{
  "Tag": [
    "geometry",
    "geometry unsolved"
  ],
  "Problem": "In triangle $ABC$, let the incircle touch side $BC$ at $P$. $R$ is an arbitrary point on side $BC$. The incircles of $ABR,ACR$ has centers $X,Y$ respectively. Show that $P$ lies on one of the internal tangent of the two circles.",
  "Solution_1": "It's sufficient to show that $\\angle XPY=90$. To prove that we must show that $XRPY$ is cyclic. Let K,L be points of tangency of incinters of $ACR$, $ABR$.  By easy couting we can show that $RL=KP$.",
  "Solution_2": "[quote=\"wojto111\"]It's sufficient to show that $\\angle XPY=90$. [/quote]\r\n\r\nEh, why?",
  "Solution_3": "Take on $BC$ point $T$ such that lie on internal tanget to this circles. Mark points of tangency to the circles $ACR$,$ABR$ by $R$,$S$. Draw this and you will see :P.",
  "Solution_4": "Okay!  :P  :)"
}
{
  "Tag": [],
  "Problem": "Dakle... \u0161to vam se naj\u010de\u0161\u0107e vrti u CD playeru? (ili gramofonu, kazetofonu, mp359 playeru ili \u010demu ve\u0107)\r\n\r\nJa bih izdvojio Beatlese i Stonese, iako znam poslu\u0161ati i balkanskije stvari tipa Rundek, Gibonni, Bala\u0161evi\u0107, Azra, Indexi, Britney itd...\r\n\r\n :lol:",
  "Solution_1": "Meni ne treba klasi\u010dna poezija jer, da citiram njema\u010dkog matemati\u010dara Dehna, \u0422\u0435\u043e\u0440\u0438\u0458\u0441\u043a\u0430 \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 \u0458\u0435 \u043f\u043e\u0435\u0437\u0438\u0458\u0430 \u043b\u043e\u0433\u0438\u0447\u043d\u0438\u0445 \u0438\u0434\u0435\u0458\u0430. (\"Pokradeno\" s sajta Dru\u0161tva matemati\u010dara Srbije, hence the cyrillic.)\r\n\r\nUz neke iznimke dosta naginjem doma\u0107im stvarima (naravno, ne folku  :noo:  :help:  :diablo: ), Bala\u0161evi\u0107 mi je for the win, a jo\u0161 povremeno slu\u0161am (puno rje\u0111e nego Bala\u0161evi\u0107a, naravno  :D ) Zakloni\u0161\u010de prepeva, Die Toten Hosen, Hladno pivo, Zabranjeno pu\u0161enje, Edo Maajka, Slon in Sade\u017e (dobro, i Rundek, Gibonni, \u0160erbed\u017eija, Bijelo dugme, Van Gogh, Mar\u010delo, Dubioza Kolektiv... u tragovima). E, da, Orthodox Celts su isto for the win, thanks Jelena.  :lol: \r\n\r\nEto, sva\u0161ta, uglavnom.\r\n\r\nBtw, ba\u0161 razmi\u0161ljam da napravim neku listu svih estradnih nadriumjetnika koje naprosto ne mogu podnijeti i onda, kad pri\u010dam s nekim, a on mi, recimo, ovla\u0161 spomene kako mu je na\u0161 \"najve\u0107i domoljubni pjesnik\" (znate ga sigurno, fin jedan gospodin, prozvao se po automatskom oru\u017eju) srcu drag ili kako je upravo bio na nastupu godine nekog od najrenomiranijih turbofolk izvo\u0111a\u010da (u narodu poznatom kao \"no\u0107 dugih no\u017eeva\"), ja samo ka\u017eem \"OK, ispri\u010daj me sada, idem zaroniti glavu u akvarij... bit \u0107e mi ugodnije pri\u010dati s ribama.\" [size=75]No, dobro, ne valja biti isklju\u010div, a i morao bih dosta \u010desto dr\u017eati glavu u akvariju.[/size]",
  "Solution_2": "Nije fer... Da sad napi\u0161em \u0161ta slu\u0161am, izgledalo bi kao quote kolege Ornika...  :lol: \r\n\r\nNo, mo\u017eda uspijem \u0161togod dodati na listu: dakle, osim pobrojanih u postu iznad, u moj player zalutaju i slijede\u0107i izvo\u0111a\u010di, pobrojani nasumi\u010dnim redom: [b]Miladin \u0160obi\u0107[/b] (umjesto pitanja: (t)ko je ovaj koje vam se upravo mota po glavi, navratite na dzuboks.de i skinite par pjesama  :lol:  ), [b]Letu \u0161tuke[/b], [b]Bajaga[/b], stare stvari od [b]\u010corbe[/b], [b]Salvatore Adamo[/b], [b]Weird Al[/b], [b]Panic(!) at the disco[/b], [b]Nick Cave[/b], [b]Guns'n'Roses, Iron Maiden, Pink Floyd[/b]... razvu\u010de se ovaj spisak, izgleda da slu\u0161am sve \u0161to mi do\u0111e pod ruku i svidi mi se...\r\n\r\nAh da, zaboravio sam sve nabrojane ispod:\r\n\r\n[quote=\"Bora \u0110or\u0111evi\u0107\"]Na priredbi kod Frankfurta\npeva Kurta, peva Murta\nSinan, Hasan, Jasar, Saban\nNino, Seki i jos neki i naravno Dzej\n\nMala Maja Marijana\nAnci, Brzi i Dajana\nKeba, Kemis, Rodja, Jami\nsvi zajedno u galami i naravno Dzej\n\nIs\u2019o sam kod oni stranci\nda im otmem njini Franci\nimam jednu zelju zarku\nda im uzmem koju marku\n\nMica, Cica i Zorica\nZoki, Boki, Rodic Roki\nSladja, Kobra, Stela, Nela\nslatka mala Karamela i naravno Dzej\n\nLepa Brena, Ekstra Nena\nLepa Lana i Cakana\nLepi Bora, Lepi Gagi\nLepi Mica, svi su dragi, i naravno Dzej\n\nEra, Gedza, Bekic Beki\ni naravno Super Sneki\ni sa njima zmija Veki\nViki, Siki, Viki, Riki i naravno Dzej\n\nPeva Sinko i Marinko\nMitar Miric, Skoric Mira\ndok muzika grozno svira\npeva hitic Mile Kitic i naravno Dzej\n[/quote]",
  "Solution_3": "Ispri\u010davam se zbog nepoznavanja srpske kulture ali tko bi bio gore vi\u0161e puta spomenuti gospodin Dzej? :D\r\n\r\nNapisao je \u010corba jo\u0161 puno dobrih stvari, kao ona [i]Neko mi je ukrao biciklo[/i] ili [i]Ve\u010de ne miri\u0161e na rakove i \u0161koljke[/i] (ili je to bio Bajaga? genijalna pjesma)\r\n\r\nI ako netko voli punk-rock, preporu\u010dam album Marquee Moon benda Television koji je stvarno genijalan i ujedno zahvaljujem Bodanu \u0161to me upoznao s njim. :)",
  "Solution_4": "D\u017eej Ramadanovski: Omaleni pjeva\u010d romskog podrijetla, s evidentnim manjkom kose, hiperaktivan na nastupima  :lol: \r\n\r\n\u0160to se pjesme \"Kad hoda\u0161 (Ve\u010de ne miri\u0161e na rakove i \u0161oljke)\" ti\u010de, da, napisao ju je Bajaga, tada\u0161nji \u010dlan Riblje \u010corbe.",
  "Solution_5": "Ne mogu vjerovati da sam zaboravio Milana Manojlovi\u0107a - Mancea (http://www.last.fm/music/mance - ima tamo i free download, pa poslu\u0161ajte). Ima mnogo impresivnih pjesama, a svakako preporu\u010dam Studijski je zapis lo\u0161, Kafi\u0107 bezveze i, recimo, Javorova grana. No, posebno je interesantna njegova pjesma Dva, \u010diji tekst \u0107ete na\u0107i dolje:\r\n\r\n[quote=\"Mance\"]\nIma jedan broj\nA djeljiv je sa pet\nU nizu brojeva\nDjeljivih sa \u0161est\n\nSvaki novi broj\nPodjeljen sa sedam\nUmanjen za tri\nPomno\u017een sa pet\n\nTo je dva, dva\ndva je labud moj\ndva, dva, dva je tajni broj\n\nPomno\u017een sa dvanaest\nUmanjen sa tri\nIma jedan broj\nBroj tramvajski\n\nA to je dva, dva\nDva do \u017ditnjaka\nDva, dva, dva do neba sad\n\nIma mnogo brojeva\nDjeljivih sa pet\nNeki pak su djeljivi\nI sa brojkom \u0161est, sa \u0161est\n\nAli dvaaaa, dvaaa\nDva do \u017ditnjaka\nDvaaaaaaaaaaaaa do neba\n\nDva je tajni broj\nDva je labud moj\nDvaaa, Dvaaa\nDva u dvanaest \u010detiri\n[/quote]\r\n\r\nI ne mogu vjerovati, \u010ditaju\u0107i Harunov popis, da sam zaboravio navesti Bajagu i Weird Ala.  :("
}
{
  "Tag": [],
  "Problem": "1. 2004 can be expressed as the sum of distinct positive numbers  of the same digits. For example 725 + 752 + 527 = 2004 or 617 + 671 + 716 = 2004 or 509 + 590 + 905 = 2004.  2003 can also be expressed as the sum of positive numbers with the same digits. What are they?\r\n\r\n2. A string is wrapped around the earth's equator (wow! that's a long string) and the two ends just touch.  Suppose that another string is tied to the original string so it becomes 30.48 m longer.  If this new string is placed around the equator and pulled tight so that it is suspended in air, how high would the string be above the ground?",
  "Solution_1": "please clarify what the radius of the earth is\r\n\r\ni think it is 6.38*10^6 meters",
  "Solution_2": "In the Flemish Mathematical Olympiad 2003, problem 3 was:\r\n\r\nLet N be a positive integer which consists of 3 distinct digits, such that the sum of the other five positive integer which can be formed with the digits of N equals 2003.\r\nFind all possible values of N.\r\n\r\n[hide]127 + 172 + 271 + 712 + 721 = 2003[/hide]",
  "Solution_3": "The circumference of the earth is approximately 24, 900 miles, then how do you solve it?",
  "Solution_4": "[quote=\"Arne\"]In the Flemish Mathematical Olympiad 2003, problem 3 was:\n\nLet N be a positive integer which consists of 3 distinct digits, such that the sum of the other five positive integer which can be formed with the digits of N equals 2003.\nFind all possible values of N.\n\n[hide]127 + 172 + 271 + 712 + 721 = 2003[/hide][/quote]\r\n\r\n  WOW, thats one of the easiest olympiad problems ive ever seen.......",
  "Solution_5": "I know. The hardest questions of the Flemish olympiad are even easy for the Pre-Olympiad section... It's sad. :)",
  "Solution_6": "2 is a mess of number and units, you know the circumfrence is changed by 30.48, how much must the diameter change?",
  "Solution_7": "http://www.google.com/search?hl=en&q=radius+of+the+earth\r\n\r\n[hide]\n$r=6 378.1 \\text { km } = 6,378,100 \\text { m.}$\n\nI don't understand the question though :huuh:\n\nEDIT: :blush: oh i see now\n[/hide]",
  "Solution_8": "[hide]Using chess64's kind given measurements...  :D \n\nCircumference = 20037.39 km. \nNew string = 20067.87 km. \nIts circle would measure 6387.80, \n\n6387.80-6387.1 = 0.70 km, \nDivide that by two (see why?) to get 350 meters. [/hide]",
  "Solution_9": "It doesn't matter what the radius of the earth is. For any circle, when you increase the circumference by 30.48 m, the radius is increased by $\\frac{30.48}{2\\pi}$ which is approximately 4.85.",
  "Solution_10": "[quote=\"mathnerd314\"]It doesn't matter what the radius of the earth is. For any circle, when you increase the circumference by 30.48 m, the radius is increased by $\\frac{30.48}{2\\pi}$ which is approximately 4.85.[/quote]\r\n\r\nNice observation, so now I don't know how I got my answer....",
  "Solution_11": "[quote=\"guile\"]\n2. A string is wrapped around the earth's equator (wow! that's a long string) and the two ends just touch.  Suppose that another string is tied to the original string so it becomes 30.48 m longer.  If this new string is placed around the equator and pulled tight so that it is suspended in air, how high would the string be above the ground?[/quote]\r\n\r\nOH this problem looks like a joke problem I have met before. It is also about a string is wrapped around the earth's equator, and the two ends just touch. If the string is 1m longer, there will be a hole. The question is if a normal cat will go through that hole  :rotfl: ."
}
{
  "Tag": [],
  "Problem": "Given that $5r+4s+3t+6u =100$, where $r\\ge s\\ge t\\ge u\\ge 0.$ are real numbers,\r\nfind, with proof, the maximum and minimum possible values of $r+s+t+u$.",
  "Solution_1": "[hide]$min=\\frac{172}{9}$ and $max=25$??[/hide]",
  "Solution_2": "[hide=correct bound]we have $5r+4s+3t+6u=100$\\\\\nnotice $u-t\\le 0$ and $-2t \\le 0$ so we have \\\\\n$5(r+s+t+u)+u-s-2t\\le 5(r+s+t+u)$\\\\\nwhich gives $r+s+t+u \\ge \\frac{100}{5}=20$\\\\\nalso we have $r-t \\ge 0$ and $2u\\ge 0$ so $r-t+2u \\ge 0\\implies 4(r+s+t+u)+r-t+2u\\ge 4(r+s+t+u)\\implies \\frac{100}{4}=25 \\ge (r+s+t+u)$\\\\\nso we get $\\boxed{20\\le r+s+t+u\\le 25}$ $\\blacksquare$[/hide]"
}
{
  "Tag": [
    "conics",
    "parabola"
  ],
  "Problem": "Points $ A$ and $ B$ are on the parabola $ y \\equal{} 4x^2 \\plus{} 7x \\minus{} 1$, and the origin is the midpoint of $ \\overline{AB}$. What is the length of $ \\overline{AB}$?\r\n\r\n$ \\textbf{(A)}\\ 2\\sqrt5 \\qquad\r\n\\textbf{(B)}\\ 5\\plus{}\\frac{\\sqrt2}{2} \\qquad\r\n\\textbf{(C)}\\ 5\\plus{}\\sqrt2 \\qquad\r\n\\textbf{(D)}\\ 7 \\qquad\r\n\\textbf{(E)}\\ 5\\sqrt2$",
  "Solution_1": "Let the 2 points be $ (a,b)$ and $ (\\minus{}a,\\minus{}b)$ (since their midpoint is the origin), with $ a \\ge 0$.\r\n\r\n$ b\\equal{}4a^2\\plus{}7a\\minus{}1$\r\n$ \\minus{}b\\equal{}4a^2\\minus{}7a\\minus{}1$\r\n\r\nSumming these 2, we have $ 8a^2\\equal{}2$, so $ a\\equal{}\\frac{1}{2}$, so $ b\\equal{}\\frac{7}{2}$. Therefore, the 2 points are $ (\\frac{1}{2},\\frac{7}{2}$ and $ (\\minus{}\\frac{1}{2},\\minus{}\\frac{7}{2})$. The distance between the 2 by the point-point distance formula is $ \\sqrt{1^2\\plus{}7^2}\\equal{}\\sqrt{50}\\equal{}2\\sqrt{5}$\r\n\r\nTherefore, the answer is A.",
  "Solution_2": "[quote=\"rofler\"]$ \\sqrt {50} \\equal{} 2\\sqrt {5}$\n\nTherefore, the answer is A.[/quote]\r\n\r\nA minor correction: $ \\sqrt{50}\\equal{}5\\sqrt{2}$, making the answer E.  :)",
  "Solution_3": "[b]Solution:[/b] We know that the points have to be $(x, y)$ and $(-x, -y)$ so both of the equations: $y=4x^2 + 7x - 1$ and $y=-4x^2 + 7x + 1$ must be true. Adding this together gives us $2y = 14x \\Rightarrow y = 7x$. Substituting this into the first equation, we get that $x = \\frac{1}{2}$ and therefore $y = \\frac{7}{2}$. This makes the distance between the points $\\sqrt{7^2 + 1^2} = \\sqrt{50} = 5\\sqrt{2} = \\boxed{E}$."
}
{
  "Tag": [],
  "Problem": "My four-digit code is 0235. Alice can't choose a code that is the same as mine in three or more of the four digit-positions, nor that is the same as mine except for switching the positions of two digits (so 3205 and 2035, for example, are forbidden, but 3025 is fine). Alice can otherwise choose any four-digit code. How many codes are available for Alice?",
  "Solution_1": "hmm this problem seems similar to a problem i did not so long ago  :rotfl:  :rotfl:  :rotfl: \r\nNow there are four digits available, or 10000 different codes.\r\nSo which ones are not available?\r\nThose with 3 digits same as 0235, or any set that is in the form\r\n023x\r\n02x5\r\n0x35\r\nx235\r\nexcept that for the set being 0235, we have 4 values that overlap into one.\r\nThis gives us 10+10+10+10-3 or 37 different codes.\r\nMoving on to the second condition, we have that there are 6 pairs of numbers (4 C 2) that can be switched around, or 6 different codes that are forbidden.\r\n\r\nNow, we can sum the forbidden codes, 37+6=43;\r\nThen subtracting this from our start point, we have\r\n10000-43=[b]9957[/b] available codes!",
  "Solution_2": "To be more transparent in the second part:\r\n\r\nTo get a code the same in three places, we would pick any spot and exchange it for one of the other nine digits.  Total $ 9\\plus{}9\\plus{}9\\plus{}9\\equal{}36$"
}
{
  "Tag": [
    "geometry"
  ],
  "Problem": "Find the shortace distance from the origin to the line segment containing (2,0) and (0,3)",
  "Solution_1": "[hide=\"a quickie spoiler DONT LOOK\"]2 times area of triangle divided by hypotenuse :D\naww told you not to look :P[/hide]",
  "Solution_2": "[hide]\n$x/2+y/3=1$\n$3x+2y=6$\n\nBy Cauchy-Schwarz,\n$(x^{2}+y^{2})(3^{2}+2^{2})\\geq (3x+2y)$\n$(x^{2}+y^{2})(3^{2}+2^{2})\\geq 6^{2}$\n$(x^{2}+y^{2})\\geq \\frac{6^{2}}{3^{2}+2^{2}}$\n$\\sqrt{(x^{2}+y^{2})}\\geq \\sqrt{\\frac{6^{2}}{3^{2}+2^{2}}}=\\boxed{6\\sqrt{13}/13}$[/hide]",
  "Solution_3": "[quote=\"contradictory\"][hide]\n$x/2+y/3=1$\n$3x+2y=6$\n\nBy Cauchy-Schwarz,\n$(x^{2}+y^{2})(3^{2}+2^{2})\\geq (3x+2y)$\n$(x^{2}+y^{2})(3^{2}+2^{2})\\geq 6$\n$(x^{2}+y^{2})\\geq \\frac{6}{3^{2}+2^{2}}$\n$\\sqrt{(x^{2}+y^{2})}\\geq \\sqrt{\\frac{6}{3^{2}+2^{2}}}=\\boxed{\\sqrt{78}/13}$[/hide][/quote]\r\n\r\nFor the RHS of the first step of Cauchy, you forgot to square it. :wink:",
  "Solution_4": "[quote=\"robinhe\"][quote=\"contradictory\"][hide]\n$x/2+y/3=1$\n$3x+2y=6$\n\nBy Cauchy-Schwarz,\n$(x^{2}+y^{2})(3^{2}+2^{2})\\geq (3x+2y)$\n$(x^{2}+y^{2})(3^{2}+2^{2})\\geq 6$\n$(x^{2}+y^{2})\\geq \\frac{6}{3^{2}+2^{2}}$\n$\\sqrt{(x^{2}+y^{2})}\\geq \\sqrt{\\frac{6}{3^{2}+2^{2}}}=\\boxed{\\sqrt{78}/13}$[/hide][/quote]\n\nFor the RHS of the first step of Cauchy, you forgot to square it. :wink:[/quote]\r\n\r\nwoops.",
  "Solution_5": "oh so altitude is the shortest distance. oooh\r\n\r\n\r\ni thought it was the median, no wonder i messed up  :lol:"
}
{
  "Tag": [
    "limit",
    "trigonometry",
    "calculus",
    "calculus computations"
  ],
  "Problem": "Calculate:\r\n\r\n$ \\lim_{x\\rightarrow0}\\frac {e^{x^2} \\minus{} e^{3x}}{\\sin(\\frac {x^2}{2}) \\minus{} \\sin x}$",
  "Solution_1": "hello :) !!\r\n\r\ni thing that $ lim \\equal{} 3$",
  "Solution_2": "the prove !!!\r\n\r\n$ e^{x^2} \\sim_0 1\\plus{}x^2$ and $ e^{3x}\\sim_0 1\\plus{}3x$ and $ sin(\\frac{x^2}{2}) \\sim_0 \\frac{x^2}{2}$ and $ sin(x) \\sim_0 x$\r\n\r\nthan:\r\n\r\n$ f(x)\\equal{}\\frac{e^{x^2}\\minus{} e^{3x}}{sin(\\frac{x^2}{2})\\minus{} sin(x)} \\sim_0 \\frac{x^2\\minus{}3x}{\\frac{x^2}{2} \\minus{} x}$\r\n\r\nthan $ if  \\ x\\to 0$ than $ f(x) \\to 3$",
  "Solution_3": "Plugging in, we get our favorite indeterminate form, $ \\frac{0}{0}$, so we attack this thing with L'Hopital's rule:\r\n\r\n$ \\lim_{x\\to 0}\\frac {e^{x^2} \\minus{} e^{3x}}{\\sin(\\frac {x^2}{2}) \\minus{} \\sin x}\\equal{}\\lim_{x\\to 0}\\frac{2xe^{x^2}\\minus{}3e^x}{2x\\cos\\frac{x^2}{2}\\minus{}\\cos x}$.\r\n\r\nPlug in 0 to get $ \\frac{\\minus{}3}{\\minus{}1}\\equal{}\\boxed{3}$.\r\n\r\n[b]mathema*[/b], what do you mean by $ \\sin\\frac{x^2}{2}\\sim_0\\frac{x^2}{2}$, and the others?",
  "Solution_4": "I believe [b]mathema*[/b] meant: $ a(x)\\sim_0 b(x)$ is equivalent to \"$ a(x)$ is asymptotic to $ b(x)$ when the variable (i.e. $ x$) tends to zero\".",
  "Solution_5": "Thanks guys"
}
{
  "Tag": [],
  "Problem": "A cell phone plan costs $ \\$50$ per calendar month. At this rate, how\nmany dollars does it cost for 10 years?",
  "Solution_1": "For every month, it's $50. 12 months in a year. do 50*12*10=6000."
}
{
  "Tag": [
    "\\/closed"
  ],
  "Problem": "I just realized that the quotes that were on the bottom of the page before are gone.\r\nWhy is the quotes gone?",
  "Solution_1": "Many people complained that they were hard to read, and they slowed down the site.  So we removed them.\r\n\r\nIf you enjoy mathematical quotes, there are many of them at http://www-groups.dcs.st-and.ac.uk/~history/Quotations/index.html"
}
{
  "Tag": [
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "3$\\frac{ab+bc+ca}{a+b+c} \\le \\sum_{a,b,c}\\frac{a^3}{b^2-bc+c^2}$\r\nI think SOS is the best for solving this problem",
  "Solution_1": "[quote=\"NC7\"]3$\\frac{ab+bc+ca}{a+b+c} \\le \\sum_{a,b,c}\\frac{a^3}{b^2-bc+c^2}$\nI think SOS is the best for solving this problem[/quote]\r\nAre you mean to $a>0,$ $b>0,$ $c>0$ $?$\r\nIf so, I don't agree with you,since\r\n$\\sum_{cyc}\\frac{a^3}{b^2-bc+c^2}\\geq\\sum_{cyc}a\\geq\\frac{3(ab+bc+ca)}{a+b+c}$ is easier. ;)",
  "Solution_2": "[quote]I don't agree with you,since\n$ \\sum_{cyc}\\frac{a^{3}}{b^{2}\\minus{}bc\\plus{}c^{2}}\\geq\\sum_{cyc}a\\geq\\frac{3(ab\\plus{}bc\\plus{}ca)}{a\\plus{}b\\plus{}c}$ is easier. ;)[/quote]\r\nSo how can you prove this without SOS or Schur",
  "Solution_3": "See\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=3221 ;\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=6551 .\r\n\r\n  darij",
  "Solution_4": "Actually, Arqady posed a good question: It seems that for any [i]reals[/i] $ a$, $ b$, $ c$, we have\r\n\r\n$ \\left(a\\plus{}b\\plus{}c\\right)\\left(\\frac{a^{3}}{b^{2}\\minus{}bc\\plus{}c^{2}}\\plus{}\\frac{b^{3}}{c^{2}\\minus{}ca\\plus{}a^{2}}\\plus{}\\frac{c^{3}}{a^{2}\\minus{}ab\\plus{}b^{2}}\\right)\\geq 3\\left(bc\\plus{}ca\\plus{}ab\\right)$.\r\n\r\nAny proof or counterexample?\r\n\r\n  darij",
  "Solution_5": "Sorry, but what you called SOS, i can't catch meaning of this word.",
  "Solution_6": "[quote=\"Zubr\"]Sorry, but what you called SOS, i can't catch meaning of this word.[/quote]\r\n\r\nSee http://www.mathlinks.ro/viewtopic.php?t=80127 and http://www.mathlinks.ro/viewtopic.php?t=65385 for different aspects of the method.\r\n\r\n  Darij",
  "Solution_7": "Thank you."
}
{
  "Tag": [],
  "Problem": "(This is a continuation of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17222]another thread[/url], but it seemed appropriate to discuss this in a separate, considering all the flameage that had taken place there.)\r\n\r\n[quote=white_horse_king88]Ahh... mathfanatic. You are in Exeter already? How's exeter - I will never make it, but I am curious![/quote]\r\n\r\nExeter is nice.  Admittedly, it's a bit different from what I expected and how it's probably been portrayed to you, but I'm glad to be here.\r\n\r\nClasses are...well, classes.  Some of them are really good (I thoroughly enjoy my Physics and European History classes, which involve a lot more experimentation/thinking and discussion, respectively, than occurs in most schools).  Some of them are decent (my Spanish class is too slow, although that may be more of a placement issue than a teaching issue).  There's generally a fair amount of work, but it's not overwhelming.  Generally, I have 5-6 hours of classes per day and 2-3 hours of homework.\r\n\r\nOutside of classes, there is generally a lot to do.  Sports are meant to take up some time (particularly if you're on a team...club sports are far less time-consuming) but Exeter sets aside about 2 hours each day for them.  There are over 100 clubs, from debate to drumming to philosophy to math team (or, as a friend calls it, \"varsity math\").  For math team, we usually work on a set of 12 problems/proofs on wednesdays, then discuss solutions on sundays.  \r\n\r\nRegarding the admissions process, I think many of you (judging by your work on AoPS) would make it.  I know plenty of not-very-qualified people who made it in; it's not as tough as you think.  If you're interested in Exeter, give it a shot and apply.",
  "Solution_1": "Hey mathfanatic, what were some of Exeter's Varsity Football team's recent scores?   \n\n\n\nBack on topic here, yeah, it's not too difficult to make it. I mean, I know this really lazy bum at my old school who almost never did his homework until the morning of classes, even flunked a test or two, and never knew when a quiz or test was, but he still got accepted to Exeter.\n\n\n\n[hide]go to choate.go to choate.go to choate.go to choate.go to choate.go to choate.go to choate.go to choate.go to choate.go to choate.go to choate.go to choate.go to choate.go to choate.go to choate.[/hide]",
  "Solution_2": "what's choate? and if your sig is true, how much are you like the \"lazy bum\"?",
  "Solution_3": "Choate is a utopia, a place of perfect harmony between everybody there. There is only total bliss and happiness, none of the evils that plague much of our world. At Choate, you'll find a place where there is no fighting or killing or name-calling. You will only find the pureness and soul-purging nature of your life.\r\n\r\nAs for that bum, let's just say we're so close, we're almost brothers.",
  "Solution_4": "[quote=\"yif man12\"]Hey mathfanatic, what were some of Exeter's Varsity Football team's recent scores?  :D [/quote]\r\n\r\nMeh.  Oooh, CRH is better than us at acting like gorillas, throwing irregularly-shaped balls over long distances, and making each other trip.\r\n\r\nI'll believe the total-bliss-and-happiness line when I see it.",
  "Solution_5": "[quote=\"mathfanatic\"][quote=\"yif man12\"]Hey mathfanatic, what were some of Exeter's Varsity Football team's recent scores?  :D [/quote]\n\nMeh.  Oooh, CRH is better than us at acting like gorillas, throwing irregularly-shaped balls over long distances, and making each other trip.[/quote]\r\n\r\nDarn right we are!!!  :D",
  "Solution_6": "I also go to Choate. I really, really like it, and would encourage you all to apply. Choate is located in Wallingford, Connecticut, a fairly small town that is close to the city of New Haven and about half an hour from Hartford. The student body is a perfect size, 850, which (correct me if I am wrong, mathfanatic) is a few hundred less than Exeter. Also, it's a lot colder at Exeter...and you don't want that.\r\n\r\nOur sports accomplishments speak for themselves (see yif man's above post). We've got a very successful math team as well as excellent debate and economics teams. There are many, many, many other clubs, including Indian Club, Chinese Club, Japanese Club, Korean Club, Russian Club, Chess Club, Asian Student Association, P.P.P.P.P.P.P.P.P.P.P.P.P.P. Club, Young Democrats Club, Young Republicans Club, Interact (Community Service), and far more, too many to name. Also, if you have are interested in starting a club that is not already in existence, you are encouraged to.\r\n\r\nClasses are excellent...students are motivated, teachers are encouraging. Material is covered at a rapid pace, but if you are unsure of something you are encouraged to seek extra help from your teacher, who will always be willing to help you out.\r\n\r\nAll in all, Choate's a great school. I encourage everyone here who is considering private schools to apply!",
  "Solution_7": "Wow. My school sponsors all those clubs too, but we aren't nearly as successful. I would really care to learn more because my school is very slow paced. I often fall asleep in class - Figuratively speaking of course. Then again, private schools are meant to have higher standards.",
  "Solution_8": "You can find more information at [url=http://www.choate.edu]Choate's website[/url] and at [url=http://crh.choate.edu]Choate's student website[/url].",
  "Solution_9": "Correct me if I'm wrong, but Exeter is a highschool right?\r\nAnd where is it located?\r\n\r\nOkay I'm pretty sure it's a highschool.  I'm looking at Choate's website, and it seems like an awesome school to go to.  Way better than my public school...by the way, how much does a school like that cost?\r\n\r\nIgnore that question too.  In fact, ignore them all."
}
{
  "Tag": [
    "AoPS Books"
  ],
  "Problem": "What's the current edition of the AoPS books? (i.e., What edition will I get if I order them?) Neither the information pages on the books nor the ordering page seems to give this info. I hope you'll update that. (I've heard there's a 5th edition. If there's a new edition, [i][u]is the number of pages is the same[/i][/u]?)",
  "Solution_1": "We're on around the 6th edition now, but the books haven't materially changed since the 2nd edition (and there wasn't that great a change from the 1st to the 2nd).  \r\n\r\n(I saw 'around 6th edition' because some the of the books are now in their 5th, some in the 6th - as I noted, there haven't been major changes in many editions.)"
}
{
  "Tag": [
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "$ a,b,c>0$,prove that $ \\frac{a^3}{b^2}\\plus{}\\frac{b^3}{c^2}\\plus{}\\frac{c^3}{a^2}\\ge a\\plus{}b\\plus{}c$.",
  "Solution_1": "very easy;only using AM-GM:\r\n    $ \\frac {a^3}{b^2} \\plus{} b \\plus{} b\\geq3a$",
  "Solution_2": "[quote=\"math10\"]very easy;only using AM-GM:\n    $ \\frac {a^3}{b^2} \\plus{} b \\plus{} b\\geq3a$[/quote]\r\n\r\nThanks,it's a nice solution.\r\n\r\nI didn't think of it!  :o \r\n\r\nI use generalized Cauchy's-Schwarz here \r\n\r\nbut I think my solution is not so nice,\r\n\r\nso I post the problem here to find some better solution.",
  "Solution_3": "Does it not follow directly from the Rearragement inequality?",
  "Solution_4": "[quote=\"kcn2rivers\"]Does it not follow directly from the Rearragement inequality?[/quote]\r\nYou are right!",
  "Solution_5": "Holder;\r\n$ \\left(\\frac{a^3}{b^2}\\plus{}\\frac{b^3}{c^2}\\plus{}\\frac{c^3}{a^2}\\right)(b\\plus{}c\\plus{}a)^2\\geq (a\\plus{}b\\plus{}c)^3$\r\nthen $ \\frac{a^3}{b^2}\\plus{}\\frac{b^3}{c^2}\\plus{}\\frac{c^3}{a^2} \\geq a\\plus{}b\\plus{}c$"
}
{
  "Tag": [],
  "Problem": "A bag contains hundreds of glass marbles, each one coloured either red, orange, green or blue. There are more than $ 2$ marbles of each colour.\r\nMarbles are drawn randomly from the bag, one at a time, and not replaced.\r\nHow many marbles must be drawn from the bag in order to ensure that at least three marbles of the same colour are drawn?",
  "Solution_1": "The worst case scenario is 2 red marbles, 2 orange, 2 green, and then 2 blue. Whatever marble is drawn afterwards will be a third, so we have 9."
}
{
  "Tag": [
    "limit",
    "geometric series",
    "calculus",
    "calculus computations"
  ],
  "Problem": "Evaluate:\r\n\r\n$ \\lim_{n \\to \\infty}{\\sum_{k\\equal{}1}^n \\frac{k}{2^k}}$.",
  "Solution_1": "$ \\sum_{k\\equal{}1}^{\\infty} k p^k \\equal{}  \\frac{p}{(1\\minus{}p)^2}$ for nice (what do I mean here?) $ p$.",
  "Solution_2": "But he actually asked for the partial sums.\r\n\r\nThis works pretty well on a \"try it and see\" basis. Explicitly compute $ \\sum_{k\\equal{}1}^n\\frac k{2^k}$ for $ n\\equal{}1,2,3,4,5,6.$ Is a pattern becoming apparent? Assume that the pattern you see is the correct formula, and try proving its validity by induction.",
  "Solution_3": "[hide]Let for $ |x| < 1$ we have: $ f \\equal{} \\sum_{k \\equal{} 1}^{n}x^k \\equal{} x\\cdot\\frac {1 \\minus{} x^n}{1 \\minus{} x}$\n\nThen $ \\sum_{k \\equal{} 1}^{n}kx^{k} \\equal{} x\\cdot f_{x}'$\n\nEasy to calculate that $ x\\cdot f_{x}' \\equal{} x\\cdot\\frac {1 \\minus{} (n \\plus{} 1)x^n \\plus{} nx^{n \\plus{} 1}}{(1 \\minus{} x)^2}$ and we get [b]Tyl's[/b] result when $ n$ tends to infinity.[/hide]",
  "Solution_4": "${ \\lim_{n\\to\\infty} \\sum_{k = 1}^n \\frac {k}{2^k} = \\lim_{n\\to\\infty} (\\frac {k + 1}{2^{k - 1}} - \\frac {k + 2}{2^k}) = \\lim_{n\\to\\infty} (\\frac {2}{2^0} - \\frac {n + 2}{2^n}}) = \\boxed{2}$.",
  "Solution_5": "uh, kunny --- since $ \\frac12\\plus{}\\frac24\\plus{}\\frac38$ is already larger than $ 1,$ I don't think $ 1$ is the sum of the infinite series. Time to recheck the details.\r\n\r\nOf course the next to last item on the line is correct: it essentially says that \r\n\\[ \\sum_{k\\equal{}1}^n\\frac{k}{2^k}\\equal{}2\\minus{}\\frac{n\\plus{}2}{2^n}\\]\r\nWhich can be gotten from what Yuriy Solovyov said, or could be found in the way I suggested earlier.",
  "Solution_6": "This is a series commonly known as the arithmetico-geometric series. Its closed form is fairly easy (though somewhat painstaking) to derive.\r\n\r\nSimply substitute that closed form, then let $ n$ tend to infinity.",
  "Solution_7": "[quote=\"Kent Merryfield\"]uh, kunny --- since $ \\frac12 \\plus{} \\frac24 \\plus{} \\frac38$ is already larger than $ 1,$ I don't think $ 1$ is the sum of the infinite series. Time to recheck the details.\n\nOf course the next to last item on the line is correct: it essentially says that\n\\[ \\sum_{k \\equal{} 1}^n\\frac {k}{2^k} \\equal{} 2 \\minus{} \\frac {n \\plus{} 2}{2^n}\n\\]\nWhich can be gotten from what Yuriy Solovyov said, or could be found in the way I suggested earlier.[/quote]\r\n\r\nYou are right. :blush: \r\n\r\nWe have explained how to solve such kind of infinite series, haven't you?,  as for me at least I have answered 5 times, this time I will copy it.\r\n\r\nAnyway, first of all we should teach beginners the way $ S_n \\minus{} rS_n$. :wink:\r\n\r\nTo Japanese high school students, you need to prove $ \\lim_{n\\to\\infty} \\frac {n}{2^n} \\equal{} 0$ in the university entrance examination. :P",
  "Solution_8": "[quote=\"grn_trtle\"]This is a series commonly known as the arithmetico-geometric series. Its closed form is fairly easy (though somewhat painstaking) to derive.[/quote]\r\nReally?  There are several straightforward ways depending on what you consider straightforward. \r\n\r\n1.  Differentiation.  You just need to deal with a finite geometric series instead of an infinite one.\r\n2.  Multiplying and taking differences, which amounts to summation by parts.\r\n3.  The combinatorial proof [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=220351]here[/url]."
}
{
  "Tag": [
    "logarithms",
    "calculus",
    "derivative",
    "integration",
    "function",
    "search",
    "projective geometry"
  ],
  "Problem": "[quote][b]Theorem:[/b] If $ G$ is either the whole plane $ \\bf C$ or some open disk. then if $ u: G \\to \\bf{R}$ is harmonic, then $ u$ has a harmonic conjugate.\n[/quote]\r\n\r\nThis happens to fail for $ u(z) \\equal{} \\log |z|$ on $ G \\equal{} \\bf{C} \\minus{} \\{0\\}$\r\n\r\n(because we can't define an analytic branch for the logarithm on $ G$)\r\n\r\nI don't really see what goes wrong in the proof of the theorem though. They seem to use Leibniz' rule of advanced calculus (Differentiation under the integral sign) with $ u: (a,R) \\times (b,R) \\to \\bf{R}$ continuous and assuming the open disk $ G$ is taken $ G \\equal{} B(z,R)$ with $ z \\equal{} a \\plus{} bi$.\r\n\r\nSo Leibniz' rule is supposably to go wrong if we don't have a continuous function from a nice open interval or what is the main issue actually?",
  "Solution_1": "Leibniz rule is really a technicality and not the point here. Differentiating under the integral sign is supposed to feel natural.\r\n\r\nThe proof goes wrong because what is your well defined harmonic conjugate? The construction doesn't work when you've punched out a hole (because your lines of integration start intersecting the hole). Of course, locally around each point the construction works fine, so it has a harmonic conjugate locally, but this doesn't give a harmonic conjugate globally.\r\n\r\nThe construction should feel pretty natural, if you think about it; the construction should give the only possible harmonic conjugate up to an additive constant, if there is one at all. We simply must attempt to define our harmonic conjugate as we do.\r\n\r\nThe problem is closely connected to simple connectivity and Cauchy's theorem. Note that the things I said above could be said about Cauchy's theorem for a disc instead, but instead of a harmonic conjugate we're trying to define a primitive of an analytic function. Indeed, if you've done Cauchy's theorem on the course, this leads to a quick proof of this business:\r\n\r\nSuppose there is a harmonic conjugate $ v$, so that $ f \\equal{} u \\plus{} iv$ is analytic in the disc. Then it follows that $ f' \\equal{} \\frac{\\partial u}{\\partial x} \\minus{} i\\frac{\\partial u}{\\partial y}$. So now, let us define $ h \\equal{} \\frac{\\partial u}{\\partial x} \\minus{} i\\frac{\\partial u}{\\partial y}$. Check that it is analytic. The function $ f$ we search for MUST be a primitive of $ h$. According to Cauchy's theorem such a primitive exists, call it $ f \\equal{} u_1 \\plus{} iv_1$. $ u_1$ has the derivatives as $ u$, so they only differ by a constant. It follows that $ v_1$ is a harmonic conjugate for $ u$",
  "Solution_2": "[quote]\n (because your lines of integration start intersecting the hole)\n[/quote]\r\nI\"m not with you on this one....",
  "Solution_3": "What is your suggested construction?"
}
{
  "Tag": [
    "calculus",
    "integration",
    "logarithms",
    "calculus computations"
  ],
  "Problem": "Evaluate $\\sum_{n=0}^{\\infty}\\frac{1}{(2n+1)2^{2n+1}}.$",
  "Solution_1": "Use the following formula:\r\n\r\n$\\frac{1}{2}\\ln \\frac{1+a}{1-a}=\\sum_{n=0}^{\\infty}\\frac{a^{2n+1}}{2n+1}$ for $|a|<1$.\r\n\r\nFor $a=\\frac{1}{2}$ we get that the value of your sum is $\\frac{1}{2}\\ln 3$.",
  "Solution_2": "Your answer is correct, didilica. :) \r\n\r\nCan anyone solve in using integral? :lol:",
  "Solution_3": "$\\sum_{n=0}^{\\infty}\\frac{1}{2^{2n+1}(2n+1)}=\\sum_{n=0}^{\\infty}\\int_{0}^{1/2}x^{2n}dx=\\int_{0}^{1/2}\\sum_{n=0}^{\\infty}x^{2n}dx$\r\n\r\n$=\\int_{0}^{1/2}\\frac{dx}{1-x^{2}}=\\frac{1}{2}\\int_{0}^{1/2}(\\frac{1}{1-x}+\\frac{1}{1+x})dx=\\frac{1}{2}\\ln(3)$",
  "Solution_4": ":)"
}
{
  "Tag": [
    "Stanford",
    "college"
  ],
  "Problem": "Does anyone know any colleges that are good for computer science (i'm looking closer to the west coast, so colleges situated mostly on the west coast are most appreciated). Thanks. :)",
  "Solution_1": "You want to consider a whole range of selectivity, right? I frequently hear Stanford and University of California Berkeley mentioned as strong computer science colleges, and Caltech should be fine too. I know less about colleges that are perhaps easier to get into but that also are west coast colleges with strong programs in computer science. \r\n\r\nWhat do the rest of you suggest?"
}
{
  "Tag": [
    "logarithms"
  ],
  "Problem": "if \\[ \\log_{12}24\\equal{}b  ,  \\log_{7}12 \\equal{} a\\]\r\n what is  $ \\log_{54}168$?",
  "Solution_1": "There is probably a simpler final expression, but \r\nSince $ 168=7*24$,   $ \\log_{12}168=\\log_{12}7+\\log_{12}24$\r\n$ \\log_{12}7=\\frac {1}{\\log_{7}12} = 1/a$\r\n$ \\log_{12}168=b+1/a$\r\nSince $ 54=\\frac {12^8}{24^5}$,   $ \\log_{12}54=8-5b$\r\n\r\n$ \\log_{54}168=\\frac{log_{12}168}{\\log_{12}54}$\r\n\r\n$ \\log_{54}168=\\frac{ b+1 a }{8-5b}$"
}
{
  "Tag": [],
  "Problem": "What is the integer value of $ x$ in the arithmetic sequence $ 3^2, x, 3^4$?",
  "Solution_1": "Call the common difference $ d$. We then have, \r\n\r\n$ 2d\\equal{}3^4\\minus{}3^2 \\implies 2d\\equal{}72 \\implies d\\equal{}36$.\r\n\r\nWe can then see that $ x\\equal{}9\\plus{}36\\equal{}\\boxed{45}$.",
  "Solution_2": "$ x \\equal{} \\frac{3^2 \\plus{} 3^4}{2} \\equal{} \\frac{9(1\\plus{}9)}{2} \\equal{} \\boxed{45}$"
}
{
  "Tag": [
    "blogs",
    "LaTeX"
  ],
  "Problem": "Hello,\r\nHow do I enable my blog or website to allow latex to appear as it does here? I will appreciate your help.",
  "Solution_1": "If you use an AoPS blog then it will already be able to use LaTeX. Otherwise, for WordPress blogs there is the LatexRender plugin at [url=http://sixthform.info/steve/wordpress/index.php]Using LaTeX in WordPress[/url] and for all other blogs and websites use [url=http://www.mayer.dial.pipex.com/tex.htm]LatexRender[/url].\r\n\r\nLatexRender is for PHP programs and you will need LaTeX, GhostScript and ImageMagick already installed on the server. If that's not possible and you can run cgi programs then there are instructions for using [url=http://www.forkosh.com/mimetex.html]Mimetex[/url] in the download. Mimetex is not quite as flexible or give such good quality as LaTeX but it's an excellent alternative if you cannot run LaTeX.",
  "Solution_2": "hmm is it possible to get LaTeX on another forum? like invisionfree? i asked on their help boards but no one answered for like a month now",
  "Solution_3": "You can use LatexRender with any PHP program [i]provided[/i] LaTeX etc is installed on the server hosting the forum. You just to need to know where in the files to add a couple of lines.",
  "Solution_4": "[url=http://www.hsn.uk.net/forum/index.php]HSN.uk.net[/url] is an Invision Board which has LatexRender installed."
}
{
  "Tag": [
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Suppose that $ f: \\mathbb R\\to\\mathbb R$ is not bounded on $ [a,b]$. Prove that there exists $ c\\in[a,b]$ such that $ f$ is not bounded on any open interval containing $ c$.",
  "Solution_1": "[quote=\"ifai\"]Suppose that $ f: \\mathbb R\\to\\mathbb R$ is not bounded on $ [a,b]$. Prove that there exists $ c\\in[a,b]$ such that $ f$ is not bounded on any open interval in $ [a,b]$ containing $ c$.[/quote]\r\n\r\nApply compactness.",
  "Solution_2": "Assume the contrary that for any $ c\\in[a,b]$, $ f$ is bounded on some open intervals containing $ c$. That is,\r\n\r\n$ (\\forall c\\in[a,b])\\,(\\exists U(c)\\subset\\mathbb R)\\,(\\exists M_c\\in\\mathbb R)\\,(\\forall x\\in U(c))\\, [|f(x)|\\le M_0],$\r\n\r\nwhere $ U(c)$ is an open intervals containing $ c$. Clearly, we have\r\n\r\n$ [a,b]\\subset\\bigcup_{c\\in[a,b]} U(c).$\r\n\r\nBy Heine-Borel Theorem, there exist $ c_1, c_2,\\cdots, c_n\\in[a,b]$ such that\r\n\r\n$ [a,b]\\subset\\bigcup_{i \\equal{} 1}^n U(c_i).$\r\n\r\nHence, for any $ x\\in[a,b]$, we have $ x\\in U(c_k)$ for some $ k$ with $ 1\\leq k\\leq n$; and so\r\n\r\n$ f(x)\\leq M_{c_k}\\leq\\max_{1\\leq i\\leq n} M_{c_i}$ for all $ x\\in[a,b]$.\r\n\r\nHowever, it violates that $ f$ is not bounded on $ [a,b]$. The result follows.",
  "Solution_3": "How about a less-fancy version of compactness - a nested interval argument?\r\n\r\nLet $ a_0\\equal{}a$ and $ b_0\\equal{}b.$ $ f$ must be unbounded on at least one of the intervals $ \\left[a_0,\\frac{a_0\\plus{}b_0}2\\right]$ or $ \\left[\\frac{a_0\\plus{}b_0}2,b_0\\right].$ Choose one of the sides for which $ f$ is unbounded and call that interval $ [a_1,b_1].$\r\n\r\nRepeat and iterate. $ f$ is unbounded on at least one of the intervals $ \\left[a_n,\\frac{a_n\\plus{}b_n}2\\right]$ or $ \\left[\\frac{a_n\\plus{}b_n}2,b_n\\right].$ Chose a half on which $ f$ is unbounded and call that $ [a_{n\\plus{}1},b_{n\\plus{}1}].$\r\n\r\nThere is a unique point $ c$ such that $ c\\in[a_n,b_n]$ for all $ n.$ Since any neighborhood of $ c$ must contain at least one of the intervals $ [a_n,b_n],$ then $ f$ is unbounded in any neighborhood of $ c.$",
  "Solution_4": "Or how about sequential compactness?\r\n\r\nIf $ f$ is not bounded (WLOG, above) on $ [a, b]$ then for every $ n$ there exists $ x_n$ such that $ f(x_n) > n$.  The sequence $ (x_n)$ takes values in a compact set, hence has a convergent subsequence that converges to $ c \\in [a, b]$.  Any open interval about $ c$ contains infinitely many elements of $ (x_n)$, hence $ f$ is not bounded on such an interval.",
  "Solution_5": "So, like the proof of t0rajir0u, we can use the theorem of accumulation points (Bolzano-Weierstrass) to prove it."
}
{
  "Tag": [
    "calculus",
    "analytic geometry",
    "graphing lines",
    "slope",
    "derivative"
  ],
  "Problem": "Suppose f(x) = x2\u22123x+1: Find all values of c so that the tangent line to the graph of f(x) at the point (c; f(c)) intersects the point (\u22123; 0)",
  "Solution_1": "Elementary calculus gives a shorter solution, but it isn't necessary:\r\n\r\n[hide=\"Hint\"]Take the general equation of a line through $(-3,0)$ and put the condition that it should have exactly one intersection point with the function.[/hide]",
  "Solution_2": "Is it $c=-3 \\pm \\sqrt{19}$ :?:",
  "Solution_3": "[quote=\"M4RI0\"]Is it $c=-3 \\pm \\sqrt{19}$ :?:[/quote]\r\n\r\nI would say those are the slopes of the two tangents, but you still have to find the touching points.",
  "Solution_4": "[quote=\"wigianto\"]Suppose $f(x) = x^{2}-3x+1$: Find all values of $c$ such that the tangent line to the graph of $f(x)$ at the point $(c; f(c))$ intersects the point $(-3; 0)$.[/quote]\r\n[hide=\"Calculus Solution\"] $f'(c)=2c-3$. The equation of a line through $(3,0)$ is $f(x)=mx+3m$. Plugging in $m=2c-3$, $x=c$ and $f(x)=c^{2}-3c+1$. \n\n$c^{2}-3c+1=(2c-3)c+3(2c-3)$\n$c^{2}-3c+1=2c^{2}-3c+6c-9$\n$c^{2}+6c-10=0$\n$c=\\frac{-6\\pm\\sqrt{36+40}}{2}=\\frac{-6\\pm2\\sqrt{19}}{2}=\\boxed{-3\\pm\\sqrt{19}}$. [/hide]\r\n\r\nM4RIO did have the correct answer. :D",
  "Solution_5": "[quote=\"lotrgreengrapes7926\"][quote=\"wigianto\"]Suppose $f(x) = x^{2}-3x+1$: Find all values of $c$ such that the tangent line to the graph of $f(x)$ at the point $(c; f(c))$ intersects the point $(-3; 0)$.[/quote]\n[hide=\"Calculus Solution\"] $f'(c)=2c-3$. The equation of a line through $(3,0)$ is $f(x)=mx+3m$. Plugging in $m=2c-3$, $x=c$ and $f(x)=c^{2}-3c+1$. \n\n$c^{2}-3c+1=(2c-3)c+3(2c-3)$\n$c^{2}-3c+1=2c^{2}-3c+6c-9$\n$c^{2}+6c-10=0$\n$c=\\frac{-6\\pm\\sqrt{36+40}}{2}=\\frac{-6\\pm2\\sqrt{19}}{2}=\\boxed{-3\\pm\\sqrt{19}}$. [/hide]\n\nM4RIO did have the correct answer. :D[/quote]\r\n\r\nI stand corrected then. Haven't dealt with the problem in detail :)",
  "Solution_6": "[hide=\"Non-Calculus Solution\"] $f(x)=mx+b$\n$0=-3m+b$\n$b=3m$\n$f(x)=mx+3m$\n$f(x)=x^{2}-3x+1$\n$x^{2}-3x+1=mx+3m$\n$x^{2}+(-3-m)x+(1-3m)=0$\n\nFor $f(x)=mx+3m$ to be a tangent line, the discriminant must be $0$.\n$(-m-3)^{2}-4(1-3m)=0$\n$m^{2}+18m+5=0$\n$m=\\frac{-18\\pm\\sqrt{324-20}}{2}=-9\\pm\\frac{\\sqrt{304}}{2}=-9\\pm2\\sqrt{19}$.\n\n$x^{2}+(-3-9\\pm2\\sqrt{19})x+(1-3(-9\\pm2\\sqrt{19}))=0$\nSince the discriminant is $0$, $x=\\frac{3-9\\pm2\\sqrt{19}}{2}=\\frac{-6\\pm2\\sqrt{19}}{2}=\\boxed{-3\\pm\\sqrt{19}}$. [/hide]",
  "Solution_7": "[quote=\"lotrgreengrapes7926\"][hide=\"Non-Calculus Solution\"] $f(x)=mx+b$\n$0=-3m+b$\n$b=3m$\n$f(x)=mx+3m$\n$f(x)=x^{2}-3x+1$\n$x^{2}-3x+1=mx+3m$\n$x^{2}+(-3-m)x+(1-3m)=0$\n\nFor $f(x)=mx+3m$ to be a tangent line, the derivative must be $0$.\n$(-m-3)^{2}-4(1-3m)=0$\n$m^{2}+18m+5=0$\n$m=\\frac{-18\\pm\\sqrt{324-20}}{2}=-9\\pm\\frac{\\sqrt{304}}{2}=-9\\pm2\\sqrt{19}$.\n\n$x^{2}+(-3-9\\pm2\\sqrt{19})x+(1-3(-9\\pm2\\sqrt{19}))=0$\nSince the derivative is $0$, $x=\\frac{3-9\\pm2\\sqrt{19}}{2}=\\frac{-6\\pm2\\sqrt{19}}{2}=\\boxed{-3\\pm\\sqrt{19}}$. [/hide][/quote]\r\n\r\nI must say that using derivatives in a \"non-calculus\" solution is a controversial practice :)",
  "Solution_8": "[quote=\"Farenhajt\"][quote=\"lotrgreengrapes7926\"][hide=\"Non-Calculus Solution\"] $f(x)=mx+b$\n$0=-3m+b$\n$b=3m$\n$f(x)=mx+3m$\n$f(x)=x^{2}-3x+1$\n$x^{2}-3x+1=mx+3m$\n$x^{2}+(-3-m)x+(1-3m)=0$\n\nFor $f(x)=mx+3m$ to be a tangent line, the derivative must be $0$.\n$(-m-3)^{2}-4(1-3m)=0$\n$m^{2}+18m+5=0$\n$m=\\frac{-18\\pm\\sqrt{324-20}}{2}=-9\\pm\\frac{\\sqrt{304}}{2}=-9\\pm2\\sqrt{19}$.\n\n$x^{2}+(-3-9\\pm2\\sqrt{19})x+(1-3(-9\\pm2\\sqrt{19}))=0$\nSince the derivative is $0$, $x=\\frac{3-9\\pm2\\sqrt{19}}{2}=\\frac{-6\\pm2\\sqrt{19}}{2}=\\boxed{-3\\pm\\sqrt{19}}$. [/hide][/quote]\n\nI must say that using derivatives in a \"non-calculus\" solution is a controversial practice :)[/quote]\r\n\r\nhe meant discriminant",
  "Solution_9": "Of course, that much is obvious... but it still should be corrected :)",
  "Solution_10": "Oops.  :blush: Sorry, I editted it."
}
{
  "Tag": [
    "function",
    "limit",
    "calculus",
    "calculus computations"
  ],
  "Problem": "Consider the sequence of function fn(x) = x^n.\r\nDoes it converges uniformly?",
  "Solution_1": "The question is meaningless as stated. Uniform convergence is tied to a set; the set is an inescapable part of the definition.\r\n\r\nDoes $ f_n(x)\\equal{}x^n$ converge uniformly on $ [0,.9]?$ Yes, it does.\r\n\r\nDoes $ f_n(x)\\equal{}x^n$ converge uniformly on $ [0,1)?$ No, it does not.",
  "Solution_2": "I forgot to mention that it is on the open interval (-1,1)",
  "Solution_3": "We have $ \\lim_{x\\to 1 \\minus{} 0}x^n \\equal{} 1$ for every $ n\\in N$.\r\n\r\nHence, for every $ \\epsilon > 0$ exists $ \\delta > 0$ sutch that\r\n \r\n$ x^n > 1 \\minus{} \\epsilon$ whenever $ 1 \\minus{} x < \\delta$ for every $ n\\in N$.\r\n\r\nThat means we can get $ x^n$ as close to $ 1$ as we want on open interval $ ( \\minus{} 1,1)$, no metter how large $ n$ is, hence $ x^n$ doesn't converge uniformly on $ ( \\minus{} 1,1)$."
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "sphere"
  ],
  "Problem": "Is it possible to have a triangle with two 90 degree angles, and the\r\nother two legs from the connected 90 degree angles meet?  Where would\r\nyou find such a triangle?  I thought maybe if the triangle is on a\r\nsphere, but then the lines aren't straight.",
  "Solution_1": "In Euclidean Geometry, the sum of the angles of any triangle is $180^{\\circ}$. If you have two $90^{\\circ}$ angles, the third angle must be $0^{\\circ}$, which is impossible.\r\n\r\nOn a sphere, I think it is possible to have a triangle with 3 $90^{\\circ}$ angles.",
  "Solution_2": "right. consider a point on the north pole of the Earth. then draw two lines south perpendicular to each other to the equator. The north pole and the intersections at the equator make a triangle of 3 90 degree angles"
}
{
  "Tag": [],
  "Problem": "Let $p$ be a fixed prime number. If $a(n)$ denotes the exponent of $p$ in the prime factorisation of $n$, determine the sum\\[S(n)=a(1)+a(2)+...+a(p^n)\\]",
  "Solution_1": "Exactly $p^{n-1}$ numbers will be divisible by $p$, $p^{n-2}$ numbers divisible by $p^2$, $p^{n-3}$ numbers divisible by $p^3$ and so forth.\r\n\r\nSo the sum will be $1+p+p^2+p^3+...+p^{n-1}$ or $\\displaystyle\\frac{p^n-1}{p-1}$."
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "tetrahedron",
    "sphere",
    "parallelogram",
    "vector",
    "perimeter"
  ],
  "Problem": "IMO 1973 ShortList:",
  "Solution_1": "[b][size=150]IMO 1973 ShortListed Problems: (Posted in Forums)[/size][/b]\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2026247#p2026247]1. (BUL 6)[/url] Let a tetrahedron $ABCD$ be inscribed in a sphere $S$. Find the locus of points $P$ inside the sphere $S$ for which the equality\n\\[\\frac{AP}{PA_1}+\\frac{BP}{PB_1}+\\frac{CP}{PC_1}+\\frac{DP}{PD_1}=4\\]\nholds, where $A_1,B_1, C_1$, and $D_1$ are the intersection points of $S$ with the lines $AP,BP,CP$, and $DP$, respectively.\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2026254#p2026254]2. (CZS 1)[/url] Given a circle $K$, find the locus of vertices $A$ of parallelograms $ABCD$ with diagonals $AC \\leq BD$, such that $BD$ is inside $K$.\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=357923&#p357923]3. (CZS 6)[/url] [b]IMO1[/b] Prove that the sum of an odd number of vectors of length 1, of common origin $O$ and all situated in the same semi-plane determined by a straight line which goes through $O,$ is at least 1.\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2026257#p2026257]4. (GBR 1)[/url] Let $P$ be a set of $7$ different prime numbers and $C$ a set of $28$ different composite numbers each of which is a product of two (not necessarily different) numbers from $P$. The set $C$ is divided into $7$ disjoint four-element subsets such that each of the numbers in one set has a common prime divisor with at least two other numbers in that set. How many such partitions of $C$ are there ?\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2026258#p2026258]5. (FRA 2)[/url] A circle of radius 1 is located in a right-angled trihedron and touches all its faces. Find the locus of centers of such circles.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=357905&#p357905]6. (POL 2)[/url][b]IMO2[/b] Establish if there exists a finite set $M$ of points in space, not all situated in the same plane, so that for any straight line $d$ which contains at least two points from M there exists another straight line $d'$, parallel with $d,$ but distinct from $d$, which also contains at least two points from $M$.\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2026262#p2026262]7. (POL 3)[/url] Given a tetrahedron $ABCD$, let $x = AB \\cdot CD$, $y = AC \\cdot BD$, and $z = AD \\cdot  BC$. Prove that there exists a triangle with edges $x, y, z.$\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2026267#p2026267]8. (ROM 1)[/url] Prove that there are exactly $\\binom{k}{[k/2]}$ arrays $a_1, a_2, \\ldots , a_{k+1}$ of nonnegative integers such that $a_1 = 0$ and $|a_i-a_{i+1}| = 1$ for $i = 1, 2, \\ldots , k.$\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2026273#p2026273]9. (ROM 2)[/url] Let $Ox, Oy, Oz$ be three rays, and $G$ a point inside the trihedron $Oxyz$. Consider all planes passing through $G$ and cutting $Ox, Oy, Oz$ at points $A,B,C$, respectively. How is the plane to be placed in order to yield a tetrahedron $OABC$ with minimal perimeter ?\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=357934#p357934]10. (SWE 3)[/url] [b]IMO6[/b] Let $a_1, \\ldots, a_n$ be $n$ positive numbers and $0 < q < 1.$ Determine $n$ positive numbers $b_1, \\ldots, b_n$ so that:\n\n[i]a.)[/i] $ a_{k} < b_{k}$ for all $k = 1, \\ldots, n,$ \n[i]b.)[/i] $q < \\frac{b_{k+1}}{b_{k}} < \\frac{1}{q}$ for all $k = 1, \\ldots, n-1,$\n[i]c.)[/i] $\\sum \\limits^n_{k=1} b_k < \\frac{1+q}{1-q} \\cdot \\sum \\limits^n_{k=1} a_k.$\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=357937&#p357937]11. (SWE 4)[/url] [b]IMO3[/b] Determine the minimum value of $a^{2} + b^{2}$ when $(a,b)$ traverses all the pairs of real numbers for which the equation \\[ x^{4} + ax^{3} + bx^{2} + ax + 1 = 0 \\] has at least one real root.\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2026275#p2026275]12. (SWE 6)[/url]Consider the two square matrices\n\\[A=\\begin{bmatrix} +1 & +1 &+1& +1 &+1 \\\\+1 &+1 &+1&-1 &-1 \\\\ +1 &-1&-1 &+1& +1 \\\\ +1 & -1 & -1 & -1 & +1 \\\\ +1 &+1&-1 &+1&-1  \\end{bmatrix} \\quad \\text{ and } \\quad B=\\begin{bmatrix} +1 & +1 &+1& +1 &+1 \\\\+1 &+1 &+1&-1 &-1 \\\\ +1 &+1&-1& +1&-1 \\\\ +1 &-1& -1& +1& +1 \\\\ +1 & -1& +1&-1 &+1 \\end{bmatrix}\\]\n\nwith entries $+1$ and $-1$. The following operations will be called elementary:\n\n(1) Changing signs of all numbers in one row;\n\n(2) Changing signs of all numbers in one column;\n\n(3) Interchanging two rows (two rows exchange their positions);\n\n(4) Interchanging two columns.\n\nProve that the matrix $B$ cannot be obtained from the matrix $A$ using these operations.\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2026289#p2026289]13. (YUG 4)[/url] Find the sphere of maximal radius that can be placed inside every tetrahedron that has all altitudes of length greater than or equal to $1.$\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=357950&#p357950]14. (YUG 5)[/url] [b]IMO4[/b] A soldier needs to check if there are any mines in the interior or on the sides of an equilateral triangle $ABC.$ His detector can detect a mine at a maximum distance equal to half the height of the triangle. The soldier leaves from one of the vertices of the triangle. Which is the minimum distance that he needs to traverse so that at the end of it he is sure that he completed successfully his mission?\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2026280#p2026280]15. (CUB 1)[/url] Prove that for all $n \\in  \\mathbb N$ the following is true:\n\\[2^n \\prod_{k=1}^n \\sin \\frac{k \\pi}{2n+1} = \\sqrt{2n+1}\\]\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2026281#p2026281]16. (CUB 2)[/url] Given $a, \\theta \\in \\mathbb R, m \\in \\mathbb N$, and $P(x) = x^{2m}- 2|a|^mx^m \\cos \\theta +a^{2m}$, factorize $P(x)$ as a product of $m$ real quadratic polynomials.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=357910&#p357910]17. (POL 1)[/url] [b]IMO5[/b] $G$ is a set of non-constant functions $f$. Each $f$ is defined on the real line and has the form $f(x)=ax+b$ for some real $a,b$. If $f$ and $g$ are in $G$, then so is $fg$, where $fg$ is defined by $fg(x)=f(g(x))$. If $f$ is in $G$, then so is the inverse $f^{-1}$. If $f(x)=ax+b$, then $f^{-1}(x)= \\frac{x-b}{a}$. Every $f$ in $G$ has a fixed point (in other words we can find $x_f$ such that $f(x_f)=x_f$. Prove that all the functions in $G$ have a common fixed point."
}
{
  "Tag": [
    "AMC",
    "AIME"
  ],
  "Problem": "Given $x$, $y$, and $z$ are real numbers that satisfy\r\n$x=\\sqrt{y^{2}-\\frac{1}{16}}+\\sqrt{z^{2}-\\frac{1}{16}}$\r\n$y=\\sqrt{z^{2}-\\frac{1}{25}}+\\sqrt{x^{2}-\\frac{1}{25}}$\r\n$z=\\sqrt{x^{2}-\\frac{1}{36}}+\\sqrt{y^{2}-\\frac{1}{36}}$\r\nand that $x+y+z=\\frac{m}{\\sqrt{n}}$, where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime, find $m+n$.",
  "Solution_1": "[hide=\"Solution\"]The equations imply $x,y,z>0$. From the first equation we get\n\n\\begin{eqnarray*}x-\\sqrt{y^{2}-{1\\over 16}}=\\sqrt{z^{2}-{1\\over 16}&\\iff& x^{2}-2x\\sqrt{y^{2}-{1\\over 16}}+y^{2}-{1\\over 16}=z^{2}-{1\\over 16}\\\\ &\\iff& 2x\\sqrt{y^{2}-{1\\over 16}}=x^{2}+y^{2}-z^{2}}\\end{eqnarray*}\n\nThat gives\n\n$2x\\left(y+\\sqrt{y^{2}-{1\\over 16}}\\right)=x^{2}+2xy+y^{2}-z^{2}=(x+y+z)(x+y-z)$\n$2x\\left(y-\\sqrt{y^{2}-{1\\over 16}}\\right)=z^{2}-x^{2}+2xy-y^{2}=(-x+y+z)(x-y+z)$\n\nMultiplying those two we get\n\n$4x^{2}\\cdot{1\\over 16}=(x+y+z)(x+y-z)(x-y+z)(-x+y+z)\\qquad(*)$\n\nSimilarly we find\n\n$4y^{2}\\cdot{1\\over 25}=(x+y+z)(x+y-z)(x-y+z)(-x+y+z)$\n$4z^{2}\\cdot{1\\over 36}=(x+y+z)(x+y-z)(x-y+z)(-x+y+z)$\n\nHence ${x^{2}\\over 4}={4y^{2}\\over 25}={z^{2}\\over 9}=t^{2}$ for some real positive $t$, giving\n\n$x=2t, y={5t\\over 2}, z=3t$.\n\nPlugging that into $(*)$ we get\n\n\\begin{eqnarray*}t^{2}&=&{15t\\over 2}\\cdot{3t\\over 2}\\cdot{5t\\over 2}\\cdot{7t\\over 2}\\\\ &=& \\frac{225\\cdot 7t^{4}}{16}\\end{eqnarray*}\n\nwhich gives $t^{2}=\\frac{16}{225\\cdot 7}\\implies t=\\frac{4}{15\\sqrt{7}}$\n\nNow $x+y+z=2t+{5t\\over 2}+3t={15t\\over 2}={15\\over 2}\\cdot\\frac{4}{15\\sqrt{7}}={2\\over\\sqrt{7}}$\n\nHence $x+y+z={2\\over\\sqrt{7}}\\implies m+n=9$[/hide]",
  "Solution_2": "[quote=\"Totally Zealous\"]Given $x$, $y$, and $z$ are real numbers that satisfy\n$x=\\sqrt{y^{2}-\\frac{1}{16}}+\\sqrt{z^{2}-\\frac{1}{16}}$\n$y=\\sqrt{z^{2}-\\frac{1}{25}}+\\sqrt{x^{2}-\\frac{1}{25}}$\n$z=\\sqrt{x^{2}-\\frac{1}{36}}+\\sqrt{y^{2}-\\frac{1}{36}}$\nand that $x+y+z=\\frac{m}{\\sqrt{n}}$, where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime, find $m+n$.[/quote]\r\n\r\ninterpreting it geometrically might give you a shorter solution :wink:",
  "Solution_3": "[quote=\"D.P.L\"][quote=\"Totally Zealous\"]Given $x$, $y$, and $z$ are real numbers that satisfy\n$x=\\sqrt{y^{2}-\\frac{1}{16}}+\\sqrt{z^{2}-\\frac{1}{16}}$\n$y=\\sqrt{z^{2}-\\frac{1}{25}}+\\sqrt{x^{2}-\\frac{1}{25}}$\n$z=\\sqrt{x^{2}-\\frac{1}{36}}+\\sqrt{y^{2}-\\frac{1}{36}}$\nand that $x+y+z=\\frac{m}{\\sqrt{n}}$, where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime, find $m+n$.[/quote]\n\ninterpreting it geometrically might give you a shorter solution :wink:[/quote]\r\nIndeed, this is #15 of the 2006 AIME, and a solution is given in the math jam.",
  "Solution_4": "Then give a geometric interpretation or post a link to the mathjam solution.",
  "Solution_5": "[quote=\"Farenhajt\"]Then give a geometric interpretation or post a link to the mathjam solution.[/quote]\r\n[url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=145#fifteen]Click this link for the solution[/url]"
}
{
  "Tag": [],
  "Problem": "There are eight furlongs in a mile. There are two weeks in a\nfortnight. The British cavalry traveled 2800 furlongs in a\nfortnight. How many miles per day did the cavalry average?",
  "Solution_1": "If they go $ 2800$ furlongs in $ 14$ days, then in one day they go $ 200$ furlongs.\r\n\r\nSince $ 200$ furlongs is $ 25$ miles, the armyu goes $ 25$ miles a day.",
  "Solution_2": "[quote=\"BOGTRO\"]\nSince $ 200$ furlongs is $ 25$ miles, the armyu goes $ 25$ miles a day.[/quote]\r\n\r\nThis is because $ \\frac{200}{8} \\equal{} 25$"
}
{
  "Tag": [
    "geometry",
    "trapezoid"
  ],
  "Problem": "What is the area in square inches of the pentagon shown?\n\n[asy]draw((0,0)--(8,0)--(8,18)--(2.5,20)--(0,12)--cycle);\nlabel(\"8''\",(1.3,16),NW);\nlabel(\"6''\",(5.2,19),NE);\nlabel(\"18''\",(8,9),E);\nlabel(\"8''\",(4,0),S);\nlabel(\"12''\",(0,6),W);\ndraw((1,0)--(1,1)--(0,1));\ndraw((7,0)--(7,1)--(8,1));[/asy]",
  "Solution_1": "Assuming that the top angle is a right angle, we can easily partition it to find an area of $ \\frac12\\cdot6\\cdot8\\plus{}\\frac12(12\\plus{}18)\\cdot8\\equal{}24\\plus{}120\\equal{}\\boxed{144}$.",
  "Solution_2": "Think of this pentagon as a triangle on top of a trapezoid. The area of the pentagon would be the area of the two parts added together. The area of the triangle is $ \\frac {1}{2}(6)(8) \\equal{} 24$. The area of the trapezoid is $ \\frac {1}{2}(12 \\plus{} 18)(8) \\equal{} 120$. Adding these two areas up, we find that the area of the pentagon is $ 24 \\plus{} 120 \\equal{} \\fbox{144}$. \r\n\r\nEDIT: Argggggh. Got beaten again."
}
{
  "Tag": [],
  "Problem": "Find all integer numbers $b$ and $d$ satisfying\r\n\\[ b^2 - d^2 +bd + 1 = 0 \\]\r\n\r\n[color=red]Edited by moderator[/color]",
  "Solution_1": "Crucial observation is probably that if (a,b) is solution then so is (a+b,a+2b) and so is (2a-b,b-a).\r\n\r\nif (a,b) is solution with $a,b > 0$ then it'a easy to see that $a < b < 2a$, so starting with (a,b) solution, it's possible to build a strictly \"smaller\" solution, by infinite descent like argument, we end with (0,1).\r\n\r\nSimilar argument if $a < 0$.\r\n\r\nSo solutions are : \r\n- $(x_n,y_n)$  with $x_0 = 1, y_1=1, x_{n+1} = x_n+y_n, y_{n+1} = x_n + 2 y_n$\r\n- $(-x_{n+1},y_n)$  \r\n- $(x_{n+1},-y_n)$  \r\n- $(-x_n,-y_n)$\r\n\r\n\r\n---------\r\nHigh Algebra is an art of counting to 20 without taking off shoes or mittens  ;)"
}
{
  "Tag": [
    "MIT",
    "college",
    "LaTeX"
  ],
  "Problem": "Hallo, kann mir bitte jmd. sagen, wie sich die Aufgabe 441144 l\u00f6sen l\u00e4sst? :blush: \r\n\r\nMit Q(n) sei die Quersumme der nat\u00fcrlichen Zahl n bezeichnet. Man beweise, dass dann Q(Q(Q(2005^2005)))=7 gilt.\r\n\r\n(Ich habe hier auf Latex verzichtet, weil man 2005^2005 ja auch gerade noch so versteht  :wink:  )\r\n\r\nDanke!",
  "Solution_1": "Hat das keiner gelesen oder will das keiner l\u00f6sen?",
  "Solution_2": "Sorry, da\u00df ich erst jetzt antworte - hatte beim ersten fl\u00fcchtigen Lesen deines Posts nur das Wort \"Quersumme\" bemerkt und mich gleich weggeklickt...\r\n\r\nHab die Aufgabe selber mal in http://www.mathlinks.ro/Forum/viewtopic.php?t=36891 gepostet, die eigentliche L\u00f6sungsidee befindet sich auf http://www.mathlinks.ro/Forum/viewtopic.php?t=31409 .\r\n\r\n  gr\u00fc\u00dfe,\r\n  darij\r\n\r\n[b]EDIT:[/b] 4444 Posts... wenn das nicht zu der Aufgabe passt :D",
  "Solution_3": "okay, danke.\r\n(wieso klicktst du gleich weg, wenn du Quersumme liest)?",
  "Solution_4": "[quote=\"Blender\"](wieso klicktst du gleich weg, wenn du Quersumme liest)?[/quote]\r\n\r\nSind meist die h\u00e4sslichsten Aufgaben von Wettbewerben...\r\nnaja, Zahlentheorie in Wettbewerben ist eh meist h\u00e4sslich.\r\n\r\n  darij",
  "Solution_5": "Zahlentee in IMO: meist recht gut.\r\nZahlentee in QEDMO: no comment ;)\r\nZahlentee in DeMO: spam.\r\nZahlentee auf ML: je nach Author oft auch sehr nett.\r\n\r\nIch hab diesen Thread erst heute \u00fcberhaupt bemerkt...",
  "Solution_6": "[quote=\"ZetaX\"]Zahlentee in IMO: meist recht gut.[/quote]\n\nFindest du da\u00df die Zahlentee letztes Jahr gut war?\n\n[quote=\"ZetaX\"]Zahlentee in QEDMO: no comment ;)[/quote]\n\n... ;)\n\n[quote=\"ZetaX\"]Zahlentee auf ML: je nach Author oft auch sehr nett.[/quote]\r\n\r\nBlo\u00df da\u00df die netten Zahlenteeaufgaben auf ML meist \u00fcber Wettbewerbsniveau angesiedelt sind.\r\n\r\n  darij"
}
{
  "Tag": [],
  "Problem": "The question is:\r\n\r\nAt the same time, two buses leave a depot and travel in opposite directions on a straight road. The first bus averages 5 mi/h more than the second. If they are 142.5 mi apart after 1 1/2 hours, how fast is each bus going?\r\n\r\nOkay, I realize that the distance is the same, so the equations can have an equal sign in between.\r\n\r\nAlso one is going 5 mi more per hour than the other, so x+5. I'm not sure where the 1.5 hours comes in and do you need to have it on both sides of the equation. Since the buses travel in miles per HOUR, and it's an hour and a half, do you need to add 30 mins to get the answer? \r\n\r\nI'm confused! Please help!  He needs this by Monday. I have others too.\r\n\r\n\r\nTwo Planes leave simultaneously from an airport, one flying East and the other West. They are 870 miles apart after 3/4 hour.  If the Eastbound plane averages 120 mi/h more than the Westbound lane, at what rate is each plane flying. I know we need to know the rate, and the distance is the same, so the equations should be equal to each other.  I am thrown off again by the 3/4 hour.  Help again!\r\n\r\nThanks so much!",
  "Solution_1": "You already solved it....\r\n\r\nD=RT...plug in values for D, R, and T.  If it's unknown, use a variable.",
  "Solution_2": "Maybe I'm an idiot, but I still don't understand how the 1.5 hours and the 3/4 hours come into play.  Can you help me with a table or how to do it? My son needs to show his work with a table that has RxT=D on the top and I'm not sure when to set the distances as equal or when to combine numbers on one side... it's been 30 yrs since I did algebra. :)\r\n\r\nThanks!",
  "Solution_3": "Oh dang...tables.  I never learned to solve it that way...\r\n\r\nI can help you with the 1 1/2 and the 3/4 hours part though.\r\n\r\nIf you are traveling at $ x$ miles per [b]hour[/b], then use hours in your equation.  So it would be D=x*3/4 or d=x*3/2.",
  "Solution_4": "Thank you.... the kid has to use the tables, the teacher demands it, and he's frustrated and doesn't know where to put what... the way he learned doesn't make any sense to me... but I guess this is the way it is. I just can't do it without just reasoning it out in my head.  He has to have all this funky stuff going on. Thanks for your help. :)",
  "Solution_5": "I remember this from weighted averages. (The spacing hee is a little off, but you should be able to read it by lining up the signs.)\r\n\r\n[u]D=r*t[/u]\r\n1.5x=x*1.5 (bus 1)\r\n142.5-1.5x=(x-5)*1.5 (bus 2)\r\n-------------------------------\r\n142.5=(2x-5)*1.5 (add equations)\r\n\r\n142.5=3x-7.5\r\n135=3x\r\n45=x\r\n\r\nSo bus 1 is traveling at 45 mph, and since x-5=45-5=40, bus 2 is traveling at 40 mph.\r\n\r\nTry doing the second one on your own. (And if the time throws you off, think of this. When you buy apples, and it says $ \\$$3/lb, does that mean you have to buy exactly 1lb, or 2lb, etc? No! Just use a decimal instead. The same applies to time. Notice how I converted 1 1/2 hours to 1.5.)\r\n\r\n[hide=\"#2\"]\n[hide=\"dont click here unless you tried to solve it\"]\nOkay.\n\n[u]D=r*t[/u]\n.75x=x*.75 (east)\n870-.75x=(x-120)*.75 (west)\n-------------------------\n870=(2x-120)*.75 (adding)\n\n870=1.5x-90\n960=1.5x\n1440=x\n\nSo the east plane is going at 1440 mph, and since x-120=1440-120=1320, the west one is going at 1320 mph.[/hide][/hide]",
  "Solution_6": "[quote=\"lifeinhim61\"]Maybe I'm an idiot, but I still don't understand how the 1.5 hours and the 3/4 hours come into play.  Can you help me with a table or how to do it?[b] My son needs to show his work with a table that has RxT=D on the top and I'm not sure when to set the distances as equal or when to combine numbers on one side... it's been 30 yrs since I did algebra. :)[/b]\n\nThanks![/quote]\r\n\r\nThey need tables....",
  "Solution_7": "Tables aren't a good idea, in general.  They just waste time.",
  "Solution_8": "... \"the kid has to use the tables, the teacher demands it\". Sorry for your kid! Lots of teachers kill creativity!",
  "Solution_9": "[quote=\"lifeinhim61\"]Thank you.... the kid has to use the tables, the teacher demands it, and he's frustrated and doesn't know where to put what... the way he learned doesn't make any sense to me... but I guess this is the way it is. I just can't do it without just reasoning it out in my head.  He has to have all this funky stuff going on. Thanks for your help. :)[/quote]\r\nif its not a HUGE assignment just do it the equation way cause the tables are unnecessary work.",
  "Solution_10": "Well, I made \"tables\", but they're in plain text...\r\n\r\nEssentially, just line up the signs, and then insert lines where they belong...(I don't really know how to do it here...)",
  "Solution_11": "Maybe like for the 2nd question:\r\n\r\n\\begin{tabular}{|c|c|}\r\nSymbol &Meaning\\\\\r\n\\hline x&Speed of westbound plane\\\\\r\n\\hline P_e&Eastbound plane\\\\\r\n\\hline P_w&Westbound plane\\\\\r\n\\hline\r\n\\end{tabular}\r\n\r\n\\begin{tabular}{|c|c|c\\parallel{}c|}\\hline Time& P_e & P_w &Distance Apart\\\\\r\n\\hline 0 min& 0 & 0 & 0\\\\\r\n\\hline 15 min& (x+120)/4 & x/4 & (2x+120)/4 \\\\\r\n\\hline 30 min& (x+120)/4 +(x+120)/2 & x/4+x/2 & (6x+360)/4\\\\\r\n\\hline 45 min& (x+120)/4 +(x+120)/2+3(x+120)/4 & x/4+x/2+3x/4 & (12x+720)4\\\\\r\n\\hline \\end{tabular}\r\n\r\n$ \\dfrac{12x+720}{4}=3x+180=870$\r\n\\begin{eqnarray*}370&=&3x+180\\\\290&=&3x\\\\\\dfrac{290}{3}&=&x\\end{eqnarray*}\r\n\r\nI probably got that wrong but maybe something like that?",
  "Solution_12": "[quote=\"lifeinhim61\"]\n\nI'm confused! Please help!  He needs this by Monday. I have others too.\n\n\n[/quote]\r\n\r\nGuess what?  This was posted last Saturday..."
}
{
  "Tag": [
    "geometry",
    "calculus",
    "3D geometry",
    "probability",
    "circumcircle"
  ],
  "Problem": "These are some medium and challenging target round questions.\r\n\r\n1. Let ABDE be a rectangle. Let C be a point on BD . If BC=11 cm, CD=9 cm and ACE is a right angle, what is the number of centimeters, to the nearest whole number, in the length of CE ? \r\n\r\n2.  A school organization consists of 5 teachers, 7 parents and 8 students. A subcommittee of this group must be formed by choosing 2 teachers, 3 parents and 5 students. How many different subcommittees can be formed?\r\n\r\n3. An equilateral triangle is inscribed in a circle with diameter 9 cm . How many square centimeters are in the area of the triangle? Express your answer as a decimal to the nearest tenth.\r\n\r\n4. At the grading of the Advanced Placement Calculus exams, 390 math teachers graded 120000 tests in 50 hours. Each test had 6 problems. What is the average number of minutes it took one teahcer to grade 25 problems? Express your answer to the nearest minute.\r\n\r\n5. Two number cubes (number cube=dice), each with the digits 1-6 on the six faces, are rolled. What is the probability that the product of the numbers on the top faces will be greater than 8 ? Express your answer as a common fraction in lowest terms. \r\n\r\nGood luck! :)",
  "Solution_1": "[quote=\"goldendomer\"]These are some medium and challenging target round questions.\n\n1. Let ABDE be a rectangle. Let C be a point on BD . If BC=11 cm, CD=9 cm and ACE is a right angle, what is the number of centimeters, to the nearest whole number, in the length of CE ? \n\n2.  A school organization consists of 5 teachers, 7 parents and 8 students. A subcommittee of this group must be formed by choosing 2 teachers, 3 parents and 5 students. How many different subcommittees can be formed?\n\n3. An equilateral triangle is inscribed in a circle with diameter 9 cm . How many square centimeters are in the area of the triangle? Express your answer as a decimal to the nearest tenth.\n\n4. At the grading of the Advanced Placement Calculus exams, 390 math teachers graded 120000 tests in 50 hours. Each test had 6 problems. What is the average number of minutes it took one teahcer to grade 25 problems? Express your answer to the nearest minute.\n\n5. Two number cubes (number cube=dice), each with the digits 1-6 on the six faces, are rolled. What is the probability that the product of the numbers on the top faces will be greater than 8 ? Express your answer as a common fraction in lowest terms. \n\nGood luck! :)[/quote]\r\n\r\n[hide=\"1\"]$CE=\\sqrt{180}$\nAE=BD=20. AB=DE=x. BC=11, CD=9. AC=$\\sqrt{x^2+121}$, CE=$\\sqrt{x^2+81}$. \n$x^2+121+x^2+81=20^2$\n$2x^2+202=400$\n$x^2+101=200$\n$x=\\sqrt{99}$\n$CE=\\sqrt{99+81}$\n$CE=\\sqrt{180}$\n[/hide]\n\n[hide=\"2\"]$binom{5}{2}\\cdot \\binom{7}{3} \\cdot \\binom{8}{5}=10\\cdot 35 \\cdot56 = 19600$(?)[/hide]\n[hide=\"3\"]\nsince circumradius = $\\frac{abc}{4A}$, $\\frac{4a}{\\sqrt{3}}=4.5$, $a=\\frac{9\\sqrt{3}}{8}$, area is $\\frac{243\\sqrt{3}}{256}$[/hide]\n\n[hide=\"4\"]total = 720000 problems, each teacher grades 480/13 problems in 1 hour, $\\frac{25}{x}=\\frac{\\frac{480}{13}}{60}$, $x=40.625$, nearest minute = 41[/hide]\n[hide=\"5\"]\n# less than  or = 8 are 16, greater than eight are 36-16=20, 20/36=$\\frac{5}{9}$[/hide]"
}
{
  "Tag": [
    "search",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Consider the sequence of numbers (u_n) defined by $ u_n=n^{2^n} $ for all n=1,2,...\r\n Put $ x_n = \\frac{1}{u_1}+\\frac{1}{u_2} + ... + \\frac{1}{u_n} $ \r\n PRove that the sequence (x_n) has a limit when n tends to infinity and the limit is an irrational",
  "Solution_1": "If you put $t_1=u_1,t_n=\\frac{u_n}{u_{n-1}},\\ \\forall n\\ge 2$, then we see that $(t_n)_n$ is strictly increasing, and it's a well known fact that $\\frac 1{t_1}+\\frac 1{t_1t_2}+\\ldots$ is irrational. The proof is pretty much the same as the usual proof that $e=1+\\frac 1{1!}+\\frac 1{2!}+\\ldots$ is irrational.",
  "Solution_2": "[quote=\"grobber\"]it's a well known fact that $\\frac 1{t_1}+\\frac 1{t_1t_2}+\\ldots$ is irrational.[/quote]\r\nOnly if $t_j$ are integers, which is not the case here  ;)",
  "Solution_3": "Oh, sorry about that :). Now it looks to me like this number contains arbitrarily long runs of $0$'s in its decimal expansion, but I'm too lazy and/or tired to complete the estimations. Sorry if this is, again, wrong :).",
  "Solution_4": "I think this problem was solved before by Moubinool but I can't remember the topic",
  "Solution_5": "Now, Let's try it again !",
  "Solution_6": "No solution ? :(",
  "Solution_7": "[quote=\"romano\"]No solution ? :([/quote]\r\n\r\nromano,\r\n\r\nTo ask for the solution of a problem three times in a row sometimes is considered spamming, and by the way, to be that insistent probably will have the oposite efect, that is, people just stop looking at the topic because it does not have much sence to see someone asking once and again the same question. Be patient, probably someone will solve it eventually.\r\n\r\nThe convergence of the series is very trivial (even when nobody has mention it). It is obvius that $2^n<n^{2^n}$ for $n>1$, then, by comparation test, or basically any test that you can imagine, it trivially convergence.\r\n\r\nThe irrationality does not seem to be easy at all. Probably you can prove (or attempt to) that if $x_n\\rightarrow x$ then \r\n$\\left|x-\\frac{p}{q}\\right|<\\frac{1}{q^\\mu}$ have no solution for $p$ and $q$ integers if $\\mu<2$, and by the irrationality measure theorem it turns to be irrational automatically.\r\n\r\nBy the way, about the Moubinool post that you mention, well, I couldn't find it, but there are search tools, just in case!\r\n\r\nBest,",
  "Solution_8": "The following lemma should help.  Let $x$ be a real number, and let $p$ and $q$ be integers such that $p \\ne qx$.  Then x is not a rational number with denominator at most $\\frac{1}{|p - qx|}$.\r\n\r\nFor this problem, let $x$ be the limit.  Let $n$ be a large integer.  Let $p$ and $q$ be the numerator and denominator of $x_n$.  Then an estimate shows that $|p - qx|$ approaches 0.  By the lemma, $x$ must be irrational.\r\n\r\nThe basic (paradoxical) idea: To show that a number is irrational, show that it is very close (but not equal) to a rational number.  That works because two different rational numbers (with bounded denominators) can't be too close to each other.",
  "Solution_9": "[quote=\"djimenez\"]\n\n\n\nBy the way, about the Moubinool post that you mention, well, I couldn't find it, but there are search tools, just in case!\n[/quote]\r\n\r\nProbably the post is in solved problem, I did not find it  :("
}
{
  "Tag": [],
  "Problem": "The spindle speed of a CD player is 300 revolutions per minute.  How many revolutions does the spindle make while playing a song that lasts exactly 4 minutes?",
  "Solution_1": "[hide]1200[/hide]",
  "Solution_2": "[hide]1200[/hide]",
  "Solution_3": "[quote=\"MCrawford\"]The spindle speed of a CD player is 300 revolutions per minute.  How many revolutions does the spindle make while playing a song that lasts exactly 4 minutes?[/quote]\r\n[hide=\"solution\"]\n300 x 4 = [b]1200[/b]\n[/hide]",
  "Solution_4": "[hide]\n300 revolutions per minute. 4 minutes.\n300*4=1200[/hide]",
  "Solution_5": "[hide]1200\nu can also use proportions\nwhen it gets harder![/hide]",
  "Solution_6": "Well I think technically you are using proportions no matter how you solve it.",
  "Solution_7": "Yep, you just don't think that way because you think it's really easy.",
  "Solution_8": "[quote=\"MCrawford\"]The spindle speed of a CD player is 300 revolutions per minute.  How many revolutions does the spindle make while playing a song that lasts exactly 4 minutes?[/quote]\r\n\r\nBy the way, you know that it is not true that the CDs always revolute to the same speed. Look at it. It goes faster at the beginning of the CD than at the end!  :D .",
  "Solution_9": "Well, now find the formula for that, then find the answer :lol:  :lol:  :D  :D  :D :lol: lol",
  "Solution_10": "[hide]\n3*400\n1200[/hide]"
}
{
  "Tag": [
    "inequalities",
    "function",
    "calculus",
    "derivative",
    "inequalities unsolved"
  ],
  "Problem": "[color=red][size=200]Let a,b,c non negatve numbers with a+b+c=3 and p>=1\nprove that                            (1+a^p-a)(1+b^p-b)(1+c^p-c)>=1[/size][/color]",
  "Solution_1": "If I am not wrong, it belongs to Zhaobin and is unsolved in VASC's book.",
  "Solution_2": "I will try to solve this problem:\r\nlet us consider function:\r\n$f(x)=1+x^{p}-x$ where $p\\geq 1$ minimal valua hold's when $x=(\\frac{1}{p})^{\\frac{1}{p-1}}$ therefore we have \r\n$g=f(x)f(y)f(z)$ where $x+y+z=3$ for wich we know when hol'd minimal for each function $f(x),f(y),f(z)$(here I must was more correct )  therefore (I think somewhere I have seen this statment, if you want I'll try to prove it, it's not hard) minimal hold's when $x=y=z$ when $p=1$ after is obviusly.",
  "Solution_3": "Let $f(a,\\ b,\\ c,\\ \\lambda)=(1+a^{p}-a)(1+b^{p}-b)(1+c^{p}-c)+\\lambda (a+b+c-3)$,\r\n\r\nUsing the method of Lagrange multipliers, taking partial derivative this for $a,\\ b,\\ c,\\ \\lambda$ and set those as 0, we obtain $a=b=c=1$.",
  "Solution_4": "[quote=\"kunny\"]Let $f(a,\\ b,\\ c,\\ \\lambda)=(1+a^{p}-a)(1+b^{p}-b)(1+c^{p}-c)+\\lambda (a+b+c-3)$,\n\nUsing the method of Lagrange multipliers, taking partial derivative this for $a,\\ b,\\ c,\\ \\lambda$ and set those as 0, we obtain $a=b=c=1$.[/quote]\r\nTry it, kunny! :wink:"
}
{
  "Tag": [
    "floor function",
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "sr for asking but I searched the solution of this 4th problem(4th round) in this linkhttp://www.mathlinks.ro/Forum/resources.php?c=80&cid=101&year=1998 but have not found ... can anyone solve it >\"< ?",
  "Solution_1": "[quote=\"long14893\"]sr for asking but I searched the solution of this 4th problem(4th round) in this linkhttp://www.mathlinks.ro/Forum/resources.php?c=80&cid=101&year=1998 but have not found ... can anyone solve it >\"< ?[/quote]\r\nA lemma:\r\n$ \\frac{1}{n} > \\frac{1}{10n\\plus{}0} \\plus{} \\frac1{10n\\plus{}1} \\plus{} \\frac{1}{10n\\plus{}2} \\plus{} ... \\plus{} \\frac1{10n\\plus{}9}$ (Which is quite obvious...)\r\n\r\n$ \\sum_{i} \\frac{1}{n_i}$ can reach $ 1 \\plus{} \\frac12 \\plus{} .. \\plus{} \\frac19$ when $ n_i\\equal{}i, i\\equal{}1,2,3..,9$. Suppose another configuration $ m_i$ is such that $ \\sum_{i} \\frac{1}{m_i} > \\sum_{i} \\frac{1}{n_i}$ with the same constraints, and that the largest $ m_i$ contains no more than $ k \\ge 2$ digits. Further more let $ m_i$ be that sequence with largest $ \\sum_{i} \\frac{1}{m_i}$. (If there are more than one then pick an abritary)\r\n\r\nLet $ m_r$ be the largest number, and hence the number with the most digits.(Obviously $ m_r\\ge10$) Let $ t \\equal{} \\lfloor \\frac{m_r}{10} \\rfloor$. Then there is at most $ 10$ numbers from $ m_i$(Including $ m_r$) on the form $ \\overline{td}$, where $ d$ is a digit. The sum of the reciprocicals of these are strichtly less than $ \\frac{1}{t}$. Hence there exist another sequence where all the numbers on the form $ \\overline{td}$ is replaced with $ t$, which still contains no element with more than $ k$ digits, but whos sum of reciprocicals is strictly bigger than $ m_i$'s. Since there are only finitly many numbers with $ k$ digits - less than $ 10^{k\\plus{}1}$ at least - we obtain a contradiction. Hence there doesn't exist such a sequence.\r\n\r\nNow we have proved the problem when $ n_i$ has finitly many elements. How can we expand it to infinitly many elements?\r\nFurther more: The proof get's a bit clumsy.. Using $ \\frac{1}{n} \\ge \\frac{1}{\\overline{n0}}\\plus{}\\frac{1}{\\overline{n2}}\\plus{}...\\plus{}\\frac{1}{\\overline{n9}}$ would do the job, but would be hard to argument with, if there were infinitly many elements..",
  "Solution_2": "[quote=\"Mathias_DK\"][quote=\"long14893\"]sr for asking but I searched the solution of this 4th problem(4th round) in this linkhttp://www.mathlinks.ro/Forum/resources.php?c=80&cid=101&year=1998 but have not found ... can anyone solve it >\"< ?[/quote]\nA lemma:\n$ \\frac {1}{n} > \\frac {1}{10n \\plus{} 0} \\plus{} \\frac1{10n \\plus{} 1} \\plus{} \\frac {1}{10n \\plus{} 2} \\plus{} ... \\plus{} \\frac1{10n \\plus{} 9}$ (Which is quite obvious...)\n\n$ \\sum_{i} \\frac {1}{n_i}$ can reach $ 1 \\plus{} \\frac12 \\plus{} .. \\plus{} \\frac19$ when $ n_i \\equal{} i, i \\equal{} 1,2,3..,9$. Suppose another configuration $ m_i$ is such that $ \\sum_{i} \\frac {1}{m_i} > \\sum_{i} \\frac {1}{n_i}$ with the same constraints, and that the largest $ m_i$ contains no more than $ k \\ge 2$ digits. Further more let $ m_i$ be that sequence with largest $ \\sum_{i} \\frac {1}{m_i}$. (If there are more than one then pick an abritary)\n\nLet $ m_r$ be the largest number, and hence the number with the most digits.(Obviously $ m_r\\ge10$) Let $ t \\equal{} \\lfloor \\frac {m_r}{10} \\rfloor$. Then there is at most $ 10$ numbers from $ m_i$(Including $ m_r$) on the form $ \\overline{td}$, where $ d$ is a digit. The sum of the reciprocicals of these are strichtly less than $ \\frac {1}{t}$. Hence there exist another sequence where all the numbers on the form $ \\overline{td}$ is replaced with $ t$, which still contains no element with more than $ k$ digits, but whos sum of reciprocicals is strictly bigger than $ m_i$'s. Since there are only finitly many numbers with $ k$ digits - less than $ 10^{k \\plus{} 1}$ at least - we obtain a contradiction. Hence there doesn't exist such a sequence.\n\nNow we have proved the problem when $ n_i$ has finitly many elements. How can we expand it to infinitly many elements?\nFurther more: The proof get's a bit clumsy.. Using $ \\frac {1}{n} \\ge \\frac {1}{\\overline{n0}} \\plus{} \\frac {1}{\\overline{n2}} \\plus{} ... \\plus{} \\frac {1}{\\overline{n9}}$ would do the job, but would be hard to argument with, if there were infinitly many elements..[/quote]thx ... nice solution :D",
  "Solution_3": "[quote=\"Mathias_DK\"][quote=\"long14893\"]sr for asking but I searched the solution of this 4th problem(4th round) in this linkhttp://www.mathlinks.ro/Forum/resources.php?c=80&cid=101&year=1998 but have not found ... can anyone solve it >\"< ?[/quote]\nA lemma:\n$ \\frac {1}{n} > \\frac {1}{10n \\plus{} 0} \\plus{} \\frac1{10n \\plus{} 1} \\plus{} \\frac {1}{10n \\plus{} 2} \\plus{} ... \\plus{} \\frac1{10n \\plus{} 9}$ (Which is quite obvious...)\n\n$ \\sum_{i} \\frac {1}{n_i}$ can reach $ 1 \\plus{} \\frac12 \\plus{} .. \\plus{} \\frac19$ when $ n_i \\equal{} i, i \\equal{} 1,2,3..,9$. Suppose another configuration $ m_i$ is such that $ \\sum_{i} \\frac {1}{m_i} > \\sum_{i} \\frac {1}{n_i}$ with the same constraints, and that the largest $ m_i$ contains no more than $ k \\ge 2$ digits. Further more let $ m_i$ be that sequence with largest $ \\sum_{i} \\frac {1}{m_i}$. (If there are more than one then pick an abritary)\n\nLet $ m_r$ be the largest number, and hence the number with the most digits.(Obviously $ m_r\\ge10$) Let $ t \\equal{} \\lfloor \\frac {m_r}{10} \\rfloor$. Then there is at most $ 10$ numbers from $ m_i$(Including $ m_r$) on the form $ \\overline{td}$, where $ d$ is a digit. The sum of the reciprocicals of these are strichtly less than $ \\frac {1}{t}$. Hence there exist another sequence where all the numbers on the form $ \\overline{td}$ is replaced with $ t$, which still contains no element with more than $ k$ digits, but whos sum of reciprocicals is strictly bigger than $ m_i$'s. Since there are only finitly many numbers with $ k$ digits - less than $ 10^{k \\plus{} 1}$ at least - we obtain a contradiction. Hence there doesn't exist such a sequence.\n\nNow we have proved the problem when $ n_i$ has finitly many elements. How can we expand it to infinitly many elements?\nFurther more: The proof get's a bit clumsy.. Using $ \\frac {1}{n} \\ge \\frac {1}{\\overline{n0}} \\plus{} \\frac {1}{\\overline{n2}} \\plus{} ... \\plus{} \\frac {1}{\\overline{n9}}$ would do the job, but would be hard to argument with, if there were infinitly many elements..[/quote]\r\n\r\nBy the way , can u help me the  2nd problem of Day 1 of the VMO 1997 :http://www.mathlinks.ro/resources.php?c=186&cid=40&year=1997   I searched the solution and could'nt fine it as well ...............>\"<"
}
{
  "Tag": [
    "function",
    "parameterization",
    "analytic geometry",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Is there a differentiable function $f:[0,1]^2->R^3$ whose image is a Moebius band?",
  "Solution_1": "Hi, Harazi,\r\n\r\nHere is the parametric function (I first saw it in the Thorpe's book, but it's quite straightforward): (t, u) -> ( (1 + t*cos(u/2))*cos(u), (1 + t*cos(u/2))*sin(u), t*sin(u/2)), -1/4 < t < 1/4. To understand geometrical meaning you can fix one of the parameters and see what coordinate lines.\r\nI bet it can't be represented as a graph of a regular smooth function.\r\n\r\nStan"
}
{
  "Tag": [
    "function",
    "induction",
    "algebra proposed",
    "algebra"
  ],
  "Problem": "Find functions $ f: \\mathbb{N} \\rightarrow \\mathbb{N}$, such that $ f(x^2 \\plus{} f(y)) \\equal{} xf(x) \\plus{} y$, for $ x,y \\in \\mathbb{N}$.",
  "Solution_1": "[hide=\"Solution\"]Put in $ x \\equal{} y \\equal{} 1$ to get that $ f(1\\plus{}f(1)) \\equal{} 1 \\plus{} f(1)$. So we know a fixed point of $ f$ exists. Call it $ c$. Putting in $ x \\equal{} 1, y \\equal{} c$ gives $ f(1\\plus{}c) \\equal{} f(1) \\plus{} c$. By induction, $ f(k\\plus{}c) \\equal{} kf(1) \\plus{} c$ for all positive integers $ k$. Now we get that $ a^2f(1) \\plus{} c \\equal{} f(a^2 \\plus{} c) \\equal{} af(a) \\plus{} c$, and so $ af(1) \\equal{} f(a)$. By this identity $ f(x^2 \\plus{} f(y)) \\equal{} x^2f(1) \\plus{} yf(1)^2$ and $ xf(x) \\plus{} y \\equal{} x^2f(1) \\plus{} y$. Setting these two equal gets $ yf(1)^2 \\equal{} y$ so $ f(1) \\equal{} 1$ and $ f(x) \\equal{} x$ for all naturals $ x$.[/hide]",
  "Solution_2": "[quote=\"MellowMelon\"][hide=\"Solution\"]Put in $ x \\equal{} y \\equal{} 1$ to get that $ f(1 \\plus{} f(1)) \\equal{} 1 \\plus{} f(1)$. So we know a fixed point of $ f$ exists. Call it $ c$. Putting in $ x \\equal{} 1, y \\equal{} c$ gives $ f(1 \\plus{} c) \\equal{} f(1) \\plus{} c$. By induction, $ f(k \\plus{} c) \\equal{} kf(1) \\plus{} c$ for all positive integers $ k$. Now we get that $ a^2f(1) \\plus{} c \\equal{} f(a^2 \\plus{} c) \\equal{} af(a) \\plus{} c$, and so $ af(1) \\equal{} f(a)$. By this identity $ f(x^2 \\plus{} f(y)) \\equal{} x^2f(1) \\plus{} yf(1)^2$ and $ xf(x) \\plus{} y \\equal{} x^2f(1) \\plus{} y$. Setting these two equal gets $ yf(1)^2 \\equal{} y$ so $ f(1) \\equal{} 1$ and $ f(x) \\equal{} x$ for all naturals $ x$.[/hide][/quote]\r\nHow do you induct ?  :wink: .\r\nHere is my solution . \r\nTake $ y \\equal{} x \\equal{} 1$ then $ f(f(1) \\plus{} 1) \\equal{} f(1) \\plus{} 1$\r\n$ a \\equal{} f(1) \\plus{} 1$ \r\nTake $ x \\equal{} 1$ then \r\n$ f(f(y) \\plus{} 1) \\equal{} y \\plus{} f(1)$\r\nTake $ y\\to f(y) \\plus{} 1$ then \r\n$ f(y \\plus{} a) \\equal{} f(y) \\plus{} a$ \r\nInduction we have $ f(x \\plus{} ka) \\equal{} f(x) \\plus{} ka$\r\nBecause $ f(a) \\equal{} a$ so $ f(ka) \\equal{} ka,\\forall k\\in N$\r\nNow take $ y \\equal{} a$ then \r\n$ f(x^2) \\equal{} xf(x)$\r\nTake $ x\\to a \\plus{} 1$ then \r\n$ f(a^2 \\plus{} 2a \\plus{} 1) \\equal{} (a \\plus{} 1)f(a \\plus{} 1)$\r\n$ \\Leftrightarrow a^2 \\plus{} 2a \\plus{} f(1) \\equal{} (a \\plus{} 1)(a \\plus{} f(1))$\r\n$ \\Leftrightarrow f(1) \\equal{} 1$ \r\n So $ a \\equal{} 2$ \r\nWe have $ f(x \\plus{} 2) \\equal{} f(x) \\plus{} 2$\r\nand $ f(2) \\equal{} 2,f(1) \\equal{} 1$\r\nInduction we have $ f(n) \\equal{} n,\\forall n\\in N$\r\nSolution of this function equation :$ f(n)\\equiv n$",
  "Solution_3": "[quote=Ahiles]Find functions $ f: \\mathbb{N} \\rightarrow \\mathbb{N}$, such that $ f(x^2 \\plus{} f(y)) \\equal{} xf(x) \\plus{} y$, for $ x,y \\in \\mathbb{N}$.[/quote]\n\nP(0,y) => $f(f(y))=y$(*)\nP(f(x),y) with (*)=>$f(f(x)^2+f(y))=xf(x)+y$\n=>$f(f(x)^2+f(y))=f(x^2+f(y))$\nAnd using (*) => $f(x)^2=x^2=>f(x)=x$"
}
{
  "Tag": [],
  "Problem": "At 12:00 noon, Anne, Beth and Carmen begin running laps around a circular track of length $300$ meters, all starting from the same point on the track. Each jogger maintains a constant speed in one of the two possible directions for an indefinite period of time. Show that if Anne's speed is different from the other two speeds, then at some later time Anne will be at least $100$ meters from each of the other runners.  (Here, distance is measured along the  shorter of the two arcs separating two runners.)",
  "Solution_1": "[quote]\nLet $M=300$ and let us define by $f: [0, \\infty) \\rightarrow [0, \\frac{M}{2}]$ the distance between Anne and Beth and $g: [0, \\infty) \\rightarrow [0, \\frac{M}{2}]$ the distance between Anne and Carmen. $f$ and $g$ are not constant functions by hypothesis. $f$ and $g$ are periodic functions and let $S$ and $T$  be their periods respectively. Since $f$ and $g$ are not constant, $S>0, T>0$. Suppose $S\\leq T$.\n\nMoreover, by hypothesis $f$ and $g$ are linear functions on any interval of the form $[k\\frac{S}{2}, (k+1)\\frac{S}{2}]$ and $[k\\frac{T}{2}, (k+1)\\frac{T}{2}]$, $k\\geq0$ respectively. Hence $f$ and $g$ are two saw functions of the following form:\n$f(x)=\\{ \\begin{array}{c}{11} f_{2k}(x), x\\in[2k\\frac{S}{2}, (2k+1)\\frac{S}{2}]\\ f_{2k+1}(x), x\\in[(2k+1)\\frac{S}{2}, (2k+2)\\frac{S}{2}]\\ \\end{array} $\n, $k\\geq0$\n$g(x)=\\{ \\begin{array}{c}{11} g_{2k}(x), x\\in[2k\\frac{T}{2}, (2k+1)\\frac{T}{2}]\\ g_{2k+1}(x), x\\in[(2k+1)\\frac{T}{2}, (2k+2)\\frac{T}{2}]\\ \\end{array} $\n, $k\\geq0$.\n\nWe have to prove that $(\\exists) \\alpha\\geq0$ s.t. $f(\\alpha)\\geq\\frac{M}{3}$ and $g(\\alpha)\\geq\\frac{M}{3}$.\n\nWe shall prove that by analyzing how $g_0$ and $g_1$ intersect different $f_k$.\nLet $k_0$ be the maximum $k\\geq0$ for which $g_0\\cap f_{k_0}\\neq\\oslash$. Such a $k_0$ exists because we supposed that $S\\leq T$. This obviously implies that $g_1\\cap f_{k_0+1}\\neq\\oslash$. Let ${P_0(x_0, y_0)=g_0\\cap f_{k_0}}$ and $P_1(x_1, y_1)=g_1\\cap f_{k_0+1}$. Let us notice that if $y_0\\geq\\frac{M}{3}$ or $y_1\\geq\\frac{M}{3}$ than $x_0$ or $x_1$ are the searched $\\alpha$ so suppose that (1) $y_0<\\frac{M}{3}$ and $y_1<\\frac{M}{3}$.\nAfter solving the two equations ${P_0(x_0, y_0)=g_0\\cap f_{k_0}}$ and $P_1(x_1, y_1)=g_1\\cap f_{k_0+1}$ we get that $P_0$ and $P_1$ are defined by $P_0((k_0+1)\\frac{ST}{S+T}, (k_0+1)\\frac{SM}{S+T})$ and $P_1((k_0+2)\\frac{ST}{S+T}, M\\frac{T-(k_0+1)S}{S+T})$.\nFrom this and (1) we get that \n$\\{ \\begin{array}{c}{11} (k_0+1)\\frac{SM}{S+T}<\\frac{M}{3}\\ M\\frac{T-(k_0+1)S}{S+T})<\\frac{M}{3}\\ \\end{array}  \\Leftrightarrow \\{ \\begin{array}{11} k_0+1<\\frac{S+T}{3S}\\ k_0+1>\\frac{2T-S}{3S}\\ \\end{array}  \\Leftrightarrow \\{ \\begin{array}{11} k_0<\\frac{T-2S}{3S}\\ k_0>\\frac{2T-4S}{3S}\\ \\end{array} $ (3).\nFrom this we get that $\\frac{2T-4S}{3S}<\\frac{T-2S}{3S}$ which is equivalent with (4) $T<2S$.\nBut (3) and (4) imply that $k_0<0$ which is a contradiction and this completes the proof.\n\n[/quote]",
  "Solution_2": "For a simpler proof, consider the frame of reference of Anne, and do cases for Beth and Carmen.",
  "Solution_3": "[quote=Altheman]For a simpler proof, consider the frame of reference of Anne, and do cases for Beth and Carmen.[/quote]\n\nI have also solved using the topic of reference point.Seems ur a physics student"
}
{
  "Tag": [
    "function",
    "linear algebra",
    "matrix",
    "LaTeX"
  ],
  "Problem": "you know in defining piece wise function you do stuff like:\r\n\r\n$ |x| = \\{_{-x, x < 0}^{ x, x \\ge 0}$\r\n\r\nBut of course that first bracket looks ugly.  I tried to use \\left which fixes it, but then I need a \\right.  I can just put the \\right and ignore the closing delimiter, and it works fine, but gives me an annoying error message.  Any other cleaner and smarter ways?\r\n\r\nedit: hmm I know the words look tiny small too, but for my tex adding \\displaystyle fixes it, but I guess on AoPS they automatically remove those stuffs??",
  "Solution_1": "Use \\phantom\r\n\r\n$ |x| = \\{\\begin{array}{l}x,\\ x\\geq0\\\\-x,\\ x<0\\end{array}\\phantom$",
  "Solution_2": "Better yet, use the \\cases environment (you need the amsmath package)\r\n\r\n$ |x|=\\begin{cases}x&\\text{if }x\\ge0\\\\-x&\\text{if }x<0 \\end{cases}$\r\n[code]$|x|=\\begin{cases}\nx&\\text{if }x\\ge0\\\\\n-x&\\text{if }x<0\n\\end{cases}$[/code]",
  "Solution_3": "[quote=\"pkerichang\"]I tried to use \\left which fixes it, but then I need a \\right.  I can just put the \\right and ignore the closing delimiter, and it works fine, but gives me an annoying error message.  Any other cleaner and smarter ways?[/quote]Do follow roadnottaken's advice for such functions but, in another situation, if you ever need a left bracket but no right bracket then use[color=red]\\right.[/color] with the [b][color=red].[/color][/b] at the end. So you can have, for example, $ \\left[\\frac{1}{2}\\right.$ and using \\left. you get $ \\left.\\frac{1}{2}\\right)$ \n[quote]edit: hmm I know the words look tiny small too, but for my tex adding \\displaystyle fixes it, but I guess on AoPS they automatically remove those stuffs??[/quote]In fact the forum adds \\displaystyle to your formula so you don't need to use it.",
  "Solution_4": "also you can try the following-gives a slightly improved look for inline(the one using cases seems to be taller):[code]\n\\left|x\\right| = \n\\left\\{\n\\begin{matrix}\n   x & \\text{if } x \\geq 0  \\\\ \n   -x & \\text{if } x < 0  \n\\end{matrix}\n\\right.[/code]",
  "Solution_5": "\\left|x\\right| =\r\n\\left\\{\r\n\\begin{matrix}\r\n   x & \\text{if } x \\geq 0  \\\\\r\n   -x & \\text{if } x < 0 \r\n\\end{matrix}\r\n\\right.\r\n\r\n\r\n(Testing how to get the forum to render \\LaTeX code?????\r\n\r\nOk how is latex code rendered inline...is there a special tag???",
  "Solution_6": "Use dollar signs like you would in a tex document. [code]$|x|=\\begin{cases} \nx&\\text{if }x\\ge0\\\\ \n-x&\\text{if }x<0 \n\\end{cases}$[/code]$ |x|=\\begin{cases}x&\\text{if }x\\ge0\\\\-x&\\text{if }x<0 \\end{cases}$"
}
{
  "Tag": [],
  "Problem": "Se considera multimea\r\n\\[A=\\{x\\in\\mathbb{N}| x=(p^{2}-q)(q^{2}-p), p, q\\in\\mathbb{N}\\}. \\]\r\nSa se arate ca:\r\na) $2006\\in{A}$;\r\nb) $A$ contine o infinitate de patrate perfecte;\r\nc) nu exista nicio multime nevida $B$ astfel incat $|B|=p^{2}-q$; $|P(B)|=q^{2}-p$ cu $p,q\\in\\mathbb{N}$.",
  "Solution_1": "[quote=\"Marius Damian\"]Se considera multimea\n\\[A=\\{x\\in\\mathbb{N}| x=2(p^{2}-q)(q^{2}-p), p, q\\in\\mathbb{N}\\}. \\]\nSa se arate ca:\na) $2006\\in{A}$;\nb) $A$ contine o infinitate de patrate perfecte;\nc) nu exista nicio multime nevida $B$ astfel incat $|B|=p^{2}-q$; $|P(B)|=q^{2}-p$ cu $p,q\\in\\mathbb{N}$.[/quote]\r\n\r\n[color=blue]Am corectat enuntul.[/color]"
}
{
  "Tag": [
    "inequalities proposed",
    "inequalities"
  ],
  "Problem": "Put $ a;b;c \\ge 0$ and $ a\\plus{}b\\plus{}c\\equal{}1$. Prove that:\r\n$ 4 \\plus{} 15abc \\plus{} (a \\minus{} b)(b \\minus{} c)(c \\minus{} a) \\ge\\ 5(a^2b \\plus{} b^2c \\plus{} c^2a) \\plus{} 12(ab \\plus{} bc \\plus{} ca)$",
  "Solution_1": "[quote=\"onlylove_math\"]Put $ a;b;c \\ge 0$ and $ a \\plus{} b \\plus{} c \\equal{} 1$. Prove that:\n$ 4 \\plus{} 15abc \\plus{} (a \\minus{} b)(b \\minus{} c)(c \\minus{} a) \\ge\\ 5(a^2b \\plus{} b^2c \\plus{} c^2a) \\plus{} 12(ab \\plus{} bc \\plus{} ca)$[/quote]\r\nIt is really easy ,my friend.This ineq is the mixing of two ineq\r\nThe first $ 2\\sum a^3 \\plus{} 3abc\\geq 3(a^2b \\plus{} b^2c \\plus{} c^2a)$  (vey well-know)\r\nAnd $ ab^2 \\plus{} bc^2 \\plus{} ca^2\\geq 3abc$\r\nIt is nice",
  "Solution_2": "[quote=\"Allnames\"][quote=\"onlylove_math\"]Put $ a;b;c \\ge 0$ and $ a \\plus{} b \\plus{} c \\equal{} 1$. Prove that:\n$ 4 \\plus{} 15abc \\plus{} (a \\minus{} b)(b \\minus{} c)(c \\minus{} a) \\ge\\ 5(a^2b \\plus{} b^2c \\plus{} c^2a) \\plus{} 12(ab \\plus{} bc \\plus{} ca)$[/quote]\nIt is really easy ,my friend.This ineq is the mixing of two ineq\nThe first $ 2(\\sum a^3 \\plus{} 3abc\\geq 3(a^2b \\plus{} b^2c \\plus{} c^2a)$  (vey well-know)\nAnd $ ab^2 \\plus{} bc^2 \\plus{} ca^2\\geq 3abc$\nIt is nice[/quote]\r\nI don't understand your proof \r\nCan you rewrite detail of your proof? please  :)",
  "Solution_3": "[quote=\"bunhiacovski\"][quote=\"Allnames\"][quote=\"onlylove_math\"]Put $ a;b;c \\ge 0$ and $ a \\plus{} b \\plus{} c \\equal{} 1$. Prove that:\n$ 4 \\plus{} 15abc \\plus{} (a \\minus{} b)(b \\minus{} c)(c \\minus{} a) \\ge\\ 5(a^2b \\plus{} b^2c \\plus{} c^2a) \\plus{} 12(ab \\plus{} bc \\plus{} ca)$[/quote]\nIt is really easy ,my friend.This ineq is the mixing of two ineq\nThe first(1) $ 2\\sum a^3 \\plus{} 3abc\\geq 3(a^2b \\plus{} b^2c \\plus{} c^2a)$  (vey well-know)\nAnd $ ab^2 \\plus{} bc^2 \\plus{} ca^2\\geq 3abc$\nIt is nice[/quote]\nI don't understand your proof \nCan you rewrite detail of your proof? please  :)[/quote]\r\nIt is quite simple.onlylove_math's ineq <=>$ 4\\sum a^3 \\plus{} 3abc\\plus{}ab^2 \\plus{} bc^2 \\plus{} ca^2\\geq 6(a^2b \\plus{} b^2c \\plus{} c^2a)$\r\nand use two ineq that i said we have a proof ((1) is a ineq in Secrets in ineq of PKH  or you can prove it by SS or pqr)",
  "Solution_4": "[quote=\"Allnames\"][/quote][quote=\"bunhiacovski\"][quote=\"Allnames\"][quote=\"onlylove_math\"]Put $ a;b;c \\ge 0$ and $ a + b + c = 1$. Prove that:\n$ 4 + 15abc + (a - b)(b - c)(c - a) \\ge\\ 5(a^2b + b^2c + c^2a) + 12(ab + bc + ca)$[/quote]\nIt is really easy ,my friend.This ineq is the mixing of two ineq\nThe first(1) $ 2\\sum a^3 + 3abc\\geq 3(a^2b + b^2c + c^2a)$  (vey well-know)\nAnd $ ab^2 + bc^2 + ca^2\\geq 3abc$\nIt is nice[/quote]\nI don't understand your proof \nCan you rewrite detail of your proof? please  :)[/quote][quote=\"Allnames\"]\nIt is quite simple.onlylove_math's ineq <=>$ 4\\sum a^3 + 3abc + ab^2 + bc^2 + ca^2\\geq 6(a^2b + b^2c + c^2a)$\nand use two ineq that i said we have a proof ((1) is a ineq in Secrets in ineq of PKH  or you can prove it by SS or pqr)[/quote]\r\nThank Allnames; It is my proof.\r\nBut with my proof i dont use $ ab^2 + bc^2 + ca^2\\geq 3abc$ . I have used\r\n$ a^3 + b^3 + c^3 + ab^2 + bc^2 + ca^2 \\ge\\ 2(a^2b + b^2c + c^2a)$"
}
{
  "Tag": [],
  "Problem": "Today I just won my 100th game. That means you get the 5th win thingy (click on the user and see how many win thingies they have). Just wondering, has anyone had trouble getting that?",
  "Solution_1": "The 5th win badge is many orders of magnitude easier to earn than the 5th rating badge, and infinite orders of magnitude easier to earn than the 5th game badge. Needless to say, most users have it.",
  "Solution_2": "What?\r\nI don't quite understand this...",
  "Solution_3": "@cheze___\r\n\r\nWhenever you win a certain amount of games, you are awarded with badges.  When you win enough games, you will have 5 badges when someone(even yourself) clicks on your name on the left bar.  You also get to win not only win badges, but also rating and game badges.  There are a total of 14 badges that you can have: 5 ratings, 5 wins, and 4 games.\r\n\r\nI don't know when you get the win and game badges, but after the rating badges in order are when you have achieved a \"record\" rating of 1200, 1350, 1500, 1750, and 2000(not very many people have this...i only know jongy, tcoas, aime15, chaus, and 1337rox)"
}
{
  "Tag": [
    "trigonometry",
    "geometry",
    "geometric transformation",
    "rotation",
    "dilation",
    "ratio",
    "circumcircle"
  ],
  "Problem": "Give sharp triangle $ ABC$,erect on the outside of $ BC,CA,AB$ in turn the squares with center are $ M,N,P$.\r\nProve that:\r\n\r\n$ \\bigtriangleup{BMN}\\equal{}\\bigtriangleup{MCP}$",
  "Solution_1": "Please help me with this problem,it's hard with me :( .Thank you!",
  "Solution_2": "Let $ BC\\equal{}a$, $ AC\\equal{}b$, $ AB\\equal{}c$, $ \\angle A\\equal{}\\alpha$, $ \\angle B\\equal{}\\beta$,\r\n$ \\angle C\\equal{}\\gamma$. Let us prove that $ BN\\equal{}PM$. Considering\r\n$ \\triangle BAN$ we get\r\n\\[ BN^{2}\\equal{}c^2\\plus{}\\frac{b^2}{2}\\minus{}bc\\sqrt{2}\\cos(\\alpha\\plus{}\\frac{\\pi}{4}) \\equal{}\r\nc^2 \\plus{}\\frac{b^2}{2} \\minus{}bc(\\cos\\alpha \\minus{}\\sin\\alpha).\\] Considering\r\n$ \\triangle PBM$ we get \\[ PM^{2}\\equal{}\\frac{c^2}{2}\r\n\\plus{}\\frac{a^2}{2}\\plus{}ac\\sin\\beta .\\] Since $ b\\sin\\alpha \\equal{} a\\sin \\beta$\r\nwe have $ bc\\cos\\alpha \\equal{} bc\\cos\\alpha \\minus{}bc\\sin\\alpha \\plus{}ac\\sin\\beta$,\r\nwhence $ \\frac{c^2}{2}\\plus{}\\frac{b^2}{2}\\minus{}\\frac{a^2}{2}\\equal{}bc(\\cos\\alpha\r\n\\minus{}\\sin\\alpha)\\plus{}ac\\sin\\beta ,$ and thus\r\n\\[ c^2 \\plus{}\\frac{b^2}{2}\\minus{}bc(\\cos\\alpha \\minus{}\\sin\\alpha) \\equal{} \\frac{c^2}{2}\\plus{}\r\n\\frac{a^2}{2} \\plus{}ac\\sin\\beta .\\] So we have proved that $ BN\\equal{}PM$.\r\nAnalogously $ NM\\equal{}CP.$ Obviously $ BM\\equal{}MC.$\r\n\r\nSo $ \\triangle BMN\\equal{}\\triangle MCP. \\     \\Box$",
  "Solution_3": "Well, unlike other people here, I do not like boring calculations. My goal is to prove that AM and PN are equal and perpendicular, similarly for the other 2 pairs.\r\nSo, let's note D - midpoint of BC, whilst AE and AF being sides of the squares, E and F the opposite vertices of C and B respectively.\r\nA $ 90^\\circ$ rotation about A will map B to F and E to C, hence BE and CF are equal and perpendicular; the same are PD and DN, as midlines in the triangles BCF and BCE, consequently $ PN \\equal{} PD \\cdot \\sqrt {2}$ ( 1 ).\r\nA $ 45^\\circ$ rotation about B followed by a dilation of $ \\sqrt 2$ ratio will map P to A and D to M, while a $ 45^\\circ$ rotation about C followed by a dilation of $ \\sqrt 2$ ratio will map N to A and D to M, hence $ AM \\equal{} PD \\cdot \\sqrt 2$ ( 2 ) and $ \\angle (PD, AM) \\equal{} 45^\\circ$ ( 3 ).\r\nFrom (1) and (2) we get AM = PN, and, with (3) we get $ AM \\perp PN$.\r\nSimilarly, BN = MP, CP = BN and, with BM = MC, $ \\triangle BMN \\equal{} \\triangle MCP$, q.e.d.\r\n\r\nBest regards,\r\nsunken rock",
  "Solution_4": "Proving that $ AM$ and $ PN$ are perpendicular will suffice . (Since after that, we can prove that the two triangles are similar and since $ BM \\equal{} MC$, we'll be done)\r\n[img]http://img41.imageshack.us/img41/1326/10561676.png[/img]\r\n\r\n\r\nThe circumcircles of two of the erected squares are shown. This is equivalent to showing that $ M$ lies on the radical axis of the circles. \r\nLet $ MB$ and $ MC$ intersect the circles at $ X$ and $ Y$. $ BU$ and $ CV$ are diameters. Observe that $ \\angle BUX \\equal{} B$ and $ \\angle CVY \\equal{} C$. We easily see that $ BX \\equal{} CY$ ($ BUsinB \\equal{} CVsinC$ iff $ ABsinB \\equal{} ACsinC$)\r\nSo,  $ MC \\equal{} MB$ and $ MX \\equal{} MY$ $ \\rightarrow MB.MX \\equal{} MC.MY$. So, power of $ M$ w.r.t. the two circles are equal and we're done."
}
{
  "Tag": [
    "vector",
    "topology",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "E is a normed vector space , and M is a submanifold of finite dimension of E;\r\nI am facing the following confusion: since M is of finite dimension , it is topologically isomorphic to Rn , which is complete, so M is cmplete\r\nbut since the interior of M is empty,so it contains no spheroid, and then M is of first Baire category which contrdict its cmpleteness.\r\nThanks.",
  "Solution_1": "[quote=\"senouy\"]E is a normed vector space , and M is a submanifold of finite dimension of E;\nI am facing the following confusion: since M is of finite dimension , it is topologically isomorphic to Rn , which is complete, so M is cmplete\nbut since the interior of M is empty,so it contains no spheroid, and then M is of first Baire category which contrdict its cmpleteness.\nThanks.[/quote]The same argument could be made to argue that $ \\mathbb R \\sim  \\mathbb R \\times \\{ 0 \\} \\subset \\mathbb R^2$ is not complete. Think about it.",
  "Solution_2": "thanks , but i can not understand what you read",
  "Solution_3": "Basically, your argument is nonsense. That's not what the Baire theorem says, and the completeness of $ M$ depends on the choice of metric, not just a topology. Is the open unit disk complete? It's homeomorphic to $ \\mathbb{R}^2$.",
  "Solution_4": "[quote]the interior of M is empty,so it contains no spheroid, and then M is of first Baire category which contrdict its cmpleteness[/quote]\r\nThe Baire category theorem asserts that a complete metric space is of second category [b]in itself[/b]. This is not related to the fact that $ M$ is of first category in $ E$."
}
{
  "Tag": [
    "floor function",
    "algebra",
    "functional equation",
    "algebra proposed"
  ],
  "Problem": "Find $f:\\mathbb{N}\\to\\mathbb{N}$ satisfying : \n   \\[f(m+f(n))= f(f(m)) +f(n)\\]",
  "Solution_1": "This one was problem 3 in IMO 1996.\r\n\r\nI'll assume that $0\\in \\mathbb{N}$ (as IMO 1996). Substituting $m=n=0$ we get $f(0)=0$. Plugging in $n=0$, one gets $f(m) = f(f(m))$. One solution is $f(n) = 0$ for all $n\\in \\mathbb{n}$.\r\n\r\nThe functional equation becomes $f(m+f(n)) = f(m) + f(n)$.\r\n\r\nAssume that there is $n$ such that $f(n)\\neq 0$. Thus there is at least one nonzero fixed point ($f(n)$, for example). Let $a>0$ be the minimum fixed point of $f$ and let $f(i) = a_i$ for $0 < i < a$. Since $f(m+a) = f(m+f(a)) = f(m) + f(a) = f(m) + a$ inductively we have $f(m+qa) = f(m) + qa$ for all nonnegative integers $q$. Thus if $n=qa+r$, $q,r$ integers and $0\\leq r<a$ then $f(n) = qa + f(r) = qa + a_r$. So, defined $f(0)=0,f(1),\\ldots,f(a)=a$, we have $f(n) = \\left\\lfloor\\frac{n}{a}\\right\\rfloor a + f(n\\bmod{a})$.\r\n\r\nI didn't think about the problem if $0\\notin \\mathbb{N}$, but it seems to be harder..."
}
{
  "Tag": [
    "trigonometry",
    "geometry",
    "angle bisector"
  ],
  "Problem": "SMO 2007, Open #17\r\n\r\nIn $\\Delta ABC$, $AC > AB$, the internal angle bisector of $\\angle A$ meets $BC$ at $D$, and $E$ is the foot of the perpendicular from $B$ onto $AD$. Suppose $AB = 5$, $BE = 4$ and $AE = 3$. Find the value of the expression \\[(\\frac{AC+AB}{AC-AB})ED\\].",
  "Solution_1": "We have $AE=c\\cos{A/2}\\longrightarrow \\cos{A/2}=\\frac{3}{5}$\r\nthen we know that $AD=\\frac{2bc\\cos{A/2}}{b+c}=\\frac{6b}{b+5}$ $\\longrightarrow ED=AD-3=\\frac{3b-15}{b+5}$ and \r\nnow :$(\\frac{b+c}{b-c})ED=3$",
  "Solution_2": "[quote=\"khashi70\"]$AD=\\frac{2bc\\cos{A/2}}{b+c}$[/quote]\r\n\r\nHow did you get this?",
  "Solution_3": "[quote=\"tjhance\"][quote=\"khashi70\"]$AD=\\frac{2bc\\cos{A/2}}{b+c}$[/quote]\n\nHow did you get this?[/quote]\r\nwith $\\sin$ law in $\\triangle ABD$ we get that $\\frac{AD}{\\sin B}=\\frac{BD}{\\sin{\\frac{A}{2}}}$\r\nbut we know that $\\frac{BD}{DC}=\\frac{AB}{AC}$ so we get that $\\frac{BD}{a}=\\frac{c}{b+c}$ so we get that $BD=\\frac{ac}{b+c}$ so we get that $\\frac{AD}{\\sin B}=\\frac{ac}{(b+c)\\sin{\\frac{A}{2}}}$\r\nbut we know that $\\frac{b}{\\sin B}=\\frac{a}{\\sin A}$ so we get that:\r\n$AD=\\frac{bc\\sin A}{(b+c)\\sin{\\frac{A}{2}}}$\r\nbut with $\\sin A=2\\sin{\\frac{A}{2}}\\cos{\\frac{A}{2}}$ we get that:\r\n$AD=\\frac{2bc\\cos{\\frac{A}{2}}}{b+c}$..."
}
{
  "Tag": [
    "conics",
    "ellipse",
    "linear algebra",
    "matrix",
    "ratio",
    "integration",
    "parabola"
  ],
  "Problem": "[color=darkred]First of all, I'd like to say sorry for making this urgent as it is despised in this board. However, I'm having maths board exams 14th and I'd be very grateful if you can try to help me in the following questions ASAP. Those questions are very important ones, taken from last years papers....\n\nThank you in advance.\n\nGuys wish me luck for exams!\n[/color]\r\n\r\n[b][u]Questions which I don't know how to solve:[/u][/b]\r\n\r\n\r\n1. Find the equation of tangent to the ellipse $ \\frac {x^2}{25} + \\frac {y^2}{16} = 1$ on the point $ (\\frac {5}{2},2\\sqrt {3})$\r\n\r\n[b]\u2713[/b] [hide]2. Solve the equation sin2x+cosx=0 for general solution. [/hide] [b]Got this, ty![/b]\n\n3. If $ C_0, C_1, C_2 ... C_n$ are binomial coefficients in the expanion of $ (1 + x)^n$, prove that $ C_0 + C_2 + C_4 + ... = 2^n^ - ^1$\n\n4. Find the sum of the following infinite series $ 1 + \\frac {1}{3}(\\frac {1}{2})^2 + \\frac {1}{5}(\\frac {1}{2})^4 + ...$\n\n5. If $ e^x^ + ^y = xy$, show that $ \\frac {dy}{dx} = \\frac {y(1 - x)}{x(y - 1)}$\n\n6. Using properties of determinants, prove that$ \\lvert A\\rvert = \\begin{vmatrix} a - b - c & 2a & 2a \\\\\n2b & b - c - a & 2b \\\\\n2c & 2c & c - a - b \\end{vmatrix} = (a + b + c)^3$\n\n7. Find $ n$ if $ {}^2{}^nC_3: ^nC_3 = 44: 3$\n\n8. Prove that $ (2n)! = 2^n.n![1.3.5....(2n - 1)]$\n\n9. Find the sum to n terms of the following series: $ \\frac {1}{1.3} + \\frac {1}{3.5} + \\frac {1}{5.7} + ....$             [color=olive][PS: (dot) is the multiplication and not the decimal][/color]\n\n10. If $ C_0, C_1, C_2,...C_n$ are the binomial coefficients in the expansion of $ (1 + x)^n$. Prove that $ C_0+C_1+C_2+C_3+...+C_n=2^n$\n\n11. Find the ratio in which the line joingin $ ( - 5, 1)$ and $ (1, - 3)$ divides the join of the points $ (3, 4)$ and $ (7, 8)$.\n\n12. Evaluate $ \\int{\\frac {3x^2}{\\sqrt {9 - 16x^6}}}dx$\n\n[b]\u2713[/b] [hide]13. Solve $ 2cos^2\\theta + 3sin\\theta = 0$.[/hide] [b]Got this, ty![/b]\n\n14. Find the vertex, axis, focus and latus rectum of the parabola $ y^2 + 4x - 2y + 3 = 0$\n\n15. Show that the derivative of $ tan^ - ^1{\\frac {\\sqrt {1 + x^} - 1}{x}} w.r.t. sin^ - ^1{\\frac {2x}{1 + x^2}} is \\frac {1}{4}.$\n\n16. Evaluate: $ \\int^\\pi_0{\\frac {xsinx}{1 + cos^2x}}dx$\nAns: $ \\int^\\pi_0{\\frac {xsinx}{1 + cos^2x}}dx$=$ \\int^\\pi_0{\\frac {xsinx}{sin^x}}dx$\n$ = \\int^\\pi_0{\\frac {x}{sinx}}dx$? [color=olive][Just gotta know how to integrate this atleast, then I know how to put the pi and 0 values there][/color]\n\n17. Solve the equation cos 3x=sin2x for x.\nMy work: $ 4cos^3x-3cosx=2cosxsinx$\n$ 4cos^2x-3-2sinx=0$\n$ 4(1-sin^2x)-3-2sinx=0$\n$ 4sin^2x+2sinx-1=0$\n$ 42sinx(2sinx+1)-1=0$ [color=olive][<<<What to do next??][/color]\n\n------------------------------------------------------------------------------------\n\n[b][u]For correction on my work:[/u][/b]\n\n[color=darkred]Following are my works....which I'm not sure if it's correct... so..plz check the following works and point out the fault if any....[/color]\n\nA. If $ y = \\frac {e{}^2{}^x.cosx}{xsinx}$, find $ \\frac {dy}{dx}$\nAns: $ \\frac {dy}{dx} = \\frac {\\frac {d(e{}^2{}^x.cosx)}{dx}xsinx - \\frac {d(xsinx)}{dx}(e{}^2{}^x.cosx)}{x{}^2sin{}^2x}$\n$ = \\frac {2x e{}^2{}^xcosxsinx - xe{}^2{}^xsin{}^2x - xe{}^2{}^xcos{}^2x - e{}^2{}^xsinxcosx}{x{}^2sin{}^2x}$\n$ = \\frac {2x - xe{}^2{}^x}{x{}^2sin{}^2x}$\n\nB. Evaluate: $ \\lim_{x \\to - 1}(\\frac {x + 1}{\\sqrt {6x^2 + 3} + 3x})$\nAns: $ \\lim_{x \\to - 1}(\\frac {x + 1}{\\sqrt {6x^2 + 3} + 3x}) = \\lim_{x\\to - 1}\\frac {x + 1}{x\\sqrt {6 + 3} + 3x}$\n$ = \\lim_{x \\to - 1}\\frac {x + 1}{6x}$\n$ = 0$\n\nC.  Solve the differential equation $ \\frac {dy}{dx} = 1 + x + y + xy$ given that y=0 when x=0.\nAns: $ \\frac {dy}{dx} = 1 + 0 + 0 + 0.0$\n$ \\int{dy} = \\int{dx}$\n$ y = x + C$\n$ = 0 + C = C$\n\n[b]\u2713 Understood this 1, ty![/b]\n[hide]D. Evaluate $ \\sqrt {13 - 20\\sqrt {3}i}$\nAns: Let $ \\sqrt {13 - 20\\sqrt {3}i} = x + yi$\n$ 13 - 20\\sqrt {3}i = (x + yi)^2 = x^2 - y^2 + 2xyi$\n\n$ 13 = x^2 - y^2$                                      .............(1)\n$ - 20\\sqrt {3}i = 2xyi$\n\n$ (x^2 + y{}^2){}^2 = (x^2 - y{}^2){}^2 + (2xy)^2$\n$ (x^2 + y{}^2){}^2 = 169 - 20\\sqrt {3}$\n$ x^2 + y^2 = \\sqrt {169 - 20\\sqrt {3}}$                     ...................(2)\n\nSubtracting (1) and (2), I get:\n$ y = \\sqrt {\\frac {\\sqrt {169 - 20\\sqrt {3}} - 13}{2}}$\n\nAlso, $ x = \\sqrt {\\frac {\\sqrt {169 - 20\\sqrt {3}} - 13}{2}}$\n\nTherefore, $ \\sqrt {13 - 20\\sqrt {3}i}$=$ \\sqrt {\\frac {\\sqrt {169 - 20\\sqrt {3}} - 13}{2}} + \\sqrt {\\frac {\\sqrt {169 - 20\\sqrt {3}} - 13}{2}}i$\n=$ \\sqrt {\\frac {\\sqrt {169 - 20\\sqrt {3}} - 13}{2}} - \\sqrt {\\frac {\\sqrt {169 - 20\\sqrt {3}} - 13}{2}}$=0[/hide]\n\n------------------------------------------------------------------------------------\n\n[b][u]For Explanation:[/u][/b]\n\n[color=darkred]Need your explanation for the following questions with solution from the text books....also plz let me know whatever formula is used....[/color]\n\n[b]\u2713 Understood these hiddens....ty![/b]\n[hide]a. The equation of the chord of the circle $ x^2 + y^2 - 2ax = 0$ is y=mx. Find the equation of the circle described on this chord as diameter.\nAns: The coordinates of the pints of intersection of the circle and the given chord are, (0, 0) and $ (\\frac {2a}{1 + m^2},\\frac {2ma}{1 + m^2})$    [color=olive]<<<How did they get this??[/color]\n[color=darkblue][size=75]------rest of the solution censored....sorry no time to type out------[/size][/color][/hide]\n\n[hide]b. Evaluate: $ \\lim_{x \\to 0}(\\frac {1 - cos x}{2x^2})$\nAns: $ \\lim_{x \\to 0}(\\frac {1 - cos x}{2x^2}) = \\lim_{x \\to 0}(\\frac {2sin\\frac {x}{2}}{2x})^2$[color=olive]<<<How did they get this??\n[/color]\n[color=darkblue][size=75]------rest of the solution censored------[/size][/color][/hide]\r\n\r\nc. Find the Mean Deviation (MD) and Standard Deviation (SD) of the AP: a, a+d, a+2d,...,a+2nd.\r\nAns: The number of items in the A.P. is (n+1)\r\n$ %Error. \"mean\" is a bad command.\n=a+nd$\r\nMean deviation about the mean\r\n$ =\\frac{1}{(2n+1)}\\displaystyle\\sum_{k=1}^n{2n}{r=0} |(a+rd)-(a+nd)|$\r\n$ =\\frac{1}{(2n+1)}.2[nd+(n-1)d+ .... +d]$\r\n$ =\\frac{2}{(2n+1)}[1+2+ ... +(n-1)+n]d$\r\n$ =\\frac{2n(n+1)}{(2n+1)2}.d=\\frac{n(n+1)d}{(2n+1)}$\r\n$ \\sqrt{Variance}=\\frac{1}{(2n+1)}\\displaystyle\\sum_{k=1}^n[(a+rd)-(a+nd)]^2$\r\n$ =\\frac{2n^2}{(2n+1)}[n^2+(n-1)^2+...+2^2+1^2]$\r\n$ =\\frac{2d^2}{(2n+1)}\\frac{n(n+1)(2n+1)}{6}$\r\n$ =\\frac{n(n+1)d^2}{3}$\r\n$ variance=d.\\sqrt{(\\frac{n(n+1)}{3})}$",
  "Solution_1": "[hide=\"#1\"]Differentiating implicitly we have\n$ \\frac{dy}{dx}\\equal{}\\minus{}\\frac{16x}{25y}\\equal{}\\minus{}\\frac{4}{5\\sqrt{3}}$\n$ \\Leftrightarrow \\ell_T : 4x\\plus{}5\\sqrt{3}y\\equal{}40$[/hide]\n[hide=\"#2\"]$ \\sin{2x}\\plus{}\\cos{x}\\equal{}0$\n$ \\displaystyle \\sin{2x}\\equal{}\\sin{\\left(\\minus{}\\frac{\\pi}{2}\\minus{}x\\right)}$\n$ x\\equal{}\\minus{}\\frac{\\pi}{6}\\pm 2\\pi n$\n\n$ 2\\sin{x}\\cos{x}\\equal{}\\minus{}\\cos{x}$\n$ \\sin{x}\\equal{}\\minus{}\\frac{1}{2}$\n$ x\\equal{}\\frac{7\\pi}{6}\\pm2\\pi n$\nNot 100% on those, but oh well.\n[/hide]",
  "Solution_2": "#3\r\nwe know that..\r\n$ (1 \\plus{} x)^n \\equal{} \\sum_{i \\equal{} 0}^n C_i^n 1^i x^{n \\minus{} i}$\r\n\r\nthus we can easily put $ x \\equal{} 1$ and $ x \\equal{} \\minus{} 1$ to get the \"part\" $ x^{n \\minus{} i}$ always $ 1$\r\n$ (1 \\plus{} 1)^n \\equal{} C_0^n \\plus{} C_1^n \\plus{} C_2^n \\plus{} C_3^n \\plus{} C_4^n.....$\r\n$ (1 \\minus{} 1)^n \\equal{} C_0^n \\minus{} C_1^n \\plus{} C_2^n \\minus{} C_3^n \\plus{} C_4^n.....$\r\n\r\n\r\nadd the 2 parts.. we have\r\n$ 2^n \\equal{} 2(C_0^n \\plus{} C_2^n \\plus{} C_4^n....)$\r\n\r\nthat's why $ C_0^n \\plus{} C_2^n \\plus{} C_4^n.... \\equal{} 2^{n \\minus{} 1}$",
  "Solution_3": "[hide=\"5\"]\nTake the derivative of both sides:\n$ (1\\plus{}y')e^{(x\\plus{}y)} \\equal{} y\\plus{}xy'$, where the chain rule was used on the left side and the product rule was used on the right side.\nWe know $ e^{x\\plus{}y}\\equal{}xy$, so:\n$ (1\\plus{}y')xy \\equal{} y\\plus{}xy'$\n$ xy\\plus{}xyy' \\equal{} y\\plus{}xy'$\n$ xyy'\\minus{}xy' \\equal{} y\\minus{}xy$\n$ x(y\\minus{}1)y' \\equal{} y(1\\minus{}x)$\n$ y' \\equal{} \\frac{y(1\\minus{}x)}{x(y\\minus{}1)}$[/hide]\n\n[hide=\"12\"]\nLet $ u\\equal{}4x^3 \\implies \\frac{du}{dx} \\equal{} 12x^2 \\implies du \\equal{} 12x^2 dx$\n\nThen, $ \\frac{1}{4} \\int \\frac{12x^2 dx}{\\sqrt{9 \\minus{} 16 x^6}} \\equal{} \\frac{1}{4} \\int \\frac{du}{\\sqrt{9 \\minus{} u^2}}$, which you may recognize as $ \\frac{1}{4} \\sin^{\\minus{}1} \\frac{4u}{3} \\plus{}C \\equal{} \\frac{1}{4} \\sin ^{\\minus{}1} \\frac{16 x^3}{3} \\plus{}C$[/hide]\n\n[hide=\"C\"]They do not tell you that $ y\\equal{}0$ and $ x\\equal{}0$. They tell you that $ y(0)\\equal{}0$, where $ y$ is treated as a function of $ x$.\n\nThe correct approach is to factor the equation as $ \\frac{dy}{dx} \\equal{} (x\\plus{}1)(y\\plus{}1)$, and then use separation of variables to get $ \\frac{1}{y\\plus{}1} dy \\equal{} (x\\plus{}1)dx$. Integrate to get $ \\ln |y\\plus{}1| \\equal{} \\frac{x^2}{2} \\plus{} x \\plus{} C$, and then exponentiate to get $ |y\\plus{}1| \\equal{} e^{\\frac{x^2}{2} \\plus{} x \\plus{} C} \\equal{} e^c e^{\\frac{x^2}{2} \\plus{} x}$.\n\n$ e^C$ is positive, but we get the $ \\pm$ from the absolute value, so the general solution is $ y \\equal{} K e^{\\frac{x^2}{2} \\plus{} x}\\minus{}1$\nfor some $ K$.\n\nNow, plug in $ (x,y) \\equal{} (0,0)$, which is just one point on the function, to get $ 0\\equal{}K\\minus{}1 \\implies K\\equal{}1$, so $ y \\equal{} e^{\\frac{x^2}{2} \\plus{} x}\\minus{}1$[/hide]\n\n[hide=\"D\"]If you squareroot, you must assume positive and negative values, and then find the right one.\n\n$ x^2\\minus{}y^2 \\equal{} 13$ and $ xy \\equal{} \\minus{} 10 \\sqrt{3} \\implies x^2y^2 \\equal{} 300$\n\nLet $ (u,v) \\equal{} (x^2,y^2)$. Then, $ u\\minus{}v \\equal{} 13$ and $ uv \\equal{} 300$, thus $ u(u\\minus{}13) \\equal{} 300 \\implies u \\equal{} \\minus{}12 \\; \\; or \\; \\; u \\equal{} 25$\n\nClearly, $ u^2 > 0$ so $ u \\equal{} 25 \\implies x \\equal{} \\pm 5$\n\nSubstituting, $ u\\minus{}v \\equal{} 13 \\implies v \\equal{} 12 \\implies y \\equal{} \\pm 2 \\sqrt{3}$\n\nFinally, you have that the simplified answer must be $ \\pm 5 \\pm 2 \\sqrt{3} i$.\n\nNote that if the signs are the same, then you get $ 13 \\plus{} 20 \\sqrt{3} i$ on the inside of the squareroot, so they must be different. And graphing in the complex plane makes it clear that we have $ 5 \\minus{} 2 \\sqrt{3} i$ [/hide]\n\n[hide=\"a\"]$ x^2 \\plus{} m^2 x^2 \\minus{} 2ax \\equal{} 0$ by direct substitution of $ y\\equal{}mx$.\n\nNow, we may either have $ x\\equal{}0$, in which case $ y\\equal{}0$ as well, giving $ (0,0)$\n\nOtherwise, we may divide by $ x$ to get $ x\\plus{}m^2 x \\minus{} 2a \\equal{} 0 \\implies (m^2\\plus{}1) x \\equal{} 2a \\implies x \\equal{} \\frac{2a}{m^2\\plus{}1}$\n\nSince $ y\\equal{}mx$, we have the associated $ y$ value of $ \\frac{2ma}{m^2\\plus{}1}$, thus $ \\left( \\frac{2a}{m^2\\plus{}1}, \\; \\frac{2ma}{m^2\\plus{}1} \\right)$ is the other intersection point.\n[/hide]",
  "Solution_4": "Q9\r\n\r\n[hide]\\[ \\displaystyle\\sum_{k\\equal{}1}^n \\frac{1}{k(k\\plus{}2)}\\equal{}\\displaystyle\\sum_{k\\equal{}1}^n \\frac{1}{2k}\\minus{}\\displaystyle\\sum_{k\\equal{}1}^n \\frac{1}{2(k\\plus{}2)}\\] Start writing out terms and it telescopes.[/hide]\r\n\r\n\r\nYour Q10 doesn't seem to be complete.",
  "Solution_5": "for part b it's just the half-angle identity $ \\sin^2(\\frac{x}2)\\equal{}\\frac{1\\minus{}\\cos(x)}2.$",
  "Solution_6": "[hide=\"Problem 13\"]\n$ 2cos^2 \\theta \\plus{}3 sin \\theta \\equal{} 0 \\Leftrightarrow 2(1\\minus{}sin^2 \\theta)\\plus{}3sin \\theta\\equal{}0 \\Leftrightarrow 2sin^2 \\theta\\minus{}3sin \\theta\\minus{}2\\equal{}0$\n$ \\star \\ \\ sin\\theta \\equal{}2$, impossible \n$ \\star \\ \\ sin \\theta\\equal{}\\minus{}0.5 \\Rightarrow \\theta\\equal{}\\frac{\\minus{}\\pi}{6}\\plus{}k\\cdot 2\\pi$ or $ \\theta\\equal{}\\frac{7\\pi}{6}\\plus{}k\\cdot 2\\pi$ where $ k \\in Z$\n[/hide]\n\n[hide=\"Problem 16\"]\nSet $ x \\equal{} \\pi \\minus{} t \\Rightarrow dx \\equal{} \\minus{} dt$\n$ I \\equal{} \\int_{0}^{\\pi}{\\frac {x sin x}{1 \\plus{} cos^2 x}dx} \\equal{} \\int_{ \\minus{} \\pi}^{0}{\\frac {(\\pi \\minus{} t) sin t}{1 \\plus{} cos^2 t}( \\minus{} dt)} \\equal{} \\int_{0}^{\\pi}{\\frac {(\\pi \\minus{} t) sin t}{1 \\plus{} cos^2 t}dt}$\n$ \\Rightarrow I \\equal{} \\frac {\\pi}{2}\\int_{0}^{\\pi}{\\frac {sin t}{1 \\plus{} cos^2 t}dt} \\equal{} \\frac {\\pi}{2}\\int_{0}^{\\pi}{\\frac {d_{cos t}}{1 \\plus{} cos^2 t}} \\equal{} \\frac { \\minus{} \\pi}{2}.\\left[arctan\\ \\cos t\\right]_{0}^{\\pi} \\equal{} \\left(\\frac {\\pi}{2}\\right)^2$\n[/hide]",
  "Solution_7": "Ty for your help!\r\n\r\nNow what about the other questions?\r\n[b]\u2713-ed[/b]\\hidden ones means I have understood it..."
}
{
  "Tag": [
    "calculus"
  ],
  "Problem": "Prove that: f(n)= n2 + 3n is O(n^2)\r\n\r\n\r\nThis what I have:\r\nn^2 + 3n <= n^2 + 3n^2 <= 4n^2 <= n^2",
  "Solution_1": "This really belongs in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=296]Calculus Computations and Tutorials[/url] section of the forum, you'll have more luck there.\r\n\r\nFor $ n \\geq 3$, we have $ n^2 \\leq f(n) \\leq n^2\\plus{}3n^2\\equal{}4n^2 \\implies 1 \\leq \\frac{f(n)}{n^2} \\leq 4$\r\n\r\nSo, as $ n$ gets arbitrarily large, $ \\frac{f(n)}{n^2}$ is bounded."
}
{
  "Tag": [
    "trigonometry",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "Let $A=\\{x|2^{1+x}+2^{1-x}=a\\}, \\ B=\\{\\sin\\theta|\\theta\\in\\mathbb R\\}$, and $A\\cap B$ be a singleton. Find the range of $a$.",
  "Solution_1": "still no one do it?",
  "Solution_2": "Obv. $B=[-1,1]$.\r\nThe equation \r\n(1)   $2^{1+x}+2^{1-X}=a$\r\nshould have a single solution in the interval $[-1,1]$.\r\nIf $x_0$ is a solution in $[-1,1]$, then $-x_0$ is also a solution in $[-1,1]$, since\r\n$2^{1+x_0}+2^{1-x_0}=a \\mbox { \\Rightarrow }2^{1-x_0}+2^{1+x_0}=a$\r\nand $x_0 \\in [-1,1] \\mbox { \\Rightarrow }-x_0\\in [-1,1]$.\r\nHence, the only case in which equation (1) has a single solution $x_0$in $[-1,1]$ is that when $x_0=0$.\r\nSubsituting $x=x_0=0$ in $(1)$ we get $a=4$ and this is also sufficient for equation $(1)$ to have a single solution."
}
{
  "Tag": [
    "percent",
    "probability",
    "expected value"
  ],
  "Problem": "Would you rather have a 50 percent chance of winning 1,000,000 dollars or a 20 percent chance of winning 3,000,000?",
  "Solution_1": "[hide=\"I highly doubt that either situation will occur any time soon\"]The second choice, because it has an expected value of $ \\$$600,000 as opposed to $ \\$$500,000 for the first one[/hide]",
  "Solution_2": "Personally, I'd rather take the first one.  Expected value is only a meaningful comparison if the event is going to happen a large number of times relative to the probabilities involved, which this one isn't.\r\n\r\nExample:\r\n\r\nI'll let you toss a coin until you get tails.  If you get $ n$ heads, I will give you $ 2^{n}$ pennies.  \r\n\r\nThis game has [i]infinite[/i] expected value, but would you ever actually play it if you had to pay, say, a dollar every time?",
  "Solution_3": "[quote=\"t0rajir0u\"]would you ever actually play it if you had to pay, say, a dollar every time?[/quote]\r\n\r\nOnly if I got to play a hell of a lot of times :)",
  "Solution_4": "[quote=\"t0rajir0u\"]Personally, I'd rather take the first one.  Expected value is only a meaningful comparison if the event is going to happen a large number of times relative to the probabilities involved, which this one isn't.\n\nExample:\n\nI'll let you toss a coin until you get tails.  If you get $ n$ heads, I will give you $ 2^{n}$ pennies.  \n\nThis game has [i]infinite[/i] expected value, but would you ever actually play it if you had to pay, say, a dollar every time?[/quote]\r\n\r\nThe people who wrote the book probably aren't expecting you to actually think about that. It's just supposed to be an exercise in expected value, and you're supposed to just plug in the numbers and take that answer. This has been mathnerd314's cynical ranting. Thank you."
}
{
  "Tag": [
    "analytic geometry",
    "graphing lines",
    "slope"
  ],
  "Problem": "The line $ y\\equal{}3\\minus{}2x$ contains points in how many quadrants of\nthe Cartesian coordinate plane?",
  "Solution_1": "[asy]draw((-25,0)--(25,0));\ndraw((0,25)--(0,-25));\ndraw((-11,25)--(14,-25),BeginArrow,EndArrow);\nlabel(\"$x$\",(25,0),E);\nlabel(\"$y$\",(0,25),N);\nlabel(\"$25$\",(25,0),S);\nlabel(\"$25$\",(0,25),E);\nlabel(\"$-25$\",(-25,0),S);\nlabel(\"$-25$\",(0,-25),E);[/asy]\r\n\r\nThus, the answer is $ \\boxed{3}$.",
  "Solution_2": "Q1: x>0, y>0\r\nQ2: x<0, y>0\r\nQ3: x<0, y<0\r\nQ4: x>0, y<0\r\n\r\nExamples:\r\nQ1: y=2, x=1/2\r\nQ2: y=5, x=-1\r\nQ3: Not possible\r\nQ4: y=-1, x=2.\r\n\r\n3 Quadrants",
  "Solution_3": "you dont have to graph it...\r\n\r\na line will pass through 3 quadrants unless the slope is 0 or undefined or b=0 where $ y\\equal{}ax\\plus{}b$",
  "Solution_4": "Don't forget the case where the line passes through the origin, it then only passes through two quadrants."
}
{
  "Tag": [
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "Each of the players in a tennis tournament played one match against each of the others. If every player won at least one match, show that there are three players $A$, $B$, and $C$ such that $A$ beats $B$, $B$ beats $C$, and $C$ beats $A$. Such a triple of player is called a [b]cycle[/b]. Determine the number of maximum [b]cycles [/b]such a tournament can have.",
  "Solution_1": "The first half is equivalent to the statement, \"Every acyclic tournament is totally ordered, i.e. there is a player who wins all matches, a player who loses only once, etc., and in particular a player who loses all matches.\"  This is classical.  I suspect the second half has been on the forum before, and recommend a search."
}
{
  "Tag": [
    "function",
    "geometry"
  ],
  "Problem": "This is an SAT question and I'm not sure if it goes here...\r\n\r\nA function f(c) is defined as the area of the circle with circumference c. For what value of z is f(z) equal to f(3)+f(4)?\r\n\r\nThis was a grid in so there is no answer choices any help is great. Thanks.",
  "Solution_1": "[hide=\"Hint\"]$ c\\equal{}2\\pi r \\Rightarrow r\\equal{} \\frac{c}{2\\pi}$. So $ f(c) \\equal{} \\pi\\left(\\frac{c}{2\\pi}\\right)^2 \\equal{} \\frac{c^2}{4\\pi}$[/hide]\n[hide=\"Solution\"]$ f(3)\\plus{}f(4) \\equal{} \\frac{3^2\\plus{}4^2}{4 \\pi} \\equal{} \\frac{5^2}{4 \\pi} \\equal{} f(5)$. So $ z\\equal{}5$.[/hide]",
  "Solution_2": "[hide=\"Solution\"]\nThe area of a circle is $ r^2\\pi$, the circumference of a circle is $ 2\\pi r$ where $ r$ is the radius.\n\nSo if $ f(c)$ is the area of a circle with circumference $ c$, then $ f(c) \\equal{} \\dfrac{c^2}{4\\pi}$\n\nNext we want to find a $ z$ such that $ f(z) \\equal{} f(3) \\plus{} f(4)$, so we calculate the R.H.S:\n\n$ f(3) \\plus{} f(4) \\equal{} \\dfrac{3^2}{4\\pi} \\plus{} \\dfrac{4^2}{4\\pi} \\equal{} \\dfrac{25}{4\\pi} \\equal{} \\dfrac{5^2}{4\\pi} \\equal{} f(5)$.\n\nA side note: If $ f(a)\\plus{}f(b)\\equal{}f(c)$ then triangle $ abc$ is a right triangle.\n[/hide]\r\n\r\nEdit: Beaten..",
  "Solution_3": "Thanks for the help guys I thought I got 5 too but I didn't think it was right"
}
{
  "Tag": [
    "mathleague.org",
    "\\/closed"
  ],
  "Problem": "At the request of several members, we've built a tool that will allow you to include a little window into the Forum from your own site.  As you'll see, it's highly customizable, allowing you to choose which forums you can see, window sizes, colors, fonts, font styles, etc.\r\n\r\nTo learn how to use it click here:\r\n\r\n[url]http://www.artofproblemsolving.com/Forum/marquee.php?help=1[/url]\r\n\r\nPlease let me know if you encounter any problems with it (but please read the instructions closely first if you have a problem!)\r\n\r\n\r\n(Special thanks to Tim Sanders of [url=http://www.mathleague.org]mathleague.org[/url] for making a suggestion that led to this being built).",
  "Solution_1": "Nifty! Also good for advertising AoPS on a banner ad.",
  "Solution_2": "Awesome! Lots of good uses for this. Now it's easier for me to monitor topics in different forums. Thanks.  :first:",
  "Solution_3": "Nice thing... but I think you should at least check permissions. Now anyone can check moderator forum, parents forum, etc... ;)",
  "Solution_4": "Have you confirmed that you can look into these forums?  Please show me a sample if so - I thought I had cut that off.",
  "Solution_5": "I logged out and could still view forum 373. (just replacing the forums with 373 in the example link)\r\n\r\nMaybe it's just one forum forgotten on the list if you made sort of a blocklist :)",
  "Solution_6": "Try it from a browser in which you are not logged in to the site - you have permission to see that forum, so you can get it in the marquee.  (Let me know if that fails.)",
  "Solution_7": "No, I don't think you can. I tried 373 from a friends account and 301 from mine, and couldn't get it. Though if those are the only forums listed on your account, it would be nice if it had a Permission denied message instead of blankness.",
  "Solution_8": "Which websites can this work in?\r\n\r\nIt doesn't work for my xanga, freewebs, or myspace.",
  "Solution_9": "I'm guessing they don't allow the 'iframe' tag, which would make it harder for them to control what sort of content appears on their site.",
  "Solution_10": "Um....there's a bug. A topic is shown that it is authored by a certain person, but it actually shows the person who last responded to it as the author. :huh:",
  "Solution_11": "great!!!!!! this is good for Internet explorer  :lol:"
}
{
  "Tag": [],
  "Problem": "does any one know how to apply for aptitude test which is for B.Arch \r\ni filled my form online and there was no such option",
  "Solution_1": "You can submit it only after 26th May 09 when JEE results are out\r\nSee this link\r\nhttp://jee.iitd.ac.in/aptitude.htm",
  "Solution_2": "can anyone suggest some books for the aptitude test for barch that i should go through",
  "Solution_3": "somebody ,please help"
}
{
  "Tag": [
    "geometry",
    "rectangle"
  ],
  "Problem": "Each point in a plane is colored either red or blue.  Show that there exists a rectangle whose all vertices are of the same color.",
  "Solution_1": "[hide=\"An Approach\"]\nConsider a rectangular array with 3 rows and 9 columns.\n\nEach column has 3 points, thus there are 2^3 = 8 possible different columns. Thus if there are 9 columns in the array, two must be the same.\n\nAlso, since there are 3 points in each column, at least two must be of the same color.\n\nTaking two points of like color from each of two identical columns will result in a rectangle with similarly colored vertexes\n[/hide]"
}
{
  "Tag": [
    "inequalities",
    "geometry",
    "perimeter",
    "inradius",
    "circumcircle",
    "inequalities unsolved"
  ],
  "Problem": "[b]Q.1.[/b] For triangle ABC,\r\n\r\n$ (a^2 \\plus{} b^2 \\plus{} c^2) \\geq \\frac{36}{35}(s^2 \\plus{} \\frac{abc}{s})$\r\n\r\n\r\n\r\n[b]Q. 2.[/b] For an acute angled triangle ABC prove that\r\n\r\n$ s^2 \\leq \\frac{27R^2}{27R^2\\minus{}8r^2} (2R \\plus{} r)^2$\r\n\r\n  s \u2013 semi-perimeter, r \u2013 inradius, R - circumradius",
  "Solution_1": "[color=red][b]Anybody got a clue ?[/b][/color]\r\n\r\n\r\n\r\n\r\n[color=blue][i].......[/i][/color]",
  "Solution_2": "Both the inequalities are easily solvable using the substitutions $ a\\equal{}y\\plus{}z, b\\equal{}z\\plus{}x, c\\equal{}x\\plus{}y$ where $ x,y,z$ are positive reals.\r\nTHe first one is really simple. Just keep collecting like terms and prove $ LHS\\minus{}RHS \\ge 0$.\r\nAs for the second one, use $ R\\ge 2r$. $ R\\equal{}\\frac{(x\\plus{}y)(y\\plus{}z)(z\\plus{}x)}{4\\sqrt{xyz(x\\plus{}y\\plus{}z)}}$. $ r\\equal{}\\sqrt{\\frac{xyz}{x\\plus{}y\\plus{}z}}$",
  "Solution_3": "hello, for the first one we have \r\n$ {a^2+b^2+c^2\\geq\\frac{4}{3}s^2=\\frac{36}{35}\\left(s^2+(\\frac{2}{3}s)^3\\cdot\\frac{1}{s}\\right)=\\frac{36}{35}\\left(s^2+(\\frac{a+b+c}{3})^3\\cdot\\frac{1}{s}\\right)\\geq\\frac{36}{35}\\left(s^2+\\frac{abc}{s}\\right)}$, \r\nwe have used that\r\n$ a^2+b^2+c^2\\geq\\frac{1}{3}(a+b+c)^2$\r\nand\r\n$ \\left(\\frac{1}{3}(a+b+c)\\right)^3 \\geq abc$.\r\nSonnhard."
}
{
  "Tag": [
    "number theory open",
    "number theory"
  ],
  "Problem": "Find all positive integers $ n > 1$ such that for any two co prime factors\r\n$ a, b$ of $ n$ the number $ a \\plus{} b \\minus{} 1$ is also a factor of $ n$.\r\n\r\n\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=267713]click[/url]",
  "Solution_1": "Obviously $ n\\equal{}p^s$ satisfies for any prime $ p$ and any natural $ s$.\r\nNow assume $ n$ has at least two different prime factor and $ p$ is the smallest.\r\nWe write $ n\\equal{}p^sN$, where $ (N,p)\\equal{}1$ and $ N$ has only greater-than-p factor.\r\n\r\nCase 1) $ p>2$.\r\nSince $ N\\plus{}p\\minus{}1|n$, $ (N\\plus{}p\\minus{}1,p)>1$ and/or $ (N\\plus{}p\\minus{}1,N)>1$.\r\n$ (N\\plus{}p\\minus{}1,N)\\equal{}(p\\minus{}1,N)<p$. But $ N$ has only greater-than-p factor. So $ (N\\plus{}p\\minus{}1,N)\\equal{}1$\r\nNow $ (N\\plus{}p\\minus{}1,p)>1$ implies \r\n$ N\\plus{}p\\minus{}1\\equal{}p^t$ with $ 2\\le t\\le s$. (*)\r\nSince $ s\\ge2$, we have $ N\\plus{}p^2\\minus{}1|n$. \r\nSince $ p^2\\minus{}1\\equal{}(p\\plus{}1)(p\\minus{}1)$ has no greater-than-p factor, $ (N\\plus{}p^2\\minus{}1,N)\\equal{}1$.\r\nSo $ N\\plus{}p^2\\minus{}1\\equal{}p^u$ with $ 2\\le u\\le s$. (**)\r\nNow (*) and (**) give us\r\n$ p^2\\minus{}p\\equal{}p^u\\minus{}p^t\\equal{}p^t(p^{u\\minus{}t}\\minus{}1)\\Rightarrow t\\equal{}1,u\\equal{}2$.\r\nThen $ N\\equal{}1$. Contradiction!\r\n\r\nCase 2) $ p\\equal{}2$.\r\n$ (N\\plus{}1,N)\\equal{}1$ implies $ N\\plus{}1\\equal{}2^w$ with $ 2\\le w\\le s$\r\nSince $ s\\ge2$, $ N\\plus{}2^2\\minus{}1\\equal{}N\\plus{}3|n$.\r\n$ (N\\plus{}3,N)\\equal{}(3,N)$ implies $ N\\plus{}3\\equal{}3^u2^v$ with $ 1\\le v\\le s$.\r\n\r\n$ 3^u2^v\\minus{}2^w\\equal{}2$ implies $ v\\equal{}1$ and thus\r\n$ 3^u\\minus{}2^{w\\minus{}1}\\equal{}1$ (***)\r\n\r\nThe only possible solution for (***) is $ (u,w)\\equal{}(1,2)$ or $ (u,w)\\equal{}(2,4)$\r\nThen $ N\\equal{}2^w\\minus{}1$ is either $ 3$ or $ 15$.\r\nIn other words, $ n\\equal{}2^s*3$ or $ n\\equal{}2^s*15$, where $ s\\ge2$.\r\nFor $ n\\equal{}2^s*3$, if $ s\\ge3$, $ 8\\plus{}3\\minus{}1\\equal{}10|n$. Contradiction!\r\nFor $ n\\equal{}2^s*15$, $ 4\\plus{}15\\minus{}1\\equal{}18|n$. Contradiction!\r\n\r\nSo $ n\\equal{}12$ is the only solution other than power of primes."
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "perimeter"
  ],
  "Problem": "The diagonal of a cube is represented by 2x. Find the total surface of the cube in terms of x.",
  "Solution_1": "if its space diagonal, with the side of the cube s, then 2x=Sqrt(3)*s. then s=2x/sqrt(3), and 6s^2=8x^2.\r\nso the answer is 8x^2.",
  "Solution_2": "first of all, you posted the same problem twice.\r\nsecond, maybe you should post all of your questions under one thread.\r\nthird, the diagonal of a cube with side $ s$ is always $ s\\sqrt{3}$. therefore, the side length of the cube is $ \\frac{2x}{\\sqrt{3}}$. the surface area of a cube with side $ s$ is always $ 6s^2$, so the surface area is $ 6(\\frac{2x}{\\sqrt{3}})^2\\equal{}6(\\frac{4x^2}{3})\\equal{}8x^2$",
  "Solution_3": "We can use the Pythag. Theorem to figure this out. \r\n$ a^2\\plus{}b^2\\equal{}c^2$\r\n$ a^2\\plus{}b^2\\equal{}2x$\r\nIn this case, $ a\\equal{}b$ because we have a cube in which all sides are equal length.\r\n$ a^2\\plus{}a^2\\equal{}2x$\r\n$ 2(a^2)\\equal{}2x$\r\nWe divde both sides by 2...\r\n$ a^2\\equal{}x$\r\nWe \"sqaure root-ify\" both sides\r\n$ \\sqrt{a^2}\\equal{}\\sqrt{x}$\r\n$ a\\equal{}\\sqrt{x}$\r\nThe surface area of a cube is Perimeter times height\r\n$ [4(\\sqrt{x})][\\sqrt{x}]$\r\nWhich is $ 5\\sqrt{x}$\r\n[b]I feel like I made a mistake along the way but I can't spot it... What did I do?![/b]",
  "Solution_4": "He said the space diagonal. The space diagonal is a line that starts from one corner that runs to the opposite corner running through and inside the cube.",
  "Solution_5": "Oh... Hah... my bad.... But still, If you were to use it, what would you find?"
}
{
  "Tag": [
    "function",
    "logarithms",
    "puzzles"
  ],
  "Problem": "Solomon Golomb's self\u00addescribing sequence $ < f(1), f(2), \\dots>$ is the only non\u00addecreasing sequence of positive integers with the property that it contains exactly $ f(k)$ occurrences of $ k$ for each $ k$. Find the sequence.",
  "Solution_1": "We want $ f(k)$ to be 1 more than $ f(k \\minus{} 1)$ if and only if $ f(k \\minus{} 1)$ has already appeared $ f(f(k \\minus{} 1))$ times in the sequence.  Another way to word this is, we want $ f(k)$ to be 1 more than $ f(k \\minus{} f(f(k \\minus{} 1)))$:\r\n\\[ \\begin{cases} f(k) \\equal{} 1 \\plus{} f(k \\minus{} f(f(k \\minus{} 1))) \\\\\r\nf(1) \\equal{} 1 \\end{cases}\r\n\\]\r\nI don't know if I could find a closed form for this, but at first glance it has me stumped.  Though you could probably get some asymptotic bounds.  Any ideas?",
  "Solution_2": "isn't it kind of obvious?\r\nlet f(1)=1.\r\nthen the minimum value of f(2)=2, since it cannot be 1.\r\nthe minimum value of f(3)=2\r\nthen f(4)=3 and f(5)=3\r\nthen f(6)=f(7)=f(8)=4 and f(9)=f(10)=f(11)=5\r\nand keep constructing the sequence like so.",
  "Solution_3": "Right, which is the recurrence I posted above.  But I imagine the OP was looking for a closed form for $ f(k)$ as a function of $ k$, or at least for asymptotic bounds, i.e. a function $ \\phi(k)$ so that $ \\frac {f(k)}{\\phi(k)}\\to 1$ as $ k\\to \\infty$.",
  "Solution_4": "There is a closed form which involves exponential function or logarithm, but I can't remember......",
  "Solution_5": "Well I managed to prove $ \\frac{f(k)}{\\phi(k)}\\to 1$ as $ k\\to \\infty$ where\r\n\\[ \\phi(k)\\equal{}\\left(\\frac{1\\plus{}\\sqrt{5}}{2}\\right)^\\frac{3\\minus{}\\sqrt{5}}{2}k^{\\frac{\\sqrt{5}\\minus{}1}{2}}\\]\r\nCan't say too much about the remainder though, except that it decays pretty slowly."
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "trigonometry",
    "angle bisector"
  ],
  "Problem": "An acute triangle $ ABC$ with $ AB \\neq AC$ is given. Let $ V$ and $ D$ be the feet of the altitude and angle bisector from $ A$, and let $ E$ and $ F$ be the intersection points of the circumcircle of $ \\triangle AVD$ with sides $ AC$ and $ AB$, respectively. Prove that $ AD$, $ BE$ and $ CF$ have a common point.",
  "Solution_1": "[quote=\"April\"]An acute triangle $ ABC$ with $ AB \\neq AC$ is given. Let $ V$ and $ D$ be the feet of the altitude and angle bisector from $ A$, and let $ E$ and $ F$ be the intersection points of the circumcircle of $ \\triangle AVD$ with sides $ AC$ and $ AB$, respectively. Prove that $ AD$, $ BE$ and $ CF$ have a common point.[/quote]\r\nAn easy problem .\r\nWe have $ DE \\perp AC,DF\\perp AB$ and $ AD$ is bisector of $ \\angle {ABC}$  so \r\n$ AE \\equal{} AF$ \r\nOther $ CF \\equal{} CD.\\cos {\\angle {ACB}},BE \\equal{} BD.\\cos {\\angle {ABC}}$ \r\n $ VB \\equal{} AB.\\cos {\\angle {ABC}},VC \\equal{} AC.\\cos {\\angle {ACB}}$\r\nFollow that : \r\n$ \\frac {VB}{VC}.\\frac {FC}{FA}.\\frac {EA}{EB} \\equal{} \\frac {AB}{AC}.\\frac {CD}{BD} \\equal{} 1$\r\nSo $ AV,BE,CF$ concurrent (Ceva's theorem )",
  "Solution_2": "[quote=\"April\"][color=darkred] An acute triangle $ ABC$ with $ AB \\neq AC$ is given. Let $ V$ and $ D$ be the feet of the altitude and angle bisector from $ A$, and let $ E$ and $ F$ be the intersection points of the circumcircle of $ \\triangle AVD$ with sides $ AC$ and $ AB$, respectively. Prove that $ AV$, $ BE$ and $ CF$ have a common point. [/color][/quote]\n\n[color=darkblue]Here is an easy extension which generates and another interesting problem ![/color]\n\n[quote=\"Virgil Nicula\"][color=darkred]Let $ ABC$ be an acute triangle with $ c < b$ and let $ M\\in (BC)$ , $ N\\in (MC)$ so that $ \\{\\begin{array}{c} x = m(\\widehat {AMC}) \\\\\n\\ y = m(\\widehat {ANB})\\end{array}$ .\n\nThe circumcircle of the triangle $ AMN$ cut again the sides $ [AB]$ , $ [AC]$ in the points $ F$ , $ E$ respectively. \n\nThen the lines $ BE$ , $ CF$ , $ AM$ are concurrently $ \\Longleftrightarrow$ $ \\cos (C - B + x - 2y) + \\cos A\\cos x = 0$ .[/color][/quote]\n\n[color=darkblue][b][u]Particular cases.[/u] [/b]\n\n$ 1\\blacktriangleright\\ x: = 90^{\\circ}$ , i.e. $ AM\\perp BC$  . In this case, $ \\{\\begin{array}{c} C - B + 90^{\\circ} - 2y = \\pm 90^{\\circ} \\\\\n\\ (\\ C\\ < \\ B\\ )\\end{array}$ $ \\Longleftrightarrow$ $ y = C + \\frac A2$ $ \\Longleftrightarrow$ $ \\widehat {NAB}\\equiv\\widehat {NAC}$ .\n\n$ 2\\blacktriangleright\\ A: = 90^{\\circ}$ . In this case obtain the following nice problem :[/color]\n\n[quote=\"Virgil Nicula\"][color=darkred]Let $ ABC$ be a $ A$ - right triangle, i.e. $ AB\\perp AC$ with $ c < b$ . Choose two points  $ \\{\\begin{array}{c} M\\in (BC) \\\\\n\\ N\\in (MC)\\end{array}$ . \n\nThe circumcircle $ w$ of the triangle $ AMN$ cut again the sides $ [AB]$ , $ [AC]$ in the points $ F$ , $ E$ respectively. \n\nThen the lines $ BE$ , $ CF$ , $ AM$ are concurently $ \\Longleftrightarrow$ $ m(\\widehat {AMC}) = 2\\cdot m(\\widehat {ANC})$ .[/color][/quote]\n\n[color=darkblue]$ 3\\blacktriangleright$ Here is a very easy problem which, for $ A = 90^{\\circ}$, is equivalently with the previous problem :[/color]\n\n[quote=\"Virgil Nicula\"][color=darkred]Let $ ABC$ be a triangle. For a point  $ M\\in (BC)$ denote the points \n\n$ \\{\\begin{array}{c} F\\in (AB)\\ ,\\ \\widehat {FMA}\\equiv\\widehat {FMB} \\\\\n \\\\\nE\\in (AC)\\ ,\\ \\widehat {EMA}\\equiv\\widehat {EMC}\\end{array}$ . Then the lines $ BE$ , $ CF$ , $ AM$ are concurently.[/color][/quote]",
  "Solution_3": "We note that\n\n$AE=AF$\n\n$CE=DCcosC$\n\n$BF=BDcosB$\n\nSo by converse of Ceva,done.......\nBut can anyone give a solution using the fact that the radical axes of three circles taken pairwise are concurrent....\n\n\nMaths is the doctor of science\n\nSayantan",
  "Solution_4": "$\\angle AVF = \\angle AEF = \\angle AFE = \\angle AVE \\implies (VA, VB; VE, VF) = -1$.",
  "Solution_5": "well an easy problem and as everyone has pointed,it can be easily done by Ceva's converse.\nMeanwhile there is a typo in the main problem it is written that prove $AD,BE and CF$ are concurrent.I think it would be $AV,BE and CF$.",
  "Solution_6": "Since $ AF=AE $ we need to prove that $ \\frac{BV}{VC}\\frac{CE}{BF}=1 $ By the power of a point theorem $ BV=\\frac{BF}{BD}BA $ so it remains to prove that $ \\frac{CE}{CV}\\frac{BA}{BD} =1 $ Note that $ \\frac{CE}{CV}=\\frac{CD}{CA} $ and now $ \\frac{CD}{CA}\\frac{BA}{BD}=\\frac{sin \\angle DAC}{sin \\angle ADC} \\frac{sin \\angle BDA}{sin \\angle BAD}=1 $",
  "Solution_7": "Very easy using Ceva! :D\n$\\angle FAD = \\angle EAD \\wedge DF \\perp AF, DE \\per AE \\wedge AD$ common side $\\implies \\triangle AFD \\cong \\triangle AED$.\n$\\therefore ED = FD$ and $AF = AE$.\nNow, $\\frac{AF}{FB} \\frac{BV}{VC} \\frac{CE}{EA} = \\frac{BV}{VC} \\frac{ED/ tan \\angle C}{FD/ tan \\angle B} = \\frac{BV}{VC} \\frac{AV/ tan \\angle C}{AV/ tan \\angle B}=1$.\n$\\therefore AV, BE, CF$ concur!",
  "Solution_8": "Dear Mathlinkers,\na proof with harmonic division is also possible...\nSincerely\nJean-Louis",
  "Solution_9": "[hide=\"Solution using harmonic division!\"]\n\nLet $BC \\cap EF = P$ and $AV \\cap EF = Q$.\n$AE=AF$ and $AFVE$ cyclic. So we must have $VA$ is the angle bisector of $\\angle FVE$ and $AV \\perp BC$. So $PFQE$ is a harmonic division.\n$\\therefore (P, V; B, C) \\stackrel{A}{=} (P, Q; F, E) = -1, i.e.,$ it's also harmonic!\nThus $AV, BE, CF$ must concur.[/hide]",
  "Solution_10": "Any solution using Desargues theorem? ",
  "Solution_11": "[url=https://artofproblemsolving.com/community/c6h1612634p10075334]Korea National Olympiad 1996 P8[/url]"
}
{
  "Tag": [
    "ratio"
  ],
  "Problem": "If $ \\frac{a}{b}\\equal{}\\frac{3}{4}, \\frac{b}{c}\\equal{}\\frac{8}{9}$ and $ \\frac{c}{d}\\equal{}\\frac{2}{3}$, what is the value of $ \\frac{ad}{b^2}$   Express your answer as a common fraction.",
  "Solution_1": "ok, let's make each of the letters the same number\r\n\r\nthe third fraction, $ \\frac {2} {3} \\equal{} \\frac {c} {d}$ and the second fraction, $ \\frac {8} {9} \\equal{} \\frac {b} {c}$, have a c in common. let's make the two c's the same value\r\nwe multiply the numerator and denominator of the third fraction by 9, and the numerator and denominator of the second fraction by 2, to make both c's the same\r\n\r\n$ \\frac {2 \\times 9} {3 \\times 9} \\equal{} \\frac {18} {27}$\r\n\r\n$ \\frac {8 \\times 2} {9 \\times 2} \\equal{} \\frac {16} {18}$\r\n\r\nnow let's make the two b values the same. the b value in the second fraction is now equal to 16, so we multiply the numerator and denominator of the first fraction by 4\r\n\r\n$ \\frac {3 \\times 4} {4 \\times 4} \\equal{} \\frac {12} {16}$\r\n\r\nso now we have the values, \r\n$ \\frac {a} {b} \\equal{} \\frac {12} {16}, \\frac {b} {c} \\equal{} \\frac {16} {18}, \\frac {c} {d} \\equal{} \\frac {18} {27}$\r\n\r\nwe have the values for a, b, c, and d now, so we can solve our problem\r\n\r\n$ a \\equal{} 12, b \\equal{} 16, c \\equal{} 18, d \\equal{} 27$\r\n\r\nwe plug it into our equation\r\n$ \\frac {ad} {b^2} \\equal{} \\frac {12 \\times 27} {16^2} \\equal{} \\frac {324} {256} \\equal{} \\frac {81} {64}$\r\n\r\ntherefore, our final answer is $ \\frac {81} {64}$\r\n\r\n\r\n\r\nif anyone sees any mistakes in that, plz tell me",
  "Solution_2": "Vallon put the ratios so that they are in relation with each other.\r\n\r\n$ A : B : C : D \\equal{} 12 : 16 : 18 : 27$.\r\n\r\nThis may count as spam.  Not sure."
}
{
  "Tag": [
    "logarithms",
    "trigonometry",
    "calculus",
    "calculus computations"
  ],
  "Problem": "can you sum  :D \r\n\r\n$ \\dfrac{1}{e^2 \\plus{} \\pi^2} \\plus{} \\dfrac{1}{2^2e^2 \\plus{} \\dfrac{\\pi^2}{2^2}} \\plus{} \\dfrac{1}{3^2e^2 \\plus{} \\dfrac{\\pi^2}{3^2}} \\plus{} \\dfrac{1}{4^2e^2 \\plus{} \\dfrac{\\pi^2}{4^2}} \\plus{} \\dfrac{1}{5^2e^2 \\plus{} \\dfrac{\\pi^2}{5^2}} \\plus{} \\dfrac{1}{6^2e^2 \\plus{} \\dfrac{\\pi^2}{6^2}} \\plus{} \\dfrac{1}{7^2e^2 \\plus{} \\dfrac{\\pi^2}{7^2}} \\plus{} \\ldots$",
  "Solution_1": "Does it converge? ;)",
  "Solution_2": "According to Mathematica: No, it doesn't converge. Changing the exponent 2 for e and pi doesn't affect that fact. But it is interesting to look up the series where 2 is substituted with n (the same number which multiplies e and divides pi). That one converges.",
  "Solution_3": "this sum $ \\sum_{1}^{N} ...$  is like something $ \\approx A\\plus{}B\\ln N$",
  "Solution_4": "sorry... fixed !  :(  difficult to concentrate with little kids running around you know.  :oops:",
  "Solution_5": "With the formula in this [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=200404]topic[/url] we can derive the two formulas\r\n\r\n$ \\sum_{k\\in\\Bbb{Z}} \\frac{\\frac{\\alpha^2}{2}}{k^4\\plus{}\\left(\\frac{\\alpha^2}{2}\\right)^2}\\equal{}\\frac{\\pi}{\\alpha}\\, \\frac{\\sinh\\alpha\\pi\\plus{}\\sin\\alpha\\pi}{\\cosh\\alpha\\pi\\minus{}\\cos\\alpha \\pi}$\r\n\r\nand \r\n\r\n$ \\sum_{k\\in\\Bbb{Z}} \\frac{k^2}{k^4\\plus{}\\left(\\frac{\\alpha^2}{2}\\right)^2}\\equal{}\\frac{\\pi}{\\alpha}\\, \\frac{\\sinh\\alpha\\pi\\minus{}\\sin\\alpha\\pi}{\\cosh\\alpha\\pi\\minus{}\\cos\\alpha \\pi}$ .\r\n\r\nYour sum $ S\\equal{}\\sum_{k\\equal{}1}^\\infty \\frac{1}{k^2 e^2\\plus{}\\frac{\\pi^2}{k^2}}$ equals\r\n\r\n$ \\frac12 \\sum_{k\\in\\Bbb{Z}} \\frac{k^2}{k^4 e^2\\plus{}\\pi^2}\\equal{}\\frac{1}{2e^2} \\sum_{k\\in\\Bbb{Z}} \\frac{k^2}{k^4\\plus{}\\left(\\frac{\\sqrt{\\frac{2\\pi}{e}}^2}{2}\\right)^2}$ .\r\n\r\nSo $ S\\equal{}\\frac{1}{2e^2}\\, \\frac{\\pi}{\\sqrt{\\frac{2\\pi}{e}}}\\,\r\n \\frac{\\sinh \\sqrt{\\frac{2\\pi^3}{e}}\\minus{}\\sin \\sqrt{\\frac{2\\pi^3}{e}}}{\\cosh \\sqrt{\\frac{2\\pi^3}{e}}\\minus{}\\cos \\sqrt{\\frac{2\\pi^3}{e}}}$ ."
}
{
  "Tag": [
    "vector",
    "geometry",
    "3D geometry",
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "Let v<sub>i</sub>, 1 \\leq i \\leq 95, be vectors in 3D space, all contained in the cube\r\nC = {(x, y, z) \\in R<sup>3</sup> || |x|, |y|, |z| \\leq 1}.  \r\nFor all vectors of the vorm\r\n\\sum<sub>1 \\leqk \\leq 95</sub>s<sub>k</sub>v<sub>k</sub>\r\nwhere v<sub>i</sub> \\in {-1, 1}, for 1 \\leq i \\leq 95,\r\nfind the smallest d such that there MUST exist a vector (a, b, c) of that form satisfying\r\na<sup>2</sup> + b<sup>2</sup> + c<sup>2</sup> \\leq d.\r\n\r\nAUSTRIA _ POLAND 1995",
  "Solution_1": "Moderators: add to resources section!!",
  "Solution_2": "[quote=\"Arne\"]Let $v_i$, $1 \\leq i \\leq 95$, be vectors in 3D space, all contained in the cube\n$C = {(x, y, z) \\in R^3 || |x|, |y|, |z| \\leq 1}$.  \nFor all vectors of the form\n$\\sum_{1 \\leqq \\leq 95}{s_k v_k}$\nwhere $v_i \\in {-1, 1}$, for $1 \\leq i \\leq 95$,\nfind the smallest $d$ such that there MUST exist a vector $(a, b, c)$ of that form satisfying\n$a^2 + b^2 + c^2 \\leq d$.\n\nAUSTRIA _ POLAND 1995[/quote]\n\nI see this problem in other place and the original says that $d=48$, anyone can help me solving this?, thank you."
}
{
  "Tag": [
    "Vieta",
    "trigonometry"
  ],
  "Problem": "Let $\\alpha$ and $\\beta$ be two differrent values of $x$ satisfying the equation:\r\n$acosx+bsinx=c$ with $a^{2}+b^{2}\\neq 0$\r\nProve that:\r\n1) $4cos^{2}\\frac{\\alpha}{2}cos^{2}\\frac{\\beta}{2}=\\frac{(a+c)^{2}}{a^{2}+b^{2}}$\r\n2) $sin(\\alpha+\\beta)=\\frac{2ab}{a^{2}+b^{2}}$\r\n3) $cos^{2}\\frac{\\alpha-\\beta}{2}=\\frac{c^{2}}{a^{2}+b^{2}}$",
  "Solution_1": "what is c?\r\n\r\njust a constant?!",
  "Solution_2": "When i have posted this problem, i have idea to solve part a of problem:\r\nFrom hyppothesis of it, we have:\r\n$acosx+bsinx=c\\Leftrightarrow acosx-c=-bsinx \\Leftrightarrow( acosx-c)^{2}=(-bsinx)^{2}\\Leftrightarrow (a^{2}+b^{2})cos^{2}x-2acosx+c^{2}-b^{2}=0$\r\nInfer $cos\\alpha$ and $cos\\beta$ are two roots of the equation with variable $t$:\r\n$(a^{2}+b^{2})t^{2}-2act+c^{2}-b^{2}=0$\r\nFollow Vieta' theorem: $cos\\alpha+cos\\beta=\\frac{2ac}{a^{2}+b^{2}}$ and $cos\\alpha.cos\\beta=\\frac{c^{2}-b^{2}}{a^{2}+b^{2}}$\r\nHence:\r\n$4cos^{2}\\frac{\\alpha}{2}.cos^{2}\\frac{\\beta}{2}=(1+cos\\alpha)(1+cos\\beta)=1+(cos\\alpha+cos\\beta)+cos\\alpha.cos\\beta=1+\\frac{2ac}{a^{2}+b^{2}}+\\frac{c^{2}-b^{2}}{a^{2}+b^{2}}=\\frac{(a+c)^{2}}{a^{2}+b^{2}}$",
  "Solution_3": "Hien, the correct way to make the trig expressions is \"\\cos\" not \"cos\". Like this : $\\cos$ versus $cos$. The first is much neater!",
  "Solution_4": "I will prove part 3 of problem:\r\nBy applying similarly the way of part 1 we have: $\\sin\\alpha.\\sin\\beta=\\frac{c^{2}-a^{2}}{a^{2}+b^{2}}$\r\nHence:\r\n$\\cos^{2}\\frac{\\alpha-\\beta}{2}=\\frac{1}{2}(1+\\cos(\\alpha-\\beta))$\r\n$=\\frac{1}{2}(1+\\cos\\alpha.\\cos\\beta-\\sin\\alpha.\\sin\\beta)$\r\n$=\\frac{1}{2}\\left(1+\\frac{c^{2}-b^{2}}{a^{2}+b^{2}}+\\frac{c^{2}-a^{2}}{a^{2}+b^{2}}\\right)=\\frac{c^{2}}{a^{2}+b^{2}}$\r\n\r\nWho can help me solve remain part of one? Part 2!",
  "Solution_5": "Hien . I can help you!!\r\n\r\nSo, we have from part 1\r\n$\\cos \\alpha \\cos \\beta = \\frac{{c^{2}-b^{2}}}{{a^{2}+b^{2}}}(I)$\r\n\r\nand from part 3\r\n\r\n$\\sin \\alpha \\sin \\beta = \\frac{{c^{2}-a^{2}}}{{a^{2}+b^{2}}}(II)$\r\n\r\n\r\ndoing\r\n$II-I$\r\n\r\nwe get\r\n$\\cos (\\alpha+\\beta ) = \\frac{{c^{2}-a^{2}}}{{a^{2}+b^{2}}}-\\frac{{c^{2}-b^{2}}}{{a^{2}+b^{2}}}= \\frac{{b^{2}-a^{2}}}{{a^{2}+b^{2}}}$\r\nnow using the well-known \r\n\r\n$\\sin^{2}(\\alpha+\\beta )+\\cos^{2}(\\alpha+\\beta ) = 1$\r\n\r\n\r\n$\\sin (\\alpha+\\beta ) = \\sqrt{1-\\left({\\frac{{b^{2}-a^{2}}}{{a^{2}+b^{2}}}}\\right)^{2}}= \\frac{{2ab}}{{a^{2}+b^{2}}}$",
  "Solution_6": "Dear [b]felipesa[/b]\r\nIt is very simple, however i don't think that. Thank a lot!"
}
{
  "Tag": [
    "AMC"
  ],
  "Problem": "Define $ x\\otimes y \\equal{} x^3 \\minus{} y$. What is $ h\\otimes (h\\otimes h)$?\r\n\r\n$ \\textbf{(A) } \\minus{} h\\qquad \\textbf{(B) } 0\\qquad \\textbf{(C) } h\\qquad \\textbf{(D) } 2h\\qquad \\textbf{(E) } h^3$",
  "Solution_1": "[hide]\n$h \\otimes (h \\otimes h)$\n$= h \\otimes (h^3 - h)$\n$= h^3 - (h^3 - h)$\n$= h$\n$\\boxed{C}$\n[/hide]",
  "Solution_2": "$h^3-(h^3-h)=\\boxed{h}$"
}
{
  "Tag": [],
  "Problem": "[b]\u0391\u03b3\u03b1\u03c0\u03b7\u03c4\u03bf\u03af \u03c6\u03af\u03bb\u03bf\u03b9 , \u03c0\u03c1\u03bf\u03c3\u03bf\u03c7\u03ae !!![/b]\r\n\r\n \u038c\u03c4\u03b1\u03bd \u03c3\u03c4\u03ad\u03bb\u03bd\u03b5\u03c4\u03b5 links \u03b3\u03b9\u03b1 \u03ba\u03b1\u03bb\u03ac \u03b2\u03b9\u03b2\u03bb\u03af\u03b1 , [color=red]\u03a0\u039f\u03a4\u0395 \u039d\u0391 \u039c\u0397\u039d \u0393\u03a1\u0391\u03a6\u0395\u03a4\u0395 \u03a4\u039f\u039d \u0391\u0393\u0393\u039b\u0399\u039a\u039f \u03a4\u0399\u03a4\u039b\u039f [/color]\u03c3\u03c4\u03b7\u03bd \u03b5\u03c0\u03b9\u03ba\u03b5\u03c6\u03b1\u03bb\u03af\u03b4\u03b1 \u03c4\u03bf\u03c5 \u03bc\u03b7\u03bd\u03cd\u03bc\u03b1\u03c4\u03bf\u03c2 , \u03bf\u03cd\u03c4\u03b5 \u03c3\u03c4\u03bf \u03ba\u03c5\u03c1\u03af\u03c9\u03c2 \u03ba\u03b5\u03af\u03bc\u03b5\u03bd\u03bf !\r\n \u03a4\u03bf\u03bd \u03b2\u03bb\u03ad\u03c0\u03bf\u03c5\u03bd \u03ba\u03b1\u03b9 \u03bf\u03b9 \u03ba\u03b1\u03c4\u03ac\u03c3\u03ba\u03bf\u03c0\u03bf\u03b9 \u03c4\u03c9\u03bd \u03b5\u03ba\u03b4\u03bf\u03c4\u03ce\u03bd \u03ba\u03b1\u03b9 \u03b1\u03c0\u03b1\u03b9\u03c4\u03bf\u03cd\u03bd \u03b1\u03c0\u03cc \u03c4\u03bf\u03bd Vornucu  \u03ae \u03b1\u03c0\u03cc \u03c4\u03bf\u03c5\u03c2 moderators \u03bd\u03b1 \u03c4\u03b1 \u03b4\u03b9\u03b1\u03b3\u03c1\u03ac\u03c8\u03bf\u03c5\u03bd.\r\n \u03a7\u03b8\u03b5\u03c2 \u03b4\u03b5\u03bd \u03bc\u03c0\u03ae\u03ba\u03b1 \u03c3\u03c4\u03bf \u03b5\u03bb\u03bb\u03b7\u03bd\u03b9\u03ba\u03cc \u03c6\u03cc\u03c1\u03bf\u03c5\u03bc , \u03b1\u03bb\u03bb\u03ac \u03bc\u03bf\u03c5 \u03b5\u03af\u03c0\u03b5 \u03c6\u03af\u03bb\u03bf\u03c2 \u03cc\u03c4\u03b9 [color=red]\u03b1\u03c0\u03b1\u03af\u03c4\u03b7\u03c3\u03b1\u03bd \u03b1\u03c0\u03cc \u03bc\u03ad\u03bd\u03b1 [/color]\u03bd\u03b1  \u03b4\u03b9\u03b1\u03b3\u03c1\u03ac\u03c8\u03c9 \u03c4\u03bf \u03c0\u03c1\u03bf\u03b7\u03b3\u03bf\u03cd\u03bc\u03b5\u03bd\u03bf \u03bc\u03ae\u03bd\u03c5\u03bc\u03b1 \u03c4\u03bf\u03c5 Michailk. \u0395\u03c0\u03b5\u03b9\u03b4\u03ae \u03b4\u03b5\u03bd  \u03bc\u03c0\u03ae\u03ba\u03b1, \u03b4\u03b5\u03bd \u03b5\u03af\u03b4\u03b1 \u03b1\u03c5\u03c4\u03ae \u03c4\u03b7\u03bd \u03c0\u03c1\u03bf\u03c4\u03c1\u03bf\u03c0\u03ae \u03ba\u03b1\u03b9 \u03b1\u03c6\u03bf\u03cd \u03b4\u03b5\u03bd \u03b4\u03b9\u03ad\u03b3\u03c1\u03b1\u03c8\u03b1 \u03b5\u03b3\u03ce \u03c4\u03bf \u03bc\u03ae\u03bd\u03c5\u03bc\u03b1 , \u03c4\u03bf \u03ad\u03ba\u03b1\u03bd\u03b1\u03bd \u03ac\u03bb\u03bb\u03bf\u03b9 \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd \u03b1\u03bc\u03b5\u03c1\u03b9\u03ba\u03ae.\r\n\r\n \u039b\u03bf\u03b9\u03c0\u03cc\u03bd , \u03bd\u03b1 \u03b1\u03ba\u03bf\u03bb\u03bf\u03c5\u03b8\u03ae\u03c3\u03bf\u03c5\u03bc\u03b5 \u03cc\u03bb\u03bf\u03b9 \u03bc\u03b1\u03c2 \u03b1\u03c5\u03c4\u03cc\u03bd \u03c4\u03bf\u03bd \u03ba\u03b1\u03bd\u03cc\u03bd\u03b1 , \u03ce\u03c3\u03c4\u03b5 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03bc\u03b1\u03c2 \u03b4\u03b9\u03b1\u03b3\u03c1\u03ac\u03c6\u03bf\u03c5\u03bd \u03c4\u03bf\u03c5\u03c2 \u03c3\u03c5\u03bd\u03b4\u03ad\u03c3\u03bc\u03bf\u03c5\u03c2 \u03b3\u03b9\u03b1 \u03c4\u03b1 \u03ba\u03b1\u03bb\u03ac \u03b7\u03bb\u03b5\u03ba\u03c4\u03c1\u03bf\u03bd\u03b9\u03ba\u03ac \u03b2\u03b9\u03b2\u03bb\u03af\u03b1.\r\n\r\n [u]\u039c\u03c0\u03ac\u03bc\u03c0\u03b7\u03c2[/u]",
  "Solution_1": "K\u03c5\u03c1\u03b9\u03b5 \u039c\u03c0\u03b1\u03bc\u03c0\u03b7,\u03b5\u03c7\u03b5\u03c4\u03b5 \u03b1\u03c0\u03bf\u03bb\u03c5\u03c4\u03bf \u03b4\u03b9\u03ba\u03b9\u03bf.\r\n\u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03b2\u03c1\u03bf\u03cd\u03bc\u03b5 \u03ad\u03bd\u03b1 \u03c4\u03c1\u03cc\u03c0\u03bf \u03bd\u03b1 \u03b1\u03c0\u03bf\u03ba\u03c1\u03cd\u03c8\u03bf\u03c5\u03bc\u03b5 \u03b1\u03c5\u03c4\u03ad\u03c2 \u03c4\u03b9\u03c2 \u03c0\u03bb\u03b7\u03c1\u03bf\u03c6\u03bf\u03c1\u03af\u03b5\u03c2\r\n\u03b1\u03c0\u03bf \u03c4\u03bf\u03c5\u03c2 ''\u03bc\u03c0\u03b1\u03bc\u03c0\u03bf\u03cd\u03bb\u03b5\u03c2'' \u03c4\u03bf\u03c5 \u03b5\u03af\u03b4\u03bf\u03c5\u03c2,\u03b3\u03b9\u03b1\u03c4\u03b9 \u03b5\u03b9\u03bd\u03b1\u03b9\r\n\u03ba\u03c1\u03af\u03bc\u03b1 \u03b2\u03b9\u03b2\u03bb\u03af\u03b1 \u03cc\u03c0\u03c9\u03c2 \u03c4\u03bf Compendiu*,\u03bd\u03b1 \u03b4\u03b9\u03b1\u03b3\u03c1\u03ac\u03c6\u03bf\u03bd\u03c4\u03b1\u03b9!",
  "Solution_2": "\u039a\u03cd\u03c1\u03b9\u03b5 \u03a3\u03c4\u03b5\u03c1\u03b3\u03af\u03bf\u03c5 \u03c0\u03bf\u03bb\u03cd \u03ba\u03b1\u03bb\u03ae \u03b7 \u03b5\u03c0\u03b9\u03c3\u03ae\u03bc\u03b1\u03bd\u03c3\u03b7 \u03c3\u03b1\u03c2.\r\n\u0398\u03b1 \u03ae\u03b8\u03b5\u03bb\u03b1 \u03bd\u03b1 \u03c1\u03c9\u03c4\u03ae\u03c3\u03c9 \u03ba\u03b9 \u03b5\u03b3\u03ce \u03ba\u03ac\u03c4\u03b9 \u03b5\u03c0\u03af \u03c4\u03bf\u03c5 \u03b8\u03ad\u03bc\u03b1\u03c4\u03bf\u03c2 \u03c4\u03c9\u03bd \u03b7\u03bb\u03b5\u03ba\u03c4\u03c1\u03bf\u03bd\u03b9\u03ba\u03ce\u03bd books.\r\n\r\n\u03a3\u03c4\u03bf \u03b4\u03af\u03c3\u03ba\u03bf \u03c4\u03bf\u03c5 \u03c5\u03c0\u03bf\u03bb\u03bf\u03b3\u03b9\u03c3\u03c4\u03ae \u03bc\u03bf\u03c5 \u03ad\u03c7\u03c9 \u03c3\u03c5\u03b3\u03ba\u03b5\u03bd\u03c4\u03c1\u03ce\u03c3\u03b5\u03b9 \u03ba\u03ac\u03c0\u03bf\u03c5 \u03c3\u03c4\u03b1 2000 \u03b2\u03b9\u03b2\u03bb\u03af\u03b1 \u03c3\u03b5 \u03bc\u03bf\u03c1\u03c6\u03ae pdf \u03b3\u03b9\u03b1 \u03c4\u03bf\u03c5\u03c2 \u03ba\u03bb\u03ac\u03b4\u03bf\u03c5\u03c2 \u03c4\u03b7\u03c2 \u0391\u03bd\u03ac\u03bb\u03c5\u03c3\u03b7\u03c2,\u0393\u03b5\u03c9\u03bc\u03b5\u03c4\u03c1\u03af\u03b1\u03c2,\u0386\u03bb\u03b3\u03b5\u03b2\u03c1\u03b1\u03c2,\u039c\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03ae\u03c2 \u03a6\u03c5\u03c3\u03b9\u03ba\u03ae\u03c2,\u03a3\u03c5\u03bd\u03b4\u03c5\u03b1\u03c3\u03c4\u03b9\u03ba\u03ae\u03c2,\u03a4\u03bf\u03c0\u03bf\u03bb\u03bf\u03b3\u03af\u03b1\u03c2,\u039c\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03ae\u03c2 \u039b\u03bf\u03b3\u03b9\u03ba\u03ae\u03c2,\u03b8\u03b5\u03c9\u03c1\u03af\u03b1\u03c2 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ce\u03bd,\u039a\u03c9\u03b4\u03b9\u03ba\u03bf\u03c0\u03bf\u03af\u03b7\u03c3\u03b7\u03c2,\u0398\u03b5\u03c9\u03c1\u03b9\u03b1\u03c2 \u03a7\u03ac\u03bf\u03c5\u03c2,\u03a0\u03b1\u03b9\u03b3\u03bd\u03af\u03c9\u03bd \u03b1\u03bb\u03bb\u03ac \u03ba\u03b1\u03b9 \u0395\u03c6\u03b1\u03c1\u03bc\u03bf\u03c3\u03bc\u03ad\u03bd\u03c9\u03bd \u039c\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03ce\u03bd.\r\n\u03a4\u03bf \u03c3\u03c5\u03bd\u03bf\u03bb\u03b9\u03ba\u03cc \u03c4\u03bf\u03c5\u03c2 \u03bc\u03ad\u03b3\u03b5\u03b8\u03bf\u03c2 \u03be\u03b5\u03c0\u03b5\u03c1\u03bd\u03ac \u03c4\u03b1 5 GB.\r\n\r\n\u03a4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03c0\u03c1\u03bf\u03c6\u03b1\u03bd\u03ce\u03c2 \u03b4\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bd \u03bd\u03b1 \u03b4\u03b9\u03b1\u03b2\u03b1\u03c3\u03c4\u03bf\u03cd\u03bd \u03cc\u03bb\u03b1 \u03b1\u03c0\u03bf \u03b5\u03bc\u03ad\u03bd\u03b1,\u03b1\u03bb\u03bb\u03ac \u03b4\u03b5\u03bd \u03b3\u03bd\u03c9\u03c1\u03af\u03b6\u03c9 \u03ba\u03b1\u03b9 \u03b1\u03bd \u03bc\u03c0\u03bf\u03c1\u03ce \u03bd\u03b1 \u03c4\u03b1 \u03b4\u03b9\u03b1\u03b8\u03b5\u03c3\u03c9 \u03c3\u03c4\u03bf\u03bd \u03ba\u03cc\u03c3\u03bc\u03bf.\r\n\u0394\u03b5\u03bd \u03ad\u03c7\u03c9 \u03b5\u03bb\u03b5\u03b3\u03be\u03b5\u03b9 Copyrights \u03ba\u03b1\u03b9 \u03c3\u03ba\u03ad\u03c6\u03c4\u03bf\u03bc\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03af\u03c3\u03c9\u03c2 \u03bc\u03b9\u03b1 \u03c4\u03ad\u03c4\u03bf\u03b9\u03b1 \u03b4\u03b9\u03ac\u03b8\u03b5\u03c3\u03b7 (\u03b1\u03bd\u03ad\u03b2\u03b1\u03c3\u03bc\u03b1 \u03c3\u03c4\u03bf\u03bd Rapidshare \u03b3\u03b9\u03b1 \u03c0\u03b1\u03c1\u03ac\u03b4\u03b5\u03b9\u03b3\u03bc\u03b1) \u03b8\u03b1 \u03ae\u03c4\u03b1\u03bd \u03c0\u03b1\u03c1\u03ac\u03bd\u03bf\u03bc\u03b7.\r\n\r\n\u039c\u03c0\u03bf\u03c1\u03b5\u03af \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03c2 \u03bd\u03b1 \u03bc\u03b5 \u03b4\u03b9\u03b1\u03c6\u03c9\u03c4\u03af\u03c3\u03b5\u03b9 \u03b5\u03c0\u03af \u03c4\u03bf\u03c5 \u03b8\u03ad\u03bc\u03b1\u03c4\u03bf\u03c2;  :|",
  "Solution_3": "\u039d\u03ad\u03b5 \u03bc\u03bf\u03c5 \u03c6\u03af\u03bb\u03b5, \u03ba\u03b1\u03bb\u03ce\u03c2 \u03cc\u03c1\u03b9\u03c3\u03b5\u03c2 \u03c3\u03c4\u03b7\u03bd \u03c0\u03b1\u03c1\u03ad\u03b1!\r\n \u039b\u03bf\u03b9\u03c0\u03cc\u03bd , \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 \u03c4\u03bf \u03c0\u03b1\u03c1\u03cc\u03bd \u03c6\u03cd\u03bb\u03bb\u03b1\u03be\u03b5 \u03ba\u03b1\u03bb\u03ac \u03b1\u03c5\u03c4\u03ac \u03c4\u03b1 \u03b1\u03c1\u03c7\u03b5\u03af\u03b1. \u03a3\u03b9\u03b3\u03ac \u03c3\u03b9\u03b3\u03ac \u03b8\u03b1 \u03b2\u03c1\u03bf\u03cd\u03bc\u03b5 \u03c4\u03c1\u03cc\u03c0\u03bf \u03bd\u03b1 \u03c4\u03b1 \u03b1\u03c0\u03bf\u03b8\u03b7\u03ba\u03b5\u03cd\u03c3\u03bf\u03c5\u03bc\u03b5 \u03ba\u03ac\u03c0\u03bf\u03c5 \u03bc\u03b5 \u03c3\u03c7\u03b5\u03c4\u03b9\u03ba\u03ae \u03b1\u03c3\u03c6\u03ac\u03bb\u03b5\u03b9\u03b1 , \u03ce\u03c3\u03c4\u03b5 \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b4\u03b9\u03b1\u03b8\u03ad\u03c3\u03b9\u03bc\u03b1 \u03c3\u03b5 \u03cc\u03bb\u03bf\u03c5\u03c2.\r\n\r\n \u03a3\u03b5 \u03b1\u03c5\u03c4\u03cc \u03c4\u03bf \u03c7\u03ce\u03c1\u03bf \u03af\u03c3\u03c9\u03c2 \u03b4\u03b7\u03bc\u03b9\u03bf\u03c5\u03c1\u03b3\u03ae\u03c3\u03c9 \u03b1\u03c1\u03b3\u03cc\u03c4\u03b5\u03c1\u03b1 \u03ad\u03bd\u03b1 \u03c5\u03c0\u03bf\u03c6\u03cc\u03c1\u03bf\u03c5\u03bc \u03bc\u03b5 \u03c6\u03bf\u03c1\u03c4\u03c9\u03bc\u03ad\u03bd\u03b1 \u03c4\u03ad\u03c4\u03bf\u03b9\u03b1 \u03b2\u03b9\u03b2\u03bb\u03af\u03b1. \u0398\u03b1 \u03b2\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bb\u03cd\u03c3\u03b7. \u0391\u03c1\u03ba\u03b5\u03af \u03bd\u03b1 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b1 \u03b2\u03b9\u03b2\u03bb\u03af\u03b1 !\r\n\r\n \u039c\u03c0\u03ac\u03bc\u03c0\u03b7\u03c2"
}
{
  "Tag": [
    "modular arithmetic",
    "induction",
    "number theory proposed",
    "number theory"
  ],
  "Problem": "For any positive integer n, let $f(n)$ be the number of positive divisors of $n$ which equal \u00b11 mod 10, and let g(n) be the number of positive divisors of n which equal $\\pm 3$  mod 10. Show that $f(n) \\geq g(n)$.",
  "Solution_1": "see http://www.mathlinks.ro/Forum/viewtopic.php?t=134074\r\nAll problem from VMO 1992 I posted :D",
  "Solution_2": "$\\pm 1,\\pm 3$ are all the coprime residues modulo $10$,\r\nand for $n=n_{1}n_{2}$ with $\\gcd (n_{1},n_{2})=1$ we have\r\n$\\{d\\in\\mathbb{N}: d|n\\}=\\{d_{1}d_{2}: d_{1},d_{2}\\in\\mathbb{N},d_{1}|n_{1},d_{2}|n_{2}\\}$\r\nand $(\\pm 1)\\cdot(\\pm 1)\\equiv (\\pm 3)\\cdot(\\pm 3)\\equiv \\pm 1 \\pmod{10}$ and $(\\pm 1)\\cdot(\\pm 3)\\equiv (\\pm 3)\\cdot(\\pm 1)\\equiv \\pm 3 \\pmod{10}$.\r\nSo $f(n)=f(n_{1})f(n_{2})+g(n_{1})g(n_{2})$ and $g(n)=f(n_{1})g(n_{2})+g(n_{1})f(n_{2})$\r\nand $f(n)-g(n)=(f(n_{1})-g(n_{1}))(f(n_{2})-g(n_{2}))$.\r\nSo by induction it's enough to show $f(n)\\geq g(n)$ for $n=p^{k}$ with $p$ prime and $k\\in\\mathbb{N}_{0}$.\r\nBut $f(p^{k})>0=g(p^{k})$ for $p=2,5$ or $p\\equiv \\pm 1\\pmod{10}$\r\nand $f(p^{k})=\\left[{k+2\\over 2}\\right]\\geq \\left[{k+1\\over 2}\\right]=g(p^{k})$ for $p\\equiv \\pm 3\\pmod{10}$.\r\n\r\nAlso $f(n)=g(n)\\Leftrightarrow n$ has some prime factor $p\\equiv\\pm 3\\pmod{10}$ with odd multiplicity\r\nand $f(n)=g(n)+1\\Leftrightarrow n$ has no prime factor $p\\equiv\\pm 1\\pmod{10}$ and all its prime factors $p\\equiv\\pm 3\\pmod{10}$ with even multiplicity."
}
{
  "Tag": [
    "\\/closed"
  ],
  "Problem": "When I first registered my username, rofler, I didn't realize how much I would actually go on the forum.\r\nI hate that my name isn't capitalized, could any admin just change it to Rofler?\r\nI know it is alot to ask, and I am not expecting much, but it would be nice :)\r\n\r\nThanks in advance,\r\n\r\nrofler",
  "Solution_1": "Done.",
  "Solution_2": "You rock! Thanks a bunch."
}
{
  "Tag": [
    "Ring Theory",
    "superior algebra",
    "superior algebra solved"
  ],
  "Problem": "Could anyone explain why in $R=Z[x]$ that $(2,x)$ is not a princial ideal?",
  "Solution_1": "I will do an attempt\r\n\r\nyou do mean $R=\\mathbb{Z}[x]$ right?\r\nwell note that the only units are 1 and -1\r\n\r\nso suppose $(2,x)$ is a principal ideal, generated by $a$\r\n\r\nwell $a$ must divide $2$ and $x$ then\r\nit is thus up to sign $1$ or $2$\r\nhowever the latter does not divide $x$ , to see this explicitly, if $x=2 p(x)$ you immediately see a problem with highest power coefficient\r\n\r\nso it is 1 , but that means the ideal is the full ring, and thus it also contains 1 \r\n\r\n$2 p(x) + q(x) x =1$ is not possible either however, as the left hand side is even for evalution with $x=0$ and the right hand side is not",
  "Solution_2": "I posted more general result in Solved Problems Section"
}
{
  "Tag": [
    "Support"
  ],
  "Problem": "Apparently I finished all the problems in the database and have 0 wrong and all that, but in the Report, I see there are topics I haven't gotten any problems from at all, yet it says \"No more problems available\".",
  "Solution_1": "Do u think thats a glitch?",
  "Solution_2": "A-ha! Now I remember why we originally disabled the view of any topics with 0 correct, 0 incorrect, and 0 given up on. Sadly, this produced a problem which is explained by MathMom in another post.\r\n\r\nThere are no questions in those subjects. They are merely placeholder subjects and parent subjects of other subjects.\r\n\r\nI'll have to put in some kind of ugly hack to not display these.",
  "Solution_3": "Wow isabella 2296, you finished all the problems!!!! I'm not even close....."
}
{
  "Tag": [
    "linear algebra",
    "matrix"
  ],
  "Problem": "There are n*n complex number $ a_{ij} (i \\equal{} 1,2,3,...,n)(j \\equal{} 1,2,3,...,n)$,and there are n complex number $ k_{j} (j \\equal{} 1,2,3,...,n)$ ,and $ \\sum_{j \\equal{} 1}^n|k_{j}|! \\equal{} 0$,and $ \\sum_{i \\equal{} 1}^n|\\sum_{j \\equal{} 1}^nk_{j}*a_{ij}| \\equal{} 0$\r\nProve:there are n complex number $ l_{i} (i \\equal{} 1,2,3,...,n)$ and $ \\sum_{i \\equal{} 1}^n|l_{i}|! \\equal{} 0$ to make $ \\sum_{j \\equal{} 1}^n|\\sum_{i \\equal{} 1}^nl_{i}a_{ij}| \\equal{} 0$",
  "Solution_1": "\\cdot for a multiplication dot, \\neq for the not-equals sign.  Anyhow, this is a linear algebra problem dressed up in silly clothes: it asserts that if a (complex) square matrix has row space of less-than-full dimension then it also has column space of less-than-full dimension, which is substantially weaker than the standard result that the dimensions of the row and column spaces are actually equal.",
  "Solution_2": "But can anybody prove it without the base of the linear algebra?"
}
{
  "Tag": [],
  "Problem": "\u039c\u03b9\u03b1\u03c2 \u03ba\u03b1\u03b9 \u03c0\u03b9\u03ac\u03c3\u03b1\u03bc\u03b5 \u03c3\u03cd\u03bd\u03bf\u03bb\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03c9\u03bd \u03c3\u03c4\u03bf \u03b5\u03c0\u03af\u03c0\u03b5\u03b4\u03bf\r\n\r\n\u039d\u03b1 \u03b4\u03b5\u03b9\u03c7\u03b8\u03b5\u03af \u03cc\u03c4\u03b9 \u03ba\u03ac\u03b8\u03b5 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03bf \u03c3\u03cd\u03bd\u03bf\u03bb\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03c9\u03bd \u0391 \u03c4\u03bf\u03c5 \u03b5\u03c0\u03b9\u03c0\u03ad\u03b4\u03bf\u03c5 \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03b3\u03c1\u03b1\u03c6\u03c4\u03b5\u03af \u03c9\u03c2 \u03ad\u03bd\u03c9\u03c3\u03b7 \u03b4\u03cd\u03bf \u03c3\u03c5\u03bd\u03cc\u03bb\u03c9\u03bd \u0392 \u03ba\u03b1\u03b9 \u0393 \u03ce\u03c3\u03c4\u03b5\r\n\u03b1) \u039a\u03ac\u03b8\u03b5 \u03bf\u03c1\u03b9\u03b6\u03cc\u03bd\u03c4\u03b9\u03b1 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ae \u03c4\u03bf\u03c5 \u03b5\u03c0\u03b9\u03c0\u03ad\u03b4\u03bf\u03c5 \u03c4\u03ad\u03bc\u03bd\u03b5\u03b9 \u03c4\u03bf \u0392 \u03c3\u03b5 \u03c0\u03b5\u03c0\u03b5\u03c1\u03b1\u03c3\u03bc\u03ad\u03bd\u03bf \u03b1\u03c1\u03b9\u03b8\u03bc\u03cc \u03c3\u03b7\u03bc\u03b5\u03af\u03c9\u03bd\r\n\u03b2) \u039a\u03ac\u03b8\u03b5 \u03ba\u03ac\u03b8\u03b5\u03c4\u03bf\u03c2 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ae \u03c4\u03bf\u03c5 \u03b5\u03c0\u03b9\u03c0\u03ad\u03b4\u03bf\u03c5 \u03c4\u03ad\u03bc\u03bd\u03b5\u03b9 \u03c4\u03bf \u0393 \u03c3\u03b5 \u03c0\u03b5\u03c0\u03b5\u03c1\u03b1\u03c3\u03bc\u03ad\u03bd\u03bf \u03b1\u03c1\u03b9\u03b8\u03bc\u03cc \u03c3\u03b7\u03bc\u03b5\u03af\u03c9\u03bd\r\n\r\n\u03a4\u03b9 \u03c3\u03c5\u03bc\u03b2\u03b1\u03af\u03bd\u03b5\u03b9 \u03b1\u03bd \u03c4\u03bf \u0391 \u03b4\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03bf;",
  "Solution_1": "[quote=\"Demetres\"]\u039c\u03b9\u03b1\u03c2 \u03ba\u03b1\u03b9 \u03c0\u03b9\u03ac\u03c3\u03b1\u03bc\u03b5 \u03c3\u03cd\u03bd\u03bf\u03bb\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03c9\u03bd \u03c3\u03c4\u03bf \u03b5\u03c0\u03af\u03c0\u03b5\u03b4\u03bf\n\n\u039d\u03b1 \u03b4\u03b5\u03b9\u03c7\u03b8\u03b5\u03af \u03cc\u03c4\u03b9 \u03ba\u03ac\u03b8\u03b5 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03bf \u03c3\u03cd\u03bd\u03bf\u03bb\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03c9\u03bd \u0391 \u03c4\u03bf\u03c5 \u03b5\u03c0\u03b9\u03c0\u03ad\u03b4\u03bf\u03c5 \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03b3\u03c1\u03b1\u03c6\u03c4\u03b5\u03af \u03c9\u03c2 \u03ad\u03bd\u03c9\u03c3\u03b7 \u03b4\u03cd\u03bf \u03c3\u03c5\u03bd\u03cc\u03bb\u03c9\u03bd \u0392 \u03ba\u03b1\u03b9 \u0393 \u03ce\u03c3\u03c4\u03b5\n\u03b1) \u039a\u03ac\u03b8\u03b5 \u03bf\u03c1\u03b9\u03b6\u03cc\u03bd\u03c4\u03b9\u03b1 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ae \u03c4\u03bf\u03c5 \u03b5\u03c0\u03b9\u03c0\u03ad\u03b4\u03bf\u03c5 \u03c4\u03ad\u03bc\u03bd\u03b5\u03b9 \u03c4\u03bf \u0392 \u03c3\u03b5 \u03c0\u03b5\u03c0\u03b5\u03c1\u03b1\u03c3\u03bc\u03ad\u03bd\u03bf \u03b1\u03c1\u03b9\u03b8\u03bc\u03cc \u03c3\u03b7\u03bc\u03b5\u03af\u03c9\u03bd\n\u03b2) \u039a\u03ac\u03b8\u03b5 \u03ba\u03ac\u03b8\u03b5\u03c4\u03bf\u03c2 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ae \u03c4\u03bf\u03c5 \u03b5\u03c0\u03b9\u03c0\u03ad\u03b4\u03bf\u03c5 \u03c4\u03ad\u03bc\u03bd\u03b5\u03b9 \u03c4\u03bf \u0393 \u03c3\u03b5 \u03c0\u03b5\u03c0\u03b5\u03c1\u03b1\u03c3\u03bc\u03ad\u03bd\u03bf \u03b1\u03c1\u03b9\u03b8\u03bc\u03cc \u03c3\u03b7\u03bc\u03b5\u03af\u03c9\u03bd\n\n\u03a4\u03b9 \u03c3\u03c5\u03bc\u03b2\u03b1\u03af\u03bd\u03b5\u03b9 \u03b1\u03bd \u03c4\u03bf \u0391 \u03b4\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03bf;[/quote]\r\n\r\n\u03a0\u03ac\u03c1\u03b1 \u03c0\u03bf\u03bb\u03cd \u03c9\u03c1\u03b1\u03af\u03b1 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7.\r\n\r\n\u0391\u03c2 \u03ba\u03ac\u03bd\u03bf\u03c5\u03bc\u03b5 \u03c0\u03c1\u03ce\u03c4\u03b1 \u03bc\u03b9\u03ac \u03b5\u03b9\u03b4\u03b9\u03ba\u03ae \u03c0\u03b5\u03c1\u03af\u03c0\u03c4\u03c9\u03c3\u03b7, \u03c0\u03bf\u03c5 \u03c0\u03b5\u03c1\u03b9\u03ad\u03c7\u03b5\u03b9 \u03c4\u03b7\u03bd \u03af\u03b4\u03b9\u03b1 \u03b4\u03c5\u03c3\u03ba\u03bf\u03bb\u03af\u03b1.\r\n\r\n\u0391\u03bd \u03c4\u03b1 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u03c0\u03bf\u03c5 \u03bc\u03b1\u03c2 \u03ad\u03b4\u03b9\u03bd\u03b1\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03cc\u03bb\u03bf \u03c4\u03b1 \u03b1\u03ba\u03ad\u03c1\u03b1\u03b9\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 (m, n) \u03c4\u03bf\u03c5 \u03b5\u03c0\u03b9\u03c0\u03ad\u03b4\u03bf\u03c5 \u03c4\u03cc\u03c4\u03b5: \r\n\r\n\u0395\u03be\u03b5\u03c4\u03ac\u03b6\u03bf\u03c5\u03bc\u03b5 \u03b4\u03cd\u03bf \u03b5\u03c5\u03b8\u03b5\u03af\u03b5\u03c2 \u03cc\u03c0\u03c9\u03c2 \u03c4\u03b7\u03bd y = x \u03ba\u03b1\u03b9 \u03c4\u03b7\u03bd y = -x.\r\n\r\n\u03a3\u03c4\u03bf \u0392 \u03b2\u03ac\u03b6\u03bf\u03c5\u03bc\u03b5 \u03cc\u03bb\u03b1 \u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u03c0\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03ac\u03bd\u03c9 \u03ba\u03b1\u03b9 \u03b1\u03c0\u03cc \u03c4\u03b9\u03c2 \u03b4\u03cd\u03bf \u03b5\u03c5\u03b8\u03b5\u03af\u03b5\u03c2 \u03ae \u03c0\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ba\u03ac\u03c4\u03c9 \u03ba\u03b1\u03b9 \u03b1\u03c0\u03cc \u03c4\u03b9\u03c2 \u03b4\u03cd\u03bf, \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae\r\n\u0392 = {(m, n) / |m| < = |n| } , \u03ba\u03b1\u03b9 C \u03c4\u03b1 \u03c5\u03c0\u03cc\u03bb\u03bf\u03b9\u03c0\u03b1. \u039c\u03b9\u03ac \u03b6\u03c9\u03b3\u03c1\u03b1\u03c6\u03b9\u03ac \u03b4\u03b5\u03af\u03c7\u03bd\u03b5\u03b9 \u03cc\u03c4\u03b9 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03c4\u03bf \u03b6\u03b7\u03c4\u03bf\u03cd\u03bc\u03b5\u03bd\u03bf.\r\n\r\n\u0393\u03b5\u03bd\u03b9\u03ba\u03ac \u03c4\u03ce\u03c1\u03b1. \u03a4\u03b1 \u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u03b2\u03c1\u03af\u03c3\u03ba\u03bf\u03bd\u03c4\u03b1\u03b9 \u03c3\u03b5 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03bf \u03c0\u03bb\u03ae\u03b8\u03bf\u03c2 \u03bf\u03c1\u03b9\u03b6\u03cc\u03bd\u03c4\u03b9\u03c9\u03bd \u03b5\u03c5\u03b8\u03b5\u03b9\u03ce\u03bd \u03ba\u03b1\u03b9, \u03b5\u03c0\u03af\u03c3\u03b7\u03c2, \u03c3\u03b5 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03bf \u03c0\u03bb\u03ae\u03b8\u03bf\u03c2 \u03ba\u03ac\u03b8\u03b5\u03c4\u03c9\u03bd. \u0394\u03b7\u03bb\u03b1\u03b4\u03ae \u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u03bc\u03b1\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf\u03b9 \u03ba\u03cc\u03bc\u03b2\u03bf\u03b9 \u03c3\u03b5 \u03ad\u03bd\u03b1 \u03c0\u03bb\u03b5\u03b3\u03bc\u03b1 \u03c0\u03bf\u03c5 \"\u03bc\u03bf\u03b9\u03ac\u03b6\u03b5\u03b9\"  \u03bc\u03b5 \u03c4\u03bf \u03c0\u03b1\u03c1\u03b1\u03c0\u03ac\u03bd\u03c9 \u03c4\u03c9\u03bd \u03b1\u03ba\u03ad\u03c1\u03b1\u03b9\u03c9\u03bd \u03c3\u03b7\u03bc\u03b5\u03af\u03c9\u03bd. \u039a\u03b1\u03b9 \u03bb\u03bf\u03b9\u03c0\u03ac, \u03b1\u03ba\u03bf\u03bb\u03bf\u03c5\u03b8\u03ce\u03bd\u03c4\u03b1\u03c2 \u03c4\u03b7\u03bd \u03af\u03b4\u03b9\u03b1 \u03b9\u03b4\u03ad\u03b1.\r\n\r\n\u03a6\u03b9\u03bb\u03b9\u03ba\u03ac. \r\n\r\n\u039c\u03b9\u03c7\u03ac\u03bb\u03b7\u03c2 \u039b\u03ac\u03bc\u03c0\u03c1\u03bf\u03c5\r\n\r\n\u0394\u03b7\u03bc\u03ae\u03c4\u03c1\u03b7: \u03c4\u03ce\u03c1\u03b1 \u03c3\u03c5\u03bd\u03b5\u03b9\u03b4\u03b7\u03c4\u03bf\u03c0\u03bf\u03b9\u03ce \u03cc\u03c4\u03b9 \u03b5\u03af\u03c3\u03b1\u03b9 \u03bf \u03af\u03b4\u03b9\u03bf\u03c2 Demetres \u03c0\u03bf\u03c5 \u03b3\u03c1\u03ac\u03c6\u03b5\u03b9 \u03c4\u03cc\u03c3\u03bf \u03b5\u03be\u03b1\u03b9\u03c1\u03b5\u03c4\u03b9\u03ba\u03ac \u03c3\u03c4\u03bf http://www.mathematica.gr.\r\n\u039f \u03af\u03b4\u03b9\u03bf\u03c2 \u03b3\u03c1\u03ac\u03c6\u03c9 \u03b5\u03ba\u03b5\u03af, \u03bf\u03c0\u03cc\u03c4\u03b5 \u03c3\u03c4\u03bf \u03b5\u03b4\u03ce \u039a\u0391\u03a4\u0391\u03a0\u039b\u0397\u039a\u03a4\u0399\u039a\u039f \u03c6\u03cc\u03c1\u03bf\u03c5\u03bc \u03bc\u03c0\u03b1\u03af\u03bd\u03c9 \u03c3\u03c0\u03ac\u03bd\u03b9\u03b1. \u0394\u03c5\u03c3\u03c4\u03c5\u03c7\u03ce\u03c2 \u03b4\u03b5\u03bd \u03c4\u03b1 \u03c0\u03c1\u03bf\u03bb\u03b1\u03b2\u03b1\u03af\u03bd\u03c9 \u03cc\u03bb\u03b1!",
  "Solution_2": "[quote=\"Demetres\"]\u039c\u03b9\u03b1\u03c2 \u03ba\u03b1\u03b9 \u03c0\u03b9\u03ac\u03c3\u03b1\u03bc\u03b5 \u03c3\u03cd\u03bd\u03bf\u03bb\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03c9\u03bd \u03c3\u03c4\u03bf \u03b5\u03c0\u03af\u03c0\u03b5\u03b4\u03bf\n\n\u039d\u03b1 \u03b4\u03b5\u03b9\u03c7\u03b8\u03b5\u03af \u03cc\u03c4\u03b9 \u03ba\u03ac\u03b8\u03b5 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03bf \u03c3\u03cd\u03bd\u03bf\u03bb\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03c9\u03bd \u0391 \u03c4\u03bf\u03c5 \u03b5\u03c0\u03b9\u03c0\u03ad\u03b4\u03bf\u03c5 \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03b3\u03c1\u03b1\u03c6\u03c4\u03b5\u03af \u03c9\u03c2 \u03ad\u03bd\u03c9\u03c3\u03b7 \u03b4\u03cd\u03bf \u03c3\u03c5\u03bd\u03cc\u03bb\u03c9\u03bd \u0392 \u03ba\u03b1\u03b9 \u0393 \u03ce\u03c3\u03c4\u03b5\n\u03b1) \u039a\u03ac\u03b8\u03b5 \u03bf\u03c1\u03b9\u03b6\u03cc\u03bd\u03c4\u03b9\u03b1 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ae \u03c4\u03bf\u03c5 \u03b5\u03c0\u03b9\u03c0\u03ad\u03b4\u03bf\u03c5 \u03c4\u03ad\u03bc\u03bd\u03b5\u03b9 \u03c4\u03bf \u0392 \u03c3\u03b5 \u03c0\u03b5\u03c0\u03b5\u03c1\u03b1\u03c3\u03bc\u03ad\u03bd\u03bf \u03b1\u03c1\u03b9\u03b8\u03bc\u03cc \u03c3\u03b7\u03bc\u03b5\u03af\u03c9\u03bd\n\u03b2) \u039a\u03ac\u03b8\u03b5 \u03ba\u03ac\u03b8\u03b5\u03c4\u03bf\u03c2 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ae \u03c4\u03bf\u03c5 \u03b5\u03c0\u03b9\u03c0\u03ad\u03b4\u03bf\u03c5 \u03c4\u03ad\u03bc\u03bd\u03b5\u03b9 \u03c4\u03bf \u0393 \u03c3\u03b5 \u03c0\u03b5\u03c0\u03b5\u03c1\u03b1\u03c3\u03bc\u03ad\u03bd\u03bf \u03b1\u03c1\u03b9\u03b8\u03bc\u03cc \u03c3\u03b7\u03bc\u03b5\u03af\u03c9\u03bd\n\n\u03a4\u03b9 \u03c3\u03c5\u03bc\u03b2\u03b1\u03af\u03bd\u03b5\u03b9 \u03b1\u03bd \u03c4\u03bf \u0391 \u03b4\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03bf;[/quote]\r\n\r\n\u0395\u03af\u03c7\u03b1 \u03b1\u03c6\u03ae\u03c3\u03b5\u03b9 \u03b1\u03bd\u03b1\u03c0\u03ac\u03bd\u03c4\u03b7\u03c4\u03bf \u03c4\u03bf \u03b5\u03c1\u03ce\u03c4\u03b7\u03bc\u03b1 \u03b3\u03b9\u03b1 \u03bc\u03b7 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03b1. \r\n\r\n\u0391\u03c0\u03bf\u03b4\u03b5\u03b9\u03ba\u03bd\u03cd\u03b5\u03c4\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03b4\u03b5\u03bd \u03b3\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03c0\u03ac\u03bd\u03c4\u03b1 \u03b7 \u03c0\u03b1\u03c1\u03b1\u03c0\u03ac\u03bd\u03c9 \u03b4\u03b9\u03b1\u03bc\u03ad\u03c1\u03b9\u03c3\u03b7. \r\n\r\n\u03a0.\u03c7. \u03c3\u03c4\u03bf [0, 1]x[0, 1] \u03b7 \u03c7\u03b1\u03c1\u03b1\u03ba\u03c4\u03b7\u03c1\u03b9\u03c3\u03c4\u03b9\u03ba\u03ae \u03c4\u03bf\u03c5 \u03c3\u03c5\u03bd\u03ac\u03c1\u03c4\u03b7\u03c3\u03b7 f(., .) \u03ad\u03c7\u03b5\u03b9 \u03b4\u03b9\u03c0\u03bb\u03cc \u03bf\u03bb\u03bf\u03ba\u03bb\u03ae\u03c1\u03c9\u03bc\u03b1 1 (= \u03c4\u03bf \u03b5\u03bc\u03b2\u03b1\u03b4\u03cc\u03bd). \r\n\r\n\u0397 \u03c0\u03b1\u03c1\u03b1\u03c0\u03ac\u03bd\u03c9 \u03c5\u03c0\u03cc\u03b8\u03b5\u03c3\u03b7 \u03b1) \u03bc\u03b5\u03c4\u03b1\u03c6\u03c1\u03ac\u03b6\u03b5\u03c4\u03b1\u03b9 \u03c3\u03c4\u03bf \u03cc\u03c4\u03b9 \u03b3\u03b9\u03b1 \u03ba\u03ac\u03b8\u03b5 \u03c3\u03c4\u03b1\u03b8\u03b5\u03c1\u03cc $ y_0$ \u03b7 \u03c7\u03b1\u03c1\u03b1\u03ba\u03c4\u03b7\u03c1\u03b9\u03c3\u03c4\u03b9\u03ba\u03ae g \u03c4\u03bf\u03c5 \u0392 \u03ad\u03c7\u03b5\u03b9 $ g(x , y_0)$  = 0 \u03b5\u03ba\u03c4\u03cc\u03c2 \u03b1\u03c0\u03cc \u03c0\u03b5\u03c0\u03b5\u03c1\u03b1\u03c3\u03bc\u03ad\u03bd\u03bf \u03c0\u03bb\u03ae\u03b8\u03bf\u03c2 \u03c4\u03b9\u03bc\u03ce\u03bd \u03c4\u03bf\u03c5 x. \u0386\u03c1\u03b1 \u03ad\u03c7\u03b5\u03b9 \u03bf\u03bb\u03bf\u03ba\u03bb\u03ae\u03c1\u03c9\u03bc\u03b1 0. \u0391\u03bd \u03c4\u03bf \u03ba\u03ac\u03bd\u03bf\u03c5\u03bc\u03b5 \u03b1\u03c5\u03c4\u03cc \u03b3\u03b9\u03b1 \u03cc\u03bb\u03b1 \u03c4\u03b1 $ y_0$ \u03ba\u03b1\u03b9 \u03bf\u03bb\u03bf\u03ba\u03bb\u03b7\u03c1\u03ce\u03c3\u03bf\u03c5\u03bc\u03b5, \u03b5\u03c1\u03c7\u03cc\u03bc\u03b1\u03c3\u03c4\u03b5 \u03c3\u03b5 \u03b1\u03bd\u03c4\u03af\u03c6\u03b1\u03c3\u03b7 \u03bc\u03b5 \u03c4\u03bf \u03b8\u03b5\u03ce\u03c1\u03b7\u03bc\u03b1 Fubini (\u03b5\u03bd\u03b1\u03bb\u03bb\u03b1\u03b3\u03ae \u03c3\u03b5\u03b9\u03c1\u03ac\u03c2 \u03bf\u03bb\u03bf\u03ba\u03bb\u03ae\u03c1\u03c9\u03c3\u03b7\u03c2 \u03b4\u03af\u03bd\u03b5\u03b9 \u03c4\u03bf \u03af\u03b4\u03b9\u03bf \u03b1\u03c0\u03bf\u03c4\u03ad\u03bb\u03b5\u03c3\u03bc\u03b1, \u03ba\u03b1\u03b9 \u03bc\u03ac\u03bb\u03b9\u03c3\u03c4\u03b1 \u03cc\u03c3\u03bf \u03c4\u03bf \u03b1\u03c1\u03c7\u03b9\u03ba\u03cc \u03b4\u03b9\u03c0\u03bb\u03cc \u03bf\u03bb\u03bf\u03ba\u03bb\u03ae\u03c1\u03c9\u03bc\u03b1). \r\n\u0397 \u03af\u03b4\u03b9\u03b1 \u03c4\u03b5\u03c7\u03bd\u03b9\u03ba\u03ae \u03b4\u03b5\u03af\u03c7\u03bd\u03b5\u03b9 \u03cc\u03c4\u03b9 \u03b4\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bf\u03cd\u03c4\u03b5 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03bf (\u03b1\u03bd\u03c4\u03af \u03b1\u03c0\u03bb\u03ac \u03c0\u03b5\u03c0\u03b5\u03c1\u03b1\u03c3\u03bc\u03ad\u03bd\u03bf) \u03c0\u03bb\u03ae\u03b8\u03bf\u03c2 \u03c3\u03c4\u03bf\u03b9\u03c7\u03b5\u03af\u03c9\u03bd \u03c3\u03b5 \u03ba\u03ac\u03b8\u03b5 \u03bf\u03c1\u03b9\u03b6\u03cc\u03bd\u03c4\u03b9\u03b1 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ae.\r\n\r\n\u0394\u03b7\u03bc\u03ae\u03c4\u03c1\u03b7, \u03c0\u03c1\u03bf\u03c3\u03c0\u03ac\u03b8\u03b7\u03c3\u03b1 \u03bd\u03b1 \u03bb\u03cd\u03c3\u03c9 \u03c4\u03b7\u03bd \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03c7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03ce\u03bd\u03c4\u03b1\u03c2 \u03bc\u03cc\u03bd\u03bf \u03c0\u03bb\u03b7\u03b8\u03ac\u03c1\u03b9\u03b8\u03bc\u03bf\u03c5\u03c2, \u03c7\u03c9\u03c1\u03af\u03c2 \u0398\u03b5\u03c9\u03c1\u03af\u03b1 \u039c\u03ad\u03c4\u03c1\u03bf\u03c5. \u0394\u03b5\u03bd \u03c4\u03b1 \u03ba\u03b1\u03c4\u03ac\u03c6\u03b5\u03c1\u03b1. \u0388\u03c7\u03b5\u03b9\u03c2 \u03c4\u03ad\u03c4\u03bf\u03b9\u03bf \u03c4\u03c1\u03cc\u03c0\u03bf; \u0398\u03b1 \u03c7\u03b1\u03c1\u03ce \u03bd\u03b1 \u03c4\u03bf\u03bd \u03b1\u03ba\u03bf\u03cd\u03c3\u03c9. \r\n\r\n\r\n\r\n\u03a6\u03b9\u03bb\u03b9\u03ba\u03ac, \r\n\r\n\u039c\u03b9\u03c7\u03ac\u03bb\u03b7\u03c2 \u039b\u03ac\u03bc\u03c0\u03c1\u03bf\u03c5",
  "Solution_3": "\u039a\u03b5 \u039b\u03ac\u03bc\u03c0\u03c1\u03bf\u03c5, \u03b5\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03ce \u03c0\u03bf\u03bb\u03cd \u03b3\u03b9\u03b1 \u03c4\u03b1 \u03ba\u03b1\u03bb\u03ac \u03c3\u03b1\u03c2 \u03bb\u03cc\u03b3\u03b9\u03b1.\r\n\r\n\u0397 \u03bb\u03cd\u03c3\u03b7 \u03bc\u03bf\u03c5 \u03c3\u03c4\u03bf (\u03b1) \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03ba\u03c1\u03b9\u03ce\u03c2 \u03b7 \u03af\u03b4\u03b9\u03b1 \u03bc\u03b5 \u03c4\u03b7\u03bd \u03b4\u03b9\u03ba\u03ae \u03c3\u03b1\u03c2. \u03a4\u03bf (\u03b2) \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03b1\u03c6\u03ce\u03c2 \u03c0\u03b9\u03bf \u03b4\u03cd\u03c3\u03ba\u03bf\u03bb\u03bf \u03b5\u03c1\u03ce\u03c4\u03b7\u03bc\u03b1. \u0398\u03b1 \u03b2\u03ac\u03bb\u03c9 \u03c4\u03b7\u03bd \u03bb\u03cd\u03c3\u03b7 \u03bc\u03bf\u03c5 \u03b1\u03c1\u03b3\u03cc\u03c4\u03b5\u03c1\u03b1 (\u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03b8\u03c5\u03bc\u03b7\u03b8\u03ce \u03ba\u03ac\u03c0\u03bf\u03b9\u03b5\u03c2 \u03bb\u03b5\u03c0\u03c4\u03bf\u03bc\u03ad\u03c1\u03b5\u03b9\u03b5\u03c2.) \u039c\u03c0\u03bf\u03c1\u03b5\u03af\u03c2 \u03c0\u03ac\u03bd\u03c4\u03c9\u03c2 \u03bd\u03b1 \u03b1\u03c0\u03bf\u03c6\u03cd\u03b3\u03b5\u03b9\u03c2 \u03c4\u03b7\u03bd \u03b8\u03b5\u03c9\u03c1\u03af\u03b1 \u03bc\u03ad\u03c4\u03c1\u03bf\u03c5.\r\n\r\n\u039a\u03ac\u03c4\u03b9 \u03cc\u03bc\u03c9\u03c2 \u03bc\u03b5 \u03ad\u03c7\u03b5\u03b9 \u03bc\u03c0\u03b5\u03c1\u03b4\u03ad\u03c8\u03b5\u03b9 \u03c3\u03c4\u03b7\u03bd \u03bb\u03cd\u03c3\u03b7 \u03c3\u03b1\u03c2. \u0395\u03bd\u03ce \u03bc\u03bf\u03c5 \u03c6\u03b1\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03b1\u03c0\u03cc\u03bb\u03c5\u03c4\u03b1 \u03bf\u03c1\u03b8\u03ae, \u03b1\u03bd \u03b1\u03bb\u03bb\u03ac\u03be\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf \u03c0\u03b5\u03c0\u03b5\u03c1\u03b1\u03c3\u03bc\u03ad\u03bd\u03bf \u03c3\u03b5 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03bf \u03b7 \u03b1\u03c0\u03ac\u03bd\u03c4\u03b7\u03c3\u03b7 \u03b1\u03bb\u03bb\u03ac\u03b6\u03b5\u03b9! (\u0393\u03b9\u03b1 \u03bd\u03b1 \u03b1\u03ba\u03c1\u03b9\u03b2\u03bf\u03bb\u03bf\u03b3\u03ce, \u03b3\u03b9\u03b1 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03b1 \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03b3\u03af\u03bd\u03b5\u03b9 \u03b1\u03bd \u03ba\u03b1\u03b9 \u03bc\u03cc\u03bd\u03bf \u03b1\u03bd \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03b7 continuum hypothesis.)",
  "Solution_4": "[quote=\"Lambrou\"]\n\u0394\u03b7\u03bc\u03ae\u03c4\u03c1\u03b7, \u03c0\u03c1\u03bf\u03c3\u03c0\u03ac\u03b8\u03b7\u03c3\u03b1 \u03bd\u03b1 \u03bb\u03cd\u03c3\u03c9 \u03c4\u03b7\u03bd \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03c7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03ce\u03bd\u03c4\u03b1\u03c2 \u03bc\u03cc\u03bd\u03bf \u03c0\u03bb\u03b7\u03b8\u03ac\u03c1\u03b9\u03b8\u03bc\u03bf\u03c5\u03c2, \u03c7\u03c9\u03c1\u03af\u03c2 \u0398\u03b5\u03c9\u03c1\u03af\u03b1 \u039c\u03ad\u03c4\u03c1\u03bf\u03c5. \u0394\u03b5\u03bd \u03c4\u03b1 \u03ba\u03b1\u03c4\u03ac\u03c6\u03b5\u03c1\u03b1. \u0388\u03c7\u03b5\u03b9\u03c2 \u03c4\u03ad\u03c4\u03bf\u03b9\u03bf \u03c4\u03c1\u03cc\u03c0\u03bf; \u0398\u03b1 \u03c7\u03b1\u03c1\u03ce \u03bd\u03b1 \u03c4\u03bf\u03bd \u03b1\u03ba\u03bf\u03cd\u03c3\u03c9. \n[/quote]\r\n\r\n\u03a5\u03c0\u03bf\u03b8\u03ad\u03c4\u03c9 \u03cc\u03c4\u03b9 \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03c4\u03ad\u03c4\u03bf\u03b9\u03bf\u03c2 \u03b4\u03b9\u03b1\u03bc\u03b5\u03c1\u03b9\u03c3\u03bc\u03cc\u03c2 \u03c4\u03bf\u03c5 $ \\mathbb{R}^2$.   \r\n\r\n\u0391\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 $ A_n \\equal{} \\{x: (n,x)\\in B\\}$. \u03a4\u03cc\u03c4\u03b5 $ \\bigcup_nA_n$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03bf. \u03a0\u03b1\u03af\u03c1\u03bd\u03c9 $ x \\notin \\bigcup_nA_n$. \u03a4\u03cc\u03c4\u03b5 $ (n,x) \\in C$ \u03b3\u03b9\u03b1 \u03ba\u03ac\u03b8\u03b5 n, \u03ac\u03c4\u03bf\u03c0\u03bf.\r\n\r\n\u039a\u03ac\u03c4\u03b9 \u03c0\u03ac\u03bd\u03c4\u03c9\u03c2 \u03c7\u03ac\u03bd\u03c9 \u03c3\u03c4\u03b7\u03bd \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03ae \u03c3\u03b1\u03c2. \u0394\u03bf\u03c5\u03bb\u03b5\u03cd\u03b5\u03b9 \u03b7 \u03af\u03b4\u03b9\u03b1 \u03c4\u03b5\u03c7\u03bd\u03b9\u03ba\u03ae \u03b1\u03bd \u03b1\u03bb\u03bb\u03ac\u03be\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf \u03c0\u03b5\u03c0\u03b5\u03c1\u03b1\u03c3\u03bc\u03ad\u03bd\u03bf \u03c3\u03b5 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03bf \u03ba\u03b1\u03b9 \u03c3\u03c4\u03b9\u03c2 \u03bf\u03c1\u03b9\u03b6\u03bf\u03bd\u03c4\u03b9\u03b5\u03c2 \u03ba\u03b1\u03b9 \u03c3\u03c4\u03b9\u03c2 \u03ba\u03ac\u03b8\u03b5\u03c4\u03b5\u03c2 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03c2 \u03ba\u03b1\u03b9 \u03b1\u03bd \u03cc\u03c7\u03b9 \u03b3\u03b9\u03b1\u03c4\u03af \u03cc\u03c7\u03b9;",
  "Solution_5": "[quote=\"Demetres\"][quote=\"Lambrou\"]\n\u0394\u03b7\u03bc\u03ae\u03c4\u03c1\u03b7, \u03c0\u03c1\u03bf\u03c3\u03c0\u03ac\u03b8\u03b7\u03c3\u03b1 \u03bd\u03b1 \u03bb\u03cd\u03c3\u03c9 \u03c4\u03b7\u03bd \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03c7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03ce\u03bd\u03c4\u03b1\u03c2 \u03bc\u03cc\u03bd\u03bf \u03c0\u03bb\u03b7\u03b8\u03ac\u03c1\u03b9\u03b8\u03bc\u03bf\u03c5\u03c2, \u03c7\u03c9\u03c1\u03af\u03c2 \u0398\u03b5\u03c9\u03c1\u03af\u03b1 \u039c\u03ad\u03c4\u03c1\u03bf\u03c5. \u0394\u03b5\u03bd \u03c4\u03b1 \u03ba\u03b1\u03c4\u03ac\u03c6\u03b5\u03c1\u03b1. \u0388\u03c7\u03b5\u03b9\u03c2 \u03c4\u03ad\u03c4\u03bf\u03b9\u03bf \u03c4\u03c1\u03cc\u03c0\u03bf; \u0398\u03b1 \u03c7\u03b1\u03c1\u03ce \u03bd\u03b1 \u03c4\u03bf\u03bd \u03b1\u03ba\u03bf\u03cd\u03c3\u03c9. \n[/quote]\n\n\u03a5\u03c0\u03bf\u03b8\u03ad\u03c4\u03c9 \u03cc\u03c4\u03b9 \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03c4\u03ad\u03c4\u03bf\u03b9\u03bf\u03c2 \u03b4\u03b9\u03b1\u03bc\u03b5\u03c1\u03b9\u03c3\u03bc\u03cc\u03c2 \u03c4\u03bf\u03c5 $ \\mathbb{R}^2$.   \n\n\u0391\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 $ A_n \\equal{} \\{x: (n,x)\\in B\\}$. \u03a4\u03cc\u03c4\u03b5 $ \\bigcup_nA_n$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03bf. \u03a0\u03b1\u03af\u03c1\u03bd\u03c9 $ x \\notin \\bigcup_nA_n$. \u03a4\u03cc\u03c4\u03b5 $ (n,x) \\in C$ \u03b3\u03b9\u03b1 \u03ba\u03ac\u03b8\u03b5 n, \u03ac\u03c4\u03bf\u03c0\u03bf.\n\n\u039a\u03ac\u03c4\u03b9 \u03c0\u03ac\u03bd\u03c4\u03c9\u03c2 \u03c7\u03ac\u03bd\u03c9 \u03c3\u03c4\u03b7\u03bd \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03ae \u03c3\u03b1\u03c2. \u0394\u03bf\u03c5\u03bb\u03b5\u03cd\u03b5\u03b9 \u03b7 \u03af\u03b4\u03b9\u03b1 \u03c4\u03b5\u03c7\u03bd\u03b9\u03ba\u03ae \u03b1\u03bd \u03b1\u03bb\u03bb\u03ac\u03be\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf \u03c0\u03b5\u03c0\u03b5\u03c1\u03b1\u03c3\u03bc\u03ad\u03bd\u03bf \u03c3\u03b5 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03bf \u03ba\u03b1\u03b9 \u03c3\u03c4\u03b9\u03c2 \u03bf\u03c1\u03b9\u03b6\u03bf\u03bd\u03c4\u03b9\u03b5\u03c2 \u03ba\u03b1\u03b9 \u03c3\u03c4\u03b9\u03c2 \u03ba\u03ac\u03b8\u03b5\u03c4\u03b5\u03c2 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03c2 \u03ba\u03b1\u03b9 \u03b1\u03bd \u03cc\u03c7\u03b9 \u03b3\u03b9\u03b1\u03c4\u03af \u03cc\u03c7\u03b9;[/quote]\r\n\r\n \u039d\u03b1\u03b9, \u03b4\u03bf\u03c5\u03bb\u03b5\u03cd\u03b5\u03b9 \u03b3\u03b9\u03b1 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03b1 \u03b7 \u03c4\u03b5\u03c7\u03bd\u03b9\u03ba\u03ae: \u0388\u03c3\u03c4\u03c9 \u03b3\u03b9\u03b1 \u03ba\u03ac\u03b8\u03b5 \u03c3\u03c4\u03b1\u03b8\u03b5\u03c1\u03cc $ y_0$ \u03b7 \u03c7\u03b1\u03c1\u03b1\u03ba\u03c4\u03b7\u03c1\u03b9\u03c3\u03c4\u03b9\u03ba\u03ae g \u03c4\u03bf\u03c5 \u0392 \u03ad\u03c7\u03b5\u03b9 $ g(x, y_0)$ = 0 \u03b5\u03ba\u03c4\u03cc\u03c2 \u03b1\u03c0\u03cc \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03bf \u03c0\u03bb\u03ae\u03b8\u03bf\u03c2 \u03c4\u03b9\u03bc\u03ce\u03bd \u03c4\u03bf\u03c5 x. \u03a4\u03cc\u03c4\u03b5 \u03c4\u03bf \u03bf\u03bb\u03bf\u03ba\u03bb\u03ae\u03c1\u03c9\u03bc\u03b1 \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 x, \u03b3\u03b9\u03b1 \u03c3\u03c4\u03b1\u03b8\u03b5\u03c1\u03cc $ y_0$, \u03b5\u03af\u03bd\u03b1\u03b9 0. \r\n\u0394\u03b5\u03bd \u03c7\u03c1\u03b5\u03b9\u03ac\u03b6\u03b5\u03c4\u03b1\u03b9 \u03b7 \u03c5\u03c0\u03cc\u03b8\u03b5\u03c3\u03b7 \u03c4\u03bf\u03c5 \u03c3\u03c5\u03bd\u03b5\u03c7\u03bf\u03cd\u03c2. \u0397 \u03c3\u03c4\u03ac\u03bd\u03c4\u03b1\u03c1 \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03b5\u03af\u03bd\u03b1\u03b9 \u03cc\u03c4\u03b9, \u03b3\u03b9\u03b1 \u03ba\u03ac\u03b8\u03b5 \u03b5>0 \u03c3\u03c4\u03b1\u03b8\u03b5\u03c1\u03cc, \u03ba\u03b1\u03bb\u03cd\u03c0\u03c4\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03bf $ (a_n)$ \u03bc\u03b5 \u03b4\u03b9\u03b1\u03c3\u03c4\u03ae\u03bc\u03b1\u03c4\u03b1 \u03ba\u03ad\u03bd\u03c4\u03c1\u03bf\u03c5 $ a_n$ \u03bc\u03ae\u03ba\u03bf\u03c5\u03c2  $ \\frac {\\epsilon}{2^n}$. \u03a4\u03bf \u03c3\u03c5\u03bd\u03bf\u03bb\u03b9\u03ba\u03cc \u03bc\u03ae\u03ba\u03bf\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b5 \u03ba\u03b1\u03b9 \u03bb\u03bf\u03b9\u03c0\u03ac. \r\n\r\n\u0388\u03c7\u03c9 \u03ba\u03b1\u03b9 \u03b5\u03b3\u03ce \u03bc\u03af\u03b1 \u03b5\u03c1\u03ce\u03c4\u03b7\u03c3\u03b7: \u03a3\u03af\u03b3\u03bf\u03c5\u03c1\u03b1 \u03c4\u03bf $ A_n \\equal{} \\{x: (n,x)\\in B\\}$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03bf (\u03ae \u03c0\u03b5\u03c0\u03b5\u03c1\u03b1\u03c3\u03bc\u03ad\u03bd\u03bf) ;  \u038a\u03c3\u03c9\u03c2 \u03b4\u03b5\u03bd \u03c4\u03bf \u03b2\u03bb\u03ad\u03c0\u03c9 \u03b3\u03b9\u03b1\u03c4\u03af \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c1\u03b3\u03ac. \u0398\u03b1 \u03c4\u03bf \u03c3\u03ba\u03b5\u03c6\u03c4\u03ce... \r\n\r\n\u039a\u03b1\u03bb\u03b7\u03bd\u03cd\u03c7\u03c4\u03b1.\r\n\r\n\u039c\u03b9\u03c7\u03ac\u03bb\u03b7\u03c2 \u039b\u03ac\u03bc\u03c0\u03c1\u03bf\u03c5",
  "Solution_6": "[quote=\"Lambrou\"] \n\u039d\u03b1\u03b9, \u03b4\u03bf\u03c5\u03bb\u03b5\u03cd\u03b5\u03b9 \u03b3\u03b9\u03b1 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03b1 \u03b7 \u03c4\u03b5\u03c7\u03bd\u03b9\u03ba\u03ae: \u0388\u03c3\u03c4\u03c9 \u03b3\u03b9\u03b1 \u03ba\u03ac\u03b8\u03b5 \u03c3\u03c4\u03b1\u03b8\u03b5\u03c1\u03cc $ y_0$ \u03b7 \u03c7\u03b1\u03c1\u03b1\u03ba\u03c4\u03b7\u03c1\u03b9\u03c3\u03c4\u03b9\u03ba\u03ae g \u03c4\u03bf\u03c5 \u0392 \u03ad\u03c7\u03b5\u03b9 $ g(x, y_0)$ = 0 \u03b5\u03ba\u03c4\u03cc\u03c2 \u03b1\u03c0\u03cc \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03bf \u03c0\u03bb\u03ae\u03b8\u03bf\u03c2 \u03c4\u03b9\u03bc\u03ce\u03bd \u03c4\u03bf\u03c5 x. \u03a4\u03cc\u03c4\u03b5 \u03c4\u03bf \u03bf\u03bb\u03bf\u03ba\u03bb\u03ae\u03c1\u03c9\u03bc\u03b1 \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 x, \u03b3\u03b9\u03b1 \u03c3\u03c4\u03b1\u03b8\u03b5\u03c1\u03cc $ y_0$, \u03b5\u03af\u03bd\u03b1\u03b9 0. \n\u0394\u03b5\u03bd \u03c7\u03c1\u03b5\u03b9\u03ac\u03b6\u03b5\u03c4\u03b1\u03b9 \u03b7 \u03c5\u03c0\u03cc\u03b8\u03b5\u03c3\u03b7 \u03c4\u03bf\u03c5 \u03c3\u03c5\u03bd\u03b5\u03c7\u03bf\u03cd\u03c2. \u0397 \u03c3\u03c4\u03ac\u03bd\u03c4\u03b1\u03c1 \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03b5\u03af\u03bd\u03b1\u03b9 \u03cc\u03c4\u03b9, \u03b3\u03b9\u03b1 \u03ba\u03ac\u03b8\u03b5 \u03b5>0 \u03c3\u03c4\u03b1\u03b8\u03b5\u03c1\u03cc, \u03ba\u03b1\u03bb\u03cd\u03c0\u03c4\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03bf $ (a_n)$ \u03bc\u03b5 \u03b4\u03b9\u03b1\u03c3\u03c4\u03ae\u03bc\u03b1\u03c4\u03b1 \u03ba\u03ad\u03bd\u03c4\u03c1\u03bf\u03c5 $ a_n$ \u03bc\u03ae\u03ba\u03bf\u03c5\u03c2  $ \\frac {\\epsilon}{2^n}$. \u03a4\u03bf \u03c3\u03c5\u03bd\u03bf\u03bb\u03b9\u03ba\u03cc \u03bc\u03ae\u03ba\u03bf\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b5 \u03ba\u03b1\u03b9 \u03bb\u03bf\u03b9\u03c0\u03ac. \n[/quote]\n\n\u0391\u03c5\u03c4\u03cc \u03ba\u03b1\u03c4\u03b1\u03bb\u03ac\u03b2\u03b1\u03b9\u03bd\u03b1 \u03ba\u03b1\u03b9 \u03b5\u03b3\u03ce. \u0391\u03bb\u03bb\u03ac \u03b5\u03b4\u03ce \u03b1\u03c1\u03c7\u03af\u03b6\u03b5\u03b9 \u03c4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03bc\u03bf\u03c5. \u03a0\u03c9\u03c2 \u03c4\u03bf \u03b5\u03be\u03b7\u03b3\u03ae\u03c4\u03b5 \u03b1\u03c5\u03c4\u03cc\n\nSierpinski Decomposition: \u03a4\u03bf \u03b5\u03c0\u03af\u03c0\u03b5\u03b4\u03bf \u03b5\u03af\u03bd\u03b1\u03b9 \u03ad\u03bd\u03c9\u03c3\u03b7 \u03b4\u03cd\u03bf \u03c3\u03c5\u03bd\u03cc\u03bb\u03c9\u03bd \u0391 \u03ba\u03b1\u03b9 \u0392 \u03ce\u03c3\u03c4\u03b5 \u03c4\u03bf \u0391 \u03c4\u03ad\u03bc\u03bd\u03b5\u03b9 \u03cc\u03bb\u03b5\u03c2 \u03c4\u03b9\u03c2 \u03bf\u03c1\u03b9\u03b6\u03cc\u03bd\u03c4\u03b5\u03b9\u03c2 \u03ba\u03b1\u03b9 \u03c4\u03bf \u0392 \u03cc\u03bb\u03b5\u03c2 \u03c4\u03b9\u03c2 \u03ba\u03ac\u03b8\u03b5\u03c4\u03b5\u03c2 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03c2 \u03c3\u03b5 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03b1 \u03c3\u03cd\u03bd\u03bf\u03bb\u03b1.\n\n\u0397 \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03c7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03b5\u03af Axiom of choice \u03ba\u03b1\u03b9 Continuum Hypothesis. \u03a4\u03b1 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03b1 \u03c3\u03cd\u03bd\u03bf\u03bb\u03b1 \u03c0\u03bf\u03c5 \u03c4\u03ad\u03bc\u03bd\u03bf\u03c5\u03bd \u03bf\u03b9 \u03bf\u03c1\u03b9\u03b6\u03cc\u03bd\u03c4\u03b5\u03c2/\u03ba\u03ac\u03b8\u03b5\u03c4\u03b5\u03c2 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03c2 \u03c6\u03c4\u03b9\u03ac\u03c7\u03bd\u03bf\u03bd\u03c4\u03b1\u03b9 \u03bc\u03b5 \u03b1\u03be\u03af\u03c9\u03bc\u03b1 \u03b5\u03c0\u03b9\u03bb\u03bf\u03b3\u03ae\u03c2. \n \n\u03a0\u03bf\u03c5 \u03b2\u03c1\u03af\u03c3\u03ba\u03b5\u03c4\u03b1\u03b9 \u03bb\u03bf\u03b9\u03c0\u03cc\u03bd \u03c4\u03bf \u03bb\u03ac\u03b8\u03bf\u03c2; \u03a4\u03bf \u03bc\u03cc\u03bd\u03bf \u03c0\u03bf\u03c5 \u03bc\u03c0\u03bf\u03c1\u03ce \u03bd\u03b1 \u03c3\u03ba\u03b5\u03c6\u03c4\u03ce \u03b5\u03af\u03bd\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03bf\u03b9 \u03c7\u03b1\u03c1\u03b1\u03ba\u03c4\u03b7\u03c1\u03b9\u03c3\u03c4\u03b9\u03ba\u03ad\u03c2 \u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03ae\u03c3\u03b5\u03b9\u03c2 \u03c4\u03c9\u03bd \u0391 \u03ba\u03b1\u03b9 \u0392 (\u03b1\u03c2 \u03c0\u03b5\u03c1\u03b9\u03bf\u03c1\u03b9\u03c3\u03c4\u03bf\u03cd\u03bc\u03b5 \u03c0.\u03c7. \u03c3\u03c4\u03bf [0,1]^2) \u03b4\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf\u03bb\u03bf\u03ba\u03bb\u03b7\u03c1\u03ce\u03c3\u03b9\u03bc\u03b5\u03c2. \u0391\u03bb\u03bb\u03ac \u03b3\u03b9\u03b1\u03c4\u03b9; \u0394\u03b5\u03bd \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03cc\u03c4\u03b9 \u03ba\u03ac\u03b8\u03b5 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03bf \u03c3\u03cd\u03bd\u03bf\u03bb\u03bf \u03ad\u03c7\u03b5\u03b9 \u03bc\u03ad\u03c4\u03c1\u03bf 0; \u0391\u03ba\u03cc\u03bc\u03b7 \u03ba\u03b1\u03b9 \u03bc\u03b5 \u03b1\u03be\u03af\u03c9\u03bc\u03b1 \u03b5\u03c0\u03b9\u03bb\u03bf\u03b3\u03ae\u03c2 \u03bd\u03b1 \u03c4\u03bf \u03c6\u03c4\u03b9\u03ac\u03be\u03c9, \u03bc\u03b5 \u03c4\u03bf \u03af\u03b4\u03b9\u03bf \u03b1\u03be\u03af\u03c9\u03bc\u03b1 \u03b4\u03b5\u03af\u03c7\u03bd\u03c9 \u03cc\u03c4\u03b9 \u03ad\u03c7\u03b5\u03b9 \u03bc\u03ad\u03c4\u03c1\u03bf 0. \u03a0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03c3\u03c4\u03b7\u03bd \u03c7\u03c1\u03ae\u03c3\u03b7 \u03c4\u03bf\u03c5 Fubini, \u03b4\u03b5\u03bd \u03b2\u03bb\u03ad\u03c0\u03c9. \u0395\u03b9\u03bb\u03b9\u03ba\u03c1\u03b9\u03bd\u03ac \u03b4\u03b5\u03bd \u03be\u03ad\u03c1\u03c9 \u03c4\u03b9 \u03c3\u03c5\u03bc\u03b2\u03b1\u03af\u03bd\u03b5\u03b9.  :?:\n\n\n[quote=\"Lambrou\"] \n\u0388\u03c7\u03c9 \u03ba\u03b1\u03b9 \u03b5\u03b3\u03ce \u03bc\u03af\u03b1 \u03b5\u03c1\u03ce\u03c4\u03b7\u03c3\u03b7: \u03a3\u03af\u03b3\u03bf\u03c5\u03c1\u03b1 \u03c4\u03bf $ A_n \\equal{} \\{x: (n,x)\\in B\\}$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03bf (\u03ae \u03c0\u03b5\u03c0\u03b5\u03c1\u03b1\u03c3\u03bc\u03ad\u03bd\u03bf) ;  \u038a\u03c3\u03c9\u03c2 \u03b4\u03b5\u03bd \u03c4\u03bf \u03b2\u03bb\u03ad\u03c0\u03c9 \u03b3\u03b9\u03b1\u03c4\u03af \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c1\u03b3\u03ac. \u0398\u03b1 \u03c4\u03bf \u03c3\u03ba\u03b5\u03c6\u03c4\u03ce... \n[/quote]\r\n\r\n\u0397 \u03b1\u03c0\u03ac\u03bd\u03c4\u03b7\u03c3\u03b7 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c0\u03bb\u03ae. \u03a0\u03b1\u03c1\u03cc\u03bb\u03bf \u03c0\u03bf\u03c5 \u03c4\u03bf \u03b4\u03b9\u03c0\u03bb\u03bf\u03ad\u03bb\u03b5\u03b3\u03be\u03b1 \u03b4\u03b9\u03cc\u03c4\u03b9 \u03ae\u03bc\u03bf\u03c5\u03bd \u03c3\u03af\u03b3\u03bf\u03c5\u03c1\u03bf\u03c2 \u03c0\u03c9\u03c2 \u03b8\u03b1 \u03ad\u03ba\u03b1\u03bd\u03b1 \u03bb\u03ac\u03b8\u03bf\u03c2 \u03bc\u03c0\u03ad\u03c1\u03b4\u03b5\u03c8\u03b1 \u03c4\u03b9\u03c2 \u03bf\u03c1\u03b9\u03b6\u03cc\u03bd\u03c4\u03b9\u03b5\u03c2 \u03bc\u03b5 \u03c4\u03b9\u03c2 \u03ba\u03ac\u03b8\u03b5\u03c4\u03b5\u03c2 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03c2. :wallbash:"
}
{
  "Tag": [
    "set theory",
    "superior algebra",
    "superior algebra solved"
  ],
  "Problem": "hello\r\nsorry if this question is not really algebra, it's more set theory\r\n\r\nsuppose A is a set with subset B\r\nsuppose C contains D\r\n\r\nlet there be bijections between B and C, and between A and G\r\nso roughly : A kinda contains C and C kinda contains A\r\n\r\nis there then a bijection between A and C\r\n\r\nsounds logic but not quite, it would allow me to prove that there is a bijection between\r\n[0,1] and ]0,1[\r\n\r\nfred",
  "Solution_1": "There are some ambiguities here, because you use $G$ once and $D$ once, and I think they are one and the same. Anyway, the well known Cantor-Bernstein Theorem states that if you have an injection from $A$ to $B$ and another injection from $B$ to $A$, then there is a bijection between them, so we can define an order relation on the cardinal numbers.",
  "Solution_2": "It is not so easy to define relation between cardinal numbers. ;)",
  "Solution_3": "[quote=\"fredbel6\"]sounds logic but not quite, it would allow me to prove that there is a bijection between\n[0,1] and ]0,1[[/quote]\r\nHa! You can easily construct bijection without Cantor-Bernstein!",
  "Solution_4": "Sure, we can, for example, construct a bijection between the rationals of the two sets (I won't construct this one :)), and put $f(x)=x$ if $x$ is irrational.",
  "Solution_5": "I am even sure it is a too complicated construction!\r\nExplicit bijection:\r\n$0\\to 1/3$\r\n$1/3\\to 1/9$\r\n...\r\n$1\\to 1/2$\r\n$1/2\\to 1/4$\r\n...\r\n$x\\to x$ for all remaining $x$.  :D"
}
{
  "Tag": [
    "limit",
    "function",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "If $\\lim_{n\\to \\infty}f_{n}= f$ and the functions $f_{n}$ are all monotone increasing, must $f$ be monotone increasing?\r\nWhat happens if $f_{n}$ are all strictly increasing?",
  "Solution_1": "For the first part: just use the fact that if $a_{n}\\le b_{n}$ for all $n,$ and the sequences have limits, then $\\lim a_{n}\\le \\lim b_{n}.$\r\n\r\nFor the second part: if you're hoping for a conclusion that includes \"strictly increasing,\" then you can't have that. Just look at $f_{n}(x)=\\frac{x}{n}.$",
  "Solution_2": "thank you for the help"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "number theory open",
    "number theory"
  ],
  "Problem": "Let r(n) be the number of distinct prime divisors of n \r\nr(50)=2 (as 50=2*5*5 and has 2 distinct prime factors 2 and 5) \r\n\r\nThen prove or disprove:\r\n\r\nThere exists a polynomial f(n) such that following holds good.\r\nThe minimum k (integer) satisfying \r\nr(f(k))=i (i >=2 is an integer) \r\nis a prime number.",
  "Solution_1": "this is not true as the number of elements in the set $ \\{f(k)\\}$ as k goes from 1 to n is approximately n. :wink:"
}
{
  "Tag": [
    "search",
    "function",
    "AoPS Books"
  ],
  "Problem": "Which book  should I probably read first? The Aops books or the Art and Craft of Promblem solving by Paul Zeitz for those of you that have read it?",
  "Solution_1": "This issue has been discussed here several times before.  I recommend you do a seach of the forum for \"Art and Craft\" or \"Zeitz\" or some such thing using the search function near the top of the page.  Also, this really isn't a getting started topic -- questions like this should go in one on the discussion forums, not one of the problem solving forums.",
  "Solution_2": "Well, it's definitely a newbie question. Part of the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=7254]owner-designated scope of this forum[/url] is asking newbie questions about problem-solving. JBL makes the good point that now this site has an extensive database of questions that have been asked and answered before, and it is an excellent idea to use the site's search function to look for answers to what are probably Frequently Asked Questions. \r\n\r\nThe short answer is that the prerequisites assumed by AoPS volume 1 are MUCH simpler than those assumed by AoPS volume 2, which in turn are (mostly) simpler than those assumed by the Zeitz book. I own all of those books, and I would work through them in the order indicated by the previous sentence. Do the search suggested by JBL for a more detailed rationale."
}
{
  "Tag": [
    "number theory",
    "Diophantine equation",
    "number theory proposed"
  ],
  "Problem": "For fixed k,n positive numbers, with (k,n) = 1, and they are both greater or equal 1.Find the number of solutions of equation : 1/x + 1/y = k/n.\r\n\r\nNote that, equation has solutions, x,y >=1 iff exits a,b .=1 such that: \r\na | n, b | n, (a,b) = 1 and k | (a+b). \r\n\r\nWhen k = 2, the problem is not hard. But can you generalize this to other k?[/code]",
  "Solution_1": "You may write the given equation as $(n-kx)(n-ky) = n^2$.\r\nThe problem is not find the number of divisors of $n^2$ which are equals to $n$ mod[k].\r\n\r\nPierre.",
  "Solution_2": "positive numbers = positive integers?\r\nand you want integer solutions right ?\r\n\r\nanyway simply working out gives $n\\cdot (x+y) = k\\cdot xy$\r\n\r\ngo on and you get $y = \\frac{n\\cdot x}{k\\cdot x-n}$ integer. Which is a well-known diophantine equation I think?"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "algebra proposed"
  ],
  "Problem": "Prove that there exist no polyniomials $f(x)$ and $g(x)$ with integer coefficients, such that $f(x)g(x) = (x-a_1)^2(x-a_2)^2.....(x-a_n)^2 +1.$\r\n   Ashwath",
  "Solution_1": "My solution is \r\nLet $P(x) = (x-a_1)^2(x-a_2)^2.....(x-a_n)^2 +1.$. Then $P(a_k)=1=f(x)g(x)$ so\r\n\r\n$f(a_k)=g(a_k)=1$ or $f(a_k)=g(a_k)=-1$ Also degree$[P(x)]=2n=$degree$[f(x)]$+degree$[g(x)]$.If one of the degrees of g(x) and f(x) is smaller than $n$ then the respect polynomial will be  $1$ or $-1$ ,contradiction Now we suppose that degree$[f(x)]$=degree$[g(x)]=n$. The coefficients of the biggest terms must have prod$1$. So degree$[f(x)-g(x)]\\leq n-1$and because$f(a_k)-g(a_k)=0$we have that$f(x)= g(x)$. So $(x-a_1)^2(x-a_2)^2.....(x-a_n)^2 +1=[f(x)]^2$ and if $z$ integer then \r\n\r\n$(z-a_1)^2(z-a_2)^2.....(z-a_n)^2 +1=[f(z)]^2$ so $[f(z)+(z-a_1)....(z-a_n)][f(z)-(z- a_1)....(z-a_n)]=1$  and $(z-a_1)....(z-a_n)=0$, $f(z)=\\pm 1$ for each $z\\in Z$, contradiction :)",
  "Solution_2": "Good one Silouan!",
  "Solution_3": "Just one word about the terms of the problem : the $a_i$ must be all different. It was used by silouan to prove $f(x)= g(x)$. \r\n\r\nFurthermore $(x-1)^2 \\times (x-1)^2 \\times (x-1)^2 + 1$ is divisible by $(x-1)^2 + 1$"
}
{
  "Tag": [
    "inequalities",
    "geometry",
    "trigonometry",
    "inequalities unsolved"
  ],
  "Problem": "Let $A_{i}B_{i}C_{i},\\ i=1,2$ be two triangles of sides $(a_{i},b_{i},c_{i})_{i=1,2}$, areas $S_{1,2}$ and of radii $(R_{i},r_{i})_{i=1,2}$ where $R_{i},r_{i}$ denote the lenghts of the radii of the circumscribed and inscribed circles for $A_{i}B_{i}C_{i}$, respectively. Then we have \\[8R_{1}R_{2}+4r_{1}r_{2}\\geq a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}\\geq 36 r_{1}r_{2}.\\]",
  "Solution_1": "This looks like an old AMM problem. Is it?",
  "Solution_2": "Here is my solution to this cool inequality\r\n\r\n[i]Lemma [/i] Given triangle $ABC$. Then $a^{2}+b^{2}+c^{2}\\leq 8R^{2}+4r^{2}$, or equivalently $sin^{2}A+sin^{2}B+sin^{2}C \\leq 2+16\\prod sin^{2}\\frac{A}{2}$\r\n\r\n[i]Proof[/i] Using the well-known identity $a^{2}+b^{2}+c^{2}= 2(p^{2}-r^{2}-4Rr)$ and $cosA+cosB+cosC = 1+\\frac{r}{R}$, we obtain that the inequality is equivalent to $sinAsinB+sinBsinC+sinCsinA \\leq (cosA+cosB+cosC)^{2}$. \r\n\r\nBy replacing $sinAsinB = cosC+cosAcosB$ and $\\sum cos^{2}A = 1-2cosAcosBcosC$, we obtain that it is sufficient to show $(1-cosA)(1-cosB)(1-cosC) \\geq cosAcosBcosC$. If $ABC$ is obtuse, then it is obvious. If not, then since $tgAtgBtgC \\geq cotg{\\frac{A}{2}}cotg{\\frac{B}{2}}cotg{\\frac{C}{2}}$, the inequality is true. Thus the lemma is completely proved.\r\n\r\nNotice that this lemma is a special case of the inequality we are trying to prove when two triangles are congruent.\r\n\r\nBefore proceeding to show the inequality, I will rewrite the problem a little bit.\r\n\r\n[b]Inequality[/b] Given two triangles $ABC$ and $DEF$. Show that\r\n\r\n$36rr' \\leq ad+be+cf \\leq 8RR'+4rr'$\r\n\r\n[b]Solution[/b]\r\n\r\nThe left most inequality is quite straightforward, followed immediately from the fact $abc \\geq 6\\sqrt{3}r^{3}$ and $def \\geq 6\\sqrt{3}r^{3}$\r\n\r\nThe rightmost inequality is much harder. According to sine law, it follows that we need to show $\\sum sinAsinD \\leq 2+16 \\prod sin \\frac{A}{2}sin \\frac{D}{2}$. By replacing $sinAsinD = sin^{2}{\\frac{A+D}{2}}-sin^{2}{\\frac{A-D}{2}}$, it follows that\r\n\r\n$LHS = \\sum (sin^{2}{\\frac{A+D}{2}}-sin^{2}{\\frac{A-D}{2}})$\r\n\r\nLet $X = \\frac{A+D}{2}, Y = \\frac{B+E}{2}, Z = \\frac{C+F}{2}$. Then $XYZ$ is a triangle. By applying the lemma, we obtain that\r\n\r\n$LHS = \\sum sin^{2}X-\\sum sin^{2}{\\frac{A-D}{2}}\\geq 2+16 \\prod sin^{2}\\frac{X}{2}-\\sum sin^{2}{\\frac{A-D}{2}}$\r\n\r\n$= 2+16 \\prod sin^{2}\\frac{A+D}{4}-\\sum sin^{2}{\\frac{A-D}{2}}$\r\n\r\nWe shall now show that\r\n\r\n$16 \\prod sin^{2}\\frac{A+D}{4}-\\sum sin^{2}{\\frac{A-D}{2}}\\leq 16 \\prod sin{\\frac{A}{2}}sin{\\frac{D}{2}}$.\r\n\r\nReplacing $sin{\\frac{A}{2}}sin{\\frac{D}{2}}= sin^{2}{\\frac{A+D}{4}}-sin^{2}{\\frac{A-D}{4}}, ...$, the inequality is equivalent to \r\n\r\n$16 \\prod sin^{2}\\frac{A+D}{4}-\\sum sin^{2}{\\frac{A-D}{2}}\\leq 16\\prod (sin^{2}{\\frac{A+D}{4}}-sin^{2}{\\frac{A-D}{4}})$\r\n\r\nor equivalently\r\n\r\n$\\sum sin^{2}\\frac{A-D}{2}+16\\sum sin^{2}\\frac{A+D}{4}sin^{2}\\frac{B-E}{4}sin^{2}\\frac{C-F}{4}-16\\sum sin^{2}\\frac{A-D}{4}sin^{2}\\frac{B+E}{4}sin^{2}\\frac{C+F}{4}-16\\prod sin^{2}{\\frac{A-D}{4}}\\geq 0 (*)$ \r\n\r\nIt looks terrible, but is in fact not hard.  :) \r\n\r\nNotice easily that $16\\sum sin^{2}\\frac{A+D}{4}sin^{2}\\frac{B-E}{4}sin^{2}\\frac{C-F}{4}\\geq 16\\prod sin^{2}{\\frac{A-D}{4}}$. \r\n\r\nWe will now show that $sin^{2}\\frac{A-D}{2}\\geq 16sin^{2}\\frac{A-D}{4}sin^{2}\\frac{B+E}{4}sin^{2}\\frac{C+F}{4}$. If we can prove it, then the whole thing is done.\r\n\r\nWe have $cotg{\\frac{Y}{2}}+cotg{\\frac{Z}{2}}\\geq 2cotg{\\frac{Y+Z}{4}}\\geq 2cotg45^{0}= 2$. Hence $cos{\\frac{X}{2}}\\geq 2sin{\\frac{Y}{2}}sin{\\frac{Z}{2}}$. This implies that\r\n\r\n$sin^{2}{\\frac{A-D}{2}}= 4sin^{2}{\\frac{A-D}{4}}cos^{2}{\\frac{A-D}{4}}\\geq 4sin^{2}{\\frac{A-D}{4}}cos^{2}{\\frac{A+D}{4}}\\geq 16sin^{2}{\\frac{A-D}{4}}sin^{2}{\\frac{B+E}{4}}sin^{2}{\\frac{C+F}{4}}$.\r\n\r\nTherefore the inequality $(*)$ is proved, and we are done."
}
{
  "Tag": [
    "conics",
    "geometry",
    "3D geometry",
    "sphere",
    "ellipse",
    "ratio",
    "geometric series"
  ],
  "Problem": "Suppose we have an (infinite) cone $ \\mathcal C$ with apex $ A$ and a plane $ \\pi$. The intersection of $ \\pi$ and $ \\mathcal C$ is an ellipse $ \\mathcal E$ with major axis $ BC$, such that $ B$ is closer to $ A$ than $ C$, and $ BC \\equal{} 4$, $ AC \\equal{} 5$, $ AB \\equal{} 3$. Suppose we inscribe a sphere in each part of $ \\mathcal C$ cut up by $ \\mathcal E$ with both spheres tangent to $ \\mathcal E$. What is the ratio of the radii of the spheres (smaller to larger)?",
  "Solution_1": "[hide=\"Diagram for Solution\"](may take a moment)\n[img]http://img.photobucket.com/albums/v194/mukmasterxps/hmmtguts12.png[/img][/hide]\n[hide=\"Solution\"][b]Inspiration:[/b] [i]AMC 10A 2006, number 16[/i]\n\nConsider the projection of the diagram onto the plane containing points A, B, and C.\n\nIt is easy to see that ABC is a 3-4-5 right triangle, where in this projection, the ellipse degenerates into segment BC. The projection of the sphere in the \"inside\" part of the cone is the incircle of triangle ABC, and has radius 1, and thus diameter 2.\n\nThe larger sphere is tangent to BC as well, but on the other side.\n\nNote that if we cut the cone with another plane parallel to $ \\pi$ which is tangent to the other side of the insphere of the \"inner\" cone part, and construct a smaller sphere in this new apical piece, we get a scaled-down version of the same problem. If we iterate this infinitely, we get a geometric series.\n\nConsider the projections of these spheres. Considering the diameters of each circle which are parallel to side BA, it is easy to see that the sum of the diameters of this geometric series of circles is AB=3.\n\nSince the radius of the given circle is $ 2$, if we let the common ratio be $ r$, we have $ \\frac {2}{1 \\minus{} \\frac {1}{r}} \\equal{} 3 \\implies r \\equal{} \\frac {1}{3}$.\n\nBack-extrapolating this pattern to the large circle in question, we see that the ratio here from smaller to larger is $ \\boxed{r \\equal{} \\frac {1}{3}}$[/hide]",
  "Solution_2": "Of course the cross section is the most important first step. Once you get that you can note that the ratio is $ \\frac {s \\minus{} a}{s} \\equal{} \\frac {6 \\minus{} 4}{6} \\equal{} \\frac {1}{3}$..."
}
{
  "Tag": [
    "AMC",
    "USA(J)MO",
    "USAMO",
    "modular arithmetic"
  ],
  "Problem": "Determine all non-negative integer solutions to\r\n$ \\ n^{4}_1 \\plus{} n^{4}_2 \\plus{} n^{4}_3 \\plus{} .... \\plus{} n^4_{14} \\equal{} 1599$\r\nI don't get how to solve this.\r\nSolutions ne1?",
  "Solution_1": "Hint: Consider a mod based approach\r\n\r\n[hide=\"Bigger hint\"]\nTake it all mod 16\n[/hide]",
  "Solution_2": "aha i see now thx mewmew",
  "Solution_3": "[hide=\"Full solution\"]\nTaking both sides modulo $ 16,$ we get \\[ n_1^4\\plus{}n_2^4\\plus{}n_3^4\\plus{}\\dots\\plus{}n_{14}^4\\equiv15\\pmod{16}.\\] $ 4$th powers modulo $ 16$ are all either $ 0,1,$ so the maximum sum is $ 14,$ thus we can't reach $ 15$ and there are no solutions.[/hide]\r\n\r\nWhich year did this come from?",
  "Solution_4": "Don't know what year, but it's probably older than you and me combined.",
  "Solution_5": "1979 (Problem 1).  I took the USAMO that year.  Back then, I think I solved this problem with a lot of casework.  :oops:"
}
{
  "Tag": [
    "inequalities proposed",
    "inequalities"
  ],
  "Problem": "for$a_{1}, ..a_{n}$ is real positive numbers such that : \r\n$(a_{1}^{2}-1)(a_{2}^{2}-1)..(a_{n}^{2}-1)=3^{n}.$\r\ndisprove or prove that : $\\frac{1}{a_{1}}+\\frac{1}{a_{2}}+...+\\frac{1}{a_{n}}\\geq \\frac{n}{2}$\r\n i sure it true with$n =3$ and $n=2$",
  "Solution_1": "[quote=\"Hong Quy\"]for$a_{1}, ..a_{n}$ is real positive numbers such that : \n$(a_{1}^{2}-1)(a_{2}^{2}-1)..(a_{n}^{2}-1)=3^{n}.$\ndisprove or prove that : $\\frac{1}{a_{1}}+\\frac{1}{a_{2}}+...+\\frac{1}{a_{n}}\\geq \\frac{n}{2}$\n i sure it true with$n =3$ and $n=2$[/quote]\r\nFor $n=2$ it's true and nice enough.\r\nFor $n=3$ it's wrong: try $a=\\sqrt7,$ $b=\\sqrt{\\frac{157}{7}}$ and $c=1.1.$ :wink:"
}
{
  "Tag": [
    "limit",
    "absolute value",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Prove that by a suitable rearrangement of terms, a conditionally convergent series may be made to converge to any desired value, or to diverge.\r\n\r\nProve that no matter how the elements of an absolutely convergent series are ordered, the sum stays the same.",
  "Solution_1": "So, what about the two questions?",
  "Solution_2": "Is this a Riemann theorem?",
  "Solution_3": "[quote=\"anik_andrew\"]Is this a Riemann theorem?[/quote]\r\n\r\nYes it is, \r\n\r\nhttp://mathworld.wolfram.com/RiemannSeriesTheorem.html\r\n\r\nmaybe knowing the name will help you in your search.",
  "Solution_4": "I posted this question some time ago, and got no real response. I think it's because I ask so many questions that the best people here get tired of me  :P \r\n\r\nAnyway, I have thought out a proof for a positive series that I think is right. Let the \"original\" positive series have partial sums $S_n$. Let a rearrangement of this series have partial sums $M_n$. Choose a $k$. By what we must mean by a rearrangement there exists a $k' > k$ such that every term in $S_k$ is also in $M_{k'}$. Since every term is positive we get that $S_k \\le M_{k'}$ . However, in the exact same way there will exist a $k'' > k' > k$ so that $S_k \\le M_{k'} \\le S_{k''}$. By reasoning similar to the proof of the \"squeeze theorem\" this shows that $\\lim_{k\\to\\infty}M_{k} = \\lim_{k\\to\\infty}S_{k}$.\r\n\r\nI have not yet really thought about how this extends to proving the same thing for absolutely convergent series.",
  "Solution_5": "Though this theorem looks like trivial, but the proof isn't too obvious.  :P",
  "Solution_6": "[quote=\"Kalle\"]I have not yet really thought about how this extends to proving the same thing for absolutely convergent series.[/quote]\r\n\r\nAbsolutely convergent means that $\\sum |a_n|$ is convergent. You have (nicely) proved the result for nonnegative series. Isn't that enough, or I'm missing something?",
  "Solution_7": "Well, no, not that I see. We have shown that if $\\sum b_n$ is a rearrangement of $\\sum a_n$ then $\\sum |b_n| = \\sum |a_n|$. I don't see how it follows that $\\sum b_n = \\sum a_n$",
  "Solution_8": "Fix $\\varepsilon>0$. Since $\\sum |a_n|$ converges, there is some $N$ such that $\\sum_{i=m}^{m+k}|a_i|<\\varepsilon$, for all $m\\ge N$ and all $k\\ge 0$. This means that for every finite set of naturals $\\mathcal F$ contained in $\\{N,N+1,\\ldots\\}$, we have $\\sum_{i\\in\\mathcal F}|a_i|<\\frac\\varepsilon 2$.  \r\n\r\nNow let $\\sigma$ be a permutation of the set of natural numbers, and take $M\\ge N$ so large that $\\{\\sigma(1),\\ldots,\\sigma(M)\\}$ contains $1,2,\\ldots,N$. Then, for all $m\\ge M$, you have $\\left|\\sum_{i=1}^ma_i-\\sum_{j=1}^ma_{\\sigma(j)}\\right|=\\left|\\sum_{i\\in\\mathcal F}a_i-\\sum_{j\\in\\mathcal F'}a_j\\right|\\ (*)$, where $\\mathcal F,\\mathcal F'$ are the finite sets $\\{1,2,\\ldots,m\\}\\setminus\\{1,2,\\ldots,N-1\\}$ and $\\{\\sigma(1),\\ldots,\\sigma(m)\\}\\setminus\\{1,2,\\ldots,N-1\\}$. The RHS of $(*)$ is now $\\le\\sum_{i\\in\\mathcal F}|a_i|+\\sum_{j\\in\\mathcal F'}|a_j|<\\varepsilon$. \r\n\r\nThe above says that given $\\varepsilon$, we can find a natural $M$ such that $\\left|\\sum_{i=1}^ma_i-\\sum_{j=1}^ma_{\\sigma(j)}\\right|<\\varepsilon$ for all $m\\ge M$, and this finishes it.",
  "Solution_9": "[quote=\"Kalle\"]Well, no, not that I see. We have shown that if $\\sum b_n$ is a rearrangement of $\\sum a_n$ then $\\sum |b_n| = \\sum |a_n|$. I don't see how it follows that $\\sum b_n = \\sum a_n$[/quote]\r\n\r\nI haven't thought about this at first, but I think this works \\[ \\sum_{n=1}^{\\infty} a_n = \\sum_{n=1}^{\\infty} (a_n+|a_n|) - \\sum_{n=1}^{\\infty} |a_n| .  \\]\r\n\r\nWell, does it?",
  "Solution_10": "[quote=\"perfect_radio\"][quote=\"Kalle\"]Well, no, not that I see. We have shown that if $\\sum b_n$ is a rearrangement of $\\sum a_n$ then $\\sum |b_n| = \\sum |a_n|$. I don't see how it follows that $\\sum b_n = \\sum a_n$[/quote]\n\nI haven't thought about this at first, but I think this works \\[ \\sum_{n=1}^{\\infty} a_n = \\sum_{n=1}^{\\infty} (a_n+|a_n|) - \\sum_{n=1}^{\\infty} |a_n| .  \\]\n\nWell, does it?[/quote]\r\nI would like to believe that yes, this equality does indeed complete the solution, since both series in that are positive. If my original proof is correct, then this is probably correct too. It is always easy to overlook minor details, though. See for example all the time I wasted in one of anik_andrew's thread. So, I'm just going to say that it would be nice if my simple proof combined with your equality would indeed work as a proof.\r\n\r\nGrobber: Thanks!",
  "Solution_11": "[quote=\"perfect_radio\"]Prove that by a suitable rearrangement of terms, a conditionally convergent series may be made to converge to any desired value, or to diverge.[/quote]\r\n\r\nSplit the set of non-zero terms in two sequences, $A=\\{a_n>0\\ |\\ n\\in\\mathbb N^*\\},\\ B=\\{b_m<0\\ |\\ m\\in\\mathbb N^*\\}$, such that $A$ and $B$ are ordered in the decreasing order of the absolute values of the $a_n$'s and $b_m$'s respectively. Now, you can prove just as we proved that other theorem you quoted that $S_A=\\sum a_n,\\ S_B=\\sum b_m$ do not depend on the order in which the summation is taken. The important thing here is that because our series is conditionally convergent, $S_A=\\infty,S_B=-\\infty$ (a simple exercise). Now let's find a rearrangement of $A\\cup B$ whose series converges to $0$. An entirely analogous process works for every other real.\r\n\r\nLet $T=\\max(\\max |a_n|,\\max |b_m|)$. We start adding terms to our sequence as follows. We add $a_1,\\ldots,a_{t_1}$ such that $a_1+\\ldots+a_{t_1}\\le T,\\ a_1+\\ldots+a_{t_1+1}>T$. Then we add $b_1,\\ldots,b_{s_1}$ such that $a_1+\\ldots+a_{t_1}+b_1,\\ldots,b_{s_1}\\ge -T$, but $a_1+\\ldots+a_{t_1}+b_1+\\ldots+b_{s_1}+b_{s_1+1}<-T$. Then we add $a_{t_1+1},\\ldots+a_{t_2}$ such that the sum of the terms which we have up until now is $\\le T$, but if we add one more $a$ we get a sum $>T$, and so on, until the maximum of the absolute value of all the $a$'s and $b$'s that are left is $\\le\\frac T2$. Then, we repeat the procedure, with $\\frac T2$ instead of $T$, then with $\\frac T3$ instead of $\\frac T2$, and so on. For every $n$, there is a large enough index $N_n$ such that all partial sums of the series we're constructing with index larger than $N_n$ lie in the interval $\\left[-\\frac Tn,\\frac Tn\\right]$, so the series converges to $0$.\r\n\r\nIf you look carefully, you'll see that we did use the fact that $S_A=\\infty,S_B=-\\infty$ in order to conclude that we always reach both sides of $0$. Otherwise the process described above could yield a series going to $\\infty$ if $S_B>-\\infty$ or $-\\infty$ if $S_A<\\infty$."
}
{
  "Tag": [
    "MATHCOUNTS",
    "geometry",
    "AMC 8",
    "AMC"
  ],
  "Problem": "I'm goinh to 9th grade in September, so I'm thinking about getting an AoPS book to prepare for AMC. Does anyone have a suggeation as to which book I should get? Thanks.\r\n(I have taken AMC 8 this year, other than that, I have also worked on old AMC problems. But I don't really feel that this is enough. I am looking for a book that would explain several topics rather than just one topic.)",
  "Solution_1": "AMC? AoPS should be enough...\r\n\r\nlater, you'll want Engel's Problem Solving Strategies and Zeitz's Art and Craft of Problem Solving (this is what I glean from these \"which book?\" topics)",
  "Solution_2": "Sorry if there are other similar posts around, I couldn't find any.\r\nThe problem with AoPS(actually it's my problem) is that I might not have access to a computer this summer.\r\nJust wondering, if anyone has [i]The Art of Problem solving, Volume 1: the Basics[/i], how helpful would it be for AMC?",
  "Solution_3": "When he said AoPS, he meant the book  :wink:  For the AMC, you'll want volume 2; volume 1 is a little too easy, although I guess the material is enough for the most part.",
  "Solution_4": "[quote=\"Phelpedo\"]When he said AoPS, he meant the book  :wink:  For the AMC, you'll want volume 2; volume 1 is a little too easy, although I guess the material is enough for the most part.[/quote]\r\n\r\nI don't think this assessment is quite accurate. From what I remember, especially from writers of the book (Mr. Rusczyk), volume I is quite sufficient for problems on the AMC. Mastering both volumes of material is essentially a large portion of what you need in order to do well on the AIME.",
  "Solution_5": "[quote=\"Phelpedo\"]When he said AoPS, he meant the book  :wink:  For the AMC, you'll want volume 2; volume 1 is a little too easy, although I guess the material is enough for the most part.[/quote]\r\nOops :oops: . I am not very good at communicating so this is defanitely not the first time I had misunderstand something like this. Thanks for the suggestion though. By the way, is there a connection between volume 1 and volume 2 (more specificly, can you understand volume 2 without doing volume 1 first?) I think my Dad will only allow me to buy one book unless I somehow convince him...",
  "Solution_6": "I read volume two without first reading volume one, and it has worked well for the most part (my AMC score jumped by over 15 points).",
  "Solution_7": "Yeah, they sometimes refer to \"we did blah in Vol. I\" but it's not necessary, really. \r\nThere is some good stuff in Vol. I, though, don't ignore it.",
  "Solution_8": "Thanks guys!!! Looks like I'm getting a Volume 2.",
  "Solution_9": "Disclaimer: Do not get volume 2 three weeks before the test and try to get through it. Believe me. I know from experience; it doesn't work, you just end up burning out.",
  "Solution_10": "Unless you're routinely scoring in the 120s or higher on the AMC 10/12, I'd recommend starting with Volume 1 before going on to Volume 2.",
  "Solution_11": "[quote=\"rrusczyk\"]Unless you're routinely scoring in the 120s or higher on the AMC 10/12, I'd recommend starting with Volume 1 before going on to Volume 2.[/quote]\r\nI really didn't see that much in volume 1 (from looking at the table of contents) that I did not learn from either MathCounts or my standard Alg 2 class (my school's curriculum is pretty advanced).",
  "Solution_12": "Ahem, by looking at the table of contents??\r\nSheesh. I mean, the practice problems in the later chapters...some of them were actually challenging.\r\nMeh, can't give an example though, I lost my copy.  :oops:",
  "Solution_13": "[quote=\"worthawholebean\"][quote=\"rrusczyk\"]Unless you're routinely scoring in the 120s or higher on the AMC 10/12, I'd recommend starting with Volume 1 before going on to Volume 2.[/quote]\nI really didn't see that much in volume 1 (from looking at the table of contents) that I did not learn from either MathCounts or my standard Alg 2 class (my school's curriculum is pretty advanced).[/quote]\r\n\r\nAnd you're consistently getting over 120s on the AMC10.",
  "Solution_14": "I've never done AMC 10/12 before (my school doesn't participate in them), so I guess maybe I'll try a few old AMC 10/12 contests first.\r\nAnother question - would it be easy to master the topics covered in volume 1 by simply studying from the Internet and library books? I don't think my Dad would buy me volume 2 if he does buy me volume 1. Even though I might not be good enough for vol. 2 at this moment, I am pretty sure (or at least I hope :) ) that I would be a year from now.",
  "Solution_15": "I don't get it. Everyone always says that AoPS volume 1 is extraordinarily easy.. but I had decent (not great, but  not bad) AMC scores, did ok in mathcounts, learned a lot last year-- and it still takes me time to get through the geo chapters of vol 1.",
  "Solution_16": "[quote=\"cat810930\"]I've never done AMC 10/12 before (my school doesn't participate in them), so I guess maybe I'll try a few old AMC 10/12 contests first.\nAnother question - would it be easy to master the topics covered in volume 1 by simply studying from the Internet and library books? I don't think my Dad would buy me volume 2 if he does buy me volume 1. Even though I might not be good enough for vol. 2 at this moment, I am pretty sure (or at least I hope :) ) that I would be a year from now.[/quote]\r\n\r\nHere's a thought: talk to your library about their getting the books.   Explain how top students and homeschoolers use these books, etc.  As I understand, some libraries are particularly responsive to the needs of homeschoolers, and are definitely interested in getting more non-adults using libraries.",
  "Solution_17": "[quote=\"rrusczyk\"][quote=\"cat810930\"]I've never done AMC 10/12 before (my school doesn't participate in them), so I guess maybe I'll try a few old AMC 10/12 contests first.\nAnother question - would it be easy to master the topics covered in volume 1 by simply studying from the Internet and library books? I don't think my Dad would buy me volume 2 if he does buy me volume 1. Even though I might not be good enough for vol. 2 at this moment, I am pretty sure (or at least I hope :) ) that I would be a year from now.[/quote]\n\nHere's a thought: talk to your library about their getting the books.   Explain how top students and homeschoolers use these books, etc.  As I understand, some libraries are particularly responsive to the needs of homeschoolers, and are definitely interested in getting more non-adults using libraries.[/quote]\r\nThanks, I might give it a try. Though I don't see how it will happen since my library is mostly used by people who need help with English rather than people who are interested in areas like math... sometimes I feel as though I'm the only one visting the math section (which, even though is large, focuses mainly on the basics).\r\nLuckily, the high school I'm going to specializes in math and science, so I would probably have more access to valuable resources this September. :)",
  "Solution_18": "Another idea: After you get AoPS 1, solve as many problems as you can, collecting your solutions in a notebook.  When you are done, with presumably more than 100 pages of solutions, show your work to your dad, and tell him how much you would like the second volume.  As a father myself, I would never say no in such a circumstance.",
  "Solution_19": "That's actually a great idea since my dad likes to say' \"If I buy you the book, you not even going to do it\" (which is not true) as a reason for not getting me the book.\r\nThanks, Ravi B! :lol:"
}
{
  "Tag": [],
  "Problem": "suppose a set exists in which it has integers that contain the digits 1-9 exactly once and all elements of S are prime.  For example, (7,83,421,569) would be considered one of these sets as 1-9 are all used exatly once and all 4 of the numbers are prime.\r\n\r\nFind the smallest possible total sum for all the elements for all sets that fit this rule.\r\n\r\nFor example, the sum of (7,83,421,569) is 7+83+421+569.",
  "Solution_1": "[quote=\"the future\"]suppose a set exists in which it has integers that contain the digits 1-9 exactly once and all elements of S are prime.  For example, (7,83,421,569) would be considered one of these sets as 1-9 are all used exatly once and all 4 of the numbers are prime.\n\nFind the smallest possible total sum for all the elements for all sets that fit this rule.\n\nFor example, the sum of (7,83,421,569) is 7+83+421+569.[/quote]\r\nSince two digit numbers will always be better than three digit numbers, the set will require two digit and one digit numbers, Also, since even digits must be in the tens digits and odds must be in the ones, then the smallest set would have a sum of:\r\n$ 10(2\\plus{}4\\plus{}6\\plus{}8)\\plus{}1\\plus{}3\\plus{}5\\plus{}7\\plus{}9\\equal{}225$",
  "Solution_2": "I am not sure how to attack this, but I got 5,89,241,367, which is 702.\r\n\r\nI left 2,3,5,7 in the front, and left the even numbers in even slot, like tenths and thousandth.",
  "Solution_3": "[quote=\"DonkeyKong\"]I am not sure how to attack this, but I got 5,89,241,367, which is 702.\n\nI left 2,3,5,7 in the front, and left the even numbers in even slot, like tenths and thousandth.[/quote]\r\nQuestion is, why would you use three digit numbers when you can use only two digit numbers?",
  "Solution_4": "[hide=\"hint\"]Maximize the number of one digit numbers. You can beat ZhangPeijin's number.[/hide]",
  "Solution_5": "[quote=\"ZhangPeijin\"][quote=\"the future\"]suppose a set exists in which it has integers that contain the digits 1-9 exactly once and all elements of S are prime.  For example, (7,83,421,569) would be considered one of these sets as 1-9 are all used exatly once and all 4 of the numbers are prime.\n\nFind the smallest possible total sum for all the elements for all sets that fit this rule.\n\nFor example, the sum of (7,83,421,569) is 7+83+421+569.[/quote]\nSince two digit numbers will always be better than three digit numbers, the set will require two digit and one digit numbers, Also, since even digits must be in the tens digits and odds must be in the ones, then the smallest set would have a sum of:\n$ 10(2 \\plus{} 4 \\plus{} 6 \\plus{} 8) \\plus{} 1 \\plus{} 3 \\plus{} 5 \\plus{} 7 \\plus{} 9 \\equal{} 225$[/quote]\r\n\r\nThats what I got at first too, but its the second smallest, not the smallest.\r\n\r\nANd you can use 3 digit numbers, but its not a good idea, the sum will be very large.",
  "Solution_6": "[quote=\"ZhangPeijin\"][quote=\"DonkeyKong\"]I am not sure how to attack this, but I got 5,89,241,367, which is 702.\n\nI left 2,3,5,7 in the front, and left the even numbers in even slot, like tenths and thousandth.[/quote]\nQuestion is, why would you use three digit numbers when you can use only two digit numbers?[/quote]\r\nI think I misunderstood the question.\r\n\r\nMy second try gives me with the set(2,5,7,89,61,43) with the sum of [b]207[/b].",
  "Solution_7": "[quote=\"DonkeyKong\"][quote=\"ZhangPeijin\"][quote=\"DonkeyKong\"]I am not sure how to attack this, but I got 5,89,241,367, which is 702.\n\nI left 2,3,5,7 in the front, and left the even numbers in even slot, like tenths and thousandth.[/quote]\nQuestion is, why would you use three digit numbers when you can use only two digit numbers?[/quote]\nI think I misunderstood the question.\n\nMy second try gives me with the set(2,5,7,89,61,43) with the sum of [b]207[/b].[/quote]\r\n\r\nya thats correct (there are actually 3 sets but they all give the sum 207), any reason why you came to that conclusion?\r\n\r\nDO you always minimize the number of digits?",
  "Solution_8": "Oh..\r\nI forgot\r\nDangit! \r\nTwo is prime...\r\nSo it becomes:\r\n$ 10(4\\plus{}6\\plus{}8)\\plus{}1\\plus{}2\\plus{}3\\plus{}5\\plus{}7\\plus{}9$"
}
{
  "Tag": [
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "find all of the answers of p^k + 1 =q^m that :\r\nk,p,q>1 and p is a coprime number",
  "Solution_1": "Posted before and bad title, too...",
  "Solution_2": "[quote=\"farshid_lk_park\"]find all of the answers of p^k + 1 =q^m that :\nk,p,q>1 and p is a coprime number[/quote]\r\nLook up Catalan-Mihailescu"
}
{
  "Tag": [
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "If f absolutely continuous, then it is of bounded variation on [a,b].\r\n\r\nlet the set of pairwise disjoint subintervals be just the interval [a, b] then l(a,b)= |a-b|?\r\n\r\nThis seems like it should be simple but I don't know where to start? Maybe start by saying that if it's not b.v.it can't possibly be a.c.?\r\n\r\nDon't tell me the answer yet!",
  "Solution_1": "Hint: If the variation is infinite, then for all $ \\epsilon > 0$, there is a subinterval of length $ < \\epsilon$ on which the variation is $ >1$.",
  "Solution_2": "Thank you so much for your help!"
}
{
  "Tag": [
    "linear algebra",
    "matrix",
    "algebra",
    "polynomial",
    "complex numbers"
  ],
  "Problem": "a) Show that $\\forall n \\in \\mathbb{N}_0, \\exists A \\in \\mathbb{R}^{n\\times n}: A^3=A+I$.\r\nb) Show that $\\det(A)>0, \\forall A$ fulfilling the above condition.",
  "Solution_1": "a) Take $x_1$ the real root of $x^3 -x - 1=0$ and the matrix $A_{nxn}=x_1I_{nxn}$\r\n this matrix satisfies $A^3 - A -I = 0$ for all n natural.... right?\r\n\r\nb) It\u00b4s enough to prove that $x_1>0$ because $Det(A)=$ product of eigenvalues and all of them satisfies $x^3 -x - 1=0$ (just multiply by a eigenvector), and the product of all complex eigenvalues is positive. So the signal of $Det(A)$ depends only of $x_1$. But the real root $x_1>0$(bolzano).",
  "Solution_2": "peter for a you used m and n\r\nshould  m=n?",
  "Solution_3": "Yes, indeed. There is no other m in the problem anyways :)",
  "Solution_4": "For part $b)$, suppose that $v \\in N(A)$, obviously $v \\in N(A^{3})$, so we have $A^{3}v=0 \\Leftrightarrow (A+ I)v=0 \\Leftrightarrow Av+v=0 \\Leftrightarrow v=0$,  $Q.E.D$.",
  "Solution_5": "Usetobe, that only proves that $\\det(A) \\neq 0$.\n\nFor part (a), take $A = \\ddag \\{L_1, ..., L_{\\lfloor {\\frac{n}{3}} \\rfloor}, O_{n-3 \\lfloor {\\frac{n}{3}} \\rfloor} \\}$, where each $L_i$ is the companion matrix of polynomial $p(x) = x^3-x-1$:\n\n$L_i = \\begin{bmatrix} 0 & 0 & 1 \\\\ 1 & 0 & 1 \\\\ 0 & 1 & 0 \\end{bmatrix}$.\n\nFor part (b), observe that p(x) has one real root $\\alpha > 0$, and two complex conjugate roots $\\beta, \\bar{\\beta}$. Therefore, det(A) is of the form $\\det(A) = \\alpha^k (|\\beta|^2)^{(n-k)/2} > 0$.",
  "Solution_6": "i dont think the soln for part (a) in post #6 is correct. because A contains a zero matrix as a submatrix means that A is singular. but A is invertible.\n\nanother approach for part (b)\n$A(A^2-I)=I$\n$A^3=A+I$\n$A=(A-I)(A^2+A+I)$\n$det(A-I)det(A+I)=det(A^2-I) \\sim det(A) \\sim det(A^3) \\sim det(A+I)$ so $det(A-I)>0$\n\n$det(A) , det(A^2-I) , det(A+I) , det(A^2+A+I)$ all have the same signs (either positive or negative).\nsince $det(A^2+A+I)>0$ , thus $det(A)>0$.",
  "Solution_7": "$A=A^3 + I = (A+I)(A+jI_n)(A+j^2I_n)$ where $j=\\exp(2i\\pi/3)$\n\n$det(A)=det(A+I_n)|det(A+jI_n)|^2 >0$",
  "Solution_8": "Others questions \n\nCompute $\\exp(A)$ ",
  "Solution_9": "in post #8, i dont understand why $det(A+I_n)>0$",
  "Solution_10": "i) moubi: Write $A=A^3-I$.\nii) $e^A\\approx 1.1766433548295131458I_n+0.55184482841941241016A^2+1.2199282716787921209A$.\niii) More generally. \nLet $n\\geq 2$, $A\\in M_n(\\mathbb{R})$ and  $P\\in \\mathbb{R}[x]$ s.t. $P(A)=0$ and $P(0)\\not= 0$.\nThen $\\det(A)\\not= 0$ but the signum of $\\det(A)$ is not fixed, except when all the real roots of $P$ are positive; in this last case, $\\det(A)$ is always positive.",
  "Solution_11": "I have same query why in post 8, $det(A+I)>0$"
}
{
  "Tag": [
    "function"
  ],
  "Problem": "Hello I was wondering how you would evaluate:\r\n\r\n$ \\sum_{x=3}^{\\infty} x (\\frac{5}{6})^{x-3}$\r\n\r\nI'm only familiar with geometric and arithmetic sequences, and this is neither.",
  "Solution_1": "Doesn't this = $ {\\infty}$",
  "Solution_2": "I went from 3 to 200 and got 48. Did the same from 3 to 400 and got 48. Also, I needed to get this to equal 48 exactly for my answer to a problem being correct. So, I'm pretty sure $ \\displaystyle\\sum_{x \\equal{} 3}^{\\infty} x \\left( \\frac {5}{6} \\right)^{x \\minus{} 3}\\equal{}48$, I just don't know why.",
  "Solution_3": "Let $ S \\equal{} \\sum^\\infty_{x\\equal{}3}x\\left(\\frac56\\right)^{x\\minus{}3}$.\r\nWhat is $ \\frac56 S$? Match up the terms of $ S$ and $ \\frac56 S$ by the powers and subtract to get a familiar series.",
  "Solution_4": "Consider the function $ G(x)\\equal{}\\sum_{k\\equal{}0}^\\infty (k\\plus{}3)x^k\\equal{}3\\plus{}4x\\plus{}5x^2\\plus{}6x^3\\plus{}\\dots$ We have $ xG(x)\\equal{}3x\\plus{}4x^2\\plus{}5x^3\\plus{}\\dots$, so $ (1\\minus{}x)G(x)\\equal{}G(x)\\minus{}xG(x)\\equal{}3\\plus{}x\\plus{}x^2\\plus{}x^3\\plus{}\\dots\\equal{}3\\plus{}\\frac x{1\\minus{}x}$, so $ G(x)\\equal{}\\frac 3{1\\minus{}x}\\plus{}\\frac x{(1\\minus{}x)^2}$. Finally, plugging in $ x\\equal{}\\frac 56$ yields $ 3\\cdot 6\\plus{}\\frac 56 \\cdot 36\\equal{}18\\plus{}30\\equal{}\\boxed{48}$.",
  "Solution_5": "[hide=\"Solution\"]Note that above expression is equal to \n\n${{ 3((\\frac {5}{6})^0 + (\\frac {5}{6}})^1 + (\\frac {5}{6}})^2 + \\cdots) + ((\\frac {5}{6})^1 + (\\frac {5}{6})^2 + \\cdots) + ((\\frac {5}{6})^2 + \\cdots)$$ + \\cdots$\n\nFirst parenthesis is equal to $ 3\\cdot\\frac {1}{1 - \\frac {5}{6}} = 18$.\n\nSecond one is equal to $ \\frac {\\frac {5}{6}}{1 - \\frac {5}{6}}$, and so on.\n\nContinuing, what we want is $ 18 + \\frac {\\frac {5}{6}}{\\frac {1}{6}} + \\frac {(\\frac {5}{6})^2}{\\frac {1}{6}} + \\cdots$.\n\n$ 18 + 6(\\frac {5}{6} + (\\frac {5}{6})^2 + \\cdots) = 18 + 6\\cdot5 = 48$.\n\nAnswer: 48[/hide]",
  "Solution_6": "Thanks everyone. Now I found a general formula with FantasyLover's solution, $ \\sum_{x=3}^{\\infty} x r^{x-3}$when $ |r|<1.$ It is$ \\, \\, \\frac{3-2r}{(1-r)^2}$\r\n\r\nHowever, I was wondering about $ \\sum_{x=3}^{\\infty} x^{n} r^{x-3}$ when $ |r|<1.$\r\n\r\nTrying $ n=2$, I saw $ \\sum_{x=3}^{\\infty} x^{2} r^{x-3} =$\r\n\r\n$ 9+16r+25r^2+36r^3+.....=9(1+r+r^2+...)+[7(r+r^2+r^3+...)+9(r^2+r^3+r^4+...)+11(r^3+r^4+r^5+...)+...] =$\r\n\r\n$ \\frac{9}{1-r}+\\frac{7r}{1-r}+\\frac{9r^2}{1-r}+\\frac{11r^3}{1-r}+\\frac{13r^4}{1-r}+\\frac{15r^5}{1-r}+...=$\r\n\r\n\\[ \\frac{9+\\sum_{q=3}^{\\infty} (2q+1) r^{q-2}}{1-r}.\\]\r\n\r\nThis then simply boils down to the $ n=1$ case. However, I'm stumped with $ n=3,4 , ...$. Is there an easier way to solve this?",
  "Solution_7": "well here's a \"strong-inductive\" way to find a closed-form formula, but im afraid i dont know how to \"automatize\" it:\r\nif you define \\[ f_n=\\sum_{x=0}^\\infty x^nr^x\\]then \\[ f_{n+1}=\\sum_{x=0}^\\infty x^{n+1}r^x=\\sum_{x=0}^\\infty (x+1)^{n+1}r^{x+1}\\]so we have\\begin{eqnarray*}f_{n+1}-rf_{n+1}&=&\\sum_{x=0}^\\infty r^{x+1}((x+1)^{n+1}-x^{n+1})\\\\\r\n&=&\\sum_{x=0}^\\infty r^{x+1}\\sum_{k=0}^n \\binom{n+1}{k}x^k\\\\\r\n&=&r\\sum_{k=0}^n \\binom{n+1}{k}\\sum_{x=0}^\\infty x^kr^x\\\\\r\n&=&r\\sum_{k=0}^n \\binom{n+1}{k}f_k\\\\\r\nf_{n+1}&=&\\frac{r}{r-1}\\sum_{k=0}^n \\binom{n+1}{k}f_k\\end{eqnarray*}",
  "Solution_8": "Idk why he just reposted a previous post to advertise.."
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "ratio",
    "AMC"
  ],
  "Problem": "Actually not an AMC problem, but similar:\r\n\r\n(from VWO (Flanders Math Olympiad) 2000 #24):\r\n\r\n9 white and 18 black cubes with edge 1 are used to form one big cube with edge 3.  The partial area of the cube that is white is at most:\r\n\r\n(A) 1/2\r\n(B) 13/27\r\n(C) 25/54\r\n(D) 4/9\r\n(E) 1/3",
  "Solution_1": "[hide]If we want to get the largest possible proportion of white on the surface of the cube, we want to use as many white cubes as possible for corners, and then use the rest for edges. Note that there are 8 corner cubes (3 faces showing), 12 edge cubes (2 faces showing), 6 face cubes (1 face showing), and 1 invisible center cube (0 faces showing). Thus we would be best off putting 8 white cubes in the corners and 1 white cube on an edge. That comes out to 8*3+1*2=26 white faces showing. On the entire cube, there are 9 faces showing per face of the large cube * 6 faces of the big cube = 54 faces. Thus we can get a maximum of 26/54=13/27 partial area being white.[/hide]",
  "Solution_2": "Right! Very nice, easy to understand answer.",
  "Solution_3": "This was the only one out of the 3 that I could solve...\n\n[hide]Practically the problem asks for the ratio of whites on the surface of the cube at most, so you have to \"show\" as many whites on the outside as possible.\n\nThe corners will be the best (3 sides showing), so place 8 white cubes on all the corners and there will be one cube left.\n\nThe next to best place will be the edges (2 sides showing) so place the remaining white cube there, and fill the rest with black cubes.\n\nThus (3*8) white faces show on the corners (24) and two on the edge, showing the total of 26 white faces. The total amount of faces showing would be (3*3*6=54), so the answer is 26/54 or 13/27 (B).[/hide]",
  "Solution_4": "Quote:9 white and 18 black cubes with edge 1 are used to form one big cube with edge 3. The partial area of the cube that is white is at most: \n\n(A) 1/2 \n(B) 13/27 \n(C) 25/54 \n(D) 4/9 \n(E) 1/3\n\n\n\n\n\n\n[hide]well, i was about to put E,  1/3. but then i reread the question as i usually do, and i saw that it said \"area\". so then i had to do the problem again. but the 1/3 answer barely took a few seconds. so no biggie. \n\n\n\n\n\n\n\nwell. to get the most area out of the 9 white cubes. i put them at the 8 corners of the cube. at each corner, a cube gets 3 square units of surface are. so, that's already 24 square units, but we still have one more white cube to deal with. if you put the cube in the center of a side of a square, then you get 25/54. but there is another option. to put it on the middle left, top right, middle right, or middle bottom of a side of a square. this way, the white cube gets 2 square units of surface area. adding 2/54+24/54=26/54=13/27. so my answer is choice B[/hide]\n\n\n\nif i helped a questioning aopser just a bit, my job is done",
  "Solution_5": "[hide]8*3+1*2=26\n\n\n\n26/(6*3^2) = 26/54 = 13/27.  (B)[/hide]",
  "Solution_6": "all i ask is \"why?\"",
  "Solution_7": "why what?",
  "Solution_8": "why \"ElectricSnowman\" would post an answer. oh the irony.",
  "Solution_9": "[hide]\n\nWell to maximize the surface area, you want one cube to take up as much area as possible. The most area a partial cube can take in a bigger one is 3, being the 8 corners. 8*3 = 24 and 1 more cube needs to be used.  The next biggest area that can be covered is 2, on one of the edges. 24+2=26 over the total surface area which is 54. So therefore I got 13/27.\n\n[/hide]",
  "Solution_10": "[quote=\"Syntax Error\"]why \"ElectricSnowman\" would post an answer. oh the irony.[/quote]\r\nThink about it.\r\nTwo equations:\r\nmathfanatic not equal answer (he made the question)\r\nElectricSnowman equal answer\r\n...?",
  "Solution_11": "spoiler. although, so many people already solved it...\n\n\n\n[hide]so there are 9 white cubes... 3 of your faces count into the total SA of the cube when you're at the corner, and there are 8 corners, so 8x1^2x3 = 24... and the last one you fit somewhere along the edges, so you get 1^2 x2 = 2... so then you get 24+2=26 (duh) so the answer is 26/3^2x6=26/54=13/27, so B is my answer[/hide]",
  "Solution_12": "yah, like you peoples."
}
{
  "Tag": [
    "Princeton",
    "college"
  ],
  "Problem": "Hi guys!!\r\n\r\nDoes anybody know the test dates of TOEFL iBT  in the 2007????\r\n\r\nThank you for your time!!! :)",
  "Solution_1": "[quote=\"Tiks\"]Hi guys!!\n\nDoes anybody know the test dates of TOEFL iBT  in the 2007????\n\nThank you for your time!!! :)[/quote]\r\nI don't see them on the TOEFL website, but I found the 2006 dates [url=http://www.ets.org/portal/site/ets/menuitem.1488512ecfd5b8849a77b13bc3921509/?vgnextoid=f368af5e44df4010VgnVCM10000022f95190RCRD&vgnextchannel=e207197a484f4010VgnVCM10000022f95190RCRD]here[/url], maybe they'll update soon.",
  "Solution_2": "isn't toefl that test that has to do with grammar for foreigners?\r\n\r\nI've taken it before  :P",
  "Solution_3": "[quote=\"Xevarion\"][quote=\"Tiks\"]Hi guys!!\n\nDoes anybody know the test dates of TOEFL iBT  in the 2007????\n\nThank you for your time!!! :)[/quote]\nI don't see them on the TOEFL website, but I found the 2006 dates [url=http://www.ets.org/portal/site/ets/menuitem.1488512ecfd5b8849a77b13bc3921509/?vgnextoid=f368af5e44df4010VgnVCM10000022f95190RCRD&vgnextchannel=e207197a484f4010VgnVCM10000022f95190RCRD]here[/url], maybe they'll update soon.[/quote]\r\nYes,Xevarion,I wasn't able to find the test dates in their site too.\r\nI am just afraid too be late with taking that test for my applications.You see I am not sure that there will be a test date in January and January is the last month to take standartized tests.",
  "Solution_4": "Is it true that taking TOEFL is benefiting American-born students from recent immigrant families?\r\nAfter all, even though their parents might be successfull professionals speaking English at work etc., these kids can't compete in English with non-recent immigrant families (all Americans are immigrants except Native Americans, of course). Because their first language is not English, parents do not speak English at home (normally), their English scores interpreted in this context would look better?\r\nOf course, taking some tests in your parent's native language probably makes sense too, to show your foreign language proficiency, especially since some Math programs (like Princeton) have Russian (French/German) requirements?",
  "Solution_5": "Yeah,guys the ETS organization at last replyed me and sent me all the test dates in the 2007  for Armenia :P (the test dates mostly the same all around the world :wink: ).The first test date in the next year is in 19 of January(in fact it is quit comfortable goes with my colleg application deadlines :lol: ).If someone wish I can send him(or her) the other test dates.Just PM me :wink: ."
}
{
  "Tag": [
    "modular arithmetic",
    "function",
    "number theory",
    "least common multiple",
    "totient function"
  ],
  "Problem": "Find the minimum value of c where:\r\n\r\n$10^{c}\\equiv 4\\ mod\\ 39$",
  "Solution_1": "[quote=\"janieluvsmusic\"]Find the minimum value of c where:\n\n$10^{c}\\equiv 4\\ mod\\ 39$[/quote]\r\n\r\n[hide] $10^{38}\\equiv1\\pmod{39}$ by Fermat's Theorem.\n\n$10^{38}\\equiv40\\pmod{39}\\Longrightarrow10^{37}\\equiv4\\pmod{39}$.\n\n$\\boxed{37}$. [/hide]",
  "Solution_2": "Are you sure that's right? It doesn't look like it.",
  "Solution_3": "[quote=\"lotrgreengrapes7926\"][quote=\"janieluvsmusic\"]Find the minimum value of c where:\n\n$10^{c}\\equiv 4\\ mod\\ 39$[/quote]\n\n[hide] $10^{38}\\equiv1\\pmod{39}$ by Fermat's Theorem.\n\n$10^{38}\\equiv40\\pmod{39}\\Longrightarrow10^{37}\\equiv4\\pmod{39}$.\n\n$\\boxed{37}$. [/hide][/quote]\r\n\r\nthat's invalid as 39 isn't prime",
  "Solution_4": "So how do you get the answer?",
  "Solution_5": "[quote=\"janieluvsmusic\"]Find the minimum value of c where:\n\n$10^{c}\\equiv 4\\ mod\\ 39$[/quote]\r\n[hide]\n$10^{\\phi(39)}\\equiv 1\\mod 39$ since $\\gcd(10,39)=1$\n\n$10^{24}\\equiv 1\\mod 39$\n\n$10^{23}\\equiv 4\\mod 39$\n\n$\\boxed{c=23}$[/hide]",
  "Solution_6": "23 works but it's not the minimum value.",
  "Solution_7": "It's necessary to take remainders modulo prime factors of $39$...",
  "Solution_8": "I think we actually will now have to look for a pattern.\r\n\r\nBut I won't work for now.\r\n\r\nI doubt any theorems work now.",
  "Solution_9": "[quote=\"janieluvsmusic\"]Find the minimum value of c where:\n\n$10^{c}\\equiv 4\\ mod\\ 39$[/quote]\r\n\r\nis the answer 5?",
  "Solution_10": "Euler's totient function (and as a special case, Fermat's Little Theorem) tell you an exponent that works.  They don't tell you it's the least such exponent.\r\n\r\nBeing nit-picky: you should ask for the least positive integer $c$ such that etcetcetc.\r\n\r\n[quote=\"D.P.L\"][quote=\"janieluvsmusic\"]Find the minimum value of c where:\n\n$10^{c}\\equiv 4\\ mod\\ 39$[/quote]\n\nis the answer 5?[/quote]\r\n\r\nYes -- why don't you give a solution?",
  "Solution_11": "[quote=\"JBL\"]\r\n\r\nBeing nit-picky: you should ask for the least positive integer $c$ such that etcetcetc.[quote]\r\n\r\nI was going to, but that would take more time to type. All I need is the solution now.",
  "Solution_12": "If you knew  $40= 39+1 = 4*10$  and   $39+10=49=7^{2}$ the answer would be a millisecond away. :) \r\n\r\nYes the answer is 5 but if you want to  dot the \"i's and cross the \"t's\" ..\r\n\r\n[hide=\"click here\"]First we notice $(10)^{-1}\\equiv 4 \\bmod 39$ \n[hide=\"becuase\"] after all $10 \\cdot 4 = 40 \\equiv 1 \\bmod 39$ [/hide]\n$-1$ is nice but you want a positive integer so .. we add 6 to it \n\n[hide=\"becuase\"]Having known how to factor  we know $39=3 \\cdot 13$ \nFeramat's   uvacuh:  $7^{12}\\equiv 1 \\bmod 13$ \nHe also says             $7^{2}\\equiv 1 \\bmod 3$ \nShowing of skills to take LCM of 2 and 12 we can say. \n$7^{12 }\\equiv 1 \\bmod 39$  \nbut $7^{2}\\equiv 10 \\bmod 39$ so we know $10^{6}\\equiv 1 \\bmod 39$ \n\n(we can quickly check  for factors of 6, eg $10^{2}$ and  $10^{3}$ are not  $1 \\bmod 39$ so 6 is the  index or the smallest number such that $10^{this number}$ is 1. \nQED\n\nPS \"uvach\" is what you use when you want to show off your knowledge of \nSanskrit :) \n\n\n[/hide][/hide]"
}
{
  "Tag": [
    "linear algebra",
    "linear algebra theorems"
  ],
  "Problem": "Hi\r\n\r\nI am looking for an exact statement on expanding a determinant by using one or more rows\r\nIt is well known how to calculate determinants by picking one row and then working with cofactors\r\n\r\nBut what if you, for instance, pick two rows?  I know it comes down then to calculating determinants of 2*2 matrices, and (n-2)*(n-2) matrices, and adding up their products, but what about the signs? \r\n\r\nCan anyone help me?  Or give me a url or something?",
  "Solution_1": "Let $A=(a_{ij})_{i,j=\\overline{1,n}}$ be a $n\\times n$ matrix. Let $0<k<n$.\r\n\r\nFor arbitrary $1\\leq i_i<i_2<\\ldots<i_k\\leq n$ and $1\\leq j_i<j_2<\\ldots<j_k\\leq n$, define \r\n\\[ d_{(i_1,\\ldots,i_k)}^{(j_1,\\ldots,j_k)}=\\det(a_{i_s,j_t})_{s,t=\\overline{1,k}}  \\]\r\nand\r\n\\[ c_{(i_1,\\ldots,i_k)}^{(j_1,\\ldots,j_k)}=\\det(a_{i'_s,j'_t})_{s,t=\\overline{1,n-k}}  \\], where $i'$s and $j'$s are chosen such that $\\{i_1,\\ldots,i_k,i'_1,\\ldots,i'_{n-k}\\}=\\{1,2,\\ldots,n\\}$ and $\\{j_1,\\ldots,j_k,j'_1,\\ldots,j'_{n-k}\\}=\\{1,2,\\ldots,n\\}$.\r\n\r\nFix $1\\leq i_i<i_2<\\ldots<i_k\\leq n$. \r\n\r\n[b]Laplace's Formula[/b]:\r\n\\[ \\det A=\\sum_{1\\leq j_1<\\ldots<j_k\\leq n}(-1)^{i_1+\\ldots+i_k+j_1+\\ldots+j_k}\\cdot d_{(i_1,\\ldots,i_k)}^{(j_1,\\ldots,j_k)}\\cdot c_{(i_1,\\ldots,i_k)}^{(j_1,\\ldots,j_k)}.  \\]"
}
{
  "Tag": [
    "geometry",
    "geometric transformation",
    "reflection",
    "ratio",
    "cyclic quadrilateral",
    "geometry unsolved"
  ],
  "Problem": "Let $X,Y$ is two point in circle $(O)$ , let $H$ is arbitrary point in line $XY$ .One line $(l)$ arbitrary pass through $H$ and meet $(O)$ at $M,N$ . Let $MX,NY$ meet $(O)$ at $P,Q$ respectively. Prove that $PQ$ pass through a fix point when $(l)$ change",
  "Solution_1": "Let $\\mathcal I_{H}$ be an inversion with center H, which carries the circle (O) into itself. If H is outside of (O), then (O) has to be perpendicular to the inversion circle and the inversion has positive power $r^{2}$ (square of the inversion circle radius). If H is inside of (O), then (O) has to cut the inversion circle in diametrically opposite points and the inversion has negative power $-r^{2},$ which is equivalent to inversion with a positive power $r^{2}$ followed by a reflection in the inversion center H.\r\n\r\nLet the line $(k) \\equiv XY$ meet the circle (O) at A, B. The sides MP, NQ and the diagonal MN of the cyclic quadrilateral MPNQ meet the line (k) at X, Y, H. Let the diagonal PQ meet (k) at G. Let $\\mathcal I_{X},\\ \\mathcal I_{Y},\\ \\mathcal I_{H},\\ \\mathcal I_{G}$ be inversions with centeres X, Y, H, G, which carry the circle (O) into itself. Inversion $\\mathcal I_{X}$ carries M into P, inversion $\\mathcal I_{G}$ carries P into Q, inversion $\\mathcal I_{Y}$ carries Q into N, and inversion $\\mathcal I_{H}$ carries N into M. Thus the Moebius transformation $\\mathcal M = \\mathcal I_{H}\\circ \\mathcal I_{Y}\\circ \\mathcal I_{G}\\circ \\mathcal I_{X}$ carries the point M to itself. In addition, $\\mathcal M$ obviously carries both A and B into themselves, hence it has 3 fixed points. But a Moebius transformation with 3 fixed points is an identity transformation. For example, the general form of a Moebius transformation in complex plane is $z' = \\frac{az+b}{cz+d},$ where $ad-bc \\neq 0.$ For a fixed point z' = z, we have $cz^{2}+(d-a)z-b = 0.$ If this has more than 2 roots, then $c = 0,\\ d = a \\neq 0,\\ b = 0$ and z' = z for all z.\r\n\r\nLet a line (l') through H different from (l) and (k) meet the circle (O) at M', N'. The sides M'P', N'Q' and the diagonal M'N' of the cyclic quadrilateral M'P'N'Q' meet (k) at the fixed points X, Y, H. Let the diagonal P'Q' meet (k) at G'. In exactly the same way as above, we find that $\\mathcal M' = \\mathcal I_{H}\\circ \\mathcal I_{Y}\\circ \\mathcal I_{G'}\\circ \\mathcal I_{X}$ is also an identity transformation. It follows that $\\mathcal I_{G'}= \\mathcal I_{G}$ and $G' \\equiv G,$ i.e., G is the fixed point of PQ.",
  "Solution_2": "Let's prove a stronger assertion:\r\n\r\n[color=blue][b]Extended Problem.[/b] Let X, Y, H be three collinear points and k be a circle. For any point M on the circle k, denote by N the point of intersection of the line MH with the circle k (apart from M), and denote by P and Q the points of intersection of the lines MX and NY with the circle k (apart from M and N), respectively. Prove that the line PQ intersects the line XY at a point independent on the position of the point M on the circle k.[/color]\r\n\r\n[i]Solution.[/i] We have to prove that for any two positions of the point M on the circle k, the corresponding lines PQ intersect the line XY at the same point.\r\n\r\nIn other words: Let $M_{1}$ and $M_{2}$ be two points on the circle k. Denote by $N_{1}$ the point of intersection of the line $M_{1}H$ with the circle k (apart from $M_{1}$), and denote by $P_{1}$ and $Q_{1}$ the points of intersection of the lines $M_{1}X$ and $N_{1}Y$ with the circle k (apart from $M_{1}$ and $N_{1}$), respectively. Similarly, denote by $N_{2}$ the point of intersection of the line $M_{2}H$ with the circle k (apart from $M_{2}$), and denote by $P_{2}$ and $Q_{2}$ the points of intersection of the lines $M_{2}X$ and $N_{2}Y$ with the circle k (apart from $M_{2}$ and $N_{2}$), respectively. We have to prove that the lines $P_{1}Q_{1}$ and $P_{2}Q_{2}$ intersect the line XY at the same point.\r\n\r\nApplying the Pascal theorem to the cyclic hexagon $Q_{1}M_{2}N_{2}Q_{2}M_{1}N_{1}$, we see that the points $Q_{1}M_{2}\\cap Q_{2}M_{1}$, $M_{2}N_{2}\\cap M_{1}N_{1}$, $N_{2}Q_{2}\\cap N_{1}Q_{1}$ are collinear. But the points $M_{2}N_{2}\\cap M_{1}N_{1}=H$ and $N_{2}Q_{2}\\cap N_{1}Q_{1}=Y$ both lie on the line XY; hence, the point $Q_{1}M_{2}\\cap Q_{2}M_{1}$ must also lie on the line XY.\r\n\r\nNow, apply the Pascal theorem to the cyclic hexagon $M_{1}P_{1}Q_{1}M_{2}P_{2}Q_{2}$. We obtain that the points $M_{1}P_{1}\\cap M_{2}P_{2}$, $P_{1}Q_{1}\\cap P_{2}Q_{2}$, $Q_{1}M_{2}\\cap Q_{2}M_{1}$ are collinear. Since the points $M_{1}P_{1}\\cap M_{2}P_{2}=X$ and $Q_{1}M_{2}\\cap Q_{2}M_{1}$ both lie on the line XY, it thus follows that the point $P_{1}Q_{1}\\cap P_{2}Q_{2}$ must also lie on the line XY. In other words, the lines $P_{1}Q_{1}$ and $P_{2}Q_{2}$ intersect the line XY at the same point. Proof complete.\r\n\r\nYetti's solution is really beautiful! I just posted mine to show that there is a more elementary way. Another solution of the Extended Problem is easy using cross-ratios on circles: Denote by U and V the (real or complex!) points of intersection of the line XY with the circle k. Let G be the point on the line XY such that (U; V; X; H) = (U; V; Y; G). Show that this point G lies on the line PQ for every position of the point M.\r\n\r\n  darij"
}
{
  "Tag": [
    "function",
    "probability",
    "integration",
    "calculus",
    "absolute value",
    "probability and stats"
  ],
  "Problem": "Let $ X$ be a random variable having the probability density $ f(x)\\equal{}\\frac12 e^{\\minus{}|x|}.$ Find the moment-generating function of $ X$.",
  "Solution_1": "$ m(t)\\equal{}\\int_{\\minus{}\\infty}^{\\infty}f(x)e^{tx}\\,dx$\r\n\r\n$ \\equal{}\\int_{\\minus{}\\infty}^0\\frac12e^xe^{tx}\\,dx\\plus{}\\int_0^{\\infty}\\frac12e^{\\minus{}x}e^{tx}\\,dx$\r\n\r\n$ \\equal{}\\frac1{2(1\\plus{}t)}\\plus{}\\frac1{2(1\\minus{}t)}\\equal{}\\frac1{1\\minus{}t^2}$\r\n\r\nThe characteristic function would be $ \\chi(t)\\equal{}\\frac1{1\\plus{}t^2},$ by a very similar computation.\r\n\r\n$ m(t)\\equal{}\\frac1{1\\minus{}t^2}\\equal{}1\\plus{}t^2\\plus{}t^4\\plus{}t^6\\plus{}\\cdots,$ so\r\n\r\n$ E(X^n)\\equal{}\\left(\\frac{d}{dt}\\right)^nm(0)\\equal{}\\begin{cases}n!&t\\text{ even}\\\\0&t\\text{ odd}\\end{cases}.$",
  "Solution_2": "[quote=\"Kent Merryfield\"]$ m(t) \\equal{} \\int_{ \\minus{} \\infty}^{\\infty}f(x)e^{tx}\\,dx$\n\n$ \\equal{} \\int_{ \\minus{} \\infty}^0\\frac12e^xe^{tx}\\,dx \\plus{} \\int_0^{\\infty}\\frac12e^{ \\minus{} x}e^{tx}\\,dx$\n\n...\n[/quote]\r\nThank you, Prof. Kent Merryfield.\r\nIn fact, the question I would like to ask is: How to calculate the integral $ \\int_0^{\\infty} e^{(t\\minus{}1)x}\\,dx$?\r\nAnd, is it true for every real $ t$?",
  "Solution_3": "Not for every real $ t.$ This particular distribution's moment generating function is only defined for $ \\minus{}1<t<1.$ The integral you wrote diverges for $ t\\ge 1,$ while the integral on the other side diverges for $ t\\le \\minus{}1.$\r\n\r\nThat's one of the theoretical advantages of the characteristic function, which is defined as $ \\chi(t)\\equal{}E(e^{itX}),$ which would equal $ \\int_{\\minus{}\\infty}^{\\infty}f(x)e^{itx}\\,dx$ for a distribution with a density. The characteristic function is always defined and finite (in fact, no larger than $ 1$ in absolute value) for all real $ t.$ The moment generating function might only be defined on some interval in $ t$ as in this example, or in the worst case might only be defined for $ t\\equal{}0$ and nowhere else.\r\n\r\nBut when the moment generating function is defined on some neighborhood of zero, it and the characteristic function each contain essentially the same information. \r\n\r\nThe moment generating function is a Laplace transform (with some sign changes); the characteristic function is a Fourier transform."
}
{
  "Tag": [
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "f derivabile neconstant on $ [0.1]$ s.t f(0)=0 .\r\nProve that exist $ c \\in [0,1]$ s.t $ f(c) <f'(c)$",
  "Solution_1": "solve the differential equation $ f'(x)\\minus{}f(x)\\equal{}\\minus{}h(x)$ where $ h(x)>0$"
}
{
  "Tag": [
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "x1^{2}+d*y1^{2}=p,x2^{2}+d*y2^{2}=p*a,where p is a prime number\r\nprove there exist x3,y3 that x3^{2}+d*y3^{2}=a",
  "Solution_1": "http://www.mathlinks.ro/viewtopic.php?p=821532#821532"
}
{
  "Tag": [],
  "Problem": "The following people have passed to the COP Round 2.\r\n\r\nTreething\r\nLynnelleYe\r\nwhitehorseking\r\nRC-7th\r\nalan\r\nsam 3.14\r\nh_s_potter 2002\r\n\r\nThe round 2 questions will be available very soon and the answers to the round one will be available also. (On the COP main website).\r\n\r\nCongratulation!",
  "Solution_1": "Awesome... I look forward to the next round.",
  "Solution_2": "Yay. Time to make hundreds of careless errors :P ."
}
{
  "Tag": [
    "function",
    "real analysis",
    "advanced fields",
    "advanced fields unsolved"
  ],
  "Problem": "Let p_t(x) be the projection of x on the t direction.t lies on [0,pi) (x lies on the plane).Let K be some compact subset of the plane\r\nHow can we show that the function t-- >l(p_t(K))(l the lebesgue measure) is Borel for t in [0,pi)?",
  "Solution_1": "The set of points where our function is less than $M$ is open for every $M$. To see it, just note that the projection to every direction is again compact set. If, for some particular direction $t$, its measure is less than $M$, then there exists a finite union of open intervals $I_j$ of total length less than $M$ containing our projection. Therefore, the set $K$ itself is contained in finitely many rectangular boxes $Q_j$ with bases $I_j$. But the projection of each box $Q_j$ to the directions close to $t$ is only slightly larger than $|I_j|$."
}
{
  "Tag": [
    "LaTeX",
    "modular arithmetic",
    "number theory"
  ],
  "Problem": "What are the positive integer values of $ n$ that make $ 2^{n} \\minus{}1$ divisible by $ 7$?",
  "Solution_1": "We consider three cases:\r\n\r\n   [u][b]Case 1[/b][/u] : $ n\\equal{}3k$ for k integer.\r\n\r\nThe we can see that: $ 2^n\\minus{}1\\equal{}(2^3)^k\\minus{}1\\equal{}8^k\\minus{}1\\equal{}7m$ where m is integer. Hence all the numbers of the form $ n\\equal{}3k$ satisfy\r\nthe condition of the problem.\r\n\r\n   [u][b]Case 2[/b][/u] : $ n\\equal{}3k\\plus{}1$\r\n\r\nThen we can write: \r\n   \r\n     $ 2^n\\minus{}1\\equal{}2^{3k\\plus{}1}\\minus{}1\\equal{}2(2^3)^k\\minus{}2\\plus{}1\\equal{}2(7m)\\plus{}1\\equal{}7(2m)\\plus{}1$ which is not a multiple of 7,hence this case is rejected.\r\n\r\n    [u][b]Case 3[/b][/u]: $ n\\equal{}3k\\plus{}2$\r\nThen we have:\r\n\r\n     $ 2^n\\minus{}1\\equal{}2^{3k\\plus{}2}\\minus{}1\\equal{}4(2^3)^k\\minus{}4\\plus{}3\\equal{}4(7m)\\plus{}3\\equal{}7(4m)\\plus{}3$ which again is not a multiple of 7.\r\n\r\nTherefore the number $ 2^n\\minus{}1$ is divisible by $ 7$ if and only if $ n\\equal{}3k$ for $ k$ integer.",
  "Solution_2": "[hide=\"Much faster\"]List out powers of 2 mod 7:\n\n$ 2^0\\equiv 1$\n$ 2^1\\equiv 2$\n$ 2^2\\equiv 4$\n$ 2^3\\equiv 1$\n\nIt clearly repeats after that, so we need the power to be $ 0\\bmod 3$ to get $ 1\\minus{}1\\equiv 0\\bmod 7$.\n\nSo any integer divisible by 3, or $ \\boxed{3k,\\ k\\in\\mathbb{Z}_{\\geq0}}$.[/hide]",
  "Solution_3": "[hide=\"Just as Fast\"]\n$ 2^n\\minus{}1 is 0 mod 7$\n$ 2^n\\equal{}1 mod 7$\nBy Fermat's Little Theorem, every multiple of 6 automatically works.\n\nIn addition, we notice that 2^3=8=1 mod 7, so we also include every multiple of 3.\n\nThen, we're done.\n\n[/hide]",
  "Solution_4": "[quote=\"zapi2007\"][hide=\"Solution with fixed LaTeX\"]\n$ 2^n \\minus{} 1 \\equiv 0 \\bmod 7$\n$ 2^n \\equiv 1 \\bmod 7$\n\nBy Fermat's Little Theorem, every multiple of 6 automatically works.\n\nIn addition, we notice that $ 2^3\\equiv 8\\equiv1 \\bmod 7$, so we also include every multiple of 3.\n\nThen, we're done.\n\n[/hide][/quote]\r\n\r\nFixed your $ \\LaTeX$. You'll want to use \\equiv for the equivalence symbol, and \\bmod {m} or \\pmod {m} for the mod part (\\pmod adds parentheses).\r\n\r\nIn my opinion, my solution is a little more straightforward than yours :wink:"
}
{
  "Tag": [
    "Putnam",
    "Harvard",
    "college",
    "Princeton",
    "MIT",
    "Stanford",
    "Duke"
  ],
  "Problem": "I just got my Putnam results envelope.\r\n\r\nCSULB Team Rank = 42. Yay!!\r\n\r\nI've got a class to teach right now. I'll be back to post more results later this afternoon or evening.",
  "Solution_1": "Does Caltech have any Putnam fellows?  :lol: (maybe someone whose surname is Lawrence?? :D )",
  "Solution_2": "The short answer to Albanian Eagle's questions: Yes, and Yes.\r\n\r\nTeam ranks:\r\n\r\n1. Harvard\r\n2. Princeton\r\n3. MIT\r\n4. Stanford\r\n5. Duke\r\n\r\nHonorable mention teams (in alphabetical order):\r\n\r\nCaltech\r\nCarnegie Mellon\r\nHarvey Mudd\r\nToronto\r\nWaterloo\r\n\r\nPutnam Fellows:\r\n\r\nJason Bland, Caltech\r\nBrian Lawrence, Caltech\r\nAaron Pixton, Princeton\r\nQingchun Ren, MIT\r\nXuancheng Shao, MIT\r\nArnav Tripathy, Harvard\r\n\r\nGroup N1:\r\n\r\nJustin Bae, Harvard\r\nRichard Biggs, Carnegie Mellon\r\nOleg Golberg, MIT\r\nYungcheng Lin, MIT\r\nTiankai Liu, Harvard\r\nKonstantin Matveev, Toronto\r\nAlison Miller, Harvard (Elizabeth Lowell Putnam Prize)\r\nAndrei Negut, Princeton\r\nLingren Zhang, Duke\r\nYufei Zhao, MIT\r\n\r\nGroup N2:\r\n\r\nZachary Abel, Harvard\r\nLurie Boreico, Harvard\r\nAnand Deopurkar, MIT\r\nAnjie Guo, Waterloo\r\nJunehyuk Jung, Chicago\r\nAnders Kaseorg, MIT\r\nJ\u00e1nos Kram\u00e1r, Toronto\r\nShrenik Shah, Harvard\r\n\r\nThere are about 50 Honorable Mention names: right now I don't have time to type all of those names.\r\n\r\nMy institution (CSULB) placed one student (Joshua Lampkins) in the top 200, and two others in the top 1000, accounting for our best-ever team rank.",
  "Solution_3": "Nice!!\r\n\r\nCongratulations to every one on that list of course!",
  "Solution_4": "The Putnam Fellows scored: 110, 97, 91, 90, 82, 82.\r\n\r\n(The question of which Fellow got which score is not, and will not be, public information.)\r\n\r\nThe minimum score needed for group N1 was 71.\r\n\r\nThe minimum score needed for group N2 was 64.\r\n\r\nThe minimum score needed for Honorable Mention was 45.\r\n\r\nThe minimum score needed for Group I was 42. (rank $ \\le 94$)\r\n\r\nThe minimum score needed for Group II was $ 30.3.$ (rank $ \\le 205$)\r\n\r\nThe minimum score needed for the top scores mailing was $ 21.$ (rank $ \\le 431.5$)",
  "Solution_5": "I see the MIT curse of picking the \"wrong\" team is alive and well.\r\nCaltech was clearly hurt by the lack of a strong third member.\r\nNo word yet at UW.",
  "Solution_6": "[quote=\"jmerry\"]\nCaltech was clearly hurt by the lack of a strong third member.\n[/quote]\r\n\r\nActually, I don't think Jason Bland was on the Caltech team. (He's a freshman, and I heard Caltech's team was just Brian Lawrence (duh) and the two top scores from previous years.) So they were hurt by the lack of strong second [i]and[/i] third members.",
  "Solution_7": "[quote=\"Kent Merryfield\"]Lurie Boreico, Harvard[/quote]\r\n\r\nAm I safe to assume this is [b]Iurie[/b] Boreico (Iura)?\r\n\r\n  darij",
  "Solution_8": "I guess so darij ;)\r\n\r\nYup jason was not on the Caltech team :) too bad...",
  "Solution_9": "The Caltech team was Lozan Ivanov, Brian Lawrence, and Doo Sung Park.\r\n\r\nThe official mailing spelled it \"Lurie.\" I assume they just misread it.\r\n\r\nWho are some people with many, many AoPS/Mathlinks posts? I see Xevarion in Group II (Rank 103-205). Sly Si is also Group II.\r\n\r\nHighest ranking Southern California ARML alum is Tedrick Leung in Group I (rank 78-94).",
  "Solution_10": "Honorable Mention by school:\r\n\r\nHarvard:\r\n\r\nZhou Fan\r\nRishi Gupta\r\n\r\nMIT:\r\n\r\nAleksandr Arkhipov\r\nThomas Belulovich\r\nGabriel Bujokas\r\nHansheng Diao\r\nNate Ince\r\nHyun Soo Kim\r\nSungyoon Kim\r\nThomas Mildorf\r\nKevin Modzelewski\r\nThanasin Nampaisarn\r\nEric Price\r\nJason Trigg\r\nAlexandr Zamorzaev\r\nBohua Zhan\r\n\r\nYale: Dimiter Ostrev\r\n\r\nPrinceton:\r\n\r\nPeter Diao\r\nJohn Pardon\r\nAaron Potechin\r\nAndrei Ungureanu\r\nAdrian Zahariuc\r\n\r\nColumbia: Dmytro Karabash\r\n\r\nCornell: Wei Quan Julius Poh\r\n\r\nSt. Joseph's (Philadelphia):  Dmytro Yeroshkin\r\n\r\nU. Maryland: Henry Scher\r\n\r\nJohns Hopkins: Kihyuk Hong\r\n\r\nU. Virginia: Wuttisak Trongsiriwat\r\n\r\nDuke:\r\n\r\nTirasan Khandawit\r\nAaron Pollack\r\n\r\nGrinnell: Rolf Hoyer\r\n\r\nCarleton: Jonah Ostroff\r\n\r\nU. Chicago:\r\n\r\nHarry Altman\r\nZhiwei Calvin Lin\r\n\r\nNorth Texas: David Simmons\r\n\r\nCaltech:\r\n\r\nEvan Dummit\r\nTimothy Black\r\nDoo Sung Park\r\nGjergji Zaimi (Albanian Eagle)\r\n\r\nHarvey Mudd:\r\n\r\nGregory Minton\r\nTheodore Spaide\r\n\r\nStanford:\r\n\r\nSerin Hong\r\nYoung Hun Jung\r\nNathan Pflueger\r\nJeffrey Wang (Paladin8)\r\n\r\nUC Berkeley: Sean Eberhard\r\n\r\nWaterloo:\r\n\r\nMichael Lipnowski\r\nYin Zhao\r\n\r\nU. Alberta: Steven Taschuk\r\n\r\nU. British Columbia: Cedric Lin",
  "Solution_11": "A comparison to other recent Putnam results. It seems to stack up a a middling-difficulty exam. One influence on the percentage of zeros: it seems to have been fairly easy to get 1 or 2 points on A1. In fact, I think some credit might have been given for answers that failed to find all four roots.\r\n\r\n$ \\begin{array}{cccccc}\\text{Year}&\\text{No. taking}&\\text{Mean}&\\text{Median}&\\text{pct zero}&\\text{Top 500}\\\\ 1998&2573&14.8&10&30.3&28\\\\ 1999&2900&6.3&0&60.2&11\\\\ 2000&2818&5.3&0&57.7&11\\\\ 2001&2954&8.9&1&44.9&20\\\\ 2002&3349&11.0&3&34.7&24\\\\ 2003&3615&7.1&1&27.8&18\\\\ 2004&3733&8.4&0&53.6&22\\\\ 2005&3545&7.9&1&46.7&20\\\\ 2006&3640&6.2&0&62.6&14\\\\ 2007&3753&7.7&1&42.5&21\\end{array}$",
  "Solution_12": "Yay Aaron Pollack at Duke! He is known at my school, since his dad is a professor there. He kind of grew up in the department.",
  "Solution_13": "does anyone know who the harvard team was? probably Tiankai, Alison, and ?\r\n\r\ncongratulations to all students",
  "Solution_14": "Is there any Rice students on the list?",
  "Solution_15": "We are not in the same high school, but we are in the same county and we've went out for ARML together, during which we talked a lot about math, music and different kind of language. I feel that he is a nice guy so I have some concern with his progress. Anyway, it's really nothing important for me if I don't know that (as you say, it is not worth discussing).  :lol:",
  "Solution_16": "[quote=\"jimhu\"]anyone know how well UCSD (San diego) or berkeley did? probably quite sucky[/quote]\r\nWhat I know and what I don't know:\r\n\r\nUC Berkeley had an H, two II's and two more in the top 500. UCSD had no one in the top 500.\r\n\r\nHere's the entire list for California public universities (UC's and CSU's both):\r\n\r\nH: UC Berkeley (1)\r\n\r\nI: -\r\n\r\nII: UC Berkeley (2), CSULB (1)\r\n\r\nother top 455: UCI (2), UC Berkeley (2), UCSC (1), SJSU (2), CSU Chico (1)\r\n\r\nI don't know anyone's team rank, other than the top 10, my own school, and two others where I've specifically been told. But UCSD is almost certainly behind our #42 placement. As for Berkeley: their test supervisor plays the identity of the team pretty close to the vest. The H (honorable mention) is a current high school student and almost certainly not on the Berkeley team; pkerichang, who is one of the II's, doesn't know whether or not he was on the team.",
  "Solution_17": "What is a \"II\"?\r\n\r\nAlso, how did you find out the results for UC Berkeley?  Is there a list posted somewhere?",
  "Solution_18": "A \"II\" is a score in the range $ [30.2,41],$ with a rank in the range $ [103,205].$ Roughly speaking, top 200 but not top 100.\r\n\r\nMy information comes from a list mailed to all Putnam supervisors (and probably a few other people) entitled \"Top Participants in the 68th Annual William Lowell Putnam Mathematical Competition.\" I've been treating this list as public information (except for the addresses printed on it) but I don't think that it's posted on the Web anywhere.\r\n\r\nThe one person who would have more detail than that regarding UC Berkeley scores would be Prof. Kahan.",
  "Solution_19": "Can someone tell me how DePaul did or where I can find the team/individual scores?",
  "Solution_20": "Information about teams outside the top 10 is not publicly released- we can tell you about any individuals in the top 500, but nothing more. For any more details, you'll have to ask the sponsor at DePaul.",
  "Solution_21": "Then were there any individuals from DePaul in the top 500?",
  "Solution_22": "No, DePaul had no one in the top \"500\". For the Chicago metropolitan area, U. Chicago had an N2, two H, one II, and three top 500; Northwestern had two top 500; IIT had two top 500.",
  "Solution_23": "[quote=\"Kent Merryfield\"]Rice? Have I mentioned that I was a Rice undergraduate myself?\n\nRice had two people in category I, which is to say rank 78-94:\n\nHo Chung Siu (from Hong Kong - is that you, Soarer?)\nShaui Ye\n\nAnd two more Rice students in the rank 221-431.5 category.\n\n----\n\nThe Harvard team was Tiankai, Alison, and Zachary Abel. So Harvard didn't pick the best team (yes, it would have been stronger with you) but won anyway.[/quote]\r\n\r\n\r\nSorry,I must point out the mistake...The name is Shuai Ye\r\nI know him.He is a freshman at Rice.",
  "Solution_24": "i am interested in who from UC Berkeley is in the top 500.\r\n\r\ncould someone with the list of top 500 participants be kind enough to post this information please?  thanks!",
  "Solution_25": "UC Berkeley:\r\n\r\nHonorable mention: Sean Eberhard (high school student)\r\n\r\nGroup II: Eric Chang (SoCal ARML alum), Joon Seok Lee\r\n\r\nOther top 500: Roman Vaisberg, Lawrence Wang",
  "Solution_26": "I wanna know who ranks the first?(score 110)",
  "Solution_27": "[quote=\"galoisj\"]I wanna know who ranks the first?(score 110)[/quote]\r\nI don't know that. The person who got the 110 doesn't know that. The professors at the university of that person don't know that. You may \"wanna know\" but you're not going to find out.",
  "Solution_28": "Wow, so the fellows do not get to know their own scores? Doesn't that have some data protection issues associated with it?",
  "Solution_29": "I wanna know the results of U of Minnesota"
}
{
  "Tag": [
    "algebra",
    "function",
    "domain",
    "group theory",
    "abstract algebra",
    "topology",
    "advanced fields"
  ],
  "Problem": "It is well-known that the torus can be constructed by using action groups; namely, $ Z^2$ acting on $ R^2$.    I know the classical model to construct the Klein Bottle, i.e., by identifying the edges, but I am figuring out if I can find some group which acts on $ R^2$ such that the equivalence classes that we get form a model as the classical one.\r\n\r\nI hope you can understand me and if not please let me know it and I can try to explain it again.\r\n\r\nRegards,\r\n\r\nAdriana",
  "Solution_1": "I think you're looking for the group generated by the following two transformations of $ \\mathbb{R}^2$:\r\n\r\n$ f(x,y) \\equal{} (x\\plus{}1,y)$\r\n\r\n$ s(x,y) \\equal{} (1\\minus{}x, y\\plus{}1)$\r\n\r\nConsider the action of these on the unit square, and notice how they identify the edges. You can conclude that unit square is a fundamental domain and the quotient space is the Klein bottle. \r\n\r\nYou can further analyze the group by studying the relations between $ f$ and $ s$. You'll find that\r\n\r\n$ f^{\\minus{}1}(x,y) \\equal{} (x\\minus{}1,y)$\r\n\r\n$ s^{\\minus{}1}(x,y) \\equal{} (1\\minus{}x,y\\minus{}1)$\r\n\r\nso\r\n\r\n$ sfs^{\\minus{}1}(x,y) \\equal{} sf(1\\minus{}x,y\\minus{}1) \\equal{} s(2\\minus{}x,y\\minus{}1) \\equal{} (x\\minus{}1,y) \\equal{} f^{\\minus{}1}(x,y)$\r\n\r\nTherefore the fundamental group $ G \\equal{} \\pi_1(K)$ is a nonabliean group having a normal subgroup $ H \\equal{} \\langle f \\rangle \\cong \\mathbb{Z}$ with $ G/H$ also isomorphic to $ \\mathbb{Z}$. Another way to say this is that $ G$ is a semidirect product of $ \\mathbb{Z} \\equal{} \\langle f \\rangle$ by $ \\mathbb{Z} \\equal{} \\langle s \\rangle$ with a non-trivial action of $ s$ on $ f$ given by $ sfs^{\\minus{}1} \\equal{} f^{\\minus{}1}$. This is sometimes called the infinite dihedral group (the finite dihedral group is what you get if you replace $ f$ by an element of finite order)."
}
{
  "Tag": [],
  "Problem": "Let $ n$ denote the minimum number of squares that must be removed from a 4 by 4 checkerboard so that there does not remain a 1 by 3 configuration of squares. What is $ n$?",
  "Solution_1": "I think it's one. If you take one from one of the middle four squares... I'm not sure. I mean like you take one from the middle and then it's four, two, one, four, four. I mean by the rows. Then by columns, it's basically the same thing. So there is no there by one configuration.",
  "Solution_2": "[geogebra]5535a01293428801f0dcb28a119d8e0e6097b7b0[/geogebra] \r\n\r\nThe ones with the dots are the ones taken out. From that, we see that we must take out $ \\boxed{5}$ squares.",
  "Solution_3": "Can't we just do this? Or do they mean like ANYWHERE there can't be a 1 by3 configuration? If that's the case then you can just do this... sorry if I reposted an answer...  :|",
  "Solution_4": "hm let's see...\r\nyou're picture has $ \\boxed{12}$ 1 x 3 configurations ;)",
  "Solution_5": "Can't we do\r\n\r\n [geogebra]c54adb2f5919a10ead7a605793353848e5480642[/geogebra] \r\n\r\nFirst we do the square (a2,2)--(a2,4)--(a4,2)--(a4,4) and then take (a1--3) and (a3--1). That's 3 squares? Or can't we even just take the whole board?",
  "Solution_6": "Well if we can do it YOUR way, then we can just take one square- the whole entire board. I doesn't state that they have to be a certain size either...",
  "Solution_7": "[quote=\"BOGTRO\"][..] Or can't we even just take the whole board? [..][/quote]\r\n\r\nI said that. So I guess the answer would be 1?",
  "Solution_8": "No, the FTW answer is 5"
}
{
  "Tag": [
    "probability"
  ],
  "Problem": "what are the four quantum numbers for iodine?",
  "Solution_1": "I'm not sure what you mean. Quantum numbers, assuming we are talking about the same thing, are used to specify where electrons are in an atom (i.e. $ n, l ,m_l,$ and of course spin).",
  "Solution_2": "That's right. As stated, that question does not make any sense. Remember that quantum numbers are a product of quantum theory, and are used to describe orbital's energy, form, and spatial orientation, and electrons. Let's recall briefly here those concepts.\r\n\r\nAccording to quantum theory, an orbital is a region around the nucleous of an atom where there is a high probability of finding one or two electrons. How are orbitals described? By three numbers that we call \"quantum numbers\". The principal quantum number, [i]n[/i], takes on natural values (1,2,3, ...) and is related with the orbital's energy. The bigger [i]n[/i] is, more energetic will be the orbital. Observe that by \"orbital's energy\" we really mean \"the energy of an electron when it \"lies\" on that orbital\". What energy is this? Well, an electron is negatively charged and the nucleous is positively charged, so there is a potential energy \"between\" them. As you know, this electric potential energy is inversely proportional to the distance between charges, so an orbital further apart from nucleous will have an increasing energy. That's the significance of [i]n[/i]: the greater it is, bigger will be the energy of an electron in that orbital because of the greater distance from nucleous.\r\nThe second quantum number (or angular momentum quantum number), [i]l[/i], describes the tridimensional form of the orbital, and for a given [i]n[/i] the only possible values of [i]l[/i] are 0, 1, 2, ..., n-1. Each value of [i]l[/i] corresponds to a given type of orbitals: orbitals with [i]l[/i] = 0 are named s orbitals, and have spherical simmetry; orbitals with [i]l[/i]= 1 are the p orbitals; those with [i]l[/i] = 2 are the d orbitals, and those with [i]l[/i] = 3 are the f orbitals.\r\nThe third quantum number (or magnetic quantum number), $ m_l$, describes the spatial orientation of orbitals. For a given [i]l[/i], the possible values of $ m_l$ are -l, -l+1, ..., 0, 1, ..., l-1, l.\r\nThose three quantum numbers are needed to describe completely a given orbital. However, to describe a given electron on a given orbital, a fourth quantum number is needed: the spin quantum number $ m_s$, whose only possible values are $ \\pm \\frac12$. These are related with the fact that an electron behaves as if it was rotating about an axis that passes through its center (just like the Earth does) - it spins -, and there are only two possible directions of rotation.",
  "Solution_3": "Thanks for the explanation! I meant to ask about the valence electron of the highest energy in the ground-state Iodine atom.",
  "Solution_4": "Iodine is in block p, period 5, group 17, and the \"previous\" inert gas is Kr. Therefore, the electron configuration of iodine is $ [Kr]5s^25p^5$, with the 5p electrons being the most energetic. Therefore, a possible set of quantum numbers for one of such electrons is $ \\left(5,1,\\minus{}1, \\frac12\\right)$.",
  "Solution_5": "so there is no set rule for whether it is $ \\pm \\frac 12$? and for the $ m_l$ quantum number, does it go from left to right on the periodic table from positive to negative?",
  "Solution_6": "I don't think there is any fixed rule about $ m_s$ being $ \\plus{}\\dfrac{1}{2}$ or  $ \\minus{}\\dfrac{1}{2}$.\r\nBut I have read that an electron with $ m_s \\equal{} \\plus{}\\dfrac{1}{2}$ (clockwise spin) has a lower energy, and thus greater stability than an electron with $ m_s \\equal{} \\minus{}\\dfrac{1}{2}$ (counter-clockwise spin).\r\n\r\nSo I [i][b]think[/b][/i] the convention is to use $ \\plus{}\\dfrac{1}{2}$.",
  "Solution_7": "so, for iodine, according to stability, it should be $ \\minus{}\\frac 12$ right? since its not as stable as a full octet.",
  "Solution_8": "I am not sure, but i think it's $ \\plus{} \\dfrac{1}{2}$. \r\n\r\nThe atom as such is not as stable as a noble gas. But that doesn't say much abt the quantum no. of the electron.\r\n\r\nUsually, an electron (or any particle for that matter) will always prefer to be in a state of lower energy(greater stability) rather than in a state of greater energy(lesser stability).\r\n\r\nWhen filling the orbitals first (using Hund's rule) with electrons of parallel spins, i guess u need to use clockwise spins first ($ \\plus{} \\dfrac{1}{2}$). Only when u are trying to pair the spins(Pauli's exclusion principle), u need to use a counterclockwise spin ( $ \\minus{} \\dfrac{1}{2}$). \r\n\r\nSince, in an Iodine atom, the last p-electron is a lone electron (unpaired), i think u should use  $ \\plus{} \\dfrac{1}{2}$. Atom's stablity as a whole ned not be considered, because that is essentially because of a lack of fully filled configuration and has nothing to do with the spin :maybe: \r\n\r\nI might be wrong though",
  "Solution_9": "@ hurdler:\r\n\r\nI am really sorry.   :oops_sign: \r\n\r\nWhat I have been saying all this while is true only if the system is placed in an external magnetic field. Only then does the electron with $ m_s \\equal{} \\plus{}\\dfrac{1}{2}$  (clockwise spin) have a lower energy than an electron with $ m_s \\equal{} \\minus{}\\dfrac{1}{2}$(counter clockwise spin).  :oops: \r\n\r\nI apologise for my mistake....",
  "Solution_10": "oh ok, thanks anyways for your help."
}
{
  "Tag": [
    "abstract algebra",
    "group theory",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "Show that if G is a finitely generated group, then every quotient group of G is finitely generated.\r\n\r\nShow that if $1 \\rightarrow N \\rightarrow G \\rightarrow Q \\rightarrow 1$ is a short exact sequence of groups, and if both N and Q are finitely generated (resp. finitely presented), then so is G.",
  "Solution_1": "The proof of the first statement is a one-liner: the images of the generators generate the quotient.",
  "Solution_2": "First I assume that the groups are abelian. This is because I borrow the definition of finite presentation from modules, and the $\\mathbb{Z}$-modules                 \r\nare exactly the abelian groups.                                                                                                                                       \r\nWe have the following diagram in which the rows and columns are exact:                                                                                                \r\n\\[ \\begin{array}{ccccccccc} \\ &\\ &\\mathbb{Z}^{n_2}&\\ &\\ &\\ &\\mathbb{Z}^{q_2}&\\ &\\ \\\\ \\ &\\ &d_2\\downarrow &\\ &\\ &\\ &\\downarrow\\partial_2&\\ &\\ \\\\ \\ &\\ &\\mathbb{Z}^{n_1}&\\ &\\ &\\ &\\mathbb{Z}^{q_1}&\\ &\\ \\\\ \\ &\\ &d_1\\downarrow &\\ &\\ &\\ &\\downarrow\\partial_1&\\ &\\ \\\\ 1&\\to &N&\\stackrel{i}{\\to}&G&\\stackrel{p}{\\to}&Q&\\to&1\\\\ \\ &\\ &\\downarrow &\\ &\\ &\\ &\\downarrow &\\ &\\ \\\\ \\ &\\ &1 &\\ &\\ &\\ &1&\\ &\\ \\\\ \\end{array}  \\]                                                                                                                                                         \r\nThis just comes from the definitions.                                                                                                                                 \r\nWe upgrade it to:                                                                                                                                                     \r\n\\[ \\begin{array}{ccccccccc} 1&\\to&\\mathbb{Z}^{n_2}&\\stackrel{i_{n_2}}{\\to} &\\mathbb{Z}^{n_2}\\oplus\\mathbb{Z}^{q_2}&\\stackrel{p_{q_2}}{\\to} & \\mathbb{Z}^{q_2}&\\to&1 \\\\ \\ &\\ &d_2\\downarrow &\\searrow^ {i_{n_1}\\circ d_2}&\\downarrow\\rho_2 &\\swarrow f_2&\\downarrow\\partial_2&\\ &\\ \\\\ 1&\\to&\\mathbb{Z}^{n_1}&\\stackrel{i_{n_1}}{\\to} &\\mathbb{Z}^{n_1}\\oplus\\mathbb{Z}^{q_1}&\\stackrel{p_{q_1}}{\\to} &\\mathbb{Z}^{q_1}&\\to&1 \\\\ \\ &\\ &d_1\\downarrow &\\searrow^ {i\\circ d_1}&\\downarrow\\rho_1 &\\swarrow f_1&\\downarrow\\partial_1&\\ &\\ \\\\ 1&\\to &N&\\stackrel{i}{\\to}&G&\\stackrel{p}{\\to}&Q&\\to&1\\\\ \\ &\\ &\\downarrow &\\ &\\downarrow &\\ &\\downarrow &\\ &\\ \\\\ \\ &\\ &1 &\\ &1 &\\ &1&\\ &\\ \\\\ \\end{array}  \\]                                                                                                                                                         \r\nThis diagram is constructed as follows:                                                                                                                               \r\n\\par The maps on the first and second row are natural injections and projections. With these construction it is obvious that now all the rows are short               \r\nexact sequences. Notice that on the first two rows we also have reverse maps because of the direct sum property.                                                      \r\n\\par We use the following lemma to construct $f_1$ and $f_2$:                                                                                                         \r\n\\par\\textbf{Lemma:} Let $p: G\\to Q$ be a surjective group homomorphism. Assume there exists $n\\in\\mathbb{N}$ and $\\pi: \\mathbb{Z}^n\\to Q$ a surjective                  \r\ngroup homomorphism. Then there exists a group homomorphism $\\bar\\pi: \\mathbb{Z}^n\\to G$ such that $p\\circ\\bar\\pi=\\pi$.                                                 \r\n\\par\\textbf{Proof of lemma:} Just choose $\\bar\\pi(e_i)\\in f^{-1}(\\pi(e_i))\\ \\forall i=\\overline{1,n}$, and extend by                                                  \r\nlinearity.\\eop\\\\                                                                                                                                                      \r\n\\par We apply the universality property of the direct sum to find maps $\\rho_1: \\mathbb{Z}^{n_1}\\oplus\\mathbb{Z}^{q_1}\\to G$ and                                       \r\n$\\rho_2: \\mathbb{Z}^{n_2}\\oplus\\mathbb{Z}^{q_2}\\to\\mathbb{Z}^{n_1}\\oplus\\mathbb{Z}^{q_1}$ such that every square in the diagram is commutative.                        \r\n\\par It follows from the snake lemma that $\\rho_1$ is surjective.                                                                                                     \r\n\\par From the commutativity of the diagram, $\\rho_1\\circ\\rho_2=i\\circ d_1\\circ d_2\\circ p_{n_2}=i\\circ 1\\circ p_{n_2}=1$.                                             \r\n\\par Also from the snake lemma it follows that there is an exact sequence                                                                                             \r\n$coker d_2\\stackrel{\\bar i}{\\to} coker\\rho_2\\stackrel{\\bar p}{\\to} coker\\partial_2\\to 1$. By exactness of the first and third column,                                 \r\n$coker d_2=N$ and $coker\\partial_2=Q$. The maps $\\bar i$ and $\\bar p$ are maps induced by $i_{n_1}$ and $p_{q_1}$ i.e. $\\bar i(\\hat a)= \\widehat{i_{n_1}(a)}$ etc. The identifications $coker d_2=N$ and $coker\\partial_2=Q$ allow us to define $\\bar i: N\\to coker\\rho_2$ by                                  \r\n$\\bar i(d_1(a))=\\widehat{i_{n_1}(a)}\\ \\forall a\\in\\mathbb{Z}^{n_1}$ and $\\bar p: coker\\rho_2\\to Q$ by                                                                  \r\n$\\bar p(\\hat b)=\\partial_1(p_{q_1}(b))\\ \\forall b\\in\\mathbb{Z}^{n_1}\\oplus\\mathbb{Z}^{q_1}$.                                                                          \r\n\\par Let $A=\\mathbb{Z}^{n_1}\\oplus\\mathbb{Z}^{q_1}$. We have a surjection $\\rho_1: A\\to G$ and a canonical projection that I will call                                 \r\n$\\alpha: A\\to coker\\rho_2$. It is clear that $\\ker\\alpha=Im\\rho_2$. $\\rho_1\\circ\\rho_2=1\\Rightarrow \\ker\\rho_1\\supseteq Im\\rho_2$. By the universality                 \r\nproperty of the factor group, there is a group homomorphism $\\beta: coker\\rho_2\\to G$. The snake lemma again proves that $\\beta$ is an isomorphism.                   \r\nThis proves $G\\simeq A/Im\\rho_2$. But $\\rho_1$ is surjective so $A/\\ker\\rho_1\\simeq G$. This and the inclusion $Im\\rho_2\\subseteq\\ker\\rho_1$ prove                    \r\n$Im\\rho_2=\\ker\\rho_1$ i.e. the exactness of the middle column.       \r\n-----------------------------------------------------------------------------------------------------------------------\r\n\r\nThe non-abelian case is a bit unclear to me.\r\nIt works very much the same up to a point by replacing $\\mathbb{Z}^n$ by the free group $\\mathbb F_n$, and the direct sum of abelian groups by the free product of groups. \r\nI'm not sure that the definition $G$ is finitely presented if and only if there exists an exact sequence $\\mathbb F^n\\to \\mathbb F^m\\to G\\to 1$ is equivalent to $G$ is a given by a finite set of generators and relations, but it seens plausible.\r\n\r\nThe problem that appears is the existence of $coker$ in the snake lemma."
}
{
  "Tag": [
    "Euler",
    "analytic geometry"
  ],
  "Problem": "Is there some specific equation or formula that could find the equation of the Euler Line of a triangle, given all coordinates each vertex on the triangle?",
  "Solution_1": "try this:[url]http://en.wikipedia.org/wiki/Euler_line[/url][/url]"
}
{
  "Tag": [
    "trigonometry",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "Let $ 0\\leq a,b,c\\leq \\frac{\\pi}{2}$ be such that\r\n\r\n$ \\sin a+\\sin b+\\sin c=1$ and $ \\sin a\\cos 2a+\\sin b\\cos 2b+\\sin c\\cos 2c=-1$. \r\n\r\nFind all possible values of $ \\sin^{2}a+\\sin^{2}b+\\sin^{2}c$.",
  "Solution_1": "I denote $ x=sina$ $ y=sinb$ $ z=sinc$ \r\nUsing the formula $ cos(2\\alpha)=1-2sin^{2}(\\alpha)$ the second relation is equivalent to:\r\n$ \\sum x(1-2x^{2})=-1$ $ \\Leftrightarrow$ $ \\sum x-2\\sum x^{3}=-1$ $ \\Leftrightarrow$ $ \\sum x^{3}=1$ ,because $ \\sum x=1$ so we have the system:\r\n$ x+y+z=1$\r\n$ x^{3}+y^{3}+z^{3}=1$ \r\n\r\nFrom $ (x+y+z)^{3}=x^{3}+y^{3}+z^{3}+3(x+y)(y+z)(x+z)$ $ \\Rightarrow$ $ y+z=0$ $ \\Rightarrow$ $ x=1$ \r\nMoreover because $ y\\geq 0$ ,$ z\\geq 0$ and $ y+z=0$ $ \\Rightarrow$ $ y=z=0$ ,therefore $ x^{2}+y^{2}+z^{2}=1$"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "function",
    "calculus",
    "derivative",
    "trigonometry",
    "algorithm"
  ],
  "Problem": "Hi everyone,\r\n\r\nI'm having some difficulty grasping the need for determining Taylor's polynomials when expanding functions by the Taylor's series. I realise that the formula foir evaluating Taylor's polynomials can be extended to generate the Taylor's series for that particular function. In that case, why evaluate Taylor's polynomials ? What is their significance ? \r\n\r\nThe reasoning usually adopted to determine the polynomial involves determining a tangent to the curve f(x), where f(x) is your function (for which you require the Taylor's series). The tangent equation is said to provide the best \"local approximation\" of the required function. I don't see how this is so ? I scoured the internet for some information and I came across this statement \"The fundamental idea in differential calculus is that a function can be 'locally' approximated by its tangent line.\" Now, this gets more and more confusing. Fundamental idea in differential calculus ????\r\n\r\nI'd appreciate and be grateful for some help on this topic.\r\n\r\nThanks, Siddharth",
  "Solution_1": "Taylor polynomials are useful because they (some of them, at least) are useful in approximating other differentiable functions, up to some degree of accuracy.  For example,\r\n$ e^{x}\\equal{}1\\plus{}x\\plus{}\\frac{x^{2}}{2!}\\plus{}\\cdots\\plus{}\\frac{x^{n}}{n!}\\plus{}R_{n}$\r\nwhere $ R_{n}$ is the remainder, or error term.  By taking derivatives over and over (assuming we can, there is some deep analysis behind the validity of differentiation), we can approximate a function more and more accurately.",
  "Solution_2": "Your calculator computes values of sines and cosines using Taylor polynomials.  Using [url=http://en.wikipedia.org/wiki/Taylor%27s_theorem]Taylor's theorem[/url], it is possible to compute exactly what degree of Taylor polynomial you need to achieve $ 13$-digit accuracy and then stop there.\r\n\r\nMore generally, studying the sequence of Taylor polynomials is a useful way to get some grip on their limit - the Taylor series itself.  We cannot really examine the Taylor series [i]except[/i] as a limit of the Taylor polynomials, and Taylor series are very useful when they have the nice properties we want them to have (i.e. convergence).",
  "Solution_3": "JRav and t0rajir0u have been speaking from a point of view that assumes that the function you're interested in is real analytic - that is, infinitely differentiable and given locally by its Taylor series. But what if the function you're interested in is not analytic, or you don't know that it is? What if all you know is that it has three derivatives, say, or four derivatives? Then the Taylor polynomial captures our knowledge.\r\n\r\nThe field of numerical analysis has many algorithms for approximating things: roots of functions, integrals, solutions of differential equations. Always, we want to know both whether the method converges and how fast it converges. And all of those estimates usually have some hypothesis of differentiability - $ C^1,C^2,C^4,\\dots$ - and an estimate of the size of some derivative is contained in the error estimate. All of those error estimating theorems use either Taylor polynomials or interpolating polynomials.\r\n\r\nIf I want to discuss finding and classifying the extrema of a smooth function of several variables, I can best explain what is going on by looking at the second order Taylor polynomial about the point in question.",
  "Solution_4": "There are certain \"bump\" functions such that the derivatives always exist, but when you calculate the Taylor series, you actually get that the series is 0, despite the fact that the function is clearly not the 0-function.",
  "Solution_5": "Hmmm, I've got a better understanding of Taylor's series now. I do have a more fundamental question. In the exponential series, if I were to substitue 'j\u0398' in place of 'x', by Euler's formula I would be able to get infinite series for the cos \u0398 and sin \u0398 terms respectively. An interesting fact over here is that all the even powers correspond to the cosine series, while the odd powers move to the sine series. I think these are Taylor's seriesfor the cosine and sine functions respectively (real-analytic functions) and I wanted to know how the interesting split of terms between the cosine and sine series was done.\r\n\r\nI was reading up some litreature on SOS Math, when I came up with doubts on Taylor's polynomials.\r\n\r\nComing to my questions:\r\na) Am I right in assuming the two infinte series generated are Taylor's series ?\r\nb) If so, could someone point as to why only even (or odd) powers of \u0398 remain in the series ?\r\n\r\nP.S: My office has blocked out certain features of Mathlinks. So, I cannot put up better images of theta,etc. Bloody *****.[/code]",
  "Solution_6": "[quote=\"Siddharth Rao\"]In the exponential series, if I were to substitue 'j\u0398' in place of 'x', by Euler's formula I would be able to get infinite series for the cos \u0398 and sin \u0398 terms respectively. An interesting fact over here is that all the even powers correspond to the cosine series, while the odd powers move to the sine series. I think these are Taylor's seriesfor the cosine and sine functions respectively (real-analytic functions) and I wanted to know how the interesting split of terms between the cosine and sine series was done.[/quote]\r\nYou mean [url=http://en.wikipedia.org/wiki/Euler%27s_formula]Euler's formula[/url]?  That is one of several \"proofs.\"  I prefer the interpretation in terms of the [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=216343]matrix exponential[/url].",
  "Solution_7": "The infinite series generated are indeed Taylor series, but you must justify the re-arrangement of terms in your series.  That is, you can't just move all of the even powers to one and side and the odd powers to the other side for any series.  The series must have a certain property (which is...?).\r\n\r\nTo your second question, what do you mean \"only even (or odd) powers\" remain?  Both even and odd powers of $ \\theta$ remain in the series, or else you could not get both sin and cosine.\r\n\r\nJust like with real numbers, you can think of a complex number as a point on a circle.  It's \"length\" is determined by the radius of the circle, and it's precise real and imaginary coordinates are determined by $ \\theta$.  So any complex number $ z$ can be written $ z\\equal{}r(\\cos{\\theta}\\plus{}i\\sin{\\theta})$.",
  "Solution_8": "While many high school students glibly work with sin x, most mathematicians feel more comfortable having its Taylor series. As far as Taylor polynomials go, they are a never ending source of computational pain to college multivariable calculus students and allegedly useful (and extremely painful) when it comes to finding extrema of functions.\r\n\r\nIn short it is nice for theoretical interest, but, as with most other things you learn in regular math classes, mathematica can do the work for you without you ever knowing the nuts and bolts. I've given up trying to memorize the different forms of the remainder, for example; I've decided that's what reference books are for.",
  "Solution_9": "I still think Weierstrass' Approximation Theorems are pretty useful.  I would much rather work with a polynomial than some crazy function I can't fathom.",
  "Solution_10": "For theoretical manipulations of Taylor series, this characterization is nice:\r\nSuppose $ f$ is $ n$ times differentiable at $ a$. The Taylor polynomial $ p$ of degree $ n$ is the unique polynomial of degree $ \\le n$ satisfying $ \\lim_{x\\to a}\\frac{f(x)\\minus{}p(x)}{x^n}\\equal{}0$.\r\n\r\nThis allows us to easily prove that a polynomial which is a good approximation is a Taylor polynomial, as might appear with $ e^{\\minus{}x^2}\\approx 1\\minus{}x^2\\plus{}\\frac{x^4}{2}\\plus{}\\cdots$ and similar substitutions.\r\n\r\nIf you can actually calculate the derivatives easily, this isn't worth it. $ \\sin$ and $ \\cos$ are the absolute best cases for the error terms in Taylor's theorem."
}
{
  "Tag": [
    "inequalities",
    "function",
    "inequalities unsolved"
  ],
  "Problem": "If $ 0<x\\leq 1$ and $ o<y\\leq 1$ prove or disprove that:\r\n\r\n$ {\\rm{(1 \\minus{} x)}}^{1 \\minus{} y}  \\plus{} (1 \\minus{} y)^{1 \\minus{} x}  \\ge 2 \\plus{} 2xy \\minus{} 2x \\minus{} 2y$",
  "Solution_1": "[quote=\"delegat\"]If $ 0 < x\\leq 1$ and $ o < y\\leq 1$ prove or disprove that:\n\n$ {\\rm{(1 \\minus{} x)}}^{1 \\minus{} y} \\plus{} (1 \\minus{} y)^{1 \\minus{} x} \\ge 2 \\plus{} 2xy \\minus{} 2x \\minus{} 2y$[/quote]\r\nSet $ a\\equal{}1\\minus{}x,b\\equal{}1\\minus{}y, 0 \\le a,b <1$\r\nThe inequality becomes:\r\n$ a^b\\plus{}b^a \\ge 2ab$\r\nBecause the function $ f(t)\\equal{}t^m, m\\in R$ is decreasing on $ [0,1)$\r\nSo $ a^b\\plus{}b^a \\ge a^1\\plus{}b^1 \\ge a^2\\plus{}b^2 \\ge 2ab$"
}
{
  "Tag": [
    "inequalities",
    "logarithms",
    "inequalities proposed"
  ],
  "Problem": "For $ a,b,c,x,y,z$ positive reals\r\n\r\n\r\n\r\n$ a^2 \\plus{} b^2 \\plus{} c^2 \\geq \\sqrt {\\frac {2x}{x \\plus{} y}}ab \\plus{} \\sqrt {\\frac {2y}{y \\plus{} z}}bc \\plus{} \\sqrt {\\frac {2z}{z \\plus{} x}}ca$",
  "Solution_1": "[quote=\"manlio\"]For $ a,b,c,x,y,z$ positive reals\n\n\n\n$ a^2 \\plus{} b^2 \\plus{} c^2 \\geq \\sqrt {\\frac {2x}{x \\plus{} y}}ab \\plus{} \\sqrt {\\frac {2y}{y \\plus{} z}}bc \\plus{} \\sqrt {\\frac {2z}{z \\plus{} x}}ca$[/quote]\r\nBy buffalo's way we need to prove that $ \\sum_{cyc}(x^4z^2\\minus{}x^2y^2z^2)\\geq0,$ which is obvious.",
  "Solution_2": "@arqady: what's buffalo's method?",
  "Solution_3": "[quote=\"FelixD\"]@arqady: what's buffalo's method?[/quote]\r\nIf $ a > 0$ and $ b^2 \\minus{} 4ac\\leq0$ then $ ax^2 \\plus{} bx \\plus{} c\\geq0$ for all value of $ x.$ :wink:",
  "Solution_4": "[quote=\"arqady\"][quote=\"manlio\"]For $ a,b,c,x,y,z$ positive reals\n\n\n\n$ a^2 \\plus{} b^2 \\plus{} c^2 \\geq \\sqrt {\\frac {2x}{x \\plus{} y}}ab \\plus{} \\sqrt {\\frac {2y}{y \\plus{} z}}bc \\plus{} \\sqrt {\\frac {2z}{z \\plus{} x}}ca$[/quote]\nBy buffalo's way we need to prove that $ \\sum_{cyc}(x^4z^2 \\minus{} x^2y^2z^2)\\geq0,$ which is obvious.[/quote]\r\n\r\nDear arqady,\r\n\r\nnice and neat solution, but can you please explain better your computation? Thank you very much. :)",
  "Solution_5": "Here is my proof:\r\n$ a^2 \\plus{} b^2 \\plus{} c^2 \\geq \\sqrt {\\frac {2x}{x \\plus{} y}}ab \\plus{} \\sqrt {\\frac {2y}{y \\plus{} z}}bc \\plus{} \\sqrt {\\frac {2z}{z \\plus{} x}}ca\\Leftrightarrow$\r\n$ \\Leftrightarrow a^2 \\minus{} \\left(b\\sqrt {\\frac {2x}{x \\plus{} y}} \\plus{} c\\sqrt {\\frac {2z}{z \\plus{} x}}\\right)a \\plus{} b^2 \\plus{} c^2 \\minus{} bc\\sqrt {\\frac {2y}{y \\plus{} z}}\\geq0.$\r\nHence, it remains to prove that\r\n$ \\left(b\\sqrt {\\frac {2x}{x \\plus{} y}} \\plus{} c\\sqrt {\\frac {2z}{z \\plus{} x}}\\right)^2 \\minus{} 4b^2 \\minus{} 4c^2 \\plus{} 4bc\\sqrt {\\frac {2y}{y \\plus{} z}}\\leq0,$\r\n which is equivalent to\r\n$ \\frac {(x \\plus{} 2y)b^2}{x \\plus{} y} \\plus{} \\frac {(z \\plus{} 2x)c^2}{x \\plus{} z} \\minus{} 2bc\\left(\\sqrt {\\frac {2y}{y \\plus{} z}} \\plus{} \\sqrt {\\frac {xz}{(x \\plus{} y)(x \\plus{} z)}}\\right)\\geq0,$ \r\nfor which we need to prove that\r\n$ \\left(\\sqrt {\\frac {2y}{y \\plus{} z}} \\plus{} \\sqrt {\\frac {xz}{(x \\plus{} y)(x \\plus{} z)}}\\right)^2 \\minus{} \\frac {(x \\plus{} 2y)(z \\plus{} 2x)}{(x \\plus{} y)(x \\plus{} z)}\\leq0.$\r\nBut $ \\left(\\sqrt {\\frac {2y}{y \\plus{} z}} \\plus{} \\sqrt {\\frac {xz}{(x \\plus{} y)(x \\plus{} z)}}\\right)^2 \\minus{} \\frac {(x \\plus{} 2y)(z \\plus{} 2x)}{(x \\plus{} y)(x \\plus{} z)}\\leq0\\Leftrightarrow$\r\n$ \\Leftrightarrow x^2z \\plus{} y^2x \\plus{} z^2y \\plus{} xyz\\geq\\sqrt {2xyz(x \\plus{} y)(x \\plus{} z)(y \\plus{} z)}\\Leftrightarrow$\r\n$ \\Leftrightarrow\\sum_{cyc}(x^4z^2 \\minus{} x^2y^2z^2)\\geq0.$",
  "Solution_6": "Thank you very much, Arqady. :)  Your proof is very nice.\r\n\r\nI tried to find a proof by AM-GM and Cauchy-Schwarz, but I failed. :(",
  "Solution_7": "[quote=\"manlio\"]Thank you very much, Arqady. :)  Your proof is very nice.\n\nI tried to find a proof by AM-GM and Cauchy-Schwarz, but I failed. :([/quote]\r\nYou can use AM-GM in that way: (But I think it's the same with [b]arqady[/b])\r\nUse AM-GM inequalitiy we have\r\n$ \\sqrt {\\frac{{2x}}{{x \\plus{} y}}} ab \\plus{} \\sqrt {\\frac{{2z}}{{z \\plus{} x}}} ca \\equal{} a\\left( {\\sqrt {\\frac{{2x}}{{x \\plus{} y}}} b \\plus{} \\sqrt {\\frac{{2z}}{{z \\plus{} x}}} c} \\right)$\r\n                 $ \\le{a^2} \\plus{} \\frac{1}{4}{\\left( {\\sqrt {\\frac{{2x}}{{x \\plus{} y}}} b \\plus{} \\sqrt {\\frac{{2z}}{{z \\plus{} x}}} c} \\right)^2}$\r\nSo it's suffficies to show that\r\n$ {b^2} \\plus{} {c^2} \\ge \\frac{1}{4}{\\left( {\\sqrt {\\frac{{2x}}{{x \\plus{} y}}} b \\plus{} \\sqrt {\\frac{{2z}}{{z \\plus{} x}}} c} \\right)^2} \\plus{} \\sqrt {\\frac{{2y}}{{y \\plus{} z}}} bc$\r\n$ \\Leftrightarrow {b^2}\\left( {4 \\minus{} \\frac{{2x}}{{x \\plus{} y}}} \\right) \\plus{} {c^2}\\left( {4 \\minus{} \\frac{{2y}}{{y \\plus{} z}}} \\right) \\ge 4bc\\left( {\\sqrt {\\frac{{2y}}{{y \\plus{} z}}}  \\plus{} \\sqrt {\\frac{{xz}}{{\\left( {x \\plus{} y} \\right)\\left( {z \\plus{} x} \\right)}}} } \\right)$\r\nNotice that $ 4\\minus{}\\frac{2x}{x\\plus{}y}>0$ and $ 4\\minus{}\\frac{2z}{z\\plus{}x}>0$ then we can use AM-GM ineq in here, so we just have to prove that\r\n$ \\left( {4 \\minus{} \\frac{{2x}}{{x \\plus{} y}}} \\right)\\left( {4 \\minus{} \\frac{{2y}}{{y \\plus{} z}}} \\right) \\ge 4{\\left( {\\sqrt {\\frac{{2y}}{{y \\plus{} z}}}  \\plus{} \\sqrt {\\frac{{xz}}{{\\left( {x \\plus{} y} \\right)\\left( {z \\plus{} x} \\right)}}} } \\right)^2}$\r\nThis inequality is equivalent to\r\n$ \\frac{x}{{x \\plus{} y}} \\plus{} \\frac{y}{{y \\plus{} z}} \\plus{} \\frac{z}{{z \\plus{} x}} \\plus{} \\sqrt {\\frac{{2xyz}}{{\\left( {x \\plus{} y} \\right)\\left( {y \\plus{} z} \\right)\\left( {z \\plus{} x} \\right)}}}  \\le 2$\r\n$ \\Leftrightarrow \\frac{y}{{y \\plus{} x}} \\plus{} \\frac{z}{{z \\plus{} y}} \\plus{} \\frac{x}{{x \\plus{} z}} \\ge 1 \\plus{} \\sqrt {\\frac{{2xyz}}{{\\left( {x \\plus{} y} \\right)\\left( {y \\plus{} z} \\right)\\left( {z \\plus{} x} \\right)}}}$ [b](I)[/b]\r\nThis inequality is a well-known.\r\nP/s: To prove[b] (I)[/b] you can see here (in page 19)",
  "Solution_8": "[quote=\"leedt26\"]\n$ \\frac {y}{{y \\plus{} x}} \\plus{} \\frac {z}{{z \\plus{} y}} \\plus{} \\frac {x}{{x \\plus{} z}} \\ge 1 \\plus{} \\sqrt {\\frac {{2xyz}}{{\\left( {x \\plus{} y} \\right)\\left( {y \\plus{} z} \\right)\\left( {z \\plus{} x} \\right)}}}$ [b](I)[/b]\nThis inequality is a well-known.\nP/s: To prove[b] (I)[/b] you can see here (in page 19)[/quote]\r\nProblem 3 (from the download)\r\nShow that for any positive real numbers $ a,$ $ b$ and $ c$ we have\r\n\\[ a^3 \\plus{} b^3 \\plus{} c^3 \\plus{} 6abc\\geq\\sqrt [3]{abc}(a \\plus{} b \\plus{} c)^2\\]\r\nMy proof:\r\nLet $ a \\plus{} b \\plus{} c \\equal{} 3u,$ $ ab \\plus{} ac \\plus{} bc \\equal{} 3v^2$ and $ abc \\equal{} w^3.$\r\nHence, $ a^3 \\plus{} b^3 \\plus{} c^3 \\plus{} 6abc\\geq\\sqrt [3]{abc}(a \\plus{} b \\plus{} c)^2\\Leftrightarrow f(v^2)\\geq0,$ where\r\n$ f(v^2) \\equal{} \\minus{} 3uv^2 \\plus{} 3u^3 \\minus{} u^2w \\plus{} w^3.$\r\nBut $ f$ gets a minimal value when $ v^2$ gets a maximal value, which happens  when two numbers from $ \\{a,b,c\\}$  are equal ($ a,$ $ b$ and $ c$ are real roots of the equation $ x^3 \\minus{} 3ux^2 \\plus{} 3v^2x \\minus{} w^3 \\equal{} 0$ :wink: ).\r\nThus, for the proof enough to check only one case: $ b \\equal{} c \\equal{} 1.$\r\nIn this case the original inequality is equivalent to $ g(a)\\geq0,$ where\r\n$ g(a) \\equal{} \\ln(a^3 \\plus{} 6a \\plus{} 2) \\minus{} \\frac {1}{3}\\ln a \\minus{} 2\\ln(a \\plus{} 2).$\r\nBut $ g'(a) \\equal{} \\frac {2(a \\minus{} 1)(a^3 \\plus{} 9a^2 \\minus{} 3a \\plus{} 2)}{3a(a^3 \\plus{} 6a \\plus{} 2)(a \\plus{} 2)},$ \r\nwhich gives $ a_{min} \\equal{} 1$ and our inequality is proven.",
  "Solution_9": "[quote=\"arqady\"][quote=\"leedt26\"]\n$ \\frac {y}{{y \\plus{} x}} \\plus{} \\frac {z}{{z \\plus{} y}} \\plus{} \\frac {x}{{x \\plus{} z}} \\ge 1 \\plus{} \\sqrt {\\frac {{2xyz}}{{\\left( {x \\plus{} y} \\right)\\left( {y \\plus{} z} \\right)\\left( {z \\plus{} x} \\right)}}}$ [b](I)[/b]\nThis inequality is a well-known.\nP/s: To prove[b] (I)[/b] you can see here (in page 19)[/quote]\nProblem 3 (from the download)\nShow that for any positive real numbers $ a,$ $ b$ and $ c$ we have\n\\[ a^3 \\plus{} b^3 \\plus{} c^3 \\plus{} 6abc\\geq\\sqrt [3]{abc}(a \\plus{} b \\plus{} c)^2\\]\nMy proof:\nLet $ a \\plus{} b \\plus{} c \\equal{} 3u,$ $ ab \\plus{} ac \\plus{} bc \\equal{} 3v^2$ and $ abc \\equal{} w^3.$\nHence, $ a^3 \\plus{} b^3 \\plus{} c^3 \\plus{} 6abc\\geq\\sqrt [3]{abc}(a \\plus{} b \\plus{} c)^2\\Leftrightarrow f(v^2)\\geq0,$ where\n$ f(v^2) \\equal{} \\minus{} 3uv^2 \\plus{} 3u^3 \\minus{} u^2w \\plus{} w^3.$\nBut $ f$ gets a minimal value when $ v^2$ gets a maximal value, which happens  when two numbers from $ \\{a,b,c\\}$  are equal ($ a,$ $ b$ and $ c$ are real roots of the equation $ x^3 \\minus{} 3ux^2 \\plus{} 3v^2x \\minus{} w^3 \\equal{} 0$ :wink: ).\nThus, for the proof enough to check only one case: $ b \\equal{} c \\equal{} 1.$\nIn this case the original inequality is equivalent to $ g(a)\\geq0,$ where\n$ g(a) \\equal{} \\ln(a^3 \\plus{} 6a \\plus{} 2) \\minus{} \\frac {1}{3}\\ln a \\minus{} 2\\ln(a \\plus{} 2).$\nBut $ g'(a) \\equal{} \\frac {2(a \\minus{} 1)(a^3 \\plus{} 9a^2 \\minus{} 3a \\plus{} 2)}{3a(a^3 \\plus{} 6a \\plus{} 2)(a \\plus{} 2)},$ \nwhich gives $ a_{min} \\equal{} 1$ and our inequality is proven.[/quote]\r\nwe can prove the stronger $ 8(a^3\\plus{}b^3\\plus{}c^3)\\plus{} 54abc \\ge 9\\sqrt [3]{abc}.(a \\plus{} b \\plus{} c)^2$\r\nBy shur,we have $ 8(a^3 \\plus{} b^3 \\plus{} c^3) \\plus{} 30abc \\ge 2(a \\plus{} b \\plus{} c)^3$\r\nand from AM-GM\r\n$ (a \\plus{} b \\plus{} c)^3 \\plus{} (a \\plus{} b \\plus{} c)^3 \\plus{} 27abc \\ge 9\\sqrt [3]{abc}.(a \\plus{} b \\plus{} c)^2$\r\nFrom it,we get\r\n$ 8(a^3 \\plus{} b^3 \\plus{} c^3) \\plus{} 57abc \\ge 9\\sqrt [3]{abc}.(a \\plus{} b \\plus{} c)^2$\r\nThis problem post by tranquocluat in maths.vn",
  "Solution_10": "In fact, the stronger inequality holds:\r\n\r\nIf $ a,$ $ b,$ $ c$ are positive real numbers such that $ abc\\equal{}1,$ then\r\n$ 3(a^3\\plus{}b^3\\plus{}c^3)\\plus{}27 \\ge 4(a\\plus{}b\\plus{}c)^2.$",
  "Solution_11": "[quote=\"can_hang2007\"]In fact, the stronger inequality holds:\n\nIf $ a,$ $ b,$ $ c$ are positive real numbers such that $ abc \\equal{} 1,$ then\n$ 3(a^3 \\plus{} b^3 \\plus{} c^3) \\plus{} 27 \\ge 4(a \\plus{} b \\plus{} c)^2.$[/quote]\r\nI think the maximal $ k$ for which the inequality\r\n$ a^3\\plus{}b^3\\plus{}c^3\\plus{}(9k\\minus{}3)abc\\geq k\\sqrt[3]{abc}(a\\plus{}b\\plus{}c)^2$\r\nholds for all non-negative $ a,$ $ b$ and $ c$ is\r\n$ k\\equal{}\\min_{a>0}\\frac{1}{18}\\left(a^2\\plus{}9a\\plus{}15\\plus{}\\frac{2}{a}\\right)\\equal{}1.316...$",
  "Solution_12": "[quote=\"arqady\"][quote=\"can_hang2007\"]In fact, the stronger inequality holds:\n\nIf $ a,$ $ b,$ $ c$ are positive real numbers such that $ abc \\equal{} 1,$ then\n$ 3(a^3 \\plus{} b^3 \\plus{} c^3) \\plus{} 27 \\ge 4(a \\plus{} b \\plus{} c)^2.$[/quote]\nI think the maximal $ k$ for which the inequality\n$ a^3 \\plus{} b^3 \\plus{} c^3 \\plus{} (9k \\minus{} 3)abc\\geq k\\sqrt [3]{abc}(a \\plus{} b \\plus{} c)^2$\nholds for all non-negative $ a,$ $ b$ and $ c$ is\n$ k \\equal{} \\min_{a > 0}\\frac {1}{18}\\left(a^2 \\plus{} 9a \\plus{} 15 \\plus{} \\frac {2}{a}\\right) \\equal{} 1.316...$[/quote]\r\nI only prove in the case $ k \\ge \\frac {3}{2}$\r\nLet abc=1.\r\n$ f(a,b,c) \\equal{} a^3 \\plus{} b^3 \\plus{} c^3 \\minus{} k(a \\plus{} b \\plus{} c)^2$\r\n$ f(a,b,c) \\minus{} f(a,\\sqrt {bc},\\sqrt {bc}) \\equal{} (\\sqrt {b} \\minus{} \\sqrt {c})^2[(b \\plus{} c \\plus{} \\sqrt {bc})^2 \\minus{} 2ka \\minus{} k(\\sqrt {b} \\plus{} \\sqrt {c})^2]$\r\n$ a \\equal{} min(a,b,c)$.It is easy to show that $ bc \\ge 1$\r\nWe have\r\n$ (b \\plus{} c \\plus{} \\sqrt {bc})^2 \\minus{} 2ka \\minus{} k(\\sqrt {b} \\plus{} \\sqrt {c})^2 \\ge \\frac {9(\\sqrt {b} \\plus{} \\sqrt {c})^4)}{16} \\minus{} k(\\sqrt {b} \\plus{} \\sqrt {c})^2 \\minus{} 2k \\ge 4.(\\frac {9}{4} \\minus{} k) \\minus{} 2k \\equal{} 9 \\minus{} 6k \\ge 0$\r\nFinally,we only need to show that in the case b=c,which is easy.",
  "Solution_13": "[quote=\"arqady\"]\nI think the maximal $ k$ for which the inequality\n$ a^3 \\plus{} b^3 \\plus{} c^3 \\plus{} (9k \\minus{} 3)abc\\geq k\\sqrt [3]{abc}(a \\plus{} b \\plus{} c)^2$\nholds for all non-negative $ a,$ $ b$ and $ c$ is\n$ k \\equal{} \\min_{a > 0}\\frac {1}{18}\\left(a^2 \\plus{} 9a \\plus{} 15 \\plus{} \\frac {2}{a}\\right) \\equal{} 1.316...$[/quote]\r\nNow I see that it's wrong.\r\nWe can say only that the maximal value of $ k$ grater than $ 1.316...$\r\nIndeed, your inequality is true, Can.",
  "Solution_14": "[quote=\"manlio\"]For $ a,b,c,x,y,z$ positive reals\n\n\n\n$ a^2 \\plus{} b^2 \\plus{} c^2 \\geq \\sqrt {\\frac {2x}{x \\plus{} y}}ab \\plus{} \\sqrt {\\frac {2y}{y \\plus{} z}}bc \\plus{} \\sqrt {\\frac {2z}{z \\plus{} x}}ca$[/quote]\r\nnice inequality,it generalize an old and hard Vasc's one.\r\nit's easy to show that the inequality is equivalent with,\r\n$ a^2 \\plus{} b^2 \\plus{} c^2 \\geq \\sqrt {\\frac {2}{1 \\plus{} x^2}}ab \\plus{} \\sqrt {\\frac {2}{1 \\plus{} y^2}}bc \\plus{} \\sqrt {\\frac {2}{1 \\plus{} z^2}}ca$\r\nwhere $ x,y,z$ are non negative real numbers such that $ xyz \\equal{} 1$. this inequality is symetric for $ x,y,z$ assume then that $ z \\equal{} max\\{x,y,z\\}$ then $ xy \\equal{} min\\{xy,yz,zx\\}\\leq 1$\r\nusing Cauchy Schwarz we have,\r\n$ \\sqrt {\\frac {2}{1 \\plus{} x^2}}ab \\plus{} \\sqrt {\\frac {2}{1 \\plus{} y^2}}bc \\equal{} b\\left(a\\sqrt {\\frac {2}{1 \\plus{} x^2}} \\plus{} c\\sqrt {\\frac {2}{1 \\plus{} y^2}}\\right)\\leq b\\sqrt {2(a^2 \\plus{} c^2)}.\\sqrt {\\frac {1}{1 \\plus{} x^2} \\plus{} \\frac {1}{1 \\plus{} y^2}}$\r\n$ \\equal{} b\\sqrt {2(a^2 \\plus{} c^2)}.\\sqrt {1 \\plus{} \\frac {1 \\minus{} x^2y^2}{(1 \\plus{} x^2)(1 \\plus{} y^2)}}\\leq b\\sqrt {a^2 \\plus{} c^2}.\\sqrt {1 \\plus{} \\frac {1 \\minus{} xy}{1 \\plus{} xy}}$\r\n$ \\equal{} b.X\\sqrt {\\frac {2z}{1 \\plus{} z}}$ where $ X \\equal{} \\sqrt {2(b^2 \\plus{} c^2)}$, then it's suffice to prove that,\r\n$ b.X\\sqrt {\\frac {2z}{1 \\plus{} z}} \\plus{} ac\\sqrt {\\frac {2}{1 \\plus{} z^2}} \\leq a^2 \\plus{} b^2 \\plus{} c^2$\r\nusing Cauchy-Schwarz we get $ \\sqrt {\\frac {2}{1 \\plus{} z^2}} \\leq \\frac {2}{1 \\plus{} z}$ then it's suffice to prove that:\r\n$ b^2 \\minus{} bX\\sqrt {\\frac {2z}{1 \\plus{} z}} \\plus{} a^2 \\plus{} c^2 \\minus{} \\frac {2ac}{1 \\plus{} z}\\geq 0$\r\n$ \\Delta \\equal{} \\frac {2zX^2}{z \\plus{} 1} \\minus{} 4\\left(a^2 \\plus{} c^2 \\minus{} \\frac {2ac}{z \\plus{} 1}\\right)$\r\n$ \\equal{} \\frac {4z(a^2 \\plus{} c^2)}{z \\plus{} 1} \\plus{} \\frac {8ac}{z \\plus{} 1} \\minus{} \\frac {4(a^2 \\plus{} c^2)(z \\plus{} 1)}{z \\plus{} 1}$\r\n$ \\equal{} \\minus{} \\frac {4(a \\minus{} c)^2}{z \\plus{} 1}$ if $ a\\neq c$ then $ \\Delta < 0$ and then we deduct our inequality, therefore it's suffice to check the case of $ a \\equal{} c$ in that case, the inequality is equivalent with,\r\n$ 2ab\\sqrt {\\frac {2z}{z \\plus{} 1}} \\plus{} \\frac {2a^2}{1 \\plus{} z}\\leq b^2 \\plus{} 2a^2$\r\n$ \\Leftrightarrow b^2 \\plus{} \\frac {2a^2z}{z \\plus{} 1}\\geq 2ab\\sqrt {\\frac {2z}{z \\plus{} 1}}$\r\nwhich is juste an application of Am-Gm inequality, that complete the proof.\r\nequality holds if and only if $ x \\equal{} y \\equal{} z$",
  "Solution_15": "[quote=\"manlio\"]For $ a,b,c,x,y,z$ positive reals\n\n\n\n$ a^2 \\plus{} b^2 \\plus{} c^2 \\geq \\sqrt {\\frac {2x}{x \\plus{} y}}ab \\plus{} \\sqrt {\\frac {2y}{y \\plus{} z}}bc \\plus{} \\sqrt {\\frac {2z}{z \\plus{} x}}ca$[/quote]\r\nBy AM-GM: $ \\frac{1}{2}\\sqrt{\\frac{2x}{x\\plus{}y}}(a^2\\plus{}b^2) \\ge \\sqrt{\\frac{2x}{x\\plus{}y}}ab$\r\nBy Cauchy-Schwarz: $ c\\sqrt{\\frac{2y}{y\\plus{}z} \\plus{} \\frac{2z}{z\\plus{}x}}\\sqrt{b^2\\plus{}a^2} \\ge \\sqrt {\\frac {2y}{y \\plus{} z}}bc \\plus{} \\sqrt {\\frac {2z}{z \\plus{} x}}ca$\r\nLet $ 2x^2 \\equal{} a^2\\plus{}b^2$. Then it is enough to prove:\r\n$ 2x^2\\plus{}c^2 \\ge \\sqrt{\\frac{2x}{x\\plus{}y}}x^2 \\plus{} \\sqrt{\\frac{4y}{y\\plus{}z} \\plus{} \\frac{4z}{z\\plus{}x}}cx \\iff$\r\n$ x^2 \\left ( 2 \\minus{} \\sqrt{\\frac{2x}{x\\plus{}y}} \\right ) \\plus{} c^2 \\ge \\sqrt{\\frac{4y}{y\\plus{}z} \\plus{} \\frac{4z}{z\\plus{}x}}cx$\r\nAnd therefore it is enough to prove:\r\n$ 4\\left ( 2 \\minus{} \\sqrt{\\frac{2x}{x\\plus{}y}} \\right ) \\ge \\frac{4y}{y\\plus{}z} \\plus{} \\frac{4z}{z\\plus{}x} \\iff$\r\n$ 1 \\minus{} \\frac{y}{y\\plus{}z} \\minus{} \\frac{z}{z\\plus{}x} \\ge \\sqrt{\\frac{2x}{x\\plus{}y}} \\minus{} 1 \\iff$\r\n$ \\frac{z(x\\minus{}y)}{(x\\plus{}z)(y\\plus{}z)} \\ge \\frac{x\\minus{}y}{\\sqrt{2x(x\\plus{}y)} \\plus{} x\\plus{}y} \\iff$\r\n$ (x\\minus{}y)(z\\sqrt{2x(x\\plus{}y)} \\minus{} z^2\\minus{}xy ) \\ge 0 \\iff$\r\n$ (x\\minus{}y)(2z^2x^2 \\minus{} z^4 \\minus{}x^2y^2) \\ge 0$\r\nSince the original inequality was cyclic in $ x,y,z$ it is enough to prove the two cases $ x \\ge z \\ge y$ and $ y \\ge z \\ge x$.\r\n\r\nCase $ x \\ge z \\ge y$ gives $ x\\minus{}y \\ge 0$ and $ x^2z^2 \\ge z^4, x^2z^2 \\ge x^2y^2$ and we are done.\r\nCase $ y \\ge z \\ge x$ gives $ x\\minus{}y \\le 0$ and $ z^2x^2 \\le z^4,z^2x^2 \\le y^2x^2$ and we are done.\r\n\r\nSo the proof is finished :)",
  "Solution_16": "[quote=\"peine\"]Then inequality is equivalent with,\n$ a^2 \\minus{} a\\left(b\\sqrt {\\frac {2x}{x \\plus{} y}} \\plus{} c\\sqrt {\\frac {2z}{z \\plus{} x}}\\right) \\plus{} b^2 \\plus{} c^2\\geq 0$\n[/quote]\r\nI think it should be\r\n$ a^2 \\minus{} a\\left(b\\sqrt {\\frac {2x}{x \\plus{} y}} \\plus{} c\\sqrt {\\frac {2z}{z \\plus{} x}}\\right) \\plus{} b^2 \\plus{} c^2 \\minus{} bc \\sqrt{\\frac{2y}{y\\plus{}z}} \\ge 0$",
  "Solution_17": "[quote=\"Mathias_DK\"][quote=\"peine\"]Then inequality is equivalent with,\n$ a^2 \\minus{} a\\left(b\\sqrt {\\frac {2x}{x \\plus{} y}} \\plus{} c\\sqrt {\\frac {2z}{z \\plus{} x}}\\right) \\plus{} b^2 \\plus{} c^2\\geq 0$\n[/quote]\nI think it should be\n$ a^2 \\minus{} a\\left(b\\sqrt {\\frac {2x}{x \\plus{} y}} \\plus{} c\\sqrt {\\frac {2z}{z \\plus{} x}}\\right) \\plus{} b^2 \\plus{} c^2 \\minus{} bc \\sqrt {\\frac {2y}{y \\plus{} z}} \\ge 0$[/quote]\r\nthank's a lot for your remark,   :!:  . I edited my message, and I posted a new solution, which is too nice, hope it's true."
}
{
  "Tag": [
    "geometry",
    "algebra proposed",
    "algebra"
  ],
  "Problem": "Numbers $p,q,r$ lies in the interval $(\\frac{2}{5},\\frac{5}{2})$ nad satisfy $pqr=1$. Prove that there exist two triangles of the same area, one with the sides $a,b,c$ and the other with the sides $pa,qb,rc$.",
  "Solution_1": "We know that if triangle $ABC$ has length of sides are $a,b,c$ then its area equal to $\\frac{2a^{2}b^{2}+2b^{2}c^{2}+2c^{2}a^{2}-a^{4}-b^{4}-c^{4}}{4}$. And if $a,b,c$ are positive reals such that $2a^{2}b^{2}+2b^{2}c^{2}+2c^{2}a^{2}-a^{4}-b^{4}-c^{4}>0$ then $a,b,c$ are lengths of sides of a triangle.\r\n\r\nNow, WLOG we can assume that $a=1$. We need prove that : Exist positive reals $b,c$ such that $0<2b^{2}+2b^{2}c^{2}+2c^{2}-1-b^{4}-c^{4}=2p^{2}q^{2}b^{2}+2q^{2}r^{2}b^{2}c^{2}+2p^{2}r^{2}c^{2}-p^{4}-q^{4}b^{4}-r^{4}c^{4}.$\r\n\r\nAt this time, that is all what i know.",
  "Solution_2": "Are you kidding me? \nThe problem does not require $p, q $ and $ r$ to be distinct. So let $p = q = r = 1$ and $1$ lies in the interval $ (\\frac{2}{5},\\frac{5}{2}) $. The two triangles are congruent because $pa = a, qb = b, rc = c$ and have the same area.  Done and totally acceptable!\n\nOn the other hand, if $p, q $ and $ r$ are distinct, arbitrarily let $r$ = 1 and $ \\frac{1}{p} = q$, the two triangles to be $ABC$ and $A'B'C'$ with $AB$ = $a$, $BC$ = $c$, $CA$ = $b$,  and $A'B'$ = $pa$, $B'C'$ = $rc$ = $c$, $C'A'$ = $qb$. We can also let $ \\angle BAC$ =  $ \\angle B'A'C'$ and $Sx$ and $Sy$ be the areas of $ABC$ and $A'B'C'$, respectively.\n\nWe have $Sx$ = $ \\frac {1}{2}ab sin\\angle BAC$ = $ \\frac {1}{2}paqb sin\\angle B'A'C'$ = $Sy$. \nActually, the interval range in the problem is not really important; to be nice you can use two numbers that are inverses of each other."
}
{
  "Tag": [
    "number theory proposed",
    "number theory"
  ],
  "Problem": "$(a+5)^2(b+5)^2=20ab(a+b+5)$",
  "Solution_1": "I thought you want a integer solution... :lol: \r\nLet x=ab, y=a+b, the equation becomes that\r\n$(x+y+20)^2=20xy$\r\nSince 20 is a divisor of a perfect square, the square must have the divisor 100,\r\nSo 5|xy and 10|x+y+20 that is 10|x+y, and 5|x or 5|y\r\nSuppose that 5|x,it's easy to get 5|y. It's clear that 5|a and 5|b.\r\nLet $x=5x_1, y=5y_1$. We get $20x_1y_1=(x_1+y_1+4)^2$\r\nSimilarly $5|x_1y_1, 10|x_1+y_1+4$\r\nBy5|a, 5|b we have 25|x, that is $5|x_1, x_1=5x_2$, and we get $y_1\\equiv 1(mod 5), y_1=5y_2+1$\r\nso $4x_2(5y_2+1)=(x_2+y_2+1)^2 \\Rightarrow 4x_2\\equiv(x_2+1)^2(mod y_2)$\r\nand $(y_2+1)^2\\equiv 0(mod x_2)$ these are $x_2|(y_2+1)^2, y_2|(x_2-1)^2$\r\nwhat's more, $x_2\\pm y_2$ is odd. \r\nLet $tx_2=(y_2+1)^2$ put it into the other equation \r\nwe get $y_2|2-2t, 2t\\equiv 2(mod y_2)$ we can get $x_2\\equiv 2(mod y_2), x_2-2|y_2$\r\nsimilarly $x_2|2+2r, 2r\\equiv 2(mod x_2)$ we can get $y_2\\equiv -2(mod x_2), y_2+2|x_2$\r\nso $y_2+2|x_2-y_2-2, x_2-2|y_2-x_2+2$\r\nFrom first we know $x_2-y_2-2$ is odd,but one of the 2 divisors must be even.Contradiction.",
  "Solution_2": "sorry, I edited them. :blush:",
  "Solution_3": "[quote=\"Skyward_Sea\"]I thought you want a integer solution... :lol: \r\nLet x=ab, y=a+b, the equation becomes that\r\n$(x+y+20)^2=20xy$\r\n\r\n\r\nI don't think so.\r\n\r\nit will be $(x+5y+25)^2=20x(y+5)$",
  "Solution_4": "[quote=\"Ghang Hwan\"][quote=\"Skyward_Sea\"]I thought you want a integer solution... :lol: \nLet x=ab, y=a+b, the equation becomes that\n$(x+y+20)^2=20xy$\n\n\nI don't think so.\n\nit will be $(x+5y+25)^2=20x(y+5)$[/quote][/quote]\r\n\r\nSorry I made such a stupid mistake... :P \r\nA new method here!\r\nYou equation is equal to $(ab-5a-5b-25)^2=0$\r\nso ab-5a-5b-25=0 that is (a-5)(b-5)=50\r\nthe steps below are easy now, I think .",
  "Solution_5": "Yeah, $(ab-5a-5b-25)^2=0$ seems to be the best rearrangement.\r\n\r\nPerhaps a more natural variation on your initial substitution would have been the best way to get it - rather than $y=a+b$, let $y=a+b+5$, and you get $(x+5y)^2=20xy$, which rearranges quite neatly into $(x-5y)^2=0$."
}
{
  "Tag": [
    "set theory",
    "advanced fields",
    "advanced fields theorems"
  ],
  "Problem": "Does anyone know what a large capital E means in set theory papers written in the 1930s? This is a large italic E not a sigma, and it is not backwards like an existential symbol.",
  "Solution_1": "Just a guess, but it might be an old existential quantifier.",
  "Solution_2": "Or in might mean union (E for \"ensemble.\") Or maybe it's not safe to guess.",
  "Solution_3": "I also saw an old paper where it was used for $\\in$ once. If you give us a few examples of usage, I think everything will become clear at once. Without that, any of our guesses seems to be as good (bad) as any other.",
  "Solution_4": "It's something like that:\r\n\r\n$E_{x}(x>0)$?\r\n\r\nIt's old notation for $\\{x| x>0\\}$."
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "This of Ductrung is so nice that I think it must have its own topic\r\n\r\n\\[ ab\\plus{}bc\\plus{}ca\\le\\frac{a^{3}(b\\plus{}c)}{a^{2}\\plus{}bc}\\plus{}\\frac{b^{3}(c\\plus{}a)}{b^{2}\\plus{}ca}\\plus{}\\frac{c^{3}(a\\plus{}b)}{c^{2}\\plus{}ab}\\le a^{2}\\plus{}b^{2}\\plus{}c^{2}\\]",
  "Solution_1": "I solved them by SOS method and Vornicu-Shur. Have you a more direct solution?\r\n\r\nThank you very much.",
  "Solution_2": "Hello manlio,\r\n\r\nI used Vornicu-Schur for both inequalities in my first solution. Since then I also found Cauchy-Schwarz-only solutions (for both) but I still think the Vornicu-Schur based are the most \"direct\" and natural.\r\n\r\nNote that\r\n\\[ \\sum_{cycl}\\frac{a^{3}(b\\plus{}c)}{a^{2}\\plus{}bc}\\plus{}\\sum_{cycl}\\frac{abc(b\\plus{}c)}{a^{2}\\plus{}bc}\\equal{}\\sum_{cycl}\\frac{a(b\\plus{}c)(a^{2}\\plus{}bc)}{a^{2}\\plus{}bc}\\equal{}2(ab\\plus{}bc\\plus{}ca)\\]\r\nThe given inequality is therefore equivalent to\r\n\\[ \\sum_{cycl}\\frac{abc(b\\plus{}c)}{a^{2}\\plus{}bc}\\le ab\\plus{}bc\\plus{}ca\\le\\frac{1}{2}\\sum_{cycl}\\left( a^{2}\\plus{}\\frac{abc(b\\plus{}c)}{a^{2}\\plus{}bc}\\right)\\ \\ (*)\\]\r\nThe left-hand side is a well-known inequality by Hoo-joo Lee [Crux 2580]; the right-hand side is equivalent to\r\n\\[ 0\\le\\sum_{cycl}\\left( a^{2}\\plus{}\\frac{abc(b\\plus{}c)}{a^{2}\\plus{}bc}\\minus{}a(b\\plus{}c)\\right)\\equal{}\\sum_{cycl}\\frac{a^{2}(a\\minus{}b)(a\\minus{}c)}{a^{2}\\plus{}bc}\\]\r\nwhich is true by Vornicu-Schur lemma.\r\n\r\nBTW: the left-hand side of (*) can also be obtained as a special case of (which I posted at [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=159311[/url]):\r\nFor any real number $ n\\ge 1$ and any positive real numbers $ a,b$ and $ c$,\r\n\\[ \\frac{a^{2}(b^{n}\\plus{}c^{n})}{a^{2}\\plus{}bc}\\plus{}\\frac{b^{2}(c^{n}\\plus{}a^{n})}{b^{2}\\plus{}ca}\\plus{}\\frac{c^{2}(a^{n}\\plus{}b^{n})}{c^{2}\\plus{}ab}\\le a^{n}\\plus{}b^{n}\\plus{}c^{n}\\]\r\n(by the change of variables $ (a,b,c)\\rightarrow (a^{\\minus{}1},b^{\\minus{}1},c^{\\minus{}1})$) or of:\r\nIf $ k\\ge 0$, $ p\\ge 1$ and $ a,b,c$ are positive real numbers, then\r\n\\[ \\frac{b\\plus{}c}{a^{k}(a^{2}\\plus{}pbc)}\\plus{}\\frac{c\\plus{}a}{b^{k}(b^{2}\\plus{}pca)}\\plus{}\\frac{a\\plus{}b}{c^{k}(c^{2}\\plus{}pab)}\\le\\frac{2}{p\\plus{}1}\\left(\\frac{1}{a^{k\\plus{}1}}\\plus{}\\frac{1}{b^{k\\plus{}1}}\\plus{}\\frac{1}{c^{k\\plus{}1}}\\right)\\]",
  "Solution_3": "Dear Ductrung, youd said it is possible to prove \r\n\r\n$ \\sum\\frac{a^{3}(b\\plus{}c)}{a^{2}\\plus{}bc}\\leq\\sum a^{2}$\r\n\r\nusing only Cauchy, but I don't see that. I tried it in two ways but I failed. :oops: \r\n\r\n$ 1st)$\r\n\r\nUsing your trick I rewrote the inequality in an equivalent form\r\n\r\n$ \\sum ab\\le\\frac{1}{2}(\\sum(a^{2}\\plus{}\\frac{abc(b\\plus{}c)}{a^{2}\\plus{}bc})$\r\n\r\nThen I used Cauchy in this form\r\n\r\n$ \\sum\\frac{(b\\plus{}c)^{2}}{(a^{2}\\plus{}bc)(b\\plus{}c)}\\geq\\frac{2(\\sum a)^{2}}{\\sum ab(a\\plus{}b)}$\r\n\r\nand putting in our inequality after some calculations we have\r\n\r\n$ \\sum ab(a\\plus{}b)(a\\minus{}b)^{2}\\geq abc\\sum (a\\minus{}b)^{2}$\r\n\r\nbut now we have to use SOS criterio to prove it (so we need not only Cauchy)\r\n\r\n$ 2nd)$\r\n\r\n$ \\sum\\frac{a^{3}(b\\plus{}c)}{a^{2}\\plus{}bc}\\leq\\sum a^{2}$\r\n\r\nI rewrote \r\n\r\n$ \\sum\\frac{a^{3}(b\\plus{}c)}{a^{2}\\plus{}bc}\\equal{}\\sum\\frac{a^{4}(b\\plus{}c)^{2}}{(a^{2}\\plus{}bc)a(b\\plus{}c)}$\r\n\r\nBut $ (a^{2}\\plus{}bc)a(b\\plus{}c)\\equal{}ab(a^{2}\\plus{}c^{2})\\plus{}ac(a^{2}\\plus{}b^{2})$ so by Cauchy we have\r\n\r\n$ \\frac{a^{4}(b\\plus{}c)^{2}}{(a^{2}\\plus{}bc)a(b\\plus{}c)}\\equal{}\\frac{a^{4}(b\\plus{}c)^{2}}{ab(a^{2}\\plus{}c^{2})\\plus{}ac(a^{2}\\plus{}b^{2})}\\leq\\frac{a^{3}b}{a^{2}\\plus{}c^{2}}\\plus{}\\frac{a^{3}c}{a^{2}\\plus{}b^{2}}$\r\n\r\nFinally we remain with\r\n\r\n$ \\sum\\frac{(a^{3}\\plus{}b^{3})c}{a^{2}\\plus{}b^{2}}\\leq\\sum a^{2}$\r\n\r\nwhich I cannot say if it is false or true. It is only easy to prove that\r\n\r\n$ \\sum\\frac{(a^{3}\\plus{}b^{3})c}{a^{2}\\plus{}b^{2}}\\geq\\sum ab$\r\n\r\nDuctrung, can you please give me a  hint for a only Cauchy solution? :) \r\n\r\nThank you very much.",
  "Solution_4": "[quote=\"manlio\"]I solved them by SOS method and Vornicu-Shur. Have you a more direct solution?\n\nThank you very much.[/quote]\r\nDear manlio,could you post your sos solution???I am a newbie who is learning sos.\r\nThanks. :)",
  "Solution_5": "Hi Manlio,\r\n\r\n[quote=\"manlio\"]\n$ \\sum ab(a\\plus{}b)(a\\minus{}b)^{2}\\geq abc\\sum (a\\minus{}b)^{2}$\nbut now we have to use SOS criterio to prove it (so we need not only Cauchy)\n[/quote]\nthe above can be proved without full expansion, without appealing to any theorem but purely algebraic manipulation using some known identity.\n\n[quote=\"manlio\"]\n$ \\sum\\frac{(a^{3}\\plus{}b^{3})c}{a^{2}\\plus{}b^{2}}\\leq\\sum a^{2}$\n[/quote]\r\nThe above is not true!",
  "Solution_6": "For stephen38:\r\n\r\n$ \\sum (ab(a\\plus{}b)\\minus{}abc)(a\\minus{}b)^{2}\\geq 0$\r\n\r\n$ \\sum C(a\\minus{}b)^{2}\\geq 0$\r\n\r\nwhere $ C\\equal{} ab(a\\plus{}b)\\minus{}abc$ ,$ B\\equal{} ca(c\\plus{}a)\\minus{}abc$,$ A\\equal{} bc(b\\plus{}c)\\minus{}abc$\r\n\r\nThen it is very easy to prove (WLOG assume $ a\\geq b\\geq c$) that \r\n\r\n$ B\\geq 0$ , $ A\\geq 0$ , $ B\\plus{}C\\geq 0$ which is third criterium of SOS method.\r\n\r\n\r\nDuctrung, can you please give me the known identity used to avoid SOS method.\r\n\r\nThank you very much.",
  "Solution_7": "[quote=\"manlio\"]\n$ \\sum ab(a\\plus{}b\\minus{}c)(a\\minus{}b)^{2}\\geq 0\\ \\ (*)$\n[/quote]\r\nWe know the transform between SOS form and Vornicu-Schur form:\r\n$ \\sum_{cycl}x(b\\minus{}c)^{2}\\equal{}\\sum_{cycl}(x\\plus{}y)(c\\minus{}a)(c\\minus{}b)$\r\nThus (*) is equivalent to\r\n$ 0\\le\\sum_{cycl}[ c^{2}(a\\plus{}b)\\plus{}c(a\\minus{}b)^{2}](c\\minus{}a)(c\\minus{}b) \\equal{}\\sum_{cycl}c^{2}(a\\plus{}b)(c\\minus{}a)(c\\minus{}b)\\ \\ (**)$\r\nThe last equality is due to the fact that the sum\r\n\\[ \\sum_{cycl}c(a\\minus{}b)^{2}(c\\minus{}a)(c\\minus{}b)\\]\r\nis vanished. To prove (**), assume $ a\\ge b\\ge c$. Then\r\n$ c^{2}(a\\plus{}b)(c\\minus{}a)(c\\minus{}b)\\ge 0$\r\nFurther,\r\n$ a^{2}(b\\plus{}c)(a\\minus{}b)(a\\minus{}c)\\plus{}b^{2}(c\\plus{}a)(b\\minus{}a)(b\\minus{}c) \\equal{} (a\\minus{}b)\\left[ a^{2}(b\\plus{}c)(a\\minus{}c)\\minus{}b^{2}(c\\plus{}a)(b\\minus{}c)\\right]$\r\n$ \\ge (a\\minus{}b)\\left[ a^{2}(b\\plus{}c)(b\\minus{}c)\\minus{}b^{2}(c\\plus{}a)(b\\minus{}c)\\right]$\r\n$ \\equal{} (a\\minus{}b)\\left[ (b\\minus{}c)[ a^{2}(b\\plus{}c)\\minus{}b^{2}(c\\plus{}a) ]\\right]$\r\n$ \\equal{} (a\\minus{}b)\\left[ (b\\minus{}c)[ ab(a\\minus{}b)\\plus{}c(a^{2}\\minus{}b^{2}) ]\\right]$\r\n$ \\equal{} (a\\minus{}b)^{2}\\left[ (b\\minus{}c)(ab\\plus{}bc\\plus{}ca)\\right]\\ge 0$\r\nand we are done. OK, I must confess that it is faster to invoke Vornicu-Schur lemma on (**).\r\nBut what about the following alternative: To prove\r\n\\[ ab\\plus{}bc\\plus{}ca\\le\\frac{1}{2}\\sum_{cycl}\\left( a^{2}\\plus{}\\frac{abc(b\\plus{}c)}{a^{2}\\plus{}bc}\\right)\\]\r\nas you noted, by Cauchy-Schwarz inequality, it suffices to show that\r\n\\[ 2(ab\\plus{}bc\\plus{}ca)\\le a^{2}\\plus{}b^{2}\\plus{}c^{2}\\plus{}\\frac{2abc(a\\plus{}b\\plus{}c)^{2}}{a(b^{2}\\plus{}c^{2})\\plus{}b(c^{2}\\plus{}a^{2})\\plus{}c(a^{2}\\plus{}b^{2})}\\]\r\nLet $ X \\equal{} a^{2}\\plus{}b^{2}\\plus{}c^{2}\\minus{}2(ab\\plus{}bc\\plus{}ca)$. The above then can be rewritten as\r\n$ 0\\le X\\sum_{cycl}a(b^{2}\\plus{}c^{2})\\plus{}2abc(a\\plus{}b\\plus{}c)^{2}$\r\n$ \\equal{} X\\sum_{cycl}a(b\\minus{}c)^{2}\\plus{}2abc[ (a\\plus{}b\\plus{}c)^{2}\\plus{}3X ]$\r\n$ \\equal{} X\\sum_{cycl}a(b\\minus{}c)^{2}\\plus{}2abc[ 4(a^{2}\\plus{}b^{2}\\plus{}c^{2}\\minus{}ab\\minus{}bc\\minus{}ca) ]$\r\n$ \\equal{} X\\sum_{cycl}a(b\\minus{}c)^{2}\\plus{}4abc\\sum_{cycl}(b\\minus{}c)^{2}$\r\n$ \\equal{}\\sum_{cycl}a(b\\minus{}c)^{2}[ X\\plus{}4bc ]$\r\n$ \\equal{}\\sum_{cycl}a(b\\minus{}c)^{2}(b\\plus{}c\\minus{}a)^{2}$\r\nwhich is obviously true.",
  "Solution_8": "Thank you very much, Ductrung. \r\nYour reasoning is very interesting. :)"
}
{
  "Tag": [
    "LaTeX"
  ],
  "Problem": "Whenever I make a small equation and compile it in $ \\text{\\LaTeX}$, the result I get is a page with my formula and a lot of blank space. Is there a way I can resize the page so that I don't have any of that blank space (like the formulas you see in this website)?",
  "Solution_1": "LaTeX is for writing documents so if you only want one equation, that's what you'll get on a page. This site extracts an image of the equation and puts it in a post. If you want a document then there is no reason why you can't put lots more on the page but if you want an image on its own then use the TeXer, where you can copy the image."
}
{
  "Tag": [
    "logarithms",
    "inequalities"
  ],
  "Problem": "What is the smallest integer value of $ n$ such that $ 2^{\\minus{}n} &lt; \\frac{1}{1000}$?",
  "Solution_1": "hello, taking the logarithm of both sides your inequality we have\r\n$ \\minus{}n\\ln(2)<\\ln\\left(\\frac{1}{1000}\\right)\\equal{}\\ln(1)\\minus{}\\ln(1000)\\equal{}\\ln(1000)$\r\nThen we get\r\n$ \\minus{}n<\\frac{\\ln(1000)}{\\ln(2)}$\r\nBy multiplying with minus one\r\n$ n>\\frac{\\ln(1000)}{\\ln(2)} \\approx 9,965784...$\r\nSo we have finally $ n\\geq10$.\r\nSonnhard.",
  "Solution_2": "This can be rewritten as $ \\frac {1}{2^n} < \\frac {1}{1000}$\r\nBoth numbers are positive, so we can cross multiply to get\r\n\r\n$ 1000 < 2^n$\r\n\r\nI'm betting you guys probably know your powers of 2 pretty well, and that $ 2^{10}$=1024 (the infamous kilobyte).\r\n\r\n9 is too small, hence $ \\boxed{10}$\r\n\r\nWhile logs are relevant, it's probably overkill for this one."
}
{
  "Tag": [
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Given real numbers $a,\\alpha,\\beta, \\sigma \\ and \\ \\varrho$ s.t. $\\sigma, \\varrho > 0$ and $\\sigma \\varrho = \\frac{1}{16}$, prove that there exist integers $x$ and $y$ s.t. \r\n\r\n\\[ - \\sigma \\leq (x+\\alpha_(ax + y + \\beta ) \\leq \\varrho  \\]",
  "Solution_1": "whats the complete question\nare they asking the min of them or max or is the comma actually a bracket\n"
}
{
  "Tag": [
    "MIT",
    "college",
    "vector",
    "linear algebra"
  ],
  "Problem": "dudes, MIT is over.. Im so happy. no more physics nor chem for 1 month. though i'll miss the math i have to admitt",
  "Solution_1": "yeaaa\r\n\r\nAre you going to be there for IAP?  If you are you should do the mystery hunt with the Simmons Hall team :) \r\n\r\nAlso, I'm going to get more events going with our lounge next semester assuming everyone signs up again.  Hopefully I can get a weekly (or every two weeks, or something) where we do a couple practice problems and, of course, get free food.\r\n\r\nI love the idea of IAP.  It better be as purely awesome as it sounds.",
  "Solution_2": "im not staying for iap man..\r\nhey i havent met u. u live in simmons right/ who are ya? do i know u",
  "Solution_3": "Yeah, I live in Simmons.  But we haven't met yet. :(",
  "Solution_4": "aw ok i'll look for u in Feb.\r\ni live in 7c.. the best floor ever",
  "Solution_5": "joml88, are tu taking 100 B next semester?\r\ni think i am taking 100 C instead. one of those 2... you should take it too",
  "Solution_6": "I'm going take two of 18.03, 18.06, and 18.100.  Not sure yet.  Maybe 18.03 and 18.100C...",
  "Solution_7": "if u take 18.03 and 18.06 you will get bored ! take 100b.. i think ill take 100b, because 100c is too much work and i can clear out the CI-M requirement with seminars that are easier ( no psets, no tests) and many of them look interesting",
  "Solution_8": "is 18.06 linear algebra?\r\nwith strang?\r\nbecause i need lots of help for that.\r\nwill someone explain vectors and graphing because i am really simple-minded",
  "Solution_9": "Postnikov is teaching it in the Spring.  I suggest asking about vectors in the pre-olympiad forum.",
  "Solution_10": "THANKS!  i never thought about that. is Prof. Strang staying at MIT?",
  "Solution_11": "I think vectors are a little too advanced for a middle schooler, although I have done physics vectors stuff and linear algebra in the summer before in class.",
  "Solution_12": "I don't think Strang's going anywhere(?).  Professors don't necessarily teach the same classes each semester.",
  "Solution_13": "I know but I thought Strang was THE LINEAR ALGEBRA GUY. Partially because he wrote the book I use.\r\n\r\nAND HOW DO YOU KNOW I AM A MIDDLE SCHOOLER MAYBE I AM SOME HACK WHO LIKES TO PRETEND THEY ARE!!!!\r\n\r\nOk I am so wtf I gave too much info on the net.",
  "Solution_14": "[quote=\"funcia\"]I know but I thought Strang was THE LINEAR ALGEBRA GUY. Partially because he wrote the book I use.\n\nAND HOW DO YOU KNOW I AM A MIDDLE SCHOOLER MAYBE I AM SOME HACK WHO LIKES TO PRETEND THEY ARE!!!!\n\nOk I am so wtf I gave too much info on the net.[/quote]\r\n\r\n\r\nI meant I AM A MIDDLE SCHOOLER  :lol:",
  "Solution_15": "phew too many people know everything aboubt me *ggrrrr*** well too bad i revealed(or did i) that i am a middle schooler",
  "Solution_16": "postnikov is a good friend of mine!\r\n\r\nhes a combinatorialist...\r\n\r\ni dont know why he is teaching  a basic math class.. hes a great mathematician.. but he is hard to follow.. i mean, its wierd for me thinking of him giving a computational class like 18.06, that so many non-math people take.",
  "Solution_17": "a paradox would be that he teaches basic math class to have time to youthfully study other more complicated activities in more spare time.  :P"
}
{
  "Tag": [
    "ratio"
  ],
  "Problem": "After one-fifth of the girls left a school dance, the ratio of girls to boys was 2:3. Then, 44 boys left and the ratio of boys to girls was 2:5. How many students remained at the dance??",
  "Solution_1": "I have seen this SO many times. \r\n[hide] Write the ratios, with g = number of girls, b = number of boys\n$\\frac{g}{b}=\\frac{2}{3}$, $\\frac{b-44}{g}=\\frac{2}{5}$\nMultiply the two equations:\n$\\frac{b-44}{b}=\\frac{4}{15} \\Longrightarrow 15b-660=4b \\Longrightarrow 11b=660 \\Longrightarrow b=60$\n$\\frac{g}{60}=\\frac{2}{3} \\Longrightarrow g=40$\n$g+b-44=40+60-44=\\boxed{56}$[/hide]",
  "Solution_2": "[quote=\"ch1n353ch3s54a1l\"]After one-fifth of the girls left a school dance, the ratio of girls to boys was 2:3. Then, 44 boys left and the ratio of boys to girls was 2:5. How many students remained at the dance??[/quote]\r\n[hide]First equation\n$.8g/b=2/3 2.4g=2b b=1.2g$\nnext equation\n$(.8g)/(b-44)=5/2 1.6g=5b-220.$\nSubstitution yields\n$4.4g=220. g=50 b=60.$\nThe answer is [b]56[/b][/hide]",
  "Solution_3": "A simpler solution :P  :D \r\n[hide]\nFlip both fractions so that they are boys over girls. Find a common denominator since the number of girls doesn't change (in the last part of the problem). The two fractions are 15/10, 4/10. 15-4=11 <- change in boys. 44/11=4 so there are 4 people per unit in the fractions. 4*4+10*4=56.\n[/hide]",
  "Solution_4": "Huge bump but surely this is better than starting a new thread when this question has been posted so many times.\nSo I get it that we can find the ratios of boys to girls before and after. Wouldn't you set $\\frac{11}{10}=\\frac{44}{x}$? I doubt that is the right proportion and I get so lost after here..",
  "Solution_5": "Call the number of girls $g$ and boys $b$.\n\nThen we set the proportion $\\frac{0.8g}{b}=\\frac{2}{3}$. From this we get $6g=5b$.\n\nNow, $\\frac{0.8g}{b-44}=\\frac{5}{2}$. From this we get $1.6g=5b-220$. But we know from our previous result that $5b=6g$, hence the substitution $\\rightarrow 4.4g=220$ and $g=50, b=60$.\n\nHowever, only 16 boys and 40 girls stayed $\\rightarrow \\boxed{56}$."
}
{
  "Tag": [],
  "Problem": "This is for people who went to nationals last year. Which state do you think had the coolest pins? Also, which had the stupidest? (i think i know that answer)\r\nMy opinion, best- Virgin Islands (also a strange story to go along w/ it)\r\nWorst- my home, connecticut",
  "Solution_1": "Nah Illinois was probably the worst.  Even though Illinois got 2nd two years ago, our pin was simply a plastic one in the shape of illinois (probably with \"illinois\" written on it)... very creative... \r\n\r\nNo wonder no one liked ours, but I suppose the previous (or previous previous, however you want to look at it) finish paid off, as people still were willing to trade for it :-O\r\n\r\nMaybe we AoPS people can arrange a place to meet and trade pins or something...",
  "Solution_2": "at least yours didn't say something stupid, like \"Connecticut, we're full of surprises\"",
  "Solution_3": "If I recall correctly, the worst pin was Mississippi.  It was a plastic pin that said \"Mississippi\".  No image of the state, no duck (a la Minnesota), no flag, etc.  Just the word \"Mississippi\".\r\n\r\nIncidently, I ended up with about 40 Mississippi pins because I traded my ESPN Zone card to some guy from Mississippi who didn't care about pin trading.  I was gonna use those pins as extra firepower to get all the states, but some states didn't bring pins.  So now I have two almost-complete pin collections from '02 and '03.",
  "Solution_4": "I think pin-trading is too tiring, scrambling about trying to find those possessing the rarest pins.  I propose some strategies:\r\n\r\n1.  Work as a team so that no extra pins will be leaking out, ensuring the value of your team's pins.\r\n\r\n2.  Don't trade until later to raise demand.\r\n\r\n3.  Forget pin-trading... I'd rather eat my meal peacefully and perhaps get a few exchanges while I'm at it.",
  "Solution_5": "[quote=\"probability1.01\"]I think pin-trading is too tiring, scrambling about trying to find those possessing the rarest pins.  I propose some strategies:\n\n1.  Work as a team so that no extra pins will be leaking out, ensuring the value of your team's pins.\n\n2.  Don't trade until later to raise demand.\n\n3.  Forget pin-trading... I'd rather eat my meal peacefully and perhaps get a few exchanges while I'm at it.[/quote]\r\nlol, what i did last year was just the opposite. i traded away 4 of my state's pins at once, and getting many extras for other states. of course, then my team got really pissed at me and stole from me. You know who you are....... :evil:",
  "Solution_6": "I believe that the purpose of pin-trading is to get to know the other guys.  It's not like we actually hold value in the pins (although knowing that we're math geeks, maybe we do...).  That's why this year I'm not even going to try (ok, I won't be able to resist temptation to do it a little).  I'll mostly chat w/ the other people around."
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "Let $ a,b,c\\geq 0$.Prove that:\r\n\r\n$ \\frac {1}{(a \\plus{} b)^2} \\plus{} \\frac {1}{(b \\plus{} c)^2} \\plus{} \\frac {1}{(c \\plus{} a)^2} \\geq \\frac {9}{(a \\plus{} 3b \\plus{} 2c)^2} \\plus{} \\frac {9}{(b \\plus{} 3c \\plus{} 2a)^2} \\plus{} \\frac {9}{(c \\plus{} 3a \\plus{} 2b)^2}$\r\n\r\n[hide]\nI think that the following inequality is also true:\n\n$ \\frac {1}{(a \\plus{} 3b \\plus{} 2c)^2} \\plus{} \\frac {1}{(b \\plus{} 3c \\plus{} 2a)^2} \\plus{} \\frac {1}{(c \\plus{} 3a \\plus{} 2b)^2} \\leq \\frac {1}{4(ab \\plus{} bc \\plus{} ca)}$\n\nfor $ a,b,c \\geq 0$[/hide]",
  "Solution_1": "Nice problem, my friend  :) . Using the lenma: $ \\frac {1}{x^2} \\plus{} \\frac {1}{y^2} \\plus{} \\frac {1}{z^2} \\geq\\ \\frac {27}{(x \\plus{} y \\plus{} z)^2}$\r\nAnd I think the stronger is not trues  :maybe: (but I haven't checked it yet)",
  "Solution_2": "[quote=\"nguoivn\"] Using the lenma: $ \\frac {1}{x^2} \\plus{} \\frac {1}{y^2} \\plus{} \\frac {1}{z^2} \\geq\\ \\frac {27}{(x \\plus{} y \\plus{} z)^2}$[/quote]\n\nI would like to see how you've proceeded then.\n\nMy proof is based on the fact that : $ \\frac {2}{3(a \\plus{} b)^2} \\plus{} \\frac {1}{3(b \\plus{} c)^2} \\geq \\frac {9}{(c \\plus{} 3b \\plus{} 2a)^2}$ ; we sum cyclically and we get the desired result.\n\n[quote=\"nguoivn\"]And I think the stronger is not trues  :maybe: (but I haven't checked it yet)[/quote]\r\nHowever,I can't find a counterexample and It seems true :maybe:",
  "Solution_3": "hello \" rachid\" I think that this lemme can be used like that:\r\n\r\n$ \\frac {1}{x^2} \\plus{} \\frac {1}{y^2} \\plus{} \\frac {1}{y^2}\\geq\\frac {27}{(x \\plus{} 2y)^2}$\r\nand we sum cyclically and we get \r\n$ \\sum\\frac {1}{x^2}\\geq\\sum{\\frac {9}{(x \\plus{} 2y)^2}}$\r\nand let $ a \\plus{} b \\equal{} x$ and $ b \\plus{} c \\equal{} y$ and  $ c \\plus{} a \\equal{} z$\r\nwe get the result\r\nthank you for the lemma",
  "Solution_4": "[quote=\"Evariste-Galois\"]hello \" rachid\" I think that this lemme can be used like that:\n\n$ \\frac {1}{x^2} \\plus{} \\frac {1}{y^2} \\plus{} \\frac {1}{y^2}\\geq\\frac {27}{(x \\plus{} 2y)^2}$\nand we sum cyclically and we get \n$ \\sum\\frac {1}{x^2}\\geq\\sum{\\frac {9}{(x \\plus{} 2y)^2}}$\nand let $ a \\plus{} b \\equal{} x$ and $ b \\plus{} c \\equal{} y$ and  $ c \\plus{} a \\equal{} z$\nwe get the result\nthank you for the lemma[/quote]\nI don't think so my friend,I guess that nguoivn's Idea is different than mine,since what you've done is also my Idea posted before  :wink:  :)\n\nNow let's try to prove/disprove the stronger:\n\n[quote=\"rachid\"]I think that the following inequality is also true:\n\n$ \\frac {1}{(a \\plus{} 3b \\plus{} 2c)^2} \\plus{} \\frac {1}{(b \\plus{} 3c \\plus{} 2a)^2} \\plus{} \\frac {1}{(c \\plus{} 3a \\plus{} 2b)^2} \\leq \\frac {1}{4(ab \\plus{} bc \\plus{} ca)}$\n\nfor $ a,b,c \\geq 0$[/quote]",
  "Solution_5": "hello Morocco.....here a solution with only AM-GM:\r\nlet\r\n$ x=a+b$ and  $ y=b+c$ and $ z=c+a$\r\nyour inequality is equivalent to:\r\n\\[{ \\frac{1}\r\n{9}\\left( {\\sum\\limits_{cyc} {\\frac{1}\r\n{{{x^2}}}} } \\right) \\geqslant \\sum\\limits_{cyc} \\frac{1}\r\n{{{{(x + 2y)}^2}}}}\\]\r\nwe have:\r\n\\[ x + 2y = x + y + y \\geqslant 3\\root 3 \\of {x{y^2}}\\]\r\n\\[ \\sum\\limits_{cyc} {\\frac{1}\r\n{{{{(x + 2y)}^2}}}}  \\leqslant \\frac{1}\r\n{9}\\sum {\\frac{1}\r\n{{\\root 3 \\of {{x^2}{y^4}} }}}\\]\r\nwe must prove that:\r\n\\[ \\sum\\limits_{cyc} {\\frac{1}\r\n{{{{x}^2}}}}\\geq\\ \\sum {\\frac{1}\r\n{{\\root 3 \\of {{x^2}{y^4}} }}}\\]\r\nso we have:\r\n\\[ \\frac{1}\r\n{{2{x^2}}} + \\frac{1}\r\n{{2{y^2}}} + \\frac{1}\r\n{{2{y^2}}} \\geqslant 3\\root 3 \\of {\\frac{1}\r\n{{8{x^2}{y^4}}}}\\]\r\nso\r\n\\[ \\sum {\\left( {\\frac{1}\r\n{{2{x^2}}} + \\frac{1}\r\n{{2{y^2}}} + \\frac{1}\r\n{{2{y^2}}}} \\right)}  = \\frac{3}\r\n{2}\\sum {\\frac{1}\r\n{{{x^2}}}}  \\geqslant \\frac{3}\r\n{2}\\sum {\\frac{1}\r\n{{\\root 3 \\of {{x^2}{y^4}} }}}\\]\r\nconclusion :)  :)",
  "Solution_6": "[quote=\"rachid\"]Let $ a,b,c\\geq 0$.Prove that:\n\n$ \\frac {1}{(a \\plus{} b)^2} \\plus{} \\frac {1}{(b \\plus{} c)^2} \\plus{} \\frac {1}{(c \\plus{} a)^2} \\geq \\frac {9}{(a \\plus{} 3b \\plus{} 2c)^2} \\plus{} \\frac {9}{(b \\plus{} 3c \\plus{} 2a)^2} \\plus{} \\frac {9}{(c \\plus{} 3a \\plus{} 2b)^2}$\n\n[hide]\nI think that the following inequality is also true:\n\n$ \\frac {1}{(a \\plus{} 3b \\plus{} 2c)^2} \\plus{} \\frac {1}{(b \\plus{} 3c \\plus{} 2a)^2} \\plus{} \\frac {1}{(c \\plus{} 3a \\plus{} 2b)^2} \\leq \\frac {1}{4(ab \\plus{} bc \\plus{} ca)}$\n\nfor $ a,b,c \\geq 0$[/hide][/quote]\r\nusing :\r\n$ \\frac {1}{(a \\plus{} b)^2} \\plus{} \\frac {1}{(b \\plus{} c)^2} \\plus{} \\frac {1}{(b \\plus{} c)^2} \\geq \\frac {27}{(a \\plus{} 3b \\plus{} 2c)^2}$",
  "Solution_7": "[quote=\"iqtimo\"][quote=\"rachid\"]Let $ a,b,c\\geq 0$.Prove that:\n\n$ \\frac {1}{(a \\plus{} b)^2} \\plus{} \\frac {1}{(b \\plus{} c)^2} \\plus{} \\frac {1}{(c \\plus{} a)^2} \\geq \\frac {9}{(a \\plus{} 3b \\plus{} 2c)^2} \\plus{} \\frac {9}{(b \\plus{} 3c \\plus{} 2a)^2} \\plus{} \\frac {9}{(c \\plus{} 3a \\plus{} 2b)^2}$\n\n[hide]\nI think that the following inequality is also true:\n\n$ \\frac {1}{(a \\plus{} 3b \\plus{} 2c)^2} \\plus{} \\frac {1}{(b \\plus{} 3c \\plus{} 2a)^2} \\plus{} \\frac {1}{(c \\plus{} 3a \\plus{} 2b)^2} \\leq \\frac {1}{4(ab \\plus{} bc \\plus{} ca)}$\n\nfor $ a,b,c \\geq 0$[/hide][/quote]\nusing :\n$ \\frac {1}{(a \\plus{} b)^2} \\plus{} \\frac {1}{(b \\plus{} c)^2} \\plus{} \\frac {1}{(b \\plus{} c)^2} \\geq \\frac {27}{(a \\plus{} 3b \\plus{} 2c)^2}$[/quote]\r\nYes, that's thing I want to mention  :)",
  "Solution_8": "The second inequality is false . Try $ a\\equal{}b\\equal{}c\\rightarrow 0$ .",
  "Solution_9": "[quote=\"alex2008\"]The second inequality is false . Try $ a \\equal{} b \\equal{} c\\rightarrow 0$ .[/quote]\r\nYou're wrong alex2008  :) .[b]However[/b],remind that no two of the variables are equal to zero.",
  "Solution_10": "But they are not!!",
  "Solution_11": "............!",
  "Solution_12": "[quote=\"FelixD\"]But they are not!![/quote]\nI know,it's just for precision.\n\nJust see that for $ a \\equal{} b \\equal{} c$ we get the equality case,wich contradicts what alex2008 says.\n\nDo you have any counterexample ?\n\n[quote=\"Evariste-Galois\"]Big mistake \"Rachid\"  :wink:  :wink:[/quote]\r\nBe sure of what you write before posting anything  :wink: :D",
  "Solution_13": ":rotfl:  :rotfl:  what do you say !!!!!!",
  "Solution_14": "I sorry for starting this \"quarell\" :( . It is my bad for miscalculating something  :D  . When $ a\\equal{}b\\equal{}c$ we have equality case , rachid was right . And when i said $ \\rightarrow 0$ i didn't reffered one of the variables is zhero , i just reffered it is very close to zhero , but even so the inequality is true . Now , i think that rachid's inequality is true  :)",
  "Solution_15": "[quote=\"rachid\"]\n[quote=\"rachid\"]I think that the following inequality is also true:\n\n$ \\frac {1}{(a \\plus{} 3b \\plus{} 2c)^2} \\plus{} \\frac {1}{(b \\plus{} 3c \\plus{} 2a)^2} \\plus{} \\frac {1}{(c \\plus{} 3a \\plus{} 2b)^2} \\leq \\frac {1}{4(ab \\plus{} bc \\plus{} ca)}$\n\nfor $ a,b,c \\geq 0$[/quote][/quote]\r\nYes, it's trues and very nice  :lol: . My solution is only Am-Gm and it's based on: $ 3(a \\plus{} b \\plus{} c)^2 \\geq\\ 2\\sum\\ a^2 \\plus{} 7\\sum\\ ab$\r\n @rachid: Can you give me your name? Indeed, I really want to have this problem in our book, And we want to write exactly the author's name. Thank you very much  :)",
  "Solution_16": "BTW,I've proved that the following inequality is true :\r\n\r\n$ \\frac {1}{(a \\plus{} 3b \\plus{} 2c)^2} \\plus{} \\frac {1}{(b \\plus{} 3c \\plus{} 2a)^2} \\plus{} \\frac {1}{(c \\plus{} 3a \\plus{} 2b)^2}$    $ \\geq$  $ \\frac {9}{8(a^2\\plus{}b^2\\plus{}c^2) \\plus{} 28(ab\\plus{}bc\\plus{}ca)}$\r\n\r\nfor $ a,b,c \\geq 0$\r\n\r\n[quote=\"nguoivn\"]\n[quote=\"rachid\"]I think that the following inequality is also true:\n\n$ \\frac {1}{(a \\plus{} 3b \\plus{} 2c)^2} \\plus{} \\frac {1}{(b \\plus{} 3c \\plus{} 2a)^2} \\plus{} \\frac {1}{(c \\plus{} 3a \\plus{} 2b)^2} \\leq \\frac {1}{4(ab \\plus{} bc \\plus{} ca)}$\n\nfor $ a,b,c \\geq 0$[/quote]\nYes, it's trues and very nice  :lol: . My solution is only Am-Gm and it's based on: $ 3(a \\plus{} b \\plus{} c)^2 \\geq\\ 2\\sum\\ a^2 \\plus{} 7\\sum\\ ab$[/quote]\r\nI would like to see your AM-GM's proof  :) \r\n\r\n@nguoivn : see your PM :)",
  "Solution_17": "Yes, I sent it to you. How do you feel with my proof?  :)",
  "Solution_18": "Very nice and elementary proof,nguoivn  :lol: .Thank you !\r\n\r\nI'll respect your want to not reveal the proof for now.",
  "Solution_19": "Now, I can post it ^^!\r\n\r\nThe first, consider the case $ a \\le b \\le c$, using Am-Gm Inequalily with the form $ \\dfrac{1}{(x \\plus{} y)^2} \\le \\dfrac{1}{4xy},$ we have $ \\frac {1}{{{{(3a \\plus{} 2b \\plus{} c)}^2}}} \\le \\frac {1}{{4(a \\plus{} 2b)(c \\plus{} 2a)}}.$\r\nSo, we only need to prove that $ \\frac {1}{{(a \\plus{} 2b)(c \\plus{} 2a)}} \\plus{} \\frac {1}{{(b \\plus{} 2c)(a \\plus{} 2b)}} \\plus{} \\frac {1}{{(c \\plus{} 2a)(b \\plus{} 2c)}} \\le \\frac {1}{{ab \\plus{} bc \\plus{} ca}},$\r\n\r\nOr $ (a \\plus{} 2b)(b \\plus{} 2c)(c \\plus{} 2a) \\ge 3(a \\plus{} b \\plus{} c)(ab \\plus{} bc \\plus{} ca).$\r\nEquivalent to  $ (a \\minus{} b)(b \\minus{} c)(c \\minus{} a) \\ge 0$ (obviously trues with $ a \\le b \\le c$).\r\nConsider the last case is $ a \\ge b \\ge c$.\r\nWe also use Am-Gm but with another hint like the following $ \\dfrac{1}{(3a \\plus{} 2b \\plus{} c)^2} \\le \\dfrac{1}{4(a \\plus{} b \\plus{} c)(2a \\plus{} b)}$. \r\nWith this estimation, easily too see that the proof'll complete if we can show that $ \\frac {1}{{a \\plus{} b \\plus{} c}}\\left( {\\frac {1}{{2a \\plus{} b}} \\plus{} \\frac {1}{{2b \\plus{} c}} \\plus{} \\frac {1}{{2c \\plus{} a}}} \\right) \\le \\frac {1}{{ab \\plus{} bc \\plus{} ca}},$\r\n\r\nEquivalent to $ (2a \\plus{} b)(2b \\plus{} c)(2c \\plus{} a)(a \\plus{} b \\plus{} c) \\ge (ab \\plus{} bc \\plus{} ca)[2({a^2} \\plus{} {b^2} \\plus{} {c^2}) \\plus{} 7(ab \\plus{} bc \\plus{} ca)].$\r\n\r\nBecause $ a \\ge b \\ge c$, so $ (2a \\plus{} b)(2b \\plus{} c)(2c \\plus{} a) \\ge 3(a \\plus{} b \\plus{} c)(ab \\plus{} bc \\plus{} ca),$ \r\nAnd we bring the problem to prove the stronger $ 3{(a \\plus{} b \\plus{} c)^2} \\ge 2({a^2} \\plus{} {b^2} \\plus{} {c^2}) \\plus{} 7(ab \\plus{} bc \\plus{} ca).$ \r\nBut this last ineq is obviously trues by Am-Gm again. We have done!\r\nEquality occurs if and only $ a \\equal{} b \\equal{} c.$ \r\nPS: It's also a nice problem in our first book \"Bat dang thuc va nhung loi giai hay\" (V.Q.B.Can - T.Q.Anh, Ha Noi Publishing House 2009)  :) \r\nThanks again Rachid Benchlikha - the author of this very nice problem  :)"
}
{
  "Tag": [
    "function",
    "limit",
    "integration",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Let $n\\in\\mathbb{N}, \\ n\\geq 2$  and the function $f_{n}(x)=\\sum_{k=0}^{n}{|x-k|}$ and $F_{n}: \\mathbb{R}\\to{R}$ a primitive of $f_{n}$. Compute: \r\n\r\n(a) $\\lim_{n\\to\\infty}\\frac{F_{n}(1)-F_{n}(0)}{n^{2}}$\r\n\r\n(b) $\\lim_{x\\to\\infty}\\frac{F_{n}(x)-F_{n}(0)}{x^{2}}$\r\n\r\nSince $f_{n}$ is continous using Leiniz-Newton formula we have (a) $\\lim_{n\\to\\infty}\\frac{F_{n}(1)-F_{n}(0)}{n^{2}}=\\lim_{n\\to\\infty}\\int_{0}^{1}f(x)dx$ right? and for (b) same thing... Well i don't know if this helps  :blush:",
  "Solution_1": "[quote=\"Slizzel\"]Let $n\\in\\mathbb{N}, \\ n\\geq 2$  and the function $f_{n}(x)=\\sum_{k=0}^{n}{|x-k|}$ and $F_{n}: \\mathbb{R}\\to{R}$ a primitive of $f_{n}$. Compute: \n\n(a) $\\lim_{n\\to\\infty}\\frac{F_{n}(1)-F_{n}(0)}{n^{2}}$\n\n(b) $\\lim_{x\\to\\infty}\\frac{F_{n}(x)-F_{n}(0)}{x^{2}}$\n\nSince $f_{n}$ is continous using Leiniz-Newton formula we have (a) $\\lim_{n\\to\\infty}\\frac{F_{n}(1)-F_{n}(0)}{n^{2}}=\\lim_{n\\to\\infty}\\int_{0}^{1}f(x)dx$ right? and for (b) same thing... Well i don't know if this helps  :blush:[/quote]\r\nFor a):\r\nUse Stolz's Theorem twice:\r\n$\\lim_{n\\to\\infty}\\frac{F_{n}(1)-F_{n}(0)}{n^{2}}=\\lim_{n\\to\\infty}\\frac{\\int_{0}^{1}f_{n}(x)dx}{n^{2}}=\\lim_{n\\to\\infty}\\frac{\\int_{0}^{1}f_{n}(x)-f_{n-1}(x)dx}{2n-1}$\r\n$=\\lim_{n\\to\\infty}\\frac{\\int_{0}^{1}|x-n|dx}{2n-1}$\r\n$=\\lim_{n\\to\\infty}\\frac{\\int_{0}^{1}|x-n|-|x-n+1|dx}{2}=\\frac{1}{2}$\r\n\r\nFor b),Use L'hosptial's Rule:\r\n$\\lim_{x\\to\\infty}\\frac{F_{n}(x)-F_{n}(0)}{x^{2}}=\\lim_{x\\to\\infty}\\frac{\\int_{0}^{x}f_{n}(t)dt}{x^{2}}=\\lim_{x\\to\\infty}\\frac{f_{n}(x)}{2x}$\r\n$=\\lim_{x\\to \\infty}\\frac{\\sum_{k=0}^{n}{|x-k|}}{2x}=\\frac{n+1}{2}$\r\nFor why we can use the two theorems is obvious,I think.",
  "Solution_2": "Very good! Thanks!  :yup:"
}
{
  "Tag": [
    "geometry",
    "conics",
    "parabola",
    "calculus",
    "integration",
    "algebra proposed",
    "algebra"
  ],
  "Problem": "There is a certain parabola, which opens up, that has no unit lattice points lying on it, and it is increasing\r\non 1 < x < 5.  Seven unit lattice points, grand total, lie (below it and above the x-axis) for 1 < x < 5.  These seven internal points lie on the lines x = 2, x = 3, and x = 4.  Three of the seven points are vertically below the parabola a distance of 0.1 unit apiece.  \r\n\r\nFind the area, to the nearest half of a square unit, for the region bounded by the curve/lines:  the certain parabola, the x-axis, x = 1, and x = 5.",
  "Solution_1": "Here's a solution without integral calculus.\r\n\r\nUnder the conditions, the seven points would have to be these:\r\n\r\n(2,1), (3,1), (3,2), (4,1), (4,2), (4,3), (4,4)\r\n\r\nThen the three points that can be 0.1 unit below (apiece) from the parabola are (2,1), (3,2), and (4,4), making (2,1.1), (3,2.1),\r\nand (4,4.1) known points on the parabola.   A QuadReg selection on a TI series graphics model calculator (for instance)\r\napplied to these gives an equation of y = 0.5x^2 - 1.5x + 2.1\r\n\r\nFor the endpoints, when x = 1, y = 1.1, and when x = 5, y = 7.1\r\n\r\nI will use Pick's Theorem:  (1/2)b + i - 1, where b is the # of\r\nboundary points and i is the # of interior points of a polygon, where\r\nit is on a grid of unit lattice points (in this case).\r\n\r\nPick's Theorem is meant for use on polygons, so I am using the section of the parabola where it is relatively close to unit lattice points, together with x = 1, x = 5, and y = 0 for the polygon.  \r\n\r\nAll the 16 boundary points are:  (1,0), (2,0), (3,0), (4,0), (5,0), (1,1), (2,1), (3,2), (4,4), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (5,7)\r\n\r\nAll of the 4 interior points are:  (3,1), (4,1), (4,2), (4,3)\r\n\r\n(1/2)16 + 4 - 1 = 11\r\n\r\n11 square units........Answer\r\n\r\n\r\nIf I were to check by integrating  (0.5x^2 - 1.5x + 2.1)dx from\r\nx = 1 to x = 5, I would get\r\n\r\napproximately  11.067"
}
{
  "Tag": [
    "function",
    "limit"
  ],
  "Problem": "The \"Inverse\" function is defined recursively as follows:\r\nThe nth number is the inverse of the sum of the previous n-1 terms.\r\n\r\nDoes the sum of the elements in this infinite sequence converge?  If so, what to?",
  "Solution_1": "Please explicate what you mean by \"the inverse of the previous n-1 terms\"",
  "Solution_2": "Sorry.  I meant inverse of the sum.  I changed it.",
  "Solution_3": "[hide=\"Solution\"]We will prove that the series diverges. \nLet the terms of the sequence be $a_{1},a_{2},\\ldots$ and $s_{n}: =a_{1}+\\ldots+a_{n}$. We have $a_{n}= \\frac{1}{s_{n-1}}$, so $s_{n}= s_{n-1}+a_{n}= s_{n-1}+\\frac{1}{s_{n-1}}$. Let us assume that $\\left\\{s_{n}\\right\\}_{n=1}^{\\infty}$ converges and $s: =\\lim_{n\\to\\infty}s_{n}$. Then $s=s+\\frac{1}{s}$, so $\\frac{1}{s}=0$ - a contradiction. Thus the sequence $\\left\\{s_{n}\\right\\}_{n=1}^{\\infty}$ diverges and the series $\\sum_{n=1}^{\\infty}a_{n}$ also diverges.[/hide]",
  "Solution_4": "The Inverse-Inverse function is defined as follows: the nth term is the inverse of the sum of the inverses of the previous terms in the sequence.  Does this sum converge?  If so, what to?\r\n\r\n[hide=\"Solution\"]\nConsider the \"Inverse-Inverse-Inverse\" function or the sequence whose terms ore the reciprocals of the corresponding term in this sequence.  The nth term of that sequence is the sum of the previous n-1 terms.  Therefore, the sum of the first n elements is half the sum of the first n+1 elements.  If the sum doubles each term, each term after the 2nd one must be twice the previous one.  This means the sequence goes 1, 1, 2, 4, 8, 16....  The inverse of this is $1, 1, \\frac{1}{2}, \\frac{1}{4}...$ which as everybody knows, converges to 3.\n[/hide]"
}
{
  "Tag": [
    "trigonometry"
  ],
  "Problem": "sin2x = tgx - cos2x\r\n\r\nWhat's the value of six-cosx?\r\nPlease help me",
  "Solution_1": "$ \\sin 2x \\equal{} \\mathrm{tg} x \\minus{} \\cos 2x\\Leftrightarrow$ $ \\sin x \\minus{} \\cos x \\plus{} 2\\sin^2 x\\cos x \\minus{} 2\\sin x\\cos^2 x \\equal{} 0\\Leftrightarrow$ $ (\\sin x \\minus{} \\cos x)(1 \\plus{} 2\\sin x \\cos x)\\equal{}0\\Leftrightarrow \\sin x \\minus{} \\cos x \\equal{} 0\\vee 1 \\plus{} 2\\sin x \\cos x \\equal{} 0$\r\n\r\n$ 1 \\plus{} 2\\sin x \\cos x \\equal{} 0\\Leftrightarrow 2\\sin x\\cos x \\equal{} \\minus{} 1\\Leftrightarrow$ $ \\cos^2 x \\minus{} 2\\sin x \\cos x \\plus{} \\sin^2 x \\equal{} 2\\Leftrightarrow |\\sin x \\minus{} \\cos x| \\equal{} \\sqrt {2}\\Leftrightarrow$ $ \\sin x \\minus{} \\cos x\\in \\{ \\minus{} \\sqrt {2}\\ ,\\ 0\\ ,\\ \\sqrt {2}\\}$"
}
{
  "Tag": [
    "inequalities",
    "function",
    "inequalities unsolved"
  ],
  "Problem": "If $ a_1,a_2,\\cdots,a_n$ are real positive numbers , prove that:\r\n\r\n\\[ \\sqrt{\\frac{a^2_1\\plus{}a^2_2\\plus{}\\cdots\\plus{}a^2_n}{n}} \\ge \\frac{a_1\\plus{}a_2\\plus{}\\cdots\\plus{}a_n}{n}\\]",
  "Solution_1": "[quote=\"Obel1x\"]If $ a_1,a_2,\\cdots,a_n$ are real positive numbers , prove that:\n\\[ \\sqrt {\\frac {a^2_1 \\plus{} a^2_2 \\plus{} \\cdots \\plus{} a^2_n}{n}} \\ge \\frac {a_1 \\plus{} a_2 \\plus{} \\cdots \\plus{} a_n}{n}\\]\n[/quote]\r\n\r\nthis ineq holds for any [b]reals[/b] $ a_i$.\r\n\r\nby CS, we have \r\n\r\n$ n(a^2_1 \\plus{} a^2_2 \\plus{} \\cdots \\plus{} a^2_n)$\r\n\r\n$ \\equal{}(1\\plus{}1\\plus{}...\\plus{}1)(a^2_1 \\plus{} a^2_2 \\plus{} \\cdots \\plus{} a^2_n)$\r\n\r\n$ \\ge (a_1 \\plus{} a_2 \\plus{} \\cdots \\plus{} a_n)^2$\r\n\r\nso we get \r\n\r\n$ \\sqrt {\\frac {a^2_1 \\plus{} a^2_2 \\plus{} \\cdots \\plus{} a^2_n}{n}}$\r\n\r\n$ \\ge \\frac {\\left| a_1 \\plus{} a_2 \\plus{} \\cdots \\plus{} a_n \\right|}{n}$\r\n\r\n$ \\ge \\frac {a_1 \\plus{} a_2 \\plus{} \\cdots \\plus{} a_n}{n}.$",
  "Solution_2": "Function $ f(k)=\\left(\\frac{a_1 ^k +...+a_n ^k}{n}^\\right)^{\\frac{1}{k}}$ is increasing since $ f'(k)\\geq 0$ ,then the conclusion follows ."
}
{
  "Tag": [
    "geometry",
    "rectangle"
  ],
  "Problem": "SMO 2004 Open #24\r\n\r\nFind the number of ways to tile a $5 \\times 2$ grid with blue $1 \\times 1$ tiles, red $2 \\times 1$ tiles and green $2 \\times 2$ tiles.\r\n\r\nHint: Use recursion.",
  "Solution_1": "[hide=\"Probably Wrong\"]\nCall the number of ways to tile an $n\\times 2$ rectangle $T_{n}$.  There are six ways to cover the $n$th column: two blue tiles, a red tile, half of two red tiles, half a green tile, and half a red tile and a blue tile in two different configurations.  \n\nThe first two can be obtained from an $n-1\\times 2$ rectangle tiling, and the last four from an $n-2\\times 2$ tiling.  Thus $T_{n}=2T_{n-1}+4T_{n-2}$.  \n\nWe can easily see that $T_{1}=2$ and $T_{2}=8$.  Thus $T_{3}=24$, $T_{4}=80$, and $T_{5}=256$.\n[/hide]",
  "Solution_2": "It's greater because for the last two configurations, there's another space which could be filled by a blue tile (which it looks like you assume), or a red tile.\r\n[hide=\"if I did not mess up...\"]\nLet $a_{n}$ be the number of ways to tile an $n\\times 2$ grid, and $b_{n}$ be the number of ways to tile an $n\\times 2$ grid with a corner removed.\n\nFor a recursion for $b_{n}$, note that green tiles can't fill in the corner adjacent to the removed one. If it is filled blue, then $a_{n-1}$ ways to fill in the rest, and if filled red, $b_{n-1}$ ways to fill in the rest. So we get\n\\[b_{n}=a_{n-1}+b_{n-1}\\]\nNow for a recursion for $a_{n}$, if we align the grid such that there are 2 rows, consider what space fills in the upper right corner. If blue, there are $b_{n}$ ways. If green, there are $a_{n-2}$ ways. If red, either the red tile is horizontal or vertical. If vertical, there are $a_{n-1}$ ways. If horizontal, the lower right corner can be blue for $b_{n-1}$ ways, or red for $a_{n-2}$ ways. This means\n\\[a_{n}=b_{n}+a_{n-1}+b_{n-1}+2a_{n-2}=2a_{n-1}+2a_{n-2}+2b_{n-1}\\]\nThen $b_{n-1}=\\frac{a_{n}}{2}-a_{n-1}-a_{n-2}$, so $b_{n-2}=\\frac{a_{n-1}}{2}-a_{n-2}-a_{n-3}$. Subtracting,\n\\[a_{n-2}=\\frac{a_{n}}{2}-\\frac{3a_{n-1}}{2}+a_{n-3}\\]\n\\[\\Rightarrow a_{n}=3a_{n-1}+2a_{n-2}-2a_{n-3}\\]\nWe find $a_{1}=2$ and $a_{2}=8$. Also, $b_{1}=1$, so for purposes of applying the recursive formula, $a_{0}=1$. Now we just compute:\n\\[a_{3}=26,\\ a_{4}=90,\\ a_{5}=\\boxed{306}\\]\n[/hide]",
  "Solution_3": "I think Scorpius119 has it  :)  Very good consideration about the corner being removed there."
}
{
  "Tag": [
    "induction",
    "number theory proposed",
    "number theory"
  ],
  "Problem": "Let $A$ be a set of positive integers such that\r\na) if $a \\in A$, the all the positive divisors of $a$are also in $A$;\r\nb) if $a,b \\in A$, with $1<a<b$, then $1+ab \\in A$. \r\nProve that if $A$ has at least 3 elements, then $A$ is the set of all positive integers.",
  "Solution_1": "hi Moroccan citizen how re u?\r\nu re from ??\r\ni m from rabat! i ve passed the IMOs USA2001 and UK2002\r\nso about your problem:\r\nBY INDUCTION :roll: \r\nif we assume that,giving a n>=5, for all k between 1 and n : k is in A!!!!\r\n{\r\nif n is  odd\r\nlet's show that n+2 is in A\r\nn(1+k)+1=k(n+2)\r\nwith k=(n+1)/2#n (coz n>3)\r\n(k+1) in A\r\nand n in A\r\nso k(n+2) in A and finally n+2 in A\r\nn(n+2)+1=(n+1)\u00b2 in A so n+1 in A\r\n}\r\nelse n is even=2k\r\nand n+1=2k+1 in A if k>2 (n>=5)\r\nNow!!\r\nfor the start values:\r\n1 in A (divisor of an element in A)\r\n2 in A \r\n3 in A\r\nso 7=(2*3)+1 in A\r\n2*7+1=15 in A\r\nand 5 in A\r\n(3*5)+1=16 so 4 in A\r\nwe may start with {1,2,3,4,5}",
  "Solution_2": "It is the same problem with Romanian Junior Balkan TSTs 2004, Test 5, problem 3, posted here: http://www.mathlinks.ro/Forum/viewtopic.php?t=5892\r\nContinue discussion there."
}
{
  "Tag": [
    "integration",
    "logarithms",
    "Putnam",
    "limit",
    "trigonometry",
    "real analysis",
    "LaTeX"
  ],
  "Problem": "[color=blue][i]\u03a3\u03c4\u03bf \u03c7\u03ce\u03c1\u03bf \u03b1\u03c5\u03c4\u03cc \u03bf\u03b9 \u03c6\u03bf\u03b9\u03c4\u03b7\u03c4\u03ad\u03c2 \u03bc\u03b1\u03c2 \u03ad\u03c7\u03bf\u03c5\u03bd \u03c4\u03b7 \u03b4\u03c5\u03bd\u03b1\u03c4\u03cc\u03c4\u03b7\u03c4\u03b1 \u03bd\u03b1 \u03b1\u03bd\u03c4\u03b1\u03bb\u03ac\u03c3\u03c3\u03bf\u03c5\u03bd \u03b9\u03b4\u03ad\u03b5\u03c2 , \u03c0\u03c1\u03bf\u03b2\u03bb\u03ae\u03bc\u03b1\u03c4\u03b1  \u03ae \u03ba\u03b1\u03b9 \u03bd\u03b1 \u03c3\u03c4\u03ad\u03bb\u03bd\u03bf\u03c5\u03bd  \n\n\u03c4\u03b9\u03c2 \u03b1\u03c0\u03bf\u03c1\u03af\u03b5\u03c2 \u03ae \u03c4\u03b9\u03c2 \u03bb\u03cd\u03c3\u03b5\u03b9\u03c2 \u03c4\u03bf\u03c5\u03c2 \u03c3\u03b5 \u03c0\u03c1\u03bf\u03c4\u03b5\u03b9\u03bd\u03cc\u03bc\u03b5\u03bd\u03b5\u03c2 \u03b1\u03c3\u03ba\u03ae\u03c3\u03b5\u03b9\u03c2 \u03ac\u03bb\u03bb\u03c9\u03bd \u03c6\u03bf\u03b9\u03c4\u03b7\u03c4\u03ce\u03bd.\n\n \u039c\u03b5 \u03b1\u03c5\u03c4\u03cc \u03c4\u03bf\u03bd \u03c4\u03c1\u03cc\u03c0\u03bf , \u03cc\u03bb\u03b1 \u03c4\u03b1 \u03b8\u03ad\u03bc\u03b1\u03c4\u03b1 \u03b8\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c5\u03b3\u03ba\u03b5\u03bd\u03c4\u03c1\u03c9\u03bc\u03ad\u03bd\u03b1 \u03ba\u03b1\u03b9 \u03b8\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b5\u03cd\u03ba\u03bf\u03bb\u03b7 \u03b7 \u03c0\u03c1\u03cc\u03c3\u03b2\u03b1\u03c3\u03b7, \u03c7\u03c9\u03c1\u03af\u03c2 \u03c7\u03c1\u03bf\u03bd\u03bf\u03c4\u03c1\u03b9\u03b2\u03ae.[/i][/color]\r\n\r\n[b] \u039a\u03b1\u03bb\u03ad\u03c2 \u03b5\u03bc\u03c0\u03bd\u03b5\u03cd\u03c3\u03b5\u03b9\u03c2 !!![/b]",
  "Solution_1": "[color=red][b]\u0391\u03a3\u039a\u0397\u03a3\u0397 1[/b][/color]\r\n\r\n\u039d\u03b1 \u03c5\u03c0\u03bf\u03bb\u03bf\u03b3\u03b9\u03c3\u03c4\u03b5\u03af \u03c4\u03bf \u03b3\u03b5\u03bd\u03b9\u03ba\u03b5\u03c5\u03bc\u03ad\u03bd\u03bf $ \\displaystyle\\int^1_0 \\frac {lnx}{1 \\plus{} x}\\,dx$\r\nA) \u03bc\u03b5 \u03bc\u03b9\u03b3\u03b1\u03b4\u03b9\u03ba\u03ae \u03b1\u03bd\u03ac\u03bb\u03c5\u03c3\u03b7 \u03b2) \u03bc\u03b5 \u03c3\u03b5\u03b9\u03c1\u03ad\u03c2 Fourier \u03b3) \u03bc\u03b5 \u03c4\u03af\u03c0\u03bf\u03c4\u03b5 \u03b1\u03c0\u03cc \u03c4\u03b1 \u03b4\u03c5\u03cc \u03c0\u03c1\u03bf\u03b7\u03b3\u03bf\u03cd\u03bc\u03b5\u03bd\u03b1 . \u0391\u03bd \u03c5\u03c0\u03ac\u03c1\u03be\u03b5\u03b9 \u03b5\u03bd\u03b4\u03b9\u03b1\u03c6\u03ad\u03c1\u03bf\u03bd  \u03b8\u03b1 \u03b4\u03ce\u03c3\u03c9 \u03bb\u03cd\u03c3\u03b7 \u03b3\u03b9\u03c3 \u03c4\u03b7\u03bd \u03b3) \u03c0\u03b5\u03c1\u03af\u03c0\u03c4\u03c9\u03c3\u03b7 \u03c3\u03b5 \u03bb\u03af\u03b3\u03b5\u03c2 \u03bc\u03ad\u03c1\u03b5\u03c2",
  "Solution_2": "r_boris, \u03bc\u03c0\u03bf\u03c1\u03b5\u03b9\u03c2 \u03bd\u03b1 \u03b3\u03c1\u03b1\u03c8\u03b5\u03b9\u03c2 \u03c4\u03b7\u03bd \u03bb\u03c5\u03c3\u03b7 \u03c4\u03c9\u03c1\u03b1?  :roll:",
  "Solution_3": "M\u03b7\u03b7\u03b7\u03b7, \u03cc\u03c7\u03b9 \u03bb\u03cd\u03c3\u03b7, \u03b1\u03ba\u03cc\u03bc\u03b1 \u03c4\u03b7\u03bd \u03c0\u03b1\u03bb\u03b5\u03cd\u03bf\u03c5\u03bc\u03b5 \u03b5\u03b4\u03ce \u03c0\u03ad\u03c1\u03b1!!!\r\n\r\n\u039a\u03b1\u03bb\u03ac, \u03c0\u03c1\u03bf\u03c6\u03b1\u03bd\u03ce\u03c2 \u03b1\u03c3\u03c4\u03b5\u03b9\u03b5\u03cd\u03bf\u03bc\u03b1\u03b9, \u03ad\u03c4\u03c3\u03b9? :) :) :) \u03a0\u03ac\u03bd\u03c4\u03c9\u03c2, \u03b1\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03bd\u03b1 \u03bc\u03c0\u03b5\u03b9 \u03b7 \u03bb\u03cd\u03c3\u03b7 \u03b1\u03c0\u03cc \u03c4\u03ce\u03c1\u03b1, \u03b3\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03bd\u03b1 \u03bc\u03c0\u03b5\u03b9 \u03c3\u03b5 spoiler \u03ae \u03c3\u03b5 attachment \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03c6\u03b1\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03bc\u03b5 \u03c4\u03bf \u03c0\u03bf\u03c5 \u03bc\u03c0\u03b1\u03af\u03bd\u03b5\u03b9 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03c2 \u03c3\u03c4\u03bf topic?\r\n\r\n\u03a0\u03c1\u03bf\u03c6\u03b1\u03bd\u03ce\u03c2 \u03ba\u03ac\u03bd\u03c9 \u03ba\u03ac\u03c4\u03b9 \u03bb\u03ac\u03b8\u03bf\u03c2 \u03b3\u03b9\u03b1\u03c4\u03af \u03bc\u03b5 \u03ad\u03bd\u03b1\u03bd \u03c4\u03c1\u03cc\u03c0\u03bf \u03b2\u03b3\u03ac\u03b6\u03c9 $ \\minus{}\\frac{\\pi^2}{2}$ \u03ba\u03b1\u03b9 \u03bc\u03b5 \u03ac\u03bb\u03bb\u03bf\u03bd \u03c4\u03c1\u03cc\u03c0\u03bf \u03b2\u03b3\u03ac\u03b6\u03c9 $ \\minus{}\\frac{\\pi^2}{12}$ (\u03c0\u03bf\u03c5 \u03c4\u03bf \u03b5\u03bc\u03c0\u03b9\u03c3\u03c4\u03b5\u03cd\u03bf\u03bc\u03b1\u03b9 \u03c0\u03b9\u03bf \u03c0\u03bf\u03bb\u03cd \u03b3\u03b9\u03b1\u03c4\u03af \u03ad\u03c7\u03b5\u03b9 \u03bb\u03b9\u03b3\u03cc\u03c4\u03b5\u03c1\u03b5\u03c2 \u03c0\u03c1\u03ac\u03be\u03b5\u03b9\u03c2).\r\n\r\nCheerio,\r\n\r\nDurandal 1707",
  "Solution_4": "[quote=\"Durandal\"] \u03ba\u03b1\u03b9 \u03bc\u03b5 \u03ac\u03bb\u03bb\u03bf\u03bd \u03c4\u03c1\u03cc\u03c0\u03bf \u03b2\u03b3\u03ac\u03b6\u03c9 $ \\minus{} \\frac {\\pi^2}{12}$ (\u03c0\u03bf\u03c5 \u03c4\u03bf \u03b5\u03bc\u03c0\u03b9\u03c3\u03c4\u03b5\u03cd\u03bf\u03bc\u03b1\u03b9 \u03c0\u03b9\u03bf \u03c0\u03bf\u03bb\u03cd \u03b3\u03b9\u03b1\u03c4\u03af \u03ad\u03c7\u03b5\u03b9 \u03bb\u03b9\u03b3\u03cc\u03c4\u03b5\u03c1\u03b5\u03c2 \u03c0\u03c1\u03ac\u03be\u03b5\u03b9\u03c2).\n[/quote]\r\n\r\n\u039a\u03b1\u03bb\u03ac \u03ba\u03ac\u03bd\u03b5\u03b9\u03c2 \u03ba\u03b1\u03b9 \u03c4\u03bf \u03b5\u03bc\u03c0\u03b9\u03c3\u03c4\u03b5\u03cd\u03b5\u03c3\u03b1\u03b9  :wink:",
  "Solution_5": "\u03a3\u03c9\u03c3\u03c4\u03cc \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf $ \\minus{}\\frac{\\pi^2}{12}$ \u03b1\u03bb\u03bb\u03ac \u03c0\u03c9\u03c2? \u03b3\u03b9\u03b1\u03c4\u03af \u03c0\u03b9\u03c3\u03c4\u03b5\u03cd\u03c9 \u03c0\u03c9\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b8\u03b1\u03c5\u03bc\u03ac\u03c3\u03b9\u03b1 \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7",
  "Solution_6": "r_boris, \u03bc\u03c0\u03bf\u03c1\u03b5\u03b9\u03c2 \u03bd\u03b1 \u03b3\u03c1\u03b1\u03c8\u03b5\u03b9\u03c2 \u03c4\u03b7\u03bd \u03bb\u03c5\u03c3\u03b7 \u03c3\u03bf\u03c5 \u03ba\u03b1\u03b9 \u03bd\u03b1 \u03c4\u03b7\u03bd \u03b2\u03b1\u03bb\u03b5\u03b9\u03c2 \u03c3\u03b5 spoiler \u03bf\u03c0\u03c9\u03c2 \u03b5\u03b9\u03c0\u03b5 \u03ba\u03b1\u03b9 \u03bf Durandal. \u0394\u03b7\u03bb\u03b1\u03b4\u03b7 \u03b1\u03bd\u03b1\u03bc\u03b5\u03c3\u03b1 \u03c3\u03b5 tags  [hide][/hide]",
  "Solution_7": "Heh, Silouan thanks \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03c8\u03ae\u03c6\u03bf \u03b5\u03bc\u03c0\u03b9\u03c3\u03c4\u03bf\u03c3\u03cd\u03bd\u03b7\u03c2 :) \u0395\u03c3\u03cd \u03c0\u03ce\u03c2 \u03c4\u03bf \u03ad\u03bb\u03c5\u03c3\u03b5\u03c2 \u03ba\u03b1\u03b9 \u03ae\u03c3\u03bf\u03c5\u03bd \u03c3\u03af\u03b3\u03bf\u03c5\u03c1\u03bf\u03c2 \u03b3\u03b9\u03b1 \u03c4\u03bf \u03b1\u03c0\u03bf\u03c4\u03ad\u03bb\u03b5\u03c3\u03bc\u03b1?\r\n\r\n\r\n\u039f\u03b9 \u03b4\u03cd\u03bf \u03b4\u03b9\u03ba\u03bf\u03af \u03bc\u03bf\u03c5 \u03c4\u03c1\u03cc\u03c0\u03bf\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf\u03b9 \u03b5\u03be\u03ae\u03c2:\r\n\r\n\r\ni. Brute force \u03bc\u03b9\u03b3\u03b1\u03b4\u03b9\u03ba\u03ae \u03b1\u03bd\u03ac\u03bb\u03c5\u03c3\u03b7 (r_boris, \u03bf \u03c4\u03c1\u03cc\u03c0\u03bf\u03c2 \u03bc\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bc\u03ac\u03bb\u03bb\u03bf\u03bd \u03ac\u03ba\u03bf\u03bc\u03c8\u03bf\u03c2 \u03bf\u03c0\u03cc\u03c4\u03b5, \u03b1\u03bd \u03bc\u03c0\u03bf\u03c1\u03b5\u03af\u03c2, \u03b3\u03c1\u03ac\u03c6\u03b5\u03b9\u03c2 \u03ba\u03b1\u03b9 \u03c4\u03b7 \u03b4\u03b9\u03ba\u03ae \u03c3\u03bf\u03c5 \u03bc\u03b9\u03b3\u03b1\u03b4\u03b9\u03ba\u03ae \u03bb\u03cd\u03c3\u03b7 \u03c3\u03c4\u03bf \u03c4\u03ad\u03bb\u03bf\u03c2?)\r\n[hide]$ \\int_0^1 \\frac {\\log x}{x \\plus{} 1} \\textrm{d}x \\equal{} \\minus{} \\int_0^\\infty \\frac {s}{1 \\plus{} e^s} \\textrm{d}s \\equal{} \\minus{} \\int_{ \\minus{} \\infty}^\\infty \\frac {e^u}{1 \\plus{} e^{e^u}}\\textrm{d}u$\n\u03a3\u03c4\u03bf \u03c4\u03b5\u03bb\u03b5\u03c5\u03c4\u03b1\u03af\u03bf \u03bf\u03bb\u03bf\u03ba\u03bb\u03ae\u03c1\u03c9\u03bc\u03b1 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03b1\u03c0\u03bb\u03cc \u03c0\u03cc\u03bb\u03bf \u03c3\u03c4\u03bf $ \\log\\pi \\plus{} i\\frac {\\pi}{2}$, \u03c5\u03c0\u03bf\u03bb\u03bf\u03b3\u03af\u03b6\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf \u03c5\u03c0\u03cc\u03bb\u03bf\u03b9\u03c0\u03bf, \u03c0\u03b1\u03af\u03c1\u03bd\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b7\u03bd \u03ba\u03bb\u03b1\u03c3\u03b9\u03ba\u03ae \u03b7\u03bc\u03b9\u03ba\u03c5\u03ba\u03bb\u03b9\u03ba\u03ae \u03b4\u03b9\u03b1\u03b4\u03c1\u03bf\u03bc\u03ae, \u03b1\u03b3\u03ba\u03bf\u03bc\u03b1\u03c7\u03ac\u03bc\u03b5 \u03bb\u03af\u03b3\u03bf \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03b2\u03b3\u03ac\u03bb\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b1 \u03c6\u03c1\u03ac\u03b3\u03bc\u03b1\u03c4\u03b1 \u03c3\u03c4\u03bf \u03b7\u03bc\u03b9\u03ba\u03cd\u03ba\u03bb\u03b9\u03bf, \u03ba\u03ac\u03bd\u03bf\u03c5\u03bc\u03b5 \u03c0\u03bf\u03bb\u03bb\u03ac\u03b1\u03b1\u03b1\u03b1\u03b1\u03b1\u03b1 \u03bb\u03ac\u03b8\u03b7 \u03c3\u03c4\u03b9\u03c2 \u03c0\u03c1\u03ac\u03be\u03b5\u03b9\u03c2, \u03ba\u03b1\u03b9 \u03c4\u03b5\u03bb\u03b9\u03ba\u03ac \u03b5\u03bb\u03c0\u03af\u03b6\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03b2\u03b3\u03ac\u03bb\u03b5\u03b9 \u03c3\u03c9\u03c3\u03c4\u03ac $ \\minus{} \\frac {\\pi^2}{12}$.\n\n\u0391\u03bd \u03c4\u03ce\u03c1\u03b1 \u03ba\u03ac\u03c0\u03bf\u03c5 \u03ad\u03c7\u03c9 \u03ba\u03ac\u03bd\u03b5\u03b9 \u03bb\u03ac\u03b8\u03bf\u03c2 \u03c3\u03c4\u03b9\u03c2 \u03c0\u03c1\u03ac\u03be\u03b5\u03b9\u03c2 \u03ba\u03b1\u03b9 \u03b1\u03c0\u03bb\u03ce\u03c2 \u03bf\u03b4\u03b7\u03b3\u03bf\u03cd\u03c3\u03b1 \u03c4\u03bf \u03c7\u03ad\u03c1\u03b9 \u03bc\u03bf\u03c5 \u03b5\u03ba\u03b5\u03af \u03c5\u03c0\u03bf\u03c3\u03c5\u03bd\u03b5\u03af\u03b4\u03b7\u03c4\u03b1, apologies... \u03bf\u03b9 \u03c0\u03c1\u03ac\u03be\u03b5\u03b9\u03c2 \u03ba\u03b9 \u03b5\u03b3\u03ce \u03b5\u03af\u03bc\u03b1\u03c3\u03c4\u03b5 \u03c3\u03b5 \u03b4\u03b9\u03ac\u03c3\u03c4\u03b1\u03c3\u03b7 \u03c4\u03b1 \u03c4\u03b5\u03bb\u03b5\u03c5\u03c4\u03b1\u03af\u03b1 15 \u03c7\u03c1\u03cc\u03bd\u03b9\u03b1 ;)[/hide]\n\n\nii. \u039c\u03b5 \u03b4\u03b9\u03c0\u03bb\u03ae \u03bf\u03bb\u03bf\u03ba\u03bb\u03ae\u03c1\u03c9\u03c3\u03b7\n[hide]\u039e\u03b5\u03ba\u03b9\u03bd\u03ac\u03bc\u03b5 \u03bc\u03b5 \u03c4\u03b7\u03bd \u03c0\u03b1\u03c1\u03b1\u03c4\u03ae\u03c1\u03b7\u03c3\u03b7: $ \\int_0^1 \\frac {\\log x}{x \\plus{} 1} \\textrm{d}x \\equal{} \\log x\\cdot \\log(x \\plus{} 1)\\Big|_0^1 \\minus{} \\int_0^1 \\frac {\\log(x \\plus{} 1)}{x} \\textrm{d}x$.\n\u03a4\u03ce\u03c1\u03b1, \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03b5\u03c0\u03af\u03c3\u03b7\u03c2 $ \\frac {\\log(x \\plus{} 1)}{x} \\equal{} \\int_0^1 \\frac {1}{1 \\plus{} xy} \\textrm{d}y$, \u03bf\u03c0\u03cc\u03c4\u03b5 \u03c4\u03bf \u03b4\u03b5\u03cd\u03c4\u03b5\u03c1\u03bf \u03bf\u03bb\u03bf\u03ba\u03bb\u03ae\u03c1\u03c9\u03bc\u03b1 \u03b8\u03b1 \u03b3\u03c1\u03ac\u03c6\u03b5\u03c4\u03b1\u03b9: $ \\int_0^1\\int_0^1 \\frac {1}{1 \\plus{} xy} \\textrm{d}y\\,\\textrm{d}x$. :!:  \u0391\u03c5\u03c4\u03cc \u03b5\u03af\u03bd\u03b1\u03b9 \u03c9\u03c1\u03b1\u03af\u03b1 \u03bc\u03bf\u03c1\u03c6\u03ae \u03b5\u03c0\u03b5\u03b9\u03b4\u03ae $ \\frac {1}{1 \\plus{} xy} \\equal{} 1 \\minus{} xy \\plus{} (xy)^2 \\plus{} \\cdots$ \u03ba\u03b1\u03b9, \u03ba\u03b1\u03c4\u03ac \u03c3\u03c5\u03bd\u03ad\u03c0\u03b5\u03b9\u03b1:\n$ \\int_0^1\\int_0^1\\frac {1}{1 \\plus{} xy} \\textrm{d}x\\,\\textrm{d}y \\equal{} \\sum_{n \\equal{} 0}^\\infty\\int_0^1\\int_0^1 ( \\minus{} 1)^n(xy)^n \\textrm{d}x\\,\\textrm{d}y \\equal{} \\minus{} \\sum_{n \\equal{} 1}^\\infty \\frac {( \\minus{} 1)^n}{n^2}$.\n\n\u038c\u03bc\u03c9\u03c2 $ \\sum_{n \\equal{} 1}^\\infty \\frac {1}{n^2} \\equal{} \\frac {\\pi^2}{6}$ \u03bf\u03c0\u03cc\u03c4\u03b5 \u03bb\u03af\u03b3\u03b1 rearrangements \u03b4\u03af\u03bd\u03bf\u03c5\u03bd $ \\sum_{n \\equal{} 1}^\\infty \\frac {( \\minus{} 1)^n}{n^2} \\equal{} \\minus{} \\frac {\\pi^2}{12}$\n\nTa daaah...    :)\n\nr_boris, \u03b1\u03c0\u03bb\u03ce\u03c2 \u03c0\u03b1\u03bd\u03ad\u03bc\u03bf\u03c1\u03c6\u03b7 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7! \u0391\u03bd \u03b4\u03b5\u03bd \u03b5\u03af\u03c7\u03b1 \u03b4\u03b5\u03b9 \u03c0\u03b1\u03bb\u03b9\u03ac \u03ad\u03bd\u03b1\u03bd \u03c5\u03c0\u03bf\u03bb\u03bf\u03b3\u03b9\u03c3\u03bc\u03cc \u03c4\u03bf\u03c5 $ \\zeta(2)$ \u03bc\u03b5 \u03ad\u03bd\u03b1 \u03b4\u03b9\u03c0\u03bb\u03cc \u03bf\u03bb\u03bf\u03ba\u03bb\u03ae\u03c1\u03c9\u03bc\u03b1, \u03b4\u03b5 \u03b8\u03b1 \u03c3\u03ba\u03b5\u03c6\u03c4\u03cc\u03bc\u03bf\u03c5\u03bd \u03c0\u03bf\u03c4\u03ad \u03c4\u03c9\u03bd \u03c0\u03bf\u03c4\u03ce\u03bd \u03c4\u03b1 \u03c0\u03b1\u03c1\u03b1\u03c0\u03ac\u03bd\u03c9!\n[/hide]\r\n\r\n\u03a5\u03c0\u03ad\u03c1\u03bf\u03c7\u03b7 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7!\r\n\r\nCheerio,\r\n\r\nDurandal 1707",
  "Solution_8": "\u039f\u03bd\u03c4\u03c9\u03c2 \u03c4\u03b5\u03bb\u03b5\u03b9\u03b1 \u03b1\u03c3\u03ba\u03b7\u03c3\u03b7 \u03ba\u03b1\u03b9 \u03c4\u03b5\u03bb\u03b5\u03b9\u03b1 \u03bb\u03c5\u03c3\u03b7 \u03c4\u03bf\u03c5 Durandal  :)\r\n\r\n\u039d\u03b1 \u03c1\u03c9\u03c4\u03b9\u03c3\u03c9 \u03bf\u03bc\u03c9\u03c2 \u03ba\u03b1\u03c4\u03b9. \u0393\u03b9\u03b1\u03c4\u03b9 \u03b5\u03c7\u03b5\u03b9\u03c2 \u03c4\u03bf logx \u03ba\u03b1\u03b9 \u03bf\u03c7\u03b9 lnx? \u039a\u03b1\u03b9 \u03c0\u03c9\u03c2 \u03c0\u03c1\u03bf\u03ba\u03c5\u03c0\u03c4\u03b5\u03b9 \u03bf\u03c4\u03b9 $ \\frac{log(x\\plus{}1)}{x}\\equal{}\\int_0^1 \\frac{1}{1\\plus{}xy}dy$? \u0395\u03c0\u03b9\u03c3\u03b7\u03c2 \u03c4\u03bf \u03b1\u03c1\u03c7\u03b9\u03ba\u03bf $ logxlog(x\\plus{}1)$ \u03c0\u03bf\u03c5 \u03c0\u03b7\u03b3\u03b5?",
  "Solution_9": "\u03a4\u03cc\u03c3\u03bf \u03b2\u03b3\u03b1\u03af\u03bd\u03b5\u03b9 \r\n\r\nMathematica rulez :P",
  "Solution_10": "[quote=\"mailo\"]\u039d\u03b1 \u03c1\u03c9\u03c4\u03b9\u03c3\u03c9 \u03bf\u03bc\u03c9\u03c2 \u03ba\u03b1\u03c4\u03b9. \u0393\u03b9\u03b1\u03c4\u03b9 \u03b5\u03c7\u03b5\u03b9\u03c2 \u03c4\u03bf logx \u03ba\u03b1\u03b9 \u03bf\u03c7\u03b9 lnx? \u039a\u03b1\u03b9 \u03c0\u03c9\u03c2 \u03c0\u03c1\u03bf\u03ba\u03c5\u03c0\u03c4\u03b5\u03b9 \u03bf\u03c4\u03b9 $ \\frac {log(x \\plus{} 1)}{x} \\equal{} \\int_0^1 \\frac {1}{1 \\plus{} xy}dy$? \u0395\u03c0\u03b9\u03c3\u03b7\u03c2 \u03c4\u03bf \u03b1\u03c1\u03c7\u03b9\u03ba\u03bf $ logxlog(x \\plus{} 1)$ \u03c0\u03bf\u03c5 \u03c0\u03b7\u03b3\u03b5?[/quote]\r\n\r\n$ \\log x, \\ln x$, \u03b8\u03ad\u03bc\u03b1 \u03c3\u03c5\u03bd\u03ae\u03b8\u03b5\u03b9\u03b1\u03c2 ;)\r\n\r\n\u03a4\u03bf \u03bf\u03bb\u03bf\u03ba\u03bb\u03ae\u03c1\u03c9\u03bc\u03b1 \u03c0\u03bf\u03c5 \u03c1\u03c9\u03c4\u03ac\u03c2, \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c0\u03bb\u03ce\u03c2 \u03ad\u03bd\u03b1\u03c2 \u03c5\u03c0\u03bf\u03bb\u03bf\u03b3\u03b9\u03c3\u03bc\u03cc\u03c2: \u03b8\u03b5\u03ce\u03c1\u03b7\u03c3\u03b5 \u03c4\u03bf $ x$ \u03c3\u03b1\u03bd \u03c3\u03c4\u03b1\u03b8\u03b5\u03c1\u03ac (\u03b1\u03c6\u03bf\u03cd \u03bf\u03bb\u03bf\u03ba\u03bb\u03b7\u03c1\u03ce\u03bd\u03b5\u03b9\u03c2 \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 $ y$) \u03ba\u03b1\u03b9 \u03b2\u03b3\u03b1\u03af\u03bd\u03b5\u03b9.\r\n\r\n\u03c4\u03bf $ \\log x \\log(x\\plus{}1)$ \u03b5\u03af\u03bd\u03b1\u03b9 $ 0$ \u03b3\u03b9\u03b1 $ x\\equal{}1$ \u03ba\u03b1\u03b9 \u03c4\u03b5\u03af\u03bd\u03b5\u03b9 \u03c3\u03c4\u03bf $ 0$ \u03ba\u03b1\u03b8\u03ce\u03c2 $ x\\to 0$.\r\n\r\nCheerio,\r\n\r\nDurandal 1707",
  "Solution_11": "[quote=\"Durandal\"]\n\n\u03a4\u03bf \u03bf\u03bb\u03bf\u03ba\u03bb\u03ae\u03c1\u03c9\u03bc\u03b1 \u03c0\u03bf\u03c5 \u03c1\u03c9\u03c4\u03ac\u03c2, \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c0\u03bb\u03ce\u03c2 \u03ad\u03bd\u03b1\u03c2 \u03c5\u03c0\u03bf\u03bb\u03bf\u03b3\u03b9\u03c3\u03bc\u03cc\u03c2: \u03b8\u03b5\u03ce\u03c1\u03b7\u03c3\u03b5 \u03c4\u03bf $ x$ \u03c3\u03b1\u03bd \u03c3\u03c4\u03b1\u03b8\u03b5\u03c1\u03ac (\u03b1\u03c6\u03bf\u03cd \u03bf\u03bb\u03bf\u03ba\u03bb\u03b7\u03c1\u03ce\u03bd\u03b5\u03b9\u03c2 \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 $ y$) \u03ba\u03b1\u03b9 \u03b2\u03b3\u03b1\u03af\u03bd\u03b5\u03b9.\n[/quote]\r\n\r\n\u039c\u03bf\u03c5 \u03b2\u03b3\u03b1\u03b9\u03bd\u03b5\u03b9 $ \\int_0^1 \\frac {1}{1 \\plus{} xy}dy\\equal{}ln(x\\plus{}1)$  :roll:",
  "Solution_12": "[size=150][b][color=red]\u0391\u03a3\u039a\u0397\u03a3\u0397 2[/color][/b][/size]\r\n\r\n\u039d\u03b1 \u03c5\u03c0\u03bf\u03bb\u03bf\u03b3\u03b9\u03c3\u03c4\u03b5\u03af \u03c4\u03bf: $ \\int_{ \\minus{} \\infty}^\\infty\\cdots\\int_{ \\minus{} \\infty}^\\infty \\frac {\\textrm{d}x_1\\cdots \\textrm{d}x_n}{(1 \\plus{} x_1^2 \\plus{} \\cdots \\plus{} x_n^2)^\\frac {n \\plus{} 1}{2}}$\r\n\r\nCheerio,\r\n\r\nDurandal 1707\r\n\r\nPS: Mailo, \u03c0\u03c1\u03cc\u03c3\u03b5\u03c7\u03b5 \u03cc\u03c4\u03b1\u03bd \u03c0\u03b1\u03c1\u03b1\u03b3\u03c9\u03b3\u03af\u03b6\u03b5\u03b9\u03c2 $ \\frac{d}{dy}\\log(1 \\plus{} xy) \\equal{} \\frac {x}{1 \\plus{} xy}$.",
  "Solution_13": "\u0397 \u03c0\u03b1\u03c1\u03ac\u03b3\u03bf\u03c5\u03c3\u03b1 \u03c4\u03bf\u03c5  $ \\frac {1}{1 \\plus{} xy}$  \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 $ \\frac {1}{x}\\ln(1 \\plus{} xy)$ \u03b5\u03c6\u03cc\u03c3\u03bf\u03bd \u03c0\u03b1\u03c1\u03b1\u03b3\u03c9\u03b3\u03af\u03b6\u03bf\u03c5\u03bc\u03b5 \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 y  \u03ba\u03b1\u03b9 \u03c4\u03bf \u03c7 \u03b8\u03b5\u03c9\u03c1\u03b5\u03af\u03c4\u03b1\u03b9  \u03c3\u03c4\u03b1\u03b8\u03b5\u03c1\u03ac.\r\n\r\n\r\n\u03a0\u03b1\u03bd\u03b1\u03b3\u03b9\u03ce\u03c4\u03b7 \u03bc\u03b5 \u03c0\u03c1\u03cc\u03bb\u03b1\u03b2\u03b5\u03c2 ... :lol:",
  "Solution_14": "\u039d\u03b1 \u03ba\u03b1\u03bd\u03c9 \u03bc\u03b9\u03b1 \u03b5\u03c1\u03c9\u03c4\u03b7\u03c3\u03b7. \u039f\u03c4\u03b1\u03bd \u03b5\u03c7\u03bf\u03c5\u03bc\u03b5 \u03bc\u03b9\u03b1 \u03c1\u03b7\u03c4\u03b7 \u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03b7\u03c3\u03b7 $ F(x)\\equal{}\\frac{f(x)}{g(x)}$ \u03c4\u03b7\u03c2 \u03bf\u03c0\u03bf\u03b9\u03b1 \u03bf \u03b2\u03b1\u03b8\u03bc\u03bf\u03c2 \u03c4\u03bf\u03c5 \u03c0\u03b1\u03c1\u03bf\u03bd\u03bf\u03bc\u03b1\u03c3\u03c4\u03b7\u03c2 $ g(x)$ \u03b5\u03b9\u03bd\u03b1\u03b9 \u03bc\u03b5\u03b3\u03b1\u03bb\u03c5\u03c4\u03b5\u03c1\u03bf\u03c2 \u03c4\u03bf\u03c5 \u03b2\u03b1\u03b8\u03bc\u03bf\u03c5 \u03c4\u03bf\u03c5 \u03b1\u03c1\u03b9\u03b8\u03bc\u03b7\u03c4\u03b7 $ f(x)$ \u03ba\u03b1\u03b9 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b4\u03b5\u03c5\u03c4\u03b5\u03c1\u03bf\u03c5 \u03b2\u03b1\u03b8\u03bc\u03bf\u03c5 \u03bc\u03b5 \u03b1\u03c1\u03bd\u03b7\u03c4\u03b9\u03ba\u03b7 \u03b4\u03b9\u03b1\u03ba\u03c1\u03b9\u03bd\u03bf\u03c5\u03c3\u03b1, \u03c4\u03bf\u03c4\u03b5 \u03c0\u03c9\u03c2 \u03b1\u03c5\u03c4\u03b7 \u03b7 \u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03b7\u03c3\u03b7 \u03b1\u03bd\u03b1\u03bb\u03c5\u03b5\u03c4\u03b1\u03b9 \u03c3\u03b5 \u03b1\u03b8\u03c1\u03bf\u03b9\u03c3\u03bc\u03b1 \u03b1\u03c0\u03bb\u03c9\u03bd \u03ba\u03bb\u03b1\u03c3\u03bc\u03b1\u03c4\u03c9\u03bd?\r\n\r\n\u0391\u03bd \u03b5\u03c7\u03bf\u03c5\u03bc\u03b5 \u03b8\u03b5\u03c4\u03b9\u03ba\u03b7 \u03b4\u03b9\u03b1\u03ba\u03c1\u03b9\u03bd\u03bf\u03c5\u03c3\u03b1 \u03c4\u03bf\u03c4\u03b5 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b1\u03c0\u03bb\u03b1 \u03c4\u03b1 \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b1, \u03b2\u03c1\u03b9\u03c3\u03ba\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b9\u03c2 \u03c1\u03b9\u03b6\u03b5\u03c2 $ r_1,r_2$ \u03c4\u03bf\u03c5 $ g(x)$ \u03ba\u03b1\u03b9 \u03b5\u03c4\u03c3\u03b9 \u03b7 $ F(x)$ \u03b3\u03b9\u03bd\u03b5\u03c4\u03b1\u03b9 $ F(x)\\equal{}\\frac{f(x)}{g(x)}\\equal{}\\frac{f(x)}{a(x\\minus{}r_1)(x\\minus{}r_2)}\\equal{}\\frac{A}{x\\minus{}r_1}\\plus{}\\frac{B}{x\\minus{}r_2}$",
  "Solution_15": "Ummm.. \u03bc\u03ac\u03bb\u03bb\u03bf\u03bd elementary \u03bc\u03bf\u03c5 \u03c6\u03b1\u03af\u03bd\u03b5\u03c4\u03b1\u03b9.\r\n\r\n[hide]\u0398\u03ad\u03c4\u03b5\u03b9\u03c2 $ t\\equal{}e^{\\minus{}u^2}\\minus{}1$ \u03c3\u03c4\u03bf \u03b4\u03b5\u03cd\u03c4\u03b5\u03c1\u03bf \u03bf\u03bb\u03bf\u03ba\u03bb\u03ae\u03c1\u03c9\u03bc\u03b1 \u03ba\u03b1\u03b9 \u03b5\u03bd\u03b1\u03bb\u03bb\u03ac\u03c3\u03c3\u03b5\u03b9\u03c2 \u03c4\u03b7 \u03c3\u03b5\u03b9\u03c1\u03ac \u03c4\u03b7\u03c2 \u03bf\u03bb\u03bf\u03ba\u03bb\u03ae\u03c1\u03c9\u03c3\u03b7\u03c2 \u03c3\u03c4\u03bf \u03b4\u03b9\u03c0\u03bb\u03cc \u03bf\u03bb\u03bf\u03ba\u03bb\u03ae\u03c1\u03c9\u03bc\u03b1 \u03c0\u03bf\u03c5 \u03c0\u03c1\u03bf\u03ba\u03cd\u03c0\u03c4\u03b5\u03b9.\n\n\u0391\u03bd \u03b4\u03b5\u03bd \u03ba\u03ac\u03bd\u03c9 \u03bb\u03ac\u03b8\u03bf\u03c2, \u03c4\u03bf \u03c4\u03b5\u03bb\u03b9\u03ba\u03cc \u03b1\u03c0\u03bf\u03c4\u03ad\u03bb\u03b5\u03c3\u03bc\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 $ \\minus{}\\frac{1}{2e}$? \u0389 \u03ba\u03ac\u03c4\u03b9 \u03c4\u03ad\u03c4\u03bf\u03b9\u03bf?[/hide]\r\n\r\n\u03a7\u03ac\u03bd\u03c9 \u03ba\u03ac\u03c4\u03b9?\r\n\r\nCheerio,\r\n\r\nDurandal 1707",
  "Solution_16": "N\u03b1\u03b9, elementary \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf \u03c4\u03c1\u03cc\u03c0\u03bf\u03c2 \u03bb\u03cd\u03c3\u03b7\u03c2, \u03b1\u03c0\u03bb\u03ce\u03c2 \u03bd\u03cc\u03bc\u03b9\u03b6\u03b1 \u03c0\u03c9\u03c2 \u03bf \u03c4\u03c1\u03cc\u03c0\u03bf\u03c2 \u03bc\u03b5 \u03c4\u03bf\u03bd \u03cc\u03c0\u03bf\u03b9\u03bf \u03c4\u03bf \u03ba\u03b1\u03c4\u03b1\u03c3\u03ba\u03b5\u03cd\u03b1\u03c3\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ba\u03b1\u03bb\u03ac \u03ba\u03c1\u03c5\u03bc\u03bc\u03ad\u03bd\u03bf\u03c2 (\u03c4\u03b7\u03bd \u03c4\u03cd\u03c6\u03bb\u03b1 \u03bc\u03bf\u03c5 :P)\r\n\r\n\u0397 \u03b9\u03b4\u03ad\u03b1 \u03bc\u03bf\u03c5 \u03ae\u03c4\u03b1\u03bd \u03cc\u03c4\u03b9:\r\n\r\n\u0391\u03c1\u03c7\u03b9\u03ba\u03ac \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9:\r\n\r\n$ \\int_{0}^{x}f(t)dt \\plus{} \\int_{0}^{f(x)}f^{ \\minus{} 1}(t)dt \\equal{} x\\cdot f(x)$\r\n\r\n\u038c\u03bb\u03b1 \u03b1\u03c5\u03c4\u03ac \u03b8\u03b5\u03c9\u03c1\u03ce\u03bd\u03c4\u03b1\u03c2 \u03c9\u03c2 \u03c3\u03c5\u03bd\u03ac\u03c1\u03c4\u03b7\u03c3\u03b7 \u03c4\u03b7\u03bd :\r\n\r\n$ f(x) \\equal{} e^{ \\minus{} x^2} \\minus{} 1$\r\n\r\n\u03cc\u03c0\u03bf\u03c5 guess \u03c0\u03bf\u03b9\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03c1\u03bf\u03c6\u03b7 \u03ba\u03b1\u03b9 \u03c0\u03c9\u03c2 \u03b1\u03c0\u03bb\u03bf\u03c5\u03c3\u03c4\u03b5\u03cd\u03b5\u03b9 \u03b1\u03c5\u03c4\u03cc \u03c4\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03c4\u03bf\u03c5 \u03c0\u03c1\u03bf\u03b2\u03bb\u03ae\u03bc\u03b1\u03c4\u03bf\u03c2!\r\n\r\n\r\n\r\n\u039c\u03ac\u03bb\u03bb\u03bf\u03bd \u03c6\u03b1\u03b9\u03ac \u03bf\u03c5\u03c3\u03af\u03b1, \u03bc\u03b9\u03b1\u03c2 \u03ba\u03b1\u03b9 \u03bc\u03bf\u03c5 \u03c0\u03ae\u03c1\u03b5 \u03b1\u03c1\u03ba\u03b5\u03c4\u03ae \u03ce\u03c1\u03b1 \u03bd\u03b1 \u03c4\u03b7 \u03c6\u03c4\u03b9\u03ac\u03be\u03c9.\r\n\r\n\r\n\u03a3\u03c4\u03ad\u03bb\u03b9\u03bf\u03c2",
  "Solution_17": "[quote=\"AlexandrosG\"][b]\u0391\u03c3\u03ba\u03b7\u03c3\u03b7 4[/b]\n\n\u0391\u03bd $ F_n$ \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b7 \u03b1\u03ba\u03bf\u03bb\u03bf\u03c5\u03b8\u03b9\u03b1 Fibonacci \u03bc\u03b5 $ F_0 \\equal{} 1$ , $ F_1 \\equal{} 1$ , $ F_{n \\plus{} 1} \\equal{} F_n \\plus{} F_{n \\minus{} 1}$ \u03bd\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03be\u03b5\u03c4\u03b5 \u03bf\u03c4\u03b9\n\n$ F_{m \\plus{} n \\plus{} 1} \\equal{} F_{m \\plus{} 1}\\cdot F_{n \\plus{} 1} \\plus{} F_m\\cdot F_n$  m,n \u03c6\u03c5\u03c3\u03b9\u03ba\u03bf\u03b9\n\n\u0395\u03bd\u03b4\u03b9\u03b1\u03c6\u03b5\u03c1\u03bf\u03bd \u03b8\u03b1 \u03b5\u03b9\u03c7\u03b5 \u03bc\u03b9\u03b1 \u03bb\u03c5\u03c3\u03b7 \u03bc\u03b5 \u03b5\u03c0\u03b1\u03b3\u03c9\u03b3\u03b7 \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 m \u03ba\u03b1\u03b9 n. \u0397 \u03bb\u03c5\u03c3\u03b7 \u03c0\u03bf\u03c5 \u03b5\u03b9\u03c7\u03b5 \u03b5\u03ba\u03b5\u03b9 \u03c0\u03bf\u03c5 \u03b2\u03c1\u03b7\u03ba\u03b1 \u03c4\u03bf \u03c0\u03c1\u03bf\u03b2\u03bb\u03b7\u03bc\u03b1 \u03c4\u03bf \u03ba\u03b1\u03bd\u03b5\u03b9 \u03ba\u03b1\u03c4\u03b1\u03bb\u03bb\u03b7\u03bb\u03bf \u03b3\u03b9\u03b1 \u03c4\u03bf topic \u03ba\u03b1\u03b9 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b1\u03c1\u03ba\u03b5\u03c4\u03b1 \u03bf\u03bc\u03bf\u03c1\u03c6\u03b7 \u03ba\u03b1\u03b9 \u03c3\u03c5\u03bd\u03c4\u03bf\u03bc\u03b7.\n\n\u0395\u03ba\u03c4\u03bf\u03c2 \u03b1\u03c0\u03bf \u03b1\u03c5\u03c4\u03b7 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03bc\u03b5\u03c1\u03b5\u03c2 \u03b1\u03bb\u03c5\u03c4\u03b5\u03c2 \u03ba\u03b1\u03b9 \u03bf\u03b9 \u03b1\u03c3\u03ba\u03b7\u03c3\u03b5\u03b9\u03c2 2 \u03ba\u03b1\u03b9 3. \u039f\u03c0\u03bf\u03b9\u03bf\u03c2 \u03bc\u03c0\u03bf\u03c1\u03b5\u03b9 \u03b1\u03c2 \u03b2\u03b1\u03bb\u03b5\u03b9 \u03c4\u03b9\u03c2 \u03bb\u03c5\u03c3\u03b5\u03b9\u03c2 \u03c4\u03bf\u03c5\u03c2.   :)[/quote]\r\n\r\n\u0392\u03b3\u03b1\u03af\u03bd\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03bc\u03b5 \u03b5\u03c0\u03b1\u03b3\u03c9\u03b3\u03ae \u03c3\u03c4\u03bf $ n$. \u0393\u03b9\u03b1 $ n\\equal{}1$ \u03c4\u03bf \u03b4\u03bf\u03c3\u03bc\u03ad\u03bd\u03bf \u03b5\u03cd\u03ba\u03bf\u03bb\u03b1 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03b3\u03b9\u03b1 \u03ba\u03ac\u03b8\u03b5 $ m$. \u0395\u03c3\u03c4\u03c9 \u03cc\u03c4\u03b9 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03b3\u03b1\u03b9 $ n$, \u03b8\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03b3\u03b9\u03b1 $ n\\plus{}1$. \r\n\r\n$ F_{m\\plus{}n\\plus{}1\\plus{}1}\\equal{}F_{m\\plus{}1\\plus{}n\\plus{}1}\\equal{}F_{m\\plus{}2}\\cdot F_{n\\plus{}1}\\plus{}F_{m\\plus{}1}\\cdot F_n\\equal{}$\r\n\r\n$ \\equal{}(F_m\\plus{}F_{m\\plus{}1})F_{n\\plus{}1}\\plus{}F_{m\\plus{}1}(F_{n\\plus{}2}\\minus{}F_{n\\plus{}1})\\equal{}$\r\n\r\n$ \\equal{}F_{m\\plus{}1}\\cdot F_{n\\plus{}2}\\plus{}F_m\\cdot F_{n\\plus{}1}$ \r\n\r\nQ.E.D.",
  "Solution_18": "[quote=\"mostel\"]N\u03b1\u03b9, elementary \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf \u03c4\u03c1\u03cc\u03c0\u03bf\u03c2 \u03bb\u03cd\u03c3\u03b7\u03c2, \u03b1\u03c0\u03bb\u03ce\u03c2 \u03bd\u03cc\u03bc\u03b9\u03b6\u03b1 \u03c0\u03c9\u03c2 \u03bf \u03c4\u03c1\u03cc\u03c0\u03bf\u03c2 \u03bc\u03b5 \u03c4\u03bf\u03bd \u03cc\u03c0\u03bf\u03b9\u03bf \u03c4\u03bf \u03ba\u03b1\u03c4\u03b1\u03c3\u03ba\u03b5\u03cd\u03b1\u03c3\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ba\u03b1\u03bb\u03ac \u03ba\u03c1\u03c5\u03bc\u03bc\u03ad\u03bd\u03bf\u03c2 (\u03c4\u03b7\u03bd \u03c4\u03cd\u03c6\u03bb\u03b1 \u03bc\u03bf\u03c5 :P)\n\n\u0397 \u03b9\u03b4\u03ad\u03b1 \u03bc\u03bf\u03c5 \u03ae\u03c4\u03b1\u03bd \u03cc\u03c4\u03b9:\n\n\u0391\u03c1\u03c7\u03b9\u03ba\u03ac \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9:\n\n$ \\int_{0}^{x}f(t)dt \\plus{} \\int_{0}^{f(x)}f^{ \\minus{} 1}(t)dt \\equal{} x\\cdot f(x)$\n\n\u038c\u03bb\u03b1 \u03b1\u03c5\u03c4\u03ac \u03b8\u03b5\u03c9\u03c1\u03ce\u03bd\u03c4\u03b1\u03c2 \u03c9\u03c2 \u03c3\u03c5\u03bd\u03ac\u03c1\u03c4\u03b7\u03c3\u03b7 \u03c4\u03b7\u03bd :\n\n$ f(x) \\equal{} e^{ \\minus{} x^2} \\minus{} 1$\n\n\u03cc\u03c0\u03bf\u03c5 guess \u03c0\u03bf\u03b9\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03c1\u03bf\u03c6\u03b7 \u03ba\u03b1\u03b9 \u03c0\u03c9\u03c2 \u03b1\u03c0\u03bb\u03bf\u03c5\u03c3\u03c4\u03b5\u03cd\u03b5\u03b9 \u03b1\u03c5\u03c4\u03cc \u03c4\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03c4\u03bf\u03c5 \u03c0\u03c1\u03bf\u03b2\u03bb\u03ae\u03bc\u03b1\u03c4\u03bf\u03c2!\n\n\n\n\u039c\u03ac\u03bb\u03bb\u03bf\u03bd \u03c6\u03b1\u03b9\u03ac \u03bf\u03c5\u03c3\u03af\u03b1, \u03bc\u03b9\u03b1\u03c2 \u03ba\u03b1\u03b9 \u03bc\u03bf\u03c5 \u03c0\u03ae\u03c1\u03b5 \u03b1\u03c1\u03ba\u03b5\u03c4\u03ae \u03ce\u03c1\u03b1 \u03bd\u03b1 \u03c4\u03b7 \u03c6\u03c4\u03b9\u03ac\u03be\u03c9.\n\n\n\u03a3\u03c4\u03ad\u03bb\u03b9\u03bf\u03c2[/quote]\r\n\u039c\u03b9\u03b1 \u03bc\u03b9\u03ba\u03c1\u03b7 \u03b4\u03b9\u03cc\u03c1\u03b8\u03c9\u03c3\u03b7 \u03c3\u03c4\u03bf\u03bd \u03c4\u03cd\u03c0\u03bf $ \\int_{0}^{x}f(t)dt \\plus{} \\int_{f(0)}^{f(x)}f^{ \\minus{} 1}(t)dt \\equal{} x\\cdot f(x)$ \u0395\u03b4\u03ce \u03b2\u03ad\u03b2\u03b1\u03b9\u03b1  \u03b5\u03af\u03bd\u03b1\u03b9 $ f(0)\\equal{}0$ \u03ba\u03b1\u03b9 \u03c0\u03b9\u03bf \u03b3\u03b5\u03bd\u03b9\u03ba\u03ac $ \\int_{y}^{x}f(t)dt \\plus{} \\int_{f(y)}^{f(x)}f^{ \\minus{} 1}(t)dt \\equal{} x\\cdot f(x)\\minus{}y\\cdot f(y)$",
  "Solution_19": "\u0392\u03b1\u03b6\u03c9 \u03c4\u03b7 \u03bb\u03c5\u03c3\u03b7 \u03c0\u03bf\u03c5 \u03b5\u03bb\u03b5\u03b3\u03b1\r\n\r\n\u0395\u03c3\u03c4\u03c9 M={{1,1},{1,0}}\r\n\r\n\u039c\u03b5 \u03b5\u03c0\u03b1\u03b3\u03c9\u03b3\u03b7 \u03b4\u03b5\u03b9\u03c7\u03bd\u03bf\u03c5\u03bc\u03b5 \u03bf\u03c4\u03b9 $ M^n$={{$ F_{n \\plus{} 1}$,$ F_n$},{$ F_n$,$ F_{n \\minus{} 1}$}}\r\n\u03a4\u03c9\u03c1\u03b1 \u03b1\u03c0\u03bf \u03c4\u03b7\u03bd \u03b9\u03b4\u03b9\u03bf\u03c4\u03b7\u03c4\u03b1 $ M^{m \\plus{} n}$=$ M^n*M^m \\equal{} M^m*M^n$ \u03b3\u03c1\u03b1\u03bc\u03bc\u03b5\u03bd\u03b7 \u03c3\u03b5 \u03bc\u03bf\u03c1\u03c6\u03b7 \u03c0\u03b9\u03bd\u03b1\u03ba\u03c9\u03bd \u03ba\u03b1\u03b9 \u03b1\u03c0\u03bf \u03c4\u03bf \u03c0\u03b1\u03bd\u03c9 \u03b1\u03c1\u03b9\u03c3\u03c4\u03b5\u03c1\u03b1 \u03c3\u03c4\u03bf\u03b9\u03c7\u03b5\u03b9\u03bf\r\n\u03c0\u03b1\u03b9\u03c1\u03bd\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b7\u03bd  $ F_{m \\plus{} n \\plus{} 1} \\equal{} F_{m \\plus{} 1}*F_{n \\plus{} 1} \\plus{} F_m*F_n$",
  "Solution_20": "\u03b5\u03c3\u03c4\u03c9 \u03b1 \u03ac\u03c1\u03b7\u03c4\u03bf\u03c2 \u03ba\u03b1\u03b9 \u03c4\u03bf \u03c3\u03cd\u03bd\u03bf\u03bb\u03bf \u0391={ma+n|m,n \u03b1\u03bd\u03ae\u03ba\u03bf\u03c5\u03bd \u03c3\u03c4\u03bf Z}.\r\n\u03bd\u03b4\u03bf \u03c4\u03bf \u0391 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03c5\u03ba\u03bd\u03cc \u03c3\u03c4\u03bf R",
  "Solution_21": "[b][color=red][size=150]\u0391\u03a3\u039a\u0397\u03a3\u0397 5[/size][/color][/b] (Dimitris T)\r\n\r\n\u0391\u03bd $ a$ \u03ac\u03c1\u03c1\u03b7\u03c4\u03bf\u03c2, \u03bd\u03b4\u03bf \u03c4\u03bf \u03c3\u03cd\u03bd\u03bf\u03bb\u03bf $ A \\equal{} \\{ma \\plus{} n: m,n\\in\\mathbb{Z}\\}$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03c5\u03ba\u03bd\u03cc \u03c3\u03c4\u03bf $ \\mathbb{R}$.\r\n\r\n\u039a\u03bb\u03b5\u03af\u03bd\u03c9 \u03c4\u03b7\u03bd \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03c3\u03b5 spoiler \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03c4\u03b7 \u03c7\u03b1\u03bb\u03ac\u03c3\u03c9 \u03c3\u03b5 \u03cc\u03c3\u03bf\u03c5\u03c2 \u03c4\u03b7 \u03b4\u03bf\u03c5\u03bb\u03b5\u03cd\u03bf\u03c5\u03bd \u03b1\u03ba\u03cc\u03bc\u03b1 :)\r\n\r\n[hide]\u039a\u03b1\u03c4'\u03b1\u03c1\u03c7\u03ac\u03c2, \u03b8\u03b5\u03c9\u03c1\u03ce\u03bd\u03c4\u03b1\u03c2 \u03bc\u03cc\u03bd\u03bf \u03c4\u03bf \u03b4\u03b5\u03ba\u03b1\u03b4\u03b9\u03ba\u03cc \u03bc\u03ad\u03c1\u03bf\u03c2, \u03b1\u03c1\u03ba\u03b5\u03af \u03bd\u03b1 \u03b4\u03b5\u03af\u03be\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03c4\u03bf \u03c3\u03cd\u03bd\u03bf\u03bb\u03bf $ \\{m a: m\\in\\mathbb{Z}\\}$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03c5\u03ba\u03bd\u03cc \u03c3\u03c4\u03bf $ [0,1]$. \u0389, \u03c0\u03b9\u03bf \u03ba\u03bf\u03bc\u03c8\u03ac, \u03b4\u03bf\u03c5\u03bb\u03b5\u03cd\u03bf\u03bd\u03c4\u03b1\u03c2 $ \\mod \\mathbb{Z}$, \u03b1\u03c1\u03ba\u03b5\u03af \u03bd\u03b1 \u03b4\u03b5\u03af\u03be\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03c4\u03bf \u03c3\u03cd\u03bd\u03bf\u03bb\u03bf $ C \\equal{} \\{e^{2\\pi i m a}: m\\in\\mathbb{Z}\\}$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03c5\u03ba\u03bd\u03cc \u03c3\u03c4\u03bf\u03bd \u03bc\u03bf\u03bd\u03b1\u03b4\u03b9\u03b1\u03af\u03bf \u03ba\u03cd\u03ba\u03bb\u03bf $ S^1 \\equal{} \\mathbb{R}/\\mathbb{Z}$.\n\n\u03a0\u03c1\u03bf\u03c6\u03b1\u03bd\u03ce\u03c2, \u03b5\u03c6\u03cc\u03c3\u03bf\u03bd \u03bf $ a$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03ac\u03c1\u03c1\u03b7\u03c4\u03bf\u03c2, \u03c4\u03bf \u03c3\u03cd\u03bd\u03bf\u03bb\u03bf $ C$ \u03b8\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ac\u03c0\u03b5\u03b9\u03c1\u03bf \u03ba\u03b1\u03b9 \u03ac\u03c1\u03b1 \u03b8\u03b1 \u03ad\u03c7\u03b5\u03b9 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf \u03bf\u03c1\u03b9\u03b1\u03ba\u03cc \u03c3\u03b7\u03bc\u03b5\u03af\u03bf $ \\xi$ \u03c3\u03c4\u03bf \u03c3\u03c5\u03bc\u03c0\u03b1\u03b3\u03ae \u03ba\u03cd\u03ba\u03bb\u03bf $ C$. \u03a4\u03cc\u03c4\u03b5, \u03b1\u03bd $ \\epsilon > 0$ \u03b8\u03b1 \u03c5\u03c0\u03ac\u03c1\u03c7\u03bf\u03c5\u03bd $ n_{1,2}$ \u03bc\u03b5 $ n_1\\neq n_2$ \u03ce\u03c3\u03c4\u03b5 $ |e^{2\\pi i n_{1(2)} a} \\minus{} \\xi| < \\frac {\\epsilon}{2}\\implies 0 < |e^{2\\pi i n_1 a} \\minus{} e^{2\\pi i n_2 a}| < \\epsilon$ (\u03b1\u03c0\u03cc \u03b5\u03c6\u03b1\u03c1\u03bc\u03bf\u03b3\u03ae \u03c4\u03c1\u03b9\u03b3\u03c9\u03bd\u03b9\u03ba\u03ae\u03c2 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1\u03c2; \u03b8\u03c5\u03bc\u03b7\u03b8\u03b5\u03af\u03c4\u03b5 \u03ba\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03bf $ a$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03ac\u03c1\u03c1\u03b7\u03c4\u03bf\u03c2). \u0391\u03c5\u03c4\u03cc \u03cc\u03bc\u03c9\u03c2 \u03b4\u03b5\u03af\u03c7\u03bd\u03b5\u03b9 \u03cc\u03c4\u03b9 \u03b3\u03b9\u03b1 $ k \\equal{} n_2 \\minus{} n_1$, \u03b8\u03b1 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 $ 0 < |e^{2\\pi i k a} \\minus{} 1| < \\epsilon$ (\u03b4\u03b7\u03bb\u03b1\u03b4\u03ae \u03ba\u03b1\u03b9 \u03c4\u03bf $ 1$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf\u03c1\u03b9\u03b1\u03ba\u03cc \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03c4\u03bf\u03c5 $ C$).\n\nT\u03ce\u03c1\u03b1, \u03ad\u03c3\u03c4\u03c9 $ \\theta \\equal{} 2k\\pi a$ \u03bf\u03c0\u03cc\u03c4\u03b5 $ 0 < |e^{i\\theta} \\minus{} 1| < \\epsilon$. \u0388\u03c4\u03c3\u03b9, \u03b1\u03c1\u03ba\u03b5\u03af \u03bd\u03b1 \u03b4\u03b5\u03af\u03be\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03b3\u03b9\u03b1 \u03ba\u03ac\u03b8\u03b5 $ w\\in S^1$ \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf $ n\\in\\mathbb{Z}$ \u03ce\u03c3\u03c4\u03b5 $ |e^{i n\\theta} \\minus{} w| < \\epsilon$. \u0391\u03c5\u03c4\u03cc \u03cc\u03bc\u03c9\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c0\u03bb\u03cc \u03bd\u03b1 \u03c4\u03bf \u03b4\u03b5\u03b9 \u03ba\u03b1\u03bd\u03b5\u03af\u03c2: \u03bf\u03b9 \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03af $ e^{i\\theta}, e^{2i\\theta}\\ldots e^{i n\\theta}\\ldots$ \u03c0\u03c1\u03bf\u03c7\u03c9\u03c1\u03ac\u03bd\u03b5 \u03ba\u03b1\u03bb\u03cd\u03c0\u03c4\u03bf\u03bd\u03c4\u03b1\u03c2 \u03c3\u03b9\u03b3\u03ac-\u03c3\u03b9\u03b3\u03ac \u03c4\u03bf\u03bd \u03ba\u03cd\u03ba\u03bb\u03bf \u03bc\u03b5 \u03b2\u03ae\u03bc\u03b1\u03c4\u03b1 $ < \\epsilon$. \u0386\u03c1\u03b1, \u03b3\u03b9\u03b1 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf $ n$, \u03b8\u03b1 \u03b2\u03c1\u03b5\u03b8\u03bf\u03cd\u03bd \u03c3\u03b5 \u03b1\u03c0\u03cc\u03c3\u03c4\u03b1\u03c3\u03b7 \u03c4\u03bf \u03c0\u03bf\u03bb\u03cd $ \\epsilon$ \u03b1\u03c0\u03cc \u03c4\u03bf $ w$, QED.\n\nNote: \u03a4\u03bf \u03c0\u03b1\u03c1\u03b1\u03c0\u03ac\u03bd\u03c9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03ba\u03b1\u03c4\u03b1\u03bb\u03b1\u03b2\u03b1\u03af\u03bd\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b9 \u03b3\u03af\u03bd\u03b5\u03c4\u03b1\u03b9. \u0393\u03b9\u03b1 \u03cc\u03c3\u03bf\u03c5\u03c2 \u03b8\u03ad\u03bb\u03bf\u03c5\u03bd \u03bc\u03af\u03b1 \u03c0\u03b9\u03bf \"\u03b1\u03c0\u03bf\u03c3\u03c4\u03b5\u03b9\u03c1\u03c9\u03bc\u03ad\u03bd\u03b7\" \u03b4\u03b9\u03b1\u03c4\u03cd\u03c0\u03c9\u03c3\u03b7, \u03b1\u03c5\u03c4\u03cc \u03c0\u03bf\u03c5 \u03bb\u03ad\u03c9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03bf\u03b9 \u03c0\u03b5\u03c1\u03b9\u03bf\u03c7\u03ad\u03c2 $ \\{z: |z \\minus{} e^{in\\theta}| < \\epsilon\\}, n\\in\\mathbb{Z}$ \u03b1\u03c0\u03bf\u03c4\u03b5\u03bb\u03bf\u03cd\u03bd \u03bc\u03af\u03b1 \u03b1\u03bd\u03bf\u03b9\u03c7\u03c4\u03ae \u03ba\u03ac\u03bb\u03c5\u03c8\u03b7 \u03c4\u03b7\u03c2 \u03c3\u03c5\u03bc\u03c0\u03b1\u03b3\u03bf\u03cd\u03c2 $ S^1$ \u03ba\u03b1\u03b9 \u03c4\u03b1 \u03c5\u03c0\u03cc\u03bb\u03bf\u03b9\u03c0\u03b1 \u03ba\u03c5\u03bb\u03ac\u03bd\u03b5 \u03bf\u03bc\u03b1\u03bb\u03ac :) [/hide]\r\n\r\n\u03a9\u03c1\u03b1\u03af\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 :) \u0395\u03c0\u03b9\u03c4\u03ad\u03bb\u03bf\u03c5\u03c2, \u03c4\u03ce\u03c1\u03b1 \u03c3\u03b9\u03b3\u03ac-\u03c3\u03b9\u03b3\u03ac \u03be\u03b1\u03bd\u03b1\u03b2\u03c1\u03af\u03c3\u03ba\u03c9 \u03bb\u03af\u03b3\u03bf \u03c7\u03c1\u03cc\u03bd\u03bf \u03b3\u03b9\u03b1 \u03c4\u03bf forum...\r\n\r\nCheerio,\r\n\r\nDurandal 1707",
  "Solution_22": "Poli oraia, \r\nTi lete gia ena bima parapera :\r\n\r\nAn $ a$ arritos kai $ P(x)\\equal{}a\\cdot x^n \\plus{}a_{n\\minus{}1}\\cdot x^{n\\minus{}1}\\plus{}\\ldots\\plus{}a_0$ einai pragmatiko polionimo,  n.d.o to sinolo $ A\\equal{}\\{P(m)\\plus{}n: m,n \\in Z\\}$  einai pikno sto $ R$.\r\n\r\nStergios",
  "Solution_23": "[b][color=red]\u03a4\u03b1 \u03bd\u03ad\u03b1 \u03c0\u03c1\u03bf\u03b2\u03bb\u03ae\u03bc\u03b1\u03c4\u03b1 \u03ba\u03b1\u03b9 \u03c4\u03b9\u03c2  \u03bb\u03cd\u03c3\u03b5\u03b9\u03c2  \u03c0\u03c1\u03bf\u03c4\u03b5\u03af\u03bd\u03c9 \u03bd\u03b1 \u03c4\u03b1 \u03b3\u03c1\u03ac\u03c6\u03b5\u03c4\u03b5 \u03cc\u03c7\u03b9 \u03c0\u03b9\u03b1 \u03c3\u03c4\u03bf \u03c3\u03c4\u03b9\u03ba\u03c5 , \u03b1\u03bb\u03bb\u03ac \u03c3\u03c4\u03bf\u03bd \u03ba\u03b1\u03bd\u03bf\u03bd\u03b9\u03ba\u03cc \u03c7\u03ce\u03c1\u03bf. \u039c\u03b5 \u03b1\u03c5\u03c4\u03cc\u03bd \u03c4\u03bf\u03bd \u03c4\u03c1\u03cc\u03c0\u03bf \u03b8\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03b9\u03bf \u03b5\u03cd\u03ba\u03bf\u03bb\u03bf \u03bd\u03b1 \u03b2\u03bb\u03ad\u03c0\u03b5\u03c4\u03b5 \u03c4\u03b9\u03c2 \u03bd\u03ad\u03b5\u03c2 \u03b1\u03c3\u03ba\u03ae\u03c3\u03b5\u03b9\u03c2 \u03ba\u03b1\u03b9 \u03c4\u03b9\u03c2 \u03bb\u03cd\u03c3\u03b5\u03b9\u03c2 .\u0391\u03bd \u03c7\u03c1\u03b5\u03b9\u03b1\u03c3\u03c4\u03b5\u03af\u03c4\u03b5 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf \u03c3\u03c4\u03b9\u03ba\u03c5 \u03b3\u03b9\u03b1 \u03ac\u03bb\u03bb\u03b7 \u03c7\u03c1\u03ae\u03c3\u03b7 ,\u03b5\u03ba\u03c4\u03cc\u03c2 \u03b1\u03c3\u03ba\u03ae\u03c3\u03b5\u03c9\u03bd ,  \u03bd\u03b1 \u03bc\u03bf\u03c5 \u03c4\u03bf \u03b6\u03b7\u03c4\u03ae\u03c3\u03b5\u03c4\u03b5 .[/color][/b].\r\n\r\n\r\n \u0395\u03bd\u03bd\u03bf\u03b5\u03af\u03c4\u03b1\u03b9 \u03c0\u03c9\u03c2 \u03b7 \u03b2\u03bf\u03ae\u03b8\u03b5\u03b9\u03ac \u03c3\u03b1\u03c2 [u]\u03ba\u03b1\u03b9[/u] \u03c3\u03c4\u03bf\u03c5\u03c2 \u03c7\u03ce\u03c1\u03bf\u03c5\u03c2 \u03c4\u03c9\u03bd \u03bf\u03bb\u03c5\u03bc\u03c0\u03b9\u03ac\u03b4\u03c9\u03bd \u03ae \u03c4\u03c9\u03bd \u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03ce\u03bd \u03b1\u03c3\u03ba\u03ae\u03c3\u03b5\u03c9\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03bf\u03bb\u03cd\u03c4\u03b9\u03bc\u03b7. \r\n\r\n[u] \u039c\u03c0\u03ac\u03bc\u03c0\u03b7\u03c2[/u]",
  "Solution_24": "[quote=\"silouan\"]\u039d\u03b1 \u03c5\u03c0\u03bf\u03bb\u03bf\u03b3\u03b9\u03c3\u03c4\u03b5\u03af \u03c4\u03bf \u03bf\u03bb\u03bf\u03ba\u03bb\u03af\u03c1\u03c9\u03bc\u03b1 \n\n$ \\int\\frac {x \\plus{} 1}{x(e^x \\plus{} 1)}$[/quote]\r\n\r\nSilouan \u03b8\u03b1 \u03ae\u03b8\u03b5\u03bb\u03b1 \u03b1\u03bd \u03bc\u03c0\u03bf\u03c1\u03b5\u03af\u03c2 \u03bd\u03b1 \u03b2\u03ac\u03bb\u03b5\u03b9\u03c2 \u03c3\u03b5 hide \u03c4\u03b7\u03bd \u03bb\u03cd\u03c3\u03b7 \u03b3\u03b9\u03b1\u03c4\u03af \u03c4\u03b7\u03bd \u03c0\u03b1\u03bb\u03b5\u03cd\u03c9 \u03b5\u03b4\u03ce \u03ba\u03b1\u03b9 \u03bb\u03af\u03b3\u03b5\u03c2 \u03bc\u03ad\u03c1\u03b5\u03c2 \u03c4\u03b7\u03bd \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03ba\u03b1\u03b9 \u03b4\u03b5\u03bd \u03ad\u03c7\u03c9 \u03ba\u03b1\u03c4\u03b1\u03c6\u03ad\u03c1\u03b5\u03b9 \u03ba\u03ac\u03c4\u03b9....\r\n\u0395\u03af\u03bd\u03b1\u03b9 \u03ba\u03b1\u03b9\u03c1\u03cc \u03ac\u03bb\u03c5\u03c4\u03b7 \u03bf\u03c0\u03cc\u03c4\u03b5 \u03b4\u03b5\u03bd \u03c0\u03b9\u03c3\u03c4\u03b5\u03cd\u03c9 \u03bd\u03b1 \u03b4\u03b9\u03b1\u03c6\u03c9\u03bd\u03ae\u03c3\u03b5\u03b9 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03c2..",
  "Solution_25": "Katarxas sorry gia ta greeklish alla grafw apo kinhto... Exw ena provlhmataki se mia askhsh pithanothtwn:\r\nSe ena kentro minhmatwn ftanoun minhmata ipshlhs mesaias kai xamhlhs proteraiothtas me sixnothta emfanishs 50% , 30% kai 20% anristoixa. Ta minhmata ipshlhs proteraiothtas kostizoun 5 monades, ta mesaias 3 monades kai ta xamhlhs 1 monada... Poia h pithanothta mia seiras 100 minhmatwn na exei kostos ligotero apo 200 monades?\r\n\r\nTo exw palepsei poli alla den mporw na vrw akrh :(",
  "Solution_26": "ok. Lithike... Skeftomouna pio poliplokes diadikasies alla den htan kati telika",
  "Solution_27": "\u039a\u03b1\u03bb\u03b7\u03bc\u03ad\u03c1\u03b1. :) \u039c\u03ae\u03c0\u03c9\u03c2 \u03b3\u03bd\u03c9\u03c1\u03af\u03b6\u03b5\u03b9 \u03ba\u03b1\u03bd\u03b5\u03af\u03c2 \u03c0\u03ce\u03c2 \u03bb\u03ad\u03b3\u03b5\u03c4\u03b1\u03b9 \u03c3\u03c4\u03b1 \u0391\u03b3\u03b3\u03bb\u03b9\u03ba\u03ac \u03b7 \u03bc\u03ad\u03b8\u03bf\u03b4\u03bf\u03c2 \u03c4\u03bf\u03c5 \u03ba\u03b1\u03c4\u03bf\u03c0\u03c4\u03c1\u03b9\u03c3\u03bc\u03bf\u03cd; :) \u0388\u03c7\u03c9 \u03ba\u03bf\u03bb\u03bb\u03ae\u03c3\u03b5\u03b9 \u03c3\u03b5 \u03ad\u03bd\u03b1 \u03c0\u03c1\u03bf\u03b2\u03bb\u03b7\u03bc\u03b1\u03c4\u03ac\u03ba\u03b9 \u03c0\u03c1\u03bf\u03c3\u03b4\u03b9\u03bf\u03c1\u03b9\u03c3\u03bc\u03bf\u03cd \u03b5\u03bd\u03cc\u03c2 \u03b4\u03b9\u03b1\u03bd\u03c5\u03c3\u03bc\u03b1\u03c4\u03b9\u03ba\u03bf\u03cd \u03c0\u03b5\u03b4\u03af\u03bf\u03c5 \u03ba\u03b1\u03b9 \u03ad\u03c7\u03c9 \u03b4\u03bf\u03ba\u03b9\u03bc\u03ac\u03c3\u03b5\u03b9 \u03cc,\u03c4\u03b9 \u03ad\u03c7\u03c9 \u03c3\u03ba\u03b5\u03c6\u03c4\u03b5\u03af \u03c3\u03b5 mirror \u03ae reflection \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03b2\u03c1\u03c9 \u03c0\u03bb\u03b7\u03c1\u03bf\u03c6\u03bf\u03c1\u03af\u03b5\u03c2 \u03c3\u03c4\u03bf google \u03b1\u03bb\u03bb\u03ac \u03bc\u03ac\u03c4\u03b1\u03b9\u03b1. :/\r\n\r\n\r\n\r\n(\u0398\u03b5\u03ce\u03c1\u03b7\u03c3\u03b1 \u03c0\u03b5\u03c1\u03b9\u03c4\u03c4\u03cc \u03bd' \u03b1\u03bd\u03bf\u03af\u03be\u03c9 \u03bd\u03ad\u03bf topic \u03b3\u03b9' \u03b1\u03c5\u03c4\u03cc \u03c4\u03bf \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03ac\u03ba\u03b9, \u03b5\u03bb\u03c0\u03af\u03b6\u03c9 \u03bd\u03b1 \u03c4\u03bf '\u03b3\u03c1\u03b1\u03c8\u03b1 \u03c3\u03c4\u03bf \u03c3\u03c9\u03c3\u03c4\u03cc \u03bc\u03ad\u03c1\u03bf\u03c2)",
  "Solution_28": "[quote=\"modulo\"]\u039a\u03b1\u03bb\u03b7\u03bc\u03ad\u03c1\u03b1. :) \u039c\u03ae\u03c0\u03c9\u03c2 \u03b3\u03bd\u03c9\u03c1\u03af\u03b6\u03b5\u03b9 \u03ba\u03b1\u03bd\u03b5\u03af\u03c2 \u03c0\u03ce\u03c2 \u03bb\u03ad\u03b3\u03b5\u03c4\u03b1\u03b9 \u03c3\u03c4\u03b1 \u0391\u03b3\u03b3\u03bb\u03b9\u03ba\u03ac \u03b7 \u03bc\u03ad\u03b8\u03bf\u03b4\u03bf\u03c2 \u03c4\u03bf\u03c5 \u03ba\u03b1\u03c4\u03bf\u03c0\u03c4\u03c1\u03b9\u03c3\u03bc\u03bf\u03cd; :) \u0388\u03c7\u03c9 \u03ba\u03bf\u03bb\u03bb\u03ae\u03c3\u03b5\u03b9 \u03c3\u03b5 \u03ad\u03bd\u03b1 \u03c0\u03c1\u03bf\u03b2\u03bb\u03b7\u03bc\u03b1\u03c4\u03ac\u03ba\u03b9 \u03c0\u03c1\u03bf\u03c3\u03b4\u03b9\u03bf\u03c1\u03b9\u03c3\u03bc\u03bf\u03cd \u03b5\u03bd\u03cc\u03c2 \u03b4\u03b9\u03b1\u03bd\u03c5\u03c3\u03bc\u03b1\u03c4\u03b9\u03ba\u03bf\u03cd \u03c0\u03b5\u03b4\u03af\u03bf\u03c5 \u03ba\u03b1\u03b9 \u03ad\u03c7\u03c9 \u03b4\u03bf\u03ba\u03b9\u03bc\u03ac\u03c3\u03b5\u03b9 \u03cc,\u03c4\u03b9 \u03ad\u03c7\u03c9 \u03c3\u03ba\u03b5\u03c6\u03c4\u03b5\u03af \u03c3\u03b5 mirror \u03ae reflection \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03b2\u03c1\u03c9 \u03c0\u03bb\u03b7\u03c1\u03bf\u03c6\u03bf\u03c1\u03af\u03b5\u03c2 \u03c3\u03c4\u03bf google \u03b1\u03bb\u03bb\u03ac \u03bc\u03ac\u03c4\u03b1\u03b9\u03b1. :/\n\n\n\n(\u0398\u03b5\u03ce\u03c1\u03b7\u03c3\u03b1 \u03c0\u03b5\u03c1\u03b9\u03c4\u03c4\u03cc \u03bd' \u03b1\u03bd\u03bf\u03af\u03be\u03c9 \u03bd\u03ad\u03bf topic \u03b3\u03b9' \u03b1\u03c5\u03c4\u03cc \u03c4\u03bf \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03ac\u03ba\u03b9, \u03b5\u03bb\u03c0\u03af\u03b6\u03c9 \u03bd\u03b1 \u03c4\u03bf '\u03b3\u03c1\u03b1\u03c8\u03b1 \u03c3\u03c4\u03bf \u03c3\u03c9\u03c3\u03c4\u03cc \u03bc\u03ad\u03c1\u03bf\u03c2)[/quote]\r\n\r\n\u0394\u03b5 \u03bd\u03bf\u03bc\u03af\u03b6\u03c9 \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ba\u03b1\u03b9 \u03c0\u03bf\u03bb\u03cd \u03c7\u03c1\u03ae\u03c3\u03b9\u03bc\u03bf \u03bc\u03b5\u03c4\u03ac \u03b1\u03c0\u03cc 2 \u03bc\u03ae\u03bd\u03b5\u03c2 :( \u03b1\u03bb\u03bb\u03ac \u03bc\u03ac\u03bb\u03bb\u03bf\u03bd \u03b8\u03ad\u03bb\u03b5\u03b9\u03c2 \u03c4\u03b7 \u03bc\u03ad\u03b8\u03bf\u03b4\u03bf \u03c4\u03c9\u03bd   [url=http://en.wikipedia.org/wiki/Method_of_image_charges]image charges[/url]...\r\n\r\nCheerio,\r\n\r\nDurandal 1707",
  "Solution_29": "\u0398\u03b1 \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03c3\u03b1\u03c4\u03b5 \u03bd\u03b1 \u03bc\u03b5 \u03b2\u03bf\u03b7\u03b8\u03ae\u03c3\u03b5\u03c4\u03b5 \u03bc\u03b5 \u03c4\u03bf\u03bd \u03c5\u03c0\u03bf\u03bb\u03bf\u03b3\u03b9\u03c3\u03bc\u03cc \u03c4\u03bf\u03c5 \u03bf\u03bb\u03bf\u03ba\u03bb\u03b7\u03c1\u03ce\u03bc\u03b1\u03c4\u03bf\u03c2 \u03c4\u03bf\u03c5 {(1+\u03c7^1/2)^1/3}*\u03c7^-1\r\n\r\n\u0395\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03ce \u03b5\u03ba \u03c4\u03c9\u03bd \u03c0\u03c1\u03bf\u03c4\u03ad\u03c1\u03c9\u03bd"
}
{
  "Tag": [
    "geometry",
    "MATHCOUNTS",
    "number theory",
    "modular arithmetic"
  ],
  "Problem": "I feel that one area that I am weak in is modular arithmetic. Are there any good places where I could learn it? I know I could buy the Introduction to Number Theory, and it would be in it, but are there any free ones? Also, I am not looking for a definition, like wikipedia, but rather an explanation.",
  "Solution_1": "I'm afraid it's too complicated to explain in one post. I suppose the only option is to buy Introduction to Number Theory, or read certain chapters in the classics.",
  "Solution_2": "Well, I just ordered the classics, and in the table of contents, I only see one chapter about Number Theory in Volume 2. Also, I understand most of the Number Theory book, it is just the last few chapters that are about modular arithmetic. Correct me if I am wrong.\r\n\r\nI was looking for links to some good website resources, like how coolmath.com has lessons for algebra.",
  "Solution_3": "Stick with AoPS volume 1. It's got a whole chapter dedicated to number theory.(chapter 5)",
  "Solution_4": "Other resources include UCSMP math in addition to AoPS",
  "Solution_5": "If you wanted, you could also have a look at this:\r\n[url]http://www-history.mcs.st-and.ac.uk/~edmund/lnotes/node5.html[/url]",
  "Solution_6": "In AoPS volume 1, one of the first few chapters explains basic modular arithmetic very well. In AoPS volue 2, it get a LOT more in depth, so read the chapter in volume 1 before reading the chapter in volume 2.",
  "Solution_7": "This link is also decent, though I'd recommend either buying Intro to Number Theory or borrowing it from a library/teacher/friend. \r\n[url]http://www.math.rutgers.edu/~erowland/modulararithmetic.html[/url]",
  "Solution_8": "Introduction to Number Theory from AoPS is better than the free links here (though obviously it's not free).",
  "Solution_9": "Santos's dispences are the best in my hopinion",
  "Solution_10": "What is Santos's dispences?",
  "Solution_11": "Look at this one http://www.openmathtext.org/lecture_notes/number_theory_book.pdf  :wink:",
  "Solution_12": "The link doesn't work, unfortunately.",
  "Solution_13": "I personally feel that intro. to number theory explains the best, AOPS vol.1, doesn't have that many chapters dedicated to that topic.\r\nWhat helped me was the advanced MATHCOUNTS/AMC 8 course. I learned modular arithmetic, base numbers much better for Mathcounts preprartion. I suggest you to try that!"
}
{
  "Tag": [
    "calculus",
    "integration",
    "calculus computations"
  ],
  "Problem": "Hi everyone, \r\n\r\nCould someone please explain to me how to solve the integral x^3-x^2-x+1/x^+9 \r\n\r\nI know that I have to do long divition first. \r\n\r\nAfter doing this I got x-1-10x+10/x^2+9, but the correct answer for this step is x-1-8x-1/x^2+9. Does anyone see where I went wrong? \r\n\r\nAfter that I am supposed to use a table and the correct answer is \r\n\r\n1/2x^2-x-8(1/2ln(x^2+9))+8(1/3tan^-1(x/3))+c \r\n\r\nCan someone explain to me how they got this as their final answer, please? \r\n\r\nThank you very much",
  "Solution_1": "That's incomprehensible. Use parentheses!\r\n\r\nDo you mean $ \\int\\frac{x^{3}\\plus{}x^{2}\\plus{}x\\plus{}1}{x^{2}\\plus{}9}\\,dx$?",
  "Solution_2": "Yes\r\n\r\nexcept it's -x^2 and -x\r\n\r\nx^3-x^2-x+1/(x^2+9)\r\n\r\nI then did long division and got x-1-(10x+10)/(x^2+9)\r\n\r\nI said that x^2+9 goes into x^3-x^2 X times. Then, after subtracting the first row from the top row I got -x^2-10x+1 and said that that went into the original equation -1 times. I then got -10x+10 as a remainder. Do you see the error?\r\n\r\nThank you",
  "Solution_3": "You're still not using enough parentheses- but I think it's actually meant to be $ \\int\\frac{x^{3}\\minus{}x^{2}\\plus{}x\\minus{}1}{x^{2}\\plus{}9}\\,dx.$\r\nThat would account for the difference between the 8 and the 10.",
  "Solution_4": "Well, to kill the remaining, make a long division, split the integral and use the following properties\r\n\r\n$ \\int\\frac{f'(x)}{f(x)}\\,dx\\equal{}\\ln|f(x)|\\plus{}k$ & $ \\int\\frac{1}{a^{2}\\plus{}u^{2}}\\,du\\equal{}\\frac{1}a\\arctan\\frac{u}{a}\\plus{}k$\r\n\r\nOf course, find them the constraints.",
  "Solution_5": "$ \\frac{x^{3}\\minus{}x^{2}\\plus{}x\\minus{}1}{x^{2}\\plus{}9}\\equal{}\\frac{(x\\minus{}1)(x^{2}\\plus{}1)}{x^{2}\\plus{}9}\\equal{}\\frac{(x\\minus{}1)(x^{2}\\plus{}1\\plus{}8\\minus{}8)}{x^{2}\\plus{}9}$\r\n\r\n$ \\equal{}\\frac{(x\\minus{}1)(x^{2}\\plus{}9)}{x^{2}\\plus{}9}\\minus{}\\frac{8(x\\minus{}1)}{x^{2}\\plus{}9}\\equal{} x\\minus{}1\\minus{}4\\frac{2x}{x^{2}\\plus{}9}\\plus{}\\frac{8}{3}\\frac{\\frac{1}{3}}{1\\plus{}\\left(\\frac{x}{3}\\right)^{2}}$.",
  "Solution_6": "Thank you very much everyone\r\n\r\nRegards"
}
{
  "Tag": [
    "probability"
  ],
  "Problem": "Let $ S$ be the set of all five-digit numbers such that the sum of their digits is 43. What is the probability that a number randomly selected from set $ S$ will be divisible by 11? Express your answer as a common fraction.",
  "Solution_1": "What is meant by the first set of numbers and the second set of numbers in the solution?",
  "Solution_2": "All the five-digit numbers with the sum of the digits 43 are either 99997 or 99988 or a permutation of those.  there are 3 ways to arrange them so that ther'e divisible by 11.  15 total, so 3/15=1/5"
}
{
  "Tag": [
    "function",
    "integration",
    "probability and stats"
  ],
  "Problem": "Let $ Y$ be a real random variable and its characteristic function is $ \\varphi_Y(t) \\equal{} e^{\\minus{}|t|^p}$, where $ p \\in (0, 2]$. Find all $ p$ such that $ \\mathbb{E}|Y| < \\infty$.",
  "Solution_1": "If $ E(|Y|)$ were finite (that is, $ Y\\in L^1$), then $ \\phi_Y$ would be in $ C^1$ - continuously differentiable. For these functions, the only question about differentiability would be at $ t\\equal{}0.$ Examining that, we see that $ e^{\\minus{}|t|^p}\\in C^1$ if and only if $ p>1.$\r\n\r\nThat's a necessary condition: is it sufficient? In this case, yes. Since $ \\phi_Y(t)$ decays rapidly at $ \\infty,$ the density $ f(y)$ is very smooth. That means the only issue will be the rate of decay as $ y\\to infty,$ and that should look like a power of $ y$ that depends in a predictable fashion on $ p.$\r\n\r\nNote that whenever the mean exists at all, we have $ E(Y)\\equal{}0.$\r\n\r\nThe borderline case is a failure. If $ p\\equal{}1,$ then $ \\phi_Y(t)\\equal{}e^{\\minus{}|t|}$ and the density of $ Y$ is $ \\frac1{\\pi(1\\plus{}y^2)}$ on $ (\\minus{}\\infty,\\infty)$ and $ E(|Y|)\\equal{}\\infty.$",
  "Solution_2": "Thanks for your reply and an idea how to deal with the case $ p \\in (1, 2]$ (the fact that it is necessay condition is clear). However, I can't do it. I only know that the density is (since $ \\varphi_Y$ is integrable) $ g(x) \\equal{} \\frac{1}{2\\pi}\\int_\\mathbb{R}e^{\\minus{}itx}e^{\\minus{}|t|^p}dt$. But how to prove that $ E|Y| \\equal{} \\int_\\mathbb{R} |y|g(y)dy < \\infty$. Could you post more hints?"
}
{
  "Tag": [
    "inequalities",
    "algebra",
    "polynomial",
    "inequalities proposed"
  ],
  "Problem": "Let $ x$, $ y$ and $ z$ be real numbers such that $ x^{2}\\plus{}y^{2}\\plus{}z^{2}\\equal{}1$. Prove that:\\[ 7(x^{4}\\plus{}y^{4}\\plus{}z^{4})\\plus{}8(x\\plus{}y\\plus{}z)(xyz)\\ge 4(xy\\plus{}yz\\plus{}zx)\\plus{}1\\]",
  "Solution_1": "Ok . Nice and easy problem .  :) \r\nThe inequality is equaivalent to $ 7\\sum x^{4}\\plus{}8xyz(x\\plus{}y\\plus{}z)\\minus{}4\\sum xy\\cdot\\sum x^{2}\\minus{}(\\sum x^{2})^{2}\\geq 0$ \r\nor $ 6(\\sum x^{4})\\plus{}4xyz(x\\plus{}y\\plus{}z)\\geq 4\\sum (xy(x^{2}\\plus{}y^{2}))\\plus{}2\\sum x^{2}y^{2}$ which true since \r\n\r\n$ \\sum x^{4}\\plus{}xyz(x\\plus{}y\\plus{}z)\\geq\\sum xy(x^{2}\\plus{}y^{2})$ from schur and the trivial $ x^{4}\\plus{}y^{4}\\plus{}z^{4}\\geq x^{2}y^{2}\\plus{}y^{2}z^{2}\\plus{}z^{2}x^{2}$",
  "Solution_2": "Shoot...I don't like when problems get 'schur-ed'...My solution was that the inequality was equivalent to:\r\n\r\n$ \\sum (x^{2}\\plus{}y^{2}\\minus{}xy\\minus{}z^{2})^{2}\\ge 0$\r\n\r\nI wonder if it is true that any symmetric inequality that is equivalent a sum of a bunch of squares of polynomials can be proven by schur/muirhead."
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "Prove that for all $a,b,c>0$ we have\r\n$abc(a+2c)(b+2a)(c+2b)\\le(a^{2}+2bc)(b^{2}+2ca)(c^{2}+2ab)$",
  "Solution_1": "It seems to me that\r\n\r\n$(a^{2}+2bc)(b^{2}+2ca)(c^{2}+2ab) \\ge (ab+2bc)(bc+2ca)(ca+2ab)$\r\n\r\nFollows from the well-known $a^{2}+b^{2}+c^{2}\\ge ab+bc+ca$ and a few variations (in other words, from Rearrangement.)",
  "Solution_2": "Don't quite see how it is rearrangement  :| \r\nAre you mistaking something with $a+b+c\\geq x+y+z\\implies abc\\geq xyz$? thats not true either...",
  "Solution_3": "[quote=\"t0rajir0u\"]It seems to me that\n\n$(a^{2}+2bc)(b^{2}+2ca)(c^{2}+2ab) \\ge (ab+2bc)(bc+2ca)(ca+2ab)$\n\nFollows from the well-known $a^{2}+b^{2}+c^{2}\\ge ab+bc+ca$ and a few variations (in other words, from Rearrangement.)[/quote]Very Good!",
  "Solution_4": "[quote=\"The soul of rock\"]Prove that for all $a,b,c>0$ we have\n$abc(a+2c)(b+2a)(c+2b)\\le(a^{2}+2bc)(b^{2}+2ca)(c^{2}+2ab)$[/quote]\r\nHere is my solution:\r\nLet $x=\\frac{a}{c},y=\\frac{c}{b},z=\\frac{b}{a}$, then $x,y,z>0,xyz=1$. The inequality becomes\r\n\\[(x+2)(y+2)(z+2) \\le \\left( \\frac{2x}{y}+1 \\right) \\left( \\frac{2y}{z}+1 \\right) \\left( \\frac{2z}{x}+1 \\right) \\]\r\n\r\n\\[\\Leftrightarrow 4\\sum{x}+2\\sum{xy}\\le 4\\sum\\frac{x}{z}+2\\sum\\frac{x}{y}\\]\r\nThis inequality follows from these inequalities\r\n\\[\\sum\\frac{x}{y}\\ge \\sum{x}\\tag{1}\\\\ \\sum\\frac{x}{z}\\ge \\sum{xy}\\tag{2}\\]\r\nYou only need to use AM - GM Inequality to prove.\r\n\r\nWe are done!\r\n\r\nHave fun! :)",
  "Solution_5": "Wow, a nine proof. But, please send your solution for two inequalities",
  "Solution_6": "[quote=\"me@home\"]Don't quite see how it is rearrangement  :| \nAre you mistaking something with $a+b+c\\geq x+y+z\\implies abc\\geq xyz$? thats not true either...[/quote]\r\nIt occurs when $(x,y,z)\\succeq(a,b,c)$ and $x,y,z,a,b,c>0$ which follow by Karamata's inequality\r\n@toanhocmuonmau: Nice proof",
  "Solution_7": "[quote=\"me@home\"]Don't quite see how it is rearrangement  :| [/quote]\r\n\r\nNearly all inequalities of this form can be solved by expanding them out and using Rearrangement  :rotfl:"
}
{
  "Tag": [
    "induction",
    "strong induction"
  ],
  "Problem": "Find all permutations of 2, 3, 4, ... , 102 in which the nth number is divisible by n for each n.",
  "Solution_1": "[hide][b]Claim:[/b] the only such permutation is $ 2,3,...,102$, i.e., the identity.\nLet $ \\sigma$ be a permutation on $ 2,3,...,102$. Denote the number of prime factors, counting multiplicity, of $ n$ as $ s(n)$. We prove our claim by strong induction on $ s(n)$.\n\n[u]Base:[/u] $ s(n) \\equal{} 1$. Since $ \\sigma(n)|n$, $ \\sigma(n) \\equal{} 1, n$. But $ 1$ is not an element. Therefore $ \\sigma(n) \\equal{} n$.\n[u]Inductive Step:[/u] suppose it holds for $ s(n) \\equal{} 1,...,k \\minus{} 1$. If $ s(n) \\equal{} k$, for any $ d|n$ with $ s(d) < k$ and $ d\\neq 1$, we have $ \\sigma(d) \\equal{} d < n$. Since $ \\sigma(n)|n$, $ \\sigma(n) \\equal{} 1$ or $ s(\\sigma(n)) \\equal{} k$. But $ 1$ is not an element. Therefore $ s(\\sigma(n)) \\equal{} k$, $ \\sigma(n) \\equal{} n$. QED\n[/hide]\r\n\r\nEdited: I add my claim to clarify my intent. Correct?\r\n\r\nNevermind. The above is remained as a testament for my stupidity.",
  "Solution_2": "I don't understand what you are proving. :maybe:  The problem asks you to find all permutations, not only to find the number of permutations.",
  "Solution_3": "[b]@ timwu:[/b] The numbers are intentionally offset by 1...\r\n\r\n[hide]Partition the set into two subsets: divisors of 102 and nondivisors of 102. Non-divisors must be in the position of their own number. Now figure out what to do with the divisors. For example, 102, 2, 3, 4, ... 101 works. You can also switch 102 with any of its divisors, and it'll still work. Say you switched it with 51. Then you can in turn switch 51 with one of its divisors and it will still work, etc. Basically, it's the number of ways to arrange the divisors.\n\nA useful approach is to start from the 101st position and work backwards. Each time, there are only two numbers greater than or equal to the position number of the position you want to fill: that number, and 102. We can obviously only use 102 in a position that is a divisor of 102. Whichever number we displace goes into the first position because any number is divisible by 1.[/hide]",
  "Solution_4": "by the way, the original formulation of this problem asks you to find the number of permutations. :D \r\n\r\nps: i'm sorry that i've posted this twice. i forgot that i've posted this. :oops:",
  "Solution_5": "[hide=\"Solution\"]There are a total of 101 numbers. The 52nd through 101st numbers are 52-101, respectively. So we have positions 1-51, with numbers 2-51, and 102.\n\n102 can only go into spots which are divisors of 102, and those are 1, 2, 3, 6, 17, 34, and 51. Likewise, the divisors of 102 can only take up those spots. Now we take the first spot that isn't taken up by a divisor of 102: 50. The only number that can go there is 50. The next spot can only be filled with 49, etcetera, and we skip all divisors of 102. Thus the $ n$th spot is filled with $ n$ for all numbers that are not divisors of 102. Thus the only permutation happens in spaces 1, 2, 3, 6, 17, 34, and 51.\n\nNow the number of the space that 102 fills is either 1, a prime, or a product of two distinct primes. If 102 fills the first spot, then 51 fills spot 51, 34 fills spot 34, and so on. That is one case. If 102 fills a prime spot, the prime must go in spot 1. That is three more cases. If 102 fills a product of primes, the product of primes can only go in one of the two prime slots, and the prime-numbered slot that it goes in must be in the first slot. That is 6 cases.\n\n1+3+6=10[/hide]\r\n\r\nForgive any un-rigor, this is my first combinatorics solution in the pre-olympiad section."
}
{
  "Tag": [
    "function",
    "induction",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "Let $f$ be a function from the set of positive integers to the set of non-negative integers such that $f(1)=0$ and for all $n \\geq 2$ : $f(n)= \\max \\{f(i)+f(n-i)+i\\}$\r\nDetermine $f(2005)$",
  "Solution_1": "$f(n)=\\frac{n(n-1)}{2}$ by induction:\r\n\r\n$f(1) = 0 = \\frac{1 \\cdot 0}{2}$ is obviously true.\r\n\r\nNow assume the above formula is true for all $i \\in {1, 2, ..., n-1}$. Then\r\n\r\n\\begin{eqnarray*}\r\n{max \\left(f(i) + f(n-i) + i\\right)&=&max \\left(\\frac{i(i-1)}{2} + \\frac{(n-i)(n-i-1)}{2} + i\\right)\\\\\r\n&=&max \\left(\\left(i-\\frac{n-1}{2}\\right)^2-\\left(\\frac{n-1}{2}\\right)^2+\\frac{n^2-n}{2}\\right) = \\\\\r\n&=&max \\left(\\left(n-1-\\frac{n-1}{2}\\right)^2-\\left(\\frac{n-1}{2}\\right)^2+\\frac{n^2-n}{2}\\right) = \\frac{n(n-1)}{2}\r\n}\\end{eqnarray*}.",
  "Solution_2": "I think that the real problem comes from Taiwan 2000, and the condition is max for $i$ between $1$ and $n/2$. In this case the answer is much more cool, and the problem is less trivial than it looked without it."
}
{
  "Tag": [
    "symmetry"
  ],
  "Problem": "Let $ABC$ be a triangle. For a fixed point $D\\in (BC)$ construct the position of the point $L\\in AD$ for which $\\widehat {DLB}\\equiv \\widehat {DLC}\\ .$",
  "Solution_1": "[b]Construction.[/b] Find the point $B'$ such that $B$ and $B'$ are symmetric about the line $AD$. The desired point $L$ is the intersection point of the lines $AD$ and $CB'.$\r\n\r\n[b]Proof.[/b] Because of the symmetry, we have $\\angle DLB=\\angle DLB'$ (both $D$ and $L$ are on the symmetrial axis), but as the points $C, B', L$ are collinear, $\\angle DLB'=\\angle DLC\\implies \\angle DLB=\\angle DLC$.\r\n\r\n[b]Discussion.[/b]\r\n1) If $B'\\equiv C$, any line passing through $C$ gives a valid solution for $L$, hence in that case there are infinitely many solutions. That will happen if and only if $\\triangle ABC$ is an isosceles with $AD$ being its symmetrial axis, and any point of the line $AD$ satisfies the given condition.\r\n2) If $CB'$ is parallel to $AD$, there will be no solution. In that case let $M=BB'\\cap AD$. It's obvious that $\\triangle BB'C$ and $\\triangle BMD$ are both right, therefore $M$ being the midpoint of $BB'$ and $DM\\parallel CB'$ imply that $D$ is the midpoint of $BC$. We conclude that if $AD$ is the median of $\\triangle ABC$ where $AB\\neq AC$, there's no solution.\r\n3) In all the other cases the line $CB'$ is uniquely determined and intersects the line $AD$, hence we'll have a unique solution for $L.$\r\n\r\n[b]Note.[/b] Obviously, the same result is obtained if we find $C'$ symmetric to $C$ and then intersect the line $BC'$ with the line $AD.$",
  "Solution_2": "The case $(b=c)\\ \\vee\\ (DB=DC)$ is evidently. Indeed, if $b=c$ then there are a $\\infty$ solutions (the all points of the line $AD$) and if $b\\ne c$ and $DB=DC$ then there is no solution ($\\emptyset$). Suppose $(b\\ne c)\\ \\wedge\\ (DB\\ne DC)$. Construct the harmonical conjugate point $E$ of the point $D$ w.r.t. the points $B$ and $C$. Then the appreciated point $L$ is the projection of the point $E$ on the line $AD\\ !$",
  "Solution_3": "Virgil, are you saying that my construction is wrong, or you're just offering another solution?",
  "Solution_4": "No, the your solution is clearly and shorterly !\r\nYou must construct only a symmetrical point of the point w.r.t. a line.\r\n\r\nI presented a another more pretentious solution !\r\nI must construct only a perpendicular from a point on a line \r\n(the construction of the conjugate is made without compasses).\r\n\r\nUnderstood it you ? $\\mathrm {Score\\ : \\ 1\\ -\\ 1\\ .}$ Sorry, I joke !",
  "Solution_5": "done similarly to Farenhajt , drawing the perpendicular from B to AD and trying to find the circumference with center L using simmetry",
  "Solution_6": "[quote=\"Virgil Nicula\"]No, the your solution is clearly and shorterly !\nYou must construct only a symmetrical point of the point w.r.t. a line.\n\nI presented a another more pretentious solution !\nI must construct only a perpendicular from a point on a line \n(the construction of the conjugate is made without compasses).\n\nUnderstood it you ? $\\mathrm {Score\\ : \\ 1\\ -\\ 1\\ .}$ Sorry, I joke ![/quote]\r\n\r\nYes, Virgil, all clear :) Thanks"
}
{
  "Tag": [
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Find all pairs $ (a,b)$ of positive integers such: $ a^{{b}^{2}}=b^{a}$",
  "Solution_1": "[quote=\"santosguzella\"]Find all pairs $ (a,b)$ of positive integers such: $ a^{{b}^{2}}=b^{a}$[/quote]\r\n\r\nFrom  $ a^{b^{2}}=b^{a}$, we get  $ a=b^{\\frac{a}{b^{2}}}$ and so $ a=b^{\\frac{p}{q}}$ for some positive coprime integers $ p$ and $ q$\r\n\r\nSo $ b^{\\frac{p}{q}b^{2}}=b^{b^{\\frac{p}{q}}}$ and so $ \\frac{p}{q}b^{2}=b^{\\frac{p}{q}}$  or $ \\frac{p}{q}=b^{\\frac{p-2q}{q}}$\r\n\r\nSince RHS is a rational number, we need $ b=c^{q}$ for some positive integer $ c$ and so $ \\frac{p}{q}=c^{p-2q}$ and so either $ p=1$ or $ q=1$\r\n\r\n$ 1).$ Case $ p=1$ : $ q=c^{2q-1}$ which gives immediately :\r\n$ q=1$ and $ c=1$\r\nAnd if $ q>1$, either $ c=1$ and equality is wrong, either $ c>1$ and $ c^{2q-1}>q$\r\n\r\n$ 2).$ Case $ q=1$ : $ p=c^{p-2}$ which gives immediately :\r\n$ c=1$ $ \\implies$ $ p=1$\r\n$ c=2$ $ \\implies$ $ p=4$\r\n$ c=3$ $ \\implies$ $ p=3$\r\n$ c>3$ $ \\implies$ $ c^{-1}\\neq 1$, $ c^{0}\\neq 2$, $ c^{1}\\neq 3$ and $ c^{p-2}>p$ $ \\forall p>3$\r\n\r\nSo we just have three different solutions :\r\n$ p=1$, $ q=1$, $ c=1$ $ \\implies$ $ a=1$ and $ b=1$ : $ 1^{1^{2}}=1^{1}$\r\n$ p=4$, $ q=1$, $ c=2$ $ \\implies$ $ a=16$ and $ b=2$ : $ 16^{2^{2}}=2^{16}$\r\n$ p=3$, $ q=1$, $ c=3$ $ \\implies$ $ a=27$ and $ b=3$ : $ 27^{3^{2}}=3^{27}$",
  "Solution_2": "Problem 5 of IMO 1997\r\n[url]http://www.mathlinks.ro/resources.php?c=1&cid=16&year=1997[/url]"
}
{
  "Tag": [],
  "Problem": "hi, i'm trying to self-study AP psych for this year.  I known I only have like 6 weeks but besides review book, what is a good text book?  any recommendations?\r\n\r\n\r\nThanks in advance",
  "Solution_1": "My previous high school used [i]Psychology[/i] by Peter Gray for the AP course."
}
{
  "Tag": [
    "calculus",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "let $f: R\\to R$ such that \r\n     $|f(x+y)-f(x)-f(y)|\\leq 1$\r\n prove that there exist $g: R\\to R$ st\r\n      $|g(x)-f(x)|\\leq 1$ and $g(x+y)=g(x)+g(y)$",
  "Solution_1": "what do you think of  it?\r\nmaybe it is very hard ;)",
  "Solution_2": "This is a calculus problem.",
  "Solution_3": "could you post your solution :?:"
}
{
  "Tag": [
    "search",
    "inequalities proposed",
    "inequalities"
  ],
  "Problem": "Let $ a,b,c$ be positive real numbers, and let $ x \\equal{} \\frac {2a}{b \\plus{} c}$, $ y \\equal{} \\frac {2b}{c \\plus{} a}$, $ z \\equal{} \\frac {2c}{a \\plus{} b}$. Find the greatest value of $ p$ such that \r\n$ \\frac 1{x^2} \\plus{} \\frac 1{y^2} \\plus{} \\frac 1{z^2} \\minus{} 3 \\ge p(x \\plus{} y \\plus{} z \\minus{} 3)$",
  "Solution_1": "[quote=\"Vasc\"]Let $ a,b,c$ be positive real numbers, and let $ x \\equal{} \\frac {2a}{b \\plus{} c}$, $ y \\equal{} \\frac {2b}{c \\plus{} a}$, $ z \\equal{} \\frac {2c}{a \\plus{} b}$. Find the greatest value of $ p$ such that \n$ \\frac 1{x^2} \\plus{} \\frac 1{y^2} \\plus{} \\frac 1{z^2} \\minus{} 3 \\ge p(x \\plus{} y \\plus{} z \\minus{} 3)$[/quote]\r\nLet $ a\\equal{}b\\equal{}1,c\\equal{}\\sqrt{3}\\plus{}1$, we have $ p \\le \\frac{5}{2}\\plus{}\\frac{3\\sqrt{3}}{2}$, we can prove it by using SOS or LCF-RCF method. :)",
  "Solution_2": "[quote=\"can_hang2007\"][quote=\"Vasc\"]Let $ a,b,c$ be positive real numbers, and let $ x \\equal{} \\frac {2a}{b \\plus{} c}$, $ y \\equal{} \\frac {2b}{c \\plus{} a}$, $ z \\equal{} \\frac {2c}{a \\plus{} b}$. Find the greatest value of $ p$ such that \n$ \\frac 1{x^2} \\plus{} \\frac 1{y^2} \\plus{} \\frac 1{z^2} \\minus{} 3 \\ge p(x \\plus{} y \\plus{} z \\minus{} 3)$[/quote]\nLet $ a \\equal{} b \\equal{} 1,c \\equal{} \\sqrt {3} \\plus{} 1$, we have $ p \\le \\frac {5}{2} \\plus{} \\frac {3\\sqrt {3}}{2}$, we can prove it by using SOS or LCF-RCF method. :)[/quote]\r\nWhy are you know $ a\\equal{}b\\equal{}1,c\\equal{}\\sqrt{3}\\plus{}1$?\r\nWhat's LCF - RCF method?\r\nPlease, !\r\nThanks!",
  "Solution_3": "[quote=\"can_hang2007\"] Let $ a \\equal{} b \\equal{} 1,c \\equal{} \\sqrt {3} \\plus{} 1$, we have $ p \\le \\frac {5}{2} \\plus{} \\frac {3\\sqrt {3}}{2}$, we can prove it by using SOS or LCF-RCF method. :)[/quote]\r\nThis is, indeed. :wink:",
  "Solution_4": "[quote=\"vipCD_A1\"][quote=\"can_hang2007\"][quote=\"Vasc\"]Let $ a,b,c$ be positive real numbers, and let $ x \\equal{} \\frac {2a}{b \\plus{} c}$, $ y \\equal{} \\frac {2b}{c \\plus{} a}$, $ z \\equal{} \\frac {2c}{a \\plus{} b}$. Find the greatest value of $ p$ such that \n$ \\frac 1{x^2} \\plus{} \\frac 1{y^2} \\plus{} \\frac 1{z^2} \\minus{} 3 \\ge p(x \\plus{} y \\plus{} z \\minus{} 3)$[/quote]\nLet $ a \\equal{} b \\equal{} 1,c \\equal{} \\sqrt {3} \\plus{} 1$, we have $ p \\le \\frac {5}{2} \\plus{} \\frac {3\\sqrt {3}}{2}$, we can prove it by using SOS or LCF-RCF method. :)[/quote]\nWhy are you know $ a \\equal{} b \\equal{} 1,c \\equal{} \\sqrt {3} \\plus{} 1$?\nWhat's LCF - RCF method?\nPlease, !\nThanks![/quote]\r\n[color=green]Have look at Vasc'book or Search in the foum.[/color]"
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "Let $ x_1,x_2,\\cdots,x_n,k$are positive numbers. prove that\r\n$ \\frac {1}{(x_1 \\plus{} k)^n} \\plus{} \\frac {1}{(x_2 \\plus{} k)^n} \\plus{} \\cdots \\plus{} \\frac {1}{(x_n \\plus{} k)^n}\\ge\\frac {n}{2^{n \\minus{} 1}(x_1x_2\\cdots x_n \\plus{} k^n)}$\r\n\r\nIt is a generalization of $ \\frac {1}{(x \\plus{} z)^2} \\plus{} \\frac {1}{(y \\plus{} z)^2}\\ge\\frac {1}{xy \\plus{} z^2}$.\r\nBut I think this inequality is so weak. Do you have idea which a nice(and strong) generalization of it?\r\nI am really want to know more nice generalization.  :blush:",
  "Solution_1": "[quote=\"NextPeace\"]Let $ x_1,x_2,\\cdots,x_n,k$are positive numbers. prove that\n$ \\frac {1}{(x_1 \\plus{} k)^n} \\plus{} \\frac {1}{(x_2 \\plus{} k)^n} \\plus{} \\cdots \\plus{} \\frac {1}{(x_n \\plus{} k)^n}\\ge\\frac {n}{2^{n \\minus{} 1}(x_1x_2\\cdots x_n \\plus{} k^n)}$\n\nIt is a generalization of $ \\frac {1}{(x \\plus{} z)^2} \\plus{} \\frac {1}{(y \\plus{} z)^2}\\ge\\frac {1}{xy \\plus{} z^2}$.\nBut I think this inequality is so weak. Do you have idea which a nice(and strong) generalization of it?\nI am really want to know more nice generalization.  :blush:[/quote]\r\nNice inequality, NextPeace. This inequality is a nice corollary of my old result:\r\n\\[ \\sum^{n}_{i \\equal{} 1} \\frac {x^{n \\minus{} 1}_{i \\equal{} 1}}{\\prod_{j\\ne i} (x_i \\plus{} x_j)} \\ge \\frac {n}{2^{n\\minus{}1}}.\r\n\\]\r\nThis inequality is my generalization for the PHONGVAN's Inequality:\r\n\\[ \\frac {a^3}{(a \\plus{} b)(a \\plus{} c)(a \\plus{} d)} \\plus{} \\frac {b^3}{(b \\plus{} c)(b \\plus{} d)(b \\plus{} a)} \\plus{} \\frac {c^3}{(c \\plus{} d)(c \\plus{} a)(c \\plus{} b)} \\plus{} \\frac {d^3}{(d \\plus{} a)(d \\plus{} b)(d \\plus{} c)} \\ge \\frac {1}{2},\r\n\\]\r\nfor $ a,b,c,d > 0.$\r\n:)",
  "Solution_2": "$ x_{i} \\equal{} e^{y_{i}}$\ub77c\uace0 \ud558\uba74 \uacf1\uc774 \ud569\uc774\ub418\uace0\r\n$ k \\equal{} e^{t}$\uc778 $ t$\uac00 \uc874\uc7ac\ud558\uace0, \uc800\ub7f0\uc2dd\uc73c\ub85c \uce58\ud658\ud558\uba74 \uc820\uc13c\uc73c\ub85c \uc548\ub418\ub098?\r\n\r\n\ud760-_-;\uc548\ud574\ubd10\uc11c \ub420\uc9c0\ub294...",
  "Solution_3": "[quote=\"can_hang2007\"][quote=\"NextPeace\"]Let $ x_1,x_2,\\cdots,x_n,k$are positive numbers. prove that\n$ \\frac {1}{(x_1 \\plus{} k)^n} \\plus{} \\frac {1}{(x_2 \\plus{} k)^n} \\plus{} \\cdots \\plus{} \\frac {1}{(x_n \\plus{} k)^n}\\ge\\frac {n}{2^{n \\minus{} 1}(x_1x_2\\cdots x_n \\plus{} k^n)}$\n\nIt is a generalization of $ \\frac {1}{(x \\plus{} z)^2} \\plus{} \\frac {1}{(y \\plus{} z)^2}\\ge\\frac {1}{xy \\plus{} z^2}$.\nBut I think this inequality is so weak. Do you have idea which a nice(and strong) generalization of it?\nI am really want to know more nice generalization.  :blush:[/quote]\nNice inequality, NextPeace. This inequality is a nice corollary of my old result:\n\\[ \\sum^{n}_{i \\equal{} 1} \\frac {x^{n \\minus{} 1}_{i \\equal{} 1}}{\\prod_{j\\ne i} (x_i \\plus{} x_j)} \\ge \\frac {n}{2^n}.\n\\]\nThis inequality is my generalization for the PHONGVAN's Inequality:\n\\[ \\frac {a^3}{(a \\plus{} b)(a \\plus{} c)(a \\plus{} d)} \\plus{} \\frac {b^3}{(b \\plus{} c)(b \\plus{} d)(b \\plus{} a)} \\plus{} \\frac {c^3}{(c \\plus{} d)(c \\plus{} a)(c \\plus{} b)} \\plus{} \\frac {d^3}{(d \\plus{} a)(d \\plus{} b)(d \\plus{} c)} \\ge \\frac {1}{2},\n\\]\nfor $ a,b,c,d > 0.$\n:)[/quote]\r\n\r\nYes, can_hang. My solution is same with you  :)\r\n\r\n*oh, sorry, I have a mistake :oops: . My solution is not same with you. Can you post your solution, can_hang?\r\n\r\nTo. Heebeen, Yang\r\n\ud574\ubcf4\uc138\uc694 \u314b\u314b",
  "Solution_4": "[quote=\"NextPeace\"][quote=\"can_hang2007\"][quote=\"NextPeace\"]Let $ x_1,x_2,\\cdots,x_n,k$are positive numbers. prove that\n$ \\frac {1}{(x_1 \\plus{} k)^n} \\plus{} \\frac {1}{(x_2 \\plus{} k)^n} \\plus{} \\cdots \\plus{} \\frac {1}{(x_n \\plus{} k)^n}\\ge\\frac {n}{2^{n \\minus{} 1}(x_1x_2\\cdots x_n \\plus{} k^n)}$\n\nIt is a generalization of $ \\frac {1}{(x \\plus{} z)^2} \\plus{} \\frac {1}{(y \\plus{} z)^2}\\ge\\frac {1}{xy \\plus{} z^2}$.\nBut I think this inequality is so weak. Do you have idea which a nice(and strong) generalization of it?\nI am really want to know more nice generalization.  :blush:[/quote]\nNice inequality, NextPeace. This inequality is a nice corollary of my old result:\n\\[ \\sum^{n}_{i \\equal{} 1} \\frac {x^{n \\minus{} 1}_{i \\equal{} 1}}{\\prod_{j\\ne i} (x_i \\plus{} x_j)} \\ge \\frac {n}{2^n}.\n\\]\nThis inequality is my generalization for the PHONGVAN's Inequality:\n\\[ \\frac {a^3}{(a \\plus{} b)(a \\plus{} c)(a \\plus{} d)} \\plus{} \\frac {b^3}{(b \\plus{} c)(b \\plus{} d)(b \\plus{} a)} \\plus{} \\frac {c^3}{(c \\plus{} d)(c \\plus{} a)(c \\plus{} b)} \\plus{} \\frac {d^3}{(d \\plus{} a)(d \\plus{} b)(d \\plus{} c)} \\ge \\frac {1}{2},\n\\]\nfor $ a,b,c,d > 0.$\n:)[/quote]\n\nYes, can_hang. My solution is same with you  :)\n\n*oh, sorry, I have a mistake :oops: . My solution is not same with you. Can you post your solution, can_hang?\n\nTo. Heebeen, Yang\n\ud574\ubcf4\uc138\uc694 \u314b\u314b[/quote]\r\nMy solution is the followiong:\r\nFor any $ 1 \\le i \\le n$, the Holder Inequality yields\r\n\\[ (x_i\\plus{}k)^n \\equal{}\\left( \\sqrt[n]{x_1x_2\\cdots x_n} \\prod_{j \\ne i} \\sqrt[n]{\\frac{x_i}{x_j}} \\plus{}\\sqrt[n]{k^n} \\prod_{j \\ne i}  \\sqrt[n]{1}\\right)^n \\le (x_1x_2\\cdots x_n \\plus{}k^n) \\prod_{j \\ne i} \\left( \\frac{x_i}{x_j}\\plus{}1\\right) .\\]\r\nUsing this inequality, we can reduce our inequality to\r\n\\[ \\sum^{n}_{i\\equal{}1} \\frac{1}{\\prod_{j \\ne i}\\left( \\frac{x_i}{x_j}\\plus{}1\\right)} \\ge \\frac{n}{2^{n\\minus{}1}}.\\]\r\nSetting $ y_i\\equal{}\\frac{1}{x_i},$ this inequality becomes\r\n\\[ \\sum^{n}_{i\\equal{}1} \\frac{y^{n\\minus{}1}_{i}}{\\prod_{j \\ne i} (y_i\\plus{}y_j)} \\ge \\frac{n}{2^{n\\minus{}1}},\\]\r\nwhich is just the inequality I have posted. :)\r\n\r\nP/s: I have mistake when typing my inequality. It should be $ \\frac{n}{2^{n\\minus{}1}}$, I have edited it.",
  "Solution_5": "[quote=\"can_hang2007\"][/quote][quote=\"NextPeace\"][quote=\"can_hang2007\"][quote=\"NextPeace\"]Let $ x_1,x_2,\\cdots,x_n,k$are positive numbers. prove that\n$ \\frac {1}{(x_1 + k)^n} + \\frac {1}{(x_2 + k)^n} + \\cdots + \\frac {1}{(x_n + k)^n}\\ge\\frac {n}{2^{n - 1}(x_1x_2\\cdots x_n + k^n)}$\n\nIt is a generalization of $ \\frac {1}{(x + z)^2} + \\frac {1}{(y + z)^2}\\ge\\frac {1}{xy + z^2}$.\nBut I think this inequality is so weak. Do you have idea which a nice(and strong) generalization of it?\nI am really want to know more nice generalization.  :blush:[/quote]\nNice inequality, NextPeace. This inequality is a nice corollary of my old result:\n\\[ \\sum^{n}_{i = 1} \\frac {x^{n - 1}_{i = 1}}{\\prod_{j\\ne i} (x_i + x_j)} \\ge \\frac {n}{2^n}.\n\\]\nThis inequality is my generalization for the PHONGVAN's Inequality:\n\\[ \\frac {a^3}{(a + b)(a + c)(a + d)} + \\frac {b^3}{(b + c)(b + d)(b + a)} + \\frac {c^3}{(c + d)(c + a)(c + b)} + \\frac {d^3}{(d + a)(d + b)(d + c)} \\ge \\frac {1}{2},\n\\]\nfor $ a,b,c,d > 0.$\n:)[/quote]\n\nYes, can_hang. My solution is same with you  :)\n\n*oh, sorry, I have a mistake :oops: . My solution is not same with you. Can you post your solution, can_hang?\n\nTo. Heebeen, Yang\n\ud574\ubcf4\uc138\uc694 \u314b\u314b[/quote][quote=\"can_hang2007\"]\nMy solution is the followiong:\nFor any $ 1 \\le i \\le n$, the Holder Inequality yields\n\\[ (x_i + k)^n = \\left( \\sqrt [n]{x_1x_2\\cdots x_n} \\prod_{j \\ne i} \\sqrt [n]{\\frac {x_i}{x_j}} + \\sqrt [n]{k^n} \\prod_{j \\ne i} \\sqrt [n]{1}\\right)^n \\le (x_1x_2\\cdots x_n + k^n) \\prod_{j \\ne i} \\left( \\frac {x_i}{x_j} + 1\\right) .\n\\]\nUsing this inequality, we can reduce our inequality to\n\\[ \\sum^{n}_{i = 1} \\frac {1}{\\prod_{j \\ne i}\\left( \\frac {x_i}{x_j} + 1\\right)} \\ge \\frac {n}{2^{n - 1}}.\n\\]\nSetting $ y_i = \\frac {1}{x_i},$ this inequality becomes\n\\[ \\sum^{n}_{i = 1} \\frac {y^{n - 1}_{i}}{\\prod_{j \\ne i} (y_i + y_j)} \\ge \\frac {n}{2^{n - 1}},\n\\]\nwhich is just the inequality I have posted. :)\n\nP/s: I have mistake when typing my inequality. It should be $ \\frac {n}{2^{n - 1}}$, I have edited it.[/quote]\r\n\r\nThanks for your posting, can_hang  :lol: \r\nMy proof is similar with yours, but I was proof \r\n$ sum^{n}_{i = 1} \\frac {1}{\\prod_{j \\ne i}\\left( \\frac {x_i}{x_j} + 1\\right)} \\ge \\frac {n}{2^{n - 1}}$ directly(with out transposition).\r\nBut your transposition is very nice! Thanks  :)"
}
{
  "Tag": [
    "AMC",
    "AIME",
    "email",
    "AMC 12",
    "AMC 12 B",
    "probability",
    "AIME II"
  ],
  "Problem": "Hey y'all,\r\n\r\nI was just wondering if these e-mails should have already been received by whatever math teacher was designated when they signed up for the AMC. The reason why I'm concerned is that last year two students that had qualified for the AIME weren't notified until after the test was administered.\r\n\r\nAlso, what steps should I (or can I?) take if the e-mails should have already arrived but the math teacher hasn't received them?\r\n\r\nThanks!",
  "Solution_1": "Yes, the teacher in charge of the AMC at your school should already know your score and therefore, whether or not you qualified for AIME (you should already know your unofficial score if you copied your answers onto the test booklet). If noone at your school knows, it probably is not even administering the AIME, but it wouldn't hurt to email the AMC director.",
  "Solution_2": "Well, it won't hurt, but it won't do any good, because it is much easier, much more sensible,  and more polite to ask your school contest manager.  Your school contest manager has a report (in most cases) on about 30-60 students, while I have about 250,000.\r\n\r\nI will delete any such requests, you have much better avenues for finding out your score.\r\n\r\nSteve Dunbar\r\nDirector, American Mathematics Competitions",
  "Solution_3": "Yes, just to clear things up, I didn't intend for any discourtesy  :blush: . I meant that you should email the AMC only if no one at your school knew anything, which isn't very likely. Sorry if I was unclear.",
  "Solution_4": "Well, we've been having technological issues at our school recently (overload caused all of the information that teachers had on their computers to be wiped and the internet wasn't working for a while). I'm in constant contact with our school AMC coordinator (because I'm heavily involved in the math team and she's the sponsor for that) so I'm assuming that if she had gotten the e-mail, she'd have reminded me (and I prescored my AMC12B as a 109.5, so yeah..), but I'll ask her tomorrow if she can check. The problem I think lies in the fact that she can only check her e-mail at school and I'm not quite sure if the internet has been fixed, yet.",
  "Solution_5": "The postal mail follow-up, with the AIME contests and a computer print-out (identical to the information in the email) should be in the possession of your teacher.  We do not rely solely on email.  Ask, your teacher has the information.\r\n\r\nSteve Dunbar\r\nAmerican Mathematics Competitions",
  "Solution_6": "Mk, like I expected, she said that she hadn't gotten the information yet (had a math competition yesterday, so I asked her on the way there).\r\n\r\nSo.. what do I do?",
  "Solution_7": "Bump...\r\n\r\nWhat a bummer if I can't take the AIME...",
  "Solution_8": "In order to help you, we need some basic information, for instance the school contest manager, your school name, your school CEEB, your city and state, whether your school took the A-date contests or the B-date contests, etc.\r\n\r\nHave your teacher call the AMC office,  we have an 800 number published on everything we print.  Check on the results for your school, check on the package tracking if available, check to see who signed for the package when it arrived, and we can work from there.  \r\n\r\nThe AIME Alternate date is still available.  \r\n\r\nSteve Dunbar\r\nDirector, American Mathematics Competitions",
  "Solution_9": "Ok, well she hasn't been at school for the last 3 days (according to my current math teacher, the math competition took a lot out of her, and the death of her mother recently isn't helping, I'm sure) and I don't know what to do. I know that you seem to be against students finding out what they got, but is there any way that I could find out if I qualified for the AIME. I mean, the probability that I will be able to take the AIME II is slim b/c I think we'll be on spring break, but I just want to know.\r\n\r\nHonestly, there should be a way for students to find out. Not everyone has a coordinator that's up to date (like I said, last year she caused 2 people to not be able to take the AIME, and it looks as if I'll be in the same position this year).\r\n\r\nOh well, it's life, right?",
  "Solution_10": "I suggest you talk to:\r\n1.  another math or science teacher in the school, perhaps the department chair\r\n2.  a principal or vice-principal,\r\n3.  a counselor.\r\n\r\nI believe that somewhere in your school the package with the results is waiting to be opened.\r\n\r\nIf you identify your school, your school CEEB number or some other identifier we can check our records to determine when and to whom that package was delivered.  Without that information, I cannot help you determine your scores.\r\n\r\nSteve Dunbar\r\nDirector, American Mathematics Competitions"
}
{
  "Tag": [
    "logarithms",
    "limit",
    "trigonometry",
    "calculus",
    "derivative",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "\\[ \\frac{{(1 \\plus{} x)^{\\frac{1}{x}}  \\minus{} e}}{x}\r\n\\]\r\n\r\nCalculate without derivates",
  "Solution_1": "You can use $ (1\\plus{}x)^\\frac{1}{x}\\equal{}e\\minus{}\\frac{e}{2}x\\plus{}\\frac{11e}{24}x^2\\plus{}...$, so\r\n$ \\frac{(1\\plus{}x)^\\frac{1}{x}\\minus{}e}{x}\\equal{}\\frac{\\minus{}\\frac{e}{2}x\\plus{}...}{x}\\to \\minus{}\\frac{e}{2}$ for $ x\\to 0$\r\n\r\nBut to calculate the first expansion, you actually use derivatives.",
  "Solution_2": "thanks\r\ni`m still waiting for a method to do it without derivates",
  "Solution_3": "i make it easier a little, if i didn`t missed any sign or number it should look like this :\r\n\r\n\\[ x \\cdot \\frac{{\\ln (1 \\plus{} x) \\minus{} x}}{{x^2 }}\r\n\\]",
  "Solution_4": "If you are allowed to use the facts $ e^x = 1+x+\\mathcal{O}\\left(x^2\\right)$ and $ \\log (1 + x) = x-\\frac{x^2}{2}+\\mathcal{O}\\left(x^3\\right)$ then,\r\n\r\n\\begin{eqnarray*} \\lim_{x\\to 0}\\frac {(1 + x)^{\\frac {1}{x}} - e}{x} & = & \\lim_{x\\to 0}\\frac {e\\left(e^{\\frac {\\log (1 + x)}{x} - 1} - 1\\right)}{x} \\\\\r\n& = & \\lim_{x\\to 0}\\frac {e}{x}\\left(e^{ - \\frac {x}{2} + \\mathcal{O}\\left(x^2\\right)} - 1\\right) \\\\\r\n& = & \\lim_{x\\to 0}\\frac {e}{x}\\left( - \\frac {x}{2} + \\mathcal{O}\\left(x^2\\right)\\right) \\\\\r\n& = & - \\frac {e}{2} \\end{eqnarray*}",
  "Solution_5": "no, i can`t use it :|\r\nyou started using derivates near the end",
  "Solution_6": "No, I didn't differentiate anything; I used only the assumptions stated at the top.  Well if you aren't allowed to use those two facts, both can be proven from definitions without using any calculus.  But what's the point of reinventing the wheel?",
  "Solution_7": "[quote=\"JoeBlow\"]But what's the point of reinventing the wheel?[/quote]\r\nI'm in agreement with JoeBlow on this. In both this topic, and to an even greater extent on night's earlier topic, I've found the request to do it \"without using derivates\" to be entirely artificial and perhaps pointless.\r\n\r\nPut another way: any method whatsoever for finding $ \\lim_{x\\to0}\\frac{\\sin x\\minus{}x}{x^3}$ has in effect computed the third derivative of the sine at zero; it doesn't matter what methods were used.",
  "Solution_8": "that was the teacher request, with derivates is very easy"
}
{
  "Tag": [
    "calculus",
    "integration",
    "logarithms",
    "trigonometry",
    "calculus computations"
  ],
  "Problem": "Prove that \r\n$ \\int _{0}^{1/4\\,\\pi }\\!{\\frac {\\ln  \\left( 1\\plus{}\\tan \\left( u \\right) \\right) }{\\cos \\left( u \\right) \\sin \\left( u \\right) }}{du}\\equal{}1/12\\,{\\pi }^{2}$\r\nand\r\n$ \\int _{0}^{1/4\\,\\pi }\\!{\\frac { \\left( \\ln  \\left( 1\\plus{}\\tan \\left( u \\right)  \\right)  \\right) ^{2}}{\\cos \\left( u \\right) \\sin \\left( u \\right) }}{du}\\equal{}1/4\\,\\zeta  \\left( 3 \\right)$",
  "Solution_1": "First one:\r\n\r\nSince for $ x \\in \\left[0, \\frac{\\pi}{4} \\right]$ we have $ 0 \\leq \\tan x \\leq 1$, we can write:\r\n\r\n$ I \\equal{} \\int ^\\frac{\\pi}{4} _ 0 \\frac{\\ln \\left(1 \\plus{} \\tan x\\right)}{\\sin x \\cos x} \\, dx \\equal{} \\int ^\\frac{\\pi}{4} _ 0 \\frac{1}{\\cos ^2 x} \\sum ^\\infty _ { n \\equal{} 1} \\frac{(\\minus{}1)^{n\\plus{}1}}{n}\\tan^{n\\minus{}1} x \\,dx$\r\n\r\nNow, due to absolute convergence of the sum, we can write:\r\n\r\n$ I \\equal{}  \\sum ^\\infty _ { n \\equal{} 1} \\frac{(\\minus{}1)^{n\\plus{}1}}{n}  \\int ^\\frac{\\pi}{4} _ 0 \\frac{\\tan^{n\\minus{}1}x }{\\cos ^2 x} \\,dx$\r\n\r\nThe integral can be easily computed since we have:\r\n\r\n$ \\int ^\\frac{\\pi}{4} _ 0 \\frac{\\tan^{n\\minus{}1}x }{\\cos ^2 x} \\,dx \\equal{} \\int ^\\frac{\\pi}{4} _ 0 \\tan^{n\\minus{}1}x \\frac{d}{dx}\\left(\\tan x\\right) \\,dx \\equal{} \\frac{\\tan ^n x}{n} \\Big| ^\\frac{\\pi}{4} _ 0 \\equal{} \\frac{1}{n}$\r\n\r\nPlugging that in, we have:\r\n\r\n$ I \\equal{} \\sum ^\\infty _ { n \\equal{} 1} \\frac{(\\minus{}1)^{n\\plus{}1}}{n^2} \\equal{} \\frac{ \\zeta(2)}{2} \\equal{} \\frac{\\pi ^2}{12}$"
}
{
  "Tag": [
    "function",
    "search",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "Prove or disprove there exists a continuous function which is nowhere differentiable.",
  "Solution_1": "Did you try googling that? You'll find lots about Weierstrass' example. You can also search the forum, since I'm sure people have posted before on the subject.",
  "Solution_2": "There is a famous example by Weierstrass showing that there are continuous nowhere differentiable functions, or we may use the Baire Category Theorem.  Would you actually like to see an example?",
  "Solution_3": "Thanks for letting me to know this is actually very famous. I just searched in mathworld and found [url=http://mathworld.wolfram.com/WeierstrassFunction.html]this[/url].",
  "Solution_4": "That function is still *somewhere* differentiable.  There are functions that are nowhere differentiable.\r\n\r\nPut $\\varphi(x)=|x|$ for $-1\\leq x\\leq 1$, and make it periodic by setting $\\varphi(x+2)=\\varphi(x)$.\r\n\r\nNow define $f(x)$ as\r\n\r\n$f(x)=\\sum^{\\infty}_{i=0} \\left(\\frac {3}{4}\\right)^i\\varphi(4^ix)$\r\n\r\nThis is continuous but nowhere differentiable."
}
{
  "Tag": [],
  "Problem": "I was wondering if someone could name a few math competitions that satisfy the following:\r\n\r\n[list]\n[*] over the Internet\n[*] free\n[*] does not require to sign up a school[/list]\r\n\r\napart from the iTest.\r\n\r\nThanks in advance!",
  "Solution_1": "http://purplecomet.org/"
}
{
  "Tag": [
    "inequalities",
    "calculus",
    "algebra",
    "function",
    "domain"
  ],
  "Problem": "for reals a, b > 0 prove that ((a+1)/(b+1))<sup>b+1</sup> :ge: (a/b)<sup>b</sup>",
  "Solution_1": "I'm curious if there are non-calculus methods:\n\n\n\n[hide]Everything that follows is in reference to the positive real numbers as our domain for x.\n\nRe-write our equation as b<sup>b</sup>/(b + 1)<sup>b + 1</sup> :ge: a<sup>b</sup>/(a + 1)<sup>b + 1</sup>.  This needs to be true for all a not equal to b, so we need to show that f(x) = x<sup>b</sup>/(x + 1)<sup>b + 1</sup> has a global maximum at b.  To do this, we can just examine f'(x).\n\nf'(x) = (b*x<sup>b - 1</sup>*(x + 1)<sup>b + 1</sup> - x<sup>b</sup>*(b + 1)*(x + 1)<sup>b</sup>)/(x + 1)<sup>2b + 2</sup>.  The denominator is always positive, so we need only examine the numerator, b*x<sup>b - 1</sup>*(x + 1)<sup>b + 1</sup> - x<sup>b</sup>*(b + 1)*(x + 1)<sup>b</sup> = x<sup>b</sup>*(x + 1)<sup>b</sup>*(b(x + 1) - x*(b + 1)) = <sup>b</sup>*(x + 1)<sup>b</sup>*(b - x).  This is positive when x < b, negative when x > b, and zero when x = b, which gives us a global maximum at x = b, which proves our theorem.[/hide]",
  "Solution_2": "((a+1)/(b+1))^(b+1)   :ge: (a/b)^b\n\n \n\n[hide] (b+1)log(b)(a+1)-(b+1)log(b)(b+1) :ge: blog(b)a-blog(b)b\n\n\n\n I'm going to simplify the right and left and sides seperately\n\n\n\nFirst the left\n\n\n\n(b+1)log(b)(a+1)=(b+1)(log(b)a+log(b)1) = (b+1)(log(b)a)\n\n\n\nand -(b+1)log(b)(b+1)=-(b+1)log(b)b-(b+1)log(b)1=-b+1\n\n\n\nSo we have the left side equal to \n\n\n\n(b+1)(log(b)a)-b+1 :ge: blog(b)a-blog(b)b\n\n\n\nthe right hand side is blog(b)a-b\n\n\n\nadding b to both sides equals\n\n\n\n(b+1)(log(b)a)+1 :ge: blog(b)a\n\n\n\ndividing by log(b)a (which must be positive, therefore no sign change is necessary)\n\n\n\nequals\n\n\n\n(b+1)+log(a)b  :ge: b\n\n\n\nSubtracting (b+1)\n\n\n\nlog(a)b :ge: -1\n\n\n\nWhich is obviously true[/hide]\n\nThis is what I got, is it wrong?",
  "Solution_3": "I believe that this is a viable non-calculus solution.",
  "Solution_4": "log<sub>b</sub>(a+1) = log<sub>b</sub>a+log<sub>b</sub>1  is not true!  log(a*b) = log(a) + log(b).  log(a + b) isn't anything!",
  "Solution_5": "yeah oops... lol",
  "Solution_6": "thanks for the sol'n Joel, that question was on a contest i took wednesday, and I thought I could do it by fixing b and varying a, but that did not work out for me.",
  "Solution_7": "Well, that's essentially what I did, but I made it look nice first :).  Was this a contest where calculus was expected?",
  "Solution_8": "Cornell Freshmen mathematics contest",
  "Solution_9": "I don't know what that means in terms of my question.\r\n\r\nBy the way, if you're a freshman at Cornell interested in math, you wouldn't happen to know Konstantin Getmanchuck (not sure of spelling there), would you?  Skinny Russian guy from Brooklyn?",
  "Solution_10": "oops, i misread your original post thinking you asked what contest it was.\r\n\r\nAnyway, i do not know Konstantin, did he go to Stuy with you?",
  "Solution_11": "Here are two more proofs.  Actually, both proofs are basically one proof in different words.  I think they qualify as non-calculus proofs, but you tell me.\n\n\n\nFirst proof: [hide]We will use the weighted AM-GM inequality, which goes like this.  Let x and y be nonnegative numbers.  Let p and q be nonnegative numbers such that p + q = 1.  Then\n\npx + qy :ge: x<sup>p</sup>y<sup>q</sup>.\n\n\n\nLet's plug in x = a/b, y = 1, p = b/(b+1), and q = 1/(b+1).  Then we have\n\n(a+1)/(b+1) = px + qy :ge: x<sup>p</sup>y<sup>q</sup> = (a/b)<sup>b/(b+1)</sup>.\n\nRaising both sides to the power b+1 gives the desired inequality.\n\n[/hide]\n\n\n\nSecond proof: [hide]We will use the concavity of the log function.  Note that\n\n(a+1)/(b+1) = b/(b+1) (a/b) + 1/(b+1) (1).\n\nBy concavity, we have\n\nln((a+1)/(b+1)) :ge: b/(b+1) ln(a/b) + 1/(b+1) ln(1) = b/(b+1) ln(a/b).\n\nTaking exponentials on both sides gives\n\n(a+1)/(b+1) :ge: (a/b)<sup>b/(b+1)</sup>.\n\nAgain, raise both sides to the b+1 power to finish the proof.\n\n[/hide]",
  "Solution_12": "Konstantin went to Brooklyn Tech, but I know him from math stuff (NYC math team, CCNY math camp).\r\n\r\nI thought about trying to use the shape of the log function but couldn't figure out how.  Nice proofs!",
  "Solution_13": "hmm, i actually think i know who you are talking about, there is this kid that I see around who fits the description and sometimes wears a nysml shirt, could be the same person."
}
{
  "Tag": [
    "analytic geometry"
  ],
  "Problem": "Find all points on the y-axis that are 5 units from the \r\npoint (-4, 4).\r\n\r\nWhat about if the question asked for all points from the same point given above but this time from the x-axis?",
  "Solution_1": "[hide]So you have one point $(-4,4)$. The distance has to be 5 units and on the y axis. In order to do this problem we do the distance formula backwards. Lets call the other point $(0,x)$. Plugging it in to the distance formula we have $5=\\sqrt{(-4)^{2}+(4-x)^{2}}$. Solving for x we have two solutions. $x=1,7$. So the two points are $(0,1)$ and $(0,7)$[/hide]\r\n\r\nIn order to do this for the x axis we have to set the $(x,0)$ and the distance fromula backwards from there.",
  "Solution_2": "So, definitely use the distance formula.\r\n\r\nGot it.\r\n\r\nThanks.",
  "Solution_3": "before you do it for the x-axis, notice that (-4,4) is the same distance from the y-axis as it is from the x-axis.  all you have to do now flip the whole cartesian coordanates around the line y=-x."
}
{
  "Tag": [
    "LaTeX",
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "Let a,b,c be positive numbers. Prove that \r\na(b+c)/(b^2+bc+c^2)+b(a+c)/(a^2+ac+c^2)+c(a+b)/(a^2+ab+b^2)>=2.\r\n( I am sorry because I don't know how to use latex now ) :blush:",
  "Solution_1": "use SOS :)",
  "Solution_2": "[quote=\"CHICKEN\"]Let a,b,c be positive numbers. Prove that \n$ \\frac {a(b \\plus{} c)}{b^2 \\plus{} bc \\plus{} c^2} \\plus{} \\frac {b(a \\plus{} c)}{a^2 \\plus{} ac \\plus{} c^2} \\plus{} \\frac {c(a \\plus{} b)}{a^2 \\plus{} ab \\plus{} b^2}\\geq2.$\n[/quote]\r\nBy Cauchy-Schwartz we obtain:  \r\n$ \\sum_{cyc}\\frac {a(b \\plus{} c)}{b^2 \\plus{} bc \\plus{} c^2}\\sum_{cyc}\\frac {a(b^2 \\plus{} bc \\plus{} c^2)}{b \\plus{} c}\\geq(a \\plus{} b \\plus{} c)^2.$\r\nId est, it remains to prove that $ (a \\plus{} b \\plus{} c)^2\\geq2\\sum_{cyc}\\frac {a(b^2 \\plus{} bc \\plus{} c^2)}{b \\plus{} c},$ which equivalent to\r\n$ \\sum_{sym}(a^4b \\minus{} a^3b^2)\\geq0.$\r\nSee also here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=25553"
}
{
  "Tag": [],
  "Problem": "Three of the following test scores are Cyprian's and the other three\nare Margaret's: 85, 87, 92, 93, 94, 98. Cyprian's mean score is\n90. What is Margaret's mean score?",
  "Solution_1": "The scores total $ 85 \\plus{} 87 \\plus{} 92 \\plus{} 93 \\plus{} 94 \\plus{} 98 \\equal{} 549$\r\n\r\nCyprian's total $ 90 \\cdot 3 \\equal{} 270$\r\n\r\nMargaret's total $ 549 \\minus{} 270 \\equal{} 279$\r\n\r\nHer mean score is $ \\frac {279}{3} \\equal{} 93$\r\n\r\nOR\r\n\r\nThe scores total $ 85 \\plus{} 87 \\plus{} 92 \\plus{} 93 \\plus{} 94 \\plus{} 98 \\equal{} 549$\r\n\r\nThe average score is $ \\frac{549}{6} \\equal{} 91.5$\r\n\r\nBoth took three tests, so Margaret's average score is $ 91.5 \\plus{} (91.5 \\minus{} 90) \\equal{} 93$",
  "Solution_2": "we first find the mean of all six scores.\r\nthat mean will also be the mean of cyprian's and margaret's scores.\r\nwe can use it to find margaret's mean.\r\nmean of 85, 87, 92, 93, 94, 98 is\r\n$ 90\\plus{}\\frac{\\minus{}5\\plus{}\\minus{}3\\plus{}2\\plus{}3\\plus{}4\\plus{}8}{6}\\equal{}90\\plus{}\\frac{9}{6}\\equal{}90\\plus{}\\frac32$\r\nso since it's $ \\frac32$ greater than cyprian's mean, it's $ \\frac32$ less than margaret's making margaret's mean $ 90\\plus{}3$\r\n93.\r\npreview edit: cyprian's total is $ 90\\cdot3$, since each person has 3 scores.",
  "Solution_3": "Let x be margaret's mean score.\r\nWe have:\r\n(90)(3) + 3x = 85+87+92+93+94+98\r\n270 + 3x = 549\r\n3x = 279\r\nx = 93\r\n\r\nMission's accomplished! :clap: Thank you. :clap2: Thank you  :winner_second:",
  "Solution_4": "An easier way is the guess and check (quite ironic). Notice that Margaret must have three scores such that the difference from 90 adds to 0. So 85 would be -5 (90-85), 92 would be 2 (92-90). So you see that 85, 92, and 93 have differences that add up to zero. Just average 87, 94, and 98 and get $\\boxed{93}$"
}
{
  "Tag": [
    "trigonometry",
    "integration",
    "calculus",
    "calculus computations"
  ],
  "Problem": "For real $a, b$ define the sequence $x_{0}, x_{1}, x_{2}, ...$ by $x_{0}= a, x_{n+1}= x_{n}+b \\sin x_{n}$. If $b = 1$, show that the sequence converges to a finite limit for all $a$. If $b > 2$, show that the sequence diverges for some $a$.",
  "Solution_1": "$x_{n+1}-x_{n}$ we can mean $x'(n)$. then $\\int_{0}^{n}\\frac{dx(n)}{sinx(n)}$~$\\int_{0}^{n}bdx$ $\\Rightarrow$ $ln(|tan\\frac{x(n)}{2}|)|_{0}^{n}$~$bn$  $\\Rightarrow$ $ln(tan(x_{n}/2))-ln(tan(a/2))$ ~ $nb$, this is idea and then is simpl. I think so."
}
{
  "Tag": [
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "Sequence ${a_{n}}$ is:\r\n$a_{1}=2$\r\n$a_{n}=2(n+a_{n-1})$, for all $x\\geq 2$\r\nProve that $a_{n}\\leq 2^{n+2}$, for all $n$, who $n$ is natural numbers.\r\n :wink:",
  "Solution_1": "$\\ a_{n}=2n+2a_{n-1}$ and $\\ a_{n-1}=2(n-1)+2a_{n-2}$\r\n\r\n$\\ \\to a_{n}=2n+2^{2}(n-1)+2^{2}a_{n-2}$\r\n\r\nIf we continue we find that $\\ a_{n}=2n+2^{2}(n-1)+2^{3}(n-2)+...$\r\n\r\n$\\ \\to  a_{n}=\\sum_{l=0}^{n-1}2^{l+1}(n-l)$\r\n\r\n$\\ \\to  a_{n}=2n(2^{n}-1)-2(2^{n}n-2^{n+1}+2)$\r\n\r\n$\\ \\to  \\boxed{a_{n}=2^{n+2}-2n-4\\ <2^{n+2}}$"
}
{
  "Tag": [
    "inequalities",
    "AMC",
    "USA(J)MO",
    "USAMO",
    "inequalities proposed"
  ],
  "Problem": "if a,b,c>0 prove the following inequality:\r\n  \r\n       (1+a/b)(1+b/c)(1+c/a)>=2(1+(a+b+c)/2 3:sqrt(abc)). I hope tou do understand what I wrote.",
  "Solution_1": "It's not USAMO, it's APMO, but the author is common: Titu Andreescu. Why is it tricky? Direct application of AM-GM?  :?",
  "Solution_2": "can you post the solution?",
  "Solution_3": "Of course I can. One solution is based on the fact that the inequality is homogenous and we can take $abc=1$. Then use the fact that $\\frac{x}{y}+\\frac{y}{z}+\\frac{z}{x}\\geq x+y+z$ when $xyz=1$, which is trivial. Use this twice and you have the result. Another solution follows from the fact that $\\frac{a}{b}+\\frac{a}{c}+1\\geq\\frac{3a}{\\sqrt[3]{abc}}$. Add them and use AM-GM once again.",
  "Solution_4": "Easy:\r\n[tex]a(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}) \\ge \\frac{3a}{\\sqrt[3]{abc}}[/tex]\r\n\r\n :lol:",
  "Solution_5": "[quote=\"ehku\"]Easy:\n[tex]a(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}) \\ge \\frac{3a}{\\sqrt[3]{abc}}[/tex]\n\n :lol:[/quote]\r\n\r\nThat's just GM-HM..."
}
{
  "Tag": [
    "inequalities",
    "inequalities solved"
  ],
  "Problem": "I created this problem and I want to know if it's not too easy.\r\n\r\nLet ABC be a triangle with standard notation. Prove that:\r\n\r\n  \\sqrt 3S/2 \\leq (ma*ma+mb*mb+mc*mc)/(ma+mb+mc)",
  "Solution_1": "I used the following identity:\r\n3S=4sqrt[(ma+mb)(ma+mc)(mb+mc)(ma+mb+mc)] and I got the following inequality which I can't figure out what to do. Perhaps this is not a good idea!\r\n\r\nI got to:\r\n[(ma^2+mb^2+mc^2)/(ma+mb+mc)]^4>=4(ma+mb)(ma+mc)(mb+mc)(ma+mb+mc) which I cannot prove! Perhaps someone could! \r\n\r\ntschuess",
  "Solution_2": "Hey!Nobody knows to solve this problem.Is someone interested to see the solution?Shall I put the solution?I am waiting your opinion.\r\n  \r\n\r\n\r\n\r\n\r\n\r\n\r\n   Regards!",
  "Solution_3": "I would love to see a solution! This one has bugged be enough! I think we should see a solution! :D",
  "Solution_4": "4 Lagrangia\r\n    I am going to put a solution of my problem\r\n    We know that ma^2+mb^2+mc^2=3(a^2+b^2+c^2)/4\r\n    After we use the ineq \\sum a^2>= \\sum ab,we use the fact that ab>2S\r\nbecause sinC<1 Idem for bc,ac.\r\n    So we have  \\sum ab>6S That means  \\sum ma^2>9S/2 (1)\r\n    Then from Cauchy we have  \\sum ma^2>=( \\sum ma)^2/3 (2)\r\n     Now we will (1)*(2) and we will have what we are loking for.\r\n    \r\n\r\n     I am still waiting other solutions!",
  "Solution_5": "well.. has anyone tried this one using the identity I gave? Cause with that identity we get an inequality only in ma,mb and mc which should be easier to solve ( I think :D).. though I can't see how..I  tried denoting ma+mb+mc=x, sum(ma*mb)=y and prod(ma)=z but got nowhere! perhaps someone will find something! \r\ncheers!",
  "Solution_6": "[quote]I created this problem and I want to know if it's not too easy. \r\n\r\nLet ABC be a triangle with standard notation. Prove that: \r\n\r\n\\sqrt 3S/2 \\leq (ma*ma+mb*mb+mc*mc)/(ma+mb+mc) [quote]\r\n  Here is my solution:\r\n   ma^2+mb^2+mc^2=3*(a^2+b^2+c^2)/4\\geq 3*4*3^(1/2)*S/4=3^(3/2)*S\r\n   By using CS  ma+mb+mc\\leq  sqrt(3*(ma^2+mb^2+mc^2))\r\n   Then  (ma*ma+mb*mb+mc*mc)/(ma+mb+mc)\\geq  sqrt(ma^2+mb^2+mc^2)*3(-1/2)\\geq sqrt(S)*3^(1/4)",
  "Solution_7": "lagrangia, can u post a proof of your identity?",
  "Solution_8": "Lagrangia you used that fact [u] ab>2S[/u] So in that problem that you gave there is no equality sign."
}
{
  "Tag": [
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "f: $ R\\minus{}{1} \\to R$\r\n[size=34]$ f(x \\plus{} 2f( \\frac {x \\plus{} 2001}{x \\minus{} 1}) \\equal{} 4013 \\minus{} x$[/size]\r\nf(2003)=?",
  "Solution_1": "In your formula where do you shut the door ?  \")\""
}
{
  "Tag": [
    "limit",
    "function",
    "calculus",
    "calculus computations"
  ],
  "Problem": "How do I get $\\lim_{x\\rightarrow 1}\\frac{x^{p}-1}{x^{q}-1}$ $=$ $\\frac{p}{q}$ without using L'Hopital's rule? ($p$ and $q$ being integers)",
  "Solution_1": "$\\frac{x^{p}-1}{x^{q}-1}=\\frac{(x-1)(x^{p-1}+x^{p-2}+...+1)}{(x-1)(x^{q-1}+x^{q-2}+...+1}=\\frac{x^{p-1}+x^{p-2}+...+1}{x^{q-1}+x^{q-2}+...+1}$.\r\n\r\n$\\lim_{x\\rightarrow1}{\\frac{x^{p-1}+x^{p-2}+...+1}{x^{q-1}+x^{q-2}+...+1}}=\\frac{p}{q}$.\r\n\r\nIf ${p<0}\\wedge{q<0}$\r\n\r\n$\\frac{x^{p}-1}{x^{q}-1}=\\frac{x^{q'}(x^{p'-1}+x^{p'-2}+...+1)}{x^{p'}(x^{q'-1}+x^{q'-2}+...+1)}$.\r\n$p'=-p, q'=-q$\r\n\r\nIf ${p<0}\\wedge{q>0}$\r\n\r\n$\\frac{x^{p}-1}{x^{q}-1}=-\\frac{x^{p'-1}+x^{p'-2}+...+1}{x^{p'}(x^{q-1}+x^{q-2}+...+1)}$.\r\n$p'=-p$\r\n\r\nIf ${p>0}\\wedge{q<0}$\r\n\r\n$\\frac{x^{p}-1}{x^{q}-1}=-\\frac{x^{q'}(x^{p-1}+x^{p-2}+...+1)}{x^{q'-1}+x^{q'-2}+...+1}$.\r\n$q'=-q$",
  "Solution_2": "easiest way is to use the substitution $y=x-1$, and then apply the binomial theorem... \r\n\r\n$\\lim_{x\\to 1}\\frac{x^{p}-1}{x^{q}-1}=\\lim_{y\\to 0}\\frac{(1+y)^{p}-1}{(1+y)^{q}-1}=\\lim_{y\\to 0}\\frac{(1+py+\\textbf o(y^{2})+\\cdots)-1}{(1+qy+\\textbf o(y^{2})+\\cdots)-1}$\r\n\r\n$=\\lim_{y\\to 0}\\frac{p+\\textbf o(y)+\\cdots}{q+\\textbf o(y)+\\cdots}=\\frac{p}{q}$",
  "Solution_3": "$\\lim_{x\\rightarrow 1}\\frac{x^{p}-1}{x^{q}-1}=\\lim_{x\\rightarrow 1} \\frac{\\frac{x^{p}-1}{x-1}}{\\frac{x^{q}-1}{x-1}}$\r\n\r\n$=\\lim_{x\\rightarrow 1}\\frac{(x^{p})'|_{x=1}}{(x^{q})'|_{x=1}}=\\frac{p}{q}$, which can be obtained by the difinition of differential coefficient.",
  "Solution_4": "[quote=\"misan\"]\n$ \\lim_{x\\to 1}\\frac{x^{p}-1}{x^{q}-1}=\\lim_{y\\to 0}\\frac{(1+y)^{p}-1}{(1+y)^{q}-1}=\\lim_{y\\to 0}\\frac{(1+py+\\textbf o(y^{2})+\\cdots)-1}{(1+qy+\\textbf o(y^{2})+\\cdots)-1}$\n[/quote]\nI agree with your answer and method, but I must tell some correction\n\n\nI note that \n[quote=\"misan\"]\n$ \\lim_{x\\to 1}\\frac{x^{p}-1}{x^{q}-1}=\\lim_{y\\to 0}\\frac{(1+y)^{p}-1}{(1+y)^{q}-1}=\\lim_{y\\to 0}\\frac{(1+py+\\textbf o(y^{2})+\\cdots)-1}{(1+qy+\\textbf o(y^{2})+\\cdots)-1}$\n[/quote]\r\n(... about defination)\r\nhere : \r\n$ (1+y)^{p}-1=1+py+\\textbf o(y^{2})+\\cdots)-1$ this is not correct. \r\n. but this \r\n$ (1+y)^{p}-1=1+py+p(p-1)y^{2}/2+\\textbf o(y^{2})+\\cdots)-1$\r\nis correct.\r\nfor example:\r\nlet's consider function \r\n$ f(x)=x^{2}$ in point $ 0$\r\nI can say that \r\n$ f(x)=o(x)$ in point $ 0$ because \r\n$ \\lim_{x\\to 0}\\frac{f(x)}{x}=0$\r\nbut in you case  \r\nyou say \r\n$ f(x)=o(x^{2})$ (this is not true because $ \\lim_{x\\to 0}\\frac{f(x)}{x^{2}}=1$)\r\nbut you can say  $ f(x)=O(x^{2})$ because it's mean $ |f(x)|<Cx^{2}$ for the some constant $ C$)\r\n\r\nAnd in our case \r\n$ \\lim_{y\\to 0}\\frac{(y+1)^{p}-1}{(y+1)^{q}-1}=\\frac{1+yp+o(y)-1}{1+qy+o(y)-1}=\\frac{p+\\frac{o(y)}{y}}{q+\\frac{o(y)}{y}}=\\frac{p}{q}$"
}
{
  "Tag": [
    "inequalities",
    "triangle inequality",
    "inequalities proposed"
  ],
  "Problem": "Let a,b,c denote the sides of a triangle . Prove that the quantity $(a/(b+c)) + (b/(c+a)) + (c/(a+b))$ lies between  2 and  3/2.Does equality hold at either limit?",
  "Solution_1": "even if a,b,c are not sides of triangles, it's still true. This has been posted for many times already.",
  "Solution_2": "No, for one half of the inequality, \\[ a/(b + c) + b/(c + a) + c/(a + b) < 2, \\] a, b and c must be sides of a triangle, I think...",
  "Solution_3": "oops i messed up with another problem. Sorry.  :blush:",
  "Solution_4": "It holds for $a,b,c>0.$ ;)",
  "Solution_5": "What do you mean Cezar?",
  "Solution_6": "If $a, b, c>0$ then the following inequality holds:\r\n\r\n       $\\dfrac{a}{b+c}+\\dfrac{b}{c+a}+\\dfrac{c}{a+b} <2$ ;)",
  "Solution_7": "[quote=\"cezar lupu\"]If $a, b, c>0$ then the following inequality holds:\n\n       $\\dfrac{a}{b+c}+\\dfrac{b}{c+a}+\\dfrac{c}{a+b} <2$ ;)[/quote]\r\n\r\n$(a,b,c)=(1,1,\\infty)$..",
  "Solution_8": "Sorry, Siuhochung is right here :oops: . Actually,\r\n\r\n$\\dfrac{a}{a+b}+\\dfrac{b}{b+c}+\\dfrac{c}{c+a} <2$ ;)",
  "Solution_9": "the solution i gave when i saw by the first time this inequality consisted in a brute-force multiplication and then using the triangle inequality... although there is a very simple one.\r\n\r\nby the triangle inequality we have that $\\dfrac{a}{b+c}<\\dfrac{2a}{a+b+c}$, then taking the cyclic sum gives the result...  :D",
  "Solution_10": "yeah I know this was easy! I'll come up with something better next time!"
}
{
  "Tag": [
    "vector"
  ],
  "Problem": "Prove that there are exactly [b]7[/b] planes that are equidistant from 4 non-coplanar points in space.",
  "Solution_1": "You might want to refine what you mean by \"equidistant from 4 points\".\r\n\r\nIf the definition for [u]a plane equidistant from 4 points[/u] is [u]the 4 perpendicular segments dropped from the given 4 points on the plane are all the same length[/u] (hence \"equidistant\"), then there are indeed 7 such planes. I'll leave this for someone to have fun with :)",
  "Solution_2": "Come on now. This shouldn't be that difficult.\r\n[hide=\"a hint\"] \nAny three points lie on the same plane. \nNow consider 3 points in space. No matter where the 4th point is chosen, the 4 points are co-planar. {why?}\nTherefore any 3 points should not be collinear.\nNow with 4 such points, we can form 4 triangles with one point not in the plane on which the triangle lies. \nConstruct a line perpendicular to the plane and passing through this point. The plane passing through the midpoint of this line segment and parallel to the plane on which the triangle lies is a plane that satisfies our condition. \nThus we have found 4 such planes.\n\nNow find the other three.\n[/hide] \r\n10000th User's interpretation is correct.",
  "Solution_3": "[hide=\"Continuation from above\"]\nThe other 3 occur when the plane divides space in half with two points in either half. There's $ \\frac{1}{2}\\binom{4}{2} \\equal{} 3$ ways that can happen.  :) \n[/hide]",
  "Solution_4": "[hide=\"Solution\"][b]3 points on one side and 1 on the other:[/b]\nThe three points define a plane. Let this plane be P, and let the point be Q. Let Q' be the projection of Q onto P, and let M be the midpoint of QQ'. There is a unique plane L passing through M that is parallel to P.\n\nThus, this case results in four possible planes -- one for each choice of P.\n\n[b]2 points on each side:[/b]\nLet P and Q be two points, which we will put on one side of the plane, and let R and S be two points on the other side of the plane. Let v be the direction vector of PQ and u be the direction vector of RS. Then, the family of planes with normal vector uxv is parallel to both line PQ and line RS.\n\nLet A and B be points on PQ and RS, respectively, such that AB is the unique such line segment that is perpendicular to both PQ and RS (AB has direction vector uxv). Let M be the midpoint of AB. Then, the plane with direction vector uxv that passes through M is the unique plane equidistant from P Q R and S.\n\nThis case results in three possible planes -- one for each way of dividing four points into two groups.[/hide]"
}
{
  "Tag": [
    "function",
    "calculus",
    "derivative",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "[b][size=150]Let f(x)=ax^2 - 4ax + b, (a>0) be defined in 1<=x<=5. \nSuppose the average of the maximum value and the minimum value of the function is 14, and the difference between the maximum value and minimum value is 18. Find the values of a (a>0) and b.[/b][/size] ;) \r\n\r\n[/b]",
  "Solution_1": "[quote=\"willydavidjr\"][b][size=150]Let f(x)=ax^2 - 4ax + b, (a>0) be defined in 1<=x<=5. \nSuppose the average of the maximum value and the minimum value of the function is 14, and the difference between the maximum value and minimum value is 18. Find the values of a (a>0) and b.[/b][/size] ;) \n\n[/b][/quote]\r\nLet $\\displaystyle f(x)=ax^2-4ax+b, (a>0)$ be defined in $\\displaystyle 1\\leq x\\leq 5$\r\n\r\nSuppose the average of the maximum value and the minimum value of the function is 14, and the difference between the maximum value and minimum value is 18. Find the values of $a$ and $b$.\r\n\r\nLaTeXed for easier reading.",
  "Solution_2": "Solution: \r\ny=ax^2-4ax+b, a>0\r\nfirst derivative=2ax-4a=0, x=2\r\nfor x>2, f(x) is an increasing function\r\nfor x<2, f(x) in a decreasing function\r\n\r\nso your max value is f(5)=5a+b\r\nyour min value=f(1)=-3a+b\r\n(max value+ min value)/2=14, a+b=14   (i)\r\n\r\nmax value - min value=18, a=9/4\r\nput x=9/4 into the equation (i), to get the value of b:\r\nb=47/4"
}
{
  "Tag": [
    "geometry",
    "geometric transformation",
    "rotation"
  ],
  "Problem": "A man of mass 80 kg mass is standing on the rim of a circular platform rotating about its axis. The platform with the man on it rotates at 12 r.p.m. How will the system rotate, if the man moves to the platform's centre? What work will the man perform in changing his position? The mass of the platform is 200 kg and the radius is 1.2 m.",
  "Solution_1": "Apply angular momentum conservation to find the new angular velocity.\r\nThen find the angular kinetic energies of new system and old system and then find the work done. ??",
  "Solution_2": "as the man moves to the axis of rotation the disk starts moving faster and faster..........\r\nthis is due to the law of conservation of angular momentum........"
}
{
  "Tag": [
    "number theory",
    "prime factorization"
  ],
  "Problem": "You may of heard of this problem before, so if you already know the answer please give others a chance.\r\n\r\nA surveyor goes to a special recipient and asks, \"what are the ages of your children?\"\r\nShe replies,\"Well, the product of my daughters'  ages is 36.\"\r\nthe surveyor says, \"Please tell me more.\"\r\nShe replies, \"It wouldn't matter if I told you the sum anyway, it won't do much good.\"\r\nThe surveyor said, \"Well, can you tell me the smallest bit more of information?\"\r\nShe replies, \"Oh no! I have to go. I need to pick up my daughter Annie from swimming lessons, and I have to pick up the twins from daycare.\"\r\n\r\nWhat are the ages of the children?",
  "Solution_1": "[hide=\"answer\"]\nI have seen this before but worded differently,\nLet's say that the age of the twins is $ x$, and the age of the daughter is $ y$.\n\nSo that $ x^2\\cdot y\\equal{}36$, now we find the prime factorization of $ 36$ which is $ 2^2\\cdot 3^2$.\n\nMeaning that $ x^2$ has to be either $ 1$,$ 4$,$ 9$,$ 36$ and the $ y$ values are $ 36$,$ 9$,$ 4$,$ 1$ respectively.\n\nNow we cross out the impossible $ y$ values like $ 36$,$ 4$,$ 1$. Because first she wouldn't be picking up a $ 36$ year old and normally $ 4$ year olds and $ 1$ year olds do not take swimming lessons. Then the two twins are each $ 2$ years old and the daughter is $ 9$ years old. I believe this answer fits best with the description of the children.\n\nMight be wrong, might be right. I am putting my solution based on a previous question like this.\n[/hide]",
  "Solution_2": "[hide] This is how you have to solve these problems : Because she says she has to pick up her twins, meaning that they have the same age. Let the age of 1 twin be A and the age of Annie be B. A x A x B = 36; A^2 x B = 36. So we have to find two numbers that multiply to 36. so lets write the factors of 36 as pairs where 1 pair is a perfect square in the form (A,B) {(36,1) (1,36) (4,9) (9,4)} So the answer has to be one of these. The lady says that her twins are at a daycare and the older daughter is at a swimming lesson. so it cannot be the pair (36,1) because 6 year-olds don't go to day care and a 1 year old doesnt go to swimming lessons alone. It can't be the pair (1,36) because you don't leave 1 year olds at daycare when they are so young. You are left with (4,9) and (9,4).\nThe swimmer must be older than the twin because swimmers are older than kids atdaycare. Therefore the answer is {2,2,9} [/hide]",
  "Solution_3": "Or we note that by the second clue, (2,2,9) and (1,6,6) are the possible solutions, since we must have two solutions to the first clue summing to the same value.\r\n\r\nUsually the problem implies that there is an oldest child. However, in this case 6 year olds don't attend daycare (actually, I'm not sure about that). So (2,2,9).",
  "Solution_4": "What would the answer be if the twins consisted of one boy and one girl?\r\nThe problem states that the product of the daughters' ages is 36. Then the answer would be xy=36",
  "Solution_5": "What if the twins were boys!!!\r\nThen, the daughter's age would be 36!!!  :D",
  "Solution_6": "To rts2007:\r\n   That's a good point you brought up, but the way the problem is stated, it tells you the product of the daughters' ages not the daughter's age. That tells me that there must be at least two daughters, but the problem does not tell you what constitutes the pair of twins. So it is possible that there could be only one girl.",
  "Solution_7": "If there were a boy and a girl, there would be multiple possibilites.",
  "Solution_8": "I agree with both bogtro and rts.a major reason.",
  "Solution_9": "There is nothing in the statement of the problem that gives you the idea that the solution is limited to only one answer.",
  "Solution_10": "Guys, the chlildren are all females.\r\n\r\nAnd the information requires a bit of logic.",
  "Solution_11": "BOGTRO quotes \"Or we note that by the second clue, (2,2,9) and (1,6,6) are the possible solutions, since we must have two solutions to the first clue summing to the same value.\"\r\nIsn't it possible that the twin girl could be 4 and the older daughter could be 9?\r\nDo you see anywhere in the problem where it says, without a doubt, that there are three girls?",
  "Solution_12": "Without doubt, this is an interesting problem!\r\n\r\n-rts2007"
}
{
  "Tag": [
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "prove that if $ [x \\plus{} y ] \\equal{} [na \\plus{} b]$for every $ \\in N* ; a,b,x,y \\in R$, then $ \\equal{} a , y \\equal{} b$",
  "Solution_1": "[quote=\"long14893\"]prove that if $ [x \\plus{} y ] \\equal{} [na \\plus{} b]$for every $ \\in N* ; a,b,x,y \\in R$, then $ x\\equal{} a , y \\equal{} b$[/quote]\r\nno one interested :(?"
}
{
  "Tag": [
    "geometry",
    "inradius",
    "circumcircle"
  ],
  "Problem": "The circumradius and the inradius of a certain right triangle are 17/2 and 3, respectively. Find the length of the shortest height. Express your answer as a mixed fraction.\r\n\r\n$ \\text {(A) } 5 \\frac{1}{7} \\qquad \\text {(B) } 7 \\frac{1}{17} \\qquad \\text {(C) } 7 \\frac{2}{19} \\qquad \\text {(D) } 8\\frac{1}{14} \\qquad \\text {(E) } 10\\frac{1}{4}$\r\n\r\nI am $ 97\\frac{1}{13}\\%$ sure that the AMC's do not have this problem in their archive.",
  "Solution_1": "Well, since its a right triangle, the circumradius  is 289\r\n\r\nand also (a+b-17)/2=3\r\n\r\nso we have a+b=23\r\nand a^2+b^2=289\r\n\r\nSquaring the first equation you get 289+2ab=529\r\n\r\nso 2ab=230 and ab=120, area is 60 (thats irrelevent)\r\n\r\nsince its a right triangle, ab=120=17*h, so h=7 1/17 B"
}
{
  "Tag": [],
  "Problem": "Wich of the following represents the largest # of moles\r\n\r\na) 10g NH3\r\nb) 10g N2\r\nc)10g CU\r\nd)10g Br2\r\ne) All these contain the same number of moles\r\n\r\n??????",
  "Solution_1": "find the molar mass of each, devide the gram by molar mass to find the MOLE"
}
{
  "Tag": [
    "Duke",
    "college",
    "Gauss"
  ],
  "Problem": "i was wondering what kind of grades/scores/etc. it takes to get a merit-based scholarship, especially when applying to an out of state college. for instance, i am planning to apply to florida and possibly duke (with clemson being my in-state choice).  would the following get a free-ride?\r\n\r\n-gpa: 4.9 weighted, 3.95 unweighted\r\n-sat: 1310 - 720 math, 590 english (only taken once and i will have to take the new one, which will hopefully be easier)\r\n-member of math team and i will join mu alpha theta my senoir year.\r\n-not too many volunteer activities\r\n-a few notable acedemic awards",
  "Solution_1": "what grade are you in?\r\ni just took the SAT and i'm in 9th right now, got 1330(760 on math)( i hate verbal)\r\n\r\ni believe you can get in to Clemson very easy(i lived in SC for 2 yrs so i'm a big clemson fan!!!)\r\n\r\nyou can probably get into FL if you get a higher score on the new SAT(which is harder).\r\n\r\ni don't think you can't get into duke(my #1 choice) because of the few extracurricular activites(i joined math, academic, beta, and plan to join Key club, and mu alpha beta and business clubs next yr)\r\n\r\nyour GPA is very very good, my is only 3.7\r\n\r\nduke is very competitive(20-25%)",
  "Solution_2": "Currently, you almost certainly wouldn't get accepted to Duke, let alone a free ride.  Certainly not with that SAT and the lack of volunteering/extracurriculars.  But you have time to bump that stuff up, so start soon.",
  "Solution_3": "Did you take the PSAT your junior year?\r\nIf you qualify for national merit scholar you can get (basically) full rides to Arizona State Univ. and Oklahoma Univ.\r\nFlorida has a pretty nice NMS package too.",
  "Solution_4": "The most sensible way to answer your questions is to call up the admissions departments of the schools in question.  Now that admissions are nearly done, they probably will be at a relative ebb of work, and I'm sure there is an admissions officer there who wouldn't mind talking about this to you.",
  "Solution_5": "You're a junior now, right? Having higher SAT scores would help IMMENSELY in making merit scholarships available to you.",
  "Solution_6": "Furious, I know plenty of people from South Carolina who have gotten into Duke with SAT scores similar to yours. So don't let anyone discourage you from applying.\r\n\r\nTalk to your guidance couselor and keep a look out for outside sources of funding and scholarships. Whether you can get a \"free-ride\" or not - who knows. Just apply to where you want to go and fill out as many scholarships applications as you can.",
  "Solution_7": "[quote=\"gauss202\"]Furious, I know plenty of people from South Carolina who have gotten into Duke with SAT scores similar to yours. So don't let anyone discourage you from applying.[/quote]\r\n\r\nGauss, the last thing I would want to do is be discouraging.  Last year's SAT 1 midrange (25-75%) at Duke was 1350-1510.  Considering that a large number of students in the lower 25% are athletes and whatnot, and also considering that he's said that his extracurriculars are relatively weak, that does not bode well - even given that admissions are marginally less selective when it comes to the Carolinas.\r\n\r\nNow, he does have plenty of time to fix the lack of extracurriculars and to up his SAT by 100 or even 150 points, both of which would help him a lot.  In particular, I think that a boost of 100 in the verbal score would be quite welcome.  I'm just saying that realistically, admissions would be a reach right now, and a full ride would be completely out of the question.",
  "Solution_8": "All I'm saying is that none of us are admissions officers. We don't really know what a school will decide on any particular student. I know of a student last year from South Carolina who got admitted to Duke with an SAT score very similar to his, and grades that were probably not quite as good. So admissions decisions can be based on a lot of factors that are not transparent to us.\r\n\r\nI'm not giving him a gaurauntee that he will get admitted, I'm just saying that its certainly possible. So he should continue to work on his scores and application profile, and then apply when the time comes. It is unlikely that he (or anyone) will get a \"full-ride\" solely from the University - but if he gets in and he's really interested in going, there are always other financial aid options and outside scholarships that he can apply for.",
  "Solution_9": "I know furious personally and I don't think it's unrealistic for him to get accepted to Duke.  I can almost guarantee that he will finish as either 1st or 2nd in his class.  He took AP US History last and got a 5.  And he will most likely get 5's on AP Euro, AP Stats and AP Bio this year.  Next year he will be taking AP Chem, AP Gov,  and AP Calc and will probably get 5's on these too.  So AP wise he is looking very strong.  He can definitely raise his SAT score.  An 800 should be easy for him to attain and if he practices his verbal that will go up also.  Don't know about writing but I think he should do decent on that section.  His extracurricular record isn't spectacular but does have math team for that.  He will have been on the math team all 4 years.  His accomplishments are very nice (compared to average people not AoPSers ;) ).  His biggest weak spot is volunteering.  I don't think has any volunteer work yet.",
  "Solution_10": "For this thread, on the subject of volunteering, do a lot of you AoPSers look at oppportunities to tutor math as a volunteer activity? The inner city schools in my town seem always to be looking for tutors (or so they say in radio public service announcements), and I wonder if that would be suitable for a teenager to do as a community service activity, or if schools are mostly looking for adult tutors.",
  "Solution_11": "And to the OP I would say, definitely build a plan to READ, READ, READ, and READ to increase your verbal score, which ought to be possible in the amount of time you still have.",
  "Solution_12": "As of right now, tutoring is my [i]only[/i] form of community service.  I tutor for about 2 and a half to 3 hrs a week.  I didn't have to look for opportunities rather they came to me.  I should have about 30 hrs at the end of this year.  This sounds kind of superficial but what is a good number to have for a transcript?  I also have a bunch of hours from helping out at church when I was in middle school.",
  "Solution_13": "thanks for all the comments.\r\n\r\n-i am a junior\r\n-i do feel my sat will go up on the new one. first of all, i expect to make an 800 math and my verbal will go up since analogies are gone. i was in the 25th percentile for analogies (yuck).\r\n-my psat score was only a 206, so i won't be a national merit scholar.\r\n-i will probably be 2nd in my graduating class, i doubt i can pass  joml88.\r\n-i did student volunteer for a year (it's a class where you help a teacher) and volunteered at a marathon once. \r\n-i don't see me picked up tutoring cuz i am not good at it as i do many things (especially math) different from the way i should be tutoring it. for not tutoring, i had to drop out of the National Honor Society (or as i call it National Tutoring Society as all you did was tutor the athletes if you didn't have anyone else to tutor).\r\n\r\nis a full-ride to florida out of the question?",
  "Solution_14": "I think you are in full-tuition \"merit scholarship\" range for some state universities with your current stats, especially now that I have seen your classmate's description of your course load. I'm not sure which private universities would offer \"merit\" money at your range, but some of those are better at offering \"financial need\" discounts than the public universities are, so you should apply to both kinds of schools and see what you are offered.",
  "Solution_15": "Duke is way to hard to get into with the score of 1310, you can hope and maybe get lucky(one or two lucky ones get in) but don't bet on it,  you have to raise it up 100 pts at least.\r\nbut the your APs and GPA and Class rank very help a lot\r\neverybody going to duke are all A students and are the best in their own school, that's why colleges care so much about EXTRA activities, like commuity services(at least 4 contuines yrs at one place), clubs, interests, they want well-rounded students\r\ntry to find volunteer work at the hospital, theatre, sports clubs.etc",
  "Solution_16": "[quote=\"tokenadult\"]For this thread, on the subject of volunteering, do a lot of you AoPSers look at oppportunities to tutor math as a volunteer activity? The inner city schools in my town seem always to be looking for tutors (or so they say in radio public service announcements), and I wonder if that would be suitable for a teenager to do as a community service activity, or if schools are mostly looking for adult tutors.[/quote]\r\n\r\nFWIW, I can say that I had a very small amount of 'real' volunteering under my belt when I applied to universities - probably less than 20 hours.  What I did have was sustained involvement in a variety of activities - I was editor in chief of my school's newspaper, I created my school's math club (an activity I continue to be involved in), I had been involved with student council in the past, I was a (founding) member of a mentorship program for younger students - heck, I was even a pretty competitive table tennis player.  To these causes, I gave hundreds of hours.\r\n\r\nI considered tutoring as a volunteer activity, but tutoring was my main source of income during high school - I was tutoring up to 8 hours a week at one point and making about 200 dollars a week, income that I was unwilling to give up for the sake of resume padding."
}
{
  "Tag": [
    "IMC",
    "college contests"
  ],
  "Problem": "Nearly 2 weeks left to the great competition and seems like everybody is so busy with preparation except me, that nobody is starting the topic  :D \r\n\r\nWho knows, may be we will get the answer to the topic after IMC  :lol:\r\n\r\n[b]Edit:[/b] National University of Uzbekistan - me  :oops: (no team leader)",
  "Solution_1": "I am. Won't be much of a trip, though :P\r\nAm guessin' i'm gonna go by bike..",
  "Solution_2": "There should be a team from Ukraine.\r\n\r\nStudents:\r\n1. FESHCHENKO Ivan (Kyiv Taras Shevchenko National University) - 1st prize at IMC-2008, gold medal at SEEMOUS-2008, 1st prize at IMC-2007 \r\n2. RADCHENKO Danylo (Kyiv Taras Shevchenko National University) - 1st prize at IMC-2008, gold medal at SEEMOUS-2008, gold medal at IMO-2007, bronze medal at IMO-2006\r\n3. SLOBODIANIUK Sergii (Kyiv Taras Shevchenko National University) - 1st prize at IMC-2008, silver medal at Iran ISMO-2008, 1st prize at IMC-2007, 2nd prize at IMC-2006, gold medal at IMO-2005\r\n4. KRAVETS Oleksandr (Kyiv Taras Shevchenko National University) - 1st prize at IMC-2008, silver medal at Iran ISMO-2008, 1st prize at IMC-2007, 2nd prize at IMC-2006, bronze medal at IMO-2005\r\n5. TYMOSHKEVYCH Larysa (Kyiv Taras Shevchenko National University) - 1st prize at IMC-2008, bronze medal at Iran ISMO-2008, 2nd prize at IMC-2007, 2nd prize at IMC-2006\r\n6. SHYSHATSKYI Iurii (Kyiv Taras Shevchenko National University) - gold medal at IMO-2008, bronze medal at IMO-2007\r\n7. IURCHENKO Ivan (Kyiv Taras Shevchenko National University) - 1st prize at IMC-2008, gold medal at SEEMOUS-2008, 1st prize at IMC-2007\r\n8. TANTSIURA Maksym (Kyiv Taras Shevchenko National University) - 2nd prize at IMC-2008, silver medal at SEEMOUS-2008\r\n9. LIVINSKYI Ivan (Kyiv Taras Shevchenko National University) - 2nd prize at IMC-2008, 2nd prize at IMC-2007, 2nd prize at IMC-2006\r\n10.KOVALENKO Oleg (Dnipropetrovsk Oles Gonchar National University)\r\n11.APYSHKOV Mykhailo (Donetsk National University)\r\n12.DZHULGAKOV Dmytro (National Technical University \"Kharkiv Polytechnic Institute\")\r\n\r\nLeaders:\r\n1.BONDARENKO Andrii (Kyiv Taras Shevchenko National University)\r\n2.MITIN Dmytro (Kyiv Taras Shevchenko National University)\r\n3.RIABYCH Nataliia (Ministry of Education and Science of Ukraine)\r\n\r\n\r\nAlso there should be a team from Odessa I.I.Mechnikov National University (2 students and 1 leader) and National Technical University of Ukraine \"Kyiv Polytechnic Institute\" (2 students and 1 leader).",
  "Solution_3": "This will most likely be the team from Zagreb, there are ten of us this year (if we, unlike before Vojtech, manage not to lose a couple of guys on illnesses or, believe it or not, burglaries):\r\n\r\nNikola Adzaga, second year\r\nGoran Drazic, third year\r\nNikola Grubisic, fourth year\r\nZeljko Kereta, third year\r\nRudi Mrazovic, fourth year\r\nMarin Misur, second year\r\nMelkior Ornik, first year\r\nPetar Sirkovic, second year\r\nSasa Stanko, second year\r\nLuka Zunic, second year\r\n\r\nOur team leader will be Matija Basic.",
  "Solution_4": "Seen a lot of ppl out there. Gonna be tough))\r\nTomorrow is a first day. Good luck everyone!\r\n\r\nP.S.: my room number is 449."
}
{
  "Tag": [
    "quadratics",
    "algebra",
    "quadratic formula",
    "special factorizations",
    "polynomial",
    "sum of roots"
  ],
  "Problem": "Please help me with this:\r\nThe sum of all the integers up to n (including n) is equal to 136. what is n? (you cannot use calculators)\r\n\r\n[hide=\"how far i got\"]First i know that $ \\frac{n(n\\plus{}1)}{2}$ is the sum all integers including n. So, we have $ \\frac{n(n\\plus{}1)}{2}\\equal{}136$ but once i solve i get $ n^2\\plus{}n\\equal{}272$ which i can't solve because if i use the quadratic formula i can't do the square root part so can someone please help me with a completing the square approach?[/hide]",
  "Solution_1": "you got to $ n^2 \\plus{} n \\equal{} 272 \\implies n^2 \\plus{} n \\minus{} 272 \\equal{} 0$\r\n\r\nsum of the roots of $ ax^2 \\plus{} bx \\plus{} c \\equal{} 0$ is $ \\minus{} \\frac {b}{a}$.\r\n\r\na=1, b=1.\r\n\r\nanswer : -1\r\n\r\nEDIT : \r\n\r\noh meh sry i misread the prob\r\n\r\nwhered -1 come from",
  "Solution_2": "First of all, memorize your perfect squares. That makes it really REALLY obvious that $ n \\equal{} 16$ (just by estimation).\r\n\r\nSecondly, I don't see why you cant do the quadratics.\r\n\r\n$ \\frac {(n)(n \\plus{} 1)}{2} \\equal{} 136$\r\n\r\n$ (n)(n \\plus{} 1) \\equal{} 272$\r\n\r\n$ n^2 \\plus{} n \\minus{} 272 \\equal{} 0$\r\n\r\n$ n \\equal{} \\frac { \\minus{} 1\\pm\\sqrt {1^2 \\plus{} 4\\times 272}}{2}$\r\n\r\n$ n \\equal{} \\frac { \\minus{} 1\\pm\\sqrt {1089}}{2}$\r\n\r\n$ n \\equal{} \\frac { \\minus{} 1\\pm33}{2}$\r\n\r\nThe answer's obviously positive, so the root is obviously $ 16$.\r\n\r\nEDIT: Lol fantasy, sum of first n whole numbers, not sum of roots  :)\r\n\r\nEDIT2: Completing the square is too beast and takes too long for a problem like this  :wink:",
  "Solution_3": "But Fantasy, since there are 2 solutions you don't want the sum, you want each of them individually.\r\n\r\nEDIT: Sorry, tex. I didn't see your post. But I stand by what I said. I think Fantasy just misread it.  :lol:",
  "Solution_4": "$ \\sqrt{1089}\\neq 3$. It equals 33.",
  "Solution_5": "Yeh, I figured that out... 3 mins after my post  :wink:\r\n\r\nEDIT: I'm sure editing a lot... It's ok super  :)",
  "Solution_6": "the method asked for was completeing the square\r\n\r\n$ n^2\\plus{}n\\equal{}272$\r\ni hope you already know this, but to complete the square $ x^2\\plus{}bx$, add $ \\frac{b^2}{4}$\r\ni will not explain it here, but you can ask someone else\r\n\r\nso we add $ \\frac14$ to both sides \r\n\r\n$ n^2\\plus{}n\\plus{}\\frac14\\equal{}\\frac{1089}{4}$\r\nthe left side may be simplified to (n+1/2)^2\r\n$ (n\\plus{}\\frac12)^2\\equal{}\\frac{1089}{4}$\r\n\r\nnow taking the square root of 1089 either requires memorization, a calculator (which we cant use) or a slight bit of work\r\nto take the square root of 1089 (assuming it is a natural number), first note that the square root is between 30 and 40 since it is between $ 30^2$ and $ 40^2$\r\nthen, since it ends in a \"9\", the square root ends in 3 or 7\r\nthen test 33 and 37 and you find it is 33\r\n\r\nnow taking the square root of both sides in the previous equation...\r\n$ n\\plus{}\\frac12\\equal{}\\pm \\frac{33}{2}$\r\n\r\nwhich yields solutions 16 or -17 (but -17 is extraneous), so 16 is the answer",
  "Solution_7": "For more on completing the square, click [url=http://www.artofproblemsolving.com/Forum/weblog.php?ppage=3&sep=%2F&w=1172&start=39]here[/url]. See \"Quadratics Part 3\".",
  "Solution_8": "i know how to complete the square  :D \r\nanyway thanks to every single one who posted here!\r\nmods, you can safely delete this if you think its getting too spammy but i think it isn't",
  "Solution_9": "[quote=\"Poincare\"]Please help me with this:\nThe sum of all the integers up to n (including n) is equal to 136. what is n? (you cannot use calculators)\n\n[hide=\"how far i got\"]First i know that $ \\frac {n(n \\plus{} 1)}{2}$ is the sum all integers including n. So, we have $ \\frac {n(n \\plus{} 1)}{2} \\equal{} 136$ but once i solve i get $ n^2 \\plus{} n \\equal{} 272$ which i can't solve because if i use the quadratic formula i can't do the square root part so can someone please help me with a completing the square approach?[/hide][/quote]\r\n\r\nI have another solution:\r\n\r\n$ 1 \\plus{} 2 \\plus{} 3 \\plus{} .... \\plus{} (n \\minus{} 1) \\plus{} n \\equal{} 136$\r\n\r\n$ \\begin{array}{l}\r\n  \\leftrightarrow \\frac{{n(n \\plus{} 1)}}{2} \\equal{} 136 \\leftrightarrow n^2  \\plus{} n \\minus{} 272 \\equal{} 0 \\\\ \r\n  \\leftrightarrow (n \\minus{} 16)(n \\plus{} 17) \\equal{} 0 \\leftrightarrow \\left[ \\begin{array}{l}\r\n n \\equal{} 16* \\\\ \r\n n \\equal{}  \\minus{} 17 \\\\ \r\n \\end{array} \\right. \\\\ \r\n \\end{array}$\r\n\r\nWith $ n\\equal{}16$ The sum of all the integers up to n (including n) is equal to 136.",
  "Solution_10": "ALTERNATIVE SOLUTION!\r\n\r\nWe can divide out 2 repeatedly from 134 and find it factors into $ 8\\cdot{17}$\r\n\r\nThat becomes equal to $ \\frac{16\\cdot{17}}{2}$ which means our answer is $ 16$",
  "Solution_11": "what mewto said gave me an idea to find the square root quickly, prime factorize  :|"
}
{
  "Tag": [
    "number theory",
    "relatively prime"
  ],
  "Problem": "Let a,b,c,d be relatively prime to m=ad-bc. Prove that the pairs of integers (x,y) for which ax+by is a multiple of m are identical with those for which cx+dy is a multiple of m.",
  "Solution_1": "[hide=\"One line\"] $ (x, y) \\equal{} (dt\\minus{}bs, as\\minus{}ct)$ [/hide]"
}
{
  "Tag": [
    "calculus",
    "calculus computations"
  ],
  "Problem": "[b]You can see now an interesting Romanian site \nhttp://www.mateforum.ro with the moderator [u]Cezar LUPU[/u] ![/b]",
  "Solution_1": "Congratulations on your opening site! :) \r\n\r\nkunny",
  "Solution_2": "[b]Thank you ![/b] [u]You are the first my quest ![/u]"
}
{
  "Tag": [
    "inequalities proposed",
    "inequalities"
  ],
  "Problem": "For $ a>b>c>0$, prove :\r\n\r\n$ \\frac{\\sqrt{a}}{(a\\minus{}b)(a\\minus{}c)}\\plus{}\\frac{\\sqrt{b}}{(b\\minus{}c)(b\\minus{}a)}\\plus{}\\frac{\\sqrt{c}}{(c\\minus{}a)(c\\minus{}b)}$\r\n\r\n$ <0<\\frac{1}{\\sqrt{a}(a\\minus{}b)(a\\minus{}c)}\\plus{}\\frac{1}{\\sqrt{b}(b\\minus{}c)(b\\minus{}a)}\\plus{}\\frac{1}{\\sqrt{c}(c\\minus{}a)(c\\minus{}b)}$",
  "Solution_1": "I doubt that we could possibly factorize these expressions in the nicer form. Nevertheless, set $ x \\equal{} \\sqrt {a}$,            $ y \\equal{} \\sqrt {b}$,             $ z \\equal{} \\sqrt {c}$.\r\n\r\nFrom the given condition, it follows $ x > y > z$\r\n\r\nBy the way, it is advisable, not compulsory to give $ x > y > z$ because after the factorization, we get the following \r\n\r\n$ \\sum_{cyc} \\frac{\\sqrt{a}}{(a\\minus{}b)(a\\minus{}c)} \\equal{}\\minus{}\\frac{1}{(\\sqrt{a} \\plus{} \\sqrt{b})(\\sqrt{b}\\plus{} \\sqrt{c})(\\sqrt{c}\\plus{}\\sqrt{a})}$\r\n\r\nthe same method for the second inequality.\r\n\r\nAs we can see, the condition given is not critically necessary even though the fulfilled constrain must be \r\n[i]x,y, z are different from each other.[/i]"
}
{
  "Tag": [
    "inequalities",
    "function",
    "induction",
    "inequalities proposed"
  ],
  "Problem": "Let $f$ be  convex function ,then for $0,a_1,...,a_n \\in D_f$\r\n\r\n\r\n$f(a_1)+f(a_2)+...+f(a_n) \\le f(a_1+a_2+...+a_n)+(n-1)f(0)$",
  "Solution_1": "I solved it by a simple induction.\r\nThe step of induction is trivial so what remains is just prove that:\r\n\r\nLet $f$ be a convex function on $[a,b]$.\r\nfor $x,y,0 \\in [a,b]$\r\n\r\n$f(x)+f(y) \\le f(x+y)+f(0)$"
}
{
  "Tag": [
    "limit",
    "integration",
    "calculus",
    "Support",
    "function",
    "Euler",
    "real analysis"
  ],
  "Problem": "Find the limit:\r\n\r\n$\\lim_{n\\rightarrow \\infty}\\frac{\\sum_{j=1}^{\\infty}\\frac{j^{n}}{e^{j}}}{n!}$.",
  "Solution_1": "${\\frac{\\sum_{j=1}^{\\infty }\\frac{j^{n}}{e^{j}}}{n!}=\\frac{\\text{Li}_{-n}\\left(\\frac{1}{e}\\right)}{n!}=\\frac{\\text{Li}_{-n}\\left(\\frac{1}{e}\\right)}{\\Gamma (n+1)}}$\r\n\r\nMAking approximations I think the result is $1$.\r\n\r\nDon't have a proof. :)",
  "Solution_2": "My first thought:\r\n\r\n$n!=\\int_{0}^{\\infty}x^{n}e^{-x}\\,dx=\\sum_{j-1}^{\\infty}j^{n}e^{-j}+\\int_{0}^{\\frac12}x^{n}e^{-x}+\\sum_{j=1}^{\\infty}\\int_{j-\\frac12}^{j+\\frac12}\\left(x^{n}e^{-x}-j^{n}e^{-j}\\right)$\r\n\r\nThe integral from zero to $\\frac12$ is trivial, of course. I would conjecture that the sum of integrals is small compared to $n!$, which would support yagaron's conjecture that the overall limit is 1. I don't yet have an argument for that that I'm happy with.",
  "Solution_3": "The function $f(t) = t^{n}e^{-t}$ increases on the segment $[1,n]$; and it decreases on $[n,+\\infty)$. Therefore,\r\n\\[\\int_{1}^{n}f(t) dt+f(1) \\leq \\sum_{j=1}^{n}j^{n}e^{-j}\\leq \\int_{1}^{n}f(t) dt+f(n). \\]\r\nNote that $f(n) = (n/e)^{n}= O(n!/\\sqrt{n})$. Hence\r\n\\[\\sum_{j=1}^{n}j^{n}e^{-j}= \\int_{1}^{n}t^{n}e^{-t}dt+o(n!). \\]\r\nSimilarly,\r\n\\[\\sum_{j=n+1}^{\\infty}j^{n}e^{-j}= \\int_{n}^{\\infty}t^{n}e^{-t}dt+o(n!). \\]\r\nCombining these two estimates we get\r\n\\[\\sum_{j=1}^{\\infty}j^{n}e^{-j}= \\int_{1}^{\\infty}t^{n}e^{-t}dt+o(n!) = (\\Gamma(n+1)-O(1))+o(n!) = n! (1+o(1)). \\]",
  "Solution_4": "${\\frac{1}{\\Gamma (n+1)}\\sum_{j=1}^{\\infty }\\frac{j^{n}}{e^{j}}\\approx \\frac{1}{\\Gamma (n+1)}\\int_{1}^{\\infty }\\frac{j^{n}}{e^{j}}\\, dj=\\frac{\\Gamma (n+1,1)}{\\Gamma (n+1)}\\approx 1}$\r\n\r\nAs $n$ get larger the approximation turn in a equality.\r\n\r\nwe can use this, don't we?",
  "Solution_5": "Nice.\r\n\r\nYury's solution is elegant and it is  based on , or probably I should say simillar to, the method of studying Euler-like sequences.",
  "Solution_6": "Answer is $\\frac{1}{2}$",
  "Solution_7": "Please, take time to read the previous posts in the thread before posting yourself. Do you see any flaws in Yury's proof? If not, you'd better check your own argument before posting a different answer. Also, just posting an unsupported answer after somebody posted a full solution is hardly ever appropriate :spam:",
  "Solution_8": ":lol: \r\nOK\r\n$\\int_{0}^{x}t^{x}e^{-t}dt$~$\\frac{1}{2}\\Gamma(1+x)$\r\nand $n!=\\Gamma(1+n)$\r\nwhen $x\\to+oo$ and $n\\to+oo$",
  "Solution_9": "This would correspond to $\\sum_{j=1}^{n}\\frac{j^{n}e^{-j}}{n!}$, not to the sum $\\sum_{j=1}^\\infty$."
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "trigonometry"
  ],
  "Problem": "In a triangle, the sum of the length of 2 sides is 10, and the height of the third side is 3.\r\nFind the maximum value of the radius of the circumcircle of the triangle.",
  "Solution_1": "Let $ a\\plus{}b\\equal{}10, \\ h_c\\equal{}3$\r\n\r\n$ \\frac{a}{\\sin A}\\equal{}\\frac{b}{\\sin B}\\equal{}\\frac{a\\plus{}b}{\\sin A\\plus{}\\sin B}\\equal{}2R$\r\n\r\nBut $ \\sin A\\equal{}\\frac{h_c}{b}\\equal{}\\frac{3}{b}, \\ \\sin B\\equal{}\\frac{h_c}{a}\\equal{}\\frac{3}{a}$\r\n\r\nThen $ R\\equal{}\\frac{1}{2}\\cdot\\frac{a\\plus{}b}{\\frac{3}{a}\\plus{}\\frac{3}{b}}$\r\n\r\n$ R\\equal{}\\frac{1}{6}ab\\equal{}\\frac{1}{6}a(10\\minus{}a)\\equal{}\\minus{}\\frac{1}{6}a^2\\plus{}\\frac{5}{3}a$\r\n\r\n$ R_{max}\\equal{}\\frac{25}{6}$"
}
{
  "Tag": [
    "calculus",
    "integration",
    "trigonometry",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Evaluate: \\[ \\int\\frac{\\cos 2x}{8\\left(\\sin^6x\\plus{}\\cos^6x\\right)\\plus{}5\\sin 4x}\\]",
  "Solution_1": "[hide]As an exercise, try to show that   $ \\sin^6x \\plus{} \\cos^6x \\equal{} \\frac{5 \\plus{} 3 \\cos(4x)}{8}$. Therefore,\n\n$ P \\equal{} \\int\\frac{\\cos(2x)}{5 \\plus{} 3 \\cos(4x) \\plus{} 5 \\sin(4x)}\\,dx$.[/hide]"
}
{
  "Tag": [],
  "Problem": "The expression $\\frac{7n + 18}{2n + 3}$ takes integer values for certain integer values of $n$. What is the sum of all such values of the expression?\r\n\r\nSource: Australian Math Competition, S28, 2000",
  "Solution_1": "[hide]$2n + 3 \\mid 7n + 18$ $\\Longrightarrow$ $2n + 3 \\mid 15$.\nThat gives 8 values of $n$: 6, 1, 0, -1, -9, -4, -3, -2. \nSo the sum should be $28$. [/hide]",
  "Solution_2": "Nope... the question asks for the sum of all the possible values of $\\frac{7n + 18}{2n + 3}$ which are integers. And there are more than 4 values as well. :)",
  "Solution_3": "[quote=\"Andreas\"]$2n + 3 \\mid 7n + 18$ $\\Longrightarrow$ $2n + 3 \\mid 15$. [/quote]\r\nUmm...how did you get this? :?  :blush:",
  "Solution_4": "$2n + 3 \\mid 7n + 18$\r\n$2n + 3 \\mid 14n + 36$\r\n$2n + 3 \\mid 14n + 36 - 14n - 21$\r\n$2n + 3 \\mid 15$.",
  "Solution_5": "Err, one possible integer value of the expression occurs when $n = -2$. Then $\\frac{7n + 18}{2n + 3} = \\frac{-14 + 18}{-4 + 3} = -4$ ,which is an integer. We want to find all the sum of all such expressions which are integers when $n$ is an integer.",
  "Solution_6": "I've edited my previous post. Please, look it.  :)",
  "Solution_7": "Ooh, didn't see that.  :blush:  It's absolutely correct.  :)"
}
{
  "Tag": [
    "vector",
    "geometry",
    "geometric transformation",
    "topology",
    "group theory",
    "abstract algebra"
  ],
  "Problem": "How to show that the collection (group) of orthogonal matrices in R^n is orientable?",
  "Solution_1": "Every Lie group is orientable as you can choose a local orientation in an arbitrary point and then move the orientation to each other point by the map on homology induced by multiplication.",
  "Solution_2": "I think I get the general idea of extending a local orientation using the multiplication,\r\nhowever, it's not quite clear to me how to do it formally.\r\ncould you give me a more detailed example?\r\n\r\nand I'm not familiar with the term Homology...",
  "Solution_3": "no need to know anything about homologies to solve the problem.\r\nlet $ G$ be a lie group, and consider a basis $ \\{E_1,\\dots,E_n\\}$ for $ T_eG$ (the tangent space to $ G$ in the identity), and define $ n$ vector fields by $ (X_i)_g \\equal{} dL_gE_i$ for $ i\\equal{}1,\\dots,n$, where $ L_g$ is the left translation in $ G$. since $ \\{(X_i)_g\\}$ is linearly independent for every $ g\\in G$, it follows that the tangent space is trivial (as a vector bundle), in particular $ G$ is orientable.\r\n\r\nnow, the fact that $ O(n)$ is a lie group should be standard.",
  "Solution_4": "[quote=\"ma_go\"]let $ G$ be a lie group, and consider a basis $ \\{E_1,\\dots,E_n\\}$ for $ T_eG$ (the tangent space to $ G$ in the identity), and define $ n$ vector fields by $ (X_i)_g \\equal{} dL_gE_i$ for $ i \\equal{} 1,\\dots,n$, where $ L_g$ is the left translation in $ G$. since $ \\{(X_i)_g\\}$ is linearly independent for every $ g\\in G$, it follows that the tangent space is trivial (as a vector bundle), in particular $ G$ is orientable.[/quote]\nin effect, what we're doing is choosing (in a smooth fashion) a basis at each point on the manifold?\n and is that the meaning of \"trivial tangent space\"?\nor is it that $ G$ is diffeomorphic to $ TG$? \n\n[quote=\"ma_go\"]now, the fact that $ O(n)$ is a lie group should be standard.[/quote]\r\nobviously",
  "Solution_5": "[quote=\"Gilad.R\"]in effect, what we're doing is choosing (in a smooth fashion) a basis at each point on the manifold?\nand is that the meaning of \"trivial tangent space\"?\nor is it that $ G$ is diffeomorphic to $ TG$? [/quote]\r\n\r\nin order:\r\n1. no, we're doing more: we're finding linearly independent sections in any point. finding a basis at each point \"in a smooth fashion\" is orientability: finding such a basis should be the triviality of the principal $ O(n)$-bundle associated to the tangent bundle of the manifold. for a (tough) review on this, you can take a look at either kobayashi-nomizu's classical book (first volume, i don't remember the title) or steenrod's \"fiber bundles\", for something easier (and of course less complete), try wikipedia.\r\n\r\n2. a bundle $ F\\to E \\stackrel{\\pi}{\\to} B$ is trivial if it's isomorphic to the trivial bundle $ F\\to F\\times B \\stackrel{\\pi_2}{\\to} B$, e.g. if there's an homeomorphism $ \\phi: E\\to F\\times B$ between total spaces commuting with projections, that is $ \\pi \\equal{} \\pi_2\\circ\\phi$. in the simpler case of vector bundles (which is what we're dealing with), being trivial is equivalent to have $ {\\rm rk}\\, E \\equal{} \\dim F$ linearly independent (at each point) sections: this is a good exercise to get a handle on definitions.\r\n\r\n3. $ G$ can't be diffeomorphic to $ TG$ unless $ \\dim G \\equal{} 0$, by a simple consideration on dimensions.",
  "Solution_6": "3. I meant $ G\\times T_{1}G$ is diff. to $ TG$...\r\n\r\nanyways, thanks!"
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "sphere",
    "analytic geometry",
    "symmetry"
  ],
  "Problem": "Given a cube of unit side. Let $A$ and $B$ be two opposite vertex. Determine the radius of the sphere, with center inside the cube, tangent to the three faces of the cube with common point $A$ and tangent to the three sides with common point $B$.",
  "Solution_1": "By \"sides\", do you mean the edges or the faces?",
  "Solution_2": "Sorry, I mean the edges.",
  "Solution_3": "well, you could do a coordinate plane, and intersect the sphere with the lines from B with a random radius, and note the symmetry that the intersections have; if it is tangent to one, it is tangent to all of them, we know that it is in the form $(x-r)^2+(y-r)^2+(z-r)^2=r^2$, if A is the origin, and it is tangent to the positive planes",
  "Solution_4": "Not completely correct. A far more concise method would be to take a sphere and both inscribe and circumscribe cubes. We can easily find the sidelength of each cube (the radius of the sphere is half the sidelength of one and half the long diagonal of the other). Now we halve the difference between the sidelengths and add this value to the sidelength of the smaller cube. This is the sidelength of the desired cube. \r\n\r\nThe inverse operation is used to go from the cube to the sphere.  :)  Hope that wasn't to wordy.",
  "Solution_5": "i don't really understand how that works because when you circumscribe of inscrible a cube in a sphere, you never have the sphere tangent to the edges, just tangent at the verticies or centers of the faces...?",
  "Solution_6": "[quote=\"Altheman\"]i don't really understand how that works because when you circumscribe of inscrible a cube in a sphere, you never have the sphere tangent to the edges, just tangent at the verticies or centers of the faces...?[/quote]\r\n\r\nThis problem is that it is tangent to three of the faces intersecting at one point, and the three edges intersecting at the opposite point.  Nevertheless, seethelight's solutions it not correct...\r\nThe sphere would be tangent to three faces of the cube intersecting at one vertex.  However, it would pass through the opposite vertex of the cube, rather than being tangent to the three edges.",
  "Solution_7": "Whoops, I meant that the first cube's sidelength was equal to the diameter of the cube, and that the second cube's shorter diagonal was also equal to the diameter. It works then. :blush:"
}
{
  "Tag": [
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "Very easy inequality with AM-GM",
  "Solution_1": "[quote]Let $ a,b,c$ be positive real numbers. Prove that\n\\[ a \\plus{} b \\plus{} c \\plus{} \\frac {1}{a} \\plus{} \\frac {1}{b} \\plus{} \\frac {1}{c} \\ge 4\\left( {\\frac {a}{{1 \\plus{} ab}} \\plus{} \\frac {b}{{1 \\plus{} bc}} \\plus{} \\frac {c}{{1 \\plus{} ca}}} \\right).\\][/quote]\r\n\r\n\r\n$ a \\plus{} \\frac {1}{b} \\equal{} a\\left(1 \\plus{} \\frac {1}{ab}\\right) \\ge \\frac {4a}{1 \\plus{} ab}.$",
  "Solution_2": "Thank you. Your solution is very nice!"
}
{
  "Tag": [
    "MATHCOUNTS",
    "quadratics",
    "algebra",
    "polynomial",
    "AMC",
    "AIME",
    "HMMT"
  ],
  "Problem": "This is, in my opinion, one of the best MathCounts problems ever.\r\n\r\nIf $ x^{2} \\minus{} 5x \\plus{} 8 \\equal{} 1$, find $ x^{4} \\minus{} 10x^{3} \\plus{} 25x^{2} \\minus{} 9$.",
  "Solution_1": "i think i got this at the competition, but i forgot if i did. i know that it's based on some clever multiplication/factoring, but i dont remember. can anyone please post a solution?",
  "Solution_2": "ugh, how did I not get this at the competition?\r\n\r\n[hide]\nWe can change this equation to $ x^2\\minus{}5x\\equal{}\\minus{}7$, and squaring both sides gives us $ x^3\\minus{}10x^3\\plus{}25x^2\\equal{}49$.  Should be obvious that the thing we're looking for is $ \\boxed{40}$\n[/hide]",
  "Solution_3": "[hide]\nWe try to complete the square for the first equation.\n\nSo we have $ (x - 2.5)^2 = - 0.75$. Yeah, $ x$ is nonreal, deal with it.\n\nNote that the second equation is $ x^2(x - 5)^2 - 9$.\n\nLet $ y = x - 2.5 = - 0.75$. Now our answer is\n\n\\begin{align*} & (y - 6.25)^2 - 9 \\\\\n= & ( - 7)^2 - 9 \\\\\n= & \\fbox{40} \\\\\n\\end{align*}\n\n[/hide]\r\n\r\nBeaten. With a simpler, although typoed, solution. :)",
  "Solution_4": "[hide=\"Hint for alternate but essentially the same solution\"] Write the desired expression as $ (x^{2}\\minus{}5x\\plus{}a)(x^{2}\\minus{}5x\\plus{}b)\\plus{}c$. Do you see the motivation for this?[/hide]",
  "Solution_5": "[quote=\"CatalystOfNostalgia\"][hide=\"Hint for alternate but essentially the same solution\"] Write the desired expression as $ (x^{2} \\minus{} 5x \\plus{} a)(x^{2} \\minus{} 5x \\plus{} b) \\plus{} c$. Do you see the motivation for this?[/hide][/quote]\r\n\r\nOHH YEAH! That's how I solved it at states. I forgot how exactly I did it.",
  "Solution_6": "how is this one of the best mathcounts problems ever =P?  This is in essence similar to a countdown round problem in 2005 with Patricia Li vs. David Benjamin, where she squared some expression to get 49, while David did the problem and got 56. xD...\r\n\r\nNote that since the $ x$ term is not in the second expression and that $ 10\\equal{}2*5$ while $ 25\\equal{}5^2$, you should definitely go for $ (x^2\\minus{}5x)^2$.\r\n\r\nOr note that if you do $ ((x^2\\minus{}5x)\\plus{}8)((x^2\\minus{}5x)\\minus{}8)$, you also cancel out the $ x$ terms.\r\n\r\nThis gives us $ x^4\\minus{}10x^3\\plus{}25x^2\\minus{}64\\equal{}1*(1\\minus{}16)\\equal{}\\minus{}15$\r\n\r\nThus to get the desired expression, we add $ 55$ to come up with $ 40$.",
  "Solution_7": "aaaaaaaaaaaaaaaaaa",
  "Solution_8": "Okay, somehow, I misread a sign, did the exact same thing, but put -40, even though the sign thing shouldn't have made a difference at all.\r\n\r\nI suck.",
  "Solution_9": "For the friendly moderator who altered the subject line, it was State.",
  "Solution_10": "[hide=\"HUGE HINT\"]\n\nsquare the quadratic  :P \n\n[/hide]",
  "Solution_11": "my method is a little similar to mathcrazed's. \r\n\r\ni used polynomial long division and got\r\n\r\n$ x^4 \\minus{} 10x^3 \\plus{} 25x^2 \\minus{} 9$ = $ (x^2 \\minus{} 5x \\plus{} 8)(x^2 \\minus{} 5x \\minus{} 8) \\plus{} 55$\r\n\r\n= and we know that $ (x^2 \\minus{} 5x \\plus{} 8) \\equal{} 1$ and $ (x^2 \\minus{} 5x \\minus{} 8)$ is 16 less cuz the constant on the first one is 8 and on the second one is -8 making a diff. of 16 since all the other terms are the same in both. \r\n\r\nso its just $ 1\\times(\\minus{}15)$ + $ 55$ = $ 55\\minus{}15$ = $ \\fbox{040}$  :D there i made it an AIME problem.",
  "Solution_12": "[quote=\"cognos599\"]my method is a little similar to mathcrazed's. \n\ni used polynomial long division and got\n\n$ x^4 \\minus{} 10x^3 \\plus{} 25x^2 \\minus{} 9$ = $ (x^2 \\minus{} 5x \\plus{} 8)(x^2 \\minus{} 5x \\minus{} 8) \\plus{} 55$\n\n= and we know that $ (x^2 \\minus{} 5x \\plus{} 8) \\equal{} 1$ and $ (x^2 \\minus{} 5x \\minus{} 8)$ is 16 less cuz the constant on the first one is 8 and on the second one is -8 making a diff. of 16 since all the other terms are the same in both. \n\nso its just $ 1\\times( \\minus{} 15)$ + $ 55$ = $ 55 \\minus{} 15$ = $ \\fbox{040}$  :D there i made it an AIME problem.[/quote]\r\nHow did you use polynomial long division to factor the 4th degree polynomial? :huh:\r\nEDIT: nevermind. Just being out of my mind, oops :oops:  :mad: I know how to do long division!",
  "Solution_13": "Yeh, I thought this was a really slick problem, with, ofcourse, a really slick solution. The complex root may have tricked some people. One of the best.. definitely.",
  "Solution_14": "Although you have motivation that the quadratic factors of the quartic start with x^2-5x, a general way of factoring quartics is just writing it as $ (x^{2}\\plus{}ax\\plus{}b)(x^{2}\\plus{}cx\\plus{}d)$, equating coefficients, and solving a system of equations. The system of equations can get messy, but if you know you're looking for integers, you can usually reduce it to a small number of cases.\r\n\r\nTry this on HMMT Algebra 2006...it totally kills two of the harder problems  :P"
}
{
  "Tag": [
    "ratio",
    "limit",
    "geometry",
    "geometry proposed"
  ],
  "Problem": "Calculate ${\\lim_{\\theta \\to \\frac{3 \\pi }{4}}\\, \\frac{\\text{CF}}{\\text{CD}}}$\r\n\r\nthis problem is very hard, but very beautiful :)\r\n\r\nIt's my own. i spent some time making this :)\r\n\r\n\r\nPS: This problem has 2 solutions. because the limit of the rigth and the left is different.\r\n\r\nI put just one limit to simplificate. but actually is 2",
  "Solution_1": "i don't see why it is a limit... the problem is solved by putting in that value of theta, making DE and AF parallel, then constructing point C \r\n\r\nbasically, <CED=60, <FEA=90, < DEA=135,\r\n\r\nsince CD and CF are chords, it is trivial now to compute the lenghts since we have their arc measures",
  "Solution_2": "[quote]PS: This problem has 2 solutions. because the limit of the rigth and the left is different.[/quote]\r\n\r\nthe limit is because we have a triangle, not parallel lines.\r\n\r\nand there is 2 possible solutions calculated from the limit",
  "Solution_3": "Yagaron, you know the solution of your own problem and such problems should be posted in the section [url=http://www.mathlinks.ro/Forum/index.php?f=47]Geometry Proposed & Own Problems[/url].",
  "Solution_4": "hey dude, i know that.\r\ni clicked on the wrong link and didnt noticed it\r\n\r\na commom mistake"
}
{
  "Tag": [
    "AMC",
    "AMC 8",
    "AIME",
    "USA(J)MO",
    "USAMO",
    "AMC 10"
  ],
  "Problem": "Thanks for voting. Good luck on the AMC10/12/beyond!",
  "Solution_1": "Phailed.  24.",
  "Solution_2": "[quote=\"numbermaster\"]Thanks for voting. Good luck on the AMC10/12/beyond![/quote]\r\n\r\nHmm... I think you should remove the labels on the scores because to most AoPSers, 24 is fail...",
  "Solution_3": "Yeah.  Even to me, 24 is fail.  And I'm not that smart.\r\n\r\nPythag, what did you get?",
  "Solution_4": "[quote=\"AIME15\"]Yeah.  Even to me, 24 is fail.  And I'm not that smart.\n\nPythag, what did you get?[/quote]\r\n\r\nI thought I failed.. but it turns out that I got 25.\r\n\r\nAnyways, some people got the wedge problem wrong because they thought the wedge was a cone...",
  "Solution_5": "[quote=\"pythag011\"][quote=\"numbermaster\"]Thanks for voting. Good luck on the AMC10/12/beyond![/quote]\n\nHmm... I think you should remove the labels on the scores because to most AoPSers, 24 is fail...[/quote]\r\n\r\n!!!!!!!! :)",
  "Solution_6": "im so angry because i got the most simple questions wrong\r\n\r\nso i got a 20-21",
  "Solution_7": "HOPEFULLY i got a 25, but i cant rememer my answers...:(\r\n\r\n\r\nWhat # was the wedge problem?",
  "Solution_8": "[quote=\"abacadaea\"]HOPEFULLY i got a 25, but i cant rememer my answers...:(\n\n\nWhat # was the wedge problem?[/quote]\r\n\r\n#21\r\n\r\ncut a cylinder into half, answer is 1/2 of the volume of cylinder :D",
  "Solution_9": "I would vote 25, but I did not officially take it. (I took it only because I was co-proctoring and my coach was taking it too - we finished at about the same time, which was 23 minutes in, 17 minutes remaining!)",
  "Solution_10": "Tell me the wedge thing was C please",
  "Solution_11": "IDK, but the numeric answer was 151.",
  "Solution_12": "Thank god...",
  "Solution_13": "25 BOGTRO?  GJ!",
  "Solution_14": "I brought new meaning to the word epic-fail. You know #8? The one with the bar graph where you had to find the average sales? Yeah. 24/25. After a perfect score last year.  :( :sob:  :(",
  "Solution_15": "Hey, I don't actually know what my score was...maybe you guys can help me figure it out. (Note: I had never taken the AMC before and I was just happy to qualify for the AIME, so my score is probably shameful by your standards  :oops: ) Okay, so I emailed the AMC people with my information and requested my scores, and this is what they sent back:\r\n\r\n[i]American Mathematics Competitions \n2008 AIME Results \n\nStudent Name AMC 12 AMC 10 AIME \nINDEX*\n(My Name) 97.5 A 4 \n137.5 [/i]\r\n\r\nWhat does that mean? I took the AMC 12 if that helps with the interpretation at all...\r\nThanks!",
  "Solution_16": "I believe that means that you scored a 97.5 on the AMC 12 and a 4 on the AIME, making your USAMO index 97.5+4*10=137.5 :)",
  "Solution_17": "Thank you!",
  "Solution_18": "Strangely, that's pretty much the exact same score I got last year...97.5 and 6 on AIME.",
  "Solution_19": "LOL i missed #2",
  "Solution_20": "I got a 25!!!!!!!\r\nfinally!\r\n4th: 17\r\n5th: 22\r\n6th: 25 YES\r\n\r\n25 IS THE NEW 42!\r\n\r\n25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25",
  "Solution_21": "Congratulations! :) It's not very often that a 6th grader gets 25.",
  "Solution_22": "uh when i was in 6th grade i took the amc 8 unofficially and got a 25 :D",
  "Solution_23": "[quote=\"BOGTRO\"]I got a 25!!!!!!!\n[/quote]\r\n\r\n[sarcasm] that was completely unexpected [/sarcasm]  :rotfl:",
  "Solution_24": "24.... fail...\r\nI got the second to last one wrong somehow. The one with the 10 cards and the dice and multiplying for a perfect square.... so embarrassed...\r\n\r\nYea when I was in 6th grade I got like a 15 and I got a 22 last year so...",
  "Solution_25": "5th grade = 22",
  "Solution_26": "Ok so...\r\n\r\n6th grade: 13 (O_O how'd a get a 13???  :rotfl: )\r\n7th grade: 22\r\n8th grade: not taken, 2009",
  "Solution_27": "I got a 16. I have no clue how i got SUCH a bad score.",
  "Solution_28": "I got 24 because I numbered BEST OF LUCK like:\r\n1234556789\r\n\r\nI can actually see that's the work I did on the packet.",
  "Solution_29": "yep thats exactly what I did, hope I don't do it on the amc 10 too.",
  "Solution_30": "So when do the perfect scorer people's names come out?\r\n\r\nI want to see how badly my middle school phailed this year after all of our smart people left.\r\n\r\nBut I heard it was reaaaaaaaaallllllllllllyyyyyyy easy, compared to that of last year, and even during last year's, I was like, oh this difficulty is game over.",
  "Solution_31": "25. hmm.. this year's was a lot easier though.. maybe because calculators weren't allowed this year..",
  "Solution_32": "When does the AMC 8 list the perfect scorers on the website?",
  "Solution_33": "there's a 6th grade in my school that is crazy smart\r\nhe got a 25 on AMC8\r\n\r\nanyway, i thought that this one was a lot easier than last year's\r\nbut i don't think its because of the calcs\r\nweren't they first banned in 2006?",
  "Solution_34": "No, I think this was the first year they were banned, because people used them last year.",
  "Solution_35": "Why on the answers on the site for #12 it gives 3 answers?  :huh:  :huh:  :huh:",
  "Solution_36": "I beleive that when the test first came out calculators were not allowed.  I think the test was a lot easier than past tests.  I thought it was like a school competition in Mathcounts.  My kids had 7 perfects including one 6th grader.",
  "Solution_37": "success!!!\r\n25/25 :bruce:  :bruce:  :bruce:"
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "\\[ \\triangle ABC\\Longrightarrow \\sum \\frac{(b+c)^2}{b+c-a}\\le \\frac{2R}{r}\\cdot (a+b+c). \\]",
  "Solution_1": "Levi, your inequality follows immediately from this one: \\[ (b+c)^{2}(a+b-c)(a+c-b)\\leq 4a^{2}bc.  \\] ( this is GL.18 from my book  :D )\r\nSo, from here we have \\[ \\frac{(b+c)^{2}}{b+c-a}\\leq abc\\cdot\\frac{a}{2pr^{2}}.  \\]\r\nWriting the analogous onesand by summing them up, we get the desired inequality. ;)",
  "Solution_2": "[quote=\"Virgil Nicula\"]$ \\triangle ABC\\Longrightarrow$\n\n$\\sum \\frac{(b+c)^2}{b+c-a}\\le \\frac{2R}{r}\\cdot (a+b+c). $[/quote][quote=\"Cezar Lupu\"]Levi, your inequality follows immediately from this one: \\[ (b+c)^{2}(c+a-b)(a+b-c)\\leq 4a^{2}bc.  \\] [/quote]$4a^{2}bc-(b+c)^{2}(c+a-b)(a+b-c)=(b-c)^2\\left[(b+c)^2-a^2\\right]\\geq 0.$",
  "Solution_3": "[quote=\"Ji Chen\"][quote=\"Virgil Nicula\"]$ \\triangle ABC\\Longrightarrow$\n\n$\\sum \\frac{(b+c)^2}{b+c-a}\\le \\frac{2R}{r}\\cdot (a+b+c). $[/quote][quote=\"Cezar Lupu\"]Levi, your inequality follows immediately from this one: \\[ (b+c)^{2}(c+a-b)(a+b-c)\\leq 4a^{2}bc.  \\] [/quote]$4a^{2}bc-(b+c)^{2}(c+a-b)(a+b-c)=(b-c)^2\\left[(b+c)^2-a^2\\right]\\geq 0.$[/quote]\n\nMr.Chen,I really want to know how can you put every inequalities into sum of squares.I read you book about\nInequalities, that is fairly good.Also, I am a student in Jiangsu Province.\nCan you tell me some skills in it?Thank you very much.",
  "Solution_4": "[quote=\"edcwconan\"][quote=\"Ji Chen\"][quote=\"Virgil Nicula\"]$ \\triangle ABC\\Longrightarrow$\n\n$\\sum \\frac{(b+c)^2}{b+c-a}\\le \\frac{2R}{r}\\cdot (a+b+c). $[/quote][quote=\"Cezar Lupu\"]Levi, your inequality follows immediately from this one: \\[ (b+c)^{2}(c+a-b)(a+b-c)\\leq 4a^{2}bc.  \\] [/quote]$4a^{2}bc-(b+c)^{2}(c+a-b)(a+b-c)=(b-c)^2\\left[(b+c)^2-a^2\\right]\\geq 0.$[/quote]\n\nMr.Chen,I really want to know how can you put every inequalities into sum of squares.I read you book about\nInequalities, that is fairly good.Also, I am a student in Jiangsu Province.\nCan you tell me some skills in it?Thank you very much.[/quote]\n\n\n$4a^{2}bc-(b+c)^{2}(c+a-b)(a+b-c) = 4a^{2}bc - (b+c)^2[a^2-(b-c)^2] = (b-c)^2(b+c)^2 +a^2[(b+c)^{2}-4bc] = (b-c)^2[(b+c)^2-a^2]$"
}
{
  "Tag": [],
  "Problem": "Hello everyone. Just became a member and this was the closest sub-forum i could find that relates to music and mathematics. Anyways, I've been highly interested in the relationship between math and music (particulary piano performance) ever since the Mozart sonata study in 1993 (mozart effect). I didnt really research it until the mid-90's but i heard about it when the study was published. Just had a few questions:\r\n\r\nAre there any composers (ie:bach, greek composer yanni) that you feel or think have an effect on mathematical performance? if so, why?\r\n\r\nI've been playing the piano on and off since i was 9 (now in my 20's) and since i dont have time to dedicate myself to a huge repertoire, i focus my playing to mozart and bach.\r\n\r\nif you play the piano or know alot about this topic, please chime in. thx \r\n\r\n-david",
  "Solution_1": "I've never really thought about that before. But I have noticed there are some pieces you can just relax and play, and some you really have to think about as you play. I think maybe Bach has as an effect, because when I play inventions, I have to think about which hand has the melody in each measure, and how it will fit in the the big picture.\r\n\r\nThanks for this topic, I will be thinking about it more now.  :read:",
  "Solution_2": "I'm not exactly sure what you mean, but I know that different styles and periods of time require a different way of playing.\r\n\r\nBach, for example, is very metronomical\r\nwhen I play Bach, the notes fall either right on the beat or off the beat\r\nthe tempo is very fixed (not sure if that's the best way to explain.)\r\n\r\nHowever, composers like Chopin are different\r\nWhen I play Chopin, I feel as if I can strech out the beat a bit\r\nit is not as metronomical as Bach's pieces\r\n\r\nIs this what you mean by mathematical performance?",
  "Solution_3": "Well, of course, scientists have had an expiriment where they \r\nfound out that piano players do better on tests\r\nthan those who don't.\r\n\r\nSome people say that Mozart and Bach make you smarter.\r\nI think maybe that's because they're very organized.\r\nThey have a exact rhythm. \r\nHowever, sometimes you they make you kinda bored.\r\n\r\nSome Romantic styles make you really peaceful, which\r\nis why I guess they call it the Romantic period.\r\nI love listening to those pieces. \r\nThere's a lot of those pieces too!\r\n(Chopin being my favorite)\r\n\r\nThe Contemporary music is either really crazy (lots\r\nof flats, sharps, polyrhythms), or similar to the \r\nRomantic period. If you listen to modern pieces, \r\nthey make your blood pressure go up.\r\n\r\nHope I made it make more sense!  :D",
  "Solution_4": "i actually listen to alot of romantic piano music (mainly chopin) but since i usually only have about half hour to play piano a day. i focus on mozart/bach. i am actually looking into pieces of music that are conducive to learning. i love music and have audiophile headphones/great music collection/excellent mp3 player,etc.  any more suggestions would help. thx everyone",
  "Solution_5": "I've also been noticing things when I listen to Mozart/Bach because I really listen to how all the melodies combine into the piece.",
  "Solution_6": "Hm. I'd never really thought about that before, but I guess I think better doing my homework after playing a piece where I have to concentrate on many melodies at the same time.",
  "Solution_7": "I have very good rhythm, and my teacher says I must be a good math student. I usually laugh, but I really think that might have to do with it. Somehow, math and piano go together."
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "Gauss"
  ],
  "Problem": "Let [tex]p[/tex] be an odd prime. Show that the polynomial\r\n\r\n[tex]x^{p-1}+\\cdots+x+1[/tex]\r\n\r\ncan not be factored into a product of two polynomials of degree > 0 with integer coefficients.",
  "Solution_1": "Make the linear changle x = y + 1. Then use Eisenstein.",
  "Solution_2": "[quote=\"koopa\"]Make the linear changle x = y + 1. Then use Eisenstein.[/quote]\r\n\r\nA specialized theorem like Eisenstein's Criterion should be quoted when used. Such theorems often slip people's memories.\r\n\r\n[b]Eisenstein's Criterion[/b]\r\n\r\nSuppose we have a polynomial with integer coefficients, [tex]f(x) = a_nx^n+a_{n-1}x^{n-1}+\\cdots+a_2x^2+a_1x+a_0[/tex]. If there exists a prime number [tex]p[/tex] that divides all [tex]a_i [/tex] except [tex]a_n[/tex], and [tex]p^2[/tex] does not divide [tex]a_0[/tex], then [tex]f(x)[/tex] is irreducible.\r\n\r\nErin Schram",
  "Solution_3": "An excellent theorem that many of you can probably prove yourself, by the way.",
  "Solution_4": "This is not a [b]challenge[/b]! It is a pretty standard exercise in algebra. My intention was to provide high school students something for thoughts.\r\nI believe that one of the best ways to learn math is proving theorems before looking at the proofs in the book. So I am restating the exercise in several steps as follows:\r\n\r\n1. Prove the [b]Eisenstein's Criterion[/b].\r\n\r\n[b]Eisenstein's Criterion[/b] Suppose we have a polynomial with integer coefficients, [tex]f(x) = a_nx^n+a_{n-1}x^{n-1}+\\cdots+a_2x^2+a_1x+a_0[/tex]. If there exists a prime number [tex]p[/tex] that divides all [tex]a_i [/tex] except [tex]a_n[/tex], and [tex]p^2[/tex] does not divide [tex]a_0[/tex], then [tex]f(x)[/tex] is irreducible.\r\n\r\n2. Let [tex]p[/tex] be an odd prime, show that p divides [tex]\\displaystyle {p\\choose k}[/tex] for [tex]1\\le k \\le p-1[/tex].\r\n\r\n3. Using [b]Eisenstein's Criterion[/b] (with a variable change) to prove the following:\r\n\r\nLet [tex]p[/tex] be an odd prime. Show that the polynomial [tex]x^{p-1}+\\cdots+x+1[/tex] is irreducible over [tex]\\mathbb{Z}[/tex].",
  "Solution_5": "[quote=\"JBL\"]An excellent theorem that many of you can probably prove yourself, by the way.[/quote]\r\n\r\nGood point! I posted [b]Gauss Lemma[/b] for primitive polynomials last night. I removed it after I saw the replies in this thread. The proofs of [b]Gauss Lemma[/b] and [b]Eisenstein's Criterion[/b] are pretty elementary and accesible for high school students (although I learned them in college :-))."
}
{
  "Tag": [],
  "Problem": "Does anyone have experience conducting an orchestra, or at least has been taking conducting lessons for a while? Care to give me some tips before I start? Some specific questions I have:\r\n\r\n-How do you practice? About how long do you practice a day?\r\n-What is considered a good piece for starters? Or good composers in general?\r\n-What other musical talents are necessary/helpful for conducting? I don't play the piano, and I'm afraid that it might make conducting hard for me. I do play the violin, though.\r\n-How long does it take on average to be able to conduct an orchestra and perform?\r\n-How much competition is there for young conductors?\r\n\r\nThanks in advance!",
  "Solution_1": "I assistant-conduct my school orchestra.\r\n\r\n[quote=\"zephyredx\"]-How do you practice? About how long do you practice a day?[/quote]\n\nI usually practice conducting for maybe 10 -15 minutes at a time. I conduct by taking piano reductions of my cello concerti, listening to the orchestral parts, and cuing as neccessary. Sometimes, I conduct music from youtube. It also helps to watch famous conductors, such as Herbert von Karajan. You can imitate them and develop your own style. There are also many good books on conducting that you can buy off of amazon.\n\n[quote=\"zephyredx\"]-What is considered a good piece for starters? Or good composers in general?[/quote]\n\nFor me, I started conducting my easy school music. You can find easy music (although somewhat modern and unknown) at http://www.jwpepper.com. An easy, yet more well known composer would be Vivaldi.\n\n[quote=\"zephyredx\"]-What other musical talents are necessary/helpful for conducting? I don't play the piano, and I'm afraid that it might make conducting hard for me. I do play the violin, though.\n[/quote]\n\nPlaying any instrument usually helps. However, you must have a deep understanding of the songs you plan to conduct. Personally, I play Cello+Piano. You should have basic knowledge of the instruments in your symphony or orchestra though. Conducting is not all about flapping your arms, jumping around, and making speeches in front of audiances. You must have leadership skills to pull your orchestra or symphony through.\n\n[quote=\"zephyredx\"]-How long does it take on average to be able to conduct an orchestra and perform?[/quote]\n\nI cannot answer you question as I am only an assistant conductor. Usually, my school takes around 2~3 months to prepare 3~4 songs.\n\n[quote=\"zephyredx\"]-How much competition is there for young conductors?[/quote]\r\n\r\nIt really depends on the city. My city has only one youth symphony and that's it. And the assistant conductor there is 18. I'm sure that you can probably find some school or youth symphony if you are in your early teens though."
}
{
  "Tag": [
    "IMO Shortlist"
  ],
  "Problem": "saludos \r\nlquisiera saber si se tienen las longlist , pues las deart of solving problem  no entra \r\nsi alguien las tiene agradeceria que las colgaran \r\nadan",
  "Solution_1": "si le das a \"resources\"  en la esquina derecha, te saldran ahi los longlists y shortlists",
  "Solution_2": "Aqui estan los links:\r\n\r\n-  [url=http://www.artofproblemsolving.com/Forum/resources.php?cid=17]IMO SHORTLIST[/url]\r\n\r\ny tambien:\r\n\r\n- [url=http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=18]IMO LongList[/url]\r\n\r\neste ultimo aun no cuenta con material ....\r\npero me parece que si se tiene el longlist de la IMO 1967, no se porque no sta aqui..\r\n\r\n\r\nCarlos Bravo :)\r\nLima-PERU"
}
{
  "Tag": [
    "abstract algebra",
    "group theory",
    "number theory",
    "least common multiple",
    "superior algebra",
    "superior algebra solved"
  ],
  "Problem": "If an abelian group has subgroups of order m & n then it will have a subgroup of order which is the lcm of m & n.\r\n\r\nPlease donot use the properties of finite abelian groups then the entire flavor of he problem will be lost.Just use the elementary concepts( i mean the defn & stuff about groups).\r\n\r\nP.S. I have encountered many solutions of this problem including a few by myself but all of them are wrong.Can anybody here do it.....",
  "Solution_1": "[quote=\"the game\"]If an abelian group has subgroups of order m & n then it will have a subgroup of order which is the lcm of m & n.\n\nPlease donot use the properties of finite abelian groups then the entire flavor of he problem will be lost.Just use the elementary concepts( i mean the defn & stuff about groups).\n\nP.S. I have encountered many solutions of this problem including a few by myself but all of them are wrong.Can anybody here do it.....[/quote]\r\n\r\nIs this the problem where in the later editions, Herstein says that he received more correspondence about this problem than all the others in the book?",
  "Solution_2": "Yup thats I called it famous.But theres no harm in trying.",
  "Solution_3": "[quote=\"the game\"]If an abelian group has subgroups of order m & n then it will have a subgroup of order which is the lcm of m & n.\n\nPlease donot use the properties of finite abelian groups then the entire flavor of he problem will be lost.Just use the elementary concepts( i mean the defn & stuff about groups).\n\n[/quote]\r\n\r\nAlthough you didn't want this (:)), you can consider it just a harmless comment.\r\n\r\nUsing the structure theorem for finite abelian groups, it's trivial to show that for any divisor $d$ of the order of the group, the group has a subgroup of order $d$ (and the converse also holds, of course). From the hypothesis we find that $m,n$ divide the order of the group, so their least common multiple also divides the order of the group, and that's that.",
  "Solution_4": "Here's how to do it without the heavy tools- just some basic stuff and a bit of number theory:\r\n(We treat the abelian group as a $\\mathbb{Z}$-module)\r\nLemma: Suppose $m$ and $n$ are relatively prime. Then $G$ has a subgroup of order $mn$ if and only if it has a subgroup of order $m$ and a subgroup of order $n$.\r\nProof. If $H$ has order $m$ and $K$ has order $n$, $H+K$ has order $mn.$\r\nIf $H$ has order $mn$, $mH$ has order $n$ and $nH$ has order $m$.\r\n\r\nThis allows us to reduce to prime powers. Once there, choose the larger one at each prime, then recombine to form the desired group of order $\\mathrm{lcm}(m,n).$",
  "Solution_5": "Wow that is one hell of a solution .Thts something ingenious i wanted.",
  "Solution_6": "I don't quite follow: what are $mH$ & $nH$?",
  "Solution_7": "In the $\\mathbb{Z}$-module view, they're the submodules $\\{mx: x\\in H\\}$ and $\\{nx: x\\in H\\}$. In the group picture, they're the subgroups of $m$th and $n$th powers of elements in $H$ respectively.",
  "Solution_8": "How to count order of mH & nH ?\r\nI don't quite follow."
}
{
  "Tag": [
    "algebra",
    "partial fractions",
    "algebra unsolved"
  ],
  "Problem": "(i) The coefficients in the series\r\n\r\n$ S\\equal{}\\frac{1}{3}x\\plus{} \\frac{1}{6}x^{2} \\plus{}\\frac{1}{12}x^{3} \\plus{} ... \\plus{} a_{r}x^{r}\\plus{}...$\r\n\r\nsatisfy a recurrence relation of the form $ a_{r\\plus{}1} \\plus{} pa_{r}\\equal{}0$. Find the value of $ p$.\r\n\r\nBy considering $ (1\\plus{}px)S$, find an expression for the sum to infinity of $ S$ (assuming that it exists). Find also expressions for the sum of the first $ n\\plus{}1$ terms of $ S$.\r\n\r\n(ii) The coefficients in the series\r\n\r\n$ T\\equal{}2\\plus{}8x\\plus{}18x^{2}\\plus{}37x^{3} \\plus{} ... \\plus{} a_{r}x^{r} \\plus{} ...$\r\n\r\nsatisfy a recurrence relation of the form $ a_{r\\plus{}2} \\plus{} pa_{r\\plus{}1} \\plus{} qa_{r} \\equal{} 0$. Find an expression for the sum to infinity of $ T$ (assuming thatit exists). By expressing $ T$ in partial fractions, or otherwise, find an expression for the sum of the first $ n\\plus{}1$ terms of $ T$.",
  "Solution_1": "(i) find $ p$ by solving the equation $ a_{r\\plus{}1}\\plus{}pa_r\\equal{}0$ for two particular consecutive terms. then the fact that $ a_{r\\plus{}1}\\plus{}pa_r\\equal{}0$ implies that $ (1\\plus{}px)S\\equal{}\\frac{1}{3}x$, giving $ S\\equal{}\\frac{x}{3(1\\plus{}px)}$. the sum of the first $ n$ terms of $ S$ is found in a similar way.\r\n\r\n(ii) this is done in a similar fashion to (i). here you'll need two equations to solve for $ p$ and $ q$. then consider $ (1\\plus{}px\\plus{}qx^2)S$"
}
{
  "Tag": [],
  "Problem": "Are they moon-quake fault lines?\r\n\r\n[img]http://www.pinktentacle.com/images/moon_kaguya_3.jpg[/img]",
  "Solution_1": "Nice inference. :lol:\r\nActually this \"lines\" are \"rays\" made of the dust the was ejected from the impacts that caused the craters. If these rays are present around a crater, it can be assumed that it is a relatively new crater. Of course, there are exceptions, like what material the rays are made of.  Different materials have different albedos.  :wink: \r\n\r\n[hide=\"Image\"][img]http://apod.nasa.gov/apod/image/0503/moon8_mandel_big.jpg[/img][/hide]\r\n[url=http://en.wikipedia.org/wiki/Ray_system]Wikipedia - Ray Systems[/url].\r\n\r\nEDIT Since the picture was added to the original post.\r\nThose still look like rays from a crater, the material from the impact was just large enough to create more defined streaks like that.\r\nEDITTED EDIT The moon does not have crust plates like the Earth so, they could not be quake lines.",
  "Solution_2": "The \"ray\" theory seems plausible to explain the rays outwards around the craters.\r\n\r\nHowever, it does not seem to explain the lines in the image of the original post. \r\nif the tracks were created by ejected big rocks, then we should be able to see that at the end of each track, there should be a big boulder. \r\n\r\nAlso, we cannot completely rule out the moon-quake fault line idea. Even though there are no tectonic plates on the moon due to its cold core, \r\nbut each meteorite impact (which caused those huge craters) would have caused a big quake on the moon in the distant past. \r\nSince there is no weathering on the moon, the fault lines have survived for billions of years, to be seen today. \r\nA close look in the image of the original post shows that the lines are not straight, some resemble cracks instead of rays.",
  "Solution_3": "About the quake idea, then where are the assumedly big craters from those big impacts? The Moon's suface does remain unchanged over time unless there is an impact, like you said, so the craters should be there still also.\r\nAnother thing is that this \"cracks\" [i]in the surface [/i]do not look like they are more than [i]on the surface [/i]of the Moon.  The mare on the Moon are cause by large impacts, where the molten layer beneath flowed onto the surface and hardened.\r\n\r\nThe rays of a crater could still create this \"lines\", the material that created it does not have to be large just needs to be moving fast enough, the distance of the picture does not do justice to seeing the \"possible boulders\" that would be at the end of the trails. :wink:"
}
{
  "Tag": [],
  "Problem": "\u039d\u03b1 \u03b2\u03c1\u03b5\u03af\u03c4\u03b5 \u03cc\u03bb\u03bf\u03c5\u03c2 \u03c4\u03bf\u03c5\u03c2 \u03b8\u03b5\u03c4\u03b9\u03ba\u03bf\u03cd\u03c2 \u03b1\u03ba\u03b5\u03c1\u03b1\u03af\u03bf\u03c5\u03c2 , \u03bf\u03b9 \u03bf\u03c0\u03bf\u03af\u03bf\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03af\u03c3\u03bf\u03b9 \u03bc\u03b5 \u03c4\u03bf $ 13 \\minus{}$ \u03c0\u03bb\u03ac\u03c3\u03b9\u03bf \u03c4\u03bf\u03c5 \u03b1\u03b8\u03c1\u03bf\u03af\u03c3\u03bc\u03b1\u03c4\u03bf\u03c2 \u03c4\u03c9\u03bd \u03c8\u03b7\u03c6\u03af\u03c9\u03bd \u03c4\u03bf\u03c5\u03c2.\r\n\r\n \u039c\u03c0\u03ac\u03bc\u03c0\u03b7\u03c2",
  "Solution_1": "\u03c8\u03b1\u03c7\u03bd\u03c9\u03bd\u03c4\u03b1\u03c2 \u03b2\u03c1\u03b7\u03ba\u03b1 \u03b1\u03c5\u03c4\u03bf \u03c4\u03bf \u03c0\u03bf\u03bb\u03c5 \u03c9\u03c1\u03b1\u03b9\u03bf \u03be\u03b5\u03c7\u03b1\u03c3\u03bc\u03b5\u03bd\u03bf \u03c0\u03c1\u03bf\u03b2\u03bb\u03b7\u03bc\u03b1 :) \r\n\u03b7 \u03bb\u03c5\u03c3\u03b7 \u03bc\u03bf\u03c5 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b7 \u03b5\u03be\u03b7\u03c2:\r\n\u03b5\u03c3\u03c4\u03c9 \u03bf\u03c4\u03b9 \u03bf \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03c2 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03bf $ xy...zk = x10^{n - 1} + y10^{n - 2} + .... + k = 13(x + y + ....z + k)$\u03bc\u03b5 $ 9 > = x,y,z....,k > = 0$\u03ba\u03b1\u03b9 n \u03bf \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03c2 \u03c4\u03c9\u03bd \u03c8\u03b7\u03c6\u03b9\u03c9\u03bd \u03c4\u03bf\u03c5.\r\n\u03b5\u03c7\u03bf\u03c5\u03bc\u03b5 \u03c9\u03c2 \u03b4\u03b5\u03b4\u03bf\u03bc\u03b5\u03bd\u03bf \u03bc\u03b5 \u03c0\u03b1\u03c1\u03b1\u03b3\u03bf\u03bd\u03c4\u03bf\u03c0\u03bf\u03b9\u03b7\u03c3\u03b7 \u03bf\u03c4\u03b9 \r\n$ (10^{n - 1} - 13)x + (10^{n - 2} - 13)y + .... - 12k = 0$\r\n$ = > k = \\frac {(10^{n - 1} - 13)x + (10^{n - 2} - 13)y + ....z10^{n - (n - 1)}}{12}\\leq9$\r\n$ < = >$\r\n$ (10^{n - 1} - 13)x + (10^{n - 2} - 13)y + ....z10^{n - (n - 1)}\\leq108$\r\n\u03bf\u03bc\u03c9\u03c2 \u03b3\u03b9\u03b1 $ n\\geq4 = > (10^{n - 1} - 13)x\\geq987x\\geq987 > 108$\u03c0\u03b1\u03b9\u03c1\u03bd\u03c9\u03bd\u03c4\u03b1\u03c2 \u03b4\u03b5\u03b4\u03bf\u03bc\u03b5\u03bd\u03bf \u03bf\u03c4\u03b9 \u03c4\u03bf \u03c7 \u03b4\u03b5\u03bd \u03b5\u03b9\u03bd\u03b1\u03b9 \u03bc\u03b7\u03b4\u03b5\u03bd\r\n\u03b1\u03c1\u03b1 \u03b1\u03c4\u03bf\u03c0\u03bf.\u03b1\u03c1\u03b1 $ n\\leq3$ \r\n\u03b4\u03b9\u03b1\u03c0\u03b9\u03c3\u03c4\u03c9\u03bd\u03bf\u03c5\u03bc\u03b5 \u03b5\u03c0\u03b9\u03c3\u03b7\u03c2 \u03bf\u03c4\u03b9 $ 13/x10^{n - 1} + y10^{n - 2} + .... + k = > x10^{n - 1} + y10^{n - 2} + .... + k\\geq13$\r\n\u03b1\u03bb\u03bb\u03b1 \u03ba\u03b1\u03b9 \u03bf\u03c4\u03b9 \r\n$ x10^{n - 1} + y10^{n - 2} + .... + k = 13(x + y + ....z + k)\\leq117n$\r\n\u03c4\u03bf\u03c4\u03b5 \u03c0\u03b1\u03b9\u03c1\u03bd\u03bf\u03c5\u03bc\u03b5 $ n\\geq\\frac {1}{9} = > n\\geq1$\r\n\u03b7 \u03c0\u03b5\u03c1\u03b9\u03c0\u03c4\u03c9\u03c3\u03b7 n=1 \u03ba\u03b1\u03b8\u03b9\u03c3\u03c4\u03b1 \u03b1\u03b4\u03c5\u03bd\u03b1\u03c4\u03bf \u03c4\u03bf\u03bd \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf \u03bd\u03b1 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c0\u03bf\u03bb/\u03c3\u03bf \u03c4\u03bf\u03c5 13 \u03ba\u03b1\u03b9\r\n\u03b3\u03b9\u03b1 n=2\r\n\u03c0\u03b1\u03b9\u03c1\u03bd\u03bf\u03c5\u03bc\u03b5 $ 10x + y = 13x + 13y < = > - 3x = 12y$=>x,y eterosimoi.\u03b1\u03c4\u03bf\u03c0\u03bf\r\n\u03b5\u03c0\u03bf\u03bc\u03b5\u03bd\u03c9\u03c2 $ n = 3$\r\n\u03b3\u03b9\u03b1 n=3,\r\n$ 100x + 10y + z = 13x + 13y + 13z(1) < = > y = 29x - 4z\\leq9$\r\n$ < = > x\\leq\\frac {9 + 4z}{29}\\leq\\frac {45}{29} = > x = 1$\r\n\u03b3\u03b9\u03b1 \u03c7=1 \u03b7 (1) \u03b3\u03c1\u03b1\u03c6\u03b5\u03c4\u03b1\u03b9 $ y + 4z = 29$\r\n\u03b4\u03b9\u03bf\u03c6\u03b1\u03bd\u03c4\u03b9\u03ba\u03b7 \u03bc\u03b5 \u03bb\u03c5\u03c3\u03b5\u03b9\u03c2 $ y = 1 - 4t,z = 7 + t$,t \u03b1\u03ba\u03b5\u03c1\u03b1\u03b9\u03bf\u03c2\r\n\u03c0\u03c1\u03b5\u03c0\u03b5\u03b9$ 0 < z,y\\leq9$\r\n$ = > t = 0, - 1, - 2 = > (y,z) = {(1,7),(5,6),(9,5)}$\r\n\u03bf\u03b9 \u03b6\u03b7\u03c4\u03bf\u03c5\u03bc\u03b5\u03bd\u03bf\u03b9 \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03b9 \u03bc\u03b1\u03c2 \u03bb\u03bf\u03b9\u03c0\u03bf\u03bd \u03b5\u03b9\u03bd\u03b1\u03b9 \u03bf\u03b9 $ 117,156,195$  :D QED"
}
{
  "Tag": [
    "trigonometry",
    "function",
    "complex analysis",
    "complex analysis unsolved"
  ],
  "Problem": "Let $ f_1(z)\\equal{}\\sin{z},f_n(z)\\equal{}f_{n\\minus{}1}(\\sin{z})$. Is the family of holomorphic functions $ f_n,n\\equal{}1,2,3,\\ldots$ normal in $ \\{z \\in \\mathbb{C} : |z|<1 \\}$ ?",
  "Solution_1": "No, $ f_n(0)\\equal{}0$ but $ f_n(0.9i)\\to\\infty$",
  "Solution_2": "Could you please explain the following : take a sequence $ f_n(z)\\equal{}z\\plus{}n$. It is clear that it diverges locally uniformly yo $ \\infty$ in $ \\mathbb{C}$. That i understand.\r\nIn the book they claim that it neither converges nor diverges locally uniformly in $ \\bar{\\mathbb{C}}$\r\n\r\nWhy is it so ? Why doesn't it converge locally uniformly to the constant function, which is $ \\infty$ everywhere.",
  "Solution_3": "[quote=\"eugene\"]in $ \\bar{\\mathbb{C}}$[/quote]\r\nThe sequence $ z\\plus{}n$ does not go to $ \\infty$ uniformly in any neighborhood of $ \\infty$.",
  "Solution_4": "Oh, yeas, indeed. For some stupid reason i forgot to look at the neighborhood of $ \\infty$.\r\nYou are saying that if $ |f_n(z)| > M(R)$ for all $ n > N(R)$ for some $ N(R),M(R)$ and all $ |z| > R$. But of course this neighborhood contains $ z$ with $ z\\equal{}\\minus{}n$ say with big enough $ n$, which is a contradiction.",
  "Solution_5": "How about everywhere black in the plot below?\r\n\r\n[img]http://img409.imageshack.us/img409/9944/iteratedsinzng6.jpg[/img]",
  "Solution_6": "Julia set of $ f(z)\\equal{}\\lambda \\sin z$ has been studied quite a bit. There is even a Youtube video. \r\n[youtube]-oJaE-iBIvA[/youtube]"
}
{
  "Tag": [
    "geometry",
    "FTW",
    "LaTeX"
  ],
  "Problem": "If you choose \"other,\" please post what the \"other\" is.\r\n\r\n[color=indigo][moved from bay area: please reserve threads like these to the fun factory][/color]",
  "Solution_1": "Where's 1337?",
  "Solution_2": "My favorite = 16",
  "Solution_3": "Mine is coming up on my post count 127!",
  "Solution_4": "Wheres 56? or 56154? or 1511558921? or 1337?",
  "Solution_5": "Sadly, it only allowed me to put nine options.",
  "Solution_6": "does anyone know the meaning of \"69\"?",
  "Solution_7": "COUGHGAGWHEEZE\r\n\r\n\r\nim sorry, please, do continue.\r\n\r\nand if you dont know its best to not know",
  "Solution_8": "please?\r\n\r\ntellz me plz\r\n\r\nplz\r\n\r\nmy 10th grade friends started laughing at \"69\"\r\n\r\ni feel left out :(\r\n\r\nplz tell me",
  "Solution_9": "ummmmm.......68<69<70!\r\n\r\n\r\ntheres always wikipedia",
  "Solution_10": "explain thouruohly plz\r\n\r\nplz?\r\n\r\nlike the actual meaning\r\n\r\ni cant go on wikipedia\r\n\r\nor on google",
  "Solution_11": "[quote=\"Joe10112\"]explain thouruohly plz\n\nplz?\n\nlike the actual meaning\n\ni cant go on wikipedia\n\nor on google[/quote]\r\n\r\nIt is seriously a very sick, disgusting joke and inappropriate for anyone here. Let's not spread the contamination.",
  "Solution_12": "eww please don't talk about that. it's gross especially when your bio teacher refers to it *chokegagdie*\r\n\r\nanway... on a more pure note... my fave number is $ 6$ =] because a lot of things in my life relate to $ 6$ =] although i'm not a devil child... the devil stole my number. he's the copy cat  :D  :rotfl:",
  "Solution_13": "I like $ 6$ too. I also like $ 56$ a lot. $ 15$, $ 56$, and $ 89$ are the three most awesome two-digit numbers, with $ 13$, $ 21$, $ 22$, $ 82$, and $ 88$ being the next-coolest two-digit numbers. If you ask for a cool triplet of two-digit numbers, $ 21$, $ 55$, and $ 89$ form a symmetrical pattern on the number pad while being Fibonacci numbers at the same time, and one of them, $ 89$, is one of those three epicly-awesome numbers. Isn't that cool? \r\n\r\n(I'm sounding even more nerdy than I normally am, right?)",
  "Solution_14": "okayz\r\n\r\ni wont ask then\r\n\r\ni will not ask 4 \"69\"",
  "Solution_15": "I like $ \\boxed{16.7}$.  Hey, my favorite number isn't a whole number!  I'M A TRUE AoPSER!  :lol:",
  "Solution_16": "Well, gee, you're leaving out an awful lot of numbers, don't you think? There's 1, and 1.1, and 1.01, and 1.001, and 1.91, and a whole lot more. My favorite number is $ \\phi$.",
  "Solution_17": "hence the answer choice \"Other\"\r\n\r\n\r\nmy favorite number is\r\n\r\n198389692832016689128025814051186435469808931027259980194805041212767924492279648804437095653839742535006120000819629040274718649969=1337^42 :)",
  "Solution_18": "$ i$.\r\n\r\nObviously.",
  "Solution_19": "I beg to differ about r15s11z55w89y21's (beg me if I spelled it wrong o_0) \"15 and not 42\" image....\r\n\r\nFavorite number = 842310 GO FIGURE!",
  "Solution_20": "[quote=\"mousy\"]\n\nFavorite number = 842310[/quote]\r\n\r\nIs that the game number that you got your highest score?",
  "Solution_21": "$ \\text{My favorite number would have to be }{\\color{cyan}\\pi \\color{black}+\\color{red}i}$",
  "Solution_22": "My favorite numbers are roots of $ x^7\\minus{}1$.",
  "Solution_23": "$ \\minus{}146e^{\\pi i}$",
  "Solution_24": "[quote=\"AIME15\"]Where's 1337?[/quote]\r\nexactly, where is 1337? And what about pi?!",
  "Solution_25": "42 ftw...   ",
  "Solution_26": "[quote=\"dragon96\"][quote=\"mousy\"]\n\nFavorite number = 842310[/quote]\n\nIs that the game number that you got your highest score?[/quote]\r\n\r\nI don't even think we got to that number yet...",
  "Solution_27": "Favorite number: lim x->(+0), 1/x\r\n\r\nShoot, that converges towards infinity.",
  "Solution_28": "[quote=\"r15s11z55y89w21\"][quote=\"dragon96\"][quote=\"mousy\"]\n\nFavorite number = 842310[/quote]\n\nIs that the game number that you got your highest score?[/quote]\n\nI don't even think we got to that number yet...[/quote]\r\n\r\nYou know, besides 42, none of the numbers are really favorites of AoPSers. If I made a poll, I would have:\r\n\r\n42\r\npi\r\n1337\r\ni\r\ne\r\n56154\r\n56\r\n15\r\n96 (7th graders only)\r\nother",
  "Solution_29": "get rid of 96 and add i",
  "Solution_30": "[quote=\"eggylv999\"]$ \\minus{} 146e^{\\pi i}$[/quote]\r\n\r\nWhich would be 146?\r\nWhile you're at it, why not write $ \\minus{}146e^{\\pi e^{\\frac{\\pi}{2}i}}$? :D",
  "Solution_31": "My favorite number is pi, which is not a interger. :lol:",
  "Solution_32": "$ \\pi$.\r\n\r\nI'm surprised no one has said pi until TPAM.\r\n\r\nOh man... 69... I don't want to hear that number again after what happened on the bus last year.",
  "Solution_33": "1337 ftw!!!!!!",
  "Solution_34": "$ \\pi \\cdot e \\cdot \\phi$",
  "Solution_35": "11235813, for obvious reasons.",
  "Solution_36": "Fibonnaci?",
  "Solution_37": "If you're looking for numbers comprised of Fibonacci numbers, $ 558921$ pwns all.",
  "Solution_38": "Okay. But why 11->THEnumber?",
  "Solution_39": "Mine is 6 for some weird reason.....It was like that ever since I was about 2 years old.",
  "Solution_40": "6 is the 1st perfect number and the only number such that its proper factors multiply to the number itself",
  "Solution_41": "5 is coool   thats why it's in my username",
  "Solution_42": "2. The only even prime. The only number where $ n^n=n\\times n=n+n$. When writing square roots, 2 is so special that it doesn't need to be written for someone to understand its power. The base that computers work in.",
  "Solution_43": "My favourite number is the sum of all real numbers that are not expressible as the sum of arbitrarily but finitely many algebraic numbers.\r\n\r\n[quote=\"PowerOfPi\"]$ n^n = n\\times n = n + n$.[/quote]\r\nLaTeX \\fail.",
  "Solution_44": "4 ;]\r\njust cuz."
}
{
  "Tag": [
    "geometry proposed",
    "geometry"
  ],
  "Problem": "Suppose that $ABC$ is a triangle and $H \\in BC, D \\in AB, E \\in AC$. Also $AH,CD$ and $BE$ are concurrent and $AH$ is an altitude of $\\triangle ABC$. Prove that $\\angle EHA=\\angle AHD$.",
  "Solution_1": "$DE$ intersects $AH$ and $BC$ at $M,N$. We have $(NDME)=-1$ ( it's well-known ). Also $HM \\perp HN$ so $HM$ is the bisector of $\\angle MHN$"
}
{
  "Tag": [
    "limit",
    "calculus",
    "calculus computations"
  ],
  "Problem": "How can I show that $\\lim_{x\\to\\infty}{\\frac{x}{2x^{2}+1}}= 0$?\r\n\r\nI can't just state it. I need to show it using $\\delta, \\epsilon$, or the like. How can I do this?\r\n\r\nTHanx",
  "Solution_1": "$\\\\ 2x^{2}+1 > x^{2}\\Rightarrow \\frac{x}{2x^{2}+1}< \\frac{x}{x^{2}}= \\frac{1}{x}$ and now you can take $\\delta = \\frac{1}{ \\epsilon }$ .",
  "Solution_2": "Ramanujan, could you explain why we choose $\\delta = \\frac{1}{\\epsilon}$ in more detail? Thanx.",
  "Solution_3": "[quote=\"bythecliff\"]How can I show that $\\lim_{x\\to\\infty}{\\frac{x}{2x^{2}+1}}= 0$?\n\nI can't just state it. I need to show it using $\\delta, \\epsilon$, or the like. How can I do this?\n\nTHanx[/quote]\r\n\r\n$\\lim_{x\\to\\infty}{\\frac{x}{2x^{2}+1}}= 0$ if and only if\r\n$\\forall\\epsilon>0\\, \\exists\\delta >0\\mbox{ suchthat }|x|>\\delta\\Rightarrow \\left| \\frac{x}{2x^{2}+1}-0\\right| <\\epsilon$\r\n\r\nchoose $\\delta =\\frac{1}{\\epsilon}$ so that $|x|>\\delta\\Rightarrow \\left| \\frac{x}{2x^{2}+1}-0\\right| =\\frac{x}{2x^{2}+1}< \\frac{1}{x}=\\epsilon$ QED"
}
{
  "Tag": [
    "quadratics"
  ],
  "Problem": "A cyclist has won a competition. Then the reporters asked him what was the speed of it, then, he replied: \r\n\r\n\"If I had been walking 4 km / ha more than I would have returned an hour earlier than when he returned, knowing that the distance is 120 km.\" \r\n\r\nSo what was the speed of the cyclist?",
  "Solution_1": "(120/x+4)+1=120/x\r\nx^2+4x-480=0\r\nx=+20 or -24.\r\nSo his speed was 20",
  "Solution_2": "Thanks, my answer was 20 too.\r\nThanks man ;)"
}
{
  "Tag": [
    "combinatorics proposed",
    "combinatorics"
  ],
  "Problem": "In a party, each person knew exactly $ 22$ other persons. For each two persons $ X$ and $ Y$, if $ X$ and $ Y$ knew each other, there is no other person who knew both of them, and if $ X$ and $ Y$ did not know each other, there are exactly $ 6$ persons who knew both of them. Assume that $ X$ knew $ Y$ iff $ Y$ knew $ X$. How many people did attend the party?\r\n[i]Yudi Satria, Jakarta[/i]",
  "Solution_1": "The answer is $ n \\equal{} 100$.\r\n\r\nConsider an arbitrary person $ A$. Let $ F(A)$ be the set of friends of $ A$ and $ S(A)$ be the set of strangers of $ A$. Then $ |F(A)| \\equal{} 22$ and $ |S(A)| \\equal{} n \\minus{} 23$. Consider a person $ S\\in S(A)$. Since $ S$ does not know $ A$, $ S$ and $ A$ must have $ 6$ friends in common. Thus, each person $ S\\in S(A)$ knows exactly $ 6$ people in $ F(A)$. Hence, the number of the relationship $ F(A) \\minus{} S(A)$ is $ 6(n \\minus{} 23)$. On the other hand, every two people in $ F(A)$ do not know each other (since they all know $ A$), therefore, every people in $ F(A)$ knows exactly $ 21$ people in $ S(A)$. This means that the number of relationship $ F(A) \\minus{} S(A)$ is $ 21\\cdot 22$. From the above argument, we deduce that $ 6(n \\minus{} 23) \\equal{} 21\\cdot 22$, which yields $ n \\equal{} 100$, QED.",
  "Solution_2": "Any construction for $100$ people?"
}
{
  "Tag": [],
  "Problem": "By using identity cot x - cot 2x = cosec 2x, prove that\r\ncot x - cot 4x > 2 for 0 < x <  :pi:/4",
  "Solution_1": "studing $x \\mapsto \\cot{x} - \\cot{4x} -2$ no?  :blush:",
  "Solution_2": "[hide]\n\n$\\cot{x} - \\cot{4x} = (\\cot{x}-\\cot{2x}) + (\\cot{2x}-\\cot{4x}) = \\csc{2x} + \\csc{4x}.$ \nBut since $\\csc{x} \\ge 1$, we know that the expression is at least 2. We also know that $\\csc{x}=1$ at $x=\\frac{\\pi}{2}, \\frac{3\\pi}{2}$ so $\\csc{2x}$ is never 1 on the interval. Then suppose $\\csc{4x}=1$. So $x=\\frac{\\pi}{8}$. Then $\\csc{2x} > 1$ so the expression is still greater than 2.[/hide]"
}
{
  "Tag": [
    "function",
    "trigonometry",
    "limit",
    "calculus",
    "calculus computations"
  ],
  "Problem": "I am stuck with a)\r\n\r\nConsider the function $ f(x) =\\sin(1/x)$\r\n\r\na) Find a sequence of x-values that approach $ 0$ such that $ \\sin(1/x) = 0$.\r\n[Hint: Use the fact that $ \\sin(\\pi) =\\six(2\\pi) =\\sin(3\\pi) = ... =\\sin(n\\pi) = 0.$]. \r\n\r\nb) Find a sequence of x - values that approach $ 0$such that $ \\sin(1/x) = 1$. [Hint: Use the fact that $ \\sin(n\\pi/2) = 1$ if $ n = 1,5,9,...$].\r\n\r\nc) Find a sequence of x-values that approach 0 such that $ \\sin(1/x) =-1$.\r\n\r\nd) Explain why your answers to any two parts (a)-(c) show that $ \\lim_{x\\to 0}\\sin(1/x)$ does not exist.",
  "Solution_1": "a) can be seen easily if you think a little bit. $ \\sin k\\pi \\equal{} 0$ for all $ k\\in\\mathbb{Z}$.  Now just solve the equation $ \\frac{1}{x}\\equal{} k\\pi$ with respect to $ x$ and prove that this value goes towards zero if $ k\\to\\infty$.",
  "Solution_2": "So\r\n\r\na)\r\n\r\n$ x \\equal{}\\frac{1}{k\\pi}$\r\n\r\n$ \\sin\\left(\\frac{1}{\\pi}\\right) \\equal{} 0.313$\r\n$ \\sin\\left(\\frac{1}{2\\pi}\\right) \\equal{} 0.158$\r\n$ \\sin\\left(\\frac{1}{3\\pi}\\right) \\equal{} 0.106$\r\n\r\nSo all of these values are approaching $ 0$.\r\n\r\nDid I do it right? \r\n\r\nb)\r\n\r\n$ x \\equal{}\\frac{n\\pi}{2}\\equal{}\\frac{1}{x}$\r\n$ x \\equal{}\\frac{2}{n\\pi}$.\r\n\r\n$ \\sin\\left(\\frac{2}{\\pi}\\right) \\equal{} 0.59$\r\n$ \\sin\\left(\\frac{2}{5\\pi}\\right) \\equal{} 0.127$\r\n$ \\sin\\left(\\frac{2}{9\\pi}\\right) \\equal{} 0.0706$\r\n\r\nSo these also approach $ 0$.\r\n\r\n???",
  "Solution_3": "For (a) you don't want the value of the function to go to $ 0$, you want it to always be zero. Look at where $ x$ is. If $ x \\equal{}\\frac{1}{k\\pi}$ then $ f(x) \\equal{}\\sin(k\\pi)$, not $ \\sin\\left(\\frac{1}{k\\pi}\\right)$. The thing you want to go to $ 0$ is the $ x$ value. So the first thing you wrote is correct, but plugging it back into the function you flipped it upside down.\r\n\r\nLikewise for (b) you want the value of the function to always be $ 1$, yet the sequence you pick goes to zero.",
  "Solution_4": ":( I still don't have a sequence of x values. is it $ \\pi, 2\\pi, 3\\pi$?",
  "Solution_5": "$ f(x)\\equal{}\\sin\\left(\\frac{1}{x}\\right)$\r\n\r\nUsing the sequence of $ x$ values $ x\\equal{}\\frac{1}{k\\pi}$. (Technically they should probably be notated as a sequence $ x_{k}\\equal{}\\frac{1}{k\\pi}$.\r\n\r\nYou get a sequence of $ f(x)$ values $ f\\left(\\frac{1}{k\\pi}\\right)\\equal{}\\sin\\left(\\frac{1}{1/k\\pi}\\right)\\equal{}\\sin(k\\pi)\\equal{}0$ for all $ k$."
}
{
  "Tag": [
    "geometry"
  ],
  "Problem": "Haley has enlarged a 3\" by 5\" picture so that both the lenght and width are tripled.  By what number must the original are of the pricture be multiplied to get the enlarged area?",
  "Solution_1": "[hide]\n$x(3\\cdot5)=3\\cdot3\\cdot3\\cdot5$\n$x=\\frac{3\\cdot3\\cdot3\\cdot5}{3\\cdot5}$\n$x=9$[/hide]",
  "Solution_2": "[hide]9 times. Each side is tripled so the area is $3^2$ times more (9). i didnt really need to calculate the actual area  :D[/hide][/hide]",
  "Solution_3": "9 times. is it correct?",
  "Solution_4": "[quote=\"risinglegend\"][hide=\"im hiding your answer\"]9 times[/hide]. is it correct?[/quote] Yes it is correct but you are supposed to tell us why its correct and HIDE your answer so other people can have a chance at the problem unless someone is asking for help.\r\n\r\nIt is useless to just post an answer and not a solution to a problem other people have already solved and explained. IT's not useless if you post a solution"
}
{
  "Tag": [
    "geometry proposed",
    "geometry"
  ],
  "Problem": "Let O be the center if the excircle of triangle ABC opposite to A. Let M be the midpoint of AC and let P be the intersection of lines MO and BC. Prove that if <BAC=2<ACB, AB=BP.",
  "Solution_1": "\u041d\u0435\u0440\u0435\u0430\u043b\u044c\u043d\u043e \u043e\u043d\u0430\u0439 \u0433\u043e\u0439!!!",
  "Solution_2": "[quote=\"Ulanbek_Kyzylorda KTL\"]Let O be the center if the excircle of triangle ABC opposite to A. Let M be the midpoint of AC and let P be the intersection of lines MO and BC. Prove that if <BAC=2<ACB, AB=BP.[/quote]\r\n\r\nLet's $ \\angle ACB \\equal{} \\gamma$. $ AO \\cap BC \\equal{} A_1$\r\nWe have \r\n $ AO\\equal{}\\frac{AC \\cdot sin(90^o\\plus{}\\gamma)}{sin(90^o\\minus{}3\\gamma)}$\r\n $ A_1O\\equal{}\\frac{A_1C \\cdot sin(90^o\\minus{}\\gamma)}{sin(90\\minus{}3\\gamma)}$\r\nSo\r\n$ \\frac{A_1O}{AO}\\equal{}\\frac{A_1C}{AC}\\equal{}\\frac{AA_1}{AC}$\r\n\r\nBy Menelaus on $ \\triangle AA_1C$ and line $ OPM$ we get $ \\frac{CM}{AM} \\cdot \\frac{A_1P}{CP} \\cdot \\frac{AO}{A_1O}\\equal{}1$\r\nor \r\n$ \\frac{A_1O}{AO}\\equal{}\\frac{A_1P}{CP}$\r\nTherefore \r\n$ \\frac{A_1P}{CP}\\equal{}\\frac{AA_1}{AC}$\r\ntherefore $ AP$ is bisector of $ \\angle A_1AC$ $ \\rightarrow$ $ \\angle BAP\\equal{}3\\gamma$\r\n$ \\angle APB \\equal{}180^o\\minus{}\\angle PAB \\minus{}\\angle ABP\\equal{}3\\gamma$ \r\nSO $ AB\\equal{}PB$",
  "Solution_3": "[color=darkred][size=125][b]Nice problem ! Thank you.[/b][/size] [/color]\r\n\r\n[quote=\"Ulanbek_Kyzylorda KTL\"]Let $ \\triangle\\ ABC$ , the midpoint $ M$ of $ [AC]$ and $ P\\in MI_a\\cap BC\\ .$ Prove that $ A \\equal{} 2C\\ \\Longleftrightarrow\\ BP \\equal{} AB\\ .$[/quote]\r\n\r\nDenote $ D\\in \\overline {AII_a}\\cap BC$ . Apply the [b]Menelaus[/b]' [u]theorem[/u] to the transversal $ \\overline {I_aPM}$ for $ \\triangle\\ ADC$ and obtain \r\n\r\n$ \\frac {PD}{PC} \\equal{} \\frac {I_aD}{I_aA} \\equal{} \\frac {ID}{IA}\\ .$ Thus, $ A \\equal{} 2C\\Longleftrightarrow\\ IP\\parallel AC\\ \\Longleftrightarrow\\ \\triangle ABI\\equiv\\triangle IPD\\Longleftrightarrow AB \\equal{} BP\\ .$"
}
{
  "Tag": [
    "calculus",
    "integration",
    "function",
    "calculus computations"
  ],
  "Problem": "Let $f(x)$ be the function such that $f(x)>0$ at $x\\geq 0$ and $\\{f(x)\\}^{2006}=\\int_{0}^{x}f(t) dt+1.$\r\n\r\nFind the value of $\\{f(2006)\\}^{2005}.$",
  "Solution_1": "I think this is the right way to do it, correct me if I am wrong.\r\nFirst differentiate both side, so by FTC we have\r\n${f(x)}^{2006}=\\int_{0}^{x}f(t)dt+1 \\\\ \\implies 2006{f(x)}^{2005}f'(x)=f(x)\\\\ \\implies{f(x)}^{2004}f'(x)=1/2006$\r\nNow, it is a first-order seperable differential equation. with $f(x)=1$, for $x=0$ \r\nso by integrate both side, we have\r\n$\\frac{{f(x)}^{2005}}{2005}=\\frac{x}{2006}+C\\\\ \\implies f(x)=(\\frac{2005x}{2006}+C)^{\\frac{1}{2005}}$\r\nBy subsititute the initial condition in, we have $C=1$\r\nso $f(x)=)=(\\frac{2005x}{2006}+1)^{\\frac{1}{2005}}$\r\nso $f(2006)=(2006)^{\\frac{1}{2005}}$",
  "Solution_2": "Your answer is correct. :) \r\nThe answer is $2006.$",
  "Solution_3": "But we first need to ensure that f(x) is differentiable.This isn't provided in the conditions!",
  "Solution_4": "[quote=galacta]But we first need to ensure that f(x) is differentiable.This isn't provided in the conditions![/quote]\nNo. We need only continuity of $f$. From there, it follows from the fact the integral of any continuous function is differentiable. As the RHS is differentiable, so is the LHS. :)\n\u2014\u2014\u2014\u2014\u2014\u2014\u2014\nCorrect me if I\u2019m wrong.",
  "Solution_5": "[quote=galacta]But we first need to ensure that f(x) is differentiable.This isn't provided in the conditions![/quote]\n\n$f(x)^{2006}=\\int_0^x f(t) \\ dt+1\\quad (*)$\n\nThe RHS is a continuous function of $x$ by fundamental theorem of calculus (FTC). Hence the LHS is continuous too. Now $f$ is positive on $[0,\\infty).$ Therefore continuity of $f(x)^{2006}$ implies that $f(x)$ is continuous. Since $f$ is continuous, applying FTC once again, we get that RHS of $(*)$ is differentiable. Hence $f(x)^{2006}$ is differentiable. Now, $f(x)>0$ for all $x\\ge 0.$ Hence $\\log (f(x)^{2006})=2006\\log f(x)$ is differentiable. Thus, $f(x)=e^{\\log f(x)}$ is differentiable too. :)",
  "Solution_6": "[quote=integrated_JRC][quote=galacta]But we first need to ensure that f(x) is differentiable.This isn't provided in the conditions![/quote]\nNo. We need only continuity of $f$. From there, it follows from the fact the integral of any continuous function is differentiable. As the RHS is differentiable, so is the LHS. :)\n\u2014\u2014\u2014\u2014\u2014\u2014\u2014\nCorrect me if I\u2019m wrong.[/quote]\n\nthe LHS is the 2006th power of f(x),so in elementary calculus,we first need to solve out f(x) by take the 2006th root on every side.Then we see the differentiability of f(x) by the chain rules.I THINK there is also a theorem talking about the differentiability of implicit function which can used in this problem directly instead of solve f(x) out.But we need more advanced tool to prove this theorem."
}
{
  "Tag": [
    "function",
    "integration",
    "calculus",
    "calculus computations"
  ],
  "Problem": "Last 3 weeks I gave the following problem to my students and  :huh: no one can solve it correctly.\r\n\r\nDoes there exist continuous function $ f : [0,1] \\to \\mathbb R$ satisfying\r\n\\[ \\int_0^1 {f\\left( x \\right) \\,dx }  = \\frac{1}{3} + \\int_0^1 {\\left( {f\\left( {x^2 } \\right)} \\right)^2  \\,dx } .\r\n\\]\r\n\r\n[hide]$ f(x) = \\sqrt{x}$.[/hide]",
  "Solution_1": "wouldn't f(x)=a, where $ a^2\\minus{}a\\plus{}1/3\\equal{}0$ work?",
  "Solution_2": "Putting $ x^2 \\equal{} y$ in the last integral, we get that $ \\int_0^1 \\left( \\frac {f(x)}{\\sqrt 2 \\sqrt [4]{x}} \\minus{} \\frac {\\sqrt [4] x}{\\sqrt 2} \\right)^2 \\, dx \\equal{} 0$ :)\r\n\r\nIt's worth mentioning that I'm talking improper integrals here, i.e. as $ a \\to 0$, $ \\int_a^1 \\frac {f^2 (y)}{2 \\sqrt y} \\, dy$ approaches the second integral from below.",
  "Solution_3": "[quote=\"0714446459923\"]wouldn't f(x)=a, where $ a^2 \\minus{} a \\plus{} 1/3 \\equal{} 0$ work?[/quote]\r\n\r\nYes, if that equation had real solutions ;)",
  "Solution_4": "JBL:Yes, if that equation had real solutions.\r\n\r\nBut that equation had no real solutions.The solution perfect_radio given is the exclusive solution.",
  "Solution_5": "how do you prove that that is the only solution?",
  "Solution_6": "Sorry. The solution perfect_radio given is the exclusive real and continuous solution.\r\nThe proof Perfect_radio had given 1n the 3th floor.If you want more simple or evident explain,I suggest that you may putting x=y^2 in the first integral.",
  "Solution_7": "[quote=\"chernfey\"]I suggest that you may putting x=y^2 in the first integral.[/quote]\r\n\r\nOh, one can easily see that but how to proceed further  :roll:",
  "Solution_8": "[quote=\"bookworm_vn\"][quote=\"chernfey\"]I suggest that you may putting x=y^2 in the first integral.[/quote]\n\nOh, one can easily see that but how to proceed further  :roll:[/quote]\r\n\r\nHello,bookworm_vn,I think you really need to see perfect_radio's proof seriously.\r\nf>=0,continuous ,in [a,b],and int[a,b]f(x)dx=0  => f=0 in [a,b].\r\nIf you still intangible,I all say nothing.",
  "Solution_9": "The full solution, just for the sake of completeness:\r\n\r\n[hide]Set $ x \\equal{} y^2, dx \\equal{} 2y\\, dy$ in the first integral, so we have\n\\[ \\int_0^1 2y \\cdot f( y^2 ) \\,dy \\equal{} \\frac {1}{3} \\plus{} \\int_0^1 {\\left( {f( {y^2 } )} \\right)^2 \\,dy }.\n\\]\nRewriting $ \\frac {1}{3} \\equal{} \\int_{0}^{1} y^2 \\,dy$ and bringing everything over to the right gives\n\\[ \\int_0^1 f(y^2)^2 \\minus{} 2yf(y^2) \\plus{} y^2 \\,dy \\equal{} 0,\n\\]\nor if we set $ g(y) \\equal{} f(y^2) \\minus{} y$\n\\[ \\int_0^1 g(y)^2 \\,dy \\equal{} 0.\n\\]\nSince $ f$ is continuous, $ g^2$ is a continuous, nonnegative function.  It's a standard exercize in a first proof-based calculus class to show that the integral of a nonnegative continuous function over some interval can be zero if and only if the function is uniquely zero.  Thus $ g(y)^2 \\equal{} 0$ and so $ f(y^2) \\minus{} y \\equal{} 0$ for all $ y \\in [0, 1]$.  It follows that $ f(x) \\equal{} \\sqrt {x}$ is the unique function satisfying the desired equation.[/hide]"
}
{
  "Tag": [
    "inequalities"
  ],
  "Problem": "If $ \\frac{1}{a} \\plus{} \\frac{1}{b} \\equal{} \\frac{2}{b}$ and $ ac > 0$.Prove that $ \\frac{a\\plus{}b}{2a\\minus{}b}\\plus{}\\frac{c\\plus{}b}{2c\\minus{}b}\\ge4$",
  "Solution_1": "Please edit.",
  "Solution_2": "if it is in that way that is written here ... than this inequality is equivalent to :\r\n\r\n$ \\frac {c \\plus{} b}{2c \\minus{} b}\\ge2$     \r\n\r\nwhere $ a\\equal{}b$\r\n\r\nand then it is very easy  :)  :D",
  "Solution_3": "[quote=\"Obel1x\"]If $ \\frac {1}{a} \\plus{} \\frac {1}{b} \\equal{} \\frac {2}{b}$ and $ ac > 0$.Prove that $ \\frac {a \\plus{} b}{2a \\minus{} b} \\plus{} \\frac {c \\plus{} b}{2c \\minus{} b}\\ge4$[/quote]\r\n\r\n\r\nIm sorry its : If $ \\frac {1}{a} \\plus{} \\frac {1}{c} \\equal{} \\frac {2}{b}$ and $ ac > 0$.Prove that $ \\frac {a \\plus{} b}{2a \\minus{} b} \\plus{} \\frac {c \\plus{} b}{2c \\minus{} b}\\ge4$\r\n\r\ni made this problem the easiest one in mathlinks.ro  :(  :oops:",
  "Solution_4": "We have $ b\\equal{}\\frac{2ac}{a\\plus{}c}$; substituting, the given inequality is equivalent to $ \\frac{a\\plus{}3c}{2a} \\plus{} \\frac{c\\plus{}3a}{2c} \\geq 4 \\Leftrightarrow \\frac{c}{a}\\plus{}\\frac{a}{c} \\geq 2$, which follows from homogeneity (WLOG $ a,c > 0$) and AM-GM."
}
{
  "Tag": [
    "calculus",
    "integration",
    "number theory",
    "Diophantine equation"
  ],
  "Problem": "Hey guys,\r\n\r\nI'm stuck on the following problem:\r\n\r\nFind the integral solutions to the Diophantine equation 317x+241y=9.\r\n\r\nI subtracted 317x to get 241y=9+317x and then put the equation in mod 317 to get 241x is congruent to 9(mod 317). \r\n\r\nThat's pretty much where I get confused. Is there a way to find a multiple of 217 that is congruent to 9(mod 317) without random guessing and checking?\r\n\r\nThanks,\r\nmathgeek93",
  "Solution_1": "hello, i hope this will help you\r\nhttp://www-groups.dcs.st-and.ac.uk/~martyn/teaching/1003/1003linearDiophantine.pdf\r\nSonnhard.",
  "Solution_2": "Using Theorem $ 4.6$ in the reference given by Sonnhard and finding lots of multiples of $ 317$ and $ 241$ on a calculator, I get $ 317 \\times 35 \\plus{} 241 \\times ( \\minus{} 46) \\equal{} 9$ and so $ x_{0} \\equal{} 35, y_{0} \\equal{} \\minus{} 46$ and the general solution is \r\n\r\n$ x \\equal{} 35 \\plus{} 241t, y \\equal{} \\minus{} 46 \\minus{} 317t$ for all $ t \\in \\mathbb{Z}$."
}
{
  "Tag": [
    "inequalities",
    "probability",
    "expected value",
    "algebra proposed",
    "algebra"
  ],
  "Problem": "Show that , for any $z_1,z_2...z_n\\in \\mathbb{C}$, there are $a_1,a_2...a_n\\in\\{\\pm 1\\}$such that:\r\n\\[|\\sum(a_iz_i)|^2\\leq(\\geq)\\sum (|z_i|^2)\\]",
  "Solution_1": "The probabilistic method works well here.  Let $a_1$, ..., $a_n$ be independent, random elements of $\\{ \\pm 1 \\}$.  Let's compute the expected value of the left-hand side of the desired inequality: \r\n[tex]\\begin{eqnarray*} \r\nE({\\bigl\\vert \\sum_i a_i z_i \\bigr\\vert}^2) \r\n  & = & E(\\sum_{i, j} a_i z_i a_j \\overline{z_j}) \\\\\r\n  & = & \\sum_{i, j} z_i \\overline{z_j} E(a_i a_j) \\\\\r\n  & = & \\sum_i {\\left\\vert z_i \\right\\vert}^2.\r\n\\end{eqnarray*}\r\n[/tex]\r\nHence, there exists a specific choice of $a_i$ such that \r\n[tex]\r\n{\\bigl\\vert \\sum_i a_i z_i \\bigr\\vert}^2 \r\n  \\geq  \\sum_i {\\left\\vert z_i \\right\\vert}^2 \r\n   =  \\sum_i {\\left\\vert a_i z_i \\right\\vert}^2.\r\n[/tex]\r\nSimilarly, there exists a specific choice of $a_i$ such that\r\n[tex]\r\n{\\bigl\\vert \\sum_i a_i z_i \\bigr\\vert}^2 \r\n  \\leq  \\sum_i {\\left\\vert z_i \\right\\vert}^2 \r\n  =  \\sum_i {\\left\\vert a_i z_i \\right\\vert}^2.\r\n[/tex]",
  "Solution_2": "Could you please tell more about this probabilistic method. Is it only based on the fact that you sum up all possible sums, or is it more than that(not necessary on this problem)?",
  "Solution_3": "The version of the problem I was given used $\\epsilon$ instead of $a$ and assumed WLOG that $\\sum |z_i|^2 =1$. \n\nLet $z_k=a_k+b_ki$ for all $1\\le k\\le n$. Now, notice that the problem condition is equivalent to \n\\[(a_1^2+a_2^2+\\dots+a_n^2)+(b_1^2+b_2^2+\\dots+b_n^2)=1,\\]\nand the problem is asking us to prove that there exist $\\epsilon_1$, $\\epsilon_2$, $\\dots$, $\\epsilon_n$ such that \n\\[ \\left|\\sum_{k=1}^n \\epsilon_k z_k \\right| \\le 1,\\]\nand squaring both sides of the equation, this is equivalent to proving that \n\\[(\\epsilon_1a_1+\\epsilon_2a_2+\\dots+\\epsilon_na_n)^2+(\\epsilon_1b_1+\\epsilon_2b_2+\\dots+\\epsilon_nb_n)^2\\leq 1.\\]\nAnd since $\\epsilon_i^2a_i^2$ is always equal to $a_i^2$, we can re-express this sum as \n\\[\\sum a_i^2 + 2\\sum \\epsilon_i\\epsilon_ja_ia_j+\\sum b_i^2 + 2\\sum \\epsilon_i\\epsilon_jb_ib_j,\\]\nand since we're given that $(a_1^2+a_2^2+\\dots+a_n^2)+(b_1^2+b_2^2+\\dots+b_n^2)=1$, this gives us that this is equal to \n\\[1+2\\sum \\epsilon_i\\epsilon_ja_ia_j+2\\sum \\epsilon_i\\epsilon_jb_ib_j.\\]\nNow note that if we sum this over all possible $n$-tuples of the different $\\epsilon$'s, the coefficient of $a_ia_j$, $\\epsilon_i\\epsilon_j$ is $-1$ with $\\frac{1}{2}$ probability and $1$ with $\\frac{1}{2}$ probability, meaning that the average value of $\\epsilon_i\\epsilon_ja_ia_j$ over all $n$-tuples of the $\\epsilon$'s for all $(i,j)$ is $0$. Similarly, the average value of $\\epsilon_i\\epsilon_jb_ib_j$ over all $n$-tuples of the $\\epsilon$'s for all $(i,j)$ is $0$. \\\\\\\\\nThis gives us that the average value of \n\\[\\left| \\sum_{k=1}^n \\epsilon_k z_k \\right|^2=1+2\\sum \\epsilon_i\\epsilon_ja_ia_j+2\\sum \\epsilon_i\\epsilon_jb_ib_j,\\]\nis $1+0$, or $1$. This implies that \n\\[\\left|\\sum_{k=1}^n \\epsilon_k z_k \\right| \\le 1,\\]\nfor some $n$-tuple of $\\epsilon$'s, which means that \n\\[\\left|\\sum_{k=1}^n \\epsilon_k z_k \\right| \\le 1,\\]\nfor the same $n$-tuple of $\\epsilon$'s, finishing the problem.",
  "Solution_4": "The square of the LHS is just\n\\begin{align*}\n& \\left(\\sum_{k=1}^n \\varepsilon_k z_k\\right)\\cdot\\overline{\\left(\\sum_{k=1}^n \\varepsilon_k z_k\\right)} \\\\\n=& \\sum_{i=1}^n\\sum_{j=1}^n \\varepsilon_i\\varepsilon_j z_i \\overline{z_j} \\\\\n=& \\sum_{i=1}^n \\varepsilon_i \\varepsilon_i |z_i|^2+\\sum_{i=1}^n\\sum_{j\\ge 1, j\\ne i} \\varepsilon_i\\varepsilon_j z_i \\overline{z_j} \\\\\n=& 1+\\sum_{i=1}^n\\sum_{j\\ge 1, j\\ne i} \\varepsilon_i\\varepsilon_j z_i \\overline{z_j}.\n\\end{align*}\nBy randomly choosing $\\varepsilon_i$, the expected value of the second sum is $0$, so some choice works. $\\blacksquare$",
  "Solution_5": "De-homogenize so that $\\sum |z_i|^2 = 1$, and use $\\epsilon_i$ instead of $a_i$. Then the LHS equals\n\\begin{align*}\n\\left| \\sum_{k = 1}^n \\epsilon_kz_k \\right| ^2 &= \\left(\\sum_{k = 1}^n \\epsilon_kz_k \\right) \\left( \\sum_{k = 1}^n \\epsilon_k \\overline{z_k} \\right) \\\\\n&= \\sum_{k = 1}^n \\epsilon_k^2 |z_k|^2 + \\sum_{i < j} (\\epsilon_i \\epsilon_j z_i \\overline{z_j} + \\epsilon_i \\epsilon_j \\overline{z_i}z_j) \\\\\n&= 1 +  \\sum_{i < j} \\epsilon_i \\epsilon_j (z_i \\overline{z_j} + \\overline{z_i}z_j).\n\\end{align*}\nWhen we sum this over all $2^n$ possible combinations of $\\epsilon_1, \\epsilon_2, \\ldots, \\epsilon_n$, each of the $\\epsilon_i$ is $-1$ half the time and $1$ the other half of the time so the $\\sum_{i < j} \\epsilon_i \\epsilon_j (z_i \\overline{z_j} + \\overline{z_i}z_j)$ terms will \"vanish\" and the sum is just $2^n$. Hence for some choice of $\\epsilon_i$, the sum is at most $1$."
}
{
  "Tag": [
    "AMC",
    "USA(J)MO",
    "USAMO",
    "inequalities solved",
    "inequalities"
  ],
  "Problem": "Let a, b, c, d are real number such that $\\large a^2+b^2+c^2+d^2 \\leq 1$ . Find maximum of : \r\n\r\n$\\large A=(a+b)^4+(a+c)^4+(a+d)^4+(b+c)^4+(b+d)^4+(c+d)^4$",
  "Solution_1": "nobody solve ???? Please help me !!!  :(  :(",
  "Solution_2": "[quote=\"Chocolate M&M\"]Let a, b, c, d are real number such that $\\large a^2+b^2+c^2+d^2 \\leq 1$ . Find maximum of : \n\n$\\large A=(a+b)^4+(a+c)^4+(a+d)^4+(b+c)^4+(b+d)^4+(c+d)^4$[/quote]\r\n\r\nWe have $\\large (a+b)^4 \\leq (a-b)^4+(a+b)^4=2(a^4+b^4+6a^2b^2)$\r\nSimilarly, we have \r\n$\\large A \\leq 6(a^4+b^4+c^4+d^4+2a^2b^2+2a^2c^2+2a^2d^2+2b^2c^2+2b^2d^2+2c^2d^2)=6(a^2+b^2+c^2+d^2)^2 \\leq 6$\r\nthat equality holds when $\\large a=b=c=d=\\frac{1}{2}$   :D",
  "Solution_3": "This seems to be an USAMO problem",
  "Solution_4": "This seems to be an USAMO problem"
}
{
  "Tag": [
    "probability",
    "logarithms"
  ],
  "Problem": "Professor Gamble buys a lottery ticket, which requires that he pick six different integers from 1 through 46, inclusive. He chooses his numbers so that the sum of the base-ten logarithrms of his six numbers is an integer. It so happens that the integers on the winning ticket have the same property - the sum of the base-ten logarithrms is an integer. What is the probability that Professor Gamble holds the winning ticket?\r\n\r\n$\\text{(A)} \\ 1/5 \\qquad \\text{(B)} \\ 1/4 \\qquad \\text{(C)} \\ 1/3 \\qquad \\text{(D)} \\ 1/2 \\qquad \\text{(E)} \\ 1$",
  "Solution_1": "[hide]Let $x_i$ be the numbers on his ticket and $y_i$ be the numbers on the winning ticket. We must have $\\log x_1 + \\log x_2 + \\cdots + \\log x_6 = \\log (x_1x_2x_3x_4x_5x_6) \\in \\mathbb Z$ and $\\log (y_1y_2y_3y_4y_5y_6) \\in \\mathbb Z$ as well. This means that both $x_1x_2\\cdots x_6$ and $y_1y_2\\cdots y_6$ must be powers of 10. The highest power of ten possible is $10^6$. We can also get $10^5$ (in 2 ways) and $10^4$, but it is impossible to get $10^3$, $10^2$, $10^1$, or $10^0$, so our answer is $\\frac{1}{4} \\implies \\boxed {B}$.[/hide]"
}
{
  "Tag": [
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Find all natural numbers $ n$ with exactly $ 8$ natural divisors which sum is equal to $ 3780$ (including 1 and $ n$).",
  "Solution_1": "$ n$ has exactly $ 8$ divisors $ \\Leftrightarrow n\\equal{}pqr, p^3 q, p^7$, where $ p,q,r$ are different prime numbers.Also $ 3780\\equal{}2^2*3^3*5*7$\r\n[b]Case 1:[/b]$ n\\equal{}pqr$, $ (p\\plus{}1)(q\\plus{}1)(r\\plus{}1)\\equal{}2^2*3^3*5*7$, then we have one of them, say $ p\\equal{}2$ as $ 8 \\nmid R.H.S.$\r\nSo $ (q\\plus{}1)(r\\plus{}1)\\equal{}2^2*3^2*5*7$, we have $ q, p\\equal{}1(mod \\;\\; 4)$ as $ 8 \\nmid R.H.S.$\r\nNow let $ q<r$, then $ q\\plus{}1<\\sqrt{2^2*3^2*5*7}$,$ q \\le 34$, by trial (as possible values of $ q$ are $ 5,13,17,29$), we get $ n\\equal{}2*13*89\\equal{}2314$ or $ n\\equal{}2*29*41\\equal{}2378$\r\n\r\n[b]Case 2:[/b]$ n\\equal{}p^3 q$, so $ (p^3\\plus{}p^2\\plus{}p\\plus{}1)(q\\plus{}1)\\equal{}2^2*3^3*5*7$ , in this case we get $ p<(2^2*3^2*5*7)^{\\frac{1}{3}}$, so $ p \\le 10$, by trial we get $ n\\equal{}2^3*251\\equal{}2008$\r\n\r\n[b]Case 3:[/b]$ n\\equal{}p^7$, then $ p^7<2^2*3^3*5*7$, so $ p\\equal{}2 or 3$ which is impossible as $ 2,3|3780$,\r\n\r\nSo the only solutions are $ n\\equal{}2008, 2314, 2378$"
}
{
  "Tag": [],
  "Problem": "If $\\displaystyle\\sum_{n=1}^{k}\\displaystyle \\left(\\frac{1}{\\sqrt{2n-1}}-\\frac{1}{\\sqrt{2n+1}}\\right)=\\frac{10}{11}$, what is the positive integer value of $k$?",
  "Solution_1": "[hide]Just from writing down the first few terms, we notice that we have a telescoping series.  The sum of the first $k$ terms is thus:\n\n$1-\\frac{1}{\\sqrt{2k+1}}=\\frac{10}{11}$\n$\\sqrt{2k+1}=11$\n$2k+1=121$\n$k=\\boxed{60}$[/hide]"
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "Prove that, \\[ (a\\plus{}3b)(b\\plus{}4c)(c\\plus{}2a) \\geq 60abc \\]for all real numbers $ 0\\leq a\\leq b\\leq c$.",
  "Solution_1": "[quote=\"Moonmathpi496\"]Prove that,\n\\[ (a \\plus{} 3b)(b \\plus{} 4c)(c \\plus{} 2a) \\geq 60abc\n\\]\nfor all real numbers $ 0\\leq a\\leq b\\leq c$.[/quote]\r\n$ (a \\plus{} 3b)(b \\plus{} 4c)(c \\plus{} 2a) \\equal{} 25abc \\plus{} 2a^2b \\plus{} 6ab^2 \\plus{} 3b^2c \\plus{} 12bc^2 \\plus{} 4c^2a \\plus{} 8ca^2 \\ge 60abc \\iff$\r\n$ ab(2a\\plus{}6b) \\plus{} bc(3b\\plus{}12c) \\plus{} ca(4c\\plus{}8a) \\ge 35abc \\iff$\r\n$ 2\\frac{a}{c}\\plus{}6\\frac{b}{c} \\plus{} 3\\frac{b}{a}\\plus{}12\\frac{c}{a} \\plus{} 4\\frac{c}{b}\\plus{}8\\frac{a}{b} \\ge 35\\iff$\r\nBy AM-GM:\r\n$ LHS \\ge 35 \\sqrt[35]{\\frac{a^{10}b^9c^{16}}{a^{15}b^{12}c^8}} \\ge 35 \\equal{} RHS \\iff$\r\n$ c^8 \\ge a^5b^3$ which is obvious since $ c \\ge a,b$\r\nEquality iff $ a\\equal{}b\\equal{}c$.",
  "Solution_2": "Apply AM-GM to $ \\frac{a\\plus{}b\\plus{}b\\plus{}b}{4}$, $ \\frac{b\\plus{}c\\plus{}c\\plus{}c\\plus{}c}{5}$ and $ \\frac{c\\plus{}a\\plus{}a}{3}$ to get \r\n\r\n$ (a\\plus{}3b)(b\\plus{}4c)(c\\plus{}2a) \\ge 60a^{\\frac{11}{12}}b^{\\frac{19}{20}}c^{\\frac{17}{15}}$.\r\n\r\nThen $ c^{\\frac{2}{15}}\\ge a^{\\frac{1}{12}}b^{\\frac{1}{20}}$ since $ \\frac{1}{12} \\plus{} \\frac{1}{20}\\equal{}\\frac{2}{15}$, and so $ a^{\\frac{11}{12}}b^{\\frac{19}{20}}c^{\\frac{17}{15}} \\ge abc$.",
  "Solution_3": "I am interested to know if there is any easy and efficient method to find the coefficient to apply AM-GM. I know that there is a method named Balancing Coefficients. \r\n\r\nDid you use balancing coefficient method?",
  "Solution_4": "[quote=\"Moonmathpi496\"]I am interested to know if there is any easy and efficient method to find the coefficient to apply AM-GM. I know that there is a method named Balancing Coefficients. \n\nDid you use balancing coefficient method?[/quote]\r\nI don't know what the balancing coefficient method is, since i've never heard about it :blush: The trick is to guess that the ineq only has equality when $ a\\equal{}b\\equal{}c$. I thought there was at least one more, so that was why I multiplied out. But you should just have the equality in mind in this case. So while $ 2a \\plus{} b \\ge 2\\sqrt{2ab}$ is true, it won't help us much since $ 2a\\plus{}b \\neq 2\\sqrt{2ab}$ when $ a\\equal{}b$. So you would choose $ 2 \\cdot a \\plus{} b \\ge (2\\plus{}1) \\sqrt[2\\plus{}1]{a \\cdot a\\cdot b} \\equal{} 3\\sqrt[3]{a^2b}$. Hope you understood :)"
}
{
  "Tag": [
    "quadratics"
  ],
  "Problem": "Let a, b, c be real numbers satisfying a < b < c, a + b + c = 6 and ab + bc + ca = 9. \r\nProve that 0 < a < 1 < b < 3 < c < 4.",
  "Solution_1": "well just by some fill-in you can easily find that they're all positive so the 0< part is pretty easy\r\n\r\nthen a<b<c is also given, so no need worrying bout that.\r\n\r\nSo we got: 0 < a < ? < b < ? < c < ?\r\n\r\nAt least 2 of them must be lower than 3 because of ab + bc + ca = 9\r\n\r\nso 0 < a < ? < b < 3 < c < ? or 0 < a < ? < b < ? < c < 3\r\n\r\nsuppose c<3, then a+b>3, then ab + c.(b+a) = 9 results us that ab < 0, which cannot be, so 0 < a < ? < b < 3 < c < ? is the correct one.\r\n\r\nIf one of them (c thus) was bigger than 4, then the other two have sum lower than 2, which makes ab + c.(b+a) = 9, and which ends us up in a quadratic equitation saying a=b=1 (which contradicts a<b<c\r\n\r\nso 0 < a < ? < b < 3 < c < 4\r\n\r\nnow last thing to prove is that a < 1 < b\r\n\r\nsuppose they were both smaller:\r\nthen c > 4, contradiction\r\n\r\nsuppose both larger: \r\nsay a = 1+d, b= 1+e\r\nd+e+c = 4      (d+e<1 as c>3)\r\nso work the second one out: \r\ned+dc+ec+c = 5\r\nand this is obviously impossible with 1 > d,e > 0 and 4 > c > 3\r\ncontradiction\r\n\r\nthus it is possible no other way then that 0 < a < ? < b < 3 < c < 4",
  "Solution_2": "I solved it this way:\r\n\r\nLet t = abc. Then a, b, c are the roots (= zeros) of f(x) = x<sup>3</sup> - 6x<sup>2</sup> + 9x - t = 0.\r\nWe have f'(x) = 3(x - 1)(x - 3) so f(3) < 0 < f(1). Hence a < 1 < b < 3 < c.\r\nSince f(1) = 4 - t > 0, t = abc < 4, and since f(3) = -t < 0, abc > 0.\r\nAs b, c are positive, a > 0.\r\nAlso, f(x) is strictly increasing on (3, +oo) and f(4) = f(1) = 4 - p > 0 so c < 4.",
  "Solution_3": "that's shorter indeed :D"
}
{
  "Tag": [
    "complex numbers"
  ],
  "Problem": "If $x+y+z=x^{-1}+y^{-1}+z^{-1}=0$, show that \\[\\frac{x^{6}+y^{6}+z^{6}}{x^{3}+y^{3}+z^{3}}=xyz\\]",
  "Solution_1": "[hide]\nRecall the factorization\n\\[x^{3}+y^{3}+z^{3}-3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx). \\]\nSince $x+y+z=0$, we have\n\\[x^{3}+y^{3}+z^{3}= 3xyz. \\]\nAlso, note that\n\\begin{align*}x^{2}+y^{2}+z^{2}&= (x+y+z)^{2}-2(xy+yz+zx)\\\\ &=-2xyz\\left(\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}\\right)\\\\ &= 0\\\\ \\end{align*}\nso $x^{6}+y^{6}+z^{6}= 3\\left(xyz\\right)^{2}$. Thus\n\\[\\frac{x^{6}+y^{6}+z^{6}}{x^{3}+y^{3}+z^{3}}= \\frac{3(xyz)^{2}}{3xyz}= xyz, \\]\nas desired.\n[/hide]",
  "Solution_2": "I don't think that\r\n$x+y+z=\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=0$\r\n\r\nis possible.\r\n\r\ncan anyone think of a example of (x,y,z) that satisfy the equation?",
  "Solution_3": "You'll have to go to the complex numbers, but there do indeed exist solutions.  (In fact, they're all just scalings and permutations of a single solution $(1, \\omega, \\omega^{2})$ where $\\omega$ is one of the complex third-roots of 1.",
  "Solution_4": "[quote=\"Tim_Lou\"]I don't think that\n$x+y+z=\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=0$\n\nis possible.\n\ncan anyone think of a example of (x,y,z) that satisfy the equation?[/quote]\r\n\r\nThe third roots of unity work. :)"
}
{
  "Tag": [
    "limit",
    "calculus",
    "calculus computations"
  ],
  "Problem": "It's pretty easy proving limits when having for example $ \\lim_{x\\to4}(2x\\plus{}1)\\equal{}9.$\r\n\r\nBut I'm getting stuck when tryin' to prove that $ \\lim_{x\\to2}x^2\\equal{}4.$\r\n\r\nWe have $ 0<|x\\minus{}2|<\\delta$ & $ |x^2\\minus{}4|<\\epsilon\\implies|x\\plus{}2\\parallel{}x\\minus{}2|<\\epsilon.$ What's the next step?",
  "Solution_1": "Create an upper bound for $ |x\\plus{}2|$, say $ M$, so that $ |x\\plus{}2|<M$. Choose $ \\delta\\equal{}\\frac{\\epsilon}{M}$ :)",
  "Solution_2": "Here's what I would do\r\n\r\n$ |x^2 \\minus{} 4| < \\epsilon$\r\n\r\n$ \\minus{} \\epsilon < x^2 \\minus{} 4 < \\epsilon$\r\n\r\n$ \\minus{} \\epsilon \\plus{} 4 < x^2 < \\epsilon \\plus{} 4$\r\n\r\nsince all three are greater than zero for sufficiently small epsilon\r\n$ \\sqrt { \\minus{} \\epsilon \\plus{} 4} < x < \\sqrt {\\epsilon \\plus{} 4}$\r\n\r\n$ \\sqrt { \\minus{} \\epsilon \\plus{} 4} \\minus{} 2 < x \\minus{} 2 < \\sqrt {\\epsilon \\plus{} 4} \\minus{} 2$\r\n\r\nSince the magnitude of lower bound of x-2 is greater than the magnitude of the upper bound we select the lower bound as our largest window\r\n\r\n$ |x \\minus{} 2| < 2 \\minus{} \\sqrt { \\minus{} \\epsilon \\plus{} 4}$\r\n\r\nand there you go\r\n\r\n$ \\delta \\equal{} 2 \\minus{} \\sqrt { \\minus{} \\epsilon \\plus{} 4}$ for any sufficiently small epsilon",
  "Solution_3": "That works as long as you're not building up foundations of calculus. (Typically, showing square roots exist uses the continuity of $ x^2$.)\r\n\r\nSo you want to show $ |x\\minus{}2\\parallel{}x\\plus{}2|<\\epsilon$ for all $ x$ satisfying $ |x\\minus{}2|<\\delta$, for some positive $ \\delta$ depending on $ \\epsilon$.\r\n\r\nIf $ \\epsilon$ is small, $ |x\\minus{}2|$ will close to 2, so $ |x\\plus{}2|$ will be close to 4. In particular, we should be able to find an open interval around 2 for which $ |x\\plus{}2|<5$. We can: $ |x\\minus{}2|<1$ works.\r\n\r\nSo choosing $ \\delta \\equal{}\\min \\left\\{ 1,\\frac{\\epsilon}{5}\\right\\}$ satisfies $ |x\\minus{}2\\parallel{}x\\plus{}2|<\\epsilon$ for all $ x$ such that $ |x\\minus{}2|<\\delta$.",
  "Solution_4": "[quote=\"scorpius119\"]Typically, showing square roots exist uses the continuity of $ x^2$.[/quote]\r\nNot really, it's easy to define each decimal in the root"
}
{
  "Tag": [],
  "Problem": "An object is propelled upward with an initial velocity of 20 meters per second.  The distance s (in meters) of the object from the ground after t seconds is s = -4.9t^2 + 20t.\r\n\r\n(a) What is the maximum height?\r\n\r\n(b) What is the minimum height?",
  "Solution_1": "$s=-4.9(t^{2}-\\frac{200t}{49})$\r\n$=-4.9\\left(t^{2}-\\frac{200t}{49}+\\left(\\frac{100}{49}\\right)^{2}-\\left(\\frac{100}{49}\\right)^{2}\\right)$\r\n$=-4.9\\left(t-\\frac{100}{49}\\right)^{2}+\\frac{1000}{49}$\r\n\r\nTherefore the maximum height is (a) $\\boxed{\\frac{1000}{49}}$..and the minimum is obviously (b) $\\boxed{0}$",
  "Solution_2": "QUESTION:\r\n\r\nWhere did you get 200/49 and 100/49?",
  "Solution_3": "$\\frac{200}{49}\\rightarrow{\\frac{20}{4.9}=\\frac{200}{49}}$\r\n\r\n$\\frac{100}{49}\\rightarrow{\\frac{1}{2}}\\cdot\\frac{200}{49}=\\frac{100}{49}$\r\n\r\nya I converted the decimal into a fraction and all I did there was complete the square..",
  "Solution_4": "[hide=\"solution\"]its just $\\frac{-b}{2a}$ or $\\frac{-20}{(2)(-4.9)}$ which equals $\\frac{200}{(2)(49)}=\\frac{100}{49}$\n\nthe minimum is obviously $0$[/hide]",
  "Solution_5": "Good replies here.",
  "Solution_6": "shouldnt this be in classroom math?  this is an algebra 1 thing.",
  "Solution_7": "no, algebra 1 canbe in hsb if its easy enuf"
}
{
  "Tag": [
    "function",
    "calculus",
    "calculus computations"
  ],
  "Problem": "Find $ f''(x)\\minus{}5f'(x)\\plus{}3f(x)$, if    $ f: \\mathbb{R}\\rightarrow\\mathbb(\\minus{}\\frac{\\pi}2, \\frac{\\pi}2)$,   $ f(x)\\equal{}arctgx$",
  "Solution_1": "$ f'(x) \\equal{} \\frac{1}{1\\plus{}x^2}$\r\n\r\n$ f''(x) \\equal{} \\frac{\\minus{}2x}{(1\\plus{}x^2)^2}$"
}
{
  "Tag": [
    "inequalities proposed",
    "inequalities"
  ],
  "Problem": "Let $a,b,c$ are positive number . Prove:\r\n$\\frac{a}{b+c+1}+\\frac{b}{a+c+1}+\\frac{c}{a+b+1}\\leq (a+b+c)$\r\n\r\nYeah. Here is what I come up with [edit according to Albanian Eagle]\r\nLet $a+b+c=1$ Find min and max of:\r\n$\\frac{a^{2}}{b+c+1}+\\frac{b^{2}}{b+c+1}+\\frac{c^{2}}{a+c+1}$\r\n\r\n$\\frac{a^{3}}{b+c+1}+\\frac{b^{3}}{b+c+1}+\\frac{c^{3}}{a+c+1}$\r\n\r\n$\\frac{a^{4}}{b+c+1}+\\frac{b^{4}}{b+c+1}+\\frac{c^{4}}{a+c+1}$\r\n\r\n:(",
  "Solution_1": "there should be a misprint because this is trivial... :maybe:",
  "Solution_2": "Er,I still don't understand....\r\n$\\sum \\frac{a^{k}}{b+c+1}\\leq \\sum a^{k}\\leq \\sum a=1$ when $k\\geq 1$\r\nand a=1,b=0,c=0 the equality holds.\r\nIs there any mistake?",
  "Solution_3": "[quote=\"Gibbenergy\"]Let $a,b,c$ are positive number . Prove:\n$\\frac{a}{b+c+1}+\\frac{b}{a+c+1}+\\frac{c}{a+b+1}\\leq (a+b+c)$\n\nYeah. Here is what I come up with [edit according to Albanian Eagle]\nLet $a+b+c=1$ Find min and max of:\n$\\frac{a^{2}}{b+c+1}+\\frac{b^{2}}{c+a+1}+\\frac{c^{2}}{a+b+1}$\n\n$\\frac{a^{3}}{b+c+1}+\\frac{b^{3}}{c+a+1}+\\frac{c^{3}}{a+b+1}$\n\n$\\frac{a^{4}}{b+c+1}+\\frac{b^{4}}{c+a+1}+\\frac{c^{4}}{a+b+1}$\n\n:([/quote]"
}
{
  "Tag": [
    "geometry",
    "3D geometry"
  ],
  "Problem": "Help me figure out how many different possible combinations of a 20x20x20 Rubik's Cube",
  "Solution_1": "[hide]\nWe have to look at the different types of pieces that make up the cube\n\nCorner Pieces\nEdge Pieces\nCenter Pieces\n\nCorner Pieces are straightforward: there are $ 8$\n\nEdge pieces can be subdivided into 9 subcategories based on how close to the corner they are. (20 sided cube, minus two for corners, and symmetry).\n\nCenter pieces can be subdivided into 45 categories of pieces (due to symmetries, this value is 1/8 of the total 18^2 adjusting for diagonal pieces)\n9 are diagonal pieces, and the other 36 are other center types\n\nNow we must enumerate each type\n\nAmount of pieces\nCorner: 8\nEdge Subcategory type: 12 x 2 = 24  (12 edges to a cube. 2 on each edge)\nCenter Diagonal type: 6 x 4 = 24 (6 faces, four on each face)\nCenter reg type: 6 x 8 = 48\n\nNumber of orientations:\nCorner: 3\nEdge: 2\nCenter: 1\n\nthrow it into one long expression, and you have\n\n$ (8!\\cdot3^{8})(24!\\cdot2^{24})^{9}(24!\\cdot 1^{24})^{9}(48!\\cdot 1^{48})^{36}$\n\nIf you put a picture on each face, suddenly the orientations on the center face skyrockets to 4, and you have to adjust accordingly.\n[/hide]\r\n\r\nOr we could just say that there are a ****load of possible combinations.",
  "Solution_2": "In case anyone's wondering, that's equal to\r\n\r\n\r\n12427622692135360803136378828873041098174068758578\r\n96219664638519956496970156525568897786278822573518\r\n08756562975039318258829386203920855055196896447399\r\n16936324369683807309514523753881151764910926848884\r\n12099433595969628146818702724787271841888520444634\r\n44787834943711079279640572128804306764501815043251\r\n91215359625426254103675803061589480946924907274758\r\n89179049596884200759728687085181396833366098609494\r\n83272234679682352756845200010610606150337513565852\r\n09351315461094802623432084972090252261808619806795\r\n39799332425284994191579614479081680754139217425753\r\n47734951458996762539104061619595430948880006839140\r\n96573448846886244058768973111097259333136495896845\r\n62671333506245256229314259392792073764809564243326\r\n71489183508388208731685125523141465701439518477740\r\n03035460389335280238306724947715547416680082524062\r\n73680169970569540147577732768982057744332067424884\r\n72185788302958725403552505635586230692911431225720\r\n25068872713855453511993670091326261239754660093007\r\n49377372479188990339733063340730083064000615206277\r\n81698389886522899711524463445574875772120721107170\r\n82844037525292964890101206413225316333725350021023\r\n47811479187536656319035892969783611542329399031785\r\n53008458195081387683443083638934288245488915046614\r\n53437229198737179405899368396971522584122424967098\r\n39640315778834201230283344508192345406829699751006\r\n80253101459624833603234527400034545685253730472795\r\n50207853026928105102474029348269477116544261366756\r\n41123479788100888450586748510605885518828493125000\r\n50562401182748138061947265844484818828507848696917\r\n55913076613688085146688390917479758629457728789371\r\n28284864470481578881303694748240352243586278795013\r\n34893419014214488749154277363261076704053383785152\r\n11003775474678459618688308961680470408592291451642\r\n21243133320891492823381067675313605723372725428117\r\n71003403201771057982874333584034173294552343298884\r\n50856199297973000781270410442251929865384465611503\r\n20726630091057750358114519769477662961913551850369\r\n72954597911297262421766756805928656437336024691076\r\n72069060647440367777293484637582413620397049656881\r\n90564090390506132273972592104070114402209328633909\r\n22430727499943450403478465451810048924412240389889\r\n32796675945363909336045692741111639487305822451449\r\n71777467237562187783948651667534001624156712813427\r\n12196324913310741602438515758779676412545758020286\r\n41823407630261944320000000000000000000000000000000\r\n00000000000000000000000000000000000000000000000000\r\n00000000000000000000000000000000000000000000000000\r\n00000000000000000000000000000000000000000000000000\r\n00000000000000000000000000000000000000000000000000\r\n00000000000000000000000000000000000000000000000000\r\n00000000000000000000000000000000000000000000000000\r\n00000000000000000000000000000000000000000000000000\r\n00000000000000000000000000000000000000000000000000\r\n00\r\n\r\nwhere each line has 50 digits (except the last)\r\n\r\nIt is roughly equal to $ e^{6219.5}$, or $ 10^{2701.94}$",
  "Solution_3": "That number is not correct, you don't take into account that there are (a lot of) combinations that are impossible to reach."
}
{
  "Tag": [
    "USAMTS",
    "LaTeX",
    "Asymptote",
    "function"
  ],
  "Problem": "Hi\r\n\r\nI just started USAMTS\r\nWhat software do you prove questions?\r\nAnd what software do you use to create images?\r\nThanks :D",
  "Solution_1": "*What software do you use to format your proofs?",
  "Solution_2": "[url=http://www.texniccenter.org]TeXnicCenter[/url] can turn LaTeX into PDFs. \r\n[url=http://www.wolfram.com]Mathematica[/url] could potentially help solve problems.",
  "Solution_3": "Use a $ \\text{\\LaTeX}$ editor. You see the sidebar (its to your left  :D) where it says LaTeX? That's a quick tutorial of LaTeX. A common one here is TeXnicCenter (google it.)\r\n\r\nTo make images, anything will work. Even microsoft paint (just save it as a jpg file). I like using Microsoft Visio 2003 because of the easy access to all the shapes and even variables (letters in Monotype Corsiva font  :P). Then, at the top, write\r\n\r\n\\usepackage{graphicx}\r\n\r\nand where you want to insert the picture\r\n\r\n\\includegraphics[scale=(insert scale)]{insert file name}",
  "Solution_4": "which types of files can you put (.doc, .gif,...) and how would you put in an asymptote code?",
  "Solution_5": "You should preferably use some sort of Integrated $ \\text{\\LaTeX}$ Environment to format your proofs. These will typically output your $ \\text{\\LaTeX}$ into .pdf or other files by themselves and help you out with accessing the function library. Also, to use asymptote or other add-ons, you simply type in the path to the asymptote info you download from somewhere (I think, can't remember).\r\n\r\nI know I was terribly vague in explaining, but I use [url=http://kile.sourceforge.net/]Kile[/url] for my $ \\text{\\LaTeX}$ needs. It only works for the KDE environment though (Linux, Mac)."
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "Gauss"
  ],
  "Problem": "$ AA_{1}$, $ BB_{1}$, $ CC_{1}$ are altitudes of an acute triangle $ ABC$. A circle passing through $ A_{1}$ and $ B_{1}$ touches the arc $ AB$ of its circumcircle at $ C_{2}$. The points $ A_{2}$, $ B_{2}$ are defined similarly. Prove that the lines $ AA_{2}$, $ BB_{2}$, $ CC_{2}$ are concurrent.",
  "Solution_1": "Let $ BC\\cap B_{1}C_{1}=X, CA\\cap C_{1}A_{1}=Y, AB\\cap A_{1}B_{1}=Z$.\r\nIt is easy to show that $ BCB_{1}C_{1}$ is cyclic, so radical center of $ (ABC), (BCB_{1}C_{1}), (B_{1}C_{1}A_{2})$ is $ X$.\r\n$ \\Rightarrow$ $ XA_{2}$ is tangent to $ (ABC)$ at $ A_{2}$.\r\nWTS : $ \\frac{BA_{2}}{A_{2}C}\\frac{CB_{2}}{B_{2}A}\\frac{AC_{2}}{C_{2}B}=1$.\r\nBut $ \\frac{BA_{2}}{A_{2}C}=\\frac{XA_{2}}{XC}=\\frac{\\sqrt{XB\\cdot XC}}{XC}=\\sqrt{\\frac{XB}{XC}}$,\r\n$ \\frac{BX}{XC}\\frac{CB_{1}}{B_{1}A}\\frac{AC_{1}}{C_{1}B}=1 \\Rightarrow \\frac{XB}{XC}=\\frac{B_{1}A}{CB_{1}}\\frac{C_{1}B}{AC_{1}}$.\r\n$ \\Rightarrow \\frac{BA_{2}}{A_{2}C}\\frac{CB_{2}}{B_{2}A}\\frac{AC_{2}}{C_{2}B}=\\frac{AC_{1}}{C_{1}B}\\frac{BA_{1}}{A_{1}C}\\frac{CB_{1}}{B_{1}A}=1$.\r\n\r\nIs it right?\r\n\r\nBTW, Problems in this olympiad are very interesting for me. Is there any official homepage of it?",
  "Solution_2": "It can be also solved with the following nice lemma:\r\nIn the triangle $ ABC$ let $ D, E, F$ be the points of tangency of the incircle with $ BC, CA, AB$. Let $ X, Y, Z$ be the points on incircle. Then $ XD, YE, ZF$ are concurrent iff $ AX, BY, CZ$ are concurrent.\r\nProof is an application of a trigonometric Ceva, I can post it in case anybody is interested, it was probably on the forum already.\r\n\r\nLet $ l, k, m$ be the tangents at $ A_{2}, B_{2}, C_{2}$ and we denote by $ P, Q, R$ intersection of lines $ k, m$, $ m, l$, $ l, k$ . As Little Gauss has already mentioned  line $ l$, line $ B_{1}C_{1}$ and line $ BC$ concurr at one point and similary for $ B_{2}$, $ C_{2}$. Since $ AA_{1}, BB_{1}, CC_{1}$ concurr in one point from the Desaurges theorem the intersections $ B_{1}C_{1}\\cap BC$, $ C_{1}A_{1}\\cap CA$, $ A_{1}B_{1}\\cap AB$ are collinear (one can omit the Desaurges here and see that these intersections lie on radical axis of circumcircle of $ ABC$ and circumcircle of $ A_{1}B_{1}C_{1}$). Once more from the Desaurges theorem we conclude that lines $ AP, BQ, CR$ concurr in one point. But note that the circumcircle of $ ABC$ is an incircle of $ PQR$ so from the lemma we are done.",
  "Solution_3": "Actually, this lemma is called [b]Steinbart's Theorem[/b]  :) !",
  "Solution_4": "Nice!\n\nLet $A_0B_0C_0$ be a triangle whose intouch triangle is $ABC$. Let $T_C=A_1B_1 \\cap AB$ and define other analogues of it. Clearly, $\\ell=T_AT_BT_C$ is the tripolar of $H$ and these three points are collinear. Now, by the radical axis theorem, one can see that $T_CC_2$ is tangent to $(ABC)$. Therefore, the pole of $\\ell$ in the circumcircle $(O)$ of $ABC$ is the intersection point of the three lines $C_0C_2,B_0B_2,A_0A_2$. We see that by the Steinbart theorem, the lines $AA_2,BB_2,CC_2$ are concurrent. The result holds.",
  "Solution_5": "[quote=TomciO]It can be also solved with the following nice lemma:\nIn the triangle $ ABC$ let $ D, E, F$ be the points of tangency of the incircle with $ BC, CA, AB$. Let $ X, Y, Z$ be the points on incircle. Then $ XD, YE, ZF$ are concurrent iff $ AX, BY, CZ$ are concurrent.[/quote]\nShouldn't this lemma have some more assumptions,? Let's take three cevians concurrent in $P$ inside the incircle and not lying on any of $AD$, $BE$, $CF$. Let $AP$ intersect incircle in $X$ and $X'$. Let one of intersection points of $BP$ with incircle be $Y$ and of $CP$ be $Z$. Now triples $AX $, $BY $, $CZ $ and $AX' $, $BY $, $CZ $ can't be both concurrent.",
  "Solution_6": "[quote=Little Gauss]Let $ BC\\cap B_{1}C_{1}=X, CA\\cap C_{1}A_{1}=Y, AB\\cap A_{1}B_{1}=Z$.\nIt is easy to show that $ BCB_{1}C_{1}$ is cyclic, so radical center of $ (ABC), (BCB_{1}C_{1}), (B_{1}C_{1}A_{2})$ is $ X$.\n$ \\Rightarrow$ $ XA_{2}$ is tangent to $ (ABC)$ at $ A_{2}$.\nWTS : $ \\frac{BA_{2}}{A_{2}C}\\frac{CB_{2}}{B_{2}A}\\frac{AC_{2}}{C_{2}B}=1$.\n[/quote]\n\nHello, how does  $\\frac{BA_{2}}{A_{2}C}\\frac{CB_{2}}{B_{2}A}\\frac{AC_{2}}{C_{2}B}=1$ prove the concurrency? Isn't Ceva supposed to be on the sides?",
  "Solution_7": "Dear Mathlinkers,\n\nfor the Steinbart's theorem, you can see and more on\n\nhttp://jl.ayme.pagesperso-orange.fr/Docs/Les%20points%20de%20Steinbart%20et%20de%20Rabinowitz.pdf\n\nSincerely\nJean-Louis\n\n",
  "Solution_8": "[quote=jayme]Dear Mathlinkers,\n\nfor the Steinbart's theorem, you can see and more on\n\nhttp://jl.ayme.pagesperso-orange.fr/Docs/Les%20points%20de%20Steinbart%20et%20de%20Rabinowitz.pdf\n\nSincerely\nJean-Louis[/quote]\n\ndo you have this article in english version???",
  "Solution_9": "[quote=pohoatza]$ AA_{1}$, $ BB_{1}$, $ CC_{1}$ are altitudes of an acute triangle $ ABC$. A circle passing through $ A_{1}$ and $ B_{1}$ touches the arc $ AB$ of its circumcircle at $ C_{2}$. The points $ A_{2}$, $ B_{2}$ are defined similarly. Prove that the lines $ AA_{2}$, $ BB_{2}$, $ CC_{2}$ are concurrent.[/quote]\n\n[hide=Solution]\n[b]Lemma[/b] Let $A_1,B_1,C_1$ be points on $\\odot{ABC}$ of a triangle $ABC$.Then $AA_1,BB_1,CC_1$ concur iff $\\left(\\dfrac{BA_1}{A_1C}.\\dfrac{CB_1}{B_1A}.\\dfrac{AC_1}{C_1B}\\right)=1$\n\n[i]Proof[/i]   Follows from Trig Ceva.$\\square$\n               --------------------------------------------\nLet $T_C=A_1B_1\\cap AB$ and $T_A,T_B$ defined similarly.Note that $TC_2$ is tangent to $\\odot{ABC}$ via radical axis theorm on $\\odot{ABC},\\odot{ABA_1B_1},\\odot{A_1B_1C_2} \\implies T_CAC_2\\sim T_CC_2B \\Rightarrow \\dfrac{C_2A}{C_2B}=\\dfrac{T_CA}{T_CC_2}=\\left(\\sqrt{\\dfrac{T_CA}{T_CB}}\\right)$.Thus since $T_AT_BT_C$ is the orthic axis so $T_A,T_B,T_C$ collinear hence by our lemma and Menelaus we are done.$\\blacksquare$\n\n[/hide]",
  "Solution_10": "[quote=Little Gauss]Let $ BC\\cap B_{1}C_{1}=X, CA\\cap C_{1}A_{1}=Y, AB\\cap A_{1}B_{1}=Z$.\nIt is easy to show that $ BCB_{1}C_{1}$ is cyclic, so radical center of $ (ABC), (BCB_{1}C_{1}), (B_{1}C_{1}A_{2})$ is $ X$.\n$ \\Rightarrow$ $ XA_{2}$ is tangent to $ (ABC)$ at $ A_{2}$.\nWTS : $ \\frac{BA_{2}}{A_{2}C}\\frac{CB_{2}}{B_{2}A}\\frac{AC_{2}}{C_{2}B}=1$.\nBut $ \\frac{BA_{2}}{A_{2}C}=\\frac{XA_{2}}{XC}=\\frac{\\sqrt{XB\\cdot XC}}{XC}=\\sqrt{\\frac{XB}{XC}}$,\n$ \\frac{BX}{XC}\\frac{CB_{1}}{B_{1}A}\\frac{AC_{1}}{C_{1}B}=1 \\Rightarrow \\frac{XB}{XC}=\\frac{B_{1}A}{CB_{1}}\\frac{C_{1}B}{AC_{1}}$.\n$ \\Rightarrow \\frac{BA_{2}}{A_{2}C}\\frac{CB_{2}}{B_{2}A}\\frac{AC_{2}}{C_{2}B}=\\frac{AC_{1}}{C_{1}B}\\frac{BA_{1}}{A_{1}C}\\frac{CB_{1}}{B_{1}A}=1$.\n\nIs it right?\n\nBTW, Problems in this olympiad are very interesting for me. Is there any official homepage of it?[/quote]\n\nYa its correct"
}
{
  "Tag": [
    "number theory proposed",
    "number theory"
  ],
  "Problem": "Let $ A \\equal{} \\{1;2;...;25\\}$. Prove that: for all $ B\\subset A$ in which $ B$ have 17 elements always exit $ a\\neq b$ in $ B$ satisfy $ ab$ is a square number.",
  "Solution_1": "Obviously $ B\\equal{}\\{2,3\\}$ is a counterexample.",
  "Solution_2": "Sorry zetaX! I have fixed this problem!",
  "Solution_3": "We can except $ \\{7,14,21,11,22,13,17,19,23\\}$ (9 terms) and have at least 8 terms in B form $ 2^k3^m5^n$.\r\nB had not terms $ 2*3*5\\equal{}30, (k,m,n)\\not \\equal{}(1,1,1)\\mod 2$, therefore 8 terms in these form had 2, suth that $ k_1\\equal{}k_2\\mod 2, m_1\\equal{}m_2\\mod 2, n_1\\equal{}n_2\\mod 2$.",
  "Solution_4": "OK Rust! Your solution is better than me!"
}
{
  "Tag": [
    "AMC"
  ],
  "Problem": "Is there a file with pictures for this test? I found a pdf by anirudh on the wiki for 2006 10A. I was wondering about if the 10B has another file with pictures, not just words saying external image.",
  "Solution_1": "As I have said many times before, this is in violation of the AMC's copyright.  We have removed the link from the wiki.\r\n\r\nHowever, since the file is not hosted on our site, there's not much we can do about it."
}
{
  "Tag": [],
  "Problem": "$ x,y,z > 0$ . Prove that: $ \\sum \\frac {\\sqrt {y}}{\\sqrt {x \\plus{} y}} \\le \\frac {3}{\\sqrt {2}}$",
  "Solution_1": "[hide=\"Hint\"]$ \\sum_{\\mathbf{cyc}} \\frac {1}{\\sqrt {1 \\plus{} \\frac{x}{y}}}$\n\nReplace with $ \\frac{x}{y} \\equal{} a$, $ \\frac{y}{z} \\equal{} b$, $ \\frac{z}{x} \\equal{} c$, $ abc\\equal{}1$\n\nThen our sum becomes $ \\frac{1}{\\sqrt{1\\plus{}a}} \\plus{} \\frac{1}{\\sqrt{1\\plus{}b}} \\plus{} \\frac{1}{\\sqrt{1\\plus{}c}}$\n\nDefine $ P_{[n]} [ \\ldots]$ be the nth power mean of the contained list. Define $ L\\equal{} (1\\plus{}a,1\\plus{}b,1\\plus{}c)$, and let $ S$ be our sum. We can write:\n\n$ S \\equal{} \\frac{3}{\\sqrt{P_{\\left[ \\minus{}\\frac 12 \\right]}[L]}}$\n\nPower Mean kills it[/hide]"
}
{
  "Tag": [
    "geometry",
    "conics",
    "parabola",
    "analytic geometry",
    "graphing lines",
    "slope"
  ],
  "Problem": "$F$ is the focus of a parabola $y=ax^2$. A line through $F$ meets the parabola at $P$ and $Q$. If $PF=p$, $QF=q$, then find the value of $\\frac{1}{p}+\\frac{1}{q}$.",
  "Solution_1": "[hide]The focus must be at $(0, \\frac{1}{4a})$.  Given that that value is constant, I know it's $4a$, but here's a proof.\n\nSuppose a line of slope $m$ passes through the focus.  Then, it has equaiton $y=mx+\\frac{1}{4a}$  At the intersections, we also have $y=ax^2$.  Hence, \\[ ax^2=mx+\\frac{1}{4a} \\implies ax^2-mx-\\frac{1}{4a}=0 \\] \\[ \\implies x=\\frac{m \\pm \\sqrt{m^2+1}}{2a} \\implies ax^2=\\frac{2m^2+1 \\pm 2m\\sqrt{m^2+1}}{4a} \\]  The distance from a point on the parabola to the focus is the same as the distance to the directrix.  The directrix has equation $y=-\\frac{1}{4a}$.  Thus the distance is just the y-coordinate plus $\\frac{1}{4a}$.  So the two distances to the directrix are then \\[ \\frac{2m^2+2 \\pm 2m\\sqrt{m^2+1}}{4a}=\\frac{m^2+1\\pm\\sqrt{m^2+1}}{2a} \\]  We know $\\frac{1}{p}+\\frac{1}{q}=\\frac{p+q}{pq}$.  Thus, \\[ \\frac{p+q}{pq}=\\frac{\\frac{2m^2+2}{2a}}{\\frac{(m^2+1)^2-(m^2(m^2+1))}{4a^2}}=\\frac{m^2+1}{a}\\frac{4a^2}{m^2+1}=\\boxed{4a} \\] [/hide]"
}
{
  "Tag": [
    "function",
    "algebra",
    "domain",
    "inequalities",
    "calculus",
    "calculus computations"
  ],
  "Problem": "Given a function $ f: I \\to R$ defined on an interval I, let $ f*(t) \\equal{} \\mathop {\\sup }\\limits_{x \\in I} (tx \\minus{} f(x))$.\r\n\r\nShow that the set $ I*$ of $ t \\in R$ for which $ f*(t) \\ne \\infty$ is either empty or consists of a single point, or is an interval of the line, and in this last case the function $ f*(t)$ is convex on $ I*$",
  "Solution_1": "can we get some continuity of f?\r\n\r\nif $ f(x_n) \\to \\minus{}\\infty$ for some $ (x_n)$, then $ f * (t) \\equal{} \\sup_{x \\in I} tx\\minus{}f(x) \\equal{} \\infty \\forall t \\to I*$ empty\r\n\r\ntherefore if I* is not empty, we must have at least $ \\inf_{x\\in I} f(x) > \\minus{} \\infty$\r\n\r\nlooking at the statement i feel like it might not be true?",
  "Solution_2": "The statement of the problem doesn't say anything about f being continuous (it's worded rather vaguely, actually).\r\n\r\nSo what of the case where f is bounded below? We can give an upper bound for f*(t) in  this case - provided I is bounded, which I assume. When is I* degenerate? Whether f is bounded above or not seems to be irrelevant.\r\n\r\nThere seems to be something [url=http://www.cc.gatech.edu/~lebanon/notes/convexFunctions.pdf]here[/url]under the name 'Fenchel conjugate' and also [url=http://books.google.com/books?id=byF4Xb1QbvMC&pg=PA106&lpg=PA106&dq=convex+conjugate+supremum&source=web&ots=zHpFHrPVem&sig=u274cKPo50ZMpXaU--J_eLL98mc&hl=en&sa=X&oi=book_result&resnum=10&ct=result#PPA107,M1]here[/url].\r\n\r\nSo linear transformations in R are an example of functions with Legendre transforms defined on a degenerate set - PROVIDED we take 'an interval' to mean R. It appears these transforms are usually defined on R (with f taking on the value infinity outside its domain) so let's work with that understanding. I wish there was an expert on this though.",
  "Solution_3": "okay i guess, assume now that it is not empty and is not a single point\r\n\r\nthen $ \\sup_{x \\in I} (t_1x \\minus{} f(x)) \\equal{}f*(t_1) < \\infty, \\sup_{x \\in I} (t_2x \\minus{} f(x)) \\equal{}f*(t_2)< \\infty, t_1<t_2$\r\n\r\n$ t_1<t<t_2 \\to t_1x \\minus{} f(x)< tx \\minus{} f(x) < t_1x \\minus{} f(x)$\r\n\r\n$ \\to \\sup_{x \\in I\\cap R^\\plus{}} \\left(t_1x \\minus{} f(x)\\right)\\le \\sup_{x \\in I\\cap R^\\plus{}} \\left(tx \\minus{} f(x)\\right) \\le\\sup_{x \\in I\\cap R^\\plus{}} \\left(t_2x \\minus{} f(x)\\right)$\r\n\r\nnot sure how to control the negative x that are in I though",
  "Solution_4": "You are trying to show connectedness, I presume. For -x, wouldn't you reverse the inequalities? (Can we say something about whether I* is connected or closed).\r\n\r\nI edited my post while you were writing to suggest a change in how we define f*.",
  "Solution_5": "$ \\sup_{x \\in I\\cap R^\\minus{}} \\left(t_0x \\minus{} f(x)\\right)\\ge \\sup_{x \\in I\\cap R^\\minus{}} \\left(t_1x \\minus{} f(x)\\right) \\ge\\sup_{x \\in I\\cap R^\\minus{}} \\left(t_2x \\minus{} f(x)\\right)$\r\n\r\ndefining $ f_i^\\minus{} \\equal{}\\sup_{x \\in I\\cap R^\\minus{}}\\left(t_ix \\minus{} f(x)\\right),f_i^\\plus{} \\equal{}\\sup_{x \\in I\\cap R^\\plus{}}\\left(t_ix \\minus{} f(x)\\right)$\r\n\r\n$ f*(t_1)\\equal{}\\max (f_1^\\minus{},f_1^\\plus{}) \\le \\max (f_0^\\minus{},f_2^\\plus{}) < \\infty$\r\n\r\nwhich makes $ f*(t)$ an interval. so at least the first claim is done.\r\n\r\nshowing convexity requires something from what you said above i thinks"
}
{
  "Tag": [
    "function",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Let $f: [0,1]\\to R$ be continuous over $[0,1]$ and $f(0)=f(1)=0$. Prove that exist convex continuous function $g: [0,1]\\to R$ such that $g(0)=g(1)=0$ and $g(x)\\geq f(x)$ for all $x\\in [0,1]$.",
  "Solution_1": "Shouldn't that be $g(x)\\le f(x)?$\r\n\r\nHere's a construction:\r\n\r\nFor each $m\\in\\mathbb{R},$ let $h(m)=\\min_{x\\in[0,1]}(f(x)-mx).$\r\n\r\nNow define $g(x)=\\sup_{m\\in\\mathbb{R}}(mx+h(m)).$\r\n\r\nSince $mx+h(m)\\le f(x)$ for each $m,$ it follows that $g(x)\\le f(x).$\r\n\r\nNow pick two points $x$ and $y$ and some $t\\in(0,1).$\r\n\r\nFor each $m,$\r\n\r\n$(1-t)(mx+h(m))+t(my+h(m))=m[(1-t)x+ty]+h(m)$\r\n\r\nNow take the sup over $m$ of both sides. The sup of the right hand side is $g[(1-t)x+ty],$ by definition. On the left hand side.\r\n\r\n$\\sup_{m}[(1-t)(mx+h(m))+t(my+h(m))]$\r\n\r\n$\\le \\sup_{m}(1-t)(m(x)+h(m))+\\sup_{m}t(m(y)+h(m))$\r\n\r\n$=(1-t)\\sup_{m}(m(x)+h(m))+t\\sup_{m}(m(y)+h(m))=(1-t)g(x)+tg(y).$\r\n\r\nHence $g[(1-t)x+ty]\\le(1-t)g(x)+tg(y)$ and $g$ is convex.",
  "Solution_2": "Wow :)  a realy nice problem and a beautiful solution Kent Merryfield\r\nIs the original statement also true?",
  "Solution_3": "[quote=\"ciprian\"]Is the original statement also true?[/quote]\r\nOf course not. Just let $f$ be some function with $f(c)>0$ for some $0<c<1.$ To have $g(x)\\ge f(x),$ we'd have to have $g(0)=0, g(c)>0,g(1)=0,$ and that violates convexity.",
  "Solution_4": "Of course g must be <=0 in [0,1]:) Sorry for my nonsense :(\r\nBut concave? The same argument works? just switching min with max and sup with inf?"
}
{
  "Tag": [],
  "Problem": "1)Explain which compound is the weaker base:\r\n\r\n[hide]Since the lone pair of the first is in conjugation so I suppose it should be the weaker base but if we consider acid/base wise , then the second one is more acidic hence less basic.[/hide]\n\n2) Rank the following amines in the increasing basic nature;\n[hide]\nDoubtful about para and ortho isomers. Also I want to know what will happen if the substituent groups attached are Cl and NO2 respectively instead of CH3[/hide]",
  "Solution_1": "1. The strongest acid is oxalic acid, and so it is the weakest base. Remember, in dicarboxylic acids the second proton is always more relutant to leave the molecule.\r\n\r\n2. The strongest base is cyclohexylmethylamine: there is no ressonance possiblity for the nitrogen lone pair.\r\n\r\nWith an electron withdrawing group like -NO2 in the ring, basicity will decrease, since the nitrogen lone pair will be more conjugated with the ring.",
  "Solution_2": "Ans 1 -edit , overlooked carcul's post :D\r\n\r\nAns -2 \r\nis it ->\r\n4>3>2>1\r\n\r\n4 - no resonance\r\n3 - +M, +I\r\n2 - +I\r\n1 - Ortho Effect"
}
{
  "Tag": [
    "Divisibility Theory"
  ],
  "Problem": "Prove that $2n \\choose n$ is divisible by $n+1$.",
  "Solution_1": "The identity $(n+1) \\binom {2n}{n-1} = n \\binom {2n}n$ shows that $(n+1)|n \\binom {2n}n$ and, since $\\gcd(n, n+1) = 1, (n+1)|\\binom {2n}n.$",
  "Solution_2": "We know that $\\frac 1{n+1} \\binom{2n}n$ is the number of triangularisations of a $n$-gon (the $n$-th Catalan number).\r\nThis number is trivially an integer.",
  "Solution_3": "[quote=\"Peter\"]Prove that $ 2n\\choose n$ is divisible by $ n\\plus{}1$.[/quote]\r\n\r\nOne may slightly generize the argument of mathmanman.\r\n\r\nProposition. Let $ N$ and $ k$ are positive integers with $ N\\geq k$ and $ \\gcd(N\\plus{}1, k\\plus{}1)\\equal{}1$. Then, $ N\\choose k$ is divisible by $ k\\plus{}1$.\r\n\r\nProof. It is well-known that\r\n\\[ (N\\plus{}1){N\\choose k}\\equal{} (k\\plus{}1){{N\\plus{}1}\\choose{k\\plus{}1}}.\\]\r\nHence, $ (N\\plus{}1){N\\choose k}$ is divisible by $ k\\plus{}1$. Since  $ \\gcd(N\\plus{}1, k\\plus{}1)\\equal{}1$, this implies that $ N\\choose k$ is divisible by $ k\\plus{}1$.",
  "Solution_4": "As you know,$C_n=\\frac{1}{n+1}\\binom{2n}{n}$ is called as [b]Catalan number[/b].\n\n$\\binom{2n}{n}-\\binom{2n}{n-1}=\\frac{(2n)!}{n!n!}-\\frac{(2n)!}{(n-1)!(n+1)!}$\n$=\\frac{(2n)!}{n!(n+1)!}=\\frac{1}{n+1}\\binom{2n}{n}$ is positive integer.The proof is completed.$\\blacksquare$ :coolspeak:"
}
{
  "Tag": [
    "MATHCOUNTS",
    "geometry",
    "Euler",
    "AMC",
    "AIME",
    "probability",
    "algebra"
  ],
  "Problem": "Does anyone have a list from easiest to hardest for chapters, states, and nats?\r\n-jorian",
  "Solution_1": "[quote=\"jhredsox\"]Does anyone have a list from easiest to hardest for chapters, states, and nats?\n-jorian[/quote]\r\nThe chapters from hardest to easiest. would be in order 04, 05, 03, 02, 06, 01 in my opinion.\r\nFor states it gets a little more complicated. Looking through my old ones I have come up with this order from hardest to easiest:\r\n06, 05(killer target) ,99, 04, 01, 03, 02, 96, 00, 98, 93, 94, 97.\r\nFor nationals, I think this is pretty accurate:\r\n05 (killer overall), 06, 04, 03, 01, 02, 96(last ten sprints), 97, 00, 98, 99 (easier than state of this year), 95, 94, 93, 91, 92, 90,",
  "Solution_2": "What about 85???",
  "Solution_3": "[quote=\"bpms\"][quote=\"jhredsox\"]Does anyone have a list from easiest to hardest for chapters, states, and nats?\n-jorian[/quote]\nThe chapters from hardest to easiest. would be in order 04, 05, 03, 02, 06, 01 in my opinion.\nFor states it gets a little more complicated. Looking through my old ones I have come up with this order from hardest to easiest:\n06, 05(killer target) ,99, 04, 01, 03, 02, 96, 00, 98, 93, 94, 97.\nFor nationals, I think this is pretty accurate:\n05 (killer overall), 06, 04, 03, 01, 02, 96(last ten sprints), 97, 00, 98, 99 (easier than state of this year), 95, 94, 93, 91, 92, 90,[/quote]\r\n\r\n'06 States was TONS easier than '05.\r\n\r\n'99 was a lot harder than most of the nationals. I got a 17 on it while I got much higher on 05.",
  "Solution_4": "[quote=\"Ignite168\"][quote=\"bpms\"][quote=\"jhredsox\"]Does anyone have a list from easiest to hardest for chapters, states, and nats?\n-jorian[/quote]\nThe chapters from hardest to easiest. would be in order 04, 05, 03, 02, 06, 01 in my opinion.\nFor states it gets a little more complicated. Looking through my old ones I have come up with this order from hardest to easiest:\n06, 05(killer target) ,99, 04, 01, 03, 02, 96, 00, 98, 93, 94, 97.\nFor nationals, I think this is pretty accurate:\n05 (killer overall), 06, 04, 03, 01, 02, 96(last ten sprints), 97, 00, 98, 99 (easier than state of this year), 95, 94, 93, 91, 92, 90,[/quote]\n\n'06 States was TONS easier than '05.\n\n'99 was a lot harder than most of the nationals. I got a 17 on it while I got much higher on 05.[/quote]\r\nThere were some good questions on the 06 sprint, and I got a 20 on the 05 sprint when I did it officially in sixth grade so It must've been easy (though I got 3 right on the target that year)...",
  "Solution_5": "well thank you bpms for all of your hard work in this area\r\n\r\nand ignite\r\n\r\nbpms, u went to nats, right?\r\n-jorian",
  "Solution_6": "04 nats was WAY harder than 06.",
  "Solution_7": "why are all these new tests harder than the old ones?\r\n-jorian",
  "Solution_8": "[quote=\"13375P34K43V312\"]04 nats was WAY harder than 06.[/quote]\r\nThey are of comparable difficulty but IMO 06 was slightly harder.\r\nTo jhredsox, I think the caliber of the students has increased, or mathcounts was getting too high scores. Perhaps a combination...\r\nEDIT: Yes, I went to nats.\r\nEDIT 2: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=98871",
  "Solution_9": "ty for this\r\nso....they generally have been getting harder?\r\n-jorian",
  "Solution_10": "89 was the hardest year, although it [i]was[/i] in the old format :wink:\r\n\r\n90 was by far the easiest(1st one in current format)",
  "Solution_11": "[quote=\"bpms\"][quote=\"13375P34K43V312\"]04 nats was WAY harder than 06.[/quote]\nThey are of comparable difficulty but IMO 06 was slightly harder.\nTo jhredsox, I think the caliber of the students has increased, or mathcounts was getting too high scores. Perhaps a combination...\nEDIT: Yes, I went to nats.\nEDIT 2: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=98871[/quote]\r\n\r\nIMO 2006??\r\n\r\nI think 2006 state was harder than 2004 state, but the 2004 nat test was harder.\r\n\r\nThe 2001 State was...ugh...#5 and #6 on target could easily have come on nationals. I found 01 nat easier than 01 state.\r\n\r\nI think this year (2007), the countdown cutoff will be around 40, with the average top 57-er having a score of like 24/12 or so. It will be a little easier than last year's.\r\n\r\n\r\n\r\nAnyway, the 1984 nat masters round was really hard.",
  "Solution_12": "[quote=\"13375P34K43V312\"][quote=\"bpms\"][quote=\"13375P34K43V312\"]04 nats was WAY harder than 06.[/quote]\nThey are of comparable difficulty but IMO 06 was slightly harder.\nTo jhredsox, I think the caliber of the students has increased, or mathcounts was getting too high scores. Perhaps a combination...\nEDIT: Yes, I went to nats.\nEDIT 2: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=98871[/quote]\n\nIMO 2006??\n\nI think 2006 state was harder than 2004 state, but the 2004 nat test was harder.\n\nThe 2001 State was...ugh...#5 and #6 on target could easily have come on nationals. I found 01 nat easier than 01 state.\n\nI think this year (2007), the countdown cutoff will be around 40, with the average top 57-er having a score of like 24/12 or so. It will be a little easier than last year's.\n\n\n\nAnyway, the 1984 nat masters round was really hard.[/quote]\r\n\r\nIMO=in my opinion (in this instance)\r\n\r\nSomehow, I think this year will be uber hard, like harder than 2005, dunno why",
  "Solution_13": "I had that feeling too, but this one's stronger.",
  "Solution_14": "I don't see how your feelings could have any basis at all.",
  "Solution_15": "Yeah, I guess we just have to see how everything plays out.\r\n\r\nI personally am hoping for a hard test, since I modestly say that I believe that I know more \"real\" math than most mathcounts national competitors, and I might be able to use higher-level shortcuts (such as Fermat-Euler or Vandermonde) on a hard Mathcounts test.",
  "Solution_16": "[quote=\"13375P34K43V312\"]Yeah, I guess we just have to see how everything plays out.\n\nI personally am hoping for a hard test, since I modestly say that I believe that I know more \"real\" math than most mathcounts national competitors, and I might be able to use higher-level shortcuts (such as Fermat-Euler or Vandermonde) on a hard Mathcounts test.[/quote]\r\nFermat Euler=Euler's Totient?\r\nAnyays, Euler's Totient only applies to the cases where you want the last 2 or maybe 3 digits and the binomial theorem dominates for those. For example, Euler's Totient would help with an old problem find the last three digits of $9^{2004}$, but the binomial theorem dominates. As for vandermonde, it is powerful, but I don't think I have ever seen it used before in mathcounts.",
  "Solution_17": "[quote=\"bpms\"]Looking through my old ones I have come up with this order from hardest to easiest:\n06, 05(killer target) ,99, 04, 01, 03, 02, 96, 00, 98, 93, 94, 97.\n[/quote]\r\n\r\nYou know this is just 93-2006 -{95} in skewed order.",
  "Solution_18": "[quote=\"Phelpedo\"][quote=\"bpms\"]Looking through my old ones I have come up with this order from hardest to easiest:\n06, 05(killer target) ,99, 04, 01, 03, 02, 96, 00, 98, 93, 94, 97.\n[/quote]\n\nYou know this is just 93-2006 -{95} in skewed order.[/quote]\r\nWell how else could it be organized :roll:. The only one I am missing is 95...\r\nEDIT: Now that I look at it, vandermonde is completely useless...",
  "Solution_19": "Yeah you're right, since vandermonde is proven using a combinatorial argument + casework, it would probably just be easier to go through the casework  :D \r\n\r\n\r\nthere have been recent last3/last2 digit problems, however.",
  "Solution_20": "[quote=\"13375P34K43V312\"]Yeah you're right, since vandermonde is proven using a combinatorial argument + casework, it would probably just be easier to go through the casework  :D \n\n\nthere have been recent last3/last2 digit problems, however.[/quote]\nDid you read what I wrote\n[quote=\"bpms\"]Fermat Euler=Euler's Totient?\nAnyays, Euler's Totient only applies to the cases where you want the last 2 or maybe 3 digits and the binomial theorem dominates for those. For example, Euler's Totient would help with an old problem find the last three digits of $9^{2004}$, but the binomial theorem dominates. As for vandermonde, it is powerful, but I don't think I have ever seen it used before in mathcounts. [/quote]",
  "Solution_21": "How would you use the binomial theorem for that problem?",
  "Solution_22": "To perfect 628, How did you do at nats last year?",
  "Solution_23": "Hard tests are WAY better in my opinion, and 07 shoulkd be one of the hardest ever. I hope it's THE hardest ever.",
  "Solution_24": "wait. why do u hope its the hardest ever?\r\n-jorian",
  "Solution_25": "[quote=\"jhredsox\"]wait. why do u hope its the hardest ever?\n-jorian[/quote]\r\nCareless mistakes have less weight and can be afforded more.",
  "Solution_26": "Yeah, with an easy test, it's just a matter of who makes the least careless mistakes but with a harder test, you can better determine who has the most knowledge.",
  "Solution_27": "thats what i was thinking in the back of my head. though, i hope there's not too much geometry on the test.\r\n\r\nwhat's the hardest geometry area mathcounts will go into?\r\n-jorian",
  "Solution_28": "[quote=\"jhredsox\"]thats what i was thinking in the back of my head. though, i hope there's not too much geometry on the test.\n\nwhat's the hardest geometry area mathcounts will go into?\n-jorian[/quote]\r\nI love geometry at the mathcounts level :). The hardest mathcounts problems aren't what you should be worried about if you want to get better at geometry. What you should be looking at is medium level AIME problems.",
  "Solution_29": "ummmmmmmmmmm.........can u tell me where i find these?\r\n-jorian",
  "Solution_30": "Contests tab at the top.",
  "Solution_31": "for me the geometry problems are much easier than other problems, especially probability, and anything to do with counting/permutation.  and those show up a heck lot on MC",
  "Solution_32": "[quote=\"star99\"]for me the geometry problems are much easier than other problems, especially probability, and anything to do with counting/permutation.  and those show up a heck lot on MC[/quote]\r\n\r\nYeah. Geometry on mathcounts, or even in general, is much easier than counting/permutation/probability.",
  "Solution_33": "bpms: Can you please stop offering advice like that? I talk to Jorian outside of AoPS and he hasn't taken Geometry yet. MathCouts Nationals problem he doesn't get most of the time and comments like that will only discourage him more.",
  "Solution_34": "@ignite:  what you just said would probably hurt him a lot more.  \r\n\r\neveryone one on aops should know that there are tons of people on the forum better than they are at math, and thats why they are here, to learn to be that good.",
  "Solution_35": "Sorry, but I was referring to the whole, \"Do problems that are really hard for you to get better\". I'd say the hardest mathcounts problem on geometry ever period would be\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=11092\r\nI have no clue how someone would get that in 3 min.\r\nOr even 6 for that matter.",
  "Solution_36": "wow, bpms, that's a real hard question\r\n\r\nand btw, the geometry portion for me is really hard compared to the probability portion and the algebra\r\n-jorian",
  "Solution_37": "I hate that question. I've seen it so many times and I still keep getting different answers for that one.",
  "Solution_38": "[quote=\"pianoforte\"]I hate that question. I've seen it so many times and I still keep getting different answers for that one.[/quote]I still don't know how one would get it in 3 min. Read Ravi B's solution it's good (not surprising seeing as he is a beast at problems :D)",
  "Solution_39": "Thanks for the compliment.  While I was preparing to coach the NY team last year, I solved that question in 3 minutes 20 seconds.  (I record my times to measure the difficulty of questions.)  It probably would have taken me a lot longer except that I had seen the question a couple of years ago when I posted that solution.",
  "Solution_40": "isn't that an inefficient way to test difficulty?\r\n-jorian",
  "Solution_41": "Maybe.  Do you know of a better method?  I like to solve the problems anyway to be well prepared to coach, so I might as well time myself doing so.",
  "Solution_42": "a good way is to do what u do now, and then record the percentage of people who got it right (don't do sprint!, they might not get to it)\r\n-jorian",
  "Solution_43": "[quote=\"Ravi B\"]Thanks for the compliment.  While I was preparing to coach the NY team last year, I solved that question in 3 minutes 20 seconds.  (I record my times to measure the difficulty of questions.)  It probably would have taken me a lot longer except that I had seen the question a couple of years ago when I posted that solution.[/quote]\r\nReally? Well, any solution under 5 minutes is okay seeing as the pair to that problem can be done in under a minute with a calculator.",
  "Solution_44": "Well you could interpret its \"difficulty\" many ways.\r\n\r\nThere could be 29 ultra easy questions, which take like 10 minutes total (i mean for sprint), then 1 uberly hard question that only like the top people at nationals can solve. \r\n\r\nThen, there could be a test with 30 questions that are all like State #15 on the sprint."
}
{
  "Tag": [
    "geometry",
    "geometric transformation",
    "reflection",
    "geometry solved"
  ],
  "Problem": "give us procedure to draw 3 cevaian line with equal size in the traingle .(with compass and ruler).",
  "Solution_1": "That's the h3ll of a problem! Actually, in [url=http://forumgeom.fau.edu/FG2003volume3/FG200307index.html]Paul Yiu, [i]On the Fermat lines[/i], Forum Geometricorum 3 (2003) p. 73-81[/url], paragraph 6, item (14), you can find a construction of the two focii of the Steiner inellipse of triangle ABC. If you reflect the centroid G of triangle ABC in these two focii, you get two new points, which are the focii of the Steiner circumellipse of triangle ABC. These two new focii have the property that the cevians of each of these focii have equal length. And these points are the only points with this property.\r\n\r\nPlease don't ask me about the proof of this construction. See also [url=http://groups.yahoo.com/group/Hyacinthos/messages/211?viscount=100&expand=1]the Hyacinthos messages #211, #212, #215, #217, #222, etc.[/url].\r\n\r\n  Darij"
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "a,b,c>0    a+b+c=1\r\n\r\n1/(ab+c)+1/(bc+a)+1/(ca+b) >= 27/4",
  "Solution_1": "[quote=\"Maverick\"]If a, b, c are positive numbers with a + b + c = 1, then prove\n\n[tex]\\frac{1}{bc+a}+\\frac{1}{ca+b}+\\frac{1}{ab+c}\\geq \\frac{27}{4}[/tex].[/quote]\r\n\r\nIndeed easy, but very nice (I have TeXed the formula in order to better appreciate its beauty  :D ).\r\n\r\n[tex]\\frac{1}{bc+a}+\\frac{1}{ca+b}+\\frac{1}{ab+c}=\\frac{1}{bc+a\\left( a+b+c\\right) }+\\frac{1}{ca+b\\left( a+b+c\\right) }+\\frac{1}{ab+c\\left( a+b+c\\right) }[/tex]\r\n[tex]=\\frac{1}{\\left( bc+ca+ab\\right) +a^{2}}+\\frac{1}{\\left( bc+ca+ab\\right) +b^{2}}+\\frac{1}{\\left( bc+ca+ab\\right) +c^{2}}[/tex]\r\n[tex]\\overset{\\text{by HM-AM}}{\\geq }9\\cdot \\frac{1}{\\left( \\left( bc+ca+ab\\right) +a^{2}\\right) +\\left( \\left( bc+ca+ab\\right) +b^{2}\\right) +\\left( \\left( bc+ca+ab\\right) +c^{2}\\right) }[/tex]\r\n[tex]=9\\cdot \\frac{1}{3\\left( bc+ca+ab\\right) +\\left( a^{2}+b^{2}+c^{2}\\right) }=\\frac{27}{4}\\cdot \\frac{4}{9\\left( bc+ca+ab\\right) +3\\left( a^{2}+b^{2}+c^{2}\\right) }[/tex]\r\n[tex]\\overset{\\text{since }bc+ca+ab\\leq a^{2}+b^{2}+c^{2}}{\\geq }\\frac{27}{4}\\cdot \\frac{4}{8\\left( bc+ca+ab\\right) +4\\left( a^{2}+b^{2}+c^{2}\\right) }[/tex]\r\n[tex]=\\frac{27}{4}\\cdot \\frac{4}{4\\left( a+b+c\\right) ^{2}}=\\frac{27}{4}\\cdot \\frac{4}{4}=\\frac{27}{4}[/tex].\r\n\r\nEquality if and only if $a=b=c=\\frac13$.\r\n\r\n  Darij",
  "Solution_2": "Or alternatively, by Cauchy,\r\n\r\n$\\sum \\frac{1}{bc+a} \\geq \\frac{9}{a+b+c+ab+bc+ca} \\geq \\frac {9}{a+b+c+\\frac{(a+b+c)^2}{3}} = \\frac{27}{4}$ with equality iff $a=b=c=\\frac{1}{3}$.",
  "Solution_3": "Whoa... I made the exact same inequality a while ago   :shock: \r\n\r\nIt's posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=13175]here[/url] with even more solutions.",
  "Solution_4": "Alternatively,\r\n\r\n$\\displaystyle \\sum \\frac{1}{bc+a}  = \\sum \\frac{1}{bc+a(a+b+c)}= \\sum \\frac{1}{(a+b)(a+c)} $\r\n\r\n$\\displaystyle  = \\frac{\\sum (b+c)}{(a+b)(b+c)(c+a)} = \\frac{2}{(a+b)(b+c)(c+a)}$\r\n\r\nAnd by AM-GM, $\\displaystyle {(a+b)(b+c)(c+a)} \\leq \\left(\\frac{(a+b)+(b+c)+(c+a)}{3}\\right)^3 = \\left(\\frac{2}{3}\\right)^3 = \\frac{8}{27}$",
  "Solution_5": "Yet another solution,\r\n\r\n[tex] \\displaystyle \\sum \\frac{1}{bc+a} = \\sum \\frac{1}{bc+a(a+b+c)} = \\sum \\frac{1}{(a+b)(a+c)} [/tex]\r\n\r\nFrom Holder,\r\n\r\n[tex] \\displaystyle \\left( \\sum \\frac{1}{(a+b)(a+c)} \\right)^{1/3} \\left( \\sum (a+b) \\right)^{1/3} \\left( \\sum (a+c) \\right)^{1/3} \\geq (1+1+1) [/tex],\r\n\r\nand since [tex]\\displaystyle \\sum (a+b) = \\sum (a+c) = 2[/tex], this is equivalent to the original inequality.",
  "Solution_6": "bc+a=bc+1-b-c=(1-b)(1-c)\r\n\r\n\r\nso \r\nthe sum = 2/(1-b)(1-c)(1-a)>=2/(3-a-b-c/3)^3=27/4"
}
{
  "Tag": [
    "\\/closed"
  ],
  "Problem": "There seems to be NO link to the Problem of the day from the Index page. Even on the resources page, there is no link to it. However, there is one on the Mathlinks Homepage.  Please see to this.",
  "Solution_1": "It is an experimental feature so for now the links are only presented there where they are now.",
  "Solution_2": "OK. Dont forget to add a prominent one when you decide to (if) make it permanent!  :lol:"
}
{
  "Tag": [
    "ratio"
  ],
  "Problem": "Triangles $ABC$ and $XYZ$ are similar, with $A$ corresponding to $X$ and $B$ to $Y$. If $AB=3$, $BC=4$, and $XY=5$, then $YZ$ is:\n\n$ \\textbf{(A)}\\ 3\\frac 3 4 \\qquad \\textbf{(B)}\\ 6 \\qquad \\textbf{(C)}\\ 6\\frac 1 4 \\qquad \\textbf{(D)}\\ 6\\frac 2 3 \\qquad \\textbf{(E)}\\ 8$",
  "Solution_1": "[hide]Becuase ABC and XYZ are similar, their respective sides must be proportional.  Therefore,\n\n$\\frac{AB}{XY} = \\frac{BC}{YZ}$\n$\\frac 35 = \\frac4{YZ}$\n$3(YZ) = 20$\n$YZ = \\frac{20}3$ = D[/hide]",
  "Solution_2": "answer\r\n\r\n--[hide] (D) [i]yz/bc=xy/ab[/i]\n[/hide]\r\n\r\n\r\n\r\n\r\n[/i]",
  "Solution_3": "[hide=\"we just learned this today...\"]\ncreate a proportion:\nab/xy=bc/yz\nsubstitute in the values:\n3/5=4/yz\ncross multiply:\n3(yz)=20\nyz=6 2/3\nor \nD\n[/hide]",
  "Solution_4": "[hide]\n$\\frac{AB}{XY}=\\frac{BC}{YZ}$\nPlug in the values:\n$\\frac{3}{5}=\\frac{4}{YZ}$\n$3(YZ)=20$\nSo $YZ=6\\frac{2}{3}$\n$\\boxed{\\boxed{D}}$[/hide]",
  "Solution_5": "[hide]\n$\\frac{3}{5}=\\frac{4}{YZ}$\n$3(YZ)=20$\n$YZ=\\frac{20}{3}=D$[/hide]",
  "Solution_6": "[hide]\nThe anwer is D. A corresponds to X and B corresponds to Y. therefore AB corresponds to XY. If AB=3, and XY=5, then we know the ratio between the two are 3:5. \n\ntherefore\n\n4 x 5/3 = 20/3 = 6+2/3\n[/hide]",
  "Solution_7": "[hide]i got $D$[/hide]",
  "Solution_8": "I got D, too.",
  "Solution_9": "Yay! :lol: \r\nI also got D.",
  "Solution_10": "[hide]As the length of AB corresponds to the length of XY, we find that the ratio of sides in triangle ABC to triangle XYZ is 1:5/3. BC corresponds to YZ, so 4*5/3=20/3, or [b]D)6 2/3[/b].[/hide]\r\n\r\nThat was an AMC 12 problem?",
  "Solution_11": "3/5=4/X\r\n\r\n3/5X=4\r\n\r\nX=4/(3/5)\r\n\r\nX=4*5/3\r\n\r\nX=20/3. (D).",
  "Solution_12": "[quote=\"easyas3.14159...\"]That was an AMC 12 problem?[/quote]\r\n\r\nWell it's number 2, and the first 5 are usually a piece of cake (usually...). You'd be surprised how many people get these types of problems wrong.",
  "Solution_13": "[quote=\"4everwise\"]Triangles $ABC$ and $XYZ$ are similar, with $A$ corresponding to $X$ and $B$ to $Y$. If $AB=3$, $BC=4$, and $XY=5$, then $YZ$ is:\n\n\\[ \\text{(A)}\\ 3\\frac 3 4 \\qquad \\text{(B)}\\ 6 \\qquad \\text{(C)}\\ 6\\frac 1 4 \\qquad \\text{(D)}\\ 6\\frac 2 3 \\qquad \\text{(E)}\\ 8  \\][/quote]\r\n\r\n[hide]without drawing the pic, you can tell that ab and xy correspond\n3/5=4/x \n20=3x\nx=20/3= 6 2/3\n[/hide]"
}
{
  "Tag": [
    "trigonometry",
    "analytic geometry",
    "ratio",
    "quadratics",
    "email",
    "geometry",
    "3D geometry"
  ],
  "Problem": "It seems that problem solving marathon is a good way to keep a forum active, especially for this one. So, here we go!\r\n\r\n[b]Problem One[/b]\r\n\r\nThe initial kinetic energy of the ball is 6J,  :theta: =30, Find the kinetic energy of the ball when it touches the incline again.",
  "Solution_1": "ok this problem is either VERY good, and challenging., in that case i like it.\r\nor it's very simple and you 'va forgotten to give more info about the problem--because nothing about the hight differenct is known, if it was, then the problem would be only simple case of conservation of energy\r\nplease state wether the problem is correct...or not. :)",
  "Solution_2": "assuming the problem is correct, is the answer:\r\n$6(1+\\sqrt{3})$?",
  "Solution_3": "Edit: I deleted this because my answer was completely off.",
  "Solution_4": "here is what i did:\r\nwe know that if we launch a projectile motion with an angle $\\theta$ on the horizantal surface, the time it takes the projectile to reach the ground would be:\r\n$t=\\frac{2v_0\\sin{\\theta}}{g}$. but since the are not on a horizantal surfact the effective $g$ would be $g\\sin{\\theta}$ therefor the time it takes for the ball to reach the surface again, in our problem would be:\r\n$t=\\frac{2v_0\\sin{\\theta}}{g\\sin{\\theta}}=\\frac{2v_0}{g}$\r\nif the diffrence of hieghts of the ball before and after moving is: $h$. then based on similar tirangles, the ball would move a lentgh of $2h$. (if this is confusing. let the point that ball left the ramp be $A$ and the point that came back be $B$, then $AB=2h$)\r\nalso we know this:\r\n$2h=v_0\\cos{\\theta}. t=\\frac{2v_0^2\\cos{\\theta}}{g}$\r\nnow based on conservation of energy:\r\n$\\frac{1}{2}mv_0^2+mgh=\\frac{1}{2}mv_2^2$ first thing you can do is to cancell out $m$'s, after substituting for $h$, you'll get:\r\n$\\cos{\\theta}v_0+\\frac{1}{2}v_0=\\frac{1}{2}v_2^2$\r\nwe know that $\\theta=\\frac{\\pi}{6}$\r\nso we get:\r\n$\\frac{\\sqrt{3}}{2}v_0+\\frac{1}{2}v_0=\\frac{1}{2}v_2^2$\r\ntherfore you get: $(1+\\sqrt{3})v_0^2=v_2^2$ therefore\r\n$E_{k2}=(1+\\sqrt{3})E_{k1}=6(1+\\sqrt{3})$\r\n\r\ni did all this, but i have this feeling that something is wrong ;)",
  "Solution_5": "there is a trick to solve this problem...\r\n[hide]\n14J[/hide]",
  "Solution_6": "If you define your coordinate system's x-axis to be along the inclined plane the problem becomes much easier, I'll post my full solution later.",
  "Solution_7": "[quote=\"shobber\"]there is a trick to solve this problem...\n[hide]\n14J[/hide][/quote]\r\ncould you explain. ? ;)",
  "Solution_8": "[quote=\"amirhtlusa\"][quote=\"shobber\"]there is a trick to solve this problem...\n[hide]\n14J[/hide][/quote]\ncould you explain. ? ;)[/quote]\r\n$\\tan{\\alpha}=2\\tan{\\beta}$",
  "Solution_9": ":huh: \r\nok, can you explain, so we can move on...? :|\r\nsince there is nothing given about the hight difference, first thing you have to do is to find the height diffrence based on the given values...\r\nthen you have to use the conservation of energy to solve the problem.\r\nhow the heck, did you get 14???? :ninja:",
  "Solution_10": "There is no need for any dirty tricks, I think. I did exatcly what amirhtlusa described:\r\n\r\n[quote]since there is nothing given about the hight difference, first thing you have to do is to find the height diffrence based on the given values...\nthen you have to use the conservation of energy to solve the problem. [/quote]\r\nand I got 14 Joules:\r\n\r\n[hide=\"step by step\"]\nRegular xy coordinate system, but y is positive as we go downwards. Denote v the original velocity of the ball. Then:\n\nx = v.t\ny = (1/2).g.t^2\n\nBall hits the incline when (at time t):\n\ny/x = tan 30\nsqrt(3).y = x\n(sqrt(3)/2).g.t^2 = v.t\nt = (v/g).(2/sqrt(3))\n\nThat makes height difference h = y(t) - y(0)\n\nh = (1/2).g.t^2 = (1/2).g.((v/g).(2/sqrt(3)))^2\nh = (2/3).(v^2)/g\n\nWhich makes potential energy decreacse\n\ndeltaE = m.g.h\ndeltaE = (2/3).m.(v^2) = 4/3 * ((1/2).m.v^2)\n\nBut the term ((1/2).m.v^2) is the original kinetic energy of the ball! That implies kinecit energy increase of (4/3) * 6 J = 8 J, total kinetic energy is therefore 6 + 8 = 14 Joules.\n\n(sorry for not using the TEX typesetting)\n[/hide]",
  "Solution_11": "For a body doing a projectile motion, there is always:\r\n\\[\\tan{\\alpha}=2\\tan{\\beta}\\]",
  "Solution_12": "[quote=\"kubus\"]There is no need for any dirty tricks, I think. I did exatcly what amirhtlusa described:\n\n[quote]since there is nothing given about the hight difference, first thing you have to do is to find the height diffrence based on the given values...\nthen you have to use the conservation of energy to solve the problem. [/quote]\nand I got 14 Joules:\n\n[hide=\"step by step\"]\nRegular xy coordinate system, but y is positive as we go downwards. Denote v the original velocity of the ball. Then:\n\n[b]x = v.t\ny = (1/2).g.t^2 [/b]\n\nBall hits the incline when (at time t):\n\ny/x = tan 30\nsqrt(3).y = x\n(sqrt(3)/2).g.t^2 = v.t\nt = (v/g).(2/sqrt(3))\n\n\n[/hide][/quote]\r\nthat is not true...",
  "Solution_13": "What is not true?\r\n\r\nIf you mean the x and y coordinate calculations - they are independent, there is no acceleration in x, therefore x = v.t. There is no initial velocity in y direction, but acceleration of magnitude g downwards (i.e. towards positive y), that means velocity vy = g.t and position (vertical position) y = (1/2).g.(t^2) = g.t.t/2.\r\n\r\nOr is it something else which is not clear?",
  "Solution_14": "shouldn't be:\r\n$x=v\\cos{30}t$\r\n$y=\\frac{1}{2}g\\sin{30}t^2$\r\nactually it depends on the coordinates, \r\nwhat was your coordinate system??",
  "Solution_15": "[hide=\"my solution\"]\nThe first ball (M) bounces up with velocity v. Looking from the reference frame of M, m is approaching with speed 2v. Since it is much lighter, it just bounces back with velocity 2v. That is 3v seen from the ground. Now m sees mu approaching with speed of 4v. It also bounces back with 4v, that is 3v+4v = 7v seen from the ground. 7 times larger velocity means 49 times larger kinetic energy = 49 times largegr potential energy at the top of the trajectory = maximum height of 49H.\n\nThis is the same as when you have a ball bounce from the earth. Why don't you always calculate the energy and momentum conservation laws for the ball and earth but just assume the ball bounces back with exactly the same velocity? In fact, the earth DOES get an impulse, but since its mass is so large, the velocity is too small and it takes up almost no energy (which depends on v squared).\n[/hide]\r\n\r\nPS. I'm not going to be online for about two weeks, have fun here ... I'll leave you another one:\r\n\r\nThere are 3 dogs (A, B, C) standing in vertices of an equilateral triangle with sides of length $a$. In one moment, they all start to chase each other: A runs exactly in the direction towards B, B towards C and C towards A. They all have constant magnitude of velocity $v$. How long does it take before they all meet?",
  "Solution_16": "wow. thats actually a really good way to do the problem",
  "Solution_17": "Won't the dogs never meet or am I missing something.",
  "Solution_18": "[quote=\"mna851\"]Won't the dogs never meet or am I missing something.[/quote]\r\nThey do meet. The problem says that [b]runs exactly in the direction towards[/b]. Which means, the direction of the velocity of every dog changes. They are not doing motion along a straight line path.",
  "Solution_19": "this way isnt very elegant.. but i think it might work:\r\n\r\nfirst.. notice that the dogs are always in an equilateral triangle wrt eachother when they are running.\r\n\r\nnow look at the picture.. this is a picture of what happens in time $dt$. the dogs go $dx$, and the sides of the triangle decreases size by $ds = s_1-s_2$\r\n\r\nif you know $ds$ in terms of $dx$, then you know $\\frac{ds}{dt}$ in terms of $v$.\r\n\r\nthen its easy to do..\r\n\r\nhowever.. i havent found a nice way to find $ds$. any seggustions?",
  "Solution_20": "OOOO now that I understand the problem ill attempt it.",
  "Solution_21": "[b]Try to solve it without calculus!!![/b]\r\nThere [b]is[/b] a way to do without calculus, but very very difficult. I remember my teacher solving this problem but i cannot recall how he did it...\r\nSo please try it.",
  "Solution_22": "instead of looking at $ds$ for $\\frac{ds}{dt}$, maybe looking at $dr$ would work?\r\n\r\nI conjecture that if one dog moves $dx$ towards another dog, it moves $dr = dx \\cos(30)$ towards the center, the point they all wind up at.  See the picture for my reasoning. It uses a 90-90-0 right triangle :) , so I'm not confident that it works. \r\n\r\nIf this is the case, then $\\frac{dr}{dt} = \\frac{dx}{dt} \\cos(30) = v \\cos(30) = v \\frac{\\sqrt{3}}{2}$\r\n\r\nThe dogs are travelling a distance $\\frac{a}{\\sqrt{3}}$, at a velocity of $v\\frac{\\sqrt{3}}{2}$, so the answer should be $\\frac{2a}{3v}$, right?",
  "Solution_23": "maybe it will help:\r\nassume one dog remains rest.",
  "Solution_24": "Due to the symmetry of the system, we assume that all dogs meet at the same time. So, $t$ can be the time required for dog A and dog B to meet.\r\n\r\nLet us imagine the whole action in the reference frame of dog A. If we sketch $v_{B} - v_{A}$ for different times, dog B follows a somewhat arc-like (of a circle) path towards A (this is an assumption), where the arc is inside the triangle and AB is a chord. (see sketch)\r\n\r\nLooking at the angles of the resultant vector at the beginning and the ending points, $\\angle XAB = \\angle XBA = 30 \\deg$.\r\nThus, $\\angle OAB = \\angle OBA = \\angle AOB = 60 \\deg$, where O is the centre of the circle.\r\nThus triangle AOB is equilateral.\r\n\r\nLet $s = AB = OB \\mbox{(radius)}$\r\nThen $arc AB = \\frac{1}{3} \\pi s$ (B's path)\r\n\r\nThe magnitude of the velocity of dog B is always the same and can be calculated. (since angle between $v_{B}$ and $v_{A}$ is $120 \\deg $)\r\n\\[v_{R} = \\sqrt{v^{2} + v^{2} - 2v^{2}cos120 \\deg}\\]\r\n\\[= \\sqrt{3}v\\]\r\nSo, $t = \\frac{s}{v} = \\frac{\\frac{1}{3} \\pi s}{\\sqrt{3}v} = \\frac{\\pi s}{3 \\sqrt{3} v}$\r\nwhere $s$ is the side of the triangle, and $v$ is the velocity of the dogs.\r\nWell, it seems I have some unexplained assumptions but they just make sense. Could anybody explain (prove) the assumptions or, otherwise, give a better solution?.",
  "Solution_25": "I understand that my solution above is wrong.\r\n\r\nAs the bugs move, they make an equilateral triangle with the same centroid at all times as repeated above. In the reference frame of the rotating triangle, the velocity component towards the centre is always $v cos 30$ (Think of the angle between the direction of a dog and the segment joining it with the centroid.)\r\nAt the same time, the displacement in this direction is also $s cos 30$. Thus, \r\n\\[ t = \\frac{s cos 30}{v cos 30}\\]\r\n\\[ t = \\frac{s}{v}\\]",
  "Solution_26": "Let's revive the Marathon!!!!\r\n\r\n\r\nProblem: Hydrogen gas is stored at a high pressure in a small, spherical container. The gas is introduced into a light balloon and its pressure becomes equal to the external atmospheric pressure. Is it possible that the balloon could lift the container in its final state? Assume that the temperature of the gas remains constant.\r\n\r\n\r\n\r\n[color=orange](P115)[/color]",
  "Solution_27": "[b]NO ![/b]\r\n\r\n[b]Rushil:[/b] Please, as a moderator you should not encourage this type of\r\nthread! It is horrible so post problems after each other like this. It gets to messy.\r\n\r\nActually one should split each problem into a new thread.\r\nAnd in the future only post one problem per thread.\r\nIt makes it easier to search the forum, and the members can easily be more active\r\nand up to date, especially new members.",
  "Solution_28": "[b]color[/b], what's the point of posting comments like this one on forsaken two-year old threads? We have established a marathon with rules, so I don't see why this old thread has any relevance...",
  "Solution_29": "Ok sorry, I am quite new in the section.\r\nI was searching for resistors and saw this thread, and that's why i did not\r\nnotice its date.\r\n\r\nSo if you already have adapted its good  :lol: .\r\nBut why continue with the marathon form?"
}
{
  "Tag": [
    "geometry",
    "ratio",
    "AMC"
  ],
  "Problem": "On circle $ O$, points $ C$ and $ D$ are on the same side of diameter $ \\overline{AB}$, $ \\angle AOC \\equal{} 30^\\circ$, and $ \\angle DOB \\equal{} 45^\\circ$. What is the ratio of the area of the smaller sector $ COD$ to the area of the circle?\r\n[asy]unitsize(6mm);\ndefaultpen(linewidth(0.7)+fontsize(8pt));\n\npair C = 3*dir (30);\npair D = 3*dir (135);\npair A = 3*dir (0);\npair B = 3*dir(180);\npair O = (0,0);\ndraw (Circle ((0, 0), 3));\nlabel (\"$C$\", C, NE);\nlabel (\"$D$\", D, NW);\nlabel (\"$B$\", B, W);\nlabel (\"$A$\", A, E);\nlabel (\"$O$\", O, S);\nlabel (\"$45^\\circ$\", (-0.3,0.1), WNW);\nlabel (\"$30^\\circ$\", (0.5,0.1), ENE);\ndraw (A--B);\ndraw (O--D);\ndraw (O--C);[/asy]$ \\textbf{(A)}\\ \\frac {2}{9} \\qquad \\textbf{(B)}\\ \\frac {1}{4} \\qquad \\textbf{(C)}\\ \\frac {5}{18} \\qquad \\textbf{(D)}\\ \\frac {7}{24} \\qquad \\textbf{(E)}\\ \\frac {3}{10}$",
  "Solution_1": "[hide=\"Solution\"]$ \\frac{180\\minus{}45\\minus{}30}{360} \\equal{} \\frac{105}{360} \\equal{} \\frac{7}{24} \\implies \\textbf{(D)}$[/hide]"
}
{
  "Tag": [
    "\\/closed"
  ],
  "Problem": "It would seem like there would be some people that would want to do both. Ah, well, there is always multi-tasking! I do that every class anyway, so nothing new here...",
  "Solution_1": "You can indeed sit in both classrooms.  Open a second browser, log in, go to the classroom.  There is a link just below the classroom window that you'll have to click in the second browser to get into a second classroom."
}
{
  "Tag": [
    "inequalities"
  ],
  "Problem": "Let $ a,b,c>0$. Prove that:\r\n\r\n$ \\dfrac{a}{2a\\plus{}b}\\plus{}\\dfrac{b}{2b\\plus{}c}\\plus{}\\dfrac{c}{2c\\plus{}a}\\leq 1$",
  "Solution_1": "[hide]Let $ \\frac {b}{a} \\equal{} x$, $ \\frac {c}{b} \\equal{} y$, and $ \\frac {a}{c} \\equal{} z$, then $ xyz \\equal{} 1$. We have to show\n$ \\frac {1}{2 \\plus{} x} \\plus{} \\frac {1}{2 \\plus{} y} \\plus{} \\frac {1}{2 \\plus{} z} \\le 1$.\nClearing denominators, this reduces to\n$ xy \\plus{} yz \\plus{} xz \\ge 3$\nWhich is true by AM-GM.\nOr we could clear denominators right away and get\n$ \\frac{c}{a}\\plus{}\\frac{a}{b}\\plus{}\\frac{b}{c} \\ge 3$ true by AM-GM.\n[/hide]",
  "Solution_2": "First of all, see easily that $\\displaystyle\\sum_{cyc} \\dfrac{b}{2a+b}=\\displaystyle\\sum_{cyc} \\dfrac{b^2}{2ab+b^2}\\ge \\dfrac{(a+b+c)^2}{a^2+b^2+c^2+2ab+2bc+2ca}=1$.\nThen we have $\\displaystyle\\sum_{cyc}\\dfrac{a}{2a+b}=\\dfrac{1}{2}\\displaystyle\\sum_{cyc}\\dfrac{2a}{2a+b}=\\dfrac{1}{2}\\displaystyle\\sum_{cyc}\\left(1-\\dfrac{b}{2a+b}\\right)\\le \\dfrac{1}{2}(3-1)=1$.",
  "Solution_3": "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=100583&hilit=Czech+Slovak]Czech and Slovak1999:[/url]\nFor arbitrary positive real numbers  $ a,b,c$ prove the inequality\\[ \\frac{a}{b\\plus{}2c}\\plus{}\\frac{b}{c\\plus{}2a}\\plus{}\\frac{c}{a\\plus{}2b}\\ge 1.\\]",
  "Solution_4": "[quote=sqing][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=100583&hilit=Czech+Slovak]Czech and Slovak1999:[/url]\nFor arbitrary positive real numbers  $ a,b,c$ prove the inequality\\[ \\frac{a}{b\\plus{}2c}\\plus{}\\frac{b}{c\\plus{}2a}\\plus{}\\frac{c}{a\\plus{}2b}\\ge 1.\\][/quote]\n\n$\\sum_{cyc}\\frac{a}{b+2c}\\geq 1$\n$\\iff$\n$\\sum_{cyc}\\frac{a^2}{ab+2ca}\\geq 1$\n\nBy Titu,\n\n$\\sum_{cyc}\\frac{a^2}{ab+2ca}\\geq \\frac{(a+b+c)^2}{3(ab+bc+ca)}$\n\nBy Rearrangement,\n\nThis is $\\geq 1$ :D"
}
{
  "Tag": [
    "analytic geometry",
    "graphing lines",
    "slope"
  ],
  "Problem": "What is the slope of a line parallel to $ 2x\\plus{}4y\\equal{}\\minus{}17$? Express\nyour answer as a common fraction.",
  "Solution_1": "lets just ignore -17.\r\n\r\nthan the equation becomes 4y=-2x, and y=$ \\minus{}\\frac{1}{2}$.\r\n\r\n$ \\minus{}\\frac{1}{2}$ * something is -1, so the answer is 2.\r\n\r\nanswer : 2",
  "Solution_2": "@FantasyLover, I think you did this problem in thinking that the problem asked for the slope of the line perpendicular to the line of $ 2x+4y=-17 $, which is why you set the slope times \"something\" equal to $-1$.\n\nWe know that the slope line $A$ perpendicular to line $B$ is the same as the slope of line $B$, all we have to do is find the slope of line $B$ by turning its equation into slope-intercept form $y=mx+b$ where $m$ is the slope of the line:\n\n$ 2x+4y=-17 $ \n\n$4y=-2x-17$\n\n$y=\\frac{-2}{4}x-17$\n\nTherefore, the slope of this line is $-\\frac{2}{4}$, or $\\boxed{-\\frac{1}{2}}$."
}
{
  "Tag": [
    "vector",
    "trigonometry",
    "linear algebra",
    "matrix",
    "linear algebra unsolved"
  ],
  "Problem": "Let $E$ be a $n$-dimensional vector space over $\\mathbb{C}$. Prove that any $f \\in L(E)$ we can represent as a linear combination of no more then $4$ unitary matrices.",
  "Solution_1": "eugene wich oral, I mean wich concours. \r\n\r\nLemma:  ($h\\in L(E)$, hermitian $h=h^{*}$, $|||h|||\\leq 1$ )\r\nis equivalent \r\n(there exist $u\\in U(E)$, $h=\\frac{1}{2}(u+u^{*})$)\r\n\r\nU(E) is unitary group",
  "Solution_2": "I think I've seen it before, but I don't think that it was on the forum. \r\n\r\nWrite $A$ (I replaced your $f$ with $A$) as $\\frac{A+A^*}2+\\frac{A-A^*}2$ (Moubinool's lemma was basically a hint given in that problem :)). $\\frac{A+A^*}2$ is a Hermitian matrix. Assume it's diagonal. Some real multiple $t\\frac{A+A^*}2,\\ t\\in\\mathbb R\\setminus\\{0\\}$ of it will have norm $\\le 1$, so its diagonal entries will be real numbers in the interval $[0,1]$. We can denote them by $\\cos\\alpha_i,\\ i=\\overline{1,n}$. Now let $U$ be the diagonal matrix with entries $\\cos\\alpha_i+i\\sin\\alpha_i$. It's unitary, and $\\frac{U+U^*}2=t\\frac{A+A^*}2$, which means that $\\frac{A+A^*}2=\\frac{U+U^*}{2t}$ (note that this means that $\\frac{A+A^*}2$ is a linear combination of two unitary matrices; it doesn't matter whether it's diagonal or not; I just assumed it was because it seemed easier to write down). \r\n\r\nWe do something similar for $s\\frac{A-A^*}2,\\ s\\in\\mathbb R\\setminus\\{0\\}$, except that this time we denote its eigenvalues (which are purely imaginary) by $i\\sin\\beta_i,\\ i=\\overline{1,n}$. We then have $s\\frac{A-A^*}2=\\frac{V-V^*}2$, where $V$ is the diagonal matrix with entries $\\cos\\beta_i+i\\sin\\beta_i$.",
  "Solution_3": "One can better do: 2 unitary matrices are enough.\r\n  Let A in Mn(C).By polar decomposition A=UH where U is an unitary\r\nmatrix and H is an hermitian nonnegative matrix.There exists an unitary matrix P and a real >=0 diagonal matrix D s.t. H=P*DP.\r\nA=UP*DP; it remains to show the result for D.\r\n  Let u=max(Djj).For all j there exists a real Sj s.t. Djj=ucos(Sj)=\r\nu/2exp(iSj)+u/2exp(-iSj).\r\nLet U1=diag(exp(iSj)) and U2=diag(exp(-iSj)).  U1 and U2 are unitary matrices such that D=u/2U1+u/2U2."
}
{
  "Tag": [
    "function",
    "limit",
    "pigeonhole principle",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "Let $ N$ denote the set of natural numbers. Let $ f$ takes $ N$ to $ N$ be a bijective funtion and assume \r\n\r\nthat there exist a finite limit \r\n                   \r\n$ \\displaystyle\\lim_{n\\to \\infty }\\frac {f(n)}{n}\\equal{}L$\r\n\r\nWhat are the possible values of $ L$?",
  "Solution_1": "Nice one.\r\n\r\n[hide=\"Hint\"]\nIf $ L \\equal{} 0.9$, then for $ n$ large enough we must have $ f(n)/n < 1$. Is this possible?\n[/hide]\n\n[hide=\"Solution\"]\nWe can certainly have L=1 by taking f(n)=n. We prove that $ L$ cannot take any other value.\n\nIf $ f(n)/n \\to L < 1$, then by the definition of a limit we must have $ f(n) / n < 1$ for all $ n$ larger than some $ N$. So we only have a finite number of $ k$'s such that $ f(k) > k$: the only candidates are $ k$'s such that $ k < N$. \n\nWe can therefore find an $ M$ so large so that for any $ k$ with $ f(k)>k$ we have $ M > f(k)$. We can also assume $ M > N$. But now the interval $ [1,2,\\ldots,M]$ is mapped onto itself by $ f$, and $ f(M\\plus{}1)$ must also belong to that interval since $ M>N$ and anything larger than $ N$ becomes smaller when $ f$ is applied. This contradicts the pigeonhole principle. \n\nThe case $ L > 1$ can be handled similarly, or by observing that replacing $ f$ with $ f^{\\minus{}1}$ (which is also a bijection) causes $ L$ to be replaced by $ 1/L$. \n[/hide]"
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "geometry proposed"
  ],
  "Problem": "Let (O) be a semircircumference. diameter BC . A a variable point on (O) , let H be the foot of the perpendicular from A to BC, Q belows to BA (in the extension though A) such that AQ=AH . show that the loci of Q is a circumference . Identify the center and the lenght of the radius.",
  "Solution_1": "Are you sure it is a circle?",
  "Solution_2": "Sorry planegeometry I confused the problem with another one, here is the correct enuntiation , the problem is very easy\r\n\r\nLet (O) be a semircircumference. diameter BC . A a variable point on (O) , Q belows to BA (in the extension though A) such that AQ=AC . show that the loci of Q is a circumference . Identify the center and the lenght of the radius",
  "Solution_3": "loci of Q is a circumcircle  with D its center ,where D is the midpoint of arcBC  the radius^2=DB^2=DC^2=1/2BC^2",
  "Solution_4": "[quote=\"Tiger100\"]Let $ B$ , $ C$ be two fixed point on a given circle $ w$ and a mobile point $ A$ which belongs to one of the arcs which are \n\ndefined in the circle $ w$ by the chord $ [BC]$ . Define the point $ Q\\in AB$ for which $ A\\in BQ$ and $ AQ\\equal{}AC$ . Show that \n\nthe loci of the point $ Q$ is a circle with the center $ D$ in the midpoint of the choosen arc and its radius is equally to $ DC$ .[/quote]",
  "Solution_5": "[quote=\"Virgil Nicula\"][quote=\"Tiger100\"]Let $ B$ , $ C$ be two fixed point on a given circle $ w$ and a mobile point $ A$ which belongs to one of the arcs which are \n\ndefined in the circle $ w$ by the chord $ [BC]$ . Define the point $ Q\\in AB$ for which $ A\\in BQ$ and $ AQ \\equal{} AC$ . Show that \n\nthe loci of the point $ Q$ is a circle with the center $ D$ in the midpoint of the choosen arc and its radius is equally to $ DC$ .[/quote][/quote]\r\n\r\nThanks for your nice extension mr Virgil, I see that the proof is exactly the same."
}
{
  "Tag": [
    "function",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Prove that we cannot find a continuous function $f:RxR->R$ such that for any continuous function $g:R->R$ there is $t$ with $g(x)=f(t,x)$ for all $x$. What if we replace $R$ with $[0,1]$.",
  "Solution_1": "For the first part:\r\n\r\nGiven a continuous function $f:\\mathbb R^2\\to\\mathbb R$, define $g$ to be continuous on the set $\\bigcup_{n\\in\\mathbb Z}[2n,2n+1]$ s.t. $\\min_{[2n,2n+1]}g>\\max_{[n,n+1]\\times[2n,2n+1]}f$, and then extend it to the whole real line (this is obviously possible).\r\n\r\nAs for the second question, could you please phrase it exactly as it should be? Which instance of $\\mathbb R$ do we replace with $[0,1]$? Is $f$ defined on $[0,1]^2$, or $\\mathbb R\\times [0,1]$, or does $g$ take values in $[0,1]$, or what? :)",
  "Solution_2": "Just replace each time the symbol $R$ with $[0,1]$. :)",
  "Solution_3": "For both parts define $g(x) = f(x,x)+1$.\r\nSo if $g(x) = f(x,t)$, then $g(t) = f(t,t) = f(t,t)+1$.",
  "Solution_4": "[quote=\"ilyasu1\"]For both parts define $g(x) = f(x,x)+1$.\nSo if $g(x) = f(x,t)$, then $g(t) = f(t,t) = f(t,t)+1$.[/quote]\r\n\r\nVery nice solution.",
  "Solution_5": "Except that for the second question it doesn't work, since both $g$ and $f$ must take values in $[0,1]$ :).",
  "Solution_6": "Indeed, this was a very silly mistake  :blush: .\r\n\r\nBut what about this approach:\r\n\r\nlet $t_n$ be a number such that $f(x,t_n) = \\sin(2\\pi n x)$. \r\nSo $t_n \\in [0,1]$. So $t_{n_j} \\to t$, i.e. there's a converging subsequence.\r\nConsider $f(0,t)$. $f(0,t_{n_j}) \\to f(0,t) = 0$, and yet $f(\\pi / 4n_j, t_{n_j}) \\to f(0,t) = 1$,\r\nsince both $(0,t_{n_j})$ and $(\\pi /4n_j, t_{n_j})$ converge to $(0,t)$.\r\n\r\nI hope that this time there are no errors."
}
{
  "Tag": [
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "Is it clear to understand?",
  "Solution_1": "Use latex... :wink: \r\n\r\nPierre."
}
{
  "Tag": [],
  "Problem": "A square-table of dimensions $ n\\times n$, where $ n\\in N^*$, is filled arbitrarly with the numbers $ 1,2,...,n^2$ such that every number appears on the table exactly one time. From each row of the table\r\nis chosen the least number and then denote by $ x$ the biggest number from the numbers chosen. From each column of the table is chosen the least number and then denote by $ y$ the biggest number from the numbers chosen. The table is called [i]balanced [/i]iff $ x \\equal{} y$. How many balanced tables we can obtain?",
  "Solution_1": "The problem is from BMO 2001 Shortlist.\r\n\r\nPlease someone solve this..........."
}
{
  "Tag": [],
  "Problem": "Let $ABC$ be a triangle with $\\angle A=120$. The angle bisectors of $\\angle A, \\angle B, \\angle C$ meet the sides $BC,CA,AB$ at $D,E,F$. Show that $\\angle EDF = 90$.",
  "Solution_1": "posted before: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=114998]foot of angle besector[/url]"
}
{
  "Tag": [
    "\\/closed"
  ],
  "Problem": "After I post a message, a screen appears asking me whether I want to see my message or go back to the forum. After about 5 seconds, it automatically goes to my message. Is this the same with everybody, or is it just my computer?",
  "Solution_1": "It's probably the same for all.\r\n\r\nIs there any problem with it? I consider it as a good option, because it actually does something, rather than asking you still. And if you want to go to the forum, you can do it with one click.",
  "Solution_2": "Ok, thanks. I don't mind it, but I thought maybe my computer is messed up or something.",
  "Solution_3": "[quote=\"PowerOfPi\"]After I post a message, a screen appears asking me whether I want to see my message or go back to the forum. After about 5 seconds, it automatically goes to my message. Is this the same with everybody, or is it just my computer?[/quote]\r\nThat doesn't happen with me.",
  "Solution_4": "Neither with me.  This probably won't work, but try clearing your cookies.  It works for everything :P"
}
{
  "Tag": [
    "algorithm",
    "induction"
  ],
  "Problem": "here are 1000 men standing in a circle numbere 1 to 1000 on their back. the person with number 1 is given a sword. he kills 2 and gives sword to 3. 3 kills 4 and passes the sword to 5. the process continues so on....................THE PROCESS CONTINUES TILL ONLY 1 PERSON IS LEFT. wat is the number of the last remaining person.",
  "Solution_1": "Is it 961?",
  "Solution_2": "I am assuming that at the end every \"round\", person #1 starts off the \"next round\" again. By this is is pretty clear that person 1 will be the last one left. Or am I going wrong anywhere?",
  "Solution_3": "Well, that's true for the first few rounds, but do it yourself, and you'll quickly see that that is not the case.",
  "Solution_4": "Ok, so how about this: if #1 starts first every round, it is impossible to it to be out that round, isn't that true? Or is it like they are all like arranged in a circle?\r\n\r\nEDIT: never mind...I over-read the question for the word circle.  :blush:",
  "Solution_5": "What happens is, 1 doesn't start all of the rounds. \r\n\r\n1 will suddenly cease to start the round when the number of people in the previous round is odd.I have reason to belive that this will happen; the reason is that 1000 is not a power of 2.",
  "Solution_6": "I tested this out for 2, 4, 6, 8, and 10 and all of them ended with the first person as the last one standing, so I'm going to say it is the first person. :P",
  "Solution_7": "What a violent problem!!! Involves killing and violence....hmmm....\r\n\r\nAnyway, since person 1 kills person 2, and person 3 kills person 4...etc etc, person 1 starts the round, so he is the only one still alive :)",
  "Solution_8": "[quote=\"TheDeity\"]I tested this out for 2, 4, 6, 8, and 10 and all of them ended with the first person as the last one standing, so I'm going to say it is the first person. :P[/quote]\r\n\r\nWell, I tested 10, and I ended up with 5 as the last one standing, so...\r\n\r\nIn that scenario, 2, 4, 6, 8, 10, 3, 7 die in that iorder, and 9 has the sword, and 1 is right next to him, so...\r\n\r\n@the cliu: person 1 doesn't always start the round, as we pointed out, so your logic is incorrect.",
  "Solution_9": "no the first person does not always start...\r\n\r\nhave a close observation and u find tha he gets killed in the 4th round.\r\n\r\nand fishythefish ,  961  is wrong.",
  "Solution_10": "I guess I'll throw out a guess.  When I first read this problem, I thought the answer was in the form $ 2^n \\plus{}1$.  So is it $ \\boxed{513}$?",
  "Solution_11": "Well, I wasn't expecting too much. I used Microsoft Excel to list all the people, then knocked out all of the evens, and so on. In the cases where the first person was killed before they started the round, I took every other person starting with the second guy. Anyways, I ended up getting 961, but I didn't have too much faith in that answer. (Does this attempt make sense?)",
  "Solution_12": "I copied fishthefish's idea with Excel, but I think that 961 gets killed in the same round that 1 gets killed.\r\nIt got confusing after a couple of rounds though so I might have messed up the numbers.\r\nI eventually got 977.",
  "Solution_13": "yes you are right tornado.adv4\r\n\r\n\r\nactualy this is called as josephus problem.   u guys can wiki it if u want some more information on it",
  "Solution_14": "1 because it will start each round",
  "Solution_15": "Who says that 1 starts each round?  Number one is not anything special, other than the fact that he starts the entire process.  Then he is as normal as number 527.\r\n\r\nAnyway, for the Josephus problem for every 2, the answer lies in the binary system:  Write the number in binary.  Then take the first digit and move it to the end.  It's that simple!\r\n\r\nFor example, 10 in binary is 1010, then move the 1 to the end you get 0101 or 101 or 5 in decimal.\r\n\r\nFor 1000, in binary this is 1111101000.  Move the first digit to the end to get 1111010001, which in decimal is 977.",
  "Solution_16": "What do you mean by \"every 2\"?",
  "Solution_17": "Every second person is killed, versus every third or fourth, ...",
  "Solution_18": "[quote=\"tenniskidperson3\"]\nAnyway, for the Josephus problem for every 2, the answer lies in the binary system:  Write the number in binary.  Then take the first digit and move it to the end.  It's that simple!\n[/quote]\r\n\r\nCan you explain why this works?\r\n\r\nThanks!",
  "Solution_19": "so if every third person was killed, you write it in trinary (whatever) and move the first digit to the end?\r\n\r\nAnd if every third person is killed and you have a number like 123 in trniary, then you get 231, which is obviously not right?\r\n\r\nEDIT: Hm never mind. This will only work consistently for every second person. So then whats the strategy if its every third person? And how do you prove the binary thing?",
  "Solution_20": "No, it doesn't generalize; it's just written that way to make it easy to remember.  I have worked on finding a strategy for every 3, and so far it has been unfruitful.\r\n\r\nTo prove that it does work, we show that it works for each power of 2 and then use induction.\r\n\r\nFor each power of 2, we use induction.  It works for 1, because the algorithm gives 1.  To do the inductive step, with $ 2^{k \\plus{} 1}$ people, the first people to be eliminated are all even numbered people, and 1 ends up with the knife again.  Thus relabel all the people that are left from $ 2n \\minus{} 1$ to $ n$.  Thus 1 has the knife and there are $ 2^k$ people left, use inductive step.  Obviously the algorithm gives 1 for each power of 2, so this case is true.\r\n\r\nNow we tackle the case when the number of people is not a power of 2.  Call when the person who has the knife killing the person ahead of him and passing on the knife a step.  Now suppose there are $ 2^k \\plus{} p$ people where $ 0\\leq p\\leq 2^k \\minus{} 1$ and right now, person 1 has the knife.  The algorithm gives that the person to survive is $ 2p \\plus{} 1$.\r\n\r\nSo do induction on $ p$.  It works for $ p \\equal{} 0$, as we've already shown.  So now we need to show that if it works for $ p$, then it works for $ p \\plus{} 1$.  Suppose that for $ p$, person $ 2p \\plus{} 1$ survives.  What happens first is that person 1 kills person 2 then gives the knife to person 3.  If we relabel all the numbers $ m$ as $ m \\minus{} 2$ except 1 is relabeled as $ p$, we have a situation where person 1 has the knife and there are $ 2^k \\plus{} p$ people left.  So obviously person $ 2p \\plus{} 1$ will survive.  Now this person, with the old labeling, is actually person $ 2p \\plus{} 1 \\plus{} 2 \\equal{} 2p \\plus{} 3 \\equal{} 2(p \\plus{} 1) \\plus{} 1$.  This works for all numbers $ 0\\leq p\\leq 2^k \\minus{} 2$.  For $ p \\equal{} 2^k \\minus{} 1$, we've already shown that $ 2^{k \\plus{} 1}$ works, so the inductive step is complete and we're done."
}
{
  "Tag": [
    "inequalities",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Find all positive integers $ n$ such that for all odd integers $ a$, if $ a^2<\\equal{}n$ then $ a$ divides $ n$.",
  "Solution_1": "We claim that the solution space is $ \\{1,2,\\ldots,9,12,15,18,21,24,30,45\\}$. As for the proof:\r\n\r\nChoose the unique $ m\\geq 0$ such that $ (2m\\plus{}1)^2\\leq n<(2m\\plus{}3)^2$ holds, i.e. $ n$ has the odd integer divisors $ 1,3,\\ldots,2m\\plus{}1$. This yields the inequality: $ (2m\\plus{}3)^2>\\prod_{i\\equal{}1}^m(2i\\plus{}1)$. This requires $ m<3$, otherwise we would conclude $ (2m\\plus{}3)^2>(2m\\plus{}1)(2m\\minus{}1)(2m\\minus{}3)$ yielding the contradiction $ 0>8m^3\\minus{}14m\\minus{}16m^2\\minus{}6\\equal{}(8m^2\\minus{}14)(m\\minus{}2)\\minus{}34\\geq (72\\minus{}14)(3\\minus{}2)\\minus{}34>0$. \r\n\r\n$ m\\equal{}2$ yields $ 25\\leq n <49$ and $ 15|n$, i.e. the solutions $ n\\equal{}30,45$\r\n$ m\\equal{}1$ yields $ 9\\leq n<25$ and $ 3|n$, i.e. the solutions $ n\\equal{}9,12,15,18,21,24$.\r\n$ m\\equal{}0$ yields $ 1\\leq n<9$ and all such $ n$ are solutions."
}
{
  "Tag": [
    "trigonometry",
    "geometry",
    "3D geometry"
  ],
  "Problem": "Let $ a$ be a real constant. Removing $ \\theta$ in the equations we have:\r\n\r\n$ \\begin{cases}x\\sin\\theta \\plus{} y\\cos\\theta \\equal{} 2a\\sin\\theta \\\\\r\nx\\cos \\theta \\minus{} y\\sin\\theta \\equal{} a\\cos\\theta\\end{cases}$\r\n\r\nWe have:\r\n\r\n$ a)\\; (x \\plus{} y)^{\\frac {2}{3}} \\plus{} (x \\minus{} y)^{\\frac {2}{3}} \\equal{} 2a^{\\frac {2}{3}} \\\\\r\nb)\\; (x \\minus{} y)^{\\frac {2}{3}} \\minus{} (x \\plus{} y)^{\\frac {2}{3}} \\equal{} 2a^{\\frac {2}{3}} \\\\\r\nc)\\; (x \\plus{} y)^{\\frac {2}{3}} \\plus{} (y \\minus{} x)^{\\frac {2}{3}} \\equal{} a^{\\frac {2}{3}} \\\\\r\nd)\\; (x \\plus{} y)^{\\frac {2}{3}} \\plus{} (x \\minus{} y)^{\\frac {2}{3}} \\equal{} \\frac {a^{\\frac {2}{3}}}{2} \\\\\r\ne)\\;\\text{None}$",
  "Solution_1": "Apparently, this is too easy for everyone else, so I figured I'd take a stab at it.\r\n\r\n$ (x \\minus{} 2a)\\sin\\theta \\equal{} \\minus{}y \\cos\\theta$\r\n$ (x \\minus{} a)\\cos\\theta \\equal{} y\\sin\\theta$\r\n$ (x \\minus{} 2a)(x \\minus{} a) \\equal{} \\minus{}y^2$\r\n$ 2a^2 \\minus{} 3ax \\plus{} x^2 \\plus{} y^2 \\equal{} 0$\r\n\r\nThe answer has a lot of cube roots and we don't have any cube roots. We plug in $ x \\equal{} 4$ and $ y \\equal{} 1$ and that gets rid of all the answer choices. So the answer is E.",
  "Solution_2": "I don't think this is too easy for everyone else, because I have posted really easy problems which always came with replies.\r\n\r\nDiffer, first of all the idea is to [i]solve[/i] the problem, even though in the exam we would not care about solving but care about getting it right. Here in the forums I am thinking the purpose of me posting the problem is because I want to see how to reach the correct alternative. Second of all, it seems you are wrong, for that is not the correct answer.",
  "Solution_3": "You are correct, of course. I should have solved the problem completely. There must be a restriction hidden within the first two equations that precludes $ x \\equal{} 4$ and $ y \\equal{} 1$; however, I cannot find it.",
  "Solution_4": "Could anybody please help me with this?",
  "Solution_5": "There must be something wrong with the statement (i.e. with the first four choices OR with the initial system). For $ \\theta\\equal{}{\\pi\\over 4}$, the system becomes $ x\\plus{}y\\equal{}2a\\land x\\minus{}y\\equal{}a$, but those values don't satisfy either of the four given formulas.",
  "Solution_6": "What's the source of the problem?",
  "Solution_7": "ITA (Aeronautic Technological Institute) Exam, here in Brasil. I don't have the exact year though.\r\n\r\nIs this problem wrong?"
}
{
  "Tag": [
    "abstract algebra",
    "group theory",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "Prove that a transitive abelian subgroup of $S_n$ has exactly $n$ elements.\r\n\r\n\r\nP.S.\r\n\r\nThis is somewhat related to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=46589]this other problem[/url].",
  "Solution_1": "anyone a solution? :) seems simple and elegant.",
  "Solution_2": "since nontrivial element $G$ ,transitive abelian subgroup,do not have fixed points we conclude ,cardinal of $G$ doesnt exceed $n$ .(should be exactly $n$ ,as stated in problem)",
  "Solution_3": "Take $f$ in $G$ and suppose that there exists some $a \\in [[1;n]]$ such that $f(a)=a$.\r\nWe know that for any $g \\in G$, $f=gfg^{-1}$. So if $f=c_1c_2..c_r$ is the cycle decomposition of $f$, any $g \\in G$ maps the element of $c_i$ to the elements of $c_j$. And because $G$ is transitive, this shows that all the $c_i$ have the same length. Hence $f$ is the identity: any non trivial element of $G$ has no fixed point.\r\n\r\nFrom here the conclusion is easy. Take any $a \\in [[1;n]]$ : its orbit under the action of $G$ is $[[1;n]]$. Hence $n=\\frac{Card(G)}{Card(Stab(a)))}=Card(G)$.",
  "Solution_4": "It's so easy :-)\r\n \r\nLet $G \\leq S_{n}$ be transitive and abelian. Consider\r\n\r\n$\\alpha : G \\to \\{1,...,n\\}, g \\to  g(1)$\r\n \r\nSince $G$ is transitive, $\\alpha$ is surjective. To show that $\\alpha$ is injective, assume $g \\neq h$ in $G$. There exists $j \\in \\{1,...,n\\}$ s.t. $g(j) \\neq h(j)$. Since $G$ is transitive, there exists $f \\in G$ s.t. $j=f(1)$. Now, $G$ is abelian, hence\r\n \r\n$f(g(1)) = g(f(1)) \\neq h(f(1))=f(h(1)) \\Rightarrow g(1) \\neq h(1)$"
}
{
  "Tag": [],
  "Problem": "Tamara knows that the arithmetic mean of her five quiz scores is 95%. However, she has misplaced one of these quizzes. The ones she can find have scores of 100%, 100%, 99% and 98%. What is her score on the misplaced quiz?",
  "Solution_1": "If the mean is 95, then the sum of the five scores must be $ 95\\times 5\\equal{}475$.  Now subtract off the other 4: $ 475\\minus{}100\\minus{}100\\minus{}99\\minus{}98\\equal{}78$.  She must have messed up bad. :D",
  "Solution_2": "for these problems, i dont like the \"normal\" method since i dont like multiplying :P . so this is how i do these \"mean\"(no pun intended) problems\r\n\r\n\r\nsince her first score is 100, and her mean is 95, you have to subtract 5 to get to 95. same thing for the second quiz. for the third quiz, you have to subtract 4.\r\n\r\ncontinuing in this fashion, you get -5,-5,-4,-3. add it all up and you get -17.\r\n\r\nso now u do 95(her average) and -17. that get you 78 :wink:",
  "Solution_3": "That method is a nice method, but only when the numbers are easily worked with.  For most problems of this nature, you should probably stick to the usual method."
}
{
  "Tag": [
    "limit",
    "integration",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Let us denote $ p={10}^{10}\\;  ,\\; \\;  X_n=\\displaystyle \\frac{1}{p} \\left(1+\r\n2\\sum\\limits_{k=1}^{n+p}\\exp{\\left(-\\frac{k^2}{p}\\right)}\\right)^2\\;\r\n. $  [b] Prove or disprove that [/b]    $\\; \\; \\lim\\limits_{n\\to \\infty}n\\cdot\\sin(X_n)=0\\; . $",
  "Solution_1": "It is obvious that $ X_{n}(p)$ converges and i found that it extemely close to $ \\pi$ for big $ n$ and $ p$. And i think that $ \\lim_{p\\to\\infty}{\\lim_{n\\to\\infty}{X_{n}(p)}} \\equal{} \\pi$.\r\n\r\nBut $ \\lim_{n\\to\\infty}{X_{n}(p)} \\equal{} X(p)\\ne\\pi$.\r\n\r\n$ p$ is very big, but not $ \\infty$, so:\r\n\r\n$ \\lim\\limits_{n\\to \\infty}n\\cdot\\sin(X_n) \\equal{} \\infty$\r\n\r\nWe should find a good approximation for the sum $ \\sum\\limits_{k \\equal{} 1}^{n \\plus{} p}\\exp{\\left( \\minus{} \\frac {k^2}{p}\\right)}$. \r\n\r\nFor big $ n$ it should be $ \\sum\\limits_{k \\equal{} 1}^{n \\plus{} p}\\exp{\\left( \\minus{} \\frac {k^2}{p}\\right)}\\approx\\frac {\\sqrt {\\pi p} \\minus{} 1}{2}$.\r\n\r\nBy Euler's summation formula: $ \\sum\\limits_{k \\equal{} 1}^{n \\plus{} p}e^{ \\minus{} \\frac {k^2}{p}} \\equal{} \\int_{1}^{n \\plus{} p}e^{ \\minus{} \\frac {x^2}{p}}\\;dx \\minus{} \\frac {1}{2}\\left(e^{ \\minus{} \\frac {(n \\plus{} p)^2}{p}} \\minus{} e^{ \\minus{} \\frac {1}{p}}\\right) \\plus{} R$, where $ R$ is small."
}
{
  "Tag": [
    "articles"
  ],
  "Problem": "People generally say that Phenyl group has greater migratory aptitude than H.. Is there any specific reason for this or is this just reaction dependent???",
  "Solution_1": "Here is a nice link. Actually relates to Pardesi's thread on stability (Pinacol-Pinacolone Rearrangement):\r\n\r\nhttp://www.cem.msu.edu/~reusch/VirtualText/rearrang.htm",
  "Solution_2": "actually there is a wonderful article on this in Peter Sykes...which clearly explains the various cases where migration aptitude is involved and how to predict which si the best :)",
  "Solution_3": "Actually H has greater migratory aptitude: the Bayer-Villiger rearrangement is a good example, where one cannot obtain formates from aldehydes, but get acids instead.",
  "Solution_4": "Phenyl still has the brgeater migratory apptitude coz in the transition state for migration , the phenyl carbon retains its tetravalency whereas in case of alkyl group it does not .... so phenyl grp has greatest migratory apptitude",
  "Solution_5": "@valerium nothing in general can be told about the migratory aptitudes it depemds on the reaction conditions and on steric factors also..",
  "Solution_6": "But we can still tell abt the transtion state during the transfer pardesi....\r\nThe product which is formed via the most stable transition state will be the most stable product...",
  "Solution_7": "what if the phenyl  group approaches already present two phenyl groups??",
  "Solution_8": "Then that only causes some steric hindrance...\r\nand moreover u should not consider those special cases..."
}
{
  "Tag": [
    "Asymptote",
    "LaTeX",
    "\\/closed"
  ],
  "Problem": "In the wiki, if you type asymptote code such as this :\r\n[code]<asy>\nsize(200pt);\ndefaultpen(1);\npair A=(0,0), B=(220,0);\nlabel(\"A\",A,SW);\nlabel(\"$B$\",B,SE);\n</asy>[/code]\nyou will get a correct image in preview, like this :\n[img]http://alt2.mathlinks.ro/Forum/latexrender/pictures/e/7/5/e75cbb71f340fd97e702d56a7764a4c9b3eab535.png[/img]\nBut when you save the page, you will see this :\n[img]http://alt1.mathlinks.ro/Forum/latexrender/pictures/2/9/0/290eb104186d0bd324d7275de0d2f0b9c349d012.png[/img]\nApparently, the dollar signs are converted to <math></math> tags, and in text mode in LaTeX, < and > become $ \\text{ < }$ and $ \\text{ > }$, so what appears is something like $ \\text{ < math > B < /math > }$, which is what you get with this code :\n[code]$\\text{<math>B</math>}$[/code]\r\nIf you replace the $ \\$$. . .$ \\$$ with \\( . . . \\) or \\begin{math} . . . \\end{math}, the picture works fine, though.",
  "Solution_1": "I've previously posted this problem, but no one responded...",
  "Solution_2": "Fixed."
}
{
  "Tag": [
    "national olympiad"
  ],
  "Problem": "1. http://www.mathlinks.ro/viewtopic.php?t=5830\r\n2. http://www.mathlinks.ro/viewtopic.php?t=5831\r\n3. http://www.mathlinks.ro/viewtopic.php?t=5832\r\n4. http://www.mathlinks.ro/viewtopic.php?t=5833\r\n5. http://www.mathlinks.ro/viewtopic.php?t=5834\r\n6. http://www.mathlinks.ro/viewtopic.php?t=5835\r\n\r\nThanks.",
  "Solution_1": "no, thank you for sharing the problems with us. What did you do in the contest? :)",
  "Solution_2": "Well I'm very lucky in this contest, and I'll participate 2004 IMO in Greece.\r\nSorry for my poor English in the above postings.. :blush:",
  "Solution_3": "don't worry, your english is fine :) all of us make small mistakes from time to time, but we don't apologize all the time for that :) once is enough :P \r\n\r\nanyway, which other members of ML from Korea will be joining in Greece? :)",
  "Solution_4": "I have spent some time to make this pdf a collection of problems from this TST. I have a word for it: I would like to know if problem #4 is translated the way it's supposed to be. If there is anything else that is also wrong, please tell me so I can correct the pdf file. Please enjoy.",
  "Solution_5": "Well I'm not sure that there is another member in ML, because this homepage isn't famous well in Korea. sorry.\r\n\r\nAnd thanks 3X.lich for making PDF type of Korean TST! I think your word isn't wrong in #4. Thanks a lot :)"
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "Prove that if $ a,b,c,d\\geq0$ then\r\n\\[ \\sum_{\\rm cyclic}\\frac{d^{3}\\plus{}abc}{(d\\plus{}a)(d\\plus{}b)(d\\plus{}c)}\\geq1.\\]",
  "Solution_1": "Once we make a full expansion (In this case with realitively little computation) it reduces to $ [3,2,1,0]$ majorises $ [2,2,1,1]$\r\nbut I guess you have a nicer solution... :lol:",
  "Solution_2": "$ \\frac{x}{x\\plus{}y}\\plus{}\\frac{y}{x\\plus{}y}\\equal{}1$ gives proof. :D",
  "Solution_3": "See:http://www.irgoc.org/bbs/dispbbs.asp?boardid=12&id=2696&star=2#2696",
  "Solution_4": "Hello fjwcls. From the linke, I saw some of your inequalities for the first time. I am interested in the following inequality\r\n\\[ \\sum_{\\rm cyclic}(a^{3}\\minus{}bcd)(a\\minus{}b)(a\\minus{}c)(a\\minus{}d)\\geq0\\]\r\nas well as your method to solve such kinds. It seems to me that you use software. Otherwise, what?",
  "Solution_5": "$ \\sum_{cyclic}\\frac{a^{3}\\plus{}bcd}{(a\\plus{}b)(a\\plus{}c)(a\\plus{}d)}\\equal{}\\sum_{cyclic}\\left\\{1\\minus{}\\left(\\frac{a}{a\\plus{}b}\\plus{}\\frac{a}{a\\plus{}c}\\plus{}\\frac{a}{a\\plus{}d}\\right)\\plus{}\\frac{a^{2}}{(a\\plus{}b)(a\\plus{}c)}\\plus{}\\frac{a^{2}}{(a\\plus{}c)(a\\plus{}d)}\\plus{}\\frac{a^{2}}{(a\\plus{}d)(a\\plus{}b)}\\right\\}$\r\n$ \\geq 4\\minus{}6\\plus{}\\frac{9(a\\plus{}b\\plus{}c\\plus{}d)^{2}}{3(a\\plus{}b\\plus{}c\\plus{}d)^{2}}\\equal{} 1$.",
  "Solution_6": "Very nice proof, kunny! :lol:",
  "Solution_7": "Thanks! :lol:",
  "Solution_8": "Another less direct approach is to show\r\n\\[ \\sum_{\\rm cyclic}\\frac{d^{3}}{(d\\plus{}a)(d\\plus{}b)(d\\plus{}c)}\\geq\\frac{1}2,\\]\r\nwhich was discussed already in this forum.\r\n\r\nNext, we show that\\[ \\sum_{\\rm cyclic}\\frac{abc}{(d\\plus{}a)(d\\plus{}b)(d\\plus{}c)}\\geq\\frac{1}2,\\]\r\nwhich is turned into the above one by substituitions $ a\\equal{}1/x,$ ..."
}
{
  "Tag": [
    "function",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "Find all $f: (1, \\infty ) \\to R$ such that : $f(x)-f(y)=(y-x)f(xy)$",
  "Solution_1": "$f(1)-f(y)=(y-1)f(y) \\Rightarrow f(y)=\\frac{f(1)}{y}\\vee [ f(1)=0 \\wedge f(y)=0] \\;\\; \\forall y \\in \\mathbb{R}^{+}$\r\n$f(1)=0 \\Rightarrow 0-f(y)=(y-1)f(y) \\Rightarrow f(y)=0 \\Rightarrow f(y)=\\frac{f(1)}{y}\\;\\;\\forall y \\in \\mathbb{R}^{+}$\r\n$f(x)=\\frac{f(1)}{x}\\Rightarrow f(1)(\\frac{1}{x}-\\frac{1}{y})=f(1)\\frac{y-x}{xy}\\Rightarrow f(x)-f(y)=(y-x)f(xy) \\;\\;$\r\n$\\forall x, y \\in \\mathbb{R}^{+}$.\r\n\r\nNo problemo. \u00bf?",
  "Solution_2": "I have corrected the problem !",
  "Solution_3": "[quote=\"Redriver\"]Find all $f: (1, \\infty ) \\to R$ such that : $f(x)-f(y)=(y-x)f(xy)$[/quote]\r\nLet $g(x)=f(x)-\\frac{2f(2)}{x},\\ \\forall x>1$\r\n$\\rightarrow g(2)=0,\\ g(x)-g(y)=(y-x)g(xy),\\ \\forall x,y>1$  (1)\r\nin (1), let $y=2\\rightarrow g(x)=(2-x)g(2x)$\r\nlet $y=xz\\rightarrow g(x)-g(xz)=(xz-x)(x^{2}z)$\r\n$\\leftrightarrow (x-z)g(x)-g(z)+g(x)=(x-z)(xz-x)g(x^{2}y)$\r\n$\\leftrightarrow (x^{2}-z)(x-z)g(x)-(x^{2}-z)g(z)+(x^{2}-z)g(x)=(x^{2}-z)(x-z)(xz-x)g(x^{2}y)$\r\n$\\leftrightarrow (x^{2}-z)(x-z)g(x)-(x^{2}-z)(g(z)-g(x))=(x-z)(xz-x)(g(z)-g(x^{2}))$ (2)\r\nin (2), let $x=2\\rightarrow g(z)(2z^{2}-5z)=(2z^{2}-6z+4)g(4)$\r\n$g(z)=\\frac{(2z^{2}-6z+4)g(4)}{z(2z-5)}, \\forall z>1,\\ z\\neq\\frac{5}{2}$ (3)\r\n$\\rightarrow g(2z)=\\frac{(8z^{2}-12z+4)g(4)}{2z(4z-5)}, \\forall z>1,\\ z\\neq\\frac{5}{4}$ (4)\r\n(3) and (4) $\\rightarrow g(2z)=\\frac{(2z^{2}-6z+4)g(4)}{z(2-z)(2z-5)}, \\forall z>1,\\ z\\neq\\frac{5}{2},\\ z\\neq 2$ (5)\r\n.cause $g(4)\\neq 0$\r\n(4) and (5), $\\rightarrow \\frac{2z-1}{4z-5}=\\frac{1}{5-2z},\\forall z>1,\\ z\\neq\\frac{5}{2},\\ z\\frac{5}{4}\\rightarrow z=2$ unreasonableness\r\n$\\rightarrow g(4)=0\\rightarrow g(z)=0,\\ \\forall z>1\\rightarrow f(x)=\\frac{2f(2)}{x}$. :D"
}
{
  "Tag": [
    "search",
    "function"
  ],
  "Problem": "In how many ways can 7 identical marbles be placed in a red bag, a white bag, and a blue bag if there must be at least one marble in each bag?",
  "Solution_1": "This is the same as distributing 4 identical marbles to the three bags, then placing 1 marble into each bag. By that balls and urns argument thing, there are $ \\binom{6}{2}\\equal{}15$ ways to do this.",
  "Solution_2": "1+2+5 x6\r\n1+1+6 x3\r\n1+3+3 x3\r\n2+2+3 x3\r\n\r\n= 6 + 3 +3 +3 = 15 ways\r\n\r\nBrute Force can u explain your balls and urns argument?",
  "Solution_3": "[hide]Basically, we have $ 7$ marbles in a row, and we want to place two partitions in between them to separate them into three different bags. However, since each bag must have at least one marble, we remove one marble from each bag, and we are left with $ 4$ marbles to work with, with no restrictions. By Balls and Urns, the amount of ways to do it is $ \\displaystyle \\binom{6}{2}\\equal{}\\boxed{15}$.\n\nOne example of a balls and urns arrangement is {1, 2, 4}:\n*|**|****\nIn this case, the * are the balls, and the | are the dividers. The diagram would be interpreted as 1 ball in the red bag, 2 in the white bag, and 4 in the blue bag.[/hide]",
  "Solution_4": "I understand about the partitions, but when u say balls and urns make it 6C2 is that a specific theorem and if so can you show the general rule and derivation, because I dont understand how the number of ways to put 4 balls into 3 different bags is 6C2",
  "Solution_5": "The search function is useful here:\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1603770934&t=137177",
  "Solution_6": "What if you were able to put one bag into another?!\r\n\r\n(This should be interesting...)"
}
{
  "Tag": [
    "geometry",
    "rectangle",
    "similar triangles"
  ],
  "Problem": "Heres a fun problem. Not sure how difficult it is..\r\n\r\nYou have a straight edge that can:\r\na) Connect two points\r\nb) Construct the perpendicular to a line from a given point on that line.\r\n\r\nShow how to construct the perpendicular to a line from a given point *not* on that line.",
  "Solution_1": "P is the point we are trying to find the perpendicular to. See attached.\r\n\r\nEdit: How can I put a picture up in non-spoiler mode?\r\n\r\nEdit: Meh, I'll just leave it for someone else to do anyways.\r\n\r\n-- James",
  "Solution_2": "I think I'm missing something, but:\n\n\n\n[hide]construct any perpendicular to a perpendicular, giving a parallel.  Let that parallel go beyond the original line.  Then we can construct a perpendicular to the parallel beyond the original line.  Or maybe if this is allowed you can extend the original line...[/hide]\n\n\n\nEdit: never mind this post; I misunderstood the problem.",
  "Solution_3": "I don't get what you are saying.. how do you make that go through the given point?\r\n\r\nedit - maybe you misunderstood the question. You have a given line (line, not line segment - lines are infinitely long) and a given point somewhere thats not on the line. You want to construct the line that goes through that point and is perpendicular to the line. I don't see how what you have written works.. you haven't mentioned the point anywhere.",
  "Solution_4": "My solution hinges on the fact that we can construct midpoints, as follows.\r\n\r\nFrom segment AB, draw perpendiculars at A and B, choose a point on one of these, and draw another perpendicular to create rectangle ABCD. Draw diagonals AC and BD, intersecting at E. Draw the line through A perpendicular to AC, and draw the line through B perpendicular to BD, and let these two lines intersect at F. Let segment EF intersect AB at M. M is the midpoint of AB. The proof that this works is straightforward.\r\n\r\nSo, here's the construction. From the point P, draw a line that intersects the line m at A, and draw the line perpendicular to PA through P, and let it intersect line m at B. (If this line doesn't intersect line m, choose a different A). Let C, D, E be the midpoints of AP, PB, and AB respectively, and let F be the midpoint of CP. Let lines CE and DF intersect at G. Draw PG. Draw the perpendicular to PG through P. This is the desired line.\r\n\r\nFor the proof that it works, just use similar triangles to prove that PG is parallel to CD, and so it is parallel to line m.",
  "Solution_5": "A slight variation, using only one midpoint:\r\n\r\nConstructs perpendiculars m and n of AB, with A on m and B on n. Construct line AP and BP, meeting m and n at C and D, respectively. Suppose that |m| < |n|. Construct rectangle ABED (with E on n), thereby constructing the midpoint M of segment AB. Join EM and extend to meet the extension of m at R. Join CR. CR meets AB at S. PS is the desired line.\r\n\r\nProof:\r\n\r\nSuppose that PQ is perpendicular to AB with foot Q. Then, PQ/AD = BQ/BA, BC/PQ = AB/AQ. Multiply to yield BC/AD = BQ/AQ.\r\n\r\nJoin DS. We have DA = EB (rectangle) and EB = AR (EBM :cong: RAM); thus DA = AR. Then, DAS :cong: RAS. Since RAS ~ CBS, we must have DAS ~ CBS. Thus BC/AD = BS/AS.\r\n\r\nThus BQ/AQ = BS/AS; we must conclude that Q and S are the same point, and thus PS is perpendicular to AB.",
  "Solution_6": "P is the point we are trying to find the perpendicular to. See attached.",
  "Solution_7": "[quote=\"Osiris\"]A slight variation, using only one midpoint:\n\nConstructs perpendiculars m and n of AB, with A on m and B on n. Construct line AP and BP, meeting m and n at C and D, respectively. Suppose that |m| < |n|. Construct rectangle ABED (with E on n), thereby constructing the midpoint M of segment AB. Join EM and extend to meet the extension of m at R. Join CR. CR meets AB at S. PS is the desired line.\n[/quote]\r\n\r\nYou can find the midpoint without breaking the rules.  Start with just P and A.  Draw line AP.  Draw a perpendicular to AP at P; call its intersection with the original line B.  Proceed with Osiris's construction until the midpoint is needed.  Now, draw perpendiculars to AP at A and BP at B; they meet at point F.  AFBP is a rectangle; its diagonals bisect each other.  Draw line FP, and call its intersection with AB point M.  Continue with the construction.",
  "Solution_8": "Yeah. I just took the midpoint construction from zabelman's post.",
  "Solution_9": "JamesMatthews, thats a really cool solution!",
  "Solution_10": "[quote=\"zabelman\"]JamesMatthews, thats a really cool solution![/quote]\r\n\r\nHehe, thanks. :)",
  "Solution_11": "My bad.  I only skimmed the posts before replying.  Kudos to JamesMatthews for that simple solution."
}
{
  "Tag": [
    "calculus",
    "integration",
    "probability"
  ],
  "Problem": "Prob. 7  How many ordered integral triples, (a,b, c) have the propery that each number is the product of the other two? I have the answer (which I can pm you) but need to know how it is arrived at, please.\r\n\r\n Prob. 8  Randomly choosing two numbers from the set (1,3, 5, 7, 9) with replacement, what is the probability that the product is greater than 40? Express your answer as a percent. (Again, I have answer, but need \"how to\" solution.",
  "Solution_1": "[quote=\"guyinPA\"]Prob. 7  How many ordered integral triples, (a,b, c) have the propery that each number is the product of the other two? I have the answer (which I can pm you) but need to know how it is arrived at, please.\n\n Prob. 8  Randomly choosing two numbers from the set (1,3, 5, 7, 9) with replacement, what is the probability that the product is greater than 40? Express your answer as a percent. (Again, I have answer, but need \"how to\" solution.[/quote]\r\nThose are from the State test, not the Nat test.\r\n\r\n\r\n[hide=\"#7\"]\n(0,0,0)\n(1,1,1)\n(1,-1,-1) and its variations, 3 in total\n\nSo 5 ORDERED pairs[/hide]\n\n[hide=\"#8\"]\nThere are 5x5 = 25 possible pairs.  The ones with a product greater than 40 are : 9x9, 9x7, 9x5, 7x9, 7x7, 5x9\n\nThats 6 out of 25 or 24%[/hide]\r\n\r\nhope that helps, \r\n\r\nfrost",
  "Solution_2": "I got number 8.\r\n\r\n[hide]24%.  The denominator is 25 because there are 5 numbers with replacement.  For the numerators of problems like these, I always draw a little chart.  So I just did a 5 by 5 grid, numbered the sides 1, 3, 5, 7, and 9, and put the products in the spaces, and then I shaded in the boxes with products higher than 40.  There were 6 (5 x 9, 7 x 7, 9 x 5, 9 x 7, 7 x 9, and 9 x 9), so the answer is 6/25, or 24%.[/hide]\r\n\r\nThis might not be the type of explanation you're looking for, and it might not even be right.  I did the problem really fast. :roll:",
  "Solution_3": "oops, problems were from the state test - sorry for the error -- but thanks for the help in finding the solution!"
}
{
  "Tag": [
    "function",
    "ratio",
    "vector",
    "algebra",
    "rational function",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "Prove or disprove that $ \\mathbb{Z}[[x]] \\cap \\mathbb{C}(x) \\subset \\mathbb{Z}(x)$.  In other words, if $ A(x) \\equal{} a_0 \\plus{} a_1 x \\plus{} a_2 x^2 \\plus{} ...$ is the generating function for an integer sequence and $ A(x)$ is also a rational function, then it must be the ratio of two integer polynomials.",
  "Solution_1": "[quote=\"t0rajir0u\"]Prove or disprove that $ \\mathbb{Z}[[x]] \\cap \\mathbb{C}(x) \\subset \\mathbb{Z}(x)$.  In other words, if $ A(x) = a_0 + a_1 x + a_2 x^2 + ...$ is the generating function for an integer sequence and $ A(x)$ is also a rational function, then it must be the ratio of two integer polynomials.[/quote]\r\nWe effectively have $ \\mathbb{Z}[[x]] \\cap \\mathbb{C}(x) \\subset \\mathbb{Z}(x)$. Indeed, let us write $ A(x) = \\sum_{n \\geq 0}{a_n x^n} = \\frac {P(x)}{Q(x)}$, where $ P,Q \\in \\mathbb{C}[X]$. We write $ Q(x) = \\sum_{n = 0}^d{q_n x^n}$. Let $ (\\omega_1,\\dots,\\omega_k)$ be a base of the vector space $ \\mathbb{Q}(q_0,q_1,\\dots,q_d)$ over the field $ \\mathbb{Q}$. $ \\forall i \\in \\{0,\\dots,d\\}$, we write $ q_i = \\sum_{j = 1}^k{\\lambda_{i,j}\\omega_j}$.\r\nNow, if $ D$ is the degree of $ P$, we know that, $ \\forall n \\geq D+1, \\sum_{i = 0}^d{a_{n + i}q_{d - i}} = \\sum_{j = 1}^k\\left(\\sum_{i = 0}^d{a_{n + i}\\lambda_{d - i,j}}\\right)\\omega_j = 0$, which assures that $ \\sum_{i = 0}^d{a_{n + 1}\\lambda_{d - i,0}} = 0$.\r\n\r\nThen, it follows that, if $ Q_0(x) = \\sum_{i \\geq 0}{\\lambda_{i,0}x^i} \\neq 0$ and $ P_0 = A Q_0 \\in \\mathbb{C}[[x]]$, $ P_0$ is in fact of degree at most $ d + D + 1$, which shows that $ P_0 \\in \\mathbb{C}[x]$. Moreover, $ A$ and $ Q_0$ have rational coefficients, and therefore $ P_0 \\in \\mathbb{Q}[X]$.\r\nTherefore, $ A = \\frac {P_0}{Q_0} \\in \\mathbb{Q}(X) = \\mathbb{Z}(X)$."
}
{
  "Tag": [
    "ratio",
    "geometry",
    "trigonometry",
    "modular arithmetic",
    "trig identities",
    "Law of Cosines",
    "number theory"
  ],
  "Problem": "Ok, the first one I got and think is kinda nice, but the second one.. I have no idea how to approach.\r\n\r\n1.  Quadrilateral $ABCD$ is inscribed in a circle with $AB=8, BC=9, CD=12$, and $DA=5$.  Diagonal $AC$ divides $ABCD$ into two triangles.  Find the ratio of the area of the smaller of the two triangles to the area of the larger one.\r\n\r\n\r\n2.  (source: 2004 senior NSML) A sequence {$a_n$} is defined as follows: $a_n$ is the $n$th digit after the decimal point in the infinite decimal representation of the rational number $\\frac{1}{n}$.  For example, $a_1=0$ since $\\frac{1}{1} = 1.00000...$, and $a_6 = 6$ since $\\frac{1}{6} = .1666666...$, and $a_8 = 0$ since $\\frac{1}{8} = .125000000...$.\r\nThe first ten terms of this sequence are: $0,0,3,0,0,6,1,0,1,0$\r\nHow many of the first 20 terms of this sequence are non-zero?\r\n\r\nThe answer is [hide]8, but I don't know how to arrive at this logically.[/hide]",
  "Solution_1": "Just find the next ten terms for #2. It seems much easier to chug out a little bit of division for a minute or two than to try to find some sort of all-encompasing pattern.",
  "Solution_2": "For number 1, all you have to do is simultaneously apply the law of cosines to find AC, then use Heron's. ;)",
  "Solution_3": "The answer to your quesiton is a little hairy.\r\n\r\nA fact is that the multiplicative order of $10\\pmod{n}$, where $n$ is an integer, is equal to the repeating places in the decimal expansion of the reciprocal of $n$.\r\n\r\nEver notice that a repeating decimal, for example, $0.2525\\ldots = \\frac{25}{99}$, or such that the repeating $n$ digits can be made into a fraction by placing them over $10^n - 1$.\r\n\r\nSo we have the multiplicative order being the smallest $e$ s.t. $10^e \\equiv 1\\pmod{n}$, where $n$ is given. In this case, $n$ is the number which we must find the repeating sequence for $\\frac{1}{n}$.\r\n\r\nBut you would have to know the modular orders of all those number that unless you know them, I forget how you would derive them.\r\n\r\nIf you are desperate enough, just divide the number $n$ into a bunch of repeating 9's and the quotient is the multiplicative order. LOL.",
  "Solution_4": "[quote=\"Elemennop\"]Just find the next ten terms for #2. It seems much easier to chug out a little bit of division for a minute or two than to try to find some sort of all-encompasing pattern.[/quote]\r\n\r\n\r\nLOL. The multiplicative order of 17 is 16, so unless you divide alot! It's going to take alot of scrap paper. And for 19, it's 18. So raw division won't work out too nicely.",
  "Solution_5": "[hide=\"1\"]I get $\\boxed{\\frac{5}{6}}$, by using the law of cosines from both triangles on $AC$ to find $\\cos x$, then using that to find $\\sin x$, then using the formula $\\frac{1}{2}ab\\sin C$ to find the areas of the two triangles (plus the fact that the angles are supplementary and that $\\sin(180^{\\circ}-x)=\\sin x$.  Then divide the two areas to get the answer.  May have screwed up computation somewhere though.[/hide]",
  "Solution_6": "[quote=\"eryaman\"][hide=\"1\"]I get $\\boxed{\\frac{5}{6}}$, by using the law of cosines from both triangles on $AC$ to find $\\cos x$, then using that to find $\\sin x$, then using the formula $\\frac{1}{2}ab\\sin C$ to find the areas of the two triangles (plus the fact that the angles are supplementary and that $\\sin(180^{\\circ}-x)=\\sin x$.  Then divide the two areas to get the answer.  May have screwed up computation somewhere though.[/hide][/quote]\r\n\r\nThat's right, but keeping in mind that $\\sin(180^{\\circ}-x)=\\sin x$ can you see a much quicker way through it just using ratios instead of calculating the areas?\r\n\r\n\r\nAbout the second one, I guess 1/17 and 1/19 are the only ones that are terrible, and the rest of the contest was pretty easy that year. . . but if that's the way they intended it to be solved, I'd say that's a pretty bad problem.",
  "Solution_7": "[quote=\"white_horse_king88\"][quote=\"Elemennop\"]Just find the next ten terms for #2. It seems much easier to chug out a little bit of division for a minute or two than to try to find some sort of all-encompasing pattern.[/quote]\n\n\nLOL. The multiplicative order of 17 is 16, so unless you divide alot! It's going to take alot of scrap paper. And for 19, it's 18. So raw division won't work out too nicely.[/quote]\r\n\r\nSure, but which do you think will generally end up being faster -- sitting for a couple minutes trying to find some distinct pattern that might point in the direction of a quicker way to handle this, or just flatout dividing in these numbers (and dividing 19 and 17 isn't terribly bad either)? I rarely get the most elegant approach my first time through a problem, especially during something like the AMC. There just isn't enough time to waste working on elegance when the solution can be brute forced just as quick or quicker most of the time, and I'm sure most other high scorers would agree.",
  "Solution_8": "[quote=\"djshowdown2\"][quote=\"eryaman\"][hide=\"1\"]I get $\\boxed{\\frac{5}{6}}$, by using the law of cosines from both triangles on $AC$ to find $\\cos x$, then using that to find $\\sin x$, then using the formula $\\frac{1}{2}ab\\sin C$ to find the areas of the two triangles (plus the fact that the angles are supplementary and that $\\sin(180^{\\circ}-x)=\\sin x$.  Then divide the two areas to get the answer.  May have screwed up computation somewhere though.[/hide][/quote]\n\nThat's right, but keeping in mind that $\\sin(180^{\\circ}-x)=\\sin x$ can you see a much quicker way through it just using ratios instead of calculating the areas?[/quote]Oh I see, you can just use the $\\frac{1}{2}ab\\sin x$ on both triangles (since $\\sin(180^{\\circ}-x)=\\sin x$, and divide the two, in which the $\\sin x$ part cancels out, leaving you with your ratio.",
  "Solution_9": "[quote=\"Elemennop\"][quote=\"white_horse_king88\"][quote=\"Elemennop\"]Just find the next ten terms for #2. It seems much easier to chug out a little bit of division for a minute or two than to try to find some sort of all-encompasing pattern.[/quote]\n\n\nLOL. The multiplicative order of 17 is 16, so unless you divide alot! It's going to take alot of scrap paper. And for 19, it's 18. So raw division won't work out too nicely.[/quote]\n\nSure, but which do you think will generally end up being faster -- sitting for a couple minutes trying to find some distinct pattern that might point in the direction of a quicker way to handle this, or just flatout dividing in these numbers (and dividing 19 and 17 isn't terribly bad either)? I rarely get the most elegant approach my first time through a problem, especially during something like the AMC. There just isn't enough time to waste working on elegance when the solution can be brute forced just as quick or quicker most of the time, and I'm sure most other high scorers would agree.[/quote]\r\n\r\nI'm pretty sure there's a way to calculate it with primitive roots, but I don't know if I would sit there and go through 18 divisions for 1/19. I mean, I guess one could do both in about 5 minutes, but you can't make an error either.",
  "Solution_10": "for any numbers in the sequence to be zero $n$ must be be divisible by only $2$ and $5$ so the numbers $2, \\ 4, \\ 5, \\ 8, \\ 10, \\ 16, \\ 20$ end with a $0$ in the sequence. The rest are non-zero numbers.\r\n\r\nThe only exception is $1$",
  "Solution_11": "[quote=\"ritchjp\"]for any numbers in the sequence to be zero $n$ must be be divisible by only $2$ and $5$ so the numbers $2, \\ 4, \\ 5, \\ 8, \\ 10, \\ 16, \\ 20$ end with a $0$ in the sequence. The rest are non-zero numbers.\n\nThe only exception is $1$[/quote]\r\n\r\nThis isn't true. Please read above.",
  "Solution_12": "I did a little more investigating and fortunately I had access to a calculator that went out to about 30 decimal places.\r\n\r\nWhat I found out was rather interesting.\r\n\r\nIn my last post I commented on the numbers that will yield a $0$ and I came up with numbers that were divisible by $2$ and $5$ only. That was only partially true.\r\n\r\nAdditional numbers I found that yielded a $0$ in the sequence are $11, \\ 13, \\ 17, \\ 19$ \r\n\r\nThe numbers with non-zero results were $12, \\ 14, \\ 15, \\ 18$\r\n\r\nDo you see a pattern?\r\n\r\nThe numbers that had non-zero results have two or more prime factors in them while the numbers that had results of $0$ in the sequence were unfactorable numbers. That may be why $1$ yields a $0$ in the sequence.\r\n\r\nSo  there are 8 non-zero numbers in the sequence.",
  "Solution_13": "[quote=\"ritchjp\"]I did a little more investigating and fortunately I had access to a calculator that went out to about 30 decimal places.\n\nWhat I found out was rather interesting.\n\nIn my last post I commented on the numbers that will yield a $0$ and I came up with numbers that were divisible by $2$ and $5$ only. That was only partially true.\n\nAdditional numbers I found that yielded a $0$ in the sequence are $11, \\ 13, \\ 17, \\ 19$ \n\nThe numbers with non-zero results were $12, \\ 14, \\ 15, \\ 18$\n\nDo you see a pattern?\n\nThe numbers that had non-zero results have two or more prime factors in them while the numbers that had results of $0$ in the sequence were unfactorable numbers. That may be why $1$ yields a $0$ in the sequence.\n\nSo  there are 8 non-zero numbers in the sequence.[/quote]\r\n\r\nThis pattern can't be taken on faith - it requires an explanation, which is relatively simple.  For prime numbers $> 10$ the first digit in the decimal expansion is $0$, and the $n^{th}$ digit will always be congruent to the $1^{st}$ digit $\\bmod \\varphi(p)$ (by 'congruent' I mean they occupy the same place in the repeating decimal expansion.)\r\n\r\nFor composite numbers $> 10$ it is untrue that $1 \\equiv n \\bmod \\varphi(n)$ and hence there is not a guarantee for the digit to be $0$ (however, I suspect it's still possible - check $\\frac{1}{34}$)."
}
{
  "Tag": [],
  "Problem": "Ok, muchos me han preguntado cuando iba a volver: La respuesta esta a la vuelta de la esquina, el reto de la semana empieza de nuevo inmediatamente que reciba 15 posts de personas diferentes en este topic diciendo que participaran!!!\r\n\r\nEspero que quienes posteen se comprometan a enviar sus soluciones!!!\r\n\r\nLas reglas son sencillas: Cuelgo 4 problemas un dia, tienen una semana para enviarme sus soluciones a pascual2k1@hotmail.com en cualquier formato y luego de esa semana cuelgo otro conjunto de problemas, los resultados que llevaremos en un ranking y discutimos las soluciones...\r\n\r\nLas inscripciones estan abiertas, para participar solo envie un mail con el nombre : Reto de la semana y sus respectivas soluciones especificando que problema resolvera.\r\n\r\nSoluciones parciales seran calificadas tambien!!!\r\n\r\nOk, espero esas 15 posts que encesito para asegurarme que esta vez no sera un fracaso como ediciones pasadas en las que tuvimos solo 1 participante... Si no las hay el reto de la semana no se hara!!! Si no se ve reflejada una alta participacion el reto de la semana se detendra, esas son las reglas!!!\r\n\r\nEso es todo, espero ver respuestas...",
  "Solution_1": "Aprovechare la semana de descanso para participar :D .",
  "Solution_2": "Yo tambien participo.",
  "Solution_3": "faltan 13",
  "Solution_4": "agora faltam 12  :D",
  "Solution_5": "Vale yo me anoto, una buena oportunidad para entrenar.",
  "Solution_6": "Yo tambi\u00e9n me les uno. Est\u00e1 perfecto pera no perder el ritmo",
  "Solution_7": "No hay limite de edad! asi que vamos a aprender todos de todos!\r\n\r\nSolo va a haber un nivel....",
  "Solution_8": "[quote=\"Pascual2005\"]\nSolo va a haber un nivel....[/quote]\r\n\r\nSi es as\u00ed, yo tambi\u00e9n.\r\n\r\n$Tipe$",
  "Solution_9": "yo tambi\u00e9n me uno...  :D",
  "Solution_10": "Sale, yo tambien entro",
  "Solution_11": "yo tambien",
  "Solution_12": "tengo poca experiencia pero de todas manera hare el intento  :blush:",
  "Solution_13": "Va, yo le entro... Ya no soy ol\u00edmpico pero necesito ver problemas, porque soy entrenador  :agent:  :D",
  "Solution_14": "Ola, yo tambien entro.\r\n \r\nFABY   :roll:",
  "Solution_15": "Yo tambien, pero hay que trabajar enserio $\\bigstar$",
  "Solution_16": "Hola a todos,\r\nsaludo la buena iniciativa .. Pascual ;)\r\n\r\nveo q ya van 13 personas ...\r\nsolo hacen faltan  2  m\u00e1s .... \r\n\r\ny si me permiten ... yo tambien me agrego ..\r\n(solo faltaria uno :) )\r\n\r\nCarlos Bravo\r\nLima  -PERU",
  "Solution_17": "Ok, ma\u00f1ana pondre la primera lista de problemas tratare de seleccioanrlos cuidadosmanente para que sean lo mas interesantes posibles y saquemos todo el provecho de ellos!!!\r\n\r\nChao y nos vemos ma\u00f1ana!!!",
  "Solution_18": "ok, los problemas ya estan posteados en la pagina de Olimpiadas de matem\u00e1ticas!!! Suerte!!!",
  "Solution_19": "AKI EL ENLACE:\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=103444\r\n\r\n\r\nSALUDOS,"
}
{
  "Tag": [
    "inequalities proposed",
    "inequalities"
  ],
  "Problem": "Prove that\r\n$\\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a}+\\frac{9}{2}\\sqrt{\\frac{ab+bc+ca}{a^{2}+b^{2}+c^{2}}}\\geq\\frac{9}{2}+3$\r\nwhere $a,b,c$ are positive real numbers",
  "Solution_1": "The following is sharper\r\n\\[\\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a}+\\frac{14}{3}\\sqrt{\\frac{ab+bc+ca}{a^{2}+b^{2}+c^{2}}}\\ge \\frac{23}{3}\\]\r\n:)"
}
{
  "Tag": [
    "Putnam",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Hi, \r\nhere's a problem I think was not yet proposed here:\r\n\r\nEvaluate\r\n\r\n$\\sum_{n=1}^{\\infty}\\sum_{m=1}^{\\infty}\\frac{m^2n}{3^m(3^m n+3^n m)}$\r\n\r\n\r\nIt has a pretty solution.\r\n\r\nRef: Putnam, 199?",
  "Solution_1": "Let ai=3 <sup>i</sup>/i. \r\nWe must evaluate  S=\\sum 1/ai(ai+aj).\r\nWe have S= \\sum 1/aj(1/ai-1/(ai+aj))= \\sum 1/(ai*aj)- \\sum 1/(aj*(ai+aj))[b](*)[/b]= \\sum 1/(ai*aj)-S, so 2*S= \\sum 1/(ai*aj)= \\sum ij/3 <sup>i+j</sup>[b](*)[/b].\r\nBy grouping the terms with the same i+j [b](*)[/b], we get\r\n2*S= \\sum {n>1} (n+1)n(n-1)/(6*3 <sup>n</sup>), so\r\n108*S= \\sum (n+1)n(n-1)/3 <sup>n-2</sup>=f'''(1/3), where\r\nf(x)= 1+x+x <sup>2</sup> +...=1/(1-x) for x in (-1,1).\r\nf'''(x)=6/(1-x) <sup>4</sup>, so we have 108S=3 <sup>5</sup>/8, so S=9/32.\r\n\r\nLets prove that  T=\\sum {i+j=n} ij=n(n+1)(n-1)/6.\r\nWe have T= \\sum i(n-i)= n*\\sum i- \\sum i <sup>2</sup> , and the result easily follows.\r\n\r\nI used some uncertain facts. For them to be true I need to know that the sum is convergent. I hope it is. I don't know if I can prove it because the writing  \\sum {ij} is not explicit. I need something like  \\sum {i} \\sum {j} or  \\sum {i+j=n}.\r\n\r\nIf the sum is not known to be convergent then I can not group terms in the sum. It is a well known result of Riemman that states that is a series is such that it is not absolutely convergent then we can permute the terms of the series and obtain any sum for the series.     \r\nWhere I used (*), the result depends on the convergence of the series.\r\n\r\nI think that my proof holds if we have the sum in the form  \\sum {i+j \\geq 2} ..."
}
{
  "Tag": [
    "inequalities",
    "rearrangement inequality",
    "inequalities proposed"
  ],
  "Problem": "Hey guys, i came across this one recently and and am drawing a blank. Ideas?\r\nLet a1, a2, . . . , an be strictly positive real numbers. Show that\r\n\r\na1 + a2 + \u00b7 \u00b7 \u00b7 + an\u22121 + an <= ((a1)^2)/a2 + ((a2)^2)/a3) +...+ ((an)^2)/a1\r\n\r\nhave fun!",
  "Solution_1": "Since the sequences $ (a_1^2,a_2^2,\\dots,a_n^2)$ and $ \\left(\\frac{1}{a_1},\\frac{1}{a_2}\\dots,\\frac{1}{a_n}\\right)$ are oppositely sorted, from Rearrangement inequality we get\r\n\\[ \\frac{a_1^2}{a_2}\\plus{}\\frac{a_2^2}{a_3}\\plus{}\\dots\\plus{}\\frac{a_n^2}{a_1}\\ge\\frac{a_1^2}{a_1}\\plus{}\\frac{a_2^2}{a_2}\\dots\\plus{}\\frac{a_n^2}{a_n}\\equal{}a_1\\plus{}a_2\\plus{}\\dots\\plus{}a_n\\]",
  "Solution_2": "$ (\\frac{a_1^2}{a_2}\\plus{}\\frac{a_2^2}{a_3}\\plus{}\\dots\\plus{}\\frac{a_n^2}{a_1})(a_2\\plus{}a_3\\plus{}\\dots\\plus{}a_n\\plus{}a_1)\\geq(a_1\\plus{}a_2\\plus{}a_3\\plus{}\\dots\\plus{}a_n)^2$"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "search",
    "quadratics",
    "algebra unsolved"
  ],
  "Problem": "The real polynomials p(x), q(x), r(x) have degree 2, 3, 3 respectively and satisfy p(x) <sup>2</sup>  + q(x) <sup>2</sup>  = r(x) <sup>2</sup> . Show that either q(x) or r(x) has all its roots real.",
  "Solution_1": "As far as I remember, I have posted a solution to a more general problem posted by Namdung. Try to search for it....",
  "Solution_2": "Since $\\deg r=3$ we conclude $r(x)$ has a root $a$, and it follows $p$ and $q$ also have this root. \r\nSuppose the contrary that $q(x)=(x-a)Q(x)$ and $r(x)=(x-a)R(x)$ where $Q$ and $R$ are positive quadratic polynomials. Then $p(x)=c(x-a)(x-b)$ and $c^{2}(x-b)^{2}=(R-Q)(Q+R)$. As $\\deg (Q+R)=2$ we conclude $\\deg (Q-R)=0$, so $c^{2}(x-b)^{2}$ positive OR negative on the $R$. Contradiction.",
  "Solution_3": "http://www.mathlinks.ro/Forum/viewtopic.php?highlight=roots+real+number&t=2857"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "algebra proposed"
  ],
  "Problem": "Find all $P(x)\\in R[x]$ such that\r\n$P(x)P(x+1) = P(x^2)$",
  "Solution_1": "If $x$ is a root of the polynomial (as a complex polynomial if needed) then $x^2$ and $(x-1)^2$ are also roots.\r\nSince there are a finite number of roots, either P is constant (and $P=0$ or$P=1$) either the only possible roots are $0$ and $1$ (by looking for instance to the norm of the complex roots)\r\n\r\n$P = X^n (X-1)^p$ then it's easy to see that $n=p$\r\n\r\n$0$, $1$, $P = X^n (X-1)^n$ are the solutions."
}
{
  "Tag": [
    "rotation",
    "geometry",
    "geometric transformation",
    "trigonometry",
    "trig identities",
    "Law of Cosines"
  ],
  "Problem": "$P$ is inside an equalateral triangle $ABC$ such that $PA=2$, $PB=2\\sqrt{3}$, $PC=4$.\r\nFind the side length of $\\triangle{ABC}$.",
  "Solution_1": "[hide=\"steps\"]1. rotate about a vertex, \n2. look at smaller equilateral triangle formed by the interior point, center of rotation and its image, \n3. look for a right triangle with the lengths of the sides given in the problem(one side from the previously mentioned triangle, one from the original triangle, 1 from the image)\n4. use the law of cosines to finish it\n[/hide]\n\n[hide=\"answer\"]after doing this, i got $2\\sqrt{7}$[/hide]",
  "Solution_2": "I just realized I made a mistake, forget it.",
  "Solution_3": "[quote=\"FMako\"]For all triangles there is a convenient formula:\n\\[ 1+\\frac{r}{R}= \\sum \\cos{A} \\]\n[/quote]\r\nIn this problem, $A=B=C=60^o$.",
  "Solution_4": "I was thinking somehow I could apply that to the interior angles, that sum to 360.",
  "Solution_5": "I'll write a complete solution for this problem based on [b]Altheman's[/b] idea.\r\n\r\nWe make a rotations transformation $R: (C, 60^o)$ so that $P \\to P'$. Connect $PP'$, $PC$ and $PB$.\r\n\r\nSince $\\angle{P'CA}=60^o-\\angle{ACP}=\\angle{BCP}$, $BC=AC$, $CP=CP'$, so $\\triangle{CBP} \\cong \\triangle{CP'A}$, which gives $PB=P'A$. Now observe $\\triangle{CPP'}$, we find that CP=CP', $\\angle{PCP'}=60^o$, hence $\\triangle{CPP'}$ is an equilateral triangle. Therefore $PP'=CP=4$.\r\n\r\nIn triangle APP', we have $PA=2$, $AP'=BP=2\\sqrt{3}$. Thus $\\triangle{APP'}$ is an right triangle with $\\angle{AP'P}=30^o$. Since $\\angle{CP'P}=60^o$, we get that $\\angle{CP'A}=90^o$. Therefore in right triangle CP'A, $CP'=4$, $AP'=2\\sqrt{3}$, thus:\r\n\\[ AC=\\sqrt{P'C^2+P'A^2}=\\sqrt{16+12}=2\\sqrt{7} \\]",
  "Solution_6": "I was thinking of manipulating the fact that if D, E, and F are feet of altitudes from P on BC, AC, AC respectively, $|PD| + |PE| + |PF| = \\sqrt{3} |AB|$, and use Pythagoras and/or Apollonius to find $|AF| + |BF|$, etc. to find $|AB|$, $|AC|$ and $|BC|$.\r\n\r\nI'll come back to it later."
}
{
  "Tag": [
    "advanced fields",
    "advanced fields unsolved"
  ],
  "Problem": "1. Self-reference statements are meaningless.\r\n\r\n2. There is one or more true statements.\r\n\r\nIf 1 is true then 2 can not be true. If 2 were true then it would be self-reference. If 2 can not be true then 1 can not be true. If 1 were true then there would be a true statement. That would mean 2 is true. If 1 is true then 2 can not be true.\r\n\r\n1 CAN NOT BE TRUE.",
  "Solution_1": "First, the word in English is \"self-referential.\"  Second, it looks to me from your posting that you don't really understand what this means.  What do you think the definition of self-referential is?",
  "Solution_2": "If 2 says \"There is one or more true statements.\"  and 2 is itself a true statement then 2 is a self reference statement.",
  "Solution_3": "I will explain it better.  2 refers to the group of all true statements.  2 says this group contains 1 or more members.  If 2 is a true statement then 2 is a member of the group it is referring to.  I came up with this paradox to show that if self-reference statements are meaningless then a contradiction occurs if there are any true statements (including 1).  On one hand you have a true statement on the other hand 2 is not true.\r\nThe purpose of 1 was to save the truth.  1 prevents there from being any truths.  You can replace 1 with \"no statement about all members of a group can be true\"  The same problem will occur.\r\n\r\n\"no statement about all members of a group can be true\"\r\n\r\nThe above statement is about all members of the group of statements about all members of a group. The statement itself is a statement about all members of a group. It is a member of the group it is referring to. It is a self-reference statement that declares itself to be untrue.",
  "Solution_4": "by your explaination, statement 2 is not self-referential.(my own oppinion^!^) it says something about this \"group\" but itself. And (even) if self-referential can be defined in this \"group\", statement 1 is not self-referential too.\r\n\r\nI think You should make a difference between \"our nature language\" and \"logic language\".",
  "Solution_5": "I can write it   \" statements 1,2,3... are true statements.  1,2,3... are all the true statements.  If this statement is true then it has a number. It is declaring itself true.\r\n\r\nI can write it \" 1 or more true statements exist\"  If it is true then it says it exists.  This is the same statement I wrote.",
  "Solution_6": "sorry\r\nI can write it \" statements 1,2,3... are true statements.\" 1,2,3... are all the true statements. If this statement is true then it has a number. It is declaring itself true. \r\n\r\nNo 1 does not refer to itself.  I wish it were that easy.  It took along time for me to come up with a problem with 1.  I understand my problem will not end all disputes.  If 1 is  \" no statement about all members of a group can be true\" then 1 is self reference.\r\n\r\na.no statement about all members of a group can be true.\r\nb.self-reference statements are meaningless.\r\nIf a is true then b is not.  if b is true then a is not.",
  "Solution_7": "What do you mean? It seems to me that you think others thought self-reference statements are meaningless, you disagree and point out how this can be \"proved\"?\r\n\r\nWhether a sentence is self-reference depends on \"us\", i.e depends on how do you make the connection between the structure and the theory. You could not tell (via syntactic knowledge) whether a Turing machine use its own code, you could not tell (via syntactic knowledge) whether a turing machine is solving 3-SAT or other NP-complete problem, because the content of the problem is explained by \"us\", i.e another theory (connected with the structure in some way) .",
  "Solution_8": "I found this problem was already discovered it is the Russell-Myhill paradox."
}
{
  "Tag": [
    "linear algebra",
    "matrix",
    "group theory",
    "abstract algebra",
    "linear algebra solved"
  ],
  "Problem": "Let $G$ be a finite subgroup of $\\mathrm{GL}_n(\\mathbb C)$ (this is the group of all $n\\times n$ matrices with complex coefficients with determinant different from $0$).\n\nThe sum of the matrices $g$ in $G$ is zero iff the sum of their traces $\\mathrm{tr}(g)$ is zero.",
  "Solution_1": "Suppose that $A=\\sum_{g\\in G}g$ has trace zero. Since $Ag=A$ for any $g\\in G$, summation yields $A^{2}=|G|A$. Therefore, each eigenvalue of $A$ is equal to either $0$ or $|G|$. But the trace of $A$ is zero, hence all eigenvalues are $0$, which means that $A$ is nilpotent. Since also $A^{n}=|G|^{n-1}A$, we conclude that $A$ is the zero matrix.",
  "Solution_2": "can some explain why for all g $ Ag\\equal{}A$ ? and why if A has trace 0 then his all eingen value are 0?  :blush:",
  "Solution_3": "$ A$ is the sum of all elements of $ G$. Multiplying each element of a finite group by $ g$ on the right, you get the same elements written in a different order. In other words, multiplication by $ g$ amounts to permutation of group elements. \r\n\r\nKnowing that (i) each eigenvalue of $ A$ is either $ 0$ or $ |G|$, and\r\n(ii) the sum of all eigenvalues (i.e. the trace) is $ 0$,\r\nit should not be hard to see that each eigenvalue is $ 0$."
}
{
  "Tag": [],
  "Problem": "agha inam az nazarie adaad...\r\nmikhaym ye shoori be pa konim...hamaeye masale haye khafano inja ba ham hal konim,...\r\nfaghat man hamin tor ke goftam momkene ye 2,3 rooz dige ta ye zamaane namaloom beram va on nasham (be khaatere masaaele siyasi :rotfl:  :ninja: )...\r\nbedrood",
  "Solution_1": "bashe masalato begoo bad boro!!\r\nbaba siasi",
  "Solution_2": "first problem:\r\n1)farz konid yeki az jomalaate tasaaodi hesaabi va namotanahi az adaade tabii,morabbae kaamel bashad.saabet konid dar in tasaaod bi nahaayat jomle vojood daarad ke morabbae kamel and.",
  "Solution_3": "[quote=\"pirhoosh\"]bashe masalato begoo bad boro!!\nbaba siasi[/quote]\r\nagha zire pato negah koni ma ro mibini...\r\nmokhlese hamegiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii..iiiiiii.iiii....iiiii....(sookhtam tamoom shod :rotfl: )\r\nta hafateye bad khodahaafez...",
  "Solution_4": "[quote=\"mehdi\"]first problem:\n1)farz konid yeki az jomalaate tasaaodi hesaabi va namotanahi az adaade tabii,morabbae kaamel bashad.saabet konid dar in tasaaod bi nahaayat jomle vojood daarad ke morabbae kamel and.[/quote]\r\ninam az javab:\r\nfarz konim $K^2$ avalin morabae kamel dar donbale bashad va $d$ ghadr nesbate donbale bashad pas:$K^2+d(2Km+dm^2)=(K+md)^2$ be ezaye tamame adde tabiyie $m$ ham morabae kamel ast va ham ozvi az donbale. ;)",
  "Solution_5": "inam az masaleye ba'd:",
  "Solution_6": "[color=cyan][size=150] Problem 2 [/color][/size]\r\ninam ye masaleye sade az short list:\r\n$s(n)$ majmoe maghsoomo alah haye adade tabiyiye $n$ minamim.\r\nsabet konid namotanahi $n$ vojood darad be goonei ke $\\frac{s(n)}{n}\\geq \\frac{s(m)}{m}$ be ezaye tamame $m$ haye tabi'ye kochektar az $n$.",
  "Solution_7": "[quote=\"mehdi\"]first problem:\n1)farz konid yeki az jomalaate tasaaodi hesaabi va namotanahi az adaade tabii,morabbae kaamel bashad.saabet konid dar in tasaaod bi nahaayat jomle vojood daarad ke morabbae kamel and.[/quote]\r\nin soal ro man hal kardam ke kheili sakht bood albate shoma nemitoonin halesh konin chon bayad mese man herfei bashin  :mad:    :arrow:  :idea:  :huh:  :P  :blush: hjfhgjhjg:",
  "Solution_8": "[quote=\"doogh khore herfei\"][quote=\"mehdi\"]first problem:\n1)farz konid yeki az jomalaate tasaaodi hesaabi va namotanahi az adaade tabii,morabbae kaamel bashad.saabet konid dar in tasaaod bi nahaayat jomle vojood daarad ke morabbae kamel and.[/quote]\nin soal ro man hal kardam ke kheili sakht bood albate shoma nemitoonin halesh konin chon bayad mese man herfei bashin  :mad:    :arrow:  :idea:  :huh:  :P  :blush: hjfhgjhjg:[/quote]\r\nbror baba az key ta hala kombooze oomade ghati miveha?\r\nage ye zare nega koni mibini ke soalesh hamchinam sakht nabood va dar zemn hame gheyr az to mitoonan hallesh konan va dar zemn mohammad hosein hallesh karde tamoomm.\r\nbadesh ham enghadr alaki khali bandi nakon ke man badam miyad az khali bandi.  :mad:",
  "Solution_9": "[quote=\"mehdi\"][quote=\"doogh khore herfei\"][quote=\"mehdi\"]first problem:\n1)farz konid yeki az jomalaate tasaaodi hesaabi va namotanahi az adaade tabii,morabbae kaamel bashad.saabet konid dar in tasaaod bi nahaayat jomle vojood daarad ke morabbae kamel and.[/quote]\nin soal ro man hal kardam ke kheili sakht bood albate shoma nemitoonin halesh konin chon bayad mese man herfei bashin  :mad:    :arrow:  :idea:  :huh:  :P  :blush: hjfhgjhjg:[/quote]\nbror baba az key ta hala kombooze oomade ghati miveha?\nage ye zare nega koni mibini ke soalesh hamchinam sakht nabood va dar zemn hame gheyr az to mitoonan hallesh konan va dar zemn mohammad hosein hallesh karde tamoomm.\nbadesh ham enghadr alaki khali bandi nakon ke man badam miyad az khali bandi.  :mad:[/quote]\r\nman ke khali nabastam.nemidoonam chera migi bastam.badesh ham ba to naboodam ke bem begi.badesh ham man moaddealam shod 17.30 pas ba man injoori harf nazan chon ke man kheili ba hoosham.vali chon midoonam ke to tazekari bet chizi nemigam.fghzgfhfghfhzfhghfhfhfhfh",
  "Solution_10": ":rotfl:  :rotfl:        \r\nye nokta ro eshare nakardi azizam...\r\nto tooye doogh khordan ham kheili maharat dari :rotfl: \r\nkheili ham bahooshi chon ke hatman emtehan varzeshi chizi az 15 balatar shode :rotfl: \r\nbadesh ham age oghde type kardan dari boro vaghti computer khamooshe bezan rooye keyboard :lol: \r\ndige ham inja chertopert nanvesi vagarna be moderator migam ke bokonadet biroon.\r\nchon inja jaye bahse elmie na injoor araajif :mad:",
  "Solution_11": "doostan behtare bepardaazim be soale agha hosein"
}
{
  "Tag": [
    "AMC",
    "USA(J)MO",
    "USAMO",
    "AIME"
  ],
  "Problem": "Hi, \r\nI'm currently living in Canada, but I'm technically American (cause I was born in California...)\r\n\r\nSo, I was wondering, am I eligible for the USA's IMO team as long as I do good on the AMC, AIME and the USAMO??\r\n\r\nOr, do I have to be LIVING in the USA to be eligible?\r\n\r\n :?:  :?:",
  "Solution_1": "There was  Canadian at MOP in my time; he was a citizen of both countries, and later competed on Canada's IMO team. I think you're eligible, although you're likely also eligible for the Canadian team, and that one's easier to reach.",
  "Solution_2": "The only place that I could find that answers this question is wikipedia, however they don't cite any sources for it\r\n\r\n[quote]Only U.S. residents and [b]citizens[/b] may join the American IMO team[/quote]"
}
{
  "Tag": [
    "geometry",
    "parallelogram",
    "analytic geometry",
    "graphing lines",
    "slope",
    "AMC",
    "USA(J)MO"
  ],
  "Problem": "If you are given any quadrilateral and you connect the midpoints of each of its sides, a parallelogram is formed. Prove why this phenomenon works.\r\n\r\n\r\nI will give my solution after people have posted theirs.",
  "Solution_1": "Proving's kinda hard\r\n\r\nbut my explanation is \" isn't it obivious\" ?\r\n\r\nif you have a quadrialteral, and connect each sides' midpoints you get a shape of diamond....\r\n( wow it's really hard to explain... arg!! )\r\n\r\nI drew a picture instead\r\n\r\nundcerstand this",
  "Solution_2": "I don't know the proof, so I'll make one up...\r\n\r\nYou have 4 points, $(a_1,a_2),(a_3,a_4),(a_5,a_6),(a_7,a_8)$... By connecting the midpoints, you obtain these points:\r\n\r\n$(\\frac{a_1+a_3}{2},\\frac{a_2+a_4}{2}),(\\frac{a_3+a_5}{2},\\frac{a_4+a_6}{2}),(\\frac{a_5+a_7}{2},\\frac{a_6+a_8}{2}),(\\frac{a_7+a_1}{2},\\frac{a_8+a_2}{2})$\r\n\r\nTo make it a paralellogram, both pairs of opposite sides must be congruent and parallel... \r\n\r\nSince using the distance formula would be a little messy here, I will find the slopes...\r\n\r\nSlope between first and second point above:\r\n\r\n$\\frac{\\frac{a_6-a_2}{2}}{\\frac{a_5-a_1}{2}}=\\frac{a_6-a_2}{a_5-a_1}$\r\n\r\nSlope between third and fourth point above:\r\n\r\n$\\frac{\\frac{a_6-a_2}{2}}{\\frac{a_5-a_1}{2}}=\\frac{a_6-a_2}{a_5-a_1}$\r\n\r\nAnd what do you know? It's the same slope!\r\n\r\nNow let's try doing the other pair of opposite sides:\r\n\r\nThe first pair (midpoint of 2nd and 3rd point):\r\n\r\n\r\n$\\frac{\\frac{a_8-a_4}{2}}{\\frac{a_7-a_3}{2}}=\\frac{a_8-a_4}{a_7-a_3}$\r\n\r\nNow for the other pair:\r\n\r\n\r\n$\\frac{\\frac{a_8-a_4}{2}}{\\frac{a_7-a_3}{2}}=\\frac{a_8-a_4}{a_7-a_3}$\r\n\r\nYay!!!\r\n\r\nBoth pairs of opposite sides are parallel!!! That means we have ourselves a parallelogram!\r\n\r\n:coolspeak:",
  "Solution_3": "This is actually a theorem called Varignon's Theorem (1654-1722).\r\n\r\nThere are many ways and one easy way I can think is using mass point. But that's way above this level so I'll not post the solution using it.",
  "Solution_4": "i like bob's proof",
  "Solution_5": "Refer to shinwoo's picture, because I'm too lazy to draw one myself. Label the four vertices A, B, C, and D in order (aren't I creative?). Draw auxiliary segments AC and BD. Now in triangles ABC and BCD, the side of the new quadrilateral connects the midpoints of the two sides. Therefore we should probably use the midpoint connector theorem (I wonder why...) and we can prove that the opposite sides of the small quadrilateral are both parallel to auxiliary segment AC, and likewise for the other pair. Therefore the small quadrilateral is a parallelogram, QED.\r\n\r\nI don't know if that helps. I could have cleared that up considerably if I had drawn a picture with labeled points.",
  "Solution_6": "[quote=\"mathnerd314\"]I don't know if that helps. I could have cleared that up considerably if I had drawn a picture with labeled points.[/quote]\r\n\r\ndo it!",
  "Solution_7": "Here's the pic. I still didn't label the points  :D  :lol:  :blush:  :mad:",
  "Solution_8": "thanks...im pretty sure i get it now...\r\n\r\ndo u  guys just draw things in paint?",
  "Solution_9": "Ok, here's how I proved it. First, i connected the opposite midpoints, which formed two diagonals. Then, I used the sides of the \"parallelogram\" as midsegments of triangles. Eventually I proved that each of them was parallel to the diagonal and using the transitive property, they are congruent to each other.\r\n\r\nI draw my figures with Macromedia Fireworks.  :lol:",
  "Solution_10": "just wondering",
  "Solution_11": "[quote=\"mathnerd314\"]Here's the pic. I still didn't label the points  :D  :lol:  :blush:  :mad:[/quote]\r\n\r\nI did it the exact same way as you.  Its the easiest since it doesn't involve messy algebra",
  "Solution_12": "y did u quote that?",
  "Solution_13": "DOn't know.  Can someone tell me what \"QED\" means and what its for?",
  "Solution_14": "[quote=\"nat mc\"]DOn't know.  Can someone tell me what \"QED\" means and what its for?[/quote]\r\n\r\nIt means Quantum Electrodynamics. Just kidding, most mathematicians would most likely think of 'Quod erat demonstrandum,' \"which was to be proved\" in Latin. You write it at the end of proofs.",
  "Solution_15": "ok\r\n\r\ni was wondering y that is in ur siggy",
  "Solution_16": "Do people put QED at the end of their USAMO proofs?",
  "Solution_17": "You don't have to, but you can.",
  "Solution_18": "I will next year",
  "Solution_19": "He quoted it cause he can't quote the image.",
  "Solution_20": "[quote]Ok, here's how I proved it. First, i connected the opposite midpoints, which formed two diagonals. Then, I used the sides of the \"parallelogram\" as midsegments of triangles. Eventually I proved that each of them was parallel to the diagonal and using the transitive property, they are congruent to each other.\n\nI draw my figures with Macromedia Fireworks. Very Happy[/quote]\r\n\r\nthat's not actually a proof\r\n\r\nthat's how it works\r\n\r\nwhat Bob said is a full proof",
  "Solution_21": "Of course its a proof, just because it doesn't have algebra doesn't mean its not a proof"
}
{
  "Tag": [
    "trigonometry",
    "integration",
    "real analysis",
    "probability",
    "function",
    "expected value",
    "absolute value"
  ],
  "Problem": "version of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=162601]counterexample to law of large numbers[/url] without details of $ X$.\r\n\r\nLet $ X_1, X_2, \\cdots$ be I.I.D. random variables and $ S_n \\equal{} \\sum_{j\\equal{}1}^{n} X_j$ and $ \\overline{X}_n \\equal{} S_n / n$ for $ n \\ge 1$. \r\nIf $ \\mathbb{E}(|X_1|) \\equal{} \\infty$ then $ \\limsup |\\overline{X}_n| \\equal{} \\infty$ almost surely.",
  "Solution_1": "A little programming fun: try simulating this. One possibility: assume that $ \\text{rand}()$ calls a random number generator that gives a uniform $ (0,1)$, and is independent each time called (well, maybe not, but we're going to act like that's what it does). Let $ X_i: \\equal{}\\tan(\\pi(\\text{rand}()\\minus{}.5)).$\r\n\r\nThis gives $ X_i$ the distribution with density $ \\frac{1}{\\pi(x^2\\plus{}1)}$ on $ \\mathbb{R}$ - I think it's called the Cauchy distribution, although I prefer to think of it as the Poisson kernel. Just run the simulation with various values of $ n$ and print out a selection of values of $ \\overline{X}_n.$",
  "Solution_2": "Let $ c>0$. Using $ \\sum P(|X_n| > cn) \\ge E(|X_1|) /c \\equal{} \\infty$ and Borel-Cantelli, we have $ |X_n| > cn$ infinitely often a.e. This means $ \\limsup \\frac{|X_n|}{n} \\equal{} \\infty$. Since $ \\frac{S_{n\\minus{}1}}{n} \\plus{} \\frac{X_n}{n} \\equal{} \\frac{S_n}{n}$ we also have $ \\limsup \\frac{|S_n|}{n} \\equal{} \\infty$",
  "Solution_3": "Since $ \\frac{S_{n\\minus{}1}}{n} \\plus{} \\frac{X_n}{n} \\equal{} \\frac{S_n}{n}$ we also have $ \\limsup \\frac{|S_n|}{n} \\equal{} \\infty$.\r\n\r\nCan you justify this?",
  "Solution_4": "[quote=\"limsup\"]Since $ \\frac {S_{n \\minus{} 1}}{n} \\plus{} \\frac {X_n}{n} \\equal{} \\frac {S_n}{n}$ we also have $ \\limsup \\frac {|S_n|}{n} \\equal{} \\infty$.\n\nCan you justify this?[/quote]\r\n\r\nClaim : The event [$ \\limsup \\frac {|S_n|}{n} < \\infty$] implies the event [$ \\limsup \\frac{|X_n|}{n} < \\infty$].\r\nProof : Let $ c$ be a positive number. The event [$ \\limsup \\frac {|S_n|}{n} \\le c$] implies the event [$ \\limsup \\frac{|S_{n\\minus{}1}|}{n} \\equal{} \\limsup \\frac{|S_{n\\minus{}1}|}{n\\minus{}1} \\cdot \\frac{n\\minus{}1}{n} \\le c$ ] and hence also implies the event [$ \\limsup \\frac {|X_n|}{n} \\le 2c$] because $ \\frac {S_{n \\minus{} 1}}{n} \\plus{} \\frac {X_n}{n} \\equal{} \\frac {S_n}{n}$.",
  "Solution_5": "Thanks Redkimchi. Mathlinks really helping me very much.",
  "Solution_6": "[quote=\"Kent Merryfield\"] Let $ X_i: \\equal{} \\tan(\\pi(\\text{rand}() \\minus{} .5)).$This gives $ X_i$ the distribution with density $ \\frac {1}{\\pi(x^2 \\plus{} 1)}$ on $ \\mathbb{R}$ [/quote]\r\nCould you please explain this?",
  "Solution_7": "compute $ E[\\phi(\\tan(\\pi(U\\minus{}\\frac{1}{2})))] \\equal{} \\int_{[0;1]} \\phi(\\tan(\\pi(x\\minus{}\\frac{1}{2}))) dx$ where $ \\phi$ is an arbitrary function. You should find that this is equal to $ \\int_{\\mathbb{R}} \\phi(x) \\frac{dx}{\\pi(1\\plus{}x^2)}$",
  "Solution_8": "alekk:\r\nYes I agree that $ \\tan(\\pi(U\\minus{}\\frac{1}{2}))$ follows a cauchy distribution. But I just want to know the motivation for that particular expression, I think $ \\tan(\\pi(U))$ would also follow Cauchy distribution. Isn't it?\r\n\r\nalekk, one more question: What for Prof Kent Merryfield's programming intended?",
  "Solution_9": "[quote=\"limsup\"]But I just want to know the motivation for that particular expression, I think $ \\tan(\\pi(U))$ would also follow Cauchy distribution.[/quote]\r\n\r\nwhere is $ x\\to\\tan(\\pi x)$ discontinuous?",
  "Solution_10": "blahblahblah and limsup: I admit that $ \\tan(\\pi(U\\minus{}1/2))$ and $ \\tan(\\pi U)$ have exactly the same distribution, as would $ \\tan(\\pi(5U\\minus{}.374)).$ Hitting the discontinuity is a event of probability zero; we may safely disregard it. I used the first form primarily because I thought it was the most intuitive.\r\n\r\n[quote=\"alekk\"]compute $ E[\\phi(\\tan(\\pi(U \\minus{} \\frac {1}{2})))] \\equal{} \\int_{[0;1]} \\phi(\\tan(\\pi(x \\minus{} \\frac {1}{2}))) dx$ where $ \\phi$ is an arbitrary function. You should find that this is equal to $ \\int_{\\mathbb{R}} \\phi(x) \\frac {dx}{\\pi(1 \\plus{} x^2)}$[/quote]\r\nThere are distinct limitations on \"arbitrary function\" here - this only makes sense if that expected value makes sense. In particular, it does not make sense for $ \\phi(x)\\equal{}x,$ since $ \\frac{x}{\\pi(1\\plus{}x^2)}$ is not absolutely integrable over $ \\mathbb{R}.$\r\n\r\nAnd that was the point of my programming exercise. It was an illustration of the principle that redkimchi started this thread with. The Cauchy distribution is a readily accessible distribution whose expectation is undefined (because the expectation of the absolute value is infinite). If you run various samples of that and print out various sample means, you should find that those sample means jump around in an unpredictable fashion. \r\n\r\nThe simulated example would probably not by itself convince you of redkimchi's result - that the sample means can become arbitrarily large - but you would at least suspect that something fishy is going on with those sample means.",
  "Solution_11": "blahblahblah: Yes you are right!\r\n\r\nTo Kent Merryfield: Is my following program correct?\r\n\r\n[b]n=100;\ns=0;\nfor i=1:n\n    x=rand(1,1);\n    x=tan(pi*(x-0.5));\n    s=s+x;\n    s1(i)=(1/i)*abs(s);\nend\nplot(s1)  [/b]",
  "Solution_12": "I don't know the syntax of the language you're using so I don't know whether \"rand(1,1)\" is the right way to say that. And I don't know what syntax you need to close a loop is this actual code in some language or system or is it \"pseudocode?\" Also: why say \"abs(s)\"? You can let $ s$ be negative as well as positive.\r\n\r\nI might push the numbers a little harder - say, the following.\r\n\r\nn=100;\r\nm=1000;\r\ns=0;\r\nfor i=1:n\r\nfor j=1:m\r\nx=rand(1,1);\r\nx=tan(pi*(x-0.5));\r\ns=s+x;\r\n(next  j? something to close the loop?)\r\ns1(i)=(1/(m*i))*s;\r\n(next i?)\r\nend\r\nplot(s1)"
}
{
  "Tag": [
    "induction",
    "combinatorics proposed",
    "combinatorics"
  ],
  "Problem": "let $ n$ is positive integer number and $ n\\geq 2$ and $ a_{1},a_{2},...,a_{n}$ are positive integers number and for any $ k \\equal{} 1,2,..,n$ we have:$ 0 < a_{k}\\leq k$ and $ a_{1} \\plus{} a_{2} \\plus{} ... \\plus{} a_{n}$ is a even number\r\nprove that: We put $ \\plus{}$ or $ \\minus{}$ in order that $ \\pm a_{1}\\pm a_{2}\\pm ...\\pm a_{n} \\equal{} 0$",
  "Solution_1": "We can use induction, no?\r\n\r\nTake $ a_{n\\minus{}1},a_n$, replace it with $ |a_n\\minus{}a_{n\\minus{}1}|$."
}
{
  "Tag": [
    "number theory",
    "prime factorization",
    "number theory unsolved"
  ],
  "Problem": "Show that the number of Pythagorean triples x, y, z (x^2+y^2=z^2) with a fixed integer x is \r\n1) ( tau(x^2) - 1  ) / 2 if x is odd\r\n2) ( tau(x^2/4) - 1  ) / 2 if x is even.\r\n( tau(n) being the the number of divisors of n )\r\n\r\nThis must be not too difficult, but I can't find the solution and I have more than 200 problems to do in 2 days.\r\nSo could someone give me a hand? thank you.",
  "Solution_1": "factor: $ x^2 \\equal{} (z\\minus{}y)(z\\plus{}y)$",
  "Solution_2": "Can you elaborate please...\r\n\r\nif x is odd, gcd(x,y,z)=d\r\nx = d(m^2-n^2) = d(m+n)(m-n)\r\nm>n\r\nm+n=a, m-n=b\r\nd , a, b are all odd.\r\na = 1 (mod 4) and b = 3 (mod 4) or a = 3 (mod 4) and b = 1 (mod 4)\r\nSo a>b, the prime factorization of a has an odd number of prime factors of which remainders are 3 modulo 4 and\r\nthe prime factorization of b has an even number of primes of which remainders are 3 modulo 4, or vice versa.\r\nIs this a right approach?[/code]",
  "Solution_3": "$ x^2 \\equal{} (z \\minus{} y)(z \\plus{} y)$. so we're looking to write $ x^2 \\equal{} mn$ for integers $ m$, $ n$, such that $ m \\equal{} z \\minus{} y$, $ n \\equal{} z \\plus{} y$. the restrictions on $ m,n$ are that $ m < n$ (b/c $ z \\minus{} y < z \\plus{} y$) and $ m \\plus{} n$ be even (b/c $ m \\plus{} n \\equal{} 2z$). and notice that any such $ m,n$ will give you a solution $ (y,z)$ when you solve the system $ m \\equal{} z \\minus{} y$, $ n \\equal{} z \\plus{} y$. so you need to find the number of ways to factor $ x^2$ as $ mn$, where $ m < n$, $ m \\plus{} n$ even. \r\n\r\n1. if $ x$ is odd, any factorization will have $ m \\plus{} n$ be even b/c $ m,n$ are odd in this case. so you only need to make sure $ m < n$. $ m < n$, $ mn \\equal{} x^2$ mean that $ m < x$. the number of factors of $ x^2$ that are less than $ x$ is $ \\frac {\\tau(x^2) \\minus{} 1}{2}$. \r\n\r\n2. if $ x$ is even, at least one factor must be even, so in light of $ m \\plus{} n$ having to be even, we have to make sure *both* $ m,n$ are even. that is, $ m \\equal{} 2M,n \\equal{} 2N$. your conditions become $ x^2 \\equal{} 4MN$, $ M < N$. (the parity of $ m \\plus{} n$ is taken care of.) $ \\frac {x^2}{4} \\equal{} MN$, $ M < N$, mean that $ M < \\frac {x}{2}$. and the number of factors of $ \\frac {x^2}{4}$ less than $ \\frac {x}{2}$ is $ \\frac {\\tau(\\frac {x^2}{4}) \\minus{} 1}{2}$."
}
{
  "Tag": [
    "ratio",
    "direct proportion"
  ],
  "Problem": "I don't understand these kind of problems. Can someone explain them to me?\r\n\r\nThe annoyance level of a particular room is directly proportional to the frequency of Wei's comments and inversely proportional to Ashley's whining. The annoyance level is 660 when Wei talks 550 times per minute and Ashley's whining intensity is $ \\frac{5}{6}$. What is the annoyance level if Wei talks 550 times per minute and Ashley's whining intensity is .05?",
  "Solution_1": "first of all you don't have to swear to get our attention. \r\n\r\nthe basics for inverse and direct proportion problems are these: when two values are inversely proportional they have a constant product, so as one increases by a certain ratio, the other value has to decrease by the same ratio to maintiain a constant product-- $ xy\\equal{}k$ such that k is a constant\r\ntwo directly proportional values are as such $ \\dfrac{x}{y}\\equal{}k$ for a constant k. as the top or bottom increases by a certain ration the top or bottom has to increase by that same ratio as well.\r\n\r\nthat should help with the problem.",
  "Solution_2": "[quote=\"mihail911\"]first of all you don't have to swear to get our attention. \n[/quote]\r\nHow is crap a swear word? ...."
}
{
  "Tag": [
    "summer program",
    "MathPath",
    "analytic geometry"
  ],
  "Problem": "Source:MathPath From Dr. Zeitz\r\n\r\n3 frogs are placed on 3 vertiecs of a sqrare. Every minute, one frog leaps over another frog, in such a way that teh \"leapee\" is at the midpoint of the line segment whose endpoints are the starting and ending position of the \"leaper\". Will a frog ever occupy the vertex of the square that was originally unoccupied.\r\n\r\nThe Names of the the frogs are Albatross, Bobert, and Cattlebot.\r\n\r\n [hide]Use Parity, [/hide] :P",
  "Solution_1": "ok....i don't understand the problem, can someone explain it differently?",
  "Solution_2": "O.K\r\n\r\nthere are 3 frogs on the verticies of a square like this.\r\nf       f\r\n\r\nf       ()\r\n\r\nevery minute, a frog jumps over another frog and goes the same distance on both sides.\r\n\r\nso a sample move would be:\r\n()      f       f\r\nf       ()\r\n\r\nNow we need to prove that the one vertex of the original triangle cannot be landed on by a frog (occupied)\r\n\r\nIn this case it is the lower right vertex that we need to prove that cannot be landed on.",
  "Solution_3": "I think that version is not the same as the original. Intrepid said that the \"leapee\" was at the midpoint of the side on which the leaper leaps (so many leaps!). I also think that the frogs are not meant to go out of the square.\r\n\r\nIn any case, for the first leap, the frog on the top left corner can choose 2 directions, right or down, while the other 2 frogs are restricted to one direction each, left and up. A frog must be able to get to either the bottom or right side of the square in order to get to the unused vertex, which poses a sort of impossibility. Maybe it has something to do with combinations? After all, 3C2 or ( 3 2 ) = 3. But how can you prove that the unused vertex is the missing one?",
  "Solution_4": "[quote=\"behemoth571\"]I think that version is not the same as the original. Intrepid said that the \"leapee\" was at the midpoint of the side on which the leaper leaps (so many leaps!). I also think that the frogs are not meant to go out of the square.\n\nIn any case, for the first leap, the frog on the top left corner can choose 2 directions, right or down, while the other 2 frogs are restricted to one direction each, left and up. A frog must be able to get to either the bottom or right side of the square in order to get to the unused vertex, which poses a sort of impossibility. Maybe it has something to do with combinations? After all, 3C2 or ( 3 2 ) = 3. But how can you prove that the unused vertex is the missing one?[/quote]\r\n\r\nThe leaper goes over one of the frogs.\r\nThe frogs can leave the square.\r\nIf we put this on coordinates with the lower left frog being the origin, and the frogs making a unit square, if it jumps over the lower right frog, it will go to the poin (2,0).\r\n\r\nOne more hint (yes, coordinates are the insight), use the parity of the points.\r\n\r\nI hope that this helps."
}
{
  "Tag": [
    "geometry",
    "perimeter",
    "inequalities",
    "triangle inequality",
    "AMC"
  ],
  "Problem": "In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is 15.  What is the greatest possible perimeter of the triangle?\r\n\r\n$ \\textbf{(A) } 43 \\qquad \\textbf{(B) } 44 \\qquad \\textbf{(C) } 45 \\qquad \\textbf{(D) } 46 \\qquad \\textbf{(E) } 47$",
  "Solution_1": "[hide=\"Answer\"]By triangle inequality, $15$ has to have the middle length. The largest value of $x$ which satisfies $x+15>3x$ is $x=7$, so we have $7+15+21=43\\Rightarrow \\boxed{A}$.[/hide]",
  "Solution_2": "I can't believe I forgot that two sides added must be greater than the third...ahhhh stupid mistake!! on such an easy question",
  "Solution_3": "[hide=Solution]\nLet the side lengths be $x$ and $3x$. By Triangle Inequality, $15+x>3x$ so $x\\le 7$. $7+21+15=\\boxed{\\textbf{(A)} 43}$\n[/hide]"
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "rotation",
    "complex numbers"
  ],
  "Problem": "Prove that if $n$ can be written as $n=a^2+ab+b^2$, then also $7n$ can be written that way.",
  "Solution_1": "As a matter of interest, what is the QEDMO. Suitable name for a maths contest!",
  "Solution_2": "[quote=\"ZetaX\"]Prove that if $n$ can be written as $n=a^2+ab+b^2$, then also $7n$ can be written that way.[/quote]\r\n$\\{a,b\\}\\subset\\math N?$",
  "Solution_3": "Yes",
  "Solution_4": "[hide]Let $x=2a+3b$, $y=a-2b$. Then $x^2=4a^2+12ab+9b^2$, $y^2=a^2-4ab+4b^2$, $xy=2a^2-ab-6b^2$, so that $x^2+xy+y^2=7a^2+7ab+7b^2=7n$.[/hide]",
  "Solution_5": "Or $7(a^2+ab+b^2)=(a+3b)^2+(a+3b)(2a-b)+(2a-b)^2,$ where $a-b>0$\r\nand $7(a^2+ab+b^2)=(3a+b)^2+(3a+b)(2b-a)+(2b-a)^2,$ where $a-b\\leq0.$",
  "Solution_6": "FWIW  a  related Question.  Why Would  any number of  the form $a^2+3b^2$ , instead of $7=2^2+3\\cdot 1^2$ would work here?",
  "Solution_7": "More generally:\r\nLet $A$ be the set of integers of type $a^2+ab+b^2$ and $B$ be the set of integers of type $a^2+3b^2$.\r\nProve for $x,y \\in A ; w,z \\in B$:\r\n$xy \\in A ; wz \\in B , xz \\in A,B$\r\nor more exactly: prove $A=B$",
  "Solution_8": "[quote=\"scorpius119\"][hide]Let $x=2a+3b$, $y=a-2b$. Then $x^2=4a^2+12ab+9b^2$, $y^2=a^2-4ab+4b^2$, $xy=2a^2-ab-6b^2$, so that $x^2+xy+y^2=7a^2+7ab+7b^2=7n$.[/hide][/quote]\r\n\r\nHow could you take $x=2a+3b$ and others by yourself. Is it some kind of a rule or something like that?",
  "Solution_9": "For such problems, trying linear combinations $x=ua+vb$ and similar for $y$ work pretty well.\r\nBut there is a more general method using complex numbers.",
  "Solution_10": "[quote]How could you take $x=2a+3b$  and others by yourself. Is it some kind of a rule or something like that? [/quote]\r\n\r\nOne useful (related to  multiplication of complex numbers) generalization of:\r\n$(a^2+b^2)(c^2+d^2) = (ac +bd)^2 +(ad-bc)^2$ (     Can also be written as  (   $(ac-bd)^2+(ad+bc)^2$    ) \r\n\r\nIs\r\n$(a^2+nb^2)(c^2+nd^2) = (ac+nbd)^2 +n (ad-bc)^2$ (    Can also be written as ( $(ac-nbd)^2+n(ad+bc)^2$ ) \r\n\r\nWhat this means, is product of two numbers of the form $(a^2+nb^2)$ can also be written  in the same form. \r\n\r\nOf course, $(a^2+ab+b^2)$ fits in this form . (this is nothing but $( (\\frac{a-b} 2)^2 + 3 (\\frac{a+b} 2)^2 ))$ \r\n\r\n(This is why I  wanted to point out  that $7=2^2+3\\cdot 1^2$ etc ... )\r\n\r\n(You write the product of two terms of  the form $x^2+3y^2$ as $x^2+3y^2$ etc ..)\r\n\r\nHope this  helps.",
  "Solution_11": "Alternately, $a^2+ab+b^2=|a+\\omega b|^2$, where $\\omega$ is a cube root of unity. The numbers of the form $a+\\omega b$ with $a,b$ integers are closed under addition and multiplication.\r\nTo get to the $a^2+3b^2$ form, rotate $60^\\circ$."
}
{
  "Tag": [
    "integration",
    "calculus",
    "trigonometry",
    "calculus computations"
  ],
  "Problem": "Find  $\\int ln(\\sqrt{x}+\\sqrt{x+1})dx$\r\n\r\n       $\\int \\frac{1}{x-\\sqrt{1-x^{2}}}dx$",
  "Solution_1": "For the second one first make the substitution x=siny and satisfy yourself thet you now need to evaluate the integral of (cosy)/(siny-cosy). To do this let I be the integral I've just mentioned and let J be the integral of (siny)/(siny-cosy). Then simply evaluate J+I and J-I and solve simultaneously for I and J. (Both I+J and I-J are easily evaluated)",
  "Solution_2": "no one know the answer",
  "Solution_3": "For the first one, change variables according to $\\sqrt{x}= t$ and then integrate by parts. The answer should then be\r\n\r\n$(x+\\frac{1}{2})\\ln|\\sqrt{x}+\\sqrt{1+x}|-\\frac{\\sqrt{x}\\sqrt{1+x}}{2}+C.$\r\n\r\nFor the second one, start by multiplying and dividing by $x+\\sqrt{1-x^{2}}$. Then split in two terms, and in the second one make $x=\\sin t$. The final answer should be easy to reach.\r\n\r\nCarcul"
}
{
  "Tag": [
    "AMC",
    "AIME",
    "geometry",
    "ARML",
    "3D geometry",
    "tetrahedron",
    "AMC 12"
  ],
  "Problem": "Let the difficulty rating of the 2004 A  AIME be 8.\r\nFrom there, we may build upon the others. So for example, if a difficulty rating of an AIME is 10, a person can expect to solve 2 more problems on the 2004 AIME  than that AIME. So lets start it off:\r\n1983 AIME: 9\r\n1984 AIME: 5\r\n1985 AIME: 4\r\n.\r\n.\r\n.\r\n\r\n\r\n\r\nKeep the ratings coming.",
  "Solution_1": "If you ask me, those last two ratings were way too low.  I find that AIMEs are fairly consistent in difficulty, except the AIME I last year which seemed exceptionally easy (I got 13 on a practice ????).",
  "Solution_2": "It's impossible to rate an AIME for the following two reasons: \r\n\r\nSay some AIME have easier #1-5s but harder #11-15s, that AIME would've been easier for people who are aiming for a 4 or 5 but harder for a person who are aiming for a 10+. \r\n\r\nAlso, say 1996 AIME, which had a lot of counting questions and very few geometry problems, it would've been an advantage for people who are good at counting but bad at geometry, but a disadvantage for people who are good at geometry but bad at counting(me).\r\n\r\nbtw I disagree with your rating for 1983, 84, 85.",
  "Solution_3": "2003 was very easy.",
  "Solution_4": "1994 was very hard. At least in my opinion, it was the hardest of all.",
  "Solution_5": "The system of classification seems a little bit counterintuitive to me. An easier way to codify it (from my POV) might be just a plain 1-10 scale, without any of the precision problem equivalence in the scale.\r\n\r\nA question I had: How would you guys rank the 1987 AIME? I just got 5/6 in 50 minutes, and had all but one case of the 1 OK. This is my first year doing the AIME; from what others at my school said, my goal was going to be a nonzero score. So I was wondering: is 1987 just one of those beginning easy, end hard AIME's that beta mentioned? Overall, how is the difficulty compared to more recent AIMES?\r\nThanks",
  "Solution_6": "First of all, [b]your goal is not to get a nonzero score[/b].  My first year taking the AIME, I could pretty consistently solve the first 7 or so.  So it's not a fluke.  I usually finished all the problems that I could do (relatively) quickly with an hour or so left.  Then I worked on the rest, usually just helping to fry my brain but occasionally squeezing out one more correct answer :).",
  "Solution_7": "Really, I usually check my answers for the last hour.",
  "Solution_8": "I ought to, but I tend to get bored that way :D.  Instead, I try to check often while finding the answer and once again right after I bubble in.",
  "Solution_9": "I'm still working frantically by the end of the three hours.",
  "Solution_10": "I usually work for about 1-2 hours and manage to solve 8-10 of them, i can usually squeeze in one of the 12-15 in the last hour.",
  "Solution_11": "Usually, I do 6 or 7 problems in the first hour. In the second hour I can do like 3 problems. Third hour I pull off one more problem, and rest of the time checking. \r\n\r\n\r\n\r\nIf I get better at counting/probability and cut down # of careless mistakes, I can do pretty well on AIME. Are these two goals achievable in the next month? How should I do them?",
  "Solution_12": "I'll hopefully do 6-7 in the first hour, then 3-4 in the next hour, and 1 in the next 30 minutes. Then I'll check for the last 30 minutes. The maximum I'll answer is probably 12 because I won't know how to do at least three of them. Consequently, when you consider stupid mistakes, I'm hoping to get a 10+.\r\n\r\nI noticed that almost always one of the last 5 problems is counting (permutations of something). I find those very annoying and can never get them, either. Grr.",
  "Solution_13": "No one else does the suicidal last hour squeeze-out-one-more-problem? I unfortunately do not have the patience to spend time checking answers, so I just try to do #14 or #15 for the last hour and fail miserably. :D\r\n\r\n[quote=\"beta\"]If I get better at counting/probability and cut down # of careless mistakes, I can do pretty well on AIME. Are these two goals achievable in the next month? How should I do them?[/quote]\r\n\r\nCut down careless mistakes is easy if you become a paranoid person like me who checks every step 3-4 times on tests like this.  For getting better at counting/probability, I feel that I tend to improve slowly but steadily, naturally gaining a slightly deeper understanding every year.  Your best bet I think is to read solutions to hard counting/probability problems and try to learn to get into the right mindset for these problems.  Also, don't be afraid to do a little casework; all elegance is defenestrated in the mindless score-oriented test taking mode for AIME :).",
  "Solution_14": "1999 was brutal. I solved like ten of them and had a stupid mistake on [i]each[/i] one. Score: 0. Lovely eh? so happy. go lucky. little angry. hey, happy rhymes with angry.",
  "Solution_15": "On the aime, and when i take it in march, im hoping that the problems are good, like geometry type (i like geometry), and three- dimensional geometry stuff like tetrahedrons and soccer ball shapes.  If there is not atleast 4 geometry problems on the aime, then i dont think i'll make the United States of America Mathematical Olympiad.  BTW, how do people say AIME? Do people say each letter A-I-M-E, or do they just say like you would say,\" my [i]aim[/i] is to make the USAMO.\"  [/i]",
  "Solution_16": "I most often hear it said \"Amy\" and occasionally \"Ay Me\" but nobody I know says each letter.\r\n\r\nUgh... geometry. I hope there are as few geometry problems in the last 5 as possible. Anything before that will probably be solvable for me (I hope).",
  "Solution_17": "LOL, 10 stupid mistakes on 10 problems...that requires SKILL.",
  "Solution_18": "I got skill. ANd they were the really stupid mistakes like I added m and n wrong in the end or wrote down a case but forgot to include it. I need to organize my work.",
  "Solution_19": "Hey recently I thought 2004/6 = 338\r\n\r\nJust wondering, was 1988 AIME generally considered the easiest AIME ever? Because I hit my record 11 on it...",
  "Solution_20": "In eight grade, right before the AMC 8, I kept on dividing 11/2 and getting 6.5. Really weird how many times I get it wrong before it sinks in.",
  "Solution_21": "[quote=\"beta\"]Hey recently I thought 2004/6 = 338\n\nJust wondering, was 1988 AIME generally considered the easiest AIME ever? Because I hit my record 11 on it...[/quote]\r\n\r\nI don't distinctly remember, but again, I don't think it has to do with the test itself.  You probably improved or were in the right mood (this makes a pretty big difference sometimes).  Except for the 2004 AIME I.  That was weirdly easy...  :?",
  "Solution_22": "[quote=\"probability1.01\"]\nI don't distinctly remember, but again, I don't think it has to do with the test itself.  You probably improved or were in the right mood (this makes a pretty big difference sometimes).  Except for the 2004 AIME I.  That was weirdly easy...  :?[/quote]\r\n\r\nRight mood? What's the \"right mood\"? I guess 1988 has only two counting questions, one of them was a easy #1, so it didn't hit my weakness too bad. \r\n\r\nBut still I'm not sure about the \"right mood\", and I know I definitely didn't improve.",
  "Solution_23": "Eh, I found the 2004 AIME to be about average, maybe a little easier than average. I got a 7 on it, and I've gotten 6-9 on most of the AIMEs I've practiced with since. So...",
  "Solution_24": "Ah I got a 10 on 1992 AIME in practice (I still made 2 careless mistakes :( ), was that year really easy too?\r\n\r\nWait I think I got a 10, kalva doesn't have an answer for #14, can someone confirm the answer to #14, I believe the answer is 094. \r\n\r\n14.  ABC is a triangle. The points A', B', C' are on sides BC, CA, AB and AA', BB', CC' meet at O. Also AO/A'O + BO/B'O + CO/C'O = 92. Find (AO/A'O)(BO/B'O)(CO/C'O).",
  "Solution_25": "[quote=\"beta\"]Hey recently I thought 2004/6 = 338\n\n[/quote]\r\n\r\n\r\nwhoaa... at arml i divided 2004 by 6 and ALSO kept getting 338! hence... i got the wrong answer.  :( freaky, eh?",
  "Solution_26": "Maybe I'm crazy, but I do better on #8-#15 than #1-#7.",
  "Solution_27": "[quote]1999 was brutal.[/quote]\r\nFrom what I recall, the scores in 1999 ran lower than they usually do. In particular, I track Southern California scores, and the second highest from that half of the state was a 7.",
  "Solution_28": "what do i need on the aime to move on if im a sophmore?\r\nand, if I took the amc12 but i am a sophmore, does it matter?",
  "Solution_29": "Where do you all find these AIMEs to take as practice tests?  Is there somewhere online or something?  It would be kinda useless to have the problems without an answer key.",
  "Solution_30": "[quote=\"everyday847\"]Maybe I'm crazy, but I do better on #8-#15 than #1-#7.[/quote]\r\n\r\nYES you are crazy.",
  "Solution_31": "http://www.kalva.demon.co.uk/aime.html\r\nLots of AIME problems, answers with solution outlines too."
}
{
  "Tag": [
    "Support",
    "induction",
    "number theory",
    "prime factorization"
  ],
  "Problem": "Here are a couple of random problems that I need major assistance on:\r\n\r\n(Well, the fact that this browser doesn't support Microsoft Equation Editor makes the next question hard to read but, try your best)\r\n\r\n1. Prove that [(N+k) Choose (k)]*(1/2^k)=2^N, where k is an integer starting from zero and increasing until N. (Its like sigma notation, where N is the limit, [(N+k) Choose (k)]*(1/2^k) is the conditional, and k starts from zero to N.\r\n ***well, I know you're supposed to use induction for this, and the base case is easy enough, but the inductive step is really killing me. Any sugestions?***\r\n\r\n2. Given m,n,p are postive integers, and m<=n<=p, and that \r\n   2mnp=(m+2)(n+2)(p+2). Find Max(P).\r\n\r\n***I have no idea.....***\r\n\r\n3.Let T(N) equal the number of ways to represent a positive integer N as the sum of one or more consecutive integers. Let \"p\" and \"q\" equal distinct odd primes, and let \"a\" and \"b\" represent positive integers. Prove:\r\n   a) T(p)=2\r\n   b)T(2^a)=1\r\n   c)T(p^a)=a+1\r\n   d)T(p^a*q^b)=T(p^a)*T(q^b)\r\n\r\n***OK. Parts a) and b) are easily proved. However, I haven't been able to get very far with parts c) and d). Perhaps given the properties of the first two parts, you can prove the last two parts. Any suggestions?",
  "Solution_1": "you do know that if you press the \"view more emoticons\" button on the left you bring up mathematical symbols and such. thats all i can help.\r\n\r\nhappy solving",
  "Solution_2": "I got #2:\r\n\r\nRewrite the equation as (m+2)/m * (n+2)/n * (p+2)/p = 2. We want (p+2)/p to be as small as possible. So we try a greedy approach: since (m+2)/m >= (n+2)/n >= (p+2)/p, we assume that (m+2)/m is as large as possible. So we try m=3. Then we need n>= 13. But (3, 13) doesn't give a solution, and neither does (3, 14). But (3, 15) does: it gives p=50. But we have no guarantee that it's the largest. So we try m=4. Then n>=7. With n=7, we get p=54, which is better. Continuing, we get (5, 5, 98). We must continue until (m+2)/m<= 2^(1/3), so we get (6, 6, 16), m=7 doesn't work with n<=7, so it must be worse, and if m=8, then (m+2)/m < 2^(1/3). Thus the optimal solution is (5, 5, 98). I'm not sure if that argument is particularly comprehensible, but it should be right if I didn't do anything horribly wrong.",
  "Solution_3": "For #3 I think you mean the sum of consecutive [b]positive[/b] integers. Otherwise some of them are false. (That's why I got confused at first.) I think it's easier to look at the number of ways to write N as the sum of one or more consecutive integers that need not be positive. Let's call that U(N). If we write N=2^a*3^b*5^c... (the prime factorization), then U(N)=2*(b+1)*(c+1)*(d+1)... (see if you can prove it; I think it's easier than tackling (c) and (d) directly). Remember that the average of the consecutive numbers must be an integer if there are an odd number of them and either an integer or a half-integer if there are an even number of them.",
  "Solution_4": "yea, ur rite. My mistake. Its the sum of consecutive postivie integers. Sorry about that.",
  "Solution_5": "Oops, COmplexZeta, Im afraid your answer is incorrect. I did the problem this morning, and I found out that the answer is 130. \r\n\r\n2mnp=(m+2)(n+2)(p+2)\r\n2mnp=mnp+2(mn+np+mp)+4(m+n+p)+8\r\nmnp-2(mn+np+mp)-4(m+n+p)-8=0.\r\n\r\nmnp-2(mn+np+mp)+4(m+n+p)-8=8(m+n+p)\r\n(m-2)(n-2)(p-2)=8(m+n+p)\r\n\r\nLet m-2=a\r\n     n-2=b\r\n     p-2=c\r\n\r\nabc=8(a+s+b+2+c+2)\r\nabc=8a+8b+8c+48\r\nc=[8(a+b+6)]/(ab-8)\r\n \r\nThus, in order to maximize c, we must minimize ab. Since m,n,p are all natural numbers (sorry bout that, i wrote they were integers before), the minimum ab-8 can be is 1. Therefore, (a,b)=(1,9) or (3,3). Plugging that back in, we see that c is greater when (a,b)=(1,9), which is 128. Thus, p is equal to 130. \r\n\r\nBut does anyone have ideas on the other two problems?",
  "Solution_6": "Oops. When I did the case m=3, I got (n+2)/n<6/5, which would imply n>10, not n>12. So that's why mine was wrong. Good catch.",
  "Solution_7": "[color=cyan]For the T(n) question, it is worth considering that for odd n, n|T(n) and for even n n|2T(n) always.\n\nI also have this distinct memeory that number 1 isn't that hard, but I'll have to go and actually try it and I don't have the time right now, sorry.  Try using simple things, like nCr + nC(r+1) = (n+1)C(r+1).[/color]"
}
{
  "Tag": [
    "function",
    "algebra",
    "functional equation"
  ],
  "Problem": "If you have a function $ f(x)$, and $ f(0)\\equal{}1$, is it always true that $ f(x\\plus{}y)\\equal{}f(x) \\cdot f(y)$. If so, why?",
  "Solution_1": "Er... try $ f(x) \\equal{} x \\plus{} 1$.  Or... any function passing through $ (0, 1)$ that isn't of the form $ e^{f(x)}$ where $ f(x \\plus{} y) \\equal{} f(x) \\plus{} f(y)$ (in other words, $ f(x)$ is a solution to the [url=http://en.wikipedia.org/wiki/Cauchy_functional_equation]Cauchy functional equation[/url]).",
  "Solution_2": "[quote=\"t0rajir0u\"]Er... try $ f(x) \\equal{} x \\plus{} 1$.  Or... any function passing through $ (0, 1)$ that isn't of the form $ e^{f(x)}$ where $ f(x \\plus{} y) \\equal{} f(x) \\plus{} f(y)$ (in other words, $ f(x)$ is a solution to the [url=http://en.wikipedia.org/wiki/Cauchy_functional_equation]Cauchy functional equation[/url]).[/quote]\r\n\r\nheh. sorry. sometimes I ask a question that I don't even look for the simplest in-my face solution. for some reason, someone in a class used f(0)=1 as sufficient proof that f(x) has to be exponential and therefore followed f(x+y)=f(x)f(y) like f(x)=2^x.",
  "Solution_3": "[quote=\"mihail911\"]someone in a class used f(0)=1 as sufficient proof that f(x) has to be exponential[/quote]\r\n\r\nIf I was a car mechanic, this would be the point where I stop, pick my head out from under your hood, and yell:\r\nYep, that's your problem right there. That'll be five hundred, please."
}
{
  "Tag": [
    "geometry",
    "parallelogram",
    "geometry solved"
  ],
  "Problem": "In the thetrahedron ABCD the sum of the areas of faces which share the same side AB equals to the sum of areas of faces which share the same side CD.Prove that the midpoints of BC,AD,AC,BD lie on a plane that contains the inecenter of the inscribed sphere.",
  "Solution_1": "Let K, L,M and N be the midpoints of BC,AD,AC,BD respectively and let I be the center of the inscribed sphere.\r\n\r\nFirst of all is it immediate to see that K,L,M and N are the vertex of a parallelogram and then they lie on the same plain. This plain is, of course, parallel and at the same  distance wrt AB and DC. It holds that ABLNKM and DCLNKM have the same volume.  Indeed as (CMK)=(CAB)/4 and (NKB)=(BCD)/4, (DMKC) = (ANKB)=(ABCD)/4 and, as A and D are at the same distance from LNKM, (ALNKM) = (DLNKM).\r\n\r\nSince, from hypothesis, (IBCD)+(IACD) = (ICAB)+(IDAB), and  as 4(ICKM) = 4(IABKM)/3 = (IABC) and similar relations hold, then it results that (DCLNKMI) = (ABLNKMI) and this implies that I lies on LNKM."
}
{
  "Tag": [
    "trigonometry"
  ],
  "Problem": "$sin x+cos x+sin xcos x=1$",
  "Solution_1": "$(\\sin(x)+\\cos(x))^{2}=(1-\\sin(x)\\cos(x)^{2}$\r\n\r\n$1+2\\sin(x)\\cos(x)=1-2\\sin(x)\\cos(x)+\\sin^{2}(x)\\cos^{2}(x)$\r\n\r\n$4\\sin(x)\\cos(x)=\\sin^{2}(x)\\cos^{2}(x)$  now for $\\sin(x)\\cos(x)\\neq{0}$ we have\r\n\r\n$4=\\sin(x)\\cos(x)\\Longrightarrow 8=\\sin(2x)$ so there are no solutons with $x\\in\\mathbb{R}$. If you want complex solutions then solve $8=\\frac{e^{2ix}-e^{-2ix}}{2i}$.\r\n\r\nnow if $\\sin(x)\\cos(x)=0$ we have that $x=\\frac{n\\pi}{2}$ for $n\\in\\mathbb{Z}$",
  "Solution_2": "Let $\\sin x+\\cos x =t,$ we can rewrite as $t+\\frac{t^{2}-1}{2}=1\\Longleftrightarrow t^{2}+2t-3=0\\Longleftrightarrow t=1,-3.$Since$(\\sin x+\\cos x)^{2}+(\\sin x-\\cos x)^{2}=2\\Longleftrightarrow t^{2}\\leq 2.$Thus$t=1.$that is to say,$\\sin x+\\cos x=1$Solving this by unit circle, yielding$x=2n\\pi , \\frac{\\pi}{2}+2n\\pi\\ (n: integer).$",
  "Solution_3": "[quote=\"maokid7\"]$(\\sin(x)+\\cos(x))^{2}=(1-\\sin(x)\\cos(x)^{2}$\n\n$1+2\\sin(x)\\cos(x)=1-2\\sin(x)\\cos(x)+\\sin^{2}(x)\\cos^{2}(x)$\n\n$4\\sin(x)\\cos(x)=\\sin^{2}(x)\\cos^{2}(x)$  now for $\\sin(x)\\cos(x)\\neq{0}$ we have\n\n$4=\\sin(x)\\cos(x)\\Longrightarrow 8=\\sin(2x)$ so there are no solutons with $x\\in\\mathbb{R}$. If you want complex solutions then solve $8=\\frac{e^{2ix}-e^{-2ix}}{2i}$.\n\nnow if $\\sin(x)\\cos(x)=0$ we have that $x=\\frac{n\\pi}{2}$ for $n\\in\\mathbb{Z}$[/quote]\r\n$sinxcosx=0\\rightarrow sinx+cosx=1$\r\nSo the answer must be $x=2k\\pi$ or $x=\\frac{\\pi}{2}+2k\\pi$ with $k\\in\\mathbb Z$"
}
{
  "Tag": [
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "If $x,y,z>0$  and $xyz=1$ .then prove that $\\frac{x}{1+xy}+\\frac{y}{1+zy}+\\frac{z}{1+xz}\\le \\frac{3}{2}$",
  "Solution_1": "[quote=\"VANCHANH123\"]If $x,y,z>0$  and $xyz=1$ .then prove that $\\frac{x}{1+xy}+\\frac{y}{1+zy}+\\frac{z}{1+xz}\\le \\frac{3}{2}$[/quote]\r\nIt's wrong: $x=y=2$ and $z=\\frac{1}{4}.$ :wink:"
}
{
  "Tag": [
    "conics",
    "parabola",
    "limit"
  ],
  "Problem": "Suppse we have an ideal vase and we do the following:\r\nat 1minute befor 1 o'clock we  put the numbers from 1 to 10 in the vase and take out the number 1;\r\nat half a minute before 1 o'clock we put the numbers from 11 to 20 and take out 2\r\nat 1/3 min befor 1 o'clock we put numbers 21-30 and take out 3...\r\nAND SO ON...\r\n\r\nQUESTION:\r\nHOW MANY NUMBERS WILL THERE BE IN THE VASE AT EXACTLY 1 O'CLOCK?",
  "Solution_1": "i just want to point out a minor error\r\n[quote]and take out the number 1; [/quote]\r\nshould be [b]and take out 1 number;[/b]\r\nhave fun!",
  "Solution_2": "No, the original version is-for the first step-take out the number one, for the second-take out the number two...     but 3X.lich's version seems more interesting!!!",
  "Solution_3": ":? I would say 0 of course, same for both versions :?\r\n\r\n(as when it's like 1/20 before 1, you'll put in 10 and take out 20, and so on",
  "Solution_4": "Hehe, this is classic. It's been heavily discussed on the sci.math newsgroup, if anybody else hangs around there.\r\n\r\n[BTW, I hope I'm understanding it right. My impression was that on the third step we take out the number 3, not 3 numbers, as Peter VDD (and 3x.lich) are assuming, because if we do it their way, there would soon be a negative number of numbers!]\r\n\r\n[edit: didn't read carefully enough. I was right, the original intention, as clarified by xxxxtt himself, was that you take out 1 number each turn. Thus, all of the below is relevant. I'll do another reply on the other method, though.]\r\n\r\nNow, if we do it xxxxtt's way, there'll be no numbers left. This may seem a bit odd, but consider it like this: try to tell me one number that is still there at the end. No matter what number N you say, that number was taken out on the Nth turn, 1/N minutes before 1 o'clock. So there are no numbers in the vase. There are an infinite number of steps, so there are an infinite number of numbers taken out.\r\n\r\nNow, let's consider another way, which is more general--you get to choose which number to take out. Now it depends on what number you chose to take out. If each time you decide to take out the lowest number, than it is the same as the original. However, say that each time you chose to take out the highest number. Then, the numbers 1-9 will be left after the first step, 11-19 after the second step, 21-29 after the third step, etc. Since there are still an infinite number of steps and there is a net gain of 9 numbers after each step, there will be an infinite number of balls left.\r\n\r\nYou can also make it so that there will be, say, exactly 5 numbers left. On the first step, take out the number 6. On the second step, take out the number 7, and so on. The only numbers left will be 1-5.\r\n\r\nThe dangers of infinity...\r\n\r\n.dzhonatan",
  "Solution_5": "The first impression that I saw was that 'take out 2' meant literally taking out two numbers but of course it meant taking out '2' because after I replied I realized that you can't take more than k numbers out after k steps, for some large k... As for the solution of swiftstream, it looks definitely like that story of a race between Achilles and the tortoise... Umm! paradoxes are fun. :D",
  "Solution_6": "The other method...\r\n\r\nThis is an analysis of the method 3x.lich and Peter VDD are suggesting, where on the first turn you take out 1 number, the second turn you take out 2 numbers, etc.\r\n\r\nNow, after N turns, the total number of numbers is  \\sum (from 1 to N) of (10-N). Or, (10-1)+(10-2)+(10-3)+...+(10-N), i.e., 10N- \\sum (from 1 to N) of (N), or, applying the sum formula,\r\n\r\n10N-(N^2+N)/2\r\n\r\n=-N^2/2+19N/2\r\n\r\nwhich is a parabola opening downward,  \\lim N-->(infinity)=(-infinity), so that after infinity steps there will be negative infinity numbers left, which just doesn't make much sense :-)\r\n\r\n.dzhonatan\r\n\r\nGrrr... ascii is frustrating. I look forward to the day when you will be able to post on forums in true mathematical form, whether it be handwritten scrawl (now that would be cool :-) or just a good typesetting interface.",
  "Solution_7": "Not quite like achilles and the tortoise, because that has an understandable resolution with limits and infinitesimals.\r\n\r\nThis is a bit wacky, because it all depends on which number you take out. It doesn't seem like that should make a difference. A lot of people just can't accept it at all...\r\n\r\nAnyway, though, I'm going to go make a thread about paradoxes, because I agree, they are fun :-)\r\n\r\n.dzhonatan",
  "Solution_8": "ya I know, then it gets negative after a while. That's also why i found it that easy :D\r\n\r\nbut still, the question is confusing :? so I'm not surprised to see more misunderstandings around this."
}
{
  "Tag": [
    "calculus",
    "derivative",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "By looking at a graphing calculator, is there a way to find inflection points for complex Calculus problems? or is there a more simplistic way of solving difficult second derivatives, besides setting the second derivative to 0?",
  "Solution_1": "I can't think of one.  You're better off just doing it by hand anyway since often times, you won't be able to use a calculator."
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "partial fractions"
  ],
  "Problem": "Show that if $f(x)$ is a polynomial whose degree is less than $n$, then the fraction \r\n$\\frac{f(x)}{(x-x_{1})(x-x_{2})...(x-x_{n})}$, \r\nwhere $x_{1},x_{2}, ...,x_{n}$ are $n$ distinct numbers, can be represented as a sum of $n$ partial fractions \\[\\frac{A_{1}}{x-x_{1}}+\\frac{A_{2}}{x-x_{2}}+\\ldots+\\frac{A_{n}}{x-x_{n}}\\] Hint: use [url=http://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html]Lagrange[/url]\r\n\r\n[hide=\"solution(please tell me if it is incorrect)\"]\nBy Lagrange we know that $f(x)$ can be expressed as $\\sum \\frac{(x-x_{2})(x-x_{2})...(x-x_{n})}{(x_{1}-x_{2})(x_{1}-x_{3})...(x_{1}-x_{n})}y_{1}+...$\nWhere $x_{i}, y_{i}\\in \\mathbb{R}$. Thus we write\n\n$\\frac{f(x)}{(x-x_{1})(x-x_{2})...(x-x_{n})}= \\frac{ \\sum y_{1}\\frac{(x-x_{2})(x-x_{2})...(x-x_{n})}{(x_{1}-x_{2})(x_{1}-x_{3})...(x_{1}-x_{n})+...}}{(x-x_{1})(x-x_{2})...(x-x_{n})}$\n\nWe can divide by $\\sum_{i=1}^{n}x-x_{i}$ to get \n\n$\\frac{y_{1}}{(x-x_{1})(x_{1}-x_{2})(x_{1}-x_{3})...(x_{1}-x_{n})}+...$\n\nSince $y_{i}, x_{i}$ are arbitrary numbers, we have found the desired partial fractions.\n[/hide]",
  "Solution_1": "So Lagrange is just the closed form of the nth degree polynomial passing through n+1 specific points...",
  "Solution_2": "Sorry, I can't tell if your post is just a comment or if you're trying to point out something wrong with my solution.",
  "Solution_3": "Looks OK to me if you mean $\\prod_{i=2}^{n}(x-x_{i})$ instead of $\\sum_{i=1}^{n}x-x_{i}$,and if you put\r\n$A_{1}=\\frac{y_{1}}{(x_{1}-x_{2})(x_{1}-x_{3})\\hdots(x_{1}-x_{n})}$\r\n\r\nEDIT: you have to divide both numerator and denominator by\r\n$\\prod_{i\\neq k}^{n}(x-x_{i})$ \r\nfor the k$^{th}$ term."
}
{
  "Tag": [],
  "Problem": "What is the maximum number of points of intersection when $ 5$ lines intersect each other?",
  "Solution_1": "There are $ 4\\plus{}3\\plus{}2\\plus{}1\\equal{}\\boxed{10}$ points of intersection.",
  "Solution_2": "Assuming no 2 lines are parallel, we need 2 lines to determine an intersection point, so $ \\binom{5}{2}\\equal{}10$",
  "Solution_3": "Ok, thanks AwesomeToad",
  "Solution_4": "@usernameyourself, please, [b]please[/b], [i]please[/i] stop bumping up random threads to not say anything useful",
  "Solution_5": "There should be an award for that 10 year bump though"
}
{
  "Tag": [
    "\\/closed"
  ],
  "Problem": "I'm in the UK and the problem solving book looks perfect for me... will it be shipped to the UK though? I know the site is American (or I guessed it is, no offence intended if it isn't), ad all the prices are in dollars which is probably the reason it looks so expensive at first glance. (I think $40 = about 25?)\r\n\r\nAlso, what is the solutions book? I guessed it was solutions to problems, but I didn't know what problems it was the solutions to? Perhaps there are problems in the book... this uncertainty is why I ask. :cool: \r\n\r\nThanks, I really want to get betetr at these maths problems but my only book \"The Mathematical Olympiad Handbook\" (for BMO-British maths olympiads) has a very breif teaching session and assumes I am already pretty good at it.",
  "Solution_1": "AoPS is American.  Well now that they have joined with mathlinks it's more international.  If you get the books you will really want the solution books.  The books consist of about 30 chapters each.  In the chapters they teach you the stuff you need to know to do the problems.  They have a few examples of problems in each chapter where they show how to do a question.  Then they give you exercises to do which help your understanding of each section within a chapter.  Finally at the end of each chapter they have a bunch of problems to do that pertain in some way to what was taught in the previous chapter.  The solution manual is very useful for these problems at the end of each chapter.",
  "Solution_2": "Ok, that sounds very help ful. Thanks.",
  "Solution_3": "We are able to ship the Art of Problem Solving books to any country.  There is an additional charge for international shipping that is not listed on the online order form ([url=http://artofproblemsolving.com/Books/AoPS_B_Order.php]click here[/url]).  There are instructions at the bottom of the order form that tell you how to find out how much the shipping charges are.\r\n\r\nI would also encourage you to buy the solutions along with the books.  Most of the time when people just buy the book they end up coming back for the solutions.  The solution manual are valuable teaching tools in that they show how to solve each problem, not just the answer."
}
{
  "Tag": [],
  "Problem": "1. There are 27 candidates in elections and $ n$ citizens that vote for them. \r\nIf a candidate gets $ m$ votes, then $ 100m/n\\le m\\minus{}1$. What is the smallest possible value of $ n$?  \r\n[hide=\"answer\"]134[/hide]\n\n2. The set A consists of $ m$ consecutive integers with sum $ 2m$. The set B consists of $ n$ consecutive integers with sum $ 2n$. The difference between the largest elements of $ m$ and $ n$ is 99. Find $ m$. \n[hide=\"answer\"]201[/hide]",
  "Solution_1": "[quote] 2. The set A consists of m consecutive integers with sum 2m. The set B consists of n consecutive integers with sum 2n. The difference between the largest elements of m and n is 99. Find m.  [/quote]\r\n\r\n[color=darkblue] Hey! Are the above two problems inter-linked or are they independent? [/color]\r\n\r\n\r\n\"Find m\", Does this mean we have to calculate the number of elements of set 'A' which conforms to the rules as stated in the problem, right? \r\n\r\n\r\nIf it's so, here's a solution:\r\n\r\nConsider the set A as follows : A={-1, 0 , 1 , 2 , 3 , 4 , 5}\r\n\r\nHere set A consists of \"seven\" consecutive integers with sum =14.  \r\n\r\nSo, [b] m =7 [/b] and [b] sum =2m = 14 [/b]\r\n\r\n\r\nNow, coming to the second part of the problem. \r\n\r\nConsider the set B as follows: B= {-100,-99,-98,..0...+100,101,102,103,104}\r\n\r\nIn set B, \"n\" i.e. the number of consecutive integers is [b] 205 [/b] and the sum (2n) of  these integers is [b] 410 [/b].\r\n\r\n(The number of consecutive integers in set B is: 205)\r\n\r\n-100 to 0 --> No of digits (including 0) is: [b] 101 [/b]\r\n\r\n1 to 104 --> No of digits (excluding 0, since it has already been counted) is :[b]104 [/b]\r\n\r\nThe largest number of set 'A' is: 5, and the largest number of set 'B' is :104 (Hence, the difference between the largest numbers of 'A' and 'B' is 99).\r\n\r\nHence, [b] 'm' is = '7' [/b] or if we consider it the other way round that is -- Set A= (-100,-99,-98.....104} and Set B = {-1, 0 , 1, 2, 3, 4, 5} then [b] 'm' = 205 [/b]",
  "Solution_2": "[quote=\"mathwizarddude\"]\n2. The set A consists of $ m$ consecutive integers with sum $ 2m$. The set B consists of $ n$ consecutive integers with sum $ 2n$. The difference between the largest elements of $ m$ and $ n$ is 99. Find $ m$. \n[hide=\"answer\"]201[/hide][/quote]\r\n\r\nWouldn't any set of consecutive integers centered around 2 work?\r\n$ 2\\equal{}2(1)$\r\n\r\n$ 1\\plus{}2\\plus{}3\\equal{}2(3)$\r\n\r\n$ 0\\plus{}1\\plus{}2\\plus{}3\\plus{}4\\equal{}2(5)$\r\n\r\nSo, couldn't M be any positive odd number?",
  "Solution_3": "This problem was taken from the imomath test [url]http://test.imomath.com/iop_logovanje.php?ImModerator=t&req=gradeTest&subjwName=gen[/url]\r\nonly the answer is given but without solution to this problem and that's why I posted it here (I'm baffled by the multiple answers, too) but does the first problem make sense?",
  "Solution_4": "For the 2nd problem if the two sequences of consecutive integers start with $ a,b$, then $ |a\\minus{}b|\\equal{}99$.\r\nDoes anyone have a solution for the first problem?\r\n(@zserf: yours doesn't work because 4-3 does not equal to 99 :P )",
  "Solution_5": "But because the largest term can be any odd number, M can be any odd number that is atleast 199.\r\n\r\nIf N is 1\r\nM is 199\r\n\r\nIf N is 3\r\nM is 201\r\n\r\nand so on.\r\n\r\n$ M\\equal{}N\\plus{}198$"
}
{
  "Tag": [
    "logarithms"
  ],
  "Problem": "Compute the number of  3 digit positive integers $b$, such that $\\log_8 b$ is a rational number.\r\n\r\nThis is how I solved it:\r\n\r\n$8= 2^3$\r\n\r\nTherefore, the number has to be multiples of $2^3$, so:\r\n\r\n$(2^3)^3 =  512$\r\n$(2^3*2)^2 = 256$\r\n$(2^3*3)^2 = 576$\r\n\r\nWhich means the only numbers that $b$ can be are $512$, $256$ or $576$. However, I got this through guess and check, and I was wondering if there were any other better, more \"mathematical\" ways of getting this answer, rather than randomly guessing.",
  "Solution_1": "Are you sure log  (base 8) of 576 is rational?  [hide](Log 9 to the base 8 is not rational, and  you still will not have a rational number if you   you add 2 to it..)\n\nNow 128,  256, 512 are another matter ..\nhint: any power of 2 would do..\n.[/hide]",
  "Solution_2": "Yep, I just realized that 128, 256 and 512 are the only values that would work, because the other two are not rational, thanks for pointing this out :blush:"
}
{
  "Tag": [
    "geometry proposed",
    "geometry"
  ],
  "Problem": "Construct a triangle ABC given the  three radios of the excircles.....",
  "Solution_1": "$ \\frac{a}{S} \\equal{} \\frac{2}{h_a} \\equal{} \\frac{1}{r_b} \\plus{} \\frac{1}{r_c}$ and cyclic exchange. Let $ A,Y,Z$ be points on a ray $ (Ox,$ such that $ OA \\equal{} r_a, OY \\equal{} r_b, OZ \\equal{} r_c.$ Inversions with center $ O$ and powers $ \\plus{}OA^2, \\minus{}OA^2$ take $ Y, Z$ into $ Y_1, Z_1$ and $ Y_2, Z_2,$ respectively. Denote $ c' \\equal{} AY_2.$ Circles $ (A), (Y_2)$ with radii $ b' \\equal{} AZ_2, a' \\equal{} Y_2Z_1$ intersect at $ Z_3$ and $ \\triangle AY_2Z_3 \\sim \\triangle ABC.$ Construct any exradius of $ \\triangle AY_2Z_3,$ say $ r_a'$ against $ A,$ and scale $ \\triangle AY_2Z_3$ by coefficient $ \\frac{r_a}{r_a'}.$",
  "Solution_2": "Thanks for the solution :lol:"
}
{
  "Tag": [
    "probability"
  ],
  "Problem": "This is from 2004-2005 handbook warmup 18\r\nThere are four cookies and four chips. If she distributes the chips randomly into the four cookies, what is the probability that there are no more than two chips in any one cookie?",
  "Solution_1": "[hide]\nI'm sorry I can't find a more algebraic way to do it, but you can take all ones, and then a two, two ones, and a 0 and rearrange them $4!/2!$ times, or 12. Finally, take two zeroes and two 2s and rearrange them $\\frac{4!}{2!2!}$ times, or 6. Add together to get $19$ ways. I hope I did that right.\n[/hide]",
  "Solution_2": "The answer is 51/64, but I don't know how to get it. Can someone help?",
  "Solution_3": "[hide]\nLet's count the ways to put 3 or 4 chips in each cookie.\n\nFirst of all, there are $4^{4}$ ways to put 4 chips in 4 cookies.\n\nThere are 4 ways to put 4 chips in one cookie.\n\nFor 3 chips in one cookie and 1 chip in another, there are 4 ways to choose the cookie with 3 chips in it and 3 ways to choose the cookie with one chip in it, for a total of 12 ways to choose the cookies. There are $\\binom{4}{3}$ ways to choose which 3 chips go into the cookie with three chips in it. So there are a total of 48 ways for this case.\n\nSo there are $256-4-48$ ways, for a probability of $\\frac{204}{256}= \\frac{51}{64}$[/hide]",
  "Solution_4": "Thank you !"
}
{
  "Tag": [
    "geometry",
    "rectangle"
  ],
  "Problem": "[img]http://img525.imageshack.us/img525/4653/augquizwv3.jpg[/img]\r\n\r\nsolve this and mensa will accept you.",
  "Solution_1": "Let $ AB \\equal{} x,\\ AD \\equal{} y, EC \\equal{} a,\\ FC \\equal{} b$, we have $ [CEF] \\equal{} 3\\Longleftrightarrow ab \\equal{} 6\\ \\cdots [1]$, $ [ABE] \\equal{} 4\\Longleftrightarrow x(y \\minus{} a) \\equal{} 8\\ \\cdots [2]$,\r\n$ \\ [ADF] \\equal{} 5\\Longleftrightarrow y(x \\minus{} b) \\equal{} 10\\ \\cdots [3]$. From $ [2],\\ [3]$, substituting $ a \\equal{} \\frac {xy \\minus{} 8}{x},\\ b \\equal{} \\frac {xy \\minus{} 10}{y}$ gives to $ [1]$, yielding $ (xy)^2 \\minus{} 24xy \\plus{} 80 \\equal{} 0\\Longleftrightarrow xy \\equal{} 4,\\ 20$. \r\n$ \\therefore [AEF] \\equal{} xy \\minus{} (3 \\plus{} 4 \\plus{} 5) \\equal{} xy \\minus{} 12 \\equal{} 20 \\minus{} 12 \\equal{} 8$.",
  "Solution_2": "Are you sure that Mensa will accept anyone for solving such an easy problem. I'm not sure...",
  "Solution_3": "How do you solve it?",
  "Solution_4": "Exactly like yours! Nice solution :lol: ! But still i think it's easy...",
  "Solution_5": "[quote=\"Bugi\"]Exactly like yours! Nice solution :lol: ! But still i think it's easy...[/quote]Oh really? There is a non-algebraic and shorter solution, much faster than kunny's, it's a slick solution and it took me a while to find it. I challenge you to find it, if you think it's easy...\r\n\r\nI'm waiting.",
  "Solution_6": "wat is your solution??",
  "Solution_7": "[hide=\" :rotfl: \"]Well, this didn't take long but...\nset EC=3, FC=2, AB=4, BE=2.  This gives the desired areas\n\nThen, rectangle-triangles  :rotfl: [/hide]\r\n\r\n[color=blue][u]click smiley to see hidden content[/u][/color]",
  "Solution_8": "The problem can be reduced to the figure where x is the area of the rectangle.\r\nNow there are two possibilities.\r\na) (x-16) + (6) >= (24- x)+ (x-14) then\r\n    (x-16) >= (24-x) gives x >= 20 \r\n    and 6>= (x-14) gives x<=20. \r\n    \r\nb) (x-16) + (6) <= (24- x)+ (x-14) then\r\n    (x-16) <= (24-x) so x <= 20\r\n    and 6<= (x-14) x>=20. \r\n    \r\n\r\n    The area of the rectangle is 20."
}
{
  "Tag": [
    "integration",
    "function",
    "calculus",
    "calculus computations"
  ],
  "Problem": "Does the series $\\sum_{n=1}^{\\infty}(-1)^{n}\\int_{2n}^{2n+1}\\frac{\\ln(1+3t)}{t}dt$ converges?",
  "Solution_1": "It converges. You show this is a Leibniz series.\r\n\r\nLet $a_{n}=\\int_{2n}^{2n+1}\\frac{\\ln(1+3t)}{t}dt=\\frac{\\ln(1+3c)}{c}$ for some $2n<c<2n+1$.\r\n\r\nNow it is simply to see that $a_{n}\\rightarrow 0$ as $n \\rightarrow \\infty$.\r\n\r\nIt remains to prove that $a_{n}$ is a decreasing sequence. It decreases in view of the fact that it converges to 0, however it can be proven to be decreasing by elementary calculations.\r\n\r\n$a'(n)=2\\left(\\frac{\\ln(1+3(2n+1))}{2n+1}-\\frac{\\ln(1+3(2n))}{2n}\\right)$, hence it sufficient to show that\r\n\r\n$x\\rightarrow \\frac{\\ln(1+3x)}{x}$ is a decreasing function, which is.",
  "Solution_2": "ok, my solution is the same"
}
{
  "Tag": [],
  "Problem": "Just wondering how fast you guys run...\r\nList your times for mile/100yd dash or whatever else you do.\r\n\r\nI run a 7:50 mile as of May/June , but I should be able to bring that down to atleast 6:40ish cuz I'm doing cross country this year(my first time-I'm in 7th grade).\r\nMy 100yd dash  was 15.9 secs I think as of May/June.\r\nI'll update in about two weeks when we run in gym.",
  "Solution_1": "[quote=\"nonie\"]Just wondering how fast you guys run...\nList your times for mile/100yd dash or whatever else you do.\n\nI run a 7:50 mile as of May/June , but I should be able to bring that down to atleast 6:40ish cuz I'm doing cross country this year(my first time).\nMy 100yd dash  was 15.9 secs I think as of May/June.\nI'll update in about two weeks when we run in gym.[/quote]\r\n\r\nMile time: 6:20\r\n\r\n100 METER dash time: 14 sec.",
  "Solution_2": "mile:   6:12\r\n\r\ni dont know the other one\r\n\r\ntheres this guy in my grade who can run 5:53",
  "Solution_3": "I ran a 5:01 mile last track season.",
  "Solution_4": "wow. i just got my mile under 6:00 this spring, and i'm a high school senior...\r\nthen again, i only ever did a season of indoor track in 9th grade, then quit.",
  "Solution_5": "My mile time is about 5:50. I prolly could do better but I haven't timed myself for a while.\r\nI forget my time for 100 yd dash though",
  "Solution_6": "my mile time is about 6:20",
  "Solution_7": "Don't laugh at mine, please \r\n\r\n10:45 :blush: :blush:\r\n\r\nI'm serious. I'm not proud.",
  "Solution_8": "i got 9ish... :P",
  "Solution_9": "I'm at 6:08, but I should be faster since I'm running cross country, and I've never ran the 100m dash, but I did it once as a trial and got 17.2, or was it 16 something?",
  "Solution_10": "[quote=\"236factorial\"]I'm at 6:08, but I should be faster since I'm running cross country, and I've never ran the 100m dash, but I did it once as a trial and got 17.2, or was it 16 something?[/quote]\r\nhmm...\r\nI've never done the 100 METER dash only the 100 YD one(and only like twice in gym)\r\nany idea how big the difference is in seconds?i think 100 m is 9 yds more than 100yds rite?",
  "Solution_11": "[quote=\"nonie\"]I've never done the 100 METER dash only the 100 YD one(and only like twice in gym)\nany idea how big the difference is in seconds?i think 100 m is 9 yds more than 100yds rite?[/quote]\r\n\r\nYeah, it's a little more than 9 yards more. \r\n\r\nI think you add about 1 second to your yards time to obtain meters time.",
  "Solution_12": "In 6th grade I ran 6:27...I havent done the mile run ever since. Im in 9th grade right now.",
  "Solution_13": "i dont know my endurance is horrible but my speed is good",
  "Solution_14": "Wow and I thought my 7 minute mile was fast...! Oh well, I'm only 13 & I don't train or anything. running is great though! I burn so many calories and I get to eat that extra slice of cheesecake at the end of the day :D",
  "Solution_15": "theres a guy at my friends school that can run a mile in 4:30! :o hes fast. I can only run it in like 7 minutes",
  "Solution_16": "the fastest i've seen anyone run a mile is 5:20.",
  "Solution_17": "[quote=\"jhollenbeck\"]theres a guy at my friends school that can run a mile in 4:30! :o hes fast. I can only run it in like 7 minutes[/quote]\r\n\r\n4:30??? That's crazy! He'd be... 400 meters ahead. me and my crazy math skills :D",
  "Solution_18": "Around 6:30 for the first mile.\r\n\r\nI don't run speed."
}
{
  "Tag": [
    "limit",
    "calculus",
    "calculus computations"
  ],
  "Problem": "Prove that $ b^{x\\plus{}y} \\equal{} b^{x}b^{y}$ for all real $ x,y$.\r\n\r\nSo $ b^{x}b^{y} \\equal{} \\sup B(x) \\sup B(y) \\equal{} \\sup(B(x)B(y))$ where $ B(x) \\equal{} \\{b^{t}| 0< t \\leq x \\}$.\r\n\r\n\r\nIs this correct?",
  "Solution_1": "Again, what is your definition of exponentiation?",
  "Solution_2": "We can always find a monotone  secuence of rationals $ \\{a_{n}\\}$ so that this secuence have  $ x$ (real) as limit,  we define $ b^x$ as the $ \\lim_{n\\rightarrow \\infty} b^{a_{n}}$. Finally if we do the same with $ y$ is easily to prove that $ b^{xy}\\equal{}b^{x\\plus{}y}$",
  "Solution_3": "Obviously you should take the sup over $ \\{b^t | t\\le x, t\\in \\mathbb{Q}\\}$.."
}
{
  "Tag": [
    "function",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "Solve the function $ (x \\plus{} y)f(yf(x)) \\equal{} x^2f(f(x) \\plus{} f(y))$ $ f: R^ \\plus{} \\to R^ \\plus{}$ for all x,y>0",
  "Solution_1": "[quote=\"StefanS\"]Solve the function $ (x \\plus{} y)f(yf(x)) \\equal{} x^2f(f(x) \\plus{} f(y))$ $ f: R^ \\plus{} \\to R^ \\plus{}$ for all x,y>0[/quote]\r\nLet $ P(x,y)$ be the assertion $ (x\\plus{}y)f(yf(x))\\equal{}x^2f(f(x)\\plus{}f(y))$\r\n\r\nIf $ f(a)\\equal{}f(b)$ : dividing $ P(a,y)$ by $ P(b,y)$, we get $ \\frac{a\\plus{}y}{b\\plus{}y}\\equal{}\\frac{a^2}{b^2}$ and so $ (a\\minus{}b)(ay\\plus{}by\\plus{}ab)\\equal{}0$ $ \\forall y$ and so $ a\\equal{}b$ and so $ f(x)$ is injective.\r\n\r\nThen $ P(\\frac{1\\plus{}\\sqrt 5}2,1)$ $ \\implies$ $ \\frac{3\\plus{}\\sqrt 5}2f(f(\\frac{1\\plus{}\\sqrt 5}2))$ $ \\equal{}\\frac{3\\plus{}\\sqrt 5}2f(f(\\frac{1\\plus{}\\sqrt 5}2)\\plus{}f(1))$ and so, since $ f(x)$ is injective :\r\n\r\n$ f(\\frac{1\\plus{}\\sqrt 5}2)\\equal{}f(\\frac{1\\plus{}\\sqrt 5}2)\\plus{}f(1)$ and so $ f(1)\\equal{}0$, which is impossible since $ f: \\mathbb R^\\plus{}\\to\\mathbb R^\\plus{}$\r\n\r\nSo no solution.\r\n\r\n(where is this silly problem coming from ?)"
}
{
  "Tag": [
    "linear algebra",
    "linear algebra solved"
  ],
  "Problem": "Let $A,B \\in M_n(\\mathbb C)$ such that $AB=A+B$ \r\nProve that $rank(A)=rank(B)$\r\nby Constantin Cocea si Mihai Piticari\r\n------------------------------------------------------------------------------------------\r\nI would have another solution here mine with mathematic in  MPSI/MP*  \r\n\r\nThe relation $AB=A+B$ => $A, B$ commutes, so there are simultaneous\r\ntrigonalisable in $M_n(C)$.They have the same rank.",
  "Solution_1": "The condition $AB=A+B$ can be rewritten as $(A-I)(B-I)=I.$\r\nSince matrices commute with their inverses, this also means that $(B-I)(A-I)=I.$\r\n\r\nSuppose $Av=0$. Then,\r\n$(A-I)v=-v$\r\nMultiply both sides of this by $B-I$:\r\n$v=-(B-I)v$\r\n$(B-I)v=-v$\r\n$Bv=0$.\r\nSo the null space of $A$ is the same as the null space of $B$; of course that means they have the same rank.",
  "Solution_2": "Nice solution Kent.",
  "Solution_3": "Another quick solution (although it uses the well-known fact that if $X$ is invertible, then $rank(AX)=rank(A)$, which I won't prove here).\r\n\r\nLet $X=I-A,\\ X^{-1}=I-B$. We have $rank(A)=rank(I-X)=rank(X^{-1}(I-X))$, which is equal to $rank(X^{-1}-I)=rank(I-X^{-1})=rank(B)$.",
  "Solution_4": "thx grobber",
  "Solution_5": "I have another methode.\n\nScince $AB=A+B$, we have $B=A(B-I)$, and then :\n\\[{\\text{rank}}(B)={\\text{rank}}\\left(A(B-I)\\right)\\leq\\min\\left({\\text{rank}}A,{\\text{rank}}(B-I)\\right)\\leq{\\text{rank}}(A).\\]\nNow we have $A=(A-I)B$, and thus :\n\\[{\\text{rank}}(A)={\\text{rank}}\\left((A-I)B\\right)\\leq\\min\\left({\\text{rank}}(A-I),{\\text{rank}}B\\right)\\leq{\\text{rank}}(B).\\]\nTherefore ${\\text{rank}}(A)={\\text{rank}}(B)$.\n\nSorry for bad English."
}
{
  "Tag": [
    "abstract algebra",
    "linear algebra",
    "matrix",
    "Ring Theory",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "Denote by $K(R)$ the [url=http://planetmath.org/encyclopedia/TotalRingOfFractions.html]total ring of fractions[/url] of a reduced ring $R$. Prove that every prime ideal of $K(R)$ is maximal.\r\nEDIT: If it helps, you may suppose, that $R$ is noetherian",
  "Solution_1": "I had posted a proof for the weaker problem (when $R$ is Noetherian), but it was flawed. I hope this one does work. \r\n\r\nInstead of working with the total ring of fractions, let's just assume the extra condition that all elements of $R$ are either invertible or zero divisors. This characterizes total rings of fractions. We can show that with these stronger hypotheses, $R$ must actually be a finite direct product of fields.\r\n\r\nConsider a primary decomposition $(0)=\\bigcap_{i=1}^n\\frak q_i\\ (*)$ of the zero ideal of $R$, and let $\\frak p_i$ be the radical of $\\frak q_i$. Because the ring is reduced, the radical of $(0)$ is $(0)$, and passing to radicals in $(*)$ we get another primary decomposition $(0)=\\bigcap_{i=1}^r\\frak p_i$, where $\\frak p_i,\\ i\\in\\overline{1,r}$ are minimal prime ideals. The set of zero divisors in $R$ is exactly $\\bigcup_{i=1}^r\\frak p_i$, and since any maximal ideal $\\frak m$ contains only non-invertible elements and is thus a subset of the set of zero divisors, each such ideal $\\frak m$ coincides with one of the minimal prime ideals $\\frak p_i$. In particular, every maximal ideal is a minimal prime ideal, and we've obtained what we initially wanted: every prime ideal is maximal. \r\n\r\nIt follows from the above that every localization $R_\\frak m$ at a maximal ideal $\\frak m$ is a field (for a commutative ring this is equivalent to the ring being reduced and of dimension zero). Now take $a\\in R$, and let $\\overline R=R/\\mbox{Ann}(a)$. $\\overline R$ is still reduced and of dimension zero, so again, each localization $\\overline R_\\frak m$ is a field. If $\\bar a$ (the image of $a$ in $\\overline R$) is not invertible, then we can find $\\frak m$ which contains it. Then on the one hand $\\frac{\\bar a}1$ is not zero in $\\overline R_\\frak m$, and on the other hand it's not invertible, and we have a contradiction (because $\\overline R_\\frak m$ is a field). This means that $\\bar a$ must be invertible, i.e. there is an $x\\in R$ such that $a=xa^2$. In other words, $R$ is a von Neumann regular ring. \r\n\r\nBeing von Neumann regular and Noetherian, it's semisimple, so by the Wedderburn-Artin Theorem it's a finite direct product of matrix rings $M_{n_i}(D_i)$ over division rings $D_i$. From the fact that our ring is commutative it follows that all $n_i$ are $1$ and all $D_i$ are in fact fields, so we're done.",
  "Solution_2": "I think I have a counterexample for the initial question (in which the ring was not assumed to be Noetherian). \r\n\r\nLet $k$ be any field, and consider the ring $S$ of formal power series over $k$ in indeterminates $x_{i},\\ i\\ge 1$ subject to $x_{i}x_{j}=0,\\ \\forall i\\ne j$. The ring $R$ will be the subring of $S$ consisting of those power series in which only finitely many of the $x_{i}$ appear. \r\n\r\nThere are now a couple of things one can easily check: (a) $R$ is reduced; (b) $R$ is local, and its maximal ideal consists of the series with no free term. By (b), a non-invertible element $f$ must be a finite sum of series of the form $a_{1}x_{i}+a_{2}x_{i}^{2}+\\ldots$ for various $i$ (remember that $x_{i}x_{j}=0,\\ \\forall i\\ne j$). If we multiply $f$ by some $x_{j}$ different from all $x_{i}$ appearing in $f$, we get $0$, so every non-invertible element is a zero divisor. Equivalently, $R$ is its own total ring of fractions. However, it cannot be true that every prime ideal is maximal, since for a local reduced ring this is equivalent to being a field. Of course, this is not the case here."
}
{
  "Tag": [
    "videos",
    "Gauss"
  ],
  "Problem": "Hi, I have a random question.\r\n\r\nIf you could meet with any person in history, who would it be, what would you say, and what is it about them that you relate to? \r\n\r\nI'm interested in your responses.",
  "Solution_1": "mmm putting this in the college forum\r\nmust be teasing me...\r\ni just had a practice SAT CR section about who someone would want to meet from history, it was saying how they wouldn't want to meet a famous person, but rather someone who was there to observe lots of them\r\n\r\ni don't really have any of my own ideas about who i'd like to meet heh :D\r\n\r\nprobably should move this to round table",
  "Solution_2": "It will surely be either Einstein or Feynman, two of the most interesting people in history. Can't decide on Erdos though ...",
  "Solution_3": "[quote=\"Koda\"]It will surely be either Einstein or Feynman, two of the most interesting people in history. Can't decide on Erdos though ...[/quote]\r\n\r\nUmm interesting, of all people in history you chose all scientists? I agree it must have been very interesting to be acquainted with both of them but I certainly don't think they are the ones who have most to teach about history. About physics sure, in the case of Erdos about math but about history...\r\n\r\nIt is an interesting question. I find myself unable to decide. So I've made a chronological list\r\n\r\nBefore Greece, Siddharta Gautama then, I'd like to know Socrates, Alexander the Great, Hadrian and Marcus Aurelius, Jesus, and Mary Magdalene if she did exist, Muhammad, da Vinci and that's nearly nine hundred years without one person that really interests me, no wait, Marco Polo is in between, there are a lot of great people in the Renaissance so I'll leave it, Newton, Fouch\u00e9, Christopher Colombus, this is becoming too lengthy I know, Lincoln, Bolivar, Lenin, Goebbels, Mandela...\r\n\r\n[color=red]Edit[/color] I forgot to say why. Gautama, Muhammad and Jesus, because they are icons of some of the greatest religions of our time and I'm interested about their true personality, time and history surely distort somehow their humanity\r\n\r\nIn the case of Mary Magdalene I'm intrigued about the truth in the Bible. This has been recently put forward in a dubious way by things like the Da Vinci Code phenomena but it is a subject that indeed has a lot of misterious.\r\n\r\nHadrian and Marcus Aurelius were good roman emperors, and I'm impressed by their achievements when Rome got to know its greatest time. And history puts them as surprisingly advanced minds for their time which makes me interested.\r\n\r\nMarco Polo and Christopher Colombus are the kind of person I would like to be in some other life. Curious explorer of strange lands.\r\n\r\nNewton and da Vinci because it must be an overwhelming experience to meet a mind of such calibre.\r\n\r\nFouch\u00e9 is a guy who has to tell us about French history at its apex, at the time it was strongly influencing world history, when there was monarchy, revolution, empire and all in thirty years or so. Fouche lived and played a role in all epochs so it must be interesting. Not a good guy it appears but wait til Goebbels.\r\n\r\nLincoln and Bolivar for their major role in American history, north and south respectively. And because they seem to have been great honourable men.\r\n\r\nGoebbels to understand how nazism could have occured to humanity. I want to hear first hand how you convince many millions of such a horrible 'truth'. What is there behind it?\r\n\r\nLenin because he marks the beginning of a world with two poles, I wan't to know what was the communism he planned, how he would have done things if he wouldn't have died.\r\n\r\nEn fin, Mandela, because he represents the modern sacrifice for a just cause, because he decided not to be a ruler and fall in the webs of power and because I find him a particularly nice old man\r\n\r\nSorry about the length :)\r\n\r\nUrsula",
  "Solution_4": "I can only pick da Vinci from that least.\r\nWhat about Puskas?",
  "Solution_5": "I would pick Hannibal, because he was such a great leader. I wonder what he REALLY lookd like, and how he spoke. I've only seen videos and stories in books about the Punic Wars and how he almost helped Carthage win the 2nd Punic war.\r\n\r\n\r\nit would be great to have a chance to meet someone who lived in a completely different era.",
  "Solution_6": "History seems to convey war heroes to most people.\r\nWell, let me tell you, war is no adventure.",
  "Solution_7": "nostradamus, cause he knew the msot about what was going to go down in history at the time",
  "Solution_8": "[quote=\"Koda\"]It will surely be either Einstein or Feynman, two of the most interesting people in history. Can't decide on Erdos though ...[/quote]\r\n\r\n\r\nfeynman would be my choice too  :)",
  "Solution_9": "Yeah it should be the natural choice.\r\nBy the way, Ignite, Nostradamus knew nothing about the future. :roll:",
  "Solution_10": "I will select two different kinds of people:\r\nGroup I: Carroll, Wodehouse, Douglas Adams\r\nGroup II: Marx, Lenin, Stalin, Mao, Ho, Che",
  "Solution_11": "[quote=\"koda\"]It will surely be either Einstein or Feynman, two of the most interesting people in history. Can't decide on Erdos though ...\n[/quote]\n\n\n[quote=\"bubka\"]I will select two different kinds of people:\nGroup I: Carroll, Wodehouse, Douglas Adams\nGroup II: Marx, Lenin, Stalin, Mao, Ho, Che[/quote]\r\n\r\n\r\nwhy i find this strange? :huh:",
  "Solution_12": "About Feynman...\r\n[quote=\"Koda\"]Yeah it should be the natural choice.\n[/quote]\n\nWhy? What is it about Feynman that is natural to pick him?\n\n[quote=\"bubka\"]\nI will select two different kinds of people: \nGroup I: Carroll, Wodehouse, Douglas Adams \nGroup II: Marx, Lenin, Stalin, Mao, Ho, Che\n[/quote]\r\n\r\nBoy, I'm not sure the people in your first group will approve people in your second, ja! But I'm interested, why do you want to meet this guys?\r\n\r\n\r\nUrsula",
  "Solution_13": "[quote=\"ursula\"]About Feynman...\nWhy? What is it about Feynman that is natural to pick him?\n[/quote]\r\n\r\n\r\ngreat man. and did you read his famous: feynman lectures on physics?",
  "Solution_14": "Yeah, Even I would go for groups-\r\nGroup I-Wodehouse, Shakespeare, Asimov, and Christie\r\nGroup II-Einstein and Ramanujan \r\nGroup III-Only Indians may understand this one- Akbar, and if u would allow me to go to  really really ancient history, then each of the Dashavatharams(10 incarnations of God :D ).\r\n :lol: \r\nWho knows, maybe someone will come up with some kind of weird time machine and we may get to do this!! :)",
  "Solution_15": "Reimann, Gauss, or Yang Mills in Mathematics.\r\n\r\n\r\nIn political issues, Ronald Reagan.\r\n\r\n\r\nIn philosophy, Socrates, Plato, or Aristotle.\r\n\r\n\r\nIn writing, Edgar Allen Poe.\r\n\r\n\r\n\r\na lot of people.....",
  "Solution_16": "[quote=\"ursula\"]About Feynman...\n[quote=\"Koda\"]Yeah it should be the natural choice.\n[/quote]\n\nWhy? What is it about Feynman that is natural to pick him?\n\n[quote=\"bubka\"]\nI will select two different kinds of people: \nGroup I: Carroll, Wodehouse, Douglas Adams \nGroup II: Marx, Lenin, Stalin, Mao, Ho, Che\n[/quote]\n\nBoy, I'm not sure the people in your first group will approve people in your second, ja! But I'm interested, why do you want to meet this guys?\n\nUrsula[/quote]\r\n\r\nThey are respectively the greatest in the fields of nonsense/farce and communism/revolution, my two true passions.",
  "Solution_17": "[quote=\"deimos\"][quote=\"ursula\"]About Feynman...\nWhy? What is it about Feynman that is natural to pick him?\n[/quote]\n\n\ngreat man. and did you read his famous: feynman lectures on physics?[/quote]\n\nYeah, as a matter of fact, I recommend them to every non physicist person I know but still with an interest for physics. But Feynman's great insight and physical intuition and even greater capacity for teaching does not make him the \"natural\" person to meet among all the other great people in history, at least not for me. I also read the book that apparently has made him so widely known to the general public, \"Surely, you're joking Mr Feynman\". And I also feel admiration for him. But I happen to know also that Feynman and others celebrated the Hiroshima bombing, he must have seen the thing as the logical conclusion to their scientific efforts in Los Alamos. Nevertheless other scientists involved, particularly Oppenheimer, were already regretting to have been contributors to project, in  Op's case the scientific leader, and warning about the catastrophic consequences. Indeed Einstein never agreed to participate in the project. So what is Feynman to me? Certainly an outstanding physicist with a good bunch of charisma and a whole lot of other amazing qualities unrelated to physics nonetheless but among other geniuses, or other outstanding people he does not occupies, for me an special place. Sure, if I list a hundred people Feynman will make it but since I studied theoretical physics [i]his[/i] contribution to mankind is familiar to me. That's why I didn't put Einstein in the list either, another modern great mind.\n\n[quote=\"nutz\"]\nReimann, Gauss, or Yang Mills in Mathematics.\n[/quote]\nYang and Mills are two different guys, you know this, do you? And I don't know about their work in mathematics but I think they are more widely known for their work in physics. They proposed for the first time one model of the kind of theory we now call Yang-Mills.\n\n[quote=\"bubka\"]\nThey are respectively the greatest in the fields of nonsense/farce and communism/revolution, my two true passions.\n[/quote]\r\nYes, I see you have a passion for revolution and communism. I myself do, but I don't think Stalin or Mao have been the greatest in this field. Probably they have been the greatest in destroying revolutions and certainly the greatest in killing innocent people. You remind me of Ignite  except for you are in the opposite corner of the ring. But I'm again interested, although this probably deserves another thread, have you ever tried to work out a 'revolutionary' world that works? Oh make another thread if you are going to answer seriously.\r\n\r\nUrsula",
  "Solution_18": "[quote=\"ursula\"]Yes, I see you have a passion for revolution and communism. I myself do, but I don't think Stalin or Mao have been the greatest in this field. Probably they have been the greatest in destroying revolutions and certainly the greatest in killing innocent people. You remind me of Ignite  except for you are in the opposite corner of the ring. But I'm again interested, although this probably deserves another thread, have you ever tried to work out a 'revolutionary' world that works? Oh make another thread if you are going to answer seriously.\n\nUrsula[/quote]\r\n\r\nDoes this mean that I am the creator of revolutions and the greatest in saving the innocent?",
  "Solution_19": "she was referring to me in that sentence and stalin/mao in the previous one, so your joke is annulled. better luck next time.\r\n\r\nhaving said that, i must add that i hope to change her ideas about these two stalwarts of communism.",
  "Solution_20": "Uh-hem.  I'd meet a relation of Nancy Reagan, or at least someone who'd say for once and for all whether she was born in 1923, like she says in her autobiography, or in 1921, which Wikipedia says.\r\n\r\nWikipedia?  Or herself?  Really there's no contest, but still.",
  "Solution_21": "Hypatia because there was very small number of woman in science/philosophy and she was one of them.\r\n\r\nTolkien because he created new world and much more than that.\r\n\r\nGalileo and Newton because I admire them.",
  "Solution_22": "[quote=\"Nienna\"]Hypatia because there was very small number of woman in science/philosophy and she was one of them.\n[/quote]\r\n\r\nBeautiful, I inmediately thought after my post that there was actually only one woman in my list and that it was very sad. I didn't editted it nonetheless because I thought that if they didn't came to my mind at first, it would be because objectively the most of the contributions were done before by men. Luckily, we're coming out of that epoch. Some other women that produce me admiration are Emmy Noether, Marie Curie, Anais Nin, Plotina(wife of Trajan, romam emperor).\r\n\r\nUrsula"
}
{
  "Tag": [
    "trigonometry",
    "analytic geometry",
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "In the Cartesian plane, let $C$ be a unit circle with center at origin $O$. For any point $Q$ in the plane distinct from $O$, define $Q'$ to be the intersection of the ray $OQ$ and the circle $C$. Prove that for any $P\\in C$ and any $k\\in\\mathbb{N}$ there exists a lattice point $Q(x,y)$ with $|x|=k$ or $|y|=k$ such that $PQ'<\\frac{1}{2k}$.",
  "Solution_1": "Consider the case when $OP$ makes an angle $\\theta$ such that $0\\leq\\theta\\leq\\frac{\\pi}{4}$.  The argument then extends to all of $C$ by symmetry.  Let $k$ be given.  Consider all the lattice points of the form $Q(k,y)$ with $0\\leq y\\leq k$.  Then by drawing all the rays $OQ$, we separate the eighth-circle into $k$ arcs.  The length of the $j^{th}$ arc is $\\tan^{-1}\\left(\\frac{j}{k}\\right)-\\tan^{-1}\\left(\\frac{j-1}{k}\\right)$ by the formula $s=r\\theta$, with $r=1$ in this case.  Since $\\tan x$ is concave down, i.e. $\\frac{d^{2}}{dx^{2}}\\tan x \\leq 0$, when $x\\geq 0$, the longest arc will be the first one, between the first and second rays, formed by the origin and $(k,0)$ and $(k,1)$, respectively.  This arc has length $\\tan^{-1}\\left(\\frac{1}{k}\\right)-\\tan^{-1}\\left(\\frac{1-1}{k}\\right)=\\tan^{-1}\\left(\\frac{1}{k}\\right)$.  Since $\\tan^{-1}x<x$ for all $x>0$, $\\tan^{-1}\\left(\\frac{1}{k}\\right)<\\frac{1}{k}$.  So all the arcs have length less than $\\frac{1}{k}$.  This means that no matter what arc $P$ lies in, the length of the arc between $P$and the closest $Q'$ must be less than $\\frac{1}{2k}$, which implies $PQ'<\\frac{1}{2k}$, as required."
}
{
  "Tag": [
    "MATHCOUNTS",
    "probability",
    "geometry",
    "3D geometry",
    "tetrahedron"
  ],
  "Problem": "[b]Another Mock MATHCOUNTS competition [/b]\r\nWill consist of Sprint (40 min) and Target (24 min)\r\n\r\nCompetition is over, but If you want you can still submit answers as long as you dont cheat.\r\n\r\n[b][u]Please reply if you are going to do it[/b][/u]",
  "Solution_1": "I want to compete",
  "Solution_2": "me 2",
  "Solution_3": "Umm, I think i can make it though i won't do very good. (Didn't participate in MATHCOUNTS but hoping to next year)...\r\n\r\nEveryone remember to turn their clocks tomorrow!\r\n\r\nAnd if you do have countdown round, how does it work?",
  "Solution_4": "He'll probably have everybody invited to some AIM chatroom, tell the competitors to mute everybody but himself and the other competitor, and then ask the question. I don't know if the person will have to \"buzz in\" or just type the answer... but that's what I'm guessing...",
  "Solution_5": "[quote=\"tarquin\"]He'll probably have everybody invited to some AIM chatroom, tell the competitors to mute everybody but himself and the other competitor, and then ask the question. I don't know if the person will have to \"buzz in\" or just type the answer... but that's what I'm guessing...[/quote]Good explanation. It will be an AIM chatroom, and you will have to buzz. Same rules as normal, but I might do It national style (bracket vs. ladder).",
  "Solution_6": "Ahh... In that case,  i won't participate in countdown. No AIM but i have an account. I might be able to go on AIM Express.",
  "Solution_7": "[quote=\"DarkKnight\"]Ahh... In that case,  i won't participate in countdown. No AIM but i have an account. I might be able to go on AIM Express.[/quote]Its free to download. I can send a link.",
  "Solution_8": "Umm... I'm not allowed to have AIM but I think my parents may let me on AIM Express to do the Countdown round.",
  "Solution_9": "What is AIM express?? If you can join a chatroom its fine",
  "Solution_10": "If there are more than 10, then the top 10 would make countdown round. Of course if you have less, the cutoff can be somewhat higher I guess.",
  "Solution_11": "[quote=\"tarquin\"]If there are more than 10, then the top 10 would make countdown round. Of course if you have less, the cutoff can be somewhat higher I guess.[/quote]The cutoff will probably be 12 if we have that many. Else everybody",
  "Solution_12": "[quote=\"biffanddoc\"]What is AIM express?? If you can join a chatroom its fine[/quote]\r\n\r\nI'm not sure if you can join a chatroom. I'll see.\r\n\r\nYes you can. I'll ask my parents if i can use AIM Express tomorrow once we're sure we have enough people.",
  "Solution_13": "3 so far...",
  "Solution_14": "Haha. Yeah, we're going to need a whole lot more. ;)\r\n\r\nTURN YOUR CLOCKS. We don't want to have anyone an hour late.",
  "Solution_15": "1\r\nYou can tell that the intercepted arcs of AC and AB are all 60 degrees, so there's an equilateral triangle. $\\frac 16$ of the circle minus the equilateral triangle with a side of 7 times four plus two times the area of the equilateral triangle with side 7 is the answer.\r\n\r\n6\r\nOnly way that the sum can be odd is if 2 is in it because 2 is the only even prime and an even plus an odd is the only way to get an odd sum. The number of ways to pick 6 numbers is $\\frac {18!}{12!6!}$. Since you know there has to be a 2 in there, you find the number of ways to pick the other 5 numbers, $\\frac {17!}{5!12!}$ and the final answer is $\\frac {\\frac {17!}{5!12!}}{\\frac {18!}{12!6!}} = \\frac 13$.",
  "Solution_16": "Thanks tarquin! :D",
  "Solution_17": "cool i'm a hodge conjecture now  :)",
  "Solution_18": "How to do #3\r\nIs the intersection of the perpendicular bisectors the center of the circle?",
  "Solution_19": "Yes that is the center of the circle. \r\nAlso note that the perpendicular bisectors of the sides of a triangle are concurrent at the center of its circumscribed circle.",
  "Solution_20": "[quote=\"tarquin\"]8\n$\\eqnarray 4\\pi r^2 = 256 \\\\\n& = & \\pi r^2 = 64 \\\\\n& = & r = \\frac {8}{\\sqrt \\pi} \\\\\n& = & r^3 = \\frac {512}{\\pi * \\sqrt \\pi} \\\\\n& = & \\frac 43 \\pi r^3 = \\frac {2048}{3\\pi \\sqrt \\pi} * \\pi$\nRationalize..\n$\\frac 43 \\pi r^3 = \\frac {2048 \\sqrt \\pi}{3 \\pi}$\n\n12 \nFive vowels, twenty-one consanants.\nVCVC\n$5 * 21 * 4 * 20$\nTimes four again because you can switch vowels with eachother and consanants with eachother... $8400 * 4 = 33600$\n\n13\n$(5, 2) ( -3 , 4) ( 8 , -2)$\nScalefactor of $\\frac 25$ ...\n$2 + \\frac 45 - \\frac 65 + \\frac 85 - \\frac {16}{5} - \\frac 45 = \\frac {28}{5}$\n\n18\nCommon fraction... $\\frac {3}{25}$\n\n21\nProbability for prime $\\frac 35$. Has to get in this order PRIME - NOPRIME or NOPRIME - PRIME, so $\\frac 35 * \\frac 25 * 2 = \\frac {12}{25}$\n\n22\nPossible points are $(3,3)(3,2)(3,1)(2,2)(2,1)(1,1)$.\nPossible midpoints are $(3,2)(2,1)(2,2)$.\nMust choose two points for every mid point backwards or forwards so $\\frac 13 * \\frac 15 * 2 * 3 = \\frac 25$\n\n28\nVolume of tetrahedron is $528 * \\frac 12 * \\frac 13 = 88$\n\nI messed the test up horribly...[/quote]\r\n#28, Let x be the side of the orginal cube, the the volume of the tetrahedron is:\r\n(1/2) * (x/2) * (x/2) * (x/2) * (1/3)= x^3/48=528/48=11. \r\n\r\n#22, {{{1, 1}, {1, 3}}, {{1, 1}, {3, 1}}, {{1, 1}, {3, 3}}, {{1, 2}, {3, 2}}, {{1, 3}, {1, 1}}, {{1, 3}, {3, 1}}, {{1, 3}, {3, 3}}, {{2, 1}, {2, 3}}. \r\nThere are 8 of them.\r\nThere are (9 choose 2)=36 total possiblities.\r\nthe answer is: 8/36 = 2/9. \r\n\r\n#21. It can be prime-prime or noprime-noprime.\r\n\r\n13 and 18 are messed up. Ill change scores. :blush:  :blush:",
  "Solution_21": "Scores are updated.\r\n\r\nTarquin-30\r\nXantos C. Guin-29\r\nChess64-10\r\nSolfidefarms-4",
  "Solution_22": "#21\r\nSays [b]exactly one[/b] prime.\r\n\r\nRest are right, sorry about that :) I'm an idiot.",
  "Solution_23": "i will",
  "Solution_24": "i will",
  "Solution_25": "i only did the target round - i got 10  :blush:",
  "Solution_26": "[quote=\"tarquin\"]#21\nSays [b]exactly one[/b] prime.\n\nRest are right, sorry about that :) I'm an idiot.[/quote]\r\nOoh youre right about exactly one prime. But, The probability of Prime-noprime is $ \\frac 25 * \\frac {12}{19} = \\frac {24}{95}$ and the probability of noprime-prime is  $ \\frac 35 * \\frac {8}{19} = \\frac {24}{95}$. Add together and get $ \\frac {48}{95}$ Am I not right?",
  "Solution_27": "I think it said with replacement.\r\nAnd how did you get 19's and 20's? I see only 5 numbers there.",
  "Solution_28": "[quote=\"tarquin\"]I think it said with replacement.\nAnd how did you get 19's and 20's? I see only 5 numbers there.[/quote]\r\nO you're right about replacement. When I did the problems, I wrote the wrong thing. Ill give you credit.",
  "Solution_29": "Lol for sprint theres 31 downloads and 2 submissions?? As long as you dont cheat, You can still submit"
}
{
  "Tag": [],
  "Problem": "A man was born in the early part of the 19th century was x years old in the year $x^{2}$, while a woman born in the later part of the 19th century was y years old in the year $y^{2}$. How many years apart were they born?",
  "Solution_1": "[quote=\"ckck\"]A man was born in the early part of the 19th century was x years old in the year $x^{2}$, while a woman born in the later part of the 19th century was y years old in the year $y^{2}$. How many years apart were they born?[/quote]\r\n[hide]Well, the only perfect square in the 1800s is 1849 (43) so the man was born in 1806. The next perfect square is 1936 so the lady was born in 1892. Our answer is 86.[/hide]"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "trigonometry",
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "Let x,y be real number such that $x^{2}+xy+y^{2}=3$. Find the greatest value of $M=x^{2}y+x-y$",
  "Solution_1": "$x^{2}+xy+y^{2}-3=0\\Leftrightarrow x=$ $-\\frac{1}{2}y\\pm \\frac{1}{2}\\sqrt{\\left(-3y^{2}+12\\right) }$\r\n\t\r\n$M=\\left(-\\frac{1}{2}y\\pm \\frac{1}{2}\\sqrt{\\left(-3y^{2}+12\\right) }\\right)^{2}y+\\left(-\\frac{1}{2}y\\pm \\frac{1}{2}\\sqrt{\\left(-3y^{2}+12\\right) }\\right)-y=-\\frac{1}{2}y^{3}-\\frac{1}{2}y^{2}\\sqrt{\\left(-3y^{2}+12\\right) }+\\frac{3}{2}y+\\frac{1}{2}\\sqrt{\\left(-3y^{2}+12\\right) }$\r\n\t\r\nAnd it's become polynomial in $y$\r\n\r\nSorry about my English.I will improve it :rotfl:",
  "Solution_2": "We can set $x=\\sqrt{3}\\cos \\theta-\\sin \\theta ,\\ y=2\\sin \\theta$.",
  "Solution_3": "For these kinds of problems, I think Lagrangian Multipliers are best.",
  "Solution_4": "Kunny, can you explain how did you get to that substitution?",
  "Solution_5": "$x^{2}+xy+y^{2}=3\\Longleftrightarrow \\left(x+\\frac{y}{2}\\right)^{2}+\\left(\\frac{\\sqrt{3}}{2}y\\right)^{2}=3$.",
  "Solution_6": "This one stumped me for a while but I managed to attack it with the substitution Kunny proposed.\r\n\r\n$x^{2}+xy+y^{2}=3$\r\n$\\left(x+\\frac{y}{2}\\right)^{2}+\\left(\\frac{\\sqrt{3}}{2}y\\right)^{2}=3$\r\n$\\left(x+\\frac{y}{2}\\right)^{2}=3\\cos^{2}\\phi$, $\\left(\\frac{\\sqrt{3}}{2}y\\right)^{2}=3\\sin^{2}\\phi$\r\n$x=\\sqrt{3}\\cos\\phi-\\sin\\phi$, $y=2\\sin\\phi$\r\n$x^{2}y+x-y=\\sqrt{3}\\cos\\phi-\\sin\\phi+4\\sin\\phi\\cos\\phi(\\cos\\phi-\\sqrt{3}\\sin\\phi)$\r\n\r\nNow we have to 'factor' somehow those summands. Notice that $\\sqrt{3}=\\tan\\frac{\\pi}{3}$.\r\n\r\n\\begin{eqnarray*}\\sqrt{3}\\cos\\phi-\\sin\\phi &=& \\tan\\frac{\\pi}{3}\\cos\\phi-\\sin\\phi \\\\ &=& \\frac{1}{\\cos\\frac{\\pi}{3}}\\left(\\sin\\frac{\\pi}{3}\\cos\\phi-\\cos\\frac{\\pi}{3}\\sin\\phi\\right) \\\\ &=& 2\\sin\\left(\\frac{\\pi}{3}-\\phi\\right) \\end{eqnarray*}\r\n\r\n\\begin{eqnarray*}\\cos\\phi-\\sqrt{3}\\sin\\phi &=& \\cos\\phi-\\tan\\frac{\\pi}{3}\\sin\\phi \\\\ &=& \\frac{1}{\\cos\\frac{\\pi}{3}}\\left(\\cos\\phi\\cos\\frac{\\pi}{3}-\\sin\\phi\\sin\\frac{\\pi}{3}\\right) \\\\ &=& 2\\cos\\left(\\phi+\\frac{\\pi}{3}\\right) \\end{eqnarray*}\r\n\r\n\\begin{eqnarray*}x^{2}y+x-y &=& 2\\sin\\left(\\frac{\\pi}{3}-\\phi\\right)+8\\sin\\phi\\cos\\phi\\cos\\left(\\phi+\\frac{\\pi}{3}\\right) \\\\ &=& 2\\sin\\left(\\frac{\\pi}{3}-\\phi\\right)+4\\sin2\\phi\\cos\\left(\\phi+\\frac{\\pi}{3}\\right) \\\\ &=&-2\\sin\\left(\\phi-\\frac{\\pi}{3}\\right)+2\\sin\\left(3\\phi+\\frac{\\pi}{3}\\right)+2\\sin\\left(\\phi-\\frac{\\pi}{3}\\right) \\\\ &=& 2\\sin\\left(3\\phi+\\frac{\\pi}{3}\\right)\\leq\\boxed{2}\\end{eqnarray*}\r\n\r\nSo, maximum of M is 2. Correct?"
}
{
  "Tag": [],
  "Problem": "When Helen jogs, every $ 20$ minutes she checks the total distance she traversed. One day, the first time she checked, Helen had jogged a distance of $ 2.5$ miles, the second time she checked, she had jogged a total distance of $ 4.8$ miles, and the third and final time that she checked, Helen found out that she jogged a total distance of $ 6.8$ miles. What is the difference between her average speeds, in miles per hour, during the first $ 20$ minutes of jogging and the last $ 20$ minutes of jogging?",
  "Solution_1": "The first $ 20$ minutes, Helen jogged $ 2.5$ miles, so her mph is $ 7.5$. The last $ 20$ minutes, she jogged $ 6.8\\minus{}4.8\\equal{}2$ miles, so her mph is $ 6$. So the difference in mph is $ 1.5$."
}
{
  "Tag": [
    "quadratics",
    "logarithms",
    "algebra",
    "calculus",
    "calculus computations"
  ],
  "Problem": "Let $ \\alpha ,\\ \\beta$\u3000be the real roots of the quadratic equation with respect to $ x$ : $ x^2 \\plus{} x\\log_8 a \\plus{} (\\log_8 a)^2 \\plus{} \\log_8 a \\minus{} 1 \\equal{} 0$. Find the range for which $ \\alpha ^3 \\plus{} \\beta ^3$\u3000can be valued.",
  "Solution_1": "I still don't understand what the \"question\" is..\r\nfor which $ \\alpha^3\\plus{}\\beta^3$ can be valued,, so $ \\alpha$ and $ \\beta$ must be valued too..(is that \"real\" according to the problem?)\r\n\r\nlet $ ^8\\log{a}\\equal{}p$ sorry it's not commonly viewed,, but in my country we write like that..\r\n\r\n$ x^2\\plus{}px\\plus{}(p^2\\plus{}p\\minus{}1)\\equal{}0$\r\nif the roots are real,, so $ D\\geq0 \\rightarrow p^2\\minus{}4(p^2\\plus{}p\\minus{}1)\\equal{}\\minus{}3p^2\\minus{}4p\\plus{}4\\geq0$\r\nso that's why $ 3p^2\\plus{}4p\\minus{}4\\leq0 \\rightarrow (3p\\minus{}2)(p\\plus{}2)\\leq0 \\rightarrow \\minus{}2\\leq p\\leq \\frac{2}{3}$\r\n\r\nso $ \\minus{}2\\leq ^8\\log{a}\\leq \\frac{2}{3} \\rightarrow \\frac{1}{64} \\leq a \\leq4$\r\n\r\nsorry if I'm mistaken..",
  "Solution_2": "ah,,yeah..\r\nI was mistaken in understanding the problem..\r\n\r\n[continue from above..]\r\n$ \\alpha\\plus{}\\beta\\equal{}\\minus{}p$\r\n$ \\alpha\\beta\\equal{}p^2\\plus{}p\\minus{}1$\r\n$ \\alpha^3\\plus{}\\beta^3\\equal{}(\\alpha\\plus{}\\beta)(\\alpha^2\\minus{}\\alpha\\beta\\plus{}\\beta^2)\\equal{}(\\alpha\\plus{}\\beta)[(\\alpha\\plus{}\\beta)^2\\minus{}3\\alpha\\beta]\\equal{}(\\minus{}p)(\\minus{}2p^2\\minus{}3p\\plus{}3)\\equal{}(2p^3\\plus{}3p^2\\minus{}3p)$\r\n\r\nokay,,subtitude the vallue of the previous $ p$ to $ \\alpha^3\\plus{}\\beta^3$\r\nif I'm not mistaken.. $ 2\\leq \\alpha^3\\plus{}\\beta^3\\leq\\frac{14}{3}$\r\n\r\n\r\nsorry if my solution is wrong.. :blush:  :blush:  :blush:",
  "Solution_3": "[quote=\"mhar_teens\"]\n\nlet $ ^8\\log{a} \\equal{} p$ sorry it's not commonly viewed,, but in my country we write like that..\n\n$ x^2 \\plus{} px \\plus{} (p^2 \\plus{} p \\minus{} 1) \\equal{} 0$\nif the roots are real,, so $ D\\geq0 \\rightarrow p^2 \\minus{} 4(p^2 \\plus{} p \\minus{} 1) \\equal{} \\minus{} 3p^2 \\minus{} 4p \\plus{} 4\\geq0$\nso that's why $ 3p^2 \\plus{} 4p \\minus{} 4\\leq0 \\rightarrow (3p \\minus{} 2)(p \\plus{} 2)\\leq0 \\rightarrow \\minus{} 2\\leq p\\leq \\frac {2}{3}$\n\nso $ \\minus{} 2\\leq ^8\\log{a}\\leq \\frac {2}{3} \\rightarrow \\frac {1}{64} \\leq a \\leq4$\n\n$ \\alpha \\plus{} \\beta \\equal{} \\minus{} p$\n$ \\alpha\\beta \\equal{} p^2 \\plus{} p \\minus{} 1$\n$ \\alpha^3 \\plus{} \\beta^3 \\equal{} (\\alpha \\plus{} \\beta)(\\alpha^2 \\minus{} \\alpha\\beta \\plus{} \\beta^2) \\equal{} (\\alpha \\plus{} \\beta)[(\\alpha \\plus{} \\beta)^2 \\minus{} 3\\alpha\\beta] \\equal{} ( \\minus{} p)( \\minus{} 2p^2 \\minus{} 3p \\plus{} 3) \\equal{} (2p^3 \\plus{} 3p^2 \\minus{} 3p)$\n[/quote]\r\n\r\nThat's correct so far.",
  "Solution_4": "when we get $ \\minus{}2 \\leq p \\leq \\frac23$,\r\n\r\nwe need to plot the graph of $ y \\equal{} 3p^3 \\plus{} 3p^2 \\minus{}3p$ to get the range of y when $ \\minus{}2 \\leq p \\leq \\frac23$\r\n\r\nI used the graph program to help me out... \r\n\r\nthen i get $ \\minus{}0.5981 \\leq \\alpha^3 \\plus{} \\beta^3 \\leq 4.5981$\r\n\r\n...my two cents.",
  "Solution_5": "[quote=\"kunny\"][quote=\"mhar_teens\"]\n\nlet $ ^8\\log{a} \\equal{} p$ sorry it's not commonly viewed,, but in my country we write like that..\n\n$ x^2 \\plus{} px \\plus{} (p^2 \\plus{} p \\minus{} 1) \\equal{} 0$\nif the roots are real,, so $ D\\geq0 \\rightarrow p^2 \\minus{} 4(p^2 \\plus{} p \\minus{} 1) \\equal{} \\minus{} 3p^2 \\minus{} 4p \\plus{} 4\\geq0$\nso that's why $ 3p^2 \\plus{} 4p \\minus{} 4\\leq0 \\rightarrow (3p \\minus{} 2)(p \\plus{} 2)\\leq0 \\rightarrow \\minus{} 2\\leq p\\leq \\frac {2}{3}$\n\nso $ \\minus{} 2\\leq ^8\\log{a}\\leq \\frac {2}{3} \\rightarrow \\frac {1}{64} \\leq a \\leq4$\n\n$ \\alpha \\plus{} \\beta \\equal{} \\minus{} p$\n$ \\alpha\\beta \\equal{} p^2 \\plus{} p \\minus{} 1$\n$ \\alpha^3 \\plus{} \\beta^3 \\equal{} (\\alpha \\plus{} \\beta)(\\alpha^2 \\minus{} \\alpha\\beta \\plus{} \\beta^2) \\equal{} (\\alpha \\plus{} \\beta)[(\\alpha \\plus{} \\beta)^2 \\minus{} 3\\alpha\\beta] \\equal{} ( \\minus{} p)( \\minus{} 2p^2 \\minus{} 3p \\plus{} 3) \\equal{} (2p^3 \\plus{} 3p^2 \\minus{} 3p)$\n[/quote]\n\nThat's correct so far.[/quote]\r\n\r\nWell I found the local maxima and minima of $ (2p^3 \\plus{} 3p^2 \\minus{} 3p)$ and obtained that it happens for $ p \\equal{} \\frac{\\pm\\sqrt{3} \\minus{} 1}{2}$\r\n\r\nSubstituting these gives:\r\n\r\n$ \\minus{}0.59807 < \\alpha^3 \\plus{} \\beta^3 < 4.59807$\r\n\r\nIs there any alternative clever way kunny?  :ninja:"
}
{
  "Tag": [
    "calculus",
    "analytic geometry",
    "graphing lines",
    "slope",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Hello,\r\n\r\nI am seeking for different solutions to a simple problem.  Now, I am seeking for a calculus solution to the problem below:\r\n\r\nThe length of the legs of a certain right traingle are 18 and 63.  What is the radius of the circle which is tangent to both legs of the traingle and has its centre on the hypotenuse?",
  "Solution_1": "Let $r$ be the desired radius, we have $\\frac{r}{63-r}=\\frac{18}{63}$, yielding $r=14.$",
  "Solution_2": "Yes, but I am trying to solve it using calculus, just to see some different solutions. :)",
  "Solution_3": "yo kunny where does the $\\frac{r}{63-r}=\\frac{18}{63}$ come from?  Not sure of the identity used here.  For the calulus solution it will be a differential equation or so i think.",
  "Solution_4": "Similar traingles and ratios.",
  "Solution_5": "i may have misunderstood the question though, but for both legs of the triangle to be tangents of the circle, they have to be at the half way points of the circle, where the slope is zero and where it approaches infinity.  so the triangle's legs have to be equal to go through the center.\r\n\r\nif the corners of the triangle have to touch the circle, and the slopes of the leg and circle didnt releate i wouldnt know how to apporach it with calculus.\r\n\r\n\r\nif the triangle wasnt a right triangle, i would make a system, one equation for the point where one leg touches, one where the other leg touches, an equation for the slope each intersection and possibly another one to put the hyponaneous through the center.  but that'd be a mess and i may have misunderstood the question"
}
{
  "Tag": [],
  "Problem": "Solve the following system of equation:\r\n\r\n//edit (sorry i had a typo): \\[x^{2}-5x-y^{2}-5y = 0\\]  \\[x^{2}+xy+y^{2}= 7\\]",
  "Solution_1": "The first equation can be written as $(x+y)(x-y-5)=0$. \r\nFirst we take $x=-y$. When we substitute this in the second equation, we get $x^{2}= 7$. So we have already 2 solutions : $x=\\sqrt 7$, $y=-\\sqrt7$ and $x=-\\sqrt7$, $y=\\sqrt 7$.\r\nThen we have $x-y=5$. We write the second equation like this : $(x-y)^{2}+3xy=7$. So we have $x-y=5$ and $x(-y)=6$. We concider $x$ and $-y$ as the solutions of the equation $a^{2}-5a+6=0$ and we find $a=2$ or $a=3$. So the other solutions are $x=3$, $y=-2$ and $x=2$, $y=-3$",
  "Solution_2": "[quote=\"Aralc\"]First we take $x=-y$. When we substitute this in the second equation, we get $x^{2}= 7$. [/quote]\nI think you mistook the constant of the $x^{2}$ term of the second equation. It is $0$, not $1$:\n[quote=\"akimatsu\"]$\\left\\{ \\matrix{ 0 \\hfill \\cr x^{2}+xy+y^{2}= 7 \\hfill \\cr}\\right\\}$\n[/quote]",
  "Solution_3": "No, there was intended to be a break between $0$ and $x^{2}$.",
  "Solution_4": "I think you mistook the constant of the  term of the second equation. It is 0, not 1:\r\n\r\nrewrite $x^{2}+xy+y^{2}=7$ as $(x+y)^{2}-xy=7$ Now if x=-y then\r\n\r\n$x^{2}=7$\r\n$x=\\pm 7$",
  "Solution_5": "[quote=\"Aralc\"]The first equation can be written as $(x+y)(x-y-5)=0$. \nFirst we take $x=-y$. When we substitute this in the second equation, we get $x^{2}= 7$. So we have already 2 solutions : $x=\\sqrt 7$, $y=-\\sqrt7$ and $x=-\\sqrt7$, $y=\\sqrt 7$.\nThen we have $x-y=5$. We write the second equation like this : $(x-y)^{2}+3xy=7$. So we have $x-y=5$ and $x(-y)=6$. We concider $x$ and $-y$ as the solutions of the equation $a^{2}-5a+6=0$ and we find $a=2$ or $a=3$. So the other solutions are $x=3$, $y=-2$ and $x=2$, $y=-3$[/quote]\r\nHow do you know $x(-y)=6$?",
  "Solution_6": "Well, we know that $x-y=5$ and $x^{2}+xy+y^{2}= 7$, so $25 = (x-y)^{2}= x^{2}-2xy+y^{2}= (x^{2}+xy+y^{2})-3xy = 7-3xy$, so $-xy=6$."
}
{
  "Tag": [
    "geometry",
    "AMC",
    "AMC 10",
    "MATHCOUNTS",
    "inequalities",
    "induction",
    "trigonometry"
  ],
  "Problem": "What sort of geometry and algebra do I need to know for the AMC10?",
  "Solution_1": "[quote=\"rnwang2\"]What sort of geometry and algebra do I need to know for the AMC10?[/quote]\r\nThis is the geometry you need for Mathcounts and AMC 8/ 10:\r\nPythagorean\r\n30-60-90\r\nIdentifying similar and congruent triangles\r\nMeasures of angles inscribed in circles and power of a point\r\nTriangle Inequality\r\nShoestring (maybe)\r\nThis is absolutely everything I can think of. Almost any problem can crushed by cleverly combining these methods/formulas.",
  "Solution_2": "What is shoestring?",
  "Solution_3": "[quote=\"rnwang2\"]What is shoestring?[/quote]\r\nIt's a really complicated formula that is one of the few that I have no idea how to prove. It gives area of a polygen when you know the coordinates.",
  "Solution_4": "It's not that complicated at all\r\n\r\nand I think it's induction off of Pick's theorem or something\r\nor maybe cross product induction",
  "Solution_5": "[quote=\"Elemennop\"]It's not that complicated at all\n\nand I think it's induction off of Pick's theorem or something\nor maybe cross product induction[/quote]\r\nIt's complicated to explain. I always thought it was from matrices or something.",
  "Solution_6": "what exactly does the theorem state?\r\n\r\nI tried looking it up but didn't find anything.",
  "Solution_7": "http://staff.imsa.edu/math/journal/volume2/articles/Shoelace.pdf\r\n\r\nYou should get the Introduction to Geometry book from the AoPS bookstore.",
  "Solution_8": "You don't need \"shoestring\" at all.",
  "Solution_9": "[quote=\"Phelpedo\"]You don't need \"shoestring\" at all.[/quote]\r\nIs there a shoestring? I always call the shoelace the shoestring...",
  "Solution_10": "[quote=\"Phelpedo\"]You don't need \"shoestring\" at all.[/quote]\r\n\r\nWell, you don't NEED shoestring, but that's like saying you don't NEED Heron's formula... after all, Heron's formula doesn't necessarily have to be used, right? You can just use the Law of Cosines to get an angle and find the area of the triangle.  :wink: \r\n\r\nLikewise, shoestring is particularly useful when trying to blast out a geometry problem on the AMC (and I believe I have found 2 or 3 AIME problems...) with analytic geometry. Shoestring is quite helpful if you know how to apply it properly.  :)",
  "Solution_11": "Knowing\r\n\r\nquadratic formula and synethetic division helps too\r\n\r\nalthough both are easily proven.",
  "Solution_12": "[quote=\"Karth\"][quote=\"Phelpedo\"]You don't need \"shoestring\" at all.[/quote]\n\nWell, you don't NEED shoestring, but that's like saying you don't NEED Heron's formula... after all, Heron's formula doesn't necessarily have to be used, right? You can just use the Law of Cosines to get an angle and find the area of the triangle.  :wink: \n\nLikewise, shoestring is particularly useful when trying to blast out a geometry problem on the AMC (and I believe I have found 2 or 3 AIME problems...) with analytic geometry. Shoestring is quite helpful if you know how to apply it properly.  :)[/quote]\r\nActually you don't even need trig for heron's. Heron's is derived by dropping an altitude of a triangle, and setting up equations. For example, drop the altitude to 14 in a 13,14,15 triangle. Call everything to its left x and the stuff to the right 14-x. Then call the altitude h and use pythagorean twice. It is only slightly more work than heron's itself. For shoestring, unless the numbers are pretty, it takes way too long by any other method.",
  "Solution_13": "It's easier for me just to take the absolute value of \\[\\frac{1}{2}\\left|\\begin{matrix}x_{1}& y_{1}& 1\\\\ x_{2}& y_{2}& 1\\\\ x_{3}& y_{3}& 1\\end{matrix}\\right|.\\]",
  "Solution_14": "[quote=\"chess64\"]It's easier for me just to take the absolute value of\n\\[\\frac{1}{2}\\left|\\begin{matrix}x_{1}& y_{1}& 1\\\\ x_{2}& y_{2}& 1\\\\ x_{3}& y_{3}& 1\\end{matrix}\\right|. \\]\n[/quote]\r\nBut that only works for triangles. Shoestring is for all polygons.",
  "Solution_15": "I never actually knew the name for \"shoestring\" until this topic.\r\n\r\nIt's pretty easy to prove using Green's theorem for line integrals though...\r\n(although it may suck up too much of your time on AMC)."
}
{
  "Tag": [],
  "Problem": "Four pens and three pencils cost $ \\$2.24$. Two pens and five pencils cost $ \\$1.54$. No prices include tax. In cents, what is the cost of a pencil?",
  "Solution_1": "$ 4a \\plus{} 3b \\equal{} 2.24$\r\n\r\n($ 2a \\plus{} 5b \\equal{} 1.54$)2\r\n___________________(-)\r\n\r\n$ \\minus{} 7b \\equal{} \\minus{} 0.84$\r\n\r\n$ \\boxed{b \\equal{} 12 \\ cents}$"
}
{
  "Tag": [],
  "Problem": "Solve for real $ x$\r\n\r\n$ 2001^{x}\\plus{}2004^{x}\\equal{}2002^{x}\\plus{}2003^{x}$",
  "Solution_1": "[hide=\"Answer\"]$ x \\equal{} 0$ or $ x\\equal{}1$[/hide]",
  "Solution_2": "[hide]... or $ x\\equal{}1$.[/hide]",
  "Solution_3": "It is too easy to give such a answer like that, isn't it?  :wink:",
  "Solution_4": "Anyone, please?"
}
{
  "Tag": [],
  "Problem": "A high school's football stadium has 11 rows of 199 seats each.  One night, several groups, each not exceeding 39 people, attend a game.  If people from the same group sit in the same row, what is the maximum number of tickets that can be sold so that everyone is guaranteed a seat?",
  "Solution_1": "$\\frac{199}{39}=5.1...$, so $(5)(39)(11)=2145$ tickets can be sold to guarantee that each group sits together."
}
{
  "Tag": [
    "absolute value"
  ],
  "Problem": "How old are you? (AS OF OCTOBER 2006)\r\n\r\nI'm 13, probably of the younger people on the forum.",
  "Solution_1": "I'm 13, too.",
  "Solution_2": "I'm about $\\frac{2e^{\\pi}}{\\pi}$",
  "Solution_3": "[quote=\"junggi\"]I'm about $\\frac{2e^{\\pi}}{\\pi}$[/quote]\r\nIs that supposed to be 14?",
  "Solution_4": "14.73182248... usually rounds to 15...",
  "Solution_5": "[quote=\"Klebian\"]14.73182248... usually rounds to 15...[/quote]\r\nwell convert the decimal to months",
  "Solution_6": "14 years old here",
  "Solution_7": "rofl I'm 19... not many active posters here older than me besides [b]fredbel6[/b] (in this subforum I mean -- I'm a youngster in College Playground :D)",
  "Solution_8": "I am 73 [color=white]...ok, i'm 14[/color]",
  "Solution_9": "[quote=\"pianoforte\"]I am 73 [color=red]...ok, i'm 14[/color][/quote]",
  "Solution_10": "I will be 19 in 14.12.2006 :P",
  "Solution_11": "[quote=\"Tiks\"]I will be 19 in 14.12.2006 :P[/quote]\r\nDecember 14? Why don't people in Europe and USA decide on common things? (Metric vs. Imperial, mm/dd/yy vs. dd/mm/yy, etc.)",
  "Solution_12": "[quote=\"i_like_pie\"][quote=\"Tiks\"]I will be 19 in 14.12.2006 :P[/quote]\nDecember 14? Why don't people in Europe and USA decide on common things? (Metric vs. Imperial, mm/dd/yy vs. dd/mm/yy, etc.)[/quote]\r\nbecause the US thinks that theyre THE BEST and dont want to change theirs to europeans.\r\n\r\nps i just noticed, why is there a category for 65-120(120?wtf?)",
  "Solution_13": "[quote=\"junggi\"][quote=\"i_like_pie\"][quote=\"Tiks\"]I will be 19 in 14.12.2006 :P[/quote]\nDecember 14? Why don't people in Europe and USA decide on common things? (Metric vs. Imperial, mm/dd/yy vs. dd/mm/yy, etc.)[/quote]\nbecause the US thinks that theyre THE BEST and dont want to change theirs to europeans.\n\nps i just noticed, why is there a category for 65-120(120?wtf?)[/quote]\r\nPeople live to be that old... However that is very rare and even if someone is 120+, I don't think they're a member of AoPS. But you should always consider all possibilities. :wink:",
  "Solution_14": "[quote=\"i_like_pie\"][quote=\"Tiks\"]I will be 19 in 14.12.2006 :P[/quote]\nDecember 14? Why don't people in Europe and USA decide on common things? (Metric vs. Imperial, mm/dd/yy vs. dd/mm/yy, etc.)[/quote]\r\nYes, you are right,December 14.Or maybe you think that I was born in 12 of 14th month :D .",
  "Solution_15": "[quote=\"K81o7\"][quote=\"not_trig\"]lol\n\nI'm a certain prime < 20 (guess which one)[/quote]\n\nI'm gonna take a stab at [hide]13[/hide][/quote]\r\nI'll guess 17, which would be my age :)",
  "Solution_16": "Take a stupid guess: 2 :!: :!: :D",
  "Solution_17": "I guess it's 19.",
  "Solution_18": "[quote=\"goldendomer\"]I guess it's 19.[/quote]\r\n\r\nditto.",
  "Solution_19": "One right answer, 4 wrong. NOT 2.",
  "Solution_20": "O I guess my age then.",
  "Solution_21": "[quote=\"not_trig\"]One right answer, 4 wrong. NOT 2.[/quote]\r\nI say you're 13",
  "Solution_22": "[quote=\"i_like_pie\"][/quote][quote=\"junggi\"][quote=\"i_like_pie\"][quote=\"Tiks\"]I will be 19 in 14.12.2006 :P[/quote]\nDecember 14? Why don't people in Europe and USA decide on common things? (Metric vs. Imperial, mm/dd/yy vs. dd/mm/yy, etc.)[/quote]\nbecause the US thinks that theyre THE BEST and dont want to change theirs to europeans.\n\nps i just noticed, why is there a category for 65-120(120?wtf?)[/quote][quote=\"i_like_pie\"]\nPeople live to be that old... However that is very rare and even if someone is 120+, I don't think they're a member of AoPS. But you should always consider all possibilities. :wink:[/quote]\r\n\r\nmaybe u should make it til 124, a french lady once lived til 123 (record)\r\n-jorian",
  "Solution_23": "as of october 2006 my age is  :rotfl: -2 :rotfl: \n\nthe coffin dance emoji:  :rotfl:  :stretcher:  :rotfl:",
  "Solution_24": "out of the womb, into the tomb",
  "Solution_25": "lol bumping ancient threads",
  "Solution_26": "in 2006, my age was -2... wow ancient threads ",
  "Solution_27": "same...\nthis is a pretty old thread...",
  "Solution_28": "[quote=nkouevda]I'm 13, too.[/quote]\n\nWow!! Right now you're 27!!!! :wow:",
  "Solution_29": "[quote=13375P34K43V312]How old are you? (AS OF OCTOBER 2006)\n\nI'm 13, probably of the younger people on the forum.[/quote]\n\nYou're not young! You're 27!!"
}
{
  "Tag": [
    "calculus",
    "derivative",
    "function",
    "limit",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Find such function $ f(x)$ for which the n-th derivative exists for $ \\forall n \\in N$ and the limit $ \\lim_{n\\to\\infty} f^n(x) \\equal{} g(x)$ converges.\r\n\r\n($ e^x$ does not count)\r\n\r\nLecturer asked me this question after the lecture about power series, but not power series function will do as well.",
  "Solution_1": "Do constant functions count?  :wink:",
  "Solution_2": "Maybe I'm missing something, but what's wrong with $ f(x)\\equal{}0$?",
  "Solution_3": "$ e^{\\alpha x}$ works for any $ |\\alpha| < 1$; the limit is $ 0$ everywhere.  Actually, I think any function $ \\sum f_n \\frac {x^n}{n!}$ where $ f_n \\equal{} O(\\alpha^n)$ for some $ |\\alpha| < 1$ should also work.",
  "Solution_4": "[quote=\"taxofonas\"]Find such function $ f(x)$ for which the n-th derivative exists for $ \\forall n \\in N$ and the limit $ \\lim_{n\\to\\infty} f^n(x) \\equal{} g(x)$ converges.\n\n($ e^x$ does not count)\n\nLecturer asked me this question after the lecture about power series, but not power series function will do as well.[/quote]You can get all nontrivial analytical limits by considering power series $ f(x) \\equal{} \\sum_{k\\equal{}0}^\\infty a_k \\frac{x^k}{k!}$ for which $ \\lim_k a_k \\equal{} \\alpha$ exists. Then $ f^{(n)}(x) \\equal{}  \\sum_{k\\equal{}0}^\\infty a_{k\\plus{}n} \\frac{x^k}{k!}$. The limiting function will be $ g(x) \\equal{} \\alpha e^x$. \r\n\r\nIn particular, the radius of convergence must be infinite.",
  "Solution_5": "Thanks.\r\n\r\nSo far looks like, regardless of constant functions, some sort of exponential ones (but expressed as power series) dominates.  Funny thing those exponentials in the world of derivatives..."
}
{
  "Tag": [
    "advanced fields",
    "advanced fields unsolved"
  ],
  "Problem": "A and B are open sets. Denote by frA the boudary of A , and by * the intersection and by + th union of sets;\r\nShow that :\r\n( A *frB) +( B*frA) is in fr(A*B) which is in  ( A *frB) +( B*frA)+(frA*frB)",
  "Solution_1": "In standard notation:\r\n$ (A\\cap\\partial B)\\cup (B\\cap\\partial A) \\subseteq \\partial (A\\cap B) \\subseteq (A\\cap\\partial B) \\cup (B\\cap\\partial A)\\cup (\\partial A\\cap \\partial B)$\r\nJust do it case by case: if $ x\\in A$ and $ x \\in \\partial B$, then any sufficiently small neighborhood of $ x$ is contained in $ A$ and includes the points from $ B$ and from its complement. Hence $ x\\in\\partial (A\\cap B)$. And so on... the whole problem is about understanding the definitions of boundary points and of open sets."
}
{
  "Tag": [
    "geometry",
    "incenter",
    "circumcircle",
    "parallelogram",
    "trigonometry",
    "function",
    "geometry unsolved"
  ],
  "Problem": "Consider a triangle $ ABC$ and its circumcircle. Let $ I$ be the incenter of this triangle, let $ B'$ be a point on the circumcircle, s.t. $ BB'$ is a diameter in this circle. Let $ M$ be the tangency point of the incircle with the side $ CA$. Consider $ K\\in (AB)$ and $ L\\in (BC)$, s.t. $ AM\\equal{}BL$ and $ MC\\equal{}BK$. Prove that $ B'I\\perp KL$.",
  "Solution_1": "KD/KE=(c-a)/2(p-c)sin(B/2)  where p=1/2(a+b+c)\r\nRS=B\u2019S=tan(B/2)   IR=2Rsin(B/2)-2Rsin(C-A)/2*tan(B/2)  \r\nB\u2019R=B\u2019S/cos(B/2)=B\u2019S/sin(A+C/2)\r\nSo KD/KE= B\u2019R/ IR=[2sin(C-A)/2]/[sinA+sinB-sinC]\r\n \u25b3KDE\u223d\u25b3B\u2019RI\r\nDONE!",
  "Solution_2": "[quote=\"plane geometry\"]KD/KE=(c-a)/2(p-c)sin(B/2)  where p=1/2(a+b+c)\nRS=B\u2019S=tan(B/2)   IR=2Rsin(B/2)-2Rsin(C-A)/2*tan(B/2)  \nB\u2019R=B\u2019S/cos(B/2)=B\u2019S/sin(A+C/2)\nSo KD/KE= B\u2019R/ IR=[2sin(C-A)/2]/[sinA+sinB-sinC]\n \u25b3KDE\u223d\u25b3B\u2019RI\nDONE![/quote]\r\n\r\nI'm sorry, but could you give more explanations in your solution? Starting with your notations, what are the points $ D,E,R$ and so on in your draw? Thank you very much.",
  "Solution_3": "DE=DK=MC  DKLE is a parallelogram     \r\nR is the intersection of B'C and DI",
  "Solution_4": "The problem is rather easy, we get it by direct calculations without trig functions: all we have to prove is $ B^'K^2 - B^'L^2 = IK^2 - IL^2$.\r\n\r\nBest regards,\r\nsunken rock"
}
{
  "Tag": [
    "function",
    "real analysis",
    "real analysis solved"
  ],
  "Problem": "Let $f:\\mathbb R\\to\\mathbb R$  be a decreasing function and $F$ a primitive of $f$. Prove that $F$ is concave.",
  "Solution_1": "If $F$ is not concave there exist $a<c<b$ such that $F(c)<xF(a)+(1-x)F(b)$ where $c=xa+(1-x)b$.So $x(F(c)-F(a))<(1-x)(F(b)-F(c))$.There exist also $e1$ in $(a,c)$ and $e2$ in $(c,b)$ such that $f(e1)=\\frac {F(c)-F(a)}{c-a}$ and $f(e2)=\\frac{F(b)-F(c)}{b-c}$.So $xf(e1)(c-a)<(1-x)f(e2)(b-c)$ $=>$ $f(e1)<f(e2)$ for $e1<e2$.This is a contradiction since f is decreasing.",
  "Solution_2": "[quote=\"Killer\"] $xf(e1)(c-a)<(1-x)f(e2)(b-c)$ $=>$ $f(e1)<f(e2)$ for $e1<e2$[/quote]\r\n\r\n\r\n\r\nCan someone give the details of this implication ?",
  "Solution_3": "Nice, killer has just shown to me his solution to this very easy problem, so I am able to justify the implication. It follows from the fact that x is pozitive, c>a and $ x(c-a)=(1-x)(b-c)$.",
  "Solution_4": "Thank you harazi"
}
{
  "Tag": [
    "linear algebra",
    "linear algebra unsolved"
  ],
  "Problem": "We have the surface defined by the equation \\[ x^/a^2 \\plus{} y^2/b^2 \\minus{} z^2/c^2\\equal{}1\\].\r\n\r\nProve that given a point X on this surface there exist two lines that pass through X and are completely contained in the surface.\r\n\r\n  i hope it is appropiate to post this problem here.",
  "Solution_1": "If such lines exist, they will be tangent to the surface and therefore they will be contained in the tangent plane. Find the equation of tangent plane at $ (x_0,y_0,z_0)$ and then find its intersection with the surface.",
  "Solution_2": "$ \\frac {x^2}{a^2} \\plus{} \\frac {y^2}{b^2} \\minus{} \\frac {z^2}{c^2} \\equal{} 1 \\Longleftrightarrow \\left( \\frac {x}{a} \\minus{} \\frac {z}{c} \\right) \\left( \\frac {x}{a} \\plus{} \\frac {z}{c} \\right) \\equal{} \\left( 1 \\minus{} \\frac {y}{b} \\right) \\left( 1 \\plus{} \\frac {y}{b} \\right)$. So you can find two different sheafs of lines completely contained in the surface by chosing\r\n\r\n$ \\left\\{\\begin{array}{l} \\frac {x}{a} \\minus{} \\frac {z}{c} \\equal{} \\alpha \\left( 1 \\minus{} \\frac {y}{b} \\right) \\\\\r\n\\frac {x}{a} \\plus{} \\frac {z}{c} \\equal{} \\frac {1}{\\alpha} \\left( 1 \\plus{} \\frac {y}{b} \\right) \\end{array} \\right.$\r\n\r\nor\r\n\r\n$ \\left\\{\\begin{array}{l} \\frac {x}{a} \\minus{} \\frac {z}{c} \\equal{} \\alpha \\left( 1 \\plus{} \\frac {y}{b} \\right) \\\\\r\n\\frac {x}{a} \\plus{} \\frac {z}{c} \\equal{} \\frac {1}{\\alpha} \\left( 1 \\minus{} \\frac {y}{b} \\right) \\end{array} \\right.$\r\n\r\nwith $ \\alpha \\in \\mathbb{R} \\setminus \\{0\\}$\r\nNow I think it's not difficult to proof that for each point X on the surface you can find two different lines passing through X, one from the first sheaf and one from the second",
  "Solution_3": "hello\r\n\r\n sorry ,but what do you mean by sheafs? would you mind explaining the proof with al ittle more detail. thanks",
  "Solution_4": "maybe my english is bad, with sheaf I mean a family of lines, for example $ y \\equal{} mx$, $ m \\in \\mathbb{R}$ is a sheaf of lines in the plane passing through $ (0,0)$. The system\r\n\r\n$ \\left\\{\\begin{array}{l} \\frac {x}{a} \\minus{} \\frac {z}{c} \\equal{} \\alpha \\left( 1 \\minus{} \\frac {y}{b} \\right) \\\\\r\n\\frac {x}{a} \\plus{} \\frac {z}{c} \\equal{} \\frac {1}{\\alpha} \\left( 1 \\plus{} \\frac {y}{b} \\right) \\end{array} \\right.$\r\n\r\nrepresent, for any fixed $ \\alpha$, the intersection of 2 planes in the $ \\mathbb{R}^3$ space, that is a line. Those lines are completely contained in your surface, because each point $ (x_0,y_0,z_0)$ that satisfies\r\n\r\n$ \\left\\{\\begin{array}{l} \\frac {x_0}{a} \\minus{} \\frac {z_0}{c} \\equal{} \\alpha \\left( 1 \\minus{} \\frac {y_0}{b} \\right) \\\\\r\n\\frac {x_0}{a} \\plus{} \\frac {z_0}{c} \\equal{} \\frac {1}{\\alpha} \\left( 1 \\plus{} \\frac {y_0}{b} \\right) \\end{array} \\right.$\r\n\r\nmust satisfy their product, too and so $ \\frac {x_0^2}{a^2} \\plus{} \\frac {y_0^2}{b^2} \\minus{} \\frac {z_0^2}{c^2} \\equal{} 1$\r\n\r\nConversely, each point $ (x_0,y_0,z_0)$ that is on the surface satisfies the equation $ \\left( \\frac {x_0}{a} \\minus{} \\frac {z_0}{c} \\right) \\left( \\frac {x_0}{a} \\plus{} \\frac {z_0}{c} \\right) \\equal{} \\left( 1 \\minus{} \\frac {y_0}{b} \\right) \\left( 1 \\plus{} \\frac {y_0}{b} \\right)$, and so you can choose an $ \\alpha$ such that this point satisfies one of the 2 systems I wrote."
}
{
  "Tag": [
    "trigonometry"
  ],
  "Problem": "In $\\triangle{ABC},A+C=2B, \\frac{1}{\\cos{A}}+\\frac{1}{\\cos{C}}=-\\frac{\\sqrt{2}}{\\cos{B}}.$ Find the value of $\\cos{\\frac{A-C}{2}}.$",
  "Solution_1": "[hide]\nSince $A+C=2B$ and $A+C+B=180$, we get $B=60$ and $A+C=120$\n\nafter multiplying through by $\\cos A \\cos B \\cos C$, you get\n\n$\\cos B (\\cos A + \\cos C) = \\sqrt{2} \\cos A \\cos C$\nsince $\\cos B=\\frac1{2}$, we have $\\cos A + \\cos C = -2\\sqrt2 \\cos A \\cos C$\n\nusing product to sum and sum to product formulas, we get\n\n$2 \\cos {60} \\cos {\\frac{A-C}{2}}=-2\\sqrt{2} \\cdot \\frac1{2} (\\cos {120}+\\cos {A-C})$\nsimplifying and letting $x=\\frac{A-B}{2}$ and also using the double angle formula for cos, \n\nwe get $\\cos x = \\frac{-\\sqrt2}2 - \\sqrt2 (2 \\cos^2 x -1)$\n\n\n\nsolving for cos x, we get, $\\cos{\\frac{A-C}{2}} = \\frac{\\sqrt{2}}2$  (used quad equation)\n\nso A=105, C=15\n\n\n\n\n\n\n\n[/hide]",
  "Solution_2": "$A + B + C = 3B = 180^o$, $B = \\frac{A + C}{2} = 60^o$, $A + C = 120^o$\r\n\r\n$\\cos B = \\cos{\\frac{A + C}{2}} = \\frac 1 2$, $\\cos{(A + B)} = -\\frac 1 2$\r\n\r\n$\\frac{1}{\\cos A} + \\frac{1}{\\cos C} = -2\\sqrt 2$\r\n\r\n$\\cos A + \\cos C = -2\\sqrt 2 \\cos A \\cos C$\r\n\r\n$\\cos A + \\cos C = 2 \\cos{\\frac{A + C}{2}} \\cos{\\frac{A - C}{2}} = \\cos{\\frac{A - C}{2}}$\r\n\r\n$2 \\cos A \\cos C = \\cos{(A + C)} + \\cos{(A - C)} = -\\frac 1 2 + 2\\cos^2{\\frac{A - C}{2}} - 1$\r\n\r\nCombining:\r\n\r\n$\\cos{\\frac{A - C}{2}} = -\\sqrt 2 \\left(2\\cos^2{\\frac{A - C}{2}} - \\frac 3 2\\right)$\r\n \r\n$\\cos^2{\\frac{A - C}{2}} + \\frac{\\sqrt 2}{4} \\cos{\\frac{A - C}{2}} - \\frac 3 4 = 0$ \r\n\r\n$\\cos{\\frac{A - C}{2}} = \\frac 1 2 \\left( -\\frac{\\sqrt 2}{4} \\pm \\sqrt{\\frac{2}{16} + 3}\\right) = \\frac 1 2 \\left( -\\frac{\\sqrt 2}{4} \\pm \\sqrt{\\frac{50}{16}}\\right) = \\frac 1 2 \\left( -\\frac{\\sqrt 2}{4} \\pm \\frac{5\\sqrt 2}{4}\\right) =$\r\n\r\n$= \\frac{\\sqrt 2}{2}$ [color=white].[/color] or [color=white].[/color] $-\\frac{3\\sqrt 2}{2}$\r\n\r\nThe negative root is less than -1, i.e., not acceptable or a cosine.\r\n\r\n$\\cos{\\frac{A - C}{2}} = \\frac{\\sqrt 2}{2}$, ${\\frac{A - C}{2}} = \\pm 45^o$, $A - C = \\pm 90^o$\r\n\r\nEither $A + C = 120^o$, $A - C = +90^o$, $A = 105^o$, $C = 15^o$\r\n[color=white].....[/color] or $A + C = 120^o$, $A - C = -90^o$, $A = 15^o$, $C = 105^o$\r\n\r\nCheck:\r\n$\\cos{15^o} = \\sqrt{\\frac{1 + \\cos{30^o}}{2}} = \\frac{\\sqrt{2 + \\sqrt 3}}{2}$\r\n\r\n$\\cos{105^o} = -\\sin{15^o} = -\\sqrt{\\frac{1 - \\cos{30^o}}{2}} = -\\frac{\\sqrt{2 - \\sqrt 3}}{2}$\r\n\r\n$\\frac{1}{\\cos{15^o}} + \\frac{1}{\\cos{105^o}} = \\frac{2}{\\sqrt{2 + \\sqrt 3}} - \\frac{2}{\\sqrt{2 - \\sqrt 3}} = 2\\left(\\sqrt{2 - \\sqrt 3} - \\sqrt{2 + \\sqrt 3}\\right) \\doteq$\r\n\r\n$\\doteq 2 \\times (0.517638 - 1.931852) \\doteq -2 \\times 1.414214 \\doteq -2 \\sqrt 2$"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "Galois Theory",
    "algebra unsolved"
  ],
  "Problem": "Solve the equation:\r\n\\[ \\frac{1}{x^{2}}+\\frac{1}{(4-\\sqrt{3}x)^{2}}=1\\]",
  "Solution_1": "$ (4-\\sqrt{3}x)^{2}=1/ (1-1/x^{2})$\r\n$ 16-8 \\sqrt{3}x+3 x^{2}= x^{2}/(x^{2}-1)$\r\n$ 16x^{2}-16-8\\sqrt{3}x^{3}+8\\sqrt{3}x+3x^{4}-3x^{2}=x^{2}$\r\n\r\n$ 3 x^{4}-8 \\sqrt{3}x^{3}+12x^{2}+8 \\sqrt{3}x-16=0$\r\nin order to avoid $ \\sqrt{3}$ let $ y=x/\\sqrt{3}$\r\n\r\n$ \\frac{1}{3}y^{4}-\\frac{8}{3}y^{3}+4 y^{2}+8 y-16 = 0$ \r\n$ y^{4}-8 y^{3}+12y^{2}+24 y-48=0$\r\n$ (y-2) (y^{3}-6 y^{2}+24)=0$\r\n$ y=2$ or $ y^{3}-6 y^{2}+24=0$",
  "Solution_2": "The beginning is as mszew did:\r\n$ (-3x^{3}+6x^{2}\\sqrt{3}-8\\sqrt{3})(-3x+2\\sqrt{3})=0$. We obtain the obvious solution $ x=\\frac{2}{\\sqrt{3}}$; now let $ x=\\frac{2t}{\\sqrt{3}}$; we obtain $ p(t)=-t^{3}+3t^{2}-3=0$. The $ p$-Galois group is $ \\mathbb{Z}_{3}$; thus we may think that if $ a$ is a $ p$-root then $ \\exists{n}$ s.t. ${ \\mathbb{Q}(a)=\\mathbb{Q}(cos(\\frac{2\\pi}{n}}))$. Necessarily $ \\phi(n)=2.3=6$, thus $ n\\in\\{7,9,14,18\\}$. That works with $ n=18$: the ${ cos(\\frac{\\pi}{9}})$-minimal polynomial,  $ q(y)=8y^{3}-6y-1$, has the roots ${{{ cos(\\frac{\\pi}{9}}),cos(\\frac{7\\pi}{9}}),cos(\\frac{5\\pi}{9}})$. Moreover $ p(1-2y)=q(y)$.\r\nConclusion: the $ p$-roots are: ${{{ 1-2cos(\\frac{\\pi}{9}}),1-2cos(\\frac{7\\pi}{9}}),1-2cos(\\frac{5\\pi}{9}})$."
}
{
  "Tag": [
    "number theory",
    "greatest common divisor",
    "modular arithmetic"
  ],
  "Problem": "Ok! this is a game ~If not proper to the rules of which i cant find i think then just lock please\r\n\r\nRules:\r\nDo not spam!\r\nCheck if the same guess you have made is posted or not!\r\nBe nice!\r\n[i]Also, If you guess if it is true then please explain how you came to that conclusion[/i]\r\n\r\nObject:\r\nthe object of this game is to figure out if a large number (really really large) is a prime number or not, and what it is divisable by if it is not\r\n[i]You must find the GCD if it is not a prime number[/i]\r\n[b]If you get it right post a new number! and if you do you get to tell someone if they are right =)[/b]\r\n\r\nFirst number, is 1111151141\r\nGuess away~ =)",
  "Solution_1": "$1111151141=19^{3}\\cdot161999$\r\n\r\nDo I post another number now?",
  "Solution_2": "Yep, And nice one. But you came to it quite quick =[ lol",
  "Solution_3": "I'll do one: $987654312$",
  "Solution_4": "it can be devided by 2\r\n\r\nnew one: 9999978347",
  "Solution_5": "Eugh... $461|9999978347$\r\n\r\n3657500101\r\n\r\nThis is cool\r\n[hide=\" :) \"]\nIf $p(N)$ represents the Nth prime, then, this is $p^{13}(1)$\n\n[/hide]",
  "Solution_6": "[quote=\"The Zuton Force\"]Eugh... $461|9999978347$\n\n3657500101\n\nThis is cool\n[hide=\" :) \"]\nIf $p(N)$ represents the Nth prime, then, this is $p^{13}(1)$\n\n[/hide][/quote]\r\n\r\ncorrect, whatever the hidden part means,\r\nalso can people please make sure to check back here to say if something is correct =)",
  "Solution_7": "BLECH. Cant figure yours out @_@",
  "Solution_8": "3657500101 is prime. :)\r\n\r\nNext: 1234567891",
  "Solution_9": "1234567891 is prime.\r\n\r\n\r\nNext:8787132021",
  "Solution_10": "[quote=\"hunter34\"]\n\nNext:8787132021[/quote]\r\n\r\nDivisible by 3\r\n\r\nGot it in like 10 sec.\r\n\r\n1234567897",
  "Solution_11": "$ 3|1234567897$\r\n\r\n$ 4.3643763431126323468034568537 \\cdot 10^{28}$",
  "Solution_12": "I think you've stumped us all, if not... WOW @ AoPS'ers",
  "Solution_13": "$ 3\\mid4.3643763431126323468034568537 \\cdot 10^{28}$\r\n\r\nNext: $ 37492875034937$",
  "Solution_14": "Wow                                              ?",
  "Solution_15": "See http://math.about.com/library/bldivide.htm , snee.\r\n\r\n$ 0 \\equiv 37492875034937 \\pmod {7}$\r\n\r\n2069384750923453",
  "Solution_16": "[quote=\"i_like_pie\"]$ 3\\mid4.3643763431126323468034568537 \\cdot 10^{28}$\n\nNext: $ 37492875034937$[/quote]\r\n\r\n7|53561250049991\r\n\r\nNext:[b]$ 35698562455953541691$[/b]",
  "Solution_17": "[quote=\"LevatorNasolabialis\"][quote=\"i_like_pie\"]$ 3\\mid4.3643763431126323468034568537 \\cdot 10^{28}$\n\nNext: $ 37492875034937$[/quote]\n\n7|53561250049991\n\nNext:[b]$ 35698562455953541691$[/b][/quote]\r\n\r\n...uh, you skipped mine.",
  "Solution_18": "[quote=\"Temperal\"][quote=\"LevatorNasolabialis\"][quote=\"i_like_pie\"]$ 3\\mid4.3643763431126323468034568537 \\cdot 10^{28}$\n\nNext: $ 37492875034937$[/quote]\n\n7|53561250049991\n\nNext:[b]$ 35698562455953541691$[/b][/quote]\n\n...uh, you skipped mine.[/quote]\r\nsorry as I was writing it you did a new one",
  "Solution_19": "(987654321!*9984234.6!^3958203!)^573038!\r\n\r\nIncase this is impossible (my scientific calc gave me some number ending in 'e'. What does that mean?) try:\r\n781633287*2531273^48",
  "Solution_20": "Um, both of those are obviously not prime, since they're products of factorials. \r\n\r\nAnd $ a \\text{ \\tiny{E} } b \\equal{} a\\cdot 10^b$\r\n\r\n\r\nTry 1732743457.",
  "Solution_21": "[quote=\"Temperal\"]$ 3|1234567897$[/quote]\r\n\r\nYou are wrong.\r\n\r\n\r\n :diablo:",
  "Solution_22": "Eh, I meant $ 1234567897|17$.",
  "Solution_23": "You people are too smart!\r\n\r\n987586323423",
  "Solution_24": "$ 3\\mid987586323423$\r\n\r\nNext: $ 123454321$",
  "Solution_25": "e = ^ in calculators.",
  "Solution_26": "[quote=\"i_like_pie\"]$ 3\\mid987586323423$\n\nNext: $ 123454321$[/quote]\r\ndivisible by 3.\r\n\r\nNext:234564799867587 :wink:",
  "Solution_27": "[quote=\"pingala\"][quote=\"i_like_pie\"]Next: $ 123454321$[/quote]\ndivisible by 3.[/quote]\r\nUmm...\r\n\r\n$ 1 \\plus{} 2 \\plus{} 3 \\plus{} 4 \\plus{} 5 \\plus{} 4 \\plus{} 3 \\plus{} 2 \\plus{} 1 \\equal{} 25\\equiv1\\mod3$",
  "Solution_28": "[quote=\"snee\"]e = ^ in calculators.[/quote]\r\n\r\n....no. That's totally different.\r\n\r\n$ a \\text{\\tiny{E} } b = a\\cdot 10^b$, like I said. $ a\\^\\, b = a^b$\r\n\r\n\r\nas for the other number pingala gave an incorrect answer to...\r\n$ 11111|123454321$\r\n\r\nRemember that old trick? \r\n\r\n164327",
  "Solution_29": "99999999999999999999999999999999999999999999999991"
}
{
  "Tag": [
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "The side of $\\Delta ABC$ are a,b,c. Prove that:\r\n$1)a^{3}b+b^{3}c+c^{3}a+2a^{2}b^{2}c^{2}\\geq ab^{3}+bc^{3}+ca^{3}+2abc(a+b+c)$\r\n$2)a^{3}b+b^{3}c+c^{3}a+a^{2}b^{2}c^{2}\\geq ab^{3}+bc^{3}+ca^{3}+abc(a+b+c)$\r\nAnd problem is: \r\nFind k max  \r\n $\\frac{ab^{3}+bc^{3}+ca^{3}-a^{3}b-b^{3}c-c^{3}a}{a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}-abc(a+b+c)}\\leq k$\r\nWith $\\forall \\Delta ABC$",
  "Solution_1": "the first two inequalities 1) and 2) are not valid as we take the [i][u]equilateral triangle with all sides have length 1.[/u][/i]____________________________________________________________",
  "Solution_2": "Hi!!\r\n$1)a^{3}b+b^{3}c+c^{3}a+2a^{2}b^{2}c^{2}\\geq ab^{3}+bc^{3}+ca^{3}+2abc(a+b+c)$\r\n$\\Leftrightarrow \\sum b(a+b-c)(a-c)^{2}\\geq 0$;\r\n               \r\n$2)a^{3}b+b^{3}c+c^{3}a+a^{2}b^{2}c^{2}\\geq ab^{3}+bc^{3}+ca^{3}+abc(a+b+c)$\r\n  In $\\Delta ABC$  $a^{2}=b^{2}+c^{2}-2bccosA$;\r\n  $\\Rightarrow a^{3}b+b^{3}c+c^{3}a=ab(b^{2}+c^{2}-2bccosA)+bc(c^{2}+a^{2}-2cacosA)+ca(a^{2}+b^{2}-2abcosA=ab^{3}+bc^{3}+ca^{3}+abc(a+b+c)-2abc(bcosA+cCosB+acosC)$\r\nI will prove that:\r\n  $2abc(bcosA+ccosB+acosC \\leq (a^{2}+b^{2}+c^{2})$;\r\n   $x=ab;t=bc;z=ca$;\r\n  $\\Rightarrow 2abc(bcosA+cCosB+acosC)=2(abbccosA+bccacosB+caabcosC)=2(xycosa+yzcosB+zxcosC) \\leq x^{2}+y^{2}+z^{2}$;\r\n  It's true.     :P \r\nAnd problem 3) k min =1; it's hard :lol:",
  "Solution_3": "[quote=\"Denbox\"]The side of $\\Delta ABC$ are a,b,c. Prove that:\n$1)a^{3}b+b^{3}c+c^{3}a+2a^{2}b^{2}c^{2}\\geq ab^{3}+bc^{3}+ca^{3}+2abc(a+b+c)$\n$2)a^{3}b+b^{3}c+c^{3}a+a^{2}b^{2}c^{2}\\geq ab^{3}+bc^{3}+ca^{3}+abc(a+b+c)$\n[/quote]\r\nMaybe \r\n$1)a^{3}b+b^{3}c+c^{3}a+6a^{2}b^{2}c^{2}\\geq ab^{3}+bc^{3}+ca^{3}+2abc(a+b+c)$\r\n$2)a^{3}b+b^{3}c+c^{3}a+3a^{2}b^{2}c^{2}\\geq ab^{3}+bc^{3}+ca^{3}+abc(a+b+c)$ $?$"
}
{
  "Tag": [
    "inequalities"
  ],
  "Problem": "Hello!!!\r\n\r\nLet S(m,n) be the set of all integers k such that m mod k + n mod k >= k. What is S(7,9)?\r\n\r\nI tried to make it more general solving as following:\r\n\r\nA mod B = A - B[A/B], where [x] is the floor function.\r\n\r\nm mod k + n mod k >= k\r\nm - k[m/k] + n - k[n/k] >= k\r\nm/k - [m/k] + n/k - [n/k] >= 1 (1)\r\n\r\nA/B - [A/B] results in a value in the range [0,1) as can be seen in the example below:\r\n\r\n2.3 - [2.3] = 2.3 - 2 = 0.3\r\n-2.3 - [-2.3] = -2.3 + 3 = 0.7\r\n\r\nSo, the left side of (1) is in the range [0,2*0.999...)\r\n\r\nBut I don't know how to solve the inequality in order to find S(m,n).\r\n\r\nThanks!!!",
  "Solution_1": "[hide=\"Answer\"]If $ S (m, n) = \\{k \\in \\mathbb  Z^{+}\\ |\\ k \\leq m\\;(mod\\ k) + n\\;(mod\\ k)\\}$, then\n\n$ S (7, 9) = \\boxed{\\left\\{2, 4, 5, 8, 10, 11, 12, 13, 14, 15, 16\\right\\}}$.\n\nIt's not asking you to generalize; it's asking you to find the elements of the set for a given $ m$ and $ n$ (in this case, $ 7$ and $ 9$)[/hide]"
}
{
  "Tag": [
    "ratio",
    "geometry",
    "inradius",
    "Vieta",
    "probability",
    "algebra",
    "polynomial"
  ],
  "Problem": "Ok. math marathon! just continue. Like, just post a solution and a new problem. Btw, this is sorta to help you in mathcounts.\nK.\nI made this one up:\nThe angles of a triangle are in the ratio of 2:4:6. What is the measure of the smallest angle in the triangle?\n\n[/hide]",
  "Solution_1": "The numbers add to twelve, so $ 12x \\equal{} 180$, and $ x \\equal{} 15$. Thus, the smallest angle is 30.\r\n\r\n2. How many factors of $ 2^{105}$ are bigger than 1,000,000?",
  "Solution_2": "$ 2^{20} \\equal{} (2^{10})^2 \\equal{} 1048576$, and every integer through $ 105$ can be made after that. Thus, the number of factors is $ 105 \\minus{} 20 \\plus{} 1 \\equal{} \\boxed{86}$.\r\n\r\n3. Find the area of a triangle with side lengths $ 5$, $ 6$ and $ 7$.",
  "Solution_3": "A 5,6,7 is not a right triangle so that doesn't work.",
  "Solution_4": "Restarting from 3.\r\nWe are given the sides of a triangle, and what better to use than Heron's right now?\r\nWe have\r\n\r\n$ s = \\frac{a+b+c}{2} = \\frac{5+6+7}{2} = 9$\r\n\r\n$ = \\sqrt{s*(s-a)(s-b)(s-c)} = {\\sqrt{9*(9-5)(9-6)(9-7)}}$\r\n\r\n$ \\boxed{6\\sqrt{6}}$\r\n\r\n4. Find the inradius of a triangle with sides $ 10$, $ 12$, and $ 13$.",
  "Solution_5": "We have that $ rs\\equal{}\\sqrt{s(s\\minus{}a)(s\\minus{}b)(s\\minus{}c)}$\r\n\r\n$ \\implies r(\\frac{35}{2})\\equal{}\\sqrt{\\frac{35}{2}(\\frac{35}{2}\\minus{}10)(\\frac{35}{2}\\minus{}12)(\\frac{35}{2}\\minus{}13)}$.\r\n\r\n$ \\implies r\\approx3.257$.\r\n\r\nNP: Find the sum of the roots of $ x^4\\plus{}8x^3\\minus{}16x^2\\plus{}39x\\minus{}5\\equal{}0$.",
  "Solution_6": "5. We have this cubic equation in the form $ ax^4 \\plus{} bx^3 \\plus{} cx^2 \\plus{} dx \\plus{} e \\equal{} 0$, and by Vieta's Formula, we have the sum of the roots equal to $ \\frac { \\minus{} b}{a}$, so therefore, the sum of the roots is $ \\frac{\\minus{} 8}{1} \\equal{} \\minus{}8$.\r\n\r\nProblem 6: What is the probability that when I flip $ 10$ coins, $ 3$ or less coins come up as heads?",
  "Solution_7": "$ \\frac {\\binom{10}{0} \\plus{} \\binom{10}{1} \\plus{} \\binom{10}{2} \\plus{} \\binom{10}{3}}{2^{10}} \\implies \\boxed{\\frac {11}{64}}$.\r\n\r\nNew Problem (7): If a polynomial $ 20x^3 \\minus{} 9x^2 \\plus{} ax \\plus{} b$ is divisible by $ x^2 \\plus{} 4$, find the sum of $ c$ and $ d$ such that $ ax^3 \\plus{} bx^2 \\plus{} cx \\plus{} d$ is divisible by $ 40x^2 \\minus{} 58x \\plus{} 40$.\r\n\r\nEDIT: Oops, didn't see the MathCounts part...\r\n\r\nNew Problem (7A): What is a general ratio of surface area of a cube with side $ r$ to the volume of a cube with side $ r$?  Express your answer in $ a: b$ form, where $ a$ and $ b$ are in simplest terms and reduced.",
  "Solution_8": "The surface area is given by $ 6r^2$, and the volume by $ r^3$, so $ \\frac{6r^2}{r^3}\\equal{}\\boxed{\\frac{6}{r}}$. \r\n\r\n8. Find the largest prime factor of $ 2^{15}\\minus{}1$.",
  "Solution_9": "8. We see that $ 2^{15} \\minus{} 1$ is not a Mersenne Prime, so therefore, we start off using the long way. $ 2^{15} \\minus{} 1$, in numerical form, is 32767. We check using our divisibility rules, and see that it is divisible by $ 7$, leaving $ 4681$. Now, through tedious work, we check by dividing the prime numbers up to $ \\sqrt {4681}$, and find that it's divisible by 31. Now, we are left with $ 151$. $ 151$ is not divisible by anything as we see by testing the primes up to $ \\sqrt{151}$, so therefore, $ 151$ is a prime number. Thus, the largest prime factor is $ \\boxed{151}$.\r\nThere seems to be an easier way to factor and find the prime by factoring, but I'm not sure how.\r\n\r\n9. I dip a cube in red, such that all the exposed surfaces are all colored red. Then, I separate this cube into $ 125$ different cubes. Then, I roll a cube. The top face is red. What is the probability that the bottom face is also red?",
  "Solution_10": "8. difference of cubes gives us 2^15-1=(2^5-1)(2^10+2^5+1) so 2^15-1 is divisible by 31.  using difference of 5th powers, we find that 2^15-1 is also divisble by 7. so 2^15-1 is divisible by lcm(31,7)=31*7.  divide.",
  "Solution_11": "[quote=\"xxrxxhxx\"]9. I dip a cube in red, such that all the exposed surfaces are all colored red. Then, I separate this cube into $ 125$ different cubes. Then, I roll a cube. The top face is red. What is the probability that the bottom face is also red?[/quote]\r\n\r\nIn math terms...0\r\n\r\nIn science terms...not enough information. \r\n\r\n1) If you don't cut the cube at all, and take 124 nanosized particles, who knows?\r\n2) The cube itself is made of atoms. Not a cube, because it has air spaces (which causes friction) between them.\r\n3) Cutting the cube will cause some extremely small pieces to chip off. Therefore, you cannot ensure that they are cubes.\r\n\r\n10. Find the sum of the coefficients of $ (x \\plus{} y)^{96}$.",
  "Solution_12": "Expanding by the binomial theorem, the coefficients are $ \\binom{96}{0}\\plus{}\\binom{96}{1}\\plus{}\\cdots\\plus{}\\binom{96}{95}\\plus{}\\binom{96}{96}\\equal{}\\boxed{2^{96}}$\r\n\r\n11.\r\nFind the distance from the incenter and circumcenter of a triangle with side lengths 3,4,5."
}
{
  "Tag": [
    "AMC",
    "AIME",
    "USA(J)MO",
    "USAMO",
    "blogs",
    "geometry",
    "AMC 10"
  ],
  "Problem": "AMC, AIME, USAMO target different skills. You can read more in my blog: \r\nhttp://blog.tanyakhovanova.com/?p=116\r\n\r\nI suggest the following: USAMO participants are allowed to go to next year's AIME no matter what their AMC scores are. USAMO winners are allowed to go to the next year's USAMO no matter what their AIME results are. \r\n\r\nThis way kids who have proved that they are great swimmers do not need to compete in running again. By swimmers I mean kids who are good at proofs, not accurate speed calculation.",
  "Solution_1": "That is wrong:\r\nIf someone was a USAMO qualifier, then he/she has demonstrated that they are capable of the speed and accuracy not only of making AIME, but having a high enough AMC score to make the USAMO. \r\nIf someone was a USAMO winner one year, they have demonstrated that they have the necessary skills to make USAMO and do well on it. \r\nThis proposition would be superfluous and reward slackers.",
  "Solution_2": "Even Olympic Gold Medalists have to go through the Olympic Trials again (atleast for the U.S.)",
  "Solution_3": "While I do agree that somebody who is capable of acheiving a high score on the USAMO should concentrate only on the USAMO, the proposed idea applies to too few people.\r\n\r\nSomebody who has made the USAMO should make the AIME. And if s/he doesn't make the AIME and is still allowed to compete, the hit on the index because of the low AMC score would almost certainly eliminate that person from USAMO qualification.\r\n\r\nSomebody who has been a USAMO winner probably won't screw up AIME bad enough to drop the USAMO index low enough to not qualify. It does happen, but at most one or two students (probably just none) who have a bad day on the AIME do not require a change of rules.\r\n\r\nWe are more concerned with those on the borderline instead of those who have proven themselves more than capable for making the USAMO.",
  "Solution_4": "What if only about 80 12th graders make it one year? That would create 420 already made spots, and 80 ones for the new 9th-graders.",
  "Solution_5": "Actually, now that I think about it, dont they have a similar procedure for the USNCO.  Its something like if you make the camp, you make it next year too?  Correct me if I am wrong.",
  "Solution_6": "That the AMC/AIME/USAMO target different skills is definitely something I agree with. Not sure about the proposal itself. For one thing, if a USAMO qualifier doesn't qualify for the AIME the year after, their qualification for the USAMO was more likely some fluke then them actually deserving to take AIME. If a USAMO winner doesn't qualify for the USAMO, something is [i]really[/i] messed up. Of all of MOP (a much bigger group), I think it's only like one or two reds, if anyone, that miss USAMO the next year. So I'm not sure how this would change things.\r\n\r\nI see why you might want to change things, based on your original point of them testing different things. However I think you might underestimate how much improving proof-writing skills and olympiad prowess makes you better at the AMCs/AIMEs too. There is still a speed element in olympiads, as only the best can do three problems that difficult in only(!) 4.5 hours, and there are still a few olympiad-level problems that involve computation.",
  "Solution_7": "Although they target different skills, isn't that the point? So that people are being tested not just in one area but all? \r\nAnyways, like simo14 said, if they've already made the usamo, haven't they demonstrated that they're capable of these tests and the different skills required for each? plus, if you really deserve to be there, you have to be good enough too & i think these tests do a good job at assessing that.",
  "Solution_8": "It's not the point. Someone who can do really well on the IMO might not even be able to pass the AIME (he can pass AMC10 through USAMTS). The goal of the AMC series is to select an IMO team, not test how quickly you can solve easy problems (as far as I know).",
  "Solution_9": "If you can succeed at IMO, you can pass AIME. Let's not be ridiculous.",
  "Solution_10": "I don't think that's a reasonable assumption at all. The AIME is EXTREMELY heavy on computation. I would say about half of the time content of the AIME is spent on computation. It is entirely feasible to assume that someone who would do quite well at IMO would be unable to do this (especially without making lots of careless mistakes). In fact, I know people who have gone to IMO who consistently just BARELY made USAMO. Doesn't that imply that there exist people who should go to IMO who consistently just barely did NOT make USAMO?",
  "Solution_11": "There are people that always don't \"barely\" make the AIME, USAMO, AIME I mean all you have to do is look on this site and see everyone going \"epic phaillll .5 away from making aime >< \" or something around the lines of that and no matter how it changes, there's always going to be people out there who don't make it, barely. So, keeping the current qualifications and system is just fine.",
  "Solution_12": "[quote=\"worthawholebean\"]If you can succeed at IMO, you can pass AIME. Let's not be ridiculous.[/quote]\r\n\r\nYes.  And if you can't pass AIME, you probably don't have the skills to do IMO or USAMO.  Although AIME does have a lot of computation, someone would have to really suck at arithmetic, and I mean [b]really[/b] suck in order to not be able to finish the AIME.  Arithmetic is easy.  If you can't do arithmetic, you probably can't do olympiad math.",
  "Solution_13": "people are overexaggerating computation time. the most time-consuming part of the aime is finding solutions and experimenting. a person who succeeds at the olympiad level should be able to find quick solutions to aime problems which would give him/her time to do the \"heavy arithmetic\" that people point to (although I find it extremely unprobable that a usamo winner can't do computation...).",
  "Solution_14": "[quote=\"jb05\"]I don't think that's a reasonable assumption at all. The AIME is EXTREMELY heavy on computation. I would say about half of the time content of the AIME is spent on computation. It is entirely feasible to assume that someone who would do quite well at IMO would be unable to do this (especially without making lots of careless mistakes). In fact, I know people who have gone to IMO who consistently just BARELY made USAMO. Doesn't that imply that there exist people who should go to IMO who consistently just barely did NOT make USAMO?[/quote]\r\n\r\nI would say computation takes me 10 minutes or less, and I attempt at least 12 of the problems (which means I'm not a newb who just does the first 5 problems then gives up).  Though I am somewhat fast at computation (I did well in Countdown Rounds back in the days of Mathcounts for me), I think \"half of the time content\" is not a \"reasonable assumption at all\" (I'm just quoting your words).\r\nAlso, I don't think worthawholebean's statement is an assumption (but a fact based on the history of these contests), and I think you're lying.  Name the people who have gone to IMO and just barely made USAMO.  The scores of previous participants do exist on the AMC website, so I can check (assuming that it was in, say, the 2000's, because the difficulty level of AMC/AIME/USAMO fluctuates if you look before the 21st century and so an argument using such information is not valid), or perhaps I can ask the AMC people for records if needed.",
  "Solution_15": "[quote=\"the future\"]Actually, now that I think about it, dont they have a similar procedure for the USNCO.  Its something like if you make the camp, you make it next year too?  Correct me if I am wrong.[/quote]\r\n\r\nI believe it's the other way around ... if you get a silver or gold at IChO, you can't go to the study camp again.",
  "Solution_16": "I agree that someone can be able to solve USAMO- and IMO-type problems without being good at the \"fast and furious\" style of the AMC (and the AIME, to a slightly lesser extent). However, don't forget that \"the main purpose of the AMC 10/12 is to spur interest in mathematics and to develop talent through solving challenging problems in a timed multiple-choice format\" (as stated on the AMC website). Selection of the IMO team is secondary. As a result, the contest must be accessible to the average test-taker. Around 200,000 students (a rough estimate) take the AMC 10 or AMC  12 each year. Of those, only around 500 qualify for the USAMO. There are certainly many more who would like to take the USAMO, but those interested in a proof-style contest still constitute a small portion of the 200,000 taking the AMC. Making the lower levels of the AMC ladder more like the USAMO would be certain to scare off the vast majority of participants. Such a change is also infeasible; every USAMO solution must be read by an actual person (or actual people), not just a machine. There are indeed tests with many thousands of participants that have essay-type questions, like AP tests. Unlike questions in AP tests, however, USAMO questions are very in-depth and require graders to closely scrutinize solutions. They can't just say, \"okay, they got A: plus 1. They got B: plus 2...\" Also, I imagine few would pay $80+ to take the USAMO, so not as many scorers could be hired as are used for AP tests to grade thousands more papers (although I'm not sure exactly how the USAMO grading is carried out).\r\n\r\nYour proposal, I'm afraid, would be of little help in ensuring that those who are good at proof-type questions and not as good at AMC-type tests would have a better chance at qualifying for the USAMO. As others in the topic have mentioned, someone who qualifies for the USAMO is unlikely to fail to qualify for the AIME the next year if they attempt to (I don't have any evidence that supports this, and there are certainly people who might qualify for the USAMO one year and not qualify for the AIME the next, but I don't imagine there are many). Someone who is in the top 12 of all USAMO participants is probably not likely to miss the USAMO cutoff the next year (again, I'm only speculating, and there may be exceptions).",
  "Solution_17": "being in the top 12 of the usamo demands a level of creativity and intelligence very few of us have. a person like that would certainly not miss the usamo index.",
  "Solution_18": "If you qualify for AIME by way of USAMTS, what is your AMC score as part of your index? By the way, I have to lean towards the view that aime is computation based, as I solved 10 problems in the first hour and a half, spending some of that time cheking my work, then spent anouther hour checking, found 5 mistakes, then solved two more problems. Although you could make the case that if you know the material well enough you should have time to check. Although I've heard many MOPers just barely qualify for USAMO.",
  "Solution_19": "personally i think it's much harder to make aime by usamts than amc, so it's kind of redundant.\r\nand you get a 100 amc 12.",
  "Solution_20": "OK, people who are good at Olympiad level problems might not be as good at the AIME - but surely in 3 hours they can be expected to solve 6 problems correctly, which has been the floor the past couple of years?",
  "Solution_21": "[quote=\"Phelpedo\"]OK, people who are good at Olympiad level problems might not be as good at the AIME - but surely in 3 hours they can be expected to solve 6 problems correctly, which has been the floor the past couple of years?[/quote]\nGetting the floor score does not guarantee AIME qualification. Last year, 3070 people scored 6 or higher on the AIME. 505 people qualified for the USAMO. That's less than a sixth of the people who had the floor score or higher.\n\n[quote=\"simo14\"]personally i think it's much harder to make aime by usamts than amc, so it's kind of redundant.[/quote]\r\nTo borrow the TC's analogy, if the AIME is to the USAMO as running is to swimming, then the AMC is to the USAMTS as a 100m dash is to a marathon. The two styles are so different that I'm certain there are many who consider getting 100+ on the AMC 12 to be harder than qualifying for the AIME through the USAMTS (which typically requires around 70 points after the first three rounds IIRC). I happen to agree that qualification through the AMC is easier, but our personal evaluations of the two contests' relative difficulties are of no consequence.",
  "Solution_22": "I am not saying that I agree with the the original post but I think it is very possible for someone who is capable of being a USAMO winner or IMO level participant to not do well on the AIME.\r\n\r\nI attended the AwesomeMath Summer Program the first year that it was held and there was a student there from Columbia who had been 7th on the Columbian TST, so basically one spot away from going to IMO. I believe he was in the Columbian equivalent of 9th or 10th grade. We had a test at the beginning of AMSP to place us into groups based on our mathematical abilities. The test was composed of an AIME format test and a USAMO format test. This particular student did very well on the USAMO format test, but did horribly on the AIME format test for whatever reason. He was placed in the 2nd lowest group (out of 5) because the AIME test was weighted heavily enough for his cumulative score to suffer severely. I also was placed in the same group that he was in and after the first day of classes, when he immediately solved every proof problem that Zuming wrote on the board in a matter of minutes, he was told to go to the highest group from then on.\r\n\r\nSo yeah, it can happen."
}
{
  "Tag": [
    "percent"
  ],
  "Problem": "In a class of 20 students, 8 girls were absent due to a basketball team trip. Fifty percent of the students present are girls. Given that no boys are absent, what percent of the total class is boys?",
  "Solution_1": "12 students are in class, and we are told that half of them are girls, so there are 6 girls currently in class and 14 total.  So 6 students are boys, giving $ 6/20\\equal{}.3$, or $ 30\\%$."
}
{
  "Tag": [
    "number theory",
    "least common multiple",
    "greatest common divisor",
    "prime factorization"
  ],
  "Problem": "Find the smallest positive integer $n$ for which there exist $n$ different positive integers $a_{1}, a_{2}, \\cdots, a_{n}$ satisfying [list] [*] $\\text{lcm}(a_1,a_2,\\cdots,a_n)=1985$,[*] for each $i, j \\in \\{1, 2, \\cdots, n \\}$, $gcd(a_i,a_j)\\not=1$, [*] the product $a_{1}a_{2} \\cdots a_{n}$ is a perfect square and is divisible by $243$, [/list] and find all such $n$-tuples $(a_{1}, \\cdots, a_{n})$.",
  "Solution_1": "The problem doesn't make sense with $ 1985$, since the first and third conditions can't be satisfied simultaneously. Judging from the topic description, $ 1995$ was meant, and here's the solution assuming that.\r\n\r\n$ n \\equal{} 7$: $ 15,21,57,105,285,399,665$ is the only sequence that satisfies every condition up to permutation.\r\n\r\nFirst, since the least common multiple of all the numbers is $ 1995 \\equal{} 3 \\cdot 5 \\cdot 7 \\cdot 19$, every $ a_i$ must be the product of some subset of $ \\{3,5,7,19\\}$. By the second condition, the subset must be nonempty. The third condition requires that the product of all elements of the sequence is divisible by $ 3^5$, but since it also must be a perfect square it must actually be divisible by $ 3^6$. Since each element of the sequence can have at most one $ 3$ in the prime factorization, this gives $ n \\equal{} 6$ as a lower bound (for now).\r\n\r\nSuppose $ n \\equal{} 6$ works. All elements must then be divisible by $ 3$, so the second condition is trivially satisfied. There are now $ 8$ possible numbers to choose for the $ 6$ elements of the sequence: The product of any subset of $ \\{5,7,19\\}$ times $ 3$. We notice that the product of all $ 8$ of these elements is a perfect square, so in order for the product of the $ 6$ elements of the sequence to be a square we must omit two distinct elements whose product is a perfect square. But since every prime has an exponent of $ 0$ or $ 1$ in these $ 8$ numbers, the product of two of them is a perfect square if and only if they are the same number. Contradiction. $ n \\equal{} 6$ is not possible.\r\n\r\n$ n \\equal{} 7$ is, however. In this case we have $ 6$ elements divisible by $ 3$. The final element is not divisible by $ 3$, since if it were the product would have $ 3^7$ in the prime factorization and not be a perfect square. Let $ a_7$ be the number not divisible by $ 3$. If $ a_7$ is prime, then the second condition requires that all numbers be divisible by $ a_7$, and then the product of the sequence has $ a_7^7$ in the prime factorization, so it can't be a square. So $ a_7$ has to be a product of at least two elements of $ \\{5,7,19\\}$.\r\n\r\nSuppose $ a_7$ is the product of two of these, which we will call $ q$ and $ r$, and is not divisible by one of them, which we will call $ p$. That is, $ a_7 \\equal{} qr$. Of the other $ 6$ elements divisible by $ 3$, we can choose four that are divisible by $ p$: $ 3p, 3pq, 3pr, 3pqr$. But including one of them, $ 3p$, contradicts the second condition. So we only have three choices of numbers divisible by $ p$. Because the product of all sequence elements must be a perfect square, we can only select two of the numbers. Now we need $ 4$ more numbers divisible by $ 3$ in the sequence. We can only choose from $ 3, 3q, 3r, 3qr$, so we must have them all. But including $ 3$ and $ qr$ in the sequence contradicts the GCD condition. Thus, $ a_7$ must be $ 5 \\cdot 7 \\cdot 19$.\r\n\r\nSo $ a_7 \\equal{} 5 \\cdot 7 \\cdot 19$. For the other $ 6$ elements of the sequence, we can only select the product of any subset of $ \\{5,7,19\\}$ times $ 3$ as before. But notice that one of these, $ 3$, contradicts the GCD condition since $ a_7$ is not divisible by $ 3$. So we only can select from seven elements. The product of these seven elements and $ a_7$ is $ 3^7 5^5 7^5 19^5$. So for this to be a perfect square we have to divide out $ 1995 \\equal{} 3 \\cdot 5 \\cdot 7 \\cdot 19$, and this is the element we omit. So the other $ 6$ elements must be $ 3 \\cdot 5,\\ 3 \\cdot 7,\\ 3 \\cdot 19,\\ 3 \\cdot 5 \\cdot 7,\\ 3 \\cdot 5 \\cdot 19,\\ 3 \\cdot 7 \\cdot 19$. This gives the unique sequence described in the first paragraph, and we're done.\r\n\r\n(edit: typo)",
  "Solution_2": "Yes, I think you're right on the 1995."
}
{
  "Tag": [
    "inequalities proposed",
    "inequalities"
  ],
  "Problem": "\\[ \\ a,b,c>0;a\\plus{}b\\plus{}c\\equal{}3.\\]Prove that:\\[ \\frac{a^2}{b\\plus{}c^3}\\plus{}\\frac{b^2}{c\\plus{}a^3}\\plus{}\\frac{c^2}{a\\plus{}b^3} \\geq \\frac{3}{2}.\\] :blush:  :blush:  :blush:",
  "Solution_1": "[quote=\"mai quoc thang\"]\n\\[ \\ a,b,c > 0;a \\plus{} b \\plus{} c \\equal{} 3.\n\\]\nProve that:\n\\[ \\frac {a^2}{b \\plus{} c^3} \\plus{} \\frac {b^2}{c \\plus{} a^3} \\plus{} \\frac {c^2}{a \\plus{} b^3} \\geq \\frac {3}{2}.\n\\]\n[/quote]\r\n$ \\sum_{cyc}\\frac {a^2}{b \\plus{} c^3}\\equal{}\\sum_{cyc}\\frac{a^4}{a^2b\\plus{}c^3a^2}\\geq\\frac{(a^2\\plus{}b^2\\plus{}c^2)^2}{\\sum(a^2b\\plus{}a^3b^2)}.$\r\nThus, it remains to prove that $ 2(a^2\\plus{}b^2\\plus{}c^2)^2\\geq3\\sum_{cyc}(a^2b\\plus{}a^3b^2).$\r\nBut $ 2(a^2\\plus{}b^2\\plus{}c^2)^2\\geq3\\sum_{cyc}(a^2b\\plus{}a^3b^2)\\Leftrightarrow$\r\n$ \\Leftrightarrow2(a\\plus{}b\\plus{}c)(a^2\\plus{}b^2\\plus{}c^2)^2\\geq(a\\plus{}b\\plus{}c)^2(a^2b\\plus{}b^2c\\plus{}c^2a)\\plus{}9(a^3b^2\\plus{}b^3c^2\\plus{}c^3a^2)\\Leftrightarrow$\r\n$ \\Leftrightarrow\\sum_{cyc}(2a^5\\plus{}a^4b\\plus{}2a^4c\\minus{}7a^3b^2\\plus{}3a^3c^2\\minus{}2a^3bc\\plus{}a^2b^2c)\\geq0.$\r\nBut by AM-GM $ \\sum_{cyc}(a^4c\\plus{}a^2b^2c\\minus{}2a^3bc)\\geq0$ is true and\r\n$ \\sum_{cyc}(2a^5\\plus{}a^4b\\plus{}a^4c\\minus{}7a^3b^2\\plus{}3a^3c^2)\\equal{}\\sum_{cyc}(a^5\\plus{}a^4b\\minus{}7a^3b^2\\plus{}3a^2b^3\\plus{}ab^4\\plus{}b^5)\\equal{}$\r\n$ \\equal{}\\sum_{cyc}(a\\minus{}b)(a^4\\plus{}2a^3b\\minus{}5a^2b^2\\minus{}2ab^3\\minus{}b^4)\\equal{}$\r\n$ \\equal{}\\sum_{cyc}\\left((a\\minus{}b)(a^4\\plus{}2a^3b\\minus{}5a^2b^2\\minus{}2ab^3\\minus{}b^4)\\plus{}a^5\\minus{}b^5\\right)\\equal{}$\r\n$ \\equal{}\\sum_{cyc}a(a\\minus{}b)^2(2a^2\\plus{}5ab\\plus{}b^2)\\geq0.$ :)",
  "Solution_2": "[quote=\"arqady\"][quote=\"mai quoc thang\"]\n\\[ \\ a,b,c > 0;a \\plus{} b \\plus{} c \\equal{} 3.\n\\]\nProve that:\n\\[ \\frac {a^2}{b \\plus{} c^3} \\plus{} \\frac {b^2}{c \\plus{} a^3} \\plus{} \\frac {c^2}{a \\plus{} b^3} \\geq \\frac {3}{2}.\n\\]\n[/quote]\n$ \\sum_{cyc}\\frac {a^2}{b \\plus{} c^3} \\equal{} \\sum_{cyc}\\frac {a^4}{a^2b \\plus{} c^3a^2}\\geq\\frac {(a^2 \\plus{} b^2 \\plus{} c^2)^2}{\\sum(a^2b \\plus{} a^3b^2)}.$\nThus, it remains to prove that $ 2(a^2 \\plus{} b^2 \\plus{} c^2)^2\\geq3\\sum_{cyc}(a^2b \\plus{} a^3b^2).$\nBut $ 2(a^2 \\plus{} b^2 \\plus{} c^2)^2\\geq3\\sum_{cyc}(a^2b \\plus{} a^3b^2)\\Leftrightarrow$\n$ \\Leftrightarrow2(a \\plus{} b \\plus{} c)(a^2 \\plus{} b^2 \\plus{} c^2)^2\\geq(a \\plus{} b \\plus{} c)^2(a^2b \\plus{} b^2c \\plus{} c^2a) \\plus{} 9(a^3b^2 \\plus{} b^3c^2 \\plus{} c^3a^2)\\Leftrightarrow$\n$ \\Leftrightarrow\\sum_{cyc}(2a^5 \\plus{} a^4b \\plus{} 2a^4c \\minus{} 7a^3b^2 \\plus{} 3a^3c^2 \\minus{} 2a^3bc \\plus{} a^2b^2c)\\geq0.$\nBut by AM-GM $ \\sum_{cyc}(a^4c \\plus{} a^2b^2c \\minus{} 2a^3bc)\\geq0$ is true and\n$ \\sum_{cyc}(2a^5 \\plus{} a^4b \\plus{} a^4c \\minus{} 7a^3b^2 \\plus{} 3a^3c^2) \\equal{} \\sum_{cyc}(a^5 \\plus{} a^4b \\minus{} 7a^3b^2 \\plus{} 3a^2b^3 \\plus{} ab^4 \\plus{} b^5) \\equal{}$\n$ \\equal{} \\sum_{cyc}(a \\minus{} b)(a^4 \\plus{} 2a^3b \\minus{} 5a^2b^2 \\minus{} 2ab^3 \\minus{} b^4) \\equal{}$\n$ \\equal{} \\sum_{cyc}\\left((a \\minus{} b)(a^4 \\plus{} 2a^3b \\minus{} 5a^2b^2 \\minus{} 2ab^3 \\minus{} b^4) \\plus{} a^5 \\minus{} b^5\\right) \\equal{}$\n$ \\equal{} \\sum_{cyc}a(a \\minus{} b)^2(2a^2 \\plus{} 5ab \\plus{} b^2)\\geq0.$ :)[/quote]\r\nWe have a harder is\r\n\\[ \\sum \\frac{a^2}{b\\plus{}3c^3} \\ge \\frac{3}{4}\\]\r\n:)"
}
{
  "Tag": [],
  "Problem": "Simplify $ \\frac{3\\minus{}5(3\\minus{}2)\\minus{}1^2}{3\\minus{}2^2}$",
  "Solution_1": "The denominator 3-2^2=3-4= -1.\r\nThe numerator reduces to 3-5-1=-3.\r\n-3/-1=3.[/b]"
}
{
  "Tag": [
    "function",
    "algebra",
    "polynomial",
    "quadratics",
    "complex numbers"
  ],
  "Problem": "In these exercises use the given zero to find all the zeros of the function\r\n\r\n\r\n56. g(x) = 4x^3 + 23x^2 + 34x - 10                     \r\n-3+[i]i[/i]\r\n\r\n\r\n57. h(x) = 3x^3 - 4x^2 + 14x + 20                       \r\n1-radical (3) [i]i[/i]\r\n\r\nplease... i need it as soon as possible",
  "Solution_1": "hello, $g(x)=0$ has the solutions $-3-i,-3+i$ and $\\frac{1}{4}$.\r\nAre you sure, that $h(x)=3x^{3}-4x^{2}+14x+20$ is right?\r\nSonnhard.",
  "Solution_2": "o dang\r\n\r\num\r\nfor the second one\r\n\r\nit is h(x) = 3x^3 - 4x^2 + 8x + 8\r\nmy bad\r\n\r\nwith one solution being 1 - radical 3 i",
  "Solution_3": "hello, if you have the solution $1-\\sqrt{3}i$ then must be\r\n$1+\\sqrt{3}i$ the other one, additionally $-\\frac{2}{3}$ ist a solution.\r\nSonnhard.",
  "Solution_4": "The point of these exercises is usually to show that for a polynomial with [i]real[/i] coefficients, if a complex number is a root then its complex conjugate is a root. (i.e. $P(n \\text{ cis}[\\theta])=0 \\Longrightarrow P(n \\text{ cis}[-\\theta])=0$, for $P(x)$ with real coefficients. )\r\n\r\nUsually, after you factor out the two complex roots (by synthetic or long division, or any other method you want) there will remain something easily factorable ,usually by inspection or by the quadratic formula.",
  "Solution_5": "[quote=\"cincodemayo5590\"]The point of these exercises is usually to show that for a polynomial with [i]real[/i] coefficients, if a complex number is a root then its complex conjugate is a root. (i.e. $P(n \\text{ cis}[\\theta])=0 \\Longrightarrow P(n \\text{ cis}[-\\theta])=0$, for $P(x)$ with real coefficients. )\n\nUsually, after you factor out the two complex roots (by synthetic or long division, or any other method you want) there will remain something easily factorable ,usually by inspection or by the quadratic formula.[/quote]\r\n\r\nall you do is take the conjugate of P(z)=0, and using the conjugate rules, you get P(z conjugate)=0, so it is also a root"
}
{
  "Tag": [
    "analytic geometry",
    "probability"
  ],
  "Problem": "[b]1. A point (x,y) is randomly selected from a rectangular grid in the coordinate plane with vertices at (0,0), (8,0), (0,4), and (8,4). What is the probability that $ x^2 \\plus{} y^2\\leq 4$?\n2. In a single elimination tournament that consisted of three rounds, the better player always won. The winner of Round 3 was the champion, and the loser was runner-up. 8 players were randomly assigned slots in Round 1. What is the probability that the runner-up was not second best player in the tournament?\n\n3. Point P is 9 units from the center of a circle of radius 15. How many different chords of the circle contain P and have integer lengths?\n\n\n[/b]",
  "Solution_1": "Hints:\r\n\r\n[hide=\"1\"]quarter circle[/hide]\n\n[hide=\"2\"]How would the players be needed to be placed so that #2 player does not make it to the finals?[/hide]\n\n[hide=\"3\"]Find the minimum and maximum lengths.  Note symmetry.[/hide]",
  "Solution_2": "[quote=\"mihail911\"][b]1. A point (x,y) is randomly selected from a rectangular grid in the coordinate plane with vertices at (0,0), (8,0), (0,4), and (8,4). What is the probability that $ x^2 \\plus{} y^2\\leq 4$?\n2. In a single elimination tournament that consisted of three rounds, the better player always won. The winner of Round 3 was the champion, and the loser was runner-up. 8 players were randomly assigned slots in Round 1. What is the probability that the runner-up was not second best player in the tournament?\n\n3. Point P is 9 units from the center of a circle of radius 15. How many different chords of the circle contain P and have integer lengths?\n\n\n[/b][/quote]\r\n\r\n\r\n[hide]\n\n1.find y and x intercepts, they are both 2. So it is a quarter-circle with radius 2.\n\npi*2^2/4 so 4pi/4 or pi total possible is 32 so 32/pi\n\n2. The ways that the 2nd best team didn't get 2nd would be if they were on the same bracket side as the 1st place one. so 3/7.\n\n3. longest chord possible is 30 (diameter) shortest is 24. So 24, 25,26,27,28,29,30. But there can be two of each not including biggest and smallest. So 24,25,25,26,26,27,27,28,28,29,29,30, or 12.[/hide]",
  "Solution_3": "How is shortest chord 24?",
  "Solution_4": "i understand where i went wrong on the geometric probability one but im still not sure about the team probability. 3/7? i know that's the right answer, but not sure how you guys got it.",
  "Solution_5": "[quote=\"mihail911\"]i understand where i went wrong on the geometric probability one but im still not sure about the team probability. 3/7? i know that's the right answer, but not sure how you guys got it.[/quote]\r\n\r\nok, think about it this way. Since there are 3 single elimination rounds, there are 8 different players in the tournament. The top (best player) is in one side of the bracket. So that means that the 2nd best player (not necessarily 2nd place) can be anywhere in those remaining 7 spots. So, the spots that the 2nd best player can be to make it so he faces off against the best player before the finals are when he is in the same side of the bracket as him. i.e. either playing him in first round or in 2nd round. There are 3 available spots on that side, so 3 out of 7."
}
{
  "Tag": [
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "Let $ a,b,c\\in\\mathbb R^\\plus{}$ then under what condition does the following inequality $ a\\plus{}b\\plus{}c\\ge ab\\plus{}bc\\plus{}ac$ hold?",
  "Solution_1": "[quote=\"Null\"]Let $ a,b,c\\in\\mathbb R^ \\plus{}$ then under what condition does the following inequality $ a \\plus{} b \\plus{} c\\ge ab \\plus{} bc \\plus{} ac$ hold?[/quote]\r\n\r\n\r\nSuppose $ {a,b,c}\\in{(0,k]},$ then\r\n\r\n  $ a \\plus{} b \\plus{} c\\ge{ab \\plus{} bc \\plus{} ca}$ \r\n\r\nholds iff $ k\\in{(0,1]}.$"
}
{
  "Tag": [],
  "Problem": "A salesperson can choose one of these two monthly salary plans.  With $ \\$25,\\!000$ of  sales this month, what is the positive difference in earnings under the two plans?\\\\\t\tA basic salary of $ \\$500$ plus a $ 10\\%$ commission on sales.\t\\\\\nA basic salary of $ \\$1000$ plus a $ 5\\%$ commission on sales.",
  "Solution_1": "well, let's find out which one gets you more money with a $ \\$25000$ of sales\r\n\r\nusing the first plan, we find that $ \\$500$ plus $ 10\\%$ would be $ 500 \\plus{} 0.1 \\times 25000 \\equal{} 3000$\r\nusing the second plan, we find that $ \\$1000$ plus $ 5\\%$ would be $ 1000 \\plus{} 0.05 \\times 25000 \\equal{} 2250$\r\n\r\ntherefore, we find that the positive difference in earnings between the two plans is $ 3000 \\minus{} 2250 \\equal{} 750$"
}
{
  "Tag": [
    "inequalities",
    "algebra",
    "polynomial"
  ],
  "Problem": "Let $ a$, $ b$, and $ c$ be positive real numbers. Show that\r\n$ (abc)^{2}(a \\plus{} b \\plus{} c)^{3}( \\minus{} a \\plus{} b \\plus{} c)(a \\minus{} b \\plus{} c)(a \\plus{} b \\minus{} c) \\geq (a^{2} \\plus{} b^{2} \\plus{} c^{2})^{3}( \\minus{} a^{2} \\plus{} b^{2} \\plus{} c^{2})(a^{2} \\minus{} b^{2} \\plus{} c^{2})(a^{2} \\plus{} b^{2} \\minus{} c^{2})$.",
  "Solution_1": "Nobody? \r\n\r\nBTW, why my latex code does NOT work ?   :( +:mad: \r\n\r\nPlus, an Idea: It will be nice if somebody create a simple program which will be called \"Muirhead-Schur\". I mean, if we input a homogeneous polynomial inequality, then, the output is the result of the proof by muirhead's theorem and Schur's inequality. Someone can do it ?",
  "Solution_2": "shold say more....",
  "Solution_3": "[quote=ideahitme]Let $ a$, $ b$, and $ c$ be positive real numbers. Show that\n$ (abc)^{2}(a \\plus{} b \\plus{} c)^{3}( \\minus{} a \\plus{} b \\plus{} c)(a \\minus{} b \\plus{} c)(a \\plus{} b \\minus{} c) \\geq (a^{2} \\plus{} b^{2} \\plus{} c^{2})^{3}( \\minus{} a^{2} \\plus{} b^{2} \\plus{} c^{2})(a^{2} \\minus{} b^{2} \\plus{} c^{2})(a^{2} \\plus{} b^{2} \\minus{} c^{2})$.[/quote]\nWe can prove it by hand without computer, even without calculator. \nLet $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.\nHence, we need to prove that $f(w^3)\\geq0$, where\n$f(w^3)=(3u^2-2v^2)^3(729u^6-1458u^4v^2+648u^2v^4+216u^3w^3-144uv^2w^3+8w^6)-u^3(27u^3-36uv^2+8w^3)w^6$\nBut $f'(w^3)=(3u^2-2v^2)^3(216u^3-144uv^2+16w^3)-54u^6w^3+72u^4v^2w^3-24u^3w^6\\geq$\n$\\geq u^6\\cdot88u^3-54u^6w^3>0$, \nwhich says that $f$ is an increasing function and it's enough to prove our inequality for a minimal value of $w^3$.\n$a$, $b$ and $c$ are positive roots of the equation $(x-a)(x-b)(x-c)=0$ or $w^3=x^3-3ux^2+3v^2x$.\nThus, a line $y=w^3$ and graph of $y=x^3-3ux^2+3v^2x$ have three common points and $w^3$ gets a minimal value in the following cases.\n1. $w^3\\rightarrow0^+$.\nLet $c\\rightarrow0^+$ and $b=1$. We obtain $(a^2+1)^4(a^2-1)^2\\geq0$;\n2. $b=c=1$, which gives $(a-1)^2(a^5+2a^4+7a^3+12a^2+16a+16)a^5\\geq0$.\nDone!",
  "Solution_4": "[img]http://s9.sinaimg.cn/middle/006ptkjAzy75NN5V0N208&690[/img]",
  "Solution_5": "[quote=luofangxiang][img]http://s9.sinaimg.cn/middle/006ptkjAzy75NN5V0N208&690[/img][/quote]\n\ndone , \u6321\u4ec0\u4e48\uff1f\nz>0 ?????????????????? \u54c8\u54c8",
  "Solution_6": "z<=0 \u663e\u7136\u6210\u7acb\uff08Clearly established\uff09",
  "Solution_7": "Use $UVW$ method",
  "Solution_8": "[quote=ideahitme]Let $ a$, $ b$, and $ c$ be positive real numbers. Show that\n$ (abc)^{2}(a \\plus{} b \\plus{} c)^{3}( \\minus{} a \\plus{} b \\plus{} c)(a \\minus{} b \\plus{} c)(a \\plus{} b \\minus{} c) \\geq (a^{2} \\plus{} b^{2} \\plus{} c^{2})^{3}( \\minus{} a^{2} \\plus{} b^{2} \\plus{} c^{2})(a^{2} \\minus{} b^{2} \\plus{} c^{2})(a^{2} \\plus{} b^{2} \\minus{} c^{2})$.[/quote]\n\nBy computer,we have\n\n\\[LHS-RHS=2(a^2+b^2+c^2)\\sum_{cyc}{a^3b^3(a-b)^2(a+b-c)^2}\\]\n\\[+\\sum_{cyc}{ab(2a^4+a^3b+6a^2b^2+ab^3+9abc^2+2b^4)(a-b)^2(a^2+b^2-c^2)^2}\\]\n\\[+\\frac{1}{2}\\sum_{cyc}{ab(3ab+ac+bc+3c^2)(a-b)^4(a+b-c)^4}\\]\n\\[+\\frac{1}{2}\\sum_{cyc}{(a-b)^4(a^2+b^2-c^2)^4}\\]\n\\[+\\frac{1}{2}\\prod{(a-b)^2(a+b-c)^2}\\]\n\\[\\ge{0}\\]\n",
  "Solution_9": "[quote]By computer,we have[/quote]\nIf not a secret, what program can do that ? ",
  "Solution_10": "Dear luofanxiang,\n\nvery nice your proof but I cannot prove\n\n$9m \\geq 2(mn)^2 +2mn +3n$\n\nCan you please tell me if your prove is by full expansion?\n\nThank you very much",
  "Solution_11": "[quote=szl6208][quote=ideahitme]Let $ a$, $ b$, and $ c$ be positive real numbers. Show that\n$ (abc)^{2}(a \\plus{} b \\plus{} c)^{3}( \\minus{} a \\plus{} b \\plus{} c)(a \\minus{} b \\plus{} c)(a \\plus{} b \\minus{} c) \\geq (a^{2} \\plus{} b^{2} \\plus{} c^{2})^{3}( \\minus{} a^{2} \\plus{} b^{2} \\plus{} c^{2})(a^{2} \\minus{} b^{2} \\plus{} c^{2})(a^{2} \\plus{} b^{2} \\minus{} c^{2})$.[/quote]\n\nBy computer,we have\n\n\\[LHS-RHS=2(a^2+b^2+c^2)\\sum_{cyc}{a^3b^3(a-b)^2(a+b-c)^2}\\]\n\\[+\\sum_{cyc}{ab(2a^4+a^3b+6a^2b^2+ab^3+9abc^2+2b^4)(a-b)^2(a^2+b^2-c^2)^2}\\]\n\\[+\\frac{1}{2}\\sum_{cyc}{ab(3ab+ac+bc+3c^2)(a-b)^4(a+b-c)^4}\\]\n\\[+\\frac{1}{2}\\sum_{cyc}{(a-b)^4(a^2+b^2-c^2)^4}\\]\n\\[+\\frac{1}{2}\\prod{(a-b)^2(a+b-c)^2}\\]\n\\[\\ge{0}\\][/quote]\n\n\u8bc4\u6ce8\uff1a\n1.\u8fd9\u4e2a\u4e0d\u7b49\u5f0f\u6709\u4e00\u4e2a\u7279\u70b9\uff0c\u4f60\u4eec\u89c9\u5f97\u5b83\u662f\u4ec0\u4e48\uff1f\n[hide]\u4e24\u4e2a\u7b26\u53f7\u4e0d\u786e\u5b9a\u7684\u91cf\u6784\u6210\u4e86\u4e0d\u7b49\u5f0f\u3002\u8fd9\u7c7b\u4e0d\u7b49\u5f0f\u76ee\u524d\u770b\u5230\u7684\u597d\u8c61\u662f\u201c\u975e\u5e38\u7a00\u5c11\u201d[/hide]\n\n2.\u4ece\u5f0f\u5dee\u7684\u8868\u8fbe\u5f0f\u53ef\u4ee5\u770b\u51fa\uff0c\u8fd9\u4e2a\u4e0d\u7b49\u5f0f\u4ecd\u542b\u6709\u8f83\u591a\u7684\u6ce5\u5df4\u3002\n\u5982\u4f55\u8ba9\u5b83\u66f4\u5f3a\u5462\uff1f\n\namount of symbol uncertainty:\n=================================\n$x,y,z>0$,if $f(x,y,z)\\geq 0$ is not true,and $f(x,y,z)\\le 0$ is not true,then call $f(x,y,z)$ is amount of symbol uncertainty.\n\nwith $M$ and $N$ are amount of symbol uncertainty,then call inequality \n\\[M\\geq N\\]\n\nis amount of symbol uncertainty inequality."
}
{
  "Tag": [
    "inequalities",
    "trigonometry",
    "geometry",
    "inradius",
    "circumcircle",
    "triangle inequality",
    "area of a triangle"
  ],
  "Problem": "In the triangle $ABC$ $AM$ is a median. Prove that : $\\sin{\\angle{MAB}}\\ge{\\frac{2r}{R}}$.\r\nHereby r is inradius, R is circumradius.\r\nPlease post your solution very quickly!!",
  "Solution_1": "Very nice problem, but it has a typo, and moreover, please don't write things like \"Please post your solution very quickly!!\" since people tend to wonder, when seeing such requests, whether the problem is from some current mathematical olympiad.\r\n\r\nHere is the corrected version of the problem:\r\n\r\n[color=blue][b]Problem.[/b] Let M be the midpoint of the side BC of a triangle ABC. Prove that $\\sin\\measuredangle MAB>\\frac{r}{2R}$, where r is the inradius and R is the circumradius of the triangle ABC.[/color]\r\n\r\n[i]Solution.[/i] The Sine Law in triangle BAM yields\r\n\r\n$\\sin\\measuredangle MAB=\\sin\\measuredangle ABM\\cdot\\frac{BM}{AM}$.\r\n\r\nBut < ABM = B and $BM=\\frac{a}{2}$ (since the point M is the midpoint of the side BC of triangle ABC). Thus, we get\r\n\r\n$\\sin\\measuredangle MAB=\\sin B\\cdot\\frac{\\left(\\frac{a}{2}\\right)}{AM}=\\frac{a\\sin B}{2\\cdot AM}$.\r\n\r\nNow, the area $\\Delta$ of triangle ABC can be expressed in the form $\\Delta=\\frac12 ca\\sin B$ on the one hand, while, on the other hand, it can be found from the formula $\\Delta=\\frac12\\left(a+b+c\\right)r$ (this is just the well-known fact that the area of a triangle equals the semiperimeter times the inradius). Thus, $\\frac12 ca\\sin B=\\frac12\\left(a+b+c\\right)r$, what simplifies to $ca\\sin B=\\left(a+b+c\\right)r$, and thus $a\\sin B=\\frac{\\left(a+b+c\\right)r}{c}$. Since a + b > c by the triangle inequality, we have $a\\sin B=\\frac{\\left(a+b+c\\right)r}{c}>\\frac{\\left(c+c\\right)r}{c}=2r$. Thus,\r\n\r\n$\\sin\\measuredangle MAB=\\frac{a\\sin B}{2\\cdot AM}>\\frac{2r}{2\\cdot AM}=\\frac{r}{AM}$.\r\n\r\nFinally, we have AM < 2R; in fact, since the point A lies on the circumcircle of triangle ABC, and the point M, being the midpoint of the side BC, lies inside this circumcircle, the segment AM is contained in the circumcircle of triangle ABC, and thus it is shorter than the diameter 2R of this circumcircle. From AM < 2R, it follows that $\\frac{r}{AM}>\\frac{r}{2R}$; altogether,\r\n\r\n$\\sin\\measuredangle MAB>\\frac{r}{AM}>\\frac{r}{2R}$,\r\n\r\nand the problem is solved.\r\n\r\n  darij",
  "Solution_2": "Thank you, darij.I'll do after as you asked.",
  "Solution_3": "And what about this? \r\n$\\sin{AMB}>\\frac{2r}{R}$.",
  "Solution_4": "[quote=\"Armo\"]And what about this? \n$\\sin{AMB}>\\frac{2r}{R}$.[/quote]\r\n\r\nI think it suffices to prove this inequality:\r\nIn triangle $ABC$, $AM$ is the median, then\r\n\\[2R \\ge \\frac{bc}{AM} \\ge 4r.\\]\r\nHowever I am not sure if it is really true.  :?",
  "Solution_5": "[quote=\"Armo\"]And what about this? \n$\\sin{AMB}>\\frac{2r}{R}$.[/quote]\n\nI guess you mean $\\sin\\measuredangle AMB\\geq\\frac{2r}{R}$. And this can be proven as follows:\n\nIf H is the foot of the altitude of triangle ABC from the vertex A, then the triangle AHM is right-angled, so that $\\sin\\measuredangle AMH = \\frac{AH}{AM}$. But we have either < AMH = < AMB, or < AMH = 180\u00b0 - < AMB; in both of these cases, sin < AMH = sin < AMB, and thus we get $\\sin\\measuredangle AMB = \\frac{AH}{AM}$. Hence, the inequality in question, $\\sin\\measuredangle AMB\\geq\\frac{2r}{R}$, rewrites as $\\frac{AH}{AM}\\geq\\frac{2r}{R}$. This is clearly equivalent to $\\frac{R}{2r}\\geq\\frac{AM}{AH}$. And this can be written in the form $\\frac{R}{2r}\\geq\\frac{m_a}{h_a}$, where $m_a=AM$ is the median and $h_a=AH$ is the altitude of triangle ABC from the vertex A. In this form, the inequality was (ingeniously!) proven by Grobber in http://www.mathlinks.ro/Forum/viewtopic.php?t=2558 post #2.\n\n[quote=\"yptsoi\"]I think it suffices to prove this inequality:\nIn triangle $ABC$, $AM$ is the median, then\n\\[2R \\ge \\frac{bc}{AM} \\ge 4r.\\][/quote]\r\n\r\nYes, this is true, although I don't see why you need the left part of this inequality (actually, only the right part, $\\frac{bc}{AM}\\geq 4r$, is equivalent to $\\sin\\measuredangle AMB\\geq\\frac{2r}{R}$). The left part is almost trivial, by the way (we have $2R=\\frac{bc}{h_a}$ and $h_a\\leq AM$).\r\n\r\n  Darij"
}
{
  "Tag": [
    "function",
    "real analysis",
    "Ramsey Theory",
    "real analysis unsolved"
  ],
  "Problem": "Does there exist  a function $ f$ such that $ f$ is not measurable and  \r\n$ f(\\frac{x\\plus{}y}{2})\\leq \\frac{1}{2}(f(x)\\plus{}f(y))$.",
  "Solution_1": "Yes. Take one of the monsters that satisfies the equality $ g(x\\plus{}y)\\equal{}g(x)\\plus{}g(y)$ and add any convex function.",
  "Solution_2": "A lot of thanks!\r\nPlease give any  of the monsters that satisfies the equality:$ g(x\\plus{}y)\\equal{}g(x)\\plus{}g(y)$",
  "Solution_3": "You can't give them explicitly, as constructions rely on a Hamel basis for $ \\bf R$ over $ \\bf Q$.",
  "Solution_4": "Any time you see \"not measurable\" (in the Lebesgue sense), you should immediately accept that you're not going to see an explicit construction or example. Constructing stuff that's not Lebesgue-measurable requires the Axiom of Choice.",
  "Solution_5": "I'm pretty sure there's a model of analysis without the AC that has all sets measurable. It isn't really what we want mathematics to be; we just give up way too much.",
  "Solution_6": "[quote=\"jmerry\"]I'm pretty sure there's a model of analysis without the AC that has all sets measurable. It isn't really what we want mathematics to be; we just give up way too much.[/quote]\r\nThis doesn't contradict my statement. :) I can say anything I want about non-measurable sets in a model where they don't exist! \r\n\r\nI used to be interested in that question of how much you can do without AC, but a lot of people whom I respect think that not having AC is such a loss that there's no point in even considering it, so I've thought about it less of late. An interesting thing I noticed recently was that the ergodic proof of Szemeredi's theorem as given by Furstenberg requires AC, but a couple of years ago Terence Tao published a paper which uses ergodic theory but not AC to prove that result."
}
{
  "Tag": [
    "function",
    "calculus",
    "derivative",
    "logarithms",
    "limit",
    "topology",
    "real analysis"
  ],
  "Problem": "Let $f: [a,b] \\to R$ be a continuous function , $\\exists f'(x)$ $\\forall x \\in (a,b)$ and $\\exists c \\in (a,b)$ such that $f'(c)=0$.\r\nProve that $\\forall k \\in R^{+}, \\exists d \\in (a,b)$ such that $f'(d)=k(f(d)-f(a))$",
  "Solution_1": "Assume additionally that $f$ is not constant; the statement is trivially true in that case.\r\n\r\nWLOG, $f(a)=0$. The quotient $\\frac{f'(x)}{f(x)}$ is the derivative of $g(x)=\\log |f(x)|$ (on the open set where $f$ is nonzero), so it has the intermediate value property on each of its intervals of definition.\r\nLet $P$ be a point in $(a,b)$ with $f'(P)=0$, $f(P)\\neq 0$; if there is no such point, we have some $x$ with $f(x)=f'(x)=0$, and the statement is trivially true. Let $Q$ be the greatest $x$ less than $P$ with $f(Q)=0$\r\n\r\nNow, consider $g$ on $(Q,P]$. The limit $\\lim_{x\\to Q^{+}}g(x)$ is $-\\infty$, so $g'$ must be unbounded above. Since $g'(P)=0$ as well, $g'$ takes all positive values by its intermediate value property."
}
{
  "Tag": [
    "LaTeX",
    "function"
  ],
  "Problem": "The set of all numbers x for which \\[x+\\sqrt{x^{2}+1}-\\frac{1}{x+\\sqrt{x^{2}+1}}\\] is a rational number is the set of all:\n$\\textbf{(A)}\\ \\text{ integers } x \\qquad\n\\textbf{(B)}\\ \\text{ rational } x \\qquad\n\\textbf{(C)}\\ \\text{ real } x\\qquad\n\\textbf{(D)}\\ x \\text{ for which } \\sqrt{x^2+1} \\text{ is rational} \\qquad\n\\textbf{(E)}\\ x \\text{ for which } x+\\sqrt{x^2+1} \\text{ is rational }$",
  "Solution_1": "Put dollar signs ($\\$$) before and after your LaTeX codes to make it look nicer ;) \r\n\r\n[quote=\"Zootieroaz\"]The set of all numbers $x$ for which $x+\\sqrt{x^{2}+1}-\\frac{1}{x+\\sqrt{x^{2}+1}}$ is a rational number is the set of all:\na, integers $x$\nb, rational $x$\nc, $x$ for which $x+\\sqrt{x^{2}+1}$ is rational \nd, real $x$\ne, $x$ for which $x+\\sqrt{x^{2}+1}$ is irrational[/quote]LaTeXed :D",
  "Solution_2": "[quote=\"Zootieroaz\"]The set of all numbers x for which $x+\\sqrt{x^{2}+1}-\\frac{1}{x+\\sqrt{x^{2}+1}}$ is a rational number is the set of all:\na, integers x\nb, rationa x\nc, x for which $x+\\sqrt{x^{2}+1}$ is rational \nd, real x\ne, x for which $x+\\sqrt{x^{2}+1}$ is irrational[/quote]\r\n\r\n[hide][size=200][b]C?[/b][/size][/hide]",
  "Solution_3": "[hide=\"hint\"]rationalize the denominator of the second term and see what happens... \n\n[hide=\"answer\"]it simplfies to $2x$, so $B$[/hide]\n\n[/hide]",
  "Solution_4": "Sorry, al, I think you made a tiny mistake when doing that. ;)\r\n\r\n[hide=\"Answer\"]$x+\\sqrt{x^2+1}-\\frac{1}{x+\\sqrt{x^2+1}}$\nRationalizing the denominator of the fraction, we have:\n$\\frac{1}{x+\\sqrt{x^2+1}}\\left(\\frac{x-\\sqrt{x^2+1}}{x-\\sqrt{x^2+1}}\\right)=x-\\sqrt{x^2+1}$\nTherefore, we have $x+\\sqrt{x^2+1}-(x-\\sqrt{x^2+1})=2\\sqrt{x^2+1}$.\nSo, our answer is $\\boxed{D}$.[/hide]",
  "Solution_5": "[quote=\"JesusFreak197\"]\n$\\frac{1}{x+\\sqrt{x^2+1}}\\left(\\frac{x-\\sqrt{x^2+1}}{x-\\sqrt{x^2+1}}\\right)=x-\\sqrt{x^2+1}$\n[/quote]\r\n\r\nSorry, but the denominator would be $(x + \\sqrt{x^2 + 1}) (x - \\sqrt{x^2 + 1}) = x^2 - (\\sqrt{x^2 + 1})^2 = (x^2) - (x^2 +1) = -1$, not $+1$.\r\n\r\nSo that expression is equal to $\\sqrt{x^2+1} - x$, and simplifies the way Altheman said it does.  The answer is, indeed, $\\boxed{B}$.  (A calculator verifies this quickly.)",
  "Solution_6": ":oops: I guess that's what happens when you don't bother to fully calculate it out. :roll:",
  "Solution_7": "i know...i should write out the problems more...",
  "Solution_8": "Sorry - I'm a bit confused.  Doesn't the expression simplify to\r\n\\[ x\\plus{}\\sqrt{x^2\\plus{}1}\\minus{}((x \\plus{} \\sqrt{x^2 \\plus{} 1}) (x \\minus{} \\sqrt{x^2 \\plus{} 1}) \\equal{} x^2 \\minus{} (\\sqrt{x^2 \\plus{} 1})^2 \\equal{} (x^2) \\minus{} (x^2 \\plus{}1) \\equal{} \\minus{}1)\\equal{}2x\r\n\\]\r\n\r\nIf so, then wouldn't whether or not $ x$ is real depend on $ \\sqrt{x^2\\plus{}1}$ being real?  Did I do something wrong?\r\n\r\nThanks in advance!",
  "Solution_9": "The problem contains the implicit assumption that $ x$ is real; otherwise, the square root function is multi-valued and the problem is not well-defined."
}
{
  "Tag": [
    "geometry",
    "incenter",
    "circumcircle",
    "IMO 1981",
    "similar triangles",
    "IMO"
  ],
  "Problem": "Three circles of equal radius have a common point $O$ and lie inside a given triangle. Each circle touches a pair of sides of the triangle. Prove that the incenter and the circumcenter of the triangle are collinear with the point $O$.",
  "Solution_1": "Let A, B, C be the centers of the 3 congruent circles (A), (B), (C) intersecting at the point O. Since the point O is equidistant from the centers A, B, C, it is the circumcenter of the triangle $\\triangle ABC$. Let a, b, c be the common external tangents of the circle pairs (B), (C); (C), (A); (A), (B), neither of them intersecting the remaining circle and let the lines a, b, c intersect at points $A' \\equiv b \\cap c,\\ B' \\equiv c \\cap a,\\ C' \\equiv a \\cap b$, forming a triangle $\\triangle A'B'C'$. Let O' be the circumcenter of this new triangle. Since the common external tangent of 2 congruent circles is parallel to their center line, $a \\equiv B'C' \\parallel BC,\\ b \\equiv C'A' \\parallel CA,\\ c \\equiv A'B' \\parallel AB$. Thus the triangles $\\triangle A'B'C' \\sim \\triangle ABC$ are centrally similar, having the corresponding sides parallel. The lines A'A, B'B, C'C connecting the corresponding vertices of the 2 triangles meet at their homothety center. But the lines A'A, B'B, C'C are the bisectors of the angles $\\angle A', \\angle B', \\angle C'$ (and also of the angles $\\angle A, \\angle B, \\angle C$), hence, the homothety center is the common incenter I of the triangles $\\triangle A'B'C',\\ \\triangle ABC$. The circumcenters O', O are the corresponding points of these 2 centrally similar triangles, hence, the line O'O also passes through the homothety center I.",
  "Solution_2": "[quote=\"yetti\"]Let A, B, C be the centers of the 3 congruent circles (A), (B), (C) intersecting at the point O. Since the point O is equidistant from the centers A, B, C, it is the circumcenter of the triangle $\\triangle ABC$. Let a, b, c be the common external tangents of the circle pairs (B), (C); (C), (A); (A), (B), neither of them intersecting the remaining circle and let the lines a, b, c intersect at points $A' \\equiv b \\cap c,\\ B' \\equiv c \\cap a,\\ C' \\equiv a \\cap b$, forming a triangle $\\triangle A'B'C'$. Let O' be the circumcenter of this new triangle. Since the common external tangent of 2 congruent circles is parallel to their center line, $a \\equiv B'C' \\parallel BC,\\ b \\equiv C'A' \\parallel CA,\\ c \\equiv A'B' \\parallel AB$. Thus the triangles $\\triangle A'B'C' \\sim \\triangle ABC$ are centrally similar, having the corresponding sides parallel. The lines A'A, B'B, C'C connecting the corresponding vertices of the 2 triangles meet at their homothety center. But the lines A'A, B'B, C'C are the bisectors of the angles $\\angle A', \\angle B', \\angle C'$ (and also of the angles $\\angle A, \\angle B, \\angle C$), hence, the homothety center is the common incenter I of the triangles $\\triangle A'B'C',\\ \\triangle ABC$. The circumcenters O', O are the corresponding points of these 2 centrally similar triangles, hence, the line O'O also passes through the homothety center I.[/quote]\r\n\r\nWhy do A'A, B'B, C'C meet at the incentre?",
  "Solution_3": "Because A', B', C' are equidistant from the sides so they all lie on the respective angle bisectors."
}
{
  "Tag": [
    "trigonometry"
  ],
  "Problem": "Prove that $ \\csc\\frac{180^{\\circ}}{7}=\\csc\\frac{360^{\\circ}}{7}+\\csc\\frac{540^{\\circ}}{7}$.",
  "Solution_1": "[hide=\"A Start\"]\n$ \\theta = \\frac{180^{\\circ}}{7}$\n$ \\frac{1}{\\sin \\theta}= \\frac{1}{\\sin 2 \\theta}+\\frac{1}{\\sin 3 \\theta}$\n$ \\frac{1}{\\sin \\theta}= \\frac{1}{2 \\sin \\theta \\cos \\theta}+\\frac{1}{3 \\sin \\theta-4 \\sin^{3}\\theta}$\n$ \\sin \\theta = \\frac{2 \\sin \\theta \\cos \\theta+3 \\sin \\theta-4 \\sin^{3}\\theta }{(2 \\sin \\theta \\cos \\theta) (3 \\sin \\theta-4 \\sin^{3}\\theta )}$.\n[/hide]",
  "Solution_2": "I think up to your second equation it was good, then it got a little crazy. \r\n\r\nInstead, try writing it as $ \\sin 2\\theta\\sin 3\\theta=\\sin\\theta(\\sin 2\\theta+\\sin 3\\theta)$ and notice that $ \\sin 3\\theta = \\sin 4\\theta$."
}
{
  "Tag": [
    "algebra",
    "function",
    "domain",
    "calculus",
    "derivative",
    "conics",
    "parabola"
  ],
  "Problem": "how would the graph of square root(X^4-16X^2) look like?\r\nwhere's the rel. extrema and intercepts?",
  "Solution_1": "$ f(x) \\equal{} \\sqrt{x^4\\minus{}16x^2}$\r\nintercepts: \r\n\r\n$ \\sqrt{x^4\\minus{}16x^2} \\equal{} 0$\r\n$ x^4\\minus{}16x^2 \\equal{} 0$\r\n$ x^2(x\\minus{}4)(x\\plus{}4)\\equal{}0$\r\n\r\n$ x\\equal{} \\pm 4, 0$\r\n\r\nThese x-intercepts are also going to be extrema because $ f(x)$'s domain is $ (\\minus{}\\infty, \\minus{}4] \\cup [0] \\cup [4,\\infty)$.  The other extrema can be found by differentiation:\r\n\r\n$ y^2 \\equal{} x^4\\minus{}16x^2$\r\n$ 2yy' \\equal{} 4x^3\\minus{}32x \\equal{} 4x(x^2\\minus{}8)$\r\nSo $ y'\\equal{}0$ when $ x\\equal{}\\pm 2 \\sqrt{2}$ (both outside our domain) and $ x\\equal{}0$",
  "Solution_2": "so how would the graph look?",
  "Solution_3": "Well, notice that $ \\sqrt {x^4 \\minus{} 16 x^2} \\approx \\sqrt {x^4 \\minus{} 16 x^2 \\plus{} 64} \\equal{} x^2 \\minus{} 8$ for $ |x| > > 0$\r\n\r\nSo you can draw that parabola in with a dotted line or whatever. Now, notice that the smaller $ |x|$ is, the lower $ f(x)$ is compared to $ x^2 \\minus{} 8$, because the difference of 64 under the squareroot makes a bigger and bigger difference. Finally, when it gets to $ |x| \\equal{} 4$, the graph, of $ f(x)$ hits $ 0$, and is undefined for $ |x| < 4$ (except at the origin).\r\n\r\nThe one additional piece of information that you might want is the points of inflection. These are the point at which the graph transitions from being locally hyperbola-like (shooting straight up near the intercept and flattening out) to being parabola like (being convex).\r\n\r\nTo do this, we find the second derivative and equate it to $ 0$, and solve:\r\n\r\n$ (\\sqrt {x^4 \\minus{} 16x^2})' \\equal{} \\minus{} \\frac {\\left( \\minus{} 32 x \\plus{} 4 x^3\\right)^2}{4 \\left( \\minus{} 16 x^2 \\plus{} x^4\\right)^{3/2}} \\plus{} \\frac { \\minus{} 32 \\plus{} 12 x^2}{2 \\sqrt { \\minus{} 16 x^2 \\plus{} x^4}} \\equal{} 0 \\implies x \\equal{} \\pm 2 \\sqrt {6}$\r\n\r\nSo, $ f$ is concave for $ 4 < |x| < 2 \\sqrt {6}$, and convex for $ |x| > 2 \\sqrt {6}$\r\n\r\nThe final graph is attached! I've circled the intercept and the point of inflection on the right half of the graph, and also included the parabola \"guide\", to show how the two are related, since you would probably plot the parabola first if you were drawing it by hand."
}
{
  "Tag": [
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Suppose that $ A$ is a compact subset of $ \\mathbb{R}$ and $ f: A \\rightarrow \\mathbb{R}$ is continuous. Prove that the set $ \\{x \\in A | 0 \\leq f(x) \\leq 1 \\}$ is a compact subset of $ \\mathbb{R}$.\r\n\r\nI know that the set $ \\{x \\in A | 0 \\leq f(x) \\leq 1 \\}$ is bounded because $ A$ is bounded, so I just need to prove that $ \\{x \\in A | 0 \\leq f(x) \\leq 1 \\}$ is closed. I was thinking to take a sequence and show that it converges in the set. This is confusing me. Thanks for any help in advance.\r\n\r\n[b]Solved it.  :lol: [/b]",
  "Solution_1": "$ (x_n\\rightarrow L) \\ \\Longrightarrow \\ (f(x_n)\\rightarrow f(L))\\ \\Longrightarrow \\ (0\\leq f(L) \\leq 1)\\  \\Longrightarrow \\ (L\\in A)$\r\n :cool:",
  "Solution_2": "It is closed because $ \\{x\\in A\\}|0\\leq f(x)\\leq 1\\}\\equal{}f^{\\minus{}1}([0,1])$ is the preimage of a closed set. Because $ f$ is continuous this preimage is closed.\r\n\r\nMark"
}
{
  "Tag": [
    "calculus",
    "integration",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Hello Math professionals,\r\n\r\nActually I am a student in the 12 grade in germany, with the age of only 17\r\n( => Your level of math is far beyond my imagination  :D )\r\n\r\nWhile doing my homework ( don't laugh at me please  :blush: )\r\nI found the following issue:\r\n\r\n$ \\int \\!{x}^{4}{{\\rm e}^{x}}{dx} \\equal{} \\left( {x}^{4} \\minus{} 4\\,{x}^{3} \\plus{} 12\\,{x}^{2} \\minus{} 24\\,x \\plus{} 24 \\right) {{\\rm e}^{x}}$\r\n\r\nCan you see the system?\r\n\r\nBasically I tried to make a formula which now looks like this:\r\n$ \\int \\!{x}^{n}t \\left( x \\right) {dx} \\equal{} {x}^{n}\\int \\!t \\left( x \\right) {dx} \\minus{} {{\\it nx}}^{n \\minus{} 1}\\int \\!\\!\\!\\int \\!t \\left( x \\right) {dx }\\,{dx} \\plus{} \\left( n \\minus{} 1 \\right) {x}^{n \\minus{} 2}\\int \\!\\!\\!\\int \\!\\!\\!\\int \\!t \\left( x \\right) {dx}\\,{dx}\\,{dx} \\minus{} ...$\r\n\r\nThis seems to always be true, but of course I cannot proove it.\r\nNow I ask you because if I tell my math teacher I solved it that way he will ask me for a proof  :)\r\nand of course because i thought this can be interesting for you. \r\n\r\nThank you for your answers\r\nListing",
  "Solution_1": "hello\r\nI m sorry but  :( R u not just using integration by parts .What to prove in this?\r\nthank u",
  "Solution_2": "Yes, sorry my mistake. I didn't see it  :blush:",
  "Solution_3": "Its OK . :)"
}
{
  "Tag": [
    "function",
    "parameterization",
    "parametric equation"
  ],
  "Problem": "The solution I saw was really elegant, but required very clever substitutions.  Surely, however, there are other solutions.\r\n\r\nFind a general solution for\r\nx^y=y^x",
  "Solution_1": "I have thought about this problem for a very long time. What do you mean by general solution. Do you mean of the form y=f(x)? The farthest I've gone to solving it is finding x^(1/x)=y^(1/y) and that 2,4 is a solution.",
  "Solution_2": "I believe Magnara is asking for a parametric solution.  That is, x = f(t) and y = g(t).  As I recall, there is one.",
  "Solution_3": "It seems my solution only works for x, y > 0.\n\n\n\n[hide]Let y = x<sup>a</sup>, where a is a real number.\n\n\n\nThen x<sup>x<sup>a</sup></sup> = (x<sup>a</sup>)<sup>x</sup>\n\n\n\nRS = x<sup>ax</sup>\n\n\n\nSince LS and RS have equal bases, their exponents must be equal as well; thus\n\n\n\nx<sup>a</sup> = ax\n\n\n\nx<sup>a - 1</sup> = a\n\n\n\nThus x = a<sup>1 / (a - 1)</sup> and y = a<sup>a / (a - 1)</sup>. I've tested it for a = 2 and 3 and it seems to work.[/hide]",
  "Solution_4": "Osiris: Your parametric equation does not include the case where x=y (and thus a=1), because then you would be dividing by 0 when you divide by a-1. It's a really neat solution though. I think that's the only missing case.",
  "Solution_5": "According to your solution, \n\n[hide]\n\nx = a<sup>1 / (a - 1)</sup> and y = a<sup>a / (a - 1)</sup>[/hide]\n\nwhen x = y, a = 1 then it becomes undefined since you are dividing by 0?  Does it show there are infinite solutions when x=y?  (I know it's obvious that x=y works)\n\n\n\nEdit: Oops, I must be slow in typing my reply then.",
  "Solution_6": "Ah yes, I forgot to take into account the trivial case.",
  "Solution_7": "Can you use Lambert's W function to express y in terms of x?  \r\nDoes that count as a valid solution?",
  "Solution_8": "What's lambert's W function?",
  "Solution_9": "I meant a parametric solution.\r\n\r\nI don't know what Lambert's W function is, but I saw a solution that gives an elementary form for x and y.",
  "Solution_10": "Here is a description of Lambert's W function:\r\n[url]http://en.wikipedia.org/wiki/Lambert_W_function[/url]\r\n\r\nI just learned about it from one of my friends.  I showed him this problem and he wrote the equation as \r\ny = x...  (with a Lambert's W function)\r\nHe even gave me a graph of that function.\r\nBut I guess it's not really useful if you write it as a Lambert's W function, pretty ugly and hard to evaluate by hand.",
  "Solution_11": "The solution I saw involved substituting k*x=y.  Can anyone get it with that?",
  "Solution_12": "Ya, I tried the substitution kx=y a while ago, and you get EXACTLY the same thing as x^k=y (do you see why?), so I didnt bother to post it.",
  "Solution_13": "I thought about this problem for about eight hours yesterday when I was coming home from college. My best attempt was to try to find a function satisfying\r\n\r\n[tex]\\displaystyle f(x)^2=x+f(2x),[/tex]\r\n\r\nbut I couldn't think of any finite ones. Clearly it must have a [tex]\\sqrt{x}[/tex] term, but then I couldn't find a way to make the smaller terms work out, so I was only able to find\r\n\r\n[tex]\\displaystyle f(x)=\\sqrt{x+\\sqrt{2x+\\sqrt{4x+\\sqrt{8x+\\cdots}}}},[/tex]\r\n\r\nwhich isn't very useful.",
  "Solution_14": "[quote=\"ComplexZeta\"]I thought about this problem for about eight hours yesterday ...[/quote]\r\n\r\nWrong thread?",
  "Solution_15": "Hm you're right. Sleep deprivation is a bad thing..."
}
{
  "Tag": [],
  "Problem": "A polyomino is a connected collection of unit squares where the unit squares create a continuous path from the first square to the last square.  Consecutive squares in the path must share a side.  The smallest polyomino that has one hole uses $ 7$ unit squares, as shown.  How many unit squares are in the smallest polyomino that has two holes? (A hole is a unit square that is not part of the polyomino but is surrounded on all sides by squares that are part of the polyomino.)\n\n[asy]defaultpen(linewidth(0.8));\nsize(3cm,3cm);\n\nfill(((0,0)--(3,0)--(3,1)--(0,1)--cycle),lightgray);\n\nfill(((0,1)--(0,3)--(2,3)--(2,2)--(1,2)--(1,1)--cycle),lightgray);\n\nfill(((2,1)--(2,2)--(3,2)--(3,2)--(3,1)--cycle),lightgray);\n\ndraw((0,0)--(3,0)--(3,1)--(0,1)--(0,0));\n\ndraw((1,0)--(1,1));\n\ndraw((2,0)--(2,1));\n\ndraw((0,1)--(0,3)--(2,3)--(2,2)--(1,2)--(1,1));\n\ndraw((0,2)--(1,2));\n\ndraw((1,3)--(1,2));\n\ndraw((2,1)--(2,2)--(3,2)--(3,2)--(3,1));[/asy]",
  "Solution_1": "You can start from the corner without a square, go 3 squares right, 2 squares down, and 1 squares left. So you add 6 to the 7 already there and get $ \\boxed{13}$",
  "Solution_2": "OR\r\nyou can start from the top middle square\r\ngo one square up, two squares across, and one square down\r\nnow, we only add 4 to the original 7, to get an answer of $ \\boxed{11}$",
  "Solution_3": "Your right. My answer isn't the smallest.",
  "Solution_4": "how do you show that 11 is the smallest answer, though?"
}
{
  "Tag": [
    "inequalities",
    "logarithms",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "$ p_k$ - k-th prime\r\nconsider the inequality $ p_{n \\plus{} 1}\\ln n > p_n\\ln(n \\plus{} 1)$\r\nis it true for every $ n > 1$ ?\r\nif not, is it true for sufficiently large $ n$ ?",
  "Solution_1": "Your inequality equavalent to\r\n$ p_{n\\plus{}1}\\minus{}p_n>\\frac{p_nln(1\\plus{}\\frac 1n)}{ln(n)}<\\frac{p_n}{nln(n)}<2$. Therefore it is true for $ n>4$."
}
{
  "Tag": [
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "Can you please explain to me what this means\r\n\r\n$\\sum_{k=0}^{n}3^{k}* k * \\binom{n}{k}$",
  "Solution_1": "$\\sum_{k=0}^{n}3^{k}* k * \\binom{n}{k}$\r\n\r\nwouldn't that mean the sum of the expression where k starts at 0 and goes up to n\r\n\r\nsuch$3^{0}* 0* \\binom{n}{0}+3^{1}* 1 * \\binom{n}{1}....$\r\n\r\nall the way until you get to \"n\" which is the last value of \"k\"\r\nsay n=5\r\n$3^{5}* 5 * \\binom{5}{5}$\r\n\r\nIs that what you are asking?\r\n\r\nso if n=5\r\n$3^{0}* 0* \\binom{5}{0}+3^{1}* 1 * \\binom{5}{1}+3^{2}* 2 * \\binom{5}{2}+3^{3}* 3 * \\binom{5}{3}+3^{4}* 4* \\binom{5}{4}+3^{5}* 5 * \\binom{5}{5}$",
  "Solution_2": "so basically you're telling me that \r\n\r\n$\\sum_{k=0}^{n}3^{k}*k*\\binom{n}{k}$ = $3^{0}*0 *\\binom{n}{0}+...+3^{n}*n*\\binom{n}{n}$",
  "Solution_3": "i edited to put my final result :lol:\r\noh and oops that is what i am saying have in mind i am human so i could have made a mistake",
  "Solution_4": "hello, it's\r\n$\\sum_{k=0}^{n}3^{k}\\cdot k\\cdot\\binom{n}{k}=3\\cdot n\\cdot4^{n-1}$.\r\nSonnhard.",
  "Solution_5": "how did you just simplify it like that? :?: \r\ni want to know...",
  "Solution_6": "[quote=\"Dr Sonnhard Graubner\"]hello, it's\n$\\sum_{k=0}^{n}3^{k}\\cdot k\\cdot\\binom{n}{k}=3\\cdot n\\cdot4^{n-1}$.\nSonnhard.[/quote]\r\n\r\nHi sonnhard\r\n\r\ncan you tell me how you got to that result pls ?",
  "Solution_7": "Yes! Please explain.",
  "Solution_8": "$\\sum_{k=1}^{n}3^{k}\\cdot k\\binom{n}{k}$\r\n$=\\sum_{k=1}^{n}3^{k}\\cdot n\\binom{n-1}{k-1}$\r\n$=3n\\sum_{k=1}^{n}3^{k-1}\\cdot\\binom{n-1}{k-1}$\r\n$=3n\\sum_{k=0}^{n-1}\\binom{n-1}{k}\\cdot 3^{k}$\r\n$=3n(3+1)^{n-1}$",
  "Solution_9": "why did you write k=1 ? if k=0 in the first condition ?",
  "Solution_10": "that one doesn't count because it is multiplied by $0$\r\nbut i am not sure how to simplify further...",
  "Solution_11": "[quote=\"funcia\"]that one doesn't count because it is multiplied by $0$\nbut i am not sure how to simplify further...[/quote]\r\n\r\nHmm.. didn't OHO explain clearly how to simplify?",
  "Solution_12": "[quote=\"Lena\"]Can you please explain to me what this means\n\n$\\sum_{k=0}^{n}3^{k}* k * \\binom{n}{k}$[/quote]\r\n\r\nLet us recall $(x+1)^{n}=\\sum_{k=0}^{n}{n \\choose k}x^{k}$.Lets modify this to get your result. \r\nWe have $\\frac{d}{dx}(x+1)^{n}=n(x+1)^{n-1}$,but also \r\n$\\frac{d}{dx}(\\sum_{k=0}^{n}{n \\choose k}x^{k})=\\sum_{k=0}^{n}{n \\choose k}kx^{k-1}$ You can see now that your sum goes from $k=1$ ( as someone mentioned).Multiply it by x ,then we have ;\r\n$xn(x+1)^{n-1}=\\sum_{k=1}^{n}{n \\choose k}kx^{k}$\r\n\r\nRegards",
  "Solution_13": "I think I can explain, how did Dr Sonnhard got his result.\r\n\r\nLet's have n lepperchauns. We want to give them hat's in 4 colours (white, red, green and blue) and also a bag of gold, to one of the lepperchauns in either red green or blue hat (we don't want to give the bag of gold to a lepperchauns in white hats). How can we do it? Well let's say we want at first to choose lepperchauns wchich we will give not-white hats, then give them hats, and then give the bag of gold to one of them. The not choosen lepperchauns will have white hats. So we do it. We choose k lepperchauns in $\\binom{n}{k}$ ways, choose the one which we will give the bag of gold to in k ways, and then give the choosen lepperchauns hats in $3^{k}$ ways. As we should assume every possible value of k, we get by this way of counting:\r\n$\\sum 3 \\cdot k\\binom{n}{k}$.\r\n\r\nThe other way to count ,i s to firstly choose the lepperchaun which will get the bag of gold, then give him a hat in three ways (he cannot heve a white one!) and then give the other lepperchauns hats in $4^{n-1}$ ways.\r\n\r\nSo we have $\\sum 3 \\cdot k\\binom{n}{k}= 3 \\cdot n \\cdot 4^{n-1}$.\r\n\r\nI hope it is clear now. And sorry for my awful english  :P",
  "Solution_14": "it is wonderfully understandable thanks"
}
{
  "Tag": [
    "function",
    "geometric series",
    "calculus",
    "calculus computations"
  ],
  "Problem": "I have to find power series for the following functions and I'm a little confused about it, so any explanations would be greatly appreciated!\r\n\r\n1) g(x) = 1/(1+x)\r\n2) g(x) = x/(1-x)\r\n3) g(x) = the square root of e^x",
  "Solution_1": "The first two do not even require any calculus.  Recall that the sum of a geometric series is given by\r\n\r\n$a+ar+ar^{2}+... = \\frac{a}{1-r}$\r\n\r\nThis gives us the power series\r\n\r\n$f(x) = \\frac{1}{1-x}= 1+x+x^{2}+...$\r\n\r\nWhich we can transform to obtain the following:\r\n\r\n[b]1.[/b]  $f(-x) = \\frac{1}{1+x}= 1-x+x^{2}-x^{3}+x^{4}+...$\r\n\r\n[b]2.[/b]  $x f(x) = x+x^{2}+x^{3}+x^{4}+...$\r\n\r\nAs for the third,\r\n\r\n$\\sqrt{ e^{x}}= \\left( e^{x}\\right)^{ \\frac{1}{2}}= e^{ \\frac{x}{2}}$\r\n\r\nRecall that $e^{x}= 1+x+\\frac{x^{2}}{2!}+\\frac{x^{3}}{3!}+...$.  Then\r\n\r\n[b]3.[/b]  $e^{ \\frac{x}{2}}= 1+\\frac{x}{2}+\\frac{x^{2}}{2^{2}2!}+\\frac{x^{3}}{2^{3}3!}+...$"
}
{
  "Tag": [
    "geometry",
    "3D geometry"
  ],
  "Problem": "i'm curious as to how one can find optimal solutions to the generalized fifteen problem. It kind of came up in myhead, and I'm really curious. I mean, how many moves will it take, how can we find them etc...?",
  "Solution_1": "I don't have any ways to compute the least number of moves it will take, but my strategy is to: \r\n1. Solve most of it layer by layer until I get down to the last two rows. \r\n2. Solve each successive \"column\" of two squares (from left to right) until I get down to the last three squares.\r\n3. Shift the last three around and around until I get it.\r\nI don't really know any other strategies other than solving it layer by layer until you get down to the last layer, then PLL. (permuting the last layer).",
  "Solution_2": "[quote=\"Benjamin Hu\"]I don't have any ways to compute the least number of moves it will take, but my strategy is to: \n1. Solve most of it layer by layer until I get down to the last two rows. \n2. Solve each successive \"column\" of two squares (from left to right) until I get down to the last three squares.\n3. Shift the last three around and around until I get it.\nI don't really know any other strategies other than solving it layer by layer until you get down to the last layer, then PLL. (permuting the last layer).[/quote]\r\n\r\nAgain, don't bring back old topics. This is the general method to solving a n by n cube. Solve the centers, pair the dedges and then solve like a 3x3. Parity will be same, I believe.",
  "Solution_3": "[quote=\"moogra\"]This is the general method to solving a n by n cube.[/quote]  Are you aware that this thread has nothing to do with Rubik's cube?",
  "Solution_4": "And as I said in another thread, there is nothing bad with bringing back old topics!"
}
{
  "Tag": [
    "algebra",
    "function",
    "domain",
    "LaTeX",
    "inequalities"
  ],
  "Problem": "y = x(x^2 - 1)(x^2 - 4)(x^2 - 9)/sqrt(x^2 - 16)\r\n\r\nPlease give the (real) domain in interval notation, please.",
  "Solution_1": "The only way for the denominator to be zero is $ x \\equal{} \\pm4$, so the domain is $ \\boxed{x\\in( \\minus{} \\infty, \\minus{} 4)\\cup( \\minus{} 4,4)\\cup(4, \\plus{} \\infty)}$.\r\n\r\nEdit: OK...then it's just all real solutions to $ x^2>16\\implies\\boxed{x\\in( \\minus{} \\infty, \\minus{} 4)\\cup(4, \\plus{} \\infty)}$.",
  "Solution_2": "I believe by \"real\" AyT meant \"giving a real number as output\", hence your interval is incorrect.",
  "Solution_3": "It's not defined when $ x^2 - 16\\leq 0\\implies x^2\\leq 16\\implies - 4\\leq x\\leq 4$.\r\n\r\nSo the domain is $ ( - \\infty, - 4)\\cup(4,\\infty)$.\r\n\r\nEdit: Beaten by math154  :mad: Shouldn't have spent so much time working out the $ \\text{\\LaTeX}$.",
  "Solution_4": "Here is the domain:\r\n\r\n[hide]{-3}U{-2}U{-1}U{0}U{1}U{2}U{3}\n\nFor example, 0/sqrt(-16) = 0\n\n(-3,0), (-2,0), (-1,0), (0,0), (1,0), (2,0), and (3,0)\nare isolated points.\n\n[/hide]",
  "Solution_5": "That's not the domain...those are the zeroes, roots, solutions, etc. of $ \\frac{x(x^2\\minus{}1)(x^2\\minus{}4)(x^2\\minus{}9)}{\\sqrt{x^2\\minus{}16}}\\equal{}0$.",
  "Solution_6": "[quote=\"math154\"]That's not the domain...those are the zeroes, roots, solutions, etc. of $ \\frac {x(x^2 \\minus{} 1)(x^2 \\minus{} 4)(x^2 \\minus{} 9)}{\\sqrt {x^2 \\minus{} 16}} \\equal{} 0$.[/quote]\r\nWrong.  Zeroes are not listed as I gave them.\r\n\r\nThe domain is the union of all of the smallest x's to the largest x's,\r\nwhich I showed in that hidden answer two (or more) posts back.",
  "Solution_7": "[quote=\"Arrange your tan\"]The domain is the union of all of the smallest x's to the largest x's,\nwhich I showed in that hidden answer two (or more) posts back.[/quote]\r\n\r\nWhere did you get this definition?\r\n\r\nAccording to [url=http://www.artofproblemsolving.com/Wiki/index.php/Domain_(function)]this[/url], [url=http://en.wikipedia.org/wiki/Domain_(mathematics)]this[/url], and [url=http://mathworld.wolfram.com/Domain.html]this[/url], [b]the domain is the set of input values on which that function is defined[/b]. :wink:\r\n\r\nI suppose that you could include $ 0,\\pm1,\\pm2,\\pm3$ in the domain, but it's sort of iffy (i.e. depending upon what you want).",
  "Solution_8": "The question (as I have interpreted it) calls for a domain where the function is defined in $ \\mathbb R$, so you wouldn't include 0, 1, etc in this case.",
  "Solution_9": "hello, for $ y(x)\\equal{}\\frac{x(x^2\\minus{}1)(x^2\\minus{}4)(x^2\\minus{}9)}{\\sqrt{x^2\\minus{}16}}$ we have the condition $ x^2\\minus{}16>0$, so the domain of definition is $ \\minus{}\\infty<x<\\minus{}4 \\cup 4<x<\\plus{}\\infty$.\r\nSonnhard.",
  "Solution_10": "This would be better if this was an inequality, I think it wouldn't cause much problem then.",
  "Solution_11": "Because all real values of x < -4 as well as all real values of x > 4 \r\nwork fine to produce real values for y, then the domain must be\r\ncorrected to:\r\n\r\n(-oo, -4) U {-3} U {-2} U {-1} U {0} U {1} U {2} U {3} U (4, oo)"
}
{
  "Tag": [
    "limit",
    "ratio",
    "calculus",
    "calculus computations"
  ],
  "Problem": "Determine which of the series diverges:\r\n\r\nthe sum of 1/square root of (n^2(n+2)) from n 1 to infinity\r\nthe sum of (((-1)^(n-1))*2^n)/n^2 from n 1 to infinity\r\nthe sum of e^-n from n 0 to infinity\r\nthe sum of (0,1)^n/n! from n 0 to infinity\r\n\r\nwhat is the limit as n goes to infinity of n/((e*n)-e)",
  "Solution_1": "1.$\\sum_{n=1}^{\\infty}\\frac{1}{\\sqrt{n^{2(n+2)}}}=\\sum_{n=1}^{\\infty}\\frac{1}{n^{n+2}}<\\sum_{n=1}^{\\infty}\\frac{1}{n^{n}}<\\sum_{n=1}^{\\infty}\\frac{1}{n^{2}}<1+\\sum_{n=2}^{\\infty}\\frac{1}{(n-1)n}=2$\r\n2.We can't find $\\lim_{n\\rightarrow\\infty}\\frac{(-1)^{n-1}2^{n}}{n^{2}}$.   So $\\sum_{n=1}^{\\infty}\\frac{(-1)^{n-1}2^{n}}{n^{2}}$ is diverges.\r\n3.$\\sum_{n=0}^{\\infty}e^{-n}=1+\\frac{1}{e}+\\frac{1}{e^{2}}+...=\\frac{1}{e-1}$\r\n4.Hence $e^{x}=\\sum_{n=0}^{\\infty}\\frac{x^{n}}{n!}\\rightarrow\\therefore\\sum_{n=0}^{\\infty}\\frac{i^{n}}{n!}=e^{i}$\r\n5.$\\lim_{n\\rightarrow\\infty}\\frac{n}{e\\times n-e}=\\frac{1}{e}\\lim_{n\\rightarrow\\infty}\\frac{n}{n-1}=\\frac{1}{e}\\left(1+\\lim_{n\\rightarrow\\infty}\\frac{1}{n-1}\\right)=\\frac{1}{e}$\r\n :D",
  "Solution_2": "1. was probably meant to be $\\sum_{n=1}^{\\infty}\\frac{1}{\\sqrt{n^{2}(n+2)}}$. It converges by comparison with $\\sum_{n=1}^{\\infty}\\frac{1}{n^{3/2}}$.\r\n\r\n3. $\\sum_{n=0}^{\\infty}e^{-n}=\\frac{1}{1-1/e}=\\frac{e}{e-1}$.\r\n\r\n4. I don't think that $(0,1)$ stands for $i$. The comma is probably a [url=http://en.wikipedia.org/wiki/Decimal_separator]decimal separator[/url]. The series converges by the Ratio Test."
}
{
  "Tag": [],
  "Problem": "x+2=5\r\n\r\n(x+2)^2=25, 9x+18=45                                                  Square, multiply by 9\r\nx^2+4x+4=25, 9x-2=25                                                  Expand, subtract 20\r\nx^2+4x+4=9x-2                                                                   Set equal\r\nx^2-5x+6=0                                                                     Subtract 9x-2\r\n(x-2)(x-3)=0                                                                         Factor\r\nx-2=0                                                                              Divide by x-3\r\nx=2                                                                                      Add 2\r\n2+2=x+2=5                                                                     Subsitiute 2 for x\r\n\r\nBy using the transitive property in the last equation, 2+2=5.\r\n[size=200]2+2=5??!!",
  "Solution_1": "[hide=\"The Big No-No\"] Dividing by $ x\\minus{}3$: From your first statement, $ x\\equal{}3$ so you'd be dividing by 0, which is a no-no.[/hide]",
  "Solution_2": "Uhmm... divided by 0..."
}
{
  "Tag": [
    "ARML"
  ],
  "Problem": "OK- apart from the stuff we HAVE to bring to ARML, who wants to play Bughouse (like 2007 NYSML) or that-game-I-cannot-mention-sometimes-involving-POOs? Set and just plain playing cards, or mmaybe Risk is also good. Also, meet-up. Who wants to go to the Chinese restaurant or the pizzeria?",
  "Solution_1": "Off topic, but I love your thing about Groups, Rings, and Fields... Haha!",
  "Solution_2": "If all goes as planned, you'll see at ARML Penn State...",
  "Solution_3": "i voted chinese",
  "Solution_4": "Are you even from Upstate NY?!",
  "Solution_5": "Oh, ha ha ha Matt.\r\nBTW, did you know Matte means dull?  :lol:",
  "Solution_6": "chez toi does not mean what you said it means in french. i believe that it means \"at your house.\"",
  "Solution_7": "I know it doesn't, the implication being that we're technically living at ARML for 2 or 3 days.",
  "Solution_8": "[quote=\"tan90=3/0\"]chez toi does not mean what you said it means in french. i believe that it means \"at your house.\"[/quote]\nHi,   \n\n\nI actually mean \"with you\".\n\n\n\n-nbute"
}
{
  "Tag": [
    "vector",
    "trigonometry"
  ],
  "Problem": "Is there any formula $ g*sin\\alpha \\equal{} a$?[/hide]",
  "Solution_1": "When a material point is moving on an inclined plane which is inclined in the angle $ \\alpha$ to the horizon, the $ mg$ force can be separated in two vectors: one's intensity is $ mg \\sin \\alpha$ and the other's $ mg \\cos \\alpha$. The first one is the component of gravity pulling the body down, and the other is the \"orthogonal\" component (the one you use when calculating friction). If you take first one and divide it with mass, you get your formula for acceleration of a material point on inclined plane.\r\n\r\n(my \"physical\" English is even worse than my \"mathematical\" English, so forgive me possible misinterpretation of some terms used here).",
  "Solution_2": "Thanks a lot  !\r\n\r\nYour english is \"udersandable\" \r\n\r\nForeigners understand each other.   :)"
}
{
  "Tag": [],
  "Problem": "If $y = 2x$ and $z = 2y$, then $x + y + z$ equals\r\n\\[ \\text{(A)}\\ x \\qquad \\text{(B)}\\ 3x \\qquad \\text{(C)}\\ 5x \\qquad \\text{(D)}\\ 7x \\qquad \\text{(E)}\\ 9x \\]",
  "Solution_1": "[hide=\"The ansewr\"]$7x$ [/hide]",
  "Solution_2": "[hide]x + 2x + 2(2x) = 7x\n(D) 7x[/hide]",
  "Solution_3": "[hide]\n$x+2x+2(2x)=\\framebox{7x}$ or $\\framebox{D}$[/hide]",
  "Solution_4": "[hide=\"to show work to the people who don't get it\"]y=2x and z=2y so substituting x in terms of y you get z = 4x so x + 2x + 4x = 7x  D[/hide]"
}
{
  "Tag": [
    "email"
  ],
  "Problem": "Three friends have a total of 6 identical pencils, and each one has at\nleast one pencil. In how many ways can this happen?",
  "Solution_1": "If guy A has 4 pencils, then guys B & C can have the following: (1,1)\r\nIf guy A has 3 pencils, then guys B & C can have the following: (1,2) (2,1)\r\nIf guy A has 2 pencils, then guys B & C can have the following: (1,3) (2,2) (3,1)\r\nIf guy A has 1 pencil, then guys B & C can have the following: (1,4) (2,3) (3,2) (4,1)\r\n\r\nadd up the solutions, and the answer would come out to be [b]10[/b]",
  "Solution_2": "Notcie we give 3possibilities for the last 3 pencils (to A, to B, or to C)\r\n\r\nso with repeated combinations forumla (i think thats what its called  :blush: )\r\n\r\nWe have $ {3\\plus{}3\\minus{}1}\\choose{3}$=$ 5\\choose{3}$=$ 10$",
  "Solution_3": "Wow mewto, I sent you a WHOLE EMAIL regarding this after states.\r\n\r\nIt's called Balls and Urns...\r\n\r\nbtw, how's my avatar.xD",
  "Solution_4": "[quote=\"math154\"]Wow mewto, I sent you a WHOLE EMAIL regarding this after states.\n\nIt's called Balls and Urns...\n\nbtw, how's my avatar.xD[/quote]\r\n\r\nlol... is that a mewto?\r\n\r\ni wasnt really good with memming my pokemon  :P",
  "Solution_5": "LOL if that is a mewtwo im a monkeys uncle.",
  "Solution_6": "LOL its like a mesed up abomasnow",
  "Solution_7": "Isn't that a character from this odd PBS show for little kids? I forget their name, it began with a \"B\" i think...",
  "Solution_8": "No, the new one looks like the Michelin Tires guy.\r\n\r\n@mewto55555- You [b]ARE[/b] a monkey's uncle.  :rotfl:  :rotfl:  :rotfl:",
  "Solution_9": "ah-ha!  its a boobah!\r\n\r\nERNIE HAS SOLVED THE MYSTERY YET AGAIN!"
}
{
  "Tag": [
    "modular arithmetic",
    "induction",
    "algebra",
    "binomial theorem",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "1) Prove that if $ g$ is a primitive root mod $ p$ ($ p$ an odd prime) and $ g^{p\\minus{}1}\\not\\equiv 1\\pmod{p^{2}}$, then $ g$ is a primitive root mod $ p^{m}$.\r\n\r\nI already know that if $ g$ is a primitive root mod $ p$ and $ g^{p\\minus{}1}\\not\\equiv 1\\pmod{p^{2}}$, then $ g^{(p\\minus{}1)p^{m\\minus{}2}}\\not\\equiv 1\\pmod{p^{m}}$, and if $ g$ is a primitive root mod $ p$, then $ g$ belongs to the exponent $ h$ mod $ p^{m}$, where $ h\\equal{}(p\\minus{}1)p^{r}$ for some $ r$.  I'm not quite sure how to combine the two, though...\r\n\r\n2) Prove that some odd numbers are primitive roots mod $ p^{m}$ for each odd prime $ p$ and each positive integer $ m$.",
  "Solution_1": "[hide=\"1\"]\nInduction of $ m$. Assume $ g$ is a generator for $ p^{m\\minus{}1}$.\n\nSince $ g^{(p\\minus{}1)p^{m\\minus{}2}}\\not\\equiv 1\\mod p^{m}$ (as you stated) but $ g^{(p\\minus{}1)p^{m\\minus{}2}}\\equiv 1\\mod p^{m\\minus{}1}$ and $ g$ is a generator mod $ p^{m\\minus{}1}$. Then $ g^{(p\\minus{}1)p^{m\\minus{}2}}\\equiv p^{m\\minus{}1}r \\plus{} 1$ for some $ r\\ne 0$. \n\nIf $ g^k\\equiv 1\\mod p^m$, then $ k$ is divisible by $ (p\\minus{}1)p^{m\\minus{}2}$, so $ k\\equal{}n(p\\minus{}1)p^{m\\minus{}2}$.\n\n$ g^{n(p\\minus{}1)p^{m\\minus{}2}}\\equal{}(g^{(p\\minus{}1)p^{m\\minus{}2}})^n\\equiv (p^{m\\minus{}1}r \\plus{} 1)^n\\equiv np^{m\\minus{}1}r \\plus{} 1$\n\n(The last step is expanding with the binomial theorem and cancelling all terms divisible by $ p^{m}$). For $ np^{m\\minus{}1}r \\plus{} 1$ to be $ 1\\mod p^m$, $ n$ must be divisible by $ p$, so the smallest exponent $ k$ is $ (p\\minus{}1)p^{m\\minus{}2}p\\equal{}(p\\minus{}1)p^{m\\minus{}1}$.\n[/hide]\n\n[hide=\"2\"]\nIf we prove that there is some generator $ g'$ such that $ g'^{p\\minus{}1}\\not\\equiv 1\\mod p^2$, which is the condition for part 1, then $ g'$ is a generator mod $ p^m$.\n\nLet $ g$ be a generator mod p, and suppose $ g^{p\\minus{}1}\\equiv 1\\mod p^2$ (Otherwise $ g'\\equal{}g$ and we're done). Then expanding\n\n$ (g\\plus{}p)^{p\\minus{}1}$ by the binomial theorem and cancelling terms divisible by $ p^2$ gives\n\n$ (p\\minus{}1)g^{p\\minus{}2}p \\plus{} g^{p\\minus{}1}\\equiv (p\\minus{}1)g^{p\\minus{}2}p \\plus{} 1\\not\\equiv 1\\mod p^2$.\n\nThen we can say $ g'\\equal{}g\\plus{}p$, and we're done.\n[/hide]",
  "Solution_2": "Thanks a lot.  However, what is a generator?  I have seen the term used but no one actually defined it.  However, I've noticed that when people use that term, they tend to not use \"primitive root.\"  So are they the same thing?",
  "Solution_3": "In this context, yes.  \"Generator\" is more general and applies to any cyclic group."
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "Let $n \\geq 2$ and $x_1, \\dots, x_n>0$ such that $x_1x_2....x_n=1$. Prove that \\[ \\sum_{i=1}^n\\left(\\frac{6}{x_i^2+4x_i+1}\\right)^2\\geq 2.  \\]",
  "Solution_1": "[quote=\"silouan\"]\n$\\sum_{i=1}^n\\frac{6^2}{(x_i+4x_i+1)^2}$[/quote]\r\n?",
  "Solution_2": "Now it is better. But what you didn't understand ?",
  "Solution_3": "I guess there is a typo in the denominator.",
  "Solution_4": "[quote=\"Arne\"]I guess there is a typo in the denominator.[/quote]\r\nyes it was a typo at the denominator, sorry. But now I corrected and the problem is correct",
  "Solution_5": "The inequality is not true for $n>36$.",
  "Solution_6": "Are you sure . Could you give me a counerexample ?",
  "Solution_7": "[quote=\"silouan\"]Are you sure . Could you give me a counerexample ?[/quote]\r\nLet $x_1\\rightarrow 0$ and $x_2=...=x_n \\rightarrow \\infty$. The inequality becomes $36 \\geq n$.",
  "Solution_8": "But we have $x_1x_2x_3...x_n=1$ . do you forget it ?",
  "Solution_9": "Vasc is correct. He is just taking the limit when $x_2, x_3, \\cdots , x_n$ tend to infinity and $x_1=\\frac{1}{x_2x_3\\cdots x_n}$.",
  "Solution_10": "Because the first is not true, could you prove the following\r\n[quote=\"silouan\"]Let $n \\geq 2$ and $x_1, \\dots, x_n>0$ such that $x_1x_2....x_n=1$. Prove that \\[ \\sum_{i=1}^n\\left(\\frac{6}{x_i^2+4x_i+1}\\right)^2\\geq 2.  \\][/quote]",
  "Solution_11": "Noone ?I think it is an interesting problem with a new idea.For a hint I will give the above. You can work as a harazi's problem. :lol:",
  "Solution_12": "Unsolved till now ?Why ?I think now it is correct .",
  "Solution_13": "now I think we should just prove it when:$n=2$(not hard)\r\nfor $n>2$ we can WLOG $x_1 x_2 \\leq 1$\r\nthen it is true.",
  "Solution_14": "yes zhaobin nice job. Please post your solution,thanks",
  "Solution_15": "my proof is ugly.\r\njust prove $18(x_1^2+4x_1+1 )^2+18(x_2^2+4x_i+1)^2 \\geq (x_1^2+4x_1+1 )^2 (x_2^2+4x_i+1)^2$\r\njust expland."
}
{
  "Tag": [
    "modular arithmetic",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Let $a,b,n$ be positive integers such that $2^n - 1 =ab$. Let $k \\in \\mathbb N$ such that $ab+a-b-1 \\equiv 0 \\pmod {2^k}$ and $ab+a-b-1 \\neq 0 \\pmod {2^{k+1}}$. Prove that $k$ is even.",
  "Solution_1": "$2^{k}$ is the highest power of two that divides $ab + a - b - 1$.\r\nWe have that $a = \\frac{2^{n} - 1}{b}$ $\\Longrightarrow$ $\\frac{2^{n}b - 2b + 2^{n} - 1 - b^2}{b} = \\frac{(2^{n} - 1 - b)(b + 1)}{b} \\equiv 0 \\bmod 2^{k}$. Let $2^{h}$ be the highest power of two that divides $b + 1$. It is also the highest power of two that divides $2^{n} - 1 - b$. So $h < n$. And the highest power of two that divides $ab + a - b - 1$ is $2^{2h}$ cause $2^{h}$ comes from both factors. Then $2^{k} = 2^{2h}$.",
  "Solution_2": "If $ a \\equal{} 1$ then $ ab \\plus{} a \\minus{} b \\minus{} 1 \\equal{} 0$ and there does not exist such a $ k$.\r\nLet $ a \\minus{} 1 \\equal{} x$, $ b \\plus{} 1 \\equal{} y$, so $ x \\ge 1$, $ y \\ge 2$. Then $ k$ is the highest power of $ 2$ dividing $ xy$, and we have\r\n$ 2^n \\equal{} xy \\minus{} x \\plus{} y \\equal{} x(y \\minus{} 1) \\plus{} y \\ge x \\plus{} y$.\r\nLet $ 2^m$ be the highest power of $ 2$ dividing $ x$. Then since $ 2^n \\ge x \\plus{} y$, $ m < n$. Let $ x \\equal{} 2^mr$, where $ r$ is odd. Substituting,\r\n$ 2^n \\equal{} 2^mry \\minus{} 2^mr \\plus{} y$. Thus, $ 2^m$ divides $ y$, so $ y \\equal{} 2^mj$. Substituting,\r\n$ 2^{n \\minus{} m} \\equal{} 2^mrj \\minus{} r \\plus{} j$\r\n$ 2^{n \\minus{} m} \\plus{} r \\equal{} j(2^m \\plus{} 1)$\r\nSo the left side is odd since $ r$ is odd, but it also divisible by $ j$, so $ j$ is odd. So $ x$ and $ y$ both have $ m$ factors of $ 2$, so $ k \\equal{} 2m$ is even.",
  "Solution_3": "easy to prove that $a,b$ are odd;and $ab+a-b-1=2^n+a-b-2=2^n+\\frac{2^n-1}{b}-b-2=2^n+\\frac{2^n}{b}-(b+1)^2$\nif $2^n|a-b-2$,thern it's easy to prove $a=b+2$hence $2^n=(b+1)^2,2|n$\notherwice,$2^k||(b+1)^2$\nhence $2|k$",
  "Solution_4": "[hide=Solution][quote=Balkan MO 2001 P1]Let $a,b,n$ be positive integers such that $2^n - 1 =ab$. Let $k \\in \\mathbb N$ such that $ab+a-b-1 \\equiv 0 \\pmod {2^k}$ and $ab+a-b-1 \\neq 0 \\pmod {2^{k+1}}$. Prove that $k$ is even.[/quote]\n[b][color=#000]Solution:[/color][/b] We know, $v_2(ab+a-b-1)=k$. Clearly, $a,b$ are odd. Note, $ab+a-b-1=(a-1)(b+1)$, Let\n$$\\begin{cases}v_2(a-1)=x \\\\ v_2(b+1)=y \\\\ x+y=k \\end{cases} \\implies \\begin{cases} a-1=k_12^x \\implies a=k_12^x+1 \\\\ b+1=k_22^y \\implies b=k_22^y-1 \\end{cases}$$\nSo,\n$$2^n-~1=ab \\implies 2^n=1~+~(k_12^x+~1)(k_22^y-~1) = k_1k_22^{x+y}+~k_22^y-~k_12^x $$ $$\\implies 2^n+k_12^x=k_22^y(k_12^x+1)$$\nNow, maximum power of $2$ dividing $\\text{ RHS }$ is $y$. If $x \\geq n$ $\\implies$ $y=n$, but then, we would have: $1+k_12^{x-n}=k_2(1+k_12^x)$, which is a clear contradiction! Hence, $n \\geq x$ $\\implies$ $x=y$ $\\implies$ $2 \\mid k$ $\\qquad \\blacksquare$\n[/hide]",
  "Solution_5": "You can easily fing that $ v_2(a-1)$$=$$v_2(b+1)$",
  "Solution_6": "$v_2(ab+a-b-1)=v_2(a-1)+v_2(b+1)$ \n$2^n-1-b=b(a-1)$\nTherefore $a-1\\mid 2^n+1-b$ and $v_2(b(a-1))=v_2(a-1)$ since $b$ is odd, and $v_2(2^n+1-b)=min\\{n, v_2(b+1)\\}$ therefore $v_2(a-1)=min\\{n, v_2(b+1)\\}=v_2(b+1)$  then $v_2(a-1)+v_2(b+1=2v_2(a-1)$ and this is even"
}
{
  "Tag": [
    "MATHCOUNTS"
  ],
  "Problem": "What is the value of:\r\n\r\n1998+2(1+2+3+4+5+......1997)?",
  "Solution_1": "[hide]3992004[/hide]\r\n\r\nThis should be in the MathCounts section not here",
  "Solution_2": "All right, but how do you move it?",
  "Solution_3": "[quote=\"hwenterprise\"]All right, but how do you move it?[/quote]\r\n\r\nYou cant, but the mods might. Im not saying this problem is too hard for the Basics section, it just looks more like a mathcounts problem",
  "Solution_4": "[hide]\n3992004\n1998*1997+1998=3992004\n[/hide]",
  "Solution_5": "DarkKnight, thanks for showing your works\r\n\r\nthe concept he used is adding backwards....\r\n\r\n1 + 2 + 3\r\n3 + 2 + 1\r\n-----------\r\n4 + 4 + 4\r\n\r\n(4+4+4)/ 2\r\n\r\nlike this",
  "Solution_6": "okay, I get the answer, but not the formula. What's the formula?",
  "Solution_7": "[quote=\"shinwoo\"]DarkKnight, thanks for showing your works\n\nthe concept he used is adding backwards....\n\n1 + 2 + 3\n3 + 2 + 1\n-----------\n4 + 4 + 4\n\n(4+4+4)/ 2\n\nlike this[/quote]\r\n\r\nI think he used the sum of consecutive positive numbers formula- \r\nn(n+1)/2\r\n\r\nUnless thats what you mean by adding backwards",
  "Solution_8": "[hide]I found the middle number of 1 and 1997 (1998/2)  then multiplied that by 1997.  That finds the average of all numbers in the parenthsies, then makes it all of the numbers.  Multiply that by 2 and add 1998 to get 3992004[/hide]",
  "Solution_9": "It is a arithmetic sequence.",
  "Solution_10": "[hide]3992004 [/hide][/hide]",
  "Solution_11": "I'll move this",
  "Solution_12": "[quote=\"GoBraves\"][quote=\"shinwoo\"]DarkKnight, thanks for showing your works\n\nthe concept he used is adding backwards....\n\n1 + 2 + 3\n3 + 2 + 1\n-----------\n4 + 4 + 4\n\n(4+4+4)/ 2\n\nlike this[/quote]\n\nI think he used the sum of consecutive positive numbers formula- \nn(n+1)/2\n\nUnless thats what you mean by adding backwards[/quote]\r\n\r\nYes, I did use the n(n+1)/2 formula but it's like the same thing.",
  "Solution_13": "i always wondered how u would do this.  thank for showing me.",
  "Solution_14": "1998 + 2(1997*1998)/2\r\n1*1998 + 1997*1998\r\n1998<sup>2</sup> \r\n3992004",
  "Solution_15": "Thank you to all of you who showed your work...I appreciate it. It was a great help.",
  "Solution_16": "Then what is (1995-1993)(1991-1989)...(3-1)\r\nin terms of 2^x?",
  "Solution_17": "[hide]3992004. Simply apply the formula   1/2n(n+1)[/hide]"
}
{
  "Tag": [
    "function",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "$\\sum_{i=1}^{n}\\ e^{^}(x^{i})$\r\n\r\nAnyone know a closed form?",
  "Solution_1": "If you are looking for an elementary function $F(x,n)$ of two variables such that this sum is equal to $F(x,n)$ for all real $x$ and integer $n$, I believe there is none. If you insist, we can even try to prove it but that may take a lot of time and considerable effort unless someone has a bright idea of how it can be done right away."
}
{
  "Tag": [
    "real analysis",
    "real analysis solved"
  ],
  "Problem": "let (x<sub>n</sub>) be a sequence of real numbers such that x<sub>0</sub> and x<sub>1</sub>>0 and x<sub>n+2</sub> = (x<sub>n+1</sub>+2)/(x<sub>n</sub>+2) for all n \\geq 0. Proove that x<sub>n</sub> is convergent\r\n[i]source i think ONM ROMANIA 1997[/i]",
  "Solution_1": "please edit your post and complete the problem's statement.",
  "Solution_2": "this is all what i have Valentin. The problem had smth else to?",
  "Solution_3": "well, if you look here, you can see that it can't be from ONM 2997\r\n\r\nhttp://www.mathlinks.ro/viewtopic.php?t=2605",
  "Solution_4": "Can anyone tell me what is so important if the problem was  in ONM 1997 or not? Try to solve it!",
  "Solution_5": "I think I've found a solution, but I'm not sure:\r\n    First of all, we have 2x_n+2<=x_n+1+2. Thus, 2^(k+2)x_k+2-2^k+1*x_k+1<=2^(k+2). Sum upp these and you will find that there exists k such that for all n>k we have x_n<3. Thus, for all n>k we have x_(n+2)>(2+x_(n+1))/5>2/5. So, for n>k+2 we have x_n>2/5. Then, for n>k+2 we have |x_(n+2)-1|=|x_(n+1)-x_n|/(2+x_(n+2)|<=(|x_(n+1)-1|+|x_n-1|)/(2+2/5). Take a_n=x_n-1. We have a_n+2<(a_n+1+a_n)/(2+2/5) for all n>k+2. Thus, since a_n is bounded and a_(n+2)<(a_n+a_(n+1))/2, a_n must be convergent. Let x be its limit. Since a_(n+2)<(a_n+a_(n+1))/(2+2/5), we must have x<=0. Since x>=0 ( clear), we must have x=0 and thus x_n tends to 1."
}
{
  "Tag": [
    "function",
    "floor function",
    "number theory",
    "prime numbers",
    "number theory theorems"
  ],
  "Problem": "$ \\textbf{Conjecture: }$ \r\n\r\nIf $ p, q, r$ are consecutive prime numbers such that $ p > q > r$ and if $ r$ is different from $ 2,3,5,7,11,13,19,23,31,47,83,89,113,199,1327$ then $ \\left\\lfloor\\frac{pq}{r}\\right\\rfloor \\equal{} p\\plus{}q\\minus{}r$.",
  "Solution_1": "Let $ p\\equal{}r\\plus{}a,q\\equal{}r\\plus{}b,a>b>0$, then $ [\\frac{pq}{r}]\\equal{}p\\plus{}q\\minus{}r$ equavalent to $ ab<r$. \r\nIf exist $ 0<c<1$, suth that $ p_{n\\plus{}1}<p_{n}\\plus{}c\\sqrt{p_{n}}$, then it followed for $ p>N(c)$."
}
{
  "Tag": [
    "complex analysis",
    "integration",
    "calculus"
  ],
  "Problem": "If ${ A=\\{z}\\in \\mathbb{C}: \\Re(z)>0\\}$, and ${ f: A\\rightarrow\\mathbb{C}}$, with $ f(z)=\\int_0^{+\\infty} \\!{e^{-zt^2}}\\, dt$, prove that $ f$ is holomorphic.",
  "Solution_1": "Well, $ f'(z)\\equal{}\\int_0^\\infty \\minus{}t^2e^{\\minus{}zt^2}\\,dt$. That's an absolutely convergent integral, and everything works.\r\n\r\nA variation on a standard theorem for use here:\r\nSuppose $ f(z)\\equal{}\\int_I g(z,t)\\,dt$, where $ g$ is differentiable in $ z$. If in addition $ |\\frac{\\partial g}{\\partial z}(z,t)|\\le h(t)$ for some integrable $ h$ on $ I$, $ f$ is differentiable and $ f'(z)\\equal{}\\int_I \\frac{\\partial g}{\\partial z}(z,t)\\,dt.$\r\n\r\nWe can't cover the whole half-plane at once with this, but differentiability is local. This method gives differentiability on each half-plane $ \\text{Re}(z)>c$ for each $ c>0$, and taking the union gives the desired result."
}
{
  "Tag": [
    "calculus",
    "derivative",
    "trigonometry",
    "function",
    "calculus computations"
  ],
  "Problem": "At what points on the curve y - sin x + cos x, 0 </= x </= 2pi, is the tangent line horizontal. \r\n\r\nI'm just wondering, for this question, would you take the derivative first and then get the critical numbers. And if so, how do you get critical numbers in these type of trig functions?[/code]",
  "Solution_1": "I'll assume you meant the curve $y = \\sin x+\\cos x,\\, \\, 0 \\leq x \\leq 2\\pi$ and yes, do what you imagined to do, namely take the derivative w.r.t. x and set equal to zero, i.e. write $y^\\prime = \\cos x-\\sin x =0$ so solve $\\cos x =\\sin x\\Rightarrow \\tan x =1$ or $x=\\tan^{-1}(1)= \\frac{\\pi}{4}+k\\pi,\\, \\, k=0,\\pm 1, \\pm 2,\\ldots$, but we wanted $0 \\leq x \\leq 2\\pi$ so we need only the solutions corresponding to $k=0,1$ i.e. $x=\\frac{\\pi}{4},\\frac{5\\pi}{4}$."
}
{
  "Tag": [],
  "Problem": "Determine el minimo valor de $(a+1)(b+1)(c+1)(d+1),$ dados $a,b,c,d \\geq 0$ y\r\n\r\n $\\frac{1}{a+1} + \\frac{1}{b+1} + \\frac{1}{c+1} + \\frac{1}{d+1} = 1.$\r\n\r\n(Sugerencia)\r\n\r\n[hide=\"Hint\"] Un inteligente cambio de variables y usar MA-MG[/hide]\r\n\r\n\r\nCarlos Bravo :)\r\nLima -PERU",
  "Solution_1": "[hide]Sup. que $(a+1)(b+1)(c+1)(d+1) = x$\nExpandiendo la suma de las fracciones tenemos que:\n$\\sum_{sym} {(a+1)(b+1)(c+1)} = x$\npor $MA - MG$ se tiene que \n$x\\geq  4 \\sqrt [4] {x^3}$ entonces se tiene que $x\\geq 4^4$ elevando todo a la $4$ y dividendo entre $x^3$\nLa igualdad se da cuando $a=b=c=d=3$[/hide]"
}
{
  "Tag": [
    "limit",
    "trigonometry",
    "LaTeX"
  ],
  "Problem": "in ham yek hadde rahat vali ziba:\r\n$\\lim_{x \\to \\frac{\\pi}{3}} \\frac{\\sqrt{\\sin{(x)}}+\\sqrt{\\sin{(2x)}}}{\\sin{(x)}+\\sqrt{3}}$\r\n\r\n[color=red]Latex edited by Mod.[/color]",
  "Solution_1": "here is the answer",
  "Solution_2": "[quote=\"ashwinrk_jain\"]here is the answer[/quote]\r\ntanx.(vali age mineveshti kheili behtar bood.) :D"
}
{
  "Tag": [
    "Diophantine equation",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "prove that:\r\n$a^2+b^2+c^2+d^2=abcd$ has infty solution in $\\mathbb{N}$",
  "Solution_1": "I claim that there are already infinity many solutions with $c=d=2$:\r\n\r\nto prove:\r\n$a^2+b^2+8=4ab$ has infinity many solutions in integers.\r\nBut this can be written as $(a-2b)^2 - 3b^2=-4$ or after substitution $e^2-3b^2=-4$ which is a generalised Pell equation with trivial solution $e=4, b=2$, so it has infinity many, which was to be proven."
}
{
  "Tag": [],
  "Problem": "\u0395\u03b1\u03bd $ a,b,c \\in \\mathbb R^\\plus{}$ \u03bc\u03b5 $ a\\plus{}b\\plus{}c\\equal{}1$.N\u03b4\u03bf:\r\n\r\n$ \\frac{1}{1\\plus{}a}\\plus{}\\frac{1}{1\\plus{}b}\\plus{}\\frac{1}{1\\plus{}c}\\plus{}6(ab\\plus{}bc\\plus{}ca) \\le \\frac{17}{4}$.\r\n\r\n[hide=\"\u03b3\u03b9\u03b1 \u03c4\u03bf\u03c5\u03c2 \u03b3\u03bd\u03c9\u03c1\u03af\u03b6\u03bf\u03bd\u03c4\u03b5\u03c2\"]\n\u03a1\u03b5 \u03c0\u03b1\u03b9\u03b4\u03b9\u03ac \u03bf Pham Kim Hung \u03c3\u03c4\u03b1 mathlinks \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf hungkhtn????[/hide]",
  "Solution_1": "\u03a3\u03c6\u03b9\u03c7\u03c4\u03ae (\u03c0\u03b1\u03c1\u03b1\u03c4\u03b7\u03c1\u03b5\u03af\u03c3\u03c4\u03b5 \u03cc\u03c4\u03b9 \u03c0\u03b9\u03ac\u03bd\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf \u03bc\u03ad\u03b3\u03b9\u03c3\u03c4\u03bf \u03c3\u03c4\u03bf \u03ba\u03ad\u03bd\u03c4\u03c1\u03bf \u03c4\u03bf\u03c5 \u03c4\u03c1\u03b9\u03b3\u03ce\u03bd\u03bf\u03c5), \u03b1\u03bb\u03bb\u03ac \u03bc\u03ac\u03bb\u03bb\u03bf\u03bd \u03b5\u03cd\u03ba\u03bf\u03bb\u03b7 (\u03b2\u03b3\u03b1\u03af\u03bd\u03b5\u03b9 \u03bc\u03b5 brute force \u03b2\u03b1\u03c3\u03b9\u03ba\u03ac):\r\n\r\n[hide]\u0393\u03b9\u03b1 \u03bd\u03b1 \u03bc\u03b7 \u03b2\u03ac\u03bb\u03c9 Lagrange, \u03b1\u03c1\u03ba\u03b5\u03af \u03bd\u03b1 \u03b2\u03c1\u03bf\u03cd\u03bc\u03b5 \u03c4\u03bf \u03bc\u03ad\u03b3\u03b9\u03c3\u03c4\u03bf \u03c4\u03b7\u03c2 \u03c3\u03c5\u03bd\u03ac\u03c1\u03c4\u03b7\u03c3\u03b7\u03c2:\n$ f(x,y) \\equal{} \\frac {1}{1 \\plus{} x} \\plus{} \\frac {1}{1 \\plus{} y} \\plus{} \\frac {1}{2 \\minus{} x \\minus{} y} \\plus{} 6\\left(x \\plus{} y \\plus{} xy \\minus{} (x \\plus{} y)^2\\right)$\n\u03bc\u03b5 $ 0\\leq x,y \\leq 1, x \\plus{} y\\leq 1$.\n\n\u039c\u03b5 \u03bc\u03af\u03b1 \u03c0\u03b1\u03c1\u03b1\u03b3\u03ce\u03b3\u03b9\u03c3\u03b7 \u03ba\u03b1\u03b9 \u03bc\u03af\u03b1 \u03b1\u03c6\u03b1\u03af\u03c1\u03b5\u03c3\u03b7, \u03c0\u03c1\u03bf\u03ba\u03cd\u03c0\u03c4\u03b5\u03b9 \u03cc\u03c4\u03b9 \u03c4\u03b1 \u03c3\u03c4\u03ac\u03c3\u03b9\u03bc\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u03c0\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c4\u03bf \u03b5\u03c3\u03c9\u03c4\u03b5\u03c1\u03b9\u03ba\u03cc \u03c4\u03bf\u03c5 \u03c0.\u03bf\u03c1\u03b9\u03c3\u03bc\u03bf\u03cd \u03b8\u03b1 \u03b9\u03ba\u03b1\u03bd\u03bf\u03c0\u03bf\u03b9\u03bf\u03cd\u03bd \u03c4\u03b7 \u03c3\u03c7\u03ad\u03c3\u03b7: $ \\frac {1}{(1 \\plus{} y)^2} \\minus{} \\frac {1}{(1 \\plus{} x)^2} \\equal{} 6(x \\minus{} y)$ \u03b7 \u03bf\u03c0\u03bf\u03af\u03b1 \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03bc\u03cc\u03bd\u03bf \u03cc\u03c4\u03b1\u03bd $ x \\equal{} y$ (\u03b3\u03b9\u03b1\u03c4\u03af?). \u0388\u03c4\u03c3\u03b9, \u03b1\u03c1\u03ba\u03b5\u03af \u03bd\u03b1 \u03ba\u03bf\u03b9\u03c4\u03ac\u03be\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b7 \u03c3\u03c5\u03bd\u03ac\u03c1\u03c4\u03b7\u03c3\u03b7 $ g(x) \\equal{} f(x,x)$ \u03b7 \u03bf\u03c0\u03bf\u03af\u03b1 \u03ba\u03b1\u03b9 \u03ad\u03c7\u03b5\u03b9 \u03bc\u03bf\u03bd\u03b1\u03b4\u03b9\u03ba\u03cc \u03c3\u03c4\u03ac\u03c3\u03b9\u03bc\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03c3\u03c4\u03bf $ x \\equal{} 1/3$.\n\n\u0391\u03c0\u03cc \u03ba\u03c1\u03b9\u03c4\u03ae\u03c1\u03b9\u03bf \u03b2' \u03c0\u03b1\u03c1\u03b1\u03b3\u03ce\u03b3\u03bf\u03c5 \u03b3\u03b9\u03b1 $ x,y \\equal{} \\frac {1}{3}$, \u03c0\u03c1\u03bf\u03ba\u03cd\u03c0\u03c4\u03b5\u03b9 \u03cc\u03c4\u03b9 \u03c4\u03bf \u03b5\u03bd \u03bb\u03cc\u03b3\u03c9 \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03b5\u03af\u03bd\u03b1\u03b9 \u03bc\u03ad\u03b3\u03b9\u03c3\u03c4\u03bf (\u03ba\u03b1\u03b9 \u03b4\u03af\u03bd\u03b5\u03b9 \u03c4\u03b7 \u03b6\u03b7\u03c4\u03bf\u03cd\u03bc\u03b5\u03bd\u03b7 \u03c4\u03b9\u03bc\u03ae $ \\frac {17}{4}$). \u03a4\u03bf \u03bc\u03cc\u03bd\u03bf \u03c0\u03bf\u03c5 \u03b1\u03c0\u03bf\u03bc\u03ad\u03bd\u03b5\u03b9 \u03bd\u03b1 \u03b5\u03bb\u03ad\u03b3\u03be\u03bf\u03c5\u03bc\u03b5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03c3\u03c4\u03bf \u03c3\u03cd\u03bd\u03bf\u03c1\u03bf \u03c4\u03bf\u03c5 \u03c0.\u03bf\u03c1\u03b9\u03c3\u03bc\u03bf\u03cd \u03b4\u03b5\u03bd \u03b5\u03bc\u03c6\u03b1\u03bd\u03af\u03b6\u03b5\u03c4\u03b1\u03b9 \u03bc\u03b5\u03b3\u03b1\u03bb\u03cd\u03c4\u03b5\u03c1\u03b7 \u03c4\u03b9\u03bc\u03ae, \u03c0\u03c1\u03ac\u03b3\u03bc\u03b1 \u03c4\u03bf \u03bf\u03c0\u03bf\u03af\u03bf \u03b5\u03af\u03bd\u03b1\u03b9 \u03b5\u03cd\u03ba\u03bf\u03bb\u03bf.[/hide]\r\n\r\nCheerio,\r\n\r\nDurandal 1707"
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "probability"
  ],
  "Problem": "Mr. Go has a weighted hexahedraon (cube) die. The number 1 is the least likely number to come up. The number 2 comes up twice as often as the number 1, the number 3 comes up three times as often as the number 1, and so on, up to the number 6 coming up 6 times as often as the number 1.\r\n\r\n\r\nHow often does the number 3 come up? Express your answer as a common fraction.",
  "Solution_1": "[hide]haha...\n\nlet the prob that 1 comes up be x.\n\nx+2x+...+6x=1\n\nx=1/21\n\n3x=3/21=1/7[/hide]",
  "Solution_2": "you get $\\frac{3}{1+2+3+4+5+6}$=probability of 3 coming up.  $\\frac{3}{21}=\\frac{1}{7}$ :idea:  our answer",
  "Solution_3": "Boy would Monopoly be fun! :D \r\n\r\nAnyway, the probability of getting a 1 is $\\frac{1}{x}$.  The probability of getting a 2 is $\\frac{2}{x}$.  That means that the sum of the probabilities of all possible outcomes is $\\frac{21}{x}=1$.  That means that x=21 and the answer is $\\frac{3}{21}=\\frac{1}{7}$.",
  "Solution_4": "[hide]roll a one, then roll two 2s, three 3s... how many dice rolls will u get?\n\n1 + 2 + 3 + 4 + 5 + 6 = 21.\n\n3 comes up thrice, so 3/21 = 1/7[/hide]",
  "Solution_5": "[hide]1/7[/hide]"
}
{
  "Tag": [],
  "Problem": "Find n to satisfy $3^{n}+63 =k^{2}$ (k $\\in N$ )",
  "Solution_1": "[hide]\n$n=0$\n$3^{0}+63 =k^{2}\\rightarrow 64 = k^{2}$\n[/hide]",
  "Solution_2": "[hide=\"solution\"]For n=1mod2, that expression might just be a square.\n\nFor n=0mod2, it isn't a square since it's 2 mod 4.\n\nThe only n that worked(that I found) is 1.[/hide]",
  "Solution_3": "[hide]Works for $n = 4 \\rightarrow 3^{4}+63 = 12^{2}$[/hide]",
  "Solution_4": "here is my solution \r\nif n is odd , we easily see that $LHS \\equiv 2(mod 4)$\r\nso n is even , take $n=2m$\r\nwe get $63=(k-3^{m})(k+3^{m})$\r\nnotice $k-3^{m}\\equiv k+3^{m}\\equiv 0 (mod 3)$\r\nso we obtain $k-3^{m}=3$ and $k+3^{m}=21$ , hence $m=2 ;n=4; k=12$",
  "Solution_5": "HTA:\r\n\r\nYour second to last line does not hold if $m=0$\r\nYou lost $n=0$ as a second solution",
  "Solution_6": "oh , that's mistake , i didnt notice :D , thanks you",
  "Solution_7": "[hide]If n is odd, then $(-1)^{odd}+3\\equiv 2\\equiv k^{2}$ is untrue, no solutions.\nIf n is even, then $k=3k'$, and $3^{n-2}+7=k'^{2}\\implies 7=k'^{2}-3^{n-2}\\implies 7=(k'+3^{(n-2)/2})(k'-3^{(n-2)/2})$. Only solutions is $(n-2)/2=1, k'=4 \\implies n=4, k=12$.\n\nEDIT:  :oops: [quote=\"chickendude\"]Your second to last line does not hold if $m=0$\nYou lost $n=0$ as a second solution[/quote]\n\n$3^{n-2}+7=k'^{2}$ is not true iff n=0, so testing that seperately gives another solution, n=0, k=8.\n[/hide]"
}
{
  "Tag": [
    "function",
    "limit"
  ],
  "Problem": "Can someone please give me instructions on how to find limits of basic functions like (x^2-4)/(x+5) quickly(10 secs or so)?",
  "Solution_1": "(assuming you meant limit as x goes to infinity, otherwise theres not always the simplest rule:)\r\nDivide top and bottom by the highest power of x you can find in that expression. Then when x goes to infinity, everything you don't care about cancels out to 0.",
  "Solution_2": "For this example, divide both the bottom and the top of (x <sup>2</sup> -4)/(x+5) by x <sup>2</sup> . Thus, you have (1-(4/x <sup>2</sup> ))/((1/x)+(5/x <sup>2</sup> ). The bottom approaches 0 as x goes to infinity, so anything over 0 is infinity, so the limit of this function as x goes to infinity is infinity.",
  "Solution_3": "\"Highest powers\" is too narrow a rule. To deal with limits quickly, what you really need to know is what grows (or shrinks) much faster than what else.\r\n\r\nIn Chinaboy's original problem, $\\lim_{x\\to\\infty}\\frac{x^2-4}{x+5}$, we see that $x^2$ gets much larger than 4 and $x$ gets much larger than 5, so we only need to look at $x^2$ and $x$. But since $x^2$ grows much faster than $x$, the limit is $\\infty.$\r\n\r\nBut powers of $x$ and going to infinity aren't the only story. What about the following limits? You can do all of them quickly, if you have a clear understanding of what's bigger than what else.\r\n\r\n1. $\\lim_{x\\to\\infty}x^4e^{-x}.$\r\n\r\n2. $\\lim_{n\\to\\infty}\\frac{n^4-4n^2-2^n}{n^3+n+2^n}.$\r\n\r\n3. $\\lim_{x\\to0^+}\\frac1{x^2}-\\frac1x.$\r\n\r\n4. $\\lim_{n\\to\\infty}\\frac{\\log(n^2)}{\\sqrt{n}}.$"
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "a,b,c are the lenghts of triangle dimensions,\r\nprove that:\r\n                    3(ab+ac+bc)=<(a+b+c)\u00b2<4(ab+ac+bc)",
  "Solution_1": "[quote] $ a,b,c$ are sides of a triangle. Prove that $ 3(ab \\plus{} bc \\plus{} ca) \\leq (a \\plus{} b \\plus{} c)^{2} < 4(ab \\plus{} bc \\plus{} ca)$[/quote]\r\n\r\n[hide=\" Solution\"]\n$ 3(ab \\plus{} bc \\plus{} ca) \\leq (a \\plus{} b \\plus{} c)^{2} \\iff a^{2} \\plus{} b^{2} \\plus{} c^{2} \\geq ab \\plus{} bc \\plus{} ca$ which is true by Rearrangement Inequality.\n\n$ (a \\plus{} b \\plus{} c)^{2} < 4(ab \\plus{} bc \\plus{} ca) \\iff (a \\plus{} b \\minus{} c)(b \\plus{} c \\minus{} a) \\plus{} (a \\plus{} b \\minus{} c)(c \\plus{} a \\minus{} b) \\plus{} (b \\plus{} c \\minus{} a)(c \\plus{} a \\minus{} b) > 0$ which is true since $ a,b,c$ are sides of a triangle.  [/hide]",
  "Solution_2": "Other sollution :\r\n$ (a\\plus{}b\\plus{}c)^2\\,\\geq\\,3(ab\\plus{}ac\\plus{}bc)$\r\n\r\n$ \\iff \\, (a\\plus{}b\\plus{}c)^2\\minus{}3(ab\\plus{}ac\\plus{}bc)\\,\\geq\\,0$\r\n$ \\iff \\, \\frac{1}{2}[(a\\minus{}b)^2\\plus{}(b\\minus{}c)^2\\plus{}(c\\minus{}a)^2]\\,\\geq\\,0$\r\nwich is true.\r\n\r\nand we have :\r\n$ |a\\minus{}b|\\,\\prec\\,c$ and  $ |b\\minus{}c|\\,\\prec\\,a$   and   $ |c\\minus{}a|\\,\\prec\\,b$\r\n\r\nso:\r\n\r\n$ 4(ab\\plus{}ac\\plus{}bc)\\minus{}(a\\plus{}b\\plus{}c)^2\\equal{}c^2\\minus{}(a\\minus{}b)^2\\plus{}a^2\\minus{}(b\\minus{}c)^2\\plus{}b^2\\minus{}(c\\minus{}a)^2$\r\n\r\nwish is also true.",
  "Solution_3": "good.easy inequality :lol:"
}
{
  "Tag": [
    "ratio",
    "geometry",
    "trigonometry"
  ],
  "Problem": "In right triangle $ \\triangle ACE$, we have $ AC \\equal{} 12$, $ CE \\equal{} 16$, and $ EA \\equal{} 20$. Points $ B$, $ D$, and $ F$ are located on $ \\overline{AC}$, $ \\overline{CE}$, and $ \\overline{EA}$, respectively, so that $ AB \\equal{} 3$, $ CD \\equal{} 4$, and $ EF \\equal{} 5$. What is the ratio of the area of $ \\triangle DBF$ to that of $ \\triangle ACE$?\n[asy]\nsize(200);defaultpen(linewidth(.8pt)+fontsize(8pt));\ndotfactor=3;\n\npair C = (0,0);\npair E = (16,0);\npair A = (0,12);\npair F = waypoint(E--A,0.25);\npair B = waypoint(A--C,0.25);\npair D = waypoint(C--E,0.25);\n\ndot(A);dot(B);dot(C);dot(D);dot(E);dot(F);\n\nlabel(\"$A$\",A,NW);label(\"$B$\",B,W);label(\"$C$\",C,SW);label(\"$D$\",D,S);label(\"$E$\",E,SE);label(\"$F$\",F,NE);\n\nlabel(\"$3$\",midpoint(A--B),W);\nlabel(\"$9$\",midpoint(B--C),W);\nlabel(\"$4$\",midpoint(C--D),S);\nlabel(\"$12$\",midpoint(D--E),S);\nlabel(\"$5$\",midpoint(E--F),NE);\nlabel(\"$15$\",midpoint(F--A),NE);\n\ndraw(A--C--E--cycle);\ndraw(B--F--D--cycle);[/asy]$ \\textbf{(A)}\\ \\frac {1}{4}\\qquad \\textbf{(B)}\\ \\frac {9}{25}\\qquad \\textbf{(C)}\\ \\frac {3}{8}\\qquad \\textbf{(D)}\\ \\frac {11}{25}\\qquad \\textbf{(E)}\\ \\frac {7}{16}$",
  "Solution_1": "[hide]Just subtract the parts away.\n$ [BDF] \\equal{} [ACE] \\minus{} [BCD] \\minus{} [DEF] \\minus{} [ABF]$\n$ [BDF] \\equal{} \\dfrac {(16)(12)}{2} \\minus{}\\dfrac{(9)(4)}{2} \\minus{}  \\dfrac {(5)(12)sin\\theta}{2} \\minus{} \\dfrac {(3)(15)sin(90\\minus{}\\theta)}{2}$\n$ [BDF] \\equal{} 96 \\minus{} 18 \\minus{} 30sin\\theta \\minus{} \\dfrac{45sin(90\\minus{}\\theta)}{2}$\n$ [BDF] \\equal{} 96 \\minus{} 18 \\minus{} \\dfrac{(30)(3)}{5} \\minus{} \\dfrac{(45)(4)}{(5)(2)}$\n$ [BDF] \\equal{} 96 \\minus{} 18 \\minus{}18 \\minus{} 18$\n$ [BDF] \\equal{} 42$\n$ \\dfrac {42}{96} \\equal{} \\dfrac {7}{16}$\n[/hide]",
  "Solution_2": "Ooooooo that's good.\r\n\r\nI was using the distance formula to find the length of each side of BDF and then using Heron's, but it took a long time. In the end I still came up with 7/16 though.",
  "Solution_3": "[hide=\"A Somewhat Longer Solution w/o Trig\"]\nWe know that $ [DBF] \\equal{} [ACE] \\minus{} [DFE] \\minus{} [BCD] \\minus{} [ABF]$. If we draw the altitudes of the three triangles we need to subtract, we find that we have similar triangles.\n\n$ [DBF] \\equal{} \\frac{12\\cdot 16}{2} \\minus{} \\frac{3\\cdot (16\\cdot \\frac{3}{4})}{2} \\minus{} \\frac{12\\cdot (12\\cdot \\frac{1}{4})}{2} \\minus{} \\frac{9\\cdot 4}{2}\\\\\n\\equal{} \\frac{12(10) \\minus{} 36}{2}\\\\\n\\equal{} \\frac{84}{2}$\n\n$ \\frac{[DBF]}{[ACE]} \\equal{} \\frac{\\frac{84}{2}}{\\frac{192}{2}}\\\\\n\\equal{}\\fbox{7/16 (E)}$\n[/hide]",
  "Solution_4": "hey sinhera, could you possibly elaborate on the similar triangles part please",
  "Solution_5": "Seriously, just altitudes doesn't prove similar triangles...",
  "Solution_6": "Sure.\r\n\r\n[hide=\"Details of the Similarity\"]\nDraw the altitude of $ \\triangle AFB$ and $ \\triangle DFE$, like so.\n\n[asy]defaultpen(linewidth(.8pt)+fontsize(8pt));\ndotfactor=3;\n\npair C = (0,0);\npair E = (16,0);\npair A = (0,12);\npair F = waypoint(E--A,0.25);\npair B = waypoint(A--C,0.25);\npair D = waypoint(C--E,0.25);\n\ndot(A);dot(B);dot(C);dot(D);dot(E);dot(F);\n\nlabel(\"$A$\",A,NW);label(\"$B$\",B,W);label(\"$C$\",C,SW);label(\"$D$\",D,S);label(\"$E$\",E,SE);label(\"$F$\",F,NE);\n\nlabel(\"$3$\",midpoint(A--B),W);\nlabel(\"$9$\",midpoint(B--C),W);\nlabel(\"$4$\",midpoint(C--D),S);\nlabel(\"$12$\",midpoint(D--E),S);\nlabel(\"$5$\",midpoint(E--F),NE);\nlabel(\"$15$\",midpoint(F--A),NE);\n\ndraw(A--C--E--cycle);\ndraw(B--F--D--cycle);\n\ndraw((foot(F,A,C))--F,linetype(\"4 4\"));\ndraw((foot(F,C,E))--F,linetype(\"4 4\"));[/asy]\n\nIt is clear that the altitudes are parallel to the legs of the right triangles (they are perpedicular to one leg and thus parallel to the other). By parallel lines we have equal angles so we get AA Similarity. You should be able to work on it from there.\n[/hide]",
  "Solution_7": "From NICE's Areas document.\n\n[hide=Solution]We use coordinates. Denote $C=(0,0)$. We immediately have $B=(0,9)$ and $D=(4,0)$. To find $F$, we drop an altitude from $F$ to $DE$, calling the contact point $G$. Note that by AA Similarity we get $\\triangle EFG \\sim \\triangle EAC$, so $FG=3$ and $EG=4$. This means that $GD=8$ and hence $CG=12$. Therefore, $G=(12,3)$. By the Shoelace Formula, the area of $\\triangle BDF$ is $42$. The area of $\\triangle EAC$ is simply $\\frac{(4+12)(9+3)}{2}=96$, so the answer is $\\frac{42}{96}=\\frac{7}{16} \\implies \\boxed{\\text{(E)}}$.[/hide]",
  "Solution_8": "Hold up, isn\u2019t this Wooga Looga Theorem?\n[hide=sol]\nWe proceed using Wooga Looga Theorem. The ratio between the segments of the sides is consistent and it is $\\frac{1}{3}$. The formula is $\\frac{r^2-r+1}{(r+1)^2}$. Plugging in $r=\\frac{1}{3}$ we get the answer is 7/16",
  "Solution_9": "Wait what is the Wooga Looga theorem again ",
  "Solution_10": "The definition of the Wooga Looga theorem:\n\"If there is $\\triangle ABC$ and points $D,E,F$ on the sides $BC,CA,AB$ respectively such that $\\frac{DB}{DC}=\\frac{EC}{EA}=\\frac{FA}{FB}=r$, then the ratio $\\frac{[DEF]}{[ABC]}=\\frac{r^2-r+1}{(r+1)^2}$.\"",
  "Solution_11": "Oh, okay. Didn't realize it was actually legit.",
  "Solution_12": "[quote=TachyonPulse]In right triangle $ \\triangle ACE$, we have $ AC \\equal{} 12$, $ CE \\equal{} 16$, and $ EA \\equal{} 20$. Points $ B$, $ D$, and $ F$ are located on $ \\overline{AC}$, $ \\overline{CE}$, and $ \\overline{EA}$, respectively, so that $ AB \\equal{} 3$, $ CD \\equal{} 4$, and $ EF \\equal{} 5$. What is the ratio of the area of $ \\triangle DBF$ to that of $ \\triangle ACE$?\n[asy]\nsize(200);defaultpen(linewidth(.8pt)+fontsize(8pt));\ndotfactor=3;\n\npair C = (0,0);\npair E = (16,0);\npair A = (0,12);\npair F = waypoint(E--A,0.25);\npair B = waypoint(A--C,0.25);\npair D = waypoint(C--E,0.25);\n\ndot(A);dot(B);dot(C);dot(D);dot(E);dot(F);\n\nlabel(\"$A$\",A,NW);label(\"$B$\",B,W);label(\"$C$\",C,SW);label(\"$D$\",D,S);label(\"$E$\",E,SE);label(\"$F$\",F,NE);\n\nlabel(\"$3$\",midpoint(A--B),W);\nlabel(\"$9$\",midpoint(B--C),W);\nlabel(\"$4$\",midpoint(C--D),S);\nlabel(\"$12$\",midpoint(D--E),S);\nlabel(\"$5$\",midpoint(E--F),NE);\nlabel(\"$15$\",midpoint(F--A),NE);\n\ndraw(A--C--E--cycle);\ndraw(B--F--D--cycle);[/asy]$ \\textbf{(A)}\\ \\frac {1}{4}\\qquad \\textbf{(B)}\\ \\frac {9}{25}\\qquad \\textbf{(C)}\\ \\frac {3}{8}\\qquad \\textbf{(D)}\\ \\frac {11}{25}\\qquad \\textbf{(E)}\\ \\frac {7}{16}$[/quote]\n\nThe inside triangle isn't right...."
}
{
  "Tag": [
    "LaTeX"
  ],
  "Problem": "can any one tell me how can I add some mathematical statements to a .TEX file?\r\n\r\nThanx for help.",
  "Solution_1": "i didn't really understand your question... what exactly do you mean by adding mathematical statements to a $\\text{\\TeX}$ file ? ( Normally, all mathematical statements in $\\text{\\TeX}$ files show up once the files been compiled to dvi or pdf... )",
  "Solution_2": "now how can I creat a $\\text{\\TeX}$ file ?",
  "Solution_3": "Shouldn't this be moved to this forum?\r\n[url]http://www.mathlinks.ro/Forum/index.php?f=123[/url] :maybe: \r\n\r\nAmir.S, AOPS/Mathlinks has a site explaining lots of stuff about (La)TeX, I suggest you check it out :\r\n[url]http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php[/url]",
  "Solution_4": "[quote=\"Amir.S\"]now how can I creat a $\\text{\\TeX}$ file ?[/quote]\r\n\r\nYou could use some software like texnic center....  If you know tex well, though you could even write out your $\\text{\\TeX}$ code on notepad and save it as a $\\text{\\TeX}$ file.",
  "Solution_5": "[quote=\"karthik_k\"]\n....  If you know tex well, though you could even write out your $\\text{\\TeX}$ code on notepad and save it as a $\\text{\\TeX}$ file.[/quote]\r\n\r\n :) How can I find these codes?",
  "Solution_6": "This thing on AoPS itself should do for the basics.... :D \r\n\r\nhttp://www.artofproblemsolving.com/LaTeX/AoPS_L_BasicFirst.php"
}
{
  "Tag": [
    "geometry theorems",
    "geometry"
  ],
  "Problem": "[b]Theorem (Ceva\u2019s Theorem for Chords).[/b] Let $A, B, C, D, E,$  and $F$  be six consecutive points around the circumference of a circle.  Then chords $AD, BE, CF$ concur  if and only if\r\n\\[ \\begin{array}{|c|} \\hline \\\\ ||AB|| \\cdot|| CD|| \\cdot ||EF|| = || BC|| \\cdot || DE || \\cdot || FA|| \\\\ \\\\ \\hline \\end{array}\\; .  \\]\r\n [b]References:[/b]\r\n[1] H. S. M. Coxeter, [i]Problem 527[/i], The Mathematics Student, 28(1980)3. \r\n[2] Stanley Rabinowitz, [i]The Seven Circles Theorem[/i],  Pi Mu Epsilon Journal, 8(1987) 441--449.",
  "Solution_1": "One implication is basically proven [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=60418]here[/url], and the converse follows immediately from it."
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "geometry unsolved"
  ],
  "Problem": "Let $A,\\ B$ and $C$ be given points on a circumference $K$ such that the triangle $\\triangle{ABC}$ is acute. Let $P$ be a point in the interior of $K$. $X,\\ Y$ and $Z$ be the other intersection of $AP, BP$ and $CP$  with the circumference. Determine the position of $P$ such that $\\triangle{XYZ}$ is equilateral.",
  "Solution_1": "One intersection P of the 3 Apollonius circles of the triangle $\\triangle ABC$ (the 1st isodynamic point) always lies in the triangle interior, even if the triangle is obtuse. Invert the vertices A, B, C in a circle (P) centered at P and with arbitrary radius r. Then $A'B' = AB \\cdot \\frac{r^2}{PA \\cdot PB},$  $B'C' = BC \\cdot \\frac{r^2}{PB \\cdot PC},$ $C'A' = CA \\cdot \\frac{r^2}{PC \\cdot PA}.$  Since P is on the B-Appolonius circle, $\\frac{AB}{BC} = \\frac{PA}{PC}$ and $\\frac{A'B'}{B'C'} = \\frac{AB}{BC} \\cdot \\frac{PB \\cdot PC}{PA \\cdot PB} = 1.$ Similarly, since P is on the C-Apollonius circle, $B'C' = C'A'.$ Thus the inverted triangle $\\triangle A'B'C'$ is equilateral. But for any point P (not on the triangle circumcircle), the circumcevian triangle $\\triangle XYZ$ of P, the pedal triangle $\\triangle DEF$ with respect to P, and the triangle $\\triangle A'B'C'$ inverted in an arbitrary circle centered at P are similar (for example, see  [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=70444]www.mathlinks.ro/Forum/viewtopic.php?t=70444[/url]).",
  "Solution_2": "i think there's a little easier solution. since $\\angle ABP+\\angle ACP=\\angle YXZ=60$, it follows that the isogonals of $AP$ and $BP$ form an angle of $120$. same for $AP$ and $CP$, and $CP$ and $BP$, then the isogonal conjugate of $P$ must be $F$ (the Fermat-Torricelli point). we conclude that $P$ is the isogonal conjugate of $F$. the converse is also easy to prove.",
  "Solution_3": "As $XYZ$ is equilateral it follows that $m\\angle APC=60+A$. Let $P'$ be the isogonal conjugate of $P$ wrt $\\triangle ABC$ we easily get that $m\\angle APC+m\\angle AP'C=180+A$ (look my lemma at [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=109112[/url] post 4)so $m\\angle AP'C=120$. Similary for the other vertices we get that $P'$ is the fermat point of $\\triangle ABC$ so that $P$ is the first isodynamic point of $\\triangle ABC$."
}
{
  "Tag": [],
  "Problem": "estw a,b,c > 0 kai abc=1 na apodeixtei oti :\r\na^3 + b^3 + c^3 + (ab)^3 + (bc)^3 + (ca)^2 >= 2(a^2*b + b^2*g + g^2*a)\r\n\r\nto thema to brhka se ena biblio toy mpamph stergioy kai moy arese idiaitera.",
  "Solution_1": "[quote=\"Athinaios\"]estw $ a,b,c > 0$ kai $ abc\\equal{}1$ na apodeixtei oti :\n$ a^3 \\plus{} b^3 \\plus{} c^3 \\plus{} (ab)^3 \\plus{} (bc)^3 \\plus{} (ca)^3 \\geq 2(a^2b \\plus{} b^2c \\plus{} c^2a)$\n\nto thema to brhka se ena biblio toy mpamph stergioy kai moy arese idiaitera.[/quote]\r\n\r\n\u0395\u03c7\u03bf\u03c5\u03bc\u03b5 \u03b1\u03c0\u03cc AM-GM: $ a^3\\plus{}a^3\\plus{}b^3\\geq 3a^2b$ \u03ba\u03b1\u03b9 \u03c0\u03c1\u03bf\u03c3\u03b8\u03ad\u03c4\u03c9\u03bd\u03c4\u03b1\u03c2 \u03ba\u03c5\u03ba\u03bb\u03b9\u03ba\u03ac \u03c0\u03c1\u03bf\u03ba\u03cd\u03c0\u03c4\u03b5\u03b9\r\n\r\n$ a^3\\plus{}b^3\\plus{}c^3\\geq a^2b\\plus{}b^2c\\plus{}c^2a$  (1)\r\n\r\n\u0395\u03c0\u03af\u03c3\u03b7\u03c2 \u03b1\u03c0\u03cc AM-GM \u03c0\u03b1\u03af\u03c1\u03bd\u03bf\u03c5\u03bc\u03b5:\r\n\r\n$ (ab)^3\\plus{}(ab)^3\\plus{}(ca)^3\\geq 3a^3b^2c\\equal{}3a^2b$ \u03ba\u03b1\u03b9 \u03c0\u03c1\u03bf\u03c3\u03b8\u03ad\u03c4\u03c9\u03bd\u03c4\u03b1\u03c2 \u03ba\u03c5\u03ba\u03bb\u03b9\u03ba\u03ac \u03c0\u03c1\u03bf\u03ba\u03cd\u03c0\u03c4\u03b5\u03b9:\r\n\r\n$ (ab)^3 \\plus{} (bc)^3 \\plus{} (ca)^3\\geq a^2b\\plus{}b^2c\\plus{}c^2a$ (2)\r\n\r\n\u0391\u03c0\u03cc (1)+(2) \u03c4\u03bf \u03b6\u03b7\u03c4\u03bf\u03cd\u03bc\u03b5\u03bd\u03bf :wink:",
  "Solution_2": "\u03a0\u03c9\u03c2 \u03c0\u03c1\u03bf\u03c3\u03b8\u03b5\u03c4\u03bf\u03c5\u03bc\u03b5 \u03ba\u03c5\u03ba\u03bb\u03b9\u03ba\u03b1?  :huh:",
  "Solution_3": "ennoei oti kanei to antisoixo gia b^2 + b^2 + c^2>=3b^2c(me AM-GM) kai opws katalavaineis k meta prosthetei kata meli.wraia lysi pantws, an k sto mialo m eixa mia alli.  :)",
  "Solution_4": "\u039f \u0394\u03b7\u03bc\u03ae\u03c4\u03c1\u03b7\u03c2 \u03b5\u03bd\u03bd\u03bf\u03b5\u03af :\r\n\r\n$ a^3 \\plus{} a^3 \\plus{} b^3 \\ge 3a^2b$  (1)\r\n$ b^3 \\plus{} b^3 \\plus{} c^3 \\ge 3b^2c$  (2)\r\n$ c^3 \\plus{} c^3 \\plus{} a^3 \\ge 3c^2a$  (3)\r\n\r\n\u03a0\u03c1\u03cc\u03c3\u03b8\u03b5\u03c3\u03b7 \u03ba\u03b1\u03c4\u03ac \u03bc\u03ad\u03bb\u03b7 \u03ba\u03b1\u03b9 ...  tsup!\r\n\r\n$ 3a^2 \\plus{} 3b^3 \\plus{} 3c^3 \\ge 3a^2b \\plus{} 3b^2c \\plus{} 3c^2a$  \r\n\r\n$ a^3 \\plus{} b^3 \\plus{} c^3 \\ge a^2b \\plus{} b^2c \\plus{} c^2a$ ,     ...\u03cc\u03bc\u03bf\u03b9\u03b1 \u03ba\u03b1\u03b9 \u03b3\u03b9\u03b1 \u03c4\u03b9\u03c2 \u03ac\u03bb\u03bb\u03b5\u03c2 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b5\u03c2.",
  "Solution_5": "nai eis tin triti ennoousa alla panw stin viasini m egrapsa ^2.. :blush: sorry...",
  "Solution_6": "\u03a4\u03bf \u03cc\u03c4\u03b9 $ a^3\\plus{}b^3\\plus{}c^3>\\equal{}a^2b\\plus{}b^2c\\plus{}c^2a$ \u03c0\u03c1\u03bf\u03ba\u03cd\u03c0\u03c4\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd Holder \u03c9\u03c2 \u03b5\u03be\u03ae\u03c2:\r\n\r\n$ a^3\\plus{}b^3\\plus{}c^3\\equal{}(a^3\\plus{}b^3\\plus{}c^3)^{2/3}(b^3\\plus{}c^3\\plus{}a^3)^{1/3}>\\equal{}a^2b\\plus{}b^2c\\plus{}c^2a$\r\n[\u03b3\u03b9\u03b1 \u03b5\u03bd\u03b1\u03bb\u03bb\u03b1\u03ba\u03c4\u03b9\u03ba\u03ae \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03b7 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03b1\u03bd\u03b1\u03b4\u03b9\u03ac\u03c4\u03b1\u03be\u03b7\u03c2!]\r\n\r\n\u03ba\u03b1\u03b9 \u03b5\u03c0\u03af\u03c3\u03b7\u03c2 \r\n$ (ab)^3\\plus{}(bc)^3\\plus{}(ca)^3\\equal{}((ab)^3\\plus{}(bc)^3\\plus{}(ca)^3)^{2/3}((ac)^3\\plus{}(ba)^3\\plus{}(cb)^3)^{1/3}\\equal{}((ab)^3\\plus{}(bc)^3\\plus{}(ca)^3)^{2/3}(1/b^3\\plus{}1/c^3\\plus{}1/a^3)^{1/3}>\\equal{}a^2b\\plus{}b^2c\\plus{}c^2a$\r\n\r\n\u03bc\u03b5 \u03c0\u03c1\u03cc\u03c3\u03b8\u03b5\u03c3\u03b7 \u03b1\u03c5\u03c4\u03ad\u03c2 \u03bf\u03b9 2 \u03b4\u03af\u03bd\u03bf\u03c5\u03bd \u03c4\u03bf \u03b6\u03b7\u03c4\u03bf\u03cd\u03bc\u03b5\u03bd\u03bf!",
  "Solution_7": "\u038c\u03c7\u03b9 \u03bc\u03cc\u03bd\u03bf \u03b1\u03c0\u03bf \u03c4\u03b7 Holder \u03b1\u03bb\u03bb\u03ac \u03ba\u03b1\u03b9 \u03b1\u03c0\u03bf \u03c4\u03b7\u03bd \u03b1\u03bd\u03bf\u03c3\u03cc\u03c4\u03b7\u03c4\u03c4\u03b1 \u03c4\u03b7\u03c2 \u03b1\u03bd\u03b1\u03b4\u03b9\u03ac\u03c4\u03b1\u03be\u03b7\u03c2. \u0388\u03c3\u03c4\u03c9 $ a\\leq b\\leq c\\Leftrightarrow a^{2}\\leq b^{2}\\leq c^{2}$. \u0393\u03b9\u03b1 \u03c4\u03b7 \u03bc\u03b5\u03c4\u03ac\u03b8\u03b5\u03c3\u03b7 $ (b,c,a)$ \u03c4\u03c9\u03bd $ a,b,c$ \u03c3\u03cd\u03bc\u03c6\u03c9\u03bd\u03b1 \u03bc\u03b5 \u03c4\u03b7\u03bd \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03c4\u03b7\u03c2 \u03b1\u03bd\u03b1\u03b4\u03b9\u03ac\u03c4\u03b1\u03be\u03b7\u03c2 \u03c0\u03c1\u03bf\u03ba\u03cd\u03c0\u03c4\u03b5\u03b9 \u03cc\u03c4\u03b9 $ a^{3} \\plus{} b^{3} \\plus{} c^{3}\\geq a^{2}b \\plus{} b^{2}c \\plus{} c^{2}a$",
  "Solution_8": "[quote][\u03b3\u03b9\u03b1 \u03b5\u03bd\u03b1\u03bb\u03bb\u03b1\u03ba\u03c4\u03b9\u03ba\u03ae \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03b7 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03b1\u03bd\u03b1\u03b4\u03b9\u03ac\u03c4\u03b1\u03be\u03b7\u03c2!] \n[/quote]\r\n\r\n\u0391\u03c5\u03c4\u03cc \u03b5\u03bd\u03bd\u03bf\u03bf\u03cd\u03c3\u03b1 \u03b5\u03b4\u03ce! \u039a\u03b1\u03b9 \u03b5\u03c0\u03af\u03c3\u03b7\u03c2 \u03bc\u03b5 \u03c4\u03b7\u03bd \u03b1\u03bd\u03b1\u03b4\u03b9\u03ac\u03c4\u03b1\u03be\u03b7\u03c2 \u03b2\u03b3\u03b1\u03af\u03bd\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03c4\u03bf \u03b4\u03b5\u03cd\u03c4\u03b5\u03c1\u03bf \u03bc\u03ad\u03c1\u03bf\u03c2 \u03c4\u03b7\u03c2 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1\u03c2!",
  "Solution_9": "kai egw aytin tn lysi eixa ipopsin m me tn anadiataksi. :wink: ..nice"
}
{
  "Tag": [
    "AMC",
    "AIME",
    "USA(J)MO",
    "USAMO",
    "articles",
    "AIME I"
  ],
  "Problem": "This seems like a stupid question, but:\r\n\r\nAm I allowed to bring and consume food and drink while taking the AIME?  I can get a good boost during practices from an energy drink, but I'm not sure if that counts as an unfair advantage or if its allowed. It doesn't seem to be mentioned in the FAQ on the MAA website.",
  "Solution_1": "I think my friend had coffee with him when he was taking the AMC 12.\r\nDepends on where you take it, I guess.",
  "Solution_2": "I think you are allowed to; my friend brought a potato chip with him on AIME :D",
  "Solution_3": "This question has been asked a few times before; the answer that AMCDirector gives is that there's no AMC restriction on it.\r\n\r\nOn an interesting and slightly related note, does anyone foresee potential problems with \"performance-enhancing\" (mainly focus) drugs like Modafinil?",
  "Solution_4": "Thanks for the information.  I guess that means I'm bringing my energy drink which I doubt contains modafinil, but worthawholebean makes an interesting observation, but of course any drugs come with serious risk.  I doubt any people good enough to qualify for USAMO would want to put their minds at jeopardy by taking drugs.  Still, it would be nice to see people care so much about math to take drugs for an exam.",
  "Solution_5": "[quote=\"worthawholebean\"]This question has been asked a few times before; the answer that AMCDirector gives is that there's no AMC restriction on it.\n\nOn an interesting and slightly related note, does anyone foresee potential problems with \"performance-enhancing\" (mainly focus) drugs like Modafinil?[/quote]\r\nWell I read in Discover that in a survey of college kids, 80% said they used drugs like Ritalin and Modafinil on tests before to help increase concentration and stuff...\r\n\r\nI drink caffeine for AMC contests but I don't know how helpful it is, seeing that I feel like puking sometimes after too much of it :O I've aced some tests after drinking energy drinks, and I think most of the time it is that way, but I've also bombed a few tests while on caffeine too. My main motivation is that I am pretty much always sleepy due to chronic lack of sleep :(\r\n\r\nHowever that same article said when they pulled random people and gave them stimulants and others placebos and had them take tests, ones who had taken the stimulant scored significantly better. Since drugs like those are so widely available, maybe the AMC would consider them a integrity threat? You can't exactly test for them though, so I suppose there isn't any way to stop it.",
  "Solution_6": "- You're full of adrenaline anyway. What's the point of adding outside stimulants on top of that?\r\n- Stimulants can backfire. I recall one orchestra practice I had after having a lot of caffeine at dinner; I suddenly dropped off and lost all ability to focus about an hour in.\r\n\r\nNow, I heartily recommend sugar and food in general, along with getting enough sleep before the test. Eating during the test is a distraction, so limit that.",
  "Solution_7": "I know I had a lot of caffeine one time before a contest for which I was really tired - I completely lost my ability to concentrate.",
  "Solution_8": "Well I think walking in calm and composed is best. Walking in anxious about results completely killed my AIME this year (although I still made USAMO most likely). I find that given I am calm, caffeine greatly improves my ability to think. If I am nervous, I start shaking, sweating, and feeling a burning sensation from taking caffeine.\r\n\r\nOf course the natural threshold for your performance is set by how well you sleep and how well you eat, so those are priority (although I never sleep well...hmmm I think I see a correlation between sleep and contest scores).\r\n\r\nEating during say, the USAMO, I think could possibly be helpful as a breather when you're stuck on a problem, since it relaxes you somewhat. It certainly helps me get ideas when I go snacking.",
  "Solution_9": "I wouldn't advocate caffiene or stimulants before a test, especially a relatively long one like the AIME.  It's hard to predict when the stimulant will wear off and leave you with little energy left.  I use an energy drink as a boost once I get to around #11-12 for some more energy on the harder problems.",
  "Solution_10": "Like jmerry said, two things are what you need the most: good night sleep and mild breakfast (not heavy please).\r\n\r\nAnother thing to note is for people like me who don't drink coffee or caffeine at all (excluding soda), trying on those drinks will be extremely harmful on first couple times. So... I'm sure we are all smart to avoid it but if you are weak on caffeine, don't drink too much on your first AMC.  :rotfl:"
}
{
  "Tag": [
    "floor function",
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "How many ways are there of coloring the vertices of a regular $2n$-gon with n colors, such that each vertex is given one color, and every color is used for two non-adjacent vertices? Colorings are regarded as the same if one is obtained from the other by rotation.",
  "Solution_1": "An equivalent question  has been posted (but not answered) in the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=85715]Pre-Olympiad forum[/url].  I could have sworn I've solved it (or at least looked it up on OEIS) recently, but I can't find it, so either I'm wrong or it wasn't on the forum.  I'll get back to you (or someone else will solve it first ;) ).",
  "Solution_2": "This problem (under the form posted in that Pre-Olympiad topic) is quite well-known in combinatorics (due to Lucas?).\r\nIt can be solved with the principle of inclusion-exclusion, after using [b]Kaplansky's lemma[/b] :\r\nThe numbers $1, 2, ..., m \\ \\ (m \\geq 3)$ are placed around a circle. For $1 \\leq k \\leq \\lfloor m/2 \\rfloor,$ let $a(k)$ be the number of subsets with k elements of $\\{1, 2, ..., m\\}$ which don't contain two adjacent elements on the circle. We then have :\r\n\\[a(k) = \\frac{m}{k}\\binom{m-k-1}{k-1}. \\]",
  "Solution_3": "this problem has been solved in:\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=4110"
}
{
  "Tag": [
    "geometry",
    "3D geometry"
  ],
  "Problem": "The complete outside including the bottom of a wooden 4inch cube is painted red. The painted cube is then cut into one inch cubes. How many cubes have absolutely no red paint on any face?",
  "Solution_1": "[hide]The only cubes that have no red are the cubes that you can't see.  That section is a 2x2x2 cube, so the answer is 8.[/hide]",
  "Solution_2": "[quote=\"bryantacademy\"]The complete outside including the bottom of a wooden 4inch cube is painted red. The painted cube is then cut into one inch cubes. How many cubes have absolutely no red paint on any face?[/quote]\r\n[hide]\nThe number of cubes you have are 64. Only the cubes in the middle have no paint. Remove the border of paint. Thus you have a 2x2x2 square. I believe that it is 8?[/hide]",
  "Solution_3": "anything on the inside, so (4-2)^3, or 8",
  "Solution_4": "its a 4 cube by 4 cube square, that is impossible!!! :furious: If you remove all the outer cubes, you must surely get a 2 cube by 2 cube square",
  "Solution_5": "how is it impossible? It needs to be a 2 by 2 by two or less!",
  "Solution_6": "[hide]\n2x2x2=8[/hide]",
  "Solution_7": "take out outer 4...\r\n\r\ninside cube = 2x2x2 = 8",
  "Solution_8": "2x2x2=8 cubes :)",
  "Solution_9": "[hide]Well, only the 2*2*2 part is going to be unpainted, so 8.[/hide]"
}
{
  "Tag": [
    "MATHCOUNTS"
  ],
  "Problem": "What is the sum of the prime factors of the number represented by:\r\n2^12-2^11+2^10-2^9+...+2:^2:-2\r\n\r\n\r\nthis was a target round, so i guess you could use a calc",
  "Solution_1": "I think the answer is... [hide]101[/hide]",
  "Solution_2": "MY bad\n\n\n\nthe correct answer is : [hide]30[/hide]",
  "Solution_3": "[hide]\n\nHmm, I did it a slow and tedious way, but got 30\n\n\n\n[/hide]",
  "Solution_4": "sorry, i meant to say no to the answer of 101. yes [hide]30[/hide] is correcto",
  "Solution_5": "How did you do it? This was my method..\n\n\n\n[hide]\n\nI noted that \n\n\n\nx^n -1/x-1 = x^n-1 + ..... 1\n\n\n\nAnd found out what the number was, then factored it.\n\n\n\nHow did you guys do it? Is there a faster way?\n\n\n\n[/hide]",
  "Solution_6": "k...one second, maybe a few minutes",
  "Solution_7": "SouthPaw: although your method is of course the better and requires more skill, most MATHCOUNTS contestants would have simply put it into their calcs, gotten 2730 out of it and then prime factored that....",
  "Solution_8": "Syntax Error wrote:What is the sum of the prime factors of the number represented by:\n2^12-2^11+2^10-2^9+...+2:^2:-2\n\n\n\n\n\n\n[hide]okay, so that equals 2^11+2^9+2^7+...+2=2(2^10+2^8+2^6+...+1)=2(4^5+4^4+4^3+4^2+4^1+1)\n\n\n\nthen i factored from there. it didnt take more than 20 seconds to get there (i think). but i dunno how far you could have factored before that. [/hide]",
  "Solution_9": "it would really help if you had a ti-89 to factor that i guess.",
  "Solution_10": "sigh, i guess its faster to factor after youve done the first step (2^11+2^9+2^7+...+2). but still, good to know how to get there, to make it easier, if you dont have that kind of...whatever",
  "Solution_11": "just wondering...syntax, did you get all this stuff from me? because I thought I was the only one weird enough to use 4's and all....hmmm...",
  "Solution_12": "im perdy sure that i told you that i had done it that way the first time i took the test in april/may\r\n\r\nanyways, mystic terminator did it the same way. yah for him. and for the rest of us!\r\n\r\nbut i really didnt copy mystic. well i did, but i had done it the same way. YAH FOR EVERYONE!",
  "Solution_13": "I don't know. For a target round questions, I think it's a lot easier and guarantees no mistakes if u use a calc.\r\n\r\nAfter all, 2^12 is pretty small, i've got it memorized to 2^16, but if u don't have it memorized, at least have 2^10 down good.",
  "Solution_14": "2+3+5+7+13 = 30"
}
{
  "Tag": [
    "algebra",
    "polynomial"
  ],
  "Problem": "Find the sum of the coefficients in the polynomial that results when $(4x^2-3x-2)^{2001}$ is completely multiplied out.\r\n\r\nEh, am I missing something really obvious?",
  "Solution_1": "Just plug in $x=1$, right?",
  "Solution_2": "I ought to gag myself...   :blush: \r\n\r\nI knew... that. :? \r\n\r\nThanks Joe.",
  "Solution_3": "Hmm why do we just need to plug in $x=1$ to find the solution",
  "Solution_4": "Take a polynomial $a_1x^{n}+a_2x^{n-1}+a_3x^{n-2}+\\dots$\r\n\r\nIf you plug in one, we get $a_1+a_2+a_3+\\dots$\r\n\r\nWhich is the sum of the coefficients."
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "Let $ a,b,c$ be none-gative numbers ,no two of which are zero.Prove that:\r\n$ \\frac{36(a^2\\plus{}b^2\\plus{}c^2)}{(a\\plus{}b\\plus{}c)^2}\\plus{}\\frac{5abc}{(a\\plus{}b)(b\\plus{}c)(c\\plus{}a)} \\ge \\frac{49}{4}\\plus{}(\\frac{a}{a\\plus{}b})^3\\plus{}(\\frac{b}{b\\plus{}c})^3\\plus{}(\\frac{c}{c\\plus{}a})^3$\r\nI hope you will like it :)",
  "Solution_1": "I am again doubtful..... Again the inequyality seems to be unchanged by replacing a, b, c respectively with ka, kb, kc so $ abc\\equal{}1$ is true.... AM I RIGHT??\r\n\r\nPS: Dear quy, your inequalities are always nice , but I cannot solve MOST of them.... :rotfl:",
  "Solution_2": "[quote=\"Potla\"]I am again doubtful..... Again the inequyality seems to be unchanged by replacing a, b, c respectively with ka, kb, kc so $ abc \\equal{} 1$ is true.... AM I RIGHT??\n\nPS: Dear quy, your inequalities are always nice , but I cannot solve MOST of them.... :rotfl:[/quote]\r\nMy friend,you can assume $ abc\\equal{}1$,but I think it isn't nececsarry  with this inequality :)",
  "Solution_3": "[quote=\"Potla\"]I am again doubtful..... Again the inequyality seems to be unchanged by replacing a, b, c respectively with ka, kb, kc so $ abc \\equal{} 1$ is true.... AM I RIGHT??\n\nPS: Dear quy, your inequalities are always nice , but I cannot solve MOST of them.... :rotfl:[/quote]\r\nWith $ abc\\equal{}1$,you can solve it? :wink: Please post solution :oops:",
  "Solution_4": "I don't think it's a strong ineq.\r\nWe can prove easily the stronger:\r\n$ \\frac{81(a^2\\plus{}b^2\\plus{}c^2)}{4(a\\plus{}b\\plus{}c)^2} \\plus{} \\frac{5abc}{(a\\plus{}b)(b\\plus{}c)(c\\plus{}a)} \\geq\\ 7 \\plus{} \\sum\\ \\frac{a^3}{(a\\plus{}b)^3}$\r\nquykhtn-qa1, can you solve this stronger ?  :)",
  "Solution_5": "[quote=\"nguoivn\"]I don't think it's a strong ineq.\nWe can prove easily the stronger:\n$ \\frac {81(a^2 \\plus{} b^2 \\plus{} c^2)}{4(a \\plus{} b \\plus{} c)^2} \\plus{} \\frac {5abc}{(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)} \\geq\\ 7 \\plus{} \\sum\\ \\frac {a^3}{(a \\plus{} b)^3}$\nquykhtn-qa1, can you solve this stronger ?  :)[/quote]\r\nI have a stronger inequality:\r\n$ \\frac{9(a^2\\plus{}b^2\\plus{}c^2)}{(a\\plus{}b\\plus{}c)^2}\\plus{}\\frac{5abc}{(a\\plus{}b)(b\\plus{}c)(c\\plus{}a)} \\ge \\frac{13}{4}\\plus{} \\sum\\ \\frac {a^3}{(a\\plus{}b)^3}$ :) \r\nAnd,with my first inequality,I have a nice solution :)",
  "Solution_6": "[quote=\"quykhtn-qa1\"][quote=\"nguoivn\"]I don't think it's a strong ineq.\nWe can prove easily the stronger:\n$ \\frac {81(a^2 \\plus{} b^2 \\plus{} c^2)}{4(a \\plus{} b \\plus{} c)^2} \\plus{} \\frac {5abc}{(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)} \\geq\\ 7 \\plus{} \\sum\\ \\frac {a^3}{(a \\plus{} b)^3}$\nquykhtn-qa1, can you solve this stronger ?  :)[/quote]\nI have a stronger inequality:\n$ \\frac {9(a^2 \\plus{} b^2 \\plus{} c^2)}{(a \\plus{} b \\plus{} c)^2} \\plus{} \\frac {5abc}{(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)} \\ge \\frac {13}{4} \\plus{} \\sum\\ \\frac {a^3}{(a \\plus{} b)^3}$ :) \nAnd,with my first inequality,I have a nice solution :)[/quote]\r\nYes, indeed, I had a mistake in my calculus. The best result is same to you:\r\n$ \\frac{9(a^2\\plus{}b^2\\plus{}c^2)}{(a\\plus{}b\\plus{}c)^2} \\geq\\ \\frac{9}{4} \\plus{} \\sum\\ \\frac{a^3\\plus{}b^3}{(a\\plus{}b)^3}$\r\nOf course, SOS solved it easily  :)",
  "Solution_7": "[quote=\"nguoivn\"][quote=\"quykhtn-qa1\"][quote=\"nguoivn\"]I don't think it's a strong ineq.\nWe can prove easily the stronger:\n$ \\frac {81(a^2 \\plus{} b^2 \\plus{} c^2)}{4(a \\plus{} b \\plus{} c)^2} \\plus{} \\frac {5abc}{(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)} \\geq\\ 7 \\plus{} \\sum\\ \\frac {a^3}{(a \\plus{} b)^3}$\nquykhtn-qa1, can you solve this stronger ?  :)[/quote]\nI have a stronger inequality:\n$ \\frac {9(a^2 \\plus{} b^2 \\plus{} c^2)}{(a \\plus{} b \\plus{} c)^2} \\plus{} \\frac {5abc}{(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)} \\ge \\frac {13}{4} \\plus{} \\sum\\ \\frac {a^3}{(a \\plus{} b)^3}$ :) \nAnd,with my first inequality,I have a nice solution :)[/quote]\nYes, indeed, I had a mistake in my calculus. The best result is same to you:\n$ \\frac {9(a^2 \\plus{} b^2 \\plus{} c^2)}{(a \\plus{} b \\plus{} c)^2} \\geq\\ \\frac {9}{4} \\plus{} \\sum\\ \\frac {a^3 \\plus{} b^3}{(a \\plus{} b)^3}$\nOf course, SOS solved it easily  :)[/quote]\r\nYES,THANKS :) \r\n@nguoivn:[hide] anh gioi qua',hinh nhu anh sap viet sach?,bao gio xong a.?' [/hide]"
}
{
  "Tag": [
    "function",
    "limit"
  ],
  "Problem": "Prove that \r\n$ f: \\mathbb {R}\\rightarrow \\mathbb {R}$ \r\n$ f(x)\\equal{}x^5\\minus{}2x^2\\plus{}x$ is a surjective function, onto mapping.",
  "Solution_1": "[hide]$ f$ is continuous on $ \\mathbb R$, and $ \\lim_{x\\to \\plus{} \\infty}f(x) \\equal{} \\plus{} \\infty$, $ \\lim_{x\\to \\minus{} \\infty}f(x) \\equal{} \\minus{} \\infty$. Therefore $ f$ is surjective.[/hide]"
}
{
  "Tag": [],
  "Problem": "The sum \u2211(from n = 1 to \u221e) arctan [n/(n^4 - 2n^2 + 2)] is equal to:\r\nA) (arctan 2 + arctan 3)/4\r\nB) 4. (arctan 1)\r\nC) -\u03c0/16\r\nD) 3\u03c0/16\r\n\r\nMore than one options may be correct.",
  "Solution_1": "hello, do you mean $ \\sum_{n\\equal{}1}^\\infty\\arctan\\left(\\frac{n}{n^4\\minus{}2n^2\\plus{}2}\\right)$?\r\nSonnhard,",
  "Solution_2": "It does mean $ \\sum_{n\\equal{}1}^\\infty\\arctan\\left(\\frac{n}{n^4\\minus{}2n^2\\plus{}2}\\right)$\r\n\r\nI am eager to know the solution. :)  :)",
  "Solution_3": "It does mean that. But I dont know the solution. Anyone on AoPS???",
  "Solution_4": "[hide=\"Not a solution but may help\"]Let's just assume for a moment the numerator is $ 4n$ instead of $ n$.\n\n$ \\begin{array}{rcl} \\arctan \\left( \\frac{4n}{n^4 \\minus{} 2n^2 \\plus{} 2} \\right) &\\equal{}& \\arctan \\left( \\frac{ (n\\plus{}1)^2 \\minus{} (n\\minus{}1)^2 }{ 1 \\plus{} (n^2\\minus{}1)^2 } \\right) \\\\ &\\equal{}& \\arctan ((n\\plus{}1)^2) \\minus{} \\arctan ((n\\minus{}1)^2) \\\\ \\end{array}$\n\nThis telescopes getting $ \\minus{}\\arctan 0 \\minus{} \\arctan 1 \\equal{} \\minus{}\\pi/4$, which is $ 4$ times one of the answers. Unfortunately you can't just yank $ 4$'s out of $ \\arctan$ like that. I'm not sure how you'd introduce it either.[/hide]",
  "Solution_5": "Ain't helping.Sigh...",
  "Solution_6": "hello, a numerical experiment with Maple ( i have calculated the sum to 100,\r\ni.e. $ \\sum_{n\\equal{}1}^{100}\\arctan\\left(\\frac{n}{n^4\\minus{}2n^2\\plus{}2}\\right)$ )shows that the result is nearly $ 1.07220308423251547552748148278$. But neither the answers A, B, C nor D show approximately this result.\r\nSonnhard.",
  "Solution_7": "[b]MNrule[/b], are you sure you have stated the problem correctly?\r\n\r\n[b]MellowMelon[/b], aren't you forgetting the limit?"
}
{
  "Tag": [
    "calculus",
    "integration",
    "calculus computations"
  ],
  "Problem": "Let $ a,\\ b,\\ c$ be positive real numbers. Prove the following inequality.\r\n\\[ \\int_1^e \\frac {x^{a \\plus{} b \\plus{} c \\minus{} 1}[2(a \\plus{} b \\plus{} c) \\plus{} (c \\plus{} 2a)x^{a \\minus{} b} \\plus{} (a \\plus{} 2b)x^{b \\minus{} c} \\plus{} (b \\plus{} 2c)x^{c \\minus{} a} \\plus{}(2a \\plus{} b)x^{a \\minus{} c} \\plus{} (2b \\plus{} c)x^{b \\minus{} a} \\plus{} (2c \\plus{} a)x^{c \\minus{} b}]}{(x^a \\plus{} x^b)(x^b \\plus{} x^c)(x^c \\plus{} x^a)}\\geq a \\plus{} b \\plus{} c.\\]\r\n\r\n\r\nI have just posted 500 th post. \r\n\r\n[color=blue]Thank you for your cooperations, mathLinkers and AOPS users.[/color]\r\n\r\nI will keep posting afterwards.\r\n\r\nJapanese Communities Modeartor\r\n\r\nkunny",
  "Solution_1": "Chikaya-san, i think there's a bit of a typo-error..anyway, it doesn't matter."
}
{
  "Tag": [
    "parameterization",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "prove that we can write any rational $r \\in (0,1)$ with odd denominator in this form$\\frac{xyz}{x^2+y^2+z^2}$",
  "Solution_1": "I think I saw it in Komal.\r\n\r\nAnyway, it seems tough. I only have some trivial remarks about it. First of all, it's enough to prove it for numbers of the form $\\frac 1{2n+1},\\ n\\in \\mathbb N^*$, because then we can take $px,py,pz$ and prove it holds for $\\frac p{2n+1}$ as well, so the problem is to show that for $n\\in \\mathbb N^*$, the equation $x^2+y^2+z^2=(2n+1)xyz$ has (non-trivial) solutions. This is equivalent to saying that for each $n\\ge 1$ we can find $y,z\\ne 0$ s.t. $(2n+1)^2y^2z^2-4y^2-4z^2$ is a perfect square. I have no idea how to do this..\r\n\r\nNote that for $n=1$ this is the famous Markov equation $x^2+y^2+z^2=3xyz$. I've searched the Web it an attempt to find refrences to the problem we're interested in (we could call it a Markov equation with generalized parameter), but to no avail. For $n=1$ we can find an infinite set of solutions: $(x,y,z)=(F_{2n-1},F_{2n+1},1)$, where $F_n$ are the Fibonacci numbers (starting with $F_0=0,F_1=1$).",
  "Solution_2": "I dont think so.\r\nIt is not difficult to prove that the equation $x^2+y^2+z^2 = kxyz$ has a solution if and only if $k=1$ or $k=3$.\r\nSo, I think there must be another way.\r\n\r\nPierre.",
  "Solution_3": "I also found this fact when reading grobber's post, but unfortunately a friendly message \" please come back again later, since the RAM memmory dosn't suffice\" didn't let me to do this. But doesn't this imply that th problem is wrong?",
  "Solution_4": "I am agree with harazi.",
  "Solution_5": "You seem to be right. \r\nOr maybe $x,y,z$ are supposed to be rationals and not integers?\r\n\r\nSo, Sam? What is your opinion about it?\r\n\r\nPierre.",
  "Solution_6": "No, the correct problem is that any such rational is the fractional part of a number of the form $\\frac{xyz}{x^2+y^2+z^2} $ with x,y,z integers."
}
{
  "Tag": [
    "real analysis",
    "function",
    "integration",
    "advanced fields",
    "advanced fields unsolved"
  ],
  "Problem": "Let A be some subset of the line with positive lebesgue measure,is there any lipschitz function f such that f(A) is an interval?",
  "Solution_1": "a point is an intervalle isn't it ?",
  "Solution_2": "I should have add ...non constant function.",
  "Solution_3": "Yes, for example $f(x)=\\int_{0}^{x}\\chi_{A}(t)\\,dt$.\r\n\r\nOne can ask the same question in higher dimensions: given a set $A\\subset\\mathbb R^{n}$ of positive measure, is there a Lipschitz mapping $f\\colon\\mathbb R^{n}\\to\\mathbb R^{n}$ such that $f(A)$ is a (non-trivial) ball? This has been proved in dimension $n=2$ (Alberti-Csornyei-Preiss, not yet published). Remains open for $n>2$.",
  "Solution_4": "Your function does not work for arbitrary sets,take the irrationals for example and you ll have f(Ir)=Ir,its of course correct for compact sets. \r\nI found out there is a proof in Laczkovich's paper 'Paradoxical decompositions using lipschitz functions',but its not of exercise level."
}
{
  "Tag": [
    "search",
    "\\/closed"
  ],
  "Problem": "On another forum I saw that once you get past a certain amount of posts and have joined for a certain period of time, you get to make up your own ranking (under your post count ranking) through the profiles page. Have no idea how that's done. \r\n\r\nAlso, I've heard that moderators used to have their own rankings like administrators. What happened to that. \r\n\r\nThis has been mentioned, but we should also have a forum where posts don't count towards your post count (although they still can searched). On that other forum, it was the home to all the games. This can also solve the \"post count deficits\" which is when you post count says x posts, but when you click \"find all posts by yyyyy\" you usually get \"search found x - k matches\" where k is non-negative. \r\n\r\nJust some ideas  :)",
  "Solution_1": "[quote=\"236factorial\"]On another forum I saw that once you get past a certain amount of posts and have joined for a certain period of time, you get to make up your own ranking (under your post count ranking) through the profiles page. Have no idea how that's done. \n[/quote]\r\n\r\nBah. Elitist :P .\r\n\r\n\r\nOn the other hand, I like your other suggestions.",
  "Solution_2": "These ideas have been discussed several times.\r\n\r\nWe don't care about post counts.  Why should they concern us at all?\r\n\r\nThe ranks are just for fun.  In our case, we add a mathematical theme -- the Millenium problems.  We're not about to spend time horsing around with them.  We have about a ginormozillion hours worth of educational projects ahead of us.  Users should remember that as entertaining as some aspects of this site might be, our goal is to build better and better educational resources for students.",
  "Solution_3": "That's why I put small ideas, they're not of high importance (because we can live without them). It's just when some of the administrators have too much free time (which I hope they do - life can't be all work and no play?  :D ), they can consider these. \r\n\r\nI just noticed something.... all of these were about post counts  :?"
}
{
  "Tag": [],
  "Problem": "A man drives along a main highway on which a regular service of buses is in operation. He notices that every three minutes he  meets a bus and that every six minutes a bus overtakes him. How often does a bus leave the terminal station at one end of the route ?",
  "Solution_1": "let $ b$ be the speed of a bus and $ m$ the man's speed. since the man meets buses twice as fast as they overtake him, we get $ b \\plus{} m \\equal{} 2(b \\minus{} m)$, which gives $ b \\equal{} 3m$. when they travel in the same direction, the speed of the buses relative to the man is therefore $ 3m \\minus{} m \\equal{} 2m$, while the actual speed of the bus is $ 3m$, so if the man were stationary, buses would pass him every $ 6\\cdot \\frac {2}{3} \\equal{} 4$ minutes. \r\n\r\nthis problem's been posted before; I believe it's an old Russian problem.",
  "Solution_2": "[quote=\"pleurestique\"]\n\nthis problem's been posted before; I believe it's an old Russian problem.[/quote]\r\n\r\n :maybe: \r\nIt is an old Britain problem   :rotfl:  from [i]\"Great book of math puzzles\" by Philip Heafford[/i]\r\nOriginally published in Great Britain under the title \"Mathematics for fun\" 1959\r\nThe solution of these problem in the  book is arithmetic solution. \r\nHow high level  has be an old math text books !",
  "Solution_3": "sloution because every three min he sees a bus then every six min he is overtaken, then \r\num every two min a bus leaves the terminal because 6/3 is two",
  "Solution_4": "[quote=\"treesamc-8\"]sloution because every three min he sees a bus then every six min he is overtaken, then \num every two min a bus leaves the terminal because 6/3 is two[/quote]\r\n\r\nsilly, the answer is 4.",
  "Solution_5": "[quote=\"pleurestique\"]let $ b$ be the speed of a bus and $ m$ the man's speed. since the man meets buses twice as fast as they overtake him, we get $ b \\plus{} m \\equal{} 2(b \\minus{} m)$, which gives $ b \\equal{} 3m$. when they travel in the same direction, the speed of the buses relative to the man is therefore $ 3m \\minus{} m \\equal{} 2m$, while the actual speed of the bus is $ 3m$, so if the man were stationary, buses would pass him every $ 6\\cdot \\frac {2}{3} \\equal{} 4$ minutes. \n\nthis problem's been posted before; I believe it's an old Russian problem.[/quote]\r\n\r\n\r\n\r\nvery nice explanation!"
}
{
  "Tag": [
    "geometry",
    "logarithms",
    "integration",
    "calculus",
    "rotation",
    "calculus computations"
  ],
  "Problem": "Let R be the region bounded by the graph of $ y\\equal{}\\frac{lnx}{x}$, the $ x\\minus{}axis$, and the line $ x\\equal{}e$\r\n\r\na. find the area of the region R.\r\nb. find the volume of the solid formed by revolving the region about the $ y\\minus{}axis$",
  "Solution_1": "By parts: $ u\\equal{}\\ln x$, $ du\\equal{}\\frac{dx}{x}$, $ v\\equal{}\\ln x$, $ dv\\equal{}\\frac{dx}{x}$.  So $ \\ln^{2} x\\minus{}\\int \\frac{\\ln x}{x}dx\\equal{}\\int \\frac{\\ln x}{x}dx$.  Move the integral on the left over and divide so $ \\int \\frac{\\ln x}{x}dx\\equal{}\\frac{\\ln^{2} x}{2}$.  The limits of integration are $ x\\equal{}1$ and $ x\\equal{}e$ so plugging these in, we get $ A\\equal{}\\frac{1}{2}$.",
  "Solution_2": "[quote=\"JRav\"]By parts: $ u = \\ln x$, $ du = \\frac {dx}{x}$, $ v = \\ln x$, $ dv = \\frac {dx}{x}$.  So $ \\ln^{2} x - \\int \\frac {\\ln x}{x}dx = \\int \\frac {\\ln x}{x}dx$.  Move the integral on the left over and divide so $ \\int \\frac {\\ln x}{x}dx = \\frac {\\ln^{2} x}{2}$.  The limits of integration are $ x = 1$ and $ x = e$ so plugging these in, we get $ A = \\frac {1}{2}$.[/quote]\r\n\r\nThis is my answer for part b, could someone check if this is correct?\r\n\r\nit rotates around y-axis\r\nuse shell dx\r\n$ V=\\int_1^e {2{\\pi}rh}dx = 2{\\pi}\\int_1^e {rh}dx$\r\nwhere $ r=(x-1)$and $ h=f(x)=\\frac {\\ln {x}}{x}$\r\nso: $ 2{\\pi}\\int_1^e {(x-1)(\\frac {\\ln {x}}{x})}dx$\r\n=$ 2{\\pi}\\int_1^e {\\ln {x}-\\frac {\\ln {x}}{x}}dx$\r\n=$ 2{\\pi}\\int_1^e {\\ln {x}}$$ -{2{\\pi}\\int_1^e {\\frac {\\ln{x}}{x}}}$\r\n=$ 2{\\pi}[xlnx-x]_1^e - 2{\\pi}[\\frac{(\\ln {x})^2}{2}]_1^e$\r\n=$ 2{\\pi}[e\\ln {e} -e-\\ln {1} +1] - {2{\\pi}[\\frac {\\ln {e^2}}{2}-\\frac {\\ln {1^2}}{2}]}$\r\n=$ 2{\\pi}[e-e+1] - 2{\\pi}[1/2]$=$ 2{\\pi}-{\\pi}$=$ {\\pi}$\r\n\r\n^does this process seem correct?\r\n\r\n=]",
  "Solution_3": "The process is correct, but you made an error early that you carried through the computation (and which made you do some extra work):\r\n[quote=\"sneezywalrus\"]$ r \\equal{} (x \\minus{} 1)$[/quote]\r\nOther than that, everything was fine.  (Well, you dropped a couple of $ dx$s, but your actual computations were correct.)"
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "trigonometry",
    "trig identities",
    "Law of Sines"
  ],
  "Problem": "Let $ ABC$ be a triangle. Prove that $ \\angle A\\equal{}60^{\\circ} \\Longleftrightarrow s\\equal{}\\sqrt{3}(R\\plus{}r)$ ($ s$-semiperimeter, $ R$-radius of the circumcircle, $ r$-radius of the incircle).",
  "Solution_1": "[hide=\"Solution\"]\n\nLet's trig it! \n\n$ s\\equal{}\\frac{a\\plus{}b\\plus{}c}{2}\\equal{}R(\\sin A\\plus{}\\sin B\\plus{}\\sin C)$\n$ \\sqrt3(R\\plus{}r)\\equal{}\\sqrt3R(\\cos A\\plus{}\\cos B\\plus{}\\cos C)$\n\nSo, it's left to prove $ \\sin A\\plus{}\\sin B\\plus{}\\sin C\\equal{}\\sqrt3(\\cos A\\plus{}\\cos B\\plus{}\\cos C)$. As $ A\\equal{}60^\\circ$, it's left to prove $ \\sin B\\plus{}\\sin C\\equal{}\\sqrt3(\\cos B\\plus{}\\cos C)$,with $ B\\plus{}C\\equal{}120^\\circ$\n\nwhich is easy if I didn't make a mistake. [/hide]",
  "Solution_2": "[hide=\"Solution 2\"]\nThe ecuation $ f(x)\\equal{} s\\cdot x^3\\minus{}(4R\\plus{}r)\\cdot x^2\\plus{}s\\cdot x\\minus{}r\\equal{}0$ has the solutions $ \\left\\{\\ \\tan\\frac A2\\ ,\\ \\tan\\frac B2\\ ,\\ \\tan\\frac C2\\ \\right\\}$.\n\nHence $ A\\equal{}60^{\\circ}\\ \\Longleftrightarrow\\ f\\left(\\frac{\\sqrt 3}{3}\\right)\\ \\equal{}\\ 0\\ \\Longleftrightarrow\\ \\boxed{s\\equal{}\\sqrt{3}(R\\plus{}r)}$\n[/hide]",
  "Solution_3": "Does anyone have a non-brute-forcing way of proving $ 1\\plus{}r/R\\equal{}\\cos A\\plus{}\\cos B\\plus{}\\cos C$ (or maybe derive it)? I can only prove it by expressing everything in terms of the side lengths. \r\nAlso Mateescu Constantin how did you come up with that nice identity (involving tangents)? Thanks.",
  "Solution_4": "[quote=\"Mateescu Constantin\"][hide=\"Solution 2\"]\nThe ecuation $ f(x) \\equal{} s\\cdot x^3 \\minus{} (4R \\plus{} r)\\cdot x^2 \\plus{} s\\cdot x \\minus{} r \\equal{} 0$ has the solutions $ \\left\\{\\ \\tan\\frac A2\\ ,\\ \\tan\\frac B2\\ ,\\ \\tan\\frac C2\\ \\right\\}$.\n\nHence $ A \\equal{} 60^{\\circ}\\ \\Longleftrightarrow\\ f\\left(\\frac {\\sqrt 3}{3}\\right)\\ \\equal{} \\ 0\\ \\Longleftrightarrow\\ \\boxed{s \\equal{} \\sqrt {3}(R \\plus{} r)}$\n[/hide][/quote]\r\n\r\nCan someone explain how he did that ?",
  "Solution_5": "[quote=\"Mateescu Constantin\"]Let $ ABC$ be a triangle. Prove that $ \\angle A\\equal{}60^{\\circ} \\Longleftrightarrow s\\equal{}\\sqrt{3}(R\\plus{}r)$ ($ s$-semiperimeter, $ R$-radius of the circumcircle, $ r$-radius of the incircle).[/quote][b][u]Remarks[/u].[/b] \n\n$\\blacktriangleright$ The equivalence $\\boxed{A=60^{\\circ}\\iff s=(R+r)\\sqrt 3}$ is false because the implication \n\n$\\boxed{A=60^{\\circ}\\implies s=(R+r)\\sqrt 3}$ is true and $\\boxed {s=(R+r)\\sqrt 3\\implies A=60^{\\circ}}$ is false.\n\n$\\blacktriangleright$ This equivalence $\\boxed{60^{\\circ}\\in\\{A,B,C\\}\\iff s=(R+r)\\sqrt 3}$ is true. Indeed, prove easily that \n\n$\\left\\{\\begin{array}{c}\n\\tan\\frac A2+\\tan\\frac B2+\\tan\\frac C2=\\frac {4R+r}{s}\\\\\\\\\n\\tan\\frac A2\\tan\\frac B2+\\tan\\frac A2\\tan\\frac B2+\\tan\\frac A2\\tan\\frac B2=1\\\\\\\\\n\\tan\\frac A2\\tan\\frac B2\\tan\\frac C2=\\frac rs\\end{array}\\right\\|\\implies$ $s\\cdot \\prod\\left(1-\\sqrt 3\\cdot\\tan\\frac A2\\right)=4\\left[s-(R+r)\\sqrt 3\\right]$ .\n\n[b][u]Other method[/u].[/b] Prove easily that $\\boxed{x+y+z=0\\implies \\sum\\sin x=-4\\prod\\sin\\frac x2}\\ (*)$ . Therefore, $s=(R+r)\\sqrt 3\\iff$ \n\n$\\frac sR=\\sqrt 3\\cdot\\left(1+\\frac rR\\right)\\iff$ $\\sum\\sin A=\\sqrt 3\\cdot\\sum\\cos A\\iff$ $\\sum\\sin\\left(A-60^{\\circ}\\right)=0$ . For  $\\left\\{\\begin{array}{c}\nx:=A-60^{\\circ}\\\\\\\ny:=B-60^{\\circ}\\\\\\\nz:=C-60^{\\circ}\\end{array}\\right\\|$ , where \n\n$x+y+z=0$ apply the identity $(*)$ . In conclusion, $s=(R+r)\\sqrt 3\\iff$ $\\prod\\sin\\left(\\frac A2-30^{\\circ}\\right)=0\\iff$  $60^{\\circ}\\in\\{A,B,C\\}$ .",
  "Solution_6": "[hide=\"Another way\"]\nPut:\n$y+z=BC, x+z=AC, x+y=AB$\nwhere $x,y,z$ are some positive real numbers. Then we must have:\n$y+z=2R\\sin 60 = R\\sqrt 3$ by the law of sines and $\\tan 30 = \\frac{1}{\\sqrt 3}= \\frac{r}{x}$. \nSolving for $r$ and $R$ and noting that $x+y+z=s$ we get exactly the desired relationship.\n[/hide]"
}
{
  "Tag": [],
  "Problem": "How many positive integer solutions exist for $ 3(x\\minus{}5)\\le 7$",
  "Solution_1": "If 3(x-5) is less than or equal to 7, then divide by 3 and get x-5 is less than or equal to 7/3.  Add 5 to both sides and get x is less than or equal to 7/3+5=7/3+15/3=22/3.  The possible values of x are 1,2,3,4,5,6 and 7, so the answer is [b]7.[/b]"
}
{
  "Tag": [
    "LaTeX"
  ],
  "Problem": "Hi,\r\n\r\nI have been trying to create a letter using LaTeX and have come across a few problems. The code I have used is:\r\n\r\n[code]\\documentclass{letter}\n\\address{Something Road \\\\ England \\\\ M12 2PL}\n\\signature{Beans}\n\n\\begin{document}\n\\begin{letter}{md}\n\\opening{Dear Mr. Wolf}\n\nBlah \n\n\\closing{Yours sincerely}\n\\end{letter}\n\\end{document}[/code]\n\nThe problem I face is that the \"Yours sincerely\" part is centred and I want it towards the left--can I do this?\n\nAlso, I don't want to write the senders address, but if I don't put anything down then my own address won't get printed! Help!\n\n(I tried googling templates for letters, but alas I didn't come up with anything better).[/code]",
  "Solution_1": "Hi, Beans\r\n\r\n1. Don't bother with \\closing just type Yours sincerely. Similarly you don't have to use \\signature if you don't want to.\r\n2. Instead of \\begin{letter}{md} use \\begin{letter}{} \r\n\r\nIncidentally, the author of the TeX FAQ [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=letterclass]Letters and the like[/url] also has had problems with the letter class so you are in good company. However it's worth sticking with for a normal bog-standard letter.",
  "Solution_2": "[quote=\"stevem\"]Hi, Beans\n\n1. Don't bother with \\closing just type Yours sincerely. Similarly you don't have to use \\signature if you don't want to.\n2. Instead of \\begin{letter}{md} use \\begin{letter}{} \n\nIncidentally, the author of the TeX FAQ [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=letterclass]Letters and the like[/url] also has had problems with the letter class so you are in good company. However it's worth sticking with for a normal bog-standard letter.[/quote] Thanks a lot--it worked a treat. \r\n\r\nI was wondering if there was any latex package/template that you would recommend for creating a CV? (The joys of looking for a summer job...)",
  "Solution_3": "There are a number of CV packages on [url=http://tug.ctan.org/search.html#byDescription]CTAN[/url]. Unfortunately that's down at the moment but have a look when it's back. Do be aware that different countries tend to use different styles for their CVs.",
  "Solution_4": "CTAN is back again and you'll find a number of CV packages at http://tug.ctan.org/cgi-bin/search.py?metadataSearch=vitae or http://texcatalogue.sarovar.org/bytopic.html#applications",
  "Solution_5": "Thanks Steve! I hadn't thought about that and instead had been using Google (unsuccessfully might I add) and found some not very nice packages."
}
{
  "Tag": [
    "function",
    "abstract algebra",
    "group theory",
    "Ring Theory",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "Recently, I have come up with an algebraic structure that I would like to call \"Partial rings\" with some interesting properties. They are similar to rings, as will be shown later. The rules for this structure are:\r\n\r\n1) Group under addition\r\n2) Another function, multiplication \r\n3) mutiplication is distributive\r\n4) multiplication is associative\r\n\r\n5 (optional)) Noncommutative group under addition.\r\n\r\nSome properties already found:\r\n1) The identity under addition annihilates under multilplication (clearly)\r\n2) The additive order of the product a*b divides the additive order of each a and b\r\n3) The set of products is commutative\r\n4) The set of products a*r, r is in R is a commutative group\r\n5) If the order of the \"partial ring\" is squarefree, then the set of products is cyclic\r\n\r\nExample:\r\n[code]+    | e    a    2a   b    b+a  b+2a\n_______________________________________\ne    | e    a    2a   b    b+a  b+2a\na    | a    2a   e    b+2a b    b+a\n2a   | 2a   e    a    b+a  b+2a b\nb    | b    b+a  b+2a e    a    2a\nb+a  | b+a  b+2a b    2a   e    a\nb+2a | b+2a b    b+a  a    2a   e\n\n*    | e    a    2a   b    b+a  b+2a\n______________________________________\ne    | e    e    e    e    e    e\na    | e    e    e    e    e    e\n2a   | e    e    e    e    e    e\nb    | e    e    e    b    b    b\nb+a  | e    e    e    b    b    b\nb+2a | e    e    e    b    b    b[/code]\n\nProof of properties:\n\n1) 0*a = 0*a + 0*a + -(0*a) = (0+0)*a + -(0*a) = 0*a + -(0*a) = 0\n2) ab+ab+ab+...+ab (o(a) times) = (a+a+a+a+...+a)b = eb = e, so o(ab)|o(a). Similar argument for b.\n3) -ac + (a+b)(c+d) + -bd = -ac + (a+b)c + (a+b)d + -bd= -ac + ac + bc + ad + bd + -bd = bc + ad = -ac + a(c+d) + b(c+d) + -bd = -ac + ac + ad + bc + bd + -bd = ad + bc.\n4) Commutative has already been shown. All we need show is closure and inverse to show it is a group. a*r1 + a*r2 = a*(r1+r2), so it is closed. a*r1 + a*-r1 = a*(r1 + -r1) = a*e = e, so we have inverse and therefore identity, so a*r is a group.\n5) This one is long. It shall be proven later.\n\nOther properties:\nThe 0th and 1st Isomorphism theorem works in these partial rings, that is:\n0th: The kernel of any \"partial ring\" homomorphism is a partial normal ideal, that is, it is a normal subgroup under addition and is an ideal under multiplication. Any partial normal ideal N of R also has a homomorphism to R/N with kernel N.\nThe definition of a \"partial ring\" homomorphism is that it takes sums to sums and products to products.\n1st: If you have an onto ring homomorphism from R to R' with kernel N, then R/N is isomorphic to R'.\nI believe that this also implies the second and third isomorphism theorems, namely, H+N/N is isomorphic to H/H intersect N for any subring H and normal partial ideal N of R, and also that G/N1 = (G/N2)/(N1/N2) where N2 and N1 are normal partial ideals of G, and N2 is in N1.\nThings found so far:\nAn 8-element partial ring with noncommutative addition and multiplication\nA 32-element ring without identity such that the set of products is not a group (not closed under addition)\nCan you guys think of anything else about it?\nNote: I am uncool on scienceforums.net, so if you see it there, it's mine.\nbmsiv[/code]",
  "Solution_1": "[quote=\"bmsiv\"]The rules for this structure are:\n\n1) Group under addition\n2) Another function, multiplication \n3) mutiplication is distributive\n4) multiplication is associative[/quote]\r\nIsn't this just the definition of a (noncommutative) ring (without unity).",
  "Solution_2": "[quote=\"ZetaX\"][quote=\"bmsiv\"]The rules for this structure are:\n\n1) Group under addition\n2) Another function, multiplication \n3) mutiplication is distributive\n4) multiplication is associative[/quote]\nIsn't this just the definition of a (noncommutative) ring (without unity).[/quote]\r\n\r\nNot quite addition is not necessarily commutative.",
  "Solution_3": "[quote=\"qleak\"][quote=\"ZetaX\"][quote=\"bmsiv\"]The rules for this structure are:\n\n1) Group under addition\n2) Another function, multiplication \n3) mutiplication is distributive\n4) multiplication is associative[/quote]\nIsn't this just the definition of a (noncommutative) ring (without unity).[/quote]\n\nNot quite addition is not necessarily commutative.[/quote]\r\nCorrect, as shown by the group that I displayed (which is S3 under addition, which is noncommutative). \r\nI believe that each ring can be simplified as follows:\r\nLet N = {a: a*r = r*a = e for all r in R}. Then R/N is a ring (that is, R/N is commutative). This occurs for any partial ring that is a whole ring to begin with, and for any partial ring with order pq, p and q prime, p =/= q.\r\n=Uncool-",
  "Solution_4": "There are currently two major theorems left that I have learned that deal with rings that do not include unity. The first one is that all of them can be embedded in a ring with unity. This clearly is not true for this structure, as if it were embedded, it would be commutative, as shown above. Therefore, let us look at the other theorem that I know of - the Chinese Remainder Theorem.\r\n\r\nThe Chinese Remainder Theorem, over rings, states the following: \r\n(from wikipedia on chinese remainder theorem)\r\nIf R is a ring (with unity) and I1, ..., Ik are two-sided ideals of R which are pairwise coprime (meaning that Ii + Ij = R whenever i \u2260 j), then the product I of these ideals is equal to their intersection, and the quotient ring R/I is isomorphic to the product ring R/I1 x R/I2 x ... x R/Ik\r\n\r\nThe proof uses the fact that R/A  x R/B is isomorphic to R/AnB, and that AnB is equal to AB, where the product is defined as the set of all finite sums of products of elements of the two rings.\r\n\r\nThe proof that R/A x R/B is isomorphic to R/AnB is the same. However, we can see that even in rings without unity, there are examples where AnB =/= AB. An example of this can be seen in the ring 6Z. Let A = B = 6Z. Then AnB = 6Z, but AB = 36Z. Therefore, all we have for both rings without unity and partial rings is that R/AnB = R/A x R/B. This is still a substantial result.\r\n\r\nThe only other proposition that may be important is how to get all of these from rings without unity. I have already explained how I believe you can.\r\n=Bmsiv-",
  "Solution_5": "Finally found how to do it.\r\nLet N = {x + y - x - y}. Then for any n in N, an = ax + ay - ax - ay = ax - ax + ay - ay = 0. One of the problems in the book by Herstein, this is a normal group. Clearly, it is a partial normal ideal, as all products are 0, which is in the group.\r\n\r\nR/N is commutative, as for any a + N and b + N: a+N + b + N = (a+b)+N = (a + b - a - b + b + a) + N = (a+b-a-b)+N + (b+a)+N = N + (b+a) + N = N.\r\n\r\nTherefore, we have that R can be created using a ring (without unity) and a normal group such that all the products of the normal group are 0.",
  "Solution_6": "A slight amendment to above, replace N with the commutator subgroup. After that, the partial ring is made trivial."
}
{
  "Tag": [
    "blogs",
    "\\/closed"
  ],
  "Problem": "I went to page 26 of the blogs section, and the attached picture shows what was displayed. \r\n\r\nAre there blogs that some users cannot see? Or is this a bug?",
  "Solution_1": "Ask one of the administrators. (For me, it says the same thing).",
  "Solution_2": "My guess - the \"754 blogs\" it says enumerates all blogs, including blogs with the \"Viewable only to friends/contribs/self\" set. (Why someone would create a blog just to view themself is beyond me, but whatever.)",
  "Solution_3": "[quote=\"james4l\"]My guess - the \"754 blogs\" it says enumerates all blogs, including blogs with the \"Viewable only to friends/contribs/self\" set. (Why someone would create a blog just to view themself is beyond me, but whatever.)[/quote]\r\nAn online journal.\r\n\r\nWell, I don't see anything at all using the MathLinks board style."
}
{
  "Tag": [],
  "Problem": "Find positive integers a and b such that: \r\n(cuberoota + cuberootb - 1) <sup>2</sup>  = 49 + 20cuberoot6",
  "Solution_1": "[hide]48, 288[/hide]"
}
{
  "Tag": [
    "real analysis",
    "limit",
    "real analysis solved"
  ],
  "Problem": "$a_n \\; b_n$ reals sequences s.t. $\\lim a_n = +\\infty$\r\n\r\nProve that in any $]a;b[$ nonempty open  interval there exixt $x_0$ \r\n\r\ns.t. $\\limsup_{n\\rightarrow +\\infty} \\cos(a_nx_0+b_n)=1$",
  "Solution_1": "Nice problem, I hope the solution is good. Suppose the contrary and take the closed sets $ A_{k,p}={x\\in (a,b), cos(a_nx+b_n)\\leq 1-\\frac{1}{p}}$ if ${n\\geq k}$. Here k,p are natural numbers. The union of these sets contains an interval and this union is numerable thus by Baire's theorem one of them contains an interval. Thus there exists an interval non trivial I and two numbers k,p such that $ cos(a_nx+b_n)\\leq 1-\\frac{1}{p}$ for all $n\\geq k$. This means that the length of the interval $ a_n I+b_n$ is always smaller than $2\\pi$, which is impossible for large n."
}
{
  "Tag": [
    "probability",
    "MATHCOUNTS"
  ],
  "Problem": "Two integers are randdomly selected from the set of integers greater than or equal to -5 and less than or equal to 5.  The two numbers need not be different.  What is the probability that the sum of the two integers is less than the product?",
  "Solution_1": "[quote=\"ccu\"]Two integers are randdomly selected from the set of integers greater than or equal to -5 and less than or equal to 5. The two numbers need not be different. What is the probability that the sum of the two integers is less than the product?[/quote][hide]\n\nThey must either be 1.both negative 2. both positive 3. negative and zero\n\nso -5 -4 -3 -2 -1 0 1 2 3 4 5\n\nThe denominator is 55.\n\nThe top part is 1.10 2.10 3.5\n\nSo it is 15/55 which simplifies to 3/11. If I didn't make a mistake.[/hide]",
  "Solution_2": "I just did this problem, it was number 30 of some sprint round, and I got [hide]$\\frac{50}{121}$[/hide] which was correct. :D [hide]The way you solve this is:\n$11^2=121$, this is the denominator.  You get 25 possible ways to solve if they are all negative, 20 negative/positive/zero, and 5 only positive, or something like that.   $20+25+5=50, \\frac{50}{121}=Answer$[/hide]",
  "Solution_3": "That's strange as I wouldn't put the denominator as 121 as that would count pairs like 4,5 twice (once as 5,4, once as 4,5) as well as any other pair like that. I would put a denominator of 66 as there are 11 pairs which only count once so subtract 11 from 121 and divide by 2 to get 55. Add the 11 back on and you have 66. Another looooooong painstaking way to get 66 is there are 11 pairings with -5, 10 with -4 (As -4,-5 is already counted), 9 with -3 (-3,-4 is already counted, so is -3,-5) etc. and if you add that up, you get 66. \r\n[hide]\nI got 29/66 as my answer. \n6 combinations of the 11 pairings work out with -5, 5 combinations of the 10 pairings with -4, 4 of the 9 with -3, 3 of the 8 with -2, and 2 of the 7 with -1. That adds up to 20. Now none work with 0 (product = 0, sum = n), none work with 1. (product is always n, sum is always n+1) Only 3, 4, and 5 work for 2 giving you 23 total so far. All of the options of 3, 4, and 5, work giving you 6 more for a total of 29 making 29/66.\n[/hide]",
  "Solution_4": "Well, the mathcounts answer key said that I was right...and they had my solution also... :D  EDIT:  You actually have to double count some pairs, as they represent different outcomes.  Example...picking 1 first and then -2 is different that picking -2 first and then 1.   There are therefore $11^2$  Possible outcomes",
  "Solution_5": "[hide]\nThere are 11 numbers, and thus $66$ combos (11+10+...+1)\n\nYou just have to list them out\n\n-5 works with\n-5,-4,-3,-2,-1,0\n\n-4 works with\n-4,-3,-2,-1,0\n\n-3 works with\n-3,-2,-1,0\n\n-2 works with\n-2,-1,0\n\n-1 works with \n-1,0\n\n0 works with none\n\n1 works with none\n\n2 works with \n3,4,5\n\n3 works with \n3,4,5\n\n4 works with\n4,5\n\n5 works with \n5\n\nTotal to get 29\n$\\frac{29}{66}$[/hide]",
  "Solution_6": "ccy, could you post the correct answer?  We have 2 different answers so far.",
  "Solution_7": "jhollenbeck, I am fairly sure that your answer is incorrect, because you said that there are 11*11=121 different combinations.\r\n\r\nThat would pair each number with each other number, but you would count each twice.\r\nFor example, for -5 you pair with -5,-4,-3,-2,-1,0,1,2,3,4,5\r\nFor -4, you pair with -5,-4,-3,-2,-1,0,1,2,3,4,5\r\n\r\nAs you can see, you pair -4 with -5 twice above.\r\n\r\nSo, you have to do it such that you pair -5 with -4,-3,-2,-1,0,1,2,3,4,5\r\nYou pair -4 with -4,-3,-2,-1,0,1,2,3,4,5\r\nYou pair -3 with -3,-2,-1,0,1,2,3,4,5\r\n...\r\nTo remove doubles",
  "Solution_8": "actually, according to the answersheet, jhollenbeck's right. It was like 15 of some state sprint."
}
{
  "Tag": [
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "Consider a set $ S$ of $ 2006$ distinct numbers.A subset $ T$ of $ S$ is called stubborn if for every $ u,v$ (not necessarily distinct) in $ T$ ,the number $ u\\plus{}v$ does not belong to $ T$. Prove that If $ S$ consists of $ 2006$ arbitrary positive integers then there exists a stubborn subset $ T$ of $ S$ having $ 669$ elements",
  "Solution_1": "http://www.mathlinks.ro/viewtopic.php?p=20031#20031 .",
  "Solution_2": "Any solutions please   :blush:  :blush: ??",
  "Solution_3": "Please download following file:"
}
{
  "Tag": [
    "search"
  ],
  "Problem": "cSplash program held at NYU in March, where for a day they offer math and mathematical science courses to high school students free of charge at NYU. \r\n[quote=\"[url=http://www.cims.nyu.edu/~csplash/index.php]cSplash[/url]\"]\nCourant Splash! is a one-day festival of classes in the mathematical and computer sciences, designed and taught by enthusiastic graduate and undergraduate students, faculty, and others associated with the Courant Institute. cSplash! is an opportunity to get your feet wet in subjects completely new to you. During the program, you can attend as many talks as you want on topics in pure math, applied math, computer science, and related fields. The program is free and is open to all students in grades 9-12 (or with equivalent mathematical background) with an interest in mathematical and computer sciences. Spaces are limited and are allocated on a first-come, first-serve basis so please register early to guarantee a spot. \n[/quote]\r\n\r\nFound out about it last year and went to it, it was quite fun.  :)",
  "Solution_1": "It was briefly mentioned in the NYC forum [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=54583303&t=75941]last year[/url].",
  "Solution_2": "Since no one is really posting here, I though I would mention that registration for classes starts in February."
}
{
  "Tag": [
    "geometry",
    "geometry open"
  ],
  "Problem": "I have a couple of questions about circle squaring sorry if this is in the wrong section.\r\n\r\nI understand that it has been proven impossible with a pencil ruler and compass, but I wanted to find the simplest way to get as close as possible.\r\n\r\nI think I can draw a square to match the area of a given circle as accuratly as I can draw another circle with the same area.\r\n\r\nI found that if you draw a square within a circle the circle takes up 57% more area than the square, so would it not be possible to simple draw  a square with 57% more area? matching the area of the circle.\r\n\r\nSo you would square the circles radius, divide by two and square root the result, double this and square it to give you the area of the square drawn within the circle, just pythagoras really.\r\n\r\nThen if this is a then (a/100*57 ) + a is the total area of the square matching the circle,\r\n\r\nDepending on how many decimal places you want to work to you can draw a square with the same area as a given circle as accuratly as you can redraw another circle with the same area.\r\n\r\nI have put together a very crude spreadsheet to demonstrate this.\r\n\r\nhttp://www.mediafire.com/?zq3xtqnkm5m\r\n\r\nThe only problem is that it shows small differences between the area of the square produced and the circle is this due to the computer only working to a limited number of decimal places? \r\n\r\nAny comments are welcome,\r\n\r\n:)",
  "Solution_1": "I think this may be in the wrong section please could a moderator move it to the right place?"
}
{
  "Tag": [
    "inequalities proposed",
    "inequalities"
  ],
  "Problem": "Find the greast value of $\\frac{a}{c}+\\frac{b}{d}+\\frac{c}{a}+\\frac{d}{b}-\\frac{abcd}{(ab+cd)^{2}}$ , where $a,b,c,d$ are distinct reals satisfying $\\frac{a}{b}+\\frac{b}{c}+\\frac{c}{d}+\\frac{d}{a}=4$ and $ac=bd$.",
  "Solution_1": "My solution.",
  "Solution_2": "[quote=\"toanhocmuonmau\"]My solution.[/quote]\r\nCh\u01b0a h\u1ebft h\u1ea1n g\u1eedi b\u00e0i .. n\u00ean \u0111\u1eebng th\u1ea3o lu\u1eadn\r\n\r\n[b]thay 4 b\u1edbi k...............[/b]",
  "Solution_3": "[quote=\"VANCHANH123\"][quote=\"toanhocmuonmau\"]My solution.[/quote]\nCh\u01b0a h\u1ebft h\u1ea1n g\u1eedi b\u00e0i .. n\u00ean \u0111\u1eebng th\u1ea3o lu\u1eadn\n\n[b]thay 4 b\u1edbi k...............[/b][/quote]\r\nNo, It is exceed the limit  :wink: . I not so, I will not post."
}
{
  "Tag": [
    "geometry",
    "geometric transformation",
    "dilation",
    "ratio"
  ],
  "Problem": "I need help on the following question and advice on how to approach similar types of the questions shown below or recommended sources to use. I could not seem to identify some of the substances. \r\n\r\nDilute hydrochloric acid was added to a metallic looking compound A (molecular weight 90.756). A colorless gas B with a characteristic, unpleasant odour was formed together with a pale green solution of the cation C. \r\n\r\n   The gas B was burned in air to give another colourless gas D that turned yellow dichromate paper green. Mixing B and D gave a yellow solid element E. Depending on the mole ratios, E reacted with chlorine gas to give two chlorides, F and G in addition to hydrogen chloride (HCl). \r\n\r\n    Addition of zinc metal to a sample of the green cation solution C gave a metal H (with electron configuration of [Ar]3d84s2. \r\n\r\nIdentify each of the substances and write balanced chemical equation for each reaction. \r\n\r\nThanks in advance.",
  "Solution_1": "[quote=\"Victornumber\"]Addition of zinc metal to a sample of the green cation solution C gave a metal H (with electron configuration of [Ar]3d84s2.[/quote]\n\nFrom the electron configuration we may conclude that H is nickel, and so $C = Ni^{2+}$. The observed reaction of C with zinc is in accordance with the reduction potentials of these two metals: $E^{0}Ni^{2+}/Ni =-0.23V$, $E^{0}Zn^{2+}/Zn =-0.76V$. This is also in accordance with the fact that Ni2+ ions in aqueous solution form the complex ions $[Ni(H_{2}O)_{6}]^{2+}$, which are responsable for the green color of their solution.\n\n[quote=\"Victornumber\"]Dilute hydrochloric acid was added to a metallic looking compound A (molecular weight 90.756). A colorless gas B with a characteristic, unpleasant odour was formed together with a pale green solution of the cation C.[/quote]\n\nThe unpleasant odour of the gas suggests that it might contains sulphur, and so compound A is NiS, nickel sulphide, which is in accordance with the given molecular weight. When this compound reacts with acid, there is the formation of sulphidric acid gas, $H_{2}S$, which is B.\n\n[quote=\"Victornumber\"]The gas B was burned in air to give another colourless gas D that turned yellow dichromate paper green. Mixing B and D gave a yellow solid element E.[/quote]\n\nThat sequence of reactions is known as [i]Claus Process[/i], which is used in oil industry to remove H2S from natural gas or from some destilation fractions of crude, turning it into elemental sulphur. The first reaction, where H2S (B) is burned in air, is\n\n$2H_{2}S(g)+2O_{2}(g) \\longrightarrow SO_{2}(g)+2H_{2}O(g)+S(l)$,\n\nand so D is SO2, sulphur dioxide. The second reaction, where H2S reacts with SO2 (usually in the presence of a catalyst) is,\n\n$2H_{2}S(g)+SO_{2}(g) \\longrightarrow 3S(l)+2H_{2}O(g)$.\n\nSo E = sulphur. The \"gas D that turned yellow dichromate paper green.\" part means that SO2 was able to reduce dichromate to the green ions Cr3+. Write this reaction as an exercise.\n\n[quote=\"Victornumber\"]Depending on the mole ratios, E reacted with chlorine gas to give two chlorides, F and G in addition to hydrogen chloride (HCl).[/quote]\r\n\r\nSulphur reacts with chlorine to form disulphur dichloride, $S_{2}Cl_{2}$, a yellow liquid used in the vulcanization of rubber. When S2Cl2 reacts with more chlorine, sulphur dichloride $SCl_{2}$ is formed as a red liquid. This last one was used in the production of [i]mustard gas[/i] by reaction with ethene.",
  "Solution_2": "[quote=\"Victornumber\"]I need help on the following question and advice on how to approach similar types of the questions shown below or recommended sources to use.[/quote]\r\n\r\nFor example:\r\n\r\n1. Knowledge of electron configurations and how to relate them with the Periodic Table.\r\n2. Knowledge of redox reactions and how to balance them.\r\n3. Common knowledge that, in general, sulphur compounds have bad odours."
}
{
  "Tag": [
    "trigonometry",
    "algebra",
    "functional equation",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "solve the differential equation:\r\nf:R->R, \r\nf(x)=f'(f(x))\r\n\r\n f(x)=x is valid, but i can't prove that :?",
  "Solution_1": "[quote=\"mrazvano\"] f(x)=x is valid[/quote]\r\nNo, it's not. If $f(x)=x$, then $f'(x)=1,$ so $f'(f(x))=1,$ which is not $f(x).$\r\n\r\nFor a solution that is obviously valid, try $f(x)=\\frac{x^2}2.$ But that's not the only solution. This is really more of a functional equation than a typical differential equation.",
  "Solution_2": "If $f$ is differentiable at $Im(f) = I$ an interval then:\r\n\r\n a. If $I$ is one point then $f = 0$\r\n b. If not for $u \\in I$ we have $u= f'(u)$ then $f = \\frac{u^2}{2} +c$ on $I$\r\n\r\n So, define $f$ as in b on an interval $I$.  The solution then depends on the way we expand $f$ contiuously on $\\mathbb{R}$ and differentiably at the end points of $I$ so that $Im(f) = I$( but not very arbitrarily as it sounds )\r\n\r\n- For example :\r\n  $f(x)$ = $x^2$ for $x \\geq 0$\r\n  $f(x)$ = $x^4$  ( or $\\sin^2(x)$ )for $x \\leq 0$ \r\n   $I$ = $[0, \\infty)$\r\n\r\n But it is quite hard to consider all cases!!!",
  "Solution_3": "I think, it is clear that $I\\neq (a,b)$, there $a,b\\in \\mathbb{R}$, isn't it? Otherwise $Im(f)\\neq I$."
}
{
  "Tag": [
    "geometry"
  ],
  "Problem": "RMO result of Chhattisgarh are out. I am selected. Thanks to mathlinks for making me prepare well.",
  "Solution_1": "Congratulations!!! :D",
  "Solution_2": "from where did you get to know of it....??? any site??? \r\n\r\n\r\nAnd of course.... Congrats...",
  "Solution_3": "Which class are u in?",
  "Solution_4": "Even the Delhi Results are out !! I have also been selected.. Thanks to all of you for helping me, and clearing all my doubts  !!",
  "Solution_5": "Which site??????????????????????????????????????????????",
  "Solution_6": "Well they Called me up and informed that i have been selected and have to attend a camp next weekends. I don't think that the results have been put up on an y site....BY THE WAY ARE YOU FROM DELHI REGION???\r\n\r\n\r\n\r\n\r\n\r\n*** INMO IS ON  JAN 18 ,2009",
  "Solution_7": "oh yes....... but never mind... I didnt make it......",
  "Solution_8": "@amangupta,I telephoned MP RMO co=ordinator and he said that due to counting of votes, results will take another 8-10 days. SO, can u tell me what is ur expected score seeing the official solutions!!",
  "Solution_9": "[quote=\"fbaggins\"]@amangupta,I telephoned MP RMO co=ordinator and he said that due to counting of votes, results will take another 8-10 days. SO, can u tell me what is ur expected score seeing the official solutions!![/quote]\r\nHi fbaggins. Just in case you were wondering the actual RMO scores you get does not depend on whether your solution is matching to the official solution. All that matters is whether your proofs are rigorous and you get the same final answer (matching with the official answer). Since  you got all atleast 3 question fully right don't be so anxious. (I understand the tension you must be  feeling though to know if you cleared RMO but there is a 99% chance that you will clear RMO, becuase you are in ninth grade.)",
  "Solution_10": "@befuddlers;\r\n\r\nThanks very much for motivating me but i feel more depressed now because I got only 2 fully right!!!!!! Q5 and Q6!!!!!!\r\n\r\nNow plz tell if i have any chance!\r\n\r\nBTW , it is my personal opinion that being in class IX doesn't help very much in clearing RMO!!",
  "Solution_11": "fbaggins, how does the counting of votes have relationship with the RMO results? \r\nSee befuddlers, it is not like 9th stds have an edge. It is the realtive performance. If the other 9th stds have given a mediocre performance and u have solved 2 with excellent presentation(I suppose-fbaggins) , U will surely get through.It is like tht. If the other 9thers have done horribly and 1 medium kinda performance, he has a better chance. Its like that. :D   :rotfl:",
  "Solution_12": "well! @ aravind,\r\nwell i was also in doubt as to what is the connection of counting of votes with RMO?But my dear frnd,\r\nvotes are counted by the teachers of the sise institute where RMO papers are checked!So u see.............",
  "Solution_13": "[quote=\"fbaggins\"]@amangupta,I telephoned MP RMO co=ordinator and he said that due to counting of votes, results will take another 8-10 days. SO, can u tell me what is ur expected score seeing the official solutions!![/quote]\r\n\r\nI was able to get 3.5 questions correct .however i attempted 4.5 questions.I am not sure how much scores i would have scored . \r\nBy the way, being able to solve 2/6 question while still being in ninth grade surely separates you from the crowd.\r\n\r\nAravind has rightly said, it is all about relative performance.There's always hope and chances are that you too might get selected  .moreover i am sure that you will do much much better the next time.\r\n :wink:  :)",
  "Solution_14": "Aravind , can you please suggest me some books / sites etc. form where i could prepare for INMO . Coz' if i am not mistaken , you are an INMO Awardee and must be aware of it . SO , can you please  help me ???",
  "Solution_15": "[quote=\"Akashnil\"]There's no pressure if you don't care about marks, they won't give assignments etc. too much.You have to study yourself. If you don't your marks will be bad that's all.[/quote]\r\n\r\n:Wow: thats really supercool man!!! :blush:",
  "Solution_16": "[quote=\"amangupta2\"] Yes I do. I have joined Vidyamandir Classes ,Delhi (You [/quote]\r\n\r\nEven i am in vidyamandir... Aim iit Group A....",
  "Solution_17": "[quote=\"anurags92\"][quote=\"amangupta2\"] Yes I do. I have joined Vidyamandir Classes ,Delhi (You [/quote]\n\nEven i am in vidyamandir... Aim iit Group A....[/quote]\r\nWow !! That's Great , I am in Regular JEE 2010 Batch 2 !! Great to Hear this .\r\n\r\nBye the way do you all know when will be the Results of ZIO 2009 declared !! They said that they will be out by 15th dec.???",
  "Solution_18": "I am not worried about zio at all coz i know i wopnt qualify at all... :|    :|",
  "Solution_19": "Well , there's some good news , I have cleared The ZIO 2009 from Delhi.....The results were out today.....I dIdn't Expect things to be this way...Bu it's Good",
  "Solution_20": "Hello! @amangupta,\r\n\r\nfirst RMO then ZIO, keep rocking!!!!!!!!!!!! :lol:  :D\r\n\r\nwell ! I am interested in ZIO but it is not held anywhere near! :(  \r\n\r\nSO.......................\r\n\r\n\r\nAnyway,  congrats again!!!!\r\n\r\nBTW, anyone else Qualified in this forum?",
  "Solution_21": "@fbaggins Thanks !!! \r\n\r\n Sorry !! I Do not Have any idea , The Result have been declared today itself  and not much people have posted since then.....\r\n\r\nBye  the way , which region do you live in ??\r\n :wink:",
  "Solution_22": "Gaurav Patil has got through.\r\nI haven't got through.",
  "Solution_23": "ny1 who knows whether or not rajasthan RMO results have been declared?",
  "Solution_24": "@amangupta,\r\nwell i live in a very very remote place called vindhyanagar in dist.-singrauli in madhya pradesh",
  "Solution_25": "@fbaggins , Oh Yes !! I have studies about Singrauli in 10th , If I am not wrong , it is famous for rich mineral deposits  , right???\r\n\r\n@rituraj , This  time In delhi INMO TC will be held for 4 days but In Two Distinct Phases.\r\n\r\n\r\nDoes anyone here know , whether any camp is organised before the INOI (like INMO TC for Maths) , If Yes, Then What all is taught there , and when is this camp organised ?????",
  "Solution_26": "well,correct amangupta,  Asia's LARGEST THERMAL POWER PLANT is located here,NTPC(National Thermal Power Corporation)!!!!",
  "Solution_27": "Wow That's Great !!",
  "Solution_28": "hey what is this zio u all are talking about?",
  "Solution_29": "ZIO stands for Zonal Informatics olympiad. for more info, go to http://www.iarcs.org.in"
}
{
  "Tag": [
    "\\/closed"
  ],
  "Problem": "look at the title",
  "Solution_1": "We're not sure.  It won't be soon, but hopefully within a year."
}
{
  "Tag": [
    "algorithm",
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "There are $ 63$ points arbitrarily on the circle $ \\mathcal{C}$ with its diameter being $ 20$. Let $ S$ denote the number of triangles whose vertices are three of the $ 63$ points and the length of its sides is no less than $ 9$. Fine the maximum of $ S$.",
  "Solution_1": "Lemma: consider a graph with $ 63$ vertices, each two connected by (one) blue or red edge. Suppose there are no [i]blue[/i] $ 7$-cliques in the graph. Define\n\n$ S_1 \\equal{} \\{V_1, V_2, ... V_{11}\\}$\n$ S_2 \\equal{} \\{V_{12}, V_{13}, ..., V_{22}\\}$\n$ S_3 \\equal{} \\{V_{23}, V_{24}, ..., V_{33}\\}$\n$ S_4 \\equal{} \\{V_{34}, V_{35}, ..., V_{43}\\}$\n$ S_5 \\equal{} \\{V_{44}, V_{45}, ..., V_{53}\\}$\nand\n$ S_6 \\equal{} \\{V_{54}, V_{55}, ..., V_{63}\\}$\n\nThen the maximimum number of [i]blue[/i] triangles is attained when two vertices are connected by a blue edge if and only if they lie in different $ S_i$'s (this distribuition is the same in Turan's theorem).\n\nProof: take a graph $ G$ and an arbitrary red clique $ C$. Take in $ C$ the vertex that lies in more blue triangles and call it $ V$. We can change our graph by making each vertex in $ C$ be linked with the vertices already linked with $ V$ (and only with this vertices). We are going to prove that this does not make a blue $ 7$-clique appear in the graph and does not decrease the number of triangles. First, notice that a blue triangle can obviously have at most one vertex in $ C$. Thus, since a blue $ 7$-clique with vertices in $ G \\minus{} C$ cannot have appeared (the edges in this graph are the same), any blue $ 7$-clique has exactly one vertex in $ C$. However, if we change this vertex by $ V$, the colors of the edges will be the same (by construction), therefore we had a $ 7$-clique before. Contradiction! On the other hand, the number of blue triangles with no vertices in $ C$ didn't change, and the number of blue triangles having a given vertex of $ C$ became the number of blue triangles having the vertex $ V$. By the maximality of $ V$, we get the result (there is no decrease).\n\nNow we are going to show an algorithm which makes any graph become the described graph (or a graph with less blue triangles), based on the operation $ O$ above. Take a red clique $ R$ (with at least $ 2$ vertices) and make the operation $ O$. If some vertex in $ R$ is connected with some vertex $ A \\in G \\minus{} R$ by a red edge, then all the vertices in $ R$ are. Thus, $ R \\bigcup \\{A\\}$ is a red clique. Make the operation $ O$ and repeat this procedure, until we get a red clique $ R_1$ such that any edge between $ R_1$ and $ G \\minus{} R_1$ is blue. Now we make the same thing with $ G \\minus{} R_1$, getting a red clique $ R_2$ and another graph $ G \\minus{} R_1 \\minus{} R_2$. Now there are no red edges between $ R_1$ and $ R_2$ or $ R_1$ and $ G \\minus{} R_1 \\minus{} R_2$ or $ R_2$ and $ G \\minus{} R_1 \\minus{} R_2$. We can make this until we get $ R_6$, at most (if we had $ 7$ red cliques as we said, taking a vertex in each of them, we would get a blue $ 7$-clique). Our algorithm only stops when all the remaining edges are blue (in particular, if $ |G \\minus{} R_1 \\minus{} ...| \\le 1$, i.e., if there are no edges in the remaining graph). In any case, in the end we will have a partition $ G \\equal{} R_1 \\bigcup ... \\bigcup R_k \\bigcup B$, where $ B$ is a (maybe empty) blue clique and each $ R_i$ is a red clique (with at least two vertices). Case $ |B| \\plus{} k \\ge 7$, it is obvious that there is a blue $ 7$-clique, which we are assuming false. Else, we can think the vertices in $ B$ as unitary red cliques, therefore having a partition $ G \\equal{} R_1 \\bigcup R_2 \\bigcup ... R_n$, with $ n \\le 6$. Moreover, if now we accept empty cliques, we may assume $ n \\equal{} 6$.\n\nNow the result is trivial! We just need to prove that, if $ a \\equal{} |R_1| > |R_2| \\plus{} 1 \\equal{} b \\plus{} 1$, moving a vertex from $ R_1$ to $ R_2$ is going to increase the number of blue triangles. This would imply that, in the maximum, the best distribuition for the points is the most balanced one. Moving a vertex $ X$ from $ R_1$ to $ R_2$ makes the blue triangles with vertices in $ X$ and some point in $ R_2$ disappear and the triangles with vertices in $ X$, some point in $ R_1$ and some point in $ R_3$, $ R_4$, $ R_5$ or $ R_6$ become blue. Since $ a \\minus{} 1 > b$, we have\n\n$ (a \\minus{} 1)(\\sum_{i \\equal{} 3, 4, 5, 6} |R_i|) > b(\\sum_{i \\equal{} 3, 4, 5, 6} |R_i|)$,\n\nid est, the number of triangles becomes greater.\n\n\n\nBack to the problem. Two vertices of the $ 63$-agon are going to be connected with a blue edge if and only if the distance between them is $ \\le 9$. The other pairs of vertices are going to be connected by a red edge. Suppose now we have a blue $ 7$-clique, say $ P_1P_2...P_7$, with $ P_i$ between $ P_{i \\minus{} 1}$ and $ P_{i \\plus{} 1}$. We know that an arc has always a length greater than the corresponding chord. Then the circle has length $ \\widehat{P_1P_2} \\plus{} \\widehat{P_2P_3} \\plus{} ... \\plus{} \\widehat{P_7P_1} > 9 \\plus{} 9 \\plus{} ... \\plus{} 9 \\equal{} 63 > 20\\pi$, absurd! It follows that the maximum number of triangles is obtained when $ P_1 \\approx ... P_{11} \\approx A$, $ P_{12} \\approx ... P_{22} \\approx B$, $ P_{23} \\approx... P_{33} \\approx C$, $ P_{34} \\approx ... P_{43} \\approx D$, $ P_{44} \\approx ... P_{53} \\approx E$ and $ P_{54} \\approx ... P_{63} \\approx F$ and $ ABCDEF$ is a regular hexagon (its side must be the radius of $ \\mathcal{C}$, that is $ 10$, that is $ > 9$, which means the maximum is really attained). It is easy to see that this number is $ 11^3 \\plus{} 9.11^2.10 \\plus{} 9.11.10^2 \\plus{} 10^3 \\equal{} 23121$.",
  "Solution_2": "Did somebody read my post?\r\n\r\nI want to know if it is right...",
  "Solution_3": "Hi Renan, I think that your proof is not right. I have the number 25515 to be the answer. The example would be when the 63 points are divided into 7 groups of 9 points each(the points in a group are closely together), 6 form a regular hexagon and the other is the center of the hexagon. It is easy to see that the example verify.\r\n\r\nYou made a great mistake when saying that the arc is greater to the corresponding chord leads to an absurd.\r\n\r\nMy proof is actually quite similar to that of the Turan Theorem. I will post it later (it is 0:00 and I am sleepy).",
  "Solution_4": "No,Cuenca,it says 63 points ON the circle.",
  "Solution_5": "But i think Feliz should connect a blue edge if and only if the distance >=9"
}
{
  "Tag": [],
  "Problem": "It is given $ A_1 A_2 A_3...A_n$ a polygon and it's diagonals such as it is imposible to find 3 diagonals that concur.Find the number of triangles having only one vertice of the polygon in their construction.(explain it)",
  "Solution_1": "Do you mean, \"Find the number of triangles whose edges belong to the diagonals of the $ n$-gon and exactly one of whose vertices is a vertex of the $ n$-gon\"?",
  "Solution_2": "[Deleted by Farenhajt]",
  "Solution_3": "I thought there were $ \\dbinom{n}{2} \\minus{} n \\equal{} \\frac {n(n \\minus{} 3)}{2}$ diagonals, and the number of the intersection points is $ \\dbinom{n}{4}$, so I do not think your result is correct.",
  "Solution_4": "It's absolutely incorrect, since I didn't put almost any thought to it. Deleted.",
  "Solution_5": "and now you can give a final response and as an extension: find in how many triangles a $ n\\minus{}$gon can be divided with help of its diagonals",
  "Solution_6": "[hide=\"Hopefully correct now\"]From each vertex (call it $ A$) we choose $ 2$ diagonals as two sides of our triangle (call those $ AB$ and $ AC$). Now we must \"cut\" those two diagonals by the third one, and that one may not contain either $ B$ or $ C$. Obviously, all triangles obtained in such a way are unique (no overcounting).\n\nThere are $ n$ vertices. There are $ \\binom{n\\minus{}3}2$ ways to choose first $ 2$ diagonals. The third must connect any two vertices but $ A, B, C$, hence there are $ \\binom{n\\minus{}3}2$ of those. So our final count is $ N\\equal{}n{\\binom{n\\minus{}3}2}^2$.[/hide]"
}
{
  "Tag": [
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "If $ a,b$ and $ c$ be any real numbers where $ a\\plus{}b\\plus{}c \\neq 0$ then prove that\r\n\r\n$ \\sum \\frac{(b\\plus{}c\\minus{}a)^2}{a^2\\plus{}(b\\plus{}c)^2}\\geq \\frac{3}{5}$",
  "Solution_1": "[quote=\"pankajsinha\"]If $ a,b$ and $ c$ be any real numbers where $ a \\plus{} b \\plus{} c \\neq 0$ then prove that\n\n$ \\sum \\frac {(b \\plus{} c \\minus{} a)^2}{a^2 \\plus{} (b \\plus{} c)^2}\\geq \\frac {3}{5}$[/quote]\r\n\r\nThis problem is similar with Japan 1997, and (I think) solution is similar, too.\r\nSee here, http://www.mathlinks.ro/viewtopic.php?p=537 ."
}
{
  "Tag": [
    "calculus",
    "integration",
    "logarithms",
    "function",
    "calculus computations"
  ],
  "Problem": "Somehow my calculus textbook doesn't mention a bit on $\\displaystyle\\int_{-x}^{x}ln x$.\r\nIf anyone knows the answer, plz respond.\r\nAlso, I'm only a 4th grader and trying to study calculus, so I don't know much about it.",
  "Solution_1": "Can you calculate $\\int \\ln x\\ dx?$",
  "Solution_2": "It is possible to calculate $\\displaystyle\\int\\ln{x} dx$, but you must use a technique called integration by parts.\r\n\r\n\r\nSee [url]http://en.wikipedia.org/wiki/Integration_by_parts[/url].\r\n\r\nIntegration by parts: $\\displaystyle\\int u dv = uv - \\displaystyle\\int v du$\r\nwhere $u$ is a differential function and $dv$ is an integrable function.\r\n\r\n\r\nIn this example:\r\nLet $u = \\ln{x}$, $du = \\frac{1}{x} dx$\r\n Let $v = x$, $dv = 1 dx$.\r\n\r\nIt simplifies out to $x \\ln{x} - x + c$.\r\n\r\nAlso, since $\\ln{x}$ is not defined over negative values, $\\displaystyle\\int_{-x}^{x}\\ln{x} dx$ doesn't exist.",
  "Solution_3": "It's also better not to have the same variable in your bounds as your variable of integration. Is it the same $x$? If so, the expression is hard to interpret -- how can you evaluate between bounds that aren't fixed? If it's not the same $x$, why call them the same thing?",
  "Solution_4": "I agree with you, Xevarion. I add to say that we can't define the definition for which the logarithm is defined, that is to say the condition of anti-logarithm.",
  "Solution_5": "[quote]Integration by parts:  $\\int u dv = uv - \\int v du$\nwhere $u$  is a differential function and $dv$ is an integrable function. \n\n\nIn this example:\nLet $u = ln x$ , $du = \\frac{1}{x} dx$. \nLet $v=x$ , $dv= 1 dx$ .[/quote]\r\n\r\nThe standard presentation of integration by parts bothers me.  I think it's unrigorous.  What is \"$dx$\"?  Is it a number?  If it's a number, how large is it?  (Some people try to say $dx$ is an \"infinitely small real number\", which is nonsense.)  If $dx$ is not a number, then what is it?\r\n\r\nBack in the old unrigorous days, people used to loosely speak of \"infinitesimal\" quantities, like $dx$ and $dy$.  When mathematicians got more careful, \"infinitesimals\" were banished from Calculus because the idea doesn't make sense, and everything was explained by taking limits of finite quantities and never referring to an \"infinitesimal.\"  And yet, after all this time, infinitesimals still linger here in the standard presentation of Integration by Parts.\r\n\r\n(Sure, the idea of an \"infinitesimal\" may have been brought back and made precise in some parts of advanced math, but that's advanced math and not basic Calculus.  And in any case, if $dx$ is going to be referred to and manipulated as its own entity, then it needs to be defined precisely, which is never done in a basic Calculus course.)\r\n\r\nIt's easy to modify the standard presentation of integration by parts to make it rigorous:\r\n\r\n$\\int u(x)v'(x) dx = u(x)v(x) - \\int u'(x)v(x) dx$.\r\n\r\nIn this example:\r\nLet $u(x) = ln(x)$, $u'(x)= \\frac{1}{x}$.\r\nLet $v(x)=x$, $v'(x)=1$.\r\n\r\nThen $\\int ln(x) dx = \\int u(x)v'(x)dx = u(x)v(x) - \\int u'(x)v(x) dx$\r\n\r\n= $x ln(x) - \\int \\left(\\frac{1}{x} \\right)x dx = x ln(x) - \\int 1 dx = x ln(x) - x + C$.\r\n\r\n(By the way, integration by parts is just the product rule in reverse.)"
}
{
  "Tag": [
    "function",
    "algebra proposed",
    "algebra"
  ],
  "Problem": "Find all functions $f : R \\longmapsto R$ such that \r\nf(x+y) = f(x)f(y)f(xy) for all real x and y.",
  "Solution_1": "The solutions are $f(x) \\equiv -1$ ; $f(x) \\equiv 0$ or $f(x) \\equiv 1$\r\nBut the proof seems to be very long  :D",
  "Solution_2": "f(x+y)=f(x)f(y)f(xy)\r\nlet f(0)=a then f(x)=f(x) a^2 for all x. so either f(x)=0 or a^2=1\r\nLet y=x/(x-1)\r\nThen f(x^2/x-1) =f(x)f(x/x-1) f(x^2/x-1)\r\nSo again either f(x)=0 or f(x) f(x/(x-1)) =1. (*)\r\nAlso Let f(1)=b \r\nThen f(x)= b [f(x-1)]^2 (**)\r\nSo f(x) has the same sign as b, and therefore f(x) has one definite sign.\r\nCase1. a=1, and therefore f(x)>0.\r\nFrom equation (*) we get that f(2)=1. from (**) then we get b^3 = 1 so b=1.\r\nhence f(0)=f(1)=f(2)=1. This together with (**) implies that f(z)=1 for all integers z.\r\nNow f(1/(x-1))^2=f(1)f(1/x-1)^2 = f(1+1/x-1) = f(x/x-1) =1/f(x) (***)\r\nAlso f(x/(x-1) -1) =f(x/x-1)^2 = 1/f(x)^2\r\nSo f(1/x-1) = 1/f(x)^2 = f(1/x-1)^4 [ from equation (***)] \r\nThus f(x) =1 for all x.\r\ncase 2. a=-1 can be handeled in the same manner, resulting in f(x)=-1 as the solution.\r\n\r\n-Ali",
  "Solution_3": "$f(x+y) = f(x) f(y) f(xy)$\r\n\r\nIf there exists $y \\in \\mathbb{R}$ such that $f(y) = 0$, then $f(x+y) = 0, (\\forall x)$, so f is identically $0$. Now, let us assume that f takes only nonzero values.\r\n\r\n$f(x+y+z) = f(x + (y+z)) = f(x)f(y+z)f(xy+xz) = $ \r\n$ f(x)f(y)f(z)f(yz)f(xy+xz) = f(x)f(y)f(z)f(yz)f(xy)f(xz)f(x^2 yz)$\r\n\r\nIn a similar manner, we can prove that, for example:\r\n\r\n$f(x+y+z) = f((x+y) + z) = \\cdots = f(x)f(y)f(z)f(xy)f(xz)f(yz)f(xyz^2)$\r\n\r\nComparing the last two relations we get:\r\n\r\n$ f(x) f(y) f(z) f(xy) f(xz) f(yz) f(xyz^2) = f(x) f(y) f(z) f(xy) f(xz) f(yz) f(x^2 yz)$, so dividing by $f(x)f(y)f(z)f(xy)f(xz)f(yz) \\neq 0$ we have:\r\n\r\n$f(xyz^2) = f(x^2yz)$\r\n\r\nNow let's take $y = 1$ and two real non-zero numbers $u, v$. The system $x^2 z = u$ and $x z^2 = v$ has the solution $x = \\frac{u}{\\sqrt[3]{uv}}, z = \\frac{v}{\\sqrt[3]{uv}}$. So plugging in these values for $x$ and $z$ we get $f(u) = f(v)$. So, $f$ is constant over all non-zero reals. Let $f(x) = c$ for $x\\neq 0$. Then $c^3 = c$. Plug in the initial equation $x=-y$ for $x\\neq 0$ and we get $f(0) = f(x)f(-x)f(-x^2) = c^3 = c$. Thus, $f$ is constant. It is easy to see that $c\\in \\{-1,0,1\\}$, but we have already ruled out the case $c=0$, so $f(x) = -1, (\\forall)x$ or $f(x) = 1, (\\forall) x$.  It is easy to see that all three solutions verify the initial equation, so we are done."
}
{
  "Tag": [
    "\\/closed"
  ],
  "Problem": "When I log in, I enter my username and password, and check the \"Log me in automatically for future visits\" box. But for some reason, that box doesn't work. Every single time I access the site, I have to log in all over again. Any ideas why?",
  "Solution_1": "Check your cookies.\r\n\r\nA cookie was probably not created to hold your username and password.",
  "Solution_2": "try clearing your cookies."
}
{
  "Tag": [
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "Is it true that there exist 100 lines in the plane, no three concurrent, such that they intersect in exactly 2002 points?",
  "Solution_1": "this sort of problem has been posted earlier and solved as well(possibly similar to what i am going to say).\r\nthe basic idea is to note that if we have $r$ parallel classes of lines with the $i$th parallel class having $k_i$ lines, then the number of intersection points(since there are no three lines concurrent) is simply $\\sum_{i<j} k_ik_j$. So all we need to do is to see if we can solve for $\\sum_{i=1}^r k_i=100,\\sum_{i<j} k_ik_j=2002$. we can change this second equation to the more simple looking $\\sum_{i=1}^r k_i^2=(100)^2-2\\cdot 2002=5996$. Now by fiddling around one cn find that $5996=77^2+7^2+2^2+14,100=77+7+2+14$. so the answer is, yes."
}
{
  "Tag": [
    "abstract algebra",
    "induction",
    "vector",
    "analytic geometry",
    "modular arithmetic",
    "linear algebra",
    "group theory"
  ],
  "Problem": "Let $(G, \\cdot)$ be a group so that $x^{2}=e$ for any $x \\in G$.  (It's easy to see that this group is comutative from this property) Prove that if $G$ is finite then:\r\n\r\na) $|G| = 2^{n}$.\r\n\r\nb) $G \\simeq \\mathbb{Z}_{2} \\times \\mathbb{Z}_{2} \\times \\cdots \\times \\mathbb{Z}_{2}$ (n terms)\r\n\r\nthank you very much!",
  "Solution_1": "Use Cauchy theorem and the representation of a commutative group as the direct product of its p-Sylows (each of them written as the product of cyclic $p$-groups).",
  "Solution_2": "[quote=\"julien_santini\"]Use Cauchy theorem and the representation of a commutative group as the direct product of its p-Sylows (each of them written as the product of cyclic $p$-groups).[/quote]\r\nA total overkill... :roll:",
  "Solution_3": "p-Sylows  .... why not using directly decomposition of finite abelian groups  :P \r\n\r\n\r\nYou can also consider $G$ as a $\\mathbb{Z}/2\\mathbb{Z}$ linear space \r\n\r\nor :\r\n\r\n$G = < x_1, .., x_n >$  with $n$ minimal and $f : \\mathbb{Z}/2\\mathbb{Z}^n \\rightarrow G$, $(a_1, ..., a_n) \\rightarrow (x_1^{a_1}, ..,x_n^{a_n})$. Then $f$ is well defined (must be proved), surjective by definition and injective. So $f$ is an isomorphism\r\n\r\nor :\r\n\r\nIt's also possible to proceed by induction assuming that $G$ has an element of order $2$ (by Cauchy or better by hypothesis ;) )",
  "Solution_4": "[quote=\"t\u00b5t\u00b5\"]It's also to proceed by induction assuming that $G$ has an element of order $2$ (by Cauchy)[/quote]\r\nWhy you all try to use Cauchy\u00bf It's [u]given[/u] that all nontrivial elements have order $2$!",
  "Solution_5": ":rotfl: you're right Zetax  :rotfl:",
  "Solution_6": "OK, here's my solution (actually, it's the induction suggested by some of you):\r\n\r\nFirst, let's prove that $G$ is abelian: $x^2y^2 = (xy)^2 = e\\iff xxyy = xyxy \\iff xy = yx$.\r\n\r\nSuppose $G \\neq \\{e\\}$. Pick $x_1 \\neq e$ and consider $\\langle x_1\\rangle$. Since $x_1^2 = e$, $\\langle x_1\\rangle$ has two elements, namely $e$ and $x_1$. If there are no more elements on $G$ then $|G| = 2$.\r\n\r\nElse, pick a element $x_2 \\in G\\setminus \\langle x_1\\rangle$ and consider $\\langle x_1, x_2\\rangle$. There are two types of elements in this set: the ones that have $x_2$ as factor and the ones that don't have. There are, of course, two elements of the latter type, namely the elements of $\\langle x_1\\rangle$. But, since $G$ is abelian and $x_2^2 = e$, either you have one factor $x_2$ or you don't have it at all. So there is a bijection between numbers of the two types: just multiply the elements of $\\langle x_1\\rangle$ by $x_2$. Thus $\\langle x_1, x_2\\rangle$ has $2\\cdot 2 = 4$ elements.\r\n\r\nNow it's not hard to see that we can proceed by induction: suppose we already found $x_1,x_2,\\ldots,x_k$ and $X_k = \\langle x_1,x_2,\\ldots,x_k\\rangle$, $|X_k| = 2^k$. If $G\\setminus X_k =\\emptyset$, then $|G| = 2^k$. Else, consider $x_{k+1} \\in G \\setminus X_k$ and do pretty much the same we did in the last paragraph: classify each element of $X_{k+1} = \\langle x_1,x_2,\\ldots,x_k,x_{k+1}\\rangle$ in $X_k$ and $Y_k = X_{k+1} \\setminus X_k$. We have again a bijection between $X_k$ and $Y_k$ (guess what bijection?) and then $|X_{k+1}| = 2|X_k| = 2^{k+1}$.\r\n\r\nSince $G$ is finite, this procedure will stop at some moment, so $|G| = 2^n$. Furthermore, we even have found the generators of $G$: $G = \\langle x_1,x_2,\\ldots,x_n\\rangle$. And now the isomorphism between $G$ and $Z_2^n$ is clear: $\\sigma\\colon G \\to Z_2^n$, $\\sigma(x) = (a_1,a_2,\\ldots,a_n)$, where $x = x_1^{a_1}x_2^{a_2}\\cdots x_n^{a_n}$.\r\n\r\nIf I'm not mistaken (and I may be), this same reasoning proves that if $G$ is abelian and finite, them $G$ has a finite number of generators and $|G|$ is equal to the product of the orders of these generators and, thus, $G$ is isomorphic to a product of $Z_m$'s.",
  "Solution_7": "[quote=\"cyshine\"] (guess what bijection?)[/quote]\r\n\r\n\r\nCan you please post which is the bijection?  :blush: \r\n\r\nThank you very much!",
  "Solution_8": "cyshine ,from what i see you give the longest solution, but the most straightforward elementary one (it's either that or overkill :blush: )\r\nis my assumption true that with some modifications you have proven that any group where ever element has order p with p prime, and abelian, is such a direct product of cyclic groups of order p?",
  "Solution_9": "[quote=\"Mitzah\"][quote=\"cyshine\"] (guess what bijection?)[/quote]\n\n\nCan you please post which is the bijection?  :blush: \n\nThank you very much![/quote]\r\n\r\nHi, and sorry from replying so late. Believe it or not, I've been busy these days (I will give classes to the national olympiad winners, so I'm preparing them, plus some material).\r\n\r\nThe bijection is the analogous to case $k=1$: multiply each element of $\\langle x_1,x_2,\\ldots,x_k\\rangle$ by $x_{k+1}$. The proof that this is indeed a bijection is also analogous to the case $k=1$: since $G$ is abelian and $x_2^2 = e$, either you have one factor $x_{k+1}$ or you don't have it at all. And since the operation \"multiply by $x_{k+1}$\" is inversible, we have a bijection.",
  "Solution_10": "[quote=\"fredbel6\"]cyshine ,from what i see you give the longest solution, but the most straightforward elementary one (it's either that or overkill :blush: )\nis my assumption true that with some modifications you have proven that any group where ever element has order p with p prime, and abelian, is such a direct product of cyclic groups of order p?[/quote]\r\n\r\nYour assumption is true.\r\n\r\nI wanted to give an \"straightforward elementary\" solution (as you wrote) because a lot of people don't know Cauchy's theorem or the theorem on representation of finite abelian groups. And these theorems have, as far as I know, long proofs (that is, if you want to start from first principles of group theory). And, most incisively, one doesn't need to know these theorems to solve the problem (I didn't remember them myself). And I actually just expanded t\u00b5t\u00b5's induction idea.",
  "Solution_11": "Then again, the most straightforward solution is probably linear algebra: since $G$ is abelian, it is naturally a module over $\\mathbb{Z}$. Multiplication by 2 gives zero, so, $G$ is a $\\mathbb{Z}/2\\mathbb{Z}$-module. That quotient is a field, and $G$ is a vector space over the field with 2 elements. Since $G$ is finite, it is spanned by finitely many elements and is finite-dimensional.\r\nGoing back to the theorems everyone learned in linear algebra, every finite-dimensional vector space over a field $F$ has a basis and is isomorphic to $F^n$, with coordinates chosen with respect to this basis. The additive structure of $F^n$ is the product of $n$ copies of the additive group of $F$, and we have the desired result.\r\n\r\nIn that first linear algebra course, the field was implicitly assumed to be $\\mathbb{R}$- but none of the proofs used anything other than field operations, so they all apply to (finite-dimensional) vector spaces over an arbitrary field. My first paragraph uses some relatively advanced language, but we could skip right to defining the vector space operations once we know that $G$ is abelian:\r\n$x+y=x*y$, where $*$ is the group operation. $0=e$. $0\\cdot x=0$, $1\\cdot x=x$. That's it for scalar multiplication, since there aren't any other scalars. The axioms to check:\r\nThe space is an abelian group under $+$: Already done.\r\n$ab\\cdot x=a\\cdot(b\\cdot x)$: All four cases are obvious.\r\n$0\\cdot x=0,1\\cdot x=x$: True by definition.\r\n$a\\cdot x+b\\cdot x=(a+b)\\cdot x$: Three cases are trivial because 0 is an identity. If $a=b=1$, we have $a+b=0$ and $x+x=0$, so this checks.\r\n$a\\cdot x+a\\cdot y=a\\cdot(x+y)$: Both cases are obvious.\r\n\r\nThere's nothing wrong with invoking a few standard theorems from an earlier course.",
  "Solution_12": "Hi,\r\n\r\nThis solution is correct, but exactly the construction of a basis of a general finite-dimensional vector space $V$ is essentially inductive.\r\n\r\nThis proof is from Herstein, Topics in Algebra. The definition of 'finite-dimensional' is that there exists a finite subset $S$ of $V$ that span $V$ ('span' is a technical term: this means that every element of $V$ is a linear combination of a finite number of elements from $S$). Let $S = \\{v_1,v_2,\\ldots, v_n\\}$. If these $n$ vectors are linear independent, $S$ is a basis of $V$. If not, weed out the first $v_j$, which is a linear combination of its predecessors. This set still span $V$: just replace $v_j$ by the corresponding linear combination. And continue until there are no more $v_k$'s that are linear combination of its predecessors. Then this set is a basis of $V$.\r\n\r\nLet me stress that jmerry's solution is OK but I still think one would need to write some additional steps, since the proof above (which is instructive itself) is, in my humble opinion, not significantly easier than my solution. I just feel that this solution is like proving that $2^6 \\equiv 1\\pmod 7$ using Fermat's theorem. It's correct, but I don't feel quite right proving a result using another that is not easier to prove.\r\n\r\nI agree that there's nothing wrong in using facts from earlier courses. But, strangely enough, I studied (I mean, with proofs) group theory before linear algebra (I studied using Herstein's book), so group theory is, to me, and maybe, only to me, a earlier course."
}
{
  "Tag": [
    "AMC",
    "AIME",
    "geometry",
    "geometric transformation",
    "rotation",
    "linear algebra",
    "matrix"
  ],
  "Problem": "When do we often see to use them in math contests?",
  "Solution_1": "Some AMC and AIME problems contain rotation matrices in disguise and ask you to identify some large iteration of the rotation.  Matrices can also be useful for doing computations with sequences such as the Fibonacci sequence, as well as for solving certain types of combinatorics problems.",
  "Solution_2": "I see. The system of equations problem in this forum would be an example, I guess.\r\nHave a sample problem on you? (I have absolutely no practice with them except how to find the determinant).",
  "Solution_3": "I don't think I've ever seen a problem that explicitly required the use of matrices.  As t0rajir0u pointed out, [url=http://en.wikipedia.org/wiki/Rotation_matrix]rotation matrices[/url] are useful in determining what happens when a point is reflected through a line (which is really just a change of basis)."
}
{
  "Tag": [
    "geometry"
  ],
  "Problem": "Hello ppl,\r\nI have been a devoted zune user for several months but just recently my dad got a free 8gig touch when he bought a macbook. I want to be able to move my files from the zune to my computer but I'm not sure how. I think the sync might work but I have no idea how to copy and paste the songs from the zune onto the hard drive. Any help would be appreciated. Even if you jsut have a suggestion. O yah, and just fyi I am planning to jailbreak my touch so any advice on that area would help too. \r\n\r\nThank you very much\r\n~zhang~ :D",
  "Solution_1": "1. Go to my computer\r\n2. Click on the Zune device, assuming its already connected.\r\n3. Copy the files to your desktop or somewhere.\r\n4. Your done :)"
}
{
  "Tag": [],
  "Problem": "Check this out: 1.e4 e5 2.Nf3 d6 3.Bc4 Nc6 4.Nc3 Bg5 5.d3 Bh5 6.d3 h6 7.b3 g5. What is White's best move?",
  "Solution_1": "Um... it seems to me that at move 4, you can't move a bishop to g5... because none of the black bishops have moved yet...\r\n\r\nAlso:\r\n[quote] 5.d3 Bh5 6.d3 h6 [/quote]\r\nHow does a pawn move from d3 to d3??",
  "Solution_2": "[quote=\"mathelete\"]Check this out: 1.e4 e5 2.Nf3 d6 3.Bc4 Nc6 4.Nc3 Bg5 5.d3 Bh5 6.d3 h6 7.b3 g5. What is White's best move?[/quote]\r\nThis is what I meant:\r\n\r\nCheck this out: 1.e4 e5 2.Nf3 d6 3.Bc4 Nc6 4.Nc3 Bg4 5.d3 Bh5 6.h3 h6 7.b3 g5. What is White's best move?"
}
{
  "Tag": [
    "trigonometry"
  ],
  "Problem": "Find the angle $ \\theta$ between the two lines 2x and 3x.\r\n\r\nI made this into a really messy deal with tangents and what not...\r\n\r\nIs their a more efficient way without a calculator? \r\n\r\nThanks in Advance",
  "Solution_1": "[quote=\"mathemagician1729\"]Find the angle $ \\theta$ between the two lines 2x and 3x.\n\nI made this into a really messy deal with tangents and what not...\n\nIs their a more efficient way without a calculator? \n\nThanks in Advance[/quote]\r\n[hide]The line $ y\\equal{}2x$ makes a $ \\tan^{\\minus{}1} 2$ with the x-axis and the line $ y\\equal{}3x$ makes a $ \\tan^{\\minus{}1} 3$ with the x-axis. Thus the angle between these two lines is $ \\tan^{\\minus{}1} 3\\minus{}\\tan^{\\minus{}1} 2$, setting this to $ \\theta$ and taking the tangent of both sides results in $ \\tan\\theta\\equal{}\\frac{1}{7}$.\nI don't know what else you could do without a calculator other then putting it in the form $ \\theta\\equal{}\\tan^{\\minus{}1} \\frac{1}{7}$. On the other hand, using a calculator gives $ \\theta$ to be approximately $ 8.13^{\\circ}$[/hide]",
  "Solution_2": "The reason I asked was because they wanted you to be able to find the range of theta as between 0 and 10 degrees without a calculator."
}
{
  "Tag": [
    "geometry proposed",
    "geometry"
  ],
  "Problem": "circle $c_1$ and $c_2$ are externally tangent at $T$. A line passing through point $T$ cuts $c_1$ at $A$ and   $c_2$ at $B$. the line tangent to $c_1$ at $A$ cuts $c_2$ at $D$ and $E$.\r\n\r\nIf $D$ $\\in$ $|AE|$, $|TA|$=a $|TB|$=b find $|BE|$",
  "Solution_1": "Let centres of $c1$ and $c2$ be $P$ and $Q$. Let angle $TAD$ be $r$, and angle $ATD$ be $s$. Because $AE$ is tangent to $c1$ then angle $APT$ is $2r$. $T$ is pt at which two circles are tangent so angle $PTD$ = angle $BTQ$. Hence angle $BQT$ is also $2r$. Then angle $TDB$ is $r$.Then also angle $BET$ is $r$.\r\nIf angle $ATD$ is $s$, then angle $BED$ is $s$. \r\nAngle $TDE$ is equal to $r+s$. But angle $BDT$ is $r$ so angle $BDE$ is $s$. Hence $DB=BE$. \r\nIn triangle $TDB$ have $DB/sin(180-s)=TB/sin(r)$ or $DB/sin(s)=TB/sin(r)$.\r\nIn triangle $ABE$ have $BE/sin(r)=AB/sin(s)$.\r\nMultiplying the 2 equations we get above, and cancelling out $sin(s)$ and $sin(r)$ we get:\r\n$DB*BE=AB*TB$. Subst $DB=AE$, $AB=a+b$, $TB$=b get:\r\n$BE^2=(a+b)b$.\r\nSo $BE = \\sqrt{(a+b)b}$ and we are done  :)"
}
{
  "Tag": [
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Is the following true?: for each positive integer $ k$ there exists a positive integer $ m$ such that all the numbers $ m,2m,3m,...,m^2$ have exactly $ k$ nonzero digits.",
  "Solution_1": "What makes you think it might be true?  Can you give one example of such an $ m$ for $ k\\geq 2$?  \r\n\r\nNot only is it not true for every $ k$, unless I've misunderstood your problem at first glance I don't think it's true for [i]any[/i] $ k\\geq 2$."
}
{
  "Tag": [
    "inequalities",
    "function",
    "integration",
    "advanced fields",
    "advanced fields theorems"
  ],
  "Problem": "We all know Jensen's inequality.\r\n\r\nWe have got a measure space $ (X,A,\\mu)$ with $ \\mu(X)\\equal{}1$, and a real-valued function $ f$ that is $ \\mu$-integrable.\r\nSo if $ \\phi$ is a measurable convex function (with $ \\phi : \\mathbb{R} \\rightarrow \\mathbb{R}$), then\r\n\r\n$ \\displaystyle \\phi\\left(\\int_X f d\\mu\\right) \\leq \\int_X (\\phi \\circ f) d\\mu$\r\n\r\nBut when does equality occur?",
  "Solution_1": "A typical proof involves an affine function $ \\psi\\colon\\mathbb R\\to\\mathbb R$ such that $ \\psi\\le \\phi$ and $ \\phi(a)\\equal{}\\psi(a)$ where $ a\\equal{}\\int_X f\\,d\\mu$. \r\nFrom this we find: equality occurs if and only if there exists $ \\psi$ as above such that $ \\phi(f(x))\\equal{}\\psi(f(x))$ for a.e. $ x$.\r\nIf $ \\phi$ is strictly convex, the above simplifies to $ f(x)\\equal{}a$ for a.e. $ x$."
}
{
  "Tag": [
    "LaTeX"
  ],
  "Problem": "En liten presentasjon av Latex  :) \r\nDet skal ikke mye til for \u00e5 bruke Latex p\u00e5 Mathlinks (ML).\r\nN\u00e5r du skal skrive en matteformel skal du sette to dollartegn rundt uttrykket.\r\nFor eksempel:   $\\$$ $a + b = c$ $\\$$. (dollar er ctrl+alt+4)\r\nFra n\u00e5 av skal jeg bare skrive det som st\u00e5r mellom disse dollartegnene.\r\nDe mest brukte ting:\r\n\r\n   $x^n$ --- du skriver  x^n  og $x^{123}$  ---  x^{123}\r\n(skal du ha mer enn 1 siffer eller bokstav n\u00e5r du bruker funksjoner m\u00e5 du ha uttrykket i slike paranteser: {}  )\r\n\r\n   $a_1$ og ${AC}_{bc}$ ---  a_1          og             {AC}_{bc}\r\n\r\n   $a \\geq b$, $b \\leq c$ og $m \\equiv n$ ----   a \\geq b   ,          b \\leq c          og        m\\equiv n\r\n\r\n   $a \\not= b$ -----           a \\not= b\r\n\r\n   $\\frac{a^2+3}{b^2+c}$ -------      \\frac {a^2+3}{b^2+c}\r\n\r\n   $\\sum_{i=0}^{n}x^i$   -------   \\sum_{i=0}^{n}x^i\r\n\r\n   $\\prod _{j=1}^{n} \\frac{j+1}{j}$ ------    \\prod _{j=1}^{n} \\frac{j+1}{j}\r\n\r\n   $f''(x+y)$ ------     f''(x+y)",
  "Solution_1": "Denne er av og til nyttig og: $3\\cdot5^3$-------3\\cdot5^3  .\r\n\r\nHer finn ein ei st\u00f8rre oversikt over latex-spr\u00e5ket. Men det er lett \u00e5 g\u00e5 seg vill, mesteparten treng ein ikkje til skriving p\u00e5 Mathlinks\r\n[url]http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php[/url]"
}
{
  "Tag": [
    "geometry",
    "complex numbers",
    "absolute value",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "A,B,C are complex numbers such that absolute value of them are equal 1\r\nand A+B+C =0 Find the area of triangle ABC in the complex graph",
  "Solution_1": "Hint No. 1: What does the absolute value of those three points imply (got something to do with the unit circle, doesn't it)?\r\nHint No. 2: Can $ A\\plus{}B\\plus{}C\\equal{}0$ tell you something about the shape of the triangle?\r\n\r\n[hide]BTW, I'm not really sure where this topic belongs... Maybe Intermediate topics?[/hide]"
}
{
  "Tag": [
    "trigonometry"
  ],
  "Problem": "let $ s\\equal{}sin{x}$\r\n\r\nHow many solutions does the equation \r\n\r\n$ 2\\equal{} s\\plus{}{s}^2 \\plus{}{s}^3 \\plus{}{s}^4\\plus{}...$\r\n\r\nhave in the range 0 (less than or equal to) x (less than 2pi)?\r\n\r\nWould this be 0 solutions because the series must be less than one?",
  "Solution_1": "[hide=\"hint\"]$ |\\sin x| \\le 1$, think of a geometric series.[/hide]\n\n[hide=\"solution\"]\nSince $ |\\sin x| \\le 1$ the series converges. \n\nIn this case $ a \\equal{} 1$, $ r \\equal{} \\sin x$\nSo $ 1 \\plus{} s \\plus{} s^{2} \\plus{} s^{3} \\plus{} ... \\equal{} \\frac {1}{1 \\minus{} \\sin x} \\equal{} 3$\n$ 1 \\equal{} 3 \\minus{} 3\\sin x$\n$ \\sin x \\equal{} \\frac {2}{3}$\nThere are two solutions, one in quadrant 1, and one in quadrant II.[/hide]",
  "Solution_2": "Where have you gotten the 3 from and why does the series suddenly start with 1 as opposed to sinx?",
  "Solution_3": "[quote=\"JasonSullivan\"]Where have you gotten the 3 from and why does the series suddenly start with 1 as opposed to sinx?[/quote]\r\nhe has added one to both sides :)",
  "Solution_4": "[quote=\"JasonSullivan\"]Where have you gotten the 3 from and why does the series suddenly start with 1 as opposed to sinx?[/quote]\n\nYou can start with $ \\sin x$ if you wanted to:\n\n$ \\sin x \\plus{} \\sin^2 x \\plus{} \\sin^3 x \\plus{} \\dots \\equal{} 2$\n$ \\implies \\frac {\\sin x}{1 \\minus{} \\sin x} \\equal{} 2$\n$ \\implies 3\\sin x \\equal{} 2$\n$ \\implies \\sin x \\equal{} \\frac {2}{3}$\n\n[quote=\"JasonSullivan\"]\nWould this be 0 solutions because the series must be less than one?[/quote]\r\n\r\nJust because the terms are all $ \\leq 1$ doesn't mean the series must be -- consider:\r\n$ 1\\plus{}\\frac{1}{2}\\plus{}\\frac{1}{4}\\plus{}\\frac{1}{8}\\plus{}\\dots \\equal{} 2$"
}
{
  "Tag": [
    "inequalities",
    "binomial coefficients",
    "rearrangement inequality"
  ],
  "Problem": "Help me please :\r\n\r\nLet $n$ be a positive integer. Determine the smallest possible value of the sum \\[a_1b_1+a_2b_2+...a_{2n+2}b_{2n+2}\\] , where $(a_1,a_2,...a_{2n+2})$ and $(b_1,b_2,...b_{2n+2})$ are rearrangements of the binomial coefficients \\[\\left(\\begin{array}{cc}{2n+1}\\\\0\\end{array}\\right) , \\left(\\begin{array}{cc}{2n+1}\\\\1\\end{array}\\right) , ... , \\left(\\begin{array}{cc}{2n+1}\\\\2n+1\\end{array}\\right)\\]",
  "Solution_1": "Well, by the rearrangement inequality (what a coincidence), you just pair up the biggest with the smallest, 2nd biggest with 2nd smallest, etc.\r\n\r\nSo one possible arrangement is $a_k = \\binom{2n+1}{k-1}$ and $b_k = \\binom{2n+1}{n-k+1}$ if $k \\le n$ or $b_k = \\binom{2n+1}{3n+2-k}$ if $k \\ge n+1$ (I think that works)."
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "algebra proposed"
  ],
  "Problem": "If the equality $ x^k\\plus{}a_1 x^{k\\minus{}1} \\plus{}...\\plus{}a_k\\equal{}0$ has $ k$ distinct real roots then prove that \r\n$ a_1^2 >\\frac{2ka_2}{k\\minus{}1}$",
  "Solution_1": "[quote=\"pradmath\"]the equality $ x^k \\plus{} a_1 x^{k \\minus{} 1} \\plus{} ... \\plus{} a_k \\equal{} 0$ has $ k$ distinct real roots then prove that \n$ a_1^2 > \\frac {2ka_2}{k \\minus{} 1}$[/quote]\r\n\r\nHere is my solution:\r\n\r\nAssume that the equality $ x^k \\plus{} a_1 x^{k \\minus{} 1} \\plus{} ... \\plus{} a_k \\equal{} 0$ has $ k$ distinct real roots $ x_1,x_2,...,x_k$. \r\n\r\nBy Viete theorem, we have:\r\n$ a_1 \\equal{} (\\minus{}1)^{k \\minus{} 1}.(\\sum_{i \\equal{} 1}^k x_i)$ (1)\r\n$ a_2 \\equal{} \\sum_{i \\neq j}^k x_ix_j$\r\n\r\nBy squaring, (1) becomes $ a_1^2 \\equal{} (\\sum_{i \\equal{} 1}^k x_i)^2$\r\n\r\nNow we must prove\r\n\r\n$ a_1^2 > \\frac {2ka_2}{k \\minus{} 1}$\r\n\r\n$ \\leftrightarrow (\\sum_{i \\equal{} 1}^k x_i)^2 > \\frac{2k}{k \\minus{} 1}\\sum_{i \\neq j}^k x_ix_j$\r\n\r\n$ \\leftrightarrow (k \\minus{} 1)\\sum_{i \\equal{} 1}^k x_i^2 > 2\\sum_{ i \\neq j}^k x_ix_j$\r\n\r\n$ \\leftrightarrow \\sum_{i \\neq j}^k (x_i \\minus{} x_j)^2 > 0$\r\n\r\nwhich is clearly true because $ x_1,x_2,...,x_k$ are distinct real numbers."
}
{
  "Tag": [
    "inequalities proposed",
    "inequalities"
  ],
  "Problem": "Prove that $\\forall a,b,c>0$and $k\\ge 0$ we have:\r\na)$\\sum\\frac{a^2-bc}{b^2+c^2+ka^2}.\\frac{b^2-ca}{c^2+a^2+kb^2}\\le 0$ :)  :) \r\nb)$\\sum\\frac{a^2-bc}{\\sqrt{b^2+c^2+ka^2}}.\\frac{b^2-ca}{\\sqrt{c^2+a^2+kb^2}}\\le 0$ :)  :)",
  "Solution_1": "[quote=\"nthd\"]Prove that $\\forall a,b,c>0$and $k\\ge 0$ we have:\na)$\\sum\\frac{a^2-bc}{b^2+c^2+ka^2}.\\frac{b^2-ca}{c^2+a^2+kb^2}\\le 0$ :)  :) [/quote]\r\n\r\nLet\r\n$f(a,b,c)=\\sum\\frac{a^2-bc}{b^2+c^2+ka^2}.\\frac{b^2-ca}{c^2+a^2+kb^2}$\r\nThen\r\n$f(1,1,3)=\\frac{4(k-78)}{(k+10)^2(9k+2)} > 0$ for $k>78$.\r\nSo where am I wrong?",
  "Solution_2": "[quote=\"ductrung\"][quote=\"nthd\"]Prove that $\\forall a,b,c>0$and $k\\ge 0$ we have:\na)$\\sum\\frac{a^2-bc}{b^2+c^2+ka^2}.\\frac{b^2-ca}{c^2+a^2+kb^2}\\le 0$ :)  :) [/quote]\n\nLet\n$f(a,b,c)=\\sum\\frac{a^2-bc}{b^2+c^2+ka^2}.\\frac{b^2-ca}{c^2+a^2+kb^2}$\nThen\n$f(1,1,3)=\\frac{4(k-78)}{(k+10)^2(9k+2)} > 0$ for $k>78$.\nSo where am I wrong?[/quote]I think you are wrong because with $k=0$ it's true.",
  "Solution_3": "[quote=\"nthd\"][quote=\"ductrung\"][quote=\"nthd\"]Prove that $\\forall a,b,c>0$and $k\\ge 0$ we have:\na)$\\sum\\frac{a^2-bc}{b^2+c^2+ka^2}.\\frac{b^2-ca}{c^2+a^2+kb^2}\\le 0$ :)  :) [/quote]\n\nLet\n$f(a,b,c)=\\sum\\frac{a^2-bc}{b^2+c^2+ka^2}.\\frac{b^2-ca}{c^2+a^2+kb^2}$\nThen\n$f(1,1,3)=\\frac{4(k-78)}{(k+10)^2(9k+2)} > 0$ for $k>78$.\nSo where am I wrong?[/quote]I think you are wrong because with $k=0$ it's true.[/quote]\r\nBut doesn't it say \"for all $a,b,c>0$ and $k\\ge 0$\"? What if we fix $k > 78$, then put $a=b=1$ and $c=3$?\r\n\r\nOr do you mean for all $a,b,c>0$, there exists $k\\ge 0$ such that\r\n$\\sum\\frac{a^2-bc}{b^2+c^2+ka^2}.\\frac{b^2-ca}{c^2+a^2+kb^2}\\le 0$ ?",
  "Solution_4": "[quote=\"ductrung\"][/quote][quote=\"nthd\"][quote=\"ductrung\"][quote=\"nthd\"]Prove that $\\forall a,b,c>0$and $k\\ge 0$ we have:\na)$\\sum\\frac{a^2-bc}{b^2+c^2+ka^2}.\\frac{b^2-ca}{c^2+a^2+kb^2}\\le 0$ :)  :) [/quote]\n\nLet\n$f(a,b,c)=\\sum\\frac{a^2-bc}{b^2+c^2+ka^2}.\\frac{b^2-ca}{c^2+a^2+kb^2}$\nThen\n$f(1,1,3)=\\frac{4(k-78)}{(k+10)^2(9k+2)} > 0$ for $k>78$.\nSo where am I wrong?[/quote]I think you are wrong because with $k=0$ it's true.[/quote][quote=\"ductrung\"]\nBut doesn't it say \"for all $a,b,c>0$ and $k\\ge 0$\"? What if we fix $k > 78$, then put $a=b=1$ and $c=3$?\n\nOr do you mean for all $a,b,c>0$, there exists $k\\ge 0$ such that\n$\\sum\\frac{a^2-bc}{b^2+c^2+ka^2}.\\frac{b^2-ca}{c^2+a^2+kb^2}\\le 0$ ?[/quote]\r\nI think your transformation is wrong! :)",
  "Solution_5": "I used a computer algebra system to check before I posted - but as you insist: ok, the software may be buggy though! I will check my calculation an extra time and be more careful.\r\n\r\nOh, BTW, I didn't do any transformation. I just put $a=b=1$, $c=3$ and, say $k=80$.",
  "Solution_6": "Oh, ductrung, I think you want to say $k \\le 78$ but you wrote $k>78$ :lol:",
  "Solution_7": "[quote=\"nthd\"]Oh, ductrung, I think you want to say $k \\le 78$ but you wrote $k>78$ :lol:[/quote]\r\n\r\nWhat a lame excuse! No matter which I wrote: $k\\le 78$ or $k>78$, the expression\r\n$f(1,1,3)=\\frac{4(k-78)}{(k+10)^2(9k+2)}$\r\nshould have rung a bell.",
  "Solution_8": "[quote=\"ductrung\"][quote=\"nthd\"]Oh, ductrung, I think you want to say $k \\le 78$ but you wrote $k>78$ :lol:[/quote]\n\nWhat a lame excuse! No matter which I wrote: $k\\le 78$ or $k>78$, the expression\n$f(1,1,3)=\\frac{4(k-78)}{(k+10)^2(9k+2)}$\nshould have rung a bell.[/quote]\r\nBecause you type wrong $k>78$ so I didn't understand you. :mad: .I don't sorry. :)",
  "Solution_9": "[quote=\"nthd\"]Because you type wrong $k>78$ so I didn't understand you.\n[/quote]\nWhere did I type wrong?\n\n[quote=\"nthd\"]\n :mad: .I don't sorry. :)[/quote]\r\nI don't expect so :-)",
  "Solution_10": "[quote=\"ductrung\"][quote=\"nthd\"]Because you type wrong $k>78$ so I didn't understand you.\n[/quote]\nWhere did I type wrong?\n\n[quote=\"nthd\"]\n :mad: .I don't sorry. :)[/quote]\nI don't expect so :-)[/quote]\r\nI think I don't understnad you. :)",
  "Solution_11": "[quote=\"nthd\"]I think I don't understnad you. :)[/quote]\r\n\r\nThat's no crime :-)\r\n\r\nSo are you going to rephrase your problem? Something like \"find all $k$ such that ...\"?"
}
{
  "Tag": [],
  "Problem": "The directions are to find all real solutions of the equation. \r\n\r\n$\\dfrac{x}{2x+7} - \\dfrac{x+1}{x+3} = 1$\r\n\r\nI found the LCD and got \r\n\r\n-x + 12x + 6 = 0. \r\n\r\nThen I found that the discriminate is 168. \r\n\r\nx = 6 - sqare root of 42 \r\n\r\nx - 6 + square root of 42 \r\n\r\nAre these right? I checked them, and they didn't work, so did i do something wrong? I know it's supposed to have at least two solutions, right? Please help me.",
  "Solution_1": "I'm just going to give the work, line by line.\r\n\r\n$\\dfrac{x}{2x+7}-\\dfrac{x+1}{x+3}=1$\r\n\r\n$x(x+3)-(x+1)(2x+7)=(2x+7)(x+3)$\r\n\r\n$x^2+3x-2x^2-9x-7=2x^2+13x+21$\r\n\r\n$-x^2-6x-7=2x^2+13x+21$\r\n\r\n$3x^2+19x+28=0$\r\n\r\n$(3x+7)(x+4)=0$\r\n\r\n$x=-\\dfrac{7}{3}$ or $x=-4$",
  "Solution_2": "Thanks, I forgot to muliply the 1 by the least common denominator :blush:"
}
{
  "Tag": [
    "real analysis",
    "integration",
    "calculus",
    "function",
    "real analysis unsolved"
  ],
  "Problem": "How to prove that the $\\int_{0}^{1}\\frac{1}{\\sqrt{x}}$ is Lebesgue integrable? And how about $\\int_{0}^{1}\\frac{cos(1/x)}{\\sqrt{x}}$?\r\n\r\nIf someone could kindly point me to some reference for this kind of problems, it will be greatly appreciated.\r\n\r\nThanks a lot!",
  "Solution_1": "It's Lebesgue integrable because it's absolutely integrable. Evaluate the improper integral; since the function is positive, that also works Lebesgue.\r\n\r\nIn the second case, use comparison.",
  "Solution_2": "Hi, could you please tell me why the function $1/\\sqrt{x}$ is absolute integrable? Somehow our prof and textbook never mention this concept.\r\n\r\nThanks!",
  "Solution_3": "Chapter 10 of Rudin's \"Principles of Mathematical Analysis\" covers Lebesgue integration. If $f\\ge 0$, then $\\int_{0}^{1}f(x)dx$ (in the Lebesgue sense) is the supremum of $\\int_{0}^{1}g(x)dx$, where $g$ is a simple function and $0\\le g\\le f$. In general, $f$ is integrable iff both $f_{+}$ and $f_{-}$ are integrable. \r\n\r\nSince $f_{+}\\le |f|$ and $f_{-}\\le |f|$, it follows that $f$ is integrable whenever $|f|$ is. This can be used to reduce your second question to the first one, because $\\left|\\frac{\\cos(1/x)}{\\sqrt{x}}\\right|\\le \\frac{1}{\\sqrt{x}}$. \r\n\r\nTo answer the first question, you need a theorem: if $f\\ge 0$ and $f$ is Riemann integrable, then $f$ is Lebesgue integrable, with the same value of the integral.",
  "Solution_4": "Hi mlok, thank you very much for the reply. Now I get that \"if $f$ is  measurable, then $|f|$ integrable => $f$ integrable\".\r\n\r\nHowever, for the case that $f = \\frac{1}{\\sqrt{x}}$ is integrable, since it is a improper Riemann integral on [0,1], I am not sure whether I can apply the Riemann integration result directly. Anyhow, now I am able to construct an integrable function $f_{n}$ that monotonically converges to $f$, I think that can prove that $f$ is Lebesgue integrable, where $f_{n}= \\frac{1}{\\sqrt{x}}$ if $1/n^{2}< x \\leq 1$, or $f_{n}= n$ when $0 \\leq x \\leq 1/n^{2}$."
}
{
  "Tag": [
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Hello all.. I'm 15-years old Korean student.\r\n\r\nI have some questions for you...\r\n\r\nPlease solve these problems...\r\n\r\nThank you..  Problems are here...\r\n\r\n\r\n(1) If ax+by=3,  a x^2 + b y^2  = 7,  ax^3 + by^3 = 16,  ax^4 + by^4 = 42, then find the value of ax^5 + by^5.\r\n\r\n(2) Find all x, y, z that satisfied (x+y)^3 = z,  (y+z)^3 = x, (z+x)^3 = y.\r\n\r\n(3) Find all x, y, z that satisfied x+[y]+{z}=1.1,  [x]+{y}+z=2.2,  {x}+y+[z]=3.3.\r\n      ( [n] is a integer part of n, and {n}=n-[n] )\r\n\r\n(4) Find all x, y, z that satisfied x^3 -9(y^2 -3y+3)=0, y^3 -9(z^2 -3z+3)=0,  z^3 -9(x^2 -3x+3)=0.\r\n\r\n(5) What is the nearest value of  \\sqrt 65 -  \\sqrt 63?\r\n     1: 0.12    2: 0.13    3: 0.14    4: 0.14    5: 0.15\r\n\r\n(6) Find (x,y) that satisfied 1!+2!+3!+...+x! = y^2    (x, y \\in Z)",
  "Solution_1": "??? why until 3pm tomorrow? Is it homework?\r\n\r\nPiere.",
  "Solution_2": "not only that, aren't these problems the same ones you posted to another forum (i think i have seen these somewhere else) the same ones, that you mentioned were homework...and that you needed to have done for you? I am pretty sure that is what you did :(  \r\n\r\n\r\nas you will notice, and other ppl might say this too, but this is not a homework solving site but mainly an olympiad problem solving site!\r\n\r\n-cheers\r\n:D :D",
  "Solution_3": "No, It' s not a homework.\r\n\r\nI mean it ask you to post until 3 p.m. is just I want to know the solution early!!\r\n\r\nI think it for over 2 hours.\r\n\r\nSo, I found some solutions about that but I can't certain that is correct.\r\n\r\nPlease Please.... just post its solution...\r\n\r\nThen I'll check it and solve again myself..\r\n\r\nWhy are you thinking that ways??. :(",
  "Solution_4": "first of all no-one says like solve these problems untill 3pm unless there is a deadline of some sort and you need them by then. \r\n\r\nsplintercell is right, you cannot ask people to do your job instead. asking for help or advice is one, but what you did is something else. \r\n\r\nnext time, please post one problem at a time, in it's apropiate subforum, without giving any deadlines. \r\n\r\npeople will gladly help you if you do not force them to."
}
{
  "Tag": [
    "LaTeX",
    "calculus",
    "derivative",
    "function",
    "Support",
    "advanced fields",
    "advanced fields unsolved"
  ],
  "Problem": "I was looking at the exercises for the first chapter of this book, and I saw:\r\n\r\n1.6. Let $u \\in D'(R^{n})$ [$u$ is a distribution] have the property that $\\langle u, \\phi \\rangle \\ge 0$ for all real valued nonnegative $\\phi \\in C_{c}^\\infty(R^{n})$. Show that $u$ is of order 0. \r\n\r\nI read the chapter, but somehow I didn't notice and now can't find what is the order of a distribution! Can someone please explain (use this example for an explanation or show me how to do it, I'll probably learn it either way). \r\n\r\n\r\nAlso, how do you make that curly D in LaTeX, the one they use for the space of distributions?",
  "Solution_1": "I'm not sure, but I think that, intuitively, the order is the number of derivatives that the distribution \"sees.\"  For example, a 5th order distribution can be evaluated on any C^5 function, but it cannot be evaluated on a function that is only C^4.  Delta function is zero order, while $\\delta'$ would be first order.  I don't remember the precise definition, but there are probably several equivalent ways to say it.\r\n\r\nAs for the curly D, perhaps you mean $\\mathcal{D}.$",
  "Solution_2": "No, \\mathcal{D} ($\\mathcal{D}$) isn't what Xevarion is looking for. I don't know how to get the font he wants.\r\n\r\nOne could find and quote the topological technicalities behind yenlee's explanation, but in this case I'd start with the intuition, just as yenlee expressed it.",
  "Solution_3": "Well now I understand what is the [i]definition[/i], but I have no idea how to approach the problem from the text. I mean, intuitively, it seems like if we allow $u$ to \"see\" the derivative, then we can pick a test function that has arbitrarily large negative derivative on some interval (tho still satisfying the original condition) and contradict the condition on $\\langle u, \\phi \\rangle$. But I can't tell what information I get from assuming $u$ is of order $\\ge 1$ that allows me to pick an appropriate test function for the contradiction... \r\n\r\nAny suggestions? Thanks again for your help so far! :)",
  "Solution_4": "If you have a test-function $\\varphi$ supported by a compact $K$, then $|\\langle u,\\varphi\\rangle|\\le C(K)\\max|\\varphi|$. Indeed, just take some test-function $\\psi$ that is non-negative everywhere and identically $1$ on $K$ and consider the functions $M\\psi\\pm\\varphi$ where $M=\\max|\\varphi|$. They are both non-negative whence we can take $C(K)=\\langle u,\\psi\\rangle$. How to finish depends on the definition of order you use.",
  "Solution_5": "fedja, I'm sorry, but you seem to have mistaken me for somebody who understands math. It's probably not your fault, but I really have absolutely no idea what you said (except at least I understand what are $\\varphi, \\psi, K$, little help though that may be). Don't worry too much about explaining it, I'm getting tempted to give up and try to practice basic analysis more before I try to go adventuring into this Advanced Topics forum again.",
  "Solution_6": "OK, let me try to explain it in more detail.\r\n[b]Claim:[/b] For every compact $K$ there exists a finite positive constant $C(K)$ depending on $K$ such that, for every test-function $\\varphi$ supported by $K$, we have $|\\langle u,\\varphi\\rangle|\\le C(K)\\max|\\varphi|$.\r\nI don't know what definition of order you have in your book, but one of them says that this property is exactly what it means that $u$ has order $0$ and all other reasonable definitions should be equivalent to it. \r\n[b]Proof:[/b]. Fix any $C_{0}^\\infty$ function $\\psi$ that is non-negative everywhere and that is identically $1$ on $K$. Let $M=\\max|\\varphi|$. The function $M\\psi-\\varphi$ is non-negative. Hence $\\langle u,M\\psi-\\varphi\\rangle=M\\langle u,\\psi\\rangle-\\langle u,\\varphi\\rangle\\ge 0$, i.e., $\\langle u,\\varphi\\rangle\\le \\langle u,\\psi\\rangle\\max|\\varphi|$. Similarly, the function $M\\psi+\\varphi$ is non-negative. Hence $\\langle u,M\\psi+\\varphi\\rangle=M\\langle u,\\psi\\rangle+\\langle u,\\varphi\\rangle\\ge 0$, i.e., $\\langle u,\\varphi\\rangle\\ge-\\langle u,\\psi\\rangle\\max|\\varphi|$. Thus, $|\\langle u,\\varphi\\rangle|\\le \\langle u,\\psi\\rangle\\max|\\varphi|$ and we can put $C(K)=\\langle u,\\psi\\rangle$.\r\n\r\nIs it clearer now?",
  "Solution_7": "OH! Well, I guess it helps to understand when I know what you're actually trying to prove. So yes, I totally understand what you did. Thanks. However! I still have absolutely no idea how that could possibly be equivalent to the thing about derivatives...  :( The reason I couldn't understand before is that I thought you were trying to suggest that there was one trivial step left to get to THAT definition of order.  :maybe: \r\n\r\nAlso, using your way of defining order, what do the definitions of orders greater than 1 look like? I guess $|\\langle u, \\phi \\rangle| \\le C(K_{0})\\max|\\varphi|+C(K_{1})\\max|\\varphi'|+...$ where $C(K_{i})$ is a constant depending on the support of the $i^{th}$ derivative (can the derivative of a $C^\\infty$ function of compact support have different support from that of the original function? I shouldn't have to think about that. This is why I should go back and actually learn analysis before I try this stuff...)? Can we always separate an order $n$ distribution into $n$ parts each of which depends only on the $i^{th}$ derivative? \r\n\r\nSorry, I'm just trying to see what's going on. You don't need to answer all the questions, the only essential one is how yenlee/professor Merryfield's definition is equivalent to yours.",
  "Solution_8": "[quote=\"Xevarion\"] How is yenlee/professor Merryfield's definition  equivalent to yours?[/quote]\r\nIt is not completely trivial but not very hard either. Indeed, if we have the property that $|\\langle u,\\varphi\\rangle|\\le C(K)\\max|\\varphi|$, then we can extend the functional $u$ to a continuous linear functional on the space $C_{0}$ of continuous functions with compact support. To do it, just take any $f\\in C_{0}$ and approximate it in the uniform norm by infinitely smooth functions $\\varphi_{k}$ such that all the supports of $\\varphi_{k}$ are contained in some compact $K$ (possibly a bit larger than $\\mathrm{supp\\,}\\varphi$). Then the sequence $\\langle u,\\varphi_{k}\\rangle$ is a Cauchy sequence and, therefore, it has a limit. We define $\\langle u,f\\rangle$ to be that limit. One needs to check that it doesn't depend on the approximating sequence and that it is a continuous linear functional on $C_{0}$ but this is a fairly straightforward exercise, so try to do it yourself. Similarly, if $u$ has order $k$, it can be extended to a continuous linear functional on $C^{k}_{0}$. \r\nAs to the last question, the answer is again \"Yes\" in the following sense: if $u$ is of order $k$, then there are distributions $u_{0},\\dots,u_{k}$ of order $0$ such that $\\langle u,\\varphi\\rangle=\\sum_{j=0}^{k}\\langle u_{j},\\varphi^{(j)}\\rangle$ for every test function $\\varphi$, though this representation is by no means unique.",
  "Solution_9": "Cool! Thanks  :)"
}
{
  "Tag": [
    "function",
    "absolute value",
    "arithmetic sequence",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "Let f be a function from the reals to reals s.t. absolute value of f(x)-f(y) is smaller than or equal to the basoltue value of x-y. Prove that if the sequence x,f(x),f(f(x)),... is an arithmetic progression for any real x , then f(x)=x+a for some real a.\r\n\r\nBomb",
  "Solution_1": "This one already appeared on ML and was solved by Myth.\r\n\r\nPierre."
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "geometry unsolved"
  ],
  "Problem": "in $ \\Delta$ $ ABC$, we have $ AC\\equal{}2BC$. the tangents at $ A$ and $ C$ to the circumcircle of $ \\Delta$ $ ABC$ meets at $ P$.\r\nprove that the line $ BP$ bissects the arc $ BAC$ of the circumcircle.",
  "Solution_1": "[quote=\"saif\"]in $ \\Delta$ $ ABC$, we have $ AC \\equal{} 2BC$. the tangents at $ A$ and $ C$ to the circumcircle of $ \\Delta$ $ ABC$ meets at $ P$.\nprove that the line $ BP$ bissects the arc $ BAC$ of the circumcircle.[/quote]\r\n\r\ndo u mean the arc ABC or AIC  or BAC  :read: ???",
  "Solution_2": ":o  you see that i write $ BAC$ so it's $ BAC$ and it means the biggest arc $ BC$.",
  "Solution_3": "[quote=\"saif\"]in $ \\Delta$ $ ABC$, we have $ AC \\equal{} 2BC$. the tangents at $ A$ and $ C$ to the circumcircle of $ \\Delta$ $ ABC$ meets at $ P$.\nprove that the line $ BP$ bissects the arc $ BAC$ of the circumcircle.[/quote]\r\n\r\nok  mr saif\r\nthis is  a geometrical construction to the question and realy i want to know how will the line PB bisects the arc  BAC  :huh: \r\nthanks again for reply\r\nalmonkez : \r\n [url=http://www.servimg.com/image_preview.php?i=313&u=12899977][img]http://i89.servimg.com/u/f89/12/89/99/77/geomet10.gif[/img][/url]",
  "Solution_4": "you see now that the arc $ CBA$ is divided into to equal arcs so the result still hold  :wink:",
  "Solution_5": "sorry for the mistake i mean $ AC\\equal{}2AB$ .",
  "Solution_6": "[quote=\"saif\"]sorry for the mistake i mean $ AC \\equal{} 2AB$ .[/quote]\r\noh , now my point of view is right mr saif :mad: \r\nok we had now to redraw and solve\r\nthanks for all\r\nmr : mohamed\r\nalmonkez",
  "Solution_7": "$ PB$ is the symmedian .thus , if $ M$ is the midpoint of $ AC$ and $ K\\equal{}(PB,BC)$,  then  $ \\angle MBC\\equal{} \\angle LBA\\equal{} \\angle LCA$.it remains to prove that $ \\angle LBM\\equal{}\\angle MCB \\implies \\angle MBA\\minus{}\\angle LBA\\equal{}\\angle MCB,\\angle LBA\\equal{}\\angle MBC \\implies  \\angle MBA\\minus{} \\angle MBC\\equal{}\\angle MCB \\implies \\angle MBA \\equal{}\\angle MCB\\plus{} \\angle MBC,\\angle MCB\\plus{}\\angle MBC\\equal{}\\angle BMA \\implies \\angle BMA\\equal{}\\angle ABM$.\r\nwhich is directly deduced from the fact that $ AB\\equal{}AM$.",
  "Solution_8": "[quote=\"saif\"][color=darkred]Let $ w$ be the circumcircle of $ \\Delta ABC$ for which $ AC \\equal{} 2\\cdot AB$. Denote \n\n$ P\\in AA\\cap CC$ and $ \\{B,D\\} \\equal{} PB\\cap w$ . Prove that $ DB \\equal{} DC$ .[/color] [/quote]\n\n[color=darkblue][b][u]Proof[/u].[/b] Denote $ S\\in PB\\cap AC$ and the midpoint $ M$ of $ [AC]$ . Is well-known that the ray $ [BS$ is the $ B$-symmedian in $ \\Delta ABC$ . Therefore,\n\n$ AM \\equal{} AB$ and $ \\widehat {AMB}\\equiv\\widehat {ABM}\\equiv\\widehat {CBS}$ , i.e. $ \\widehat {AMB}\\equiv\\widehat {DBC}$ . Since $ \\widehat {BAM}\\equiv\\widehat {CDB}$ obtain $ \\Delta AMB\\stackrel{(a.a.)}{\\sim}\\Delta DBC$ , i.e. $ DB \\equal{} DC$ .[/color]\n\n[quote=\"Virgil Nicula\"][color=darkred][b][u]An easy extension[/u].[/b] Let $ ABC$ be a triangle inscribed in a circle $ w$ . Denote \n\n$ P\\in AA\\cap CC$ and $ \\{B,D\\} \\equal{} PB\\cap w$ . Prove that $ \\frac {DB}{DC} \\equal{} \\frac {2\\cdot AB}{AC}$ . [/color][/quote]\r\n\r\n[color=darkblue][b][u]Proof[/u].[/b]. The quadrilateral $ ABCD$ is harmonically, i.e. $ AB\\cdot DC \\stackrel{(*)}{ \\equal{} } BC\\cdot AD$ . From the [u][b]Ptolemeu[/b]'s theorem[/u] \n\nobtain $ AC\\cdot DB \\equal{} AB\\cdot DC \\plus{} BC\\cdot AD$ , i.e. $ AC\\cdot DB \\equal{} 2\\cdot AB\\cdot DC\\ \\implies\\ \\frac {DB}{DC} \\equal{} \\frac {2\\cdot AB}{AC}$ .\n\n[b][u]Remark[/u].[/b] Observe that $ 2\\cdot AB \\equal{} AC\\ \\implies\\ DB \\equal{} DC$ (the proposed problem).\n\nWe can obtain easily the relation $ (*)$ . Indeed, $ \\left\\|\\begin{array}{ccc} PAB\\sim PDA & \\implies & \\frac {PA}{PD} \\equal{} \\frac {AB}{DA} \\\\\n \\\\\nPCB\\sim PDC & \\implies & \\frac {PC}{PD} \\equal{} \\frac {CB}{DC}\\end{array}\\right\\|\\ \\stackrel{(PA \\equal{} PC)}{\\implies}\\ (*)$ .[/color]"
}
{
  "Tag": [
    "inequalities",
    "trigonometry",
    "triangle inequality",
    "trig identities",
    "Law of Cosines",
    "IMO Shortlist",
    "inequalities proposed"
  ],
  "Problem": "For  $a,b,c$ positive reals such that $\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c} =1$ prove that\r\n\r\n$\\sum \\sqrt{a^4-a^2b^2+b^4} \\geq \\frac{\\sqrt{3}}{2}abc $",
  "Solution_1": "i have a question.. why do you use $\\sum$ in this case? \r\ndo you mean \r\n\r\n\r\n$\\sqrt{a^4-a^2b^2+b^4} \\geq \\frac{\\sqrt{3}}{2}abc $",
  "Solution_2": "Hmmm...looks strange....where did I miss something?\r\n\r\n$a^4 - a^2b^2 + b^4 \\geq a^2b^2$ so that the lhs is at least $ab+bc+ca$, and this last term is $abc$.\r\nSince $\\frac {\\sqrt {3}} 2 < 1$, we have a stronger result, and best possible since equality occurs if and only if $a=b=c = 3$.\r\n\r\nPierre.",
  "Solution_3": "I am very,very,very stupid to post this inequality :( .\r\n\r\nThis was my bad reasoning.\r\n\r\nAt first I proved that\r\n\r\n$ \\sqrt{x^2+xy+y^2} +\\sqrt{x^2+xz+z^2 } \\geq \\sqrt{(x+y)^2 +(x+z)^2 +(x+y)(x+z)} $\r\n\r\nthen I made the substitution $x+y=c,x+z=b,x=a$ to obtain\r\n\r\n$\\sqrt{a^2-ab+b^2}+\\sqrt{b^2-bc+c^2}+\\sqrt{c^2-ca+a^2} \\geq\\frac{1}{2}(\\sqrt{a^2+ab+b^2}+\\sqrt{b^2+bc+c^2}+\\sqrt{c^2+ca+a^2})$\r\n\r\nNow I used\r\n\r\n$(\\sqrt{a^2+ab+b^2}+\\sqrt{b^2+bc+c^2}+\\sqrt{c^2+ca+a^2}$$ \\geq \\sqrt{3ab} +\\sqrt{3bc} +\\sqrt{3ca}  =\\sqrt{3}\\sqrt{abc}(\\frac{1}{\\sqrt{a}}+\\frac{1}{\\sqrt{b}}+\\frac{1}{\\sqrt{c}})$\r\n\r\nand finally I have substituted $a,b,c$ with$a^2,b^2,c^2$ and imposed the condition $ \\frac{1}{a} +\\frac{1}{b}+\\frac{1}{c} =1 $\r\n\r\nbut in my reasoning there is a big mistake :blush: !!!!!!!.\r\nIt was better to post my initial inequality\r\n\r\n[quote]$\\sqrt{a^2-ab+b^2}+\\sqrt{b^2-bc+c^2}+\\sqrt{c^2-ca+a^2} \\geq\\frac{1}{2}(\\sqrt{a^2+ab+b^2}+\\sqrt{b^2+bc+c^2}+\\sqrt{c^2+ca+a^2})$[/quote]\r\n\r\nwhich I hope it is not banal.\r\n\r\nThank you very much Pierre to have pointed it.\r\n\r\nSorry very much again :blush: .",
  "Solution_4": "[quote=\"manlio\"]\n$\\sqrt{a^2-ab+b^2}+\\sqrt{b^2-bc+c^2}+\\sqrt{c^2-ca+a^2} \\geq\\frac{1}{2}(\\sqrt{a^2+ab+b^2}+\\sqrt{b^2+bc+c^2}+\\sqrt{c^2+ca+a^2})$\n[/quote]\r\n\r\nThis follows from the triangle inequality:\r\n\r\n$\\sqrt{a^2-ab+b^2}+\\sqrt{b^2-bc+c^2} \\geq \\sqrt{c^2+ca+a^2}$ ---- (*)\r\n\r\nYou could see (*) by Minkowski, Law of Cosines or complex numbers.\r\n\r\nManlio, what is the original IMO SL problem?",
  "Solution_5": "It is problem g.2 of IMO shortlist 2002 you can find at\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=15588&highlight=shortlist"
}
{
  "Tag": [
    "calculus",
    "integration",
    "trigonometry",
    "logarithms",
    "calculus computations"
  ],
  "Problem": "I'm not sure about answer.It looks very strange.\r\n$ \\int_{1}^{e}\\frac {dx}{x\\sqrt {1 \\plus{} ln^2x}}$\r\n\r\nfor u=lnx-->u'=1/x\r\n\r\n$ \\int \\frac {du}{\\sqrt {1 \\plus{} u^2}}$\r\n\r\nsubstituting $ u \\equal{} tan\\theta$\r\n \r\n$ \\equal{} \\int \\frac {d\\theta}{cos\\theta} \\equal{} ln|sec\\theta \\plus{} tan\\theta|$\r\n\r\n$ \\int_{1}^{e}\\frac {dx}{x\\sqrt {1 \\plus{} ln^2x}} \\equal{} ln|\\sqrt { \\minus{} 1}|$",
  "Solution_1": "Let $ w \\equal{} lnx$, then $ xdw \\equal{} dx$\r\nWe get:\r\n\r\n$ \\int_{1}^{e}\\frac {dx}{x\\sqrt {1 \\plus{} ln^2x}} \\equal{} \\int_{0}^{1} \\frac {1}{\\sqrt {1 \\plus{} w^{2}}}dw$\r\nWith $ w \\equal{} \\tan(\\theta)$ and $ dw \\equal{} \\sec^{2}(\\theta)d\\theta$\r\n\r\nWe get:\r\n$ \\int_{0}^{\\frac{\\pi}{4}} \\frac {\\sec^{2}(\\theta)d\\theta}{\\sqrt {1 \\plus{} \\tan^{2}(\\theta)}} \\equal{} \\int_{0}^{\\frac {\\pi}{4}} \\sec\\theta d\\theta$",
  "Solution_2": "$ ln|sec\\frac{\\pi}{4}\\minus{}sec\\frac{\\pi}{4} tan\\frac{\\pi}{4}|\\minus{}ln|sec0\\minus{}sec0tan0|$\r\n$ ln|\\sqrt{2}\\minus{}\\sqrt{2}|\\minus{} ln|1\\minus{}0|\\equal{}ln0$:oops_sign:",
  "Solution_3": "$ \\int_0^{\\pi/4}\\sec \\theta\\,d \\theta \\equal{} [\\ln|\\sec \\theta \\plus{} \\tan \\theta|]_0^{\\pi/4} \\equal{} \\ln(\\sqrt{2} \\plus{} 1) \\minus{} \\ln(1 \\plus{} 0) \\equal{} \\ln(\\sqrt{2} \\plus{} 1)$."
}
{
  "Tag": [],
  "Problem": "Hello, This is a stupid question but I don't really know the answer:\r\nHow is 12,988,816 pronounced? :maybe: \r\nThanks!",
  "Solution_1": "I would pronounce it as:\r\n\r\n\"Twelve million, nine hundred eighty-eight thousand, eight hundred sixteen\"\r\n\r\nwhere commas indicate pauses.",
  "Solution_2": "At least in American English, there are also optional \"ands\" that isabella omitted.  Including them gives, \"Twelve million nine hundred and eighty-eight thousand, eight hundred and sixteen.\"",
  "Solution_3": "My teachers always told me to never say \"and\" in a number. But I guess that was their opinion."
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "vector",
    "geometry solved"
  ],
  "Problem": "Let $ABCDEF$ be a convex hexagon with $AB = BC, CD = DE$ and $EF = FA$. Prove that the lines through $C,E,A$ perpendicular to $BD,DF, FB$ are concurrent.",
  "Solution_1": "Obviously, the lines through $B,D,F$ perpendicular to $AC,CE,EA$ respectively, are concurrent (in the circumcenter of $ACE$). This means that the triangles $ACE,BDF$ are [url=http://mathworld.wolfram.com/OrthologicTriangles.html]orthologic[/url], i.e. the desired conclusion holds: the lines through $A,C,E$ to $FB,BD,DF$ are concurrent.",
  "Solution_2": "Thanks Grobber for quick response: actually I've come up with my solution just after posting - one simple lemma solves all:\r\n\r\nIn convex quadriteral $ABCD$ we have: diagonals are perpendicular if and only if the following holds:\r\n\r\n$|AB|^2+|CD|^2=|BC|^2+|AD|^2$\r\n\r\nsorry for bothering",
  "Solution_3": "Megus,can u elaborate? :?",
  "Solution_4": "Suppose we have proven lemma  :D  (actually proof of it is not hard but I dont like it - it's just some calculations on vectors) \r\n\r\nLet $X$ be the point of concurrency of perpendiculars from $A$ and $C$. So we have to prove that $EX$ is perpendicular to $FD$. Now our lemma come handy. Consider quadriterals $FXAB$ and $BXDC$ - use lemma to them and you obtain as a result an equality (after using equality of some sides) which induce that quadriteral $EFXD$ has perpendicular diagonals. Sorry for my messy language above - I tried to elaborate a bit and avoid calculations simultanously  :lol:",
  "Solution_5": "[quote=\"Megus\"]Let $ABCDEF$ be a convex hexagon with $AB = BC, CD = DE$ and $EF = FA$. Prove that the lines through $C,E,A$ perpendicular to $BD,DF, FB$ are concurrent.[/quote]\r\n\r\nI know this is hyper-useless now, but let me note that the same problem has been discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=1240 .\r\n\r\n  darij"
}
{
  "Tag": [
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "Let $a_{k},b_{k}, c_{k} >0$, $ \\forall k=1,n$. Prove that:\r\n\r\n$ \\frac{ \\sum_{k=1}^{n} a_{k}^2} { \\sum_{k=1}^{n} a_{k}} +  \\frac{ \\sum_{k=1}^{n} b_{k}^2} { \\sum_{k=1}^{n} b_{k}} + \\frac{ \\sum_{k=1}^{n} c_{k}^2} { \\sum_{k=1}^{n} c_{k}} \\geq \\frac{ \\sum_{k=1}^{n} (a_{k}+b_{k}+c_{k})^2} { \\sum_{k=1}^{n} (a_{k}+b_k+c_k)}$\r\n\r\nmade by Lascu M.\r\n\r\ncheers! :D :D  :cool:",
  "Solution_1": "Beautiful inequality, thanks Lagrangia!\r\n\r\nThe inequality can be generalized to n positive sequences (and since I am a masochist, I write it up using the $\\sum$ notation):\r\n\r\nIf $a_{1,1},$ $a_{1,2},$ $...,$ $a_{1,n},$ $a_{2,1},$ $a_{2,2},$ $...,$ $a_{2,n},$ $...,$ $a_{u,1},$ $a_{u,2},$ $...,$ $a_{u,n}$ are positive numbers, then\r\n\r\n$\\Huge \\sum_{i=1}^{u}\\frac{\\sum_{k=1}^{n}a_{i,k}^{2}}{\\sum_{k=1}^{n}a_{i,k}}\\geq \\frac{\\sum_{k=1}^{n}\\left( \\left( \\sum_{i=1}^{u}a_{i,k}\\right) ^{2}\\right) }{\\sum_{k=1}^{n}\\sum_{i=1}^{u}a_{i,k}}$.\r\n\r\nMy [i]proof[/i] uses two lemmata:\r\n\r\n[b]Lemma 1 (Cauchy-Schwartz inequality in the Arthur Engel form).[/b] If $b_{1},$ $b_{2},$ $...,$ $b_{u}$ are arbitrary reals, and $c_{1},$ $c_{2},$ $...,$ $c_{u}$ are positive reals, then\r\n\r\n$\\Huge \\sum_{i=1}^{u}\\frac{b_{i}^{2}}{c_{i}}\\geq \\frac{\\left(\\sum_{i=1}^{u}b_{i}\\right) ^{2}}{\\sum_{i=1}^{u}c_{i}}$.\r\n\r\n[i]Proof.[/i] This is quite an unspectacular application of the Cauchy-Schwartz inequality:\r\n\r\n$\\Huge \\sum_{i=1}^{u}\\frac{b_{i}^{2}}{c_{i}}\\sum_{i=1}^{u}c_{i}=\\sum_{i=1}^{u}\\left( \\left( \\frac{b_{i}}{\\sqrt{c_{i}}}\\right) ^{2}\\right) \\sum_{i=1}^{u}\\left( \\left( \\sqrt{c_{i}}\\right) ^{2}\\right) \\geq \\left( \\sum_{i=1}^{u}\\left( \\frac{b_{i}}{\\sqrt{c_{i}}}\\sqrt{c_{i}}\\right) \\right)^{2}=\\left( \\sum_{i=1}^{u}b_{i}\\right) ^{2}$.\r\n\r\n[b]Lemma 2 (Minkowski inequality).[/b] If $a_{1,1},$ $a_{1,2},$ $...,$ $a_{1,n},$ $a_{2,1},$ $a_{2,2},$ $...,$ $a_{2,n},$ $...,$ $a_{u,1},$ $a_{u,2},$ $...,$ $a_{u,n}$ are arbitrary real numbers, then\r\n\r\n$\\Huge \\sum_{i=1}^{u}\\sqrt{\\sum_{k=1}^{n}a_{i,k}^{2}}\\geq \\sqrt{\\sum_{k=1}^{n}\\left( \\left( \\sum_{i=1}^{u}a_{i,k}\\right) ^{2}\\right) }$.\r\n\r\n[i]Proof[/i] (for the sake of completeness)[i].[/i] By squaring, our inequality becomes\r\n\r\n$\\Huge \\left( \\sum_{i=1}^{u}\\sqrt{\\sum_{k=1}^{n}a_{i,k}^{2}}\\right) ^{2}\\geq \\sum_{k=1}^{n}\\left( \\left( \\sum_{i=1}^{u}a_{i,k}\\right) ^{2}\\right)$,\r\n\r\nUsing the formula\r\n\r\n$\\Huge \\left( \\sum_{i=1}^{u}x_{i}\\right)^{2}=\\sum_{i=1}^{u}x_{i}^{2}+2\\cdot\\sum_{1\\leq i<j\\leq u}\\left( x_{i}x_{j}\\right)$,\r\n\r\nthis is rewritten as\r\n\r\n$\\Huge \\sum_{i=1}^{u}\\sum_{k=1}^{n}a_{i,k}^{2}+2\\cdot \\sum_{1\\leq i<j\\leq u}\\sqrt{\\sum_{k=1}^{n}a_{i,k}^{2}\\cdot \\sum_{k=1}^{n}a_{j,k}^{2}}\\geq \\sum_{k=1}^{n}\\left( \\sum_{i=1}^{u}a_{i,k}^{2}+2\\cdot \\sum_{1\\leq i<j\\leq u}\\left( a_{i,k}a_{j,k}\\right) \\right)$,\r\n\r\nor\r\n\r\n$\\Huge \\sum_{i=1}^{u}\\sum_{k=1}^{n}a_{i,k}^{2}+2\\cdot \\sum_{1\\leq i<j\\leq u}\\sqrt{\\sum_{k=1}^{n}a_{i,k}^{2}\\cdot \\sum_{k=1}^{n}a_{j,k}^{2}}\\geq \\sum_{k=1}^{n}\\sum_{i=1}^{u}a_{i,k}^{2}+2\\cdot \\sum_{k=1}^{n}\\sum_{1\\leq i<j\\leq u}\\left( a_{i,k}a_{j,k}\\right)$.\r\n\r\nHowever, obviously\r\n\r\n$\\Huge \\sum_{i=1}^{u}\\sum_{k=1}^{n}a_{i,k}^{2}=\\sum_{k=1}^{n}\\sum_{i=1}^{u}a_{i,k}^{2}$,\r\n\r\nso that it remains to prove\r\n\r\n$\\Huge \\sum_{1\\leq i<j\\leq u}\\sqrt{\\sum_{k=1}^{n}a_{i,k}^{2}\\cdot \\sum_{k=1}^{n}a_{j,k}^{2}}\\geq \\sum_{k=1}^{n}\\sum_{1\\leq i<j\\leq u}\\left( a_{i,k}a_{j,k}\\right)$,\r\n\r\nor, equivalently,\r\n\r\n$\\Huge \\sum_{1\\leq i<j\\leq u}\\sqrt{\\sum_{k=1}^{n}a_{i,k}^{2}\\cdot \\sum_{k=1}^{n}a_{j,k}^{2}}\\geq \\sum_{1\\leq i<j\\leq u}\\sum_{k=1}^{n}\\left( a_{i,k}a_{j,k}\\right)$.\r\n\r\nBut actually, for any i and j, we have\r\n\r\n$\\Huge \\sqrt{\\sum_{k=1}^{n}a_{i,k}^{2}\\cdot \\sum_{k=1}^{n}a_{j,k}^{2}}\\geq \\sum_{k=1}^{n}\\left( a_{i,k}a_{j,k}\\right)$\r\n\r\nby the Cauchy-Schwartz inequality. Hence Lemma 2 is proven.\r\n\r\nNow to the proof of the main inequality: By Lemma 1, we have\r\n\r\n$\\Huge \\sum_{i=1}^{u}\\frac{\\sum_{k=1}^{n}a_{i,k}^{2}}{\\sum_{k=1}^{n}a_{i,k}}=\\sum_{i=1}^{u}\\frac{\\left( \\sqrt{\\sum_{k=1}^{n}a_{i,k}^{2}}\\right) ^{2}}{\\sum_{k=1}^{n}a_{i,k}}\\geq \\frac{\\left( \\sum_{i=1}^{u}\\sqrt{\\sum_{k=1}^{n}a_{i,k}^{2}}\\right) ^{2}}{\\sum_{i=1}^{u}\\sum_{k=1}^{n}a_{i,k}}$.\r\n\r\nBut Lemma 2 yields\r\n\r\n$\\Huge \\left( \\sum_{i=1}^{u}\\sqrt{\\sum_{k=1}^{n}a_{i,k}^{2}}\\right) ^{2}\\geq \\sum_{k=1}^{n}\\left( \\left( \\sum_{i=1}^{u}a_{i,k}\\right) ^{2}\\right)$,\r\n\r\nso that we get\r\n\r\n$\\Huge \\sum_{i=1}^{u}\\frac{\\sum_{k=1}^{n}a_{i,k}^{2}}{\\sum_{k=1}^{n}a_{i,k}}\\geq \\frac{\\left( \\sum_{i=1}^{u}\\sqrt{\\sum_{k=1}^{n}a_{i,k}^{2}}\\right) ^{2}}{\\sum_{i=1}^{u}\\sum_{k=1}^{n}a_{i,k}}\\geq \\frac{\\sum_{k=1}^{n}\\left( \\left( \\sum_{i=1}^{u}a_{i,k}\\right) ^{2}\\right) }{\\sum_{i=1}^{u}\\sum_{k=1}^{n}a_{i,k}}=\\frac{\\sum_{k=1}^{n}\\left( \\left( \\sum_{i=1}^{u}a_{i,k}\\right) ^{2}\\right) }{\\sum_{k=1}^{n}\\sum_{i=1}^{u}a_{i,k}}$,\r\n\r\nand the inequality is proven.\r\n\r\nNow I hope to have convinced everybody of the superiority of the $\\sum$ notation :-)\r\n\r\n  Darij"
}
{
  "Tag": [
    "vector",
    "linear algebra"
  ],
  "Problem": "Can someone please explain the concept of why the dot product of the normals of two planes is 0 if they are perpendicular? Our teacher told us that it is so, but he can't explain why.",
  "Solution_1": "The intersection of two planes that aren't parallel is a line, and it can be parameterized.  You can find information on this in any introductory textbook on linear algebra; we can, for example, set $ z \\equal{} t$ which gives $ x \\equal{} 5 \\plus{} 3t, y \\equal{} 5 \\minus{} 4t$.",
  "Solution_2": "[quote=\"t0rajir0u\"]The intersection of two planes that aren't parallel is a line, and it can be parameterized.  You can find information on this in any introductory textbook on linear algebra; we can, for example, set $ z \\equal{} t$ which gives $ x \\equal{} 5 \\plus{} 3t, y \\equal{} 5 \\minus{} 4t$.[/quote]\r\n\r\noops, sorry, i already knew the answer, so i changed the question. I didn't know someome would post immediately.",
  "Solution_3": "[hide=\"Hint\"]The dot-product of two vectors is zero iff. they are perpendicular.[/hide]\n[hide=\"Off Topic\"] :wink: Of course, I won't draw any premature conclusions, I'm just pickin'\n[quote=\"guile in another thread\"][quote=\"JBL\"]1) Do your own homework.[/quote]\n\nThis is not a homework. Im studying linear algebra on my own in preparation for the next semester. Im just answering exercises on the lecture notes i have borrowed from my friends who have already taken linear algeba.  :D[/quote]\n[quote=\"guile in this thread\"]Our teacher told us that it is so, but he can't explain why.[/quote][/hide]"
}
{
  "Tag": [
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "Given a natural sequence $ \\{a_n\\}$ such that\r\n$ (a_{2n}\\plus{}a_{2n\\plus{}1}\\plus{}1)(a_{2n\\plus{}1}\\minus{}a_{2n}\\minus{}1)\\equal{}3(1\\plus{}2a_n)$\r\n$ a_{2n} \\geq a_n$ with all $ n \\in N$\r\nFind all $ n \\in N$ such that $ a_n \\leq 2009$",
  "Solution_1": "posted before $ a_{2n}=3a_n,a_{2n+1}=a_{2n}+2$. Therefore if $ a_1=a, n=a_0+a_1*2+...+a_k*2^k$, then ${ a_n=2[a_0+a(a_1*3+...+a_k*3^k)}.$\r\nMaximal numbers $ a_n<2009$, when $ a_1=a=1$ and $ 108$.",
  "Solution_2": "[quote=\"hcccccccccc\"]Given a natural sequence $ \\{a_n\\}$ such that\n$ (a_{2n} \\plus{} a_{2n \\plus{} 1} \\plus{} 1)(a_{2n \\plus{} 1} \\minus{} a_{2n} \\minus{} 1) \\equal{} 3(1 \\plus{} 2a_n)$\n$ a_{2n} \\geq a_n$ with all $ n \\in N$\nFind all $ n \\in N$ such that $ a_n \\leq 2009$[/quote]\r\n\r\nIt seems this is a problem from the still running  MYM 45th anniversary contest.\r\nsee http://www.mathlinks.ro/Forum/viewtopic.php?p=1568836#1568836\r\n\r\nIf so, posting the problem to find solution from others is a rather unfair behaviour, IMHO."
}
{
  "Tag": [
    "function",
    "integration",
    "real analysis",
    "real analysis solved"
  ],
  "Problem": "Given a sequence of continuous functions $f_n:[0,1],\\ n\\ge 1$ s.t. $\\int^1_0f_i(x)f_j(x)dx=1$ if $i=j$ and $0$ otherwise and $\\sup\\{f_n(x)|x\\in[0,1],n\\ge 1\\}<\\infty$, show that it has no pointwise convergent subsequence.",
  "Solution_1": "What is $ \\sup\\{f_n(x)|x\\in[0,1],n\\ge 1\\}<\\infty$ in your opinion?",
  "Solution_2": "I have just seen official source. I can't understand why such obvious problem was used for IMC as B6 for 20 points?",
  "Solution_3": "Yeah, it does seem a bit obvious. Might as well give an answer: the condition of uniform boundedness is more then enough to invoke the Dominated Convergence Theorem, so a pointwise convergent subsequence would also converge in $L^2.$ But every function in an orthonormal set must be at a distance of $\\sqrt{2}$ from every other function in that set, and hence an orthonormal set can have no convergent subsequences."
}
{
  "Tag": [],
  "Problem": "This is my first game, and these are the rules.\r\nThe first person will post a word in BOLD in a certain topic, like board games (I will post the topic), and the second player will post a word with the 1st letter being the same as the last letter of the previous post. \r\n\r\nExample:\r\nPlayer 1: Monopoly\r\nPlayer 2: Yahtzee\r\nPlayer 3: Ummm..... \r\nSo since Player 2 got 3 stumped, he/she gains 2 points. \r\nIf Player 3 has any points (which is not the case), he/she loses 3 points because he/she couldn't think of a word fitting the topic. There is no penalty if the player has 0 points. If you post with a word, then you get 1 point.\r\n\r\nScoreboard:\r\nPlayer 1: 1\r\nPlayer 2: 2\r\n\r\nPlease post the scoreboard.  Are the rules clear enough?\r\n\r\nI don't get any points for starting us off. \r\n\r\nThe topic is..... food. (Every once in a while I'll change the topic.) \r\n\r\nOk..... \r\n[b]Hamburger[/b]\r\n\r\n[u]Scoreboard[/u]\r\nNobody yet...",
  "Solution_1": "*SIGH*..... Somebody please delete this.",
  "Solution_2": "remember: DO NOT GIVE UP HOPE,\r\n\r\nespecially within 30 minutes of posting your game.\r\n\r\nAfter all, these games require time, LOTSA time.\r\n\r\n-Raspberries-\r\n\r\nScoreboard:\r\n\r\nlouiethechewy123-0\r\n\r\ntcjy-1",
  "Solution_3": "Uhhh, the game is in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=283787]this[/url] topic.",
  "Solution_4": "Yeah, randomly caught the FAMOUS Virus called copied posting. It breaks the \"Cannot post consecutively for 10 min\" law, too."
}
{
  "Tag": [],
  "Problem": "I'm just curious about the heights of everyone.\r\n\r\nI'm only 5'4\"-ish...i'm pretty short for an 8th grader.... :(",
  "Solution_1": "1.87 meters :)",
  "Solution_2": "sorry I have no idea what everyone's height is\r\nin fact, I doubt anybody has\r\nit's not something you commonly take pains to ask and note down, you see",
  "Solution_3": "6'11\"",
  "Solution_4": "7'2. I'm tall for my age.\r\n\r\n\r\n\r\nok, ok. I'm 5'2.",
  "Solution_5": "5'8 or so?",
  "Solution_6": "4'6\r\n\r\n[color=white]The message is too small. Please make the message longer before submitting.[/color]",
  "Solution_7": "around 5'3 or so",
  "Solution_8": "[quote=\"1=2\"]7'2. I'm tall for my age.\n\n\n\nok, ok. I'm 5'2.[/quote]\r\nyippeee i think now i am not abnormally short....wait, how old were you again?\r\n :(",
  "Solution_9": "im 5 feet 3 inches...\r\nif it makes any of you feel better my grandmother is 4 feet 3 inches",
  "Solution_10": "[quote=\"funcia\"]\nyippeee i think now i am not abnormally short....wait, how old were you again?\n :([/quote]He's 11.",
  "Solution_11": "*curses under breath* and i'm almost 14(in june hahahaha got june 93)",
  "Solution_12": "About 5'8''. i'm tall and thin compared to those in my school.",
  "Solution_13": "5'7/5'8.  something like that",
  "Solution_14": "6'1ish ",
  "Solution_15": "lol. everyone thinks that BOB's actually short even though he's not. BOB YOU WERE SHORT LAST YEAR. I told him that at a nats practice last year and he got really p^$\\$\\$$ed.\r\n\r\nOh ya and I'm 5'11\"\r\n\r\nyou're still safe from stalkers BOB",
  "Solution_16": "5'7\" about\r\n\r\nman, eating cheese will make u tall or what? or is it that only white ppl live in cheese country cause we all know azns are short",
  "Solution_17": "1.74 meters",
  "Solution_18": "About 5'11\"...",
  "Solution_19": "5'3\"-5'4\" ish\r\n\r\nFuncia, don't be so sad. Aren't you a girl? (terribally sorry if I am wrong)\r\nGirls are naturally shorter.",
  "Solution_20": "[quote=\"pianoforte\"]5'7\" about\n\nman, eating cheese will make u tall or what? or is it that only white ppl live in cheese country cause we all know azns are short[/quote]\r\n\r\nMilk is good for your bones :P . There are some asians in Madison, but I don't think that there are many farther north.",
  "Solution_21": "1.81 m , but I can jump high enough.",
  "Solution_22": "[quote=\"pianoforte\"]5'7\" about\n\nman, eating cheese will make u tall or what? or is it that only white ppl live in cheese country cause we all know azns are short[/quote]\r\nwell i guess maybe if i ate cheese i could be taller but cheese is umm really icky\r\nall dairy products are errr like cheese :what?: \r\n\r\nand yeah i'm a girl but i still am really short",
  "Solution_23": "There was a thread about this a long time ago. I remember because Ashegh said he was taller than me! :o\r\n\r\nI'm 6'5\" $\\approx$ 197 cm.",
  "Solution_24": "wow you tall! I'm around Funcias height.",
  "Solution_25": "[quote=\"spix\"]1.81 m , but I can jump high enough.[/quote]Can you dunk a regular basketball? :D Cheese doesn't help you: I ate lots of cheese while I was young, but I'm still at a measly 5'10'' (1.78m) :(",
  "Solution_26": "[quote=\"asmallgoat\"]wow you tall! I'm around Funcias height.[/quote]\r\nyou forget to add the 1 & 1/2 years younger part  :(",
  "Solution_27": "6'6'' counting my hair.",
  "Solution_28": "[quote=\"mustafa\"][quote=\"pianoforte\"]5'7\" about\n\nman, eating cheese will make u tall or what? or is it that only white ppl live in cheese country cause we all know azns are short[/quote]\n\nMilk is good for your bones :P . There are some asians in Madison, but I don't think that there are many farther north.[/quote]\r\n\r\n??I'm azn and i live farther north!",
  "Solution_29": "around 5 feet last time i measured i was 4 11",
  "Solution_30": "never mind, I just measured myself. I'm 6'7'' counting my hair. Without my hair, I'm a little shorter.",
  "Solution_31": "1.89 m . high tower  :!:",
  "Solution_32": "like 6'4\" if i stick my hair straight up?\r\n\r\nsadly i'm only 5'11\" normally",
  "Solution_33": "[quote=\"footballrocks41237\"]\nMilk is good for your bones :P . There are some asians in Madison, but I don't think that there are [b][size=150]many[/size][/b] farther north.[/quote]\n\n??I'm azn and i live farther north![/quote]\r\n\r\nAhem."
}
{
  "Tag": [
    "conics",
    "ellipse"
  ],
  "Problem": "For a ellipse\r\n\r\n$\\frac{x^{2}}{4}+y^{2}$ = 1 \r\n\r\nhow do you know that the auxillary circle has eqn $x^{2}+y^{2}= 4$\r\n\r\n--------------------",
  "Solution_1": "The radius of the auxiliary circle would be the same as the semimajor axis of the ellipse. The ellipse has axes  length 4 (x-direction) and length 2 (y-direction), So the major axis is 4 and the semimajor axis is $\\frac{4}{2}= 2$. Thus the circle is $x^{2}+y^{2}= 2^{2}$.\r\n\r\nIn general, for an ellipse with equation $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}= 1$, we get an auxiliary circle of $x^{2}+y^{2}= \\max\\{a^{2},b^{2}\\}$"
}
{
  "Tag": [
    "search",
    "Ring Theory",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "Show that the ring $\\mathbb Z \\left[ \\tfrac{1+\\sqrt{-19}}{2}\\right]$ is not euclidian to any norm.\r\n\r\n[hide=\"hint\"]prove first it's not euclidian to complex norm. what do we know about the ring of units?[/hide]",
  "Solution_1": "is it easy? :o \r\nI think it was an ariticle. :ninja: \r\ndo you know an easy proof?\r\nI am eagerly waiting for it.",
  "Solution_2": "It was a class exercice, so it's definitely not unsolvable or a research article. But I admit I haven't found its solution yet.",
  "Solution_3": "No one? :)",
  "Solution_4": "As my friend has mentioned, this is due to [i]J.C Wilson[/i], A principal ideal ring that is not a Euclidean ring, Math. Mag $46, 34-8 (1973)$. Do a search on the web and you'll certainly find lots of related links.",
  "Solution_5": "check out this \"instruction\" :-)\r\n\r\nhttp://math.berkeley.edu/~gbergman/grad.hndts/nonEucPID.ps"
}
{
  "Tag": [
    "Sequence",
    "algebra",
    "algebraic identities",
    "IMO",
    "IMO 1967"
  ],
  "Problem": "Let $a_1,\\ldots,a_8$ be reals, not all equal to zero. Let \n\\[ c_n = \\sum^8_{k=1} a^n_k\\] \nfor $n=1,2,3,\\ldots$. Given that among the numbers of the sequence $(c_n)$, there are infinitely many equal to zero, determine all the values of $n$ for which $c_n = 0.$",
  "Solution_1": "Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)",
  "Solution_2": "$c_n$ must be zero for all odd $n$. \r\n\r\nProof:\r\nWLOG suppose that $a_1 \\geq a_2 \\geq ... \\geq a_8$. If $a_1+a_8 > 0$ then for sufficiently high odd $n$, $c_n$ will be dominated by $a_1$ alone i.e. it will always be positive. Similarly if $a_1+a_8 < 0$; hence $a_1=-a_8$. Now for odd $n$ these terms cancel, so we can repeat for the remaining values. Now all the terms cancel for all odd $n$. Since some $a_i$ is nonzero $c_n > 0$ for even $n$.",
  "Solution_3": "QED\r\nMany early IMO problems are simply jokes in comparing with present problems.",
  "Solution_4": "$n$ can't be even since that would mean that all $a_1,...,a_8=0$ which goes against the conditions.\nIf n is even.\nNow we must prove that for all odd n the statement is true.\nAssume the opposite that:\n$a_1>=a_2>=.....>=a_8$\nCase 1: $a_1+a_8>0$\nThen we would have that $c_n$ is infinite for huge n and then there wouldn't be infinitely many $c_n$ which are equal to 0.\nCase 2: $a_1+a_8<0$\nThen we would have that $c_n$ is infinite for huge n and then there wouldn't be infinitely many $c_n$ which are equal to 0.\nTherefore, $a_1+a_8=0$.\nNow we can repeat this process for the pairs."
}
{
  "Tag": [
    "inequalities proposed",
    "inequalities"
  ],
  "Problem": "Let $x,y,z \\geq 0$.Prove that\r\n$(2x+y+z)(2y+x+z)(2z+x+y)xyz \\leq (x+y)^{2}(y+z)^{2}(z+x)^{2}$\r\n\r\nP.S.I have a solution with opening brackets,does anyone have an elegant solution? :maybe:",
  "Solution_1": "RHS- LHS  $= (x+y+z) \\left(\\sum x^{2}(y+z)(y-z)^{2}\\right) \\geq 0$\r\nNot hard ! :lol:",
  "Solution_2": "Since the inequality is homogenous we can assume that $x+y+z=3.$ Then the inequality turns to $(3+x)(3+y)(3+z)xyz\\leq (3x+yz)(3y+zx)(3z+xy)$ $\\iff$ $(3+x)(3+y)(3+z)\\leq (3+\\frac{yz}{x})(3+\\frac{zx}{y})(3+\\frac{xy}{z}).$ By $Cauchy-Schwarz \\ inequality$ we have $(3+\\frac{yz}{x})(3+\\frac{zx}{y})\\geq (3+z)^2.$ \nMultiplying similar inequalities we get the result.\n\n[color=#0f0]Note:[/color] $(x+y)^2(y+z)^2(z+x)^2=(x+y)(y+z)\\cdot (y+z)(z+x)\\cdot (z+x)(x+y)=((y(x+y+z)+zx)((z(x+y+z)+xy)(x(x+y+z)+yz)=(3x+yz)(3y+zx)(3z+xy).$\n\n\n",
  "Solution_3": "[quote=henderson]Since the inequality is homogenous we can assume that $x+y+z=3.$ Then the inequality turns to $(3+x)(3+y)(3+z)xyz\\leq (3x+yz)(3y+zx)(3z+xy)$ $\\iff$ $(3+x)(3+y)(3+z)\\leq (3+\\frac{yz}{x})(3+\\frac{zx}{y})(3+\\frac{xy}{z}).$ By $Cauchy-Schwarz \\ inequality$ we have $(3+\\frac{yz}{x})(3+\\frac{zx}{y})\\geq (3+z)^2.$ \nMultiplying similar inequalities we get the result.\n\n[color=#0f0]Note:[/color] $(x+y)^2(y+z)^2(z+x)^2=(x+y)(y+z)\\cdot (y+z)(z+x)\\cdot (z+x)(x+y)=((y(x+y+z)+zx)((z(x+y+z)+xy)(x(x+y+z)+yz)=(3x+yz)(3y+zx)(3z+xy).$[/quote]\n\nI cannot understand your method. How can we assume the sum of x,y,z ?No such statement is given in the question?",
  "Solution_4": "[quote=Gibbenergy]RHS- LHS  $= (x+y+z) \\left(\\sum x^{2}(y+z)(y-z)^{2}\\right) \\geq 0$\nNot hard ! :lol:[/quote]\n\nVery long",
  "Solution_5": "[quote=cefer]Let $x,y,z \\geq 0$.Prove that\n$(2x+y+z)(2y+x+z)(2z+x+y)xyz \\leq (x+y)^{2}(y+z)^{2}(z+x)^{2}$\nP.S.I have a solution with opening brackets,does anyone have an elegant solution? :maybe:[/quote]\nPut $p=a+b+c,\\,q=ab+bc+ca$ and $r=abc$ inequality become\n\\[(pq-r)^2 \\geqslant r(2p^3+pq+r),\\]\nor\n\\[pq^2 \\geqslant (2p^2+3q)r.\\]\nWhich is true because $r \\leqslant \\frac{q^2}{3p},$ and \n\\[pq^2-(2p^2+3q) \\cdot \\frac{q^2}{3p} = \\frac{q^2(p^2-3q)}{3p} \\geqslant 0.\\]\nThe proof is complete.",
  "Solution_6": "[quote=cefer]Let $x,y,z \\geq 0$.Prove that\n$(2x+y+z)(2y+x+z)(2z+x+y)xyz \\leq (x+y)^{2}(y+z)^{2}(z+x)^{2}$\nP.S.I have a solution with opening brackets,does anyone have an elegant solution? :maybe:[/quote]\nSuppose $x \\geqslant y \\geqslant z,$ apply AM-GM Inequality, we have \n\\[xyz(2x+y+z)(2y+x+z)(2z+x+y) \\leqslant \\frac{1}{16}[4zx(2y+z+x)+y(2z+x+y)(2x+y+z)]^2.\\]\nWe need to prove\n\\[4(x+y)(y+z)(z+x) \\geqslant 4zx(2y+z+x)+y(2z+x+y)(2x+y+z),\\]\nequivalent to\n\\[y(2x-y-z)(x+y-2z) \\geqslant 0.\\]\nWhich is true.",
  "Solution_7": "[quote=tarzanjunior][quote=henderson]Since the inequality is homogenous we can assume that $x+y+z=3.$ Then the inequality turns to $(3+x)(3+y)(3+z)xyz\\leq (3x+yz)(3y+zx)(3z+xy)$ $\\iff$ $(3+x)(3+y)(3+z)\\leq (3+\\frac{yz}{x})(3+\\frac{zx}{y})(3+\\frac{xy}{z}).$ By $Cauchy-Schwarz \\ inequality$ we have $(3+\\frac{yz}{x})(3+\\frac{zx}{y})\\geq (3+z)^2.$ \nMultiplying similar inequalities we get the result.\n\n[color=#0f0]Note:[/color] $(x+y)^2(y+z)^2(z+x)^2=(x+y)(y+z)\\cdot (y+z)(z+x)\\cdot (z+x)(x+y)=((y(x+y+z)+zx)((z(x+y+z)+xy)(x(x+y+z)+yz)=(3x+yz)(3y+zx)(3z+xy).$[/quote]\n\nI cannot understand your method. How can we assume the sum of x,y,z ?No such statement is given in the question?[/quote]\n\nLet $x=kx_1, y=ky_1$ and $z=kz_1$ such that $x_1+y_1+z_1=3.$ Then substituting $x,y,z$ in the given inequality we have to prove:\n\n$(2x_1+y_1+z_1)(2y_1+x_1+z_1)(2z_1+x_1+y_1)x_1y_1z_1 \\leq (x_1+y_1)^{2}(y_1+z_1)^{2}(z_1+x_1)^{2}$ where $x_1+y_1+z_1=3.$ (you can observe that this is same with the given inequality with additional condition $x+y+z=3.$) So, we can assume that $x+y+z=3.$\n[color=#f00]Note:[/color] If the inequality is homogenous, we can also give any certain value to $x+y+z,$ $xyz,$ $xy+yz+zx,$ ...",
  "Solution_8": "[quote=cefer]Let $x,y,z \\geq 0$.Prove that\n$(2x+y+z)(2y+x+z)(2z+x+y)xyz \\leq (x+y)^{2}(y+z)^{2}(z+x)^{2}$\n\nP.S.I have a solution with opening brackets,does anyone have an elegant solution? :maybe:[/quote]\n\n[img]http://s1.sinaimg.cn/middle/006ptkjAzy77zpNEEUw40&690[/img]",
  "Solution_9": "[quote=cefer]Let $x,y,z \\geq 0$.Prove that\n$(2x+y+z)(2y+x+z)(2z+x+y)xyz \\leq (x+y)^{2}(y+z)^{2}(z+x)^{2}$\n\nP.S.I have a solution with opening brackets,does anyone have an elegant solution? :maybe:[/quote]\n\n\noneline down",
  "Solution_10": "Nice inequality.",
  "Solution_11": "It is well-known that in triangle holds sin<A+sin<B+sin<C<=3sqrt(3)/2. Use that for a=x+y... then calcul all sines in function of a,b,c.",
  "Solution_12": "[quote=cefer]Let $x,y,z \\geq 0$.Prove that\n$(2x+y+z)(2y+x+z)(2z+x+y)xyz \\leq (x+y)^{2}(y+z)^{2}(z+x)^{2}$\n\nP.S.I have a solution with opening brackets,does anyone have an elegant solution? :maybe:[/quote]\n\nstronger:\n\n\\[{\\frac { \\left( y+z \\right) ^{2} \\left( z+x \\right) ^{2} \\left( x+y\n \\right) ^{2}}{xyz \\left( 2\\,y+z+x \\right)  \\left( 2\\,z+x+y \\right) \n \\left( z+2\\,x+y \\right) }}\\geq {\\frac {27}{64}}\\,{\\frac { \\left( y+z\n \\right) ^{2} \\left( z+x \\right) ^{2} \\left( x+y \\right) ^{2}}{\n \\left( x+y+z \\right) ^{3}xyz}}\\geq 1\\]\n\nLeft is am-gm.\nplease nice do right.",
  "Solution_13": "[quote=xzlbq]\n\nstronger:\n\n\\[{\\frac { \\left( y+z \\right) ^{2} \\left( z+x \\right) ^{2} \\left( x+y\n \\right) ^{2}}{xyz \\left( 2\\,y+z+x \\right)  \\left( 2\\,z+x+y \\right) \n \\left( z+2\\,x+y \\right) }}\\geq {\\frac {27}{64}}\\,{\\frac { \\left( y+z\n \\right) ^{2} \\left( z+x \\right) ^{2} \\left( x+y \\right) ^{2}}{\n \\left( x+y+z \\right) ^{3}xyz}}\\geq 1\\]\n\nLeft is am-gm.\nplease nice do right.[/quote]\n\n\n$ \\left( x+y \\right) ^{2} \\left( y+z \\right) ^{2} \\left( z+x \\right) ^{\n2}-{\\frac {64}{27}}\\, \\left( x+y+z \\right) ^{3}xyz$\n$={\\it ssgm} \\left( \n \\left( y-z \\right) ^{2} \\left( {x}^{4}+5/7\\,{x}^{3}y+5/7\\,{x}^{3}z\n \\right)  \\right) +{\\it ssgm} \\left( {\\frac {83}{189}}\\, \\left( {x}^{2\n}-yz \\right) ^{2}yz \\right) +{\\it ssgm} \\left( {\\frac {25}{189}}\\,\n \\left( 3\\,{x}^{2}-xy-zx-yz \\right) ^{2}yz \\right)\\geq 0 $\n\nssgm=sum.\n",
  "Solution_14": "it equals with\n$$\\prod \\left(\\frac{1}{x+y}+\\frac{1}{y+z}\\right) \\leq \\frac{1}{xyz}$$\n\nand just AG",
  "Solution_15": "by ABC meathod...:D"
}
{
  "Tag": [
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "Show that for all positive reals $ x_i$ we have $ \\frac {1}{x_1} \\plus{} \\frac {2}{x_1 \\plus{} x_2} \\plus{} ... \\plus{} \\frac {n}{x_1 \\plus{} x_2 \\plus{} ... \\plus{} x_n} < \\frac {4}{x_1} \\plus{} \\frac {4}{x_2} \\plus{} ... \\plus{} \\frac {4}{x_n}$.\r\n\r\n[color=red]Edited by N.T.TUAN[/color]",
  "Solution_1": "Here is a stronger one  :wink: http://www.mathlinks.ro/viewtopic.php?t=173813",
  "Solution_2": "[quote=\"Myszon11\"]Here is a stronger one  :wink: http://www.mathlinks.ro/viewtopic.php?t=173813[/quote]\r\n\r\nThank you very much.  :D"
}
{
  "Tag": [
    "real analysis",
    "function",
    "topology",
    "advanced fields",
    "advanced fields unsolved"
  ],
  "Problem": "Today I heard that it is possible to find two sets $A$ and $B$ so that \r\n$A \\cup B = \\mathbb{R}$\r\n$A \\cap B$ is empty\r\n$\\mu(A) = 0$ (lebesgue measure)\r\n$B$ is of Baire category 1\r\n\r\nHow to construct such sets? :D",
  "Solution_1": "Let $(q_{n})_{n\\ge 1}$ be an enumeration of all the rationals, $A_{n}=\\bigcup_{k\\ge 1}I_{k}^{n}$, where $I_{k}^{n}$ is an open interval of length $\\frac 1{2^{n+k}}$ around $q_{n}$, and $B_{n}=\\mathbb R\\setminus A_{n}$. Now take $A=\\bigcap_{n}A_{n},\\ B=\\bigcup_{n}B_{n}$. \r\n\r\n$A_{n}$ is an open set with measure going to $0$ as $n\\to\\infty$, so the intersection $A$ of all $A_{n}$ is a $G_\\delta$ set of measure zero. $B_{n}$ is a nowhere dense closed set, so the union $B$ of all $B_{n}$ is an $F_\\sigma$ set of category $1$, and it's clear that they form a partition of $\\mathbb R$.",
  "Solution_2": "This is a rather indirect argument from real one-variable analysis. Try to fill in the details if you like. \r\n\r\nFact #1:  every increasing function is differentiable everywhere except for a set of measure zero. \r\n\r\nFact #2: if a function is differentiable on a set of second category, then it is Lipschitz on some interval. (Being Lipschitz means that $|f(x)-f(y)|\\le L|x-y|$ for some $L$ independent of $x$ and $y$.)\r\n\r\nFact #3: there exists an increasing function that is not Lipschitz on any (nontrivial) interval.",
  "Solution_3": "There is a book named \"Measure and Category\" in Graduate Texts in Mathematics,",
  "Solution_4": "[quote=\"cyqwhu\"]There is a book named \"Measure and Category\" in Graduate Texts in Mathematics,[/quote]\r\n\r\nThe complete reference is\r\nJ. C. Oxtoby, Measure and Category, Springer-Verlag, 1980.",
  "Solution_5": "Note that the set $A$ in grobber's example has Hausdorff dimension $0$. Here is an obvious generalization: Suppose that $\\{q_{n}\\colon n\\ge 1\\}$ is dense in $\\mathbb R$, and $\\sum_{n\\ge 1}c_{n}$ is a convergent series with positive terms. Then the set $A=\\bigcap_{m\\ge 1}\\bigcup_{n\\ge 1}\\left(q_{n}-\\frac{c_{n}}{m},q_{n}+\\frac{c_{n}}{m}\\right)$ has measure zero and $\\mathbb R\\setminus A$ is of first category.\r\n\r\nIs it possible to choose $q_{n}$ and $c_{n}$ so that $\\mathrm{dim}\\,A>0$? Or even $\\mathrm{dim}\\,A=1$?"
}
{
  "Tag": [
    "MATHCOUNTS",
    "geometry",
    "3D geometry"
  ],
  "Problem": "i participate in mathcounts and was wondering if there are any good sites that show shortcuts and tricks for mathcounts problems.",
  "Solution_1": "Ah yes... I remember a few sites when I was doing Mathcounts:\r\n\r\n[url]http://www.mathcounts.org/Problems/strategies01.html[/url]\r\n\r\n[url]http://www.unidata.ucar.edu/staff/russ/mathcounts/diaz.html[/url]\r\n\r\nHope these help!  :D \r\n\r\n-interesting_move",
  "Solution_2": "Yep, I remember reading Nick Diaz' mathcounts bible when I was little...well not that little, maybe 2 years ago.  But it's kind of common sense, you know...memorize squares thru 30, cubes thru 15, square roots of 2, 3, 5, 6, 10.  Memorize fractions (especially 1/7 - very helpful).  I don't know if those are [u]shortcuts[/u] but they're things to memorize.  I'll post more soon - after I find them again.\r\n\r\nGoogle is king!",
  "Solution_3": "MATHCOUNTS TRICKS AND TIPS at http://www.artofproblemsolving.com/Forum/viewtopic.php?mode=attach&id=5939 Yeah that should help"
}
{
  "Tag": [
    "geometry",
    "Olimpiada de matematicas"
  ],
  "Problem": "Veamos c\u00f3mo les va con este  :) \r\n\r\n[img]http://img49.imageshack.us/img49/2136/trianglezy7.png[/img]\r\n\r\nEn la figura, sea $\\triangle{ABC}$ acut\u00e1ngulo, donde $O$ es el centro de su circunferencia circunscrita. Sean $x,y,z$ segmentos perpendiculares a los lados, como muestra la figura. Sean $r$ y $R$ los radios de las circunferencias inscrita y circunscrita, respectivamente. Demuestre que $x+y+z=r+R$\r\n\r\nSaludos",
  "Solution_1": "Aki mi solucion:\r\n\r\nComo O es circuncentro entonces sabemos que x, y, z cortan a BC(a), AC(b) y AB(c) en sus puntos medios.\r\nAhora, aplicamos el Teorema de Ptolomeo en cada uno de los tres cuadrilateros ciclicos formados por x,y, z.\r\n\r\n$(xc)/2+(ya)/2 = (Rb)/2$   ... $(*)$\r\n$(yb)/2+(zc)/2 = (Ra)/2$  ...  $(**)$\r\n$(za)/2+(xb)/2 = (Rc)/2$   ...  $(***)$\r\n\r\nPor suma de areas tenemos:\r\n\r\n$(xa)/2+(yb)/2+(zc)/2 = (ABC)$ ...  $(****)$\r\n\r\nSumando $(*), (**), (***) y (****)$ tenemos:\r\n\r\n$(x+y+z)(a+b+c)(1/2) = R(a+b+c)(1/2)+(ABC)$\r\n$(x+y+z) = R+(ABC)(2/(a+b+c)) = R+r$\r\n\r\nLo ultimo lo deduzco de que $(ABC)= sr$ donde $s= semiperimetro$ y $r=inradio$",
  "Solution_2": "Muy buena soluci\u00f3n, bastante mejor que la m\u00eda, que es por trigonometr\u00eda  \r\n\r\nHoy en el colegio obtuve esta soluci\u00f3n, por casualidad aprend\u00ed Ptolomeo  :wink: \r\n\r\nSaludos  :P"
}
{
  "Tag": [],
  "Problem": "Find all real numbers $a$ such that the equations below have a common root.\r\n\r\n$x(x^4 - a)^2 - x^3(1-a)^2 + x^2 + 4 = 0$\r\n\r\nand\r\n\r\n$x^3 + x^2 + x + a = 0$\r\n\r\nHow do I find the roots of the second equation? I've gotten as far as $a = -rst$ and $-r-s-t = rs + st + tr = 1$ where $r$, $s$, and $t$ are roots of the second equation, but I can't see how to substitute or work that into the first equation. What do you do with the first equation to solve it?",
  "Solution_1": "It looks like a manipulation of the second equation will do the trick, \\[ -a=x^3+x^2+x.  \\]",
  "Solution_2": "[hide]Solving for $a$ in terms of $x$ in the second equation, $x^3+x^2+x=-a$. Then substitute. Since $x^3(x^3+x^2+x+1)^2=x(x^4+x^3+x^2+x)^2$, the first equation is just $x^2+4=0$.\n\nEither $2i$ or $-2i$ must be a root of the second equation. By complex conjugates, if one of them works, then both will, so just plug in $x=2i$. Then $-8i-4+2i+a=0$. But this makes $a$ imaginary. Therefore $\\boxed{\\text{no solution}}$.[/hide]"
}
{
  "Tag": [],
  "Problem": "The numbers 3, 5 and 7 are important in many aspects of Japanese life. Most positive integers can be expressed as a sum of only 3's, 5's and 7's. For example, 15 can be expressed as 5 + 5 + 5 or 3 + 5 + 7. What is the greatest even integer that cannot be expressed as a sum of 3's, 5's and 7's?",
  "Solution_1": "6=3+3\r\n8=3+5\r\n10=3+7\r\n\r\nNow you can add 3+3=6 to one of those numbers repeatedly to get any even number greater or equal to 6. Thus, the answer is 4."
}
{
  "Tag": [
    "geometry proposed",
    "geometry"
  ],
  "Problem": "let A and B is a point of circle w. and C is the point of the circle diameter. and the radius of this circle is 4\u221a7. D is the piont of the line AB. angle ADC is 90 degree, and angle CAB=60. then find the value of CD.",
  "Solution_1": "Could you, please, post a sketch?\r\nThanks!\r\n\r\nBest regards,\r\nsunken rock",
  "Solution_2": "for this picture R=2\u221a7!",
  "Solution_3": "sorry for this sketch, angle CAB=60, and angle ADC=90 find the value of CD"
}
{
  "Tag": [
    "Euler",
    "function",
    "logarithms",
    "inequalities",
    "limit"
  ],
  "Problem": "Prove that $ \\sum_{i\\equal{}1}^{\\infty} \\frac1{p_i}$ diverges, where $ p_i$ is the $ i$th prime.",
  "Solution_1": "hello, here you will find the answer\r\nhttp://everything2.com/title/Erdos%2527%2520proof%2520that%2520the%2520sum%2520of%2520the%2520reciprocals%2520of%2520the%2520primes%2520diverges\r\nSonnhard.",
  "Solution_2": "For reference, the standard proof is by the Euler product formula\r\n\r\n$ \\zeta(s) \\equal{} \\sum_{n \\ge 1} \\frac{1}{n^s} \\equal{} \\prod_{p \\in \\mathbb{P}} \\frac{1}{1 \\minus{} p^{\\minus{}s}}$.\r\n\r\nThe divergence of the product is equivalent to the divergence of the sum $ \\sum_{p \\in \\mathbb{P}} \\frac{1}{p^s}$ as $ s \\to 1^{\\plus{}}$ by a standard lemma.  One says that the zeta function has a simple pole at $ s \\equal{} 1$.",
  "Solution_3": "In Euler's proof we have $ \\prod_{i \\equal{} 1}^{\\infty}1 \\plus{} \\frac {1}{p_i \\minus{} 1}$ diverges\r\n\r\nbut $ 1 \\plus{} x\\le e^x$ so the product is bounded by $ \\exp\\left(\\sum \\frac1{p_i \\minus{} 1}\\right)$, but can this be modified to\r\n\r\n$ \\sum \\frac1{p_i}$?",
  "Solution_4": "[b]Lemma:[/b]  Let $ (a_n)$ be a sequence of real numbers such that $ \\sum_n a_n^2$ converges.  Then $ \\prod_n (1 \\plus{} a_n)$ converges if and only if $ \\sum_n a_n$ converges.\r\n\r\nThe proof is by the approximation $ \\ln (1 \\plus{} a_n) \\equal{} a_n \\plus{} O(a_n^2)$.",
  "Solution_5": "So from that lemma we have $ \\sum_{i\\equal{}1}^{\\infty}\\frac1{p_i\\minus{}1}$ diverges, but how would this result prove the divergence of $ \\sum_{i\\equal{}1}^{\\infty}\\frac1{p_i}$?\r\n\r\n(hmm if we use that lemma how would we deal with the convergence of $ \\sum a_n^2$ first?)",
  "Solution_6": "Use the inequality $ \\frac{1}{p} \\ge \\frac{1}{2(p\\minus{}1)}$.",
  "Solution_7": "[quote=\"mathwizarddude\"](hmm if we use that lemma how would we deal with the convergence of $ \\sum a_n^2$ first?)[/quote]\r\nCompare it to $ \\sum_n \\frac{1}{n^2}$.",
  "Solution_8": "The version of that lemma I prefer:\r\nLet $ a_n$ be a sequence of (real or complex) numbers. The series $ \\sum |a_n|$ converges if and only if the product $ \\prod(1\\plus{}|a_n|)$ converges.\r\nIf that series $ \\sum |a_n|$ converges, the series $ \\sum a_n$ and the product $ \\prod (1\\plus{}a_n)$ also converge*.\r\n\r\n* If $ 1\\plus{}a_k\\equal{}0$ for some $ k$, we say that $ \\prod(1\\plus{}a_n)$ converges to zero. If none of those terms is zero but $ \\lim_{N\\to\\infty}\\prod_{n\\le N}(1\\plus{}a_n)\\equal{}0$, we say that the product diverges to zero.\r\nThat technicality in the definition also matters for t0rajir0u's version, which needs to exclude the cases of $ a_k\\equal{}\\minus{}1$. It doesn't matter for application to this problem, which has $ a_n>0$ everywhere.",
  "Solution_9": "I saw a beautiful solution.I am posting it.According to PNT ,$ln n +ln ln n -1\\leq \\frac{p_n}{n}\\leq ln n+ln ln n$\nfor $n\\geq 6$ .\nNow $\\sum_{n=1}^{\\infty}\\frac {1}{p_n} > \\sum_{n=6}^{\\infty} \\frac {1}{p_n} \\geq  \\sum_{n=6}^{\\infty} \\frac {1}{n ln n+ n ln ln n} \\geq  \\sum_{n=6}^{\\infty} \\frac {1}{2n ln n}=\\infty$ \n\nfrom the well known fact that  $\\sum_{n=2}^{\\infty} \\frac {1}{n (ln n)^p}$ diverges for $p\\leq 1$ and converges for $p >1$. So done!."
}
{
  "Tag": [
    "geometry",
    "function"
  ],
  "Problem": "The set of ordered pairs below define a function  { (3,1), (-1,1), (2,4), (-2,4) }\r\n\r\nWhat is the range of the function?\r\n\r\na. { 3,4 } b. { -2,3 } c. { -2,4 } d. { 1,4 } e. NG",
  "Solution_1": "[hide] The range is just the set of $ y$ values in a function. Hence, in this case, the range is $ {(1,4)}$, so the answer is D.[/hide]"
}
{
  "Tag": [
    "inequalities"
  ],
  "Problem": "If $ a$ and $ b$ are the catetes in a right triangle and $ c$ is the hypothenuses. \r\nProve that $ c^{3}>a^{3}\\plus{}b^{3}$\r\n\r\nThe following is my solution. Is it right?Give you your own solution.\r\n\r\n$ c^{3}\\equal{}(a^{2}\\plus{}b^{2})\\sqrt{a^{2}\\plus{}b^{3}}$\r\n$ c^{6}\\equal{}(a^{2}\\plus{}b^{2})^{2}(a^{2}\\plus{}b^{2})$\r\n\r\nSince a,b,c are length is true that $ (c^{3})^2>(a^{3}\\plus{}b^{3})^{2}$\r\nso:\r\n$ a^{6}\\plus{}3a^{4}b^{2}\\plus{}3a^{2}b^{4}\\plus{}b^{6}>a^{6}\\plus{}2a^{2}b^{2}\\plus{}b^{6}$\r\n\r\n$ 3a^{2}b^{2}(a^{2}\\plus{}b^{2})>2a^{3}b^{3}$\r\n$ 2(a^{2}\\plus{}b^{2})\\plus{}a^{2}\\plus{}b^{2}\\minus{}2ab>0$\r\n$ 2c^{2}\\plus{}(a\\minus{}b)^{2}>0$\r\n\r\n\r\nThe last inequality is always true :$ 2c^{2}\\plus{}(a\\minus{}b)^{2}>0$  :)",
  "Solution_1": "[hide=\"Another solution.\"]\nFirst, start off with the statement that $ c > b$.  Then it follows that $ c > \\frac {b^3}{b^2}$ and that $ c > \\frac {b^3 \\minus{} 2ab^2 \\minus{} a^3}{a^2 \\plus{} b^2}$.\n\nFactoring out a negative sign and rewriting the expression in another way, you obtain:\n\n$ c > \\minus{} \\frac {a^2b \\plus{} ab^2 \\plus{} a^3 \\plus{} ab^2 \\minus{} a^2b \\minus{} b^3}{a^2 \\plus{} b^2}$\n\n$ c > \\minus{} \\frac {ab(a \\plus{} b) \\plus{} (a \\minus{} b)(a^2 \\plus{} b^2)}{a^2 \\plus{} b^2}$\n\n$ c > \\minus{} \\frac {ab(a \\plus{} b)}{a^2 \\plus{} b^2} \\plus{} a \\minus{} b$\n\n$ c \\minus{} a \\plus{} b > \\minus{} \\frac {ab(a \\plus{} b)}{a^2 \\plus{} b^2}$.\n\nSince $ c^2 \\equal{} a^2 \\plus{} b^2$, you can multiply both sides by $ c^2$:\n\n$ c^2(c \\minus{} a \\plus{} b ) > \\minus{} ab(a \\plus{} b)$\n$ c^3 \\minus{} c^2(a \\plus{} b) > \\minus{} ab(a \\plus{} b)$\n$ c^3 > c^2(a \\plus{} b) \\minus{} ab(a \\plus{} b)$.\n\nSubstituting $ a^2 \\plus{} b^2$ back for $ c^2$:\n\n$ c^3 > (a^2 \\minus{} ab \\plus{} b^2)(a \\plus{} b)$\n$ c^3 > [ (a \\plus{} b)^2 \\minus{} 3ab ]( a \\plus{} b )$\n$ c^3 > (a \\plus{} b)^3 \\minus{} 3ab(a \\plus{} b)$\n$ c^3 > (a \\plus{} b)^3 \\minus{} 3a^2b \\minus{} 3b^2a$\n$ c^3 > a ^ 3 \\plus{} b^3$.\n[/hide]\r\nIf I understand your solution correctly, when you expanded $ (a^3 \\plus{} b^3)^2$, the middle term should have been $ 2a^3b^3$, I believe.",
  "Solution_2": "$ (x^2\\plus{}y^2)^3\\minus{}(x^3\\plus{}y^3)^2\\equal{}x^2y^2(3x^2\\minus{}2xy\\plus{}y^2)\\equal{}x^2y^2(3(x\\minus{}y)^2\\plus{}4xy)>0$",
  "Solution_3": "Where did the square term in $ (x^3 \\plus{} y^3)^2$ come from?"
}
{
  "Tag": [
    "induction",
    "inequalities",
    "combinatorics proposed",
    "combinatorics"
  ],
  "Problem": "Let $ G=(V,E)$ be a simple graph.\n\na) Let $ A,B$ be a subsets of $ E$, and spanning subgraphs of $ G$ with edges $ A,B,A\\cup B$ and $ A\\cap B$ have $ a,b,c$ and $ d$ connected components respectively. Prove that $ a+b\\leq c+d$.\n\nWe say that subsets $ A_1,A_2,\\dots,A_m$ of $ E$ have $ (R)$ property if and only if for each $ I\\subset\\{1,2,\\dots,m\\}$ the spanning subgraph of $ G$ with edges $ \\cup_{i\\in I}A_i$ has at most $ n-|I|$ connected components.\nb) Prove that when $ A_1,\\dots,A_m,B$ have $ (R)$ property, and $ |B|\\geq2$, there exists an $ x\\in B$ such that $ A_1,A_2,\\dots,A_m,B\\backslash\\{x\\}$ also have property $ (R)$.\n\nSuppose that edges of $ G$ are colored arbitrarily. A spanning subtree in $ G$ is called colorful if and only if it does not have any two edges with the same color.\nc) Prove that $ G$ has a colorful subtree if and only if for each partition of $ V$ to $ k$ non-empty subsets such as $ V_1,\\dots,V_k$, there are at least $ k\\minus{}1$ edges with distinct colors that each of these edges has its two ends in two different $ V_i$s.\nd) Assume that edges of $ K_n$ has been colored such that each color is repeated $ \\left[\\frac n2\\right]$ times. Prove that there exists a colorful subtree.\ne) Prove that in part d) if $ n\\geq5$ there is a colorful subtree that is non-isomorphic to $ K_{1,n-1}$.\nf) Prove that in part e) there are at least two non-intersecting colorful subtrees.",
  "Solution_1": "I can't believe we are filling the Combinatorics forum with chessboard exercises while masterpieces like this remain unsolved.\n\nLet me give some hints to a).\n\na) For every subset $F$ of $E$, let $c_F$ denote the number of connected components in the spanning subgraph of $G$ with edge set $F$. Then, the problem asks us to prove that $c_A + c_B \\leq c_{A\\cup B} + c_{A\\cap B}$ for any pair $\\left(A,B\\right)$ of subsets of $E$.\n\nWe will prove this by induction over $\\left|A\\setminus B\\right|$. The base case ($\\left|A\\setminus B\\right|=0$) is obvious (in this case, $A\\subseteq B$, so that $A\\cap B=A$ and $A\\cup B=B$). So let us go for the induction step:\n\nLet $k\\in\\mathbb N$ be positive. Assume that the problem is already solved for all pairs $\\left(A,B\\right)$ with $\\left|A\\setminus B\\right|<k$. Now let $\\left(A,B\\right)$ be a pair with $\\left|A\\setminus B\\right|=k$. Let $a\\in A\\setminus B$ be arbitrary, and let $B'=B\\cup\\left\\{a\\right\\}$. Then, $\\left|A\\setminus B'\\right|=\\underbrace{\\left|A\\setminus B\\right|}_{=k}-1=k-1<k$, so that (by our induction assumption) the problem is already solved for the pair $\\left(A,B'\\right)$. In other words, $c_A + c_{B'} \\leq c_{A\\cup B'} + c_{A\\cap B'}$. Since $A\\cup B' = A\\cup B$, this rewrites as $c_A + c_{B'} \\leq c_{A\\cup B} + c_{A\\cap B'}$.\n\nIn the following, when $F$ is a subset of $E$ and $x$ is an edge in $F$, we say that [i]$x$ is a bridge in $F$[/i] if $x$ is a bridge in the spanning subgraph of $G$ with edge set $F$.\n\nNow, $c_{B'}$ is either $c_B - 1$ or $c_B$, depending on whether the edge $a$ is a bridge in $B'$ or not. Similarly, $c_{A\\cap B'}$ is either $c_{A\\cap B} - 1$ or $c_{A\\cap B}$, depending on whether the edge $a$ is a bridge in $A\\cap B'$ or not. This yields four possible cases:\n\n[i]Case 1:[/i] We have $c_{B'} = c_B - 1$ and $c_{A\\cap B'} = c_{A\\cap B} - 1$.\n[i]Case 2:[/i] We have $c_{B'} = c_B$ and $c_{A\\cap B'} = c_{A\\cap B} - 1$.\n[i]Case 3:[/i] We have $c_{B'} = c_B - 1$ and $c_{A\\cap B'} = c_{A\\cap B}$.\n[i]Case 4:[/i] We have $c_{B'} = c_B$ and $c_{A\\cap B'} = c_{A\\cap B}$.\n\nIn each of the Cases 1, 2 and 4, we can easily deduce $c_A + c_B \\leq c_{A\\cup B} + c_{A\\cap B}$ from $c_A + c_{B'} \\leq c_{A\\cup B} + c_{A\\cap B'}$, and thus the induction step is completed in these cases. So it remains to handle the Case 3. But in fact, this case cannot occur. In fact, $c_{B'} = c_B - 1$ can only hold when the edge $a$ is a bridge in $B'$. But if the edge $a$ is a bridge in $B'$, then $a$ must also be a bridge in $A\\cap B'$, so that $c_{A\\cap B'} = c_{A\\cap B} - 1$. Hence, if $c_{B'} = c_B - 1$, then $c_{A\\cap B'} = c_{A\\cap B} - 1$. This rules out Case 3, and we are done.\n\nWho can solve b) and c)? (These seem the most interesting to me.)",
  "Solution_2": "[b]a)[/b] Let $n$ denote the number of vertices. Let $G_1=G(V,A\\cap B)$, $G_2 = G(V,A)$, $G_3=G(V,B)$ and $G_4= G(V,A\\cup B)$. Also let $\\kappa_j = \\kappa (G_j)$ for $j=1,2,3,4$. WLOG $\\kappa_4 = 1$, otherwise pick a connected component of $G_4$\n\nLet $T_1\\subseteq G_1$ be a spanning tree. It has $n-\\kappa_1$ edges and $\\kappa_1$ components. Now choose two spanning trees $T_2\\subseteq G_2$ and $T_3 \\subseteq G_3$ that contain the tree $T_1$. If we let $|T_j|$ denote the number of edges then $\\kappa_2 = n-|T_2|$ and $\\kappa_3 = n-|T_3|$.\n\nFurther, $|T_2|+|T_3| \\ge n-1 + (n-\\kappa_1)$ because $G_4$ is connected. Therefore\n\n\\[\\kappa_2 + \\kappa_3 = 2n - (|T_2| + |T_3|) \\le 2n - (2n-1 -\\kappa_1 ) = \\kappa_1 + 1 = \\kappa_1 + \\kappa_4\\]\n\nFor $\\kappa_4 > 1$, summing over all components of $G_4$ gives the desired inequality $\\square$",
  "Solution_3": "[b]b)[/b] [b]Equivalent Problem:[/b] Let $G(V,E)$ be a graph such that $|V|=n$ and $E = A_1\\cup A_2 \\cup ... \\cup A_m$, where each $A_j$ is non-empty and $m\\le n-1$.  If $G(V,E)$ has property $(R)$ then there exists a subforest (collection of disjoint trees) whose edge set contains exactly one edge from each of the sets $A_j$.\n\n[i]Note: you can observe that his problem is equivalent by repeatedly applying the original statement.[/i]\n\n[i]Proof:[/i] Induction on $m$. If $m=1$ then the statement is clearly true because $A_1$ is not empty. Assume true for the case $m-1$.\n\nIf the set $E_m = \\{A_1,A_2...,A_m\\}$ has property (R) then so does the set $E_{m-1}=\\{A_1,A_2...,A_{m-1}\\}$. By induction we can find a colourful subforest of $G(V,E_{m-1})$, call it $T_0$. If there exists an edge in $A_m$ such that adding it to $T_0$ will decrease the number of components of $T_0$ then we're done. So assume the contrary.\n\nBy property (R) there must be an edge in a set $A_j$ ($j<m$) whose addition to $T_0$ will decrease the number of components. Call all such edges [i]nice edges[/i], and call sets containing a nice edge, [i]nice sets[/i]\n\nAdding an edge from $A_m$ to $T_0$ will create a cycle in $T_0$. We can then remove any edge in this cycle to form a new subforest $T_1$. We call this process a [i]switch[/i]; note that all edges of $T$ have distinct colour after a switch and the set of [i]nice edges/nice sets[/i] has not changed (*). So we wish to find a series of switches that ends in removing an edge from a nice set, because then we can add the nice edge from that set and decrease the number of components- giving us a forest with $m$ edges.\n\nWe define the family of sets $\\mathcal{E}$. If there is a series of [i]switches[/i] that starts by adding edge $x\\in A_m$ to $T_0$ and edges by removing edge $y\\in A_i$, then place $A_i \\in \\mathcal{E}$. Also let $A_m\\in\\mathcal{E}$.\n\nIf $\\mathcal{E}$ contains a [i]nice set[/i], $A_i$, then by performing that series of switches we can remove and edge $x\\in A_i$ from that set and add the corresponding [i]nice edge[/i], $y\\in A_i$. By the invariance property (*), this edge $y$ will decreases the number of components of $T$ - giving us a forest with $m$ edges. \n\nIf none of the sets in $\\mathcal{E}$ are nice then delete all edges from $T_0$ that do not belong to a set in $\\mathcal{E}$, call the new forest $T^*$. $T^*$ has $n-|\\mathcal{E}|+1$ components (because $A_m\\in\\mathcal{E}$, but no edge from $A_m\\in E(T_0)$). Adding any edge $x\\in \\mathcal{E}$ to $T^*$ will not decrease the number of components, otherwise $x$ would lie in a cycle of $T_0$ with an edge $y\\not\\in\\mathcal{E}$, contradicting the definition of $\\mathcal{E}$. So in fact the graph with edge set $\\mathcal{E}$ also has $n-|\\mathcal{E}|+1$ components. But this contradicts property (R). \n\nThis completes the inductive step $\\square$",
  "Solution_4": "[b]c) [/b] Suppose the graph has the given property, call it property (P). Let $A_1, A_2, ..., A_m$ be the edge sets of distinct colour in $G$. By applying property (P) for $k=n$ it is clear that $m\\ge n-1$. Further it is clear that the graph $G$ is connected, for if we could partition $G$ into two disconnected sets $V_1$ and $V_2$ with no edge between them, we would contradict property (P) for $n=2$.\n\nLet $F\\subseteq G$ be the [i]largest[/i] colourful forest, and let $\\mathcal{A}=\\{A_1,A_2,...,A_j\\}$ be the edge sets who have an edge in the forest. Assume for the sake of contradiction that $j < n-1$.\n\nNone of the remaining sets $\\{A_{j+1},...,A_m\\}$ contains an edge that decreases the number of components of $F$. Therefore, such edge must exist in the set $\\mathcal{A}$ because $G$ is connected. Let $\\mathcal{A}_1\\subseteq \\mathcal{A}$ be the sets that contain an edge, $x$, such that adding $x$ to $F$ will decrease the number of components\n\nFor $k > j$, adding an edge $x \\in A_k$ to $F$ will create a cycle. You can then remove any edge from that cycle and you have a new colourful forest, call this operation a [i]switch[/i] (as we did in part b).  If it was possible to remove an edge from a set $A_i\\in \\mathcal{A}_1$ by a series of switches, then you could add the edge in $y\\in A_i$ to $F$ and decrease the number of components (again like part b). This contradicts the fact that $F$ is maximum. \n\nSo it turns out that the graph with edges in the set $\\{A_1,....,A_m\\} \\backslash \\mathcal{A}_1$ has at least $n-j+\\mathcal{A}_1$ components. By placing each of these components in its own set $V_i$ and applying property (P) there are at least $n-j+\\mathcal{A}_1 -1$ other edges of distinct colour. These must come from the only remaining sets $\\mathcal{A}_1$. But $n-1-j+\\mathcal{A}_1 > \\mathcal{A}_1$ by assumption. This is a contradiction. \n\nTherefore $F$ has at least $n-1$ edges, that is, it has exactly $n-1$. So $F\\subset G$ is a colourful subtree.\n \nThis completes the proof of the [i]if[/i] direction. The [i]only if[/i] direction is more or less trivial, so the proof is complete $\\square$",
  "Solution_5": "[b]d)[/b] We will show that $G$ has property (P) from part c.\n\nIt suffices to show that if $V$ is partition into non-empty sets $V= V_1 \\cup ... \\cup V_k$, with cardinality $|V_i|=v_i$, then the number of edges between these components is at least $\\frac{n}{2}(k-2) + 1$. The result will then follow from the pigeon hole principle.\n\nSo if $v_1,...,v_k$ are positive integers with sum $n$ then\n\n\\begin{align*}\n\\sum_{i <j} v_iv_j &= \\frac{1}{2} \\left ( n^2 - v_1^2-v_2^2-...-v_k^2\\right)\\\\\n& \\ge \\frac{1}{2} \\left (n^2 - (n-k+1)^2 - 1^2 - 1^2 -...-1^2\\right)\\\\\n& = \\frac{1}{2} (k-1)(2n-k)\\\\\n& \\ge (k-1)\\frac{n}{2}\n\\end{align*}\n\nSo there are at least $k-1$ distinct coloured edges between the components $V_1,...,V_k$.",
  "Solution_6": "[b]e)[/b] [i]n is odd:[/i] If there was a colourful subtree isomorphic to $K_{1,n-1}$, then we may remove the vertex of degree $n-1$ and reduce the problem to the case of $n-1$ vertices. If we find a desired subtree in the reduced graph then we add back the vertex we removed and connect it to the tree via the remaining edge of distinct colour. \n\n[i]n is even[/i] Again, if there is a colourful subtree isomorphic to $K_{1,n-1}$ we remove the root vertex. But this time we haven't reduced the problem entirely; there are too many colours. So pick an arbitrary colour, say blue, and re-colour the blue edges with the remaining colours (each remaining colour used on one blue edge). Then the problem has been reduced to the $n-1$ vertices case.\n\nIn this manner we are done by induction if we can show the claim is true for $n=5$, but this is easily verified by hand. So the proof is complete $\\square$",
  "Solution_7": "[b]f) [/b] We know that there exists at least one colourful subtree, so find one and remove those edges from the graph. Now each colour is repeated at most $\\frac{n-1}{2}$ times, and there are $n-1$ pairs of vertices with no edges.\n\nTherefore, as in part d, we require that $\\textstyle\\sum_{i<j} v_iv_j \\ge (k-2)\\frac{n-1}{2} + 1 + (n-1)$, where the $n-1$ is accounting for the edges we removed. But we already showed in part d that the LHS is at least $(n-1)(2n-k)$ so it's left to show that\n\\[(n-1)(2n-k) \\ge (k-2)\\frac{n-1}{2} + n\\]\nWhich after some rearrangement becomes\n\\[nk \\ge (k-1)^2 + 1\\]\nThis is clearly true so the proof is complete $\\square$"
}
{
  "Tag": [],
  "Problem": "I gave a mean 10th grader this problem:\r\n$ \\frac{1}{1\\plus{}{\\frac{1}{1\\plus{}{\\frac{1}{1\\plus{}...}}}}}\\equal{}$  :?: \r\n\r\n\r\nSince he didn't know the answer, he gave this to me:\r\n$ x^{x^{x^{...}}}\\equal{}2$, and i couldn't solve it...\r\nThose are dots at the top, so the expression goes on forever.\r\n\r\nI tried doing this:\r\n$ x^{x^{x^{...}}}\\equal{}2$\r\n$ x^{x^{x^{...}}}\\equal{}\\sqrt[x]{2}$\r\n$ \\sqrt[x]{2}\\equal{}2$\r\n$ x\\equal{}1$\r\n\r\nObviously, if $ x\\equal{}1$, then the whole thing equals 1.\r\n\r\nIs there a solution to this problem? If so, then how do you get it?\r\n\r\nThanks for the help.",
  "Solution_1": "[quote=\"Kevin K.\"]I gave a mean 10th grader this problem:\n$ \\frac{1}{1\\plus{}{\\frac{1}{1\\plus{}{\\frac{1}{1\\plus{}...}}}}}\\equal{}$  :?: [/quote]\r\n\r\nThe thing with solving those infinite problems is to substitute a variable for an expression. For example, in this problem, you can substitute $ x$ for $ \\frac{1}{1\\plus{}x}$. This usually makes solving them a lot easier  :wink: .\r\n\r\nSo I believe in this problem then the set up you are looking for is (and just solve for x):\r\n\r\n$ x \\equal{}\\frac{1}{1\\plus{}x}$  :?:",
  "Solution_2": "[hide]\n$ x^{x^{x^{x^{...}}}}\\equal{} 2\\implies x^{2}\\equal{} 2\\implies x \\equal{}\\pm\\sqrt{2}$\n\n$ x\\equal{}\\frac{1}{x\\plus{}1}$\n$ x^{2}\\plus{}x\\minus{}1\\equal{}0$\n$ x\\equal{}\\frac{1\\pm\\sqrt{5}}{2}$\n[/hide]",
  "Solution_3": "I know how to do $ \\frac{1}{1\\plus{}{\\frac{1}{1\\plus{}{\\frac{1}{1\\plus{}...}}}}}$\r\n\r\nBut it's the second one that I don't get.\r\nIf $ x\\equal{}\\sqrt{2}$, then how does it make an infinite expression equal 2?\r\nI thought that an answer greater than 1 will give you $ \\infty$.",
  "Solution_4": "No, it converges in this case. Think about it. Do some of $ \\sqrt{2}^{\\sqrt{2}^{\\sqrt{2}^{\\cdots}}}$ with a computer. You'll notice the change starts getting very small. In fact, it's not hard to prove that for $ 1 < a < 2$, $ a <\\sqrt{2}^{a}< 2$. Then all you'd have to do to show it converges is that you can eventually get above any number less than 2, but that's the harder part (takes a bit of calculus).\r\n\r\nYou're right to worry about convergence though. When dealing with \"power towers\"  as they're usually called, it's a very big issue.",
  "Solution_5": "[quote=\"Kevin K.\"]I know how to do $ \\frac{1}{1\\plus{}{\\frac{1}{1\\plus{}{\\frac{1}{1\\plus{}...}}}}}$\n\nBut it's the second one that I don't get.\nIf $ x \\equal{}\\sqrt{2}$, then how does it make an infinite expression equal 2?[/quote]\n\nThe first expression is infinite and it still converges, doesn't it?  ;)\n\nWhen we talk about infinite expressions like that we are really talking about the convergence or divergence of the sequence\n\n$ a_{1}\\equal{} x$\n$ a_{n}\\equal{} x^{a_{n\\minus{}1}}$\n\nFinding the interval of convergence of $ x$ is in fact not trivial and requires a bit of calculus.\n\n[quote=\"Kevin K.\"]$ x^{x^{x^{...}}}\\equal{} 2$\n$ x^{x^{x^{...}}}\\equal{}\\sqrt [x]{2}$[/quote]\r\n\r\nThis step is not valid because the exponent on the left is not $ x$.  What I mean is that \r\n\r\n$ 3^{3^{3}}\\equal{} a$\r\n\r\nDoes [b]not[/b] imply\r\n\r\n$ 3^{3}\\equal{}\\sqrt [3]{a}$\r\n\r\nIn fact, $ \\sqrt [3]{a}\\equal{} 3^{9}$.  This is because $ 3^{3^{3}}\\equal{} 3^{27}$ and not \r\n\r\n$ (3^{3})^{3}\\equal{} 3^{9}$",
  "Solution_6": "[quote=\"AstroPhys\"][hide]\n$ x^{x^{x^{x^{...}}}}\\equal{} 2\\implies x^{2}\\equal{} 2\\implies x \\equal{}\\pm\\sqrt{2}$\n[/hide][/quote]\r\n\r\nHow exactly does $ x^{x^{x^{x^{...}}}}\\equal{} 2$ turn into $ x^{2}\\equal{} 2$?\r\n\r\nI missed the jump from one step to the other...",
  "Solution_7": "If $ x^{x^{x^{...}}}= 2$ then \r\n\r\n${ x^{x^{x^{...}}}= x^{( x^{x^{x^{...}}}}) }= x^{2}$",
  "Solution_8": "[quote=\"Icy\"][quote=\"AstroPhys\"][hide]\n$ x^{x^{x^{x^{...}}}}\\equal{} 2\\implies x^{2}\\equal{} 2\\implies x \\equal{}\\pm\\sqrt{2}$\n[/hide][/quote]\n\nHow exactly does $ x^{x^{x^{x^{...}}}}\\equal{} 2$ turn into $ x^{2}\\equal{} 2$?\n\nI missed the jump from one step to the other...[/quote]\r\nIn the first expression, cover up the bottommost x with your finger and you'll see why. :)",
  "Solution_9": "The towers of $ x$'s is equal to $ 2$ so $ x^{\\text{tower of x's}}\\equal{}2\\implies x^{2}\\equal{}2$.",
  "Solution_10": "Oh! These things are so weird...I get it now though, thanks a lot!"
}
{
  "Tag": [
    "function",
    "floor function",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "How do you prove that:\r\n\r\n$ \\sum_{n \\le x}\\left \\lfloor \\sqrt \\frac{x}{n} \\right \\rfloor \\equal{}\\sum_{n\\le\\sqrt {x}}\\left \\lfloor \\frac{x}{n^2} \\right \\rfloor$\r\n\r\nI have tried looking at lattice points (a,b) under ab=x and also looking at $ \\left \\lfloor \\sqrt \\frac{x}{n}\\right  \\rfloor$ as the number of squares less than or equal to $ \\frac{x}{n}$ but so far no luck.",
  "Solution_1": "count in two different ways the lattice points on or under the graph of $ y(t) \\equal{} \\sqrt {\\frac {x}{t}}$ between $ t \\equal{} 1$  and $ t \\equal{} x$. when you solve for $ t$, you have $ t \\equal{} \\frac {x}{y^2}$. the rest is easy...",
  "Solution_2": "We can also use $\\sum_{n \\leq x}\\lambda(n) \\left\\lfloor {\\frac{x}{n}} \\right\\rfloor=\\lfloor \\sqrt{x} \\rfloor$ where $\\lambda(n)$ is the Liouville function\nso \n$\\sum_{n \\leq x} \\left\\lfloor \\sqrt{\\frac{x}{n}} \\right \\rfloor =\\sum_{n \\leq x} \\sum_{d \\leq \\frac{x}{n}} \\lambda(d) \\left\\lfloor \\frac{x}{dn} \\right\\rfloor=\\sum_{q \\leq x} \\left\\lfloor \\frac{x}{q} \\right\\rfloor \\sum_{d|q}\\lambda(d)$ on setting q=nd\nand $\\sum_{d|q}\\lambda(d)=1$ if q a square, =0 if not, gives result."
}
{
  "Tag": [],
  "Problem": "I found this very hard problem in a book. I have worked on it for over four hours, but I am unable to solve it. If you can solve it, please tell me the logic path you took to solve it. This is similar in form to the famous \"Einstein Problem\". Here is the exact text from the book:\r\n\r\n\"In the days before it got sadly bare, old Mother Hubbard's cupboard was actually stocked quite amply. At least, it contained enough to meet her needs and those of her loyal pooch, Alphonse. She had a supply of bones, of course; and she kept anchovies, chutney, dill pickles, egg drop soup, farina, gluten, ham hocks, Irish stew, jellied eels, kidney beans, ladyfingers, marshmallows, nutcakes, and oat bran. Each of these 15 necessities was stored in one of 15 compartments of her cupboard, though if there was any method to their placement, only Mother Hubbard (or Alphonse) fathomed it. By boning up on the clues, however, you should be able to identify the compartment in which each item was placed.\"\r\n\r\n\r\nFYI, the 15 compartments are arranged in a 3 x 5 grid. 3 rows by 5 columns. Each row has 5 compartments and each column has 3 compartments.\r\n\r\n\r\n\"The Clues:\r\n\r\n1. If you scanned the cupboard shelves from left to right, starting with the top shelf and ending with the bottom shelf, you'd find the five canned items (anchovies, ham hocks, Irish stew, jellied eels, and kidney beans) in alphabetical order, with no two side by side.\r\n\r\n2. The boxed items (farina, gluten, and oat bran) were all on the same shelf, but only two of these were side by side.\r\n\r\n3. The bottled items (chutney, dill pickles, and egg drop soup) were all on different shelves, though no two were in the same vertical file.\r\n\r\n4. None of the bagged items (ladyfingers, marshmallows, and nutcakes) was to be found directly beside a canned item.\r\n\r\n5. One item was directly below the ham hocks, directly above the farina, and directly left of the ladyfingers.\r\n\r\n6. The chutney was either directly beside the marshamallows or directly above the oat bran, but not both.\r\n\r\n7. The dill pickles were either directly below the nutcakes or directly beside the gluten, but not both.\r\n\r\n8. No corner compartment contained kidney beans or oat bran.\"\r\n\r\nThank you.",
  "Solution_1": "No idea up to now. I think it's harder than Einstein's test(when I was 11, I slved it in 8 minutes :wink: ).",
  "Solution_2": "What does \"side by side\" mean here?\r\n[code]\n1   2   3   4   5\n6   7   8   9   10\n11  12  13  14  15[/code]\r\n\r\nI just want to clarify...would 1 and 6 be side to side? What about like 5 and 6? (sorry if these are stupid questions :blush:)",
  "Solution_3": "I couldn't do it.\r\n\r\nAnyways, here's a prod for people that can.\r\n\r\n[hide=\"where to start\"]\n[code]\nRow 1      1   2   3   4   5\nRow 2      6   7   8   9   10\nRow 3      11  12  13  14  15[/code]\n\nClue number five tells you that the ham hocks must be in Row 1, the farina in Row 3 directly below the ham hocks, and the ladyfingers in row 2 to the right of the ham hocks and the farina.\n\nBy clue 1, The ham hocks must be in 3, 4, or 5, and if its in 5, then there would be no place to put the ladyfingers. Thus, the ham hocks are 3 or 4, and the farina in 13 or 14, and the ladyfingers in 9 or 10. The \"mystery item\" in between them cannot be a canned item by clue 4.\n\nBy clue 2 and clue 8, because the farina is in the bottom row, the oat bran is in 12, 13, or 14, and the gluten is in the bottom row.\n\nBy clue 1, the kidney beans must be in the bottom row.\n[/hide]\r\n\r\nThat's as far as I got. I almost solved it a few times, and then reached a few contradictions. Good luck people!",
  "Solution_4": "[quote=\"Textangle\"]I couldn't do it.\n\nAnyways, here's a prod for people that can.\n\n[hide=\"where to start\"]\n[code]\nRow 1      1   2   3   4   5\nRow 2      6   7   8   9   10\nRow 3      11  12  13  14  15[/code]\n\nClue number five tells you that the ham hocks must be in Row 1, the farina in Row 3 directly below the ham hocks, and the ladyfingers in row 2 to the right of the ham hocks and the farina.\n\nBy clue 1, The ham hocks must be in 3, 4, or 5, and if its in 5, then there would be no place to put the ladyfingers. Thus, the ham hocks are 3 or 4, and the farina in 13 or 14, and the ladyfingers in 9 or 10. The \"mystery item\" in between them cannot be a canned item by clue 4.\n\nBy clue 2 and clue 8, because the farina is in the bottom row, the oat bran is in 12, 13, or 14, and the gluten is in the bottom row.\n\nBy clue 1, the kidney beans must be in the bottom row.\n[/hide]\n\nThat's as far as I got. I almost solved it a few times, and then reached a few contradictions. Good luck people![/quote]\r\n\r\nok\r\nhere's what i got\r\nsince the kidney beans and the oat bran can't be on the corners, that means the farina, kidney beans, and oat bran are somewhere in 12, 13, and 14\r\nthe gluten is either in 11, or 15\r\nif the farina is in 13, then the oat bran would have to be next to the farina, since the gluten has no way of being next to it\r\nif the farina is in 12 or 14, then the gluten could be in 11 or 15, in which case the oat bran is in 14 or 12, or the gluten is in 11 or 15, in which case the oat bran is in 13 or 13 (i made two numbers for the two cases, each of them is in their respective positions)\r\n\r\nand just a question, couldn't the ham hocks be in 2, and the anchovies be in 1?",
  "Solution_5": "I am not sure what they mean when they say things like \"beside\", \"side by side\", or \"next to\". Sides could mean horizontal only or both horizontal and vertical. That is a major problem with this problem; things are too ambiguous. They don't give definitions of the terms they use.\r\n\r\nMy teacher said she was able to solve the problem, but I think it is because she interpreted certain terms to mean horizontal only. The bagged items are a big problem. If \"beside\" meant horizontal only, the problem might be solvable. If \"beside\" meant horizontal and vertical, then I'm pretty sure the problem is unsolvable.",
  "Solution_6": "on that note...\r\ni would say that beside only means horizontal\r\n\r\nanyway, anyone get anywhere closer than me yet?",
  "Solution_7": "I got everything except the bagged items, but I reached a contradiction, so I had to start over.\r\n\r\nHam hocks can't be in 2 because\r\n\r\n[quote=\"12markkram34\"]1. If you scanned the cupboard shelves from left to right, starting with the top shelf and ending with the bottom shelf, you'd find the five canned items (anchovies, ham hocks, Irish stew, jellied eels, and kidney beans) in alphabetical order, with no two side by side. [/quote]\r\n\r\nIf the Ham hocks are in 2, anchovies are restricted to 1, which can't be true.",
  "Solution_8": "ooohhhh\r\nok\r\nanyway, i know what has to be on the bottom shelf, but i don't know where they go\r\n\r\nfarina is either in 13 or 14\r\noat bran is either in 12, 13, or 14\r\nthe kidney beans are either in 12, 13, or 14\r\nthe bluten is either in 11, or 15 but cannot be next to the farina if the oat bran is (11) or has to be next to the farina if the oat bran isn't (15)\r\none of the following three: chutney, dill pickles, and egg drop soup, are on the last place on the bottom shelf\r\n\r\nthat's what's on the bottom shelf and where they could go\r\nbut i haven't gotten futher than that yet",
  "Solution_9": "[size=150][color=blue][b]I found the solution!!![/b][/color][/size]\r\n\r\nI assumed that 1 and 6 aren't beside!\r\n\r\n[hide=\"Solution\"]\n1-Anchovies     2-Bones                3-Ham hocks      4-Egg drop soup     5-Irish stew\n6-Jellied eels    7-Chutney            8-Marshmallows  9-Ladyfingers        10-Nutcakes\n11-Gluten        12-Kidney beans   13-Farina            14-Oat bran           15-Dill pickles \n[hide=\"How I got this...\"]\n[hide=\"First\"]At first, there are two cases: Ham hocks and farina are in 4th column and the other case is that they are in 3rd column.\nTry the 4th column way and it leads to contradiction.\nSo, they are in the third column.[/hide]\n[hide=\"Second\"]\nBecause of the bag-can condition and 1.clue, kidney beans are in 3. shelf. Kidney beans and oat bran are in positions 12 and 14 because of the 8.clue. Trying cases we see that the correct one is: 12-KB, 14-OB.[/hide]\n[hide=\"Third\"]\nOat Bran is below Ladyfingers, so in clue 6 we se that chutney and marshmallows are beside so they are in positions 4-5 or 6-7 or 7-8.\nBut, Irish stew and Jelied eels can be only in 5,6 or 7 so Irish stew is in 5, Chutney in 7 and marshmallows in 8, and eels in 6.\n[/hide]\n[hide=\"Finally...\"]\nGluten is in 11 or 15, so in clue 7 we see that pickles are directly below nutcakes, so they are in 10 and 15, and gluten in 11. \nThe remaining fields are 2 and 4. As Chutney is in 2. column, egg drop soup can't be so it's in 4, and bones in 2[/hide][/hide][/hide]",
  "Solution_10": "wow!\r\nnice!\r\ni couldn't find any contradictions, so...\r\nmarkkram, do you have the answer?",
  "Solution_11": "Well,thank you very much!\r\n\r\nMost of my contradictions were e.g: bag is next to can, in clues 6 or 7 both or none holds or couldn't find place for all 5 canned items.",
  "Solution_12": "Good job. I will check when I find out the answer in the book. I guess that \"beside\" meant \"horizontal only\" after all.",
  "Solution_13": "Well, it wasn't so hard. It took me a hour to solve it. I assumed that beside is horizontal only.\r\n\r\nAlso, for those who don't know: Einstein's test:\r\nEinstein said that only 2% of people in the world can solve this test in less than 15 minutes with only paper and pencil.\r\nHere's the test:\r\n\r\n5 houses are painted in 5 different colors, in every house lives a person of different nationality, those 5 owners drink specific drink, smoke specific cigarettes, and hold specific pets, none 2 owners have same pet, cigares or drink.\r\nQuestion: Who has the fishes?\r\n[u][b]Clues[/b][/u]:\r\nBritish lives in red house\r\nSwedish has dogs\r\nDanish drinks tea\r\nGreen house is left from the white house\r\nOwner of green house drinks coffee\r\nPerson who smokes Pall-Mall has birds\r\nOwner of yellow house smokes Dunhill\r\nOwner of the middle house drinks milk\r\nPerson from Norway lives in first house\r\nPerson who smokes Blend lives beside the person who has cats\r\nThe man who has horses lives beside the person which smokes Dunhill\r\nOwner which smokes Blue Master drinks beer\r\nGerman smokes Prince \r\nPerson from Norway lives beside blue house\r\nPerson which smokes Blend has neighbour who drinks water\r\n\r\n\r\n[u][b][size=125]There is only one true solution!!![/size][/b][/u]",
  "Solution_14": "I think I should put this in my mock MC test. :D\r\n\r\nHmmm, I wish I had some graph paper.  :roll:",
  "Solution_15": "[u]Answer to Einstein Question[/u]:\r\n\r\n[hide]German owns the fish.[/hide]",
  "Solution_16": "Yes, that is correct.",
  "Solution_17": "Can you please show your solution?",
  "Solution_18": "[quote=\"isabella2296\"]Can you please show your solution?[/quote]\r\n\r\nJust draw a table and fill in things you know.\r\n\r\n:D that was a 4th grade homework assignment for the advanced math group at my elementary school lol"
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "If $x_1,x_2,...,x_n$ are non-negative numbers, then\r\n$(n-1)(x_1^{n+1}+x_2^{n+1}+...+x_n^{n+1}) \\geq (x_1+...+x_n)(x_1^n+x_2^n+...+x_n^n-x_1x_2...x_n)$\r\n\r\nor\r\n\r\nIf $x_1,x_2,...,x_n$ are non-negative numbers such that $x_1+x_2+...+x_n=n-1$, then\r\n$x_1^n(1-x_1)+x_2^n(1-x_2)+...+x_n^n(1-x_n) \\leq x_1x_2...x_n$.",
  "Solution_1": "Sorry,I can't understand the meaning of the inequality.\r\nCan you explain more,VASC?",
  "Solution_2": "Did Vasc want to say that the two form of the Inequalities is equivalent?",
  "Solution_3": "[quote=\"Hawk Tiger\"]Did Vasc want to say that the two form of the Inequalities is equivalent?[/quote]\r\nI know now,thank you :)",
  "Solution_4": "hey I have the proof for $(n-1)\\sum a_i ^{n+k}+\\prod a_i \\sum a_i ^k \\geq \\sum a_i \\sum a_i ^{n+k-1}$ where k  and $a_i$ are positive real",
  "Solution_5": "[quote=\"Albanian Eagle\"]hey I have the proof for $(n-1)\\sum a_i ^{n+k}+\\prod a_i \\sum a_i ^k \\geq \\sum a_i \\sum a_i ^{n+k-1}$ where k  and $a_i$ are positive real[/quote]\r\n\r\nis k positive real or positive integer?",
  "Solution_6": "k real positive (or maybe for some egative too but i have to check)!",
  "Solution_7": "A. Eagle, have you a nice solution for positive real or only positive integer k ?",
  "Solution_8": "$k$real\r\nand since the solution is quite similar to the basic suranyi ineq I guess it's nice...",
  "Solution_9": "Indeed, as You and Son of Skanderbeg had prove, the inequality is valid for any real k.\r\nCongratulations for this generalization. :lol:",
  "Solution_10": "Could I have question that what  the basic suranyi Ineq is,and  where  the proof is.\r\nI just proove the original inquality.",
  "Solution_11": "it's the inequality in my post above for $k=0$\r\n(I emember harazi has posted a proof somwhere in the forum...)",
  "Solution_12": "[quote=\"Vasc\"]Indeed, as You and Son of Skanderbeg had prove, the inequality is valid for any real k.\nCongratulations for this generalization. :lol:[/quote]\r\n\r\nThank you !\r\n :) \r\n\r\n(We are working now on an inequality which is a bit more general than that generalisation. :) )",
  "Solution_13": "Congratulations for this generalization from you're friends Fitim Dika! :D \r\n[b]Go for it[/b] and [b]Good for you![/b] :wink:",
  "Solution_14": "no one can ?"
}
{
  "Tag": [],
  "Problem": "A space-shuttle revolves around the Earth along an elliptical path, the smallest distance between the Earth and the shuttle is $ 6870$ km (measured from the centre of the Earth), the speed of the shuttle at this position is $ 10 km/s$. What is the least speed of the shuttle? (The radius of the Earth is $ 6370$ km)",
  "Solution_1": "Actually, is enough information given? I'm not so sure..."
}
{
  "Tag": [
    "inequalities",
    "inequalities solved"
  ],
  "Problem": "Prove  that\r\nx/(y+z) + y/(z+x) + z/(x+y) >=3/2  for  any  positive  reals  x,y,z\r\n12 th  british  mo , 1976\r\nThanks",
  "Solution_1": "Isn't-it [b]THE[/b] inequality (Nesbitt's) found in every tutorials on inequalities ??\r\n\r\n\r\nOne of the many solutions (the classical one?) is to use rearrangement\r\n\r\nx,y,z  and 1/y+z, 1/z+x, 1/x+y are ordered the same way so\r\n\r\nx/(y+z) + y/(z+x) + z/(x+y) >= x/(z+x) + y/(x+y) + z/(y+z)\r\nand\r\nx/(y+z) + y/(z+x) + z/(x+y) >= x/(y+x) + y/(z+y) + z/(x+z)\r\n\r\nAdding these two inequalites gives the wanted result.",
  "Solution_2": "Cauchy is also straightforward:\r\n\r\n\\sum x/(y + z) \\sum x(y + z) \\geq (\\sum x)<sup>2</sup>\r\n\r\nSince 3/2 \\sum x(y + z) \\leq (\\sum x)<sup>2</sup> the result follows.",
  "Solution_3": "Or\r\na=y+z, b=x+z, c=x+y\r\nThe inequality comes to\r\n(a+b-c)/2c+(a-b+c)/2b+(-a+b+c)/2c \\geq 3/2 <=>\r\na/b+b/a+a/c+c/a+b/c+c/b \\geq 6.\r\nThis is true because we know A/B+B/A \\geq 2 for all A,B>0",
  "Solution_4": "heh.. this one is really easy.. :D\r\n\r\nit is equivalent also to:\r\n\r\nsum ((x+y)(x-y)^2))>=0. which is obiuos.\r\n\r\ncheers!",
  "Solution_5": "I think that this problem has a lot of solutions .. here is another one:\r\n\r\nPut x=p-a; y=p-b; z=p-c and we have:\r\n\r\nsum((p-a)/a)>=3/2 <=> (a+b+c)(1/a+1/b+1/c)>=9 (true) :D :D",
  "Solution_6": "Are you trying to gather as many solutions as possible for this WELL-KNOWN ineq? :D\r\n\r\nWe add 1 to each fraction and it becomes equivalent to (x+y+z)(1/(x+y)+1/(y+z)+1/(z+x)) >= 9/2, which is equivalent to (a+b+c)(1/a+1/b+1/c) >= 9, where a=y+z etc."
}
{
  "Tag": [
    "number theory"
  ],
  "Problem": "Great Aunt Minnie's age plus the square of Great Uncle Paul's age is 7308. Great Uncle Paul's age plus the square of Great Aunt Minnie's age is 6974. What is the sum of Great Aunt Minnie's age and Great Uncle Paul's age?\r\n\r\nI formed an equation:\r\n[hide][b]M+P^2=7308\nP+M^2=6974\nP+M=???\n[/b][/hide]\r\n\r\nPlease help me solve this",
  "Solution_1": "[hide=\"hint\"]Solve the system of equations[/hide]",
  "Solution_2": "[hide]Well, obviously they have to be pretty old. I just picked numbers. You know that Great Uncle Paul has to be older than 80 and younger than 90. I just picked 85. \n85 <sup>2</sup> =7225 and to get 7308, you add 83 to it. When you then take and square Great Aunt Minnie's age 83 <sup>2</sup> = 6889, and 6889 + 85 = 6974. \n\nThe sum of their ages = 85 + 83 = 168. \n :)  :)  :) \n\nBTW- For those of you who didn't know this- a quick way to multiply 2 two digit numbers that both have the same 10's digit and the sum of the 1's digits add up to 10 is that the last two digits of the number are the 1's digits multitplied and the first digit/s are the first two digits multiplied, where you raise one of the numbers to the next number. So if you have 34 x 36, you would get 1224, because 4 x 6 = 24, and     3 x 4 = 12. (If the last two numbers are 1 and 9, the last two digits are 09.)\n\n :)  :) [/hide]",
  "Solution_3": "Yeah, guess and check appears to work better for this one than algebra; when I tried to solve the system I got a fourth degree equation :o",
  "Solution_4": "I've got a solution using number theory:\r\n\r\n[hide] Suppose x=Great Uncle Paul's age and y=Great Aunt Minnie's age. Thus x^2+y=7308 and y^2+x=6974. Subtracting the second from the first gives x^2-y^2=y-x=334, and factoring and rearranging gives (x+y)(x-y)+(y-x)=334. Multiply by -1 on both sides gives:\n\n(x-y)(-x-y)+(x-y)=-334, which is also (x-y)(1-x-y)=-334.\n\nMultiply by -1 again gives (x-y)(x+y-1)=334. Factoring 334 gives 2x167. Supposing x-y=2 and x+y-1=167, then solve the systems of equations, which gives x=85 and y=83. \n\n[/hide]",
  "Solution_5": "Nice work, anonymous Ohio teammate :D",
  "Solution_6": "[hide]\nEasy guess-and-check:\n\\begin{eqnarray*}\nm+p^2=7308\\\\\np+m^2=6974 \n\\end{eqnarray*}\n\nYou know that $p^2 < 7308$, so $p < 85.5...$. Try $p = 85$. Then $7308 - 85^2 = 83$, so $m = 83$. Try this in the other equation. It also works, and is reasonable. $83 + 85 = 168$.\n[/hide]",
  "Solution_7": "Yea I guess guess and check is better...",
  "Solution_8": "This was on the 1997-1998 Mathcounts Sprint Nationals round, so you aren't given a calculator. Doing it algebraically would be more reliable. \n",
  "Solution_9": "If you know the rational roots theorem, it's pretty easy to solve a fourth degree polynomial. I personally hate algebra(no offense) so I would have done guess and check. But algebra is more reliable. :help: "
}
{
  "Tag": [
    "linear algebra",
    "linear algebra unsolved"
  ],
  "Problem": "In $ R^{n\\plus{}1}$, indentify $ R^{n}$ as $ span(e_{1},...,e_{n})$. \r\nFor any fixed $ z\\in R^{n}$, define $ f: R\\to S_{n}$ by $ f(t)\\equal{}cos(\\parallel z\\parallel t)e_{n\\plus{}1}\\plus{}sin(\\parallel z\\parallel t)\\frac{z}{\\parallel z\\parallel }$\r\n\r\nHow to show $ f(t)$ is a great circle of $ S_{n}$? (maybe this is the defination of great circle).",
  "Solution_1": "Here is a [url=http://www.d-e-f-i-n-i-t-e-l-y.com/]very helpful link[/url].\r\n\r\nA great circle is the intersection of $ S^{n}$ with a two-dimensional subspace (a plane passing through the center of the sphere). The plane you have here is spanned by $ e_{n\\plus{}1}$ and $ z$. It is easy to show that a linear combination $ ae_{n\\plus{}1}\\plus{}bz$ has norm one if and only if it is equal to $ f(t)$ for some $ t$."
}
{
  "Tag": [
    "algebra proposed",
    "algebra"
  ],
  "Problem": "Find $ a$ if:\r\n\r\n$ a \\equal{} \\displaystyle\\sum_{i\\equal{}1}^\\infty {\\displaystyle(\\frac{2007n}{2007^n})}$",
  "Solution_1": "If $ \\vert x \\vert < 1$ we have that\r\n\r\n$ \\sum_{n \\equal{} 1}^{\\infty} nx^{n \\minus{} 1} \\equal{} \\frac {1}{(1 \\minus{} x)^{2}} .$\r\n\r\nHence,\r\n\r\n$ a \\equal{} 2007 \\cdot \\sum_{n \\equal{} 1}^{\\infty} \\frac {n}{2007^{n}} \\equal{} \\sum_{n \\equal{} 1}^{\\infty} \\frac {n}{2007^{n \\minus{} 1}} \\equal{} \\frac {1}{\\left(1 \\minus{} \\frac {1}{2007}\\right)^{2}} \\equal{} \\left(\\frac{2007}{2006}\\right)^{2}.$"
}
{
  "Tag": [
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Prove that the statement  \"The sum of 2 irrational numbers is irrational.\" is wrong, using straight forward prove (not a simple counter example). :(\r\nthanks in advance.",
  "Solution_1": "If x irratianal, then -x irrational $x+(-x)=0$ rational.  :D",
  "Solution_2": "thats an counter example...\r\nstraight forward prove? also for all positive irrational numbers please.",
  "Solution_3": "if x irrational and a rational then a-x irrational.(a-x)+(x)=a is rational\r\n(if a>x>0 then x,a-x are positive)"
}
{
  "Tag": [
    "geometry",
    "perimeter",
    "rotation",
    "absolute value"
  ],
  "Problem": "|x+2| + |y-3| = 1 is an equation for a square. How many units are in the length of its diagonal.\r\n\r\nPlease show ur solution",
  "Solution_1": "[hide]Graph it out. Looks like a square with sides of $\\sqrt 2$. The answer is $2$.[/hide]",
  "Solution_2": "this was a national sprint question...you cant graph it in less than 40 seconds...but thanx for the answer",
  "Solution_3": "$|x-a|+|y-b|=c$ is the generic equation for a square with a center at $(a,b)$ and a a side length of $\\sqrt{2c}$ (so a diagonal is equal to $2\\sqrt{c}$ , perimeter is equal to $4\\sqrt{2c}$ and area is equal to $2c$).",
  "Solution_4": "[quote=\"NoSoupForYou\"]$|x-a|+|y-b|=c$ is the generic equation for a square with a center at $(a,b)$ and a a side length of $\\sqrt{2c}$ (so a diagonal is equal to $2\\sqrt{c}$ , perimeter is equal to $4\\sqrt{2c}$ and area is equal to $2c$).[/quote]\r\n\r\nSo if we were to have $|x+2|+|y-3|=2$, then the side length would be $2\\sqrt{2}$? I thought it was $2c$.",
  "Solution_5": "try solving the problem using $s=2c$; I'm pretty sure the answer you get won't be 2",
  "Solution_6": "[quote=\"mathwizard91\"]|x+2| + |y-3| = 1 is an equation for a square. How many units are in the length of its diagonal.\n\nPlease show ur solution[/quote]\r\n\r\n[hide=\"Something\"]\nIt looks like each side length is 1. [b] :sqrt: 2[/b].\n[/hide]",
  "Solution_7": "The vertices are located at $(-2,2)$,$(-1,3)$,$(-2,4)$ and $(-2,3)$\r\n\r\nThe side length is $\\sqrt{2}$, so the area is 2",
  "Solution_8": "For abs(x-a)+abs(y-b)=c, it is a square with center (a,b).  Think about it., when x is a, y is b+c or b-c, and when y is b, x is a+c or a-c.  These are the four vertices.  The square is rotated 45 degrees.  That makes the diagonal 2c and the area 2c <sup>2</sup>  because dist((a,b-c),(a,b+c)) is 2c.",
  "Solution_9": "Oh yeah, I messed up. Here is what I should have put;\r\n\r\n$|x-a|+|y-b|=c$ is the generic equation for a square with a center at $(a,b)$ and a a side length of $c\\sqrt{2}$ (so a diagonal is equal to $2c$ , perimeter is equal to $4c\\sqrt{2}$ and area is equal to $2c^2$).",
  "Solution_10": "yeah that is correct",
  "Solution_11": "this is a annoying problem becuz there is no use memorizing the formula but u still need to know how to do it so just learn how to do it w/o the formula",
  "Solution_12": "You can solve the problem just by memorizing the formula.",
  "Solution_13": "I'm not very familiar with these equations of a polygon. What does the equation of a square tell me?",
  "Solution_14": "[quote=\"236factorial\"]I'm not very familiar with these equations of a polygon. What does the equation of a square tell me?[/quote]\r\n\r\nUmm, well you graph it out and you get a square.",
  "Solution_15": "how do you get the formula for a square?",
  "Solution_16": "think about it: a regular absolute value graph is a right angle up and two x values can lead to a single y value.  But for a square, you need two x values to lead to two y values, which will make another right angle and a square"
}
{
  "Tag": [
    "LaTeX"
  ],
  "Problem": "How do I make a table of contents like this one:\r\n\r\n[img]http://img78.imageshack.us/img78/8190/asdasdkg3.png[/img]",
  "Solution_1": "The command for producing a table of contents is: \\tableofcontents. To change the way the sections and subsections are numbered  (x1.x2.x3.x4), use \\setcounter{tocdepth}{x}.\r\nSameet.",
  "Solution_2": "Okay, but what could be the code to produce the result above?",
  "Solution_3": "ssn116 has just told you, \\tableofcontents.\r\nYou will need to compile your document more than once so LaTeX can go through the document, saving the details in a .toc file, and then printing the result."
}
{
  "Tag": [],
  "Problem": "\u03a3\u03c4\u03bf \u03b2\u03b9\u03b2\u03bb\u03af\u03bf \u03c4\u03bf\u03c5 \u03b1\u03c0\u03b5\u03b9\u03c1\u03bf\u03c3\u03c4\u03b9\u03ba\u03bf\u03cd 1 \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03b7 \u03b5\u03be\u03ae\u03c2 \u03ac\u03c3\u03b7\u03c3\u03b7:\r\n\u0391\u03bd \u03b1,\u03b2>0 \u03ba\u03b1\u03b9 \u03b1+\u03b2=1, \u03c4\u03cc\u03c4\u03b5 (\u03b1+1/\u03b1)^2+(\u03b2+1/\u03b2)^2>=25/2\r\n\u0395\u03b3\u03ce \u03c4\u03b1 \u03ad\u03c6\u03b5\u03c1\u03b1 \u03cc\u03bb\u03b1 \u03c3\u03c4\u03bf \u03c0\u03c1\u03ce\u03c4\u03bf \u03bc\u03ad\u03bb\u03bf\u03c2,\u03ad\u03ba\u03b1\u03bd\u03b1 \u03c0\u03c1\u03ac\u03be\u03b5\u03b9\u03c2,\u03c0\u03b1\u03c1\u03b1\u03b3\u03bf\u03bd\u03c4\u03bf\u03c0\u03bf\u03b9\u03ae\u03c3\u03b5\u03b9\u03c2 \u03ba\u03c4\u03bb \u03ba\u03b1\u03b9 \u03ba\u03b1\u03c4\u03ad\u03bb\u03b7\u03be\u03b1 \u03c3\u03c4\u03bf:\r\n(\u03b2^2)(\u03b1^2-1)^2+(\u03b1^2)(\u03b2^2-1)^2>=0 \u03c0\u03bf\u03c5 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9.\u0395\u03bb\u03b5\u03b3\u03be\u03b1 \u03be\u03b1\u03bd\u03ac \u03c4\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03ba\u03b1\u03b9 \u03b4\u03b9\u03b1\u03c0\u03af\u03c3\u03c4\u03c9\u03c3\u03b1 \u03bf\u03c4\u03b9 \u03b4\u03b5\u03bd \u03c7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03af\u03b7\u03c3\u03b1 \u03c4\u03bf \u03c3\u03c4\u03bf\u03b9\u03c7\u03b5\u03af\u03bf [u]\u03b1+\u03b2=1[/u].\r\n\u03a4\u03b9 \u03b3\u03b9\u03bd\u03b5\u03c4\u03b1\u03b9 \u03b5\u03b4\u03c9?\u0394\u03af\u03bd\u03b5\u03b9 \u03c0\u03b5\u03c1\u03b9\u03c4\u03ac \u03b4\u03b5\u03b4\u03bf\u03bc\u03ad\u03bd\u03b1 \u03b7 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7?",
  "Solution_1": "[quote=\"DIMITRIS T\"]\u03a3\u03c4\u03bf \u03b2\u03b9\u03b2\u03bb\u03af\u03bf \u03c4\u03bf\u03c5 \u03b1\u03c0\u03b5\u03b9\u03c1\u03bf\u03c3\u03c4\u03b9\u03ba\u03bf\u03cd 1 \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03b7 \u03b5\u03be\u03ae\u03c2 \u03ac\u03c3\u03b7\u03c3\u03b7:\n\u0391\u03bd \u03b1,\u03b2>0 \u03ba\u03b1\u03b9 \u03b1+\u03b2=1, \n\n\u03c4\u03cc\u03c4\u03b5 (\u03b1+1/\u03b1)^2+(\u03b2+1/\u03b2)^2>=25/2\n[/quote]\r\n\u03a4\u03bf \u03c0\u03b1\u03c1\u03b1\u03c0\u03b1\u03bd\u03c9 \u03b5\u03b9\u03bd\u03b1\u03b9: \u03c0\u03c7 \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03c0\u03c1\u03c9\u03c4\u03b7 \u03c0\u03b1\u03c1\u03b5\u03bd\u03b8\u03b5\u03c3\u03b7 (\u03b1+1/\u03b1) :\r\n(\u03b1 + (1/\u03b1)) \u03b5\u03be \u03bf\u03c1\u03b9\u03c3\u03bc\u03bf\u03c5.\r\n\r\n\u039c\u03b7\u03c0\u03c9\u03c2 \u03bf\u03bc\u03c9\u03c2 \u03b5\u03bd\u03bd\u03bf\u03b5\u03b9\u03c2 ((\u03b1+1)/\u03b1) ?\r\n\r\n\u0395\u03c4\u03c3\u03b9 \u03ae \u03b1\u03bb\u03bb\u03b9\u03c9\u03c2 \u03c0\u03b1\u03bd\u03c4\u03c9\u03c2 \u03b4\u03b5\u03bd \u03b2\u03bb\u03b5\u03c0\u03c9 \u03c0\u03c9\u03c2 \u03bc\u03b5 \u03b9\u03c3\u03bf\u03b4\u03c5\u03bd\u03b1\u03bc\u03b9\u03b5\u03c2 \u03bc\u03c0\u03bf\u03c1\u03b5\u03b9\u03c2 \u03bd\u03b1 \u03ba\u03b1\u03c4\u03b1\u03bb\u03b7\u03be\u03b5\u03b9\u03c2 \u03c3\u03b5 \u03b1\u03c5\u03c4\u03bf \u03c0\u03bf\u03c5 \u03ba\u03b1\u03c4\u03b5\u03bb\u03b7\u03be\u03b5\u03c2.",
  "Solution_2": "\u0395\u03af\u03bd\u03b1\u03b9 \u03bc\u03b9\u03b1 \u03b1\u03c0\u03bb\u03ae B. C. S. \u03b1\u03bd \u03c0\u03bf\u03bb\u03bb\u03b1\u03c0\u03bb\u03b1\u03c3\u03b9\u03ac\u03c3\u03b5\u03b9\u03c2 \u03ba\u03b1\u03b9 \u03c4\u03b1 \u03b4\u03cd\u03bf \u03bc\u03ad\u03bb\u03b7 \u03bc\u03b5 2.\r\n\u039c\u03b5\u03c4\u03ac \u03bc\u03c0\u03bf\u03c1\u03b5\u03af\u03c2 \u03bd\u03b1 \u03ba\u03ac\u03bd\u03b5\u03b9\u03c2 \u03b1\u03bd\u03c4\u03b9\u03ba\u03b1\u03c4\u03ac\u03c3\u03c4\u03b1\u03c3\u03b7 \u03ba\u03b1\u03b9 \u03bd\u03b1 \u03c7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03ae\u03c3\u03b5\u03b9\u03c2 \u03c0\u03b1\u03c1\u03b1\u03b3\u03ce\u03b3\u03bf\u03c5\u03c2 \u03b3\u03b9\u03b1 \u03c4\u03bf \u03b5\u03bb\u03bb\u03ac\u03c7\u03b9\u03c3\u03c4\u03bf \u03bc\u03b5\u03c4\u03ac \u03b1\u03c0\u03cc \u03b1\u03c5\u03c4\u03cc.",
  "Solution_3": ":oops_sign:   Sorry \u03c0\u03b1\u03b9\u03b4\u03b9\u03ac.\u0395\u03c6\u03b1\u03b3\u03b1 \u03ad\u03bd\u03b1 \u03cc\u03c1\u03bf \u03cc\u03c4\u03b1\u03bd \u03ad\u03ba\u03b1\u03bd\u03b1 \u03c0\u03b1\u03c1\u03b1\u03b3\u03bf\u03bd\u03c4\u03bf\u03c0\u03bf\u03af\u03b7\u03c3\u03b7 \u03ba\u03b1\u03b9 \u03ba\u03b1\u03c4\u03ad\u03bb\u03b7\u03be\u03b1 \u03c3\u03b5 \u03bb\u03ac\u03b8\u03bf\u03c2 \u03c3\u03c5\u03bc\u03c0\u03ad\u03c1\u03b1\u03c3\u03bc\u03b1.",
  "Solution_4": "$ 2( (a \\plus{} \\frac{1}{a})^2\\plus{}(b \\plus{} \\frac{1}{b})^2 ) \\equal{}$\r\n$ (1^2 \\plus{} 1^2)( (a \\plus{} \\frac{1}{a})^2\\plus{}(b \\plus{} \\frac{1}{b})^2 ) >\\equal{}$ (\u0391\u03c0\u03cc B. C. S.)\r\n$ ( 1(a \\plus{} \\frac{1}{a})\\plus{} 1(b \\plus{} \\frac{1}{b}) )^2 \\equal{}$\r\n$ (a \\plus{} \\frac{1}{a} \\plus{} b \\plus{} \\frac{1}{a})^2 \\equal{}$\r\n$ (a \\plus{} b \\plus{} \\frac{a \\plus{} b}{ab})^2 \\equal{}$\r\n$ (1 \\plus{} \\frac{1}{ab})^2 \\equal{} A$\r\n$ a \\plus{} b \\equal{} 1 < \\equal{} > ab <\\equal{} \\frac{1}{4} < \\equal{} >\\frac{1}{ab} >\\equal{} 4$ \u03ac\u03c1\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9\r\n$ A  >\\equal{} (1 \\plus{} 4)^2 \\equal{} 5^2 \\equal{} 25 < \\equal{} >$\r\n$ (a \\plus{} \\frac{1}{a})^2\\plus{}(b \\plus{} \\frac{1}{b})^2 >\\equal{} \\frac{25}{2}$"
}
{
  "Tag": [],
  "Problem": "Compute the least positive prime $p$ greater than $2$ such that $p^{3}+7p^{2}$ is a perfect square.",
  "Solution_1": "[hide=\"Solution\"]  If $p^{3}+7p^{2}$ is a perfect square, then $p+7$ must also be a perfect square.  Checking, we find that the smallest perfect square for which this is true is $p=29$.[/hide]"
}
{
  "Tag": [
    "integration",
    "floor function",
    "calculus",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "$\\int_{0}^{e}e^{x}d\\lfloor x\\rfloor$",
  "Solution_1": "If I remember correctly, $\\int_{a}^{b}f(x)dh(x)$ is the limit of sums like $\\sum_{i}f(x_{i}^{*})(h(x_{i})-h(x_{i-1}))$, where $x_{i}*\\in [x_{i-1},x_{i}]$. Since $h$ is piecewise constant, only subintervals about $1$ and $2$ will contribute nonzero terms to the sum. Consequently, the integral is $e+e^{2}$. \r\n\r\nNow how do we put this theory into practice?  :maybe:"
}
{
  "Tag": [
    "linear algebra",
    "matrix",
    "trigonometry",
    "function",
    "complex numbers"
  ],
  "Problem": "There are given 2 numbers $ a$ and $ b$ with $ a^2\\plus{}b^2<1$ and two rows $ (u_n)_{n\\geq 1}$ and $ (v_n)_{n\\geq 1}$ defined by \r\n\\[ u_{n\\plus{}1}\\equal{}au_n\\minus{}bv_n\\]\r\n\\[ v_{n\\plus{}1}\\equal{}bu_n\\plus{}av_n\\]\r\nProve that $ lim u_n\\equal{}lim v_n\\equal{}0$ and the series $ \\sum_{n\\equal{}1}^{\\infty} u_n$ and $ \\sum_{n\\equal{}1}^{\\infty} v_n$ are convergent and calculate them",
  "Solution_1": "[hide=\"nice problem\"]Use matrix representation of complex numbers $ a\\plus{}bi \\to \\rho \\begin{pmatrix}\\cos \\varphi&\\minus{}\\sin \\varphi\\\\ \\sin \\varphi&\\cos \\varphi\\end{pmatrix}, 0\\le \\rho < 1$ [/hide]",
  "Solution_2": "Digger, I can easily see how that statement proves both sums converge,  but could you explain how that representation allows one to actually calculate the sums?",
  "Solution_3": "[hide]Let $ \\alpha \\equal{} a \\plus{} bi$ and $ \\omega_k \\equal{} u_k \\plus{} v_k i$, then $ \\omega_k \\alpha \\equal{} au_k \\plus{} b i u_k \\plus{} a i v_k \\plus{} b i^2 v_k \\equal{} (a u_k \\minus{} b v_k ) \\plus{} (a v_k \\plus{} b u_k) i \\equal{} \\omega_{k\\plus{}1}$. So $ \\omega_k \\equal{} \\omega_1 \\alpha^{k\\minus{}1}$.[/hide]",
  "Solution_4": "[hide]I had actually gotten exactly that far, but failed to make the final connection with the geometric sum formula. Thanks anyway![/hide]",
  "Solution_5": "[quote=\"Darmani\"]Digger, I can easily see how that statement proves both sums converge,  but could you explain how that representation allows one to actually calculate the sums?[/quote]\r\n[hide=\"what i was thinking about\"]Write the given recursion in the form $ w_{n+1}=Cw_n$, where $ w_n=\\begin{pmatrix}u_n\\\\ v_n\\end{pmatrix}, \\ C=\\begin{pmatrix}a & - b \\\\ b & a\\end{pmatrix} \\Rightarrow w_n=C^nw_0$ - I prefer to start the recursion whith $ n=0$.\n\nIt is easy to see that the map $ a+bi \\rightarrow \\begin{pmatrix}a & - b \\\\ b & a\\end{pmatrix}$ preserves convergence - for this we need distance-function for matrixes of our form. It is evidently to see that $ \\parallel{}C\\parallel{} = \\sqrt{\\det C}=|a+bi|$ works.\n\nBecause complex serie $ \\sum\\limits_{n=0}^\\infty z^n$ converges for $ |z|<1$ it follows that matrix serie  $ \\sum\\limits_{n=0}^\\infty C^n$ also converges and $ \\sum\\limits_{n=0}^\\infty C^n=(I-C)^{-1}$ and\n\n$ \\sum\\limits_{n=0}^\\infty \\begin{pmatrix}u_n\\\\ v_n\\end{pmatrix}=\\sum\\limits_{n=0}^\\infty w_n=\\sum\\limits_{n=0}^\\infty C^nw_0=(I-C)^{-1}w_0=\n\\frac{1}{(1-a)^2+b^2}\\begin{pmatrix}1-a & - b \\\\ b & 1-a\\end{pmatrix}\\begin{pmatrix}u_0\\\\ v_0 \\end{pmatrix} \\Rightarrow \\sum\\limits_{n=0}^\\infty u_n = \\dots, \\ \\sum\\limits_{n=0}^\\infty v_n = \\dots$[/hide]\n\n[hide=\"how to eliminate matrixes from the above\"]\nJust write the given recursion twice:\n\n$ \\begin{pmatrix}u_{n+1} & -v_{n+1} \\\\ v_{n+1} & u_{n+1}\\end{pmatrix}=\\begin{pmatrix}a & -b \\\\ b & a\\end{pmatrix}\\begin{pmatrix}u_n & -v_n \\\\ v_n & u_n\\end{pmatrix}$, use the inverse map $ \\begin{pmatrix}a & - b \\\\ b & a\\end{pmatrix} \\rightarrow a+bi$ and get $ w_{n+1}=u_{n+1} +i v_{n+1}=(a+bi)(u_n+v_ni)=cw_n$ - what have got [b]MellowMelon[/b]\n[/hide]",
  "Solution_6": "See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1038337#1038337]here[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=187900]here[/url]."
}
{
  "Tag": [
    "function",
    "trigonometry"
  ],
  "Problem": "Find constants $ a, b, c,$ and $ d$ such that:\r\n$ 4x^3 \\minus{} 3x \\plus{} \\frac{\\sqrt3}2\\equal{}a(x\\minus{}b)(x\\minus{}c)(x\\minus{}d)$\r\n\r\nRules:\r\n\r\n#1: Your factors must be exact. No numerical approximations allowed.\r\n#2: Neither the imaginary constant, nor the square root of any negative number may appear in your answer.\r\n#3: Your answer must be simple. Nothing more complicated than a trigonometric function may appear in your answer.\r\n#4: Yes, this problem is solvable.",
  "Solution_1": "[hide=\"Hint\"] The first two constants should remind you of a trigonometric identity. [/hide]",
  "Solution_2": "Yes! I have noticed that as well. :lol:",
  "Solution_3": "hello, you have\r\n${ 4x^3-3x+\\frac{\\sqrt{3}}{2}}=ax^3+x^2(-ad-ac-ab)+x(acd+abd+abc)-abcd$\r\nand from this equation we get the equation system\r\n$ 4=a$\r\n$ 0=-ad-ac-ab$\r\n$ -3=acd+abd+abc$\r\n$ \\frac{\\sqrt{3}}{2}=-abcd$\r\nSonnhard.",
  "Solution_4": "[quote=\"Dr Sonnhard Graubner\"]we get the equation system\n$ 4 \\equal{} a$\n$ 0 \\equal{} \\minus{} ad \\minus{} ac \\minus{} ab$\n$ \\minus{} 3 \\equal{} acd \\plus{} abd \\plus{} abc$\n$ \\frac {\\sqrt {3}}{2} \\equal{} \\minus{} abcd$\nSonnhard.[/quote]\r\n\r\nHow do you solve a continuation? :D \r\n\r\n[hide=\"My Solution\"]We use $ \\cos 3 \\theta \\equal{} 4 \\cos^3 \\theta \\minus{} 3 \\cos$.\n\nLet, $ x \\equal{} \\cos \\theta$.\nSo\n$ 4x^3 \\minus{} 3x \\plus{} \\frac{\\sqrt{3}}{2} \\equal{} 4 \\cos^3 \\theta \\minus{} 3 \\cos \\plus{} \\frac{\\sqrt{3}}{2} \\equal{} \\cos 3 \\theta \\plus{} \\frac{\\sqrt{3}}{2}$\n\nTherefore, $ \\cos 3 \\theta \\plus{} \\frac{\\sqrt{3}}{2} \\equal{} 0 \\iff 3 \\theta \\equal{} \\frac{5\\pi}{6}, \\frac{7\\pi}{6}, \\frac{17\\pi}{6} \\iff \\theta \\equal{} \\frac{5\\pi}{18}, \\frac{7\\pi}{18}, \\frac{17\\pi}{18}$\n\nThus,\n$ 4x^3 \\minus{} 3x \\plus{} \\frac{\\sqrt{3}}{2} \\equal{} 4 \\left(x \\minus{} \\cos \\frac{5\\pi}{18}\\right) \\left(x \\minus{} \\cos \\frac{7\\pi}{18}\\right) \\left(x \\minus{} \\cos \\frac{18\\pi}{18}\\right)$[/hide]",
  "Solution_5": "[color=darkblue]Oh. Your problem is nice, but it's easy!\n\nWe will find a root of equation: $ 4x^3\\minus{}3x\\plus{}\\frac{\\sqrt{3}}{2}\\equal{}0$, (1)\nLet $ x \\equal{} sin\\ t$ so equation (1) $ \\Longrightarrow 4sin^3t\\minus{}3sint\\plus{}\\frac{\\sqrt{3}}{2}\\equal{}0$, (2)\n$ \\Longleftrightarrow sin3t\\equal{}\\frac{\\sqrt{3}}{2}$\n$ \\Longleftrightarrow sin3t\\equal{}sin\\frac{\\pi}{3}$\n$ \\Longrightarrow$ equation (2) has a root $ t\\equal{}\\frac{\\pi}{9}$\n$ \\Longrightarrow$ equation (1) has a root $ x\\equal{}sin\\frac{\\pi}{9}$\nHence $ 4x^3\\minus{}3x\\plus{}\\frac{\\sqrt{3}}{2}\\equal{}4(x\\minus{}sin \\frac{\\pi}{9})(x^2\\plus{}mx\\plus{}n)$\nVery easy you will get $ m$ and $ n$.\nThen OK![/color]"
}
{
  "Tag": [
    "calculus",
    "calculus computations"
  ],
  "Problem": "Use the series method, I am having problems constructing the solution to y''+ay=0, a>0\r\n\r\nThe solution should be y=Asin((a^.5)x)+Bcos((a^.5)x)\r\n\r\nIn the recursion relation I get -aCn/(n+2)(n+1) = Cn+2\r\n\r\nHowever to construct sin, I need a to have fractional power which I cannot seem to get.\r\n\r\ni.e. for odd n, Cn should have 'a' with a fractional power like n+.5 or a^(n+.5) but I can only see a^n in Cn for all n.\r\n\r\nWhere have I gone wrong?\r\n\r\nIn my series solution I have a constant C1 so if I can show C1=a^.5 then things will work out. But how do I do that? Is that what is left to do?\r\n\r\nI also have to show C0 = 1 for even n series.",
  "Solution_1": "As $C1$ is not determined, you may assume $C1 = B\\sqrt{a}$. Then you will have the desired result $C_{2n+1}= B \\frac{(-1)^{n}(\\sqrt{a})^{2n+1}}{(2n+1)!}$."
}
{
  "Tag": [
    "floor function",
    "number theory"
  ],
  "Problem": "An integer is square-free if it is not divisible by $a^2$ for any integer $a>1$. Let $S$ be the set of positive square-free integers. Determine, with justification, the value of\\[\\sum_{k\\epsilon S}\\left[\\sqrt{\\frac{10^{10}}{k}}\\right]\\]where $[x]$ denote the greatest integer less than or equal to $x$",
  "Solution_1": "Anyone knows how to do this problem?  :?:   :?",
  "Solution_2": "I have a strong suspicion that the answer is $10^{10}$.\r\n\r\nI do not, however, have a strong proof...\r\n\r\nThe main thing is that the result is independent of the number $10^{10}$ - I'm pretty sure $\\sum_{k\\in S}\\left\\lfloor\\sqrt{\\frac{a}{k}}\\right\\rfloor=\\lfloor a\\rfloor$ for all positive reals $a$.",
  "Solution_3": "Yes, you're right Diarmuid, you put me on the right track. I will prove your stronger statement for all integers $a$.\r\n\r\n[b]Lemma[/b]: if $n$, $k$ are positive integers then $\\left[\\sqrt{\\frac{n}{k}}\\right] =\\left[\\sqrt{\\frac{n - 1}{k}}\\right]$, unless $\\sqrt{\\frac{n}{k}} \\in \\mathbb{N}$, in which case we have $\\left[\\sqrt{\\frac{n}{k}}\\right] =\\left[\\sqrt{\\frac{n - 1}{k}}\\right] + 1.$\r\n\r\n[b]Proof[/b] Take $\\left[\\sqrt{\\frac{n}{k}}\\right] = m$. Then $m \\leq \\left[\\sqrt{\\frac{n}{k}}\\right] < m + 1$. Hence $m^2 - \\frac{1}{k} \\leq \\frac{n - 1}{k} < (a + 1)^2 - \\frac{1}{k}$. Hence $m - 1 = \\left[\\sqrt{m^2 - \\frac{1}{k}}\\right] \\leq \\sqrt{\\frac{n - 1}{k}} \\leq \\left[\\sqrt{(m + 1)^2 - \\frac{1}{k}}\\right] = m.$  :| Now, if $\\sqrt{\\frac{n}{k}} = m \\in \\mathbb{N}$, $n = km^2$, and $\\left[\\sqrt{\\frac{n - 1}{k}}\\right] = \\left[\\sqrt{m^2 - \\frac{1}{k}}\\right] = m - 1.$ So $\\left[\\sqrt{\\frac{n}{k}}\\right] =\\left[\\sqrt{\\frac{n - 1}{k}}\\right] + 1.$ Conversely, if $\\left[\\sqrt{\\frac{n}{k}}\\right] = m$ and $\\left[\\sqrt{\\frac{n - 1}{k}}\\right] = m - 1$, we get $\\sqrt{\\frac{n - 1}{k}} < m \\leq \\sqrt{\\frac{n}{k}}$ or $km^2 \\leq n < km^2 + 1$. Hence $n = km^2$, or $\\sqrt{\\frac{n}{k}} \\in \\mathbb{N}$, and the lemma is proven. (That was NOT a fun proof :P)\r\n\r\nNow we define $f(n) = \\sum_{k \\in S}\\left[\\sqrt{\\frac{n}{k}}\\right]$. We only need to show that $f(n) = f(n - 1) + 1$ but that is obvious from the lemma (indeed, for all positive integers $n$ there exists a unique $k \\in S$ such that $\\sqrt{\\frac{n}{k}} \\in \\mathbb{N}$.\r\n\r\nAnd we are done. :D",
  "Solution_4": "A related approach is to note that $\\lfloor \\sqrt{a/k} \\rfloor$ is the number of integers less than or equal to $a$ whose square-free part is equal to $k$.",
  "Solution_5": "[quote=\"Ravi B\"]A related approach is to note that $\\lfloor \\sqrt{a/k} \\rfloor$ is the number of integers less than or equal to $a$ whose square-free part is equal to $k$.[/quote]\r\n\r\nI see. That's neat :) it's basically the same, but the combinatorial interpretation looks much simpler.",
  "Solution_6": "Sry, how do you use the lemma to find the answer/value?",
  "Solution_7": "We have $f(n) = \\sum_{k \\in S}\\left[\\sqrt{\\frac{n}{k}}\\right]$. We only need to show that $f(n) = f(n - 1) + 1$... Now \\[f(n) - f(n - 1) = \\sum_{k \\in S}\\left(\\left[\\sqrt{\\frac{n}{k}}\\right] - \\left[\\sqrt{\\frac{n - 1}{k}}\\right]\\right).\\] By the lemma, this is equal to 1 since there is a unique $k \\in S$ such that $\\sqrt{\\frac{n}{k}} \\in \\mathbb{N}$ (and for the other $k$, we get $\\left[\\sqrt{\\frac{n}{k}}\\right] = \\left[\\sqrt{\\frac{n - 1}{k}}\\right]$, giving only zeroes).\r\n\r\nClear?",
  "Solution_8": "Very cool.  :maybe: \n\nRewrite the equation as\n\\[\\sum_{k\\in S} \\lfloor \\sqrt{\\frac{10^{10}}{k}}\\rfloor=\\sum_{k\\in S} \\sum_{ki^2\\leq 10^{10}, i\\geq 1} 1 = \\sum_{m=1}^{10^{10}} 1= \\boxed{10^{10}}\\]\nwhere $m=ki^2$, which has a unique representation as a product of a non-square number and a square number. ",
  "Solution_9": "Note that $\\left \\lfloor \\sqrt{\\frac{10^{10}}{k}}\\right \\rfloor$ counts the number of integers $m \\geq 1$ with $km^2 \\leq 10^{10}$, so we have\n$$\\sum_{k \\in S} \\left \\lfloor \\sqrt{\\frac{10^{10}}{k}}\\right \\rfloor=\\sum_{k \\in S} \\sum_{km^2 \\leq 10^{10}} 1=\\sum_{n=1}^{10^{10}} 1=10^{10},$$\nwhere the second equality comes from counting over $n=km^2$, since $k$ is squarefree."
}
{
  "Tag": [
    "percent"
  ],
  "Problem": "Mr. Green receives a $ 10 \\%$ raise every year.  His salary after four such raises has gone up by what percent?\r\n\r\n\\[ \\textbf{(A)}\\ \\text{less than }40 \\% \\qquad\r\n\\textbf{(B)}\\ 40 \\% \\qquad\r\n\\textbf{(C)}\\ 44 \\% \\qquad\r\n\\textbf{(D)}\\ 45 \\% \\qquad\r\n\\textbf{(E)}\\ \\text{More than }45 \\%\r\n\\]",
  "Solution_1": "This is simply $ 100\\% \\times (1.1)^4 \\minus{} 100\\%$, or $ 146.41 \\minus{} 100$, so $ \\boxed{\\textbf{(E)}\\ \\text{More than }45 \\%}$.",
  "Solution_2": "[hide=\"Alternate Solution\"]\nClearly, $ (1.1)^4$ has more than two digits past the decimal point, so we eliminate $ \\textbf{(B)}$, $ \\textbf{(C)}$, and $ \\textbf{(D)}$.  It is also quite clearly greater than $ 40 \\%$, so the only answer choice left is $ \\textbf{(E)}$.\n[/hide]"
}
{
  "Tag": [
    "abstract algebra",
    "group theory",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "Problem. Let $G = \\{g_1,\\ldots,g_n\\}$ be a non-abelian finite simple group of finite order $n$. For a subset $S \\subseteq G$ and integer $k \\in \\mathbb{N}$, define $S^{(k)} := \\{a^k \\, | \\, a \\in S\\}$. Consider the set \r\n\\[\r\nP_G := \\{g_{\\pi(1)} \\cdots g_{\\pi (n)} \\, | \\, \\pi \\in \\Pi_n\\},\r\n\\]\r\nwhere $\\Pi_n$ is the set of all permutations of $\\{1,\\ldots,n\\}$. Show that there exists an integer $m$ such that $P_G^{(m)} = G$.\r\n\r\n--Vesselin",
  "Solution_1": "Actually, for $G$ simple and non-commutative, we have that $P_G = G$, but I guess that would be very difficult to prove (I don't know if an elementary proof is known). So, I invite you to try to prove the weaker version above to which I am sure that there is a very short and elementary (yet beautiful) solution.\r\n\r\n--Vesselin"
}
{
  "Tag": [
    "analytic geometry",
    "geometry",
    "geometric transformation",
    "reflection"
  ],
  "Problem": "AB with endpoints A($ \\minus{}4,2$) and B(5,5) is reflected over the $ y$-axis. Then, its image is reflected over the $ x$-axis. What is the product of the coordinates of the midpoint of the final image?",
  "Solution_1": "current midpoint: (0.5,3.5)\r\nreflect over y then x = (a,b) => (-a,-b)\r\n\r\nProduct: [b]1.75[/b]",
  "Solution_2": "[quote=GameBot]AB with endpoints A($ \\minus{}4,2$) and B(5,5) is reflected over the $ y$-axis. Then, its image is reflected over the $ x$-axis. What is the product of the coordinates of the midpoint of the final image?[/quote]\n\num in ftw it says $A=(-1,2)$\n\n[quote=james4l]current midpoint: (0.5,3.5)\nreflect over y then x = (a,b) => (-a,-b)\n\nProduct: [b]1.75[/b][/quote]\n\nftw says the ans is 7"
}
{
  "Tag": [
    "probability",
    "expected value"
  ],
  "Problem": "Kesha caught 40 fish, marked them, and returned them to the pond.\nLater, she randomly netted 60 fish and found 4 marked fish in this\ngroup. What is the expected number of fish in the pond?",
  "Solution_1": "If out of 60 you find 4 marked fish, out of n you find all 40 marked fish.\r\n\r\nThis is a simple proportions problem, for which n=600 (answer)"
}
{
  "Tag": [
    "geometry"
  ],
  "Problem": "you have triangle ABC. the three medians are AX, BY, and CZ, where X,Y,Z, are on BC, AC, AB respectively\r\nthe medians intersect at point G. if AG= sqrt(5), BG= sqrt(3), \r\nCG= sqrt(2), then find the area of triangle ABC",
  "Solution_1": "[quote=\"davidlizeng\"]you have triangle ABC. the three medians are AX, BY, and CZ, where X,Y,Z, are on BC, AC, AB respectively\nthe medians intersect at point G. if AG= sqrt(5), BG= sqrt(3), \nCG= sqrt(2), then find the area of triangle ABC[/quote]\r\n[hide=\"interesting\"]Set the sides to be x, y, and z. We now know can use the formula for the length of a median to get $\\sqrt{\\frac{2a^{2}+2b^{2}-c^{2}}{2}}=1.5\\sqrt{3}$ and some other equations. Squaring our equations gives us the equations:\n$2a^{2}+2b^{2}-c^{2}=27$\n$2a^{2}+2c^{2}-b^{2}=45$\n$2b^{2}+2c^{2}-a^{2}=18$\nAdding them, we have\n$5a^{2}+5b^{2}+5c^{2}=90$\n$2a^{2}+2b^{2}+2c^{2}=36$\nNow I'm getting that $b=i\\sqrt{3}$ uh oh.[/hide]\r\nCan someone point out my mistake? I think I have the right method...",
  "Solution_2": "I am not quite sure if this is correct, but here is an idea. \r\n[hide] Well, I will use the same exact method to do this as bpms used.  Say the side opposite point A has a length of a, the side opposite point B has a length of b, and the side opposite point C has a length of c.  We have that the median to the side with length a is $\\sqrt{\\frac{2c^{2}+2b^{2}-a^{2}}{2}}=1.5(\\sqrt{5})\\Rightarrow 2c^{2}+2b^{2}-a^{2}=22.5$ (2 is also under the squareroot, bpms, this is where you made the mistake).  Therefore, we can just take all of bpm's numbers and divide them by 2.  We have: \n1. $2a^{2}+2c^{2}-b^{2}=13.5$\n2. $2a^{2}+2b^{2}-c^{2}=9$\n3. $2c^{2}+2b^{2}-a^{2}=22.5$. \nNow, when you add these all up, you have that 3a^2+3b^2+3c^2=45 (You added the variables wrong here, bpms).  We have that $a^{2}+b^{2}+c^{2}=15$.  Adding this to 1,2, and 3 respectively and dividing by 3, we get:\n4. $a^{2}+c^{2}=9.5$\n5. $a^{2}+b^{2}=8$\n6. $c^{2}+b^{2}=12.5$\nFrom here, we can solve for a, b, and c.  Then, we can use Hero's Formula to get the area.[/hide]",
  "Solution_3": "Where did you get all the square roots and stuff? I don't quite follow.",
  "Solution_4": "Oh, sorry about that; I didn't make it quite clear.  The length of the median of a triangle to the side with length a (the sides if the triange have lengths are a, b, and c) is going to be $\\sqrt{\\frac{2b^{2}+2c^{2}-a^{2}}{2}}$."
}
{
  "Tag": [
    "number theory",
    "prime numbers"
  ],
  "Problem": "Nine different two-digit numbers can be formed with the\ndigits 1, 3 and 7. How many of these numbers are prime?",
  "Solution_1": "The 9 numbers that can be formed are \r\n11,13,17,31,33,37,71,73,77.\r\nThe 2 not prime numbers are 33 and 77.\r\nThe rest are prime-the answer is 7.",
  "Solution_2": "You don't have to really list them out - you can just see that only numbers that aren't prime are 33 and 77, so our answer is $ 9 \\minus{} 2 \\equal{} 7$."
}
{
  "Tag": [
    "AMC",
    "AIME"
  ],
  "Problem": "Anyone knows the question for #10?\r\nI also find out that E, A, D are colinear, but i got a different answer.",
  "Solution_1": "http://www.artofproblemsolving.com/Forum/viewtopic.php?t=195725"
}
{
  "Tag": [],
  "Problem": "\u0395\u03af\u03bd\u03b1\u03b9 \u03b7 \u03c0\u03c1\u03ce\u03c4\u03b7 (\u03ba\u03b1\u03b9 \u03c4\u03b5\u03bb\u03b5\u03c5\u03c4\u03b1\u03af\u03b1---\u03b4\u03b5\u03bd \u03b5\u03af\u03bc\u03b1\u03b9 \u03ba\u03b1\u03b8\u03cc\u03bb\u03bf\u03c5 fan \u03c4\u03bf\u03c5 \u03c3\u03c5\u03b3\u03ba\u03b5\u03ba\u03c1\u03b9\u03bc\u03ad\u03bd\u03bf\u03c5 \u03ba\u03bb\u03ac\u03b4\u03bf\u03c5) \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03c0\u03bf\u03c5 \u03b2\u03ac\u03b6\u03c9 \u03c3\u03b5 abstract algebra. \u039c\u03bf\u03c5 \u03c4\u03b7\u03bd \u03b5\u03af\u03c7\u03b5 \u03b4\u03b5\u03af\u03be\u03b5\u03b9 \u03bf Anto \u03c0\u03c1\u03b9\u03bd \u03b1\u03c1\u03ba\u03b5\u03c4\u03cc \u03ba\u03b1\u03b9\u03c1\u03cc, \u03ba\u03b1\u03b9 \u03bd\u03bf\u03bc\u03af\u03b6\u03c9 \u03cc\u03c4\u03b9 \u03b1\u03b4\u03b9\u03ba\u03ce \u03c4\u03b7\u03bd \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03ba\u03c1\u03b1\u03c4\u03ce\u03bd\u03c4\u03b1\u03c2 \u03c4\u03b7\u03bd \u03b3\u03b9\u03b1 \u03c4\u03bf\u03bd \u03b5\u03b1\u03c5\u03c4\u03cc \u03bc\u03bf\u03c5, \u03ba\u03b1\u03b8\u03ce\u03c2 \u03b4\u03b5\u03bd \u03b5\u03ba\u03c4\u03b9\u03bc\u03ce \u03c4\u03b7 \u03b8\u03b5\u03c9\u03c1\u03af\u03b1 \u03bf\u03bc\u03ac\u03b4\u03c9\u03bd \u03cc\u03c3\u03bf (\u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03b9 \u03bb\u03ad\u03bd\u03b5 \u03cc\u03c4\u03b9) \u03b8\u03b1 \u03ad\u03c0\u03c1\u03b5\u03c0\u03b5. \u03a0\u03b9\u03c3\u03c4\u03b5\u03cd\u03c9 \u03cc\u03c4\u03b9 \u03ad\u03c7\u03c9 \u03bb\u03cd\u03c3\u03b7, \u03b1\u03bb\u03bb\u03ac \u03b8\u03b5\u03c9\u03c1\u03ce \u03cc\u03c4\u03b9 \u03ae\u03c4\u03b1\u03bd \u03ba\u03ac\u03c0\u03c9\u03c2 \u03c4\u03c5\u03c7\u03b1\u03af\u03b1, \u03bf\u03c0\u03cc\u03c4\u03b5 \u03b5\u03c5\u03b5\u03bb\u03c0\u03b9\u03c3\u03c4\u03ce \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03c2 \u03bd\u03b1 \u03b2\u03c1\u03b5\u03b9 \u03bc\u03b9\u03b1 \u03c0\u03bf\u03bb\u03cd \u03c0\u03b9\u03bf \u03b1\u03c0\u03bb\u03ae/elegant/common-sense \u03bb\u03cd\u03c3\u03b7:\r\n\r\n\u0391\u03bd \u03bf\u03b9 $ G$, $ H$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf\u03bc\u03ac\u03b4\u03b5\u03c2, \u03c4\u03ad\u03c4\u03bf\u03b9\u03b5\u03c2 \u03ce\u03c3\u03c4\u03b5:\r\n- \u0397 $ G$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03b9\u03c3\u03cc\u03bc\u03bf\u03c1\u03c6\u03b7 \u03bc\u03b9\u03b1\u03c2 \u03c5\u03c0\u03bf\u03bf\u03bc\u03ac\u03b4\u03b1\u03c2 $ H'$ \u03c4\u03b7\u03c2 $ H$.\r\n- \u0397 $ H$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03b9\u03c3\u03cc\u03bc\u03bf\u03c1\u03c6\u03b7 \u03bc\u03b9\u03b1\u03c2 \u03c5\u03c0\u03bf\u03bf\u03bc\u03ac\u03b4\u03b1\u03c2 $ G'$ \u03c4\u03b7\u03c2 $ G$.\r\n\u039c\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bd\u03b1 \u03ba\u03b1\u03c4\u03b1\u03bb\u03ae\u03be\u03bf\u03c5\u03bc\u03b5 \u03c3\u03c4\u03bf \u03cc\u03c4\u03b9 \u03bf\u03b9 $ G$, $ H$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03b9\u03c3\u03cc\u03bc\u03bf\u03c1\u03c6\u03b5\u03c2;\r\n\r\n(\u0393\u03b9\u03b1 \u03c4\u03bf\u03c5\u03c2 \u03b1\u03b3\u03b3\u03bb\u03cc\u03c6\u03c9\u03bd\u03bf\u03c5\u03c2 \u03b1\u03bd\u03b1\u03b3\u03bd\u03ce\u03c3\u03c4\u03b5\u03c2 \u03bc\u03b1\u03c2, if any: if $ G, H$ are groups such that $ G \\cong H' \\leq H$ and $ H \\cong G' \\leq G$, then does it necessarily follow that $ G \\cong H$?)",
  "Solution_1": "[quote=\"cmad\"]\u0391\u03bd \u03bf\u03b9 $ G$, $ H$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf\u03bc\u03ac\u03b4\u03b5\u03c2, \u03c4\u03ad\u03c4\u03bf\u03b9\u03b5\u03c2 \u03ce\u03c3\u03c4\u03b5:\n- \u0397 $ G$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03b9\u03c3\u03cc\u03bc\u03bf\u03c1\u03c6\u03b7 \u03bc\u03b9\u03b1\u03c2 \u03c5\u03c0\u03bf\u03bf\u03bc\u03ac\u03b4\u03b1\u03c2 $ H'$ \u03c4\u03b7\u03c2 $ H$.\n- \u0397 $ H$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03b9\u03c3\u03cc\u03bc\u03bf\u03c1\u03c6\u03b7 \u03bc\u03b9\u03b1\u03c2 \u03c5\u03c0\u03bf\u03bf\u03bc\u03ac\u03b4\u03b1\u03c2 $ G'$ \u03c4\u03b7\u03c2 $ G$.\n\u039c\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bd\u03b1 \u03ba\u03b1\u03c4\u03b1\u03bb\u03ae\u03be\u03bf\u03c5\u03bc\u03b5 \u03c3\u03c4\u03bf \u03cc\u03c4\u03b9 \u03bf\u03b9 $ G$, $ H$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03b9\u03c3\u03cc\u03bc\u03bf\u03c1\u03c6\u03b5\u03c2;[/quote]\r\n\r\nLet me see if I get this straight (\u03c0\u03bf\u03c5 \u03bc\u03b5 \u03c4\u03b9\u03c2 \u03b1\u03c3\u03ba\u03ae\u03c3\u03b5\u03b9\u03c2 \u03c4\u03bf\u03c5 Anto, I never do ;) )... \u03b9\u03c3\u03bf\u03b4\u03cd\u03bd\u03b1\u03bc\u03b1 \u03b8\u03ad\u03bb\u03bf\u03c5\u03bc\u03b5 \u03bd\u03b1 \u03b4\u03bf\u03cd\u03bc\u03b5 \u03b1\u03bd \u03b7 \u03c3\u03c7\u03ad\u03c3\u03b7 $ K\\leq H\\leq G$ \u03cc\u03c0\u03bf\u03c5 $ G,K$ \u03b9\u03c3\u03cc\u03bc\u03bf\u03c1\u03c6\u03b5\u03c2 \u03c3\u03c5\u03bd\u03b5\u03c0\u03ac\u03b3\u03b5\u03c4\u03b1\u03b9 \u03cc\u03c4\u03b9 $ H\\cong G$, \u03c3\u03c9\u03c3\u03c4\u03ac?\r\n\r\nCheerio,\r\n\r\nDurandal 1707",
  "Solution_2": "\u039d\u03b1\u03b9 \u03bc\u03c0\u03bf\u03c1\u03b5\u03af\u03c2 \u03bd\u03b1 \u03b4\u03b5\u03af\u03be\u03b5\u03b9\u03c2 \u03ba\u03b1\u03b9 \u03b5\u03ba\u03b5\u03af\u03bd\u03bf, \u03b9\u03c3\u03bf\u03b4\u03cd\u03bd\u03b1\u03bc\u03b1.\r\n\r\n\u03a5\u0393: Oops, \u03c4\u03b5\u03bb\u03b9\u03ba\u03ac \u03b2\u03c1\u03ae\u03ba\u03b1 \u03bc\u03b9\u03b1 \u03b1\u03c1\u03ba\u03b5\u03c4\u03ac \u03b1\u03c0\u03bb\u03bf\u03cd\u03c3\u03c4\u03b5\u03c1\u03b7 \u03bb\u03cd\u03c3\u03b7... \u038c\u03c0\u03c9\u03c2 \u03ba\u03b1\u03c4\u03b1\u03bb\u03b1\u03b2\u03b1\u03af\u03bd\u03b5\u03b9\u03c2, \u03b4\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03cc\u03c3\u03bf \u03b4\u03cd\u03c3\u03ba\u03bf\u03bb\u03b7 \u03cc\u03c3\u03bf \u03bf\u03b9 \u03c5\u03c0\u03cc\u03bb\u03bf\u03b9\u03c0\u03b5\u03c2 \u03b1\u03c3\u03ba\u03ae\u03c3\u03b5\u03b9\u03c2 \u03c4\u03bf\u03c5 (\u03b1\u03bb\u03bb\u03b9\u03ce\u03c2, to be honest, \u03b5\u03b9\u03b4\u03b9\u03ba\u03ac \u03b5\u03c0\u03b5\u03b9\u03b4\u03ae \u03b5\u03af\u03bd\u03b1\u03b9 \u03ac\u03bb\u03b3\u03b5\u03b2\u03c1\u03b1, \u03b4\u03b5 \u03b8\u03b1 \u03bc\u03bf\u03c5 \u03ad\u03b2\u03b3\u03b1\u03b9\u03bd\u03b5 \u03bc\u03b5 \u03c4\u03af\u03c0\u03bf\u03c4\u03b1 :oops:)",
  "Solution_3": "\u039d\u03bf\u03bc\u03af\u03b6\u03c9 \u03c0\u03c9\u03c2 \u03c4\u03ad\u03c4\u03bf\u03b9\u03b5\u03c2 \u03b1\u03c3\u03ba\u03ae\u03c3\u03b5\u03b9\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c1\u03ba\u03b5\u03c4\u03ac \u03c9\u03c1\u03b1\u03af\u03b5\u03c2. \u03a3\u03c4\u03b1 \u03c0\u03b5\u03c1\u03b9\u03c3\u03c3\u03cc\u03c4\u03b5\u03c1\u03b1 \u03c0\u03c1\u03bf\u03c0\u03c4\u03c5\u03c7\u03b9\u03b1\u03ba\u03ac \u03bc\u03b1\u03b8\u03ae\u03bc\u03b1\u03c4\u03b1 \u03bc\u03b1\u03b8\u03b1\u03af\u03bd\u03bf\u03c5\u03bc\u03b5 \u03b8\u03b5\u03c9\u03c1\u03af\u03b1 \u03bf\u03bc\u03ac\u03b4\u03c9\u03bd \u03b1\u03bb\u03bb\u03ac \u03b1\u03c3\u03c7\u03bf\u03bb\u03bf\u03cd\u03bc\u03b1\u03c3\u03c4\u03b5 \u03c3\u03c5\u03bd\u03ae\u03b8\u03c9\u03c2 \u03bc\u03cc\u03bd\u03bf \u03bc\u03b5 \u03c0\u03b5\u03c0\u03b5\u03c1\u03b1\u03c3\u03bc\u03ad\u03bd\u03b5\u03c2 \u03bf\u03bc\u03ac\u03b4\u03b5\u03c2 (\u03ac\u03bd\u03c4\u03b5 \u03ba\u03b1\u03b9 \u03bc\u03b5 \u03c4\u03bf $ \\mathbb{Z}$). \u03a3\u03c5\u03bc\u03b2\u03b1\u03af\u03bd\u03bf\u03c5\u03bd \u03cc\u03bc\u03c9\u03c2 \u03b1\u03c1\u03ba\u03b5\u03c4\u03ac \u03c0\u03b5\u03c1\u03af\u03b5\u03c1\u03b3\u03b1 \u03c3\u03c4\u03b9\u03c2 \u03bc\u03b7 \u03c0\u03b5\u03c0\u03b5\u03c1\u03b1\u03c3\u03bc\u03ad\u03bd\u03b5\u03c2 \u03bf\u03bc\u03ac\u03b4\u03b5\u03c2 \u03c0\u03bf\u03c5 \u03b7 \u03b4\u03b9\u03b1\u03af\u03c3\u03b8\u03b7\u03c3\u03ae \u03bc\u03b1\u03c2 (\u03bc\u03b5 \u03b2\u03ac\u03c3\u03b7 \u03c4\u03bf \u03c4\u03b9 \u03b3\u03bd\u03c9\u03c1\u03af\u03b6\u03bf\u03c5\u03bc\u03b5 \u03b1\u03c0\u03cc \u03c4\u03b9\u03c2 \u03c0\u03b5\u03c0\u03b5\u03c1\u03b1\u03c3\u03bc\u03ad\u03bd\u03b5\u03c2 \u03bf\u03bc\u03ac\u03b4\u03b5\u03c2) \u03bc\u03b1\u03c2 \u03bb\u03ad\u03b5\u03b9 \u03c0\u03c9\u03c2 \u03b4\u03b5\u03bd \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03c3\u03c5\u03bc\u03b2\u03b1\u03af\u03bd\u03bf\u03c5\u03bd. \r\n\r\n[hide]\n\u039a\u03bf\u03b9\u03c4\u03ac\u03be\u03c4\u03b5 \u03c3\u03c4\u03b1 free groups.\n[/hide]"
}
{
  "Tag": [
    "inequalities",
    "calculus",
    "integration",
    "derivative",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Let $ f$ differentiable, $ f(0) \\equal{} 0$ and $ 0 < f'(x)\\leq1$.\r\n\r\nProve that for all $ x\\geq 0$\r\n\r\n$ \\int_0^xf^3 \\leq (\\int_0^x f)^2$",
  "Solution_1": "[hide=\"Hint\"]Show that $ g(x) \\equal{} 2\\int_0^x f \\minus{} f^2(x)$ and $ h(x) \\equal{} (\\int_0^xf)^2 \\minus{} \\int_0^x f^3$ have non-negative derivatives with $ g(0) \\equal{} h(0) \\equal{} 0$.[/hide]"
}
{
  "Tag": [
    "conics",
    "parameterization"
  ],
  "Problem": "The equation of a circle which passes through the point $ (1,2)$, bisects the circumference of the circle $ x^2\\plus{}y^2\\equal{}9$ and cuts orthogonally the circle $ x^2\\plus{}y^2\\minus{}2x\\plus{}8y\\minus{}7\\equal{}0$ is ......\r\n\r\nans:$ x^2\\plus{}y^2\\plus{}8x\\minus{}2y\\minus{}9\\equal{}0$",
  "Solution_1": "Errr...Do you want the method?? Assume general equation of circle. Since it bisects the C1, the chord of contact passes through the centre of circle (diameter).\r\n\r\n2. Use the equation for orthogonal cutting.\r\n\r\n3. Substitute (1,2)\r\n\r\n3 Equations, 3 unknowns, solve!~",
  "Solution_2": "sorry shreyas i have got the ans now. :huh: ...but i did in a different way :) \r\n\r\nI assumed the circle as $ S \\plus{} \\lambda L \\equal{} 0$ :)",
  "Solution_3": "[quote=\"anandghegde\"]sorry shreyas i have got the ans now. :huh: ...but i did in a different way :) \n\nI assumed the circle as $ S \\plus{} \\lambda L \\equal{} 0$ :)[/quote]\r\nAtleast someone got my trick at last. :D",
  "Solution_4": "yeah chemrock. :) \r\n\r\nthanks for that idea :D",
  "Solution_5": "[quote=\"anandghegde\"]sorry shreyas i have got the ans now. :huh: ...but i did in a different way :) \n\nI assumed the circle as $ S \\plus{} \\lambda L \\equal{} 0$ :)[/quote]\r\n\r\n\r\nThis method of solving problems is very powerful and gives you answers fast.\r\nEspecially say when you want to find the conic $ \\Gamma$ that circumscibes the triangle formed by some three lines, say $ L_{i}, i\\equal{}1,2,3.$\r\nA simple method would be to assume it to be of the form\r\n$ \\Gamma \\equal{} \\lambda L_{1}L_{2} \\plus{} \\mu L_{2}L_{3} \\plus{} L_{3}L_{1}$.\r\nThen use constraints on the kinda conic you want $ \\Gamma$ to be to solve for the parameters, \r\nfor a circle, tha cross term must vanish and the square terms coeff must be equal and stuff like that."
}
{
  "Tag": [],
  "Problem": "Rectangulaer cards, 2 inches by 3 inches are cut froma  rectangular shett 2 feet by 3 feet.  What is the greatest number of cards that can be cutr from the sheet?",
  "Solution_1": "[hide] either 12 * 12= 144 or 864/6=144 [/hide]",
  "Solution_2": "[hide]There are 12 inches in a foot, so we have 12 * 12 or 144 rectangles. [/hide]",
  "Solution_3": "3f=36\r\n2f=24\r\n36*24=864\r\n2*3=6\r\n864\\6=144"
}
{
  "Tag": [
    "projective geometry",
    "geometry proposed",
    "geometry"
  ],
  "Problem": "Let ABCD be a convex quadrilateral (AC,BD are two diagonals).Let be given points E,F on AC,BD respectively.AF $\\cap BE=K,DE\\cap CF=P,AD\\cap BC=S$.prove that:S,K,P are colinear",
  "Solution_1": "Are you sure about the question?\r\nI think you mean ${P}= DE \\cap CF$.\r\n\r\nIf you do: it's literally Pappus-Pascal and a very transparent question.\r\n$B,F,D$ are collinear, $A,E,C$ are collinear.\r\nTherefore, $S,K,P$ are collinear.\r\n\r\nIf you don't: Please post the correct problem, I'm curious."
}
{
  "Tag": [
    "geometry unsolved",
    "geometry"
  ],
  "Problem": "$ \\mbox{Given a convex quadrilateral } ABCD. \\mbox{ On the segments } AB,BC,CD,DA \\mbox{ are chosen points }M,N,P,Q$ $ \\mbox{ , respectively, such that } AQ\\equal{}DP\\equal{}CN\\equal{}BM.$\r\n$ \\mbox{Show that if } MNPQ \\mbox{ is a square, then } ABCD \\mbox{ is also a square. }$  :)",
  "Solution_1": "easily proved $ AQM\\equal{}DQP \\Longrightarrow AM\\equal{}DQ ,AQ\\equal{}BM \\Longrightarrow AB\\equal{}AD$\r\n<AQM+<AMQ=<AQM+DQP=90 thus <A=90\r\nsame  as this ...ABCD is square",
  "Solution_2": "How can you prove $ \\Delta AQM\\equal{}\\Delta DQP$ ?",
  "Solution_3": "<AQM+<DQP=90 thus sin(AQM)=cos(DQP)\r\n$ \\frac{AM}{QM}\\equal{}\\frac{DQ}{PQ}$ thus AM=DQ",
  "Solution_4": "It's not right because if $ \\angle QDP<90^o$, we can construct $ D'$ on $ AD$ such that $ PD'\\equal{}PD$ ...",
  "Solution_5": "Anyone?  :)"
}
{
  "Tag": [
    "AMC"
  ],
  "Problem": "This is a fun one.\r\n\r\nSource: College of Charleston Math Meet, 2003 Lv. 2\r\n\r\nThe 2003 inhabitants of an island are divided into two groups: the truth tellers and the liars.  The former group always tells the truth, while the latter one always lies.  Each person is either a soccer player, a singer, or a fisherman (but not both or all 3).  We ask each inhabitant these three questions:\r\n1) are you a soccer player?\r\n2) are you a singer?\r\n3) are you a fisherman?\r\n\r\n1000 people answer \"yes\" to the first question.  700 answer \"yes\" to the second one, and 500 answer \"yes\" to the third one.  How many liars are there on the island?",
  "Solution_1": "spoiler\n\n[hide]\n\nEvery truth-teller says yes to one of the questions and no to two of the questions (he answers 'yes' to whichever of the three is his job and 'no' to the other ones.) Every liar answers yes to two questions and no to one ('yes' to whichever two jobs he doesn't do and no to the job that he actually has.) Straightforward algebra from there. [/hide]",
  "Solution_2": "Well, its' not THAT straightforward. THere are still several cases to consider, such as which ones the liars say yes to. BUt, still, it is pretty easy...",
  "Solution_3": "Does it matter which jobs the liars choose? \r\nI was just assuming there were 2003 real jobs, 2200 claimed jobs, and each liar answered twice, which would make 197 liars?"
}
{
  "Tag": [
    "function",
    "algebra",
    "polynomial",
    "domain",
    "ratio",
    "complex numbers",
    "complex analysis"
  ],
  "Problem": "let $f(z)=\\frac{p(z)}{q(z)}$ where $p(z)$ and $q(z)$ are polynomials, all 3 are $\\mathbb{C} \\rightarrow \\mathbb{C}$ functions, prove either prove or disprove the existence a function $f(z)$ such that $f(z)=\\overline{z}$ for all $z$ in the domain of $f(z)$",
  "Solution_1": "More simply:\r\n\r\nProve or disprove that the complex conjugate function can be expressed as a ratio of two complex polynomials.\r\n\r\n[hide=\"A first step hint\"]Consider the roots of the denominator[/hide]",
  "Solution_2": "$f(z)=\\overline{z}$\r\n\r\nin never analytcic in the complex plane.",
  "Solution_3": "You need \\overline, not \\overbar.  Anyhow, there are PreOlympiad methods (i.e. not using complex analysis) that can solve this question.",
  "Solution_4": "Just thought i was a nice way of doing it! Ill have a look for a more accessible proof"
}
{
  "Tag": [
    "linear algebra",
    "matrix",
    "induction",
    "algebra",
    "binomial theorem"
  ],
  "Problem": "The matrix $ M$ is given by $ M\\equal{} \\begin {bmatrix} 1 & 2 \\\\ 0 & 1\\end {bmatrix}$.\r\n\r\n(i) Find $ M^{2}$ and $ M^{3}$.\r\n\r\n(ii) Hence suggest a suitable form for the matrix $ M^{n}$.\r\n\r\n(iii) Use induction to prove that your answer to part (ii) is correct.\r\n\r\n(iv) Descibe fully the single transformation represented by $ M^{10}$.",
  "Solution_1": "In general, if we can write $ M\\equal{}I\\plus{}N,$ where $ N^2\\equal{}0,$ we can just use the binomial theorem for $ M^n$, and then have that binomial expansion simplify enormously.\r\n\r\nThe geometric word I would use to describe $ M^n$ for any $ n$ is \"shear.\" (And the Lay textbook that I use for an elementary course, in one of the author's few moments of levity, has a picture of a \"sheared sheep\" - you'll have to pick up a copy of the book to see it.)",
  "Solution_2": "This will help you.\r\n\r\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=308360[/url]",
  "Solution_3": "Thanks, Kent and kunny."
}
{
  "Tag": [
    "function",
    "combinatorics proposed",
    "combinatorics"
  ],
  "Problem": "prove that $\\sum_{k=0}^{\\infty}\\frac {k^n}{2^{k+1}}$ ,(n is natural )is positive integer.",
  "Solution_1": "Hi!\r\nI use gf.\r\nTake f[n](x) = your formula. I don't know latex.Sorry.\r\nNow you get Integral(f[n](x)/x) = f[n-1](x) / 2!!!\r\nSo you can get a formula for f[n](x). You have a reccurence.\r\nThen you need only f[n](1).\r\nI think, though I didn't check, that the number is a power of 2 ?",
  "Solution_2": "First of all, I think we have to show that for $\\displaystyle \\sum_{k\\geq 1}\\frac{k^n}{2^k}$, because if the exponent of $2$ is $k+1$ then for $n=0$ we get $\\frac 1{2^2}+\\frac 2{2^3}+..=\\frac 12$, which isn't an integer.\r\n\r\nLet $f(n)=\\displaystyle \\sum_{k\\geq 1}\\frac {k^n}{2^k}$. $f(0)=1\\in Z$. Now let's assume we have shown that $f(k)\\in Z,\\ \\forall k\\in \\overline{0,n}$ and let's try to show it for $f(n+1)$. \r\n\r\nWe have (a very well-known trick) $f(n+1)\\cdot \\frac 12=f(n+1)\\cdot (1-\\frac 12)=\\displaystyle \\sum_{k\\geq 1}\\frac{k^{n+1}-(k-1)^{n+1}}{2^k}$ and by decomposing $k^{n+1}-(k-1)^{n+1}$ we find that $\\frac{f(n+1)}2$ can be written as a linear combination with integer coefficients of $f(k),\\ k\\in \\overline{0,n}$. This shows that $f(n+1)$ is an even integer for $n\\geq 1$. Let's observe that the problem as you had initially posed it is true for $n>0$. When you say \"natural\" I also understand $0$, and that's why I've modified it. However, as you can see, your assertion is also true for $n\\in N^*=N-\\{0\\}$.",
  "Solution_3": "may I'd better to post the prob to combinatorics because the official idea of the problem is pure combinatorics :(",
  "Solution_4": "It is true because we have following representation:\r\n\\[k^n=\\sum_{i=0}^{n-1}(-1)^{n-1-i}C_n^i(1^i+2^i+\\dots+k^i)\\]",
  "Solution_5": "its nice solution is here:\r\nassume that $A_n$ is the number map $f:${$1,2,3,..,n$}$\\rightarrow${$1,...,n$} s.t satifies this condition: if $f^{-1}(i)$ is'nt empty then $f^{-1}(j)$ for $j=1,..,i$ isnt either .so if $a_n=|A_n|$ .and the number of map in $A_n$ s.t exactly$ l $number of $i$ exsits such that $f(i)=1$is ${l\\choose n}.a_{n-l}$so we have :\r\n$a_n=\\sum_{l=1}^{n}{l\\choose n}.a_{n-l}=\\sum {l \\choose n}a_l$we suppose that $a_0=0$and $b_n=\\frac { a_n}{n!}$so we have $b_n=\\sum_{l=0}^{n-1}\\frac {b_l}{(n-l)!}$ now we use generating function .let $f(z)=\\sum_{n=0}^{\\infty}b_n.z^n$ so we have $f(z)=1+\\sum_{n =1}^{\\infty}z^n.(\\sum_{l=0}^{n-1}\\frac {b_l}{(n-i)!})=1+(e^z-1)f(z)$ therefore $f(z)=\\frac {1}{2-e^z}$ so $f(z)=\\sum_{k=0}^{\\infty}\\frac {e^{kz}}{2^{k+1}}=\\sum \\frac {z^n}{n!}(\\sum \\frac {k^n}{2^{k+1}}$ so we are done."
}
{
  "Tag": [
    "ceiling function",
    "pigeonhole principle",
    "modular arithmetic",
    "search",
    "combinatorics proposed",
    "combinatorics"
  ],
  "Problem": "Let $A$ be the largest subset of $\\{1,\\dots,n\\}$ such that for each $x\\in A$, $x$ divides at most one other element in $A$. Prove that \\[\\frac{2n}3\\leq |A|\\leq \\left\\lceil \\frac{3n}4\\right\\rceil. \\]",
  "Solution_1": "Iran TST? I'm happy! I really like iranian problems!\r\n\r\nFirst a few observation: let us divide A in the set $A_{1}$, the set of numbers that divide exactly another element of A, and $A_{2}$, the set of numbers that divide no other element of $A_{2}$. By definition, $A_{2}$ is an antichain. But it's obvious that also $A_{1}$ is an antichain. Therefore, A is the disjoint union of two antichains.\r\nWhat is the maximum size of an antichain? $\\frac{1}{2}n$, because if you take the greatest odd divisor of each element, these must be distinct, but there are only $\\frac{1}{2}n$ odd numbers $\\le n$. An example of a maximal antichain is the set $\\frac{1}{2}n+1 ,\\frac{1}{2}n+2 \\ldots, n-1,n$.\r\n\r\nNow let's start solving the problem. Suppose that $|A| > \\frac{3n}4$. Take the greatest odd divisor of each element of A. There are no three elements with the same greatest odd divisor, and there are only $\\frac{n}{2}$ odd numbers, then, by the pigeonhole principle, there are more than $\\frac{n}{4}$ couples of elements of A with the same greatest odd divisor. Among their greatest odd divisors, take d, the greatest. Since in $[1,\\frac{n}{2}]$ there are at most $\\frac{1}{4}n$ odd numbers, it must be $d > \\frac{n}{2}$. But then, there is a pair of elements of A with d as divisor, and the bigger must be $\\ge 2d > n$, contradiction.\r\n\r\nTo prove that $|A| \\ge \\frac{3}{2}n$, consider\r\n$A = \\{ k : \\frac{1}{3}n \\le k \\le \\frac{1}{2}n \\land k \\equiv 1 \\pmod 2\\}\\cup \\{ 2k : \\frac{1}{3}n \\le k \\le \\frac{1}{2}n \\land k \\equiv 1 \\pmod 2\\}$.\r\nThe two sets of which A is union are antichains, the first because if an odd number divides another, the second is at least 3 times the first, the second set because if $2k \\mid 2j$ with both k,j odd, we have $k \\mid j$.",
  "Solution_2": "What about the case $A = \\{1\\}$, $n = 1$? Then we have $|A| = 1 > \\frac{3n}{4}$...",
  "Solution_3": "Excuse me, that must be $\\left\\lceil \\frac{3n}4\\right\\rceil$.\r\nModerators, Could you please edit this topic?",
  "Solution_4": "What's an antichain?  :huh:",
  "Solution_5": "Do you know what a search engine is?\r\n\r\nhttp://mathworld.wolfram.com/Antichain.html\r\n\r\nThe poset in question is the set of integers from 1 to $ n$ ordered by division.",
  "Solution_6": "i think this solution is a bit easier than the one presented above :) :\r\n\r\n\r\nfor the first part it is clear just by taking  $ B={k+1,k+2,......,3k+e}$ where $ n=3k+e$ and $ e = 0,1,2$ because $ |A|\\geq |B|$for the second part,let's take $ n=4k$ and for $ n=4k+1,2,3$ it can be done almost exactly the same way.we can write $ {1,2,....,4k}$ as $ 1,3,.....,4k-1; 2*1,2*3,....2*(2k-1); 4*1,4*3,.....; ..........;2^{v}*1;$ where$ 4k\\geq 2^{v}$and v is the greatest number with that property.so $ {1,2,3,....,4k}$can be partitioned in $ A_{1}={1,2*2,4*1,......,2^{v}*1}+A_{3}{3,2*3,4*3,.......}+........A_{(}2k-1)={2k-1,2*(2k-1)}+C={2k+1,2k+3,.....,4k-1}$.it is clear that from $ A_{(}2j-1)$ we can choose at most 2 numbers,for every $ j=0,1,2,....,k-1$ so we can get at most $ 2k$ numbers from there and we are left with just $ k$ numbers that we can choose ,the numbers from $ C$.so we kave at most $ 3k$ numbers in $ A$. that's all :lol:",
  "Solution_7": "[quote name=\"bodom\" url=\"/community/p882822\"]\ni think this solution is a bit easier than the one presented above :) :\n\n\nfor the first part it is clear just by taking  $ B={k+1,k+2,......,3k+e}$ where $ n=3k+e$ and $ e = 0,1,2$ because $ |A|\\geq |B|$for the second part,let's take $ n=4k$ and for $ n=4k+1,2,3$ it can be done almost exactly the same way.we can write $ {1,2,....,4k}$ as $ 1,3,.....,4k-1; 2*1,2*3,....2*(2k-1); 4*1,4*3,.....; ..........;2^{v}*1;$ where$ 4k\\geq 2^{v}$and v is the greatest number with that property.so $ {1,2,3,....,4k}$can be partitioned in $ A_{1}={1,2*2,4*1,......,2^{v}*1}+A_{3}{3,2*3,4*3,.......}+........A_{(}2k-1)={2k-1,2*(2k-1)}+C={2k+1,2k+3,.....,4k-1}$.it is clear that from $ A_{(}2j-1)$ we can choose at most 2 numbers,for every $ j=0,1,2,....,k-1$ so we can get at most $ 2k$ numbers from there and we are left with just $ k$ numbers that we can choose ,the numbers from $ C$.so we kave at most $ 3k$ numbers in $ A$. that's all :lol:\n[/quote]\n\nHow is this clear for the first part? Still confused with that",
  "Solution_8": "[hide=wrong solution]Note that \n$$ A = \\left\\{ \\left\\lfloor \\frac{n}{3} \\right\\rfloor + 1, \\ldots,n \\right\\}  $$\nworks. Hence we get \n$$|A| \\ge \\frac{2n}{3}$$\nNext assume on the contrary that \n$$ |A| \\ge \\frac{3n}{4} + 1 $$\nLet set \n$$ B = 2A := \\{2a : a \\in A\\} \\qquad , \\qquad C = A \\cap B $$\nSince \n$$ |A|,|B| \\ge \\frac{3n}{4} + 1 \\textcolor{red}{\\implies |C| \\ge \\frac{n}{2} + 2} $$\nThis means one can pick $c_1,c_2 \\in C$ such that \n$$ c_1 \\mid c_2 $$\nBut as $c_1  \\in B$, so we can pick some $a \\in A$ such that \n$$ \\frac{c_1}{a} = 2^l ~~, l \\ge 1$$\nPick the least such $a$ (i.e. $l$ is max possible). We get that $a \\notin B$ (otherwise minimality gets contradicted). Then $a \\in A$ divides both of $c_1,c_2 \\in A$, as desired. $\\blacksquare$[/hide]"
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "geometry unsolved"
  ],
  "Problem": "If $ T$ is Toricelli's point and $ O$ the circumcenter of $ \\triangle ABC$, prove that $ OT^2\\equal{}a^2\\plus{}b^2\\plus{}c^2\\minus{}4S\\sqrt{3}$(usual notations)",
  "Solution_1": "I remember I have read the formula in an articles of a fourteen boy (it seem that he is the member [b]mr.danh[/b]), but I don't have more details. \r\nMaybe, the proof proceeds from $ \\vec{OA}\\equal{}\\vec{OT}\\plus{}\\vec{TA}$"
}
{
  "Tag": [
    "geometry",
    "quadratics",
    "algebra"
  ],
  "Problem": "A circular garden is surrounded by a sidewalk with a uniform width of 25 feet. The total area of the sidewalk equals the total area of the garden. How many feet are in the diameter of the garden? Round your answer to the nearest whole number.",
  "Solution_1": "[quote=\"ch1n353ch3s54a1l\"]A circular garden is surrounded by a sidewalk with a uniform width of 25 feet. The total area of the sidewalk equals the total area of the garden. How many feet are in the diameter of the garden? Round your answer to the nearest whole number.[/quote]\r\n[hide]$(x+25)^2-x^2=x^2$\n$50x+625=x^2$\n$x^2-50x-625=0$\n$x=25+25(sq(2))$\n$60.35...$\n$60$[/hide]",
  "Solution_2": "[quote=\"pgpatel\"][quote=\"ch1n353ch3s54a1l\"]A circular garden is surrounded by a sidewalk with a uniform width of 25 feet. The total area of the sidewalk equals the total area of the garden. How many feet are in the diameter of the garden? Round your answer to the nearest whole number.[/quote]\n[hide]$(x+25)^2-x^2=x^2$\n$50x+625=x^2$\n$x^2-50x-625=0$\n$x=25+25(sq(2))$\n$60.35...$\n$60$[/hide][/quote]\n\n[hide]$25/(sqrt(2)-1)=60.355... 60.355*2=120.71$\n\n[b]121[/b]\n\nI think you want the diameter of the garden, not the radius...I am probably mistaken though...[/hide]",
  "Solution_3": "[quote=\"ehehheehee\"][quote=\"pgpatel\"][quote=\"ch1n353ch3s54a1l\"]A circular garden is surrounded by a sidewalk with a uniform width of 25 feet. The total area of the sidewalk equals the total area of the garden. How many feet are in the diameter of the garden? Round your answer to the nearest whole number.[/quote]\n[hide]$(x+25)^2-x^2=x^2$\n$50x+625=x^2$\n$x^2-50x-625=0$\n$x=25+25(sq(2))$\n$60.35...$\n$60$[/hide][/quote]\n\n[hide]$25/(sqrt(2)-1)=60.355... 60.355*2=120.71$\n\n[b]121[/b]\n\nI think you want the diameter of the garden, not the radius...I am probably mistaken though...[/hide][/quote] no you are right...\r\n :wallbash_red:  :wallbash_red: :wallbash_red: :wallbash_red: :wallbash_red: :wallbash_red: :wallbash_red: :wallbash_red: :wallbash_red: :wallbash_red:",
  "Solution_4": "[quote=\"pgpatel\"][quote=\"ch1n353ch3s54a1l\"]A circular garden is surrounded by a sidewalk with a uniform width of 25 feet. The total area of the sidewalk equals the total area of the garden. How many feet are in the diameter of the garden? Round your answer to the nearest whole number.[/quote]\n[hide]$(x+25)^2-x^2=x^2$\n$50x+625=x^2$\n$x^2-50x-625=0$\n$x=25+25(sq(2))$\n$60.35...$\n$60$[/hide][/quote]\r\n\r\nSorry...I don't really understand your first expression...",
  "Solution_5": "[quote=\"ch1n353ch3s54a1l\"][quote=\"pgpatel\"][quote=\"ch1n353ch3s54a1l\"]A circular garden is surrounded by a sidewalk with a uniform width of 25 feet. The total area of the sidewalk equals the total area of the garden. How many feet are in the diameter of the garden? Round your answer to the nearest whole number.[/quote]\n[hide]$(x+25)^2-x^2=x^2$\n$50x+625=x^2$\n$x^2-50x-625=0$\n$x=25+25(sq(2))$\n$60.35...$\n$60$[/hide][/quote]\n\nSorry...I don't really understand your first expression...[/quote]\r\n\r\nI'll explain it for pgpatel. x is the radius of the garden, so the first equation is the area of the sidewalk on the LHS and the area of the garden on the RHS (with the $\\pi$s already cancelled out.) :D",
  "Solution_6": "[hide]Let $x$ be the radius of the inner circle. You are trying to find $2x$.\n$x^2\\pi = (x+25)^2\\pi - x^2\\pi$ as the radius of the inner circle plus the walkway's length would be $x+25$. That would be radius of the larger circle so you'd use that in finding area. But, you don't want total area so you subtract the area of the inner circle to get the area of the walkway.\nExpand to get $x^2\\pi = x^2\\pi + 50x\\pi + 625\\pi - x^2\\pi$.\nSimplify to get $50x + 625 = x^2$.\nSubtract $x^2$ from both sides to get $-x^2 + 50x + 625 = 0$.\nApply quadratic equation to get $\\frac{-50 \\pm \\sqrt{2500 + 2500}}{-2}$.\nSimplify this to $25 \\pm -25\\sqrt{2}$.\nSince it obviously can't be negative and $+-$ would make negative, reject $+$ and change the double negative to $+$ to get $25+25\\sqrt{2}$.\nNow, just calculate it with a calculator and multiply the result by $2$ and round to the nearest integer to get $\\boxed {121}$.[/hide]",
  "Solution_7": "OHH... For some reason I assumed the sidewalk was a square... :read:  :huh:",
  "Solution_8": "[quote=\"ch1n353ch3s54a1l\"]OHH... For some reason I assumed the sidewalk was a square... :read:  :huh:[/quote]\r\n\r\nBut then a vertice of the square would be a larger distance away than the vertice of the midpoint of a side and the distance wouldn't be uniform. It probably would be solvable if it was a square but you'd need more information to narrow it down to one solution and it'd be harder.",
  "Solution_9": "Well the problem said \"[u]with[/u] a uniform width of 25\" so I thought the 25 was referring to the sidewalk...and I didn't know what it was talking about.",
  "Solution_10": "It is.  Basically just picture two concentric circles, and the larger circle has a radius that is 25 feet larger than the radius of the other circle.  Then the area of the larger circle is twice that of the area of the smaller circle."
}
{
  "Tag": [
    "function",
    "topology"
  ],
  "Problem": "Let X be a compact space, Y be a Hausdoff space and f a continumous mapping from X to Y. If f is bijective, prove that the inverse of f is continuous.",
  "Solution_1": "take a closed set $ F\\subset X$ which is compact then $ F$ is compact, $ f(F)$ is compact in $ Y$ which is Hausdorff, so $ f(F)$ must be closed in $ Y$, i.e $ f^{\\minus{}1}$ is continuous !"
}
{
  "Tag": [
    "geometry",
    "calculus",
    "integration",
    "calculus computations"
  ],
  "Problem": "d:1  2   3  4   5   5   4   3   2  1(in m)\r\nX:1 10 20 40 50 60 70 80 90 100(in m)\r\n\r\n\r\nHere X denotes distance in m of a pond from the starting pt.\r\nd denotes depth of the pond coresponding to \"X\"\r\n\r\nFind the area of such pond :(",
  "Solution_1": "which methods of numerical integration is your class working on?\r\n\r\nsimpsons rule, midpoint rule, trapezoidal rule?",
  "Solution_2": "simpsons rule."
}
{
  "Tag": [
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "Let -positive real numbers,\nShow that: $ a\\plus{}b\\plus{}c\\plus{}\\frac{3}{\\frac{1}{a}\\plus{}\\frac{1}{b}\\plus{}\\frac{1}{c}}>\\equal{}4\\sqrt[3] {abc}.$ thanx :wink:\n\n[color=#FF0000][Mod: Do not double post: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=229033[/url].][/color]",
  "Solution_1": "The inequality is equivalent to\r\n\r\n$ a\\plus{}b\\plus{}c \\plus{} \\frac{3abc}{ab\\plus{}bc\\plus{}ca} \\geq 4 \\sqrt[3]{abc}$.\r\n\r\nMany approach to solve. ABC might be the nicest.",
  "Solution_2": "[quote=\"ely_en\"]Let -positive real numbers,\nShow that: $ a \\plus{} b \\plus{} c \\plus{} \\frac {3}{\\frac {1}{a} \\plus{} \\frac {1}{b} \\plus{} \\frac {1}{c}} \\ge\\ 4\\sqrt [3] {abc}.$ thanx :wink:[/quote]\r\nAssum $ abc \\equal{} 1$. \r\n\r\n$ Ineq \\leftrightarrow a \\plus{} b \\plus{} c \\plus{} \\frac {3}{ab \\plus{} bc \\plus{} ca} \\ge\\ 4$\r\n\r\nFollow AM-GM:\r\n\r\n$ LHS \\ge\\ a \\plus{} b \\plus{} c \\plus{} \\frac {9}{(a \\plus{} b \\plus{} c)^2} \\ge\\ \\frac {a \\plus{} b \\plus{} c}{3} \\plus{} 3 \\ge\\ 4$"
}
{
  "Tag": [
    "inequalities",
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "Prove by combinatorial reasoning,\r\n\r\n\\[ \\sum_{i\\equal{}1}^{n} {n\\choose i}\\leq {\\bigg( \\frac{2^n\\minus{}2}{n\\minus{}1} \\bigg)}^{n\\minus{}1}\\quad \\forall n>1,n\\in \\mathbb{N}\\]",
  "Solution_1": "Did you mean\r\n\\[ \\prod_{i \\equal{} 1}^{n} {n\\choose i}\\leq {\\bigg( \\frac {2^n \\minus{} 2}{n \\minus{} 1} \\bigg)}^{n \\minus{} 1}\r\n\\]\r\n?  If so, it can be checked directly for small $ n$, and proven for $ n$ greater than $ 5$ or so by asymptotic estimates, as the left hand side is substantially smaller for sufficiently large $ n$.",
  "Solution_2": "That is simply AM-GM inequality, which in equivalent form is:\r\n\\[ \\sqrt [n \\minus{} 1]{\\prod_{i \\equal{} 1}^{n \\minus{} 1}{n\\choose i}} \\leq \\frac {\\sum_{i \\equal{} 1}^{n \\minus{} 1} {n\\choose i}}{n \\minus{} 1}\r\n\\]",
  "Solution_3": "Sorry, I meant the product.\r\n\r\nYes, I am aware of the proof through AM-GM Inequality. Can we have a counting argument for it ? or do we have a way to combinatorially interpret AM-GM Inequality ?"
}
{
  "Tag": [
    "function",
    "FTW"
  ],
  "Problem": "1. Are we allowed to program formulas?\r\n2. If I archive a program and then change out my calculators batteries, will my program be deleted?",
  "Solution_1": "1.\r\nwell im pretty sure you can program even though i dont know why because it would give people unfair advantages, but the rules say you can use any calculator that doesnt have a qwerty keyboard, so if we can use programmable calcs then im assuming we can write programs.",
  "Solution_2": "[quote=\"Lawrence Wu\"]1. Are we allowed to program formulas?\n2. If I archive a program and then change out my calculators batteries, will my program be deleted?[/quote]\r\n\r\n1. Yes, as long as you are using a non-QWERTY calculator. (Concise version of micromath's answer).\r\n2. No. But it helps to use TI Connect and backup the programs anyway.",
  "Solution_3": "thx. does anyone know how to program a bases program\r\nbtw, distracted523, according to your \"location\", you live in texas. I'll see you at states!",
  "Solution_4": "base programs? like what?",
  "Solution_5": "[quote=\"Lawrence Wu\"]thx. does anyone know how to program a bases program\nbtw, distracted523, according to your \"location\", you live in texas. I'll see you at states![/quote]\r\nwhoa, dang I'm in your chapter, distracted! Were you in the top 10?\r\n\r\nOntopic: \r\n1. yes you can use programmed calc's, I know someone who always does\r\n2. I think no but I'm not sure..",
  "Solution_6": "Programming calculators are not necessary, and they take more time, in my opinion.\r\n\r\nTwo ways to run a program:\r\n\r\n1. Type the name of the program (note the calculator doesn't implement a qwerty keyboard, so that step takes about 20 seconds)\r\n\r\n2. Go to the program directory and scroll millions of times until you find the program name you want. That takes about 10-30 seconds, depending how many programs you have :D \r\n\r\nI usually program functions that take more than 30 seconds to calculate manually...\r\n\r\nBut I got third  in my state without using any user-defined programs on my calculator.",
  "Solution_7": "Programs give you a bit of advantage in MC.\r\nAlthough it's not really worth the time it takes for you to program it unless you have a lot of free time.\r\nAnyways, for 2, it depends on what calculator you have I guess.",
  "Solution_8": "[quote=\"kthxbai\"][quote=\"Lawrence Wu\"]thx. does anyone know how to program a bases program\nbtw, distracted523, according to your \"location\", you live in texas. I'll see you at states![/quote]\nwhoa, dang I'm in your chapter, distracted! Were you in the top 10?\n[/quote]\r\n\r\nYep! I was the winner!  :lol:\r\n\r\n6th last year at state...so much pressure.",
  "Solution_9": "[quote=\"ZhangPeijin\"]Programs give you a bit of advantage in MC.\nAlthough it's not really worth the time it takes for you to program it unless you have a lot of free time.\nAnyways, for 2, it depends on what calculator you have I guess.[/quote]\r\n\r\nI'm a fast calculator programmer, but I'm not concerned about that part. It's actually finding the program under pressure that I'm concerned about.",
  "Solution_10": "Don't calculators have an archive for all the programs?\r\nUnless you have so many you can't find the one you want.",
  "Solution_11": "[quote=\"dingzhou\"]base programs? like what?[/quote]\r\na program that can convert a number from base 10 to say, base 2\r\nand vice versa",
  "Solution_12": "Hmm...I don't remember seeing many base problems, especially not in calculator rounds.",
  "Solution_13": "[quote=\"ZhangPeijin\"]Don't calculators have an archive for all the programs?\nUnless you have so many you can't find the one you want.[/quote]\r\n\r\nBut unfortunately, I have a TI-89 and it stores programs as \"files\", along with other \"files\" that were added to my calculator.",
  "Solution_14": "[quote=\"Lawrence Wu\"][quote=\"dingzhou\"]base programs? like what?[/quote]\na program that can convert a number from base 10 to say, base 2\nand vice versa[/quote]\r\n\r\n\r\noh...well...its easy enough to just do it by hands, and its unlikely that you need to convert bases in this years target...i don't even remember a question where it told you to convert bases...why dont u try to make ur own?",
  "Solution_15": "1. Yes you're allowed to program formulas.\r\n\r\n2. As long as the programs are archived. NEVER pull the batteries from your calculator. Ever. Only do it if your calculator freezes for over 5 minutes. I also find it helpful to make backups (just use the Recall function to copy and paste programs) and archive+hide them in case my memory does randomly clear.",
  "Solution_16": "[quote=\"distracted523\"][quote=\"kthxbai\"][quote=\"Lawrence Wu\"]thx. does anyone know how to program a bases program\nbtw, distracted523, according to your \"location\", you live in texas. I'll see you at states![/quote]\nwhoa, dang I'm in your chapter, distracted! Were you in the top 10?\n[/quote]\n\nYep! I was the winner!  :lol:\n\n6th last year at state...so much pressure.[/quote]\r\nOH U!\r\n...expect some messages in ftw and pm's :lol:",
  "Solution_17": "[quote=\"kthxbai\"][/quote][quote=\"distracted523\"][quote=\"kthxbai\"][quote=\"Lawrence Wu\"]thx. does anyone know how to program a bases program\nbtw, distracted523, according to your \"location\", you live in texas. I'll see you at states![/quote]\nwhoa, dang I'm in your chapter, distracted! Were you in the top 10?\n[/quote]\n\nYep! I was the winner!  :lol:\n\n6th last year at state...so much pressure.[/quote][quote=\"kthxbai\"]\nOH U!\n...expect some messages in ftw and pm's :lol:[/quote]\r\n\r\nkthxbai, are you in tx too?",
  "Solution_18": "[quote=\"Lawrence Wu\"][quote=\"dingzhou\"]base programs? like what?[/quote]\na program that can convert a number from base 10 to say, base 2\nand vice versa[/quote]\r\n\r\nIs there a remainder function? Or mod?\r\nIt's really easy if there is one. \r\nI'll try to make one\r\n\r\n-EDIT-\r\nHere we go:\r\n-> means Store (button above on) <= means less than or equal\r\n0->B\r\n0->C\r\nDisp \"What number\"\r\nPrompt A\r\nLbl 1\r\nIf A<=1\r\nThen\r\n10^B+C->C\r\nDisp C\r\nStop\r\nEnd\r\nIf (A/2-int(A/2)=.5\r\nThen\r\n10^B+C->C\r\nB+1->B\r\nA-int(A/2)->A\r\nGoto 1\r\nElse\r\nB+1->B\r\nA-int(A/2)->A\r\nGoto 1\r\nEnd"
}
{
  "Tag": [],
  "Problem": "Hello I am a Chinese student ready to receive my undergraduate education in American universities---I don't have much prominent experiences in my resume---even though I did well at school and I am a quality student---I love math---I would like to know what do you think of Math department of Purdue University West Lafayatte, University of Illinois Urban-Champaign and University of California San Diego?\r\nI would like to listen to your comments and informations---\r\nThanks a lot---",
  "Solution_1": "following is written in Chinese, please choose GBK/GB2312 encoding to read.\r\n\r\n\u548c\u4e2d\u56fd\u5927\u5b66\u6bd4\uff0c\u4e00\u822c\u7f8e\u56fd\u5927\u5b66\u90fd\u662f\u5f88\u597d\u7684\u3002\u6240\u4ee5\uff0c\u4e0d\u7ba1\u4f60\u53bb\u54ea\u4e2a\u5b66\u6821\uff0c\u90fd\u4e0d\u4f1a\u5931\u671b\u7684\u3002\u53ea\u8981\u80fd\u88ab\u5f55\u53d6\uff0c\u5176\u4ed6\u7684\u4e8b\u60c5\u6ca1\u6709\u4ec0\u4e48\u597d\u62c5\u5fc3\u7684\u3002\r\n\r\n\u4e0d\u8fc7\u8981\u6ce8\u610f\u8eab\u4f53\u662f\u5426\u9002\u5e94\u957f\u9014\u65c5\u884c\u3001\u4ee5\u53ca\u65e5\u591c\u98a0\u5012(\u5012\u65f6\u5dee)\u7684\u751f\u6d3b\u3002",
  "Solution_2": "All three of those schools have respectable math departments.  (Personally, I think UCSD is pretty good, but consult US News for the \"conventional wisdom.\")  I see little reason to chose one over the others for an undergraduate degree simply because of math.",
  "Solution_3": "Thank you for your advice---I think what you said make sense---"
}
{
  "Tag": [
    "algebra",
    "polynomial"
  ],
  "Problem": "Let the polynomial $ p(x)\\equal{}x^3\\plus{}bx^2\\minus{}(4\\plus{}3b\\plus{}2d)x\\plus{}d$ where the numers $ b,c,d$ are nonzero integers . Prove that $ p(x)$ can't have three integer roots .",
  "Solution_1": "Where's $ c$?",
  "Solution_2": "Counterexample: b=-3, d=3 (has roots 3,1,-1)\r\n\r\nYeah, silouan must have left out a $ c$.",
  "Solution_3": "There is no $ c$ . The one I missed is that $ 3\\plus{}3d\\neq \\minus{}4b$"
}
{
  "Tag": [
    "geometry",
    "quadratics",
    "algebra",
    "polynomial",
    "conics",
    "parabola",
    "analytic geometry"
  ],
  "Problem": "These are stolen from the Lehigh Univ. contest (2002 and 2003). (I don't know how to solve 1 and 4, the rest are just nice problems I came across).\r\n\r\n1. For each point (m,n) in the plane, with m and n integers, draw a circle of radius 1/(2^((absolute value of m)+1))(3^((absolute value of n) +1)). What is the total area enclosed by all these circles?\r\n\r\n2. Write a quadratic polynomial having as its two roots the numbers obtained by increasing each root of x :^2: -2x-5=0 by the reciprocal of the other. \r\n\r\n3. What is the largest number of 0s that can occur at the end of 1^n+ 2^n + 3^n + 4^n for any positive integer n?\r\n\r\n4. Suppose the parabola y= x^2+px+q is situated so that it has two arcs lying between the rays y=x and y=2x, x  :ge: 0. These two arcs are projected onto the x-axis, yielding segments SL and SR, with SR to the right of SL. What is the difference of the lengths (SR-SL)?\r\n\r\n5. In triangle ABC, D is the midpoint of AB, while E lies on BC satisfying BE= 2EC. If angle ADC= angle BAE, then how many degrees are in angle BAC?\r\n\r\n6. Find the smallest possible value of (x^2+y^2)/(x-y) for all real numbers x and y satisfying x>y>0 and xy= 3.",
  "Solution_1": "These are nice questions.  This will be edited to include my solutions, when I get any.",
  "Solution_2": "[quote=\"314159265358979323846\"]\n2. Write a quadratic polynomial having as its two roots the numbers obtained by increasing each root of x :^2: -2x-5=0 by the reciprocal of the other. \n[/quote]\r\n\r\nLet the two roots of x^2 - 2x - 5 = 0 be r and s. Then the roots of the new quadratic are r+1/s and s+1/r. Their sum (with use of Viete's formula) is r+s+1/r+1/s=r+s+(r+s)/rs=2-2/5=8/5, while their product is (r+1/s)(s+1/r)=rs+1/rs+2=-5-1/5+2=-16/5. The new quadratic is thus P(x)=x^2-(8/5)x-16/5.",
  "Solution_3": "Jennifer wrote:. For each point (m,n) in the plane, with m and n integers, draw a circle of radius 1/(2^(|m|+1))(3^(|n| +1)). What is the total area enclosed by all these circles?\n\n2. Write a quadratic polynomial having as its two roots the numbers obtained by increasing each root of x :^2: -2x-5=0 by the reciprocal of the other. \n\n3. What is the largest number of 0s that can occur at the end of 1^n+ 2^n + 3^n + 4^n for any positive integer n?\n\n4. Suppose the parabola y= x^2+px+q is situated so that it has two arcs lying between the rays y=x and y=2x, x  :ge: 0. These two arcs are projected onto the x-axis, yielding segments SL and SR, with SR to the right of SL. What is the difference of the lengths (SR-SL)?\n\n5. In triangle ABC, D is the midpoint of AB, while E lies on BC satisfying BE= 2EC. If angle ADC= angle BAE, then how many degrees are in angle BAC?\n\n6. Find the smallest possible value of (x^2+y^2)/(x-y) for all real numbers x and y satisfying x>y>0 and xy= 3.\n\n\n\n2. [hide]The roots of our new polynomial will have sum (1+1/ab)(a+b)=(1+1/-5)(2)=8/5 and product (a+1/b)(b+1/a)=ab+2+1/ab=(-5+2+1/-5)=-16/5, where a and b are the roots of the original polynomial. Thus one possible new polynomial is 5x:^2:-8x-16. Correct?[/hide]\n\n\n\n4. [hide]The lengths of SL and SR are the distances between the x-coordinates for where this parabola intersects the two lines. y=x:^2:+px+q intersects y=2x at x:^2:+(p-2)x+q=0 while it intersects y=x at x:^2:+(p-1)x+q=0. Let the points of intersection for y=2x be a and b and let the points of intersection with y=x be c and d. Then we are looking for (b-d)-(c-a)=a+b-c-d=(a+b)-(c+d)=(p-2)-(p-1)=-1. Correct?[/hide]\n\n\n\n6. [hide](x:^2:+y:^2:)/(x-y)=((x:^2:-2xy+y:^2:)+2xy)/(x-y)=\n\n(x-y)+2xy/(x-y)=a+6/a, where a is the real number x-y. a+6/a:ge:2:sqrt:(a*6/a)=2:sqrt:6 by AM-GM, which is attainable when x=y+:sqrt:6. Correct?[/hide]\n\n\n\nOdd problems forthcoming.",
  "Solution_4": "But Mystic,\r\n\r\nFor #4, the question is to find not the difference between the x-coordinates, but the difference in the x-coordinates after the arcs are projected onto x-axis (which i think means transforming the curves into lines).",
  "Solution_5": "think about it a little; i think i'm right",
  "Solution_6": "Well, I will wait for someone else to chime in, since I don't know much about projecting stuff (is this related to projective geometry?), but I just saw your statement [hide]The lengths of SL and SR are the distances between the x-coordinates for where this parabola intersects the two lines. [/hide] and noticed how it contradicted with the actual problem.",
  "Solution_7": "http://www.lehigh.edu/~dmd1/test02.pdf  - solutions\r\n\r\nAccording to the website, the solution for number 4 should be (d-b)-(a-c).. whether this is right or not, I don't know... I didn't actually figure out what an x-projection was until this morning...\r\n\r\nThe other two are correct. (It is interesting to note that none of the contestants who took this test got number 6 right...)",
  "Solution_8": "well the only one i see a solution for off the bat is #6... so..\n\n\n\n[hide] (x :^2:+y :^2:)/(x-y) = {(x-y) :^2:+2xy}/(x-y), break it up, you get (x-y) + 6/(x-y)...cuz xy=3... so let z=x-y, then you have to minimize z+6/z. by am-gm, we have (z+6/z)/2 :ge:  :sqrt: (6), multiplying both sides by two and you get z+6/z :ge: 2 :sqrt: (6)....ill update with the eq case in a few min...[/hide]",
  "Solution_9": "[color=cyan]\"Projection\" means dropping perpendiculars at each point.  Thus, the difference in the x-coordinates of the endpoints of an arc is [i]equal[/i] to the difference in the x-coordinates of the projections of the arc.  That is, the process of projection is such that it does not change the value of the x-coordinates at all.[/color]",
  "Solution_10": "Oh, so projection is NOT what this problem is refering to, right?",
  "Solution_11": "[color=cyan]The problem IS refering to projection -- the two arcs are projected onto the x-axis, meaning that you drop a perpendicular from each point of the arcs to the x-axis.  The set of the feet of the perpendiculars is the segment the problem refers to.\n\nA question, however:  why is the letter S used twice there?  I find that rather confusing.[/color]",
  "Solution_12": "[quote=\"314159265358979323846\"]1. For each point (m,n) in the plane, with m and n integers, draw a circle of radius 1/(2^(|m|+1) * 3^(|n| +1)). What is the total area enclosed by all these circles?[/quote]\r\n\r\nWell, lets take this question apart a bit -- fix n.  That is, consider it not to be variable but instead constant for a while.  Now, what we are looking for is the summation :Sigma: :pi:*[1/(2^(|m|+1) * 3^(|n| +1))]:^2:  over all integers m.  The n and the pi parts are constant, so lets pull them out front, to give us\r\n:pi:/3^(2|n| + 2) * :Sigma: 1/(2^(2|m| + 2)\r\nLet us just calculate the value of the sum first, and then multiply it by the other stuff afterwards.\r\n:Sigma: 1/(2^(2|m| + 2) = . . . 1/64 + 1/16 + 1/4 + 1/16 + 1/64 + . . .\r\n1/16 + 1/64 + 1/256 + . . . is a geometric series with first term 1/16 and common difference 1/4.  Thus, its sum is (1/16)/(1 - 1/4) = 1/12.  Thus the total sum here is 1/12 + 1/4 + 1/12 = 5/12.\r\n\r\nSo, going back to the expression we want, we have that it is\r\n:pi:/3^(2|n| + 2) * :Sigma: 1/(2^(2|m| + 2) = :pi:/3^(2|n| + 2) * 5/12\r\n= 5:pi:/12 * 1/3^(2|n| + 2), which we want to sum over all integral values of n.  Ignore the coefficient for a bit again, and just look at\r\n:Sigma: 1/3^(2|n| + 2) = . . . 1/729 + 1/81 + 1/9 + 1/81 + 1/729 + . . .\r\n1/81 + 1/729 + . . . is a geometric sequence with initial term 1/81 and common difference 1/9.  Thus, its sum is \r\n(1/81)/(1 - 1/9) = 1/72.  This makes the total value of our sum \r\n1/72 + 1/9 + 1/72 = 5/36.  Thus the sum of all of the areas is\r\n5:pi:/12 * 5/36 = 25:pi:/432.\r\n\r\nIf you see any arithmetic errors or have questions on the technique, feel free to point them out/ask, or to otherwise comment."
}
{
  "Tag": [
    "modular arithmetic",
    "quadratics",
    "number theory proposed",
    "number theory"
  ],
  "Problem": "Find all integer solutios  to  $\\frac{13}{x^2}+\\frac{1996}{y^2}=\\frac{z}{1997}.$",
  "Solution_1": "We claim that there are no possible solutions.\r\n\r\nRearranging the equation, we get $13\\cdot 1997 y^2z + 1996 \\cdot 1997x^2z = {(xyz)}^2$. Note that $1997$ divides the L.H.S., so since $1997$ is prime and ${(xyz)}^2$ is a perfect square, we have ${1997}^2$ divides R.H.S., so $1997$ must divide at least one of $x,y, z$.\r\n\r\nSuppose $1997 \\nmid z$, then taking $\\mod 1997$ to the above equation, we mst have $1997 \\mid x$ and $1997 \\mid y$. Letting $x=1997x_0, y = 1997y_0$, we have \\[ 13{y_0}^2z + 1996{x_0}^2z = 1997{x_0y_0z}^2 \\] \\[ \\Rightarrow 13{y_0}^2 + 1996{x_0}^2 = 1997{x_0}^2{y_0}^2z \\] \\[ \\Rightarrow 13{y_0}^2 - {x_0}^2 \\equiv 0 \\pmod{1997} \\] \\[ \\Rightarrow 13 \\text{ is a quadratic residue} \\mod 1997 \\] Yet $(\\frac{13}{1997})(\\frac{1997}{13}) = (\\frac{13}{1997})(\\frac{8}{13}) = 1$ and $(\\frac{8}{13}) = -1$ implies $(\\frac{13}{1997}) = -1$. (Contradiction)\r\n\r\nThus $1997\\mid z$. Letting $z=1997z_0$, we have \\[ 13y^2z_0 + 1996x^2z_0 = x^2y^2{z_0}^2 \\] \\[ \\Leftrightarrow z_0(x^2y^2z_0 - (13y^2 + 1996x^2)) = 0 \\] \\[ \\Rightarrow z_0 = 0 \\text{ or } z_0 = \\frac{13y^2+1996x^2}{x^2y^2} \\] Note that $z_0 = 0$ is not possible as $\\frac{13}{x^2} + \\frac{1996}{y^2} > 0 \\forall x,y \\in \\mathbb{R}$ , so we have $13y^2 + 1996x^2 \\equiv 0 \\pmod{x^2y^2}$. Letting $\\gcd (x,y) = d, x = dx_0, y = dy_0$, we have $\\frac{13{y_0}^2 + 1996{x_0}^2}{d^2{x_0}^2{y_0}^2} \\in \\mathbb{Z} \\Rightarrow 13{y_0}^2 + 1996{x_0}^2 \\equiv 0 \\pmod{{x_0}^2{y_0}^2}$. Taking $\\mod {x_0}^2$, we have $13{y_0}^2 \\equiv 0 \\pmod{{x_0}^2} \\Rightarrow 13 \\mid x_0$.\r\nLetting $x_0 = 13x_1$, we have $\\frac{{y_0}^2 + 1996\\cdot 13{x_1}^2}{13d^2{x_1}^2{y_0}^2} \\in \\mathbb{Z}$\r\n$\\Rightarrow {y_0}^2 + 1996\\cdot 13{x_1}^2 \\equiv 0 \\pmod{13}\\Rightarrow y_0 \\equiv 0 \\pmod{13}$. This contradicts the original assumption that $x_0, y_0$ are coprime.\r\n\r\nTherefore there are no solutions."
}
{
  "Tag": [
    "geometry",
    "perimeter"
  ],
  "Problem": "How many pythagorean triangles are there such that their perimeter equals their area?\r\n[hide=\"hint\"]Diophantine[/hide]",
  "Solution_1": "5,12,13 is the only solution, I think.",
  "Solution_2": "[quote=\"Elemennop\"]5,12,13 is the only solution, I think.[/quote]\r\ndid you use trial and error? :P \r\nIf not, please post your solution.",
  "Solution_3": "Denote the sides by $a,b$. Then we have\r\n\r\n\\begin{eqnarray*}\\frac{ab}{2}=a+b+\\sqrt{a^{2}+b^{2}}&\\iff& \\frac{ab}{2}-a-b=\\sqrt{a^{2}+b^{2}}\\\\ &\\implies& \\frac{a^{2}b^{2}}{4}+a^{2}+b^{2}-a^{2}b-ab^{2}+2ab=a^{2}+b^{2}\\\\ &\\iff& \\frac{a^{2}b^{2}}{4}-a^{2}b-ab^{2}+2ab=0\\\\ &\\iff& \\frac{ab}{4}-a-b+2=0\\\\ &\\iff& ab-4a-4b+8=0\\\\ &\\iff& (a-4)(b-4)=8\\end{eqnarray*}\r\n\r\nSince $8$ can be factorized as $1\\cdot 8,2\\cdot 4,4\\cdot 2,8\\cdot 1$, we get $4$ possibilities that reduce to $2$ different triangles: $5,12,13$ and $6,8,10$.",
  "Solution_4": "[quote=\"Farenhajt\"]Denote the sides by $a,b$. Then we have\n\\begin{eqnarray*}\\frac{ab}{2}=a+b+\\sqrt{a^{2}+b^{2}}&\\iff& \\frac{ab}{2}-a-b=\\sqrt{a^{2}+b^{2}}\\\\ &\\implies& \\frac{a^{2}b^{2}}{4}+a^{2}+b^{2}-a^{2}b-ab^{2}+2ab=a^{2}+b^{2}\\\\ &\\iff& \\frac{a^{2}b^{2}}{4}-a^{2}b-ab^{2}+2ab=0\\\\ &\\iff& \\frac{ab}{4}-a-b+2=0\\\\ &\\iff& ab-4a-4b+8=0\\\\ &\\iff& (a-4)(b-4)=8\\end{eqnarray*}\nSince $8$ can be factorized as $1\\cdot 8,2\\cdot 4,4\\cdot 2,8\\cdot 1$, we get $4$ possibilities that reduce to $2$ different triangles: $5,12,13$ and $6,8,10$.[/quote]\r\nthanks Farenhajt. :)",
  "Solution_5": "Jeez, I didn't even consider the 2*4 factorization... I'm getting lazy.",
  "Solution_6": "[quote=\"Elemennop\"]Jeez, I didn't even consider the 2*4 factorization... I'm getting lazy.[/quote]\r\nIt happens. :) \r\nthank you for your contribution.",
  "Solution_7": "This has been asked here  a few times, but remove the condition about \"pythagorean\" ..how many triangles (whose sides are integers - but  they need not be right angle) are there whose perimeter is equal to the area.\r\n\r\n[hide](For example triangle  with the sides like (9,10,17) and others in additon to the two given above.. [/hide]",
  "Solution_8": "Continuing along this topic, find all primitive pythagorean triangles whose perimeters are twice their areas. :)",
  "Solution_9": "[quote=\"Inspired By Nature\"]Continuing along this topic, find all primitive pythagorean triangles whose perimeters are twice their areas. :)[/quote]\r\n\r\n[hide=\"Answer\"]The equation would be $(a-2)(b-2)=2$ and the only solution is $3,4,5$.[/hide]",
  "Solution_10": "thats what I found, thanks Farenhajt. :D",
  "Solution_11": "[quote=\"Gyan\"]This has been asked here  a few times, but remove the condition about \"pythagorean\" ..how many triangles (whose sides are integers - but  they need not be right angle) are there whose perimeter is equal to the area.\n\n[hide](For example triangle  with the sides like (9,10,17) and others in additon to the two given above.. [/hide][/quote]\n\nwell...\n[hide]\nyou get $4s = (s-a)(s-b)(s-c)$,multiply both side by 8 we get:\n$16(a+b+c) = (b+c-a)(a+c-b)(a+b-c)$, and what now... :maybe: \n[/hide]",
  "Solution_12": "[quote=\"pkerichang\"]\n$16(a+b+c) = (b+c-a)(a+c-b)(a+b-c)$, and what now... :maybe: \n [/quote]\r\nSome Hints: \r\n1. The problem (though harder than that of pythagorean triangles) is not that hard ( I think, it can still be in \"intermediate\" or \" pre-olympiad level\") \r\n2. It may be a little easier to work with:\r\n[hide]Iif you let  $x=b+c-a,y=a+c-b,z=a+b-c$ the equation becomes $16(x+y+z)= xyz$ \n(and x,y,z all are >0)     [/hide]\r\n3. There are only a few answers.\r\n\r\nedit: removed a typo.",
  "Solution_13": "Using Gyan's hints so far I can find the following triples:\r\n$(6, 25, 29), (7, 15, 20), (9, 10, 17), \\text{ and }(14, 15, 20)$",
  "Solution_14": "Think, there is a typo, No $(14,15,20)$, all other three  are okay. \r\n[hide=\"Also\"](And if you include  the two already given before ($(6,8,10)$ and $(5,12,13)$ there are no more solutions.) [/hide]",
  "Solution_15": "If there is still interest.. here  are hints which give away the solution.\r\n\r\n[hide]1.  (x, y, z ) are all even and thus the equation can be further simplified.\nto $4(p+q+r)= pqr$ \n[hide=\"because\"]\n(If $x$ is odd, so is $y$ because $a=(x+y)/2$ is integer. .., and $z$ too  is odd because ($(x+z)/2$ is also a  side of the triangle  .. and if all are odd the product  of xyz can't be even) \n[/hide]\n2. WLOG, if you assume $p \\ge q \\ge r$ then $r \\le 3$ (In fact one can see that it is not 3, and thus only possible values  of r are 1 and 2) \n[hide=\"because\"]\n$4(p+p+p) \\ge pqr \\implies qr \\le 12 \\implies r^{2}\\le 12$ [/hide]\n\n3. Once you know r, finding p and q is trivial.  [/hide]\r\n\r\nBTW - One geometric answer makes the above solution trivial is to notice that  in-circle of  all such triangles  have radius =2.\r\nOnce you notice that (and realize one of the angle has to be more than, or equal to $60^{o}$ you can find all the answers .. just as the above logic."
}
{
  "Tag": [
    "Ring Theory",
    "algebra",
    "polynomial",
    "calculus",
    "integration",
    "function",
    "domain"
  ],
  "Problem": "I have two questions about R-Algebras.\r\nFirst I don't think that if R is a UFD, then every finitely generated R-Algebra is a UFD. Is it true? My reasoning is since the R-Algebra, say A, is an extension of A, there could be elements which are not characteristic to UFDs adjoined to R. I mean given a ring R which is not a UFD, there may be a subring which is a UFD. \r\n\r\nAlso is every finitely generated R-Algebra isomorphic over R to a quotient of a polynomial ring in many variables by a prime ideal??\r\n\r\nThanks",
  "Solution_1": "Um, im not sure what your definition of an algebra is, so I'll state mine and you can tell me if we're on the same page: \r\n\r\nA (commutative) $ R$-algebra is a ring $ A$ with a homomorphism $ \\phi: R\\rightarrow A$. Given this definition of an $ R$-algebra certainly not every algebra over a UFD is a UFD: for example, consider $ \\mathbb{Z}[\\sqrt { \\minus{} 5}]$. Then $ 6 \\equal{} (1 \\plus{} \\sqrt { \\minus{} 5})(1 \\minus{} \\sqrt { \\minus{} 5}) \\equal{} 2\\times 3$ so we don't have unique factorization. \r\n\r\nAs for the second, I'm pretty sure that its true.",
  "Solution_2": "My definition is: A finitely generated R algebra is an integral domain A, extending R, with the property : there exists a finite number of elements in A such that there is no proper subring of A containing R and the finite elements.\r\n\r\nHow would the second thing be proven?",
  "Solution_3": "Alright so I didn't know you wanted integrality (thats ok, my counter example to UFD still holds). I actually messed up in my last post cause if you didn't want integrality the ideal doesnt have to be prime so I implicitly assumed it was. my bad. So, basically your definition is the same as mine except you want the homomorphism $ R\\rightarrow A$ to be the inclusion map. The second condition that there is a finite number of elements in $ A$ such that no proper subring of $ A$ containing $ R$ and those elements means that those elements generate $ A$. \r\n\r\nAlright, so suppose our finite set of elements is $ \\{a_1, a_2, a_3... a_n\\}$. Then $ R[a_1, a_2,...a_n] \\equal{} A$. Consider the polynomial ring $ R[x_1,...x_n]$. There is an $ R$-algebra homomorphism $ R[x_1,...x_n]\\rightarrow A$ by taking $ x_i$ to $ a_i$. This map is surjective because the $ a_i$ generate $ A$. Now, call its kernel $ I$. Then $ A\\cong R[x_1,...x_n]/I$ by the first isomorphism theorem. This actually proves a stronger result than what you asked for because we don't need integrality. If indeed $ A$ happens to be integral, then we know $ I$ is a prime ideal because the product of two things cannot land in $ I$ without one of them being in $ I$.",
  "Solution_4": "The rationals are an extension of the integers which is not a $ \\mathbb{Z}$-algebra, hm? (Since the rationals are not finitely generated)",
  "Solution_5": "yeah its not finitely generated, thats why it doesn't work. I'm starting to think we might want to add finite presentation just to be safe but I think it works with just finite generation. Also if you think about it, the rational numbers are obtained by adjoining the roots of all linear polynomials to $ \\mathbb{Z}$ in the similar way that roots to polynomials are adjoined to ring extensions, so its really not that different. For example, the subset of rational numbers of the form $ a \\plus{} b/2$ with integer values for $ a$ and $ b$ is formed by $ \\mathbb{Z}[x]/(2x \\plus{} 1)$. So we can think of $ \\mathbb{Q}$ as some sort of giant binding together of all these extensions.",
  "Solution_6": "I don't know what a finite presentation is at all. So I don't think it will be necessary =)\r\nDo you have any suggested books on which I can read up on this stuff some more? Its difficult for me but I hardly want to give up XD",
  "Solution_7": "If you want a very thorough introduction to commutative algebra (which is where your inquiries seem to direct you), then i would recommend Commutative Algebra with a view toward Algebraic Geometry. (You don't have to read the second half of the book, the first half of the book should be more than enough). I would also recommend Hungerford's Algebra.",
  "Solution_8": "Your argument that $ I$ is prime is slightly more obscure than it needs to be. $ R[x_1, \\ldots, x_n]/I$ is isomorphic to an integral domain, so $ I$ is prime (any ideal such that the ring mod this ideal is an integral domain is prime).",
  "Solution_9": "Right, I was trying to prove the above statement (or at least explain it) by saying that the reason why it must be the case is because if the product of two things is congruent to zero mod $ I$, by integrality one of them must be congruent to zero mod $ I$ which means one of the must be in $ I$ which means $ I$ is prime. From your reaction, I guess my explanation was unsuccessful."
}
{
  "Tag": [
    "inequalities",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Let $(x_i)$ be a decreasing sequence of positive reals, then show that:\r\n\r\n(a) for every positive integer $n$ we have $\\sqrt{\\sum^n_{i=1}{x_i^2}} \\leq \\sum^n_{i=1}\\frac{x_i}{\\sqrt{i}}$.\r\n\r\n(b) there is a constant C for which we have $\\sum^{\\infty}_{k=1}\\frac{1}{\\sqrt{k}}\\sqrt{\\sum^{\\infty}_{i=k}x_i^2} \\le C\\sum^{\\infty}_{i=1}x_i$.",
  "Solution_1": "see http://www.mathlinks.ro/Forum/viewtopic.php?t=56175",
  "Solution_2": "[quote=\"Fermat -Euler\"]see http://www.mathlinks.ro/Forum/viewtopic.php?t=56175[/quote]No, you only write a trivial start to the problem. the real question is part (b), which is only in this topic. ;)",
  "Solution_3": "[quote=Peter]Let $(x_i)$ be a decreasing sequence of positive reals, then show that ... there is a constant [url=https://www.imc-math.org.uk/imc2000/prob_sol1.pdf]${\\it{C}}$[/url] for which we have $\\sum^{\\infty}_{k=1}\\frac{1}{\\sqrt{k}}\\sqrt{\\sum^{\\infty}_{i=k}x_i^2} \\le {\\it{C}}\\sum^{\\infty}_{i=1}x_i$.[/quote]\n\nIn fact, we have \\[\\sum_{m=1}^\\infty{\\mspace{-4mu}\\sqrt{\\frac1m\\sum_{i=m}^\\infty{\\mspace{-1mu}x_i^2}}}\\leq\\frac\\pi2\\sum_{i=1}^\\infty{\\mspace{-1mu}x_i}\\] for any decreasing sequence \\(\\left\\{x_i>0\\right\\}_{i\\geq1}\\), where the constant $\\frac\\pi2$ on the right-hand side is best possible.\nFor more details, please see [url=http://nsmath.cn/filedownload/173078]here[/url].",
  "Solution_4": "Does anybody take care of convergence of both series? What if x_i=1+1/i ? C can be in that case arbitraty small..."
}
{
  "Tag": [
    "calculus",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "For a partition P $ \\equal{} \\{ x_0, x_1,\\ldots, x_n \\equal{} b\\}.$\r\n\r\nLet $ f: [a,b]$ $ \\to$ $ \\mathbb{R}$ satisfy the Lipschitz condition i.e. for u,v, in [a,b] where $ \\mid f(u) \\minus{} f(v)\\mid \\leq c\\mid u\\minus{}v\\mid$ for a positive constant c. Prove that\r\n$ 0 \\leq$ U(f,P) $ \\minus{}$ L(f,P) $ \\leq c[b\\minus{}a]*gap P$",
  "Solution_1": "[quote=\"ayatollah85\"]For a partition P $ \\equal{} \\{ x_0, x_1,\\ldots, x_n \\equal{} b\\}.$\n\nLet $ f: [a,b]$ $ \\to$ $ \\mathbb{R}$ satisfy the Lipschitz condition i.e. for u,v, in [a,b] where $ \\mid f(u) \\minus{} f(v)\\mid \\leq c\\mid u \\minus{} v\\mid$ for a positive constant c. Prove that\n$ 0 \\leq$ U(f,P) $ \\minus{}$ L(f,P) $ \\leq c[b \\minus{} a]*gap P$[/quote]\r\n\r\nThis is straighforward. What are you having trouble with?"
}
{
  "Tag": [
    "Euler",
    "geometry unsolved",
    "geometry"
  ],
  "Problem": "Let P be the intersection of the diagonals of an inscribed quadrilateral ABCD. Prove that all four Euler lines of the triangles PAB, PBC, PCD, and PDA intersect in a single point.( the Euler line of a triangle is the line connecting its centroid and its orthocenter).",
  "Solution_1": "posted before:\r\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=107997[/url]"
}
{
  "Tag": [
    "geometry",
    "3D geometry"
  ],
  "Problem": "The length of a face diagonal of a cube is 3 cm. What is the number of square centimeters in the surface area of the cube?",
  "Solution_1": "We have that $ s\\sqrt{2}\\equal{}d$, where $ s$= the side length of the cube, and $ d$= the length of a face diagonal.\r\nSo, \r\n           $ s\\equal{}\\frac{3}{\\sqrt{2}}\\equal{}\\frac{3\\sqrt{2}}{2}$.\r\nThen, using that the surface area of a cube is $ 6s^2$,\r\n$ 6s^2\\equal{}6(\\frac{3\\sqrt{2}}{2})^2\\equal{}\\boxed{27}$."
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "advanced fields",
    "advanced fields unsolved"
  ],
  "Problem": "Let $ n\\geqslant 2$, and $ z_{1},\\ldots , z_{n}\\in\\mathbb{C}$.\r\nFind out $ P\\in\\mathbb{C}[X]$, s.t $ P\\bigl( (X\\minus{}z_{1})(X\\minus{}z_{2})\\cdots (X\\minus{}z_{n})\\bigr)\\equal{}P(X\\minus{}z_{1})P(X\\minus{}z_{2})\\cdots P(X\\minus{}z_{n})$.\r\nP.S.: I only have a partial answer, and I post it here hoping that some bigger tools can be used to crack this nut.",
  "Solution_1": "every $ P\\equal{}X^{k}$ satisfies the equation. maybe I've misunderstood something, or we want [i]all[/i] solutions?",
  "Solution_2": "And those are the only solutions that work for general $ z_{i}$. (There are no others when $ z_{1}\\equal{}\\dots\\equal{}z_{n}\\equal{}0$.)",
  "Solution_3": "-oo-, you are right, I should I've been clearer, instead of \r\n[quote=\"JC_math\"]... Find out $ P\\in\\mathbb{C}[X]$, s.t ...[/quote]\r\nI should have written \"\"... Find out [b][i]all[/i][/b] $ P\\in\\mathbb{C}[X]$, s.t ...\".\r\nIn fact $ P\\equal{}aX^{k}$ are solutions if $ a^{n\\minus{}1}\\equal{}1$, and of course $ P\\equal{}0$.\r\nIf $ n\\equal{}2$ these are the only solutions, but also if $ z_{1}, z_{2},\\ldots, z_{n}$ is a regular polygon as well. \r\nBut in general???\r\n\r\nmlok, \r\nTo answer your comment, my question was not\r\nLet $ n\\geqslant 2$, Find out all $ P\\in\\mathbb{C}[X]$, s.t for all $ z_{1},\\ldots , z_{n}\\in\\mathbb{C}$ $ P\\bigl( (X\\minus{}z_{1})(X\\minus{}z_{2})\\cdots (X\\minus{}z_{n})\\bigr) \\equal{} P(X\\minus{}z_{1})P(X\\minus{}z_{2})\\cdots P(X\\minus{}z_{n})$.\r\n\r\nP.S.: I tried to edit my previous message, but I got \"Sorry, but you cannot edit this post anymore. Ask a moderator to do it if it is really necessary.\", strange????"
}
{
  "Tag": [],
  "Problem": "The quantity of heat liberated when $ 2.00$ moles of aluminum oxide is formed under standard conditions is closest to\r\n\r\n$ \\textbf{(1)} \\ 419 \\ \\text{kJ} \\qquad \\textbf{(2)} \\ 838 \\ \\text{kJ} \\qquad \\textbf{(3)} \\ 1680 \\ \\text{kJ} \\qquad \\textbf{(4)} \\ 3350 \\ \\text{kJ}$",
  "Solution_1": "The answer is (4)\r\nHere's why:\r\nFor forming of the aluminum oxide, we have a reaction:\r\n4Al(s) + 3 O2 = 2 Al2O3\r\nHeat released= Q= 2* (-1676)= -3352 Kj.\r\nthe number -1676 is obtained from the standard delta H of alumnium oxide. This delta H for both alumnium and oxygen are zero because they are at standard state.\r\nYou can get this number from the index of a chemistry book.",
  "Solution_2": "[quote=\"i_thirst_for_knowledge\"]The answer is (4)\nHere's why:\nFor forming of the aluminum oxide, we have a reaction:\n4Al(s) + 3 O2 = 2 Al2O3\nHeat released= Q= 2* (-1676)= -3352 Kj.\nthe number -1676 is obtained from the standard delta H of alumnium oxide. This delta H for both alumnium and oxygen are zero because they are at standard state.\nYou can get this number from the index of a chemistry book.[/quote]\r\n\r\nThanks a lot!"
}
{
  "Tag": [
    "geometry",
    "ratio"
  ],
  "Problem": "I like doing these old Math League Contests because they're enjoyable: here's a good review problem for area: \r\n[b]Let there be acute triangle \u2206ABC, such that AE, BF, and CD intersect at point P (each of these are lines to the opposite side of \u2206ABC from each angle). If the area of \u2206AFP is 40, the area of \u2206CFP is 30, and the area of \u2206CEP is 35, what is the area of \u2206BPD?[/b]",
  "Solution_1": "I don't think that's really clear...\r\nYou say that the other points are on the sides opposite the angle, but if that were true, then point P would be there too, meaning that the the other three lines couldn't intersect point P...",
  "Solution_2": "Yeah. I agree with ZhangPeijin. The directions aren't very clear. Can you please accompany this with a figure or diagram?",
  "Solution_3": "Huh? Guys, there is an acute triangle ABC, and points E,F, and D on BC,AC, and D on AB, respectively.The areas of triangle AFP,CEP, and CFP Z-Coli gave.\r\nArgh, I'm really not sure in this problem.\r\n[hide]Since triangles APC and CPE have the same height, and [APC]=70,[CPE]=35, the ratio of AP to EP=2:1, inother words, AE is a median. For triangles ABE and ACE, they have the same base and altitude, so equal area;thus, the total area is 210.[/hide].\r\n\r\nYeah, probably wrong.",
  "Solution_4": "No, it's good.  I think\r\n\r\nIt's actually pretty clear, and Peijin, I don't see why point P would have to be on a side.  Just draw 2 of the cevians from 2 different angles and draw the third from the last angle to hit that intersection point.\r\n\r\ncevian=line from an angle to the opposite side\r\n\r\n\r\nAnd I'm not sure if this actually belongs in Classroom Math or not, most people probably wouldn't consider the ratio to find that one of the cevians was a median.",
  "Solution_5": "Yeah, but he said each of these are on lines opposite to the angle.\r\nMeaning that Point P is on a side of the triangle.",
  "Solution_6": "Hence why you fail English.  Read the question.  The lines [b]intersect[/b] at point P.",
  "Solution_7": "Ohh...\r\nAnyways, I didn't fail english.\r\nI just got a A-...\r\n-worst grade ever-",
  "Solution_8": "Okay, there are a couple of things which should be addressed:\r\n1) The original problem had a picture but I don't have a scanner so I had to word it.  I think it's pretty clear, but if you have specific questions about the problem, I'll answer it.\r\n2) P IS NOT NECESSARILY THE CENTROID.  That is assuming that the lines drawn all have the same properties==>however, all I said was that they intersect at point P.  Every line has a point that divides it into a 2:1 ratio (so no, 210 is not correct)",
  "Solution_9": "yeah, that's why I thought it was wrong.\r\nToo tired to do anything right now.",
  "Solution_10": "Obviously, this problem uses side ratios that correspond to area ratios.  Hint: Let [CEP]=k, and then find the remaining areas in terms of k.  Set up some equations to solve for k, and later, solve for the remaining unknown areas.",
  "Solution_11": "Are you sure this is solveable based on the information given?\r\nSeeing as the lines could reach the opposite side at any point?\r\nOr maybe it's just because I suck at Geometry.",
  "Solution_12": "Also, I don't think Math League problems go in Classroom Math",
  "Solution_13": "If it's the High school kind, then no. I don't think the middle school kind would be this hard though.",
  "Solution_14": "Yeah, this problem is from the high school Math League, but I didn't think that you guys would have trouble with it.  I guess I'll post the answer soon, then.  Seriously though, when I did it I though it wasn't too hard.  Sorry guys :(",
  "Solution_15": "Some of us are rather bad at geometry  :D",
  "Solution_16": "omg i got it!!!\r\n\r\nI will include diagrams later\r\n\r\n[hide=\"My Solution\"]If the heights of two triangles are equal, then their areas are in the ratio of their bases.  In this case, $ \\triangle AFP$ and $ \\triangle CFP$ have the same height.  We know that the ratio of the bases must be $ 4: 3$.  $ AF: CF \\equal{} 4: 3$.  So now that we have the ratio of the bases, we can assign weights to $ A$ and $ C$.  The weight of $ A$ is $ 3$ and the weight of $ C$ is $ 4$.  Now, let the weight of $ B$ be $ x$.  Therefore, $ BE: CE \\equal{} 4: x$.  We know that the areas are in the ratio of the bases so the area of $ \\triangle BEP \\equal{} \\dfrac{140}{x}$.  Since the weight of $ A$ is $ 3$ and the weight of $ B$ is $ x$, $ AD: BD \\equal{} x: 3$.  Now, let us take the big triangles, $ \\triangle ABE$ and $ \\triangle ACE$.  We know that the area of $ \\triangle ACE$ is $ 40 \\plus{} 35 \\plus{} 30 \\equal{} 105$.  We also know that the areas are in the ratio $ 4: x$.  The area of $ \\triangle ABE$ is $ 105\\times\\dfrac{4}{x} \\equal{} \\dfrac{420}{x}$.  Therefore, the area of $ \\triangle ABP$ is $ \\dfrac{280}{x}$.  The ratio of the area of $ \\triangle ADP$ to the area of $ \\triangle BDP$ is $ x: 3$.  We find that the area of $ \\triangle ADP \\equal{} \\dfrac{280}{x(x \\plus{} 3)}\\times x \\equal{} \\dfrac{280}{x \\plus{} 3} and the area of$\\triangle BDP=\\dfrac{280}{x(x+3)}\\times3=\\dfrac{840}{x(x+3)}.  Now, let us take $ \\triangle ABF$ and $ \\triangle CBF$.  The ratio of the areas of the triangles is $ 4: 3$.  Adding the areas of all the little triangles up, we get the areas of the big triangles.\n\\[ \\dfrac{4}{3}(65 \\plus{} \\dfrac{140}{x}) \\equal{} 40 \\plus{} \\dfrac{280}{x}\n\\]\n\n\\[ \\dfrac{260}{3} \\plus{} \\dfrac{560}{3x} \\equal{} \\dfrac{40x}{x} \\plus{} \\dfrac{280}{x}\n\\]\n\n\\[ \\dfrac{260x}{3x} \\plus{} \\dfrac{560}{3x} \\equal{} \\dfrac{40x \\plus{} 280}{x}\n\\]\n\n\\[ \\dfrac{260x \\plus{} 560}{3x} \\equal{} \\dfrac{120x \\plus{} 840}{3x}\n\\]\n\n\\[ 260x \\plus{} 560 \\equal{} 120x \\plus{} 840\n\\]\n\n\\[ 140x \\equal{} 280\n\\]\n\n\\[ x \\equal{} 2\n\\]\nThe area of $ \\triangle BPD$ is $ \\dfrac{840}{x(x \\plus{} 3)} \\equal{} \\dfrac{840}{2\\times5} \\equal{} \\dfrac{840}{10} \\equal{} \\boxed{84}$[/hide]",
  "Solution_17": "The answer is correct, good.  The answer and solution was already up on the high school math section, but it looks like a different solution.  Excellent job.",
  "Solution_18": "[quote=\"Z-coli\"]I like doing these old Math League Contests because they're enjoyable: [/quote]\r\nwhere do you find these problems?",
  "Solution_19": "I go to a math team every Sunday where they have tons of old math books.  Our coach give me the book, it's pretty fun."
}
{
  "Tag": [
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Find all positive integers $ x;y;z;n$ sauch that $ x!\\plus{}y!\\plus{}z!\\equal{}5n!$ where $ k!\\equal{}1*2*3*..*k$",
  "Solution_1": "Assume that $ x\\geq y\\geq z>0$ then we have: $ 5.n!\\leq 3.x!<5.x!\\Rightarrow n<x\\Rightarrow n\\plus{}1\\leq x$\r\nNow $ 5.n!>x!\\geq (n\\plus{}1)!\\Rightarrow 5>n\\plus{}1\\Rightarrow n\\equal{}1,2,3$\r\nConsider $ 3$ cases we have result!",
  "Solution_2": "Interesting solution tdl!\r\nmy solution :\r\n\r\n Note that if we take the bigest  possible solution, we get: x=y=z=n : and then \r\n3n!=5n!\r\n\r\nimpossible,so $ a\\ne b$, and $ a\\ne c$, but $ b$ can be equal to $ c$,.\r\nNow,\r\nFor n=0\r\n\r\nRHS=5,\r\n\r\nLHS=$ x! \\plus{} y! \\plus{} z!$ So the only solutions are $ (x,y,z)\\equal{}{(2,2,1);(2,2,0)}$ \r\nand  simmetricaly.\r\n\r\n$ n\\equal{}1$\r\n\r\n$ RHS\\equal{}5$,\r\nSame solution \r\n\r\n$ n\\equal{}2$\r\n\r\n$ RHS\\equal{}10$ and $ 10 \\equal{} 2\\plus{}2\\plus{}6$ the only possible partition,\r\nso  $ (x,y,z)\\equal{}{(2,2,3)}$  and permutations\r\n\r\n$ n\\equal{}3$\r\n\r\n$ RHS\\equal{}5\\cdot 3!\\equal{}30$\r\nno solutions \r\n\r\n$ n\\equal{}4$\r\n\r\n$ 5n!\\equal{}120$,\r\n$ x! \\plus{}y!\\plus{}z!\\equal{} 5!$which is $ x! \\plus{}y!\\equal{}5! \\minus{} z!$\r\nso  $ LHS\\equal{}5/( x!\\plus{}y!)$       $ RHS\\equal{}4! \\minus{}5/z!$\r\nbut since $ 4 \\geq z\\geq 0$, the result found previously is absurd.\r\nBesides if we take the bigest  possible solution, we get:$ x\\equal{}y\\equal{}n$ : and then $ 3n!\\equal{}5n!$\r\nimpossible.\r\n\r\nFor $ n\\geq 5$,  $ x! \\plus{} y! \\plus{} z!$=5n!\r\nso $ \\frac{(x!\\plus{} y! \\plus{}z!)}{5}\\equal{}x! (x\\plus{}1)\\cdot...\\cdot n$\r\n supposing that $ x> (z,y)$.\r\nDividing by x! we get $ (\\frac {1}{5})\\plus{} \\frac {(y! \\plus{}z!)}{5} \\equal{}\r\n (x\\plus{}1)\\cdot...\\cdot n$\r\nSo impossible since LHS is not an integer and RHS is.\r\nSo the only solutions are  \r\n$ (x,y,z,n)\\equal{}{ (2,2,1,0) (2,2,0,0) (2,2,1,1)(2,2,0,1),(2,2,3,2)}$ \r\nand all the permutations of $ (x,y,z)$."
}
{
  "Tag": [
    "inequalities proposed",
    "inequalities"
  ],
  "Problem": "sorry  :( \r\nEdit\r\n[quote=\"AnhHungThamLang\"][i]Let[/i] $\\large a,b,c\\ge0$ [i].prove that[/i]\n...............\n$\\large(1+\\sqrt{\\frac{a}{a+b}})(1+\\sqrt{\\frac{b}{b+c}})(1+\\sqrt{\\frac{c}{c+a}})\\leq \\frac{10+7\\sqrt{2}}{4}$\n[b]Undefined Mixing Variables[/b]  :wink: \n[b]@pvthuan:[/b]  Th\u1ea7y th\u1ea5y b\u00e0i to\u00e1n n\u00e0y th\u1ebf n\u00e0o \u1ea1!  :rotfl: Th\u1ea5y h\u00e3y nh\u1eafn l\u1ea1i cho em qua nick [b]silenthero_k09[/b] nh\u00e9. :D \n\u0110\u1ec1 n\u00e0y m\u1edbi \u0111\u00fang th\u1ea7y \u1ea1,em \u0111\u00e1nh nh\u1ea7m d\u1ea5u \u0111\u1eb3ng th\u1ee9c\n [/quote]",
  "Solution_1": "Please don't post repeat in future!\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=138945\r\nYou can edit your post in that topic."
}
{
  "Tag": [
    "Ross Mathematics Program",
    "summer program",
    "PROMYS",
    "AMC",
    "USA(J)MO",
    "USAMO",
    "AwesomeMath"
  ],
  "Problem": "What is more prestigious, the Ross Program/ PROMYS or making USAMO??",
  "Solution_1": "idk y ppl r always worrying abt prestige but i'd say usamo is harder to make",
  "Solution_2": "Don't worry about prestige just try to do what you enjoy.  Regardless of that, USAMO is definitely far more prestigious than any summer program short of MOP/RSI.",
  "Solution_3": "Are the two mutually exclusive?  \r\n\r\nSeriously, if you're worried about anything other than college applications, there's not much relation between the two.  Ross and PROMYS are [b]not training camps.[/b]  They are designed to get you interested in, and better at, [b]mathematics[/b] - not contest mathematics, but the kind of mathematics you might want to spend the rest of your life doing and go to graduate school for.  The USAMO, on the other hand, is a high school competition.  It would be fair to say that PROMYS is the reason I became less interested in the USAMO.\r\n\r\nFor what it's worth, a good letter of recommendation from PROMYS is probably going to help your application more than making USAMO, but please don't think about either in such terms.",
  "Solution_4": "They are not mutually exclusive. \r\n\r\nI enjoyed AwesomeMath and plan to apply to PROMYS, MathPath or Ross this year. Like many of AoPSers, I do math because I enjoy it. I am sure most of AoPSers also do other math competitions, such as Mandelbrot, ARML, USAMTS etc. \r\n\r\nBut, in my mind, USAMO always has a special place in our heart as this is the only way we can qualify for the USA IMO team someday. I will continue to practice math and continue to enjoy it. Once a math lover, forever a math lover, period.",
  "Solution_5": "yea i did well in all my other comps: mandelbrot, RIML, USAMTS\r\nBut FAILED the AMC's this year and did not make AIME. Very dissapointing....",
  "Solution_6": "[quote=\"mround\"]They are not mutually exclusive. [/quote]\nRhetorical question!\n\n[quote=\"mround\"] plan to apply to PROMYS, MathPath or Ross this year.[/quote]\nIsn't the PROMYS deadline in a few days?  \n\n[quote=\"mround\"]USAMO always has a special place in our heart as this is the only way we can qualify for the USA IMO team someday. I will continue to practice math and continue to enjoy it. Once a math lover, forever a math lover, period.[/quote]\r\nBut that's exactly my point!  A love of math is lifelong, but the period in which you can qualify for the IMO team is 4 years (plus or minus).  What I'm trying to say is that I got much more out of going to PROMYS as a mathematician than I got out of qualifying for the USAMO.",
  "Solution_7": "errrr no \r\npromys deadline is in like end of may.",
  "Solution_8": "[quote=\"simo14\"]idk y ppl r always worrying abt prestige but i'd say usamo is harder to make[/quote]\r\n\r\nPrestige makes you look good. It's like saying \"I go to Stanford/Top University\" and \"I go to <insert something>\"",
  "Solution_9": "And that really matters because?",
  "Solution_10": "Obviously it matters because society judges people based on their past. That said, it's obvious that the bragging rights aren't what matter, but the achievements themselves.",
  "Solution_11": "[quote=\"shelly32494\"]errrr no \npromys deadline is in like end of may.[/quote]\r\n\r\nMAY 30th to be precise.t0rajir0u you should be a counselor ^^\r\nI've finished 9 of the problems and typed them up though: 1-5, 7-10",
  "Solution_12": "I'm applying to be a counselor, but I'll be skimming the counselor deadline (which is what I was thinking of, my bad) pretty closely so that might be a problem.",
  "Solution_13": "this is not collegeconfidential.",
  "Solution_14": "Let's just say that I do not know what PROMYS is but know what USAMO is. That must mean that USAMO is more prestigious.",
  "Solution_15": "[quote=\"james4l\"]Let's just say that I do not know what PROMYS is but know what USAMO is. That must mean that USAMO is more prestigious.[/quote]\r\nHoribble logic. Most students don't know what IMO is.",
  "Solution_16": "Well let's take a poll - mods, please make one.\r\n\r\nI'm fairly certain that PROMYS is something that \"most students\" would not know.",
  "Solution_17": "That's completely irrelevant to the question being asked.  When people talk about prestige, it's usually with reference to college applications, so it's less important what students are aware of and more important what college admissions officers are aware of, but again, this is [b]still the wrong perspective[/b] to take with respect to how you decide to allot your time.",
  "Solution_18": "If you make IMO is it true that (almost) every college would accept you (or maybe even in other countries)? Just curious...",
  "Solution_19": "Not if they don't care about math!  This is both a college-specific decision and [b]still the wrong perspective.[/b]  If you're good enough to make IMO, especially from the United States, chances are you've got a lot of other good stuff on your application.",
  "Solution_20": "Also, if you are talented enough to make IMO, chances are you won't want to apply to, say, a liberal arts college, which probably won't care about math at all, but you'll probably have a better chance at an engineering/math college.  Still, torajirou has the right point: if you only want to get to IMO for college, then you should reset your priorities or just don't include math competitions in your priorities.  Furthermore, if all you care about is college, chances are you won't make IMO anyways because of stress over grades, extracurriculars, and focusing on those things that \"look good\" rather than math, which should be done because you want to."
}
{
  "Tag": [
    "AwesomeMath",
    "summer program",
    "AMC",
    "USA(J)MO",
    "USAMO",
    "AMC 8",
    "AMC 10",
    "\\/closed"
  ],
  "Problem": "In terms of AIME/USAMO preparation, which one would be more useful? Any suggestions would be greatly appreciated.",
  "Solution_1": "I think it depends on your level of math.",
  "Solution_2": "That question may be impossible to answer.  This is the first year of the AwesomeMath program and the WOOT program itself has evolved a lot over the past year.  You might be better off asking specific questions about each program or their instructors.",
  "Solution_3": "I have a question......Is USAMO Awesome Math Camp? I have been hearing many abbreviated tests and competitions lately. I am pretty confused.",
  "Solution_4": "USAMO=United States of America Mathematical Olympiad",
  "Solution_5": "AMC8, AMC10, and AMC12 have AMC=American Mathematics Competition.\r\n\r\nAIME = American Invitational Mathematics Examination\r\n\r\nUSAMO= USA Math Olympiad\r\n\r\nMOP/MOSP = Math Olympiad Summer Program. \r\n\r\nAwesomemath, Mathpath, Mathcamp = No common abbreviation (I think).\r\n\r\nW00t = Worldwide Online Olympiad Training.",
  "Solution_6": "I would think WOOT would be a better course, if only because of the quality of instruction.\r\n\r\nBut if it takes you a little while to see things (like me) then an online classroom might not be best.  However, during math jams, I can usually grasp a concept I missed by going back and reviewing transcripts.  So maybe the online classroom isn't a setback after all...",
  "Solution_7": "WOOT has a definite advantage in the fact that you can always look back at the transcripts if you need any help or want to review.  The transcripts are a very valuable resource that can help you a lot later, when the class is over.  Also, I think you get lots of good feedback from the instructor (if it is anything like the feedback you get in the other classes) that really helps.\r\n\r\nBecause the AwesomeMath program is new, the people who run it might not have had as much experience running it, so it could potentially have more problems.  However, you would have to look at the website of the people who run the AwesomeMath program to see what it is like.\r\n\r\nBasically, there is not as information about the AwesomeMath program as there is for the WOOT program.",
  "Solution_8": "[quote=\"b-flat\"]WOOT has a definite advantage in the fact that you can always look back at the transcripts if you need any help or want to review.  The transcripts are a very valuable resource that can help you a lot later, when the class is over.  Also, I think you get lots of good feedback from the instructor (if it is anything like the feedback you get in the other classes) that really helps.\n\nBecause the AwesomeMath program is new, the people who run it might not have had as much experience running it, so it could potentially have more problems.  However, you would have to look at the website of the people who run the AwesomeMath program to see what it is like.\n\nBasically, there is not as information about the AwesomeMath program as there is for the WOOT program.[/quote]\r\nI haven't taken WOOT yet, but I would say from what I saw in the Math Jam yesterday it looks like it will be a very helpful program.  :) \r\n\r\nActually, the AwesomeMath program (which is abbreviated as AMSP for AwesomeMath Summer Program) is very new, but the instructors are definitely not inexperienced; many of them are MOP instructors. Here's the faculty list if you're interested:\r\nhttp://www.awesomemath.org/staff.html\r\n\r\nHowever, WOOT is definitely less expensive than AwesomeMath, since you don't have to live there. (Or were you talking about just participating in the Year-Round program?)\r\n\r\nI would say that if you can (if your budget allows and you have enough time), the best thing would be to do both.",
  "Solution_9": "From the title of this topic, I think he's talking about [url=http://www.awesomemath.org/amyhome.html]the AwesomeMath Year-round[/url], not the Awesome Math Summer Program."
}
{
  "Tag": [
    "geometry",
    "ratio",
    "analytic geometry",
    "circumcircle",
    "power of a point"
  ],
  "Problem": "Let an arc $AB=90^\\circ$ of a circle . We produce chord $AB$ and take point $C$ , such that  $BC=AB$. Let  $CD$ the tangent segment to the arc  and $K$  the foot of the perpendiculer of  $A$  to  $BD$. Prove that  $KB=2KA$.\r\n\r\n[u]babis[/u]\r\n\r\n [i]Have you ever seen this problem before ?[/i]",
  "Solution_1": "A couple questions.\r\n1. Produce chord AB (does this mean just connect the segment AB?)\r\n2. Take point C (anywhere in the plane under the following constraint, or is it on line AB?)\r\n3. CD the tangent segment to the arc (what does this mean?)\r\n\r\nI tried to make I diagram but I coulldn't, at least not from my undestanding.",
  "Solution_2": "i think $C$ is on the line $\\overline{AB}$, and $D$ is the point on the minor arc $\\widehat{AB}$ of the given circle such that $\\overline{CD}$ is tangent to the circle.\r\n\r\n[hide]\nLet $O$ be the center of the circle and $E$ be the point opposite $A$ on the circle. Then $AC=2AB$ and $AE=2AO$ so triangles $\\triangle AOB$ and $\\triangle AEC$ are similar by SAS.\n\nApplying the side ratio to the third pair of sides, $CE=2OB=2OE$. Let $\\alpha =\\angle OCE$ and $\\beta =\\angle COE$. Then $\\angle COD=\\beta$ since triangles $\\triangle COE$ and $\\triangle COD$ are congruent.\n\nFinally, $\\angle AOD=180^\\circ-2\\beta =2\\alpha$, which gives us $\\angle ABD=\\alpha$ since the inscribed angle is half of arc $\\widehat{AD}$. Then $\\triangle AKB$ is similar to $\\triangle OEC$, so $KB=2KA$, as desired.\n[/hide]",
  "Solution_3": "[quote=\"me@home\"]A couple questions.\n1. Produce chord AB (does this mean just connect the segment AB?)\n2. Take point C (anywhere in the plane under the following constraint, or is it on line AB?)\n3. CD the tangent segment to the arc (what does this mean?)\n\nI tried to make I diagram but I coulldn't, at least not from my undestanding.[/quote]\r\n\r\n Really sorry! Probably I transleted the greek text in a direct way.\r\n[color=red]a)[/color] We extend $AB$ and on ray $AB$ we take segment $BC= AB$\r\n[color=red]b) [/color]From point $C$ we draw the tangent $CD$ \r\n[color=red]c) [/color]In (very nice !) solution by scorpius 119 , I hope you 'll clear the enounciation. \r\n[color=red]d)[/color] Try for a new solution , if there is .\r\n[u]babis[/u]",
  "Solution_4": "Alternative solution:\r\n\r\n[hide]Clearly, $\\angle ADB = \\frac{3\\pi}{4}$, so we conclude that $\\left|KD\\right|=\\left|KA\\right|$   (1)\n\nBy power of a point, we know that $\\left|CD\\right|^{2}= 2 \\left|AB\\right|^{2}$.\n\nNow, in $\\Delta ACD$, a median is $DB$. Therefore, it's length is given by $\\left|BD\\right|^{2}= \\frac{1}{4}\\left(2\\left|AD\\right|^{2}+2\\left|CD\\right|^{2}-\\left|AC\\right|^{2}\\right)= AK^{2}$\nOr $\\left|BD\\right|=\\left|AK\\right|$     (2)\n\nAdding $(1)$ and $(2)$ together, we get what we wanted.[/hide]",
  "Solution_5": "Solution with coordinates[hide]Let $O$ be the center of the circle. We put $O(0,0), A(0,1)$ and $B(1,0)$. Then we found $C(2,-1), D(\\frac{4}{5},\\frac{3}{5})$ and $K(\\frac{3}{5},\\frac{6}{5})$. Then $AK= \\frac{2}{\\sqrt10}$ and $KB=\\frac{4}{\\sqrt{10}}$. So $KB=2KA$.[/hide]\r\nIt's ugly but mechanical ... :rotfl:",
  "Solution_6": "My solution uses only some concyclic points...This is the solution I found during the competition.\r\n\r\n             [u][b]  SOLUTION[/b][/u]\r\nIt is enough to prove that $D$ is the midpoint of $KB$.Let us consider the midpoint $M$ of segment $AB$.I have to show that $\\widehat{MDB}=90^{o}$.But $\\widehat{ODB}=90^{o}$, so i have to prove that $\\widehat{MDO}=\\widehat{BDC}$, or $\\widehat{DAC}=\\widehat{MCO}$ $\\Longleftrightarrow CO//AD$.\r\nWe now cosider the point $A'$, the symmetric point of $A$ with respect to $O$ and denote $Cx$ the ray $CD$.\r\nThen we can easilly show that $\\widehat{ADx}$ $=\\widehat{AA'D}$ $=\\widehat{OCD}$ , which is enough to prove that $AD//OC$ and we are Q.E.D.\r\n",
  "Solution_7": "Nice solutions guys, Btw I understand the problem now :D \r\nhere is a diagram [img]6993[/img]\r\n\r\n\r\n[quote=\"Nick Rapanos\"]My solution uses only some concyclic points...This is the solution I found during the competition.\n\n             [u][b]  SOLUTION[/b][/u]\nIt is enough to prove that $D$ is the midpoint of $KB$.Let us consider the midpoint $M$ of segment $AB$.I have to show that $\\widehat{MDB}=90^{o}$.But $\\boxed{\\boxed{\\widehat{ODB}}}=90^{o}$, so i have to prove that $\\boxed{\\boxed{\\widehat{MDO}}}=\\widehat{BDC}$, or $\\widehat{DAC}=\\boxed{\\boxed{\\widehat{MCO}}}$ $\\Longleftrightarrow CO//AD$.\nWe now cosider the point $A'$, the symmetric point of $A$ with respect to $O$ and denote $Cx$ the ray $CD$.\nThen we can easilly show that $\\widehat{ADx}$ $=\\widehat{AA'D}$ $=\\widehat{OCD}$ , which is enough to prove that $AD//OC$ and we are Q.E.D.\n[/quote]\r\nDouble box:\r\n1. The double boxed angle should be ODC, right?\r\n2. / 3. It appears like the first and second angles are equal, as you claim, but how is this trivial? I drew in the circumcircle just in case, but still, what am I missing... it seems non obvious to me.\r\n\r\nThat's all, thank you",
  "Solution_8": "[quote=\"me@home\"]Nice solutions guys, Btw I understand the problem now :D \nhere is a diagram [img]6993[/img]\n\n\n[quote=\"Nick Rapanos\"]My solution uses only some concyclic points...This is the solution I found during the competition.\n\n             [u][b]  SOLUTION[/b][/u]\nIt is enough to prove that $D$ is the midpoint of $KB$.Let us consider the midpoint $M$ of segment $AB$.I have to show that $\\widehat{MDB}=90^{o}$.But $\\boxed{\\boxed{\\widehat{ODB}}}=90^{o}$, so i have to prove that $\\boxed{\\boxed{\\widehat{MDO}}}=\\widehat{BDC}$, or $\\widehat{DAC}=\\boxed{\\boxed{\\widehat{MCO}}}$ $\\Longleftrightarrow CO//AD$.\nWe now cosider the point $A'$, the symmetric point of $A$ with respect to $O$ and denote $Cx$ the ray $CD$.\nThen we can easilly show that $\\widehat{ADx}$ $=\\widehat{AA'D}$ $=\\widehat{OCD}$ , which is enough to prove that $AD//OC$ and we are Q.E.D.\n[/quote]\nDouble box:\n1. The double boxed angle should be ODC, right?\n2. / 3. It appears like the first and second angles are equal, as you claim, but how is this trivial? I drew in the circumcircle just in case, but still, what am I missing... it seems non obvious to me.\n\nThat's all, thank you[/quote]\r\n\r\n\r\nYes you are right its not obvious...Firstly, i wanted to write $\\widehat{ODC}$ as you understood!\r\nNow I have to prove that $\\widehat{MDO}=\\widehat{BDC}$.\r\n\r\nBut $\\widehat{BDC}=\\widehat{BAD}$ (because $CD$ is tangent to the cicle) and $\\widehat{MDO}=\\widehat{MCO}$ because of the concyclic points you have drawn in your shape.\r\n\r\nI think it is clear now!Ask me if you have any questions without any doubt!\r\nBye",
  "Solution_9": "Thanks for answering my questions, but I am still a little confused... I drew that circle in the diagram but I do not know [b]why[/b] they are concyclic, in your proof it kind of looks like you assumed it :maybe: but I am not sure, that is what I was really asking about (we know they are concyclic from the diagram but we haven't proven it yet or so it appears to me)",
  "Solution_10": "Sorry I didnt prove that these points are concyclic because it was too easy to prove...\r\nLook $\\widehat{ODC}=\\widehat{OMC}=\\widehat{OA'C}=90^{o}$",
  "Solution_11": "Sorry about that, I truly did not see that and t didn't seem \"trivial\" to me at least... thank you for pointing that out :oops:"
}
{
  "Tag": [],
  "Problem": "Suppose there is a language similar to the English language, but any letter combination forms a word.  For example, \"gdsf\" is a word. How many four letter words can be formed from the word, \"MISSISSIPPI\"?.\r\n\r\nHow many three letter words can be formed from the word, KARAT?",
  "Solution_1": "Rezult of \"MISSISSIPPI\" is: $\\frac{11!}{4!\\cdot 4!\\cdot 4!}$\r\nRezult of KARAT is $\\frac{5!}{2!}$",
  "Solution_2": "Uh, the first one gives you a non-integer.  :wink: \r\nI'm no good at these, but shouldn't it be:\r\n$\\frac{11!}{4!4!2!}$?",
  "Solution_3": "You guys are all counting the number of permutations when all the letters are used, but the questions ask when 4 letters and 3 letters are used.",
  "Solution_4": "Ah, well, isn't it:\r\n$\\frac{11!}{4!4!2!7!}$?\r\n...But that isn't a whole number; never mind, I fail.",
  "Solution_5": "I actually do not have the answer. A teacher of mine messed up while making a question for a quiz, and I just posted it here because maybe there's some tricky counting technique you could use.",
  "Solution_6": "is it 176 for the first question (the one with MISSISSIPPI and 4 letter words)",
  "Solution_7": "If someone could post a solution to the MISSISSIPPI problem as well as a general method of doing problems like this, that would be great.\r\nI admit; I'm curious.",
  "Solution_8": "I have a feeling that with the logic we've been using so far it would just be really ridiculous casework... But maybe somebody has a cool intermediate counting technique that could get it done quickly?"
}
{
  "Tag": [
    "AMC",
    "USA(J)MO",
    "USAMO",
    "AIME",
    "geometry",
    "inequalities",
    "number theory"
  ],
  "Problem": "Does anyone know of a nice book with good USAMO-level practice problems? About the level of #1 and #4 USAMO problems.\r\n\r\nThanks! :)",
  "Solution_1": "I think Mathematical Olympiad Challenges is a good place to start - it groups problems by subject to give a more thorough grounding in a variety of Olympiad topics than simply selecting random Olympiad problems could provide.",
  "Solution_2": "The USSR Olympiad Problem Book has a lot of good problems, although they may be a little difficult for just #1 and #4 USAMO level.",
  "Solution_3": "Why not Engel? It has lots of problems at lots of different levels.",
  "Solution_4": "Hm, can I get the full title? On amazon engel seems to be a fiction book. :D thanks!",
  "Solution_5": "[url=http://www.amazon.com/exec/obidos/tg/detail/-/0387982191/qid=1125807069/sr=8-1/ref=pd_bbs_1/103-6210280-3474222?v=glance&s=books&n=507846][i]Problem Solving Strategies[/i] by Arthur Engel[/url]",
  "Solution_6": "Ah, thanks! :) I actually already have that book but I have never really looked at it before... I'll go check it out now!",
  "Solution_7": "The Arbelos help quite a bit too, i heard.",
  "Solution_8": "[quote=\"xxreddevilzxx\"]The Arbelos help quite a bit too, i heard.[/quote]\r\n\r\nI've got those too, about 4 or 5 volumes or booklets or whatever. It seems kind of dense, I read them when I was preparing for the AMC 12 but I figured they wouldn't be of much use in problem solving. It probably is pretty good for the USAMO though.",
  "Solution_9": "Engel is great. If you only get one book, get that. Actually going through the WHOLE thing rigorously would put you at borderline top 12, honestly. (disclaimer: borderline top 12 depending on the strength of the year)",
  "Solution_10": "*sigh* I am going through Engel right now... it's soooo long! I'm a freshman trying to qualify for MOP. I don't know how I'm going to finish this book. x___x",
  "Solution_11": "[quote=\"mathclass\"]*sigh* I am going through Engel right now... it's soooo long! I'm a freshman trying to qualify for MOP. I don't know how I'm going to finish this book. x___x[/quote]\r\n\r\nIt's not about finishing. It's always going to be a continuous work in progress, which involves doing problems, GOING BACK AND DOING STUFF AGAIN, analyzing your strengths and weaknesses, etc. It's not a novel.",
  "Solution_12": "Oh by the way for USAMO 1's/4's, some AIME problems really aren't any easier than many 1's and 4's (sometimes significantly more time consuming), and doing stuff like Canadian, Irish, etc. Olympiads from http://www.kalva.demon.co.uk is good too.",
  "Solution_13": "Darktreb, I've gone through quite a bit of Engel but I was wondering if there are any things which are left out that would be good to look over for USAMO?\r\n\r\nAlso, I own Art and Craft of Problem Solving, Math Olympiad Challenges, USSR Olympiad Problem Book, and quite a bit of MOP notes. In our opinion do you think Engel is still best to look over when preparing for the USAMO?",
  "Solution_14": "In my opinion, Engel is a bit light on geometry and inequalities, but excellent otherwise.",
  "Solution_15": "There's no magic book ... I know a lot of people who have never used an olympiad book in their lives. Personally I had Engel as a reference but didn't really use it either.\r\n\r\nIn retrospect, I agree with Engel being slightly light on Inequalities, and shortchanging Geometry a bit, but those are both more \"trainable\" topics than some others (like hard hard Number Theory and Combinatorics). There are other places to learn those things. Inequalities have been relatively easy at the IMO lately, look at this year's #3. Or, if you look through last year's IMO shortlist, A6 was a two step Holder problem, and A6 is usually supposed to be significantly hard....",
  "Solution_16": "[quote=\"DPopov\"]Also, I own Art and Craft of Problem Solving, Math Olympiad Challenges, USSR Olympiad Problem Book, and quite a bit of MOP notes. [/quote]\r\n\r\nI have all that except MOP notes. Where did u get them? and can I get them?",
  "Solution_17": "A friend lent them to me. I don't think you could get them unless you can find someone to let you borrow them.",
  "Solution_18": "I did my whole IMO training with stuff I found on the internet (because of a lack of money and resources), and, guess what, it worked :D You can find a LOT of good things on the web if you know where to look.",
  "Solution_19": "what do you mean with\r\n[quote=\"Arne\"] and, guess what, it worked :D[/quote]\r\nwhat did you get?",
  "Solution_20": "A honourable mention the first year, a comfortable silver in my second year.\r\n\r\nSo, you see, by working hard you will eventually get there - keep in mind that I had no \"good resources\" (no books, I mean) and only 2 training weekends a year. So, I had to do everything on my own, but I believed I could do it, so...",
  "Solution_21": "Wow!!! Arne amazing job.....\r\n\r\ncan you tell me the websites...."
}
{
  "Tag": [
    "linear algebra",
    "matrix",
    "vector",
    "induction",
    "algorithm",
    "pigeonhole principle",
    "combinatorics solved"
  ],
  "Problem": "We consider graphs with vertices colored black or white. \"Switching\" a vertex means: coloring it black if it was formerly white, and coloring it white if it was formerly black.\r\n\r\nConsider a finite graph with all vertices colored white. Now, we can do the following operation: Switch a vertex and simultaneously switch all of its neighbours (i. e. all vertices connected to this vertex by an edge). Can we, just by performing this operation several times, obtain a graph with all vertices colored black?\r\n\r\n[It is assumed that our graph has no loops (a [i]loop[/i] means an edge connecting one vertex with itself) and no multiple edges (a [i]multiple edge[/i] means a pair of vertices connected by more than one edge).]",
  "Solution_1": "The answer is \"yes\". The goal is to find a subset of the set of vertices ($S$) of $G$ (the given graph) s.t. each vertex in $G$ is connected to an odd number of vertices in $S$ (assume a vertex is connected to itself). As Omid said here:\r\n\r\n[url]http://www.mathlinks.ro/viewtopic.php?t=5656[/url] (the last message), this is a very famous problem. I don't know the exact solution, but I did try some things. If $A$ is the adjacency matrix of $G$ (assume $a_{ii}=1,\\ \\forall i\\in \\overline{1,n}$) then we must show that there is a $0,1$ column-vector $v$ s.t. $Av=\\left (\\begin{array}{c}1\\\\1\\\\\\vdots\\\\1\\end{array}\\right )$. In conclusion, we must show that the system is compatible. I think (although I haven't proven it) that the compatibility is equivalent with the fact that no set of lines of $A$ with an odd number of elements adds up to the $0$ line (i.e. the line with $n$ elements which has $0$ everywhere). It's easy to show that this is indeed the case for $A$, but I haven't shown that it's equivalent to the compatibility of the system mentioned above.",
  "Solution_2": "I forgot to mention a crucial thing: I've been working with matrices which have elements in $F_2$, meaning that all integers are reduced $\\mod 2$. That's why instead of a column vector having only odd entries we have a column-vector having only $1$'s as entries.",
  "Solution_3": "That is exactly what I was trying to prove....Keep searching....\r\n\r\nPierre.",
  "Solution_4": "Pfff....I'd rather look more often in my archives.... :maybe:\r\nThe result is du to Sutner (1989).\r\n\r\nReferences :\r\nK.Sutner, 'Linear cellular automata and the garden of eden', Math. Intelligencer, vol.11, n2, (1989), p.49-53.\r\nK.Sutner, ' The sigma-game and cellular automata, A.M.M., 1990, p.24-34.\r\n\r\nPierre.",
  "Solution_5": "In another hand, I would be interested about the induction proof that Omid mentioned in the other topic.\r\n\r\nPierre.",
  "Solution_6": "I remember seeing quite a short, elementary and ingenious solution of the above TST problem by one of my friends (it used induction over the number of vertices in the graph), but I can't remember it...\r\n\r\n  Darij",
  "Solution_7": "the inducthion proof is easy.assume that the hypothesis is valid for n-1 and $A$ is n*n matrice .suppress the i th row and ith column,we get n-1*n-1 matrice with same property,then according to the hypothesis there exist some column such that their sum mod 2 equal n-1 dimention vector (1,...,1),now if we add the similar columns of A we get n dimention vector s.t all its component equals 1 just maybe ith component is'nt,if there exists i such that the number of 1 in its row is odd we are done otherwise for any$1\\leq i\\leq n$ according to above we get $v(i)=(1,1,...,0,1,1,...,1)$ assume that n is even then we have $v(1)+...+v(n)\\equiv (1,...,1) means that if we add the column which is in relation to v(i) we get (1,1,...) but in this summand maybe some column is used more than 1 ,so if some column is used even times we can omit it and or we can consider column in parityso we can get (1,1,..1) and for n=2k+1 the sol is same as above ( alittle diffrent)",
  "Solution_8": "Thus it is a linear algebra induction proof. I was expecting a graph theoretic proof (even not the ingenuous and elementary one that Darij cannot remember).\r\n\r\nPierre.",
  "Solution_9": "Here is the ingenious elementary solution I promised. It was found by Peter Scholze (thanks to Peter for reminding me of his induction argument, BTW):\r\n\r\nThe answer is \"yes\". We will prove this by induction over the number n of the vertices of the graph. For n = 1, there is nothing to prove. Now turn to the step n --> n+1:\r\n\r\nAssume that the theorem holds for an n = k. Consider a graph G with k+1 vertices, all colored white. Choose an arbitrary vertex A of this graph, and let G' = G \\ {A} be the graph consisting of the remaining k vertices. Call O the operation \"switch a vertex and simultaneously switch all of its neighbours\". Now, since the theorem is assumed true for n = k, there exists an algorithm T which consists of multiple applications of the operation O and which, applied to the graph G', colors all of its vertices black. If we apply this algorithm T to the graph G' regarded as a subgraph of G, then we know that all vertices of G' become black, but we don't know what happens to the vertex A.\r\n\r\n(In the following, \"applying the algorithm T to a vertex A of G\" will mean \"applying the algorithm T to the graph G' = G \\ {A}\".)\r\n\r\nIf after our algorithm T applied to A, the vertex A is colored black, too, then our graph G is completely black, and we are done. If the vertex A remains white after the algorithm, then start again with a white graph G and try applying the algorithm T to another vertex B of G instead of A. If this doesn't work, too, (i. e. if the vertex B remains white) then start again, choose another vertex C and apply T again, etc.. If our algorithm T doesn't work for any of our vertices, then we can draw the following conclusion: For any vertex V of the graph G, there exists an algorithm T(V) which switches all vertices of G apart from V, while V retains its initial color.\r\n\r\nNow, we will distinguish between two cases: if k+1 is even, and if k+1 is odd.\r\n\r\nIf k+1 is even, then start with the white graph G, apply the algorithm T(V) to all vertices V of G one after the other, and finally each vertex gets switched k times, and, since k is odd (as k+1 is even), this means that all vertices become black. Proof complete.\r\n\r\nNow consider the case when k+1 is odd. Then, there exists a vertex A of G with even valency (the \"valency\" of a vertex is the number of edges starting at the vertex). [This is because the sum of the valencies of all vertices is even.] Let $ B_1$, $ B_2$, ..., $ B_{n}$ be all vertices of G which are connected with A. Then, $ n$ is even (since A has even valency). Let $ C_1$, $ C_2$, ..., $ C_m$ be all vertices of G which neither coincide with A nor are connected to A; in other words, let $ C_1$, $ C_2$, ..., $ C_m$ be all vertices of G except of A and $ B_1$, $ B_2$, ..., $ B_n$. Then, $ m$ is even (since\r\n$ m\\equal{}\\left(\\text{number of vertices of }G\\text{ except of }A,B_1,B_2,...,B_n\\right)$\r\n$ \\equal{}\\left(\\text{number of vertices of }G\\right)\\minus{}\\left(n\\plus{}1\\right)\\equal{}\\left(k\\plus{}1\\right)\\minus{}\\left(n\\plus{}1\\right)$,\r\nand $ k\\plus{}1$ is odd while $ n$ is even). Apply the algorithm T to each of these vertices $ C_1$, $ C_2$, ..., $ C_m$. Then, these vertices become black (since a vertex $ C_i$ gets switched each time when T is applied to a vertex $ C_j$ with $ j \\neq i$, and there is an odd number of such vertices $ C_j$). On the other hand, each of the vertices A, $ B_1$, $ B_2$, ..., $ B_n$ was switched m times and hence remains white (since $ m$ is even). Now finally apply the operation O to the vertex A (i. e. switch A and all neighbours of A), and all of the vertices A, $ B_1$, $ B_2$, ..., $ B_n$ (and only they) become black. Again, we have colored G completely black. Proof complete.\r\n\r\nIngenious, isn't it? I believe Peter was the only one to solve the problem on the exam.\r\n\r\n  Darij",
  "Solution_10": "You are right Darij, it is a very nice proof.\r\n\r\nPierre.",
  "Solution_11": "Here's how to prove the lemma:\n\nThe system $Av=u$ where $v,u$ are column-vectors and $u$ has only $1$'s is compatible iff there's no subset of lines of $A$ with an odd number of elements which add up to the $0$ line (call this property (P)). \n\nAssume the system is compatible. Then it's clear that there's no such subset because we would have an odd number of $1$'s adding up to $0$, which is false.\n\nConversely, assume there's no subset with the mentioned property. Take a minor of $A$ of order $r$ (the rank of $A$). We have to show that if we add a column with only $1$'s and a line of the matrix getting a $r+1$ minor then it must be $0$ (the system is compatible iff adding $u$ to $A$, the new matrix doesn't have a rank different from that of $A$). Let's assume that this minor isn't $0$. If we look at its $r+1$ lines without the $1$ column, there must be some subset whose sum is $0$ because otherwise we would be able to find a non-zero minor of order $r+1$ within $A$, and that's false. If this subset has an odd number of elements, then it's ok, because the $1$'s at the end also add up to $0$, so we're done. We must thus assume that this set has an odd number of elements and get a contradiction. A very important fact is that because the $r$ minor isn't $0$, it means that the subset of lines with a null sum is unique.\n\nIf we add a column from $A$ to the $(r+1)\\times r$ matrix we're currently looking at we must get a zero determinant, so there must be some subset of lines with a $0$ sum. Bu this subset must be the same as before because otherwise in our $(r+1)\\times r$ matrix we would find more than one subset of lines with $0$ sum, and I've mentioned that this is impossible. This means that the mentioned subset of lines has a $0$ sum even when we add all the columns in $A$, and we thus get a subset of lines of $A$ with a $0$ sum and an odd number of elements, which is what we wanted. We've assumed there's no such subset, so we have a contradiction, which means that the system is compatible.\n\nI think this proves the lemma. Here's how we use it to the problem: all we must show is that the adjacency matrix has property (P). This is easy to do because if we assume that lines $L_{a_1},L_{a_2},\\cdots,L_{a_{2k+1}}$ add up to $0$ then the same happens in the minor matrix formed by these lines and the columns with the same indexes, so we have a symmetric matrix of odd order which has only $1$'s on the main diagonal  s.t. its lines add up to $0$. This is false because it implies that the matrix has an even number of $1$'s, but it has an odd number of $1$'s (an even number below and above the main diagonal and an odd number on the main diagonal).\n\nI think we're done. Can you spot any mistakes? (hope it's not too unclear..)",
  "Solution_12": "Here is a solution by my friend Janos Kramar, and it is by far the most ingenious one i have seen so far.  Use induction. assume its true for n-1.\r\n\r\nNow, there are 2^n possible states of color, and each point can be either \"switched\" or not, giving 2^n possible sequences of operations. Therefore, if we cannot get them all black, there must be a set S of k points such that switching every point in S causes no change. Now, each point in S is connected to an even number of points in S, so each point in S has an odd number of neighbours in S. So |S| is even. Now, as each point is connected to an even number of points in S, the number of white vertices in S never changes parity mod 2. Now, fix a vertex V in S. If we can get all points except for V black, then V must also be black due to our parity argument. But this is doable by induction. Hence,  result.",
  "Solution_13": "Sorry for the useless post, but this is just beautiful! :D",
  "Solution_14": "A wonderful solution, indeed! I didn't ever think it could be solved using the pigeonhole principle.\r\n\r\n  dg",
  "Solution_15": "This is a similar (but easier) problem, taken from SNS admission exam, 2001. The official solution used a technique that's very similar to Lipnowski's one.\r\n\r\nIn a graph, every vertex has an on-off button, and every vertex can be white or black. If we switch a button, then the color of its vertex and all its neighbours switch, too (but the neighbours' buttons don't switch). Suppose that in our graph, all vertices are colored white, all buttons are off, and that the only way to have all vertices colored white is keeping all buttons switched off. Then, prove that we can reach every possible color configuration.",
  "Solution_16": "I think this problem was known earlier. Indeed, there was a problem at All-Russian Olympiad-1999:\r\n\r\n[b]Problem 9.4.[/b] The numbers $1,\\ldots,1000000$ are colored in two colors --- black and white. During the move we may choose a number and recolor it and all the numbers, that don't coprime with it, into opposite color. At the beginning all the numbers were black. Is there a finite sequence of moves that recolors all the numbers into white color?\r\n\r\nAuthor of the problem 9.4: S Berlov.\r\n\r\nClearly, it is a particular case of the problem (about arbitrary graph). But there was a story concerning this problem. Only one participant solved this problem at the olympiad (ARO-1999). It was Ilya Megirov. He generalized the problem 9.4 to the arbitrary graph and suggested a solution by induction (this solution has already been posted here by darij). And the committee didn't belive that his solution was correct. And the correctness of his solution was revealed only at the appeal. So, the original problem posted by darij has already been known in 1999.\r\n\r\nP.S. The story about I. Megirov and his solution I read in the journal \"Kvant\", 1999, \u2116 5, page 50. It can be found here (in pdf format, and certainly in Russian):\r\nhttp://kvant.mccme.ru/pdf/1999/05/50.pdf .\r\n\r\nThe solution of I. Megirov was also published in the same journal, see page 61. It can be found here:\r\nhttp://kvant.mccme.ru/pdf/1999/05/61.pdf .",
  "Solution_17": "Of course the problem is Iran TST 1996, so it was know since 1996. As I know the problem is from Peled (name of person), 1992.",
  "Solution_18": "[hide = Slightly Different]\nSuppose $G$ has $n$ vertices; then this becomes equivalent to choosing some subset of the $n$ vertices to mark so that every vertex is adjacent to an odd number of marked vertices, where a vertex is adjacent to itself. Call a vertex [i]lit[/i] if it is adjacent to an odd number of marked vertices, and call it [i]unlit[/i] otherwise.\n\nWe induct on $n$; clearly this is trivial for $n=1.$ Suppose that in $G$ with $n$ vertices, $2k$ of the vertices have odd degree. Consider one of these odd vertices, $V.$ If $V$ is unmarked, we can choose some subset, which we call $L(V),$ of $G\\backslash V$ to mark so that all the vertices of $G,$ except possibly $V,$ are lit. If $V$ is lit by this process, we are done, so suppose $V$ is unlit. Repeat this for every one of the $2k$ odd vertices, each time making the assumption that the $V$ we chose is unlit by marking $L(V).$ \n\nNow take the sum set $S$ of $L(V)$ over all odd vertices $V$ using binary xor (so a vertex $W$ is in $S$ iff it appeared an odd number of times among all $L(V)$). Every odd vertex is now adjacent to an $\\text{odd}\\cdot (2k-1)+\\text{even}=\\text{odd}$ number of vertices in $S,$ and every even vertex is adjacent to an $\\text{odd}\\cdot 2k=\\text{even}$ number of vertices in $S.$ (If there are no odd vertices, $S$ is simply the empty set.) So if we choose $S$ to be our marked subset, then all the odd vertices are lit and all the even vertices are unlit. Then if we take the complement of $S$ among the vertices of $G,$ every odd remains adjacent to an odd number of marked vertices and so remains lit, but every even becomes adjacent to an odd number of marked vertices as well. Therefore all the vertices of $G$ are lit, as desired.\n[/hide]",
  "Solution_19": "Induction. Assume n-1. There\u2019s two cases. For n is even, take all n combinations of n-1 vertices and apply the inductive hypothesis to each of those combinations. Every vertex is flipped an odd number of times.\nFor n is odd, there\u2019s a vertex with even degree 2k, call it v_i. Take all 2k combinations of n-1 vertices consisting of all vertices but one of v_i neighbours. Apply hypothesis to all 2k combinations, then flip v_i. The only black vertex is v_i. Apply hypothesis to subgraph that excludes v_i (if this doesnt work, then making this move to the original graph gives the result)",
  "Solution_20": "Pure linear algebra:\n\nWe prove the following claim.\n\n[u]Claim:[/u] Let $A\\in\\mathbb{F}_2^{n\\times n}$ be symmetric, and let $d=\\mathrm{diag}A\\in\\mathbb{F}_2^n$ be the diagonal of $A$. Then, $\\mathrm{diag} A\\in\\mathrm{im} A$.\n\nThe result follows then by setting $A$ to be the adjacency matrix of $G$ plus the identity matrix.\n\n[u]Proof of Claim:[/u] We show that $d\\in(\\mathrm{null}A)^\\perp=\\mathrm{im}A$. Suppose $Ax=0$. We want to show $d\\cdot x=0$.\n\nLet $I=\\{i:x_i=1\\}$, and let $B$ be the restriction of $A$ on to $I$. We have that $B1=0$ where $1$ us the all $1$s vector. Adding up all the rows of the equation $B1=0$, we see that the off diagonal terms cancel because $B$ is symmetric and we are working mod $2$, so we see that $d'\\cdot 1=0$ where $d'$ is the restriction of $d$ on to $I$, so we have $d\\cdot x=0$, as desired. Thus, the claim, and the problem are proven. $\\blacksquare$",
  "Solution_21": "Sorry for bumping but (speaking to yayups) is there motivation for using linear algebra in this scenario?",
  "Solution_22": "Well for these kind of problems, the process can be modeled with a matrix similar to the adjacency matrix, so its natural to try to do something with that. You quickly see that the problem is then equivalent to the lemma. Proving the lemma is just some linear algebra tricks.",
  "Solution_23": "We will use induction on the number of vertices. The base case is trivial; now, suppose it holds for $n-1$, and I will prove it for $n$. By the inductive hypothesis, for every vertex $V$, there is a sequence of moves that keeps $V$ the same and toggles all the other vertices (we are supposing that each sequence that toggles $G\\backslash V$ does not toggle $V$, or else we'll be immediately done). This means that for any two vertices $W, V$, after applying each of their respective sequences, $W, V$ each are toggled once while all the other vertices are toggled twice, meaning $W, V$ are toggled and everything else stays the same.\\\\\n\nThus it is possible to toggle vertices two at a time, which means that if $n$ is even then we are automatically done. If $n$ is odd, then choose vertex $V$ with even degree (there will be such a vertex because the sum of the degrees is even), and let $S$ be the subgraph of $G$ that excludes $V$ and its neighbors. Note that $S$ has an even number of vertices, so by toggling vertices two at a time, we can make everything in $S$ red while keeping $V$ and its neighbors black. Finally, just toggle $V$ and its neighbors, and we've made everything red. \n",
  "Solution_24": "We will induct on the number of vertices, with the base case $n = 1$ being trivial. Suppose that the result holds for $n - 1$ vertices, and consider adding another black vertex $v$ to the graph, and $V$ the set of $v$ and its neighbors. For brevity, let $G$ denote the original graph and $G'$ denote the union of $G$ and $v$. By the induction hypothesis, there exists a sequence of moves toggling all of the colors of $G$, meaning that it is possible to have a graph of $n - 1$ red vertices and $1$ black vertex. This likewise holds for all graphs on $i$ vertices, where $1 \\le i \\le n - 1$, by applying the induction hypothesis for $i - 1$. Hence, it is possible for $G$ to have $1$ red vertex and $n - 2$ black vertices. Let this black vertex neighbor $v$. Repeatedly applying this process to all subgraphs with $1$ vertex being a neighbor of $v$, it is possible to turn $G \\backslash V$ into all red vertices and $V$ into a set of all black vertices. To finish, toggle $v$, which turns all of its neighbors black, completing the induction. $\\blacksquare$",
  "Solution_25": "(I used red/black instead of white/black)\n\nLet $n$ be the number of vertices. We will prove this via induction on $n$. Our base case is $n = 1$, where we just flip the lone vertex. We move on to the inductive step.\n\nAssume we can make a sequence of moves for $n-1$; we prove we can do the same for $n$. For any vertex $A$, we can change the colour of all vertices except for $A$ via our inductive step, if $A$'s colour is also changed, then we're done, otherwise, $A$ must be black. Call this a supermove on $A$. If any supermove on any vertex leaves that vertex as red, we're finished with our inductive step, so assume all supermoves leaves the vertex as black. If $n$ was even, we do a supermove on each vertex. Then, each vertex will change it's colour $n-1$ times, which means each vertex will become red. Otherwise, if $n$ was odd, then let $S$ be the set of all vertices with even degree. Let $k$ be the size of $S$, we must have $k$ as odd (since the sum of all degrees is even). Then, let's do supermoves on each vertex in $S$. Each vertex in $S$ will change it's colour $k-1$ times, resulting in black, while each vertex outside of $S$ will change it's colour $k$ times, resulting in red. Finally, if we take the complement of all moves, such that for each vertex, if it was moved on, it will not be moved on, and vice versa, observe that all vertices with even degree will change their colour, while all vertices with odd degree wil not change their colour. Thus, we can take the complement of the supermoves of vertices in $S$, which will result in the vertices outside of $S$ to retain its redness, while the vertices in $S$ will become red. Thus, if $n$ is even or odd, we can turn all it's vertices red, so our inductive step is complete. We conclude that it is possible for all graphs.\n",
  "Solution_26": "[color=#00f][b]Remark:[/b][/color] Not my favorite problem :(\n\nWe will use strong induction. The base cases of $n = 1$ and $2$ are direct. Next we suppose that for some $k \\geq 3$, it is possible to make all vertices red for graphs on $\\leq k$ vertices. We wish to show that it is also always possible for graphs on $k+1$ vertices.\n\nFirst let's consider a graph $G$ on $k+1$ vertices. By the inductive hypothesis, for all vertices $v \\in G$, it is possible to toggle $G \\symbol{92} v$ to all red. If in this toggling process, $v$ is toggled as well, then we are done, so suppose otherwise. We pick two vertices $u, w \\in G$ to apply this toggling to. Note that all vertices in $G \\symbol{92} \\{u, w\\}$ are toggled twice and $u, w$ are each toggled once, so this process actually turns both $u, v$ to red and keeps all other vertices black. \n\nThus, if we are not able to immediately finish, then we can toggle two vertices at a time, which completes the case when $k+1$ is even. Otherwise, we choose a vertex $v$ with even degree, which is known to exist since sum of degrees is even, and consider the set $S \\subseteq G$ excluding it and its neighbors. $|S|$ must be even so we can apply the double toggling trick to turn all of $S$ red, and then we toggle $v$ to turn $v$ and its neighbors red, and we are done. $\\blacksquare$",
  "Solution_27": "Quite nice!\n\nWLOG, add self-loops to each vertex. To each vertex $v$, associate a binary string $N_v$, for which each coordinate is $0$ except for those vertices which are neighbors of $v$. Represent the state of the graph with a binary string, where the $i$'th coordinate is $0$ if the $i$'th color has black, and $1$ otherwise. A move then corresponds to replacing $T$ with the XOR of $T$ and one of the $N_v$. In particular, this implies that moves are commutative. We say a binary string $L$ of length $|G|$ is \\textit{good} if there is some sequence of moves which results in $G$ having state $L$. We wish to show that the string $L$ of length $|G|$ with all entries $1$ is good.\n\nNote that if two strings $L_1$ and $L_2$ are good, so is their XOR (perform the set of operations which results in $L_1$, then perform the set of operations which results in $L_2$).\n\nWe now proceed by induction on $n = |G|$, with base case $n = 1$ trivial (toggle the only vertex).\n\nConsider a graph with $n > 1$. Consider the induced subgraph formed by removing vertex $1$. By induction, there is a sequence of moves which will cause each of the remaining vertices to be colored red. If this sequence of moves also colors vertex $1$ red in $G$, we are done. Otherwise, we find that $011\\cdots 1$ is good, where there are $n - 1$ $1$'s. Repeating the argument for all vertices of $G$, we find that all binary strings with bitcount $n - 1$ are good. \n\nIn the case that $n$ is even, we are already done. Indeed, the XOR of all of these strings is also good, implying that the string $11\\cdots 1$ is good, so we are done.\n\nNow, suppose $n$ is odd. Note that an odd number of vertices have odd degree (since each vertex has a self-loop). WLOG, suppose the vertices with odd degree are $1, \\ldots, k$. Let $X$ be the result of operating on each vertex. As $X$ is the XOR of all $T_v$,\nit follows that the $i$'th coordinate of $X$ is $1$ iff the degree of the $i$'th vertex in $G$ is odd. Now, for each $1 \\leq i \\leq k$, let $S_i$ denote the string for which the only coordinate which is $1$ is the $i$'th coordinate. Then, for each $i$, let $Y_i$ be the complement of the XOR of $S_i$ and $X$. As $Y_i$ is equal to the XOR of the complement of $S_i$ and $X$, both of which are good, $Y_i$ is good as well. In $Y_i$, the $j$'th coordinate is $1$ iff the $j$'th vertex is even or $j = i$. Now, let $Y$ be the XOR of all of the $Y_i$'s; note that $Y$ is good. For each $1 \\leq i \\leq k$, the $i$'th coordinate is $1$ in exactly one of the $Y_j$'s (namely $j = i$). The coordinates corresponding to even vertices are $1$ in every $Y_i$. Thus, as there are an odd number of $Y_i$, it follows that $Y = 11\\cdots 1$ is good. This completes the proof. $\\Box$\n",
  "Solution_28": "Label the vertices with elements of $\\mathbb{F}_2$, let black be 0 and red be 1. The [i]toggling[/i] operation adds 1 to a vertex and all its neighbors. We induct on $n$, the number of vertices of $G$.\n\n[color=#00f][b]Claim:[/color][/b]  For any vertex $v$, there is a sequence of togglings that adds 1 to all but $v$, and keeps $v$ constant.\n\n[color=#00f][i]Proof:[/i][/color]  By the inductive hypothesis, there is a sequence of togglings involving vertices not $v$ that adds 1 to all vertices not $v$, but not necessarily to $v$. If $v$ also increases by 1, then all vertices are red, so done. Else, $v$ stays constant, while all other vertices increase by 1.  $\\blacksquare$\n\n[color=#00f][b]Claim:[/color][/b]  For any two vertices $u,v$, we can add 1 to just $u$ and $v$.\n\n[color=#00f][i]Proof:[/i][/color]  Add 1 to all but $u$, then add 1 to all but $v$.  $\\blacksquare$\n\nIf all degrees are odd, then for each vertex $w$, toggling all but $w$ makes $w$ increase by $\\deg w\\equiv 1 \\pmod2$, finishing. So now consider a vertex $v$ of even degree. Add 1 to all but $v$. Now, toggle $v$, so that $v$ is 1 while every vertex in the neighbor set $S$ of $v$ is 0. Since $|S|$ is even, arbitrarily pair up the vertices of $S$ and add 1 to each. Now all vertices are 1, finishing.\n\n[hide=Remarks]For the first claim, it's actually pretty hard to see that you can not control what happens exactly to $v$, and instead realize simply that if it also adds by 1, then you're done. I was trying to use other global arguments to control what happens to $v$ for too long. After realizing that this second operation exists, there is a lot more room to get more power in terms of what you can do.[/hide]",
  "Solution_29": "Short and sweet linear algebra.\n\nWork in $(\\mathbb{F}_2)^n.$ Let $A$ be a $n \\times n$ matrix representing this graph, where the $i$-th column contains $1$'s for all of $i$'s neighbors and $i,$ and $0$'s elsewhere. $A$ is symmetric and has $1$'s on the diagonal. Let $d = (1,\\ldots,1)^T$ be the all ones vector. It is sufficient to show that there exists some $x \\in \\mathbb{F}_2^n$ such that $A x = d,$ ie. $d \\in range(A) = ker(A^T)^{\\perp} = ker(A)^{\\perp}$ since $A$ is symmetric, ie. $ker(A) \\subseteq \\langle d \\rangle^{\\perp}$\nIt suffices to show that $A y = 0$ implies $d \\cdot y = 0.$\nBut $Ay = 0$ means $y^T A y = \\sum_{i,j} a_{i,j} y_i y_j = \\sum_{i} a_{i,i} y_i^2 + 2 \\sum_{i < j} a_{i,j} y_i y_j = \\sum_{i} y_i = d \\cdot y = 0,$ since we are in $\\mathbb{F}_2.$"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "inequalities",
    "algebra proposed"
  ],
  "Problem": "Let $ P$ be a complex monic polynomial with $ |P(0)|\\equal{}1$. prove there exists another complex monic polynomial $ Q$ of the same degree with $ |Q(0)|\\equal{}1$ and all roots in the unit circle (if $ z$ is a root then $ |z|\\equal{}1$)  such that for all $ x$ with $ |x|\\equal{}1$ the following inequality holds\r\n\\[ |P(x)|\\geq|Q(x)|.\\]\r\n\r\nBTW: Not related to this problem, but does someone know theorems involving inequalities and zeros of polynomials in complex disks?  or useful facts?",
  "Solution_1": "The idea is to extend or shorten the roots till they touch the unit circle and find a relation between the new and the old distance to some point of the unit circle."
}
{
  "Tag": [
    "AMC",
    "USA(J)MO",
    "USAMO",
    "AIME",
    "ARML",
    "trigonometry",
    "probability"
  ],
  "Problem": "I think this topic would be appropriate for all math competitions. But just in case there is a number I should randomly guess all the blanks with when I'm almost out of time, which one shall it be?\r\n\r\nOne USAMO qualifier suggested 0. My other classmates in the math team put in 5 for some reason.\r\n\r\nWhat are your thhoughts?",
  "Solution_1": "I would take 0",
  "Solution_2": "Good guesses are 0, 1, and 2004. On the AIME, good guesses would be factors of 2004.",
  "Solution_3": "hmm.. what's wrong with  guessing the one which happens to be the  corrrect answer too.?\r\n\r\nI mean, if you have time enough to write an answer,  why not write the correct answer? Answers do depend on well defined logic and  all you have to do is to just follow that logic  and  then you don't even have to guess. .. after all you, for crying out loud,  are not picking a number for lotto you know...\r\n\r\nJust my thoughts... :cool:",
  "Solution_4": "Gyan, don't get smart.  Inevitably, there will be a time in your life when you have to guess; we're discussing what number would be best in such a situation.\r\n\r\nI favor 501 this year.",
  "Solution_5": "[quote]Inevitably, there will be a time in your life when you have to guess;[/quote]\r\n\r\nMay be  some time in my life I have to guess in some situations,  but  I NEVER have to do that in a  AIME type, (or any otehr math test for that matter)  contest...and you know what, that principle works. I have gotten 100% in  every math tests in my school years. and believe me it was not because  of my knwoledge of \"best number to guess with\" \r\n\r\nNo I am not being smart,   I am serious, the point is trying to find a stretagy of \"guessing\"  the best answer, IMO, is pointless.  AIME type tests are not that hard  that  one would think that investing in \"better guessing stretegy\"  has better chance of return than just stuyding some serious stuff. Besides studing may help you in your life even after AIME type tests.  \r\n\r\nOkay, not everyone gets 100% in AIME but do you really think that trying to find the \"best number to guess with\" has any merit as far as your objectives are concerned?  \r\n\r\nDin't want to flame any one .. this is just my personal opinion..\r\n\r\nBottom line:   expected value of even the \"best guess\" is not going to make any significant difference in your. \r\nJust my thoughts.",
  "Solution_6": "The only reasonable way to answer this question, of course, is to go collect thousands of math competitions and check out what answers are most frequent.  (On a random note, about four years ago a senior captain of Stuyvesant math team gave a test he had written to determine the junior teams in which every single answer was either 1 or 0.)  If this is for the AIME, though, the strategy is going to suck -- there is far too high a degree of randomness in the construction of answers to make it a significantly useful strategy.",
  "Solution_7": "You can bash guessing all you want, but if you can't solve a problem or you run out of time on the AIME or some other contest, and there isn't a penalty for wrong answers, it's not very smart to leave it blank. Therefore you might as well guess something that has what you believe to be a decent chance of being correct.",
  "Solution_8": "I'm not bashing guessing -- I'm bashing the process of choosing a reasonable number to guess :).  It very much depends on the circumstances -- remember that ARML question with the heptagon?  Anyone who guessed 1001 was absurd, because it was obvious the asnwer was going to be of order 2.  They all should have guessed 1002001, but I don't see that on anyones list of numbers to guess :)",
  "Solution_9": "happy numbers like 144 or 288, with lots of factors of 2 and maybe 3. If you can narrow it down to a range of 10 numbers or so, pick the one in that range with the most factors of 2 or 3",
  "Solution_10": "There have been 27 AIME tests to date, which means 405 questions.  \r\n\r\nAnd the winner for most popular answer is (drum roll, please...) 025, which has appeared 6 times.  The runnerup is 840, which has appeared 5 times.  There is a five-way tie for third place: The answers 005, 012, 020, 061, and 400 have each appeared 4 times.\r\n\r\nAt that rate, if you choose 025, you have about a 1.5% chance of being right...\r\n\r\nThe answers 000 and 001 have appeared once each.\r\n\r\nEnquiring minds want to know.",
  "Solution_11": "[quote=\"JBL\"]I'm not bashing guessing -- I'm bashing the process of choosing a reasonable number to guess :).  It very much depends on the circumstances -- remember that ARML question with the heptagon?  Anyone who guessed 1001 was absurd, because it was obvious the asnwer was going to be of order 2.  They all should have guessed 1002001, but I don't see that on anyones list of numbers to guess :)[/quote]\r\n\r\nHaha I guessed 1001 on that question, and it didn't take me long to realize that it was a really stupid guess. (I'm pretty sure I was certain it was wrong before they called out the answer.) Did you guess 1002001? Or did you solve it properly?",
  "Solution_12": "I did it trig-wise and got the angles in terms of 7ths.  Then I said, \"Hey, these angles add to 90 degrees and aren't terribly far away from 30 and 60,\" so I plugged in 30 and 60 and it worked out to 1002001, and when I saw that, I knew I was set :)",
  "Solution_13": "You must. Definitely. Guess. 17.",
  "Solution_14": "[quote=\"Ravi B\"]And the winner for most popular answer is (drum roll, please...) 025, which has appeared 6 times. The runnerup is 840, which has appeared 5 times. There is a five-way tie for third place: The answers 005, 012, 020, 061, and 400 have each appeared 4 times. [/quote]\r\n\r\nThanks for doing the scratch work. There is very much of a spy-counterspy aspect to making tests hard to guess correct answers for. I remember once, back when I was just finishing up law school, I took a bar exam review course that promised that topic X (a topic that was fair game, according to that state's bar examiners) would never be on the bar exam. Naturally, the day I sat for the bar exam, the very first question was on topic X, which I had had the foresight to learn even though I was promised it wouldn't be on the test. \r\n\r\nSimilarly, I am SURE that every year there are SAT I sentence completion or critical reading questions that involve the words kids memorize on SAT vocabulary lists being used with unusual meanings. Just memorizing vocabulary lists is not enough. \r\n\r\nI note for the record that at least one staff member of the AMC program, namely the program's director, reads this site from time to time, so I think the AMC staff will figure out some way to make guessing unlikely to be successful on the next AIME.",
  "Solution_15": "I mainly favor 3, 5, 12, 20 and 144. But it depends on my mood and the problem. It also helps if you have to solve the measure for an angle and there's a diagram. Even though the diagram might not be to scale, if it's kind of close, it indicates a lot towards guessing.",
  "Solution_16": "0 0 and 0",
  "Solution_17": "On one math test, I only knew about half the answers, so I guessed on the other ones.  But I had some fun with it.  For example, one of my guesses was Abraham Lincoln (we could write in answers).  For another, I wrote a long thing out involving i, e, pi, square roots, and fractions.  Somehow....they missed it and marked it right.  Either that, or someone had a really really good sense of humor.",
  "Solution_18": "Why Zero? Why not 1? or -1?\r\n\r\nAlso, be sure to guess numbers relatively prime to 2004 but less than 25...\r\n\r\n[quote=\"Magnara\"]On one math test, I only knew about half the answers, so I guessed on the other ones.  But I had some fun with it.  For example, one of my guesses was Abraham Lincoln (we could write in answers).  For another, I wrote a long thing out involving i, e, pi, square roots, and fractions.  Somehow....they missed it and marked it right.  Either that, or someone had a really really good sense of humor.[/quote]\r\n\r\nyou mean a baaaaaaaaad sense of humor...",
  "Solution_19": "Also, be sure to guess numbers relatively prime to 2004 but less than 25...\r\n\r\nWhy?",
  "Solution_20": "Ooohhh !!!  so many responses in this thread. This certainly looks like fun.\r\n:D \r\nI wonder if the authors of AoPs have seen this thread.  If some one writes a book about Art of guessing the best number -  Problem Solving the easy way, little doubt that it will become  a best-seller like  any other  get_rich_quick/cheat_the_system/get_something_for_nothing_type  book. \r\n\r\n\r\nI can see it clearly now. A typical question in the contest:\r\n[quote]\nFind the number of zeros in the end of  1000! [/quote]\r\n\r\nLet me see,   as clear as a drum roll (okay I cant see a drum roll but I can  imagine the amplitude of the  sound waves it will produce) rolling after 405 AIME problems, and I answer with  unheard of 1.5% confidence level 025. Now I dont see why this has more probability to get you a point  than  say, obvious and easily  calculable answer,  of  249.  . But hey! What do I know? I am just being two smart and am bashing  this wonderful art.  :D\r\n\r\nOne can always rush and buy Dr. Matrixs book on numerology. :D   ( may be better than  following old statistics..:D) \r\n\r\n[size=125][b]But arent you selling yourselves short and aiming too low. I mean what an extra point in AIME will get you. Now  why not use the same logic i to pick the  numbers for super-lotto.[/b].[/size] :)\r\n\r\n\r\n. Okay ducking for cover now!  \r\n\r\nP.S. Giving you best try (and that includes guessing) and entering  a number rather than leaving a  blank for an answer is one thing, but   investing energy in trying to find a best  number to guess  wont, IMVHO,   make  you a better problem solver, it may not even increase your score or even get a spot on your local TV news  How I added   the number of toes  on my cat  to the age  of my crazy uncle to guess the answer . \r\n  :D  :D",
  "Solution_21": "[quote=\"Magnara\"]For another, I wrote a long thing out involving i, e, pi, square roots, and fractions.  Somehow....they missed it and marked it right.  Either that, or someone had a really really good sense of humor.[/quote]\r\n\r\nMaybe that was the correct answer  :D.",
  "Solution_22": "Really, your probablility of guessing an answer right on AIME is 1/1000, so I don't see the point of even trying to guess.  If you want to guess, just guess randomly.\r\n\r\nOn AMC, if you are pretty sure that you did really well, guess B for #25",
  "Solution_23": "Guessing on AMC 10 or AMC 12 throws away points. Guessing on the AIME, after making an honest effort, is hoping that you'll win the lottery.",
  "Solution_24": "Humans cannot generate random numbers. We always need some reason to put the answers we put down. So why not put down an answer you think might have a better chance?",
  "Solution_25": "[quote=\"tokenadult\"]Guessing on AMC 10 or AMC 12 throws away points. Guessing on the AIME, after making an honest effort, is hoping that you'll win the lottery.[/quote]\r\n\r\nI agree, that's why I said if you think you did REALLY well and losing 2.5 point is not really going to hurt you.",
  "Solution_26": "Being a Douglas Adams fan, when I have absolutely no idea whatsoever, I like to guess 42 or its variants (4,2) and 4 :sqrt: 2.\r\n\r\nOn a different note, at ARML practice last year, we did a team round problem involving a polynomial for which f(x)=1/x for all integral x between 1 and 1997.  We needed to find the value at x=1998.  I guessed zero (figuring the answer would be nice and simple); a coach pointed out afterwards that if it has nonzero rational values at 1997 points in a row, the next one is probably a nonzero rational as well.",
  "Solution_27": "[quote]a coach pointed out afterwards that if it has nonzero rational values at 1997 points in a row, the next one is probably a nonzero rational as well.[/quote]\r\n\r\nWhy stop at that? (you need not even browse in that direction), look at the  polynomial:\r\n(x*f(x)) . Isn't its value 1 for 1 to 1997, and  isn't it is easy to guess the actual value?  (of say (x-1)*(x-2)*...(x-1997)+1)\r\n\r\nPoint is, more often than one would think, the answer is nice and  simple if you happen to see it, it may be even simpler than stopping at \"art of guessing type reasoning\"   in many cases,",
  "Solution_28": "Yes, that's how you solve it. But what if you don't see how to solve it?",
  "Solution_29": "[quote]But what if you don't see how to solve it?[/quote]Most likely it is not the end of the world (and if it is then who cares!), and you may miss a point  in the test.  \r\nBut  then that is the point of the test, to see who can see how to solve it.  All is not wasted though becuase  one can use such  learnings  in the  future.",
  "Solution_30": "I'd just leave it blank on the AMCs..."
}
{
  "Tag": [
    "function",
    "calculus",
    "derivative",
    "complex analysis",
    "topology",
    "real analysis",
    "real analysis theorems"
  ],
  "Problem": "Is a function that is analytic always continuous and differentiable?\r\n\r\nIs it possible to have a differentiable function that is not analytic?",
  "Solution_1": "The answer to the first question: yes, $ f$ analytic implies $ f\\in C^{\\infty}.$ (Asking for \"continuous and differentiable\" isn't going far enough.)\r\n\r\nThe argument goes like this: analytic means locally given by a power series with a positive radius of convergence. Any power series is differentiable in the interior of its interval of convergence, with derivative given by the term-by-term derivative of the series. (Proving that is a nice exercise). But then the derived series is still a power series with the same radius of convergence, so we can repeat the process arbitrarily many times.\r\n\r\nThe converse is false. A $ C^{\\infty}$ function need not be analytic. There's a classic example which is given in several forms, slightly differing as to the details. I usually do it this way:\r\n\\[ f(x)\\equal{}\\begin{cases}e^{\\minus{}1/x},&x>0\\\\0,&x\\le 0\\end{cases}\\]\r\nThat function is $ C^{\\infty}$ everywhere but fails to be analytic at $ x\\equal{}0.$\r\n\r\nIn fact a function can be $ C^{\\infty}$ everywhere and analytic nowhere. I know I've posted my example of that somewhere here.",
  "Solution_2": "Now, if you were thinking of the complex realm...\r\n\r\nA function from $ \\mathbb{C}$ to $ \\mathbb{C}$ which is differentiable (holomorphic) in the complex sense on some open set $ U$ is in fact analytic on that set, and equal to its power series on each disk contained in $ U$. That's a theorem, and it defines how complex analysis is so different from real analysis. Just one derivative is good enough for infinite niceness."
}
{
  "Tag": [
    "geometry",
    "geometric transformation",
    "reflection",
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "If you have m \"a\"s and n \"b\"s, where m>n>0, how many different strings of length m+n can you generate such that every non-empty prefix(substring starting at the beginning) of the string has more \"a\"s then \"b\"s? For example, the smallest case is 2 \"a\"s, and 1 \"b\", where the number of possible strings is only 1 - \"aab\". Is there a formula for the general case? I don't even know how to approach the problem.",
  "Solution_1": "the problem is equivalent to finding the number of paths (consisting of \"right\" and \"up\" moves) from (1, 0) to (m, n) that don't go above the line $ y = x - 1 $...  to compute this number, we find the number of paths that *do* go above the line $ y = x - 1 $, and then subtract from the total number of paths from (1, 0) to (m, n) (which is $ {m+n-1} \\choose {n} $).\r\n\r\neach path that *does* go above $ y = x - 1 $ must meet the line $ y = x $.  for such a path, consider the first time it meets the line $ y = x $ and *reflect* the part of the path up to that point in the line $ y = x $, leaving the rest of the path intact.  this results in a path from (0, 1) to (m, n).  conversely, for every path from (0, 1) to (m, n) there's a unique path from (1, 0) to (m, n) that goes above the line $ y = x - 1 $ (obtained by reversing the above procedure).  this determines a bijection between the set of (all) paths from (0, 1) to (m, n) and the set of paths from (1, 0) to (m, n) that go above $ y = x - 1 $.  thus, the number of such paths is $ {m+n - 1} \\choose {n - 1} $.  \r\n\r\nso your answer is $ {m+n-1 \\choose n} - {m+n -1 \\choose n - 1}$",
  "Solution_2": "Thank you. Your solution works, and I understand why.   :)",
  "Solution_3": "cool.  no problem."
}
{
  "Tag": [
    "email",
    "AMC",
    "AIME",
    "induction",
    "analytic geometry",
    "trigonometry",
    "geometry",
    "\\/closed"
  ],
  "Problem": "We have now listed all classes that we plan to have available for enrollment that start in 2004.  PreTests and PostTests will soon be available for classes that do not already have them.\r\n\r\nIf you have any questions about the courses you can ask here or email me at crawford@artofproblemsolving.com.",
  "Solution_1": "[quote=\"Rep123max\"]Once the pre test comes up I'll check it, but what are the basic standars for Olympiad Problem Solving. Is it for people who are sure they will make USAMO, or also for people who think they can (but it's not a definite)?[/quote]\r\n\r\nStudents need to be able to score a 5 on the AIME or get a significant majority of the problems on the Pre Test.  The class will be challenging and will be good preparation for the Olympiads.\r\n\r\nStudents will be expected to have very good algebraic problem solving skills, know a fair amount of number theory, be able to handle most of the Intermediate Counting/Probability material, and have a basic understanding of contradiction and induction as problem solving tools.",
  "Solution_2": "is problem solving with polar coordinates in the trig and complex numbers class?",
  "Solution_3": "Polar coordinates are discussed in the complex numbers class.  One of the goals of the class is to teach students to understand the deep inter-relations between polar coordinates, complex numbers, exponential notation, and trig functions.  Many areas which are often understood as different in a standard mathematics curriculum are actually the same the thing."
}
{
  "Tag": [
    "abstract algebra",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "$ R$ is commutative ring with unity  $ T\\equal{}\\{y \\in R: \\exists t \\in R$ so that $ yt \\in S \\}$  $ S$ is multiplicative closed .suppose M is a finite generated R-module so that $ \\forall x \\in M$ we have  $ T \\cap (0 : _R  x)  \\not\\equal{} \\phi$ show that exist $ s \\in S$ so that $ \\forall y \\in M$   $ sy\\equal{}0$\r\n\r\nI try to use this: $ T\\equal{}R\\minus{} \\cup$ $ p$  s.t.  p is prime and  $ p\\cap S \\equal{} \\phi$",
  "Solution_1": "let $ M\\equal{}x_1R\\plus{}\\cdots\\plus{}x_nR$.\r\nto  $ x_1$,  we have $ y\\in T, t\\in R$, $ s_1\\equal{}yt\\in S$ ,$ s_1x_1\\equal{}0$.\r\nlet $ s\\equal{}s_1s_2\\cdots s_n\\in S$",
  "Solution_2": "the struction of $ T$ you gave is also right..since $ T$ is also mutiplicative, and statured( if $ y_1y_2\\in T\\Rightarrow y_1\\in T$ and $ y_2\\in T$),......apply ex3.7 of atiyah'book ."
}
{
  "Tag": [],
  "Problem": "How many sets of four consecutive positive integers are there such that the product of the four integers is less than $ 100,\\!000$?",
  "Solution_1": "I asked this because I didn't really understand the solution.",
  "Solution_2": "let's experiment with some sets of four consecutive positive integers\r\n\r\n$ 14\\times15\\times16\\times17 \\equal{} 57120$\r\n$ 15\\times16\\times17\\times18 \\equal{} 73440$\r\n\r\neach of those fit the restrictions set\r\n\r\nhowever, we find that $ 17\\times18\\times19\\times20 \\equal{} 116280$, which is greater than $ 100000$\r\n\r\nthus, the sets are $ (1,2,3,4), (2,3,4,5).....(16,17,18,19)$\r\n\r\nwe can immediately see that there are $ 16$ sets that work",
  "Solution_3": "How do you do this w/o a calculator?",
  "Solution_4": "So we want to approximate 100,000^1/4 and center our set around that.\n100,000^1/4 = 10*10^1/4\n\n10^1/4 is approx 3.16^1/2 which is approx 1.775\n\n[hide]\n\nSo we end up with a center of about 17.7, so we consider the set (16,17,18,19). 16*17*18*19=93024<100,000, so we try (17,18,19,20)\n\n17*18*19*20=116280>100,000, so the sets we want are bounded by (1,2,3,4) and (16,17,18,19), inclusive. This yields 16 sets.\n\n[/hide]",
  "Solution_5": "I don't understand and I don't think that yall are conting \n1*1*1*999999\n1*1*1*999998\n1*1*1*999997\n1*1*1*999996\nand so on that is so much more!! than 16 !!! Help me!!!",
  "Solution_6": "[hide=solution]What I did was first note that product of the largest must be less than 316. So 17*18. 15*16*17*18, and there are 15 from 1*2*3*4. But since 16*17*18*19 also works, it's 17.[/hide]\n\nIs there a better solution though? Because I got it correct on second try, since you had to notice that 16*17*18*19 also works.",
  "Solution_7": "[quote=Sippi]I don't understand and I don't think that yall are conting \n1*1*1*999999\n1*1*1*999998\n1*1*1*999997\n1*1*1*999996\nand so on that is so much more!! than 16 !!! Help me!!![/quote]\n\nthese are not consecutive integers",
  "Solution_8": "Dude, he/she posted 2 years ago.",
  "Solution_9": "It seems a little ridiculous to have to manually compute $16*17*18*19$. ",
  "Solution_10": "Yes, there is nothing wrong with using a calculator, as far I know\n\n~ :heli:"
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "Let a,b,c be positive real numbers such that \r\na+b+c=3 Prove that\r\n[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/b2f603d58bdfdfafa1feb3b4506ad7b7.gif[/img] ;)",
  "Solution_1": "[quote=\"suweijie\"]Let a,b,c be positive real numbers such that \na+b+c=3 Prove that\n[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/b2f603d58bdfdfafa1feb3b4506ad7b7.gif[/img] ;)[/quote]\r\nweighted jensen on $f(x)=\\frac{1}{x^2+14x+1}$ and use $(a+b+c)^2 \\ge 3(ab+bc+ca)$.",
  "Solution_2": "First I don't understand what you meant,but I have a much easier proof. :D",
  "Solution_3": "Just use the  fact:[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/d36e36d8f23b960b10352aa37430801e.gif[/img] :D"
}
{
  "Tag": [
    "trigonometry",
    "calculus",
    "calculus computations"
  ],
  "Problem": "can any one please evaluate the inverse of this transform\r\n\r\n(s+1)/(s^2+4s+6)",
  "Solution_1": "You just have to complete the square in that denominator.\r\n\r\n$\\frac{s+1}{s^{2}+4x+6}= \\frac{(s+2)-1}{(s+2)^{2}+2}$\r\n\r\n$=\\frac{s+2}{(s+2)^{2}+2}-\\frac1{\\sqrt{2}}\\cdot\\frac{\\sqrt{2}}{(s+2)^{2}+2}$\r\n\r\nWe can now read off the inverse Laplace transform as\r\n\r\n$e^{-2t}\\left(\\cos(\\sqrt{2}t)-\\frac1{\\sqrt{2}}\\sin(\\sqrt{2}t)\\right)$",
  "Solution_2": ":D  cheers kent",
  "Solution_3": "I debated whether to move your three topics to the \"Calculus Computations and Tutorials\" section but decided to do so for two reasons.\r\n\r\n1. The questions seem to fit the spirit of \"computations and tutorials.\"\r\n\r\n2. I tend to say that differential equations (ODE's - not PDE's) are part of calculus, and then I suppose Laplace transforms are part of differential equations.\r\n\r\nBut it's a borderline case."
}
{
  "Tag": [],
  "Problem": "Prove, that for any natural $n\\geq 2$ the equation:\r\n$x_1+x_2+\\cdots+x_n=x_1x_2\\cdots x_n$ has solution is positive integers.",
  "Solution_1": "For $n\\geqslant 3$:\r\n\r\n$\\underbrace{1+1+...+1}_{n-2}+2+n=\\underbrace{1\\cdot 1\\cdot...\\cdot 1}_{n-2}\\cdot 2\\cdot n$\r\n\r\nFor $n=2$:\r\n\r\n$2+2=2\\cdot 2$",
  "Solution_2": "In fact, there are infinite solutions for each $n$.  For any given set of numbers $x_1, x_2, ...$ such that their product is greater than their sum, a solution can be found simply by adding $1$s to the sequence until the sum is equal to the product.  :)",
  "Solution_3": "That doesn't work, because you add additional terms and thus increase the size of $n$.  In fact, there are only finitely many solutions for each $n$.",
  "Solution_4": "Oops.  Righty-o.  Of course, as the $x_k \\to \\infty$ the product becomes much larger than the sum... silly me  :blush: \r\n\r\nIt remains that there are more nontrivial solutions than the ones given, though.  :P",
  "Solution_5": "How to find other non-trivial solutions :?:",
  "Solution_6": "[quote=\"Farenhajt\"]For $n\\geqslant 3$:\n\n$\\underbrace{1+1+...+1}_{n-2}+2+n=\\underbrace{1\\cdot 1\\cdot...\\cdot 1}_{n-2}\\cdot 2\\cdot n$\n\nFor $n=2$:\n\n$2+2=2\\cdot 2$[/quote]\r\n\r\nJust nice!",
  "Solution_7": "[quote=\"dondigo\"]How to find other non-trivial solutions :?:[/quote]\r\n\r\nTake any collection of integers - say, $2, 6, 10, 17$.  Call $P = 2040$ their product and $S = 35$ their sum.  Now clearly $P > S$.  So we add $P-S = 2005$ $1$s to the collection; then the sum increases to $P$ but the product does not change.  :)",
  "Solution_8": "[quote=\"t0rajir0u\"]Take any collection of integers - say, $2, 6, 10, 17$.  Call $P = 2040$ their product and $S = 35$ their sum.  Now clearly $P > S$.  So we add $P-S = 2005$ $1$s to the collection; then the sum increases to $P$ but the product does not change.  :)[/quote]\r\n\r\nI might be missing something, but as far as I can see, after adding $P-S$ the sum will be $P$ and the product $P(P-S)$. How these two can be equal for $P-S\\neq 1$?\r\n\r\nEdit: $P-S$ [u]ones[/u]... Almost missed it :)",
  "Solution_9": "he means $P-S$ unique terms each of value one.",
  "Solution_10": "See also : http://www.mathlinks.ro/Forum/viewtopic.php?t=63671. :)"
}
{
  "Tag": [
    "MATHCOUNTS"
  ],
  "Problem": "Hello,\r\n\r\nThis pretends to create a nice discussion, but it is not really a \"Serius Discussion\", hence, I believe it belongs more here (Games and Fun Factory) than to Round Table.\r\n\r\nI had a collection of currency of different countries since at least 1987, and it has grown a lot (unfortunatelly the 95% of it is three thousand miles away) and I have noticed that regularly most of the currency denominations are of the form $2^n5^m$, where most of the time $|m-n|\\leq2$, for example, the  currency denomiations in Costa Rica are 1, 5, 10, 25, 50, 100, 500, 1 000, 2 000, 5 000 and 10 000. In the US, besides the cents coins, those are 1, 2, 5, 10, 20, 50, 100, 500, 1 000, 5 000 and 10 000, but the last four are pretty difficoult to see and are not longer in print (I know that some people say that it is not true, but go to the page of the US Department of Theasury, and you will see that it is true, they exist)\r\n\r\nNevertheless, I have seen some denominations out of common. I remember some 13 pesos and other of 12 cents, from the 1920s or so, I do not remember the countries, but they exist. Does anyone knows about any other 'Odd' denominations, specially if they are currently in circulation?\r\n\r\nBest regards,",
  "Solution_1": "I have some really old english coins that are 6 pence and 1/2 penny, but I am pretty sure that they are no longer circulated.",
  "Solution_2": "Sixpence made sense pre-1971; it was half a schilling. You see, back then it was:\r\n\r\n1 pound = 20 schillings = 240 pence = 960 farthings.",
  "Solution_3": "way back in US history, there was a 1/2 cent coin, 3 cent coin, 3 dollar coin (gold) There were plans for a 4 dollar coin that was abandoned. These are all no longer in circulation, but...",
  "Solution_4": "There's a cool picture in my economics book of these people on an island who use massive, circular stones as money.",
  "Solution_5": "[quote=\"miraculouspostmaster\"]There's a cool picture in my economics book of these people on an island who use massive, circular stones as money.[/quote]\r\n\r\nIt would probably make quite difficoult to have a couple of bucks on your wallet  :D  :P",
  "Solution_6": "[quote=\"cardwacko9999\"]way back in US history, there was a 1/2 cent coin[/quote]\r\n\r\nReally?  I was under the impression that the 'mill' was never produced.\r\n\r\nIt was proposed in the Great Depression.",
  "Solution_7": "[quote=\"cardwacko9999\"]way back in US history, there was a 1/2 cent coin, 3 cent coin, 3 dollar coin (gold) There were plans for a 4 dollar coin that was abandoned. These are all no longer in circulation, but...[/quote]\n\nI don't know if you read the  [url=http://www.auburn.edu/~vestmon/Gift_of_the_Magi.html]Gift of the Magi[/url]  (O.Henry)  which starts:\n[quote] One dollar and[b] eighty-seven[/b] cents. That was all. And [b]sixty cents[/b] of it was in pennies. Pennies saved one and two at a time by bulldozing the grocer and the vegetable man and the butcher until one's cheeks burned with the silent imputation of parsimony that such close dealing implied. Three times Della counted it. One dollar and eighty- seven cents. And the next day would be Christmas.[/quote]\r\n\r\nMathcounts students always wondered .. How to have  \"One dollar and twenty-seven cents\" without having pennies...????\r\n[b]Mystry solved!![/b]\r\nFrom googling: ..The US produced a 2-cent coin for general circulation up until 1872 and a 3-cent coin until 1889. (Either of these would allow the math to work) there were also half-cent pieces produced from 1794 until 1857,..  O. Henry lived from 1862 to 1910..",
  "Solution_8": "Israel definitely has a 1-2 coin",
  "Solution_9": "Is this odd = strange or odd= not even?",
  "Solution_10": "[quote=\"Danbert\"][quote=\"cardwacko9999\"]way back in US history, there was a 1/2 cent coin[/quote]\n\nReally?  I was under the impression that the 'mill' was never produced.\n\nIt was proposed in the Great Depression.[/quote]\r\n\r\nI believed it was called a pence",
  "Solution_11": "isnt that in europe?\r\n\r\nalong with the euros, etc",
  "Solution_12": "For some clarification... :)\r\n\r\nUS started minting its own coins in 1792. Of course, they made very few that year (mostly test coins). Regular circulation coins, more like 1794, 1796, dunno. At that time, there were minted half cent, one cent, \"half dime\", dime, quarter, half dollar, dollar, and in gold 2.50, 5, 10. People sometimes cut a quarter into 2 and used each piece as 12.5 cents. (It is the reason for the name \"two bits\" for a quarter. Actually, that comes from an earlier practice of cutting mexican 8 real coins into eighths (\"pieces of eight\").) \r\n\r\nLater in the 1800s, there are some changes. Around 1857, they stop the 1/2 cent, and they change the one cent to a much smaller coin. Not long after, they start making the 3 cent coins (in copper-nickel and in silver) and change the silver half dime for the nickel. Around the same time (1850s), they also made some interesting endeavors with gold coins. They have a gold 1 dollar for a short time. The 3 dollar lasts for like 20 yrs at least (though not many are minted). They actually made a 4 dollar coin, \"Stella\", but there are VERY few (test coins, probably). Also, the 20 dollar coin started around then. \r\n\r\nIn the 1870s, for a short time there was a 20 cent coin. \r\n\r\nUnfortunately I don't know the reasons for most of these \"odd\" denominations.\r\n\r\nThe UK used their English system for a long time. They had many weird denominations. There are 12 pence to a shilling, 20 shillings to a pound. Coins were usually from 1 farthing (1/4 penny) to a pound, but there were 1/4, 1/3, 1/2 farthing in the mid 1800s (circulated in the colonies, probably because money had more buying power there). They had farthing, half penny, penny, three pence, six pence, shilling, 2 shillings (florin), 2.5 shillings (half crown), 5 shillings (crown), pound, and guinea (21 shillings). For some reason, the guinea was more popular among rich people. I think it arose as a result of some tax which was 5%. In 1887-1889, there was a failed experiment where they minted a double florin (4 shillings). I have no idea why. There were also 4 pence (aka groat, hilarious name) which stopped circulating in 1800s, and there existed 2 pence and 2.5 pence that (as far as I know) were only in Maundy sets (for collectors) in recent times (i.e. since late 1700s). \r\n\r\nI am less knowledgeable about other countries. I think I have a 9 kreuzer coin from some central European country, though. Austria, maybe? It was from the 19th century."
}
{
  "Tag": [
    "function",
    "calculus",
    "integration",
    "geometry",
    "rectangle",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Hello: I need a hint for showing the following\r\nGiven f_n:A->R a sequence of continous and integrable (in the extended sense) functions who uniformly converge in compact sets\r\nto f:A->R, show that \r\nlim_n integral over A (f_n) = integral over A (f)   [/code]",
  "Solution_1": "This is not true with $A=\\mathbb R$ and $f_{n}(x)=\\frac{1}{(x-n)^{2}+1}$. The sequence converges to $0$ uniformly on compact sets, yet the integral of $f_{n}$ is $\\pi$ for any $n$.\r\n\r\nThe conclusion is true if the sequence is [i]uniformly integrable[/i], that is, for every $\\epsilon>0$ there exists $M$ such that $\\sup_{n}\\int_{-M}^{M}|f_{n}(x)|dx<\\epsilon$",
  "Solution_2": "If the sequence were uniformly integrable how would the proof go?",
  "Solution_3": "If $A$ $\\subseteq$ $\\mathbb R^{n}$ for $n>1$ what would be the condition of uniformly integrable for the sequence? Perhaps:\r\n$\\forall$$\\epsilon>0$,$\\exists Q$, rectangle, such that\r\n$\\sup_{n}\\int^{}_{Q}\\left|f_{n}\\right|<\\epsilon$  ?"
}
{
  "Tag": [
    "quadratics",
    "algebra",
    "polynomial",
    "algebra proposed"
  ],
  "Problem": "Suppose that $ a$ and $ b$ are real numbers such that the quadratic polynomial $ f(x)\\equal{}x^2\\plus{}ax\\plus{}b$ has no nonnegative real roots. Prove that there exist two polynomials $ g,h$ whose coefficients are nonnegative real numbers such that: $ f(x)\\equal{}\\frac{g(x)}{h(x)}$ for all real numbers $ x$.",
  "Solution_1": "Even more is true actually.\r\nTake a look at http://www.mathlinks.ro/viewtopic.php?p=1219057#1219057"
}
{
  "Tag": [
    "inequalities proposed",
    "inequalities"
  ],
  "Problem": "Let $ a,b,c>0$. Prove that\r\n\\[ \\left(\\frac{a+b+c}{3\\sqrt[3]{abc}}\\right)^{3}\\le \\left(\\frac{a^{2}b+b^{2}c+c^{2}a}{3abc}\\right)^{4}\\]",
  "Solution_1": "I think that's wrong.\r\nTry $ a=1, b=3, c=2$",
  "Solution_2": "Now it is correct. Thank you very much Yulia for pointing out the mistake! :lol:",
  "Solution_3": "[quote=\"nayel\"]Let $ a,b,c>0$. Prove that\n\\[ \\left(\\frac{a+b+c}{3\\sqrt[3]{abc}}\\right)^{3}\\le \\left(\\frac{a^{2}b+b^{2}c+c^{2}a}{3abc}\\right)^{4}\\]\n[/quote]\r\n\r\n$ (\\frac{a}{c}+\\frac{b}{a}+\\frac{c}{b})^{4}\\geq 3(\\sqrt[3]{\\frac{a^{2}}{bc}}+\\sqrt[3]{\\frac{b^{2}}{ac}}+\\sqrt[3]{\\frac{c^{2}}{ab}})$\r\nLet \r\n$ x^{3}=\\frac{a}{c}$\r\n$ y^{3}=\\frac{b}{a}$\r\n$ z^{3}=\\frac{c}{b}$\r\nNote that $ xyz=1$\r\nThen we want to prove\r\n$ (x^{3}+y^{3}+z^{3})^{4}\\geq 3(\\frac{x}{y}+\\frac{y}{z}+\\frac{z}{x})^{3}=3(x^{2}z+y^{2}x+z^{2}y)^{3}$\r\n\r\n$ x^{3}+y^{3}+z^{3}\\geq 3xyz=3$, by AM-GM\r\n\r\n$ x^{3}+y^{3}+z^{3}=(\\frac{x^{3}}{3}+\\frac{x^{3}}{3}+\\frac{z^{3}}{3})+(\\frac{y^{3}}{3}+\\frac{y^{3}}{3}+\\frac{x^{3}}{3})+(\\frac{z^{3}}{3}+\\frac{z^{3}}{3}+\\frac{y^{3}}{3})\\geq  x^{2}z+y^{2}x+z^{2}y$, by AM-GM  :)",
  "Solution_4": "[quote=\"nayel\"]Let $ a,b,c>0$. Prove that\n\\[ \\left(\\frac{a+b+c}{3\\sqrt[3]{abc}}\\right)^{3}\\le \\left(\\frac{a^{2}b+b^{2}c+c^{2}a}{3abc}\\right)^{4}\\]\n[/quote]\r\n\r\n\\[ \\left(\\frac{b}{a}+\\frac{c}{b}+\\frac{a}{c}\\right)^{4}\\ge \\left(\\frac{a+b+c}{\\sqrt[3]{abc}}\\right)^{4}\\ge 3\\left(\\frac{a+b+c}{\\sqrt[3]{abc}}\\right)^{3}\\]"
}
{
  "Tag": [],
  "Problem": "Let  f(x),  x\u03b5[0,1] with f'(x)>=M>0  \r\nShow that there is a space I, in the [0,1] , with length  1/4 : |f(x)|>=M/4 for all x whose belongs to I",
  "Solution_1": "Note that f is increasing, so that it suffices to find x >= 1/4 s.t. f(x) <= -M/4 or x <= 3/4 s.t. f(x) >= M/4. Suppose that is not true... then in [1/4, 3/4] f is between -M/4 and M/4. This is contradicted by the mean value theorem, as |f(3/4) - f(1/4)|/(3/4 - 1/4) < |M/4 + M/4|/(1/2) = M, so that some value t in [1/4, 3/4] has f'(t) < M."
}
{
  "Tag": [
    "USAMTS",
    "LaTeX"
  ],
  "Problem": "Hi,\r\n\r\nI wish to type up my solutions using LaTeX this year.\r\n\r\nIn order to do so, I learned that I must download fancyhdr.sty.\r\n\r\nHowever, when I click the the link 'Click here to download fancyhdr.sty', I see a [url=http://www.artofproblemsolving.com/LaTeX/Images/fancyhdr.sty]text page[/url] instead of a file download window. (In [url=http://usamts.org/Forms/U_Forms.php]this[/url] page)\r\n\r\nCould AoPS fix this?\r\n\r\nThanks. :)",
  "Solution_1": "Well your browser must be interpreting it as text.\r\n\r\nSimply right-click on the link to fancyhdr.sty and choose \"Save Link\" or \"Save Link As\" to save it.",
  "Solution_2": "Thanks a lot. :)\r\n\r\nEDIT: Um, am I supposed to download the file the way dysfunctionalequations mentioned?\r\n\r\nI did what he said and when I compiled it, it came out as attached... :|",
  "Solution_3": "Did you also download usamts.tex",
  "Solution_4": "Yes. I downloaded both solutions.tex and usamts.tex.",
  "Solution_5": "Did you put your solutions and the fancyhdr file in the same folder?",
  "Solution_6": "Yes...\r\n\r\nHuh, when I put all three files in my folder, it says it has 16 errors, and when I put it on the desktop, it says it has 0 errors... (although it says the file does not exist)\r\n\r\nEDIT: Okay, now it just says the file does not exist when I compile it and see it...",
  "Solution_7": "This depends on what broswer your using, but when you right click and save fancyhdr, make sure you have the filename as \"fancyhdr.sty\".  For example, in google chrome it will save the file as \"fancyhdr\", and you won't get what you want without the .sty.",
  "Solution_8": "Ugh, still the same...\r\n\r\nI downloaded solution.tex, usamts.tex, and fancyhdr.sty, and it says it has 16 errors :stink:",
  "Solution_9": "Mine has errors to (about 10, but there is only 1 problem complete), but it turns out fine.  YOu could take the template for handwritten solutions and make it a word doc. and type everything up there.",
  "Solution_10": "Don't do that... that is a very bad idea. First of all, word docs are not accepted. Secondly, even if you print them out and submit them by mail, the typesetting looks... very bad. Also, submitting by mail is a pain. I guess you could possibly write the document to a PDF file, but the typesetting would still look... very bad.\r\n\r\nI would suggest that you redownload the solutions.tex file and try compiling it again. If it doesn't work, try redownloading all three files. If that doesn't work, reinstall LaTeX.",
  "Solution_11": "Reinstalled LaTex.\r\n\r\nStill doesn't work...\r\n\r\nEDIT: Phew, I got it now.\r\n\r\nThanks for the help!  :)",
  "Solution_12": "[quote=\"dysfunctionalequations\"]Don't do that... that is a very bad idea. First of all, word docs are not accepted. Secondly, even if you print them out and submit them by mail, the typesetting looks... very bad. Also, submitting by mail is a pain. I guess you could possibly write the document to a PDF file, but the typesetting would still look... very bad.[/quote]\r\nYou don't [i]have[/i] to use LaTeX. I have it, but I prefer Word. You can just use the equation editor in word, if it has one. You don't need LaTeX at all. Also, you can print it as a PDF file, and it will come out fine (the Word equations, not the LaTeX). Then you have what you need, and don't need to mail.",
  "Solution_13": "Yeah...Word's equations are much easier to type, and you can easily convert a word document to pdf and then upload it.",
  "Solution_14": "[quote=\"Math Champion\"]Yeah...Word's equations are much easier to type, and you can easily convert a word document to pdf and then upload it.[/quote]\r\n\r\nIt's up to the user, but in terms of a document dominated by equations, I find LaTeX a lot easier to use because the keyboard is so much quicker than clicking the mathematical symbols and typing, and then pressing the mouse, and then typing, etc. Also, Word's equation-rendering IMO is a bit ugly.",
  "Solution_15": "What's an IMO? I don't think you mean the math competition.",
  "Solution_16": "IMO=In My Opinion",
  "Solution_17": "Oh... :blush:"
}
{
  "Tag": [
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "Let $ a,b,c>0$.Prove that:\r\n$ \\sqrt{(a\\plus{}b\\plus{}c)(\\frac{1}{a}\\plus{}\\frac{1}{b}\\plus{}\\frac{1}{c})}\\geq 1\\plus{}\\sqrt{1\\plus{}\\sqrt{(a^2\\plus{}b^2\\plus{}c^2)(\\frac{1}{a^2}\\plus{}\\frac{1}{b^2}\\plus{}\\frac{1}{c^2})}}$.",
  "Solution_1": "See here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=82868"
}
{
  "Tag": [],
  "Problem": "1) A traveling salesman has to visit six of the ten cities in his territoriy. Cities A and B are two cities among the ten. If A and B are to be included in the itinerary what is the number of ways? In how many of these will A come before B?\r\n\r\n2) A delegation of 4 students is to be chosen from among 12 for a conference. How many ways if 2 particular students refuse to attend the meeting together? How many ways if 2 students who are married to each other will attend on if they both can go? (Note: Are these the same question??? I was thinking of 10 choose 2 = 45 ways for these two questions...)",
  "Solution_1": "1.\r\n[hide] 8C4*6! (take 4 from the remaining 8, and permute them). Half of them.[/hide]\n\n2.\n[hide] 11C4 (with one student) + 11C4 (the other student) - 10C4 (those without either is double counted)\n10C2 (if we choose the pair) + 10C4 (if we don't)[/hide]",
  "Solution_2": "[quote] 1) A traveling salesman has to visit six of the ten cities in his territoriy.\nCities A and B are two cities among the ten.\n(a) If A and B are to be included in the itinerary, what is the number of ways?\n(b)In how many of these will A come before B?[/quote][hide]\n(a) Since A and B are already selected, we must choose 4 more from the other 8 cities.\n[color=yellow]. . . [/color]There are $C(8,4) = 70$ choices for cities W,X,Y,Z.\n[color=yellow]. . . [/color]Then these six cities can be ordered in $6!= 720$ ways.\nTherefore, there are $70 \\times 720 = 50,400$ ways.\n\n(b) Half of them: $25,200$ ways.\n[color=yellow]. . . [/color][I have also worked this out \"the long way\".][/hide]\n[quote]2) A delegation of 4 students is to be chosen from among 12 for a conference.\n(a) How many ways if 2 particular students refuse to attend the meeting together?\n(b) How many ways if 2 students who are married to each other will attend only if they both can go?[/quote][hide]\nThere are $C(12,4) = 495$ possible delegations.\n\n[b](b)[/b] There are two cases:\n[color=yellow]. . . [/color][1] A and B are both selected.\n[color=yellow]. . . . . .[/color]We must select the other 2 delegates from the remaining 10 students.\n[color=yellow]. . . [/color]Hence, there are $C(10,2) = 45$ ways for both A and B to be in the delegation.\n\n[color=yellow]. . . [/color][2] Neither is selected.\n[color=yellow]. . . . . . [/color]We must select 4 delegates from the remaining 10 students.\n[color=yellow]. . . [/color]Hence, there are $C(10,4) = 210$ ways in which neither A and B is selected.\n\nTherefore, there are $45 + 210 = 255$ ways for A and B to be \"together\".\n\n\n[b](a)[/b] And therefore, there are $495 - 255 = 240$ ways for A and B to be separated.[/hide]\r\n\r\n\r\nEdit: I [b]love[/b] your one- and two-line solutions, kueh!"
}
{
  "Tag": [
    "vector",
    "linear algebra",
    "linear algebra theorems"
  ],
  "Problem": "Let $V$ and $W$ be finite dimensional vector spaces over $F$ which are dual with respect to a bilinear form $B$.\r\nIf $V_{1}$ be a subspace of $V$ and $V^{\\bot}_{1}$ be the [u]annihilator[/u] of $V_{1}$ with respect to $B$.\r\n\r\nI want to show the following result:\r\n\r\n$\\boxed{dim\\, V_{1} \\,+dim\\, V^{\\bot}_{1} \\, = dim \\, V }$\r\n\r\nIn my book we take a base for $V_{1}$, extend it to a base for $V$ and  since $W \\cong V^{*}$ we find a base of $W$ \r\ncorresponding to the dual base of $V^{*}$ relative to the base of $V$ etc.\r\n\r\nI tried the the following:\r\n\r\nThe vector spaces $V_{1}$ and $W/V^{\\bot}_{1}$ are dual with respect to the bilinear form $B_{1}$ defined by\r\n\r\n$B_{1}(v_{1},[w])=B(v_{1},w) \\, ,v_{1} \\in V_{1} \\, ,[w] \\in W/V^{\\bot}_{1}$  therefore  $dim\\, W/V^{\\bot}_{1} = dim\\,V_{1}$.\r\n\r\nBut $dim\\, W/V^{\\bot}_{1} = dim\\, W - dim\\, V^{\\bot}_{1}$ and $dim\\, W = dim\\, V$ so $dim\\, V -dim\\, V^{\\bot}_{1} = dim\\,V_{1}$.\r\n\r\nIt is wrong ?? Because i told that to my teacher and he didnt liked it  :?",
  "Solution_1": "Well i am almost sure it is correct and maybe it's trivial but i need a verification from someone who knows the subject  :)\r\n\r\nAnd if it is not correct please explain me why  :)"
}
{
  "Tag": [],
  "Problem": "\u0633\u0644\u0627\u0645 . \u0627\u0632 \u0647\u0645\u06af\u06cc \u062e\u0648\u0627\u0647\u0634 \u0645\u06cc \u06a9\u0646\u0645 \u0647\u0631 \u06a9\u06cc \u0633\u0648\u0627\u0644\u0627\u062a \u0645\u0631\u062d\u0644\u0647 \u0633\u0648\u0645 \u0627\u0645\u0633\u0627\u0644 88 \u0631\u0627 \u062f\u0627\u0631\u0647 \u0628\u0630\u0627\u0631\u0647. \u062e\u0648\u0627\u0647\u0634 \u0645\u0646\u062f\u0645 \u0648 \u0634\u062f\u06cc\u062f \u0634\u062f\u06cc\u062f \u0646\u06cc\u0627\u0632 \u062f\u0627\u0631\u0645 . \u0633\u0648\u0627\u0644\u0627\u062a \u062e\u0644\u0627\u0642\u06cc\u062a \u0648 \u062a\u0631\u06a9\u06cc\u0628\u06cc\u0627\u062a \u0648 \u0647\u0646\u062f\u0633\u0647 \u0648 \u062c\u0628\u0631 \u0648 \u062a\u0648\u067e\u0648\u0644\u0648\u0698\u06cc. \u0628\u0627 \u062a\u0634\u06a9\u0631",
  "Solution_1": "mitoni beri to resources sayt!",
  "Solution_2": "hendese ro khodam post kardam ([url=http://www.mathlinks.ro/viewtopic.php?t=299851]inja[/url]).\r\nnazariye adad ham ghablan post shode ([url=http://www.mathlinks.ro/viewtopic.php?t=295800]inja[/url]).\r\njabr ro ham type kardam [url=http://www.irysc.com/ftopict-1587.html]inja[/url] (bazam mazerat mikham chon behtar nashode).\r\nmimune tarkibiat ke type shode hast va baz ham sharmande ke behtar az in nashod...\r\nbe tartibe safahat:\r\nuntitled , untitled 1 , untitled 2, untitled 3...",
  "Solution_3": "mamnoon !! dastet dard nakone . :lol:",
  "Solution_4": "khahesh mikonam, inam akharin safhe ke forsat nashode bud bezaram..."
}
{
  "Tag": [],
  "Problem": "Who do think is going to win?\r\nI say MENS: Roddick, Federer, or Safin if he plays well.\r\n        WOMENS: Henin-hardene or sharapova",
  "Solution_1": "MENS: Federer or Hewitt\r\nWOMAN:H-H or Williams, if she comes back.\r\n\r\nRoddick has to prove that he is out of the major-less slump.... or whether he could be the player he was when he used to be No.1 in the world.\r\n\r\nNadal has to prove that he is better than what people think he is [b] outside [/b] of clay.\r\n\r\nSafin has to prove that he can control himself.\r\n\r\nAs for the woman's side:\r\n\r\nSharapova needs to prove that she can beat H-H or Serena if she were to face them before finals, or even before semi.\r\n\r\nPersonally, I dont think Davenport deserves to be No. 1, and she would have to prove that she is, by winning the Wimbledon.\r\n\r\n\r\nBut my picks still stands as it is.",
  "Solution_2": "MEN: Federer... I don't even think I need another name here, but we'll see. Next would be Roddick or Hewitt.\r\nWOMEN: Henin-Hardenne or Sharapova\r\n\r\nFederer is just unstoppable on grass and, frankly, he's due for a Grand Slam now :P. I'd like to see Roddick make it to the semis at least since grass is one of his better surfaces, and Hewitt is just fast so he's always a possibility imo. Anyway, Federer just took out Safin in a tournament, so I'd say Safin has relatively little chance unless some fluke knocks Federer out before he plays Safin. I doubt either of the two other players I mentioned could beat Federer either, so yeah, we'll see...\r\n\r\nOn the women's side, I'd go for Sharapova but she's really young, so who knows. Henin-Hardenne has done pretty well coming back, but I don't know how she'll fair on grass. Then again, I don't pay attention to the women as much so I wouldn't be one to call it.",
  "Solution_3": "When does it start/when does the coverage of it start?",
  "Solution_4": "[quote=\"WindSlicer\"]One does it start/when does the coverage of it start?[/quote]\r\n\r\nJune 20, so about a week from now... yay! :D",
  "Solution_5": "Awesome, can't wait. Just started playing tennis, and when I watch them play I improve a lot, like after the French Open I played tons better.\r\n\r\nOn a side note, I think it's funny how sometimes we type stuff depending on how sometimes it sounds in our head.. i.e. how I put \"one\" instead of \"when\". *shrugs* heh",
  "Solution_6": "Hmm...someone with more tennis experience could probably shed more light on this, but in my opinion Sharapova is not experienced enough to be reliably consistent in major tournaments like Wimbledon.\r\n\r\nOn the other side, Federer is probably my pick for most likely to win. A few of my friends got his autograph when they were playing in Houston. :-) He seems like a nice fellow actually.",
  "Solution_7": "sharapova is young but she is more than fit and perfect for W.\r\nand she super hot",
  "Solution_8": "hopefully, federer will be able to tackle the grass surface better than he did at the French open. personally, i want roddick to win cause he hasn't won a major in a while, but federer will probably win it all.",
  "Solution_9": "[quote=\"plokoon51\"]hopefully, federer will be able to tackle the grass surface better than he did at the French open. personally, i want roddick to win cause he hasn't won a major in a while, but federer will probably win it all.[/quote]\r\n\r\nFederer's game is way better on grass than on clay, or at least compared to his competitors",
  "Solution_10": "An absurdly optimistic prediction:\r\n\r\nHenman will win!",
  "Solution_11": "[quote=\"cadge_nottosh\"]An absurdly optimistic prediction:\n\nHenman will win![/quote]\r\n\r\nGod save the queen!",
  "Solution_12": "ANDY RODDICK! err, that's how you spell it right? but dude rafael-something is really good too for a beginner. but he's not always a great player.",
  "Solution_13": "Rafael Nadal is only good on clay, he stinks on grass!",
  "Solution_14": "wow,serena wiiliams lost to Craybas, who is no. 85 in the world.  Did any of you guys see thr shot Federer madeb when he was playing Kiefer?  It was in the last game of the match, Kiefer was at the net, and Federer hit an amazing crosscourt backhand.  After the match, he said that the shot was the best shot of his life.",
  "Solution_15": "[quote=\"math92\"]If they played i think Roddick would win in 5 sets, but hewitt is 6-1 agains roddick[/quote]\r\n\r\nWhich is exactly why Lleyton would win..because he's better. Roddick is overrated a bit too much.",
  "Solution_16": "federer is just non human..",
  "Solution_17": "[quote=\"GoBraves\"][quote=\"math92\"]If they played i think Roddick would win in 5 sets, but hewitt is 6-1 agains roddick[/quote]\n\nWhich is exactly why Lleyton would win..because he's better. Roddick is overrated a bit too much.[/quote]\r\n\r\nHe's overrated, yeah, but not that much. There's just a lot of hype about him because he's American. I think he deserves his #4 spot in the world right now. A match between Roddick and Hewitt might actually be interesting, instead of both being owned by Federer.",
  "Solution_18": "The only one that can beat federer is marat safin!!! he's the guy with most ability on tour! but he is insane! \r\nbtw, i hit yesterday with a guy that beat andy 2 times in boy's 16 !!!",
  "Solution_19": "[quote=\"manuel\"]The only one that can beat federer is marat safin!!! he's the guy with most ability on tour! but he is insane! \nbtw, i hit yesterday with a guy that beat andy 2 times in boy's 16 !!![/quote]\r\n\r\n :o thats awespme!!! \r\n\r\ntoo bad safin isn't very dedicated to practicing tennis. he has won one grand slam tournament already and could (should) do better in wimbledon.",
  "Solution_20": "[quote=\"plokoon51\"][quote=\"manuel\"]The only one that can beat federer is marat safin!!! he's the guy with most ability on tour! but he is insane! \nbtw, i hit yesterday with a guy that beat andy 2 times in boy's 16 !!![/quote]\n\n :o thats awespme!!! \n\ntoo bad safin isn't very dedicated to practicing tennis. he has won one grand slam tournament already and could (should) do better in wimbledon.[/quote]\r\n\r\nSafin has 2",
  "Solution_21": "[quote=\"juicybooty911\"][quote=\"plokoon51\"][quote=\"manuel\"]The only one that can beat federer is marat safin!!! he's the guy with most ability on tour! but he is insane! \nbtw, i hit yesterday with a guy that beat andy 2 times in boy's 16 !!![/quote]\n\n :o thats awespme!!! \n\ntoo bad safin isn't very dedicated to practicing tennis. he has won one grand slam tournament already and could (should) do better in wimbledon.[/quote]\n\nSafin has 2[/quote]\r\n\r\nYeah he has One U.S open and an Australian, but he's made the finals tons of times. I wonder why he isnt doing well in U.S Open anymore, when he won he played amazingly.",
  "Solution_22": "[quote=\"GoBraves\"][/quote][quote=\"juicybooty911\"][quote=\"plokoon51\"][quote=\"manuel\"]The only one that can beat federer is marat safin!!! he's the guy with most ability on tour! but he is insane! \nbtw, i hit yesterday with a guy that beat andy 2 times in boy's 16 !!![/quote]\n\n :o thats awespme!!! \n\ntoo bad safin isn't very dedicated to practicing tennis. he has won one grand slam tournament already and could (should) do better in wimbledon.[/quote]\n\nSafin has 2[/quote][quote=\"GoBraves\"]\n\nYeah he has One U.S open and an Australian, but he's made the finals tons of times. I wonder why he isnt doing well in U.S Open anymore, when he won he played amazingly.[/quote]\r\n\r\noops...sorry :blush: i was talking about this year. i forgot about the US Open win over Sampras (how could i forget!!!)",
  "Solution_23": "Safin is inconsistent and seems to be bad on grass (at least he seriously complained about it after last year's Wimbledon.)",
  "Solution_24": "safin can be the best in the world when he is on fire and he can be the worst when he is angry.",
  "Solution_25": "yeah... but i thik he will win like a couple of  grand slams more...",
  "Solution_26": "maybe...i dunno...possibly one more.\r\nbut for Nadal i see 4 more.",
  "Solution_27": "[quote=\"math92\"]maybe...i dunno...possibly one more.\nbut for Nadal i see 4 more.[/quote]\r\n\r\nI dont see anymore for Nadal...",
  "Solution_28": "well there is this really good 15 year old named Donald Young who might have 4 slams.",
  "Solution_29": "i think nadal can become fairly competitive on grass in a few years time. dont forget, he's only 19. however, i doubt anyone's gonna beat federer anytime soon."
}
{
  "Tag": [
    "trigonometry",
    "function"
  ],
  "Problem": "8*sin(20)*sin(70)*sin(10)*sin(50) = ?\r\ni got it down to \r\n8*sin(20)*cos(20)*sin(10)*sin(50)\r\n\r\ndont know what to do from there.",
  "Solution_1": "Just use a bunch of co-function and double angle identities.\r\n\r\n[hide]\n\\begin{align*}\n\\ 8\\sin(20)\\sin(70)\\sin(10)\\sin(50) &=8\\sin(20)\\cos(20)\\sin(10)\\cos(40) \\\\\n\\ &=4\\sin(40)\\cos(40)\\sin(10) \\\\\n\\ &=2\\sin(80)\\cos(80) \\\\\n\\ &= \\sin(160) \\\\ \n\\ &= \\sin(20).\n\\end{align*}\n[/hide]",
  "Solution_2": "this does not go in intermediate, more like HSB."
}
{
  "Tag": [
    "MATHCOUNTS",
    "spam"
  ],
  "Problem": "Has the jenks team been chosen yet?\r\nIf so, who is on the team this year?",
  "Solution_1": "Hi Basm? Is dat how u spell ur name? yeah.. This is the team: \r\nJonathan-Me/7th\r\nCheeho -8th\r\nEric-8th (me bro)\r\nCallie-7th\r\nArnav-8th\r\nMitchell-8th\r\nCaleb-8th\r\nShawn-8th\r\n\r\nBTW: Regionals already over and I won!!!",
  "Solution_2": "Yo. State are me, carphead, aznkid, mrk, and shortstack",
  "Solution_3": "What about your \"brother\" jonathanchou711?",
  "Solution_4": "same person as aznkid. duh. i don't have multi tho.",
  "Solution_5": "azn kid711\r\n[size=200][b]BASEM[/b][/size]\r\nnot \r\n[size=200][b]BASM[/b][/size]\r\n\r\nE is very important\r\n$ e\\equal{}mc^2$ you know\r\n(or 2.\t7182818284590452353602874713526624977572470936999595749669676277240766303535475945713821785251664274 depending on your defintion)",
  "Solution_6": "Whoa dude! I've never met you! Dang... \r\n\r\nI get it now! \r\n\r\n[size=200]Basem!!! [/size]",
  "Solution_7": "thank you!!!\r\nand kill arnav for starting the \"[b]basem[/b]ent\" joke\r\nplease",
  "Solution_8": "Basement? It's okay. I already beat him up. Despite his size, I beat him up with disses... :) jk jk jk. I'm only 4 ft 10 though.... :(  :|"
}
{
  "Tag": [
    "geometry",
    "trapezoid"
  ],
  "Problem": "I am trying to write a conjecture for an isosceles trapezoid without using an altitude. I hope you can help me.",
  "Solution_1": "do you mean construct?\r\n\r\notherwise, i have not idea what your question is",
  "Solution_2": "yes, construct and write the proof",
  "Solution_3": "draw a circle, draw a line that intersects the circle 2 times (and does not go through the center of the circle), connect these 2 intersections and the center of the circle, this is an isosceles triangle because 2 sides are congruent because they are radii of a circle",
  "Solution_4": "He wants an isosceles trapezoid, not triangle.  I think you could do it similarly though.  Draw a secant that doesn't go thru the center.  Draw the radii to the intersection points.  Now draw another secant that intersects both radii that is parallel to the first secant.  Then you have an isosceles trapezoid.",
  "Solution_5": "Thanks, how would you write the proof for this?"
}
{
  "Tag": [
    "search"
  ],
  "Problem": "I found these a bit tough. It would be great if you could provide the steps long with the answer. Thanks in advance!  :P \r\n\r\n1. A swimming pool can be filled either using a pipe, a hose, or both. Using the pipe alone takes 12 hours. Using both it takes 60/7 hours. How long does it take using the hose alone?\r\n\r\n2. Wayne drove 3 hours on a freeway at 55 mi/h and then drove 10 miles in the city at 35 mi/h. What was his average speed? \r\n\r\n3. Donna drove half the distance of a trip at 40 mi/h. At what speed would she have to drive for the rest of the distance so that the aberage speed for the entire trip would be 45 mi/h? \r\n\r\n4. At what time after 4:00 will the minute hand and the hour hand of a clock first be in the same position? \r\n\r\n\r\nI solved a bunch of these types of questions, and for some reason, I kept getting the incorrect answers for these.",
  "Solution_1": "[quote=\"apriori\"]1. A swimming pool can be filled either using a pipe, a hose, or both. Using the pipe alone takes 12 hours. Using both it takes 60/7 hours. How long does it take using the hose alone?[/quote]\n\n[hide=\"Solution\"]Okay, so in one hour, using the pipe would fill up $ \\frac1{12}$th of the pool.  Using both, it would fill up $ \\frac {7}{60}$ths of the pool.  So to find how long it would take to use the hose, we take the reciprocal of $ \\frac {7}{60} \\minus{} \\frac1{12} \\equal{} \\frac {2}{60} \\implies \\frac {1}{30} \\implies \\boxed{30 \\text{ hours}}$[/hide]\n\n[quote=\"apriori\"]2. Wayne drove 3 hours on a freeway at 55 mi/h and then drove 10 miles in the city at 35 mi/h. What was his average speed?[/quote]\n\n[hide=\"Solution\"]Wayne drove $ 3 \\times 55$ or $ 165$ miles on the highway.  Thus, we must do a weighted average:\n\n$ \\frac{165 \\text{ miles} \\times 55 \\frac{\\text{miles}}{\\text{hour}} \\plus{} 10 \\text{ miles} \\times 35 \\frac{\\text{miles}}{\\text{hour}}}{175 \\text{ miles}} \\implies \\frac{(9075 \\plus{} 350) \\frac{\\text{miles}^2}{\\text{hour}}}{175 \\text{ miles}} \\implies \\frac{9425 \\frac{\\text{miles}^2}{\\text{hour}}}{175 \\text{ miles}} \\implies \\boxed{53 \\frac67 \\text{ miles}/\\text{hour}}$[/hide]",
  "Solution_2": "[quote=\"apriori\"]I found these a bit tough. It would be great if you could provide the steps long with the answer. Thanks in advance!  :P \n\nI solved a bunch of these types of questions, and for some reason, I kept getting the incorrect answers for these.[/quote]\r\n\r\napriori, you see what I highlighted?\r\n\r\nYou should post at least some of your work on all of these problems, along with \r\nyour final answer attempt, so we can see your thinking, and then we not doing \r\nyour work.",
  "Solution_3": "[hide=\"3\"]\n$ \\frac {d}{2} \\equal{} 40t$\n$ t \\equal{} \\frac {d}{80}$\nLet $ r$ be the rate for the second half of the distance.\nThen, $ t \\equal{} \\frac {d}{2r}$.\n\nThen, we have:\naverage $ r \\equal{} \\frac {total distance}{total time} \\equal{} 45$\n\naverage $ r \\equal{} \\frac {d}{\\frac {d}{80} \\plus{} \\frac {d}{2r}} \\equal{} \\frac {1}{\\frac {1}{80} \\plus{} \\frac {1}{2r}} \\equal{} 45$\nSolve for $ r$.\n[/hide]",
  "Solution_4": "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=548908835&t=172357]Hint for Number 4.[/url]"
}
{
  "Tag": [
    "inequalities"
  ],
  "Problem": "This question comes from The Art and Craft of Problem Solveing.\r\n\r\nFind the minimum value of $xy+yz+xz$, given that $x$, $y$, $z$, are real and $x^2+y^2+z^2=1$.",
  "Solution_1": "oi... this is pretty easy with lagrange multipliers, which we just learned to use...\r\nbut that's calc, so i'm guessing you want a more basic approach?",
  "Solution_2": "$(x-y)^2+(y-z)^2+(z-x)^2=.....$ :)",
  "Solution_3": "$2x^2+2y^2+2z^2-2(xy+xz+yz)$\r\nand since $x^2+y^2+z^2=1$ we have\r\n$2-2(xy+xz+yz)=2(1-xy-yz-xz)$\r\n\r\nsorry but i do not see how this helps. :(",
  "Solution_4": "[hide]$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)=1+2(xy+yz+zx)\\geq 0\\Rightarrow xy+yz+zx\\geq -\\frac{1}{2}$\n\nEquality is attained when $x=\\frac{\\sqrt{2}}{2},\\ y=-\\frac{\\sqrt{2}}{2},\\ z=0$. Therefore minimum is $-\\frac{1}{2}$.[/hide]",
  "Solution_5": "[quote=\"maokid7\"]$2x^2+2y^2+2z^2-2(xy+xz+yz)$\nand since $x^2+y^2+z^2=1$ we have\n$2-2(xy+xz+yz)=2(1-xy-yz-xz)$\n\nsorry but i do not see how this helps. :([/quote]\r\n\r\nSorry, I was mistaken about the Maximum :blush:  value!",
  "Solution_6": "o i see now. i feel so stupid now. i tried AM-GM and other inequality tactics on $(x+y+z)^2$. but never thought about the trivial inequality.  :oops:  :blush:"
}
{
  "Tag": [
    "inequalities",
    "Galois Theory"
  ],
  "Problem": "I was wondering what cyclic and symmetric sums were, and possibly some of their common uses.\r\n\r\nCheers,\r\nThe Idiot without a village",
  "Solution_1": "Basically cyclic sums are when we have some formula where relabelling the variables doesn't change anything. For example ab+bc+ca, or a:^2:+b:^2:+c:^2:+2ab+2bc+2ca+abc.\r\n\r\nAs to what you'd use them for.. apart from just simplfying how much you need to write (for the first thing above you could just write cyc  :Sigma: ab (not sure if this is official language or anything but I think its usable).. they just often pop up all over the place because they are 'nice'.  For example most inequalities use cyclic sums. Can't really think of any special use of them though..",
  "Solution_2": "Cyclic sums are of outmost importance in some part of Algebra, the Galois theory for instance.\r\n\r\n\r\nJust to give a small example :\r\n\r\nLet's consider an equation  :\r\n\r\n123 * x^56 + 456 * x^12 + 789 * x^5 + 123456789 = 0\r\n\r\n\r\nThere are 56 roots x_i (not necessary all different) to this horrible equation.\r\nI have NO idea of its roots but by Viete relations I can state that :\r\n\r\nsum x_i = 456/123\r\nsum x_i * x_j (i < j) = 0\r\nsum x_i * x_j * x_k (i < j < k) = 0\r\n....\r\n\r\nprod x_i = 123456789/123.",
  "Solution_3": "What about symmetric sums?  What are they?",
  "Solution_4": "I'm pretty sure they're just the same thing.",
  "Solution_5": "There's a slight difference. For example, assume we have the variables a, b, and c. Then the symmetric sum of a^2*b is a^2*b+a^2*c+b^2*a+b^2*c+c^2*a+c^2*b. But the cyclic sum of a^2*b is a^2*b+b^2*c+c^2*a. Sorry, but I can't think of a way to explain it better. I hope this is sufficient.",
  "Solution_6": "Are there any specific links or theorems you may provide me with so that I may view exactly where these cyclic/symmetric sums occur?\r\n\r\nAlso are there any links that contains elementary mathematics topics not coverred in the High School Curricula throughout the nation.\r\n\r\nThank you, I appreciate your aid on this matter,\r\nThe Idiot without a village",
  "Solution_7": "Is it safe to assume that this has nothing to do with cyclic quadrilaterals?",
  "Solution_8": "Why yes Pierre [if that is indeed your real name]... I do believe that it is safe to assume that this has nothing to do with cyclic quadrilaterals... see ya tommorow in class..\r\n\r\nCyclic quads taste yummy...  but what really turn's me on is modelling photosynthetic biochemistry via PDEs.\r\n\r\nPeace from the Idiot...\r\n\r\nI also like it when people rub my tummy.",
  "Solution_9": "Just 15 years late but; Newton's Theorem or something like that about symmetric sums",
  "Solution_10": "[quote=achen29]Just 15 years late but; Newton's Theorem or something like that about symmetric sums[/quote]\n\nIf you had a question, you could start a new topic. Just necroing a 15 year old post with no major contribution other than \"Hey just wanted say Newton's Theorem\" is bad, especially since the OP's question already got answered long long LONG time ago. I remember a bunch of these guys when we were in high school. I'm sure they all graduated college by now...."
}
{
  "Tag": [
    "\\/closed"
  ],
  "Problem": "If you take [url=http://www.artofproblemsolving.com/Forum/index.php?f=271]this[/url] link, and \"close\" the Vietnam tab, India, Iran, Pakistan, Tajikistan become hidden as well.\r\n\r\nJust something small I'd like to mention.",
  "Solution_1": "how come India, Iran, Pakistan, & Tajikistan are part of Vietnam now? :P"
}
{
  "Tag": [
    "search",
    "Ross Mathematics Program",
    "inequalities",
    "IMO Shortlist",
    "graph theory",
    "\\/closed"
  ],
  "Problem": "Hi,  :) \r\n\r\nI would like to know your opinion. Do you think we should mention the source to all problems we are posting ? One of the advantages of not mentioning the sources: Nobody will be seduced to the search the web for the problem, e.g. kalva and others.\r\nSo if we want to mention the sources. Do you think we should edit the old postings, e.g. from the unsolved section or even the solved ones (this would mean to unlock all the problems first) ? It also should be nice if others will tell us about the source where they have seen the problem as well. If the source is unknown, it should be indicated as unknown. Otherwise we would assume that you forgot to mention the source. Another advantage is that we can remember a problem easier though we might have forgotten about the problem's statement. Moreover we get to know about the preferences, meaning which olympiads they like, of the mathlinks users.",
  "Solution_1": "searching the web for problems doesn't seem like a suitable idea for those pursuing math (olympiad :D) knowledge. \r\n\r\nso my guess is that we should try to post the sources when we can. :) for quicker indexing and searching of the problems.",
  "Solution_2": "So my idea is the following one: Please indicate the source of the problem below the problem statement in Italic letters. If you want to prevent others from looking up the problem somewhere, e.g. Kalva. In this case you might post the source when the problem is solved. Therefore we don't try to lock/remove the problems quickly so that you have time to add the source.\r\n\r\nSuggestions for citing sources are given at: http://www.mathlinks.ro/phpBB/viewtopic.php?t=1053",
  "Solution_3": "Honestly, I don't know why anybody would want to prevent somebody else from looking up a problem on the web.  After all, it does seem like \"cheating\", so I don't think anybody who really wants to do the problem seriously would consider the idea.  It's much simpler to state the source when posting the problem.\r\n\r\nI think, more importantly, since most questions are not our own, it's only correct that we acknowledge where we got the problem from (so if you forgot, just say so).  And of course, it is convenient for future reference.\r\n\r\nOrlando, regarding your idea of including the year, stage, class, problem no., I think a simple way to get around the issue of people omitting certain information would be to put a letter in front of each number e.g. 2003 S4 C13 P3.",
  "Solution_4": "Many thanks for your suggestion Valiowk. I hope everybody will keep to the rule. Otherwise I will always ask for the source. Maybe we should publish a document on the site where you can read about how your posting should look like, meaning some general summarized rule. Every new member should be informed about the document when subscribing to mathlinks.",
  "Solution_5": "well I have made some annoucements earlier concerning that. I am planning to gather them up ... thanks for the sugestions, and keep up the good work orl :)",
  "Solution_6": "In my opinion it is a good idea to indicate the source of the problem only after its solution is posted.\r\nA question. I have in my notebook a lot of problems of which I don't remember the source so shall I write  \"source unknown\" or \"source forgotten\" ?",
  "Solution_7": "Why is it so important to know the source of a problem  :?  Problems exist by them-selves, don't they ?",
  "Solution_8": "Each week hundreds of new problems emerge on the site. already some of them are repeatting themselves. It is a good ideea to be able to search for specific problems to see if they are posted. :)",
  "Solution_9": "[quote=\"Valentin Vornicu\"]Each week hundreds of new problems emerge on the site.[/quote]\r\nSure they do. :D \r\nLets not exaggerate. This forum has yet to grow to reach 100 new topics a week.",
  "Solution_10": "Maybe you could even add some dropdown/text inputbox in the \"post a reply\" window?",
  "Solution_11": "I am not sure what you mean mindspa. maybe you can ellaborate a little bit ? :)",
  "Solution_12": "I think it is very important to indicate the source. There are three kinds of sources:\r\n\r\n(1) self-created  - you created the problem by yourself and you are not aware that it has already been proposed anywhere else\r\n(2) \"competition\" - tell the source in the way valiowk suggested it\r\n(3)  unknown       - in the case you don't know the source\r\n\r\nMaybe it is possible for (2) and (3) to say where you have found the problem, meaning which journal, book etc. There are several problem which are used in competitions again, sometimes known by the guys proposing problems and sometimes not. Anyway I feel it is an appreciation for a certain problem to emerge in many sources.\r\n\r\nPlease have a look at the following postings:\r\n\r\n(a)  http://www.mathlinks.ro/phpBB/viewtopic.php?t=1985\r\n(b)  http://www.mathlinks.ro/phpBB/viewtopic.php?t=1028\r\n\r\nTo provide an example, problem (b) was used in many sources and I even did not remember every source back then. So I try to mention all sources I am right now aware of:\r\n\r\n(I)   \"early\" Chinese mathematical olympiad\r\n(II)  \"early\" AMM volume which covered Chinese math olympiad problems from (I)\r\n(III) Honsberger, Ross; Mathematical morsels; MAA, 1978\r\n(IV)  Richard K. Guy, The strong law of small numbers, American Mathematical Monthly, 95: 697-712, 1988\r\n(V)   E. A. Morozova, I. S. Petrakov, V.A. Skvortov; International Mathematical Olympiads; Moscow\r\n      (comment: the problem in this source is taken from an early IMO shortlist)\r\n(VI)  Paul Zeitz, The Art and Craft of Problem Solving, John Wiley & Sons, 1999\r\n(VII) I. Tomescu, R. A. Melter; Problems in Combinatorics and Graph Theory; John Wiley & Sons, 1985\r\n\r\nAnother reason might be the Valentin's idea to browse the forum easier. Maybe it is possible to establish a menu for each problem where the user can write down the source of a problem. Other users may edit the menu. But they can only add different sources where they run across the problem.\r\n\r\nPlease also indicate the soure below the problem statement though the subject (of your created) might be something like \"country\" + \"year\".",
  "Solution_13": "I also think that mentioning problem sources is important. Maybe even mention similar problems or generalizations etc (with sources). And to clarify my earlier post I was thinking about adding some extra fields to the \"Post a reply\" page, and of course connecting these fields with the search engine. Some plausible fields would be a drop down box which would have a list on the majority of olympiads, another one with the year, another one with problem number etc etc, those last could be entered manually though. I would also like to stress the importance, according to me of couse -- I'm sure many don't agree, in naming problem entries unambigously, giving them some distinct name that enlightens the problems nature or statement. Having 59 entries named inequality xx doesn't describe the problem that well, but as I said before, this are only my thoughts.",
  "Solution_14": "Technically, the search engine already searches the entire post, so it would not make that much of a difference to link the separate parts to the search engine.  The drop down box is a good idea though; of course, there should always be the option labelled \"others\".",
  "Solution_15": "I was mostly refering to that the search output could create tabulated information about the source of the problems."
}
{
  "Tag": [
    "algebra",
    "system of equations"
  ],
  "Problem": "I'm wondering if there's a simpler solution to this problem than what I have.\r\n\r\nThe lengths of the three medians of a given triangle are 9, 12, 15. Find the length of the side which the median of length 15 is drawn to.\r\n\r\n[hide=\"outline of my solution\"]I let the lengths of the three sides be $2x,2y,2z$, then wrote three equations involving $x,y,z$ using Stewart's Theorem on all three medians, I ended up with pretty symmetric system of equations, and solved for the side I wanted.[/hide][/i]",
  "Solution_1": "i think that is pretty much what you do...just to note though...your systems should be very simple...linear if you do m=x^2, n=y^2, p=z^2"
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "calculus",
    "integration",
    "perimeter",
    "function",
    "inequalities"
  ],
  "Problem": "ARO (= All-Russian Olympiad): \r\n\r\n[b]problems:[/b] http://www.mccme.ru/olympiads/vmo/31/zadachi.htm\r\n[b]results:[/b] http://www.mccme.ru/olympiads/vmo/31/protocol.htm\r\n\r\nThe proposers of the problems are in the post below. \r\n\r\nHere are the links of the problems. Thanks to [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=270]Fedor Petrov[/url] for providing the translation. :) \r\n\r\n\r\n[hide=\"9th grade problem\"]\n\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35296]9.1[/url] Given parrallelogram $ABCD\\ (AB<BC)$. Prove that the circumcircles of triangles $APQ$, where points $P$ and $Q$ are chosen on sides $BC$ and $CD$ respectively such that $CP=CQ$, have a common point, different from $A$.\n\n[i] proposed by T. Emelyanova[/i]\n\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35325]9.2[/url] Lesha put numbers from 1 to $22^{2}$ into cells of $22\\times 22$ board. Can Oleg always choose two cells, adjacent by the side or by vertex, the sum of numbers in which is divisible by 4?\n\n[i] proposed by O. Podlipsky[/i]\n\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35324]9.3[/url] Given three  reals $a_{1},\\,a_{2},\\,a_{3}>1,\\,S=a_{1}+a_{2}+a_{3}$. Provided ${a_{i}^{2}\\over a_{i}-1}>S$ for every  $i=1,\\,2,\\,3$ prove that $\\frac{1}{a_{1}+a_{2}}+\\frac{1}{a_{2}+a_{3}}+\\frac{1}{a_{3}+a_{1}}>1.$\n\n[i] proposed by S. Berlov[/i]\n\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35323]9.4[/url] Given 365 cards, in which distinct numbers are written. We may ask for any three cards, the order of numbers written in them. Is it always possible to find out the order of all 365 cards by 2000 such questions?\n\n[i] proposed by M. Garber[/i]\n\n\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35322]9.5[/url] Ten mutually distinct non-zero reals are given such that for any two, either their sum or their product is rational. Prove that squares of all these numbers are rational.\n\n[i] proposed by O. Podlipsky[/i]\n\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35321]9.6[/url] Find the number of subsets  $A\\subset M=\\{2^{0},\\,2^{1},\\,2^{2},\\dots,2^{2005}\\}$ such that equation $x^{2}-S(A)x+S(B)=0$ has integral roots, where $S(M)$ is the sum of all elements of $M$, and $B=M\\setminus A$ ($A$ and $B$ are not empty).\n\n[i] proposed by Nazar-Bashi, I. Bogdanov[/i]\n\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35298]9.7[/url]  = [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35298]10.6[/url] \n\n[i] proposed by A. Akopyan[/i]\n\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35320]9.8[/url] 100 people from 50 countries, two from each countries, stay on a circle. Prove that one may partition them onto 2 groups in such way that neither no two countrymen, nor three consecutive people  on a circle, are in the same group.\n\n[i] proposed by S. Berlov[/i] \n[/hide]\n\n[hide=\"10th grade problem\"] \n\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35319]10.1[/url] Find the least positive integer, which may not be represented as ${2^{a}-2^{b}\\over 2^{c}-2^{d}}$, where $a,\\,b,\\,c,\\,d$ are positive integers.\n\n[i] proposed by V. Senderov[/i]\n\n [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35304]10.2[/url] Positive numbers are written in cells of a board $2\\times n$ such that the sum of numbers in each of $n$ columns equals 1. Prove that   it is possible to select one number in every  column such that the sum of selected numbers in each of two rows does not exceed ${n+1\\over 4}$.\n\n[i] proposed by E. Kulikov[/i]\n\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35318]10.3[/url] Given 2005 distinct numbers $a_{1},\\,a_{2},\\dots,a_{2005}$. By one question, we may take three different indices $1\\le i<j<k\\le 2005$ and find out the set of numbers $\\{a_{i},\\,a_{j},\\,a_{k}\\}$ (unordered, of course).  Find the minimal number of questions, which are necessary to find out all numbers $a_{i}$.\n\n[i] proposed by I. Bogdanov[/i]\n\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35317]10.4[/url] $w_{B}$ and $w_{C}$ are excircles of a triangle $ABC$. The circle $w_{B}'$ is symmetric to $w_{B}$ with respect to the midpoint of $AC$, the circle $w_{C}'$ is symmetric to $w_{C}$ with respect to the midpoint of $AB$. Prove that the radical axis of $w_{B}'$ and $w_{C}'$ halves the perimeter of $ABC$.\n\n[i] proposed by P. Kozhevnikov[/i]\n\n\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35306]10.5[/url] Some 16 cells are chosen on a chessboard. Find the minimal number of pairs of cells, which are in the same row or column. \n\n[i] proposed by E. Kulikov[/i]\n\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35298]10.6[/url] Given acute triangle $ABC$, $AA'$ and $BB'$ are its altitudes.  A point $D$ is chosen on an arc $ACB$ of a circumcircle of $ABC$. Lines $AA'$ and $BD$ meet in $P$,  lines $BB'$ and $AD$ meet in $Q$.  Prove that the line $A'B'$ contains a midpoint of $PQ$\n\n[i] proposed by A. Akopyan[/i]\n\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35316]10.7[/url] Positive integers $x>1$ and $y$  satisfy an equation $2x^{2}-1=y^{15}$. Prove that 5 divides $x$.\n\n[i] proposed by V. Senderov[/i]\n\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35315]10.8[/url] A white plane is partitioned onto cells (in a usual way). A finite number of cells are coloured black. Each black cell has an even (0, 2 or 4) adjacent (by the side) white cells. Prove that one may colour each white cell in green or red such that every black cell will have equal number of red and green adjacent cells.\n\n[i] proposed by A. Glebov, D. von der Flaas[/i]\n\n [/hide]\n\n[hide=\"11th grade problem\"] \n\n\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35314]11.1[/url] Find the maximal possible  finite number of roots of the equation  $|x-a_{1}|+\\dots+|x-a_{50}|=|x-b_{1}|+\\dots+|x-b_{50}|$, where $a_{1},\\,a_{2},\\,\\dots,a_{50},\\,b_{1},\\dots,\\,b_{50}$ are distinct reals.\n\n[i] proposed by I. Rubanov[/i]\n\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35318]11.2[/url] = [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35318]10.3[/url] \n\n[i] proposed by I. Bogdanov[/i]\n\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35313]11.3[/url] Let $A',\\,B',\\,C'$ be points, in which excircles touch corresponding sides of triangle $ABC$. Circumcircles of triangles $A'B'C,\\,AB'C',\\,A'BC'$ intersect a circumcircle of $ABC$ in points  $C_{1}\\ne C,\\,A_{1}\\ne A,\\,B_{1}\\ne B$ respectively. Prove that a triangle $A_{1}B_{1}C_{1}$ is similar to a triangle, formed by points, in which incircle of $ABC$ touches its sides.\n\n[i] proposed by L. Emelyanov[/i]\n\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35312]11.4[/url] Integers $x>2,\\,y>1,\\,z>0$ satisfy an equation $x^{y}+1=z^{2}$. Let $p$ be a number of different prime divisors of $x$, $q$ be a number of different prime divisors of $y$. Prove that $p\\ge q+2$.\n\n[i] proposed by V. Senderov[/i]\n\n\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35311]11.5[/url] Do there exist a bounded function $f: R\\to R$ such that $f(1)>0$ and $f(x)$ satisfies an inequality $f^{2}(x+y)\\ge f^{2}(x)+2f(xy)+f^{2}(y)$?\n\n[i] proposed by Nazar-Bashi[/i]\n\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35310]11.6[/url] Do there exist 12 rectangular parallelepipeds  $P_{1},\\,P_{2},\\dots,P_{12}$ with edges parallel to coordinate axes $OX,\\,OY,\\,OZ$ such that $P_{i}$ and $P_{j}$ have a common point iff $i\\ne j\\pm 1$ modulo 12?\n\n[i] proposed by A. Akopyan[/i]\n\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35309]11.7[/url] A quadrilateral $ABCD$ without parallel sides is circumscribed around a circle with centre $O$. Prove that $O$ is a point of intersection of middle lines of quadrilateral $ABCD$ (i.e. barycentre of points $A,\\,B,\\,C,\\,D$) iff $OA\\cdot OC=OB\\cdot OD$.\n\n[i] proposed by A. Zaslavsky, M. Isaev, D. Tsvetov[/i]\n\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35308]11.8[/url] 100 people from 25 countries, four from each countries, stay on a circle. Prove that one may partition them onto 4 groups in such way that neither no two countrymans, nor two neighbours will be  in the same group.\n\n[i] proposed by S. Berlov[/i]\n\n [/hide]",
  "Solution_1": "Enjoy. I am in haste, so the translation is even worse then usually.\r\nI do not know how define normal capital R for reals. Probelms of 9 klass are in Russian, ifa anybody wants, he may translate them, else please wait a little.\r\n\r\n\\q{9.1.} Given parrallelogram $ABCD\\ (AB<BC)$.\r\nProve that the circumcircles of triangles\r\n$APQ$, where points $P$ and $Q$ are chosen on sides\r\n$BC$ and $CD$ respectively such that\r\n$CP=CQ$, have a common point, different from\r\n$A$.\r\n\r\n\\q{9.2.} Lesha put numbers from 1 to $22^2$\r\ninto cells of $22\\times 22$ board.\r\nCan Oleg always choose two cells,\r\nadjacent by the side or by vertex, the\r\nsum of numbers in which is divisible by 4?\r\n\r\n\\q{9.3.} Given three  reals $a_1,\\,a_2,\\,a_3>1,\\,S=a_1+a_2+a_3$\r\nProvided ${a_i^2\\over a_i-1}>S$ for every \r\n$i=1,\\,2,\\,3$ prove that\r\n$\\frac{1}{a_1+a_2}+\\frac{1}{a_2+a_3}+\\frac{1}{a_3+a_1}>1.$\r\n\r\n\\q{9.4.} Given 365 cards, in which distinct numbers are written.\r\nWe may ask for any three cards, the order of numbers\r\nwritten in them. Is it always possible to find out\r\nthe order of all 365 cards by 2000 such questions?\r\n\r\n\\smallskip\r\n\\q{9.5.} Ten mutually distinct non-zero\r\nreals are given such that for any two, either\r\ntheir sum or their product is rational.\r\nProve that squares of all these numbers\r\nare equal.\r\n\r\n\\q{9.6.} Find the number of subsets \r\n$A\\subset M=\\{2^0,\\,2^1,\\,2^2,\\dots,2^{2005}\\}$\r\nsuch that equation $x^2-S(A)x+S(B)=0$ has integral root,\r\nwhere $S(M)$ is the sum of all elements of $M$,\r\nand $B=M\\setminus A$ ($A$ and $B$ are not empty).\r\n\r\n\\q{9.7.} see 10.6.\r\n\r\n\r\n\r\n\\q{9.8.} 100 people from 50 countries,\r\ntwo from each countries, stay on a circle.\r\nProve that one may partition them\r\nonto 2 groups in such way that neither no two\r\ncountrymans, nor thre consecutive people \r\non a circle are in the same group.\r\n\r\n\r\n\\q{10.1.} Find the least positive integer,\r\nwhich may not be represented as ${2^a-2^b\\over 2^c-2^d}$,\r\nwhere $a,\\,b,\\,c,\\,d$ are positive integers.\r\n\r\n\\q{10.2.} Positive numbers are written in cells\r\nof a board $2\\times n$ such that the sum of numbers\r\nin each of $n$ columns equals 1. Prove that  \r\nit is possible to select one number in every \r\ncolumn such that the sum of selected numbers in each of\r\ntwo rows does not exceed ${n+1\\over 4}$.\r\n\r\n\\q{10.3.} Given 2005 distinct numbers $a_1,\\,a_2,\\dots,a_{2005}$.\r\nBy one question, we may take three different\r\nindecies $1\\le i<j<k\\le 2005$ and find out the set of numbers\r\n$\\{a_i,\\,a_j,\\,a_k\\}$ (unordered, of course). \r\nFind the minimal number\r\nof questions, which are necessary to find\r\nout all numbers $a_i$.\r\n\r\n\\q{10.4.} $w_B$ and $w_C$ are excircles of a \r\ntriangle $ABC$. The circle $w_B'$ is symmetric to $w_B$\r\nwith respect to the midpoint of $AC$,\r\nthe circle $w_C'$ is symmetric to $w_C$\r\nwith respect to the midpoint of $AB$.\r\nProve that the radical axis of $w_B'$ and $w_C'$\r\nhalves the perimeter of $ABC$.\r\n\r\n\\q{10.5.} Some 16 cells are chosen on a chessboard.\r\nFind the minimal number of pairs of cells, which are\r\nin the same row or column. \r\n\r\n\\q{10.6.}\r\nGiven acute triangle $ABC$,\r\n$AA'$ and $BB'$ are its altitudes. \r\nA point $D$ is chosen on an arc\r\n$ACB$ of a circumcircle of $ABC$. \r\nLines $AA'$ and $BD$ meet in $P$, \r\nlines $BB'$ and $AD$ meet in $Q$. \r\nProve that the line $A'B'$ contains\r\na midpoint of $PQ$.\r\n\r\n\\q{10.7.} Positive integers $x>1$ and $y$ \r\nsatisfy an equation $2x^2-1=y^{15}$.\r\nProve that 5 divides $x$.\r\n\r\n\\q{10.8.} A white plane is partioned onto cells\r\n(in a usual way). A finite number of cells\r\nare coloured black. Each black cell\r\nhas an even (0, 2 or 4) adjacent (by the side)\r\nwhite cells. Prove that one may colour each white cell\r\nin green or red such that every black cell\r\nwill have equal number of red\r\nand green adjacent cells.\r\n\r\n\\q{11.1.} Find the maximal possible number\r\nof roots of an equation \r\n$|x-a_1|+\\dots+|x-a_{50}|=|x-b_1|+\\dots+|x-b_50|$,\r\nwhere $a_1,\\,a_2,\\,\\dots,a_50,\\,b_1,\\dots,\\,b_50$\r\nare distinct reals.\r\n\r\n\\q{11.2.} see 10.3\r\n\r\n\\q{11.3.} Let $A',\\,B',\\,C'$ be points,\r\nin which excircles touch corresponding\r\nsides of triangle $ABC$. Circumcircles of triangles\r\n$A'B'C,\\,AB'C',\\,A'BC'$ intersect \r\na circumcircle of $ABC$ in points \r\n$C_1\\ne C,\\,A_1\\ne A,\\,B_1\\ne B$ respectively.\r\nProve that a triangle $A_1B_1C_1$ is similar\r\nto a triangle, formed by points, in which\r\nincircle of $ABC$ touches its sides.\r\n\r\n\\q{11.4.} Integers $x>2,\\,y>1,\\,z>0$\r\nsatisfy an equation $x^y+1=z^2$.\r\nLet $p$ be a number of different prime divisors of\r\n$x$, $q$ be a number of different prime divisors of\r\n$y$. Prove that $p\\ge q+2$.\r\n \r\n\r\n\\q{11.5.} Do there exist a \r\nbounded function $f:R\\to R$ such that $f(1)>0$\r\nand $f(x)$ satisfies an inequality \r\n$f^2(x+y)\\ge f^2(x)+2f(xy)+f^2(y)$?\r\n\r\n\\q{11.6.} Do there exist 12 rectangular parallelepipeds \r\n$P_1,\\,P_2,\\dots,P_{12}$ with edges parallel to\r\ncoordinate aces $OX,\\,OY,\\,OZ$ such that $P_i$\r\nand $P_j$ have a common point iff $i\\ne j\\pm 1$\r\nmodulo 12?\r\n\r\n\\q{11.7.} A quadrilateral $ABCD$ without\r\nparallel sides is circumscribed arounda circle\r\nwith centre $O$. Prove that $O$ is a poiint\r\nof intersection of middle lines of quadrilateral\r\n$ABCD$ (i.e. barycentre of points $A,\\,B,\\,C,\\,D$)\r\niff $OA\\cdot OC=OB\\cdot OD$.\r\n\r\n\\q{11.8.} 100 people from 25 countries,\r\nfour from each countries, stay on a circle.\r\nProve that one may partition them\r\nonto 4 groups in such way that neither no two\r\ncountrymans, nor two neighbours will be \r\nin the same group.",
  "Solution_2": "My students got the following results:\r\n\r\n11th grade\r\nBelousov Kirill - I-st Diploma  :first:  :first:  :first: \r\n\r\n10th grade\r\nSmotrov Dmitrii - II-nd Diploma  :D :D :D\r\n\r\nAlso one students from my school, Dmitrii Yarushin, took II-nd Diploma in the 9th grade.\r\n\r\nI congratulate them!!!\r\n\r\nUnfortunately, despite of I-st diploma, Kirill wasn't included into the russian IMO team (because of Winter TSTs).",
  "Solution_3": "[quote=\"Myth\"]Unfortunately, despite of I-st diploma, Kirill wasn't included into the russian IMO team (because of Winter TSTs).[/quote]I am really sorry to hear that :( \r\n\r\nBut anyway it's great to be the first in the Russian MO!!! :first: :first: Congrats!",
  "Solution_4": "[quote=\"Myth\"]My students got the following results:\n\n11th grade\nBelousov Kirill - I-st Diploma  :first:  :first:  :first: \n\n10th grade\nSmotrov Dmitrii - II-nd Diploma  :D :D :D\n\nAlso one students from my school, Dmitrii Yarushin, took II-nd Diploma in the 9th grade.\n\nI congratulate them!!!\n\nUnfortunately, despite of I-st diploma, Kirill wasn't included into the russian IMO team (because of Winter TSTs).[/quote]\r\n\r\nCongrats to teacher Myth! You are great teacher!!!",
  "Solution_5": "Did any bulgarians participated?\r\nCongratulations,Myth,I wish I had such a teacher like you!",
  "Solution_6": "did Chinese students participate?? \r\nWhat's their results?",
  "Solution_7": "Bulgaria\r\n\r\n11 klass\r\n\r\nKralev Rosen\t65774761\t43\t1diploma\r\nDimitrov Ivan\t07777770\t42\t2diploma\r\nLishkov Alexandr\t31774070\t29\t3diploma\r\nDimitrov Veselin\t01077000\t15             no diploma\r\n\r\n10 klass\r\n\r\nBaronov Emil\t77607770\t41\t2diploma\r\nSimeonov Dimitar\t76077711\t36\t2diploma\r\n\r\n\r\nChina (I apologize for transcription of names from Russian)\r\n\r\n10 klass\r\n\r\nLyu Chziyu                77777747   53\t1diploma\r\nVan Czenang\t55777710\t 39\t2diploma\r\nCzan Tau\t                71177770\t 37\t2diploma\r\nIng Cze\t\t57107710\t 28\t3diploma\r\n\r\n9 klass\r\n\r\nIng Li                        77707770   42            2diploma\r\nVang Yuli                 70707670   34            3diploma",
  "Solution_8": "[quote=\"Myth\"]My students got the following results:...\nI congratulate them!!!\n[/quote]\r\nCongratulations, Mikhail! I wish you that your students do even better the next year!  :)",
  "Solution_9": "[quote=\"Fedor Petrov\"]Bulgaria\r\n\r\n11 klass\r\n\r\nKralev Rosen\t65774761\t43\t1diploma\r\nDimitrov Ivan\t07777770\t42\t2diploma\r\nLishkov Alexandr\t31774070\t29\t3diploma\r\nDimitrov Veselin\t01077000\t15             no diploma\r\n\r\n\r\n1diploma means first place in Russia?"
}
{
  "Tag": [
    "AMC"
  ],
  "Problem": "Should we create a new thread to link together all the mock tests?\r\n\r\nMathfanatics thread hasn't been updated in a while.\r\n\r\nIf you're planning to provide difficulty ratings for the tests, I can provide input, as I've taken quite a bit of them.",
  "Solution_1": "Not really, since they can just be added to the AoPSWiki.  There's already a section on mock amc stuff."
}
{
  "Tag": [
    "linear algebra",
    "matrix",
    "Gauss",
    "algebra",
    "system of equations",
    "linear algebra unsolved"
  ],
  "Problem": "Let $ E$ be linear equations system with all integer coefficients. Let us suppose that $ E$ has only one solution $ (x_{1}, ..., x_{n})$ (all real numbers). Prove that all $ x_{i}$ are rational.",
  "Solution_1": "Suppose your system of equations is $ Ax \\equal{} b$, where $ A$ is an $ m\\times n$ matrix. Obviously $ n\\leq m$ and $ n \\equal{} \\text{rank}(A)$. Without loss of generality, assume that the first $ n$ rows of $ A$ are linearly independent and $ A \\equal{} \\begin{bmatrix}C \\\\\r\nD\\end{bmatrix}, b \\equal{} \\begin{bmatrix}b_1 \\\\\r\nb_2\\end{bmatrix}$, where $ C$ is an invertible $ n\\times n$ submatrix of $ A$, and $ b_1\\in \\mathbb{Z}^n$. Since $ Cx \\equal{} b$ has a unique solution, which is clearly in $ \\mathbb{Q}^n$, the hypothetical solution to your original system should also be in $ \\mathbb{Q}^n$.",
  "Solution_2": "[quote=\"Invariant\"]Let $ E$ be linear equations system with all integer coefficients. Let us suppose that $ E$ has only one solution $ (x_{1}, ..., x_{n})$ (all real numbers). Prove that all $ x_{i}$ are rational.[/quote]\r\nUse Gauss Method for solving this problem. After transforming $ E$ you will have equations system with rational coefficients.",
  "Solution_3": "[quote=\"Kirill\"]After transforming $ E$ you will have equations system with rational coefficients.[/quote]\r\nYes. But you haven't pointed out the way this fact is related to the solutions of the given system.",
  "Solution_4": "[quote=\"The Captain\"]\nBut you haven't pointed out the way this fact is related to the solutions of the given system.[/quote]\r\nI think, it's easy, and there is no reason to extend this idea to full solution.",
  "Solution_5": "[quote=\"Kirill\"][quote=\"Invariant\"]Let $ E$ be linear equations system with all integer coefficients. Let us suppose that $ E$ has only one solution $ (x_{1}, ..., x_{n})$ (all real numbers). Prove that all $ x_{i}$ are rational.[/quote]\nUse Gauss Method for solving this problem. After transforming $ E$ you will have equations system with rational coefficients.[/quote]\r\n\r\nAccording to your idea - For me it is not so easy. It seems it is interesting problem for begginers and I'f like someone to post a complete solution, because this question suffices it. \r\n\r\nThank you",
  "Solution_6": "[b]Problem.[/b] Suppose $ Ax\\equal{}b$ is a consistent system of equations, where $ A, b$ have rational entries only. Then there exists a solution to the aforementioned system whose components are all rational numbers.\r\n\r\n[b]Solution.[/b] (Note that the set of rational numbers is closed under addition and multiplication.)\r\nApply the Gauss-Jordan elimination method to the augmented matrix $ [A|b]$. It is clear that the resulting matrix has only rational entries, therefore you can assign arbitrary rational values to the free variables, if they exist of course, and get some rational values for the pivot variables. $ \\blacksquare$\r\n\r\nPS. Note how the original problem follows from this one."
}
{
  "Tag": [],
  "Problem": "Johnny starts writing down numbers starting with 1, then 2, then 3, ... He gets tired and stops. If the last number he wrote was 2006, how many digits did peter write in all?",
  "Solution_1": "[hide]Well you know that the # of digits between 1-9 is 9. Then 10-99 is 180. And 100-999 is 2700. 1000-1999 is 4000. 2000-2006 is 28. Adding all this up gets 6917.[/hide]",
  "Solution_2": "[hide=\"solution\"]\n\n9  one digit numbers\n90 2 digit numbers\n900 3 digit numbers\n1007 4 digit numbers\n\n9+90*2+900*3+1007*4=6917[/hide]",
  "Solution_3": "[quote=\"xxxxx\"]Johnny starts writing down numbers starting with 1, then 2, then 3, ... He gets tired and stops. If the last number he wrote was 2006, how many digits did peter write in all?[/quote]\r\nTechnically, you cannot figure out how many Peter has written because we are given information about Johnny!",
  "Solution_4": "[hide]Peter wrote 0 digits[/hide]"
}
{
  "Tag": [
    "articles",
    "LaTeX",
    "\\/not_monitored"
  ],
  "Problem": "Please post here any problems you may have and we shall try to fix them as soon as possible. \r\n\r\nThe source field problem is solved.",
  "Solution_1": "No, it's not supposed to do that... strangely. I'll look into it.",
  "Solution_2": "since the site merger i'm no longer able to log in automatically. I select the log me in automatically box, but i still have to log in manually every time.",
  "Solution_3": "The automatic log in seems to work fine for me..",
  "Solution_4": "[quote=\"jeffrey\"]since the site merger i'm no longer able to log in automatically. I select the log me in automatically box, but i still have to log in manually every time.[/quote]\r\n\r\nStrange... it works for me.",
  "Solution_5": "can anyone give me insight into this problem? its not that inconvenient, but it would still be great if i could fix it.",
  "Solution_6": "The only thing that I can think of is to clear your cookies.  It works for me and since noone else has complained that leads me to believe that it is just you.",
  "Solution_7": "[quote=\"jeffrey\"]can anyone give me insight into this problem? its not that inconvenient, but it would still be great if i could fix it.[/quote] This is usually a problem associated with phpBB's management of cookies. Deleting the AoPS/MathLinks cookie is the simplest way to solve the problem. Then log in checking the box with the automatical login and you are done.",
  "Solution_8": "Thanks. It's fixed now.",
  "Solution_9": "this isnt so much of a problem as it is an inconvenience- every time i log in, i i always check new stuff on the site by looking at \"see posts since last visit\"- the problem is, it will say \"seach found 164 matches\" (or something to that effect) when in reality, there have only been about 10-20 posts since i last logged in. never had this problem previously, anything you can do about it? its not a big huge problem, but you know, if you could fix it, that would be nice too.",
  "Solution_10": "One thing you can do for now is hit 'Mark all forums read' (at the top of the forum on the index page) when you're done with a visit.",
  "Solution_11": "The old AoPS site did not require login to read the \"public\" forums.  However, there are some forums on the new, combined site that *do* require login.  For example, under \"Advanced --> Algebra\", forums #37 and #38 require login to view the posts.  [I don't know how to get an exhaustive list.  I've just noticed these two.]\r\n\r\nIs this an intentional change or is this an oversight?",
  "Solution_12": "Thanks- I tried that, didn't work... Hmm... very odd. But its a very small prob/inconvenience, really. I wouldn't worry too much about it.",
  "Solution_13": "[quote=\"rcv\"]The old AoPS site did not require login to read the \"public\" forums.  However, there are some forums on the new, combined site that *do* require login.  For example, under \"Advanced --> Algebra\", forums #37 and #38 require login to view the posts.  [I don't know how to get an exhaustive list.  I've just noticed these two.]\n\nIs this an intentional change or is this an oversight?[/quote]\r\n\r\nLooking in to it. . .",
  "Solution_14": "Here's an interesting one.  [I'm guessing it's probably because some forums are not openly readable.]  It's amusing, anyway.\r\n\r\n- I just log on to AoPS, after having been off the computer for hours.\r\n\r\n- I happen across this post:  [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=101506&highlight=#101506[/url]\r\n\r\n- I'm interested in reading the referenced article, which points to [url]http://www.mathlinks.ro/viewtopic.php?t=14167[/url].\r\n\r\n- I am taken to the Mathlinks login screen.\r\n\r\n- I login to Mathlinks.\r\n\r\n- I am taken to the \"Page does not exist\" page.\r\n\r\n- After clicking \"back\" twice, I get to see article 14167.  Yeah!\r\n\r\n\r\n\r\nBy the way, when I use more modern browsers, the security icons light up like Christmas trees, as http://www.artofproblemsolving.com and http://www.mathlinks.ro try to share cookies between each other.  I presume you folks know it isn't Kosher (with modern security recommendations) to share cookies across domains.  I'm wondering if some of the cookie problems and different behaviors folks are seeing might be due to differences in security settings.  Cheers!",
  "Solution_15": "We've solved (I think) the links in posts problem.  As you'll see in the post above, the links to topics now both point to whatever server you're coming from.  I'll have to look into more detail for the cookies issue.  Please let me know if you see instances where this little fix didn't work out.",
  "Solution_16": "[quote=\"rcv\"]The old AoPS site did not require login to read the \"public\" forums.  However, there are some forums on the new, combined site that *do* require login.  For example, under \"Advanced --> Algebra\", forums #37 and #38 require login to view the posts.  [I don't know how to get an exhaustive list.  I've just noticed these two.]\n\nIs this an intentional change or is this an oversight?[/quote]It is an intentional change, or rather a remain from the old mathlinks site. The forums that required users to be logged in are the solved problems forums. We will remove this soon.",
  "Solution_17": "I'm having troubles on the forums. First of all, numbers in Latex don't show up. There's just a white square with a red X in it followed by the latex in normal text. Also, on the top where Forum, Math Jams, and My Classes should be are squares with red X's. Same thing goes with the side headings like Enroll and Resources. I have HTML and everything else enabled so I don't know what the problem is.",
  "Solution_18": "I see no one else complaining about this, so I assume it must be a particular thing. \r\nIf you can answer to these questions maybe we can help: what url do you use and does not work ? and what browser (name and version) & os ?",
  "Solution_19": "URL?  Isn't that like http://www.yahoo.com?  I just type in artofproblemsolving.com\r\nMy browser is Internet Explorer but I'm not sure which version, and my operating system is Microsoft 98.",
  "Solution_20": "[quote=\"NightFlarer\"]My browser is Internet Explorer but I'm not sure which version, and my operating system is Microsoft 98.[/quote]\r\n\r\nClick the \"Help\" menu and click \"About\".  It should tell you which version you are using.",
  "Solution_21": "A comment: If you login on mathlinks and look through all the forums and stuff, then login on artofproblemsolving, it will still say that all the stuff you looked at is unread.",
  "Solution_22": "well apparently there's still a very very small flaw in the new site: http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=4\r\n\r\nit doesn't disturb me, but sometimes these 'peeks' into your source code can lead to hackers easily finding security holes in it, which can lead to eum... unwished effects :D (not by me, all I now know is that there must be more forums out there than you actually display)\r\n\r\ncould easily be fixed by first checking if there's any user with that user_id before going on (cuz now the user_id apparently gets lost somehow if there is no such user :?)",
  "Solution_23": "I'm still seeing two distinct, weird things.  I suspect they are both related to cookie handling.\r\n\r\nTwice, in the last 7 - 10 days, I have gotten an error that reports a cookie was too long for the Apache Web Server.  [I think the error occured when logging on, and I think it said the Apache limit was 8KB.]  The solution was to delete the cookies.  I note your recent instruction for users to delete their cookies.  Perhaps this issue will never occur again.  Or perhaps my cookies will grow gradually, until I hit this limit again.\r\n\r\nThe other issue occurs only when I'm using MS XP SP 1, with IE 6.0.28.  I go to the Logon screen, and enter my Userid and Password.  I am then taken to the Forums page, without any errors, but I do not appear to be logged on.  [I don't have a link to view postings since last visit, I cannot post, and the Logon button is still active.]  So, I go through the logon process a second time.  The second time, the logon really succeeds, but it reports that my last logon was ~1 minute ago.  I deleted all IE cookies yesterday, but this problem bit me again this morning.  [I have never observed this behavior when using IE 5, Netscape, or Mozilla.]",
  "Solution_24": "I hope that first error is dead now - let me know if it happens again.\r\n\r\nWill think about that other one - I have no immediate idea what's causing it.",
  "Solution_25": "could be the append_sid() being forgotten in a page redirection that was used. \r\n\r\ncan you post exactly what happens and how we can reproduce it?",
  "Solution_26": "when i was browsing the forums(not logged in) things appeared in yellow saying there was a new message.\r\nIs this supposed to happen?",
  "Solution_27": "What things appeared in yellow? A window? A pop up? ... How many times did it happen to you? \r\n\r\nAre you sure you weren't logged in?",
  "Solution_28": "[quote=\"Valentin Vornicu\"]What things appeared in yellow? A window? A pop up? ... How many times did it happen to you? \n\nAre you sure you weren't logged in?[/quote]\r\nUm it says there are new messsages.\r\nand the login icon was on the side",
  "Solution_29": "Do you use a pop-up blocker? Maybe that's what's causing your problem. Everytime you receive a private message a small pop-up appears when you loggin announcing you of that event. If you have a pop-up blocker, most probably this will not be shown, but instead you will see a warning message, usually colored in yellow, to attract attention. That's all ;)",
  "Solution_30": "[quote=\"Valentin Vornicu\"]Do you use a pop-up blocker? Maybe that's what's causing your problem. Everytime you receive a private message a small pop-up appears when you loggin announcing you of that event. If you have a pop-up blocker, most probably this will not be shown, but instead you will see a warning message, usually colored in yellow, to attract attention. That's all ;)[/quote]\r\ni dont have a popup blocker"
}
{
  "Tag": [
    "limit",
    "MATHCOUNTS"
  ],
  "Problem": "what is $\\lim_{n\\to\\infty}{\\sum_{x=1}^n{\\frac1x}}$?",
  "Solution_1": "didn't this come up in mathcounts a while ago? \r\n\r\ngo [url=http://www.rocketcitymath.org/]here[/url], click on archives, and go to apollo round 2. They have answers.",
  "Solution_2": "so basically you are looking for the sum of the harmonic series.\r\nit divergest to infinity.\r\nthere are many different ways to prove it.",
  "Solution_3": "or look here: [url=http://mathworld.wolfram.com/HarmonicSeries.html]mathworld.wolfram.com/HarmonicSeries.html[/url]"
}
{
  "Tag": [
    "function",
    "induction",
    "algebra proposed",
    "algebra"
  ],
  "Problem": "Find all continues functions $ f: \\mathbb R\\to\\mathbb R$ such that for all $ x,y,z\\in\\mathbb R$,\n\\[f(x\\plus{}f(y\\plus{}f(z)))\\equal{}f(x)\\plus{}f(f(y))\\plus{}f(f(f(z))).\\]",
  "Solution_1": "[quote=\"hana1122\"]Find all continues functions $ f: \\mathbb R\\to\\mathbb R$ such that for all $ x,y,z\\in\\mathbb R$,\n$ f(x \\plus{} f(y \\plus{} f(z))) \\equal{} f(x) \\plus{} f(f(y)) \\plus{} f(f(f(z)))$[/quote]\r\n\r\nLet $ P(x,y,z)$ be the assertion $ f(x \\plus{} f(y \\plus{} f(z))) \\equal{} f(x) \\plus{} f(f(y)) \\plus{} f(f(f(z)))$\r\n\r\n$ P(0,0,0)$ $ \\implies$ $ f(0) \\plus{} f(f(0)) \\equal{} 0$ and so either $ f(0) \\equal{} f(f(0)) \\equal{} 0$, either $ f(0)$ and $ f(f(0))$ have opposite signs and $ \\exists u$ between them such that $ f(u) \\equal{} 0$\r\n\r\nSo $ \\exists u$ such that $ f(u) \\equal{} 0$\r\n\r\n$ P(x,y,u)$ $ \\implies$ $ f(x \\plus{} f(y)) \\equal{} f(x) \\plus{} f(f(y)) \\plus{} f(f(0))$ and so $ g(x \\plus{} f(y)) \\equal{} g(x) \\plus{} g(f(y))$ where $ g(x) \\equal{} f(x) \\plus{} f(f(0)) \\equal{} f(x) \\minus{} f(0)$. So :\r\n\r\n$ g(x \\plus{} y) \\equal{} g(x) \\plus{} g(y)$ $ \\forall x\\in\\mathbb R,\\forall y\\in f(\\mathbb R)$ and also (using $ x \\minus{} y$ instead of $ x$) :\r\n$ g(x \\minus{} y) \\equal{} g(x) \\minus{} g(y)$ $ \\forall x\\in\\mathbb R,\\forall y\\in f(\\mathbb R)$\r\n\r\nIf $ f(x) \\equal{} c$ constant, then $ P(x,y,z)$ $ \\implies$ $ c \\equal{} 3c$ and so $ f(x) \\equal{} 0$ $ \\forall x$, which indeed is a solution.\r\n\r\nIf $ f(x)$ is not constant, then $ \\exists b\\ne0$ and $ y$ such that $ g(y) \\equal{} b\\ne 0$ and an immediate induction gives :\r\n$ g(x \\plus{} ny) \\equal{} g(x) \\plus{} nb$ $ \\forall x,\\forall n\\in\\mathbb N$\r\n$ g(x \\minus{} ny) \\equal{} g(x) \\minus{} nb$ $ \\forall x,\\forall n\\in\\mathbb N$\r\n\r\nand so, since $ g(x)$ is continuous,  $ g(\\mathbb R) \\equal{} \\mathbb R$ and so $ f(\\mathbb R) \\equal{} \\mathbb R$ and so :\r\n\r\n$ g(x \\plus{} y) \\equal{} g(x) \\plus{} g(y)$ $ \\forall x,y$\r\n\r\nAnd so $ g(x) \\equal{} ax$ (continuous solutions of Cauchy) and $ f(x) \\equal{} ax \\plus{} b$\r\n\r\nPlugging this back in the original equation, we get $ b(a \\plus{} 2) \\equal{} 0$ hence the solutions :\r\n\r\n$ \\boxed{f(x) \\equal{} ax}$ and $ \\boxed{f(x) \\equal{} b \\minus{} 2x}$"
}
{
  "Tag": [
    "trigonometry",
    "algebra",
    "system of equations"
  ],
  "Problem": "If $\\sin a+\\cos a=.2$, what is $\\sin a-\\cos a$?",
  "Solution_1": "[hide]$\\sin a-\\cos a=\\frac{-\\cos 2a}{\\sin a+\\cos a}$\n[/hide]",
  "Solution_2": "If sin a+ cos a=.2, then (sin a)^2+(cos a)^2+2sinacosa=0.04, so 2sina*cosa=-0.96 so -2sinacosa=0.96\r\n\r\nso then sina-cosa=x, (sina)^2+(cosa)^2-2sinacosa=x^2, so we have x^2=1+0.96, so x^2=1.96, so $x=1.4$.",
  "Solution_3": "[hide=\"In other words\"] $(\\sin a+\\cos a)^{2}+(\\sin a-\\cos a)^{2}= 2$ [/hide]\n[hide=\"Why?  Because...\"] $\\sin^{2}a+\\cos^{2}a+2 \\sin a \\cos a+\\sin^{2}a+\\cos^{2}a-2 \\sin a \\cos a = 2 (\\sin^{2}a+\\cos^{2}a) = 2$ [/hide]\n[hide=\"But also because...\"] $\\sin a+\\cos a = \\sqrt{2}\\cos (a-\\frac{\\pi}{4})$ \n$\\sin a-\\cos a = \\sqrt{2}\\cos (a-\\frac{3\\pi}{4}) =-\\sqrt{2}\\sin (a-\\frac{\\pi}{4})$ [/hide]",
  "Solution_4": "This can be transformed into an interesting system of equations problem: let $\\sin x=a$ and $\\cos x=b$, then we have that:\r\n\r\n$a+b=\\frac{1}{5}$\r\n$a^{2}+b^{2}=1$\r\n\r\nDetermine $t=a-b$.\r\n\r\nI guess this might seem contrived, but it usually helps to square when you have $a+b$ and $a-b$.\r\n\r\n$t^{2}+\\frac{1}{25}=(a+b)^{2}+(a-b)^{2}=a^{2}+2ab+b^{2}+a^{2}-2ab+b^{2}=2(a^{2}+b^{2})=2$.\r\n\r\nFrom this we determine that $|t|=\\frac{7}{5}$."
}
{
  "Tag": [
    "probability",
    "search",
    "floor function",
    "limit",
    "number theory",
    "prime numbers",
    "number theory proposed"
  ],
  "Problem": "Let $ n$ be a number in the form $ 0.abcd\\cdots$ for example $ \\frac{\\pi}{10}$\r\nIt's true that this number can be written in the form $ 0.(p_1)(p_2)(p_3)\\cdots$? ($ p_i$ is a prime number) for the same example above:\r\n\r\n$ 0.(314159)(2)(6535897)(93238462643)(3832795028841971693)\\cdots$\r\n\r\neach number in () is a prime.\r\n\r\nthe answer is probably a yes, since each digit in pi is random (it really is?), and the probability of forming a finite prime of any length is non-zero.\r\n\r\nBut the real question is the following.\r\nlet $ p_1$ the biggest prime number possible to form by this \"sequence\", no matter how further you search, you can't form another prime number. (althought each digit of pi is random, will exist a prime numbers in this exact form? will it grow infinitely? i assume it will not, and this biggest prime in this form exist)\r\nSo you stop there and start searching for $ p_2$ and etc.\r\n\r\nThe questions I have are:\r\na) there is this biggest $ p_1$ prime? If no, ends. if yes, there is $ p_2$? etc.?\r\n\r\nI made some search in this, but the density of prime numbers found start to get really thin, so the purpose of the question.\r\n\r\nthanks",
  "Solution_1": "By the conjectured normality of $ \\pi$, you would be able to find arbitrary long primes in it's decimal expansion.\r\nMore exactly, there are arbitrary long primes in any such \"prime decomposition\" of $ \\pi$: if there would be a maximal prime of length $ n$ in one such sequence, assuming normality, we can find $ 2n$ consecutive $ 0$'s (or alternatively and easier to find: just even digits; or ones divisible by $ 3$ or $ 5$ or $ 7$) in $ \\pi$. Then any way to write $ \\pi$ as such a sequence of primes must get over this subsequence, which can only be done if there is a prime of length at least $ n\\plus{}1$ (note that no prime can be completely in that $ 2n$ digits). Contradiction.",
  "Solution_2": "I think there is a little bit confusion on this.\r\nThe question I proposed can be formuled as \"How many primes there are in a specific form\"\r\nThe answer is not just that.\r\nHow many primes start with $ n$ specific digits?\r\nIf it were like that you could prove that there is infinitely many fermat numbers, because it is isomorphic to this problem, isn't it?\r\n\r\nCorrect if I am wrong, but I don't belive in your \"proof\", there are infinitely many primes, but the existence or not of this maximum is not yet prooven or disprooved",
  "Solution_3": "I don't see how Fermat numbers should come into the game. And by the prime number theorem, infinitely many primes start with $ n$ specific digits (note that we can have as many digits at the end as we like).\r\nWell, normality is definitely not proven, but it it's necessary for randomness (a random number is normal almost surely). So is this the only doubt of yours\u00bf",
  "Solution_4": "Consider the number $ 0.1111111111111...$ in binary.\r\nThere exist this prime decomposition.\r\nSuppose there is a maximum $ p_1$ of length $ n$.\r\nTo overcome the series of $ m$ $ 1$'s next, there must exist a primer number greater than $ n\\plus{}1$. And there is.\r\nSo there is no maximum.\r\nAnd this prime numers are only fermat numbers, So there is infinitely many fermat numbers?",
  "Solution_5": "I was talking about $ \\pi$ as an example where you don't have maximal primes for sure (as you have in any number that contains arbitrary long sequences of $ 0$. And you should reread my proof as we don't need to overcome $ 1$'s but $ 0$'s (a big difference).",
  "Solution_6": "So, Since there is no maximum prime. We can say that:\r\n$ \\lfloor 10^n \\pi \\rfloor$ is prime for infinitely many $ n$ and\r\n$ \\lim_{x \\to \\infty} \\lfloor 10^n \\pi \\rfloor$ is a prime number? (is infinity,but you can think as a infinity prime, since there is no prime upper bond)",
  "Solution_7": "It was that shown any \"subdivision\" of $ \\pi$ into primes must contain arbitrary long primes. This doesn't show (and is in some sense unrelated to) the possible length of a prime at the beginning, which is probably infinite, but somewhat more difficult.\r\nAnd I don't see any usefull concept to define an \"infinity prime\" in this meaning (that limit is somewhat strange as even if the first statement is true, still not all these numbers are prime (in fact, only a very small fraction of them is)).",
  "Solution_8": "So for any normal number $ x$: $ S \\equal{} \\left\\{\\left\\lfloor x 10^n\\right\\rfloor : n\\in \\mathbb{Z}\\land \\left\\lfloor x 10^n\\right\\rfloor \\in \\mathbb{P}\\right\\}$, is a incontable set.\r\n\r\nThis is somewhat similar to mill's constant. (but is not truel for all $ n$, but for some (and not so related at all))\r\n\r\nI saw my stupid dumb mistake now.\r\nSorry."
}
{
  "Tag": [
    "ratio",
    "geometry",
    "circumcircle",
    "vector",
    "symmetry",
    "search"
  ],
  "Problem": "Could anyone show that the medians intersect at a point known as the centroid and that this centroid splits the medians into a $2: 1$ ratio?",
  "Solution_1": "Medians?? I though it was the Perpendicular Bisectors?",
  "Solution_2": "[quote=\"fireforce\"]Medians?? I though it was the Perpendicular Bisectors?[/quote]\r\nThe perpendicular bisectors are concurrent at the circumcenter.",
  "Solution_3": "Perpendicular bisectors intersect at the circumcenter, not the centroid.",
  "Solution_4": "There's a really short proof using areal co-ordinates:\r\n\r\nLet $AD$ and $BE$ be two medians. The co-ordinates of $A$ and $B$ are $(1, 0, 0)$ and $(0, 1, 0)$.  $D$ and $E$ have co-ordinates $(0, \\frac{1}{2}, \\frac{1}{2})$ and $(\\frac{1}{2}, 0, \\frac{1}{2})$. From this we can see the point they intersect at $G$ must have co-ordinates are $(\\frac{1}{3}, \\frac{1}{3}, \\frac{1}{3})$. $\\overrightarrow{AG} = \\frac{2}{3} \\overrightarrow{AD}$. $\\blacklozenge$",
  "Solution_5": "[hide]\nIn $\\triangle ABC$, let $D$, $E$ and $F$ be the midpoints of $AB$, $BC$ and $CA$. Let $G$ be the centroid.\n\nNote that $DE\\parallel AC$ and $DE=\\frac{1}{2} AC$ (Midline theorem).\n$\\triangle DEG\\sim \\triangle CAG$, so $DG: GC=EG: GA=DE: CA=1: 2$.\n[/hide]",
  "Solution_6": "See i learnt how to prove it using vectors, but i'm not a big fan of vectors so i was just wondering if there was a better proof.",
  "Solution_7": "Let $\\triangle ABC$ have midpoints $D$, $E$, and $F$ on $AB$, $BC$, and $CA$, respectively. \r\n\r\n--\r\nLemma:\r\n$DC$, $EA$, and $FB$ are concurrent\r\nby ceva, we have that\r\n$\\frac{AD}{DB}*\\frac{BE}{EC}*\\frac{CF}{FA}=1$ using the midpoints, so they are concurrent. Say this point is $G$\r\n--\r\n\r\nAssign a mass of $1$ to point $A$ or $w_A=1$, then to balance along the midpoints, $w_B=w_C=1$, then using mass points again,\r\n$w_A+w_B=w_D=2$, symmetrically, we have $w_E=w_F=2$, then\r\n\r\n$w_A*AG=w_E*EG$\r\n$\\frac{AG}{EG}=2$\r\nthe other results follow by symmetry\r\n\r\n\r\nP.S. if you don't know mass points search \"mass points\"\r\nthis question has been asked a million times\r\n\r\n----\r\n\r\nanother solution\r\n\r\nassuming that they are concurrent, apply menelaus:\r\n\r\n$\\frac{AD}{DB}*\\frac{BC}{CE}*\\frac{EG}{GA}=1$\r\n$1*2*\\frac{EG}{GA}=1$\r\n$\\frac{AG}{GE}=2$ as desired\r\n\r\n\r\n----\r\n\r\nanother...\r\n\r\n--\r\nLemma:\r\nIf $A$, $D$,  and $B$ are collinear, and $P$ is a point not on that line, then $\\frac{[ADP]}{[BDP]}=\\frac{AD}{DB}$ (where $[ABC]$ is the area of $\\triangle ABC$)\r\n\r\nconsider the height from $P$ to $AB$, this is the height for both triangles, so dividing the area formula $[ADP]=\\frac{1}{2}*AD*h$, we get the result\r\n--\r\n\r\nthen using the midpoints, $[AGF]=[CGF]$ and $[AFB]=[CFB]$, subtract those to get\r\n\r\n$[AGB]=[CGB]$ (1)\r\n\r\nthen we have that $[CGE]=[BGE]$ and $[AGD]=[BGD]$, so put that into (1) to get:\r\n\r\n$2[CGE]=2[AGD]$\r\n\r\nusing transivity, we have that all the \"little triangles\" are equal in area\r\n\r\nthen $\\frac{[ABG]}{[EBG]}=\\frac{AG}{GE}$\r\n\r\n$[ABG]=2[EBG]$ by the little triangles all having the same area, so\r\n\r\n$\\frac{AG}{GE}=2$"
}
{
  "Tag": [
    "logarithms"
  ],
  "Problem": "Find all positive integers $ x, y ,z $ s.t.\r\n\r\n$ x^ {y^z} . y^ {z^x} . z^ {x^y} =\\ 5 xyz $",
  "Solution_1": "hmmm are you sure this is possible?",
  "Solution_2": "It's from a book!!!! And the answer satisfies!!!",
  "Solution_3": "Rushil, are you sure, all of the variables are distinct :? ?",
  "Solution_4": "Maybe log it?\r\n\r\n$\\log{x^{y^z}\\cdot{y}^{z^x}\\cdot{z}^{x^y}}=\\log{5xyz}$ \r\n\r\n$\\log{x^{y^z}}+\\log{y^{z^x}}+\\log{z^{x^y}}=\\log5+\\log{x}+\\log{y}+\\log{z}$\r\n\r\n$y^z\\log{x}+z^x\\log{y}+x^y\\log{z}=\\log5+\\log{x}+\\log{y}+\\log{z}$\r\n\r\nDoes that help any?",
  "Solution_5": "Actually you'll have to make some cases since we require integer solutins!!!",
  "Solution_6": "[quote=\"nat mc\"]Maybe log it?\n\n$\\log{x^{y^z}\\cdot{y}^{z^x}\\cdot{z}^{x^y}}=\\log{5xyz}$ \n\n$\\log{x^{y^z}}+\\log{y^{z^x}}+\\log{z^{x^y}}=\\log5+\\log{x}+\\log{y}+\\log{z}$\n\n$y^z\\log{x}+z^x\\log{y}+x^y\\log{z}=\\log5+\\log{x}+\\log{y}+\\log{z}$\n\nDoes that help any?[/quote]\r\n\r\nthat's exactly what I did nat mc, however, I was kind of lost after that, and didn't know exactly what to do :?",
  "Solution_7": "it is obvious that WLOG $x=5k$",
  "Solution_8": "[quote=\"peeta\"]it is obvious that WLOG $x=5k$[/quote]\r\nWhat do you mean. How does that help any?",
  "Solution_9": "Has nobody got any solution. Should I post the one which I have??",
  "Solution_10": "Here's how I got my solution.\r\n\r\nStarting with the original equation:\r\n\r\n$x^{y^{z}}y^{z^{x}}z^{x^{y}} = 5xyz$\r\n\r\ndivide both sides by $xyz$ to get:\r\n\r\n$x^{y^z-1}y^{z^x-1}z^{x^y-1}=5$\r\n\r\nsince 5 is prime, we have to have 2 factors of 1, and 1 factor of 5. If $z=1$, we know that the last two factors will always be 1. So, we have to find $x$ and $y$ such that $x^{y-1} = 5$ Since we want integer solutions, we have to make the exponent 1 and the base 5. So:\r\n\r\n$x=5$\r\n$y=2$\r\n\r\nSo we have the solution (5,2,1). Since the exponents always go in the same order, the positions of the numbers don't matter, so all the solutions (that i found) are:\r\n\r\n(5,2,1)\r\n(2,1,5)\r\n(1,5,2)",
  "Solution_11": "[quote=\"Hamster1800\"]Here's how I got my solution.\n\nStarting with the original equation:\n\n$x^{y^{z}}y^{z^{x}}z^{x^{y}} = 5xyz$\n\ndivide both sides by $xyz$ to get:\n\n$x^{y^z-1}y^{z^x-1}z^{x^y-1}=5$\n\nsince 5 is prime, we have to have 2 factors of 1, and 1 factor of 5. If $z=1$, we know that the last two factors will always be 1. So, we have to find $x$ and $y$ such that $x^{y-1} = 5$ Since we want integer solutions, we have to make the exponent 1 and the base 5. So:\n\n$x=5$\n$y=2$\n\nSo we have the solution (5,2,1). Since the exponents always go in the same order, the positions of the numbers don't matter, so all the solutions (that i found) are:\n\n(5,2,1)\n(2,1,5)\n(1,5,2)[/quote]\r\nvery nice, and those are the only integer solutions"
}
{
  "Tag": [
    "ratio",
    "geometry",
    "angle bisector",
    "geometry proposed"
  ],
  "Problem": "(O) is a circle which touches side AB,BC,CD,DA of the quadrilateral ABCD. The diagonal BD meets (O) at E and F. H is the projection of O on AC. Prove that $ \\angle BHE\\equal{}\\angle CHF$",
  "Solution_1": "Should be: Prove that $ \\angle BHE \\equal{} \\angle DHF.$\n\nThe diagonals $ AC, BD$ meet at $ X.$ $ AB, BC, CD, DA$ touch $ (O)$ at $ P, Q, R, S.$ $ OH$ cuts $ (O)$ at $ M, N$ on the same sides as $ B, D.$ $ BD, QR, SP, MN$ meet at the pole $ K$ of $ AC$ and $ OK \\equal{} \\frac {r^2}{OH}$, where $ r$ is radius of $ (O)$ $ \\Longrightarrow$ the cross ratio $ G(K, H, M, N) \\equal{} \\minus{} 1$ is harmonic. $ QR, SP$ meet $ AC$ at $ U, V.$ Since $ MQ, NR$ and $ MP, NS$ meet on the polar $ AC$ of $ K,$ the cross ratios  $ (K, X, E, F) \\equal{} (K, H, M, N) \\equal{} \\minus{} 1$ and $ (K, X, B, D) \\equal{} (K, V, P, S) \\equal{} (K, H, M, N) \\equal{} \\minus{} 1$ are also harmonic. Since the angle $ \\angle KHX$ is right, $ HX \\equiv AC$ is the common angle bisector of the angles $ \\angle BHD, \\angle EHF.$",
  "Solution_2": "Dear yetti,\r\nCan you detail your solution?",
  "Solution_3": "Could you first confirm that $ \\angle BHE \\equal{} \\angle DHF$ is the correct proposition ? You typed $ \\angle BHE \\equal{} \\angle CHF$ twice, they are not equal and I am guessing.",
  "Solution_4": "What is the [quote=\"mr.danh\"]projection of $ O$ on $ AC$[/quote]?????"
}
{
  "Tag": [
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "Let $\\Sigma$ and $\\Sigma '$ be cuboids with $\\Sigma$ lying inside $\\Sigma '$. Suppose $\\Sigma$ has dimensions $a \\times b \\times c$ and $\\Sigma '$ has dimensions $a' \\times b' \\times c'$. Must we have $a + b + c \\leq a' + b' + c'$?",
  "Solution_1": "i can remember the answer... but not know how to prove it"
}
{
  "Tag": [
    "geometry",
    "perimeter",
    "ratio",
    "trigonometry",
    "analytic geometry",
    "similar triangles"
  ],
  "Problem": "Delroy\u2019s sailboat has two sails that are similar triangles. The larger sail has\r\nsides of 10 feet, 24 feet, and 26 feet. If the shortest side of the smaller sail\r\nmeasures 6 feet, what is the perimeter of the smaller sail?\r\n(1) 15 ft     (3) 60 ft\r\n(2) 36 ft     (4) 100 ft",
  "Solution_1": "[hide]choice 2: 36 is the answer. Use similar triangles. [/hide]",
  "Solution_2": "How would you set up similar triangles to solve this question?",
  "Solution_3": "[hide]Shortest and longest sides correspond in similar triangles, so the ratio of the smaller sail to the bigger one is $3: 5$. Perimeters also have the same ratio, so $p=\\frac{3}{5}\\cdot(10+24+26)=36\\Longrightarrow(2)$.[/hide]",
  "Solution_4": "i_like_pie said:\r\n\r\n\"Perimeters also have the same ratio.\"\r\n\r\nDo you mean all perimeters or just questions dealing with similar triangles?\r\n\r\nAlso, does this site answer trigonometry questions and geometric proof questions using the famous high school statement vs reason chart?",
  "Solution_5": "In similar triangles, everything has the same ratio: corresponding sides, heights, medians, perimeters, etc.\r\n\r\nIf the ratio of corresponding sides is $3: 5$, then the ratio of the perimeters will also be $3: 5$.",
  "Solution_6": "[quote=\"i_like_pie\"]In similar triangles, everything has the same ratio: corresponding sides, heights, medians, perimeters, etc.\n\nIf the ratio of corresponding sides is $3: 5$, then the ratio of the perimeters will also be $3: 5$.[/quote]\r\n\r\nnot EVERYTHING\r\n\r\nthe ratio of the area is the ratio of the sides (or the perimeter or altitudes or whatever) squared.\r\n\r\nso if the ratio of perimeters is $\\frac{3}{5}$, the ratio of the areas would be $\\frac{9}{25}$\r\n\r\n(also, if the figure was 3-D, then the ratio of volumes would be the ratio of sides cubed)",
  "Solution_7": "[quote=\"Interval\"]\nAlso, does this site answer trigonometry questions and geometric proof questions using the famous high school statement vs reason chart?[/quote]\r\n\r\nPlease understand that this site is a [u]forum[/u], [b]NOT[/b] a tutoring site; so it does not usually act as a collective. Thus, the usage of the \"statement vs. reason chart\" will vary depending on who is responding; but in most cases, the chart is unnecessarily tedious to write out, as everything which can be put onto the chart can also be put into plain, intelligible English sentences. I do not like the chart as I see it as unwieldy, but also since it has given me many a wrist cramp in freshman year geometry :P .",
  "Solution_8": "I think the \"statement vs reason charts\" does not help high school students.  What it creates is more confusion and fear in the lives of students who are so afraid of math and math teachers trying to intimidate rather than teach numbers clearly.",
  "Solution_9": "[quote=\"Interval\"]I think the \"statement vs reason charts\" does not help high school students.  What it creates is more confusion and fear in the lives of students who are so afraid of math and math teachers trying to intimidate rather than teach numbers clearly.[/quote]\r\n\r\nWell for some it is easier to organize their thoughts in that manner; for me, a paragraph proof works just as well, usually even better, than that organizational plan.",
  "Solution_10": "Can coordinate geometry or analytical geometry be used to solve geometric proofs?\r\n\r\nIf so, do you have an example?"
}
{
  "Tag": [
    "trigonometry",
    "conics",
    "parabola",
    "calculus",
    "derivative",
    "function",
    "vector"
  ],
  "Problem": "Find the maximum and minimum value of\r\n\r\n$ \\sin^2\\theta\\plus{}\\sin\\theta\\plus{}\\cos\\theta$.",
  "Solution_1": "[hide=\"Maximum\"]Write as $ x^2 \\plus{} x \\plus{} \\sqrt {1 \\minus{} x^2}$ with $ \\minus{} 1\\le x \\le 1$. Differentiating gives the answer easily: Solve $ \\minus{} \\frac {x}{\\sqrt {1 \\minus{} x^2}} \\plus{} 2x \\plus{} 1 \\equal{} 0$ and check that the value is attainable.[/hide]\n\n[hide=\"Minimum\"]Write as $ x^2 \\plus{} x \\minus{} \\sqrt {1 \\minus{} x^2}$ with $ \\minus{} 1\\le x \\le 1$. Use the same technique as above and check that the value is attainable.[/hide]",
  "Solution_2": "[quote=\"Temperal\"][hide=\"Maximum\"]Write as $ x^2 \\plus{} x \\plus{} \\sqrt {1 \\minus{} x^2}$ with $ \\minus{} 1\\le x \\le 1$. Differentiating gives the answer easily: Solve $ \\minus{} \\frac {x}{\\sqrt {1 \\minus{} x^2}} \\plus{} 2x \\plus{} 1 \\equal{} 0$ and check that the value is attainable.[/hide]\n\n[hide=\"Minimum\"]Write as $ x^2 \\plus{} x \\minus{} \\sqrt {1 \\minus{} x^2}$ with $ \\minus{} 1\\le x \\le 1$. Use the same technique as above and check that the value is attainable.[/hide][/quote]\r\n\r\n\r\nOther than Differentiating Method, anybody has any idea?",
  "Solution_3": "Anybody has new idea?",
  "Solution_4": "[quote=\"vinskman\"]Find the maximum and minimum value of\n\n$ \\sin^2\\theta \\plus{} \\sin\\theta \\plus{} \\cos\\theta$.[/quote]\r\n\r\nWe are to find the condition such that the parabola $ x\\plus{}\\left(y\\plus{}\\frac{1}{2}\\right)^2\\equal{}k\\plus{}\\frac{1}{4}$ has intersection point with the unit circle $ x^2\\plus{}y^2\\equal{}1$.",
  "Solution_5": "[quote=\"kunny\"][quote=\"vinskman\"]Find the maximum and minimum value of\n\n$ \\sin^2\\theta \\plus{} \\sin\\theta \\plus{} \\cos\\theta$.[/quote]\n\nWe are to find the condition such that the parabola $ x \\plus{} \\left(y \\plus{} \\frac {1}{2}\\right)^2 \\equal{} k \\plus{} \\frac {1}{4}$ has intersection point with the unit circle $ x^2 \\plus{} y^2 \\equal{} 1$.[/quote]\r\n\r\nwould you please post the details ?\r\n\r\n :o  :)",
  "Solution_6": "If $ \\sin ^ 2 \\theta \\plus{}\\sin \\theta \\plus{}\\cos \\theta \\Longleftrightarrow \\left(\\sin \\theta \\plus{}\\frac{1}{2}\\right)^2\\plus{}\\cos \\theta \\minus{}\\frac{1}{4}$ takes some value $ k$,\r\n\r\nthe parabpla $ x\\plus{}\\left(y\\plus{}\\frac{1}{2}\\right)^2\\equal{}k\\plus{}\\frac{1}{4}$ has more than one intersection points with the unit circle $ x^2\\plus{}y^2\\equal{}1$.",
  "Solution_7": "[quote=\"kunny\"]If $ \\sin ^ 2 \\theta \\plus{} \\sin \\theta \\plus{} \\cos \\theta \\Longleftrightarrow \\left(\\sin \\theta \\plus{} \\frac {1}{2}\\right)^2 \\plus{} \\cos \\theta \\minus{} \\frac {1}{4}$ takes some value $ k$,\n\nthe parabpla $ x \\plus{} \\left(y \\plus{} \\frac {1}{2}\\right)^2 \\equal{} k \\plus{} \\frac {1}{4}$ has more than one intersection points with the unit circle $ x^2 \\plus{} y^2 \\equal{} 1$.[/quote]\r\n\r\n\r\nMay be I stupid, why are their points of intersection related to the max. & min. values of $ \\sin ^ 2 \\theta \\plus{} \\sin \\theta \\plus{} \\cos \\theta$?",
  "Solution_8": "\uff29\uff46 $ k$ could take value of $ \\frac {3}{4} \\plus{} \\frac {\\sqrt {3}}{2}$, can't it be possible to exist the values of $ x,\\ y$ in $ x^2 \\plus{} y^2 \\equal{} 1$?\r\n\r\nThe answer is Yes. For $ x \\equal{} \\frac {\\sqrt {3}}{2},\\ y \\equal{} \\frac {1}{2}$, we can make a confirmation the existence of the point in unit circle, that is to say, \r\nthe parabola and the unit circle have more than one points of the intersection.",
  "Solution_9": "[quote=\"kunny\"]\uff29\uff46 $ k$ could take value of $ \\frac {3}{4} \\plus{} \\frac {\\sqrt {3}}{2}$, can't it be possible to exist the values of $ x,\\ y$ in $ x^2 \\plus{} y^2 \\equal{} 1$?\n\nThe answer is Yes. For $ x \\equal{} \\frac {\\sqrt {3}}{2},\\ y \\equal{} \\frac {1}{2}$, we can make a confirmation the existence of the point in unit circle, that is to say, \nthe parabola and the unit circle have more than one points of the intersection.[/quote]\r\n\r\nJust you mean that under the constraints $ x^2\\plus{}y^2\\equal{}1$ ,the the two values of k by counting the points of intersections is justified as max. & min. values for $ \\sin^2\\theta\\plus{}\\sin\\theta\\plus{}\\cos\\theta$? \r\n\r\nIs it?",
  "Solution_10": "Yes. :)",
  "Solution_11": "How would you find the maximum and minimum vales of $ x\\plus{}\\sqrt{1\\minus{}x^2}$ without calculus or trig?",
  "Solution_12": "[quote=\"kunny\"]Yes. :)[/quote]\r\n\r\nThen , as your examples to raise the value of $ k \\equal{}\\frac{3}{4}\\plus{}\\frac{\\sqrt 3}{2}$\r\njust by your obseveration nad not by solving equations?",
  "Solution_13": "Just by observation.",
  "Solution_14": "[quote=\"kunny\"]How would you find the maximum and minimum vales of $ x \\plus{} \\sqrt {1 \\minus{} x^2}$ without calculus or trig?[/quote]\r\n\r\nSorry , I have tried , but in vain.",
  "Solution_15": "$ x \\plus{} \\sqrt {1 \\minus{} x^2}$ would take some value of $ k$\r\n\r\n$ \\Longleftrightarrow$ there exists some real value $ x\\ ( \\minus{} 1\\leq x\\leq 1)$ such that $ x \\plus{} \\sqrt {1 \\minus{} x^2} \\equal{} k$\r\n\r\n$ \\Longleftrightarrow$ there exists some real value $ x\\ ( \\minus{} 1\\leq x\\leq 1)$ such that $ \\sqrt {1 \\minus{} x^2} \\equal{} \\minus{} x \\plus{} k$\r\n\r\n$ \\Longleftrightarrow$ the graph of $ y \\equal{} \\sqrt {1 \\minus{} x^2}$ has more than one points of intersection with the line $ y \\equal{} \\minus{} x \\plus{} k$.\r\n\r\nWe are to maximize and minimize the $ y$ intercept $ k$.",
  "Solution_16": "[quote=\"kunny\"]$ x \\plus{} \\sqrt {1 \\minus{} x^2}$ would take some value of $ k$\n\n$ \\Longleftrightarrow$ there exists some real value $ x\\ ( \\minus{} 1\\leq x\\leq 1)$ such that $ x \\plus{} \\sqrt {1 \\minus{} x^2} \\equal{} k$\n\n$ \\Longleftrightarrow$ there exists some real value $ x\\ ( \\minus{} 1\\leq x\\leq 1)$ such that $ \\sqrt {1 \\minus{} x^2} \\equal{} \\minus{} x \\plus{} k$\n\n$ \\Longleftrightarrow$ the graph of $ y \\equal{} \\sqrt {1 \\minus{} x^2}$ has more than one points of intersection with the line $ y \\equal{} \\minus{} x \\plus{} k$.\n\nWe are to maximize and minimize the $ y$ intercept $ k$.[/quote]\r\n\r\n\r\nJust you say,\r\n\r\nfirstly , we find out all the points of intersection between the unit circle and parabola.\r\nsecondly , we test all such points in the ways you mentioned y-intercept k to jusify the points for max.and min. value for the trigo. expressions, is it?\r\n :D",
  "Solution_17": "[quote=\"vinskman\"][quote=\"kunny\"]$ x \\plus{} \\sqrt {1 \\minus{} x^2}$ would take some value of $ k$\n\n$ \\Longleftrightarrow$ there exists some real value $ x\\ ( \\minus{} 1\\leq x\\leq 1)$ such that $ x \\plus{} \\sqrt {1 \\minus{} x^2} \\equal{} k$\n\n$ \\Longleftrightarrow$ there exists some real value $ x\\ ( \\minus{} 1\\leq x\\leq 1)$ such that $ \\sqrt {1 \\minus{} x^2} \\equal{} \\minus{} x \\plus{} k$\n\n$ \\Longleftrightarrow$ the graph of $ y \\equal{} \\sqrt {1 \\minus{} x^2}$ has more than one points of intersection with the line $ y \\equal{} \\minus{} x \\plus{} k$.\n\nWe are to maximize and minimize the $ y$ intercept $ k$.[/quote]\n\n\nJust you say,\n\nfirstly , we find out all the points of intersection between the unit circle and parabola.\nsecondly , we test all such points in the ways you mentioned y-intercept k to jusify the points for max.and min. value for the trigo. expressions, is it?\n :D[/quote]\r\n\r\nFor which problem?",
  "Solution_18": "[quote=\"kunny\"]If $ \\sin ^ 2 \\theta \\plus{} \\sin \\theta \\plus{} \\cos \\theta \\Longleftrightarrow \\left(\\sin \\theta \\plus{} \\frac {1}{2}\\right)^2 \\plus{} \\cos \\theta \\minus{} \\frac {1}{4}$ takes some value $ k$,\n\nthe parabpla $ x \\plus{} \\left(y \\plus{} \\frac {1}{2}\\right)^2 \\equal{} k \\plus{} \\frac {1}{4}$ has more than one intersection points with the unit circle $ x^2 \\plus{} y^2 \\equal{} 1$.[/quote]\r\n\r\n\r\nLet's consider that if k becomes a variable then the vertex of parabola will shift along positive /negative x-direction.When the parabola just touches the unit circle, just may count the max./min.  value of k. Is it?",
  "Solution_19": "That may be true.",
  "Solution_20": "[quote=\"kunny\"]That may be true.[/quote]\n\n[quote=\"Temperal\"][hide=\"Maximum\"]Write as $ x^2 \\plus{} x \\plus{} \\sqrt {1 \\minus{} x^2}$ with $ \\minus{} 1\\le x \\le 1$. Differentiating gives the answer easily: Solve $ \\minus{} \\frac {x}{\\sqrt {1 \\minus{} x^2}} \\plus{} 2x \\plus{} 1 \\equal{} 0$ and check that the value is attainable.[/hide]\n\nMinimum]Write as $ x^2 \\plus{} x \\minus{} \\sqrt {1 \\minus{} x^2}$ with $ \\minus{} 1\\le x \\le 1$. Use the same technique as above and check that the value is attainable.[/quote]\r\n\r\nBy the way,\r\nBy using differentiating method , can you solve the derivative equation:\r\n$ \\minus{} \\frac {x}{\\sqrt {1 \\minus{} x^2}} \\plus{} 2x \\plus{} 1 \\equal{} 0$",
  "Solution_21": "The calculation of the differentiation is wrong.",
  "Solution_22": "[quote=\"kunny\"]The calculation of the differentiation is wrong.[/quote]\r\n\r\nwhy not? :)",
  "Solution_23": "$ f(x)\\equal{}x^2\\plus{}x\\minus{}\\sqrt{1\\minus{}x^2}$, we have $ f'(x)\\equal{}2x\\plus{}1\\plus{}\\frac{x}{\\sqrt{1\\minus{}x^2}}$",
  "Solution_24": "[quote=\"kunny\"]$ f(x) \\equal{} x^2 \\plus{} x \\minus{} \\sqrt {1 \\minus{} x^2}$, we have $ f'(x) \\equal{} 2x \\plus{} 1 \\plus{} \\frac {x}{\\sqrt {1 \\minus{} x^2}}$[/quote]\r\n\r\nI think he is typo ,in fact $ f(x) \\equal{} x^2 \\plus{} x \\plus{}\\sqrt {1 \\minus{} x^2}, \\sin\\theta\\equal{}x$, \r\n\r\n$ \\implies f'(x) \\equal{} 2x \\plus{} 1 \\minus{} \\frac {x}{\\sqrt {1 \\minus{} x^2}}$\r\n\r\nif it is so , how to solve it when $ f'(x)\\equal{}0$",
  "Solution_25": "We can use slightly more straightforward math if we use Lagrange Multipliers instead:\r\n[hide]$ x \\equal{} \\sin \\theta, \\; \\; y \\equal{} \\cos \\theta$. We want to optimize $ f \\equal{} x^2 \\plus{} x \\plus{} y$ subject to $ g \\equal{} x^2 \\plus{} y^2 \\minus{} 1 \\equal{} 0$.\n$ (2x \\plus{} 1,1) \\equal{} \\lambda (2x,2y)$\n$ \\implies \\frac {2x \\plus{} 1}{x} \\equal{} \\frac {1}{y} \\iff 2 \\equal{} \\frac {x \\minus{} y}{xy}$\n$ \\implies 4 \\equal{} \\frac {x^2 \\plus{} y^2 \\minus{} 2xy}{x^2 y^2} \\equal{} \\frac {1 \\minus{} 2xy}{x^2 y^2}$\n$ \\implies 4 (xy)^2 \\plus{} 2(xy) \\minus{} 1 \\equal{} 0$\n$ \\implies xy \\equal{} \\frac { \\minus{} 1 \\pm \\sqrt {5}}{4}$\n\n$ (x \\plus{} y)^2 \\equal{} 1 \\plus{} 2xy \\equal{} \\frac {1 \\pm \\sqrt {5}}{2} \\implies x \\plus{} y \\equal{} \\sqrt {\\frac {1 \\pm \\sqrt {5}}{2}}$\n\nClearly we cannot take the squareroot of a negative, therefore we change the $ \\pm$'s into $ \\plus{}$\n\nWe thus have that $ x,y$ are the roots of the equation:\n\n$ t^2 \\minus{} \\sqrt {\\frac {1 \\plus{} \\sqrt {5}}{2}} \\plus{} \\frac { \\minus{} 1 \\plus{} \\sqrt {5}}{4}$[/hide]",
  "Solution_26": "[quote=\"TZF\"]We can use slightly more straightforward math if we use Lagrange Multipliers instead:\n[hide]$ x \\equal{} \\sin \\theta, \\; \\; y \\equal{} \\cos \\theta$. We want to optimize $ f \\equal{} x^2 \\plus{} x \\plus{} y$ subject to $ g \\equal{} x^2 \\plus{} y^2 \\minus{} 1 \\equal{} 0$.\n$ (2x \\plus{} 1,1) \\equal{} \\lambda (2x,2y)$\n$ \\implies \\frac {2x \\plus{} 1}{x} \\equal{} \\frac {1}{y} \\iff 2 \\equal{} \\frac {x \\minus{} y}{xy}$\n$ \\implies 4 \\equal{} \\frac {x^2 \\plus{} y^2 \\plus{} 2xy}{x^2 y^2} \\equal{} \\frac {1 \\plus{} 2xy}{x^2 y^2}$\n$ \\implies 4 (xy)^2 \\plus{} 2(xy) \\minus{} 1 \\equal{} 0$\n$ \\implies xy \\equal{} \\frac { \\minus{} 1 \\pm \\sqrt {5}}{4}$\n\n$ (x \\plus{} y)^2 \\equal{} 1 \\plus{} 2xy \\equal{} \\frac {1 \\pm \\sqrt {5}}{2} \\implies x \\plus{} y \\equal{} \\sqrt {\\frac {1 \\pm \\sqrt {5}}{2}}$\n\nClearly we cannot take the squareroot of a negative, therefore we change the $ \\pm$'s into $ \\plus{}$\n\nWe thus have that $ x,y$ are the roots of the equation:\n\n$ t^2 \\minus{} \\sqrt {\\frac {1 \\plus{} \\sqrt {5}}{2}} \\plus{} \\frac { \\minus{} 1 \\plus{} \\sqrt {5}}{4}$[/hide][/quote]\r\n\r\n$ \\implies \\frac {2x \\plus{} 1}{x} \\equal{} \\frac {1}{y} \\iff 2 \\equal{} \\frac {x \\minus{} y}{xy}$\r\n$ \\implies 4 \\equal{} \\boxed{\\frac {x^2 \\plus{} y^2 \\plus{} 2xy}{x^2 y^2} }\\equal{}\\boxed{ \\frac {1 \\plus{} 2xy}{x^2 y^2}}$\r\n\r\nIs it typo? :lol:",
  "Solution_27": "Right, it should be $ \\minus{}2xy$ in those; I copied it off my paper incorrectly! But I copied the remaining steps correctly so nothing else changes  :)",
  "Solution_28": "[quote=\"TZF\"]Right, it should be $ \\minus{} 2xy$ in those; I copied it off my paper incorrectly! But I copied the remaining steps correctly so nothing else changes  :)[/quote]\r\n\r\nsorry ,I am unfamiliar with applying for Largrange Multiplier in your proof, even though after reading \"http://en.wikipedia.org/wiki/Lagrange_multipliers\"\r\n\r\nCould you please introduce it definitely?",
  "Solution_29": "Suppose you wanted to minimize $ f \\equal{} 3x\\plus{}4y$, given that $ y\\equal{}x^2$. What would you do, visually?\r\n\r\nYou would plot the curve $ y\\equal{}x^2$ and all lines of the form $ 3x\\plus{}4y\\equal{}k$ (we call these \"level curves\"), and find the line with the least possible $ k$ that still intersects $ y\\equal{}x^2$. In doing this, you would notice that at the point of optimization, the line $ 3x\\plus{}4y\\equal{}k$ is tangent to the curve $ y\\equal{}x^2$. So, for this two-variable case, we can take a shortcut and find the place where $ 3x\\plus{}4y\\equal{}k$ is tangent to $ y\\equal{}x^2$.\r\n\r\nWe can extend this notion into multiple dimensions. Suppose we want to optimize a function $ f(x \\ldots)$ given a constraint $ g(x \\ldots)\\equal{}0$. Optimization will occur if $ f$ and $ g$ are tangent, which happens where $ f$ and $ g$ have a [i]gradient[/i] (normal vector) in the same direction. Since vectors in the same direction are scalar multiples, we say that:\r\n\r\n$ \\nabla f \\equal{} \\lambda \\nabla g$\r\n\r\nwhere $ \\lambda$ is the scale factor and $ \\nabla f$ denotes the gradient of $ f$. This is the essence of the method of lagrange multipliers.\r\n\r\nIn this case, we have $ f \\equal{} x^2\\plus{}x\\plus{}y$, which we want to optimize subject to the constraint $ g\\equal{}x^2\\plus{}y^2\\minus{}1 \\equal{} 0$\r\n\r\n$ \\nabla f \\equal{} \\left( \\frac{\\partial}{\\partial x} f , \\; \\frac{\\partial}{\\partial y} f  \\right) \\equal{} (2x\\plus{}1, \\; 1)$\r\n\r\n$ \\nabla g \\equal{} \\left( \\frac{\\partial}{\\partial x} g , \\; \\frac{\\partial}{\\partial y} g  \\right) \\equal{} (2x, \\; 2y)$\r\n\r\nThus, there exists a scalar $ \\lambda$ such that:\r\n\r\n$ (2x\\plus{}1 , \\; 1) \\equal{} \\lambda (2x, \\; 2y)$\r\n\r\nWe can write this as a system:\r\n\r\n$ 2x\\plus{}1 \\equal{} \\lambda 2x \\implies \\lambda \\equal{} \\frac{2x\\plus{}1}{2x}$\r\n$ 1 \\equal{} \\lambda 2y \\implies \\lambda \\equal{} \\frac{1}{2y}$\r\n\r\nEquating the two expressions for $ \\lambda$, we find that $ \\frac {2x \\plus{} 1}{2x} \\equal{} \\frac {1}{2y}$",
  "Solution_30": "[quote=\"TZF\"]Suppose you wanted to minimize $ f \\equal{} 3x \\plus{} 4y$, given that $ y \\equal{} x^2$. What would you do, visually?\n\nYou would plot the curve $ y \\equal{} x^2$ and all lines of the form $ 3x \\plus{} 4y \\equal{} k$ (we call these \"level curves\"), and find the line with the least possible $ k$ that still intersects $ y \\equal{} x^2$. In doing this, you would notice that at the point of optimization, the line $ 3x \\plus{} 4y \\equal{} k$ is tangent to the curve $ y \\equal{} x^2$. So, for this two-variable case, we can take a shortcut and find the place where $ 3x \\plus{} 4y \\equal{} k$ is tangent to $ y \\equal{} x^2$.\n\nWe can extend this notion into multiple dimensions. Suppose we want to optimize a function $ f(x \\ldots)$ given a constraint $ g(x \\ldots) \\equal{} 0$. Optimization will occur if $ f$ and $ g$ are tangent, which happens where $ f$ and $ g$ have a [i]gradient[/i] (normal vector) in the same direction. Since vectors in the same direction are scalar multiples, we say that:\n\n\n\n\n$ \\nabla f \\equal{} \\lambda \\nabla g$\n\nwhere $ \\lambda$ is the scale factor and $ \\nabla f$ denotes the gradient of $ f$. This is the essence of the method of lagrange multipliers.\n\nIn this case, we have $ f \\equal{} x^2 \\plus{} x \\plus{} y$, which we want to optimize subject to the constraint $ g \\equal{} x^2 \\plus{} y^2 \\minus{} 1 \\equal{} 0$\n\n$ \\nabla f \\equal{} \\left( \\frac {\\partial}{\\partial x} f , \\; \\frac {\\partial}{\\partial y} f \\right) \\equal{} (2x \\plus{} 1, \\; 1)$\n\n$ \\nabla g \\equal{} \\left( \\frac {\\partial}{\\partial x} g , \\; \\frac {\\partial}{\\partial y} g \\right) \\equal{} (2x, \\; 2y)$\n\nThus, there exists a scalar $ \\lambda$ such that:\n\n$ (2x \\plus{} 1 , \\; 1) \\equal{} \\lambda (2x, \\; 2y)$\n\nWe can write this as a system:\n\n$ 2x \\plus{} 1 \\equal{} \\lambda 2x \\implies \\lambda \\equal{} \\frac {2x \\plus{} 1}{2x}$\n$ 1 \\equal{} \\lambda 2y \\implies \\lambda \\equal{} \\frac {1}{2y}$\n\nEquating the two expressions for $ \\lambda$, we find that $ \\frac {2x \\plus{} 1}{2x} \\equal{} \\frac {1}{2y}$[/quote]\r\n\r\n \r\nHello, TZF\r\nThank you very much!!\r\nYou are nice and great...\r\n :P  :clap:",
  "Solution_31": "Here is the solution for Japanese high school students, because they don't study Lagrage's Multipliers.\r\n\r\n$ f(\\theta) \\equal{} \\sin ^2 \\theta \\plus{} \\sin \\theta \\plus{} \\cos \\theta\\ \\Longrightarrow f'(\\theta) \\equal{} 2\\sin \\theta \\cos \\theta \\plus{} \\cos \\theta \\minus{} \\sin \\theta$.\r\nSet $ \\cos \\theta \\minus{} \\sin \\theta \\equal{} t\\Longleftrightarrow \\cos \\theta \\plus{} \\sin \\theta \\equal{} \\pm\\sqrt {2 \\minus{} t^2} \\equal{} \\pm\\sqrt {1 \\minus{} t}\\ \\because t^2 \\minus{} t \\minus{} 1 \\equal{} 0$, we have $ f'(\\theta) \\equal{} \\minus{} (t^2 \\minus{} t \\minus{} 1)$ for $ \\minus{} \\sqrt {2}\\leq t\\leq \\sqrt {2}$. Let $ \\alpha \\equal{} \\frac {1 \\minus{} \\sqrt {5}}{2},\\ \\beta \\equal{} \\frac {1 \\plus{} \\sqrt {5}}{2}$, $ f(\\theta)\\equiv g(t)$ has a local minimum $ g(\\alpha)$ and a local maximum $ g(\\beta)$.   Since $ \\sin \\theta \\equal{} \\frac {\\pm\\sqrt {1 \\minus{} t} \\minus{} t}{2}$, remark that $ \\alpha ^ 2 \\minus{} \\alpha \\minus{} 1 \\equal{} 0$, we have $ g(\\alpha)$ is either \r\n\r\n$ \\left(\\frac {\\sqrt {1 \\minus{} \\alpha} \\minus{} \\alpha}{2}\\right)^2\\pm\\sqrt {1 \\minus{} \\alpha} \\equal{} \\frac {1}{2}(1 \\minus{} \\alpha \\sqrt {1 \\minus{} \\alpha}\\pm \\sqrt {1 \\minus{} \\alpha})$ or \r\n\r\n$ \\left(\\frac { \\minus{} \\sqrt {1 \\minus{} \\alpha} \\minus{} \\alpha}{2}\\right)^2\\pm\\sqrt {1 \\minus{} \\alpha} \\equal{} \\frac {1}{2}(1 \\plus{} \\alpha \\sqrt {1 \\minus{} \\alpha}\\pm \\sqrt {1 \\minus{} \\alpha})$\r\n\r\nwe also get the similar result for $ g(\\beta)$.\r\n\r\nFinally there wasn't a solution without calculus. :wink:",
  "Solution_32": "[quote=\"kunny\"]Here is the solution for Japanese high school students, because they don't study Lagrage's Multipliers.\n\n$ f(\\theta) \\equal{} \\sin ^2 \\theta \\plus{} \\sin \\theta \\plus{} \\cos \\theta\\ \\Longrightarrow f'(\\theta) \\equal{} 2\\sin \\theta \\cos \\theta \\plus{} \\cos \\theta \\minus{} \\sin \\theta$.\nSet $ \\cos \\theta \\minus{} \\sin \\theta \\equal{} t\\Longleftrightarrow \\cos \\theta \\plus{} \\sin \\theta \\equal{} \\pm\\sqrt {2 \\minus{} t^2} \\equal{} \\pm\\sqrt {1 \\minus{} t}\\ \\because t^2 \\minus{} t \\minus{} 1 \\equal{} 0$, we have $ f'(\\theta) \\equal{} \\minus{} (t^2 \\minus{} t \\minus{} 1)$ for $ \\minus{} \\sqrt {2}\\leq t\\leq \\sqrt {2}$. Let $ \\alpha \\equal{} \\frac {1 \\minus{} \\sqrt {5}}{2},\\ \\beta \\equal{} \\frac {1 \\plus{} \\sqrt {5}}{2}$, $ f(\\theta)\\equiv g(t)$ has a local minimum $ g(\\alpha)$ and a local maximum $ g(\\beta)$.   Since $ \\sin \\theta \\equal{} \\frac {\\pm\\sqrt {1 \\minus{} t} \\minus{} t}{2}$, remark that $ \\alpha ^ 2 \\minus{} \\alpha \\minus{} 1 \\equal{} 0$, we have $ g(\\alpha)$ is either \n\n$ \\left(\\frac {\\sqrt {1 \\minus{} \\alpha} \\minus{} \\alpha}{2}\\right)^2\\pm\\sqrt {1 \\minus{} \\alpha} \\equal{} \\frac {1}{2}(1 \\minus{} \\alpha \\sqrt {1 \\minus{} \\alpha}\\pm \\sqrt {1 \\minus{} \\alpha})$ or \n\n$ \\left(\\frac { \\minus{} \\sqrt {1 \\minus{} \\alpha} \\minus{} \\alpha}{2}\\right)^2\\pm\\sqrt {1 \\minus{} \\alpha} \\equal{} \\frac {1}{2}(1 \\plus{} \\alpha \\sqrt {1 \\minus{} \\alpha}\\pm \\sqrt {1 \\minus{} \\alpha})$\n\nwe also get the similar result for $ g(\\beta)$.\n\nFinally there wasn't a solution without calculus. :wink:[/quote]\r\n\r\n\r\nThanks for your instructions and sorry to waste your much time to find the details.\r\n :blush:"
}
{
  "Tag": [],
  "Problem": "In how many ways can a president and a 2-person committee be chosen from a group of 8 people (where the order in which we choose the 2 people doesn't matter)?",
  "Solution_1": "Firstly, the president can be chosen in eight ways. The two person committee, can be chosen from the 7 people, but the order doesnt matter. \r\nCommittee=7*6/2=21\r\n\r\n8*21=168\r\n\r\n(i think that's right...)\r\n\r\nANSWER:168"
}
{
  "Tag": [
    "LaTeX"
  ],
  "Problem": "Factor completely:\r\n\r\n[size=150]1. (a-b)^3+(b-c)^3+(c-a)^3\n2. (x+1)(y+1)(xy+1)+xy[/size]\r\n\r\nMaybe these are relatively easy for intermediate, but anyway, have fun! :)",
  "Solution_1": "1.\r\n\r\n$\\sum (a-b)^3 = \\sum a^3 - b^3 + 3ab(a-b) = 3[ a^2b + b^2c + c^2a - ab^2 - bc^2 - ca^2 ] = 3(b-a)(c-b)(a-c)$",
  "Solution_2": "[quote=\"Singular\"]1.\n\n$\\sum (a-b)^3 = \\sum a^3 - b^3 + 3ab(a-b) = 3[ a^2b + b^2c + c^2a - ab^2 - bc^2 - ca^2 ] = 3(b-a)(c-b)(a-c)$[/quote]\r\n\r\ni got $-3(b-a)(c-b)(a-c)$... and i plugged in values in a calculator and it worked  ;) you might wanna check your expansion of $(a-b)^3$ i think",
  "Solution_3": "You're right, pkothari13 :)",
  "Solution_4": "1. Let $a-b=x,b-c=y,c-a=$,....\r\n\r\nor you can use the eqality $A^3+B^3+C^3-3ABC=$....\r\n\r\n2. Regard $y$ as constant number and arrange with respect to $x$.",
  "Solution_5": "shouldnt it be $-xy$ in the $2nd$ one? :?",
  "Solution_6": "[quote=\"riddler\"]shouldnt it be $-xy$ in the $2nd$ one? :?[/quote]\r\nNo, it's not -xy.",
  "Solution_7": "(x+1)(y+1)(xy+1)+xy\r\n=(xy+x+y+1)(xy+1)+xy\r\n\r\n=(x^2)(y^2)+(x^2*y) + (x*y^2)+2xy+xy+x+y+1\r\n\r\n=[(x^2)(y^2) + 2xy+1] + [(x^2*y)+ x] + [(x*y^2) + y] + xy\r\n\r\n=(xy+1)^2   +    x(xy+1)  +  y(xy+1) + xy\r\nlet xy+1 be a\r\n\r\na^2 + (x+y)a + xy\r\n(a+x)(a+y)\r\n\r\n(xy+x+1)(xy+y+1)\r\nwell is it correct?",
  "Solution_8": "Yep, that's correct ;)",
  "Solution_9": "One well known trick which make  things easier:\r\n\r\n[hide]One can easily check that if  $a=b$ LHS is zero .\n[hide]Because  it becomes  $0^3 +(a-c)^3 +(c-a)^3 $ [/hide]\n\nso $(a-b)$ is a factor, similarly $(b-c)$ and $(c-a)$ are also   factors;\nso LHS = :\n\n$(a-b)^3+(b-c)^3+(c-a)^3  = K(a-b)(b-c)(c-a)$\nwhere K is a constant.   (Because you have 3rd degree on both sides)\nput   a=0, b=1, c=-1  (or any simple value to plug in) and you get K=3) [/hide]\n\nAnother easy way  if you happen to remember, and it is nice identity to remember:\n[hide]$x^3+y^3+z^3 - 3 xyz = (x+y+z) (x^2+y^2+z^2-xy-yz-zx) $ \n[hide]if $x+y+z=0 $ then $x^3+y^3+z^3 = 3xyz $ \nand here x+y+z  becomes 0 if x=a-b, y=b-c and z=c-a :) [/hide][/hide]",
  "Solution_10": "You can also use the fact that $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)$ to prove AM-GM for three variables. Can anyone do it?",
  "Solution_11": "[quote=\"DPopov\"]You can also use the fact that $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)$ to prove AM-GM for three variables. Can anyone do it?[/quote]\r\nI'm not sure if this is right, but...\r\n[hide]Let  A=a^(1/3), B=b^(1/3), C=c^(1/3).\n\nThen, (a+b+c)/3 - (abc)^(1/3)\n=(A^3+B^3+C^3-3ABC)/3\n=(A+B+C)(A^2+B^2+C^2-AB-BC-CA)/3\n=(A+B+C)((A-B)^2+(B-C)^2+(C-A)^2)/6\n\nGiven that a,b,c are positive numbers, A,B,C are also positive. (--> A+B+C is positive)\nSo, the expression above is positive.\nTherefore, (a+b+c)/3 >= (abc)^(1/3) (for a,b,c >=0)\n[/hide]\r\nSorry, it looks really messy without using LaTeX :( \r\nI definately should learn LaTeX...",
  "Solution_12": "You are right, frt.\r\n\r\nGood luck, for Tex. Here is my first Tex.[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=16677$highlight=[/url]\r\n\r\nIf you make a mistake, someone would correct like this kindly! :)  First I didn't know how to use edit,but it's the very\r\n\r\nPractice makes perfect!\r\n\r\nKunihiko"
}
{
  "Tag": [
    "inequalities",
    "algebra",
    "polynomial",
    "function",
    "limit",
    "quadratics",
    "real analysis"
  ],
  "Problem": "Let $x$, $y$, $z$ be positive reals satisfying $\\left(x+y+z\\right)^{3}=32xyz$\r\nFind the minimum and the maximum of $P=\\frac{x^{4}+y^{4}+z^{4}}{\\left(x+y+z\\right)^{4}}$",
  "Solution_1": "$\\frac{383-165 \\sqrt{5}}{256}$ to be the minimum ?!?!?!\r\n$\\frac{x^4+y^4+z^4}{(x+y+z)^4} \\geq \\frac{xyz(x+y+z)}{(x+y+z)^3(x+y+z)}=\\frac{xyz(x+y+z)}{32xyz(x+y+z)}=\\frac{1}{32}$",
  "Solution_2": "Your inequalities prove nothing, Gohan, since you cannot have equality. Now I remember, indeed I put it in Old and New Inequalities. But I'm not going to post the solutions here, since there are too many formulae and I hate Latex.",
  "Solution_3": "I have a sketch of a solution, but it's horrendously messy.\r\n\r\nLet $s_1=x+y+z=1$, $s_2=xy+yz+zx=a$ and $s_3=xyz=\\frac{1}{32}$.\r\n\r\nThen I get $P=2(s_2-1)^2-\\frac{15}{16}$.\r\n\r\nWe can determine the range of values of $s_2$ subject to the condition that $x,y$ and $z$ all be real by finding the range of values of $a$ for which the polynomial discriminant of $x^3-x^2+ax-\\frac{1}{32}$ is less than or equal to zero.\r\n\r\nThe discriminant will be $\\left(\\frac{3a+1}{9}\\right)^3+\\left(\\frac{1}{32}+\\frac{2}{27}-\\frac{a}{3}\\right)^2$, so the solution could be completed by finding where this is positive and negative.  I just couldn't be bothered to go through all that mess...\r\n\r\nSilouan, do you have a nicer way to finish off the proof once you have the form of $P$?  Do you have a nice way to find what values of $s_2$ are valid?",
  "Solution_4": "I think you should use Horner in a point.",
  "Solution_5": "My solution but  in Greek because I have not time to write it to English.",
  "Solution_6": "[hide=\"A hint\"]Assume that $x+y+z=1,xy+yz+zx=p$, then $xyz=\\frac{1}{32}$. Called $D$ is set of values of $p$. \n1) What is $D$?\n2) Write $P=f(p)$.\n3) Use real analysis for $f(p)$.[/hide]",
  "Solution_7": "I did something similar (maybe a little bit easier): \r\n\r\n[hide=\"A hint\"] W.l.o.g. $x+y+z=4$, then $xyz=2$. W.l.o.g. $x \\ge y,z$ , \n0) By AM-GM applied to $y$ and $z$ we have $x \\le 2$, so $4/3 \\le x \\le 2$ \n1) Express $y^{4}+z^{4}$ as a function of $z+y = 4-x$ and $zy = 2/x$.\n2) Write $P=f(x)$.\n3) Use real analysis to show $f(x) \\le f(2)$.[/hide]",
  "Solution_8": "silouan, I have just solved this problem completely. You can have the solution by sea mail now. Greece is far from Japan. :lol: \n\n[quote=\"silouan\"]Let $x$, $y$, $z$ be positive reals satisfying $\\left(x+y+z\\right)^{3}=32xyz$\nFind the minimum and the maximum of $P=\\frac{x^{4}+y^{4}+z^{4}}{\\left(x+y+z\\right)^{4}}$[/quote]\n\n\nHere is my solution:\n\n\nLet $\\frac{x}{x+y+z}=a,\\ \\frac{y}{x+y+z}=b,\\ \\frac{z}{x+y+z}=c\\ (a>0,\\ b>0,\\ c>0)$, the given condition gives $abc=\\frac{1}{32}$ and $a+b+c=1$.\n\nSet $ab+bc+ca=k$, since $a,\\ b,\\ c$ are the positive roots of the cubic equation with respect to $t$, $t^{3}-t^{2}+kt-\\frac{1}{32}=0\\ \\cdots [*]$.By $t>0$,\n\nRewrite this as $k=-t^{2}+t+\\frac{1}{32t}: =f(t)$. Now we are to find the condition of $t$ for which $[*]$ has more than two positive roots including double root,\n\nthat is to say, the condition such that two graphs of the curve $y=f(t)$ and the line $y=k$ have more than two intersection points at $t>0$.\n\n$f'(t)=-2t+1-\\frac{1}{32t^{2}}=\\frac{(1-4t)(16t^{2}-4t-1)}{32t^{2}}$, so $f'(t)=0\\ (t>0)\\Longleftrightarrow t=\\frac{1}{4},\\ \\frac{1+\\sqrt{5}}{8}$.\n\nLet $t_{0}=\\frac{1+\\sqrt{5}}{8}$,the behabiour of $f(t)$ is $\\lim_{t\\rightarrow+0}f(t)=+\\infty$, which is decreasing at $0<t<\\frac{1}{4},\\ t_{0}<t$ and \n\nincreasing at $\\frac{1}{4}<t<t_{0}$ and $\\lim_{t\\rightarrow \\infty}f(t)=-\\infty,\\ \\lim_{t\\rightarrow \\infty}\\{f(t)-(-t^{2}+t)\\}=0$, yielding the local minimum \n\n$f\\left(\\frac{1}{4}\\right)=\\frac{5}{16}$, since $t_{0}$ satisfies $16t_{0}^{2}-4t_{0}-1=0\\Longleftrightarrow \\frac{1}{32t_{0}}=t_{0}-2t_{0}^{2}$, \n\nwe have the local maximum $f(t_{0})=-t_{0}^{2}+t_{0}+\\frac{1}{32t_{0}}=2t_{0}-3t_{0}^{2}$\n\n$=2t_{0}-\\frac{3}{16}(4t_{0}+1)=\\frac{1}{16}(20t_{0}-3)=\\frac{5\\sqrt{5}-1}{32}\\because 16t_{0}^{2}-4t_{0}-1=0$. \n\nTherefore the range of $k$ for which $k$ should be taken is $\\frac{5}{16}\\leq k\\leq \\frac{5\\sqrt{5}-1}{32}$.\n\nThen $P=a^{4}+b^{4}+c^{4}=(a^{2}+b^{2}+c^{2})^{2}-2(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})$\n\n$=\\{(a+b+c)^2-2(ab+bc+ca)\\}^2-2\\{(ab+bc+ca)^2-2abc(a+b+c)\\}$\n\n$=(1-2k)^{2}-2\\left(k^{2}-\\frac{1}{16}\\right)=2k^{2}-4k+\\frac{9}{8}$\n\n$=2(k-1)^{2}-\\frac{7}{8}: =g(k)$ where $\\frac{5}{16}\\leq k\\leq \\frac{5\\sqrt{5}-1}{32}$.\n\nFor this interval, the quadratic function $y=g(k)$ is decreasing and concave down, so $g\\left(\\frac{5\\sqrt{5}-1}{32}\\right)\\leq P\\leq g\\left(\\frac{5}{16}\\right)$.\n\nConsequently the desired answer is  \n\nminimum value         $\\frac{383-165\\sqrt{5}}{256}$ for $a+b+c=1,\\ ab+bc+ca=\\frac{5\\sqrt{5}-1}{32},\\ abc=\\frac{1}{32}$\n\nmaximum value        $\\frac{9}{128}$ for $a+b+c=1,\\ ab+bc+ca=\\frac{5}{16},\\ abc=\\frac{1}{32}$\n\nkunny",
  "Solution_9": "Without losing of generality, we let $ x + y + z = 4 $ and then $ abc = 2 . $\nLet $ f(x) = x^4 , $and then $$ f'(x) = 4x^3 ,  f''(x) = 12x^2 > 0, $$ \nwhich means $ f $ is convex on the interval $ ( 0 , + \\infty ) . $\nUsing [b]EV[/b] method, we can let $ y = z $, then there's nothing left.\n"
}
{
  "Tag": [
    "trigonometry",
    "geometry",
    "geometry solved"
  ],
  "Problem": "Can be solved without using trigonometry?\r\n\r\nLet $P$ a point within $\\triangle{ABC}$ equilateral, such that $\\overline{PA}= 5$, $\\overline{PB}= 7$ and $\\overline{PC}= 8$. Find the measure of $\\overline{AB}$\r\n\r\n :)",
  "Solution_1": "If $x$ is the length of the side of the triangle then $x$ satisfies $3(5^{4}+7^{4}+8^{4}+x^{4}) = (5^{2}+7^{2}+8^{2}+x^{2})^{2}$. This is equivalent to $x^{4}-138x^{2}+1161 = 0$. Thus $x = 9$ or $x= 129$. $x=129$ is not admissible. Therefore $x=9$.",
  "Solution_2": "Thank you\r\n\r\nI put $x=\\sqrt{129}$ in geometry program and it works.\r\n\r\nCan you tell me why $x$ satisfies that equation.\r\n\r\n :)",
  "Solution_3": "http://www.artofproblemsolving.com/Forum/viewtopic.php?t=67254"
}
{
  "Tag": [],
  "Problem": "Need help with the mechanism",
  "Solution_1": "This a cyclic acetal, and the carbon \"bearing\" the two oxygens have one hydrogen, so the aqueous acid product must have two -OH groups and an aldehyde group, -CHO.\r\n\r\n[hide=\"Answer\"]$ [HO \\minus{} (CH_2)_3]_2 \\minus{} CH \\minus{} CHO$[/hide]\r\n\r\nThe mechanism is essentialy:\r\n\r\n1) Protonation of one oxygen;\r\n\r\n2) Attack of one water molecule on the carbon bearing the two oxygens;\r\n\r\n3) Deprotonation of oxygen \"from\" water, and protonation of the other oxygen;\r\n\r\n4) Departure of HO-... assisted by one of \"water oxygen\" 's lone pair, followed by deprotonation of the \"water oxygen\"."
}
{
  "Tag": [],
  "Problem": "\u039b\u03c5\u03c3\u03c4\u03b5 \u03c4\u03b9\u03c2 \u03b4\u03b9\u03bf\u03c6\u03b1\u03bd\u03c4\u03b9\u03ba\u03b5\u03c2 \u03c3\u03c4\u03bf\u03c5\u03c2 \u03c6\u03c5\u03c3\u03b9\u03ba\u03bf\u03c5\u03c2 \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03c5\u03c2\r\n$ 1) 2^{x \\minus{} 1} \\plus{} 1 \\equal{} y^{z \\plus{} 1}$\r\n$ 2)2^{x \\plus{} 1} \\minus{} 1 \\equal{} y^{z \\plus{} 1}$",
  "Solution_1": "\u0391\u03bd \u03b8\u03b5\u03bb\u03b5\u03b9 \u03bd\u03b1 \u03c4\u03b9\u03c2 \u03c0\u03c1\u03bf\u03c3\u03c0\u03b1\u03b8\u03b7\u03c3\u03b5\u03b9 \u03ba\u03b1\u03c0\u03bf\u03b9\u03bf\u03c2 \u03b1\u03c2 \u03c4\u03bf \u03ba\u03b1\u03bd\u03b5\u03b9....\u0391\u03be\u03b9\u03b6\u03bf\u03c5\u03bd \u03c4\u03bf\u03bd \u03ba\u03bf\u03c0\u03bf....\r\n\u0391\u03bb\u03bb\u03b9\u03c9\u03c2 \u03b1\u03c5\u03c1\u03b9\u03bf \u03b8\u03b1 \u03c0\u03bf\u03c3\u03c4\u03b1\u03c1\u03c9 \u03c4\u03b9\u03c2 \u03bb\u03c5\u03c3\u03b5\u03b9\u03c2 \u03c4\u03bf\u03c5\u03c2 \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03bc\u03b5\u03bd\u03bf\u03c5\u03bd \u03b1\u03bb\u03c5\u03c4\u03b1 \u03c0\u03c1\u03bf\u03b2\u03bb\u03b7\u03bc\u03b1\u03c4\u03b1 \u03c3\u03c4\u03bf forum...."
}
{
  "Tag": [
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "[u][i]Problem 1[/i][/u] [i]Let a,b,c be positive real numbers such that[/i] $ abc \\equal{} 1$\r\n[i]   Prove that[/i]$ \\sqrt{\\frac{a}{b\\plus{}3}} \\plus{} \\sqrt{\\frac{b}{c\\plus{}3}}\\plus{}\\sqrt{\\frac{c}{a\\plus{}3}}\\geq \\frac{3}{2}$\r\n\r\n[u][i] Problem 2\n [/i][/u]\r\n [i] Let a,b,c be positive real numbers such that[/i] $ a\\plus{}b\\plus{}c \\equal{} \\frac{1}{a}\\plus{} \\frac{1}{b}\\plus{} \\frac{1}{c}$\r\n  [i]and[/i] $ a \\leq b \\leq c$\r\n[i] Prove that\n [/i]\r\n               $ ab^2c^3 \\geq 1$",
  "Solution_1": "Problem 2\r\n\r\nReplacing $ x\\equal{}bc$ and so on, the condition becomes $ xy\\plus{}yz\\plus{}zx\\equal{}x\\plus{}y\\plus{}z$. Also, $ a\\le b\\le c$ implies $ x\\ge y\\ge z$. Now we need to prove $ x^2 y\\ge 1$.\r\n\r\nSince $ (x\\plus{}y\\plus{}z)^2 \\ge 3(xy\\plus{}yz\\plus{}zx) \\equal{} 3(x\\plus{}y\\plus{}z)$, we have $ x\\plus{}y\\plus{}z\\ge 3$ and $ xy\\plus{}yz\\plus{}zx \\ge 3$. Hence $ x\\ge 1$, $ xy\\ge 1$ and the conclusion follows.",
  "Solution_2": "[quote=\"nkht-tk14\"][u][i]Problem 1[/i][/u] [i]Let a,b,c be positive real numbers such that[/i] $ abc \\equal{} 1$\n[i]   Prove that[/i]$ \\sqrt {\\frac {a}{b \\plus{} 3}} \\plus{} \\sqrt {\\frac {b}{c \\plus{} 3}} \\plus{} \\sqrt {\\frac {c}{a \\plus{} 3}}\\geq \\frac {3}{2}$\n[/quote]\r\nSee here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=142431",
  "Solution_3": "I think i can give a nicer solution :D.\r\n$ abc \\equal{} 1\\rightarrow S \\equal{} a \\plus{} b \\plus{} c\\geq3\\rightarrow \\sum\\sqrt {\\frac {a}{b \\plus{} 3}}\\geq\\sum\\sqrt {\\frac {a}{a \\plus{} 2b \\plus{} c}} \\equal{} \\sum\\sqrt {\\frac {a}{S \\plus{} a}}.$\r\nLet $ f(x) \\equal{} \\sqrt {\\frac {x}{S \\plus{} x}}$, and $ f$ is convex, then from Jensen we have that $ \\sum\\sqrt {\\frac {a}{S \\plus{} a}}\\geq3f(\\frac {S}{3}) \\equal{} \\frac {3}{2}$.\r\nThe equality occurs when $ a \\equal{} b \\equal{} c \\equal{} 1$.",
  "Solution_4": "Unfortunately, your $ f$ is concave for positive $ x$.",
  "Solution_5": "yes it is not so nice any more :))"
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "Let $ a,b,c,x,y,z$ are real numbers. If $ |x| \\plus{} |z|\\geq |y|$, prove the inequality\r\n\r\n\r\n$ x^2(a \\minus{} b)(a \\minus{} c) \\plus{} y^2(b \\minus{} c)(b \\minus{} a) \\plus{} z^2(c \\minus{} a)(c \\minus{} b)\\geq 0$",
  "Solution_1": "You need not only $ \\left|x\\right| \\plus{} \\left|z\\right|\\geq\\left|y\\right|$, but also $ \\left|y\\right| \\plus{} \\left|z\\right|\\geq\\left|x\\right|$ and $ \\left|x\\right| \\plus{} \\left|y\\right|\\geq\\left|z\\right|$. In other words, you need $ \\left|x\\right|$, $ \\left|y\\right|$ and $ \\left|z\\right|$ to be sides of a triangle.\r\n\r\nThis is one of the cases of the Vornicu-Schur inequality; in my note [url=http://www.cip.ifi.lmu.de/~grinberg/VornicuS.pdf]The Vornicu-Schur inequality and its variations[/url], it is Theorem 1 [b]h)[/b]. It is not due to Vornicu and neither it is to Schur, but it was not invented by me either - if you are looking for the source, here is an equivalent version:\r\n\r\n[quote=\"[url=http://www.mathlinks.ro/viewtopic.php?p=117296#117296]http://www.mathlinks.ro/viewtopic.php?t=4357 post #5[/url]\"][b]Kvant Problem M1166, solved in Kvant 11/1989, page 26:[/b]\n\nProve the inequality $ a^2pq \\plus{} b^2qr \\plus{} c^2rp \\leq 0$, where a, b, c are the sidelengths of a triangle, and p, q, r are three real numbers satisfying p + q + r = 0.[/quote]\r\n\r\nI forgot the author, but you can look up in Kvant.\r\n\r\n  Darij",
  "Solution_2": "I am sorry. This inequality is true if $ a\\geq b\\geq c$ or $ a\\leq b\\leq c$. So, I think it is stronger than Vornicu-schur because it is true for all real numbers a,b,c,x,y,z. And, can you prove it? or any idea? :)",
  "Solution_3": "[quote=\"manlio\"]Let $ a,b,c,x,y,z$ are real numbers, and $ |z| \\plus{} |x|\\geq |y|$ etc, prove the inequality\n\n$ x^2(a \\minus{} b)(a \\minus{} c) \\plus{} y^2(b \\minus{} c)(b \\minus{} a) \\plus{} z^2(c \\minus{} a)(c \\minus{} b)\\geq 0$[/quote][quote=\"thegod277\"]This inequality is true if $ b \\equal{} {\\rm mid}\\{a,b,c\\}$. I think it is stronger. And, can you prove it? [/quote]$ x^2(a \\minus{} b)(a \\minus{} c) \\plus{} y^2(b \\minus{} c)(b \\minus{} a) \\plus{} z^2(c \\minus{} a)(c \\minus{} b)$\n\n$ \\equal{} \\left[|x|a \\minus{} \\left(|z| \\plus{} |x|\\right)b \\plus{} |z|c\\right]^2 \\plus{} \\left[\\left(|z| \\plus{} |x|\\right)^2 \\minus{} y^2\\right](a \\minus{} b)(b \\minus{} c)\\geq 0$.\n[quote=\"darij grinberg\"]if you are looking for the source, here is[/quote]A. Oppenheim and Roy O. Davies, Inequalities of Schur's Type,  [url=http://www.m-a.org.uk/resources/periodicals/the_mathematical_gazette/]The Mathematical Gazette[/url], Vol. 48 (1964), pp. 25-27."
}
{
  "Tag": [
    "linear algebra",
    "matrix",
    "linear algebra unsolved"
  ],
  "Problem": "Let $\\zeta$ be a primitive $n$-th root of unity.\r\n\r\nProve that all eigenvalues of the matrix $\\left(\\begin{matrix} 1 & 1 & 1 & \\cdots & 1\\\\ 1 & \\zeta & \\zeta^2 & \\cdots & \\zeta^{n-1}\\\\ 1 & \\zeta^2 & \\zeta^4 & \\cdots & \\zeta^{2(n-1)}\\\\ \\vdots & \\vdots & \\vdots & \\ddots & \\vdots\\\\ 1 & \\zeta^{n-1} & \\zeta^{2(n-1)} & \\cdots & \\zeta^{(n-1)^2} \\end{matrix}\\right)$ have modulus $\\sqrt{n}$.",
  "Solution_1": "[hide=\"Edit - ignore this\"]Terminology issue: I've seen this matrix given several names, but but it is clearly [b]not[/b] a Vandermonde matrix.[/hide]\r\nCall the matrix $F$. (F is for Fourier - this is the matrix on which the discrete Fourier transform is based.)\r\n\r\nProve that $F^*F=nI.$ That is, $F$ is a constant multiple of a unitary matrix. This will lead to Peter's result.",
  "Solution_2": "sorry for terminology... but it is a vandermonde matrix i think, though not a general one.",
  "Solution_3": "Sorry for asking, but if that is not a Vandermonde... then what`s a Vandermonde?",
  "Solution_4": "Wait a minute - now I realize that it is a Vandermonde matrix, albeit a very specific and special one."
}
{
  "Tag": [
    "limit",
    "logarithms",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "Solve: $ x.\\ln(x\\plus{}1)\\minus{}(x\\plus{}1).\\ lnx\\equal{}0$",
  "Solution_1": "hello, let $ f(x)=x\\ln(x+1)-(x+1)\\ln(x)$,then we have $ \\lim_{x\\to 0+}f(x)=+\\infty$, $ \\lim_{x\\to \\infty}f(x)=+\\infty$, further we get $ f^'(x)=\\ln\\left(\\frac{x+1}{x}\\right)-\\frac{2x+1}{x(x+1)}<0 \\forall x>0$, so it exist an uniquely determined intersection point with the $ x$ axis.\r\nSonnhard."
}
{
  "Tag": [
    "function",
    "number theory theorems",
    "number theory"
  ],
  "Problem": "Hello.. Does anyone know if there is a formula for $ \\displaystyle \\sum_{i\\equal{}1}^{n} \\phi(i)$?\r\n\r\n(here $ \\phi$ is the number of positive integers coprime with n and less than n)",
  "Solution_1": "You can describe it asymptotically. Should be $ \\sim \\frac{3}{\\pi^2} n^2$.",
  "Solution_2": "Will you please prove your asymptotic formula?",
  "Solution_3": "See here:\r\n\r\nhttp://en.wikipedia.org/wiki/Proofs_involving_the_totient_function#Average_order_of_the_totient"
}
{
  "Tag": [],
  "Problem": "let $ m \\in N$ and $ m>4$ and it isn't a prime  number.\r\nprove that  $ m$ divides $ (m\\minus{}1)!$",
  "Solution_1": "if $ m\\equal{}a^2$ and $ m>4$ \r\nthen $ 2a<m$\r\nif $ m$ is not a square and not prime then $ m\\equal{}ab$ with $ a$ different than $ b$ and $ a,b<m$"
}
{
  "Tag": [
    "geometry"
  ],
  "Problem": "Hi,\r\n\r\nThis doesn't have to do anything with [i]100 Math Problems[/i]. It just something that I got from geometry worksheet that I get answer but not quite sure.\r\n\r\nGiven: BD is perpendicular to plane M, AB :cong: BC\r\n\r\nProve:  :tri: ABD = :tri: CBD\r\n\r\n-----------------------------------------------------------------------------------\r\n1. BD :perp: plane M, AB :cong: BC | 1. Given\r\n2. BD :perp: BC, BD :perp: AB | 2. If line is :perp: to plane, the line is also :perp: to other lines in same plane\r\n3. :ang: DBC :cong: DBA | 3. Def. :perp: lines\r\n4. DB = DB |4. Reflexive Prop.\r\n5. :tri: ABD = :tri: CBD |5. SAS\r\n\r\nHope this is right.",
  "Solution_1": "Looks good to me.\n\n[hide]If line is perpendicular to a plane, it is perpendicular to all lines in the plane that pass through its foot[/hide]",
  "Solution_2": "Proof is basic, simple, and sound. Getting started stuff...\r\nBTW Rep123max, nice avatar, I love the show [i]Family Guy[/i] :lol:"
}
{
  "Tag": [
    "calculus",
    "integration",
    "logarithms",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "an   integral i just made up, LOL  :D \r\n\r\nshow\r\n\r\n$ \\int_0^1\\;\\dfrac{\\textbf dx}{\\sqrt {\\; \\ln\\;\\left(\\ln\\; \\dfrac{1}{x}\\; \\right)\\; }}\\; \\equal{} \\;\\boxed{\\sqrt \\pi}$",
  "Solution_1": "Now tell me why the RHS of your formula contains no imaginary part.",
  "Solution_2": "Isn't it  $ \\int_{0}^{1} \\frac{dx}{\\sqrt{ln(\\frac{1}{x})}}\\equal{}\\sqrt{\\pi}$ ?",
  "Solution_3": "If we put here $ \\frac {1}{x} \\equal{} e^t$, we get: \r\n$ \\int_{0}^{1} \\frac {\\textbf dx}{\\sqrt {\\ln (\\frac {1}{x})}} \\equal{} \\int_{0}^{\\infty}{e^{ \\minus{} t}t^{ \\minus{} \\frac {1}{2}}\\textbf dx} \\equal{} \\Gamma\\left(\\frac {1}{2}\\right) \\equal{} \\sqrt {\\pi}$\r\n\r\nand if $ x>\\frac{1}{e}$ then $ \\ln \\ln\\left(\\frac{1}{x}\\right)<0$",
  "Solution_4": "oh ! sorry....  :( \r\n\r\nyou see, the computer which i use to connect to the internet is   old, and i sometimes see things on the screen that looks OK after i type (preview window), but apparently it is coming out  wrong !\r\n\r\nyou are correct, the integral i wanted to pose was\r\n\r\n$ \\int_0^1\\dfrac{1}{\\sqrt {\\minus{}\\,\\ln\\,x } } \\;\\textbf dx$\r\n\r\n(maybe i typed in an extra $ \\ln$ who knows...   :maybe: )\r\n\r\nanyway, i was looking at $ \\int_0^1x^m\\cdot \\left(\\minus{}\\,\\ln\\,x\\right)^n\\;\\textbf dx\\;\\equal{}\\;\\dfrac{\\Gamma(n\\plus{}1)}{(m\\plus{}1)^{n\\plus{}1}}$\r\n\r\n( where $ m,n>\\minus{}1$ )\r\n\r\nand thought i would ask that question. well, you guys got it right even with the wrong question.  :lol:",
  "Solution_5": "[quote=\"misan\"]\nyou are correct, the integral i wanted to pose was\n\n$ \\int_0^1\\dfrac{1}{\\sqrt { \\minus{} \\,\\ln\\,x } } \\;\\textbf dx$\n\n(maybe i typed in an extra $ \\ln$ who knows...   :maybe: )[/quote]\nAfter substitution $ z \\equal{} \\minus{} \\ln x$ yields the well-known gaussian integral and the result clearly follows.\n\n[quote=\"misan\"]anyway, i was looking at $ \\int_0^1x^m\\cdot \\left( \\minus{} \\,\\ln\\,x\\right)^n\\;\\textbf dx\\; \\equal{} \\;\\dfrac{\\Gamma(n \\plus{} 1)}{(m \\plus{} 1)^{n \\plus{} 1}}$\n\n( where $ m,n > \\minus{} 1$ )[/quote]\r\nAfter makin' $ z \\equal{} \\minus{} \\ln x$ the integral becomes $ \\int_0^\\infty {e^{ \\minus{} z(m \\plus{} 1)} z^n \\,dz} \\equal{} \\frac {{n!}} {{(m \\plus{} 1)^{n \\plus{} 1} }}.\\quad\\blacksquare$"
}
{
  "Tag": [
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Let $ p_i \\geq 2$, $ i \\equal{} 1,2, \\cdots n$ be $ n$ integers such that any two of them are relatively prime. Let:\r\n\r\n\\[ P \\equal{} \\{ x \\equal{} \\sum_{i \\equal{} 1}^{n} x_i \\prod_{j \\equal{} 1, j \\neq i}^{n} p_j \\mid x_i \\text{is a non \\minus{} negative integer}, i \\equal{} 1,2, \\cdots n \\}\r\n\\]\r\nProve that the biggest integer $ M$ such that $ M \\not\\in P$ is greater than $ \\displaystyle \\frac {n \\minus{} 2}{2} \\cdot \\prod_{i \\equal{} 1}^{n} p_i$, and also find $ M$.",
  "Solution_1": "very easy problem for china tst",
  "Solution_2": "And the answer is $$\\prod ^{n}_{i=1}p_{i}\\cdot \\left( n-1-\\sum ^{n}_{k=1}\\dfrac {1}{p_{i}}\\right) $$",
  "Solution_3": "This is just a (pretty straightforward) generalization of problem IMO 1983/3.\n(The same argument works for both problems.)\n\nhttp://www.artofproblemsolving.com/community/c6h60800p366618"
}
{
  "Tag": [],
  "Problem": "\u0391\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03c4\u03b5 \u03cc\u03c4\u03b9 \u03b3\u03b9\u03b1 \u03c4\u03bf\u03c5\u03c2 \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03bf\u03cd\u03c2 \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03cd\u03c2 $ a,b,c$ \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \r\n\r\n$ 4\\left( \\sum_{cyclic}a^{2}b^{2}\\minus{}abc\\sum_{cyclic}a\\right)\\left( \\sum_{cyclic}a^{4}\\minus{}\r\n\\sum_{cyclic}a^{2}b^{2}\\right)\\geq 3\\left( \\sum_{cyclic} a^{3}b\\minus{}abc\\sum_{cyclic}a\\right)^{2}$",
  "Solution_1": "\u0394\u03b5\u03bd \u03c4\u03b7\u03bd \u03ad\u03c7\u03c9 \u03ba\u03b1\u03bb\u03bf\u03ba\u03bf\u03b9\u03c4\u03ac\u03be\u03b5\u03b9, \u03b1\u03bb\u03bb\u03ac \u03bd\u03bf\u03bc\u03af\u03b6\u03c9 \u03cc\u03c4\u03b9 \u03b2\u03b3\u03b1\u03af\u03bd\u03b5\u03b9 \u03bc\u03b5 Moirhead \u03ba\u03b1\u03b9 Holder.",
  "Solution_2": "\u0391\u03bd \u03c6\u03ad\u03c1\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf \u03b1\u03c1\u03b9\u03c3\u03c4\u03b5\u03c1\u03cc \u03bc\u03ad\u03bb\u03bf\u03c2 \u03c3\u03c4\u03bf \u03b4\u03b5\u03be\u03af \u03b1\u03c1\u03ba\u03b5\u03af \u03bd\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03b7 \u03b4\u03b9\u03b1\u03ba\u03c1\u03af\u03bd\u03bf\u03c5\u03c3\u03b1 \u03c4\u03bf\u03c5 \u03c4\u03c1\u03b9\u03c9\u03bd\u03cd\u03bc\u03bf\u03c5  \r\n$ T\\equal{}3(\\sum x^2y^2\\minus{}xyz\\sum x)r^2\\minus{}3(\\sum x^3y\\minus{}xyz\\sum x)r\\plus{}\\sum x^4\\minus{}\\sum x^2y^2$  \u0391\u03c1ke\u03af \u03bd\u03b4\u03bf $ T\\geq 0$\r\n\r\n\u0395\u03af\u03bd\u03b1\u03b9 $ 3(\\sum x^2y^2\\minus{}xyz\\sum x)\\equal{}\\frac{1}{2}\\sum (xy\\minus{}2yz\\plus{}zx)^2$  ,\r\n\r\n$ 3(\\sum x^3y\\minus{}xyz\\sum x)r\\equal{}\\sum (x^2\\minus{}y^2)(xy\\minus{}2yz\\plus{}zx)$ \u03ba\u03b1\u03b9 $ \\sum x^4\\minus{}\\sum x^2y^2\\geq\\frac{1}{2}\\sum (x^2\\minus{}y^2)^2$ \r\n\r\n\u03ac\u03c1\u03b1 \u03b1\u03c1\u03ba\u03b5\u03af \u03bd\u03b1 \u03b4\u03b5\u03af\u03be\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 $ \\frac{1}{2}\\sum (xy\\minus{}2yz\\plus{}zx)^2 r^2\\minus{}r\\sum (x^2\\minus{}y^2)(xy\\minus{}2yz\\plus{}zx)\\plus{}\\frac{1}{2}\\sum (x^2\\minus{}y^2)^2\\geq 0$ (1)\r\n\r\n\u03a4\u03ce\u03c1\u03b1 \u03c0\u03b1\u03c1\u03b1\u03c4\u03b7\u03c1\u03b5\u03af\u03c3\u03c4\u03b5 \u03cc\u03c4\u03b9 \u03c4\u03bf \u03bc\u03b5\u03c3\u03b1\u03af\u03bf \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf \u03b4\u03b9\u03c0\u03bb\u03ac\u03c3\u03b9\u03bf \u03b3\u03b9\u03bd\u03cc\u03bc\u03b5\u03bd\u03bf \u03b3\u03b9\u03bd\u03cc\u03bc\u03b5\u03bd\u03bf \u03c4\u03c9\u03bd \u03b1\u03ba\u03c1\u03b1\u03af\u03c9\u03bd \u03ac\u03c1\u03b1 \u03b7 (1) \u03b5\u03af\u03bd\u03b1\u03b9 \u03b9\u03c3\u03bf\u03b4\u03cd\u03bd\u03b1\u03bc\u03b7 \u03bc\u03b5 \r\n$ \\frac{1}{2}\\sum (x^2\\minus{}y^2\\minus{}rxy\\plus{}2ryz\\minus{}rzx)^2\\geq 0$ , \u03c4\u03bf \u03bf\u03c0\u03bf\u03af\u03bf \u03b5\u03af\u03bd\u03b1\u03b9 \u03ba\u03b1\u03b9 \u03c4\u03bf \u03b6\u03b7\u03c4\u03bf\u03cd\u03bc\u03b5\u03bd\u03bf .\r\n\r\n\u0391\u03c0\u03cc \u03c4\u03b7\u03bd \u03c0\u03b1\u03c1\u03b1\u03c0\u03ac\u03bd\u03c9 \u03bb\u03cd\u03c3\u03b7 \u03b2\u03bb\u03ad\u03c0\u03bf\u03bd\u03c4\u03b1\u03c2 \u03c4\u03b7\u03bd \u03c0\u03b5\u03c1\u03af\u03c0\u03c4\u03c9\u03c3\u03b7 \u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1\u03c2 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03c2 \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03b2\u03b3\u03ac\u03bb\u03b5\u03b9 \u03b1\u03c0\u03b5\u03c5\u03b8\u03b5\u03af\u03b1\u03c2 \u03cc\u03c4\u03b9 \u03b7 \u03b4\u03bf\u03c3\u03bc\u03ad\u03bd\u03b7 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b9\u03c3\u03bf\u03b4\u03cd\u03bd\u03b1\u03bc\u03b7 \u03bc\u03b5 \r\n$ (\\sum x^2(xy\\plus{}yz\\minus{}2zx))^2\\geq 0$    :wink:",
  "Solution_3": "[quote=\"silouan\"]\n\u0391\u03c0\u03cc \u03c4\u03b7\u03bd \u03c0\u03b1\u03c1\u03b1\u03c0\u03ac\u03bd\u03c9 \u03bb\u03cd\u03c3\u03b7 \u03b2\u03bb\u03ad\u03c0\u03bf\u03bd\u03c4\u03b1\u03c2 \u03c4\u03b7\u03bd \u03c0\u03b5\u03c1\u03af\u03c0\u03c4\u03c9\u03c3\u03b7 \u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1\u03c2 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03c2 \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03b2\u03b3\u03ac\u03bb\u03b5\u03b9 \u03b1\u03c0\u03b5\u03c5\u03b8\u03b5\u03af\u03b1\u03c2 \u03cc\u03c4\u03b9 \u03b7 \u03b4\u03bf\u03c3\u03bc\u03ad\u03bd\u03b7 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b9\u03c3\u03bf\u03b4\u03cd\u03bd\u03b1\u03bc\u03b7 \u03bc\u03b5 \n$ (\\sum x^2(xy \\plus{} yz \\minus{} 2zx))^2\\geq 0$    :wink:[/quote]\r\n\r\n :P wow!ton elegxo ton ekanes esu sil i i mathematica??",
  "Solution_4": "\u0394\u03b5\u03bd \u03ad\u03ba\u03b1\u03bd\u03b1 \u03ad\u03bb\u03b5\u03b3\u03c7\u03bf  :P  \u0388\u03bb\u03c5\u03c3\u03b1 \u03c4\u03bf \u03c3\u03cd\u03c3\u03c4\u03b7\u03bc\u03b1 \r\n\r\n$ \\sum (x^2(kxy \\plus{} kyz \\minus{} lzx))^2 \\equal{} 4(\\sum a^2b^2 \\minus{} abc\\sum a)(\\sum a^4 \\minus{} \\sum a^2b^2) \\minus{} 3(\\sum a^3b \\minus{} abc\\sum a)^2$\r\n\r\n\u0391\u03bd \u03be\u03ad\u03c1\u03b5\u03b9 \u03ba\u03b1\u03bd\u03b5\u03af\u03c2 \u03bd\u03b1 \u03c4\u03b1 \u03b1\u03bd\u03bf\u03af\u03b3\u03b5\u03b9 \u03c3\u03c5\u03bc\u03bc\u03b5\u03c4\u03c1\u03b9\u03ba\u03ac (\u03b2\u03bb\u03ad\u03c0\u03b5 \u03bb\u03cd\u03c3\u03b7 silouan IMO 2008) \u03bf\u03b9 \u03c0\u03c1\u03ac\u03be\u03b5\u03b9\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03bf\u03bb\u03cd \u03c3\u03cd\u03bd\u03c4\u03bf\u03bc\u03b5\u03c2  :wink:",
  "Solution_5": "[quote=\"silouan\"]\u0394\u03b5\u03bd \u03ad\u03ba\u03b1\u03bd\u03b1 \u03ad\u03bb\u03b5\u03b3\u03c7\u03bf  :P  \u0388\u03bb\u03c5\u03c3\u03b1 \u03c4\u03bf \u03c3\u03cd\u03c3\u03c4\u03b7\u03bc\u03b1 \n\n$ \\sum (x^2(kxy \\plus{} kyz \\minus{} lzx))^2 \\equal{} 4(\\sum a^2b^2 \\minus{} abc\\sum a)(\\sum a^4 \\minus{} \\sum a^2b^2) \\minus{} 3(\\sum a^3b \\minus{} abc\\sum a)^2$\n\n\u0391\u03bd \u03be\u03ad\u03c1\u03b5\u03b9 \u03ba\u03b1\u03bd\u03b5\u03af\u03c2 \u03bd\u03b1 \u03c4\u03b1 \u03b1\u03bd\u03bf\u03af\u03b3\u03b5\u03b9 \u03c3\u03c5\u03bc\u03bc\u03b5\u03c4\u03c1\u03b9\u03ba\u03ac (\u03b2\u03bb\u03ad\u03c0\u03b5 \u03bb\u03cd\u03c3\u03b7 silouan IMO 2008) \u03bf\u03b9 \u03c0\u03c1\u03ac\u03be\u03b5\u03b9\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03bf\u03bb\u03cd \u03c3\u03cd\u03bd\u03c4\u03bf\u03bc\u03b5\u03c2  :wink:[/quote]\r\n\r\nexeis dikio...alla thelei ligi eksaskisi ..right?",
  "Solution_6": "[quote=\"silouan\"]\u0394\u03b5\u03bd \u03ad\u03ba\u03b1\u03bd\u03b1 \u03ad\u03bb\u03b5\u03b3\u03c7\u03bf  :P  \u0388\u03bb\u03c5\u03c3\u03b1 \u03c4\u03bf \u03c3\u03cd\u03c3\u03c4\u03b7\u03bc\u03b1 \n\n$ \\sum (x^2(kxy \\plus{} kyz \\minus{} lzx))^2 \\equal{} 4(\\sum a^2b^2 \\minus{} abc\\sum a)(\\sum a^4 \\minus{} \\sum a^2b^2) \\minus{} 3(\\sum a^3b \\minus{} abc\\sum a)^2$\n\n\u0391\u03bd \u03be\u03ad\u03c1\u03b5\u03b9 \u03ba\u03b1\u03bd\u03b5\u03af\u03c2 \u03bd\u03b1 \u03c4\u03b1 \u03b1\u03bd\u03bf\u03af\u03b3\u03b5\u03b9 \u03c3\u03c5\u03bc\u03bc\u03b5\u03c4\u03c1\u03b9\u03ba\u03ac (\u03b2\u03bb\u03ad\u03c0\u03b5 \u03bb\u03cd\u03c3\u03b7 silouan IMO 2008) \u03bf\u03b9 \u03c0\u03c1\u03ac\u03be\u03b5\u03b9\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03bf\u03bb\u03cd \u03c3\u03cd\u03bd\u03c4\u03bf\u03bc\u03b5\u03c2  :wink:[/quote]\r\n\r\n\u0394\u03b5\u03bd \u03ba\u03b1\u03c4\u03b1\u03bb\u03b1\u03b2\u03b1\u03b9\u03bd\u03c9 \u03c0\u03c9\u03c2 \u03b1\u03c0\u03bf \u03c4\u03b7 \u03bb\u03c5\u03c3\u03b7 \u03ba\u03b1\u03c4\u03b1\u03bb\u03b1\u03b2\u03b1\u03b9\u03bd\u03bf\u03c5\u03bc\u03b5 \u03bf\u03c4\u03b9 \u03b7 \u03b4\u03b9\u03b1\u03ba\u03c1\u03b9\u03bd\u03bf\u03c5\u03c3\u03b1 \u03b8\u03b1 \u03b5\u03c7\u03b5\u03b9 \u03b1\u03c5\u03c4\u03b7 \u03c4\u03b7 \u03bc\u03bf\u03c1\u03c6\u03b7. \u0399\u03c3\u03bf\u03c2 \u03bd\u03b1 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c0\u03c1\u03bf\u03c6\u03b1\u03bd\u03b5\u03c2 \u03b1\u03bb\u03bb\u03b1 \u03b5\u03c7\u03c9 \u03bc\u03c0\u03b5\u03c1\u03b4\u03b5\u03c5\u03c4\u03b5\u03b9...\r\n\u039c\u03b7\u03c0\u03c9\u03c2 \u03bc\u03c0\u03bf\u03c1\u03b5\u03b9\u03c2 \u03bd\u03b1 \u03c4\u03bf \u03b5\u03be\u03b7\u03b3\u03b7\u03c3\u03b5\u03b9\u03c2 \u03bb\u03b9\u03b3\u03bf \u03b1\u03bd \u03b5\u03c7\u03b5\u03b9\u03c2 \u03c7\u03c1\u03bf\u03bd\u03bf?",
  "Solution_7": "\u03a3\u03b1\u03c2 \u03b5\u03bc\u03c0\u03bb\u03b5\u03be\u03b1 \u03bb\u03af\u03b3\u03bf  :maybe:  \u039b\u03bf\u03b9\u03c0\u03cc\u03bd \u03c0\u03b1\u03c1\u03b1\u03c0\u03ac\u03bd\u03c9 \u03ad\u03c7\u03c9 \u03c0\u03b1\u03c1\u03bf\u03c5\u03c3\u03b9\u03ac\u03c3\u03b5\u03b9 \u03b4\u03cd\u03bf \u03b5\u03bd\u03c4\u03b5\u03bb\u03ce\u03c2 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03ad\u03c2 \u03c0\u03c1\u03bf\u03c3\u03b5\u03b3\u03b3\u03af\u03c3\u03b5\u03b9\u03c2 . \u0397 \u03c0\u03b9\u03bf \u03bd\u03bf\u03c1\u03bc\u03ac\u03bb \u03ba\u03b1\u03b9 \u03b5\u03bd\u03c4\u03b5\u03bb\u03ce\u03c2 \u03c6\u03c5\u03c3\u03b9\u03bf\u03bb\u03bf\u03b3\u03b9\u03ba\u03ae \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c5\u03c4\u03ae \u03bc\u03b5 \u03c4\u03b7 \u03b4\u03b9\u03b1\u03ba\u03c1\u03af\u03bd\u03bf\u03c5\u03c3\u03b1 . \u0397 \u03b4\u03b5\u03cd\u03c4\u03b5\u03c1\u03b7 \u03ba\u03ac\u03bd\u03b5\u03b9 \u03c7\u03c1\u03ae\u03c3\u03b7 \u03c4\u03b7\u03c2 \u03ac\u03bb\u03bb\u03b7\u03c2 \u03c4\u03b1\u03c5\u03c4\u03cc\u03c4\u03b7\u03c4\u03b1\u03c2 . \u039a\u03b1\u03b9 \u03bb\u03ad\u03c9 \u03cc\u03c4\u03b9 \u03c4\u03b7\u03bd \u03c4\u03b1\u03c5\u03c4\u03cc\u03c4\u03b7\u03c4\u03b1 \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bd\u03b1 \u03c4\u03b7\u03bd \u03c0\u03ac\u03c1\u03bf\u03c5\u03bc\u03b5 \u03bc\u03b5 \u03b4\u03cd\u03bf \u03c4\u03c1\u03cc\u03c0\u03bf\u03c5\u03c2 . \u039f \u03c0\u03c1\u03ce\u03c4\u03bf\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bd\u03b1 \u03bb\u03cd\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf \u03c3\u03cd\u03c3\u03c4\u03b7\u03bc\u03b1 \u03bc\u03b5 \u03c4\u03b1 $ k,l$ \u03ba\u03b1\u03b9 \u03bf \u03b4\u03b5\u03cd\u03c4\u03b5\u03c1\u03bf\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bc\u03b5 \u03c0\u03bf\u03bd\u03b7\u03c1\u03b9\u03ac (\u03b4\u03b7\u03bb\u03b1\u03b4\u03ae \u03ba\u03bf\u03b9\u03c4\u03ac\u03bc\u03b5 \u03c4\u03b7\u03bd \u03c0\u03c1\u03ce\u03c4\u03b7 \u03bb\u03cd\u03c3\u03b7 , \u03b2\u03bb\u03ad\u03c0\u03bf\u03c5\u03bc\u03b5 \u03c0\u03bf\u03c5 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03ba\u03b1\u03b9 \u03c4\u03bf\u03c5\u03c2 \u03c3\u03c5\u03bd\u03c4\u03b5\u03bb\u03b5\u03c3\u03c4\u03ad\u03c2 \u03c4\u03b7\u03c2 \u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1\u03c2 \u03c4\u03bf\u03c5\u03c2 \u03c7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03bf\u03cd\u03bc\u03b5 \u03c9\u03c2 \u03cc\u03c1\u03bf\u03c5\u03c2 \u03b5\u03bd\u03cc\u03c2 \u03b1\u03b8\u03c1\u03bf\u03af\u03c3\u03bc\u03b1\u03c4\u03bf\u03c2 \u03c4\u03bf \u03bf\u03c0\u03bf\u03af\u03bf \u03c4\u03bf \u03c5\u03c8\u03ce\u03bd\u03bf\u03c5\u03bc\u03b5 \u03cc\u03bb\u03bf \u03c3\u03c4\u03bf \u03c4\u03b5\u03c4\u03c1\u03ac\u03b3\u03c9\u03bd\u03bf ) \r\nHope it is more clear now  :)   . \u0391\u03bb\u03bb\u03ac \u03b8\u03b1 \u03ad\u03bb\u03b5\u03b3\u03b1 \u03cc\u03c4\u03b9 \u03b7 \u03c0\u03c1\u03ce\u03c4\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c5\u03c4\u03ae \u03c0\u03bf\u03c5 \u03b1\u03be\u03af\u03b6\u03b5\u03b9 . \u039c\u03b5 \u03c4\u03b7\u03bd \u03ba\u03b1\u03c4\u03b1\u03c3\u03ba\u03b5\u03c5\u03ae \u03c4\u03bf\u03c5 \u03c4\u03c1\u03b9\u03c9\u03bd\u03cd\u03bc\u03bf\u03c5 \u03ba\u03b1\u03b9 \u03c4\u03b7\u03bd \u03c0\u03bf\u03bb\u03cd \u03bb\u03bf\u03b3\u03b9\u03ba\u03ae \u03bc\u03ac\u03b6\u03b5\u03c8\u03b7 \u03c4\u03bf\u03c5 \u03c4\u03b5\u03c4\u03c1\u03b1\u03b3\u03ce\u03bd\u03bf\u03c5  :wink:",
  "Solution_8": "\u039d\u03b1\u03b9, \u03bf\u03bc\u03c9\u03c2 \u03b4\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03c9 \u03bd\u03b1 \u03ba\u03b1\u03c4\u03b1\u03bb\u03b1\u03b2\u03c9 \u03c3\u03c4\u03bf\u03bd 1\u03bf \u03c4\u03c1\u03bf\u03c0\u03bf \u03b3\u03b9\u03b1\u03c4\u03b9 \u03bb\u03c5\u03bd\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf \u03c3\u03c5\u03b3\u03ba\u03b5\u03ba\u03c1\u03b9\u03bc\u03b5\u03bd\u03bf \u03c3\u03c5\u03c3\u03c4\u03b7\u03bc\u03b1... \u0393\u03bd\u03c9\u03c1\u03b9\u03b6\u03bf\u03c5\u03bc\u03b5 \u03c3\u03c4\u03bf \u03c0\u03b5\u03c1\u03b9\u03c0\u03bf\u03c5 \u03c4\u03b7 \u03bc\u03bf\u03c1\u03c6\u03b7 \u03c0\u03bf\u03c5 \u03b8\u03b1 \u03b5\u03c7\u03b5\u03b9 \u03b7 \u03c0\u03b1\u03c1\u03b1\u03c3\u03c4\u03b1\u03c3\u03b7 \u03bf\u03c4\u03b1\u03bd \u03c4\u03b7 \u03b3\u03c1\u03b1\u03c8\u03bf\u03c5\u03bc\u03b5 \u03b5\u03c4\u03c3\u03b9 \u03c9\u03c3\u03c4\u03b5 \u03bd\u03b1 \u03c6\u03b1\u03b9\u03bd\u03b5\u03c4\u03b1\u03b9 \u03bf\u03c4\u03b9 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b8\u03b5\u03c4\u03b9\u03ba\u03b7 \u03ba\u03b1\u03b9 \u03b8\u03b5\u03c9\u03c1\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf \u03ba\u03b1\u03c4\u03b1\u03bb\u03b7\u03bb\u03bf \u03c3\u03c5\u03c3\u03c4\u03b7\u03bc\u03b1?",
  "Solution_9": "\u03a4\u03ce\u03c1\u03b1 \u03ba\u03b1\u03c4\u03b1\u03bb\u03ac\u03b2\u03b1 \u03c4\u03b9 \u03c1\u03ce\u03c4\u03b7\u03c3\u03b5\u03c2 ... \u039b\u03bf\u03b9\u03c0\u03cc\u03bd \u03ba\u03b1\u03bd\u03bf\u03bd\u03b9\u03ba\u03ac \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03c0\u03ac\u03c1\u03b5\u03b9\u03c2 \u03ba\u03ac\u03b8\u03b5 \u03cc\u03c1\u03bf \u03c4\u03b7\u03c2 \u03bc\u03bf\u03c1\u03c6\u03ae\u03c2 $ x^3y$ \u03bc\u03b5 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03bf\u03cd\u03c2 \u03c3\u03c5\u03bd\u03c4\u03b5\u03bb\u03b5\u03c3\u03c4\u03ad\u03c2 \u03c4\u03bf \u03ba\u03b1\u03b8\u03ad\u03bd\u03b1 \u03b1\u03c0\u03cc \u03bc\u03c0\u03c1\u03bf\u03c3\u03c4\u03ac , \u03b1\u03bb\u03bb\u03ac \u03b5\u03b4\u03ce \u03bb\u03cc\u03b3\u03c9 \u03ba\u03c5\u03ba\u03bb\u03b9\u03ba\u03cc\u03c4\u03b7\u03c4\u03b1\u03c2 \u03bf\u03b9 \u03c3\u03c5\u03bd\u03c4\u03b5\u03bb\u03b5\u03c3\u03c4\u03ad\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03af\u03b4\u03b9\u03b9\u03bf\u03b9 \u03ba\u03b1\u03b9 \u03b1\u03bd \u03b4\u03b5\u03b9\u03c2 \u03c0\u03bf\u03b9\u03bf\u03b9 \u03cc\u03c1\u03bf\u03b9 \u03bc\u03ad\u03bd\u03bf\u03c5\u03bd \u03c4\u03bf \u03b1\u03c0\u03bb\u03bf\u03c0\u03bf\u03b9\u03b5\u03af\u03c2 \u03c3\u03b9\u03b3\u03ac \u03c3\u03b9\u03b3\u03ac",
  "Solution_10": "[quote=\"silouan\"]\u03a4\u03ce\u03c1\u03b1 \u03ba\u03b1\u03c4\u03b1\u03bb\u03ac\u03b2\u03b1 \u03c4\u03b9 \u03c1\u03ce\u03c4\u03b7\u03c3\u03b5\u03c2 ... \u039b\u03bf\u03b9\u03c0\u03cc\u03bd \u03ba\u03b1\u03bd\u03bf\u03bd\u03b9\u03ba\u03ac \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03c0\u03ac\u03c1\u03b5\u03b9\u03c2 \u03ba\u03ac\u03b8\u03b5 \u03cc\u03c1\u03bf \u03c4\u03b7\u03c2 \u03bc\u03bf\u03c1\u03c6\u03ae\u03c2 $ x^3y$ \u03bc\u03b5 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03bf\u03cd\u03c2 \u03c3\u03c5\u03bd\u03c4\u03b5\u03bb\u03b5\u03c3\u03c4\u03ad\u03c2 \u03bf \u03ba\u03b1\u03b8\u03ad\u03bd\u03b1 \u03b1\u03c0\u03cc \u03bc\u03c0\u03c1\u03bf\u03c3\u03c4\u03ac , \u03b1\u03bb\u03bb\u03ac \u03b5\u03b4\u03ce \u03bb\u03cc\u03b3\u03c9 \u03ba\u03c5\u03ba\u03bb\u03b9\u03ba\u03cc\u03c4\u03b7\u03c4\u03b1\u03c2 \u03bf\u03b9 \u03c3\u03c5\u03bd\u03c4\u03b5\u03bb\u03b5\u03c3\u03c4\u03ad\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03af\u03b4\u03b9\u03b9\u03bf\u03b9 \u03ba\u03b1\u03b9 \u03b1\u03bd \u03b4\u03b5\u03b9\u03c2 \u03c0\u03bf\u03b9\u03bf\u03b9 \u03cc\u03c1\u03bf\u03b9 \u03bc\u03ad\u03bd\u03bf\u03c5\u03bd \u03c4\u03bf \u03b1\u03c0\u03bb\u03bf\u03c0\u03bf\u03b9\u03b5\u03af\u03c2 \u03c3\u03b9\u03b3\u03ac \u03c3\u03b9\u03b3\u03ac[/quote]\r\n\r\n\u0391 \u03bf\u03ba, \u03ba\u03b1\u03c4\u03b1\u03bb\u03b1\u03b2\u03b1.\r\nThanx"
}
{
  "Tag": [
    "MATHCOUNTS",
    "rate problems"
  ],
  "Problem": "when im practicing, im missing a lot of problems like \r\n\r\nA jet airplane departing on, flying between two airports at an average speed of 540 miles per hour arrives 8 minutes late. departing on time and flying at an average of 480 miles per hour arrives 53 minutes late. what is the number of miles between the two airsports?\r\n\r\nhow do you solve these?",
  "Solution_1": "Let $ s$ miles be the distance between two airports, and $ t$ minutes be the expected time of travel. We now have the following simultaneous equations:\r\n\r\n$ s \\equal{}\\frac{540}{60}*(t\\plus{}8)$\r\n$ s \\equal{}\\frac{480}{60}*(t\\plus{}53)$\r\n\r\nHence we have:\r\n$ \\frac{540}{60}*(t\\plus{}8) \\equal{}\\frac{480}{60}*(t\\plus{}53)$\r\n$ \\Rightarrow 540*(t\\plus{}8) \\equal{} 480*(t\\plus{}53)$\r\n$ \\Rightarrow 1.125(t\\plus{}8) \\equal{} t\\plus{}53$\r\n$ \\Rightarrow 1.125t\\plus{}9 \\equal{} t\\plus{}53$\r\n$ \\Rightarrow t \\equal{} 352$\r\n\r\nTherefore we have:\r\n$ s \\equal{}\\frac{540}{60}*(352\\plus{}8)$\r\n$ \\Rightarrow s \\equal{} 3240$ miles",
  "Solution_2": "ok i worded that teribly. i suck at those types of problems. thanks thought minicon!\r\n\r\nok let me try these\r\n\r\nlet x be the rate and y be the time\r\nwe have\r\n$ 120\\equal{}xy$\r\n$ 120\\equal{}(x\\plus{}8)(y\\minus{}30)$\r\nso\r\nwe need two numbers whose difference is 38 and whose procut is 120...\r\nummmm what did i do wrong?",
  "Solution_3": "[hide=\"Solution for 1\"]\n$ RT \\equal{} 120$\n$ R \\equal{} 120/T$\n\n$ (R\\plus{}8)(T\\minus{}.5)\\equal{}120$\n$ (120/T\\plus{}8)(T\\minus{}.5)\\equal{}120$\n$ (120\\plus{}8T)(T\\minus{}.5) \\equal{} 120T$\n$ 120T\\minus{}60\\plus{}8T^{2}\\minus{}4T\\equal{}120T$\n$ 2T^{2}\\minus{}T\\minus{}15\\equal{}0$\n$ (2T\\plus{}5)(T\\minus{}3)\\equal{}0$\n$ T\\equal{}3\\text{ hours}$\n[/hide]",
  "Solution_4": "I suck at those kind of problems, too.\r\n\r\nMinicon, what's the solution for #2?\r\n\r\nI got 12 minutes.\r\n\r\nI am not confident, so can you explain it your way??",
  "Solution_5": "you can do that too. that saves a lot of time. thanks",
  "Solution_6": "person A and B are biking 1200 meters apart. they start at the same time. A travels 25 kmph, and B travels 30 kmph. what is the difference in time between when A finishes and B finishes?",
  "Solution_7": "1.200/30=.04. thats 2.4 minutes\r\n1.200/25=.048. thats 2.88 minutes.\r\nthey are .48 minutes apart. thats 28.8 seconds",
  "Solution_8": "they r all based on the d=rt formula, with x variables and other algebra.  Try doing practice problems to improve.  For each problem, know wat u wanna find, and wat u have as x variable (sometimes they're not the same).  Also, clearly know for each equation wat is time and wats speed",
  "Solution_9": "I was wondering about that too, but in a MathCounts competition, if the times are like $30$, $40$, $45$, etc. minutes, I would just bash it like in minicon's solution, since trying relative speeds may take some time to think about. However, a time like $53$ minutes, probably would be quicker using relative speeds."
}
{
  "Tag": [
    "quadratics",
    "algebra",
    "number theory",
    "modular arithmetic",
    "quadratic formula"
  ],
  "Problem": "Find all $n \\leq 100$ such that $n^{3}+n+31$ is divisible by $43$. \r\n\r\n\r\n\r\n\r\n\r\n\r\nI have often heard that you should not use the quadratic equation in modular arithmetic, but Thomas Mildorf uses it on page 3 of http://www.tjhsst.edu/~tmildorf/math/NumberTheory.pdf. Is it only applicable to prime moduli?",
  "Solution_1": "[quote=\"Phelpedo\"]Find all $n \\leq 100$ such that $n^{3}+n+31$ is divisible by $43$. [/quote]\r\nI think you meant $n^{2}$. And I'm pretty sure you're able to use the quadratic formula, at least in prime moduli. I don't know about other moduli.\r\n[hide=\"Solution, without the quadratic formula\"]\n\\[n^{2}+n+31 \\equiv n^{2}+n-12 \\equiv (n+4)(n-3) \\bmod 43 \\implies n \\in \\{3,39,46,82,89\\}\\]\n[/hide]"
}
{
  "Tag": [
    "symmetry"
  ],
  "Problem": "Which of the following lettered items possesses\r\nline symmetry?  Consider vertical and/or horizontal symmetry.\r\n\r\na. the letter F \r\n\r\nb. an  equilateral triangle\r\n\r\nc. the letter X\r\n\r\nd. a square",
  "Solution_1": "[quote=\"Interval\"]Which of the following lettered items possesses\nline symmetry?  Consider vertical and/or horizontal symmetry.\n\na. the letter F \n\nb. an  equilateral triangle\n\nc. the letter X\n\nd. a square[/quote]\r\n[hide=\"a\"]No F does not have one.[/hide]\n[hide=\"b\"]Yes, an equilateral triangle split in half is a 30-60-90 triangle.[/hide]\n\n[hide=\"c\"]Yes X has one straight through it.[/hide]\n\n[hide=\"d\"]Yes a square has one . A square split along its diagonal makes two 45-45-90 triangles.[/hide]",
  "Solution_2": "[quote=\"Interval\"]Which of the following lettered items possesses\nline symmetry?  Consider vertical and/or horizontal symmetry.\n\na. the letter F \n\nb. an  equilateral triangle\n\nc. the letter X\n\nd. a square[/quote]\r\n\r\nB,C,D do\r\nFor the triangle, any perpendicular line that go through a corner of the triangle works\r\nFor the X, either go straight down the middle or straight across the side\r\nSquare have 4 lines of symmetry"
}
{
  "Tag": [
    "calculus",
    "integration",
    "derivative",
    "calculus computations"
  ],
  "Problem": "When using integration by parts, is there any advantage to using the du and dv?  To me, it seems much simpler to just use u, v, u', and v' (so that the formula is $ \\int uv' \\equal{} uv \\minus{} \\int u'v )$.  However, my teacher and most websites seem to prefer to use the du and dv, which appears to me to be more work and more confusing than is necessary.  So is there any reason to use du and dv instead of u' and v'?",
  "Solution_1": "It's just a matter of pleasure.\r\n\r\nI do use $ \\int f(t)g'(t)\\,dt \\equal{} f(t)g(t) \\minus{} \\int f'(t)g(t)\\,dt,$ (*) because for me (at least) is a faster way on doing the integration by parts and saves a lot of time.\r\n\r\nIf your professor likes using $ du$ and $ dv,$ it's probably 'cause he(she) has taught so many times IBP in the same fashion that he(she) doesn't have time on changing the way on teaching it, or perhaps he(she) doesn't realize that using (*) is faster.",
  "Solution_2": "I personally find it less confusing rather than more confusing to use $ du$ and $ dv,$ primarily because it is the same language and the same thought processes used in making substitution, and because it creates a consistent level of bookkeeping. In my mind it's a notation that forces you to do things the right way, and that's a good thing. And that is the way I teach, because it works for me and it works for my students.\r\n\r\nA way, way, down the road thought: Once you've defined the Riemann-Stieljes integral, then provided $ f$ and $ g$ satisfy some reasonable hypotheses,\r\n\r\n$ \\int_a^bf(x)\\,dg(x)\\equal{}f(b)g(b)\\minus{}f(a)g(a)\\minus{}\\int_a^bg(x)\\,df(x).$",
  "Solution_3": "On a side note: I sometimes see other students having trouble remembering the \"formula\" for integration by parts.\r\n\r\nI then remind them that it is really just the product rule for differentiation, integrated:\r\n\r\n$ \\frac{d}{dx} (f g) \\equal{} f ' g \\plus{} g' f$, so $ f g' \\equal{} \\frac{d}{dx} ( f g ) \\minus{} f' g$, which gives:\r\n\r\n\r\n$ \\int f g'  \\equal{}  \\int \\frac{d}{dx} ( f g ) \\minus{} \\int f' g \\equal{}  f g \\minus{} \\int f' g$.",
  "Solution_4": "\"it is really just the product rule for differentiation, integrated\"\r\n\r\nThat's why I prefer to use u' and v' rather than du and dv; to me; the former notation makes the connection between integration by parts and the product rule clear and obvious, while the latter notation does not.",
  "Solution_5": "You should never write down an integral without specifying the variable you're integrating with respect to.  The notation $ \\int uv'$ is ambiguous.\r\n\r\n[b]Edit:[/b]  More precisely, you are supposed to assume that the variable you're taking the derivative of $ v$ with respect to and the variable you're integrating with respect to are the same.  This is ambiguous, and the better notation $ \\int u \\, dv$ resolves the ambiguity by pointing out a more important truth: that as long as those two variables are the same variable, it doesn't matter (up to some smoothness requirement) what it is.",
  "Solution_6": "I agree with t0rajir0u and feel that I probably should have included that as part of my previous post.",
  "Solution_7": "[quote=\"t0rajir0u\"]You should never write down an integral without specifying the variable you're integrating with respect to.  The notation $ \\int uv'$ is ambiguous.[/quote]\r\n\r\nOkay, so why not say that $ \\int uv' dx \\equal{} uv \\minus{} \\int u'v dx$ ?",
  "Solution_8": "Because it's still not clear whether the derivatives are with respect to $ x$.  What you should write is $ \\int u \\frac {dv}{dx} \\, dx$, which, of course, you should really write $ \\int u \\, dv$.\r\n\r\nIt is maybe not clear to you why specifying variables is so important, but if you start doing a lot of changing of variables or working with more than one variable you will quickly come to see the benefit of this notation.",
  "Solution_9": "[quote=\"t0rajir0u\"] It is maybe not clear to you why specifying variables is so important, but if you start doing a lot of changing of variables or working with more than one variable you will quickly come to see the benefit of this notation.[/quote]\r\n\r\nI'll take your word on this, as I have no experience with multivariable calculus/advanced calculus (and you and Kent obviously do).  Thanks for the help and insight.",
  "Solution_10": "It's not even a matter of preparing yourself for multivariable calculus.  You just don't want to get into the habit of always thinking of $ x$ as the independent variable, since it could also be $ y$ or $ t$ or anything else, depending on the problem."
}
{
  "Tag": [
    "articles",
    "trigonometry"
  ],
  "Problem": "I registered for a hosting service this summer for a free math league our high school is coordinating ([url]http://www.rocketcitymath.org[/url]). Well, I still have about a gigabyte left, so I was hoping to put that space to good use by constructing an open content math website. Anyone who wants to can contribute articles, mock contests, proofs, etc., that will help other avid math students. If there is enough interest, I'll post an E-mail address dedicated to contributions. Thank you.",
  "Solution_1": "is this school math or olympiad math? :)",
  "Solution_2": "It's tougher than school for sure.",
  "Solution_3": "I'm not sure what you are asking, but if you're talking about the Rocket City Math League, it's probably around the level of normal math competitions for Algebra II / Trigonometry students or lower and is not as tough as the Olympiad. An archive of past tests is available at the website if you would like to view some of the past contests. As for contributing articles and such, I think students would be happy to see a variety of math levels covered."
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "perpendicular bisector",
    "projective geometry",
    "geometry solved"
  ],
  "Problem": "Given 6 arbitrary points [tex]A_1,A_2,B_1,B_2,C_1,C_2[/tex]. [tex]M_A, M_B, M_C[/tex] are the midpoints of [tex]A_1A_2, B_1B_2, C_1C_2[/tex] respectively. The perpendicular bisector of [tex]A_1A_2[/tex] and [tex]B_1B_2[/tex] meet at [tex]P_{AB}[/tex]. The same applies for [tex]P_{BC}[/tex] and [tex]P_{CA}[/tex].\r\nNow choose two midpoints and the intersection point of the respective perpendicular bisectors and draw a circle through these three points. Do this three times, so that you'll get three circles. Proof that there's a common point of intersection",
  "Solution_1": "Unfortunately, this is just a complicated application of the [b]Miquel theorem[/b]:\r\n\r\n[i]If ABC is a triangle, and X, Y, Z are three arbitrary points on its sidelines BC, CA, AB, then the circumcircles of triangles AYZ, BZX, CXY have a common point[/i].\r\n\r\nIn your case, the triangle is $P_{BC}P_{CA}P_{AB}$ and the points on its sidelines are $M_A$, $M_B$, $M_C$, respectively.\r\n\r\nThe proof of the Miquel theorem above is a simple application of the chordal angle theorem. [You can find a proof together with some further discussion at http://cut-the-knot.com/proofs/Pivot.shtml , Statement B.]\r\n\r\n  Darij",
  "Solution_2": "Thanks Darij! \r\nMy poor geometric knowledge is gradually increasing. Yesterday: Carnot's theorem;Today: Miquel's Theorem; Tomorrow: Duck's Theorem! \r\n :D",
  "Solution_3": "[quote=\"Christian Hirsch\"]Tomorrow: Duck's Theorem![/quote]\r\n\r\nThanks for your interest, but just to mention it, this is not the established name of that fact ;)  others would call it a particular case of Pascal's theorem with the Pascal line being the line at infinity :D\r\n\r\n  darij"
}
{
  "Tag": [
    "function",
    "algebra proposed",
    "algebra"
  ],
  "Problem": "a) Prove that there does not exist any function $f: \\mathbb{R}\\rightarrow \\mathbb{R}$ so that $(f\\circ f)(x)=\\{\\begin{array}{cc}{II}\\sqrt{2007}&\\textrm{if}\\;x\\in\\mathbb{Q}\\2007,&\\textrm{if}\\;x \\in\\mathbb{R}-\\mathbb{Q}\\end{array}$\r\nb) Prove that there exist an infinity of functions $g: \\mathbb{R}\\rightarrow \\mathbb{R}$  so that $(g\\circ g)(x)=\\{\\begin{array}{cc}{II}2007&\\textrm{if}\\;x\\in \\mathbb{Q}\\\\sqrt{2007},&\\textrm{if}\\;x \\in\\mathbb{R}-\\mathbb{Q}\\end{array}$",
  "Solution_1": "a) Suppose $x \\in \\mathbb{Q}$ and $f(x) \\in \\mathbb{Q}$.\r\nThen:\r\n- $f(x) \\= q \\in \\mathbb{Q}$\r\n- $\\sqrt{2007}= f(q)$\r\n- $f(\\sqrt{2007}) = f(f(q) = \\sqrt{2007}$\r\n- $f(f(\\sqrt{2007})) = 2007 = f(\\sqrt{2007})$.\r\nThe last two lines are a contradiction.\r\n\r\nSuppose $f(x) = r \\in \\mathbb{R}-\\mathbb{Q}$. Then:\r\n- $f(x) = r$\r\n- $f(f(x)) = \\sqrt{2007}= f(r)$\r\n- $f(\\sqrt{2007}) = f(f(r)) = 2007$\r\n- $f(f(\\sqrt{2007})) = 2007 = f(2007)$\r\n- $f(2007) = f(f(2007) = \\sqrt{2007}$\r\nThe last two lines are a contradiction.\r\n\r\nb) Let r a rational number. Define f as:\r\n$f(r) = 2007$\r\n$f(2007) = 2007$\r\n$f(\\mathbb{Q}-\\{r; 2007\\}) = r$\r\n\r\n$f(\\mathbb{R}-\\mathbb{Q}$.\r\n\r\nThen, if x is rational:\r\n- if x is r or 2007, then $f(f(x)) = f(2007) = 2007 \\in \\mathbb{Q}$\r\n- otherwise, $f(f(x)) = f(r) = 2007 \\in \\mathbb{Q}$\r\nIf x is irrational:\r\n- $f(f(x)) = f(\\sqrt{2007}) = \\sqrt{2007}$\r\n\r\nYou can choose r in infinte ways."
}
{
  "Tag": [
    "trigonometry"
  ],
  "Problem": "Demonstrati ca $\\ h_{a}+h_{b}+h_{c}\\leq\\ 9r+\\frac{a^{2}+b^{2}+c^{2}-ab-bc-ca}{p}$, notatiile fiind cele obisnuite.",
  "Solution_1": "Ce imbunatatire!!!Acuma sa o demonstram.Avem ca $h_{a}=\\frac{2S}{a}$ si analoagele.Trebuie sa demonstram ca \r\n $ S \\left[(\\sum a)\\left(\\sum\\frac{1}{a}\\right)-9\\right]\\leq \\sum a^{2}-\\sum ab$.Acuma folosim substitutiile Ravi $a=y+z,b=x+z,c=x+y$.Avem de demonstrat astfel $\\sqrt{xyz\\sum x}\\cdot \\frac{2\\sum x^{3}-\\sum xy(x+y)}{\\prod xy(x+y)}\\leq \\sum x^{2}-\\sum xy$.Acuma o rescriem $ 2\\sqrt{xyz\\sum x}\\sum_{cyc}(x-y)^{2}(x+y)\\leq \\prod (x+y)\\sum (x-y)^{2}$ echivalent cu $\\sum (x-y)^{2}(x+y)[z(x+y+z)+xy-2\\sqrt{xyz\\sum x}] \\geq 0$ echivalent cu $\\sum (x-y)^{2}(x+y)(\\sqrt{z(x+y+z)}-\\sqrt{xy})^{2}\\geq 0$.(QED) :lol:",
  "Solution_2": "Din $ah_{a}=bh_{b}=ch_{c}=2S$ si $r=\\frac{S}{p}=\\frac{2S}{a+b+c},$ avem:\r\n$\\ h_{a}+h_{b}+h_{c}\\leq\\ 9r+\\frac{a^{2}+b^{2}+c^{2}-ab-bc-ca}{p}\\Longleftrightarrow$ \r\n$\\Longleftrightarrow S(a+b+c)\\bigg(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\bigg)\\leq 9S+(a^{2}+b^{2}+c^{2}-ab-bc-ca)\\Longleftrightarrow$\r\n$\\Longleftrightarrow S\\bigg[\\bigg(\\frac{a}{b}+\\frac{b}{a}-2\\bigg)+\\bigg(\\frac{b}{c}+\\frac{c}{b}-2\\bigg)+\\bigg(\\frac{c}{a}+\\frac{a}{c}-2\\bigg)\\bigg]\\leq \\frac{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}{2}\\Longleftrightarrow$\r\n$\\Longleftrightarrow S\\bigg[\\frac{(a-b)^{2}}{ab}+\\frac{(b-c)^{2}}{bc}+\\frac{(c-a)^{2}}{ca}\\bigg]\\leq \\frac{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}{2}\\Longleftrightarrow$\r\n$\\Longleftrightarrow \\sum \\bigg[\\frac{(a-b)^{2}}{ab}\\bigg(S-\\frac{ab}{2}\\bigg)\\bigg]\\leq 0$, inegalitate evidenta daca tinem cont ca $S=\\frac{ab\\sin C}{2}\\leq \\frac{ab}{2}.$",
  "Solution_3": "[color=darkblue][b][u]Demonstratia de mai jos este oarecum similara cu demonstratia lui Marius Damian.[/u][/b]\n\nInmultind cu $4Rp$, inegalitatea $\\sum h_{a}\\le 9r+\\frac{1}{p}\\cdot(\\sum a^{2}-\\sum bc)$ devine : $2p\\cdot\\sum 2Rh_{a}\\le 9\\cdot 4Rpr+2R\\cdot(2\\sum a^{2}-2\\sum bc)$.\n\nFolosind formulele $\\{\\begin{array}{c}2Rh_{a}=bc\\\\\\\\ 4Rpr=abc\\end{array}$ si identitatile cunoscute $\\{\\begin{array}{c}2\\sum a^{2}-2\\sum bc=\\sum (b-c)^{2}\\\\\\\\ \\sum a\\cdot\\sum bc=9abc+\\sum a(b-c)^{2}\\end{array}$ inegalitatea devine : \n\n$\\sum a\\sum bc\\le 9abc+2R\\cdot\\sum (b-c)^{2}$ $\\Longleftrightarrow$ $9abc+\\sum a(b-c)^{2}\\le 9abc+2R\\cdot\\sum (b-c)^{2}$ $\\Longleftrightarrow$\n\n$\\sum a(b-c)^{2}\\le 2R\\cdot\\sum (b-c)^{2}$ $\\Longleftrightarrow$ $\\sum (2R-a)(b-c)^{2}\\ge 0$, ceea ce este adevarat. Avem egalitate daca si numai daca $a=b=c$.[/color]"
}
{
  "Tag": [
    "trigonometry"
  ],
  "Problem": "Solve.\r\n\r\n$ sin(2\\theta)\\equal{}\\sqrt{2}cos(\\theta)$",
  "Solution_1": "\\[ \\sin 2x \\equal{} \\sqrt{2} \\cos x \\iff 2 \\sin x \\cos x \\equal{} \\sqrt{2} \\cos x.\\]\r\n\r\nIf $ \\cos x \\neq 0$,\r\n\\[ \\sin x \\equal{} \\frac{1}{\\sqrt{2}} \\iff x \\equal{} \\frac{1}{4}\\pi, \\frac{3}{4}\\pi.\\]\r\n\r\nIf $ \\cos x \\equal{} 0$,\r\n\\[ \\cos x \\equal{} 0 \\iff x \\equal{} \\minus{}\\frac{1}{2}\\pi, \\frac{1}{2}\\pi.\\]\r\n\r\nThus $ x \\equal{} \\frac{1}{4}\\pi, \\frac{3}{4}\\pi, \\pm \\frac{1}{2}\\pi$."
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "geometry proposed"
  ],
  "Problem": "Given triangle $ ABC$, with circumcenter $ O$, let the $ A$-symmedian hit $ BC$ at $ S$.  Let the parallel through $ S$ to $ AO$ hit the $ A$-altitude at $ X$.  Prove that $ OX\\perp AS$.",
  "Solution_1": "Let the A-symmedian cut  the circumcircle again at $ T.$ Tangents to $ (O)$ at $ A,T$ meet at a point $ V$ on $ BC,$ i.e. the center of the A-Apollonius circle of $\\triangle ABC.$ Let $ OY \\perp AS$ and define $ X'\\equiv OY \\cap AH_a.$  From the cyclic quadrilaterals $ AYH_aV$ and $ YSH_aX'$ we get  $ \\angle YSX' \\equal{} \\angle YH_aA \\equal{} \\angle YVA \\equal{} \\angle OAY $ $ \\Longrightarrow$ $ OA \\parallel SX'$ $\\Longrightarrow$ $ X \\equiv X'.$",
  "Solution_2": "My solution:\n\nLet $ Z $ be the projection of $ X $ on $ AS $ .\nLet the tangent of $ \\odot (ABC) $ through $ A $ intersect $ BC $ at $ Y $ .\n\nIt suffices to prove $ O \\in XZ $ .\n\nFrom $ SX \\parallel AO \\Longrightarrow SX \\perp AY $ ,\nso $ X $ is the orthocenter of $ \\triangle AYS \\Longrightarrow Y \\in XZ $ .\n\nSince $ AS $ is the polar of $ Y $ WRT $ \\odot (O) $ ,\nso we get $ YO \\perp AS \\Longrightarrow O, X, Y, Z $ are collinear .\n\nQ.E.D",
  "Solution_3": "An extension\n\nLet $O$ be circumcenter of triangle $ABC$. $P$ is a point on $OA$. $(K)$ is circumcircle of triangle $PBC$. $PQ$ is diameter of $(K)$. $AQ$ cuts $BC$ at $S$. $T$ is a point such that $AT\\perp BC$ and $ST\\parallel AO$. Prove that $PT$ and $AS$ intersect on $(K)$.",
  "Solution_4": "[quote=\"buratinogigle\"]An extension\n\nLet $O$ be circumcenter of triangle $ABC$. $P$ is a point on $OA$. $(K)$ is circumcircle of triangle $PBC$. $PQ$ is diameter of $(K)$. $AQ$ cuts $BC$ at $S$. $T$ is a point such that $AT\\perp BC$ and $ST\\parallel AO$. Prove that $PT$ and $AS$ intersect on $(K)$.[/quote]\n\n\nReformulation\n\nLet $O$ be circumcenter of triangle $ABC$. $S$ is a point on $BC$;  $T$ is a point such that $AT\\perp BC$ and $ST\\parallel AO$ ; $P$ a point on $AO$  sucht that $PT \\bot AS$ and $ PT \\cap AS \\doteq F$.  Prove that  $B,C,F$ and $P$ are cyclic\n\nProof \n\n$ BC \\cap TP \\doteq Z $ and  $T$ is orthocenter of $\\Delta AZS$ then $TS \\bot AZ$ ie. $ AO \\bot AZ \\Rightarrow ZP.ZF \\doteq AZ^2$ and $AZ$ is tangent to circumcircle of $ \\Delta ABC$ ie. $ZB.ZC \\doteq AZ^2$ \n$ ZP.ZF \\doteq ZB.ZC \\doteq AZ^2 \\Rightarrow B,C,F$ and $P$ are cyclic.\nDone!",
  "Solution_5": "[quote=\"buratinogigle\"]An extension\n\nLet $O$ be circumcenter of triangle $ABC$. $P$ is a point on $OA$. $(K)$ is circumcircle of triangle $PBC$. $PQ$ is diameter of $(K)$. $AQ$ cuts $BC$ at $S$. $T$ is a point such that $AT\\perp BC$ and $ST\\parallel AO$. Prove that $PT$ and $AS$ intersect on $(K)$.[/quote]\nMy solution:\n\n[b]Lemma[/b]:\n\nLet $ O $ be the circumcenter of $ \\triangle ABC $ and $ P $ be a point on $ AO $ .\nLet $ Q $ be the antipode of $ P $ in $ \\odot (PBC) $ and $ M $ be the projection of $ P $ on $ BC $ .\n\nThen $ AM, AQ $ are isogonal conjugate of $ \\angle BAC $ .\n\n[b]Proof[/b]:\n\nLet $ R $ be the isogonal conjugate of $ Q $ WRT $ \\triangle ABC $ .\nLet $ Y, Z $ be the projection of $ P $ on $ AC, AB $, respectively .\nLet $ B^*, C^* $ be the projection of  $ R $ on $ AC, AB $, respectively .\n\nIt suffices to prove $ A, M, R $ are collinear .\n\nSince $ PM, PQ $ are isogonal conjugate of $ \\angle BPC $ ,\nso we get $ \\angle C^*BR=\\angle QBC=\\angle BPM=\\angle BZM \\Longrightarrow MZ \\parallel RB $ .\nSimilarly we can prove $ MY \\parallel RC $ .\n\nFrom $ P \\in AO $ we get $ YZ \\parallel CB $ ,\nso $ \\triangle MYZ $ and $ \\triangle RCB $ are homothetic with center $ A \\Longrightarrow R \\in AM $ .\n____________________________________________________________\n[b]Back to the main problem[/b]:\n\nLet the tangent of $ \\odot (ABC) $ through $ A $ intersect $ BC $ at $ X $ .\nLet $ Y $ be the projection of $ T $ on $ AS $ and $ M $ be the projection of $ P $ on $ BC $ .\n\nFrom $ ST \\parallel AO \\Longrightarrow ST \\perp AX $ ,\nso $ T $ is the orthocenter of $ \\triangle AXS \\Longrightarrow Y \\in XT $ .\nSince $ A, P, M, X $ lie on a circle with diameter $ XP $ ,\nso combine with the lemma we get $ \\angle MXY=\\angle TAS=\\angle MAP=\\angle MXP $ ,\nhence $ P, T, X, Y $ are collinear $ \\Longrightarrow \\angle QYP=90^{\\circ} $ . i.e. $ Y \\equiv AS \\cap PT \\in \\odot (K) $ \n\nQ.E.D"
}
{
  "Tag": [
    "trigonometry",
    "search",
    "complex numbers"
  ],
  "Problem": "Find the value of the following expression \r\n$ \\sin\\frac{\\pi}{14}\\sin\\frac{3\\pi}{14}\\sin\\frac{5\\pi}{14}....\\sin\\frac{13\\pi}{14}$",
  "Solution_1": "[hide=\"Hint\"]$ m$ is integer numbers such that $ m\\ge2$.\n\\[ \\prod_{i \\equal{} 1}^{m \\minus{} 1}\\sin\\frac {i \\pi}{m} \\equal{} \\frac {m}{2^{m \\minus{} 1}}\n\\]\n[/hide]",
  "Solution_2": "[quote=\"RAGHAV GROVER\"]Find the value of the following expression \n$ \\sin\\frac {\\pi}{14}\\sin\\frac {3\\pi}{14}\\sin\\frac {5\\pi}{14}....\\sin\\frac {13\\pi}{14}$[/quote]\r\n\r\n[hide]\nSince $ \\sin \\theta=\\cos({\\pi\\over2}-\\theta)$\n\nThen $ \\sin\\frac {k\\pi}{14}=\\cos({7\\pi\\over14}-\\frac {k\\pi}{14})$;$ \\quad k=1,3,5\\cdots13$\n\n$ \\therefore \\sin\\frac {\\pi}{14}\\sin\\frac {3\\pi}{14}\\sin\\frac {5\\pi}{14}....\\sin\\frac {13\\pi}{14}=\\cos\\frac {3\\pi}{7}\\cos\\frac {2\\pi}{7}\\cos\\frac {\\pi}{7}....\\cos\\frac {-3\\pi}{7}$\n$ =\\left(\\cos\\frac {3\\pi}{7}\\cos\\frac {2\\pi}{7}\\cos\\frac {\\pi}{7}\\right)^2$\n\nAnd\n\n$ \\cos\\frac {3\\pi}{7}\\cos\\frac {2\\pi}{7}\\cos\\frac {\\pi}{7}$\n\n$ ={\\cos\\frac {3\\pi}{7}\\cos\\frac {2\\pi}{7}\\cdot 2\\sin \\frac {\\pi}{7}\\cos\\frac {\\pi}{7}\\over 2\\sin \\frac {\\pi}{7}}$\n\n$ ={\\cos\\frac {3\\pi}{7}\\cos\\frac {2\\pi}{7}\\sin \\frac {2\\pi}{7}\\over 2\\sin \\frac {\\pi}{7}}$\n\n$ ={\\cos\\frac {3\\pi}{7}\\cdot2\\cos\\frac {2\\pi}{7}\\sin \\frac {2\\pi}{7}\\over 2^2\\sin \\frac {\\pi}{7}}$\n\n$ ={\\cos\\frac {3\\pi}{7}\\sin \\frac {4\\pi}{7}\\over 2^2\\sin \\frac {\\pi}{7}}$\n\n$ ={\\cos\\frac {3\\pi}{7}\\sin \\frac {3\\pi}{7}\\over 2^2\\sin \\frac {\\pi}{7}}$\n\n$ ={2\\cos\\frac {3\\pi}{7}\\sin \\frac {3\\pi}{7}\\over 2^3\\sin \\frac {\\pi}{7}}$\n\n$ = {\\sin \\frac {6\\pi}{7}\\over 2^3\\sin \\frac {\\pi}{7}}={\\sin \\frac {\\pi}{7}\\over 2^3\\sin \\frac {\\pi}{7}}= {1\\over 2^3}$\n\n$ \\therefore \\sin\\frac {\\pi}{14}\\sin\\frac {3\\pi}{14}\\sin\\frac {5\\pi}{14}....\\sin\\frac {13\\pi}{14}=  \\left({1\\over 2^3}\\right)^2= {1\\over 2^6}$\n\n[/hide]",
  "Solution_3": "yes i also solved by the same method..",
  "Solution_4": "[quote=\"Kouichi Nakagawa\"][hide=\"Hint\"]$ m$ is integer numbers such that $ m\\ge2$.\n\\[ \\prod_{i \\equal{} 1}^{m \\minus{} 1}\\sin\\frac {i \\pi}{m} \\equal{} \\frac {m}{2^{m \\minus{} 1}}\n\\]\n[/hide][/quote]\n[hide=\"Using formula\"]\n$ \\prod_{i \\equal{} 1}^{13}\\sin\\frac {i\\pi}{14} \\equal{} \\left(\\prod_{i \\equal{} 1}^{6}\\sin\\frac {i\\pi}{7}\\right) \\times \\left(\\prod_{i \\equal{} 1}^{7}\\sin\\frac {(2i \\minus{} 1)\\pi}{14}\\right)$\n\nSo,\n$ \\prod_{i \\equal{} 1}^{13}\\sin\\frac {i\\pi}{14} \\equal{} \\frac {14}{2^{13}}$\n$ \\prod_{i \\equal{} 1}^{6}\\sin\\frac {i\\pi}{7} \\equal{} \\frac {7}{2^{6}}$\n\nThus, $ \\prod_{i \\equal{} 1}^{7}\\sin\\frac {(2i \\minus{} 1)\\pi}{14} \\equal{} \\left(\\prod_{i \\equal{} 1}^{13}\\sin\\frac {i\\pi}{14}\\right)/\\left(\\prod_{i \\equal{} 1}^{6}\\sin\\frac {i\\pi}{7}\\right) \\equal{} \\frac {\\frac {14}{2^{13}}}{\\frac {7}{2^{6}}} \\equal{} \\frac {2}{2^7} \\equal{} \\frac {1}{2^6} \\equal{} \\boxed{\\frac {1}{64}}$.[/hide]",
  "Solution_5": "[quote=\"Kouichi Nakagawa\"][hide=\"Hint\"]$ m$ is integer numbers such that $ m\\ge2$.\n\\[ \\prod_{i \\equal{} 1}^{m \\minus{} 1}\\sin\\frac {i \\pi}{m} \\equal{} \\frac {m}{2^{m \\minus{} 1}}\n\\]\n[/hide][/quote]\r\n\r\nThanks for ur help ,ur solution is very cute and short can u plz tell me the same  formula for cosine also :) \r\n\r\nWhat is the name of this formula , i have never seen this one ,plz tell me so that i can search for its proof.. :!:",
  "Solution_6": "Use complex numbers! Although, it may not look like it, it can easily be evaluated using complex numbers. \r\n[hide]Hint: z^14=1.[/hide]",
  "Solution_7": "Hey,Kouichi Nakagawa can u evaluate the following by the \r\nformula,you have stated\r\n$ \\sin\\frac{5\\pi}{14}\\sin\\frac{3\\pi}{14}\\sin\\frac{\\pi}{14}$. :roll: \r\n\r\nThanks in advance!!",
  "Solution_8": "I'm not Kouichi Nakagawa, but, here's what I'd do...\r\n\r\n[hide]Let's say $ x \\equal{}  \\sin\\frac{5\\pi}{14}\\sin\\frac{3\\pi}{14}\\sin\\frac{\\pi}{14}$. Note that since $ \\sin y \\equal{} \\sin(\\pi \\minus{} y)$, we get that $ x^2 \\sin{\\frac{7\\pi}{14}} \\equal{} x^2 \\equal{} \\sin\\frac{\\pi}{14}\\sin\\frac{3\\pi}{14}\\sin\\frac{5\\pi}{14}....\\sin\\frac{13\\pi}{14} \\equal{} \\frac{1}{64}$. $ x$ must be positive, so $ x \\equal{} \\frac{1}{8}$. \n\nAnd Google calculator agrees with me. =D\n[/hide]",
  "Solution_9": "$ z^m=1 \\iff z=1, e^{\\frac{2\\pi i}{m}}, \\ e^{\\frac{4\\pi i}{m}}, \\ \\cdots, \\ e^{\\frac{2(m-1)\\pi i}{m}}$.\r\nSo,\r\n\\[ z^m-1=(z-1)(z-e^{\\frac{2\\pi i}{m}})(z-e^{\\frac{4\\pi i}{m}})\\cdots(z-e^{\\frac{2(m-1)\\pi i}{m}})\\]\r\nAnd,\r\n\\[ \\left.\\frac{z^m-1}{z-1}\\right|_{z=1}=m=(1-e^{\\frac{2\\pi i}{m}})(1-e^{\\frac{4\\pi i}{m}})\\cdots(1-e^{\\frac{2(m-1)\\pi i}{m}})\\]\r\n$ (\\because \\frac{z^m-1}{z-1}=1+z+\\cdots+z^{m-1})$\r\nAnd,\r\n\\begin{eqnarray*}\r\n& & \\overline{m}=\\overline{(1-e^{\\frac{2\\pi i}{m}})(1-e^{\\frac{4\\pi i}{m}})\\cdots(1-e^{\\frac{2(m-1)\\pi i}{m}})} \\\\\r\n& \\iff & m=(1-e^{-\\frac{2\\pi i}{m}})(1-e^{-\\frac{4\\pi i}{m}})\\cdots(1-e^{-\\frac{2(m-1)\\pi i}{m}})\r\n\\end{eqnarray*}\r\n\r\nWe use $ (1-e^{\\frac{2k\\pi i}{m}})(1-e^{-\\frac{2k\\pi i}{m}})=2-2\\cos(\\frac{2k\\pi}{m})$.\r\nSo,\r\n\\[ m^2=2^{m-1}\\left(1-\\cos\\frac{2\\pi}{m}\\right)\\left(1-\\cos\\frac{4\\pi}{m}\\right)\\cdots\\left(1-\\cos\\frac{2(m-1)\\pi}{m}\\right)\\]\r\nAnd, $ 1-\\cos\\frac{2\\pi}{m}=2\\sin^2 \\frac{k\\pi}{m}$.\r\nTherefore,\r\n\\[ m^2=2^{2m-2}\\sin^2 \\frac{\\pi}{m}\\sin^2 \\frac{2\\pi}{m}\\cdots\\sin^2 \\frac{(m-1)\\pi}{m}\\]\r\n\r\nThus,\r\n\\[ \\prod_{i=1}^{m-1} \\sin \\frac{i\\pi}{m}=\\frac{m}{2^{m-1}}\\]",
  "Solution_10": "See here. [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=819011&search_id=1630616319#819011[/url]",
  "Solution_11": "Kunny is there a same type of formula for cosine also??",
  "Solution_12": "I'm sure you can do it easily with complex numbers too. Anyone want to give it a shot?",
  "Solution_13": "Can anyone tell me the link on which the above formula is stated (except AOPS)??",
  "Solution_14": "[quote=\"Kouichi Nakagawa\"]$ z^m = 1 \\iff z = 1, e^{\\frac {2\\pi i}{m}}, \\ e^{\\frac {4\\pi i}{m}}, \\ \\cdots, \\ e^{\\frac {2(m - 1)\\pi i}{m}}$.\nSo,\n\\[ z^m - 1 = (z - 1)(z - e^{\\frac {2\\pi i}{m}})(z - e^{\\frac {4\\pi i}{m}})\\cdots(z - e^{\\frac {2(m - 1)\\pi i}{m}})\n\\]\nAnd,\n\\[ \\left.\\frac {z^m - 1}{z - 1}\\right|_{z = 1} = m = (1 - e^{\\frac {2\\pi i}{m}})(1 - e^{\\frac {4\\pi i}{m}})\\cdots(1 - e^{\\frac {2(m - 1)\\pi i}{m}})\n\\]\n$ (\\because \\frac {z^m - 1}{z - 1} = 1 + z + \\cdots + z^{m - 1})$\nAnd,\n\\begin{eqnarray*} & & \\overline{m} = \\overline{(1 - e^{\\frac {2\\pi i}{m}})(1 - e^{\\frac {4\\pi i}{m}})\\cdots(1 - e^{\\frac {2(m - 1)\\pi i}{m}})} \\\\\n& \\iff & m = (1 - e^{ - \\frac {2\\pi i}{m}})(1 - e^{ - \\frac {4\\pi i}{m}})\\cdots(1 - e^{ - \\frac {2(m - 1)\\pi i}{m}}) \\end{eqnarray*}\nWe use $ (1 - e^{\\frac {2k\\pi i}{m}})(1 - e^{ - \\frac {2k\\pi i}{m}}) = 2 - 2\\cos(\\frac {2k\\pi}{m})$.\nSo,\n\\[ m^2 = 2^{m - 1}\\left(1 - \\cos\\frac {2\\pi}{m}\\right)\\left(1 - \\cos\\frac {4\\pi}{m}\\right)\\cdots\\left(1 - \\cos\\frac {2(m - 1)\\pi}{m}\\right)\n\\]\nAnd, $ 1 - \\cos\\frac {2\\pi}{m} = 2\\sin^2 \\frac {k\\pi}{m}$.\nTherefore,\n\\[ m^2 = 2^{2m - 2}\\sin^2 \\frac {\\pi}{m}\\sin^2 \\frac {2\\pi}{m}\\cdots\\sin^2 \\frac {(m - 1)\\pi}{m}\n\\]\nThus,\n\\[ \\prod_{i = 1}^{m - 1} \\sin \\frac {i\\pi}{m} = \\frac {m}{2^{m - 1}}\n\\]\n[/quote]\r\n\r\nUse this, we obtain :\r\n\r\n$ \\boxed{\\sin \\frac{\\pi}{2n}\\sin \\frac{3\\pi}{2n}\\cdots \\sin \\frac{(2n-1)\\pi}{2n}=\\frac{1}{2^{n-1}}}$",
  "Solution_15": "[quote=\"electron-proton\"]Can anyone tell me the link on which the above formula is stated (except AOPS)??[/quote]\r\n\r\nYou can look this book inside: Challenging Mathematical Problems With Elementary Solutions vol.2 written by A.M. Yaglom & I.M. Yaglom\r\n\r\nSee Page 23-24!\r\n\r\n[url]http://www.flipkart.com/challenging-mathematical-problems-elementary-solutions/0486655377-gyw3fulx8d#previewbook[/url]",
  "Solution_16": "How to solve \r\n$ \\sin\\frac{\\pi}{30}.\\sin\\frac{7\\pi}{30}.\\sin\\frac{11\\pi}{30}.\\sin\\frac{17\\pi}{30}$\r\nby the above mentioned formula..\r\n\r\nI am not posting the problem in the new topic because this question is directly linked to these discussions..",
  "Solution_17": "[quote=\"RAGHAV GROVER\"]Kunny is there a same type of formula for cosine also??[/quote]\r\n\r\nYes, here is:\r\n\r\n$ \\boxed{\\prod_{j \\equal{} 1}^{n \\minus{} 1} \\cos \\frac {2\\pi}{n}j \\equal{} \\frac {\\sqrt{n}}{2^{n \\minus{} 1}}}$"
}
{
  "Tag": [],
  "Problem": "Determine the number of ways to arrange the letters of the word ELEVEN",
  "Solution_1": "The solution needs a little more for those wanting to understand why, something like:\r\n\"If we treat the three E's as distinct, there are 6x5x4x3x2x1 ways to rearrange the six letters.  In each arrangement there are 3x2x1 ways to arrange the three E's, so to correct for overcounting we divide 6! by 3!.\"",
  "Solution_2": "Isn't that the same solution...?",
  "Solution_3": "Well the official solution just says There are $ 6!$ ways to arrange the letters, but we divide by $ 3!$ for overcounting to get $ \\frac{6!}{3!}\\equal{}120$ or something like that.  Mr. Batterson showed WHY you divide by $ 3!$."
}
{
  "Tag": [
    "geometry",
    "rectangle",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Ok. This is a simple question, yet it managed to get me thinking for a couple of days.\r\nI hope someone here might be able to solve it.\r\n\r\nLet $ (X_i,\\mathcal{M}_i), i\\equal{}1,2,3$ be three measurable spaces. We can define the product measure in more than on way. For instance, $ \\mathcal{M}_1 \\otimes \\mathcal{M}_2 \\otimes \\mathcal{M}_3$ is the $ \\sigma$-algebra generated by the mesurable rectangles $ A_1 \\times A_2 \\times A_3$ where $ A_i \\in \\mathcal{M}_i, i\\equal{}1,2,3$. Another way would be to define $ \\left( \\mathcal{M}_1 \\otimes \\mathcal{M}_2 \\right) \\otimes \\mathcal{M}_3$ as the $ \\sigma$-algebra generated by measurable rectangles $ B \\times A$ where $ B \\in \\mathcal{M}_1 \\otimes \\mathcal{M}_2$ and $ A \\in \\mathcal{M}_3$.\r\nNow Folland claims that $ \\mathcal{M}_1 \\otimes \\mathcal{M}_2 \\otimes \\mathcal{M}_3 \\equal{} \\left( \\mathcal{M}_1 \\otimes \\mathcal{M}_2 \\right) \\otimes \\mathcal{M}_3$.\r\nI can see why $ \\mathcal{M}_1 \\otimes \\mathcal{M}_2 \\otimes \\mathcal{M}_3 \\subseteq \\left( \\mathcal{M}_1 \\otimes \\mathcal{M}_2 \\right) \\otimes \\mathcal{M}_3$, since any mesurable rectangles $ A_1 \\times A_2 \\times A_3$ where $ A_i \\in \\mathcal{M}_i, i\\equal{}1,2,3$ is trivially in $ \\left( \\mathcal{M}_1 \\otimes \\mathcal{M}_2 \\right) \\otimes \\mathcal{M}_3$.\r\nI can't even start to imagine the other direction.",
  "Solution_1": "Well, just consider the sets $ A\\subset X_1\\times X_2$ such that $ A\\times B\\subset M_1\\otimes M_2\\otimes M_3$ for all $ B\\in M_3$. It is a sibma-algebra and conatins all rectangles. ;)",
  "Solution_2": "Hi fedja,\r\nI have thought along similar lines, but was unable to show that it was closed under complements.\r\nBut to tell you the truth, I think the family you have devised is easily shown to be a $ \\sigma$-algebra.\r\n\r\nThank you for the help.",
  "Solution_3": "[quote=\"karitaru\"]\nI have thought along similar lines, but was unable to show that it was closed under complements.[/quote]\r\n$ [(X_1\\times X_2)\\setminus A]\\times B\\equal{}(X_1\\times X_2\\times B)\\setminus (A\\times B)$",
  "Solution_4": "Thank you so much, it worked fine."
}
{
  "Tag": [
    "geometry",
    "geometric transformation",
    "rotation",
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "A chessboard is the set $ C \\equal{} \\{(i,j),1\\leq i,j\\leq 8\\}$. We put $ 21$ $ 3\\times 1$ tiles on chessboard such that no two tiles overlaps. Determine those points in $ C$ that can not be covered by some tile.",
  "Solution_1": "Would you give one example of placing 21 of $ 3\\times1$ dominoes ? ( I believe we is allowed to use  $ \\pi/2$ rotation of it too ie., $ 1\\times 3$ domino). I could not find one and I feel there does not exist one such arrangement.",
  "Solution_2": "[quote=\"srikanth\"]Would you give one example of placing 21 of $ 3\\times1$ dominoes ?[/quote]\nGood point! I haven't found a proof that it is possible or impossible. That should be proven first.\n[quote=\"srikanth\"]( I believe we is allowed to use  $ \\pi/2$ rotation of it too ie., $ 1\\times 3$ domino).[/quote]\r\nYes. You can use also $ 1\\times 3$ dominoes.",
  "Solution_3": "There exists...",
  "Solution_4": "Is the following approach correct? One can number the squares of the board as\r\n12312312\r\n31231231\r\n23123123\r\n12312312\r\n31231231\r\n23123123\r\n12312312\r\n31231231\r\nNow every tile covers three different numbers. So after putting $ 21$ tiles on the board we have one square with number $ 1$ uncovered. On the other hand, one can cover the board as\r\n21321321\r\n13213213\r\n32132132\r\n21321321\r\n13213213\r\n32132132\r\n21321321\r\n13213213\r\nAgain one $ 1$ stays uncovered. Now the result is those position $ a_{i,j}$ on the board which has $ 1$ in both numberings.",
  "Solution_5": "[quote=\"MeKnowsNothing\"]Is the following approach correct? One can number the squares of the board as\n12312312\n31231231\n23123123\n12312312\n31231231\n23123123\n12312312\n31231231\nNow every tile covers three different numbers. So after putting $ 21$ tiles on the board we have one square with number $ 1$ uncovered. On the other hand, one can cover the board as\n21321321\n13213213\n32132132\n21321321\n13213213\n32132132\n21321321\n13213213\nAgain one $ 1$ stays uncovered. Now the result is those position $ a_{i,j}$ on the board which has $ 1$ in both numberings.[/quote]\r\n\r\nThis is what I did.  In total there are four squares which cannot be covered.\r\n\r\nGeneralization:  For each integer $ n>3$ with $ n$ not a multiple of 3, find the number of ways to deleta a square from an $ n^2$ chessboard so we can tile it completely with $ 1$ x $ 3$ tiles."
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "Give a,b,c >0\r\nShow that \r\n$ \\[\\frac{a}{b} + \\frac{b}{c} + \\frac{c}{a} \\ge \\frac{{9({a^2} + {b^2} + {c^2})}}{{{{(a + b + c)}^2}}}\\] $",
  "Solution_1": "Give a,b,c >0\r\nShow that \r\n$ \\frac{a}{b} \\plus{} \\frac{b}{c} \\plus{} \\frac{c}{a} \\ge \\frac{{9({a^2} \\plus{} {b^2} \\plus{} {c^2})}}{{{{(a \\plus{} b \\plus{} c)}^2}}}$",
  "Solution_2": "[quote=\"Fabiano\"]Give a,b,c >0\nShow that \n$ \\frac {a}{b} \\plus{} \\frac {b}{c} \\plus{} \\frac {c}{a} \\ge \\frac {{9({a^2} \\plus{} {b^2} \\plus{} {c^2})}}{{{{(a \\plus{} b \\plus{} c)}^2}}}$[/quote]\r\n\r\nWe have stronger Inequality\r\n$ 2(\\frac {a}{b} \\plus{} \\frac {b}{c} \\plus{} \\frac {c}{a})\\plus{}1 \\ge \\frac{21(a^2\\plus{}b^2\\plus{}c^2}{(a\\plus{}b\\plus{}c)^2}$\r\nSOS can slove it :)"
}
{
  "Tag": [
    "MIT",
    "college",
    "Harvard",
    "Princeton"
  ],
  "Problem": "I am interested in studying Mathematics and Computer Science.\r\nDoes anyone know if Caltech has a good math and computer science department?",
  "Solution_1": "Yes, it does. Ordinarily, Caltech is just outside the top five ratings for computer science, but that is partly because of the very theoretical orientation of the department. Caltech's CS department is academic, not job training.",
  "Solution_2": "And what are the top ratings for mathematics instead?\r\n \r\nI guess MIT, Caltech, Harvard, Princeton, Stanford. Others?\r\n\r\nThanks for the answer.",
  "Solution_3": "UC Berkeley's graduate math program is consistently in the top 5, if not number one on a some occasions",
  "Solution_4": "[quote=\"XT\"] \nI guess MIT, Caltech, Harvard, Princeton, Stanford. Others?\n[/quote]\r\nChicago",
  "Solution_5": "[quote=\"tokenadult\"]Yes, it does. Ordinarily, Caltech is just outside the top five ratings for computer science, but that is partly because of the very theoretical orientation of the department. Caltech's CS department is academic, not job training.[/quote]\r\n\r\nAs an alum of Caltech CS:\r\n\r\nEverything at Caltech is theoretical.  It's not that good if you plan to work in industry (though a lot of cs grads end up at Microsoft/Yahoo/Google/nVidia/etc), but I find that it teaches you a lot of more general skills that let you pick up things very quickly, so you end up being able to outpace the standard software engineers and have a deeper understanding of the problems.  Of course, Caltech's also so small that there aren't any co-op programs, so most CS grads got their job offers as a result of summer internships.  (I believe)\r\n\r\nAs far as math goes, I don't think there's a huge difference in the quality among the top schools.  I'm willing to bet their curricula are similar, since the math is likely to be theoretical anyway.  The only major negative I would mention about Caltech is that because it's so small, you won't get a diverse selection of faculty working on all sorts of problems, there might only be one or two professors working in any particular subfield.  This could potentially be a factor depending on what you want to study.\r\n\r\nI would say that most people who end up at Caltech chose because the environment is unique, not because Caltech is a top-school.  (I don't think many people slip by our admissions who can't handle any other top school.)"
}
{
  "Tag": [
    "number theory",
    "number theory unsolved"
  ],
  "Problem": "Find integer roots of the equation : $x^{10} + y^{10} -z^{10} = 1999$",
  "Solution_1": "Hmmm... I presume you meant $x^{10}+y^{10} - z^{10} = 1999$, am I wrong?\r\n\r\nPierre.",
  "Solution_2": "According to Fermat's little theorem, $n^{10} = 0$ or $1$ modulo $11$.\r\nTherefore $x^{10}+y^{10}-z^{10} = 0,1,2,10 \\mod [11]$. Since $1999 = 8 \\mod [11]$ there is no solution at the given equation.\r\n\r\nPierre."
}
{
  "Tag": [
    "\\/closed"
  ],
  "Problem": "Ok.. the way I see it, the main problem with this merger is that it makes the forum selection page abhorrently convoluted. \r\nMy suggestions on how to correct it:\r\nTake away all the links to the subforums under advanced and college.\r\nThen on the next page, where the links to the subforums are, take away all those subforum links. Now, everything on that topic will be in that topic, and not in 5 other possible places.\r\nNow get rid of cultural exchange. Aops has had that in the round table (in addition to the hundred billion posts on the current administration).\r\nNext, lets get rid of the news flashes area and the polls section. Thats quite an extra feature that was already used everywhere else on the site.\r\nAlso, I think it would be a better design if instead of posting a bunch of separate continents in that one section (the one with Non-US Americas, Africa/Australia, Asia, Europe), it was just combined into \"Other Continents\". \r\nAnd a shout-out to all those people who have a knack for picking on others' english grammar, hold back on it for a while.\r\nI think thats pretty much it. If I think of anything else, I'll post it.",
  "Solution_1": "Well,  they were fine combined into one before AoPS merged.  But now we have all kinds of people from different countries.  I don't think that it is fair if we combine it into one.  I think rep's idea is fine.  Don't take each one away but maybe make subforums.  I don't know how many people are from which continent so it also kind of depends on that.",
  "Solution_2": "[quote=\"jelyman\"]My suggestions on how to correct it:\nTake away all the links to the subforums under advanced and college.\nThen on the next page, where the links to the subforums are, take away all those subforum links. Now, everything on that topic will be in that topic, and not in 5 other possible places.[/quote]\n\nI would rather propose a systematic subdivision of the maths sections into \"Getting Started\", \"Intermediate Level\" and \"Advanced Corner\". Each of these sections should be subdivided into the subforums \"Algebra\", \"Combinatorics\", \"Geometry\", etc. etc.. Actually, this is because many of us are beginners in one field and professionals in another one, hence it is better if everybody has the appropriate category for each field. The material currently crowded in the \"Advanced Problems (Temporary)\" forum should be moved into the other categories, namely each thread into the appropriate subforum (I know this is a lot of work, but it should be worth doing it).\n\n[quote=\"jelyman\"]Next, lets get rid of the news flashes area and the polls section.[/quote]\n\nIndeed this is right.\n\n[quote=\"jelyman\"]Also, I think it would be a better design if instead of posting a bunch of separate continents in that one section (the one with Non-US Americas, Africa/Australia, Asia, Europe), it was just combined into \"Other Continents\".[/quote]\r\n\r\nYes. Take the subforum \"National Olympiads\" out of the \"IMO\" forum, and join it will the Americas, Africa/Australia, Asia, Europe etc. olympiad ones.\r\n\r\n  Darij",
  "Solution_3": "One idea that we have discussed is letting students select which forums show up on their main page when they log in.  This ways students can select to see only the 4 or 10 or 15 they normally look at, which they can click a button to bring up the full list when they feel like it.\r\n\r\nIt is still undetermined how much work this would take, but any feedback on this idea would be appreciated.",
  "Solution_4": "Hmm. It seems as if I worded that section wrong. I didnt meant to say get rid of all of the subforums completely. What I meant was that on the main forum page, get rid of those little subforums sections that u can click on in each of the advacned and college boxes. Then on the next page (when u click either the advanced or college section), dont get rid of all those links, but get rid of all of those sub-sub forums. Just combine all of the double-subs into one sub. Get it? Heh.",
  "Solution_5": "it's just my opinion, but i like these sub-subforum links. they don't take much place and if you've got a modem which has just 56kbit and every site needs about 5 or 10 seconds to load, it is quite useful you don't have to click through the thousands of subforums, but can go directly to the forum you want to go to.\r\n\r\nPeter",
  "Solution_6": "Yes I like the sub-subforums too. Cuz I also have a 56 kbit connection. And BTW I would like to have the text mode back (even if it's not the most important thing).\r\n\r\nConcerning the old AoPS ranks, I don't know what they were like, but if they were better than the MathLinks rank then I have nothing against using them.\r\n\r\n  Darij",
  "Solution_7": "Yeah, that'd definitely be nice. I think most community members don't use all the forums, so it would be very useful.",
  "Solution_8": "[quote=\"Rep123max\"]That would be awesome to let us choose what forums show up when we log on. Cause I really only look at about 6 forums.[/quote]\r\n\r\nWe're looking into this.  Not clear yet how hard it will be to do.",
  "Solution_9": "[quote=\"jelyman\"]Hmm. It seems as if I worded that section wrong. I didnt meant to say get rid of all of the subforums completely. What I meant was that on the main forum page, get rid of those little subforums sections that u can click on in each of the advacned and college boxes. Then on the next page (when u click either the advanced or college section), dont get rid of all those links, but get rid of all of those sub-sub forums. Just combine all of the double-subs into one sub. Get it? Heh.[/quote] You can do that if you use the MathLinks skin (profile -> style -> MathLinks) and then hit the options and you can customize anything. \r\nThese options will soon be available on the AoPS skin soon (Richards' working on them).",
  "Solution_10": "Here's what I'd like to see (or something similar)\r\n\r\n[u]Problem Solving[/u]\r\n[b]-Getting Started[/b] (as is)\r\n[b]-Intermediate Topics[/b] (as is)\r\n[b]-Advanced Problems[/b]\r\n[i]Perhaps leave the sub-forums under advanced problems, but I don't think the sub-sub-forum classification is necessary; I feel it creates entirely too much clicking and is confusing. Instead of having Unsolved Problems, Proposed & Own Problems, Open Questions, Solved Problems, Theorems and Formulas sub-sub-forums, just combine it all under the respective sub-forums (i.e. once someone clicked on \"algeba,\" it would look just like the rest of the forums, with all the \"algebra\" topics listed normally) The old \"Advanced Problems\" could be filed under those topics[/i]\r\n[b]-College Playground[/b]\r\n[i]Again, leave the sub-forums, ditch the sub-sub-forums. There also appears to be a \"main index\" link with a globe next to it that takes you back to the forum index. That seems unnecessary. Mathematics Contests for Undergraduate Students should perhaps be moved to \"Other Problem Solving Topics[/i]\r\n[b]-Other Problem Solving Topics[/b] (as is)\r\n\r\n[u]US Contests and Programs[/u] (as is)\r\n\r\n[u]International Contests & Programs[/u]\r\n[i]This can be consolidated a lot. The Europe, Asia, and Africa & Australia forums have one post each. The Americas outside the US forum has some posts, but they could probably be moved to \"Other Problem Solving Topics\" or something similar. As for the Mathlinks Contest and Download forums, I think that perhaps a better idea would be to have a seperate webpage for these forums. The mathlinks contest could be on a page similar to the \"Resources\" page on AoPS, with a link off to the far left. It doesn't really need to be a forum at all, probably. The download forum could be similar, on a seperate page listing the olympiads, etc that are available for download (in a civilized, ordered manner), along with a way for members to send new files to the site maintainers to add to the page. It wouldn't have to be a forum either. The idea sounds like a lot of work, but it seems like it should be possible to script some PHP that would help out considerably in the process. (Unfortunately, I don't know PHP, so I can't help.) Doing those things would allow you to eliminate this entire section, without losing any content.[/i]\r\n----\r\nWhat happens to the rest of the sections doesn't matter much, although the Announcements section seems pointless.\r\n\r\nI hope this post was of some help.",
  "Solution_11": "[quote=\"Valentin Vornicu\"] These options will soon be available on the AoPS skin soon (Richards' working on them).[/quote]\r\n\r\nDone.  Try the Options button above - you can set it to see no subforums listed.",
  "Solution_12": "I just noticed a subforum under the LaTeX forum. That seems like it can go, since it serves the same purpose as the LaTeX forum and the preview button, essentially.",
  "Solution_13": "[quote=\"confuted\"]Here's what I'd like to see (or something similar)\n\nPerhaps leave the sub-forums under advanced problems, but I don't think the sub-sub-forum classification is necessary; I feel it creates entirely too much clicking and is confusing. Instead of having Unsolved Problems, Proposed & Own Problems, Open Questions, Solved Problems, Theorems and Formulas sub-sub-forums, just combine it all under the respective sub-forums (i.e. once someone clicked on \"algeba,\" it would look just like the rest of the forums, with all the \"algebra\" topics listed normally) The old \"Advanced Problems\" could be filed under those topics\n[/quote] \n\nThat would be a really pain-in-the-ass for people searching only to solve the (yet) unsolved problems. Experience shows that for Olympiad problems that's the best way to do it. \n[quote=\"confuted\"] [i]Again, leave the sub-forums, ditch the sub-sub-forums. There also appears to be a \"main index\" link with a globe next to it that takes you back to the forum index. That seems unnecessary. Mathematics Contests for Undergraduate Students should perhaps be moved to \"Other Problem Solving Topics[/i]\n[/quote]\n\nThe main index link, which you don't see unless you go into the problem solving category, is there because most of the Forum links on MathLinks skin are designed to get you there (skip all the other parts) and then if still users want to get into the other places, they would use that link. \n\n[quote=\"confuted\"] This can be consolidated a lot. The Europe, Asia, and Africa & Australia forums have one post each. The Americas outside the US forum has some posts, but they could probably be moved to \"Other Problem Solving Topics\" or something similar. \n[/quote] At the beginning the US Contest & Programs had only one post per forum. It's not a real criterion, having more or less posts (at least for now). \n\n[quote=\"confuted\"]As for the Mathlinks Contest and Download forums, I think that perhaps a better idea would be to have a seperate webpage for these forums. The mathlinks contest could be on a page similar to the \"Resources\" page on AoPS, with a link off to the far left. It doesn't really need to be a forum at all, probably. The download forum could be similar, on a seperate page listing the olympiads, etc that are available for download (in a civilized, ordered manner), along with a way for members to send new files to the site maintainers to add to the page. It wouldn't have to be a forum either. The idea sounds like a lot of work, but it seems like it should be possible to script some PHP that would help out considerably in the process. (Unfortunately, I don't know PHP, so I can't help.) Doing those things would allow you to eliminate this entire section, without losing any content. [/quote]You can post yourself any materials you want. That's the purpose of those files being in the forum. Many people use them already. It's also a good way to track what people are downloading. \n\n[quote=\"confuted\"] What happens to the rest of the sections doesn't matter much, although the Announcements section seems pointless.\n[/quote] It only seems. \n\n[quote=\"confuted\"] I hope this post was of some help.[/quote] Glad to be of assistance in clearing some of the issues. :)"
}
{
  "Tag": [
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Isn't it true that the limit of the sequence a_n=n^2*|sinn| doesn't exist? But how do you prove that?",
  "Solution_1": "It is really hard if we want to show n^2.|sin n| does not coverge to infinity. \r\nThe question n^2|sin n| has not finite limit is triavial, because for every N, we can find n > N such that |sinn| > 1/2."
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "geometric transformation",
    "analytic geometry",
    "homothety",
    "symmetry"
  ],
  "Problem": "Let $O$ be the circumcenter of an acute-angled triangle $ABC$, let $T$ be the circumcenter of the triangle $AOC$, and let $M$ be the midpoint of the segment $AC$. We take a point $D$ on the side $AB$ and a point $E$ on the side $BC$ that satisfy $\\angle BDM = \\angle BEM = \\angle ABC$. Show that the straight lines $BT$ and $DE$ are perpendicular.",
  "Solution_1": "Assume the circumcircle of $\\triangle AOC$ intersects the sides $AB$ and $BC$ again at $A'$ and $C'$ respectively. The foot of the perpendicular from $T$ to $BC$ is the midpt of $CC'$. It's easy to show (quick angle chase) that $AC'||DM||A\\\"C$, where $A\\\"$ is the reflection of $A$ in $D$. All of this means that $DM$ cuts $BC$ in the midpoint of $CC'$, which is also the foot of the perpendicular from $T$ to $BC$. In the same way we show that $EM$ cuts $AB$ in the midpt of $AA'$, which is the foot of the perpendicular from $T$ to $AB$.\r\n\r\nLet the perpendicular from $T$ to $AB$ cut $BC$ in $C_1$. Similarly we define $A_1$. Let $X$ and $Y$ be the midpts of $AA'$ and $CC'$ respectively. The figures $BXEC_1$ and $BYDA_1$ are obviously similar, so $A_1C_1||DE$ (#).\r\n\r\nIn $\\triangle BA_1C_1$ the point $T$ is the orthocenter, so $BT\\perp A_1C_1$ (##).\r\n\r\nFrom (#) and (##) we get the desired conclusion.",
  "Solution_2": "Let N be the midpoint of BC and P be the midpoint of AC. Let D' be the intersection of the perpendicular bisector of MN with AB. It is easy to show that D' is the midpoint of BD. Let E' be the intersection of the perpendicular bisector of MP with BC. E' is the midpoint of BE.\n\nSet up a coordinate system with the center in O and such that the circumcenter of ABC is the unit circle. Then, by a quick and easy computation we obtain : (we have denoted by lowercase leters the affixes of points denoted by uppercase letters)\n\nd = 2a + b + c - ab/c\ne = a + b + 2c - bc/a\nt = ac/(a+c)\n\nWe must prove that (d-e)/(b-t) is an imaginary complex number. But :\n\nd-e = (ab+bc-ac)(c-a)/ac\nb-t = (ab+bc-ac)/(a+c)\n\nTherefore z = (d-e)/(b-t) = (c-a)(c+a)/ac. It is easy to see that if z' is the conjugate of z , z+z'=0. Whence the conclusion.",
  "Solution_3": "we have BE=RsinA+2RsinCcosB\r\n             BD=RsinC+2sinAcosB\r\n DE <sup>2</sup> =R <sup>2</sup> [(sinC) <sup>2</sup>+6sinAsinCcosB +(sinC) <sup>2</sup> -8sinAsinC(cosB)^3]\r\n then[ 1/(sinDEB) <sup>2</sup>] -1=\r\n[(cos(A-B)) <sup>2</sup> /(sinC+2sinAcosB) <sup>2</sup> ]\r\n then tg TBC=[cosBcos(A-B)]/[sin(A-B)cosB-sinA] then tg TBC=cotgDEB => we prove",
  "Solution_4": "Consider a homothety $h$ with center in centroid of $\\triangle ABC$, s.t. $h(M)=B$. Denote\r\n$\\angle ABC=\\alpha$\r\n$F=h(D)$ and $G=h(E)$.\r\nIt is easy that \r\n$\\angle BFC=\\angle BCF=\\angle BGA=\\angle BAG=\\angle TCO=\\angle TOC=\\angle TAO=\\angle TOA=\\alpha$.\r\nThus\r\n$\\triangle CFB \\sim \\triangle COT$ and $\\triangle AGB \\sim \\triangle AOT$, so\r\n$\\triangle CFO \\sim \\triangle CBT$ and $\\triangle AGO \\sim \\triangle ABT$. Hence\r\n$OF=\\frac{TB\\cdot OC}{TC}=\\frac{TB\\cdot OA}{TA}=OG$ and\r\n$\\angle FOG=\\angle OCT+\\angle OAT=2\\alpha$.\r\nLet $k$ be a bisector of $\\angle FOG$. We have\r\n$\\angle(OF, BT)=\\angle OCT=\\alpha=\\angle(OF, k)$.\r\nThus $BT \\parallel k$ and $k\\perp FG$ because $\\triangle OFG$ is isocles  \r\ntriangle, so $BT\\perp FG$. By homothety $DE\\parallel FG$ and in the end $BT\\perp DE$. :)   :D",
  "Solution_5": "[quote=\"Thanhliem\"]we have BE=RsinA+2RsinCcosB\n             BD=RsinC+2sinAcosB\n DE <sup>2</sup> =R <sup>2</sup> [(sinC) <sup>2</sup>+6sinAsinCcosB +(sinC) <sup>2</sup> -8sinAsinC(cosB)^3]\n then[ 1/(sinDEB) <sup>2</sup>] -1=\n[(cos(A-B)) <sup>2</sup> /(sinC+2sinAcosB) <sup>2</sup> ]\n then tg TBC=[cosBcos(A-B)]/[sin(A-B)cosB-sinA] then tg TBC=cotgDEB => we prove[/quote]\r\n\r\nCan you involve a little bit more detail?",
  "Solution_6": "[quote=\"orl\"][color=darkred]Let $ C(O)$ , $ C(T)$ be the circumcircles of an acute $ \\triangle ABC$ and $ \\triangle AOC$ respectively. Let $ M$ be the midpoint of $ [AC]$ . \n\nConsider the points $ D\\in [AB]$ and $ E\\in [BC]$ which satisfy $ \\widehat {BDM}\\equiv\\widehat {BEM}\\equiv\\widehat {ABC}$ . Show that $ BT\\perp DE$ .[/color][/quote]\n\n[color=darkblue][b][u]Proof[/u][/b] (similar with the nice [b]Grobber[/b]'s proof). I\"ll use the [b]Grobber[/b]'s notations. Denote the points $ \\left\\{\\begin{array}{c} A'\\in BA\\ ,\\ A'B = A'C \\\\\n \\\\\nC'\\in BC\\ ,\\ C'B = C'A\\end{array}\\right\\|$ .\n\nObserve that $\\left \\{\\begin{array}{c} m(\\widehat {AOC}) = 2B \\\\\n \\\\\nm(\\widehat {AA'C}) = m(\\widehat {AC'C}) = 180^{\\circ} - 2B\\end{array}\\right\\|$ , i.e. the points $ A'$ , $ C'$ belong to the circle $ C(T)$ .\n\nDenote the midpoints $ X$ , $ Y$ of the segments $ [AA']$ , $ [CC']$ respectively. Observe that $ \\left\\{\\begin{array}{c} MX\\parallel A'C\\implies E\\in XM \\\\\n \\\\\nMY\\parallel C'A\\implies D\\in MY\\end{array}\\right\\|$ .\n\n$\\left \\{\\begin{array}{c} TX\\perp BA \\\\\n \\\\\nTY\\perp BC\\end{array}\\right\\|\\implies$ the quadrilateral $ BXTY$ is ciclically (the circumcenter is the midpoint of the segement $ [BT]$ ).\n\nFrom the relations $ m(\\widehat {XDY}) = m(\\widehat {XEY}) = 180^{\\circ} - B$ obtain that the quadrilateral $ DEYX$ is cyclically. \n\nThus, the perpendicular line $ d$ from the point $ B$ to the line $ DE$ passes through the center of the circumcircle of $ \\triangle BXY$ \n\n(and of the quadrilateral $ BXTY$ ) , i.e. the midpoint of the segment $ [BT]$ . Thus, $ d\\equiv BT$ . In conclusion, $ BT\\perp DE$ .[/color]",
  "Solution_7": "[hide=\"Solution\"]\nLet $ \\angle ABC\\equal{}\\alpha$.  Letting $ DM$ and $ EM$ meet $ AB$ and $ BC$ at $ X$ and $ Y$ respectively, we see that $ \\angle XDB\\equal{}\\angle XBD\\equal{}\\alpha$.  Thus, $ \\angle MXA\\equal{}\\angle XDB\\plus{}\\angle XBD\\equal{}2\\alpha\\equal{}\\angle AOC\\equal{}\\frac{\\angle ATC}{2}\\equal{}\\angle ATM$.  It follows that $ TXAM$ is cyclic, so $ TX\\perp AB$ since $ TM\\perp AC$.  Similarly, $ TY\\perp BC$, so $ BYTX$ is cyclic.  Now, notice that $ \\angle TYM\\equal{}\\angle TCA\\equal{}90\\minus{}\\angle AOC\\equal{}90\\minus{}2\\alpha$.  Letting $ \\angle TYX\\equal{}\\beta$, we see that $ \\angle TBX\\equal{}\\beta$, so $ \\angle TBY\\equal{}\\alpha\\minus{}\\beta$.  Furthermore, $ \\angle XDY\\equal{}\\alpha\\equal{}\\angle YEX$, so $ YXED$ is cyclic, so $ \\angle XDE\\equal{}\\angle XYM\\equal{}\\angle TYX\\plus{}\\angle TYM\\equal{}90\\minus{}2\\alpha\\plus{}\\beta$.  Yet, $ \\angle XDB\\equal{}\\alpha$, so $ \\angle BDE\\equal{}\\angle XDB\\plus{}\\angle XDE\\equal{}90\\minus{}\\alpha\\plus{}\\beta$.  It follows that $ \\angle TBD\\equal{}\\alpha\\minus{}\\beta\\equal{}90\\minus{}(90\\minus{}\\alpha\\plus{}\\beta)\\equal{}90\\minus{}\\angle BDE$, so $ BT\\perp DE$. [/hide]",
  "Solution_8": "[quote=\"iandrei\"]Let N be the midpoint of BC and P be the midpoint of AC. Let D' be the intersection of the perpendicular bisector of MN with AB. It is easy to show that D' is the midpoint of BD. Let E' be the intersection of the perpendicular bisector of MP with BC. E' is the midpoint of BE.\n\nSet up a coordinate system with the center in O and such that the circumcenter of ABC is the unit circle. Then, by a quick and easy computation we obtain : (we have denoted by lowercase leters the affixes of points denoted by uppercase letters)\n\nd = 2a + b + c - ab/c\ne = a + b + 2c - bc/a\nt = ac/(a+c)\n\nWe must prove that (d-e)/(b-t) is an imaginary complex number. But :\n\nd-e = (ab+bc-ac)(c-a)/ac\nb-t = (ab+bc-ac)/(a+c)\n\nTherefore z = (d-e)/(b-t) = (c-a)(c+a)/ac. It is easy to see that if z' is the conjugate of z , z+z'=0. Whence the conclusion.[/quote]\nIt's a pretty minor thing, but I think it's worth pointing out: it should be\n\\[2d=2a+c-\\frac{ab}{c}\\]and\n\\[2e=a+2c-\\frac{bc}{a}.\\]",
  "Solution_9": "[hide=\"Solution\"]\nWe do not use isometric coordinates.\n\nLet $BD\\cap EM=N$ and $BE\\cap DM=P$.  By trivial angle chasing, $\\angle BNE=\\angle ATM=\\angle MTC=\\angle BPD$.  Then $\\angle ANT=\\pi-\\angle TMA=\\dfrac{\\pi}{2}$, and $\\angle CPT=\\pi-\\angle CMT=\\dfrac{\\pi}{2}$.  Thus, quadrilateral $BNTP$ is cyclic.  Since $\\angle NDP=\\pi-\\angle ABC=\\angle NEP$, quadrilateral $NDEP$ is also cyclic.  Thus, we have $\\angle DBF+\\angle BDF=\\angle NBT+\\angle BPN=\\angle NBT+\\angle NTB=\\dfrac{\\pi}{2}$, so $BT \\perp DE$ as desired.\n\nQED\n[/hide]",
  "Solution_10": "Let $N,P,L$ be midpoints of $BC,BA$ and the center of the 9-point circle of $\\triangle ABC$. $S$ is the foot of perpendicular of $B$ on $AC$. Since the composition of inversion with center $B$ and radius $\\frac{BA\\cdot BC}{2}$ and symmetry WRT the internal bisector of $\\angle ABC$ pictures $S,N,P$ to $O,A,C$ then it pictures $\\odot SNP$ to $\\odot AOC$ so $\\angle CBT=\\angle ABL$(1). Note that $MN||BD$ and $\\angle MDB=\\angle NBD$ then $MNBD$ is an isosceles trapezoid so since $LM=LN$ then $LB=LD$ Likewise $LB=LE$ so $L$ is the circumcenter of $\\triangle BDE$. So from (1) $BT\\perp DE$",
  "Solution_11": "Let the circumcircle of AOC intersects AB and CB at S and P,respectively.Now,we easy have that MD/ME=BS/BP=BC/BA(Let R and Q be the midpoints of BC and BA,then MD=MQ and MR=ME and the conclusion follows).Now,it is easy by angle chasing that <STP=2<ABC,now let point N be such that BR=BN and <RBN=<STP and <SBN=3<ABC.Now,SBN is similar to MDE,so we need to prove that(this is reduced by trivial angle chase) <PNS=<PBT,but this is obvious aplying spiral homothety with center P,so BPN is similar to STP,from which we have that SPN is similar to BPT,so we are finished.",
  "Solution_12": "Let $F$ be the point such that $\\triangle FBA \\sim \\triangle FCB$. It is well-known that $BF$ is a symmedian in triangle $ABC$ and points $A,F,O,C$ lie on a circle. Let $E'$ be the intersection of lines $AB$ and $CF,$ and $D'$ be the intersection of lines $AF$ and $BC$.\n\nApply inversion at $B$ of radius $\\sqrt{\\frac{1}{2}\\cdot AB\\cdot BC}$ followed by reflection in the internal bisector of angle $ABC$. As the circle $(BDM)$ passes through the midpoint of $BC$ and $M \\rightarrow F,$ we conclude that $D \\rightarrow D'$. Similarly, $E \\rightarrow E'$ and we obtain $D'E' \\parallel DE$. \n\nFinally, we note that $$\\angle D'BF=\\angle D'CB \\Longrightarrow D'B^2=D'F\\cdot D'C.$$ Similarly, $E'B^2=E'F\\cdot E'A$. Thus, the line $D'E'$ is the radical axis of $B$ and the circle $(AFC)$. The conclusion $BT \\perp DE$ follows.\n\n[b]Note:[/b] The fact that $B$ acts as a \"point circle\" was crucial to finish the proof.",
  "Solution_13": "Solved with [b]p_square, Aayuu[/b]\n\nLet $Z$ denote the reflection of $O$ across $AC$. Observe that $AMZD$ and $CMZE$ are cyclic. So, it suffices to show $BZ$ and $BT$ are isogonal since $BZ$ is the diameter of $(BDZE)$.\n\nLet $T'$ be the isogonal conjugate of $T$ in $\\triangle ABC$, so it suffices to show $Z,B,T'$ are collinear or that $BT'$ goes through the circumcenter of $\\triangle AHC$, or show that $T'$ lies on the euler liner of $\\triangle AHC$.\n\nSo we have reduced the problem to the following. \n\n[quote=Rephrased problem]\nLet $ABDC$ be a quadrilateral with $\\angle A = \\angle B = \\angle C$. Show that $D$ lies on the Euler line of $\\triangle ABC$.\n[/quote]\n\nTo do this, let $X = AB \\cap CD$ and $Y = BD \\cap AC$. Let $P,Q$ denote midpoints of $AB,AC$ and let $G$ be the centroid. Note that $YB = YA$ and $XA = XC$.\n\nBlack magic, by Pappus on $XBP$ and $YCQ$, $D$ lies on $OG$, the Euler line, done. $\\blacksquare$",
  "Solution_14": "Let rays $BA,BC$ meet $\\odot (AOC)$ again at $S,R$ respectively; $DM\\cap BC=N,DE\\cap BA=K.$ Observe that $RA\\parallel MD$ (analogously $SC\\parallel ME$), since $\\angle RAB=\\pi-\\angle ABR-\\angle ARB=\\angle ABR=\\angle MDB.$ Hence by the Thales $N,K$ are midpoints of $CR,AS.$ Next $\\angle BKT=90^{\\circ}=\\angle BNT,\\angle KDN=\\angle KEN,$ so $BNTK,DENK$ are cyclic. \nThus the diameter $BT$ of $\\odot (BNK)$ is perpendicular to $DE,$ as desired.",
  "Solution_15": "Let $P$ and $Q$ be the feet of the altitudes from $M$ to $\\overline{BA}$ and $\\overline{BC}$ respectively. Then it is clear that $D$ is the reflection of the midpoint of $\\overline{BA}$ over $P$, and a similar fact is true for $E$. Now use complex numbers with $(ABC)$ as the unit circle, setting $A=a,B=1,C=c$, so $M=\\tfrac{a+c}{2}$. We can compute $P=\\tfrac{3}{4}a+\\tfrac{1}{4}c-\\tfrac{a}{4c}+\\tfrac{1}{4}$ and $Q=\\tfrac{3}{4}c+\\tfrac{1}{4}a-\\tfrac{c}{4a}+\\tfrac{1}{4}$, hence $D=a+\\tfrac{1}{2}c-\\tfrac{a}{2c}$ and $E=c+\\tfrac{1}{2}a-\\tfrac{c}{2a}$. Furthermore, the circumcenter formula yields $t=\\tfrac{ac}{a+c}$. We wish to prove that\n$$\\frac{2(d-e)}{t-1}=\\frac{a-c-\\frac{a}{c}+\\frac{c}{a}}{\\frac{ac-a-c}{a+c}}=\\frac{(a^2-c^2)(1-\\frac{a+c}{ac})}{ac-a-c}=\\frac{a^2-c^2}{ac}$$\nis imaginary, which is clear since its conjugate is its negative."
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "Rational Root Theorem"
  ],
  "Problem": "solve the natural solution  system   $ \\left\\{\\begin{array}{l}\r\n\r\n  x^4\\plus{}y\\equal{}17 \\\\\r\n\r\n  x\\plus{}y^4\\equal{}3\r\n\r\n\\end{array}\\right.$  :roll:   :wink:",
  "Solution_1": "I dnt know how to formally show this is the only real solution, but the only positive integer solution is (2,1)",
  "Solution_2": "hello, setting $ y=17-x^4$ in the second equation we get\r\n$ (17-x^4)^4+x-3=0$\r\nfactoring the equation we get\r\n$ ( x-2 )  ( {x}^{15}+2\\,{x}^{14}+4\\,{x}^{13}+8\\,{x}^{\r\n12}-52\\,{x}^{11}-104\\,{x}^{10}-208\\,{x}^{9}-416\\,{x}^{8}+902\\,{x}^{7}$$ +1804\\,{x}^{6}+3608\\,{x}^{5}+7216\\,{x}^{4}-5220\\,{x}^{3}-10440\\,{x}^{2}-20880\\,x-41759 )=0$\r\nby a numerical method i get three solutions\r\n$ -2.073948198, -1.984374742, 2., 2.059332877$\r\nSonnhard.",
  "Solution_3": "When you're solving polynomials of a high degree for integer roots, it is often useful to use the:\r\n[hide=\"One of many possible hints\"]Rational Root Theorem[/hide]\n[hide=\"One of many possible solutions\"]$ x \\equal{} 3 \\minus{} y^4 \\implies (3 \\minus{} y^4)^4 \\plus{} y \\minus{} 17 \\equal{} 0$\n\nSubstituting $ y \\equal{} 0$ into the LHS reveals that the constant term is $ 64$, so $ y|64$.\n\nThus $ 17 \\minus{} x^4 \\equal{} y \\geq \\minus{} 64 \\implies x^4 \\leq 81 \\implies |x| \\leq 3$\n\nTesting these seven values, the only solution is $ (x,y) \\equal{} (2,1)$.[/hide]"
}
{
  "Tag": [
    "probability"
  ],
  "Problem": "The probability of wind tommorrow is 2/3, and the probability of rain tommorow is 2/5. If they are independent, what is the probability of wind but not rain tommorrow. \r\n\r\nhow would u do this problem?",
  "Solution_1": "If the probability of an event is p, then the probability of that event [b]not[/b] occuring is 1-p. So the probability of there being no rain is 1- 2/5.\r\n\r\nThe probability of two independent results occuring= probability of one * probability of the other.\r\n\r\n[hide=\"solution\"]Probability of no rain = 3/5, so probability of wind but no rain = 3/5 * 2/3 = 6/15 = 2/5[/hide]",
  "Solution_2": "So you want wind, which has probability $\\frac{2}{3}$. You don't want rain, which has probability $1-\\frac{2}{5}$ = $\\frac{3}{5}$. We have to multiply these since they are independent:\r\n$(\\frac{2}{3})(\\frac{3}{5})=\\frac{2}{5}$\r\nThe answer is $\\frac{2}{5}$.",
  "Solution_3": "Could you please hide your answers?  Thank You\r\n\r\n[hide]The answer is\n\n$\\frac{2}{3}*(1-\\frac{2}{5})=\\frac{2}{5}$[/hide]"
}
{
  "Tag": [
    "Euler",
    "geometry",
    "3D geometry",
    "tetrahedron",
    "analytic geometry",
    "geometry unsolved"
  ],
  "Problem": "For any tetrahedron, Is following true?\r\n\r\nLet $R$ be the radii of circumsphere, and $r$ be the radii of insphere of a tetrahedron. \r\n\r\nThen $R^2 - 3Rr = OI^2$",
  "Solution_1": "No!!\r\nFor instance, consider an isoseles tetrahedron, the coordinates of the vertices being\r\n(a,0,0); (0,b,0); (0,0,c); (a,b,c)\r\nWe have O = I and (R/r)^2 = (a^2+b^2+c^2)(1/a^2+1/b^2+1/c^2)",
  "Solution_2": "Oh, You're right. \r\n\r\nThen Could you show me the extention form of Euler theorem in a tetrahedron?  :D",
  "Solution_3": "I think that a relation involving only r, R and OI cannot exist.",
  "Solution_4": "Proof, that $R^2-9r^2\\geq OI^2$."
}
{
  "Tag": [],
  "Problem": "\u0633\u0644\u0627\u0645 \u062f\u0648\u0628\u0627\u0631\u0647.\n\u0627\u0645\u0631\u0648\u0632 \u0631\u0648 \u0686\u0647 \u0643\u0631\u062f\u064a\u062f\u061f\n\u0645\u0646 \u0628\u062f \u0646\u062f\u0627\u062f\u0645.\n20 ta zadam ke fekr konam 2tash ghalate.\n\nKilid ro kasi nadare?",
  "Solution_1": "salam man ke kelid nadaram...\r\nvali motaassefane gand zadam (be onvane yek 3-vomi) va faghat 20 ta ro zadam...\r\nkode daftarcham 2 hast...\r\n1-d,2-b,3-a,4-( :D ,fekr konam befahmid manzuram chie ...[hide=\"click konid ta befahmid\"] nazadam  :rotfl: [/hide])\r\n5-c,6-e,7-b,8-ghalat zadam (soale be in asuni ro  :rotfl: ),10-a,11-a,12-a,13-e,14-e,15-d,16-fekr mikonam soal ghalat bashe manzuram ine ke javab tuye gozineha nist badan tozih midam chera (age mishe kasi ke javab dade bege chijuri daravorde)\r\n17-c,18-a,19-a,23-c,24-b,25-(nazadam vali fekr konam a beshe chon hatman katar az 20 hast)\r\n28-b,29-c\r\nmibinid cheghadr gand zadam  :) ...",
  "Solution_2": "[quote=\"shoki\"]salam man ke kelid nadaram...\nvali motaassefane gand zadam (be onvane yek 3-vomi) va faghat 20 ta ro zadam...\nkode daftarcham 2 hast...\n1-d,2-b,3-a,4-( :D ,fekr konam befahmid manzuram chie ...[hide=\"click konid ta befahmid\"] nazadam  :rotfl: [/hide])\n5-c,6-e,7-b,8-ghalat zadam (soale be in asuni ro  :rotfl: ),10-a,11-a,12-a,13-e,14-e,15-d,16-fekr mikonam soal ghalat bashe manzuram ine ke javab tuye gozineha nist badan tozih midam chera (age mishe kasi ke javab dade bege chijuri daravorde)\n17-c,18-a,19-a,23-c,24-b,25-(nazadam vali fekr konam a beshe chon hatman katar az 20 hast)\n28-b,29-c\nmibinid cheghadr gand zadam  :) ...[/quote]\r\n\r\n\r\nthx.\r\n\r\nrasti soale kashia ro ($ 1388 \\times 1388$) che joori hal mikonid?",
  "Solution_3": "hal nakardam har chand ke sare emtehan fahmidam taye jahayisho vali ....(hesssesh nabud :D )\r\nbaraye un soal ta jayi ke midunam faghat kafiye in mozu ro esbat konid ke do ta az char ta adadi ke avale kar mizarim va gharare majmueshun bar 6 bakhshpazir bashe barabar hastan va baraye do taye dige ham ye khassiyate dige darim(be peymaneye 3 hamneheshtand) ...\r\nvali be har hal...\r\nsoalate marhale aval (nemigam hamash vali) 99% asunand (manzuram ine ke chize fogholaddei nadarand ) vali baraye man... nemidunam chera.... vali vaght kam miyaram :rotfl:  :rotfl:\r\nkelid ro ham tu site imo09 ta ye chand saate dige mizaran",
  "Solution_4": "Salam !! be nazare man ke kheyli sakht nabood. man [b]63%[/b] zadam .(az ye dovommi bishtar entezar nadashte bashin  :!: ) \r\nfek mikonam kaf az [b]30%[/b] kamtar nabashe (khodam doost daram kamtar bashe chon chand ta az doostam bad dadan  :( ) vali az [b]40%[/b] ghatan bishtar nist.\r\nBye !",
  "Solution_5": "na . soal ghalat nist .\r\njavabesh mishe 116 . halesh ham ine ke avalan $ x \\plus{} y | 5x$ va $ x \\plus{} y | 5y$ . \r\nhala $ d \\equal{} (x,y)$ va $ d$ ro az 2 taraf sade kon natije mide $ x` \\plus{} y` | 5$ pas 2 halat darim :\r\n1 - $ x` \\equal{} 1$ va $ y` \\equal{} 4$ dar in soorat chon $ y < 101$ yani $ d \\leq 25$ yani $ 25$ ta javab darim  ke be elate vojode halat $ x` \\equal{} 4$ va $ y` \\equal{} 1$ dar ye 2 zarb mishe . \r\n2 - $ x` \\equal{} 2$  va $ y` \\equal{} 3$ mesle ghabli to in halat $ 33$ ta javab darim ke dar 2 zarb mishe va mishe $ 66$ . \r\npas javab e soal mishe $ 50 \\plus{} 66 \\equal{} 116$",
  "Solution_6": "[quote=\"navidj\"]na . soal ghalat nist .\njavabesh mishe 116 . halesh ham ine ke avalan $ x \\plus{} y | 5x$ va $ x \\plus{} y | 5y$ . \nhala $ d \\equal{} (x,y)$ va $ d$ ro az 2 taraf sade kon natije mide $ x` \\plus{} y` | 5$ pas 2 halat darim :\n1 - $ x` \\equal{} 1$ va $ y` \\equal{} 4$ dar in soorat chon $ y < 101$ yani $ d \\leq 25$ yani $ 25$ ta javab darim  ke be elate vojode halat $ x` \\equal{} 4$ va $ y` \\equal{} 1$ dar ye 2 zarb mishe . \n2 - $ x` \\equal{} 2$  va $ y` \\equal{} 3$ mesle ghabli to in halat $ 33$ ta javab darim ke dar 2 zarb mishe va mishe $ 66$ . \npas javab e soal mishe $ 50 \\plus{} 66 \\equal{} 116$[/quote]\r\nEyval manam haminjoori hal kardam  :lol: ... \r\nbebakhshid [b]navidj[/b] shoma chandomi ?",
  "Solution_7": "Man 3vomam . \r\nChetor mage ???",
  "Solution_8": "[quote=\"navidj\"]Man 3vomam . \nChetor mage ???[/quote]\r\netela`at omoomi !! (hamoon fooozoooli)",
  "Solution_9": "injoor ke maloomev hamatoon balaye 60,70% zadin. mane badbakht....\r\n\r\nagha man 55% zadam.torokhoda begid ghabool misham ya na,daram az esteres mimiram.\r\n\r\nalbate bachehamoon bazia kheili kharab kardan,manam omidvaram kaf paiin bashe.",
  "Solution_10": "man ke injuri halesh kardam be dast miarim $ x\\plus{}y |5y$\r\npas yani $ x\\plus{}y\\equal{}\\frac{5y}{k}$\r\nchon $ k$ hadeaksar $ 4$ mitune bashe (va agar na $ x$ gheyre tabii mishe) $ 4$ ta halat darim\r\n1-$ k\\equal{}1$ be ma midahad $ (y,4y) ; (4y,y)$ ya be ebarate digar $ 50$ ta zoj.\r\n2-$ k\\equal{}2$ be ma midahad $ (y,\\frac{3}{2}y) ; (\\frac{3}{2}y,y)$ ke kolan yani $ 66\\equal{}2 \\times 33$.dalil an ast ke avalan $ \\frac{3}{2}y \\le 100$ ke natije midahad $ y \\le 66$ va saniyan $ y$ bayad zoj bashad yani az beyne adade zoje az $ 1$ ta $ 66$ entekhab shavad pas $ 33$ entekhab baraye $ y$ mojud hast pas dar kol $ 66$ zoj be dast midahad.\r\n3-$ k\\equal{}3$ ke ba estedlali moshabeh zojhaye $ (y,\\frac{2}{3}y) ; (\\frac{2}{3}y,y)$ ra be ma midahad yani $ 66\\equal{}33 \\times 2$ ta zoj midahad.\r\n4-$ k\\equal{}4$ ke be dast midahad $ (y,\\frac{y}{4}) ; (\\frac{y}{4},y)$ ama agar yekami aghl ra dar inja be kar andazim mifahmium ke in zojha dar halate 1 yani $ k\\equal{}1$ she morde shodeand.\r\npas dar kol $ 66\\plus{}66\\plus{}50$ ta zoj darim yani $ 182$ ta zoj.\r\nmontazere javabetun hastam ke be man begin koja eshtebah kardam mamnun misham...",
  "Solution_11": "Oon soale ke ye 4zeli mohaddab bood javabesh chi mishe ? man sabet kardam $ 6$ mishe vali bazia migoftan etela`at kafi nist !!\r\nBayad sabet koni $ B,C,D$ rooye dayere ee be markaze $ A$ hastan . Dalilesh ine ke $ \\frac{110}{2} \\plus{} 125\\equal{}180$",
  "Solution_12": "[quote=\"mahanmath\"][quote=\"navidj\"]Man 3vomam . \nChetor mage ???[/quote]\netela`at omoomi !! (hamoon fooozoooli)[/quote]\r\n :rotfl:   :rotfl:   :rotfl: \r\nJavabe jalebi bood ... . \r\nBe nazaretoon kaf emsal chande ?\r\nMan ke fekr nemikonam kaf riazi emsal balaye 12 ya 12.5 bere nazare shoma chiye ?",
  "Solution_13": "[quote=\"mahanmath\"]Oon soale ke ye 4zeli mohaddab bood javabesh chi mishe ? man sabet kardam $ 6$ mishe vali bazia migoftan etela`at kafi nist !!\nBayad sabet koni $ B,C,D$ rooye dayere ee be markaze $ A$ hastan . Dalilesh ine ke $ \\frac {110}{2} \\plus{} 125 \\equal{} 180$[/quote]\r\n\r\n\r\njavabesh mishe 6.yani bachehaye ma migan 6.",
  "Solution_14": "vaaaaaaaaaayyyyyyyyyyyyyyyy fahmidam koja eshtebah kardam halate 2,3 yeki bud khakkkkkkkkkkkkk bar saram ajab eshtebahiiiiiiiiiiiiiiiiiiiiii :rotfl:  :rotfl:",
  "Solution_15": "[quote=\"amparvardi\"][quote=\"navidj\"][quote=\"amparvardi\"]vali age kaf 11 bashe ke hame ghaboolan.mage kasi zire 11 ham zade?[/quote]\n\nParsal ham emtahan oonghadr sakht nabood ke KAF BESHE 8 !!!!!!!!!!!!!!!!!!!!!!!!!!!!!![/quote]\n\n\nHARFETOON kamelan matine.\nrasti shoma parsal chand ta zadin?(va ghabool shodin?)\nemsaal chi?\nyeki az bachehaye ma dasht az esteres mimord.behesh begam kaf 11e?[/quote]\r\n\r\nman parsal 16.75 shodam (18 ta zadam 1 ghalat) . Emsal ham 17.75 shodam (19 ta zadam 1 ghalat) .  \r\n(Man kolan Test Olympiad ro khob nemizanam  :D ) \r\nraje be doostetoon hanooz nemishe goft kaf 11 e . vali dar hamoon hoodood hast .",
  "Solution_16": "vaaaaaaaaaaaaaaaayyyyyyyyyyy eeeeeeyyyyyyyyyyyyy       khhhhhhhhhhhhhhhhhooooooooodaaaaaaaaaaaaaa ye ghalate dige  :bomb: \r\najab sutii dadam.... \r\nmaaaaaaaaaaamannnnnnnnnnnnnnnnnn che ghadr bad dadam........ :wallbash_red:",
  "Solution_17": "[quote=\"shoki\"]vaaaaaaaaaaaaaaaayyyyyyyyyyy eeeeeeyyyyyyyyyyyyy       khhhhhhhhhhhhhhhhhooooooooodaaaaaaaaaaaaaa ye ghalate dige  :bomb: \najab sutii dadam.... \nmaaaaaaaaaaamannnnnnnnnnnnnnnnnn che ghadr bad dadam........ :wallbash_red:[/quote]\r\n\r\n\r\nna baba chera roohie mANFI midin?\r\n\r\n,migan kaf 11e.shomam ke sad dar sad bishtar az 15 ta dorost zadid.\r\ndige ghametoon chie??\r\n\r\nbe nazaretoon kaf hala 11 hast??\r\n\r\nmanam ye ghalate tablo daram ke nemidoonam chera eshtebaah zadam.\r\nsoale  5 kode1,hamoon 100 ta danesh amooze alef o b.\r\nnemidoonam chera javabesh mishe 49,man zadam 25!!!\r\nage momkene rahetoon ro begid.",
  "Solution_18": "man ke osulan kaf ro khub nemitunam takhmin bezanam(behtare begam ke hich vaght takhmini nazadam).\r\nkhob kafiye masalan $ 49$ ta $ 51$ begirid baraye kelase b(manzuram kelase ba miyangine kamtare)va yeki ham 0 va 50 ta 50 begirid baraye kelase a.\r\nkhahid did ke dorost dar miyad.",
  "Solution_19": "[quote=\"shoki\"]man ke osulan kaf ro khub nemitunam takhmin bezanam(behtare begam ke hich vaght takhmini nazadam).\nkhob kafiye masalan $ 49$ ta $ 51$ begirid baraye kelase b(manzuram kelase ba miyangine kamtare)va yeki ham 0 va 50 ta 50 begirid baraye kelase a.\nkhahid did ke dorost dar miyad.[/quote]\r\n\r\ney lanat.\r\n\r\nvelesh dige mohem nist.ba ye soal be jaii nemiresim.",
  "Solution_20": "Dustan guya kasi soale akharo fek nakarde!\r\nkasy hast hal karde bashe?\r\nhaddeaksar tedade davayer........",
  "Solution_21": "[quote=\"stvs_f\"]Dustan guya kasi soale akharo fek nakarde!\nkasy hast hal karde bashe?\nhaddeaksar tedade davayer........[/quote]\r\n\r\n\r\njavabesh mishe 4.vali mese in ke bachehaye ma ba 6 ta ham dar avorde boodan.\r\nnemidoonam.\r\nmiporsam :!:  :!:  :!:",
  "Solution_22": "[quote=\"stvs_f\"]Dustan guya kasi soale akharo fek nakarde!\nkasy hast hal karde bashe?\nhaddeaksar tedade davayer........[/quote]\r\n\r\njavabesh gozineye akhar fekr konam beshe.",
  "Solution_23": "[quote=\"amparvardi\"]migam in soale 23 daftarche 1 (hamoon chand jomleii ba daraje 5) chi mishe/?\nman har chi fekr kardam nafahmidam bayad che konam.[/quote]\r\n\r\nmishe 48 vali man eshtebah zadam :blush:",
  "Solution_24": "hi to all \r\non soale daraje 5 kheili ghashang bud!\r\nman halidam\r\ntaghsimesh ke mikoni bedast miad zaribr x^5 ba zaribe x^2 va zaribe x^4 ba x va x^3 ham ba x^0 barabare ka ba halat bandie adade 1 ta 6 kolan 48 joor mishe ja gozari kard!\r\nsoale akharo begid chi mishe??",
  "Solution_25": "salam.soal akhar mishe binahyat masalan mitunid khat ro dayere be shoae binahayat dar nazar begirid(omidvaram manzooramo resoonde basham :!: )",
  "Solution_26": "[quote=\"MJ GEO\"]salam.soal akhar mishe binahyat masalan mitunid khat ro dayere be shoae binahayat dar nazar begirid(omidvaram manzooramo resoonde basham :!: )[/quote]\r\n\r\nMan ke nafahmidam !!( kheily khengam  :blush: ) age mishe bishtar tozih bede.\r\n\r\namma mishod injoori in masalaro hal kard ke ye kore ro dar nazar begirid bad dayere haye azimasho rasm konid ...  :!:",
  "Solution_27": "[quote=\"mahanmath\"][quote=\"MJ GEO\"]salam.soal akhar mishe binahyat masalan mitunid khat ro dayere be shoae binahayat dar nazar begirid(omidvaram manzooramo resoonde basham :!: )[/quote]\n\nMan ke nafahmidam !!( kheily khengam  :blush: ) age mishe bishtar tozih bede.\n\namma mishod injoori in masalaro hal kard ke ye kore ro dar nazar begirid bad dayere haye azimasho rasm konid ...  :!:[/quote]\r\n\r\n\r\ndayerehaye chisho rasm konim??",
  "Solution_28": "[quote=\"amparvardi\"][quote=\"mahanmath\"][quote=\"MJ GEO\"]salam.soal akhar mishe binahyat masalan mitunid khat ro dayere be shoae binahayat dar nazar begirid(omidvaram manzooramo resoonde basham :!: )[/quote]\n\nMan ke nafahmidam !!( kheily khengam  :blush: ) age mishe bishtar tozih bede.\n\namma mishod injoori in masalaro hal kard ke ye kore ro dar nazar begirid bad dayere haye azimasho rasm konid ...  :!:[/quote]\n\n\ndayerehaye chisho rasm konim??[/quote]\r\ndayere haye azime (mesle ostova too koreye zamin) ...(azime = \u0639\u0638\u06cc\u0645\u0647)",
  "Solution_29": "Dar har soorat mamnoon.\r\n\r\n[url]http://www.IMO09.blogfa.com[/url]  javabha ro zade,age mikhaid berid bebinid.bahal javab dadan."
}
{
  "Tag": [
    "analytic geometry",
    "geometry",
    "conics",
    "perpendicular bisector",
    "radical axis",
    "geometry proposed",
    "BritishMathematicalOlympiad"
  ],
  "Problem": "Let $ \\Delta ABC$ be a triangle and $ (O)$ be a circle which intersects the sides $ BC,CA,AB$ at $ \\{A_1,A_2\\},\\{B_1,B_2\\},\\{C_1,C_2\\}.$ Also the intersection of the tangents at $ A_1,A_2$ to $ (O)$ is $ A_3.$ The two other points ($ B_3,C_3$) are defined similarly. Prove the concurrency of $ AA_3,BB_3,CC_3.$ I know a solution which is really ugly and I wanted to know if there exists other solutions. (Sorry if it has been posted before, I don't know the source).",
  "Solution_1": "[b]Theorem:[/b] $(O)$ is an arbitrary circle in the plane of $ \\triangle ABC.$ $P,Q,R$ are the poles of the sidelines $ BC,CA,AB$ with respect to $ (O).$ Then $AP,BQ,CR$ concur.\n\n[b]Proof:[/b] $ \\delta(P,a)$ stands for the distance from $ P$ to the sideline $ BC$  and $ \\delta(P,b),$  $ \\delta(P,c)$ are defined similarly. By [url=http://mathworld.wolfram.com/SalmonsTheorem.html]1st Salmon's theorem[/url] we have the following relations:\n\n$ \\frac {\\overline{RO}}{\\overline{PO}} = \\frac {\\delta(R,a)}{\\delta(P,c)} \\ , \\ \\frac {\\overline{QO}}{\\overline{PO}} = \\frac {\\delta(Q,a)}{\\delta(P,b)} \\ \\Longrightarrow \\ \\frac {\\delta(P,c)}{\\delta(P,b)} = \\frac {\\delta(R,a)}{\\delta(Q,a)} \\cdot \\frac {\\overline{QO}}{\\overline{RO}} \\ \\ (1)$\n\nBy cyclic exchange we have the relations:\n\n$ \\frac {\\delta(Q,a)}{\\delta(Q,c)} = \\frac {\\delta(P,b)}{\\delta(R,b)} \\cdot \\frac {\\overline{RO}}{\\overline{PO}} \\ . \\ (2) \\ , \\ \\frac {\\delta(R,b)}{\\delta(R,a)} = \\frac {\\delta(Q,c)}{\\delta(P,c)} \\cdot \\frac {\\overline{PO}}{\\overline{QO}} \\  \\ (3) $   \n\nMultiplying $ (1),(2)$ and $ (3)$ together yields:\n\n$ \\frac {\\delta(P,c)}{\\delta(P,b)} \\cdot \\frac {\\delta(Q,a)}{\\delta(Q,c)} \\cdot \\frac {\\delta(R,b)}{\\delta(R,a)} = \\frac {\\delta(R,a)}{\\delta(Q,a)} \\cdot \\frac {\\delta(P,b)}{\\delta(R,b)} \\cdot \\frac {\\delta(Q,c)}{\\delta(P,c)}\\cdot \\frac {\\overline{QO}}{\\overline{RO}} \\cdot \\frac {\\overline{RO}}{\\overline{PO}} \\cdot \\frac {\\overline{PO}}{\\overline{QO}}$\n\n$ \\frac {\\delta(P,c)}{\\delta(P,b)} \\cdot \\frac {\\delta(Q,a)}{\\delta(Q,c)} \\cdot \\frac {\\delta(R,b)}{\\delta(R,a)}  = 1$ $ \\Longrightarrow$ $AP,BQ,CR$ concur by Ceva's theorem.",
  "Solution_2": "I didn't check the computations but your idea was really NICE\nALL OF MY CONGRATS :D  :D \nThat was a really nice idea to use this theorem (and what a useful theorem it is! :) )",
  "Solution_3": "Thanks for your kind words dear shoki, indeed it is a very useful theorem but surprinsingly unknown as well. The problem is really nice. I was working on a solution since several days ago when a friend of mine sent me the generalization via private message and then I realized that you submitted it earlier.",
  "Solution_4": "We have an interesting concurrency when the subject circle concides with the 9-point circle.\n\nGiven a triangle $ \\triangle ABC$ and its 9-nine point circle $ (N).$ $ P,Q,R$  are the poles of the sidelines $ BC,CA,AB$  with respect to $ (N).$ Then $ AP,BQ,CR$ concur at the isotomic conjugate of $ X_{1078}.$",
  "Solution_5": "[quote]Given a triangle $ \\triangle ABC$ and its 9-nine point circle $ (N).$ $ P,Q,R$  are the poles of the sidelines $ BC,CA,AB$  with respect to $ (N).$ Then $ AP,BQ,CR$ concur at the isotomic conjugate of $ X_{1078}.$[/quote]\nThe pole $ P$ of $ BC$ WRT $ (N)$ is the intersection of the tangent of $ (N)$ at the midpoint of $ BC$ and the perpendicular dropped from $N$ to $ BC.$ Its anticomplement $P'$ is then the intersection of the tangent $ \\tau_a$ of $ (O)$ at $ A$ and the perpendicular bisector $ \\ell_a$ of $ BC.$ Using barycentric coordinates with respect to $ \\triangle ABC,$ we have:\n\n$ \\ell_a \\equiv (b^2 \\minus{} c^2)x \\plus{} a^2y \\minus{} a^2z \\equal{} 0 \\ , \\ \\tau_a \\equiv c^2y \\plus{} b^2z \\equal{} 0$ \n\n$ P' \\equiv \\ell_a \\cap \\tau_a \\equiv{} (a^2(b^2 \\plus{} c^2): b^2(c^2 \\minus{} b^2): c^2(b^2 \\minus{} c^2))$\n\nTherefore, $ P(2b^2c^2 \\minus{} b^4 \\minus{} c^4: a^2b^2 \\plus{} a^2c^2 \\plus{} b^2c^2 \\minus{} c^4: a^2b^2 \\plus{} a^2c^2 \\plus{} b^2c^2 \\minus{} b^4)$ \n\nCoordinates of $ Q,R$ are found  by cyclic permutation. Thus, $ AP,BQ,CR$ concur at\n\n$ \\left (\\frac {1}{b^2c^2 \\plus{} c^2a^2 \\plus{} a^2b^2 \\minus{} a^4}: \\frac {1}{b^2c^2 \\plus{} c^2a^2 \\plus{} a^2b^2 \\minus{} b^4}:\\frac {1}{b^2c^2 \\plus{} c^2a^2 \\plus{} a^2b^2 \\minus{} c^4} \\right )$\n\nWhich is the isotomic conjugate of $ X_{1078} \\equiv (a^2b^2 \\plus{} a^2c^2 \\plus{} b^2c^2 \\minus{} a^4)$",
  "Solution_6": "[quote=\"Luis Gonz\u00e1lez\"][b]Theorem:[/b] $(O)$ is an arbitrary circle in the plane of $ \\triangle ABC.$ $P,Q,R$ are the poles of the sidelines $ BC,CA,AB$ with respect to $ (O).$ Then $AP,BQ,CR$ concur.\n[/quote]\nHere is a synthetic solution for this theorem:-\nIts enough to prove that intersection of the polars of $ A,B,C $ wrt $ (O) $ with $ BC,CA,AB $ are collinear. Suppose, $ A',B',C' $ are the inverse points of $ A,B,C $ wrt $ (O) $. $ A_1,B_1,C_1 $ are located on $ BC,CA,AB $ such that $ A_1A'\\perp OA $ and similar for the others. We have to prove that $ A_1,B_1,C_1 $ are collinear. Suppose, $ AA_2,BB_2,CC_2 $ are the altitudes of $ ABC $. Consider the circles $ \\odot AA'A_2,\\odot BB'B_2,\\odot CC'C_2 $. Clearly, the orthocenter $ H $ of $ ABC $ wrt these circles has same power and so has $ O $. So these circles are co-axial. Their centers are midpoints of $ AA_1,BB_1,CC_1 $ and so they are collinear and this implies $ A_1,B_1,C_1 $ are collinear. So done.",
  "Solution_7": "A friend of mine from New Zealand( Malcolm Granville,  a former participant of IMO) remarked to me that he discovered this theorem while working on a special case of Nine Point Circle which appeared in some Romanian TST, as also remarked by Luis in his second post! The generalized statement, as he puts, is the following:\n\nGiven a triangle $ ABC $ and a conic $ \\omega $. Let the poles of the lines of the sides be $ A' $, $ B' $ and $ C'  $ corresponding  to $ BC $  , $ AB $ and $ AC $. Then, the lines $ AA' $ , $ BB' $ and $ CC' $ are concurrent.\n\nProof: Let the points   $ AB \\cap A'B' =   Z  $ ,  $ AA'  \\cap BB'  = P $ ,      $ AB \\cap   CC'  =   U  $ and   $  A' B'  \\cap CC'   =  U'    $  be in the projective plane.\n\nWe have  $  ( A, B  ; Z , U ) = ( CA  , CB  ; CZ  ,  CU ) $ and taking the poles of the lines in the last bundle, we have  $  ( CA, CB ; CZ ,  CU )  =  (  B' , A' ; U' , Z  )  = ( A' , B' , Z , U' )  =  ( PA', PB' , PZ, PU' ) $ and so the perspectivity from $ P $ sends $ A $, $ B $ , $ Z $ and $ U $ to $ A' $ , $ B' $ , $ Z $ and $ U ' $  respectively or $ P $, $ U $ and $ U' $  are  collinear which means $ P \\in  CC' $.\n\n\n\nThe existence of Gergonne and Symmedian Point can be proved by this theorem, as observed by him.",
  "Solution_8": "[quote=\"RSM\"]Clearly, the orthocenter $ H $ of $ ABC $ wrt these circles has same power and so has $ O $. So these circles are co-axial.[/quote]\nI'm not understanding this part. $H$ can  be the radical center of those three circles but how can you claim that those three circles are co axial?",
  "Solution_9": "I wrote that, both $ H $ and $ O $ have same power wrt the 3 circles. So clearly, they are co-axial.",
  "Solution_10": "[quote=\"RSM\"]I wrote that, both $ H $ and $ O $ have same power wrt the 3 circles. So clearly, they are co-axial.[/quote]\nOpps. Sorry i didn't notice that before :oops_sign:  :oops:",
  "Solution_11": "Is there any suloution without polar??",
  "Solution_12": "We can prove a stronger result when $A, B, C$ are inscribed in a conic: \n\nLet $\\triangle ABC$ be inscribed in a conic $\\mathcal{C}$ and let $M$ be a variable point on $\\mathcal{C}.$ Let $P, Q, R$ be the poles of $BC, CA, AB$ WRT $\\mathcal{C}$, and let $\\tau$ be the trilinear polar of $M$ WRT $\\triangle ABC.$ Then $AP, BQ, CR, \\tau$ concur.\n\n[b]Proof:[/b] Let $\\triangle M_AM_BM_C$ be the cevian triangle of $M$ WRT $\\triangle ABC$ and denote $D \\equiv BC \\cap M_BM_C.$ Let $Y, Z$ be the poles of $BQ, CR$ and denote $X \\equiv QR \\cap YZ.$\n\n[asy]\n /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */\nimport graph; size(15cm); \nreal labelscalefactor = 0.5; /* changes label-to-point distance */\npen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ \npen dotstyle = black; /* point style */ \nreal xmin = 956.2083418698593, xmax = 1112.9566474270691, ymin = 963, ymax = 1045;  /* image dimensions */\npen efefef = rgb(0.9372549019607843,0.9372549019607843,0.9372549019607843); pen wqwqwq = rgb(0.3764705882352941,0.3764705882352941,0.3764705882352941); pen ffqqff = rgb(1.,0.,1.); \n\nfilldraw((999.3440507142487,1013.1722703375563)--(996.4641665206781,1002.5120551177166)--(1017.7478969694971,1002.5756519218465)--cycle, efefef, wqwqwq); \nLabel laxis; laxis.p = fontsize(10); \nxaxis(xmin, xmax, Ticks(laxis, Step = 5., Size = 2, NoZero),EndArrow(6), above = true); \nyaxis(ymin, ymax, Ticks(laxis, Step = 5., Size = 2, NoZero),EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */ \n /* draw figures */draw(shift((1007.0127467743117,1006.7411397507293))*rotate(-0.6412881891215791)*xscale(12.441839109267029)*yscale(8.104044558855877)*unitcircle); \ndraw((1027.6066477315999,996.8991526492077)--(970.382782794154,1029.8476707254933), red); \ndraw((970.382782794154,1029.8476707254933)--(987.4245541699095,969.0509133891155)); \ndraw((978.0605681438306,1002.457064285911)--(1050.734592358769,1039.0449272681938), blue); \ndraw((970.382782794154,1029.8476707254933)--(1007.3605573895838,991.0916659454263), blue); \ndraw((1050.734592358769,1039.0449272681938)--(987.4245541699095,969.0509133891155), blue); \ndraw((992.2044439283167,986.7442128854926)--(1047.7081837536498,1002.6651746911193)); \ndraw((999.3440507142487,1013.1722703375563)--(1012.8734702434173,999.5545572630017), dotted); \ndraw((996.4641665206781,1002.5120551177166)--(1027.6066477315999,996.8991526492077), dotted); \ndraw((999.3440507142487,1013.1722703375563)--(987.4245541699095,969.0509133891155), red); \ndraw((996.4641665206781,1002.5120551177166)--(1047.7081837536498,1002.6651746911193), red); \ndraw((1017.7478969694971,1002.5756519218465)--(992.2044439283167,986.7442128854926), dotted); \n /* dots and labels */\ndot((999.3440507142487,1013.1722703375563),linewidth(3.pt) + dotstyle); \nlabel(\"$A$\", (998.3603474663743,1014.4745810022042), NE * labelscalefactor); \ndot((996.4641665206781,1002.5120551177166),linewidth(3.pt) + dotstyle); \nlabel(\"$B$\", (994.0022587521582,1000.4000977776053), N * labelscalefactor); \ndot((1017.7478969694971,1002.5756519218465),linewidth(3.pt) + dotstyle); \nlabel(\"$C$\", (1018.0931364871405,1000.9714333139119), S * labelscalefactor); \ndot((1012.8734702434173,999.5545572630017),linewidth(3.pt) + dotstyle); \nlabel(\"$M$\", (1012.2775017635865,997.7851142999051), S * labelscalefactor); \ndot((990.5279178417685,1008.7337730195741),linewidth(3.pt) + dotstyle); \nlabel(\"$R$\", (990.5729430426113,1009.6163837470127), N * labelscalefactor); \ndot((1050.734592358769,1039.0449272681938),linewidth(3.pt) + dotstyle); \nlabel(\"$Q$\", (1049.800083109579,1040.6231132874996), N * labelscalefactor); \ndot((1009.8952754145088,1002.5521879132139),linewidth(3.pt) + dotstyle); \nlabel(\"$M_A$\", (1008.6482945950152,1004.0437457189989), N * labelscalefactor); \ndot((1027.6066477315999,996.8991526492077),linewidth(3.pt) + dotstyle); \nlabel(\"$M_B$\", (1026.866642842959,994.3272426676403), SE * labelscalefactor); \ndot((992.2044439283167,986.7442128854926),linewidth(3.pt) + dotstyle); \nlabel(\"$M_C$\", (988.8582851878379,986.182726398442), N * labelscalefactor); \ndot((1007.3605573895838,991.0916659454263),linewidth(3.pt) + dotstyle); \nlabel(\"$P$\", (1007.3623012039351,988.1831605623444), E * labelscalefactor); \ndot((978.0605681438306,1002.457064285911),linewidth(3.pt) + dotstyle); \nlabel(\"$X$\", (976.3555716634478,1000.614430009452), W * labelscalefactor); \ndot((970.382782794154,1029.8476707254933),linewidth(3.pt) + dotstyle); \nlabel(\"$Y$\", (970.6400454808695,1030.2637220815766), NE * labelscalefactor); \ndot((987.4245541699095,969.0509133891155),linewidth(3.pt) + dotstyle); \nlabel(\"$Z$\", (985.500413555573,966.6784933003938), SE * labelscalefactor); \ndot((1047.7081837536498,1002.6651746911193),linewidth(3.pt) + dotstyle); \nlabel(\"$D$\", (1047.7282048683942,1000.2571010820653), S * labelscalefactor); \nclip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); \n /* end of picture */\n[/asy]\n\nBy Brokard's Theorem for quadrilateral $ABCM$, $\\triangle M_AM_BM_C$ is self-polar WRT $\\mathcal{C}.$ Hence the poles of $P, M_B, M_C$ concur (at $M_A$), so $P, M_B, M_C$ are collinear.\n\nNote that the polar of $D$ is the intersection of the polars of $BC$ and $M_BM_C$, which is precisely line $PM_A.$ As $D \\in \\tau$, we get that $PM_A$ passes through the pole $\\mathcal{T}$ of $\\tau$ WRT $\\mathcal{C}.$ Similarly, $PM_A, QM_B, RM_C$ are concurrent at $\\mathcal{T}.$ From \\[ Q(X, Y; Z, M_B) = (A, Y; C, M_B) \\stackrel{P}{=} (A, B; Z, M_C) = R(X, Y; Z, M_C), \\] we get that $\\mathcal{T} \\equiv QM_B \\cap RM_C \\in YZ.$ Hence, the polars $BQ, CR, \\tau$ of $Y, Z, \\mathcal{T}$ are concurrent. Similarly we get $AP, BQ, CR, \\tau$ are concurrent. $\\square$",
  "Solution_13": "Essentially the same solution as posted by multiple users, but I still feel like posting it :)\n\nSolution: Let $O$ be the centre of the circle $(A_1B_1C_1)$. Clearly, $A_3$ lies on the perpendicular bisector of $A_1A_2$. Thus, the perpendiculars from $A_3$ to $BC$, and the like, are concurrent at $O$. Notice that $A$ lies on the polar $B_3$ and $C_3$ since $A=B_1B_2\\cap C_1C_2$ and by duality, $B_3C_3$ is the polar line of $A$ wrt $(O)$. Thus, perpendiculars from $A$ to $B_3C_3$ and other lines defined similarly, are concurrent at point $O$. Hence, triangles $ABC$ and $A_3B_3C_3$ are orthologic and both the centers of orthology coincide. By the Sondat's theorem, the above pair of triangles is perspective and the result follows.\n\n[b]Comment[/b] The result also holds when the circle is replaced by any conic. The proof goes by applying a projective transformation sending the conic to a circle and proceeding as above."
}
{
  "Tag": [
    "analytic geometry",
    "trigonometry"
  ],
  "Problem": "The letters x and y represent rectangular coordinates. Write each equation using polar coordinates (r, theta).\r\n\r\n(1) x^2 + y^2 = x\r\n\r\n(2) 4x^2y = 1\r\n\r\n(3) y = -3",
  "Solution_1": "Do your own homework.",
  "Solution_2": "A = theta\r\n\r\n1. x^2 + y^2 = x\r\nr^2 = r cos A\r\nr = cos A\r\n\r\n2. 4x^2*y = 1 (is this what you meant by 4x^2y?)\r\nr^2*cos^2 A = 1/(r sin A)\r\nr^3*cos^2 A = 1/sin A\r\nr^3 = 1/sin A/cos^2 A.\r\nr^3 = 1/(cos^2 A * sin A) = sec^2 A *csc A\r\n\r\n(I'm not too sure if this one's right)\r\n\r\n3. y = -3\r\nr sin A = -3\r\nsin A = -3/r\r\n\r\nSorry if this is wrong--haven't done trig for a while"
}
{
  "Tag": [
    "search"
  ],
  "Problem": "With 4 digits  5, 5, 5  and 1, and the operator + - *  /  can you present the number 24 ? Remeber $5*5-1^{5}$ is not allowed. Also you can't have $55$ in your solution.\r\n\r\n  Good luck  :D  :D  :D",
  "Solution_1": "[hide]\n$5\\cdot\\left(5-\\frac{1}{5}\\right)$[/hide]",
  "Solution_2": "[quote=\"rnwang2\"][hide]\n$5\\cdot\\left(5-\\frac{1}{5}\\right)$[/hide][/quote]\r\nis there any method to do this or is it guess and check?\r\nJW",
  "Solution_3": "[quote=\"funcia\"][quote=\"rnwang2\"][hide]\n$5\\cdot\\left(5-\\frac{1}{5}\\right)$[/hide][/quote]\nis there any method to do this or is it guess and check?\nJW[/quote]\r\nIt's basically just guess and check.",
  "Solution_4": "[quote=\"rnwang2\"][hide]\n$5\\cdot\\left(5-\\frac{1}{5}\\right)$[/hide][/quote]\r\n\r\n Very nice sols. I think it is the only way too.  :D",
  "Solution_5": "[quote=\"Gibbenergy\"][quote=\"rnwang2\"][hide]\n$5\\cdot\\left(5-\\frac{1}{5}\\right)$[/hide][/quote]\n\n Very nice sols. I think it is the only way too.  :D[/quote]\n\n*plays clip of Dr. Cox singing \"wrong wrong wrong wrong...\"*\n\n[hide]$5-5+5-1$[/hide]",
  "Solution_6": "[quote=\"lingomaniac88\"][quote=\"Gibbenergy\"][quote=\"rnwang2\"][hide]\n$5\\cdot\\left(5-\\frac{1}{5}\\right)$[/hide][/quote]\n\n Very nice sols. I think it is the only way too.  :D[/quote]\n\n*plays clip of Dr. Cox singing \"wrong wrong wrong wrong...\"*\n\n[hide]$5-5+5-1$[/hide][/quote]\r\nwhat the heck?  that equals 4 :maybe:",
  "Solution_7": "[quote=\"lingomaniac88\"][quote=\"Gibbenergy\"][quote=\"rnwang2\"][hide]\n$5\\cdot\\left(5-\\frac{1}{5}\\right)$[/hide][/quote]\n\n Very nice sols. I think it is the only way too.  :D[/quote]\n\n*plays clip of Dr. Cox singing \"wrong wrong wrong wrong...\"*\n\n[hide]$5-5+5-1$[/hide][/quote]\r\nuhh that's 4",
  "Solution_8": "I have a disease.  I can't see the number 2.  :rotfl: \r\n\r\nHuh?  Did I type something after the word \"number\"?  What is it?  I don't see it.  I swear I did, but I can't see it!",
  "Solution_9": "Another way of doing it is this\r\n\r\n$(5-\\frac{1}{5})(5)$",
  "Solution_10": "[quote=\"mathgeniuse^ln(x)\"]Another way of doing it is this\n\n$(5-\\frac{1}{5})(5)$[/quote]\r\nUh that's the same as rnwang's solution.",
  "Solution_11": "[quote=\"Gibbenergy\"][quote=\"rnwang2\"][hide]\n$5\\cdot\\left(5-\\frac{1}{5}\\right)$[/hide][/quote]\n\nVery nice sols. I think it is the only way too.  :D[/quote]\r\n\r\nThe following webpage can search for the solutions in this type of questions: :lol: \r\n[url]http://www.mikekong.net/Maths/Quizzes/24game.php[/url]\r\n\r\nI have entered (5,5,5,1), and the only outcome is the solution given by rnwang2, so I believe that (5,5,5,1) is the only way. :)"
}
{
  "Tag": [
    "inequalities",
    "function",
    "integration",
    "calculus",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "let $f,g,h: \\mathbb{R}\\to \\mathbb{R}^{+}$ be $3$ positive continuous functions and $\\lambda \\in [0;1]$. Suppose that for every $x,y \\in \\mathbb{R}$ the following inequality holds:\r\n$h(\\lambda x+(1-\\lambda ) y) \\geq f^{\\lambda}(x) g^{1-\\lambda}(y)$\r\nShow that\r\n$\\int_{\\mathbb{R}}h \\geq (\\int_{\\mathbb{R}}f)^{\\lambda}(\\int_{\\mathbb{R}}g)^{1-\\lambda}$",
  "Solution_1": "This inequality is true for function from $\\mathbb R$ to $\\mathbb R$. For 1-dimensional situation it's collorary of Brunn-Minkowski inequality. \r\nActually wlog $\\| f \\|_{\\infty}= \\| g \\|_{\\infty}= 1$. For an arbitrary $a \\in [0,1)$ we have relation\r\n$\\{ h \\geq a \\}\\supset \\lambda \\{ f \\geq a \\}+(1-\\lambda) \\{ g \\geq a \\}$. Therefore $\\mu \\{ h \\geq a \\}\\geq \\lambda \\mu \\{ f \\geq a \\}+(1-\\lambda) \\mu \\{ g \\geq a \\}$. The last step is integration over $a \\in [0,1]$."
}
{
  "Tag": [
    "probability"
  ],
  "Problem": "Erika, who is 14 years old, flips a fair coin whose sides are labeled 10 and 20, and then she adds the number on the top of the flipped coin to the number she rolls on a standard die. What is the probability that the sum equals her age in years? Express your answer as a common fraction.",
  "Solution_1": "The only way to get 14 is to get 10 on the coin and 4 on the die.  This happens with probability $ \\frac{1}{2}\\cdot \\frac{1}{6}\\equal{}\\boxed{\\frac{1}{12}}$",
  "Solution_2": "Yup! Nice job getting to the solution~"
}
{
  "Tag": [
    "geometry",
    "research"
  ],
  "Problem": "So here's a problem I've been thinking about, but I've not been able to settle yet. Let's say you have a convex polygon with n sides, where n >= 4 and then we cover that polygon with a smaller homothetic copy. Is it true that the non-covered region will always be nonconvex? Prove whether this is true or not.",
  "Solution_1": "If the smaller copy does not have to be in the interior of the original larger polygon, then no, the statement is not true and the non-covered region can be convex. One example is the pentagon with just one edge of the smaller copy lying inside the original; this can be done homothetically.\r\n\r\nIf the smaller copy must be contained in the larger original, then the statement is true. Since the smaller copy can share at most two (consecutive) edges with the original, the copy will have at least two consecutive edges, neither one which shares edges with the original. Since the polygon is convex, the region outside of those two edges is concave.",
  "Solution_2": "Cover a hexagon (with appropraite shape, say regular hexagon) with 4 copy at four coner with two opposite coner ignored, then remained areas are 2 triangulars and 3 quadrangles."
}
{
  "Tag": [],
  "Problem": "lithium heptylide with 1-bromo 3-chloro propane ofcourse with reasoning :)",
  "Solution_1": "[hide=\"Answer\"]Sn2 reaction, with bromide being a better leaving group than chloride.[/hide]",
  "Solution_2": "yes that's perfect but i am 'worried' about the aswer pretty much the reason we were discussing in the other thread today...\r\nwhy doesn't the strong base heptylide capture the acidic $ \\beta$ hydrogen $ \\beta$ to both chlorine and bromine :?:",
  "Solution_3": "Because that's a secondary carbon from which the base would have to approach (with also the two big halogens encumbering the way). SN2 at a primary and electropositive carbon is much faster.",
  "Solution_4": "would the answer be the same if i had acetylide ions :maybe:",
  "Solution_5": "Yes, it would.",
  "Solution_6": "can we take this as a thumb rule that the alkenylide ions mostly behave as nucleophilles with halides,tosylates that is Saytzeff Leaving Groups :)",
  "Solution_7": "Yes, if they are primary. Otherwise, elimination will occur.",
  "Solution_8": "I think this is corey-houser reaction. :)",
  "Solution_9": "yes somewhat but not corey house itself i suppose corey house specifically refers to alkane chain expansion :maybe:",
  "Solution_10": "The Corey-Posner, Whitesides-House reaction refers to a completely different thing: the reaction between Gilman reagents and organic halides."
}
{
  "Tag": [],
  "Problem": "the maximum deceleration of a car on a dry road is about 8.0 m/s2. if two cars are moving head on toward each other at 88 kph, and thier drivers apply thier brakes when they are 85 m apart, will they collide? if so, at what relative speed? if not. how far apart will they be when they stop?",
  "Solution_1": "Take each car separately.  First, you have $v_{0}=\\frac{220}{9}\\,\\textrm{m}/\\textrm{s}$ and $a=-8\\,\\textrm{m}/\\textrm{s}^{2}$.  Now use the equation \\[v=v_{0}+at=0,\\] solve for $t$, then plug this into the equation \\[d=v_{0}t+\\frac{1}{2}at^{2}\\] to get the distance travelled.  Doing the same thing for both cars will give you $d_{1}$ and $d_{2}$.  If $d_{1}+d_{2}<85$ then the cars will not collide.  Otherwise they will.",
  "Solution_2": "[quote=\"arlene de la cruz\"]the maximum deceleration of a car on a dry road is about 8.0 m/s2. if two cars are moving head on toward each other at 88 kph, and thier drivers apply thier brakes when they are 85 m apart, will they collide? if so, at what relative speed? if not. how far apart will they be when they stop?[/quote]\r\n\r\nApply the formula: $v_{f}^{2}= v_{i}^{2}+2a\\Delta x$ in the following way.\r\n\r\n$v_{f}= 0,$\r\n\r\n$v_{i}= 88 \\times 10^{3}\\text{ m/hr}\\times\\frac{\\text{ 1 hour}}{3.6 \\times 10^{3}\\text{ seconds}}= 24.45 \\text{ m/s}$\r\n\r\n$a =-8.0\\, m/s^{2}$\r\n\r\n$\\therefore \\Delta x = \\frac{(0-(24.45)^{2})}{2(-8.0)}= 37.35\\text{ m}$\r\n\r\n${\\Delta x < \\frac{85 \\text{ meters}}{2}}$, so yes the cars will stop before they collide.\r\n\r\n$\\text{separation}= 85 \\text{ m}-2(37.35\\text{ m}) = 10.3 \\text{ m}$"
}
{
  "Tag": [
    "geometry",
    "probability",
    "AMC",
    "AIME",
    "USA(J)MO",
    "USAMO",
    "algebra",
    "\\/closed"
  ],
  "Problem": "How many people from WOOT are taking it next year?",
  "Solution_1": "I'm trying to convince my parents to let me take WOOT this year...I think I'll probably be able to, but I have to convince them that it's worth the money.",
  "Solution_2": "I want to know if WOOT will be more difficult and intense than last year before I decide whether I want to join or not. The problem sets last year were short and many had too many easyish problems. Also does anyone know about the difficulty of the problems that will be done in class? I hope WOOT is more difficult this year and also I think there is some repetition which isn't nice.",
  "Solution_3": "We do not intend for there to be any repetition of problems in 2009-10 from 2008-09.  As for the difficulty of the problems, I think you're in the minority -- we had more students complaining about the problems being too hard than too easy.  That said, there will be some more very hard problems, as we will attempt to increase the variance of the difficulty of the material.",
  "Solution_4": "I should add that we will re-use some handouts from 2008-09, but these will be in addition to the 8+ handouts that we will offer for 2009-10 that were not part of 2008-09 WOOT.  (That is, we'll have more than 8 handouts for the course.)",
  "Solution_5": "Overall will there be more message board problems this year because there wasn't enough challenging message board problems (there was like one or two really hard ones I couldn't get but I didn't really like that.)",
  "Solution_6": "We'll make sure there are plenty of challenging message board problems.",
  "Solution_7": "Will there be Olympiad classes this upcoming year? I think those are generally more challenging than the WOOT lectures. If not will WOOT be more challenging this year? Last year many of the classes only did easy problems while the message board problems were often hard.",
  "Solution_8": "I will be going to WOOT, probably",
  "Solution_9": "[quote=\"thing1\"]Will there be Olympiad classes this upcoming year? I think those are generally more challenging than the WOOT lectures. If not will WOOT be more challenging this year? Last year many of the classes only did easy problems while the message board problems were often hard.[/quote]\r\n\r\nAs you'll note from the WOOT pages, we have restructured the course somewhat, by grouping classes roughly by subject area.  Our intent here is to allow us to drill deeper into each subject area, pushing more of the introductory material into handouts, and allowing us to do more of the challenging material in class.  Of course, this only helps students who will actually read the handouts and do some of the problems therein!\r\n\r\nWe might offer the Olympiad Geometry class next spring, as we did this year, but we haven't yet decided.  Most of the material from our old Olympiad Problem Solving course has been integrated into WOOT and the Intermediate Counting & Probability course.",
  "Solution_10": "Is there any way we can buy the course materials or take the course without feedback for cheaper? I would still be willing to pay reasonably.",
  "Solution_11": "No, we don't offer that option.",
  "Solution_12": "So does the Intermediate Counting and Probability class cover a lot of AIME/USAMO-level material?  Is it also targeted toward WOOT-level people?  Does it cover material not included in WOOT?",
  "Solution_13": "Yes, it has a lot of AIME - beginning USAMO material.  It is targeted towards a similar crowd as beginners in WOOT.  It has a little topical overlap with WOOT, but will not use the same problems.",
  "Solution_14": "Some of the sample problems like all the polynomial ones are really simple or stuff thats in the AOPS text. Same goes for many other sections but the graph theory has one from 2008 USAMO so I'm wondering what the level of difficulty of WOOT will be for this year? Will the classes be as challenging as the advanced part of the practice olympiads?",
  "Solution_15": "Yes, some will.  That is part of the reason we re-structured WOOT the way we have, to allow more depth in fewer areas each year rather than just touching on many areas.  We'll still give some exposure to areas outside the lectures, through handouts and problems on the message board, but rather than try to have classes on 12-16 different areas, we'll have classes on fewer areas, but cover them to greater depth (i.e. to greater complexity and difficulty) by having more than one class per area for most areas we study.",
  "Solution_16": "Hmm...\r\nWill WOOT still be offered to all MOPpers for free next year?",
  "Solution_17": "I've got a question. I REALLY REALLY want to take WOOT next year, because all the topics seem interesting, and I definitely want to improve my proof-writing skills.\r\n\r\nThere are three things though:\r\n\r\n1. My average score on the AIME is precisely 5 now, and I am unsure whether this will be enough for WOOT (because I have not seen any incredibly increase in score)\r\n\r\n2. Put simply, I find it extremely difficult to write a proof. Usually, what ends up happening is that I think I have this great solution, but when I try and put on paper, it either doesn't work, or there is some case that I missed and renders the whole thing useless.\r\n\r\n3. My problem solving and mathematical ability...unfortunately is not that high. I am mostly good at finding patterns and trying something out for a while before I can come to a legitimate conclusion. This type of thinking is also very difficult to write down on paper in the form of a proof. \r\nIn other words, I can solve problems requiring trivial observations, but for something more complex (like nearly any olympiad problem), something always goes wrong (to this day, I haven't solved one).\r\n\r\nDo you still think WOOT would be a valid choice for me? One reason I want to take it next year is because that is 10th grade for me, and it will be the year I can put most time into it, as Junior year is 3 times more hectic.",
  "Solution_18": "[quote=\"blackbelt14253\"]Hmm...\nWill WOOT still be offered to all MOPpers for free next year?[/quote]\r\n\r\nThe 2009 MOPpers will be invited to 2009-2010 WOOT for free.",
  "Solution_19": "[quote=\"veezbo\"]I've got a question. I REALLY REALLY want to take WOOT next year, because all the topics seem interesting, and I definitely want to improve my proof-writing skills.\n\nThere are three things though:\n\n1. My average score on the AIME is precisely 5 now, and I am unsure whether this will be enough for WOOT (because I have not seen any incredibly increase in score)\n\n2. Put simply, I find it extremely difficult to write a proof. Usually, what ends up happening is that I think I have this great solution, but when I try and put on paper, it either doesn't work, or there is some case that I missed and renders the whole thing useless.\n\n3. My problem solving and mathematical ability...unfortunately is not that high. I am mostly good at finding patterns and trying something out for a while before I can come to a legitimate conclusion. This type of thinking is also very difficult to write down on paper in the form of a proof. \nIn other words, I can solve problems requiring trivial observations, but for something more complex (like nearly any olympiad problem), something always goes wrong (to this day, I haven't solved one).\n\nDo you still think WOOT would be a valid choice for me? One reason I want to take it next year is because that is 10th grade for me, and it will be the year I can put most time into it, as Junior year is 3 times more hectic.[/quote]\r\n\r\nIt depends on what you do between now and September.  If you do nothing, then you would likely be better off with the Intermediate classes than WOOT.  If you spend the next 2-3 months working on math, then you will likely be ready for WOOT.  Some resources to consider: from AoPS, Interm Algebra, Interm Counting & Probability, AoPS Vol 2.  From Amazon: Art and Craft of Problem Solving by Paul Zeitz, Problem Solving Strategies by Arthur Engel.  (I recommend the latter two after you've done either both Interm books or AoPS Vol 2.)  You don't have to finish all of these books prior to WOOT, but you should spend a significant amount of time with them.\r\n\r\nAs for proof writing, we will be including mathematical writing as part of the curriculum in WOOT this year."
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "Let $ x,y,z$ be the non-negative real number satisfying $ (x\\plus{}y\\plus{}z)^2\\plus{}xy\\plus{}yz\\plus{}zx\\equal{}2$.Prove that $ \\frac{x\\plus{}y}{\\sqrt{z^2\\plus{}xy\\plus{}1}}\\plus{}\\frac{y\\plus{}z}{\\sqrt{x^2\\plus{}yz\\plus{}1}}\\plus{}\\frac{z\\plus{}x}{\\sqrt{y^2\\plus{}zx\\plus{}1}}$\u2265$ \\frac{3\\sqrt2}{2}$",
  "Solution_1": "I really like your inequality.\r\nwe can let:a=x+y,b=y+z,c=z+x\r\nthen we have ab+bc+ca=2\r\nand we have to prove that:\r\n$ \\sum_{cyc}\\frac{a}{\\sqrt{a^2\\plus{}3bc}} \\geq \\frac{3}{2}$(1)\r\nbut i can't solve (1) .\r\ncan you or somebody help me,pleasezzz>?\r\nthanks\r\n :blush:",
  "Solution_2": "[quote=\"tuandokim\"]I really like your inequality.\nwe can let:a=x+y,b=y+z,c=z+x\nthen we have ab+bc+ca=2\nand we have to prove that:\n$ \\sum_{cyc}\\frac {a}{\\sqrt {a^2 \\plus{} 3bc}} \\geq \\frac {3}{2}$(1)\nbut i can't solve (1) .\ncan you or somebody help me,pleasezzz>?\nthanks\n :blush:[/quote]\r\nYou have a right idea.Your idea is same to mine.We have if a,b,c  are 3 sides of a triangle,then\r\n$ \\sum_{cyc}\\frac {a}{\\sqrt {a^2 \\plus{} 3bc}} \\geq \\frac {3}{2}$(1)\r\nThis ineq is not hard to prove.\r\n$ aGA^2\\plus{}bGB^2\\plus{}cGC^2\\equal{}(a\\plus{}b\\plus{}c)IG^2\\plus{}abc$\r\n=>$ 2ab(a\\plus{}b)\\plus{}2bc(b\\plus{}c)\\plus{}2ca(c\\plus{}a)$\u2265$ a^3\\plus{}b^3\\plus{}c^3\\plus{}9abc$(**)\r\nNow use Cauchy-Schwarzt ineq:\r\n$ \\sum_{cyc}\\frac{a}{a^2\\plus{}3bc}$\u2265$ \\frac{(a\\plus{}b\\plus{}c)^2}{\\sum_{cyc}a\\sqrt{a^2\\plus{}3bc}}\\equal{}\\frac{(a\\plus{}b\\plus{}c)^2}{\\sum_{cyc}\\sqrt{a}\\sqrt{a^3\\plus{}3abc}}$\u2265$ \\frac{(a\\plus{}b\\plus{}c)^2}{\\sqrt{a\\plus{}b\\plus{}c}\\sqrt{a^3\\plus{}b^3\\plus{}c^3\\plus{}9abc}}$\r\nSo it remains to prove that $ \\frac{(a\\plus{}b\\plus{}c)^2}{\\sqrt{a\\plus{}b\\plus{}c}\\sqrt{a^3\\plus{}b^3\\plus{}c^3\\plus{}9abc}}$\u2265$ \\frac{3}{2}$\r\nThis is equivalent to $ 4(a\\plus{}b\\plus{}c)^3$\u2265$ 9(a^3\\plus{}b^3\\plus{}c^3\\plus{}9abc)$\r\n<=> $ 12\\sum_{cyc}ab(a\\plus{}b)$\u2265$ 5(a^3\\plus{}b^3\\plus{}c^3)\\plus{}57abc$\r\nFrom (**),we have\r\n$ 12\\sum_{cyc}ab(a\\plus{}b)$\u2265$ 6(a^3\\plus{}b^3\\plus{}c^3)\\plus{}54abc$\u2265$ 5(a^3\\plus{}b^3\\plus{}c^3)\\plus{}57abc$\r\nThe ineq hold iff $ a\\equal{}b\\equal{}c$,this leads to $ x\\equal{}y\\equal{}z\\equal{}\\frac{\\sqrt6}{6}$",
  "Solution_3": "[quote=\"tuandokim\"]I really like your inequality.\nwe can let:a=x+y,b=y+z,c=z+x\nthen we have ab+bc+ca=2\nand we have to prove that:\n$ \\sum_{cyc}\\frac {a}{\\sqrt {a^2 \\plus{} 3bc}} \\geq \\frac {3}{2}$(1)\nbut i can't solve (1) .\ncan you or somebody help me,pleasezzz>?\nthanks\n :blush:[/quote]\r\nIt can solve by Cauchy-schwarz and SOS :)",
  "Solution_4": "quykhtn-qa1 are u telling about solving the original inequality by Cauchy Schwarz and SOS???\r\nI would like to see it...........",
  "Solution_5": "[quote=\"tuandokim\"]\n$ \\sum_{cyc}\\frac {a}{\\sqrt {a^2 \\plus{} 3bc}} \\geq \\frac {3}{2}$(1)\n[/quote]\r\nJust use Holder, let the LHS be S, then by Holder\r\n$ S^2(\\sum a(a^2 \\plus{} 3bc)) \\ge (a \\plus{} b \\plus{} c)^3$\r\nSo we have to show $ 4(a \\plus{} b \\plus{} c)^3 \\ge 9(a^3 \\plus{} b^3 \\plus{} c^3 \\plus{} 9abc)$ which is obvious by Muirhead",
  "Solution_6": "That last inequality that you want to show is false... b=c=0 gives 4a^3>=9a^3 which is obviously false...",
  "Solution_7": "Whoops, I realized that right after you posted  :oops:",
  "Solution_8": "[quote=\"Altheman\"]That last inequality that you want to show is false... b=c=0 gives 4a^3>=9a^3 which is obviously false...[/quote]\r\nWhy is that false? Isn't $ a,b,c$ sides of a triangle, since tuandokim has set $ a \\equal{} x \\plus{} y, b \\equal{} y \\plus{} z, c \\equal{} z \\plus{} x$ ? :)\r\n\r\nI get $ 4(a \\plus{} b \\plus{} c)^3 \\minus{} 9(a^3 \\plus{} b^3 \\plus{} c^3 \\plus{} 9abc) \\equal{}$ $ (7b \\plus{} 7c \\minus{} 5a)(a \\minus{} b)(a \\minus{} c) \\plus{} (19a \\minus{} 5b \\minus{} 5c)(b \\minus{} c)^2 \\ge 0$, which is obvious if we assume $ a \\ge b,c$ :)",
  "Solution_9": "[quote=\"Agr_94_Math\"]quykhtn-qa1 are u telling about solving the original inequality by Cauchy Schwarz and SOS???\nI would like to see it...........[/quote]\r\nI will must prove that:\r\n$ 2(ab(a\\plus{}b)\\plus{}bc(b\\plus{}c)\\plus{}ca(c\\plus{}a)) \\ge a^3\\plus{}b^3\\plus{}c^3\\plus{}9abc$\r\n$ <\\equal{}> (3c\\minus{}a\\minus{}b)(a\\minus{}b)^2\\plus{}(3a\\minus{}b\\minus{}c)(b\\minus{}c)^2\\plus{}(3b\\minus{}a\\minus{}c)(c\\minus{}a)^2 \\ge 0$  (*)\r\nSuppose $ a \\ge b \\ge \\ge c$\r\nWe have: $ 3b\\minus{}a\\minus{}c \\ge 3b\\minus{}c\\minus{}(b\\plus{}c) \\equal{}2(b\\minus{}c) \\ge 0 ,,(c\\minus{}a)^2 \\ge (a\\minus{}b)^2\\plus{}(b\\minus{}c)^2$\r\nthen$ LHS \\ge (3c\\minus{}a\\minus{}b\\plus{}3b\\minus{}a\\minus{}c)(a\\minus{}b)^2\\plus{}(3b\\minus{}a\\minus{}c\\plus{}3a\\minus{}b\\minus{}c)(b\\minus{}c)^2  \\ge 0$\r\nWe have Q.E.D \r\n :wink: :)"
}
{
  "Tag": [
    "Ross Mathematics Program"
  ],
  "Problem": "ok, so i think i can handle the math problems but the application is a no-no  :P ... how does mathematics fit in my special interests? how do u answer a question like that? \"ummmm... i like math?\"  :blush: lol... & describe ur experiences with mathematics... i guess i could list out competitions and math fairs and all that but thats not gonna be too long & its more of a list than an essay  :? ... *sigh*  i need help, any Ross veterans out there, please share ur thoughts!!!!!!!!  ;)",
  "Solution_1": "Well I've been going for the past two years, and am returning as a counselor...\r\n\r\nI wouldn't worry too much about these questions, just write what comes to mind.\r\n\r\nyou should probably focus on why you enjoy mathematics, and what makes you want to pursue it."
}
{
  "Tag": [
    "integration",
    "logarithms",
    "calculus",
    "calculus computations"
  ],
  "Problem": "Evaluate $ \\int_{0}^{1}\\frac {\\log ( 1 \\plus{} x ) }{x} dx .$",
  "Solution_1": "[b]log(1+x) = x-(x^2)/2+(x^3)/3...........\nexpansion of log(1+X) AND THEN INTEGRATE[/b]",
  "Solution_2": "But how do i integrate infinite terms . Can u please work it out ?? And i don't think we can do that.",
  "Solution_3": "If you do so, you would get an infinite series $ \\sum_{k\\equal{}1}^{\\infty}\\frac{(\\minus{}1)^{k\\plus{}1}}{k^{2}}$ which equals to $ \\frac{\\pi^{2}}{12}$.",
  "Solution_4": "Oh ya thats correct ,\r\n\r\nAny other ideas without using the series expansions of $ log (1\\plus{}x)$ ??",
  "Solution_5": "I think they aren't others, there's no way to get the result by not using series.\r\n\r\nBy the way to justify its convergence, it's enough to see that the integrand has limit as $ x\\to0$ ?",
  "Solution_6": "Yes, another method can be: \\[ J\\left( \\theta  \\right) \\equal{} \\int_0^1 {\\frac{{\\log \\left( {1 \\plus{} \\theta x} \\right)}}\r\n{x}} dx\\underbrace  \\Rightarrow _{Leibniz}J'\\left( \\theta  \\right) \\equal{} \\int_0^1 {\\frac{{dx}}\r\n{{\\left( {1 \\plus{} \\theta x} \\right)}}}\\]\r\n\r\nAnd then integrate."
}
{
  "Tag": [
    "geometry"
  ],
  "Problem": "Can you please provide a brief discription of the math courses below?\r\n\r\n(1) Differential Equations\r\n\r\n(2) Applied Mathematics\r\n\r\n(3) Mathematical Logic \r\n\r\n(4) Axiomatic Geometry\r\n\r\n(5) Game Theory and Linear Programming\r\n\r\nThanks",
  "Solution_1": "You can find information about your recent posts in Wikipedia or other websites.",
  "Solution_2": "Okay, I will do a research online.  I just wanted the insight of most tutors regarding other areas of math not common to people who did not major in math.\r\n\r\nThanks anyway."
}
{
  "Tag": [
    "geometry",
    "geometric transformation",
    "reflection",
    "abstract algebra",
    "superior algebra",
    "superior algebra solved"
  ],
  "Problem": "Find all $n\\in \\mathbb{N}$ for which there is an epimorphism from $D_{2n}$ to $C_n$. \r\n[i](in our notation, $D_{2n}$ has $2n$ elements)[/i]",
  "Solution_1": "Let such a epimorphism be given.\r\nThere are $n$ reflections in $D_{2n}$, all their images having order $1$ or $2$ in $C_n$.\r\nNow any of the other elements can be written as product of two reflections and their images have, since $C_n$ is abelian,  also order $1$ or $2$. But thats just possible for $n=1$ or $2$, since $C_n$ is by definition cyclic."
}
{
  "Tag": [
    "modular arithmetic",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Find the remainder of $ 5^{100}$ when divided by $ 1001$",
  "Solution_1": "$ 1001 \\equal{} 7\\cdot 11 \\cdot 13$. find the remainders of $ 5^{100}$ modulo $ 7, 11, 13$, using fermat's little theorem. then use these to find the remainder modulo 1001...",
  "Solution_2": "is the ans 396",
  "Solution_3": "I got $ 5^{100}\\equiv 2 \\pmod 7$, $ 5^{100}\\equiv 1 \\pmod {11}$, $ 5^{100}\\equiv 1 \\pmod {13}$, and finally $ 5^{100}\\equiv 716 \\pmod {1001}$."
}
{
  "Tag": [],
  "Problem": "During one season from May to September, a pair of mosquitoes can become parents and ancestors to $ 1.7\\times 10^{21}$ mosquitoes. If a large swamp in Florida contains $ 10^{10}$ pairs of mosquitoes, how many mosquitoes can be produced in one season. Express your answer in scientific notation.",
  "Solution_1": "Because each pair can be parents and ancestors to 1.7 x 10^{21} mosquitoes, the answer is simply $ 1.7 * 10^{21} * 10^{10} \\equal{} \\boxed{1.7 * 10^{31}}$\r\nSomeone please check this..."
}
{
  "Tag": [
    "combinatorics proposed",
    "combinatorics"
  ],
  "Problem": "Find all $(m,n) \\in \\mathbb{Z}^2$ that we can color each unit square of $m \\times n$ with the colors black and white that for each unit square number of unit squares that have the same color with it and have at least one common vertex (including itself) is even.",
  "Solution_1": "so, is the problem to find all m,n for which an mxn rectangle's unit squares can be colored W/B, so that every black squares has vertices touching to 0,2 or 4 (a square number) of other black squares and vice versa for white?\r\n\r\nBut that cannot be, since a chessboard always fulfils the condition [always 0]... please clarify yourself.",
  "Solution_2": "Peter, I have no idea what you find hard to understand here.. :?\r\n\r\nFor a small square, call all the squares which share at least a vertex with it (including itself) its neighbours. We have to find those $m,n$ for which it's possible to color the squares of an $m\\times n$ board in two colors, s.t. each square has an even number of neighbours colored in the same color as itself.\r\n\r\nI believe the answer is: those $m,n$ for which $2|mn$ (i.e. at least one of $m,n$ is even). If $m$ is even, for example, we color the first two lines in white, the next two in black, and so on.\r\n\r\nConversely, let's assume we can find such a coloring for our $m\\times n$ board. We look at the graph having the black squares as vertices and in which two different squares are connected iff they are neighbours. The hypothesis tells us that in this graph, each vertex has an odd degree, so it has an even number of vertices. The same goes for the graph formed by the white squares, so the board has an even number of squares, and this is what we wanted to show: $mn$ (the number of squares) is even.",
  "Solution_3": "For $m \\times n$ table that $m$ is even, Divide the table in $\\frac m2$ of $ n \\times 2$ tables.\r\nAnd color the tables black and white",
  "Solution_4": "That's what I said, but i guess my post looked too long to be worth reading :).",
  "Solution_5": "Is that long? :? I'm still waiting for Pierre's first post of that length :D\r\n\r\nAs for my misunderstanding: I skipped the last line and thus thought that the number of neigbours not including itself needed to be even... and then a chessboard pattern does the job :)"
}
{
  "Tag": [
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "3(p^q+q^p)=n! p,q are prime numbers and n is natural number. Find these unknown values",
  "Solution_1": "[b]Case 1:[/b] $ p \\neq q$ and $ p,q \\neq 3$, then $ p,q>n$ or one of $ p,q |n!$, which is impossible. Now $ 3(p^q\\plus{}q^p)>6n^n>n!$, contradiction.\r\n[b]Case 2:[/b] one of $ p,q$ , say $ p\\equal{}3$, then the equation becomes $ 3(3^q\\plus{}q^3)\\equal{}n!$. As $ 3 \\nmid 3^q\\plus{}q^3$ , the only possible values of $ n$ are $ 3,4,5$. After checking we see that they are impossible also.\r\n[b]Case 3:[/b] $ p\\equal{}q$, then the equation becomes $ 6(p^p)\\equal{}n!$. For $ n \\ge 4$, we have $ 5!|n!\\equal{}6(p^p) \\Rightarrow (2^2 \\times 5|p^p$, which is impossible .So the only possible values of $ n$ are $ 3$ and $ 4$. After checking, $ n\\equal{}4, p\\equal{}q\\equal{}2$ is the only solution."
}
{
  "Tag": [
    "MATHCOUNTS",
    "geometry",
    "trapezoid"
  ],
  "Problem": "The acute angles of a particular isosceles trapezoid each measure 40 degrees. What is the measure, in degrees, of each obtuse angle in the trapezoid?",
  "Solution_1": "[hide]There are two acute angles and two obtuse angles in an isosceles trapezoid; the two acute angles are congruent, and the two obtuse angles are also congruent. The angle sum of any quadrilateral is 360 degrees, so the two obtuse angles sum to $ 360 \\minus{} 2 \\times 40 \\equal{} 280$ degrees, so one obtuse angle is 140 degrees.[/hide]",
  "Solution_2": "[quote=\"math_explorer\"][hide]There are two acute angles and two obtuse angles in an isosceles trapezoid; the two acute angles are congruent, and the two obtuse angles are also congruent. The angle sum of any quadrilateral is 360 degrees, so the two obtuse angles sum to $ 360 \\minus{} 2 \\times 40 \\equal{} 280$ degrees, so one obtuse angle is 140 degrees.[/hide][/quote]\r\n\r\n\r\nNice explanation.  :idea:",
  "Solution_3": "All you need to do is that the consecutive angles are supplemantary. So the other angle must be 140 degrees."
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "pyramid"
  ],
  "Problem": "Hi!!\r\nI have a problem to solve by tomorrow, please help me :!: \r\nYou have a cube whose sides measure 1 cm and you section it with a plan which passes in the vertexs A, B,C\r\n(look at the attachment). I need to find out the volume of the piramyd it originates from sectioning the cube.\r\nthanks for your answers :)",
  "Solution_1": "[hide=\"EDIT Answer\"]\n$V= \\frac{ah}{3}$\nArea of the base\n$a= \\frac{\\sqrt{2}^{2}}{2}\\rightarrow a=1$\nHeight of Pyramid\n$h=1^{2}-l^{2}$\n$l= \\frac{{\\sqrt{2}}^{2}-\\left( \\frac{\\sqrt{2}}{2}\\right)^{2}}{2}\\rightarrow l=\\frac{3}{4}$\n$h=\\frac{7}{16}$\nVolume\n$V=\\frac{1 \\cdot \\frac{7}{16}}{3}\\rightarrow V= \\frac{7}{48}$\n[/hide]",
  "Solution_2": "thanks!!\r\ncould you explain me why?",
  "Solution_3": ":|  :blush: I am sorry I did it wrong.\r\nI will post a correction.",
  "Solution_4": "Do your own homework."
}
{
  "Tag": [],
  "Problem": "Last month sue made $\\$$200 by baby sitting for 12 hours and tutoring for 20 hours.  This month she worked the same number of hours at each job as she did last muchm but she increased her babysitting fee by $\\$$2 per hour and her turtoring fee by $\\$$1 per hour.  How much did se earn this month?",
  "Solution_1": "[hide]$12x + 20y = 200$\n$12(x + 2) + 20(y + 1) = 12x + 24 + 20y + 20 = 200 + 24 + 20 = 244$[/hide]",
  "Solution_2": "$244$\r\n\r\nsorry idonthave time toprove it :( .",
  "Solution_3": "[hide]12(x+2)+20(y+1)= \n12x+24+20y+20=\n200+24+20 = \n244[/hide]",
  "Solution_4": "sorry I can't understand muchm or se\r\n[hide]200 + 12 x 2 + 20 x 1\n= 200 + 24 + 20\n= [b]244[/b][/hide]",
  "Solution_5": "[hide]\n244, 12*2=24 is the extra that she got for baby sitting and 20*1=20 for tutoring.\n200+24+20=244[/hide]"
}
{
  "Tag": [
    "invariant",
    "modular arithmetic",
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "1994 lamps which can be switched single threaded are arranged in a circle and initially switched on. At random the lamps' state is changed (on-> off, or off -> on) by the following two rules: \r\n\r\n1.) four consecutive lamps,\r\n\r\n2.) among five lamps all states are changed except the middle one.\r\n\r\nIs it possible to reach a state where all lamps are switched out ?",
  "Solution_1": "please explain the rules again\r\nu haven't said nothing about those 4\r\nthose five, how are they chosen, and what does middle mean.",
  "Solution_2": "I think it's pretty clear: the moves you can perform are the following:\r\n\r\na) You can choose any four lamps in consecutive positions on the circle and change switch all of them (turn a lamp on if it's off and vice-versa);\r\nb) You can chose any five lamps in consecutive positions on the circle and switch the first two and the last two (the \"middle\" remains unchanged).\r\n\r\nYou choose these groups of lamps randomly (orl actually mentioned that).",
  "Solution_3": "Number the lamps starting from one of them and going around the circle. Now regard the indices $\\pmod 4$. For a certain configuration of lamps, consider $s$ to be the sum of the numbers of the lamps which are on $\\pmod 4$. We can see that $s$ is always odd, so we can't have $0$ lamps which are on.",
  "Solution_4": "One sees that an operation repeated twice cancels each other\r\nConsider the following \r\nxxxx\r\n..xxxx\r\n.xxxx=\r\nx.xx.x\r\nand\r\nxx.xx\r\n.xx. xx=\r\nx.xx.x\r\n\r\nSo we see that two neighbour operations of the second type can be replaced by operations of the first type\r\nSince two operations cancel each other we see\r\n$O_iO_j=O_iO_{i+1} O_{i+1}O_{i+2} \\cdots O_{j-1}O_{j}$.\r\n$O_i$ is the operation of second type  applied with lamp #i \r\nHence every two operations of the second type can be replaced by operation of the first type.\r\nHence we must so\\\\look just at two cases:\r\n1. All operations are of first type\r\n2. All are of the first type execept one.\r\n\r\nTo solve the first case denote the opqrations $a_1\\cdots a_n$ where $a_i$ is the \r\nleftmost square changed at operation #i.\r\nSince every lamp is changed odd number of times one gets \r\n$a_i$ and $a_i-1$ are changed the same # of times which means $a_i-4$ also belong to $a_1, \\cdots .. an$. Now we see that the operations cover $4k$ lamps which cannot be true since $1994=4k+2$.\r\nAnalogously one can handle the second case.",
  "Solution_5": "Here is a brief outline of my solution, it shares some resemblance with grobber's:\r\n\r\nLabel each lamp Ai (i=1,2,...,1994), now assign Ai=0 if lamp is on and Ai=1 if lamp is off. After each operation (either i) or ii)), calculate S=Sum (A_i after - A_i before)^2 over all 1994 Ai. Observe that S changes by 4 after each operation. Initially, S is, by definition, 0 (multiple of 4). So an easy contradiction argument follows because 1994 is not divisible by 4.\r\n\r\nNext time, why not change 1994 to 2004 for some modernization, eh?  :D",
  "Solution_6": "But with the present conditions, that problem (with $2004$) would be trivial :).",
  "Solution_7": "Yes. In the way stated there it's a quite stupid problem. Either 4 $|$ n, then it's trivially possible, either 4 $\\not|$ n, then it's trivially impossible",
  "Solution_8": "I see that al.M.V.'s solution is better than mine: for me it was crucial that $1994=2 \\pmod 4$.",
  "Solution_9": "grobber, your solution is just as equally effective.  :)"
}
{
  "Tag": [
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "Let $x,y,z>0$ such that:$x+y+z=1$.Prove that:\r\n\\[\\frac{xy}{1+z}+\\frac{yz}{1+x}+\\frac{xz}{1+y}\\leq \\frac{1}{4}\\]",
  "Solution_1": "It resembles Singapore TST 2004.\r\nxy/(1+z)=xy/((x+z)+(y+z)) now by AM-HM :1/a+1/b>=4/(a+b) putting a=x+y,b=y+z we get \r\nxy/((x+z)+(y+z))=<1/4*xy*(1/(x+z)+1/(y+z))\r\nyz/((x+y)+(x+z))=<1/4*yz*(1/(x+y)+1/(x+z)) \r\nzx/((y+z)+(y+x))=<1/4*xy*(1/(y+z)+1/(y+x)) \r\nnow adding these inequalities LHS=<1/4*(x+y+z)."
}
{
  "Tag": [],
  "Problem": "I have 300 or so of these so if you want more help improving your logic, PM me and I'll type up some more problems.\r\n\r\nOn the show Quiz Kids, five contestants (Males: Caleb, Dean, and Paul; Females: Megan and Shirley) were questioned about pop music each from a different decade (40s, 50s, 60s, 70s, or 80s). In the next round, they each answered questions about a different 20th century figure. (Albert Einstein, Charles Lindbergh, Dwight Eisenhower, Pablo Picasso, and Sigmund Freud) Each contestant ended up with a different # of points and a different prize (Bomber Jacket, Cutlery, Dinner for Two, Snowboard, and a Vacation) Determine the decade and 20th century figure each contestant answered questions about, the number of points, and the prize they went home with.\r\n\r\nClues:\r\n1. The three men are the ones who answered questions about Albert Einsten, the one who won the snowboard, and the one who scored exactly 5 points. The two women are the one who answered questions about 80s music and the one who won a vacation.\r\n2. The three people who answered questions about Charles Lindbergh, Dwight Eisenhower, and Pablo Picasso are the one who discussed music of the 40s, the one who won a set of cutlery (who scored 30 points), and Shirley, in some order.\r\n3. The sum created by adding 40 points to Paul's score is equal to either Megan's or Shirley's score. The difference created by subtracting Paul's score from 40 is equal to either Megan's or Shirley's score. Paul scored exactly 10 fewer points that the one who answered questions about 60s music.\r\n4. Shirley didn't answer questions about 70s music. Caleb (who isn't the one who answered questions about Sigmund Freud) is not the one who won dinner for two at a swanky downtown restaurant.\r\n5. The one who discussed 70s music (who didn't win a snowboard) ended the game with exactly 20 fewer points that the one who answered questions about Lindberg. The one who won the bomber jacket didn't answer questions about Picasso.",
  "Solution_1": "Suggestion: slap Hide tags around your clues.",
  "Solution_2": "Suggestion: the clues are necessary to solve the problem - without all of them, you can't solve it"
}
{
  "Tag": [
    "ratio",
    "geometry",
    "trigonometry",
    "quadratics",
    "algebra",
    "quadratic formula"
  ],
  "Problem": "Find $k$ if P,Q,R, and S are points on the sides of quadrilateral ABCD so that $\\frac{AP}{PB}=\\frac{BQ}{QC}=\\frac{CR}{RD}=\\frac{DS}{SA}=k$, and the area of quadrilateral PQRS is exactly 52% of the area of quadrilateral ABCD.",
  "Solution_1": "[hide=\"A (well-known, maybe) formula says that\"]\n\\[\\frac{ [PQRS] }{ [ABCD]}= \\frac{k^{2}+1}{(k+1)^{2}}\\]\n\n[b]Try to prove this on your own.[/b] The solution is gives is $\\boxed{k= \\frac{2}{3}}$ or equivalently $\\boxed{k= \\frac{3}{2}}$. [/hide]",
  "Solution_2": "Never seen that formula before. :D Here's how I solved it.\r\n\r\n[hide=\"Areas\"]Let the area of the quadrilateral $ABCD$ be $x.$ Also, let $AB=a,$ $BC=b,$ $CD=c,$ and $DA=d.$ It follows that $AP=\\frac{ka}{k+1},$ $PB=\\frac{a}{k+1},$ and etc.\n\nNote that\\[ad\\sin A+bc\\sin C= 2x\\]\nand\\[ab\\sin B+bc\\sin D=2x.\\] \n\nAdditionally,\n\\begin{eqnarray*}2(48\\%) &=& (AP)(AS)\\sin A+(BP)(BQ)\\sin B+(CQ)(CR)\\sin C+(DR)(DS)\\sin D \\\\ &=& \\frac{ka}{k+1}\\frac{d}{k+1}\\sin A+\\frac{kb}{k+1}\\frac{a}{k+1}\\sin B+\\frac{kc}{k+1}\\frac{b}{k+1}\\sin C+\\frac{kd}{k+1}\\frac{c}{k+1}\\sin D \\\\ &=& \\frac{k}{(k+1)^{2}}(ad\\sin A+ab\\sin B+bc\\sin C+cd\\sin D)\\\\ &=&\\frac{k}{(k+1)^{2}}(4x). \\end{eqnarray*}\nHence, $\\frac{96}{100}=\\frac{k}{(k+1)^{2}}(4x)$ and simple quadratic formula yields $k=\\frac{3}{2}, \\frac{2}{3}.$\n\n[/hide]",
  "Solution_3": "The way I solved this when I first saw out of the AoPS book used a little cheating... since it holds for any quadrilateral, it holds for a square.\r\nLet the big square have area 25 so the little has area 13. Clear, the sides of the big are 5 by 5, and we find that AS to SD is 3 to 2 since 3^2+2^2=13.\r\nThe result follows :P",
  "Solution_4": "uhh...that formula on teh noah sheets was basically copying the problem here...so i suppose if you mean this is a well-known problem...which i think it is, then fine, but this is obviously a result that you would just prove on a contest",
  "Solution_5": "I think it's also pretty easy to use vector/cross products? You should get the 'well known formula' mentioned earlier."
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "sphere"
  ],
  "Problem": "I'm just wondering, do people at AOPs play trading card games like Magic, I played magic back in middle school with the same group of people I did math team with, but then we all kinda drifted away from it.  What about you people?",
  "Solution_1": "I have pokemon which I play with my sister and a couple friends.",
  "Solution_2": "Wow! I play Magic. Used to play Pokemon and Yu-gi-oh",
  "Solution_3": "What is the point of these pieces of paper that have pictures on them?",
  "Solution_4": "lol, what's the point of those large rectangular pieces of grass with lines painted on them?  or those strange boards with black and white squares on them and funny pieces on top. or for that matter --- does it matter if we can prove that if you have 6 points on a plane (no 4 coplanar) that you can connect 3 with one color and the other 3 with another color and one passes thru the interior of the other.\r\n\r\nSeriously tho CTG's expand your mind in a way no other game does, besides the fact that they're SO much more fun then a limiting game like chess or a physical game like football. ;)",
  "Solution_5": "I consider TGC's very addictive games.",
  "Solution_6": "[quote=\"NAS\"]lol, what's the point of those large rectangular pieces of grass with lines painted on them?  or those strange boards with black and white squares on them and funny pieces on top. or for that matter ---does it matter if we can prove that if you have 6 points on a plane (no 4 coplanar) that you can connect 3 with one color and the other 3 with another color and one passes thru the interior of the other.\n\nSeriously tho CTG's expand your mind in a way no other game does, besides the fact that they're SO much more fun then a limiting game like chess or a physical game like football. ;)[/quote]\r\n\r\nThen how come chess has survived so long. And so many books, strategies been developed.",
  "Solution_7": "I used to play Pokemon, switched over to Yu-gi-oh!\r\n\r\n[i]Moderator note: Watch the language, Tare.[/i]",
  "Solution_8": "Realistically, CCGs are far more limiting than chess, as (in most cases), given two CCG players with roughly the same experience, the one with the more expensive deck will win. For example, assuming you're into magic... can you imagine a Storm deck with power 9 losing to a $5 stompy? Such a situation would never happen in chess, where money doesn't inevitably create an unfair playing ground. In general a good game of chess requires more thinking, as does finding a solution to a math proof. \nI'm being quite hypocritical here since I've spent about $300 between magic in middle school and pokemon in fifth grade... But I was mostly undefeatable in both games against other players with years more experience who hadn't spent as much on their decks, which further proves my point.",
  "Solution_9": "I have to disagree, (I have to say I love chess, and math obviously) the key difference between chess and magic is that in magic half the fun (at least for me) is that building the decks.  The other part is the 'on your feet thinking' you draw seven cards, sure you have (or should have) a very good idea what kind of cards you're going to get but you never know.  You also (at least in the beginning) have no idea what kind of cards your opponent has.  In short, it's the before game creativity, and the in game randomness.  With chess, on the other hand, everything is out in the open and predictable aka in every chess game you play, you have exactly 20 opening moves.\r\nNot to be too poetic, but the reason I like chess is the reason I like this site.  The reason I like Magic is the reason I like music.\r\n\r\n[quote] can you imagine a Storm deck with power 9 losing to a $5 stompy [/quote]\r\nSure, it's certainly not very likely but that's the thing about Magic that I love(also the formats).  The Stompy deck has a better chance of winning that game by a person that's never played magic before then someone that's never played chess playing with Bobby Fischer.",
  "Solution_10": "Do you people who play magic have like really, really good cards? If you got something really good, please post it. My cards aren't very good.",
  "Solution_11": "Umm... probably the rarest card I have is an Alpha Sierra Angel and/or and alpha Nightmare",
  "Solution_12": "My bro's rarest card is planeshift Draco which isn't that rare but it's pretty good.",
  "Solution_13": "Alpha Serra Angel!!!!!!!!!!!!!!!!!!!!!!! WHOA!!!!!!!!!!!!!!!!!\r\nMine is Akroma, Angel of Wrath.\r\n\r\nMTG rocks!!!\r\n\r\nI have a counter control deck and a seismic assault deck which throws lands - how about you guys?",
  "Solution_14": "I agree with PI(up there aka 3.14159265358979 etc). Money is a major factor in Magic -  my friend spends lots on Magic and builds a great deck, but I don't and I have a mediocre deck. However, there is the point that the win loss is 50/50.\r\n\r\nEveryone that plays magic w/me is also on my MC team - strange. It must take smart people then to build a good deck with little money.\r\nWith a few thousand different cards - there is always a good undiscovered strategy awaiting.",
  "Solution_15": "I used to play magic in the fifth grade....but its way to expensive so I dropped out in the sixth.  Chess limited! Pah! Magic is more limited because you actually need to spend money on it.  Its more like a \"who has the most money to spend\" game than chess where everyone is equal and takes lots of thinking to beat your opponents.",
  "Solution_16": "WoW!! Akroma is good!\r\n\r\nAnybody have a dragon deck?\r\n\r\nUhh... I got an Elf, a Mercanary and a spectrum deck. They're pretty bad but oh well...",
  "Solution_17": "MAGIC rules!!!\r\nI have several decks, and I luv arcbound creatures-DArksteel\r\nCheap combination-Broodstar, Mycosynth Lattice, and DARKSTEEL FORGE!!!!=\r\nSuperpowered indestructible, giant bashing BROODSTAR",
  "Solution_18": "I got 4 Darksteel Forges, 3 Memnarchs, a Phage the Untouchable, a Broodstar, 5 Mycosinth lattices, and a whole ton of other rares.\r\nI build stomp decks that are semitribal with lots of support. AKA Darksteel Forge: Artifacts that you control are indestructible and Mycosynth Lattice:  All permanents become artifacts in addition to other types. All spells are colorless. Players may spend mana as though it were mana of any color and Broodstar: Power and toughness equal to the number of artifacts you control. Has flying too!\r\nArcbound creatures rule with Fangrens.",
  "Solution_19": "I have a pro-black (turns everything black, courtesy of Darkest hour, then gives me super pro black- Multiple \"Sphere of Grace\"s\r\n, an Arcbound(modular) deck as mentioned above, and  black red nightmare burn deck.",
  "Solution_20": "Dragon decks are good if you use a lot of squirrel tokens and day of the dragons!",
  "Solution_21": "I had a beta/alpha (can't remember) Ancestral Recall..... I have a whole bunch of cards because I used to play it in 6th grade.  I guess the best deck I ever made never consisted of expensive cards, just cheap ones.\r\n\r\nHorseshoe Crab + Hermetic Study",
  "Solution_22": "wow...did u sell the ancestral recall?",
  "Solution_23": "I believe trading card games are fun and I would love to play Magic if I had my own deck, but I'm not willing to spend my own money on them. Pokemon battling is fun too but doesn't require as much ingenuity and deck modifying skills as Magic.  Yet, since I have thousands of Pokemon in my closet, I would use them to battle if there were some Pokemaniacs out there who aren't afraid to admit Pokemadness.",
  "Solution_24": "Well the ancestral recall was in fair condition... I bought it for 20 dollars one day and I got 50 dollars the next! Ka ching!",
  "Solution_25": "Wow. Thats lucky. :lol:",
  "Solution_26": "pokemon OWNS! the game is pretty good, and its not \"in fasion\" so the cards are really really cheap and you can have a good deck without paying as much (those cost a dollar) as \"in fasion\" yugioh and magic decks. (those cost like 12 dollars). \r\nsum1 already voted in my account.  \r\n\r\nwhats wierd about tcgs are that popular tcgs are always denounced by some other group",
  "Solution_27": "i used to play magic, pretty seriously actually about 7 years ago, pokemon was either shortly before or shortly after (I think before)... I might still enjoy playing them, except I have more important things that consume my time these days.",
  "Solution_28": "I dont see the point in these games but I do know how to play Yu-gi-oh and pokemon for what its worth. I dont know any of the cards and Im not very good at it one of my best friends lil brothers taught me how to play. I still see no point why not just play regular cards there is so much more u can do with them. Or why not have baseball trading cards which totally rock. My brother collects football, baseball, and nascar trading cards why cant we just go back to those?",
  "Solution_29": "Does anyone who plays MTG have a Serra Avatar? Those things are like $30.",
  "Solution_30": "14 year BOOOMP\nAwesome! Another 3",
  "Solution_31": "[quote=mikimoto12]14 year BOOOMP\nAwesome! Another 3[/quote]\n\naaaaaaaarggggghhhhh"
}
{
  "Tag": [
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "Let $\\large a_0;a_1;. . . ;a_n$ be a sequence satisfying: $\\large a_0=1; a_{n}=a_{[7n/9]}+a_{[n/9]}$ for all n=1;2;. . .\r\nProve that there exist a positive integer k with $\\large a_k < \\frac{k}{2001!}$",
  "Solution_1": "The expectation is that $a_n$ is within a small (e.g., bounded or logarithmic) factor of ${n^c}$ for some $c < 1$.   There is only one possible value of $c$, that is the one satisfying  $9^c = 7^c + 1$.  Proving it seems like a difficult problem about uniform distribution."
}
{
  "Tag": [
    "ceiling function",
    "graph theory",
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "Given a complete graph on $ k\\ge 3$ vertices whose edges are colored in black and white s.t. every triangle has an even number of black edges (either $ 0$ or $ 2$). Prove that we can find a complete subgraph on $ \\left\\lceil\\frac {k}{2}\\right\\rceil$ vertices all of whose edges are white .\r\nit was in [url]http://www.mathlinks.ro/viewtopic.php?p=1652182#1652182[/url] but i couldn`t understand the proof .\r\n(sorry for my bad English) :D",
  "Solution_1": "grobber wrote\r\n[quote]It\u2019s easy to see that the partial graph generated by all the vertices and the black edges is a complete bipartite graph. If $ A,B$ is its bipartition, the subgraphs induced on $ A,B$ are both complete graphs with white edges only, and at least one of $ |A|,|B|$ is $ \\ge\\left\\lceil\\frac k2\\right\\rceil$, so we\u2019re done.[/quote]\nTake two vertices $ x, y$, joined by a black edge, and put them in two classes $ A,B$. Any third vertex $ z$ is joined to exactly one of $ x,y$ by a black edge - put it in the opposite class. Now check each class cannot contain black edges, so they are complete subgraphs with white edges only.\n\nBy the way, you made a typo when copying. You wrote\n[quote]even number of black edges (either $ 0$ or $ 1$)[/quote]\r\n$ 1$ is odd. It should be (either $ 0$ or $ 2$), like in the original.",
  "Solution_2": "thank you very much !!! (I edited my typo)"
}
{
  "Tag": [
    "calculus",
    "integration",
    "calculus computations"
  ],
  "Problem": "I just need help getting started mainly with problem # 1.  However if you find this problem to be a challenge go ahead and have at it.\r\n\r\n\r\nAir resistance is a force that acts in the direction opposite to the motion and increases in magnitude as velocity increases, let us assume at least initially that air resistance r is proportional to the velocity: r = pv, where p is a negative constant. suppose a ball of mass m is thrown upward from the ground. The net force f on the ball is F = r \u2013 mg (the direction of the force r is downward (negative) when the ball is traveling upward and the direction of r is upward when the ball is traveling upward.)\r\n\r\n1. Use the net force equation and the fact that F = ma to write a differential equation for the ball\u2019s velocity.\r\n\r\n2. Assume m = .5kg and p = .1. Make a direction field for the differential equation and sketch a solution of the initial value problem v(0) = 50 m/s.\r\n\r\n3. Solve the initial value problem algebraically. Hint: be sure to take the constant of integration into account.\r\n\r\n4. Find an equation of the height of the ball at time t.\r\n\r\n5. When does the ball reach the apex of its trajectory? When does the ball land?\r\n\r\n6. Does it take the ball longer to come up or come down?",
  "Solution_1": "From $ma=pv-mg$ it follows that $\\frac{dv}{dt}=\\frac{p}{m}v-g$."
}
{
  "Tag": [
    "ratio",
    "LaTeX",
    "AMC"
  ],
  "Problem": "Points $ B$ and $ C$ lie on $ \\overline{AD}$. The length of $ \\overline{AB}$ is $ 4$ times the length of $ \\overline{BD}$, and the length of $ \\overline{AC}$ is $ 9$ times the length of $ \\overline{CD}$. The length of $ \\overline{BC}$ is what fraction of the length of $ \\overline{AD}$?\r\n\r\n$ \\textbf{(A)}\\ \\frac{1}{36} \\qquad\r\n\\textbf{(B)}\\ \\frac{1}{13} \\qquad\r\n\\textbf{(C)}\\ \\frac{1}{10} \\qquad\r\n\\textbf{(D)}\\ \\frac{5}{36} \\qquad\r\n\\textbf{(E)}\\ \\frac{1}{5}$",
  "Solution_1": "[hide]Since it doesn't matter the length since they are just ratios i took the length to be 10. and so BD = 2. and CD = 1 so BC = 1 and so BC/AD = $ \\frac{1}{10}$ $ \\fbox{\\mbox{C}}$ sry i am too tired to do it in Latex now.[/hide]",
  "Solution_2": "[hide]Let BD = x and CD = y.\n\nThus,\nAB = 4x and AC = 9y\nAD = AB + BD = 4x + x = 5x\nAD = AC + CD = 10y\nAD = 10y = 5x\nx=2y\n\nBC = AC - AB = 9y - 4x = 9y - 8y = y\nBC/AD = y/(10y) = 1/10[/hide]"
}
{
  "Tag": [
    "calculus",
    "integration",
    "trigonometry",
    "calculus computations"
  ],
  "Problem": "Find $ \\int_0^{2\\pi}\\frac{d\\,x}{\\sqrt{25\\plus{}24\\cos x}}$.",
  "Solution_1": "$ \\int_{x\\equal{}0}^{2\\pi} \\frac{dx}{\\sqrt{25\\plus{}24\\cos x}} \\equal{} \\frac{2\\pi}{7} \\sum_{k\\equal{}0}^\\infty \\left(\\frac{(2k\\minus{}1)!!}{(2k)!!}\\right)^2 \\left(\\frac{48}{49}\\right)^k$."
}
{
  "Tag": [
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "let  $\\mu^{*}$ be an outer measure and $\\mu^{-}$ be the  restriction\r\nof $\\mu^{*}$ on $\\mu^{*}$ measurable sets. prove that $\\mu^{-}$ is\r\nsaturated.",
  "Solution_1": "I don't know what a saturated measure is. Maybe you mean \"complete\" (that would be \"$\\mu^{-}(A)=0\\implies$ all subsets of $A$ are $\\mu^{-}$ measurable\").",
  "Solution_2": "let$(X,\\mathbf{B},\\mu)$be a measure space\r\n we say  $E$ subset of $X$  is locally measurable\r\n  if for each $B\\in\\mathbf{B}$ with $\\mu(B)<+\\infty$ we have  $E \\cap B\\in\\mathbf{B}$\r\n\r\n\r\nwe say $\\mu$ is saturated if each locally measurable set be measurable\r\n ."
}
{
  "Tag": [
    "trigonometry",
    "geometry",
    "angle bisector",
    "geometry unsolved"
  ],
  "Problem": "$ D$ is a point on the edge $ BC$ of triangle $ ABC$ such that $ AD\\equal{}\\frac{BD^2}{AB\\plus{}AD}\\equal{}\\frac{CD^2}{AC\\plus{}AD}$. $ E$ is a point such that $ D$ is on $ [AE]$ and $ CD\\equal{}\\frac{DE^2}{CD\\plus{}CE}$. Prove that $ AE\\equal{}AB\\plus{}AC$.",
  "Solution_1": "Who know the solution of this problem?  :?:",
  "Solution_2": "[color=darkred][size=117][b]A nice and difficult (for who doesn't know the lower lemma) metrical problem ![/b][/size][/color]\r\n\r\n[quote=\"anonymous1173\"][color=darkred]$ D$ is a point on the edge $ BC$ of triangle $ ABC$ such that $ AD = \\frac {BD^2}{AB + AD} = \\frac {CD^2}{AC + AD}$. Denote \n\nthe point $ E$ is a point such that $ D\\in [AE]$ and $ CD = \\frac {DE^2}{CD + CE}$. Prove that $ AE = AB + AC$.[/color][/quote]\n[color=darkblue][b][u]Proof.[/u][/b] The relations from hypothesis suggest me to apply the well-known property :[/color]\n\n[quote][color=darkred][b][u]Lemma.[/u][/b] Let $ ABC$ be a triangle. Denote the point $ D\\in BC$ for which $ AD\\perp BC$ . \n\nSuppose $ D\\in (BC)$ . Then $ b^2 = c(a + c)\\Longleftrightarrow B = 2C\\Longleftrightarrow DC = DB + BA$ .[/color]\n\n[color=darkblue][b][u]Proof.[/u][/b] $ \\boxed {\\ b^2 = c(a + c)\\ }\\Longleftrightarrow b^2 - c^2 = ac\\Longleftrightarrow$ $ DC^2 - DB^2 = ac\\Longleftrightarrow$ $ \\boxed {\\ DC = DB + BA\\ }\\Longleftrightarrow$ \n\n$ b\\cos C = c + c\\cdot \\cos B$ $ \\Longleftrightarrow$ $ \\sin B\\cos C = \\sin C(1 + \\cos B)$ $ \\Longleftrightarrow$ $ \\tan\\frac B2 = \\tan C$ $ \\Longleftrightarrow$ $ \\boxed {\\ B = 2C\\ }$ .[/color][/quote]\r\n\r\n[color=darkblue]Denote $ \\|\\begin{array}{c} m(\\widehat {ABC}) = x \\\\\n \\\\\nm(\\widehat {ACB}) = y \\\\\n \\\\\nm(\\widehat {CED}) = z\\end{array}$ . Apply the upper lemma to the triangles :\n\n$ \\|\\begin{array}{cccc} \\triangle\\ ABD\\ : & BD^2 = AD\\cdot (AB + AD) & \\implies & m(\\widehat {BAD}) = 2\\cdot\\widehat {ABD}) = 2x \\\\\n \\\\\n\\triangle\\ ACD\\ : & CD^2 = AD\\cdot (AC + AD) & \\implies & m(\\widehat {CAD}) = 2\\cdot m(\\widehat {ACD}) = 2y \\\\\n \\\\\n\\triangle\\ CDE\\ : & DE^2 = CD\\cdot (CE + CD) & \\implies & m(\\widehat {DCE}) = 2\\cdot m(\\widehat {CED}) = 2z\\end{array}$ .\n\nObserve that $ A + B + C = 180^{\\circ}$ $ \\Longleftrightarrow$ $ (2x + 2y) + x + y = 180^{\\circ}$ $ \\Longleftrightarrow$ $ \\boxed {\\ x + y = 60^{\\circ}\\ }$ ,\n\n$ m(\\widehat {ABD}) + m(\\widehat {BAD}) = m(\\widehat {CED}) + m(\\widehat {DCE})$ $ \\Longleftrightarrow$ $ x + 2x = z + 2z$ $ \\Longleftrightarrow$ $ \\boxed {\\ z = x\\ }$ and \n\n$ ABEC$ is cyclically. Thus, $ AE = AB + AC$ $ \\Longleftrightarrow$ $ \\sin \\widehat {ACE} = \\sin\\widehat {ACB} + \\sin\\widehat {ABC}$ $ \\Longleftrightarrow$\n\n$ \\sin (2x + y) = \\sin y + \\sin x$ $ \\Longleftrightarrow$ $ \\sin (2x + y) - \\sin y = \\sin x$ $ \\Longleftrightarrow$\n\n$ 2\\sin x\\cos (x + y) = \\sin x$ $ \\Longleftrightarrow$ $ \\cos (x + y) = \\frac 12$ $ \\Longleftrightarrow$ $ x + y = 60^{\\circ}$ , what is truly.[/color]",
  "Solution_3": "We have $ AD=\\frac{BD^2}{AB+AD}\\Rightarrow AB\\cdot AD = BD^2 - AD^2 $.\n\nLet the circle with center $D$ and radius $DA$ cut $BA$ at $X$. \nThe power of $B$ is $BD^2 - DA^2 = BA \\cdot BX$.\nSo $BX = AD = XD \\Rightarrow \\angle BAD = 2\\cdot \\angle ABD = 2\\alpha$.\nSimilarly, $\\angle DAC = 2\\cdot \\angle ACD = 2\\beta$.\nWe have $\\alpha + \\beta + 2\\alpha + 2\\beta = 180^\\circ \\Rightarrow \\alpha + \\beta = 60^\\circ$.\nAgain similarly, and since $\\angle EDC = \\angle BDA$, $\\angle DCE = 2\\cdot\\angle DEC = 2\\alpha$.\nSince, $\\angle BCE = \\angle BAE = 2\\alpha$, $B,A,C,E$ are concyclic.\n\nLet $H$ be foot of altitude from $B$ to $EC$.\nLet $BH$ and $AE$ meet at $F$.\nSince $\\angle BAC = 120^\\circ$, $\\angle BEH = 60^\\circ$.\nLet $G$ be on $EC$ such that $\\triangle EBG$ is equilateral.\nLet $BG$ meet $AE$ at $I$. \nSince $F$ is on the angle bisector of $\\triangle EBG$, $EF=FG$ and $\\angle IFG = 2\\angle FEG = 2\\alpha$. So $FG \\parallel AB$.\n\nBy Sin Law, $BI/IG = AB/AC$. \nIf you want synthetic proof: $EI$ cuts $(BEG)$ at $J$. $\\triangle JBG \\sim \\triangle ABC$. $JI$ is angle bisector of $\\angle BJG$. So $AB/AC=BJ/JG=BI/IG$\n\n$AB \\parallel FG \\Rightarrow AB/FG = BI/IG = AB/AC \\Rightarrow FG = AC = EF$.\nWe have $\\angle EFH = \\angle BFA = 90^\\circ - \\angle AEC = 90^\\circ - \\alpha$ and also $\\angle BAF = 2\\alpha$. So $AB=AF$.\n$AE=AF+FE=AB+AC$. $\\blacksquare$"
}
{
  "Tag": [
    "modular arithmetic"
  ],
  "Problem": "A collection S of integers is defined by the following three rules: (I) 2 is in S: (II)for every x in S, 3x and x+7 are also in S; (III) no integers except those defined by rules I, II are in S. What is the smallest integer greater than 2004 which is NOT in S?",
  "Solution_1": "What is the cardinality of S?",
  "Solution_2": "Since the title of this topic is \"Another Hard Question\", I think I screwed this up...\r\n\r\n[hide]Well first, you can notice that 2004 is one of the numbers that fall under the category of \"x+7\". (since $ 2\\plus{}7\\cdot286\\equal{}2004$)\nThen, 2005 Doesn't fall under the \"x+7\" category. Also, $ \\dfrac{2005}{3}\\equal{}\\text{Integer}$. Hence, $ \\boxed{2005}$ is the answer?[/hide]",
  "Solution_3": "If we look at this mod 7, we quickly see that all numbers of the form $ 2 \\mod 7$ fit are in $ S$. Also if $ 2$ is in $ S$, then $ 6$ is in $ S$ and all numbers in the form $ 6 \\mod 7$ are in $ S$ as well. Then all numbers in the form $ 4 \\mod 7$ ($ 18 \\equal{} 4 \\pmod 7$) are in $ S$, and so are $ 5 \\mod 7$, and $ 1 \\mod 7$, and $ 3 \\mod 7$, and back to $ 2 \\mod 7$. So we see that all numbers are in $ S$ except some smaller numbers and eventually just the multiples of 7. So our answer is the smallest multiple of 7 greater than 2004, which is $ \\boxed {2009}$."
}
{
  "Tag": [
    "inequalities",
    "inequalities unsolved",
    "acute"
  ],
  "Problem": "if a,b,c are the sides of a right triangle prove that $ c^n > a^n \\plus{} b^n$.$ n>2$",
  "Solution_1": "I suppose $ c$ is the hypothenuse of the triangle. :maybe: \r\n\r\nIs $ n>2$? For $ n\\equal{}1$ or $ n\\equal{}2$ for example your inequality does not hold...",
  "Solution_2": "[quote=\"aadil\"]if a,b,c are the sides of a right triangle prove that $ c^n > a^n \\plus{} b^n$[/quote]\r\n\r\n\r\nfirst  we have $ p > 1$ \r\n $ (x \\plus{} y)^p > x^p \\plus{} y^p$\r\n\r\n$ x^p \\plus{} y^p \\equal{} (x \\plus{} y)^p \\minus{} (x \\plus{} y)(x \\plus{} y)^{p \\minus{} 1} \\plus{} x^p \\plus{} y^p \\equal{} (x \\plus{} y)^p \\minus{} x((x \\plus{} y)^{p \\minus{} 1} \\minus{} x^{p \\minus{} 1}) \\minus{} y((x \\plus{} y)^{p \\minus{} 1} \\minus{} y^{p \\minus{} 1}) < (x \\plus{} y)^p$\r\n\r\nTherefore,\r\n\r\n$ c^n \\equal{} (c^2)^{\\frac {n}{2}} \\equal{} (a^2 \\plus{} b^2)^{\\frac {n}{2}} > a^n \\plus{} b^n$\r\n\r\nwhile $ n > 2$ and n is positive reals",
  "Solution_3": "$ n>2$.sorry i didnt mention before"
}
{
  "Tag": [],
  "Problem": "How many positive integers $k$ leave a remainder of 2 when 101 is divided by $k$?",
  "Solution_1": "[hide]If the remainder when divided by $k$ is 2, then $k$ must be a factor of $101-2=99$.  The factors of 99 are: ${1,3,9,11,33,99}$.  However, 1 does not work, so the only values of $k$ which satisfy are ${3,9,11,33,99}.$\n\nThe number of values of $k$ which satisfy are $\\boxed{5}$\n[/hide]"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "calculus",
    "derivative",
    "induction"
  ],
  "Problem": "How can we prove that a root a of a polynomial f(x) has multiplicity m iff f(a) = f'(a) = f''(a) = f'''(a) = ... = (m-1)th derivative of f(x) = 0 ?",
  "Solution_1": "One way is to use Taylor's theorem, which for polynomials goes like this.  Let f be a polynomial of degree n, and let a and x be numbers.  Then f(x) equals the sum (for k from 0 to n) of f<sup>(k)</sup>(a) (x - a)<sup>k</sup> / k!.\r\n\r\nDo you see how Taylor's theorem implies what you want?",
  "Solution_2": "[quote]Do you see how Taylor's theorem implies what you want?[/quote]\n\nNot really...but that might simply be because I am not too familiar with the Taylor Series and am missing something obvious.\n\n[quote]Then f(x) equals the sum (for k from 0 to n) [/quote]\r\n\r\nIsn't it from 0 to infinity (at least this is what my calculus textbook says)?",
  "Solution_3": "Well, you could think of the sum as an infinite sum, but for a polynomial of degree n, all the derivatives past the nth derivative are 0.",
  "Solution_4": "Hint:  Prove that if g'(x) has multplicty j of root k, and g(k)=0, then g has j+1 roots.",
  "Solution_5": "[quote]Well, you could think of the sum as an infinite sum, but for a polynomial of degree n, all the derivatives past the nth derivative are 0.[/quote]\n\nYes, but aren't we looking to show the opposite? We must show that f(a) = f'(a) = f''(a) = f'''(a) = ... = (m-1)th derivative of f(x) = 0, not that (m+1)th derivative of f(x) = (m+2)th derivative of f(x)  = ... = 0, which is what your statement suggested, no?\n\n[quote]Prove that if g'(x) has multplicty j of root k, and g(k)=0, then g has j+1 roots.[/quote]\r\n\r\nSo induction, eh?",
  "Solution_6": "You've asked two separate questions.  One was your original post.  The second was why my statement of Taylor's theorem had a finite sum.  My last post, about the derivatives of a polynomial being 0 past a certain point, was meant to answer your second question, not your first.",
  "Solution_7": "Oh, I see, you were commenting on my question \"Isn't it from 0 to infinity (at least this is what my calculus textbook says)?\" Anyways, thanks for the responses."
}
{
  "Tag": [
    "real analysis",
    "summer program",
    "Mathcamp"
  ],
  "Problem": "I got one point off for Round 3 Problem 5 because the grader said that I did not prove the link between the Dirichlet series and the Phi Function. But in the solution, I proved it:\r\n\r\n[img]http://img263.imageshack.us/img263/8661/prooffp3.jpg[/img]",
  "Solution_1": "Possibly the grader wanted you to prove that $ \\sum_{d | h} \\varphi(d) \\equal{} h$.",
  "Solution_2": "This looks copied from Wikipedia.",
  "Solution_3": "Don't worry, you're not the only one who lost a point for citing wikipedia...",
  "Solution_4": "But I'm not just citing it...I'm showing the proof, which is completely valid.",
  "Solution_5": "Yeah, but I easily could have copy-pasted it from Wikipedia as well, since I'm assuming you didn't come up with that yourself.",
  "Solution_6": "I did the exact same thing as you and got 1 point off. In my case, I didn't use Wikipedia at all, but used knowledge about Dirichlet convolutions that I learned at Mathcamp last year. I figured that that would be common knowledge, but apparently it wasn't  :o",
  "Solution_7": "Did you remember cite Wikipedia? I copied the proof and cited the URL, and I got a 5.",
  "Solution_8": "I forgot to cite wikipedia...but the grader didn't cite that as the reason for taking points off...he said i didn't prove it.",
  "Solution_9": "The rubric specifically said that citing wikipedia should result in taking off a point.",
  "Solution_10": "I got a commended 5 and I didn't even go through all of those steps.  I jumped from the end of 123456789's line 1 to the end of line 3 by just saying that we sum along a different index.  I did show a simple proof of Gauss's lemma (that is, $ \\sum_{d|k} \\phi(d) \\equal{} k$).  It's really quite easy if we consider reducing the fractions $ \\frac {1}{k}, \\frac {2}{k}, \\cdots, \\frac {k}{k}$.  But then again, I did come up with the \"Dirichlet convolution\" idea independently of an outside source."
}
{
  "Tag": [],
  "Problem": "[b]The Sagebrush student council has 6 boys and 6 girls as class representatives. Two committees, each consisting of 2 boys and 2 girls, are to be created. If no student can serve on both committees, how many different combinations of committees are possible?[/b]",
  "Solution_1": "[hide]$ \\binom{6}{2}\\cdot\\binom{6}{2}\\cdot\\binom{4}{2}\\cdot\\binom{4}{2}$\nThis equals $ 15*15*6*6$ or $ 8100$. Now we divide this by $ 2!$ and get $ \\boxed{4050}$. [/hide]",
  "Solution_2": "I got a completely diff answer\r\n\r\n[hide]\n\nso I got\n\n(6) *  (6)\n(2)     (2)\n\nso it is n!/(n-k)!k!\n\nor 6!/(6-2)!2!\n\nor 720/48\n\nthis simplifies to 15 and 15*15=\n\n[225][/hide]\r\n\r\nEdit: I realized that each committee had 2 boys and two girls, not one with 2 boys and one with 2 girls.",
  "Solution_3": "[quote=\"lokesh\"][hide]$ \\binom{6}{2}\\cdot\\binom{6}{2}\\cdot\\binom{4}{2}\\cdot\\binom{4}{2}$\nThis equals $ 15*15*6*6$ or $ 8100$. Now we divide this by $ 2!$ and get $ \\boxed{4050}$. [/hide][/quote]\r\n\r\nI agree with you on $ \\binom{6}{2}\\cdot\\binom{6}{2}\\cdot\\binom{4}{2}\\cdot\\binom{4}{2}$.\r\n\r\nBut why do you divide it by $ 2!$?",
  "Solution_4": "In the original count, the committee group |AB|CD| was counted as a different committee than |CD|AB|.",
  "Solution_5": "[quote=\"alanchou\"]In the original count, the committee group |AB|CD| was counted as a different committee than |CD|AB|.[/quote]\r\nOh, okay. Thanks."
}
{
  "Tag": [
    "USAMTS",
    "email"
  ],
  "Problem": "As much as I hate to admit it, I found myself scrambling to solve my final unsolved problem late last night (solution had eluded me for a good week at that point).  The good news:  I solved it.  The bad news:  I had roughly 3 minutes until the deadline for submissions, and, being overly hasty, sent an e-mail titled simply \"Round 3 Solutions,\" as opposed to \"Round 3 - username - ID#.\"  Looking at the submission guidelines afterwards (wonderful idea, right?), I saw that my speed-titling was certainly a bad idea, but I also did not want to submit a second set of solutions, since that could potentially make the situation worse.  My question is simply... am I screwed (perhaps rightly so, for wasting someone's _free_ time on something like this), or is there something that I should do to correct this?  My assumption is that whoever opens the e-mail will see the lack of information in the title and check the attached PDF (which takes time, and thus, I should have done for them, obviously).  However, if a script runs through this stuff, then I'm in trouble.\r\n\r\nWhatever happens, thanks for your time in advance.",
  "Solution_1": "My advice is to reply to your own e-mail, send it to the solutions address again, and attach the full title, explaining in the body your mistake.  I don't think you'll need to attach another set of solutions, as they'll see that your second e-mail is coming from the same e-mail address.  I've made this mistake before, and they've had no problem with that way of correcting it.",
  "Solution_2": "Humans go through it, not a script.  Don't worry about it unless you don't have a green checkmark next to Round 3 on the Scores page at the end of the week.",
  "Solution_3": "One thing you can always count upon in these forums are solid, quick responses to questions.  Thanks a bunch!",
  "Solution_4": "i always send with subject \"Round # Solutions\"\r\n\r\nis that bad?  :blush:  :blush:",
  "Solution_5": "It's not 'bad', but we would like you to follow the instructions regarding including identifying information.  You can find them here:\r\n\r\nhttp://www.usamts.org/Problems/U_ProblemsSoln.php",
  "Solution_6": "Not bad as in doomsday bad, as evidenced by the responses from Richard and Zanttrang, but I believe that it does make it easier on the one who runs through all of the e-mails if you have your ID number and name in the topic, as they ask for you to send them in that format on the submitting solutions page (http://www.usamts.org/Problems/U_ProblemsSoln.php).",
  "Solution_7": "ah ok.\r\n\r\ni had my USAMTS ID in the body of the email, but I'll change next time :)"
}
{
  "Tag": [],
  "Problem": "Alfreda hiked half her planned trip the first day;tiring she only covered one-third the remaining distance the second day; rain slowed her to one-fourth the remaining distance the third day, and the fourth day she hiked three miles, which was one-fifth of what she had left. How many miles did she hike during the four days?",
  "Solution_1": "[quote=\"ckck\"]Alfreda hiked half her planned trip the first day;tiring she only covered one-third the remaining distance the second day; rain slowed her to one-fourth the remaining distance the third day, and the fourth day she hiked three miles, which was one-fifth of what she had left. How many miles did she hike during the four days?[/quote]\r\n\r\n[hide]On the first day she hiked 1/2, the second she hiked 1/6, and the third she hiked 1/12. This leaves 1/4 left. \n3=x/20\nx=60 total miles\n\nSo she hiked a total of 48 miles.[/hide]"
}
{
  "Tag": [],
  "Problem": "For what value of $ x$ does $ \\frac89 \\equal{} \\frac{x}{x\\plus{}\\frac{x}{x\\plus{}x}}$?",
  "Solution_1": "[quote=\"GameBot\"]For what value of $ x$ does $ \\frac89 \\equal{} \\frac {x}{x \\plus{} \\frac {x}{x \\plus{} x}}$?[/quote]\r\n\r\n$ \\frac{8}{9}\\equal{}\\frac{x}{x\\plus{}\\frac{x}{x\\plus{}x}}\\equal{}\\frac{x}{x\\plus{}\\frac{1}{2}}$\r\n\r\n$ 8x\\plus{}4\\equal{}9x$\r\n\r\n$ x\\equal{}4$.",
  "Solution_2": "Go BOGTRO!"
}
{
  "Tag": [
    "inequalities proposed",
    "inequalities"
  ],
  "Problem": "Let a,b,c >0, a<1+b , b+c<1 . Prove\r\n$ 1\\ge 27abc(1\\plus{}b\\minus{}a)(1\\plus{}b\\minus{}c)(1\\plus{}a\\plus{}c)$",
  "Solution_1": "[quote=\"dragon1\"]Let a,b,c >0, a<1+b , b+c<1 . Prove\n$ 1\\ge 27abc(1 \\plus{} b \\minus{} a)(1 \\plus{} b \\minus{} c)(1 \\plus{} a \\plus{} c)$[/quote]\r\nIt 's wrong .Try $ a\\equal{}b\\equal{}c\\equal{}\\frac{1}{3}$ :wink:",
  "Solution_2": "[quote=\"quykhtn-qa1\"][quote=\"dragon1\"]Let a,b,c >0, a<1+b , b+c<1 . Prove\n$ 1\\ge 27abc(1 \\plus{} b \\minus{} a)(1 \\plus{} b \\minus{} c)(1 \\plus{} a \\plus{} c)$[/quote]\nIt 's wrong .Try $ a \\equal{} b \\equal{} c \\equal{} \\frac {1}{3}$ :wink:[/quote]\r\n\r\nSorry\r\n\r\nLet a,b,c >0, a<1+b , b+c<1 . Prove\r\n$ 1\\ge 27abc(1 \\plus{} b \\minus{} a)(1 \\minus{} b \\minus{} c)(1 \\plus{} a \\plus{} c)$"
}
{
  "Tag": [
    "limit",
    "integration",
    "function",
    "calculus",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Assume $ 1\\le p < \\infty.$ Suppose $ f_k,f\\in L^p$ such that $ f_k\\to f$ a.e. and $ \\|f_k\\|_p\\to\\|f\\|_p.$ Prove that $ \\|f_k \\minus{} f\\|_p\\to0.$",
  "Solution_1": "Here's a proof I have for $ p \\equal{} 1$ that I don't see how to make work for $ p > 1.$\r\n\r\nDecompose $ f$ and $ f_n$ into positive and negative parts (or positive and negative parts of real and imaginary parts). For the rest of the problem, assume that $ f\\ge 0$ and $ f_k\\ge 0.$\r\n\r\nWe know that $ \\lim \\int (f_k \\minus{} f) \\equal{} 0.$ Write $ \\int (f_k \\minus{} f) \\equal{} \\int (f_k \\minus{} f)^ \\plus{} \\minus{} \\int (f_k \\minus{} f)^ \\minus{} .$\r\n\r\nNote that $ (f_k \\minus{} f)^ \\minus{} \\le f.$ Therefore, we can use $ f$ as a dominating function and use the DCT to show that $ \\lim\\int(f_k \\minus{} f)^ \\minus{} \\to0.$\r\n\r\n(*) Since $ \\int (f_k \\minus{} f)^ \\plus{} \\equal{} \\int (f_k \\minus{} f) \\plus{} \\int (f_k \\minus{} f)^ \\minus{}$ and the limits both integrals on the right are zero, we have that $ \\lim\\int (f_k \\minus{} f)^ \\plus{} \\equal{} 0.$\r\n\r\nTherefore $ \\lim \\int|f_k \\minus{} f| \\equal{} \\lim\\int(f_k \\minus{} f)^ \\plus{}\\plus{} \\lim\\int(f_k \\minus{} f)^ \\minus{} \\equal{} 0.$\r\n\r\nIt's the step marked (*) that doesn't seem to extend to $ p > 1.$",
  "Solution_2": "Let $ h_{k} \\equal{} (f_{k} \\minus{} f)^{p}$. Then \r\n\r\n$ |h_{k}|\\leq \\left(|f_{k}| \\plus{} |f|\\right)^{p}\\leq 2^{p \\minus{} 1}\\left(|f_{k}|^{p} \\plus{} |f|^{p}\\right) \\equal{} g_{k}$\r\n\r\n\r\nOn the other hand, $ g_{k}\\rightarrow 2^{p}|f|^{p} \\equal{} g$, which is an integrable function and $ \\int g \\equal{} \\lim\\int g_{k}$ (this follows based on the fact that $ \\parallel{}f_{k}\\parallel{}\\rightarrow \\parallel{}f\\parallel{}$.)\r\n\r\nAlso $ h_{k}\\rightarrow 0$.\r\n\r\nThus, $ \\lim\\int h_{k} \\equal{} \\int\\limits \\lim h_{k}\\equal{}\\int 0 \\equal{} 0$.\r\n\r\nNote that I have used the so called generalized Lebesque Convergence Theorem.",
  "Solution_3": "[quote=\"didilica\"]Note that I have used the so called generalized Lebesque Convergence Theorem.[/quote]\r\nWhat is the statement of this? Because I'm not seeing how the very last line follows.",
  "Solution_4": "Theorem: Let $ g_{n}$ be a sequence of integrable functions which converges a.e to an integrable function g. \r\nLet $ f_{n}$ be a sequence of measurable functions such that $ |f_{n}|\\leq g_{n}$ and $ f_{n}$ converges to $ f$ a.e. \r\nIf \\[ \\int g\\equal{}\\lim\\int g_{n}\\] then $ \\int f\\equal{}\\lim\\int f_{n}$.",
  "Solution_5": "You can apply Fatou's lemma to $ g_k \\minus{} |h_k|$ and it falls right out",
  "Solution_6": "for the case $ 1<p<\\infty$, we could replace the hypothesis \"$ \\|f_n\\|_p\\to \\|f\\|_p$\" by \"$ \\limsup\\limits_{n\\to\\infty}\\|f_n\\|_p\\leq \\|f\\|_p$\""
}
{
  "Tag": [],
  "Problem": "\u03a3\u03b5 \u03ad\u03bd\u03b1 \u03b2\u03b9\u03b2\u03bb\u03af\u03bf \u03b2\u03c1\u03ae\u03ba\u03b1 \u03c4\u03bf \u03c0\u03b1\u03c1\u03b1\u03ba\u03ac\u03c4\u03c9 \u03b5\u03bd\u03b4\u03b9\u03b1\u03c6\u03ad\u03c1\u03bf\u03bd \u03bb\u03ae\u03bc\u03bc\u03b1 \u03b1\u03bb\u03bb\u03ac \u03c7\u03c9\u03c1\u03af\u03c2 \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7, \u03b3\u03b9' \u03b1\u03c5\u03c4\u03cc \u03b8\u03b1 \u03c0\u03b1\u03c1\u03b1\u03ba\u03b1\u03bb\u03bf\u03cd\u03c3\u03b1 \u03cc\u03c0\u03bf\u03b9\u03bf\u03bd \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03c3\u03b5 \u03bd\u03b1 \u03b2\u03ac\u03bb\u03b5\u03b9 \u03c4\u03b7\u03bd \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03ae \u03c4\u03bf\u03c5\r\n\r\n[color=red][u]\u039b\u03ae\u03bc\u03bc\u03b1[/u][/color]\r\n\r\n\u03a3\u03b5 \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03bf $ ABC$ \u03ad\u03c3\u03c4\u03c9 $ I$ \u03c4\u03bf \u03ad\u03b3\u03ba\u03b5\u03bd\u03c4\u03c1\u03bf, \u03ba\u03b1\u03b9 $ a,b,c$ \u03c4\u03b1 \u03bc\u03ae\u03ba\u03b7 \u03c4\u03c9\u03bd \u03c0\u03bb\u03b5\u03c5\u03c1\u03ce\u03bd  $ BC,CA,AB$ \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03bf\u03b9\u03c7\u03b1. \u03a4\u03cc\u03c4\u03b5 \u03b3\u03b9\u03b1 \u03ba\u03ac\u03b8\u03b5 \u03c3\u03b7\u03bc\u03b5\u03af\u03bf $ M$ \u03c4\u03bf\u03c5 \u03b5\u03c0\u03b9\u03c0\u03ad\u03b4\u03bf\u03c5 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9:\r\n\r\n$ a*MA^2 \\plus{} b*MB^2 \\plus{} c*MC^2 \\equal{} abc \\plus{} (a \\plus{} b \\plus{} c)*MI^2$",
  "Solution_1": "\u038a\u03c3\u03c9\u03c2 \u03b2\u03b3\u03b1\u03af\u03bd\u03b5\u03b9 \u03b5\u03cd\u03ba\u03bf\u03bb\u03b1 \u03bc\u03b5 \u03b4\u03b9\u03b1\u03bd\u03cd\u03c3\u03bc\u03b1\u03c4\u03b1 \u03ba\u03b1\u03b9 \u03bc\u03b9\u03b3\u03b1\u03b4\u03b9\u03ba\u03bf\u03cd\u03c2. \u03a4\u03bf \u03b4\u03bf\u03ba\u03af\u03bc\u03b1\u03c3\u03b5\u03c2 \u03ad\u03c4\u03c3\u03b9; \r\n\r\n\u03a3\u03c4\u03ad\u03bb\u03b9\u03bf\u03c2",
  "Solution_2": "bilot4 \u03bb\u03b9\u03b3\u03bf \u03b6\u03bf\u03c1\u03b9\u03ba\u03bf \u03c4\u03bf \u03bb\u03b7\u03bc\u03bc\u03b1!... \u03ba\u03b1\u03c4\u03b1 \u03c4\u03c5\u03c7\u03b7 \u03ba\u03b1\u03c4\u03b5\u03b2\u03b1\u03c3\u03b1 \u03b1\u03c5\u03c4\u03bf \u03c4\u03bf pdf \u03c0\u03b1\u03bd\u03c9 \u03c3\u03b5 \u03b1\u03bd\u03b9\u03c3\u03bf\u03c4\u03b7\u03c4\u03b5\u03c2 \u03b1\u03c0\u03bf \u03c4\u03bf \u0391\u03bfPS  \u03c4\u03bf\u03c5 Hojoo Lee \u03c0\u03bf\u03c5 \u03c0\u03b5\u03c1\u03b9\u03b5\u03c7\u03b5\u03b9 \u03c4\u03b7 \u03bb\u03c5\u03c3\u03b7 (\u03c3\u03b5\u03bb.9-10) :)"
}
{
  "Tag": [
    "email",
    "AMC"
  ],
  "Problem": "I have received 4 protests on 3 separate problems.  Each was received through the regular communication channels we publish on every piece of information we produce.  After consultation with the Contest Chairs, I have responded personally and in detail to each communication.\r\n\r\nAs for the 4 protests on 3 problems, obviously two protests were about one problem, but it was not the number 11 on the 2009 AMC 12B.  \r\n\r\nSteve Dunbar\r\nMAA Director, American Mathematics Competitions",
  "Solution_1": "Perhaps I am mistaken, but I believe I filed my protest through the correct channel (amcinfo@maa.org), and I am currently looking at my Inbox that is without a response.  The only communication I have regarding my protest is: [quote]I will forward you concern onto the appropriate people.\n\nThank you[/quote]",
  "Solution_2": "[quote=\"AMCDirector\"]I have responded personally and in detail to each communication[/quote]\r\n\r\nDoes that mean all protests have been rejected :D",
  "Solution_3": "[quote]perhaps I am mistaken, but I believe I filed my protest through the correct channel (amcinfo@maa.org), and I am currently looking at my Inbox that is without a response. The only communication I have regarding my protest is:[/quote]\r\n\r\namcinfo@maa.org received a message concerning this problem at:\r\nSun, 01 Mar 2009 23:53:20 -0500 \r\nThe message was relayed to me, and I responded:\r\n03/04/2009 11:54 AM\r\nafter consultation with the Contest Chairs.\r\nThe message was sent to an aol.com address.\r\nOne of the contest chairs informed me this weekend that he had seen the content of the message, so I know it was sent.\r\nI have not received a bounce-back informing me the message was not deliverable.\r\n\r\nHave you checked your spam filter?\r\n\r\n\r\nSteve Dunbar\r\nMAA Director, American Mathematics Competitions",
  "Solution_4": "[quote]The message was sent to an aol.com address[/quote]\r\n\r\nI assume this is the problem considering I sent the email from my one, and only, email address that is a Gmail account.  Would there be any way to have this resent?",
  "Solution_5": "[quote]is a Gmail account[/quote]\r\n\r\nSend me a message via the Private Messages here on AoPS with your email address, I will resend.\r\n\r\nSteve Dunbar\r\nMAA Director, American Mathematics Competitions",
  "Solution_6": "Were any additional answers accepted?",
  "Solution_7": "No additional answers were accepted."
}
{
  "Tag": [
    "\\/closed"
  ],
  "Problem": "I could not post in the problems thread, so I started this one.  In addition, the graphics are inversed and I can not seem to fix it.",
  "Solution_1": "What do you mean the graphics are inversed?\r\n\r\nAlso, what do you mean you couldn't post?  Try again - I've been messing around with the site all day as people identify little problems; you may have tried posting as one of the files was being replaced.",
  "Solution_2": "When I clicked on Post, which I did several times, it took me to a blank page, and never loaded anything.  By inverted colors, I mean that everything is yellow, green and gray instead of red yellow and blue.",
  "Solution_3": "As for the colors, I am seeing the same, although I think that such is intended (that the color scheme was changed).",
  "Solution_4": "yes the colors are lyk that for me too.... so i agree with fierytycoon. but im able to post normally",
  "Solution_5": "No duh, I'm shocked, I mean wow. </no offence, but that was kind of obvious :)>\r\n\r\nAnyways the posting I have no problems with...well, I've had that once but as he says it was during when he was tweaking the phpBB so yeah, things happen, but they should be alright by the time everything's sorted out."
}
{
  "Tag": [
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Prove that for all $ k \\in  N$, there exist $ A$ is positive integer, p is a prime number such that $ ORD_p (A) \\equal{}k$\r\n$ N \\equal{}{1,2,3....}$\r\n\r\n\r\nWHO CAN ?",
  "Solution_1": "In other words, there exists a prime of the form $ nk \\plus{} 1$.  In fact, there are infinitely many: see http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=155461 ."
}
{
  "Tag": [
    "probability",
    "number theory",
    "relatively prime"
  ],
  "Problem": "When rolling a certain unfair six-sided die with faces numbered 1,\n2, 3, 4, 5, and 6, the probability of obtaining face $ F$ is greater\nthan $ 1/6$, the probability of obtaining the face opposite face $ F$\nis less than $ 1/6$, the probability of obtaining each of the other\nfaces is $ 1/6$, and the sum of the numbers on each pair of opposite\nfaces is 7. When two such dice are rolled, the probability of\nobtaining a sum of 7 is 47/288. Given that the probability of\nobtaining face $ F$ is $ m/n$, where $ m$ and $ n$ are relatively prime\npositive integers, find $ m\\plus{}n$.",
  "Solution_1": "Well. To get 7 there are 3 ways. Letting the faces be a,b,c,d each with probability 1/6, and F and f with probability 1/6+p and 1/6-p.\r\n\r\nTo get 7:\r\n\r\nab:1/18\r\n\r\ncd:1/18\r\n\r\nFf:2(1/36-p^2)=1/18-2p^2\r\n\r\nThus 1/18+1/18+1/18-2p^2=1/6-2p^2=47/288\r\n\r\nTherefore 2p^2=1/288\r\np^2=1/576\r\n\r\np=1/24\r\n\r\nThus the probability of rolling F is 1/6+1/24=5/24, so m+n=29"
}
{
  "Tag": [
    "inequalities",
    "algebra",
    "function",
    "domain"
  ],
  "Problem": "Let  [tex] a,b,c>0[/tex] .Prove that  [tex]\\frac{ab}{ab+c^2}+\\frac{bc}{bc+a^2}+\\frac{ca}{ca+b^2}\\geq\\frac{3}{2}[/tex]",
  "Solution_1": "[quote=\"nickolas\"]Let  [tex] a,b,c>0[/tex] .Prove that  [tex]\\frac{ab}{ab+c^2}+\\frac{bc}{bc+a^2}+\\frac{ca}{ca+b^2}\\geq\\frac{3}{2}[/tex][/quote]\r\nWill AM-HM help?",
  "Solution_2": "[quote=\"leepakhin\"]Let $x=\\frac{a^2}{bc}$ and so on, $xyz=1$\nBy AM-HM inequality,\n$\\frac{1}{1+x}+\\frac{1}{1+y}+\\frac{1}{1+z}\\ge 3(\\frac{3}{1+x+1+y+1+z})$\nIt remains to prove $\\frac{3}{3+x+y+z}\\ge \\frac{3}{2}$\n$\\Leftrightarrow x+y+z\\le 3$\nThis is true by AM-GM inequality.[/quote]\r\n\r\nActually, it remains to prove $\\frac{3}{3+x+y+z}\\ge \\frac{1}{2}$\r\n\r\nand $x+y+z \\le 3$ is not AMGM ineq, because $x + y + z \\ge 3 \\sqrt[3]{xyz} = 3$ is AMGM.",
  "Solution_3": "[quote=\"Singular\"]But $(1,1,0)$ majorizes $(4/3, 1/3, 1/3)$, so by muirhead we are done.[/quote]\r\n\r\nIt does?\r\n\r\nAnyways, the inequality cannot be true either way. Letting $(a,b,c) \\to (0, 1, 1)$ gives $LHS \\to 1$ and $(a,b,c) \\to (0, 0, 1)$ gives $LHS \\to 2$.",
  "Solution_4": "Bill:  a,b,c must be greater than zero. :)  So those do not work.",
  "Solution_5": "Of course they serve as counterexamples. Take anything close to it and try it out (like $(0.001, 1, 1)$).\r\n\r\nIt's just easier to see why the inequality it's wrong when we extend the domain to include 0.\r\n\r\nOr if you want to be more formal, we can let $(a,b,c) = (\\epsilon, 1, 1)$ and let $\\epsilon \\to 0$.",
  "Solution_6": "[quote=\"billzhao\"][quote=\"Singular\"]But $(1,1,0)$ majorizes $(4/3, 1/3, 1/3)$, so by muirhead we are done.[/quote]\n\nIt does?\n\nAnyways, the inequality cannot be true either way. Letting $(a,b,c) \\to (0, 1, 1)$ gives $LHS \\to 1$ and $(a,b,c) \\to (0, 0, 1)$ gives $LHS \\to 2$.[/quote]\r\n\r\n\r\nmy bad, you are right.  i was rushing lol",
  "Solution_7": "Oops, my bad, I understand what you mean now (a limit).",
  "Solution_8": "Another way to realize that the inequality is wrong...  :P \r\n\r\n$\\sum \\frac{ab}{ab+c^2} \\geq \\frac{3}{2} \\iff \\sum \\frac{c^2}{ab+c^2} \\leq \\frac{3}{2}$\r\n\r\nwhich after setting $x = c^2/ab$, etc or $p = ab/c^2$, etc becomes\r\n\r\n$\\sum \\frac{1}{1+x} \\geq \\frac{3}{2} \\iff \\sum \\frac{1}{1+p} \\leq \\frac{3}{2}$\r\n\r\nunder the conditions $xyz = pqr = 1$... prove one and the other's false."
}
{
  "Tag": [],
  "Problem": "Two different numbers are chosen from the sequence of consecutive odd integers 51, 53, 55, \u2026,95, 97, 99. Which of the following could not be their sum?\r\na. 161 b. 168 c. 186 d. 196 e. NG",
  "Solution_1": "[hide]Well, all the numbers are odd.\nAn odd number plus an odd number is always an even number.\nThe only odd answer choice is [b](A) 161[/b][/hide]"
}
{
  "Tag": [
    "geometry",
    "rectangle",
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "Let $n$ be a positive integer. Nancy is given a rectangular table in which each entry is a positive integer.  She is permitted to make either of the following two moves:\r\n\r\n(1)  select a row and multiply each entry in this row by $n$;\r\n\r\n(2)  select a column and subtract $n$ from each entry in this column.\r\n\r\nFind all possible values of $n$ for which the following statement is true:\r\n\r\nGiven any rectangular table, it is possible for Nancy to perform a finite sequence of moves to create a table in which each entry is $0$.",
  "Solution_1": "May I clear up another old unsolved problem?\r\n\r\nClearly $n=1$ is not going to work if the entries in a column are not equal.\r\n\r\nIf $n>2$, then the rectangle consisting just of one column of a $2$ and a $1$ cannot be changed to $0$s- it is impossible to make the two values congruent $\\mod (n-1)$.\r\n\r\nTo prove that it works for $n=2$:\r\n\r\nClearly if we can do it for one column, we can do it for any number of columns, because once a column is all zeros we can move on to another column. Also, we only need consider columns with two numbers in them because once we have made two numbers in a column equal, we can just treat them together and move on to the rest of the column.\r\nSo we only need consider $2 \\times 1$ rectangles. \r\nWe can double both numbers in this $2 \\times 1$ rectangle, and then subtract $2$ repeatedly, until one of the numbers is $2$. Then we double that to $4$, subtract $2$ again until both numbers are $2$. Then we can move onto the rest of the rectangle, if it is larger than a $2 \\times 1$.\r\n\r\nHence $n=2$ is the only possibility."
}
{
  "Tag": [
    "pigeonhole principle",
    "linear algebra",
    "matrix",
    "modular arithmetic",
    "linear algebra unsolved"
  ],
  "Problem": "Let $n\\geq 2$ and $A\\in\\mathcal M_n(Z)$ with $\\det(A)=\\pm 1$. Prove that in the sequence $A,A^2,A^3,...$ there exists at least one matrix with the property that each of it`s elements is congruent modulo $n$ with the element in the same postion in the corresponding inverse.",
  "Solution_1": "Yes, pigeonhole :). There is, however, something I don't quite get: that $n$ modulo which we have to work need not be the same $n$ as the dimension of the matrix. Was it a typo of yours, or did the problem really look like this? I'll replace the former with $m$. \r\n\r\nAnyway, let $\\widetilde A$ be the matrix reduced modulo $m$. Since $\\mathcal M_n(\\mathbb Z/m\\mathbb Z)$ has only finitely many elements, we have $\\widetilde A^k=\\widetilde A^p$ for some $0\\le k<p$. Since $\\widetilde A$ is invertible, we have $\\widetilde A^{p-k}=I\\Rightarrow \\left(\\widetilde A^{p-k}\\right)^2=I$, which means that $\\widetilde A^{p-k}\\equiv\\left(\\widetilde A^{p-k}\\right)^{-1}\\pmod m$, and this is exactly what we want.",
  "Solution_2": "In the original problem the $n$ with which we must modulate is the same as the dimension of the matrix, however since you generalised the problem, it does not matter any longer :) .( The problem is from a book of Gheorghe Andrei or C.Caragea about algebra.)"
}
{
  "Tag": [
    "email",
    "USAMTS",
    "LaTeX",
    "\\/closed"
  ],
  "Problem": "Today morning, for about 2 hours ET, the AoPS Forum gave a message:\r\n\r\nCritical Error\r\n\r\nCannot connect to the database.\r\n\r\nWhat happened? :maybe:",
  "Solution_1": "It happened for me for 4 hours (Some time around 10GMT to 3:30GMT). Bizarre, though I guess the admins knew about the problem (and fixed it)",
  "Solution_2": "I emailed Mr. Vornicu about it...",
  "Solution_3": "i was wondering why i was the only one other than bots at 10 :ninja:",
  "Solution_4": "When I just went to http://artofproblemsolving.com or http://mathlinks.ro, it gave me:\r\n\r\nThis problem is temporary for 5 minutes. If the problem persists, email webmaster@artofproblemsolving.com or http://mathlinks.ro respectively\r\n\r\nor something to that extent",
  "Solution_5": "Great that you did. This most likely appeared due to an automatic system update and some paths were screwed up :)",
  "Solution_6": "Strange. I was on fine. :maybe:",
  "Solution_7": "The same thing happened to me. I was checking the USAMTS website too, and I got the same message.",
  "Solution_8": "Aren't like the Mathlinks and USAMTS website on teh same server? :maybe:",
  "Solution_9": "I know this is totally not the place to report a new, distinct, problem, but whenever I try to go to the forum (by going to mathlinks.ro, or artofproblemsolving.com and clicking on forum, or by clicking a link to any subforum) I get an error message as such:\r\n\r\nSQL requests not achieved\r\n\r\nDEBUG MODE\r\n\r\nSQL Error: 1030 Got error 134 from storage engine\r\n\r\nSQL Request: SELECT COUNT(DISTINCT session_ip) AS count_session_ip FROM phpbb_sessions WHERE session_user_id = -1 AND session_time >= 1166987998\r\n\r\nLine : 86\r\nFile : class_stats.php\r\n\r\nHowever, all the header/background is still there, and I can click on \"view my posts\" and click on the topics I have posted in (hence this topic). Just about anywhere else I try to go in the forum, I get the same error. Is this a bug/temporary problem/only my problem? (Using IE7 on XP).\r\n\r\nAgain, sorry for reposting here but it is the only thing I can do at the moment.",
  "Solution_10": "Same here. I think it is some huge bug.",
  "Solution_11": "It definitely is... emailed Mr. Vornicu again. He's probably working on it though...",
  "Solution_12": "I am getting this message:\r\n[quote]\nSQL requests not achieved\n\nDEBUG MODE\n\nSQL Error: 1030 Got error 134 from storage engine\n\nSQL Request: SELECT COUNT(DISTINCT session_ip) AS count_session_ip FROM phpbb_sessions WHERE session_user_id = -1 AND session_time >= 1166991021\n\nLine : 86\nFile : class_stats.php\n[/quote]\r\nFor like the last two hours i got the other one",
  "Solution_13": "It appears one of our tables was messed up; I think it is fixed now.  Is anyone still having problems?",
  "Solution_14": "That message came since today morning for me.",
  "Solution_15": "No, its OK now, thanks.",
  "Solution_16": "[quote=\"anirudh\"]That message came since today morning for me.[/quote]\r\n\r\nIs it still happening?",
  "Solution_17": "Always on X-Mas something happens :)",
  "Solution_18": "[quote=\"Valentin Vornicu\"]Always on X-Mas something happens :)[/quote]\r\n\r\nYeah, I'm becoming a real Scrooge.",
  "Solution_19": "I call it the Christmas Bug",
  "Solution_20": "Yes. The Forums work. But the classroom doesn't work. :(",
  "Solution_21": "The classroom does not work for me either  :maybe:",
  "Solution_22": "It gives some kind of error.\r\n\r\nCannot connect to server\r\n\r\nor something to that extent",
  "Solution_23": "Classroom should be fixed now.",
  "Solution_24": "Yes. the classroom works now.\r\n\r\nAlso, sometimes, latex just crashes in the classroom sometimes...\r\n\r\nand only:\r\n\r\n[code];$x$[/code]\r\n\r\nworks.\r\n\r\nNothing else works.\r\n\r\nBut it works fine now :) \r\n\r\nSometimes, though, when in discussion in the classroom, not in the class, it crashes such.",
  "Solution_25": "This is a little unrelated (and I guess I might be bumping this post... its been almost a month)\r\n\r\nbut after reading this, I remembered what I was going to mention - I use firefox, and I always have problems logging in. I will be working on a post and try to submit and it will say, \"please log in\" or something. Sometimes, every 15 minutes or so I get automatically logged out. Especially when I'm editing a post of mine or something.\r\n\r\nJust wondering if someone could look into it r if other people have similar experiences."
}
{
  "Tag": [
    "algebra proposed",
    "algebra"
  ],
  "Problem": "Calculate $ \\displaystyle \\sum_{k\\equal{}0}^m  \\left( \\begin{array} {c}  s\\minus{}1 \\\\ m\\minus{}k  \\end{array} \\right) \\left( \\begin{array} {c}  n\\minus{}s \\\\ k\\minus{}1  \\end{array} \\right)$, where $ n>s>m$.",
  "Solution_1": "The result is $ \\binom{n}{m}$.\r\nThis is the number of ways to choose $ m$ balls from an urn which contains $ n$ balls numbered from $ 1$ to $ n$.\r\nIn another hand, for each $ k$, the $ k^{th}$ summand in the sum is the number of ways to choose $ m$ of balls from this urn such that the $ k^{th}$ greatest number of the chosen balls is $ s$.\r\n\r\nPierre."
}
{
  "Tag": [
    "irrational number"
  ],
  "Problem": "I know that it's possible for an irrational number multiplied by an irrational number can equal a whole number, (i.e. take a 1 foot line bend it into a circle multiply the diameter times pi [both are irrational] and you should get 1 foot) but how can we apply this in real life? How can we tell when it will equal a whole number and how to do the problem?",
  "Solution_1": "This really shouldn't be here....\r\n\r\nWell, we have $ \\sqrt{2} \\cdot \\sqrt{2}\\equal{}2$.",
  "Solution_2": "[quote=\"mathblitz\"]How can we tell when it will equal a whole number[/quote]\r\n\r\nTake $ \\sqrt{a} \\times \\sqrt{b}$.  If $ ab$ is a square number, or if $ a\\equal{}b$, or if $ a$ or $ b$ is $ 0$, it will equal a whole number.",
  "Solution_3": "[quote=\"mathblitz\"]how can we apply this in real life? How can we tell when it will equal a whole number and how to do the problem?[/quote]\r\nThat depends on the problem.  The algebra of irrational numbers is exactly the same as the algebra of rational numbers, so nothing changes until you have to do numerical computations - is that what you're asking about?  Do you need help understanding roots or irrational numbers in general?  It would be helpful if you were more specific."
}
{
  "Tag": [
    "probability"
  ],
  "Problem": "Players A and B play a (fair) dice game. \"A\" deposits one coin and they take turns rolling a single dice, \"B\" rolling first.\r\n\r\nIf \"B\" rolls an even number, he collects a coin from the pot. If he rolls an odd number, he put a coin (coins with same values always).  If \"A\" (plays and) rolls an even number, he collects a coin but if he rolls an odd number, he does NOT add a coin.\r\n\r\nThe game continues until the pot is exhausted.\r\n\r\nQuestion: what is the probability that \"A\" wins this game (that is, exhaust the pot) ?\r\n\r\n(Show solution, not just answer.)",
  "Solution_1": "You could solve this by getting ever-closer approximations by looking at the probability that the game ends after n rounds.  The probability that b wins after the first round is $\\frac12$, which immediately sets an upper bound on the probability that a will win.\r\n\r\nOf the remaining $\\frac12$, there is a $\\frac12$ probability that a will remove one of the two coins in the pot.  B might then also remove a coin, lowering the upper bound by $\\frac38$.\r\n\r\nThere is also a $\\frac18$ probability that a will not remove a coin, but b will.  Then a will have an immediate shot at winning, yielding a lower bound of $\\frac1{16}$.\r\n\r\nIf, in the second round, b instead places a coin in the pot, there is no chance for a to win in the second round.\r\n\r\nThis is obviously not the most elegant solution to this problem, and really requires writing a program to get a decent approximation in a decent time interval, but it's the only thing I can think of.",
  "Solution_2": "P(i): The probability that a wins with i in the pot at the start of B's turn.\r\n\r\nP(i)=(1/4)(P(i+1)+P(i)+P(i-1)+P(i-2))\r\nP(i)=(1/3)(P(i+1)+P(i)+P(i-1)+P(i-2))\r\n\r\nP(0)=1\r\nP(1)=(1/3)(P(2)+1+1)=(2/3)+(1/3)P(2)\r\nP(2)=(1/3)(P(3)+P(1)+1)=(1/9)(3P(3)+2+P(2)+3) so P(2)=(1/8)(3P(3)+5)\r\nP(3)=(1/3)(P(4)+P(2)+P(1))=\r\n(1/120)(40P(4)+15P(3)+25+16+8P(2)=\r\n(1/120)(40P(4)+15P(3)+8P(2)+41)\r\nP(3)=(1/105)(40P(4)+8P(2)+41)=(1/105)(40P(4)+3P(3)+46)=\r\n(1/102)(40P(4)+46)=(1/51)(20P(4)+23)\r\n\r\nA lower bound is P(3)=(23/51) & P(2)=(23+5*17)/(8*17)=27/34 and P(1)=2/3+9/34=(27+68)/102=95/102.\r\n\r\nWait a second. The probability is 1. A 1-dimensional random walk (and this walk is actually down-leaning) ALWAYS returns to 1.",
  "Solution_3": "roadnottaken:  Good approach!\r\n\r\nI realized the A:B winning probabilities is NOT 1:1 using the very same (start of your) reasoning: \r\n-------------\r\nThe probability that B wins after the first round is... sets an upper bound on the probability that A will win \r\n-------------\r\nand the begining of your argument is more-than-strong enough to garantee the probability asked is NOT 1, too...\r\n\r\n\r\nBut... I am still looking for an analytical-only solution and I will try a bit harder on the \"total probability theorem\" and related ideas. \r\nNow it seems to me that my first difficulties are not silly, anyway. (I guess I had tried unecessary-hard paths, I mean.)\r\n\r\nLet us wait other contributions. For myself, I will keep my word to come back to bring some options -- if not a solution, who knows? -- in the near future (that is, in some days)!\r\n\r\nBest regards,\r\nfskilnik.\r\n\r\nP.S.: have you seen the \"Dead Poets Society\" movie? (A.Frost poem, beautiful!)",
  "Solution_4": "It said \"A wins if the coins are exhausted\".\r\n\r\nIf you want: B wins if B gets to exhaust the pile before A gets any, then it gets a litle more complicated.\r\n\r\nBasically let R(i) be the probability of a random walk returning to 0 after2i steps. \\[sum_{i=1}^{\\inf}R(i)2^{1-2i}\\]\r\n\r\nAnyone know R(i)?",
  "Solution_5": "Sorry, Bill.\r\n\r\nWhen I wrote \r\n\r\nwhat is the probability that \"A\" wins this game (that is, exhaust the pot) ? \r\n\r\n\r\nI should have written \r\n\r\nwhat is the probability that \"A\" wins this game (that is, A exhaust the pot) ?",
  "Solution_6": "Not a realistic situation, because B could lose money and still win the game. Whatever.\r\n\r\n\r\nThe pot is 1 before A's turn:\r\nA has a 1/2 chance. B has a 1/4 chance. There is a 1/4 chance of going up to 2.\r\n\r\nThe pot is 2 before A's turn:\r\nB has a 1/4 chance. There is a 1/4 chance of it staying the same. There is a 1/4 chance of it going down to 1. There is a 1/4 chance of it going up to 3.\r\n\r\nThe pot is 3 before A's turn:\r\nThere is a 1/4 chance, respectively, of, 1, 2, 3, and 4\r\n\r\nTherefore at the start $B$'s chance is $1/2+P(2)/2=2/3+P(1)/6+P(3)/6=2/3+1/24+P(2)/24+P(1)/18+P(2)/18+P(4)/18 =17/24+P(1)/18+7P(2)/72+P(4)/18 =17/24+1/72+P(2)/72+7/216+7P(3)/216+7P(1)/216+[lots]/18$\r\n\r\nMinimum (for B): 163/216$,$53/216<3/8$Maximum (for B): 192/216=24/27=8/9$ $1/9>1/16$\r\nMy range is now $29/216=13.4%\n$"
}
{
  "Tag": [
    "modular arithmetic"
  ],
  "Problem": "Calculate $\\left(2^{100}-2\\right) \\text{ mod } 100$.",
  "Solution_1": "[hide=\"solution\"]\n$2^{100}\\text{mod} 100 = (2^{10}\\text{mod} 100)^{10}\\text{mod} 100 = 24^{10} \\text{mod} 100 = ((24^2) \\text{mod} 100)^5 \\text{mod} 100 = 76^5 \\text{mod} 100 = 76$ since $76^2=5776$.  \nSo $\\left(2^{100}-2\\right)$ $mod$ $100 = 74$\n[/hide]\r\n\r\nedit: added some {}",
  "Solution_2": "[hide]we know that $(2^10)$ $mod$ $100=48$ and $(48^2)$ $mod$ $100=4$ then we know #$(4^5$ $mod$ $100=48$ $\\rightarrow$ $(2^100- 2 )$ $mod$ $100=48-2=46$[/hide]",
  "Solution_3": "[quote=\"jensen\"]we know that $(2^{10})$ $mod$ $100=48$[/quote]\r\n\r\n :roll: \r\n$2^{10} = 1024$",
  "Solution_4": "[hide]mod 100 means finding the last two digits. \n\nOkay then..... \n\n$2^1=02$\n$2^2=04$\n$2^3=08$\n$2^4=16$\n$2^5=32$\n$2^6=64$\n$2^7=28$ \n\n..... \n\n*feels bored, says K.W.M.A.N's solution is way better. Uses calculator because of sleepiness* \n\n$2^{21}=52$\n$2^{22}=04$ \n$2^{23}=08$ \n\nHey a pattern, $2^2$, $2^22$, etc. have the same last digit. That means $2^{102}$ must also end with four, implying that $2^{100}$ ends with 76. \n\n$2^{100}-2\\equiv74\\pmod{100}$. [/hide]"
}
{
  "Tag": [
    "function",
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "Find the number of lattice walks from $ (0, 0)$ to $ (n, n)$, \r\nconsisting of up and right steps of length one only, \r\nsuch that paths do not cross the lines $ y \\minus{} x \\equal{} k$ and $ y \\minus{} x \\equal{} \\minus{}m$.",
  "Solution_1": "If I'm not mistaken, [url=http://www.research.att.com/~njas/sequences/A080934]A080934[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url] gives the number of paths from $ (0, 0)$ to $ (n, n)$ lying above $ y \\equal{} x$ and below $ y \\equal{} x \\plus{} k$ (encoded in some not-entirely-obvious form), a natural version of restricted Dyck paths.  One can write down recursions for these sequences and maybe generating functions for these, but they aren't overwhelmingly elegant.  Anything that will solve this case can probably be modified for $ y \\equal{} x \\minus{} m$ for $ m > 0$, as it's not all that different then the case where you're bounded between $ y \\equal{} x$ and $ y \\equal{} x \\plus{} k \\plus{} m$."
}
{
  "Tag": [],
  "Problem": "Calculate: $ 6 \\plus{} 4 \\times 3$.",
  "Solution_1": "Order of operations\r\nMultiplication comes first\r\n4x3=12\r\n12+6=18",
  "Solution_2": "easy, oreder of operations, 4 times 3 and then add 12 and six",
  "Solution_3": "hello, or so: $ 6\\plus{}4\\times3\\equal{}2\\times(3\\plus{}2\\times3)\\equal{}6\\times(1\\plus{}2)\\equal{}18$\r\nSonnhard.",
  "Solution_4": "ans is 6+4*3\r\n12+6=18\r\nso it is just using pemdas"
}
{
  "Tag": [
    "geometry",
    "combinatorics proposed",
    "combinatorics"
  ],
  "Problem": "Let $ABC$ be a given triangle .\r\nProve that if we color the plane with 2 colors(Black and white) then \r\nEither we have 2 white points with distance of unit or  we have 3  black points \r\nsuch that they will form a triangle with equal area to $ABC$",
  "Solution_1": "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=chromatic+number&t=26773]Here it is[/url]. In fact, that seems to be a stronger version (the triangles are congruent, not merely equivalent)."
}
{
  "Tag": [
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "A claw is a graph consisting of 4 vertices and three edges, one vertex of degree three and three vertices of degree one (one vertex adjacent to other 3 that are pairwise non-adjacent). A graph is claw free if it does not cotained a induced claw as a subgraph.\r\n\r\nProve that if G is 4-connected and claw free then G is Hamiltonian.",
  "Solution_1": "[quote=\"manuel\"]A graph is claw free if it does not cotained a induced claw as a subgraph.[/quote]\r\nWhat does it mean? There are no vertices A,B, C, D s.t. A is connected to B, C, and D, and there are no edges between B, C and D. Right?",
  "Solution_2": "Yes Myth",
  "Solution_3": "Any hints from your teacher?  ;)",
  "Solution_4": "no.. my teacher is out of the country  :(",
  "Solution_5": "Manuel, i know you already explined me this, but what is k-conected? thanks!",
  "Solution_6": "Pascual here are some definitions of graph connectivity that maybe useful to you:\r\nA [b]separating set [/b]of a graph G is a subset S of the set of vertices of G ( written V(G) ) such that G - S disconnectes the graph. The [b]connectivity[/b] of G, sometimes written k(G), is the minimum size of a separating set in G. A graph G is [b]k-connected[/b] if its connectivity is at least k. In oder words G is [b]k-connected [/b]iff when you remove any k-1 vertices the graph remains connected. \r\nI also told you some theorems regarding this that if you are interested you can ask me... ;)"
}
{
  "Tag": [
    "blogs",
    "MATHCOUNTS"
  ],
  "Problem": "ok...im making a puzzle type thing where ill give you some clues and after you figure out one clue, it leads to another and another untill it stops (kind of like national treasurer)  The only thing you will need is you smarts and the AoPS website (i.e all the clues will be on the website...)  Too hard?  ask for a hint.  NO PMING OTHER PEOPLES\r\n\r\nREAD THIS PASSAGE:\r\n\r\nThere once was a man named bob.  he was a garbaje cllector.  he always wen on AoPS.  one day, he goot fired.  he had nothing do do so he went on the AoPS foru for hours.  his favorite was the communiti section.  Bobb friend was Molly.  one day moly died.  So bob started smamming in the getting started frum.  the moderators got really mad.  One moderator was Geroge.  geore was nice\r\n\r\nTHE END :D",
  "Solution_1": "the answer is someones username\r\n\r\npm me the answer when you are done",
  "Solution_2": "why dont u just edit your first post",
  "Solution_3": "becuase i thought that if people already read it they wouldnt know i had edited it\r\n\r\njeez",
  "Solution_4": "I don't understand this... :huh:",
  "Solution_5": "its like notpron (in a way)\r\n\r\nlook for clues in the passage which will lead you to the next clue and so on\r\n\r\nHINT: [hide]look for spelling errors[/hide]\n\nHINT2: [hide]carreers in math...wondering...hmm[/hide]\n\nHINT3: [hide]line-letter[/hide]\n\nHINT4: [hide]porblem, getting started[/hide]",
  "Solution_6": "I understood what you are trying to do here but I can't figure it out. What's next after the spelling errors?",
  "Solution_7": "have you decoded the message?",
  "Solution_8": "I'm working on it now :weightlift: Sorry if my brain is so slow :)",
  "Solution_9": "post somthing when you get there",
  "Solution_10": "Does this username do all of the things you described?",
  "Solution_11": "i would like to congradulate 236! for being the first to solve the puzzle...using TONS of hints though\r\n\r\nfirst...look for spelling errors which spell out\"go to my blog\"\r\n\r\nnext...i posted something called \"careers in math\" and said ii was WONDERING about careers in math.  \r\n\r\ngo to the careers in math foruma and click wondering\r\n\r\nall the pairs of numbers stand for a letter in the first post\r\n\r\nit comes out as \"go to mathcounts\" or something\r\n\r\ngo to mathcounts and click on \"problem\"\r\n\r\nBOB:lets go to the getting started forum...i like their MODERATORS\r\n\r\nlater on in the post after the problem was stated...\r\n\r\nHINT:the answer is less than [b]10000[/b]\r\n\r\ncan you get the answer now?",
  "Solution_12": "[quote=\"jli\"]There once was a man named bob.  he was a garbaje cllector.  he always wen on AoPS.  one day, he goot fired.  he had nothing [b]do[/b] do so he went on the AoPS foru for hours.  his favorite was the communiti section.  Bobb friend was Molly.  one day moly died.  So bob started [b]smamming[/b] in the getting started frum.  the moderators got really mad.  One moderator was Geroge.  geore was nice[/quote]\r\n :?\r\nI get GOTODMYBLMOG.",
  "Solution_13": "sorry...its supposed to say go to my blog",
  "Solution_14": "ahh, I sort of get it now..."
}
{
  "Tag": [
    "calculus",
    "integration",
    "trigonometry",
    "conics",
    "ellipse",
    "search",
    "calculus computations"
  ],
  "Problem": "Please find (if its possible):\r\n\r\n\\[{ I=\\int^{0}_{\\frac{\\pi}{2}}\\sqrt{a^{2}\\sin^{2}x+b^{2}\\cos^{2}x}}\\,dx\\]",
  "Solution_1": "This calculation seems to be the arc length for the ellipse of a quarter,but  generally the calculation is impossible.",
  "Solution_2": "This is a classical example of something called [i]elliptic integrals[/i]. It turns out that finding the circumfurence of an ellipse isn't an easy problem. It leads to an non-elementary integral. Which is why a math reference book will write the circumfurence of an ellipse is $ \\approx\\pi(a\\plus{}b)/2$, which is just an approximation. If you consider $ \\bold{r}(t) \\equal{} a\\cos(t)\\bold{i}\\plus{}b\\sin(t)\\bold{j}$ for $ 0\\leq t\\leq 2\\pi$ this is an ellipse. Its length is given by:\r\n$ \\int_{0}^{2\\pi}\\sqrt{a^{2}\\sin^{2}(t)\\plus{}b^{2}\\cos^{2}(t)}dt$.\r\nWhich [b]isnt[/b] doable if $ a,b>0$ and $ a\\not \\equal{} b$.\r\n\r\nBecause of appearaces of this integrals in mathematical physics some work on tables and identities have been developed. You can find more [url=http://mathworld.wolfram.com/EllipticIntegral.html]here[/url] including your integral which is formula #108.",
  "Solution_3": "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1082482064&t=55166]This topic[/url] discusses the situation and includes a discussion of various approximations."
}
{
  "Tag": [
    "inequalities",
    "function",
    "vector",
    "induction",
    "inequalities theorems"
  ],
  "Problem": "What  is the  general  form  of  Holder's  inequality??\r\nThanks  Harazi !",
  "Solution_1": "But I really don't know!",
  "Solution_2": "sum(i=1,n) p_i*a_i*b_i<=(sum(i=1,n) p_i*a_i^x)^(1/x)*(sum(i=1,n) p_i*b_i^y)^(1/y).\r\n\r\nwhere 1/x+1/y=1 and x,y,pi,ai,bi>0 and n>2.\r\n\r\nI hope this is what u wanted",
  "Solution_3": "I knew this one! I'm so glad. I thought he was asking about the most general form of his inequality and I don't know that. Anyway, open question-as far as I know- means a problem that was not solved by the one who thought about it and is proposed for the others to think about it. So, this question is not an open question.",
  "Solution_4": "This one should go to formulas and theorems corner! :D",
  "Solution_5": "Well  i  knew  that  one , but  i  was  wondering  if  there  is  a  more  general  form  like  schure's  inequality...Thanks",
  "Solution_6": "btw I have extracted the generalisation I gave for schur and put it into this folder if you want to take a look at it nickolas.",
  "Solution_7": "I've only seen it with the sets $A=\\left\\{a_{1},a_{2},...,a_{n}\\right\\}$ and $B=\\left\\{b_{1},b_{2},...,b_{n}\\right\\}$. I think though it can be extended to $k$ sets of numbers where $v^{j}$ denotes the $jth$ set. So for $v^{1}=\\left\\{(v^{1})_{1},(v^{1})_{2},...,(v^{1})_{n}\\right\\}, v^{2}=\\left\\{(v^{2})_{1},(v^{2})_{2},...,(v^{2})_{n}\\right\\},...,v^{k}=\\left\\{(v^{k})_{1},(v^{k})_{2},...,(v^{k})_{n}\\right\\}$, and for $\\displaystyle\\ p_{1},p_{2},...,p_{k}>1$, given $\\displaystyle\\sum_{i=1}^{k}\\frac{1}{p_{i}}=1$, I believe this inequality holds:\r\n\r\n$\\displaystyle\\sum_{i=1}^{n}\r\n\\left|(v^{1})_{i}(v^{2})_{i}...(v^{k})_{i}\\right|\\leq\\left(\\sum_{i=1}^{n}\\left|(v^{1})_{i}\\right|^{p_{1}}\\right)^{\\frac{1}{p_{1}}}\\left(\\sum_{i=1}^{n}\\left|(v^{2})_{i}\\right|^{p_{2}}\\right)^{\\frac{1}{p_{2}}}...\\left(\\sum_{i=1}^{n}\\left|(v^{k})_{i}\\right|^{p_{k}}\\right)^{\\frac{1}{p_{k}}}$\r\n\r\nCan anyone verify/disprove my statement?",
  "Solution_8": "Let $a_{1,1},a_{1,2},...a_{1,n},a_{2,1},a_{2,2}...a_{2,n...}a_{k,n} > 0$ and  $\\frac{1}{p_1}+\\frac{1}{p_2}+...+\\frac{1}{p_k}=\\frac{1}{s}$. Than\r\n$(a_{1,1}^{p_1}+a_{1,2}^{p_1}+...+a_{1,n}^{p_1})^{\\frac{1}{p_1}} (a_{2,1}^{p_2}+a_{2,2}^{p_2}+...+a_{2,n}^{p_2})^{\\frac{1}{p_2}}....(a_{k,1}^{p_k}+a_{k,2}^{p_k}+...+a_{k,n}^{p_k})^{\\frac{1}{p_k}}\\geqslant ((a_{1,1}a_{2,1}...a_{k,1})^s+(a_{1,2}a_{2,2}...a_{k,2})^s+...+(a_{1,n}a_{2,n}...a_{k,n})^s)^{\\frac{1}{S}}$\r\ngeneralization by jalowiec ( http://www.matematyka.org/forum/viewtopic.php?t=1893 - the site is in Polish)\r\n\r\nproof 1 (by me)\r\n\r\nlet $\\mathbb{R}_+^n \\ni \\vec{A_j} = [ \\ a_{j,1} \\  , \\  a_{j,2} \\  , \\  a_{j,3} \\  , \\  ... \\  , \\  a_{j,n} \\ ]$\r\nlet us define $\\Vert \\vec{A_j} \\Vert_m = (a_{j,1}^m + a_{j,2}^m + ... + a_{j,n}^m)^{1/m}$\r\nand a realtion - $\\ell$-multiplication - denoted as: it's a function $\\mathcal{M}_\\ell : \\mathbb{R}_+^{n \\ell} \\mapsto \\mathbb{R}_+^n$ ($\\ell$ vectors give one vector) and\r\n$\\mathcal{M}_\\ell (\\vec{A_1} , \\vec{A_2} , ... , \\vec{A_\\ell}) = [ \\  a_{1,1} a_{2,1} ... a_{\\ell, 1} \\  , \\  a_{1,2} a_{2,2} ... a_{\\ell , 2} \\  , \\  ... \\  , \\  a_{1,n} a_{2,n} ... a_{\\ell , n} \\  ]$ (analogically to usual vector addition, but with multiplication instead). of course $\\ell$-multiplication is commutative, you can randomly change the order of the factors and the result is always the same due to commutativity of regular multiplication. $\\ell$-multiplication has also this property:\r\n$\\mathcal{M}_2 (\\mathcal{M}_r (\\vec{A_1} , ... , \\vec{A_r}) , \\mathcal{M}_w (\\vec{A_{r+1}} , ... , \\vec{A_{r+w}}) ) = \\mathcal{M}_{r+w} (\\vec{A_1} , ... , \\vec{A_{r+w}})$ - r-multiplication times w-multiplication is r+w-multiplication\r\nfirst let us prove that $\\Vert \\vec{X} \\Vert_t \\Vert \\vec{Y} \\Vert_t \\geqslant \\Vert \\mathcal{M}_2 (\\vec{X} , \\vec{Y} ) \\Vert_t$ , ($\\vec{X} = [ \\  x_1 \\  , \\ x_2 \\  , \\  ... \\  , \\  x_n \\  ] \\  , \\  \\vec{Y} = [ \\  y_1 \\  , \\  y_2 \\  , \\  ... \\  , \\  y_n \\  ]$)\r\nplotting it to our definition of norm we get\r\n$(\\sum_{i=1}^n x_i^t)^{1/t} (\\sum_{i=1}^n y_i^t)^{1/t} \\geqslant (\\sum_{i=1}^n (x_i y_i)^t)^{1/t}$\r\nit's visible that after putting both sides to the power $t$ and multiplicating the left side, we will get there products $x_i x_j$ with both $i=j$ and $i \\neq j$ , while on the right side there will only be $i=j$.\r\nabove all this, the Holder inequality states that \r\n${1 \\over p} + {1 \\over q} = 1 \\  , \\  a_i , b_i \\geqslant 0 \\Rightarrow (\\sum_{i=1}^n a_i^p)^{1/p}(\\sum_{i=1}^n b_i^q)^{1/q} \\geqslant \\sum_{i=1}^n a_i b_i$\r\nsetting ${p' \\over s} = p$ i ${q' \\over s} = q$ we get\r\n$(\\sum_{i=1}^n a_i^{p'/s})^{s/p'}(\\sum_{i=1}^n b_i^{q'/s})^{s/q'} \\geqslant \\sum_{i=1}^n a_i b_i$\r\nputting both sides to the power $1 \\over s$ and substtuting $c_i = a_i^{1/s} , d_i = b_i^{1/s}$ we obtain\r\n$(\\sum_{i=1}^n c_i^{p'})^{1/p'}(\\sum_{i=1}^n d_i^{q'})^{1/q'} \\geqslant (\\sum_{i=1}^n (c_i d_i)^s)^{1/s}$ and ${1 \\over p'} + {1 \\over q'} = {1 \\over s}$\r\nand this is equivalent to\r\n$\\Vert \\vec{C} \\Vert_{p'} \\Vert \\vec{D} \\Vert_{q'} \\geqslant \\Vert \\mathcal{M}_2 (\\vec{C} , \\vec{D}) \\Vert_{(p'^{-1} + q'^{-1})^{-1}}$\r\nwe'll call this a \"generalized Holder inequality for two sequences/vectors\"\r\nand now returning to the proper inequality:\r\nwe have to prove that:\r\n$\\Vert \\mathcal{M}_k (\\vec{A_1} , \\vec{A_2} , ... , \\vec{A_k}) \\Vert_{(\\sum_{i=1}^k p_i^{-1})^{-1}} \\leqslant \\Vert \\vec{A_1} \\Vert_{p_1} \\Vert \\vec{A_2} \\Vert_{p_2} ... \\Vert \\vec{A_k} \\Vert_{p_k}$\r\nwe'll do this with induction (variable - $k$). the latter has to imply in\r\n$\\Vert \\mathcal{M}_{k+1} (\\vec{A_1} , \\vec{A_2} , ... , \\vec{A_k} , \\vec{A_{k+1}}) \\Vert_{(\\sum_{i=1}^k p_i^{-1} + p_{k+1}^{-1})^{-1}} \\leqslant \\Vert \\vec{A_1} \\Vert_{p_1} \\Vert \\vec{A_2} \\Vert_{p_2} ... \\Vert \\vec{A_k} \\Vert_{p_k} \\Vert \\vec{A_{k+1}} \\Vert_{p_{k+1}}$\r\ndue to the second given by me property of $\\ell$-multiplication we have:\r\n$\\Vert \\mathcal{M}_2 ( \\mathcal{M}_k (\\vec{A_1} , \\vec{A_2} , ... , \\vec{A_k} ) , \\vec{A_{k+1}}) \\Vert_{(\\sum_{i=1}^k p_i^{-1} + p_{k+1}^{-1})^{-1}} \\leqslant \\Vert \\vec{A_1} \\Vert_{p_1} \\Vert \\vec{A_2} \\Vert_{p_2} ... \\Vert \\vec{A_k} \\Vert_{p_k} \\Vert \\vec{A_{k+1}} \\Vert_{p_{k+1}}$\r\nlet us substitute $\\vec{M} = \\mathcal{M}_k (\\vec{A_1} , \\vec{A_2} , ... , \\vec{A_k} ) , \\vec{\\alpha} = \\vec{A_{k+1}} , s^{-1} = \\sum_{i=1}^k p_i^{-1}$ because i've had enough writing so far :)\r\nhence we get\r\n$\\Vert \\mathcal{M}_2 (\\vec{M} , \\vec{\\alpha}) \\Vert_{(s^{-1} + p_{k+1}^{-1})^{-1}} \\leqslant \\Vert \\vec{A_1} \\Vert_{p_1} \\Vert \\vec{A_2} \\Vert_{p_2} ... \\Vert \\vec{A_k} \\Vert_{p_k} \\Vert \\vec{\\alpha} \\Vert_{p_{k+1}}$\r\nwe can substitute $\\Vert \\vec{M} \\Vert_s$ for $\\Vert \\vec{A_1} \\Vert_{p_1} \\Vert \\vec{A_2}\\Vert_{p_2} ... \\Vert \\vec{A_k} \\Vert_{p_k}$ because due to inductive assumption the latter is larger, thus we make the right hand side smaller.\r\nhence:\r\n$\\Vert \\mathcal{M}_2 (\\vec{M} , \\vec{\\alpha}) \\Vert_{(s^{-1} + p_{k+1}^{-1})^{-1}} \\leqslant \\Vert \\vec{M} \\Vert_s \\Vert \\vec{\\alpha} \\Vert_{p_{k+1}}$\r\nand this is true due to generalized Holder inequality for two sequences/vectors ($\\vec{C} = \\vec{M} , \\vec{D} = \\vec{\\alpha} , p' = s , q' = p_{k+1}$)\r\n\r\nproof 2 (by jalowiec)\r\n\r\nlet us denote $(a_{1,1}a_{2,1}...a_{k,1})^s+(a_{1,2}a_{2,2}...a_{k,2})^s+...+(a_{1,n}a_{2,n}...a_{k,n})^s=c$ and write the inequality this way:\r\n$(\\frac{(a_{1,1}a_{2,1}...a_{k,1})^s}{c}\\cdot \\frac{a_{1,1}^{p_1}}{(a_{1,1}a_{2,1}...a_{k,1})^s}+\\frac{(a_{1,2}a_{2,2}...a_{k,2})^s}{c}\\cdot \\frac{a_{1,2}^{p_1}}{(a_{1,2}a_{2,2}...a_{k,2})^s}+...+\\frac{(a_{1,n}a_{2,n}...a_{k,n})^s}{c}\\cdot \\frac{a_{1,n}^{p_1}}{(a_{1,n}a_{2,n}...a_{k,n})^s})^\\frac{1}{p_1}\\cdot$ $\\cdot(\\frac{(a_{1,1}a_{2,1}...a_{k,1})^s}{c}\\cdot \\frac{a_{2,1}^{p_2}}{(a_{1,1}a_{2,1}...a_{k,1})^s}+\\frac{(a_{1,2}a_{2,2}...a_{k,2})^s}{c}\\cdot \\frac{a_{2,2}^{p_2}}{(a_{1,2}a_{2,2}...a_{k,2})^s}+...+\\frac{(a_{1,n}a_{2,n}...a_{k,n})^s}{c}\\cdot \\frac{a_{2,n}^{p_2}}{(a_{1,n}a_{2,n}...a_{k,n})^s})^\\frac{1}{p_2}\\cdot...$ $...\\cdot(\\frac{(a_{1,1}a_{2,1}...a_{k,1})^s}{c}\\cdot \\frac{a_{k,1}^{p_k}}{(a_{1,1}a_{2,1}...a_{k,1})^s}+\\frac{(a_{1,2}a_{2,2}...a_{k,2})^s}{c}\\cdot \\frac{a_{k,2}^{p_k}}{(a_{1,2}a_{2,2}...a_{k,2})^s}+...+\\frac{(a_{1,n}a_{2,n}...a_{k,n})^s}{c}\\cdot \\frac{a_{k,n}^{p_k}}{(a_{1,n}a_{2,n}...a_{k,n})^s})^\\frac{1}{p_k}\\geq 1$ \r\nlet us now use the inequality $q_1x_1+q_2x_2+...+q_nx_n\\geq x_1^{q_1}x_2^{q_2}...x_n^{q_n}$ ($\\sum_{j=1}^n q_j = 1$) for $q_j = \\frac{(a_{1,j}a_{2,j}...a_{k,j})^s}{c}$ in each of the \"wider\" brackets. it substitutes each bracket with \r\n$((\\frac{a_{1,j}^{p_1}}{(a_{1,j}a_{2,j}...a_{k,j})^s})^\\frac{1}{p_1}(\\frac{a_{2,j}^{p_2}}{(a_{1,j}a_{2,j}...a_{k,j})^s})^\\frac{1}{p_2}...(\\frac{a_{k,j}^{p_k}}{(a_{1,j}a_{2,j}...a_{k,j})^s})^\\frac{1}{p_k})^\\frac{(a_{1,j}a_{2,j}...a_{k,j})^s}{c}$, which is equal to $1$\r\nmultiplying the latter from $j=1$ to $j=n$ we obtain the initial inequality."
}
{
  "Tag": [
    "search"
  ],
  "Problem": "Someone asked if there's a relationship between math and musical skill.  Let's see!\r\n\r\n\r\n\r\nBy the way, I play piano and viola.",
  "Solution_1": "I am a percussionist. woohoo",
  "Solution_2": "violin.",
  "Solution_3": "i used to play clarinet...",
  "Solution_4": "I play trumpet but I'm not good at it",
  "Solution_5": "I'm a pianist.",
  "Solution_6": "[quote=\"Treething\"]I'm a pianist.[/quote]\r\nSame here.",
  "Solution_7": "cello",
  "Solution_8": "[quote=\"Danbert\"]Someone asked if there's a relationship between math and musical skill.  Let's see![/quote]\r\n\r\nThere's no \"Don't play\" choice there. So how do we say we don't play an instrument?",
  "Solution_9": "Piano since I was 5... so I've been playing for about 7 years.",
  "Solution_10": "played piano, hated it, gave it up after maybe 2 years....",
  "Solution_11": "piano since i was 5",
  "Solution_12": "[quote]There's no \"Don't play\" choice there. So how do we say we don't play an instrument?[/quote]\r\n\r\nOops.  Just check 'Other,' I guess.  \r\n\r\n[i]Are[/i] there any other instruments?",
  "Solution_13": "a 'don't play' option has been added",
  "Solution_14": "Kazoo.  :D \r\n\r\nSeriously, piano since I was 7. Violin since I was 10, but seriously thinking about giving it up this year. Didn't join school orchestra, might just play by myself a little bit.",
  "Solution_15": "I'm an asian girl. Asian girls plays strings, woodwind and piano, it's like a law.",
  "Solution_16": "i play lots of things",
  "Solution_17": "i have played the trumpet before (didn't work out so well) and i've played the piano for 10 years",
  "Solution_18": "I used to play sax but stopped and now I play mandoline :)",
  "Solution_19": "I play the piano, and I sing Indian music a bit. I can play a few songs on the guitar too! I have to admit, guitarists are very talented. :D",
  "Solution_20": "heh piano here... almost every asian guy at my school plays the piano or the violin lol",
  "Solution_21": "Vibraphone!!! (percussion)\r\nGo Cadets of Bergen County",
  "Solution_22": "i play accordian. \r\nBTW, which section does it belong to? Piano?",
  "Solution_23": "Trumpet!   :)",
  "Solution_24": "Piano",
  "Solution_25": "i used to play recorder, then i began playing guitar and now i play piano and drum... :)",
  "Solution_26": "I play both violin and piano. Piano for like 6 years and violin for like 1.",
  "Solution_27": "Alto Saxophone!\r\n\r\nAnd learning the piano.",
  "Solution_28": "Piano since I was five...\r\n\r\nPecussionist\r\n\r\nTrumpet Player\r\n\r\nHarmonica (Hey this actually does take some skill)\r\n\r\nGuitar...Kind of...  It a bit complicated.\r\n\r\n :P",
  "Solution_29": "Piano for ten years, singing in the school chorus for 3 years, trombone for school for 5 years.\r\n\r\nMusic is one of the single most important activities one can take part in. Music and sports. And...competetive math :)"
}
{
  "Tag": [],
  "Problem": "A car-rental company charges $ \\$50$ for the first 500 miles driven and $ \\$0.25$ for each additional mile. Given that Meghan was charged $ \\$239.25$ for total mileage, how many miles did she drive?",
  "Solution_1": "We first compute the number of miles she drove past $ 500$. Since she spent $ \\$189.25$, on those miles, we have that $ \\frac{189.25}{.25}\\equal{}757$. Thus, the total number of miles is $ 757\\plus{}500\\equal{}\\boxed{1257}$."
}
{
  "Tag": [
    "Euler"
  ],
  "Problem": "Ai bi\u1ebft \u0111\u01b0\u1eddng th\u1eb3ng \u01a0-le cho t\u1ee9 gi\u00e1c v\u00e0 n-gi\u00e1c kh\u00f4ng ch\u1ec9 cho e v\u1edbi :?:",
  "Solution_1": "M\u1ecdi ng\u01b0\u1eddi d\u1ecbch h\u1ed9 e c\u00e1i n\u00e0y v\u1edbi\r\n[url]http://forumgeom.fau.edu/FG2006volume6/FG200634.pdf[/url]\r\nD\u1ecbch trang 2,3,4 th\u00f4i nha.\r\nC\u1ea3m \u01a1n nh\u00ecu! :)",
  "Solution_2": "[quote=\"Kyoshiro\"]M\u1ecdi ng\u01b0\u1eddi d\u1ecbch h\u1ed9 e c\u00e1i n\u00e0y v\u1edbi\n[url]http://forumgeom.fau.edu/FG2006volume6/FG200634.pdf[/url]\nD\u1ecbch trang 2,3,4 th\u00f4i nha.\nC\u1ea3m \u01a1n nh\u00ecu! :)[/quote]\r\nKh\u00f4ng m\u1edf \u0111\u01b0\u1ee3c em \u00e0!!! :(",
  "Solution_3": "E v\u1eabn m\u1edf \u0111\u01b0\u1ee3c m\u00e0.\r\nAnh th\u1eed link n\u00e0y xem [url]http://209.85.175.104/search?q=cache:qCSjtWryIaUJ:forumgeom.fau.edu/FG2006volume6/FG200634.pdf+the+Euler+line+of+quadrilateral&hl=vi&ct=clnk&cd=6&gl=vn[/url]"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "trigonometry",
    "binomial theorem"
  ],
  "Problem": "please don't use binomial theorem as I have not learnt it yet\r\n\r\n------------------------------\r\n\r\nUse De Moivre's theorem to show that $ cos3\\theta \\equal{} 4cos^{3}\\theta\\minus{}3cos\\theta.$ Hence solve $ 8x^{3}\\minus{}6x\\minus{}1 \\equal{} 0.$Deduce that $ cos\\frac{\\pi}{9}\\equal{} cos\\frac{2\\pi}{9}\\plus{}cos\\frac{4\\pi}{9}$\r\n\r\nI done the show $ cos3\\theta\\minus{}3cos\\theta$ bit. So don't waste time on that.\r\nHow do I solve the eqn now?",
  "Solution_1": "$ cos{3\\phi}= Re[(\\cos\\phi+i\\sin\\phi)^{3}]$\r\n$ \\Rightarrow\\cos{3\\phi}=\\cos^{3}{\\phi}-3\\cos{\\phi}\\sin^{2}{\\phi}$\r\n$ \\Rightarrow\\cos{3\\phi}=\\cos^{3}{\\phi}-3\\cos{\\phi}(1-\\cos^{2}{\\phi})$\r\n$ \\Rightarrow\\cos{3\\phi}=\\cos^{3}{\\phi}-3\\cos{\\phi}+3\\cos^{3}{\\phi})$\r\n$ \\Rightarrow\\cos{3\\phi}= 4\\cos^{3}{\\phi}-3\\cos{\\phi}$\r\n\r\nLet $ x =\\cos{\\phi}$\r\n$ 8x^{3}-6x-1 = 0$\r\n$ \\Rightarrow 2(4\\cos^{3}{\\phi}-3\\cos{\\phi})-1 = 0$\r\n$ \\Rightarrow\\cos{3\\phi}=\\frac{1}{2}$\r\n$ \\Rightarrow 3\\phi =\\frac{\\pi}{3},\\frac{5\\pi}{3},\\frac{7\\pi}{3}$\r\n$ \\Rightarrow\\phi =\\frac{\\pi}{9},\\frac{5\\pi}{9},\\frac{7\\pi}{9}$\r\n$ \\Rightarrow x =\\cos{\\frac{\\pi}{9},\\cos{\\frac{5\\pi}{9},\\cos{\\frac{7\\pi}{9}}}}$",
  "Solution_2": "oh I see it now.\r\n\r\nthanks for your help. Appreciate it.",
  "Solution_3": "I have just figured out the last part of the question.\r\n\r\n$ 8x^{3}-6x-1 = 0$\r\n$ \\Rightarrow x^{3}-\\frac{3x}{4}-\\frac{1}{8}= 0$\r\n\r\nand we have:\r\n$ (x-\\cos{\\frac{\\pi}{9})(x-\\cos{\\frac{5\\pi}{9})(x-\\cos{\\frac{7\\pi}{9}) = 0}}}$\r\n$ \\Rightarrow (x-\\cos{\\frac{\\pi}{9})(x+\\cos{\\frac{4\\pi}{9})(x+\\cos{\\frac{2\\pi}{9}) = 0}}}$\r\n(since $ -\\cos{\\frac{5\\pi}{9}=\\cos{\\frac{4\\pi}{9}}}$ and $ -\\cos{\\frac{7\\pi}{9}=\\cos{\\frac{2\\pi}{9}}}$\r\n\r\nConsider the coefficient of the term $ x^{2}$ of $ x^{3}-\\frac{3x}{4}-\\frac{1}{8}= 0$, then we have:\r\n$ -\\cos{\\frac{\\pi}{9}+\\cos{\\frac{2\\pi}{9}+\\cos{\\frac{4\\pi}{9}= 0}}}$\r\n$ \\Rightarrow\\cos{\\frac{\\pi}{9}=\\cos{\\frac{2\\pi}{9}+\\cos{\\frac{4\\pi}{9}}}}$",
  "Solution_4": "[quote=\"IrinaSharova\"]Let $ x \\equal{}\\cos{\\phi}$\n$ 8x^{3}\\minus{}6x\\minus{}1 \\equal{} 0$\n$ \\Rightarrow 2(4\\cos^{3}{\\phi}\\minus{}3\\cos{\\phi})\\minus{}1 \\equal{} 0$\n[/quote]\r\n\r\nI have a small question about this substitution. Wouldn't this restrict $ x\\in [\\minus{}1, 1]$? So, hypothetically, if one were given another polynomial with a root that is $ <\\minus{}1$ or $ > 1$, then this substitution would be ineffective, correct?\r\n\r\nAnd also, how would one prove that all the roots in a polynomial is $ \\in [\\minus{}1, 1]$?",
  "Solution_5": "[quote=\"p4fn2w\"][quote=\"IrinaSharova\"]Let $ x \\equal{}\\cos{\\phi}$\n$ 8x^{3}\\minus{}6x\\minus{}1 \\equal{} 0$\n$ \\Rightarrow 2(4\\cos^{3}{\\phi}\\minus{}3\\cos{\\phi})\\minus{}1 \\equal{} 0$\n[/quote]\n\nI have a small question about this substitution. Wouldn't this restrict $ x\\in [\\minus{}1, 1]$? So, hypothetically, if one were given another polynomial with a root that is $ <\\minus{}1$ or $ > 1$, then this substitution would be ineffective, correct?\n\nAnd also, how would one prove that all the roots in a polynomial is $ \\in [\\minus{}1, 1]$?[/quote]\r\nWell, doesn't it increase/decrease without bound in both directions?",
  "Solution_6": "[quote=\"archimedes1\"]\nWell, doesn't it increase/decrease without bound in both directions?[/quote]\r\n\r\nI'm not sure I follow... can you please elaborate more?",
  "Solution_7": "[quote=\"p4fn2w\"][quote=\"IrinaSharova\"]Let $ x \\equal{}\\cos{\\phi}$\n$ 8x^{3}\\minus{}6x\\minus{}1 \\equal{} 0$\n$ \\Rightarrow 2(4\\cos^{3}{\\phi}\\minus{}3\\cos{\\phi})\\minus{}1 \\equal{} 0$\n[/quote]\n\nI have a small question about this substitution. Wouldn't this restrict $ x\\in [\\minus{}1, 1]$?[/quote]\r\n\r\nActually, when $ \\phi$ is allowed to be complex $ \\cos$ can take on any value in $ \\mathbb{C}$.  In particular, often it is desirable to [b]define[/b] the cosine via the complex exponential\r\n\r\n$ \\cos z \\equal{}\\frac{e^{iz}\\plus{}e^{\\minus{}iz}}{2}$\r\n\r\nso that, for example, $ \\cos i \\equal{}\\frac{e\\plus{}\\frac{1}{e}}{2}> 1$."
}
{
  "Tag": [],
  "Problem": "Adam grew rabbits. Adam grew lots of rabbits. Adam started with 229 rabbits that he bought from various stores in 1999. It is now 2006 and adam has 1727 rabbits. assuming that the rabbit population grew exponentially find,\r\n\r\na)The formula for the growth in (t) years\r\n\r\n\r\n\r\nb) the year the rabbit population grew by 50%",
  "Solution_1": "I don't think there's enough info... :?",
  "Solution_2": "[hide]Although a few more numbers would be nice, I think we can manage.\n\n$y=a\\cdot b^t$\n$229=a \\cdot b^0$\n$229=a$\n\n$1727=229 \\cdot b^7$\n$b\\approx 1.33$\n\n$y=229\\cdot 1.33^t$\n\n^Is assuming that the rabbits grew exponentially [b]per year[/b]\n\n$\\frac{229\\cdot 1.33^{t+1}}{229\\cdot 1.33^t} \\neq 1.5$ (I think...)\n\nOr, if you mean when the rabbits were 50% then the original level.. it's\n\n$229\\cdot 1.5=229\\cdot 1.33^t$\n$t\\approx1.42$\nSo approximately 1.42 years after 1999.\n[/hide]",
  "Solution_3": "surge got it right, if you round to 1.3346 you get a little more but that's O.K. otherwise, good job!",
  "Solution_4": "I was going to do that, but isn't the more general form of exponential growth:\r\n\\[ P = P_o * C ^{a*t + b} \\]\r\nor\r\n\\[ P = P_o * C ^{a*t + b} + d \\]"
}
{
  "Tag": [
    "geometry",
    "perimeter",
    "number theory",
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "Good afternoon\r\nIm having a trouble with this :(\r\nLet $ n\\geq 3$ be an integer. Suppose that a convex $ n$-gon is drawn in the plane, and then each pair of nonadjacent corner points is joined by a straight line through the interior. Suppose that no $ 3$ of these lines through the interior meet at a common point in the interior. Let $ f_n$ be the number of regions into which the interior of the $ n$-gon is divided. $ f_3 \\equal{} 1, f_4 \\equal{} 4$. Use Euler's Formula to find $ f_n$.\r\n\r\nMy attempt has been thus: Euler's formula is $ p \\minus{} q \\plus{} r \\equal{} c \\plus{} 1$, with $ p$ vertices, $ q$ edges, $ r$ faces and $ c$ components.\r\nSince the graph is connected, $ c \\equal{} 1$. Now the convex $ n$-gon has p vertices...plus the number of intersections of lines in the interior. I've figured that taking any $ 4$ points and drawing the diagonals we get $ 1$ intersection.\r\nSo the number of vertices we have is actually $ n \\plus{} \\binom{n}{4}$. I did this so that the graph becomes planar to use Euler's Formula...if I am on the right track.\r\nSo we have $ r \\equal{} f_n \\equal{} q \\minus{} (n \\plus{} \\binom{n}{4}) \\plus{} 1$. I don't remember why I substracted $ 1$ from the $ 2$. >_<\r\nAnyways my problem is I cannot find a closed expression for the number of edges :( \r\nThanks for any help =)\r\n\r\nEdit: I subtracted 1 because the outer region is not a region considered for the problem =)",
  "Solution_1": "Well, let's see... every interior point is an endpoint for exactly four edges, while every point on the perimeter is an endpoint for exactly $ n\\minus{}1$.  So $ 4(p\\minus{}n) \\plus{} (n\\minus{}1)n \\equal{} 2q$.  Combining this with $ p\\equal{}n \\plus{} {n\\choose 4}$ gives $ q \\equal{} {n\\choose 2} \\plus{} 2{n\\choose 4}$.\r\n\r\nFinally, Euler's formula gives $ r \\equal{} 1\\minus{}p\\plus{}q \\equal{} 1 \\minus{} n \\plus{} {n\\choose 2} \\plus{} {n\\choose 4}$.\r\n\r\nNow try it for the [url=http://www-math.mit.edu/~poonen/papers/ngon.pdf]regular n-gon[/url] :)",
  "Solution_2": "Hmm yea I have read that paper before. Michael Rubenstein, one of the co-authors, is a professor at my school (Waterloo). He teachers most of the number theory courses (analytic, etc..)"
}
{
  "Tag": [],
  "Problem": "When the measure of $ \\angle A$ is added to the measure of the complement of $ \\angle B$, their sum is $ 120^\\circ$. $ \\angle A$ measures $ 86^\\circ$. How many degrees are in the measure of $ \\angle B$?",
  "Solution_1": "$ 86 \\plus{} (90 \\minus{} B) \\equal{} 120$\r\n\r\n$ 176 \\minus{} B \\equal{} 120$\r\n\r\n$ B \\equal{} 56$ degrees"
}
{
  "Tag": [
    "geometry",
    "incenter",
    "geometric transformation",
    "rotation",
    "trapezoid",
    "perpendicular bisector",
    "angle bisector"
  ],
  "Problem": "On the internal bisector of the angle <ACB of the isosceles triangle ABC with <BAC = 100 degs take the point D so that CD = BC.\r\nFind the angle <BAD.\r\n\r\nSorry to still bother with this kind of problems, but I just found two nice synthetical solutions...\r\n\r\nBest regards,\r\nsunken rock",
  "Solution_1": "Denote $ m(\\widehat {BAD})\\equal{}x$ , $ P\\in AB\\cap CD$ and $ R\\in AC\\cap BD$ . Thus, the point $ P$ is the incenter \r\n\r\nof the triangle $ EBC$ and $ AEDP$ is cyclic. In conclusion, $ x\\equal{}60^{\\circ}\\minus{}x$ , i.e. $ m(\\widehat{BAD})\\equal{}30^{\\circ}$ .",
  "Solution_2": "Sometimes the proof you have in mind drives you to suppose the problem was more difficult as it actually is...\r\nI was thinking the problem was college level.\r\n\r\nYour E must be common to CA and BD.\r\n\r\nI took as the easiest method the following:\r\nA 20 degs rotation around C brings B to D and A to F, on (BA, hence DF = AC. Take now Q on (BF) so that FQ = FD; the triangles DFQ and ACF are congruent, thence DQ = AF and it's well known that the angle <ADF = 10 degs, which gives <BAD = 30 degs.\r\n\r\n[i] Remark: In the triangle BDF (20-40-120 degs), if take A on (BF) s.t. AB = DF, then <ADF = 10 degs [/i]\r\n\r\nA 2nd method:\r\nThe perpendicular bisector of AC intersects CD at M, then take N the symmetrical of M about the perpendicular bisector of BC, the triangle ANM is equilateral, the trapezoid BCMN isosceles with BM angle bisector of <NBC = 20 degs, hence BD = CM = AM and \r\n<DBM = <AMB = 70 degs, thence DBMA is an isosceles trapezoid, with <(AB, DM) = 60 degs.\r\n\r\nBest regards,\r\nsunken rock"
}
{
  "Tag": [
    "linear algebra",
    "matrix",
    "parameterization",
    "graph theory",
    "linear algebra unsolved"
  ],
  "Problem": "a strongly regular graph $srg(v,k,\\lambda,\\mu)$\r\n\r\nis a graph with v vertices, each vertex has k neighbours, two neigbours have $\\lambda$ common neighbours\r\ntwo nonneighbours have $\\mu$ common neighbours\r\n\r\nYou can prove that $(v-1-k)\\mu =k(k-1-\\lambda)$ (ask if you want this)\r\n\r\nlet us assume they are not disconnected or have a complement that is disconnected\r\n\r\nConsider its adjacency matrix\r\nOne can prove that if $2 k\\neq (v-1)(\\mu-\\lambda)$ , the eigenvalues are integers (just write out the eigenvalues)\r\n\r\nProve that all strongly regular graphs with noninteger eigenvalues are conference graphs : \r\n\r\n$srg(2 k+1,k,\\frac{k}{2}-1,\\frac{k}{2})$\r\n\r\nI find it weird that two equations on four parameters imply this.\r\n\r\nApologies if this is the wrong place.  I placed it here because of the eigenvalues.  What is in general the best place for graph theory?",
  "Solution_1": "Maybe a better place is the advanced forum?  :? \r\nbecause i do realise it's more graph theory than matrices",
  "Solution_2": "i think i got it, for the interested :\r\n\r\n $2 k\\neq (v-1)(\\mu-\\lambda)$\r\n\r\nif the graph is not complete :$v-1>k$\r\n\r\nthis implies $\\mu-\\lambda<2$ but it has to divide $2 k$ and therefore \r\n\r\n$\\mu-\\lambda=1$\r\n\r\nthis implies $v=2 k+1$\r\n\r\nuse the other equation i gave, you quickly find everything\r\n\r\napparently type I strongly regular graphs are also used for conference graphs"
}
{
  "Tag": [
    "function",
    "algebra",
    "polynomial"
  ],
  "Problem": "f is an even function, g and h are odd functions, all 3 are polynomials. given: f(1)=0, f(2)=1, f(3)=-5, g(1)=-1, g(-3)=2, g(5)=3, h(1)=3, h(3)=5, and h(5)=1. \r\nFind f(g(h(1)))+g(h(f(3)))+h(f(g(-1)))\r\n\r\n[hide=\"Question\"]For the last one I simplified it down to h(0) and put unsolvable because in the previous given statements, h(0) was not given. However looking at the solution key, it stated(or assumed) that h(0)=0...can you assume such a statement in this case, or is that wrong?[/hide]",
  "Solution_1": "$ h$ is odd, so $ h(0) \\equal{} h(\\minus{}0) \\equal{} \\minus{}h(0)$, which is true if and only if $ h(0) \\equal{} 0$.  This is true of any odd function.",
  "Solution_2": "An odd function means it's symmetric across the origin. An even is symmetric across the y-axis. For it to be symmetric across the origin, doesn't that mean it has to go through the origin? Which means for any odd function, f(0)=0.",
  "Solution_3": "[quote]An odd function means it's symmetric across the origin. An even is symmetric across the y-axis. For it to be symmetric across the origin, doesn't that mean it has to go through the origin? Which means for any odd function, f(0)=0.[/quote]\r\n\r\nConsider $ f(x)\\equal{}\\frac{1}{x}$",
  "Solution_4": "Perhaps I should rephrase that. If an odd function is defined at x=0, then it goes through the origin. Nice catch grn_trtle!\r\n\r\nBut in this question, as they are polynomials, they are all defined at x=0.",
  "Solution_5": "I think that holds for all polynomials. Notice that if some we have some function $ p(x)\\equal{}a_0x^1\\pm a_1x^3\\pm a_2x^5\\pm \\cdots\\pm a_nx^{2n\\plus{}1}$, clearly odd and continuous, then $ p(0)\\equal{}0$. Basically, an odd polynomial cannot have a y-intercept not equal to 0 without losing its origin symmetry.\r\n\r\nOr something like that.",
  "Solution_6": "\"Odd function\" implicitly means \"odd function on $ \\mathbb{R}$,\" and $ f(x) \\equal{} \\frac{1}{x}$ is not defined on all of $ \\mathbb{R}$.  The only value to which it can be set at zero to make it odd is $ 0$.  The condition that the functions are polynomial is not necessary (and neither is the weaker condition that they are continuous)."
}
{
  "Tag": [
    "linear algebra",
    "matrix",
    "algebra",
    "function",
    "domain",
    "linear algebra unsolved"
  ],
  "Problem": "S is called a twisting subsituation if $S^2$ is indentically equal .Prove that :All of subsituation can be expressed form product of the twisting situations",
  "Solution_1": "... just wondering, do you mean idempotent matrix? every matrix can be written as a product of idempotent matrices? also what domain are you working on?"
}
{
  "Tag": [
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "Let triangle $ ABC$. Prove\r\n$ a^4b^2\\plus{}b^4c^2\\plus{}c^4a^2\\geq a^3c^2b\\plus{}b^3a^2c\\plus{}c^3b^2a$\r\nQuickly!",
  "Solution_1": "[quote=\"mufc\"]Let triangle $ ABC$. Prove\n$ a^4b^2 \\plus{} b^4c^2 \\plus{} c^4a^2\\geq a^3c^2b \\plus{} b^3a^2c \\plus{} c^3b^2a$\nQuickly![/quote]\r\nIt's true for all reals $ a,$ $ b$ and $ c.$ :wink:",
  "Solution_2": "[quote=\"mufc\"]Let triangle $ ABC$. Prove\n$ a^4b^2 \\plus{} b^4c^2 \\plus{} c^4a^2\\geq a^3c^2b \\plus{} b^3a^2c \\plus{} c^3b^2a$\nQuickly![/quote]\r\nInequality follows from these three am-gm inequalities.\r\n$ a^4b^2\\plus{}b^4c^2 \\geq 2b^3a^2c$\r\n$ a^4b^2\\plus{}c^4a^2\\geq 2a^3c^2b$\r\n$ b^4c^2\\plus{}c^4a^2\\geq 2c^3b^2a$.",
  "Solution_3": "$ (4,4,4)  >> (3,3,3)$ and $ (6,6,6) >> (5,5,5)$,  the end : $ (6,6,6)\\equal{}(6,6,6)$",
  "Solution_4": "[quote=\"Hong Quy\"]$ (4,4,4) > > (3,3,3)$ and $ (6,6,6) > > (5,5,5)$,  the end : $ (6,6,6) \\equal{} (6,6,6)$[/quote]\r\n\r\nWhatever you mean by this...\r\n\r\n  dg"
}
{
  "Tag": [
    "invariant",
    "combinatorics proposed",
    "combinatorics"
  ],
  "Problem": "A regular $ (5 \\times 5)$-array of lights is defective, so that toggling the switch for one light causes each adjacent light in the same row and in the same column as well as the light itself to change state, from on to off, or from off to on. Initially all the lights are switched off. After a certain number of toggles, exactly one light is switched on. Find all the possible positions of this light.",
  "Solution_1": "label the light by its position (i;j) $1\\leq\\i\\leq\\ 5;1\\leq\\j\\leq\\ 5$\r\ncall (i;j) good if there is an algorithm make all lights but light(i;j) off\r\nnotice that ;if (i;j) is good then the light symmetric to it through 2 diagonal is also good\r\nnow we define an invariant\r\nconsider set S (1;1);(1;2);(1;4);(1;5);(3;1);(3;2);(3;4);(3;5);(5;1);(5;2);(5;1);(5;3)\r\nthe parity of M,the number of light on in S is invariant.A light not in S has even adjacent light in S(0 or 2);so toggle it will make M unchanged,or reduce 2 or increase 2,so parity is unchanged.A light in S has one adjacent light in S,so toggle it also make the parity of M unchanged\r\nInitially,M is 0,so a good position (i;j) cannot in S\r\nalso (1;3);(2;1);(4;1);(2;5);(4;5) is not good since it is symmetric to a position in S\r\nso there are 5 possible good position and we can check that all of it works\r\neasy problem,right?",
  "Solution_2": "The parity in M is not invariant just operate on 2;5. However if u consider a square bound by 1;1, 1;5, 4;1, 4;5 the parity of on lights in that region is invariant and if we reflect this region we conclude that there is no light that can be on while all others are off.",
  "Solution_3": "My progress(yet to reach the whole solution :blush: )\nLet me define \"good\" bulb; (1,1), (1,5), (2,2), (2,4), (3,3), (4,2), (4,4), (5,1), (5,5) are good.\nLet me call the other \"bad\".\nThen, the number of bad bulb on will be always even.\nSo, the group of good bulbs contains the solution.\nI'm not a contestant, but if I were a contestant, could I get partial credit? :D",
  "Solution_4": "[quote=\"peregrinefalcon88\"]The parity in M is not invariant just operate on 2;5. However if u consider a square bound by 1;1, 1;5, 4;1, 4;5 the parity of on lights in that region is invariant and if we reflect this region we conclude that there is no light that can be on while all others are off.[/quote]\nIncorrect; operating on any of a lot of lights already tell that it's wrong. Try (3,3).\n\nThe invariant is actually the set dhthstn said, only with a few typos there. The parity of the number of turned on lights in the set (1,1), (1,2), (1,4), (1,5), (3,1), (3,2), (3,4), (3,5), (5,1), (5,2), (5,4), (5,5) is an invariant. So the bulbs can only be in the complementary region. Reflecting through the main diagonal, we get that the set (1,1), (1,3), (1,5), (2,1), (2,3), (2,5), (4,1), (4,3), (4,5), (5,1), (5,3), (5,5) also have the same invariant. Thus the lights that are in the complement of both regions are the only candidates; these are (2,2), (2,4), (3,3), (4,2), (4,4). Now for the construction...",
  "Solution_5": "Another way of looking at the invariants above is to think of the lights' statuses as a collection of $25$ equations $\\pmod 2$, where $a_{i,j} \\equiv 1 \\pmod 2$ if and only if the light at $(i,j)$ is on (otherwise it is $0$) and $a_{i,j} \\equiv a_{i,j} + a_{i-1,j} + a_{i+1,j} + a_{i,j-1} + a_{i,j+1} \\pmod 2$ (with some of the RHS terms missing if $(i,j)$ is an edge/corner square). Then if a collection $S$ of squares which, when all toggled (starting from no lights on), leave all lights off, then no square in $S$ can be a lone illuminated square since summing the corresponding equations $\\pmod 2$ give a LHS of $0 \\pmod 2$ and a RHS of $1 \\pmod 2$ (a sum of $0$s along with a single $1$ corresponding to the one illuminated square).\n\nSuch sets include, as partly mentioned above:\n\n$\\{(1,1), (1,2), (1,4), (1,5), (3,1), (3,2), (3,4), (3,5), (5,1), (5,2), (5,4), (5,5)\\}$, implying that the lone illuminated square cannot be a corner square\n\n$\\{(1,2), (1,3), (1,4), (2,1), (2,3), (2,5), (3,1), (3,2), (3,4), (3,5), (4,1), (4,3), (4,5), (5,2), (5,3), (5,4)\\}$, implying that the lone illuminated square must be on a diagonal.\n\nThe remaining five squares are achievable. For the center square, toggle $(1,1), (1,2), (2,3), (3,1), (3,3), (3,4), (4,1), (4,5), (5,2), (5,3), (5,5)$. To achieve only $(4,4)$ illuminated, toggle $(1,4), (2,3), (2,4), (2,5), (3,2), (4,1), (4,2), (4,4), (4,5), (5,2), (5,4)$ (rotate this set to get the other three of $(2,2), (2,4), (4,2)$).",
  "Solution_6": "[quote=dhthstn]label the light by its position (i;j) $1\\leq\\i\\leq\\ 5;1\\leq\\j\\leq\\ 5$\ncall (i;j) good if there is an algorithm make all lights but light(i;j) off\nnotice that ;if (i;j) is good then the light symmetric to it through 2 diagonal is also good\nnow we define an invariant\nconsider set S (1;1);(1;2);(1;4);(1;5);(3;1);(3;2);(3;4);(3;5);(5;1);(5;2);(5;1);(5;3)\nthe parity of M,the number of light on in S is invariant.A light not in S has even adjacent light in S(0 or 2);so toggle it will make M unchanged,or reduce 2 or increase 2,so parity is unchanged.A light in S has one adjacent light in S,so toggle it also make the parity of M unchanged\nInitially,M is 0,so a good position (i;j) cannot in S\nalso (1;3);(2;1);(4;1);(2;5);(4;5) is not good since it is symmetric to a position in S\nso there are 5 possible good position and we can check that all of it works\neasy problem,right?[/quote]\n\nI think(if understands correctly) if (1,2),(3,2) is on, (5,2) is off, then operates on (2,2) make M alternates the parity.",
  "Solution_7": "Is there any other invariant? what about bigger fields?",
  "Solution_8": "I wonder how to determine a binary matrix(relation matrix) is invertible? Or restrict to some symmetric case like this problem.  Can anyone recommend some theoretical or computational reference ! :) ",
  "Solution_9": " The construction took me more time than the solution itself :) \nMy solution is very similar to dhthstn's, so I'll just post the construction.\nIn the construction, black squares are the lights which we will toggle:\n[hide=First construction] Costruction for the square $(2,4)$: (and $(2,2),(4,4),(4,2)$ which follows by symmetry)\n$$\\begin{bmatrix}\n \\square& \\blacksquare  & \\square & \\blacksquare & \\square\\\\ \n \\blacksquare& \\blacksquare & \\square & \\blacksquare &\\blacksquare \\\\ \n \\square& \\square & \\square & \\blacksquare & \\square\\\\ \n \\blacksquare & \\blacksquare & \\blacksquare & \\square & \\square\\\\ \n \\square& \\blacksquare & \\square &\\square  & \\square\n\\end{bmatrix}$$[/hide]\n[hide=Second construction]Construction for the square $(3,3)$:\n$$\\begin{bmatrix}\n\\blacksquare & \\square & \\blacksquare & \\blacksquare & \\square\\\\ \n\\blacksquare & \\square & \\square & \\square & \\blacksquare\\\\ \n\\square & \\blacksquare & \\blacksquare & \\square & \\blacksquare\\\\ \n\\square & \\square & \\blacksquare & \\square & \\square\\\\ \n\\square & \\square & \\square & \\blacksquare & \\blacksquare\n\\end{bmatrix}$$[/hide]"
}
{
  "Tag": [
    "group theory",
    "abstract algebra",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "Let $ G$ be an additive subgroup of $ (R, \\plus{})$. Show that either $ G$ is dense in $ R$or it is discrite (\u0434\u0438\u0441\u043a\u0440\u0435\u0442\u043d\u0430\u044f).",
  "Solution_1": "Sketch:\r\nIf there is a smallest positive element in that subgroup, this element generates the whole group. If such an element doesn't exist, then there are arbitrary small elements. This implies density."
}
{
  "Tag": [
    "function",
    "limit",
    "calculus",
    "derivative",
    "calculus computations"
  ],
  "Problem": "If $ f, g$ are non negative real valued functions, where $ f$ is continuous,  is it true that $ \\inf_{x\\in E}f(g(x))\\equal{}f\\left(\\inf_{x\\in E}g(x)\\right)$?",
  "Solution_1": "No, it is not true. Take for example $ f(x)\\equal{}|lnx|$, $ g(x)\\equal{}x^2\\plus{}\\frac{1}{2}$, $ E\\equal{}\\mathbb R^\\plus{}$.",
  "Solution_2": "The question is same as if $ E$ is set of non negative reals then $ \\inf f(E) \\equal{} f\\left(\\inf{E}\\right)$.\r\n\r\nI hope the result would be true for monotone increasing continuous $ f$.\r\nHere's a proof:\r\n\r\n$ f\\left(\\inf{E}\\right)\\le f(x)$ for all $ x\\in E$ ( since $ f$ is increasing)\r\nTherefore $ f\\left(\\inf{E}\\right)\\le \\inf f(E)$\r\n\r\nOn the other hand, \r\nLet $ x_n$ be sequence in $ E$ s.t. $ x_n\\to \\inf E$\r\n\r\nNow $ \\inf f(E)\\le f(x_n)$ for every $ n$\r\n\r\nSo, $ \\inf f(E)\\le \\lim_{n\\to \\infty}f(x_n)$\r\n$ \\equal{} f\\left(\\lim_{n\\to \\infty}x_n\\right)$   (using continuity of $ f$)\r\n$ \\equal{} f\\left(\\inf{E}\\right)$\r\n\r\nFor monotone decreasing continuous $ f$ I think the result would be $ \\sup f(E) \\equal{} f\\left(\\sup{E}\\right)$.",
  "Solution_3": "It seems fine.  Note that if you are assuming that $ f$ is monotone, not only do you get continuity a.e., you get a finite derivative a.e. (see [url=http://en.wikipedia.org/wiki/Lebesgue_differentiation_theorem]Lebesgue's Differentiation Theorem[/url] for the second one).  And if you assume $ f$ is surjective, you're guaranteed continuity on the interval."
}
{
  "Tag": [
    "MIT",
    "college",
    "Harvard",
    "Princeton",
    "geometry",
    "Stanford",
    "email"
  ],
  "Problem": "Hi all!\r\n\r\n  I'm going to take 3 SAT subject test in December 2; 2 math and a physics \r\ntest.I am taking SAT test fo the first time and I would like someone who \r\nhave already taken it to describe the atmospher during the test.I think it \r\nwill help me to score batter if I know what to expect.\r\n\r\n  So some things I would like to know.Can I choose the order of the exams?\r\nI mean if I want to take Math 1 then Physics and in the and Math \r\n2 can I do that or I have to write the exams in that order in which they will tell me on \r\nthe test day?Also, can I make notes on the test form( I mean it may be \r\nhelpful to note the questions that you miss).At last I will be gratefull to \r\nhear any advice that you think can help me during the test.     \r\n\r\nThank you!\r\n\r\nTigran",
  "Solution_1": "Hrmm well in response to your questions, Yes you can choose whatever order you want to take the exams. All the tests will come in a big book and you just flip to a page to take the exams. You can mark up the test as much as you want and it won't be graded.\r\n\r\nMy advice is to get a good nights sleep because even though it is slightly shorter than the SAT I's, it is still quite demanding on concentration. Cramming usually doesn't work and I don't know about other states but I believe Math II is the only recommended one to take since Math II is like the Math I plus more.",
  "Solution_2": "Thanks for the reply and advice.And congratulations with the first post in  Mathlinks  :lol: .",
  "Solution_3": "As far as i know, if you are going to take Math 2, i heard that you don't have to take math 1. (my school recommends to take only Math 2...) BTW, I'm taking Physics on Dec. 2 too..(i should've taken last year... :( )",
  "Solution_4": "[quote=\"kimby_102\"]As far as i know, if you are going to take Math 2, i heard that you don't have to take math 1. (my school recommends to take only Math 2...) [/quote]\r\nAre you sure about it? :maybe: \r\nI think some of my friends hwo are students already took both math1 ant math 2.So,anyway I 'm going to take them both because I have already registered :) .But I don't think I will have any problems with math,the difficult part for me is physics :( .",
  "Solution_5": "I took all 3 (not at the same time).  Taking Math I and II is good.  When applying to college, I found some schools asked for either, while some asked for one specifically.  They're both easy, though.\r\n\r\nOn the physics, there's a ton of leeway.  You can get quite a few wrong and still get 800.  One thing to look out for- when I took it there were a couple of questions on topics which weren't part of the AP Physics B curriculum (especially relating to sound).  I specifically remember a question about timbre.  So look out for that, but it's nothing major.\r\n\r\nAnd good luck.  Taking 3 is exhausting.",
  "Solution_6": "[quote=\"Tiks\"][quote=\"kimby_102\"]As far as i know, if you are going to take Math 2, i heard that you don't have to take math 1. (my school recommends to take only Math 2...) [/quote]\nAre you sure about it? :maybe: \nI think some of my friends hwo are students already took both math1 ant math 2.So,anyway I 'm going to take them both because I have already registered :) .But I don't think I will have any problems with math,the difficult part for me is physics :( .[/quote]\r\n\r\nafter you register you can change, it doesnt matter what you registered for. on the test day they give you a big booklet, just do the test that you WANT to do. Dont even do math 1, just do math 2.",
  "Solution_7": "i took both math 1 and 2 and 2 is a tiny little harder than 1 but has a way better curve so your scores comes out better.",
  "Solution_8": "[quote=\"ShoeFactory\"][quote=\"Tiks\"][quote=\"kimby_102\"]As far as i know, if you are going to take Math 2, i heard that you don't have to take math 1. (my school recommends to take only Math 2...) [/quote]\nAre you sure about it? :maybe: \nI think some of my friends hwo are students already took both math1 ant math 2.So,anyway I 'm going to take them both because I have already registered :) .But I don't think I will have any problems with math,the difficult part for me is physics :( .[/quote]\n\nafter you register you can change, it doesnt matter what you registered for. on the test day they give you a big booklet, just do the test that you WANT to do. Dont even do math 1, just do math 2.[/quote]\r\nLook guys,in the colleg application requirements it is written that you should take three subject tests,and I think math 1 and math 2 are different tests.So,in my opinion I need to write both Math 1 and math2,and I 'm not worring about them. :P",
  "Solution_9": "You may wish to ask the college whether Math 1 and Math 2 count as two different subject tests.",
  "Solution_10": "[quote=\"Ravi B\"]You may wish to ask the college whether Math 1 and Math 2 count as two different subject tests.[/quote]\r\n\r\nThat would be my advice too: check the college admission Web pages for what each college recommends.",
  "Solution_11": "[quote=\"Tiks\"][Look guys,in the colleg application requirements it is written that you should take three subject tests,and I think math 1 and math 2 are different tests.So,in my opinion I need to write both Math 1 and math2,and I 'm not worring about them. :P[/quote]\r\n\r\nHonestly, I doubt a college is going to fall for that. I think if you take math 2, they are not going to look at Math 1, unless you do a lot better on Math 1, in which case, they might be kind enough to ignore your math 2. \r\n\r\nThose tests are really designed for everyone to be able to take a math SAT II, yet, still allow different math levels. I can see how some people may end up with both scores, but to take it on the same day may make you look like a Grade Hog, and that is a bad thing to be. A Grade Hog (or someother word) is someone who's actions are only taken to look good on the app, not out of a pursuit of excellence. \r\n\r\nI agree with the others. Check with the school before you go too far on this plan.",
  "Solution_12": "I am rather sure that the colleges will admitt math 1 and math 2 as 2 subject tests,because my two friends Iurie Boreico(Iura)(twice perfect scorer at the IMO :P ) and Dimitar Semionov(imortal) (silver medalist from Bulgarian team)took math1,math2 and Physics as three subject tests for their application and now Dimitar is a freshman at MIT,Iura has already applied to Harvard :) .So,I don't think it is a problem to take math1 and math 2 as different subject tests.However,I will try to find out more about that!  :ninja:",
  "Solution_13": "[quote=\"Tiks\"]I am rather sure that the colleges will admitt math 1 and math 2 as 2 subject tests,because my two friends Iurie Boreico(Iura)(twice perfect scorer at the IMO :P ) and Dimitar Semionov(imortal) (silver medalist from Bulgarian team)took math1,math2 and Physics as three subject tests for their application and now Dimitar is a freshman at MIT,Iura has already applied to Harvard :) .So,I don't think it is a problem to take math1 and math 2 as different subject tests.However,I will try to find out more about that!  :ninja:[/quote]\r\n\r\nMIT only requires 2 SAT2s - one in Math 1 or 2 and one in a science. Not sure about Harvard. In any case, standardized test scores are a bit negotiable. I know that Harvey Mudd wants something life 5 or 6 from homeschooled students. I wrote to the admissions and nicely asked them if they are on something. I made it clear my son would not be applying if that was the case. They admitted they would back down on that as long as everything else is good. \r\n\r\nsome other examples...\r\n\r\nCaltech also requires only 2, but wants Math 2, not Math 1, and a science SAT. \r\n\r\nPrinceton wants 3, and doesnt specify. Presumably they want 3 to see how you have mastered 3 different subjects. If your app is very strong in other areas, they might ignore the fact you havent really shown this, and that you may be trying to do something bogus.\r\n\r\nStanford only wants 2, and with a preference for Math 2. \r\n\r\nHere is a cautionary tale told to me by a UCLA prof who used to be at MIT. Some kid has 3 SAT 1 scores. The first one is 1590. He takes it again, and again, gets 1590, but this time it is reversed, so he does have 800s in both M and V. This is not good enough for him. He takes it a 3rd time and gets 1600. Although he looked pretty good in other areas, the 3 SAT scores tipped the balance against him - he appeared highly strung and not beyond doing anything to make his app look good, thus suspect.\r\n\r\nGood luck on your apps.",
  "Solution_14": "I would never retake a 1590, but I would like to know why an admissions officer would deny a kid who tried to strive for perfection... and accomplished it.\r\n\r\nI would hope there were other gaps in the app to justify the decision.. poor kid.",
  "Solution_15": "[quote=\"WindSlicer\"]I would never retake a 1590, but I would like to know why an admissions officer would deny a kid who tried to strive for perfection... and accomplished it.\n\nI would hope there were other gaps in the app to justify the decision.. poor kid.[/quote]\r\n\r\nSATs are not that important. Not worth the time to retake. Striving for perfection on an SAT shows a lack of maturity. Striving for perfection on your coursework or extracurricular activities looks good.",
  "Solution_16": "[quote=\"WindSlicer\"]I would never retake a 1590, but I would like to know why an admissions officer would deny a kid who tried to strive for perfection... and accomplished it.\n\nI would hope there were other gaps in the app to justify the decision.. poor kid.[/quote]\r\n1600 isn't perfection anyway. At my school, when I took SAT, there were 4 1600s. Only one of them was actually a perfect paper. ;)\r\n\r\nAnyway I don't have much respect for standardized tests as a way to determine the best from among good students (obviously it is good for determining the good students from the average ones). Just like IQ -- the IQ test gets less and less accurate for people farther from the mean.\r\n\r\n@ Tiks: I will try to email an admissions officer and find out what their opinion is about the Math I, Math II thing.",
  "Solution_17": "If instead of thinking about it as \"getting rejected on account of retaking the SAT to get a 1600\" we think of it as \"not getting accepted on account of retaking the SAT to get a 1600\" I think it seems more reasonable for this to possibly hurt the chances of the applicant.  Admissions officers (at the more selective schools, at least) usually look at applicants holistically.  If they look at your application, see that you spent a lot of time on the SAT (as indicated by the number of times you took it and the progression of scores) then they may very well think that the chances you're just a tool are higher...at least, that's what I would think.  Of course, if you aren't a total tool, and your application shows that, then this impression can be overriden.  Anyway, the whole idea of someone getting rejected for the sole reason that they took the SAT $x$ times, and scored $y_{i}$ on them is dubious.  It's the [i]whole[/i] application that gets someone in a school and, likewise, it's (generally) the whole application that gets someone rejected.",
  "Solution_18": "[quote=\"Tiks\"]I am rather sure that the colleges will admitt math 1 and math 2 as 2 subject tests,because my two friends Iurie Boreico(Iura)(twice perfect scorer at the IMO :P ) and Dimitar Semionov(imortal) (silver medalist from Bulgarian team)took math1,math2 and Physics as three subject tests for their application and now Dimitar is a freshman at MIT,Iura has already applied to Harvard :) .So,I don't think it is a problem to take math1 and math 2 as different subject tests.However,I will try to find out more about that!  :ninja:[/quote]\r\nPerhaps going to the IMO had something to do with getting in.",
  "Solution_19": "[quote=\"WatsonLadd\"][quote=\"Tiks\"]I am rather sure that the colleges will admitt math 1 and math 2 as 2 subject tests,because my two friends Iurie Boreico(Iura)(twice perfect scorer at the IMO :P ) and Dimitar Semionov(imortal) (silver medalist from Bulgarian team)took math1,math2 and Physics as three subject tests for their application and now Dimitar is a freshman at MIT,Iura has already applied to Harvard :) .So,I don't think it is a problem to take math1 and math 2 as different subject tests.However,I will try to find out more about that!  :ninja:[/quote]\nPerhaps going to the IMO had something to do with getting in.[/quote]\r\n\r\nYeah,I went to the IMO too(hmm...4 times :blush: and it is not the end :D  )."
}
{
  "Tag": [
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "let a, b, c, d  be positive numbers such that abcd=1. prove  that :!:  :!: \r\n\\[ \\sum{\\frac{1}{{1+a+a^{2}+a^{3}}}}\\ge 1 \\]",
  "Solution_1": "$ \\sum \\frac{1}{1+a+a^{2}+a^{3}}\\geq \\sum \\left(-\\frac{3}{8}\\left( a-1\\right)+\\frac{1}{4}\\right) =-\\frac{3}{8}\\left( a+b+c+d\\right)+\\frac{5}{2}\\geq-\\frac{3}{8}.4\\sqrt[4]{abcd}+\\frac{5}{2}=1$",
  "Solution_2": "[quote=\"Vorelin Tancu\"]$ \\sum \\frac{1}{1+a+a^{2}+a^{3}}\\geq \\sum \\left(-\\frac{3}{8}\\left( a-1\\right)+\\frac{1}{4}\\right) =-\\frac{3}{8}\\left( a+b+c+d\\right)+\\frac{5}{2}\\geq-\\frac{3}{8}.4\\sqrt[4]{abcd}+\\frac{5}{2}=1$[/quote]\r\n\r\nbut i think the latest inequality is incorrect!!!  :blush:",
  "Solution_3": "Another studios... :huh: \r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=17246\r\n\r\n  darij"
}
{
  "Tag": [
    "geometry",
    "geometry solved"
  ],
  "Problem": "Consider a triangle ABC and two different points P and Q.\r\nThe parallel to BC through Q intersects the line AP at A*.\r\nThe parallel to AP through Q intersects the line BC at A#.\r\nDefine the points B*, B#, C* and C# similarly.\r\nThen prove that the triangles A*B*C* and A#B#C# are perspective.\r\n\r\n  Darij",
  "Solution_1": "I think the projective generalization is also true: replace the line at infinity with a line $\\ell$ and instead of the parallel through $Q$ to $BC$ (for example) draw the line through $Q$ which is concurrent with $BC,\\ell$ and so on.\r\n\r\nThis is important because we might be able to use some projective transformations to make the problem easier. For example, I'm trying to solve it in the case $PQ=\\ell_{\\infty}$ (I haven't succeeded, though, so maybe this isn't such a slick approach :D).",
  "Solution_2": "Apparently, some very interesting projective properties are related to the configuration I mentioned. Let me resume it:\r\n\r\nI took $PQ=\\ell_{\\infty}$, so instead of lines through $P,Q$ we have lines parallel to two given directions $p,q$. Let $\\ell$ be the line replacing $\\ell_{\\infty}$ from the initial problem. We obtain points $A^*,A\\#$ and the others in the obvious way, extending the definitions from Darij's post. Then I've noticed that this very interesting thing happens: no matter how we move $\\ell$, the point $A^*A\\#\\wedge B^*B\\#$ always stays on the same line parallel to $q$, and the same can be said about the other two points (we want to show that these points are, in fact, one and the same). Then if $A^*A\\#,B^*B\\#,C^*C\\#$ are concurrent for a position of $\\ell$, then they're always concurrent, so all we need to do is choose a particular position of $\\ell$ and prove the concurrence. We may, for example, choose $\\ell||p$. I don't know if it works out Ok, but it looks like we're only dealing with some similarities (we can choose [b]any[/b] of the lines $\\ell||p$).\r\n\r\nNote that we can't choose $\\ell||q$ because in that case the three concurrence points would trivially be one and the same, because the three lines parallel to $q$ which are their loci concur on the infinity line ($\\ell_{\\infty}$), but we want to show that they concur at a finite point.",
  "Solution_3": "Now I have a proof. Admittedly, it uses none of your ideas, Grobber, I haven't managed to complete them; it just uses Ceva and Menelaos (well, Menelaos for quadrilaterals). You can find the proof in the note \"On the paracevian perspector\" on [url=http://www.cip.ifi.lmu.de/~grinberg/]my geometry website[/url]. (BTW, I have replaced the notations A#, B#, C#, A*, B*, C* by U, V, W, U', V', W' for typographical reasons.)\r\n\r\n  Darij",
  "Solution_4": "Dear Mathlinkers,\r\nanother synthetic proof can be see on \r\nhttp://perso.orange.fr/jl.ayme   vol. 4  the paracevian perspector (in French)\r\nSincerely\r\nJean-Louis",
  "Solution_5": "Also, two other solutions have been posted in the topic [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=126138]Perpendicular feet[/url], of [b]barasawala[/b].\r\n\r\nKostas Vittas."
}
{
  "Tag": [
    "ratio"
  ],
  "Problem": "The four angles of a quadrilateral are in the ratio of $ 1: 2: 3: 4$. What is the number of degrees in the measure of the smallest angle?",
  "Solution_1": "So, you first know that a quadrilateral has $ 360$ degrees in it. Then, you add up the ratios. So, $ 1 \\plus{} 2 \\plus{} 3 \\plus{} 4 \\equal{} 10$  So now you can do $ \\dfrac{360}{10}$ and get $ 36$ for each unit. The $ 1$ is obviously the smallest so the answer is $ \\boxed{36}$",
  "Solution_2": "A little clarification:\r\n\r\nThe reason we added up the ratios and divided, is because we know there exists some number $ x$ such that $ 1x \\plus{} 2x \\plus{} 3x \\plus{} 4x \\equal{} 360$. So all we need to do is solve that equation for $ x$ and then multiply by the respective number (1, 2, 3, or 4) to find the angle we wanted. In this case, we wanted the smallest angle, so we took $ x \\equal{} 36$ and multiplied it by 1, which is just $ \\boxed{36}$."
}
{
  "Tag": [
    "logarithms",
    "floor function"
  ],
  "Problem": "Why is the number of digits in any integer 1 + log(n)?\r\n\r\nCan someone show me a derivation or give me an explanation?",
  "Solution_1": "it's not $ 1\\plus{}\\log n$. It's $ 1\\plus{}\\lfloor \\log n \\rfloor$; since we write numbers in base 10, a number with $ n$ digits can be written as $ 10^{n\\minus{}1}\\plus{}\\text{stuff less than }10^n$ (test it for small values of $ n$ to see why). Just rewrite that in logarithmic form."
}
{
  "Tag": [
    "inequalities",
    "algebra",
    "binomial theorem"
  ],
  "Problem": "Let $x,y$ be positive reals such that $\\frac{1}{x}+\\frac{1}{y}=1$, and let $n$ be a positive integer. Prove that $(x+y)^n + 2^{n+1} \\geq x^n + y^n + 2^{2n}$.",
  "Solution_1": "[hide=\"Solution\"]\n$x + y = xy$\n$(x-1)(y-1) = 1$\n\nLet $a = x-1$.  Then $x = a+1, y = \\frac{1}{a} + 1$ and the problem condition becomes\n\n$(a + 2 + \\frac{1}{a})^n + 2^{n+1} \\ge a^n + \\frac{1}{a^n} + 2^{2n}$\n\nLet $b = \\sqrt{a}$.  We can rewrite the problem as\n\n$\\left( b + \\frac{1}{b} \\right)^{2n} - b^{2n} - \\frac{1}{b^{2n}} \\ge 2^{2n} - 2^{n+1}$\n\nLet $s_k = b^k + \\frac{1}{b^k}$.  It is fairly clear from binomial theorem that\n\n$s_1^{2n} - s_{2n} = {2n \\choose 1} s_{2n-2} + {2n \\choose 2} s_{2n-4} + ... + {2n \\choose n}$\n\nNow, we have the inequality $s_k \\ge 2$.  So:\n\n$s_1^{2n} - s_{2n} \\ge {2n \\choose 1} 2 + {2n \\choose 2} 2 + ... + {2n \\choose n} = 2^{2n} - 2$\n\nWhich seems to be much stronger than the initial statement...  :huh: \n[/hide]",
  "Solution_2": "[quote=\"t0rajir0u\"]\nWhich seems to be much stronger than the initial statement...  :huh: \n[/quote]\r\n\r\nIf $x=y=2$, then the equality holds.",
  "Solution_3": "[quote=\"t0rajir0u\"]\n$x + y = xy$\n$(x-1)(y-1) = 1$\n\nLet $a = x-1$.  Then $x = a+1, y = \\frac{1}{a} + 1$ and the problem condition becomes\n\n$(a + 2 + \\frac{1}{a})^n + 2^{n+1} \\ge a^n + \\frac{1}{a^n} + 2^{2n}$\n\n[/quote]\r\n\r\nOh, I see where your mistake is. It should be\r\n\r\n$(a + 2 + \\frac{1}{a})^n + 2^{n+1} \\ge (a+1)^n + (\\frac{1}{a}+1)^n + 2^{2n}$",
  "Solution_4": "Edit: -deleted for mistaken the question-",
  "Solution_5": "[quote=\"t0rajir0u\"][hide=\"Solution\"]\n\n[/hide][/quote]\n\nYour solution is not correct. Would you like to try again? :-) Here is a little hint:\n\n[hide=\"hint\"]\nPairing $x^ky^{n-k}$ and $x^{n-k}y^k$ in the binomial expansion of $(x+y)^n$.\n[/hide]",
  "Solution_6": "[hide=\"AM GM\"]\nBy AM-GM, we have $x+y=xy\\ge 2\\sqrt{xy}\\implies xy\\ge 4\\implies (x+y)^n \\ge 2^{2n}$.\n\nBut also by AM-GM, $x^n+y^n\\ge 2(\\sqrt{xy})^n \\ge 2(2)^n = 2^{n+1}$.\n\nAdding these equations, we get $(x+y)^n+(x^n+y^n) \\ge 2^{2n}+2^{n+1}$.\n[/hide]\r\n\r\nErr oops what I did wrong? :P",
  "Solution_7": "[quote=\"chess64\"][hide=\"AM GM\"]\nBy AM-GM, we have $x+y=xy\\ge 2\\sqrt{xy}\\implies xy\\ge 4\\implies (x+y)^n \\ge 2^{2n}$.\n\nBut also by AM-GM, $x^n+y^n\\ge 2(\\sqrt{xy})^n \\ge 2(2)^n = 2^{n+1}$.\n\nAdding these equations, we get $(x+y)^n+(x^n+y^n) \\ge 2^{2n}+2^{n+1}$.\n[/hide]\n\nErr oops what I did wrong? :P[/quote]\r\n\r\nNice try :-)\r\n\r\nYou didn't do anything wrong except the thing you proved is a little different from the original inequality :-)"
}
{
  "Tag": [
    "calculus",
    "integration",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "If x is a root of an equation x^m+C1*x^(m-1)+.....+Cm where C1,....,Cm are integers then x is either integral or irrational",
  "Solution_1": "this is a special case of the rational root theorem. the proof of the theorem is simple: seee it at http://planetmath.org/encyclopedia/ProofOfRationalRootTheorem.html"
}
{
  "Tag": [
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Let $ p$ be a prime number with $ p > 2$ such that $ 3$ divides $ p - 2$. Let $ S$ be a set such that $ S$ = {$ y^2 - x^3 - 1$ and $ x,y$ integers with ${ 0<= x, y<= p - 1}$}\r\nShow that $ p$ divides at most $ p - 1$ members of the set $ S$.",
  "Solution_1": "http://www.mathlinks.ro/viewtopic.php?p=495342#495342",
  "Solution_2": "i didnt get it. could you explain it ???? :)",
  "Solution_3": "we take $ A \\equal{} \\{\\overline{x^3}|x\\in[|0,p \\minus{} 1|]\\}$ and $ B \\equal{} \\{\\overline{y^2 \\minus{} 1}|y\\in[|0,p \\minus{} 1|]\\}$\r\nwe have $ a^3\\equiv b^3[p]$ ==> $ a^{3}(a^3)^{\\frac {p \\minus{} 2}{3}}\\equiv b^3(b^3)^{\\frac {p \\minus{} 2}{3}}[p]$ ==> $ a\\equiv b[p]$.\r\nthen $ A \\equal{} \\{\\overline{x^3}|x\\in[|0,p \\minus{} 1|]\\} \\equal{} \\{\\overline{k}|\\ k\\in [|0,p \\minus{} 1|]\\}$\r\nthen $ S \\equal{} A \\plus{} B$\r\nif $ \\exists (a_1^3 \\plus{} b_1,a_2^3 \\plus{} b_2,...,a_{\\frac {p \\plus{} 1}{2}}^3 \\plus{} b_{\\frac {p \\plus{} 1}{2}})\\in S^{\\frac {p \\plus{} 1}{2}}$ such that for $ i \\equal{} 1..p: \\ p|a_i^3 \\plus{} b_i,\\ (a_i^3,b_i)\\in A\\times B$\r\nthen exists $ i\\neq j$ such that $ b_i\\equiv b_j[p]$ because $ |B| \\equal{} \\frac {p \\minus{} 1}{2} < \\frac {p \\plus{} 1}{2}$\r\nso $ a_i^3\\equiv a_j^3[p]$ ==> $ a_i\\equiv b_i$\r\nthen $ |\\{X\\in S: \\ p|X\\}|\\le\\boxed{\\frac {p \\plus{} 1}{2}}\\le p$"
}
{
  "Tag": [
    "inequalities",
    "AMC",
    "USA(J)MO",
    "USAMO",
    "inequalities solved"
  ],
  "Problem": "I need a problems, which are solved with Bernoulli's inequality. Can you help me to find them! Bernoulli's inequality says: \\[(1+x)^n\\geqq 1+nx,\\quad x\\geqq -1\\in\\mathbb{R},\\quad n\\in\\mathbb{N}\\]\r\nI just want a problems and their decisions.  :)",
  "Solution_1": "I believe #4 of 1991 USAMO first introduced Bernoulli to me...\r\n\r\nhttp://www.kalva.demon.co.uk/usa/usa91.html",
  "Solution_2": "our imo team has solved many problems using bernulli s ineq.\r\ni will post them later ;)",
  "Solution_3": "Here is a problem using Bernoulli's inequality:\r\n\r\n   Solve the equation: $2^x+3^x=3x+2$. If you want more I have some hard one's."
}
{
  "Tag": [
    "inequalities",
    "inequalities solved"
  ],
  "Problem": "Let a,b,c,d \\in (0,\\infty). Prove that:\r\n(a/a+b)<sup>2</sup> + (b/b+c)<sup>2</sup> + (c/c+d)<sup>2</sup> + (d/d+a)<sup>2</sup>  \\geq 1.",
  "Solution_1": "it suffices to prove that \\sum a/a+b >2\r\n\\sum a/a+b =\\sum 1/(1+(b/a))\r\nso we need to show that \\sum 1/(1+x) >2 where xyzt=1\r\nwhich is equivalent to prove that \\sum x> \\sum 1/x.\r\nconsider r.h.s of this last inequality to be proved:\r\n\\sum 1/x=xyz+yzt+ztx+txy =xy(z+t)+zt(x+y)\r\nconsider (x+y)(1-zt)+(z+t)(1-xy)\r\nobviously one of xy and zt is less than 1. let zt<=1.\r\nconsider the change x to root xy and y to root xy.\r\nthe new variables still satisfy xyzt=1 \r\nbut now since x+y>=2 times root xy. therefore the expression is decreased. \r\nhence the minimum occurs when x=y. \r\nexpression now equals: (2x)(1-zt)+(z+t)(1-x^2)=(zt-1)(1/z+1/t -2x) \r\nwhere x^2 z t =1 \r\nbut 1/z +1/t>= 2/root(zt) =2x hence expression >=0\r\nhence \\sum x \\geq \\sum 1/x.   \\",
  "Solution_2": "The above solution is wrong. The inequality sum 1/(1+x) >=2 isn't always true: for example, it fails when y,z,t do to infinity and x tends to zero so that xyzt=1. I've seen this inequality before and , if I'm right, it belongs to Vasile Cartoaje. I managed to solve it some time ago, but I've tried know to solve it and I didn't manage.",
  "Solution_3": "Just a little help: we have to prove that sum 1/(1+x)^2 is at least 1 when xyzt=1. Make the substitutions 1/(1+x)=a, etc and you will obtain the following inequality proposed by the same author at a GM contest ( in 2000 or 2001) : if a,b,c,d>=0 and abcd=(1-a)(1-b)(1-c)(1-d) , then a^2 +b^2+c^2+d^2>=1.",
  "Solution_4": "Harazi, have you a solution for the inequality :\r\n\r\nIf a,b,c,d>=0 and abcd=(1-a)(1-b)(1-c)(1-d) , then \r\n\r\n                    a^2 +b^2+c^2+d^2>=1.\r\n \r\nI cannot solve it.\r\n\r\nThank you, very much.",
  "Solution_5": "No, I don't have it, but maybe a romanian who has a book with that GM contest would be so kind to post a solution.",
  "Solution_6": "well, up to my knowledge, there is no book with the solutions of those problems ( GM contest 2000-2001) and neither the GM presents the solutions, so I guess we have to solve the problem :)",
  "Solution_7": "Isn't the problem like this: if a <sup>2</sup> +b <sup>2</sup> +c <sup>2</sup> +d <sup>2</sup> =1 then prove that abcd \\geq (1-a)(1-b)(1-c)(1-d)?",
  "Solution_8": "The inequality can be obtained from the product of the following inequalities:\r\n  (1 - a)(1 - b)(a<sup>2</sup> + b<sup>2</sup>) \\geq ab(c<sup>2</sup> + d<sup>2</sup>)    (1)\r\n  (1 - c)(1 - d)(c<sup>2</sup> + d<sup>2</sup>) \\geq cd(a<sup>2</sup> +b<sup>2</sup>)     (2).\r\nStarting with the identity\r\n  2(1 - a)(1 - b) = (a + b - 1)<sup>2</sup> +c<sup>2</sup> + d<sup>2</sup>\r\nwe have:\r\n  2(1 - a)(1 - b)(a<sup>2</sup> + b<sup>2</sup>)-2ab(c<sup>2</sup> +d<sup>2</sup>) = [(a + b - 1)<sup>2</sup> + c<sup>2</sup> + d<sup>2</sup>](a<sup>2</sup> + b<sup>2</sup> - 2ab(c<sup>2</sup> +d<sup>2</sup>) = (a + b - 1)<sup>2</sup> (a <sup>2</sup> + b<sup>2</sup>) + (c<sup>2</sup> + d<sup>2</sup>)(a - b)<sup>2</sup>  \\geq 0.",
  "Solution_9": "Yeah, right. All I can say about that solution is that it is very nice, but not at all natural.",
  "Solution_10": "May anyone explain to me why the inequality:\r\n\r\nif a,b,c,d>=0 and abcd=(1-a)(1-b)(1-c)(1-d) , then \r\na^2 +b^2+c^2+d^2>=1.\r\n\r\nis equivalent to\r\n\r\nif a,b,c,d>=o and  a^2 +b^2 +c^2 +d^2 =1 , then \r\nabcd \\geq (1-a)(1-b)(1-c)(1-d)\r\n\r\nhow pcalin said.\r\nSorry, but I don't understand.",
  "Solution_11": "Sorry, pcalin, I have misunderstood the meaning of the adjective \"like\". \r\nIt means \"similar\" and not, as I did, \"equivalent\".\r\n\r\nSo this beautiful inequality is again an open challenge.",
  "Solution_12": "Hi all,\r\n\r\nI'm agree with Manlio. \r\n\r\nTwo problem \"if a,b,c,d>=0 and abcd=(1-a)(1-b)(1-c)(1-d) , then \r\na^2 +b^2+c^2+d^2>=1\" \r\n\r\nand \r\n\r\n\"if a,b,c,d>=0 and a^2 +b^2 +c^2 +d^2 =1 , then \r\nabcd  (1-a)(1-b)(1-c)(1-d)\". \r\n\r\nare not equivalent. (At least I dont know how to reduce one from the other).  \r\n\r\nSo, the problem is open (at least in this forum). I have a solution, not very nice, but contains many interesting ideas. I will post it in future.\r\n\r\nNamdung",
  "Solution_13": "By dividing all the fractions with a, b, c, d and b/a = x, c/d = y, d/c = z, a/d = t we have:\r\n1/(a+x)<sup>2</sup> + 1/(1+y)<sup>2</sup> -1/(1+xy) = [(xy-1)<sup>2</sup> + xy(x-y)<sup>2</sup>]/[(1+x)<sup>2</sup> (1+y)<sup>2</sup>(1+xy)<sup>2</sup>] => 1/(1+x)<sup>2</sup> + 1/(1+y)<sup>2</sup> \\geq 1/(1+xy) = a/(a+c) also 1/(1+z)<sup>2</sup> + 1/(1+t)<sup>2</sup> \\geq 1/(1+zt) = c/(a+c).\r\nFrom the sum of the two inequalities we get our inequality for a=b=c=d.",
  "Solution_14": "[quote=\"Gil\"]By dividing all the fractions with a, b, c, d and b/a = x, c/d = y, d/c = z, a/d = t we have:\n1/(1+x)<sup>2</sup> + 1/(1+y)<sup>2</sup> -1/(1+xy) = [(xy-1)<sup>2</sup> + xy(x-y)<sup>2</sup>]/[(1+x)<sup>2</sup> (1+y)<sup>2</sup>(1+xy)<sup></sup>] => 1/(1+x)<sup>2</sup> + 1/(1+y)<sup>2</sup> \\geq 1/(1+xy) = a/(a+c) also 1/(1+z)<sup>2</sup> + 1/(1+t)<sup>2</sup> \\geq 1/(1+zt) = c/(a+c).\nFrom the sum of the two inequalities we get our inequality for a=b=c=d.[/quote]\r\n============\r\nVery nice and neat solution (I corected some misprint). So we can forget my solution.\r\n\r\nNamdung"
}
{
  "Tag": [
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "$ a,b,c,x,y,z > 0$such that $ abc \\equal{} m^3$ and $ xyz \\equal{} n^3$,where $ m,n$are given const.Find the max and the min of the following expression.\r\n\\[ \\frac {1}{1 \\plus{} a \\plus{} x} \\plus{} \\frac {1}{1 \\plus{} b \\plus{} y} \\plus{} \\frac {1 }{1\\plus{} c \\plus{} z}\\]",
  "Solution_1": ":blush: Anyone want to give a solution?",
  "Solution_2": ":maybe: one and a half month has passed....still no one have a solution?????? :(",
  "Solution_3": "I think there isn't any minimum constant value. We can approach that expression to $ 0$.\r\n\r\nIf there is a constant $ k>0$ such that the expression $ \\geq k$. Let's take $ x,y,c>\\frac{3}{k}\\minus{}1$ is a contradiction. :wink:"
}
{
  "Tag": [
    "LaTeX",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Given two positive integers $m>n$,denote $A=\\{a|(a,n)=1,1\\le a\\le m\\}$,show that $\\sum_{a \\in A}\\frac{1}{a}$is not an integer.\r\n\r\n[i]moderator edit: no, the problem is not hard; choose better titles next time![/i]",
  "Solution_1": "It is not correct. For example $if \\ P(m)|n, \\ P(m)=p_1*p_2* \\dots *p_s \\ - \\ product \\ all \\ primes \\ \\le m.$ $For \\ another \\ cases \\ it \\ is \\ true, \\ and \\ may \\ be \\ proved \\ by \\ consider \\ maximal \\ prime \\ p \\lem: (p,n)=1$.",
  "Solution_2": "Rust: please don't write everything in $\\LaTeX$ now, especially the text ;)\r\nBut I don't see what your condition implies...\r\n\r\nIndeed, a full prove (I just need $m$ being greater than the smallest prime not dividing $n$):\r\nTake the smallest $p>1$ with $\\gcd(n,p)=1$. $p$ will be prime then since otherwiese every prime divisor of $p$ is smaller and fulfills the conditions, too.\r\nNow to kill this prime $p$ from the denominator, you will need another number having this denominator; let $pq$ be the next multiple of $p$ with $\\gcd(q,n)= \\gcd(pq,n) =1$.\r\nSince $p$ is minimal, $q \\geq p \\implies p=q$, thus the next multiple of $p$ (occuring as denominator) is already $p^2$.\r\nNow to kill $p^2$, a $p$ is not enough, you need a denominator that is a multiple of $p^2$ this time, and continuing as before, the next such multiple is $p^3$, the same problems occure again and so on, thus you will never get an integer."
}
{
  "Tag": [
    "group theory",
    "abstract algebra",
    "induction",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "(I)Let $G$ be a group of order $2^{m}k$,where $k$ is odd. Prove that if $G$ contains an element of order $2^{m}$,then the set of all odd order in $G$ is a (normal) subgroup of $G$.\r\n(Hint:Consider $G$ as permutations via Cayley\u2019s theorem , and show that it contains an odd permutation.)\r\n(II)Show that a finite simple group of even order must have order divisible by 4.",
  "Solution_1": "We have a homomorphism $\\varphi: G\\to S_{G}$, taking an element $g\\in G$ to the permutation $\\sigma_{g}$ of $G$, where $\\sigma_{g}(x) = gx$.  Suppose $g$ has order $2^{m}$.  Then $\\sigma_{g}$ is a permutation of $|G|$ elements, of order $2^{m}$, with no fixed points.  It must therefore consist of $k$ disjoint $2^{m}$-cycles, so it is an odd permutation.  It follows that $sign\\circ\\varphi : G\\to \\{\\pm 1\\}$ is surjective, and that $N=\\ker(sign\\circ\\varphi)$ is a subgroup of $G$ of order $2^{m-1}k$.  Clearly the odd order elements of $N$ and of $G$ are the same.  By induction (since $\\sigma^{2}$ has order $2^{m-1}$), these elements form a subgroup, which is obviously characteristic, hence normal.\r\n\r\nNow (II) follows from the fact that a group of even order has an element of order 2 (Cauchy)."
}
{
  "Tag": [],
  "Problem": "A recent national survey found that high school students watched an average\r\n(mean) of 6.8 DVDs per month. A random sample of 36 college students\r\nrevealed that the mean number of DVDs watched last month was 6.2, with a\r\nstandard deviation of 0.5. At the .05 significance level, can we conclude\r\nthat college students watch fewer DVDs a month than high school students?",
  "Solution_1": "[quote=\"Interval\"]A recent national survey found that high school students watched an average\n(mean) of 6.8 DVDs per month. A random sample of 36 college students\nrevealed that the mean number of DVDs watched last month was 6.2, with a\nstandard deviation of 0.5. At the .05 significance level, can we conclude\nthat college students watch fewer DVDs a month than high school students?[/quote]\r\nYou do realize that .5 and .05 are different, right?",
  "Solution_2": "You asked:\r\n\r\nYou do realize that .5 and .05 are different, right?\r\n\r\nMy reply: Yes, of course.",
  "Solution_3": "[hide] Since the survey said that college students watch 6.2 DVDs per month with a standard deviation of .5, we can say that the range of the possible average DVDs per month that are watched by all college students is more than 6.2-.5=5.7 and less than 6.2+.5=6.7.  However, 6.8 is more than the highest possible average, which is 6.7.  This means that the average DVDs watched by the high schoolers is greater than the average DVD watched by the college students. [/hide]"
}
{
  "Tag": [],
  "Problem": "Find A+B, if A and B are positive, rational numbers and\r\n\r\n$ A\\plus{}B\\sqrt{7}\\equal{}\\sqrt{6\\sqrt{28}\\plus{}\\frac{2}{8\\plus{}3\\sqrt{7}}}$",
  "Solution_1": "[hide] We can simplify some of the terms in the RHS to get $ \\sqrt{12\\sqrt{7}\\plus{}16\\minus{}6\\sqrt{7}}\\equal{}\\sqrt{6\\sqrt{7}\\plus{}16}$. So,\n\n$ A\\plus{}B\\sqrt{7}\\equal{}\\sqrt{16\\plus{}6\\sqrt{7}}$\n\n$ \\implies A^2\\plus{}7B^2\\plus{}2AB\\sqrt{7}\\equal{}16\\plus{}6\\sqrt{7}$.\n\nComparing like terms,\n\n$ A^2\\plus{}7B^2\\equal{}16$\n\n$ 2AB\\equal{}6$\n\nThe solution of this is $ A\\equal{}\\pm3$, $ B\\equal{}\\pm1$. However, we want the positive values (since square roots are positive), so $ A\\equal{}3$, $ B\\equal{}1$.\n\nTherefore, $ A\\plus{}B\\equal{}3\\plus{}1\\equal{}\\boxed{4}$.[/hide]"
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "MATHCOUNTS"
  ],
  "Problem": "Find the sum $\\sqrt[3]{5+2\\sqrt{13}}+\\sqrt[3]{5-2\\sqrt{13}}$.\r\n\r\nA. $\\frac{3}{2}$\r\nB. $\\frac{\\sqrt[3] {65}}{4}$\r\nC. $\\frac{1 + \\sqrt[6] {13}}{2}$\r\nD. $\\sqrt[3] {2}$\r\nE. None of these",
  "Solution_1": "[hide]The answer is $1$.[/hide]",
  "Solution_2": "HIDE YOUR ANSWER! And also, how do I know you didn't just use a calculator?? :)",
  "Solution_3": "This is a very popular question.  I have seen it on [i]so[/i] many competitons :D",
  "Solution_4": "[quote=\"joml88\"]This is a very popular question.  I have seen it on [i]so[/i] many competitons :D[/quote]\r\n\r\nYou have? I haven't...but then again the only competitions I have ever participated in are MATHCOUNTS and Math League round 1... :D",
  "Solution_5": "This problem is 1980 AHSME #27.\r\n\r\nchess64, can you edit your post so it has just the problems and the choices? I want to link this thread to Olympiad Resources page.\r\n\r\nHere are choices:\r\n\r\nA. $\\frac{3}{2}$\r\nB. $\\frac{\\sqrt[3] {65}}{4}$\r\nC. $\\frac{1 + \\sqrt[6] {13}}{2}$\r\nD. $\\sqrt[3] {2}$\r\nE. None of these",
  "Solution_6": "Let $x = \\sqrt[3]{5+2\\sqrt{13}}$ and $y = \\sqrt[3]{5-2\\sqrt{13}}$\r\n$x^3+y^3 = (x+y)(x^2-xy+y^2) = (x+y)((x+y)^2-3xy) = 10$\r\n$xy = \\sqrt[3]{5+2\\sqrt{13}}\\sqrt[3]{5-2\\sqrt{13}} = \\sqrt[3]{-27} = -3$\r\nLet $z = x+y$\r\n$z^3+9z -10 = (z-1)(z^2+z+10) = 0$\r\n$\\sqrt[3]{5+2\\sqrt{13}}+\\sqrt[3]{5-2\\sqrt{13}} = 1$",
  "Solution_7": "[hide=\"Answer\"]$x^3+y^3=(x+y)(x^2-xy+y^2)$\n$xy=\\sqrt[3]{(5+2\\sqrt{13})(5-2\\sqrt{13})}=\\sqrt[3]{25-52}=\\sqrt[3]{-27}=-3$\n$x^2+y^2=(x+y)^2-2xy\\Rightarrow (x+y)(x^2-xy+y^2)=(x+y)((x+y)^2-3xy)=(x+y)((x+y)^2+9)=(x+y)^3+9(x+y)$\n$5+2\\sqrt{13}+5-2\\sqrt{13}=10=(x+y)^3+9(x+y)\\Rightarrow x+y=1\\Rightarrow \\boxed{E}$[/hide]",
  "Solution_8": "My own solution...\r\n[hide]\n[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=313[/img]\n[/hide]",
  "Solution_9": "[quote=\"chess64\"]HIDE YOUR ANSWER! And also, how do I know you didn't just use a calculator?? :)[/quote]\r\n\r\nIn Japan, there is no custom  to use calculator in examination.\r\n\r\nAnyway in solving your problem, we are to use the next inequality.\r\n\r\nIf $a+b+c=0$, then $a^3+b^3+c^3=3abc$. :)",
  "Solution_10": "or just cube the whole thing\r\n$(x+y)^3=x^3+y^3+3xy(x+y)$",
  "Solution_11": "Which, if you'll notice, is what I did. ;)"
}
{
  "Tag": [
    "MATHCOUNTS",
    "LaTeX",
    "probability",
    "AMC",
    "AIME"
  ],
  "Problem": "I love the MATHCOUNTS competition. \r\nI would like to start a new and fresh round of a [b]Mock Mathcounts Competition[/b]\r\n\r\nIt will start out with a mock school round.\r\nEveryone will then take the mock chapter round.\r\nThose doing well on the chapter will qualify and take the mock state round.\r\nThe top scorers on the state will take the mock national round.\r\nThe winners will come to FTW2 for an unrated round of CountDown.\r\nThe ranking will be posted.  :) \r\n\r\nThose interested are welcome to sign-up!\r\n\r\nI would also appreciate a few helpers to write the problems.\r\n[i]Note: Helpers cannot also take the test.[/i]\r\n\r\nI hope this works smoothly, have fun!",
  "Solution_1": "I'll participate :lol: !!\r\n\r\n[hide=\"Participants\"]the cliu[/hide]\r\n\r\nPlease copy and paste this list, thanks.",
  "Solution_2": "No problem.\r\n\r\n[hide=\"Participants(1)\"]\nthe cliu\n[/hide]\r\n\r\nThanks!",
  "Solution_3": "I would like to help write the problems\r\n(I'm KingSmasher3's sister)\r\n\r\nThanks!",
  "Solution_4": "i would like to be a helper also.\r\n\r\nsince i am in high school now, it would be more helpful to myself to write problems than solve them",
  "Solution_5": "I would like to sign up for this mock-mathcounts competition. :D",
  "Solution_6": "count me in!!!  :ninja:",
  "Solution_7": "No Problem.\r\n\r\n[hide=\"Participants(3)\"]\nthe cliu\nStreet\nBarbieRocks\n[/hide]\n\n[hide=\"Helpers(2)\"]\n15!!!\nriggabamboo\n[/hide]\r\n\r\nI will let everyone know when we are ready.  :) \r\n\r\nThanks!",
  "Solution_8": "I will join. I hope i dont fail like i did PoP's :oops:",
  "Solution_9": "I'll join. Warning: might forget I'm in this :P",
  "Solution_10": "UPDATE:\r\n\r\n[hide=\"Participants(6)\"]\nthe cliu \nStreet \nBarbieRocks \ndragon96\nTwinDracula\nmath rulz\n[/hide]\n\n[hide=\"Helpers(2)\"]\n15!!!\nriggabamboo\n[/hide]\r\n\r\nThanks all! I will PM all of you regarding the Competition!",
  "Solution_11": "I join.     .",
  "Solution_12": "Cool I'm in.",
  "Solution_13": "About how hard is it going to be? Like, what grade level? :fool:",
  "Solution_14": "I'll join.\r\n\r\n[hide=\"Participants (7)\"]the cliu\nStreet\nBarbieRocks\ndragon96\nTwinDracula\nmath rulz \nPowerOfPi[/hide]\n[hide=\"Helpers (2)\"]\n15!!!\nriggabamboo[/hide]",
  "Solution_15": "@Kingsmasher3:\r\nCan I still participate in the State round and pm you the answers but my score will not be counted?",
  "Solution_16": "[quote=\"exmath89\"]@Kingsmasher3:\nCan I still participate in the State round and pm you the answers but my score will not be counted?[/quote]\r\n\r\nOf Course.",
  "Solution_17": "can you send me tehh questions ?\r\nof course, im not joinging",
  "Solution_18": "Look at the previous pages.",
  "Solution_19": "Can we sign up? This is getting sort of inactive, because Nats is in 6 days.",
  "Solution_20": "Yes, I noticed.  :)  \n\nI know that so many people are getting PUMPED for nats.\nI'm sorry I havent gotten around to grading all the tests and keeping the tournament going. I'm preparing too.  :P \n\nI am surely going to pick up after nats is over on the 10th.\nThanks!",
  "Solution_21": "So can we sign up? You siggy says so.",
  "Solution_22": "Yes, it would be great for you to sign up. Copy and paste your name to the most recent list and you may find the School, Chapter, and State Problems on the previous Pages.\n\nThanks!\nI am waiting for a few more answers to be turned in before I grade them\n[color=#0040BF]THANKS ALL![/color]",
  "Solution_23": "What's the current deadline for State? I'll try to fit it in my schedule, but the earliest available day is Friday, so...",
  "Solution_24": "[hide=\"Participants(A lot)\"]AdithyaGanesh\nkelleyzhao\nthe cliu\nStreet\nBarbieRocks\ndragon96\nTwinDracula\nmath rulz\nPowerOfPi\nMaybach\npercy314\nslayer42\nDrepix\nbbgun34\ncheeseyicecream\nCRICKET229\nyaofan\nAwesomeToad\nAIME15USAMO\n\\LaTeX\nbasketball9\nMinamoto\npi31415926\nThe Hobbit\nPutneni07\nTheMan1998\nSonyWii\njerry125\norigamiguy777\nbluecarneal\nCDKing129\nJust_Beginner\nAIME15\nbiblioman\nmonkeygirl13\npolicecap\ndiger\nhahagooman\nhi how are you doing toda\ndwei1019\nbarcelona\nIggy Iguana\nDemosthenes\n141matho\ne,i,pi\nThe Archer\nliltobe9\nathunder\nmathiscool888\nmatho141\namparvardi\ntheoneforce\nvjnmath\nksun48\nhli\ncolinhy\ndwade884\nphysicsguymark\njoshim5\nshangdevin\nzhanga\njonathanchou711\nksun48[/hide]",
  "Solution_25": "[quote=\"theoneforce\"]What's the current deadline for State? I'll try to fit it in my schedule, but the earliest available day is Friday, so...[/quote]\n\nYou may turn in your answers for the State round any time before the end of the competition.  :)",
  "Solution_26": "Can I still do the state round?  Sorry I've been wrapped up in other things lately.",
  "Solution_27": "Ugh.\nMe too.\nI have had an extremely busy summer.  :dry: \n\nI'll my best to keep this going.\n(If not there will be next year)\n\n@Maybach: Of course. The Q's are open to everyone. You are also welcome to Pm me your answers\n\nTHANKS ALL!",
  "Solution_28": "Ok!\nI have set a date for the release of the final test for this year's round. The national round will be posted on July 20th. Feel free to turn in your answers anytime befor the end of August!\nSorry about the delay of the grading of the state answers I will get them back to you as soon as possible.\nI hope things will be better next year!\n\nThanks!",
  "Solution_29": "[quote=AwesomeToad][quote=\"stargroup\"]oh oops I didn't realize I posted in a mock MATHCOUNTS competition\n\nnvm I'm out haha[/quote]\n\nSometimes it would be helpful to read the title of the thread you're posting in maybe? :lol:[/quote]\n\nHI GUYS! I JUST WANTED TO SAY HI BECAUSE  I DON'T KNOW ANYBODY HERE. I'M PRETTY GOOD AT MATH AND I THINK Y'ALL CAN GET PERFECT SCORES. EVERYBODY CAN BE SMART IF THEY ACTUALLY TRY AND DO THEIR BEST. GOOD LUCK!\n\n[color=#f00]Welcome to AoPS!  But please try not to revive ten year old threads to add discussion that isn't relevant to the original topic :) ~dj[/color]"
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "sphere",
    "parameterization",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "[hide]As you know one can compute the order of contact of two curves at a certain point $ P$ by comparing the terms of their Tailor series, by definition.\n\nthere is another way, when we want to compute the order of contact between a curve and it's osculating circle at point $ P$:[/hide]\r\n\r\nassume that $ O(s)$ be the origin of osculating circle to curve at point $ s$ for a [b]fixed point[/b] $ s$. define: $ r(t): \\equal{}O(s)\\minus{}C(t)$ where $ C(t)$ is an arbitrary parametrization for that curve. It is obvious that the first non-zero term of the Tailor series of $ r(t)$ around $ s$ determines the order of contact.\r\n\r\nnow there is a question: can we extend this method in order to compute the order of contact between a curve and it's osculating [b]sphere[/b]?",
  "Solution_1": "That's already stated for spheres, in any number of dimensions.\r\n\r\nIf you talk about surfaces instead of curves, there isn't an osculating sphere in general.",
  "Solution_2": "It's alright. but how can I prove it?\r\nOr this is definition? if yes, can you say where could I find it?"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "number theory proposed",
    "number theory"
  ],
  "Problem": "Prove that for any integer $ n$, there exists a unique polynomial $ Q$\r\nwith coefficients in {$ 0, 1, ...,9$} such that $ Q( \\minus{} 2)$ $ \\equal{}$ $ Q( \\minus{} 5)$ $ \\equal{}$ $ n$.",
  "Solution_1": "A tough one:\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?p=343873#343873"
}
{
  "Tag": [
    "AMC",
    "USA(J)MO",
    "USAMO"
  ],
  "Problem": "Suppose that the set $ \\{1,2,\\cdots, 1998\\}$ has been partitioned into disjoint pairs $ \\{a_i,b_i\\}$ ($ 1\\leq i\\leq 999$) so that for all $ i$, $ |a_i \\minus{} b_i|$ equals $ 1$ or $ 6$. Prove that the sum\r\n\\[ |a_1 \\minus{} b_1| \\plus{} |a_2 \\minus{} b_2| \\plus{} \\cdots \\plus{} |a_{999} \\minus{} b_{999}|\r\n\\]\r\nends in the digit $ 9$.\r\n\r\n\r\nHere is one proof. I do not think it's very well written, especially the lemma (did I even need a lemma??) but here it is. (Sorry, I also didn't know how to make the congruence sign, and I was in a hurry, so I just put equal signs.)\r\n\r\n[hide=\"lemma\"]\nDefine $ x$ as the number of pairs where $ \\[ |a_{i} \\minus{} b_{i}| \\equal{} 6$, and similarly define $ y$ as the number of pairs where $ \\[ |a_{i} \\minus{} b_{i}| \\equal{} 1$.\n\nWe can prove that $ a \\equal{} 0 (mod 2)$.\n\nAssume a was not even. Then, we would have some interval, $ (m, m \\plus{} 2n)$, where $ m$ and $ n$ are positive integers, where the middle numbers weren't any $ a_i$ or $ b_i$. We could not assign pairs $ |a_{i} \\minus{} b_{i}| \\equal{} 1$ to these integers.\n\nSo, a must be even.\n[/hide]\n\n[hide=\"main proof\"]\nUsing definitions of $ a$ and $ b$ from the lemma, we know that \n$ a \\plus{} b \\equal{} 999$.\nUsing our results from the lemma,\n\n$ a \\equal{} 0 (mod 2)$\n$ \\Leftrightarrow 5a \\equal{} 0 (mod 10)$\n$ \\Leftrightarrow 5a \\plus{} 999 \\equal{} 999 (mod 10)$\n$ \\Leftrightarrow 5a \\plus{} 999 \\equal{} 9 (mod 10)$\n\nUsing the equation $ a \\plus{} b \\equal{} 999$, we substitute:\n\n$ 5a \\plus{} a \\plus{} b \\equal{} 9 (mod 10)$\n$ \\Leftrightarrow 6a \\plus{} b \\equal{} 9 (mod 10)$\n\nWe notice that 6a + b represents the total sum, and it is congruente to 9 modulo 10, i.e. its last digit is nine.\n\nQED\n\n [/hide]",
  "Solution_1": "$ \\text{USAMO} \\not\\subset \\text{Intermediate Topics}$.  Choose more appropriate fora in the future.  Unrelatedly, you have some notational errors in your solution you might want to fix.",
  "Solution_2": "http://www.artofproblemsolving.com/Forum/viewtopic.php?p=343865#343865\r\n\r\nPlease do not double post."
}
{
  "Tag": [
    "function",
    "geometry",
    "3D geometry",
    "LaTeX",
    "algebra",
    "domain",
    "logarithms"
  ],
  "Problem": "Show that the cube root of z is an analytic function.\r\n(Sorry about the lack of symbols but I am trying to get Latex up and running)",
  "Solution_1": "What do you mean, \"the\" cube root? You have to choose a branch.\r\nAlso, the origin is a bad point- whatever region you choose as a domain will have to exclude it, as well as some curve between zero and $ \\infty$.\r\n\r\nSome possible approaches:\r\n- Write down the derivative.\r\n- Write down a power series (the binomial series).\r\n- Write it as a composition involving $ \\log$ and $ \\exp$."
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "number theory",
    "greatest common divisor"
  ],
  "Problem": "For how many integers $N$ between 1 and 1990 is the improper fraction $\\frac{N^2+7}{N+4}$ NOT in lowest terms?\r\n\r\nA. 0\r\nB. 86\r\nC. 90\r\nD. 104\r\nE. 105",
  "Solution_1": "[hide=\"Solution\"]\n$N+4$ and $N^2 + 7$ must have some common factor $p$ in order for the expression not to be in lowest terms.  Thus, $(N+4)^2 - (N^2 + 7) - 8(N+4) = 23$ must also share this common factor $p$, which must therefore be $23$.  Hence we want $N+4 = 23k < 1990$.  We find that there are $86$ values of $N$ for which this is true.  Hence, our answer is $\\boxed{B}$\n[/hide]",
  "Solution_2": "[hide]\nAnswer: B\nGiven (N^2+7)/(N-4), we can use synthetic division to get\nN-4+23/(N-4)\nSince N-4 is always an integer, we only need to ensure that the term 23/(N-4) is in the lowest terms. In this, it is important to consider that 23 is a prime number, rendering it impossible to reduce the numerator. As a result, the only denominator that can be less than 23 so that the resulting fraction can be reduced is 1, which is impossible since then N would equal -3, outside the given range. Also, as a result, N-4 must be a multiple of 23. Between 1 and 1990, there are 86 such multiples of 23. Now, if 4 is added to the greatest multiple in order to solve for N, the greatest value of N remains less than 1990, and thus within the given range. Therefore, the answer is [b]86[/b][/hide]",
  "Solution_3": "[quote=\"t0rajir0u\"]Thus, $(N+4)^2 - (N^2 + 7) - 8(N+4) = 23$ must also share this common factor $p$, which must therefore be $23$. [/quote]\r\n\r\nWhere did that come from?",
  "Solution_4": "This seems way too hard for a 19 out of 30. :huh:",
  "Solution_5": "Maybe because you guys are making it way too complicated... :roll:\r\n\r\n[hide=\"Solution\"]\nUse polynomial long division to find $\\frac{N^2+7}{N+4}=N-4+\\frac{23}{N+4}$. If this is in lowest terms, then $\\frac{23}{N+4}$ must be in lowest terms. The only way that $\\text {gcd}(N+4,23)\\ne 1$ and $1\\le N\\le 1990$ is if $N\\in \\{ 19, 42, 65, \\ldots, 23\\cdot 86-4\\}$, so there are $\\boxed {86}$ such $N$.\n[/hide]",
  "Solution_6": ":wallbash:\r\n\r\nI hate it when I do that.",
  "Solution_7": "[quote=\"chess64\"]Maybe because you guys are making it way too complicated... :roll:\n\n[hide=\"Solution\"]\nUse polynomial long division to find $\\frac{N^2+7}{N+4}=N-4+\\frac{23}{N+4}$. If this is in lowest terms, then $\\frac{23}{N+4}$ must be in lowest terms. The only way that $\\text {gcd}(N+4,23)\\ne 1$ and $1\\le N\\le 1990$ is if $N\\in \\{ 19, 42, 65, \\ldots, 23\\cdot 86-4\\}$, so there are $\\boxed {86}$ such $N$.\n[/hide][/quote]\r\n\r\nAhhhh... that makes me feel a lot better  :P",
  "Solution_8": "[quote=\"236factorial\"][quote=\"t0rajir0u\"]Thus, $(N+4)^2 - (N^2 + 7) - 8(N+4) = 23$ must also share this common factor $p$, which must therefore be $23$. [/quote]\n\nWhere did that come from?[/quote]\r\n\r\nWell, I had considered the synthetic division solution, but thought this one was a little purer, so what I did was to try to show that a constant had to have $p$ as one of its factors, so I considered $(N+4)^2 - (N^2 + 7) = 8N + 9$, which also has to have this factor, but I wanted to eliminate $N$ entirely, so, reasoning that $8(N+4)$ must have this factor, I subtracted and, er, took absolute value.   :D"
}
{
  "Tag": [
    "articles",
    "LaTeX"
  ],
  "Problem": "Hi,\r\nI tried to use latex for my resume. Need to add an article class with .cls..\r\nhow to add it (where to save the file I downloaded in my computer)? Thanks!",
  "Solution_1": "What do you mean by a new article class? Exactly what did you download and from where? Does it require other packages? What LaTeX distribution are you using? What is the class called? Is it already there? What happens if you use \\documentclass{[i]name of class[/i]} in your document?\r\nIn other words you must provide much more information about your system and what you've done to answer your question.",
  "Solution_2": "I downloaded a resume document in .tex. I found that TexnicCenter could use the usepackage directly. (sorry, somehow, I made a mistake for using \"article class\" instead of \"documentclass\" :( )."
}
{
  "Tag": [],
  "Problem": "In regular pentagon $ PQRST$, $ X$ is the midpoint of segment $ ST$. What is the measure of angle $ XQS$?\n\n[asy]size(101);\npicture p;\npair P  = (0,0); pair Q = (-cos(108),sin(108)); pair R = Q + expi(36*pi/180);  pair T = (1,0); pair S1 = T + (cos(108),sin(108));\ndraw(p,(P--Q--R--S1--T--cycle),linewidth(1)); pair X = (S1+T)/2;\nlabel(p,rotate(20)*\"$P$\",P,rotate(20)*W,fontsize(10pt)); label(p,rotate(20)*\"$Q$\",Q,rotate(20)*NW,fontsize(10pt)); label(p,rotate(20)*\"$R$\",R,rotate(20)*NE,fontsize(10pt)); label(p,rotate(20)*\"$S$\",S1,rotate(20)*E,fontsize(10pt)); label(p,rotate(20)*\"$T$\",T,rotate(20)*S,fontsize(10pt));label(p,rotate(20)*\"$X$\",X,rotate(20)*SE,fontsize(10pt));\ndot(p,X);\nadd(rotate(-20)*p);[/asy]",
  "Solution_1": "Triangle RQS is an isosceles triangle, so <RSQ is (180-108)/2=36 degrees. Triangle XQS is a right triangle, so <XQS is 180-<QXS-<XSQ=180-90-(108-36)=90-72=18 degrees."
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "power of a point",
    "radical axis",
    "geometry proposed"
  ],
  "Problem": "Let $ ABC$ be a triangle inscribed in circumcircle $ (O)$. Denote $ A_1,B_1,C_1$ respectively be the projections of $ A$,$ B,$ $ C$ onto $ BC,$ $ CA,$ $ AB$. Let $ A_2,B_2,C_2$ respectively be the intersections of $ AO,BO,CO$ with $ BC,$ $ CA,$ and $ AB$. A circle $ \\Omega_a$ passes through $ A_1$, $ A_2$ and tangents to the arc of $ BC$ that doesn't contain $ A$ of $ (O)$ at $ T_a$. The same definition holds for $ T_b,T_c$. Prove that $ AT_a$, $ BT_b$ and $ CT_c$ are concurrent.\r\n\r\n[i]Best regard,\n[b]mathVNpro[/b] [/i]",
  "Solution_1": "Since the circle $ \\odot(AA_1A_2)$ is tangent to $ (O)$ at $ A,$ then the tangent $ \\ell_1$ to $ (O)$ at $ A$ is the radical axis of $ \\odot(AA_1A_2)$ and $ (O).$ The sideline $ BC$ is the radical axis of $ \\odot(AA_1A_2) , \\Omega_a$ and the tangent $ \\ell_2$ to $ (O)$ at $ T_a$ is the radical axis of $ \\Omega_a$ and $ (O).$ Hence,  lines $ BC ,$ $ \\ell_1 ,$ $ \\ell_2$ concur at the radical center $ O_a$ of circles $ (O) ,\\Omega_a$ and $ \\odot(AA_1A_2).$ It's known that $ O_a \\equiv \\ell_1 \\cap BC$ is the center of the A-Apollonius circle of $\\triangle ABC,$ which is orthogonal to $ (O)$ $ \\Longrightarrow$ $ AT_a$ is the A-symmedian. By similar reasoning we conclude that lines $ AT_a,BT_b,CT_c$ concur at the symmedian point of $ \\triangle ABC.$",
  "Solution_2": "The generalization of this problem also holds:(instead of $ H$ is orthocenter and $ O$ is circumcenter we have $ H,O$ are arbitary isogonal conjugates points).  :wink: \r\nProof:Let $ (AA_1A_2)$ cut $ AB$ at $ A_3$ and $ AC$ at $ A_4$.Then $ \\angle{A_3A_4A_2} \\equal{} \\angle{A_4A_3A_1} \\rightarrow$ $ A_1A_2A_3A_4$ is isoceles trapezium.From that $ A_3A_4 \\| BC \\rightarrow$ $ (AA_3A_4)$ is tangent to $ (ABC)$.The back of this proof is same of luisgeometria's proof."
}
{
  "Tag": [
    "induction",
    "modular arithmetic",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Show that every integer greater than $ 89.315$ is the sum of two abundant numbers.",
  "Solution_1": "Well, this is not a solution to your problem, but I will sketch a weaker result:\r\n\r\n\"Every large integer is a sum of two abundant numbers\". More precisely, exists a number $ N$ such that every integer larger than $ N$ is a sum of two abundants.\r\n\r\nLet $ S(k)$ the sum of all positive divisors of $ k$. If $ S(k)>2k$ then $ k$ is abundant.\r\n\r\n[b]Lemma 1:[/b]\r\nIf $ k\\equal{}{p_1}^{e_1}{p_2}^{e_2}\\ldots{p_i}^{e_i}$ is the canonical factoring of $ k$ then \r\n\\[ S(k)\\equal{}\\frac{{p_1}^{e_1\\plus{}1}\\minus{}1}{p_1\\minus{}1} \\cdot \\frac{{p_2}^{e_2\\plus{}1}\\minus{}1}{p_2\\minus{}1} \\ldots \\frac{{p_i}^{e_i\\plus{}1}\\minus{}1}{p_i\\minus{}1}\\]\r\n\r\nThis can be proved with double counting techniques or with a simple induction (and even can be generalized : try to calculate the sum of powers of divisors!).\r\n\r\n[b]Lemma 1.1:[/b]\r\n\\[ \\frac{S(k)}{k}\\equal{} \\prod_{1 \\leq j \\leq i}\\left({\\sum_{0 \\leq t \\leq e_j}{p_j^{\\minus{}t}}}\\right)\\]\r\n\r\n[b]Lemma 2:[/b]\r\nFor any $ m \\geq 1$ and $ k \\geq 1$, $ S(km)/km \\geq S(m)/m$ with equality if and only if $ k\\equal{}1$.\r\nThis can be proved by a two-parallel-way induction:\r\n\r\n1) If $ k$ has no common prime factors with $ m$ then it adds another primes to the $ km$ factoring, and so it adds other $ {\\sum_{0 \\leq t \\leq e_j}{p_j^{\\minus{}t}}}$ factors to the product above;\r\n\r\n2) If $ k$ has some common prime factor with $ m$, some of existing factors {\\sum_{0 \\leq t \\leq e_j}{p_j^{-t}}} gain an extra term $ p_i^t$.\r\n\r\nIn every case, the only way to make none of adds above is $ k\\equal{}1$.\r\n\r\n[b]Lemma 2.1:[/b]\r\nIf $ m$ is perfect or abundant then $ Km$ is abundant for $ K>1$.\r\n\r\n[b]Lemma 2.1.1:[/b]\r\nEvery multiple of 6, greater than 6, is abundant!\r\n\r\nIf $ N \\equiv t \\pmod 6$ then $ N\\equal{}6M\\plus{}t$, and we can force $ M>1$ just in the case. But we need to guarrantee that $ t$ is abundant. Or, better, we need to arrange a residue class of 6, containing only abundant numbers.\r\nIf we find an abundant $ N \\equiv 1 \\pmod 6$ then our residue class can be just $ \\{6N,1N,2N,3N,4N,5N\\}$; and if we can find an abundant $ N \\equiv \\minus{}1 \\pmod 6$ it is just fine too (the residue class is reverted!$ \\{6N,5N,4N,3N,2N,1N\\}$).\r\n\r\nEvery prime (except 2 and 3) leaves a residue $ 1$ or $ 5 \\equiv \\minus{}1$ modulo 5. Our idea is to find an abundant number of form $ N\\equal{}5 \\cdot 7 \\cdot 11 \\cdot 13 \\cdot 17 \\ldots$. You can verify that $ 5 \\cdot 7 \\cdot 11 \\cdot 13 \\cdot 17 \\cdot 19 \\cdot 23 \\cdot 29 \\cdot 31$ works!\r\n\r\nMaybe adapting this idea and using greater exponents instead I can obtain better estimates... But, in fact, as far as I know, the lower limit is very large! 89315=5*17863 appears to be very small...\r\n\r\nObs.: The limit for \"even as sum of two abundants\" is far less."
}
{
  "Tag": [
    "trigonometry"
  ],
  "Problem": "two boats A and B lives a port C at the same time on different routes.B travelledon a bearing of 150degrees and A travelled north of B.when A had travelled 8km,B had travelled 10km and the distance between them was 12km.Calculate the bearing of A's route from C.",
  "Solution_1": "Those sides call for the Law of Cosines.\r\n\r\nWe are looking for the center angle at C, right? So use\r\n$ c^2 \\equal{} a^2 \\plus{} b^2 \\minus{} 2ab \\cos C$\r\nand $ 160 \\cos C \\equal{} 64 \\plus{} 100 \\minus{} 144$\r\n$ \\cos C \\equal{} \\frac{1}{8}$\r\nSolving with a calculator yields 82.819244218541718676961004373851 degrees. Sadly, I don't remember the way bearings turn anymore."
}
{
  "Tag": [
    "IMO Shortlist",
    "geometry unsolved",
    "geometry"
  ],
  "Problem": "The circles $ k_1$ and $ k_2$ with respective centers $ O_1$ and $ O_2$ are externally tangent at point $ C$, while the circle $ k$ with center $ O$ is externally tangent to $ k_1$ and $ k_2$. Let $ l$ be the common tangent of $ k_1$ and $ k_2$ at the point $ C$ and let $ AB$ be the diameter of $ k$ perpendicular to $ l$. Assume that $ O$ and $ A$ lie on the same side of $ l$. Show that the lines $ AO_2,BO_1,l$ have a common point.",
  "Solution_1": "Well, it's nice to hearing that this problem is from Bulgaria 1996. It was also proposed for IMO Shortlist 2006, you can find some nice solutions here: http://www.mathlinks.ro/viewtopic.php?p=875026#875026"
}
{
  "Tag": [
    "number theory",
    "relatively prime"
  ],
  "Problem": "Prove that the congruence x^37\u2261x(mod 13) is valid for any integer x.",
  "Solution_1": "[hide]\n\nIf x is relatively prime to 13 then by Fermat's Little Theorem\n\nx^12 = 1 (mod 13)\n\n(x^12)^3 = x^36 = 1 (mod 13)\n\nx^37 = x (mod 13)\n\n\n\nOtherwise if 13 | x\n\nx^37 = x = 0 (mod 13)\n\n[/hide]",
  "Solution_2": "[quote=\"Iris Aliaj\"]Prove that the congruence x^37\u2261x(mod 13) is valid for any integer x.[/quote]\r\n\r\nI will prove FLT:\r\n\r\nModulo p, [0, 1, .... , p-1] is congruent to [0, a , 2a , ..... a(p-1)]\r\n\r\nHence:\r\n\r\n(p-1)! = a^(n-1)(p-1)! mod p\r\n\r\nBut (p-1)! = -1 mod p -->\r\n\r\n1 = a^(n-1) mod p as desired."
}
{
  "Tag": [
    "Pascal\\u0027s Triangle"
  ],
  "Problem": "Help me please!\r\n\r\nWhat's the maximum number of regions formed by lines in the plane?\r\n\r\nThis is obviously $ \\binom{n\\plus{}1}{2}\\plus{}1$.  But what if $ k$ are parallel?  Or $ c$ are concurrent?  What about planes in space?",
  "Solution_1": "Generally, in the $ n^{th}$ dimension, the maximum number of regions that $ k$ things of $ (n \\minus{} 1)^{th}$ dimension can cut it into is the sum of the first $ n \\plus{} 1$ numbers in the $ k^{th}$ row of Pascal's Triangle, where the zeroth row is $ 1$, the first row is $ 1,1$, etc., i.e. $ R_n \\equal{} \\left\\{\\dbinom nj\\right\\}_{j \\equal{} 0}^n$, or equivalently, $ \\sum_{i \\equal{} 0}^n\\dbinom ki$, where if $ k < i$, we have $ \\dbinom ki \\equal{} 0$.",
  "Solution_2": "[quote=\"AIME15\"]\n\n\n\nThis is > obviously < $ \\binom{n \\plus{} 1}{2} \\plus{} 1$. [/quote]\r\nThis is a silly statement.  There is no \"obviously\" about it.\r\n\r\nYou just state \"This is $ \\binom{n \\plus{} 1}{2} \\plus{} 1$.\"",
  "Solution_3": "[quote=\"AIME15\"]But what if $ k$ are parallel?  Or $ c$ are concurrent?[/quote]\r\n\r\nThen we haven't attained the maximum number of regions formed by lines in the plane.\r\n\r\n...I must be missing something.  :maybe:",
  "Solution_4": "I think he means the maximum possible given that $ k$ lines are parallel, $ c$ lines are concurrent (probably not too clear, but something of that nature), or both."
}
{
  "Tag": [
    "MATHCOUNTS",
    "email"
  ],
  "Problem": "No, I am pretty certain that you cannot go as an individual if you do not make your school's team.",
  "Solution_1": "I'm pretty sure you can sign up as \"homeschooled\".  I don't think you actually have to be homeschooled.",
  "Solution_2": "Probably the smartest thing to do, instead of asking random AoPSers who have not been in a similar situation before and can do nothing more than make educated guesses, would be to find your chapter coordinater's email and ask him specifically.",
  "Solution_3": "[quote=\"athunder\"]If one does not make the school team for whatever reason, is he/she aloud to go and still take the mathcounts test officially?\n\nI am in a situation like this. \n\nI know you can sign up as an individual if your school does not offer it, but what if it does?\n\nThanks.[/quote]\r\n\r\nDo you mean that if you didn't make the top 4? Because, register your school can 4 team members and 4 individuals. Or, do you mean that you didn't make the top 8? If that happens, you can be an alternate in chapter and practice with the team. Thats up to your ranking in the school competition and the discretion of the teacher."
}
{
  "Tag": [
    "number theory",
    "greatest common divisor",
    "relatively prime",
    "prime factorization"
  ],
  "Problem": "If $ \\frac{1}{a}\\plus{}\\frac{1}{b}\\equal{}\\frac{1}{c}$, where a, b, c are positive integers with no common factor, prove that $ (a \\plus{} b)$ is the square of an integer.",
  "Solution_1": "[hide=\"solution\"]$ \\dfrac{a \\plus{} b}{ab} \\equal{} \\dfrac{1}{c}$\n\n$ ab \\minus{} ca \\minus{} cb \\equal{} 0$\n\n$ ab \\minus{} ca \\minus{} cb \\plus{} c^2 \\equal{} c^2$\n\n$ (a \\minus{} c)(b \\minus{} c) \\equal{} c^2$\n\n\nSince $ a$, $ b$, and $ c$ have no common factors with each other other than 1, $ a \\minus{} c$ and $ b \\minus{} c$ may not have any common factors with $ c^2$ other than 1, but that's impossible if $ c^2$ is not equal to 1. Therefore, $ c \\equal{} 1$, and $ a \\equal{} b \\equal{} 2$, but that's impossible.[/hide]\r\n\r\nAm I missing something?",
  "Solution_2": "I think he means there is no factor which all three have in common.",
  "Solution_3": "[hide]\nRearranging and factoring gives $ (a\\minus{}c)(b\\minus{}c) \\equal{} c^2$. If this has integer solutions we must have $ a\\minus{}c\\equal{}k$ and $ b\\minus{}c \\equal{} c^2/k$ where $ k$ is some factor of $ c^2$. Then $ a\\plus{}b \\equal{} \\frac{1}{k}(k\\plus{}c)^2$. It remains only to show that $ k$ must be a perfect square. Probably something to do with the relatively prime condition.\n[/hide]",
  "Solution_4": "I think 1=2 makes a good point.\r\n\r\n$ \\frac {1}{a} \\plus{} \\frac {1}{b} \\equal{} \\frac {1}{c}$\r\n\r\nTherefore\r\n\r\n$ \\frac {a \\plus{} b}{ab} \\equal{} \\frac {1}{c}$\r\n\r\nTherefore $ a \\plus{} b \\equal{} \\frac {ab}{c}$\r\n\r\nBut since $ gcd(c,ab) \\equal{} 1$ if $ a \\plus{} b$ is an integer, $ c \\equal{} 1$\r\n\r\nTherefore it is as 1=2 suggests\r\n\r\n[b]Related Problem[/b]\r\nBMO 2: 1998 Q3\r\n\r\nSuppose x, y, and z are integers satisfying\r\n\\[ \\frac {1}{x} \\minus{} \\frac {1}{y} \\equal{} \\frac {1}{c}\r\n\\]\r\nLet h be the highest common factor of x, y, and z.\r\n\r\nProve\r\n\r\ni) hxyz is a perfect square\r\nii) h(y-x) is a perfect square",
  "Solution_5": "[b]@1=2:[/b] Why is $ (a,b,c) \\equal{} (2,2,1)$ impossible? Also consider $ \\frac {1}{a} \\plus{} \\frac {1}{a(a \\minus{} 1)} \\equal{} \\frac {1}{a \\minus{} 1}$, where clearly the GCD is 1 since $ a$ and $ c \\equal{} a \\minus{} 1$ are co-prime.\r\n\r\n[b]@SimonM:[/b] I think the thing here is that we don't necessarily have $ \\text{GCD}[a,b] \\equal{} \\text{GCD}[b,c] \\equal{} \\text{GCD}[c,a] \\equal{} 1$, only $ \\text{GCD}[a,b,c]$\r\n\r\n\r\nNote that we must have $ a|bc \\; \\; \\; b|ac \\; \\; \\; c|ab$",
  "Solution_6": "[hide=\"my solution\"] \nSay $ c \\equal{} a \\minus{} d$. \n\n$ \\frac {1}{a} \\plus{} \\frac {1}{b} \\equal{} \\frac {1}{a \\minus{} d} \\\\\n\\Rightarrow \\frac {ab}{a \\plus{} b} \\equal{} a \\minus{} d \\\\\n\\Rightarrow d \\equal{} a \\minus{} \\frac {ab}{a \\plus{} b} \\\\\n\\Rightarrow d \\equal{} \\frac {a^2}{a \\plus{} b} \\\\\n\\Rightarrow d(a \\plus{} b) \\equal{} a^2$ \n\nConsider a prime factor $ p$ of $ d$. It is clear that $ p|a$, so $ a$ and $ d$ have p in common, and therefore $ a$ and $ c$ have $ p$ in common. Follows that $ p$ doesn't divide $ b$, so neither doess it divide $ a \\plus{} b$. We conclude that $ p$ occurs equally many times in $ d$ as in $ a^2$. We do this for every prime factor of $ d$ and so we see that $ d$ is a perfect square. Therefore also $ (a \\plus{} b)$ is a perfect square.\n[/hide]",
  "Solution_7": "Yeah, that's what I was thinking, [b]Mat(h)s[/b].",
  "Solution_8": "Nice proof, [b]Mat(h)s[/b].\r\n\r\nHere is my conjecture of a stronger proposition:\r\n\r\nGiven $ \\gcd(a,b,c)\\equal{}1$, if $ (a,b,c)$ satisfies $ \\frac1a\\plus{}\\frac 1b\\equal{}\\frac1c$, then $ (a,b,c)\\equal{}\\left(p(p\\plus{}q),q(p\\plus{}q),pq\\right)$ for $ \\gcd(p,q)\\equal{}1$.\r\n\r\nTrue or false?",
  "Solution_9": "[quote=\"The Zuton Force\"]\n[b]@SimonM:[/b] I think the thing here is that we don't necessarily have $ \\text{GCD}[a,b] \\equal{} \\text{GCD}[b,c] \\equal{} \\text{GCD}[c,a] \\equal{} 1$, only $ \\text{GCD}[a,b,c]$\n[/quote]\r\n\r\nAh, it didn't read like that when I first read it.",
  "Solution_10": "[hide=\"Solution\"]We have $ bc\\plus{}ac\\equal{}ab$.  Note that if any prime $ p$ divides one of $ a,b,c$, it must divide exactly one of the other two.  Furthermore, suppose for instance $ p^{k}$ strictly divides $ a$.  Then $ p^{k}$ divides $ bc$, so $ p^{k}$ divides either $ b$ or $ c$.  But if $ p^{k\\plus{}1}$ divides either $ b$ or $ c$, then $ a$ must be divisible by $ p^{k\\plus{}1}$ by the same logic, a contradiction.  Therefore, if $ p^{k}$ strictly divides one of $ a,b,c$, then $ p^{k}$ strictly divides another of $ a,b,c$, and the third is not divisible by $ p$.\n\nNow let $ x\\equal{}(a,b)$, $ y\\equal{}(b,c)$, and $ z\\equal{}(c,a)$.  Every power of a prime $ p_{i}^{a_{i}}$ that is part of the prime factorization of $ a$ is also part of the prime factorization of exactly one of $ b,c$.  Therefore, $ a\\equal{}xz$, and similarly $ b\\equal{}xy$ and $ c\\equal{}yz$.  Substituting, we get $ xy\\cdot yz\\plus{}xz\\cdot yz\\equal{}xz\\cdot xy$.  Dividing through by $ xyz$, we get $ y\\plus{}z\\equal{}x$.\n\nThen $ a\\plus{}b\\equal{}xz\\plus{}xy\\equal{}x(y\\plus{}z)\\equal{}x^{2}$ as desired.[/hide]",
  "Solution_11": "Oh, thank for all! I don't think we can have many proofs as above.",
  "Solution_12": "[quote=\"NguyenDungTN\"]If $ \\frac {1}{a} \\plus{} \\frac {1}{b} \\equal{} \\frac {1}{c}$, where a, b, c are positive integers with no common factor, prove that $ (a \\plus{} b)$ is the square of an integer.[/quote]\r\n\r\nSorry for reviving this thread...but i didnt see a satisfactoryanswer so i decided to post mine.\r\n\r\n\r\nsolving the equation gives us \r\n$ ab\\minus{}cb\\minus{}ca\\equal{}0$\r\n$ \\equal{}> (a\\minus{}c)(a\\minus{}b)\\equal{}c^2$  (*)\r\n\r\n$ (a,b,c)\\equal{}1$ $ \\equal{}> (a\\minus{}c,b\\minus{}c,1)\\equal{}1$       (**)\r\nfact * and ** together imply that $ a\\minus{}c\\equal{}y^2$ and  $ b\\minus{}c\\equal{}z^2$\r\n\r\nthis means that $ c \\equal{} yz$ \r\n$ a\\plus{}b \\equal{} y^2\\plus{}z^2\\plus{} 2c\\equal{}y^2\\plus{}z^2\\plus{}2yz\\equal{}(y\\plus{}z)^2$ . Hence we are done"
}
{
  "Tag": [
    "inequalities",
    "induction",
    "inequalities unsolved"
  ],
  "Problem": "[color=indigo][b][n+1)/2]^n * [(2n+1)/3]^n > (n ! ) ^ 2[/b][/color]",
  "Solution_1": "Hey,\r\n\r\nWe want to prove the inequality: $ (\\frac{n\\plus{}1}{2})^n(\\frac{2n\\plus{}1}{3})^n > (n!)^2$.\r\n\r\nNotice that the following equality holds for each positive integer $ n$ (it can be proved by induction on n):\r\n\r\n     $ 1^2\\plus{}2^2\\plus{}...\\plus{}n^2\\equal{}\\frac{n(n\\plus{}1)(2n\\plus{}1)}{6}$\r\n\r\nTherefore considering the left part of our desired inequality,we have:\r\n\r\n $ (\\frac{n\\plus{}1}{2})^n(\\frac{2n\\plus{}1}{3})^n\\equal{}(\\frac{(n\\plus{}1)(2n\\plus{}1)}{6})^n\\equal{}(\\frac{1^2\\plus{}2^2\\plus{}...n^2}{n})^n>1^22^2...n^2\\equal{}(n!)^2$\r\n\r\nbecause of the AM-GM.Notice that the inequality is strict!! \r\n\r\n~Kostas",
  "Solution_2": "Do not double post:\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=306724[/url]"
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "vector",
    "LaTeX",
    "calculus",
    "linear algebra",
    "matrix"
  ],
  "Problem": "Find the angle between the diagonal of cube and  another diagonal of the\r\ncube. \r\n\r\nI can visualize it, but how do you express the vector for the 2nd diagonal? I'm using a unit cube so $ u \\equal{} <1,1,1>$ is my first diagonal but I don't know how to express the other diagonal in component form.",
  "Solution_1": "$ \\text{\\LaTeX}$ hint: use \\langle and \\rangle for that kind of bracket, not < and >.\r\n\r\nYour vector $ \\langle 1,1,1\\rangle$ is the displacement from the point $ (0,0,0)$ to $ (1,1,1).$ That's two of the 8 vertices of the cube. Now, pick two [i]other[/i] diametrically opposed vertices; it shouldn't matter which. What's another vertex? Say, $ (1,0,0).$ Now find what's opposite to that by just changing ones to zeros and zeros to ones. That gives us $ (0,1,1).$ The displacement vector would then be $ \\langle -1,1,1\\rangle$ - or maybe that in the other direction, $ \\langle 1,-1,-1\\rangle.$ You can try both, and report the angle you get as the acute angle.\r\n\r\nVaguely related chemistry question: What is the H-C-H bond angle in methane?",
  "Solution_2": "[quote=\"Kent Merryfield\"]$ \\text{\\LaTeX}$ hint: use \\langle and \\rangle for that kind of bracket, not < and >.\n\nYour vector $ \\langle 1,1,1\\rangle$ is the displacement from the point $ (0,0,0)$ to $ (1,1,1).$ That's two of the 8 vertices of the cube. Now, pick two [i]other[/i] diametrically opposed vertices; it shouldn't matter which. What's another vertex? Say, $ (1,0,0).$ Now find what's opposite to that by just changing ones to zeros and zeros to ones. That gives us $ (0,1,1).$ The displacement vector would then be $ \\langle - 1,1,1\\rangle$ - or maybe that in the other direction, $ \\langle 1, - 1, - 1\\rangle.$ You can try both, and report the angle you get as the acute angle.\n\nVaguely related chemistry question: What is the H-C-H bond angle in methane?[/quote]\r\n\r\nAh, thank you. I didn't think of having negative coordinates. I did get the same as the bond angle, which is 109.5. Coincidentally, that happens to be #45 in my book :]\r\n\r\nI have a question about the section though. I am doing this problem from 11.1 of Strang's Calculus. I always thought vectors were a part of calculus, so why are they in the Linear Algebra section? I know you can express a vector as a column matrix, but I didn't know they played a big role in matrix algebra. Guess I have a lot more to learn lol!",
  "Solution_3": "I'd probably report the answer to this particular problem as $ \\cos^{\\minus{}1}\\left(\\frac13\\right)\\approxeq 70.5^{\\circ}$ - the acute angle rather than the obtuse angle. But that's not particularly important.\r\n\r\nI'm the one who moved the problem. Yes, I am very much aware that this comes from a calculus course and there's a chapter in most calculus books containing exactly this sort of material. It's just that I wanted to make the point that not everything taught in such courses is actually calculus - if it's actually vector algebra or vector geometry, I think it fits a little better in this forum.",
  "Solution_4": "I have one more question.\r\n\r\n27 The force F = 3i - 4k acts at the point (1,2,2). How much\r\nforce pulls toward the origin? How much force pulls vertically\r\ndown? Which direction does a mass move under the force F?\r\n\r\nFor vertically downward, the answer is stated as 4. How is that computed?\r\n\r\nThe vertically downward vector should be $ \\langle 1,2,2\\rangle \\minus{} \\langle 1,0,2 \\rangle$ correct? That gives a vector of $ \\langle 0,2,0 \\rangle$, but the magnitutde of the projection comes out to be zero.",
  "Solution_5": "Not correct.  You're asked to find the downwards component of the force vector, not the $ z$-coordinate of the position.  (Note that in three dimensions, the third coordinate is the vertical coordinate, not the second.)  For the first part of the question, you want to find the component of the force in the direction that joins the origin to the current position.",
  "Solution_6": "I got the first and third part, but I'm not sure how to compute the second part."
}
{
  "Tag": [
    "calculus",
    "integration",
    "trigonometry",
    "Putnam",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Compute \r\n\\[ \\int_0^1\\frac{ln(1+x)}{1+x^2}dx. \\]",
  "Solution_1": "You can set $x=\\tan \\theta$",
  "Solution_2": "more details [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=23498]http://www.mathlinks.ro/Forum/viewtopic.php?t=23498[/url]",
  "Solution_3": "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=64436]Putnam 2005 A5[/url]",
  "Solution_4": "See and the exercise $8.1.$ from http://www.mathlinks.ro/Forum/viewtopic.php?t=66197"
}
{
  "Tag": [
    "vector",
    "abstract algebra",
    "Functional Analysis",
    "advanced fields",
    "advanced fields unsolved"
  ],
  "Problem": "Let $X,Y$ be Banach spaces, and let $T: X\\to Y$ be a continuous linear map such that the linear space $Y/T(X)$ is finite-dimensional. \r\n\r\nProve that the range of $T$ (that is, $T(X)$) is a closed subspace of $Y$.",
  "Solution_1": "I'll try this:  Suppose not, if necessary by replacing the spaces we can assume that $cl(TX)=Y$.  Let a vector $v$ in $Y$ representing some vector in $Y/X$.  Then we can approximate v in TX by $Tx_{n}$, which I can choose to be a linearly independent set, here I use that v is not in the image of X.  We define $A_{n}\\in Y^{*}$ as $A_{n}(v)=1, A_{n}(Tx_{i})=0, \\forall i<n$ and extend them each by Hahn-Banach, here I use the linearly independence.  By Banach-Alaoglu(mispelling?) theorem, exist $A\\in Y^{*}$ s.t. $A_{n}$ weak* converges to A.  By definition that means $ATx_{n}=0, Av=1$, contradiction.\r\n\r\nBut I still kinda feel that there's some problem in the proof, I would be very appreciative if someone would like to point it out.  Finite dimension is not used, but I am not concious enough to see where it should be.  I'll sleep first... :lol:",
  "Solution_2": "Banach-Alaoglu says that the unit ball of $Y^{*}$ is weak-$*$ compact. In order to apply it, you need a boundedness condition on your sequence of functionals $(A_{n})_{n}$, which you don't have.\r\n\r\n\r\nP.S.\r\n\r\nI just wanted to point out that (a) as you noticed, you did not use the fact that $Y/TX$ is finite-dimensional, and there are plenty of coutnerexamples which show that it's a crucial condition, and (b) you did not use the fact that $TX$ is the range of a continuous operator $T$; that too is crucial, since there are finite-codimensional non-closed subspaces of $Y$ (the kernel of a discontinuous linear functional, for example). \r\n\r\nThere's nothing redundant in the hypothesis.",
  "Solution_3": "First, we can suppose $T$ is injective, otherwise we replace $T$ with $T/\\mathrm{ker}T$ wich has the same range.\r\n\r\n(i) By the hypotesis, there exist linearly independent $y_{1},...,y_{n}$ s.t. $Y=\\langle y_{1},...,y_{n}\\rangle+T(X)$. Moreover, using linear independence of the $y_{i}$, it can be shown that the $+$ above is in fact a $\\oplus$.\r\n\r\n(ii) Define a map $\\phi: X\\times\\mathbb{R}^{n}\\to Y$, sending $(x,t_{1},...,t_{n})$ in $t_{1}y_{1}+...t_{n}+y_{n}+T(x)$. This is linear, continuous and surjective by construction (hence an open map) and injective by assumption. So its inverse is continuous.\r\n\r\n(iii) Take $\\{y_{k}\\}$ in $T(X)$ which converge to $y\\in Y$. Then $\\phi^{-1}(z_{k})=: (x_{k},0,...,0)\\to\\phi^{-1}(y)$. So $\\phi^{-1}(y)=(x,0,...,0)$ for some $x\\in X$, that is, $y=T(x)$."
}
{
  "Tag": [
    "calculus",
    "derivative",
    "integration",
    "logarithms",
    "calculus computations"
  ],
  "Problem": "If $ x^{3}\\minus{}3a^{2}x\\equal{}a^{2}y$ show that \r\n$ \\frac{dx}{\\sqrt{x^{2}\\minus{}4a^{2}}}\\equal{}\\frac{dy}{3\\sqrt{y^{2}\\minus{}4a^{2}}}$\r\nand hence obtain Cardan's formula for the solution of a cubic.\r\n\r\nProving the differential equation is just a straightforward exercise of implicit differentiation, but I can't seem to get the solution of a cubic from it, integrating both sides results in a rather large mess...Any suggestions?",
  "Solution_1": "The formula is messy no matter what; you'll need to get the integral in terms of logarithms, and do some complicated algebra.\r\n\r\nActually, this method is a lot nicer if we look at the case where $ y$ is small and there are three real roots- it's an angle trisection."
}
{
  "Tag": [],
  "Problem": "Suppose that $4^{x}-4^{y}=48$ and $2^{x+y}=32$. Find $3x-4y$.",
  "Solution_1": "obviously x=3, y=2",
  "Solution_2": "[quote=\"surshu\"]obviously x=3, y=2[/quote]\r\n\r\nYou must always show your work surshu...\r\n\r\n[hide]\n$2^{2x}-2^{2y}=48$\n$2^{x}\\times2^{y}=32$\nLet $X=2^{x}$ and $Y=2^{y}$\n$(X-Y)(X+Y)=48$ and $XY=32$\nFor X and Y to have a natural value in the first equation, there are only 3 possible pairs: (X-Y,X+Y)=(6,8),(2,24),(4,12)\n\nIf you solve for X and Y, only (4,12) gives you perfect powers of 2. $X=8$ and $Y=4$. This fits into the second equation well.\n\nFinally, we see that $x=3$ and $x=2$.\n$3x-4y=9-8=1$\n[/hide]",
  "Solution_3": "i did it a different way.  \r\n$2^{5}=32$\r\n$x+y=5$\r\n$4^{3}-4^{2}=48$\r\n$3x-4y=1$",
  "Solution_4": "[quote=\"Andrew Kim\"]\nFor X and Y to have a natural value in the first equation[/quote]\r\nAlthough this works, you don't know that they're natural"
}
{
  "Tag": [
    "number theory solved",
    "number theory"
  ],
  "Problem": "$p_1,p_2,...,p_{25}$ is primes which is less than  $2004$Find the maximal integer $T$ such that all of positive integer which is not greater than $T$ can be represented as sum of distinct divisor of  $(p_1.p_2...p_{25})^{2004}$",
  "Solution_1": "Check: http://www.mathlinks.ro/Forum/viewtopic.php?p=347239"
}
{
  "Tag": [
    "geometry unsolved",
    "geometry"
  ],
  "Problem": ":(  :(  :(",
  "Solution_1": "$ \\frac{s^2}{2p}\\equal{}r$ then is equal to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=215261]this[/url]",
  "Solution_2": ":clap:  :wink:"
}
{
  "Tag": [
    "limit",
    "logarithms",
    "integration",
    "calculus",
    "calculus computations"
  ],
  "Problem": "Calculate       $\\lim_{n\\to\\infty}\\frac{n}{\\sqrt[n]{n!}}$",
  "Solution_1": "$\\log \\left ( \\frac{\\sqrt[n]{n!}}{n}\\right ) = \\log \\left ( \\frac{\\prod_{i=1}^{n}i^{\\frac{1}{n}}}{n}\\right ) = \\sum_{i=1}^{n}\\log \\left ( \\frac{i^{\\frac{1}{n}}}{n}\\right ) = \\frac{1}{n}\\sum_{i=1}^{n}\\log \\left ( \\frac{i}{n}\\right )$\r\n\r\nTherefore,\r\n$\\lim_{n \\to \\infty}\\log \\left (\\frac{n}{\\sqrt[n]{n!}}\\right ) = \\lim_{n \\to \\infty}\\log \\left (\\frac{1}{\\frac{\\sqrt[n]{n!}}{n}}\\right ) =-\\lim_{n \\to \\infty}\\log \\left ( \\frac{\\sqrt[n]{n!}}{n}\\right )$\r\n$=-\\lim_{n \\to \\infty}\\left ( \\frac{1}{n}\\sum_{i=1}^{n}\\log \\left ( \\frac{i}{n}\\right ) \\right ) =-\\int_{0}^{1}\\log x \\ dx = 1$\r\n\r\nThus, \r\n$\\lim_{n \\to \\infty}\\frac{n}{\\sqrt[n]{n!}}= e^{1}= e$",
  "Solution_2": "[b]Theorem:[/b] Let $(s_{n})$ be a sequence of non-zero terms and $\\lim \\left| \\frac{s_{n+1}}{s_{n}}\\right| = L$ (note $L$ can be finite or infinite) then $\\lim |s_{n}|^{1/n}= L$.\r\n----\r\n\r\nWith above theorem it is simple. Consider the sequence $(s_{n}) = \\frac{n^{n}}{n!}$ then $\\frac{s_{n+1}}{s_{n}}= \\frac{(n+1)^{n+1}}{(n+1)n!}\\cdot \\frac{n!}{n^{n}}$.\r\nWhich is,\r\n$=(1+1/n)^{n}$\r\nSo,\r\n$\\lim \\frac{s_{n+1}}{s_{n}}= e$.\r\n\r\nWhich means that,\r\n$\\lim \\left| \\frac{n^{n}}{n!}\\right|^{1/n}= \\lim \\frac{n}{\\sqrt[n]{n!}}= e$",
  "Solution_3": "[quote=\"Kouichi Nakagawa\"]$ \\log \\left ( \\frac {\\sqrt [n]{n!}}{n}\\right ) = \\log \\left ( \\frac {\\prod_{i = 1}^{n}i^{\\frac {1}{n}}}{n}\\right ) = \\sum_{i = 1}^{n}\\log \\left ( \\frac {i^{\\frac {1}{n}}}{n}\\right ) = \\frac {1}{n}\\sum_{i = 1}^{n}\\log \\left ( \\frac {i}{n}\\right )$[/quote]\r\n\r\nSorry :( \r\nI'm mistake.\r\n\r\n\\begin{eqnarray*}\r\n\\log \\left( \\frac{\\sqrt[n]{n!}}{n} \\right)\r\n& = & \\log \\left( \\frac{\\prod_{i=1}^n i^{\\frac{1}{n}}}{n} \\right) \\\\\r\n& = & \\log \\left( \\frac{\\prod_{i=1}^n i^{\\frac{1}{n}}}{\\prod_{i=1}^n n^{\\frac{1}{n}}} \\right) \\\\\r\n& = & \\log \\left( \\prod_{i=1}^n \\frac{i^{\\frac{1}{n}}}{n^{\\frac{1}{n}}} \\right) \\\\\r\n& = & \\log \\left( \\prod_{i=1}^n \\left( \\frac{i}{n} \\right)^{\\frac{1}{n}} \\right) \\\\\r\n& = & \\sum_{i=1}^n \\log \\left( \\left( \\frac{i}{n} \\right)^{\\frac{1}{n}} \\right) \\\\\r\n& = & \\frac{1}{n} \\sum_{i=1}^n \\log \\left( \\frac{i}{n} \\right) \\\\\r\n\\end{eqnarray*}",
  "Solution_4": "$ n! \\approx \\sqrt{2 \\pi n} \\frac{n^n}{e^n}$\r\n\r\nso\r\n\r\n$ \\frac{n}{\\sqrt[n]{n!}}\\approx e \\frac{1}{\\sqrt[\\frac{n}{2}]{2 \\pi n} } \\Rightarrow \\lim_{n \\rightarrow \\infty}\\frac{n}{\\sqrt[n]{n!}}=e$"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "geometry",
    "3D geometry"
  ],
  "Problem": "The equation $x^{3}+ax^{2}+bx+c = 0$ has the root $\\sqrt[3]{(\\sqrt{33}+5)}-\\sqrt[3]{(\\sqrt{33}-5)}$. Find the ordered triple $(a,b,c)$ if $a,b,$ and $c$ are integers.",
  "Solution_1": "Let $p=\\sqrt[3]{\\sqrt{33}+5},\\ q=\\sqrt[3]{\\sqrt{33}-5}$ and $x=p-q$, we have $x^{3}=(p-q)^{3}=p^{3}-q^{3}-3pq(p-q)$, yielding $x^{3}=10-3\\cdot 2\\cdot x\\Longleftrightarrow x^{3}+6x-10=0$. The answer is $(a,\\ b,\\ c)=(0,\\ 6,-10)$.",
  "Solution_2": "hmm thats cool whats the motivation behind this?",
  "Solution_3": "Thanks. $A+B+C=0\\Longrightarrow A^{3}+B^{3}+C^{3}=3ABC$",
  "Solution_4": "well the motivation is that you have seen the problem before...like i suppose you can be like hey lets just cube it, that gets rid of stuff...then i guess you just need to be creative to see the recursive substitution"
}
{
  "Tag": [
    "calculus",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "Find all real solutions of the equation [quote]$ x^5  \\plus{} 10x^3  \\plus{} 20x \\minus{} 18 \\equal{} 0$[/quote]",
  "Solution_1": "[quote=\"xdanielxedgex\"]Find all real solutions of the equation [quote]$ x^5 \\plus{} 10x^3 \\plus{} 20x \\minus{} 18 \\equal{} 0$[/quote][/quote]\r\n\r\n$ f(x)\\equal{}x^5 \\plus{} 10x^3 \\plus{} 20x \\minus{} 18$ $ \\implies$ $ f'(x)\\equal{}5x^4\\plus{}30x^2\\plus{}20>0$ and so $ f(x)$ is increasing and has exactly one real root.\r\n\r\nThen numeric calculus gives $ x\\equal{}0.711111084051342...$",
  "Solution_2": "Hi, Can you explain how you obtain this result?",
  "Solution_3": "[quote=\"xdanielxedgex\"]Hi, Can you explain how you obtain this result?[/quote]\r\n\r\nThe numeric result ?\r\n\r\nThere exists a lot of method to numerically solve equations. In this case, I used a spreadsheet and a basic Newton method : \r\n\r\n$ u_0\\equal{}0$\r\n$ u_{n\\plus{}1}\\equal{}u_n\\minus{}\\frac{f(u_n)}{f'(u_n)}$"
}
{
  "Tag": [
    "\\/closed"
  ],
  "Problem": "I think there's a typo at the bottom of the screen, when it says there have been 8250144700 users online in the last 24 hours. ([i]has[/i] should be [i]have[/i])",
  "Solution_1": "[quote=\"zabelman\"]I think there's a typo at the bottom of the screen, when it says there have been 8250144700 users online in the last 24 hours. ([i]has[/i] should be [i]have[/i])[/quote]\r\n\r\nThere's a mistake in your comment above.  The set of all possible users is less than 8 billion.  And we won't have the attention of all of them for another few years at least.",
  "Solution_2": "The language error is fixed :)",
  "Solution_3": "[quote=\"MCrawford\"][quote=\"zabelman\"]I think there's a typo at the bottom of the screen, when it says there have been 8250144700 users online in the last 24 hours. ([i]has[/i] should be [i]have[/i])[/quote]\n\nThere's a mistake in your comment above.  The set of all possible users is less than 8 billion.  And we won't have the attention of all of them for another few years at least.[/quote]\r\n\r\nYou've worked with zableman enough to know he's always thinking ahead."
}
{
  "Tag": [
    "conics",
    "ellipse",
    "analytic geometry",
    "graphing lines",
    "slope",
    "geometry",
    "geometric transformation"
  ],
  "Problem": "Let the major axis of an ellipse be $AB$, let $O$ be its center, and let $F$ be one of its foci. $P$ is a point on the ellipse, and $CD$ a chord through $O$, such that $CD$ is parallel to the tangent of the ellipse at $P$. $PF$ and $CD$ intersect at $Q$. Compare the lengths of $PQ$ and $OA$.",
  "Solution_1": "$\\left(\\forall\\right)P\\in \\Sigma\\ ,\\ PQ=OA\\ .$. [b]Very nice and interesting problem ![/b][hide=\"Here is the proof.\"][color=darkblue]Let $b^2x^2+a^2y^2=a^2b^2$ be the equation of the ellipse $\\Sigma$, where $b^2+c^2=a^2$ and let $F(\\epsilon c,0)$ be a focus, where $\\epsilon ^2=1$. Let $P(au,bv)\\in \\Sigma$ be a mobile point, where $u^2+v^2=1$. Then the slope of the tangent $PP$ in the point $P\\in \\Sigma$ to the ellipse $\\Sigma$ is $m_{PP}=-\\frac{bu}{av}$. The equation of the line $CD$ is $y=-\\frac{bu}{av}\\cdot x$ and the equation of the line $CF$ is $y=\\frac{bv}{au-\\epsilon c}\\cdot (x-\\epsilon c)$. Thus, for the intersection point $Q\\in CD\\cap PF$ we have $x_Q=\\frac{\\epsilon acv^2}{a-\\epsilon cu}$, $y_Q=-\\frac{\\epsilon bcuv}{a-\\epsilon cu}$ and $\\left(\\frac{a-\\epsilon cu}{a}\\right)^2\\cdot PQ^2=$ $\\left[u(a-\\epsilon cu)-\\epsilon cv^2\\right]^2+b^2v^2=$ $(au-\\epsilon c)^2+b^2v^2=$ $a^2u^2+b^2(1-u^2)+c^2-2\\epsilon acu=$ $(a^2-b^2)u^2+a^2-2\\epsilon acu=$ $c^2u^2+a^2-2\\epsilon acu=$ $(a-\\epsilon cu)^2\\Longrightarrow PQ=a$, i.e. $PQ=OA$.[/color][/hide]",
  "Solution_2": "Let the foci be $F_1$ and $F_2$, and let $Q_1 = PF_1 \\cap CD$ and $Q_2 = PF_2 \\cap CD$.  Let $M$, $R_1$, and $R_2$ be the projections of $P$, $F_1$, and $F_2$ onto $CD$, respectively.\r\n\r\nFirst, we claim that triangles $PMQ_1$ and $PMQ_2$ are congruent.  We are given that $CD$ is parallel to the tangent at $P$, so by the reflection property of ellipses, $\\angle MPQ_1 = \\angle MPQ_2$.  Also, $\\angle PMQ_1 = \\angle PMQ_2 = 90^\\circ$, and both triangles share side $PM$, so they are congruent.  Therefore, $PQ_1 = PQ_2$.\r\n\r\nNext, we claim that triangles $F_1 R_1 Q_1$ and $F_2 Q_2 R_2$ are congruent.  Since $CD$ passes through the centre of the ellipse, foci $F_1$ and $F_2$ are equidistant from $CD$, i.e. $F_1 R_1 = F_2 R_2$.  By the argument above, $\\angle R_1 F_1 Q_1 = \\angle R_2 F_2 Q_2$, and $\\angle F_1 R_1 Q_1 = \\angle F_2 R_2 Q_2 = 90^\\circ$, so the two triangles are congruent.  Therefore, $F_1 Q_1 = F_2 Q_2$.\r\n\r\nBut, $PF_1 + PF_2 = AB$, and\r\n\\begin{eqnarray*} PF_1 + PF_2 &=& F_1 Q_1 + Q_1 P + PQ_2 - Q_2 F_2 \\\\ &=& Q_1 P + PQ_2 \\\\ &=& 2 PQ_1 = 2PQ_2, \\end{eqnarray*}\r\nso $PQ_1 = PQ_2 = AB/2 = OA$.\r\n\r\n[img]http://www.artofproblemsolving.com/Admin/latexrender/pictures/f044e9673a5d3b9e6f446ce1be0c9dbb.png[/img]"
}
{
  "Tag": [
    "Euler",
    "algorithm",
    "real analysis",
    "real analysis solved"
  ],
  "Problem": "prove that { \\emptyset (n)/n} is dense in [0,1]?\r\n( \\emptyset (n) means totient of euler and {x} means the decimal part of x) excuse me I don't know if I send it to right topic or not?",
  "Solution_1": "Let 2=p1<p2<... be the sequence of all prime numbers.\r\n\r\n \\emptyset (p)/p=1-1/p ->1 for p prime, p-> \\infty.\r\n\r\n \\emptyset (p1*p2*...*pk)/(p1*...*pk)=(1-1/p1)...(1-1/pk)=1/(1+1/p1+1/p1 <sup>2</sup> +...)...(1+1/pk+1/pk <sup>2</sup> +...)=1/(\\sum 1/x :x is a number that has only prime divisors int he set p1...pk). It is easy to see that the limit of this, when k ->\\infty is 0.\r\n\r\nFor a number a \\in (0,1) use the following algorithm:\r\n\r\n(00) for k:=1 to \\infty do if  (1-1/pk)>a then goto (01);\r\n(01) n:=pk;\r\n(02) for i:=k+1 to \\infty do if  (\\emptyset (n)/n)(1-1/pi)>a then goto (03);\r\n(03) k:=i; while ( \\emptyset(n)/n)(1-1/pk)>a do {n:=n*pk; k:=k+1;} goto (02);    \r\n\r\nThis alghorithm, chained indefinitely, constructs an increasing sequence of n's such as  \\emptyset (n)/n decreases to a.\r\n\r\nI hope you don't find it that hard to prove that the sequence actually converges to a. You must use: (1-1/pk) increases to 1 and (1-1/p1)*...*(1-1/pk) decreases to 0. You can a;so use this to prove that the alghorithm doesn't stop at some point.\r\n----------------------------------------------------------------------------------\r\n{} - the decimal part, it is superfluous for the problem because  0 \\leq \\emptyset(n) <n"
}
{
  "Tag": [
    "geometry",
    "LaTeX"
  ],
  "Problem": "Does anyone have a link to a [b]free[/b] geometry rendering program?",
  "Solution_1": "I'm using TeXgraph, but there is a lot of programs for that: gnuplot,....\r\n\r\nGoogle is your friend :lol:",
  "Solution_2": "try graphmatica, from download.com. If you need geometrical shapes, then do what i did. Use a mixture of ppt and paint. :P"
}
{
  "Tag": [
    "counting",
    "distinguishability"
  ],
  "Problem": "How many ways are there to put 4 distinguishable balls into 2 distinguishable boxes?",
  "Solution_1": "Each ball has $ 2$ choices, and there are $ 4$ balls, so this is $ 2 \\times 2 \\times 2 \\times 2\\equal{}2^4\\equal{}\\boxed{16}$.",
  "Solution_2": "It appeared that 16 is correct. I am wondering, however, if the correct answer should be 14 instead. It is quite explicitly mentioned that the 4 balls should be put into 2 boxes. Therefore, putting all balls in one box should not be counted as a valid solution. Using the same reasoning, putting all balls in box 2 is also invalid, which results in $16-2=14$ valid ways to put the 4 distinguishable balls into 2 distinguishable boxes.",
  "Solution_3": "4 choose 0 + 4 choose 1 + 4 choose 2 + 4 choose 3 + 4 choose 4 = 2^4 = 16  :P ",
  "Solution_4": "$ 2 \\times 2 \\times 2 \\times 2=2^4=\\boxed{16}$",
  "Solution_5": "[quote=blijkijker]It appeared that 16 is correct. I am wondering, however, if the correct answer should be 14 instead. It is quite explicitly mentioned that the 4 balls should be put into 2 boxes. Therefore, putting all balls in one box should not be counted as a valid solution. Using the same reasoning, putting all balls in box 2 is also invalid, which results in $16-2=14$ valid ways to put the 4 distinguishable balls into 2 distinguishable boxes.[/quote]\n\nThere is no restriction in the problem that said \"putting all balls in one box should not be counted as a valid solution\" You can imagine it as putting 0 balls in box 1 and 4 balls in box 2. \n\nI know this user probably won't reply as their post was made a few years ago, but can someone else confirm my reasoning.",
  "Solution_6": "The generalization is there are $m^n$ ways to put $n$ distinguishable balls in $m$ distinguishable boxes.",
  "Solution_7": "[quote=Nutty_Nitty][quote=blijkijker]It appeared that 16 is correct. I am wondering, however, if the correct answer should be 14 instead. It is quite explicitly mentioned that the 4 balls should be put into 2 boxes. Therefore, putting all balls in one box should not be counted as a valid solution. Using the same reasoning, putting all balls in box 2 is also invalid, which results in $16-2=14$ valid ways to put the 4 distinguishable balls into 2 distinguishable boxes.[/quote]\n\nThere is no restriction in the problem that said \"putting all balls in one box should not be counted as a valid solution\" You can imagine it as putting 0 balls in box 1 and 4 balls in box 2. \n\nI know this user probably won't reply as their post was made a few years ago, but Can someone else confirm my reasoning.[/quote]\n\nYeah that is how I interpreted it too. :)"
}
{
  "Tag": [
    "calculus",
    "integration",
    "function",
    "trigonometry",
    "limit",
    "derivative",
    "real analysis"
  ],
  "Problem": "This one's been eating me up for quite some time. Any help is welcome! :blush: \r\nI found in a paper the following equality:\r\n$ I_0(r,z)\\equal{}rz\\int_z^r\\frac {1}{p\\sqrt {r^2 \\minus{} p^2}\\sqrt {p^2 \\minus{} z^2}}\\,dp \\equal{} \\frac {\\pi}{2}$\r\n$ 0 < z < r < 1$\r\nBut it doesn't say how they got there nor has any helpful reference.\r\nSo here I am, asking someone who can solve this and show me how to do it, please!\r\n(There's also $ I_1$ and $ I_n$ in the bag, but I'll be glad to solve this one first  :wink: )",
  "Solution_1": "http://en.wikipedia.org/wiki/Elliptic_integral",
  "Solution_2": "[quote=\"JRav\"]http://en.wikipedia.org/wiki/Elliptic_integral[/quote]\r\nThanks, but my doubt comes from page 596 of [url=http://books.google.com/books?id=MtU8uP7XMvoC&pg=PA596&lpg=PA587&dq=abramowitz&output=html&sig=i1Wvt7QhvO_-Qf5KjOmcNOLLT7I]Abramowitz and Stegun's handbook of mathematical functions[/url]. They have a table with some integrals similar to mine:\r\n17.4.43\r\n$ a\\int_b^x\\frac{dt}{\\sqrt{(a^2\\minus{}t^2)(t^2\\minus{}b^2)}}$\r\nand\r\n17.4.44\r\n$ a\\int_x^a\\frac{dt}{\\sqrt{(a^2\\minus{}t^2)(t^2\\minus{}b^2)}}$\r\n\r\nBut I don't know how to use this thing. I don't know anything about elliptic integrals. That's why I'm asking for help. How can I use this? What are those 'cn', 'sc', 'nd', etc functions?\r\n\r\nAlso, I have an extra $ p$ in the denominator, so I can't just apply the formula from the book. There's some extra brain crunching to do, but I feel unprepared for such a momentous task.\r\nOn a bright note, I put this integral on Mathematica and, after substituting $ y\\equal{}p^2,\\,dy\\equal{}2p\\,dp$ it was able to calculate it and gave a correct result. So my theory is: if the computer can do it, then a human can do it. I just need to learn.\r\nPlease, if anyone has any ideas, fell free! :blush:",
  "Solution_3": "That extra $ p$ is exactly what you need for elementary methods. Substitute $ p^2$, and then combine the square roots into the square root of a quadratic. Trig substitution works.",
  "Solution_4": "[quote=\"pocaracas\"] \n$ I_0(r,z) \\equal{} rz\\int_z^r\\frac {1}{p\\sqrt {r^2 \\minus{} p^2}\\sqrt {p^2 \\minus{} z^2}}\\,dp \\equal{} \\frac {\\pi}{2}$\n  [/quote]\r\n\r\nthis is an [b]easy[/b] integral, why on earth r u associating it with elliptic integrals  ?   :?: \r\n\r\n  divide both num and den by $ rz\\,p^3$ to get\r\n\r\n $ I_0(r,z) \\equal{} \\int_z^r \\frac {\\frac {1}{p^3}\\,dp } {\\sqrt {\\frac {1}{p^2} \\minus{} \\frac {1}{r^2} }\\sqrt {\\frac {1}{z^2} \\minus{} \\frac {1}{p^2}} } \\equal{} \\minus{} \\frac {1}{2}\\int_z^r \\frac {d\\left(\\frac {1}{p^2}\\right) } {\\sqrt {\\frac {1}{p^2} \\minus{} \\frac {1}{r^2} }\\sqrt {\\frac {1}{z^2} \\minus{} \\frac {1}{p^2}} }$\r\n\r\nletting $ x \\equal{} \\frac {1}{p^2}$ gives (notice $ 1 > r > z > 0 )$\r\n\r\n$ I_0(r,z) \\equal{}\\frac {1}{2}\\int_{\\frac {1}{r^2}}^{\\frac {1}{z^2}} \\frac {dx}{\\sqrt {x \\minus{} \\frac {1}{r^2}}\\sqrt {\\frac {1}{z^2} \\minus{} x}} \\equal{}\\frac {1}{2}\\;\\left\\{ \\minus{} \\tan^{ \\minus{} 1} \\left( \\; \\frac { \\frac {1}{2r^2} \\plus{} \\frac {1}{2z^2} \\minus{}x}{\\sqrt {x \\minus{} \\frac {1}{r^2} }\\sqrt {\\frac {1}{z^2} \\minus{} x}} \\;\\right)\\;\\bigg|_{\\frac {1}{r^2}}^{\\frac {1}{z^2}}\\right\\}$\r\n\r\n\r\n$ \\equal{}  \\frac {1}{2}\\left\\{  \\;\\lim_{p_0\\to z}  \\; \\minus{} \\tan^{ \\minus{} 1} \\left( \\frac{\\frac {1}{2r^2} \\minus{} \\frac {1}{2z^2} }{\\sqrt { \\frac {1}{p_0^2} \\minus{} \\frac {1}{r^2} }\\sqrt {\\frac {1}{z^2}  \\minus{} \\frac {1}{p_0^2} }} \\right) \\plus{} \\;\\lim_{p_1\\to r}  \\;  \\tan^{ \\minus{} 1}  \\left( \\frac{\\frac {1}{2z^2} \\minus{} \\frac {1}{2r^2} }{\\sqrt { \\frac {1}{p_1^2} \\minus{} \\frac {1}{r^2} }\\sqrt {\\frac {1}{z^2}  \\minus{} \\frac {1}{p_1^2} }} \\right) \\right\\}$\r\n\r\n\r\n since $ 1 > r > z > 0,$ we have\r\n\r\n\r\n $ I_0(r,z) \\equal{} \\frac {1}{2} \\; \\left\\{ \\minus{}  \\tan^{ \\minus{} 1}  \\left( \\frac{ something \\quad negative} { something \\quad going \\quad to \\quad 0} \\right) \\plus{}  \\tan^{ \\minus{} 1} \\left( \\frac{ something \\quad positive} { something \\quad going \\quad to \\quad 0}\\right) \\right\\}$\r\n\r\n$ \\equal{}\\frac {1}{2} \\;\\lim_{R\\to\\infty } \\; \\left(\\minus{} \\tan^{ \\minus{} 1} ( \\minus{} R) \\plus{}  \\tan^{ \\minus{} 1} R  \\right)\\equal{} \\frac {1}{2} \\; \\left( \\minus{} \\left( \\minus{}\\frac {\\pi}{2}\\right) \\plus{} \\frac {\\pi}{2}\\right) \\equal{} \\frac {\\pi}{2}$\r\n\r\n$ \\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}$\r\n\r\nPS: to solve the integral, i used the  fact \r\n\r\n$ d(\\tan^{ \\minus{} 1} \\; X) \\equal{} \\frac {1}{1 \\plus{} X^2}$\r\n\r\n can you identify what my $ X$ was ?",
  "Solution_5": "[quote=\"jmerry\"]That extra $ p$ is exactly what you need for elementary methods. Substitute $ p^2$, and then combine the square roots into the square root of a quadratic. Trig substitution works.[/quote]\r\nThanks, jmerry, but it's not so obvious to me... I must have burned some neurons after graduating!  :wink:",
  "Solution_6": "[quote=\"misan\"]\n... can you identify what my $ X$ was ?[/quote]\r\nOk, \r\nI got as far as\r\n$ I_0(r,z) \\equal{} \\minus{} \\frac {1}{2}\\int_{\\frac {1}{z^2}}^{\\frac {1}{r^2}} \\frac {dx}{\\sqrt {x \\minus{} \\frac {1}{r^2}}\\sqrt {\\frac {1}{z^2} \\minus{} x}}$\r\nno prob!\r\nBut I can't see how you did the next step\r\n\r\nUsing $ d(\\tan^{ \\minus{} 1} \\; X) \\equal{} \\frac {1}{1 \\plus{} X^2}$, one would be led to assume that $ 1\\plus{}X^2 \\equal{} \\sqrt {x \\minus{} \\frac {1}{r^2}}\\sqrt {\\frac {1}{z^2} \\minus{} x}$, but that can't be right since it leads to $ \\sqrt{\\minus{}1}$... I think.\r\nThe limit afterwards just blew me away, completely. Can you explain how you get there, please?",
  "Solution_7": "Ok! I see how the limit works. thanks!\r\nBut this leap:\r\n$ I_0(r,z) \\equal{}\\frac {1}{2}\\int_{\\frac {1}{r^2}}^{\\frac {1}{z^2}} \\frac {dx}{\\sqrt {x \\minus{} \\frac {1}{r^2}}\\sqrt {\\frac {1}{z^2} \\minus{} x}} \\equal{}\\frac {1}{2}\\;\\left\\{ \\minus{} \\tan^{ \\minus{} 1} \\left( \\; \\frac { \\frac {1}{2r^2} \\plus{} \\frac {1}{2z^2} \\minus{}x}{\\sqrt {x \\minus{} \\frac {1}{r^2} }\\sqrt {\\frac {1}{z^2} \\minus{} x}} \\;\\right)\\;\\bigg|_{\\frac {1}{r^2}}^{\\frac {1}{z^2}}\\right\\}$\r\nis still unclear.\r\n\r\nHow do you go from $ \\frac {1}{\\sqrt {x \\minus{} \\frac {1}{r^2}}\\sqrt {\\frac {1}{z^2} \\minus{} x}}$ to $ \\frac{1}{1\\plus{}X^2}$\r\nwhere, appearently, $ X$ is $ \\frac { \\frac {1}{2r^2} \\plus{} \\frac {1}{2z^2} \\minus{}x}{\\sqrt {x \\minus{} \\frac {1}{r^2} }\\sqrt {\\frac {1}{z^2} \\minus{} x}}$?",
  "Solution_8": "I found it!\r\nmisan, you must have done something wrong to get the right result.\r\nBut I found equation 3.3.27 of [url=http://books.google.com/books?id=MtU8uP7XMvoC&pg=PA12&lpg=PA5&dq=abramowitz&output=html&sig=4Z9EC21rwOhDxOsClgtxRaOPM1s]Abramowitz[/url] which gives a ready made formula to find this integral:\r\n$ \\int\\frac{dx}{\\sqrt{(a\\plus{}bx)(c\\plus{}dx)}} \\equal{} \\minus{}\\frac{1}{\\sqrt{\\minus{}bd}}\\arcsin\\left(\\frac{2bdx\\plus{}ad\\plus{}bc}{bc\\minus{}ad}\\right),\\,b>0, d<0$\r\n\r\nSo,\r\n$ I_0(r,z) \\equal{} \\frac{1}{2}\\int_{\\frac{1}{r^2}}^{\\frac{1}{z^2}}\\frac{dx}{\\sqrt{(x\\minus{}\\frac{1}{r^2})(\\frac{1}{z^2}\\minus{}x)}} \\equal{}\\frac{1}{2}\\left[\\minus{}\\frac{1}{\\sqrt{\\minus{}(\\minus{}1)}}\\arcsin\\left(\\frac{2(\\minus{}1)x\\plus{}(\\minus{}\\frac{1}{r^2})(\\minus{}1)\\plus{}\\frac{1}{z^2}}{\\frac{1}{z^2}\\minus{}\\frac{1}{r^2}}\\right)\\right]_{\\frac{1}{r^2}}^{\\frac{1}{z^2}}$\r\n$ \\equal{}\\minus{}\\frac{1}{2}[\\arcsin(\\minus{}1)\\minus{}\\arcsin(1)] \\equal{} \\minus{}\\frac{1}{2}(\\minus{}\\frac{\\pi}{2}\\minus{}\\frac{\\pi}{2}) \\equal{} \\frac{\\pi}{2}$",
  "Solution_9": "ok, misan, I double checked your result and deriving the $ \\tan^{\\minus{}1}$ you present one does get to $ \\frac {1}{\\sqrt {x \\minus{} \\frac {1}{r^2}}\\sqrt {\\frac {1}{z^2} \\minus{} x}}$.\r\nOn my first check it didn't give that result.\r\nI'm [b]sorry[/b] for saying you were wrong.\r\n\r\nOne other thing that made me say it was wrong was that $ \\frac{1}{1\\plus{}X^2}$, with $ X\\equal{}\\frac { \\frac {1}{2r^2} \\plus{} \\frac {1}{2z^2} \\minus{}x}{\\sqrt {x \\minus{} \\frac {1}{r^2} }\\sqrt {\\frac {1}{z^2} \\minus{} x}}$, I arrive at $ \\minus{}\\frac{4r^2(r^2x\\minus{}1)z^2(z^2x\\minus{}1)}{(r^2\\minus{}z^2)^2}$ which is considerably different from $ \\frac {1}{\\sqrt {x \\minus{} \\frac {1}{r^2}}\\sqrt {\\frac {1}{z^2} \\minus{} x}}$ in the degree of $ x$. :ninja: \r\nSo... I'm guessing you didn't really use that formula for the derivative of the $ \\tan^{\\minus{}1}$, did you?",
  "Solution_10": "So do it the long way. $ \\frac1p\\equal{}\\frac{p}{p^2}$; now you have $ p\\,dp$ and can substitute $ p^2$. There are square roots of linear factors in the new denominator; combine them, complete the square, and perform the appropriate trig substitution. This leaves a rational combination of trigonometric functions, which you may need to use the $ \\tan\\frac{\\theta}{2}$ rationalizing substitution on. It's long, but it's guaranteed to work and find an elementary antiderivative."
}
{
  "Tag": [
    "function",
    "induction",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "Let $f: \\Bbb{R} \\rightarrow \\Bbb{R}$ be a function such that\r\n(i) For all $x,y \\in \\Bbb{R}$, \\[ f(x)+f(y)+1 \\geq f(x+y) \\geq f(x)+f(y) \\]\r\n(ii) For all $x \\in [0,1)$, $f(0) \\geq f(x)$,\r\n(iii) $-f(-1) = f(1) = 1$.\r\n\r\nFind all such functions $f$.",
  "Solution_1": "by plugging $x=-1$, $y=1$ we get $f(0)\\geq0$, by plugging $(0,0)$ we get $f(0) \\leq 0$ so $f(0)=0$. You can easy prove that for all $x \\in [0,1)$ we have $f(x)=0$ (it's a key step  :) ) Now we'll prove by induction that $f(x)=[x]$. Our inductive sentence $T_n$ is: $f(x)=[x]$ for all $x<n$. Suppose we have proved it for some $n$. Now we have for $n-1 \\leq x < n$ and $\\epsilon$ such that $x+\\epsilon<n$:\r\n\r\n$[x+1]=[x]+1=f(x+\\epsilon)+f(1-\\epsilon)+1 \\geq f(x+1) \\geq f(x)+1=[x]+1=[x+1]$. So we have proved $T_{n+1}$.\r\n\r\nQED",
  "Solution_2": "My solution from WOOT:\n\nLet $P(x,y)$ denote the assertion of $(\\text i)$.\n$P(0,0),P(1,-1)\\Rightarrow f(0)=0$\n$P(x,1)\\Rightarrow f(x+1)\\ge f(x)+1$\n$P(x+1,-1)\\Rightarrow f(x)\\ge f(x+1)-1$\nThen equality must hold, so $f(x+1)=f(x)+1$.\nFor $x\\in[0,1)$, $f(x)\\le0$. Then:\n$P(x,1-x)\\Rightarrow f(x)+f(1-x)\\ge0$, but since $f(x)+f(1-x)\\le0+0$ equality holds for each and $f(x)=0$ for $x\\in[0,1)$. Then by induction on $f(x+1)=f(x)+1$ we obtain $\\boxed{f(x)=\\lfloor x\\rfloor}$, which fits.",
  "Solution_3": "[hide=Solution]\nSubstituting $x=y=0$, we get $f(0)\\geq2f(0)$, so $f(0)\\leq0$. Substituting $x=1$ and $y=-1$, we get $f(0)\\geq f(1)+f(-1)=0$. Therefore, $f(0)=0$. Substituting $x=1$, we get $f(y+1)\\geq f(y)+1$. Substituting $x=-1$, we get $f(y-1)\\geq f(y)-1$, so $f(y)\\geq f(y+1)-1$. Substituting this into the first inequality, we get $f(y+1)\\geq f(y)+1\\geq f(y+1)$, so $f(y+1)=f(y)+1$. Substituting $y=1-x$, we get $f(x)+f(1-x)\\geq0$. If $0\\leq x<1$, then $f(x)\\leq0$, so $f(x)=0$. Therefore, we have $f(x)=\\lfloor x\\rfloor$. We will show that this works. The left inequality is true since $\\lfloor x\\rfloor+\\lfloor y\\rfloor+1-\\lfloor x+y\\rfloor=-\\{x\\}-\\{y\\}+1+\\{x+y\\}>-1$ and it is an integer, so it is nonnegative. The right inequality is equivalent to $\\{x\\}+\\{y\\}\\geq\\{x+y\\}$, which is also true. Since $\\lfloor -1\\rfloor=-1$ and $\\lfloor 1\\rfloor=1$, this means that the third condition is satisfied. Since $\\lfloor x\\rfloor=0$ when $0\\leq x<1$, this means that the second condition is also true.\n\nTherefore, the only solution is $\\boxed{f(x)=\\lfloor x\\rfloor}$.",
  "Solution_4": "As everybody else is posting solutions, I'll also post mine:\n\nWe claim that $f(x)=\\lfloor x \\rfloor$ is the only solution. It clearly works.\n\n[b][color=#f00]Claim:[/color][/b] $f(x+1)=f(x)+1$ for all real $x$.\nNotice that plugging in $x=1$ gives $$1+f(y)\\le f(1+y)\\le 2+f(y).$$ On the other hand, plugging in $x=-1$ gives $$f(y)-1\\le f(y-1)\\le f(y)\\Rightarrow f(y-1)\\le f(y) \\le f(y-1)+1\\Rightarrow f(y)\\le f(y+1)\\le f(y)+1.$$ The result follows from these two inequalities. $\\blacksquare$\n\nThus it suffices to find the values of $f(x)$ for $0\\le x<1$; then we can use the claim to find the rest of the function. Using the claim, we get $f(0)=0$. Let $0<a<1$ and plug $x=a, y=1-a$ to get $$f(a)+f(1-a)\\le 1\\le 1+f(a)+f(1-a).$$ In particular, we get $0\\le f(a)+f(1-a)$. However, from condition (ii) we get $0\\ge f(a), f(1-a)$, thus we have $f(a)=f(1-a)=0$. Hence, for all $0\\le x<1$, $f(x)=0$. Using the claim, we get $f(x)=\\lfloor x \\rfloor$, and we are done.",
  "Solution_5": "The answer is $f(x) = \\lfloor x \\rfloor$, which works. \n\n[b][color=blue]Step 1: Extracting $f(0)$.[/b][/color] We claim that $f(0)=0$. First, by setting $x=y=0$, $$2f(0)+1 \\geq f(0) \\geq 2f(0) \\iff -1 \\leq f(0)\\leq 0.$$ On the other hand, letting $x=1$ and $y=-1$, $$f(-1)+f(1)+1=1\\geq f(0) \\geq 0.$$ Combining these two inequalities gives $f(0)=0$.\n\n[b][color=blue]Step 2: Getting $f$ over $[0, 1)$.[/b][/color] Now, we claim that $f(x)=0$ for all $x \\in [0, 1)$. The idea here is similar to the idea in the first step: by condition (ii), we know that $f(x) \\leq 0$ for $x \\in [0, 1)$, but on the contrary, letting $y=1-x$ gives us $$f(x)+f(1-x)+1\\geq f(1) = 1 \\iff f(x)+f(1-x) \\geq 0.$$ For $x \\in [0, 1)$, this is only possible if $f(x)=f(1-x)=0$, as required.\n\n[b][color=blue]Step 3: Inducting upwards/downwards.[/b][/color] Finally, we are ready to finish. Letting $y=1$ gives us $$f(x+1) \\geq f(x)+1$$ for all $x$, and setting $y=-1$ and $x=x+1$ gives us $$f(x+1)+f(-1)\\leq f(x) \\iff f(x+1)\\leq f(x)+1.$$ Therefore, $f(x+1)=f(x)+1$ for all $x$. Inducting upwards and downwards gives us the result.",
  "Solution_6": "Let $P(x,y)$ denote the given assertion. \n\nClaim: $f(0)=0$. \nProof: $P(0,0): 2f(0)+1\\ge f(0)\\ge 2f(0)\\implies -1\\le f(0)\\le 0$. \n\n$P(1,-1): f(0)\\ge 0$, so $f(0)=0$. \n\nClaim: $f(k)=0\\forall 0\\le k<1$. \nProof: Note that $f(k), f(1-k)\\le 0$. \n\n$P(k,1-k): f(k)+f(1-k)+1\\ge 1\\implies f(k)+f(1-k)\\ge 0\\implies f(k)=f(1-k)=0$. \n\nClaim: $f(x+1)=f(x)+1$. \nProof: $P(x,1): f(x)+2\\ge f(x+1)\\ge f(x)+1$. \n\n$P(x+1,-1): f(x)\\ge f(x+1)-1\\implies f(x)+1\\ge f(x+1)$. \n\nThis proves our claim. We can also find that $f(x)-1=f(x-1)$. \n\nFrom this claim, $f(x)=\\lfloor x \\rfloor +\\{x\\}\\implies \\boxed{f(x)=\\lfloor x\\rfloor}$, which works."
}
{
  "Tag": [
    "geometry",
    "ratio",
    "Pythagorean Theorem"
  ],
  "Problem": "The area of the semicircle in Figure A is half the area of the circle in Figure B. The area of a square inscribed in the semicircle, as shown, is what fraction of the area of a square inscribed in the circle? Express your answer as a common fraction.\n\n[asy]defaultpen(linewidth(0.8));\n\nsize(5cm,5cm);\n\ndraw((0,0)..(1,1)..(2,0)--(0,0));\n\ndraw((0.5,0)--(0.5,0.87)--(1.5,0.87)--(1.5,0));\n\ndraw(Circle((4,0),1));\n\npair A,B,C,D;\n\nA=(3.3,0.7);\nB=(3.3,-0.7);\nD=(4.7,0.7);\nC=(4.7,-0.7);\n\ndraw(A--B--C--D--A);\n\nlabel(\"Figure A\",(1,1.3));\nlabel(\"Figure B\",(4,1.3));[/asy]",
  "Solution_1": "hello\r\nArea of square in semicircle=$ \\frac{4}{5}r^2$\r\nArea of square in circle=$ 2r^2$\r\nso the ratio=$ \\frac{2}{5}$\r\nthank u",
  "Solution_2": "could you please explain why the area of the square in the semicircle is   (4/5)r^2?\r\n\r\n\r\nsorry for the fact that I can't type in latex.",
  "Solution_3": "[quote=\"covertcognomen\"]could you please explain why the area of the square in the semicircle is   (4/5)r^2?\n\n\nsorry for the fact that I can't type in latex.[/quote]\nLetting the radius of the semicircle be $r$, then the line segment from one corner of the the square in Figure A to the center of the semicircle makes a right triangle.  Let the side of the square in Figure A be $x$ units in length, then by the Pythagorean Theorem, we have:\n\\begin{align*}x^2+\\left(\\frac{x}{2}\\right)^2&=r^2 \\\\ x^2+\\frac{x^2}{4}&=r^2 \\\\ \\frac{5x^2}{4}&=r^2 \\\\ x^2&=\\frac{4r^2}{5} \\\\ x&=\\frac{2r}{\\sqrt5}\\end{align*}\nSince the area of a square is the square of its side, the area of the square in Figure A is:\n\\[\\left(\\frac{2r}{\\sqrt5}\\right)^2 = \\frac{4r^2}{5}\\]\nWhich is what kabi got."
}
{
  "Tag": [
    "inequalities",
    "function",
    "calculus",
    "derivative",
    "logarithms",
    "rearrangement inequality",
    "inequalities proposed"
  ],
  "Problem": "[color=darkred]Let $ m,n\\in \\mathbb{N}$, $ n\\ge 2$ and numbers $ a_i > 0$, $ i \\equal{} \\overline{1,n}$, such that $ \\sum a_i \\equal{} 1$. Prove that\n\n$ \\small{\\dfrac{a_1^{2 \\minus{} m} \\plus{} a_2 \\plus{} ... \\plus{} a_{n \\minus{} 1}}{1 \\minus{} a_1} \\plus{} \\dfrac{a_2^{2 \\minus{} m} \\plus{} a_3 \\plus{} ... \\plus{} a_n}{1 \\minus{} a_1} \\plus{} ... \\plus{} \\dfrac{a_n^{2 \\minus{} m} \\plus{} a_1 \\plus{} ... \\plus{} a_{n \\minus{} 2}}{1 \\minus{} a_1}\\ge n \\plus{} \\dfrac{n^m \\minus{} n}{n \\minus{} 1}}$[/color]",
  "Solution_1": "Ahiles,\r\nDid you solve it?",
  "Solution_2": "Oops... I did a mistake.... it should be\r\n\r\n$ \\small{\\dfrac{a_1^{2 \\minus{} m} \\plus{} a_2 \\plus{} ... \\plus{} a_{n \\minus{} 1}}{1 \\minus{} a_1} \\plus{} \\dfrac{a_2^{2 \\minus{} m} \\plus{} a_3 \\plus{} ... \\plus{} a_n}{1 \\minus{} a_2} \\plus{} ... \\plus{} \\dfrac{a_n^{2 \\minus{} m} \\plus{} a_1 \\plus{} ... \\plus{} a_{n \\minus{} 2}}{1 \\minus{} a_n}\\ge n \\plus{} \\dfrac{n^m \\minus{} n}{n \\minus{} 1}}$",
  "Solution_3": "[color=darkblue]ok... if no one is gonna write.. I will post the offcial solution\n\n$ \\sum_{i=1}^n \\dfrac{a_i^{2-m}+a_{i+1}+...+a_{i-2}}{1-a_i}=\\sum_{i=1}^n \\dfrac{a_i^{2-m}+1-a_{i-1}-a_i}{1-a_i}=$\n\n$ =\\sum_{i=1}^n \\bigg(\\dfrac{a_i^{2-m}-a_{i-1}}{1-a_i}+1\\bigg)=\\sum_{i=1}^n \\dfrac{a_i^{2-m}-a_{i-1}}{1-a_i}+n$\n\nSo our inequality is equivalent to\n\n$ \\sum_{i=1}^n \\dfrac{a_i^{2-m}-a_{i-1}}{1-a_i} \\ge \\dfrac{n^m-n}{n-1}$\n$ \\underline{\\phantom{xxxxxxxxxxx}}$[/color]\r\n[color=darkred]\n[b]Case 1.[/b]   $ m=0$\n\n$ \\sum_{i=1}^n \\dfrac{a_i^{2}-a_{i-1}}{1-a_i}=\\sum_{i=1}^n \\dfrac{(a_i^{2}-1)+(1-a_{i-1})}{1-a_i}=\\sum_{i=1}^n \\dfrac{1-a_{i-1}}{1-a_i}+\\sum_{i=1}^n (a_i+1)=$\n\n$ =\\sum_{i=1}^n \\dfrac{1-a_{i-1}}{1-a_i}+n+1 >-1=\\dfrac{n^0-n}{n-1}$\n$ \\underline{\\phantom{xxxxxxxxxxx}}$\n\n[b]Case 2.[/b]   $ m=1$\n\n$ \\sum_{i=1}^n \\dfrac{a_i-a_{i-1}}{1-a_i}=\\sum_{i=1}^n \\dfrac{(a_i-1)+(1-a_{i-1})}{1-a_i}=\\sum_{i=1}^n \\dfrac{1-a_{i-1}}{1-a_i}-n\\ge$\n\n$ \\ge n\\sqrt[n] {\\dfrac{\\prod_{i=1}^n (1-a_{i-1})}{\\prod_{i=1}^n (1-a_{i-1})}}-n =n-n=0=\\dfrac{n-n}{n-1}$\n$ \\underline{\\phantom{xxxxxxxxxxx}}$\n\n[b]Case 3.[/b]   $ m=2$\n$ \\sum_{i=1}^n \\dfrac{1-a_{i-1}}{1-a_i}\\ge n\\sqrt[n] {\\dfrac{\\prod_{i=1}^n (1-a_{i-1})}{\\prod_{i=1}^n (1-a_{i-1})}}=n=\\dfrac{n^2-n}{n-1}$\n\n$ \\underline{\\phantom{xxxxxxxxxxx}}$\n\n[b]Case 4.[/b]   $ m>2$\n$ \\sum_{i=1}^n \\dfrac{a_i^{2-m}-a_{i-1}}{1-a_i}=\\sum_{i=1}^n \\dfrac{\\dfrac{1}{a_i^{m-2}}-a_{i-1}}{1-a_i}=\\sum_{i=1}^n \\dfrac{\\big(\\dfrac{1}{a_i^{m-2}}-1\\big)+(1-a_{i-1})}{1-a_i}=$[/color]\r\n\r\n${ =\\sum_{i=1}^n \\dfrac{1-a_{i-1}}{1-a_i}+\\sum_{i=1}^n \\dfrac{1}{a_i^{m-2}} \\cdot  \\dfrac{1-a_{i}^{m-2}}{1-a_i}\\ge n+\\sum_{i=1}^n \\dfrac{1}{a_i^{m-2}} (1+a_i+...+a_i^{m-3})}=$\r\n\r\n$ = n+ \\sum_{i=1}^n \\bigg(\\dfrac{1}{a_i^{m-2}} +\\dfrac{1}{a_i^{m-1}}+...+\\dfrac{1}{a_i} \\bigg) = \\sum_{j=1}^{m-2}\\sum_{i=1}^{n}  \\dfrac{1}{a_i^{j}}$\r\n[color=darkblue]\nBut from AM-GM and Cauchy we get\n\n$ \\dfrac{1}{n}\\sum_{i=1}^{n}  \\dfrac{1}{a_i^{j}} \\ge \\bigg( \\dfrac{1}{n} \\sum_{i=1}^{n}\\dfrac{1}{a_i} \\bigg)^j\\ge\n\\bigg( \\dfrac{1}{n} \\sum_{i=1}^{n}\\dfrac{n^2}{\\sum_{i=1}^{n}a_i} \\bigg)^j=n^j$\n\nTherefore\n\n$ \\sum_{j=1}^{m-2}\\sum_{i=1}^{n}  \\dfrac{1}{a_i^{j}} \\ge \\sum_{j=1}^{m-2} n^{j+1}=\\dfrac{n^2(n^{m-2}-1)}{n-1}$\n\nAnd finally\n\n$ \\sum_{i=1}^n \\dfrac{a_i^{2-m}-a_{i-1}}{1-a_i}\\ge n+ \\dfrac{n^2(n^{m-2}-1)}{n-1}= \\dfrac{n^{m}-n}{n-1}$\n\n[/color]",
  "Solution_4": "I solved the problem looking the function $ F(m) \\equal{} \\sum_{i \\equal{} 1}^{n}\\frac {a_i^{2 \\minus{} m}}{1 \\minus{} a_i} \\minus{} \\frac {n^m}{n \\minus{} 1}$ with first derivative - $ F'(m) \\equal{} \\minus{} \\sum_{i \\equal{} 1}^{n}\\frac {a_i^{2 \\minus{} m}}{1 \\minus{} a_i}\\ln a_i \\plus{} \\frac {n^m}{n \\minus{} 1}\\ln\\frac {1}{n}\\ge 0$. :roll: \r\n\r\nHence $ LHS \\minus{} \\frac {n^m}{n \\minus{} 1}\\ge \\sum_{i \\equal{} 1}^n\\frac {1 \\minus{} a_i}{1 \\minus{} a_{i \\plus{} 1}} \\minus{} \\frac {n}{n \\minus{} 1}\\ge n\\minus{}\\frac{n}{n\\minus{}1}$ and therefore $ LHS\\ge RHS$  :wink:  \r\nHowever, it's obvious but very ugly and because of that I didn't write anything :oops:",
  "Solution_5": "my solution:\r\n\r\nFirst step is the same as the official solution\r\nour inequality is equivalent to $ \\sum_{i\\equal{}1}^{n}\\frac{a_{i}^{2\\minus{}m}\\minus{}a_{i\\minus{}1}}{1\\minus{}a_{i}}\\ge\\frac{n^{m}\\minus{}n}{n\\minus{}1}$\r\n\r\nthen using the rearrangement inequality we have\r\n\r\n$ \\sum_{i\\equal{}1}^{n}\\frac {a_{i\\minus{}1}}{1\\minus{}a_i} \\le \\sum _{i\\equal{}1}^{n}\\frac {a_i}{1\\minus{}a_i}$\r\n\r\nso it suffices to prove\r\n\r\n$ \\sum_{i\\equal{}1}^{n}\\frac {a_{i}^{2\\minus{}m}\\minus{}a_i}{1\\minus{}a_i} \\ge \\frac{n^{m}\\minus{}n}{n\\minus{}1}$\r\n\r\nthe case of $ m\\equal{}0,1,2$ is obvious because $ LHS$ is just equal to $ RHS$\r\n\r\nfor $ m \\ge 3$\r\n\r\n$ \\frac {a_{i}^{2\\minus{}m}\\minus{}a_i}{1\\minus{}a_i}\\equal{}\\sum_{k\\equal{}0}^{m\\minus{}2} \\frac {1}{a_{i}^k}$\r\n\r\nso it remains to prove \r\n\r\n$ \\sum_{k\\equal{}0}^{m\\minus{}2} \\sum_{i\\equal{}1}^{n} \\frac {1}{a_{i}^k} \\ge \\frac{n^{m}\\minus{}n}{n\\minus{}1}$\r\n\r\nnow applying the Holder inequality we have \r\n\r\n$ \\sum_{i\\equal{}1}^{n} \\frac {1}{a_{i}^k}\\equal{}(\\sum_{i\\equal{}1}^{n} \\frac {1}{a_{i}^k}) ( \\sum_{i\\equal{}1}^{n} a_i )^k \\ge n^{k\\plus{}1}$, for $ k\\equal{}0,1,...,m\\minus{}2$\r\n\r\nso \r\n\r\n$ \\sum_{k\\equal{}0}^{m\\minus{}2} \\sum_{i\\equal{}1}^{n} \\frac {1}{a_{i}^k} \\ge \\sum_{k\\equal{}0}^{m\\minus{}2} n^{k\\plus{}1}\\equal{}\\frac{n^{m}\\minus{}n}{n\\minus{}1}$\r\n\r\nQ.E.D.",
  "Solution_6": "[quote=\"linboll\"]my solution:\n\nFirst step is the same as the official solution\nour inequality is equivalent to $ \\sum_{i \\equal{} 1}^{n}\\frac {a_{i}^{2 \\minus{} m} \\minus{} a_{i \\minus{} 1}}{1 \\minus{} a_{i}}\\ge\\frac {n^{m} \\minus{} n}{n \\minus{} 1}$\n\nthen using the rearrangement inequality we have\n\n$ \\sum_{i \\equal{} 1}^{n}\\frac {a_{i \\minus{} 1}}{1 \\minus{} a_i} \\le \\sum _{i \\equal{} 1}^{n}\\frac {a_i}{1 \\minus{} a_i}$\n\nso it suffices to prove\n\n$ \\sum_{i \\equal{} 1}^{n}\\frac {a_{i}^{2 \\minus{} m} \\minus{} a_i}{1 \\minus{} a_i} \\ge \\frac {n^{m} \\minus{} n}{n \\minus{} 1}$\n\nthe case of $ m \\equal{} 0,1,2$ is obvious because $ LHS$ is just equal to $ RHS$\n\nfor $ m \\ge 3$\n\n$ \\frac {a_{i}^{2 \\minus{} m} \\minus{} a_i}{1 \\minus{} a_i} \\equal{} \\sum_{k \\equal{} 0}^{m \\minus{} 2} \\frac {1}{a_{i}^k}$\n\nso it remains to prove \n\n$ \\sum_{k \\equal{} 0}^{m \\minus{} 2} \\sum_{i \\equal{} 1}^{n} \\frac {1}{a_{i}^k} \\ge \\frac {n^{m} \\minus{} n}{n \\minus{} 1}$\n\nnow applying the Holder inequality we have \n\n$ \\sum_{i \\equal{} 1}^{n} \\frac {1}{a_{i}^k} \\equal{} (\\sum_{i \\equal{} 1}^{n} \\frac {1}{a_{i}^k}) ( \\sum_{i \\equal{} 1}^{n} a_i )^k \\ge n^{k \\plus{} 1}$, for $ k \\equal{} 0,1,...,m \\minus{} 2$\n\nso \n\n$ \\sum_{k \\equal{} 0}^{m \\minus{} 2} \\sum_{i \\equal{} 1}^{n} \\frac {1}{a_{i}^k} \\ge \\sum_{k \\equal{} 0}^{m \\minus{} 2} n^{k \\plus{} 1} \\equal{} \\frac {n^{m} \\minus{} n}{n \\minus{} 1}$\n\nQ.E.D.[/quote]\r\n\r\nsee my solution here [url]http://ddbdt.co.cc/forum/showthread.php?p=969#post969[/url][/url]"
}
{
  "Tag": [
    "calculus",
    "integration"
  ],
  "Problem": "If you have a repelling force in which the strength varies with the inverse square of the distance (i.e. Columb's Law for like charges), imagine having one object fixed. Then, if you have the other start some distance away from this fixed object, obviously the strength of the force diminishes as this object gets further and further away. Say you let the forces act until they get d (distance) away from each other. Is there a way to calculate the speed of the non-fixed object as we let d go to infinity?\r\n\r\nHere is the exact wording of a problem I got:\r\nAn electron starts from rest 72.5 cm from a fixed point charge with Q=-0.125 microC (1 microC = 1 x 10^-6 C). How fast will the electron be moving when it is very far away?\r\n\r\nThe constants used are the charge on an electron is 1.60 x 10^-19 C, and its mass is 9.11 x 10^-31 kg. \r\n\r\nThanks in advance!",
  "Solution_1": "applying the conservation of energy:\r\n$\\frac{1}{4\\pi \\varepsilon_{0}}\\frac{eQ}{d}=\\frac12 mv^{2}$\r\nfrom here the velocity:\r\n$v=\\sqrt{\\frac{eQ}{2\\pi \\varepsilon_{0}m d}}$\r\n($e$ is the charge of the electron, $\\varepsilon_{0}$ is the dielectrical constant, $d$ is the distance between the charge and the electron at the beginning.)",
  "Solution_2": "Using calculus:\r\nthe equation of motion:\r\n$\\frac{1}{4\\pi \\varepsilon_{0}}\\frac{eQ}{s^{2}}=m \\frac{dv}{dt}$\r\nusing:$v=\\frac{ds}{dt}$\r\nwe get:\r\n$\\frac{1}{4\\pi \\varepsilon_{0}}\\frac{eQ}{m}\\frac{ds}{s^{2}}=vdv$\r\nif at beginning the electron was at $d$ distance from the charge and it goes very very far:\r\n$\\frac{eQ}{4\\pi \\varepsilon_{0}m}\\int_{d}^{\\infty}\\frac{ds}{s^{2}}=\\int_{0}^{v}vdv$\r\nfrom  here:\r\n$v=\\sqrt{\\frac{eQ}{2\\pi \\varepsilon_{0}m d}}$\r\nAnd this is just the same as before.",
  "Solution_3": "Thanks. That really helped me figure it out. :)"
}
{
  "Tag": [
    "AMC",
    "AIME",
    "geometry",
    "USA(J)MO",
    "USAMO",
    "\\/closed"
  ],
  "Problem": "Hard question to answer.\r\n\r\nDepends on your comfort with counting.  If you don't find the post-test for the counting class pretty easy, I'd recommend the counting class (followed up by the intermediate counting class, which we'll offer again sometime in 2004).  Otherwise, I'd recommend the AMC/AIME class.  \r\n\r\nThe counting class should give you virtually all the tools to tackle counting problems on MATHCOUNTS/AMC (and a little on the AIME) - it's a pretty thorough exploration of fundamental counting problems.  If you generally tend to make mistakes on those problems now, this class will help with that.  If you already cover those problems very easily, then go for the AMC/AIME class.",
  "Solution_1": "[quote=\"Rep123max\"]If I had to choose between the AMC/AIME problem series class and the MATHCOUNTS/AMC Counting class which would you recommend?[/quote]\r\n\r\nWhat are your goals?\r\n\r\nThe AMC/AIME class will examine the last ten problems from AMC/AHSME exams and many AIME problems of all levels.  It is a class built for the kinds of students who can already score 2-10 problems on the AIME.  Dozens of topics and problem ideas will be discussed.\r\n\r\nOur subject classes are designed to give a more thorough view of each individual subject -- something good for any student who wants to build a comprehensive base in each of the subject areas that are covered on the USAMO, though the subject classes are certainly also very helpful with AMC and AIME problems."
}
{
  "Tag": [
    "geometry",
    "parallelogram"
  ],
  "Problem": "attatchment contains all information.............",
  "Solution_1": "would u all mind shifting it to an appropriate  forum. i am curious to know the answer",
  "Solution_2": "What does RTP mean?"
}
{
  "Tag": [
    "Support"
  ],
  "Problem": "On my Get Problem page, the info isnt showing up, I have tried clicking refresh, but it didnt work. please reply someone.",
  "Solution_1": "Have you possibly ran out of [url=http://www.artofproblemsolving.com/Alcumus/Instructions.php#karmapoints]karma[/url]?",
  "Solution_2": "no i havent, i fixed it, sorry for your time. :blush:"
}
{
  "Tag": [
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "find the least value and the greatest value of the expression \r\n $P= \\frac{x+y}{1+z}+ \\frac{z+y}{1+x}+ \\frac{x+z}{1+y}$ where x,y,z are real numbers belonging to the segment $[1/2;1]$",
  "Solution_1": "we have $P+3= (x+y+z+1)[1/(x+1)+1/(y+1)+1/(z+1)]$=> $P \\geq 2$\r\nhence : $min P=2$      \r\n  i guess that $max P=3$",
  "Solution_2": "The given expression is convex in each of the variables, thus it reaches its maximum at the endpoint of the given interval...\r\n\r\nPierre."
}
{
  "Tag": [
    "function",
    "calculus",
    "integration",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "let I=[a,b] and f be riemann integrable function from I to R satisfying \r\n  /f(x)/<=integral /f/ on [a,x] for all x in I.show f(x)=0 for all x in I.\r\n  P.S. /X/=mod x. :huh:",
  "Solution_1": "Let $g: [a,b]\\to R$$g(x)=|f(x)|$. We have that g is positive and $g(x) \\leq \\int_{a}^{x}g(t)dt$ $\\forall$ $x\\in [a,b]$. We need to prove that g is constant 0.\r\n\r\nLet $M\\geq 0$ such that $g(x)\\leq M \\forall x\\in [a,b]$. Than $g(x)\\leq M(x-a)$. We substitute x with t and reintegrating we get that $g(x)\\leq M\\frac{(x-a)^{2}}{2}$. After we reiterate this n times we get:\r\n$0\\leqq(x)\\leq M\\frac{(x-a)^{n}}{n!}$ $\\forall x\\in [a,b]$. Taking the limit we get that g(x)=0 $\\forall x\\in [a,b]$.\r\n\r\nThis problem also appeared in a strange form with a different solution $\\int_{a}^{b}f^{n}(t)dt\\leq (\\int_{a}^{b}f(t)dt)^{n}$ but the conclusion is obvious once you know the above problem :) \r\n\r\nBTW arline, I like your problems :)  and I met most of them in the last weeks. I wonder if we were working the same book (mine is an olympiad romanian book for analysis and superior algebra...) What's your source?",
  "Solution_2": "I've also seen this in a book, but with a slightly different solution:\r\nFix $x \\in [a,b]$. Let $M = \\sup_{t \\in [a,x]}|f(t)|$. We have that $|f(t)| \\leq \\int_{a}^{x}|f(u)| \\, du \\leq (x-a) M$ for all $t \\in [a,x]$. Taking sups, we get that $M \\leq (x-a)M$. If $|x-a| < 1$, then this implies $M = 0$, so $f \\equiv 0$ on $\\left[ a, a+1 \\right)$. We can easily cover the whole interval.\r\n\r\n[quote]This problem also appeared in a strange form with a different solution $\\int_{a}^{b}f^{n}(t) \\, dt \\leq \\left( \\int_{a}^{b}f(t) \\, dt \\right)^{n}$ but the conclusion is obvious once you know the above problem :)[/quote]\r\nWhat do you mean?",
  "Solution_3": "Maybe not strange was the right word to use :oops: ...'trap problem' is more expressive :roll:"
}
{
  "Tag": [
    "inequalities",
    "function",
    "logarithms",
    "inequalities proposed"
  ],
  "Problem": "Given $ a, b, c$ and $ a \\plus{} b \\plus{} c \\equal{} 3$. Prove that: $ \\frac {1}{2 \\plus{} a^3b^3} \\plus{} \\frac {1}{2 \\plus{} b^3c^3} \\plus{} \\frac {1}{2 \\plus{} c^3a^3} \\geq\\ 1$\r\n\r\nWith this ineq, we can solve it easily by Am-Gm. But the following (of him) is harder and I still have no solution:  :oops: \r\n$ \\frac {1}{2 \\plus{} a^5b^5} \\plus{} \\frac {1}{2 \\plus{} b^5c^5} \\plus{} \\frac {1}{2 \\plus{} c^5a^5} \\geq\\ 1$\r\n(with the same condition)",
  "Solution_1": "[quote=\"nguoivn\"]\nWith this ineq, we can solve it easily by Am-Gm. But the following (of him) is harder and I still have no solution:  :oops: \n$ \\frac {1}{2 \\plus{} a^5b^5} \\plus{} \\frac {1}{2 \\plus{} b^5c^5} \\plus{} \\frac {1}{2 \\plus{} c^5a^5} \\geq\\ 1$\n(with the same condition)[/quote]\r\n\r\nMaybe ..... I have a very ugly solution for this  :blush: \r\n\r\nAssume : $ c \\equal{} min \\{a;b;c\\}$\r\n\r\nWith : $ a \\equal{} x \\plus{} c \\ ; \\ b \\equal{} y \\plus{} c \\ ; \\ x,y \\geq 0$ .\r\n\r\nThis inequality is equivalent to : \r\n\r\n$ \\minus{} 343151886824415\\,xy{c}^{28} \\plus{} 343151886824415\\,{x}^{2}{c}^{28} \\plus{} 343151886824415\\,{y}^{2}{c}^{28}$\r\n$ \\plus{} 1258223585022855\\,x{y}^{2}{c}^{27} \\plus{} 2363935220345970\\,{x}^{3}{c}^{27} \\plus{} 1258223585022855\\,{x}^{2}y{c}^{27} \\plus{} 2363935220345970\\,{y}^{3}{c}^{27}$\r\n$ \\plus{} 23677480190884635\\,{x}^{2}{y}^{2}{c} ^{26} \\plus{} 7206189623312715\\,{y}^{4}{c}^{26} \\plus{} 7206189623312715\\,{x}^{4}{c}^{ 26} \\plus{} 17500746228045165\\,{x}^{3}y{c}^{26} \\plus{} 17500746228045165\\,x{y}^{3}{c} ^{26}$\r\n$ \\plus{} 160099419197302065\\,{x}^{3}{y}^{2}{c}^{25} \\plus{} 12887259749628030\\,{y }^{5}{c}^{25} \\plus{} 61462315728995220\\,{x}^{4}y{c}^{25} \\plus{} 12887259749628030\\,{ x}^{5}{c}^{25} \\plus{} 160099419197302065\\,{x}^{2}{y}^{3}{c}^{25} \\plus{} 61462315728995220\\,x{y}^{4}{c}^{25}$\r\n$ \\plus{} 16267941301305600\\,{y}^{6}{c}^{24} \\plus{} 520637668287487425\\,{x}^{2}{y}^{4}{c}^{24} \\plus{} 16267941301305600\\,{x}^{6} {c}^{24} \\plus{} 849237373244718900\\,{x}^{3}{y}^{3}{c}^{24} \\plus{} 112286922966433575 \\,{x}^{5}y{c}^{24} \\plus{} 520637668287487425\\,{x}^{4}{y}^{2}{c}^{24} \\plus{} 112286922966433575\\,x{y}^{5}{c}^{24}$\r\n$ \\plus{} 20462019918048450\\,{x}^{7}{c}^{23 } \\plus{} 2583870161142136725\\,{x}^{3}{y}^{4}{c}^{23} \\plus{} 123598225902497625\\,{x}^ {6}y{c}^{23} \\plus{} 2583870161142136725\\,{x}^{4}{y}^{3}{c}^{23} \\plus{} 123598225902497625\\,x{y}^{6}{c}^{23} \\plus{} 1010963586572151525\\,{x}^{5}{y}^{ 2}{c}^{23} \\plus{} 20462019918048450\\,{y}^{7}{c}^{23} \\plus{} 1010963586572151525\\,{x} ^{2}{y}^{5}{c}^{23}$\r\n$ \\plus{} 100227181758694875\\,x{y}^{7}{c}^{22} \\plus{} 33771511824715575\\,{y}^{8}{c}^{22} \\plus{} 5052078366356891925\\,{x}^{3}{y}^{5} {c}^{22} \\plus{} 1300256160789639825\\,{x}^{2}{y}^{6}{c}^{22} \\plus{} 5052078366356891925\\,{x}^{5}{y}^{3}{c}^{22} \\plus{} 100227181758694875\\,{x}^{7 }y{c}^{22} \\plus{} 1300256160789639825\\,{x}^{6}{y}^{2}{c}^{22} \\plus{} 7610819359490631675\\,{x}^{4}{y}^{4}{c}^{22} \\plus{} 33771511824715575\\,{x}^{8} {c}^{22 }$\r\n$ \\plus{} 6871482379629081900\\,{x}^{3}{y}^{6}{c}^{21} \\plus{} 56307035256428700 \\,{y}^{9}{c}^{21} \\plus{} 14738655968645105250\\,{x}^{4}{y}^{5}{c}^{21} \\plus{} 118104971417942175\\,x{y}^{8}{c}^{21} \\plus{} 6871482379629081900\\,{x}^{6}{y}^{ 3}{c}^{21} \\plus{} 1211700379626022275\\,{x}^{2}{y}^{7}{c}^{21} \\plus{} 14738655968645105250\\,{x}^{5}{y}^{4}{c}^{21} \\plus{} 1211700379626022275\\,{x}^ {7}{y}^{2}{c}^{21} \\plus{} 56307035256428700\\,{x}^{9}{c}^{21} \\plus{} 118104971417942175\\,{x}^{8}y{c}^{21}$\r\n$ \\plus{} 6922333817672485950\\,{x}^{3}{y}^{ 7}{c}^{20} \\plus{} 74579755550855265\\,{x}^{10}{c}^{20} \\plus{} 6922333817672485950\\,{x }^{7}{y}^{3}{c}^{20} \\plus{} 218325092438225025\\,x{y}^{9}{c}^{20} \\plus{} 20205326773079281125\\,{x}^{6}{y}^{4}{c}^{20} \\plus{} 218325092438225025\\,{x}^{ 9}y{c}^{20} \\plus{} 20205326773079281125\\,{x}^{4}{y}^{6}{c}^{20} \\plus{} 984690578941006500\\,{x}^{2}{y}^{8}{c}^{20} \\plus{} 74579755550855265\\,{y}^{10} {c}^{20} \\plus{} 28195374156236728335\\,{x}^{5}{y}^{5}{c}^{20} \\plus{} 984690578941006500\\,{x}^{8}{y}^{2}{c}^{20}$\r\n$ \\plus{} 332431506674710425\\,{x}^{10 }y{c}^{19} \\plus{} 20703523584131542575\\,{x}^{4}{y}^{7}{c}^{19} \\plus{} 929452678951534425\\,{x}^{2}{y}^{9}{c}^{19} \\plus{} 332431506674710425\\,x{y}^{ 10}{c}^{19} \\plus{} 20703523584131542575\\,{x}^{7}{y}^{4}{c}^{19} \\plus{} 929452678951534425\\,{x}^{9}{y}^{2}{c}^{19} \\plus{} 5587211207586606525\\,{x}^{8 }{y}^{3}{c}^{19} \\plus{} 5587211207586606525\\,{x}^{3}{y}^{8}{c}^{19} \\plus{} 75157463424334950\\,{y}^{11}{c}^{19} \\plus{} 38253020811296949225\\,{x}^{6}{y}^{ 5}{c}^{19} \\plus{} 75157463424334950\\,{x}^{11}{c}^{19} \\plus{} 38253020811296949225\\,{ x}^{5}{y}^{6}{c}^{19}$\r\n$ \\plus{} 1009513729316449710\\,{x}^{10}{y}^{2}{c}^{18} \\plus{} 58564007553966660\\,{y}^{12}{c}^{18} \\plus{} 38837506435766674065\\,{x}^{7}{y}^{ 5}{c}^{18} \\plus{} 4071659563211697225\\,{x}^{9}{y}^{3}{c}^{18} \\plus{} 58564007553966660\\,{x}^{12}{c}^{18} \\plus{} 16701086221823724750\\,{x}^{8}{y}^{ 4}{c}^{18} \\plus{} 362611857207382065\\,x{y}^{11}{c}^{18} \\plus{} 50955813923280508710 \\,{x}^{6}{y}^{6}{c}^{18} \\plus{} 4071659563211697225\\,{x}^{3}{y}^{9}{c}^{18} \\plus{} 38837506435766674065\\,{x}^{5}{y}^{7}{c}^{18} \\plus{} 16701086221823724750\\,{x} ^{4}{y}^{8}{c}^{18} \\plus{} 362611857207382065\\,{x}^{11}y{c}^{18} \\plus{} 1009513729316449710\\,{x}^{2}{y}^{10}{c}^{18}$\r\n$ \\plus{} 977917657147362045\\,{x}^{ 11}{y}^{2}{c}^{17} \\plus{} 2984439815827840980\\,{x}^{10}{y}^{3}{c}^{17} \\plus{} 2984439815827840980\\,{x}^{3}{y}^{10}{c}^{17} \\plus{} 36130164566694030\\,{y}^{ 13}{c}^{17} \\plus{} 30825593044084271745\\,{x}^{8}{y}^{5}{c}^{17} \\plus{} 11308185750809556000\\,{x}^{4}{y}^{9}{c}^{17} \\plus{} 292229998302188745\\,{x}^{ 12}y{c}^{17} \\plus{} 30825593044084271745\\,{x}^{5}{y}^{8}{c}^{17} \\plus{} 11308185750809556000\\,{x}^{9}{y}^{4}{c}^{17} \\plus{} 50567705251517488230\\,{x} ^{6}{y}^{7}{c}^{17} \\plus{} 36130164566694030\\,{x}^{13}{c}^{17} \\plus{} 50567705251517488230\\,{x}^{7}{y}^{6}{c}^{17} \\plus{} 292229998302188745\\,x{y}^ {12}{c}^{17} \\plus{} 977917657147362045\\,{x}^{2}{y}^{11}{c}^{17}$\r\n$ \\plus{} 2160360452109822660\\,{x}^{3}{y}^{11}{c}^{16} \\plus{} 2160360452109822660\\,{x}^ {11}{y}^{3}{c}^{16} \\plus{} 38992364468350626915\\,{x}^{8}{y}^{6}{c}^{16} \\plus{} 748135554864262515\\,{x}^{12}{y}^{2}{c}^{16} \\plus{} 180949529841796485\\,x{y}^{ 13}{c}^{16} \\plus{} 6906323613662794935\\,{x}^{4}{y}^{10}{c}^{16} \\plus{} 6906323613662794935\\,{x}^{10}{y}^{4}{c}^{16} \\plus{} 18022409853586110\\,{y}^{ 14}{c}^{16} \\plus{} 18022409853586110\\,{x}^{14}{c}^{16} \\plus{} 748135554864262515\\,{x }^{2}{y}^{12}{c}^{16} \\plus{} 19994508314853588315\\,{x}^{5}{y}^{9}{c}^{16} \\plus{} 48631289652679821660\\,{x}^{7}{y}^{7}{c}^{16} \\plus{} 180949529841796485\\,{x}^{ 13}y{c}^{16} \\plus{} 38992364468350626915\\,{x}^{6}{y}^{8}{c}^{16} \\plus{} 19994508314853588315\\,{x}^{9}{y}^{5}{c}^{16}$\r\n$ \\plus{} 88711037987481855\\,{x}^{ 14}y{c}^{15} \\plus{} 36070087439992712715\\,{x}^{8}{y}^{7}{c}^{15} \\plus{} 88711037987481855\\,x{y}^{14}{c}^{15} \\plus{} 1393142007744442485\\,{x}^{3}{y}^{ 12}{c}^{15} \\plus{} 3960181901596608855\\,{x}^{4}{y}^{11}{c}^{15} \\plus{} 36070087439992712715\\,{x}^{7}{y}^{8}{c}^{15} \\plus{} 445465963285976835\\,{x}^{ 2}{y}^{13}{c}^{15} \\plus{} 24236016130783711395\\,{x}^{6}{y}^{9}{c}^{15} \\plus{} 11119352295374912205\\,{x}^{5}{y}^{10}{c}^{15} \\plus{} 3960181901596608855\\,{x} ^{11}{y}^{4}{c}^{15} \\plus{} 1393142007744442485\\,{x}^{12}{y}^{3}{c}^{15} \\plus{} 7395765445494270\\,{x}^{15}{c}^{15} \\plus{} 445465963285976835\\,{x}^{13}{y}^{2} {c}^{15} \\plus{} 24236016130783711395\\,{x}^{9}{y}^{6}{c}^{15} \\plus{} 7395765445494270 \\,{y}^{15}{c}^{15} \\plus{} 11119352295374912205\\,{x}^{10}{y}^{5}{c}^{15}$\r\n$ \\plus{} 2084305487797867050\\,{x}^{12}{y}^{4}{c}^{14} \\plus{} 209424571608036825\\,{x}^{ 14}{y}^{2}{c}^{14} \\plus{} 21349556612200860075\\,{x}^{7}{y}^{9}{c}^{14} \\plus{} 744857579775620475\\,{x}^{3}{y}^{13}{c}^{14} \\plus{} 12585125502138452670\\,{x}^ {10}{y}^{6}{c}^{14} \\plus{} 2084305487797867050\\,{x}^{4}{y}^{12}{c}^{14} \\plus{} 744857579775620475\\,{x}^{13}{y}^{3}{c}^{14} \\plus{} 35203478647936185\\,x{y}^{ 15}{c}^{14} \\plus{} 35203478647936185\\,{x}^{15}y{c}^{14} \\plus{} 21349556612200860075 \\,{x}^{9}{y}^{7}{c}^{14} \\plus{} 2533095274158855\\,{y}^{16}{c}^{14} \\plus{} 12585125502138452670\\,{x}^{6}{y}^{10}{c}^{14} \\plus{} 2533095274158855\\,{x}^{ 16}{c}^{14} \\plus{} 209424571608036825\\,{x}^{2}{y}^{14}{c}^{14} \\plus{} 25510966047864850275\\,{x}^{8}{y}^{8}{c}^{14} \\plus{} 5511886560598653405\\,{x}^ {5}{y}^{11}{c}^{14} \\plus{} 5511886560598653405\\,{x}^{11}{y}^{5}{c}^{14}$\r\n$ \\plus{} 322423177403419200\\,{x}^{14}{y}^{3}{c}^{13} \\plus{} 733199301973440\\,{y}^{17}{ c}^{13} \\plus{} 2456477683392122055\\,{x}^{12}{y}^{5}{c}^{13} \\plus{} 11499472852337745 \\,x{y}^{16}{c}^{13} \\plus{} 2456477683392122055\\,{x}^{5}{y}^{12}{c}^{13} \\plus{} 955075836063775050\\,{x}^{13}{y}^{4}{c}^{13} \\plus{} 79098393960389025\\,{x}^{2} {y}^{15}{c}^{13} \\plus{} 10388985093352033110\\,{x}^{7}{y}^{10}{c}^{13} \\plus{} 5632415391716311230\\,{x}^{6}{y}^{11}{c}^{13} \\plus{} 955075836063775050\\,{x}^{ 4}{y}^{13}{c}^{13} \\plus{} 10388985093352033110\\,{x}^{10}{y}^{7}{c}^{13} \\plus{} 14264412913360975275\\,{x}^{9}{y}^{8}{c}^{13} \\plus{} 733199301973440\\,{x}^{17} {c}^{13} \\plus{} 5632415391716311230\\,{x}^{11}{y}^{6}{c}^{13} \\plus{} 322423177403419200\\,{x}^{3}{y}^{14}{c}^{13} \\plus{} 14264412913360975275\\,{x}^ {8}{y}^{9}{c}^{13} \\plus{} 11499472852337745\\,{x}^{16}y{c}^{13} \\plus{} 79098393960389025\\,{x}^{15}{y}^{2}{c}^{13}$\r\n$ \\plus{} 2206418360572908315\\,{x}^{ 12}{y}^{6}{c}^{12} \\plus{} 24362825601034935\\,{x}^{16}{y}^{2}{c}^{12} \\plus{} 6477654862591944300\\,{x}^{10}{y}^{8}{c}^{12} \\plus{} 2206418360572908315\\,{x}^ {6}{y}^{12}{c}^{12} \\plus{} 112916040343061310\\,{x}^{3}{y}^{15}{c}^{12} \\plus{} 7466933476884696750\\,{x}^{9}{y}^{9}{c}^{12} \\plus{} 3134553601994010\\,x{y}^{17 }{c}^{12} \\plus{} 6477654862591944300\\,{x}^{8}{y}^{10}{c}^{12} \\plus{} 366353221781733975\\,{x}^{14}{y}^{4}{c}^{12} \\plus{} 4262032766997513990\\,{x}^{ 7}{y}^{11}{c}^{12} \\plus{} 366353221781733975\\,{x}^{4}{y}^{14}{c}^{12} \\plus{} 181249095648150\\,{y}^{18}{c}^{12} \\plus{} 3134553601994010\\,{x}^{17}y{c}^{12} \\plus{} 4262032766997513990\\,{x}^{11}{y}^{7}{c}^{12} \\plus{} 181249095648150\\,{x}^{18} {c}^{12} \\plus{} 24362825601034935\\,{x}^{2}{y}^{16}{c}^{12} \\plus{} 962390964597878745 \\,{x}^{13}{y}^{5}{c}^{12} \\plus{} 962390964597878745\\,{x}^{5}{y}^{13}{c}^{12} \\plus{} 112916040343061310\\,{x}^{15}{y}^{3}{c}^{12}$\r\n$ \\plus{} 6196930766887320\\,{x}^{17} {y}^{2}{c}^{11} \\plus{} 32196892281454305\\,{x}^{3}{y}^{16}{c}^{11} \\plus{} 38568777876360\\,{y}^{19}{c}^{11} \\plus{} 6196930766887320\\,{x}^{2}{y}^{17}{c}^ {11} \\plus{} 1496578870705849605\\,{x}^{12}{y}^{7}{c}^{11} \\plus{} 3140506203143784840 \\,{x}^{9}{y}^{10}{c}^{11} \\plus{} 721091184063480\\,x{y}^{18}{c}^{11} \\plus{} 115583970861408465\\,{x}^{4}{y}^{15}{c}^{11} \\plus{} 32196892281454305\\,{x}^{16 }{y}^{3}{c}^{11} \\plus{} 1496578870705849605\\,{x}^{7}{y}^{12}{c}^{11} \\plus{} 38568777876360\\,{x}^{19}{c}^{11} \\plus{} 2440286663280994845\\,{x}^{8}{y}^{11}{ c}^{11} \\plus{} 751829705460213795\\,{x}^{6}{y}^{13}{c}^{11} \\plus{} 721091184063480\\,{ x}^{18}y{c}^{11} \\plus{} 321388973115663375\\,{x}^{5}{y}^{14}{c}^{11} \\plus{} 2440286663280994845\\,{x}^{11}{y}^{8}{c}^{11} \\plus{} 321388973115663375\\,{x}^{ 14}{y}^{5}{c}^{11} \\plus{} 751829705460213795\\,{x}^{13}{y}^{6}{c}^{11} \\plus{} 3140506203143784840\\,{x}^{10}{y}^{9}{c}^{11} \\plus{} 115583970861408465\\,{x}^{ 15}{y}^{4}{c}^{11}$\r\n$ \\plus{} 772772734181827170\\,{x}^{12}{y}^{8}{c}^{10} \\plus{} 449889121836280395\\,{x}^{7}{y}^{13}{c}^{10} \\plus{} 29842861410108315\\,{x}^{4} {y}^{16}{c}^{10} \\plus{} 1081313759617721910\\,{x}^{9}{y}^{11}{c}^{10} \\plus{} 7534683942204330\\,{x}^{17}{y}^{3}{c}^{10} \\plus{} 141198022534590\\,{x}^{19}y{c }^{10} \\plus{} 449889121836280395\\,{x}^{13}{y}^{7}{c}^{10} \\plus{} 1211445894374410257 \\,{x}^{10}{y}^{10}{c}^{10} \\plus{} 7093025578539\\,{x}^{20}{c}^{10} \\plus{} 7534683942204330\\,{x}^{3}{y}^{17}{c}^{10} \\plus{} 1315579009511205\\,{x}^{2}{y} ^{18}{c}^{10} \\plus{} 218583663720200910\\,{x}^{6}{y}^{14}{c}^{10} \\plus{} 89469060681737367\\,{x}^{15}{y}^{5}{c}^{10} \\plus{} 1081313759617721910\\,{x}^{ 11}{y}^{9}{c}^{10} \\plus{} 1315579009511205\\,{x}^{18}{y}^{2}{c}^{10} \\plus{} 772772734181827170\\,{x}^{8}{y}^{12}{c}^{10} \\plus{} 7093025578539\\,{y}^{20}{c} ^{10} \\plus{} 141198022534590\\,x{y}^{19}{c}^{10} \\plus{} 29842861410108315\\,{x}^{16}{y }^{4}{c}^{10} \\plus{} 218583663720200910\\,{x}^{14}{y}^{6}{c}^{10} \\plus{} 89469060681737367\\,{x}^{5}{y}^{15}{c}^{10}$\r\n$ \\plus{} 308183466115653975\\,{x}^{9} {y}^{12}{c}^{9} \\plus{} 1126430533800\\,{y}^{21}{c}^{9} \\plus{} 20503667384124705\\,{x}^ {5}{y}^{16}{c}^{9} \\plus{} 1458316098172575\\,{x}^{3}{y}^{18}{c}^{9} \\plus{} 6310720192829400\\,{x}^{17}{y}^{4}{c}^{9} \\plus{} 308183466115653975\\,{x}^{12}{ y}^{9}{c}^{9} \\plus{} 114414758085012750\\,{x}^{7}{y}^{14}{c}^{9} \\plus{} 1458316098172575\\,{x}^{18}{y}^{3}{c}^{9} \\plus{} 23637607287795\\,x{y}^{20}{c}^ {9} \\plus{} 206128996224470325\\,{x}^{13}{y}^{8}{c}^{9} \\plus{} 20503667384124705\\,{x}^ {16}{y}^{5}{c}^{9} \\plus{} 1126430533800\\,{x}^{21}{c}^{9} \\plus{} 234981359117550\\,{x} ^{2}{y}^{19}{c}^{9} \\plus{} 53176745518951770\\,{x}^{6}{y}^{15}{c}^{9} \\plus{} 378024510732571755\\,{x}^{10}{y}^{11}{c}^{9} \\plus{} 378024510732571755\\,{x}^{ 11}{y}^{10}{c}^{9} \\plus{} 206128996224470325\\,{x}^{8}{y}^{13}{c}^{9} \\plus{} 6310720192829400\\,{x}^{4}{y}^{17}{c}^{9} \\plus{} 234981359117550\\,{x}^{19}{y}^ {2}{c}^{9} \\plus{} 53176745518951770\\,{x}^{15}{y}^{6}{c}^{9} \\plus{} 23637607287795\\,{ x}^{20}y{c}^{9} \\plus{} 114414758085012750\\,{x}^{14}{y}^{7}{c}^{9}$\r\n$ \\plus{} 3845533111294590\\,{x}^{17}{y}^{5}{c}^{8} \\plus{} 153604163700\\,{x}^{22}{c}^{8} \\plus{} 1096068860697600\\,{x}^{18}{y}^{4}{c}^{8} \\plus{} 153604163700\\,{y}^{22}{c}^{8 } \\plus{} 10663945457377545\\,{x}^{16}{y}^{6}{c}^{8} \\plus{} 96280138209164220\\,{x}^{12 }{y}^{10}{c}^{8} \\plus{} 3379291601400\\,x{y}^{21}{c}^{8} \\plus{} 3845533111294590\\,{x} ^{5}{y}^{17}{c}^{8} \\plus{} 45931518331885125\\,{x}^{8}{y}^{14}{c}^{8} \\plus{} 24232232576657520\\,{x}^{7}{y}^{15}{c}^{8} \\plus{} 24232232576657520\\,{x}^{15}{ y}^{7}{c}^{8} \\plus{} 234807019897500\\,{x}^{3}{y}^{19}{c}^{8} \\plus{} 3379291601400\\,{ x}^{21}y{c}^{8} \\plus{} 10663945457377545\\,{x}^{6}{y}^{16}{c}^{8} \\plus{} 45931518331885125\\,{x}^{14}{y}^{8}{c}^{8} \\plus{} 234807019897500\\,{x}^{19}{y} ^{3}{c}^{8} \\plus{} 96280138209164220\\,{x}^{10}{y}^{12}{c}^{8} \\plus{} 72834792863369850\\,{x}^{13}{y}^{9}{c}^{8} \\plus{} 35447693970690\\,{x}^{2}{y}^{ 20}{c}^{8} \\plus{} 105713054588876040\\,{x}^{11}{y}^{11}{c}^{8} \\plus{} 1096068860697600\\,{x}^{4}{y}^{18}{c}^{8} \\plus{} 72834792863369850\\,{x}^{9}{y} ^{13}{c}^{8} \\plus{} 35447693970690\\,{x}^{20}{y}^{2}{c}^{8}$\r\n$ \\plus{} 14169936212586000 \\,{x}^{9}{y}^{14}{c}^{7} \\plus{} 588626351558550\\,{x}^{18}{y}^{5}{c}^{7} \\plus{} 23797038511922220\\,{x}^{12}{y}^{11}{c}^{7} \\plus{} 17809178400\\,{y}^{23}{c}^{7 } \\plus{} 1744644537673560\\,{x}^{17}{y}^{6}{c}^{7} \\plus{} 4505722135200\\,{x}^{21}{y}^ {2}{c}^{7} \\plus{} 1744644537673560\\,{x}^{6}{y}^{17}{c}^{7} \\plus{} 588626351558550\\,{ x}^{5}{y}^{18}{c}^{7} \\plus{} 4209610166857935\\,{x}^{16}{y}^{7}{c}^{7} \\plus{} 4209610166857935\\,{x}^{7}{y}^{16}{c}^{7} \\plus{} 31505187102390\\,{x}^{3}{y}^{ 20}{c}^{7} \\plus{} 8434858561754355\\,{x}^{8}{y}^{15}{c}^{7} \\plus{} 17809178400\\,{x}^{ 23}{c}^{7} \\plus{} 31505187102390\\,{x}^{20}{y}^{3}{c}^{7} \\plus{} 156654239411700\\,{x} ^{19}{y}^{4}{c}^{7} \\plus{} 409611103200\\,{x}^{22}y{c}^{7} \\plus{} 23797038511922220\\, {x}^{11}{y}^{12}{c}^{7} \\plus{} 4505722135200\\,{x}^{2}{y}^{21}{c}^{7} \\plus{} 20019537297068490\\,{x}^{10}{y}^{13}{c}^{7} \\plus{} 20019537297068490\\,{x}^{13} {y}^{10}{c}^{7} \\plus{} 8434858561754355\\,{x}^{15}{y}^{8}{c}^{7} \\plus{} 156654239411700\\,{x}^{4}{y}^{19}{c}^{7} \\plus{} 14169936212586000\\,{x}^{14}{y} ^{9}{c}^{7} \\plus{} 409611103200\\,x{y}^{22}{c}^{7}$\r\n$ \\plus{} 41554749600\\,{x}^{23}y{c}^{ 6} \\plus{} 4301057448335610\\,{x}^{13}{y}^{11}{c}^{6} \\plus{} 230971324829805\\,{x}^{18} {y}^{6}{c}^{6} \\plus{} 1731447900\\,{x}^{24}{c}^{6} \\plus{} 4301057448335610\\,{x}^{11}{ y}^{13}{c}^{6} \\plus{} 230971324829805\\,{x}^{6}{y}^{18}{c}^{6} \\plus{} 1256797756868130\\,{x}^{8}{y}^{16}{c}^{6} \\plus{} 3504450549600\\,{x}^{3}{y}^{21 }{c}^{6} \\plus{} 3369714001320915\\,{x}^{10}{y}^{14}{c}^{6} \\plus{} 1256797756868130\\,{ x}^{16}{y}^{8}{c}^{6} \\plus{} 3504450549600\\,{x}^{21}{y}^{3}{c}^{6} \\plus{} 591834193207470\\,{x}^{7}{y}^{17}{c}^{6} \\plus{} 73279650945510\\,{x}^{19}{y}^{5 }{c}^{6} \\plus{} 591834193207470\\,{x}^{17}{y}^{7}{c}^{6} \\plus{} 477879620400\\,{x}^{2} {y}^{22}{c}^{6} \\plus{} 477879620400\\,{x}^{22}{y}^{2}{c}^{6} \\plus{} 4664636437623390 \\,{x}^{12}{y}^{12}{c}^{6} \\plus{} 2238944546574450\\,{x}^{9}{y}^{15}{c}^{6} \\plus{} 2238944546574450\\,{x}^{15}{y}^{9}{c}^{6} \\plus{} 18380931463395\\,{x}^{20}{y}^{ 4}{c}^{6} \\plus{} 3369714001320915\\,{x}^{14}{y}^{10}{c}^{6} \\plus{} 18380931463395\\,{x }^{4}{y}^{20}{c}^{6} \\plus{} 1731447900\\,{y}^{24}{c}^{6} \\plus{} 41554749600\\,x{y}^{23 }{c}^{6} \\plus{} 73279650945510\\,{x}^{5}{y}^{19}{c}^{6}$\r\n$ \\plus{} 66427655144355\\,{x}^{7 }{y}^{18}{c}^{5} \\plus{} 41554749600\\,{x}^{23}{y}^{2}{c}^{5} \\plus{} 616653376988580\\, {x}^{11}{y}^{14}{c}^{5} \\plus{} 282252157155990\\,{x}^{16}{y}^{9}{c}^{5} \\plus{} 24496286003190\\,{x}^{19}{y}^{6}{c}^{5} \\plus{} 66427655144355\\,{x}^{18}{y}^{7} {c}^{5} \\plus{} 451892854554867\\,{x}^{10}{y}^{15}{c}^{5} \\plus{} 1752225274800\\,{x}^{ 21}{y}^{4}{c}^{5} \\plus{} 3462895800\\,{x}^{24}y{c}^{5} \\plus{} 616653376988580\\,{x}^{ 14}{y}^{11}{c}^{5} \\plus{} 318586413600\\,{x}^{22}{y}^{3}{c}^{5} \\plus{} 138515832\\,{x} ^{25}{c}^{5} \\plus{} 1752225274800\\,{x}^{4}{y}^{21}{c}^{5} \\plus{} 451892854554867\\,{x }^{15}{y}^{10}{c}^{5} \\plus{} 149396846867280\\,{x}^{17}{y}^{8}{c}^{5} \\plus{} 719748561723435\\,{x}^{12}{y}^{13}{c}^{5} \\plus{} 7355859369759\\,{x}^{20}{y}^{5 }{c}^{5} \\plus{} 719748561723435\\,{x}^{13}{y}^{12}{c}^{5} \\plus{} 3462895800\\,x{y}^{24 }{c}^{5} \\plus{} 149396846867280\\,{x}^{8}{y}^{17}{c}^{5} \\plus{} 318586413600\\,{x}^{3} {y}^{22}{c}^{5} \\plus{} 282252157155990\\,{x}^{9}{y}^{16}{c}^{5} \\plus{} 24496286003190 \\,{x}^{6}{y}^{19}{c}^{5} \\plus{} 138515832\\,{y}^{25}{c}^{5} \\plus{} 7355859369759\\,{x} ^{5}{y}^{20}{c}^{5} \\plus{} 41554749600\\,{x}^{2}{y}^{23}{c}^{5}$\r\n$ \\plus{} 8879220\\,{x}^{ 26}{c}^{4} \\plus{} 132744339000\\,{x}^{22}{y}^{4}{c}^{4} \\plus{} 68602274395200\\,{x}^{ 15}{y}^{11}{c}^{4} \\plus{} 13871783425500\\,{x}^{8}{y}^{18}{c}^{4} \\plus{} 2885746500\\, {x}^{2}{y}^{24}{c}^{4} \\plus{} 13871783425500\\,{x}^{18}{y}^{8}{c}^{4} \\plus{} 230859720\\,x{y}^{25}{c}^{4} \\plus{} 2885746500\\,{x}^{24}{y}^{2}{c}^{4} \\plus{} 47164063646700\\,{x}^{10}{y}^{16}{c}^{4} \\plus{} 5840750916000\\,{x}^{7}{y}^{19} {c}^{4} \\plus{} 584075091600\\,{x}^{21}{y}^{5}{c}^{4} \\plus{} 132744339000\\,{x}^{4}{y}^ {22}{c}^{4} \\plus{} 23085972000\\,{x}^{3}{y}^{23}{c}^{4} \\plus{} 8879220\\,{y}^{26}{c}^{ 4} \\plus{} 2044262820600\\,{x}^{20}{y}^{6}{c}^{4} \\plus{} 27743566851000\\,{x}^{17}{y}^{ 9}{c}^{4} \\plus{} 2044262820600\\,{x}^{6}{y}^{20}{c}^{4} \\plus{} 47164063646700\\,{x}^{ 16}{y}^{10}{c}^{4} \\plus{} 584075091600\\,{x}^{5}{y}^{21}{c}^{4} \\plus{} 92349215532000 \\,{x}^{13}{y}^{13}{c}^{4} \\plus{} 27743566851000\\,{x}^{9}{y}^{17}{c}^{4} \\plus{} 23085972000\\,{x}^{23}{y}^{3}{c}^{4} \\plus{} 85752842994000\\,{x}^{14}{y}^{12}{c }^{4} \\plus{} 68602274395200\\,{x}^{11}{y}^{15}{c}^{4} \\plus{} 85752842994000\\,{x}^{12} {y}^{14}{c}^{4} \\plus{} 230859720\\,{x}^{25}y{c}^{4} \\plus{} 5840750916000\\,{x}^{19}{y} ^{7}{c}^{4}$\r\n$ \\plus{} 438480\\,{y}^{27}{c}^{3} \\plus{} 973458486000\\,{x}^{19}{y}^{8}{c}^{ 3} \\plus{} 11838960\\,{x}^{26}y{c}^{3} \\plus{} 5716856199600\\,{x}^{16}{y}^{11}{c}^{3} \\plus{} 389383394400\\,{x}^{7}{y}^{20}{c}^{3} \\plus{} 35398490400\\,{x}^{5}{y}^{22}{c}^{ 3} \\plus{} 438480\\,{x}^{27}{c}^{3} \\plus{} 7695324000\\,{x}^{4}{y}^{23}{c}^{3} \\plus{} 1282554000\\,{x}^{24}{y}^{3}{c}^{3} \\plus{} 1282554000\\,{x}^{3}{y}^{24}{c}^{3} \\plus{} 153906480\\,{x}^{2}{y}^{25}{c}^{3} \\plus{} 2055079026000\\,{x}^{18}{y}^{9}{c}^{3 } \\plus{} 7695324000\\,{x}^{23}{y}^{4}{c}^{3} \\plus{} 973458486000\\,{x}^{8}{y}^{19}{c}^ {3} \\plus{} 35398490400\\,{x}^{22}{y}^{5}{c}^{3} \\plus{} 389383394400\\,{x}^{20}{y}^{7}{ c}^{3} \\plus{} 8795163384000\\,{x}^{13}{y}^{14}{c}^{3} \\plus{} 5716856199600\\,{x}^{11}{ y}^{16}{c}^{3} \\plus{} 7622474932800\\,{x}^{15}{y}^{12}{c}^{3} \\plus{} 11838960\\,x{y}^{ 26}{c}^{3} \\plus{} 7622474932800\\,{x}^{12}{y}^{15}{c}^{3} \\plus{} 8795163384000\\,{x}^{ 14}{y}^{13}{c}^{3} \\plus{} 129794464800\\,{x}^{21}{y}^{6}{c}^{3} \\plus{} 2055079026000 \\,{x}^{9}{y}^{18}{c}^{3} \\plus{} 153906480\\,{x}^{25}{y}^{2}{c}^{3} \\plus{} 3699142246800\\,{x}^{10}{y}^{17}{c}^{3} \\plus{} 129794464800\\,{x}^{6}{y}^{21}{c }^{3} \\plus{} 3699142246800\\,{x}^{17}{y}^{10}{c}^{3}$\r\n$ \\plus{} 476404683300\\,{x}^{16}{y} ^{12}{c}^{2} \\plus{} 438480\\,{x}^{27}y{c}^{2} \\plus{} 15660\\,{y}^{28}{c}^{2} \\plus{} 1539064800\\,{x}^{23}{y}^{5}{c}^{2} \\plus{} 320638500\\,{x}^{4}{y}^{24}{c}^{2} \\plus{} 108162054000\\,{x}^{9}{y}^{19}{c}^{2} \\plus{} 1539064800\\,{x}^{5}{y}^{23}{c}^{2 } \\plus{} 320638500\\,{x}^{24}{y}^{4}{c}^{2} \\plus{} 15660\\,{x}^{28}{c}^{2} \\plus{} 108162054000\\,{x}^{19}{y}^{9}{c}^{2} \\plus{} 5919480\\,{x}^{2}{y}^{26}{c}^{2} \\plus{} 48672924300\\,{x}^{20}{y}^{8}{c}^{2} \\plus{} 51302160\\,{x}^{3}{y}^{25}{c}^{2} \\plus{} 438480\\,x{y}^{27}{c}^{2} \\plus{} 336285658800\\,{x}^{11}{y}^{17}{c}^{2} \\plus{} 5919480 \\,{x}^{26}{y}^{2}{c}^{2} \\plus{} 5899748400\\,{x}^{6}{y}^{22}{c}^{2} \\plus{} 586344225600\\,{x}^{15}{y}^{13}{c}^{2} \\plus{} 5899748400\\,{x}^{22}{y}^{6}{c}^{ 2} \\plus{} 628225956000\\,{x}^{14}{y}^{14}{c}^{2} \\plus{} 205507902600\\,{x}^{18}{y}^{10 }{c}^{2} \\plus{} 205507902600\\,{x}^{10}{y}^{18}{c}^{2} \\plus{} 586344225600\\,{x}^{13}{ y}^{15}{c}^{2} \\plus{} 18542066400\\,{x}^{21}{y}^{7}{c}^{2} \\plus{} 48672924300\\,{x}^{8 }{y}^{20}{c}^{2} \\plus{} 336285658800\\,{x}^{17}{y}^{11}{c}^{2} \\plus{} 476404683300\\,{ x}^{12}{y}^{16}{c}^{2} \\plus{} 18542066400\\,{x}^{7}{y}^{21}{c}^{2} \\plus{} 51302160\\,{ x}^{25}{y}^{3}{c}^{2}$\r\n$ \\plus{} 1315440\\,{x}^{3}{y}^{26}c \\plus{} 360\\,{y}^{29}c \\plus{} 171007200\\,{x}^{23}{y}^{6}c \\plus{} 7210803600\\,{x}^{19}{y}^{10}c \\plus{} 1545172200\\, {x}^{8}{y}^{21}c \\plus{} 3605401800\\,{x}^{9}{y}^{20}c \\plus{} 27921153600\\,{x}^{14}{y} ^{15}c \\plus{} 1545172200\\,{x}^{21}{y}^{8}c \\plus{} 7210803600\\,{x}^{10}{y}^{19}c \\plus{} 24431009400\\,{x}^{13}{y}^{16}c \\plus{} 146160\\,{x}^{27}{y}^{2}c \\plus{} 42751800\\,{x}^ {5}{y}^{24}c \\plus{} 561880800\\,{x}^{22}{y}^{7}c \\plus{} 146160\\,{x}^{2}{y}^{27}c \\plus{} 10440\\,{x}^{28}yc \\plus{} 3605401800\\,{x}^{20}{y}^{9}c \\plus{} 27921153600\\,{x}^{15}{y }^{14}c \\plus{} 171007200\\,{x}^{6}{y}^{23}c \\plus{} 12455024400\\,{x}^{11}{y}^{18}c \\plus{} 12455024400\\,{x}^{18}{y}^{11}c \\plus{} 1315440\\,{x}^{26}{y}^{3}c \\plus{} 18682536600\\, {x}^{12}{y}^{17}c \\plus{} 8550360\\,{x}^{4}{y}^{25}c \\plus{} 360\\,{x}^{29}c \\plus{} 8550360\\,{x }^{25}{y}^{4}c \\plus{} 10440\\,x{y}^{28}c \\plus{} 18682536600\\,{x}^{17}{y}^{12}c \\plus{} 42751800\\,{x}^{24}{y}^{5}c \\plus{} 24431009400\\,{x}^{16}{y}^{13}c \\plus{} 561880800\\,{ x}^{7}{y}^{22}c$\r\n$ \\plus{} 109620\\,{x}^{4}{y}^{26} \\plus{} 1740\\,{x}^{2}{y}^{28} \\plus{} 16240\\,{ x}^{27}{y}^{3} \\plus{} 4\\,{y}^{30} \\plus{} 109620\\,{x}^{26}{y}^{4} \\plus{} 16240\\,{x}^{3}{y}^{ 27} \\plus{} 570024\\,{x}^{25}{y}^{5} \\plus{} 120\\,x{y}^{29} \\plus{} 2375100\\,{x}^{24}{y}^{6} \\plus{} 4 \\,{x}^{30} \\plus{} 8143200\\,{x}^{23}{y}^{7} \\plus{} 120\\,{x}^{29}y \\plus{} 23411700\\,{x}^{22}{ y}^{8} \\plus{} 1740\\,{x}^{28}{y}^{2} \\plus{} 57228600\\,{x}^{21}{y}^{9} \\plus{} 345972900\\,{x}^ {12}{y}^{18} \\plus{} 120180060\\,{x}^{20}{y}^{10} \\plus{} 218509200\\,{x}^{11}{y}^{19} \\plus{} 218509200\\,{x}^{19}{y}^{11} \\plus{} 120180060\\,{x}^{10}{y}^{20} \\plus{} 345972900\\,{x} ^{18}{y}^{12} \\plus{} 57228600\\,{x}^{9}{y}^{21} \\plus{} 479039400\\,{x}^{17}{y}^{13} \\plus{} 23411700\\,{x}^{8}{y}^{22} \\plus{} 581690700\\,{x}^{16}{y}^{14} \\plus{} 8143200\\,{x}^{7} {y}^{23} \\plus{} 620470080\\,{x}^{15}{y}^{15} \\plus{} 2375100\\,{x}^{6}{y}^{24} \\plus{} 581690700\\,{x}^{14}{y}^{16} \\plus{} 570024\\,{x}^{5}{y}^{25} \\plus{} 479039400\\,{x}^{13 }{y}^{17} \\geq 0$\r\n\r\nDone ! \r\n\r\nPS : It's just for fun  :lol:\r\n\r\nPS : Do you believe i do it by hand ???  :rotfl:",
  "Solution_2": "[quote=\"Materazzi\"][quote=\"nguoivn\"]\nWith this ineq, we can solve it easily by Am-Gm. But the following (of him) is harder and I still have no solution:  :oops: \n$ \\frac {1}{2 \\plus{} a^5b^5} \\plus{} \\frac {1}{2 \\plus{} b^5c^5} \\plus{} \\frac {1}{2 \\plus{} c^5a^5} \\geq\\ 1$\n(with the same condition)[/quote]\n\nMaybe ..... I have a very ugly solution for this  :blush: \n\nAssume : $ c \\equal{} min \\{a;b;c\\}$\n\nWith : $ a \\equal{} x \\plus{} c \\ ; \\ b \\equal{} y \\plus{} c \\ ; \\ x,y \\geq 0$ .\n\nThis inequality is equivalent to : \n\n$ \\minus{} 343151886824415\\,xy{c}^{28} \\plus{} 343151886824415\\,{x}^{2}{c}^{28} \\plus{} 343151886824415\\,{y}^{2}{c}^{28}$\n$ \\plus{} 1258223585022855\\,x{y}^{2}{c}^{27} \\plus{} 2363935220345970\\,{x}^{3}{c}^{27} \\plus{} 1258223585022855\\,{x}^{2}y{c}^{27} \\plus{} 2363935220345970\\,{y}^{3}{c}^{27}$\n$ \\plus{} 23677480190884635\\,{x}^{2}{y}^{2}{c} ^{26} \\plus{} 7206189623312715\\,{y}^{4}{c}^{26} \\plus{} 7206189623312715\\,{x}^{4}{c}^{ 26} \\plus{} 17500746228045165\\,{x}^{3}y{c}^{26} \\plus{} 17500746228045165\\,x{y}^{3}{c} ^{26}$\n$ \\plus{} 160099419197302065\\,{x}^{3}{y}^{2}{c}^{25} \\plus{} 12887259749628030\\,{y }^{5}{c}^{25} \\plus{} 61462315728995220\\,{x}^{4}y{c}^{25} \\plus{} 12887259749628030\\,{ x}^{5}{c}^{25} \\plus{} 160099419197302065\\,{x}^{2}{y}^{3}{c}^{25} \\plus{} 61462315728995220\\,x{y}^{4}{c}^{25}$\n$ \\plus{} 16267941301305600\\,{y}^{6}{c}^{24} \\plus{} 520637668287487425\\,{x}^{2}{y}^{4}{c}^{24} \\plus{} 16267941301305600\\,{x}^{6} {c}^{24} \\plus{} 849237373244718900\\,{x}^{3}{y}^{3}{c}^{24} \\plus{} 112286922966433575 \\,{x}^{5}y{c}^{24} \\plus{} 520637668287487425\\,{x}^{4}{y}^{2}{c}^{24} \\plus{} 112286922966433575\\,x{y}^{5}{c}^{24}$\n$ \\plus{} 20462019918048450\\,{x}^{7}{c}^{23 } \\plus{} 2583870161142136725\\,{x}^{3}{y}^{4}{c}^{23} \\plus{} 123598225902497625\\,{x}^ {6}y{c}^{23} \\plus{} 2583870161142136725\\,{x}^{4}{y}^{3}{c}^{23} \\plus{} 123598225902497625\\,x{y}^{6}{c}^{23} \\plus{} 1010963586572151525\\,{x}^{5}{y}^{ 2}{c}^{23} \\plus{} 20462019918048450\\,{y}^{7}{c}^{23} \\plus{} 1010963586572151525\\,{x} ^{2}{y}^{5}{c}^{23}$\n$ \\plus{} 100227181758694875\\,x{y}^{7}{c}^{22} \\plus{} 33771511824715575\\,{y}^{8}{c}^{22} \\plus{} 5052078366356891925\\,{x}^{3}{y}^{5} {c}^{22} \\plus{} 1300256160789639825\\,{x}^{2}{y}^{6}{c}^{22} \\plus{} 5052078366356891925\\,{x}^{5}{y}^{3}{c}^{22} \\plus{} 100227181758694875\\,{x}^{7 }y{c}^{22} \\plus{} 1300256160789639825\\,{x}^{6}{y}^{2}{c}^{22} \\plus{} 7610819359490631675\\,{x}^{4}{y}^{4}{c}^{22} \\plus{} 33771511824715575\\,{x}^{8} {c}^{22 }$\n$ \\plus{} 6871482379629081900\\,{x}^{3}{y}^{6}{c}^{21} \\plus{} 56307035256428700 \\,{y}^{9}{c}^{21} \\plus{} 14738655968645105250\\,{x}^{4}{y}^{5}{c}^{21} \\plus{} 118104971417942175\\,x{y}^{8}{c}^{21} \\plus{} 6871482379629081900\\,{x}^{6}{y}^{ 3}{c}^{21} \\plus{} 1211700379626022275\\,{x}^{2}{y}^{7}{c}^{21} \\plus{} 14738655968645105250\\,{x}^{5}{y}^{4}{c}^{21} \\plus{} 1211700379626022275\\,{x}^ {7}{y}^{2}{c}^{21} \\plus{} 56307035256428700\\,{x}^{9}{c}^{21} \\plus{} 118104971417942175\\,{x}^{8}y{c}^{21}$\n$ \\plus{} 6922333817672485950\\,{x}^{3}{y}^{ 7}{c}^{20} \\plus{} 74579755550855265\\,{x}^{10}{c}^{20} \\plus{} 6922333817672485950\\,{x }^{7}{y}^{3}{c}^{20} \\plus{} 218325092438225025\\,x{y}^{9}{c}^{20} \\plus{} 20205326773079281125\\,{x}^{6}{y}^{4}{c}^{20} \\plus{} 218325092438225025\\,{x}^{ 9}y{c}^{20} \\plus{} 20205326773079281125\\,{x}^{4}{y}^{6}{c}^{20} \\plus{} 984690578941006500\\,{x}^{2}{y}^{8}{c}^{20} \\plus{} 74579755550855265\\,{y}^{10} {c}^{20} \\plus{} 28195374156236728335\\,{x}^{5}{y}^{5}{c}^{20} \\plus{} 984690578941006500\\,{x}^{8}{y}^{2}{c}^{20}$\n$ \\plus{} 332431506674710425\\,{x}^{10 }y{c}^{19} \\plus{} 20703523584131542575\\,{x}^{4}{y}^{7}{c}^{19} \\plus{} 929452678951534425\\,{x}^{2}{y}^{9}{c}^{19} \\plus{} 332431506674710425\\,x{y}^{ 10}{c}^{19} \\plus{} 20703523584131542575\\,{x}^{7}{y}^{4}{c}^{19} \\plus{} 929452678951534425\\,{x}^{9}{y}^{2}{c}^{19} \\plus{} 5587211207586606525\\,{x}^{8 }{y}^{3}{c}^{19} \\plus{} 5587211207586606525\\,{x}^{3}{y}^{8}{c}^{19} \\plus{} 75157463424334950\\,{y}^{11}{c}^{19} \\plus{} 38253020811296949225\\,{x}^{6}{y}^{ 5}{c}^{19} \\plus{} 75157463424334950\\,{x}^{11}{c}^{19} \\plus{} 38253020811296949225\\,{ x}^{5}{y}^{6}{c}^{19}$\n$ \\plus{} 1009513729316449710\\,{x}^{10}{y}^{2}{c}^{18} \\plus{} 58564007553966660\\,{y}^{12}{c}^{18} \\plus{} 38837506435766674065\\,{x}^{7}{y}^{ 5}{c}^{18} \\plus{} 4071659563211697225\\,{x}^{9}{y}^{3}{c}^{18} \\plus{} 58564007553966660\\,{x}^{12}{c}^{18} \\plus{} 16701086221823724750\\,{x}^{8}{y}^{ 4}{c}^{18} \\plus{} 362611857207382065\\,x{y}^{11}{c}^{18} \\plus{} 50955813923280508710 \\,{x}^{6}{y}^{6}{c}^{18} \\plus{} 4071659563211697225\\,{x}^{3}{y}^{9}{c}^{18} \\plus{} 38837506435766674065\\,{x}^{5}{y}^{7}{c}^{18} \\plus{} 16701086221823724750\\,{x} ^{4}{y}^{8}{c}^{18} \\plus{} 362611857207382065\\,{x}^{11}y{c}^{18} \\plus{} 1009513729316449710\\,{x}^{2}{y}^{10}{c}^{18}$\n$ \\plus{} 977917657147362045\\,{x}^{ 11}{y}^{2}{c}^{17} \\plus{} 2984439815827840980\\,{x}^{10}{y}^{3}{c}^{17} \\plus{} 2984439815827840980\\,{x}^{3}{y}^{10}{c}^{17} \\plus{} 36130164566694030\\,{y}^{ 13}{c}^{17} \\plus{} 30825593044084271745\\,{x}^{8}{y}^{5}{c}^{17} \\plus{} 11308185750809556000\\,{x}^{4}{y}^{9}{c}^{17} \\plus{} 292229998302188745\\,{x}^{ 12}y{c}^{17} \\plus{} 30825593044084271745\\,{x}^{5}{y}^{8}{c}^{17} \\plus{} 11308185750809556000\\,{x}^{9}{y}^{4}{c}^{17} \\plus{} 50567705251517488230\\,{x} ^{6}{y}^{7}{c}^{17} \\plus{} 36130164566694030\\,{x}^{13}{c}^{17} \\plus{} 50567705251517488230\\,{x}^{7}{y}^{6}{c}^{17} \\plus{} 292229998302188745\\,x{y}^ {12}{c}^{17} \\plus{} 977917657147362045\\,{x}^{2}{y}^{11}{c}^{17}$\n$ \\plus{} 2160360452109822660\\,{x}^{3}{y}^{11}{c}^{16} \\plus{} 2160360452109822660\\,{x}^ {11}{y}^{3}{c}^{16} \\plus{} 38992364468350626915\\,{x}^{8}{y}^{6}{c}^{16} \\plus{} 748135554864262515\\,{x}^{12}{y}^{2}{c}^{16} \\plus{} 180949529841796485\\,x{y}^{ 13}{c}^{16} \\plus{} 6906323613662794935\\,{x}^{4}{y}^{10}{c}^{16} \\plus{} 6906323613662794935\\,{x}^{10}{y}^{4}{c}^{16} \\plus{} 18022409853586110\\,{y}^{ 14}{c}^{16} \\plus{} 18022409853586110\\,{x}^{14}{c}^{16} \\plus{} 748135554864262515\\,{x }^{2}{y}^{12}{c}^{16} \\plus{} 19994508314853588315\\,{x}^{5}{y}^{9}{c}^{16} \\plus{} 48631289652679821660\\,{x}^{7}{y}^{7}{c}^{16} \\plus{} 180949529841796485\\,{x}^{ 13}y{c}^{16} \\plus{} 38992364468350626915\\,{x}^{6}{y}^{8}{c}^{16} \\plus{} 19994508314853588315\\,{x}^{9}{y}^{5}{c}^{16}$\n$ \\plus{} 88711037987481855\\,{x}^{ 14}y{c}^{15} \\plus{} 36070087439992712715\\,{x}^{8}{y}^{7}{c}^{15} \\plus{} 88711037987481855\\,x{y}^{14}{c}^{15} \\plus{} 1393142007744442485\\,{x}^{3}{y}^{ 12}{c}^{15} \\plus{} 3960181901596608855\\,{x}^{4}{y}^{11}{c}^{15} \\plus{} 36070087439992712715\\,{x}^{7}{y}^{8}{c}^{15} \\plus{} 445465963285976835\\,{x}^{ 2}{y}^{13}{c}^{15} \\plus{} 24236016130783711395\\,{x}^{6}{y}^{9}{c}^{15} \\plus{} 11119352295374912205\\,{x}^{5}{y}^{10}{c}^{15} \\plus{} 3960181901596608855\\,{x} ^{11}{y}^{4}{c}^{15} \\plus{} 1393142007744442485\\,{x}^{12}{y}^{3}{c}^{15} \\plus{} 7395765445494270\\,{x}^{15}{c}^{15} \\plus{} 445465963285976835\\,{x}^{13}{y}^{2} {c}^{15} \\plus{} 24236016130783711395\\,{x}^{9}{y}^{6}{c}^{15} \\plus{} 7395765445494270 \\,{y}^{15}{c}^{15} \\plus{} 11119352295374912205\\,{x}^{10}{y}^{5}{c}^{15}$\n$ \\plus{} 2084305487797867050\\,{x}^{12}{y}^{4}{c}^{14} \\plus{} 209424571608036825\\,{x}^{ 14}{y}^{2}{c}^{14} \\plus{} 21349556612200860075\\,{x}^{7}{y}^{9}{c}^{14} \\plus{} 744857579775620475\\,{x}^{3}{y}^{13}{c}^{14} \\plus{} 12585125502138452670\\,{x}^ {10}{y}^{6}{c}^{14} \\plus{} 2084305487797867050\\,{x}^{4}{y}^{12}{c}^{14} \\plus{} 744857579775620475\\,{x}^{13}{y}^{3}{c}^{14} \\plus{} 35203478647936185\\,x{y}^{ 15}{c}^{14} \\plus{} 35203478647936185\\,{x}^{15}y{c}^{14} \\plus{} 21349556612200860075 \\,{x}^{9}{y}^{7}{c}^{14} \\plus{} 2533095274158855\\,{y}^{16}{c}^{14} \\plus{} 12585125502138452670\\,{x}^{6}{y}^{10}{c}^{14} \\plus{} 2533095274158855\\,{x}^{ 16}{c}^{14} \\plus{} 209424571608036825\\,{x}^{2}{y}^{14}{c}^{14} \\plus{} 25510966047864850275\\,{x}^{8}{y}^{8}{c}^{14} \\plus{} 5511886560598653405\\,{x}^ {5}{y}^{11}{c}^{14} \\plus{} 5511886560598653405\\,{x}^{11}{y}^{5}{c}^{14}$\n$ \\plus{} 322423177403419200\\,{x}^{14}{y}^{3}{c}^{13} \\plus{} 733199301973440\\,{y}^{17}{ c}^{13} \\plus{} 2456477683392122055\\,{x}^{12}{y}^{5}{c}^{13} \\plus{} 11499472852337745 \\,x{y}^{16}{c}^{13} \\plus{} 2456477683392122055\\,{x}^{5}{y}^{12}{c}^{13} \\plus{} 955075836063775050\\,{x}^{13}{y}^{4}{c}^{13} \\plus{} 79098393960389025\\,{x}^{2} {y}^{15}{c}^{13} \\plus{} 10388985093352033110\\,{x}^{7}{y}^{10}{c}^{13} \\plus{} 5632415391716311230\\,{x}^{6}{y}^{11}{c}^{13} \\plus{} 955075836063775050\\,{x}^{ 4}{y}^{13}{c}^{13} \\plus{} 10388985093352033110\\,{x}^{10}{y}^{7}{c}^{13} \\plus{} 14264412913360975275\\,{x}^{9}{y}^{8}{c}^{13} \\plus{} 733199301973440\\,{x}^{17} {c}^{13} \\plus{} 5632415391716311230\\,{x}^{11}{y}^{6}{c}^{13} \\plus{} 322423177403419200\\,{x}^{3}{y}^{14}{c}^{13} \\plus{} 14264412913360975275\\,{x}^ {8}{y}^{9}{c}^{13} \\plus{} 11499472852337745\\,{x}^{16}y{c}^{13} \\plus{} 79098393960389025\\,{x}^{15}{y}^{2}{c}^{13}$\n$ \\plus{} 2206418360572908315\\,{x}^{ 12}{y}^{6}{c}^{12} \\plus{} 24362825601034935\\,{x}^{16}{y}^{2}{c}^{12} \\plus{} 6477654862591944300\\,{x}^{10}{y}^{8}{c}^{12} \\plus{} 2206418360572908315\\,{x}^ {6}{y}^{12}{c}^{12} \\plus{} 112916040343061310\\,{x}^{3}{y}^{15}{c}^{12} \\plus{} 7466933476884696750\\,{x}^{9}{y}^{9}{c}^{12} \\plus{} 3134553601994010\\,x{y}^{17 }{c}^{12} \\plus{} 6477654862591944300\\,{x}^{8}{y}^{10}{c}^{12} \\plus{} 366353221781733975\\,{x}^{14}{y}^{4}{c}^{12} \\plus{} 4262032766997513990\\,{x}^{ 7}{y}^{11}{c}^{12} \\plus{} 366353221781733975\\,{x}^{4}{y}^{14}{c}^{12} \\plus{} 181249095648150\\,{y}^{18}{c}^{12} \\plus{} 3134553601994010\\,{x}^{17}y{c}^{12} \\plus{} 4262032766997513990\\,{x}^{11}{y}^{7}{c}^{12} \\plus{} 181249095648150\\,{x}^{18} {c}^{12} \\plus{} 24362825601034935\\,{x}^{2}{y}^{16}{c}^{12} \\plus{} 962390964597878745 \\,{x}^{13}{y}^{5}{c}^{12} \\plus{} 962390964597878745\\,{x}^{5}{y}^{13}{c}^{12} \\plus{} 112916040343061310\\,{x}^{15}{y}^{3}{c}^{12}$\n$ \\plus{} 6196930766887320\\,{x}^{17} {y}^{2}{c}^{11} \\plus{} 32196892281454305\\,{x}^{3}{y}^{16}{c}^{11} \\plus{} 38568777876360\\,{y}^{19}{c}^{11} \\plus{} 6196930766887320\\,{x}^{2}{y}^{17}{c}^ {11} \\plus{} 1496578870705849605\\,{x}^{12}{y}^{7}{c}^{11} \\plus{} 3140506203143784840 \\,{x}^{9}{y}^{10}{c}^{11} \\plus{} 721091184063480\\,x{y}^{18}{c}^{11} \\plus{} 115583970861408465\\,{x}^{4}{y}^{15}{c}^{11} \\plus{} 32196892281454305\\,{x}^{16 }{y}^{3}{c}^{11} \\plus{} 1496578870705849605\\,{x}^{7}{y}^{12}{c}^{11} \\plus{} 38568777876360\\,{x}^{19}{c}^{11} \\plus{} 2440286663280994845\\,{x}^{8}{y}^{11}{ c}^{11} \\plus{} 751829705460213795\\,{x}^{6}{y}^{13}{c}^{11} \\plus{} 721091184063480\\,{ x}^{18}y{c}^{11} \\plus{} 321388973115663375\\,{x}^{5}{y}^{14}{c}^{11} \\plus{} 2440286663280994845\\,{x}^{11}{y}^{8}{c}^{11} \\plus{} 321388973115663375\\,{x}^{ 14}{y}^{5}{c}^{11} \\plus{} 751829705460213795\\,{x}^{13}{y}^{6}{c}^{11} \\plus{} 3140506203143784840\\,{x}^{10}{y}^{9}{c}^{11} \\plus{} 115583970861408465\\,{x}^{ 15}{y}^{4}{c}^{11}$\n$ \\plus{} 772772734181827170\\,{x}^{12}{y}^{8}{c}^{10} \\plus{} 449889121836280395\\,{x}^{7}{y}^{13}{c}^{10} \\plus{} 29842861410108315\\,{x}^{4} {y}^{16}{c}^{10} \\plus{} 1081313759617721910\\,{x}^{9}{y}^{11}{c}^{10} \\plus{} 7534683942204330\\,{x}^{17}{y}^{3}{c}^{10} \\plus{} 141198022534590\\,{x}^{19}y{c }^{10} \\plus{} 449889121836280395\\,{x}^{13}{y}^{7}{c}^{10} \\plus{} 1211445894374410257 \\,{x}^{10}{y}^{10}{c}^{10} \\plus{} 7093025578539\\,{x}^{20}{c}^{10} \\plus{} 7534683942204330\\,{x}^{3}{y}^{17}{c}^{10} \\plus{} 1315579009511205\\,{x}^{2}{y} ^{18}{c}^{10} \\plus{} 218583663720200910\\,{x}^{6}{y}^{14}{c}^{10} \\plus{} 89469060681737367\\,{x}^{15}{y}^{5}{c}^{10} \\plus{} 1081313759617721910\\,{x}^{ 11}{y}^{9}{c}^{10} \\plus{} 1315579009511205\\,{x}^{18}{y}^{2}{c}^{10} \\plus{} 772772734181827170\\,{x}^{8}{y}^{12}{c}^{10} \\plus{} 7093025578539\\,{y}^{20}{c} ^{10} \\plus{} 141198022534590\\,x{y}^{19}{c}^{10} \\plus{} 29842861410108315\\,{x}^{16}{y }^{4}{c}^{10} \\plus{} 218583663720200910\\,{x}^{14}{y}^{6}{c}^{10} \\plus{} 89469060681737367\\,{x}^{5}{y}^{15}{c}^{10}$\n$ \\plus{} 308183466115653975\\,{x}^{9} {y}^{12}{c}^{9} \\plus{} 1126430533800\\,{y}^{21}{c}^{9} \\plus{} 20503667384124705\\,{x}^ {5}{y}^{16}{c}^{9} \\plus{} 1458316098172575\\,{x}^{3}{y}^{18}{c}^{9} \\plus{} 6310720192829400\\,{x}^{17}{y}^{4}{c}^{9} \\plus{} 308183466115653975\\,{x}^{12}{ y}^{9}{c}^{9} \\plus{} 114414758085012750\\,{x}^{7}{y}^{14}{c}^{9} \\plus{} 1458316098172575\\,{x}^{18}{y}^{3}{c}^{9} \\plus{} 23637607287795\\,x{y}^{20}{c}^ {9} \\plus{} 206128996224470325\\,{x}^{13}{y}^{8}{c}^{9} \\plus{} 20503667384124705\\,{x}^ {16}{y}^{5}{c}^{9} \\plus{} 1126430533800\\,{x}^{21}{c}^{9} \\plus{} 234981359117550\\,{x} ^{2}{y}^{19}{c}^{9} \\plus{} 53176745518951770\\,{x}^{6}{y}^{15}{c}^{9} \\plus{} 378024510732571755\\,{x}^{10}{y}^{11}{c}^{9} \\plus{} 378024510732571755\\,{x}^{ 11}{y}^{10}{c}^{9} \\plus{} 206128996224470325\\,{x}^{8}{y}^{13}{c}^{9} \\plus{} 6310720192829400\\,{x}^{4}{y}^{17}{c}^{9} \\plus{} 234981359117550\\,{x}^{19}{y}^ {2}{c}^{9} \\plus{} 53176745518951770\\,{x}^{15}{y}^{6}{c}^{9} \\plus{} 23637607287795\\,{ x}^{20}y{c}^{9} \\plus{} 114414758085012750\\,{x}^{14}{y}^{7}{c}^{9}$\n$ \\plus{} 3845533111294590\\,{x}^{17}{y}^{5}{c}^{8} \\plus{} 153604163700\\,{x}^{22}{c}^{8} \\plus{} 1096068860697600\\,{x}^{18}{y}^{4}{c}^{8} \\plus{} 153604163700\\,{y}^{22}{c}^{8 } \\plus{} 10663945457377545\\,{x}^{16}{y}^{6}{c}^{8} \\plus{} 96280138209164220\\,{x}^{12 }{y}^{10}{c}^{8} \\plus{} 3379291601400\\,x{y}^{21}{c}^{8} \\plus{} 3845533111294590\\,{x} ^{5}{y}^{17}{c}^{8} \\plus{} 45931518331885125\\,{x}^{8}{y}^{14}{c}^{8} \\plus{} 24232232576657520\\,{x}^{7}{y}^{15}{c}^{8} \\plus{} 24232232576657520\\,{x}^{15}{ y}^{7}{c}^{8} \\plus{} 234807019897500\\,{x}^{3}{y}^{19}{c}^{8} \\plus{} 3379291601400\\,{ x}^{21}y{c}^{8} \\plus{} 10663945457377545\\,{x}^{6}{y}^{16}{c}^{8} \\plus{} 45931518331885125\\,{x}^{14}{y}^{8}{c}^{8} \\plus{} 234807019897500\\,{x}^{19}{y} ^{3}{c}^{8} \\plus{} 96280138209164220\\,{x}^{10}{y}^{12}{c}^{8} \\plus{} 72834792863369850\\,{x}^{13}{y}^{9}{c}^{8} \\plus{} 35447693970690\\,{x}^{2}{y}^{ 20}{c}^{8} \\plus{} 105713054588876040\\,{x}^{11}{y}^{11}{c}^{8} \\plus{} 1096068860697600\\,{x}^{4}{y}^{18}{c}^{8} \\plus{} 72834792863369850\\,{x}^{9}{y} ^{13}{c}^{8} \\plus{} 35447693970690\\,{x}^{20}{y}^{2}{c}^{8}$\n$ \\plus{} 14169936212586000 \\,{x}^{9}{y}^{14}{c}^{7} \\plus{} 588626351558550\\,{x}^{18}{y}^{5}{c}^{7} \\plus{} 23797038511922220\\,{x}^{12}{y}^{11}{c}^{7} \\plus{} 17809178400\\,{y}^{23}{c}^{7 } \\plus{} 1744644537673560\\,{x}^{17}{y}^{6}{c}^{7} \\plus{} 4505722135200\\,{x}^{21}{y}^ {2}{c}^{7} \\plus{} 1744644537673560\\,{x}^{6}{y}^{17}{c}^{7} \\plus{} 588626351558550\\,{ x}^{5}{y}^{18}{c}^{7} \\plus{} 4209610166857935\\,{x}^{16}{y}^{7}{c}^{7} \\plus{} 4209610166857935\\,{x}^{7}{y}^{16}{c}^{7} \\plus{} 31505187102390\\,{x}^{3}{y}^{ 20}{c}^{7} \\plus{} 8434858561754355\\,{x}^{8}{y}^{15}{c}^{7} \\plus{} 17809178400\\,{x}^{ 23}{c}^{7} \\plus{} 31505187102390\\,{x}^{20}{y}^{3}{c}^{7} \\plus{} 156654239411700\\,{x} ^{19}{y}^{4}{c}^{7} \\plus{} 409611103200\\,{x}^{22}y{c}^{7} \\plus{} 23797038511922220\\, {x}^{11}{y}^{12}{c}^{7} \\plus{} 4505722135200\\,{x}^{2}{y}^{21}{c}^{7} \\plus{} 20019537297068490\\,{x}^{10}{y}^{13}{c}^{7} \\plus{} 20019537297068490\\,{x}^{13} {y}^{10}{c}^{7} \\plus{} 8434858561754355\\,{x}^{15}{y}^{8}{c}^{7} \\plus{} 156654239411700\\,{x}^{4}{y}^{19}{c}^{7} \\plus{} 14169936212586000\\,{x}^{14}{y} ^{9}{c}^{7} \\plus{} 409611103200\\,x{y}^{22}{c}^{7}$\n$ \\plus{} 41554749600\\,{x}^{23}y{c}^{ 6} \\plus{} 4301057448335610\\,{x}^{13}{y}^{11}{c}^{6} \\plus{} 230971324829805\\,{x}^{18} {y}^{6}{c}^{6} \\plus{} 1731447900\\,{x}^{24}{c}^{6} \\plus{} 4301057448335610\\,{x}^{11}{ y}^{13}{c}^{6} \\plus{} 230971324829805\\,{x}^{6}{y}^{18}{c}^{6} \\plus{} 1256797756868130\\,{x}^{8}{y}^{16}{c}^{6} \\plus{} 3504450549600\\,{x}^{3}{y}^{21 }{c}^{6} \\plus{} 3369714001320915\\,{x}^{10}{y}^{14}{c}^{6} \\plus{} 1256797756868130\\,{ x}^{16}{y}^{8}{c}^{6} \\plus{} 3504450549600\\,{x}^{21}{y}^{3}{c}^{6} \\plus{} 591834193207470\\,{x}^{7}{y}^{17}{c}^{6} \\plus{} 73279650945510\\,{x}^{19}{y}^{5 }{c}^{6} \\plus{} 591834193207470\\,{x}^{17}{y}^{7}{c}^{6} \\plus{} 477879620400\\,{x}^{2} {y}^{22}{c}^{6} \\plus{} 477879620400\\,{x}^{22}{y}^{2}{c}^{6} \\plus{} 4664636437623390 \\,{x}^{12}{y}^{12}{c}^{6} \\plus{} 2238944546574450\\,{x}^{9}{y}^{15}{c}^{6} \\plus{} 2238944546574450\\,{x}^{15}{y}^{9}{c}^{6} \\plus{} 18380931463395\\,{x}^{20}{y}^{ 4}{c}^{6} \\plus{} 3369714001320915\\,{x}^{14}{y}^{10}{c}^{6} \\plus{} 18380931463395\\,{x }^{4}{y}^{20}{c}^{6} \\plus{} 1731447900\\,{y}^{24}{c}^{6} \\plus{} 41554749600\\,x{y}^{23 }{c}^{6} \\plus{} 73279650945510\\,{x}^{5}{y}^{19}{c}^{6}$\n$ \\plus{} 66427655144355\\,{x}^{7 }{y}^{18}{c}^{5} \\plus{} 41554749600\\,{x}^{23}{y}^{2}{c}^{5} \\plus{} 616653376988580\\, {x}^{11}{y}^{14}{c}^{5} \\plus{} 282252157155990\\,{x}^{16}{y}^{9}{c}^{5} \\plus{} 24496286003190\\,{x}^{19}{y}^{6}{c}^{5} \\plus{} 66427655144355\\,{x}^{18}{y}^{7} {c}^{5} \\plus{} 451892854554867\\,{x}^{10}{y}^{15}{c}^{5} \\plus{} 1752225274800\\,{x}^{ 21}{y}^{4}{c}^{5} \\plus{} 3462895800\\,{x}^{24}y{c}^{5} \\plus{} 616653376988580\\,{x}^{ 14}{y}^{11}{c}^{5} \\plus{} 318586413600\\,{x}^{22}{y}^{3}{c}^{5} \\plus{} 138515832\\,{x} ^{25}{c}^{5} \\plus{} 1752225274800\\,{x}^{4}{y}^{21}{c}^{5} \\plus{} 451892854554867\\,{x }^{15}{y}^{10}{c}^{5} \\plus{} 149396846867280\\,{x}^{17}{y}^{8}{c}^{5} \\plus{} 719748561723435\\,{x}^{12}{y}^{13}{c}^{5} \\plus{} 7355859369759\\,{x}^{20}{y}^{5 }{c}^{5} \\plus{} 719748561723435\\,{x}^{13}{y}^{12}{c}^{5} \\plus{} 3462895800\\,x{y}^{24 }{c}^{5} \\plus{} 149396846867280\\,{x}^{8}{y}^{17}{c}^{5} \\plus{} 318586413600\\,{x}^{3} {y}^{22}{c}^{5} \\plus{} 282252157155990\\,{x}^{9}{y}^{16}{c}^{5} \\plus{} 24496286003190 \\,{x}^{6}{y}^{19}{c}^{5} \\plus{} 138515832\\,{y}^{25}{c}^{5} \\plus{} 7355859369759\\,{x} ^{5}{y}^{20}{c}^{5} \\plus{} 41554749600\\,{x}^{2}{y}^{23}{c}^{5}$\n$ \\plus{} 8879220\\,{x}^{ 26}{c}^{4} \\plus{} 132744339000\\,{x}^{22}{y}^{4}{c}^{4} \\plus{} 68602274395200\\,{x}^{ 15}{y}^{11}{c}^{4} \\plus{} 13871783425500\\,{x}^{8}{y}^{18}{c}^{4} \\plus{} 2885746500\\, {x}^{2}{y}^{24}{c}^{4} \\plus{} 13871783425500\\,{x}^{18}{y}^{8}{c}^{4} \\plus{} 230859720\\,x{y}^{25}{c}^{4} \\plus{} 2885746500\\,{x}^{24}{y}^{2}{c}^{4} \\plus{} 47164063646700\\,{x}^{10}{y}^{16}{c}^{4} \\plus{} 5840750916000\\,{x}^{7}{y}^{19} {c}^{4} \\plus{} 584075091600\\,{x}^{21}{y}^{5}{c}^{4} \\plus{} 132744339000\\,{x}^{4}{y}^ {22}{c}^{4} \\plus{} 23085972000\\,{x}^{3}{y}^{23}{c}^{4} \\plus{} 8879220\\,{y}^{26}{c}^{ 4} \\plus{} 2044262820600\\,{x}^{20}{y}^{6}{c}^{4} \\plus{} 27743566851000\\,{x}^{17}{y}^{ 9}{c}^{4} \\plus{} 2044262820600\\,{x}^{6}{y}^{20}{c}^{4} \\plus{} 47164063646700\\,{x}^{ 16}{y}^{10}{c}^{4} \\plus{} 584075091600\\,{x}^{5}{y}^{21}{c}^{4} \\plus{} 92349215532000 \\,{x}^{13}{y}^{13}{c}^{4} \\plus{} 27743566851000\\,{x}^{9}{y}^{17}{c}^{4} \\plus{} 23085972000\\,{x}^{23}{y}^{3}{c}^{4} \\plus{} 85752842994000\\,{x}^{14}{y}^{12}{c }^{4} \\plus{} 68602274395200\\,{x}^{11}{y}^{15}{c}^{4} \\plus{} 85752842994000\\,{x}^{12} {y}^{14}{c}^{4} \\plus{} 230859720\\,{x}^{25}y{c}^{4} \\plus{} 5840750916000\\,{x}^{19}{y} ^{7}{c}^{4}$\n$ \\plus{} 438480\\,{y}^{27}{c}^{3} \\plus{} 973458486000\\,{x}^{19}{y}^{8}{c}^{ 3} \\plus{} 11838960\\,{x}^{26}y{c}^{3} \\plus{} 5716856199600\\,{x}^{16}{y}^{11}{c}^{3} \\plus{} 389383394400\\,{x}^{7}{y}^{20}{c}^{3} \\plus{} 35398490400\\,{x}^{5}{y}^{22}{c}^{ 3} \\plus{} 438480\\,{x}^{27}{c}^{3} \\plus{} 7695324000\\,{x}^{4}{y}^{23}{c}^{3} \\plus{} 1282554000\\,{x}^{24}{y}^{3}{c}^{3} \\plus{} 1282554000\\,{x}^{3}{y}^{24}{c}^{3} \\plus{} 153906480\\,{x}^{2}{y}^{25}{c}^{3} \\plus{} 2055079026000\\,{x}^{18}{y}^{9}{c}^{3 } \\plus{} 7695324000\\,{x}^{23}{y}^{4}{c}^{3} \\plus{} 973458486000\\,{x}^{8}{y}^{19}{c}^ {3} \\plus{} 35398490400\\,{x}^{22}{y}^{5}{c}^{3} \\plus{} 389383394400\\,{x}^{20}{y}^{7}{ c}^{3} \\plus{} 8795163384000\\,{x}^{13}{y}^{14}{c}^{3} \\plus{} 5716856199600\\,{x}^{11}{ y}^{16}{c}^{3} \\plus{} 7622474932800\\,{x}^{15}{y}^{12}{c}^{3} \\plus{} 11838960\\,x{y}^{ 26}{c}^{3} \\plus{} 7622474932800\\,{x}^{12}{y}^{15}{c}^{3} \\plus{} 8795163384000\\,{x}^{ 14}{y}^{13}{c}^{3} \\plus{} 129794464800\\,{x}^{21}{y}^{6}{c}^{3} \\plus{} 2055079026000 \\,{x}^{9}{y}^{18}{c}^{3} \\plus{} 153906480\\,{x}^{25}{y}^{2}{c}^{3} \\plus{} 3699142246800\\,{x}^{10}{y}^{17}{c}^{3} \\plus{} 129794464800\\,{x}^{6}{y}^{21}{c }^{3} \\plus{} 3699142246800\\,{x}^{17}{y}^{10}{c}^{3}$\n$ \\plus{} 476404683300\\,{x}^{16}{y} ^{12}{c}^{2} \\plus{} 438480\\,{x}^{27}y{c}^{2} \\plus{} 15660\\,{y}^{28}{c}^{2} \\plus{} 1539064800\\,{x}^{23}{y}^{5}{c}^{2} \\plus{} 320638500\\,{x}^{4}{y}^{24}{c}^{2} \\plus{} 108162054000\\,{x}^{9}{y}^{19}{c}^{2} \\plus{} 1539064800\\,{x}^{5}{y}^{23}{c}^{2 } \\plus{} 320638500\\,{x}^{24}{y}^{4}{c}^{2} \\plus{} 15660\\,{x}^{28}{c}^{2} \\plus{} 108162054000\\,{x}^{19}{y}^{9}{c}^{2} \\plus{} 5919480\\,{x}^{2}{y}^{26}{c}^{2} \\plus{} 48672924300\\,{x}^{20}{y}^{8}{c}^{2} \\plus{} 51302160\\,{x}^{3}{y}^{25}{c}^{2} \\plus{} 438480\\,x{y}^{27}{c}^{2} \\plus{} 336285658800\\,{x}^{11}{y}^{17}{c}^{2} \\plus{} 5919480 \\,{x}^{26}{y}^{2}{c}^{2} \\plus{} 5899748400\\,{x}^{6}{y}^{22}{c}^{2} \\plus{} 586344225600\\,{x}^{15}{y}^{13}{c}^{2} \\plus{} 5899748400\\,{x}^{22}{y}^{6}{c}^{ 2} \\plus{} 628225956000\\,{x}^{14}{y}^{14}{c}^{2} \\plus{} 205507902600\\,{x}^{18}{y}^{10 }{c}^{2} \\plus{} 205507902600\\,{x}^{10}{y}^{18}{c}^{2} \\plus{} 586344225600\\,{x}^{13}{ y}^{15}{c}^{2} \\plus{} 18542066400\\,{x}^{21}{y}^{7}{c}^{2} \\plus{} 48672924300\\,{x}^{8 }{y}^{20}{c}^{2} \\plus{} 336285658800\\,{x}^{17}{y}^{11}{c}^{2} \\plus{} 476404683300\\,{ x}^{12}{y}^{16}{c}^{2} \\plus{} 18542066400\\,{x}^{7}{y}^{21}{c}^{2} \\plus{} 51302160\\,{ x}^{25}{y}^{3}{c}^{2}$\n$ \\plus{} 1315440\\,{x}^{3}{y}^{26}c \\plus{} 360\\,{y}^{29}c \\plus{} 171007200\\,{x}^{23}{y}^{6}c \\plus{} 7210803600\\,{x}^{19}{y}^{10}c \\plus{} 1545172200\\, {x}^{8}{y}^{21}c \\plus{} 3605401800\\,{x}^{9}{y}^{20}c \\plus{} 27921153600\\,{x}^{14}{y} ^{15}c \\plus{} 1545172200\\,{x}^{21}{y}^{8}c \\plus{} 7210803600\\,{x}^{10}{y}^{19}c \\plus{} 24431009400\\,{x}^{13}{y}^{16}c \\plus{} 146160\\,{x}^{27}{y}^{2}c \\plus{} 42751800\\,{x}^ {5}{y}^{24}c \\plus{} 561880800\\,{x}^{22}{y}^{7}c \\plus{} 146160\\,{x}^{2}{y}^{27}c \\plus{} 10440\\,{x}^{28}yc \\plus{} 3605401800\\,{x}^{20}{y}^{9}c \\plus{} 27921153600\\,{x}^{15}{y }^{14}c \\plus{} 171007200\\,{x}^{6}{y}^{23}c \\plus{} 12455024400\\,{x}^{11}{y}^{18}c \\plus{} 12455024400\\,{x}^{18}{y}^{11}c \\plus{} 1315440\\,{x}^{26}{y}^{3}c \\plus{} 18682536600\\, {x}^{12}{y}^{17}c \\plus{} 8550360\\,{x}^{4}{y}^{25}c \\plus{} 360\\,{x}^{29}c \\plus{} 8550360\\,{x }^{25}{y}^{4}c \\plus{} 10440\\,x{y}^{28}c \\plus{} 18682536600\\,{x}^{17}{y}^{12}c \\plus{} 42751800\\,{x}^{24}{y}^{5}c \\plus{} 24431009400\\,{x}^{16}{y}^{13}c \\plus{} 561880800\\,{ x}^{7}{y}^{22}c$\n$ \\plus{} 109620\\,{x}^{4}{y}^{26} \\plus{} 1740\\,{x}^{2}{y}^{28} \\plus{} 16240\\,{ x}^{27}{y}^{3} \\plus{} 4\\,{y}^{30} \\plus{} 109620\\,{x}^{26}{y}^{4} \\plus{} 16240\\,{x}^{3}{y}^{ 27} \\plus{} 570024\\,{x}^{25}{y}^{5} \\plus{} 120\\,x{y}^{29} \\plus{} 2375100\\,{x}^{24}{y}^{6} \\plus{} 4 \\,{x}^{30} \\plus{} 8143200\\,{x}^{23}{y}^{7} \\plus{} 120\\,{x}^{29}y \\plus{} 23411700\\,{x}^{22}{ y}^{8} \\plus{} 1740\\,{x}^{28}{y}^{2} \\plus{} 57228600\\,{x}^{21}{y}^{9} \\plus{} 345972900\\,{x}^ {12}{y}^{18} \\plus{} 120180060\\,{x}^{20}{y}^{10} \\plus{} 218509200\\,{x}^{11}{y}^{19} \\plus{} 218509200\\,{x}^{19}{y}^{11} \\plus{} 120180060\\,{x}^{10}{y}^{20} \\plus{} 345972900\\,{x} ^{18}{y}^{12} \\plus{} 57228600\\,{x}^{9}{y}^{21} \\plus{} 479039400\\,{x}^{17}{y}^{13} \\plus{} 23411700\\,{x}^{8}{y}^{22} \\plus{} 581690700\\,{x}^{16}{y}^{14} \\plus{} 8143200\\,{x}^{7} {y}^{23} \\plus{} 620470080\\,{x}^{15}{y}^{15} \\plus{} 2375100\\,{x}^{6}{y}^{24} \\plus{} 581690700\\,{x}^{14}{y}^{16} \\plus{} 570024\\,{x}^{5}{y}^{25} \\plus{} 479039400\\,{x}^{13 }{y}^{17} \\geq 0$\n\nDone ! \n\nPS : It's just for fun  :lol:\n\nPS : Do you believe i do it by hand ???  :rotfl:[/quote]\r\n\"very nice solution\".I think only arqady,jichen and you can do it. :)",
  "Solution_3": "[quote=\"Materazzi\"][quote=\"nguoivn\"]\nWith this ineq, we can solve it easily by Am-Gm. But the following (of him) is harder and I still have no solution:  :oops: \n$ \\frac {1}{2 \\plus{} a^5b^5} \\plus{} \\frac {1}{2 \\plus{} b^5c^5} \\plus{} \\frac {1}{2 \\plus{} c^5a^5} \\geq\\ 1$\n(with the same condition)[/quote]\n\nMaybe ..... I have a very ugly solution for this  :blush: \n\nAssume : $ c \\equal{} min \\{a;b;c\\}$\n\nWith : $ a \\equal{} x \\plus{} c \\ ; \\ b \\equal{} y \\plus{} c \\ ; \\ x,y \\geq 0$ .\n\nThis inequality is equivalent to : \n\n$ \\minus{} 343151886824415\\,xy{c}^{28} \\plus{} 343151886824415\\,{x}^{2}{c}^{28} \\plus{} 343151886824415\\,{y}^{2}{c}^{28}$\n$ \\plus{} 1258223585022855\\,x{y}^{2}{c}^{27} \\plus{} 2363935220345970\\,{x}^{3}{c}^{27} \\plus{} 1258223585022855\\,{x}^{2}y{c}^{27} \\plus{} 2363935220345970\\,{y}^{3}{c}^{27}$\n$ \\plus{} 23677480190884635\\,{x}^{2}{y}^{2}{c} ^{26} \\plus{} 7206189623312715\\,{y}^{4}{c}^{26} \\plus{} 7206189623312715\\,{x}^{4}{c}^{ 26} \\plus{} 17500746228045165\\,{x}^{3}y{c}^{26} \\plus{} 17500746228045165\\,x{y}^{3}{c} ^{26}$\n$ \\plus{} 160099419197302065\\,{x}^{3}{y}^{2}{c}^{25} \\plus{} 12887259749628030\\,{y }^{5}{c}^{25} \\plus{} 61462315728995220\\,{x}^{4}y{c}^{25} \\plus{} 12887259749628030\\,{ x}^{5}{c}^{25} \\plus{} 160099419197302065\\,{x}^{2}{y}^{3}{c}^{25} \\plus{} 61462315728995220\\,x{y}^{4}{c}^{25}$\n$ \\plus{} 16267941301305600\\,{y}^{6}{c}^{24} \\plus{} 520637668287487425\\,{x}^{2}{y}^{4}{c}^{24} \\plus{} 16267941301305600\\,{x}^{6} {c}^{24} \\plus{} 849237373244718900\\,{x}^{3}{y}^{3}{c}^{24} \\plus{} 112286922966433575 \\,{x}^{5}y{c}^{24} \\plus{} 520637668287487425\\,{x}^{4}{y}^{2}{c}^{24} \\plus{} 112286922966433575\\,x{y}^{5}{c}^{24}$\n$ \\plus{} 20462019918048450\\,{x}^{7}{c}^{23 } \\plus{} 2583870161142136725\\,{x}^{3}{y}^{4}{c}^{23} \\plus{} 123598225902497625\\,{x}^ {6}y{c}^{23} \\plus{} 2583870161142136725\\,{x}^{4}{y}^{3}{c}^{23} \\plus{} 123598225902497625\\,x{y}^{6}{c}^{23} \\plus{} 1010963586572151525\\,{x}^{5}{y}^{ 2}{c}^{23} \\plus{} 20462019918048450\\,{y}^{7}{c}^{23} \\plus{} 1010963586572151525\\,{x} ^{2}{y}^{5}{c}^{23}$\n$ \\plus{} 100227181758694875\\,x{y}^{7}{c}^{22} \\plus{} 33771511824715575\\,{y}^{8}{c}^{22} \\plus{} 5052078366356891925\\,{x}^{3}{y}^{5} {c}^{22} \\plus{} 1300256160789639825\\,{x}^{2}{y}^{6}{c}^{22} \\plus{} 5052078366356891925\\,{x}^{5}{y}^{3}{c}^{22} \\plus{} 100227181758694875\\,{x}^{7 }y{c}^{22} \\plus{} 1300256160789639825\\,{x}^{6}{y}^{2}{c}^{22} \\plus{} 7610819359490631675\\,{x}^{4}{y}^{4}{c}^{22} \\plus{} 33771511824715575\\,{x}^{8} {c}^{22 }$\n$ \\plus{} 6871482379629081900\\,{x}^{3}{y}^{6}{c}^{21} \\plus{} 56307035256428700 \\,{y}^{9}{c}^{21} \\plus{} 14738655968645105250\\,{x}^{4}{y}^{5}{c}^{21} \\plus{} 118104971417942175\\,x{y}^{8}{c}^{21} \\plus{} 6871482379629081900\\,{x}^{6}{y}^{ 3}{c}^{21} \\plus{} 1211700379626022275\\,{x}^{2}{y}^{7}{c}^{21} \\plus{} 14738655968645105250\\,{x}^{5}{y}^{4}{c}^{21} \\plus{} 1211700379626022275\\,{x}^ {7}{y}^{2}{c}^{21} \\plus{} 56307035256428700\\,{x}^{9}{c}^{21} \\plus{} 118104971417942175\\,{x}^{8}y{c}^{21}$\n$ \\plus{} 6922333817672485950\\,{x}^{3}{y}^{ 7}{c}^{20} \\plus{} 74579755550855265\\,{x}^{10}{c}^{20} \\plus{} 6922333817672485950\\,{x }^{7}{y}^{3}{c}^{20} \\plus{} 218325092438225025\\,x{y}^{9}{c}^{20} \\plus{} 20205326773079281125\\,{x}^{6}{y}^{4}{c}^{20} \\plus{} 218325092438225025\\,{x}^{ 9}y{c}^{20} \\plus{} 20205326773079281125\\,{x}^{4}{y}^{6}{c}^{20} \\plus{} 984690578941006500\\,{x}^{2}{y}^{8}{c}^{20} \\plus{} 74579755550855265\\,{y}^{10} {c}^{20} \\plus{} 28195374156236728335\\,{x}^{5}{y}^{5}{c}^{20} \\plus{} 984690578941006500\\,{x}^{8}{y}^{2}{c}^{20}$\n$ \\plus{} 332431506674710425\\,{x}^{10 }y{c}^{19} \\plus{} 20703523584131542575\\,{x}^{4}{y}^{7}{c}^{19} \\plus{} 929452678951534425\\,{x}^{2}{y}^{9}{c}^{19} \\plus{} 332431506674710425\\,x{y}^{ 10}{c}^{19} \\plus{} 20703523584131542575\\,{x}^{7}{y}^{4}{c}^{19} \\plus{} 929452678951534425\\,{x}^{9}{y}^{2}{c}^{19} \\plus{} 5587211207586606525\\,{x}^{8 }{y}^{3}{c}^{19} \\plus{} 5587211207586606525\\,{x}^{3}{y}^{8}{c}^{19} \\plus{} 75157463424334950\\,{y}^{11}{c}^{19} \\plus{} 38253020811296949225\\,{x}^{6}{y}^{ 5}{c}^{19} \\plus{} 75157463424334950\\,{x}^{11}{c}^{19} \\plus{} 38253020811296949225\\,{ x}^{5}{y}^{6}{c}^{19}$\n$ \\plus{} 1009513729316449710\\,{x}^{10}{y}^{2}{c}^{18} \\plus{} 58564007553966660\\,{y}^{12}{c}^{18} \\plus{} 38837506435766674065\\,{x}^{7}{y}^{ 5}{c}^{18} \\plus{} 4071659563211697225\\,{x}^{9}{y}^{3}{c}^{18} \\plus{} 58564007553966660\\,{x}^{12}{c}^{18} \\plus{} 16701086221823724750\\,{x}^{8}{y}^{ 4}{c}^{18} \\plus{} 362611857207382065\\,x{y}^{11}{c}^{18} \\plus{} 50955813923280508710 \\,{x}^{6}{y}^{6}{c}^{18} \\plus{} 4071659563211697225\\,{x}^{3}{y}^{9}{c}^{18} \\plus{} 38837506435766674065\\,{x}^{5}{y}^{7}{c}^{18} \\plus{} 16701086221823724750\\,{x} ^{4}{y}^{8}{c}^{18} \\plus{} 362611857207382065\\,{x}^{11}y{c}^{18} \\plus{} 1009513729316449710\\,{x}^{2}{y}^{10}{c}^{18}$\n$ \\plus{} 977917657147362045\\,{x}^{ 11}{y}^{2}{c}^{17} \\plus{} 2984439815827840980\\,{x}^{10}{y}^{3}{c}^{17} \\plus{} 2984439815827840980\\,{x}^{3}{y}^{10}{c}^{17} \\plus{} 36130164566694030\\,{y}^{ 13}{c}^{17} \\plus{} 30825593044084271745\\,{x}^{8}{y}^{5}{c}^{17} \\plus{} 11308185750809556000\\,{x}^{4}{y}^{9}{c}^{17} \\plus{} 292229998302188745\\,{x}^{ 12}y{c}^{17} \\plus{} 30825593044084271745\\,{x}^{5}{y}^{8}{c}^{17} \\plus{} 11308185750809556000\\,{x}^{9}{y}^{4}{c}^{17} \\plus{} 50567705251517488230\\,{x} ^{6}{y}^{7}{c}^{17} \\plus{} 36130164566694030\\,{x}^{13}{c}^{17} \\plus{} 50567705251517488230\\,{x}^{7}{y}^{6}{c}^{17} \\plus{} 292229998302188745\\,x{y}^ {12}{c}^{17} \\plus{} 977917657147362045\\,{x}^{2}{y}^{11}{c}^{17}$\n$ \\plus{} 2160360452109822660\\,{x}^{3}{y}^{11}{c}^{16} \\plus{} 2160360452109822660\\,{x}^ {11}{y}^{3}{c}^{16} \\plus{} 38992364468350626915\\,{x}^{8}{y}^{6}{c}^{16} \\plus{} 748135554864262515\\,{x}^{12}{y}^{2}{c}^{16} \\plus{} 180949529841796485\\,x{y}^{ 13}{c}^{16} \\plus{} 6906323613662794935\\,{x}^{4}{y}^{10}{c}^{16} \\plus{} 6906323613662794935\\,{x}^{10}{y}^{4}{c}^{16} \\plus{} 18022409853586110\\,{y}^{ 14}{c}^{16} \\plus{} 18022409853586110\\,{x}^{14}{c}^{16} \\plus{} 748135554864262515\\,{x }^{2}{y}^{12}{c}^{16} \\plus{} 19994508314853588315\\,{x}^{5}{y}^{9}{c}^{16} \\plus{} 48631289652679821660\\,{x}^{7}{y}^{7}{c}^{16} \\plus{} 180949529841796485\\,{x}^{ 13}y{c}^{16} \\plus{} 38992364468350626915\\,{x}^{6}{y}^{8}{c}^{16} \\plus{} 19994508314853588315\\,{x}^{9}{y}^{5}{c}^{16}$\n$ \\plus{} 88711037987481855\\,{x}^{ 14}y{c}^{15} \\plus{} 36070087439992712715\\,{x}^{8}{y}^{7}{c}^{15} \\plus{} 88711037987481855\\,x{y}^{14}{c}^{15} \\plus{} 1393142007744442485\\,{x}^{3}{y}^{ 12}{c}^{15} \\plus{} 3960181901596608855\\,{x}^{4}{y}^{11}{c}^{15} \\plus{} 36070087439992712715\\,{x}^{7}{y}^{8}{c}^{15} \\plus{} 445465963285976835\\,{x}^{ 2}{y}^{13}{c}^{15} \\plus{} 24236016130783711395\\,{x}^{6}{y}^{9}{c}^{15} \\plus{} 11119352295374912205\\,{x}^{5}{y}^{10}{c}^{15} \\plus{} 3960181901596608855\\,{x} ^{11}{y}^{4}{c}^{15} \\plus{} 1393142007744442485\\,{x}^{12}{y}^{3}{c}^{15} \\plus{} 7395765445494270\\,{x}^{15}{c}^{15} \\plus{} 445465963285976835\\,{x}^{13}{y}^{2} {c}^{15} \\plus{} 24236016130783711395\\,{x}^{9}{y}^{6}{c}^{15} \\plus{} 7395765445494270 \\,{y}^{15}{c}^{15} \\plus{} 11119352295374912205\\,{x}^{10}{y}^{5}{c}^{15}$\n$ \\plus{} 2084305487797867050\\,{x}^{12}{y}^{4}{c}^{14} \\plus{} 209424571608036825\\,{x}^{ 14}{y}^{2}{c}^{14} \\plus{} 21349556612200860075\\,{x}^{7}{y}^{9}{c}^{14} \\plus{} 744857579775620475\\,{x}^{3}{y}^{13}{c}^{14} \\plus{} 12585125502138452670\\,{x}^ {10}{y}^{6}{c}^{14} \\plus{} 2084305487797867050\\,{x}^{4}{y}^{12}{c}^{14} \\plus{} 744857579775620475\\,{x}^{13}{y}^{3}{c}^{14} \\plus{} 35203478647936185\\,x{y}^{ 15}{c}^{14} \\plus{} 35203478647936185\\,{x}^{15}y{c}^{14} \\plus{} 21349556612200860075 \\,{x}^{9}{y}^{7}{c}^{14} \\plus{} 2533095274158855\\,{y}^{16}{c}^{14} \\plus{} 12585125502138452670\\,{x}^{6}{y}^{10}{c}^{14} \\plus{} 2533095274158855\\,{x}^{ 16}{c}^{14} \\plus{} 209424571608036825\\,{x}^{2}{y}^{14}{c}^{14} \\plus{} 25510966047864850275\\,{x}^{8}{y}^{8}{c}^{14} \\plus{} 5511886560598653405\\,{x}^ {5}{y}^{11}{c}^{14} \\plus{} 5511886560598653405\\,{x}^{11}{y}^{5}{c}^{14}$\n$ \\plus{} 322423177403419200\\,{x}^{14}{y}^{3}{c}^{13} \\plus{} 733199301973440\\,{y}^{17}{ c}^{13} \\plus{} 2456477683392122055\\,{x}^{12}{y}^{5}{c}^{13} \\plus{} 11499472852337745 \\,x{y}^{16}{c}^{13} \\plus{} 2456477683392122055\\,{x}^{5}{y}^{12}{c}^{13} \\plus{} 955075836063775050\\,{x}^{13}{y}^{4}{c}^{13} \\plus{} 79098393960389025\\,{x}^{2} {y}^{15}{c}^{13} \\plus{} 10388985093352033110\\,{x}^{7}{y}^{10}{c}^{13} \\plus{} 5632415391716311230\\,{x}^{6}{y}^{11}{c}^{13} \\plus{} 955075836063775050\\,{x}^{ 4}{y}^{13}{c}^{13} \\plus{} 10388985093352033110\\,{x}^{10}{y}^{7}{c}^{13} \\plus{} 14264412913360975275\\,{x}^{9}{y}^{8}{c}^{13} \\plus{} 733199301973440\\,{x}^{17} {c}^{13} \\plus{} 5632415391716311230\\,{x}^{11}{y}^{6}{c}^{13} \\plus{} 322423177403419200\\,{x}^{3}{y}^{14}{c}^{13} \\plus{} 14264412913360975275\\,{x}^ {8}{y}^{9}{c}^{13} \\plus{} 11499472852337745\\,{x}^{16}y{c}^{13} \\plus{} 79098393960389025\\,{x}^{15}{y}^{2}{c}^{13}$\n$ \\plus{} 2206418360572908315\\,{x}^{ 12}{y}^{6}{c}^{12} \\plus{} 24362825601034935\\,{x}^{16}{y}^{2}{c}^{12} \\plus{} 6477654862591944300\\,{x}^{10}{y}^{8}{c}^{12} \\plus{} 2206418360572908315\\,{x}^ {6}{y}^{12}{c}^{12} \\plus{} 112916040343061310\\,{x}^{3}{y}^{15}{c}^{12} \\plus{} 7466933476884696750\\,{x}^{9}{y}^{9}{c}^{12} \\plus{} 3134553601994010\\,x{y}^{17 }{c}^{12} \\plus{} 6477654862591944300\\,{x}^{8}{y}^{10}{c}^{12} \\plus{} 366353221781733975\\,{x}^{14}{y}^{4}{c}^{12} \\plus{} 4262032766997513990\\,{x}^{ 7}{y}^{11}{c}^{12} \\plus{} 366353221781733975\\,{x}^{4}{y}^{14}{c}^{12} \\plus{} 181249095648150\\,{y}^{18}{c}^{12} \\plus{} 3134553601994010\\,{x}^{17}y{c}^{12} \\plus{} 4262032766997513990\\,{x}^{11}{y}^{7}{c}^{12} \\plus{} 181249095648150\\,{x}^{18} {c}^{12} \\plus{} 24362825601034935\\,{x}^{2}{y}^{16}{c}^{12} \\plus{} 962390964597878745 \\,{x}^{13}{y}^{5}{c}^{12} \\plus{} 962390964597878745\\,{x}^{5}{y}^{13}{c}^{12} \\plus{} 112916040343061310\\,{x}^{15}{y}^{3}{c}^{12}$\n$ \\plus{} 6196930766887320\\,{x}^{17} {y}^{2}{c}^{11} \\plus{} 32196892281454305\\,{x}^{3}{y}^{16}{c}^{11} \\plus{} 38568777876360\\,{y}^{19}{c}^{11} \\plus{} 6196930766887320\\,{x}^{2}{y}^{17}{c}^ {11} \\plus{} 1496578870705849605\\,{x}^{12}{y}^{7}{c}^{11} \\plus{} 3140506203143784840 \\,{x}^{9}{y}^{10}{c}^{11} \\plus{} 721091184063480\\,x{y}^{18}{c}^{11} \\plus{} 115583970861408465\\,{x}^{4}{y}^{15}{c}^{11} \\plus{} 32196892281454305\\,{x}^{16 }{y}^{3}{c}^{11} \\plus{} 1496578870705849605\\,{x}^{7}{y}^{12}{c}^{11} \\plus{} 38568777876360\\,{x}^{19}{c}^{11} \\plus{} 2440286663280994845\\,{x}^{8}{y}^{11}{ c}^{11} \\plus{} 751829705460213795\\,{x}^{6}{y}^{13}{c}^{11} \\plus{} 721091184063480\\,{ x}^{18}y{c}^{11} \\plus{} 321388973115663375\\,{x}^{5}{y}^{14}{c}^{11} \\plus{} 2440286663280994845\\,{x}^{11}{y}^{8}{c}^{11} \\plus{} 321388973115663375\\,{x}^{ 14}{y}^{5}{c}^{11} \\plus{} 751829705460213795\\,{x}^{13}{y}^{6}{c}^{11} \\plus{} 3140506203143784840\\,{x}^{10}{y}^{9}{c}^{11} \\plus{} 115583970861408465\\,{x}^{ 15}{y}^{4}{c}^{11}$\n$ \\plus{} 772772734181827170\\,{x}^{12}{y}^{8}{c}^{10} \\plus{} 449889121836280395\\,{x}^{7}{y}^{13}{c}^{10} \\plus{} 29842861410108315\\,{x}^{4} {y}^{16}{c}^{10} \\plus{} 1081313759617721910\\,{x}^{9}{y}^{11}{c}^{10} \\plus{} 7534683942204330\\,{x}^{17}{y}^{3}{c}^{10} \\plus{} 141198022534590\\,{x}^{19}y{c }^{10} \\plus{} 449889121836280395\\,{x}^{13}{y}^{7}{c}^{10} \\plus{} 1211445894374410257 \\,{x}^{10}{y}^{10}{c}^{10} \\plus{} 7093025578539\\,{x}^{20}{c}^{10} \\plus{} 7534683942204330\\,{x}^{3}{y}^{17}{c}^{10} \\plus{} 1315579009511205\\,{x}^{2}{y} ^{18}{c}^{10} \\plus{} 218583663720200910\\,{x}^{6}{y}^{14}{c}^{10} \\plus{} 89469060681737367\\,{x}^{15}{y}^{5}{c}^{10} \\plus{} 1081313759617721910\\,{x}^{ 11}{y}^{9}{c}^{10} \\plus{} 1315579009511205\\,{x}^{18}{y}^{2}{c}^{10} \\plus{} 772772734181827170\\,{x}^{8}{y}^{12}{c}^{10} \\plus{} 7093025578539\\,{y}^{20}{c} ^{10} \\plus{} 141198022534590\\,x{y}^{19}{c}^{10} \\plus{} 29842861410108315\\,{x}^{16}{y }^{4}{c}^{10} \\plus{} 218583663720200910\\,{x}^{14}{y}^{6}{c}^{10} \\plus{} 89469060681737367\\,{x}^{5}{y}^{15}{c}^{10}$\n$ \\plus{} 308183466115653975\\,{x}^{9} {y}^{12}{c}^{9} \\plus{} 1126430533800\\,{y}^{21}{c}^{9} \\plus{} 20503667384124705\\,{x}^ {5}{y}^{16}{c}^{9} \\plus{} 1458316098172575\\,{x}^{3}{y}^{18}{c}^{9} \\plus{} 6310720192829400\\,{x}^{17}{y}^{4}{c}^{9} \\plus{} 308183466115653975\\,{x}^{12}{ y}^{9}{c}^{9} \\plus{} 114414758085012750\\,{x}^{7}{y}^{14}{c}^{9} \\plus{} 1458316098172575\\,{x}^{18}{y}^{3}{c}^{9} \\plus{} 23637607287795\\,x{y}^{20}{c}^ {9} \\plus{} 206128996224470325\\,{x}^{13}{y}^{8}{c}^{9} \\plus{} 20503667384124705\\,{x}^ {16}{y}^{5}{c}^{9} \\plus{} 1126430533800\\,{x}^{21}{c}^{9} \\plus{} 234981359117550\\,{x} ^{2}{y}^{19}{c}^{9} \\plus{} 53176745518951770\\,{x}^{6}{y}^{15}{c}^{9} \\plus{} 378024510732571755\\,{x}^{10}{y}^{11}{c}^{9} \\plus{} 378024510732571755\\,{x}^{ 11}{y}^{10}{c}^{9} \\plus{} 206128996224470325\\,{x}^{8}{y}^{13}{c}^{9} \\plus{} 6310720192829400\\,{x}^{4}{y}^{17}{c}^{9} \\plus{} 234981359117550\\,{x}^{19}{y}^ {2}{c}^{9} \\plus{} 53176745518951770\\,{x}^{15}{y}^{6}{c}^{9} \\plus{} 23637607287795\\,{ x}^{20}y{c}^{9} \\plus{} 114414758085012750\\,{x}^{14}{y}^{7}{c}^{9}$\n$ \\plus{} 3845533111294590\\,{x}^{17}{y}^{5}{c}^{8} \\plus{} 153604163700\\,{x}^{22}{c}^{8} \\plus{} 1096068860697600\\,{x}^{18}{y}^{4}{c}^{8} \\plus{} 153604163700\\,{y}^{22}{c}^{8 } \\plus{} 10663945457377545\\,{x}^{16}{y}^{6}{c}^{8} \\plus{} 96280138209164220\\,{x}^{12 }{y}^{10}{c}^{8} \\plus{} 3379291601400\\,x{y}^{21}{c}^{8} \\plus{} 3845533111294590\\,{x} ^{5}{y}^{17}{c}^{8} \\plus{} 45931518331885125\\,{x}^{8}{y}^{14}{c}^{8} \\plus{} 24232232576657520\\,{x}^{7}{y}^{15}{c}^{8} \\plus{} 24232232576657520\\,{x}^{15}{ y}^{7}{c}^{8} \\plus{} 234807019897500\\,{x}^{3}{y}^{19}{c}^{8} \\plus{} 3379291601400\\,{ x}^{21}y{c}^{8} \\plus{} 10663945457377545\\,{x}^{6}{y}^{16}{c}^{8} \\plus{} 45931518331885125\\,{x}^{14}{y}^{8}{c}^{8} \\plus{} 234807019897500\\,{x}^{19}{y} ^{3}{c}^{8} \\plus{} 96280138209164220\\,{x}^{10}{y}^{12}{c}^{8} \\plus{} 72834792863369850\\,{x}^{13}{y}^{9}{c}^{8} \\plus{} 35447693970690\\,{x}^{2}{y}^{ 20}{c}^{8} \\plus{} 105713054588876040\\,{x}^{11}{y}^{11}{c}^{8} \\plus{} 1096068860697600\\,{x}^{4}{y}^{18}{c}^{8} \\plus{} 72834792863369850\\,{x}^{9}{y} ^{13}{c}^{8} \\plus{} 35447693970690\\,{x}^{20}{y}^{2}{c}^{8}$\n$ \\plus{} 14169936212586000 \\,{x}^{9}{y}^{14}{c}^{7} \\plus{} 588626351558550\\,{x}^{18}{y}^{5}{c}^{7} \\plus{} 23797038511922220\\,{x}^{12}{y}^{11}{c}^{7} \\plus{} 17809178400\\,{y}^{23}{c}^{7 } \\plus{} 1744644537673560\\,{x}^{17}{y}^{6}{c}^{7} \\plus{} 4505722135200\\,{x}^{21}{y}^ {2}{c}^{7} \\plus{} 1744644537673560\\,{x}^{6}{y}^{17}{c}^{7} \\plus{} 588626351558550\\,{ x}^{5}{y}^{18}{c}^{7} \\plus{} 4209610166857935\\,{x}^{16}{y}^{7}{c}^{7} \\plus{} 4209610166857935\\,{x}^{7}{y}^{16}{c}^{7} \\plus{} 31505187102390\\,{x}^{3}{y}^{ 20}{c}^{7} \\plus{} 8434858561754355\\,{x}^{8}{y}^{15}{c}^{7} \\plus{} 17809178400\\,{x}^{ 23}{c}^{7} \\plus{} 31505187102390\\,{x}^{20}{y}^{3}{c}^{7} \\plus{} 156654239411700\\,{x} ^{19}{y}^{4}{c}^{7} \\plus{} 409611103200\\,{x}^{22}y{c}^{7} \\plus{} 23797038511922220\\, {x}^{11}{y}^{12}{c}^{7} \\plus{} 4505722135200\\,{x}^{2}{y}^{21}{c}^{7} \\plus{} 20019537297068490\\,{x}^{10}{y}^{13}{c}^{7} \\plus{} 20019537297068490\\,{x}^{13} {y}^{10}{c}^{7} \\plus{} 8434858561754355\\,{x}^{15}{y}^{8}{c}^{7} \\plus{} 156654239411700\\,{x}^{4}{y}^{19}{c}^{7} \\plus{} 14169936212586000\\,{x}^{14}{y} ^{9}{c}^{7} \\plus{} 409611103200\\,x{y}^{22}{c}^{7}$\n$ \\plus{} 41554749600\\,{x}^{23}y{c}^{ 6} \\plus{} 4301057448335610\\,{x}^{13}{y}^{11}{c}^{6} \\plus{} 230971324829805\\,{x}^{18} {y}^{6}{c}^{6} \\plus{} 1731447900\\,{x}^{24}{c}^{6} \\plus{} 4301057448335610\\,{x}^{11}{ y}^{13}{c}^{6} \\plus{} 230971324829805\\,{x}^{6}{y}^{18}{c}^{6} \\plus{} 1256797756868130\\,{x}^{8}{y}^{16}{c}^{6} \\plus{} 3504450549600\\,{x}^{3}{y}^{21 }{c}^{6} \\plus{} 3369714001320915\\,{x}^{10}{y}^{14}{c}^{6} \\plus{} 1256797756868130\\,{ x}^{16}{y}^{8}{c}^{6} \\plus{} 3504450549600\\,{x}^{21}{y}^{3}{c}^{6} \\plus{} 591834193207470\\,{x}^{7}{y}^{17}{c}^{6} \\plus{} 73279650945510\\,{x}^{19}{y}^{5 }{c}^{6} \\plus{} 591834193207470\\,{x}^{17}{y}^{7}{c}^{6} \\plus{} 477879620400\\,{x}^{2} {y}^{22}{c}^{6} \\plus{} 477879620400\\,{x}^{22}{y}^{2}{c}^{6} \\plus{} 4664636437623390 \\,{x}^{12}{y}^{12}{c}^{6} \\plus{} 2238944546574450\\,{x}^{9}{y}^{15}{c}^{6} \\plus{} 2238944546574450\\,{x}^{15}{y}^{9}{c}^{6} \\plus{} 18380931463395\\,{x}^{20}{y}^{ 4}{c}^{6} \\plus{} 3369714001320915\\,{x}^{14}{y}^{10}{c}^{6} \\plus{} 18380931463395\\,{x }^{4}{y}^{20}{c}^{6} \\plus{} 1731447900\\,{y}^{24}{c}^{6} \\plus{} 41554749600\\,x{y}^{23 }{c}^{6} \\plus{} 73279650945510\\,{x}^{5}{y}^{19}{c}^{6}$\n$ \\plus{} 66427655144355\\,{x}^{7 }{y}^{18}{c}^{5} \\plus{} 41554749600\\,{x}^{23}{y}^{2}{c}^{5} \\plus{} 616653376988580\\, {x}^{11}{y}^{14}{c}^{5} \\plus{} 282252157155990\\,{x}^{16}{y}^{9}{c}^{5} \\plus{} 24496286003190\\,{x}^{19}{y}^{6}{c}^{5} \\plus{} 66427655144355\\,{x}^{18}{y}^{7} {c}^{5} \\plus{} 451892854554867\\,{x}^{10}{y}^{15}{c}^{5} \\plus{} 1752225274800\\,{x}^{ 21}{y}^{4}{c}^{5} \\plus{} 3462895800\\,{x}^{24}y{c}^{5} \\plus{} 616653376988580\\,{x}^{ 14}{y}^{11}{c}^{5} \\plus{} 318586413600\\,{x}^{22}{y}^{3}{c}^{5} \\plus{} 138515832\\,{x} ^{25}{c}^{5} \\plus{} 1752225274800\\,{x}^{4}{y}^{21}{c}^{5} \\plus{} 451892854554867\\,{x }^{15}{y}^{10}{c}^{5} \\plus{} 149396846867280\\,{x}^{17}{y}^{8}{c}^{5} \\plus{} 719748561723435\\,{x}^{12}{y}^{13}{c}^{5} \\plus{} 7355859369759\\,{x}^{20}{y}^{5 }{c}^{5} \\plus{} 719748561723435\\,{x}^{13}{y}^{12}{c}^{5} \\plus{} 3462895800\\,x{y}^{24 }{c}^{5} \\plus{} 149396846867280\\,{x}^{8}{y}^{17}{c}^{5} \\plus{} 318586413600\\,{x}^{3} {y}^{22}{c}^{5} \\plus{} 282252157155990\\,{x}^{9}{y}^{16}{c}^{5} \\plus{} 24496286003190 \\,{x}^{6}{y}^{19}{c}^{5} \\plus{} 138515832\\,{y}^{25}{c}^{5} \\plus{} 7355859369759\\,{x} ^{5}{y}^{20}{c}^{5} \\plus{} 41554749600\\,{x}^{2}{y}^{23}{c}^{5}$\n$ \\plus{} 8879220\\,{x}^{ 26}{c}^{4} \\plus{} 132744339000\\,{x}^{22}{y}^{4}{c}^{4} \\plus{} 68602274395200\\,{x}^{ 15}{y}^{11}{c}^{4} \\plus{} 13871783425500\\,{x}^{8}{y}^{18}{c}^{4} \\plus{} 2885746500\\, {x}^{2}{y}^{24}{c}^{4} \\plus{} 13871783425500\\,{x}^{18}{y}^{8}{c}^{4} \\plus{} 230859720\\,x{y}^{25}{c}^{4} \\plus{} 2885746500\\,{x}^{24}{y}^{2}{c}^{4} \\plus{} 47164063646700\\,{x}^{10}{y}^{16}{c}^{4} \\plus{} 5840750916000\\,{x}^{7}{y}^{19} {c}^{4} \\plus{} 584075091600\\,{x}^{21}{y}^{5}{c}^{4} \\plus{} 132744339000\\,{x}^{4}{y}^ {22}{c}^{4} \\plus{} 23085972000\\,{x}^{3}{y}^{23}{c}^{4} \\plus{} 8879220\\,{y}^{26}{c}^{ 4} \\plus{} 2044262820600\\,{x}^{20}{y}^{6}{c}^{4} \\plus{} 27743566851000\\,{x}^{17}{y}^{ 9}{c}^{4} \\plus{} 2044262820600\\,{x}^{6}{y}^{20}{c}^{4} \\plus{} 47164063646700\\,{x}^{ 16}{y}^{10}{c}^{4} \\plus{} 584075091600\\,{x}^{5}{y}^{21}{c}^{4} \\plus{} 92349215532000 \\,{x}^{13}{y}^{13}{c}^{4} \\plus{} 27743566851000\\,{x}^{9}{y}^{17}{c}^{4} \\plus{} 23085972000\\,{x}^{23}{y}^{3}{c}^{4} \\plus{} 85752842994000\\,{x}^{14}{y}^{12}{c }^{4} \\plus{} 68602274395200\\,{x}^{11}{y}^{15}{c}^{4} \\plus{} 85752842994000\\,{x}^{12} {y}^{14}{c}^{4} \\plus{} 230859720\\,{x}^{25}y{c}^{4} \\plus{} 5840750916000\\,{x}^{19}{y} ^{7}{c}^{4}$\n$ \\plus{} 438480\\,{y}^{27}{c}^{3} \\plus{} 973458486000\\,{x}^{19}{y}^{8}{c}^{ 3} \\plus{} 11838960\\,{x}^{26}y{c}^{3} \\plus{} 5716856199600\\,{x}^{16}{y}^{11}{c}^{3} \\plus{} 389383394400\\,{x}^{7}{y}^{20}{c}^{3} \\plus{} 35398490400\\,{x}^{5}{y}^{22}{c}^{ 3} \\plus{} 438480\\,{x}^{27}{c}^{3} \\plus{} 7695324000\\,{x}^{4}{y}^{23}{c}^{3} \\plus{} 1282554000\\,{x}^{24}{y}^{3}{c}^{3} \\plus{} 1282554000\\,{x}^{3}{y}^{24}{c}^{3} \\plus{} 153906480\\,{x}^{2}{y}^{25}{c}^{3} \\plus{} 2055079026000\\,{x}^{18}{y}^{9}{c}^{3 } \\plus{} 7695324000\\,{x}^{23}{y}^{4}{c}^{3} \\plus{} 973458486000\\,{x}^{8}{y}^{19}{c}^ {3} \\plus{} 35398490400\\,{x}^{22}{y}^{5}{c}^{3} \\plus{} 389383394400\\,{x}^{20}{y}^{7}{ c}^{3} \\plus{} 8795163384000\\,{x}^{13}{y}^{14}{c}^{3} \\plus{} 5716856199600\\,{x}^{11}{ y}^{16}{c}^{3} \\plus{} 7622474932800\\,{x}^{15}{y}^{12}{c}^{3} \\plus{} 11838960\\,x{y}^{ 26}{c}^{3} \\plus{} 7622474932800\\,{x}^{12}{y}^{15}{c}^{3} \\plus{} 8795163384000\\,{x}^{ 14}{y}^{13}{c}^{3} \\plus{} 129794464800\\,{x}^{21}{y}^{6}{c}^{3} \\plus{} 2055079026000 \\,{x}^{9}{y}^{18}{c}^{3} \\plus{} 153906480\\,{x}^{25}{y}^{2}{c}^{3} \\plus{} 3699142246800\\,{x}^{10}{y}^{17}{c}^{3} \\plus{} 129794464800\\,{x}^{6}{y}^{21}{c }^{3} \\plus{} 3699142246800\\,{x}^{17}{y}^{10}{c}^{3}$\n$ \\plus{} 476404683300\\,{x}^{16}{y} ^{12}{c}^{2} \\plus{} 438480\\,{x}^{27}y{c}^{2} \\plus{} 15660\\,{y}^{28}{c}^{2} \\plus{} 1539064800\\,{x}^{23}{y}^{5}{c}^{2} \\plus{} 320638500\\,{x}^{4}{y}^{24}{c}^{2} \\plus{} 108162054000\\,{x}^{9}{y}^{19}{c}^{2} \\plus{} 1539064800\\,{x}^{5}{y}^{23}{c}^{2 } \\plus{} 320638500\\,{x}^{24}{y}^{4}{c}^{2} \\plus{} 15660\\,{x}^{28}{c}^{2} \\plus{} 108162054000\\,{x}^{19}{y}^{9}{c}^{2} \\plus{} 5919480\\,{x}^{2}{y}^{26}{c}^{2} \\plus{} 48672924300\\,{x}^{20}{y}^{8}{c}^{2} \\plus{} 51302160\\,{x}^{3}{y}^{25}{c}^{2} \\plus{} 438480\\,x{y}^{27}{c}^{2} \\plus{} 336285658800\\,{x}^{11}{y}^{17}{c}^{2} \\plus{} 5919480 \\,{x}^{26}{y}^{2}{c}^{2} \\plus{} 5899748400\\,{x}^{6}{y}^{22}{c}^{2} \\plus{} 586344225600\\,{x}^{15}{y}^{13}{c}^{2} \\plus{} 5899748400\\,{x}^{22}{y}^{6}{c}^{ 2} \\plus{} 628225956000\\,{x}^{14}{y}^{14}{c}^{2} \\plus{} 205507902600\\,{x}^{18}{y}^{10 }{c}^{2} \\plus{} 205507902600\\,{x}^{10}{y}^{18}{c}^{2} \\plus{} 586344225600\\,{x}^{13}{ y}^{15}{c}^{2} \\plus{} 18542066400\\,{x}^{21}{y}^{7}{c}^{2} \\plus{} 48672924300\\,{x}^{8 }{y}^{20}{c}^{2} \\plus{} 336285658800\\,{x}^{17}{y}^{11}{c}^{2} \\plus{} 476404683300\\,{ x}^{12}{y}^{16}{c}^{2} \\plus{} 18542066400\\,{x}^{7}{y}^{21}{c}^{2} \\plus{} 51302160\\,{ x}^{25}{y}^{3}{c}^{2}$\n$ \\plus{} 1315440\\,{x}^{3}{y}^{26}c \\plus{} 360\\,{y}^{29}c \\plus{} 171007200\\,{x}^{23}{y}^{6}c \\plus{} 7210803600\\,{x}^{19}{y}^{10}c \\plus{} 1545172200\\, {x}^{8}{y}^{21}c \\plus{} 3605401800\\,{x}^{9}{y}^{20}c \\plus{} 27921153600\\,{x}^{14}{y} ^{15}c \\plus{} 1545172200\\,{x}^{21}{y}^{8}c \\plus{} 7210803600\\,{x}^{10}{y}^{19}c \\plus{} 24431009400\\,{x}^{13}{y}^{16}c \\plus{} 146160\\,{x}^{27}{y}^{2}c \\plus{} 42751800\\,{x}^ {5}{y}^{24}c \\plus{} 561880800\\,{x}^{22}{y}^{7}c \\plus{} 146160\\,{x}^{2}{y}^{27}c \\plus{} 10440\\,{x}^{28}yc \\plus{} 3605401800\\,{x}^{20}{y}^{9}c \\plus{} 27921153600\\,{x}^{15}{y }^{14}c \\plus{} 171007200\\,{x}^{6}{y}^{23}c \\plus{} 12455024400\\,{x}^{11}{y}^{18}c \\plus{} 12455024400\\,{x}^{18}{y}^{11}c \\plus{} 1315440\\,{x}^{26}{y}^{3}c \\plus{} 18682536600\\, {x}^{12}{y}^{17}c \\plus{} 8550360\\,{x}^{4}{y}^{25}c \\plus{} 360\\,{x}^{29}c \\plus{} 8550360\\,{x }^{25}{y}^{4}c \\plus{} 10440\\,x{y}^{28}c \\plus{} 18682536600\\,{x}^{17}{y}^{12}c \\plus{} 42751800\\,{x}^{24}{y}^{5}c \\plus{} 24431009400\\,{x}^{16}{y}^{13}c \\plus{} 561880800\\,{ x}^{7}{y}^{22}c$\n$ \\plus{} 109620\\,{x}^{4}{y}^{26} \\plus{} 1740\\,{x}^{2}{y}^{28} \\plus{} 16240\\,{ x}^{27}{y}^{3} \\plus{} 4\\,{y}^{30} \\plus{} 109620\\,{x}^{26}{y}^{4} \\plus{} 16240\\,{x}^{3}{y}^{ 27} \\plus{} 570024\\,{x}^{25}{y}^{5} \\plus{} 120\\,x{y}^{29} \\plus{} 2375100\\,{x}^{24}{y}^{6} \\plus{} 4 \\,{x}^{30} \\plus{} 8143200\\,{x}^{23}{y}^{7} \\plus{} 120\\,{x}^{29}y \\plus{} 23411700\\,{x}^{22}{ y}^{8} \\plus{} 1740\\,{x}^{28}{y}^{2} \\plus{} 57228600\\,{x}^{21}{y}^{9} \\plus{} 345972900\\,{x}^ {12}{y}^{18} \\plus{} 120180060\\,{x}^{20}{y}^{10} \\plus{} 218509200\\,{x}^{11}{y}^{19} \\plus{} 218509200\\,{x}^{19}{y}^{11} \\plus{} 120180060\\,{x}^{10}{y}^{20} \\plus{} 345972900\\,{x} ^{18}{y}^{12} \\plus{} 57228600\\,{x}^{9}{y}^{21} \\plus{} 479039400\\,{x}^{17}{y}^{13} \\plus{} 23411700\\,{x}^{8}{y}^{22} \\plus{} 581690700\\,{x}^{16}{y}^{14} \\plus{} 8143200\\,{x}^{7} {y}^{23} \\plus{} 620470080\\,{x}^{15}{y}^{15} \\plus{} 2375100\\,{x}^{6}{y}^{24} \\plus{} 581690700\\,{x}^{14}{y}^{16} \\plus{} 570024\\,{x}^{5}{y}^{25} \\plus{} 479039400\\,{x}^{13 }{y}^{17} \\geq 0$\n\nDone ! \n\nPS : It's just for fun  :lol:\n\nPS : Do you believe i do it by hand ???  :rotfl:[/quote]\r\nOh  :o ,Your solutions very frightful.Thank you very much :lol:",
  "Solution_4": "[quote=\"nguoivn\"]Given $ a, b, c$ and $ a \\plus{} b \\plus{} c \\equal{} 3$. Prove that: $ \\frac {1}{2 \\plus{} a^3b^3} \\plus{} \\frac {1}{2 \\plus{} b^3c^3} \\plus{} \\frac {1}{2 \\plus{} c^3a^3} \\geq\\ 1$\n\nWith this ineq, we can solve it easily by Am-Gm. But the following (of him) is harder and I still have no solution:  :oops: \n$ \\frac {1}{2 \\plus{} a^5b^5} \\plus{} \\frac {1}{2 \\plus{} b^5c^5} \\plus{} \\frac {1}{2 \\plus{} c^5a^5} \\geq\\ 1$\n(with the same condition)[/quote]\r\nLet $ a \\plus{} b \\plus{} c \\equal{} 3u,$ $ ab \\plus{} ac \\plus{} bc \\equal{} 3v^2$ and $ abc \\equal{} w^3 \\equal{} x.$\r\nHence, $ \\sum_{cyc}\\frac {1}{2 \\plus{} a^5b^5}\\leq1\\Leftrightarrow(abc)^{10} \\plus{} (a^5 \\plus{} b^5 \\plus{} c^5)(abc)^5\\leq4\\Leftrightarrow$\r\n$ \\Leftrightarrow w^{30} \\plus{} (243u^5 \\minus{} 405u^3v^2 \\plus{} 135uv^4 \\plus{} 15(3u^2 \\minus{} v^2)w^3)u^{10}w^{15} \\minus{} 4u^{30}\\leq0$\r\n$ \\Leftrightarrow w^{30} \\plus{} 15(3u^2 \\minus{} v^2)u^{10}w^{18} \\plus{} 27(9u^4 \\minus{} 15u^2v^2 \\plus{} 5v^4)u^{11}w^{15} \\minus{} 4u^{30}\\leq0$\r\nLet $ f(x) \\equal{} x^{10} \\plus{} 15(3u^2 \\minus{} v^2)u^{10}x^6 \\plus{} 27(9u^4 \\minus{} 15u^2v^2 \\plus{} 5v^4)u^{11}x^5 \\minus{} 4u^{30}.$\r\nWe obtain: $ f''(x) \\equal{} 90x^3(x^5 \\plus{} 5(3u^2 \\minus{} v^2)u^{10}x \\plus{} 6(9u^4 \\minus{} 15u^2v^2 \\plus{} 5v^4)u^{11})\\geq0$ because \r\n$ g(x) \\equal{} x^5 \\plus{} 5(3u^2 \\minus{} v^2)u^{10}x \\plus{} 6(9u^4 \\minus{} 15u^2v^2 \\plus{} 5v^4)u^{11}$\r\n is an increasing function of $ x$ and we need to check only \r\n$ 9u^4 \\minus{} 15u^2v^2 \\plus{} 5v^4\\geq0$ for $ w^3 \\equal{} 0,$ which is obviously true.\r\nThus, $ f$ is a convex function.\r\nHence, $ f$ gets the maximal value on bounds of $ w^3.$\r\nId est, it remains to check the original inequality for $ b \\equal{} c$ and for $ c \\equal{} 0,$ which are true.",
  "Solution_5": "The case $ b\\equal{}c$ is not easy to check  :maybe: . Hope that we'll find the nicer proofs for it  :)",
  "Solution_6": "I am sure that no one has enough free time to check these stuffs  :D  properly. If this problem is proposed in IMO, it would be a nice surprise. :P",
  "Solution_7": "By Holder ,the inequality is equivalent with : \r\n\r\n$ a^3 (2\\plus{}a^3 b^3 ) \\plus{} b^3 (2\\plus{}b^3 c^3 ) \\plus{} c^3(2 \\plus{}c^3 a^3) \\geq 9$ \r\n\r\nIs this true ?  :blush:",
  "Solution_8": "[quote=\"enndb0x\"]By Holder ,the inequality is equivalent with : \n\n$ a^3 (2 \\plus{} a^3 b^3 ) \\plus{} b^3 (2 \\plus{} b^3 c^3 ) \\plus{} c^3(2 \\plus{} c^3 a^3) \\geq 9$ \n\nIs this true ?  :blush:[/quote]\r\n\r\nYou shouldn't say \"is equivalent with\" but rather \"would follow from\",  that's different  :!:",
  "Solution_9": "[quote=\"Materazzi\"][quote=\"nguoivn\"]\nWith this ineq, we can solve it easily by Am-Gm. But the following (of him) is harder and I still have no solution:  :oops: \n$ \\frac {1}{2 \\plus{} a^5b^5} \\plus{} \\frac {1}{2 \\plus{} b^5c^5} \\plus{} \\frac {1}{2 \\plus{} c^5a^5} \\geq\\ 1$\n(with the same condition)[/quote]\n\nMaybe ..... I have a very ugly solution for this  :blush: \n\nAssume : $ c \\equal{} min \\{a;b;c\\}$\n\nWith : $ a \\equal{} x \\plus{} c \\ ; \\ b \\equal{} y \\plus{} c \\ ; \\ x,y \\geq 0$ .\n\nThis inequality is equivalent to : \n\n$ \\minus{} 343151886824415\\,xy{c}^{28} \\plus{} 343151886824415\\,{x}^{2}{c}^{28} \\plus{} 343151886824415\\,{y}^{2}{c}^{28}$\n$ \\plus{} 1258223585022855\\,x{y}^{2}{c}^{27} \\plus{} 2363935220345970\\,{x}^{3}{c}^{27} \\plus{} 1258223585022855\\,{x}^{2}y{c}^{27} \\plus{} 2363935220345970\\,{y}^{3}{c}^{27}$\n$ \\plus{} 23677480190884635\\,{x}^{2}{y}^{2}{c} ^{26} \\plus{} 7206189623312715\\,{y}^{4}{c}^{26} \\plus{} 7206189623312715\\,{x}^{4}{c}^{ 26} \\plus{} 17500746228045165\\,{x}^{3}y{c}^{26} \\plus{} 17500746228045165\\,x{y}^{3}{c} ^{26}$\n$ \\plus{} 160099419197302065\\,{x}^{3}{y}^{2}{c}^{25} \\plus{} 12887259749628030\\,{y }^{5}{c}^{25} \\plus{} 61462315728995220\\,{x}^{4}y{c}^{25} \\plus{} 12887259749628030\\,{ x}^{5}{c}^{25} \\plus{} 160099419197302065\\,{x}^{2}{y}^{3}{c}^{25} \\plus{} 61462315728995220\\,x{y}^{4}{c}^{25}$\n$ \\plus{} 16267941301305600\\,{y}^{6}{c}^{24} \\plus{} 520637668287487425\\,{x}^{2}{y}^{4}{c}^{24} \\plus{} 16267941301305600\\,{x}^{6} {c}^{24} \\plus{} 849237373244718900\\,{x}^{3}{y}^{3}{c}^{24} \\plus{} 112286922966433575 \\,{x}^{5}y{c}^{24} \\plus{} 520637668287487425\\,{x}^{4}{y}^{2}{c}^{24} \\plus{} 112286922966433575\\,x{y}^{5}{c}^{24}$\n$ \\plus{} 20462019918048450\\,{x}^{7}{c}^{23 } \\plus{} 2583870161142136725\\,{x}^{3}{y}^{4}{c}^{23} \\plus{} 123598225902497625\\,{x}^ {6}y{c}^{23} \\plus{} 2583870161142136725\\,{x}^{4}{y}^{3}{c}^{23} \\plus{} 123598225902497625\\,x{y}^{6}{c}^{23} \\plus{} 1010963586572151525\\,{x}^{5}{y}^{ 2}{c}^{23} \\plus{} 20462019918048450\\,{y}^{7}{c}^{23} \\plus{} 1010963586572151525\\,{x} ^{2}{y}^{5}{c}^{23}$\n$ \\plus{} 100227181758694875\\,x{y}^{7}{c}^{22} \\plus{} 33771511824715575\\,{y}^{8}{c}^{22} \\plus{} 5052078366356891925\\,{x}^{3}{y}^{5} {c}^{22} \\plus{} 1300256160789639825\\,{x}^{2}{y}^{6}{c}^{22} \\plus{} 5052078366356891925\\,{x}^{5}{y}^{3}{c}^{22} \\plus{} 100227181758694875\\,{x}^{7 }y{c}^{22} \\plus{} 1300256160789639825\\,{x}^{6}{y}^{2}{c}^{22} \\plus{} 7610819359490631675\\,{x}^{4}{y}^{4}{c}^{22} \\plus{} 33771511824715575\\,{x}^{8} {c}^{22 }$\n$ \\plus{} 6871482379629081900\\,{x}^{3}{y}^{6}{c}^{21} \\plus{} 56307035256428700 \\,{y}^{9}{c}^{21} \\plus{} 14738655968645105250\\,{x}^{4}{y}^{5}{c}^{21} \\plus{} 118104971417942175\\,x{y}^{8}{c}^{21} \\plus{} 6871482379629081900\\,{x}^{6}{y}^{ 3}{c}^{21} \\plus{} 1211700379626022275\\,{x}^{2}{y}^{7}{c}^{21} \\plus{} 14738655968645105250\\,{x}^{5}{y}^{4}{c}^{21} \\plus{} 1211700379626022275\\,{x}^ {7}{y}^{2}{c}^{21} \\plus{} 56307035256428700\\,{x}^{9}{c}^{21} \\plus{} 118104971417942175\\,{x}^{8}y{c}^{21}$\n$ \\plus{} 6922333817672485950\\,{x}^{3}{y}^{ 7}{c}^{20} \\plus{} 74579755550855265\\,{x}^{10}{c}^{20} \\plus{} 6922333817672485950\\,{x }^{7}{y}^{3}{c}^{20} \\plus{} 218325092438225025\\,x{y}^{9}{c}^{20} \\plus{} 20205326773079281125\\,{x}^{6}{y}^{4}{c}^{20} \\plus{} 218325092438225025\\,{x}^{ 9}y{c}^{20} \\plus{} 20205326773079281125\\,{x}^{4}{y}^{6}{c}^{20} \\plus{} 984690578941006500\\,{x}^{2}{y}^{8}{c}^{20} \\plus{} 74579755550855265\\,{y}^{10} {c}^{20} \\plus{} 28195374156236728335\\,{x}^{5}{y}^{5}{c}^{20} \\plus{} 984690578941006500\\,{x}^{8}{y}^{2}{c}^{20}$\n$ \\plus{} 332431506674710425\\,{x}^{10 }y{c}^{19} \\plus{} 20703523584131542575\\,{x}^{4}{y}^{7}{c}^{19} \\plus{} 929452678951534425\\,{x}^{2}{y}^{9}{c}^{19} \\plus{} 332431506674710425\\,x{y}^{ 10}{c}^{19} \\plus{} 20703523584131542575\\,{x}^{7}{y}^{4}{c}^{19} \\plus{} 929452678951534425\\,{x}^{9}{y}^{2}{c}^{19} \\plus{} 5587211207586606525\\,{x}^{8 }{y}^{3}{c}^{19} \\plus{} 5587211207586606525\\,{x}^{3}{y}^{8}{c}^{19} \\plus{} 75157463424334950\\,{y}^{11}{c}^{19} \\plus{} 38253020811296949225\\,{x}^{6}{y}^{ 5}{c}^{19} \\plus{} 75157463424334950\\,{x}^{11}{c}^{19} \\plus{} 38253020811296949225\\,{ x}^{5}{y}^{6}{c}^{19}$\n$ \\plus{} 1009513729316449710\\,{x}^{10}{y}^{2}{c}^{18} \\plus{} 58564007553966660\\,{y}^{12}{c}^{18} \\plus{} 38837506435766674065\\,{x}^{7}{y}^{ 5}{c}^{18} \\plus{} 4071659563211697225\\,{x}^{9}{y}^{3}{c}^{18} \\plus{} 58564007553966660\\,{x}^{12}{c}^{18} \\plus{} 16701086221823724750\\,{x}^{8}{y}^{ 4}{c}^{18} \\plus{} 362611857207382065\\,x{y}^{11}{c}^{18} \\plus{} 50955813923280508710 \\,{x}^{6}{y}^{6}{c}^{18} \\plus{} 4071659563211697225\\,{x}^{3}{y}^{9}{c}^{18} \\plus{} 38837506435766674065\\,{x}^{5}{y}^{7}{c}^{18} \\plus{} 16701086221823724750\\,{x} ^{4}{y}^{8}{c}^{18} \\plus{} 362611857207382065\\,{x}^{11}y{c}^{18} \\plus{} 1009513729316449710\\,{x}^{2}{y}^{10}{c}^{18}$\n$ \\plus{} 977917657147362045\\,{x}^{ 11}{y}^{2}{c}^{17} \\plus{} 2984439815827840980\\,{x}^{10}{y}^{3}{c}^{17} \\plus{} 2984439815827840980\\,{x}^{3}{y}^{10}{c}^{17} \\plus{} 36130164566694030\\,{y}^{ 13}{c}^{17} \\plus{} 30825593044084271745\\,{x}^{8}{y}^{5}{c}^{17} \\plus{} 11308185750809556000\\,{x}^{4}{y}^{9}{c}^{17} \\plus{} 292229998302188745\\,{x}^{ 12}y{c}^{17} \\plus{} 30825593044084271745\\,{x}^{5}{y}^{8}{c}^{17} \\plus{} 11308185750809556000\\,{x}^{9}{y}^{4}{c}^{17} \\plus{} 50567705251517488230\\,{x} ^{6}{y}^{7}{c}^{17} \\plus{} 36130164566694030\\,{x}^{13}{c}^{17} \\plus{} 50567705251517488230\\,{x}^{7}{y}^{6}{c}^{17} \\plus{} 292229998302188745\\,x{y}^ {12}{c}^{17} \\plus{} 977917657147362045\\,{x}^{2}{y}^{11}{c}^{17}$\n$ \\plus{} 2160360452109822660\\,{x}^{3}{y}^{11}{c}^{16} \\plus{} 2160360452109822660\\,{x}^ {11}{y}^{3}{c}^{16} \\plus{} 38992364468350626915\\,{x}^{8}{y}^{6}{c}^{16} \\plus{} 748135554864262515\\,{x}^{12}{y}^{2}{c}^{16} \\plus{} 180949529841796485\\,x{y}^{ 13}{c}^{16} \\plus{} 6906323613662794935\\,{x}^{4}{y}^{10}{c}^{16} \\plus{} 6906323613662794935\\,{x}^{10}{y}^{4}{c}^{16} \\plus{} 18022409853586110\\,{y}^{ 14}{c}^{16} \\plus{} 18022409853586110\\,{x}^{14}{c}^{16} \\plus{} 748135554864262515\\,{x }^{2}{y}^{12}{c}^{16} \\plus{} 19994508314853588315\\,{x}^{5}{y}^{9}{c}^{16} \\plus{} 48631289652679821660\\,{x}^{7}{y}^{7}{c}^{16} \\plus{} 180949529841796485\\,{x}^{ 13}y{c}^{16} \\plus{} 38992364468350626915\\,{x}^{6}{y}^{8}{c}^{16} \\plus{} 19994508314853588315\\,{x}^{9}{y}^{5}{c}^{16}$\n$ \\plus{} 88711037987481855\\,{x}^{ 14}y{c}^{15} \\plus{} 36070087439992712715\\,{x}^{8}{y}^{7}{c}^{15} \\plus{} 88711037987481855\\,x{y}^{14}{c}^{15} \\plus{} 1393142007744442485\\,{x}^{3}{y}^{ 12}{c}^{15} \\plus{} 3960181901596608855\\,{x}^{4}{y}^{11}{c}^{15} \\plus{} 36070087439992712715\\,{x}^{7}{y}^{8}{c}^{15} \\plus{} 445465963285976835\\,{x}^{ 2}{y}^{13}{c}^{15} \\plus{} 24236016130783711395\\,{x}^{6}{y}^{9}{c}^{15} \\plus{} 11119352295374912205\\,{x}^{5}{y}^{10}{c}^{15} \\plus{} 3960181901596608855\\,{x} ^{11}{y}^{4}{c}^{15} \\plus{} 1393142007744442485\\,{x}^{12}{y}^{3}{c}^{15} \\plus{} 7395765445494270\\,{x}^{15}{c}^{15} \\plus{} 445465963285976835\\,{x}^{13}{y}^{2} {c}^{15} \\plus{} 24236016130783711395\\,{x}^{9}{y}^{6}{c}^{15} \\plus{} 7395765445494270 \\,{y}^{15}{c}^{15} \\plus{} 11119352295374912205\\,{x}^{10}{y}^{5}{c}^{15}$\n$ \\plus{} 2084305487797867050\\,{x}^{12}{y}^{4}{c}^{14} \\plus{} 209424571608036825\\,{x}^{ 14}{y}^{2}{c}^{14} \\plus{} 21349556612200860075\\,{x}^{7}{y}^{9}{c}^{14} \\plus{} 744857579775620475\\,{x}^{3}{y}^{13}{c}^{14} \\plus{} 12585125502138452670\\,{x}^ {10}{y}^{6}{c}^{14} \\plus{} 2084305487797867050\\,{x}^{4}{y}^{12}{c}^{14} \\plus{} 744857579775620475\\,{x}^{13}{y}^{3}{c}^{14} \\plus{} 35203478647936185\\,x{y}^{ 15}{c}^{14} \\plus{} 35203478647936185\\,{x}^{15}y{c}^{14} \\plus{} 21349556612200860075 \\,{x}^{9}{y}^{7}{c}^{14} \\plus{} 2533095274158855\\,{y}^{16}{c}^{14} \\plus{} 12585125502138452670\\,{x}^{6}{y}^{10}{c}^{14} \\plus{} 2533095274158855\\,{x}^{ 16}{c}^{14} \\plus{} 209424571608036825\\,{x}^{2}{y}^{14}{c}^{14} \\plus{} 25510966047864850275\\,{x}^{8}{y}^{8}{c}^{14} \\plus{} 5511886560598653405\\,{x}^ {5}{y}^{11}{c}^{14} \\plus{} 5511886560598653405\\,{x}^{11}{y}^{5}{c}^{14}$\n$ \\plus{} 322423177403419200\\,{x}^{14}{y}^{3}{c}^{13} \\plus{} 733199301973440\\,{y}^{17}{ c}^{13} \\plus{} 2456477683392122055\\,{x}^{12}{y}^{5}{c}^{13} \\plus{} 11499472852337745 \\,x{y}^{16}{c}^{13} \\plus{} 2456477683392122055\\,{x}^{5}{y}^{12}{c}^{13} \\plus{} 955075836063775050\\,{x}^{13}{y}^{4}{c}^{13} \\plus{} 79098393960389025\\,{x}^{2} {y}^{15}{c}^{13} \\plus{} 10388985093352033110\\,{x}^{7}{y}^{10}{c}^{13} \\plus{} 5632415391716311230\\,{x}^{6}{y}^{11}{c}^{13} \\plus{} 955075836063775050\\,{x}^{ 4}{y}^{13}{c}^{13} \\plus{} 10388985093352033110\\,{x}^{10}{y}^{7}{c}^{13} \\plus{} 14264412913360975275\\,{x}^{9}{y}^{8}{c}^{13} \\plus{} 733199301973440\\,{x}^{17} {c}^{13} \\plus{} 5632415391716311230\\,{x}^{11}{y}^{6}{c}^{13} \\plus{} 322423177403419200\\,{x}^{3}{y}^{14}{c}^{13} \\plus{} 14264412913360975275\\,{x}^ {8}{y}^{9}{c}^{13} \\plus{} 11499472852337745\\,{x}^{16}y{c}^{13} \\plus{} 79098393960389025\\,{x}^{15}{y}^{2}{c}^{13}$\n$ \\plus{} 2206418360572908315\\,{x}^{ 12}{y}^{6}{c}^{12} \\plus{} 24362825601034935\\,{x}^{16}{y}^{2}{c}^{12} \\plus{} 6477654862591944300\\,{x}^{10}{y}^{8}{c}^{12} \\plus{} 2206418360572908315\\,{x}^ {6}{y}^{12}{c}^{12} \\plus{} 112916040343061310\\,{x}^{3}{y}^{15}{c}^{12} \\plus{} 7466933476884696750\\,{x}^{9}{y}^{9}{c}^{12} \\plus{} 3134553601994010\\,x{y}^{17 }{c}^{12} \\plus{} 6477654862591944300\\,{x}^{8}{y}^{10}{c}^{12} \\plus{} 366353221781733975\\,{x}^{14}{y}^{4}{c}^{12} \\plus{} 4262032766997513990\\,{x}^{ 7}{y}^{11}{c}^{12} \\plus{} 366353221781733975\\,{x}^{4}{y}^{14}{c}^{12} \\plus{} 181249095648150\\,{y}^{18}{c}^{12} \\plus{} 3134553601994010\\,{x}^{17}y{c}^{12} \\plus{} 4262032766997513990\\,{x}^{11}{y}^{7}{c}^{12} \\plus{} 181249095648150\\,{x}^{18} {c}^{12} \\plus{} 24362825601034935\\,{x}^{2}{y}^{16}{c}^{12} \\plus{} 962390964597878745 \\,{x}^{13}{y}^{5}{c}^{12} \\plus{} 962390964597878745\\,{x}^{5}{y}^{13}{c}^{12} \\plus{} 112916040343061310\\,{x}^{15}{y}^{3}{c}^{12}$\n$ \\plus{} 6196930766887320\\,{x}^{17} {y}^{2}{c}^{11} \\plus{} 32196892281454305\\,{x}^{3}{y}^{16}{c}^{11} \\plus{} 38568777876360\\,{y}^{19}{c}^{11} \\plus{} 6196930766887320\\,{x}^{2}{y}^{17}{c}^ {11} \\plus{} 1496578870705849605\\,{x}^{12}{y}^{7}{c}^{11} \\plus{} 3140506203143784840 \\,{x}^{9}{y}^{10}{c}^{11} \\plus{} 721091184063480\\,x{y}^{18}{c}^{11} \\plus{} 115583970861408465\\,{x}^{4}{y}^{15}{c}^{11} \\plus{} 32196892281454305\\,{x}^{16 }{y}^{3}{c}^{11} \\plus{} 1496578870705849605\\,{x}^{7}{y}^{12}{c}^{11} \\plus{} 38568777876360\\,{x}^{19}{c}^{11} \\plus{} 2440286663280994845\\,{x}^{8}{y}^{11}{ c}^{11} \\plus{} 751829705460213795\\,{x}^{6}{y}^{13}{c}^{11} \\plus{} 721091184063480\\,{ x}^{18}y{c}^{11} \\plus{} 321388973115663375\\,{x}^{5}{y}^{14}{c}^{11} \\plus{} 2440286663280994845\\,{x}^{11}{y}^{8}{c}^{11} \\plus{} 321388973115663375\\,{x}^{ 14}{y}^{5}{c}^{11} \\plus{} 751829705460213795\\,{x}^{13}{y}^{6}{c}^{11} \\plus{} 3140506203143784840\\,{x}^{10}{y}^{9}{c}^{11} \\plus{} 115583970861408465\\,{x}^{ 15}{y}^{4}{c}^{11}$\n$ \\plus{} 772772734181827170\\,{x}^{12}{y}^{8}{c}^{10} \\plus{} 449889121836280395\\,{x}^{7}{y}^{13}{c}^{10} \\plus{} 29842861410108315\\,{x}^{4} {y}^{16}{c}^{10} \\plus{} 1081313759617721910\\,{x}^{9}{y}^{11}{c}^{10} \\plus{} 7534683942204330\\,{x}^{17}{y}^{3}{c}^{10} \\plus{} 141198022534590\\,{x}^{19}y{c }^{10} \\plus{} 449889121836280395\\,{x}^{13}{y}^{7}{c}^{10} \\plus{} 1211445894374410257 \\,{x}^{10}{y}^{10}{c}^{10} \\plus{} 7093025578539\\,{x}^{20}{c}^{10} \\plus{} 7534683942204330\\,{x}^{3}{y}^{17}{c}^{10} \\plus{} 1315579009511205\\,{x}^{2}{y} ^{18}{c}^{10} \\plus{} 218583663720200910\\,{x}^{6}{y}^{14}{c}^{10} \\plus{} 89469060681737367\\,{x}^{15}{y}^{5}{c}^{10} \\plus{} 1081313759617721910\\,{x}^{ 11}{y}^{9}{c}^{10} \\plus{} 1315579009511205\\,{x}^{18}{y}^{2}{c}^{10} \\plus{} 772772734181827170\\,{x}^{8}{y}^{12}{c}^{10} \\plus{} 7093025578539\\,{y}^{20}{c} ^{10} \\plus{} 141198022534590\\,x{y}^{19}{c}^{10} \\plus{} 29842861410108315\\,{x}^{16}{y }^{4}{c}^{10} \\plus{} 218583663720200910\\,{x}^{14}{y}^{6}{c}^{10} \\plus{} 89469060681737367\\,{x}^{5}{y}^{15}{c}^{10}$\n$ \\plus{} 308183466115653975\\,{x}^{9} {y}^{12}{c}^{9} \\plus{} 1126430533800\\,{y}^{21}{c}^{9} \\plus{} 20503667384124705\\,{x}^ {5}{y}^{16}{c}^{9} \\plus{} 1458316098172575\\,{x}^{3}{y}^{18}{c}^{9} \\plus{} 6310720192829400\\,{x}^{17}{y}^{4}{c}^{9} \\plus{} 308183466115653975\\,{x}^{12}{ y}^{9}{c}^{9} \\plus{} 114414758085012750\\,{x}^{7}{y}^{14}{c}^{9} \\plus{} 1458316098172575\\,{x}^{18}{y}^{3}{c}^{9} \\plus{} 23637607287795\\,x{y}^{20}{c}^ {9} \\plus{} 206128996224470325\\,{x}^{13}{y}^{8}{c}^{9} \\plus{} 20503667384124705\\,{x}^ {16}{y}^{5}{c}^{9} \\plus{} 1126430533800\\,{x}^{21}{c}^{9} \\plus{} 234981359117550\\,{x} ^{2}{y}^{19}{c}^{9} \\plus{} 53176745518951770\\,{x}^{6}{y}^{15}{c}^{9} \\plus{} 378024510732571755\\,{x}^{10}{y}^{11}{c}^{9} \\plus{} 378024510732571755\\,{x}^{ 11}{y}^{10}{c}^{9} \\plus{} 206128996224470325\\,{x}^{8}{y}^{13}{c}^{9} \\plus{} 6310720192829400\\,{x}^{4}{y}^{17}{c}^{9} \\plus{} 234981359117550\\,{x}^{19}{y}^ {2}{c}^{9} \\plus{} 53176745518951770\\,{x}^{15}{y}^{6}{c}^{9} \\plus{} 23637607287795\\,{ x}^{20}y{c}^{9} \\plus{} 114414758085012750\\,{x}^{14}{y}^{7}{c}^{9}$\n$ \\plus{} 3845533111294590\\,{x}^{17}{y}^{5}{c}^{8} \\plus{} 153604163700\\,{x}^{22}{c}^{8} \\plus{} 1096068860697600\\,{x}^{18}{y}^{4}{c}^{8} \\plus{} 153604163700\\,{y}^{22}{c}^{8 } \\plus{} 10663945457377545\\,{x}^{16}{y}^{6}{c}^{8} \\plus{} 96280138209164220\\,{x}^{12 }{y}^{10}{c}^{8} \\plus{} 3379291601400\\,x{y}^{21}{c}^{8} \\plus{} 3845533111294590\\,{x} ^{5}{y}^{17}{c}^{8} \\plus{} 45931518331885125\\,{x}^{8}{y}^{14}{c}^{8} \\plus{} 24232232576657520\\,{x}^{7}{y}^{15}{c}^{8} \\plus{} 24232232576657520\\,{x}^{15}{ y}^{7}{c}^{8} \\plus{} 234807019897500\\,{x}^{3}{y}^{19}{c}^{8} \\plus{} 3379291601400\\,{ x}^{21}y{c}^{8} \\plus{} 10663945457377545\\,{x}^{6}{y}^{16}{c}^{8} \\plus{} 45931518331885125\\,{x}^{14}{y}^{8}{c}^{8} \\plus{} 234807019897500\\,{x}^{19}{y} ^{3}{c}^{8} \\plus{} 96280138209164220\\,{x}^{10}{y}^{12}{c}^{8} \\plus{} 72834792863369850\\,{x}^{13}{y}^{9}{c}^{8} \\plus{} 35447693970690\\,{x}^{2}{y}^{ 20}{c}^{8} \\plus{} 105713054588876040\\,{x}^{11}{y}^{11}{c}^{8} \\plus{} 1096068860697600\\,{x}^{4}{y}^{18}{c}^{8} \\plus{} 72834792863369850\\,{x}^{9}{y} ^{13}{c}^{8} \\plus{} 35447693970690\\,{x}^{20}{y}^{2}{c}^{8}$\n$ \\plus{} 14169936212586000 \\,{x}^{9}{y}^{14}{c}^{7} \\plus{} 588626351558550\\,{x}^{18}{y}^{5}{c}^{7} \\plus{} 23797038511922220\\,{x}^{12}{y}^{11}{c}^{7} \\plus{} 17809178400\\,{y}^{23}{c}^{7 } \\plus{} 1744644537673560\\,{x}^{17}{y}^{6}{c}^{7} \\plus{} 4505722135200\\,{x}^{21}{y}^ {2}{c}^{7} \\plus{} 1744644537673560\\,{x}^{6}{y}^{17}{c}^{7} \\plus{} 588626351558550\\,{ x}^{5}{y}^{18}{c}^{7} \\plus{} 4209610166857935\\,{x}^{16}{y}^{7}{c}^{7} \\plus{} 4209610166857935\\,{x}^{7}{y}^{16}{c}^{7} \\plus{} 31505187102390\\,{x}^{3}{y}^{ 20}{c}^{7} \\plus{} 8434858561754355\\,{x}^{8}{y}^{15}{c}^{7} \\plus{} 17809178400\\,{x}^{ 23}{c}^{7} \\plus{} 31505187102390\\,{x}^{20}{y}^{3}{c}^{7} \\plus{} 156654239411700\\,{x} ^{19}{y}^{4}{c}^{7} \\plus{} 409611103200\\,{x}^{22}y{c}^{7} \\plus{} 23797038511922220\\, {x}^{11}{y}^{12}{c}^{7} \\plus{} 4505722135200\\,{x}^{2}{y}^{21}{c}^{7} \\plus{} 20019537297068490\\,{x}^{10}{y}^{13}{c}^{7} \\plus{} 20019537297068490\\,{x}^{13} {y}^{10}{c}^{7} \\plus{} 8434858561754355\\,{x}^{15}{y}^{8}{c}^{7} \\plus{} 156654239411700\\,{x}^{4}{y}^{19}{c}^{7} \\plus{} 14169936212586000\\,{x}^{14}{y} ^{9}{c}^{7} \\plus{} 409611103200\\,x{y}^{22}{c}^{7}$\n$ \\plus{} 41554749600\\,{x}^{23}y{c}^{ 6} \\plus{} 4301057448335610\\,{x}^{13}{y}^{11}{c}^{6} \\plus{} 230971324829805\\,{x}^{18} {y}^{6}{c}^{6} \\plus{} 1731447900\\,{x}^{24}{c}^{6} \\plus{} 4301057448335610\\,{x}^{11}{ y}^{13}{c}^{6} \\plus{} 230971324829805\\,{x}^{6}{y}^{18}{c}^{6} \\plus{} 1256797756868130\\,{x}^{8}{y}^{16}{c}^{6} \\plus{} 3504450549600\\,{x}^{3}{y}^{21 }{c}^{6} \\plus{} 3369714001320915\\,{x}^{10}{y}^{14}{c}^{6} \\plus{} 1256797756868130\\,{ x}^{16}{y}^{8}{c}^{6} \\plus{} 3504450549600\\,{x}^{21}{y}^{3}{c}^{6} \\plus{} 591834193207470\\,{x}^{7}{y}^{17}{c}^{6} \\plus{} 73279650945510\\,{x}^{19}{y}^{5 }{c}^{6} \\plus{} 591834193207470\\,{x}^{17}{y}^{7}{c}^{6} \\plus{} 477879620400\\,{x}^{2} {y}^{22}{c}^{6} \\plus{} 477879620400\\,{x}^{22}{y}^{2}{c}^{6} \\plus{} 4664636437623390 \\,{x}^{12}{y}^{12}{c}^{6} \\plus{} 2238944546574450\\,{x}^{9}{y}^{15}{c}^{6} \\plus{} 2238944546574450\\,{x}^{15}{y}^{9}{c}^{6} \\plus{} 18380931463395\\,{x}^{20}{y}^{ 4}{c}^{6} \\plus{} 3369714001320915\\,{x}^{14}{y}^{10}{c}^{6} \\plus{} 18380931463395\\,{x }^{4}{y}^{20}{c}^{6} \\plus{} 1731447900\\,{y}^{24}{c}^{6} \\plus{} 41554749600\\,x{y}^{23 }{c}^{6} \\plus{} 73279650945510\\,{x}^{5}{y}^{19}{c}^{6}$\n$ \\plus{} 66427655144355\\,{x}^{7 }{y}^{18}{c}^{5} \\plus{} 41554749600\\,{x}^{23}{y}^{2}{c}^{5} \\plus{} 616653376988580\\, {x}^{11}{y}^{14}{c}^{5} \\plus{} 282252157155990\\,{x}^{16}{y}^{9}{c}^{5} \\plus{} 24496286003190\\,{x}^{19}{y}^{6}{c}^{5} \\plus{} 66427655144355\\,{x}^{18}{y}^{7} {c}^{5} \\plus{} 451892854554867\\,{x}^{10}{y}^{15}{c}^{5} \\plus{} 1752225274800\\,{x}^{ 21}{y}^{4}{c}^{5} \\plus{} 3462895800\\,{x}^{24}y{c}^{5} \\plus{} 616653376988580\\,{x}^{ 14}{y}^{11}{c}^{5} \\plus{} 318586413600\\,{x}^{22}{y}^{3}{c}^{5} \\plus{} 138515832\\,{x} ^{25}{c}^{5} \\plus{} 1752225274800\\,{x}^{4}{y}^{21}{c}^{5} \\plus{} 451892854554867\\,{x }^{15}{y}^{10}{c}^{5} \\plus{} 149396846867280\\,{x}^{17}{y}^{8}{c}^{5} \\plus{} 719748561723435\\,{x}^{12}{y}^{13}{c}^{5} \\plus{} 7355859369759\\,{x}^{20}{y}^{5 }{c}^{5} \\plus{} 719748561723435\\,{x}^{13}{y}^{12}{c}^{5} \\plus{} 3462895800\\,x{y}^{24 }{c}^{5} \\plus{} 149396846867280\\,{x}^{8}{y}^{17}{c}^{5} \\plus{} 318586413600\\,{x}^{3} {y}^{22}{c}^{5} \\plus{} 282252157155990\\,{x}^{9}{y}^{16}{c}^{5} \\plus{} 24496286003190 \\,{x}^{6}{y}^{19}{c}^{5} \\plus{} 138515832\\,{y}^{25}{c}^{5} \\plus{} 7355859369759\\,{x} ^{5}{y}^{20}{c}^{5} \\plus{} 41554749600\\,{x}^{2}{y}^{23}{c}^{5}$\n$ \\plus{} 8879220\\,{x}^{ 26}{c}^{4} \\plus{} 132744339000\\,{x}^{22}{y}^{4}{c}^{4} \\plus{} 68602274395200\\,{x}^{ 15}{y}^{11}{c}^{4} \\plus{} 13871783425500\\,{x}^{8}{y}^{18}{c}^{4} \\plus{} 2885746500\\, {x}^{2}{y}^{24}{c}^{4} \\plus{} 13871783425500\\,{x}^{18}{y}^{8}{c}^{4} \\plus{} 230859720\\,x{y}^{25}{c}^{4} \\plus{} 2885746500\\,{x}^{24}{y}^{2}{c}^{4} \\plus{} 47164063646700\\,{x}^{10}{y}^{16}{c}^{4} \\plus{} 5840750916000\\,{x}^{7}{y}^{19} {c}^{4} \\plus{} 584075091600\\,{x}^{21}{y}^{5}{c}^{4} \\plus{} 132744339000\\,{x}^{4}{y}^ {22}{c}^{4} \\plus{} 23085972000\\,{x}^{3}{y}^{23}{c}^{4} \\plus{} 8879220\\,{y}^{26}{c}^{ 4} \\plus{} 2044262820600\\,{x}^{20}{y}^{6}{c}^{4} \\plus{} 27743566851000\\,{x}^{17}{y}^{ 9}{c}^{4} \\plus{} 2044262820600\\,{x}^{6}{y}^{20}{c}^{4} \\plus{} 47164063646700\\,{x}^{ 16}{y}^{10}{c}^{4} \\plus{} 584075091600\\,{x}^{5}{y}^{21}{c}^{4} \\plus{} 92349215532000 \\,{x}^{13}{y}^{13}{c}^{4} \\plus{} 27743566851000\\,{x}^{9}{y}^{17}{c}^{4} \\plus{} 23085972000\\,{x}^{23}{y}^{3}{c}^{4} \\plus{} 85752842994000\\,{x}^{14}{y}^{12}{c }^{4} \\plus{} 68602274395200\\,{x}^{11}{y}^{15}{c}^{4} \\plus{} 85752842994000\\,{x}^{12} {y}^{14}{c}^{4} \\plus{} 230859720\\,{x}^{25}y{c}^{4} \\plus{} 5840750916000\\,{x}^{19}{y} ^{7}{c}^{4}$\n$ \\plus{} 438480\\,{y}^{27}{c}^{3} \\plus{} 973458486000\\,{x}^{19}{y}^{8}{c}^{ 3} \\plus{} 11838960\\,{x}^{26}y{c}^{3} \\plus{} 5716856199600\\,{x}^{16}{y}^{11}{c}^{3} \\plus{} 389383394400\\,{x}^{7}{y}^{20}{c}^{3} \\plus{} 35398490400\\,{x}^{5}{y}^{22}{c}^{ 3} \\plus{} 438480\\,{x}^{27}{c}^{3} \\plus{} 7695324000\\,{x}^{4}{y}^{23}{c}^{3} \\plus{} 1282554000\\,{x}^{24}{y}^{3}{c}^{3} \\plus{} 1282554000\\,{x}^{3}{y}^{24}{c}^{3} \\plus{} 153906480\\,{x}^{2}{y}^{25}{c}^{3} \\plus{} 2055079026000\\,{x}^{18}{y}^{9}{c}^{3 } \\plus{} 7695324000\\,{x}^{23}{y}^{4}{c}^{3} \\plus{} 973458486000\\,{x}^{8}{y}^{19}{c}^ {3} \\plus{} 35398490400\\,{x}^{22}{y}^{5}{c}^{3} \\plus{} 389383394400\\,{x}^{20}{y}^{7}{ c}^{3} \\plus{} 8795163384000\\,{x}^{13}{y}^{14}{c}^{3} \\plus{} 5716856199600\\,{x}^{11}{ y}^{16}{c}^{3} \\plus{} 7622474932800\\,{x}^{15}{y}^{12}{c}^{3} \\plus{} 11838960\\,x{y}^{ 26}{c}^{3} \\plus{} 7622474932800\\,{x}^{12}{y}^{15}{c}^{3} \\plus{} 8795163384000\\,{x}^{ 14}{y}^{13}{c}^{3} \\plus{} 129794464800\\,{x}^{21}{y}^{6}{c}^{3} \\plus{} 2055079026000 \\,{x}^{9}{y}^{18}{c}^{3} \\plus{} 153906480\\,{x}^{25}{y}^{2}{c}^{3} \\plus{} 3699142246800\\,{x}^{10}{y}^{17}{c}^{3} \\plus{} 129794464800\\,{x}^{6}{y}^{21}{c }^{3} \\plus{} 3699142246800\\,{x}^{17}{y}^{10}{c}^{3}$\n$ \\plus{} 476404683300\\,{x}^{16}{y} ^{12}{c}^{2} \\plus{} 438480\\,{x}^{27}y{c}^{2} \\plus{} 15660\\,{y}^{28}{c}^{2} \\plus{} 1539064800\\,{x}^{23}{y}^{5}{c}^{2} \\plus{} 320638500\\,{x}^{4}{y}^{24}{c}^{2} \\plus{} 108162054000\\,{x}^{9}{y}^{19}{c}^{2} \\plus{} 1539064800\\,{x}^{5}{y}^{23}{c}^{2 } \\plus{} 320638500\\,{x}^{24}{y}^{4}{c}^{2} \\plus{} 15660\\,{x}^{28}{c}^{2} \\plus{} 108162054000\\,{x}^{19}{y}^{9}{c}^{2} \\plus{} 5919480\\,{x}^{2}{y}^{26}{c}^{2} \\plus{} 48672924300\\,{x}^{20}{y}^{8}{c}^{2} \\plus{} 51302160\\,{x}^{3}{y}^{25}{c}^{2} \\plus{} 438480\\,x{y}^{27}{c}^{2} \\plus{} 336285658800\\,{x}^{11}{y}^{17}{c}^{2} \\plus{} 5919480 \\,{x}^{26}{y}^{2}{c}^{2} \\plus{} 5899748400\\,{x}^{6}{y}^{22}{c}^{2} \\plus{} 586344225600\\,{x}^{15}{y}^{13}{c}^{2} \\plus{} 5899748400\\,{x}^{22}{y}^{6}{c}^{ 2} \\plus{} 628225956000\\,{x}^{14}{y}^{14}{c}^{2} \\plus{} 205507902600\\,{x}^{18}{y}^{10 }{c}^{2} \\plus{} 205507902600\\,{x}^{10}{y}^{18}{c}^{2} \\plus{} 586344225600\\,{x}^{13}{ y}^{15}{c}^{2} \\plus{} 18542066400\\,{x}^{21}{y}^{7}{c}^{2} \\plus{} 48672924300\\,{x}^{8 }{y}^{20}{c}^{2} \\plus{} 336285658800\\,{x}^{17}{y}^{11}{c}^{2} \\plus{} 476404683300\\,{ x}^{12}{y}^{16}{c}^{2} \\plus{} 18542066400\\,{x}^{7}{y}^{21}{c}^{2} \\plus{} 51302160\\,{ x}^{25}{y}^{3}{c}^{2}$\n$ \\plus{} 1315440\\,{x}^{3}{y}^{26}c \\plus{} 360\\,{y}^{29}c \\plus{} 171007200\\,{x}^{23}{y}^{6}c \\plus{} 7210803600\\,{x}^{19}{y}^{10}c \\plus{} 1545172200\\, {x}^{8}{y}^{21}c \\plus{} 3605401800\\,{x}^{9}{y}^{20}c \\plus{} 27921153600\\,{x}^{14}{y} ^{15}c \\plus{} 1545172200\\,{x}^{21}{y}^{8}c \\plus{} 7210803600\\,{x}^{10}{y}^{19}c \\plus{} 24431009400\\,{x}^{13}{y}^{16}c \\plus{} 146160\\,{x}^{27}{y}^{2}c \\plus{} 42751800\\,{x}^ {5}{y}^{24}c \\plus{} 561880800\\,{x}^{22}{y}^{7}c \\plus{} 146160\\,{x}^{2}{y}^{27}c \\plus{} 10440\\,{x}^{28}yc \\plus{} 3605401800\\,{x}^{20}{y}^{9}c \\plus{} 27921153600\\,{x}^{15}{y }^{14}c \\plus{} 171007200\\,{x}^{6}{y}^{23}c \\plus{} 12455024400\\,{x}^{11}{y}^{18}c \\plus{} 12455024400\\,{x}^{18}{y}^{11}c \\plus{} 1315440\\,{x}^{26}{y}^{3}c \\plus{} 18682536600\\, {x}^{12}{y}^{17}c \\plus{} 8550360\\,{x}^{4}{y}^{25}c \\plus{} 360\\,{x}^{29}c \\plus{} 8550360\\,{x }^{25}{y}^{4}c \\plus{} 10440\\,x{y}^{28}c \\plus{} 18682536600\\,{x}^{17}{y}^{12}c \\plus{} 42751800\\,{x}^{24}{y}^{5}c \\plus{} 24431009400\\,{x}^{16}{y}^{13}c \\plus{} 561880800\\,{ x}^{7}{y}^{22}c$\n$ \\plus{} 109620\\,{x}^{4}{y}^{26} \\plus{} 1740\\,{x}^{2}{y}^{28} \\plus{} 16240\\,{ x}^{27}{y}^{3} \\plus{} 4\\,{y}^{30} \\plus{} 109620\\,{x}^{26}{y}^{4} \\plus{} 16240\\,{x}^{3}{y}^{ 27} \\plus{} 570024\\,{x}^{25}{y}^{5} \\plus{} 120\\,x{y}^{29} \\plus{} 2375100\\,{x}^{24}{y}^{6} \\plus{} 4 \\,{x}^{30} \\plus{} 8143200\\,{x}^{23}{y}^{7} \\plus{} 120\\,{x}^{29}y \\plus{} 23411700\\,{x}^{22}{ y}^{8} \\plus{} 1740\\,{x}^{28}{y}^{2} \\plus{} 57228600\\,{x}^{21}{y}^{9} \\plus{} 345972900\\,{x}^ {12}{y}^{18} \\plus{} 120180060\\,{x}^{20}{y}^{10} \\plus{} 218509200\\,{x}^{11}{y}^{19} \\plus{} 218509200\\,{x}^{19}{y}^{11} \\plus{} 120180060\\,{x}^{10}{y}^{20} \\plus{} 345972900\\,{x} ^{18}{y}^{12} \\plus{} 57228600\\,{x}^{9}{y}^{21} \\plus{} 479039400\\,{x}^{17}{y}^{13} \\plus{} 23411700\\,{x}^{8}{y}^{22} \\plus{} 581690700\\,{x}^{16}{y}^{14} \\plus{} 8143200\\,{x}^{7} {y}^{23} \\plus{} 620470080\\,{x}^{15}{y}^{15} \\plus{} 2375100\\,{x}^{6}{y}^{24} \\plus{} 581690700\\,{x}^{14}{y}^{16} \\plus{} 570024\\,{x}^{5}{y}^{25} \\plus{} 479039400\\,{x}^{13 }{y}^{17} \\geq 0$\n\nDone ! \n\nPS : It's just for fun  :lol:\n\nPS : Do you believe i do it by hand ???  :rotfl:[/quote]\r\nI can't believe that you did it by hand. Blah blah.... :D",
  "Solution_10": "I think the stronger also trues: $ a^5b^5c^5(a^5\\plus{}b^5\\plus{}c^5) \\leq\\ 3$\r\nWith $ a, b, c > 0$ and $ a\\plus{}b\\plus{}c\\equal{}3$.\r\nThe general problem: Find all $ k$ for this ineq is trues (with the same condition):\r\n$ a^kb^kc^k(a^k\\plus{}b^k\\plus{}c^k) \\leq\\ 3$",
  "Solution_11": "[quote=\"nguoivn\"]I think the stronger also trues: $ a^5b^5c^5(a^5 \\plus{} b^5 \\plus{} c^5) \\leq\\ 3$\nWith $ a, b, c > 0$ and $ a \\plus{} b \\plus{} c \\equal{} 3$.\n[/quote]\r\n\r\nYou can try $ a\\equal{}1 \\ ; \\ b\\equal{}0,9 \\ ; \\ c\\equal{}1,1$\r\n\r\n~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`\r\n\r\nM\u1ea1c Th\u1ecb Thanh Hoa",
  "Solution_12": "Thank you, Blueflower. Maybe the following is trues:\r\n$ a^4b^4c^4(a^4\\plus{}b^4\\plus{}c^4) \\leq\\ 3$  :maybe: \r\nWith $ a^3b^3c^3(a^3\\plus{}b^3\\plus{}c^3) \\leq\\ 3$, I had had solution by Am-Gm.",
  "Solution_13": "[quote=\"nguoivn\"]Thank you, Blueflower. Maybe the following is trues:\n$ a^4b^4c^4(a^4 \\plus{} b^4 \\plus{} c^4) \\leq\\ 3$  :maybe: \n[/quote]\r\n\r\nTry $ a\\equal{}2 \\ ; \\ b\\equal{}c\\equal{}1$\r\n\r\nP/S : [hide]nguoivn : anh c\u00f3 bi\u1ebft anh Zi ( Materazzi ) nh\u00e0 em \u0111i \u0111\u00e2u r\u1ed3i ko \u1ea1 :(([/hide]\r\n\r\n~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\r\n\r\nM\u1ea1c Th\u1ecb Thanh Hoa",
  "Solution_14": "With the problem \r\n\r\n$ (abc)^3(a^3\\plus{}b^3\\plus{}c^3) \\le 3$ then we can actually prove by Mixing Varable although it seems to us a little bit messy",
  "Solution_15": "[quote=\"Blueflower\"][quote=\"nguoivn\"]Thank you, Blueflower. Maybe the following is trues:\n$ a^4b^4c^4(a^4 \\plus{} b^4 \\plus{} c^4) \\leq\\ 3$  :maybe: \n[/quote]\n\nTry $ a \\equal{} 2 \\ ; \\ b \\equal{} c \\equal{} 1$\n\nP/S : [hide]nguoivn : anh c\u00f3 bi\u1ebft anh Zi ( Materazzi ) nh\u00e0 em \u0111i \u0111\u00e2u r\u1ed3i ko \u1ea1 :(([/hide]\n\n~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\n\nM\u1ea1c Th\u1ecb Thanh Hoa[/quote]\r\nDear my friend,but $ a\\plus{}b\\plus{}c\\equal{}3$\r\n:)Ch\u1ecb l\u00e0 em g\u00e1i hay b\u1ea1n g\u00e1i anh materazzi :)",
  "Solution_16": "[quote=\"Blueflower\"][quote=\"nguoivn\"]Thank you, Blueflower. Maybe the following is trues:\n$ a^4b^4c^4(a^4 \\plus{} b^4 \\plus{} c^4) \\leq\\ 3$  :maybe: \n[/quote]\n\nTry $ a \\equal{} 2 \\ ; \\ b \\equal{} c \\equal{} 1$\n\nP/S : [hide]nguoivn : anh c\u00f3 bi\u1ebft anh Zi ( Materazzi ) nh\u00e0 em \u0111i \u0111\u00e2u r\u1ed3i ko \u1ea1 :(([/hide]\n\n~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\n\nM\u1ea1c Th\u1ecb Thanh Hoa[/quote]\n\nBut $ a\\plus{}b\\plus{}c\\equal{}3$, Blueflower. I think this ineq is trues.\nFrom it, we have a nice result:\nGiven $ a, b, c > 0$ and $ abc \\equal{} 1$. Prove that: $ \\frac{a\\plus{}b\\plus{}c}{3} \\geq\\ \\sqrt[16]{\\frac{a^4\\plus{}b^4\\plus{}c^4}{3}}$\nA similar result and well-known (with the same condition): $ \\frac{a\\plus{}b\\plus{}c}{3} \\geq\\ \\sqrt[10]{\\frac{a^3\\plus{}b^3\\plus{}c^3}{3}}$\n[hide]@Blueflower: Ec, sao anh bit dc. Ku Thang' len day ma` con` bi nguoi` yeu duoi?. Po tay ong em nay` lun:| [/hide]",
  "Solution_17": "[quote=\"nguoivn\"]\nFrom it, we have a nice result:\nGiven $ a, b, c > 0$ and $ abc \\equal{} 1$. Prove that: $ \\frac {a \\plus{} b \\plus{} c}{3} \\geq\\ \\sqrt [16]{\\frac {a^4 \\plus{} b^4 \\plus{} c^4}{3}}$[/quote]\r\nwhich is wrong. Try $ a\\equal{}1.44$ and $ b\\equal{}c\\equal{}\\frac{5}{6}.$",
  "Solution_18": "[quote=\"nguoivn\"]\n@Blueflower: Ec, sao anh bit dc. Ku Thang' len day ma` con` bi nguoi` yeu duoi?. Po tay ong em nay` lun:| [/quote]\r\n\r\n[hide]Ai m\u00e0 th\u00e8m y\u00eau anh \u1ea5y ch\u1ee9  :blush: \n\n~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`\n\nM\u1ea1c Th\u1ecb Thanh Hoa[/hide]",
  "Solution_19": "By the way, the following inequality is true.\r\nLet $ a,$ $ b$ and $ c$ are positive numbers such that $ abc\\equal{}1.$ Prove that\r\n\\[ \\frac {a \\plus{} b \\plus{} c}{3} \\geq\\ \\sqrt [17]{\\frac {a^4 \\plus{} b^4 \\plus{} c^4}{3}}\\]",
  "Solution_20": "[quote=\"arqady\"]By the way, the following inequality is true.\nLet $ a,$ $ b$ and $ c$ are positive numbers such that $ abc \\equal{} 1.$ Prove that\n\\[ \\frac {a \\plus{} b \\plus{} c}{3} \\geq\\ \\sqrt [17]{\\frac {a^4 \\plus{} b^4 \\plus{} c^4}{3}}\\]\n[/quote]\r\nDid you have solution for it, arqady? \r\nI think with $ abc\\equal{}1$, this ineq is trues: $ \\sqrt[10]{\\frac{a^3\\plus{}b^3\\plus{}c^3}{3}} \\geq\\ \\sqrt[17]{\\frac{a^4\\plus{}b^4\\plus{}c^4}{3}}$  :maybe:",
  "Solution_21": "[quote=\"Materazzi\"][quote=\"nguoivn\"]\nWith this ineq, we can solve it easily by Am-Gm. But the following (of him) is harder and I still have no solution:  :oops: \n$ \\frac {1}{2 \\plus{} a^5b^5} \\plus{} \\frac {1}{2 \\plus{} b^5c^5} \\plus{} \\frac {1}{2 \\plus{} c^5a^5} \\geq\\ 1$\n(with the same condition)[/quote]\n\nMaybe ..... I have a very ugly solution for this  :blush: \n\nAssume : $ c \\equal{} min \\{a;b;c\\}$\n\nWith : $ a \\equal{} x \\plus{} c \\ ; \\ b \\equal{} y \\plus{} c \\ ; \\ x,y \\geq 0$ .\n\nThis inequality is equivalent to : \n\n$ \\minus{} 343151886824415\\,xy{c}^{28} \\plus{} 343151886824415\\,{x}^{2}{c}^{28} \\plus{} 343151886824415\\,{y}^{2}{c}^{28}$\n$ \\plus{} 1258223585022855\\,x{y}^{2}{c}^{27} \\plus{} 2363935220345970\\,{x}^{3}{c}^{27} \\plus{} 1258223585022855\\,{x}^{2}y{c}^{27} \\plus{} 2363935220345970\\,{y}^{3}{c}^{27}$\n$ \\plus{} 23677480190884635\\,{x}^{2}{y}^{2}{c} ^{26} \\plus{} 7206189623312715\\,{y}^{4}{c}^{26} \\plus{} 7206189623312715\\,{x}^{4}{c}^{ 26} \\plus{} 17500746228045165\\,{x}^{3}y{c}^{26} \\plus{} 17500746228045165\\,x{y}^{3}{c} ^{26}$\n$ \\plus{} 160099419197302065\\,{x}^{3}{y}^{2}{c}^{25} \\plus{} 12887259749628030\\,{y }^{5}{c}^{25} \\plus{} 61462315728995220\\,{x}^{4}y{c}^{25} \\plus{} 12887259749628030\\,{ x}^{5}{c}^{25} \\plus{} 160099419197302065\\,{x}^{2}{y}^{3}{c}^{25} \\plus{} 61462315728995220\\,x{y}^{4}{c}^{25}$\n$ \\plus{} 16267941301305600\\,{y}^{6}{c}^{24} \\plus{} 520637668287487425\\,{x}^{2}{y}^{4}{c}^{24} \\plus{} 16267941301305600\\,{x}^{6} {c}^{24} \\plus{} 849237373244718900\\,{x}^{3}{y}^{3}{c}^{24} \\plus{} 112286922966433575 \\,{x}^{5}y{c}^{24} \\plus{} 520637668287487425\\,{x}^{4}{y}^{2}{c}^{24} \\plus{} 112286922966433575\\,x{y}^{5}{c}^{24}$\n$ \\plus{} 20462019918048450\\,{x}^{7}{c}^{23 } \\plus{} 2583870161142136725\\,{x}^{3}{y}^{4}{c}^{23} \\plus{} 123598225902497625\\,{x}^ {6}y{c}^{23} \\plus{} 2583870161142136725\\,{x}^{4}{y}^{3}{c}^{23} \\plus{} 123598225902497625\\,x{y}^{6}{c}^{23} \\plus{} 1010963586572151525\\,{x}^{5}{y}^{ 2}{c}^{23} \\plus{} 20462019918048450\\,{y}^{7}{c}^{23} \\plus{} 1010963586572151525\\,{x} ^{2}{y}^{5}{c}^{23}$\n$ \\plus{} 100227181758694875\\,x{y}^{7}{c}^{22} \\plus{} 33771511824715575\\,{y}^{8}{c}^{22} \\plus{} 5052078366356891925\\,{x}^{3}{y}^{5} {c}^{22} \\plus{} 1300256160789639825\\,{x}^{2}{y}^{6}{c}^{22} \\plus{} 5052078366356891925\\,{x}^{5}{y}^{3}{c}^{22} \\plus{} 100227181758694875\\,{x}^{7 }y{c}^{22} \\plus{} 1300256160789639825\\,{x}^{6}{y}^{2}{c}^{22} \\plus{} 7610819359490631675\\,{x}^{4}{y}^{4}{c}^{22} \\plus{} 33771511824715575\\,{x}^{8} {c}^{22 }$\n$ \\plus{} 6871482379629081900\\,{x}^{3}{y}^{6}{c}^{21} \\plus{} 56307035256428700 \\,{y}^{9}{c}^{21} \\plus{} 14738655968645105250\\,{x}^{4}{y}^{5}{c}^{21} \\plus{} 118104971417942175\\,x{y}^{8}{c}^{21} \\plus{} 6871482379629081900\\,{x}^{6}{y}^{ 3}{c}^{21} \\plus{} 1211700379626022275\\,{x}^{2}{y}^{7}{c}^{21} \\plus{} 14738655968645105250\\,{x}^{5}{y}^{4}{c}^{21} \\plus{} 1211700379626022275\\,{x}^ {7}{y}^{2}{c}^{21} \\plus{} 56307035256428700\\,{x}^{9}{c}^{21} \\plus{} 118104971417942175\\,{x}^{8}y{c}^{21}$\n$ \\plus{} 6922333817672485950\\,{x}^{3}{y}^{ 7}{c}^{20} \\plus{} 74579755550855265\\,{x}^{10}{c}^{20} \\plus{} 6922333817672485950\\,{x }^{7}{y}^{3}{c}^{20} \\plus{} 218325092438225025\\,x{y}^{9}{c}^{20} \\plus{} 20205326773079281125\\,{x}^{6}{y}^{4}{c}^{20} \\plus{} 218325092438225025\\,{x}^{ 9}y{c}^{20} \\plus{} 20205326773079281125\\,{x}^{4}{y}^{6}{c}^{20} \\plus{} 984690578941006500\\,{x}^{2}{y}^{8}{c}^{20} \\plus{} 74579755550855265\\,{y}^{10} {c}^{20} \\plus{} 28195374156236728335\\,{x}^{5}{y}^{5}{c}^{20} \\plus{} 984690578941006500\\,{x}^{8}{y}^{2}{c}^{20}$\n$ \\plus{} 332431506674710425\\,{x}^{10 }y{c}^{19} \\plus{} 20703523584131542575\\,{x}^{4}{y}^{7}{c}^{19} \\plus{} 929452678951534425\\,{x}^{2}{y}^{9}{c}^{19} \\plus{} 332431506674710425\\,x{y}^{ 10}{c}^{19} \\plus{} 20703523584131542575\\,{x}^{7}{y}^{4}{c}^{19} \\plus{} 929452678951534425\\,{x}^{9}{y}^{2}{c}^{19} \\plus{} 5587211207586606525\\,{x}^{8 }{y}^{3}{c}^{19} \\plus{} 5587211207586606525\\,{x}^{3}{y}^{8}{c}^{19} \\plus{} 75157463424334950\\,{y}^{11}{c}^{19} \\plus{} 38253020811296949225\\,{x}^{6}{y}^{ 5}{c}^{19} \\plus{} 75157463424334950\\,{x}^{11}{c}^{19} \\plus{} 38253020811296949225\\,{ x}^{5}{y}^{6}{c}^{19}$\n$ \\plus{} 1009513729316449710\\,{x}^{10}{y}^{2}{c}^{18} \\plus{} 58564007553966660\\,{y}^{12}{c}^{18} \\plus{} 38837506435766674065\\,{x}^{7}{y}^{ 5}{c}^{18} \\plus{} 4071659563211697225\\,{x}^{9}{y}^{3}{c}^{18} \\plus{} 58564007553966660\\,{x}^{12}{c}^{18} \\plus{} 16701086221823724750\\,{x}^{8}{y}^{ 4}{c}^{18} \\plus{} 362611857207382065\\,x{y}^{11}{c}^{18} \\plus{} 50955813923280508710 \\,{x}^{6}{y}^{6}{c}^{18} \\plus{} 4071659563211697225\\,{x}^{3}{y}^{9}{c}^{18} \\plus{} 38837506435766674065\\,{x}^{5}{y}^{7}{c}^{18} \\plus{} 16701086221823724750\\,{x} ^{4}{y}^{8}{c}^{18} \\plus{} 362611857207382065\\,{x}^{11}y{c}^{18} \\plus{} 1009513729316449710\\,{x}^{2}{y}^{10}{c}^{18}$\n$ \\plus{} 977917657147362045\\,{x}^{ 11}{y}^{2}{c}^{17} \\plus{} 2984439815827840980\\,{x}^{10}{y}^{3}{c}^{17} \\plus{} 2984439815827840980\\,{x}^{3}{y}^{10}{c}^{17} \\plus{} 36130164566694030\\,{y}^{ 13}{c}^{17} \\plus{} 30825593044084271745\\,{x}^{8}{y}^{5}{c}^{17} \\plus{} 11308185750809556000\\,{x}^{4}{y}^{9}{c}^{17} \\plus{} 292229998302188745\\,{x}^{ 12}y{c}^{17} \\plus{} 30825593044084271745\\,{x}^{5}{y}^{8}{c}^{17} \\plus{} 11308185750809556000\\,{x}^{9}{y}^{4}{c}^{17} \\plus{} 50567705251517488230\\,{x} ^{6}{y}^{7}{c}^{17} \\plus{} 36130164566694030\\,{x}^{13}{c}^{17} \\plus{} 50567705251517488230\\,{x}^{7}{y}^{6}{c}^{17} \\plus{} 292229998302188745\\,x{y}^ {12}{c}^{17} \\plus{} 977917657147362045\\,{x}^{2}{y}^{11}{c}^{17}$\n$ \\plus{} 2160360452109822660\\,{x}^{3}{y}^{11}{c}^{16} \\plus{} 2160360452109822660\\,{x}^ {11}{y}^{3}{c}^{16} \\plus{} 38992364468350626915\\,{x}^{8}{y}^{6}{c}^{16} \\plus{} 748135554864262515\\,{x}^{12}{y}^{2}{c}^{16} \\plus{} 180949529841796485\\,x{y}^{ 13}{c}^{16} \\plus{} 6906323613662794935\\,{x}^{4}{y}^{10}{c}^{16} \\plus{} 6906323613662794935\\,{x}^{10}{y}^{4}{c}^{16} \\plus{} 18022409853586110\\,{y}^{ 14}{c}^{16} \\plus{} 18022409853586110\\,{x}^{14}{c}^{16} \\plus{} 748135554864262515\\,{x }^{2}{y}^{12}{c}^{16} \\plus{} 19994508314853588315\\,{x}^{5}{y}^{9}{c}^{16} \\plus{} 48631289652679821660\\,{x}^{7}{y}^{7}{c}^{16} \\plus{} 180949529841796485\\,{x}^{ 13}y{c}^{16} \\plus{} 38992364468350626915\\,{x}^{6}{y}^{8}{c}^{16} \\plus{} 19994508314853588315\\,{x}^{9}{y}^{5}{c}^{16}$\n$ \\plus{} 88711037987481855\\,{x}^{ 14}y{c}^{15} \\plus{} 36070087439992712715\\,{x}^{8}{y}^{7}{c}^{15} \\plus{} 88711037987481855\\,x{y}^{14}{c}^{15} \\plus{} 1393142007744442485\\,{x}^{3}{y}^{ 12}{c}^{15} \\plus{} 3960181901596608855\\,{x}^{4}{y}^{11}{c}^{15} \\plus{} 36070087439992712715\\,{x}^{7}{y}^{8}{c}^{15} \\plus{} 445465963285976835\\,{x}^{ 2}{y}^{13}{c}^{15} \\plus{} 24236016130783711395\\,{x}^{6}{y}^{9}{c}^{15} \\plus{} 11119352295374912205\\,{x}^{5}{y}^{10}{c}^{15} \\plus{} 3960181901596608855\\,{x} ^{11}{y}^{4}{c}^{15} \\plus{} 1393142007744442485\\,{x}^{12}{y}^{3}{c}^{15} \\plus{} 7395765445494270\\,{x}^{15}{c}^{15} \\plus{} 445465963285976835\\,{x}^{13}{y}^{2} {c}^{15} \\plus{} 24236016130783711395\\,{x}^{9}{y}^{6}{c}^{15} \\plus{} 7395765445494270 \\,{y}^{15}{c}^{15} \\plus{} 11119352295374912205\\,{x}^{10}{y}^{5}{c}^{15}$\n$ \\plus{} 2084305487797867050\\,{x}^{12}{y}^{4}{c}^{14} \\plus{} 209424571608036825\\,{x}^{ 14}{y}^{2}{c}^{14} \\plus{} 21349556612200860075\\,{x}^{7}{y}^{9}{c}^{14} \\plus{} 744857579775620475\\,{x}^{3}{y}^{13}{c}^{14} \\plus{} 12585125502138452670\\,{x}^ {10}{y}^{6}{c}^{14} \\plus{} 2084305487797867050\\,{x}^{4}{y}^{12}{c}^{14} \\plus{} 744857579775620475\\,{x}^{13}{y}^{3}{c}^{14} \\plus{} 35203478647936185\\,x{y}^{ 15}{c}^{14} \\plus{} 35203478647936185\\,{x}^{15}y{c}^{14} \\plus{} 21349556612200860075 \\,{x}^{9}{y}^{7}{c}^{14} \\plus{} 2533095274158855\\,{y}^{16}{c}^{14} \\plus{} 12585125502138452670\\,{x}^{6}{y}^{10}{c}^{14} \\plus{} 2533095274158855\\,{x}^{ 16}{c}^{14} \\plus{} 209424571608036825\\,{x}^{2}{y}^{14}{c}^{14} \\plus{} 25510966047864850275\\,{x}^{8}{y}^{8}{c}^{14} \\plus{} 5511886560598653405\\,{x}^ {5}{y}^{11}{c}^{14} \\plus{} 5511886560598653405\\,{x}^{11}{y}^{5}{c}^{14}$\n$ \\plus{} 322423177403419200\\,{x}^{14}{y}^{3}{c}^{13} \\plus{} 733199301973440\\,{y}^{17}{ c}^{13} \\plus{} 2456477683392122055\\,{x}^{12}{y}^{5}{c}^{13} \\plus{} 11499472852337745 \\,x{y}^{16}{c}^{13} \\plus{} 2456477683392122055\\,{x}^{5}{y}^{12}{c}^{13} \\plus{} 955075836063775050\\,{x}^{13}{y}^{4}{c}^{13} \\plus{} 79098393960389025\\,{x}^{2} {y}^{15}{c}^{13} \\plus{} 10388985093352033110\\,{x}^{7}{y}^{10}{c}^{13} \\plus{} 5632415391716311230\\,{x}^{6}{y}^{11}{c}^{13} \\plus{} 955075836063775050\\,{x}^{ 4}{y}^{13}{c}^{13} \\plus{} 10388985093352033110\\,{x}^{10}{y}^{7}{c}^{13} \\plus{} 14264412913360975275\\,{x}^{9}{y}^{8}{c}^{13} \\plus{} 733199301973440\\,{x}^{17} {c}^{13} \\plus{} 5632415391716311230\\,{x}^{11}{y}^{6}{c}^{13} \\plus{} 322423177403419200\\,{x}^{3}{y}^{14}{c}^{13} \\plus{} 14264412913360975275\\,{x}^ {8}{y}^{9}{c}^{13} \\plus{} 11499472852337745\\,{x}^{16}y{c}^{13} \\plus{} 79098393960389025\\,{x}^{15}{y}^{2}{c}^{13}$\n$ \\plus{} 2206418360572908315\\,{x}^{ 12}{y}^{6}{c}^{12} \\plus{} 24362825601034935\\,{x}^{16}{y}^{2}{c}^{12} \\plus{} 6477654862591944300\\,{x}^{10}{y}^{8}{c}^{12} \\plus{} 2206418360572908315\\,{x}^ {6}{y}^{12}{c}^{12} \\plus{} 112916040343061310\\,{x}^{3}{y}^{15}{c}^{12} \\plus{} 7466933476884696750\\,{x}^{9}{y}^{9}{c}^{12} \\plus{} 3134553601994010\\,x{y}^{17 }{c}^{12} \\plus{} 6477654862591944300\\,{x}^{8}{y}^{10}{c}^{12} \\plus{} 366353221781733975\\,{x}^{14}{y}^{4}{c}^{12} \\plus{} 4262032766997513990\\,{x}^{ 7}{y}^{11}{c}^{12} \\plus{} 366353221781733975\\,{x}^{4}{y}^{14}{c}^{12} \\plus{} 181249095648150\\,{y}^{18}{c}^{12} \\plus{} 3134553601994010\\,{x}^{17}y{c}^{12} \\plus{} 4262032766997513990\\,{x}^{11}{y}^{7}{c}^{12} \\plus{} 181249095648150\\,{x}^{18} {c}^{12} \\plus{} 24362825601034935\\,{x}^{2}{y}^{16}{c}^{12} \\plus{} 962390964597878745 \\,{x}^{13}{y}^{5}{c}^{12} \\plus{} 962390964597878745\\,{x}^{5}{y}^{13}{c}^{12} \\plus{} 112916040343061310\\,{x}^{15}{y}^{3}{c}^{12}$\n$ \\plus{} 6196930766887320\\,{x}^{17} {y}^{2}{c}^{11} \\plus{} 32196892281454305\\,{x}^{3}{y}^{16}{c}^{11} \\plus{} 38568777876360\\,{y}^{19}{c}^{11} \\plus{} 6196930766887320\\,{x}^{2}{y}^{17}{c}^ {11} \\plus{} 1496578870705849605\\,{x}^{12}{y}^{7}{c}^{11} \\plus{} 3140506203143784840 \\,{x}^{9}{y}^{10}{c}^{11} \\plus{} 721091184063480\\,x{y}^{18}{c}^{11} \\plus{} 115583970861408465\\,{x}^{4}{y}^{15}{c}^{11} \\plus{} 32196892281454305\\,{x}^{16 }{y}^{3}{c}^{11} \\plus{} 1496578870705849605\\,{x}^{7}{y}^{12}{c}^{11} \\plus{} 38568777876360\\,{x}^{19}{c}^{11} \\plus{} 2440286663280994845\\,{x}^{8}{y}^{11}{ c}^{11} \\plus{} 751829705460213795\\,{x}^{6}{y}^{13}{c}^{11} \\plus{} 721091184063480\\,{ x}^{18}y{c}^{11} \\plus{} 321388973115663375\\,{x}^{5}{y}^{14}{c}^{11} \\plus{} 2440286663280994845\\,{x}^{11}{y}^{8}{c}^{11} \\plus{} 321388973115663375\\,{x}^{ 14}{y}^{5}{c}^{11} \\plus{} 751829705460213795\\,{x}^{13}{y}^{6}{c}^{11} \\plus{} 3140506203143784840\\,{x}^{10}{y}^{9}{c}^{11} \\plus{} 115583970861408465\\,{x}^{ 15}{y}^{4}{c}^{11}$\n$ \\plus{} 772772734181827170\\,{x}^{12}{y}^{8}{c}^{10} \\plus{} 449889121836280395\\,{x}^{7}{y}^{13}{c}^{10} \\plus{} 29842861410108315\\,{x}^{4} {y}^{16}{c}^{10} \\plus{} 1081313759617721910\\,{x}^{9}{y}^{11}{c}^{10} \\plus{} 7534683942204330\\,{x}^{17}{y}^{3}{c}^{10} \\plus{} 141198022534590\\,{x}^{19}y{c }^{10} \\plus{} 449889121836280395\\,{x}^{13}{y}^{7}{c}^{10} \\plus{} 1211445894374410257 \\,{x}^{10}{y}^{10}{c}^{10} \\plus{} 7093025578539\\,{x}^{20}{c}^{10} \\plus{} 7534683942204330\\,{x}^{3}{y}^{17}{c}^{10} \\plus{} 1315579009511205\\,{x}^{2}{y} ^{18}{c}^{10} \\plus{} 218583663720200910\\,{x}^{6}{y}^{14}{c}^{10} \\plus{} 89469060681737367\\,{x}^{15}{y}^{5}{c}^{10} \\plus{} 1081313759617721910\\,{x}^{ 11}{y}^{9}{c}^{10} \\plus{} 1315579009511205\\,{x}^{18}{y}^{2}{c}^{10} \\plus{} 772772734181827170\\,{x}^{8}{y}^{12}{c}^{10} \\plus{} 7093025578539\\,{y}^{20}{c} ^{10} \\plus{} 141198022534590\\,x{y}^{19}{c}^{10} \\plus{} 29842861410108315\\,{x}^{16}{y }^{4}{c}^{10} \\plus{} 218583663720200910\\,{x}^{14}{y}^{6}{c}^{10} \\plus{} 89469060681737367\\,{x}^{5}{y}^{15}{c}^{10}$\n$ \\plus{} 308183466115653975\\,{x}^{9} {y}^{12}{c}^{9} \\plus{} 1126430533800\\,{y}^{21}{c}^{9} \\plus{} 20503667384124705\\,{x}^ {5}{y}^{16}{c}^{9} \\plus{} 1458316098172575\\,{x}^{3}{y}^{18}{c}^{9} \\plus{} 6310720192829400\\,{x}^{17}{y}^{4}{c}^{9} \\plus{} 308183466115653975\\,{x}^{12}{ y}^{9}{c}^{9} \\plus{} 114414758085012750\\,{x}^{7}{y}^{14}{c}^{9} \\plus{} 1458316098172575\\,{x}^{18}{y}^{3}{c}^{9} \\plus{} 23637607287795\\,x{y}^{20}{c}^ {9} \\plus{} 206128996224470325\\,{x}^{13}{y}^{8}{c}^{9} \\plus{} 20503667384124705\\,{x}^ {16}{y}^{5}{c}^{9} \\plus{} 1126430533800\\,{x}^{21}{c}^{9} \\plus{} 234981359117550\\,{x} ^{2}{y}^{19}{c}^{9} \\plus{} 53176745518951770\\,{x}^{6}{y}^{15}{c}^{9} \\plus{} 378024510732571755\\,{x}^{10}{y}^{11}{c}^{9} \\plus{} 378024510732571755\\,{x}^{ 11}{y}^{10}{c}^{9} \\plus{} 206128996224470325\\,{x}^{8}{y}^{13}{c}^{9} \\plus{} 6310720192829400\\,{x}^{4}{y}^{17}{c}^{9} \\plus{} 234981359117550\\,{x}^{19}{y}^ {2}{c}^{9} \\plus{} 53176745518951770\\,{x}^{15}{y}^{6}{c}^{9} \\plus{} 23637607287795\\,{ x}^{20}y{c}^{9} \\plus{} 114414758085012750\\,{x}^{14}{y}^{7}{c}^{9}$\n$ \\plus{} 3845533111294590\\,{x}^{17}{y}^{5}{c}^{8} \\plus{} 153604163700\\,{x}^{22}{c}^{8} \\plus{} 1096068860697600\\,{x}^{18}{y}^{4}{c}^{8} \\plus{} 153604163700\\,{y}^{22}{c}^{8 } \\plus{} 10663945457377545\\,{x}^{16}{y}^{6}{c}^{8} \\plus{} 96280138209164220\\,{x}^{12 }{y}^{10}{c}^{8} \\plus{} 3379291601400\\,x{y}^{21}{c}^{8} \\plus{} 3845533111294590\\,{x} ^{5}{y}^{17}{c}^{8} \\plus{} 45931518331885125\\,{x}^{8}{y}^{14}{c}^{8} \\plus{} 24232232576657520\\,{x}^{7}{y}^{15}{c}^{8} \\plus{} 24232232576657520\\,{x}^{15}{ y}^{7}{c}^{8} \\plus{} 234807019897500\\,{x}^{3}{y}^{19}{c}^{8} \\plus{} 3379291601400\\,{ x}^{21}y{c}^{8} \\plus{} 10663945457377545\\,{x}^{6}{y}^{16}{c}^{8} \\plus{} 45931518331885125\\,{x}^{14}{y}^{8}{c}^{8} \\plus{} 234807019897500\\,{x}^{19}{y} ^{3}{c}^{8} \\plus{} 96280138209164220\\,{x}^{10}{y}^{12}{c}^{8} \\plus{} 72834792863369850\\,{x}^{13}{y}^{9}{c}^{8} \\plus{} 35447693970690\\,{x}^{2}{y}^{ 20}{c}^{8} \\plus{} 105713054588876040\\,{x}^{11}{y}^{11}{c}^{8} \\plus{} 1096068860697600\\,{x}^{4}{y}^{18}{c}^{8} \\plus{} 72834792863369850\\,{x}^{9}{y} ^{13}{c}^{8} \\plus{} 35447693970690\\,{x}^{20}{y}^{2}{c}^{8}$\n$ \\plus{} 14169936212586000 \\,{x}^{9}{y}^{14}{c}^{7} \\plus{} 588626351558550\\,{x}^{18}{y}^{5}{c}^{7} \\plus{} 23797038511922220\\,{x}^{12}{y}^{11}{c}^{7} \\plus{} 17809178400\\,{y}^{23}{c}^{7 } \\plus{} 1744644537673560\\,{x}^{17}{y}^{6}{c}^{7} \\plus{} 4505722135200\\,{x}^{21}{y}^ {2}{c}^{7} \\plus{} 1744644537673560\\,{x}^{6}{y}^{17}{c}^{7} \\plus{} 588626351558550\\,{ x}^{5}{y}^{18}{c}^{7} \\plus{} 4209610166857935\\,{x}^{16}{y}^{7}{c}^{7} \\plus{} 4209610166857935\\,{x}^{7}{y}^{16}{c}^{7} \\plus{} 31505187102390\\,{x}^{3}{y}^{ 20}{c}^{7} \\plus{} 8434858561754355\\,{x}^{8}{y}^{15}{c}^{7} \\plus{} 17809178400\\,{x}^{ 23}{c}^{7} \\plus{} 31505187102390\\,{x}^{20}{y}^{3}{c}^{7} \\plus{} 156654239411700\\,{x} ^{19}{y}^{4}{c}^{7} \\plus{} 409611103200\\,{x}^{22}y{c}^{7} \\plus{} 23797038511922220\\, {x}^{11}{y}^{12}{c}^{7} \\plus{} 4505722135200\\,{x}^{2}{y}^{21}{c}^{7} \\plus{} 20019537297068490\\,{x}^{10}{y}^{13}{c}^{7} \\plus{} 20019537297068490\\,{x}^{13} {y}^{10}{c}^{7} \\plus{} 8434858561754355\\,{x}^{15}{y}^{8}{c}^{7} \\plus{} 156654239411700\\,{x}^{4}{y}^{19}{c}^{7} \\plus{} 14169936212586000\\,{x}^{14}{y} ^{9}{c}^{7} \\plus{} 409611103200\\,x{y}^{22}{c}^{7}$\n$ \\plus{} 41554749600\\,{x}^{23}y{c}^{ 6} \\plus{} 4301057448335610\\,{x}^{13}{y}^{11}{c}^{6} \\plus{} 230971324829805\\,{x}^{18} {y}^{6}{c}^{6} \\plus{} 1731447900\\,{x}^{24}{c}^{6} \\plus{} 4301057448335610\\,{x}^{11}{ y}^{13}{c}^{6} \\plus{} 230971324829805\\,{x}^{6}{y}^{18}{c}^{6} \\plus{} 1256797756868130\\,{x}^{8}{y}^{16}{c}^{6} \\plus{} 3504450549600\\,{x}^{3}{y}^{21 }{c}^{6} \\plus{} 3369714001320915\\,{x}^{10}{y}^{14}{c}^{6} \\plus{} 1256797756868130\\,{ x}^{16}{y}^{8}{c}^{6} \\plus{} 3504450549600\\,{x}^{21}{y}^{3}{c}^{6} \\plus{} 591834193207470\\,{x}^{7}{y}^{17}{c}^{6} \\plus{} 73279650945510\\,{x}^{19}{y}^{5 }{c}^{6} \\plus{} 591834193207470\\,{x}^{17}{y}^{7}{c}^{6} \\plus{} 477879620400\\,{x}^{2} {y}^{22}{c}^{6} \\plus{} 477879620400\\,{x}^{22}{y}^{2}{c}^{6} \\plus{} 4664636437623390 \\,{x}^{12}{y}^{12}{c}^{6} \\plus{} 2238944546574450\\,{x}^{9}{y}^{15}{c}^{6} \\plus{} 2238944546574450\\,{x}^{15}{y}^{9}{c}^{6} \\plus{} 18380931463395\\,{x}^{20}{y}^{ 4}{c}^{6} \\plus{} 3369714001320915\\,{x}^{14}{y}^{10}{c}^{6} \\plus{} 18380931463395\\,{x }^{4}{y}^{20}{c}^{6} \\plus{} 1731447900\\,{y}^{24}{c}^{6} \\plus{} 41554749600\\,x{y}^{23 }{c}^{6} \\plus{} 73279650945510\\,{x}^{5}{y}^{19}{c}^{6}$\n$ \\plus{} 66427655144355\\,{x}^{7 }{y}^{18}{c}^{5} \\plus{} 41554749600\\,{x}^{23}{y}^{2}{c}^{5} \\plus{} 616653376988580\\, {x}^{11}{y}^{14}{c}^{5} \\plus{} 282252157155990\\,{x}^{16}{y}^{9}{c}^{5} \\plus{} 24496286003190\\,{x}^{19}{y}^{6}{c}^{5} \\plus{} 66427655144355\\,{x}^{18}{y}^{7} {c}^{5} \\plus{} 451892854554867\\,{x}^{10}{y}^{15}{c}^{5} \\plus{} 1752225274800\\,{x}^{ 21}{y}^{4}{c}^{5} \\plus{} 3462895800\\,{x}^{24}y{c}^{5} \\plus{} 616653376988580\\,{x}^{ 14}{y}^{11}{c}^{5} \\plus{} 318586413600\\,{x}^{22}{y}^{3}{c}^{5} \\plus{} 138515832\\,{x} ^{25}{c}^{5} \\plus{} 1752225274800\\,{x}^{4}{y}^{21}{c}^{5} \\plus{} 451892854554867\\,{x }^{15}{y}^{10}{c}^{5} \\plus{} 149396846867280\\,{x}^{17}{y}^{8}{c}^{5} \\plus{} 719748561723435\\,{x}^{12}{y}^{13}{c}^{5} \\plus{} 7355859369759\\,{x}^{20}{y}^{5 }{c}^{5} \\plus{} 719748561723435\\,{x}^{13}{y}^{12}{c}^{5} \\plus{} 3462895800\\,x{y}^{24 }{c}^{5} \\plus{} 149396846867280\\,{x}^{8}{y}^{17}{c}^{5} \\plus{} 318586413600\\,{x}^{3} {y}^{22}{c}^{5} \\plus{} 282252157155990\\,{x}^{9}{y}^{16}{c}^{5} \\plus{} 24496286003190 \\,{x}^{6}{y}^{19}{c}^{5} \\plus{} 138515832\\,{y}^{25}{c}^{5} \\plus{} 7355859369759\\,{x} ^{5}{y}^{20}{c}^{5} \\plus{} 41554749600\\,{x}^{2}{y}^{23}{c}^{5}$\n$ \\plus{} 8879220\\,{x}^{ 26}{c}^{4} \\plus{} 132744339000\\,{x}^{22}{y}^{4}{c}^{4} \\plus{} 68602274395200\\,{x}^{ 15}{y}^{11}{c}^{4} \\plus{} 13871783425500\\,{x}^{8}{y}^{18}{c}^{4} \\plus{} 2885746500\\, {x}^{2}{y}^{24}{c}^{4} \\plus{} 13871783425500\\,{x}^{18}{y}^{8}{c}^{4} \\plus{} 230859720\\,x{y}^{25}{c}^{4} \\plus{} 2885746500\\,{x}^{24}{y}^{2}{c}^{4} \\plus{} 47164063646700\\,{x}^{10}{y}^{16}{c}^{4} \\plus{} 5840750916000\\,{x}^{7}{y}^{19} {c}^{4} \\plus{} 584075091600\\,{x}^{21}{y}^{5}{c}^{4} \\plus{} 132744339000\\,{x}^{4}{y}^ {22}{c}^{4} \\plus{} 23085972000\\,{x}^{3}{y}^{23}{c}^{4} \\plus{} 8879220\\,{y}^{26}{c}^{ 4} \\plus{} 2044262820600\\,{x}^{20}{y}^{6}{c}^{4} \\plus{} 27743566851000\\,{x}^{17}{y}^{ 9}{c}^{4} \\plus{} 2044262820600\\,{x}^{6}{y}^{20}{c}^{4} \\plus{} 47164063646700\\,{x}^{ 16}{y}^{10}{c}^{4} \\plus{} 584075091600\\,{x}^{5}{y}^{21}{c}^{4} \\plus{} 92349215532000 \\,{x}^{13}{y}^{13}{c}^{4} \\plus{} 27743566851000\\,{x}^{9}{y}^{17}{c}^{4} \\plus{} 23085972000\\,{x}^{23}{y}^{3}{c}^{4} \\plus{} 85752842994000\\,{x}^{14}{y}^{12}{c }^{4} \\plus{} 68602274395200\\,{x}^{11}{y}^{15}{c}^{4} \\plus{} 85752842994000\\,{x}^{12} {y}^{14}{c}^{4} \\plus{} 230859720\\,{x}^{25}y{c}^{4} \\plus{} 5840750916000\\,{x}^{19}{y} ^{7}{c}^{4}$\n$ \\plus{} 438480\\,{y}^{27}{c}^{3} \\plus{} 973458486000\\,{x}^{19}{y}^{8}{c}^{ 3} \\plus{} 11838960\\,{x}^{26}y{c}^{3} \\plus{} 5716856199600\\,{x}^{16}{y}^{11}{c}^{3} \\plus{} 389383394400\\,{x}^{7}{y}^{20}{c}^{3} \\plus{} 35398490400\\,{x}^{5}{y}^{22}{c}^{ 3} \\plus{} 438480\\,{x}^{27}{c}^{3} \\plus{} 7695324000\\,{x}^{4}{y}^{23}{c}^{3} \\plus{} 1282554000\\,{x}^{24}{y}^{3}{c}^{3} \\plus{} 1282554000\\,{x}^{3}{y}^{24}{c}^{3} \\plus{} 153906480\\,{x}^{2}{y}^{25}{c}^{3} \\plus{} 2055079026000\\,{x}^{18}{y}^{9}{c}^{3 } \\plus{} 7695324000\\,{x}^{23}{y}^{4}{c}^{3} \\plus{} 973458486000\\,{x}^{8}{y}^{19}{c}^ {3} \\plus{} 35398490400\\,{x}^{22}{y}^{5}{c}^{3} \\plus{} 389383394400\\,{x}^{20}{y}^{7}{ c}^{3} \\plus{} 8795163384000\\,{x}^{13}{y}^{14}{c}^{3} \\plus{} 5716856199600\\,{x}^{11}{ y}^{16}{c}^{3} \\plus{} 7622474932800\\,{x}^{15}{y}^{12}{c}^{3} \\plus{} 11838960\\,x{y}^{ 26}{c}^{3} \\plus{} 7622474932800\\,{x}^{12}{y}^{15}{c}^{3} \\plus{} 8795163384000\\,{x}^{ 14}{y}^{13}{c}^{3} \\plus{} 129794464800\\,{x}^{21}{y}^{6}{c}^{3} \\plus{} 2055079026000 \\,{x}^{9}{y}^{18}{c}^{3} \\plus{} 153906480\\,{x}^{25}{y}^{2}{c}^{3} \\plus{} 3699142246800\\,{x}^{10}{y}^{17}{c}^{3} \\plus{} 129794464800\\,{x}^{6}{y}^{21}{c }^{3} \\plus{} 3699142246800\\,{x}^{17}{y}^{10}{c}^{3}$\n$ \\plus{} 476404683300\\,{x}^{16}{y} ^{12}{c}^{2} \\plus{} 438480\\,{x}^{27}y{c}^{2} \\plus{} 15660\\,{y}^{28}{c}^{2} \\plus{} 1539064800\\,{x}^{23}{y}^{5}{c}^{2} \\plus{} 320638500\\,{x}^{4}{y}^{24}{c}^{2} \\plus{} 108162054000\\,{x}^{9}{y}^{19}{c}^{2} \\plus{} 1539064800\\,{x}^{5}{y}^{23}{c}^{2 } \\plus{} 320638500\\,{x}^{24}{y}^{4}{c}^{2} \\plus{} 15660\\,{x}^{28}{c}^{2} \\plus{} 108162054000\\,{x}^{19}{y}^{9}{c}^{2} \\plus{} 5919480\\,{x}^{2}{y}^{26}{c}^{2} \\plus{} 48672924300\\,{x}^{20}{y}^{8}{c}^{2} \\plus{} 51302160\\,{x}^{3}{y}^{25}{c}^{2} \\plus{} 438480\\,x{y}^{27}{c}^{2} \\plus{} 336285658800\\,{x}^{11}{y}^{17}{c}^{2} \\plus{} 5919480 \\,{x}^{26}{y}^{2}{c}^{2} \\plus{} 5899748400\\,{x}^{6}{y}^{22}{c}^{2} \\plus{} 586344225600\\,{x}^{15}{y}^{13}{c}^{2} \\plus{} 5899748400\\,{x}^{22}{y}^{6}{c}^{ 2} \\plus{} 628225956000\\,{x}^{14}{y}^{14}{c}^{2} \\plus{} 205507902600\\,{x}^{18}{y}^{10 }{c}^{2} \\plus{} 205507902600\\,{x}^{10}{y}^{18}{c}^{2} \\plus{} 586344225600\\,{x}^{13}{ y}^{15}{c}^{2} \\plus{} 18542066400\\,{x}^{21}{y}^{7}{c}^{2} \\plus{} 48672924300\\,{x}^{8 }{y}^{20}{c}^{2} \\plus{} 336285658800\\,{x}^{17}{y}^{11}{c}^{2} \\plus{} 476404683300\\,{ x}^{12}{y}^{16}{c}^{2} \\plus{} 18542066400\\,{x}^{7}{y}^{21}{c}^{2} \\plus{} 51302160\\,{ x}^{25}{y}^{3}{c}^{2}$\n$ \\plus{} 1315440\\,{x}^{3}{y}^{26}c \\plus{} 360\\,{y}^{29}c \\plus{} 171007200\\,{x}^{23}{y}^{6}c \\plus{} 7210803600\\,{x}^{19}{y}^{10}c \\plus{} 1545172200\\, {x}^{8}{y}^{21}c \\plus{} 3605401800\\,{x}^{9}{y}^{20}c \\plus{} 27921153600\\,{x}^{14}{y} ^{15}c \\plus{} 1545172200\\,{x}^{21}{y}^{8}c \\plus{} 7210803600\\,{x}^{10}{y}^{19}c \\plus{} 24431009400\\,{x}^{13}{y}^{16}c \\plus{} 146160\\,{x}^{27}{y}^{2}c \\plus{} 42751800\\,{x}^ {5}{y}^{24}c \\plus{} 561880800\\,{x}^{22}{y}^{7}c \\plus{} 146160\\,{x}^{2}{y}^{27}c \\plus{} 10440\\,{x}^{28}yc \\plus{} 3605401800\\,{x}^{20}{y}^{9}c \\plus{} 27921153600\\,{x}^{15}{y }^{14}c \\plus{} 171007200\\,{x}^{6}{y}^{23}c \\plus{} 12455024400\\,{x}^{11}{y}^{18}c \\plus{} 12455024400\\,{x}^{18}{y}^{11}c \\plus{} 1315440\\,{x}^{26}{y}^{3}c \\plus{} 18682536600\\, {x}^{12}{y}^{17}c \\plus{} 8550360\\,{x}^{4}{y}^{25}c \\plus{} 360\\,{x}^{29}c \\plus{} 8550360\\,{x }^{25}{y}^{4}c \\plus{} 10440\\,x{y}^{28}c \\plus{} 18682536600\\,{x}^{17}{y}^{12}c \\plus{} 42751800\\,{x}^{24}{y}^{5}c \\plus{} 24431009400\\,{x}^{16}{y}^{13}c \\plus{} 561880800\\,{ x}^{7}{y}^{22}c$\n$ \\plus{} 109620\\,{x}^{4}{y}^{26} \\plus{} 1740\\,{x}^{2}{y}^{28} \\plus{} 16240\\,{ x}^{27}{y}^{3} \\plus{} 4\\,{y}^{30} \\plus{} 109620\\,{x}^{26}{y}^{4} \\plus{} 16240\\,{x}^{3}{y}^{ 27} \\plus{} 570024\\,{x}^{25}{y}^{5} \\plus{} 120\\,x{y}^{29} \\plus{} 2375100\\,{x}^{24}{y}^{6} \\plus{} 4 \\,{x}^{30} \\plus{} 8143200\\,{x}^{23}{y}^{7} \\plus{} 120\\,{x}^{29}y \\plus{} 23411700\\,{x}^{22}{ y}^{8} \\plus{} 1740\\,{x}^{28}{y}^{2} \\plus{} 57228600\\,{x}^{21}{y}^{9} \\plus{} 345972900\\,{x}^ {12}{y}^{18} \\plus{} 120180060\\,{x}^{20}{y}^{10} \\plus{} 218509200\\,{x}^{11}{y}^{19} \\plus{} 218509200\\,{x}^{19}{y}^{11} \\plus{} 120180060\\,{x}^{10}{y}^{20} \\plus{} 345972900\\,{x} ^{18}{y}^{12} \\plus{} 57228600\\,{x}^{9}{y}^{21} \\plus{} 479039400\\,{x}^{17}{y}^{13} \\plus{} 23411700\\,{x}^{8}{y}^{22} \\plus{} 581690700\\,{x}^{16}{y}^{14} \\plus{} 8143200\\,{x}^{7} {y}^{23} \\plus{} 620470080\\,{x}^{15}{y}^{15} \\plus{} 2375100\\,{x}^{6}{y}^{24} \\plus{} 581690700\\,{x}^{14}{y}^{16} \\plus{} 570024\\,{x}^{5}{y}^{25} \\plus{} 479039400\\,{x}^{13 }{y}^{17} \\geq 0$\n\nDone ! \n\nPS : It's just for fun  :lol:\n\nPS : Do you believe i do it by hand ???  :rotfl:[/quote]\r\n\r\nWell, it starts with $ \\minus{} 343151886824415xyc^{28}$. So you have to better check for it.",
  "Solution_22": "[quote=\"Stephen\"][quote=\"Materazzi\"][quote=\"nguoivn\"]\nWith this ineq, we can solve it easily by Am-Gm. But the following (of him) is harder and I still have no solution:  :oops: \n$ \\frac {1}{2 \\plus{} a^5b^5} \\plus{} \\frac {1}{2 \\plus{} b^5c^5} \\plus{} \\frac {1}{2 \\plus{} c^5a^5} \\geq\\ 1$\n(with the same condition)[/quote]\n\nMaybe ..... I have a very ugly solution for this  :blush: \n\nAssume : $ c \\equal{} min \\{a;b;c\\}$\n\nWith : $ a \\equal{} x \\plus{} c \\ ; \\ b \\equal{} y \\plus{} c \\ ; \\ x,y \\geq 0$ .\n\nThis inequality is equivalent to : \n\n$ \\minus{} 343151886824415\\,xy{c}^{28} \\plus{} 343151886824415\\,{x}^{2}{c}^{28} \\plus{} 343151886824415\\,{y}^{2}{c}^{28}$\n$ \\plus{} 1258223585022855\\,x{y}^{2}{c}^{27} \\plus{} 2363935220345970\\,{x}^{3}{c}^{27} \\plus{} 1258223585022855\\,{x}^{2}y{c}^{27} \\plus{} 2363935220345970\\,{y}^{3}{c}^{27}$\n$ \\plus{} 23677480190884635\\,{x}^{2}{y}^{2}{c} ^{26} \\plus{} 7206189623312715\\,{y}^{4}{c}^{26} \\plus{} 7206189623312715\\,{x}^{4}{c}^{ 26} \\plus{} 17500746228045165\\,{x}^{3}y{c}^{26} \\plus{} 17500746228045165\\,x{y}^{3}{c} ^{26}$\n$ \\plus{} 160099419197302065\\,{x}^{3}{y}^{2}{c}^{25} \\plus{} 12887259749628030\\,{y }^{5}{c}^{25} \\plus{} 61462315728995220\\,{x}^{4}y{c}^{25} \\plus{} 12887259749628030\\,{ x}^{5}{c}^{25} \\plus{} 160099419197302065\\,{x}^{2}{y}^{3}{c}^{25} \\plus{} 61462315728995220\\,x{y}^{4}{c}^{25}$\n$ \\plus{} 16267941301305600\\,{y}^{6}{c}^{24} \\plus{} 520637668287487425\\,{x}^{2}{y}^{4}{c}^{24} \\plus{} 16267941301305600\\,{x}^{6} {c}^{24} \\plus{} 849237373244718900\\,{x}^{3}{y}^{3}{c}^{24} \\plus{} 112286922966433575 \\,{x}^{5}y{c}^{24} \\plus{} 520637668287487425\\,{x}^{4}{y}^{2}{c}^{24} \\plus{} 112286922966433575\\,x{y}^{5}{c}^{24}$\n$ \\plus{} 20462019918048450\\,{x}^{7}{c}^{23 } \\plus{} 2583870161142136725\\,{x}^{3}{y}^{4}{c}^{23} \\plus{} 123598225902497625\\,{x}^ {6}y{c}^{23} \\plus{} 2583870161142136725\\,{x}^{4}{y}^{3}{c}^{23} \\plus{} 123598225902497625\\,x{y}^{6}{c}^{23} \\plus{} 1010963586572151525\\,{x}^{5}{y}^{ 2}{c}^{23} \\plus{} 20462019918048450\\,{y}^{7}{c}^{23} \\plus{} 1010963586572151525\\,{x} ^{2}{y}^{5}{c}^{23}$\n$ \\plus{} 100227181758694875\\,x{y}^{7}{c}^{22} \\plus{} 33771511824715575\\,{y}^{8}{c}^{22} \\plus{} 5052078366356891925\\,{x}^{3}{y}^{5} {c}^{22} \\plus{} 1300256160789639825\\,{x}^{2}{y}^{6}{c}^{22} \\plus{} 5052078366356891925\\,{x}^{5}{y}^{3}{c}^{22} \\plus{} 100227181758694875\\,{x}^{7 }y{c}^{22} \\plus{} 1300256160789639825\\,{x}^{6}{y}^{2}{c}^{22} \\plus{} 7610819359490631675\\,{x}^{4}{y}^{4}{c}^{22} \\plus{} 33771511824715575\\,{x}^{8} {c}^{22 }$\n$ \\plus{} 6871482379629081900\\,{x}^{3}{y}^{6}{c}^{21} \\plus{} 56307035256428700 \\,{y}^{9}{c}^{21} \\plus{} 14738655968645105250\\,{x}^{4}{y}^{5}{c}^{21} \\plus{} 118104971417942175\\,x{y}^{8}{c}^{21} \\plus{} 6871482379629081900\\,{x}^{6}{y}^{ 3}{c}^{21} \\plus{} 1211700379626022275\\,{x}^{2}{y}^{7}{c}^{21} \\plus{} 14738655968645105250\\,{x}^{5}{y}^{4}{c}^{21} \\plus{} 1211700379626022275\\,{x}^ {7}{y}^{2}{c}^{21} \\plus{} 56307035256428700\\,{x}^{9}{c}^{21} \\plus{} 118104971417942175\\,{x}^{8}y{c}^{21}$\n$ \\plus{} 6922333817672485950\\,{x}^{3}{y}^{ 7}{c}^{20} \\plus{} 74579755550855265\\,{x}^{10}{c}^{20} \\plus{} 6922333817672485950\\,{x }^{7}{y}^{3}{c}^{20} \\plus{} 218325092438225025\\,x{y}^{9}{c}^{20} \\plus{} 20205326773079281125\\,{x}^{6}{y}^{4}{c}^{20} \\plus{} 218325092438225025\\,{x}^{ 9}y{c}^{20} \\plus{} 20205326773079281125\\,{x}^{4}{y}^{6}{c}^{20} \\plus{} 984690578941006500\\,{x}^{2}{y}^{8}{c}^{20} \\plus{} 74579755550855265\\,{y}^{10} {c}^{20} \\plus{} 28195374156236728335\\,{x}^{5}{y}^{5}{c}^{20} \\plus{} 984690578941006500\\,{x}^{8}{y}^{2}{c}^{20}$\n$ \\plus{} 332431506674710425\\,{x}^{10 }y{c}^{19} \\plus{} 20703523584131542575\\,{x}^{4}{y}^{7}{c}^{19} \\plus{} 929452678951534425\\,{x}^{2}{y}^{9}{c}^{19} \\plus{} 332431506674710425\\,x{y}^{ 10}{c}^{19} \\plus{} 20703523584131542575\\,{x}^{7}{y}^{4}{c}^{19} \\plus{} 929452678951534425\\,{x}^{9}{y}^{2}{c}^{19} \\plus{} 5587211207586606525\\,{x}^{8 }{y}^{3}{c}^{19} \\plus{} 5587211207586606525\\,{x}^{3}{y}^{8}{c}^{19} \\plus{} 75157463424334950\\,{y}^{11}{c}^{19} \\plus{} 38253020811296949225\\,{x}^{6}{y}^{ 5}{c}^{19} \\plus{} 75157463424334950\\,{x}^{11}{c}^{19} \\plus{} 38253020811296949225\\,{ x}^{5}{y}^{6}{c}^{19}$\n$ \\plus{} 1009513729316449710\\,{x}^{10}{y}^{2}{c}^{18} \\plus{} 58564007553966660\\,{y}^{12}{c}^{18} \\plus{} 38837506435766674065\\,{x}^{7}{y}^{ 5}{c}^{18} \\plus{} 4071659563211697225\\,{x}^{9}{y}^{3}{c}^{18} \\plus{} 58564007553966660\\,{x}^{12}{c}^{18} \\plus{} 16701086221823724750\\,{x}^{8}{y}^{ 4}{c}^{18} \\plus{} 362611857207382065\\,x{y}^{11}{c}^{18} \\plus{} 50955813923280508710 \\,{x}^{6}{y}^{6}{c}^{18} \\plus{} 4071659563211697225\\,{x}^{3}{y}^{9}{c}^{18} \\plus{} 38837506435766674065\\,{x}^{5}{y}^{7}{c}^{18} \\plus{} 16701086221823724750\\,{x} ^{4}{y}^{8}{c}^{18} \\plus{} 362611857207382065\\,{x}^{11}y{c}^{18} \\plus{} 1009513729316449710\\,{x}^{2}{y}^{10}{c}^{18}$\n$ \\plus{} 977917657147362045\\,{x}^{ 11}{y}^{2}{c}^{17} \\plus{} 2984439815827840980\\,{x}^{10}{y}^{3}{c}^{17} \\plus{} 2984439815827840980\\,{x}^{3}{y}^{10}{c}^{17} \\plus{} 36130164566694030\\,{y}^{ 13}{c}^{17} \\plus{} 30825593044084271745\\,{x}^{8}{y}^{5}{c}^{17} \\plus{} 11308185750809556000\\,{x}^{4}{y}^{9}{c}^{17} \\plus{} 292229998302188745\\,{x}^{ 12}y{c}^{17} \\plus{} 30825593044084271745\\,{x}^{5}{y}^{8}{c}^{17} \\plus{} 11308185750809556000\\,{x}^{9}{y}^{4}{c}^{17} \\plus{} 50567705251517488230\\,{x} ^{6}{y}^{7}{c}^{17} \\plus{} 36130164566694030\\,{x}^{13}{c}^{17} \\plus{} 50567705251517488230\\,{x}^{7}{y}^{6}{c}^{17} \\plus{} 292229998302188745\\,x{y}^ {12}{c}^{17} \\plus{} 977917657147362045\\,{x}^{2}{y}^{11}{c}^{17}$\n$ \\plus{} 2160360452109822660\\,{x}^{3}{y}^{11}{c}^{16} \\plus{} 2160360452109822660\\,{x}^ {11}{y}^{3}{c}^{16} \\plus{} 38992364468350626915\\,{x}^{8}{y}^{6}{c}^{16} \\plus{} 748135554864262515\\,{x}^{12}{y}^{2}{c}^{16} \\plus{} 180949529841796485\\,x{y}^{ 13}{c}^{16} \\plus{} 6906323613662794935\\,{x}^{4}{y}^{10}{c}^{16} \\plus{} 6906323613662794935\\,{x}^{10}{y}^{4}{c}^{16} \\plus{} 18022409853586110\\,{y}^{ 14}{c}^{16} \\plus{} 18022409853586110\\,{x}^{14}{c}^{16} \\plus{} 748135554864262515\\,{x }^{2}{y}^{12}{c}^{16} \\plus{} 19994508314853588315\\,{x}^{5}{y}^{9}{c}^{16} \\plus{} 48631289652679821660\\,{x}^{7}{y}^{7}{c}^{16} \\plus{} 180949529841796485\\,{x}^{ 13}y{c}^{16} \\plus{} 38992364468350626915\\,{x}^{6}{y}^{8}{c}^{16} \\plus{} 19994508314853588315\\,{x}^{9}{y}^{5}{c}^{16}$\n$ \\plus{} 88711037987481855\\,{x}^{ 14}y{c}^{15} \\plus{} 36070087439992712715\\,{x}^{8}{y}^{7}{c}^{15} \\plus{} 88711037987481855\\,x{y}^{14}{c}^{15} \\plus{} 1393142007744442485\\,{x}^{3}{y}^{ 12}{c}^{15} \\plus{} 3960181901596608855\\,{x}^{4}{y}^{11}{c}^{15} \\plus{} 36070087439992712715\\,{x}^{7}{y}^{8}{c}^{15} \\plus{} 445465963285976835\\,{x}^{ 2}{y}^{13}{c}^{15} \\plus{} 24236016130783711395\\,{x}^{6}{y}^{9}{c}^{15} \\plus{} 11119352295374912205\\,{x}^{5}{y}^{10}{c}^{15} \\plus{} 3960181901596608855\\,{x} ^{11}{y}^{4}{c}^{15} \\plus{} 1393142007744442485\\,{x}^{12}{y}^{3}{c}^{15} \\plus{} 7395765445494270\\,{x}^{15}{c}^{15} \\plus{} 445465963285976835\\,{x}^{13}{y}^{2} {c}^{15} \\plus{} 24236016130783711395\\,{x}^{9}{y}^{6}{c}^{15} \\plus{} 7395765445494270 \\,{y}^{15}{c}^{15} \\plus{} 11119352295374912205\\,{x}^{10}{y}^{5}{c}^{15}$\n$ \\plus{} 2084305487797867050\\,{x}^{12}{y}^{4}{c}^{14} \\plus{} 209424571608036825\\,{x}^{ 14}{y}^{2}{c}^{14} \\plus{} 21349556612200860075\\,{x}^{7}{y}^{9}{c}^{14} \\plus{} 744857579775620475\\,{x}^{3}{y}^{13}{c}^{14} \\plus{} 12585125502138452670\\,{x}^ {10}{y}^{6}{c}^{14} \\plus{} 2084305487797867050\\,{x}^{4}{y}^{12}{c}^{14} \\plus{} 744857579775620475\\,{x}^{13}{y}^{3}{c}^{14} \\plus{} 35203478647936185\\,x{y}^{ 15}{c}^{14} \\plus{} 35203478647936185\\,{x}^{15}y{c}^{14} \\plus{} 21349556612200860075 \\,{x}^{9}{y}^{7}{c}^{14} \\plus{} 2533095274158855\\,{y}^{16}{c}^{14} \\plus{} 12585125502138452670\\,{x}^{6}{y}^{10}{c}^{14} \\plus{} 2533095274158855\\,{x}^{ 16}{c}^{14} \\plus{} 209424571608036825\\,{x}^{2}{y}^{14}{c}^{14} \\plus{} 25510966047864850275\\,{x}^{8}{y}^{8}{c}^{14} \\plus{} 5511886560598653405\\,{x}^ {5}{y}^{11}{c}^{14} \\plus{} 5511886560598653405\\,{x}^{11}{y}^{5}{c}^{14}$\n$ \\plus{} 322423177403419200\\,{x}^{14}{y}^{3}{c}^{13} \\plus{} 733199301973440\\,{y}^{17}{ c}^{13} \\plus{} 2456477683392122055\\,{x}^{12}{y}^{5}{c}^{13} \\plus{} 11499472852337745 \\,x{y}^{16}{c}^{13} \\plus{} 2456477683392122055\\,{x}^{5}{y}^{12}{c}^{13} \\plus{} 955075836063775050\\,{x}^{13}{y}^{4}{c}^{13} \\plus{} 79098393960389025\\,{x}^{2} {y}^{15}{c}^{13} \\plus{} 10388985093352033110\\,{x}^{7}{y}^{10}{c}^{13} \\plus{} 5632415391716311230\\,{x}^{6}{y}^{11}{c}^{13} \\plus{} 955075836063775050\\,{x}^{ 4}{y}^{13}{c}^{13} \\plus{} 10388985093352033110\\,{x}^{10}{y}^{7}{c}^{13} \\plus{} 14264412913360975275\\,{x}^{9}{y}^{8}{c}^{13} \\plus{} 733199301973440\\,{x}^{17} {c}^{13} \\plus{} 5632415391716311230\\,{x}^{11}{y}^{6}{c}^{13} \\plus{} 322423177403419200\\,{x}^{3}{y}^{14}{c}^{13} \\plus{} 14264412913360975275\\,{x}^ {8}{y}^{9}{c}^{13} \\plus{} 11499472852337745\\,{x}^{16}y{c}^{13} \\plus{} 79098393960389025\\,{x}^{15}{y}^{2}{c}^{13}$\n$ \\plus{} 2206418360572908315\\,{x}^{ 12}{y}^{6}{c}^{12} \\plus{} 24362825601034935\\,{x}^{16}{y}^{2}{c}^{12} \\plus{} 6477654862591944300\\,{x}^{10}{y}^{8}{c}^{12} \\plus{} 2206418360572908315\\,{x}^ {6}{y}^{12}{c}^{12} \\plus{} 112916040343061310\\,{x}^{3}{y}^{15}{c}^{12} \\plus{} 7466933476884696750\\,{x}^{9}{y}^{9}{c}^{12} \\plus{} 3134553601994010\\,x{y}^{17 }{c}^{12} \\plus{} 6477654862591944300\\,{x}^{8}{y}^{10}{c}^{12} \\plus{} 366353221781733975\\,{x}^{14}{y}^{4}{c}^{12} \\plus{} 4262032766997513990\\,{x}^{ 7}{y}^{11}{c}^{12} \\plus{} 366353221781733975\\,{x}^{4}{y}^{14}{c}^{12} \\plus{} 181249095648150\\,{y}^{18}{c}^{12} \\plus{} 3134553601994010\\,{x}^{17}y{c}^{12} \\plus{} 4262032766997513990\\,{x}^{11}{y}^{7}{c}^{12} \\plus{} 181249095648150\\,{x}^{18} {c}^{12} \\plus{} 24362825601034935\\,{x}^{2}{y}^{16}{c}^{12} \\plus{} 962390964597878745 \\,{x}^{13}{y}^{5}{c}^{12} \\plus{} 962390964597878745\\,{x}^{5}{y}^{13}{c}^{12} \\plus{} 112916040343061310\\,{x}^{15}{y}^{3}{c}^{12}$\n$ \\plus{} 6196930766887320\\,{x}^{17} {y}^{2}{c}^{11} \\plus{} 32196892281454305\\,{x}^{3}{y}^{16}{c}^{11} \\plus{} 38568777876360\\,{y}^{19}{c}^{11} \\plus{} 6196930766887320\\,{x}^{2}{y}^{17}{c}^ {11} \\plus{} 1496578870705849605\\,{x}^{12}{y}^{7}{c}^{11} \\plus{} 3140506203143784840 \\,{x}^{9}{y}^{10}{c}^{11} \\plus{} 721091184063480\\,x{y}^{18}{c}^{11} \\plus{} 115583970861408465\\,{x}^{4}{y}^{15}{c}^{11} \\plus{} 32196892281454305\\,{x}^{16 }{y}^{3}{c}^{11} \\plus{} 1496578870705849605\\,{x}^{7}{y}^{12}{c}^{11} \\plus{} 38568777876360\\,{x}^{19}{c}^{11} \\plus{} 2440286663280994845\\,{x}^{8}{y}^{11}{ c}^{11} \\plus{} 751829705460213795\\,{x}^{6}{y}^{13}{c}^{11} \\plus{} 721091184063480\\,{ x}^{18}y{c}^{11} \\plus{} 321388973115663375\\,{x}^{5}{y}^{14}{c}^{11} \\plus{} 2440286663280994845\\,{x}^{11}{y}^{8}{c}^{11} \\plus{} 321388973115663375\\,{x}^{ 14}{y}^{5}{c}^{11} \\plus{} 751829705460213795\\,{x}^{13}{y}^{6}{c}^{11} \\plus{} 3140506203143784840\\,{x}^{10}{y}^{9}{c}^{11} \\plus{} 115583970861408465\\,{x}^{ 15}{y}^{4}{c}^{11}$\n$ \\plus{} 772772734181827170\\,{x}^{12}{y}^{8}{c}^{10} \\plus{} 449889121836280395\\,{x}^{7}{y}^{13}{c}^{10} \\plus{} 29842861410108315\\,{x}^{4} {y}^{16}{c}^{10} \\plus{} 1081313759617721910\\,{x}^{9}{y}^{11}{c}^{10} \\plus{} 7534683942204330\\,{x}^{17}{y}^{3}{c}^{10} \\plus{} 141198022534590\\,{x}^{19}y{c }^{10} \\plus{} 449889121836280395\\,{x}^{13}{y}^{7}{c}^{10} \\plus{} 1211445894374410257 \\,{x}^{10}{y}^{10}{c}^{10} \\plus{} 7093025578539\\,{x}^{20}{c}^{10} \\plus{} 7534683942204330\\,{x}^{3}{y}^{17}{c}^{10} \\plus{} 1315579009511205\\,{x}^{2}{y} ^{18}{c}^{10} \\plus{} 218583663720200910\\,{x}^{6}{y}^{14}{c}^{10} \\plus{} 89469060681737367\\,{x}^{15}{y}^{5}{c}^{10} \\plus{} 1081313759617721910\\,{x}^{ 11}{y}^{9}{c}^{10} \\plus{} 1315579009511205\\,{x}^{18}{y}^{2}{c}^{10} \\plus{} 772772734181827170\\,{x}^{8}{y}^{12}{c}^{10} \\plus{} 7093025578539\\,{y}^{20}{c} ^{10} \\plus{} 141198022534590\\,x{y}^{19}{c}^{10} \\plus{} 29842861410108315\\,{x}^{16}{y }^{4}{c}^{10} \\plus{} 218583663720200910\\,{x}^{14}{y}^{6}{c}^{10} \\plus{} 89469060681737367\\,{x}^{5}{y}^{15}{c}^{10}$\n$ \\plus{} 308183466115653975\\,{x}^{9} {y}^{12}{c}^{9} \\plus{} 1126430533800\\,{y}^{21}{c}^{9} \\plus{} 20503667384124705\\,{x}^ {5}{y}^{16}{c}^{9} \\plus{} 1458316098172575\\,{x}^{3}{y}^{18}{c}^{9} \\plus{} 6310720192829400\\,{x}^{17}{y}^{4}{c}^{9} \\plus{} 308183466115653975\\,{x}^{12}{ y}^{9}{c}^{9} \\plus{} 114414758085012750\\,{x}^{7}{y}^{14}{c}^{9} \\plus{} 1458316098172575\\,{x}^{18}{y}^{3}{c}^{9} \\plus{} 23637607287795\\,x{y}^{20}{c}^ {9} \\plus{} 206128996224470325\\,{x}^{13}{y}^{8}{c}^{9} \\plus{} 20503667384124705\\,{x}^ {16}{y}^{5}{c}^{9} \\plus{} 1126430533800\\,{x}^{21}{c}^{9} \\plus{} 234981359117550\\,{x} ^{2}{y}^{19}{c}^{9} \\plus{} 53176745518951770\\,{x}^{6}{y}^{15}{c}^{9} \\plus{} 378024510732571755\\,{x}^{10}{y}^{11}{c}^{9} \\plus{} 378024510732571755\\,{x}^{ 11}{y}^{10}{c}^{9} \\plus{} 206128996224470325\\,{x}^{8}{y}^{13}{c}^{9} \\plus{} 6310720192829400\\,{x}^{4}{y}^{17}{c}^{9} \\plus{} 234981359117550\\,{x}^{19}{y}^ {2}{c}^{9} \\plus{} 53176745518951770\\,{x}^{15}{y}^{6}{c}^{9} \\plus{} 23637607287795\\,{ x}^{20}y{c}^{9} \\plus{} 114414758085012750\\,{x}^{14}{y}^{7}{c}^{9}$\n$ \\plus{} 3845533111294590\\,{x}^{17}{y}^{5}{c}^{8} \\plus{} 153604163700\\,{x}^{22}{c}^{8} \\plus{} 1096068860697600\\,{x}^{18}{y}^{4}{c}^{8} \\plus{} 153604163700\\,{y}^{22}{c}^{8 } \\plus{} 10663945457377545\\,{x}^{16}{y}^{6}{c}^{8} \\plus{} 96280138209164220\\,{x}^{12 }{y}^{10}{c}^{8} \\plus{} 3379291601400\\,x{y}^{21}{c}^{8} \\plus{} 3845533111294590\\,{x} ^{5}{y}^{17}{c}^{8} \\plus{} 45931518331885125\\,{x}^{8}{y}^{14}{c}^{8} \\plus{} 24232232576657520\\,{x}^{7}{y}^{15}{c}^{8} \\plus{} 24232232576657520\\,{x}^{15}{ y}^{7}{c}^{8} \\plus{} 234807019897500\\,{x}^{3}{y}^{19}{c}^{8} \\plus{} 3379291601400\\,{ x}^{21}y{c}^{8} \\plus{} 10663945457377545\\,{x}^{6}{y}^{16}{c}^{8} \\plus{} 45931518331885125\\,{x}^{14}{y}^{8}{c}^{8} \\plus{} 234807019897500\\,{x}^{19}{y} ^{3}{c}^{8} \\plus{} 96280138209164220\\,{x}^{10}{y}^{12}{c}^{8} \\plus{} 72834792863369850\\,{x}^{13}{y}^{9}{c}^{8} \\plus{} 35447693970690\\,{x}^{2}{y}^{ 20}{c}^{8} \\plus{} 105713054588876040\\,{x}^{11}{y}^{11}{c}^{8} \\plus{} 1096068860697600\\,{x}^{4}{y}^{18}{c}^{8} \\plus{} 72834792863369850\\,{x}^{9}{y} ^{13}{c}^{8} \\plus{} 35447693970690\\,{x}^{20}{y}^{2}{c}^{8}$\n$ \\plus{} 14169936212586000 \\,{x}^{9}{y}^{14}{c}^{7} \\plus{} 588626351558550\\,{x}^{18}{y}^{5}{c}^{7} \\plus{} 23797038511922220\\,{x}^{12}{y}^{11}{c}^{7} \\plus{} 17809178400\\,{y}^{23}{c}^{7 } \\plus{} 1744644537673560\\,{x}^{17}{y}^{6}{c}^{7} \\plus{} 4505722135200\\,{x}^{21}{y}^ {2}{c}^{7} \\plus{} 1744644537673560\\,{x}^{6}{y}^{17}{c}^{7} \\plus{} 588626351558550\\,{ x}^{5}{y}^{18}{c}^{7} \\plus{} 4209610166857935\\,{x}^{16}{y}^{7}{c}^{7} \\plus{} 4209610166857935\\,{x}^{7}{y}^{16}{c}^{7} \\plus{} 31505187102390\\,{x}^{3}{y}^{ 20}{c}^{7} \\plus{} 8434858561754355\\,{x}^{8}{y}^{15}{c}^{7} \\plus{} 17809178400\\,{x}^{ 23}{c}^{7} \\plus{} 31505187102390\\,{x}^{20}{y}^{3}{c}^{7} \\plus{} 156654239411700\\,{x} ^{19}{y}^{4}{c}^{7} \\plus{} 409611103200\\,{x}^{22}y{c}^{7} \\plus{} 23797038511922220\\, {x}^{11}{y}^{12}{c}^{7} \\plus{} 4505722135200\\,{x}^{2}{y}^{21}{c}^{7} \\plus{} 20019537297068490\\,{x}^{10}{y}^{13}{c}^{7} \\plus{} 20019537297068490\\,{x}^{13} {y}^{10}{c}^{7} \\plus{} 8434858561754355\\,{x}^{15}{y}^{8}{c}^{7} \\plus{} 156654239411700\\,{x}^{4}{y}^{19}{c}^{7} \\plus{} 14169936212586000\\,{x}^{14}{y} ^{9}{c}^{7} \\plus{} 409611103200\\,x{y}^{22}{c}^{7}$\n$ \\plus{} 41554749600\\,{x}^{23}y{c}^{ 6} \\plus{} 4301057448335610\\,{x}^{13}{y}^{11}{c}^{6} \\plus{} 230971324829805\\,{x}^{18} {y}^{6}{c}^{6} \\plus{} 1731447900\\,{x}^{24}{c}^{6} \\plus{} 4301057448335610\\,{x}^{11}{ y}^{13}{c}^{6} \\plus{} 230971324829805\\,{x}^{6}{y}^{18}{c}^{6} \\plus{} 1256797756868130\\,{x}^{8}{y}^{16}{c}^{6} \\plus{} 3504450549600\\,{x}^{3}{y}^{21 }{c}^{6} \\plus{} 3369714001320915\\,{x}^{10}{y}^{14}{c}^{6} \\plus{} 1256797756868130\\,{ x}^{16}{y}^{8}{c}^{6} \\plus{} 3504450549600\\,{x}^{21}{y}^{3}{c}^{6} \\plus{} 591834193207470\\,{x}^{7}{y}^{17}{c}^{6} \\plus{} 73279650945510\\,{x}^{19}{y}^{5 }{c}^{6} \\plus{} 591834193207470\\,{x}^{17}{y}^{7}{c}^{6} \\plus{} 477879620400\\,{x}^{2} {y}^{22}{c}^{6} \\plus{} 477879620400\\,{x}^{22}{y}^{2}{c}^{6} \\plus{} 4664636437623390 \\,{x}^{12}{y}^{12}{c}^{6} \\plus{} 2238944546574450\\,{x}^{9}{y}^{15}{c}^{6} \\plus{} 2238944546574450\\,{x}^{15}{y}^{9}{c}^{6} \\plus{} 18380931463395\\,{x}^{20}{y}^{ 4}{c}^{6} \\plus{} 3369714001320915\\,{x}^{14}{y}^{10}{c}^{6} \\plus{} 18380931463395\\,{x }^{4}{y}^{20}{c}^{6} \\plus{} 1731447900\\,{y}^{24}{c}^{6} \\plus{} 41554749600\\,x{y}^{23 }{c}^{6} \\plus{} 73279650945510\\,{x}^{5}{y}^{19}{c}^{6}$\n$ \\plus{} 66427655144355\\,{x}^{7 }{y}^{18}{c}^{5} \\plus{} 41554749600\\,{x}^{23}{y}^{2}{c}^{5} \\plus{} 616653376988580\\, {x}^{11}{y}^{14}{c}^{5} \\plus{} 282252157155990\\,{x}^{16}{y}^{9}{c}^{5} \\plus{} 24496286003190\\,{x}^{19}{y}^{6}{c}^{5} \\plus{} 66427655144355\\,{x}^{18}{y}^{7} {c}^{5} \\plus{} 451892854554867\\,{x}^{10}{y}^{15}{c}^{5} \\plus{} 1752225274800\\,{x}^{ 21}{y}^{4}{c}^{5} \\plus{} 3462895800\\,{x}^{24}y{c}^{5} \\plus{} 616653376988580\\,{x}^{ 14}{y}^{11}{c}^{5} \\plus{} 318586413600\\,{x}^{22}{y}^{3}{c}^{5} \\plus{} 138515832\\,{x} ^{25}{c}^{5} \\plus{} 1752225274800\\,{x}^{4}{y}^{21}{c}^{5} \\plus{} 451892854554867\\,{x }^{15}{y}^{10}{c}^{5} \\plus{} 149396846867280\\,{x}^{17}{y}^{8}{c}^{5} \\plus{} 719748561723435\\,{x}^{12}{y}^{13}{c}^{5} \\plus{} 7355859369759\\,{x}^{20}{y}^{5 }{c}^{5} \\plus{} 719748561723435\\,{x}^{13}{y}^{12}{c}^{5} \\plus{} 3462895800\\,x{y}^{24 }{c}^{5} \\plus{} 149396846867280\\,{x}^{8}{y}^{17}{c}^{5} \\plus{} 318586413600\\,{x}^{3} {y}^{22}{c}^{5} \\plus{} 282252157155990\\,{x}^{9}{y}^{16}{c}^{5} \\plus{} 24496286003190 \\,{x}^{6}{y}^{19}{c}^{5} \\plus{} 138515832\\,{y}^{25}{c}^{5} \\plus{} 7355859369759\\,{x} ^{5}{y}^{20}{c}^{5} \\plus{} 41554749600\\,{x}^{2}{y}^{23}{c}^{5}$\n$ \\plus{} 8879220\\,{x}^{ 26}{c}^{4} \\plus{} 132744339000\\,{x}^{22}{y}^{4}{c}^{4} \\plus{} 68602274395200\\,{x}^{ 15}{y}^{11}{c}^{4} \\plus{} 13871783425500\\,{x}^{8}{y}^{18}{c}^{4} \\plus{} 2885746500\\, {x}^{2}{y}^{24}{c}^{4} \\plus{} 13871783425500\\,{x}^{18}{y}^{8}{c}^{4} \\plus{} 230859720\\,x{y}^{25}{c}^{4} \\plus{} 2885746500\\,{x}^{24}{y}^{2}{c}^{4} \\plus{} 47164063646700\\,{x}^{10}{y}^{16}{c}^{4} \\plus{} 5840750916000\\,{x}^{7}{y}^{19} {c}^{4} \\plus{} 584075091600\\,{x}^{21}{y}^{5}{c}^{4} \\plus{} 132744339000\\,{x}^{4}{y}^ {22}{c}^{4} \\plus{} 23085972000\\,{x}^{3}{y}^{23}{c}^{4} \\plus{} 8879220\\,{y}^{26}{c}^{ 4} \\plus{} 2044262820600\\,{x}^{20}{y}^{6}{c}^{4} \\plus{} 27743566851000\\,{x}^{17}{y}^{ 9}{c}^{4} \\plus{} 2044262820600\\,{x}^{6}{y}^{20}{c}^{4} \\plus{} 47164063646700\\,{x}^{ 16}{y}^{10}{c}^{4} \\plus{} 584075091600\\,{x}^{5}{y}^{21}{c}^{4} \\plus{} 92349215532000 \\,{x}^{13}{y}^{13}{c}^{4} \\plus{} 27743566851000\\,{x}^{9}{y}^{17}{c}^{4} \\plus{} 23085972000\\,{x}^{23}{y}^{3}{c}^{4} \\plus{} 85752842994000\\,{x}^{14}{y}^{12}{c }^{4} \\plus{} 68602274395200\\,{x}^{11}{y}^{15}{c}^{4} \\plus{} 85752842994000\\,{x}^{12} {y}^{14}{c}^{4} \\plus{} 230859720\\,{x}^{25}y{c}^{4} \\plus{} 5840750916000\\,{x}^{19}{y} ^{7}{c}^{4}$\n$ \\plus{} 438480\\,{y}^{27}{c}^{3} \\plus{} 973458486000\\,{x}^{19}{y}^{8}{c}^{ 3} \\plus{} 11838960\\,{x}^{26}y{c}^{3} \\plus{} 5716856199600\\,{x}^{16}{y}^{11}{c}^{3} \\plus{} 389383394400\\,{x}^{7}{y}^{20}{c}^{3} \\plus{} 35398490400\\,{x}^{5}{y}^{22}{c}^{ 3} \\plus{} 438480\\,{x}^{27}{c}^{3} \\plus{} 7695324000\\,{x}^{4}{y}^{23}{c}^{3} \\plus{} 1282554000\\,{x}^{24}{y}^{3}{c}^{3} \\plus{} 1282554000\\,{x}^{3}{y}^{24}{c}^{3} \\plus{} 153906480\\,{x}^{2}{y}^{25}{c}^{3} \\plus{} 2055079026000\\,{x}^{18}{y}^{9}{c}^{3 } \\plus{} 7695324000\\,{x}^{23}{y}^{4}{c}^{3} \\plus{} 973458486000\\,{x}^{8}{y}^{19}{c}^ {3} \\plus{} 35398490400\\,{x}^{22}{y}^{5}{c}^{3} \\plus{} 389383394400\\,{x}^{20}{y}^{7}{ c}^{3} \\plus{} 8795163384000\\,{x}^{13}{y}^{14}{c}^{3} \\plus{} 5716856199600\\,{x}^{11}{ y}^{16}{c}^{3} \\plus{} 7622474932800\\,{x}^{15}{y}^{12}{c}^{3} \\plus{} 11838960\\,x{y}^{ 26}{c}^{3} \\plus{} 7622474932800\\,{x}^{12}{y}^{15}{c}^{3} \\plus{} 8795163384000\\,{x}^{ 14}{y}^{13}{c}^{3} \\plus{} 129794464800\\,{x}^{21}{y}^{6}{c}^{3} \\plus{} 2055079026000 \\,{x}^{9}{y}^{18}{c}^{3} \\plus{} 153906480\\,{x}^{25}{y}^{2}{c}^{3} \\plus{} 3699142246800\\,{x}^{10}{y}^{17}{c}^{3} \\plus{} 129794464800\\,{x}^{6}{y}^{21}{c }^{3} \\plus{} 3699142246800\\,{x}^{17}{y}^{10}{c}^{3}$\n$ \\plus{} 476404683300\\,{x}^{16}{y} ^{12}{c}^{2} \\plus{} 438480\\,{x}^{27}y{c}^{2} \\plus{} 15660\\,{y}^{28}{c}^{2} \\plus{} 1539064800\\,{x}^{23}{y}^{5}{c}^{2} \\plus{} 320638500\\,{x}^{4}{y}^{24}{c}^{2} \\plus{} 108162054000\\,{x}^{9}{y}^{19}{c}^{2} \\plus{} 1539064800\\,{x}^{5}{y}^{23}{c}^{2 } \\plus{} 320638500\\,{x}^{24}{y}^{4}{c}^{2} \\plus{} 15660\\,{x}^{28}{c}^{2} \\plus{} 108162054000\\,{x}^{19}{y}^{9}{c}^{2} \\plus{} 5919480\\,{x}^{2}{y}^{26}{c}^{2} \\plus{} 48672924300\\,{x}^{20}{y}^{8}{c}^{2} \\plus{} 51302160\\,{x}^{3}{y}^{25}{c}^{2} \\plus{} 438480\\,x{y}^{27}{c}^{2} \\plus{} 336285658800\\,{x}^{11}{y}^{17}{c}^{2} \\plus{} 5919480 \\,{x}^{26}{y}^{2}{c}^{2} \\plus{} 5899748400\\,{x}^{6}{y}^{22}{c}^{2} \\plus{} 586344225600\\,{x}^{15}{y}^{13}{c}^{2} \\plus{} 5899748400\\,{x}^{22}{y}^{6}{c}^{ 2} \\plus{} 628225956000\\,{x}^{14}{y}^{14}{c}^{2} \\plus{} 205507902600\\,{x}^{18}{y}^{10 }{c}^{2} \\plus{} 205507902600\\,{x}^{10}{y}^{18}{c}^{2} \\plus{} 586344225600\\,{x}^{13}{ y}^{15}{c}^{2} \\plus{} 18542066400\\,{x}^{21}{y}^{7}{c}^{2} \\plus{} 48672924300\\,{x}^{8 }{y}^{20}{c}^{2} \\plus{} 336285658800\\,{x}^{17}{y}^{11}{c}^{2} \\plus{} 476404683300\\,{ x}^{12}{y}^{16}{c}^{2} \\plus{} 18542066400\\,{x}^{7}{y}^{21}{c}^{2} \\plus{} 51302160\\,{ x}^{25}{y}^{3}{c}^{2}$\n$ \\plus{} 1315440\\,{x}^{3}{y}^{26}c \\plus{} 360\\,{y}^{29}c \\plus{} 171007200\\,{x}^{23}{y}^{6}c \\plus{} 7210803600\\,{x}^{19}{y}^{10}c \\plus{} 1545172200\\, {x}^{8}{y}^{21}c \\plus{} 3605401800\\,{x}^{9}{y}^{20}c \\plus{} 27921153600\\,{x}^{14}{y} ^{15}c \\plus{} 1545172200\\,{x}^{21}{y}^{8}c \\plus{} 7210803600\\,{x}^{10}{y}^{19}c \\plus{} 24431009400\\,{x}^{13}{y}^{16}c \\plus{} 146160\\,{x}^{27}{y}^{2}c \\plus{} 42751800\\,{x}^ {5}{y}^{24}c \\plus{} 561880800\\,{x}^{22}{y}^{7}c \\plus{} 146160\\,{x}^{2}{y}^{27}c \\plus{} 10440\\,{x}^{28}yc \\plus{} 3605401800\\,{x}^{20}{y}^{9}c \\plus{} 27921153600\\,{x}^{15}{y }^{14}c \\plus{} 171007200\\,{x}^{6}{y}^{23}c \\plus{} 12455024400\\,{x}^{11}{y}^{18}c \\plus{} 12455024400\\,{x}^{18}{y}^{11}c \\plus{} 1315440\\,{x}^{26}{y}^{3}c \\plus{} 18682536600\\, {x}^{12}{y}^{17}c \\plus{} 8550360\\,{x}^{4}{y}^{25}c \\plus{} 360\\,{x}^{29}c \\plus{} 8550360\\,{x }^{25}{y}^{4}c \\plus{} 10440\\,x{y}^{28}c \\plus{} 18682536600\\,{x}^{17}{y}^{12}c \\plus{} 42751800\\,{x}^{24}{y}^{5}c \\plus{} 24431009400\\,{x}^{16}{y}^{13}c \\plus{} 561880800\\,{ x}^{7}{y}^{22}c$\n$ \\plus{} 109620\\,{x}^{4}{y}^{26} \\plus{} 1740\\,{x}^{2}{y}^{28} \\plus{} 16240\\,{ x}^{27}{y}^{3} \\plus{} 4\\,{y}^{30} \\plus{} 109620\\,{x}^{26}{y}^{4} \\plus{} 16240\\,{x}^{3}{y}^{ 27} \\plus{} 570024\\,{x}^{25}{y}^{5} \\plus{} 120\\,x{y}^{29} \\plus{} 2375100\\,{x}^{24}{y}^{6} \\plus{} 4 \\,{x}^{30} \\plus{} 8143200\\,{x}^{23}{y}^{7} \\plus{} 120\\,{x}^{29}y \\plus{} 23411700\\,{x}^{22}{ y}^{8} \\plus{} 1740\\,{x}^{28}{y}^{2} \\plus{} 57228600\\,{x}^{21}{y}^{9} \\plus{} 345972900\\,{x}^ {12}{y}^{18} \\plus{} 120180060\\,{x}^{20}{y}^{10} \\plus{} 218509200\\,{x}^{11}{y}^{19} \\plus{} 218509200\\,{x}^{19}{y}^{11} \\plus{} 120180060\\,{x}^{10}{y}^{20} \\plus{} 345972900\\,{x} ^{18}{y}^{12} \\plus{} 57228600\\,{x}^{9}{y}^{21} \\plus{} 479039400\\,{x}^{17}{y}^{13} \\plus{} 23411700\\,{x}^{8}{y}^{22} \\plus{} 581690700\\,{x}^{16}{y}^{14} \\plus{} 8143200\\,{x}^{7} {y}^{23} \\plus{} 620470080\\,{x}^{15}{y}^{15} \\plus{} 2375100\\,{x}^{6}{y}^{24} \\plus{} 581690700\\,{x}^{14}{y}^{16} \\plus{} 570024\\,{x}^{5}{y}^{25} \\plus{} 479039400\\,{x}^{13 }{y}^{17} \\geq 0$\n\nDone ! \n\nPS : It's just for fun  :lol:\n\nPS : Do you believe i do it by hand ???  :rotfl:[/quote]\n\nWell, it starts with $ \\minus{} 343151886824415xyc^{28}$. So you have to better check for it.[/quote]\r\n\r\nWhat do you mean Stephen ??? \r\n\r\nDo you know : $ x^2 \\minus{} xy \\plus{} y^2 \\geq 0$ ???\r\n\r\n;)\r\n\r\n~~~~~~~~~~~~~~~~~~~~~~~~~~\r\n\r\nM\u1ea1c Th\u1ecb Thanh Hoa",
  "Solution_23": "[quote=\"nguoivn\"][quote=\"arqady\"]By the way, the following inequality is true.\nLet $ a,$ $ b$ and $ c$ are positive numbers such that $ abc \\equal{} 1.$ Prove that\n\\[ \\frac {a \\plus{} b \\plus{} c}{3} \\geq\\ \\sqrt [17]{\\frac {a^4 \\plus{} b^4 \\plus{} c^4}{3}}\\]\n[/quote]\nDid you have solution for it, arqady? [/quote]\nHere is my proof.\nLet $ a \\plus{} b \\plus{} c \\equal{} 3u,$ $ ab \\plus{} ac \\plus{} bc \\equal{} 3v^2$ and $ abc \\equal{} w^3.$\nHence, $ \\frac {a \\plus{} b \\plus{} c}{3} \\geq\\ \\sqrt [17]{\\frac {a^4 \\plus{} b^4 \\plus{} c^4}{3}}\\Leftrightarrow$\n$ \\Leftrightarrow u^{17}\\geq(27u^4 \\minus{} 36u^2v^2 \\plus{} 6v^4 \\plus{} 4uw^3)w^{13}\\Leftrightarrow f(x)\\leq0,$ where $ x \\equal{} w^3$ and\n$ f(x) \\equal{} 4ux^{\\frac {16}{3}} \\plus{} 3(9u^4 \\minus{} 12u^2v^2 \\plus{} 2v^4)x^{\\frac {13}{3}} \\minus{} u^{17}.$\nHence, $ f''(x) \\equal{} 4\\cdot\\frac {16}{3}\\cdot\\frac {13}{3}ux^{\\frac {10}{3}} \\plus{} 3\\cdot\\frac {13}{3}\\cdot\\frac {10}{3}(9u^4 \\minus{} 12u^2v^2 \\plus{} 2v^4)x^{\\frac {7}{3}} \\equal{}$\n$ \\equal{} \\frac {26w^7}{9}\\left(32uw^3 \\plus{} 15(9u^4 \\minus{} 12u^2v^2 \\plus{} 2v^4)\\right) \\equal{}$\n$ \\equal{} \\frac {26w^7}{9}\\left(32uw^3 \\plus{} 15\\cdot\\frac {a^4 \\plus{} b^4 \\plus{} c^4 \\minus{} 12uw^3}{9}\\right) \\equal{}$\n$ \\equal{} \\frac {26w^7}{27}(36uw^3 \\plus{} 5(a^4 \\plus{} b^4 \\plus{} c^4))\\geq0.$\nThus, $ f$ is a convex function. \nId est, $ f$ gets a maximal value on bounders of $ w^3,$ which holds when \n1) $ w^3 \\equal{} 0$ ( for homogeneous inequality ), which is true and \n2) when  $ b \\equal{} c.$\nLet $ b \\equal{} c \\equal{} 1$ in homogeneous inequality.\nHence, we need to check here $ \\left(\\frac {a \\plus{} 2}{3}\\right)^{17}\\geq\\frac {a^4 \\plus{} 2}{3}\\cdot a^{\\frac {13}{3}},$\nwhich is equivalent to $ g(a)\\geq0,$ where $ g(a) \\equal{} 17\\ln(a \\plus{} 2) \\minus{} \\ln(a^4 \\plus{} 2) \\minus{} \\frac {13}{3}\\ln a \\minus{} 16\\ln3.$\nBut $ g'(a) \\equal{} \\frac {17}{a \\plus{} 2} \\minus{} \\frac {4a^3}{a^4 \\plus{} 2} \\minus{} \\frac {13}{3a} \\equal{} \\frac {2(a \\minus{} 1)(13a^4 \\minus{} 12a^3 \\minus{} 12a^2 \\minus{} 12a \\plus{} 26)}{3a(a \\plus{} 2)(a^4 \\plus{} 2)}.$\nEasy to show that $ 13a^4 \\minus{} 12a^3 \\minus{} 12a^2 \\minus{} 12a \\plus{} 26 > 0.$ Hence, $ a_{min} \\equal{} 1.$\nHence, $ g(a)\\geq g(1) \\equal{} 0.$  Done! :) \n[quote=\"nguoivn\"]\nI think with $ abc \\equal{} 1$, this ineq is trues: $ \\sqrt [10]{\\frac {a^3 \\plus{} b^3 \\plus{} c^3}{3}} \\geq\\ \\sqrt [17]{\\frac {a^4 \\plus{} b^4 \\plus{} c^4}{3}}$  :maybe:[/quote]\r\nIt's not true. Try $ a \\equal{} 1.21$ and $ b \\equal{} c \\equal{} \\frac {10}{11}.$ :wink:",
  "Solution_24": "[quote=\"arqady\"]Hence, we need to check here $ \\left(\\frac {a \\plus{} 2}{3}\\right)^{17}\\geq\\frac {a^4 \\plus{} 2}{3}\\cdot a^{\\frac {13}{3}},$\nwhich is equivalent to $ g(a)\\geq0,$ where $ g(a) \\equal{} 17\\ln(a \\plus{} 2) \\minus{} \\ln(a^4 \\plus{} 2) \\minus{} \\frac {13}{3}\\ln a \\minus{} 16\\ln3.$[/quote]\r\nI must say that I was suprised by this one! Very nice! :)",
  "Solution_25": "Thank you, Mathias_DK!\r\n[quote=\"nguoivn\"]The case $ b \\equal{} c$ is not easy to check  :maybe: [/quote]\r\nWe need to check only that\r\n$ (3\\minus{}2b)^5\\plus{}2b^5\\plus{}2b^{10}(3\\minus{}2b)^5\\minus{}b^4(3\\minus{}2b)\\minus{}b^3(3\\minus{}2b)^2\\minus{}b^2(3\\minus{}2b)^3\\minus{}b(3\\minus{}2b)^4$\r\n is positive on $ [0,1.5]$ or maybe a root of him is small than $ 1$ or something similar.",
  "Solution_26": "Thank you, arqady with your proof  :lol: \r\nI'm trying to use Am-Gm to solve it  :)",
  "Solution_27": "[quote=\"nguoivn\"] \nWith $ a^3b^3c^3(a^3 \\plus{} b^3 \\plus{} c^3) \\leq\\ 3$, I had had solution by Am-Gm.[/quote]\r\n\r\nDear nguoivn, can you please explain better how to solve $ \\sum \\frac{1}{1\\plus{}a^3b^3}\\geq 1$ by AM-GM?\r\n\r\nThank you very much. :)",
  "Solution_28": "[quote=\"nguoivn\"]\n[hide]@Blueflower: Ec, sao anh bit dc. Ku Thang' len day ma` con` bi nguoi` yeu duoi?. Po tay ong em nay` lun:| [/hide][/quote]\n\n[hide]@nguoivn : \u0111\u00e0n em v\u00f4 t\u1ed9i  :lol: [/hide]\n[hide]@Blueflower: anh \u0111\u00e2y anh \u0111\u00e2y :( .... ch\u1ea1y theo anh l\u00ean t\u1edbi \u0111\u00e2y lu\u00f4n m\u1edbi gh\u00ea ch\u1ee9 :([/hide]"
}
{
  "Tag": [],
  "Problem": "The measure of the supplement of an angle is three times the measure of its complement. How many degrees are in the measure of the angle?",
  "Solution_1": "The supplement of an angle is always 90 degrees more than the complement. So, let x be the complement, and\r\nx+90=3x\r\nx=45\r\n\r\nSince the complement is 45 degrees, the angle is 90-45= $ \\boxed{45}$",
  "Solution_2": "Let the angle be $ x.$ The complement of angle $ x$ is $ 90\\minus{}x$ and the supplement is $ 180\\minus{}x.$ We then have the equation $ 180\\minus{}x\\equal{}3(90\\minus{}x).$Solving, we get that $ \\angle x\\equal{}45^\\circ.$"
}
{
  "Tag": [
    "LaTeX",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "1.Let $f:R\\to R$ derivable such that: for any $a,b$ real numbers with $a<b$ exists and it's unique a point\r\n $c$ in $(a,b)$ such that $f^'(c)=f(b)-f(a)/b-a$.Prove that $f$ has no inflexion points.\r\n\r\n 2. Let $f:R\\to R$ derivable such that $f^{''}(x)>=0$ and $f(x)<=c$ for every $x$ real number. \r\n     Prove that $f$ is constant.",
  "Solution_1": "I hope you do understand what I've written.",
  "Solution_2": "cezar learn latex see top of mathlinks Latex help\r\n\r\n[quote=\"cezar lupu\"]1.Let $f:R\\to R$ derivable such that: for any $a,b$ real numbers with $a<b$ exists and it's unique a point\n $c$ in $(a,b)$ such that $f^{'}(c)=\\frac{f(b)-f(a)}{b-a}$.Prove that $f$ has no inflexion points.\n\n 2. Let $f:R\\to R$ derivable such that $f^{''}(x)\\geq 0$ and $f(x)\\leq c$ for every $x$ real number. \n     Prove that $f$ is constant.[/quote]",
  "Solution_3": "Still nobody?",
  "Solution_4": "I don't understand what you're saying in 1)\r\n\r\nFor 2), it is clear that $f'$ is everywhere increasing.  Say that we have $f'(x)>0$ for some $x$.  Then there exists $a>x$ and $c\\in(x,a)$ such that $f'(c)=\\frac {f(a)-f(x)}{x-a}\\leq \\frac {c-f(x)}{a-x}<\\epsilon$ for every $\\epsilon>0$.  In particular, we can find a $c$ such that $f'(x)>f'(c)$, a contradiction.\r\n\r\nOtherwise, $f'(x)<0$ for all $x$.  Fix a point $x$.  Then there exists a $c\\in(a,x)$ such that $f'(c)=\\frac {f(x)-f(a)}{x-a}\\geq \\frac {f(x)-c}{x-a}$.  As $a\\to -\\infty$, we can find arbitrarily small $f'(c)$, which contradicts that $f'$ is everywhere increasing.",
  "Solution_5": "To understand better, we call a inflexion point a number $x$ a root for the equation $f''(x)=0$.All $x$ which verify\r\n that equality are called inflexion points. ;)"
}
{
  "Tag": [],
  "Problem": "The age of a man is the same as his wife's age with the digits reversed. The sum of their ages is 99 and the man is 9 years older than his wife. How old is the man?",
  "Solution_1": "[quote=\"tcjy\"]The age of a man is the same as his wife's age with the digits reversed. The sum of their ages is 99 and the man is 9 years older than his wife. How old is the man?[/quote]\r\n\r\nLet $ 10a \\plus{} b$ be the age of the man.  Therefore, his wife's age is $ 10b \\plus{} a$.  We get:\r\n\\[ 11a \\plus{} 11b \\equal{} 99\r\n\\]\r\n\r\n\\[ 9a \\minus{} 9b \\equal{} 9\r\n\\]\r\nSolving this system, we get $ (a,b) \\equal{} (5,4)$, so the man is $ \\boxed{54 \\text{ years}}$",
  "Solution_2": "[quote=\"007math\"][quote=\"tcjy\"]The age of a man is the same as his wife's age with the digits reversed. The sum of their ages is 99 and the man is 9 years older than his wife. How old is the man?[/quote]\n\nLet $ 10a \\plus{} b$ be the age of the man.  Therefore, his wife's age is $ 10b \\plus{} a$.  We get:\n\\[ 11a \\plus{} 11b \\equal{} 99\n\\]\n\n\\[ 9a \\minus{} 9b \\equal{} 9\n\\]\nSolving this system, we get $ (a,b) \\equal{} (5,4)$, so the man is $ \\boxed{54 \\text{ years}}$[/quote]\r\n\r\n\r\nthis is the correct answer.\r\n\r\nmy way is to guess and check.",
  "Solution_3": "umm..\r\nu just revived a month old post, if u haven't noticed... :o",
  "Solution_4": "dude this is such an old contest problem! :roll:",
  "Solution_5": "54 \r\nwhy are all of the problems that are like this have the same answer?",
  "Solution_6": "plz in the future refrain from answering the same question unless\r\n1) u have a different solution\r\n2) u think the other person is wrong \r\nthx :lol:  :wink:",
  "Solution_7": "If pi is 3.14159265358979323846, what is the square root of pi?",
  "Solution_8": "[quote=\"Twilight5331\"]If pi is 3.14159265358979323846, what is the square root of pi?[/quote]\r\nI see you are new, but if you want to post a new problem, you should post a new topic unless it is related.\r\nBTW, the answer can be found using your calculator, which is $ \\approx1.772453851$.",
  "Solution_9": "I hate to be one of those people who are always criticizing other peoples' posts...\r\n\r\n[quote=\"isabella2296\"]4. Keep in mind that each thread is only, and I repeat, only, for purposes of solving the problem shown. \n\n-- Be careful not to stray off topic. [/quote]\r\n\r\nModerators, please lock.",
  "Solution_10": "What is sum of number from 11 to 55 with reversed digits ?\r\n :roll:",
  "Solution_11": "uhh dude \r\n[quote]I see you are new, but if you want to post a new problem, you should post a new topic unless it is related. \nBTW, the answer can be found using your calculator, which is .\n[/quote]\r\nsoo don't need to post a np... :wink:",
  "Solution_12": "What is the expanded value of x(x+5)(x+5+3a)?  Good luck.",
  "Solution_13": "No offense but [b] DUDE R U BLIND! CAN YOU READ WAT I POSTED BEFORE U![/b]",
  "Solution_14": "[quote=\"tcjy\"]The age of a man is the same as his wife's age with the digits reversed. The sum of their ages is 99 and the man is 9 years older than his wife. How old is the man?[/quote]\r\nA number reversed so that it is 9 less than the original number has to have a ones digit 1 less than the 10s so you find two reasonable tens digits like 10 20 30 that when doubled is around the 90s or 100s which it should be 5 then subtract 1 and use that as your ones digit so 54 is the mans age and revearsed is 45 which 45+54=99.  Or you could use algebra which is slower",
  "Solution_15": "FOR THE LAST TIME STOP REVIVING.\r\nthere my angers out. rip now.",
  "Solution_16": "Why do you have a problem with people doing what they want with their posting privileges? If you're going to post, at least contribute to the discussion, not be a \"moderator\" Also there was never a rule against reviving.\r\n\r\n[quote=\"cgyao15\"]No offense but [b]DUDE R U BLIND! CAN YOU READ WAT I POSTED BEFORE U![/b][/quote]\r\nNot the best thing to say to get on admins' good side ;)\r\n\r\nAgain, moderators/admins, please lock this thread.",
  "Solution_17": "[i]Contents originally posted by my sister while I was logged on under the Automatic Login option, now deleted by me.[/i]",
  "Solution_18": "the wife's age= (99-9)/2=45\r\nthe man is 54"
}
{
  "Tag": [
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "the sequence $u_n$ has the following properties:\r\n\r\n$\\forall n \\exists i>n$ s.t. $u_i>0$\r\n$\\forall n \\exists j<n$ s.t. $u_j<0$\r\n$\\forall i\\in\\mathbb{Z^+}: i-j|u_i-u_j$\r\n\r\nprove every positive integer appears exactly once in $\\{u_n\\}$",
  "Solution_1": "There is a missing hypothesis, because if $\\{U_n\\}$ has the desired properties then so has $\\{kU_n \\}$ for each non-zero integer $k$, but in this last case you will only have the multiples of $k$...\r\n\r\nPierre."
}
{
  "Tag": [
    "limit",
    "integration",
    "function",
    "calculus",
    "algebra",
    "polynomial",
    "Gauss"
  ],
  "Problem": "Show that [tex]\\displaystyle\\lim_{n\\rightarrow\\infty}n\\int_0^1 x^nf(x)\\,dx = f(1)[/tex], where the function [tex]f[/tex] is continuous at 1 and Riemann integrable on the interval [tex][0,\\,1][/tex].\r\n\r\n(Edited) Thanks to JBL for pointing out an error in the statement.\r\n\r\nThis was one of my favorite exercises using [tex]\\epsilon-\\delta, \\epsilon-\\-N[/tex] language for limit problems when I was a freshman.",
  "Solution_1": "I assume that you also want the function to be integrable.  In that case, our function f must be a uniform limit of step functions [tex]f_n:[0, 1]\\to\\mathbb{R}[/tex], defined by points [tex]\\{a_0, a_1, ..., a_m\\}\\subset[0,1], a_i<a_{i+1}, a_0 = 0, a_m = 1[/tex], f constant on [tex](a_i,a_{i+1}][/tex].  Since the convergence is uniform, we may switch our limit outside the integral.  It is easy to verify the given statement in the case of step functions and the desired result follows for all integrable functions immediately.",
  "Solution_2": "Fuzzy what grade are you in?",
  "Solution_3": "[quote=\"gauss202\"]Fuzzy what grade are you in?[/quote]\r\n\r\nRetired stock trader. math is my hobby. A lot of stuff is coming from memory. Sometimes, statements in my post may not be entirely accurate. Sorry for the inconvenience if that happens.",
  "Solution_4": "[quote=\"JBL\"]I assume that you also want the function to be integrable.  In that case, our function f must be a uniform limit of step functions [tex]f_n:[0, 1]\\to\\mathbb{R}[/tex], defined by points [tex]\\{a_0, a_1, ..., a_m\\}\\subset[0,1], a_i<a_{i+1}, a_0 = 0, a_m = 1[/tex], f constant on [tex](a_i,a_{i+1}][/tex].  Since the convergence is uniform, we may switch our limit outside the integral.  It is easy to verify the given statement in the case of step functions and the desired result follows for all integrable functions immediately.[/quote]\r\n\r\nDo you need the condition that [tex]f[/tex] is continuous at 1?",
  "Solution_5": "No, that's fine. I was just wondering because the last few problems you've posted have been rather advanced for a secondary school student. They're good questions though.",
  "Solution_6": "Hmm, that's a good question.  It is definitely easy to verify for step functions ... I think the nature of the convergence and how I constructed the functions (so that each of the step functions are continuous at 1) implies that f is continuous at 1.  At least, I hope it does, because otherwise there is obviously something wrong with my proof :)\r\n\r\nPerhaps I should actually try to do it with epsilons :)",
  "Solution_7": "I think we need the condition that f is continuous at 1.  Otherwise we could change f(1) to anything we want, and falsify the equation.",
  "Solution_8": "Right, we need it.  I purported to have solved the question.  Thus, if I am correct, I must have used that property somewhere.",
  "Solution_9": "JBL,\r\n\r\nAssume that [tex]f(x)=1[/tex] for all [tex]x\\in[0,\\,1][/tex], then [tex]\\displaystyle\\lim_{n\\rightarrow\\infty}n\\int_0^1 x^nf(x)\\,dx = \\lim_{n\\rightarrow\\infty}\\frac{n}{n+1}=1[/tex]. On the other hand, if we switch the limit and the integral, we have [tex]\\displaystyle\\int_0^1\\lim_{n\\rightarrow\\infty}\\left(n x^nf(x)\\right)\\,dx = 0[/tex]. Therefore, your switching method doesn't work here.\r\n\r\nIt's time to try to do it with epsilons. :-)",
  "Solution_10": "JBL wrote:f must be a uniform limit of step functions.\n\n\n\nI don't think this is true for *all* Riemann integrable functions, only a strict sub-set (so-called \"ruled functions\"). For instance f(x) = sin(1/x), f(0)=0 is Riemann integrable but not ruled.\n\n\n\nI think this may work :\n\n\n\n[hide]Let eps > 0, there exist a > 0 such as |1-x|<=a => |f(x)-f(1)|< eps, or f(x) = f(1) + g(x) with |g(x)|< eps.\n\n\n\nLet M >= 0 a bound of |f|\n\n\n\nI(n) = int (0..1)(x^n f(x)) = int (0..a)(x^n f(x)) + int (a..1)(x^n f(x))\n\n\n\nI1(n) = | int (0..a)(x^n f(x)) | <= a^n / (n+1) M\n\n\n\nint (a..1)(x^n f(x)) = int (a..1)(x^n f(1)) + int (a..1)(x^n g(x))\n\n\n\nI2(n) = int (a..1)(x^n f(1)) = f(1) (1-a^n)/(n+1) \n\nI3(n) = | int (a..1)(x^n g(x))| <= eps (1-a^n)/(n+1) <= eps\n\n\n\nI1(n) -> 0, so for n >= N0, |I1(n)| < eps\n\nI2(n) -> f(1), so for n >= N1, |I2(n) - f(1)| < eps\n\n\n\nSo for n > max(N1,N2), | I(n) - f(1) | < 3*eps. QED.[/hide]",
  "Solution_11": "belenos, excellent job! (The main idea is right, but you missed n in the front of the integral).\r\n\r\nThe original version of this problem assume that [tex]f[/tex] is continuous on the whole interval [tex][0,\\,1][/tex]. Then I realized that it can be approximated uniformly by a sequence of polynomials (the therorem itself is more advanced than this problem!). Using the approximation (I am not sure whether it works or not) might defeat the original purpose (an exercise for epsilons :-)).\r\n\r\nI am going to ask you the same question that gauss asked me a while ago :-) in this thread. [quote]what grade are you in?[/quote]\r\nBut I am putting up the reason first so that I don't scare you off.  I am just wondering because you solved many \"Advanced \" problems posted here. By the way, you don't have to answer my question if you don't feel comfortable.",
  "Solution_12": "[quote=\"fuzzylogic\"] what grade are you in?[/quote]\r\nNo problem, see private message  :)",
  "Solution_13": "[quote=\"belenos\"][quote=\"fuzzylogic\"] what grade are you in?[/quote]\nNo problem, see private message  :)[/quote]\r\n\r\nYou are certainly smarter than me! :-)",
  "Solution_14": "[quote=\"fuzzylogic\"][quote=\"gauss202\"]Fuzzy what grade are you in?[/quote]\n\nRetired stock trader. math is my hobby. A lot of stuff is coming from memory. Sometimes, statements in my post may not be entirely accurate. Sorry for the inconvenience if that happens.[/quote]\r\n\r\n\r\nwhew. all this time I thought you were a brilliant 8th grader for some reason.",
  "Solution_15": "I actually wasn't trying to switch that limit -- if I was correct and we can write every Riemann-Integrable function as a uniform limit of step functions, then we write f as a limit of functions of x and switch that limit outside the integral.  Because that limit would be uniform, we should be able to do the switching.\r\n\r\nI had thought that Riemann-integrable functions were equivalent to ruled functions, but I could be wrong -- I don't even know explicitly what Riemann-integrable means.  Are you sure Sin(1/x) is riemann-integrable?",
  "Solution_16": "Riemann integrable : \r\n\r\nit's explained (briefly!) there : http://planetmath.org/encyclopedia/RiemannIntegral.html\r\n\r\n\r\nsin(1/x) : \r\n\r\na very strong (and beautiful) result from Lebesgue is the following :\r\n\r\nA bounded function is Riemann integrable iff the set of points at which f is discontinuous has Lebesgue measure zero.\r\n\r\nsin(1/x) is discontinuous only at 0."
}
{
  "Tag": [
    "algebra",
    "linear equation"
  ],
  "Problem": "This is a type of magic square. The eight sums from the numbers in each of the 3 rows, in each of the 3 columns, and in each of the 2 main diagonals are equal. What is the value of $ n$?\n\n[asy]for(int i = 0; i&lt;4; ++i)\n{\n\tdraw((0,i)--(3,i),linewidth(1));\n}\n\nfor(int j = 0; j&lt;4; ++j)\n{\n\tdraw((j,0)--(j,3),linewidth(1));\n}\n\nlabel(\"$n-3$\",(.5,.5));\nlabel(\"3\",(.5,1.5));\nlabel(\"$n+1$\",(.5,2.5));\n\nlabel(\"$n+2$\",(1.5,.5));\nlabel(\"$2n-9$\",(1.5,1.5));\nlabel(\"$1$\",(1.5,2.5));\n\nlabel(\"$2$\",(2.5,.5));\nlabel(\"$n$\",(2.5,1.5));\nlabel(\"$n-1$\",(2.5,2.5));[/asy]",
  "Solution_1": "We can just use two of the rows, columns, or diagonals to write a linear equation, and solve.  Using the first column and the second column yields:\r\n\r\n\\begin{align*}\r\n(n+1)+(3)+(n-3)&=(1)+(2n-9)+(n+2)\r\n\\\\2n+1&=3n-6\r\n\\\\n&=\\boxed{7}\r\n\\end{align*}"
}
{
  "Tag": [
    "group theory",
    "abstract algebra",
    "superior algebra",
    "superior algebra theorems"
  ],
  "Problem": "Hi,\r\n\r\nIs it true that if $G_1$ is isomorph to $G_2$ and the normal subgroups $H_1$ of $G_1$ and $H_2$ of $G_2$ are isomorph, the quotient groups $G_1/H_1$ and $G_2/H_2$ are isomorph ? It's true if it's the same morphism restricted but is it true in general ?\r\n\r\nThank you",
  "Solution_1": "Consider $C_4 \\times C_2 = \\langle g \\rangle \\times \\langle h \\rangle$.  The subgroups $\\langle g^2 \\rangle$ and $\\langle h \\rangle$ are isomorphic, being $C_2$, but the quotients are $C_2 \\times C_2$ and $C_4$.",
  "Solution_2": "Thanks  :)"
}
{
  "Tag": [
    "probability"
  ],
  "Problem": "1. Bachukas draw a triangle on a plane, what's the probability that a triangle is acute?\r\n(This problem is equivalent to: 'a+b+c=180; a,b,c - positive real numbers. What's the probability that a<90,b<90 and c<90?')\r\n\r\n2. Bachukas draw a hexagon on a plane, what's the probability that a hexagon is convex?\r\n\r\nI guess that answer to my first problem is $ 1/2$ and the answer to my second problem is $ 1/2^3\\equal{}1/8$, am I right?",
  "Solution_1": "For your first one (of course, I'm not sure), I'm leaning towards $ \\frac12$ as well.\r\n\r\nFor your second one, I think it may be $ \\left(\\frac12\\right)^5$ or $ \\frac1{32}$.  But I think it may be smaller.",
  "Solution_2": "Well, the question more complex than you could imagine :)\r\n\r\nhttp://www.jstor.org/pss/3616588\r\nhttp://jd2718.wordpress.com/2007/09/01/prepuzzle-puzzle-what-is-a-random-triangle/\r\nhttp://www.ratio.huji.ac.il/dp_files/dp235.doc",
  "Solution_3": "For the first one, there are $ 4$ conditions:\r\na) $ a < 90$\r\nb) $ b < 90$\r\nc) $ 180 \\minus{} a \\minus{} b < 90\\implies a \\plus{} b > 90$\r\nd) Also, we must have $ a\\plus{}b<180$, as this is a triangle.\r\n\r\nNow, on a graph with $ a$ as the x-axis and $ b$ as the y-axis, this looks like\r\n[geogebra]d122927f186efb65c75eeb127df10815db9cda85[/geogebra] \r\n\r\nYou can tell that the probability is $ \\frac14$. \r\n\r\nThe hexagon is much more complicated. :wink:\r\n\r\nEDIT: Diagram fixed. Bachukas, that's only considering integer values, which isn't \"rigorous\".",
  "Solution_4": "Thanks hsiljak, answer to my first problem is $ \\frac{1}{4}$, not $ \\frac{1}{2}$  :o \r\nI found fast way of solving that problem: \r\nWhat would be the answer, if degrees could be only positive integers?\r\n\r\nIt's $ \\frac{\\binom{92}{2}}{\\binom{182}{2}}\\equal{}\\frac{4186}{16471}\\approx\\frac{1}{4}$",
  "Solution_5": "Solution to my second problem:\r\n\r\nThe probability that a random triangle is convex - 1.\r\n\r\nThe probability that a random quadriteral is convex - 1/2 :\r\nWe choose a point in one of the triangle's side, we can move that point either inside, or outside of the triangle.\r\n\r\nThe probability that a random pentagon is convex - 1/2*1/2:\r\nWe choose a point in one of the convex quadriteral's side, we can move that point either inside, or outside of the convex quadriteral.\r\n\r\nThe probability that a random hexagon is convex - 1/2*1/2*1/2:\r\nWe choose a point in one of the convex pentagon's side, we can move that point either inside, or outside of the convex pentagon. \r\n\r\n\r\nWhat are you thinking about this solution?  :maybe:"
}
{
  "Tag": [],
  "Problem": ":D solve for x and y if x+y=5,and x^y+y^x=17",
  "Solution_1": "[quote=\"williams\"]:D solve for x and y if x+y=5,and x^y+y^x=17[/quote]\r\n[hide]$ (2,3)$\n[/hide]",
  "Solution_2": "[hide]and $ (3,2)$. You now need to show that these are the only solutions, i.e. that the curves $ x^{5 \\minus{} x}$ and $ 17 \\minus{} (5 \\minus{} x)^x$ intersect in exactly two points-it isn't hard, with or without calculus.[/hide]"
}
{
  "Tag": [
    "\\/closed"
  ],
  "Problem": "I was just curious if there was any way you could have a more structured Location thing in the Profile box... Would it be possible to have a list of countries, and if they were from the U.S.  a list of states to choose from as well?  \r\n\r\nAlso, is there any way you can allow students to put in their grade?  I think that would be nifty!\r\n\r\nLemme know.  :D",
  "Solution_1": "Won't happen soon, but we'll put it on the todo list.",
  "Solution_2": "[quote=\"Ragingg\"]Also, is there any way you can allow students to put in their grade?  I think that would be nifty![/quote]\n\nWould be quite hard I think... since there are many students who are partially in more than 1 grade :)\n\n[quote=\"Ragingg\"]Would it be possible to have a list of countries[/quote]\r\n\r\nIsn't that there? :?\r\nwhen selecting flag I could not type my country, could only select it in a list..."
}
{
  "Tag": [
    "function",
    "floor function",
    "induction",
    "algebra"
  ],
  "Problem": "Calculate the sum :\r\n\\[ \\left[\\frac {n \\plus{} 1}{2}\\right] \\plus{} \\left[\\frac {n \\plus{} 2}{2^2}\\right] \\plus{} ... \\plus{} \\left[\\frac {n \\plus{} 2^k}{2^{k \\plus{} 1}}\\right] \\plus{} ...\r\n\\]",
  "Solution_1": "I assume the square brackets denote the floor function?\r\n\r\nEdit: ... and I assume also that you mean $ \\left\\lfloor \\frac {n \\plus{} k}{2^{k \\plus{} 1}} \\right\\rfloor$? Or is it $ \\left\\lfloor \\frac {n \\plus{} 2^k}{2^{k \\plus{} 1}} \\right\\rfloor$?",
  "Solution_2": "[quote=\"worthawholebean\"]I assume the square brackets denote the floor function?[/quote]\r\n\r\nGreatest integer function!",
  "Solution_3": "[hide=\"Hint\"] Let $ S_n$ be the sum in question and compute $ S_{n\\plus{}1} \\minus{} S_n$. [/hide]",
  "Solution_4": "floor (2x)-floor(x)=floor(x+1/2)",
  "Solution_5": "[hide=\"Solution\"]Let $ n\\equal{}a_m2^m\\plus{}a_{n\\minus{}1}2^m\\minus{}1\\plus{}...\\plus{}a_0$ ; $ a_m\\equal{}1$\n\\[ a_1,...,a_{m\\minus{}1} \\in {0,1}\\]\nIf $ k>m$ $ \\boxed{\\left[\\frac{n\\plus{}2^k}{2^{k\\plus{}1}}\\right]\\equal{}0}$\nIf $ k\\equal{}m$ $ \\boxed{\\left[\\frac{n\\plus{}2^k}{2^{k\\plus{}1}}\\right]\\equal{}1\\equal{}a_m}$\nFor $ k<m$ we have the cases:\n[color=red][b]1)[/b][/color]\n$ a_k\\equal{}0$ $ \\implies$ $ n\\plus{}2^k\\equal{}a_m2^m\\plus{}...\\plus{}a_{k\\plus{}1}2^{k\\plus{}1}\\plus{}2^k\\plus{}a_{n\\minus{}1}2^{k\\minus{}1}\\plus{}...\\plus{}a_0$\n\\[ \\left[\\frac{n\\plus{}2^k}{2^{k\\plus{}1}}\\right]\\equal{}a_m2^{m\\minus{}k\\plus{}1}\\plus{}a_{m\\minus{}1}2^{m\\minus{}1\\minus{}k\\minus{}1}\\plus{}...\\plus{}a_{k\\plus{}1}.\\]\n[color=red][b]2)[/b][/color]\n$ a_k\\equal{}1$ therefore $ \\left[\\frac{n\\plus{}2^k}{2^{k\\plus{}1}}\\right]\\equal{}a_m2^{m\\minus{}k\\minus{}1}\\plus{}a_{m\\minus{}1}2^{m\\minus{}1\\minus{}k\\minus{}1}\\plus{}...\\plus{}a_{k\\plus{}1}\\plus{}1$\n\nIn both cases we have:$ \\left[\\frac{n\\plus{}2^k}{2^{k\\plus{}1}}\\right]\\equal{}a_m2^{m\\minus{}k\\minus{}1}\\plus{}a_{m\\minus{}1}2^{m\\minus{}k\\minus{}2}\\plus{}...\\plus{}a_{k\\plus{}1}\\plus{}a_k$\n\\[ \\sum_{k\\equal{}0}^{\\infty} \\left[\\frac{n\\plus{}2^k}{2^{k\\plus{}1}}\\right]\\equal{}a_m(2^{m\\minus{}1}\\plus{}2^{m\\minus{}2}\\plus{}...\\plus{}2\\plus{}1)\\plus{}a_m\\plus{}a_{m\\minus{}1}(2^{m\\minus{}2}\\plus{}2^{m\\minus{}1}\\plus{}...\\plus{}1)\\plus{}a_{m\\minus{}1}\\plus{}...\\plus{}a_2(2\\plus{}1)\\plus{}a_2\\plus{}a_1\\plus{}a_1\\plus{}a_0\\equal{}a_m(2^m\\minus{}1\\plus{}1)\\plus{}a_{m\\minus{}1}(2^{m\\minus{}1}\\minus{}1\\plus{}1)\\plus{}...\\plus{}a_2(2^2\\minus{}1\\plus{}1)\\plus{}a_1(1\\plus{}1)\\plus{}a_0\\equal{}a_m2^m\\plus{}a_{m\\minus{}1}2^{m\\minus{}1}\\plus{}...\\plus{}a_1 \\cdot 2\\plus{}a_0\\equal{}n\\]\n And this can be proved with induction![/hide]"
}
{
  "Tag": [
    "trigonometry"
  ],
  "Problem": "Solve for all $x$ \\[ \\cos^2{x}+\\cos^2{2x}+\\cos^2{3x}=1  \\]",
  "Solution_1": "[hide=\"Here are two short and complete solutions.\"][b]The first method.[/b]\n$2\\left(\\cos^2x+\\cos^22x+\\cos^23x\\right)=2\\Longleftrightarrow$\n$2\\cos^2x+(1+\\cos 4x)+(1+\\cos 6x)=2\\Longleftrightarrow$\n$2\\cos^2x+\\cos 4x+\\cos 6x=0\\Longleftrightarrow$\n$2\\cos^2x+2\\cos 5x\\cos x=0\\Longleftrightarrow$\n$\\cos x(\\cos x+\\cos 5x)=0\\Longleftrightarrow$\n$\\cos x\\cos 2x\\cos 3x=0\\Longleftrightarrow$\n$\\boxed {x\\in \\pi Z+A,\\ \\mathrm {\\ where\\ }A=\\left\\{\\frac{\\pi}{6}\\ ,\\ \\frac{\\pi}{4}\\ ,\\frac{\\pi}{2}\\ ,\\ \\frac{3\\pi}{4}\\ ,\\ \\frac{5\\pi}{6}\\right\\}}\\ .$\n[b]The second method.[/b]\nDenote $z=\\cos x+i\\cdot \\sin x$, $x\\in [0,2\\pi )$. Thus, $z\\cdot \\overline z=1$ and for any $n\\in N$,\n$z^n=\\cos nx+i\\cdot \\sin nx$. Therefore, $\\boxed {\\ \\cos nx=\\frac{z^{2n}+1}{2z^n}\\ }\\ .$\nThe our equation becomes: $\\left(\\frac{z^2+1}{2z}\\right)^2+\\left(\\frac{z^4+1}{2z^2}\\right)^2+\\left(\\frac{z^6+1}{2z^3}\\right)^2=1\\Longleftrightarrow$\n$z^{12}+z^{10}+z^8+2z^6+z^4+z^2+1=0\\ \\ \\boxed {z^2=t}\\Longleftrightarrow$\n$t^6+t^5+t^4+2t^3+t^2+t+1=0\\ \\ \\boxed {t+\\frac 1t=u}\\Longleftrightarrow$\n$u(u^2+u-2)=0\\Longleftrightarrow u\\in \\{-2,0,1\\}\\Longleftrightarrow t\\in \\left\\{-1,\\pm i,\\frac{1\\pm i\\sqrt 3}{2}\\right\\}\\Longleftrightarrow$\n$\\boxed {z\\in \\left\\{\\pm i,\\pm \\frac{\\sqrt 2}{2}(1\\pm i),\\pm \\frac 12(\\sqrt 3\\pm i)\\right\\}}$.[/hide]",
  "Solution_2": "[quote=\"Virgil Nicula\"][hide=\"Here are two short and complete solutions.\"][b]The first method.[/b]\n$2\\left(\\cos^2x+\\cos^22x+\\cos^23x\\right)=2\\Longleftrightarrow$\n$2\\cos^2x+(1+\\cos 4x)+(1+\\cos 6x)=2\\Longleftrightarrow$\n$2\\cos^2x+\\cos 4x+\\cos 6x=0\\Longleftrightarrow$\n$2\\cos^2x+2\\cos 5x\\cos x=0\\Longleftrightarrow$\n$\\cos x(\\cos x+\\cos 5x)=0\\Longleftrightarrow$\n$\\cos x\\cos 2x\\cos 3x=0\\Longleftrightarrow$\n$\\boxed {x\\in \\pi Z+A,\\ \\mathrm {\\ where\\ }A=\\left\\{\\frac{\\pi}{6}\\ ,\\ \\frac{\\pi}{4}\\ ,\\frac{\\pi}{2}\\ ,\\ \\frac{3\\pi}{4}\\ ,\\ \\frac{5\\pi}{6}\\right\\}}\\ .$\n[b]The second method.[/b]\nDenote $z=\\cos x+i\\cdot \\sin x$, $x\\in [0,2\\pi )$. Thus, $z\\cdot \\overline z=1$ and for any $n\\in N$,\n$z^n=\\cos nx+i\\cdot \\sin nx$. Therefore, $\\boxed {\\ \\cos nx=\\frac{z^{2n}+1}{2z^n}\\ }\\ .$\nThe our equation becomes: $\\left(\\frac{z^2+1}{2z}\\right)^2+\\left(\\frac{z^4+1}{2z^2}\\right)^2+\\left(\\frac{z^6+1}{2z^3}\\right)^2=1\\Longleftrightarrow$\n$z^{12}+z^{10}+z^8+2z^6+z^4+z^2+1=0\\ \\ \\boxed {z^2=t}\\Longleftrightarrow$\n$t^6+t^5+t^4+2t^3+t^2+t+1=0\\ \\ \\boxed {t+\\frac 1t=u}\\Longleftrightarrow$\n$u(u^2+u-2)=0\\Longleftrightarrow u\\in \\{-2,0,1\\}\\Longleftrightarrow t\\in \\left\\{-1,\\pm i,\\frac{1\\pm i\\sqrt 3}{2}\\right\\}\\Longleftrightarrow$\n$\\boxed {z\\in \\left\\{\\pm i,\\pm \\frac{\\sqrt 2}{2}(1\\pm i),\\pm \\frac 12(\\sqrt 3\\pm i)\\right\\}}$.[/hide][/quote]\r\n[b]Remark.[/b] \\[ \\boxed {\\ z^2=a+bi\\ ,\\ a,b\\in R\\ ,\\ b\\ne 0\\ ,\\ \\rho=\\sqrt{a^2+b^2}\\ } \\] \\[ \\Downarrow\\Downarrow\\Downarrow \\] \\[ \\boxed {\\ z_{1,2}=\\pm\\left(\\sqrt{\\frac{\\rho +a}{2}}+i\\cdot \\mathrm {sgn}b\\cdot \\sqrt{\\frac{\\rho-a}{2}}\\right)\\ } \\]",
  "Solution_3": "Nice solutions.\r\n\r\nBut what is $z_{1,2}$?",
  "Solution_4": "$z_{1,2}$ are the roots of the equation $Z^2=a+bi$, i.e. the $2$-roots of the complex number $a+bi$. Use for the above substitution $z^2=t$."
}
{
  "Tag": [
    "limit",
    "calculus",
    "calculus computations"
  ],
  "Problem": "Find $ \\lim_{n\\to\\infty}\\sum\\limits_{k\\,\\equal{}\\,1}^{n}{\\frac{k\\minus{}n^{2}}{n^{2}\\sqrt{n^{2}\\plus{}k}}}.$",
  "Solution_1": "You may use the estimation $ \\frac{n^2 \\minus{} n}{n^2 \\sqrt{n^2 \\plus{} 2n \\plus{} 1}} \\leq \\frac{n^2 \\minus{} k}{n^2 \\sqrt{n^2 \\plus{} k}} \\leq \\frac{n^2}{n^2 \\sqrt{n^2}}.$",
  "Solution_2": "Thank you, but I got a question: how do you make those estimations?",
  "Solution_3": "btw/ in case ppl are wondering, the answer is $ \\minus{}1$. at least that is what i get using riemann summation. :|"
}
{
  "Tag": [
    "geometry",
    "rotation",
    "function",
    "algebra",
    "system of equations"
  ],
  "Problem": "I found these Problems Sets from [url]http://www.chem.utoronto.ca/IChO.Ontario/problems/english/englishprob.htm[/url] and I can't find any answers so I thought it might be good if we solve them because I have solved some of them but I am not sure if I got them correct. So just feel free to post answers or thoughts on these and try to keep it in order (so set 1,2,3.....).\r\n\r\nThanks\r\n\r\nP.S. They are not very hard.",
  "Solution_1": "Problem Set 1, Problem 1\r\n\r\n\"1. Element X reacts with oxygen to produce a pure sample of X2O3. In an experiment, it is found that 1.0000 g of X produces 1.1596 g of X2O3. If the atomic weight of oxygen is known to be 15.9994 g/mol, what is the atomic weight of X?\r\n(a) 20.70 (b) 66.85 (c) 100.2 (d) 150.4 (e) 168.9\"\r\n\r\nX reacts with oxygen according to the chemical equation\r\n\r\n$4X+3O_{2}\\longrightarrow 2X_{2}O_{3}$\r\n\r\nSo, the number of moles of X, $N_{X}$, that react is double of number of moles of oxide produced, $N_{P}$. But the number of moles is given by the quotient of mass and molar mass. So:\r\n\r\n$\\frac{1.0000g(X)}{M_{X}}= 2\\frac{1.1596g(X2O3)}{2M_{X}+3M_{O}}$\r\n\r\nwhere $M_{X}$ and $M_{O}$ are the molar masses of X and oxygen. Solving this we get $M_{X}= 150.4 g/mol$, and the correct answer is (d).",
  "Solution_2": "Problem Set 1, Problem 2\r\n\r\n\"2. An ore contains 1.34% of the mineral argentite, Ag2S, by weight. How many grams of this ore would have to be processed in order to obtain 1.00 g of pure solid silver, Ag?\r\n(a) 74.6 g (b) 85.7 g (c) 107.9 g (d) 134.0 g (e) 171.4 g\"\r\n\r\nWhatever method is used to reduce the sulphide to the metal, for example\r\n\r\n$Ag_{2}S+2H^{+}\\longrightarrow 2Ag+H_{2}S$,\r\n\r\nwe always have that, from 1 mol of Ag2S we get 2 mol of Ag. As 1.00 g of silver correspond to 9.27 mmole, then we must use 4.64 mmole of sulphide, which corresponds to a mass of 1.15 g. Now, in 100 g of ore there is 1.34 g of pure sulphide, so the mass of ore that contains 1.15 g of Ag2S is 85.8 g. The correct answer is (b).",
  "Solution_3": "Problem Set 1, Problem 3\r\n\r\n\"3. When pentane, C5H12, is burned in excess oxygen, the products of the reaction are CO2(g) and H2O(l). In the balanced equation for this combustion\r\n\r\nC5H12(g) + a O2(g) \u2192 b CO2(g) + c H2O(l)\r\n\r\nThe coefficient a should be\r\n\r\n(a) 16 (b) 12 (c) 11 (d) 8 (e) 6.\"\r\n\r\nFrom the stoicheometry of the reaction we must have:\r\n\r\n$b = 5; 12 = 2c; 2a = 2b+c$.\r\n\r\nSolving this system of equations we get b = 5, c = 6, and a = 8. So the correct answer is (d).",
  "Solution_4": "Problem Set 1, Problem 4\r\n\r\n\"4. Cortisone is a molecular substance containing 21 atoms of carbon per molecule. The weight percentage of carbon in cortisone is 69.98%. What is the molecular weight of cortisone?\r\n(a) 176.5 (b) 252.2 (c) 287.6 (d) 312.8 (e) 360.4.\"\r\n\r\nAs the weight percentage of carbon in cortisone is 69.98%, that means that in 100 g of cortisone we will have 69.98 g of carbon. So in M g of cortisone (where M is the molecular weight) we will have \r\n\r\n$m_{C}= \\frac{69.98M}{100}g$\r\n\r\nof carbon. But, as each molecule of cortisone has 21 atoms of carbon, then this mass must also equal (knowing that the molar mass of carbon is 12.011 g/mol) $m_{C}= 21\\cdot12.011$. Equating these two relations we get M = 360.4 g/mol.\r\nThe correct answer is (e).",
  "Solution_5": "Problem Set 1, Problem 5\r\n\r\n\"5. Which atom has the same number of neutrons as 85Rb?\r\n\r\n(a) 85Kr (b) 87Rb (c) 85Sr (d) 86Sr (e) 86Kr.\"\r\n\r\nAn atom with an atomic number one unit greater and a massa number also one unit greater will have the same number of neutrons. So the answer is (d).\r\n\r\n\r\nProblem Set 1, Problem 6\r\n\r\n\"6. An aqueous solution of which of the following compounds will not conduct\r\nelectricity ?\r\n(a) CH3OH (b) Ba(OH)2 (c) KOH (d) Ca(CH3COO)2\r\n(e) All of compounds (a)-(d) will conduct electricity.\"\r\n\r\nCompounds (b)-(c) are ionic, and so in aquous solution will dissociate in ions. Compound (a), methanol, is a molecular compound that will not ionize appreciably in solution. So the answer is (a).\r\n\r\n\r\nProblem Set 1, Problem 7\r\n\r\n\"7. The first ionization energy of an element is the energy required to remove one electron from a gaseous atom of that element, that is, it is the energy required for the reaction X(g) \u2192 X+(g) + e\u2212 where X stands for any element. Which of the following elements would you expect to have the lowest first ionization energy?\r\n(a) Mg (b) Rb (c) Li (d) Ca (e) Be.\"\r\n\r\nThe first ionization energy decreases down a group and increases along a period (with some exceptions). So Rubidium is the atom, from the ones given, with the lowest first ionization energy. Answer (b).\r\n\r\n\r\nProblem Set 1, Problem 8\r\n\r\n\"8. The name of the SO32\u2212 ion is\r\n(a) sulfite (b) sulfate (c) sulfide (d) sulfurate (e) bisulfide.\"\r\n\r\nAnswer (a), sulfite.",
  "Solution_6": "$5.$ Which atom has the same number of neutrons as $^{85}Rb$?\r\n$(a)^{85}Kr \\qquad (b)^{87}Rb \\qquad (c)^{85}Sr \\qquad (d)^{86}Sr \\qquad (e)^{86}Kr$\r\n[hide=\"Problem Set 1, Problem 5\"]\nSo the order on he table is Krypton, Rubidium, Strontium if I remember correctly with Krypton a noble gas of course.\nLet $Rb$ have $x$ protons so this isotope has $85-x$ neutrons. The answer is either $^{86}Sr$ or $^{86}Kr$ which have $86-(x+1),86-(x-1)$ neutrons respectively so the answer is $(d)$\n[/hide]",
  "Solution_7": "Problem Set 1, Problem 9\r\n\r\n\"9. The combustion of butane, C4H10, yields only CO2(g) and H2O. A sample of butane gas at 23.8\u00b0C and 753.4 mm Hg occupies a volume of 1.85 L. How many litres of CO2(g), measured at the same temperature and pressure, will be formed on complete combustion of this C4H10?\r\n(a) 7.40 L (b) 9.25 L (c) 12.0 L (d) 16.7 L (e) 24.0 L.\"\r\n\r\nThe chemical equation is\r\n\r\n$2C_{4}H_{10}+13O_{2}\\longrightarrow 8CO_{2}+10H_{2}O$.\r\n\r\nSince we are dealing with gases we will naturally approximate their behaviour with the ideal gas law. Now, as from 2 moles of butane we get 8 of carbon dioxide, and from the gas law \r\n\r\n$n = \\frac{PV}{RT}$\r\n\r\nwe conclude that $V(CO_{2}) = 4V(C_{4}H_{10}) = 7.40 L$. The correct answer is (a).",
  "Solution_8": "Problem Set 1, Problem 10\r\n\r\n\"10. At 27\u00b0C and 1.00 atm, the density of a gaseous hydrocarbon is 1.22 g/L. The hydrocarbon is (a) CH4 (b) C2H4 (c) C2H6 (d) C3H8 (e) C3H6.\"\r\n\r\nFrom the perfect gas law we have $M = \\frac{dRT}{P}$, where M is the molar mass and d the density. Substituting by the given values, and using $R = 0.0820578 L atm K^{-1}mol^{-1}$, we get M = 30.05 g/mol. So the correct answer is (c).\r\n\r\n\r\nProblem Set 1, Problem 11\r\n\r\n\"11. A mixture of 50.0 mL of ammonia and 60.0 mL of oxygen gas reacts according to the following equation: 4 NH3(g) + 5 O2(g) \u2192 4 NO(g) + 6 H2O(g).\r\nIf all gases are at the same temperature and pressure, and the reaction continues until one of the gases is completely consumed, what volume of water vapour is produced ?\r\n(a) 48.0 mL (b) 60.0 mL (c) 72.0 mL (d) 75.0 mL (e) 110 mL.\"\r\n\r\nConsider ideal gas behaviour. Avogadro's principle tells us that, under the same conditions of temperature and pressure, the number of moles of a gas is directly proportional to its volume. So, for a given volume of oxygen, the volume of ammonia needed to react completely with the oxygen is given by \r\n\r\n$V(NH3) = \\frac{4V(O2)}{5}$.\r\n\r\nIn the present case, for V(O2) = 60.0 mL we would need V(NH3) = 48.0 mL. But we use 50.0 mL of ammonia, so the limiting reagent is oxygen and reaction continues until all the oxygen is consumed. So, if 5 moles of O2 produce 6 moles of water vapour, then the volume of water produced will be\r\n\r\n$V(H2O) = \\frac{6V(O2)}{5}= 72.0 mL$\r\n\r\nThe correct answer is (c).",
  "Solution_9": "Problem Set 1, Problem 12\r\n\r\n\"12. What is the partial pressure of SO2 if 100 g of O2 are mixed with 100 g of SO2, and the total pressure is 600 mm Hg ?\r\n(a) 500 mm Hg (b) 400 mm Hg (c) 300 mm Hg (d) 200 mm Hg\r\n(e) 100 mm Hg.\"\r\n\r\nWe have\r\n\r\n$P(SO_{2}) = x(SO_{2})P$\r\n\r\nwhere $x(SO_{2})$ is the mole fraction of SO2 and P is the total pressure. From the molar masses of the gases we get 100 g (O2) = 3.12 mol and 100 g (SO2) = 1.56 mol, and so $x(SO_{2}) = 0.333$. Finally, $P(SO_{2}) = 0.333\\cdot600 mm Hg = 200 mm Hg$. The correct answer is (d).\r\n\r\n\r\nProblem Set 1, Problem 13\r\n\r\n\"13. A 0.20-mol sample of a hydrocarbon CxHy yields, after complete combustion with excess O2 gas, 0.80 mol of CO2 and 1.0 mol of H2O. The molecular formula of CxHy is (a) C4H10 (b) C4H8 (c) C4H5 (d) C8H16 (e) C8H10.\"\r\n\r\nIn the complete combustion of an hydrocarbon CxHy, for every mole of CxHy we obtain x moles of CO2 and y/2 moles of H2O. Since the problem tells us that from 0.20 mol of CxHy we get 0.80 mol CO2 and 1.o mol H2O, then we must have x = 4 and y = 10. The answer is (a).",
  "Solution_10": "Problem Set 1, Problem 14\r\n\r\n\"14. A sample of gas is kept in a closed container of fixed volume at 250 K and 400 mmHg pressure. If the gas is heated to 375 K, its pressure increases to 600 mm Hg. By what factor will the average speed of the molecules increase ?\r\n(a) 1.22 (b) 1.50 (c) 2.25 (d) 2.00\r\n(e) There is not enough information to determine this.\"\r\n\r\nAccording to the kinetic model of gases, the average speed of molecules of gas is directly proportional to the square root of absolute temperature. So, \r\n\r\n$\\frac{v_{2}}{v_{1}}= \\sqrt{\\frac{T_{2}}{T_{1}}}= \\sqrt{\\frac{375K}{250K}}= 1.22$.\r\n\r\nThe correct answer is then (a).",
  "Solution_11": "Problem Set 2, Problem 16\r\n\r\n\"16. On the basis of the relative strengths of intermolecular forces, predict the order of increasing boiling points of the following substances. [N.B. CH3CHO is acetaldehyde or ethanal (IUPAC)]\r\n(a) CH3CH3 < CH3CHO < CH3CH2OH\r\n(b) CH3CH2OH < CH3CH3 < CH3CHO\r\n(c) CH3CHO < CH3CH2OH < CH3CH3\r\n(d) CH3CH2OH < CH3CHO < CH3CH3\r\n(e) CH3CH3 < CH3CH2OH < CH3CHO.\"\r\n\r\nThe CH3CH3 molecules (ethane) are non polar, and so between them there are weak London interactions. The acetaldehyde molecules are polar, but they cannot form hydrogen bonds between them: so the only intermolecular forces operating here are dipole-dipole interactions. Now, between the molecules of ethanol relatively strong hydrogen bonds can and do operate. So the order of increasing boiling point is ethane < ethanal < ethanol, and the correct answer is (a).",
  "Solution_12": "Problem Set 3, Problem 31\r\n\r\n\"31. Ksp for lead iodate, Pb(IO3)2, is 3.2 \u00d7 10\u221214 at a given temperature. What is the solubility of Pb(IO3)2 in mole litre\u20131?\r\n(a) 1.8 \u00d7 10\u20137 (b) 3.6 \u00d7 10\u20137 (c) 2.0 \u00d7 10\u20135 (d) 4.0 \u00d7 10\u20135 (e) 8.0 \u00d7 10\u20135.\"\r\n\r\nThe solubility equilibrium is\r\n\r\n$Pb(IO_{3})_{2}(s)+aq \\rightleftharpoons Pb^{2+}(aq)+2IO_{3}^{-}(aq)$.\r\n\r\nSo we have $[Pb^{2+}] = S$ and $[IO_{3}^{-}] = 2S$ where S is the molar solubility. So $K_{sp}= [Pb^{2+}][IO_{3}^{-}]^{2}= 4S^{3}$ from which we get $S = 2\\times10^{-5}mol/L$.",
  "Solution_13": "Problem Set 3, Problem 32\r\n\r\n\"32. A solution has the following molar concentrations: [Cl\u2013] = 1.5 \u00d7 10\u20131, [Br\u2013] = 5.0 \u00d710\u20134, [CrO42\u2013] = 1.9 \u00d7 10\u20132. A solution of AgNO3 (100% dissociated) is added to the above solution, drop by drop. Which silver salt will precipitate first ?\r\nData: Ksp(AgCl) = 1.5 x 10\u201310; Ksp(AgBr)= 5.0 x 10\u201313; Ksp(Ag2CrO4) = 1.9 x 10\u201312\r\n(a) AgCl (b) AgBr (c) Ag2CrO4 (d) AgCl and AgBr together\r\n(e) None of the above\"\r\n\r\nA given salt will start to precipitate as soon the quotient of reaction equals Ksp, and the salt that will precipitate first is the one that requires the lower concentration of silver ions in solution. The relevant equilibria are:\r\n\r\n$AgCl(s)+aq \\rightleftharpoons Ag^{+}(aq)+Cl^{-}(aq), K_{sp}= [Ag^{+}][Cl^{-}]$\r\n$AgBr(s)+aq \\rightleftharpoons Ag^{+}(aq)+Br^{-}(aq), K_{sp}= [Ag^{+}][Br^{-}]$\r\n$Ag_{2}CrO_{4}(s)+aq \\rightleftharpoons 2Ag^{+}(aq)+CrO_{4}^{2-}(aq), K_{sp}= [Ag^{+}]^{2}[CrO_{4}^{2-}]$.\r\n\r\nFor AgCl we have: $[Ag^{+}] = \\frac{K_{sp}}{[Cl^{-}]}= 1.0\\times10^{-9}M$.\r\nFor AgBr: $[Ag^{+}] = \\frac{K_{sp}}{[Br^{-}]}= 1.0\\times10^{-9}$.\r\nFor Ag2CrO4: $[Ag^{+}] = \\sqrt{\\frac{K_{sp}}{[CrO_{4}^{2-}]}}= 1.0\\times10^{-5}$.\r\nSo AgCl and AgBr will precipitate first and at the same time. The correct answer is (d).",
  "Solution_14": "Problem Set 3, Problem 33\r\n\r\n\"33. BaSO4 is a compound that aids radiographic investigation of the gastrointestinal (GI) tract because it is opaque to radiation and only slightly soluble (Ksp = 1.1 x 10\u201310). Suppose that you had added some solid BaSO4 to pure water to obtain a saturated solution and wished to reduce the Ba2+ concentration to a minimum to prepare it for ingestion by a patient uncommonly allergic to Ba2+. Which procedure might be recommended ?\r\nData: BaSO4(s) \u2192 Ba2+(aq) + SO42\u2013(aq) is endothermic\r\n(a) Heat the system\r\n(b) Add sufficient H2SO4 to give a [H+] = 1.0 molar\r\n(c) Add more solid BaSO4\r\n(d) Reduce the amount of saturated solution over the solid\r\n(e) Add enough Na2SO4 to give a [Na+] = 10\u20132 molar.\"\r\n\r\n(a) As the solubilization process is endothermic, increasing the temperature will favour the direct reaction, so more solid will dissolve and increase the concentration of Ba2+.\r\n\r\n(b) The equilibria that describe the ionization of sulphuric acid in solution are:\r\n$H_{2}SO_{4}(aq)+H_{2}O(l) \\longrightarrow HSO_{4}^{-}(aq)+H^{+}(aq), K very large$\r\n$HSO_{4}^{-}(aq)+H_{2}O(l) \\rightleftharpoons SO_{4}^{2-}(aq)+H^{+}(aq), K small$\r\n\r\nSo the concentration of sulphate ions due to sulphuric acid will be small but the concentration of H+ will be high (1.0 M as stated in the problem), and these H+ will reduce the concentration of SO42-in solution due to\r\n\r\n$SO_{4}^{2-}(aq)+H^{+}\\rightleftharpoons HSO_{4}^{-}(aq)$\r\n\r\nwhich will shift the solubility equilibrium of BaSO4 to the right, dissolving more solid and increasing the concentration of Ba2+.\r\n\r\n(c) Adding more solid will of course favour the direct reaction, dissolving more solid and increasing the concentration of Ba2+.\r\n\r\n(d) The concentration of ions in solution will remane the same, so this will have no effect.\r\n\r\n(e) Adding a salt with a common ion with the solid will shift the equilibrium to the left and diminish the concentration of Ba2+. This is the common ion effect.\r\n\r\nThe correct answer is then (e).",
  "Solution_15": "Problem Set 3, Problem 34\r\n\r\n\"34. A 10.00g sample of Ca(OH)2 (Ksp = 4.7 x 10\u22126) is added to 250.0mL of pure water. Determine the pH of the solution after just enough HCl is added to completely dissolve the Ca(OH)2. Assume that there is no change in the solution volume during the addition of HCl.\r\n(a) 10.50 (b) 11.70 (c) 12.11 (d) 11.47 (e) 10.89.\"\r\n\r\n10.00 g of Ca(HO)2 correspond to 0.135 mol, so when the solid is completely dissolved the same number of moles of Ca2+ will exist in solution, that, with a volume of 0.250 L, correspond to a concentration of 0.54 M. In the end there is no solid but the solubility equilibrium still holds, and we can use it to compute the equilibrium concentration of hydroxide:\r\n\r\n$[HO^{-}] = \\sqrt{\\frac{K_{sp}}{[Ca^{2+}]}}= 2.95\\times10^{-3}M$.\r\n\r\nThis corresponds to a pOH of 2.53 and so a pH of 11.47 - answer (d).",
  "Solution_16": "Problem Set 3, Problem 35\r\n\r\n\"35. The bicarbonate ion, HCO3\u2013, can act as a base by hydrolyzing in aqueous solution. Given only Ka1(H2CO3) and Kw, what is Kb(HCO3\u2013)? Note: Kb(HCO3\u2013) is K for the reaction of HCO3\u2013 acting as base: HCO3\u2013 + H2O \u2194 H2CO3 + OH\u2013\r\n(a) Ka1 x Kw (b) Ka1/Kw (c) Kw/Ka1 (d) Ka12/Kw\r\n(e) There is insufficient data as Ka2(H2CO3) must be given.\"\r\n\r\nWe have $H_{2}CO_{3}(aq) \\rightleftharpoons HCO_{3}^{-}(aq)+H^{+}(aq), K_{a1}$, and so\r\n\r\n$HCO_{3}^{-}(aq)+H^{+}(aq) \\rightleftharpoons H_{2}CO_{3}(aq), \\frac{1}{K_{a1}}$\r\n$H_{2}O(l) \\rightleftharpoons H^{+}(aq)+HO^{-}(aq), K_{w}$.\r\n\r\nAdding these two equilibria we get:\r\n\r\n$HCO_{3}^{-}(aq)+H_{2}O(l) \\rightleftharpoons H_{2}CO_{3}(aq)+HO^{-}(aq), \\frac{K_{w}}{K_{a1}}= K_{b}$.\r\n\r\nThe correct answer is then (c).",
  "Solution_17": "Problem Set 3, Problem 36\r\n\r\n\"36. What is the pH of a 1.0 molar sodium acetate, CH3COONa, solution, given that Ka(CH3COOH) = 1.8 x 10\u20135 and Kw = 1.0 x 10\u201314?\r\n(a) 0.0 (b) 2.4 (c) 4.8 (d) 7.0 (e) 9.4.\"\r\n\r\nWe have:\r\n\r\n$CH_{3}COONa(s)+aq \\longrightarrow CH_{3}COO^{-}(aq)+Na^{+}(aq)$\r\n$CH_{3}COO^{-}(aq)+H_{2}O(l) \\rightleftharpoons CH_{3}COOH(aq)+HO^{-}(aq), K = \\frac{K_{w}}{K_{a}}$.\r\n\r\nThe equilibrium concentrations will be [CH3COOH] = [HO-] = x, [CH3COO-] = 1.0 - x. But, as Kw/Ka is quite small, and the initial concentration of acetate is relatively large, than we can make the approximation [CH3COO-] = 1.0 in the equilibrium. So:\r\n\r\n$\\frac{x^{2}}{1.0}= \\frac{K_{w}}{K_{a}}= 5.6\\times10^{-10}\\Rightarrow  [HO^{-}] = x = 2.4\\times10^{-5}\\Rightarrow  pOH = 4.6 \\Rightarrow  pH = 9.4$.\r\n\r\nThe correct answer is (e).",
  "Solution_18": "Problem Set 3, Problem 42\r\n\r\n\"42. Which of the following statements describes Markovnikov's rule?\r\n(a) In the electrophilic addition of H\u2013Cl to an alkene, the chlorine atom becomes\r\nattached to the carbon which has the most hydrogen atoms.\r\n(b) In the nucleophilic addition of H\u2013Br to an alkene, the major product is formed from the more stable carbocation intermediate.\r\n(c) In the electrophilic addition of H\u2013Br to an alkene, the major product is formed from the more stable radical intermediate.\r\n(d) The transition state for an endothermic process will closely resemble the product.\r\n(e) In the electrophilic addition of H\u2013Cl to an alkene, the chlorine atom becomes\r\nattached to the carbon which has the least hydrogen atoms.\"\r\n\r\nThe Markovnikov's rule states that in the electrophilic addition of a compound A-B to an alkene, where A have a partial or total positive charge and B a partial or total negative charge, A will bond to the carbon atom of the double that possess the greatest number of hydrogen atoms, and B to the other atom. So the correct answer is (e).\r\n\r\n\r\nProblem Set 3, Problem 43\r\n\r\n\"43. Betamethasone is a drug widely prescribed for the treatment of rheumatoid arthritis. How many degrees of unsaturation are present within the molecule?\r\n(a) 3 (b) 4 (c) 7 (d) 8 (e) 10.\"\r\n\r\nEvery double bond and every ring contributes each with one degree of unsaturation, so counting them on the molecule we get the answer (d), 8.\r\n\r\n\r\nProblem Set 3, Problem 44\r\n\r\n\"44. The number of different constitutional isomers generated by radical\r\nmonochlorination of 1,3-diethylcyclopentane is\r\n(a) 4 (b) 5 (c) 9 (d) 12 (e) 18.\"\r\n\r\nThe radical chlorination is rather unselective, so every hydrogen on the molecule will be chlorinated. Noting that 1,3-diethylcyclopentane has a plane of simmetry, then the number of monochlorinated isomers is 5 - answer (b).",
  "Solution_19": "Problem Set 3, Problem 45\r\n\r\n\"45. The correct IUPAC name for the following compound is\r\n(a) (3E)-3,7-dimethyl-1,3,6-octatriene\r\n(b) (2E,5E)-2,6-dimethyl-2,5,7-octatriene\r\n(c) (1Z,3E,6E)-3,7-dimethyl-1,3,6-octatriene\r\n(d) (2Z)-2,6-dimethyl-2,5,7-octatriene\r\n(e) (3E,6E)-3,7-dimethyl-1,3,6-octatriene.\"\r\n\r\nThe compound given has three double bounds, but only the middle one can have cis/trans isomery. We also start to number the carbon chain from the right in order to give the lowest numbers for the double bounds, and we have two methyl groups as substituents in carbons 3 and 7. The geometry of the central double bond is trans or E, so the correct answer is (a).\r\n\r\n\r\nProblem Set 3, Problem 46\r\n\r\n\"46. A chiral compound, C5H8, on catalytic hydrogenation yields an achiral compound, C5H10. What is the original chiral compound?\r\n(a) 1-methylcyclobutene (b) 3-methylcyclobutene\r\n(c) 1,2-dimethylcyclopropene (d) cyclopentene\r\n(e) 3-ethylcyclopropene.\"\r\n\r\nFrom the compounds given, the only one that have a chiral carbon is 3-methylcyclobutene, so the answer is (b).\r\n\r\n\r\nProblem Set 3, Problem 47\r\n\r\n\"47. Which of the following compounds has an infra red absorption band at 3300 cm\u20131 but no band between 1650 and 1850 cm\u20131 ?\r\n(a) CH3COOH (b) CH3OCH3 (c) CH3CONH2 (d) CH3CH2OH (e) CH3COCH3.\"\r\n\r\nThere are three bonds that can have an IR absorption at 3300 cm-1: O-H of alcohols, N-H of amines and amides, and C-H of terminal alkynes. However, for the amide we would also have a carbonyl absorption near 1650 cm-1, so the correct answer (d).",
  "Solution_20": "Problem Set 3, Problem 48\r\n\r\n\"48. Which alkyl halides upon dehydrohalogenation give alkenes that do not show\r\ncis-trans isomerism?\r\ni) 2-chloropentane iv) 1-chloro-2-methylpropane\r\nii) 2-chlorobutane v) 2-chloro-2-methylpropane\r\niii) chlorocyclohexane\r\n(a) i, ii and iv\r\n(b) iii, iv and v\r\n(c) i, iii and iv\r\n(d) ii, iii and v\r\n(e) ii, iii and iv.\"\r\n\r\nFor an alkene to not show cis-trans isomerism it must have two equal groups attached to at least one of its carbons. So the correct answer can only be (iii), (iv), and (v) - (b).\r\n\r\n\r\nProblem Set 3, Problem 49\r\n\r\n\"49. What is the relationship between these two structures?\r\n(a) identical compounds (b) enantiomers (c) diastereomers\r\n(d) cis/trans isomers (e) constitutional isomers.\"\r\n\r\nThe structures are the same compound, since for example the structure on the right, if rotated by 180\u00ba can then be exactly superposed to the one on the left. The answer is (a).\r\n\r\n\r\nProblem Set 3, Problem 50\r\n\r\n\"50. Which one of the following statements is false?\r\n(a) The bond order of C2\u2013 is 2.5.\r\n(b) The predominant source of strain energy in a planar cyclopentane is the deviation\r\nfrom tetrahedral geometry at carbon.\r\n(c) The correct IUPAC name for the following compound is\r\n(E)-3,5,5-trimethyl-3-heptene:\r\n(d) Halogenation of an alkene takes place with anti stereochemistry\r\n(e) The following species can be described as having aromatic properties:\r\n\r\n(a) True (the configuration of C2- would be $\\sigma_{2s}^{2}\\sigma_{2s}^{*2}\\pi_{2p}^{2}\\pi_{2p}^{2}\\sigma_{2p}^{1}$, and $bond order = \\frac{number of bonding electrons-number of antibonding electrons}{2})$.\r\n(b) False.\r\n(c) True.\r\n(d) True.\r\n(e) True.",
  "Solution_21": "Problem Set 3, Problem 39\r\n\r\n\"39. Sodium thiosulfate, Na2S2O3\u00b75 H2O (commonly known as \"hypo\" from an old name, sodium hyposulfite) is used extensively in photography and volumetric analysis. Iodine reacts quantitatively with sodium thiosulfate in water and sodium tetrathionate is formed: 2 Na2S2O3 + I2 \u2192 Na2S4O6 + 2 NaI\r\nWhat are the oxidation numbers of sulfur in the thiosulfate ion, S2O3\r\n2\u2013 and the tetrathionate ion, S4O62\u2013?\r\nS2O32\u2013 S4O62\u2013\r\n(a) \u20132 \u20132.5\r\n(b) \u20130.5 \u20131\r\n(c) +0.5 +1\r\n(d) +2 +2.5\r\n(e) +3 +3.\"\r\n\r\nFor the thiosulfate ion we have 2x -6 = -2 => x = +2. For the tetrathionate ion we have 4x - 12 = -2 => x = +5/2. So the correct answer is (d).\r\n\r\n\r\nProblem Set 3, Problem 40\r\n\r\n\"40. One of the steps in the manufacture of nitric acid from ammonia, NH3, involves the disproportionation (self-oxidation and self-reduction) of nitrous acid, HNO2, according to the following unbalanced equation:\r\na HNO2 \u2192 b NO + c HNO3 + d H2O\r\nWhat are the stoichiometric coefficients when the equation is properly balanced ?\r\na b c d\r\n(a) 1 1 1 1\r\n(b) 2 2 1 1\r\n(c) 5 2 3 1\r\n(d) 6 6 3 3\r\n(e) None of the above.\"\r\n\r\nLet's start by computing the oxidation number of nitrogen, x, in each compound: HNO2, x = +3; NO, x = +2; HNO3, x = +5. So the semi equation of reduction is\r\n\r\n$HNO_{2}\\longrightarrow NO$,\r\n\r\nand the semi equation of oxidation is\r\n\r\n$HNO_{2}\\longrightarrow HNO_{3}$.\r\n\r\nNow, to balance semi equations of reduction or oxidation we follow the following steps:\r\n\r\n1) Balance the number of atoms that undergo red. or ox. in each side of the equation.\r\n2) Add the appropriate number of electrons, according to step 1, to the left if reduction and to the right if oxidation.\r\n3) Balance the number of charges so as to have the same number in each side: with H+ if we are in acid medium and with HO- in alcaline medium.\r\n4) Balance the number of hydrogen and oxygen atoms by adding enough molecules of water to the appropriate side of the equation.\r\n5) If necessary, multiply each semi equation by a number so as to have the same number of electrons transferred in both.\r\n6) Add the two semi equations.\r\n\r\nFollowing these steps we get:\r\n\r\n$HNO_{2}+e^{-}+H^{+}\\longrightarrow NO+H_{2}O$\r\n$HNO_{2}+H_{2}O \\longrightarrow HNO_{3}+2e^{-}+2H^{+}$.\r\n\r\nNow multiply the first one by 2 and sum:\r\n\r\n$3HNO_{2}\\longrightarrow HNO_{3}+2NO+H_{2}O$.\r\n\r\nSo the correct answer is (e).",
  "Solution_22": "Problem Set 3, Problem 41\r\n\r\n\"41. Calcium is the fifth most abundant element in man. The Ca2+ content in the blood of an average adult is 0.90\u20131.15 mg per 10 cm3 blood. In a typical analysis, a 10.0 cm3 blood sample from an adult male patient was diluted to 100 cm3. From a 10.0 cm3 portion of this diluted sample, the Ca2+ was precipitated as CaC2O4. After this precipitate had been dissolved in H2SO4, it required 1.00 cm3 of 0.00500 normal (1.00 x 10\u20133 mole litre\u20131) KMnO4 to titrate to an end point. What is the [Ca2+] expressed as milligrams of Ca2+ per 10.0 cm3 of blood? The equation for the titration reaction is:\r\n5 CaC2O4 + 2 KMnO4 + 8 H2SO4 \u2192 K2SO4 + 2 MnSO4 + 5 CaSO4 + 10 CO2 + 8 H2O\r\n(a) 0.100 (b) 1.00 (c) 2.00 (d) 3.20 (e) 10.0.\"\r\n\r\nThe number of moles of KMnO4 that reacted is $n(KMnO4) = 1.0\\times10^{-3}L \\cdot 1.0\\times10^{-3}mol/L = 1.0\\times10^{-6}mol$. According to the titration reaction, 5 moles of calcium oxalate react with 2 moles of KMnO4, so the number of moles of CaC2O4 that reacted is $n = 2.5\\times10^{-6}mol$, which is also the number of moles of Ca2+ present in the 10 cm3 portion of the diluted sample of 100 cm3. As the concentration is the same, then the number of moles of Ca2+ present in the diluted sample of 100 cm3 is $n(Ca^{2+}) = \\frac{100cm^{3}\\cdot2.5\\times10^{-6}mol}{10cm^{3}}= 2.5\\times10^{-5}mol$. But this is also the number of moles initially present in the blood sample of 10 cm3. Using the molar mass of calcium, we finally get that the concentration of calcium ions is $[Ca^{2+}] = 1.00 mg/10mL$ of blood. The correct answer is then (b).",
  "Solution_23": "Problem Set 4, Problem 51\r\n\r\n\"51. 1.00 mole of an ideal gas at 0oC confined initially to a volume of 22.4 litres is first allowed to expand isothermally against a constant external pressure of 0.50 atm until equilibrium is attained. The gas is then compressed isothermally and reversibly back to a volume of 22.4 litres. The total enthalpy change \u0394H of the gas as a result of the two processes is:\r\n(a) 11.2 L atm (b) 15.5 L atm (c) 4.3 L atm (d) \u22124.3 L atm (e) 0.0 L atm.\"\r\n\r\nAs H is a state function, then for every cyclic process we must have \u0394H = 0. As the problem describes a cyclic process, then the correct answer is (e)."
}
{
  "Tag": [
    "LaTeX"
  ],
  "Problem": "How to change the line spacing to double space? Thank you",
  "Solution_1": "Put [code]\\usepackage{setspace}\n\\doublespacing[/code]in the preamble - before \\begin{document}"
}
{
  "Tag": [
    "geometry",
    "perimeter",
    "ratio",
    "geometry proposed"
  ],
  "Problem": "[u][b]The author of this posting is : nttu[/b][/u]\r\n____________________________________________________________________\r\n\r\nLet a triangle ABC . M , N , P are the midpoints of BC, CA, AB .\r\na) $d_1, d_2, d_3$ are lines throughing M, N, P and dividing the perimeter of triangle ABC into halves . Prove that : $d_1, d_2, d_3$ are concurrent at K .\r\nb) Prove that : among the ratios : $\\frac{KA}{BC}, \\frac{KB}{AC}, \\frac{KC}{AB}$, there exists at least one ratio $\\geq \\frac{1}{\\sqrt3}$.",
  "Solution_1": "See [url=http://groups.yahoo.com/group/Hyacinthos/messages/9627?expand=1&viscount=8]Hyacinthos messages #9627, #9628, #9629 and some later messages[/url]. Since some people have troubles with displaying these pages, here is a copy of the relevant contents:\r\n\r\n[quote=\"Hyacinthos message #9540\"]\n[b]From:[/b] \"orl_ml\" (Orlando D\u00f6hring)\n[b]Subject:[/b] a ratio\n\nOn the sides of triangle ABC take the points M_1, N_1, P_1 such\nthat each line MM_1, NN_1, PP_1 divides the perimeter of ABC in two \nequal parts (M, N, P are respectively the midpoints of the sides BC, \nCA, AB).\n\n1/ Prove that the lines MM_1, NN_1, PP_1 are concurrent at a point K.\n\n2/ Prove that among the ratios KA/BC, KB/CA, KC/AB there exist at\nleast a ratio which is not less than 1/?\u00e33.\n\n[/quote]\n\n[quote=\"Hyacinthos message #9627\"]\n[b]From:[/b] \"ben_goss_ro\" (aka Grobber)\n[b]Subject:[/b] Re: a ratio\n\n--- In \"Hyacinthos\", \"orl_ml\" (Orlando D\u00f6hring) wrote:\n> On the sides of triangle ABC take the points M_1, N_1, P_1 such\n> that each line MM_1, NN_1, PP_1 divides the perimeter of ABC in two \n> equal parts (M, N, P are respectively the midpoints of the sides BC, \n> CA, AB).\n> \n> 1/ Prove that the lines MM_1, NN_1, PP_1 are concurrent at a point K.\n> \n> 2/ Prove that among the ratios KA/BC, KB/CA, KC/AB there exist at\n> least a ratio which is not less than 1/?\u00e33.\n\n(1): The intersection point is the centroid of the perimeter of \ntriangle ABC. It's the well-known Spieker center (X(10)). The fact \nthat it's the centroid of the perimeter is a good-enough argument to \nshow the concurrence: the centroid of the perimeter of ABC must lie \non MM_1, on NN_1 and on PP_1, so they must be concurrent.\n\nWhat's the ratio for (2)? I see gibberish, but maybe it's my \nbrowser's fault.\n\n[/quote]\n\n[quote=\"Hyacinthos message #9628\"]\n[b]From:[/b] Darij Grinberg\n[b]Subject:[/b] Re: a ratio\n\nDear \"ben_goss_ro\" (= grobber at MathLinks?),\n\nIn Hyacinthos message #9627, you wrote:\n\n>> (1): The intersection point is the centroid of\n>> the perimeter of triangle ABC. It's the\n>> well-known Spieker center (X(10)). The fact\n>> that it's the centroid of the perimeter is a\n>> good-enough argument to show the concurrence:\n>> the centroid of the perimeter of ABC must lie\n>> on MM_1, on NN_1 and on PP_1, so they must be\n>> concurrent.\n\nThis puzzles me a bit. In fact, the lines MM_1,\nNN_1, PP_1 do bisect the perimeter of triangle\nABC, but the cevians of the Nagel point (X(8))\ndo it as well; however, the former lines pass\nthrough the centroid of the perimeter of\ntriangle ABC, whereas the latter ones don't!\nI guess this is a consequence of my bad\nphysics knowledge (in fact, must a centroid of\na perimeter always lie on every perimeter\nbisector?), but I am still searching a better\nexplanation why X(10) is the centroid of the\nperimeter, while X(8) is not.\n\n>> What's the ratio for (2)? I see gibberish,\n\nActually it's because Orlando used a non-ASCII\nsign. If I understand him correctly, he means\n\"[...] a ratio which is not less than\n1/sqrt(3)\". Actually, the problem is quite\nsimple; K doesn't need to be the Spieker point\nin order to have this satisfied.\n\nWe will prove that for every point K in the\nplane of triangle ABC, at least one of the\nratios KA / BC, KB / CA, KC / AB is not less\nthan 1/sqrt(3).\n\nWell, assume the contrary: Let all ratios\nKA / BC, KB / CA, KC / AB be less than\n1/sqrt(3). If the sidelengths BC, CA, AB are\ndenoted by a, b, c, then we have\n\n  KA^2 < a^2 / 3,\n  KB^2 < b^2 / 3,\n  KC^2 < c^2 / 3.\n\nHence,\n\n  KA^2 + KB^2 + KC^2 < (a^2 + b^2 + c^2) / 3.\n\nBut the well-known Leibniz formula gives\n\n  KG^2 = (KA^2 + KB^2 + KC^2) / 3\n            - (a^2 + b^2 + c^2) / 9,\n\nwhere G is the centroid of triangle ABC.\n\n[Actually, this formula is also known in the\nform\n\n  KG^2 = (KA^2 + KB^2 + KC^2 - GA^2 - GB^2 - GC^2) / 3.\n\nBut this is equivalent to the above, as some\nformulas for the length of a median show.]\n\nSince KG^2 >= 0, we thus have\n\n  KA^2 + KB^2 + KC^2 >= (a^2 + b^2 + c^2) / 3,\n\ncontradicting our assumption. Hence, at\nleast one of the ratios KA / BC, KB / CA,\nKC / AB must be >= 1/sqrt(3). Qed..\n\n  Sincerely,\n  Darij Grinberg[/quote]\n\n[quote=\"Hyacinthos message #9629\"]\n[b]From:[/b] \"ben_goss_ro\" (aka Grobber)\n[b]Subject:[/b] Re: a ratio\n\n--- In \"Hyacinthos\", Darij Grinberg wrote:\n> Dear \"ben_goss_ro\" (= grobber at MathLinks?),\n\n> This puzzles me a bit. In fact, the lines MM_1,\n> NN_1, PP_1 do bisect the perimeter of triangle\n> ABC, but the cevians of the Nagel point (X(8))\n> do it as well; however, the former lines pass\n> through the centroid of the perimeter of\n> triangle ABC, whereas the latter ones don't!\n> I guess this is a consequence of my bad\n> physics knowledge (in fact, must a centroid of\n> a perimeter always lie on every perimeter\n> bisector?), but I am still searching a better\n> explanation why X(10) is the centroid of the\n> perimeter, while X(8) is not.\n> \n\nHi! Yes, I am grobber (:)). \n\nAbout the first part of the problem:\nI apologize for not completing it. No, not all the lines which bisect \nthe perimeter pass through Spieker's point, but these ones do because \nthey also pass through the midpoints of the sides, and here's why:\n\nLet's look at MM_1. Assume AB<AC, and take X on [AC] s.t. AB=CX. It's \nobvious that M_1 is the midpoint of AX. Now look at \nthe \"quadrilateral\" ABCX and let's try to find the centroid of its \nperimeter. First of all, we take the centroid G_1 of the system \nformed by the sides AX and BC. To do this we take the midpoints of AX \nand BC (which are M_1 and M respectively), and G_1 must lie on MM_1 \ns.t. MG_1*BC=M_1G_1*AX (basic rod equilibrium physics). Now take G_2 \nto be the centroid of the system formed by the sides AB and XC. We do \npretty much the same as before, taking G_2 on the line connecting the \nmidpts of AB and XC (call them S and T respectively) s.t. \nG_2X*AB=G_2B*XC, only this time AB=XC, so SG_2=BG_2. This means that \nG_2 is also the centroid of the system of points A,B,X,C and that \nit's also situatd on the line connecting the midpts of AX and BC, \nnamely MM_1.\n\nWe have 2 things: G_1 is on MM_1 and G_2 is on MM_1, but G_1 and G_2 \nare the centroids of 2 disjoint parts of our system (AB, BC, CX, XA) \nthe union of which is the entire system, so the centroid of the whole \nsystem must lie on G_1G_2=MM_1, so the centroid of the perimeter of \nABC is on MM_1. I'd like to stress that G_1 is never equal to G_2, \nbecause that would mean that AX=BC iff AC-AB=BC, which is clearly \nfalse (triangle inequality). X(10) is on MM_1 so it must, of course, \nbe on NN_1 and PP_1 as well, so we're done.\n\n[/quote]\r\n\r\n  Darij"
}
{
  "Tag": [
    "geometry",
    "vector",
    "analytic geometry",
    "Euler",
    "circumcircle",
    "ratio",
    "geometry solved"
  ],
  "Problem": "prove that for $n\\geq 3$ if an n-simplex has an orthocenter and if the point coincides with centroid or cercumcirle then simplex is regular.",
  "Solution_1": "(All multiplication of vectors here is the usual dot product)\r\nLemma: The n-simplex $X_1X_2X_3 \\dots X_{n+1}$ has an orthocenter in its interior if and only if a congruent simplex $A_1A_2A_3 \\dots A_{n+1}$ can be embedded into $\\Re^{n+1}$ with each $A_i$ on a coordinate axis but not at the origin.\r\n\r\nProof: If $A_1A_2A_3 \\dots A_{n+1}$ has $A_i=a_iE_i$, where $E_i$ is the i-th unit vector in the standard basis of $\\Re^{n+1}$, let $H$ be the projection of $0$ onto the hyperplane containing the simplex. Since $(A_i-0) \\cdot (A_j-0) = 0$ for all distinct $i,j$, $(A_i-0) \\cdot (A_j-A_k) =0$ for any distinct $i,j,k$ and $(A_i-H) \\cdot (A_j-A_k)= (0-H) \\cdot (A_j-A_k)=0$ by the projection, making $H$ the orthocenter of $A_1A_2A_3 \\dots A_{n+1}$.\r\nWLOG, suppose each $a_i>0$. Then the equation of the hyperplane is $\\frac{x_1}{a_1}+\\frac{x_2}{a_2}+\\dots + \\frac{x_{n+1}}{a_{n+1}}=1$ and  $H$ is $\\frac1B(\\frac1{a_1},\\frac1{a_2}, \\dots ,\\frac1{a_{n+1}})$, where $B=\\frac1{a_1^2}+\\frac1{a_2^2}+ \\dots +\\frac1{a_{n+1^2}}$. Since H has positive coordinates, it must be in the interior of the simplex.\r\n\r\nIf $X_1X_2X_3 \\dots X_{n+1}$ has an orthocenter $H$, we have $(X_i-H)\\cdot (X_j-X_k)=0$ for any distinct $i,j,k$. This gives $(X_i-H)\\cdot (X_j-H) = (X_i-H)\\cdot (X_k-H)=-C$, where C is the same for any triple of distinct indices $i,j,k$. Consider right triangle $X_iX_jF_i$, where $F_i$ is the foot of the altitude from $X_i$. $H$ is in the interior of segment $X_iF_i$, so $\\angle X_jHX_i$ is obtuse and $(X_i-H)\\cdot (X_j-H)$ is negative, making $C$ positive.\r\nThen \r\n$(X_i-X_j)^2=((X_i-H)-(X_j-H))^2 =$\r\n$ (X_i-H)^2-(X_i-H)\\cdot (X_j-H)+(X_j-H)^2-(X_i-H)\\cdot (X_j-H) = $\r\n$(X_i-H)^2+C+(X_j-H)^2+C$.\r\nLet $a_i=\\sqrt{(X_i-H)^2+C}$ and $A_i=a_iE_i$ as above. We have $A_iA_j=X_iX_j$ for all $i,j$, so $X_1X_2X_3 \\dots X_{n+1}$ and $A_1A_2A_3 \\dots A_{n+1}$ are congruent.\r\n\r\n\r\nProblem. Assume that a simplex has an orthocenter $H$ which coincides with its centroid $G$. The centroid is in its interior, so we can embed the simplex as in the lemma. The vector $H-0=G-0=\\frac1{n+1}(a_1,a_2,\\dots ,a_{n+1})$ is orthogonal to the hyperplane $\\frac{x_1}{a_1}+\\frac{x_2}{a_2}+\\dots + \\frac{x_{n+1}}{a_{n+1}}=1$, so $a_1^2=a_2^2= \\dots =a_{n+1}^2$ and the simplex is regular.\r\n As a side note, it is trivial to establish from here the generalization of the Euler line- the circumcenter $O$ is the projection of the point $\\frac12(a_1,a_2,\\dots ,a_{n+1}) $. The ratio of distances $\\frac{HG}{GO}$ is $\\frac2{n-1}$, so the same result would hold if we started with $H$ and $O$ coinciding."
}
{
  "Tag": [
    "topology",
    "advanced fields",
    "advanced fields unsolved"
  ],
  "Problem": "Hello,\r\n\r\nlet\r\n$ M \\equal{} \\left \\{([x],[y]) \\in \\mathbb{RP}^n \\times \\mathbb{RP}^m | n \\leq m, \\sum_{i\\equal{}0}^n x_i y_i \\equal{} 0 \\right \\}$\r\n\r\nProve M is a differntial manifold of dimension $ n\\plus{}m\\minus{}1$.\r\n\r\nI would like to write M as $ f^{\\minus{}1}(y)$ for $ y \\in$ some manifold Y and \"good\" map f, but I can't find such objects.\r\nany help?\r\n\r\nthx",
  "Solution_1": "Set $ f \\equal{} \\sum_{i \\equal{} 0}^n x_i y_i$"
}
{
  "Tag": [],
  "Problem": "Hi!  I'm going to be a senior, and I'm currently in AP Bio.  I know I want to major in bio in college (either molecular or cellular).  Should I take AP Chem or AP Environmental Science next year?\r\n\r\nThanks!",
  "Solution_1": "I suggest AP Chem. Chemistry is fundamental to the understanding of biological systems."
}
{
  "Tag": [
    "probability"
  ],
  "Problem": "There are 100 seats on an aeroplane. Each passenger boarding must sit in their own seat unless it is occupied in which case they may choose a seat at random. The first passenger on is a little old lady who sits in the first seat she comes to. What is the probability that the last passenger gets to sit in their own seat?\r\n\r\nI keep coming up with either 1/2, 1/100 or 100!. Any ideas please?",
  "Solution_1": "I remember having solved this problem a while ago. If you don't want me to ruin it for you, well, try this:\r\n\r\n-manually calculate it for smaller values of $n$ (here $n=100$);\r\n-try to find a recurrence between the cases for different $n$;\r\n-induction. ;)",
  "Solution_2": "Let the plane have n seats and let $p_n$ be the probability that the last passenger goes to his own seat. Assume that the old lady is assigned seat #1 and passengers #2, #3, ..., #n are assigned seats #2, #3, ..., #n.  The old lady chooses the seat #2 with probability $\\frac 1 n$ and some other seat with probability $\\frac {n - 1}{n}$. If she selects seat #2, then passenger #2 is exactly in the position of the old lady in a (n-1)-seat airplane and the probability of the last passenger getting his own seat (under the condition that she chose seat #2) is $p_{n-1}$. If she selects some other seat, then we can ignore passenger #2 and his seat, consider an (n-1)-seat airplane and again, the probability of the last passenger getting his own seat (under the condition that she did not select seat #2) is $p_{n-1}$. Hence,\r\n\r\n$p_n = \\frac 1 n p_{n-1} + \\frac{n - 1}{n} p_{n-1} = p_{n-1}$\r\n\r\nThus the probability $p_n$ does not depend on the number of seats in the airplane. It is trivial to see that in a 2-seat  plane, the probability $p_2 = \\frac 1 2$ and consequently, $p_n = \\frac 1 2$ for any $n \\ge 2$. :P"
}
{
  "Tag": [
    "AMC",
    "USA(J)MO",
    "USAMO",
    "USAJMO",
    "AMC 10",
    "AMC 10 A",
    "AMC 12"
  ],
  "Problem": "Hi,\r\nI just have a question about how to qualify for USAMO or USAJMO\r\n\r\nI am taking AMC10A through my high school, and AMC12B through another math organization\r\n\r\nFor example, if I get a 150 on 10A and 138 on 12B and a 9 on AIME (which I will take through my high school) will it still be possible for me to make USAMO, or will I only be able to make USAJMO because I took 12B through at a different place than I took AIME?",
  "Solution_1": "It's irrelevant where you take the tests",
  "Solution_2": "On the other hand, the AMC has been known to lose track of people who take the tests in multiple locations. If something like this happens and you wind up on the wrong list, contact them to correct it.",
  "Solution_3": "Fill in the same name and home address on every answer form.  We lose track of people when they take contests at different locations using nickames, initials, or if the answer form is returned with incomplete information.\r\n\r\nSteve Dunbar\r\nMAA Director, American Mathematics Competitions"
}
{
  "Tag": [
    "probability"
  ],
  "Problem": "this is from IIT JEE 96\r\n\r\na number of 5 distinct digits is written down at random using the digits 1,2,3,4,5 and 6 . What is the chance that the no. is divisible by 3 but not by 12?",
  "Solution_1": "Well , we know that a number is divisible by $3$ iff the sum of the digit is divisible by 3. We see that $1+2+3+4+5+6=21$ is divisible by three . Now you want to make a five digit numbers . Obviously the number you are going to give up is either 3 or 6 (or else the sum wont be divisible by 3) . \r\n\r\nIf you give up 3 , then the number of way of such arrangement is $5!=120$ . If it is not divisible by 12 , the number should be not divisible by 4 also . This means there are 1/4 of them is divisible by 4 . So those not divisible by 4 is 3/4 . Hence the total number not divisible by 4 is $120\\cdot 3/4 = 90$ . But you can also give up \" 6 \" .Hence th enumber of ways is $90\\cdot 2=180$ .So the probability is\r\n\r\n$\\frac{180}{240}=\\frac{3}{4}$  :)\r\n\r\nEDITTED : Thanks for pointing out FMako  :)",
  "Solution_2": "I disagree, assuming I understand what I'm doing.\r\n\r\nI agree that you have to remove only a $3$ or $6$.\r\n\r\nCase 1: $1,2,4,5,6$\r\n\r\nTotal combinations: $5! = 120$\r\n\r\nTotal combinations that are divisible by $12$, since it is divisible by $3$ already, that means it is divisible by $4$.\r\nTherefore you only need to look at the last two digits of the number.\r\nEndings of $12, 52, 24, 64, 16, 56$ satisfy divisibility by $4$, \r\ntherefore the total combinations divisible by $4$ are $3! \\cdot 6 = 36$.\r\n\r\nCase 2: $1,2,3,4,5$\r\n\r\nTotal combinations: $5! = 120$\r\n\r\n\"\": $12,32,52,24$ Thus the total combinations divisible by $4$ are $3! \\cdot 4 = 24$\r\n\r\nThus the probability that the number is divisible by $3$ but not $12$ is $\\frac{240-60}{240} = \\frac{3}{4}$\r\n\r\nI hope I did this right.",
  "Solution_3": "[quote=\"FMako\"]I disagree, assuming I understand what I'm doing.\n\nI agree that you have to remove only a $3$ or $6$.\n\nCase 1: $1,2,4,5,6$\n\nTotal combinations: $5! = 120$\n\nTotal combinations that are divisible by $12$, since it is divisible by $3$ already, that means it is divisible by $4$.\nTherefore you only need to look at the last two digits of the number.\nEndings of $12, 52, 24, 64, 16, 56$ satisfy divisibility by $4$, \ntherefore the total combinations divisible by $4$ are $3! \\cdot 6 = 36$.\n\nCase 2: $1,2,3,4,5$\n\nTotal combinations: $5! = 120$\n\n\"\": $12,32,52,24$ Thus the total combinations divisible by $4$ are $3! \\cdot 4 = 24$\n\nThus the probability that the number is divisible by $3$ but not $12$ is $\\frac{240-60}{240} = \\frac{3}{4}$\n\nI hope I did this right.[/quote]\r\n\r\nWhy divide by 240 ? The total number of ways of writing 5 digit number from 6 number is $6P5=720$  :) For the 240 , you only included the case which you didnt take 3 or 6. But what about if you didnt take 1,2,4,6 ?  :P There are another 480 cases .",
  "Solution_4": "[quote=\"eminem\"]this is from IIT JEE 96\n\na number of 5 distinct digits is written down at random using the digits 1,2,3,4,5 and 6 . What is the chance that the no. is divisible by 3 but not by 12?[/quote]\r\n\r\nThe question asks for the probability that the number formed is not divisible by 12, if the number is divisible by 3.\r\n\r\nI hope that's what the question means at least.",
  "Solution_5": "[quote=\"FMako\"][quote=\"eminem\"]this is from IIT JEE 96\n\na number of 5 distinct digits is written down at random using the digits 1,2,3,4,5 and 6 . What is the chance that the no. is divisible by 3 but not by 12?[/quote]\n\nThe question asks for the probability that the number formed is not divisible by 12, if the number is divisible by 3.\n\nI hope that's what the question means at least.[/quote]\r\n\r\nOh , I got what you mean :) . I read the first sentence only and forget the back one . Then I agree with you that it is 3/4  :P",
  "Solution_6": "correct guys :)"
}
{
  "Tag": [
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "Mykolka the numismatist possesses $ 241$ coins of the total amount $ 360$ tughriks (the Mongolian currency), where the value in tughriks of each coin is an integer. Does it necessarily follow that these coins can be divided into three heaps of equal amount?",
  "Solution_1": "http://www.mathlinks.ro/viewtopic.php?t=149608"
}
{
  "Tag": [
    "probability",
    "real analysis"
  ],
  "Problem": "Jenny and Alex flip four fair coins each. What is the probability that they get the same number of tails?",
  "Solution_1": "[hide]For each player, P(0 tails) = 4 C 0 * (1/16) = 1/16, P(1 tail) = 4 C 1 * (1/16) = 1/4, P(2 tails) = 4 C 2 * (1/16) = 3/8, P(3 tails) = 4 C 3 * (1/16) = 1/4, and P(4 tails) = 4 C 4 * (1/16) = 1/16. \n\nThen the probability of them both getting the same number of tails is just the sum of each of the probabilities squared. \n\n$ \\\\(\\frac{1}{16})^{2}+(\\frac{1}{4})^{2}+(\\frac{3}{8})^{2}+(\\frac{1}{4})^{2}+(\\frac{1}{16})^{2}\\\\ = \\frac{1}{256}+\\frac{1}{16}+\\frac{9}{64}+\\frac{1}{16}+\\frac{1}{256}\\\\ = \\frac{1+16+36+16+1}{256}\\\\ = \\frac{35}{128}$[/hide]\r\n\r\nEdit: Thanks, cinco de mayo.",
  "Solution_2": "[hide=\"Generalization\"]\n\\[ P(n)= \\sum_{k=0}^{n}\\left( \\frac{ \\binom{n}{k}}{2^{n}}\\right)^{2}= \\frac{\\sum \\binom{n}{k}^{2}}{4^{n}}= \\boxed{ \\frac{ \\binom{2n}{n}}{4^{n}}}\\] \nNote that $ \\sum_{k=0}^{n}\\binom{n}{k}^{2}= \\binom{2n}{n}$ by the [url=http://mathworld.wolfram.com/Chu-VandermondeIdentity.html]Vandermonde Convolution Formula.[/url]\n\nThis gives $ P(4)=\\frac{35}{128}$.[/hide]\r\n[b]winluo2[/b]: Check your calculations in the beginning."
}
{
  "Tag": [],
  "Problem": "Hopkins requires you to buy all your textbooks. At minimum, they would cost around $\\$$300 and that's if you buy all your books used, at the maximum and buying them new, they could cost you well over $\\$$500 and also, the weight of your backpack for all your homework is crazy! I'm like stunting my growth with the weight of my backpack. \n\nAlso, the amount of homework you get is ridiculous.",
  "Solution_1": "You guys are freshman. You don't know anything about homework.",
  "Solution_2": "[quote=\"Phelpedo\"]You guys are freshman. You don't know anything about homework.[/quote]\r\n\r\nFreshman at Hopkins' Homework > Seniors at Hamden High Homework\r\n\r\nkthx.",
  "Solution_3": "No duh seniors never have much homework. Freshman at Hopkins < most people, depending on courses.",
  "Solution_4": "[quote=\"Phelpedo\"]No duh seniors never have much homework. Freshman at Hopkins < most people, depending on courses.[/quote]\r\n\r\nSeniors should have the most homework.\r\n\r\nAnd any student > most people.\r\n\r\nkthx.",
  "Solution_5": "No no. The homework hierarchy goes like this: Senior $\\leq$ Freshman < Sophomore < Junior. Why don't seniors have any homework? BECAUSE THEY\"VE ALREADY DONE COLLEGE APPLICATIONS AND DON\"T CARE ABOUT ANYTHING ANYMORE. \r\n\r\nAlso, when I said Freshmen @ Hopkins < most people (upperclassment), I was talking about homework load. \r\n\r\nkthx.",
  "Solution_6": "Except you're just generally assuming that. It's not true at Hamden High, it's not true at Hopkins.",
  "Solution_7": "Ok. So Hamden high guys get no homework and Hopkins freshman get a little homework.",
  "Solution_8": "wrong idea",
  "Solution_9": "Ummmm... That is DEFINITELY true at Hopkins.",
  "Solution_10": "Ok the \"that\" is quite out of context now so I don't know what you're referring to. However, I will assume it to be Ignite's statement. Anyhow, I don't think you can really say until you have a better frame of reference (I mean come on, you get like 30 minutes of HW every night in middle school).  4-5 hours a night for freshman? Not unusual at all.",
  "Solution_11": "Well you can expect around 4 hours a week per subject.\r\n\r\nUp to a maximum of 7 subjects\r\n\r\n28 hours a week",
  "Solution_12": "For me, it generally ends up about\r\n\r\n6h for English\r\n4h for Math\r\n2h for the other 3\r\n\r\nSo 16h of homework\r\n\r\nAnd most students take 5 subjects.",
  "Solution_13": "[quote=\"worthawholebean\"]For me, it generally ends up about\n\n6h for English\n4h for Math\n2h for the other 3\n\nSo 16h of homework\n\nAnd most students take 5 subjects.[/quote]\r\n\r\nI think most students take 6.",
  "Solution_14": "I don't think he's including non-academic subjects.",
  "Solution_15": "No actually many people take 6 academic subjects",
  "Solution_16": "5 ACADEMIC - foreing language, social studies, sci, MATH, la",
  "Solution_17": "If you count MG as \"many people\", let it be known he is not a representative sample.\n\nFrom freshman year onward, not many people take 6 academics.",
  "Solution_18": "[quote=\"worthawholebean\"]If you count MG as \"many people\", let it be known he is not a representative sample.\n\nFrom freshman year onward, not many people take 6 academics.[/quote]\nI was talking about in the Jschool.\n\nHow do you expect me to know how many courses he takes?",
  "Solution_19": "You know him. How many courses do you think he's taking?",
  "Solution_20": "[quote=\"worthawholebean\"]You know him. How many courses do you think he's taking?[/quote]\r\nUmmm......fifteen and a half?",
  "Solution_21": "fifteen in a half....... you've gotta be kiding me.. :0",
  "Solution_22": "gee whiz\r\n\r\nthink",
  "Solution_23": "Is he taking two arts?\r\n\r\nI'm thinking of taking two sciences in my junior or senior year.",
  "Solution_24": "Or he might still be taking both Spanish and Latin \r\n\r\nI don't really know. Last year I had one class with him but not this year.",
  "Solution_25": "See, MG has time to play Ping Pong. I told you study halls aren't very necessary. I should've took Chinese and Spanish and an art and the other 5 courses.",
  "Solution_26": "I think this whole heavy book bag, and lot's of hw thing isn't that big of a problem. Hopkins is a very academic school. So you should expect to keep alot of hw and heavy books to carry home each day. :)\r\n\r\nHopkins, is a hard school and academic's are a major so heavy books and lots of hw.. shouldn't be a bir problem to you. Since you applied and learn alot about the school.. \r\n\r\n:",
  "Solution_27": "Omg I am NOT MSG. Kthanks."
}
{
  "Tag": [
    "analytic geometry",
    "number theory proposed",
    "number theory"
  ],
  "Problem": "Given a Cartesian coordinate system in the plane, a point with coordinates (x; y) is called a grid point if and only if his coordinates x and y are integers. Does there exist a circle with exactly five grid points lying on its periphery?\r\n\r\n  Darij",
  "Solution_1": "This is a problem that needs some spirit of observation ;]\r\nThere are two ideas that may bring you to its solution:\r\n1) The equation sin^2(x)+cos^2(x)=1 has only four integer solutions;\r\n2) If a grid point lies on the periphery then its symmetric for the center of the circle also lies on the periphery of the circle.\r\n\r\n1)One knows that the parametrical equations of the circle are :\r\nx=a+Rcos(x)\r\ny=b+Rsin(x),  where (a,b) are the coordinates of the center of the circle, and R is its radius.\r\nBut cos(x) and sin(x) are integer only for four points on the circle, so there exist only 4 grid points on the periphery of the circle. (this may be difficult to understand for R irrational, but when one looks at the circle and at the coordinates, everything becomes clear :))\r\n \r\n2) This is a solution only for circles with grid center points :\r\n Let M(c,d) be a point on the periphery of the circle, then its symetrical point M' (according to the center of the circle) is also grid, because its coordinates are (2a-c,2b-d). So there can be only an even number of grid points on the periphery of the circle, and because 5 is even there doesn't exist a circle with such a property.\r\n\r\nThis solution came to my mind quickly, but its generality is hard to prove.\r\n\r\n\r\n\r\n__________________________________________________________\r\n\r\n[SYSTEM OF A DOWN]\r\n\"Our days are never coming back\" (Highway Song) ;]",
  "Solution_2": "This is a theorem of A. Schinzel, which in fact states that for any n there exist a circle with has exactly n laticial point on it.",
  "Solution_3": "@ Palosh : Hmm... in fact, there does exist a circle through 5 grid points! Of course, its center is not a grid point; it is not even a point with coordinates (x/2; y/2) where x and y are integers.\r\n\r\n  Darij",
  "Solution_4": "Valentin, could you tell me that theorem... and Darij what is then the complete solution of this problem?\r\nThanx...\r\n:):):)\r\n\r\n\r\n______________________\r\n\"Our days are never coming back...\" (Highway Song -> System of a Down)",
  "Solution_5": "The complete solution is quite straightforward once you have the explicite answer: The circle with equation $x^{2}+\\left( y+\\frac{7}{4}\\right) ^{2}=\\left( \\frac{25}{4}\\right) ^{2}$ passes through the points (0; -8), (5; 2), (-5; 2), (6; 0) and (-6; 0) and through no other lattice points. About the general theorem, I think there is [url=http://mathworld.wolfram.com/SchinzelsTheorem.html]some information on the Mathworld site[/url].\n\n  Darij",
  "Solution_6": "Thanx a lot...\r\nNice problem though i've such a mistake!\r\n :):):)\r\n\r\n\r\n\r\n       Palosh alias Downito\r\n_____________________________\r\n\"Our days are never coming back... \" (Highway Song -> System of a Down)"
}
{
  "Tag": [],
  "Problem": "Do you think if your parents should be invited to join AoPS site?  Give your reason please.",
  "Solution_1": "for the most part, parents would learn quite a bit if they were in AoPS. my mom was the one who invited ME to AoPS however.",
  "Solution_2": "Aren't there alredy parents browsing the AoPS forums, along with the teachers?",
  "Solution_3": "My mother/teacher is Sola Deo Gloria.",
  "Solution_4": "well...I guess. But mine already is.  Hi Dad! :)",
  "Solution_5": "I think he should because he's a math professor, and this would be fun for him, but he hasn't joined yet."
}
{
  "Tag": [
    "AMC",
    "USA(J)MO",
    "USAMO",
    "geometry",
    "circumcircle",
    "induction"
  ],
  "Problem": "I was totally confuzzled by the problems. I suck at proofs :( \r\n\r\nHow do you tackle this:\r\n\r\n2008 USAMO #2: Let ABC be an acute, scalene triangle and let M, N, and P be the midpoints of BC, CA, and AB, respectively. Let teh perpendicular bisectors of AB and AC intersect ray AM in points D and E respectively, and let lines BD and CE intersect in point F, inside triangle ABC. Prove that points A, N, F, and P all lie on one circle.\r\n\r\nI tried drawing the figure as described and drew the circumcenter of ABC. I tried to make a circle with AO as a diameter. Knowing that angles APO and ANO are right angles, they form two semicircles so the four points and O must form a circle.\r\n\r\nDo I have good enough progres to at least get SOME credit for the problem?",
  "Solution_1": "here is another one too:\r\n\r\n4. Let P be a convex polygon with n sides, n \u00b8 3. Any set of n\u00a13 diagonals of P that do not\r\nintersect in the interior of the polygon determine a triangulation of P into n\u00a12 triangles.\r\nIf P is regular and there is a triangulation of P consisting of only isosceles triangles, \u00afnd\r\nall the possible values of n.",
  "Solution_2": "The problems are under \"Contests\" so you should not repost them.\r\n\r\nBut anyways, #2 is tricky but not too far-fetched.\r\n#4 uses more intuition than geometry knowledge.\r\n\r\nTo be honest, your #2 would probably not earn any points... it's pretty obvious that any quads with opposing right angles are always cyclic, and that doesn't demonstrate enough.\r\n\r\nTry #1. It's much, much simpler imo, and doesn't need anything strange or obscure. Try to find one such sequence (since it just says to prove that one exists); play around with the numbers, preferably without a calculator... and then\r\n\r\n\r\n[hide]\n\nuse induction[/hide]",
  "Solution_3": "But you can't use a calc for USAMO.\r\n :maybe:",
  "Solution_4": "exactly the point."
}
{
  "Tag": [
    "AMC"
  ],
  "Problem": "http://www.unl.edu/amc/e-exams/e4-amc08/e4-1-8archive/2006-8a/2006-amc8schHonorRoll.shtml",
  "Solution_1": "I get the feeling there may be other schools added.",
  "Solution_2": "like I said in the mc forum, I doubt this is from this year.\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?p=675930#675930\r\nThen again I may be wrong.",
  "Solution_3": "It is definately from this year because AMC of last year was in 2005, and the title says 2006, so it's from this year.  I need to stop repeating myself.",
  "Solution_4": "yay i got first at the place i took it and its on the honor roll :D",
  "Solution_5": "Same here! Perfect at Kennedy Junior High School, Lisle, IL.\r\nWhat's weird is that our school is in Lisle, but it's a Naperville school.",
  "Solution_6": "After the latest update, we are on the list.",
  "Solution_7": "yeahmore schools have been added."
}
{
  "Tag": [
    "factorial"
  ],
  "Problem": "Find a three digit number of the form $xyz$ such that the number $xyz = x!+y!+z!$, where $x$ and $y$ and $z$ are all between 4 and 10, inclusive.\r\n\r\nIn other words, find a three digit number, whose digits are between 4 and 10 inclusive, s.t. it equals the sum of the factorials of its digits.",
  "Solution_1": "[hide]I sort of saw the solution immediately the first time I solved this, so I don't have a very good logical explanation.\n\nOne of the digits is either 5 or 6. 6 doesn't work, so one of the digits (the last one) is 5. The first one must be a 1, which leaves the last one to be 4. The number is $\\boxed{145}$.[/hide]",
  "Solution_2": "[quote=\"i_like_pie\"][hide]I sort of saw the solution immediately the first time I solved this, so I don't have a very good logical explanation.\n\nOne of the digits is either 5 or 6. 6 doesn't work, so one of the digits (the last one) is 5. The first one must be a 1, which leaves the last one to be 4. The number is $\\boxed{145}$.[/hide][/quote]\r\nRead the question again ;)",
  "Solution_3": "Is there something like that????",
  "Solution_4": "[quote=\"i_like_pie\"][hide]I sort of saw the solution immediately the first time I solved this, so I don't have a very good logical explanation.\n\nOne of the digits is either 5 or 6. 6 doesn't work, so one of the digits (the last one) is 5. The first one must be a 1, which leaves the last one to be 4. The number is $\\boxed{145}$.[/hide][/quote]\r\nJust wondering, why does one of the digits have to be five or six?",
  "Solution_5": "the largest possible way to get xyz is 10*10*10 or 1000\r\nsince 6!=720 and 7!=5040, the numbers x, y, z could at most be 6!\r\nthen just use 6! and work your way down.",
  "Solution_6": "[quote=\"xscapezaer\"]Find a three digit number of the form $xyz$ such that the number $xyz = x!+y!+z!$, where $x$ and $y$ and $z$ are all between 4 and 10, inclusive.\n\nIn other words, find a three digit number, whose digits are between 4 and 10 inclusive, s.t. it equals the sum of the factorials of its digits.[/quote]\r\n\r\nis 10 a digit? :rotfl:",
  "Solution_7": "Oops... I didn't see they had to be bewteen 4 and 10.\r\n\r\n[hide=\"Solution\"]The maximum possible value of any of the digits is $5$ because $6!=720$, which means one of the digits has to be $7$, but $7!=5040$.\n\nThe only possible digits are $4$ and $5$, but the sum of the factorials of any combination of those digits will always be less, so there is $\\boxed{\\texttt{no solution}}$.[/hide]\r\n[quote=\"davidlizeng\"]the largest possible way to get xyz is 10*10*10 or 1000\nsince 6!=720 and 7!=5040, the numbers x, y, z could at most be 6!\nthen just use 6! and work your way down.[/quote]\r\nIt can't be 6! because then one of the digits is 7.",
  "Solution_8": "[quote=\"xscapezaer\"]Find a three digit number of the form $xyz$ such that the number $xyz = x!+y!+z!$, where $x$ and $y$ and $z$ are all between 4 and 10, inclusive.\n\nIn other words, find a three digit number, whose digits are between 4 and 10 inclusive, s.t. it equals the sum of the factorials of its digits.[/quote]\r\n\r\nThe original poster means $\\overline{xyz}= x!+y!+z!$, where $\\overline{xyz}$ is the number with digits $x,y,z$, as can be inferred from the second paragraph. ;) So yeah, 145.",
  "Solution_9": "[quote=\"chess64\"]The original poster means $\\overline{xyz}= x!+y!+z!$, where $\\overline{xyz}$ is the number with digits $x,y,z$, as can be inferred from the second paragraph. ;) So yeah, 145.[/quote]\n[quote=\"xscapezaer\"]Find a three digit number of the form $xyz$ such that the number $xyz = x!+y!+z!$, where $x$ and $y$ and $z$ [b]are all between 4 and 10, inclusive.[/b][/quote]\r\nRead my second post (the one above your post).",
  "Solution_10": "It is [hide]no solution.\n\nA digit can obviously not be above 6, since 7! is over 1000. If one of the digits is 6, then the hundreds digit would be at least 7, which is not possible. Since 3X4!=72, one of the numbers mst be 5. It is obvious that this is not possible by looking at the remaining possibilities.[/hide]"
}
{
  "Tag": [
    "calculus",
    "derivative",
    "limit",
    "function",
    "calculus computations"
  ],
  "Problem": "find the limit as x->4 of (4^x - x^4) / (x-4).\r\n\r\nWe are not allowed to use l'hopital's rule.\r\n\r\nAny help is appreciated. Thanks!",
  "Solution_1": "Are you allowed to use the definition of the derivative? \r\n(For the given limit this amounts to the same as using L'Hopital.) \r\n\r\nIn any case you should try to break up the fraction into a sum of two terms \r\nthat are difference quotients and deal with them seperately.",
  "Solution_2": "yes, we are allowed to use the definition of the derivative.\r\n\r\nI tried using algebra to try and cancel terms so that i could substitute 4 in, but i havent come up with anything. I also tried multiplying by the conjugate, using exponentials, etc.\r\n\r\nAny tips? :)",
  "Solution_3": "You're going to have to quote your knowledge of derivatives. $ \\lim_{x\\to 4}\\frac{4^4\\minus{}x^4}{x\\minus{}4}$ is doable algebraically, but you need to know the derivative of the exponential for $ \\lim_{x\\to 4}\\frac{4^x\\minus{}4^4}{x\\minus{}4}$.",
  "Solution_4": "is it basically just differentiating $ \\frac{4^{x} \\minus{} 4^{4}}{x \\minus{} 4}$ and then substituting the value 4?\r\n\r\nbut differentiating that would give a denominator of (x-4)^2 which would give a denominator of 0 when substituting 4.\r\n\r\nI'm still stuck :(",
  "Solution_5": "$ \\lim_{x\\to 4}\\frac {4^x \\minus{} 4^4}{x \\minus{} 4}$  [i]is[/i] a derivative -- can you recognize the function that it's a derivative of (and the point at which it's the derivative)?",
  "Solution_6": "oops, its $ \\lim_{x\\to 4}\\frac{4^{x}\\minus{}x^{4}}{x\\minus{}4}$ not $ \\lim_{x\\to 4}\\frac{4^{x}\\minus{}4^{4}}{x\\minus{}4}$\r\n\r\nif i let y = $ \\frac{4^{x}\\minus{}x^{4}}{x\\minus{}4}$ , and i differentiate w.r.t. x, and substitute it in at point 4, i get 240/0. Does that mean the limit is infinity?",
  "Solution_7": "No, it means that you don't know how to recognize a limit as the definition of a derivative.  What's the definition of the derivative of the function $ f(x)$ at the point $ x \\equal{} a$?\r\n\r\n(Also, jmerry and I both know that the problem we're suggesting to you is not exactly the same as the problem you asked; but it's an intermediate step in solving your problem.  Once you've figured out our problem, you'll begin to see why.)",
  "Solution_8": "$ \\frac{240}{0}$? $ 4^4$ is $ 256$, not $ 16$.\r\n\r\nIt has to be a $ \\frac{0}{0}$ form, because it's a derivative.\r\n\r\nAs for the different limits, there are two pieces in my post #4. Their sum is the original limit, and each is simpler than that original.",
  "Solution_9": "$ \\lim_{x\\to 0}\\frac{f(a\\plus{}m)\\minus{}f(a)}{m}$   ? I think thats right. Sorry if I'm a bit slow, I'm really bad at limits. Taking calculus in college and learning derivatives without learning limits in high school is a terrible experience.\r\n\r\nI did realise that I could add them together and get the limit, but I can't find the derivative. I'm going around in circles. I can't seem to apply anything I've learnt in my calc class to this. I think I'm missing a huge chunk of the concept here.",
  "Solution_10": "For the way you wrote that, you wanted to take $ \\lim_{m\\to 0},$ not $ x.$\r\n\r\nBut you might also want to consider this alternate (but algebraically equivalent) form:\r\n\r\n$ f'(a) \\equal{} \\lim_{x\\to a}\\frac {f(x) \\minus{} f(a)}{x \\minus{} a}.$",
  "Solution_11": "[quote=\"tennischan\"]Sorry if I'm a bit slow, I'm really bad at limits.[/quote]  No need to apologize :)  \r\n\r\nKent made a helpful suggestion for another way to write the same expression.  Now, the question becomes: can you recognize the limit you're given as the derivative of some function?  This is a pattern-matching exercise: you have to come up with the right choice of $ f(x)$ and $ a$ so that the definition of the derivative of $ f$ at $ a$ ends up looking like the limit you're trying to compute.",
  "Solution_12": "Okay, so I tried something I thought of, and I added the limits together by comparing the definition of the derivative for both parts and finding out what f(x) for each part was, and then finding f'(x).\r\n\r\nSo I found the derivative of 4^x and the derivative of x^4, which are 4^x (ln 4) and 4x^3 respectively.\r\n\r\nAdding them together, I got 256 ln 4 + 256 = 256 (ln 4 + 1).\r\n\r\nIs that right?",
  "Solution_13": "Almost: are you sure adding the two limits was what you wanted to do?  After you fix this, you'll be done; for purposes of your own education, though, you should definitely make sure you understand [i]why[/i] you did each step in this process and practice doing other examples so that you don't forget the important ideas  :)",
  "Solution_14": "$ \\frac {4^x \\minus{} x^4}{x \\minus{} 4} \\equal{} \\frac {4^x \\minus{} 4^4}{x \\minus{} 4} \\minus{} \\frac {x^4 \\minus{} 4^4}{x \\minus{} 4}$.\r\n\r\nAnd apply definition of derivative.",
  "Solution_15": "oops, its 256(ln 4 -1)\r\n\r\nI get it now, thanks so much for all your help! :)"
}
{
  "Tag": [
    "function",
    "inequalities",
    "logarithms",
    "inequalities proposed"
  ],
  "Problem": "Let $ x,y,z$ be positive real numbers. Prove that for any $ k > 0$ such that $ k^3 \\plus{} 3k^2 \\minus{} 6k \\minus{} 2 \\ge 0,$ then\r\n$ \\left( \\frac {x^2 \\plus{} yz}{x^2 \\plus{} kyz}\\right)^2 \\plus{} \\left( \\frac {y^2 \\plus{} zx}{y^2 \\plus{} kzx}\\right)^2 \\plus{} \\left( \\frac {z^2 \\plus{} xy}{z^2 \\plus{} kxy}\\right)^2 \\ge \\frac {12}{(k \\plus{} 1)^2}.$\r\n:)",
  "Solution_1": "It follows from the convexity of function $ f(t) \\equal{} \\left( \\frac{e^t\\plus{}1}{e^t\\plus{}k} \\right)^2$\r\n\r\nIf we let $ xyz \\equal{} 1$ then the inequality becomes:\r\n$ f(\\ln x) \\plus{} f(\\ln y) \\plus{} f(\\ln z) \\geq 3f(0)$ which is Jensen",
  "Solution_2": "Hello, thank you for your reply. But I think the function $ f(t)\\equal{}\\left( \\frac{e^t\\plus{}1}{e^t\\plus{}k}\\right)^2$ is not always convex, so I think you can't use Jensen's Inequality Here. Please try it again. BTW, the condition $ k^3\\plus{}3k^2\\minus{}6k\\minus{}2 \\ge 0$ is also the best condition such that the original inequality holds.\r\nRegards"
}
{
  "Tag": [
    "inequalities"
  ],
  "Problem": "Let {x_i} be a set of reals such that $ \\sum^n x_i \\equal{} n$. Prove that\r\n\r\n$ \\sum^n (n \\plus{} x_i \\plus{} \\frac {1}{x_i})(n \\plus{} x_i \\plus{} x_i^2) \\ge n^3 \\plus{} 4n^2 \\plus{} 4n$",
  "Solution_1": "[quote=\"4865550150\"]Let {x_i} be a set of reals such that $ \\sum^n x_i \\equal{} n$. Prove that\n\n$ \\sum^n (n \\plus{} x_i \\plus{} \\frac {1}{x_i})(n \\plus{} x_i \\plus{} x_i^2) \\ge n^3 \\plus{} 4n^2 \\plus{} 4n$[/quote]\r\n\r\nThe LHS is equivalent to $ n^3\\plus{}2n^2\\plus{}2n\\plus{}\\sum x^{3}_{i}\\plus{}(n\\plus{}1)\\sum x^{2}_{i}\\plus{}\\sum x_{i}\\sum \\frac{1}{x_i}\\geq n^3\\plus{}4n^2\\plus{}4n$.\r\n\r\nso, it suffices to prove that $ \\sum x^{3}_{i}\\plus{}(n\\plus{}1)\\sum x^{2}_{i}\\plus{}\\sum x_{i}\\sum \\frac{1}{x_i}\\geq 2n^2\\plus{}2n$.\r\n\r\nBut from Chebychev's inequality we get that $ \\sum x^{3}_{i}\\geq \\sum x^{2}_{i}$. So, we need to prove that $ (n\\plus{}2)\\sum x^{2}_{i}\\plus{}\\sum x_{i}\\sum\\frac{1}{x_i}\\geq n^2\\plus{}n^2\\plus{}2n$ which is obviously true from CS and AM-GM"
}
{
  "Tag": [
    "inequalities proposed",
    "inequalities"
  ],
  "Problem": "Let a,b,c > 0 .Prove that :\r\n \r\ni, $ \\frac{a}{a\\plus{}2b}\\plus{}\\frac{b}{b\\plus{}2c}\\plus{}\\frac{c}{c\\plus{}2a}\\leq \\frac{3(a^2\\plus{}b^2\\plus{}c^2)}{(a\\plus{}b\\plus{}c)^2}$\r\n\r\nii, $ \\frac{a^2}{a^2\\plus{}2b^2}\\plus{}\\frac{b^2}{b^2\\plus{}2c^2}\\plus{}\\frac{c^2}{c^2\\plus{}2a^2}\\leq \\frac{a^2\\plus{}b^2\\plus{}c^2}{ab\\plus{}bc\\plus{}ca}$",
  "Solution_1": "[quote=\"PHONGVAN\"]Let a,b,c > 0 .Prove that :\n \ni, $ \\frac {a}{a \\plus{} 2b} \\plus{} \\frac {b}{b \\plus{} 2c} \\plus{} \\frac {c}{c \\plus{} 2a}\\leq \\frac {3(a^2 \\plus{} b^2 \\plus{} c^2)}{(a \\plus{} b \\plus{} c)^2}$\n\n[/quote]\r\n\\[ \\Longleftrightarrow \\sum{\\left(1\\minus{}\\frac{a}{a\\plus{}2b}\\right)}\\geq 3\\minus{}\\frac{3(a^2\\plus{}b^2\\plus{}c^2)}{(a\\plus{}b\\plus{}c)^2}\\equal{}\\frac{6(ab\\plus{}bc\\plus{}ca)}{(a\\plus{}b\\plus{}c)^2}\\]\r\n\\[ \\Longleftrightarrow \\frac{b}{a\\plus{}2b}\\plus{}\\frac{c}{b\\plus{}2c}\\plus{}\\frac{a}{c\\plus{}2a}\\geq \\frac{3(ab\\plus{}bc\\plus{}ca)}{(a\\plus{}b\\plus{}c)^2}\\]\r\nBy $ Cauchy$ we get\r\n\\[ \\sum{\\frac{b}{a\\plus{}2b}}\\sum{b(a\\plus{}2b)}\\geq (a\\plus{}b\\plus{}c)^2\\]\r\nIt suffice to show that\r\n\\[ (a\\plus{}b\\plus{}c)^4\\geq 3(ab\\plus{}bc\\plus{}ca)(ab\\plus{}bc\\plus{}ca\\plus{}2a^2\\plus{}2b^2\\plus{}2c^2)\\]\r\nWithout loss of generosity,assume that $ ab\\plus{}bc\\plus{}ca\\equal{}3$,then it becomes\r\n\\[ \\left[(a\\plus{}b\\plus{}c)^2\\minus{}9\\right]^2\\geq 0\\]\r\nwhich is obvious."
}
{
  "Tag": [],
  "Problem": "Rohan plans to open a bank for his friends. Some friends will allow him to keep their money in savings, and others will borrow money from him. The charts below describe the amounts that students are saving and borrowing from him. What is the least possible value, in dollars, of the money that Rohan is currently keeping in his bank? Express your answer to the nearest whole number.\n\n\\[ \\begin{array}{cccc}\n\\multicolumn{2}{c}{\\textrm{\\underline{Amount put in for savings}}} & \\multicolumn{2}{c}{\\textrm{\\underline{Amount taken for loans}}} \\\\\n\\textrm{\\underline{Value of accounts}} & \\textrm{\\underline{Number of accounts}} &\n\\textrm{\\underline{Value of loan}} & \\textrm{\\underline{Number of loans}} \\\\\n\\$0.01\\minus{}\\$20.00 & 3 & \\$0.01\\minus{}\\$20.00 & 5\\\\\n\\$20.01\\minus{}\\$40.00 & 4 & \\$20.01\\minus{}\\$40.00 & 3 \\\\\n\\$40.01\\minus{}\\$60.00 & 6 & \\$40.01\\minus{}\\$60.00 & 1 \\\\\n\\$60.01\\minus{}\\$80.00 & 2 & \\$60.01\\minus{}\\$80.00 & 1 \\\\\n\\end{array}\\]",
  "Solution_1": "[quote=\"GameBot\"]Rohan plans to open a bank for his friends. Some friends will allow him to keep their money in savings, and others will borrow money from him. The charts below describe the amounts that students are saving and borrowing from him. What is the least possible value, in dollars, of the money that Rohan is currently keeping in his bank? Express your answer to the nearest whole number.\n\\[ \\begin{array}{cccc} \\multicolumn{2}{c}{\\textrm{\\underline{Amount put in for savings}}} & \\multicolumn{2}{c}{\\textrm{\\underline{Amount taken for loans}}} \\\\\n\\textrm{\\underline{Value of accounts}} & \\textrm{\\underline{Number of accounts}} & \\textrm{\\underline{Value of loan}} & \\textrm{\\underline{Number of loans}} \\\\\n\\$0.01 \\minus{} \\$20.00 & 3 & \\$0.01 \\minus{} \\$20.00 & 5 \\\\\n\\$20.01 \\minus{} \\$40.00 & 4 & \\$20.01 \\minus{} \\$40.00 & 3 \\\\\n\\$40.01 \\minus{} \\$60.00 & 6 & \\$40.01 \\minus{} \\$60.00 & 1 \\\\\n\\$60.01 \\minus{} \\$80.00 & 2 & \\$60.01 \\minus{} \\$80.00 & 1 \\\\\n\\end{array}\n\\]\n[/quote]\r\n\r\n$ .01 \\cdot 3\\plus{}20.01 \\cdot 4\\plus{}40.01 \\cdot 6\\plus{}60.01 \\cdot 2\\equal{}440.15$.\r\n\r\n$ 440.15\\minus{}100\\minus{}120\\minus{}60\\minus{}80\\equal{}80.15$. Rounded this is $ 80$.",
  "Solution_2": "[quote=\"BOGTRO\"][quote=\"GameBot\"]Rohan plans to open a bank for his friends. Some friends will allow him to keep their money in savings, and others will borrow money from him. The charts below describe the amounts that students are saving and borrowing from him. What is the least possible value, in dollars, of the money that Rohan is currently keeping in his bank? Express your answer to the nearest whole number.\n\\[ \\begin{array}{cccc} \\multicolumn{2}{c}{\\textrm{\\underline{Amount put in for savings}}} & \\multicolumn{2}{c}{\\textrm{\\underline{Amount taken for loans}}} \\\\\n\\textrm{\\underline{Value of accounts}} & \\textrm{\\underline{Number of accounts}} & \\textrm{\\underline{Value of loan}} & \\textrm{\\underline{Number of loans}} \\\\\n\\$0.01 \\minus{} \\$20.00 & 3 & \\$0.01 \\minus{} \\$20.00 & 5 \\\\\n\\$20.01 \\minus{} \\$40.00 & 4 & \\$20.01 \\minus{} \\$40.00 & 3 \\\\\n\\$40.01 \\minus{} \\$60.00 & 6 & \\$40.01 \\minus{} \\$60.00 & 1 \\\\\n\\$60.01 \\minus{} \\$80.00 & 2 & \\$60.01 \\minus{} \\$80.00 & 1 \\\\\n\\end{array}\n\\]\n[/quote]\n\n$ .01 \\cdot 3\\plus{}20.01 \\cdot 4\\plus{}40.01 \\cdot 6\\plus{}60.01 \\cdot 2\\equal{}440.15$.\n\n$ 440.15\\minus{}100\\minus{}120\\minus{}60\\minus{}80\\equal{}80.15$. Rounded this is $ 80$.[/quote]\n\nHow did you get the second line?",
  "Solution_3": "\\[20\\cdot5=100\\] \\[40\\cdot3=120\\] \\[60\\cdot1=60\\] \\[80\\cdot1=80\\]"
}
{
  "Tag": [],
  "Problem": "\u5982\u56fe\uff0c\u6b63\u65b9\u5f62ABCD\u7684\u8fb9\u957f\u4e3a1\uff0c\u70b9M\u3001N\u5206\u522b\u5728BC\u3001CD\u4e0a\uff0c\u4f7f\u5f97\u25b3CMN\u7684\u5468\u957f\u4e3a2\u3002\u6c42\u25b3MAN\u9762\u79ef\u7684\u6700\u5c0f\u503c\u3002",
  "Solution_1": "\u7b54\u6848\u662f$\\sqrt{2}-1$,\u53ef\u7531\u65cb\u8f6c\u53d8\u6362\u5f97\u6765.",
  "Solution_2": "\u65cb\u8f6c\u6211\u77e5\u9053\uff0c\u53ef\u8bc1\u89d2MAN=45\u5ea6\uff0c\u53ef\u5426\u544a\u77e5\u63a5\u4e0b\u6765\u7684\u63a8\u7406\u3001\u8fd0\u7b97\u8fc7\u7a0b\uff1f",
  "Solution_3": "Let $MC=a$, $NC=b$, $MN=c$. hence: $(a+b)^2 \\leq 2c^2$ and $a+b+c=2$, so $c \\leq 2\\sqrt{2}-2$.\r\n\r\n$S_{AMN}=1-\\frac12 (c+ab) = 1 - \\frac12 (2-c) \\geq \\sqrt{2}-1$."
}
{
  "Tag": [
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "For a given natural number $n > 3$, the real numbers $x_1, x_2, \\ldots, x_n, x_{n + 1}, x_{n + 2}$ satisfy the conditions $0\r\n< x_1 < x_2 < \\cdots < x_n < x_{n + 1} < x_{n + 2}$.  Find the minimum possible value of\r\n\\[\\frac{(\\sum _{i=1}^n \\frac{x_{i + 1}}{x_i})(\\sum _{j=1}^n \\frac{x_{j + 2}}{x_{j +\r\n1}})}{(\\sum _{k=1}^n \\frac{x_{k + 1} x_{k + 2}}{x_{k + 1}^2 + x_k\r\nx_{k + 2}})(\\sum _{l=1}^n \\frac{x_{l + 1}^2 + x_l x_{l + 2}}{x_l\r\nx_{l + 1}})}\\] and find all $(n + 2)$-tuplets of real numbers $(x_1, x_2, \\ldots, x_n, x_{n + 1}, x_{n + 2})$ which gives this value.",
  "Solution_1": "Anybody get this?",
  "Solution_2": "Let $a_{i}=\\frac{x_{i+1}}{x_{i}},i=1,2,...,n+1$, then \\[X=\\frac{s(s+b)}{(2s+b)(2t+c)},\\] were $s=\\sum_{i=1}^{n}a_{i}, \\ \\ t=\\sum_{i=1}^{n}\\frac{1}{a_{i}}, b=a_{n+1}-a_{1},c=\\frac{1}{a_{n+1}}-\\frac{1}{a_{1}}.$\r\nWe can chose $a_{1}=a_{n+1}$, then b=c=0 and $X=\\frac{s}{4t}$. Minimal value for X we get when $a_{i}=a$, then $X=\\frac{1}{4}a^{2}$, because $a\\ge 1$, minimal value $X=\\frac{1}{4}$.",
  "Solution_3": "If it holds for all $a_i$ equal, shouldn't the minimum be 1?\n And why can we choose $a_1=a_{n+1}$?",
  "Solution_4": "Using Rust 's notation,in fact we are looking for the minimum value of $$\\dfrac {1}{\\left( \\sum ^{n}_{k=1}\\dfrac {1}{\\dfrac {1}{a_{k}}+\\dfrac {1}{a_{k+1}}}\\right) (\\dfrac {1}{\\sum ^{n}_{l=1}a_{l}}+\\dfrac {1}{\\sum ^{n}_{l=1}a_{l+1}})}$$ for all $a_{i} > 1$.\nAnd I think it's not so easy.",
  "Solution_5": "And I think it's not so easy!",
  "Solution_6": "wow....\n\nwhat a formula!"
}
{
  "Tag": [
    "linear algebra",
    "matrix",
    "algebra",
    "polynomial",
    "linear algebra unsolved"
  ],
  "Problem": "[b]Prove that[/b] if $ k_{1},k_{2},..,k_{n-1},$ [b]are (n-1) positive different integer numbers and[/b] $ a_{1},...,a_{n}$ are [b]n positive different real numbers[/b] \r\nthen : $ \\left| \\begin {array} {LLLL} 1 & 1 & ... & 1 \\\\  a^{k_{1}}_{1} & a^{k_{1}}_{2} & ... & a^{k_{1}}_{n} \\\\ a^{k_{2}}_{1} & a^{k_{2}}_{2} & ... & a^{k_{2}}_{n} \\\\  ... & ... & ... & ... & ... \\\\ a^{k_{n-1}}_{1} & a^{k_{n-1}}_{2} & ... & a^{k_{n-1}}_{n} \\end {array} \\right| \\neq 0$\r\n Here [b]max [/b]{$ k_{1},...,k_{(n-1)}$} >n",
  "Solution_1": "This is not fundamentally different from the Vandermonde case:  the inversion of this matrix decides the solvability of the interpolation problem $ P(a_i) \\equal{} y_i$ for a polynomial $ P(x) \\equal{} c_0 \\plus{} c_1 x^{k_1} \\plus{} c_2 x^{k_2} \\plus{} ...$, which is clearly well-posed by a dimension argument.  I believe jmerry made comments to this effect in a thread a few weeks ago."
}
{
  "Tag": [
    "inequalities",
    "AMC",
    "USA(J)MO",
    "USAMO",
    "IMO Shortlist",
    "inequalities solved"
  ],
  "Problem": "For a,b,c positive real numbers such that a+b+c=1 prove\r\n\r\n1/(1+a+b)  + 1/(1+b+c) + 1/(1+c+a)  \\leq 1",
  "Solution_1": "There's definitely sth wrong with this one: what if a=b=c=1/3? The sum on the left becomes 9/5>1...",
  "Solution_2": "Sorry very much. The condition is abc=1 and not a+b+c=1.",
  "Solution_3": "Well, put x = a<sup>3</sup>, y = b<sup>3</sup>, z = c<sup>3</sup>. Then we have to prove that\r\n\r\n\\sum<sub>cyclic</sub>xyz/(x<sup>3</sup> + y<sup>3</sup> + xyz) \\leq 1.\r\n\r\nThis is exactly USAMO 1997/5. Solution at the kalva site.",
  "Solution_4": "This is a well-known trick:\r\n\r\nI used the sum Arne wrote. If we consider the fact that x^3+y^3>=xy(x+y) we get left sum <=  \\sum z/(x+y+z)=1. This was the official solution of a very similar (if not identical, I can't eally remember      :D) IMO problem.",
  "Solution_5": "You probably mean A1 of the IMO shortlist 1996 ?\r\n\r\nIf x, y, z > 0 and xyz = 1 then\r\n\r\n\\sum<sub>cyclic</sub> xy/(x<sup>5</sup> + y<sup>5</sup> + xy) \\leq 1.",
  "Solution_6": "Yeah, that's the one.",
  "Solution_7": "x=b+c y=c+a z=a+b\r\nSo we have:\r\n \\sum 1/(1+a+b)= \\sum 1/(1+x)=(3+2 \\sum x+ \\sum xy)/(1+ \\sum x+ \\sum xy+xyz)\r\n2+ \\sum x \\leq xyz\r\n2+2 \\sum a \\leq (a+b)(b+c)(c+a)\r\n2 \\sum a \\leq  \\sum (a^2b+a^2c+1)\r\na^2b+a^c+1 \\geq 3a\r\n \\sum(a^2b+a^c+1) \\geq 3 \\sum a\r\n3 \\sum a \\geq 2 \\sum a+3\r\n \\sum a \\geq 3",
  "Solution_8": "If $x,y,z$ positive real numbers such that $yz+zx+xy=3$ and $k\\geq1$, then\r\n\r\n$\\frac{1}{k+y+z}+\\frac{1}{k+z+x}+\\frac{1}{k+x+y}\\leq\\frac{3}{k+2}$,\r\n\r\nwith equality if and only if $x=y=z=1$.",
  "Solution_9": "use harazi's substitution\r\na=x/y,b=y/z,c=z/x and we are through! :roll:",
  "Solution_10": "How does that help Ji Chen's problem?"
}
{
  "Tag": [
    "search",
    "arithmetic sequence",
    "geometric sequence",
    "number theory open",
    "number theory"
  ],
  "Problem": "Does an arithmetic progression of four distinct triangular numbers exist?\r\n\r\nI've done search through all possible quadruples with the largest member smaller than $ T_{32000}$, however I didn't find anything. There exist a lot of triples, e.g. the smallest ones $ (T_1, T_7, T_{10}), (T_3, T_6, T_8)$.\r\n\r\nDoes anybody know something about this problem or have any ideas?\r\n\r\nNote: Similar problem with four triangular numbers in geometric progression (Sierpinski, Szymiczek) was proven to have no solution.",
  "Solution_1": "Note that $ T_n \\equal{} \\frac{n(n\\plus{}1)}{2} \\equal{} \\frac{(2n\\plus{}1)^2\\minus{}1}{8}$. Therefore, if triangular numbers $ T_{n_1}, T_{n_2}, T_{n_3}, T_{n_4}$ are consecutive term of an arithmetic progression, then such are the numbers $ (2n_1\\plus{}1)^2, (2n_2\\plus{}1)^2, (2n_3\\plus{}1)^2, (2n_4\\plus{}1)^2$. But it is well known (see [url=http://www.mathpages.com/home/kmath044/kmath044.htm]this page[/url], for example) that there are no 4 consecutive squares in arithmetic progression."
}
{
  "Tag": [
    "inequalities",
    "trigonometry",
    "inequalities unsolved"
  ],
  "Problem": "Prove the inequality\r\n\\[ \\frac{\\sqrt{1\\plus{}8\\cos^{2}A}}{\\sin B}\\plus{}\\frac{\\sqrt{1\\plus{}8\\cos^{2}B}}{\\sin C}\\plus{}\\frac{\\sqrt{1\\plus{}8\\cos^{2}C}}{\\sin A}\\geq 6,\\]\r\nwhere $ A$, $ B$ and $ C$ are the three angles of a triangle $ ABC$.",
  "Solution_1": "The inequality above is the corollary of this one: \r\n $ (1+8cos^2A)(1+8\\cos^2B)(1+8\\cos^2C)\\geq\\ 64sin^2Asin^2Bsin^2C $  (*)\r\nTo prove (*)  we replace $ sin^2A=1-cos^2A $ and etc. \r\nExpand the expression the inequality becomes :\r\n$  -63+72(cos^2A+cos^2B+cos^2C)+576cos^2Acos^2Bcos^2C \\geq\\ 0 $\r\nUse the indentity $ cos^2A+cos^2B+cos^2C+2cosAcosBcosC=1 $ (*) is equivalent to $ 64(cosAcosBcosC)^2 - 16(cosAcosBcosC) + 1 \\geq\\ 0 $ \r\nThe last inequality is clearly true. Therefore (*) is proved. Apply the Cauchy's inequality for the LHS of the original inequality we finish.",
  "Solution_2": "[quote=\"tranminhhoang\"]$ cos^2A+cos^2B+cos^2C+2cosAcosBcosC=1 $[/quote]\r\nIt was unknown for me.",
  "Solution_3": "[quote=\"Myth\"][quote=\"tranminhhoang\"]$ cos^2A+cos^2B+cos^2C+2cosAcosBcosC=1 $[/quote]\nIt was unknown for me.[/quote]\r\n\r\nI use this identity so frequently that suddenly it is difficult to write the proof. But anyway, i have the proof now. \r\nUse the formulae : $ cos^2A=(1+cos2A)/2 $ and etc. All we have to prove now is :\r\n$ P=cos2A+cos2B+2cos^2C+4cosAcosBcosC = 0 $\r\nWe have $cos2A+cos2B=2cos(A+B)cos(A-B)=-2cosCcos(A-B) $ Therefore it is equivalent to prove \r\n$ P=2cosC(cosC-cos(A-B)+2cosAcosB)=0 $ \r\nBut this is clearly true because $ cosC-cos(A-B)=-cos(A+B)-cos(A-B)=-2cosAcosB $"
}
{
  "Tag": [
    "inequalities",
    "combinatorics proposed",
    "combinatorics"
  ],
  "Problem": "Here's a quite easy, but elegant problem:\r\n\r\n[i]Let $ A \\equal{} (A_{i})_{1\\le i\\le n}$, $ B \\equal{} (B_{i})_{1\\le i\\le n}$, $ C \\equal{} (C_{i})_{1\\le i\\le n}$ be three partitions of a finite set $ M$.\nProve that if $ |A_{i}\\cap B_{j}|\\plus{}|B_{j}\\cap C_{k}|\\plus{}|C_{k}\\cap A_{i}|\\ge n$ holds for any $ i,j,k$, then $ |M|\\ge\\frac{n^{3}}{3}$.\n\nEquality can hold iff $ n\\equiv0($mod $ 3)$.\n[/i]",
  "Solution_1": "Summation over all $ i,j,k$ in your inequality proves the first part of your assertion \r\n\r\n$ \\begin{array}{lcl}n^{4}&\\leq &\\sum_{i,j,k=1}^{n}\\left(|A_{i}\\cap B_{j}|+|B_{j}\\cap C_{k}|+|C_{k}\\cap A_{i}|\\right)\\\\[0.1cm] & =&\\sum_{i,j=1}^{n}\\left(n|A_{i}\\cap B_{j}|+|B_{j}|+|A_{i}|\\right)\\\\[0.1cm] & = &\\sum_{i=1}^{n}\\left(n|A_{i}|+|M|+n|A_{i}|\\right)\\\\[0.1cm] & = & n|M|+n|M|+n|M|\\\\[0.1cm] & = & 3n|M|\\end{array}$ \r\n\r\nEquality obviously implies $ n$ divisible by 3. This proves one direction of the \"iff\"-statement. As for the other direction (the \"can hold\") let $ n$ be divisible by 3 and choose an arbitrary partition $ \\left(D_{r,s}\\right)_{r,s=1,\\ldots n}$ such that $ |D_{(r,s)}|=\\frac{n}{3}$ holds. We interpret the indexes $ (r,s)$ of this partition as all possible pairs of residues $ 1,\\ldots, n$ modulo $ n$. With that interpretation in mind define\r\n\r\n$ \\begin{array}{lcl}A_{i}& = &\\bigcup_{t=1}^{n}D_{(i,t)}\\\\[0.1cm] B_{j}& = &\\bigcup_{t=1}^{n}D_{(t,j)}\\\\[0.1cm] C_{k}& = &\\bigcup_{t=1}^{n}D_{(t,t+k)}\\end{array}$\r\n\r\nand observe for all indexes $ i,j,k$ the relationship\r\n\r\n$ |A_{i}\\cap B_{j}|+|B_{j}\\cap C_{k}|+|C_{k}\\cap A_{i}|=|D_{(i,j)}|+|D_{(j-k,j)}|+|D_{(i,i+k)}|=n$ verifying the last bit."
}
{
  "Tag": [],
  "Problem": "whats wron with my solution\r\n\r\nWhen Bill drives to work at 45 mph, he is 5 minutes late. when bill drives 53 mph to work, he is 5 minutes early. how fast should he drive to work if he wants to be right on time?\r\n\r\n[hide]\nset the time it takes for him to be right on time x.\n$ D \\equal{} 45(x \\plus{} 5)$\n$ D \\equal{} 53(x \\minus{} 5)$\nsubstitute\n$ 53(x \\minus{} 5) \\equal{} 45(x \\plus{} 5)$\ni solve and get 61.25\nwhich is clearly not the answer...[/hide]",
  "Solution_1": "You solved for $ x$, not the rate.  The rate would be $ \\frac {D} {x}$.  So, go redo it, and see if you end up with the correct answer.",
  "Solution_2": "[hide=\"new solution\"]\ni did what i did before, got $ x\\equal{}61.25$\ni plugged it back in, and got $ D\\equal{}2981.25$\n$ \\frac{D} {x}\\equal{}48.67$(i dont know how to make the aproximate sign)[/hide]",
  "Solution_3": "The approximate sign is \\approx. $ \\approx$"
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "Let $ a,b,c \\ge \\frac{1}{2}$ be positive real numbers such that $ abc\\equal{}1$. Prove that\r\n\\[ \\frac{1}{a\\plus{}2b}\\plus{}\\frac{1}{b\\plus{}2c}\\plus{}\\frac{1}{c\\plus{}2a} \\le 1\\]\r\n\r\nWe can prove this inequality using Cauchy Schwarz Inequality. :)",
  "Solution_1": "Can, see your problem, I remembered an inequality created by Anh Tran. \r\n[hide]Let $ a,b,c$ be positive real numbers. Prove that \n$ \\frac {b}{a \\plus{} 2b} \\plus{} \\frac {c}{b \\plus{} 2c} \\plus{} \\frac {a}{c \\plus{} 2a}\\leq 1$[/hide]",
  "Solution_2": "[quote=\"Sudoku\"]Can, see your problem, I remembered an inequality created by Anh Tran. \nLet $ a,b,c$ be positive real numbers. Prove that \n$ \\frac {b}{a \\plus{} 2b} \\plus{} \\frac {c}{b \\plus{} 2c} \\plus{} \\frac {a}{c \\plus{} 2a}\\leq 1$[/quote]\r\nYes, you are right. :)\r\nBut I think this problem was proposed first time by VasC at Romany team selection test (it is the generalization)\r\n\\[ \\overset{n}{\\underset{i\\equal{}1}{\\sum}} \\frac{1}{a_i\\plus{}n\\minus{}1} \\le 1\\]\r\nwith $ a_1,a_2,...,a_n$ are positive real numbers such that $ a_1a_2\\cdots a_n\\equal{}1$. \r\n:)"
}
{
  "Tag": [
    "inequalities",
    "topology",
    "triangle inequality",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Let A be a bounded subset of a metric space X equipped with a metric 'd'. show that A is bounded iff for every $x\\in X$ there exists r>0 such that $A\\subset B(x,r)$\r\n\r\nlet A be bounded, then there exists $y\\in X$ and R>0, so that $A\\subset B(y,R)$\r\n\r\nnow i am stuck when proving that if A is bounded, then that condn holds. here's what i tried \r\n\r\nlet us assume the contrary. ie$\\exists x\\in X$ such that however big u take r,points of A lie outside $B(x,r)$. tkae r=n, then $\\exists a_n$ such that $d(a_n,x)\\geq n$ . now using this and the bounded condition i tried to arrive at a contradiction,by triangle inequality but i'm getting stuck.is my approach on the right lines?could someone give me a hint? or suggest another approach. i dont want a detailed proof as i want to do it myself\r\n\r\np.s- by $B(x,r)$ i mean the open ball, with radius r and centre x",
  "Solution_1": "anyone? :?",
  "Solution_2": "Let $R_1 = R+\\rho(x,y)$, and look at the $B(x,R_1)$",
  "Solution_3": "ok i've got it. thanks andrew!"
}
{
  "Tag": [
    "calculus",
    "calculus computations"
  ],
  "Problem": "Does anyone know of a really good Partial Differential Equation's Book?",
  "Solution_1": "we used evan's, \"partial differntial equations\" at my school",
  "Solution_2": "how did you like the book?"
}
{
  "Tag": [],
  "Problem": "Take the figure $ 123456789$. Now insert two and only two multiplication signs ($ \\times$) anywhere inbetween.\r\n \r\nHow many even numbers will you be able to generate by this exercise ?\r\n \r\nFor example,\r\n \r\n$ 12\\times34\\times56789$ and $ 1\\times2\\times3456789$ are two such exercises which will generate even numbers.\r\n \r\nYou are not permitted to change the sequence of the digits $ 1$ thru' $ 9$.",
  "Solution_1": "[hide=\"Hint\"]Count all the cases (number of ways of havin 2 x and 7 numbers) and substract the odd ones (remembering the only you get and odd number by multipliying odd numbers only)[/hide]",
  "Solution_2": "[hide] Total ways = $ \\binom{9\\minus{}1}{2}\\equal{}28$\nTo make odd numbers, we can only put both \"X\"s after 1,3,5,7. Therefore, we have $ \\binom{4}{2}\\equal{}6$ ways\nSo the there're 22 ways. Right?[/hide]"
}
{
  "Tag": [
    "geometry",
    "incenter",
    "inradius",
    "rectangle",
    "geometry proposed"
  ],
  "Problem": "Given three parallel lines on the plane. Find the locus of incenters of triangles with vertices lying on these lines (a single vertex on each line).",
  "Solution_1": "Call $ r$ the parallel line that is contained in the stripe determined by the other two lines $ s,s'$.  Clearly, if the incenter $ I$ is contained in the stripe defined by $ r$ and $ s$, then it must be closer to $ r$ than to $ s$; otherwise, since the incircle cannot touch line $ s$ because the triangle is entirely contained in the stripe between $ s$ and $ s'$, then the inradius is less than half the distance between $ r$ and $ s$, hence it would not cross $ r$, and any ray originating in any point on $ r$ and touching the incircle could not cross $ s'$.  Similarly, if the incenter is contained in the stripe defined by $ r$ and $ s'$, then it is closer to $ r$ than to $ s'$.  We will now show that for any point strictly inside the stripe defined by the parallel lines halfway between $ r$ and $ s$, and halfway between $ r$ and $ s'$, one point may be found on each of the three given lines such that this point is the incircle of the triangle thus formed.  Take $ A\\in r$, $ B\\in s$ and $ C\\in s'$, collinear and such that $ AB\\perp r$.  Displace $ C$ along $ s'$ away from $ A,B$.  As $ C$ goes arbitrarily far, the figure tends towards an open rectangle with one side $ AB$, two sides being rays on $ r$ and $ s$, hence the incircle tends towards the circle touching $ AB$, $ r$ and $ s$, ie, the incenter tends towards a point on the line halfway between $ r$ and $ s$, getting arbitrarily close but never reaching it.  From the original position, displace $ A$ along $ s$ away from $ B,C$, the incenter tends towards a point on the line halfway between $ r$ and $ s'$, getting arbitrarily close but never reaching it.  Since $ C$ on the first displacement, and $ A$ on the second displacement, may be moved continuously, and in doing so the incenter moves continuously, for each line $ t$ parallel to $ r$ contained strictly in the stripe defined by the lines halfway from $ r$ and $ s$, and halfway from $ r$ and $ s'$, there is a position of $ A,B,C$ such that the incenter is on $ t$.  Displace now simultaneously $ A,B,C$ along $ s,r,s'$ as needed until the incenter reaches any desired point on $ t$.  The conclusion follows."
}
{
  "Tag": [
    "quadratics"
  ],
  "Problem": "Find the value of x. There are 4 solutions.\r\n(x^4) + (5 x^2) = 36",
  "Solution_1": "Are you actually supposed to do [url=http://planetmath.org/encyclopedia/QuarticFormula.html]this[/url], or is there an easier way?",
  "Solution_2": "OF COURSE NOT!!\r\n\r\nHint: try reducing it to something you can solve, such as.......................................................... a quadratic?\r\n\r\nOh yeah, and what are those stars under your name for?? :?:",
  "Solution_3": "[hide=\"Hint\"]Let $ X\\equal{}x^{2}$.[/hide]\r\n\r\nand the stars relate to how many posts you have.",
  "Solution_4": "thats what i was thinking. only i used y. so we have y^2+5y=36 so y^2+5y-36=0 (y+9)(y-4). plug in x^2. x^2+9=0 and x^2-4=0.\r\n\r\nare there imaginary numbers involved cuz i get [hide=\"my answer\"]$ 2,\\minus{}2,3i,\\minus{}3i$[/hide]",
  "Solution_5": "That is correct!",
  "Solution_6": "[quote Kevin K.\"]Find the value of x. There are 4 solutions.\r\n(x^4) + (5 x^2) = 36[/quote]\r\nSolution (by galois01)\r\n$ x^{4}\\plus{}5x^{2}\\equal{}36\\Longleftrightarrow$\r\n$ x^{4}\\plus{}5x^{2}\\minus{}36\\equal{}0 (1)$\r\nLet:\r\n$ x^{2}\\equal{}z$ (2)          $ z\\geq 0$\r\n$ (1)\\Longrightarrow z^{2}\\plus{}5z\\minus{}36\\equal{}0$\r\n$ D\\equal{}5^{2}\\minus{}4(\\minus{}36)\\equal{}169$\r\n$ z_{1,2}\\equal{}\\frac{\\minus{}5\\pm\\sqrt{169}}{2}$\r\n$ \\Longrightarrow$  $ z\\equal{}4$and $ z\\equal{}\\minus{}9$\r\nFrom (2) we get\r\n$ x^{2}\\equal{}4\\Longleftrightarrow x\\equal{}\\pm2$\r\n$ x^{2}\\equal{}\\minus{}9\\Longleftrightarrow x\\equal{}\\pm3i$\r\nSo the solutions are:\r\n$ x\\equal{}2,\\minus{}2, 3i,\\minus{}3i$",
  "Solution_7": "Oh. yeah. Oops. This is simple, sorry. I thought... \r\n\r\nHere, let me try it on my own.\r\n\r\n[hide]\n$ x^{4}\\plus{}5x^{2}\\equal{}36$\n$ n\\equal{}x^{2}$\n$ n^{2}\\plus{}5n\\equal{}36$\n$ n^{2}\\plus{}5n\\plus{}6.25\\equal{}42.25$\n$ \\pm(n\\plus{}2.5)\\equal{}6.5$\n$ n\\equal{}4,\\minus{}9$\n$ x\\equal{}2,\\minus{}2,3i,\\minus{}3i$[/hide]",
  "Solution_8": "yep, thats it",
  "Solution_9": "[hide]\nReplace $ x^{2}$ with $ y$\n$ y^{2}\\plus{}5y\\minus{}36\\equal{}0$\nFactor: $ (y\\plus{}9)(y\\minus{}4)\\equal{}0$\nSo, $ y\\equal{}\\minus{}9 , 4$\nThe 4 solutions for x are $ \\boxed{\\minus{}2 , 2 ,\\minus{}3i, 3i}$[/hide]",
  "Solution_10": "x^4+5x^2=36----> Y^2+5Y=36\r\n\r\n(X^2=Y)",
  "Solution_11": "Is there any particular reason that you have to make the substitution $ y\\equal{}x^{2}$? You can just directly factor...\r\n\r\n\\[ x^{4}\\plus{}5x^{2}\\minus{}36\\equal{}(x^{2}\\minus{}4)(x^{2}\\plus{}9)\\equal{}(x\\minus{}2)(x\\plus{}2)(x\\minus{}3i)(x\\plus{}3i)\\equal{}0\\]",
  "Solution_12": "you dont have to, it just makes it easier to see what we have to do. but yes you can do it that way to"
}
{
  "Tag": [
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Solve in natural numbers x,y,z,t the equation: $x^2+y^2+z^2=2^t$",
  "Solution_1": "Well, if t equals 0, the solution is trivial : (x,y,z)=(1,0,0) of any permutation of this solution. \r\nNow assume that $t\\geq 1$. The value of  $x^2+y^2+z^2$ must be divisible by 2. So x,y and z are all even, or exactly one of those is even. In the first case, we can put $x=2x',y=2y',z=2z'$ and the equality becomes $x'^2+y'^2+z'^2=2^{t-2}$. If t equals 2, we have our trivial case, otherwise we have one more time two cases to see. We can do the substitution again and again to get exactly one of the numbers even. Let WLOG  $x=2x'+1,y=2y'+1,z=2z'$ and the equality becomes now $2(x'^2+x'+y'^2+y'+z'^2)+1=2^{t-1}$. So $2^{t-1}$ is not even and t=1. Then obviously $x'=y'=z'=0$. Finally we have   \r\n$(x,y,z,t)\\in\\{(1,0,0,0),(0,1,0,0),(0,0,1,0),(2^k,2^k,0,2k+1),(2^k,0,2^k,2k+1),(0,2^k,2^k,2k+1)|k\\in\\mathbb{N}\\}$",
  "Solution_2": "You omitted $(2^k,0,0,2k)$ and the permutations (since you left the discussion on all the numbers even unfinished), and for $k=0$ you get the trivial solution."
}
{
  "Tag": [
    "inequalities",
    "ceiling function",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Let p(n) be the greatest odd divisor of n. Prove that\r\n                  $\\frac{1}{2^n} \\sum_{k=1}^{2^n} \\frac{p(k)}{k} > \\frac{2}{3}$",
  "Solution_1": "Solution: Let $ n,k$ be two positive integers.\r\nWe define $ v_2: \\mathbb{N} \\rightarrow \\mathbb{N}$, by $ v_2(n) = k$ iff $ 2^k \\parallel{} n$ (i.e $ 2^k$ divides $ n$ and $ 2^{k + 1}$ does not divide $ n$) \r\nWe have\r\n\r\n\\begin{eqnarray*} \\sum_{k = 1}^{2^n} \\frac {p ( k )}{k} & = & \\sum_{j = 0}^{\\infty} \\sum_{\\tmscript{\\begin{array}{c} k \\leq 2^n \\\\\r\nv_2 ( k ) = j \\end{array}}} \\frac {p ( k )}{k} \\\\\r\n& = & \\sum_{j = 0}^{\\infty} \\sum_{\\tmscript{\\begin{array}{c} k \\leq 2^n \\\\\r\nv_2 ( k ) = j \\end{array}}} \\frac {\\frac {k}{2^j}}{k} \\\\\r\n& = & \\sum_{j = 0}^{\\infty} \\frac {1}{2^j} \\sum_{\\tmscript{\\begin{array}{c} k \\leq 2^n \\\\\r\nv_2 ( k ) = j \\end{array}}} 1 \\\\\r\n& = & \\sum_{j = 0}^n \\frac {1}{2^j} \\sum_{\\tmscript{\\begin{array}{c} 2^j k \\leq 2^n \\\\\r\nk \\mbox{ odd} \\end{array}}} 1 \\\\\r\n& = & \\sum_{j = 0}^n \\frac {1}{2^j} \\sum_{\\tmscript{\\begin{array}{c} k \\leq 2^{n - j} \\\\\r\nk \\mbox{ odd} \\end{array}}} 1 \\\\\r\n& = & \\sum^n_{j = 0} \\frac {1}{2^j} \\cdot \\lceil 2^{n - j - 1} \\rceil \\\\\r\n& = & \\sum_{j = 0}^{n - 1} \\frac {1}{2^j} \\cdot 2^{n - j - 1} + \\frac {1}{2^n} \\\\\r\n& = & 2^{n - 1} \\sum_{j = 0}^{n - 1} \\left( \\frac {1}{4} \\right)^j + \\frac {1}{2^n} \\\\\r\n& = & 2^{n + 1} \\cdot \\left( \\frac {1 - \\left( \\frac {1}{4} \\right)^n}{3} \\right) + \\frac {1}{2^n} \\end{eqnarray*}\r\n\r\nHence\r\n\r\n\\begin{eqnarray*} \\frac {1}{2^n} \\sum_{k = 1}^{2^n} \\frac {p ( k )}{k} & = & 2 \\cdot \\left( \\frac {1 - \\frac {1}{4^n}}{3} \\right) + \\frac {1}{4^n} \\\\\r\n& = & \\frac {2 - \\frac {2}{4^n} + \\frac {3}{4^n}}{3} \\\\\r\n& = & \\frac {2 + \\frac {1}{4^n}}{3} > \\frac {2}{3} \\end{eqnarray*}\r\n\r\nFunny solution, huh ? :)"
}
{
  "Tag": [],
  "Problem": "Five integers have a sum of 92. What is the greatest possible product of these five numbers?",
  "Solution_1": "But we want 5 numbers  :roll:  (and I never thought of AM-GM even though it is obvious that you can use it).\r\n\r\nEDIT: Deleted post",
  "Solution_2": "[quote=\"236factorial\"]But we want 5 numbers  :roll:  (and I never thought of AM-GM even though it is obvious that you can use it).[/quote]\r\n\r\nSorry, I misread the question and caught my mistake before you posted.  I found the answer for 5 numbers, but realised that this is the middle school section, so I'll let someone else post."
}
{
  "Tag": [
    "HMMT",
    "Vieta"
  ],
  "Problem": "Give that three roots of $ f(x)\\equal{}x^4\\plus{}ax^2\\plus{}bx\\plus{}c$ are 2, -3, and 5, what is the value of $ a\\plus{}b\\plus{}c$?",
  "Solution_1": "[hide=\"solution\"]\nLet the 4 roots of $ f(x)$ be $ r_1, r_2, r_3, r_4$. Then $ r_1 = 2, r_2 = -3, r_3=5$.\n\nBy Vieta's Formulas, $ r_1 + r_2 + r_3 + r_4 = 0$ (since there is no $ x^3$ term).\n\n$ 2 - 3 + 5 + r_4 = 0 \\implies r_4 = -4$.\n\nSo $ f(x) = x^4 +ax^2 + bx + c = (x - 2)(x+3)(x-5)(x+4)$\n\nThen letting $ x=1$ we get:\n\n\\begin{align*}\nf(1) = 1^4 +a(1)^2 + b(1) + c &= (1 - 2)(1+3)(1-5)(1+4) \\\\\n1 + a + b + c &= 80 \\\\\na+b+c &= \\boxed{79}\n\\end{align*}\n[/hide]",
  "Solution_2": "well plugging in\r\n\r\n16+4a+2b+c=0\r\n81+9a-3b+c=0\r\n625+25a+5b+c=0\r\n\r\nsubtracting (2) and (1) you get 5b-5a=65, b-a=13\r\nsubtracting (3) and (1) you get 544=16a+8b or 2a+b=-68\r\n\r\nsubstituting b=13+a, you get 3a=-81, a=-27, so b=-14\r\n\r\nplugging into any of the three equations, c=120\r\n\r\nso -27-14+120=79"
}
{
  "Tag": [
    "inequalities proposed",
    "inequalities"
  ],
  "Problem": "$ S_n \\equal{} a_1 \\plus{}a_2 \\plus{} \\cdots \\plus{} a_n , a_n \\plus{}2 S_n S_{n\\minus{}1} \\equal{}0 (n \\geq 2),a_1 \\equal{} \\frac 12 ,$\r\n$ b_n \\equal{} 2 (1\\minus{}n) a_n (n \\geq 2), \\Longrightarrow b_2 ^2 \\plus{}b_3^2 \\plus{} \\cdots \\plus{}b_n^2<1.$",
  "Solution_1": "[quote=\"jcc0107\"]$ S_n \\equal{} a_1 \\plus{} a_2 \\plus{} \\cdots \\plus{} a_n , a_n \\plus{} 2 S_n S_{n \\minus{} 1} \\equal{} 0 (n \\geq 2),a_1 \\equal{} \\frac 12 ,$\n$ b_n \\equal{} 2 (1 \\minus{} n) a_n (n \\geq 2), \\Longrightarrow b_2 ^2 \\plus{} b_3^2 \\plus{} \\cdots \\plus{} b_n^2 < 1.$[/quote]\r\n\r\n$ a_{n}\\equal{}S_{n}\\minus{}S_{n\\minus{}1}$---> $ a_{n}\\plus{}2S_n S_{n\\minus{}1}\\equal{}S_{n}\\minus{}S_{n\\minus{}1}\\plus{}2S_n S_{n\\minus{}1}\\equal{}0$----> $ S_{n}\\equal{}\\frac{1}{c_{n}}$----> $ c_{n\\minus{}1}\\minus{}c_{n}\\plus{}2\\equal{}0$---> $ \\sum(c_{n\\minus{}1}\\minus{}c_{n}\\plus{}2)\\equal{}c_{1}\\minus{}c_{n}\\plus{}2(n\\minus{}1)\\equal{}0$---> $ c_{n}\\equal{}2n$ ---> $ a_{n}\\equal{}\\frac{1}{2n(1\\minus{}n)}$-----> $ b_{n}\\equal{}\\frac{1}{n}$--->\r\n\r\n$ \\sum b^{2}_{k}\\equal{}\\sum \\frac{1}{k^{2}}<\\sum \\frac{1}{k(k\\minus{}1)}\\equal{}\\sum(\\frac{1}{k\\minus{}1}\\minus{}\\frac{1}{k})\\equal{}1\\minus{}\\frac{1}{n}<1$"
}
{
  "Tag": [],
  "Problem": "How many C's are used when writing down all of the numbers from 1 to 500 using Roman numerals?\r\n\r\nI keep getting 720, but the correct answer is 750.  Can someone explain this to me?\r\n\r\nHere's a quick chart if you forgot which letters go with which numbers:\r\n\r\nI = 1\r\nV = 5\r\nX = 10\r\nL = 50\r\nC = 100\r\nD = 500",
  "Solution_1": "I don't know how to express numbers higher than like 10, so could someone explain how to express higher numbers to me?",
  "Solution_2": "Here are the rules according to the makers of the problem:\r\n\r\n[quote] As many as three of the same marks that represent $10^n$ may be placed consecutively to form other numbers:\n\n    * III is 3\n    * CCC is 300 \n\nMarks that have the value $5\\cdot10^n$ are never used consecutively.\n\nGenerally (with the exception of the next rule), marks are connected together and written in descending order to form even more numbers:\n# CCLXVIII = 100+100+50+10+5+1+1+1 = 268\n\nSometimes, a mark that represents 10^n is placed before a mark of one of the two next higher values (I before V or X; X before L or C; etc.). In this case, the value of the smaller mark is SUBTRACTED from the mark it precedes:\n\n    * IV = 4\n    * IX = 9\n    * XL = 40 \n\nThis compound mark forms a unit and may not be combined to make another compound mark (e.g., IXL is wrong for 39; XXXIX is correct).\n\nCompound marks like XD, IC, and XM are not legal, since the smaller mark is too much smaller than the larger one. For XD (wrong for 490), one would use CDXC; for IC (wrong for 99), one would use XCIX; for XM (wrong for 990), one would use CMXC. [/quote]",
  "Solution_3": "C is first used at 90 = XC\r\nOne C is used until 190 = CXC\r\nTwo C's are used until 290 = CCXC\r\nThree C's are used until 390 = CCCXC\r\nFour C's are used until 400 = CD\r\nOne C is again used until 490 = CDXC\r\nTwo C's are used until 500 = D\r\n\r\n100 + 200 + 300 + 40 + 90 + 20 = 750",
  "Solution_4": "I need help again with something similar to this (It's for a computer program).\r\n\r\nHow many X's, L's and C's are in 2974.  I need the work, since I already know the answers.  \r\n\r\nThanks.",
  "Solution_5": "Sorry, I worded it wrong.  How many of those letters do you need to write [b]all[/b] of the number [b]from[/b] 1 [b]to[/b] 2974.\r\n\r\nThe bolds are just to emphasize, not to act like I'm getting mad at you or anything.",
  "Solution_6": "Here's what I have for the number of L's for all of the numbers from 1 to 2974 so far:\r\n\r\nHere are all of the multiples of 10 from 40 to 100 that need L's:\r\n\r\nXL = 40 to 49\r\nL = 50 to 59\r\nLX = 60 to 69\r\nLXX = 70 to 79\r\nLXXX = 80 to 89\r\n\r\nThat means for every 100 numbers you wirte, you need 50 L's.  That means we have 28*50 L's from 1 to 2900.  But, we still have 74 numbers left to check.\r\n\r\nFrom 40 to 69, there are 30 L's, and from 70-74 there are 5 L's.\r\n\r\nSo in total, we have 28*50 + 30 + 5 = 1400+35 = 1435.  But the correct answer is 1485.  This is the one I'm having to most trouble with.",
  "Solution_7": "1-2900 is 29 hundreds, not 28?",
  "Solution_8": "Yeah, I forgot about the 1-100 part.  I though that it should have been 28 since 2974 is in the 2900-3000 range.\r\n\r\nSorry, but I always make stupid mistakes like that."
}
{
  "Tag": [],
  "Problem": "A house with a square floor plan is to have each of its four outer walls finished with brick, stone, stucco or wood. If no two adjacent outer walls are to have the same finish in how many ways can the house's walls be constructed? (One such way is to brick on the front and back and stone on the left and right.)",
  "Solution_1": "If all four walls have the same finish, there are $ 4!\\equal{}24$ different ways to finish the walls.\r\nIf two walls have the same finish and the other walls have different finishes, there are $ 4\\cdot 2\\cdot {3\\choose 2}\\cdot 2\\equal{}4\\cdot 2\\cdot 3\\cdot 2\\equal{}48$ different ways.\r\nIf two walls have the same finish and the other two walls also have the same finish, then there are $ {4\\choose 2}\\cdot 2\\equal{}6\\cdot 2\\equal{}12$ different ways.\r\nIn total, there are $ 24\\plus{}12\\plus{}48\\equal{}\\boxed{84}$ ways.",
  "Solution_2": "I got this with the casework solution, but why can't we just say that there are 4 choices for the front most wall. Then 3 choices for each of the side and 2 in the back to get $4 \\cdot 3^2 \\cdot 2 = 72$?",
  "Solution_3": "It gets complicated then. What if one of the sides with 3 has the same choice as the other side with 3? Then instead of 2, you have 3 for the back, etc. Those things screw up the easy count.",
  "Solution_4": "So it's then 4x3x3 if the two sides are the same and 4x3x2x2 if the two sides are different (Note 3x2 instead of 3x3 as the two sides have to be different) for a total of 84",
  "Solution_5": "What's wrong with this [hide=solution?]\nLet's call the walls $A$, $B$, $C$, and $D$ and paint them in that order. Wall $A$ has four choices, wall $B$ has three choices (because it cannot be the same as wall $A$, we aren't considering wall $C$ yet), and wall $C$ has three choices (same logic). So the ways to paint these three walls is $4\\cdot3\\cdot3=36$. Exactly one quarter of the time, $A$ and $C$ will be painted with the same material, so for that one quarter of the time ($36\\cdot\\frac{1}{4}=9$ possibilities), $D$ has three choices, so it's $9\\cdot3=27$. The other three quarters of the time ($36\\cdot\\frac{3}{4}=27$ possibilities), $A$ and $C$ have different material, so $D$ has two choices, so it's $27\\cdot2=54$. Add the two up, and the answer is $81$.\n\nBut clearly, the answer isn't 81! I don't understand why.\n[/hide]\n\nI hope this isn't considered revival, I just didn't understand, and I know that we are supposed to discuss with Gamebot. Thanks!\nhiabc",
  "Solution_6": "\uff55\uff4d\u3000\uff54\uff48\uff49\uff53\u3000\uff49\uff53\u3000\uff4d\uff49\uff53\uff50\uff4c\uff41\uff43\uff45\uff44\uff0e\u3000\uff29\uff54\u3000\uff53\uff48\uff4f\uff55\uff4c\uff44\u3000\uff42\uff45\u3000\uff49\uff4e\u3000\uff43\uff41\uff53\uff45\uff57\uff4f\uff52\uff4b\uff0c\u3000\uff4e\uff4f\uff54\u3000\uff43\uff4f\uff55\uff4e\uff54\uff49\uff4e\uff47\u3000\uff57\uff49\uff54\uff48\u3000\uff52\uff45\uff53\uff54\uff52\uff49\uff43\uff54\uff49\uff4f\uff4e\uff53",
  "Solution_7": "Hello,\n\n[hide=Alternate Casework]\nThe north side has 4 possibilities.\n[b]Case 1: East and west side have same finishing.[/b] There are 3 ways to choose the shared finishing and 3 ways to choose the south side's finishing.\n[b]Case 2: East and west side have different finishings.[/b] There are 3 ways to choose the east side's, then 2 ways to choose the west side's, then 2 ways to choose the south side's.\nAltogether, there are a total of $4 \\cdot (3 \\cdot 3 + 3 \\cdot 2 \\cdot 2) = \\boxed{84}$ choices. $\\square$\n[/hide]\n\n[b]@hiabc:[/b] [hide=Hopefully this makes sense.]Let your $A, B, C, D$ be the north, east, south, and west walls of the house, respectively. If $A = C$ then there are 4 ways to choose the shared finishing, 3 ways to choose $B,$ and 3 ways to choose $D.$ If $A \\neq C$ then there are 2 ways each to choose $B$ and $D$ finishing. This gives us an answer of $4 \\cdot 3 \\cdot 3 + 4 \\cdot 3 \\cdot 2 \\cdot 2 = \\boxed{84}$ ways. $\\square$[/hide]\n\n$\\varnothing$",
  "Solution_8": "[hide = solution]\nEquivalent to the 4th chromatic number, which is simply 3^4+1*3=84",
  "Solution_9": "[quote=SigmaPiE][hide = solution]\nEquivalent to the 4th chromatic number, which is simply 3^4+1*3=84[/quote]\n\nWhat's a \"chromatic\" number?",
  "Solution_10": "[quote=MarcusRashford][quote=SigmaPiE][hide = solution]\nEquivalent to the 4th chromatic number, which is simply 3^4+1*3=84[/quote]\n\nWhat's a \"chromatic\" number?[/quote]\n\na chromatic number is the least number of colours needed to colour a graph's vertices such that no two adjacent vertices has the same colour."
}
{
  "Tag": [
    "number theory",
    "least common multiple",
    "arithmetic sequence"
  ],
  "Problem": "The two arithmetic sequence, 1, 5, 9, 13, $ \\ldots$ and 1, 6, 11, 16, $ \\ldots$, have infinitely many terms in common. What is the sum of the first three common terms?",
  "Solution_1": "The lists are $ 1 \\plus{} 4x$ and $ 1 \\plus{} 5x$, respectively, where x >= 0.\r\nSubtract 1 from every term in the list, giving lists of $ 4x$ and $ 5x$, x>=0.\r\nSince the lcm of 4 and 5 is 20, we want to find the first 3 terms of $ 20x$, which are 0, 20, and 40.\r\nAdding those up gives 60, but we still need to add the 3 that we subtracted before.\r\nSo the sum is $ 60 \\plus{} 3 \\equal{} \\boxed{63}$."
}
{
  "Tag": [
    "function",
    "limit",
    "LaTeX",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Let $f\\text{: }[0,4]\\rightarrow\\mathbb{R}$ be a continuous function. such that $f(0)\\cdot f(4)<0$. Show that $(\\exists) c\\in(0,2)$ such that $1-<f(c^{2})+f(2c)<0$.",
  "Solution_1": "If we denote $g: [0,2]\\to\\Re,g(x)=f(x^2)+f(2x)$ then $g(0)g(4)<0$ so let $g(\\alpha)=0,\\alpha \\in (0,2)$. Suppose $g(0)<0$,$g$ is continous so $\\lim_{x\\to\\alpha}g(x)=g(\\alpha)=0$ (don`t know how to write this in $LaTeX$ but in the precent limit we consider it from te left) so that means that $\\forall \\epsilon >0 |g(x)|<\\epsilon$ iff $|x-\\alpha|<\\delta$. If we take $\\epsilon=1$ then $-1<g(x)<1$ and from the fact that $g$ is continous and $g(\\alpha)=0$ then it is clearly that exists $\\beta$ such that $-1<g(\\beta)<0$ ... Is this proof ok?",
  "Solution_2": "I think it's ok... I found the following statment:\r\n\r\nLet $f: [0,2]\\to\\mathbb{R}$ a continuous funtion that satisfyes $f(a)\\cdot f(b)\\leq0$. Then there is at least on point $c\\in[a,b]$ such that $f(c)=0$.\r\n\r\nCan this help?",
  "Solution_3": "Well ... I used that.... It`s called Cauchy-Bolzano theorem. :P",
  "Solution_4": "I thought so  :blush: . I'm new with this type of problems... need to practice more  :read:"
}
{
  "Tag": [
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "Let ABC be a triangle inscribed in a circle of radius R \r\n\r\nand a, b, c be respectly sides of the triangle ABC.\r\n\r\nProve that:\r\n\r\n   R/r >= b/c + c/b\r\n\r\nPd.  r: inradio of the triangle ABC.",
  "Solution_1": "R/r >= (b^2+c^2)/bc   (1)\r\n\r\n(abc)/4(p-a)(p-b)(p-c)  >= (b^2+c^2)/bc    (2)\r\n\r\nWe denote   x=2(p-a)        y=2(p-b)           z=2(p-c)\r\nso we have  a=(y+z)/2      b=(x+z)/2         c=(y+x)/2  \r\n\r\n(2) equivalent to\r\n\r\n(x+y)(y+z)(z+x)/4xyz   >=   ((x+z)^2 + (y+x)^2)/(x+z)(y+x)\r\n\r\n((x+y)(y+z)(z+x)-8xyz)/4xyz   >=   ((x+z)^2 + (y+x)^2)/(x+z)(y+x)\r\n\r\n(x(y-z)^2 + y(z-x)^2 +z(x-y)^2 )/4xyz   >=   (y-z)^2/(x+z)(x+y)\r\n\r\nWe have (y(z-x)^2 + z(x-y)^2)/4xyz = (z-x)^2/4xz  +  (x-y)^2/4xy \r\n>= (z-y)^2/4x(y+z)\r\n\r\nTherefore now we have to prove :\r\n\r\n1/4yz + 1/4x(y+z)  >=  1/(x+z)(x+y) (3)\r\n\r\nUse the lemma for a,b>0 1/a +1/b >=4/(a+b)  (3) is proved.\r\n\r\nAfter all, (1) is proved.  The equality holds iff ABC is the equalateral triangle"
}
{
  "Tag": [
    "geometry",
    "function",
    "MATHCOUNTS"
  ],
  "Problem": "Is everyone who posts here actually from the bay area? I was wondering if possibly there are some ppl here who don't actually live here...\r\n\r\nI really do by the way...",
  "Solution_1": "This forum isn't restricted to residents of the Bay Area. It's just that most people who don't live here wouldn't really have anything to post here, now, would they?\r\n\r\nBy the way, I live in Millbrae, which is near the SF Airport.",
  "Solution_2": "Yeah!!!!! I'm from the east bay. In a town near concord. Go campo!",
  "Solution_3": "eh whats the point of this topic\r\nSPAM SPAM\r\n\r\n\r\num\r\n\r\n\r\ndoom\r\n\r\n\r\n\r\num\r\n\r\n\r\n\r\ndio is a better singer than osbourne\r\nyea\r\n\r\n666",
  "Solution_4": "how many people here from cupertino? or cupertino schools (as in cusd or parts of fuhsd)?  i am.",
  "Solution_5": "here\r\n\r\n[size=9]This message is too small. Please make the message longer before submitting.[/size]",
  "Solution_6": "I shouldn't be posting here, as I'm from Ohio, but oh well.",
  "Solution_7": "[quote=\"xpmath\"]I shouldn't be posting here, as I'm from Ohio, but oh well.[/quote]\r\ndude.\r\nstop.",
  "Solution_8": "i'm from IN. i don't normally post here but i am a member of the usergroup and i read threads here.",
  "Solution_9": "not that you want me to, austin :P",
  "Solution_10": "WELCOME TO THE FORUMS!!!\r\n\r\nalrighty Bay Area-ers let's introduce myself.\r\n\r\nI'm Ubemaya. I suck at math. And I want Meggy-Meg.",
  "Solution_11": "OMFG stop posting that you want Meggy-Meg everywhere on the forums! It's hard now to find a place where you have posted that doesn't mention Meggy-Meg anymore...\r\n\r\nAnd you don't #$%@! suck at math!!!",
  "Solution_12": "WELL COMPARED TO PEOPLE AT MY SCHOOL I DON'T BUT COMPARED TO YOU I'M LIKE A PRESCHOOLER!!!!!",
  "Solution_13": "[quote=\"person w/o a username\"]not that you want me to, austin :P[/quote]\n\nWhat's your real name?\n\nEDIT:\n[quote=\"xscapezaer\"]OMFG stop posting that you want Meggy-Meg everywhere on the forums! It's hard now to find a place where you have posted that doesn't mention Meggy-Meg anymore...\n\nAnd you don't #%@! suck at math!!![/quote]\r\n\r\nEh, just ignore him.\r\n\r\nBesides, you are starting to sound like Ubemaya the II, anyways, so you shouldn't be talking... :rotfl: \r\nYou say that you're going to be on the C team, but I think you're going to be on the B team, and if you really work hard, you have a chance of making A team.",
  "Solution_14": "wtf?! [b][i]Me[/i][/b] sound like Ubemaya v.2?! And what about you?! Remember those \"enlightening words of wisdom\" you put on your calculator :roll: or the...well, you know what i'm talking about.\r\n\r\nEDIT: and I still don't have a chance of doing well on ARML...so shaddup.",
  "Solution_15": "Yeah, that is what the world may become.",
  "Solution_16": "[quote=\"CheeseIsGood\"]That is UNPOSSIBLE!\n\nThen the world will suck, there will be no more food or any other products.\nMaybe there would, until all that we have is used up, and then there will be nothing left, and everyone will die, since no new things are being made.[/quote]\r\n\r\nnah I'll just grow some food and farm like a farmer in those days. It's going to be hard work. I should start now. Lemme save up my apples",
  "Solution_17": "[quote=\"moogra\"]\nnah I'll just grow some food and farm like a farmer in those days. It's going to be hard work. I should start now. Lemme save up my apples\n[/quote]\r\n\r\nAwesome, I want a banana.",
  "Solution_18": "I'm sorry. I don't own a banana tree. Maybe when I get one, I'll send you a free banana.",
  "Solution_19": "Someone I know gave me this account, apparently he didn't need AoPS anymore.\r\n\r\nI go to Redwood Middle...probably going to Mitty or Harker for High School though.",
  "Solution_20": "[quote=\"happyme\"]Someone I know gave me this account, apparently he didn't need AoPS anymore.\n\nI go to Redwood Middle...probably going to Mitty or Harker for High School though.[/quote]\r\nOh hi! Were you at Mathcounts last year?",
  "Solution_21": "[quote=\"happyme\"]\nI go to Redwood Middle...probably going to [b]Mitty[/b] or Harker for High School though.[/quote]\r\nwe need more math people!",
  "Solution_22": "[quote=\"happyme\"]...apparently he didn't need AoPS anymore.[/quote]\r\nI sense arrogance.\r\n\r\nAnyway, yes, I'm in the bay area. I go to Hopkins (in Fremont), as well as lifeisacircle, superquark1123, psichilikapati, and another person who's username i dont know (the first two people are people who joined AoPS because i told them about it and pestered them until they made accounts and bought all the books, lol). bowei is from mission high, where im going in 2 years time. aznphatso is the brother of the person who's username i dont know that used to be in mission but is now in MIT. yeah.\r\n\r\nOkay.\r\n\r\nEnd of spam message.",
  "Solution_23": "[quote=\"archimedes1\"][quote=\"happyme\"]\nI go to Redwood Middle...probably going to [b]Mitty[/b] or Harker for High School though.[/quote]\nwe need more math people![/quote]\r\n\r\nyou're probably good enough for your whole high school...\r\n\r\nWait... who's happyme now?",
  "Solution_24": "Uh my friend Ho (I'm not sure if he wanted people to know his first name o.o) gave me this.\r\n\r\nI am currently nobody, and I suck at math.",
  "Solution_25": "I was just kidding. But I bet I suck more.",
  "Solution_26": "haha tony had an account? lol\r\nand now he quit and gave it to you, wow\r\nhow do you know tony?",
  "Solution_27": "I don't know who tony is.\r\n\r\no.o\r\n\r\nWhat grade are you in Mission? I know a lot of people that went to Hopkins/Mission.",
  "Solution_28": "bowei is a freshman\r\n\r\nseeing as i'm not in mission, i don't know a lot of people. but i do have a brother in mission. is tony a junior? i can remember some times when a guy called tony is mentioned.",
  "Solution_29": "oopsie lol i thought when you said ho you meant tony ho\r\nmy bad but i would have expected him to have an aops account\r\ntheres only one more person i know that goes onto aops in hopk/mission\r\nits mathclass, i dunno if he ever posted here"
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "geometry proposed",
    "moving points"
  ],
  "Problem": "Let a triangle ABC and a given line $ d$. A point M moves on d. The perp line from the midpoint of MB to $ d$ meets AB at P. The perp line from the midpoint of MC to $ d$ meets AC at Q. Prove that the perp line from M to PQ always passes through a fixed point when M moves on the line $ d$.\r\n[hide=\"Consequences\"]Two consequences ( one is Pr 7, VMO 2008 and other is my own problem ) were posted at [url=http://www.mathlinks.ro/viewtopic.php?t=185640]here[/url]. They refer to two particular cases of the above problem.[/hide]",
  "Solution_1": "Hello!\r\n\r\nYour problem can be generalized:\r\n\r\nLet $ ABC$ be a triangle, $ d$, $ g$ are two given lines $ ($ the line $ g$ is not parallel to the lines $ AB$, $ AC$ or $ d$ $ )$ and a point $ M$ moves on the line $ d$.\r\n$ S\\in MB$, $ T\\in MC$ such that $ ST\\parallel{}BC$ and $ \\frac {SM}{SB} \\equal{} k$  $ ($ $ k\\in\\Re^{*}_{ \\plus{} }$, $ k$ is a fixed number $ )$.\r\n$ X,Y\\in d$ such that $ SX\\parallel{}TY\\parallel{}g$, $ \\left\\{P\\right\\} \\equal{} SX\\cap AB$ and $ \\left\\{Q\\right\\} \\equal{} TY\\cap AC$.\r\nThe circumcircles of the triangles $ PXM$ and $ QYM$ intersect again at the point $ N$. Prove that the line $ MN$ \r\nalways passes through a fixed point when the point $ M$ moves on the line $ d$.  :)",
  "Solution_2": "Oh, you have a wonderful generalization, Petry. \r\nBut I don't have enough time to check it. If your generalization is true, I hope that you will introduce us about its proof. For my problem, I have a synthetic proof, but maybe your solution to the generalized problem is very nice and I am really curious about it. \r\nSincerely, mr.danh",
  "Solution_3": "I don't have a proof :(, but I will try  :) !",
  "Solution_4": "[quote=\"Petry\"]I don't have a proof :(, but I will try  :) ![/quote]\r\n :?:  Why can you generalize my problem?  :)  Are you sure that your generalization is true?",
  "Solution_5": "[quote=\"mr.danh\"][quote=\"Petry\"]I don't have a proof :(, but I will try  :) ![/quote]\n :?:  Why can you generalize my problem?  :)  Are you sure that your generalization is true?[/quote]\r\nAs I can see by the drawing, I think this challenge problem is true.\r\n\r\nI will check more carefully before a definite answer (probably, later today).\r\n\r\nKostas Vittas.",
  "Solution_6": "I verified the generalization with \"Compass and Ruller\" program, it seems to be true, so maybe one of the Mathlinkers will prove this problem. \r\n   Sincerely, Petry",
  "Solution_7": "@mr.danh: I have seen your solution to your extensive problem in a link. It is very nice  :lol:  \r\n@vittasko:\r\n[quote=\"vittasko\"]As I can see by the drawing, I think this challenge problem is true. \nI will check more carefully before a definite answer (probably, later today). \nKostas Vittas. [/quote]\r\nDid you solve the generalized problem of Petry? Why don't you introduce to your proof ?",
  "Solution_8": "[quote=\"thaithuan_GC\"][quote=\"vittasko\"]As I can see by the drawing, I think this challenge problem is true. \nI will check more carefully before a definite answer (probably, later today). \nKostas Vittas. [/quote]\nDid you solve the generalized problem of Petry? Why don't you introduce to your proof ?[/quote]\r\nI can only say by the drawing that it is true the problem in generalization of [b][size=100]Petry[/size][/b], but unforunately, I have not in mind any proof of this nice result dear [b][size=100]thaithan_GC[/size][/b] and sorry for the late answer.\r\n\r\nKostas Vittas.",
  "Solution_9": "Hello!\r\n\r\nFinally, I proved the generalization of the proposed problem of mr.danh.\r\nFirst we prove a generalization of the Carnot's theorem, then a aplication of this generalization and at last the generalization of the proposed problem.",
  "Solution_10": "The generalization of the Carnot's theorem:\r\n\r\nLet $ ABC$ be a triangle and the points $ X$, $ Y$, $ Z$ are the feet of the lines equally inclined to the sidelines $ BC$, $ CA$, $ AB$ from a point $ P$ \r\n($ \\angle\\left(PX;BC\\right) \\equal{} \\angle\\left(PY;CA\\right) \\equal{} \\angle\\left(PZ;AB\\right) \\equal{} \\alpha$; this angles have the same orientation). Prove that \r\n$ \\left|BX^{2} \\minus{} CX^{2} \\plus{} CY^{2} \\minus{} AY^{2} \\plus{} AZ^{2} \\minus{} BZ^{2}\\right| \\equal{} \\left|4\\cdot ctg\\alpha\\cdot\\left[ABC\\right]\\right|$.\r\nThe reciprocal is also true.\r\n\r\n$ \\left[...\\right]$ - the area of the triangle $ ...$\r\n\r\nSolution:\r\n$ PB^{2} \\equal{} PX^{2} \\plus{} BX^{2}\\mp 2\\cdot PX\\cdot BX\\cdot cos\\alpha\\Rightarrow PB^{2} \\equal{} PX^{2} \\plus{} BX^{2}\\mp 4\\cdot ctg\\alpha\\cdot\\left[PXB\\right]$    (*)\r\n$ PC^{2} \\equal{} PX^{2} \\plus{} CX^{2}\\pm 2\\cdot PX\\cdot CX\\cdot cos\\alpha\\Rightarrow PC^{2} \\equal{} PX^{2} \\plus{} CX^{2}\\pm 4\\cdot ctg\\alpha\\cdot\\left[PXC\\right]$     (**)\r\n(*), (**) $ \\Rightarrow PB^{2} \\minus{} PC^{2} \\equal{} BX^{2} \\minus{} CX^{2}\\mp 4\\cdot ctg\\alpha\\cdot\\left[PBC\\right]\\Rightarrow$ (1) \r\nSimilarly $ PC^{2} \\minus{} PA^{2} \\equal{} CY^{2} \\minus{} AY^{2}\\mp 4\\cdot ctg\\alpha\\cdot\\left[PCA\\right]$(2)\r\nand       $ PA^{2} \\minus{} PB^{2} \\equal{} AZ^{2} \\minus{} BZ^{2}\\mp 4\\cdot ctg\\alpha\\cdot\\left[PAB\\right]$(3)\r\n(1), (2), (3) $ \\Rightarrow BX^{2} \\minus{} CX^{2} \\plus{} CY^{2} \\minus{} AY^{2} \\plus{} AZ^{2} \\minus{} BZ^{2} \\equal{} \\pm 4\\cdot ctg\\alpha\\cdot\\left[ABC\\right]\\Rightarrow$\r\n$ \\Rightarrow\\left|BX^{2} \\minus{} CX^{2} \\plus{} CY^{2} \\minus{} AY^{2} \\plus{} AZ^{2} \\minus{} BZ^{2}\\right| \\equal{} \\left|4\\cdot ctg\\alpha\\cdot\\left[ABC\\right]\\right|$.",
  "Solution_11": "A aplication of the generalization of the Carnot's theorem:\r\n\r\nLet $ ABC$ be a triangle and $ d$ a line. The points $ D$, $ E$, $ F$ are the feet of the lines equally\r\ninclined to the line $ d$ from the points $ A$, $ B$, $ C$ ($ \\angle\\left(AD;d\\right) \\equal{} \\angle\\left(BE;d\\right) \\equal{} \\angle\\left(CF;d\\right) \\equal{} \\alpha$; \r\nthis angles have the same orientation, so $ AD\\parallel{}BE\\parallel{}CF$).\r\nThe points $ X$, $ Y$, $ Z$ are the feet of the lines equally inclined to the lines $ BC$, $ CA$, $ AB$ from the points $ D$, $ E$, $ F$ respectively \r\n($ \\angle\\left(DX;BC\\right) \\equal{} \\angle\\left(EY;CA\\right) \\equal{} \\angle\\left(FZ;AB\\right) \\equal{} \\minus{} \\alpha$; this angles have the same orientation).\r\nThe angles $ \\angle\\left(AD;d\\right)$ and $ \\angle\\left(DX;BC\\right)$ have different orientation. Then prove that the lines $ DX$, $ EY$, $ FZ$ are concurrent.\r\n\r\n$ \\left[...\\right]$ - the area of the triangle or quadrilateral $ ...$\r\n\r\nSolution:\r\n$ DB^{2} \\equal{} DX^{2} \\plus{} BX^{2}\\mp 2\\cdot DX\\cdot BX\\cdot cos\\alpha\\Rightarrow DB^{2} \\equal{} DX^{2} \\plus{} BX^{2}\\mp 4\\cdot ctg\\alpha\\cdot\\left[DXB\\right]$     (*)\r\n$ DC^{2} \\equal{} DX^{2} \\plus{} CX^{2}\\pm 2\\cdot DX\\cdot CX\\cdot cos\\alpha\\Rightarrow DC^{2} \\equal{} DX^{2} \\plus{} CX^{2}\\pm 4\\cdot ctg\\alpha\\cdot\\left[DXC\\right]$    (**)\r\n(*), (**) $ \\Rightarrow DB^{2} \\minus{} DC^{2} \\equal{} BX^{2} \\minus{} CX^{2}\\mp 4\\cdot ctg\\alpha\\cdot\\left[DBC\\right]\\Rightarrow$ \r\n$ \\Rightarrow BX^{2} \\minus{} CX^{2} \\equal{} DB^{2} \\minus{} DC^{2}\\pm 4\\cdot ctg\\alpha\\cdot\\left[DBC\\right]$   (1)\r\n\r\n$ DB^{2} \\equal{} DE^{2} \\plus{} BE^{2}\\pm 2\\cdot DE\\cdot BE\\cdot cos\\alpha\\Rightarrow DB^{2} \\equal{} DE^{2} \\plus{} BE^{2}\\pm 4\\cdot ctg\\alpha\\cdot\\left[DEB\\right]$ ($ \\bullet$)\r\n$ DC^{2} \\equal{} DF^{2} \\plus{} CF^{2}\\mp 2\\cdot DF\\cdot CF\\cdot cos\\alpha\\Rightarrow DC^{2} \\equal{} DF^{2} \\plus{} CF^{2}\\mp 4\\cdot ctg\\alpha\\cdot\\left[DFC\\right]$  ($ \\bullet\\bullet$)\r\n($ \\bullet$), ($ \\bullet\\bullet$) $ \\Rightarrow DB^{2} \\minus{} DC^{2} \\equal{} DE^{2} \\minus{} DF^{2} \\plus{} BE^{2} \\minus{} CF^{2}\\pm 4\\cdot ctg\\alpha\\cdot\\left(\\left[DEB\\right] \\plus{} \\left[DFC\\right]\\right)$   (2)\r\n\r\n(1),(2)$ \\Rightarrow BX^{2} \\minus{} CX^{2} \\equal{} DE^{2} \\minus{} DF^{2} \\plus{} BE^{2} \\minus{} CF^{2}\\pm 4\\cdot ctg\\alpha\\cdot\\left[BEFC\\right]$   (3)\r\nSimilarly $ CY^{2} \\minus{} AY^{2} \\equal{} EF^{2} \\minus{} ED^{2} \\plus{} CF^{2} \\minus{} AD^{2}\\mp 4\\cdot ctg\\alpha\\cdot\\left[CFDA\\right]$   (4)\r\nand $ AZ^{2} \\minus{} BZ^{2} \\equal{} FD^{2} \\minus{} FE^{2} \\plus{} AD^{2} \\minus{} BE^{2}\\mp 4\\cdot ctg\\alpha\\cdot\\left[ADEB\\right]$   (5)\r\n\r\n(3), (4), (5) $ \\Rightarrow BX^{2} \\minus{} CX^{2} \\plus{} CY^{2} \\minus{} AY^{2} \\plus{} AZ^{2} \\minus{} BZ^{2} \\equal{} \\mp 4\\cdot ctg\\alpha\\cdot\\left[ABC\\right]\\Rightarrow$\r\n$ \\Rightarrow\\left|BX^{2} \\minus{} CX^{2} \\plus{} CY^{2} \\minus{} AY^{2} \\plus{} AZ^{2} \\minus{} BZ^{2}\\right| \\equal{} \\left|4\\cdot ctg\\alpha\\cdot\\left[ABC\\right]\\right|$.\r\nFrom the reciprocal of the generalization of the Carnot's theorem we have that\r\nthe lines $ DX$, $ EY$, $ FZ$ are concurrent.",
  "Solution_12": "The generalization of the proposed problem:\r\n\r\nLet $ ABC$ be a triangle, $ d$, $ g$ are two given lines (the line $ g$ is not parallel to the lines $ AB$, $ AC$ or $ d$) and a point $ M$ moves on the line $ d$.\r\n$ S\\in MB$, $ T\\in MC$ such that $ ST\\parallel{}BC$ and $ \\frac {SM}{SB} \\equal{} k$, ($ k\\in\\Re^{*}_{ \\plus{} }$, $ k$ is a fixed number).\r\n$ X,Y\\in d$ such that $ SX\\parallel{}TY\\parallel{}g$, $ \\left\\{P\\right\\} \\equal{} SX\\cap AB$ and $ \\left\\{Q\\right\\} \\equal{} TY\\cap AC$.\r\nThe circumcircles of the triangles $ PXM$ and $ QYM$ intersect again at the point $ N$. Prove that the line $ MN$ \r\nalways passes through a fixed point when the point $ M$ moves on the line $ d$.\r\n\r\nSolution:\r\n$ \\angle\\left(g;d\\right) \\equal{} \\alpha$ and it's easy to prove that $ N\\in PQ$.\r\n\r\nConstruct $ AA'\\parallel{}g$, $ A'\\in d$, so $ AA'\\parallel{}PX\\parallel{}QY$.\r\nThe points $ D$, $ E$, $ F$ are the feet of the lines equally inclined to the lines $ PQ$, $ QA$, $ AP$ from the points $ A'$, $ X$, $ Y$ respectively \r\n($ \\angle\\left(A'D;PQ\\right) \\equal{} \\angle\\left(XE;QA\\right) \\equal{} \\angle\\left(YF;AP\\right) \\equal{} \\minus{} \\alpha$; this angles have the same orientation).\r\nThe angles $ \\angle\\left(AA';d\\right)$ and $ \\angle\\left(A'D;PQ\\right)$ have different orientation.\r\nFrom the aplication of the generalization of the Carnot's theorem we have that the lines $ A'D$, $ XE$, $ YF$ are concurrent. $ \\left\\{W\\right\\} \\equal{} A'D\\cap XE\\cap YF$.\r\nIt's easy to prove that $ MN\\parallel{}A'D$.\r\n\r\nConstruct $ BB'\\parallel{}CC'\\parallel{}g$, $ B',C'\\in d$. Through the points $ B'$, $ C'$ construct parallels to the lines $ XE$, $ YF$ respectively \r\nand let $ V'$ be their point of intersection.\r\nIt's easy to prove that the points $ M$, $ W$, $ V'$ are collinear and  $ \\frac {WM}{WV'} \\equal{} \\frac {XM}{XB'} \\equal{} \\frac {SM}{SB} \\equal{} k$. \r\n\r\n$ \\left\\{V\\right\\} \\equal{} V'A'\\cap MN$\r\nConstruct $ VB\"\\parallel{}V'B'$, $ B\"\\in d$ and $ VC\"\\parallel{}V'C'$, $ C\"\\in d$.\r\n$ \\frac {A'B\"}{A'B'} \\equal{} \\frac {A'V}{A'V'} \\equal{} \\frac {WM}{WV'} \\equal{} k$. Similarly $ \\frac {A'C\"}{A'C'} \\equal{} \\frac {A'V}{A'V'} \\equal{} \\frac {WM}{WV'} \\equal{} k$\r\n\r\nThe points $ A'$, $ B'$, $ C'$ are fixed, $ k$ is a fixed number, so the points $ B\"$, $ C\"$ are fixed. (1)\r\n$ VB\"\\parallel{}V'B'\\parallel{}XE\\Rightarrow\\angle\\left(VB\";AC\\right) \\equal{} \\alpha$ (2)\r\n$ VC\"\\parallel{}V'C'\\parallel{}YF\\Rightarrow\\angle\\left(VC\";AB\\right) \\equal{} \\alpha$ (3)\r\n\r\n(1),(2),(3)$ \\Rightarrow V$ is a fixed point.",
  "Solution_13": "Wow! Thanks Petry for your nice complete proof of the generalized problem. And I will be very glad if someone have a synthetic proof.  :)",
  "Solution_14": "Here's a proof using moving points for this beautiful problem: Let $X,Y,Z$ be the feet of perpendicular from the points $A,B,C$ respectively to $d$. Suppose the line through $Y$ perpendicular to $AC$ meets the line through $Z$ perpendicular to $AB$ at $T$, and let $T'$ be the reflection of $T$ in the point $X$. We claim that $T$ is the desired fixed point. Fix the line $d$, and the points $A,B,M,Z$ and animate $C$ on the line through $Z$ perpendicular to $d$ (call this line $\\ell$). Note that $X,Y,P$ automatically get fixed, and that there is no loss of degrees of freedom in doing this. Let $U$ be the midpoint of $CM$. Then the line $UQ$ is simply the perpendicular bisector of $MZ$. As $Z$ is fixed, so this means that $UQ$ is also fixed. Then $$C \\mapsto AC \\mapsto AQ \\mapsto Q \\mapsto PQ \\mapsto P\\infty_{\\perp_{PQ}} \\mapsto \\infty_{\\perp_{PQ}}$$ is a projective map (where the last two maps follow from rotation of angle $90^{\\circ}$ about $P$). Now, as both $AB$ and $Z$ are fixed, so the line through $Z$ perpendicular to $AB$ (i.e. $ZT$) is also fixed. This gives $$C \\mapsto AC \\mapsto \\infty_{\\perp_{AC}} \\mapsto Y\\infty_{\\perp_{AC}} \\mapsto YT \\mapsto T \\mapsto T'$$\nis also projective (where the last map is reflection in $X$). Thus, it suffices to prove that $M,T',\\infty_{\\perp_{PQ}}$ are collinear for three positions of $C$. Take $C=\\infty_{\\ell}$ and $C=AB \\cap \\ell$ for the first two positions. One can easily check that these work. Finally, take $C$ as the foot of perpendicular from $B$ to $\\ell$ (i.e. $BC \\parallel d$). As $T$ is the orthopole of $\\triangle ABC$ with respect to $d$ (by the definition of the point $T$), so the line $TX$ must be perpendicular to $BC$, or in this case, to $d$ as well. So this means that $X'$ is simply the reflection of $X$ in $d$. Now, notice that if we move the line $d$ with constant velocity, keeping it parallel to $BC$, and without changing the distances $MY$ and $MZ$, then the points $P,Q$ remain fixed (as they lie on perpendicular bisectors of $MY$ and $MZ$ respectively), while the line $MT'$ simply moves parallel to its original position. So it suffices to prove the desired conclusion for just one position of the line $d$. Take $d=BC$ (obviously :D). Then $T$ becomes the orthocenter of $\\triangle ABC$, while $T'$ is the point where the $A$-altitude meets $\\odot (ABC)$ again. Suppose $T'M$ meets $\\odot (ABC)$ again at $N$. Then $$2\\angle MNB=2\\angle T'NB=2\\angle T'AB=\\angle MPB \\Rightarrow \\text{As PM=PB, so } N \\in \\odot (P,PM)$$\nSimilarly, $N$ lies on $\\odot (Q,QM)$, which in turn gives that $T'$ lies on the radical axis of $\\odot (P,PM)$ and $\\odot (Q,QM)$. Thus, $T'M \\perp PQ$, as desired. Hence, done. $\\blacksquare$"
}
{
  "Tag": [],
  "Problem": "I was wondering how to do this without guess and check.\r\n\r\n#15\r\n15. The integers from 1 through 15 are written in numerical order in pencil going clockwise around a circle. A student begins moving clockwise around the circle erasing every third integer that has not yet been erased until only the integer 11 remains. Which integer did the student erase first?\r\n\r\n\r\n\r\n#27\r\nA rectangular band formation is a formation with m band members in each of r rows, where m and r are integers. A particular band has less than 100 band members. The director arranges them in a rectangular formation and finds that he has two members left over. If he increases the number of members in each row by 1 and reduces the number of rows by 2, there are exactly enough places in the new formation for each band member. What is the largest number of members the band could have?",
  "Solution_1": "Look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=32450337&t=249626]here[/url] for number $ 15$.\r\n\r\n[hide=\"27\"]\nLet there be some $ a$ and $ b$ so that $ ab\\plus{}2\\equal{}mr$, which satisfies the first formation. Then $ mr\\equal{}(a\\plus{}1)(b\\minus{}2)\\equal{}ab\\minus{}2a\\plus{}b\\minus{}2\\implies mr\\minus{}ab\\equal{}2\\equal{}\\minus{}2a\\plus{}b\\minus{}2\\implies b\\minus{}2a\\equal{}4$ as well. $ mr\\equal{}(a\\plus{}1)(b\\minus{}2)$ is maximized when $ b$ is maximized, since both $ a$ and $ b$ increase as $ b$ increases. Testing, $ (a,b)\\equal{}(6,16)$ is the largest pair that works, so the answer is $ mr\\equal{}(a\\plus{}1)(b\\minus{}2)\\equal{}7\\cdot14\\equal{}\\boxed{98}$.\n[/hide]",
  "Solution_2": "for #27, (m+1)(r-2)=mr+2\r\nr-2m=4\r\nfind the largest m such that mr+2<100 which happens to be m=6, r=16\r\nso the answer 6*16+2=98",
  "Solution_3": "Any way to solve #27 using mods?",
  "Solution_4": "[quote=\"math154\"]Look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=32450337&t=249626]here[/url] for number $ 15$.\n\n[hide=\"27\"]\nLet there be some $ a$ and $ b$ so that $ ab\\plus{}2\\equal{}mr$, which satisfies the first formation. Then $ mr\\equal{}(a\\plus{}1)(b\\minus{}2)\\equal{}ab\\minus{}2a\\plus{}b\\minus{}2\\implies mr\\minus{}ab\\equal{}2\\equal{}\\minus{}2a\\plus{}b\\minus{}2\\implies b\\minus{}2a\\equal{}4$ as well. $ mr\\equal{}(a\\plus{}1)(b\\minus{}2)$ is maximized when $ b$ is maximized, since both $ a$ and $ b$ increase as $ b$ increases. Testing, $ (a,b)\\equal{}(6,16)$ is the largest pair that works, so the answer is $ mr\\equal{}(a\\plus{}1)(b\\minus{}2)\\equal{}7\\cdot14\\equal{}\\boxed{98}$.\n[/hide][/quote]\n$ mr=(a+1)(b-2)=ab-2a+b-2\\implies mr-ab=2=-2a+b-2\\implies b-2a=4 $\n\nWhat's the point of the above step if in the end you're looking at $(a+1)(b-2)=mr$?",
  "Solution_5": "[quote=Aequilipse]Any way to solve #27 using mods?[/quote]\n\nSorry for the huge bump (oops 4 year bump)\n\nCan someone explain how you would do #27 with mods?\n\nI started off with something like this. We would $mr + 2 = (m + 1)(r - 2)$. Which would mean that (m + 1)(r - 2) is either $1 \\pmod{2}$ (this is when $mr$ is even, and $0 \\pmod{2}$ when $mr$ is odd. I am not sure on how to go from here, can someone explain? \n\n",
  "Solution_6": "For #15, work backwards.",
  "Solution_7": "I remember doing that sprint, #15 was a pain in the neck"
}
{
  "Tag": [
    "trigonometry",
    "logarithms",
    "complex analysis",
    "calculus",
    "calculus computations"
  ],
  "Problem": "Find the Taylor series centered at $1$ for $f(x)=\\tan^{-1}(x).$ Determine the radius of convergence of that series.\r\n\r\n\r\n[hide=\"Hint\"]Somwhere in this forum.\n\n[hide=\"OK I give up\"][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=42805]Here.[/url] [/hide][/hide]",
  "Solution_1": "Instead of explictly calculating the series, I give another way to find the radius of convergence, using Complex Analysis\r\n[hide=\"Radius\"]For the series $\\arctan(z)$ with $z \\in \\mathbb{C}$ we can calculate the radius of convergence as the shortest distance from the centre, to a singularity (Singularity or pole...I always get confused). Anyway this is the distance from 1 (the centre) to $i$, or $\\sqrt{2}$.\nHope its correct![/hide]",
  "Solution_2": "That is correct, as far as it goes. We're still looking for the series itself.\r\n\r\n[hide]The arctangent has singularities at both $i$ and $-i$ and your argument should acknowledge both of them. Of course, the distance from $1$ to either one is the same, $\\sqrt{2}.$\n\nThe singularities of the arctangent at $\\pm i$ are not poles. They are branch points. They're the same kind of singularities that the logarithm has at zero. (One can even express the arctangent in terms of logarithms.)[/hide]",
  "Solution_3": "[quote=\"Kent Merryfield\"]That is correct, as far as it goes. We're still looking for the series itself.\n\n[hide]The arctangent has singularities at both $i$ and $-i$ and your argument should acknowledge both of them. Of course, the distance from $1$ to either one is the same, $\\sqrt{2}.$\n\nThe singularities of the arctangent at $\\pm i$ are not poles. They are branch points. They're the same kind of singularities that the logarithm has at zero. (One can even express the arctangent in terms of logarithms.)[/hide][/quote]\n\nThank you kent. How can we tell we we have a pole or a branch point?\n\nBTW: The answer to the original problem is\n[hide]$\\frac{\\pi}{4}+\\frac{1}{2}(x-1)-\\frac{1}{4}(x-1)^2+\\frac{1}{12}(x-1)^3+ \\cdots $[/hide]"
}
{
  "Tag": [],
  "Problem": "Which program(s) is/are the best for drawing geometric figures?",
  "Solution_1": "our teacher uses Geometer's Sketchpad."
}
{
  "Tag": [
    "AMC 10"
  ],
  "Problem": "The mean of 20 numbers is 30, and the average of 30 other numbers is 20.  What is the average of all 50 numbers?\r\n\r\n$a.) 23 b.) 24 c.) 25 d.) 26 e.) 27$",
  "Solution_1": "[hide]$\\frac{20\\cdot30+30\\cdot20}{50}=24\\implies\\boxed B$[/hide]",
  "Solution_2": "[hide]B[/hide]",
  "Solution_3": "[hide](20*30+30*20)/(20+30)=24\nB[/hide]",
  "Solution_4": "[hide]\n$\\frac{30 \\cdot 20 + 20 \\cdot 30}{50}=\\frac{1200}{50}=24$, so B\n[/hide]\r\nI just took the AMC-12 2005 and the same problem was on it, word for word. :o  :o  :o",
  "Solution_5": "[quote=\"mathnerd314\"]I just took the AMC-12 2005 and the same problem was on it, word for word. :o  :o  :o[/quote]\r\nI think the 15 or so problems are the same on the AMC-10 and AMC-12."
}
{
  "Tag": [
    "limit",
    "calculus",
    "calculus computations"
  ],
  "Problem": "Find a series $ \\sum_{n}(\\minus{}1)^na_n$ with these properties:\r\n\r\n1) $ a_n>0$ for all sufficiently large $ n.$\r\n\r\n2) $ \\lim_{n\\to\\infty}na_n\\equal{}1.$ (That is, the terms \"look like\" $ \\frac1n.$)\r\n\r\n3) The alternating series diverges.",
  "Solution_1": "[hide]\nhow about $ \\frac {ln(n) \\plus{} ( \\minus{} 1)^n}{nln(n)}$\n\nalso, doesn't requirement 2 imply requirement 1?\n[/hide]",
  "Solution_2": "[quote=\"0714446459923\"]also, doesn't requirement 2 imply requirement 1?[/quote]\r\nTrue, that was redundant. I think I changed my mind about the exact phrasing in the middle of writing down the problem, and accidentally introduced the redundancy.\r\n\r\nNice example.\r\n\r\nOf course, what I was pushing here is that the requirement that the absolute values decrease monotonically cannot be dispensed with as a hypothesis of the alternating series test."
}
{
  "Tag": [
    "function",
    "inequalities",
    "trigonometry",
    "calculus",
    "integration",
    "real analysis",
    "complex numbers"
  ],
  "Problem": "Prove the following inequality.\r\n$\\sum_{k=1}^{1001}\\sin\\frac{k\\pi}{1002}<\\frac{2004}{\\pi}$",
  "Solution_1": "It's not very hard. The left side is less than 2 times the integral from 1 to 501 of the function sin(xPi/1002) plus 1. This expression is 2*1002*cos(Pi/1002)/Pi+1 which is more than 2004/Pi which can be proved with a calculator.\r\n\r\nMisha",
  "Solution_2": "This is a very well kown trick for summing trigs.\r\n\r\nWe have\\begin{eqnarray*}\n&& 2\\sin\\frac{\\pi}{2004} \\sum_{k=1}^{1001} \\sin\\frac{k\\pi}{1002} \\\\\r\n&=& \\sum_{k=1}^{1001} \\cos\\frac{(2k-1)\\pi}{2004} - \\cos\\frac{(2k+1)\\pi}{2004} \\\\\r\n&=& \\cos\\frac{\\pi}{2004} - \\cos\\frac{2003\\pi}{2004}\r\n&=& 2\\cos\\frac{\\pi}{2004}\r\n\\end{eqnarray*}\r\n\r\nSo \\[\r\n\\sum_{k=1}^{1001} \\sin\\frac{k\\pi}{1002} = \\cot\\frac{\\pi}{2004}\r\n\\]\r\n\r\nAnd the result come immediately from the fact that $\\tan x > x$ in $(0, \\pi/2)$.",
  "Solution_3": "Two comments:\r\n\r\nThe first is a variation on what Misha123 said. \r\n\r\n$\\frac{\\pi}{1002}\\sum_{k=1}^{1001}\\sin\\left(\\frac{k\\pi}{2002}\\right)$ is both the right-hand endpoint Riemann sum and the left-hand endpoint Riemann sum (with 1002 subintervals of equal width) for the integral $\\int_0^{\\pi}\\sin x dx$. The average of the right and left-hand Riemann sums is the trapeziodal rule estimate. Since the function is concave downward on the whole interval, the trapezoidal rule estimate is less than the actual integral. Hence,\r\n$\\frac{\\pi}{1002}\\sum_{k=1}^{1001}\\sin\\left(\\frac{k\\pi}{1002}\\right) < \\int_0^{\\pi}\\sin x dx = 2$, which proves the inequality quite directly. You could even use the error estimate for the trapezoidal rule to determine how close together the two sides must be.\r\n\r\nThe second is a comment on what billzhao said. This exact summation is both familiar and important in Fourier analysis. The exact formula can be recovered without remembering as many trigonometric identities by employing complex numbers and the formula for the sum of a finite geometric series.\r\n$\\sum_{k=1}^{n-1}\\sin\\left(\\frac{k\\pi}{n}\\right)= \\sum_{k=0}^{n-1}\\sin\\left(\\frac{k\\pi}{n}\\right)$\r\n$=Imag\\left[\\sum_{k=0}^{n-1}e^{ik\\pi/n}\\right] = Imag\\left[\\frac{1-e^{i\\pi}}{1-e^{i\\pi/n}}\\right] = Imag\\left[\\frac{2}{1-e^{i\\pi/n}}\\right]$\r\n$=Imag\\left[\\frac{2e^{\\frac{-i\\pi}{2n}}} {e^{\\frac{-i\\pi}{2n}}-e^{\\frac{i\\pi}{2n}}}\\right] = \\frac{\\cos\\left(\\frac{\\pi}{2n}\\right)}{\\sin\\left(\\frac{\\pi}{2n}\\right)}$, which is the same formula billzhao found.",
  "Solution_4": "Thank you for your comment. I used the trapezoidel rule several times indirectly but I never realized the real matture. Your proof is of course the 'perfect' proof for this problem.\r\n\r\nMisha",
  "Solution_5": "Thank you for your replies!  Sorry for my delaying reply.\r\nIndeed, my solution is the same as that of Kent Merryfield.\r\n\r\nkunny"
}
{
  "Tag": [
    "abstract algebra",
    "geometry",
    "calculus",
    "integration"
  ],
  "Problem": "Post your jokes here! Here's two to get started.\r\n\r\n\r\nOld maid's laughter--He! he! he!\r\n\r\n\r\nA flea and a fly in a flue,\r\nWere improsoned, so what could they do?\r\nSaid the fly, \"let us flee!\"\r\n\"Let us fly!\" said the flea,\r\nAnd they flew through a flaw in the flue.",
  "Solution_1": "Why are they called apartements when they are all stuck together?",
  "Solution_2": "What's nice and works for the phone company? A deferential operator!!\r\n\r\nWhat's purple and commutes? An abelian grape!!\r\n\r\nTwo wrongs don't make a right... but two Wrights make an airplane...\r\n\r\nA mathematician decides that his job as a professor isn't paying enough, and decides to be a plumber instead. However, he lies about his educational experience, saying that he dropped out of seventh grade, because he thinks that the plumbing company doesn't like educated people. Well, one evening, the company decides to round up all the plumbers and enroll them in an eighth grade level class. It just happens that the first class of the evening is math, the first formula is to find the area of a circle, and the mathematician is called up to the board to demonstrate his knowledge. The mathematician suddenly has a mental blank and forgets, so he fills the board with equations trying to derive the formula. But he keeps getting $A=-\\pi r^{2}$. Thoroughly confused, he desperately looks to the other plumbers for help, only to find them all mouthing, \"Switch the limits of the integral!\" (No offense to plumbers :D)\r\n\r\n:roll:",
  "Solution_3": "One off the top of my head, and not so good\r\n\r\nA man goes into a bar.  He notices a woman sitting alone at a table, so he goes to her.\r\n\r\nMan: Can I buy you a drink?\r\n\r\nShe replies at the top of her lungs \"NO I WILL NOT SLEEP WITH YOU TONIGHT!\"  Everyone in the bar stares as the man crawls away to a table in the corner and sits dejected.  A few minutes later, the woman goes to the man.\r\n\r\nWoman: I'm sorry about earlier.  I'm a psychologist, and I wanted to see how you'd react to that situation.\r\n\r\nMan: WHAT DO YOU MEAN TWO HUNDRED DOLLARS?!",
  "Solution_4": "[quote=\"K81o7\"]One off the top of my head, and not so good\n\nA man goes into a bar.  He notices a woman sitting alone at a table, so he goes to her.\n\nMan: Can I buy you a drink?\n\nShe replies at the top of her lungs \"NO I WILL NOT SLEEP WITH YOU TONIGHT!\"  Everyone in the bar stares as the man crawls away to a table in the corner and sits dejected.  A few minutes later, the woman goes to the man.\n\nWoman: I'm sorry about earlier.  I'm a psychologist, and I wanted to see how you'd react to that situation.\n\nMan: WHAT DO YOU MEAN TWO HUNDRED DOLLARS?![/quote]\r\n\r\n:rotfl:\r\n\r\nHe: I'm a photographer. I've been looking for a face like yours.\r\nShe: I'm a plastic surgeon. I've been looking for a face like yours.",
  "Solution_5": "-There are two groups of people in the world; those who believe that the world can be divided into two groups of people, and those who don't. \r\n\r\n-The less you know, the more you make.\r\nProof:\r\nPostulate 1: Knowledge is Power.\r\nPostulate 2: Time is Money.\r\nAs every engineer knows: Power = Work / Time\r\nAnd since Knowledge = Power and Time = Money\r\nIt is therefore true that Knowledge = Work / Money .\r\nSolving for Money, we get:\r\nMoney = Work / Knowledge\r\nThus, as Knowledge approaches zero, Money approaches infinity, regardless of the amount of Work done.\r\n\r\n-Teacher: \"Who can tell me what 7 times 6 is?\"\r\nStudent: \"It's 42!\"\r\nTeacher: \"Very good! - And who can tell me what 6 times 7 is?\"\r\nSame student: \"It's 24!\"",
  "Solution_6": "Why is the alphabet in that order?  Is it because of that song?",
  "Solution_7": "What do vampires say when they kiss each other?\r\nOuch!\r\n\r\nWhy did the monster take six hours to finish the small book?\r\nHe was full!",
  "Solution_8": "[quote=\"lingomaniac88\"]Why is the alphabet in that order?  Is it because of that song?[/quote]\r\n\r\nGoooooooooood question...\r\n\r\nPhilosophy Explained in Two-Cow Terms!!\r\n\r\nSocialism: You have two cows. You give one to your neighbor.\r\nCommunism: You have two cows. You give them to the government and the government gives you some milk.\r\nFascism: You have two cows. The government confiscates them and sells you some milk.\r\nTotalitarianism: You have two cows. The government shoots you and keeps the cows.\r\nDemocracy: You have two cows. They outvote you 2-1 to ban all meat and dairy products."
}
{
  "Tag": [
    "inequalities",
    "trigonometry",
    "inequalities unsolved"
  ],
  "Problem": "if a\u00b2+b\u00b2+c\u00b2+abc=4, prove a+b+c<=3",
  "Solution_1": "a,b,c>=0",
  "Solution_2": "You must have the condition $a,b,c\\ge 0$ \r\nThen In fact you can let:\r\n$a=2\\cos A,b=2\\cos B,c=2\\cosh$\r\n$A,B,C$ is three angles of an triangle.\r\nThe rest is easy.",
  "Solution_3": "[quote=\"Stuart\"]no one nows how to solve it[/quote]\r\n\r\nDon't generalize from yourself to others!\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=25977\r\n\r\n  darij",
  "Solution_4": "Here is my solution.\r\n$a^{2}+b^{2}+c^{2}+abc=4 \\Longrightarrow a=\\frac{-bc+\\sqrt{(4-b^{2})(4-c^{2})}}{2}$\r\nWe must prove that \r\n$\\frac{-bc+\\sqrt{(4-b^{2})(4-c^{2})}}{2}+b+c \\leq 3$\r\n$S=\\sqrt{(4-b^{2})(4-c^{2})}\\leq 6-2(b+c)+bc$\r\n$S \\leq \\frac{4-b^{2}+4-c^{2}}{2}\\leq 6-2(b+c)+bc$\r\n$\\Longrightarrow (b+c-2)^{2}\\geq 0$",
  "Solution_5": "[quote=\"Stuart\"]if a\u00b2+b\u00b2+c\u00b2+abc=4, a,b,c>=0 prove a+b+c<=3[/quote]\r\nThe following inequality is stronger:\r\nlet $a,$ $b$ and $c$ are non-negative numbers such that $a^{2}+b^{2}+c^{2}+1.5abc=4.5.$\r\nProve that $a+b+c\\leq3.$ :wink:",
  "Solution_6": "[quote=\"arqady\"][quote=\"Stuart\"]if a\u00b2+b\u00b2+c\u00b2+abc=4, a,b,c>=0 prove a+b+c<=3[/quote]\nThe following inequality is stronger:\nlet $a,$ $b$ and $c$ are non-negative numbers such that $a^{2}+b^{2}+c^{2}+1.5abc=4.5.$\nProve that $a+b+c\\leq3.$ :wink:[/quote]\r\n\r\nDear [b]arqady[/b], can you give a solution to your problem? :roll:",
  "Solution_7": "[quote=\"cefer\"]\nDear [b]arqady[/b], can you give a solution to your problem? :roll:[/quote]\r\nFor you, cefer :lol: \r\nLet $a+b+c>3,$  $a=kx,$ $b=ky$ and $c=kz$ such that $x+y+z=3.$\r\nHence, $k>1$ and $4.5=a^{2}+b^{2}+c^{2}+1.5abc=k^{2}(x^{2}+y^{2}+z^{2})+1.5k^{3}xyz>x^{2}+y^{2}+z^{2}+1.5xyz.$\r\nThus, $4.5>x^{2}+y^{2}+z^{2}+1.5xyz.$ It's contradiction since, $4.5\\leq x^{2}+y^{2}+z^{2}+1.5xyz\\Leftrightarrow\\sum_{cyc}(x^{3}-x^{2}y-x^{2}z+xyz)\\geq0.$ :)\r\nAlso see here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=123877",
  "Solution_8": "[quote=\"arqady\"]For you, cefer :lol: \n[/quote]\r\n\r\n\r\nThanks, very much"
}
{
  "Tag": [
    "geometry",
    "integration",
    "calculus",
    "calculus computations"
  ],
  "Problem": "I need help in solving this..... Please and Thank You\r\nFind the area of the region in the first quadrant bounded by the line y=8x, the line x=1 , the curve y=1/sqaureroot(x), and the x-axis.",
  "Solution_1": "Hi :) !!!\r\n\r\nare you want to say thatt?????\r\n\r\ncalcul: $ \\int_1^x \\left|8t \\minus{} \\frac{1}{\\sqrt{t}} \\right|dt$ \r\n\r\nand thanks :)",
  "Solution_2": "How to start: draw all the curves given to you.  Shade in the region described by the problem -- there's only one portion of the plane that has all four of those curves as part of its boundary.  Then figure out an integral (or sum of integrals) that gives you this area.  (mathema*'s answer isn't right -- he or she made false assumptions about what the region looks like.  Draw the picture and you won't make the same mistake.)",
  "Solution_3": "[quote=\"JBL\"]How to start: draw all the curves given to you.  Shade in the region described by the problem -- there's only one portion of the plane that has all four of those curves as part of its boundary.  Then figure out an integral (or sum of integrals) that gives you this area.  [b](mathema*'s answer isn't right -- he or she made false assumptions about what the region looks like.  Draw the picture and you won't make the same mistake.[/b])[/quote]\r\n\r\nHI JBL \r\n\r\nI didn't give any answer and i didn't understood the questions of \"dizizviet\". than i just asked (he or she) !!!!\r\n\r\nand thanks\r\n\r\nPS: i'm a man ((he))",
  "Solution_4": "Instead of \"Are you say thatt?\" use \"Is this what you mean?\" for the question you intended.  Also, \"calcul\" is not an English word -- the verb you want is \"calculate.\"\r\n\r\ndizizviet gave four curves, $ y \\equal{} 8x$, $ x \\equal{} 1$, $ y \\equal{} 0$ and $ y \\equal{} \\frac{1}{\\sqrt{x}}$.  These curves cut the plane into several regions.  One of these regions is bounded by all four curves.  dizizviet is asking for the area of this region.",
  "Solution_5": "Find where:\r\n$ 8x\\equal{}\\frac{1}{\\sqrt{x}}$\r\n$ x^\\frac{3}{2}\\equal{}\\frac{1}{8}$\r\n$ \\sqrt{x}\\equal{}\\frac{1}{2}$\r\n$ x\\equal{}\\frac{1}{4}$\r\n\r\nThus, the area under the curves is equal to:\r\n\r\n$ \\int^{\\frac{1}{4}}_08x \\plus{} \\int_{\\frac{1}{4}}^1 \\frac{1}{\\sqrt x}$\r\n$ \\equal{}4 \\left (\\frac{1}{4} \\right )^2  \\minus{}4(0)^2\\plus{}2 \\sqrt{1} \\minus{} 2 \\sqrt{\\frac{1}{4}}$\r\n$ \\equal{}\\frac{5}{4}$"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "algebra proposed"
  ],
  "Problem": "hi,\r\nwe consider the polynominal with integer coeffitients\r\n\\[ P(x) \\equal{} a_{0}x^{n} \\plus{} a_{1}x^{n\\minus{}1} \\plus{} ... \\plus{} a_{n},\\]\r\nand a prime $ p$ s.t. $ a_{n} \\not\\equiv 0 \\bmod{p}$. If there exists $ n\\plus{}1$ integer numbers $ \\alpha_{1}, ..., \\alpha_{n\\plus{}1}$ with $ P(\\alpha_{r}) \\equiv 0 \\bmod{p}, \\forall r \\equal{} 1, ..., n\\plus{}1$, prove that there exists two of the $ \\alpha_{i}, \\alpha_{j}$ s.t. $ \\alpha_{i} \\equiv \\alpha_{j} \\bmod{p}$.",
  "Solution_1": "[quote=\"doicu\"]hi,\nwe consider the polynominal with integer coeffitients\n\\[ P(x) \\equal{} a_{0}x^{n} \\plus{} a_{1}x^{n \\minus{} 1} \\plus{} ... \\plus{} a_{n},\n\\]\nand a prime $ p$ s.t. $ a_{n} \\not\\equiv 0 \\bmod{p}$. If there exists $ n \\plus{} 1$ integer numbers $ \\alpha_{1}, ..., \\alpha_{n \\plus{} 1}$ with $ P(\\alpha_{r}) \\equiv 0 \\bmod{p}, \\forall r \\equal{} 1, ..., n \\plus{} 1$, prove that there exists two of the $ \\alpha_{i}, \\alpha_{j}$ s.t. $ \\alpha_{i} \\equiv \\alpha_{j} \\bmod{p}$.[/quote]\r\nWe will prove $ P(x)$ has max $ n$ root $ mod p$(*)\r\n$ n \\equal{} 1$,easy\r\nSuppose that we have (*) for $ n \\equal{} k$\r\nWe will prove that this is true for $ n \\equal{} k \\plus{} 1$\r\nConsider $ degP(x) \\equal{} k \\plus{} 1$ and\r\n$ P(x) \\equiv 0(mod p)$(**)\r\nIf (**) hasn't root,we done\r\nIf (**) has a root is $ x_o$ then we have\r\n$ P(x) \\equiv (x \\minus{} x_o)Q(x) \\plus{} R(x)(mod p)$\r\n$ degR(x) < 1$ so $ R(x) \\equiv const \\equiv c$=>$ degQ(x) \\equal{} k$\r\n$ P(x_o) \\equiv R(x_o) \\equiv c(mod p)$ so $ p|c$\r\n=>$ P(x) \\equiv (x \\minus{} x_o)Q(x)(mod p)$\r\nThen $ P(x) \\equiv 0(mod p)$ <=>$ x \\equiv x_o(mod p)$ or $ Q(x) \\equiv 0(mod p)$\r\nOn other hand $ degQ(x) \\equal{} k$ so $ Q(x)$ has max $ k$ root $ mod p$\r\nSo $ P(x)$ has max $ k \\plus{} 1$ root $ mod p$\r\n :P \r\nSorry beacuse my English isn't good!"
}
{
  "Tag": [
    "geometry",
    "incenter",
    "geometry unsolved"
  ],
  "Problem": "In $\\triangle ABC$, $\\angle A=60^\\circ$. The incentre is $I$. The inscribed circle of $\\triangle ABC$ meets $AB$, $AC$ at $D$, $E$ respectively. Line $DE$ meets line $BI$ and line $CI$ at $F$, $G$ respectively. Show that $FG=\\frac12 BC$.",
  "Solution_1": "see http://www.mathlinks.ro/Forum/viewtopic.php?p=224997#p224997 , it's almost the same problem.",
  "Solution_2": " this is my solution:\r\neasy to see that $\\angle ADE=\\angle AED=60$ ,$\\angle BIC=120$\r\n\r\nthen $\\angle CGE=60-\\frac{c}{2}$ , $\\angle BFD=\\frac{c}{2}$ \r\nand $\\angle GDI=30$\r\nsin theo. in $\\triangle DGI$ :\r\n\r\n$\\frac{DG}{sin(\\frac{B-60}{2})}=\\frac{DI}{sin(\\frac{B}{2})}$ then\r\n\r\n$DG=\\frac{rsin(\\frac{B-60}{2})}{sin(\\frac{B}{2})}$ \r\n\r\nif we use the sin theo. in $\\triangle DIF$\r\n\r\nwe obtain $DF=\\frac{rcos(\\frac{B}{2})}{sin(\\frac{C}{2})}$\r\n\r\n$\\Rightarrow$ $DF-DG=GF=r(\\frac{sin(\\frac{B}{2})cos(\\frac{B}{2})+sin(\\frac{B-60}{2})sin(\\frac{C}{2})}{sin(\\frac{C}{2})sin(\\frac{B}{2})})$\r\nwe use the trigonometrical changes we obtain:\r\n$GF=\\frac{1}{2}\\frac{rsin(60)}{sin(\\frac{C}{2})sin(\\frac{B}{2})}$ we must prove $GF=\\frac{1}{2}BC=Rsin(60)$ \r\nLEMMA: $r=4Rsin(\\frac{C}{2})sin(\\frac{B}{2})sin(\\frac{A}{2})$\r\n(this is a famous lemma this proves easy)\r\nif we use the lemma we obtain \r\n$GF=4Rsin(\\frac{60}{2})sin(60)\\frac{1}{2}=Rsin60=BC\\frac{1}{2}$ \r\n\r\nand that is the end of the solution :lol: \r\n",
  "Solution_3": "can anyone give a pure geom soln?",
  "Solution_4": "[quote=\"modeler\"]can anyone give a pure geom soln?[/quote]\r\nSee the link at Post 2."
}
{
  "Tag": [
    "ceiling function",
    "algebra proposed",
    "algebra"
  ],
  "Problem": "Find all possible triplets that satisfy:\r\n\r\n$ xyz\\equal{}4104$\r\n\r\nand\r\n\r\n$ x\\plus{}y\\plus{}z\\equal{}77$\r\n\r\n :)",
  "Solution_1": "$ 4104|n \\ \\text{for} \\ \\left\\lceil \\frac{77\\minus{}2}{2}\\right\\rceil<n<77\\minus{}2 \\Rightarrow n\\equal{}38, \\ 54, \\ 57, \\ 72$.\r\n\r\nCase $ n\\equal{}38$\r\n$ \\frac{4104}{38}|m \\ \\text{for} \\ \\left\\lceil \\frac{77\\minus{}38\\minus{}1}{2}\\right\\rceil<m<77\\minus{}38\\minus{}1 \\Rightarrow m\\equal{}27, \\ 36$.\r\n\r\nSubcase $ n\\equal{}38, \\ m\\equal{}27$\r\n$ 38\\plus{}27\\plus{}\\frac{4104}{38\\times27}\\equal{}38\\plus{}27\\plus{}4\\equal{}69 \\neq 77$\r\n\r\nSubcase $ n\\equal{}38, \\ m\\equal{}36$\r\n$ \\mathbf{38\\plus{}36\\plus{}\\frac{4104}{38\\times36}\\equal{}38\\plus{}36\\plus{}3\\equal{}77}$\r\n\r\n\r\nCase $ n\\equal{}54$\r\n$ \\frac{4104}{54}|m \\ \\text{for} \\ \\left\\lceil \\frac{77\\minus{}54\\minus{}1}{2}\\right\\rceil<m<77\\minus{}54\\minus{}1 \\Rightarrow m\\equal{}19$.\r\n\r\nSubcase $ n\\equal{}54, \\ m\\equal{}19$\r\n$ \\mathbf{54\\plus{}19\\plus{}\\frac{4104}{54\\times19}\\equal{}54\\plus{}19\\plus{}4\\equal{}77}$\r\n\r\n\r\nCase $ n\\equal{}57$\r\n$ \\frac{4104}{57}|m \\ \\text{for} \\ \\left\\lceil \\frac{77\\minus{}57\\minus{}1}{2}\\right\\rceil<m<77\\minus{}57\\minus{}1 \\Rightarrow m\\equal{}12, \\ 18$.\r\n\r\nSubcase $ n\\equal{}57, \\ m\\equal{}12$\r\n$ 57\\plus{}12\\plus{}\\frac{4104}{57\\times12}\\equal{}57\\plus{}12\\plus{}6\\equal{}75 \\neq 77$\r\n\r\nSubcase $ n\\equal{}57, \\ m\\equal{}18$\r\n$ 57\\plus{}18\\plus{}\\frac{4104}{57\\times18}\\equal{}57\\plus{}18\\plus{}4\\equal{}79 \\neq 77$\r\n\r\n\r\nCase $ n\\equal{}72$\r\n$ \\frac{4104}{72}|m \\ \\text{for} \\ \\left\\lceil \\frac{77\\minus{}72\\minus{}1}{2}\\right\\rceil<m<77\\minus{}72\\minus{}1 \\Rightarrow m\\equal{}3$.\r\n\r\nSubcase $ n\\equal{}72, \\ m\\equal{}3$\r\n$ 72\\plus{}3\\plus{}\\frac{4104}{72\\times3}\\equal{}72\\plus{}3\\plus{}19\\equal{}94 \\neq 77$\r\n\r\n\r\nThus, $ (x, \\ y, \\ z)\\equal{}(3, \\ 36, \\ 38), \\ (4, \\ 19, \\ 54)$",
  "Solution_2": "Brilliant, you were very clear  :)"
}
{
  "Tag": [
    "Putnam",
    "calculus",
    "AMC",
    "USA(J)MO",
    "USAMO",
    "MIT",
    "college"
  ],
  "Problem": "I'm pretty sure that this is the wrong forum to post this, and I apologize in advance, so someone can move this if they are disturbed  :P \r\n\r\nWhat is the Putnam, and what is it's difficulty level? Is it much harder than the IMO? How can you prepare for this? I'm still in high school, but I guess i could start knowing what it is so when it comes, I can be prepared  :)",
  "Solution_1": "The Putnam is the most prominent under grad math competition. It includes topics that wouldn't be normally learned in a high school curriculum. I wouldn't say \"more difficult\" than the IMO, but the Putnam is with no doubt at a much faster pace. Someone in college who has taken it probably has a lot more to say...",
  "Solution_2": "As krustyteklown points out, the Putnam is an undergraduate level competition.  The Putnam tests the same problem solving skills as any olympiad such as IMO, but the topics also include differential equations, linear algebra, and multivariable calculus, since they are covered in undergraduate mathematics.\r\n\r\nYou can find previous Putnam problems at these sites:\r\n[url=http://www.kalva.demon.co.uk/putnam.html]www.kalva.demon.co.uk/putnam.html[/url]\r\n[url=http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pindex.html]www.unl.edu/amc/a-activities/a7-problems/putnam/-pindex.html[/url]\r\n\r\nThe AoPS bookstore also sells compilations of Putnam problems:\r\n[url=http://www.artofproblemsolving.com/Books/AoPS_B_CP_Putnam.php]www.artofproblemsolving.com/Books/AoPS_B_CP_Putnam.php[/url]",
  "Solution_3": "Look in the \"College Playground\" section here at AoPS. We do have a whole forum devoted to the Putnam. It's briefly active in two spurts - December (the exam itself) and March (the results are announced).\r\n\r\nThe structure of the Putnam: no qualifying rounds - whoever wants to take it can. Many colleges have study groups or even courses devoted to preparing for it. Held on the first Saturday in December. Two three-hour sessions on the same day, morning and afternoon. 6 problems each session, 12 total. (When krustyteklown said \"at a much faster pace\" he was probably referring to this 30 minute per problem time allowance, in contrast to the USAMO or IMO's hour and a half.) Usually a considerable range of difficulty. Scored 0-10 per problem but the numbers 3 through 7 are not used. About 3500 participants. Median score (out of 120 possible) usually low single digits, occasionally zero.\r\n\r\nThere is \"college-level\" material on the Putnam, but it's usually not highly technical. Students who bring a strong problem-solving and competition background with them from high school have a substantial advantage over those trying to pick it up for the first time at the college level. College freshmen can and do do well on it.\r\n\r\nLook at the forum here to see discussion of the 2005 and 2004 versions.",
  "Solution_4": "yes over 2/3 of the participants this year got a 0 out of 120\r\nhighest score was a 92",
  "Solution_5": "[quote=\"sapphyre571\"]yes over 2/3 of the participants this year got a 0 out of 120\nhighest score was a 92[/quote]\r\nAre you sure about the highest score?\r\n\r\nhttp://www.unl.edu/amc/a-activities/a7-problems/putnam/-html/putnam2005stats.html",
  "Solution_6": "Just wondering, do you have to be math major to take Putnam? I know it's hard to answer since it might differ from college to college but this is something I always wondered.",
  "Solution_7": "[quote=\"amcavoy\"][quote=\"sapphyre571\"]yes over 2/3 of the participants this year got a 0 out of 120\nhighest score was a 92[/quote]\nAre you sure about the highest score?\n\nhttp://www.unl.edu/amc/a-activities/a7-problems/putnam/-html/putnam[b]2005[/b]stats.html[/quote]\r\n\r\n;)",
  "Solution_8": "Actually, in 2005, 46.7% (1657 out of 3545) of the participants got a score of zero. The high score was 100. The median score was 1.\r\n\r\nThe percentages of particpants scoring zero:\r\n\r\n2005: 46.7%\r\n2004: 53.6%\r\n2003: 27.8%\r\n2002: 34.7%\r\n2001: 44.9%\r\n2000: 57.7%\r\n1999: 60.2%\r\n1998: 30.2%\r\n\r\nI didn't find a year in there with a top score of 92 and a 2/3 rate of zeros, although 1999 featured a top score of 74.\r\n\r\nDo you have to be a math major to take the Putnam? Almost certainly not. However, it will probably be run by a Math Dept. faculty member, and information about it will come through the Math Dept. If you're not a math major but are interested, you may need to actively seek out the information about it at your college. The entry form is due in mid-October, and each college must at that time name it's three-member \"team\" - the three individuals whose scores will count for the team score. Note that schools with a huge number of students taking the Putnam and scoring well on it - MIT and Harvard, in particular - almost never get that naming of the team \"right\" in hindsight.\r\n\r\nThere may or may not be some kind of qualifying test; if so, that's likely to be in early October so that the results could be used to name the team. Even if there is a \"qualifying\" test, any student who truly wants to take the Putnam will probably be allowed to do so.\r\n\r\nI've run the Putnam for 7 years at my school, and usually get 6 to 8 students to take it, almost all of whom take (or at least sit in on) the special problem-solving course. In those years I've had maybe 2 who weren't math majors (or at least double majors - math with something else), and one of those had been on an ARML team when he was in high school.",
  "Solution_9": "For future reference, what is the normal procedure for high school students taking the Putnam (if there is a \"normal procedure\")?",
  "Solution_10": "There isn't a normal procedure.",
  "Solution_11": "By the way, I think you get at most 4 shots to take the Putnam.  So if you take it as a high-school student, you use up one of your shots.",
  "Solution_12": "[quote=\"Ravi B\"]By the way, I think you get at most 4 shots to take the Putnam.  So if you take it as a high-school student, you use up one of your shots.[/quote]It seems this is true :  [url]http://www.cs.berkeley.edu/~wkahan/MathH90/PutnamRegs.pdf[/url]\n[quote=\"Putnam Regulations\"]This contest is open only to registered students who have not received a college degree, and who have competed fewer than four times in previous Putnam examinations.[/quote]"
}
{
  "Tag": [
    "function",
    "vector",
    "analytic geometry",
    "group theory",
    "abstract algebra",
    "absolute value",
    "graph theory"
  ],
  "Problem": "[b](i)[/b] Let $S$ be a finite set with at least three elements. Show that there exists a function $f : \\mathbb{R} \\to \\mathbb{Q}$ with the following properties: 1) $f(x) = 0$ if and only if $x=0$; and 2) for any $x \\in \\mathbb{R}$, $\\sum_{a \\in S} f(ax) = 0$.\r\n[b](ii)[/b] Does there necessarily exist such function $: \\mathbb{R} \\to \\mathbb{Z}$?",
  "Solution_1": "I have a very ambigous and ugly solution. Vess, please show us yours!\r\nWe prove the following lemma:\r\nLet $u_1,\\cdots,u_m$ be pairwise distinct vectors in $Z^n$.\r\nThen there is a function $f:Z^n \\rightarrow Z*$ with\r\n$f(x+u_1)+\\cdots +f(x+u_n)=0$ for any $x$.\r\nProof. \r\nDefine $S_x=\\{x+u_1,\\cdots,x+u_n\\}$. \r\nSublemma. The vectors in $Z^n$ can be enumerated \r\n$v_1,v_2,\\cdots,v_n,\\cdots$ in such a way that for any $k$ \r\nthere is at most one $X$ with $v_k \\in S_{X}$ and $S_{X} \\subseteq \\{v_1,v_2,\\cdots,v_k\\}$.\r\nProof. Consider a function $g: Z^n \\rightarrow Z$ defined by\r\n $g(x_1,\\cdots,x_n)=\\sum_{i=1}^n m_i x_i$ \r\nwhere $m_i \\in N / \\{0\\}$ are choosen in such a way that \r\n$g(u_i)$ are all distinct and non-zero. \r\nNote that $g$ is additive.\r\nNow define $U_k$ be the set of all vectors $v$ with $|g(v)|=k$.\r\nLet's enumerate the vectors in $Z^n$ as follows:\r\nChoose a sufficiently large number $N$.\r\nThen begin: \r\nStep $1$: Enumerate all vectors in $U_0$ having all coordinates \r\nat most $N$ by absolute value (there are finitely many of them)\r\nStep $2$: Enumerate all vectors in $U_0$ having all coordinates \r\nat most $N^2$ by absolute value and not already enumerated.\r\nStep $3$: Enumerate all vectors in $U_0$ having all coordinates \r\nat most $N^3$ by absolute value and not already enumerated,\r\nthen enumerate all vectors in $U_1$ having all coordinates \r\nat most $N$ by absolute value.\r\n$\\cdots$\r\nStep $k$: Enumerate all vectors in $U_0$ having all coordinates \r\nat most $N^k$ by absolute value and not already enumerated, \r\nthen enumerate all vectors in $U_1$ having all components \r\nat most $N^{k-2}$ by absolute value and so on.\r\n\r\nOf course every vector will be enumerated at some time.\r\nNow let's prove the conclusion of the sublemma:\r\nSuppose $v_k \\in S_X,S_Y$. WLOG $g(v_k) \\geq 0$. \r\nWe note that we must have $X=v_k-u_i$, $Y=v_k-u_j$.\r\n WLOG $g(u_i)>g(u_j)$.\r\nThen $V=v_k-u_j+u_i \\in S_Y$ and hence was alredy enumerated. \r\nBut $g(V)=g(v_k)-g(u_j)+g(u_i)>g(v_k)$ so $|g(V)|>|g(v_k)|$.\r\nNow suppose $|g(v_k)|=s,|g(v)|=t>s$ and \r\n$v_k$ was enumerated at step $r$, $V$ at step $q\\leq r$.\r\nThen by the properties of our enumeration\r\n $v_k$ has at least one component greater that $N^{r-2s-1}$ \r\nby absolute value \r\n(or else it would have been  enumerated at a previous step).\r\nBut then if $N$ is sufficiently large \r\nthen $V$ must have at least one component \r\ngreater than $N^{r-2s-2}$ by absolute value.\r\nBut this contradicts the fact that $V$ has all components\r\n smaller than $N^{q-2s}\\leq N^{r-2s-2}$ by absolute value.\r\nThe sublemma is proven.\r\n\r\nTo prove our lemma, \r\nconsider $\\left(v_i\\right)$ our enumeration as in the sublemma.\r\nNow we can define $f(v_i)$ inductively on $i$. \r\nWe can choose $f(v_i)$ at random unless \r\n$v_i \\in S_X$ with $S_{X} \\subseteq \\{v_1,v_2,\\cdots,v_i\\}$.\r\nBut then by sublemma, this $X$ is unique and then \r\nwe can choose an unique value of $f(v_i)$ to satisfy\r\n $\\sum_{v \\in S_x} f(v)=0$.\r\nMoreover, we can ensure \r\n$f(v_i)=1$ $(mod$ $n)$ at each step so $f(v_i) \\neq 0$.\r\n \r\nthe lemma is proven.\r\n\r\nNow to pass to the solution of the problem.\r\nLet $S={x_1,\\cdots,x_n}$ and \r\nwlog suppose $S$ doesn't contain $0$ (it doesn't matter). \r\nWe define $T=\\{\\prod x_i^{a_i}|a_i \\in Z\\}$. \r\nLet $u_1,u_2,u_3,\\cdots,u_m$ \r\nbe the basis of $T$ over $Q$(it has dimension at most $n$).\r\nThen every element $x$ of $T$ can be written as\r\n$\\prod u_i^{a_i}$ for $a_i \\in Q$. \r\nBut multiplying by a suitable constant \r\nwe can suppose that $a_i \\in Z$ for $x=x_i$. \r\nThen as easily seen, $a_i \\in Z$ for any $X$.\r\nNow extend $T$ to become subgroup of $(R,\\times)$\r\n (still having dimension $m$).\r\nSo we can take the group $R\\T$. \r\nBy axiom of choice  we can choose from every class $C$\r\n an element $x \\in C$. \r\nNow every element in $C$ can be expressed uniquely as \r\n$x \\prod u_i^{a_i}, a_i \\in Z$ \r\nand hence can be associated with the vector \r\n$(u_1,\\cdots,u_n)$.\r\nNow taking $w_i$ be the vector associated with $x_i$ \r\n(in $T'$ this time, not in $C$) \r\nand using the lemma we get the result.\r\n\r\nVess, am I right? I'm not sure..\r\n\r\n  [Edit: I have edited my proof to correct one possible flaw]",
  "Solution_2": "[quote=\"vess\"][b](i)[/b] Let $S$ be a finite set with at least three elements. Show that there exists a function $f : \\mathbb{R} \\to \\mathbb{Q}$ with the following properties: 1) $f(x) = 0$ if and only if $x=0$; and 2) for any $x \\in \\mathbb{R}$, $\\sum_{a \\in S} f(ax) = 0$.\n[b](ii)[/b] Does there necessarily exist such function $: \\mathbb{R} \\to \\mathbb{Z}$?[/quote]\r\n\r\nTo avoid irrelevant issues about Zorn's lemma let me rephrase the meat of the problem:\r\nYou have a finitely generated group, generated by the nonzero elements of S.\r\nDoes its Cayley graph have a labelling where the sum of labels of the neighbors \r\nof any point = 0, and all labels are nonzero?\r\n For R or Q as the labels maybe this is true for any f.g group.  For Z, I don't know."
}
{
  "Tag": [
    "limit",
    "algebra",
    "polynomial",
    "calculus",
    "calculus computations"
  ],
  "Problem": "What is the limit of\r\n\r\n(10^8 * x^5 + 10^6 * x^4 + 10^4*x^2) / (10^9 * x^6 + 10^7 * x^5 + 10^5*x^3)\r\n\r\nas x approaches positive infinity \r\n\r\nthanks",
  "Solution_1": "0\r\n\r\n[hide=\".\"]\nlook at the degree of the denominator\n[/hide]",
  "Solution_2": "im still not seeing how the limit is 0",
  "Solution_3": "Let $ L \\equal{} \\lim_{x \\to \\infty} \\frac {P(x)}{Q(x)}$, where P(x) and Q(x) are two polynomials in x with degrees m and n, respectively. Then we have:\r\n\r\nL = 0, if m < n;\r\nL = a/b, if m = n (whera a and b are the coefficients of the terms with degrees m and n);\r\nL = infinity, if m > n.\r\n\r\nTo see this write P(x)/Q(x) in following way:\r\n\r\n$ \\frac {P(x)}{Q(x)} \\equal{} \\frac {ax^m \\plus{} a_1x^{m \\minus{} 1} \\plus{} a_2x^{m \\minus{} 2} \\plus{} ... \\plus{} a_m}{bx^n \\plus{} b_1x^{n \\minus{} 1} \\plus{} b_2x^{n \\minus{} 2} \\plus{} ... \\plus{} b_n} \\equal{} \\frac {x^m\\left(a \\plus{} \\frac {a_1}{x} \\plus{} \\frac {a_2}{x^2} \\plus{} ... \\plus{} \\frac {a_m}{x^m}\\right)}{x^n\\left(b \\plus{} \\frac {b_1}{x} \\plus{} \\frac {b_2}{x^2} \\plus{} ... \\plus{} \\frac {b_n}{x^n}\\right)}$\r\n\r\n$ \\equal{} x^{m \\minus{} n} \\underbrace{\\frac {a \\plus{} \\frac {a_1}{x} \\plus{} \\frac {a_2}{x^2} \\plus{} ... \\plus{} \\frac {a_m}{x^m}}{b \\plus{} \\frac {b_1}{x} \\plus{} \\frac {b_2}{x^2} \\plus{} ... \\plus{} \\frac {b_n}{x^n}}}_{ \\equal{} T}$.\r\n\r\nNow letting x -> +oo, the limit of the fraction T is always a/b, but the limit of the term $ x^{m \\minus{} n}$ is zero if m < n (and so the total limit is also zero), is 1 if m = n (and so the total limit is a/b), and is infinite if m > n (and so the total limit is also infinite)."
}
{
  "Tag": [
    "arithmetic sequence",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "I have easy problem\r\nProve that if$a,b$be two realnumbers $\\geq2$.Provethat ${a^n-b^n}$ doesnot contain a arithmetic progression contains three terms",
  "Solution_1": "I split it to start a new topic... ;) \r\n\r\nPierre.",
  "Solution_2": "Assume that $0<b<a$ and $0 \\leq n<q<p$ are integers such that \r\n$(a^n-b^n) + (a^p-b^p) = 2(a^q-b^q)$ (which means that $a^n-b^n,a^q-b^q,a^p-b^p$ form a $3$-arithmetical progression).\r\nThen $a^n+a^p-2a^q = b^n+b^p - 2b^q .$\r\n\r\nLet $f(x) =x^n+x^p-2x^q$ defined for $x \\geq 2.$\r\nThen $f'(x) = \\frac 1 x (nx^n + px^p - 2qx^q)$.\r\nBut, since $x \\geq 2$ and $q \\leq p-1$, we have :\r\n$nx^n + px^p - 2qx^q \\geq px^p - 2qx^q > qx^p - 2qx^q \\geq 2qx^{p-1}-2qx^q \\geq 0$.\r\nThus $f'(x) >0$ and $f$ is increasing on $(2; + \\infty)$, which contradicts that $f(a)=f(b)$ with $a<b.$\r\n\r\nPierre.",
  "Solution_3": "Yes,my solution is above.I said that this problem is very easy."
}
{
  "Tag": [],
  "Problem": "Hmm check out \r\n[url=http://www.trendio.com]www.trendio.com[/url].  It's a pretty interesting site, where you \"buy\" popular words that appear often in the news - kind of like the stock market.  The price of the words fluctuates based on how often the word appears in the news, and the object is to get as much money as possible.",
  "Solution_1": "[hide=\"for me is\"]200% waste of time[/hide]",
  "Solution_2": "JESUS CHRIST JOSE, STOP PISSING PEOPLE OFF AND SPAMMING THE ROUND TABLE. u posted like 10 meaningless posts in the last 2 mins. WTF is with that?If u're gonna spam something, go to G&FF.  here are two examples:\r\n\r\n[quote=\"Jos\u00e9\"][hide=\"lol\"]superb signature![/hide][/quote]\n\n[quote=\"Jos\u00e9\"][hide=\"for me is\"]200% waste of time[/hide][/quote]",
  "Solution_3": "There's a huge problem here in MathLinks: Everything is spam.... you can't even post your opinion! come on! what's this then??? a military service??? ONLY MATH ONLY MATH... come on!",
  "Solution_4": "[quote=\"Jos\u00e9\"]There's a huge problem here in MathLinks: Everything is spam.... you can't even post your opinion! come on! what's this then??? a military service??? ONLY MATH ONLY MATH... come on![/quote]\r\n\r\nJose, look around.  No one else posts as much as you.  And yet your posts contain very little content.  They are short and devoid of insight.  You do not have to post in every single thread you read.\r\n\r\nOf course you can post your opinion.  But posting an opinion without any justification is spam (not to mention rude).\r\n\r\nRegards"
}
{
  "Tag": [
    "inequalities"
  ],
  "Problem": "Let $ a,b,c>0$ so that $ a\\plus{}b\\plus{}c\\equal{}1$. Prove that:\r\n$ \\frac{\\sqrt{a^2\\plus{}abc}}{c\\plus{}ab}\\plus{}\\frac{\\sqrt{b^2\\plus{}abc}}{a\\plus{}bc}\\plus{}\\frac{\\sqrt{c^2\\plus{}abc}}{b\\plus{}ca}\\leq\\frac{1}{2\\sqrt{abc}}$",
  "Solution_1": "I failed to find a less uglier solution  :( \r\n\r\nNote that $ \\sum\\frac{\\sqrt{a^2+abc}}{c+ab}=\\sum\\frac{\\sqrt{a(c+a)(a+b)}}{(b+c)(c+a)}$.\r\n\r\nTherefore our inequality is equivalent to\r\n\\begin{align*}&\\sum\\frac{\\sqrt{a(c+a)(a+b)}}{(b+c)(c+a)}\\le\\frac{a+b+c}{2\\sqrt{abc}}\\\\\r\n\\iff &\\sum a(a+b)\\sqrt{bc(c+a)(a+b)}\\le \\frac 12(a+b+c)(a+b)(b+c)(c+a)\\end{align*}\r\nBy AM-GM,\r\n\\begin{align*}\\sum a(a+b)\\cdot 2\\sqrt{bc(c+a)(a+b)}&\\le\\sum a(a+b)(b(c+a)+c(a+b))\\\\\r\n&=\\sum a(a+b)(ab+2bc+ca)\\end{align*}\r\nNow\r\n\\begin{align*}\\sum a(a+b)(ab+2bc+ca)&=\\sum a^2(ab+bc+ca)+\\sum a^2bc+\\sum ab(ab+bc+ca)+\\sum ab^2c\\\\\r\n&=(a^2+b^2+c^2+ab+bc+ca)(ab+bc+ca)+2abc(a+b+c)\\\\\r\n&=(a+b+c)^2(ab+bc+ca)-(ab+bc+ca)^2+2abc(a+b+c)\\\\\r\n&=(a+b+c)^2(ab+bc+ca)-(a^2b^2+b^2c^2+c^2a^2)\\\\\r\n&\\le (a+b+c)^2(ab+bc+ca)-abc(a+b+c)\\\\\r\n&=(a+b+c)(a+b)(b+c)(c+a)\\end{align*}\r\nwhich was what we wanted.",
  "Solution_2": "Beautiful proof, nayel! :lol:",
  "Solution_3": "@nayel:\r\n\r\nIn your proof, you used AM-GM to give an upper bound for the square root. In particular, you did\r\n\r\n$ 2\\sqrt{bc(c\\plus{}a)(a\\plus{}b)}\\le b(c\\plus{}a)\\plus{}c(a\\plus{}b)$\r\n\r\nI am wondering, why did you choose that particular way of breaking up the square root. For example, one could group $ bc$ and $ (c\\plus{}a)(a\\plus{}b)$ instead. What is your thinking on this?",
  "Solution_4": "Actually when using AM-GM or Cauchy-Schwarz in this sort of cases, I always try to end up with the form where the degree minorizes as much as possible. For example, if I used AM-GM for $ 4bc$ and $ (c \\plus{} a)(a \\plus{} b)$ I would get $ [2,0,0]$ and $ [1,1,0]$ terms. But if I use it on $ b(c \\plus{} a)$ and $ c(a \\plus{} b)$ I get all $ [1,1,0]$ terms, which in the long run could possibly be useful.\r\n\r\nEDIT: Thanks arqady! :)",
  "Solution_5": "Geat proof, but werent you suposed to show it was less than half of what you showed?",
  "Solution_6": "But I already multiplied both sides by $ 2$ before applying AM-GM."
}
{
  "Tag": [
    "calculus",
    "LaTeX"
  ],
  "Problem": "I'm not sure how to use the math symbols with a post like I've seen in other posts; but let me give it a shot.\r\n\r\nI'm asked to write the equation of a circle in standard form and sketch its graph.  I thought: Simple enough; I'll just work backwards and put them back in their respective places for the standard equation of a circle.  Then I realized there was no way I could tell how the orignal two were split up (because in the end they were combined)... how else can I approach this?  Any suggestions?\r\n\r\nx\u00b2+y\u00b2-2x+6y+6=0\r\n\r\nThanks,",
  "Solution_1": "BTW; The standard equation of a circle is (x-h)\u00b2+(y-k)\u00b2=r\u00b2.\r\n\r\nr = radius\r\n(x,y) = point on circle\r\n(h,k) = center",
  "Solution_2": "complete the square?",
  "Solution_3": "This forum makes me feel immensely stupid in all honesty...  maybe I should find a group of less intelligent people.  :P \r\n\r\nThank you for the quick response; I haven't done any of this stuff since Algebra II.  We are reviewing an incredible amount of old material for the first week in A.P. Calculus AB and these topics seem almost foreign to me.  However it all seems to be coming back.\r\n\r\nThanks (once again),",
  "Solution_4": "[hide]\n$x^2+y^2-2x+6y+6=0$\n$(x^2-2x)+(y^2+6y)+6=0$\n$(x^2-2x+1)-1+(y^2+6y+9)-9+6=0$\n$(x-1)^2+(y+3)^2=4$\n\nSo the center is $(1,-3)$ and the radius is $2$.\n[/hide]\r\n\r\n[quote=\"shyce\"]This forum makes me feel immensely stupid in all honesty...  maybe I should find a group of less intelligent people.  :P \n[/quote]\r\nYou can use [url=http://www.artofproblemsolving.com/Forum/index.php?f=149]Getting Started[/url] forum ;) \r\n\r\nAlso, for \"math symbols\" ($\\LaTeX$), all you have to do is to put dollar signs before and after the expression."
}
{
  "Tag": [],
  "Problem": "Prove that $ x^2-(m^2+3m+1)x+(3n^2+5n+7)=0$ has no integer root $ \\forall m,n \\in \\mathbb{Z}$",
  "Solution_1": "[size=150]It is important to use the fact  that $ x^2\\plus{}x$ is even (because $ x^2\\plus{}x\\equal{}x(x\\plus{}1)$, and one of $ x$ or $ x\\plus{}1$ is even).[/size]\r\n$ m^2\\plus{}3m\\plus{}1\\equal{}(m^2\\plus{}m)\\plus{}(2m\\plus{}1)$, which is always odd.\r\n$ 3n^2\\plus{}5n\\plus{}7\\equal{}3(n^2\\plus{}n)\\plus{}(2n\\plus{}7)$, which is always odd.\r\nIf $ x$ is even or odd, $ x^2\\minus{}x(m^2\\plus{}3m\\plus{}1)\\plus{}3n^2\\plus{}5n\\plus{}7$ is always odd, while $ 0$ is even, so it is impossible the equality.",
  "Solution_2": "[quote=\"erudito\"][size=150]It is important to use the fact  that $ x^2 \\plus{} x$ is even (because $ x^2 \\plus{} x \\equal{} x(x \\plus{} 1)$, and one of $ x$ or $ x \\plus{} 1$ is even).[/size]\n$ m^2 \\plus{} 3m \\plus{} 1 \\equal{} (m^2 \\plus{} m) \\plus{} (2m \\plus{} 1)$, which is always odd.\n$ 3n^2 \\plus{} 5n \\plus{} 7 \\equal{} 3(n^2 \\plus{} n) \\plus{} (2n \\plus{} 7)$, which is always odd.\nIf $ x$ is even or odd, $ x^2 \\minus{} x(m^2 \\plus{} 3m \\plus{} 1) \\plus{} 3n^2 \\plus{} 5n \\plus{} 7$ is always odd, while $ 0$ is even, so it is impossible the equality.[/quote]\r\n\r\nCorrect.",
  "Solution_3": "It is interesting how people sometimes are only able to see the most difficult solution, trying to calculate the both roots of the equation:\r\n$ x \\equal{} \\dfrac{m^2 \\plus{} 3m \\plus{} 1 \\plus{} \\sqrt {( \\minus{} (m^2 \\plus{} 3m \\plus{} 1))^2 \\minus{} 4(3n^2 \\plus{} 5n \\plus{} 7)}}{2}$ and $ \\dfrac{m^2 \\plus{} 3m \\plus{} 1 \\minus{} \\sqrt {( \\minus{} (m^2 \\plus{} 3m \\plus{} 1))^2 \\minus{} 4(3n^2 \\plus{} 5n \\plus{} 7)}}{2}$, trying to see when the term $ ( \\minus{} (m^2 \\plus{} 3m \\plus{} 1))^2 \\minus{} 4(3n^2 \\plus{} 5n \\plus{} 7)$ is a perfect square.\r\nThere is a problem (actually easier then this one) in which people normally procced by the most difficult solution:\r\nTwo trains in a straight line are going to crash, and both run with speed $ 60$ against the other. A bug, which flies with speed $ 80$, always keep his way in this straight line, and when arrives in the front side of a train, begin to fly exactly in the opposite way, same direction, without stoping, against the other train, and keep itself like this until both hit, then stop. Initially, they are at a distance $ 300$ one from other. How much is the bug going to fly ?\r\nPeople try to sum the progression of distances that the bug flies from one to other train each time, but realise that they are going to hit in $ 2.5$ \"hours\", and the speed of the bug is constant, so it flies $ 200$."
}
{
  "Tag": [
    "algebra proposed",
    "algebra"
  ],
  "Problem": "The following numbers are the first seven numbers obtained through a few logical tranformations involving a non-Cardinal sequnce of numbers.\r\nAll i can realy say about this term \"logical sequence\" is that i'm not sure if i can write an equation for it, but i can deffinatly write a program that outputs numbers in the sequnce, without useing conditional statments.\r\n\r\nHere are the first seven numbers produced by the formula:\r\n7,15,36,76,144,220,324\r\n\r\nThre may be more then one way to get this sequnce of numbers... i'll provide more numbers if necisary.",
  "Solution_1": "[quote=\"LaughingMan42\"]The following numbers are the first seven numbers obtained through a few logical tranformations involving a non-Cardinal sequnce of numbers.\nAll i can realy say about this term \"logical sequence\" is that i'm not sure if i can write an equation for it, but i can deffinatly write a program that outputs numbers in the sequnce, without useing conditional statments.\n\nHere are the first seven numbers produced by the formula:\n7,15,36,76,144,220,324\n\nThre may be more then one way to get this sequnce of numbers... i'll provide more numbers if necisary.[/quote]\r\n$a_{n}=\\frac{(n-2)(n-3)(n-4)(n-5)(n-6)(n-7)}{(1-2)(1-3)(1-4)(1-5)(1-6)(1-7)}\\cdot7+$\r\n$+\\frac{(n-1)(n-3)(n-4)(n-5)(n-6)(n-7)}{(2-1)(2-3)(2-4)(2-5)(2-6)(2-7)}\\cdot15+$\r\n$+\\frac{(n-1)(n-2)(n-4)(n-5)(n-6)(n-7)}{(3-1)(3-2)(3-4)(3-5)(3-6)(3-7)}\\cdot36+$\r\n$+\\frac{(n-1)(n-2)(n-3)(n-5)(n-6)(n-7)}{(4-1)(4-2)(4-3)(4-5)(4-6)(4-7)}\\cdot76+$\r\n$+\\frac{(n-1)(n-2)(n-3)(n-4)(n-6)(n-7)}{(5-1)(5-2)(5-3)(5-4)(5-6)(5-7)}\\cdot144+$\r\n$+\\frac{(n-1)(n-2)(n-3)(n-4)(n-5)(n-7)}{(6-1)(6-2)(6-3)(6-4)(6-5)(6-7)}\\cdot220+$\r\n$+\\frac{(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)}{(7-1)(7-2)(7-3)(7-4)(7-5)(7-6)}\\cdot324.$ :D"
}
{
  "Tag": [
    "geometry",
    "geometric transformation",
    "reflection",
    "incenter",
    "parallelogram",
    "circumcircle",
    "ratio"
  ],
  "Problem": "Let $ABC$ be a triangle in which no angle is $90^{\\circ}$. For any point $P$ in the plane of the triangle, let $A_1, B_1, C_1$ denote the reflections of $P$ in the sides $BC,CA,AB$ respectively. Prove that\r\n \r\n(i) If $P$ is the incenter or an excentre of $ABC$, then $P$ is the circumenter of $A_1B_1C_1$;\r\n\r\n(ii)  If $P$ is the circumcentre of $ABC$, then $P$ is the orthocentre of $A_1B_1C_1$;\r\n\r\n(iii) If $P$ is the orthocentre of $ABC$, then $P$ is either the incentre or an excentre of $A_1B_1C_1$.",
  "Solution_1": "(1) The distance from $I$ to the sides of the triangles are all equal. Therefore we have $IA_1=IB_1=IC_1$, Hence $I$ is the circumcentre of $\\triangle{A_1B_1C_1}$.\r\n\r\n(2) Since $O$ is the circumcentre, so $BC_1 || AO || CB_1$ and $BC_1=AO=CB_1$, therefore $BCB_1C_1$ is a paralellogram hence $B_1C_1 || BC$. Hence $A_1O \\perp B_1C_1$. Similarly, we can get that $O$ is the orthocentre of $\\triangle{A_1B_1C_1}$.\r\n\r\n(3) The reflections of the orthocentre of the triangle is on the circumcircle of the triangle. Also, since $H$ is the intersection of the altitudes, we have $A, H, A_1$ collinear. Thus: \\[ \\angle{B_1A_1A}=\\angle{B_1BA}=\\angle{C_1CA}=\\angle{C_1A_1A} \\]\r\nSo $AHA_1$ is the bisector of $\\angle{B_1A_1C_1}$. Similarly, we will have other two results. Therefore, $H$ is the incenter of $\\triangle{A_1B_1C_1}$.\r\n\r\nThe proof of excentre is basically the same as above.",
  "Solution_2": "Well for the incenter and circumcenter, it is very easy as shown by shobber.\r\nFor the part 3) another proof:\r\nCOnsider the orthocenter $ H$ of $ ABC$ and its reflections $ A_1,B_1,C_1$ . We know and can prove easily that reflections are concyclic with the the points $ A,B,C$ by proving specifically the concyclity for $ A_1, A,B,C$, $ B_1,A,B,C$, $ C_1,A,B,C$. So now consider triangle $ A_1B_1C_1$ with the same circumcircle as $ ABC$. Draw the pedal triangle of $ ABC$. Let it be $ DEF$. Obviosuly, $ DEF$ and $ A_1B_1C_1$ are homothetic with ratio $ \\frac{1}{2}$ center of homothety $ H$. So, now the the $ H$ of the triangle $ ABC$ is the incenter of the podar triangle and done. Similarly, we can do for excenter in the case of the original triangle being obtuse as the prthocenter of the original triangle will be the excenter of the podar triangle in that case.",
  "Solution_3": "What is reflection? How can we construct it here and what point is P ? Pls can someone explain.\n",
  "Solution_4": "My sol\n1use incentre is circumcentre of pedal triangle and homothety about p with ratio 1/2\n2use circumcentre is orthocentre of medial triangle and homothety\n3use orthocentre is incentre of pedal triangle and homthety"
}
{
  "Tag": [],
  "Problem": "What is the value of $ (\\minus{}3) \\plus{} (\\minus{}6) \\div (\\minus{}6) \\times (\\minus{}3) \\minus{} (\\minus{}6)$?",
  "Solution_1": "remember order of operations\r\n\\[ ( \\minus{} 3) \\plus{} ( \\minus{} 6) \\div ( \\minus{} 6) \\times ( \\minus{} 3) \\minus{} ( \\minus{} 6)\\]\r\n\\[ ( \\minus{} 3) \\plus{} 1 \\times ( \\minus{} 3) \\minus{} ( \\minus{} 6)\\]\r\n\\[ ( \\minus{} 3) \\plus{} ( \\minus{} 3) \\minus{} ( \\minus{} 6)\\]\r\n\\[ \\minus{} 3 \\minus{} 3 \\plus{} 6\\]\r\n\\[ 0\\]"
}
{
  "Tag": [
    "number theory proposed",
    "number theory"
  ],
  "Problem": "Find all natural numbers $a$ such that : \r\n\r\n$a = \\frac{1}{x_1}+\\frac{1}{x_2}+\\frac{1}{x_3}+...............+\\frac{1}{x_{2005}}$\r\nwhere $x_1 , ... , a_{2005}$ are natural numbers.",
  "Solution_1": "$1\\le a\\le 2005$\r\ntake $x_{i}=2006-a$ for $i\\in [1,2006-a]$ and $x_{i}=1$ for $i\\in [2007-a,2005]$ :)"
}
{
  "Tag": [
    "quadratics",
    "inequalities proposed",
    "inequalities"
  ],
  "Problem": "For $-1\\leq a_{i}\\leq 1\\ (i=1,\\ 2,\\ \\cdots ,\\ 2007)$, Prove the following inequality.\r\n\r\n$(1-{a_{1}}^{2})^{1004}+(1-{a_{2}}^{2})^{1004}+\\cdots+(1-{a_{2007}}^{2})^{1004}+2008(|a_{1}|+|a_{2}|+\\cdots+|a_{2007}|)$\r\n$\\geq 2007$.",
  "Solution_1": "[quote=\"kunny\"]For $-1\\leq a_{i}\\leq 1\\ (i=1,\\ 2,\\ \\cdots 2007)$, Prove the following inequality.\n\n$(1-a_{1}^{2})^{1004}+(1-a_{2}^{2})^{1004}+\\cdots+(1-a_{2007})^{1004}+2008(|a_{1}|+|a_{2}|+\\cdots+|a_{2007}|)$\n$\\geq 2007$.[/quote]\r\nHm... Are you sure? Please edit it!"
}
{
  "Tag": [
    "function",
    "trigonometry",
    "limit",
    "inequalities",
    "logarithms",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Prove that a function $f\\in C^{1}\\bigl((0,+\\infty)\\bigr)$ which satisfy $\\ds f'(x)=\\frac{1}{1+x^{4}+\\cos f(x)},\\, x>0,$ is bounded at $(0,+\\infty).$ (O. Nesterenko)\r\n\r\nSee all the problems from [color=orange]Kyiv Mechmat Competition, 2007[/color] [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=135749]here[/url]",
  "Solution_1": "This is an idea:\r\n\r\nObserve that $f'(x)>0$ so $f$ increases, therefore you only need to show that $\\lim_{x\\rightarrow \\infty}f(x)<\\infty$.\r\n\r\nWe have \r\n\r\n$-1\\leq \\cos f =\\frac{1}{f'}-1-x^{4}\\leq 1$, it follows that\r\n\r\n$\\frac{1}{2+x^{4}}\\leq f'\\leq \\frac{1}{x^{4}}$\r\n\r\n$(f+\\frac{x^{-3}}{3})'\\leq 0$ for all $x\\geq 1$. Thus the function \r\n\r\n$x\\rightarrow f(x)+\\frac{x^{-3}}{3}$ decreases on $(1,\\infty)$ hence\r\n\r\n$f(x)\\leq f(1)+\\frac{1}{3}-\\frac{x^{-3}}{3}$, for all $x\\geq 1$ and the desired result follows.",
  "Solution_2": "[quote=\"didilica\"]This is an idea:\nObserve that $f'(x)>0$ so $f$ increases, therefore you only need to show that $\\lim_{x\\rightarrow \\infty}f(x)<\\infty$.\n[/quote]\r\nAnd what about $\\lim_{x\\to 0+}f(x)>-\\infty$? :maybe:",
  "Solution_3": "Use the other inequality:\r\n\r\nLet $F$ be a primitive of $\\frac{1}{2+x^{4}}$. Then \r\n\r\n$(f-F)'\\geq 0$ implies that $x\\rightarrow f(x)-F(x)$ increases on $(0,\\infty)$ hence\r\n\r\n$f(x)\\geq F(x)+f(0)-F(0)$ for all $x\\in (0,\\infty)$.\r\n\r\nA calculation shows that\r\n $F(x)=\\frac{1}{8\\sqrt[4]{2}}(-2\\arctan (1-\\sqrt[4]{2}x)+2\\arctan(1+\\sqrt[4]{x})+\\ln \\frac{\\sqrt{2}x^{2}+2\\sqrt[4]{2}x+2}{\\sqrt{2}x^{2}-2\\sqrt[4]{x}+2})$\r\n\r\nNow, let $x\\rightarrow 0$ and you get that the function is bounded. I hope that there are no computational mistakes.",
  "Solution_4": "[quote=\"didilica\"]Use the other inequality:\n\nLet $F$ be a primitive of $\\frac{1}{2+x^{4}}$. Then \n\n$(f-F)'\\geq 0$ implies that $x\\rightarrow f(x)-F(x)$ increases on $(0,\\infty)$ hence\n\n$f(x)\\geq F(x)+f(0)-F(0)$ for all $x\\in (0,\\infty)$.\n\n[/quote]\r\n\r\n $f(0)$ ? \r\nyou don't know if it is defined"
}
{
  "Tag": [
    "Support"
  ],
  "Problem": "http://www.dailymail.co.uk/sciencetech/article-1031438/Pictured-The-floating-cities-day-house-climate-change-refugees.html\r\n\r\nI have a few questions: There are bound to be sea storms. If these are lilypads, then what's to keep them from floating into hurricanes??\r\n\r\nAlso, what keeps them from floating into each other, or onto debris, like reefs or rocks?",
  "Solution_1": "what exactl do u mean or???? and how is this related to aops??? :huh:",
  "Solution_2": "I completely support innovation like this, but there is so much that needs to go into this like proper safety precautions which with 50,000 inhabitants it sounds over the top. So I am somewhat against it.",
  "Solution_3": "rocket boosters.\r\n\r\njk, but they could have GPS systems and propulsion systems (probably underwater) that prevent collisions. Maybe a wide buffer zone around each lilypad? The propulsion systems won't have to be very powerful then."
}
{
  "Tag": [
    "geometry",
    "ratio",
    "geometry proposed"
  ],
  "Problem": "Let a circle with diameter $  BC $ and a point $  A $ on his plane , such that  $\\triangle ABC $ is regular. On side  $ BC $  we take a point $ D $ with $ CD = 2DB $.Line \r\n $ AD $  meets the circle [b]for second time [/b]at $  E $ . Prove that $  BE / / AC $ .\r\n\r\n Babis\r\n\r\n[i](  It could be preolympiad , but some times easier problems encourange the solvers ! So , to this problem I' d like to see as many as possible solutions)[/i]",
  "Solution_1": "are you sure the wording of the problem is correct?? i just cant get AD  to intersect the circle",
  "Solution_2": "Well , A is an exterior point of the circle , triangle$\\triangle ABC  $is equilateral and D is on  -cirle's diameter - side BC. Line AD intersects the circle first in a point interior to the triangle and then, for second time , at E. Does this help , or I must write it in an other way?\r\nBabis",
  "Solution_3": "Stergiu, you frisk with this problem. See (High School, Pre-Olimpiad, Shobber)\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?p=331011#p331011",
  "Solution_4": "Levi ,yes ! It took me more than an hour to find a geometric solution( you know that I' m not so young !!!) for Shobber ' s problem . So I changed the wording  and posted it to solvers with more experience , - you are beetween them of course.I ' m sure that I have seen a good solution in the past , but I did not write it. Believe me , this is the only reason for reposting the problem. \r\n   Cu respekt - Babis  \r\n(By the way , Tsitsifas has also an excellent geometry book in greek. )",
  "Solution_5": "It is O.K.  Try to prove the fourth criterion of equivalence and of similarity\r\n(the enunciations are in reply at Grobber's problem).",
  "Solution_6": "Consider the omotety $f$ with center D and ratio -1/2; then\r\n$f(C)=B$\r\n$f(B)=O$\r\nwhere O is the midpoint of BC.\r\nNow, $F=f(A)$ is the vertex of the equilateral triangle on BO, in the halfplane that doesn't cointain A, therefore is on the circle with diameter BC, because $BF=BC/2$ and $\\angle BOF=\\pi/3$.\r\nSo, A,D and F are collinear and F is on the circle with diameter BC, then $F\\equiv E$, but $BF\\parallel AC$ (the second is the image of the first under omotety) and this is the thesis."
}
{
  "Tag": [
    "geometry",
    "geometric transformation",
    "rotation"
  ],
  "Problem": "Any suggestions on how  to teach a five year old mathematics? I have tried, but I don't know how to explain it to her. I want my little sister to be a math nerd like me, but its kind of hard when she still does not know addition or subtraction.  :|",
  "Solution_1": "Addition and subtraction? Use Cheerios. So easy to count and sort, and when you're done, you can eat them. \r\n\r\nThis is meant as a serious suggestion.",
  "Solution_2": "When you want to teach multiplication, first teach grids, rows, and columns.  Then you can visually represent it to her.  Another concept that helps multiplication is counting by intervals.  Counting by fives also greatly helps in money and time.  Normal multiplication might be beyond her now, but you never know how soon this might become helpful...",
  "Solution_3": "I do pattern recognition with my little four year-old brother. Sometimes it's the simple ones like <square> <triangle> <square> to rotations to number progressions. They're all pretty much what comes next. It's really tough trying to get a feel for what he can understand and what he fooled you into thinking he can understand.",
  "Solution_4": "THE questions for testing understanding are \"How?\" and \"Why?\"\r\n\r\nOccasionally, when the situation allows, review topics, and get the kid to explain the topic to YOU.  If he/she can clearly phrase his/her thoughts, I, at least, think that's worth half a cookie or a shiny nickel.  \r\n\r\nSlight tangent?: Totally serious about the nickel.  Wash it and it becomes shiny and clean.  Like alot of kids, I still like shiny nickels.  You should have seen me at the nearby museum's gem exhibition.   :wink:",
  "Solution_5": "[quote=\"alexhhmun\"]Totally serious about the nickel. [/quote]\r\nOnly for a child that's old enough not to automatically put everything into his or her mouth.",
  "Solution_6": "true.  bad idea.   :(   I should have left while I was ahead.",
  "Solution_7": "Start with word problems and real life situations instead of abstractions. Say \"How many are there?\" and \"How long will it take?\" instead of \"What comes after eighty-seven?\" and \"What is $ 20 \\minus{} 4$?\". Make it amusing. Not \"How many pencils does the teacher have?\", but \"How many chocolate bars did Mr. Fat eat?\" Encourage thinking and early exposure to geometric ideas: \"What is a circle?\" \"...A round thingy?\" Draw an extremely huge elongated and twisted but round figure. Ask if it is a circle. \"...NooooooOOOO... like this...\" Pretend you don't know what a circle is. \"Oh, so smaller!\" Draw a small twisted and loopy thing. Etc.  :)"
}
{
  "Tag": [
    "inequalities"
  ],
  "Problem": "Find all real solutions to the following inequality:\r\n\r\n$ \\frac {8}{\\sqrt {x \\plus{} 6} \\minus{} \\sqrt {x \\minus{} 2}} \\leq 6 \\minus{} \\sqrt {x \\plus{} 1}$",
  "Solution_1": "Reminds me quite a bit of an IMO problem from the mid-1960s.\r\n\r\n[hide=\"First step\"]\n$ x\\minus{}2\\geq0\\implies x\\geq 2$, giving us a lower bound.\n\n$ \\text{LHS}\\cdot\\frac{\\sqrt{x\\plus{}6}\\plus{}\\sqrt{x\\minus{}2}}{\\sqrt{x\\plus{}6}\\plus{}\\sqrt{x\\minus{}2}}\\equal{}\\frac{8(\\sqrt{x\\plus{}6}\\plus{}\\sqrt{x\\minus{}2})}{4}\\equal{}2\\sqrt{x\\plus{}6}\\plus{}2\\sqrt{x\\minus{}2}$\n\n$ 2\\sqrt{x\\plus{}6}\\plus{}2\\sqrt{x\\minus{}2}<6\\minus{}\\sqrt{x\\plus{}1}$\n\nNow it seems kind of ugly, unless there's a substitution or something that makes things easier.\n[/hide]",
  "Solution_2": "hint:\r\n\r\n[hide]Think about what values of $ x$ will make the radical values real, you can find an interval for $ x$.[/hide]",
  "Solution_3": "[quote=\"grn_trtle\"]Reminds me quite a bit of an IMO problem from the mid-1960s.\n\n[hide=\"First step\"]\n$ x \\minus{} 2\\geq0\\implies x\\geq 2$, giving us a lower bound.\n\n$ \\text{LHS}\\cdot\\frac {\\sqrt {x \\plus{} 6} \\plus{} \\sqrt {x \\minus{} 2}}{\\sqrt {x \\plus{} 6} \\plus{} \\sqrt {x \\minus{} 2}} \\equal{} \\frac {8(\\sqrt {x \\plus{} 6} \\plus{} \\sqrt {x \\minus{} 2})}{4} \\equal{} 2\\sqrt {x \\plus{} 6} \\plus{} 2\\sqrt {x \\minus{} 2}$\n\n$ 2\\sqrt {x \\plus{} 6} \\plus{} 2\\sqrt {x \\minus{} 2} < 6 \\minus{} \\sqrt {x \\plus{} 1}$\n\nNow it seems kind of ugly, unless there's a substitution or something that makes things easier.\n[/hide][/quote]\r\n\r\nAre you sure that $ \\text{LHS}\\cdot\\frac {\\sqrt {x \\plus{} 6} \\plus{} \\sqrt {x \\minus{} 2}}{\\sqrt {x \\plus{} 6} \\plus{} \\sqrt {x \\minus{} 2}} \\equal{} \\frac {8(\\sqrt {x \\plus{} 6} \\plus{} \\sqrt {x \\minus{} 2})}{4}$? \r\n\r\nI think once you fix it, it should be easier.",
  "Solution_4": "[hide=\"Solution\"]It is apparent that $ x \\geq 2$. Observe that:\n\n(1) $ \\sqrt {x \\plus{} 1}$ is strictly increasing, thus $ 6 \\minus{} \\sqrt {x \\plus{} 1}$ is strictly decreasing.\n\n(2) $ \\sqrt {x \\plus{} 6} \\minus{} \\sqrt {x \\minus{} 2}$ is strictly decreasing, thus $ \\frac {8}{\\sqrt {x \\plus{} 6} \\minus{} \\sqrt {x \\minus{} 2}}$ is strictly increasing.\n\n(3) At our lower bound $ x \\equal{} 2$, the former is greater than the latter. \n\n(4) The LHS is strictly positive, but the RHS is eventually negative, so eventually, the LHS exceeds the RHS\n\nConsidering the above, we realize there must be a unique number $ r>2$ such that the solution set is $ 2 \\leq x \\leq r$.\n\nBy inspection, this number is $ r\\equal{}3$, and the solution set is $ \\boxed{2 \\leq x \\leq 3 }$[/hide]"
}
{
  "Tag": [],
  "Problem": "Simplify:\r\n$ \\frac {(x^2 \\minus{} 16)\\sqrt {x^2 \\minus{} 64} \\plus{} x^2 \\minus{} 48x \\plus{} 128}{(x^2 \\minus{} 16)\\sqrt {x^2 \\minus{} 64} \\plus{} x^2 \\minus{} 48x \\minus{} 128}$",
  "Solution_1": "Is there supposed to be another parenthesis anywhere?",
  "Solution_2": "There is not supposed to be a ) at the end of the 128s",
  "Solution_3": "ok I fixed it",
  "Solution_4": "Uh... in mixed fractions... that's equal to $ 1\\plus{}\\frac{256}{(x^2\\minus{}16)\\sqrt{x^2\\minus{}64}\\plus{}x^2\\minus{}48x\\minus{}128}$ and $ x^2\\minus{}48x\\pm128$ are both not reducible in $ \\mathbb{Z}$. Do you want to rationalize the numerator/denominator? It's a mess  :maybe:",
  "Solution_5": "It's supposed to simplify to something, I don't know what."
}
{
  "Tag": [
    "function",
    "inequalities",
    "inequalities open"
  ],
  "Problem": "I don't know if this is true or not.  I am not aware of any proof of it, but it would certainly be neat if the so-called [tex]p[/tex]-norm were a descreasing function of [tex]p[/tex].  \r\n\r\nLet [tex]n[/tex] be a positive integer.  Prove (or disprove) that [tex]\\forall p, q \\in \\mathbb{R^{+}} \\,| \\,p \\ge q; \\forall\\mathbf{x} \\in \\mathbb{R}^{n}[/tex], we have\r\n\r\n[tex]\r\n\\displaymode{\r\n||\\mathbf{x}||_{p} \\le ||\\mathbf{x}||_{q}\r\n}[/tex]\r\n\r\nWhere we define the [tex]p[/tex]-norm of [tex]\\mathbf{x} = \\,\\,<x_{1}, \\dots , x_{n}>[/tex] by\r\n\r\n[tex]\r\n\\displaymode{\r\n||\\mathbf{x}||_{p} = \\sqrt[p]{|x_{1}|^{p} + \\cdots + |x_{n}|^{p}}\r\n}[/tex]",
  "Solution_1": "Shouldn't the inequality go the other way? I think you can prove it with Holder's Inequality.",
  "Solution_2": "I think the above is the correct statement, or the following is a quick counterexample\r\n\r\n[tex]\\sqrt[3]{91} = \\sqrt[3]{3^3 + 4^3} <  \\sqrt{3^2 + 4^2} = 5[/tex].\r\n\r\nDoes it follow from the expansion of the [tex]pq[/tex]-th power of both sides?",
  "Solution_3": "Interesting. Then I think you are right. I'm not sure if you can do it by raising each side to the power of $pq$. But geometrically it might be ``obvious'' by looking at unit ``circles'' with respect to various $l_p$-norms and considering them as distance formulae. (At least that convinces me that it's true...)",
  "Solution_4": "Ok, this is actually quite a low-tech result.   :D   Since the entire inequality is homogenous in the first degree, we assume [tex]x_{1}^{p} + \\cdots + x_{n}^{p} = 1[/tex]. Let [tex]a_{i} = x_{i}^{p}[/tex].  Substituting [tex]x_{i}[/tex] with [tex]a_{i}[/tex], we want\r\n\r\n[tex]\\displaymode{\r\n1 = \\sqrt[p]{a_{1} + \\cdots + a_{n}} \\le \\sqrt[q]{a_{1}^{\\frac{q}{p}} + \\cdots + a_{n}^{\\frac{q}{p}}}[/tex]\r\n\r\nBut we have [tex]0 \\le a_{i} \\le 1[/tex], where the upper bound results from the sum of the [tex]a_{i}[/tex] being 1.  Hence, [tex]a_{i}^{\\frac{q}{p}} \\ge a_{i}[/tex] (*), and it follows immediately that\r\n\r\n[tex]\\displaymode{\r\n\\sqrt[q]{a_{1}^{\\frac{q}{p}} + \\cdots + a_{n}^{\\frac{q}{p}}} \\ge \\sqrt[q]{a_{1} + \\cdots + a_{n}} = 1}[/tex]\r\n\r\nwhich is sufficient.  As for the equality condition, equality in (*) holds iff [tex]a_{j} = 0[/tex] or [tex]a_{j} = 1[/tex].  It is now clear that equality in the [tex]p[/tex]-norm inequality holds iff [tex]x_{1} = \\cdots = x_{j-1} = x_{j+1} = \\cdots = x_{n} = 0[/tex] and [tex]x_{j} \\in \\mathbb{R}[/tex].  Moreover, [tex]||\\mathbf{x}||_{p}[/tex] is a strictly decreasing function of [tex]p[/tex] iff [tex]\\exists[/tex] at least two non-zero components of [tex]\\mathbf{x}.[/tex]"
}
{
  "Tag": [
    "ratio",
    "calculus",
    "derivative"
  ],
  "Problem": "1.A body falling from a given height H hits an inclined plane in its path at a height\r\n\r\n   'h',as a result of its impact the direction of the body becomes horizontal.For what\r\n\r\n    value of  (h/H),the body will take maximum time to reach the ground\r\n\r\n\r\n2.A police inspector in a jeep is chasing a pickpocketer in a straight road.The\r\n\r\n  jeep is going at its maximum speed v.The pickpocketer rides on the motorcycle\r\n\r\n  of a waiting friend.When the jeep is at a distance 'd' away and the motorcycle starts\r\n\r\n  with a constant acceleration 'a'.Show that pickpocketer will be caught if \r\n\r\n  $v\\ge\\sqrt{2ad}$",
  "Solution_1": "[size=150][b]$1$[/b][/size].whatever happens, the net time taken has to be [b]$\\sqrt{\\frac{2(H - h)}{g}} + \\sqrt{\\frac{2h}{g}} \\geq 2 \\sqrt{\\frac{2}{g} \\sqrt{(H - h)(h)}} \\geq 2\\sqrt{\\frac{h}{g}}$[/b]\r\n the required ratio of [b]$\\frac{h}{H} = \\frac{1}{2}$[/b]\r\n\r\n [size=150][b]$2$[/b][/size]. Consider the frame of reference of the pickpocketer.\r\n w.r.t him, the inspector is slowing down with a decelration of [b]$a$[/b] ,\r\n having started with initial velocity[b] $v$.[/b]\r\n then, we require to find the minimum [b]$v$[/b] such that the inspector covers [b]$d$[/b].\r\n using third eq^n of kinematics, \r\n [b]$v^2 = 2ad$[/b],\r\n hence the result.",
  "Solution_2": "[quote=\"madatmath\"][size=150][b]$1$[/b][/size].whatever happens, the net time taken has to be [b]$\\sqrt{\\frac{2(H - h)}{g}} + \\sqrt{\\frac{2h}{g}} \\geq 2 \\sqrt{\\frac{2}{g} \\sqrt{(H - h)(h)}} \\geq 2\\sqrt{\\frac{h}{g}}$[/b]\n the required ratio of [b]$\\frac{h}{H} = \\frac{1}{2}$[/b]\n\n [size=150][b]$2$[/b][/size]. Consider the frame of reference of the pickpocketer.\n w.r.t him, the inspector is slowing down with a decelration of [b]$a$[/b] ,\n having started with initial velocity[b] $v$.[/b]\n then, we require to find the minimum [b]$v$[/b] such that the inspector covers [b]$d$[/b].\n using third eq^n of kinematics, \n [b]$v^2 = 2ad$[/b],\n hence the result.[/quote]\r\n\r\n1.your answer is right but why is $\\sqrt{\\frac{2}{g} \\sqrt{(H - h)(h)}} \\geq 2\\sqrt{\\frac{h}{g}}$\r\nhow did you conclude $\\frac{h}{H} = \\frac{1}{2}$\r\n\r\n2.if the inspector covers the distance d with the minimum v then the pickpocketer could have covered some distance .then how could he catch the p.p",
  "Solution_3": "$1$. The minimum value of $(H - h)(h)$ is $h^2/4$and occurs at\r\n        $h = H/2$. this is easily verifiable by differentiation.\r\n \r\n$2$. Not with minimum $v$.\r\n        you see, we are talking in the pickpocketer's frame of reference,\r\n        and the pickpocketer does not move w.r.t himself. He watches the inspector being the distance $d$ away, and  then actually decelerating as he moves towards the p.p. this is becoz' in the ground frame of ref. , the p.p. is acclerating away frm the inspector.",
  "Solution_4": "[quote=\"madatmath\"]$1$. The minimum value of $(H - h)(h)$ is $h^2/4$and occurs at\n        $h = H/2$. this is easily verifiable by differentiation.\n \n$2$. Not with minimum $v$.\n        you see, we are talking in the pickpocketer's frame of reference,\n        and the pickpocketer does not move w.r.t himself. He watches the inspector being the distance $d$ away, and  then actually decelerating as he moves towards the p.p. this is becoz' in the ground frame of ref. , the p.p. is acclerating away frm the inspector.[/quote]\r\n\r\nyes.thanks for the solution",
  "Solution_5": "[quote=\"bkunchanapalli\"][quote=\"madatmath\"]$1$. The minimum value of $(H - h)(h)$ is $h^2/4$and occurs at\n        $h = H/2$. this is easily verifiable by differentiation.\n \n$2$. Not with minimum $v$.\n        you see, we are talking in the pickpocketer's frame of reference,\n        and the pickpocketer does not move w.r.t himself. He watches the inspector being the distance $d$ away, and  then actually decelerating as he moves towards the p.p. this is becoz' in the ground frame of ref. , the p.p. is acclerating away frm the inspector.[/quote]\n\nyes.thanks for the solution[/quote]\r\n\r\nwait a minute\r\ni think after differentiation it should be $H^2$/4 instead of $h^2$/4",
  "Solution_6": "true, true.\r\n my mistake. i think i forgot to hit the shift key   :oops: \r\n the correct answer is $2 \\sqrt{\\frac{H}{g}}$",
  "Solution_7": "[quote=\"madatmath\"]true, true.\n my mistake. i think i forgot to hit the shift key   :oops: \n the correct answer is $2 \\sqrt{\\frac{H}{g}}$[/quote]\r\n\r\nit's ok once again thanks for the solution"
}
{
  "Tag": [
    "function"
  ],
  "Problem": "Let $ n=2^{k-1}(2^{k}-1)$ be a natural, where $ k$ is a positive integer such that $ 2^{k}-1$ is prime.\r\n Compute the number of divisors of $ n$ and the sum of them.",
  "Solution_1": "[hide]That is the general form for all perfect numbers that we know of (all even ones): http://en.wikipedia.org/wiki/Perfect_number.  \n\nThere are $ (k-1+1)2=2k$ factors. \n\nThe sum of the factors is:\n\n$ (1+2+4...+2^{k-1})(1+2^{k}-1)$\n$ (2^{k}-1)2^{k}$\n\nAnd if we subtract the number itself we get the original number, which is the definition of a perfect number.[/hide]",
  "Solution_2": "Yes, A Wanted To Compare ...",
  "Solution_3": "[quote=\"jonysatie\"]Let $ n=2^{k-1}(2^{k}-1)$ be a natural, where $ k$ is a positive integer such that $ 2^{k}-1$ is prime.\n Compute the number of divisors of $ n$ and the sum of them.[/quote]\r\n\r\nThe number theoretic function $ \\sigma (n)$ which is the sum of the divisors is [b]weakly multiplicative[/b]. Meaning $ \\sigma (ab) = \\sigma(a) \\sigma(b)$ whenever $ \\gcd(a,b)=1$.\r\n\r\nSo, $ \\sigma \\left( 2^{k-1}(2^{k}-1) \\right) = \\sigma (2^{k-1}) \\cdot \\sigma (2^{k}-1)$\r\n\r\nNow, $ \\sigma (2^{k}-1)=2^{k}$ because it is prime.\r\n\r\nUnd, $ \\sigma (2^{k-1}) = 1+2+2^{2}+...+2^{k-1}= 2^{k}-1$\r\n\r\nSo their product is,\r\n$ 2^{k}(2^{k}-1)$\r\n\r\nWhich is TWICE as much as the number itself. So the number is perfect."
}
{
  "Tag": [
    "calculus",
    "integration",
    "trigonometry",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Compute:\r\n$ \\int_{ \\minus{} \\infty}^{ \\plus{} \\infty} \\frac {e^{ix t} }{1 \\plus{} i x} \\, \\mathbf{d}x$",
  "Solution_1": "${ \\textbf I\\;=\\;\\int_{-\\infty}^\\infty}\\;\\dfrac{e^{\\imath \\,x\\,t}}{1+\\imath\\,x}\\;\\textbf dx$\r\n\r\n${ =\\; \\int_{-\\infty}^\\infty} \\dfrac {\\left[\\cos\\,(x\\,t)+\\imath\\,\\sin\\,(x\\,t)\\right]\\cdot\\left(1-\\imath\\,x\\right) } {1+x^2}\\;\\textbf dx$\r\n\r\n${{ \\;=\\int_{-\\infty}^\\infty} \\left[ \\dfrac{\\cos\\,(x\\,t)+x\\,\\sin\\,(x\\,t)}{1+x^2}\\right]\\textbf dx\\;+\\;\\imath\\;\\int_{-\\infty}^\\infty} \\left[ \\dfrac{\\sin\\,(x\\,t)-x\\,\\cos\\,(x\\,t)}{1+x^2}\\right]\\textbf dx$\r\n\r\n${{ \\;=\\int_{-\\infty}^\\infty} \\left[ \\dfrac{\\cos\\,(x\\,t)-\\dfrac{d}{dt}\\cos\\,(x\\,t)}{1+x^2}\\right]\\textbf dx\\;+\\;\\imath\\;\\int_{-\\infty}^\\infty} \\left[ \\dfrac{\\sin\\,(x\\,t)-\\dfrac{d}{dt}\\sin\\,(x\\,t)}{1+x^2}\\right]\\textbf dx$\r\n\r\n$ =\\;\\left(\\pi\\,e^{-t} -\\dfrac{d}{dt}\\left(\\pi\\,e^{-t}\\right)\\right)+\\;\\imath\\left( 0 -\\dfrac{d}{dt} (0)\\right)$\r\n\r\n$ =\\;\\boxed{2\\pi\\,e^{-t}}$",
  "Solution_2": "Very nice :D Thank you!\r\nBut isn't it true only for t>0 ? :huh:",
  "Solution_3": "[quote=\"@petko\"]Very nice :D Thank you!\nBut isn't it true only for t>0 ? :huh:[/quote]\r\nWhen t<0, the answer is zero. When t=0, the integral does not converge, however it is reasonable to set its value $ \\pi$.\r\nYou can see the conclusion easily from the following (rough) derivation\r\n$ \\int_{ \\minus{} \\infty}^{ \\plus{} \\infty}\\frac {e^{ix t} }{1 \\plus{} i x}dx \\equal{} \\int_{ \\minus{} \\infty}^{ \\plus{} \\infty}\\int_{ \\minus{} \\infty}^t e^{i x u \\plus{} u \\minus{} t}du dx \\equal{} \\int_{ \\minus{} \\infty}^t 2\\pi\\delta(u)e^{u \\minus{} t} du$",
  "Solution_4": "[b]Buffalo[/b] could you explain in more detail the very last part of your equation, please?\r\nI mean:\r\n\r\n$ \\int\\limits_{ \\minus{} \\infty}^{ \\plus{} \\infty}\\int\\limits_{ \\minus{} \\infty}^t e^{i x u \\plus{} u \\minus{} t}du dx \\equal{} \\int\\limits_{ \\minus{} \\infty}^t 2\\pi\\delta(u)e^{u \\minus{} t} du \\stackrel{\\text{is it true that...?}}{\\Rightarrow} \\int\\limits_{ \\minus{} \\infty}^{ \\plus{} \\infty} e^{i x u} dx \\equal{} 2 \\pi \\delta(u)$\r\n\r\n:?: :huh:",
  "Solution_5": "I think that if t=0 the integral is $ 2\\pi$."
}
{
  "Tag": [
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Is there a subset of $ E\\subset [0,1]\\times [0,1] \\times [0,1]$ such that $ m(E)>0$ and $ E$ is connected, compact, and nowhere dense?",
  "Solution_1": "Let $ A$ be a \"fat\" Cantor set in $ [0,1]$ - construct it by deleting middle intervals, but arrange the sizes of the deleted intervals so that $ m(A)>0.$ Now let ${ E=\\{(x,y,z}: x\\in A \\text{ or }y\\in A\\text{ or }z\\in A\\}.$ This should be path-connected, hence connected, and compact and nowhere dense should be true as well.",
  "Solution_2": "Oh. This was what I was fiddling with, but I couldn't quite tweak it properly. I didn't think about getting connected from allowing the other two dimensions to vary at each point."
}
{
  "Tag": [
    "algorithm",
    "floor function"
  ],
  "Problem": "Show that if a is an integer and d is a positive integer then when a divides d the quotient is floor (a/d) and remainder is a - d * floor (a/d)",
  "Solution_1": "[quote=\"Christine\"]Show that if a is an integer and d is a positive integer then when a divides d the quotient is floor (a/d) and remainder is a - d * floor (a/d)[/quote]\r\n\r\n[hide=\"Not sure\"]I guess this really depends on how you define quotient. If you define it as the largest multiple of $ d$ that divides $ a$, then it is obviously $ \\lfloor a/d \\rfloor$. So:\n\n$ a/d \\equal{} \\lfloor a/d \\rfloor \\plus{} r/d$\n\nSolve for $ r$ to get $ r\\equal{}a\\minus{}d\\cdot\\lfloor a/d \\rfloor$.[/hide]",
  "Solution_2": "[hide=\"Part 2\"]The remainder is $ a \\minus{} n$, where $ n$ is the largest multiple of $ d$ less than $ a$. This is $ d\\left\\lfloor\\frac {a}{d}\\right\\rfloor$, so the remainder is $ a \\minus{} d\\left\\lfloor\\frac {a}{d}\\right\\rfloor$.[/hide]"
}
{
  "Tag": [
    "LaTeX",
    "\\/closed"
  ],
  "Problem": "I would like to copy some LaTeX code in the private message that I sent. But I can't see quote for the messages in the Sentbox, Outbox and Savebox, Why?",
  "Solution_1": "[quote=\"fuzzylogic\"]I can't see quote for the messages in the Sentbox, Outbox and Savebox, Why?[/quote]\r\n\r\nI don't know about savebox, but outbox and sentbox is stuff that you sent, so usually you ain't quoting yourself :)\r\n\r\nThe thing doesn't think of latex apparently :)",
  "Solution_2": "[quote=\"fuzzylogic\"]I would like to copy some LaTeX code in the private message that I sent. But I can't see quote for the messages in the Sentbox, Outbox and Savebox, Why?[/quote]if you are really lazy, you can send yourself that pm (works!) and then quote yourself :D",
  "Solution_3": "[quote=\"Valentin Vornicu\"][quote=\"fuzzylogic\"]I would like to copy some LaTeX code in the private message that I sent. But I can't see quote for the messages in the Sentbox, Outbox and Savebox, Why?[/quote]if you are really lazy, you can send yourself that pm (works!) and then quote yourself :D[/quote]\r\n\r\nI don't understand what you mean. I sent a message before. So there is a saved copy in the Sentbox. How do I reply that pm to myself?\r\n\r\nI am pretty sure it's very easy for the administrator to add a \"post reply\" or \"quote\" button, but they are not there right now.",
  "Solution_4": "fuzzylogic, I tried it out, and in my box there are quote buttons there. Just check it out, they're at the place where reply/newtopic buttons usually are..",
  "Solution_5": "[quote=\"fuzzylogic\"][quote=\"Valentin Vornicu\"][quote=\"fuzzylogic\"]I would like to copy some LaTeX code in the private message that I sent. But I can't see quote for the messages in the Sentbox, Outbox and Savebox, Why?[/quote]if you are really lazy, you can send yourself that pm (works!) and then quote yourself :D[/quote]\n\nI don't understand what you mean. I sent a message before. So there is a saved copy in the Sentbox. How do I reply that pm to myself?\n\nI am pretty sure it's very easy for the administrator to add a \"post reply\" or \"quote\" button, but they are not there right now.[/quote]I am one of the Administrators, and I was there :) actually I added the quote buttons today, but I didn't yet get them to work. Have a little patience :)",
  "Solution_6": "If you really need it you can just go to [u]V[/u]iew -> Sour[u]c[/u]e in your menu\r\n\r\nthere you should find the latex code somewhere ( ... alt=\"$here_the_latex_code$\"...)",
  "Solution_7": "[quote=\"darij grinberg\"]\n[quote=\"fuzzylogic\"]I strongly suggest admin to let us see the quote of our own PMs.[/quote]\n\nI agree. Note that the \"quote message\" button is there, but strangely enough it just leads back to the inbox.\n\n  Darij[/quote]\r\n\r\nAdmin, could you fix the problem?"
}
{
  "Tag": [
    "trigonometry"
  ],
  "Problem": "Show that the triangle whose angles satisfy the equality \\[\\frac{\\sin^{2}{A}+\\sin^{2}{B}+\\sin^{2}{C}}{\\cos^{2}{A}+\\cos^{2}{B}+\\cos^{2}{C}}=2\\]\r\nis a right-angled triangle .",
  "Solution_1": "Trivial by the well-known identity $\\cos^{2}A+\\cos^{2}B+\\cos^{2}C = 1-2\\cos A\\cos B\\cos C$.",
  "Solution_2": "[hide]\n$\\frac{3-(\\cos^{2}A+\\cos^{2}B+\\cos^{2}C)}{\\cos^{2}A+\\cos^{2}B+\\cos^{2}C}=2$\n$3-(\\cos^{2}A+\\cos^{2}B+\\cos^{2}C)=2(\\cos^{2}A+\\cos^{2}B+\\cos^{2}C)$\n$3=3(\\cos^{2}A+\\cos^{2}B+\\cos^{2}C)$\n$\\cos^{2}A+\\cos^{2}B+\\cos^{2}C=1$\n$1-2\\cos A\\cos B\\cos C=1$\n$\\cos A\\cos B\\cos C=0$\nSo one of those cosines has to be 0 meaning one of the angles is right.\n[/hide]"
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "geometric transformation",
    "homothety",
    "cyclic quadrilateral",
    "power of a point",
    "radical axis"
  ],
  "Problem": "Quadrilateral ABCD is inscribed in circle k centered at O. The bisector of angle ABD meets AD and k at points K and M, respectively. The bisector of angle CBD meets CD and k at points L and N, respectively. Suppose that KL ll MN.  Denote midpoint of BD with S. Prove that fopur points M, O, N and S are concyclic.",
  "Solution_1": "What you need to prove is to show that $MBND$ is harmonic.",
  "Solution_2": "Let $(P)$ be the circumcircle of the triangle $\\triangle MON$ intersecting the center line $OP$ of the circles $(O), (P)$ at another point $Q$ different from $O.$  Since $OQ$ is a diameter of $(P),$ $OM \\perp QM,\\ ON \\perp QN$ and $QM, QN$ are tangents to $(O)$ from $Q.$ Hence\r\n\r\n$\\angle QMN = \\angle QNM = \\angle MAN = \\frac{\\angle ABC}{2}$\r\n \r\n$\\angle QMN = \\angle QNM = \\frac{\\angle MON}{2}$\r\n\r\nIt follows that $\\angle MON = \\angle ABC$ and from the cyclic quadrilateral $OMQN,\\ ABCD,$ $\\angle MQN = \\angle ADC \\equiv \\angle KDL.$ Let $(O)$ meet the center line $OP$ at $U, V$ outside and inside of $(P),$ respectively, and let a parallel to $UV \\equiv OP$ through $D$ meet $(O)$ again at $T.$ Then\r\n\r\n$\\angle AOT = 180^\\circ-(2 \\angle MOV+\\angle VOD)+\\angle UOT = 180^\\circ-2 \\angle MOV$ \r\n\r\n$\\angle COT = 180^\\circ-(2 \\angle NOV-\\angle VOD)-\\angle UOT = 180^\\circ-2 \\angle NOV$ \r\n\r\nIt follows that $T$ is the midpoint of the arc $AC$ opposite to $D,$ so that $DT \\parallel OP$ bisects the angle $\\angle ADC \\equiv \\angle KDL.$  The radical axis of $(O), (P)$ is perpendicular to their center line, $MN \\perp OP.$ By the problem condition, $KL \\parallel MN.$ Combining, we get $DT \\perp KL,$ which means that the triangle $\\triangle KDL$ is isosceles with $DK = DL.$ As a result, the isosceles triangles $\\triangle KDL \\sim \\triangle MQN$ with $\\angle KDL = \\angle MQN$ are centrally similar with the similarity center $B \\equiv KM \\cap LN.$ Thus the line $QD$ connecting their remaining corresponding vertices $Q, D$ passes through the similarity center $B.$ Let $QD \\equiv BD$ meet $(P)$ again at $S.$ Since $OQ$ is a diameter of $(P),$ the angle $\\angle QSO \\equiv \\angle DSO = 90^\\circ$ is right, which means that $S$ is the midpoint of the chord $BD.$",
  "Solution_3": "Consider the homotety centered at $B$ which sends $KL$ into $MN$. \r\nIt will map the point $D$ into a point $X$ which is the intersection of the tangents to the circumcircle of $ABCD$ at $M$ and $N$. It's easy to see that $MOSNX$ is cyclic and we are done."
}
{
  "Tag": [
    "function",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "Determine all functions $ f: \\mathbb{R} \\mapsto \\mathbb{R}$ such that\r\n\\[ x f(x \\plus{} xy) \\equal{} x f(x) \\plus{} f \\left( x^2 \\right) f(y) \\quad  \\forall  x,y \\in \\mathbb{R}.\\]",
  "Solution_1": "$ x \\equal{} y \\equal{} 0$ gives $ f(0)^2 \\equal{} 0$, so $ f(0) \\equal{} 0$.\r\n\r\n$ x \\equal{} 1, y \\equal{} \\minus{}1$ gives $ f(1)f(\\minus{}1) \\equal{} \\minus{}f(1)$. Either $ f(1) \\equal{} 0$ or $ f(\\minus{}1) \\equal{} \\minus{}1$.\r\n\r\nSuppose $ f(1) \\equal{} 0$. Then $ x \\equal{} 1$ gives $ f(1\\plus{}y) \\equal{} 0$, or $ f \\equiv 0$. This is a solution. Now assume it away. $ f(\\minus{}1) \\equal{} \\minus{}1$.\r\n\r\n$ y \\equal{} \\minus{}1$ gives $ f(x^2) \\equal{} xf(x)$. Put $ x \\equal{} \\minus{}1$ in to get $ f(1) \\equal{} 1$.\r\n\r\n$ x \\equal{} 1$ gives $ f(y\\plus{}1) \\equal{} f(y)\\plus{}1$.\r\n\r\nSubstituting $ f(x^2)$ for $ xf(x)$ gives $ xf(x\\plus{}xy) \\equal{} xf(x) \\plus{} xf(x)f(y)$. Let's only allow $ x$ to be nonzero, and then the equation reduces to $ f(x(y\\plus{}1)) \\equal{} f(x) \\plus{} f(x)f(y)$, which reduces to $ f(xy) \\equal{} f(x)f(y)$ when $ y$ is shifted by $ 1$.\r\n\r\nPut in $ x \\equal{} y$ in this new equation to get $ f(x^2) \\equal{} f(x)^2$, or $ xf(x) \\equal{} f(x)^2$. So for all $ x$, $ f(x) \\equal{} 0$ or $ f(x) \\equal{} x$. If $ f(a) \\equal{} 0$ for some $ a \\neq 0$, then $ f(xa) \\equal{} f(x)f(a) \\equal{} 0$. $ xa$ can take any nonzero real, so $ f \\equiv 0$ in this case.\r\n\r\nWe end up with $ f(x) \\equal{} x$ or $ f(x) \\equal{} 0$ as our solutions.",
  "Solution_2": "[quote=MellowMelon]$ x \\equal{} y \\equal{} 0$ gives $ f(0)^2 \\equal{} 0$, so $ f(0) \\equal{} 0$.\n\n$ x \\equal{} 1, y \\equal{} \\minus{}1$ gives $ f(1)f(\\minus{}1) \\equal{} \\minus{}f(1)$. Either $ f(1) \\equal{} 0$ or $ f(\\minus{}1) \\equal{} \\minus{}1$.\n\nSuppose $ f(1) \\equal{} 0$. Then $ x \\equal{} 1$ gives $ f(1\\plus{}y) \\equal{} 0$, or $ f \\equiv 0$. This is a solution. Now assume it away. $ f(\\minus{}1) \\equal{} \\minus{}1$.\n\n$ y \\equal{} \\minus{}1$ gives $ f(x^2) \\equal{} xf(x)$. Put $ x \\equal{} \\minus{}1$ in to get $ f(1) \\equal{} 1$.\n\n$ x \\equal{} 1$ gives $ f(y\\plus{}1) \\equal{} f(y)\\plus{}1$.\n\nSubstituting $ f(x^2)$ for $ xf(x)$ gives $ xf(x\\plus{}xy) \\equal{} xf(x) \\plus{} xf(x)f(y)$. Let's only allow $ x$ to be nonzero, and then the equation reduces to $ f(x(y\\plus{}1)) \\equal{} f(x) \\plus{} f(x)f(y)$, which reduces to $ f(xy) \\equal{} f(x)f(y)$ when $ y$ is shifted by $ 1$.\n\nPut in $ x \\equal{} y$ in this new equation to get $ f(x^2) \\equal{} f(x)^2$, or $ xf(x) \\equal{} f(x)^2$. So for all $ x$, $ f(x) \\equal{} 0$ or $ f(x) \\equal{} x$. If $ f(a) \\equal{} 0$ for some $ a \\neq 0$, then $ f(xa) \\equal{} f(x)f(a) \\equal{} 0$. $ xa$ can take any nonzero real, so $ f \\equiv 0$ in this case.\n\nWe end up with $ f(x) \\equal{} x$ or $ f(x) \\equal{} 0$ as our solutions.[/quote]\n\nCan you pleas explain \" which reduces to $ f(xy) = f(x)f(y)$ when y is shifted by 1\"\nI just cant figure out what you did there. Help please ;)",
  "Solution_3": "$f(x(y+1)) = f(x) + f(x)f(y)$ is equivalent to $f(xy) = f(x) + f(x)f(y-1) = f(x)(1 + f(y-1) = f(x)f(y)$ when you replace $y+1$ with $y$.",
  "Solution_4": "Let $P(x,y)$ be the assertion into the problem statement.  \n$P(0,y)\\implies f(y)f(0)=0.$  Either $\\boxed{f(x)=0}$ for all $x$ or we have $f(0)=0$.  Assume that there is some $x$ such that $f(x)\\neq 0$; thus, we have $f(0)=0$.\n$P(1,-1)\\implies 0=f(1)+f(1)f(-1)\\implies f(1)(1+f(-1))=0.$  Either $f(1)=0$ or $f(-1)=-1.$\n\n[b]Case 1:[/b] $f(1)=0$.\n$P(1,y)\\implies f(y+1)=0$ but this doesn't satisfy the assumption that there exists some $x$ such that $f(x)\\neq 0$.  We already counted this solution before the assumption of $f(0)=0$.\n\n[b]Case 2:[/b] $f(-1)=-1$\n$P(-1,-1)\\implies 0=1-f(1)$ so $f(1)=1$. \n\nThus, $P(1,y)\\implies f(y+1)=f(y)+1$.\n$P(x,-1)\\implies xf(x)=f(x^2)$. \n Substituting these two identities into the problem yields $xf(x+xy)=xf(x)+f(x^2)f(y)=xf(x)(1+f(y))=xf(x)(f(y+1))$ so $$f(x+xy)=f(x)(f(y+1))$$ given that $x\\neq 0.$  \nLet $Q(x,y)$ be the assertion into this new function.\n$Q(x,x-1)\\implies f(x^2)=(f(x))^2$.  \nHence, $(f(x))^2=f(x^2)=xf(x)$ so $$f(x)(f(x)-x)=0.$$\nNow, we've run into the Pointwise Values Trap. We claim that the only possible solution from this if $\\boxed{f(x)=x}$ and pathological solutions do not exist. Suppose for the sake of contradiction that there exists distinct values $s$ and $t$ such that $f(s)=s$ and $f(t)=0$ for $s,t\\neq 0$(since they still can be in the same function).  Then, $f(s+st)=s+st$ or $f(s+st)=0$.  We consider the revised function $$xf(x+xy)=xf(x)(f(y)+1).$$\n\n[b]Case 2.1:[/b] $f(s)=s, f(t)=0$, and $f(s+st)=s+st$\nLet $x=s$ and $y=t$ in the revised function yields $sf(s+st)=sf(s)(f(t)+1)\\implies s^2+s^2t=s^2$ or $s^2t=0$ so we must have either $s=0$ or $t=0$, contradiction.\n\n[b]Case 2.2:[/b] $f(s)=s, f(t)=0$, and $f(s+st)=0$.\nLet $x=s$ and $y=t$ in the revised function yields $sf(s+st)=sf(s)(f(t)+1)\\implies 0=s^2$ or $s=0$, contradiction.\n\nHence, we have proven that $f(x)=x$ is the only solution that is not identically zero and our two solutions are $f(x)=x$ and $f(x)=0.$ $\\blacksquare$\n",
  "Solution_5": "all the solutions are beautiful but i found an easier solution",
  "Solution_6": "Let $P(x,y)$ the assertion:\n$P(0,0)$\n$f(0)^2=0 \\implies f(0)=0$\n$P(1,-1)$\n$-f(-1)f(1)=f(1) \\implies$ $f(-1)=-1$ or $f(1)=0$\nIf $f(-1)=0$:\n$P(1,x-1)$\n$f(x)=0$ for all $x \\in \\mathbb R$\nIf $f(-1)=-1$\n$P(x,-1)$\n$xf(x)=f(x^2)$.\n$P(x+1,x)$\n$(x+1)f((x+1)^2)=f((x+1)^2)(1+f(x)) \\implies x+1=1+f(x) \\implies f(x)=x$ for all $x \\in \\mathbb R$.\nOr also implies $f((x+1)^2)=0$ but we found $f(x)=0$ is another solution soo..\n$f(x)=x$ or $f(x)=0$ for all $x \\in \\mathbb R$",
  "Solution_7": "[quote=MathLuis]$(x+1)f((x+1)^2)=f((x+1)^2)(1+f(x)) \\implies x+1=1+f(x) \\implies f(x)=x$ for all $x \\in \\mathbb R$[/quote]\nNo.\n$(x+1)f((x+1)^2)=f((x+1)^2)(1+f(x)) $ implies :\n$\\forall x\\in\\mathbb R$, either $f((x+1)^2)=0$, either $f(x)=x$\n\n\n",
  "Solution_8": "[quote=orl]Determine all functions $ f: \\mathbb{R} \\mapsto \\mathbb{R}$ such that\n\\[ x f(x \\plus{} xy) \\equal{} x f(x) \\plus{} f \\left( x^2 \\right) f(y) \\quad  \\forall  x,y \\in \\mathbb{R}.\\][/quote]\n\nCan anyone check this proof? \nThe only solutions are $f\\equiv \\text{Id}$ and $f\\equiv 0$ which clearly works.We will now prove that indeed these are the only solutions.\n\nLet $P(x,y)$ denote the assertion $x f(x + xy) = x f(x) + f \\left( x^2 \\right) f(y)$.$\\qquad$.           We proceed in the following steps:-\n              Note that $P(0,0)$ gives $f(0)=0$.From $P(x,-1)$ we get $f(-1)f(x^2)+xf(x)=0$.       Now split into two cases; If $f(-1)=0$ then we get one of the solutions $f\\equiv 0$.Otherwise Compare $P(-1,x)$ and $P(-1,-x)$ to see $xf(x)=-xf(-x)$ which implies $f$ is odd.\n\nNow $x=1$ in the above para easily gives $f(1)=1$.Thus $f(-1)=-1$ and hence $f(x^2)=xf(x)$ and  $P(1,y)$ gives $f(y+1)=f(y)+1$.To finish up $$P(x,x-1)\\implies x^2f(x)=xf(x^2)=xf(x)+f(x^2)f(x-1)=xf(x)+f(x^2)f(x-1)=xf(x)+xf(x)(f(x)-1) \\implies f(x)=x \\forall x\\neq 0 $$ but since $f(0)=0$ hence we conclude $f(x)=x$ for all $x\\in \\mathbb{R}$.$\\square$",
  "Solution_9": "[quote=Pluto1708]$x^2f(x)=......=xf(x)+xf(x)(f(x)-1) \\implies f(x)=x \\forall x\\neq 0 $\nNo\n$x^2f(x)=xf(x)^2$ implies \n$\\forall x\\ne 0$, either $f(x)=0$, either $f(x)=x$\n\n",
  "Solution_10": "[quote=pco][quote=MathLuis]$(x+1)f((x+1)^2)=f((x+1)^2)(1+f(x)) \\implies x+1=1+f(x) \\implies f(x)=x$ for all $x \\in \\mathbb R$[/quote]\nNo.\n$(x+1)f((x+1)^2)=f((x+1)^2)(1+f(x)) $ implies :\n$\\forall x\\in\\mathbb R$, either $f((x+1)^2)=0$, either $f(x)=x$[/quote]\n\nYes @pco i made a mistake in that part i have done yet the case when $f(x)=0$ soo that doesnt make a problem i think",
  "Solution_11": "i will correct it right now\n",
  "Solution_12": "$xf(x+xy)=xf(x)+f(x^2)f(y)$\n$P(0,0)$ $\\implies$ $f(0)=0$\n$P(-1,-1)$ $\\implies$ $f(-1)=0$ or $f(1)=1$\nIf $f(-1)=0$ then $P(x,-1)$ $\\implies$ $f(x)=0$ which is a solution.\nIf $f(1)=1$ \n$P(1,x)$ $\\implies$ $f(x+1)=1+f(x)$\n$P(x,x-1)$ \n$xf(x^2)=xf(x)+f(x^2)f(x-1)=xf(x)+f(x^2)(f(x)-1)$\n$xf(x^2)=xf(x)+f(x^2)f(x)-f(x^2)$\n$f(x^2)=\\frac{xf(x)}{x+1-f(x)}$\nit is obvious that $f(x) \\ne x+1$\nLet's substitute this with original equation\n$xf(x+xy)=xf(x)+\\frac{xf(x)f(y)}{x+1-f(x)}$\nCall it's assertion $Q(x,y)$\n$Q(1,-1)$  $\\implies$ $f(-1)=-1$\n$Q(-1,x)$ $\\implies$ $f(x)=-f(-x)$\n$Q(-x,-1)$ $\\implies$ $x+1-f(x)=1$ \nso that, $f(x)=x$ \nWe have pointwise trap which is easy to get contradicition.\nSols:$f(x)=0,x$"
}
{
  "Tag": [],
  "Problem": "Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 2.5\u00d710^4 m/s when at a distance of 2.3\u00d710^11 m from the center of the sun, what is its speed when at a distance of 5.9\u00d710^10 m .",
  "Solution_1": "[hide=\"HINT\"]Conversation of angular momentum.[/hide]",
  "Solution_2": "ya thats correct!"
}
{
  "Tag": [
    "linear algebra",
    "matrix",
    "vector",
    "LaTeX",
    "linear algebra unsolved"
  ],
  "Problem": "Hello, Can someone help me on any of these problems ?\r\n\r\n1) Let A be a matrix with n columns. Prove that Null(A)=R^n if and only if A=0\r\n\r\n2) Let A and B be matrices with n columns. Prove that A=B if and only if Ax=Bx, for all x in R^n\r\n\r\n3)Let V=R^2={(a) (b)* | a,b is contain in R } be the vector space of column vectors of length 2.\r\n                                                                                               Show that W={(a) (b)* | a,b is contain/element of Z}is not a vector subspace of V\r\n\r\nI would appreciate any help and thank you in advance\r\n\r\n* the (b) should be written below the letter \"a\" but I don't know how to use the computer to do it like that. I am not very good at computers  :blush:",
  "Solution_1": "for the first and second questions ,start plugging in special vectors in $\\mathbb{R}^n$ :\r\n\r\nlike these : $v_{i}=\\left[\\begin{matrix} 0\\\\ \\vdots\\\\ 1\\\\ \\vdots \\\\0 \\\\ \\end{matrix}\\right]$ with zeroes everywhere except on the i th position\r\n\r\n\r\nI am terrible with computers yet these people here helped me  :oops:  to write in a much neater way : \r\n\r\nPress the button \"Latex Help\" on top of your screen, it will take you to this link:\r\n[url]http://www.mathlinks.ro/LaTeX/AoPS_L_About.php[/url]",
  "Solution_2": "Thanks for your reply fredbel6, I'll try to learn and use Latex next time as only Question 3 require the use of Latex anyway.\r\n\r\nMeanwhile, I really want to know what the solution is for Question 1 and Question 2, but I am more interested to know what the solution is for Question 1  than Question 2 as I can't figure it out. However, if anyone who can solve Question 2 but not Question 1, feel free to post it here as I am dying to know the solution.\r\n\r\nCan you fredbel6 or anyone  help me on any of the first two questions ? I reallly really want to know it  :) \r\n\r\n(Question 1 and Question 2 doesn't need to use Latex for it , so the question is as it is :)",
  "Solution_3": "1) The dimension of $N(A)$ (the nullspace of $A$) is $n-r$ ($r$ is the rank) so $N(A) = \\mathbb{R}^n$ iff $r = 0$, i.e. $A = 0$.\r\n\r\n2) If $A = B$, then clearly $Ax = Bx$ for all $x \\in\\mathbb{R}^n$.  If $Ax = Bx$ for all $x \\in\\mathbb{R}^n$ then $(A-B)x = 0$ for all $x \\in\\mathbb{R}^n$, i.e. $A-B$ is the zero matrix, and this is exactly what we want.\r\n\r\n3) So $V = \\{(a,b) | a,b\\in\\mathbb{R}\\}$ and $W = \\{(a,b) | a,b\\in\\mathbb{Z}\\}$.  If $W$ is a subspace of $V$, then for instance, $c(a,b) \\in W$ for any $c \\in\\mathbb{R}$ and $(a,b) \\in V$ which is absurd."
}
{
  "Tag": [
    "Princeton",
    "college",
    "geometry",
    "email",
    "2007",
    "function",
    "videos"
  ],
  "Problem": "In this thread you answer a question that's posted, and then you ask a generalized question to someone on this forum (whoever chooses to answer it), and so on. ( No math questions)\r\n ( and if you're on geobeecentral -yes i did copy this from there)\r\n\r\nI'll start---\r\n\r\nDo you like chocolate?",
  "Solution_1": "No, I'm allergic to it. \r\n\r\n(BTW, geobeecentral.com is a geography site with a forum. There are a few GBC members who are also AoPS members. )\r\n\r\nWhat is the most useless emoticon?",
  "Solution_2": ":arrow: I think it's this one....I don't know why..\r\n\r\nDo you have a sibling?",
  "Solution_3": "[img]http://www.venganza.org/forum/images/smiles/fsm7.gif[/img]\r\n\r\n\r\n\r\n[url=http://www.venganza.org/]That^[/url]\r\nWhat is the largest 20 digit prime?",
  "Solution_4": "Isn't that a math question?",
  "Solution_5": "[quote=\"nonie\"]\n\nDo you have a sibling?[/quote]\r\nyes, she's going into 6th grade.\r\n Do you have an i-pod?",
  "Solution_6": "yeah i do  :D \r\n\r\nWhat country do you live in?",
  "Solution_7": "The USA. What is your favorite kind of rodent?",
  "Solution_8": "[quote=\"nonie\"]Do you have a sibling?[/quote]\n\n[quote=\"PenguinIntegral\"][img]http://www.venganza.org/forum/images/smiles/fsm7.gif[/img]\n\n\n\n[url=http://www.venganza.org/]That^[/url][/quote]\r\n\r\nIs that your brother or your sister? :D",
  "Solution_9": "chinchillas\r\n\r\nif you are to be turned into a dessert of your choice(dont ask why), what would you be?",
  "Solution_10": "ice cream, I think..\r\n\r\n\r\nIf you could be any of the emoticons on this forum, which one would you be?",
  "Solution_11": ":spam:\r\n\r\nDo you listen to Pink Floyd?",
  "Solution_12": "No.\r\n\r\nDo you know who Bela Bartok is?",
  "Solution_13": "It sounds familiar.\r\n\r\nDo you own a pet reptile?",
  "Solution_14": "nope\r\n\r\ndo you know what Japanese guy is nicknamed \"Tsunami\"?",
  "Solution_15": "Because you PROBABLY are.. but you're never sure...\r\n\r\nwhy is  :P 's tongue pentagonular?",
  "Solution_16": "idk, my bff rose?\r\n\r\nWhy are GEICO's commercials so stupid?",
  "Solution_17": "Stupidity attracts money.\r\n\r\nIs money 1337?",
  "Solution_18": "No.\r\n\r\nWhat does 1337 mean?",
  "Solution_19": "17 15 4 1_4l\\lGl_l4G3 \\/\\/l-l3123 Y0l_l 741_1< 1_11<3 7l-l15.\r\n\r\nDoes anyone know what I just said?",
  "Solution_20": "It is a language where you talk like this.\r\n\r\nIf pro is the opposite of con, what is the opposite of Congress?",
  "Solution_21": "[quote=\"Brut3Forc3\"]It is a language where you talk like this.\n\nIf pro is the opposite of con, what is the opposite of Congress?[/quote]\r\nIt should be progress by logic, but by common sense it isn't.\r\n :huh: \r\n\\/\\/l-lY 4lVl 1 5711_1_ 741_1<1l\\lG 1_11<3 7l-l15?",
  "Solution_22": "Because you think its cool for some odd reason.\r\n\r\nHow much wood would a woodchuck chuck if a woodchuck could chuck would?",
  "Solution_23": "Who cares?\r\n\r\nWho cares?",
  "Solution_24": "The Woodchuck International Theological Society (TWITS) does.\r\n\r\nWhat question answers itself?",
  "Solution_25": "Is this a question?\r\n\r\nWhat does x = if 1=2 and 1=x?",
  "Solution_26": "Square both sides of the second equation to get x=4.\r\n\r\nMirror mirror, on t3h wall,\r\nWho's t3h 1337est of them all?",
  "Solution_27": "Your face.\r\n\r\nWhat is Zlarx?",
  "Solution_28": "xralZ backwards.\r\n\r\n:ddr:  :ninja:  :rotfl: ?",
  "Solution_29": "This thread has pretty much degenerated into spam as well.\r\nLocking this."
}
{
  "Tag": [
    "binomial coefficients"
  ],
  "Problem": "This was posted before, but I haven't seen a solution yet.\r\n\r\n\r\nProve that the Generalized Binomial Coefficients defined as:\r\n\r\n[tex]\\displaystyle {n\\choose k}_C = \\frac {\\prod^n_{i=1}C_i}{\\left(\\prod^k_{i=1}C_i\\right)\\left( \\prod^{n-k}_{i=1}C_i\\right)}[/tex] for [tex]1\\le k\\le n[/tex] are all integers, \r\n\r\nwhere [tex]\\{C_n\\}^{\\infty}_{n=1}[/tex] is a sequence of positive integers such that [tex]\\gcd(C_m,C_n)=C_{\\gcd(m,n)}[/tex].",
  "Solution_1": "let $p$ be an arbitrary prime.  for each $ i \\geq 1 $ let $ m_i $ (if it exists) be the smallest positive integer such that $ p^i | C_{m_i} $.  then if $ p^i | C_k $, where $ k = qm_i+r $, then $ p^i | (C_{m_i}, C_{qm_i+r}) = C_{(m_i, qm_i+r)} = C_{(m_i,r)} $, so $ r = 0 $ (else $(m_i, r)$ contradicts the minimality of $m_i$.  hence the only $ k $ for which $ p^i | C_k $ are the multiples of $ m_i $, and, in general, the number of $ C_j $ with $ j \\leq N $ for which $ p^i | C_j $ is $ [\\frac{N}{m_i}] $.  \r\n\r\nthis means that, in general, the highest power of $ p $ dividing $ \\prod^N_{i=1}C_i $ is $ \\sum^{\\infty}_{j = 1}[\\frac{N}{m_j}] $.  \r\n\r\nso we need to show $ \\sum^{\\infty}_{j = 1}[\\frac{n}{m_j}] \\geq \\sum^{\\infty}_{j = 1}[\\frac{k}{m_j}] + \\sum^{\\infty}_{j = 1}[\\frac{n-k}{m_j}] $\r\n\r\nthis is evident from the general fact that \r\n\r\n${ [\\frac{n}{r}}] \\geq [\\frac{k}{r}] + [\\frac{n-k}{r}] $",
  "Solution_2": "Yikes, I don't think this is pre-olympiad.",
  "Solution_3": "[quote=\"mna851\"]Yikes, I don't think this is pre-olympiad.[/quote]\r\n\r\nI disagree.  His techniques are just a fancier version of the ones you use to solve a problem like: \"Find how many zeroes are at the end of 100!\""
}
{
  "Tag": [
    "integration",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "$\\sum_{n\\geq 0}a_nz^n$ power serie with radius convergence 1\r\nwrite\r\n$f(z)=\\sum_{n\\geq 0}a_nz^n$ when $|z|<1$ \r\n\r\nsuppose\r\n\r\n$|f(z)|\\leq \\frac{1}{1-|z|}$ for any $|z|<1$\r\n\r\nProve that $|a_n|<(n+1)e$ for any $n\\geq 0$",
  "Solution_1": "Just write that $a_n=\\frac{1}{2\\cdot\\pi\\cdot r^n}{\\int_{0}^{2\\pi}{ f(re^{it})e^{-int}dt}}$ and then integrate and take $r=n/(n+1)$. I hope it's correct.",
  "Solution_2": ":first:"
}
{
  "Tag": [
    "integration",
    "abstract algebra",
    "algebra",
    "polynomial",
    "calculus",
    "derivative",
    "real analysis"
  ],
  "Problem": "Let $n , (n> 2) ,$ be a given integer and  $K\\in C[0,1]$ such that : \r\n[b]i)[/b]  $\\int_{0}^{1}K(t)\\; dt = 1$,\r\n[b]ii)[/b] there exists a system $c_{1},c_{2},...,c_{n}$ of real numbers such that $\\int_{0}^{1}K(t)f^{(n-1)}(t)\\; dt = \\sum_{j=1}^{n}c_{j}f\\left(\\frac{j}{n}\\right)\\; \\; ,\\; \\; \\forall \\; f\\in C^{n-1}[0,1] .$\r\n[i]Questions : [/i]\r\n[b]1.[/b] Prove that there exists at least one such continuous ,,[i]kernel[/i]\"  $K$   having above properties [b]i)[/b]--[b]ii)[/b] ;\r\n[b]2.[/b] Suppose that $S(n,j) , 0\\le j \\le n,$ are Stirling numbers of second kind, i.e. the coefficients from  $x^{n}=\\sum_{j=0}^{n}S(n,j)x(x-1)\\cdots (x-j+1) .$\r\nProve that for  $p \\in{\\mathbb N}$ one has \\[\\int_{0}^{1}K(t)t^{p}\\; dt = \\frac{S(n+p,n)}{n^{p}{n-1+p\\choose p}}\\; .\\] [b]3.[/b] Prove or disprove that $\\int_{\\frac{1}{n}}^{1}K(t)\\frac{dt}{t^{n}}=\\frac{n^{n}}{n!}$ and $\\int_{\\frac{1}{n}}^{1}K(t)\\frac{dt}{t^{n+1}}=\\frac{n^{n}}{n!}H_{n}$ where  $H_{n}: = \\sum_{j=1}^{n}\\frac{1}{j}\\; .$",
  "Solution_1": "I think something is wrong with the condition [b]ii)[/b].",
  "Solution_2": "[quote=\"Myth\"]I think something is wrong with the condition [b]ii)[/b].[/quote]\r\nHi Myth, many thanks ! Alex",
  "Solution_3": "Anyway, I can't believe it is true, since we can smoothly change $f$ on $[0,1/n]$, then LHS will change and RHS will remain unchanged. Am I missing something?",
  "Solution_4": "[quote=\"Myth\"]Anyway, I can't believe it is true, since we can smoothly change $f$ on $[0,1/n]$, then LHS will change and RHS will remain unchanged. Am I missing something?[/quote]\r\nFurther I consider the problem to be OK  ! Likewise, my opinion is that 0 instead 1/n it was  also good. \r\nPerhaps the reason  ist that the kernel K statisfies (after me)  the condition $ K(t)=0 $ , $ \\forall t \\in   [0,1/n] $.\r\nHINT : consider the linear functional  $ A[f] =\\frac{1}{(n-1)!}\\int\\limits_{0}^{1}K(t)f^{(n-1)}(t)\\equiv \\frac{1}{(n-1)!}\\sum\\limits_{j=1}^{n}c_j f\\left(\\frac{j}{n}\\right)  .\r\n$ Observe that $ A[h]=0 $ for all polynomials of degree $\\le n-2 . $\r\nMoreover  $ A[e_{n-1}]=1 $ where   $e_m(t)=t^m . $  Further find the explicit form of coefficients  $ c_1,c_2 ,...,c_n $ . It's  possible !  Recognize the functional $ A $,.... .\r\nAlex.",
  "Solution_5": "I don't like meaningless discussions!\r\n\r\nWe can smoothly change values of $f$ inside of intervals $(i/n,(i+1)/n)$, then LHS changes and RHS doesn't changes, therefore the equlity doesn't hold for all $f\\in C^{n-1}[0,1]$!",
  "Solution_6": "Not quite. It means only that $K$ is a polynomial of small degree between the points and satisfies some conditions at the points imposed on low order left and right derivatives (like in the formula $\\int_0^1 1\\cdot f'(t)\\,dt=f(1)-f(0)$ or in $\\int_{-1}^1 (1-|t|)f''(t)\\,dt=f(1)+f(-1)-2f(0)$).  ;)",
  "Solution_7": "I see. Indeed, I forgot about this trick, though it was in the course of numerical methods.",
  "Solution_8": "[quote=\"flip2004\"][quote=\"Myth\"]Anyway, I can't believe it is true, since we can smoothly change $f$ on $[0,1/n]$, then LHS will change and RHS will remain unchanged. Am I missing something?[/quote]\nFurther I consider the problem to be OK  ! Likewise, my opinion is that 0 instead 1/n it was  also good. \nPerhaps the reason  ist that the kernel K statisfies (after me)  the condition $ K(t)=0 $ , $ \\forall t \\in   [0,1/n] $.\nHINT : consider the linear functional  $ A[f] =\\frac{1}{(n-1)!}\\int\\limits_{0}^{1}K(t)f^{(n-1)}(t)\\equiv \\frac{1}{(n-1)!}\\sum\\limits_{j=1}^{n}c_j f\\left(\\frac{j}{n}\\right)  .\n$ Observe that $ A[h]=0 $ for all polynomials of degree $\\le n-2 . $\nMoreover  $ A[e_{n-1}]=1 $ where   $e_m(t)=t^m . $  Further find the explicit form of coefficients  $ c_1,c_2 ,...,c_n $ . It's  possible !  Recognize the functional $ A $,.... .\nAlex.[/quote]\r\nFor $ k\\in \\{1,2,...,n\\} $, $x_j=\\frac{j}{n} , 1\\le j\\le n ,$ denote ${  \\omega(x)=\\prod\\limits_{\\nu=1}^{n}(x-x_{\\nu})\\; \\; ,\\; \\; \\;\r\nl_k(x)=\\frac{\\omega(x)}{(x-x_k)\\omega^{\\prime}(x_k)}} \\; .\r\n$\r\nConsider the polynomial  $ H_k $ of degree  $\\le n-2$ defined as  $ H_k(x)= l_k(x) - \\frac{1}{\\omega^{\\prime}(x_k)}x^{n-1} . $ Because   $ l_k(x_j)=\\left\\{\\begin{array}{lcl}\r\n1 & , & j =k\\\\     \r\n0 & , & j\\ne k\r\n\\end{array}\\right. , $ and $A[e_{n-1}]=1, $ we obtain $ 0 = A[H_k]= c_k -\\frac{1}{\\omega{\\prime}(x_k)}, $ that is    $ c_k= \\frac{1}{\\omega{\\prime}(x_k)}. $ In other words  \\[  A[f]=\\frac{1}{(n-1)!}\\int\\limits_{0}^{1}K(t)f^{(n-1)}(t)\\; dt = \\frac{1}{(n-1)!}\\sum\\limits_{j=1}^{n}\\frac{f(x_j)}{\\omega^{\\prime}(x_j)} .\r\n\\]\r\n\r\n........."
}
{
  "Tag": [
    "conics",
    "ellipse",
    "hyperbola",
    "LaTeX",
    "parameterization"
  ],
  "Problem": "For the hyperbola $frac{x^{2}}{4}-\\frac{y^{2}}{12}$= 1\r\n\r\na)eccentricity\r\nb)coordinates of foci\r\nc)eqn of directrices",
  "Solution_1": "$\\LaTeX$ correction $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$\r\n[hide=\"HINT\"]\nJust use the definitions of the parameters of the hyperbola equation.\n[/hide]",
  "Solution_2": "Well the titles say Ellipse Equation so:\r\n\r\n[hide=\"Solutions\"]\n\n$E: \\frac{x^{2}}{4}+\\frac{y^{2}}{12}=1$\n\na) $e= \\sqrt{1-\\frac{4}{12}}= \\sqrt{\\frac{2}{3}}$\n\nb) $\\text{Foci}= (0, \\pm \\sqrt12 e)= (0, \\pm \\sqrt{8})$\n\nc) $y= \\pm \\sqrt{18}$\n\n[/hide]"
}
{
  "Tag": [
    "factorial"
  ],
  "Problem": "How many zeros are there at the end of the base 2 representation of $132!$?\r\n\r\nplease detail your  solution\r\n\r\nthanks!!!",
  "Solution_1": "[hide]If you multiply something in base 2 by $10_2$ (= $2_{10}$) then it adds another 0 to the end of the base 2 representation.\n\nSo, we only have to count how many powers of 2 are in 132!.\n\n132/2 = 66\n132/4 = 33\n132/8 = 16.5 (so 16 powers of 2)\n132/16 = 8.25 (so 8 powers of 2)\n132/32 = 4.125 (so 4 powers of 2)\n132/64 = 2.0625 (so 2 powers of 2)\n132/128 = 1.3125 (so 1 power of 2)\n\nSo there are 66 + 33 + 16 + 8 + 4 + 2 + 1 = 130 powers of 2 in 132!. That means that there are [b]130[/b] zeros at the end when written in base 2.\n\nIs that right...?[/hide]",
  "Solution_2": "This is Legendre formula see http://mathworld.wolfram.com/Factorial.html for details",
  "Solution_3": "When you say written in base two, do you mean in like computer code.\r\nThen the answer would be 2.  132 is divisble by 2. which correponds to 0.\r\n66 is divisble by 2. which correponds to 0.   33 is not divisble by 2. which correponds to 1.  16 is divisble by 2. which correponds to 0.  8 is divisble by 2. which correponds to 0.   4 is divisble by 2. which correponds to 0.  2 is divisble by 2. which correponds to 0.   1 is not divisble by 2. which correponds to 1.\r\n\r\nso we get 0010001.  We have to flip the numbers to put to binary.\r\n\r\nSo 132 =   1000100  in binary.  There are two zeros at the end.\r\n\r\n[size=24][/size]2",
  "Solution_4": "The problem asks for the number of trailing 0's of $132!$ in binary, not 132 ;)",
  "Solution_5": "thanks MaThWhIz2004,t\u00b5t\u00b5,zacharytmitchell for the help!!\r\n\r\nAnyone has a different solution?\r\n\r\nthanks!!",
  "Solution_6": "I was thinking about the ninth ciphering question at this year's UGA Math Meet.\r\n\r\nHow many zeroes are at the end of the base three decimal for 27!?\r\n\r\nSo, does the solution follow like...\r\n\r\n[hide=\"This\"]\n\nCount the number of threes.\n\n27 / 3 = 9\n\n27 / 9 = 3\n\n27 / 27 = 1.\n\n9 + 3 + 1 = 13.\n\n[/hide]\r\n\r\n?",
  "Solution_7": "[hide]I don't really think there are many alternate ways to do this type of problem.\n\nBasically what you're do when you're trying to find the number of zeroes at the end of a factorial is you think:  What would make a zero in this base system?\n\nIn Base 2, 2s\nIn Base 3, 3s\nIn Base 4, 2 2s\nIn Base 5, 5\nIn Base 6, 2*3\nIn Base 7, 7\nIn Base 8, 2*2*2\nIn Base 9, 3*3\nIn Base 10, 2*5\nIn Base 11, 11. \n...\nIn Base 14, 7*2\n\nBut now you start to think, which one is the limiting factor here (like in 2*5), because there is a 2 in every other number, there are many more twos than 5s.  So we're trying to count the number of 5s in base 10.  But we're talking about base 2 here so we're trying to count the number of 2s which occur.  132/2+132/4+132/8+132/16+132/32+132/64+132/128\n\nRound each individual part down\n\n66+33+16+8+4+2+1=130[/hide]",
  "Solution_8": "To find the number of terminal zeroes of a number $n!$ in base $b$, just divide powers of the largest prime factor of $b$ into $n$ until you start to get zeroes in the division.",
  "Solution_9": "[quote=\"erik-the-red\"]\nSo, does the solution follow like...\nThis\n?[/quote]\r\n\r\nYes, it does ;)\r\nPakman2012 explained why + t\u00b5t\u00b5's link and Elemennop's post."
}
{
  "Tag": [
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "$300$ people must elect one of three persons $A , B, C$. The first $200$ of them gave $90$ votes to $A$ , 60 to the second , $40$ votes to the third and  $5+5 = 10$ votes were invalid or white. To be elected from the first round, one person must have at least $127$ votes.Ih how many ways can  vote the other $100$ people , in order to elect person $A$ in the first round ? \r\n\r\n [u]Babis[/u]\r\n\r\n [i]The problem was given to students in a Greek technology Institute.[/i]",
  "Solution_1": "[hide]\n\n127-90 = 37\n\nSo, 63 of the votes are up in the air.\n\n63x4x3x2 = 1512 ways? Not sure.[/hide]"
}
{
  "Tag": [
    "inequalities"
  ],
  "Problem": "Prove that:\r\n\\[ 1+\\frac12+\\frac13+\\cdots+\\frac1{2^n-1} > \\frac{n}{2}\\ ,\\ \\forall n \\in \\mathbb{N^*}  \\]",
  "Solution_1": "Do you mean:\r\n$\\frac 1 2 + (\\frac1 3 + \\frac 1 4 ) + (\\frac 1 5 + \\frac 1 6 + \\frac 17 + \\frac 18) + ..+(\\frac 1{2^{n-1}+1}+...+\\frac 1{2^n}) \\\\ > \\frac 1 2 + 2 \\times \\frac 1 4 + 4 \\times \\frac 1 8 + ....+ 2^{n-1} \\times \\frac 1{2^n}= \\frac n 2$",
  "Solution_2": "[quote=\"luimichael\"]Do you mean:\n$\\frac 1 2 + (\\frac1 3 + \\frac 1 4 ) + (\\frac 1 5 + \\frac 1 6 + \\frac 17 + \\frac 18) + ..+(\\frac 1{2^{n-1}+1}+...+\\frac 1{2^n}) \\\\ > \\frac 1 2 + 2 \\times \\frac 1 4 + 4 \\times \\frac 1 8 + ....+ 2^{n-1} \\times \\frac 1{2^n}= \\frac n 2$[/quote]\r\n\r\nWell that proves the given problem too... since $1 > \\frac{1}{2^n}$... it is a pretty weak inequality as stated, though."
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "number theory proposed",
    "number theory"
  ],
  "Problem": "Consider a \ffinite sequence of integers $ a_1, a_2, \\ldots, a_n$. We say a polynomial $ W$ fits the sequence\r\nif $ W(i) \\equal{} a_i$ for $ i \\equal{} 1, 2,\\ldots, n$:\r\n(a) Does there necessarily (that is, for every \ffinite sequence $ a_i$) exist a polynomial $ W$ with\r\ninteger coefficients which fi\fts this sequence?\r\n(b) Does there necessarily exist a polynomial W which fits the sequence and such that $ W(n)$ is an integer for all integers $ n$?",
  "Solution_1": "a)  No.  Polynomials with integer coefficients are \"locally obstructed\" by the fact that $ m \\minus{} n | W(m) \\minus{} W(n)$.  For example, $ W(0), W(2), W(4), ...$ must all have the same parity.  (I believe this is the only such obstruction; in other words, if for every prime $ p$ the sequence $ a_i$ satisfies $ a_{i\\plus{}p} \\equiv a_i \\bmod p$, then $ W$ should exist.)\r\n\r\nb)  Yes.  If you require that the polynomial has degree $ n \\minus{} 1$, then the polynomial is unique and the fact that it only takes integer values is a corollary of some basic theorems about [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=227941]finite differences[/url].  Such a polynomial takes the form\r\n\r\n$ W(x) \\equal{} \\sum_{i \\equal{} 0}^{n \\minus{} 1} b_i {x \\choose i}$\r\n\r\nwhere the $ b_i$ are certain integers that appear in the finite difference table of $ W$."
}
{
  "Tag": [
    "function",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "Let $f: [0,1]\\rightarrow[0,1]$ a strictly increasing function such that:\r\n    * $f(0)=0$\r\n    * $\\forall x\\in[0,1]: f\\left(\\frac{x}{3}\\right)=\\frac{f(x)}{2}$\r\n    * $\\forall x\\in[0,1]: f(1-x)=1-f(x)$\r\n\r\nFind $f\\left(\\frac{18}{1991}\\right)$",
  "Solution_1": "If  $f: [0,1]\\rightarrow[0,1]$  function such that:\r\n     * $\\forall x\\in[0,1]: f\\left(\\frac{x}{3}\\right)=\\frac{f(x)}{2},$\r\nthen f(x) can be defined for all x, if we know $f(x)=g(x): [\\frac{1}{3},1]\\to [0,1]$ in interval $[\\frac{1}{3},1], g(\\frac{1}{3})=\\frac{1}{2}g(1)$\r\n (1) $f(\\frac{x}{3^{k}})=\\frac{1}{2^{k}}f(x)$. \r\nIf f(x) increase, then g(x) increase and from (1) f(0)=0.\r\nIf f(x) increase and satisfied: \r\n   $\\forall x\\in[0,1]: f(1-x)=1-f(x),$\r\nthen $f(1)=1\\Longrightarrow f(\\frac{1}{3})=\\frac{1}{2}=f(\\frac{2}{3})$, therefore\r\n$g(x)=\\frac{1}{2}\\ \\forall x\\in[\\frac{1}{3},\\frac{2}{3}].$\r\nIt give $f(x)=\\frac{1}{2^{k}}$ if $x\\in [\\frac{1}{3^{k}}, \\frac{2}{3^{k}}]$, but in interval $\\frac{2}{3^{k}},\\frac{1}{3^{k-1}}$ we can not define f(x) from (1).\r\n$f(\\frac{18}{1991})<\\frac{1}{16}f(\\frac{1458}{1991})=\\frac{1}{16}(1-f(\\frac{533}{1991}))=\\frac{1}{16}-\\frac{1}{32}f(\\frac{1599}{1991})=\\frac{1}{32}(1+f(\\frac{392}{1991}))=\\frac{3}{64}$\r\n because $\\frac{1}{3}< \\frac{3*392}{1991}< \\frac{2}{3}$."
}
{
  "Tag": [
    "probability",
    "Asymptote"
  ],
  "Problem": "I have a question.\r\n\r\nSay, when scientists do experiments, record data, and come up wtih a formula, how do they prove that the formula is correct? There will inevitably be some error so how do they prove, besides just looking at the graph, that the formula still stands in spite of the errors affecting the data? Is there some staticistical analytic way to prove the relationship they found (such as using software)? What are some ways?",
  "Solution_1": "This gets into statistics, and such formulas cannot be proven to be completely correct, only approximated very well. Using staistics, you can show that it is very very very unlikely that your formula is incorrect, but you cannot prove that it's absolutely true, and there will sometimes be slight differences between your output and the actual observations.",
  "Solution_2": "Thank you.\r\n\r\nCan you tell me (or offer me a site) how to use statistics to show that the formula is unlikely to be untrue?",
  "Solution_3": "you can do a best fit line on your graphing calculator!\r\nbut thats only more linear data\r\nfor the other ones you'll have to transfer your data\r\nlike if exponential , you log y\r\nif power log both X and Y\r\nand it goes to very complicated stuff",
  "Solution_4": "Yes. My question is, however, how can you prove mathematically or show using statistics that the formula that you come up with is unlikely to be wrong. That is, the line of best fit really exists, and the data is not completely random. The data will inevitably have errors, so you can't say that the formula stands just by finding a line of best fit.\r\n\r\nSo how do you show that the formula that you come up with, with the line of best fit and everything, is true or likely to be true?",
  "Solution_5": "For some physical laws, like Snell's Law, there is a mathematical proof of a natural phenomenon. If you are in this forum, you can probably understand the link I'll post here. This is a nicely laid out proof of Snell's Law. [url]http://scienceworld.wolfram.com/physics/SnellsLaw.html[/url]\r\nThis is an example of a mathematically derived proof of a physics law. There are many more of these, to be sure, this is just an excellent example that is laid out well.",
  "Solution_6": "Thanks a bunch.\r\n\r\nWhat about for laws that involve an experimental factor? Like, say, $F=uF_{N}$?",
  "Solution_7": "Which law is that? I might be able to explain that if I had a proper name...",
  "Solution_8": "That's the Coulomb friction equation. I'm talking about laws in which there is a constant that must be determined experimentally for each kind of material. Another example is Hooke's Law.\r\n\r\nSo for these formulae, is there a MATHEMATICAL way using statistical analysis to show that the formula that you've derived is unlikely to be false? Like, sure, I can look on the graph and see most of my plotted points are in a line, some are off, but most are on a line. But can I prove, not just by looking at the graph, but by using statistics (like standard deviation and whatever), that my formula is unlikely to be untrue?\r\n\r\nThanks.",
  "Solution_9": "Think of the probabilty definiton.\r\nIf you do an experiment correctly (assuming very slight human error) infinity times, then your result should be infinitely accurate.\r\nHave you done limits in math yet? Take the graph y = 1 / x.\r\nAs X values increase without bound, Y values decrease to the horizontal asymptote (the X-axis).\r\nBecause the more times we do an experiment, the more accurate it is, we can say that if we do an experiment an infinite amount of times, we will be infinitely accurate.\r\n\r\nThus let the previous graphs' Y values represent units of innacuracy, and let the X values represent repetitions of experiment. This way, you can statistically say how innaccurate you are after doing X amount of trials.\r\n\r\nI *think* this is how they derive laws from experimental data.",
  "Solution_10": "Thanks.\r\n\r\nSay I've been experimenting on Hooke's Law $F=-kx$ with a spring, a ruler, and some masses. There will be imprecision and random errors from the ruler. And if I collect the data, plot it, it will not assume the shape of a perfect line. So my question is: how do I know that I should fit a linear regression to it? How can I prove that it is very likely to be a linear relationship and not some other, like an exponential relationship? I look at the graph, yes, it look slinear, but how do I prove on paper or with a software mathematically that it is highly likely that this is a linear relationship?\r\n\r\nThanks a lot.",
  "Solution_11": "does your calculator give you a value for 'r'?\r\n'r' has not been well explained to me but i have been told it indicates the accuracy with which you line describes your data. I am not sure how this value is calculated... can someone explain??\r\nWhat I do know is that one way that you can show that your data is most likely linear as opposed to exponential is that the 'r' value for the linear regression is closer to one (with 1 being perfect accuracy) than the 'r' value for exponential regression. again, this is only what I have learned in my math class and an explanation of what 'r' is and how it is found woul be useful... thanks"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "special factorizations"
  ],
  "Problem": "For which c real numbers ,there can be found a line that intersects  $y=x^4+9x^3+cx^2+9x+4$ curve at four  distinct points?",
  "Solution_1": "$x^4+9x^3+cx^2+9x+4=mx+b$ has four real solutions for some m, b. $x^4+9x^3+cx^2+(9-m)x+(4-b)=0$ I can't remember the rules about numbers of real solutions for polynomials though. I think there should be an easy enough way of telling how many solutions there are. Change of signs shows how many negative/positive solutions there are? Can't remember.",
  "Solution_2": "[hide]\nWe want two distinct points of inflection.\n$y''=12x^2+54x+2c$ $y'''=24x+54$, so points of inflection\noccur at x values such that $12x^2+54x+2c=0$ and $24x+54$ is not 0\nSo, by completing the square, $x^2+\\frac{9x}{2}+\\frac{81}{16}=\\frac{81}{16}-\\frac{c}{6}$ and and x does not equal $-\\frac{9}{4}$\nIn order to have two distinct real solutions, $\\frac{81}{16}-\\frac{c}{6}>0$ and $c<\\frac{243}{8}$[/hide]",
  "Solution_3": "[hide]If $r_k$ denote the roots of the polynomial $x^4+9x^3+cx^2+(9-a)x+(4-b)$, then the $r_k$ are all real and distinct. Then $\\sum_{j<k} (r_j-r_k)^2> 0$.\n\nExpanding each, this is $3\\sum_{k} r_k^2> 2\\sum_{j<k} r_jr_k=2c$.\n\nAdding on $6c=6\\sum_{j<k} r_jr_k$ to both sides to complete the square, $3\\left( \\sum_k r_k \\right)^2>8c$. But $\\sum_k r_k=-9\\Rightarrow c<\\frac{243}{8}$.\n\nSo if $c\\geq \\frac{243}{8}$ it certainly won't work. But if $c<\\frac{243}{8}$ then the sum $\\sum_{j<k} (r_j-r_k)^2$, or $243-8c$, is positive.\n\nJust suppose that $r_1=r_2+\\alpha=r_3+2\\alpha=r_4+3\\alpha$. Then this sum is $(3\\alpha)^2+2(2\\alpha)^2+3\\alpha^2=243-8c$.\n\nThen $20\\alpha^2=243-8c$, so $\\alpha=\\sqrt{\\frac{243-8c}{20}}$.\n\nThen from $\\sum_k r_k=4r_4+6\\alpha=-9$, $r_4=-\\frac{9+6\\alpha}{4}$. So a set of $r_k$ can be constructed, which then determine the coefficients.\n\nTherefore there is a line if and only if $\\boxed{c<\\frac{243}{8}}$.[/hide]"
}
{
  "Tag": [
    "number theory theorems",
    "number theory"
  ],
  "Problem": "gcd(2^{p}-1,2^{q}-1)=1  equivalent with gcd(p,q)=1 ?  :?: : i don't work it  ,,,,please help me !\r\nthanks you !everyone",
  "Solution_1": "Stronger:  $ \\gcd(2^p \\minus{} 1, 2^q \\minus{} 1) \\equal{} 2^{\\gcd(p, q)} \\minus{} 1$.\r\n\r\nLet $ d \\equal{} \\gcd(2^p \\minus{} 1, 2^q \\minus{} 1)$; then $ 2^p \\equiv 2^q \\equiv 1 \\bmod d$.  Let $ a, b$ be integers such that $ ap \\plus{} bq \\equal{} \\gcd(a, b)$ (by [url=http://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity]Bezout's lemma[/url]); then $ 2^{ap \\plus{} bq} \\equiv 1 \\bmod d$.\r\n\r\nHence $ d | 2^{\\gcd(p, q)} \\minus{} 1$.  And it's obvious that $ 2^{\\gcd(p, q)} \\minus{} 1 | d$."
}
{
  "Tag": [
    "algorithm",
    "videos"
  ],
  "Problem": "[youtube]2bpKiwVDZP8[/youtube]\r\n\r\nA parody of Crank that Soulja boy... about math =).\r\nThere was very little choreographing done, so bear with it.\r\nI hope you like it. The lyrics are in the \"about this video\". I'll post them here too.\r\n\r\n[hide=\"lyrics\"]\nSigma boy add-em\nHey I got this new algorithm fo yall. It's called tha Sigma Boy\n\nYou gotta partition and add n times from i to infinity\n\n(Chorus){\nSigma boi up in Big O\nWatch me max the bound with Rolle\nWatch me crank that sigma boi and\nSubstitute dat ho\n\nnow sub with UUUUUUUU crank that sigma boi\nnow sub with UUUUUUUU crank that sigma boi\nnow sub with UUUUUUUU crank that sigma boi\nnow sub with UUUUUUUU crank that sigma boi\n}\n\nSigma boi up in big O\nWatch me work it with dat chalk\nSimplify and show\nThat Riemann cranks dat Robocop\nPlug and chug, now watch me calc\nCalckin' on dem hatahs mang\nWhen I do dat sigma boy\nI punch it in tha calc and watch it summate\nI'm calckin on you (calckin on you)\nAnd if we get to graphin'\nI'll be plottin' on ya (plottin on ya)\nYou catch me at yo local party\nIntegrating with yo date\nHaterz get mad cuz\nThey got F's and I got A\n\nChorus\n\nSpherical C.O.s\nI roll with Theta, phi, and rho\nIma pass it to Leibniz\nAnd he gon' calc it up fo sho\nHaterz wanna be me Sigma boy, I'm the man\nThey be lookin' at my index\nIterate through the summand (maan) (maan)\nAdd it through (Add it through)\nDance (Dance)\nLie tangent to (tangent to)\nNope, you can't add it like me\nHoe, so don't add it like me\nFolk, I see you tryin to add it like me\nMan that sum was ugly\n\nChorus\n[/hide]",
  "Solution_1": "Hmmm.\r\nI'd say nice video except lately all of the videos posted in G&FF have only played for 2 seconds then stopped but said they were playing...",
  "Solution_2": "Haha interesting  :D",
  "Solution_3": "Awesome video  :) . \r\n\r\nIf it was choreographed then that would seriously be on Youtube's favorites.",
  "Solution_4": "[quote=\"chickendude\"][youtube]2bpKiwVDZP8[/youtube]\n\nA parody of Crank that Soulja boy... about math =).\nThere was very little choreographing done, so bear with it.\nI hope you like it. The lyrics are in the \"about this video\". I'll post them here too.\n\n[hide=\"lyrics\"]\nSigma boy add-em\nHey I got this new algorithm fo yall. It's called tha Sigma Boy\n\nYou gotta partition and add n times from i to infinity\n\n(Chorus){\nSigma boi up in Big O\nWatch me max the bound with Rolle\nWatch me crank that sigma boi and\nSubstitute dat ho\n\nnow sub with UUUUUUUU crank that sigma boi\nnow sub with UUUUUUUU crank that sigma boi\nnow sub with UUUUUUUU crank that sigma boi\nnow sub with UUUUUUUU crank that sigma boi\n}\n\nSigma boi up in big O\nWatch me work it with dat chalk\nSimplify and show\nThat Riemann cranks dat Robocop\nPlug and chug, now watch me calc\nCalckin' on dem hatahs mang\nWhen I do dat sigma boy\nI punch it in tha calc and watch it summate\nI'm calckin on you (calckin on you)\nAnd if we get to graphin'\nI'll be plottin' on ya (plottin on ya)\nYou catch me at yo local party\nIntegrating with yo date\nHaterz get mad cuz\nThey got F's and I got A\n\nChorus\n\nSpherical C.O.s\nI roll with Theta, phi, and rho\nIma pass it to Leibniz\nAnd he gon' calc it up fo sho\nHaterz wanna be me Sigma boy, I'm the man\nThey be lookin' at my index\nIterate through the summand (maan) (maan)\nAdd it through (Add it through)\nDance (Dance)\nLie tangent to (tangent to)\nNope, you can't add it like me\nHoe, so don't add it like me\nFolk, I see you tryin to add it like me\nMan that sum was ugly\n\nChorus\n[/hide][/quote]\r\n\r\nlol :lol:"
}
{
  "Tag": [
    "AMC",
    "USA(J)MO",
    "USAMO"
  ],
  "Problem": "How is your GPA this year, university students?\r\nHere's mine:\r\nSemester 1:\r\nOverall 3.29\r\nMajor 3.59\r\nSemester 2:\r\nOverall 3.07  :( \r\nMajor 3.59\r\n\r\nThough I think I deserve my low major GPA for my utter laziness...\r\nNo one believes I was participating IMO last year  :D",
  "Solution_1": "uh, let's not. we hate talking about our GPA",
  "Solution_2": "i don't have a gpa yet. :D",
  "Solution_3": "huh?",
  "Solution_4": "well, at least i don't think i do.",
  "Solution_5": "Well, a lot of 8th graders don't have GPA's yet.\r\n\r\nBut it said university students.",
  "Solution_6": "senior in high school with 4.93 weighted gpa (3.94 unweighted)",
  "Solution_7": "*cough* university students, furious *cough*",
  "Solution_8": "[quote=\"bleumoose\"]*cough* university students, furious *cough*[/quote]\r\n\r\nYeah, but to be fair, The Round Table isn't exactly swarming with university students...",
  "Solution_9": "this is true. aight, im 3.93 unweighted, 4.5something weighted.",
  "Solution_10": "4.0 uw. We don't weigh or rank (I wish!)",
  "Solution_11": "Well, I don't technically have a gpa yet, since I just started high school, but the high school classes that we took in 7th/8th grade count for it.  So only counting the high school classes, I have a 4.0 (w), and am probably ranked 1/615, since only 6 people are ranked.",
  "Solution_12": "I'm a Freshman, but considering school started today, my unweighed GPA is probably either 0 or 4.  :D",
  "Solution_13": "My unweighted is 4, my weighted is 4.5. Im Freshman, and school started 3 weeks ago.",
  "Solution_14": "Community colleges count, right?\r\n\r\n4.0 @ Bellevue Community College (I'm still \"technically\" a high school student though lol) last year. :)",
  "Solution_15": "4.0, 4.5\r\n\r\nit would be higher but my high school only has 2 ap classes for freshmen...and honors english doesn't even count as a 5...",
  "Solution_16": "Junior in High School:\r\nCumulative GPA: 3.98 unweighted (#^@! A-'s), 4.75 weighted",
  "Solution_17": "I am a freshman in high school. I'm taking 1 AP class so um... I dunno.  :)",
  "Solution_18": "Wow I'd expect much higher GPA's from USAMO qualifiers... Maybe you guys spend all your time studying math? Probably becuase you school doesn't offer that many AP classes. Yes that must be it. \r\n\r\nMy GPA is low, I'm ashamed."
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "geometry proposed"
  ],
  "Problem": "Let $ ABCD$ be a cyclic quadrilateral. And let $ E \\equal{} AB\\cap CD$, $ F \\equal{} BC\\cap DA$ and $ G \\equal{} AC\\cap BD$. The circumcircles of triangles $ DAE$ and $ DCF$ intersect each other at $ D$ and $ H$. The angle bisectors of the angles $ AHB$, $ DHC$ intersect $ AB$, $ CD$ at $ I$, $ J$, respectively. Prove that $ I$, $ J$ and $ G$ are collinear.",
  "Solution_1": "Use Miguel's lemma with triangle $ BAF$ we have $ H \\in (BCE)$. From this we have $ \\Delta HAD \\sim \\Delta HBC \\Rightarrow \\frac{HA}{HB} \\equal{} \\frac{HD}{HC} \\equal{} \\frac{AD}{BC} \\Rightarrow \\frac{IA}{IB} \\equal{} \\frac{GA}{GB} , \\frac{JD}{JC} \\equal{} \\frac{GD}{GC}$ \r\nIt means that $ GI,GJ$ is the bisectors of the angles $ \\angle BGA , \\angle DGC$, so $ G,I,J$ are collinear",
  "Solution_2": "Thank you dear [b]PDatK40SP[/b] for your interest and for your solution. We can easily to prove two other nice properties of this configure that $ E$, $ H$, $ F$ are collinear and $ \\angle AHB\\equal{}\\angle CHD$  :wink:"
}
{
  "Tag": [
    "linear algebra",
    "matrix",
    "vector",
    "arithmetic sequence"
  ],
  "Problem": "(sorry for posting so many questions)\r\n\r\n How can one know what the reduced row echelon form will look like? I know that you can row reduce any matrix into something, but is there a way to tell how it will look like? (i.e. how many independent columns will it have?) I know that if a matrix A is m x n then it can only have at most m independent columns (obviously)\r\n\r\n thanks \r\n\r\nfor example, I thought \r\n\r\n1 2 3\r\n4 5 6 \r\n7 8 9\r\n\r\n could eventually get you\r\n\r\n1 0 0\r\n0 1 0 \r\n0 0 1\r\n\r\n but it doesn't. How could I have known this without just \"experimenting\"?\r\n\r\nthanks a lot",
  "Solution_1": "I think I found the answer for this one. There's a theorem that states, for scalars a1,...,an and columns c1,....,cn, if column K = a1c1 + a2c2 +..... + ancn, then in the row reduced form, it will equal to a1e1 + .... + anen where e1, .... , en are columns for the standard F^n identity matrix",
  "Solution_2": "That doesn't make sense, and can't possibly be right.\r\n\r\nThere is no simple answer; the easiest way to find the row-reduced form in general is to actually do the row reduction.",
  "Solution_3": "[quote=\"jmerry\"]That doesn't make sense, and can't possibly be right.\n\nThere is no simple answer; the easiest way to find the row-reduced form in general is to actually do the row reduction.[/quote] \r\n\r\n  it's from my textbook which states:\r\n\r\n \"Theorem 3.16: Let A be an m x n matrix of rank r, where r > 0, and let B be the reduced row echelon form of A. Then \r\n\r\n ... For each k = 1,2,....,n if column k of B is d1e1 + d2e2 +.... + drer then column K of A is  d1a1 + d2a2 +.... + drar\"\r\n\r\n Oops, I messed up the notation in my previous post, I said d1e1 + d2e2 +.... + dnen\r\n\r\nthanks for your time",
  "Solution_4": "Here's the thing I can see at a glance: if the columns (or rows) of a matrix form an arithmetic progression (which is to say, each pair of adjacent columns has the same difference), then the rank of the matrix is at most 2. I see that right away, and see that it applies to this matrix, meaning its rank is at most 2. And since the columns are not all multiples of the same thing, the rank is exactly 2. That means that your reduced row echelon form will be\r\n\\[ \\begin{bmatrix}1&0&*\\\\0&1&*\\\\0&0&0\\end{bmatrix}.\\]\r\n(And its determinant is zero, and it's not invertible, and its null space has dimension 1, and so on.)\r\n\r\nFor more details: we can see that the difference of adjacent columns is $ \\begin{bmatrix}1\\\\1\\\\1\\end{bmatrix}.$ That means that all three columns can be written as linear combinations of the first column and the difference I just stated - and if all three columns lie in the span of a set of two vectors, then the dimension of the column space (also known as the rank) of this matrix is at most 2."
}
{
  "Tag": [
    "trigonometry",
    "geometry",
    "incenter",
    "LaTeX",
    "inradius",
    "circumcircle"
  ],
  "Problem": "An equilateral triangle is inscribed in a circle of radius 6cm. Find the length of each side of the triangle...\r\nI just wanna check if everyone else does it the same way I have done it so please show ur working..",
  "Solution_1": "[hide=\"with trig\"]the incenter of the triangle = the center of the circle\n\nthus, the distance from each vertex of the triangle to the incenter = 6cm\n\nthis splits our equilateral triangle into 3 isosceles triangles ABC with:\n\n<A = 30 degrees\n<B = 30 degrees\n<C = 120 degrees\nAC = 6cm\nBC = 6cm\nAB = ?? (this is one edge of the original triangle)\n\nAB / sin120 = 6 / Sin30\nthus\nAB = 6sin120 / sin30 = 10.39cm\n\nso each side of the triangle is 10.39cm[/hide]\r\n\r\nsorry I dont know how to use latex just yet..",
  "Solution_2": "[hide=\"Without trig\"]Call the triangle ABC. The center of the triangle is also the middle of the circle since the triangle is equilateral. Draw the radius to any vertice, lets do it to A. Then from the center draw a perpendicular line to AB. Call the point where the perpendicular line X. Call the center of the circle O. Angle OAB is 30 degrees, and that means that we have a 30-60-90 triangle. Since OA equals 6, we have that AX equals $3\\sqrt{3}$. But that is half of the side of the equilateral triangle so the side length is $6\\sqrt{3}$.[/hide]",
  "Solution_3": "[b] :rotfl: \n\nInscribe triangles in a circle or in another triangle are easy to solve if you'll be given the givens you need! :lol: \nAm I right?\n\nAnd being stupid adds up a lot because the more stupid you are the least problems you'll have? because you're to stupid to know that you don't even know that you don't. :P [/b]",
  "Solution_4": "[quote=\"bikturya\"][b] \n\nInscribe triangles in a circle or in another triangle are easy to solve if you'll be given the givens you need! \nAm I right?\n\nAnd being stupid adds up a lot because the more stupid you are the least problems you'll have? because you're to stupid to know that you don't even know that you don't.  [/b][/quote]\r\n\r\nWhat the...?\r\n\r\n :spam:  :spam:  :spam:",
  "Solution_5": "[quote=\"7h3.D3m0n.117\"][quote=\"bikturya\"][b] \n\nInscribe triangles in a circle or in another triangle are easy to solve if you'll be given the givens you need! \nAm I right?\n\nAnd being stupid adds up a lot because the more stupid you are the least problems you'll have? because you're to stupid to know that you don't even know that you don't.  [/b][/quote]\n\nWhat the...?\n\n :spam:  :spam:  :spam:[/quote]\r\nYeah, I agree.\r\n I know you're new here, but some people don't know how to do it that fast. So be nice to those people.",
  "Solution_6": "you could be really cool and use A=rs...\r\n\r\nso $\\frac{\\sqrt{3}}{4}x^{2}=\\frac{3x}{2}\\cdot6$\r\n$\\frac{\\sqrt{3}}{4}x=9$\r\nso whazam $x=36/sqrt{3}=12\\sqrt{3}$\r\n\r\nwait wtf did I do something wrong?",
  "Solution_7": "[quote=\"13375P34K43V312\"]you could be really cool and use A=rs...\n\nso $\\frac{\\sqrt{3}}{4}x^{2}=\\frac{3x}{2}\\cdot6$\n$\\frac{\\sqrt{3}}{4}x=9$\nso whazam $x=36/sqrt{3}=12\\sqrt{3}$\n\nwait wtf did I do something wrong?[/quote]\r\nWhat does the x represent, the radius?",
  "Solution_8": "[quote=\"ckck\"]\nWhat does the x represent, the radius?[/quote]\r\n\r\nIdeally, it should represent the side of the equilateral triangle !!!",
  "Solution_9": "[quote=\"13375P34K43V312\"]you could be really cool and use A=rs...\n\nso $\\frac{\\sqrt{3}}{4}x^{2}=\\frac{3x}{2}\\cdot6$\n$\\frac{\\sqrt{3}}{4}x=9$\nso whazam $x=36/sqrt{3}=12\\sqrt{3}$\n\nwait wtf did I do something wrong?[/quote]\r\nI know what you did wrong. That formula is when the circle is inscribed in the triangle. Not the other way around.",
  "Solution_10": "I did it with trig  and I got the same answer...thanx!",
  "Solution_11": "[quote=\"13375P34K43V312\"]you could be really cool and use A=rs...\n\nso $\\frac{\\sqrt{3}}{4}x^{2}=\\frac{3x}{2}\\cdot6$\n$\\frac{\\sqrt{3}}{4}x=9$\nso whazam $x=36/sqrt{3}=12\\sqrt{3}$\n\nwait wtf did I do something wrong?[/quote]\r\n\r\nyes. r stands for the inradius of the triangle, which is 3, not the circumradius, 6. But once you've found the inradius, it's easier to just find the side lengths using that 30-60-90 triangle."
}
{
  "Tag": [
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "Prove that for $a,b,c>0$ is valid:\r\n\r\n$\\sum_{cyclic}{\\frac{a}{\\sqrt{b+c}}}\\geq \\sqrt{\\frac{3}2(a+b+c)}$.",
  "Solution_1": "Denoting by $LHS$ by $S$ and applying chebyshev inequality we get\r\n$3S \\geq (a+b+c)(\\frac{1}{\\sqrt{b+c}}+\\frac{1}{\\sqrt{a+c}}+\\frac{1}{\\sqrt{b+a}}) \\geq (a+b+c)(\\frac{9}{\\sqrt{b+c}+\\sqrt{a+c}+\\sqrt{b+a}}) \\geq (a+b+c)(\\frac{9}{\\sqrt{6(a+b+c)}}) = 3\\sqrt{\\frac{3}{2}(a+b+c)}$\r\n So $S \\geq \\sqrt{\\frac{3}{2}(a+b+c)}$",
  "Solution_2": "--------------------\nBy [b]Holder Inequality[/b] \\[\\left(\\frac{a}{\\sqrt{b+c}}+\\frac{b}{\\sqrt{c+a}}+\\frac{c}{\\sqrt{a+b}} \\right ) ^2\\left( a(b+c)+b(a+c)+c(a+b) \\right ) \\geq (a+b+c)^3\\] $\\longrightarrow$ \\[\\left(\\frac{a}{\\sqrt{b+c}}+\\frac{b}{\\sqrt{c+a}}+\\frac{c}{\\sqrt{a+b}} \\right) \\geq \\sqrt{\\frac{(a+b+c)^3}{2(ab+bc+ca)}}\\] So it remains to prove that \\[\\sqrt{\\frac{(a+b+c)^3}{2(ab+bc+ca)}}\\geq \\sqrt{\\frac{3}2(a+b+c)}\\] $\\longrightarrow$ $$(a+b+c)^2\\geq 3(ab+bc+ca)$$ which is obvious.\n------------",
  "Solution_3": "$\\sum_{cyc}{\\frac{a}{\\sqrt{b+c}}}\\geq \\sqrt{\\frac{3}2(a+b+c)}$\nassume:a+b+c=6,now we can prove \n$\\sum_{cyc}{\\frac{a}{\\sqrt{b+c}}}\\geq 3$\nM=$\\sum_{cyc}{\\frac {a} {\\sqrt{b+c}}}$,N=$\\sum_{cyc}{\\frac {\\sqrt{b+c}} {a}}$\nthen M+N\u22656\nand M-N=$\\sum_{cyc}{\\frac {a^2-b-c} {a\\sqrt{b+c}}}$\nso $\\sum_{cyc}{\\frac{a}{\\sqrt{b+c}}}\\geq \\sqrt{\\frac{3}2(a+b+c)}$"
}
{
  "Tag": [
    "geometry"
  ],
  "Problem": "An Isoceles triangle ABC, with base AC, has a point P that lies in BC, and a point Q that lies in AB. Given that AC=AP=QP=BQ, find the measure in degrees of the angle B.\r\n\r\n :)",
  "Solution_1": "I hand-wrote this on a tablet computer, sorry if my handwriting is a little messy.\r\n\r\nFrom my algebraic solution, I got an answer of $ \\angle B\\equal{}\\boxed{\\frac{180}{7}^{\\circ}}$.",
  "Solution_2": "Thank you gauss1181; I understood  :lol:"
}
{
  "Tag": [
    "trigonometry",
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "prove that for any triangle ABC\r\n[2r/R] <sup>2</sup>  < cos{(A-B)/2}cos{(B-C)/2}cos{(C-A)/2}\r\nin my opinion this one's really sweet ;)",
  "Solution_1": "I think I have something for this one:\r\n\r\nusing Mollweide formula we have: $\\cos (\\frac{A-B}{2})=\\frac{\\sin( \\frac{C}{2})(a+b)}{c}$ we rewrite the inequality like so: $64 \\prod (a) \\prod \\sin (\\frac{A}{2}) < \\prod(a+b)$, but $\\prod \\sin (\\frac{A}{2}) = \\frac{r}{4R}$. Thus the inequality is equivalent to:$ R \\prod(a+b) \\geq 16r \\prod(a)$ but this is the product of $R\\geq 2r$ and $ (a+b) \\geq 2\\sqrt{ab}$ (and the simmilar relations).\r\n\r\ncheers! :D  I hope it is okay. What is your sol galois?",
  "Solution_2": "nice sol,lagrangia(tho i've to go thru it fully but it seems impressive)\r\ni converted everyting in terms of half angles and then converted it into an obvious inequality in homogenous polynomials(with the variables being cot(A/2) and similar terms) using the fact that \r\ncot(A/2)+cot(B/2)+cot(C/2)=cot(A/2)cot(B/2)cot(C/2)"
}
{
  "Tag": [],
  "Problem": "I'm told that the answer to: \r\n\r\nT - xg = xg((y-x)/(x + y)) (Find T) \r\n\r\nis: \r\n\r\nT = g(2xy/(x + y)) \r\n\r\nCan anyone tell me the intermediate steps please?",
  "Solution_1": "[hide=\"Full List of Steps\"]\\begin{align*}T-xg&=xg\\cdot\\left(\\frac{y-x}{x+y}\\right)\\\\\\\\\nT&=xg+xg\\cdot\\left(\\frac{y-x}{x+y}\\right)\\\\\\\\\nT&=xg\\cdot\\left(1+\\frac{y-x}{x+y}\\right)\\\\\\\\\nT&=xg\\cdot\\left(\\frac{(x+y)+(y-x)}{x+y}\\right)\\\\\\\\\nT&=xg\\cdot\\left(\\frac{2y}{x+y}\\right)\\\\\\\\\nT&=g\\cdot\\left(\\frac{2xy}{x+y}\\right)\\end{align*}[/hide]",
  "Solution_2": "Thanks very much for the quick answer! :)"
}
{
  "Tag": [
    "logarithms"
  ],
  "Problem": "If I have a homework assignment that has 25 questions and I finish 10% of the remaining homework on any given day, how many days will it take me to finish the all but the last question?\r\n\r\n(You can finish a little bit of a question at a time)",
  "Solution_1": "[hide=\"answer\"]I multiplied 25 by powers of $\\frac{9}{10}$ until I got an answer smaller than $1$, which was $31$[/hide]",
  "Solution_2": "Since 10% of the homework is completed, 90% always remains. Thus we can write the exponential equation 25(9/10)^t. We want to know when all but the last question is answered, so set this equation equal to 1. Graph is equation on a graphing calculator and the answer should be 31 days.\r\n\r\nP.S. Could someone check this?",
  "Solution_3": "I got the same answer, but shouldn't it be $25(\\frac{9}{10})^{t}\\le 1$?  The answer would have to be a fraction for it to equal $1$ exactly",
  "Solution_4": "You really don't need to guess.  :P There will always be 9/10 of the homework remaining (you can never actually finish it), and there are 25 problems:\r\n[hide]\n((9/10)^n)25 = 1\n(9/10)^n = 0.04\nn = (log0.04)/log(0.9)\nHitting this value onto your calculator...\nn=30.5510637...\nRounding up, \nn=31[/hide]",
  "Solution_5": "[hide]Set up the equation $25\\left(\\frac{9}{10}\\right)^{x}\\le 1$\n\nRounding $x$ to an integer, we get $31$.[/hide]",
  "Solution_6": "[hide]\nEven though you aren't allowed to use logs,\n\nYou can easily find out that you could find out \nthat because this is true:\n$\\frac{9}{10}^{4}= 0.04$, that the answer of $n$ would be:\n$\\frac{\\log 0.04}{\\log 0.9}$\nThen, you use Log tables or a calculator to find out that the answer to that would be $30.5520\\dots$. I think you round up to get $31$.[/hide]",
  "Solution_7": "[hide]25*.9 all the way until it is lessthan or equal to one (im too lazy to do it)[/hide]",
  "Solution_8": "you can use logs after that, which is what I did"
}
{
  "Tag": [
    "search",
    "algebra proposed",
    "algebra"
  ],
  "Problem": "1)find f:R-->R such that\r\n$ f(f(x)\\plus{}y))\\equal{}2f(x)\\plus{}f(y)$\r\n2)find g:R-->R such that\r\n$ g(xy)\\minus{}g(x)g(y)\\plus{}g(x\\plus{}y)\\equal{}g(x)(y\\plus{}1)\\plus{}g(y)(x\\plus{}1)$\r\n3)find f:R->R such that\r\n$ f(x^2\\minus{}y^2)\\equal{}xf(x)\\minus{}yf(y)$\r\n4)find h:R->R such that\r\n$ h(x\\plus{}y)\\minus{}2h(x\\minus{}y)\\equal{}2h(y)\\minus{}h(x)$\r\nCan you help me? :(",
  "Solution_1": "[quote=\"tronghieu\"]\n\n3)find f:R->R such that\n$ f(x^2 \\minus{} y^2) \\equal{} xf(x) \\minus{} yf(y)$\n[/quote]\r\n[url]http://www.mathlinks.ro/viewtopic.php?search_id=901152022&t=138850[/url]\r\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=337857#337857[/url]"
}
{
  "Tag": [
    "number theory solved",
    "number theory"
  ],
  "Problem": "Prove that there doesn't exist any positive integer $n$ such that $2n^2+1,3n^2+1$ and $6n^2+1$ are perfect squares.",
  "Solution_1": "Very nice problem,I tried to solve it with Pell's equations till I came up with this\r\n(6n^3-3n) <sup>2</sup>  <= (2n <sup>2</sup> -1)(3n <sup>2</sup>-1 )(6n <sup>2</sup> -1) <= (6n ^3-3n+1) <sup>2</sup>",
  "Solution_2": "very nice key.thank you.I also want to solve it by Pell's equations but failed\r\nbut is that be (6n^3+3n) <sup>2</sup> <(2n <sup>2</sup>+1)(3n <sup>2</sup> +1 )(6n<sup>2</sup> +1)<(6n ^3+3n+1)  <sup>2</sup>",
  "Solution_3": "$(3n)^2 (2n^2+1)+1=18n^4+9n^2+1=(3n^2+1)(6n^2+1)$, so you would have two squares with difference 1, so its only possible for $n=0$ which is not positive.",
  "Solution_4": "The problem does not only come from the Japanese mathematical olympiad, it was also used as problem 1 in the 1st German TST 2005:\r\n\r\n[color=blue]Prove that there is no positive integer n such that all three numbers $2n^2+1$, $3n^2+1$ and $6n^2+1$ are perfect squares.[/color]\r\n\r\nMy solution uses the fact that $\\left(2n^2+1\\right)\\left(6n^2+1\\right)-\\left(2n\\right)^2\\left(3n^2+1\\right)=4n^2+1$ (now I think you can complete it for yourself). All other solutions were easier ;) .\r\n\r\n  darij",
  "Solution_5": "[quote=darij grinberg]The problem does not only come from the Japanese mathematical olympiad, it was also used as problem 1 in the 1st German TST 2005:\n\n[color=blue]Prove that there is no positive integer n such that all three numbers $2n^2+1$, $3n^2+1$ and $6n^2+1$ are perfect squares.[/color]\n\nMy solution uses the fact that $\\left(2n^2+1\\right)\\left(6n^2+1\\right)-\\left(2n\\right)^2\\left(3n^2+1\\right)=4n^2+1$ (now I think you can complete it for yourself). All other solutions were easier ;) .\n\n  darij[/quote]\n\nHow would you advance from this, for a dumb person like me?\n",
  "Solution_6": "[quote=aThousandWords][quote=darij grinberg]The problem does not only come from the Japanese mathematical olympiad, it was also used as problem 1 in the 1st German TST 2005:\n\n[color=blue]Prove that there is no positive integer n such that all three numbers $2n^2+1$, $3n^2+1$ and $6n^2+1$ are perfect squares.[/color]\n\nMy solution uses the fact that $\\left(2n^2+1\\right)\\left(6n^2+1\\right)-\\left(2n\\right)^2\\left(3n^2+1\\right)=4n^2+1$ (now I think you can complete it for yourself). All other solutions were easier ;) .\n\n  darij[/quote]\n\nHow would you advance from this, for a dumb person like me?[/quote]\n\nDifference of two squares couldn't be of kind $4k+1$."
}
{
  "Tag": [
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "Let $ x,y,z \\ge  0$.Prove that $ \\frac{\\sqrt{x}}{1\\plus{}x}\\plus{}\\frac{\\sqrt{y}}{1\\plus{}x\\plus{}y}\\plus{}\\frac{\\sqrt{z}}{1\\plus{}x\\plus{}y\\plus{}z}<\\sqrt{3}$\r\n\r\n\r\n_______________________________________\r\nAzerbaijan Land of the Fire  :P",
  "Solution_1": "[quote=\"Pirkuliyev Rovsen\"]Let $ x,y,z \\ge 0$.Prove that $ \\frac {\\sqrt {x}}{1 \\plus{} x} \\plus{} \\frac {\\sqrt {y}}{1 \\plus{} x \\plus{} y} \\plus{} \\frac {\\sqrt {z}}{1 \\plus{} x \\plus{} y \\plus{} z} < \\sqrt {3}$\n\n\n_______________________________________\nAzerbaijan Land of the Fire  :P[/quote]\r\n\r\n\r\n$ \\frac {\\sqrt {x}}{1 \\plus{} x} \\plus{} \\frac {\\sqrt {y}}{1 \\plus{} x \\plus{} y} \\plus{} \\frac {\\sqrt {z}}{1 \\plus{} x \\plus{} y \\plus{} z} < \\sqrt {3}$\r\n\r\nwe just need to prove\r\n\r\n$ \\frac {\\sqrt {x}}{1 \\plus{} x} \\plus{} \\frac {\\sqrt {y}}{1 \\plus{} y} \\plus{} \\frac {\\sqrt {z}}{1 \\plus{} z} \\le \\sqrt {3}$\r\n\r\n$ \\frac {\\sqrt {x}}{1 \\plus{} x} \\plus{} \\frac {\\sqrt {y}}{1 \\plus{} y} \\plus{} \\frac {\\sqrt {z}}{1 \\plus{} z} \\le\\frac {\\sqrt {x}}{2\\sqrt{x}} \\plus{}$ $ \\frac {\\sqrt {y}}{2\\sqrt{y}} \\plus{} \\frac {\\sqrt {z}}{2\\sqrt{z}} \\equal{}\\frac{3}{2}\\le\\sqrt {3}$\r\n\r\n :lol:",
  "Solution_2": "your solution is similary to mine :)"
}
{
  "Tag": [
    "geometry proposed",
    "geometry"
  ],
  "Problem": "Let $ ABCD$ be cyclic quadrilateral.$ P$ $ Q$ $ R$ are orthogonal projections of $ D$ on $ BC$ $ CA$ $ AB$ respectively. Prove that $ PQ \\equal{} QR$ if and only if bisectors of angles $ ABC and ADC$ intersect on $ AC$.\r\n\r\nthanks for Your help",
  "Solution_1": "$  \\begin{array}{c} |PQ|=|CD|.sin{C}=|CD|.\\frac{|AB|}{2R}\\\\\r\n|RQ|=|AD|.sin{A}=|AD|.\\frac{BC|}{2R}\\end{array}\\}\\Rightarrow \\frac{|PQ|}{|RQ|}=\\frac{|AB|.|CD|}{|AD|.|BC|}\\Leftrightarrow\\\\\r\n\\Leftrightarrow [|PQ|=|QR|\\Leftrightarrow |AB|.|CD|=|AD|.|BC|\\Leftrightarrow \\mbox{the bisectors of angles ABC and ADC intersect in (AC).}]$",
  "Solution_2": "Why Your last conclusion is equivalent to the statement from the problem( why then this bisectors intersect each other on $ AC$) thanks",
  "Solution_3": "This is IMO 2003 P4"
}
{
  "Tag": [
    "geometry"
  ],
  "Problem": "FIND AREA OF IIGM AND TRIANGLE.BD =15 \r\n\r\n\r\n\r\nANGLE( D )OF IIGM=35\r\n\r\n\r\n\r\n \r\n \r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n \r\n\r\n\r\n\r\n\r\n\r\n \r\n\r\n\r\n\r\nFIND AREA OF IIGM AND TRIANGLE.BD =15 \r\n\r\n\r\n\r\nANGLE( D )OF IIGM=35\r\n\r\n\r\n\r\n \r\n\r\n\r\n\r\n\r\n\r\n \r\n\r\n\r\n\r\n \r\n\r\n\r\n\r\n\r\n\r\n \r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n \r\n\r\n\r\n\r\n\r\n\r\n \r\n\r\n\r\n\r\nSORRY FOR DIAGRAM\r\n\r\n\r\n\r\n \r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n \r\n\r\n\r\n\r\n \r\n\r\n\r\n\r\n\r\n\r\n \r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n \r\n\r\n\r\n\r\n\r\n\r\n \r\n\r\n\r\n\r\nSORRY FOR DIAGRAM\r\n\r\n\r\n\r\n \r\n\r\n[url][/url][youtube]hide[/youtube]",
  "Solution_1": "sorry i am unable to understand the question,so please post what is in diagram in words\r\nit may help\r\n\r\n[hide]though even then i'ld not be able to answer[/hide]"
}
{
  "Tag": [
    "induction",
    "number theory",
    "relatively prime",
    "number theory proposed"
  ],
  "Problem": "Prove that the sequence $ a_n$ with \r\n\r\n$ a_n \\equal{} 6a_{n \\minus{} 1} \\minus{} a_{n \\minus{} 2}$ with  $ a_1 \\equal{} 4$ and $ a_2 \\equal{} 24$\r\n\r\ncontains no perfect squares",
  "Solution_1": "Can somebody prove this statement?I have a proof for this...Nobody? :wink:",
  "Solution_2": "Dunno, but I think this could potentially help:\r\n$ a_{n}\\equal{}\\frac{\\sqrt{2}}{2}((3\\plus{}2\\sqrt{2})^{n}\\minus{}(3\\minus{}2\\sqrt{2})^{n})$\r\nMaybe this will help...  :wink:",
  "Solution_3": "There is no help....Its only a transformation... :wink:",
  "Solution_4": "[quote]Can somebody prove this statement?I have a proof for this...Nobody? [/quote]\r\n\r\nCan you give hint for solution?",
  "Solution_5": "[quote=\"\u0391\u03c1\u03c7\u03b9\u03bc\u03ae\u03b4\u03b7\u03c2 6\"]There is no help....Its only a transformation... :wink:[/quote]\nEvery proof consists of transformations...\n\n[quote=\"\u0391\u03c1\u03c7\u03b9\u03bc\u03ae\u03b4\u03b7\u03c2 6\"]Can somebody prove this statement?I have a proof for this...Nobody? :wink:[/quote]\r\nStop posing. Either post a solution, a hint, etc., or don't post.",
  "Solution_6": "$ a_1\\equal{}4$ is a perfect square  :rotfl: \r\nA hint is to divide every term of the sequence by 4. Now you have smaller numbers and the same problem.",
  "Solution_7": "maybe $ a_i a_{i \\plus{} 1} \\plus{} 4$ would be a perfect square,,,,\r\nI just guess, not prove yet :oops:",
  "Solution_8": "divide the sequence by 4.\r\nthen a_1=1,a_2=6,a_3=35  etc\r\nand a_n=6a_n-1-a_n-2\r\nwe need to show that the sequence a_n contains no squares.\r\nLemma.1 a_n=a_k*a_(n-k+1)-a_(k-1)*a_(n-k)(easy proof by induction)\r\n(in general this holds for a_n=a_2*a_n-1-a_1*a_n-2)....\r\nLemma 2. gcd(a_n,a_n-1)=1 (easy proof by induction)\r\nusing this, we get a_2n=a_n*(a_n+1-a_n-1) and a_(2n+1)=a_(n+1)*(a_n+1-a_n)\r\nif a_2n is a square, since a_n and a_n+1-a_n-1 are coprime (transform a_n+1 into 6a_n-a_n-1 and apply lemma)\r\na_n is a square \r\nif a_2n+1 is a square, similar argument leads to a_n+1 is a square.\r\nsince a_1 is square, a_2, a_3,a_4 is not,\r\nchoosing the smallest square that is not a_1 we get contradiction\r\n\r\nsorry for not latexing..",
  "Solution_9": "Let $ (1\\plus{}\\sqrt 2 )^n\\equal{}x_n\\plus{}y_n\\sqrt 2$, then $ a_n\\equal{}y_{2n}$.\r\nBecause $ (1\\minus{}\\sqrt 2 )^n\\equal{}x_n\\minus{}y_n\\sqrt 2$ we get $ x_n^2\\minus{}2y_n^2\\equal{}(\\minus{}1)^n\\to (x_n,y_n)\\equal{}1.$\r\nIf $ n\\equal{}2^km$, m - odd, then $ y_{2n}\\equal{}2^{m\\plus{}1}x_nx_{2^{k\\minus{}1}m}...x_my_m$, were $ x_{2^i}m,i\\equal{}0,..k$ are relatively prime odd numbers and $ (x_{2^im},y_m)\\equal{}1$. If $ y_{2n}$ is square any $ x_{2^im}$ must be square. But $ x_m\\equal{}3\\mod 8$ is not square.",
  "Solution_10": "SORRY :blush: \r\n$ 1)$If $ (b_n \\minus{} 1)^2 < a_n < b_n^2$ and $ n > 6$ then:\r\n$ a)$for $ n \\equal{} 0,2,4mod5 \\minus{} > b_n \\equal{} 2b_{n \\minus{} 1} \\plus{} b_{n \\minus{} 2} \\minus{} 1$\r\n$ b)$for $ n \\equal{} 1,3mod5 \\minus{} > b_n \\equal{} 2b_{n \\minus{} 1} \\plus{} b_{n \\minus{} 2}$\r\n$ 2)lim(b_{n \\plus{} 1}/b_n)^2 \\equal{} lim(a_{n \\plus{} 1}/a_n) \\equal{} c$ and hold that $ c^2 \\minus{} 6c \\plus{} 1 \\equal{} 0$  and  $ c \\equal{} 3 \\plus{} 2\\sqrt {2}$"
}
{
  "Tag": [
    "geometry"
  ],
  "Problem": "Given a triangle $ABC$ such that $AB=2BC$ and $\\angle ABC=60^{o}$. A point $P$ lies within the triangle and satisfy the condition: $\\angle APB= \\angle BPC= \\angle CPA$. Find the area of triangle $APC$ in term of the area of triangle $ABC$",
  "Solution_1": "If you solved the above problem, try this:\r\nGiven a convex quadrangle $ABCD$ such that $AD=AB$, $\\angle DAB= \\angle DCB= \\angle AEC=90^{o}$ and $AE=5$. Find the area of quadrangle $ABCD$"
}
{
  "Tag": [
    "linear algebra",
    "matrix"
  ],
  "Problem": "Show  the next equality:\r\n$ \\begin{vmatrix} \\frac{1}{p\\plus{}1}   & \\frac{1}{p\\plus{}2}  & \\ldots   &\\frac{1}{p\\plus{}n}\\\\ \\frac{1}{p\\plus{}2}   &    \\frac{1}{p\\plus{}3}       & \\ldots &     \\frac{1}{p\\plus{}n\\plus{}1}            \\\\ \\vdots &\\vdots &\\ddots &\\vdots\\\\    \\frac{1}{p\\plus{}n}            &   \\frac{1}{p\\plus{}n\\plus{}1}   & \\ldots &     \\frac{1}{p\\plus{}2n\\minus{}1} \\end{vmatrix}\\equal{}\\frac{[1! 2! \\ldots (n\\minus{}1)! (p\\plus{}1)! (p\\plus{}2)!\\ldots (p\\plus{}n\\minus{}1)!  ] ^2 }{(p\\plus{}1)! (p\\plus{}2)!\\ldots (p\\plus{}2n\\minus{}1)!  }$.",
  "Solution_1": "Read Determinant Cauchy : http://vi.wikipedia.org/wiki/%C4%90%E1%BB%8Bnh_th%E1%BB%A9c_Cauchy",
  "Solution_2": "A Romanian probably won't get much out of a Vietnamese-language link.\r\n\r\nMy best idea here is to find an $ LL^T$ or maybe $ LDL^T$ factorization by exploiting the inner product structure on polynomials. Unfortunately, that $ p$ means we've got something meaner and not just the Legendre polynomials.",
  "Solution_3": "He probable meant the Cauchy deteminant.\r\nhttp://www.mathlinks.ro/viewtopic.php?t=290874 contains the statement and proofs.",
  "Solution_4": "First apply column operations, col i -> col i - col 1 for i = 2,...,n\r\nThen for each row, we take out the common factors and get $ \\frac{1}{(p\\plus{}1)...(p\\plus{}n)}$\r\nFor each column, we take out the common factors and get $ (\\minus{}1)^{n\\minus{}1}(n\\minus{}1)!$\r\nWe are then left with a matrix with the first column becomes all 1s.\r\nNow, apply row operations row i -> row i - row 1 for i = 2,...,n\r\nand use similar method, then proceed by induction."
}
{
  "Tag": [
    "algebra",
    "function",
    "domain",
    "trigonometry"
  ],
  "Problem": "when f(x)= 3sin(4x) the max turning point on f(x) should be (pie/2,3) right?\r\n\r\nhow would you figure out the domain and range of f(x)=-3sec(x)?",
  "Solution_1": "[hide=\"not sure if it needs to be hidden\"]\nconsidering the unit circle, the angle with the greatest sine value is $\\frac{\\pi}{2}$. However, we are looking at sin(4x). so we want $4x=\\frac{\\pi}{2}$, so $x=\\frac{\\pi}{8}$. plug it in and you get your point to be \n$(\\frac{\\pi}{8},3)$\n\n[/hide]"
}
{
  "Tag": [],
  "Problem": "At what velocity $v_{0}$ should a bullet of mass $m$ hit(and get stuck in) a block of mass $M$ resting on a flat surface for it to move a distance $d$ if the coefficient of static and kinetic friction are $\\mu_{s}$ and $\\mu_{s}$, respectively and air resistance is negligible in terms of the given variables.",
  "Solution_1": "By conservation of momentum, $mv_{0}= (M+m)v$.\r\n\r\nSince the collision is virtually instantaneous, the static friction is negligible.\r\n\r\nThus, $(M+m) \\mu g d = \\frac{1}{2}(M+m)v^{2}= \\frac{m^{2}v_{0}^{2}}{2(M+m)}$.\r\n\r\nTherefore, $v_{0}^{2}= \\frac{2 \\mu (M+m)^{2}g d}{m^{2}}$.\r\n\r\nSo we have $v_{0}= \\frac{M+m}{m}\\sqrt{2 \\mu gd}$.",
  "Solution_2": "r u sure about static friction, cuz thats y i asked this question, consider a speed not large enough to move the block, its definately greater than zero, if a marble hits an 18 wheeler, it won't move",
  "Solution_3": "The momentum lost in the collision due to static friction is $\\mu (M+m)g t$, where $t$ is the time of the collision.\r\n\r\nThus, since this collision is assumed to be nearly instantaneous, $t$, and therefore the momentum lost due to static friction, is negligible.\r\n\r\nThe reason is that these are assumed to be two solid incompressible blocks.  In an actual collision, all objects involved undergo some compression, however small, during which the impulse of the collision is applied to them.\r\n\r\nIf the problem were changed to include a spring on one of the blocks, say with spring constant $k$, then the static friction could be taken into account, since it takes a nonzero time to compress the spring.  If you wanted the blocks to still stick, you could say that after the spring was compressed as far as it would be, a latch holds the blocks together.",
  "Solution_4": "fine how would one solve the problem with a spring on the 2nd block with spring constant k"
}
{
  "Tag": [
    "function",
    "induction",
    "group theory",
    "abstract algebra",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "Here's a problem I just composed; hope you'll find it interesting!  :) \r\n\r\n[b]Problem. [/b] Let $f(n)$ denote the number of groups of order $n$.\r\n(a) Find all pairs $(m,n)$ of positive integers such that $(m,n)=1$ and $f(mn) = f(m)f(n)$.\r\n(b) Find all pairs $(m,n)$ of positive integers such that $(m,n) = 1$ and $f(mn) = f(m)f(n)  + 1$.\r\n(c)* Is it possible to characterize the pairs $(m,n)$ with $f(mn)=f(m)f(n)$ without the assumption of $m$ and $n$ being co-prime?\r\n\r\n--Vesselin",
  "Solution_1": "I think I can solve (a), but this is going to be rather long.\r\n\r\nIn [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=automorphisms&t=45849]another thread[/url], vess introduced the multiplicative function $g$ defined by $g(p^{k})=(p^{k}-1)(p^{k}-p)\\ldots(p^{k}-p^{k-1})$. I\u2019ll (try to :)) prove that if $(m,n)=1$, then $(*)\\ f(mn)=f(m)f(n)$ is equivalent to $(**)\\ (m,g(n))=(n,g(m))=1$. \r\n\r\nFirst, notice that $(*)$ is equivalent to saying that every group of order $mn$ is a direct product of a group of order $m$ and one of order $n$. Next, notice that $(**)$ implies that an element $x$ of order dividing $n$ which normalizes a group $H$ of prime power order $p^{a}|m$ actually centralizes $H$: I showed in that other thread that every prime dividing $|\\mbox{Aut}~H|$ divides $|H|\\cdot g(p^{a})$, and since $x$ acts by conjugation on $H$ and the order of $x$ is coprime to $g(p^{a})$ and hence to $|\\mbox{Aut}~H|$, the conclusion follows. I\u2019ll use this observation below. \r\n\r\n\r\n$(*)\\Rightarrow (**)$\r\n\r\nAssume $(**)$ doesn\u2019t hold because, say, a prime divisor $p$ of $m$ divides $g(q^{k})$ for a prime power $q^{k}|n$. There is then an automorphism of $(\\mathbb Z/q\\mathbb Z)^{k}$ of order $p$, so $\\mathbb Z/p\\mathbb Z$ can be made to act non-trivially on $(\\mathbb Z/q\\mathbb Z)^{k}$. This means that there is a semidirect product $H=(\\mathbb Z/q\\mathbb Z)^{k}\\rtimes\\mathbb Z/p\\mathbb Z$ which is not direct. We can take our counterexample $G$ to be the direct product between $H$ and, for example, a cyclic group of order $\\frac{mn}{pq^{k}}$. $G$ is a group of order $mn$ which is not a direct product of one of order $m$ and one of order $n$, i.e. $(*)$ does not hold.\r\n\r\n\r\n$(**)\\Rightarrow (*)$\r\n\r\nAssume first that we\u2019ve managed to prove this for the case when $m,n$ are prime powers. Then we could solve the general case by induction on the order $mn$ of $G$ as follows:\r\n\r\nAssume that $m$ has at least two prime factors, and let $p^{a}\\|m$ for some prime $p$. Clearly, $mn$ is odd, so $G$ is solvable (the Feit-Thompson theorem). I\u2019ll also use Hall\u2019s theorem, stating that if $G$ is solvable, and $u\\ |\\ |G|,\\ \\left(u,\\frac{|G|}u\\right)=1$, then $G$ has subgroups of order $u$ which are all conjugate. By Hall\u2019s theorem, $G$ has a subgroup $N$ of order $n$, and all such subgroups of $G$ are conjugates of $N$. This means that we can find subgroups $N\\le N_{1}\\le G,\\ N\\le N_{2}\\le G$ such that $|N_{1}|=\\frac{mn}{p^{a}},\\ |N_{2}|=np^{a}$. By the induction hypothesis, we can find subgroups $M_{1}\\le N_{1},\\ M_{2}\\le N_{2}$ with $|M_{1}|=\\frac m{p^{a}},\\ |M_{2}|=p^{a}$ which centralize $N$. This implies that the centralizer of $N$ contains a subgroup $M$ of order $m$, and it\u2019s clear from here that $G=M\\times N$. \r\n\r\nAll that's left now is to prove that $(**)\\Rightarrow(*)$ when $m=p^{a},n=q^{b}$ are prime powers. \r\n\r\nLet $P$ be a Sylow $p$-subgroup of $G$. Another result due to Hall states that if for every subgroup $H$ of $P$, every element $x\\in G$ of order coprime to $p$ which normalizes $H$ also centralizes $H$, then $G$ has a normal subgroup $Q$ of order $q^{b}$. This condition holds thanks to $(**)$, so we do have such a group $Q$. $(**)$ now tells us that $P$ centralizes $Q$, so we get $G=P\\times Q$, as desired.\r\n\r\n\r\nI hope it\u2019s finally ok. I\u2019ve posted several incorrect versions up until now :)."
}
{
  "Tag": [
    "algebra",
    "polynomial"
  ],
  "Problem": "If I find the formula/method/theorem in a book/journal, how do I cite it???  Say for example,  I find in a journal call \"Mathematics\" issue number 2, in November 1950, how to find the roots to a cubic polynomial, how do I cite it???",
  "Solution_1": "There are various formats, but a standard citation might be:\r\n\r\nJ. Smith, \"An interesting equation\", [i]Mathematics[/i] [b]5[/b] (1950), 129-131.\r\n\r\nAuthor, \"Title\", [i]Journal Name[/i] [b]Volume[/b] (year), pages.\r\n\r\nIf citing a theorem or result from a paper, state the result in full; do not simply say something like \"By Theorem 3 in [1], f(x) is convex\".",
  "Solution_2": "So like we have a page of \"works cited\" and we put that line in there???or we put the line in the middle of our solutions???",
  "Solution_3": "As long as it's clear to the reader, it doesn't matter."
}
{
  "Tag": [
    "combinatorics proposed",
    "combinatorics"
  ],
  "Problem": "What is the minimum number of squares that is necessary to draw on a white sheet to obtain a square grid of side $n$?",
  "Solution_1": "I think it is n^2-(n-2)^2/2 if n is even and n^2-((n-2)^2+1)/2 if n is odd, \r\nbecause every exterior square must be drawn and for the rest (n-2)*(n-2)\r\njust notice that in any 2*2 piece at least 2 squares must be drawn.",
  "Solution_2": "Have a look here :\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?highlight=square&t=1403\r\n\r\nPierre."
}
{
  "Tag": [
    "MATHCOUNTS",
    "search",
    "research"
  ],
  "Problem": "does anyone know here an excellent math forum site (intended for math undergraduates and graduates) aside from mathlinks???please tell me the link..thanks",
  "Solution_1": "you can try the mathcounts website    \r\nits like mathcounts.saab.com\r\nbut im not sure so search on google",
  "Solution_2": "Is MathCounts website for undergraduates/graduates?\r\n\r\nAnyway, I would suggest [url=http://www.mathoverflow.net]Math Overflow[/url]. I think many undergraduates/graduates can learn a lot by seeing how professional mathematicians/other graduate students think.",
  "Solution_3": "thanks guys!",
  "Solution_4": "[url]mymathforum.com[/url]"
}
{
  "Tag": [],
  "Problem": "Express the following as a simplified common fraction: \\\\\n\n$ \\frac{1^2\\plus{}2^2\\plus{}3^2\\plus{}4^2\\plus{}5^2}{\n\\sqrt{1^2} \\cdot\n\\sqrt{2^2} \\cdot\n\\sqrt{3^2} \\cdot\n\\sqrt{4^2} \\cdot\n\\sqrt{5^2}}$",
  "Solution_1": "The formula for sum of squares is $ \\frac{n(n \\plus{} 1)(2n \\plus{} 1)}{6}$ where $ n$ is the last term (this only works when starting from one). Plug in five to get fifty-five. Everything on the bottom is just $ 1\\times 2 \\times 3 \\times 4 \\times 5$. This obviously equals $ 120$. $ \\frac{55}{120} \\implies \\boxed{\\frac{11}{24}}$."
}
{
  "Tag": [
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "A line dance requires two steps forward and then one step back. If a wall is ten steps away, how many steps must the dancers take to reach the wall?",
  "Solution_1": "Should go into the middle school forum ;)"
}
{
  "Tag": [],
  "Problem": "Two equal circles in the same plane cannot have the following number of common tangents:\r\n\r\n$ \\textbf{(A)}\\ 1 \\qquad\\textbf{(B)}\\ 2 \\qquad\\textbf{(C)}\\ 3 \\qquad\\textbf{(D)}\\ 4 \\qquad\\textbf{(E)}\\ \\text{none of these}$",
  "Solution_1": "[hide=\"Click for solution\"]\nTwo equal circles in the same plane cannot have only one common tangent (draw a diagram with them intersecting, with them tangent, or with them separate).  Thus, the answer is $ \\boxed{\\textbf{(A)}}$.\n[/hide]"
}
{
  "Tag": [
    "greatest common divisor",
    "quadratics",
    "number theory proposed",
    "number theory"
  ],
  "Problem": "Solve in integers.$y(x+y)=x^3-7x^2+11x-3$",
  "Solution_1": "Multiplying both sides by 64 and adding $16x^2$ yields the equation\r\n\\[ 16x^2+64xy+64y^2=64x^3-432x^2+704x-192  \\] that is,\r\n\\[ (4x+8y)^2=(4x)^3-27(4x)^2+176(4x)-192  \\] and so we can deal instead with the equation $A^2=B^3-27B^2+176B-192$ where $A=4(x+2y)$ and $B=4x$ are also integers.\r\n\r\nNote $B=8$ is a root of the cubic on the RHS, so $A^2=(B-8)(B^2-19B+24)=(B-8)[(B-8)^2-3(B-8)-64]$.  If $B=8$ the LHS is 0 and we have the solution $(x, y)=(2, -1)$ to the original equation.  The second factor has no integer roots.  Otherwise, the product of $B-8$ and $B^2-19B+24$ is a nonzero square.  As we know that the gcf of these two numbers is a factor of 64, $B^2-19B+24$ must be either a square or twice a square.\r\n\r\nIf $B^2-19B+24=k^2$, $(2B-19)^2-(2k)^2=265=1\\times 265=5\\times 53$, so $2B-19$ is one of $\\pm\\frac{265+1}{2}=\\pm 133$ and $\\pm\\frac{53+5}{2}=\\pm 29$, and $B\\in\\{76,-57,24,-5\\}$.  In fact, as $A^2=(B-8)(B^2-19B+24)$, $B-8$ is square also, and this happens only for $B=24$.  In this case $A^2=16\\times 144$ so $A=\\pm 48$, and the two resulting solutions $(x, y)$ to the original equation are $(6,3)$ and $(6,-9)$.\r\n\r\nIf $B^2-19B+24=2k^2$, $(2B-19)^2-2(2k)^2=265$, which, checking quadratic residues mod 5, implies $2B-19$ and $2k$ are multiples of 5, a contradiction since then the RHS should be a multiple of 25, which it is not.  So there are no solutions in this case.",
  "Solution_2": "Nice solution $Agrippina$ :clap2:  :clap:",
  "Solution_3": "Very very nice thanks"
}
{
  "Tag": [
    "abstract algebra",
    "superior algebra",
    "superior algebra theorems"
  ],
  "Problem": "I just have a general question about showing a group G is Abelian. In trying to show that G is Abelian, does it suffice to show that the inverses of two arbitrary elements in the group commute? That is (a^-1)(b^-1) = (b^-1)(a^-1) for any a and b in G. Or would I have to take it one step further and show that this implies that ab=ba? Thanks for your help.",
  "Solution_1": "The statement with inverses is precisely equivalent to the original statement. You can prove this by manipulation, or see that inversion is a bijection from the set to itself."
}
{
  "Tag": [
    "function"
  ],
  "Problem": "The largest integer $k$ such that $5^k$ is a factor of $100! = (1)(2)(3)(4)...(100)$ is what?",
  "Solution_1": "[hide]100! = 1*2*3*4*5*...*98*99*100\n\nNumbers with a 5 in them:\n5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95,100\n20 total\n\nNumbers with another 5 in them:\n25,50,75,100\n4 total\n\nNumbers with a third 5 in them:\n[none]\n\n20+4=24 ANS.[/hide]",
  "Solution_2": "[hide]I got 24[/hide]",
  "Solution_3": "[hide]24\n\nthe reason i didn't show my work is because i had i a whole big page of work.  then my annoying little sister deleted it all and typed all this weird stuff.  now i can't get it back. :mad: [/hide]",
  "Solution_4": "24 is what i got too.",
  "Solution_5": "Note that this is asking the same as the question, \"What are the number of zeros at the end of $100!$?\"",
  "Solution_6": "[quote=\"Silverfalcon\"]The largest integer $k$ such that $5^k$ is a factor of $100! = (1)(2)(3)(4)...(100)$ is what?[/quote]\r\n\r\n[hide=\"Solution\"]\n20 + 4\n[b]24[/b]\n[/hide]",
  "Solution_7": "you can generalise this: (i had this problem in one of my correspondence programmes)\r\n\r\nummm...i give u a rigourous proof (yeh im aussie)\r\n\r\nProve: find an expression for the greatest power of a prime number $k$ that divides $n$!\r\n\r\nProof:\r\nlet's imagine each of the integers from 1 to n strecthed out on a line as in a continued product:\r\nn!=1.2.3.4.5....n\r\nCLEARLY each term of the sequence p, 2p, 3p, ..., kp will contribute to at least one factor of p, kp being the largest multiple of p not exceeding n.\r\n\r\nlet the greatest power of p which divides n! be x\r\nif 1 <= n <= p-1 then x=0\r\nif p <= n <= $p^2$, n! will contain one factor of p from each of the corresponding multiples of p; that is x=[$n/p$] (integer part function)\r\n\r\nif $p^2$ <= n <= $p^3$-1, n! will contain one factor for each corresponding multiple of p that is not a multiple of $p^2$ and an additional one for each factor that is a multiple of $p^2$.\r\nthat is x=[$n/p$]+[$n/p^2$]\r\n\r\nthis pattern will continue:\r\n\r\nthat is x=[$n/p$]+[$n/p^2$]+[$n/p^3$]+...+[$n/p^j$] where $p^j$ is the largest power of p that does not exceed n.\r\n\r\nplug in 100 into n, 5 into p and then you get the answer: 24",
  "Solution_8": "That about covers it all. Nice proof random_mathematician.",
  "Solution_9": "I always did it that way, random_mathematician, only I used a simple proof based on doublecounting :P .\r\n\r\n[hide=\"Answer\"]\n20 + 20/5\n20 + 4\n[b]24[/b]\n[/hide]"
}
{
  "Tag": [
    "vector",
    "induction",
    "superior algebra",
    "superior algebra solved"
  ],
  "Problem": "Let $K$ a corp of caracteristic $0$ and $E$ a vectorial space over $K$.Let $E1,....En$ vectorial subspaces in $E$ with their reunion a vectorial subspace.Then there is one which contains all the others.",
  "Solution_1": "it suffices to prove that if K is an infinite field then a vector space E over K is not a finite union of subspaces strictly included in E. You can prove it by induction on n, the number of subspace:\r\nIt's clear for n=1,n=2.\r\nsuppose n>=3 and the result proved for k<n. \r\n [code]  :cup:  [/code]",
  "Solution_2": "it suffices to prove that if K is an infinite field then a vector space E over K is not a finite union of subspaces strictly included in E. You can prove this result by induction on n=dim E, the number of subspace:\r\nIt's clear for n=1,n=2.\r\nsuppose that E=F1 U F2 U .. U Fk where Fi is a strict subspace of E. Then take an arbitrary hyperplan H of E. Then H=F'1 U F'2 U .. U F'k where F'i=(H inter Fi). But , because of the inductive assumption, H is equal to one of the Fi. -contradiction because there are infinitely many hyperplans in E ( K is infinite)-",
  "Solution_3": "I've just realized I've only proved the result for vector spaces of finite dimension  :( \r\n(for the case where K=R you can prove the result thanks to Baire's theorem and the fact that a subspace have void interior)",
  "Solution_4": "it also works in the general case. Assume that E=(F1 U F2 U..U Fn) where n is minimal. Take A in E\\(F1 U F2 U..U F(n-1) ) and B in E\\Fn and consider the line L that goes through A and B. Then this line is not included in Fi. Thus L cuts Fi in only one point. -contradiction-"
}
{
  "Tag": [
    "algebra proposed",
    "algebra"
  ],
  "Problem": "For the sequence $ a_n$ we have \r\n\r\n   $ a_{n \\plus{} 1} \\minus{} a_n \\equal{} 2n \\plus{} 1 , \\forall n \\in \\mathbb N*$ and $ a_1 \\in \\mathbb Z$\r\n\r\nProve that the product of any two consecutive  terms of the sequence is also a term of the sequense.",
  "Solution_1": "[quote=\"stergiu\"]For the sequence $ a_n$ we have \n\n   $ a_{n \\plus{} 1} \\minus{} a_n \\equal{} 2n \\plus{} 1 , \\forall n \\in \\mathbb N*$\n\nProve that the product of any two consecutive  terms of the sequence is also a term of the sequense.[/quote]\r\n\r\nWe have $ (a_{n\\plus{}1}\\minus{}(n\\plus{}1)^2)\\minus{}(a_{n}\\minus{}n^2)\\equal{}0$ and so $ a_n\\equal{}n^2\\plus{}c$\r\n\r\nThen $ a_na_{n\\plus{}1}\\equal{}(n^2\\plus{}c)((n\\plus{}1)^2\\plus{}c)$ $ \\equal{}(n^2\\plus{}n\\plus{}c)^2\\plus{}c$  $ \\equal{}a_{n^2\\plus{}n\\plus{}c}$\r\n\r\nQ.E.D.",
  "Solution_2": "Call $ D_n\\equal{}a_{n\\plus{}1}\\minus{}a_{n}$. Call $ a_1\\equal{}k$\r\n\r\nSo, $ \\sum_{i\\equal{}1}^nD_i\\plus{}k\\equal{}a_{n\\plus{}1}\\equal{}n(n\\plus{}2)\\plus{}k$\r\n\r\nSo, $ a_n\\equal{}(n\\minus{}1)(n\\plus{}1)\\plus{}k$\r\n\r\nSo,\r\n$ a_na_{n\\plus{}1}\\equal{}[(n\\minus{}1)(n\\plus{}1)\\plus{}k][n(n\\plus{}2)\\plus{}k]\\equal{}n^4\\plus{}2n^3\\plus{}(2k\\minus{}1)n^2\\plus{}(2k\\minus{}2)n\\minus{}k\\plus{}k^2\\equal{}(n^2\\plus{}n\\plus{}k\\minus{}2)(n^2\\plus{}n\\plus{}k)\\plus{}k\\equal{}a_{n^2\\plus{}n\\plus{}k\\minus{}1}$"
}
{
  "Tag": [
    "geometry",
    "calculus",
    "derivative",
    "perimeter",
    "conics",
    "parabola"
  ],
  "Problem": "Uhm...Yeah I'm in Pre-Calc. and have a test Monday. I've been absent for a week and have no idea how to do these, there are two problems:\r\n\r\nENCLOSING THE MOST AREA WITH A FENCE: A farmer with 4000 meters of fencing wants to enclose a rectangular plot that borders on a river. If the farmer does not fence the side along the river, what is the largest area that can be enclosed?\r\n\r\nyet another one like it...\r\n\r\nA farmer with 2000 meters of fencing wants to enclose a rectangular plot that borders on a straight highway. If the farmer does not fence the side along the highway, what is the largest area that can be enclosed\r\n\r\nPlease help me...\r\nAIM: KomyunisutoNeko\r\nYIM: floopyjellytinkiefishie",
  "Solution_1": "I dont think this exactly qualifies for preolympiad (maybe getting started?)",
  "Solution_2": "Ok, optimization.\r\n\r\nI've never taken pre-calc, so I'm not sure if you want to find the answer by derivatives, or basic inequalities.  :huh:",
  "Solution_3": "[quote=\"Scrambled\"]I dont think this exactly qualifies for preolympiad (maybe getting started?)[/quote]Anything with calculus goes in other problem solving topics. Definately NOT getting started.  :P",
  "Solution_4": "this can be solved using getting started techniques though (since its halfway through the year in precalc, i highly doubt theyd ask you do differentiate anything)",
  "Solution_5": "[quote=\"4everwise\"][quote=\"Scrambled\"]I dont think this exactly qualifies for preolympiad (maybe getting started?)[/quote]Anything with calculus goes in other problem solving topics. Definately NOT getting started.  :P[/quote]\r\n\r\n(cough)...yah I'm new so i didn't know where the hell to put this sorry ^^;",
  "Solution_6": "Ok, here's to start you off.\r\n\r\nIf he wants to enclose 3 sides, that means the perimeter is $2l+w=4000$ instead of $2l+2w$, where $l$ and $w$ are length and width. The area is $A=lw$.\r\n\r\nBecause $2l+w=4000$, $4000-2l=w$ and $A=l(4000-2l)$. Then just take the derivative and find the maximum and minimum values.  :) \r\n\r\nWhen working on optimization, try to get everything in terms of one variable so that you can solve it easily.",
  "Solution_7": "First of all, I also don't know where \"-\" is, so yeah. But anyways this probably belongs in intermediate? Pre-calculus $\\ne$ pre-olympiad :D.\r\n\r\nOh yeah ignore 4everwise I'm AIMing you with a much beastlier method.",
  "Solution_8": "[quote=\"probability1.01\"]Oh yeah ignore 4everwise I'm AIMing you with a much beastlier method.[/quote]What's that supposed to mean?  :mad:  :roll:",
  "Solution_9": "[quote=\"4everwise\"][quote=\"probability1.01\"]Oh yeah ignore 4everwise I'm AIMing you with a much beastlier method.[/quote]What's that supposed to mean?  :mad:  :roll:[/quote]\r\n\r\nErrmm... it means I told her* to be all like $l(4000-2l) = 1,000,000 - 2(l-1000)^2 \\le 1,000,000$.\r\n\r\n*oops I said him darn oh well",
  "Solution_10": "[quote=\"probability1.01\"][quote=\"4everwise\"][quote=\"probability1.01\"]Oh yeah ignore 4everwise I'm AIMing you with a much beastlier method.[/quote]What's that supposed to mean?  :mad:  :roll:[/quote]\n\nErrmm... it means I told him to be all like $l(4000-2l) = 1,000,000 - 2(l-1000)^2 \\le 1,000,000$.[/quote]\r\n\r\nthank youu-u-u! But it was all about parabolas (in that section of the book anyways)...\r\n\r\n((sends you jelly filled donuts in the mail))"
}
{
  "Tag": [
    "geometry",
    "perimeter",
    "rectangle"
  ],
  "Problem": "A single large banquet table is created by joining the sides of several square tables, each of which seats one person for each open side. What is the maximum number of people that could be seated at a table which uses 10 small tables.",
  "Solution_1": "I got $ \\boxed{22}$... anyone get more?",
  "Solution_2": "I think that's correct. The initial count is $ 4\\cdot10\\equal{}40$, but there are at least $ 9$ places where two tables join, so we take away $ 2$ from each of those places to get $ 40\\minus{}2\\cdot9\\equal{}\\boxed{22}$.",
  "Solution_3": "[quote=\"math154\"]I think that's correct. The initial count is $ 4\\cdot10 \\equal{} 40$, but there are at least $ 9$ places where two tables join, so we take away $ 2$ from each of those places to get $ 40 \\minus{} 2\\cdot9 \\equal{} \\boxed{22}$.[/quote]\r\n\r\nWhich means that every single formation seats $ 22$?",
  "Solution_4": "You wish to maximize the perimeter of a rectangle with area 10, which clearly occurs with a 10x1 rectangle. So it is 2x(10+1)=22.\r\n\r\nAs for math154's solution, he says there are at LEAST 9 junctions."
}
{
  "Tag": [
    "calculus",
    "derivative",
    "limit",
    "function",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "If f: R --->R is continous at 0. Determine a necessary and sufficient condition for ( f(2x) - f(x) ) / x having a limit at 0.\r\n\r\nI guess the answer should be that f is differentiable at 0, and I manage to do the forward direction. However, I am stuck in doing the reverse direction.  :oops:",
  "Solution_1": "$ \\lim_{x\\to 0}\\frac{f(2x)\\minus{}f(x)}{x}\\equal{}\\lim_{x\\to 0} \\frac{f(2x)\\minus{}f(0)\\plus{}f(0)\\minus{}f(x)}{x}\\equal{}$\r\n\r\n$ \\lim_{x\\to 0} \\left( 2\\frac{f(2x)\\minus{}f(0)}{2x}\\minus{}\\frac{f(x)\\minus{}f(0)}{x}\\right)\\equal{}f'(0)$",
  "Solution_2": "That has to be what distantstar called the \"forward direction,\" which is the easy direction. You assumed differentiability and derived that the limit in question equals the derivative. What if you start merely by assuming that the limit in question exists?\r\n\r\nNote the extra hypothesis that $ f$ is continuous at zero. Without that we could simply have an example like $ f(x) \\equal{} \\text{sgn}(x),$ in which case the limit is zero but the function is not even continuous, much less differentiable.",
  "Solution_3": "Actually, $ f$ is Lipschitz is some neighborhood of zero: indeed, for $ x$ small enough $ |f(2x) \\minus{} f(x)|\\leq C|x|$ - follows from the existence of a limit. Let $ f(0) \\equal{} 0$. We can write $ |f(x)| \\equal{} |f(x) \\minus{} f(\\frac x2) \\plus{} f(\\frac x2)|\\leq |f(\\frac x2)| \\plus{} c|\\frac x2|\\leq \\dots\\leq f(0) \\plus{} c\\sum \\frac {x}{2^k} \\equal{} c|x|$.\r\nNow it's easy to prove the existence of right and left limits and conclude that they are equal.",
  "Solution_4": "Yeah, you are right. However, we can only show that f(x) / x is bounded by some positive constant c, how do we actually know that it admits a limit?",
  "Solution_5": "If $ M \\equal{} \\lim \\frac {f(2x) \\minus{} f(x)}{x}$, apply my previous post to $ g(x) \\equal{} f(x) \\minus{} Mx$ and see what you get."
}
{
  "Tag": [
    "percent"
  ],
  "Problem": "At Pigskins High School $ 20\\%$ of the boys and $ 80\\%$ of the girls attended a football game. If $ 45\\%$ of the student body is female, what percent of the school population attended the game?",
  "Solution_1": "If $ 45\\%$ of the student body is female, $ 55\\%$ is male. To solve, we find the percentages of each gender percent that attended: \r\n\r\n$ 0.8(45\\%)\\plus{}0.2(55\\%)\\equal{}\\fbox{47}\\%$."
}
{
  "Tag": [
    "ratio"
  ],
  "Problem": "Given that:\r\n\r\n\\[\\frac {4}{y} + \\frac {3}{x} = 44\\\\\r\n\\frac {12}{y} - \\frac {2}{x} = 44\\]\r\n\r\nFind the ratio of x to y.",
  "Solution_1": "[hide]4x + 3y/xy = 44\n12x - 2y/x = 44\nso\n4x + 3y/xy = 12x - 2y/xy\nso \n4x + 3y = 12x - 2y\n5y = 8x\n\nthe ratio of x to y is 8:5 or 8/5 or 8 to 5...[/hide]",
  "Solution_2": "[color=indigo][hide] first you use substitution to combine the two equations and from there you can get x/y=5/8 :lol:  [/color][/hide]",
  "Solution_3": "My method would be ...\r\n\r\n[hide=\"My answer\"]\nBecause the two equations are equal to 44, set the variable expressions equal to each other, so that:\n\n4/y + 3/x = 12/y - 2/x\n\nMultiply by x*y.\n\n4x + 3y = 12x - 2y\n\n5y = 8x\n\ny = 8/5 * x\n\nThere's your answer.\n[/hide]",
  "Solution_4": "[quote=\"Silverfalcon\"]Given that:\n\n\\[\\frac {4}{y} + \\frac {3}{x} = 44\\\\\n\\frac {12}{y} - \\frac {2}{x} = 44\\]\n\nFind the ratio of x to y.[/quote]\r\n\r\n[hide=\"click here\"]$\\[\\frac {4}{y} + \\frac {3}{x} =\\frac {12}{y} - \\frac {2}{x}$\n\n$4x+3y=12x-2y$\n\n$8x=5y$\n\n$x:y=5:8$[/hide][/hide]",
  "Solution_5": "[hide]\n4x+3y=12x-2y\n8x=5y\nx:y=5:8[/hide]"
}
{
  "Tag": [
    "AMC",
    "USA(J)MO",
    "USAMO",
    "AIME",
    "geometry",
    "number theory",
    "AMC 10"
  ],
  "Problem": "alright.. can someone describe exactly how to get in the usamo?\r\ni know something about getting an index score of 210 or something..\r\nbut.. i need to know exactly how it will be scored and stuff...\r\ni will be a senior next year.. my chances are low.. but i and willing to work my butt off for this.. i think theres no better advantage to taking a test then knowing how its gonna be graded.. \r\nya.. no one in my school has ever gone to the usamo level so i really want this..\r\nthanks..\r\nalso.. if u guys got any material/books/websites/tips/strats.. anything to help me pass.. i'd love it..\r\n\r\nthanks\r\n\r\nkent",
  "Solution_1": "I think the cut off point is 200 points total in AMC and AIME. \r\n\r\nSuppose your score for AMC = x, AIME = y. Then you need x + 10y >= 200. \r\n\r\nWhich is not that bad nowadays. AMC is a lot easier then in past. You should at least shoot for 120+ for AMC, and then 8+ should be pretty easy if you are strong in two of the four subjects: Algebra, Number Theory, Geometry, Combinatorics. By strong I mean you can do all(almost all) the problems in your area. Also, there ought to be some easy ones in your not so strong area also. So, it's not that bad. Getting into IMO is a complete different world...",
  "Solution_2": "I believe that the USAMO floor is dependent on the year.  To get into USAMO you basically have to be in the top 250.  If for some reason 500 kids get a 200 index one year then the index will be raised.  So basically just aim to get as high a score as possible.  If you really want a good numerical goal I would say that a good goal for the USAMO would be a 130 on the AMC 12 and an 8 on the AIME.  That's a 210 index which should put you in in most circumstances.  The scoring is like what koopa said.  This website is the best site I know for preparing.  I suggest taking some of their classes and reading their books.  In Volume 1 they have a lot of problems that will help you with the AMC.  I haven't got to Volume yet but I'm thinking that it will be more of late AMC to AIME and possibly more.  You should also consider ordering past AMC and AIME test for practice.  [url]http://www.kalva.demon.co.uk/[/url] has past AIME s.  If you do most of the stuff I told you and you understand it you have a good chance.  Oh yeah don't memorize formulas or problem types.  It is a lot easier to just understand the concept and use that to figure a problem out.",
  "Solution_3": "[quote=\"kentliu\"]alright.. can someone describe exactly how to get in the usamo? . . . thanks[/quote]\r\n\r\nStart by reading the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11693]FAQ on the AMC contests[/url], and follow the links from there. Don't rely solely (or at all) on the Kalva site for the solutions to old exams; instead, buy and use the official AMC copies of old exams and study and think about their solutions after taking the tests for practice. Practice old AIMEs at least weekly between now and March 2005. \r\n\r\nAs the others have said, this AoPS site is the best Web site for AMC preparation. This site is raising the level of competition in the United States, so aim high to be one of the 120 or so test-takers after the AMC 12 and AIME next year. \r\n\r\nGood luck.",
  "Solution_4": "So is there one cutoff score each year for all grades?  Or is there a score for each grade?\r\nI'm going to be a 10th grader this year and was wondering what the cutoff score might be.  Also, the top 250 students are accepted in each grade nationwide or is it the top 250 students in all grades in a certain state?  I'm guessing it's the top 250 students in all grades nationwide that make it to the USAMO.  I live in Texas just to let you'll know, so maybe some of you know the requirements for Texas usually.\r\n\r\nIt's cool how if you get a 150 on the AMC you only have to get 5 questions right on the AIME (not that that is easy or anything, just fun to point out)",
  "Solution_5": "From the AMC website (http://www.unl.edu/amc):\r\n\r\nSelection for the USAMO will be made according to the follwing rules:\r\n\r\nThe goal is to select about 250 of the top scorers from the prior AIME and AMC 12A, AMC 12B, AMC 10A and AMC 10B contests to participate in the USAMO. \r\n\r\nSelection will be based on the USAMO index which is defined as 10 times the students AIME score plus the students score on the AMC 12 or the AMC 10. \r\n\r\nThe first selection will be the approximately 160 highest USAMO indices of students taking the AMC 12A or AMC 12B contest. \r\n\r\nThe lowest AIME score among those 160 first selected will determine a floor value. The second selection of USAMO participants will be from the highest USAMO indices among students who took the AMC 10A or AMC 10B and the AIME, and got an AIME score at least as high as the floor value. \r\n\r\nThe student with the highest USAMO index from each state, territory, or U.S. possession not already represented in the selection of the first and second groups will be invited to take the USAMO. \r\n\r\nTo adjust for variations in contest difficulty, the number of students selected from A & B contests will be proportional to the number of students who took the (A & B) Contests. \r\n\r\nThe selection process is designed to favor students who take the more mathematically comprehensive AMC 12A and AMC 12B contests.",
  "Solution_6": "Unfortunately, most years, even with 150 on the AMC 12, you will not make USAMO with a 5 on AIME. This year you needed at least 7 and last year at least 8.",
  "Solution_7": "Yes, that would be unfortunate, and I could be wrong, but what are the chances that a person who makes a 150 on the AMC 10 or 12 is going to make a 5 on the AIME?  It's possible I guess, but just not too likely, unless that person got really lucky guessing or something, which wouldn't help him/her on the AIME.  Now, I just got both AoPS volumes, and I have some serious studying to do.  Good day, and thanks for the help.",
  "Solution_8": "[quote=\"Rep123max\"]Does that mean since I will be a Freshmen next year and probably take AMC 10, if I qualify for AIME, my AMC 10 score wont matter, its just that I get whatever the floor was?[/quote]\r\n\r\nIf and only if the ninth-grade MOP program exists next year, and works the same way it did this year, AND if and only if there aren't a whole lot more ninth graders trying to get into MOP and succeeding at qualifying for AIME, then it will be as you suppose, because it was that way this year. This year people with USAMO scores of 0 went to ninth-grade MOP, and I think they got into USAMO in some cases mostly having a floor score on the AIME (which is not trivially easy to obtain) and whatever AMC 10 score got them into AIME in the first place (which anyone who has had algebra 1 and geometry ought to be able to obtain). But if next year is more competitive, all this may change, and I expect AoPS's very existence to make next year more competitive. And having ninth-grade MOP at all depends on funding, as indeed offering a broad eligibility for USAMO also does. \r\n\r\nGood luck.",
  "Solution_9": "[quote=\"Rep123max\"]So the floor is around 7 or 8? Does that mean since I will be a Freshmen next year and probably take AMC 10, if I qualify for AIME, my AMC 10 score wont matter, its just that I get whatever the floor was?[/quote]\r\n\r\nbasically yes, all you need is a aime score higher than the floor and an amc 10 score that aime-quals.  it's substantially more difficult when you're older tho...",
  "Solution_10": "oh man.. so AoPS is a new site?\r\nman.. its gonna be tough for senior like me then..",
  "Solution_11": "[quote=\"koopa\"]I think the cut off point is 200 points total in AMC and AIME. \n\nSuppose your score for AMC = x, AIME = y. Then you need x + 10y >= 200. \n\nWhich is not that bad nowadays. AMC is a lot easier then in past. You should at least shoot for 120+ for AMC, and then 8+ should be pretty easy if you are strong in two of the four subjects: Algebra, Number Theory, Geometry, Combinatorics. By strong I mean you can do all(almost all) the problems in your area. Also, there ought to be some easy ones in your not so strong area also. So, it's not that bad. Getting into IMO is a complete different world...[/quote]\r\n\r\nkoopa, u said algebra, number theory, gemetry, and combinatorics..\r\n\r\nmay i ask what topics are involved in number theory and combinatorics?",
  "Solution_12": "Here's a description from a novice:\r\n\r\nNumber theory is like really advanced arithmetic.\r\n\r\nThe permutation and combination formulas you learn in Alg II are combinatorics.  There's a lot more, but that will give you an idea.",
  "Solution_13": "They say that all the problems can be done without any math above precal.  That s not to say that anyone who has completed precal can do good.  They pretty much take seemingly easy concepts and make them a whole lot more tricky.  \r\n\r\nHere 's a sample NT problem \r\n\r\n Find the largest integer n such that n + 10 divides n^3 + 100.\r\n\r\n\r\nI would say that there is usually more algebra/ geometry on the AIME on average.",
  "Solution_14": "[hide]\n\nDividing n :^3: + 100 by n + 10 we get n :^2: - 10n + 100n + 10n :^2: - 900/(n+10).\n\n\n\nFor n + 10 to divide n :^3:  + 100, the result must be an integer, so n :^2: - 10n + 100n + 10n :^2: - 900/(n+10) must be an integer and thus 900/(n+10) must be also.\n\n\n\nThe largest number is obviously 890, when the fraction is 1[/hide]",
  "Solution_15": "[quote=\"tokenadult\"][quote=\"Rep123max\"]Does that mean since I will be a Freshmen next year and probably take AMC 10, if I qualify for AIME, my AMC 10 score wont matter, its just that I get whatever the floor was?[/quote]\n\nIf and only if the ninth-grade MOP program exists next year, and works the same way it did this year, AND if and only if there aren't a whole lot more ninth graders trying to get into MOP and succeeding at qualifying for AIME, then it will be as you suppose, because it was that way this year. This year people with USAMO scores of 0 went to ninth-grade MOP, and I think they got into USAMO in some cases mostly having a floor score on the AIME (which is not trivially easy to obtain) and whatever AMC 10 score got them into AIME in the first place (which anyone who has had algebra 1 and geometry ought to be able to obtain). But if next year is more competitive, all this may change, and I expect AoPS's very existence to make next year more competitive. And having ninth-grade MOP at all depends on funding, as indeed offering a broad eligibility for USAMO also does. \n\nGood luck.[/quote]\r\n\r\nIFF the program exists next year and works the same way?! The \"if\" part is certainly true, but I don't know about the \"only if\" part.\r\n\r\nBut if you think you'll be able to make a ~130 on AMC 12, and a ~10 on AIME, you're pretty safe, right?",
  "Solution_16": "The last three years have seen substantially easier contests and much more competition.  In 2002, the USAMO cutoff was an index of 210 (AIME floor of 6 or 7; I can't remember this one exactly). In 2003, the cutoff was a steep 226 (AIME floor of 8), but last year it came down to 210 (AIME floor of 7) again thanks mostly to slightly harder AMC-10/12 and AIME contests.\r\n\r\nPreviously, the AMC-12 had been much more difficult, with the AIME cutoff (through the AMC-12) dipping well below 100 just a couple years earlier.\r\n\r\nBottom line - I don't know if a 230 is particularly safe.  It all depends on what the folks in Lincoln have put together.  A 230 hasn't failed to reach USAMO yet, but it was pretty close in 2003."
}
{
  "Tag": [],
  "Problem": "Suppose you are given a graph of y = f(x).  How would you use that graph to sketch y = -f(x+1) + 2?\r\n\r\nIf you find Intermediate problems pretty easy, please leave this challenge problem for less experienced students to solve.\r\n\r\nSource: Alabama State Contest",
  "Solution_1": "I haven't been here in a looooooong time...\n\n\n\n[hide]\n\nFirst, the graph of y=f(x+1) would be the graph of y=f(x) but moved to the left 1 unit.  y=-f(x+1) would be f(x+1) reflected over the x axis.  Finally, y=-f(x+1) + 2 would be the graph of -f(x+1) moved up 2 units.\n\n[/hide]",
  "Solution_2": "[hide]Flip it over the x-axis, move it left one, and up 2.[/hide]"
}
{
  "Tag": [],
  "Problem": "hi my name is rami i an a new member \r\n\r\nany body want to say hi for me ?\r\n\r\n\r\n\r\n :blush: \r\n\r\n[url=http://zzrz.com/english.htm]http://zzrz.com/english.htm[/url]",
  "Solution_1": "Hello rami, welcome to this site!\r\nBy the way, you should post it in the Game and Fun Factory, there is also a self-introduction thread! :D",
  "Solution_2": "hey rami! welcome!  :D",
  "Solution_3": ":jump:  :clap:  :w00tb:  :welcomeani:",
  "Solution_4": ":) welcome! hope you like the site!"
}
{
  "Tag": [],
  "Problem": "What is the number of leap years between 1937 and 2005?",
  "Solution_1": "[hide]$68/4=17$? i think. but thats assuming that 1938 is a leap year[/hide]",
  "Solution_2": "[quote=\"ToBe\"][hide]$68/4=17$? i think. but thats assuming that 1938 is a leap year[/hide][/quote]\r\n\r\nBut 1938 wasn't a leap year...",
  "Solution_3": "[b]it must be 16 since (2000-1937)/4 +1[/b]",
  "Solution_4": "[quote=\"nvidura\"][b]it must be 16 since (2000-1937)/4 +1[/b][/quote]\r\n\r\nwould you please hide your answer next time, so other ppl can try. thx!",
  "Solution_5": "by the way, I got the same answer, same way.",
  "Solution_6": "how to find leap years:\r\n[hide]to find out whether or not a year is a leap year, you use the last two digits of the year, and if it's divisible by $4$, it's leap. The exception is when the last two digits is $00$, in that case you divide all four digits by $4$[/hide]\nanswer:\n[hide]I got $17$... i've got no clue how y'all did it: why would you do $2000-1937$? doesn't the question ask to find from $1937$ to $2005$? but you have to notice $1937$ isn't a leap year, so you have to start on $1940$, so it would be $2005-1940 = 65$, then you use $\\frac{65}{4}  +1 = 16+1 = 17$.... but I might be completely wrong here...[/hide]\r\nbtw, Kristy, you aren't really suppose to double post... but I'm only saying... :lol:"
}
{
  "Tag": [
    "geometry",
    "MATHCOUNTS",
    "algebra",
    "polynomial",
    "function",
    "AMC",
    "USA(J)MO"
  ],
  "Problem": "I got a 118.5 on AMC 12, and I really want to qualify for USAMO. That means I probably need a 10 or 11 on AIME. Right now I'm averaging 7-8, and I was wondering which topics you think are the easiest to improve on.",
  "Solution_1": "personally, Geometry was easiest to improve.  You just need to pinpoint the key points.  In one year, i went from a 16 on the AMC 8 :oops:    to a 138 on the AMC 10 :D   this year(which had lots of geometry).",
  "Solution_2": "[quote=\"indianamath\"]personally, Geometry was easiest to improve.  You just need to pinpoint the key points.  In one year, i went from a 16 on the AMC 8 :oops:    to a 138 on the AMC 10 :D   this year(which had lots of geometry).[/quote]\r\nActually, I'm going to have to say that it's number theory.\r\nNumber theory may look confusing, but once you get past the simple basics (mods and whatnot) it's smooth sailing.",
  "Solution_3": "number theory is easy to improve in, but AMCS's and MATHCOUNTS don't use Number Theory that much.",
  "Solution_4": "[quote=\"indianamath\"]number theory is easy to improve in, but AMCS's and MATHCOUNTS don't use Number Theory that much.[/quote]\r\n\r\nare you kidding?\r\n\r\nhave you ever even SEEN an AMC?\r\n\r\nit's 25 number theory questions",
  "Solution_5": "not on AMC 10's",
  "Solution_6": "I heard polynomials and functions are used in questions toward the end of the AMC 10.",
  "Solution_7": "[quote=\"indianamath\"]number theory is easy to improve in, but AMCS's and MATHCOUNTS don't use Number Theory that much.[/quote]\r\nNumber 25 on the AMC 10 was Number Theory (and Algebra).\r\nI think Number 22 on the 10 was as well, I'm not sure though. (don't exactly remember the question, but I remember thinking that there were two number theory near the end).",
  "Solution_8": "those weren't really pure number theory problems. I thought you were talking about like finding the tens digit of (blahblahblah) and stuff like that.",
  "Solution_9": "[quote=\"indianamath\"]those weren't really pure number theory problems. I thought you were talking about like finding the tens digit of (blahblahblah) and stuff like that.[/quote]\r\nMathcounts is actually very NT heavy.\r\nLet's look at past nationals.\r\n2005: 3 out of last 10\r\n2004: 3 out of last 10\r\n2003: 1 out of last 10\r\n2002: 3 out of last 10",
  "Solution_10": "ok i'm wrong, just like a lot of my answers in MATHCOUNTS. :)",
  "Solution_11": "USAMO is very number theory heavy, which is why I can never do the problems when I look at one.",
  "Solution_12": "Personally, I think number theory is good too, because it provides a basis for a lot of the other things you'll need to do...",
  "Solution_13": "The easiest subject to improve quickly in is \"Techniques for cutting back on stupid mistakes\". I'm serious; if you master this between now and the AIME your score will go up by at least two points. (or twelve in AMC terms)",
  "Solution_14": "pressure with a capital P can bring out the worst in you",
  "Solution_15": "[quote=\"randomdragoon\"]The easiest subject to improve quickly in is \"Techniques for cutting back on stupid mistakes\". I'm serious; if you master this between now and the AIME your score will go up by at least two points. (or twelve in AMC terms)[/quote]\r\n\r\nI would really like to improve on this. Do you have any tips for doing it?",
  "Solution_16": "Tip 1: [u]Underline[/u] the question being asked. Put a square around important information that you might forget to take into account later into doing a question (such as a>b, x is an integer, y is positive, etc.)\r\n\r\nTip 2: Use scrap paper liberally. [b]Write big and neat.[/b] This will avoid errors such as turning + into -1, 4 into 9, dropping - signs, etc. Also, draw [i]large[/i] diagrams for geometry problems. \r\n\r\nAlso, when doing algebra, don't fail to write out intermediate steps just because it isn't cool. Don't write out the intermediate steps if you can see how to get there quickly, but you have to think hard about what's going on you might as well write it out. Things can get muddled in your head.\r\n\r\nTip 3: [b]Organize[/b] your scrap paper. Do not turn half-used scrap paper upside down and start working on another problem; if you have ample space draw a line under the last problem and start working on the next problem under it. [b]If in doubt get another piece of scrap paper.[/b] This tip will save you time when you go back and check over your answers.\r\n\r\nTip 4: Always allot time to check your work. Unless you are a master at avoiding silly errors because of the first three tips (or screwed up following tip 5) chances are you'll gain more points catching stupid mistakes than trying to solve more problems under time pressure.\r\n\r\nExample: Many people on the AIME make 2 stupid errors. Say you have gotten to question 12 already. Which do you think is more likely? Backtracking and finding those two errors, netting you 2 points, or solving two of the last three problems?\r\n\r\nTip 5: Don't get caught up in a hard problem if there are others that you haven't even read yet. The AMCs and AIMEs try, but are never in perfect difficulty order. Just because it's problem 15/25, doesn't mean you don't know how to do it.",
  "Solution_17": "[quote=\"tjhance\"][quote=\"randomdragoon\"]The easiest subject to improve quickly in is \"Techniques for cutting back on stupid mistakes\". I'm serious; if you master this between now and the AIME your score will go up by at least two points. (or twelve in AMC terms)[/quote] \n\nI would really like to improve on this. Do you have any tips for doing it?[/quote]\r\n\r\nTry reading this:\r\n\r\nhttp://www.artofproblemsolving.com/Resources/AoPS_R_A_Mistakes.php\r\n\r\nNumber Theory is easy to improve in for MATHCOUNTS and the AMC, because most NT problems on those tests are mods and  . . . mods. :P  However, if you're looking at hard AIME and USAMO, number theory requires a lot of work to get good at.\r\n\r\nFor the AIME, I'd say the qickest to improve in would be complex numbers/trigonometry.  For some reason, the AIME often assumes that students don't know those subjects and has a problem with them at around #11-#13 that is about #5 in difficulty if you learn the basic ideas (see 2002 AIME I #12).  Unfortunately, the trig #12 on last year's AIME I was far from trivial, so maybe they've gotten away from this. :( \r\n\r\nIf you are geometrically inclined (I'm not), geometry is probably good too."
}
{
  "Tag": [
    "trigonometry"
  ],
  "Problem": "find  all the solutions of the eq.\r\n\\[cos(12x)=5sen(3x)+9tg^2x+ctg^2x\\]\r\nin the interval $[0,2\\pi]$",
  "Solution_1": "Um, I'm just starting trig, so can someone help me with what $sen$ $tg$ and $ctg$ are?",
  "Solution_2": "this is probably just a terribly misguided guess, but i'm going to guess that sen=secant, tg=tangent, and ctg=cotangent.\r\n\r\ni've never seen ctg, tg, and sn, before either >_<",
  "Solution_3": "it could be a peruvian thing...",
  "Solution_4": "It's a chinese way of saying it too.\r\n\r\nDo you know what they mean? You should.\r\n\r\nReply, and I'll tell you."
}
{
  "Tag": [
    "algebra",
    "difference of squares",
    "special factorizations",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Find all (x,y), given that\r\n$ x,y \\in Z \\\\\r\nx^2(y\\minus{}1)\\plus{}y^2(x\\minus{}1)\\equal{}1$",
  "Solution_1": "[quote=\"NikolayKaz\"]Find all (x,y), given that $ x,y \\in Z, x^2(y \\minus{} 1) \\plus{} y^2(x \\minus{} 1) \\equal{} 1$[/quote]\r\n\r\nRewriting the equation we have $ \\displaystyle xy \\plus{}2\\minus{}x\\minus{}y\\equal{} \\frac {5}{x \\plus{} y \\plus{} 2} \\in \\mathbb{Z}$. So we are lead in a finite number of cases (only 4 because 5 is prime..)",
  "Solution_2": "Wow, how did you see that? What inspired you to do that? o_O\r\n\r\n[hide=\"What I did :/\"]Let $ u \\equal{} x \\plus{} y$ and $ v \\equal{} xy$. We have $ x^2y \\minus{} x^2 \\plus{} xy^2 \\minus{} y^2 \\minus{} 1 \\equal{} 0$, or $ uv \\minus{} (u^2 \\minus{} 2v) \\minus{} 1 \\equal{} 0$, or $ u^2 \\minus{} uv \\minus{} 2v \\plus{} 1 \\equal{} 0$. Multiplying by 4, we have $ 4u^2 \\minus{} 4uv \\minus{} 8v \\plus{} 4 \\equal{} 0$, or $ (2u \\minus{} v)^2 \\minus{} v^2 \\minus{} 8v \\plus{} 4 \\equal{} 0$, that is, $ (2u \\minus{} v)^2 \\minus{} (v \\plus{} 4)^2 \\plus{} 20 \\equal{} 0$, so $ (v \\plus{} 4)^2 \\minus{} (2u \\minus{} v)^2 \\equal{} 20$. We have a difference of squares, so we again only have finitely many cases to check. \n\nGranted, many many more cases. Oh well. [/hide]",
  "Solution_3": "[quote=\"Zhero\"]Wow, how did you see that? What inspired you to do that? o_O [...] or $ u^2 \\minus{} uv \\minus{} 2v \\plus{} 1 \\equal{} 0$. [...][/quote]\r\n\r\nBut it is the same equation, only try to explicit v and reduce the other fraction  :wink:",
  "Solution_4": "[hide]\nFirst, putting y = x giving (x^2)(x-1) = 0.5 which is impossible, obviously putting unknown(s) to be zero gets no solution.\nAs x^2, y^2 >= 1,\nso we first put one of the unknown be 1 obtaining the other be 2, so (1,2) or (2,1) are part of the solutions.\nWLOG, assuming x>y,\nwith greater x, a negative y is needed for obtaining solutions.\ne.g., putting x = 3,\ny = -0.5 or -0.4. which neither are integers and both negative.\nSo putting x = a + 1, y = b + 1,\neasily we get:\n(1 + ab)(a + b)  + 4ab = 1, which L.H.S. is negative while the right is positive.\nSo the only solutions are (1,2) and (2,1).\n[/hide]",
  "Solution_5": "What's about this problem : Find all pairs of integers $ (x,y)$  and positive integer n such that :\r\n$ x^2(y^n\\minus{}1)\\plus{}y^2(x^n\\minus{}1)\\equal{}1$",
  "Solution_6": "[quote=\"TTsphn\"]What's about this problem : Find all pairs of integers $ (x,y)$  and positive integer n such that :\n$ x^2(y^n \\minus{} 1) \\plus{} y^2(x^n \\minus{} 1) \\equal{} 1$[/quote]\r\nthat's easy since u can notice $ xy(y^{n\\minus{}1}x\\plus{}x^{n\\minus{}1}y)\\equal{}1\\plus{}x^2\\plus{}y^2$ but it's known that if $ \\frac{1\\plus{}x^2\\plus{}y^2}{|xy|}$ is an integer then it's equal to 3(the cases $ x\\equal{}0$ or $ y\\equal{}0$ are trivial).it means that $ |y^{n\\minus{}1}x\\plus{}x^{n\\minus{}1}y|\\equal{}3$ so we are left with 2 cases which are trivial."
}
{
  "Tag": [],
  "Problem": "In how many ways can 4 on/off switches in a row be set so that no two adjacent switches are on?",
  "Solution_1": "There are 8 ways for there to be at lest 2 switches on:\r\n\r\n1100\r\n1101\r\n0110\r\n0011\r\n1011\r\n1110\r\n0111\r\n1111\r\n\r\nThere are 2^4 - 8 = 8 ways for there to be no adjacent on switches",
  "Solution_2": "You can also just count the number of ways not to have two adjacent switches on:\r\n\r\nAll 4(0)s,\r\n3(0)s and 1(1),\r\nand\r\n\r\n1001\r\n1010\r\n0101\r\n\r\nfor a total of $ 1 \\plus{} 3 \\plus{} 4 \\equal{} \\boxed{8}$ ways.",
  "Solution_3": "We can either have $ONON\\implies 2!\\times 2!=4\\text{ ways}$ or $NONO\\implies 4\\text{ ways}$. So $4+4=\\boxed{8\\text{ ways}}$",
  "Solution_4": "If there is 2 on switch then there is 3 ways. If there is 1 on switch we have 4 ways. If there is 0 on switch there is 1 way. Hence $3+4+1=8.$"
}
{
  "Tag": [
    "geometry"
  ],
  "Problem": "I can't figure out this question:\r\n\r\nThe radius of a particular circle is inscribed in a equilateral triangle is 2 units.  What is the perimenter of the riangle?  Express your answer in simplist radical form.\r\n\r\nThanks in advance. :lol:",
  "Solution_1": "Blah, I'll make a diagram later.\r\n\r\nBasically, make a large triangle that uses 120 degrees of the central angle. Then divide it into a 30 60 90 triangle with one side as half of the side of the triangle. You can find that half, and then multiply by 6.",
  "Solution_2": "couldn't you just draw the radius of the circle te be perpendicular to the traingle's base to form a 30 60 90 with 2 as the shorter leg the longer leg which is 1/2 of a side of a triangle to be 2root 3. So multiply by 2 and 4root 3 is the length of each side of the equilateral. the area of an equialeteral is A=(S^2root3)/4 so the area should be [b]12root3[/b]  :D \r\n\r\nAm I right Adeesh",
  "Solution_3": "[quote=\"neurosurgeon1118\"]couldn't you just draw the radius of the circle te be perpendicular to the traingle's base to form a 30 60 90 with 2 as the shorter leg the longer leg which is 1/2 of a side of a triangle to be 2root 3. So multiply by 2 and 4root 3 is the length of each side of the equilateral. the area of an equialeteral is A=(S^2root3)/4 so the area should be [b]12root3[/b]  :D \n\nAm I right Adeesh[/quote]\r\n\r\n\r\n\r\nNo fingers, but it said perimeter.",
  "Solution_4": "Thats exactly what I said...\r\n\r\n :huh:",
  "Solution_5": "[quote=\"happyme\"]Thats exactly what I said...\n\n :huh:[/quote]\r\n\r\n\r\nDon't copy of other people *cough karthik cough*",
  "Solution_6": "sorry guys i just couldn't help myself, Adeesh was this on Warm Up 13",
  "Solution_7": "[quote=\"neurosurgeon1118\"]sorry guys i just couldn't help myself, Adeesh was this on Warm Up 13[/quote]\r\n\r\nNever give names out, surgeon, its denies netiquitte",
  "Solution_8": "Uh, its fine I guess?\r\n\r\nI have no clue whats going on, this is getting slightly off-topic.\r\n\r\nI have posted problems (and solutions) to most of the early Warm-Ups, so look in the past 2 or 3 pages, you'll get a lot of answers.\r\n\r\nAnd didn't you just call the other guy Cartick or something?\r\n\r\no.o",
  "Solution_9": "yeah were friends",
  "Solution_10": "Welcome to AoPS.",
  "Solution_11": "ajain is that an f-117"
}
{
  "Tag": [
    "integration",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Suppose $ (a_n)_{n\\equal{}0}^\\infty$ is a sequence of positive numbers such that $ \\sum_{n\\equal{}0}^\\infty a_n$ diverges. What can you say about \r\n$ \\sum_{n\\equal{}0}^\\infty \\frac{a_n}{(a_0\\plus{}\\ldots\\plus{}a_n)^2}$?",
  "Solution_1": "Let $ s_n \\equal{} a_0 \\plus{} \\cdots \\plus{} a_n$. Then $ 0 < s_{n \\minus{} 1} < s_n$ and hence $ \\frac {1}{s_{n \\minus{} 1}} \\minus{} \\frac {1}{s_n} \\equal{} \\frac {a_n}{s_{n \\minus{} 1} s_n} > \\frac {a_n}{s_n^2} > 0$. Summing each sides shows that $ \\sum_{n \\equal{} 0}^{\\infty} \\frac {a_n}{s_n^2}$ is bounded, hence convergent.",
  "Solution_2": "Nice.  It seems (from the small assortment of examples I looked at) that perhaps the 2 in the exponent could be replaced with a smaller number and the result would still be true.  So, can anyone either prove this or find a sequence for which the 2 is best possible, i.e. $ \\sum_{n \\geq 0} \\frac {a_n}{s_n^{2 \\minus{} \\varepsilon}}$ diverges for all $ \\varepsilon > 0$?",
  "Solution_3": "$ \\sum_{n \\equal{} 0}^\\infty \\frac {a_n}{(a_0 \\plus{} \\ldots \\plus{} a_n)^{1\\plus{}\\epsilon}} \\approx \\int_0^{\\infty} \\frac{f'(t)}{f(t)^{1\\plus{}\\epsilon}} dt$\r\nwhere $ f(x) \\approx \\sum_0^x a_k$ is a positive  increasing function. Because $ \\frac{1}{f^{\\epsilon}(t)}$ is a primitive of $ \\frac{f'(t)}{f(t)^{1\\plus{}\\epsilon}}$, that \"shows\" that for every $ \\epsilon > 0$, the sum $ \\sum_{n \\equal{} 0}^\\infty \\frac {a_n}{(a_0 \\plus{} \\ldots \\plus{} a_n)^{1\\plus{}\\epsilon}}$ converges. (you can easily make all this a proper proof, I think ... )",
  "Solution_4": "Very nice solutions [b]sos440[/b] and [b]alekk[/b]!"
}
{
  "Tag": [
    "LaTeX",
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "prove:$r>1$,$\\frac{a}{ra+b+c}+\\frac{b}{rb+a+c}+\\frac{c}{rc+a+b}\\leq \\frac{3}{2+r}$\r\nand also what about the general case",
  "Solution_1": "easy,but I can't use LaTeX,so can't write here,Who can tell me how to use LaTeX in this forum",
  "Solution_2": "we are are chinese.So if you like,tell me your qq number.I'll tell you how to use.(but a little) :D",
  "Solution_3": "Expansion works  :blush: .\r\nExpanding all the terms we can know that this inequality is equivalent to:\r\n\\[6(r^2-r)abc \\le (r-1)\\displaystyle\\sum_{sym}a^3+(r^2-2r+1)\\displaystyle\\sum_{sym}a^2b.\\]But I think it is easy now.  :)\r\n\r\nEdit: Jensen also works after homogenization using $a+b+c=1$.",
  "Solution_4": "i correct my latex :?",
  "Solution_5": "Cause of homogenious we can assume that \r\n\\[a+b+c=1\\]\r\nand the inequality becomes:\r\n$ \\frac{a}{(r-1)a+1}+ \\frac{b}{(r-1)b+1}+ \\frac{c}{(r-1)c+1}$ <=$\\frac{3}{r+2}$\r\ncause of symetri suppose that:\r\n$a$ <= $b$ <= $c$ and as $r$  >= 1 then :\r\n$ \\frac{1}{(r-1)a+1}$ >= $ \\frac{1}{(r-1)b+1}$ >= $ \\frac{1}{(r-1)c+1}$\r\nChebetchev gives:\r\n$3(\\frac{a}{(r-1)a+1}+\\frac{b}{(r-1)b+1}+\\frac{c}{(r-1)c+1})$<=$(a+b+c)(\\frac{1}{(r-1)a+1}+\\frac{1}{(r-1)b+1}+\\frac{1}{(r-1)c+1})$=$\\frac{1}{(r-1)a+1}+\\frac{1}{(r-1)b+1}+ \\frac{1}{(r-1)c+1}$\r\n=$\\frac{((ab+ac+bc)(r-1)^2+2(r-1)+3)}{((r-1)a+1))((r-1)b+1)((r-1)c+1)}$\r\nOr $ab+ac+bc$ <= $\\frac{1}{3}$\r\nAnd$((r-1)a+1))((r-1)b+1)((r-1)c+1))$ <= $(\\frac{(r-1)(a+b+c)+3}{3})^3$ AMGM\r\nThen $3(\\frac{a}{(r-1)a+1}+\\frac{b}{(r-1)b+1}+\\frac{c}{(r-1)c+1})$ <=$9\\frac{(r-1)^2+6[r-1)+9}{(r+2)^3}$\r\nAnd we done.",
  "Solution_6": "thank you for your proof.\r\nand here is my proof:\r\nwe have:$\\frac{a}{ra+b+c}+\\frac{b}{rb+a+c}+\\frac{c}{rc+a+b}\\leq \\frac{b}{ra+b+c}+\\frac{c}{rb+a+c}+\\frac{a}{rc+a+b}$,$(1)$\r\n$\\frac{a}{ra+b+c}+\\frac{b}{rb+a+c}+\\frac{c}{rc+a+b}\\leq \\frac{c}{ra+b+c}+\\frac{a}{rb+a+c}+\\frac{b}{rc+a+b}$,$(2)$\r\nadd $(1)+(2)+r \\frac{a}{ra+b+c}+\\frac{b}{rb+a+c}+\\frac{c}{rc+a+b}$\r\n :)  :) \r\nand I'm sorry I find $AYMANE's$ solution doesn't work.check it please. :D  :D \r\nJensen won't be work(I think),maybe RCF theorem will work.",
  "Solution_7": "Ooops......you are right,isee my mistake :blush:",
  "Solution_8": "Finally, something non computational  :D . Let us prove that for $ r\\geq 1$ and $a_1,...,a_n>0$ we have \r\n  $\\sum{\\frac{a_1}{ra_1+a_2+...+a_n}}\\leq\\frac{n}{n+r-1}$. Put $S=a_1+...+a_n$ and $ x_i=\\frac{a_i}{(r-1)a_i+S}$. Then we have $n+r-1=\\sum{\\frac{1}{1-(r-1)x_i}}\\geq\\frac{n^2}{n-(r-1)(x_1+...+x_n)}$ from where we get $x_1+...+x_n\\leq\\frac{n}{n+r-1}$",
  "Solution_9": "thank you for your nice proof :) $harazi$",
  "Solution_10": "My solution. Noting the given inequality is equivalent to \r\n\\[\r\n\\sum {(1 - r)(a - b)^2 (rc + a + b)}  \\leqslant 0\r\n\\]\r\n :lol:"
}
{
  "Tag": [
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "if $ A \\subseteq {\\mathbb{R}}^n$ be compact and has measure 0 then A has content 0.",
  "Solution_1": "The only difference between Lebesgue measure and Jordan content is that when you are finding the content, you use only [b]finite[/b] covers. The rest should be clear.\r\n\r\nBy the way, it seems like I smell homeworks here :lol:."
}
{
  "Tag": [
    "trigonometry"
  ],
  "Problem": "Find the sum of all the roots contained in the interval $ [0;2\\pi]$\r\n\r\n$ \\sqrt {3}\\tan x - \\sqrt {3} \\sin2x + cos2x = 0$\r\n\r\n[hide=\"My Solution\"]Dividing both sides by $ \\sin 2x$\n$ \\sqrt {3}\\cdot\\frac {1}{2\\cos^2x} - \\sqrt {3} + \\text{cot}2x = 0 \\\\\n\\therefore\\;\\sqrt {3}\\cdot \\frac { - \\cos 2x}{2\\cos^2x} + \\frac {1 - \\text{tan}^2x}{2\\text{tan}x} = 0 \\\\\n\\therefore\\;\\sqrt {3}\\cdot\\frac {\\cos2x}{\\cos^2x} = \\frac {\\cos2x}{\\cos^2x\\cdot\\text{tan}x} \\\\\n\\therefore\\;\\text{tan}x = \\frac {\\sqrt {3}}{3}\\text{ or }\\cos2x = 0$\n\n$ \\text{tan}x = \\frac {\\sqrt {3}}{3}\\;\\;x = \\{\\frac {\\pi}{6}\\,,\\,\\frac {7\\pi}{6}\\}$\n\n$ \\cos 2x = 0\\;\\;2x = \\frac {\\pi}{2} + k\\pi\\;\\therefore\\;x = \\{\\frac {\\pi}{4}\\,,\\,\\frac {3\\pi}{4}\\,,\\,\\frac {5\\pi}{4}\\,,\\,\\frac {7\\pi}{4}\\}$[/hide]\r\n\r\nI solved it like this, but I was wondering if anybody had a more clever and faster solution for this problem. Thanks in advance.",
  "Solution_1": "hello, do you mean $ \\sqrt{3}\\tan(x)\\minus{}\\sqrt{3}\\sec(2x)\\plus{}\\cos(2x)\\equal{}0$?\r\nSonnhard.",
  "Solution_2": "No,  it is $ \\sin(2x)$",
  "Solution_3": "hello, you can rewrite your equation int the form\r\n$ 4\\sec(x)\\sin(\\pi/6\\minus{}x)\\sin(\\pi/4\\minus{}x)\\sin(\\pi/4\\plus{}x)\\equal{}0$\r\nSonnhard.",
  "Solution_4": "hello, i have found the solutions\r\n$ x\\equal{}\\frac{\\pi}{6}$\r\n$ x\\equal{}\\frac{\\pi}{4}$\r\n$ x\\equal{}\\frac{3\\pi}{4}$\r\nSonnhard.",
  "Solution_5": "Let $ t \\equal{} \\tan x\\Longrightarrow \\sin 2x \\equal{} \\frac {2t}{1 \\plus{} t^2},\\ \\cos 2x \\equal{} \\frac {1 \\minus{} t^2}{1 \\plus{} t^2}$, \r\n\r\nThe given equation can be written as $ \\sqrt {3}t^3 \\minus{} t^2 \\minus{} \\sqrt {3}t \\plus{} 1 \\equal{} 0$\r\n\r\n$ \\Longleftrightarrow (\\sqrt {3}t \\minus{} 1)(t \\plus{} 1)(t \\minus{} 1) \\equal{} 0$, which follows Dr Sonnhard Graubner's solution. :lol:"
}
{
  "Tag": [
    "geometry",
    "trigonometry",
    "number theory"
  ],
  "Problem": "I've often had this thought and was wondering if anyone else had too. Actually, I'm mainly just posting this because I thought it was interesting.\r\n\r\nOften when you are trying to solve a hard problem, you try one method of attack for a while, and end up with a completely horrible equation to solve, say. You can see absolutely no way to solve this equation, you go back to the start and eventually solve the problem in a 'nice' way.\r\nNow what I've thought is that after this, you could say \"hey - what I have just done is a way to prove that horrible equation!\"\r\nOr to give a more specific example, you have a geometry problem, try a trig bash and, as long as you don't make any mistakes, you end up with a horrible trig equation that is probably \"impossible\" to prove. However, you could prove it by saying this trig equation is true if and only if this geometry problem is true, which I can prove in a nice way.\r\n\r\nOf course, there'd be absolutely no way you'd be able to translate something horrible into a completely random geometry problem, but that idea always fascinated me..",
  "Solution_1": "Well, IMO 2001 problem 6 was a number theory problem with a horrible equation but you can solve it beautifully by using simple geometry.",
  "Solution_2": "[quote=\"mozilla\"]Well, IMO 2001 problem 6 was a number theory problem with a horrible equation but you can solve it beautifully by using simple geometry.[/quote]\r\n\r\nReally?  How?"
}
{
  "Tag": [
    "probability"
  ],
  "Problem": "While staying in a 15-story hotel, Polya plays the following game. She enters an elevator on the $ 6^{\\mathrm{th}}$ floor. She flips a fair coin five times to determine her next five stops. Each time she flips heads, she goes up one floor. Each time she flips tails, she goes down one floor. What is the probability that each of her next five stops is on the $ 7^{\\mathrm{th}}$ floor or higher? Express your answer as a common fraction.",
  "Solution_1": "There are 2^5 = 32 different ways to flip the coin 5 times.\r\n\r\nSince Polya's stops must be at the 7th floor or higher, she must flip heads first. After that, she must flip head again because flipping tails will result in going back to the 6th floor.\r\n\r\nSo Polya must go in this order: 7th floor, 8th floor, ...\r\n\r\nOn the third flip, she can flip either heads or tails. If she flips tails, she will go back to level 7. The next flip must be heads. There are 2 choices for the last flip. They are 7,8,7,8,7 and 7,8,7,8,9.\r\n\r\nIf she flips heads for the 3rd flip, she will go in this order: 7,8,9, ...\r\nThe 4th flip can be either heads or tails.\r\n\r\nHeads (two ways): 7,8,9,10,9 and 7,8,9,10,11\r\n\r\nTails (two ways): 7,8,9,8,7 and 7,8,9,8,9\r\n\r\nThere is a total of 6 ways to stay at the 7th floor or higher for each stop.\r\nSo there the probability is 6/32 which is simplified to 3/16.",
  "Solution_2": "[hide = Quicker Way]She starts at the 6th floor, so in order for the conditions to be satisfied she must go up on her first move, to the 7th floor.\n\nOn her second move, she must also go up, to the 8th floor. The probability of flipping two heads is $\\frac{1}{4}.$\n\nNow, she's on the 8th floor and has 3 moves left. If she goes up in any one of those moves, she won't be able to reach the 6th floor. The probability she flips at least one head (aka goes up at least once) is $1$ minus the probability she flips two tails, or $\\frac{1}{4}.$\n\nThe answer is $\\frac{1}{4} \\cdot \\frac{3}{4} = \\boxed{\\frac{3}{16}}.$[/hide]"
}
{
  "Tag": [],
  "Problem": "How many pairs of vertical angles are formed by five distinct lines that have a common point of intersection?",
  "Solution_1": "For any pair of lines, there are $ 2$ unique pairs of vertical angles formed, so the answer is $ 2\\binom52 \\equal{} \\boxed{20}$.\r\n\r\nEdit: Darn, I seem to have forgotten what vertical angles are. :P",
  "Solution_2": "Aren't there 2 pairs? \r\n\r\n[asy]draw((0,0)--(10,10));\ndraw((10,0)--(0,10));\nlabel(\"Pair 1\",(5,6),N);\nlabel(\"Pair 1\",(5,4),S);\nlabel(\"Pair 2\",(6,5),E);\nlabel(\"Pair 2\",(4,5),W);[/asy]\r\n\r\nSo our answer is 20.",
  "Solution_3": "NOPE, good try\n\nThe answer is 5 \n\n(Obviously. 5 lines make 5 vertical angles)",
  "Solution_4": "The accepted solution seems to be incorrect. For example's sake have the $5$ lines intersect at the center of a circle, splitting it into $10$ congruent arcs. We count $5$ pairs of vertical angles of $36^{\\circ}$, $72^{\\circ}$, $108^{\\circ}$, $144^{\\circ}$, and $180^{\\circ}$ each. Thus we have $5\\cdot5 = 25$ pairs of vertical angles.",
  "Solution_5": "I disagree about the vertical angles measuring $180^{\\circ}$. Vertical angles are formed by two lines, and you would only be able to form $180^{\\circ}$ angles using two copies of the same line (which isn't allowed). If you discount them, you end up with the same answer as above.",
  "Solution_6": "I believe the currently accepted solution is $5$.",
  "Solution_7": "I reported it. They changed it from $5$."
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "incenter",
    "inradius",
    "inequalities",
    "trigonometry",
    "symmetry"
  ],
  "Problem": "Let [tex]ABC[/tex] be triangle. I is centre of triangle's incircle. [tex]AI[/tex] line cut circumcircle at [tex]A'[/tex]. Similarly, we have [tex]B',C'[/tex]. Prove that:\r\n[tex]AA'+BB'+CC'\\leq \\frac{29}{6}R+\\frac{7}{3}r[/tex].",
  "Solution_1": "Wlog, assume that A>=B>=C. (angles)\r\nNote that AA' = 2R cos(B-C)/2 \r\nSo it suffices to show that sum cos (B-C)/2 <= 29/12 +7/6 (r/R)\r\nFinally taking into account that cosA + cosB +cosC= 1+ r/R we need to show that \r\n[sum cos(B-C)/2] + 7/6 cos(B+C)-7/6(cos B +cos C) <=5/4\r\neliminating A, we need to show that :\r\ncos(B-C)/2 [1-7/3 cos(B+C)/2] +7/6 cos(B+C) - 2cos(B-C)/4cos3/2(B+C) <= 5/4",
  "Solution_2": "[quote=\"keira_khtn\"]Let [tex]ABC[/tex] be triangle. I is centre of triangle's incircle. [tex]AI[/tex] line cuts circumcircle at [tex]A'[/tex]. Similarly, we have [tex]B',C'[/tex]. Prove that:\n[tex]AA'+BB'+CC'\\leq \\frac{29}{6}R+\\frac{7}{3}r[/tex].[/quote]\r\nLet $O, I$ be the triangle circumcenter and incenter, $R, r$ the circumradius and inradius and $\\alpha = \\angle A, \\beta = \\angle B, \\gamma = \\angle C$. For an equilateral triangle $\\triangle ABC$, $r = \\frac R 2$, $AA' = BB' = CC' = 2R$ and the inequality becomes a trivial equality:\r\n\r\n$AA' + BB' + CC'  = 6R$\r\n\r\n${\\frac{29}{6}\\ R + \\frac 7 3\\ r = \\frac{29 + 7}{6}\\ R} = 6R$\r\n\r\nLet $\\triangle ABC$ be an arbitrary triangle. Consider the triangle $\\triangle CAA'$. Since $AA'$ bisects the angle $\\angle CAB$, the angle $\\angle CAA' = \\frac{\\angle CAB}{2} = \\frac \\alpha 2$. Since the angles $\\angle AA'C, \\angle ABC$ span the same arc $AC$ of the circumcircle $(O)$, $\\angle AA'C = \\angle ABC = \\beta$ and the remaining angle of this triangle is\r\n\r\n$\\angle ACA' = 180^o - \\left(\\frac \\alpha 2 + \\beta\\right) = \\gamma + \\frac \\alpha 2$\r\n\r\nUsing the sine theorem for this triangle,\r\n\r\n$AA' = CA\\ \\frac{\\sin{\\left(\\gamma + \\frac \\alpha 2\\right)}}{\\sin{\\beta}} = 2R \\sin{\\left(\\gamma + \\frac \\alpha 2\\right)}$\r\n\r\nSimilar expressions could be obtained for $BB', CC'$, Hence,\r\n\r\n$AA' + BB' + CC' = 2R \\left[\\sin{\\left(\\gamma + \\frac \\alpha 2\\right)} + \\sin{\\left(\\alpha + \\frac \\beta 2\\right)} + \\sin{\\left(\\beta + \\frac \\gamma 2\\right)} \\right]$\r\n\r\nLet $a = BC, b = CA, c = AB$ be the triangle sides and $s$ the semiperimeter. The inradius is equal to\r\n\r\n$r = \\frac{|\\triangle ABC|}{s} = \\frac{abc}{4Rs} = 2R\\ \\frac{\\sin \\alpha \\sin \\beta \\sin \\gamma}{\\sin \\alpha + \\sin \\beta + \\sin \\gamma}$\r\n\r\nConsider the triangle $\\triangle ABC$ \"rotating\" according to Poncelet's porism. During this process, it passes through 2 different isosceles triangles, one acute and the other obtuse, 3 times through each one in one full \"rotation\". Let $\\triangle A_1B_1C_1$ the acute isosceles triangle with $A_1B_1 = A_1C_1$. Then $A_1A_1' = 2R$ is maximum. When \"rotating\" the triangle counter-clockwise into the obtuse isosceles triangle $\\triangle B_2C_2A_2$ with $B_2C_2 = B_2A_2$, the segments $A_1A_1', C_1C_1'$ both monotonously decrease into $A_2A_2', C_2C_2'$ and the segment $B_1B_1'$ monotonously increases into $B_2B_2' = 2R$. Because of symmetry of these isosceles triangles, $B_1B_1' = C_1C_1', A_2A_2' = C_2C_2'$ and obviously, $C_2C_2' < C_1C_1'$. Consequently,\r\n\r\n$A_2A_2' + B_2B_2' + C_2C_2' = 2R + 2C_2C_2' < 2R + 2C_1C_1' = A_1A_1' + B_1B_1' + C_1C_1'$\r\n\r\nbecomes a minimum. In a subsequent \"rotation\" into the acute isosceles triangle $\\triangle C_3A_3B_3 \\cong \\triangle A_1B_1C_1$ with $C_3A_3 = C_3B_3$, the sum $A_3A_3' + B_3B_3' + C_3C_3'$ again increases to its maximum. Therefore, it is sufficient to prove the desired inequality for an acute isosceles triangle $\\triangle ABC$, for example, with $AB = AC,\\ \\alpha \\le 90^o,\\ \\beta = \\gamma < 90^o$. The desired inequality then becomes\r\n\r\n$(?)\\ \\ 2R\\left(1 + 2 \\sin{\\frac{3 \\gamma}{2}}\\right) \\le \\frac{29R}{6} + \\frac{14R}{3}\\ \\frac{\\sin^2{\\gamma} \\sin{2 \\gamma}}{2 \\sin \\gamma + \\sin{2 \\gamma}} =$\r\n\r\n$= \\frac{29R}{6} + \\frac{14R}{3}\\ \\frac{\\sin^2{\\gamma} \\cos \\gamma}{1 + \\cos \\gamma} = \\frac{29R}{6} + \\frac{14R}{3}\\ (1 - \\cos \\gamma) \\cos \\gamma$\r\n\r\n$(?)\\ \\ 4\\sin{\\frac{3 \\gamma}{2}} \\le \\frac{17}{6} + \\frac{14}{3} (1 - \\cos \\gamma) \\cos \\gamma$\r\n\r\nSince both sides are obviously positive, we can square the inequality in question. The left side becomes\r\n\r\n$16\\left(\\sin^2{\\gamma} \\cos^2{\\frac \\gamma 2} + 2 \\sin \\gamma \\cos \\gamma \\sin{\\frac \\gamma 2} \\cos{\\frac \\gamma 2} + \\cos^2{\\gamma} \\sin^2{\\frac \\gamma 2}\\right) =$\r\n\r\n$16\\left(\\sin^2{\\gamma} \\frac{1 + \\cos \\gamma}{2} + \\sin^2{\\gamma} \\cos \\gamma + \\cos^2{\\gamma} \\frac{1 - \\cos \\gamma}{2}\\right) =$\r\n\r\n$8\\left[(1 - \\cos^2{\\gamma})(1 + \\cos \\gamma) + 2(1 - \\cos^2{\\gamma}) \\cos \\gamma + \\cos^2{\\gamma}(1 - \\cos \\gamma)\\right] =$\r\n\r\n$= 8 (1 + 3 \\cos \\gamma - 4 \\cos^3{\\gamma})$\r\n\r\nand the right side becomes\r\n\r\n$\\frac{1}{36} (289  + 952 \\cos \\gamma - 952 \\cos^2{\\gamma} + 784 \\cos^2{\\gamma} - 1568 \\cos^3{\\gamma} + 784 \\cos^4{\\gamma})$\r\n\r\nDenote $x = \\cos \\gamma$. Since $0 < \\gamma < 90^o$, $0 < x < 1$. Thus we have to show\r\n\r\n$288 (1 + 3x - 4x^3) \\le 289  + 952x - 168x^2 - 1568x^3 + 784x^4$\r\n\r\n$784x^4 - 416x^3 - 168x^2 + 88x + 1 \\ge 0$\r\n\r\nSince we have an equality for an equilateral triangle, this most ugly quartic polynomial must have a root $x = \\cos{60^o} = \\frac 1 2$. Moreover, if a maximum is reached at $x = \\frac 1 2$, this root has to be a double root. This can be verified by a long division:\r\n\r\n$P(x) = 784x^4 - 416x^3 - 168x^2 + 88x + 1 = (2x - 1)^2 (196x^2 + 92x + 1)$\r\n\r\nThe roots of the remaining quadratic factor $x = \\frac{-23 \\pm 4\\sqrt{30}}{98} < 0$ are both negative, i.e., not acceptable. The fact that the above quartic polynomial has a minimum for $x = \\frac 1 2$ can be verified by taking the 2nd derivative:\r\n\r\n$P'(x) = 8(392x^3 - 156x^2 - 42x + 11) = 0$ [color=white].[/color] for [color=white].[/color] $x = \\frac 1 2$\r\n\r\n$P''(x) = 48(196x^2 - 52x - 7) = 48 \\cdot 16 > 0$ [color=white].[/color] for [color=white].[/color] $x = \\frac 1 2$\r\n\r\nThus we have a single minimum in the interval $(0, 1)$ equal to zero for $x = \\cos \\gamma = \\frac 1 2$, $\\beta = \\gamma = 60^o$ and $P(x) \\ge 0$ elsewhere in this interval. As a result\r\n\r\n$AA' + BB' + CC' \\le \\frac{29}{6}\\ R + \\frac 7 3\\ r$\r\n\r\nfor an arbitrary triangle, with the equality only for an equilateral triangle.",
  "Solution_3": "Oh! It 's so long but true..\r\nI have a short solution but have to use 2 lemma...A stronger result:\r\n[tex]AA'+BB'+CC'\\leq\\ \\sqrt{\\frac{70}{3}}R+[12-2\\sqrt{\\frac{70}{3}}]r[/tex]\r\nAnd find better [tex]k[/tex] constant st:\r\n[tex]AA'+BB'+CC'\\leq\\ kR+[12-2k]r[/tex]\r\nFirst inequality is particular case when:[tex]k=\\frac{29}{6}[/tex]\r\nAnd second inequality:[tex]k=\\sqrt{\\frac{70}{3}}[/tex]"
}
{
  "Tag": [
    "Olimpiada de matematicas"
  ],
  "Problem": "Sea $ABCDEF$ un hexagono con todos sus lados opuestos paralelos tal que:\r\n$-FC \\cap AB= X_1, FC\\cap ED= X_2$\r\n$-AD\\cap EF = Y_1 ,AD\\cap BC =Y_2$\r\n$-BE \\cap CD = Z_1, BE\\cap AF=Z_2$\r\n\r\nDemuestre:\r\na)\r\n$X_1, Y_1$ y $Z_1$ son colineales y $X_2, Y_2$ y $Z_2$ son colineales\r\nb)\r\nLas lineas $X_1Y_1Z_1$ y $X_2Y_2Z_2$ son paralelas",
  "Solution_1": "[quote=\"skuge\"]Sea $ABCDEF$ un hexagono con todos sus lados opuestos paralelos tal que:\n$-FC \\cap AB= X_{1}, FC\\cap ED= X_{2}$\n$-AD\\cap EF = Y_{1},AD\\cap BC =Y_{2}$\n$-BE \\cap CD = Z_{1}, BE\\cap AF=Z_{2}$\n\nDemuestre:\na)\n$X_{1}, Y_{1}$ y $Z_{1}$ son colineales y $X_{2}, Y_{2}$ y $Z_{2}$ son colineales\nb)\nLas lineas $X_{1}Y_{1}Z_{1}$ y $X_{2}Y_{2}Z_{2}$ son paralelas[/quote]\r\nSean:\r\nG:$AB\\cap DC$\r\nH:$CD\\cap FE$\r\nI:$EF\\cap BA$\r\nlos tri\u00e1ngulos:BGC, EDH, IAF, IGH son semejantes, entonces si a,b,c son las longitudes de los segmentos:IG, GH, HI respectivamente; podemos decir que existen r,s,t tales que:BG=at, GC=bt, BC=ct, ED=ar, DH=br, AH=cr, IA=as, AF=bs, IF=cs. Con ayuda de esto, aplicando el teorema de menelao 3 veces(sobre IGH y una recta:AD, BE o  CF), podemos hallar las razones se necesitan para demostrar que $X_{1}, Y_{1}, Z_{1}$ son colineales(con el teorema rec\u00edproco de menelao, para el triangulo IGH)\r\nAnalogamente se demuestra que $X_{2}, Y_{2}, Z_{2}$ son colineales.\r\nPara probar la parte b) es suficiente probar que $Y_{1}X_{1}$ y $X_{2}Y_{2}$ son paralelas, y ello se demuestra con semejanza(depende del grafico)",
  "Solution_2": "[quote=\"Chen241290\"]\nSean:\nG:$AB\\cap DC$\nH:$CD\\cap FE$\nI:$EF\\cap BA$\nlos tri\u00e1ngulos:BGC, EDH, IAF, IGH son semejantes, entonces si a,b,c son las longitudes de los segmentos:IG, GH, HI respectivamente; podemos decir que existen r,s,t tales que:BG=at, GC=bt, BC=ct, ED=ar, DH=br, AH=cr, IA=as, AF=bs, IF=cs. Con ayuda de esto, aplicando el teorema de menelao 3 veces(sobre IGH y una recta:AD, BE o  CF), podemos hallar las razones se necesitan para demostrar que $X_{1}, Y_{1}, Z_{1}$ son colineales(con el teorema rec\u00edproco de menelao, para el triangulo IGH)\nAnalogamente se demuestra que $X_{2}, Y_{2}, Z_{2}$ son colineales.\nPara probar la parte b) es suficiente probar que $Y_{1}X_{1}$ y $X_{2}Y_{2}$ son paralelas, y ello se demuestra con semejanza(depende del grafico)[/quote]\r\n\r\nMi soluci\u00f3n es similar usando homotesias y esto lo simplifica un poco:\r\nusando la misma notaci\u00f3n, \r\n$FIA$ es homotesia de $CBG$ con centro $X_{1}$\r\n$HED$ es homotesia de $FIA$ con centro $Y_{1}$\r\n Sabemos que el producto de dos homotesias resulta en una homotesia con centro colinear a los centros anteriores:\r\n $CBG$ es homotesia de $HED$ con centro $Z_{1}$\r\npero adem\u00e1s es el producto de las dos homotesias anteriores entonces $X_{1},Y_{1}$ y $Z_{1}$ son colineares.\r\nEl resto de la soluci\u00f3n es igual a la de Chen.",
  "Solution_3": "Se pueden evitar todos los c\u00e1lculos echando mano de la geometr\u00eda proyectiva. Sean $ U_{\\infty},V_{\\infty},T_{\\infty}$ los puntos del infinito de $ FA,AB,BC$ y $ \\tau_{\\infty} \\equiv U_{\\infty}V_{\\infty}T_{\\infty}$ la recta del infinito del plano del hex\u00e1gono. Entonces por reciproco del teorema de Pascal, $ A,B,C,D,E,F$ yacen en un misma c\u00f3nica. Consiguientemente por el teorema de Pascal en el hex\u00e1gono $ BADCFE,$ las intersecciones $ X_1 \\equiv AB \\cap FC,$ $ Y_1 \\equiv AD \\cap FE$ y $ Z_1\\equiv EB \\cap DC$ son colineales. De forma an\u00e1loga en el hex\u00e1gono $ EDAFCB,$ las intersecciones $ X_2 \\equiv ED \\cap CF,$ $ Y_2 \\equiv DA \\cap CB$ y $ Z_2 \\equiv AF \\cap BE$ son colineales. En el hex\u00e1gono $ AFEBCD,$ las intersecciones $ P_1 \\equiv FA \\cap CB ,$ $P_2 \\equiv EF \\cap DC$ y $ P_3 \\equiv EB \\cap DA$ son colineales. Entonces los tri\u00e1ngulos degenerados  $ \\triangle Y_1Y_2T_{\\infty}$ y $ \\triangle Z_1Z_2U_{\\infty}$ son homol\u00f3gicos $ \\Longrightarrow$ $ P \\equiv Y_1Z_1 \\cap Y_2Z_2 \\cap \\tau_{\\infty}$ $ \\Longrightarrow$ $ P$ est\u00e1 en el infinito. Por tanto, las rectas $ X_1Y_1Z_1$ y $X_2Y_2Z_2$ son paralelas."
}
{
  "Tag": [
    "LaTeX",
    "email"
  ],
  "Problem": "I would like to make one document with LaTeX in which there would be empty input boxes for name, surname, email and so on...\r\n\r\nIs it possible to make an LaTeX document and then print it to PDF with these options?\r\n\r\nIs it possible do make PDF in which you can put these text boxes so that other persons that get this PDF can input their name, surname and email and then print the document. \r\n\r\nI know PDF isn't writeable, only readable document. But is it possible to make it only possible for someone to input this data and print out the document?\r\n\r\nThank you in advance.",
  "Solution_1": "You can use forms in PDF which will do exactly what you want. To achieve this use the hyperref package and see the [url=http://www.tug.org/applications/hyperref/manual.html#x1-160006]PDF and HTML forms[/url] section of the hyperref manual",
  "Solution_2": "Thank you."
}
{
  "Tag": [
    "vector",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "Let $n\\geq 1$ be an integer. Find all rings $(A,+,\\cdot)$ such that all $x\\in A\\setminus\\{0\\}$ satisfy $x^{2^{n}+1}=1$.",
  "Solution_1": "$1 = (-x)^{2^{n}+1}=-x^{2^{n}+1}=-1$, so $2 = 0$. Now, we note that $(t+1)^{2}= t^{2}+1$; applying this $n$ times yields $(t+1)^{2^{n}}= t^{2^{n}}+1$. We therefore find that if neither $x$ nor $x+1$ is $0$, $1 = (x+1)^{2^{n}+1}= (x+1)(x+1)^{2^{n}}= (x+1)(x^{2^{n}}+1) = x^{2^{n}+1}+x^{2^{n}}+x+1 = x^{2^{n}}+x$. Left-multiplying by $x$ yields $x = 1+x^{2}$, or $x^{2}+x+1 = 0$. Note that the condition that $x, x+1 \\ne 0$ is equivalent to $x \\ne 0, 1$. Note furthermore that this implies $x^{3}= 1$ for all nonzero $x$. Now, for $x, y$ nonzero elements, suppose $x, y \\ne 1$. We claim that either $xy = 1$ or $xy^{2}= 1$, i.e. $x = y$ or $y^{2}$. Suppose not. Then $1 \\ne xy^{2}= x(y+1) = xy+x$ and $1 \\ne xy$, so we can apply the result that $t^{2}+t+1 = 0$ for $t \\ne 1$ to find that $0 = (xy+x)(xy+x)+(xy+x)+1 = (xyxy+xy+1)+(xx+x+1)+(x^{2})y+xyx+1 = xy+y+xyx+1 = xy(x+1)+(y+1) = xyx^{2}+y+1$, so $xyx^{2}= y+1$ and right-multiplying by $x$ yields $xy = yx+x$, so $x = xy+yx$. Similarly, $y = xy+yx$, so $x = y$. This yields that besides $0, 1$, if we have any other element, the only other possible elements are its square, whereupon it is immediate that the only possible rings that satisfy this condition are the trivial ring with $0 = 1$, $\\mathbb{F}_{2}$, and $\\mathbb{F}_{4}$.",
  "Solution_2": "I'm interested in a full solution using vector spaces...I tried something similar in the contest itself, but forgot to prove one of my statements... :( ...got 2 f***ing points 4 it"
}
{
  "Tag": [
    "geometry",
    "incenter",
    "geometric transformation",
    "reflection",
    "circumcircle",
    "perpendicular bisector",
    "geometry unsolved"
  ],
  "Problem": ":(  :(  :(",
  "Solution_1": "[b]Construction[/b]. It is well-known that $ M_{a}E_{a}$ is a diameter of the nine-point circle; thus the midpoint $ N$ of $ M_{a}E_{a}$ is the nine-point center of $ ABC$. We can now construct the nine-point circle (we have its center and a point on it - $ E_{a}$ or $ M_{a}$). Now, the Feuerbach point is the tangency point of the incircle and the nine-point circle. This point obviously lies on the line $ IN$. Thus, prolong $ IN$ to intersect the nine-point center in two points; one of them is the Feuerbach point $ Fe$ (this part isn't quite exact, but at least it gives a Geogebra construction). Construct now the incircle (we have the incenter $ I$ and the $ Fe$, which we now that lies on it). Draw the tangents from $ M_{a}$ to the incircle. One of them is clearly the sideline $ BC$. Let the intersection points of the line $ BC$ with the incircle and nine-point circle be $ D$ and $ H_{a}$, respectively. Let $ M_{a}Fe$ intersect the incircle again at $ X$ and draw the perpendicular bisector of $ DX$ (this line is the bisector $ AI$). The intersection of this line $ AI$ with the perpendicular in $ H_{a}$ to $ BC$ gives the vertex $ A$. Draw the tangents from $ A$ to the incircle and intersect them with $ BC$. This gives the vertices $ B$ and $ C$.\r\n\r\nExplanation:\r\n[quote=\"I wrote\"]\nLet $ M_{a}Fe$ intersect the incircle again at $ X$ and draw the perpendicular bisector of $ DX$ (this line is the bisector $ AI$).[/quote]\r\n\r\nThis is a consequence of the following theorem due to Emelyanov.\r\n[b]\nTheorem[/b]. Let $ ABC$ be a triangle and let $ D$, $ E$, $ F$ be the tangecy points of the incircle with $ BC$, $ CA$, $ AB$, respectively. Let $ X$, $ Y$, $ Z$ be the reflections of $ D$, $ E$, $ F$ in $ AI$, $ BI$, $ CI$, respectively, and let $ M$, $ N$, $ P$ be the midpoints of $ BC$, $ CA$, $ AB$. Then, the lines $ MX$, $ NY$, $ PZ$ are concurrent at the Feuerbach point $ Fe$ of $ ABC$.\r\n\r\nA synthetic proof can be viewed here:[url]http://pagesperso-orange.fr/jl.ayme/Docs/Symetriques%20de%20OI%20par%20rapport%20au%20triangle%20de%20contact.pdf[/url].\r\n\r\nDo you have any simpler solutions on your mind, jrrbc? I find this somehow unnatural.",
  "Solution_2": "thanks  :lol:",
  "Solution_3": "Dear Cosmin,\r\nAfter constructing the incircle from Feuerbach point, we may solve in other way:\r\nDraw tangent $ d$ from $ M_a$ to the incirlce to determine the line $ BC$, let $ D$ be the tangency point. The perp line $ d'$ from $ E_a$ to $ d$ is just the A-altitude. Let $ K$ be the orthogonal projection of $ I$ on $ d'$. Then the perp line to $ d$ at $ M_a$ (perp bisector of BC) meets $ KD$ at the midpoint $ M$ of the arc $ BC$ which doesn't contain $ A$ of the circumcircle (see my post in http://www.mathlinks.ro/viewtopic.php?t=256552 ). $ IM$ meets $ d'$ is just the vertex $ A$.\r\nMaybe this solution has the same method with your solution but it doesn't need the consequence of Emelyanov's theorem.  :D"
}
{
  "Tag": [
    "geometry",
    "Pythagorean Theorem"
  ],
  "Problem": "Do any of you know what President Garfield made an original proof for?",
  "Solution_1": "[hide=\"I do!\"]The Pythagorean Theorem.[/hide]",
  "Solution_2": "Darn. I thought I was the only one who knew that.",
  "Solution_3": "[quote=\"ln(dx/dy)\"]Darn. I thought I was the only one who knew that.[/quote]\r\n\r\nReally?  This is a math website so it's a pretty reasonable to assume that for each bit of mathematical knowledge you have that at least one person on this sites also has that knowledge.  \r\n\r\n(Btw, I knew that Garfield came up with a proof for the pythagorean theorem also  :P )",
  "Solution_4": "[quote=\"joml88\"][quote=\"ln(dx/dy)\"]Darn. I thought I was the only one who knew that.[/quote]\n\nReally?  This is a math website so it's a pretty reasonable to assume that for each bit of mathematical knowledge you have that at least one person on this sites also has that knowledge.  \n\n(Btw, I knew that Garfield came up with a proof for the pythagorean theorem also  :P )[/quote]\r\n\r\nO, i really don't get your meaning.. \r\nCan you tell me who Garfield is?? Don't tell me the famour cat...",
  "Solution_5": "http://www.pbs.org/teachersource/mathline/concepts/president/activity2.shtm",
  "Solution_6": "[quote=\"shobber\"]O, i really don't get your meaning.. \nCan you tell me who Garfield is?? Don't tell me the famour cat...[/quote]\r\nA president. How do you not know that?",
  "Solution_7": "[quote=\"ln(dx/dy)\"][quote=\"shobber\"]O, i really don't get your meaning.. \nCan you tell me who Garfield is?? Don't tell me the famour cat...[/quote]\nA president. How do you not know that?[/quote]\r\n\r\nDoesn't shobber live in China?",
  "Solution_8": "It says 'President Garfield' as the thread title and in the original post though...",
  "Solution_9": "Thats really cool that someone in that kind of office would have studied math (and more, have some original results in it!)",
  "Solution_10": "[quote=\"eryaman\"][quote=\"ln(dx/dy)\"][quote=\"shobber\"]O, i really don't get your meaning.. \nCan you tell me who Garfield is?? Don't tell me the famour cat...[/quote]\nA president. How do you not know that?[/quote]\n\nDoesn't shobber live in China?[/quote]\r\n\r\nWell, i do live in China and i have never been to the USA......\r\nSo that means a president got an original solution of this theorem? That's really amazing!",
  "Solution_11": "[quote=\"tetrahedr0n\"]Thats really cool that someone in that kind of office would have studied math (and more, have some original results in it!)[/quote]\r\n\r\nWell, Michael Jordan was a math major! :D",
  "Solution_12": "Michael Jordan switched majors his junior year....\r\n\r\nIt is very cool to see people in office do math in their spare time...brings to mind Napolean and one of his servicemen Poncelet.",
  "Solution_13": "...and Dan Quayle was dumb enough to spell \"potato\" wrong. Political figures are getting dumber and dumber. Especially George Bush and his famous Bushisms :P .",
  "Solution_14": "[quote=\"zscool\"]Michael Jordan switched majors his junior year....\n\nIt is very cool to see people in office do math in their spare time...brings to mind Napolean and one of his servicemen Poncelet.[/quote]\r\n\r\nJordan?? He was a math major?? Incredible......\r\nAnd for Napolean, do you mean the geometry theorem named after him??"
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "Prove for $x,y,z>0$:\r\na) $\\frac{x^3}{x^2+xy+y^2}+\\frac{y^3}{y^2+yz+z^2}+\\frac{z^3}{z^2+zx+x^2}\\geq \\frac{x+y+z}{3}$\r\nb) $\\frac{x^3}{y^2-yz+z^2}+\\frac{y^3}{z^2-zx+x^2}+\\frac{z^3}{x^2-xy+y^2}\\geq x+y+z$",
  "Solution_1": "[quote=\"indybar\"]Prove for $x,y,z>0$:\na) $\\frac{x^3}{x^2+xy+y^2}+\\frac{y^3}{y^2+yz+z^2}+\\frac{z^3}{z^2+zx+x^2}\\geq \\frac{x+y+z}{3}$\nb) $\\frac{x^3}{y^2-yz+z^2}+\\frac{y^3}{z^2-zx+x^2}+\\frac{z^3}{x^2-xy+y^2}\\geq x+y+z$[/quote]\r\n\r\na. Inequality is not difficult. We have:\r\n$\\frac{x^3}{x^2+xy+y^2}+\\frac{y^3}{y^2+yz+z^2}+\\frac{z^3}{z^2+zx+x^2}=\\frac{1}{2}\\sum(\\frac{x^3+y^3}{x^2+xy+y^2})$\r\n :lol:  :roll:",
  "Solution_2": "I will post it in tomorrow :)",
  "Solution_3": "[quote=\"indybar\"]Prove for $x,y,z>0$:\na) $\\frac{x^3}{x^2+xy+y^2}+\\frac{y^3}{y^2+yz+z^2}+\\frac{z^3}{z^2+zx+x^2}\\geq \\frac{x+y+z}{3}$\nb) $\\frac{x^3}{y^2-yz+z^2}+\\frac{y^3}{z^2-zx+x^2}+\\frac{z^3}{x^2-xy+y^2}\\geq x+y+z$[/quote]\r\nToo easy. I think you shouldn't post them here because they are too old.\r\na. Another solution:\r\n$\\frac{x^3}{x^2+xy+y^2}\\ge\\frac{2x-y}{3} \\Leftrightarrow x^3+y^3\\ge xy^2+x^2y$\r\nb.$\\frac{x^3}{y^2-yz+z^2}+\\frac{y^3}{z^2-zx+x^2}+\\frac{z^3}{x^2-xy+y^2}\\geq \\frac{(x^2+y^2+z^2)^2}{\\sum(x^2y+x^2z)-3xyz}$\r\nAnd we have:\r\n$\\frac{(x^2+y^2+z^2)^2}{\\sum(x^2y+x^2z)-3xyz}\\ge x+y+z \\Leftrightarrow\\sum x^4+\\sum x^2yz\\ge\\sum(x^3y+x^3z)$",
  "Solution_4": "I think :\r\n\r\nWe have :$\\frac{x^3}{x^2+xy+y^2}+\\frac{y^3}{y^2+yz+z^2}+\\frac{z^3}{z^2+zx+x^2}=\\frac{1}{2}\\sum(\\frac{x^3+y^3}{x^2+xy+y^2})$\r\n\r\n$\\sum(\\frac{x^3+y^3}{x^2+xy+y^2}) \\geq \\frac{1}{3}(x+y)$\r\n\r\nbecause: ${\\frac{x^3+y^3}{x^2+xy+y^2} = (x+y)\\frac{x^2-xy+y^2}{x^2+xy+y^2}= (x+y)(1-\\frac{2xy}{x^2+xy+y^2}) \\geq \\frac{1}{3}(x+y})$\r\n(Cosi inequality)"
}
{
  "Tag": [
    "calculus",
    "integration",
    "analytic geometry",
    "graphing lines",
    "slope",
    "vector",
    "trigonometry"
  ],
  "Problem": "Does anyone know that how many questions do I have to get right to get a 5? Thanks in advance.",
  "Solution_1": "The curve for a 5 is approximately 65%. So, if you do about equally well on the FR and the MC, 65% of 45 questions is about 30.",
  "Solution_2": "I think it depends on the year.  If you actually understand the material well, you shouldn't have any trouble getting a 5.",
  "Solution_3": "Yea I think all AP tests are graded relative to everyone taking the examn, so if you pay 1 million dumb people to take the examn then a 5 is a lot easier to get. :D",
  "Solution_4": "hehe, for the easy 1 million dumb people. Well, to get a 5 in AB is defintely easy becuase I got it. Probably a bit tougher in BC, but not that tough, I think you can make a lot of mistakes and you dont need extremely advanced problem solving to solve the problems.",
  "Solution_5": "The BC curve is HUGE. If you go to the AP site, 40% of test-takers get 5 or above; compare that to about 25% for AB and 11% for like Physics.",
  "Solution_6": "umm, you're on this site, which means you should know some math! getting a 5 on BC Calculus (at least in my school) seems to be much easier than a 5 on AB Calculus, and the curve is really big, so you should be fine",
  "Solution_7": "What?!?!?! Somehow that makes zero sense....",
  "Solution_8": "Yeah, I doubt that BC is easier then AB because BC covers infinite series and more integration techinques then AB.",
  "Solution_9": "If you study well, a 100% on that exam is definitely not out of reach. (Of course, for those of us who just love making stupid mistakes, that may not always be true.)\r\n\r\nI personally think there should be a 6 to recognize those students who achieve a perfect.",
  "Solution_10": "[quote=\"MithsApprentice\"]If you study well, a 100% on that exam is definitely not out of reach. (Of course, for those of us who just love making stupid mistakes, that may not always be true.)\n\nI personally think there should be a 6 to recognize those students who achieve a perfect.[/quote]\r\n\r\n*nods in agreement* I can get a 90% basically from doing the test as fast as I can... if I really took the time, I wouldn't be surprised to get 100% at least once in a while... not that I would ever actually take the time... ;)",
  "Solution_11": "Ah, thank you guys very much!  :)",
  "Solution_12": "[quote=\"paladin8\"]The BC curve is HUGE. If you go to the AP site, 40% of test-takers get 5 or above; compare that to about 25% for AB and 11% for like Physics.[/quote]\r\n\r\nwhere can i take a look at these curves?\r\nI'm shocked that AB 5's are only 25%, and with about 4-5 days left before the exam... I thought AB was very easy. I did practice tests from years before and I could answer almost all the questions.\r\n\r\nMaybe this curve is just for last year since people didn't learn slope fields which was a new thing?",
  "Solution_13": "I got a 5 in Calc AB last year and it was easy, I am not a top aopser. The only slope fields question was in free response and you had to draw a graph and solve the differential equation by the seperabhle differential equation method.",
  "Solution_14": "[quote=\"qweretyq\"][quote=\"paladin8\"]The BC curve is HUGE. If you go to the AP site, 40% of test-takers get 5 or above; compare that to about 25% for AB and 11% for like Physics.[/quote]\n\nwhere can i take a look at these curves?\nI'm shocked that AB 5's are only 25%, and with about 4-5 days left before the exam... I thought AB was very easy. I did practice tests from years before and I could answer almost all the questions.\n\nMaybe this curve is just for last year since people didn't learn slope fields which was a new thing?[/quote]\r\n\r\nCalc AB is pretty easy.  If you can do all the problems then you should be fine.\r\n\r\nI'm a little worried about BC though...I'm self studying it this year and am still not very confident with my abilities to do vector, parametric and some of the trig stuff.  Well, one more weekend...",
  "Solution_15": "Really? I don't remember much except plugging it into my trusty TI-89 :D",
  "Solution_16": "Dont worry about Calc BC if you answered [b][u]most[/u][/b] of the questions. I mean there is a curve, and this is AOPS, so I can never tell if what you say \"bombing\" = I missed a free response part and some MC.",
  "Solution_17": "the polar one was pretty easy... ;) \r\ni mean as far as i remember, all you had  to was to find the area of the under the curve with the integral formula, and for the other part, you had to let $dr/d\\theta=0$.... ;)",
  "Solution_18": "i solved all of the MC except for one of them, which i was $70\\%$ sure i knew the right answer but i didn't answer it...\r\ni made one stupid mistake in the second part of the first question...(finding the volume of the solid...) ;)",
  "Solution_19": "I feel hella dumb\r\nBecause I didn't know how to derive or integrate that\r\n\r\nAnd, on top of that, I didn't know how to use my calculator (89, also) for it\r\n\r\nThat's what I get for slackin off 2nd semester senior year....\r\n\r\noh well, I'm expectin at least a 4, but a 5 would be nice.",
  "Solution_20": "I was glad that they had a polar rather than a vector/parametric one.  It's just those stupid velocity/acceleration vectors that always throw me off.  I know how to do them but I often times screw them up :D .  For both parts of the free response I ended up working till the very end of the time limit.  Maybe my answers were more thorough than they had to be....I thought that a little more room to write the solutions would have been nice :lol: .  Guess I'm still on USAMO mode...",
  "Solution_21": "You cant get points off on free response like MC right?\r\nYou can only [i]gain[/i] points?",
  "Solution_22": "on MC, every wrong answer is 1/4 off.  on free response, you can only gain points.",
  "Solution_23": "[quote=\"joml88\"]I was glad that they had a polar rather than a vector/parametric one.  It's just those stupid velocity/acceleration vectors that always throw me off.  I know how to do them but I often times screw them up :D .  For both parts of the free response I ended up working till the very end of the time limit.  Maybe my answers were more thorough than they had to be....I thought that a little more room to write the solutions would have been nice :lol: .  Guess I'm still on USAMO mode...[/quote]\r\n\r\nHaha. We've been practicing grading our FR's in class a TON, so I pretty much know exactly how they grade; it's completely unrigorous and extremely easy to get points on, so I didn't bother to write a lot :D",
  "Solution_24": "I was hoping for a parametric one....\r\n\r\nI'm just glad BC kids didn't get a related rates like AB had to do....\r\n\r\n\r\nIf it was up to me, free response would consist of 3 questions on parametric\r\nand then 3 questions on rotating  areas and what not....",
  "Solution_25": "Ah you guys, I was HOPING for a parametric/vector FRQ or a related rates one.....\r\n\r\ni did okay on the multiple choice but made some silly, and i mean SILLY errors on the FRQs. i messed up on parts of the polar and on parts of the taylor series one (i forgot to write down that 7 was the first term of the series! and that was given information.  :( ). but then even worse, i had:\r\n\r\n$-1<\\frac{x-2}{3}<1$\r\n\r\nfor the interval of convergence (which is correct) but i forgot my algebra and simplified it incorrectly! that sucks! i am still screaming at myself over that!\r\n\r\nI HOPE I GET A $5$!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!",
  "Solution_26": "For those who are interested, preliminary solutions to the 2005 AB and BC Free Response questions can be found here: http://clem.mscd.edu/~talmanl/APCalculus.html",
  "Solution_27": "I can't really remember my answers :!: Looking at his answers, though, I feel pretty good.  They have the questions up on collegeboard.com also: http://www.collegeboard.com/student/testing/ap/calculus_bc/samp.html?calcbc",
  "Solution_28": "oh man, i got half way through multiple choice and then suddenly realized that i had forgotten how to do related rates. i spent the rest of the time PRAYING that a related rate wouldn't be on the test. i was SO LUCKY.",
  "Solution_29": "[quote=\"gauss202\"]For those who are interested, preliminary solutions to the 2005 AB and BC Free Response questions can be found here: http://clem.mscd.edu/~talmanl/APCalculus.html[/quote]\nlooking at the answers, of some of the FR, i got all of it right, except for one part of the first problem, finding the volume one...\nand by the way..\n[quote=\"joml88\"]I know how to do them but I often times screw them up  . For both parts of the free response I ended up working till the very end of the time limit. Maybe my answers were more thorough than they had to be....[/quote]\r\nthat happened to me too, like in the last one, the series one the moment i finished the problem, time was up... ;) but my answeres were correct.... :P"
}
{
  "Tag": [
    "AMC",
    "AIME",
    "trigonometry",
    "geometry",
    "LaTeX",
    "incenter",
    "search"
  ],
  "Problem": "Do you know any AIME Level Trig and/or Geometry Problem book. I don't want other topics but just those two concentrated.\r\n\r\nThanks.",
  "Solution_1": "Well here are are from AIME 1 2005\r\n\r\n7. Quadrilateral ABCD has angle A=angle B=60 degrees (sorry i don't have latex)  BC=8, AD=10 and CD=12.  The length of AB can be expressed as m+sqrt[n] where m and n are natural numbers.  Find m+n.\r\n\r\n\r\n15. In triangle ABC, AB=20.  The incircle of the triangle (not including the incenter) trisects the median CD.  If the are of the triangle is m+sqrt[n] where m and n are natural numbers, find m+n.  (I think that's the form you have to write it in...don't quite remember...if that isn't, just write the area)",
  "Solution_2": "But are there any good books or sets of problems online??",
  "Solution_3": "Don't know any, but i'm sure you could find one in any large bookstore.",
  "Solution_4": "[quote=\"K81o7\"]Don't know any, but i'm sure you could find one in any large bookstore.[/quote]\r\nImpossible...I live in China... :(",
  "Solution_5": "You could probably search on the internet and find some...",
  "Solution_6": "For geometry, I would suggest Problems in Geometry (I am not sure if thats the right title). It is similar to Geometry Revisited but much easier.",
  "Solution_7": "Thanks I'm getting that one.\r\n\r\nI wonder what about trig. Trig is something that surely appears on AMC and if I'm good at it, it usually makes geometry problem easier. :ninja:"
}
{
  "Tag": [
    "inequalities"
  ],
  "Problem": "Let $a,b,c$ are positive such that $abc=1$. Prove\r\n$\\frac{a}{c}+\\frac{b}{a}+\\frac{c}{b}+2(a+b+c) \\geq 3 (\\frac1a+\\frac1b+\\frac1c)$",
  "Solution_1": "[hide=\"rather messy, but it works\"]\nSubstitute $a=\\frac{x}{y}$, $b=\\frac{y}{z}$, $c=\\frac{z}{x}$. After clearing denominators, it reduces to\n\\[x^{3}+y^{3}+z^{3}+2(x^{2}z+y^{2}x+z^{2}y)\\geq 3(x^{2}y+y^{2}z+z^{2}x)\\]\nBut we have $x^{3}+y^{2}x-2x^{2}y=x(x-y)^{2}$, and\n\\begin{eqnarray*}x^{2}y+y^{2}z+z^{2}x-x^{2}z-y^{2}x-z^{2}y&=&x^{2}(y-z)-y^{2}(x-z)+z^{2}(x-y)\\\\ &=&(x^{2}-y^{2})(y-z)+(z^{2}-y^{2})(x-y)\\\\ &=&(x+y)(x-y)(y-z)-(y+z)(x-y)(x-z)\\\\ &=&(x-y)(y-z)(x-z)\\\\ \\end{eqnarray*}\nso the inequality is equivalent to\n\\[x(x-y)^{2}+y(y-z)^{2}+z(z-x)^{2}\\geq (x-y)(y-z)(x-z)\\]\nWLOG let $z=\\min \\{ x,y,z \\}$. Then let $p=x-z$ and $q=y-z$, so that $x-y=p-q$ as well. Replacing $(x,y,z)$ with $(p,q,0)$ does not change any of the differences, so the right hand side is left unchanged. But as $x\\geq p$, $y\\geq q$, $z\\geq 0$, the left hand side decreases:\n\\[x(x-y)^{2}+y(y-z)^{2}+z(z-x)^{2}\\geq p(x-y)^{2}+q(y-z)^{2}=p(p-q)^{2}+q^{3}\\]\nIt remains to prove that $p(p-q)^{2}+q^{3}\\geq pq(p-q)$. If $q=0$, the inequality is trivial. Otherwise, let $r=\\frac{p}{q}$ so that it becomes equivalent to\n\\[r(r-1)^{2}+1\\geq r(r-1)\\]\nBut this follows from\n\\[r(r-1)^{2}+1-r(r-1)=r^{3}-3r^{2}+2r+1=1+(r+2)(r-1)^{2}\\geq 0\\]\n[/hide]"
}
{
  "Tag": [],
  "Problem": "Hi friends,\r\n                 I am planning to join for MSc mathematics course in India. I would like to know about the the opportunities for PhD abroad.\r\n1. What else exam other than GRE should I write ?\r\n2. What percentage of my expense will be covered by the scholarship, where do I apply to them ?\r\n3. Do I have to write entrance exam ie., math paper for each University separately ?\r\n4. What about the recommendation letters ?\r\n5. What else I can do during my MSc years like MTTS to improve my math skills and better my CV ?\r\n6. Most Important of all- Is this foreign PhD really worth, what do I get out of it ? ( question is intended both materialistically and intellectually)\r\n\r\nHoping for help,\r\nRegards,\r\nSrikanth",
  "Solution_1": "Hi srikanth ..\r\nwell i think you will get good replies if you repost it [url=http://www.mathlinks.ro/index.php?f=143]here[/url] :)",
  "Solution_2": "[quote=\"pardesi\"]Hi srikanth ..\nwell i think you will get good replies if you repost it [url=http://www.mathlinks.ro/index.php?f=143]here[/url] :)[/quote]\r\n\r\nDon't you mean\r\n[url]http://www.artofproblemsolving.com/Forum/index.php?f=143[/url]\r\n :maybe:",
  "Solution_3": "[quote=\"vishalarul\"]Don't you mean\n[url]http://www.artofproblemsolving.com/Forum/index.php?f=143[/url]\n :maybe:[/quote]\r\nyes i think that's obvious and that's where u r lead into when u click on \"here\" link",
  "Solution_4": "[quote=\"pardesi\"][quote=\"vishalarul\"]Don't you mean\n[url]http://www.artofproblemsolving.com/Forum/index.php?f=143[/url]\n :maybe:[/quote]\nyes i think that's obvious and that's where u r lead into when u click on \"here\" link[/quote]\r\nNo, when you click that link, you are lead to\r\n[url]http://www.artofproblemsolving.com[/url]",
  "Solution_5": "no it doesn't happen to me and i suppose it does't happen to srikanth since he posted in the college forum just few hours after i asked him to do so :wink:",
  "Solution_6": "Not necessarily. I get lead to the home page as well.\r\n\r\nI shouldn't even be in the forum, as I'm not Indian.  :|",
  "Solution_7": "wait are u all posting from Aops or mathlinks",
  "Solution_8": "Hope the answers are useful.\r\n\r\nDisclaimer: Accurate answers can be gotten from the universities, use these as guidelines: \r\n1. What else exam other than GRE should I write ?\r\n>>> Most  require or strongly suggest  you take GRE.   The marks are generally used as a \"cutoff\"  but other than that not much weightage.\r\n\r\n>>>2. What percentage of my expense will be covered by the scholarship, where do I apply to them ? \r\n\r\nGenerally full tuition, room and board are covered in virtually all graduate assistance. You may have to teach (TA)  (15-20 hours a week) \r\n\r\n3. Do I have to write entrance exam ie., math paper for each University separately ?\r\n\r\n>>> Most likely no. \r\n4. What about the recommendation letters ?\r\n\r\n>>.>Very important. \r\n5. What else I can do during my MSc years like MTTS to improve my math skills and better my CV ?\r\n\r\n>>> publishing papers in peer reviewed journals.\r\n6. Most Important of all- Is this foreign PhD really worth, what do I get out of it ? ( question is intended both materialistically and intellectually) \r\n\r\n>>> Depends on you..for many   the subject,  faculty  (who is the adviser .. what field .. how good your work is etc) makes one place more worthy than others...",
  "Solution_9": "Thanks Gyan"
}
{
  "Tag": [
    "articles",
    "LaTeX"
  ],
  "Problem": "I wrote an article with 5 sections. I want that each displayed equation be numbered according to that section. Like (1.1), (1.2) etc for the first section, then (2.1), (2.2) etc for the second section and so on. \r\n\r\n1. How can I make Latex to know when a section starts and ends? \r\n2. How can I make the equations be numbered as above in the output (final) document? \r\n3. How can I make the refferences to those equations change according to the above numerotation? \r\n\r\nApologize for my silly questions but I'm exhausted after typing 14 pages. \r\n\r\nThanks",
  "Solution_1": "Assuming you have amsmath loaded (which you should do anyway) then it's been made really simple for you. Just put \\numberwithin{equation}{section} at the start of your document and it's done!"
}
{
  "Tag": [
    "set theory",
    "combinatorics proposed",
    "combinatorics"
  ],
  "Problem": "The elements of the set $F$ are some subsets of $\\left\\{1,2,\\ldots ,n\\right\\}$ and satisfy the conditions:\ni) if $A$ belongs to $F$, then $A$ has three elements;\nii)if $A$ and $B$ are distinct elements of $F$ , then $A$ and $B$ have at most one common element.\nLet $f(n)$ be the greatest possible number of elements of $F$. Prove that $\\frac{n^{2}-4n}{6}\\leq  f(n) \\leq  \\frac{n^{2}-n}{6}$",
  "Solution_1": "i proved the RHS! (for all n, $f(n)\\leq \\frac{n(n-1)}{6}$).\r\n\r\nto prove this, see that all elements of the family F have 3 diferent pairs of elements, and this pais occurs exactly one time in F, because the intersection of every two elements of F have at most 1 element!.\r\n\r\nso, $3f(n) \\leq C(n,2)$, and we get the result!",
  "Solution_2": "The LHS isn't all that hard as well . Let $K_{i}$ consist of ${(a,b,c)|a+b+c \\equiv i \\bmod n, a,b,c\\in{1,2,\\ldots n}}$ Now we see very clearly that all K_i's are disjoint.\r\n  We also show that in any $K_{i}$ any two 3-sets can have atmost one element in common (as if they had two then by the congruence condition and the fact that our elements are chosen from ${1,\\ldots,n}$ the third also would be fixed). The fact that every set has 3 elements is obvious (If you don't see this .. :rotfl: ) So our set $K_{i}$ satisfies the conditions that $F$ does. Now, it is clear that all the $K_{i}$'s are non-intersecting and their union is the set of all 3-subsets which has size ${n \\choose 3}$. Since we have n such $K_{i}$'s we can say that there exists a set of size atleast$\\frac{{n\\choose 3}}{n}= \\frac{(n-1)(n-2)}{6}$ So we choose that $K_{i}$ which satisfies our condition. Now,\r\n   $\\frac{(n-1)(n-2)}{6}\\leq|K_{i}| \\leq f(n)$ thus giving our condition. Now since $\\frac{n-1)(n-2)}{6}\\geq \\frac{n^{2}-4n}{6}$ we are done. Note this gives us a stronger condition - have I gone wrong anywhere?(The idea of partitioning it $\\bmod n$ is a general heuristic and I usually try it on any nice set theory problem.)\r\n  Ashwath"
}
{
  "Tag": [
    "geometry"
  ],
  "Problem": "Do any of you know about any programs where one can be mentored by a mathematician personally? All the greats had opportunities like that. I think it could really help. Sadly, none of my teachers can solve the problems we discuss in these forums. :(  I need constant contact with a person who has alot of experience and can break hard problems down face to face. Thank you. Happy Holidays :lol:",
  "Solution_1": "There is no problem.\r\nJust contact via PMs with persons you want to teach you.",
  "Solution_2": "Maybe one of the independent studies programs listed at http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php would meet your needs?\r\n\r\nIf you want an in-person mentor, that may be harder to find depending on where you are located.   You might try a local college math department, if there is one near you, or look for programs that provide mentors from local industry.",
  "Solution_3": "go to like a college near ur area and talk to the math professors.. try to get a private college or at least a good state one.."
}
{
  "Tag": [
    "geometry",
    "calculus",
    "topology",
    "summer program",
    "Mathcamp",
    "Stanford",
    "college"
  ],
  "Problem": "All the math I know since freshman year geometry is self taught, as I skipped algebra II to take a C++ course and then continued on to self teach myself precalc and calc BC. Going into my senior year, I have a 5 on the AP Calc BC test and a decent knowledge more advanced math, i.e. Multivariable Calculus, ODE's, The calculus of differential forms, some real analysis. Right now I'm working on some point-set topology. My question is, how do I emphasize to a college to which I am applying that I have [u]self-studied[/u] all my high school math and beyond? For the stuff that's beyond, should I send my notes/exercises?\r\n\r\nAn example of a book that I'm working through: \r\n\r\n[url=http://www.amazon.com/exec/obidos/tg/detail/-/0817637079/]Advanced Calculus : A Differential Forms Approach[/url] \r\n\r\n\r\nExcellent book, recommend it to anyone with a knowledge of basic calculus.",
  "Solution_1": "I'd recommend at the very least writing about it as one of your essays.",
  "Solution_2": "That is an interesting book, and you should definitely mention in your application that you do that kind of reading for \"fun.\"",
  "Solution_3": "[quote=\"curl\"]All the math I know since freshman year geometry is self taught, as I skipped algebra II to take a C++ course and then continued on to self teach myself precalc and calc BC. Going into my senior year, I have a 5 on the AP Calc BC test and a decent knowledge more advanced math, i.e. Multivariable Calculus, ODE's, The calculus of differential forms, some real analysis. Right now I'm working on some point-set topology. My question is, how do I emphasize to a college to which I am applying that I have [u]self-studied[/u] all my high school math and beyond? For the stuff that's beyond, should I send my notes/exercises? [/quote]\r\n\r\nYour question isn't clear.   What information do you want to convey, and why?\r\n\r\nFor all but the most elite US universities, differential forms, real analysis etc are at the very limit of what an applicant might know, and having studied them in any form places you at the top of the applicant pool (in the absence of severe disqualifying \"issues\" such as a criminal record, or worse, a lack of money).\r\n\r\nAt universities where a math professor may read your application and you will not be the only applicant to have seen such material, they know that both course-taking and self-study can vary widely in the results, and that people can cite whatever they want in an application, including reams of notes and exercises.  If they really need to know the details (probably not) they can invite you for a conversation with a professor, or ask to see the notes and exercises that you allude to.  I doubt it would ever be that crucial. \r\n\r\nThere is also a difference, that savvy admissions officers will be aware of, between self-study added to your existing course work, and self-study where your advanced mathematical interests are assigned some course credit in lieu of regular high school courses.  Both demonstrate aptitude and motivation, but if it's a situation where you free up 1-2 hours during your school day for an accredited hobby, it is exaggerated to celebrate the self-study aspect (as opposed to just the learning of advanced material, by whatever means) and people familiar with such arrangements may not necessarily be bowled over by it.",
  "Solution_4": "I want to convey the effort that I have put forth to cover all of high school mathematics on my own (which is not at all a great feat, by the way), as well as my curiosity and fascination with math.",
  "Solution_5": "It almost goes without saying that learning a serious programming language, differential forms and real analysis within the first $n-1$ years of high school involves a lot of self-study.   Discussion of the effort as such, beyond just stating the fact of self-study, can backfire for many reasons, and is mostly irrelevant.  More pertinent would be to quantify or concretely indicate the mathematical level reached, (e.g., \"calculus, analysis and topology at about the third-year undergraduate level\", or \"solved all exercises in Spivak's book on manifolds\").  Others can judge the effort from such information.",
  "Solution_6": "my son was in a similar situation, needing to somehow document a bunch of stuff beyond Calc BC that he learned on his own. I don't think that sending copies of primary material like notes etc will be considered \"proof\" that you mastered the subjects you have studied. Anybody can assert anything, and few colleges have the resources to sort through massive amounts of primary material to assess the rigor. My son documented multi and linear alg by taking distance learning courses thru http://www.utexas.edu. They are self paced, so if you already know those subjects you could work thru them pretty fast and get grades on an official transcript in time to submit it with your apps. He also audited a bunch of classes at the local uni. Administration would not allow him to register, but the profs did allow him to submit all the homework and exams for grading. At the end, the profs each wrote a note with what his grade would have been if he had been allowed to register, and he included copies of those with his apps. That still left a lot of stuff that he somehow \"knew\" which could not be documented (5 years of Mathcamp will do that). But there was plenty of documented stuff to make it obvious that he is a strong math person. Success in competitions can also be viewed as documentation that you have learned a significant body of mathematical knowledge outside a standard curriculum.",
  "Solution_7": "you would have to show that you know the material well, and you didn't just skim through for the sake of saying you did \"Ind study\". I would guess there aren't AP tests for Real analysis (am I right? :)) so maybe you could go to your local community college & ask to take the Real Analysis, etc, finals. and see what you get, above a 90% or 95% I bet colleges would take that as proof that you understand the material.",
  "Solution_8": "[quote=\"mathclass\"]you would have to show that you know the material well, and you didn't just skim through for the sake of saying you did \"Ind study\". I would guess there aren't AP tests for Real analysis (am I right? :)) so maybe you could go to your local community college & ask to take the Real Analysis, etc, finals. and see what you get, above a 90% or 95% I bet colleges would take that as proof that you understand the material.[/quote]\r\n\r\n.. If your local community college offers Real Analysis (mine doesn't)",
  "Solution_9": "Worse, taking \"real analysis\" at a community college is hardly going to be at the level of the analogous course at a decent university, much less an elite one.",
  "Solution_10": "[quote=\"blahblahblah\"]Worse, taking \"real analysis\" at a community college is hardly going to be at the level of the analogous course at a decent university, much less an elite one.[/quote]\r\n\r\nEntirely true that it's hardly likely that a real analysis course at a community college, where such a thing exists, is anything at all like a real analysis course at a top university. But for the top university admissions officer sorting through applications, it is enough that a student who took the most advanced courses at his local community college showed more ambition than the student who took only high school courses. The top universities all have first-year courses that introduce challenging material even to very well prepared new admittees.",
  "Solution_11": "[quote=\"texas137\"]my son was in a similar situation, needing to somehow document a bunch of stuff beyond Calc BC that he learned on his own. I don't think that sending copies of primary material like notes etc will be considered \"proof\" that you mastered the subjects you have studied. Anybody can assert anything, and few colleges have the resources to sort through massive amounts of primary material to assess the rigor. [/quote]\r\n\r\nFew colleges have the [i]need[/i] to assess the rigor.   At the non-elite schools, it doesn't matter what the details of the study program are, just that it substantially exceeds high school.   At the elite schools, for purposes of admission the fine details of the self-study program are not important either; if it is consistent with other information (such as competitions or summer mathematical programs as you mention) the claims will probably be taken at face value as an indication of interest, ambition and high, but not precisely calibrated, ability. \r\nFor purposes of course placement or credit once at the university, the content of the self-study doesn't matter because beyond the standard year 1-2 placement exams, a student can generally enroll in whatever courses they want, and universities rarely accredit academic studies without institutional documentation.    Apart from competitions for large scholarships, the need to investigate the exact material studied will rarely arise, and if it does, the school can simply have a math professor contact the applicant.",
  "Solution_12": "It's interesting to note that at most of the top math schools there is a professor of mathematics on the admissions committee, or else a committee of the mathematics faculty to whom the main admissions committee can refer the files of applicants who appear to have strong math backgrounds. A former Stanford admissions officer, in her book Questions and Admissions, describes how Stanford adopted the latter system after noting that it was losing out on strong math students to other colleges that were better staffed to evaluate the credentials of strong math students who applied.",
  "Solution_13": "Most departments will have a liaison to the admissions office if not a fixed spot on the admissions committee.  Mathematics is unusual in that it is one of the few subjects where a (relatively) large number of applicants claim specific achievements outside the standard curriculum.",
  "Solution_14": "[quote=\"fleeting_guest\"]\nFew colleges have the [i]need[/i] to assess the rigor.   At the non-elite schools, it doesn't matter what the details of the study program are, just that it substantially exceeds high school.   At the elite schools, for purposes of admission the fine details of the self-study program are not important either; if it is consistent with other information (such as competitions or summer mathematical programs as you mention) the claims will probably be taken at face value as an indication of interest, ambition and high, but not precisely calibrated, ability. [/quote]\r\n\r\nWe're basically in agreement. Courses being used to support an application for admission do not need to be detailed to the same extent as courses being used for placement or credit. But I would only expect colleges to take a limited amount of bald assertion on face value. If a student has self-studied a lot of post-secondary math, it is in their best interest to document what they can to make the whole package plausible. They can document with grades, letters from profs who know their work, or by success in competitions. I think it is reasonable to assume that someone from the math department where a student applies will review that sort of thing, and accept a certain amount of additional, undocumented but plausible stuff based solely on a student's description of what they did. I do not think it is reasonable to assume that someone from the math department is going to sift through (hundreds of?) pages of the OP's \"notes/exercises...from Multivariable Calculus, ODE's, the calculus of differential forms, real analysis, and point-set topology\", or quiz them, and come up with some opinion about whether or not the student has actually mastered any of it. I also think it is unreasonable to assume that the OP can simply assert that he has covered those subjects with no documentation of anything beyond AP calc. It is the student's responsibility, not the math department's of every school where s/he is applying, to make his/her application a winning one. If they can document some of that stuff with a grade on a transcript or a short letter from a prof that anyone in the math dept. can understand and believe without spending a lot of time on it, I think that's the student's best route.",
  "Solution_15": "Well, again, I just want to communicate the fact that I am a very motivated student and that I am very devoted to math, not necessarily that I am a master of the entire undergraduate curriculum.\r\n\r\nThat being said, I think that an admissions officer would take my word on the level of math I have reached as much as he or she would take my word on a personal story or a life-changing event. Everything written in a college essay could be cooked up, yet officers do seem to accept what is in most people's essays because it seems reasonable and honest. I am hoping that the same logic will extend to my independent study. As a precaution, though, I will take the advice of some people here and ask a teacher that knows about my progress for a recommendation. \r\n\r\nI am actually a bit rusty in some of the areas I studied; for example I studied ODE's summer before my junior year, and now that it's my senior year, I have forgotten the bulk of the material. I may or may not try to refresh my memory on these topics and go on to take a final at a local college to get the certified credit.\r\n\r\nEither way, thanks for the advice folks.",
  "Solution_16": "Do taking courses off EPGY Stanford seem credible to universities?",
  "Solution_17": "Off topic, but I have a question.  I live around a lot of colleges, like OSU, Captiol, Otterbein, and Columbus State.  Obviously OSU is going to have harder classes that Columbus State, so would it look bad If I took college classes at Columbus State instead of OSU?",
  "Solution_18": "Not if the classes themselves cover the same material....\r\n\r\nA class being \"harder\" than another just means that the prof. assigns more work, or just requires more from his/her students by traveling at high speed through material, or perhaps with more rigorous proofs. So to answer your question, no, it wouldn't look bad if you took classes at Columbus instead of OSU so long as the courses you plan on taking cover the same material as they would have at OSU. College admissions officers aren't going to judge you based off which college you take classes at; the fact that you have taken classes at the college level at all will be impressive to them.",
  "Solution_19": "If you can only get into classes at one place and not the other, you did the best you could by getting into the place you could get into. A lot of high school students try to increase their academic challenges by self-studying or attending part-time classes at some college near their home, so do what you can, and be satisfied that at least you are challenging yourself beyond the high school curriculum.",
  "Solution_20": "[quote=\"h_s_potter2002\"]Off topic, but I have a question.  I live around a lot of colleges, like OSU, Captiol, Otterbein, and Columbus State.  Obviously OSU is going to have harder classes that Columbus State, so would it look bad If I took college classes at Columbus State instead of OSU?[/quote]\r\n\r\nI think they'll be impressed in general that you were taking classes at the college.  But.  More information is necessary.  For example, the University of Illinois at Chicago, hereafter to be referred to as UIC, is about a 25 minute train ride away from me.  It is, to put it nicely, not a very good school.  The University of Chicago, one of the strongest math schools in the country (but I'm biased, I'm actually starting my freshman year there in 6 days :)), is about an hour and a half away by public transportation.  Therefore, I went to UIC.  And evidently, the University of Chicago wasn't at all offended.  \r\n\r\nThere could be other equally valid reasons why you would go to Columbus State instead of OSU.  OSU is more expensive, OSU doesn't like high school students in its classes, and so on.  Even if you just though Columbus State would be easier, which would be the worst from the point of view of a college to which you are applying, it's entirely understandable.  You're in high school still, and balancing a college class on top of that is very time consuming.  To take the easier isn't shameful.",
  "Solution_21": "[quote=\"Magnara\"] For example, [b]the University of Illinois at Chicago[/b], hereafter to be referred to as [b]UIC[/b], is about a 25 minute train ride away from me.  It is, to put it nicely, not a very good school.  The University of Chicago, one of the strongest math schools in the country (but I'm biased, I'm actually starting my freshman year there in 6 days :)), [/quote]\r\n\r\nSure, UIC is not as good as Univ. of Chicago. However, UIC's math department is by no means weak. UIC's math department is very comparable to the math department at the Ohio State University in Columbus, OH.",
  "Solution_22": "Are you thinking of the University of Illinois at Urbana Champaign (UIUC)?",
  "Solution_23": "No, the University of Illinois at Urbana Champaign (UIUC) is better in math and engineering than both UIC and Ohio State University (OSU). However, the math departments at both UIC and OSU are not bad at all.",
  "Solution_24": "[quote=\"qvc122\"]No, the University of Illinois at Urbana Champaign (UIUC) is better in math and engineering than both UIC and Ohio State University (OSU). However, the math departments at both UIC and OSU are not bad at all.[/quote]\r\n\r\nOSU is a sort of \"geometric mean\" of UIC and UIUC; in quantity and level of researchers, it bears roughly the same relationship to UIC as the relationship between UIUC and OSU.  All three are large departments with a few people in (or near) almost any field that a student could be interested in, but whether or not there are very strong people or a research group, will depend on the school and the subject.   \r\n\r\nAt undergraduate level what will make more the difference is the level of other students, the amount of funding, the quality of life, and so on; one will not be in a position to really detect the level of the faculty."
}
{
  "Tag": [
    "LaTeX"
  ],
  "Problem": "Currently, I have tried many free word document to pdf converters which i find unsatisfactory.\r\n\r\nEvery time I try one, the $ \\text{\\LaTeX}$ images I added are blurred and it is barely legible.\r\n\r\nDoes anyone know a good pdf converter that produces clear images?\r\n\r\nThanks.",
  "Solution_1": "Have you tried this?\r\n[url]http://sourceforge.net/projects/pdfcreator/[/url]\r\n\r\nIf you have, you can google \"pdf printer\" and get a lot of programs...",
  "Solution_2": "eh LaTeX documents aren't very hard to write..",
  "Solution_3": "I know......\r\n\r\nI can use the text command, right??\r\n\r\nbut MikeTeX is screwed up on my computer for some reason and even with the fancydhr (or something like that) it can't export as a pdf file.\r\n\r\nthat's why i need a good pdf converter......",
  "Solution_4": "Did you look at the quality settings before you converted?",
  "Solution_5": "Or an alternative, print it out and scan it into a pdf.",
  "Solution_6": "[quote=\"USAMTS Rules\"]The solutions will be printed and/or photocopied in black-and-white for grading. [/quote]\r\n\r\nmy latex comes out blurry too when i convert it to pdf, but it comes out ok when i print it.\r\n\r\nbut eventually graders are going to print it so no problem."
}
{
  "Tag": [],
  "Problem": "The title is self explanatory.\r\nThanks, and see everyone soon.",
  "Solution_1": "There will be computers available. It's been 4 years since I was at Colorado, but at that time we had a group go over to the library to use the computers on a nightly basis during free time. They do not recommend bringing laptops as they can easily get broken, especially in a summer camp setting with lots of people running around."
}
{
  "Tag": [
    "calculus",
    "derivative",
    "function",
    "calculus computations"
  ],
  "Problem": "Has f(x) = exp(x) / 3x any inflection point(s)?",
  "Solution_1": "At an inflection point, the second derivative of a funciton is zero.  $\\frac{d^2}{dx^2}(\\frac{e^x}{3x})=\\frac{e^x}{3x}-\\frac{2e^x}{3x^2}+\\frac{2e^x}{3x^3}=0$  \r\nThis can be rearranged as $e^x(x^2-2x+2)=0$  Neither of these factors has any real roots and therefore, there is no point of inflection. :D"
}
{
  "Tag": [
    "function",
    "induction",
    "algebra proposed",
    "algebra"
  ],
  "Problem": "find all functions such that:($f: \\mathbb{Q}^+\\to\\mathbb{Q}^+$)\r\n$f(x+1)=f(x)+1$ and $f(x^2)={f(x)}^2$",
  "Solution_1": "I'm assuming the \"Q+\" means positive rational numbers.\r\n\r\n[hide]\n$f(1) = f(1)^2 \\rightarrow f(1) = 1$ or $0$, but $0 \\notin \\mathbb{Q}^+$, therefore $f(1) = 1$.  From there, an easy induction using the first property of the function gives $f(x) = x$ as the only possible solution.\n\n\n[/hide]",
  "Solution_2": "$f(x)=x$ for any natural namber.\r\nbut problem wants to find the she solution of this equation in $\\mathbb{Q}^+$",
  "Solution_3": "justdudxit: your solution is wrong",
  "Solution_4": "Damn.  I'm looking at $[0,1]$, seeing if it gets me anywhere",
  "Solution_5": "[quote=\"justdudxit\"]-.  I'm looking at $[0,1]$, seeing if it gets me anywhere[/quote]\r\nIts not as easy as you wrote i think.",
  "Solution_6": "my solution;\r\nfor a rational number$\\frac pq$\r\nwe suppose $f(\\frac pq)=a$\r\nthen$f(\\frac pq+q)=a+q$\r\n$f((\\frac{p+q^2)}{q})^2)=(a+q)^2$\r\n$f((\\frac{p+q^2}{q})^2)=f(\\frac{p^2+2pq^2+q^4}{q^2})=f(\\frac{p^2}{q^2}+2p+q^2)=f(\\frac{p^2}{q^2})+2p+q^2=a^2+2p+q^2$\r\nthen$f((\\frac{p+q^2}{q})^2)=(a+q)^2=a^2+2p+q^2$\r\nso $a=\\frac pq$\r\nthen$f(\\frac pq)=\\frac pq$",
  "Solution_7": "As already said $f(1)=1$\r\nand by Induction f(x)=x $\\forall x \\in N$\r\nso $f(nx)=nx=nf(x)$     $\\forall x,n \\in N$\r\nNow let $x=\\frac{a}{b}$\r\nthen $x*b=a*1$\r\nand $f(b*x)=f(a*1)$ using $f(nx)=nf(x)$\r\nwe get $f(x)=\\frac{a}{b}*f(1)=\\frac{a}{b}=x$\r\n\r\nCorrect?",
  "Solution_8": "[quote=\"sqrt4\"]As already said $f(1)=1$\nand by Induction f(x)=x $\\forall x \\in N$\nso $f(nx)=nx=nf(x)$     $\\forall x,n \\in N$\nNow let $x=\\frac{a}{b}$\nthen $x*b=a*1$\nand $f(b*x)=f(a*1)$ using $f(nx)=nf(x)$\nwe get $f(x)=\\frac{a}{b}*f(1)=\\frac{a}{b}=x$\n\nCorrect?[/quote]\r\nI don't think so.\r\nwe can't get $f(nx)=nx=nf(x)$easily by induction based on the condition$f(x+1)=f(x)+1$",
  "Solution_9": "Well I do.\r\nBy Induction:$f(x)=x$ $\\forall x \\in N$\r\n$nx$ is also natural so f(nx)=nx is correct \r\nand then nx=x+x+x.....+x=f(x)+f(x)+....+f(x)=nf(x)\r\n\r\nI don\u00b4t see the mistake.",
  "Solution_10": "[quote=\"sqrt4\"]Well I do.\nBy Induction:$f(x)=x$ $\\forall x \\in N$\n$nx$ is also natural so f(nx)=nx is correct \nand then nx=x+x+x.....+x=f(x)+f(x)+....+f(x)=nf(x)\n\nI don\u00b4t see the mistake.[/quote]\r\nwell,I see.\r\nIt's right that$f(nx)=nf(x),x \\in N$\r\nbut In your first post ,you wrote\"Let $x=\\frac{a}{b}$\r\nbutl$\\frac{a}{b}$is not a integer,except $b=1$\r\nBe careful when trying to solve the problems.",
  "Solution_11": "Let $f(\\frac{p}{q})=a$, then $f(\\frac{p}{q}+q)=a+q$.\r\nWe have \\[f((\\frac{p}{q})^{2})=a^{2},(a+q)^{2}=(f((\\frac{p}{q})^{2}+2p+q^{2})=f((\\frac{p}{q})^{2}+2p+q^{2})=2p+q^{2}+a^{2}\\Longrightarrow 2aq=2p\\Longrightarrow a=\\frac{p}{q}.\\]",
  "Solution_12": "[quote=\"Rust\"]Let $f(\\frac{p}{q})=a$, then $f(\\frac{p}{q}+q)=a+q$.\nWe have \\[f((\\frac{p}{q})^{2})=a^{2},(a+q)^{2}=(f((\\frac{p}{q})^{2}+2p+q^{2})=f((\\frac{p}{q})^{2}+2p+q^{2})=2p+q^{2}+a^{2}\\Longrightarrow 2aq=2p\\Longrightarrow a=\\frac{p}{q}.\\] [/quote]\r\nSorry,I can't understand your solution.But it seem to be like my solution and there seems to be  some typo in it.",
  "Solution_13": "Let $f(\\frac{p}{q})=a$, then $f(\\frac{p}{q}+q)=a+q,f((\\frac{p}{q}+q)^{2})=(a+q)^{2},f((\\frac{p}{q})^{2})=a^{2}$.\r\nWe have \\[a^{2}+2aq+q^{2}=(a+q)^{2}=(f((\\frac{p}{q})^{2}+2p+q^{2})=f((\\frac{p}{q})^{2}+2p+q^{2}=a^{2}+2p+q^{2}\\Longrightarrow 2aq=2p\\Longrightarrow f(\\frac{p}{q})=a=\\frac{p}{q}.\\]"
}
{
  "Tag": [],
  "Problem": "Any of your school math tecahers hate you?\r\n\r\nmy math teacher hates me",
  "Solution_1": "mine is just the opposite. she's great.",
  "Solution_2": "in 5th grade i really liked my teacher.\r\nthen in 6th grade my teacher was really mean so i hated her and she hated me\r\nin 7th and 8th grade my teacher was mean too and I hate her so she hates me"
}
{
  "Tag": [
    "inequalities",
    "absolute value"
  ],
  "Problem": "Thought it might be entertaining for you guys:\r\n\r\n Find the restrictions on $ x$ if:\r\n\r\n#1. $ |x\\minus{}5| \\leq |2x\\minus{}3|$\r\n\r\n#2. $ |2x| \\leq |x\\minus{}3|$\r\n\r\n#3. $ |x| > |2x\\minus{}6|$",
  "Solution_1": "i like to deal with absolute value problems by squaring :D \r\nsince these inequalities are preserved while squaring(both sides are positive)\r\n\r\n1.$ x^2 \\minus{} 10x \\plus{} 25 <\\equal{} 4x^2 \\minus{} 12x \\plus{} 9, so\r\n3x^2 \\minus{} 2x \\minus{} 16 >\\equal{}0$\r\n. the solution to the equation is -2,8/3, so x<-2 or x>8/3\r\n\r\nthe rest can be solved similarly",
  "Solution_2": "I am not good at these kinds of problems, but I will still give it a go. \r\n\r\n#2. Square it to get 4x^2 < or = x^2 - 6x + 9\r\n \r\n         3x^2 < or =  9 - 6x \r\n\r\n         3x^2 + 6x - 9 < or = 0 \r\n     \r\n         Solve the equation and the solutions are 1, and -3, so the solution is; \r\n         x < -3 , and x > 1 \r\n\r\n\r\nI let somebody else solve number 3.",
  "Solution_3": "3. For x<3, it becomes\r\nx>-2x+6.\r\n3x>6\r\nx>2\r\n\r\nFor x>0, it becomes\r\nx>2x-6\r\nx<6\r\n\r\nThen, answer is $ (2,6)$ (I guess :|)",
  "Solution_4": "please don't revive old threads",
  "Solution_5": "In this case it's fine, since the thread is not [i]that[/i] old, and the problems were not all finished yet (so people may still be looking for feedback on their answers / solutions)."
}
{
  "Tag": [
    "trigonometry",
    "inequalities proposed",
    "inequalities"
  ],
  "Problem": "Let x,y,z is possitive real number satisfies xy+yz+zx=3 . Prove this ineq\r\n$ \\sqrt{x^2\\plus{}3}\\plus{} \\sqrt{y^2\\plus{}3} \\plus{} \\sqrt{z^2\\plus{}3} \\geq 3\\plus{}x\\plus{}y\\plus{}z$",
  "Solution_1": "[quote=\"eagerkill\"]Let x,y,z is possitive real number satisfies xy+yz+zx=3 . Prove this ineq\n$ \\sqrt {x^2 \\plus{} 3} \\plus{} \\sqrt {y^2 \\plus{} 3} \\plus{} \\sqrt {z^2 \\plus{} 3} \\geq 3 \\plus{} x \\plus{} y \\plus{} z$[/quote]\r\nSee here http://www.artofproblemsolving.com/Forum/viewtopic.php?t=133953",
  "Solution_2": "Hi,arqady,Nice to see you :) Hope my solution is right and not so ugly.\r\nBy change variables $ x \\equal{} \\cot A,y \\equal{} \\cot B,z \\equal{} \\cot C$,Where $ ABC$ is an acute triangle.\r\n$ \\frac {1}{\\sin A} \\plus{} \\frac {1}{\\sin B} \\plus{} \\frac {1}{\\sin C} \\ge \\sqrt {3} \\plus{} \\cot A \\plus{} \\cot B \\plus{} \\cot C$\r\nBut $ \\frac {1}{\\sin x} \\minus{} \\cot x \\equal{} \\frac {1 \\minus{} \\cos x}{\\sin x} \\equal{} \\tan \\frac {x}{2}$\r\nSo it suffices to prove:\r\n$ \\tan \\frac {A}{2} \\plus{} \\tan \\frac {B}{2} \\plus{} \\tan \\frac {C}{2} \\ge \\sqrt {3}$\r\nBut this is obvious by $ \\tan x$ is convex on $ [0,\\frac{\\pi}{2})$.",
  "Solution_3": "[quote=\"zhaobin\"]Hi,arqady,Nice to see you :)[/quote]\r\nMutually! :lol:"
}
{
  "Tag": [
    "number theory proposed",
    "number theory"
  ],
  "Problem": "A natural number is a [i]palindrome[/i] when one obtains the same number when writing its digits in reverse order. For example, $481184$, $131$ and $2$ are palindromes.\r\n\r\nDetermine all pairs $(m,n)$ of positive integers such that $\\underbrace{111\\ldots 1}_{m\\ {\\rm ones}}\\times\\underbrace{111\\ldots 1}_{n\\ {\\rm ones}}$ is a palindrome.",
  "Solution_1": "I think it 's $m < 10$ or $n < 10$\r\n\r\nWhen doing the multiplication like in small classes, if $m \\geq 10$ or $n \\geq 10$ there is an extra carry which introduces a malicious $8$ on the right part.\r\n\r\nI would find it difficult to write a full and complete solution in competitions ....",
  "Solution_2": "It's not so hard to write a full solution...\r\n\r\nNote that $N = \\underbrace{111\\ldots 1}_{m\\ {\\rm ones}}\\times\\underbrace{111\\ldots 1}_{n\\ {\\rm ones}}$ has exactly $m + n - 1$ digits, since $\\underbrace{111\\ldots 1}_{m\\ {\\rm ones}}\\times\\underbrace{111\\ldots 1}_{n\\ {\\rm ones}} < 2\\cdot 10^{m-1}\\times2\\cdot 10^{n-1} = 4 \\cdot 10^{m+n-2}$. If $m, n > 9$, then considering the tenth leftmost digit, there would be a carry, thus the number consisting of the first nine digits of $N$ is bigger than the number formed by the last nine digits of $N$, in reversed order. Then if $m,n > 10$ the number $N$ is not a palindrome.",
  "Solution_3": "Would that be seen as a full solution ???"
}
{
  "Tag": [],
  "Problem": "In an office during the day the boss gives the secretary a letter to type each time putting the letter on the top of the secretary's inbox.When there is time the secretary takes the top letter off the pile and types it.If there are five letters in all and the boss delivers them in the order 1 2 3 4 5 which of the following could not be the order in which the secretary types them?\r\nA, 12345\r\nB, 24351\r\nC,32415\r\nD,45231\r\nE,54321\r\n\r\nCan anyone help me to solve this problem?",
  "Solution_1": "Is this assuming you can't just grab one out of the middle?",
  "Solution_2": "[hide]It would have to be D.\n\nYou wait until the 4th is delivered. Do it and then wait til 5th is delivered. You take that but you can't do the 2nd right after because you have to do it in order so it would be the 3rd.[/hide]",
  "Solution_3": "This problem definitely doesn't belong in this forum.\r\n\r\nHere is a thread for which this same problem was included in the Problem of the Day.\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?p=362143#p362143",
  "Solution_4": "Yes.  Please look through the forums to determine the forum in which the problems you post are most relative in difficulty."
}
{
  "Tag": [],
  "Problem": "Hi africans.\r\nCan some one send the last test of african MO?\r\nBy the way,what was the results?",
  "Solution_1": "http://www.mathlinks.ro/Forum/viewtopic.php?t=48794\r\nThey are the pan aftrican MO problems. Don't know if they are the ones you want.",
  "Solution_2": "Thanks shobber :)",
  "Solution_3": "It's also hard to find Belarussian MO, which contains many interesting problem?",
  "Solution_4": "[quote=\"shobber\"]http://www.mathlinks.ro/Forum/viewtopic.php?t=48794\nThey are the pan aftrican MO problems. Don't know if they are the ones you want.[/quote]Actually shobber here's the link with all the problems: http://www.mathlinks.ro/Forum/resources.php?c=1&cid=28 ;)"
}
{
  "Tag": [
    "linear algebra",
    "matrix",
    "vector",
    "algebra",
    "polynomial",
    "combinatorics proposed",
    "combinatorics"
  ],
  "Problem": "A graph will be called [i]even[/i] if each of its vertices has an even degree.\r\n\r\nLet $ V$ be the set of vertices of an even graph $ G$ such that $ \\left|V\\right|$ is an even number. Prove that the set $ V$ can be partitioned into two non-empty subsets $ V_{1}$ and $ V_{2}$ such that the induced subgraph of $ G$ on each of these two subsets $ V_{1}$ and $ V_{2}$ is an even graph.\r\n\r\n  darij",
  "Solution_1": "I don't know if the problem is new.  Here is a related problem from Lovasz's book [i]Combinatorial Problems and Exercises[/i].\r\n\r\nLet $ G$ be an arbitrary graph with vertex set $ V$.  Then $ V$ can be partitioned into two subsets $ V_{1}$ and $ V_{2}$ such that the induced subgraphs are even.  (This problem allows empty subsets.)\r\n\r\nAs a corollary, $ V$ can be partitioned into two sets such that one induced subgraph has all-even degrees and the other has all-odd degrees.\r\n\r\nLovasz credits Gallai (unpublished), Chen (SIAM J. Appl. Math 1971), and Posa.",
  "Solution_2": "Darij tells me to post the following in his name:\r\n\r\nThe problem with $ G$ being an arbitrary graph and $ V_{1}$ and $ V_{2}$ not having to be empty is known to him (see [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=59434[/url]), but he thinks his problem requires some other idea.",
  "Solution_3": "It does need a little more work, but the algebraic proof in that other thread (see the link in NoName's message) can be adapted. \r\n\r\nLet's start just as in the topic mentioned above: let $ A$ be the adjacency matrix of the graph, regarded as an endomorphism of $ \\mathbb F_{2}^{n}$ (with $ n$ being the number of vertices of $ G$). We want to prove that there is a non-zero vector $ x\\in\\mathbb F_{2}^{n}$, different from the all-$ 1$ vector (this, together with $ x\\ne 0$, says that the sets $ V_{i}$ are both non-empty) such that $ Ax \\equal{} 0$. In other words, what we want to show is that the rank of $ A$ cannot be $ n\\minus{}1$; we already know that it's at most $ n\\minus{}1$ because $ G$ is even and hence the sum of all the columns of $ A$ is zero.\r\n\r\nAfter casting out the $ i$'th row and column for any $ i\\in\\overline{1,n}$, the problem is reduced to showing that a symmetric $ m\\times m$ matrix $ M$ over $ \\mathbb F_{2}$ having $ 0$ on the main diagonal can't be invertible if $ m$ is odd. The simplest proof I could think of is this: write the determinant of $ M$ in the usual way, as a sum of products of elements of $ M$. The products which contain an element on the main diagonal are zero, and the other ones can be partitioned (because $ m$ is odd and $ M$ is symmetric) in equal pairs which are the transpose of one another. It follows that the sum of all the products must be $ 0$, i.e. $ \\det M \\equal{} 0$.",
  "Solution_4": "I don't have much to say because your solution is exactly the same as mine, Grobber. Only one remark:\n\n[quote=\"grobber\"]the problem is reduced to showing that a symmetric $ m\\times m$ matrix $ M$ over $ \\mathbb F_{2}$ having $ 0$ on the main diagonal can't be invertible if $ m$ is odd.[/quote]\n\nThis, in fact, holds for any field with characteristic $ 2$ instead of $ \\mathbb{F}_{2}$, and is a particular case of the following theorem:\n\n[b](*)[/b] An alternating $ m\\times m$-matrix over any field cannot be invertible if $ m$ is odd.\n\nHere, a matrix $ A$ is called [i]alternating[/i] if it satisfies $A^{T} = -A$ and all of its entries on the main diagonal are $0$. Over fields with characteristic $ \\neq 2$, a matrix $A$ is alternating if and only if it is antisymmetric (i.e., it satisfies $A^T = -A$), but this does not hold over fields of characteristic $ 2$. Instead, over fields of characteristic $ 2$, a matrix is alternating if and only if it is symmetric and all of its entries on the main diagonal are $ 0$.\n\nYou can prove the assertion [b](*)[/b] by first showing it for fields of characteristic $ \\neq 2$ (trivially: $ \\det A = \\det A^T = \\det\\left(-A\\right) = \\left(-1\\right)^{m} \\det A = -\\det A$, and thus $2 \\det A = 0$, so that $ \\det A = 0$) and then concluding that it must hold for fields of characteristic $ 2$ as well by a polynomial identity argument. The latter argument works because the determinant of an alternating matrix is a universal polynomial in its entries below the main diagonal and these entries are completely independent.  (However, the same cannot be said of an antisymmetric matrix, so that this argument would not work anymore if we replaced \"alternating\" by \"antisymmetric\" in [b](*)[/b].)\n\n  Darij"
}
{
  "Tag": [
    "inequalities proposed",
    "inequalities"
  ],
  "Problem": "$ a,b,c$ are positive reals such that $ abc=1$. Prove that\r\n$ \\frac{a}{(a+1)(b+1)}+\\frac{b}{(b+1)(c+1)}+\\frac{c}{(c+1)(a+1)}\\ge{\\frac{3}{4}}$",
  "Solution_1": "$ \\frac{a}{(a+1)(b+1)}+\\frac{b}{(b+1)(c+1)}+\\frac{c}{(c+1)(a+1)}\\ge{\\frac{3}{4}}$\r\n\r\ncan be rewritten as follows\r\n\r\n$ 4[a(c+1)+b(a+1)+c(b+1)] \\ge 3(a+1)(b+1)(c+1)$\r\n\r\n$ 4ab+4bc+4ca+4a+4b+4c \\ge 3(abc+ab+bc+ca+a+b+c+1)$\r\n\r\n$ ab+bc+ca+a+b+c \\ge 6$ since $ abc=1$\r\n\r\n$ abc=1 \\Rightarrow  ab=\\frac{1}{c}$\r\n\r\nNow, $ \\frac{1}{c}+c\\ge2$ since $ (c-1)^{2}\\ge0$\r\n\r\nThen $ ab+c\\ge2$, $ bc+a\\ge2$, $ ca+2\\ge2 \\Rightarrow ab+bc+ca+a+b+c \\ge 6$ is true.",
  "Solution_2": "[quote=\"Litlle 1000t\"]$ a,b,c$ are positive reals such that $ abc=1$. Prove that\n$ \\frac{a}{(a+1)(b+1)}+\\frac{b}{(b+1)(c+1)}+\\frac{c}{(c+1)(a+1)}\\ge{\\frac{3}{4}}$[/quote]\r\n\r\n[b] HINT[/b] \r\n\r\n Put \r\n\r\n$ a=\\frac{x}{y}, b=\\frac{y}{z}, , c = \\frac{z}{x}$ , \r\n\r\n clear the denominators and it is sufficient to prove that \r\n\r\n   $ 4[xy(x+y)+yz(y+z)+zx(z+x)] \\geq 3(x+y)(y+z)(z+x)$ , \r\n\r\n which easy becaoms\r\n\r\n $ x(y-z)^{2}+y(z-x)^{2}+z(x-y)^{2}\\geq 0$ \r\n\r\n We have equality only if      $ a=b=c=1$\r\n\r\n[u] Babis [/u]",
  "Solution_3": "I have another solution  :oops: \r\n\r\n$ \\frac{a}{(a+1)(b+1)}+\\frac{b}{(b+1)(c+1)}+\\frac{c}{(c+1)(a+1)}\\geq\\frac{3}{4}$\r\n\r\n$ \\Leftrightarrow4[a(c+1)+b(a+1)+c(b+1)]\\geq3(a+1)(b+1)(c+1)$ $ (1)$\r\n\r\nSetting $ p=a+b+c$ and $ q=ab+ac+bc$ we have :\r\n\r\n$ (1)\\Leftrightarrow4(p+q)\\geq3(p+q+2)\\Leftrightarrow p+q\\geq6$\r\n\r\nBut,\r\n\r\n$ p+q=(a+b+c)+(ab+ac+bc)\\geq6\\sqrt[6]{(abc)^{3}}=6$ from AM-GM.\r\n\r\nDone.\r\n\r\nEDIT : Oh, it's the same than stergiu  :oops_sign: , sorry, I didn't see your solution",
  "Solution_4": "see also [url]http://www.mathlinks.ro/viewtopic.php?p=521094#p521094[/url]\r\n\r\nThis problem also was in Czech Slovak M.C. (beffore then it appeared in France)",
  "Solution_5": "Ohh, this problem comes from French TST  :o",
  "Solution_6": "[quote=\"Litlle 1000t\"]$ a,b,c$ are positive reals such that $ abc=1$. Prove that\n$ \\frac{a}{(a+1)(b+1)}+\\frac{b}{(b+1)(c+1)}+\\frac{c}{(c+1)(a+1)}\\ge{\\frac{3}{4}}$[/quote]\r\n\r\nAnother solution by AM-GM:\r\n\r\n$ \\frac{a}{(a+1)(b+1)}+\\frac{a+1}{8}+\\frac{b+1}{8}\\geq\\frac{3a}{4}$\r\n\r\nSo that: \r\n\r\n$ \\sum{\\frac{a}{(a+1)(b+1)}}\\geq (\\frac{3}{4}-\\frac{1}{4})(a+b+c)-\\frac{3}{4}=\\frac{1}{2}(a+b+c)-\\frac{3}{4}\\geq\\frac{3}{2}-\\frac{3}{4}=\\frac{3}{4}$\r\n\r\nSo we are done!!^^"
}
{
  "Tag": [
    "search",
    "\\/closed"
  ],
  "Problem": "While accessing some pages,I get a message saying it might be unsafe to let an ActiveX control to interact with other parts of the page.\r\nI'm using IE 6.0\r\n\r\nWhat is this problem ?",
  "Solution_1": "Can you give us a link to the problem pages? Thanks.",
  "Solution_2": "I got the message while I accessed this page.\r\nBTW I accessed this Page from http://www.artofproblemsolving.com/Forum/search.php?search_id=egosearch\r\ni.e. I was viewing my posts and then accessed this page.\r\n\r\nHere's the exact message :\r\n\r\n\r\nI usually get the message when accessing my post from http://www.artofproblemsolving.com/Forum/search.php?search_id=egosearch\r\n\r\nElse there is no problem.",
  "Solution_3": "I think your firewall may be a little too sensitive."
}
{
  "Tag": [
    "MATHCOUNTS"
  ],
  "Problem": "It is once again that time of year for MATHCOUNTS.  Chapter competitions have already taken place across the nation.\r\n\r\nDue to the popularity of the format for our last MATHCOUNTS Math Jam we will again be running a countdown round style Math Jam on Wednesday, Feb. 18.  We will keep a tally of points based on the first three students to submit correct answers as well as the best explanations of the solutions offered.  To see how this worked last time, check out the transcript:\r\n\r\nhttp://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=20\r\n\r\nThere will be a slight change in the way the game will be run.  No split points will be awarded for best solutions.  If two or more solutions are equally nice, the student who answered the problem the quickest will be awarded the extra point.\r\n\r\nCome and challenge some of the top MATHCOUNTS students in the nation!",
  "Solution_1": "The questions will come from all levels.  I think they will be appropriate practice for the upcoming competitions.",
  "Solution_2": "darn.  the math jam is four days after our chapter competition.",
  "Solution_3": "I just checked the Math Jams Page and it doesn't list the MATHCOUNTS Jam on the 11th, but on the 18th.  Was that a typo, or have you added one and just haven't listed it yet?\r\n\r\nthanks",
  "Solution_4": "[quote=\"frost13\"]I just checked the Math Jams Page and it doesn't list the MATHCOUNTS Jam on the 11th, but on the 18th.  Was that a typo, or have you added one and just haven't listed it yet?\n\nthanks[/quote]\r\n\r\nThank you for noticing my mistake.  The Math Jam will indeed by on the 18<sup>th</sup>.  I will edit my post."
}
{
  "Tag": [],
  "Problem": "Prove that if p is prime, then $ 2^p \\minus{} 1$ is also prime. Do not be afraid to include examples if it helps your proof!",
  "Solution_1": "No, if $ p\\equal{}11$.",
  "Solution_2": "lol, 3 and 11 were my first guesses. \r\n\r\nSometimes is the answer. $ 2^3 \\minus{} 1$ is prime but $ 2^{11} \\minus{} 1$ is not (23 times 89.)",
  "Solution_3": "[quote=\"Zhero\"]lol, 3 and 11 were my first guesses. \n\nSometimes is the answer. $ 2^3 \\minus{} 1$ is prime but $ 2^{11} \\minus{} 1$ is not (23 times 89.)[/quote]\r\nOkay so I guess that does not work.\r\n\r\nHow about this?\r\n\r\n$ a_1 \\equal{} 2$, $ a_n$=2^($ {a}_{n \\minus{} 1}$)-1\r\nEvery term of this sequence is a prime.",
  "Solution_4": "Beyond the fifth term of this sequence are numbers that are not known to be prime or composite.  In particular, only [url=http://en.wikipedia.org/wiki/Mersenne_prime]46 Mersenne primes[/url] are known.",
  "Solution_5": "[quote=\"t0rajir0u\"]Beyond the fifth term of this sequence are numbers that are not known to be prime or composite.  In particular, only [url=http://en.wikipedia.org/wiki/Mersenne_prime]46 Mersenne primes[/url] are known.[/quote]\r\n\r\nMy original conjecture was an unproved conjecture, but I think that the new one I just thought of is correct."
}
{
  "Tag": [],
  "Problem": "1. On a 10 problem test, Jim will score 5 points for a correct answer, 1 point for no answer, and 0 points for an incorrect answer. What is the smallest positive score that Jim [b]cannot[/b] earn?\r\n\r\n2. If 4/5 of the students in Ms. Smith's class use 5/6 of the desks in the room, what is the least possible number of students in the class?",
  "Solution_1": "[hide=\"2\"]\nWe have $ \\frac45s\\equal{}\\frac56d\\implies d\\equal{}\\frac{24}{25}s\\implies s\\geq\\boxed{25}$.\n[/hide]\r\n\r\nFor $ 1$, do you mean smallest positive score? And can he skip questions?",
  "Solution_2": "[quote=\"mathemagician1729\"]1. On a 10 problem test, Jim will score 5 points for a correct answer, 1 point for no answer, and 0 points for an incorrect answer. What is the smallest positive score that Jim [b]cannot[/b] earn?\n\n2. If 4/5 of the students in Ms. Smith's class use 5/6 of the desks in the room, what is the least possible number of students in the class?[/quote]\r\n\r\n\r\n[hide=\"1\"]  Okay...We know he can get $ 1\\minus{}9$ since he can get $ x$ for no answer.  He can get 10-19 by getting 2 or 3 correct ones and 1-4 no answers.  He can get 20-29 by getting 4 or 5 correct and 1-4 no answer.  ON 30-39, though, he cannot attain 39 since if he has 7 correct answers, he can only have 3 no answers, therefore $ 39$ is the lowest score Jim cannot earn.[/hide]",
  "Solution_3": "1.\r\n\r\nx:  number of correct answers\r\ny:  number of no answers\r\n\r\n\r\nTotal score s = 5x + y  where   0 <= x <= 10; 0 <= y <= 10; 0 <= x + y <= 10; \r\n\r\nFor each x, the scores differ by 5, so the y has to be at least 4 to make the possible \r\nscores consecutive, therefore x = 7, y = 3 gives s = 5 * 7 + 3 = 38.  The answer is 39.\r\n\r\n\r\nI hope someone could give more elegant solutions."
}
{
  "Tag": [],
  "Problem": "How many positive integer divisors of 2007^2007 are divisible by exactly 2007 positive integers",
  "Solution_1": "Call x is a positive integer divisor of $\\ 2007^{2007}$ are divisible by exactly 2007 positive integers.\r\n   $\\ 2007=223*3^{2}$\r\n    $\\ x=223^{a}.3^{b}$ and (a+1)(b+1)=2007=223.9=669.3=2007.1\r\n  So there're 6"
}
{
  "Tag": [],
  "Problem": "My father is four times as old as me. In 20 years, he will be only twice as old as me. How old is my father and how old am I?",
  "Solution_1": "[hide]\nI made a chart:\nboy    dad in 20 years\n                  boy  dad\n8        32    28    52\n9        36    29    56\n10      40    30    60\nSo the answer is he is 10 years old[/hide]",
  "Solution_2": "[hide]x=kid\ny=dad\n\n4x=y\n2(x+20)=y\nx=20[/hide]"
}
{
  "Tag": [],
  "Problem": "Ms. Hamilton's eighth-grade class wants to participate in the annual\nthree-person-team basketball tournament.  Lance, Sally, Joy, and Fred\nare chosen for the team.  In how many ways can the three starters be chosen?",
  "Solution_1": "You have 4! ways to choose people, but you have counted every team 3! times. Take $ \\frac{4!}{3!}$ which gives $ \\boxed{4}$\r\n\r\nAlternatively, you have 4 different choices of who sits out, the other 3 will be playing.",
  "Solution_2": "Or we can do $\\binom{4}{3}=4$"
}
{
  "Tag": [],
  "Problem": "Sa se afle in cate regiuni este impartit spatiul de 10 plane aflate in pozitie generala.",
  "Solution_1": "$\\binom{10}{0}+\\binom{10}{1}+\\binom{10}{2}+\\binom{10}{3}$ :)",
  "Solution_2": "Da, intr-adevar asta este raspunsul, insa e nevoie de mai multe detalii mai ales pentru un pusti de clasa a-9-a care inca nu a facut la clasa combinari. :)",
  "Solution_3": "hai ca dau eu mai multe detalii:\r\n\r\nstim ca trei plane formeaza un varf in spatiu. deci exista $\\binom{10}{3}$ varfuri, iar fiecare este punctul cel mai de jos exact al unei parti a spatiului.\r\n\r\nasadar exista $\\binom{10}{3}$ parti care au \"cel mai de jos punct\", fiecare parte care nu are cel mai de jos punct taie un plan orizontal in una cele $X$ parti plane.\r\n\r\ndeci obtinem numaru $X+\\binom{10}{3}$, ne mai ramane sa aflam $X$, si anume numaru de parti in care poate fi impartit planul de 10 drepte.\r\n\r\ncele $\\binom{10}{2}$ puncte de intersectie ale dreptelor sunt usor de numarat, dar fiecare punct de intersectie este punctul \"cel mai de jos\" exact al unei parti. deci exista $\\binom{10}{2}$ regiuni avand un cel mai de jos punct, exista astfel $10+1=11$ regiuni care nu au un punct cel mai de jos, adika $\\binom{10}{0}+\\binom{10}{1}$.\r\n\r\ndeci adunand obtinem $X=\\binom{10}{0}+\\binom{10}{1}+\\binom{10}{2}$.\r\n\r\nacum inlocuind $X$ in ce obtinusem mai sus dam raspunsul final dat si de domnul Enescu, $\\binom{10}{0}+\\binom{10}{1}+\\binom{10}{2}+\\binom{10}{3}$\r\n\r\n\r\nPS: m-a mai intrebat Cezar de problema asta mai de demult dar ii zisesem o prostie :)",
  "Solution_4": "Solu\u0163ia e din  Arthur Engel \"Problem Solving Strategies\", tradus\u0103 de editura GIL. Nu vreau s\u0103 fac reclam\u0103, dar e o carte esen\u0163ial\u0103 pentru cei care doresc s\u0103 ajung\u0103 la baraj. Exista, p\u00e2n\u0103 acum c\u00e2teva luni, un download pe site-ul lib.org.by . Oricum, varianta \u00een limba rom\u00e2n\u0103 (a prof. Mihai B\u0103lun\u0103) e mai bun\u0103, deoarece au fost \u00eenl\u0103turate o serie \u00eentreag\u0103 de gre\u015feli din edi\u0163ia original\u0103 \u015fi au fost ad\u0103ugate solu\u0163ii.",
  "Solution_5": "dap , de acolo e rezolvarea, nu am zis ca it`s mine:)",
  "Solution_6": "[quote=\"pohoatza\"]dap , de acolo e rezolvarea, nu am zis ca it`s mine:)[/quote]\r\nNu asta am vrut s\u0103 spun...scuze dac\u0103 asta s-a \u00een\u0163eles...\r\nEu \u00eensumi am scris zeci de solu\u0163ii pe acest site pe care le \u015ftiam din diverse surse, f\u0103r\u0103 a le men\u0163iona.\r\nAm vrut doar s\u0103 atrag aten\u0163ia asupra unei c\u0103r\u0163i care este, \u00eentr-adev\u0103r, de excep\u0163ie. Acum ceva ani, un elev de clasa a 9-a, foarte talentat de altfel, m-a rugat s\u0103 \u00eel ajut \u00een preg\u0103tire. I-am dat cartea lui Engel. A parcurs-o, cu creionul pe h\u00e2rtie, \u00een 3 s\u0103pt\u0103m\u00e2ni (!). \u00cen urm\u0103torii trei ani a luat aur la OIM. Nu spun c\u0103 numai Engel a fost cauza, dar..."
}
{
  "Tag": [
    "integration",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Let $ f \\in C_{[0;1]}$. Suppose that, for all $ n \\in \\mathbb{N}$, exists $ a_n,b_n \\in \\mathbb{R}$ such that $ \\int \\limits_0^1 (f(x)-a_nx-b_n)^4 dx < \\frac{1}{n}$. Prove that $ f(x)=ax+b, \\forall x \\in [0;1]$ ($ a,b \\in \\mathbb{R}$).",
  "Solution_1": "Since $ \\inf_{(a,b) \\in \\mathbb{R}^2} \\int_0^1 (f(x) \\minus{} ax \\minus{} b)^2 dx \\equal{}0$, what is the orthogonal projection of $ f$ onto the subspace $ \\{ P_{a,b}(x) \\equal{} ax\\plus{}b : (a,b) \\in \\mathbb{R}^2 \\}$ ?"
}
{
  "Tag": [
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "construct a nonabelian group of order 21.",
  "Solution_1": "Take a look at the following link:\r\n\r\n[url]http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist;task=show_msg;msg=2573.0001[/url]\r\n\r\nRegards.",
  "Solution_2": "$ G\\equal{}\\{x^i y^j | x^7\\equal{}y^3\\equal{}xyxy^{\\minus{}1}\\equal{}e\\}$",
  "Solution_3": "[quote=\"mitchan88\"]$ G \\equal{} \\{x^i y^j | x^7 \\equal{} y^3 \\equal{} xyxy^{ \\minus{} 1} \\equal{} e\\}$[/quote]\r\nNo, $ x^{\\minus{}2}yxy^{\\minus{}1}\\equal{}1$.\r\n$ xyxy^{ \\minus{} 1}\\equal{}1$ would give $ y^2xy^{\\minus{}2}\\equal{}yx^{\\minus{}1}y^{\\minus{}1}\\equal{}x$ and $ G\\equal{}\\langle x,y^2\\rangle$  abelian.",
  "Solution_4": "More to the point, it would give $ x \\equal{} y^3 x y^{\\minus{}3} \\equal{} x^{\\minus{}1}$.  So $ x^7\\equal{}x^2\\equal{}e$, hence $ x\\equal{}e$.\r\n\r\nWe need $ yxy^{\\minus{}1} \\equal{} x^2$ or $ x^4$. (the action needs to have order 3!)"
}
{
  "Tag": [
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "Find all endomorphisms of an euclidean space $E$ which conserve the orthogonality of vectors.",
  "Solution_1": "let's prove that such an endomorphism $u$ is of the form $u=k.f$ where $k \\in R$ and $f \\in O_n(R)$.\r\nWe know that $u$ conserves orthogonality hence it maps the canonical basis $(e_1,..,e_n)$ to $(b_1,..,b_n)$ where $(b_i | b_j)=0$ for all $i \\neq j$. It suffices to show that $|b_i|=k$ for all $i$.\r\nBut, for $i \\neq j$, $(u(e_i + e_j) | u(e_i + e_j))=((b_i + b_j)|(b_i + b_j))=|b_i|^2 - |b_j|^2 = 0$ because \r\n$((e_i+e_j)|(e_i - e_j))=0$. Conclusion follows."
}
{
  "Tag": [
    "Pascal\\u0027s Triangle",
    "binomial coefficients"
  ],
  "Problem": "In row n of Pascal's triangle, let E be the sum of the binomial coefficients, n C k, for which k is even and let D be the sum of the binomial coefficients, for which k is odd. How do E and D compare?",
  "Solution_1": "[hide=\"my clumsy explanation\"]$E=D$ because if you expand $(a+b)^n$ and you let $a=1$, $b=-1$, then $(a+b)^n=0$ so the result follows[/hide]",
  "Solution_2": "Works for me!"
}
{
  "Tag": [
    "inequalities proposed",
    "inequalities"
  ],
  "Problem": "Given a positive integer $k \\geq 2$.Find the minimum value of:\r\n                   $F(x_{1},x_{2},...,x_{n})=\\frac{x_{1}^{k}+x_{2}^{k}+...+x_{n}^{k}}{x_{1}+x_{2}+...+x_{n}}$ ($x_{1},x_{2},...,x_{n}$ be distinct positive integers,$n \\geq 1$)",
  "Solution_1": "Because the sequence$\\ (x_{1},x_{2},...,x_{n})$is composed of distinct positive integers, then if we want F minimum then we most have the smallest sequence of differents integers possible that is:$\\ (x_{1},x_{2},...,x_{n})=(1,2,..,n)$\r\n\r\n$\\rightarrow$ $\\ F(x_{1},x_{2},...,x_{n})=\\frac{1^{k}+2^{k}+...+n^{k}}{1+2+...+n}$,\r\n\r\n$\\rightarrow$ $\\ F(x_{1},x_{2},...x_{n})=\\frac{(B+n+1)^{[k+1]}-B^{[k+1]}}{k+1}\\frac{2}{n(n+1)}$\r\n\r\nwhere the notation$\\ B^{[k]}$means the quantity in question is raised to the appropriate power$\\ k$, and all terms of the form $\\ B^{m}$are replaced with the corresponding Bernoulli numbers$\\ B_{m}$."
}
{
  "Tag": [
    "geometry",
    "3D geometry"
  ],
  "Problem": "A rectangular aquarium 16 inches long, 9 inches wide and 10 inches tall is filled with water to a level of 8 inches. A solid steel cube with an edge of 3 inches is dropped into the aquarium. How many inches high is the new water level in the aquarium? Express your answer as a decimal rounded to the nearest tenth.",
  "Solution_1": "The cube has a volume of 3*3*3=27 cubic inches \r\nthe base of the aquarium has an area of 9*16= 144 square inches\r\nso dropping in the cube will cause the water level to rise 27/144 = 3/16 = 0.1875 inches\r\nthe water level started at 8 inches so the final water level would be 8.1875 inches, or 8.2 inches rounded to the nearest tenth",
  "Solution_2": "Why would it rise 27/144? I don't get it...",
  "Solution_3": "It's kind of an intuition kind of thing, but it's like the Archimedes problem: An object will displace an amount of liquid equal to its volume when placed in a tank of liquid. The base area of the tank is $144$ and the cube's volume is $27$. The cube will displace $27$ cubic inches of water spread out over $144$ square inches of water, hence $\\frac{27}{144}$. $\\box[4]$"
}
{
  "Tag": [
    "function",
    "topology",
    "advanced fields",
    "advanced fields unsolved"
  ],
  "Problem": "a) Let  $f: S^{n}\\longrightarrow S^{m}$ ($n<m$) be a continious function. Prove that $f$ is nullhomotopic. (i.e. it is homotop to constant function)\r\nb) Let $X$ is subspace of $Y$ and $Y$ is simply connected, and $X$ is contractible in $Y$. Prove that each function $f: X\\longrightarrow Y$ is nullhomotopic.\r\n\r\na) is given by our teacher, and b) is my conjecture. I don't know whether it is correct or not.",
  "Solution_1": "b) is wrong: consider $Y = \\mathbb{R}^{2}\\setminus 0$, $X = S^{1}+100$, then $X \\to S^{1}$ (obvious map) is not nullhomotopic",
  "Solution_2": "Yes you are right, but now I think when $Y$ is simply connected my conjecture is correct.",
  "Solution_3": "There is a map $S^{3}\\to S^{2}$ which is not nullhomotopic (http://en.wikipedia.org/wiki/Hopf_bundle). Now take $Y = \\mathbb{R^{4}}\\setminus D^{3}$ and do the analogous thing.",
  "Solution_4": "a) is easy to prove if $f(S^{n})$ is a proper subset of $S^{m}$: pick a point $a\\in S^{m}\\setminus f(S^{n})$ and gradually push $f(S^{n})$ into the antipode of $a$. But a continuous map $f\\colon S^{n}\\to S^{m}$ may be surjective even though $n<m$. I suppose one can avoid this difficulty by constructing a \"nice\" map $\\tilde f\\colon S^{n}\\to S^{m}$ such that $|\\tilde f(x)-f(x)|<\\epsilon$ for all $x\\in S^{n}$. However, this is not so easy when $n>1$.",
  "Solution_5": "a) Sketch:\r\n\r\nThere is such a theorem. Let $(X,A)$ and $(Y,A)$ be $CW$-pairs, and let $f: (X,A)\\to (Y,B)$ be a map of pairs. Then there exists a cellular map $g: (X,A)\\to (Y,B)$ such that $g\\simeq f\\text{ rel }A$.\r\n\r\nThen, according to the theorem, $f: \\mathbb{S}^{n}\\to \\mathbb{S}^{m}$ factors up to homotopy through the inclusion $\\mathbb{S}^{m-1}\\hookrightarrow \\mathbb{S}^{m}$. Inclusions $\\mathbb{S}^{n-1}\\to \\mathbb{S}^{n}$ are nullhomotopic."
}
{
  "Tag": [
    "MATHCOUNTS",
    "ARML",
    "AMC",
    "AIME",
    "USA(J)MO",
    "USAMO",
    "algebra"
  ],
  "Problem": "You can sign up for The MATHCOUNTS Ultimo Competition. Please register before August 31. Late Registration until September 7. Late Registration means that you will/may have to work out both Round 1 and Round 2 Both together. Here are the rules:\r\n\r\n1) You must register before August 31; Late-Register before September 7. If you need an exception, please private message me.\r\n\r\n2) The format will be as described:\r\n\r\nRound 1: Mostly Sprint Questions (15 Questions, 15 points)\r\n\r\nRound 2: Extra Round. This round is not required, but will give you extra points. (5 questions, 5 points)\r\n\r\nRound 3: Target Round questions (10 Questions, 20 points)\r\n\r\nRound 4: Extra Round. Once again, this round is not required but will give you extra points. (10 Questions, 20 points)\r\n\r\nRound 5: Final Round: Hard Team Round Questions (10 questions, 40 points)\r\n\r\nTie-Breakers: Various Questions; Maximum of 2 tie-breakers. Joint winners can be declared. (1 point per question, 5 questions per set; 2 sets max)\r\n\r\nFreebie rounds: 1 question, 1 pt extra\r\n\r\nTotal points possible (extra round included): 100\r\n\r\n3) Scores will be posted.\r\n\r\n4) Questions can be downloaded from this topic. Questions for Round 1 will be posted any time after there are a minimum of 5 participants. Due dates will be posted with the scores.\r\n\r\n5) Late submissions accepted until 3 days after due date. After that, submissions will be accepted for a 5 point penalty.\r\n\r\n6) YOU ARE NOT ELIMINATED IF YOU LOSE A ROUND. YOU WILL BE CONTINUING TO THE NEXT ROUND. WINNERS WILL BE DECLARED AT END.\r\n\r\n7) Roundly Scores will be posted. Cumulative Scores will  also be posted throughout the competition, along with placements.",
  "Solution_1": "According to rcv, I don't think I will be allowed to since I am going into 9th Grade.",
  "Solution_2": "I'll join.",
  "Solution_3": "You shouldn't be in this forum neil.\r\n\r\nI'm in too.",
  "Solution_4": "ILL BE IN\r\n\r\nwen does teh competition start",
  "Solution_5": "The competition begins once there are 5 participants. The due date for Round 1 is set then. It is expected to start within the next few days.\r\n[u]\n[b]Current Participants:[/b][/u]\r\nSplashD\r\npascal_1623\r\nakalra1",
  "Solution_6": "I'll participate.",
  "Solution_7": "[u][b]Current Participants:[/b][/u]\r\nSplashD\r\npascal_1623\r\nakalra1\r\nragnarok23\r\n\r\nWe need 1 more participant to begin...",
  "Solution_8": "I'm in :D",
  "Solution_9": "I'm in. Thats 6!\r\n\r\n[quote=\"pascal_1623\"]You shouldn't be in this forum neil.\n\nI'm in too.[/quote]\r\n\r\nHe's still participating in my competition, which started when he was still in 8th grade.",
  "Solution_10": "[u][b]Current Participants:[/b][/u]\r\nSplashD\r\npascal_1623\r\nakalra1\r\nragnarok23 \r\nbpms\r\njclarkemathL314159\r\n\r\nThat's 6 people. I will now post round 1 below. The round consists of mostly sprint round with some target round questions.\r\n\r\nPeople can sign up as well as pm me the answers at the same time. This round is due September 2. All questions are permitted. Discussion of the problems is allowed if and only if the problem answer is not directly stated. [color=darkred][size=200][u][b]ONLY PROBLEM INTERPRETATION HELP CAN BE GIVEN!!! ANY DIRECT STATEMENT OF THE ANSWERS MAY RESULT IN DISQUALIFICATION.[/b][/u][/size][/color]\r\n\r\nAdditional \"Freebie rounds\" may be posted which connsist of 1 question, worth 1 point for extra bonus's.",
  "Solution_11": "I m in.\r\n\r\n\r\n\r\n\r\n\r\n\r\n",
  "Solution_12": "I'm in too! :lol:",
  "Solution_13": "count me in! :D\r\n\r\ncan we use a calculator?",
  "Solution_14": "I'm signing up, up UP, UP!!!\r\n\r\nDo we have to send solutions?",
  "Solution_15": "[quote=\"i_like_pie\"][quote=\"SplashD\"]For condition 5, you have \"the four interior regions AQPB...\" but AQ intersects PB, so it's not a valid quadrilateral[/quote]\nThat makes two triangles if I understand correctly.[/quote]\r\n\r\nwhat two triangles?",
  "Solution_16": "[quote=\"SplashD\"][quote=\"i_like_pie\"][quote=\"SplashD\"]For condition 5, you have \"the four interior regions AQPB...\" but AQ intersects PB, so it's not a valid quadrilateral[/quote]\nThat makes two triangles if I understand correctly.[/quote]\nwhat two triangles?[/quote]\r\nThese two triangles:",
  "Solution_17": "i don't think that's what the question means though.",
  "Solution_18": "that [i]is[/i] what the question means",
  "Solution_19": "[quote=\"anirudh\"]that [i]is[/i] what the question means[/quote]\r\n\r\ni have never seen the area of a shape defined like that. did you make these questions?",
  "Solution_20": "No. I got problem 1 from unknown sources... :) \r\nThe rest are original",
  "Solution_21": "[quote=\"anirudh\"]that [i]is[/i] what the question means[/quote]\r\nHow is APQB a quadrilateral then?",
  "Solution_22": "i don't know...the question is copied and pasted from saab",
  "Solution_23": "Well, I just finished the target round for my mock!!!!!!!!! Yay...",
  "Solution_24": "[quote=\"anirudh\"]i don't know...the question is copied and pasted from saab[/quote]\r\n\r\nthat's what i said. you should at least read over the problems before you post it as a contest question, sheesh.\r\n\r\nand you said it was from \"unknown sources\"  :mad:  :roll:",
  "Solution_25": "[b]NOTE: PROBLEM #1 DOES NOT COUNT!!! YOU WILL GET FULL CREDIT FOR PROBLEM 1 IF YOU GIVE AN ANSWER FOR IT!!![/b]\r\n\r\nOne person has turned in all rounds:\r\n\r\n[hide=\"SplashD\"]\n[hide=\"Round 1 Score\"]\nPerfect!\n[/hide]\n\n[hide=\"Round 2 Score\"]\nPerfect!\n[/hide]\n\n[hide=\"Round 3 Score\"]\n$\\frac{27}{30}$\nNear Perfect!\n[/hide]\n\n[hide=\"Overall Score\"]\n$\\frac{57}{60}$!\n[/hide]\n[/hide]",
  "Solution_26": "OK. By far, SplashD is winning this competition with an overall score of $\\frac{57}{60}$.",
  "Solution_27": "wait is it all right if I send in my round 3 answers tomorrow? or is that way too late?",
  "Solution_28": "Actually, I'll delay it. You can send it in tommorrow.  :)",
  "Solution_29": "OK. This competition is officially over. SplashD won, with 13375P34K43V312 as runner-up. Mods can now lock this up."
}
{
  "Tag": [
    "algebra",
    "binomial theorem",
    "number theory",
    "number theory solved"
  ],
  "Problem": "Let p be an odd prime, prove that\r\n<sub>p-1</sub>\r\n   \\sum   k<sup>2p-1</sup> == p(p+1)/2 (mod p <sup>2</sup>)\r\n<sup>k=1</sup>",
  "Solution_1": "Let a in {1,..., (p-1)/2}.\r\nThen, from the binomial theorem :\r\na <sup>2p-1</sup>  + (p-a) <sup>2p-1</sup>  = p(2p-1)a <sup>2p-2</sup>  mod[p <sup>2</sup> ]\r\n= p(2p-1)(a <sup>p-1</sup> ) <sup>2</sup> mod[p <sup>2</sup> ].\r\n\r\nIn another hand, from Fermat's little theorem, we have \r\na <sup>p-1</sup>  = 1 +p*k for some integer k, thus (a <sup>p-1</sup> ) <sup>2</sup>  = 1 + 2k*p mod[p <sup>2</sup> ], that leads to p*(a <sup>p-1</sup> ) <sup>2</sup>  = p mod[p <sup>2</sup> ].\r\nThen, if S denotes the given sum, we have :\r\n\r\nS = sum{1 \\leq a \\leq (p-1)/2} (2p-1)p mod[p <sup>2</sup> ]\r\n= p(2p-1)(p-1)/2 mod[p <sup>2</sup> ]\r\n= p(p+1)/2 mod[p <sup>2</sup> ], and we are done.\r\n\r\nPierre."
}
{
  "Tag": [
    "probability",
    "MATHCOUNTS",
    "AMC",
    "AIME",
    "geometry",
    "number theory"
  ],
  "Problem": "I have two questions. \r\n\r\nI recently did well at the Chapter Competition, and I got a AOPS gift certificate to use to buy books on this site. I was wondering which book I should buy. Which book (out of Intro to Counting and Probability and Intro to Number Theory) would help me the most in Mathcounts?\r\n\r\nAlso, my 2nd question is, are there any awards associated with getting first place in states/nats other than the trophy?",
  "Solution_1": "You get scholarships for some math programs.\r\nAnyways, you should get AoPS volume 1, it's the best for the long run.",
  "Solution_2": "I got  one of those too. Although, I'm wondering whether to use it on Intermediate Trig/complex numbers, Intermediate C+P or the AIME seminar...",
  "Solution_3": "[quote=\"Ihatepie\"](out of Intro to Counting and Probability and Intro to Number Theory) [/quote]\r\n\r\nI would love to buy the AOPS book, but it costs too much. So which one is better for MATHCOUNTS?",
  "Solution_4": "[quote=\"xpmath\"]I got  one of those too. Although, I'm wondering whether to use it on Intermediate Trig/complex numbers, Intermediate C+P or the AIME seminar...[/quote]\r\n\r\nhow much did you get xp?\r\nhopefully ill do well enough at chapter to get one too. use it on intermediate complex and trig. we'll be classmates. :-)",
  "Solution_5": "[quote=\"Ihatepie\"]I would love to buy the AOPS book, but it costs too much. [/quote]\r\n\r\nThey don't cost that much (or at least the ones I bought). Regular math textbooks cost approx $ \\$60$ or more. \r\n\r\nAoPS vol 1 is a great start for any MC participant. If you already have it, you can order a specific book, like number theory or probability, if you find you're having trouble in those areas.",
  "Solution_6": "at states and nats you can sometimes get scholarships and free TI-84+'s",
  "Solution_7": "[quote=\"mihail911\"][quote=\"xpmath\"]I got  one of those too. Although, I'm wondering whether to use it on Intermediate Trig/complex numbers, Intermediate C+P or the AIME seminar...[/quote]\n\nhow much did you get xp?\nhopefully ill do well enough at chapter to get one too. use it on intermediate complex and trig. we'll be classmates. :-)[/quote]\r\nOnly 25 dollars.\r\nI intend to use it on that, seeing the first day of the AIME seminar is on the same day as states, so I'll have to look at transcripts.",
  "Solution_8": "[quote=\"7h3.D3m0n.117\"][quote=\"Ihatepie\"]I would love to buy the AOPS book, but it costs too much. [/quote]\n\nThey don't cost that much (or at least the ones I bought). Regular math textbooks cost approx $ \\$60$ or more. \n\nAoPS vol 1 is a great start for any MC participant. If you already have it, you can order a specific book, like number theory or probability, if you find you're having trouble in those areas.[/quote]\r\nAnd the four Introduction books as well. They help a lot :)",
  "Solution_9": "I've also got $25. I'm trying to decide on the whole set year 2000, or year 1999. They both cost 24.10 Which one?",
  "Solution_10": "I'd get vol 1 if you dont already have it, then id get the intro to geometry",
  "Solution_11": "vol 1 isn't great, get vol. 2 \r\n\r\nand i have the probabitlity book \r\nand it really doesn't help me a whole lot, so dont get that",
  "Solution_12": "[quote=\"Hustla25\"]vol 1 isn't great, get vol. 2 [/quote]\r\n\r\nVolume 2's material content isn't compatible for MATHCOUNTS. You're better off with AoPS vol 1.",
  "Solution_13": "I think out of C/P and NT, C/P appears more often on the tests. But that doesn't mean NT isn't important- NT is just not used as often as C/P on MATHCOUNTS TESTS..... However, I've recently noticed that a lot of Aopsers use mod to solve lots of stuff....\r\nDon't blame me if the state test is all NT- I'm just observing from the past tests",
  "Solution_14": "what eva u decide to buy, study it!!!\r\n\r\nthat is all i can say",
  "Solution_15": "I think they give everyone at Nats a free TI-89.\r\nOr TI-89 titanium.",
  "Solution_16": "Nats = Ti 84 Silver Plus. As for the gift certificate, why didn't I get one?!?! So unfair T.T xD... But... are you sure you're only going to choose one of those two? Becuase AoPS volume 1 contains both of them, just not as in depth, but it's good enough.",
  "Solution_17": "i think ihatepie if you are already pretty good at C\\P than the introduction to C\\P will not help you, i have it and it helps me becuase im really really really bad at C\\P but if your pretty good at it id go with number theory cause it does start to show up on MATHCOUNTS at states",
  "Solution_18": "thanks all, I got C/P\r\n\r\n@mathcrazed did you get first place written round?"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "Vieta",
    "number theory"
  ],
  "Problem": "For all positive integers $ {i}$ let $ {S_i}$ be the sum of the products of $ {1,2,...,p\\minus{}1}$ taken $ {i}$ at a time,where $ {p}$ is an odd prime.Show that $ {S_1}$ is congruent to $ {S_2}$  in short all $ {S_i}$ up to $ {S_p\\minus{}_2}$  MODULO P",
  "Solution_1": "In $ \\mathbb{F}_p \\equal{} \\mathbb{Z}_p \\equal{} \\{0, 1, \\ldots, p\\minus{}1 \\}$, the remainders modulo $ p$, the polynomial $ f(x) \\equal{} x^{p\\minus{}1} \\minus{} 1$ has all the values as roots, except $ 0$, by Fermat's little theorem, hence these are all its roots. From Viete's relations, one gets precisely that all your sums $ S_i$ are congruent to $ 0$ modulo $ p$, i.e. divisible by $ p$.",
  "Solution_2": "Use vietas to explain your point plz your solution is vague",
  "Solution_3": "$ S_1$ is the sum of the roots, $ S_2$ is the sum of their products two by two , ..., $ S_{p\\minus{}2}$ is the sum of their products $ p\\minus{}2$ by $ p\\minus{}2$. But all intermediate coefficients in $ x^{p\\minus{}1} \\minus{} 1$ (except the leading coefficient, and the free term, corresponding to the product of all roots) are zero, and they correspond by Viete's relations to the sums above."
}
{
  "Tag": [
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "For every $0\\leq a,b,c<\\frac{1}{\\sqrt{3}}$, prove that\r\n\r\n$\\frac{a+b}{1-ab}+\\frac{b+c}{1-bc}+\\frac{c+a}{1-ca}\\leq 2\\frac{a+b+c-abc}{1-ab-bc-ac}$",
  "Solution_1": "We can confirm this inequality like belove\r\nLet $a=tan(\\alpha),b=tan(\\beta)$ and $c=tan(\\gamma)$ where $\\alpha,\\beta,\\gamma < 30$,and $\\alpha \\geq \\beta \\geq \\gamma$\r\nThen we must prove that\r\n$tan(\\alpha+\\beta)+tan(\\beta+\\gamma)+tan(\\gamma+\\alpha) \\leq 2.tan(\\alpha+\\beta+\\gamma)$\r\n$2.tan(\\alpha+\\beta+\\gamma)-tan(\\alpha+\\beta)-tan(\\beta+\\gamma)-tan(\\gamma+\\alpha) \\geq 0$\r\nFix $\\alpha$ and $\\beta$.And let $f(\\gamma)=2.tan(\\alpha+\\beta+\\gamma)-tan(\\alpha+\\beta)-tan(\\beta+\\gamma)-tan(\\gamma+\\alpha)$.First derivate of $f(\\gamma)$ is positive, then it gets it's minimum at $\\gamma=0$.With the same method it gets minimum when $\\beta=0$ and it is $0$"
}
{
  "Tag": [
    "trigonometry",
    "function",
    "complex analysis",
    "calculus",
    "calculus computations"
  ],
  "Problem": "A part of a question asks me to show that that $\\frac{\\sin x}{x}$ is strictly decreasing in $[0,\\pi]$ My question is why 0 is included in the interval. the intermediate value theorem requires f to be differentiable in $(0,\\pi)$ and continuous in $[0,\\pi]$. The function is not continuous in 0 (since it's not defined there), so why is the 0 included?\r\n\r\nThanks!",
  "Solution_1": "$\\frac{\\sin x}x$ has a removable discontinuity at zero. There's a possible attitude one could take - and I go along with this about half of the time - that removable discontinuities deserve to be removed.\r\n\r\nIf we take that attitude, we are dealing with the function $f(x)$ such that for $x\\ne0,\\,f(x)=\\frac{\\sin x}x$ whereas $f(0)=1.$ This is an extremely nice function - it's continuous, differentiable, anything of that sort you'd want to ask for, at every point, including zero. OK, to say that formally: it's an entire function - a function given everywhere by a power series with an infinite radius of convergence.\r\n\r\nAnd $f(x)$ is decreasing on $[0,\\pi].$"
}
{
  "Tag": [
    "geometry",
    "analytic geometry",
    "algebra",
    "polynomial",
    "vector",
    "advanced fields",
    "advanced fields open"
  ],
  "Problem": "A set of countable points in complex affine variety is not constructible set? why? :P",
  "Solution_1": "A proof would require developing some theory, but the explanation is simple.  The coordinates of points in an finite affine variety are roots of some polynomial equations, and a polynomial has only a finite number of roots, so there are a finite number of possibilities for each coordinate.  You have to prove that the algebra of coordinates of a 0-dimensional variety is a finite-dimensional vector space.\r\n(If the dimension is higher than 0, the variety has topological dimension at least 2 and is uncountable, so this case cannot arise)."
}
{
  "Tag": [
    "algorithm",
    "combinatorics proposed",
    "combinatorics"
  ],
  "Problem": "An excellent exercise on graphs would be to prove Brooks' theorem: \r\n\r\nLet $G$ be any graph in which every vertex is of order at most $\\Delta$. If $\\Delta \\geq 3$ and $G$ is not $K_{\\Delta+1}$, prove that $G$ is $\\Delta$-colorable.",
  "Solution_1": "[quote=\"vess\"]An excellent exercise on graphs would be to prove Brooks' theorem: [/quote]\r\n\r\nYou are right  :D \r\n\r\nPierre.",
  "Solution_2": "i'll prove that $\\chi(G)\\leq\\Delta(G)$ so the result will follow.\r\nwhere $\\chi(G)$ is the chromatic number of G.\r\n\r\nthe idea of the proof is the same as the one in the proof of the following easy lemma\r\nLemma: $\\chi(G)\\leq\\Delta(G) + 1$\r\nProof: Let $\\Delta(G)=k$ (maximum degree) . Now consider a ordering of vertices $v_1,...v_n$. We can color properly (no two adjacent vertices with the same color) the following way: to each $v_i$ assing the color that corresponds to the vertex with smallest index in the ordering that is not a neighbor of $v_i$. Since each vertex of the ordering has at most $k$ lower indexed neighbor the lemma follows.\r\n\r\nNow we have to order the vertices so that each has at most k-1 lower indexed neighbors. \r\n\r\ncase 1: G is not k-regular. Let x be a vertex with degree less than k. Now construct a spanning tree from x and assing indices in the decreasing order as we reach vertices. Then in our ordering $v_1,...,v_{n-1}$ each vertex has a neighbor with a higher index. Since every vertex has degree at most k, then each vertex has at most k-1 neighbors with lower index, so by the same idea in the lemma above the coloring algorithm will produce at most k colors.\r\n\r\ncase 2A: G is k -regular and G has a cut vertex, say $x$. Let $G'$ be the subgraph formed by a component of $G-x$ together with it edges to $x$.  The degree of $x$ in $G'$ is less than k, so the method used in case 1 provides a desired coloring. So we can permutate the coloring in each component so that x recieves the same color and we get a graph with a k-coloring.\r\n\r\ncase 2B:G is k-regular and 2-connected. (contains no cut vertex)\r\nSuppose that some vertex x has neighbors y and z such that yz is not an edge and $G-{y,z}$ is connected. If we have this case, we grow a spannig tree from x assinging the labels $3,...n$ so that $v_n=x$ in decreasing order so each vertex has at most k-1 neighbors with a lower index. As for vertex $x=v_n$ we color y and z with the same color so the algorithm uses at most k-1 colors in the neighbors of x and produces a k-coloring in G.\r\nNow we prove that ina 2-connected graph such triple $x,y,z$ always exist.\r\nChoose a vertex v. If $G-v$ is still  2-connected then let v=y and let z be a vertex of distance 2 from v=y. this vertex y exists because G is k-regular and it is not complete. So we have our desired triple $x,y,z$ where $G-{y,z}$ is connected.\r\nIf $G-v$ has a cut vertex then put x=v. Since $G-v$ is not 2-connected it is made up of blocks (maximal 2-connected subgraphs) .  Since x can't be a cut vertex x has a nieghbor in every block of $G-x$ and this neighbor is not a cut vertex of $G-x$ . Since $k\\geq 3$ then x has neighbors y and z in different blocks therefore of $G-x$ so yz is not an edge of G and $G-{y,z}$ is connected since y and z can't be cut vertices. We have the desired $x,y,z$ triple. \r\nDone!\r\n\r\nIf something is unclear or wrong let me know.\r\nBtw, does someone has a shorter or easier proof?",
  "Solution_3": "I read in another post of mathlinks a couple of months a go the following related problem which i haven't been able to solve.\r\n\r\n(i think it was in the open questions section)\r\n prove that $\\chi(G)\\leq\\frac {\\Delta(G)+1+\\omega(G)}{2}$\r\nwhere $\\omega(G)$ denotes the cardinality of the largest complete subgraph of G.\r\nany ideas?",
  "Solution_4": "yes this is conjecture and it is due to Reed I remember that when i had posted the thread i had read something about this conjecture for example this is proved : for some $\\lambda >0$ such that if $\\omega(G) \\geq (1-\\lambda )(\\Delta +1)$ then$ \\chi(G)\\leq \\frac12(\\Delta +1)+\\frac12\\omega(G)$.\r\nReference:Graph coloring and probablitistic method,Michael Molloy,Bruce Reed."
}
{
  "Tag": [
    "calculus",
    "integration",
    "limit",
    "logarithms",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Let $ f(x)$ be integrable on $ [a,b]$,and $ f_{in}\\equal{}f(a\\plus{}i\\frac{b\\minus{}a}{n})$,\r\nShow that\r\n$ \\lim_{n\\to\\infty}(1\\plus{}f_{1n}\\frac{b\\minus{}a}{n})(1\\plus{}f_{2n}\\frac{b\\minus{}a}{n})\\cdots(1\\plus{}f_{nn}\\frac{b\\minus{}a}{n})\\equal{}e^{\\int_a^bf(x)dx}$",
  "Solution_1": "[quote=\"wngyny\"]Let $ f(x)$ be integrable on $ [a,b]$,and $ f_{in} \\equal{} f(a \\plus{} i\\frac {b \\minus{} a}{n})$,\nShow that\n$ \\lim_{n\\to\\infty}(1 \\plus{} f_{1n}\\frac {b \\minus{} a}{n})(1 \\plus{} f_{2n}\\frac {b \\minus{} a}{n})\\cdots(1 \\plus{} f_{nn}\\frac {b \\minus{} a}{n}) \\equal{} e^{\\int_a^bf(x)dx}$[/quote]\r\ntake $ \\ln$ (also say $ \\ln(1 \\plus{} x) \\approx x$ for   $ \\minus{} 1 < x\\leq 1$) and use the fact $ \\int_{a}^{b}f(x)dx \\equal{}(b\\minus{}a) \\int_{0}^{1}f(a \\plus{} x(b \\minus{} a))dx\\minus{}>\\lim _{n\\minus{}>\\infty} \\frac{b\\minus{}a}{n}\\sum_{i}^{n} f_{in}$ :wink:"
}
{
  "Tag": [
    "LaTeX",
    "articles",
    "function"
  ],
  "Problem": "uhh where do you get all the LaTeX codes??",
  "Solution_1": "To learn the basics, start with the AoPS Wiki article: [[LaTeX]]. A more advanced tutorial can be found [url=http://www.ctan.org/tex-archive/info/Math_into_LaTeX-4/Short_Course.pdf]here[/url], and the full documentation is [url=http://www.ctan.org/]here[/url].\r\n\r\nIf you're only looking to learn the commonly used codes, I think the first link will suffice; the other two links are more focused on documents created with $ \\text{\\LaTeX}$ instead of basic functions and such.",
  "Solution_2": "ahh thanks =D\r\n\r\nalthough i found those on LaTeX-Reference..."
}
{
  "Tag": [
    "group theory",
    "abstract algebra",
    "linear algebra",
    "linear algebra unsolved"
  ],
  "Problem": "Let $A \\in \\mbox{GL}_{n}(\\mathbb{Q})$ s.t. $A^{4}=I_{n}$. \r\nShow that there exists $P\\in \\mbox{GL}_{n}(\\mathbb{Q})$ and $PAP^{-1}\\in \\mbox{GL}_{n}(\\mathbb{Z})$",
  "Solution_1": "I have shown on the forum (a problem by Alekk) that any finite subgroup of $GL_{n}(Q)$ is conjugated to a finite subgroup of $GL_{n}(Z)$. This immediately applies here, by considering the cyclic group generated by $A$.",
  "Solution_2": "[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=finite&t=27907[/url]\r\n\r\nYou even replied on that topic, JC_math.",
  "Solution_3": "True, I didn't make the link with that problem when I posted it :blush:"
}
{
  "Tag": [
    "topology",
    "abstract algebra",
    "advanced fields",
    "advanced fields unsolved"
  ],
  "Problem": "This is 14.15, on p.81 of greenberg and harper's algebraic topology book.\r\n\r\nLet $ X \\equal{} X_1 \\cup X_2$, and $ A \\equal{} X_1 \\cap X_2$. Using the exact sequences of triples, show that if the inclusion $ (X_1, A) \\rightarrow (X,X_2)$ induces an isomorphism in homology then the same holds for the inclusion $ (X_2,A) \\rightarrow (X,X_1)$.\r\n\r\nI don't really know how I can use the exact sequences of triples. Can anyone enlighten me on this?",
  "Solution_1": "I'm not sure if I can bump this yet, anyway.. can anyone help?",
  "Solution_2": "[quote=\"Soarer\"]This is 14.15, on p.81 of greenberg and harper's algebraic topology book.\n\nLet $ X = X_1 \\cup X_2$, and $ A = X_1 \\cap X_2$. Using the exact sequences of triples, show that if the inclusion $ (X_1, A) \\rightarrow (X,X_2)$ induces an isomorphism in homology then the same holds for the inclusion $ (X_2,A) \\rightarrow (X,X_1)$.[/quote]\n\nI have got something, but I don't really believe it's true - check it carefully, please.\n\nWe do not need $ X = X_1 \\cup X_2$, and $ A = X_1 \\cap X_2$. It is enough to have $ X_1 \\cup X_2\\subset X$, and $ A \\subset X_1 \\cap X_2$.\n\nHowever, I hope that by saying \"isomorphism in homology\", you mean \"isomorphism in homologies of every degree\"; else, I don't have a proof (but I haven't searched for counterexamples either).\n\nConsider the sequence $ H_n\\left(X_1,A\\right)\\to H_n\\left(X,A\\right)\\to H_n\\left(X,X_2\\right)$, where both arrows are induced by inclusions. The composition of these transformations is induced by the inclusion $ \\left(X_1,A\\right)\\to\\left(X,X_2\\right)$, and therefore (by a condition of the problem) it is an isomorphism, so that it is injective. Hence, its \"first arrow\" $ H_n\\left(X_1,A\\right)\\to H_n\\left(X,A\\right)$ must also be injective. Hence, the kernel of the arrow $ H_n\\left(X_1,A\\right)\\to H_n\\left(X,A\\right)$ is zero.\n\nThe long exact sequence for the triple $ \\left(X,X_1,A\\right)$ looks like this:\n\n$ ...\\to H_{n + 1}\\left(X,X_1\\right)\\to H_n\\left(X_1,A\\right)\\to H_n\\left(X,A\\right)\\to H_n\\left(X,X_1\\right)$\n$ \\to H_{n - 1}\\left(X_1,A\\right)\\to ...$.\n\nWe know that the kernel of the arrow $ H_n\\left(X_1,A\\right)\\to H_n\\left(X,A\\right)$ is zero here - but this kernel is the image of the preceding arrow $ H_{n + 1}\\left(X,X_1\\right)\\to H_n\\left(X_1,A\\right)$. Hence, the image of the arrow $ H_{n + 1}\\left(X,X_1\\right)\\to H_n\\left(X_1,A\\right)$ must be zero, so that this arrow is simply the zero homomorphism. Repeating this argument with $ n - 1$ instead of $ n$, we see that $ H_n\\left(X,X_1\\right)\\to H_{n - 1}\\left(X_1,A\\right)$ is the zero homomorphism. Thus, our sequence becomes\n\n$ ...\\to H_{n + 1}\\left(X,X_1\\right)\\substack{0 \\\\\n\\to} H_n\\left(X_1,A\\right)\\to H_n\\left(X,A\\right)\\to H_n\\left(X,X_1\\right)$\n$ \\substack{0 \\\\\n\\to} H_{n - 1}\\left(X_1,A\\right)\\to ...$.\n\nThis readily yields the exact sequence\n\n$ 0\\to H_n\\left(X_1,A\\right)\\to H_n\\left(X,A\\right)\\to H_n\\left(X,X_1\\right)\\to 0$.\n\nOn the other hand, the long exact sequence for the triple $ \\left(X,X_2,A\\right)$ yields\n\n$ H_n\\left(X_2,A\\right)\\to H_n\\left(X,A\\right)\\to H_n\\left(X,X_2\\right)$.\n\nAn easy diagram chase yields the following fact: If $ A$, $ B$, $ X$, $ Y$, $ U$ are (say) abelian groups such that $ 0\\to A\\to U\\to B\\to 0$ and $ X\\to U\\to Y$ are exact sequences and such that the composition of the arrows $ A\\to U$ and $ U\\to Y$ is an isomorphism, then the compositions of the arrows $ X\\to U$ and $ U\\to B$ is an isomorphism as well.\n\nThis applies to our problem, because the composition of the arrows $ H_n\\left(X_1,A\\right)\\to H_n\\left(X,A\\right)$ and $ H_n\\left(X,A\\right)\\to H_n\\left(X,X_2\\right)$ is an isomorphism (namely, the one induced by the inclusion $ \\left(X_1,A\\right)\\to\\left(X,X_2\\right)$), and thus we conclude that the composition of the arrows $ H_n\\left(X_2,A\\right)\\to H_n\\left(X,A\\right)$ and $ H_n\\left(X,A\\right)\\to H_n\\left(X,X_1\\right)$ is an isomorphism as well, i. e. the inclusion $ \\left(X_2,A\\right)\\to\\left(X,X_1\\right)$ induces an isomorphism. We are done.\n\nDid this all make sense? (I'm not asking whether this could have been done on 2 lines as well, because most likely the answer is yes.)\n\n  darij",
  "Solution_3": "Yeah it works, thanks.  :)"
}
{
  "Tag": [
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "There are $ n_i$ balls of color $ i$ , $ i \\equal{} 1,2,...,k$. Assume that $ n_1 \\plus{} n_2 \\plus{} ... \\plus{} n_k$ is even\r\nLet $ s \\equal{} \\frac {n_1 \\plus{} n_2 \\plus{} ... \\plus{} n_k}{2}$\r\nLet $ f(n_1,n_2,...,n_k)$ denotes the number of ways to arrange this collection of balls into $ s$ pairs such that each pairs consist of balls with different colors (no balls of the same color are paired up together). Find $ f(n_1,n_2,...,n_k)$\r\n\r\nEx. $ f(n,n) \\equal{} n!, f(n_1,n_2) \\equal{} 0$ if $ n_1\\not \\equal{} n_2$, \r\n$ f(n_1,n_2,...,n_k) \\equal{} 0$ if $ \\max{n_i}> s$\r\n$ f(n_1,n_2,n_3) \\equal{} \\frac {n_1!n_2!n_3!}{(s \\minus{} n_1)!(s \\minus{} n_2)!(s \\minus{} n_3)!}$ provided that $ \\max{n_i}> s$",
  "Solution_1": "I think we are looking for \"how many edges there can be in a maximum matching in a complete multipartite graph ?\"\r\n\r\nhttp://math.furman.edu/~mwoodard/fuejum/content/1996/1996paper2.pdf",
  "Solution_2": "In the paper, it doesn't talk about how many ways to partition it.\r\nThe set up is the same, but my question is different.\r\n\r\nIt can be done using brute force and the following recurrent relation,\r\n\r\n$ f(n_1,n_2,...,n_k)\\equal{}\\sum_{p\\equal{}0}^{min(n_1,n_2)}p!C^{n_1}_pC^{n_2}_pf(n_1\\plus{}n_2\\minus{}2p,n_3,...,n_k)$\r\n\r\nThe answer is explicit, but ugly. So I am looking for some smarter way to do it."
}
{
  "Tag": [
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "For what k is $ 10^k \\plus{} 1$ divisible by a square.\r\n\r\nodd powers of 11 is obvious\r\n\r\nI think for infinitely many prime p $ k \\equal{} \\frac {p(p \\minus{} 1)}{2}$ is a candidate. \r\n\r\nCould someone provide a proof? Thank you",
  "Solution_1": "It is not true. $ 10^k\\plus{}1$ never divisible $ p\\equal{}2,3,5,..$ infinetely many primes.\r\nLet $ T_p$ is minimal period, suth that $ p|10^{T_p}\\minus{}1$ and for any divisor $ k|T_p \\ \\ p\\not |10^k\\minus{}1$.\r\nThen if $ T_p$ - odd, never $ p|10^k\\plus{}1$. If $ T_p$ even $ p^2|10^k\\plus{}1$ for $ k\\equal{}\\frac{pT_p}{2}$.\r\nFor example $ T_{13}\\equal{}6$ and $ 13^2|10^{39}\\plus{}1$, but $ 13\\not| 10^{13*12/2}\\plus{}1$.",
  "Solution_2": "But, it is certainly true for odd mulitples of 11.\r\nAnd for a host of other numbers like k = 21, 136, 171, etc.\r\nAnd id it is true for a number, it is true for all odd multiples of that number. So, 63, 125 etc. also qualify"
}
{
  "Tag": [
    "limit",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Let $(x_n)_{n\\geq 0} \\in R_{+}^N$ be a sequence which tends to $0$.Let $y_n=n(x_n-x_{n+1})$.Show that the convergence of $\\sum y_n$ implies the convergence of $\\sum x_n$.",
  "Solution_1": "Is the sequence also supposed to be decreasing?",
  "Solution_2": "------------------",
  "Solution_3": "Grobber, I would say yes. :?  Anyway, from Stolz Cesaro we have $\\lim_{n\\to\\infty} n(x_{n+1}-x_{n})=\\lim_{n\\to\\infty} \\frac{x_{n}}{n}$ and from \"compare\" :?  criterion it follows the result. ;)",
  "Solution_4": "----------------",
  "Solution_5": "--------------------",
  "Solution_6": "Cimabue, do you know Cesaro-Stolz theorem? :?",
  "Solution_7": "What about you, Cezar, do you know Cesaro-Stolz? Here's a counter-example to what you have written above $x_n=1+1/2+...+1/n$.",
  "Solution_8": "Yes, but $x_n=1=\\frac{1}{2}+...+\\frac{1}{n}$ doesn't tend to $0.$ ;)",
  "Solution_9": "Cimuabue has, in effect, used Abel's summation by parts identity, which is well worth knowing.\r\n\r\nTo cope with the \"more or less\" in his first post, we can write that exactly as:\r\n\r\n$\\sum_{k=1}^nx_k=nx_n+\\sum_{k=1}^{n-1}k(x_k-x_{k+1}).$\r\n\r\nSo cimabue is completely correct: If we assume that $\\sum y_n$ converges, then having $nx_n$ converge to a limit is a necessary and sufficient condition for the convegence of $\\sum x_n.$ Suprisingly, we don't need for $x_n$ to be decreasing.",
  "Solution_10": "The are no other hypothesis apart the ones mentioned.",
  "Solution_11": "--------------",
  "Solution_12": "Cesar, read again your formula, application of Cesaro-Stolz theorem and you'll see what I wanted to say.",
  "Solution_13": "-------------------",
  "Solution_14": "-------------------",
  "Solution_15": "The convergence of $\\sum y_n$ gives the boundness of $S_n -nx_n$.So let $M$ such that $S_n-nx_n$ is bounded by $M$.From this we get that $|\\frac{S_{n-1}}{n-1}-\\frac{S_n}{n}|\\leq \\frac{M}{n(n-1)}$.So,for $n\\leq p,|\\frac{S_{n}}{n}-\\frac{S_p}{p}|\\leq M(\\frac{1}{n}-\\frac{1}{p})$.Let $p$ go to infinity and we get $S_n\\leq M$.Conclusion fallows.",
  "Solution_16": "-------------",
  "Solution_17": "Probably...but it gives what the posted problem asks for.So I dont see what is your remark doing there.My post was just a solution for the initial problem. ;)"
}
{
  "Tag": [
    "group theory",
    "abstract algebra",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "Let $ G$ be a finite group, and $ H$ a normal subgroup of $ G$ whose order is the lowest prime dividing $ |G|$. Then $ H \\subset Z(G)$",
  "Solution_1": "I tried to use the action of $ G$ on $ H$ by conjugation...",
  "Solution_2": "EDIT: Incomplete solution. (Thanks [b]LydianRain[/b] for noticing the very strange mistake I made!)",
  "Solution_3": "An alternative solution: \r\n\r\nSince $ H$ is a normal subgroup, $ (G,H)$ is a valid permutation representation with conjugation as it's action. The order of the orbit of a nontrivial element $ h$ of $ H$ has to be a divisor of $ |G|$, and since $ h$ is not conjugated to $ e$, we must have that $ h^G \\equal{} \\left\\{h\\right\\}$.",
  "Solution_4": "Thanks for both solutions!",
  "Solution_5": "[quote=\"ysharifi\"]Thus $ k^n \\equiv 1 \\mod p,$ and so $ p \\minus{} 1 \\mid n.$[/quote]\r\nWhy?\r\n\r\nYou proof cannot be correct; it would imply that a group with a non-central normal subgroup of order 5 must have order divisible by 4.  But the dihedral group of order 10 is a counterexample.\r\n\r\nJan's proof looks good."
}
{
  "Tag": [
    "geometry",
    "MATHCOUNTS"
  ],
  "Problem": "in triangle ABC, points D and E are drawn so that E is on BC and AB=BE and AE is perpendicular to BC.  if [] represents area, prove that [BDC]=[BACD]",
  "Solution_1": "[quote=\"daermon\"]in triangle ABC, points D and E are drawn so that E is on BC and AB=BE and AE is perpendicular to BC.  if [] represents area, prove that [BDC]=[BACD][/quote]\r\nProofs certainly aren't MathCounts level, so I'll move this to Intermediate.\r\nAlso, are you sure the problem statement is correct? You never gave any information about where point D was located.  :wink:",
  "Solution_2": "Where is D?",
  "Solution_3": "how can AB=BE\r\nif $BE^{2}+AE^{2}=AB^{2}$ since they are perpendicular?",
  "Solution_4": "oops.  I meant AD=DE, not AB=BE  :blush: \r\nD is inside the triangle"
}
{
  "Tag": [
    "function",
    "calculus",
    "derivative",
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "Let a,b,c>0\r\nand $ a^2\\plus{}b^2\\plus{}c^2\\equal{}3$\r\nProve that $ (2\\minus{}ab)(2\\minus{}bc)(2\\minus{}ac) \\geq 1$",
  "Solution_1": "if we can use Lagrange's method... :D \r\ntake the function $ L \\equal{} f(a,b,c) \\minus{} \\lambda(a^2 \\plus{} b^2 \\plus{} c^2 \\minus{} 3)$.where $ \\lambda \\in R$ and $ f(a,b,c) \\equal{} 7 \\plus{} 2abc(\\sum a) \\minus{} (abc)^2 \\minus{} 4(\\sum ab)$.\r\nthen we should compute the derivatives of the function $ L$  wrt $ a,b,c$ and let them be equal to $ 0$.\r\nby doing all this work we get that the extremum of this function are reached at points $ (0,0,\\sqrt {3})$ and $ (1,1,1)$ which leads to $ 8 \\ge (2 \\minus{} ab)(2 \\minus{} bc)(2 \\minus{} ac)\\geq 1$ but since $ a,b,c > 0$ then $ 8 > (2 \\minus{} ab)(2 \\minus{} bc)(2 \\minus{} ac)\\geq 1$.",
  "Solution_2": "[quote=\"nguyensilva\"]Let a,b,c>0\nand $ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} 3$\nProve that $ (2 \\minus{} ab)(2 \\minus{} bc)(2 \\minus{} ac) \\geq 1$[/quote]\r\nThis inequality is equivalent to $ w^6 \\plus{} Aw^3 \\plus{} B\\leq0,$ where\r\n$ a \\plus{} b \\plus{} c \\equal{} 3u,$ $ ab \\plus{} ac \\plus{} bc \\equal{} 3v^2,$ $ abc \\equal{} w^3$ and $ A$ and $ B$  are functions of $ u$ and $ v^2$ only.\r\nThus, $ w^6 \\plus{} Aw^3 \\plus{} B$ gets a maximal value when $ w^3$ gets an extremal value, \r\nwhich happens  when two numbers from $ \\{a,b,c\\}$ are equal or when $ w^3 \\equal{} 0.$\r\nHence, it remains to check two cases:\r\n1) $ c \\equal{} 0.$ In this case we need to prove that $ ab\\leq\\frac {7}{4},$ when $ a^2 \\plus{} b^2 \\equal{} 3,$ which is true.\r\n2) $ b \\equal{} c.$ In this case after homogenization we obtain \r\n$ 7(a^2 \\plus{} b^2 \\plus{} c^2)^3 \\minus{} 12(ab \\plus{} ac \\plus{} bc)(a^2 \\plus{} b^2 \\plus{} c^2)^2 \\plus{}$\r\n$ \\plus{} 18abc(a \\plus{} b \\plus{} c)(a^2 \\plus{} b^2 \\plus{} c^2) \\minus{} 27a^2b^2c^2\\geq0$ and we can assume here $ b \\equal{} c \\equal{} 1.$\r\n But $ 7(a^2 \\plus{} 2)^3 \\minus{} 12(2a \\plus{} 1)(a^2 \\plus{} 2)^2 \\plus{} 18a(a \\plus{} 2)(a^2 \\plus{} 2) \\minus{} 27a^2\\geq0\\Leftrightarrow$\r\n$ \\Leftrightarrow7a^6 \\minus{} 24a^5 \\plus{} 48a^4 \\minus{} 60a^3 \\plus{} 45a^2 \\minus{} 24a \\plus{} 8\\geq0\\Leftrightarrow$\r\n$ \\Leftrightarrow(a \\minus{} 1)^2(7a^4 \\minus{} 10a^3 \\plus{} 21a^2 \\minus{} 8a \\plus{} 8)\\geq0,$ which is true.\r\nBy the way, we see that your inequality is true for all reals $ a,$ $ b$ and $ c$ such that $ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} 3.$ :wink:\r\nSee also here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=136486"
}
{
  "Tag": [
    "function",
    "trigonometry",
    "algebra",
    "domain",
    "inequalities",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Let f be a continuous function defined for 0<=x<=1. Assume that f(0) is not 0 and that f takes any value a finite number of times. Then there is a value taken an odd number of times.",
  "Solution_1": "There's something I don't quite get about this: what do we need the condition f(0)<>0 for? If f is real-valued and f(0)=0 then we can prove what we want for g(x)=f(x)+1 and then go back to f and the problem is Ok for f(0)=0 as well. \r\n\r\nMaybe the text states that f can't take any value, just values in [0,1] or something like that, but I find that hard to believe as well. \r\n\r\nSo what's the deal with the condition f(0)<>0 then?",
  "Solution_2": "I just saw this problem in one of my materials, I don't have the magazine to see what is with that condition.",
  "Solution_3": "[u][b]The author of this posting is : grobber[/b][/u]\r\n____________________________________________________________________\r\n\r\nFrankly, if f is real-valued, I don't see the purpose of the condition f(0) <> 0. If f(0)=0 we solve the question for g(x)=f(x)+1 and go back to f, and it holds in the case f(0)=0 as well, so what's up with that?\r\n\r\nMaybe we also have f:[0,1] $\\longmapsto$ [0,1] or something like that, or maybe the condition f(0) <> 0 just doesn't exist and f is real-valued.",
  "Solution_4": "the condition that $f$ takes a value a finite number of times is trivial, since $f\\in C[0,1]$ , $f^{-1}(a)$ is finite for any $a\\in\\mathbb{R}$ .",
  "Solution_5": "We should show that it takes a value an [b]odd[/b] number of times.",
  "Solution_6": "yes, but harazi wrote: \r\n[quote]Assume that f(0) is not 0 and that [size=150]f takes any value a finite number of times[/size].[/quote]\r\nSince $f$ is continious it always happens.",
  "Solution_7": "[quote=\"hcast\"]the condition that $f$ takes a value a finite number of times is trivial, since $f\\in C[0,1] , f^{-1}(a)$ is finite for any $a\\in\\mathbb{R}$ .[/quote]\r\nUh - better be careful what you're saying there. let $f(x) = x\\sin(\\frac1x)$ for $x\\neq 0$ and $f(0)=0$. Then $f\\in C[0,1]$ but $f^{-1}(0)$ is an infinite set. We can say that $f^{-1}(a)$ is closed and we can say that $f^{-1}(a)$ is compact, but we can't say that $f^{-1}(a)$ is finite.",
  "Solution_8": "Even simpler:  Let\r\n\r\nf(x) = x (0 <= x < 1/3)\r\n       = 1/3 (1/3 <= x <= 2/3)\r\n       =  1 - x (2/3 < x <= 1)\r\n\r\nf^(-1)(1/3) is very infinite.  (And for no a does f(x) take on the value a an odd number of times.)",
  "Solution_9": "So the condition on the original problem is that $f\\in C[0,1]$ [b]AND[/b] $f^{-1}(a)$ is finite $\\forall a$. As Alison and I have noted, this condition is more restrictive than merely $f\\in C[0,1]$, and as Alison's example shows, the theorem is false without this additional condition. However, Alison's example is not a counterexample to the original problem.\r\n\r\nThe examples $f(x)=\\sin(4\\pi x), g(x)=\\sin((4\\pi-.01)x),$ and $h(x)=\\sin((4\\pi+.01)x)$ may provide some insight into this problem, as the conclusion is true for each of them. You might want to investigate whether or not the domain can be partitioned into finitely many intervals on which the function is monotone.",
  "Solution_10": "I'm pretty sure we can't say that the domain can be partitioned into finitely many intervals on which the function is monotone. I can't give you a specific counterexample, but think of some sort of stairway s.t. the steps are, in fact, little \"humps\" which get smaller and smaller and higher and higher as you go towards the right endpoint of the domain. The function is thus \"sort of\" increasing, havin occasional downward \"falls\".\r\n\r\nSuch a graph might not satisfy the condition that every value is taken an even number of times (actually, we have to prove that no graph satisfies the condition :)), but I don't think it's obvious.",
  "Solution_11": "I was hoping to use that method to solve this problem yesterday, but unfortunately it's false. Consider the function $f(x)=2x+x\\sin\\frac{1}{x}$.",
  "Solution_12": "Here is a counterexample:\r\n\r\nLet $f:[0,2] \\to [0,1]$ be a piecewise linear function with vertices:\r\n$(2,0), (\\frac32,1), (1,\\frac12), (\\frac12,1), (\\frac14,\\frac18), (\\frac18,\\frac12), \\dots, (\\frac1{2^{2k}},\\frac1{2^{2k+1}}), (\\frac1{2^{2k+1}},\\frac1{2^{2k-1}}), \\dots$, and $f(0)=0$. Since $\\frac12x \\le f(x) \\le 4x$ for $0 < x \\le 1$, $f$ is continuous at 0, and the piecewise linearity makes it continuous everywhere else as well.\r\n\r\n$f$ takes on each value in its image four times, except that the values 0 and 1 are only taken twice.\r\nIf you wanted to, you could rescale this to match the original conditions.",
  "Solution_13": "Perhaps that mysterious and nonsensical condition \"$f(0)\\neq 0$\" should have been the not-so-nonsensical \"$f(0)\\neq f(1)$.\" jmerry's example did have (and need) $f(0)=f(1)$ (well, $f(2)$ in his case but it's the same thing.)\r\n\r\nConsider the problem still alive and kicking, with this condition.",
  "Solution_14": "Sorry to revive such an old topic, but I remembered this problem by accident, and I realized that Kent is right: if we make the reasonable assumption that that condition was meant to be $f(0)\\ne f(1)$, then the problem becomes true:\r\n\r\nSuppose WLOG that $f(0)=0$. We want to prove that $f(1)=0$ as well. For every $M>0$, we call a point $x\\in(0,1),\\ |f(x)|=M$ an in-point if for some $\\varepsilon>0,\\ |f|>M$ on $(x-\\varepsilon,x)$ and $|f|<M$ on $(x,x+\\varepsilon)$. We call $x\\in(0,1),\\ |f(x)|=M$ an out-point if the reversed inequalities take place for small enough $\\varepsilon>0$. The only way $x$ could fail to be either an in our an out-point is if $|f|$ reaches a local extremum at $x$. This, however, is possible only for countably many $M$. Indeed, if $|f(x)|=M$ and $x$ is a local extremum for $|f|$, then $M$ is the maximal/minimal value of $|f|$ on some interval with rational endpoints, and there are only countably many such intervals. \r\n\r\nAssume now that $0<M<|f(1)|$, with $M$ chosen such that all elements of $|f|^{-1}(M)$ are either in or out-points (it is possible to find such $M$, according to the paragraph above). The largest element of $|f|^{-1}(M)$ must then be an out-point. However, when ordered increasingly, the elements of $|f|^{-1}(M)$ are alternatively in and out-points, with the first one being an out-point. Since $|f|^{-1}(M)$ has an even number of elements, the largest one must be an in-point, and we have a contradiction."
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "$a>0,$ $b>0,$ $c>0,$ $a^3+b^3+c^3=3.$ Prove that\r\n$\\frac{a^2}{b+c}+\\frac{b^2}{a+c}+\\frac{c^2}{a+b}\\geq\\frac{3}{2}.$",
  "Solution_1": "I am interested in this kind. Another idea from this kind is the following\r\n\r\nLet $x,y,z$ be positive real numbers such that $x^2+y^2+z^2=1$, find all positive rational $r$ number such that the following expression\r\n\\[ \\frac{x^r}{y+z}+\\frac{y^r}{z+x}+\\frac{z^r}{x+y} \\]\r\nhas a minimum.",
  "Solution_2": "$\\frac{a^2}{b+c}+\\frac{b^2}{a+c}+\\frac{c^2}{a+b}\\geq\\frac{3}{2}$\r\n\r\n$\\Longleftrightarrow 2(a+b+c)(abc+3)\\geq 3(a+b)(b+c)(c+a)$ ;)",
  "Solution_3": "SOS :D.... Nice problem.",
  "Solution_4": "[quote=\"arqady\"]$a>0,$ $b>0,$ $c>0,$ $a^3+b^3+c^3=3.$ Prove that\n$\\frac{a^2}{b+c}+\\frac{b^2}{a+c}+\\frac{c^2}{a+b}\\geq\\frac{3}{2}.$[/quote]\r\n\r\nAre the following inequality right?\r\n\r\n$0<a+b+c\\leq 3,\\ 0<ab+bc+ca\\leq 3.$",
  "Solution_5": "[quote=\"arqady\"]$a>0,$ $b>0,$ $c>0,$ $a^3+b^3+c^3=3.$ Prove that\n$\\frac{a^2}{b+c}+\\frac{b^2}{a+c}+\\frac{c^2}{a+b}\\geq\\frac{3}{2}.$[/quote]\r\n\r\nThis follows immediately from the following inequality (which is not hard to prove): for all positive reals $a,b$ and $c$,\r\n\\[ \\frac{a^2}{b+c}+\\frac{b^2}{c+a}+\\frac{c^2}{a+b}\\ge\\frac{3}{2}\\cdot\\frac{a^3+b^3+c^3}{a^2+b^2+c^2}  \\]",
  "Solution_6": "$\\frac{a^2}{b+c}+\\frac{b^2}{c+a}+\\frac{c^2}{a+b} \\ge \\frac{a^3+b^3+c^3}{3}\\left(\\frac{1}{a(b+c)}+\\frac{1}{b(c+a)}+\\frac{1}{c(a+b)}\\right) \\ge \\frac9{2(ab+bc+ca)} \\ge \\frac32$",
  "Solution_7": "[quote=\"Lovasz\"]$\\frac{a^2}{b+c}+\\frac{b^2}{c+a}+\\frac{c^2}{a+b} \\ge \\frac{a^3+b^3+c^3}{3}\\left(\\frac{1}{a(b+c)}+\\frac{1}{b(c+a)}+\\frac{1}{c(a+b)}\\right)$[/quote]\r\n\r\nLovasz, could you teach me how we can obtain this inequality? :?\r\n\r\nkunny",
  "Solution_8": "[quote=\"Lovasz\"]$\\frac{a^2}{b+c}+\\frac{b^2}{c+a}+\\frac{c^2}{a+b} \\ge \\frac{a^3+b^3+c^3}{3}\\left(\\frac{1}{a(b+c)}+\\frac{1}{b(c+a)}+\\frac{1}{c(a+b)}\\right)$[/quote]\r\n\r\nsign reversed !",
  "Solution_9": "Yes, it's wrong.\r\n[hide=\"to ductrung\"]Where are you from, Australia?[/hide]",
  "Solution_10": "Oh, really? I didn't notice that.\r\n\r\nCan't I proceed afterwards? Could you tell me, please?\r\n\r\n$\\frac{a^2}{b+c}+\\frac{b^2}{a+c}+\\frac{c^2}{a+b}\\geq\\frac{3}{2}$\r\n\r\n$\\Longleftrightarrow 2(a+b+c)(abc+3)\\geq 3(a+b)(b+c)(c+a)$\r\n\r\nkunny",
  "Solution_11": "[quote=\"kunny\"]Oh, really? I didn't notice that.\nCan't I proceed afterwards? Could you tell me, please?\nkunny[/quote]\nTell you about what, kunny?\n\n[quote=\"kunny\"]$\\frac{a^2}{b+c}+\\frac{b^2}{a+c}+\\frac{c^2}{a+b}\\geq\\frac{3}{2}$\n$\\Longleftrightarrow 2(a+b+c)(abc+3)\\geq 3(a+b)(b+c)(c+a)$\n[/quote]\r\nThis follows what?",
  "Solution_12": "Can we prove $2(a+b+c)(abc+3)\\geq 3(a+b)(b+c)(c+a)?$",
  "Solution_13": "[quote=\"kunny\"]Can we prove $2(a+b+c)(abc+3)\\geq 3(a+b)(b+c)(c+a)?$[/quote]\r\nI think we cannot. \r\nYour question:\r\n$2(a+b+c)(abc+3) \\geq 3(a+b)(b+c)(c+a)$\r\nLet $a=1, b \\to 0$, and you can find a counter-example base on the ineqn $3c^2-3c-6 \\ge 0$",
  "Solution_14": "Under conditions that $a>0,\\ b>0,\\ c>0,\\ a^3+b^3+c^3=3?$\r\n\r\nI am comfusing now. :?  Anyway I will show you my approach as follows.\r\n\r\nLet $\\frac{a^2}{b+c}+\\frac{b^2}{c+a}+\\frac{c^2}{a+b}=\\frac{f(a,b,c)}{(a+b)(b+c)(c+a)}.$\r\n\r\n$f(a,b,c)=a^2(c+a)(a+b)+b^2(a+b)(b+c)+c^2(b+c)(c+a)$\r\n\r\n$=a^4+b^4+c^4+(b+c)a^3+(c+a)b^3+(a+b)c^3+abc(a+b+c)$\r\n\r\n$=(a+b+c)(a^3+b^3+c^3)-a(b^3+c^3)-b(c^3+a^3)-c(a^3+b^3)$\r\n\r\n$+(b+c)a^3+(c+a)b^3+(a+b)c^3+abc(a+b+c)$\r\n\r\n$=(a+b+c)(abc+3)\\ \\because a^3+b^3+c^3=3.$\r\n\r\nTherefore we can rewrite the given inequality as $2(a+b+c)(abc+3)\\geq 3(a+b)(b+c)(c+a).$\r\n\r\nAm I something wrong?\r\n\r\nkunny",
  "Solution_15": "[quote=\"hungkhtn\"]SOS :D.... Nice problem.[/quote]\r\nThank you, hungkhtn.\r\nI think SOS here it is very ugly. (Maybe I don't see something.)\r\nA.Gladkich has found  very nice proof for an original inequality without SOS. ;)",
  "Solution_16": "[quote=\"kunny\"]Can we prove $2(a+b+c)(abc+3)\\geq 3(a+b)(b+c)(c+a)?$[/quote]\r\nYes, kunny. It is true.\r\nUse helping inequality\r\n$\\frac{2(a+b+c)}{(a+b)(b+c)(c+a)} \\geq \\frac 9{a^2+b^2+c^2+3(ab+bc+ca)}$.",
  "Solution_17": "Hello to everybody after my long absence. I have tried to solve this inequality using Vasc approach and Ductrung approach, but I cannot. Can anybody help me, please? :blush: \r\n\r\n\r\nThank you very much.",
  "Solution_18": "[quote=\"ductrung\"]\nThis follows immediately from the following inequality (which is not hard to prove): for all positive reals $a,b$ and $c$,\n\\[ \\frac{a^2}{b+c}+\\frac{b^2}{c+a}+\\frac{c^2}{a+b}\\ge\\frac{3}{2}\\cdot\\frac{a^3+b^3+c^3}{a^2+b^2+c^2}  \\][/quote]\r\n\r\n\r\n :?:  :?:",
  "Solution_19": "manlio, try the gladkich's hint:\r\n$\\frac{a}{b+c}\\geq\\frac{8a-b-c}{4(a+b+c)}.$  ;)",
  "Solution_20": "Ok,I post my solution to the problem :\r\n                $\\frac{a^2}{b+c}+\\frac{b^2}{a+c}+\\frac{c^2}{a+b} \\geq \\frac{3}{2}\\dots\\frac{a^3+b^3+c^3}{a^2+b^2+c^2}$ (*)\r\nSolution:\r\n(*) $\\Longleftrightarrow 2\\sum \\frac{a^2}{b+c}-\\sum a \\geq 3\\frac{a^3+b^3+c^3}{a^2+b^2+c^2}-\\sum a$\r\n     $\\Longleftrightarrow \\sum (2\\frac{a^2}{b+c}-a) \\geq \\frac{2(a^3+b^3+c^3)-\\sum_{sym} a^2b}{a^2+b^2+c^2}$\r\n     $\\Longleftrightarrow \\sum \\frac{a(a-b)+a(a-c)}{b+c} \\geq \\frac{\\sum (b-c)^2(b+c)}{a^2+b^2+c^2}$\r\n     $\\Longleftrightarrow \\sum (\\frac{a(a-b)}{b+c}+\\frac{b(b-a)}{a+c}) \\geq \\frac{\\sum (b-c)^2(b+c)}{a^2+b^2+c^2}$\r\n     $\\Longleftrightarrow \\sum \\frac{(a-b)^2(a+b+c)}{(a+c)(b+c)} \\geq \\frac{\\sum (b-c)^2(b+c)}{a^2+b^2+c^2}$\r\nPut:$S_{a}=\\frac{a+b+c}{(a+b)(a+c)}-\\frac{b+c}{a^2+b^2+c^2}$\r\n      $S_{b}=\\frac{a+b+c}{(a+b)(b+c)}-\\frac{a+c}{a^2+b^2+c^2}$\r\n      $S_{a}=\\frac{a+b+c}{(c+b)(a+c)}-\\frac{a+b}{a^2+b^2+c^2}$\r\nNow ne need to prove $\\sum S_{a} (b-c)^2 \\geq 0$\r\nIt approves because we have $S_{a},S_{b},S_{c} \\geq 0$\r\nSo the problem is proved :D  :D",
  "Solution_21": "Very nice proof, nhat!\r\nWhat if $a^4+b^4+c^4=3$ $?$\r\nI think that your method is not work here. :P  :)",
  "Solution_22": "[quote=\"Vasc\"][quote=\"kunny\"]Can we prove $2(a+b+c)(abc+3)\\geq 3(a+b)(b+c)(c+a)?$[/quote]\nYes, kunny. It is true.\nUse helping inequality\n$\\frac{2(a+b+c)}{(a+b)(b+c)(c+a)} \\geq \\frac 9{a^2+b^2+c^2+3(ab+bc+ca)}$.[/quote]\r\n\r\nThank you very much, [b]Vasc[/b].\r\n\r\nkunny",
  "Solution_23": "Hello manlio,\r\n\r\nMy proof is exactly the same as Nhat's. The end result is the identity\r\n$\\sum_{cycl}\\frac{a^2}{b+c}-\\frac{3}{2}\\cdot\\frac{a^3+b^3+c^3}{a^2+b^2+c^2}$\r\n$=\\frac{a^3+b^3+c^3-2abc}{2(a+b)(b+c)(c+a)(a^2+b^2+c^2)}\\sum_{cycl}(a+b)(a-b)^2$",
  "Solution_24": "Thank you very much to everybody",
  "Solution_25": "Dear arqady I use your hint this way, but I cannot solve last ineq\r\n\r\n\r\n$\\frac{a^2}{b+c}+\\frac{b^2}{a+c}+\\frac{c^2}{a+b} \\geq \\sum\\frac{8a^2-ab-ac}{4(a+b+c)} \\geq \\frac{3}{2}\\frac{a^3+b^3+c^3}{a^2+b^2+c^2}$ \r\n\r\nCan you please help me?\r\n\r\nThank you very much.",
  "Solution_26": "[quote=\"manlio\"]Dear arqady I use your hint this way, but I cannot solve last ineq\n\n\n$\\frac{a^2}{b+c}+\\frac{b^2}{a+c}+\\frac{c^2}{a+b} \\geq \\sum\\frac{8a^2-ab-ac}{4(a+b+c)} \\geq \\frac{3}{2}\\frac{a^3+b^3+c^3}{a^2+b^2+c^2}$ \n\nCan you please help me?\n\nThank you very much.[/quote]\r\nDear manlio! I meant to\r\n $\\frac{a^2}{b+c}+\\frac{b^2}{a+c}+\\frac{c^2}{a+b} \\geq \\sum_{cyc}\\frac{8a^2-ab-ac}{4(a+b+c)} \\geq \\frac{1}{2}\\cdot\\sqrt[3]{9(x^3+y^3+z^3)}=\\frac{3}{2}.$  ;)  :)\r\nBy the way, your last inequality is not true. Try $a=b=2,$ $c=1.$ ;)",
  "Solution_27": "Both this problem and older one\r\n\\[ \\frac{a^2}{b}+\\frac{b^2}{c}+\\frac{c^2}{a}\\ge 3\\sqrt[4]{\\frac{a^4+b^4+c^4}{3}} \\]\r\ncan be proved by one way. Notice that the following result implies all of them\r\n\\[ (x+y+z)^3 \\ge 3(xy+yz+zx)\\sqrt{3(x^2+y^2+z^2)} \\]",
  "Solution_28": "[quote=\"hungkhtn\"] Notice that the following result implies all of them\n\\[ (x+y+z)^3 \\ge 3(xy+yz+zx)\\sqrt{3(x^2+y^2+z^2)}  \\][/quote]\r\nIt is a special case ($n=3$) of my inequality:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=78240\r\n :)",
  "Solution_29": "Thank you very much, Arqady.\r\n\r\n\r\nNow it works  :)",
  "Solution_30": "[quote=\"arqady\"][quote=\"hungkhtn\"] Notice that the following result implies all of them\n\\[ (x+y+z)^3 \\ge 3(xy+yz+zx)\\sqrt{3(x^2+y^2+z^2)}  \\][/quote]\nIt is a special case ($n=3$) of my inequality:\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=78240\n :)[/quote]\r\n\r\nOh, actually I haven't see it before and this inequality I post was insprired from my solution for the older one. Very nice result, arqady.",
  "Solution_31": "[quote=\"hungkhtn\"]\n\\[ (x+y+z)^3 \\ge 3(xy+yz+zx)\\sqrt{3(x^2+y^2+z^2)}  \\][/quote]\n\nAfter some computation I reducede it to SOS method\n\n[quote] \\[ \\sum{[13z^3(x+y) +14xy(x^2+y^2) +x^4+y^4 +3x^2y^2 -xyz(7z + 24x +24y)](x-y)^2 \\geq 0}  \\][/quote]\r\n\r\nbut now I am in trouble with this one  :blush: \r\n\r\nHas anyone tried to solve it?\r\n\r\nIs this method OK?\r\n\r\nThank you very much.\r\n\r\nP.S. Vey nice ineq :)",
  "Solution_32": "Dear Manlio, the SOS form is not such complicated as that. Change it into the form\r\n\\[ \\frac{(a+b+c)^2}{ab+bc+ca}-3 \\ge 3(\\frac{\\sqrt{3(a^2+b^2+c^2}}{a+b+c}-1) \\] \r\nNow I think it's easier so much.",
  "Solution_33": "[quote=\"hungkhtn\"]SOS :D.... Nice problem.[/quote]\r\nThank you, hungkhtn. I think SOS not work here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=86713\r\n  :P  :)",
  "Solution_34": "Dear Aquady,\r\n\r\nI'm sure that S.O.S will work. The lemma I post will help to solve both problems, involve yours.\r\n\r\nFirst, use Holder\r\n\\[ (\\sum\\frac{a^2}{b+c})(\\sum\\frac{a^2}{b+c})(\\sum a^2(b+c)^2) \\ge (\\sum a^2)^2 \\]\r\nThen notice that\r\n\\[ \\sum a^2(b+c)^2 \\le 4\\sum a^2b^2 \\]\r\n\r\n :lol: So we are done~",
  "Solution_35": "[quote=\"hungkhtn\"]\nFirst, use Holder\n\\[ (\\sum\\frac{a^2}{b+c})(\\sum\\frac{a^2}{b+c})(\\sum a^2(b+c)^2) \\ge (\\sum a^2)^2  \\]\nThen notice that\n\\[ \\sum a^2(b+c)^2 \\le 4\\sum a^2b^2  \\]\n[/quote]\r\n\r\nNice proof, hungkhtn.\r\n\r\nBTW, I believe that the following inequality holds but I don't have a nice proof yet:\r\n\\[ \\frac{a^2}{b+c}+\\frac{b^2}{c+a}+\\frac{c^2}{a+b} \\ge \\frac{1}{2}\\sqrt[5]{27(a^5+b^5+c^5)} \\]\r\nIt does not hold if we replace 5 by 6.",
  "Solution_36": "[quote=\"ductrung\"]\nBTW, I believe that the following inequality holds but I don't have a nice proof yet:\n\\[ \\frac{a^2}{b+c}+\\frac{b^2}{c+a}+\\frac{c^2}{a+b} \\ge \\frac{1}{2}\\sqrt[5]{27(a^5+b^5+c^5)}  \\]\n[/quote]\r\nMaybe $\\frac{a^2}{b+c}+\\frac{b^2}{c+a}+\\frac{c^2}{a+b} \\ge \\frac{1}{2}\\sqrt[5]{81(a^5+b^5+c^5)}$ $?$ ;)",
  "Solution_37": "[quote=\"hungkhtn\"]\nFirst, use Holder\n\\[ (\\sum\\frac{a^2}{b+c})(\\sum\\frac{a^2}{b+c})(\\sum a^2(b+c)^2) \\ge (\\sum a^2)^2  \\]\n[/quote]\r\nMaybe $(\\sum\\frac{a^2}{b+c})(\\sum\\frac{a^2}{b+c})(\\sum a^2(b+c)^2) \\ge (\\sum a^2)^3$ $?$\r\nIf so, why $\\frac{a^2}{b+c}+\\frac{b^2}{c+a}+\\frac{c^2}{a+b} \\ge \\frac{1}{2}\\sqrt[4]{27(a^4+b^4+c^4)}$ $?$ Thank you. :)",
  "Solution_38": "[quote=\"arqady\"][quote=\"hungkhtn\"]\nFirst, use Holder\n\\[ (\\sum\\frac{a^2}{b+c})(\\sum\\frac{a^2}{b+c})(\\sum a^2(b+c)^2) \\ge (\\sum a^2)^2  \\]\n[/quote]\nMaybe $(\\sum\\frac{a^2}{b+c})(\\sum\\frac{a^2}{b+c})(\\sum a^2(b+c)^2) \\ge (\\sum a^2)^3$ $?$\nIf so, why $\\frac{a^2}{b+c}+\\frac{b^2}{c+a}+\\frac{c^2}{a+b} \\ge \\frac{1}{2}\\sqrt[4]{27(a^4+b^4+c^4)}$ $?$ Thank you. :)[/quote]\r\nThen use the Hung's Lemma:\r\n$(x+y+z)^3 \\ge 3(xy+yz+zx)\\sqrt{3(x^2+y^2+z^2)}$ :) \r\nNice solution.",
  "Solution_39": "WOW! Thank you, zhaobin. Very nice proof, hungkhtn!",
  "Solution_40": "[quote=\"arqady\"]\nMaybe $\\frac{a^2}{b+c}+\\frac{b^2}{c+a}+\\frac{c^2}{a+b} \\ge \\frac{1}{2}\\sqrt[5]{81(a^5+b^5+c^5)}$ $?$ ;)[/quote]\r\n\r\nYeah, that's what I mean. Thank you, arqady.",
  "Solution_41": "[quote=arqady]$a>0,$ $b>0,$ $c>0,$ $a^3+b^3+c^3=3.$ Prove that\n$\\frac{a^2}{b+c}+\\frac{b^2}{a+c}+\\frac{c^2}{a+b}\\geq\\frac{3}{2}.$[/quote]\n\n[img]http://s12.sinaimg.cn/middle/006ptkjAzy76sGtu0DF3b&690[/img]"
}
{
  "Tag": [
    "MATHCOUNTS"
  ],
  "Problem": "http://groups.myspace.com/mathdorksunited\r\n\r\nJoin if you can... Lets make this a huge kick butt myspace community!\r\n\r\n :D \r\n\r\nMathCounts portion of my site is down... Mad My site will be up in september again... I am waiting for my uncle to bring the server down from Massachusetts! If someone would be willing to temporarily host all of this stuff i have I would really appreciate it.",
  "Solution_1": "why did you put it in this forum?"
}
{
  "Tag": [
    "geometry",
    "geometric transformation",
    "rotation",
    "AMC"
  ],
  "Problem": "Ten chairs are evenly spaced around a round table and numbered clockwise from $ 1$ through $ 10$. Five married couples are to sit in the chairs with men and women alternating, and no one is to sit either next to or directly across from his or her spouse. How many seating arrangements are possible?\r\n\r\n$ \\textbf{(A)}\\ 240\\qquad\r\n\\textbf{(B)}\\ 360\\qquad\r\n\\textbf{(C)}\\ 480\\qquad\r\n\\textbf{(D)}\\ 540\\qquad\r\n\\textbf{(E)}\\ 720$",
  "Solution_1": "[hide]Notice that if the wife is in one seat, the husband can only be in 2 different seats by the restriction of the problem.   Now notice that there is 10 places for the first woman to be.  Then, there are only 4 places for the next one to be, since the women and men alternate.  Then there are 3 more places for the 3rd woman, 2 for the 4th, and 1 for the 5th woman.  This gives 240 ways for the women to be seated.\nFor the men, on any open place, there are only 2 men that can be there.  Given one man, the other man that can be there has only one other place to go.  So there are only 2 choices for the men for every choice of women.  This gives 2*240 or 480 choices.[/hide]",
  "Solution_2": "well it has to be a multiple of 240 since its 5!*2 for 5! for the men to sit then 2 for either all even or odd....the women i just got 2 ways they can sit down after the men had been seated...i didnt think 720 would work becuase 3 seems like a weird number due to rotations...\r\nhad 960 been a choice, i mightve put that..",
  "Solution_3": "480 i think",
  "Solution_4": "[hide=\"pretty much the same...\"]The partner of the person in seat 1 must sit in either seat 4 or seat 8. Either way, the rest of the pairs can only be placed in one way. This gives two symmetric cases. Multiplying by 5! (to account for permutations of the 5 pairs) and by 2 (placing men/women on even or odd numbered chairs) we get $ \\boxed{480\\ C}$.[/hide]",
  "Solution_5": "Sorry, I'm confused why it's 5! and not 4!. Aren't rotations the same so wouldn't you do $(5-1)!$ ?",
  "Solution_6": "Rotations aren't the same since the seats are labeled differently from each other.  If they were unlabeled then we would have to divide by 10 (not 5) to account for the rotations.  As the problem stands, we cannot account for rotations because the problem never specifies that rotations are equivalent.",
  "Solution_7": "[quote=\"hello.world\"][hide=\"pretty much the same...\"]The partner of the person in seat 1 must sit in either seat 4 or seat 8. Either way, the rest of the pairs can only be placed in one way. This gives two symmetric cases. Multiplying by 5! (to account for permutations of the 5 pairs) and by 2 (placing men/women on even or odd numbered chairs) we get $ \\boxed{480\\ C}$.[/hide][/quote]\n\nSorry, but can somebody further explain this? Why can the partner of person in seat 1 only sit in seat 4 or 8? Is this some intuition I need to develop - I'm not sure how I would see this in the heat of a competition? \n\nWhy can the rest of the pairs only be placed in one way? Can't we switch the positions of two pairs and it will still be valid? (To elaborate what I mean, if B and b sit in 2 and 7, C and c can be changed with them...no?)\n\nAlso the AMC solution involves seating the women first. is this a general good way of tackling these \"seating\" problems?",
  "Solution_8": "There isnt any \"6th sense\" type deals, here or in most cases, anywhere. Just draw the table. Then $3$ are taken out by the spouse restriction and $2$ are taken out by the gender restriction.\n\n[quote=\"Aequilipse\"]\n\nAlso the AMC solution involves seating the women first. is this a general good way of tackling these \"seating\" problems?[/quote]\n\n I am sure the AMC seats the women first because all male mathematicians are gentlemen. So, yes, generally it is a good idea to seat women first."
}
{
  "Tag": [
    "probability",
    "ratio",
    "geometric series"
  ],
  "Problem": "Hi all,\r\n\r\nI need help solving this probability problem. If anyone can help, can you write the solution in a  step by step manner? Thanks in advance. \r\n\r\n\"Xavier, Yul, and Zeb are having a 3 way paint ball duel. Xavier is a poor shot, hitting his target 1/3 of the time correctly. Yul hits the target 2/3 of the time. Zeb hits his target all the time(no miss). Each person will get one shot at a target of his choice. Once you are hit, you are out. The last person left wins. Since he is the worst, Xavier goes first, followed by Yul, then Zeb. What is Xavier's best strategy, and what is his probability of winning? \"",
  "Solution_1": "Hello tpulak.\r\n\r\nI myself keep forgetting how to do this problem. I just saw it the other day in the Intro to Counting and Probability book. Here's a link that might help....\r\n\r\nhttp://rutherglen.ics.mq.edu.au/math106S206/ex/threewayduel.pdf\r\n\r\nThe numbers are a bit different, but the theory is all the same. You just have to change the fractions around...",
  "Solution_2": "Yes, as a matter of fact, I have gotten this question from my introduction to probability book. Thanks for the comparative solution,but I would really like a solution oriented to this exact problem, as I still am learning this concept.Thanks for helping though.",
  "Solution_3": "Oh no, not again.\r\n[hide]\nAfter Xavier shoots (we don't decide who he should shoot yet), there are three possibilities: Yul is out, Zeb is out, or nobody is hit.\n\nSuppose Yul is out. Then it is Zeb's turn, so he will obviously shoot Xavier and Xavier will always get out. Chance of winning: $ 0$.\n\nSuppose Zeb is out. Yul will shoot at Xavier, hitting 2/3 of the time. Then, if Yul misses, Xavier shoots at Yul, hitting 1/3 of the time. And so on until somebody finally hits. Yul wins $ \\frac{2}{3} \\plus{} \\frac{1}{3}\\times\\frac{2}{3}\\times\\frac{2}{3} \\plus{} \\frac{1}{3}\\times\\frac{2}{3}\\times\\frac{1}{3}\\times\\frac{2}{3}\\times\\frac{2}{3} \\plus{} \\cdots$ of the time.\n\nThis is a geometric series, with first element $ \\frac{2}{3}$ and ratio $ \\frac{2}{9}$. Apply the formula for geometric series. The probability that Yul wins is $ \\frac{6}{7}$, so Xavier only wins $ \\frac{1}{7}$ of the time.\n\nFinally, suppose Xavier misses. Then Yul will shoot Zeb, since Zeb would shoot Yul if Zeb got to shoot. Then if Yul misses Zeb shoots and outs Yul. Thus, 1/3 of the time Xavier gets to shoot first against Zeb; 2/3 of the time it's against Yul.\n\nXavier versus Zeb: Xavier wins only if he can hit Zeb the first time, $ \\frac{1}{3}$.\nXavier versus Yul: Note this is different from the above annoying geometric series probability since Xavier shoots first, but we can use the result. If Xavier hits Yul, Xavier wins: $ \\frac{1}{3}$ of the time. Otherwise, $ \\frac{2}{3}$ of the time, it reduces to the above series. Xavier wins $ \\frac{1}{7}$ of the time, so his total chance against Yul is $ \\frac{1}{3} \\plus{} \\frac{2}{3}\\times\\frac{1}{7} \\equal{} \\frac{9}{21} \\equal{} \\frac{3}{7}$.\n\nSo his total chance against either Zeb or Yul, whoever he faces off, is $ \\frac{1}{3}\\times\\frac{1}{3} \\plus{} \\frac{2}{3}\\times\\frac{3}{7} \\equal{} \\frac{1}{9} \\plus{} \\frac{2}{7} \\equal{} \\frac{25}{63}$ which is the best probability compared to if he shot and outed anybody. So Xavier should deliberately miss or shoot the ground or the air or behind him or whatever.\n\nThis is actually not complete; we must demonstrate Yul and Zeb's best strategies to prove the above, but I'm going to leave that to, um, the next poster. Also, check my arithmetic. I make the most amazing mistakes.\n[/hide]",
  "Solution_4": "Xavier should fire into the air.  Yul will shoot at Zeb, because he is the biggest threat.  If he misses, then Zeb will shoot at Yul and Yul is out since Zeb is a perfect shot.  \r\n\r\nI'm not sure about the probability part though? :?:"
}
{
  "Tag": [
    "inequalities"
  ],
  "Problem": "Let $ x, y, z$ are non- negative integers number a different pair such that $ (x\\plus{}z)(z\\plus{}y)\\equal{}1$. Prove that\r\n$ \\dfrac{1}{(x\\minus{}y)^2}\\plus{}\\dfrac{1}{(x\\plus{}z)^2}\\plus{}\\dfrac{1}{(z\\plus{}y)^2}\\geq 4$.",
  "Solution_1": "If $ x, y,$ and $ z$ are non-negative integers, then $ (x\\plus{}z)(z\\plus{}y) \\equal{} 1$ gives us that $ x\\plus{}z\\equal{}1$ and $ z\\plus{}y\\equal{}1$, which gives us that $ x\\equal{}y$. However, this would be impossible since it makes the inequality undefined?",
  "Solution_2": "Maybe they're supposed to be nonnegative reals?",
  "Solution_3": "[hide=\"Assuming that x, y, z are distinct real numbers\"]Let $ x\\plus{}z\\equal{}a$.\nSince $ (x\\plus{}z)(z\\plus{}y)\\equal{}1$, we have $ z\\plus{}y\\equal{}\\frac{1}{a}$.\n\nThen, the original inequality turns into $ \\frac{1}{\\left(a\\minus{}\\frac{1}{a}\\right)^2}\\plus{}\\frac{1}{a^2}\\plus{}a^2\\ge 4$.\n\nSince $ a^2\\plus{}\\frac{1}{a^2}\\equal{}\\left(a\\minus{}\\frac{1}{a}\\right)^2\\plus{}2$, we now have to prove that $ \\frac{1}{\\left(a\\minus{}\\frac{1}{a}\\right)^2}\\plus{}\\left(a\\minus{}\\frac{1}{a}\\right)^2\\plus{}2\\ge 4$.\n\nSubstituting $ \\left(a\\minus{}\\frac{1}{a}\\right)^2\\equal{}b$, we have to prove that $ b\\plus{}\\frac{1}{b}\\ge 2$, which is trivial by AM-GM Inequality.\n\nQ.E.D.[/hide]",
  "Solution_4": "Sorry,\r\n$ x,y,z$ are nonnegative reals."
}
{
  "Tag": [
    "function",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "Does there exist a function $ f: \\mathbb{N}\\to\\mathbb{N}$ such that $ f(f(n\\minus{}1))\\equal{}f(n\\plus{}1)\\minus{}f(n)$ for each natural number $ n$?",
  "Solution_1": "Observe that f(n + 1) > f(n), and this is imediact.\r\n\r\nTake an integer x.\r\n\r\nf(x + 2) - f(x + 1) = f(f(x))\r\n\r\nLet k = f(x). We have\r\n\r\nf(x + 2) = f(x + 1) + f(k)\r\n\r\nIf k > x + 1, clearly left side < right side.. So f(x) < x + 2.\r\n\r\nBut f(2) > 0, so f(3) > 1, ..., f(n) > n - 2. Then,\r\n\r\nn - 2 < f(n) < n + 2.\r\n\r\nIt\u00b4s easy to see that the difference between f(n) and n + 2 can\u00b4t become greater if n grows. However, this difference can change just two times. So there is an z > 5 such that\r\n\r\nf(z + 1) = f(z) + 1\r\n\r\nBut f(z + 1) = f(z) + f(f(z - 1)),\r\n\r\nand we get f(f(z - 1)) = 1, which implies f(z - 1) < 3, which implies z - 1 < 4, absurd.",
  "Solution_2": "i think you might consider deleting your post feliz :)",
  "Solution_3": "Sorry. Is it better now?",
  "Solution_4": "[quote=\"feliz\"]\n\n\nIt\u00b4s easy to see that the difference between f(n) and n + 2 can\u00b4t become greater if n grows. However, this difference can change just two times. So there is an z > 5 such that\n\nf(z + 1) = f(z) + 1\n[/quote]\r\n\r\nits okay now, but could you just explain a bit more that statement..."
}
{
  "Tag": [],
  "Problem": "1) Predict the shape of $ ClF_3$\r\n\r\n2) $ N_3H$ has two possible structure with the atoms written as NNNH. Find the type of hybridisation of the N atom bonded to the hydrogen atom in each case.",
  "Solution_1": "1) Pyramidal\r\n\r\n2) $ N\\equiv N^\\plus{}\\minus{}N^\\minus{}\\minus{}H$. ----> sp3\r\n $ N^\\minus{}\\equal{}N^\\plus{}\\equal{}N\\minus{}H$ ---->sp2",
  "Solution_2": "[quote=\"hell_ever\"]1) Pyramidal[/quote]\r\n\r\nI think it's actually T-shaped.",
  "Solution_3": "edit : yeah there are lone pairs so it would be t-shaped\r\n\r\nhell_ever : in your first structure for 2), that is a dative bond from the second to the third N, so why isnt it sp2 as well?",
  "Solution_4": "There are no dative bonds in hydrazoic acid.",
  "Solution_5": "Yep my bad ... could someone explain the hybridizations?",
  "Solution_6": "What do you want to be explained? In the first ressonance structure above the nitrogen forms bonds with two atoms and there two lone pairs: so four hybrid orbitals are needed, and the appropriate hybridization scheme is therefore sp3. A similar explanation holds for the other structure."
}
{
  "Tag": [
    "number theory solved",
    "number theory"
  ],
  "Problem": "Let $p \\geq 5$ be a prime number.  Prove that there exist at least 2 distinct primes $q_1, q_2$ satisfying $1 < q_i < p - 1$ and $q_i^{p-1} \\not\\equiv 1 \\mbox{ (mod }p^2)$, for $i = 1, 2$.",
  "Solution_1": "This is strange, since if I'm not wrong it follows from the following IMO 2001 shortlisted problem:\r\n  If p>3 is a prime then we can find n<p-1 such that $ p^2$ doesn't divide $ n^{p-1}-1 $ nor $ (n+1)^{p-1}-1$. Now, suppose for a certain x $ p^2$ doesn't divide $ x^{p-1}-1$. It's quite obvious that there must be a prime q|x such that $ p^2$ does not divide $ q^{p-1}-1$. Thus we can find two different prime factors of n and n+1 with the property stated. They will be certainly different. Am I missing something?",
  "Solution_2": "Valiowk, you guys should really be studying the past ISLs right? :)",
  "Solution_3": "Well it isn't necessary to use the other shortlisted problem.  It's enough to consider $(p-1)^{p-1}, (p+1)^{p-1}, (2p-1)^{p-1}, (p+1)^{p-1}$ and by a simple argument it can be shown that there must exist $q_1, q_2$ which are factors of some of $p-1, p+1, 2p-1, 2p+1$.\r\n\r\nAnd yes, Valentin, we should have studied the past ISLs harder... :(",
  "Solution_4": "The main observation is that if $x^{p-1} \\not \\equiv 1$ (mod $p^2$), then some prime divisor $q$ of $x$ satisfies $q^{p-1} \\not \\equiv 1$ (mod $p^2$). \n\nIndeed, if this were not the case, then we could multiply some numbers which are all $1$ (mod $p^2$) to obtain $x^{p-1}$, absurd. \n\nNow, with the cases $p < 11$ being trivial, we will use this observation to solve the cases where $p\\ge 11.$ Call a prime $q$ [b] tasty [/b] if $q^{p-1} \\not \\equiv 1$ (mod $p^2$). Notice that $(p-1)^{(p-1)} \\equiv p+1$ (mod $p^2$) and $(p+1)^{(p-1)} \\equiv -p+1$ (mod $p^2$). Since these are both not $1$ (mod $p^2$), both $p-1$ and $p+1$ have a prime divisor which is tasty. If these are not both $2$, we're done since they then must be distinct. \n\nOtherwise, we have that $2$ is tasty. Now, we have that $(2p-1)^{(p-1)} \\equiv -2p(p-1) + 1 \\not \\equiv 1$ (mod $p^2$). This means that $2p-1$ has a tasty prime divisor, which is clearly in $(1, p-1)$ and not equal to $2.$ Together with $2$, we are done.\n\n$\\square$ "
}
{
  "Tag": [
    "function",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "Suppose that the function $ f: \\mathbb{R}\\rightarrow\\mathbb{R}$ has the prperty that\r\n\r\n$ f(x\\plus{}y)\\equal{}f(x)\\plus{}f(y)$ for all $ x$ and $ y \\in \\mathbb{R}$.\r\n\r\nProve that if $ f$ is continuous at some $ x_0\\in \\mathbb{R}$,then there exists a const $ m$ such that $ \\forall x\\in\\mathbb{R}$,$ f(x)\\equal{}mx$",
  "Solution_1": "[quote=\"skywalkerJ.L.\"]Suppose that the function $ f: \\mathbb{R}\\rightarrow\\mathbb{R}$ has the prperty that\n\n$ f(x \\plus{} y) \\equal{} f(x) \\plus{} f(y)$ for all $ x$ and $ y \\in \\mathbb{R}$.\n\nProve that if $ f$ is continuous at some $ x_0\\in \\mathbb{R}$,then there exists a const $ m$ such that $ \\forall x\\in\\mathbb{R}$,$ f(x) \\equal{} mx$[/quote]\r\n\r\nIt's a basic classical result.\r\n\r\nAny of the following supplementary conditions for Cauchy equation imply $ f(x) \\equal{} \\alpha x$ :\r\n\r\n$ f(x)$ continuous\r\n$ f(x)$ continuous at one point\r\n$ f(x)$ monotonous\r\n$ f(x)$ monotonous on a non empty open interval\r\n$ f(x) > m$ $ \\forall x\\in(a,b)$ for some $ b > a$\r\n$ f(x) < M$ $ \\forall x\\in(a,b)$ for some $ b > a$",
  "Solution_2": "[quote=\"pco\"][quote=\"skywalkerJ.L.\"]Suppose that the function $ f: \\mathbb{R}\\rightarrow\\mathbb{R}$ has the prperty that\n\n$ f(x \\plus{} y) \\equal{} f(x) \\plus{} f(y)$ for all $ x$ and $ y \\in \\mathbb{R}$.\n\nProve that if $ f$ is continuous at some $ x_0\\in \\mathbb{R}$,then there exists a const $ m$ such that $ \\forall x\\in\\mathbb{R}$,$ f(x) \\equal{} mx$[/quote]\n\nIt's a basic classical result.\n\nAny of the following supplementary conditions for Cauchy equation imply $ f(x) \\equal{} \\alpha x$ :\n\n$ f(x)$ continuous\n$ f(x)$ continuous at one point\n$ f(x)$ monotonous\n$ f(x)$ monotonous on a non empty open interval\n$ f(x) > m$ $ \\forall x\\in(a,b)$ for some $ b > a$\n$ f(x) < M$ $ \\forall x\\in(a,b)$ for some $ b > a$[/quote]\r\n\r\nIn the case of $ f(x)$ continuous at one point,I just have to prove $ f(nx_0\\plus{}q)\\equal{}f(nx_0)\\plus{}f(q)\\equal{}(nx_0\\plus{}q)f(1)$($ q\\in\\mathbb{Q}$)\r\n\r\nIf all the real numbers can be expressed by $ nx_0\\plus{}q$,then we are done.\r\n\r\nAm i right?"
}
{
  "Tag": [],
  "Problem": "Evaluate \\begin{align*} (5a^2 - 13a + 4)(2a - 3) \\end{align*} for $ a \\equal{} 1\\frac12$.",
  "Solution_1": "Plug in a\r\nNotice that (2a-3)=0\r\nSince anything multiplied by 0 is 0, the answer is\r\n0.",
  "Solution_2": "2a-3=0, so 0."
}
{
  "Tag": [
    "AMC",
    "AIME",
    "USA(J)MO",
    "USAMO",
    "AIME I",
    "AIME II"
  ],
  "Problem": "Seeing as my school is taking the National French Exam on the same day as AIME I, and I don't want to have two major academic competitions on the same day, I want to take the alternate AIME.\r\n\r\nSo, how do I sign up for it? Who should I talk to? Do I simply need to pay the money or do I have to give some sort of reason?\r\nAlso, I don't think any teacher at my school wants to stay there for 3 hours after school, so does anyone have any recommendations for who I should call or look for, etc? Note that I live in a suburb west of Boston, MA.",
  "Solution_1": "You should tell whoever is in charge of the AMC at your school.  (Or did you not take the AMC through your regular school?)  They can make arrangements to get the Alternate AIME for you.  It costs $\\$25$ -- this is something that you should [i]insist[/i] that the school pay for.\r\n\r\nYou're only supposed to take the Alternate AIME if you have some sort of conflict with the regular date, but clearly you do.  You don't need to provide the reason to the AMC.\r\n\r\nThe AIME is supposed to be given from 9 a.m. to Noon, unless your school has a really good reason not to do it at that time, and they're supposed to get permission from the AMC to give it at a time other than 9 a.m.  \r\n\r\nBonne chance au concours fran\u00e7ais!  (J'ai etudi\u00e9 le fran\u00e7ais dans l'\u00e9cole aussi.  :) )",
  "Solution_2": "I also need to take the AIME II, but I heard it usually harder than the AIME I. And is it possible that the better people will be taking the AIME II, trying to get more time to practice, so that even the fact that they take a percentage of AIME I and AIME II for the USAMO will not help?",
  "Solution_3": "Correct, timely information from DPatrick.  Thanks for passing it along!\r\n\r\n-- Steve Dunbar\r\nDiector, American Mathematics Competitions",
  "Solution_4": "[quote=\"123s\"]And is it possible that the better people will be taking the AIME II, trying to get more time to practice, so that even the fact that they take a percentage of AIME I and AIME II for the USAMO will not help?[/quote]\r\n\r\nNo, because it's not allowed unless you specifically have a conflict of dates for the AIME I.",
  "Solution_5": "But could there be a possibility that the aime don't arrive to my school? I'm qualified to write the aime but I don't know if my school had received the contest yet  :(",
  "Solution_6": "[quote=\"AMCDirector\"]> If we cannot receive our material by March 7, would it be possible to allow us to take the AIME II?\n\nIf you have not received materials by March 6, have your teacher contact us. We will make an alternate arrangement.[/quote]",
  "Solution_7": "Does the alternate AIME decrease your chances of qualifying for USAMO though? I've heard that it is usually more difficult.",
  "Solution_8": "Darn... people ask that a lot. The answer is search. Actually, the answer is if there really is a huge difference, the AMC will find a way to take it into account. But usually there's not that big of a difference.",
  "Solution_9": "As seen by last year's (I think), if there's a large enough difference (probably index difference of 10+ points), they'll have a separate index for the AIME II. Otherwise, it's close enough that in general it won't matter.",
  "Solution_10": "Does anyone know when deadline is to register from AIME II?\r\n\r\nThe guy in charge of AMC at my school agrees to let me take it, however he says that \"he called and e-mailed the AMC officials and has not gotten a single reply,\" suggesting that if he doesn't manage to get a response I might have to take the AIME I.",
  "Solution_11": "[quote=\"probability1.01\"]Darn... people ask that a lot. The answer is search.[/quote]\r\n\r\nOnly if everyone was like probability1.01 who knows how nice that button is.  :)",
  "Solution_12": "[quote=\"calc rulz\"]Does anyone know when deadline is to register from AIME II?\n\nThe guy in charge of AMC at my school agrees to let me take it, however he says that \"he called and e-mailed the AMC officials and has not gotten a single reply,\" suggesting that if he doesn't manage to get a response I might have to take the AIME I.[/quote]Does that mean he called and no one answered the phone?  Just to verify, the number listed in the AIME packet is (800) 527-3690.\r\n\r\nAlso, the AMC has set up a website where you can register for the Alternate AIME online.  I'm not sure that I'm supposed to give out the URL publicly (since I couldn't find it anywhere public), but the URL is listed on the AMC Score Report that you teacher should have received (on ours, it's on the page that says \"THE 2006 AIME POLICY\" near the top).",
  "Solution_13": "When registering for the Alternate Aime, do you do it yourself, or does your contest manager do it for you. Does this change if you are sending it to an alternate test location?",
  "Solution_14": "[quote=\"123s\"]When registering for the Alternate Aime, do you do it yourself, or does your contest manager do it for you.[/quote]The teacher / contest manager should do it.  [quote] Does this change if you are sending it to an alternate test location?[/quote]For this you probably need to call the AMC."
}
{
  "Tag": [],
  "Problem": "Hey When does your school start?",
  "Solution_1": "My school starts on the 24th.  I hate this summer vacation, since this is like the shortest one I've had in a long time.  Usually our school starts aorun Sept. 1 or Aug. 31, but last year it started after labor day.  Now this year, it's starting earlier than usual.  It hasn't started this early since kindergarden.  Only 2 months and a week of summer vacation for me  :( .",
  "Solution_2": "Well mine the same thing.  We got like 2 months and maybe 2 weeks of summer.  But my school district always does that.",
  "Solution_3": "Monday.   :(",
  "Solution_4": "End of the month yeah! :)",
  "Solution_5": "September 7th",
  "Solution_6": "September 9!  :lol:",
  "Solution_7": "September 7",
  "Solution_8": "September 7, the tuesday after Labour Day, for Univ. of Toronto students.",
  "Solution_9": "The first day of classes at MTU is Aug. 29, IIRC, but first-year orientation starts Aug. 21.",
  "Solution_10": "ick, aug. 22nd",
  "Solution_11": "September 7th",
  "Solution_12": "My school starts August 16th :( Last year we started September 7th so I only had two months. Plus we have yearbook/pick up book day the 10th! btw, how long are your winter and spring breaks?",
  "Solution_13": "August 18th baby!!! WOOOHOOOOOO!!!!!!!!!!!!!!!!!!!!!!!",
  "Solution_14": "Sept. 2nd\r\nSorry, what's labor day ?  :?",
  "Solution_15": "In US it is Sept 1st...but I think in most countries, it is May 1st",
  "Solution_16": "Labor day is when people just take a break off from school or work.  In the US, it's the first monday of september.",
  "Solution_17": "I start August 25th.",
  "Solution_18": "August 26th!  Can't wait!   :P",
  "Solution_19": "I start August 19th, 2004.  My enrollment is a weekish before that.",
  "Solution_20": "September 8th. Why do most of you start before labor day. Never seen a school that does it.",
  "Solution_21": "Wow... a lot of ppl start realli early... my school used to start Aug. 15, but now we're early september.. this yr is Sept. 2.",
  "Solution_22": "My school had to start early this year so I started Monday.  I have a busy schedule this year.  My classes aren't too bad compared to most people on this forum.  However, I am taking 3 AP's(4 if I can independent study stats, I have get that straightened out) and an honors, and a CP(below honors).  The CP is for Latin which they only offer in CP until you get Latin 4/AP.  What really is hard is doing cross country after school.  That means I won't get home until 5:45-6:00 everyday!  I 'm just glad I m not bored!  ;)",
  "Solution_23": "My schoos starts september 13th",
  "Solution_24": "classes don't start until the 30th, but i can move back in on the 23rd\r\n\r\nI know someone who starts on the 15th of September (yes...for college)",
  "Solution_25": "Well, in college come people might start even on September 22nd or 23rd, somewhere near there.  It just means that they are on a quarter system.  THose start late and end late compared to high school and stuff.",
  "Solution_26": "My school starts september 13th... I've just started my summer-homework...\r\nyuk....\r\n\r\nHistory, Latin, Philosofy, Italian Licterature... and a bit of trigonometry/algebra routine problems... And 13 technical paintings (bad english :maybe: )\r\n\r\nI'd love to know when chinese school starts...",
  "Solution_27": "mine starts September 5 th. Can u believe? it's Sunday, terrible  :(",
  "Solution_28": "September 10th, but registration and stuff starts september 8th (?)",
  "Solution_29": "Wow! Your school starts very early. When I was looking at the dates I thought they were for September. In Albania it started on September 20th, and here in Canada it started on September 7 (the day after Labour Day). I didn't think other countries actually started in August!  :D  When does it end though? In Canada it ends June 27th. (Unless you're doing gr. 12 IB, where it ends at the end of April and te exams are in May, but that is applicable to Ib students all over the world, not only in Canada). :)",
  "Solution_30": "We do start early in the U.S.  I don't know why though.  Most of us usually get out at the end of may or early june though.",
  "Solution_31": "Mine already started...",
  "Solution_32": "26! :D",
  "Solution_33": "YOu're a little off topic. This was last year  ;)",
  "Solution_34": "last monday in august...\r\nwhich in this year's case is the 29th  ;)",
  "Solution_35": "August 23rd",
  "Solution_36": "September 7th  :P",
  "Solution_37": "Aug. 25",
  "Solution_38": "my school starts \"late\" this year, aug 25",
  "Solution_39": "29th. cuz we always start on a monday",
  "Solution_40": "I don't start classes until September 26th (due to quarter system).  I go out for orientation the week before.  But then I don't get out until mid-June.",
  "Solution_41": "August 15th... :(",
  "Solution_42": "My school start's on none of those dates it starts on August 29th :)  :!:",
  "Solution_43": "Septtember 7th :(",
  "Solution_44": "[quote=\"236factorial\"]YOu're a little off topic. This was last year  ;)[/quote]\r\n\r\nQuite so.\r\n\r\nLocked."
}
{
  "Tag": [
    "AoPSwiki",
    "ARML",
    "search"
  ],
  "Problem": "Does anyone know where I can find past NYSML Problems?  I checked the AoPSWiki, but the [url=http://www.artofproblemsolving.com/Wiki/index.php/NYSML_Problems_and_Solutions]NYSML Page[/url] was blank.  Furthermore, my coach said that she didn't have any old NYSML exams either.  Thanks!",
  "Solution_1": "i don't know if these are helpful but...\r\n\r\nhttp://www.mathlinks.ro/viewtopic.php?t=16613",
  "Solution_2": "The recent batch of contests are being sold as a book for the quite reasonable price of 15 dollars: http://www.nysml.org/index.php?n=NYSML.ContestArchive\r\n\r\nI don't know if the old ARML-NYSML books are still in print.  For a random selection of past problems, search for \"NYSML\" on this website.",
  "Solution_3": "Do you know if there are any free resources?  I can't afford all of the contests they're selling.  :(",
  "Solution_4": "do what jbl said and try to find some tests from other people",
  "Solution_5": "You might also ask your coach or the department head if the school can purchase it.  Are you on the NYC math team?  If so, you should ask around if anyone has some you can copy, including the other coaches (e.g. Mr. C.).  And, again, lots of NYSML and ARML stuff around the forums.",
  "Solution_6": "The NYSML 2005 individual problems were posted here:\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=33106[/url].\r\n\r\nThe NYSML 2008 individual problems were posted here:\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=205255[/url].\r\n\r\nI also posted a dozen problems from the years 1989 through 1992 in this forum (NYC).  I got those problems from the book [i]ARML-NYSML Contests 1989-1994[/i] by Zimmerman and Kessler, published by MathPro Press.  I believe I purchased that book 4 years ago for about 20 dollars from here:\r\n[url]http://www.mathpropress.com[/url]."
}
{
  "Tag": [],
  "Problem": "1/2+1/3+1/42\r\n\r\n\r\n47, but it's probably not the least",
  "Solution_1": "I don't get it. Would 1/7+2/7+3/7 count for 21? :maybe:",
  "Solution_2": "UNIT FRACTIONS ARE IN THE FORM 1/x, where x is not 0.  here, x must be POSITIVE because it asks for positive fractions",
  "Solution_3": "[quote=\"asianhottie8293\"]The sum of the three different positive unit fractions is 6/7. What is the least number that could be the sum of the denominators of these fractions?[/quote]\r\n\r\n[hide=\"solution\"]We know that one of the fractions must be $\\frac{1}{2}$ because $\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{5}<\\frac{6}{7}$.  \n\nIf one of the fractions is $\\frac{1}{2}$, the sum of the other two unit fractions is $\\frac{5}{14}$.  \n\n$\\frac{5}{14}=\\frac{1}{a}+\\frac{1}{b}$\n\n$\\Rightarrow \\frac{5}{14}=\\frac{a+b}{ab}$\n\n$\\Rightarrow 5ab-14a-14b=0$\n\n$\\Rightarrow (5a-14)(5b-14)=196$\n\nThe factors of $196$ are:\n\n$1,196$\n$2,98$\n$4,49$\n$7,28$\n$14,14$\n\n$5a-14=1,2,4,7,14$ and the only integer value for $a$ is $3$\n\nThe three fractions are $\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{52}$.\n\nTherefore, $2+3+52=\\boxed{57}$.[/hide]",
  "Solution_4": "[quote=\"jli\"][quote=\"asianhottie8293\"]The sum of the three different positive unit fractions is 6/7. What is the least number that could be the sum of the denominators of these fractions?[/quote]\n\n[hide=\"solution\"]We know that one of the fractions must be $\\frac{1}{2}$ because $\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{5}<\\frac{6}{7}$.  \n\nIf one of the fractions is $\\frac{1}{2}$, the sum of the other two unit fractions is $\\frac{5}{14}$.  \n\n$\\frac{5}{14}=\\frac{1}{a}+\\frac{1}{b}$\n\n$\\Rightarrow \\frac{5}{14}=\\frac{a+b}{ab}$\n\n$\\Rightarrow 5ab-14a-14b=0$\n\n$\\Rightarrow (5a-14)(5b-14)=196$\n\nThe factors of $196$ are:\n\n$1,196$\n$2,98$\n$4,49$\n$7,28$\n$14,14$\n\n$5a-14=1,2,4,7,14$ and the only integer value for $a$ is $3$\n\nThe three fractions are $\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{52}$.\n\nTherefore, $2+3+52=\\boxed{57}$.[/hide][/quote]\r\n\r\nwell that cant be right since indiana found one that was less.",
  "Solution_5": "ravi b came up with a baller solution to this, lemme see if i can find it\r\nedit: http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=fractions+unit&t=127023",
  "Solution_6": "[quote=\"Walk Around The River\"][quote=\"jli\"][quote=\"asianhottie8293\"]The sum of the three different positive unit fractions is 6/7. What is the least number that could be the sum of the denominators of these fractions?[/quote]\n\n[hide=\"solution\"]We know that one of the fractions must be $\\frac{1}{2}$ because $\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{5}<\\frac{6}{7}$.  \n\nIf one of the fractions is $\\frac{1}{2}$, the sum of the other two unit fractions is $\\frac{5}{14}$.  \n\n$\\frac{5}{14}=\\frac{1}{a}+\\frac{1}{b}$\n\n$\\Rightarrow \\frac{5}{14}=\\frac{a+b}{ab}$\n\n$\\Rightarrow 5ab-14a-14b=0$\n\n$\\Rightarrow (5a-14)(5b-14)=196$\n\nThe factors of $196$ are:\n\n$1,196$\n$2,98$\n$4,49$\n$7,28$\n$14,14$\n\n$5a-14=1,2,4,7,14$ and the only integer value for $a$ is $3$\n\nThe three fractions are $\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{52}$.\n\nTherefore, $2+3+52=\\boxed{57}$.[/hide][/quote]\n\nwell that cant be right since indiana found one that was less.[/quote]\r\n\r\noops...it should be $\\frac{1}{42}$ so 47.",
  "Solution_7": "[quote=\"ragnarok23\"]ravi b came up with a baller solution to this, lemme see if i can find it\nedit: http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=fractions+unit&t=127023[/quote]\r\nDude, Ravi B is a beast.\r\nIt's no surprise that he came up with such a pwnage solution.\r\nIf he could be in countdown I'd be even more scared of him than splash D"
}
{
  "Tag": [
    "LaTeX",
    "set theory"
  ],
  "Problem": "This is a part of a proof my dad gave me, can anyone make anysense out of it?\r\nlet ^ be intersection (i'm bad at latex)\r\n=A-B^C+B^C-b^A-A^C+B^C\r\n\r\n=A-B^C+B^(C-A)+C^(B-A)\r\n\r\n=(A-B^C) + BU(C-A)+C^(B-A)\r\n\r\nDoes this make any sense?\r\nAnd is + a symbol you can use in set theory that you can replace for U?",
  "Solution_1": "let me try to write it out\r\n\r\n\\cup and \\cap represent union and intersection\r\neverything else you typed is the same\r\n\r\n$ A\\minus{}B\\cap C\\plus{}B\\cap C\\minus{}B\\cap A\\minus{}A\\cap C\\plus{}B\\cap C$\r\n\r\n$ A\\minus{}B\\cap C\\plus{}B\\cap (C\\minus{}A)\\plus{}C\\cap (B\\minus{}A)$\r\n\r\n$ (A\\minus{}B\\cap C)\\plus{}B\\cup (C\\minus{}A)\\plus{}C\\cap (B\\minus{}A)$\r\n\r\nGoing from that first statement to the second seems like a simple use of the distributive property\r\nIntersections distribute\r\n\r\nGoing from the 2nd to the 3rd doesn't seem to make sense. It looks like your dad just turned an intersection into a union (which you can't do as far as I know). Did you copy it wrong?"
}
{
  "Tag": [],
  "Problem": "A rational number between $ \\sqrt{2}$ and $ \\sqrt{3}$ is:\r\n\r\n$ \\textbf{(A)}\\ \\frac{\\sqrt{2} \\plus{} \\sqrt{3}}{2} \\qquad \r\n\\textbf{(B)}\\ \\frac{\\sqrt{2} \\cdot \\sqrt{3}}{2}\\qquad \r\n\\textbf{(C)}\\ 1.5\\qquad \r\n\\textbf{(D)}\\ 1.8\\qquad \r\n\\textbf{(E)}\\ 1.4$",
  "Solution_1": "[hide=\"Solution\"]\n\nOptions $ \\textbf{(A)}$ and $ \\textbf{(B)}$ are quickly discarted because they are not rational.\n\nNow we square $ 1.4$, this gives $ 1.96$ which is clearly smaller than $ 2$ so it can't be in between $ \\sqrt {2}$ and $ \\sqrt {3}$. So option $ \\textbf{(E)}$ is out. Another way is noticing that $ \\sqrt {2}$ is aproximately $ 1.414 > 1.4$\n\nNow square $ 1.8$, that will give you $ 3.24 > 3$ so option $ \\textbf{(D)}$ is out. \n\nSo by canceling out all other possibilities $ \\boxed{\\textbf{(C)}}$ is the answer.\n[/hide]"
}
{
  "Tag": [
    "function",
    "inequalities",
    "calculus",
    "derivative",
    "real analysis",
    "inequalities proposed"
  ],
  "Problem": "I noticed that the following lemma is quite useful. It is very simple with Lagrange multipliers but, maybe, somebody will offer an elementary proof.\r\n\r\nLet $x_1,\\ldots ,x_n>0$. Fix the sum $\\sum_j x_j=S$ and the product $\\prod_j x_j=P$. Suppose that $F(x)$ is a smooth function such that $F'(1/u)$ is strictly convex or strictly concave. Then the maximum and the minimum of the expression $\\sum_j F(x_j)$ are achieved when among the values of $x_j$ there are at most 2 different. (I do not know whether we can strengthen it to \"$(n-1)$ variables are equal\"; if we can, we'll probably, have something equivalent or close to $(n-1)EV$ principle of Vasc. Nevertheless, even as is, it is a nice thing to know)\r\n\r\nAs an example, let us consider the famous $\\sum \\frac1{5-abc}\\leq 1$ inequality of Vasc. If we fix the product to be $P$, then it can be written as $\\sum\\frac1{5-\\frac Pa}\\leq 1$. But the function $F(x)=\\frac 1{5-\\frac Px}$ satisfies the above property and, therefore, it is enough to consider the two-value cases. The 2-2 case (a=b=x, c=d=y) reduces to the inequality $\\frac 1{5-x^2y}+\\frac 1{5-xy^2}\\leq \\frac 12$ under the condition $x+y=2$, which can be rewritten as $(xy)^3-6xy+5=(1-xy)(5-xy-(xy)^2)\\geq 0$, which is clear because $0\\leq xy\\leq 1$. The 3-1 case (a=b=c=x, d=y) reduces to the inequality $\\frac 1{5-x^3}+\\frac 3{5-x^2 y}\\leq 1$ under the condition $3x+y=4$, which can be rewritten as $5-4x^2y-2x^3+x^5y\\geq 0$. Plugging $y=4-3x$, we obtain $5-16x^2+10x^3+4x^5-3x^6\\geq 0$. The left hand side may seem unpleasant, but it factors as $(1-x)^2(5+10x-x^2-2x^3-3x^4)$, so it is enough to check that the second factor is non-negative on $(0,4/3)$. But it is a concave function of $x$ on that interval, so it suffices to verify the inequality for the endpoints. The end.",
  "Solution_1": "Indeed Fedja, if $F'(1/x)$ is convex, then the expression $\\sum_j F(x_j)$ is maximal and minimal when $n-1$ of $x_i$ are equal.\r\n ;)",
  "Solution_2": "Actually, so called (n-1) Equal Variable Principle is the following:\r\n\r\nLet $a_1,...,a_n$ ($n \\ge 3$) be given positive numbers, let $p$ and $q$ be unzero and non-equal real numbers, and let $x_1,...,x_n$ be positive numbers such that\r\n$x_1^p+...+x_n^p=a_1^p+...+a_n^p$ and\r\n$x_1^q+...+x_n^q=a_1^q+...+a_n^q$.\r\nLet $f(x)$ be a differentiable function on $(0,\\infty)$ so that $g(x)=x^{\\frac{1-q}{p-q}}f'(x^{\\frac{1}{p-q}})$ is strictly convex. Then, the symmetric function\r\n$F_n(x_1,...,x_n)=f(x_1)+...+f(x_n)$\r\nis minimal for $x_1\\le x_2=x_3=...=x_n$, and is maximal for $x_1=x_2=...=x_{n-1}\\le x_n$.\r\nNotice that for $p \\rightarrow 0$ or $ q \\rightarrow 0$, the sums become products.\r\n\r\n[b]Attention.[/b] This primary form needs some correction. The corrected statement is the following:\r\n\r\nLet $a_1,...,a_n$ ($n \\ge 3$) be given positive numbers, let $p$ and $q$ be non-equal real numbers, and let $x_1,...,x_n$ be positive numbers such that\r\n$x_1^p+...+x_n^p=a_1^p+...+a_n^p$ and\r\n$x_1^q+...+x_n^q=a_1^q+...+a_n^q$ ($p=0$ or $q=0$ means $x_1x_2...x_n=a_1a_2...a_n$).\r\nLet $f(x)$ be a differentiable function on $(0,\\infty)$ so that $g(x)=x^{\\frac{1-q}{p-q}}f'(x^{\\frac{1}{p-q}})$ is strictly convex, and $F_n(x_1,...,x_n)=f(x_1)+...+f(x_n)$.\r\n(a) If $pq \\le 0$, then $F$ is minimal for $x_1\\le x_2=x_3=...=x_n$, and is maximal for $x_1=x_2=...=x_{n-1}\\le x_n$;\r\n(b) If $p>0$ and $q>0$, then $F$ is maximal for $x_1=x_2=...=x_{n-1}\\le x_n$;\r\n(c) If $p<0$ and $q<0$, then $F$ is minimal for $x_1\\le x_2=x_3=...=x_n$.",
  "Solution_3": "Very nice idea, Vasc! Thank you for sharing it with us :). Now let's see if somebody can provide a nice proof  ;) .",
  "Solution_4": "[quote=\"fedja\"]I noticed that the following lemma is quite useful. It is very simple with Lagrange multipliers but, maybe, somebody will offer an elementary proof.\n[/quote]\r\n\r\nVery interesting this lemma. Can you please post your solution?\r\n\r\nThank you very much, Fedja.",
  "Solution_5": "[quote=\"manlio\"]\nVery interesting this lemma. Can you please post your solution?\n[/quote]\r\nOK, but note that what I'll prove falls short of Vasc's $(n-1)EV$ principle. Using Lagrange multipliers, we see that, at the extremum point, either all $n$ variables are equal or there exist $\\mu$ and $\\nu$ such that $F'(x_j)=\\mu\\cdot 1+\\nu\\cdot\\frac1{x_j}$ for all $j$. But, due to strict convexity (concavity) of $F'(1/u)$, the equation $F'(x)=\\mu+\\nu\\frac1x$ may have at most 2 positive solutions (since the left hand side is a strictly convex (concave) function of $1/x$ and the right hand side is a linear function of $1/x$). The end. \r\nThis solution applies to the more general case considered by Vasc too, but gives only the same conclusion about just 2 different values. To prove the $(n-1)$EV principle in full, a different idea is needed. I'd like to see if somebody else finds it  ;) .",
  "Solution_6": "Thank you very much.",
  "Solution_7": "I think Fedja proof is OK  ;)\r\n\r\n\r\nNever mind if it is a calculus based solution, the important thingh is that it works.",
  "Solution_8": "It suiffices to prove Vasc's lemma for three variables. Indeed, if for any three $x$-s, two greater of them are equal, then all $x$-s except the least are equal. Fedja's argument shows that two of variables are equal, and we only have to show that the case $x_1 < x_2 = x_3$ is not a maximum case. For proving this, we redenote the above point $(x_1,x_2,x_3)$ as $ P = (a,b,c)$ and make the following: vary $x_2$ in a neighborhood of $b$, and consider $x_1$ and $x_3$ as functions of $x_2$, defined by two equations $x_1^p+x_2^p+x_3^p=a^p+2b^p$ and $x_1^q+x_2^q+x_3^q=a^q+2b^q$ ($x_1$ varies in a neighborhood of $a$, $x_3$ in a neighborhood of $c=b$). They are well-defined due to implicit function theorem. More over, we may compute their derivatives in a point $(a,b,c)$. Do it, and then study a function $G(b)=f(x_1(x_2))+f(x_2)+f(x_3(x_2))$ in a point $x_2=b$. We may assure after some computations, that $G'(b)=0$ and $G''(b)>0$, since $g'(b)>(g(b)-g(a))/(b-a)$ from convexity of Vasc's function $g(x)=x^{\\frac{1-q}{p-q}}f'(x^{\\frac{1}{p-q}})$. So, our point does not correspond to a maximum, q.e.d. \r\n\r\nComputations may be simplified, if we put $p=q+1$, which may be gotten if replace variables ($x_i=y_i^\\alpha$\r\nfor appropriate $\\alpha$).\r\n\r\nWe also have to consider a case, in which some of variables vanish (as we did not have to consider in Fedja's case), it is the work of the same kind.",
  "Solution_9": "I'm curious, what will happend if someone solves a problem in IMO 2005 using this method ?\r\nWill he get all the points? none? or what ? Will the coordination comitee and the examiners be aware of it ?\r\n\r\nthanks",
  "Solution_10": "Hey,\r\nI think you forget my question. I'm really curious , what will happend ?\r\nWill the coordination comitee accept such a proof ? what about B.M.O ? (which is gonna take place in some days... )\r\n\r\nthanks",
  "Solution_11": "I used Lagrange multipliers on IMO, and got 7. But it was in Romania. Many things depend on coordinators. In any case, you must clearly formulate the facts you use, and preferably write the sketch of the proof. After that, you have to inform your team leader of what you meant, before coordination (it is forbidden, I know).  Then he says on coordination: \"or, it's a basic thing, we in Greece learn it in primary school, our students would not imagine that this obvious statement  must be proved on a such high-level exam, as IMO\", and presents to a coordinator a bottle of the very best Greek wine.",
  "Solution_12": "I think that (almost) the same coordinators will be present this year at the BMO as they were in 1999 at the IMO in Romania. So if Fedor Petrov got 7p points using Lagrange' multipliers (correctly!) I see no reason for you to fear :)\r\n\r\nBTW, Fedor, AFAIK it is not forbidden to speak to the leader [b]after[/b] the two examinations  :?",
  "Solution_13": "Well, Firstly i won't participate in B.M.O or I.M.O . I'm a first year student at a university. (But i did participate in 2004). But i'm still intrested in such competitions.\r\nAs far as Lagrange' multipliers is conserned it is true that it's well know method for proving inequalities. So the coordination comitee should accept it. \r\nBut what about the \"n-1 E.V.M of Vasc \" ? Can they still accept a proof that uses this method ? \r\n\r\nAnd generally, what are the limits of this  ? Can the participants use any know theorem ? any theorem ? and if they can ? should they? (i meant that should the coordinators accept such proofs ? or they must put some restrictions ? )\r\n\r\nValentine you are a coordinator ? (of at least one competition  ;) ) What's your opinion?",
  "Solution_14": "There are no general rules. But it is always better to formulate correctly all not-too-much-basic things, and outline their proofs. Usually, you should ask jury on an olympiad. In Russia, combinatorial and geometrical facts are appreciated rather than analytical and algebraical.",
  "Solution_15": "[quote=\"Fedor Petrov\"]\nWe also have to consider a case, in which some of variables vanish (as we did not have to consider in Fedja's case), it is the work of the same kind.[/quote]\nActually this case can really cause some trouble. Seems like I just overlooked it when saying that the general Vasc's result can be proved the same way :blush:.  For instance, the minimum of $x^3+y^3+z^3$ over non-negative $x,y,z$ with $x+y+z=3$, $x^2+y^2+z^2=5$ is attained on the triple $(0,1,2)$ because $2(x^3+y^3+z^3)=3(x+y+z)(x^2+y^2+z^2)-(x+y+z)^3+6xyz$, in which expression everything is fixed except $xyz$, which is always non-negative and vanishes on the triple $(0,1,2)$. So, Vasc's (n-1)EV should be formulated more accurately: there is a possibility that a few variables vanish and, among the others, all but one are equal.\nAlso, some variables can escape to $+\\infty$ in the similar way (if $p,q<0$). Still, even with this correction, (n-1)EV remains very nice.\n\n[quote=\"Anto\"] And generally, what are the limits of this ? Can the participants use any know theorem ? any theorem ? and if they can ? should they? (i meant that should the coordinators accept such proofs ? or they must put some restrictions ? ) [/quote]\r\nI used to be on both sides of math. competitions and my impression is that the limits are set by coordinators just using common sense. There are no written rules for it. Of course, any theorem whose full proof is included will be accepted. Also, usually there is a pool of theorems that constitute \"common knowledge\". Again, there is no fixed list and the pool heavily depends on the particular contest, but Vasc's (n-1)EV is definitely not in that pool for any contest I know of and I doubt it'll be there soon (except, maybe, for the Mathlinks contest, where Valentin will, probably, allow it from now on, though you'd better ask him first). Everything outside that pool is usually considered requiring a proof though partial credit can be given in some border cases.",
  "Solution_16": "I wonder how many people except those from mathlinks know Vasc's invention. It is a great thing. Yet, I hope not to see applications of it in mathlinks contest, since it would be just as a homework.  ;)",
  "Solution_17": "Ops. Indeed, I was too much self- and Vasc- assured, and did not check it carefully. But I think that either minimum, or maximum may not be achieved for null values of some variables (that depends of convexity/concavity).",
  "Solution_18": "You are right fedja, the (n-1)EV theorem needs some correction. Thanks. :? \r\nThe corrected theorem ends as follows:\r\n(a) If $pq \\le 0$, then $F$ is minimal for $x_1\\le x_2=x_3=...=x_n$, and is maximal for $x_1=x_2=...=x_{n-1}\\le x_n$;\r\n(b) If $p>0$ and $q>0$, then $F$ is maximal for $x_1=x_2=...=x_{n-1}\\le x_n$;\r\n(c) If $p<0$ and $q<0$, then $F$ is minimal for $x_1\\le x_2=x_3=...=x_n$.\r\n\r\nAny new note is welcome.",
  "Solution_19": "In addition to the [b](n-1)EV Theorem[/b]:\r\n\r\n[i]For $p>0$ and $q>0$, $F$ is minimal for $0=x_1=...=x_{k-1}<x_k\\le x_{k+1}=...=x_n$, where $1\\le k\\le n$.[/i]\r\n\r\n[b]Remark.[/b] Replacing all $x_i$ by $1/x_i$, the case $p<0$ and $q<0$ reduces to the case $p>0$ and $q>0$.",
  "Solution_20": "[quote=\"fedja\"]This solution applies to the more general case considered by Vasc too, but gives only the same conclusion about just 2 different values.[/quote]\r\n\r\nSorry, but I don't understand in what variable is the RHS linear (in the general case). could you explain a little?",
  "Solution_21": "Can someone answer my question?",
  "Solution_22": "I sent a paper concerning EV-Theorem to be published in Gazeta Matematica A, in English.\r\nI hope it will appear in 3-4 months.\r\nIn Attachment is the last form of the statement.",
  "Solution_23": "thanks.Vasc :)",
  "Solution_24": "what is \"is minimal for \" means? does it nmeans that F(x) can attain its maximum and when it attain its maximum, must have n-1 equal variables ?"
}
{
  "Tag": [
    "quadratics",
    "limit",
    "algebra",
    "binomial theorem"
  ],
  "Problem": "a) Prove that the values of $x$ for which $x=(x^2+1)/198$ lie between $1/198$ and $197.99494949\\cdots$.\r\n\r\nb) Use the result of problem a) to prove that $\\sqrt{2}<1.41\\overline{421356}$.\r\n\r\nc) Is it true that $\\sqrt{2}<1.41421356$?",
  "Solution_1": "[quote=\"chess64\"]a) Prove that the values of $x$ for which $x=(x^2+1)/198$ lie between $1/198$ and $197.99494949\\cdots$.\n\nb) Use the result of problem a) to prove that $\\sqrt{2}<1.41\\overline{421356}$.\n\nc) Is it true that $\\sqrt{2}<1.41421356$?[/quote]\r\n\r\n[hide=\"a\"]\n\n$x=\\frac{(x^2+1)}{198}$\n$198x=x^2+1$\n$0=x^2-198x+1$You continue then.[/hide]\n[hide=\"c\"]$\\sqrt{2}\\equiv 1,4142135623730950488016887242097...$. Then, the statement is false.[/hide]",
  "Solution_2": "um, do you honestly think that he would post problems you can just plug in on a calculator in Pre-Olympiad? ;) \r\n\r\n[hide=\"hint\"]Extended Binomial Theorem should do the trick.[/hide]",
  "Solution_3": "[hide=\"a\"]\n\nWe see that $x^2-198x+1=0$.  \n\nThe minimum of a quadratic is at $\\frac{-b}{2a}=\\frac{198}{2}=99$.  $99^2-198*99+1=-9800$ and $\\displaystyle\\lim_{x\\rightarrow \\infty}x^2-198x+1=\\displaystyle\\lim_{x\\rightarrow -\\infty}x^2-198x+1=\\infty$, so the quadratic has two roots. \n\n$\\frac{1}{198}^2-198(\\frac{1}{198})+1=\\frac{1}{39204}>0$, so both roots must be $>\\frac{1}{198}$.  \n\n$197.9949494949\\ldots=\\frac{39203}{198}$, and $\\frac{39203}{198}^2-198(\\frac{39203}{198})+1=\\frac{39203^2}{39204}-39203+1=1-\\frac{39203}{39204}>0$, so both roots are $<197.99494949\\ldots$, as desired.\n\n  [/hide]",
  "Solution_4": "[quote]m, do you honestly think that he would post problems you can just plug in on a calculator in Pre-Olympiad?[/quote]\r\n\r\nHuh?  Does  truthfulness of $1+1=2$ depend on \"in which forum the problem is posted?\" :)  \r\n IOW:  The problem is VERY easy, easy enough for a simple calculator (or  few seconds of  simple arithmetic even without a calculator).\r\n\r\nSure,  there may be a fun or a clever way to do this  problem, but there is no reason to think  that one has to have a  particular method just because a problem is in any particular forum.",
  "Solution_5": "[quote=\"E^(pi*i)=-1\"][hide=\"a\"]\n\nWe see that $x^2-198x+1=0$.  \n\nThe minimum of a quadratic is at $\\frac{-b}{2a}=\\frac{198}{2}=99$.  $99^2-198*99+1=-9800$ and $\\displaystyle\\lim_{x\\rightarrow \\infty}x^2-198x+1=\\displaystyle\\lim_{x\\rightarrow -\\infty}x^2-198x+1=\\infty$, so the quadratic has two roots. \n\n$\\frac{1}{198}^2-198(\\frac{1}{198})+1=\\frac{1}{39204}>0$, so both roots must be $>\\frac{1}{198}$.  \n\n$197.9949494949\\ldots=\\frac{39203}{198}$, and $\\frac{39203}{198}^2-198(\\frac{39203}{198})+1=\\frac{39203^2}{39204}-39203+1=1-\\frac{39203}{39204}>1$, so both roots are $<197.99494949\\ldots$, as desired.[/hide][/quote]\r\nI believe you meant $1-\\frac{39203}{39204}>0$.",
  "Solution_6": "[quote=\"Gyan\"]but there is no reason to think  that one has to have a  particular method just because a problem is in any particular forum.[/quote]\r\n\r\nThe point, I think, is that both problems are trivialized if one bothers to actually calculate with a calculator!",
  "Solution_7": "[quote=\"t0rajir0u\"][quote=\"Gyan\"]but there is no reason to think  that one has to have a  particular method just because a problem is in any particular forum.[/quote]\n\nThe point, I think, is that both problems are trivialized if one bothers to actually calculate with a calculator![/quote]\r\n\r\nSorry, but I can solve problems in any way I wanted. The thing to solve it. I don't \"bother\" because I use methods you don't like.",
  "Solution_8": "I hope it increases your self-confidence to go around trivializing olympiad problems by using your calculator. :)",
  "Solution_9": "[quote=\"Jos\u00e9\"][quote=\"t0rajir0u\"][quote=\"Gyan\"]but there is no reason to think  that one has to have a  particular method just because a problem is in any particular forum.[/quote]\n\nThe point, I think, is that both problems are trivialized if one bothers to actually calculate with a calculator![/quote]\n\nSorry, but I can solve problems in any way I wanted. The thing to solve it. I don't \"bother\" because I use methods you don't like.[/quote]\r\n1. You have already been proven false on your calculator in another thread.\r\n2. You don't have one e.g. at IMO.",
  "Solution_10": "Jose, sure you can solve it any way you want, but honestly, what high degree of happiness did you gain by solving this problem with a calculator?  It all comes down to solving a quadratic, which is not hard to do if you know the quadratic formula...it's almost as if chess64 should restate the problem saying that someone gave him the problem(s) posted, and since he doesn't have a calculator, he need someone else who has a calculator to solve it for him (oh yes, and first one to post the answer wins!)",
  "Solution_11": "Basically, you have contributed nothing of any mathematical significance whatsoever. Before posting, Jose, please make sure that your either (a) your post contains informative value or (b) your post will interest or be useful to someone.",
  "Solution_12": "In general, the assumption on the main AoPS boards (i.e. the ones for problem solving math like GS, Intermediate, Pre Olympiad, and all the Olympiad forums) is that your solution should not invoke the use of a calculator unless the person who posted the problem specifically said you can't.\r\n\r\nWhen a problem is trivialized by the use of a calculator, you should think to look for a more elegant method.  Additionally, whenever you see the source of a problem as ---- Math Olympiad, chances are that calculators are not allowed.\r\n\r\nAs an aside, I'd like to note that historically mathematicians have been displeased by proofs which require calculators/computers (most famously...the four color theorem).",
  "Solution_13": "[quote=\"ZetaX\"]\n1. You have already been proven false on your calculator in another thread.\n2. You don't have one e.g. at IMO.[/quote]\r\n\r\n1. are you perfect???? can't you make a mistake?\r\n2. and? My life is not only an IMO\r\n\r\nwhat's wrong with you? what do you have against me, man?",
  "Solution_14": "He's just saying that it is obvious you can do it with a calculator, so we want a better way. A monkey could use a calculator. So please stop.",
  "Solution_15": "[quote=\"Jos\u00e9\"][quote=\"ZetaX\"]\n1. You have already been proven false on your calculator in another thread.\n2. You don't have one e.g. at IMO.[/quote]\n\n1. are you perfect???? can't you make a mistake?\n2. and? My life is not only an IMO\n\nwhat's wrong with you? what do you have against me, man?[/quote]\r\nZetaX just wants ppl not to decrease the quality of the forum. Consider other ppl who want to see a proof of a problem. They will not learn much if they just see a solution which requires pressing keys on a calculator instead of using mathematical logic. And ZetaX is pretty ownage in my opinion.",
  "Solution_16": "[quote=\"Jos\u00e9\"]\n1. are you perfect???? can't you make a mistake?\n2. and? My life is not only an IMO\n\nwhat's wrong with you? what do you have against me, man?[/quote]\r\n\r\nHe, I, and many other people are annoyed that you continue to mess up perfectly good threads by posting uselessly.\r\n\r\nConsider NOT POSTING AT ALL if you don't have anything useful to contribute (for your purposes, consider 'something useful' to be a complete, well written solution).",
  "Solution_17": "[quote=\"Jos\u00e9\"]1. are you perfect???? can't you make a mistake?[/quote]\nNo, I'm very far away from being perfect.\n[quote=\"Jos\u00e9\"]what's wrong with you? what do you have against me, man?[/quote]\nNothing against you, but only against your posting/problem solving habit.\n\n[quote=\"rem\"]ZetaX just wants ppl not to decrease the quality of the forum.[/quote]\r\nYes, but I failed  :(",
  "Solution_18": "Sorry for reviving an old post, I didnt know where else to put this. I found another solution for a. Could you guys see if it's good?\r\n\r\n[hide]We can manipulate the equation to yield\n$ x^2 \\minus{} 198x \\plus{} 1 \\equal{} 0$\n\nSince it's a quadratic, it must have two roots. Let's let\n$ f(x) \\equal{} x^2 \\minus{} 198x \\plus{} 1$\n\nWe know that for two numbers $ a$ and $ b$, $ a < b$, if $ f(a) < 0$, and $ f(b) > 0$, or if $ f(a) > 0$, and $ f(b) < 0$ then between $ a$ and $ b$ there must be a root.\n\n$ f(\\frac {1}{198}) > 0$. $ f(1) < 0$. So, we know there is a root of $ f(x)$ between $ \\frac {1}{198}$ and $ 1$.\n\nWe find that $ 197.994949\\cdots$ is $ 198 \\minus{} \\frac {1}{198}$. $ f(198 \\minus{} \\frac {1}{198}) < 0$.r\n\n\n$ f(198) > 0$. There must be a zero between these two also.\n\nQED\n[/hide]\r\n\r\nAny comments or suggestions would be greatly appreciated.\r\nThanks!"
}
{
  "Tag": [
    "MIT",
    "college",
    "Harvard",
    "Princeton",
    "Stanford",
    "Support",
    "Gauss"
  ],
  "Problem": "What is the strongest math program in the nation?\r\n\r\nAccording to US News Ranking it is...\r\n\r\n1. MIT\r\n2. Harvard, Princeton, Stanford, UC-Berkeley\r\n\r\nBut.. how legitimate is the provided ranking?\r\n\r\nAlthough it might be different for each field of mathematics, here we are primarily concerned with 'general' reputation of school and 'overall' image of the school.",
  "Solution_1": "It is very difficult to say which is the best. Each of these schools have different but very strong math programs. I know the most about MIT, Harvard, and Princeton (based on talking with friends who are there), so I will discuss those.\r\n\r\nHarvard has Math 55. This class is basically in a class of its own (no pun intended). Most of the people I know are Freshmen, so beyond that I don't know what they have.\r\n\r\nPrinceton, from what I have heard, probably has the best staff of any that you listed, if you are going based on contributions to super-famous modern developments in mathematics (I believe everyone involved in the proof of Fermat's Last Theorem is part of their faculty, for one thing).\r\n\r\nMIT seems to have a student culture that supports mathematics (the same can be said of any of the above schools, but from what I have heard, the culture at MIT is particularly intense among the mathy folks). They also let you take any classes you want.",
  "Solution_2": "I'm just taking a reasonable guesss here, but I'm saying that there's a difference between having the overall best program for math in the country or having the best certain course for math in the country. I'm sure if people come to a consensus that MIT is the best, that doesn't mean that all of their classes are the best in the country...I mean, I'm almost positive Harvard would have probably a few up there.\r\n\r\n-jorian",
  "Solution_3": "Hmm, I would not have expected Stanford's math program to be up there, but I'm very glad it is :P.",
  "Solution_4": "[quote=\"JSteinhardt\"]It is very difficult to say which is the best. Each of these schools have different but very strong math programs. I know the most about MIT, Harvard, and Princeton (based on talking with friends who are there), so I will discuss those.\n\nHarvard has Math 55. This class is basically in a class of its own (no pun intended). Most of the people I know are Freshmen, so beyond that I don't know what they have.\n\nPrinceton, from what I have heard, probably has the best staff of any that you listed, if you are going based on contributions to super-famous modern developments in mathematics (I believe everyone involved in the proof of Fermat's Last Theorem is part of their faculty, for one thing).\n\nMIT seems to have a student culture that supports mathematics (the same can be said of any of the above schools, but from what I have heard, the culture at MIT is particularly intense among the mathy folks). They also let you take any classes you want.[/quote]\r\n\r\nmmm, ribet is at berkeley",
  "Solution_5": "The math program at any school is only as good as the students, and the students are only as good as the effort they put in. Many schools offer accelerated classes, accomplished faculty, and so forth, and most, if not all, have a library. Unlike other sciences, high quality facilities don't mean as much. \r\n\r\nIn short, all those math programs are great and who cares which one is best. It's just between the student and the math, the math in the books, and his/her peers.",
  "Solution_6": "I wonder which people have voted here. MIT is great but I do not see why it should have a much better math program than, say Berkeley. It seems people have rather voted for the school they like best. Go to the respective course lists and compare.",
  "Solution_7": "I haven't voted yet since I don't really know...\r\n\r\norl, people can vote for a school without thinking that school is [i]much[/i] better than the others.  They may think the programs are really close yet one is slightly better.\r\n\r\nOn the other hand, there are a lot of factors which would probably need to be considered when figuring out which one is the best.  I guess mostly what should be considered are the calibre of the students in the math department, the quality of the teachers, and the classes and how well they are taught.\r\n\r\nIt seems to me that the strongest student body is at MIT.  The other factors I'm not so sure about.  But basically, I think the math programs at Harvard, MIT, and Princeton are about equivalent in terms of what math undergraduates can get out of them.  It doesn't seem to me that math majors at one of these schools have a huge advantage over math majors at another school.  If you're trying to decide between the top schools, I would place much more of an emphasis on other factors such as geography, surroundings, and the required curriculum (do you want a solid science background or an exceptional humanities one?).",
  "Solution_8": "What is the meaning of \"strongest\" in this context? I could imagine that different schools would be \"best\" for different students.\r\n\r\nI think its human nature to try to rank things -- it makes things easier to analyze. But I don't think everything (including the strengths of different math departments) has a logical ordering. Every student is different, and every student will have a different experience a given school.\r\n\r\nIf I were forced to try to make some sense out of this question though, I would probably look at what recent graduates from those school's departments have gone on to do, and see which ones have accomplished things that seem appealing to you. For example, what graduate schools did they go to and where are they working now? That information might be hard to find though. The best way to find it would probably be to talk to a current student or recent graduate of the school.",
  "Solution_9": "For strongest student body I suppose you could look at the quantity and quality of the math students at each school.  This is also hard to determine (to the accuracy needed to declare a student body as the strongest).  So even though my statement in my last post that MIT has the strongest body was an opinion, it was perhaps a little naive.",
  "Solution_10": "What exactly is Harvard math 55? I read the topic called that but it didn't really explain it.",
  "Solution_11": "It's a freshman math course at Harvard.  [url=http://www.math.harvard.edu/undergrad/Pamphlets/freshmenguide.html]Here's[/url] one site that gives a brief description of it along with other freshman math options at Harvard.",
  "Solution_12": "This debate probably won't get anywhere. In order to find out how each experience is, you'd need to have some guy attend each of those schools and take a bunch of classes in each, but even so it'd be biased because some schools have better this, some schools better that. Who cares.",
  "Solution_13": "The US News Ranking you quote there is only a ranking of the graduate programs, not of the undergraduate ones.",
  "Solution_14": "The revival of this thread from last year shows that this is a topic of continued interest.",
  "Solution_15": "[quote]I believe everyone involved in the proof of Fermat's Last Theorem is part of their faculty, for one thing[/quote]\r\n\r\nYes but Andrew Wiles, the man who is most acredited with proving the Tanyama-Shimura conjecture and thus Fermat's last theorem attended Cambridge university. He also chose to announce the proof at the Sir Isaac Newton institute in Cambridge. I would say be adventurous and travel abroad to England and follow in the footsteps of Sir Isaac Newton and in the footsteps (perhaps \"wheel-marks\") of Stephen Hawking (who holds the post of Lucasian professor that Newton once held). However, I am rather biased as I will be going to read maths at Cambridge in a month. I would have applied to Harvard if i were living in America.",
  "Solution_16": "\"These courses are not taught in any other university because no other university has the same caliber of first-year mathematicians\"\r\n\r\nI saw this in a description of math 25 and 55 on Harvard's math dept. website. Does anyone else find this a little pretentious?",
  "Solution_17": "Is it pretentious if it's true?",
  "Solution_18": "[quote=\"curl\"]\"These courses are not taught in any other university because no other university has the same caliber of first-year mathematicians\"\n\nI saw this in a description of math 25 and 55 on Harvard's math dept. website. Does anyone else find this a little pretentious?[/quote]\r\n\r\nWhere did you see this on the website and do you know whom it was written by?\r\n\r\nI happen to disagree with the statement as its written for several reasons. First of all, because I can think of several other schools who's top undergraduate students have comparable mathematical ability to Harvard's students.\r\n\r\nSecondly, I disagree with the supposition that such courses are the only ideal introduction to undergraduate mathematics for top students.",
  "Solution_19": "The quotation comes from \r\n\r\nhttp://www.math.harvard.edu/pamphlets/courses.html",
  "Solution_20": "I strongly agree with both of gauss202's points. I think this kind of language, presented by an official body, is a serious issue. The elitist intellectual attitude is very easy to develop when you are a student at a top university, and these universities should be making efforts to counter that effect, not to encourage it.",
  "Solution_21": "I heard that Princeton has a really good math program and physics program.\r\nI heard that one professor is considered one of the top 5 physicists of all time.",
  "Solution_22": "Which professor?\r\n\r\nNewton, Maxwell, Einstein already have three of the top 5 pretty easily...\r\n\r\nSchrodinger is also almost certainly in the top 5, and then there's people like Ampere, Gauss, and Planck who, if not in the top 5, are certainly contenders. And as pointed out, Stephen Hawking is at Cambridge unless that changed at some point.",
  "Solution_23": "If you mean a current professor, perhaps it's Witten (Fields medalist)? Or Taylor (Nobel laureate)? \r\nhttp://en.wikipedia.org/wiki/Edward_Witten\r\nhttp://en.wikipedia.org/wiki/Joseph_Hooton_Taylor,_Jr.",
  "Solution_24": "What is University of Chicago ranked in mathematics? I have heard good things about that school as well.",
  "Solution_25": "I wanna add that its not all about the university you go to. It has to more do with amount of work you put into your studies. Yes, Princeton is one of the best but if you are a lazy type of guy its not gonna do any good. Similarly if you go to an ordinary college but you are very serious about mathematics you can sometimes get departmental approval to let you study you own advanced stuff not taught in college. Going to a top university will help you get better but you can avoid it altogether (not that I am saying that is what you should do). That is why Karl Weierstrass is a very big inspiration to me because he never got his doctrate and yet earned the immortal title \"father or analysis\".",
  "Solution_26": "Unless Prof. Jesus Christ Superstar Einstein IV is going to be your personal mentor, why exactly is it critical to your undergraduate education if you might occasionally pass by him on campus or have him be doing research you could never understand or even teach one of your classes that any qualified math professor would grasp to the same extent?"
}
{
  "Tag": [
    "algorithm"
  ],
  "Problem": "Here is a problem that took me quite a while...\n\n\n\nThere are twelve pennies. Among the 12 pennies there is \n\none bad penny that either weighs less or more than the \n\nnormal pennies. You have a scale that tells you whether one \n\nside is heavier than the other or not. \n\nFind the ultimate solution to find the bad penny in 3 steps.\n\n(meaning using the scale only 3 times...)\n\nConsider all the possibilities...\n\n\n\nIf you have seen this problem before, and solved it\n\ngive some chance for other people.\n\n\n\n[hide] And Don't cheat...[/hide]",
  "Solution_1": "[hide]Put 6 on each side.  Set aside the six on the side that weighs more.  With the six remaining, split them into two groups of three, and weigh those.  Set aside the three that weigh more.  Take the last three, and pick two from there.  Put one on each side of the scale.  If they are equal, the one that you set aside is the light one.  If they are different, the lighter one is the light one.[/hide]",
  "Solution_2": "But you don't know whether the bad penny is lighter or heavier.",
  "Solution_3": "Magnara, I think you misunderstood the problem. The \"bad\" coin isn't necessarily lighter than the others, and you don't know if it's lighter or heavier when you start the weighings. This makes the question soooo much more fun.... and soooo much harder!",
  "Solution_4": "I think I really need to start actually reading these problems.  And if I think it's too easy for advanced, it probably is....",
  "Solution_5": "Having Fun?\n\n[hide]\n\nThere is more than 1 answer to this problem.\n\nAnd don't cheat!\n\nAnd have fun w/ it!\n\n[/hide]",
  "Solution_6": "[hide]Divide 12 coins into 4 piles a,b,c,d, each pile has 3 coins[/hide]\n\n\n\nAm I at least on the right track?",
  "Solution_7": "hmm... you might be...\r\n\r\nI've done the problem w/o any specific algorithm\r\nbut i've heard that it is possible to approach the problem \r\nmathematically.",
  "Solution_8": "This doesn't quite work out...but here's what I did.\n\n\n\n[hide]Divide into three piles of four. Weigh two piles. \n\n\n\nCase 1: If they are the same, the bad penny must be in the third pile. Weigh any two of the pennies in the last pile.\n\n\n\nSubcase 1: Weights are equal. Remove one penny and replace with any other penny in the last pile. If they are still both the same, the other penny left is the bad penny. If they are different, the one replaced is the bad penny.\n\n\n\nSubcase 2: Weights differ. Remove any one penny, and replace with another penny. If weights still differ, the penny not removed is the bad penny. Otherwise, the penny removed is the bad penny.\n\n\n\nCase 2: Two piles are of unequal weight. Err...unless I'm mistaken, this requires four tries. Can anyone come up with fewer?\n\n[/hide]",
  "Solution_9": "It might seem that unequal case needs 1 extra step but\r\nit can be done within 3 steps as the problem is stated.",
  "Solution_10": "I got four tries too using my way, but the question asks for three.\n\nIs any of us on the right track?\n\n\n\n[hide]Divide 12 coins into 4 piles a,b,c,d, each pile has 3 coins\n\nTake any a and b piles and weigh them.\n\n  If they are balanced, then none of them have the bad one.\n\nThe bad one is in pile c or d.  Take pile a and compare it with pile c.   \n\n     If they are balanced, then bad one is in pile d. Take any two of the 3 in pile d and compare them. if they balance, then good, the other one is the bad. If they don't, we  have a problem.that takes another two tries.\n\n\n\n     If a and c don't balance, means one of the 3 in pile c is bad. we only need one try by taking any 2 of 3 and compare them. we already know the bad one is heavier of lighter by comparing a, c\n\n\n\nIf a and b do not balance. compare a to c, since all 3 in pile c are normal.  if they balance, then the bad one is in b, but we do not know if it is heavier or lighter, so it takes another 2 tries\n\n if they don't we know the bad one is in pile a and it's heavier of lighter. only take one more try[/hide].",
  "Solution_11": "I did it too by weighing groups of 4 in the first step.\r\nThere is a trick involved in the case when the first measurement\r\nisn't equal.",
  "Solution_12": "[quote=\"xirtamax\"]I did it too by weighing groups of 4 in the first step.\nThere is a trick* involved in the case when the first measurement\nisn't equal.[/quote]\r\n\r\n*Sneak in a fourth weighing while no one's looking.",
  "Solution_13": "lol\r\n\r\nif we do it that way, we can get it in 2 steps.",
  "Solution_14": "I was just found this site tonight and this will be my first post on it :)  I thought up a solution but it has one step that might not be considered valid.  It is a creative step though....\n\n[hide]\n\nWeight 4 vs. 4.  If they are equal we know the bad penny is in the last group of 4.  Weigh 1 vs. 1 from the last group.  If not equal.  Weight either one of those 2 with any other penny.  If it is equal it was the one of the 2 we did not wiegh.  If not equal it was the one we did.  If that 1 vs. 1 was equal, then that tells us the same information as weighing the other 2 and them not being equal, so follow the above last step on those 2 that we know aren't.\n\n\n\nDang I need to learn to write shorter, but anyways...\n\n\n\nIf the initial 4 vs. 4 was not equal, then steal 2 pennies from Bill Gates and assume them to be normal weighted pennies.  This is the step that might not be considered valid.  Take any 3 of the penies on the left side and any 3 on the right side of the 4 vs. 4 we just did, and put them on the same side.  Then take the 4 from the group we know is right, and our stolen 2 on the other.\n\n\n\nIf this 6 vs. 6 is equal, then one of the 2 excluded - from this test is the bad one.  Just weight one of them aginst a good one, if its not equal its that one, if it is equal its the other.\n\n\n\nIf this 6 vs. 6 is not equal.  We just gained some valuable info.  We know if the bad penny is heaver or lighter than a normal penny.  So...if it is heavier, then we know the bad penny was on the side of the 4 vs. 4 test that was heavier.  So we know it is in one of those 3 pennies that we taken from the heavier side.  Weigh any 2 of those three pennies.  If one is heavier than the other, then it is the bad penny.  If they are equal, then the 3 penny is the bad one.  If from the 6 vs. 6 we determined the bad penny is lighter, then do the same steps as above but with the word lighter in place of heavier.\n\n[/hide]\n\n\n\nThe moral of the story is to try to determine if its heavy or light.  And also that one must steal.",
  "Solution_15": "What the heck.  I told it to set my font color to white on my solution.  But.....its black.  Well, if some one scrolls down this far they are proly looking for the solution anyways.",
  "Solution_16": "[quote=\"Douglas\"]If the initial 4 vs. 4 was not equal, then steal 2 pennies from Bill Gates and assume them to be normal weighted pennies.  This is the step that might not be considered valid.[/quote]\r\n\r\nYou know that the third group of 4 pennies is all normal, so you don't need to steal from Mr. Gates. :)\r\n\r\nAlso, you can use the [spoiler] tag to hide your text. Nice solution, by the way.",
  "Solution_17": "Well, since I can't steal anymore, I did this...\n\n[hide]\n\nOkay, back to if the 4 vs. 4 is not equal.  Take 3 from one side and 1 from the other and weight that aginst the 4 we know are normal in another 4 vs. 4.  If they are not equal.  We know if the bad penny is heavy or light.  If the 1 penny came from the heavy side, its that guy.  Else it is one of those other 3 and you can determine the bad penny in 1 weighing if you have 3 pennies and you know if its heavy or not.  If the 4vs. 4 was light then do same steps above with light in place of heavy.\n\n\n\nSo, what if this 4 vs. 4 is equal.  Then the bad one is either the 1 that was not in the group of 3 we took from one side, or the 3 that were not the 1 we took from the other side.\n\n\n\nTake 1 of those 3 out.  Put 1 of those 3 on the side with the 1.  Then put a normal weighted penny on the side with the 1 of the 3 that is left.  So we have a 2 vs. 2\n\n\n\nIf this is equal, it is the 1 we took out.  If the side with the 1 and the 1 from the 3 is heavy, and the side we got the 1 from was heavy, then it is that 1 that is bad.  If it was the side with the 1 from those 3 that was heavy, then it is that 1 from the 3 that is bad.  If this side is light and the side we got the 3 from was heavy, then it is the penny next to the normal penny that is bad.  And do same stuff with light in place of the word heavy in the above steps.\n\n\n\n\n\nMan...I need to learn to write shorter and clearer.  It would have proly helped if I labed pennies A1,A2,B1,B2.. instead of saying things like \"the 1 from the 3\".  And maybe call sides of the balance S1 and S2.[/hide]",
  "Solution_18": "Douglas said: \r\n[quote]\nIf the side with the 1 and the 1 from the 3 is heavy, and the side we got the 1 from was heavy, then it is that 1 that is bad. \n[/quote]\r\n\r\nHmm... You did a good job, but i think that part is wrong.\r\nThe remaining penny of the hevier side could be the bad penny, \r\nbut the penny on the left side that measured lighter could also be\r\nthe bad penny.\r\nEverything else seem to work.",
  "Solution_19": "Okay I'm about 99% sure this works.  The explanation may be confusing, but bear with me. \n\n[hide]\n\nNOTE: Nv1 is the proccess to determine between 2 that you are unsure about for your final weighing.  You have 2 that you are unsure about.  Measuring 1 of those against a normal penny will give you a definite answer.  Here's why:  If they are equal, the one you didn't measure is the faulty penny.  If they are unequal, you know that the one that is NOT normal is the faulty penny.\n\n\n\nMeasure 4 against 4 first.  That's your 1st weighing.   \n\nA)  Assume they are equal\n\n\n\nYou now know 8 that are definitely normal.  On your 2nd weighing, do 1 against 1 (of those you don't know).  If that was even, then do Nv1 with the remaining 2.  If the 1v1 was uneven, you do Nv1 with those 2.\n\n\n\nNow, if the original 4v4 was unequal, \n\n\n\nB) Measure 3v3 like this:\n\nHHL v HHL. (H = heavier, L = lighter.) This will be your second weighing.  If these are equal, You do the process of Nv1 to measure the 2 remaining pennies.\n\nIf HHL v HHL was uneven, then you are left with 3 pennies that could possibly be faulty.  \n\n\n\nHere's why:\n\nIn the 4v4, one side went down, and one side went up.  Out of those pennies, only one is faulty.  Either the penny is light, and therefore made it's side weigh less, or the penny was heavy, and made its side weigh more.  When you do HHL v HHL, (each letter representing an individual penny), one side will go down, and the other will go up.  Therefore, the only possible faulty pennies are the HH on the heavier side or the L on the lighter side.  If any other penny was faulty, it would both be heavier and lighter than the other pennies.\n\n\n\nSo, with your final weighing, you have HHL.  Do this measurement:\n\n\n\nHL v NN (N represents a Normal penny).\n\n\n\nAnd that's the genius step.\n\n\n\nIf equal, the only questionable penny is the faulty one.  If they are unequal, then you can figure out very easily which is the faulty penny.  IF the HL side goes up, L is the faulty penny.  If the HL side goes down, H is the faulty penny.\n\n[/hide]\n\nI'm done now.  That took me a while.",
  "Solution_20": "Oops, yeah, mine messes up.",
  "Solution_21": "Good Job, Green Twilight.\n\n[hide]\n\nThe trick was to use the elimination method \n\nand the second step had to be HHL and HHL\n\nor LLH and LLH because we have to have only 3 \n\nremaining suspects to complete the problem.\n\n[/hide]\n\nBut I believe there is an another solution. I only was able to\n\nsolve the problem w/ Green's solution, but there is another one.\n\n\n\n(Actually, I was too happy that i solved it, and never went back\n\nto try other methods.  [/hide]\n\n\n\nI forgot to mention, can you put the solution in spoiler.",
  "Solution_22": "[I've edited in a few spoilers. By the way, Douglas, the reason it showed up black was because you had [color=white][/color] then all the text afterwards, instead of in the middle of that].\r\n\r\nThis is a very very famous problem.. and now theres about 1000 variants of it. I've seen the solution a few times, never really tried it properly myself, I get too fed up trying :)",
  "Solution_23": "[quote=\"TripleM\"]This is a very very famous problem.. and now theres about 1000 variants of it.[/quote]\r\nIndeed.  But, now that someone has found a solution, extend the problem.  The next incremental step is to ask not only which coin is cointerfeit, but whether it is heavier or lighter than normal.  All other conditions and requirements remain the same.  Yes, it is possible.",
  "Solution_24": "rcv wrote:\r\n[quote]\nIndeed. But, now that someone has found a solution, extend the problem. The next incremental step is to ask not only which coin is cointerfeit, but whether it is heavier or lighter than normal. All other conditions and requirements remain the same. Yes, it is possible.\n[/quote]\r\n\r\nIf you look at greentwilight's solution, you'll see that \r\nwe can not only figure out the bad penny but figure \r\nwhether it's heavier or lighter than the normal ones.",
  "Solution_25": "[quote=\"xirtamax\"]If you look at greentwilight's solution, you'll see that \nwe can not only figure out the bad penny but figure \nwhether it's heavier or lighter than the normal ones.[/quote]\n\n[quote=\"GreenTwilight\"]NOTE: Nv1 ... If they are equal, the one you didn't measure is the faulty penny.[/quote]\r\n\r\nUsually, we can distinguish heavier vs lighter.  But if GreenTwighlight's first two measurements are even, he uses NV1 to determine which of the final two coins are bad.  How can you know if an unmeasured coin is heavier or lighter?",
  "Solution_26": "Yes, I think if the initial 4v4 was even, then there is a case where you don't know the weight of the penny.",
  "Solution_27": "Ok here we go:\n\n[hide]\n\nHHHH > LLLL\n\n\n\ncase 1:\n\nHHL > HHL\n\nIt's either HH on the left side or L on the right side.\n\nMeasure HL and 2 normal pennies.\n\nIf the HL is greater than it's H.\n\nIf the HL is lighter than it's L.\n\nIf the HL is equal than it's H.\n\n\n\ncase 2:\n\nHHL < HHL\n\nSame as case 1, just sides reversed.\n\n\n\ncase 3:\n\nHHL = HHL\n\nthen we measure one of the 2 remaining L and a normal penny.\n\nIf L is equal to normal than it's the other L.\n\nIf L is lighter than normal than it's that L.\n\n[/hide]",
  "Solution_28": "xirtamax wrote:Ok here we go:\n[hide]HHHH > LLLL[/hide]\n\n\n\nUnless I am not seeing something, I think you are still missing the case where the first two measurement are both in balance.",
  "Solution_29": "Oh I see. Greenlight's solution is different than what i had in mind.\n\nI just looked at the unequal case.\n\nAnd I thought that the equal case is obvious, but I guess it's a little\n\nmore than obvious. (not too hard to figure out though once you\n\nknow how to find the bad penny.) \n\n\n\n[hide]\n\nN = Normal\n\nU = Unknown\n\nH = Possibly Heavy \n\nL = Possibly Light\n\n\n\nMeasure:\n\nNNN and UUU.\n\nCase 1:\n\nNNN < UUU then UUU becomes HHH, then measure \n\nHN and HN. \n\n   If HN > HN, then H on the left is the bad penny.\n\n   If HN < HN, then H on the right is the bad penny.\n\n   If HN = HN, then the third H is the bad penny.\n\nCase 2:\n\nIf NNN > UUU then UUU becomes LLL, then measure \n\nUN and UN.\n\n   Use the same method as case 1.\n\nCase 3:\n\nNNN = UUU, then measure N and U.\n\nU will be figured out.\n\n[/hide]",
  "Solution_30": "[quote=\"xirtamax\"]Oh I see.[/quote]\r\nGood!  I think your revised solution takes care of my issues.\r\n\r\nLet me point out there is one more extension.\r\n\r\n\"There may or may not be a counterfeit coin.\"\r\n\r\nSome say starting with this extra condition actually makes the original problem easier.",
  "Solution_31": "This is a stupid small thing...but isnt\r\n\r\nHN v HN the same as H v H?\r\n\r\nAnyways, this is probably in a tie for the coolest problem I have seen."
}
{
  "Tag": [
    "analytic geometry",
    "graphing lines",
    "slope",
    "inequalities",
    "algebra",
    "linear equation"
  ],
  "Problem": "PROBLEM\r\nPlanning for the prom\r\n\r\nPam and Remy are organizing the junior prom. They plan to sell two types of tickets- individual tickets and couples tickets. The ballroom where the prom will be held holds a maximum of 400 people.\r\n\r\nPam is pretty opinionated. She believes the prom will be more successful when more people go as couples. So that at least half the people at the prom will be in couples, Pam and Remy decide that the number of individual tickets they sell should be at most twice the number of couples tickets.\r\n\r\nRemy ordered door prizes to be handed out at the prom (one prize for each ticket, even if it's a couples ticket). Unfortunately, the supplier to get more, so they can sell a total of at most 225 tickets.\r\n\r\nVersion a: Suppose individual tickets sell for 22 dollars and couples tickets sell for 30 dollars. HOw many of each type of ticket should they sell to maximize the money they take in?\r\n\r\nVersion b: ...individuals ticket 15 dollars and couples ticket 35 dollars...?\r\nVersion c: ...individuals ticket 30 dollars and couples ticket 20 dollars...?\r\n\r\nThis problem is too wordy and confusing. So there's at most 225 tickets and the number of couples ticket is x and individuals ticket is 2x... How do you find how they can maximum the take-in???\r\n\r\nThere's graphing involve but I am not quite sure how to do that.\r\n\r\nI need help with this as quick as possible. Thank You!",
  "Solution_1": "this isnt a place were you come and ask for homework help, if you needed help so badly, im sure you could find out how to do it if you cared enough, you obviously dont and you want us to do it for you...",
  "Solution_2": "Actually, I disagree - this is a great place to get homework help.  What's the point of a community if people can't get help from it?",
  "Solution_3": "i'm stupid.",
  "Solution_4": "Whenever tackling a word problem, the first thing you should do is figure out what the problem is asking for.  So let's take a look at Version A: Suppose individual tickets sell for 22 dollars and couples tickets sell for 30 dollars.  How many of each type of ticket should they sell to maximize the money they take in?\r\n\r\nSo it's an optimization problem.  First, let's define a few variables.  Let's say $x =$ the number of individual tickets they sell.  Since the maximum number of tickets they can sell is 225, the number of couples tickets they can sell is $225 - x$.  Therefore, the revenue from the sale of tickets is:\r\n\r\n$R = 22x + 30(225 - x)$\r\n\r\nSimplify this linear equation:\r\n\r\n$R = 6750 - 8x$\r\n\r\nBecause this is a line and it has a negative slope, the obvious maximum value would be at $x = 0$ since you can't sell a negative amount of tickets.  However, there is a restriction on $x$.  If $x = 0$, then there would be 225 couples tickets sold.  For every couples ticket there are two people going, meaning there would be a total of 450 people at the ball.  The maximum capacity is 400 people.  So $x$ cannot be 0.  We need to find the number of individual tickets to sell so that the number of people going is exactly 400 in order to maximize the revenue.\r\n\r\nUsing our original variable $x$, the total number of people attending would be:\r\n\r\n$P = x + 2(225 - x) = 450 - x$\r\n\r\nThe information given about maximum capacity creates the inequality $450 - x \\le 400 \\rightarrow 50\\le x$, but we want the minimum value of x, which is $x = 50$.\r\n\r\nIn order to maximize revenue, 50 individual tickets and 175 couples tickets should be sold.  This does not violate the other condition where the number of individual tickets can be at most twice the number of couples tickets sold.\r\n\r\nUsing this approach, you should be able to solve the other versions as well."
}
{
  "Tag": [
    "geometry",
    "calculus",
    "calculus computations"
  ],
  "Problem": "Hello,\r\nI am trying to graph the mapping of the transformation u=xcosy and v=xsiny where 1<=x<=2 and 0<=y<=1.\r\nI have some trouble sketching the area, basicly on the fact that i reach the step where i find that 0<=arctan(v/u)<=1. Does this affect the shape of the region that corresponds to the mapping?\r\nThanks",
  "Solution_1": "if we had $ 0 \\le y \\le 2\\pi$ we would get a washer\r\n\r\nlooks nicer when $ y\\equal{}\\theta, x\\equal{}r$\r\n\r\nso instead of the full circle, we only get part of a washer,"
}
{
  "Tag": [
    "probability"
  ],
  "Problem": "if a player has  2/5 chance of earning a \"walk\" on each up-to- bat, what is probability that earn a walk exactly once in next two plate appearances?\r\n\r\nanswer is 12/25 but how do you get that?",
  "Solution_1": "[quote=\"epatjn\"]if a player has  2/5 chance of earning a \"walk\" on each up-to- bat, what is probability that earn a walk exactly once in next two plate appearances?\n\nanswer is 12/25 but how do you get that?[/quote]\r\n\r\n[hide]Let W be a walk and N be not a walk.\n\nThese are the combinations that can happen in two at bats.\n\nW,W\nW,N\nN,W\nN,N\n\nOf those, W,N and N,W have the property that there is exactly one walk (W) in the two at bats.\n\nThe chance of getting W,N is\n$\\frac{2}{5}\\cdot \\frac{3}{5}= \\frac{6}{25}$\n\nThe chance of getting N,W is\n$\\frac{3}{5}\\cdot \\frac{2}{5}= \\frac{6}{25}$\n\nThus, the chance of getting one walk in two at bats is\n$\\frac{6}{25}+\\frac{6}{25}= \\frac{12}{25}$.\n[/hide]",
  "Solution_2": "how did you get \r\nThe chance of getting W,N is \r\n3/5?  \r\n\r\ni get the options of \r\nw,w\r\nw,n\r\nn, n\r\nn, w",
  "Solution_3": "The chance of getting W,N is $\\frac{6}{25}$.\r\n\r\nI got this by multiplying $\\frac{3}{5}\\cdot \\frac{2}{5}$,\r\nbecause this means, \"the probability that this event will happen, then this event\".\r\n\r\nIn general, if we have two probabilities, $P_{1}$ and $P_{2}$, the probability that $P_{1}$ will occur, then $P_{2}$ will occur is $P_{1}\\cdot P_{2}$."
}
{
  "Tag": [],
  "Problem": "My first ever math contet comes up next sarturday.i need tips on (Algebra,combinatorics,tringles,bearings,compoud anglesfactorisation and any othrt topic you know i have not mentioned).I realy need what they call math contest tips.",
  "Solution_1": "Reading the AoPS books would have helped but now it's too late.  Practice problems and get a good night's sleep.",
  "Solution_2": "[quote=\"21 7 15\"](Algebra,combinatorics,tringles,bearings,compoud anglesfactorisation and any othrt topic you know i have not mentioned)[/quote]\r\n\r\nOh man that's quite a list. Work over basics if you know them; if you don't, like what JRav said, it's too late to learn for the competition."
}
{
  "Tag": [
    "function",
    "calculus",
    "limit",
    "algebra proposed",
    "algebra"
  ],
  "Problem": "This one seems to be a criterion of continous function:\r\nLet f be a surjective function on [0,1] and f is increasing.Prove that f is continous on [0,1].\r\n...Any role in this one?",
  "Solution_1": "This probably belongs in the calculus-analysis forum; it's not algebra. I also assmue that the condition is that $f$ maps [0, 1] onto [0, 1] (or to some closed bounded interval). But yes, it is true.\r\n\r\nNote that since $f$ is increasing, it is injective, which means that it is bijective and has an inverse. Suppose $f(a)=b$, and suppose for the time being that $a,b\\in (0, 1)$. Choose $\\epsilon >0$ arbitrary except for the condition that it is small enough that $b+\\epsilon$ and $b-\\epsilon\\in [0,1]$. Then let $u=f^{-1}(b+\\epsilon)$ and $v=f^{-1}(b-\\epsilon)$. Since $f$ is increasing, $v<a<u$. Let $\\delta = \\min(u-a,a-v)$. Then, $|x-a|<\\delta\\implies v<x<u\\implies b-\\epsilon<f(x)<b+\\epsilon$. Thus, $f$ is continuous at $a$. A slight modification of the same argument shows that $f$ is continous at 0 and at 1.\r\n\r\nThis is a standard argument, and is in a number of textbooks. It's particularly useful for showing that the inverse of a continuous function from an interval to an interval is also continuous. This theorem is a companion (and partial converse) to the Intermediate Value Theorem.",
  "Solution_2": "I bet you can find this criterion in any analysis book you can get your hands on :).\r\n\r\nIt uses the fact that every monotonic function has lateral limits in any point, and if $\\lim_l,\\lim_r$ are the left and right limits in $x_0$ respectively, then $\\lim_l\\le f(x_0)\\le\\lim_r$. The surjectivity simply tells us that $\\lim_l=f(x_0)=\\lim_r$, which is the condition for continuity."
}
{
  "Tag": [
    "\\/closed"
  ],
  "Problem": "Do you know a site where I can find KSEG activities?",
  "Solution_1": "[quote=\"guile\"]Do you know a site where I can find KSEG activities?[/quote]\r\n\r\nThis could maybe help, but I'm not sure: http://www.mit.edu/~ibaran/kseg.html\r\nhttp://en.wikipedia.org/wiki/KSEG"
}
{
  "Tag": [],
  "Problem": "The mean, median, unique mode, and range of a collection of eight integers are all equal to 8. What is the largest interger that can be an element of the collection?\r\n\r\nSource: AMC",
  "Solution_1": "[quote=\"buzzer11\"]The mean, median, unique mode, and range of a collection of eight integers are all equal to 8. What is the largest interger that can be an element of the collection?\nSource: AMC[/quote]\r\nwhat's unique mode? I have never heard of it before. :huh:",
  "Solution_2": "some sets have two modes like 1,1,2,2",
  "Solution_3": "So nobody has gotten it yet?",
  "Solution_4": "[quote=\"buzzer11\"]The mean, median, unique mode, and range of a collection of eight integers are all equal to 8. What is the largest interger that can be an element of the collection?\n\nSource: AMC[/quote]\r\n\r\n[hide] Obviously, $8,8$ are the middle 2 numbers. Let's call the largest number $a+8$ and the smallest $a$. We want $a+8$ as large as possible, so we need lower numbers on the right hand side than on the left to balance the mean at $8$. So our 8 integers are $a,a,a,8,8,8,8,a+8$. $4a+40=64\\Longrightarrow a=6$. $\\boxed{14}$? [/hide]",
  "Solution_5": "Going down the list...is it 14?",
  "Solution_6": "[hide]Yes, the answer is 14.  :) [/hide]",
  "Solution_7": "[hide]We see that 15, is impossible but 14 is possible.\n\nWhen it is this\n\n6,6,6,8,8,8,8,14\n\n$14$[/hide]"
}
{
  "Tag": [],
  "Problem": "What is the simplified value of $ 0.75 \\div \\frac{1}{4}$?",
  "Solution_1": "3/4x4/1=3 \r\n\r\n999th post!"
}
{
  "Tag": [
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "Prove that there exist 10 distinct real number $ a_1,a_2,..,a_{10}$ such that the equation:\r\n           $ (x\\minus{}a_1)(x\\minus{}a_2)...(x\\minus{}a_{10})\\equal{}(x\\plus{}a_1)(x\\plus{}a_2)...(x\\plus{}a_{10})$\r\nhas exactly 5 different real roots.",
  "Solution_1": "For example $ a_6 \\equal{} 0,a_7 \\equal{} 1,a_8 \\equal{} \\minus{} 1,a_9 \\equal{} 2,a_{10} \\equal{} \\minus{} \\minus{} 2$. \r\nThen $ x \\equal{} 0,\\pm 1,\\pm 2$ are roots. Another roots satysfyed\r\n$ \\sigma_1 x^4 \\plus{} \\sigma_3 x^5 \\plus{} \\sigma_5 \\equal{} 0.$\r\nWere $ \\sigma_1 \\equal{} a_1 \\plus{} a_2 \\plus{} a_3 \\plus{} a_4 \\plus{} a_5, sigma_3 \\equal{} \\sum_{1\\le i < j < k\\le 5}a_ia_ja_k, \\sigma_5 \\equal{} a_1a_2a_3a_4a_5$.\r\nIt had not another real roots if $ a_i$ are positive, because roots are $ x^2 \\equal{} \\frac { \\minus{} \\sigma_3 \\plus{} \\pm \\sqrt {\\sigma_3^2 \\minus{} 4\\sigma_1\\sigma_5}}{2\\sigma_1}.$ If $ D < 0$ $ x^2,x$ are complex, if $ D\\ge 0$ $ x^2 < 0$."
}
{
  "Tag": [],
  "Problem": "Check out the poll.",
  "Solution_1": "NEVER. I don't get homework.",
  "Solution_2": "Sometimes Friday if I'm not too tired, but usually Saturday. Sunday only if I had absolutely no time before, and never Monday morning.",
  "Solution_3": "usually friday. that way i can enjoy my weekend",
  "Solution_4": "All through. :(\r\nI get a lot of homework",
  "Solution_5": "i sometimes don't get weekend homework :D too bad you guys always do :P",
  "Solution_6": "In senior year, one can expect a lot of homework.  I've only had two days of school and already I have an essay to write, two presentations to give, and lots of reading.  And I have a test in about a week (though that's not such a worry).  Also, I have college applications to worry about.  :maybe: \r\nAnd then you have to worry about running clubs, activities, groups, etc.",
  "Solution_7": "IT IS LIKE A RIGHT-SKEWED NORMAL DISTRIBUTION",
  "Solution_8": "Mostly Saturday, occasionally Sunday (i.e. this weekend, because I'm procrastinating on history homework).",
  "Solution_9": "usuallly monday, but now our homeroom is in the middle of the day, so, i probably can't do it. :(",
  "Solution_10": "my bus ride is really long because i take a bus to another school, get off, and ride that bus all around the county before it gets to a stop near my home.  \r\n\r\ni usually do my hw then, on Fridays.",
  "Solution_11": "Hmm freshman year:\r\n\r\nMath homework: Nope. I do it during the lunch period before it's due.\r\n\r\nAP bio homework: Chh-yeah I HAVE to do it during the weekend.\r\n\r\nEnglish homework: Meh. Mostly during weekends.\r\n\r\nSpanish: See math.\r\n\r\nGovernment: What homework?\r\n\r\nI think that's it.  :lol:",
  "Solution_12": "I know its a bad habit, but I usually procrastinate and tend to leave all my homework to Sunday (although never Monday)",
  "Solution_13": "im in elementary school still so i dont get a lot of homework",
  "Solution_14": "/bump\n\nThe day before I was born...",
  "Solution_15": "what is with bumping old threads lmho",
  "Solution_16": "This isn't even an interesting thread to bump"
}
{
  "Tag": [
    "group theory",
    "abstract algebra",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Let $ A$ be a set of positive integers such that no positive integer greater than 1 divides all the elements of $ A.$ Prove that any sufficiently large positive integer can be written as a sum of elements of $ A.$ (Elements may occur several times in the sum.)",
  "Solution_1": "Take some $a \\in A$. If $a=1$ we're done, otherwise factorize $a=\\prod_{k=1}^n{p_k}^{e_k}$. By the fact that the setwise $\\gcd$ of the elements of $A$ is $1$, we can find $a_k \\in A$ such that $p_k \\nmid a_k$ for all $1\\leq k \\leq n$. Take $P = \\prod_{k=1}^n p_k$, then $b_k = a_kP/p_k$ for all $1\\leq k \\leq n$, then take $b=\\sum_{k=1}^n b_k$. Since $p_k \\nmid b_k$ and $p_i \\mid b_j$ for all $1\\leq k \\leq n$ and $1\\leq i\\neq j \\leq n$, it follows that $p_k \\nmid b$ for any $1\\leq k\\leq n$, thus $\\gcd(a,b)=1$. Notice that $b$ is a sum of sums of elements of $A$, thus a sum of elements of $A$.\n\nBy Sylvester's theorem, or any other weaker and easier to prove method, there exists some $N_0$ such that any positive integer $N\\geq N_0$ can be represented as a sum of $a$'s and/or $b$'s (from Sylvester's theorem $N_0 = ab - a - b + 1$), thus as a sum of elements of $A$.\n\nIn effect, this problem shows that the additive subgroup $\\left < A \\right >$ generated by a set $A$ of positive integers with setwise $\\gcd$ of its elements $1$ contains consecutive numbers from some point on. Since for any additive subgroup $S$ with setwise $\\gcd$ of its elements $1$ we can take as generating set $S$ itself, this is a structure theorem for the additive subgroups of $\\mathbb{N}^*$ (and it immediately extends to the case when the setwise $\\gcd$ is any $d$)."
}
{
  "Tag": [],
  "Problem": "Do you know any more of those shortcuts?? {hold alt and press one or more numbers in the [b]side[/b] numberpad, then release alt}",
  "Solution_1": "it's the numberpad",
  "Solution_2": "you don't have a sig though, do you?",
  "Solution_3": "Oops...deleted it...",
  "Solution_4": "Yess theres a whole bunch of them. The number you press in the numpad is the ASCII code for the character. To find the shortcuts, go to the Start menu and press Run. Then type in \"charmap\". Choose the character you want a shortcut for and look in the lower right hand corner for the keystroke. note that you can only go up to  (alt-0255) because the characters above that are Unicode."
}
{
  "Tag": [
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "Let $ a>0, b>0, c>0$ and $ \\frac{1}{a}\\plus{}\\frac{1}{b}\\plus{}\\frac{1}{c}\\le 2009$.\r\n\r\nProve $ \\frac{1}{2008a\\plus{}b\\plus{}c}\\plus{}\\frac{1}{2008b\\plus{}c\\plus{}a}\\plus{}\\frac{1}{2008c\\plus{}a\\plus{}b}\\le \\frac{2009}{2010}$.\r\n\r\n\r\n--------------------------------------------------------\r\nMy school site: http://www.chuyenquangtrung.com.vn",
  "Solution_1": "Quite nice!!\r\n\r\n$ a\\plus{}b\\plus{}c \\geq \\frac{9}{2009}$\r\n\r\nThen\r\n\r\n$ \\sum \\frac{1}{2008a\\plus{}b\\plus{}c} \\leq \\sum \\frac{1}{2007a\\plus{} \\frac{9}{2009}} \\leq \\frac{1}{2010^2}  \\sum ( \\frac{2007}{a}\\plus{}2009) \\leq \\frac{2009}{2010}$",
  "Solution_2": "Yes! Your solution is nice. This problem can done by some way. Your solution is one of them.",
  "Solution_3": "[quote=\"thanhnam2902\"]Let $ a > 0, b > 0, c > 0$ and $ \\frac {1}{a} \\plus{} \\frac {1}{b} \\plus{} \\frac {1}{c}\\le 2009$.\n\nProve $ \\frac {1}{2008a \\plus{} b \\plus{} c} \\plus{} \\frac {1}{2008b \\plus{} c \\plus{} a} \\plus{} \\frac {1}{2008c \\plus{} a \\plus{} b}\\le \\frac {2009}{2010}$.\n\n\n--------------------------------------------------------\nMy school site: http://www.chuyenquangtrung.com.vn[/quote]\r\n$ \\frac{1}{2008a\\plus{}b\\plus{}c} \\leq  \\frac{1}{2010^2}(\\frac{2008}{a} \\plus{}\\frac{1}{b} \\plus{}\\frac{1}{c})$\r\n  hence : \r\n$ LHS  \\leq \\frac{1}{2010}(\\frac{1}{a} \\plus{}\\frac{1}{b} \\plus{}\\frac{1}{c}) \\equal{}RHS$",
  "Solution_4": "generalization problem :  $ \\sum {\\frac{1}{ma\\plus{}nb\\plus{}pc}}\\leq\\frac{1}{m\\plus{}n\\plus{}p}(\\sum {\\frac{1}{a}})$\r\nthe case of this problem is $ m\\equal{}2008,n\\equal{}p\\equal{}1$",
  "Solution_5": "OK! this is my problem! \r\n$ \\sum {\\frac{1}{ma\\plus{}nb\\plus{}pc}}\\leq\\frac{1}{m\\plus{}n\\plus{}p}(\\sum {\\frac{1}{a}})$",
  "Solution_6": "a same problem in some ebook:\r\nLet $ a,b,c>0$ and $ \\frac{1}{a}\\plus{}\\frac{1}{b}\\plus{}\\frac{1}{c}\\equal{} 4$.\r\n\r\nProve:     $ \\frac{1}{2a\\plus{}b\\plus{}c}\\plus{}\\frac{1}{2b\\plus{}c\\plus{}a}\\plus{}\\frac{1}{2c\\plus{}a\\plus{}b}\\le 1$.",
  "Solution_7": "[quote=\"thanhnam2902\"]a same problem in some ebook:\nLet $ a,b,c > 0$ and $ \\frac {1}{a} \\plus{} \\frac {1}{b} \\plus{} \\frac {1}{c} \\equal{} 4$.\n\nProve:     $ \\frac {1}{2a \\plus{} b \\plus{} c} \\plus{} \\frac {1}{2b \\plus{} c \\plus{} a} \\plus{} \\frac {1}{2c \\plus{} a \\plus{} b}\\le 1$.[/quote]\r\n similar  we have :\r\n$ A\\equal{}\\frac {1}{2a \\plus{} b \\plus{} c} \\plus{} \\frac {1}{2b \\plus{} c \\plus{} a} \\plus{} \\frac {1}{2c \\plus{} a \\plus{} b}  \\leq \\frac{1}{16} (\\frac{4}{a} \\plus{}\\frac{4}{b}\\plus{}\\frac{4}{c} \\equal{} 1$",
  "Solution_8": "Ok! Thank you."
}
{
  "Tag": [],
  "Problem": "This is the question:\r\nWhat is the sum of all the perfect squares between 5 and 30?\r\n\r\nSimple, but is there an easier way besides 9+16+25?",
  "Solution_1": "You can't factor sum of squares in any way, so it would be the quickest except if you notice $ 3^2\\plus{}4^2\\equal{}5^2$",
  "Solution_2": "The sum of the first $ n$ square numbers is $ \\frac{n(n\\plus{}1)(2n\\plus{}1)}{6}$. On this problem, you want the sum of the first $ 5$ perfect squares minus the sum of the first $ 2$ perfect squares, but with such small numbers involved, it makes more sense to just add.",
  "Solution_3": "[quote=\"erictao200500\"]This is the question:\nWhat is the sum of all the perfect squares between 5 and 30?\n\nSimple, but is there an easier way besides 9+16+25?[/quote]\r\n\r\nYou could use the formula from Discrete_math, but I'm guessing that wouldn't be as fast as noticing $ 3^2 \\plus{} 4^2 \\equal{} 5^2$, so $ (2)(50) \\equal{} 50$.  Or you could jut do 9 + 16 + 25, which is pretty simple to do in your head..not that hard...",
  "Solution_4": "I tell my kids to not use that formula unless you have a large number of perfect squares to add.",
  "Solution_5": "i think that both are equal ( 2*50 and 9+16+25)\r\nbecause it is such a small range of numbers 2*50 wouldn't be quicker by that much.",
  "Solution_6": "yikes!\r\ni meant 2*25 \r\nnot 2*50\r\nthat would be way off\r\nsorry :oops:",
  "Solution_7": "[quote=\"007math\"]You could use the formula from Discrete_math, but I'm guessing that wouldn't be as fast as noticing $ 3^2 \\plus{} 4^2 \\equal{} 5^2$, so $ (2)(50) \\equal{} 50$.  Or you could jut do 9 + 16 + 25, which is pretty simple to do in your head..not that hard...[/quote]\r\n\r\n$ 2(50)\\neq 50$  :wink: \r\n\r\nIt should be $ 2(25)$ anyway.",
  "Solution_8": "[quote=\"Dojo\"][quote=\"007math\"]You could use the formula from Discrete_math, but I'm guessing that wouldn't be as fast as noticing $ 3^2 \\plus{} 4^2 \\equal{} 5^2$, so $ (2)(50) \\equal{} 50$.  Or you could jut do 9 + 16 + 25, which is pretty simple to do in your head..not that hard...[/quote]\n\n$ 2(50)\\neq 50$  :wink: \n\nIt should be $ 2(25)$ anyway.[/quote]\r\n\r\nargh... :wallbash_red:\r\n\r\nEDIT: I remember even checking over my multiplication to see if i did it right...me is dumb  :("
}
{
  "Tag": [
    "MATHCOUNTS"
  ],
  "Problem": "Let S be the set of all natural numbers whose digits are chosen from the set {1, 3, 5, 7} such that\r\nno digits are repeated. Find the sum of the elements of S.",
  "Solution_1": "[quote=\"mdk\"]Let S be the set of all natural numbers whose digits are chosen from the set {1, 3, 5, 7} such that\nno digits are repeated. Find the sum of the elements of S.[/quote]\r\n\r\nI also managed to solve this...\r\n\r\n[hide]Look at all the numbers with first digit $ 1$:\n1357\n1375\n1537\n1573\n1735\n1753\n\nSo 1 appears in the first digit 6 times. Same for the other digits in each of the other spots. \n\nWhen added together, each digit will have a value of\n$ 6\\cdot (1+3+5+7) = 6\\cdot 4^{2}= 96$\n\nSince there are 4 digits, multiply by $ 1111$:\n$ 96\\cdot 1111 = 106656$\nTo account for $ 3$ digits, note that the possibilities are \n135\n137\n157\n357\nApplying the same technique,\n$ 18\\times 111+22\\times 111+26\\times 111+30\\times = 10656$\n\nTo account for $ 2$ digits, note that the possibilities are:\n13\n15\n17\n35\n37\n57\nApplying the same technique,\n$ 4\\times 11+6\\times 11+8\\times 11+8\\times 11+10\\times 11+12\\times 11 = 528$\n1 digit numbers are easy... $ 1+3+5+7 = 4^{2}= 16$\nAdding up,\n\\[ 106656+10656+528+16 = \\boxed{117856}\\]\n[/hide]\r\n\r\nEDIT: Maybe MathCounts words it differently. I've got to be more careful next time.",
  "Solution_2": "how did you do it so fast?",
  "Solution_3": "[quote=\"mdk\"]how did you do it so fast?[/quote]\r\n\r\nMathCounts training. :P \r\n\r\nThis is a classic type of MathCounts problem. When I trained for MathCounts, I frequently came across these types of problems, and I just got used to them.",
  "Solution_4": "[quote=\"vishalarul\"][quote=\"mdk\"]Let S be the set of all natural numbers whose digits are chosen from the set {1, 3, 5, 7} such that\nno digits are repeated. Find the sum of the elements of S.[/quote]\n\nI also managed to solve this...\n\n[hide]Look at all the numbers with first digit $ 1$:\n1357\n1375\n1537\n1573\n1735\n1753\n\nSo 1 appears in the first digit 6 times. Same for the other digits in each of the other spots. \n\nWhen added together, each digit will have a value of\n$ 6\\cdot (1+3+5+7) = 6\\cdot 4^{2}= 96$\n\nSince there are 4 digits, multiply by $ 1111$:\n$ 96\\cdot 1111 = \\boxed{106656}$[/hide][/quote]\n\nhmmmmmmmmmm... but you haven't accounted for the 3 digit, 2 digit and single digit numbers!\n\n[hide=\"solution\"]\nBy a similar method, if you place 1 at the front of the 3 digit possibilities, you get:\n\n1st digit: 1\n2nd digit: 3 choices\n3rd digit: 2 choices\n\nSo again 1 will appear at the start 6 times, and it must be the same for the other digits and the other places.\n\nSo: $ 6*(1+3+5+7) = 96$\n\nand multiply by 111, because these are the 3 digit numbers:\n\n$ 96*111 = 10656$\n\n\nFor 2 digits, each of the {1,3,5,7} digits will appear in each place 3 times, so:\n\n$ 3*(1+3+5+7)=48$\n\nand times 11:\n\n$ 48*11 = 528$\n\nFor single digits each appears once, and multiply by 1, so:\n\n$ 1+3+5+7 = 16$\n\nSo the final answer is:\n\n$ 106656+10656+528+16 = 117856$\n[/hide]"
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Is it possible for a perfect cube to start with 2002 ones in decimal notation?\r\nGuess the answer is yes but I have no idea how to show this.",
  "Solution_1": "The following problem has been discussed on the forum several times: if $a$ is not a power of $10$, then given any combination of digits, some $a^n$ begins with that combination. In your case, just take $a$ to be a cube which is not a power of $10$, and you're done.",
  "Solution_2": "[quote=\"grobber\"]The following problem has been discussed on the forum several times: if $a$ is not a power of $10$, then given any combination of digits, some $a^n$ begins with that combination. In your case, just take $a$ to be a cube which is not a power of $10$, and you're done.[/quote]\r\nWhere can I find a proof of this?",
  "Solution_3": "Just some threads under this one you can find http://www.mathlinks.ro/Forum/viewtopic.php?t=50019"
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "tetrahedron",
    "geometric inequality",
    "IMO",
    "IMO 1970"
  ],
  "Problem": "In the tetrahedron $ABCD,\\angle BDC=90^o$ and the foot of the perpendicular from $D$ to $ABC$ is the intersection of the altitudes of $ABC$. Prove that: \\[ (AB+BC+CA)^2\\le6(AD^2+BD^2+CD^2). \\] When do we have equality?",
  "Solution_1": "[hide=\"proof that also appears in the other thread ...\"]\nA problem that everyone must try..\n\nLet $H$ be the orthocentre of $ABC$. Then $AH=2R\\cos A$ etc and $BC=2R\\sin A$ etc (well-known).\n\nNote that \n$AH^2+BC^2-CH^2=4R^2(\\cos^2A+\\sin^2A-\\cos^2C)=4R^2(1-\\cos^2C)=$ $4R^2\\sin^2C=AB^2$\n\nThe following equations hold due to Pythagoras' theorem:\n\n$AH^2+HD^2=AD^2$ (1)\n$BH^2+HD^2=BD^2$ (2)\n$CH^2+HD^2=CD^2$ (3)\n$BD^2+CD^2=BC^2$ (4)\n\nNow (2)+(3) imply $2HD^2=BC^2-CH^2-BH^2$ using (4).\nTherefore (1)+(2) imply $AD^2+BD^2=AH^2+BC^2-CH^2=AB^2$.\n\nSo:\n$BD^2+CD^2=BC^2$\n$AD^2+BD^2=AB^2$\n$CD^2+AD^2=CA^2$\n\nAdding these up and multiplying by 3 gives $6(AD^2+BD^2+CD^2)=3(AB^2+BC^2+CD^2)$.\n\nAnd clearly $3(AB^2+BC^2+CD^2)\\ge (AB+BC+CD)^2$. (we have equality when $ABC$ is equilateral)\n\n[/hide]\n\nEDIT: posted later here (thanks [b]Luis Gonz\u00e1lez[/b]): http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=407847"
}
{
  "Tag": [
    "LaTeX",
    "geometry"
  ],
  "Problem": "Does anyone know is it possible to put images in a table in latex. I've tried but no luck.",
  "Solution_1": "You do it the same way as you would put graphics in anywhere http://www.artofproblemsolving.com/LaTeX/AoPS_L_PictHow.php as long as you don't use a float since that would make no sense. So use \\includegraphics{myimage.png} but not \\begin{figure}...\\end{figure}.",
  "Solution_2": "Thanks for your reply\r\n\r\nI came across the following which may be of use to anyone who has found their way here.\r\n\r\nImages in an array does the job for me.\r\n\r\n[code]\\begin{figure}[h]\n\\begin{center}$\\begin{array}{cc}\\includegraphics[width=2.5in]{image1.jpg}& \\includegraphics[width=2.5in]{image2.jpg}\\\\ \\includegraphics[width=2.5in]{image1.jpg}& \\includegraphics[width=2.5in]{image2.jpg}\\end{array}$\n\\end{center}\n\\caption{my caption}\n\\end{figure}[/code]\r\n\r\nShane",
  "Solution_3": "There's no need for the dollar signs as there is no mathematics in the array so you can use tabular instead which does the same job as array but in text mode.",
  "Solution_4": "I thought array only worked in mathmode, obviously not.\r\n\r\nThanks for you help.\r\n\r\nShane\r\n\r\nI see you are a probably a disciple of Yang-Mills theory and so I presume you are familiar with tensors. I just wanted to ask you do you know off hand of any good introductions to tensors. Something to explain in physical terms what exactly a tensor and dual space etc. really is?",
  "Solution_5": "[quote=\"ofeyrpf\"]I thought array only worked in mathmode, obviously not.[/quote]You are quite correct but what I meant was that as you didn't need maths mode you didn't need array and could use tabular instead. They are exactly the same except that tabular is for text mode.\n\n[quote=\"ofeyrpf\"]I see you are a probably a disciple of Yang-Mills theory[/quote]That's only something that AoPS puts there to indicate the number of posts that have been made. So I'm afraid tensors is not really my area of mathematics but have you looked at and followed the links at [url=http://en.wikipedia.org/wiki/Tensor]Wikipedia[/url], [url=http://mathworld.wolfram.com/Tensor.html]MathWorld[/url] and [url=http://planetmath.org/encyclopedia/Tensor.html]PlanetMath[/url]?",
  "Solution_6": "Thanks Stevem, I'll have a look."
}
{
  "Tag": [
    "number theory",
    "Diophantine equation"
  ],
  "Problem": "Determine the values of the coefficient $a$ for which the equations:\r\n\r\n$x^2 + ax + 1 = 0$\r\n$x^2 + x + a = 0$\r\n\r\nhave at least one common root.",
  "Solution_1": "Hello, Mikey\n\n\n\nQuote:Determine the values of the coefficient  for which the equations:\n\n\n\n\nhave at least one common root.\n\n[hide]\n\nLet the roots of x^2 + ax + 1 = 0 be  {p,q}.\n\nThen: [1] p + q = -a . . . [2] pq = 1\n\n\n\nLet the roots of x^2 + x + a = 0  be {q,r}.\n\nThen: [3] q + r = -1 . . . [4] qr = a\n\n\n\nSubtract [3] from [1]: [5] p - r = 1 - a\n\n\n\nDivide [2] by [4]:  pq/qr = 1/a . . . r = ap\n\n\n\nSubstitute into [5]: p - ap = 1 - a . . . (p - 1)(1 - a) = 0\n\nHence: a = 1 ... or ... p = 1\n\n\n\nIf a = 1, the equations are identical, both are x^2 + x + 1 = 0,\n\nand hence, have two common roots: x = (-1 p/m sqrt{3}i)/(2).\n\n\n\nIf p = 1, then from [2] q = 1, and from [3], r = -2\n\nAnd finally from [1], a = -2\n\n\n\nThe equations are:\n\nx^2 - 2x + 1 = 0  . . . roots: 1, 1\n\nx^2 + x - 2 = 0 . . . roots: 1, -2\n\n[/hide]",
  "Solution_2": "Let $t$ be common root,we have \r\n$t^2+at+1=0\\cdots$(1)\r\n$t^2+t+a=0\\cdots$(2)\r\n(2)-(1):$(a-1)(t-1)=0$,which follows Soroban's solutiuon.\r\n\r\nkunny",
  "Solution_3": "Hmm...\r\n\r\nThis is a form of a Diophantine equation, right\r\n\r\nx^2 + ax + 1 = x^2 + x + a\r\nax + 1 = x + a\r\nax - a - x + 1 = 0\r\n\r\n(a - 1)(x - 1) = 0\r\n\r\nOne value is 1... there's probably others, but I can't find a way to find any other...",
  "Solution_4": "[quote=\"Treething\"]Hmm...\n\nThis is a form of a Diophantine equation, right\n\nx^2 + ax + 1 = x^2 + x + a\nax + 1 = x + a\nax - a - x + 1 = 0\n\n(a - 1)(x - 1) = 0\n\nOne value is 1... there's probably others, but I can't find a way to find any other...[/quote]\r\n\r\nWell, you are basically finished.. if (a-1)(x-1) = 0, you either have a solution a=1.. or.. how else can that product be 0 :P And in that case, whats a?",
  "Solution_5": "If x is one...\r\n\r\nx^2 + ax + 1 = 0\r\n\r\n1 + a + 1 = 0\r\n\r\na = -2\r\n\r\n\r\nx^2 + x + a = 0\r\n\r\n1 + 1 + a = 0\r\n\r\na = -2"
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed",
    "algebra",
    "China",
    "Hi"
  ],
  "Problem": "$ a,b,c$ are positive reals. Prove that\r\n$ \\sqrt{\\frac{a^4 \\plus{}2b^2 c^2}{a^2 \\plus{}2bc}}\\plus{}\\sqrt{\\frac{b^4 \\plus{}2c^2 a^2}{b^2 \\plus{}2ca}}\\plus{}\\sqrt{\\frac{c^4 \\plus{}2a^2 b^2}{c^2 \\plus{}2ab}}\\geq a\\plus{}b\\plus{}c$",
  "Solution_1": "[quote=\"jedaihan\"]$ a,b,c$ are positive reals. Prove that\n$ \\sqrt {\\frac {a^4 \\plus{} 2b^2 c^2}{a^2 \\plus{} 2bc}} \\plus{} \\sqrt {\\frac {b^4 \\plus{} 2c^2 a^2}{b^2 \\plus{} 2ca}} \\plus{} \\sqrt {\\frac {c^4 \\plus{} 2a^2 b^2}{c^2 \\plus{} 2ab}}\\geq a \\plus{} b \\plus{} c$[/quote]\r\nVery nice inequality, jedaihan. :) I like it very much. :)\r\nBecause of the following identity $ (a^2 \\plus{} 2bc)(b^2 \\plus{} 2ca) \\minus{} (ab \\plus{} bc \\plus{} ca)^2 \\equal{} c(a \\minus{} b)^2(2a \\plus{} 2b \\minus{} c)$, it suggested me to solve your inequality as follow:\r\nAssume that $ a \\ge b \\ge c$, we will consider 2 cases:\r\n\r\nCase 1. If $ a \\le 2(b \\plus{} c)$, then rewriting our inequality as\r\n\\[ \\sum\\frac {a^4 \\plus{} 2b^2c^2}{a^2 \\plus{} 2bc} \\plus{} 2\\sum \\sqrt {\\frac {(a^4 \\plus{} 2b^2c^2)(b^4 \\plus{} 2c^2a^2)}{(b^2 \\plus{} 2ca)(c^2 \\plus{} 2ab)}} \\ge \\sum a^2 \\plus{} 2\\sum bc\r\n\\]\r\nNotice that in this case, the above identity and Cauchy Schwarz Inequality yield\r\n\\[ \\sum \\sqrt {\\frac {(a^4 \\plus{} 2b^2c^2)(b^4 \\plus{} 2c^2a^2)}{(b^2 \\plus{} 2ca)(c^2 \\plus{} 2ab)}} \\ge \\sum \\sqrt {\\frac {(a^2 \\plus{} 2bc)(b^2 \\plus{} 2ca)}{9}} \\ge \\sum \\sqrt {\\frac {(ab \\plus{} bc \\plus{} ca)^2}{9}} \\equal{} \\sum bc.\r\n\\]\r\nMoreover, the Chebyshev's Inequality gives us\r\n\\[ \\sum \\frac {a^4 \\plus{} 2b^2c^2}{a^2 \\plus{} 2bc} \\ge \\sum a^2.\r\n\\]\r\nTherefore, our inequality is trivial in this case. :)\r\n\r\nCase 2. If $ a \\ge 2(b \\plus{} c)$, then we see that the original inequality is just the sum of the following inequalities:\r\n\\[ \\sqrt {\\frac {a^4 \\plus{} 2b^2c^2}{a^2 \\plus{} 2bc}} \\ge a \\plus{} (1 \\minus{} \\sqrt {2})(b \\plus{} c),\\qquad (1)\r\n\\]\r\nand\r\n\\[ \\sqrt {\\frac {b^4 \\plus{} 2c^2a^2}{b^2 \\plus{} 2ca}} \\ge \\sqrt {2}c,\\quad \\sqrt {\\frac {c^4 \\plus{} 2a^2b^2}{c^2 \\plus{} 2ab}} \\ge \\sqrt {2}b. \\qquad (2)\r\n\\]\r\nThese inequalities can be easily proved if we use the hypothesis $ a \\ge 2(b \\plus{} c)$. ;) :)\r\n\r\nThe proof is completed. :)\r\n\r\nP/s: The more detailed proof, we leave it to the readers. :)",
  "Solution_2": "Yes, I also very like it. Thank you, jedaihan.\r\n@can_hang: We can prove it by only Am-Gm, my brother  :)",
  "Solution_3": "[quote=\"nguoivn\"]Yes, I also very like it. Thank you, jedaihan.\n@can_hang: We can prove it by only Am-Gm, my brother  :)[/quote]\r\n[hide=\"tonguoivn\"]@nguoivn: I know the proof by AM-GM ;), but I like the nonstandard proof more.  ;)[/hide]",
  "Solution_4": "How can you do it can_hung using only AM-GM?\r\n\r\nI can't do it :( \r\nI used\r\n$ \\sqrt{\\frac{a^{4}+2b^{2}c^{2}}{a^{2}+2bc}}=\\sqrt{a^{2}+bc-\\frac{3a^{2}bc}{a^{2}+2bc}}$\r\nor\r\n$ x=\\frac{bc}{a}, y=\\rac{ca}{b}, z=\\frac{ab}{c}$, then $ xyz=abc$ and because this inequality is homogenoeous,\r\nlet $ xyz=1$ then enough to show $ \\sum_{cyc}\\sqrt{\\frac{2x^{3}+1}{2x^{2}+x}} \\ge \\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}$\r\nand so on.\r\nI tried with using a lot of transform. But I can't -_-..",
  "Solution_5": "Thanks, can_hang. Very nice solution  :)  :) \r\n\r\nMy solution)\r\nBy Cauchy Schwarz Inequality,\r\n$ (a^4\\plus{}b^2c^2\\plus{}b^2c^2)(a\\plus{}b\\plus{}c)(a\\plus{}c\\plus{}b)\\geq (a^2\\plus{}2bc)^3$\r\nTherefore,\r\n$ \\sum{\\sqrt{\\frac{a^4\\plus{}2b^2c^2}{a^2\\plus{}2bc}}}\\geq \\sum{\\frac{a^2\\plus{}2bc}{a\\plus{}b\\plus{}c}}\\equal{}a\\plus{}b\\plus{}c$",
  "Solution_6": "[quote=\"jedaihan\"]Thanks, can_hang. Very nice solution  :)  :) \n\nMy solution)\nBy Cauchy Schwarz Inequality,\n$ (a^4 \\plus{} b^2c^2 \\plus{} b^2c^2)(a \\plus{} b \\plus{} c)(a \\plus{} c \\plus{} b)\\geq (a^2 \\plus{} 2bc)^3$\nTherefore,\n$ \\sum{\\sqrt {\\frac {a^4 \\plus{} 2b^2c^2}{a^2 \\plus{} 2bc}}}\\geq \\sum{\\frac {a^2 \\plus{} 2bc}{a \\plus{} b \\plus{} c}} \\equal{} a \\plus{} b \\plus{} c$[/quote]\r\nNice solution, jedaihan. But I think your way and nguoivn's way will not work for the following:\r\n\\[ \\sqrt{\\frac{a^4\\plus{}5b^2c^2}{a^2\\plus{}5bc}} \\plus{}\\sqrt{ \\frac{b^4\\plus{}5c^2a^2}{b^2\\plus{}5ca}}\\plus{}\\sqrt{\\frac{c^4\\plus{}5a^2b^2}{c^2\\plus{}5ab}} \\ge a\\plus{}b\\plus{}c.\\]\r\n;) :)\r\n\r\nFor more general, I think that the following inequality still holds:\r\n\\[ \\sqrt{\\frac{a^4\\plus{}kb^2c^2}{a^2\\plus{}kbc}} \\plus{}\\sqrt{ \\frac{b^4\\plus{}kc^2a^2}{b^2\\plus{}kca}}\\plus{}\\sqrt{\\frac{c^4\\plus{}ka^2b^2}{c^2\\plus{}kab}} \\ge a\\plus{}b\\plus{}c.\\]\r\nWith $ 0 \\le k \\le 12.$\r\nBut I only have nice solution (with Cauchy Schwarz) for $ 0 \\le k \\le 5.$\r\nI hope you will be interested with this inequality. :)",
  "Solution_7": "Wow! that's interesting :) \r\nI'll try it",
  "Solution_8": "[quote=\"jedaihan\"]\nMy solution)\nBy Cauchy Schwarz Inequality,\n$ (a^4 \\plus{} b^2c^2 \\plus{} b^2c^2)(a \\plus{} b \\plus{} c)(a \\plus{} c \\plus{} b)\\geq (a^2 \\plus{} 2bc)^3$\nTherefore,\n$ \\sum{\\sqrt {\\frac {a^4 \\plus{} 2b^2c^2}{a^2 \\plus{} 2bc}}}\\geq \\sum{\\frac {a^2 \\plus{} 2bc}{a \\plus{} b \\plus{} c}} \\equal{} a \\plus{} b \\plus{} c$[/quote]\r\nVery nice proof, my friend  :lol: \r\nHere is my proof:\r\nIndeed, I think this is a good problem which let us see the power of classic ineqs. Model methods is not an easy ways with it.\r\nJedaihan's proof is shortest and nicest.\r\nCan_hang's proof is also nice and strange.\r\nMy proof rather long but also only use classic ineq. \r\nNow, we'll try to solve the harder problems of can_hang. I hope that we'll have another nice proofs  :)",
  "Solution_9": "[quote=\"Heebeen, Yang\"]How can you do it can_hung using only AM-GM?\n\nI can't do it :( \nI used\n$ \\sqrt {\\frac {a^{4} + 2b^{2}c^{2}}{a^{2} + 2bc}} = \\sqrt {a^{2} + bc - \\frac {3a^{2}bc}{a^{2} + 2bc}}$\nor\n$ x = \\frac {bc}{a}, y = \\rac{ca}{b}, z = \\frac {ab}{c}$, then $ xyz = abc$ and because this inequality is homogenoeous,\nlet $ xyz = 1$ then enough to show $ \\sum_{cyc}\\sqrt {\\frac {2x^{3} + 1}{2x^{2} + x}} \\ge \\frac {1}{x} + \\frac {1}{y} + \\frac {1}{z}$\nand so on.\nI tried with using a lot of transform. But I can't -_-..[/quote]\r\nYour second approach can lead to a completed nice solution, my friend. Your normalize equivalent to normalize for $ abc=1$, the inequality becomes\r\n$ \\sum \\sqrt {\\frac {x^6 + 2}{x(x^3 + 2)}} \\ge x + y + z.$\r\nNow, using Holder Inequality, we see that\r\n$ (x^6 + 1 + 1)(x^2 + x + 1)(x^2 + 1 + x) \\ge x(x^3 + 2)^3,$\r\nwhich yields\r\n$ \\sqrt {\\frac {x^6 + 2}{x(x^3 + 2)}} \\ge \\frac {x^3 + 2}{x^2 + x + 1} = x - 1 + \\frac {3}{x^2 + x + 1}.$\r\nUsing this inequality, we can reduce our inequality to\r\n$ \\frac {1}{x^2 + x + 1} + \\frac {1}{y^2 + y + 1} + \\frac {1}{z^2 + z + 1} \\ge 1,$\r\nwhich is a well-known inequality (as I remember, it is Vasc's result :) ) and you can prove this by Cauchy Schwarz. :)",
  "Solution_10": "Wow!!!\r\n\r\nMy think was short -_-..\r\n\r\nI must train cauchy more...\r\n\r\nThank you for your nice solution^^!",
  "Solution_11": "[quote=\"jedaihan\"]$ a,b,c$ are positive reals. Prove that\n$ \\sqrt {\\frac {a^4 \\plus{} 2b^2 c^2}{a^2 \\plus{} 2bc}} \\plus{} \\sqrt {\\frac {b^4 \\plus{} 2c^2 a^2}{b^2 \\plus{} 2ca}} \\plus{} \\sqrt {\\frac {c^4 \\plus{} 2a^2 b^2}{c^2 \\plus{} 2ab}}\\geq a \\plus{} b \\plus{} c$[/quote]\r\nAnother solution is the following:\r\nSetting $ a\\equal{}x^2,b\\equal{}y^2,c\\equal{}z^2$ where $ x,y,z \\ge 0$, then using Cauchy Schwarz Inequality, we have\r\n\\[ \\sqrt {\\frac {a^4 \\plus{} 2b^2 c^2}{a^2 \\plus{} 2bc}}  \\equal{}\\frac{\\sqrt{(a^4\\plus{}2b^2c^2)(a^2\\plus{}2bc)}}{a^2\\plus{}2bc} \\ge \\frac{a^3\\plus{}2bc\\sqrt{bc}}{a^2\\plus{}2bc} \\equal{}\\frac{x^6\\plus{}2y^3z^3}{x^4\\plus{}2y^2z^2}.\\]\r\nIt suffices to prove that\r\n\\[ \\sum \\frac{x^6\\plus{}2y^3z^3}{x^4\\plus{}2y^2z^2} \\ge \\sum x^2,\\]\r\nor\r\n\\[ \\sum \\frac{yz\\minus{}x^2}{x^6\\plus{}2x^2y^2z^2} \\ge 0.\\]\r\nUsing GM-HM inequality, we have\r\n\\[ \\frac{yz\\minus{}x^2}{x^6\\plus{}2x^2y^2z^2} \\ge \\frac{\\frac{2y^2z^2}{y^2\\plus{}z^2} \\minus{}x^2}{x^6\\plus{}2x^2y^2z^2} \\equal{}\\frac{c(b\\minus{}a)\\plus{}b(c\\minus{}a)}{(b\\plus{}c)(a^3\\plus{}2abc)}.\\]\r\nThe inequality deduces to\r\n\\[ \\sum \\frac{c(b\\minus{}a)\\plus{}b(c\\minus{}a)}{(b\\plus{}c)(a^3\\plus{}2abc)} \\ge 0,\\]\r\nwhich is equivalent to\r\n\\[ \\sum c(a\\minus{}b)\\left( \\frac{1}{(a\\plus{}c)(b^3\\plus{}2abc)} \\minus{}\\frac{1}{(b\\plus{}c)(a^3\\plus{}2abc)}\\right) \\ge 0,\\]\r\nor\r\n\\[ \\sum \\frac{c(a\\minus{}b)^2[ab(a\\plus{}b)\\plus{}c(a^2\\minus{}ab\\plus{}b^2)]}{(a\\plus{}c)(b\\plus{}c)(a^3\\plus{}2abc)(b^3\\plus{}2abc)} \\ge 0,\\]\r\nwhich is trivial. :)\r\n\r\nNote: This way can help us to prove the more general with $ 0 \\le k \\le 9$. ;) :)",
  "Solution_12": "Nice solution :)  :wink: \r\nBut I can't solve your expansion when $ 0\\leq k\\leq 12$  :blush:\r\nCan you post your solution?",
  "Solution_13": "[quote=\"can_hang2007\"]Moreover, the Chebyshev's Inequality gives us\n\\[ \\sum \\frac {a^4 \\plus{} 2b^2c^2}{a^2 \\plus{} 2bc} \\ge \\sum a^2.\n\\]\n:)[/quote]\r\n\r\n\r\nMy dear friend, can you please explain how to prove it by Chebyshev. Thank you very much. :)",
  "Solution_14": "[quote=\"manlio\"][quote=\"can_hang2007\"]Moreover, the Chebyshev's Inequality gives us\n\\[ \\sum \\frac {a^4 \\plus{} 2b^2c^2}{a^2 \\plus{} 2bc} \\ge \\sum a^2.\n\\]\n:)[/quote]\n\n\nMy dear friend, can you please explain how to prove it by Chebyshev. Thank you very much. :)[/quote]\r\nMy friend, write the inequality in the form\r\n$ \\sum \\frac{bc(bc\\minus{}a^2)}{a^2\\plus{}2bc} \\ge 0,$\r\n$ \\sum \\frac{(bc\\minus{}a^2)(b\\plus{}c)}{(b\\plus{}c)(a^3\\plus{}2abc)} \\ge 0.$\r\nNow assume that $ a \\ge b \\ge c$, we can see that\r\n$ (bc\\minus{}a^2)(b\\plus{}c) \\le (ca\\minus{}b^2)(c\\plus{}a) \\le (ab\\minus{}c^2)(a\\plus{}b),$\r\nand\r\n$ \\frac{1}{(b\\plus{}c)(a^3\\plus{}2abc)} \\le \\frac{1}{(c\\plus{}a)(b^3\\plus{}2abc)} \\le \\frac{1}{(a\\plus{}b)(c^3\\plus{}2abc)}.$\r\nNow, use Cauchy Schwarz, we can get the result. :)",
  "Solution_15": "Thank you very much. :)",
  "Solution_16": "Let $ a,b,c$ are  nonnegative reals. Prove that\\[\\sqrt{\\frac{a^2 \\plus{}bc}{b^2 \\plus{}c^2}}\\plus{}\\sqrt{\\frac{b^2 \\plus{}ca}{c^2 \\plus{}a^2}}\\plus{}\\sqrt{\\frac{c^2 \\plus{}ab}{a^2 \\plus{}b^2}}\\geq 2+\\frac{\\sqrt{2}}{2}.\\]\\[\\sqrt{\\frac{a^2 \\plus{}2bc}{b^2 \\plus{}c^2}}\\plus{}\\sqrt{\\frac{b^2 \\plus{}2ca}{c^2 \\plus{}a^2}}\\plus{}\\sqrt{\\frac{c^2 \\plus{}2ab}{a^2 \\plus{}b^2}}\\geq 3.\\]\nOld? Where?",
  "Solution_17": "[quote=\"sqing\"]Let $ a,b,c$ are  nonnegative reals. Prove that\\[\\sqrt{\\frac{a^2 +bc}{b^2 +c^2}}+\\sqrt{\\frac{b^2 +ca}{c^2 +a^2}}+\\sqrt{\\frac{c^2 +ab}{a^2 +b^2}}\\geq 2+\\frac{\\sqrt{2}}{2}\\][/quote]\n\n$\\sqrt{\\frac{a^2 +bc}{b^2 +c^2}}+\\sqrt{\\frac{b^2 +ca}{c^2 +a^2}}+\\sqrt{\\frac{c^2 +ab}{a^2 +b^2}}\\geq \\sqrt{\\frac{2(a^2+b^2+c^2)}{ab+ac+bc}}+\\frac{\\sqrt{2}}{2}$ is also true.",
  "Solution_18": "[quote=\"arqady\"][quote=\"sqing\"]Let $ a,b,c$ are  nonnegative reals. Prove that\\[\\sqrt{\\frac{a^2 +bc}{b^2 +c^2}}+\\sqrt{\\frac{b^2 +ca}{c^2 +a^2}}+\\sqrt{\\frac{c^2 +ab}{a^2 +b^2}}\\geq 2+\\frac{\\sqrt{2}}{2}\\][/quote]\n\n$\\sqrt{\\frac{a^2 +bc}{b^2 +c^2}}+\\sqrt{\\frac{b^2 +ca}{c^2 +a^2}}+\\sqrt{\\frac{c^2 +ab}{a^2 +b^2}}\\geq \\sqrt{\\frac{2(a^2+b^2+c^2)}{ab+ac+bc}}+\\frac{\\sqrt{2}}{2}$ is also true.[/quote]\nWhe the equality holds?",
  "Solution_19": "$c=0$ and $a=b$.",
  "Solution_20": "[quote=sqing]Let $ a,b,c$ are  nonnegative reals. Prove that\\[\\sqrt{\\frac{a^2 \\plus{}2bc}{b^2 \\plus{}c^2}}\\plus{}\\sqrt{\\frac{b^2 \\plus{}2ca}{c^2 \\plus{}a^2}}\\plus{}\\sqrt{\\frac{c^2 \\plus{}2ab}{a^2 \\plus{}b^2}}\\geq 3.\\]\n[/quote]\n[color=#FF0000][b]Moscow MO 1999?:[/b][/color]\nLet $a,b,c$ be side lengths of a triangle . Prove that : \\[\\frac{a^2 \\plus{}2bc}{b^2 \\plus{}c^2}\\plus{}\\frac{b^2 \\plus{}2ca}{c^2 \\plus{}a^2}\\plus{}\\frac{c^2 \\plus{}2ab}{a^2 \\plus{}b^2}> 3.\\]",
  "Solution_21": "[hide]$a^2+2bc>b^2+c^2\\Leftrightarrow (a+b-c)(a+c-b)>0$\nand we're done.[/hide]",
  "Solution_22": "Let $ a,b,c$ are  nonnegative reals. Prove that\\[\\sqrt{\\frac{a^3 \\plus{}abc}{b \\plus{}c}}\\plus{}\\sqrt{\\frac{b^3 \\plus{}abc}{c \\plus{}a}}\\plus{}\\sqrt{\\frac{c^3 \\plus{}abc}{a \\plus{}b}}\\geq a+b+c.\\]\nOld? Where?\n",
  "Solution_23": "[quote=sqing]Let $ a,b,c$ are  nonnegative reals. Prove that\\[\\sqrt{\\frac{a^3 \\plus{}abc}{b \\plus{}c}}\\plus{}\\sqrt{\\frac{b^3 \\plus{}abc}{c \\plus{}a}}\\plus{}\\sqrt{\\frac{c^3 \\plus{}abc}{a \\plus{}b}}\\geq a+b+c.\\]\nOld? Where?[/quote]\nI think I saw it.\nBy the way, it's easy Holder:\n$\\left(\\sum_{cyc}\\sqrt{\\frac{a^3 \\plus{}abc}{b \\plus{}c}}\\right)^2\\sum_{cyc}\\frac{a^2(b+c)}{a^2+bc}\\geq(a+b+c)^3$.\nHence, it remains to prove that $a+b+c\\geq\\sum_{cyc}\\frac{a^2(b+c)}{a^2+bc}$, which is $abc\\sum_{cyc}(a^4-a^2b^2)\\geq0$, which is obvious.",
  "Solution_24": "I also saw it.\nIt should be an old inequality .\nThanks.",
  "Solution_25": "[img]http://s13.sinaimg.cn/middle/0018zOAxgy719RYuyAI5c&690[/img]",
  "Solution_26": "[quote=sqing]Let $ a,b,c$ are  nonnegative reals. Prove that\\[\\sqrt{\\frac{a^2 \\plus{}bc}{b^2 \\plus{}c^2}}\\plus{}\\sqrt{\\frac{b^2 \\plus{}ca}{c^2 \\plus{}a^2}}\\plus{}\\sqrt{\\frac{c^2 \\plus{}ab}{a^2 \\plus{}b^2}}\\geq 2+\\frac{\\sqrt{2}}{2}.\\]\\[\\sqrt{\\frac{a^2 \\plus{}2bc}{b^2 \\plus{}c^2}}\\plus{}\\sqrt{\\frac{b^2 \\plus{}2ca}{c^2 \\plus{}a^2}}\\plus{}\\sqrt{\\frac{c^2 \\plus{}2ab}{a^2 \\plus{}b^2}}\\geq 3.\\]\nOld? Where?[/quote]\nCrux,Problems 3419:\nLet $ a,b,c$ are  nonnegative reals. Prove that\n$$\\sqrt{\\frac{a^{2}+4bc}{b^2+c^2}}+\\sqrt{\\frac{b^{2}+4ca}{c^2+a^2}}+\\sqrt{\\frac{c^{2}+4ab}{a^2+a^2}}\\ge 2+\\sqrt{2}$$\n$$\\sqrt{\\frac{a^{2}+bc}{b^2+c^2}}+\\sqrt{\\frac{b^{2}+ca}{c^2+a^2}}+\\sqrt{\\frac{c^{2}+ab}{a^2+a^2}}\\ge 2+\\frac{1}{\\sqrt[3]{2}}$$\n[url=http://www.artofproblemsolving.com/community/c6h1115969p5108246]here[/url]",
  "Solution_27": "[quote=sqing]Let $ a,b,c$ are  nonnegative reals. Prove that\\[\\sqrt{\\frac{a^3 \\plus{}abc}{b \\plus{}c}}\\plus{}\\sqrt{\\frac{b^3 \\plus{}abc}{c \\plus{}a}}\\plus{}\\sqrt{\\frac{c^3 \\plus{}abc}{a \\plus{}b}}\\geq a+b+c.\\]\nOld? Where?[/quote]\n\n[img]http://s1.sinaimg.cn/middle/006ptkjAzy75Lkd18qs10&690[/img]",
  "Solution_28": "[img]http://s10.sinaimg.cn/middle/006ptkjAzy75LkvM8n7e9&690[/img]",
  "Solution_29": "[img]http://s9.sinaimg.cn/middle/006ptkjAzy75LkLXmCk78&690[/img]",
  "Solution_30": "[quote=sqing]Let $ a,b,c$ are  nonnegative reals. Prove that\\[\\sqrt{\\frac{a^3 \\plus{}abc}{b \\plus{}c}}\\plus{}\\sqrt{\\frac{b^3 \\plus{}abc}{c \\plus{}a}}\\plus{}\\sqrt{\\frac{c^3 \\plus{}abc}{a \\plus{}b}}\\geq a+b+c.\\]\nOld? Where?[/quote]\nIf $a,b,c>0$ [url=http://www.artofproblemsolving.com/community/c6h1220622p6101528]then prove:[/url]\n$$\\frac{a^3+abc}{b+c}+\\frac{b^3+abc}{a+c}+\\frac{c^3+abc}{b+a}\\geq a^2+b^2+c^2$$",
  "Solution_31": "[quote=sqing][hide=**][quote=sqing]Let $ a,b,c$ are  nonnegative reals. Prove that\\[\\sqrt{\\frac{a^2 \\plus{}bc}{b^2 \\plus{}c^2}}\\plus{}\\sqrt{\\frac{b^2 \\plus{}ca}{c^2 \\plus{}a^2}}\\plus{}\\sqrt{\\frac{c^2 \\plus{}ab}{a^2 \\plus{}b^2}}\\geq 2+\\frac{\\sqrt{2}}{2}.\\]\\[\\sqrt{\\frac{a^2 \\plus{}2bc}{b^2 \\plus{}c^2}}\\plus{}\\sqrt{\\frac{b^2 \\plus{}2ca}{c^2 \\plus{}a^2}}\\plus{}\\sqrt{\\frac{c^2 \\plus{}2ab}{a^2 \\plus{}b^2}}\\geq 3.\\]\nOld? Where?[/quote][/hide]\nCrux,Problems 3419:\nLet $ a,b,c$ are  nonnegative reals. Prove that\n$$\\sqrt{\\frac{a^{2}+4bc}{b^2+c^2}}+\\sqrt{\\frac{b^{2}+4ca}{c^2+a^2}}+\\sqrt{\\frac{c^{2}+4ab}{a^2+a^2}}\\ge 2+\\sqrt{2}$$\n$$\\sqrt{\\frac{a^{2}+bc}{b^2+c^2}}+\\sqrt{\\frac{b^{2}+ca}{c^2+a^2}}+\\sqrt{\\frac{c^{2}+ab}{a^2+a^2}}\\ge 2+\\frac{1}{\\sqrt[3]{2}}$$\n[url=http://www.artofproblemsolving.com/community/c6h1115969p5108246]here[/url][/quote]\nLet $ a,b,c$ are nonnegative reals. [url=http://www.artofproblemsolving.com/community/c6h186334p1042076]Prove that[/url]\n$$\\sqrt[3]{\\frac{a^{2}+bc}{b^2+c^2}}+\\sqrt[3]{\\frac{b^{2}+ca}{c^2+a^2}}+\\sqrt[3]{\\frac{c^{2}+ab}{a^2+a^2}}\\ge 2+\\frac{1}{\\sqrt[3]{2}}$$",
  "Solution_32": "[quote=jedaihan]Thanks, can_hang. Very nice solution  :)  :) \n\nMy solution)\nBy Cauchy Schwarz Inequality,\n$ (a^4\\plus{}b^2c^2\\plus{}b^2c^2)(a\\plus{}b\\plus{}c)(a\\plus{}c\\plus{}b)\\geq (a^2\\plus{}2bc)^3$\nTherefore,\n$ \\sum{\\sqrt{\\frac{a^4\\plus{}2b^2c^2}{a^2\\plus{}2bc}}}\\geq \\sum{\\frac{a^2\\plus{}2bc}{a\\plus{}b\\plus{}c}}\\equal{}a\\plus{}b\\plus{}c$[/quote]\n\ncan anyone show how this result comes from CS inequality ??",
  "Solution_33": "[quote=ISIBS1531][quote=jedaihan]Thanks, can_hang. Very nice solution  :)  :) \n\nMy solution)\nBy Cauchy Schwarz Inequality,\n$ (a^4\\plus{}b^2c^2\\plus{}b^2c^2)(a\\plus{}b\\plus{}c)(a\\plus{}c\\plus{}b)\\geq (a^2\\plus{}2bc)^3$\nTherefore,\n$ \\sum{\\sqrt{\\frac{a^4\\plus{}2b^2c^2}{a^2\\plus{}2bc}}}\\geq \\sum{\\frac{a^2\\plus{}2bc}{a\\plus{}b\\plus{}c}}\\equal{}a\\plus{}b\\plus{}c$[/quote]\n\ncan anyone show how this result comes from CS inequality ??[/quote]\n\nIt's Holder:\n$$(a^4+b^2c^2+b^2c^2)(a+b+c)(a+c+b)\\ge(\\sqrt[3]{a^6}+\\sqrt[3]{b^3c^3}+\\sqrt[3]{b^3c^3})^3$$\nThen by results:\n$$\\frac{(a^4+b^2c^2+b^2c^2)}{(a^2+bc+bc)}\\ge\\frac{(a^2+bc+bc)^2}{(a+b+c)(a+c+b)}$$\nOr\n$$\\sqrt{\\frac{(a^4+b^2c^2+b^2c^2)}{(a^2+bc+bc)}}\\ge\\frac{(a^2+bc+bc)}{(a+b+c)}$$\nHence we need to prove:\n$$\\sum\\sqrt{\\frac{(a^4+b^2c^2+b^2c^2)}{(a^2+bc+bc)}}\\ge\\sum\\frac{(a^2+bc+bc)}{(a+b+c)}$$\nThen:\n$$\\sum\\frac{(a^2+bc+bc)}{(a+b+c)}=\\sum\\frac{\\sum a^2+2\\sum ab}{a+b+c}=\\sum\\frac{(\\sum a)^2}{a+b+c}=\\sum a$$\nSo we're done...!",
  "Solution_34": "You can homogenise with $abc=1$ ,write the  terms as functions of a,b,c and apply jensen,although the convexity is very hard to prove. ",
  "Solution_35": "[quote=sqing]\nCrux,Problems 3419:\nLet $ a,b,c$ are  nonnegative reals. Prove that\n$$\\sqrt{\\frac{a^{2}+4bc}{b^2+c^2}}+\\sqrt{\\frac{b^{2}+4ca}{c^2+a^2}}+\\sqrt{\\frac{c^{2}+4ab}{a^2+b^2}}\\ge 2+\\sqrt{2}$$\n[/quote]\nLet $c=\\min\\{a,b,c\\}$.\nThus, by C-S $$\\sqrt{\\frac{a^{2}+4bc}{b^2+c^2}}+\\sqrt{\\frac{b^{2}+4ca}{c^2+a^2}}=\\frac{\\sqrt{(a^{2}+4bc)(b^2+c^2)}}{b^2+c^2}+\\frac{\\sqrt{(b^{2}+4ca)(a^2+c^2)}}{c^2+a^2}\\geq$$\n$$\\geq\\frac{ab+2\\sqrt{bc^3}}{b^2+c^2}+\\frac{ab+2\\sqrt{ac^3}}{a^2+c^2}\\geq\\frac{ab+2c^2}{b^2+c^2}+\\frac{ab+2c^2}{a^2+c^2}.$$\nAlso, $$\\sqrt2-\\sqrt{\\frac{c^{2}+4ab}{a^2+b^2}}\\leq\\sqrt2-\\sqrt{\\frac{4ab}{a^2+b^2}}=$$\n$$=\\frac{\\sqrt2(a-b)^2}{a^2+b^2+\\sqrt{2ab(a^2+b^2)}}\\leq\\frac{1.5(a-b)^2}{(a+b)^2}.$$\nId est, it's enough to prove that:\n$$\\frac{ab+2c^2}{b^2+c^2}+\\frac{ab+2c^2}{a^2+c^2}\\geq2+\\frac{1.5(a-b)^2}{(a+b)^2},$$\nwhich is obvious.",
  "Solution_36": "[quote=arqady][quote=sqing]\nCrux,Problems 3419:\nLet $ a,b,c$ are  nonnegative reals. Prove that\n$$\\sqrt{\\frac{a^{2}+4bc}{b^2+c^2}}+\\sqrt{\\frac{b^{2}+4ca}{c^2+a^2}}+\\sqrt{\\frac{c^{2}+4ab}{a^2+b^2}}\\ge 2+\\sqrt{2}$$\n[/quote]\nLet $c=\\min\\{a,b,c\\}$.\nThus, by C-S $$\\sqrt{\\frac{a^{2}+4bc}{b^2+c^2}}+\\sqrt{\\frac{b^{2}+4ca}{c^2+a^2}}=\\frac{\\sqrt{(a^{2}+4bc)(b^2+c^2)}}{b^2+c^2}+\\frac{\\sqrt{(b^{2}+4ca)(a^2+c^2)}}{c^2+a^2}\\geq$$\n$$\\geq\\frac{ab+2\\sqrt{bc^3}}{b^2+c^2}+\\frac{ab+2\\sqrt{ac^3}}{a^2+c^2}\\geq\\frac{ab+2c^2}{b^2+c^2}+\\frac{ab+2c^2}{a^2+c^2}.$$\nAlso, $$\\sqrt2-\\sqrt{\\frac{c^{2}+4ab}{a^2+b^2}}\\leq\\sqrt2-\\sqrt{\\frac{4ab}{a^2+b^2}}=$$\n$$=\\frac{\\sqrt2(a-b)^2}{a^2+b^2+\\sqrt{2ab(a^2+b^2)}}\\leq\\frac{1.5(a-b)^2}{(a+b)^2}.$$\nId est, it's enough to prove that:\n$$\\frac{ab+2c^2}{b^2+c^2}+\\frac{ab+2c^2}{a^2+c^2}\\geq2+\\frac{1.5(a-b)^2}{(a+b)^2},$$\nwhich is obvious.[/quote]\nSir how is the last inequality obvious, Is buffalo way neccessary?",
  "Solution_37": "[quote=sqing]Let $ a,b,c$ are  nonnegative reals. Prove that\\[\\sqrt{\\frac{a^2 \\plus{}2bc}{b^2 \\plus{}c^2}}\\plus{}\\sqrt{\\frac{b^2 \\plus{}2ca}{c^2 \\plus{}a^2}}\\plus{}\\sqrt{\\frac{c^2 \\plus{}2ab}{a^2 \\plus{}b^2}}\\geq 3.\\][/quote]\n[url=https://artofproblemsolving.com/community/c6h1115969p5108246]here[/url] [url=https://artofproblemsolving.com/community/c6h1646201p10405989]here[/url]\n[color=#f00]Answers are welcome.[/color]",
  "Solution_38": "[quote=sqing]Let $ a,b,c$ are  nonnegative reals. Prove that\\[\\sqrt{\\frac{a^2 \\plus{}2bc}{b^2 \\plus{}c^2}}\\plus{}\\sqrt{\\frac{b^2 \\plus{}2ca}{c^2 \\plus{}a^2}}\\plus{}\\sqrt{\\frac{c^2 \\plus{}2ab}{a^2 \\plus{}b^2}}\\geq 3.\\][/quote][size=50]p/1383596268[/size]",
  "Solution_39": "[quote=sqing]Let $ a,b,c$ are  nonnegative reals. Prove that\\[\\sqrt{\\frac{a^2 \\plus{}2bc}{b^2 \\plus{}c^2}}\\plus{}\\sqrt{\\frac{b^2 \\plus{}2ca}{c^2 \\plus{}a^2}}\\plus{}\\sqrt{\\frac{c^2 \\plus{}2ab}{a^2 \\plus{}b^2}}\\geq 3.\\][/quote]\n[url=https://artofproblemsolving.com/community/c6h3007280p27010757]h[/url] [url=https://artofproblemsolving.com/community/c6h361540p1979760]h[/url] [url=https://artofproblemsolving.com/community/c4h2464062p20573119]h[/url] [url=https://artofproblemsolving.com/community/c6h559045p3252033]h[/url] [url=https://math.stackexchange.com/questions/4040811/sqrt-fraca24bcb2c2-sqrt-fracb24aca2c2-sqrt-fracc24?noredirect=1]h[/url] [url=https://artofproblemsolving.com/community/c6h1300514p6925295]h[/url] [url=https://artofproblemsolving.com/community/c6h1823379p12187705]h[/url]",
  "Solution_40": "Let $ a,b,c$ are nonnegative reals. Prove that$$\\sqrt{\\frac{a^2 +8bc}{b^2 +c^2}}+\\sqrt{\\frac{b^2 +8ca}{c^2 +a^2}}+\\sqrt{\\frac{c^2 +8ab}{a^2 +b^2}} \\geq 4$$\n$$\\sqrt{\\frac{a^2 +18bc}{b^2 +c^2}}+\\sqrt{\\frac{b^2 +18ca}{c^2 +a^2}}+\\sqrt{\\frac{c^2 +18ab}{a^2 +b^2}} \\geq 5$$\n$$\\sqrt{\\frac{a^2 +32bc}{b^2 +c^2}}+\\sqrt{\\frac{b^2 +32ca}{c^2 +a^2}}+\\sqrt{\\frac{c^2 +32ab}{a^2 +b^2}} \\geq 6$$",
  "Solution_41": "[quote=sqing]Let $ a,b,c$ are  nonnegative reals. Prove that\\[\\sqrt{\\frac{a^2 \\plus{}bc}{b^2 \\plus{}c^2}}\\plus{}\\sqrt{\\frac{b^2 \\plus{}ca}{c^2 \\plus{}a^2}}\\plus{}\\sqrt{\\frac{c^2 \\plus{}ab}{a^2 \\plus{}b^2}}\\geq 2+\\frac{\\sqrt{2}}{2}.\\][/quote]\nLet $a$, $b$ and $c$ be positive numbers. [url=https://artofproblemsolving.com/community/c6h3090659p27934129]Prove that:[/url]\n$$\\sqrt{\\frac{a^2+bc}{b^2+c^2}}+\\sqrt{\\frac{b^2+ca}{c^2+a^2}}+\\sqrt{\\frac{c^2+ab}{a^2+b^2}}\\geq2\\left(\\sqrt{\\frac{ab}{(a+c)(b+c)}}+\\sqrt{\\frac{bc}{(b+a)(c+a)}}+\\sqrt{\\frac{ca}{(c+b)(a+b)}}\\right)$$\n[url]https://artofproblemsolving.com/community/c6h1646201p10405989[/url]\n[url]https://artofproblemsolving.com/community/c6h1115969p5108413[/url]\nIf $a,b,c\\geq 0 ,\\lambda >0.$ Then$$\\sqrt{\\frac{a^{2}+\\lambda bc}{b^{2}+c^{2}}}+\\sqrt{\\frac{b^{2}+\\lambda ca}{c^{2}+a^{2}}}+\\sqrt{\\frac{c^{2}+\\lambda ab}{a^{2}+b^{2}}}\\geq 3\\sqrt[3]{\\frac{\\lambda }{4}} $$\n$$\\sqrt{\\frac{a^{2}+ 256 bc}{b^{2}+c^{2}}}+\\sqrt{\\frac{b^{2}+ 256 ca}{c^{2}+a^{2}}}+\\sqrt{\\frac{c^{2}+ 256 ab}{a^{2}+b^{2}}}\\geq 12 $$"
}
{
  "Tag": [
    "vector",
    "LaTeX",
    "topology",
    "linear algebra",
    "linear algebra unsolved"
  ],
  "Problem": "Prove that any contraction operator C of an Euclidean space equals $\\Pi \\circ Q$ where $\\Pi$ is an orthogonal projection and $Q$ is an orthogonal operator",
  "Solution_1": "First of all this is kind of analysis more isn't it?  On the other hand there are the linear operators...\r\n\r\nNever the less : what about this map \r\n\r\n$(x,y)$--> (x/2,y/2)\r\n\r\nwhy is this not a contraction?",
  "Solution_2": "[quote=\"fredbel6\"]First of all this is kind of analysis more isn't it?  On the other hand there are the linear operators...\n\nNever the less : what about this map \n\n$(x,y)$--> (x/2,y/2)\n\nwhy is this not a contraction?[/quote]\r\nIs it true if I add the contraction is not injective",
  "Solution_3": "What do you mean by \"contraction\" anyway? :? As far as I know, it means an operator with norm $\\le 1$. However, non-null orthogonal projections have norm precisely $1$, and multiplication by an orthogonal operator does not change the norm, so all operators of that form have norm equal to $1$. You can't get anything with norm $<1$.",
  "Solution_4": "As i saw it :? , contractions f were mappings from a metric space (with a distance not necessarily induced by a norm as the space does not even need to have vector space structure) to the same space, where there was a constant k with  $0<k<1$, such that\r\n$d(f(x),f(y))\\leq k d(x,y)$ \r\nthose mappings are not even linear so how could we speak of norm? \r\nor am i just babbling?  \r\n\r\nPS : sorry about botching up in Latex the smaller or equal sign, does anyone know advice?"
}
{
  "Tag": [
    "Putnam",
    "function",
    "combinatorics proposed",
    "combinatorics"
  ],
  "Problem": "$n$ integers that sum to $n - 1$ are arranged in a circle. Show that we can choose one of the integers $x_1$ so that the other integers going around the circle are, in order, $x_2, ..., x_n$ and so that we have $\\Sigma_{i=1}^k x_i \\leq k-1$ for $k = 1, 2, ... , n$.\r\n\r\n\r\nI have a solution to this problem, but I do not understand how someone would come up with it. Maybe someone could explain how he/she thought about this problem to successfully solve it.\r\n\r\nThanks.",
  "Solution_1": "For me, it helped to have seen a similar problem before.  For example, I have seen the following problem (with clever solution) in Lovasz's book on [i]Combinatorial Problems and Exercises[/i]: \r\n\r\nAlong a speed track there are some gas stations. The total amount of gasoline available in them is equal to what our car needs for going around the track. Prove that there is a gas station such that if we start there with an empty tank, we shall be able to go around the track without running out of gasoline.",
  "Solution_2": "I remember solving a very similar problem a long time ago, which is, in fact, the same as this one. \r\n\r\nInstead of working with $x_i$, we'll replace them with $1-x_i$ (and denote the quantities $1-x_i$ also by $x_i$). The new problem reads: given $n$ integers $x-1,\\ldots,x_n$ with sum $1$ around the circle, we can find one of them s.t. the sums mentioned in the original post are all positive as we go around the circle (in a counterclockwise direction, for instance).\r\n\r\nIt's going to be a bit hard to explain, I think, but it's perfectly natural, if you don't let the definitions fool you. Just draw a picture.\r\n\r\nSuppose there is a function $f:\\{1,2,\\ldots,n\\}\\to\\{1,2,\\ldots,n\\}$ s.t. $x_k+x_{k+1}+\\ldots+x_{f(k)}\\le 0$ for all $k\\in\\overline{1,n}$, if $x_k,x_{k+1},\\ldots,x_{f(k)}$ are consecutive points on the circle in a counterclockwise direction ($x_{n+1}=x_1$, as usual). Let $S_k=\\{x_k,\\ldots,x_{f(k)}\\}$. Start with $a_1=1$, and put $a_{n+1}=f(a_n)+1$. Let $(m,p),\\ m<p$ the first pair with $a_p=a_m$. \r\n\r\nObserve now that if you add up all the numbers in $S_m,\\ldots,S_{p-1}$, you get a non-positive sum on the one hand, and, on the other hand, the multiset formed by adjoining the elements of $S_m,\\ldots,S_{p-1}$ contains each $x_i$ the same number of times, so the sum should have been positive (since all the $x_i$'s add up to $1>0$). \r\n\r\nThe contradiction shows that there is no function with the properties of $f$, and this is, in fact, precisely what we want to prove.\r\n\r\nEdit:\r\n\r\nYes, Ravi, this problem reminded me of the one you mention as well. That one has also been discussed on the forum, also in relation to the one I solved, which was posted either by Moubinool or Pierre (I cannot remember which one, unfortunately, but I suspect they both discussed it, since I have the feeling it was used in a Concours Generale :)).",
  "Solution_3": "[edit:removed]",
  "Solution_4": "well also the following can be done, consider all possible walks, and in each walk record under each number the partial sum (start each walk with sum 0). Suppose point $i$ gets the smallest recorded partial sum, then it is easy to prove starting from $i+1$ we get a good walk.",
  "Solution_5": "You guys are kind of confussing me.  Anyways, here is my solution which I think is straight forward:\r\n\r\nLet $S_j=\\sum_{i=1}^{k} x_{(i+j\\mod n)}$ (where $x_0=x_n$).\r\n\r\nNote that $S_1+S_2+\\ldots+S_n=k(n-1)$.\r\n\r\nTheir average is $k\\frac{n-1}{n}$ so there must be at least one $S_j$ such that $S_j\\leq k\\frac{n-1}{n}=k-\\frac{k}{n}$.  Since $S_j$ is an integer, $S_j\\leq k-1$.",
  "Solution_6": "There is not just one value of $k$.  You have to prove that $\\sum_{i=1}^k x_i \\le k - 1$ holds for all $k$ (between 1 and $n$) simultaneously.",
  "Solution_7": "Ah, yes, thank you.  That did seem ridicolously easy for an A4 Putnam Problem."
}
{
  "Tag": [],
  "Problem": "A dragon was tamed by Dumbledore to guard his home. The dragon guarded the home on the condition that Dumbledore feed him 4 pieces of meat for every 10 hours. If one cow provides 10 pieces of meat, what is the smallest whole number of cows  Dumbledore needs to guard his home for 1 month (30 days)?",
  "Solution_1": "Ooh...a dragon. 30 days is equiv to 720 hours, so Dumbledore would have to feed me 72 times, or 288 pieces of meat. 288 divided by 10 and rounded up is equivalent to 29 cows."
}
{
  "Tag": [
    "vector",
    "trigonometry"
  ],
  "Problem": "A boy is playing with a toy airplane attached to a string. At time=0 the toy plane is moving at 5.0m/s [N] . At t=0.15 the plane is moving at 5.0m/s[N 40 degrees  E]. What was the magnitude of the average acceleration of the plane during thist time?",
  "Solution_1": "I have actually asked this before, and you ignored me, please say what you mean by [N 40 degrees E]. Is it 40 degrees east of north, or 40 degrees north of east???",
  "Solution_2": "I assume it is $ 40^{o}$ East of North.\r\n\r\nSolution\r\n\r\nLet $ \\mathbf{i}$ and $ \\mathbf{j}$ be unit position vectors in the directions East and North respectively.\r\n\r\nInitial velocity is: $ u = 5\\mathbf{j}$\r\n\r\nFinal velocity is: $ v = 5\\cos{50^{o}}\\mathbf{i}+5\\sin{50^{o}}\\mathbf{j}= 3.214\\mathbf{i}+3.830\\mathbf{j}$\r\n\r\nChange in velocity: $ \\Delta v = v-u = 3.214\\mathbf{i}-1.17\\mathbf{j}$\r\n\r\nAverage acceleration: $ a =\\frac{\\Delta v}{t}= 21.43\\mathbf{i}-7.8\\mathbf{j}$\r\n\r\n$ |a| =\\sqrt{21.43^{2}+7.8^{2}}= 23 ms^{-2}$\r\n\r\n$ \\theta =\\arctan{\\frac{-7.8}{21.43}}=-20^{o}$\r\n\r\nAcceleration is $ 23 ms^{-2}$, $ 20$ degrees clockwise from East.\r\n\r\nThis can angle can also be obtained from purely geometric reasoning, and the acceleration's magnitude from the cosine rule.",
  "Solution_3": "[quote=\"sludgethrower\"]I have actually asked this before, and you ignored me, please say what you mean by [N 40 degrees E]. Is it 40 degrees east of north, or 40 degrees north of east???[/quote]\r\n\r\nI am so sorry. It means going 40 degrees east from north so you were right ."
}
{
  "Tag": [
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Find the solution of \r\n\r\ny\"(x) - 4y(x) = 0\r\n\r\nI simply started it off by trying to create a general form but I'm stuck.\r\n\r\na^2 - 4 = 0\r\n\r\nWhat should I do from here?",
  "Solution_1": "You are supposing some solution in the form of $ Ce^{ax}$ for the constant $ C$.\r\n\r\nGo ahead. :)",
  "Solution_2": "hello, by your ansatz you get $ a_1\\equal{}2$ or $ a_2\\equal{}\\minus{}2$, so the general solution \r\nis $ y(x)\\equal{}C_1e^{\\minus{}2x}\\plus{}C_2e^{2x}$.\r\nSonnhard."
}
{
  "Tag": [
    "function",
    "integration",
    "calculus",
    "vector",
    "trigonometry",
    "linear algebra",
    "real analysis"
  ],
  "Problem": "i had this for an exam question, im not sure what it means or what my professor is trying to show:\r\n\r\nConsider the set of functions which are continuous on the interval [a, b].\r\nDefine the two dot products between the functions $f$ and $g$ as\r\n\r\n$f \\cdot g = \\int_a^bf(x)g(x)dx$\r\n\r\nshow that the two functions $f(x) = sin(x)$ and $g(x) = cos(x)$ are orthogonal on the interval $[-\\pi, \\pi]$.  Hint: show their dot product is zero\r\n\r\n\r\ni evaluted the integral which came out to zero, and got the credit, but im clueless.  we just finished cross products, projections... and we had done a little bit with parametric equations.  right after we went into 3 dimensional parametric equations in vector form, r(t) = <f(t), g(t), h(t)>, so im thinking he was trying to set us up for that, but the above question seems like nonsense.  two things that got me were that there's an infinite number of sumations within the dot product which to me would be an infinite number of dimensions.  also there's only one value for f and one for g, which wouldnt be able to define a point in two dimensions.",
  "Solution_1": "Yes, there are infinitely many dimensions. What's wrong with that?\r\n\r\nThe definition of an inner product (dot product): A vector space $V$ has an inner product $\\langle\\,,\\,\\rangle$ if $\\langle\\,,\\,\\rangle$ satisfies these properties:\r\n1: Bilinear: $\\langle ax+by,z\\rangle=a\\langle x,z\\rangle+b\\langle y,z\\rangle$ and $\\langle z,ax+by\\rangle=a\\langle z,x\\rangle+b\\langle z,y\\rangle$.\r\n2: Symmetric: $\\langle x,y\\rangle=\\langle y,x\\rangle$\r\n3: Positive definite: $\\langle x,x\\rangle>0$ if $x\\neq0$.\r\n\r\nThe inner product can be used to define a norm: $\\|x\\|=\\sqrt{\\langle x,x\\rangle}$.\r\nHow does this relate to what you already know? On any finite-dimensional subspace, it is always possible to find an orthonormal basis, and the inner product behaves like the standard dot product with respect to that basis. In your example, you could always reduce to functions of the form $f(x)=a\\sin x+b\\cos x+c$; that's three-dimensional, and the obvious basis is already orthogonal, but not orthonormal.\r\nFor infinite-dimensional spaces, the right definitions are trickier, but this question hints at the way to Fourier series, which are based on the functions $1,\\sin x,\\cos x,\\sin 2x,\\cos 2x,\\dots$ being an orthogonal basis for the appropriate space of functions.\r\n\r\nThere is no continuation in the course, but it's worth keeping in mind that $\\mathbb{R}^n$ is not the only interesting vector space.",
  "Solution_2": "[quote=\"oooooh ya\"]... there's an infinite number of sumations within the dot product which to me would be an infinite number of dimensions ...[/quote]\r\nIndeed so. The set whose members are all of the continuous functions on a certain interval is an infinite-dimensional vector space.\r\n\r\nWhat we have here is a typical case of how mathematics abstracts and generalizes from certain concrete cases.\r\n\r\nStart with $\\mathbb{R}^2$ or $\\mathbb{R}^3.$ We talk about vectors in these spaces, and we have some specific, concrete pictures in mind for what we mean. But back off from the question of what a vector \"is\" and consider what you can do with vectors, algebraically. The essence boils down to just two things: you can add (or subtract) two vectors to get another vector, and you can multiply a scalar (a real number) by a vector to get a vector. These algebraic operations obey some (more or less) familiar laws: addition is commutative and associative; there's a zero vector (an additive identity); there are negatives (additive inverses). For scalar multiplication, every reasonable identity that reminds us of an associative or distributive law is true.\r\n\r\nThe next step is to take that list of algebraic properties as [i]axioms[/i]; we call these the axioms of a (real) vector space. What good is this? Any theorems and definitions that we base on these axioms - and there are plenty of theorems and definitions; the subject is called \"Linear Algebra\" - are true in any system, any new example that we create that satisfies these axioms. \"Dimension\" is one of the definitions from linear algebra; it's the number of elements in a minimal spanning set.\r\n\r\nAt this point, what about $C[-\\pi,\\pi],$ the set of (real-valued) continuous functions on the interval $[-\\pi,\\pi]?$ Is the sum of two continuous functions a continuous function? Is a constant times a continuous function a continuous function? The rest of the arithemetic all works; we have a vector space. But we can't find a finite spanning set - a finite sets such that we can write any contininuous function as a linear combination of elements of this set. So the $C[\\pi,\\pi]$ is an infinite-dimensional vector space, and with that observation, we lose confidence in much of our geometric intuition.\r\n\r\nThen there's the dot product. For $x$ and $y$ in $\\mathbb{R}^2,$ we can define the dot product (aka scalar product) of $x$ and $y$ so that it is a real number. Does the dot product have significant algebraic properties? Yes. We can boil them down to three. The dot product is:\r\n\r\nSymmetric: $x\\cdot y=y\\cdot x$ for all $x,y.$\r\n\r\nBilinear: $(ax+by)\\cdot z=a(x\\cdot z)+b(y\\cdot z)$ and $x\\cdot(ay+bz)=a(x\\cdot y)+b(x\\cdot z)$ for all vectors $x,y,z$ and all scalars $a,b.$\r\n\r\nPositive definite: $x\\cdot x>0$ whenever $x\\ne0.$\r\n\r\nAnd we can prove a lot of interesting things based on this, including the Cauchy-Schawrz inequality. We make the obervation that $x$ and $y$ are orthogonal whenever $x\\cdot y=0.$\r\n\r\nAnd so we abstract that; any symmetric positive definite bilinear form on a real vector space gets called an [i]inner product[/i]. What about $\\langle f,g\\rangle=\\int_{-\\pi}^{\\pi}f(x)g(x)\\,dx?$ We can show that that is symmetric, bilinear and positive definite (at least on the continuous functions.) And so we call $C[-\\pi,\\pi]$ an inner product space. \r\n\r\nBack in $\\mathbb{R}^2,$ the idea that $x\\cdot y=0$ iff $x$ and $y$ are orthogonal was an observation. In a more abstract inner product space, it's a definition, not a theorem. It's the definition of \"orthogonal.\" Hence all you did was to verify the condition of the definition. The \"hint\" you reported was in fact the only thing you could do.\r\n\r\nThis is actually a powerful tool in mathematics. But you have to accept not being bound by familiar pictures in order to use it."
}
{
  "Tag": [
    "geometry"
  ],
  "Problem": "In a triangle ABC, O is the circumcenter.\r\nsides AB and AC are extended. A circle passing through B , O and C meets AB and AC at P and Q.\r\nShow that AO when intersects PQ is perpendicular to it.",
  "Solution_1": "No one as yet?",
  "Solution_2": "Anyone???\r\nIm not able to prove BC parallel to PQ",
  "Solution_3": "[quote=\"Zed_darkx\"]Anyone???\nIm not able to prove BC parallel to PQ[/quote]Why does BC have to be parallel to PQ? After all, AO does not necessarily be perpendicular to BC in an arbitrary triangle ABC.",
  "Solution_4": "[quote=\"10000th User\"][quote=\"Zed_darkx\"]Anyone???\nIm not able to prove BC parallel to PQ[/quote]Why does BC have to be parallel to PQ? After all, AO does not necessarily be perpendicular to BC in an arbitrary triangle ABC.[/quote]\r\nThen how do we prove this thing?\r\nOr is true for some special case?"
}
{
  "Tag": [],
  "Problem": "I am trying to add a method to my source file, but I get the error \"(method name) was not declared in this scope.\"\r\nI am using Eclipse. This is my code:\r\n\r\n[code]#include <iostream>\nusing namespace std;\n\nint main()\n{\n\tcout << \"Hello World!\" << endl;\n\tprintIt();\n\treturn 0;\n}\n\nvoid printIt()\n{\n\tcout << \"Goodbye!\" << endl;\n}\n[/code]",
  "Solution_1": "I forgot the prototype! Problem fixed."
}
{
  "Tag": [
    "calculus",
    "integration",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Find all integers $x,y$ that :\r\n\\[ x^2+x+2=y^3\\]",
  "Solution_1": "Well it's easy for positive integers \r\nlook if 2+x+x^2=y^3 then 1+x+x^2=y^3-1 since for x=1 and y=1 there are no sol.   (x^3-1)/(y^3-1)=x-1 then x\\leq y. if x=y there are sol. x=y=1 if \r\nx<y  then y^3=2+x+x^2<2+y+y^2 wich is false for y>2. \r\nSo the only positive integer sol. is x=y=2. [/tex]",
  "Solution_2": "im sorry if im slow today, but why does (x^3-1)/(y^3-1)=x-1 imply that x <= y? cause if x < y then (x^3-1)/(y^3-1) is not an integer.\r\n\r\nanyway, you only need to consider positive integral solutions, because once you have a solutions (x,y) with x positive, you get the solution (-x-1,y).",
  "Solution_3": "[quote=\"zscool\"]anyway, you only need to consider positive integral solutions, because once you have a solutions (x,y) with x positive, you get the solution (-x-1,y).[/quote]\r\n\r\nyou also need to prove there are no others there ;)",
  "Solution_4": "Here is my attempt:\r\n\r\nIt's not difficult to see that if (x, y) is a solution, then (-x-1, y) is also a solution. Hence we only consider x,y>=1. We see there is no solution when x=1 or y=1, so assume x,y>=2. Now x>=y from (x^3-1)/(y^3-1)=x-1 (otherwise 0<(x^3-1)/(y^3-1)<1, impossible.) So x>=y>=2.\r\n\r\nLet a=[:sqrt:(y<sup>3</sup>-y<sup>3/2</sup>+9/4)], where [] denotes the floor function. Then (a-1)a<y^3<(a+1)(a+2). Hence x=a. Similarly, let b=[:sqrt:(y<sup>3</sup>+y<sup>3/2</sup>-7/4)], we have (b-1)b<y^3<(b+1)(b+2). Hence x=b, or, a=b. But that implies  [:sqrt:( (y<sup>3/2</sup>-1/2)<sup>2</sup>-2)]=[:sqrt:( (y<sup>3/2</sup>+1/2)<sup>2</sup>+2)]. For sufficiently large y, we see that LHS=[y<sup>3/2</sup>-1/2]=[y<sup>3/2</sup>+1/2]=RHS. Impossible. Now we can check for smaller values of y to conclude that (x, y)=(2, 2) and (x, y)=(-3, 2) are the only solutions.\r\n\r\nI know that the \"sufficiently large y\" part of the proof seems shaky, so it's still an attempt, not yet a proof. Is this correct?",
  "Solution_5": "This is not quite correct.\r\n[tex] (a-1)a<y^3<(a+1)(a+2) [/tex] implies both [tex]x=a-1[/tex] and  [tex]x=a[/tex], as well for [tex]b[/tex].\r\nThen a contradiction of this type cannot arise:\r\n[tex] [ (y^3/2-1/2)^2-2) ]=[ (y^3/2+1/2)^2-2 ] [/tex]\r\n(btw this has an infinite number of solutions)",
  "Solution_6": "OOPS! My mistake in typing! The second paragraph should have been the following:\r\n\r\n(a-1)a<y<sup>3</sup>-2<(a+1)(a+2) and (b-1)b<y<sup>3</sup>-2<(b+1)(b+2).\r\n\r\nNow it should be okay. Also, heartwork, I think you forgot the square roots inside the floor function...  :D",
  "Solution_7": "I missed the square roots in typing, but i ment that the equation:\r\n[tex] [ {\\sqrt{(y^3/2-1/2)^2-2)}} ]=[ {\\sqrt{(y^3/2+1/2)^2-2)}} ] [/tex]\r\nhas no solutions. Indeed you can check:\r\n[tex] {\\sqrt{(y^3/2+1/2)^2-2)}} - {\\sqrt{(y^3/2-1/2)^2-2)}} >1 [/tex]"
}
{
  "Tag": [
    "inequalities solved",
    "inequalities"
  ],
  "Problem": "Let $a,b,c\\in R$ and $a+b+c=1$ .Prove that \r\n$\\frac{a}{1+a^2}+\\frac{b}{1+b^2}+\\frac{c}{1+c^2}\\le\\frac{9}{10}$",
  "Solution_1": "See http://www.mathlinks.ro/Forum/viewtopic.php?t=27450 , http://www.mathlinks.ro/Forum/viewtopic.php?t=1831 and http://www.mathlinks.ro/Forum/viewtopic.php?t=50269 .\r\n\r\n  darij"
}
{
  "Tag": [
    "abstract algebra",
    "geometry",
    "parameterization",
    "Ring Theory",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "Let $ \\phi: A \\to B$ be a homomorphism of Noetherian rings, and $ P$ a prime ideal of $ B$. Then setting $ p \\equal{} P \\cap A$ we have\r\n\r\n$ ht P \\leq ht p \\plus{} dim B_P/pB_P$.\r\n\r\n[b]Proof:[/b]\r\n\r\nWe can replace $ A$ and $ B$ by $ A_p$ and $ B_P$ and assume $ (A,m)$ and $ (B,n)$ are local rings, with $ mB \\subset n$. Then the problem reduces to showing that:\r\n\r\n $ dim B \\leq dim A \\plus{} dim B/mB$. \r\n\r\nUp to here I understand the proof, but then it goes wrong :)\r\n\r\n[quote]\nTo prove this, take a system of parameters $ x_1,...,x_r$ of $ A$ and choose $ y_1,...,y_s \\in B$ such that their images in $ B/mB$ form a system of parameters in $ B/mB$. Then for $ v,u$ large enough we have $ n^v \\subset mB \\plus{} \\sum y_i B$ and $ m^u \\subset \\sum x_j A$, giving $ n^{vu} \\subset \\sum y_i B \\plus{} \\sum x_j B$. Hence $ dim B \\leq r \\plus{} s$\n[/quote]\r\n\r\nSeriously lost :( \r\n\r\n(p.s.: I have no algebraic geometry background, so my interpretation is very limited of this theorem.)",
  "Solution_1": "Let $ I\\equal{}(y_1,\\ldots,y_s)\\plus{}mB$. The assumption says that $ I/mB$ is a parameter ideal in the local ring $ (B/mB,n/mB)$, so there exists $ v$ such that $ (n/mB)^v\\subseteq I/mB$, which is equivalent to $ n^v\\subseteq I$. owk",
  "Solution_2": "What's a \"parameter ideal\"? An ideal generated by a system of parameters?\r\n\r\nFurthermore, I don't understand either why it is possible to:\r\n\r\n[quote]\ntake a system of parameters $ x_1,...,x_r$ of $ A$ and choose $ y_1,...,y_s \\in B$ such that their images in $ B/mB$ form a system of parameters \n[/quote]",
  "Solution_3": "[quote=\"Jacobson\"]What's a \"parameter ideal\"? An ideal generated by a system of parameters?[/quote]\n\nYes.\n\n[quote]\ntake a system of parameters $ x_1,...,x_r$ of $ A$ and choose $ y_1,...,y_s \\in B$ such that their images in $ B/mB$ form a system of parameters \n[/quote]\r\n\r\nI do not understand what is unclear about this. Maybe you should read the first paragraph of \u00a714 again. owk"
}
{
  "Tag": [],
  "Problem": "A semicircle is inscribed in a quadrilateral ABCD in such a way that the midpoint\r\nof BC coincides with the center of the semicircle, and the diameter of the semicircle\r\nlies along a portion of BC. If AB = 4 and CD = 5, what is BC?",
  "Solution_1": "[hide=\"solution\"]\nCall the points of tangency to AB, AD, and DC F,G,and H. We let BF=HC=n (since the center of the semicircle's on the midpoint), so DH=GD=5-n and AF=GA=4-n. Let the radius be r. We wish to find $ 2\\sqrt {r^2 \\plus{} n^2}$. Call the center of the semicircle O. Angles BOF,FOA,AOG,DOG,DOH,and HOC must add up to 180. So 2arctan(n/r)+2arctan((4-n)/r)+2arctan((5-n)/r)=180. Solving this gives $ r^2 \\plus{} n^2 \\equal{} 20$, so subbing in the answer is $ 4\\sqrt {5}$. [/hide]\r\nSince there is more than one possible figure, though, you could assume something in solving the problem, making it easier but not fully proving the result. By the way, I just edited my solution to make it easier.",
  "Solution_2": "There are 689 Theorems on cut-the-knot and I was just randomly clicking on them on and the second one I clicked was similar to this problem! (I'm not kidding!). It's called the see-saw theorem. [url]http://www.cut-the-knot.org/Curriculum/Geometry/SeeSaw.shtml[/url]. It's not the same as this but I think there must be a way to generalize it to this, seeing that both have the answer as the geometric mean."
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "geometry unsolved"
  ],
  "Problem": "Two circles $ \\Gamma$ and $ \\Gamma '$ intersect at $ A$ and $ D$. A line is tangent to $ \\Gamma$ and $ \\Gamma '$ at $ E$ and $ F$ , respectively. The line $ BC$ contains the point $ D$ parallel to the line $ EF$ such that $ C$ in $ \\Gamma$ and $ B$ in $ \\Gamma '$. Show that the circumcircles of $ \\triangle BDE$ and $ \\triangle CDF$ intersect again on the line $ AD$.",
  "Solution_1": "Consider the positive inversion through center $ D$ and arbitrary radius. Let  $ B',C',E',F'$ be the images of $ B,C,E,F.$ The circumferences $ \\odot(BDE)$ and $ \\odot(CDF)$ are taken into the lines $ B'E'$ and $ C'F'$ intersecting at the inverse image $ A'$ of $ A$ and the circles $ \\Gamma$ and $ \\Gamma'$ are taken  into the lines $ A'C'$ and $ A'B'.$ Since $ BC \\parallel EF,$ the common tangent $ EF$ of $ \\Gamma, \\Gamma'$ is transformed into the incircle $ \\omega$ of $ \\triangle A'B'C'$ tangent to $ B'C',$ $C'A',$ $A'B'$ at $ D,E',F'.$ The lines $ B'E',C'F'$ and $ A'D$ concur at the Gergonne point of $ \\triangle A'B'C',$ thus the primitive figures, namely the circles  $ \\odot(BDE) , \\odot(CDF)$ and the double  line $ AD$ concur at another point besides $ D.$",
  "Solution_2": "another solution is to prove that $ O_1S_1\\equal{}O_2S_2$ and then obviously we r done!\r\n($ O_1,S_1,O_2,S_2$ are the centers of $ \\Gamma,(CDF),\\Gamma',(BDE)$)",
  "Solution_3": "[quote=\"luisgeometria\"]\nThe circumferences $ \\odot(BDE)$ and $ \\odot(CDF)$ are taken into the lines $ B'E'$ and $ C'F'$ intersecting at the inverse image $ A'$ of $ A$[/quote]\r\n\r\nI think the intersection of $ B'E'$ and $ C'F'$ is not always $ A'$. Check the image.\r\nIf I'm wrong, I want to know why?",
  "Solution_4": "[quote=\"mathson\"]I think the intersection of $ B'E'$ and $ C'F'$ is not always $ A'.$ Check the image.[/quote]\nDear mathson, I did not claim that the lines B'E' and C'F' meet at A', but they meet at the Gergonne point of  A'B'C'.",
  "Solution_5": "I'm really sorry, I didn't understand you at the first time.\n\n[quote]\nthe common tangent $ EF$ to $ \\Gamma$ and $ \\Gamma'$ is transformed into the incircle $ \\omega$ of $ \\triangle A'B'C'$ \n[/quote]\nIs there any generalization for this? Or theorem?"
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "For $ a,b,c$ positive reals prove that\r\n\r\n$ \\sum \\frac{a}{b} \\plus{} 3 \\geq 2(\\sum \\frac{a\\plus{}b}{b\\plus{}c})$",
  "Solution_1": "after expanding It's equivalent to prove that \r\n\\[ \\sum_{cyc}a^3b^3\\plus{}\\sum_{cyc}a^4c^2\\plus{}\\sum_{cyc}a^3bc^2 \\geq \\sum_{cyc}a^4bc\\plus{}\\sum_{cyc}a^3b^2c\\plus{}3a^2b^2c^2\\]\r\nbut I don't know how to prove it. :(",
  "Solution_2": "[quote=\"manlio\"]For $ a,b,c$ positive reals prove that\n\n$ \\sum \\frac {a}{b} \\plus{} 3 \\geq 2(\\sum \\frac {a \\plus{} b}{b \\plus{} c})$[/quote]\r\nLet $ a\\equal{}0.1,b\\equal{}1.2,c\\equal{}0.2$, the inequality reversed. :(\r\nHence, the desired inequality is wrong. Actually, it holds for $ a,b,c$ are the side lengths of a triangle.",
  "Solution_3": "[quote=\"can_hang2007\"][quote=\"manlio\"]For $ a,b,c$ positive reals prove that\n\n$ \\sum \\frac {a}{b} \\plus{} 3 \\geq 2(\\sum \\frac {a \\plus{} b}{b \\plus{} c})$[/quote]\nLet $ a\\equal{}0.1,b\\equal{}1.2,c\\equal{}0.2$, the inequality reversed. :(\nHence, the desired inequality is wrong. Actually, it holds for $ a,b,c$ are the side lengths of a triangle.[/quote]\r\nNow, let us prove the inequality in the case $ a,b,c$ are the side lengths of a triangle. Rewrite it as\r\n\\[ \\sum \\frac {a}{b} \\plus{} \\sum\\frac {2b}{a \\plus{} b} \\ge \\sum \\frac {2a}{b \\plus{} c} \\plus{} 3\r\n\\]\r\n\r\n\\[ \\Leftrightarrow (\\sum ab)(\\sum \\frac {a}{b} \\plus{} \\sum\\frac {2b}{a \\plus{} b}) \\ge (\\sum \\frac {2a}{b \\plus{} c} \\plus{} 3)(\\sum ab)\r\n\\]\r\n\r\n\\[ \\Leftrightarrow \\sum \\frac {ca^2}{b} \\plus{} 2\\sum \\frac {ab^2}{a \\plus{} b} \\ge \\sum a^2 \\plus{} 2abc\\sum \\frac {1}{b \\plus{} c}\r\n\\]\r\nSince $ a,b,c$ are the side lengths of a triangle, by VasC's Inequality (Moldova 2006), we have\r\n\\[ \\sum \\frac {ca^2}{b} \\ge \\sum a^2\r\n\\]\r\nIt suffices to prove that\r\n\\[ \\sum \\frac {ab^2}{a \\plus{} b} \\ge \\sum \\frac {abc}{b \\plus{} c}\r\n\\]\r\n\r\n\\[ \\Leftrightarrow \\sum \\frac {ab(b \\minus{} c)}{a \\plus{} b} \\ge 0\r\n\\]\r\n\r\n\\[ \\Leftrightarrow \\sum ab(b^2 \\minus{} c^2)(a \\plus{} c) \\ge 0\r\n\\]\r\n\r\n\\[ \\Leftrightarrow \\sum a^2b^3 \\ge \\sum a^2b^2c\r\n\\]\r\nwhich is trivial by the AM-GM. Equality holds if and only if $ a \\equal{} b \\equal{} c.$ :)"
}
{
  "Tag": [],
  "Problem": "Find all four digit numbers of the form $aabb$ such that they are squares.\r\n\r\nMasoud Zargar",
  "Solution_1": "[hide]Let $\\ aabb = c^2$\n=> $\\ a.1100+b.11= c^2$\nSo $\\ c^2\\equiv\\ 0\\mod11$ so $\\ c\\equiv\\ 0\\mod11$ \nLet c =11d(d belong to Z)We have:\n$\\ a.1100+b.1100= 121d^2$\n$\\ a.100+b= 11d^2$\nSo $\\ a.100+b\\equiv\\ 0\\mod11$ => $\\ a+b\\equiv\\ 0\\mod11$\nAnd a+b<20 so a+b=11\nTry with all 2299, 3388, 4477, 5566, 6655, 7744, 8833, 9922.We only have 7744[/hide] :play_ball: \r\n\r\n\r\n\r\nMath is my life :!:",
  "Solution_2": "[quote=\"TRAN THAI HUNG\"][hide]Let $\\ aabb = c^2$\n=> $\\ a.1100+b.11= c^2$\nSo $\\ c^2\\equiv\\ 0\\mod11$ so $\\ c\\equiv\\ 0\\mod11$ \nLet c =11d(d belong to Z)We have:\n$\\ a.1100+b.1100= 121d^2$\n$\\ a.100+b= 11d^2$\nSo $\\ a.100+b\\equiv\\ 0\\mod11$ => $\\ a+b\\equiv\\ 0\\mod11$\nAnd a+b<20 so a+b=11\nTry with all 2299, 3388, 4477, 5566, 6655, 7744, 8833, 9922.We only have 7744[/hide] :play_ball: \n\n\n\nMath is my life :!:[/quote]\r\nProbably easier if you just checked the possibilities after finding out c=11d since 2<d<10",
  "Solution_3": "\"bb\" must be 0 or 1 modulo 4, implying $3b$ is 0 or 1 mod 4, implying $b = 0, 4, 8, 3, 7$ is possible.\r\n\r\nThe number is divisible by 11.  Therefore it is divisible by 121, implying $100a + b = 0 \\mod 11$, or $a+b = 0 \\mod 11$.  From $b = 0,4,8,3,7$ we have at most 5 numbers to check.",
  "Solution_4": "$11(100a+b)=p^2 \\implies 100a+b=11\\cdot x^2$\r\n\r\nNow $100a+b\\leq909$ otherwise, $11(100a+b)>9999$\r\n\r\n$\\implies x\\leq 9$\r\n\r\ncheck all possibilties $\\implies 100a+b=11\\cdot8^2=704$\r\n\r\nonly soln is $11\\cdot 704=7744$"
}
{
  "Tag": [
    "MATHCOUNTS"
  ],
  "Problem": "[color=#00CCFF]MATHCOUNTS Scores by jda12.8[/color]\r\n\r\n[color=#FF3300]6th Grade - 2004\n\nChapter competition 2004 - 32/46 (7th)\nState competition 2004 - 22/46 (9th)[/color]\r\n\r\n[color=#00CC00]7th Grade - 2005\n\nChapter competition 2005 - 41/46 (1st)\nState competition 2005 - 37/46 (1st)\nNational competition 2005 - ??/46 (99/228th)[/color]\r\n\r\n[color=#6600FF]8th Grade - 2006\n\nChapter competition 2006 - 43/46 (1st)\nState competition 2006 - 44/46 (1st)\nNational competition 2006 - 31/46 (41/228th, Arkansas team 12/57th)[/color]\r\n\r\n[color=#000000]mario_jda128[/color] :D",
  "Solution_1": "what is the purpose of this post?",
  "Solution_2": "[quote=\"funcia\"]what is the purpose of this post?[/quote]\r\nI think he wants us to post our scores.\r\nMaybe not"
}
{
  "Tag": [
    "inequalities",
    "geometry unsolved",
    "geometry"
  ],
  "Problem": "$a,b,c$ are sides of a triangle and $r$ is radius of interior circle. Prove that ; \r\n\r\n$\\frac{1}{a^{2}}+\\frac{1}{b^{2}}+\\frac{1}{c^{2}}\\leq\\frac{1}{4r^{2}}$",
  "Solution_1": "[hide]Substitute $a=x+y,b=y+z,c=z+x,s=x+y+z$ where $x,y,z$ are positive real numbers. \n$[ABC]=\\sqrt{s(s-a)(s-b)(s-c)}=\\sqrt{xyz(x+y+z)}=rs\\implies r^{2}=\\frac{xyz}{x+y+z}$\nThe inequality becomes\n$\\sum\\frac{1}{(x+y)^{2}}\\le \\frac{x+y+z}{4xyz}=\\sum\\frac{1}{4xy}$\nwhich is true since\n$\\frac{1}{(x+y)^{2}}\\le\\frac{1}{4xy}\\implies 4xy\\le x^{2}+2xy+y^{2}\\implies 0\\le(x-y)^{2}$\n[/hide]"
}
{
  "Tag": [
    "quadratics",
    "function",
    "algebra unsolved",
    "algebra",
    "inequalities"
  ],
  "Problem": "Let $n \\geq 2, n \\in \\mathbb{N}$, $a,b,c,d \\in \\mathbb{N}$, $\\frac{a}{b} + \\frac{c}{d} < 1$ and $a + c \\leq n,$ find the maximum value of $\\frac{a}{b} + \\frac{c}{d}$ for fixed $n.$",
  "Solution_1": "It is easy. Let $s=\\frac{a}{b}+\\frac{c}{d}=\\frac{ad+bc}{bd}$. \r\nIf b=d then $s\\le 1-\\frac{1}{b}\\le 1-\\frac{1}{2n}$. \r\nIf $b\\not =d$ then $s\\le \\frac{bd-1}{bd}=1-\\frac{1}{bd}.$ From $a+c\\le n$ we get $b,d\\le 2n$.\r\n We have $\\frac{n-1}{n}+\\frac{1}{n+1}=1-\\frac{1}{n(n+1)}$.",
  "Solution_2": "[quote=\"Rust\"]\nFrom $a+c\\le n$ we get $b,d\\le 2n$.\nWe have $\\frac{n-1}{n}+\\frac{1}{n+1}=1-\\frac{1}{n(n+1)}$.[/quote]\r\n\r\nI don't understand that, can you help me?",
  "Solution_3": "Let $s=\\frac{a}{b}+\\frac{c}{d}=\\frac{ad+bc}{bd}<1$. \r\nIf b=d then $s=\\frac{a+c}{b}$, maximum when $b=a+c+1$, because $a+c\\le n$, $s\\le 1-\\frac{1}{n+1}$. \r\nIf $b\\not =d$ then $s\\le \\frac{bd-1}{bd}=1-\\frac{1}{bd}.$\r\nIf $b\\ge n+1,c\\ge n+1$, then $s\\le 1-\\frac{1}{n+1}$. Therefore if $s>1-\\frac{1}{n+1}$ one of $b,d$ less than n+1. Let $b\\le n$. If $d\\le n$, then $s\\le 1-\\frac{1}{n^{2}}$. If $d>n+2$, then $s=1-\\frac{b-a}{b}+\\frac{c}{d}\\le 1-\\frac{b-a}{b}+\\frac{n-a}{n+2}<1-\\frac{1}{n(n+1)}$. \r\n We have $\\frac{n-1}{n}+\\frac{1}{n+1}=1-\\frac{1}{n(n+1)}$. Therefore it give maximum s.",
  "Solution_4": "Not true unfortunately.\nFor n=5,for example, we have 2/9+3/4=35/36 > 29/30.\nThe exact answer is a bit complicated. The maximum equals 1-1/t where\nt(3p)=4p^3+4p^2+3p+1\nt(3p+1)=4p^3+8p^2+7p+2\nt(3p+2)=4p^3+12p^2+14p+6\n\nSo t is not quadratic,but of third power wrt. n",
  "Solution_5": "can you give detail solution alexheinis?",
  "Solution_6": "I included the details in a pdf-file, I only had the solution in Dutch so I translated.\nBye.",
  "Solution_7": "Setting $b=a+x,d=c+y$ the problem gets easier, because we have\n${a\\over {a+x}}+{c\\over {c+y}}<1\\iff xy>ac$. We assume $a\\le c$.\nKeeping $\\pi:=xy$ fixed, we will show that the form ${a\\over {a+x}}+{c\\over {c+y}}$ \nis maximal for\n$x=\\pi,y=1$. We have ${a\\over {a+\\pi}}+{c\\over{c+1}}\\ge \n{a\\over {a+x}}+{c\\over {c+y}}\\iff ay+cx\\le a+c\\pi$.\nUsing the convex function $f(x)=cx+{{a\\pi}\\over x}$ one sees that the maximum on $[1,\\pi]$\nis attained in one of the extremities and, in this case, when $x=\\pi$.\nHence, if $xy=\\pi$, then the form is at most ${a\\over {a+\\pi}}+{c\\over {c+1}}$.\nThis form is maximal when $\\pi=ac+1$, substitution yields $1-{1\\over t}$, where\n$t:=(ac+a+1)(c+1)$. We want to maximise $t$, it is increasing in $a,c$, hence we may assume $a+c=n$.\nWith $\\gamma:=c+1,\\sigma:=n+1$ we find $\\gamma(a\\gamma+1)$ and we consider\n$g(x)=x+x^2(\\sigma-x)$ on $[{\\sigma\\over 2},\\sigma]$.\nThen $g$ is concave on this interval with a maximum in $x={{\\sigma+\\sqrt{\\sigma^2+3}}\\over 3}$.\nNotice ${{2\\sigma}\\over 3}<x<{{2\\sigma+1}\\over 3}$ and because we consider integers,\nwe split up with regards to $n$ modulo 3.\nIf $n=3p+1$ we have $\\sigma=3p+2$ and $2p+1<x<2p+2$. We need to plug in $x=2p+1,2p+2$\nand check which $x$ yields the bigger value. Similarly for the other cases modulo 3.\nWe find \\[t(3p)=4p^3+4p^2+3p+1,t(3p+1)=4p^3+8p^2+7p+2,t(3p+2)=4p^3+12p^2+14p+6\\]",
  "Solution_8": "Let $a,b,c,d \\in \\mathbb{N}$ and $\\frac{a}{b}+\\frac{c}{d}<1$. Prove that $$\\frac{a}{b}+\\frac{c}{d}<1-\\frac{1}{(a+c)^3}.$$",
  "Solution_9": "\u53ef\u4ee5\u7528n\u7684\u4ee3\u6570\u5f0f\u8868\u793a\u3002"
}
{
  "Tag": [
    "geometry"
  ],
  "Problem": "Hi, \r\nThis may sound like an obvious question, but why do you want to go to college?",
  "Solution_1": "I think this is a very good question, and I think the answers will be interesting and thought-provoking.",
  "Solution_2": "a) I love knowledge.\r\nb) College opens up all kinds of opportunites.  There are more job choices and more opportunities for internships etc.\r\nPi) I loved high school so much!  Seriously though, high school was a huge part of me-in particular cross country and math team.  In college, I still get to be a part of the cross country team.  After college I just wont have many options for running competitively with a group of people so close-knit.\r\n3) It's an adventurous, CHALLENGING experience that I can't pass up.\r\n$\\omega$) It's been indoctrinated in me for almost as long as I've been alive that I will/should go to college.",
  "Solution_3": "[quote=\"joml88\"]$\\omega$) It's been indoctrinated in me for almost as long as I've been alive that I will/should go to college.[/quote]\r\nThis obviously does not apply to joml88, but I will say it anyway: [u]make sure this is not your only reason[/u].",
  "Solution_4": "College is overrated.\r\n\r\nGarbage men in NY make $60,000 to$80,000 a year.",
  "Solution_5": "For me, it's trying to help me figure out what the heck I want to do in life.",
  "Solution_6": "[quote=\"neelnal\"]College is overrated.\n\nGarbage men in NY make $60,000 to$80,000 a year.[/quote]\r\n\r\nIf you want to be a garbage men, by all means, go ahead. Nobody is stopping you.\r\n\r\nWhy? To learn, of course, and expand my opportunities for getting a job doing something I love doing.",
  "Solution_7": "[quote=\"paladin8\"][quote=\"neelnal\"]College is overrated.\n\nGarbage men in NY make $60,000 to$80,000 a year.[/quote]\n\nIf you want to be a garbage men, by all means, go ahead. Nobody is stopping you.\n\nWhy? To learn, of course, and expand my opportunities for getting a job doing something I love doing.[/quote]\r\n\r\ni was kidding man. i wanna do a ph.d in math",
  "Solution_8": "I want to go to college because I know that I wont get anywhere without a college education.  I have all sorts of ideas and theories that I want to share in college, and hopefully turn the ideas into reality.  I want to show the world the impossible is indeed possible.  A solid college education will help me show my goal.",
  "Solution_9": "[hide] I definetely want to go to college.I'm thinking of going to BC and getting a bachelors degree. I'm currently in middle school and I am taking Geometry. How well do i have to do on my SATs to get a scolarship in BC?\n[/hide]",
  "Solution_10": "[quote=\"neelnal\"]College is overrated.\n\nGarbage men in NY make $60,000 to$80,000 a year.[/quote]\r\n\r\nwhat are you serious? WHOA, that's like 40K dollars more than the average us income",
  "Solution_11": "I wanna go to get a good job and actually get a degree in something.  :rotfl:",
  "Solution_12": "Why do I want to go to college?\r\n\r\n I want to go to college because I will earn more money, meet people with similar interests, and be the recipient of higher education. I also want to go to college because I believe that it will make me more self confident, have greater knowledge of governance, be less likely to become a criminal, be emotionally and financially secure, and have a deeper understanding of human nature. It is not just for the money. A college education makes you a rounded person. It shapes your communication skills, expands your knowledge base, and exposes you to a whole new world of learning.",
  "Solution_13": "[quote=\"Iversonfan2007\"][quote=\"neelnal\"]College is overrated.\n\nGarbage men in NY make $60,000 to$80,000 a year.[/quote]\n\nwhat are you serious? WHOA, that's like 40K dollars more than the average us income[/quote]\r\n\r\nYea hahaa. Depends on the area of NYcity you pick the trash up in. It is because the demand for the job is very low due to the labor and dirtiness of the job."
}
{
  "Tag": [
    "analytic geometry"
  ],
  "Problem": "I'm just curious: with respect to a 2D coordinate plane, does anyone know why on a 3D coordinate plane the x and y axes are moved around?",
  "Solution_1": "can you clarify what you mean by the $x$ and $y$ axes are moved around? When you \"draw\" a three dimensional space on a piece of paper... sometimes they're \"moved around\" to help with perspective and that allows you to \"see\"",
  "Solution_2": "i know that it is kind of confusing because the y-axis goes towards you in 2-d, but in 3-d, it runs parallel to you",
  "Solution_3": "How would simply moving the 3 axes around help one to see 3D on paper?  :? And to answer dangvh's question, when drawing a 2D coordinate plane, the x axis is horizontal and the y vertical (on the paper). However, on a 3D coordinate plane (drawn on the paper), the x axis is diagonal and y is horizontal. That's what I mean by \"moving around\"."
}
{
  "Tag": [
    "probability",
    "expected value",
    "geometric series"
  ],
  "Problem": "We construct a base 2 number as follows. First we flip a coin. If we get tails, we write a $0$ and our number is complete. If instead we get heads we write a $1$ and then flip the coin again. If tails this time, our number is complete. If heads, we continue to the next digit. For the next digit we do the exact same thing, keeping in mind that we add the digit to the right of the number. If we continue this process, what is the expected base 10 value of our number?",
  "Solution_1": "I get 1. Is that the answer? If I'm right I'll post soln, but it's kinda long.",
  "Solution_2": "That's close to what I found, but not exactly what I have. One thing to make sure you read carefully: there is an intermediate flip between the flip that determines a digit and the flip that determines the next digit. A lot of people overlook that. But post your solution if you don't see an error (maybe i made one  ;) )",
  "Solution_3": "To clarify: we always stop after the first tails, but half of the time it adds a 0 at the end, and half of the time it doesn't?",
  "Solution_4": "[hide] I also got 1. The probability of getting $0$ is $\\frac{1}{2}$ (with T), getting $1$ is $\\frac{1}{4}$ (with HT), getting $10_2=2$ is $\\frac{1}{8}$ (with HHT), etc. So the expected value is $\\frac{1}{4}+\\frac{2}{8}+\\frac{3}{16}+...=\\boxed{1}$. [/hide]",
  "Solution_5": "Yeah, lotrgreengrapes has the same solution as waht i got. Nicely explained.",
  "Solution_6": "Yeah, if we get tails on the very first flip then we stop. In general, half the time a tails will mean to end in a $0$ and half the time a tails will indicate that we don't move to the next digit.\r\n\r\nOk your series is correct for the first three values shown, but you incorrectly assumed that all the following terms would be of the form $\\frac{k}{2^{k+1}}$. The possible numbers are $0$,$1$,$10$,$11$,$110$,$111$,$1110$,$1111$,... Notice that when we go from 2 digit numbers to 3 digit numbers we skip from $3$ to $6$. To be more precise, the values listed above are $0$,$1$,$2$,$3$,$6$,$7$,$14$,$15$,... in base 10.",
  "Solution_7": "It was a bit awkward, but that's what I thought you meant.  Thanks for the clarification.\r\n\r\nI get [hide]4/3[/hide]",
  "Solution_8": "that's what i got yoober  :lol: \r\ngood job. post an explanation",
  "Solution_9": "Solution:\r\n\r\n[hide]\nExtending the pattern, I got:\n0/2+1/4+2/8+3/16+6/32+7/64+14/128+15/256+30/512+31/1024+..\nReducing the odd terms:\n1/4+1/4+3/16+3/16+7/64+7/64+..\nAdd and reducing:\n= 1/2+3/8+7/32+15/64+...\n= sum((2^n-1)/(2^(2n-1))) for n=1 to inf\nSplitting the fraction:\n= sum(2^(n-1)) - sum(1/(2^(2n-1)))\n\nEach is geometric with r=1/2 and with r=1/4:\n= 1/(1/2) - (1/2)/(3/4) = 2 - 2/3 = 4/3[/hide]",
  "Solution_10": "Here's my whole solution (it arises very easily if you draw a tree of the outcomes):\r\nFirst flip is 0,1 stay if 0, flip to stay or continue for 1 and so forth (bolded steps are where it stops, giving a final number).\r\n[b]0[/b]         /1\\\r\n       /1\\         [b]1[/b]\r\n  [b]10[/b]      /11\\\r\n      /11\\      [b]11[/b]\r\n [b]110[/b]    .../111\\...\r\nThis gives $0*\\frac12 + 1*\\frac14 + 2*\\frac18 + 3*\\frac{1}{16} + 6*\\frac{1}{32} + \\cdots$. Looking at adjacent terms, they're the same so we get rid of every other term and multiply the sum by 2 giving $2(\\frac14 + \\frac{3}{16} + \\frac{7}{64} + \\cdots)$. We notice that the numerators are one less than powers of 2, and the denominators are powers of 4, so rewriting in summation notation to simplify things, it becomes $\\displaystyle 2*\\sum \\frac{2^{n+1}-2^n}{4^{n+1}}$. Separating the fraction gives two geometric series which are easily evaluated: $\\frac{2^{n+1}}{4^{n+1}} - \\frac{2^n}{4^{n+1}} = (\\frac12)^{n+1} - (\\frac12)^{n+2}=1-\\frac12=\\frac12$. Now, multiplying the 2 which we took out before gives a final sum of $\\boxed{1}$.\r\n\r\nIt looks like we got the same series, but summed them differently. Does anyone see an error that I made?",
  "Solution_11": "Hokkage, just check the way you wrote your sum. \r\nYou wrote $\\sum \\frac{2^{n+1}-2^n}{4^{n+1}}=\\sum \\frac{2^n}{4^{n+1}}$. However, you said \"the numerators are one less than powers of 2\" so it should be $\\sum \\frac{2^{n+1}-1}{4^{n+1}}$",
  "Solution_12": "That was a dumb mistake on my part  :blush: \r\nI must have been tired or something.\r\nYes, in light of my obvious mistake I agree with the answer of $\\frac{4}{3}$."
}
{
  "Tag": [
    "LaTeX"
  ],
  "Problem": "I put Hyperlink on Word and made that to PDF by CutePDF. However, Hyperlink doesn't work. Any idea?",
  "Solution_1": "[quote=\"Silverfalcon\"]I put Hyperlink on Word and made that to PDF by CutePDF. However, Hyperlink doesn't work. Any idea?[/quote]\r\n\r\nWhat it is CutePDF?",
  "Solution_2": "CutePDF is a program that converts the regular document into PDF. \r\n\r\nAnyone has an idea for this? Also, is there anyway that I can put hyperlink in PDF then? I know it's possible (duh) but I don't know how to do it whenever I convert regular documents into PDFs."
}
{
  "Tag": [
    "geometry",
    "parallelogram",
    "analytic geometry",
    "geometry unsolved"
  ],
  "Problem": "Describe a construction of quadrilateral $ABCD$ given:\r\n\r\n(i) the lengths of all four sides;\r\n\r\n(ii) that $AB$ and $CD$ are parallel;\r\n\r\n(iii) that $BC$ and $DA$ do not intersect.",
  "Solution_1": "Let AB=a, BC=b, CD=c and DA=d, w.l.o.g.  a<c (if a=c then ABCD is a parallelogram and the construction is easy). Construct a triangle BCX, where  BC=b, BX=d and CX=c-a, then take a point D on the ray CX such that CD=c. Draw the segment AB parallel to DC such that AB=a. Since AB and DX are parallel and have the same length, then ABXD is a parallelogram and AD=BX=d. Finally, is easy to see that the cuadrilateral ABCD fullfills the conditions. \r\n\r\n$Tipe$",
  "Solution_2": "Answer of Reynaldo Gil Pons:\r\nI do it by coordinate geometry.\r\nLet $ AB \\equal{} a$, $ BC \\equal{} b$, $ CD \\equal{} c$ and $ AD \\equal{} d$.\r\nLet $ A(\\minus{}a;0)$, $ B(0;0)$, $ C(x_C;y_C)$ and $ D(x_D;x_D)$.\r\nBut $ y_C\\equal{}y_D$ because $ AB$ is parallel to $ CD$.\r\nAs $ BC\\equal{}b$, then:\r\n$ b^2\\equal{}x_C^2\\plus{}y_C^2\\hspace{1cm}(1)$\r\nAs $ AD\\equal{}d$, then:\r\n$ d^2\\equal{}(x_D\\plus{}a)^2\\plus{}y_C^2\\hspace{1cm}(2)$\r\nAs $ CD\\equal{}c$, then:\r\n$ c^2\\equal{}(x_C\\minus{}x_D)^2$\r\nAs $ AD$ and $ BC$ do not intersect:\r\n$ c\\equal{}x_C\\minus{}x_D\\hspace{1cm}(3)$\r\nReplacing $ (3)$ on $ (1)$:\r\n$ b^2\\equal{}(x_D\\plus{}c)^2\\plus{}y_C^2\\hspace{1cm}(4)$\r\nSubtracting $ (4)$ from $ (2)$:\r\n$ b^2\\minus{}d^2\\equal{}2x_D(c\\minus{}a)\\plus{}c^2\\minus{}a^2$\r\nAnd: $ x_D\\equal{}\\frac{b^2\\minus{}d^2\\minus{}c^2\\plus{}a^2}{2(c\\minus{}a)}$\r\nThen we can calculate $ x_C$ and $ y_C$ in terms of $ a$, $ b$, $ c$, $ d$. However the problem is solved."
}
{
  "Tag": [
    "calculus",
    "integration",
    "complex analysis"
  ],
  "Problem": "Evaluate $ \\frac{1}{2\\pi i} \\oint \\frac{1}{(z\\plus{}2)(z\\minus{}i)^{2}} \\ dz$ around $ |z\\minus{}1| \\equal{} 2$.\r\n\r\nThis has two singularities at $ z_1\\equal{} \\minus{}2$ and $ z_2 \\equal{} i$. $ z_1$ is on $ C$ which $ z_2$ is in $ C$. In the answer, it just works with the Extended Cauchy Integral Formula for $ z_2 \\equal{} i$. What about $ z_1$? That is also a singularity, so why can't we just apply the regular Cauchy Integral Formula?",
  "Solution_1": "There is a difference between singularities inside the loop and singularities outside the loop.",
  "Solution_2": "but one singularity is [b]on[/b] the loop while the other singularity is [b]in[/b] the loop. Neither is outside the loop. So for the singularity which is [b]in[/b] the loop we use the Extended Cauchy Integral Formula. While for the singularity [b]on[/b] the loop, can we use the Cauchy Integral Formula?\r\n\r\nAnd please keep this question in this section. Because no one visits the Complex Analysis section.",
  "Solution_3": "I'm not a moderator- but a moderator [b]will[/b] move this.\r\n\r\nIt's not on the loop. $ |1\\minus{}(\\minus{}2)|>2$.",
  "Solution_4": "Then how come when I draw it, it is on the loop? I have the circle centered at $ z_0 \\equal{} 1$. Ok I get it."
}
{
  "Tag": [],
  "Problem": "Is it a better idea to get a PhD in Math or to get a MS in financial engineering if you want to become a quant? Also, how long does it usually take to get those degrees and are there any exams that you need to pass in order to quality as a quant? \r\nAny help will he appreciated.",
  "Solution_1": "I'd say go for the pure sciences degree.",
  "Solution_2": "[quote=\"satyam\"]Is it a better idea to get a PhD in Math or to get a MS in financial engineering if you want to become a quant? Also, how long does it usually take to get those degrees and are there any exams that you need to pass in order to quality as a quant? \nAny help will he appreciated.[/quote]\r\n\r\nNo specific exams to 'qualify' as a quant like there are for the actuaries.\r\n\r\nPhD in math takes a lot longer than an MS in financial engineering.  I don't know much about the latter; it's a fairly new degree.  Also, it depends on where you want to work.  Hedge funds seem to go more for the folks with pure science degrees.  That said, they aim more at the people with applied knowledge (e.g. physics) than at the pure math people.  \r\n\r\nWhatever route you go, learn how to program.",
  "Solution_3": "so if i am a pure math major...that takes classes in physics and other applied fields and programming as well i should be fine right?",
  "Solution_4": "[quote=\"maokid7\"]so if i am a pure math major...that takes classes in physics and other applied fields and programming as well i should be fine right?[/quote]\r\n\r\nIf you do well in them, etc., then you should be ok.  Try to get an internship after your junior year, if not earlier.",
  "Solution_5": "do the top hedge fund managers tend to have math and science degrees rather than degrees in economics or finance as well, or is it mainly in the employees that they look for such people?"
}
{
  "Tag": [
    "induction",
    "algebra proposed",
    "algebra"
  ],
  "Problem": "let the sequence $ a_n$:\r\n$ a_1\\equal{}1$ and $ a_2\\equal{}7$ and $ a_{n\\plus{}2}\\equal{}1/a_n (a_{n\\plus{}1}^2\\minus{}1)$.\r\nprove that $ 1\\plus{}9a_{n\\plus{}1}a_n$ is a square of integer.",
  "Solution_1": "It's easy to check the fomular $ a_{n\\plus{}2}\\equal{}7a_{n\\plus{}1}\\minus{}a_{n}$\r\nNext, prove by induction $ 9a_{n}a_{n\\plus{}1}\\plus{}1\\equal{}(a_n\\plus{}a_{n\\plus{}1})^2$",
  "Solution_2": "$ a_{n} \\equal{} \\frac{F_{4n}}{3}$ and $ 9 a_{n} a_{n\\plus{}1} \\plus{} 1 \\equal{} (F_{4n\\plus{}2})^2$\r\n($ F_{n}$ is Fibonacci sequence)",
  "Solution_3": "you are right,hjen! :wink: \r\nhere is my full solution:\r\n$ a_{n\\plus{}2}\\equal{}7a_{n\\plus{}1}\\minus{}a_{n}$ (->easy to see it)\r\nwe have:\r\n$ a_{n\\plus{}2}\\minus{}3/2.a_{n\\plus{}1}\\equal{}3/2.a_{n\\plus{}1}\\minus{}a_{n}$\r\nthen we have:\r\n$ a_{n\\plus{}1}^2\\minus{}3.a_{n}.a_{n\\plus{}1}\\plus{}a_{n}^2\\equal{}a_{1}^2\\minus{}3.a_{1}.a_{0}\\plus{}a_{0}^2$\r\nIt is enough for every body! :oops:",
  "Solution_4": "[quote=\"Math pro\"]\\we have:\n$ a_{n \\plus{} 2} \\minus{} 3/2.a_{n \\plus{} 1} \\equal{} 3/2.a_{n \\plus{} 1} \\minus{} a_{n}$\nthen we have:\n$ a_{n \\plus{} 1}^2 \\minus{} 3.a_{n}.a_{n \\plus{} 1} \\plus{} a_{n}^2 \\equal{} a_{1}^2 \\minus{} 3.a_{1}.a_{0} \\plus{} a_{0}^2$\nIt is enough for every body! :oops:[/quote]\r\nWhy ??.the proof whichis used Fibonaci sequence is nice and quite natural but but with yours ,i cant understand those lines"
}
{
  "Tag": [
    "calculus",
    "integration",
    "derivative",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "can you do  :maybe: \r\n\r\n$ \\int_{0}^{1}\\;\\;\\sqrt{\\;1+e^{x}+e^{2x}\\;}\\;\\;\\textbf dx$",
  "Solution_1": "Derivative of $ \\sqrt{1+e^{x}+e^{2x}}$ is given by\r\n\r\n$ \\frac{d}{dx}\\sqrt{1+e^{x}+e^{2x}}\\; = \\; \\frac{\\sqrt{1+e^{x}+e^{2x}}}{2}+\\frac{e^{2x}-1}{2\\sqrt{1+e^{x}+e^{2x}}}.$\r\n\r\nSo\r\n\r\n\\begin{eqnarray*}\\int \\sqrt{1+e^{x}+e^{2x}}\\; dx & = & 2\\sqrt{1+e^{x}+e^{2x}}-\\int \\frac{e^{2x}-1}{\\sqrt{1+e^{x}+e^{2x}}}\\; dx\\\\ & = & 2\\sqrt{1+e^{x}+e^{2x}}-\\int e^{\\frac{x}{2}}\\frac{e^{x}-e^{-x}}{\\sqrt{e^{x}+e^{-x}+1}}\\; dx\\\\ & = & 2\\sqrt{1+e^{x}+e^{2x}}-\\int 2e^{t}\\frac{e^{2t}-e^{-2t}}{\\sqrt{e^{2t}+e^{-2t}+1}}\\; dt \\quad \\quad (x = 2t)\\\\ & = & 2\\sqrt{1+e^{x}+e^{2x}}-\\int \\frac{8\\sinh^{2}t \\cosh t}{\\sqrt{4\\sinh^{2}t+3}}\\; dt-\\int \\frac{8\\sinh t \\cosh^{2}t}{\\sqrt{4 \\cosh^{2}t-1}}\\; dt\\\\ \\end{eqnarray*}\r\n\r\nLast two integrals can be computed using substitutions $ 2 \\sinh t = \\sqrt{3}\\sinh u$ and $ 2 \\cosh t = \\cosh v$, respectively. But I will not proceed this calculation.\r\n\r\n\r\np.s. I tried to solve it using series expansion for more than 2 hours, only realizing that it just yields terribly messy formulas.",
  "Solution_2": "$ \\int \\sqrt{1+e^{x}+e^{2x}}\\, dx = \\int e^{-x}\\sqrt{1+e^{x}+e^{2x}}\\, (e^{x}dx) = \\int \\frac{1}{z}\\sqrt{1+z+z^{2}}\\, dz =$\r\n\r\n$ =\\int \\frac{1+z+z^{2}}{z \\sqrt{1+z+z^{2}}}\\, dz =$\r\n\r\n$ = \\int \\frac{dz}{z \\sqrt{1+z+z^{2}}}+\\int \\frac{dz}{\\sqrt{1+z+z^{2}}}+\\frac{1}{2}\\int \\frac{(1+2z)-1}{\\sqrt{1+z+z^{2}}}\\, dz =$\r\n\r\n$ = \\int \\frac{dz/z^{2}}{\\sqrt{1/z^{2}+1/z+1}}+\\frac{1}{2}\\int \\frac{dz}{\\sqrt{(z+1/2)^{2}+3/4}}+\\int \\frac{d(1+z+z^{2})}{2 \\sqrt{1+z+z^{2}}}=$\r\n\r\n$ =-\\int \\frac{d(1/z)}{\\sqrt{(1/z+1/2)^{2}+3/4}}+\\frac{1}{2}\\arg \\sinh \\frac{z+1/2}{\\sqrt 3/2}+\\sqrt{1+z+z^{2}}=$\r\n\r\n$ =-\\arg \\sinh \\frac{1/z+1/2}{\\sqrt 3/2}+\\frac{1}{2}\\arg \\sinh \\frac{2z+1}{\\sqrt 3}+\\sqrt{1+z+z^{2}}+C =$\r\n\r\n$ =-\\arg \\sinh \\frac{2+z}{z \\sqrt 3}+\\frac{1}{2}\\arg \\sinh \\frac{2z+1}{\\sqrt 3}+\\sqrt{1+z+z^{2}}+C =$\r\n\r\n$ =-\\arg \\sinh \\frac{2+e^{x}}{e^{x}\\sqrt 3}+\\frac{1}{2}\\arg \\sinh \\frac{2e^{x}+1}{\\sqrt 3}+\\sqrt{1+e^{x}+e^{2x}}+C$"
}
{
  "Tag": [
    "geometry",
    "analytic geometry",
    "geometry unsolved"
  ],
  "Problem": "Let $ P$ be an interior point of a triangle with sides $ a,b,c$ in which lines parallel to the sides of the triangle are traced. THe segments of the parallels delimited by the sides of the triangle have the same length. Find its value.\r\n\r\n\r\n\r\n[hide=\"Answer\"]\n$ \\frac{2abc}{ab\\plus{}bc\\plus{}ac}$\n[/hide]",
  "Solution_1": "Let the desired segment be $ d$, and let the perpendiculars from $ P$ to the sides be $ n_a,n_b,n_c$ respectively. Then\r\n\r\n$ {d\\over a} \\equal{} {h_a \\minus{} n_a\\over h_a} \\equal{} 1 \\minus{} {n_a\\over h_a}$\r\n\r\nSumming those up, we get $ d\\left(\\frac 1 a \\plus{} \\frac 1 b \\plus{} \\frac 1 c\\right) \\equal{} 3 \\minus{} \\left({n_a\\over h_a} \\plus{} {n_b\\over h_b} \\plus{} {n_c\\over h_c}\\right)$\r\n\r\nThe expression in the parentheses on the RHS is equal to $ 1$, which is easily proven if $ h_x$ is substituted by $ {2A\\over x}$, where $ x\\in\\{a,b,c\\}$ and $ A$ is the area of the triangle. After that, the result follows.",
  "Solution_2": "Let $(u:v:w)$ be the normalized  barycentric coordinates of $P$ WRT $\\triangle ABC.$ Then lengths of the paralell sections from $P$ to $BC,CA,AB$ are given by\n\n$d_a= a(1-u) \\ \\ , \\ \\ d_b=b(1-v) \\ \\ , \\ \\ d_c=c(1-w)$\n\nSolving $a(1-u)=b(1-v)=c(1-w),$  together with $u+v+w=1$ gives\n\n$P \\equiv \\left( \\frac{1}{b}+\\frac{1}{c}-\\frac{1}{a}:\\frac{1}{a}+\\frac{1}{c}-\\frac{1}{b}:\\frac{1}{a}+\\frac{1}{b}-\\frac{1}{c} \\right) \\Longrightarrow$\n\n$X_{192} \\equiv 3 \\cdot X_2-2 \\cdot {X_1}^{-1} \\equiv \\left (\\frac{1}{b}+\\frac{1}{c}-\\frac{1}{a}:\\frac{1}{a}+\\frac{1}{c}-\\frac{1}{b}:\\frac{1}{b}+\\frac{1}{a}-\\frac{1}{c} \\right)$",
  "Solution_3": "Thanks a lot guys, that is very helpful.\n\nluis, I am not accustomed to this notation you use but I would like to understand how you solved this problem.\n\nIs this really a simple solution that I would understand if I understood the notation or does it involve more than basic concepts?"
}
{
  "Tag": [
    "inequalities",
    "induction",
    "parameterization",
    "combinatorics proposed",
    "combinatorics"
  ],
  "Problem": "For a graph $G$, let $\\nu(G)$ be the maximal number of edges s.t. no two of them have a common endpoint, and let $\\rho(G)$ be the minimal number of edges to cover all the vertices (an edge covers a vertex if that vertex is one of the endpoints of the edge). If the simple undirected graph $G$ has $n$ vertices which are not isolated, show that $\\nu(G)+\\rho(G)=n$.\r\n\r\nI'm sure you people will find this easy, but I thought it was a rather nice fact.",
  "Solution_1": "isn't it  just $v(G)=p(G)=\\frac{n}{2}$ ? Anyway it's a nice fact",
  "Solution_2": "It's definitely not always like that (there are situations when they're not both $\\frac n2$).. Had it been like that, it would have been really stupid of me to post it :D.",
  "Solution_3": "ok - I've posted it almost without being careful. But I think it is $v(G)=p(G)=\\frac{n}{2}$ when $n\\equiv0 \\ (mod \\ 2)$ and $v(G)=[\\frac{n}{2}]$   $p(G)=[\\frac{n}{2}] + 1$ when $n\\equiv1 \\ (mod \\ 2)$",
  "Solution_4": "No Megus : Consider the graph with 4 vertices $A,B,C,D$ and edges $AB,AC,AD$.\r\nThen $\\nu (G) = 1$ and $p(G) = 3$.\r\n\r\nPierre.",
  "Solution_5": "Megus! You aren't realizing how much you wrong. ;) \r\n$\\nu(G)=n/2$ corresponds to complete matching in graph $G$. Why do you think Hall invented his theorem?  :)",
  "Solution_6": "Holy **** ! Now I'm realizing how wrong was I  :blush:  - eh - next time I'll think twice  :maybe:",
  "Solution_7": "[color=red][[b]Moderator edit:[/b] In this posting, $p$ means what grobber denoted by $\\rho$ in the original post.][/color]\n\nLet $G$ be a graph with $n$ vertices.\nLet $\\nu (G) = k$ and consider a maximal set $E$ of pairwise disjoint $k$ edges. Thus, these edges uses exactly $2k$ vertices from $G$. Since no vertex is isolated and from the maximality of $k$, it left $n-2k$ vertices, each of which are connected by an edge to at least one of the endpoint of an edge belonging to $E$. Thus, choosing one of these edges per vertex among the $n-2k$ and using the $k$ edges from $E$, we cover all the vertices. The minimality of $p(G)$ leads to $p(G) \\leq (n-2k)+k = n-k$ so that $p(G)+ \\nu (G) \\leq n$.\n\nNow, we prove the reverse inequality by induction on the number $a$ of edges in $G$.\nSince no vertex is isolated, we have $a \\geq [ \\frac {n+1} 2]$.\nIf $a = [ \\frac {n+1} 2]$ then for even $n$ it means that each vertex belongs to exactly one edge, so that $p(G) = a = \\nu (G)$, so that $p(G) + \\nu(G) = 2a = n$. And, for odd $n$, it means that each vertex except one, say $v$, belongs to exactly one edge and $v$ belongs to two edges. Thus, $p(G) = a = \\frac {n+1} 2 $ and $\\nu (G) = a-1$ so that $p(G) + \\nu (G) = 2a-1 = n$. Therefore, we are done in the case where $a = [ \\frac {n+1} 2]$.\n\nNow assume the desired inequality holds for each graph with $n$ vertices (none isolated) and at most $a-1$ edges. Let $G$ be such a graph with $a$ edges.\nLet $F$ be a set of $p(G)$ edges which cover each vertex from $G$.\n- If $p(G) = a$ then we have to choose all the edges so as to cover all the vertices. Moreover, with the edges from $F$ we cannot form a cycle otherwise we could omit one edge from the cycle without losing the covering, which would contradicts the minimality of $p(G)$. It follows that $G$ is a forest, which means that each of its connex components is a tree (with at least two vertices in each component, since no vertex is isolated). If there are at least two components, then we may use the induction hypothesis on each of the components, and sum all the correspondant inequalities (clearly $p(G)$ (resp. $\\nu (G)$) is the sum of the analog parameter for each component). If there is only one component, then $G$ itself is a tree, so that $a=n-1$. Since, we obviously have $\\nu (G) \\geq 1$, we are done again.\n- If $p(G) < a$ then consider the graph $G'$ whith the same vertices than $G$ but only the edges from $F$. Since each vertex is covered, there is no isolated vertex. Then we may use the induction hypothesis to deduce that $p(G') + \\nu (G') \\geq n$.\nBut, from the minimality of $p(G)$ we have $p(G') = p(G)$, and clearly $\\nu(G') \\leq \\nu(G)$ and the conclusion follows.\nThis ends the induction step.\nTherefore $p(G) + \\nu (G) \\geq n$, and we are done.\n\nPierre.\n\nP.S. Please, Grobber stop apologize when posting a graph problem ;)",
  "Solution_8": "Sorry for apologizing. :D:D\r\n\r\nI seem to have done almost the same thing. Thanks for the solution.",
  "Solution_9": ":?  Oh...I was hoping that there was a direct proof of the equality (not using the two inequalities).\r\n\r\nPierre.",
  "Solution_10": "[quote]\nthen we may use the induction hypothesis on each of the components, and sum all the correspondant inequalities\n[/quote]\r\n\r\n\r\nHmmm...to be rigourous, I have to use a double induction in the above proof :\r\nFirst on $n$  : the case $n = 2$ is trivial, then we assume ($H_1$) that the inequality holds for all $i \\leq n-1$.\r\nThen, on $a$ as done above ($H_2$). The induction hypothesis in quote is then $H_1$. And the one used for $G'$ is $H_2$.\r\n\r\nPierre.",
  "Solution_11": "Well, the problem is much easier.\r\n\r\nFor any matching with $k$ edges, choose for the $n-2k$ unoccupied vertices some edge to cover them. This gives a covering with $k+(n-2k)=n-k$ edges, thus proving one inequality.\r\n\r\nFor any covering with $k$ edges, choose a maximal submatching(that means all edges of the matching are edges of the covering and no more edges from the covering may be added to give a matching); let it have $l$ edges. Then the $n-2l$ uncovered vertices all belong to some edge of the covering. Choosing for every uncovered vertex such an edge, no edge can be taken twice by the maximality of the matching. Thus $k\\geq l+(n-2l)=n-l$ thus proving the other inequality.\r\n\r\nPeter"
}
{
  "Tag": [
    "geometry unsolved",
    "geometry"
  ],
  "Problem": "Let $ A_0 A_1 ... A_{n\\minus{}1}$ be a regular $ n$-gon inscribed in a circle with radius $ r$. Prove that for every point P of the circle and every natural number $ m<n$ we have:\r\n$ \\sum_{k\\equal{}0}^{n\\minus{}1} PA_k^{2m}\\equal{}\\binom{2m}{m}nr^{2m}$",
  "Solution_1": "Anyone can help me with this? Please try it."
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "advanced fields",
    "advanced fields theorems"
  ],
  "Problem": "What would constitute the easiest proof of this theorem ? I would like to have a look since I am using it (implicitely) to determine Galois groups of certain irreducible polynomials of $\\mathbb{Q}[X]$.",
  "Solution_1": "I have read a proof which is a mixture of Artin's reciprocity law, Chebotarev's proof for the Frobenius conjecture, and Deuring's reduction to the cyclic case. The proof is quite a long, 15 pages, but rather elementary."
}
{
  "Tag": [
    "number theory proposed",
    "number theory"
  ],
  "Problem": "hallo,\r\nLet $a, b \\in N$ s.t. $\\frac{a+1}{b}+\\frac{b+1}{a} \\in N$. Let $x=gcd(a, b)$. Prove:\r\n\\[ x^2 \\le a+b. \\] \r\n\r\n :)",
  "Solution_1": "[hide]Suppose that $gcd(a,b)=x$ then $a=a_1x$ , $b=b_1x$ where $gcd(a_1,b_1)=1$[/hide]",
  "Solution_2": "$ab \\mid a^2 + b^2 + a + b = (a + b)^2 - 2ab + a + b$ $\\Longrightarrow$ $ab \\mid (a + b)(a + b + 1)$.\r\nLet $a = xh$ and $b = xk$. Then $x^2hk \\mid (a + b)(a + b + 1)$. \r\nIf $x = 1$ it's trivial. So suppose $x > 1$. Then $x^2 \\mid a + b$ from which $x^2 \\leq a + b$."
}
{
  "Tag": [
    "vector",
    "algebra",
    "polynomial",
    "integration",
    "calculus",
    "function",
    "derivative"
  ],
  "Problem": "Let V be the vector space of all polynomials over R with scalar product defined as:  $ \\displaystyle \\langle f(t); g(t) \\rangle \\equal{} \\int_{0}^{1}{f(t)g(t)dt}$ Prove that the operator D defined as $ D(f) \\equal{} \\frac {df}{dt}$ has no adjoint operator",
  "Solution_1": "can someone solve this problem... :( ?",
  "Solution_2": "If $ D$ had an adjoint operator $ D^*$, we would have the following:\r\n\r\n$ \\langle f,g'\\rangle \\equal{} \\langle D^*f,g\\rangle.$\r\n\r\nBut we actually have this (by integration by parts):\r\n\r\n$ \\langle f,g'\\rangle \\equal{} \\minus{}\\langle f',g\\rangle\\plus{}f(1)g(1)\\minus{}f(0)g(0).$\r\n\r\nLet's be very particular: let $ f(x)\\equal{}x.$ Then $ \\langle x,g'\\rangle \\equal{}\\minus{}\\int_0^1g\\plus{}g(1).$\r\n\r\nSo, hypothetically, if $ h\\equal{}D^*(f)$ where $ f(x)\\equal{}x,$ then for all $ g$\r\n\r\n$ \\int_0^1hg\\equal{}\\minus{}\\int_0^1g\\plus{}g(1),$ or $ \\int_0^1(h\\plus{}1)g\\equal{}g(1).$\r\n\r\nNow let $ g(x)\\equal{}x^n.$ We have\r\n\r\n$ \\left|\\int_0^1(h\\plus{}1)g\\right|^2\\le\\int_0^1(h\\plus{}1)^2\\cdot\\int_0^1x^{2n}\\equal{}\r\n\\frac1{2n\\plus{}1}\\int_0^1(h\\plus{}1)^2.$\r\n\r\nThis can be made arbitrarily small by letting $ n$ be sufficiently large. But at the same time this had to be equal $ (g(1))^2\\equal{}1.$ Since we can't do that, then there is no such (polynomial) function $ h$ and thus $ D^*$ can't be defined.\r\n\r\nIf the vector space were polynomials such that $ f(0)\\equal{}f(1)\\equal{}0,$ then the derivative operator would have an adjoint.",
  "Solution_3": "wow very nice solution :D ! thank you"
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "probability",
    "sphere",
    "rectangle",
    "rotation",
    "modular arithmetic"
  ],
  "Problem": "1. Three large cubes with sides 1m, 3 m, and 5m, are glued together at their faces. What is the least possible surface area of the resulting solid figure?\r\n\r\n2. You are given six 2-meter wide walls to fully surround the lion, You are tasked to enclose him with the largest possible area without breaking the walls. What is the largest possible area?\r\n\r\n3. Which positive integer becomes 57 times smaller when removing the leftmost digit?\r\n\r\n4. Let A be the set of all integers from 23 to 99. If S is any six-element subset of A, how many possible values are there for the sum of the elements in S?\r\n\r\n5. Mr. Lee awards extra credit to his students with quiz grades that are higher than the class mean. Given that 87 students take the same quiz, what is the largest number of students who can be awarded extra credit?\r\n\r\n6. Tom randomly selects two distinct numbers from the set {1, 2, 3, 4, 5} and Steve randomly selects a number from the set {1, 2, 3,...., 9, 10}. What is the probability that Steve's number is larger than the sum of the two numbers selcted by Tom?\r\n\r\n7. Two equal squares ABCD and DEFG have vertex D in common. The angle of CDG is 60. Find angle CFE.\r\n\r\n8. The price of gum has skyrocketed in recent years. Each year for the last 7 years the price has increasedm and the new price is the sum of the prices of the two previous years. Last year, a piece of gum costs 60 cents. How much does a price of gum cost today?\r\n\r\n9. A cylindrical barrel 50cm high with a radius of 18cm is filled with water up to 1cm from the brim.A solid sphere with radius 9cm is placed into the water so that the water level rises and some water spills out. If the sphere is half submerged underwater, how much water has spilled out?\r\n\r\n10. Rectangle PQRS lies in plane with PQ=RS=2 and QR=SP=2. The rectangle is rotated 90 degrees about R, then rotated 90 degrees clockwise about the point that S moved after the first rotation. What is the length of the path travled by point P?\r\n\r\n11. During a chicken-spaghetti day, people go to the restaurant to eat either chicken, spaghetti,or both chicken and spaghetti. If 1/3 of those who ate spaghetti also ate chicken, and 1/5 of those who ate chicken also ate spaghetti, what fraction of the people ate spaghetti only?\r\n\r\n12. John can cross a river on his boat in 5 min. James can do it in 7 min. If they go back and forth from one side of the river to the other, and if they simultaneously cross the river, how many times would they meet in 35 min?\r\n\r\n13. A, B, and C each had a bag of candies. A gave B candies equal to half of what C had. Then B gave C candies equal to half of what A now had. B ended up with 17 candies. A now had a number of candies equal to what C originally had, while C now had a number of candies equal to what A originally had. If there were 57 candies all in all, how many candies did C originally have?\r\n\r\n14. Suppose that today, the last Friday of this month is on the 24th day of that month. The first day of the next month is a Thursday. What month is it now?",
  "Solution_1": "[quote=\"orangefronted\"]\n3. Which positive integer becomes 57 times smaller when removing the leftmost digit?[/quote]\r\n\r\n[hide=\"Solution\"]\n=====\nLet the resulting number after the leftmost digit is removed be $ n$. Clearly $ n\\neq 1$, so $ 57n$ has at least three digits. Suppose $ 57n$ has exactly three digits. Then $ n \\equiv 57n \\pmod{100} \\implies 56n \\equiv 0 \\pmod{100} \\implies 14n \\equiv 0 \\pmod{25}$. Since $ (14,25)\\equal{}1$, $ 25|n$. But for $ n\\geq 25$, $ 57n$ has more than three digits. So suppose $ 57n$ has exactly four digits. Similar to above, we find $ 56n \\equiv 0 \\pmod{1000} \\implies 7n \\equiv 0 \\pmod{125}$, so $ 125|n$. Trying $ n\\equal{}125$, we find $ 57n\\equal{}\\boxed{7125}$, which works.\n=====\n[/hide]",
  "Solution_2": "14. Suppose that today, the last Friday of this month is on the 24th day of that month. The first day of the next month is a Thursday. What month is it now?\r\n\r\nToday is the 24th, and the month ends in five days (Saturday-Wednesday is a five-day period).  Therefore, the month has a total of 29 days.  The only month with 29 days is $ \\boxed{February}$ (during leap years).",
  "Solution_3": "Do these belong in this forum?",
  "Solution_4": "Yes I think so.. They are not that simple though.\r\n\r\nThanks for the answers...12 more to go... :o  :o",
  "Solution_5": "1.the reduction of surface area can happen only when the surfaces of two different cubes touch. Between any two cube at maximum one surface of each cubes can touch each other, so the area reduction is at most twice the area of one of the surface of the smaller cube. So total area is at least 6*(25+9+1)-2*(1+1+9)=188. Obviously this can be achieved.",
  "Solution_6": "[hide=\"1\"]the total surface area is $ 6\\cdot 1^2 + 6\\cdot 3^2 + 6\\cdot 5^2 = 210 m^2$.  Now we wish to attach these cubes together to cover the most amount of surface area.  Gluing the 5m cube to the 3m cube covers $ 2 \\cdot 3^2 = 18 m^2$ of surface area.  Now we can attach the 1m cube at any point.  We want to cover as many of its surfaces as possible, so we glue it in the corner of the 3m and 5m cube, covering an additional $ 4m^2$ of surface area.  Thus the minimum area is $ 210-18-4= 188$ square meters.[/hide]\n\n[hide=\"2\"]a regular shape always gives the greatest area for perimeter, for a given number of sides.  A circle gives the greatest area for its \"perimeter\" (circumference).\n\nSo we arrange the walls into a regular hexagon.  the area of a regular hexagon is the are of $ 6$ equilateral triangles with the same side length, so we have $ 6 \\cdot \\frac{(2)^2\\sqrt{3}}{4} = 6\\sqrt{3}$ square meters.[/hide]\n\n[hide=\"4\"]Claim: we can obtain any value between $ 153$ and $ 579$ for a sum of a six element subset.\n\nTake the first 6 consecutive integers.  Then we have $ 23+24+25+26+27+28 = 153$.  We can then change the $ 28$ to $ 29$, $ 30$, $ 31$, $ 32$, and $ 33$.  Then we can take the set of consecutive integers starting with $ 24$, giving a sum one larger than we just had with $ 23+24+25+26+27+33$.  Continuing this process, we can get all the way up to $ 89+90+91+92+93+99$.  Then we can increase the penultimate number by 1, and then the 4th number by one, etc down to the first number, and then go back and increase the penultimate number again, and continue in this fashion until reaching $ 94+95+96+97+98+99 = 579$\n\nSo the sum of any 6 element subset in A can be any integer in $ [153,579]$ for a total of $ 579-153+1 = 427$ possible values.\n[/hide]\n\n[hide=\"5\"]\nThe \"optimal\" case is when the distribution is skewed left so the mean is lower than the median.  The best way to drag the mean down is to have as many kids as possible get $ 0$'s on the quiz.  But at the same time we want as many as possible to beat the mean.\n\nSo suppose some people get 0's on the quiz.  If even one person got a zero, and the other $ 86$ students got the same score, then the mean is slightly lower than the score that these $ 86$ students got.  So the maximum number of students getting the extra credit is $ 86$.  Though this is possible, it is highly unlikely to actually happen.\n[/hide]\n\n[hide=\"6\"]\nTom has the possible choices: (the sums are a set of chosen numbers)\n$ 1+2=3$\n$ 1+3 = 4$\n$ 1+4=2+3=5$\n$ 1+5=2+4=6$\n$ 2+5=3+4=7$\n$ 3+5=8$\n$ 4+5=9$\n\nSo Tom has $ 10$ possible choices.  If Tom chooses a sum of $ n$, then steve has a $ 10-n$ ways of choosing a number larger than the sum of tom's numbers.  Let $ p(n)$ be the numbe of ways for Tom to chose a sum of $ n$.  Then the probability we want is:\n\n$ \\sum_{n=3}^{9}\\frac{p(n)}{10}\\cdot \\frac{10-n}{10} = 1/10 \\cdot 7/10 + 1/10 \\cdot 6/10 + 2/10 \\cdot 5/10 + 2/10\\cdot 4/10 + 2/10 \\cdot 3/10 + 1/10 \\cdot 2/10 + 1/10 \\cdot 1/10$\n$ = \\boxed{2/5}$\n[/hide]\n\n[hide=\"7\"]\nThe squares are congruent, not equal.  Anyway, this tells us that they have the same side length.  $ \\angle CDG \\cong 60^\\circ$ tells us that $ CG$ is the same length as the sides of the squares; thus, $ \\triangle CDG$ is equilateral, so $ \\angle CGD = 60^\\circ$.  Now $ \\angle DGF = 90^\\circ$ so $ \\angle CGF = 60+90 = 150^\\circ$.  Since $ CG =FG$, $ \\angle CFG = \\angle GCF$.  Since the sum of the angles in a triangle is $ 180^\\circ$, using $ \\triangle CGF$ we see that $ \\angle GFC = (180-150)/2 = 15^\\circ$.  Now $ \\angle GFE = \\angle GFC + \\angle CFE$.  $ \\angle GFE=90^\\circ$, so $ \\angle CFE = 90-15 =  \\boxed{75^\\circ}$\n[/hide]\n\n[hide=\"8\"]\nLet the price of gum $ 8$ and $ 9$ years ago be $ x$.  Then the price of gum $ 7$ years ago was $ 2x$, $ 6$ years ago was $ 3x$, $ 5$ years ago was $ 5x$, $ 4$ years ago was $ 8x$, $ 3$ years ago was $ 13x$, $ 2$ years ago was $ 21x$, $ 1$ year ago was $ 34x$, and now is $ 55x$.  So we have $ 34x=60 \\implies x=60/34$ and the price of gum is now $ 55\\cdot60/34$ which is approximately $ 97$ cents.\n\nI'm not totally sure about this... I'm not sure if I started off the problem correctly.  But either way we are finding $ 60$ time some ratio of the fibonacci numbers, and this ratio approaches $ \\frac{1+\\sqrt{5}}{2}$ quick enough that the gum price only changes a small fraction of a cent if we start from the wrong year.\n[/hide]\n\n[hide=\"9\"]\nFirst, we need to know how much volume half the sphere is.  The volume of a sphere is$ V=4/3 \\pi r^3$, so we have $ 1/2 \\cdot 4/3 \\pi (9)^3 = 486 \\pi$ cubic centimeters of space taken up.  This is no different than adding $ 54 \\pi$ cubic centimeters of water to the barrel.  Now the barrel can hold an additional $ v=\\pi r^2 h = \\pi (18)^2 \\cdot 1 = 324 \\pi$ cubic centimeters of water.  Therefore, $ 486\\pi-324\\pi = 162\\pi \\approx 508.938$ mL of water spill out (1 cubic centimeter is 1 mL).\n[/hide]\n\n[hide=\"10\"]\nFirst of all, this rectangle is really a square.\n\nThe direction of the rotation does not matter.  In each rotation, $ P$ travels along an arc of a circle, whose radius is determined by the distance of $ P$ from the point we are rotating around, whose measure is the measure of the rotation.  Both rotations are $ \\pi/2$ radians (90 degrees); $ PR = 2\\sqrt{2}$ and $ PS = 2$.  The formula for arc length is $ s=r\\theta$ with $ \\theta$ in radians.  So $ P$ has moved $ 2\\sqrt{2} \\cdot \\pi/2 + 2 \\cdot \\pi/2 = \\pi (\\sqrt{2}+1)$.\n[/hide]\n\n[hide=\"11\"]\nIf the number of people who ate both is $ x$, then $ 4x$ ate only chicken and $ 2x$ ate only spaghetti, for a total of $ 7x$ people.  Therefore, $ \\boxed{2/7}$ of the people only ate spaghetti.\n[/hide]\n\nwhat does \"meet\" mean in #12? do you mean end up on the same side of the river? because this clearly only happens at the beginning and at the end (they both cross an odd number of times).  Anyway I'll assume you mean \"pass\" when you say \"meet\".  Anyway, I'm not sure If I interpreted this correctly.\n[hide=\"12\"]\n\nJohn would cross the river a total of $ 7$ times, and would meet Jack every time he crossed (except when they started together) for a total of $ 6$ meetings.\n[/hide]\n\n[hide=\"13\"]\nLet $ a$, $ b$, and $ c$ be the number of candies that A, B and C respectively start with.  Now lets keep score:\n\n1st step: A gave B candies equal to half of what C had.\nA: $ a-1/2 c$\nB: $ b+1/2c$\nC: $ c$\n\n2nd step: B gave C candies equal to half of what A now had.\nA: $ a-1/2 c$\nB: $ b+1/2c - 1/2(a-1/2c)$\nC: $ c + 1/2(a-1/2c)$\n\nNow we have the system:\nA: $ a-1/2 c  = c$\nB: $ b+1/2c - 1/2a+1/4c= 17$\nC: $ 3/4c + 1/2a = a$\n$ a+b+c = 57$\n\nfrom C we obtain $ 3/2c = a$\nfrom A we obtain the same.\nfrom B we obtain $ b-1/2a+3/4c = 17 \\implies 4b-2a+3c = 68$.\n$ 4b-2(3/2 c)+3c = 68 \\implies 4b=68 \\implies b=17$\n\nTherefore $ a+b+c = 57 \\implies a+c = 40$\n$ 3/2c+c = 40 \\implies 5/2 c=40$\n$ c=\\boxed{16}$\n[/hide]\r\n\r\n:)"
}
{
  "Tag": [
    "induction"
  ],
  "Problem": "I cannot figure out how to do this problem at all. I've been working on it, and for some reason I can't solve it.\r\n\r\nUse mathematical induction to prove for n >= 1 and n an integer, n <sup>2</sup> +n is divisible by 2.\r\n\r\nThe method we were taught was by setting n=k and then assume true for n=k+1 .... but I can't figure this one out!\r\n\r\nPlease!",
  "Solution_1": "well... $n^2+n=n(n+1)$\r\n\r\nso we start with the base case, when $n=1$.\r\n$1\\cdot2 = 2$, so it is true.\r\n\r\nNow, let's say it works for some number $k$. Then $k(k+1)$ is an even number\r\nWe need to prove that it also holds true for $n=k+1$. That means we need to prove that $(k+1)(k+2)$ is diivisble by 2.\r\n\r\nLet's go back. $k(k+1) = k^2+k$\r\nthen $(k+1)(k+2) = k^2+3k+2 = (k^2+k)+(2k+2) = (k^2+k)+2(k+1)$ \r\n\r\nSince both $k^2+k$ and $2(k+1)$ are divisible by 2, we are done. QED.",
  "Solution_2": "jenybeny, I'd just like to add something.  reddevilz's solution is impeccable, but I'm not sure you understand exactly what mathematical induction is.  \r\n\r\nYou said [quote]The method we were taught was by setting n=k and then assume true for n=k+1 .... but I can't figure this one out! [/quote]\r\n\r\nNo, you don't assume it's true for $n=k+1$\r\n\r\nWhat you do is the following: First, you show it works for $n=1$.  \r\n\r\nThen, you assume it works for $n=k$ (called the INDUCTION HYPOTHESIS) and show it works for $n=k+1$\r\n\r\nThis works because you've shown it works for $n=1$ in step 1, so by step 2, it works for $n=2$.  Then, by step 2 again, it works for $n=3$.  And then, by step 2 again, it works for $n=4$ ad infinitum (is that the right expression, anyone?).  \r\n\r\nThus, by going through just these two steps, you've shown it works for all natural numbers n.",
  "Solution_3": "[quote=\"bubala\"]What you do is the following: First, you show it works for $n=1$. [/quote]\r\n\r\nIt is called the base case, and you dont have to use $1$. In some cases it might be easier to use other number, simply because of simplifying thingy.",
  "Solution_4": "o yea, and by the way, I dont think this belongs here. Probably Intermediate Section.",
  "Solution_5": "[quote=\"xxreddevilzxx\"]well... $n^2+n=n(n+1)$[/quote]You can stop there. Either n or n+1 is always even.",
  "Solution_6": "[quote=\"Peter VDD\"][quote=\"xxreddevilzxx\"]well... $n^2+n=n(n+1)$[/quote]You can stop there. Either n or n+1 is always even.[/quote]\r\n\r\nNot if what is requested is a proof by induction, Peter.",
  "Solution_7": "It works for 1: 1(2) is even\r\nNow assume that it works for n.  Then either n or n+1 has to be even.\r\nNow let's see if it works for n+1.  The number in question is then (n+1)(n+2).\r\nIf n was originally even, then we now have n+2 again, which must be even because an even number was added.\r\nIf n+1 was originally even, then the new number is even still.",
  "Solution_8": "[quote=\"Jenybeny27\"]I cannot figure out how to do this problem at all. I've been working on it, and for some reason I can't solve it.\n\nUse mathematical induction to prove for n >= 1 and n an integer, n <sup>2</sup> +n is divisible by 2.\n\nThe method we were taught was by setting n=k and then assume true for n=k+1 .... but I can't figure this one out!\n\nPlease![/quote]\r\n\r\nWell, to use induction, I guess you have to go far but I agree with Peter on general idea.\r\n\r\n[hide=\"What I think..\"]\n\nSince $n^2+n = n(n+1)$ and no matter what $n$ is, $n+1$ and $n$ are in different parity and this means that one is even and other one is odd (doesn't matter which is which) so it's going to be divisible by 2.[/hide]"
}
{
  "Tag": [],
  "Problem": "Consider O = {1, 3, 5, 7, . . .} the set of odd integers.  Define an operation ~ on O by the following rule:\r\na~b = the odd part of (a + b).  For example,\r\n7~5 = the odd part of 12 = 3\r\n3~11 = the odd part of 14 = 7\r\n3~5 = the odd part of 8 = 1\r\n\r\n1) Prove  a*(b~c) = (a*b)~(a*c)\r\n2) Is ~ commutative?  That is, does a~b = b~a?\r\n3) Is ~ associative?  That is, does a~(b~c) = (a~b)~c?\r\n4) Consider any odd numbers a and b.  Prove that there exists an odd number x such that a~x = b.\r\n5) Define f(x) = x~1.  Is it true that for any odd number x, if we continually apply f we will enventually get 1?\r\nFor example:  x = 9\r\nf(9) = 5, f(5) = 3, f(3) = 1, done.\r\nf(7) = 1, done.\r\nf(97) = 49, f(49) = 25, f(25) = 13, f(13) = 7, f(7) = 1, done.\r\nWill this always happen?",
  "Solution_1": "Sol'ns for 1 and 2; 3 forthcoming when I wake up.\n\n\n\n[hide]Let D (n) represent the \"odd part\" - i.e. the greatest odd divisor of n. Then a ~ b = D (a + b).\n\nObserve that if a = 2<sup>m</sup> * k where n :ge: 1, k is an positive integer (not necessarily odd), then D (a) = D (k).\n\n\n\nThen,\n\n\n\n1) Observe that a, b and c must all be odd in order for the question to make sense. (a * b) ~ (a * c) = D (a * (b + c)). Let b + c = 2<sup>m</sup> * k, k is odd. Then D (a * (b + c)) = D (ak * 2<sup>m</sup>) = D (ak) = ak.\n\n\n\na * (b ~ c) = a * D (b + c) = a * D (2<sup>m</sup> * k) = ak.\n\n\n\n2) This is obvious (D (a + b) = D (b + a)).\n\n\n\nTo be continued...[/hide]",
  "Solution_2": "[hide]Yes. Since the answer is always odd and always less than x, the numbers will gradually converge on the least odd number, 1. There are no cycles of more than 1 number, since a cycle would involve the numbers involved increasing, and the greastest odd divisor is never greater than the number in question. The number will never repeat until 1 is reached, since that would require the greatest odd divisor of x+1 to be x. Only possible for 2.[/hide]",
  "Solution_3": "Rocking so far . . . how about 3 and 4?",
  "Solution_4": "Solution  to 3 in spoiler:[hide] The case a=1 b=3 c=5 is a contradiction.\n\n1~3=1\n\n1~5=3\n\n\n\nBUT\n\n3~5=1\n\n1~1=1\n\n\n\n3 isn't 1.\n\n\n\nSo no.[/hide]\n\nSolution to 4, also in spoiler:[hide]The problem is basically saying \"If you have 2 odd numbers, a & b, is there always a number\" (x) \"that you can add to a to make b a's greatest odd divisor?\" There will always be a number x to add to a so that a=2 (or 3 or whatever, until you get the greatest odd divisor) b. Then, b is a's greatest odd divisor.[/hide]"
}
{
  "Tag": [
    "inequalities",
    "logarithms"
  ],
  "Problem": "Let $ a_1,a_2,\\ldots a_{2004}\\in [4; 5,6]$ and let $ x_1, x_2, \\ldots x_{2004}$ be a permutation of the numbers $ a_1, a_2,\\ldots a_{2004}$. Prove that:\r\n\r\n$ 1603 < \\log_{a_1}x_1 + \\log_{a_2}x_2 + \\ldots + \\log_{a_n}x_n < 2505$",
  "Solution_1": "For the LHS, Since $ x_i$ is a permutation shouln't it be $ \\ge 2004$ by AM-GM."
}
{
  "Tag": [
    "AMC",
    "AIME",
    "AIME I",
    "Science Olympiad"
  ],
  "Problem": "Unfortunately, there is big conflict between AIME I and NJ Science Olympiad State Competition.   What are you going to do if you plan to take AIME and you are in NJ Science Olympiad State Competition?",
  "Solution_1": "Take AIME II :wink:",
  "Solution_2": "Do you think AIME II is harder than AIME I?",
  "Solution_3": "In theory, they should be the same level.",
  "Solution_4": "[quote=\"spse\"]Do you think AIME II is harder than AIME I?[/quote]No -- they are intended to be the same level of difficulty.",
  "Solution_5": "They will average the same level of difficulty - the AMC makes up two essentially equal-difficulty tests, the AIME A and the AIME B, and flips a coin to determine which will be the I and II.",
  "Solution_6": "I highly suggest that you read the rules on AMC Pamphlet for this year and carry out the procedures for alternate testing so you do not end up missing the AIME! Good luck on both of exams."
}
{
  "Tag": [
    "abstract algebra",
    "calculus",
    "integration",
    "calculus computations"
  ],
  "Problem": "We know that the Dirichlet kernel $ D_n(y) \\equal{} \\frac {1}{2\\pi}\\sum_{k \\equal{} \\minus{} n}^{n}e^{iky}$. The integral $ \\int_{ \\minus{} \\pi}^{\\pi}D_n(y)\\ dy$ \"should\" be 1. But my calculations tell me that it is $ 0$.\r\nHere what I did:\r\n\\[ \\int_{ \\minus{} \\pi}^{\\pi}D_n(y)\\ dy \\equal{} \\int_{ \\minus{} \\pi}^{\\pi}\\frac {1}{2\\pi}(\\sum_{k \\equal{} \\minus{} n,k\\neq 0}^{n}e^{iky})\\ dy \\equal{}\r\n\\]\r\n\r\n\\[ \\frac {1}{2\\pi}\\sum_{k \\equal{} \\minus{} n,k\\neq 0}^{n}([\\frac {sin(ky)}{k}]_{ \\minus{} \\pi}^{\\pi} \\minus{} i[\\frac {cos(ky)}{k}]_{ \\minus{} \\pi}^{\\pi}) \\equal{} 0\r\n\\]\r\nWhat do I do wrong ?  :|",
  "Solution_1": "[quote=\"Amazigh\"] Dirichlet kernel $ D_n(y) \\equal{} \\frac {1}{2\\pi}\\sum_{k \\equal{} \\minus{} n,k\\neq 0}^{n}e^{iky}$. |[/quote]\r\n\r\nWhy did you remove the term $ k\\equal{}0$ from the sum? As I recall, that term makes your integral $ 1$ :wink: .",
  "Solution_2": "Haha, thanks :lol:"
}
{
  "Tag": [],
  "Problem": "Two objects with masses 6.00 kg and 2.00 kg, respectively, hang 1.00m above the floor from the ends of a cord 4.00 m long passing over a frictionless pulley.Both objects start from rest. Find the maximum height reached by 2.00 kg-object",
  "Solution_1": "First, calculate the velocity of the 6kg box just before it hits the ground. We have four forces: $ m_1g,m_2g,F_{T1} \\equal{} m_2g,F_{T2} \\equal{} m_1g$. Since $ m_1 \\equal{} 6$ and $ m_2 \\equal{} 2$, the net force on the 6kg box will be $ 4g$ down, and thus the 6kg box will accelerate at $ \\frac 23 g\\ \\mathrm{m/s^2}$.\r\n\r\nThe velocity of the 6kg box just before it hits the ground is $ \\sqrt {\\frac 43 g} \\equal{} \\frac 23 \\sqrt {3g}$ down because $ v^2 \\equal{} 2ax,a \\equal{} g,x \\equal{} 1$. Therefore, the velocity of the 2kg box will also be $ \\frac 23 \\sqrt {3g}\\ \\mathrm{m/s}$ up. At this point, there will be no more tension in the string, so the only force acting on the 2kg box will be gravity, $ 2g\\ \\mathrm{m/s^2}$ down. (Also, the 2kg box is 2 meters in the air.) Now, we just use $ s \\equal{} \\minus{} \\frac 12 gt^2 \\plus{} v_0t \\plus{} s_0$, plug and chug, we get the maximum height is $ s_0 \\plus{} \\frac 23$, and since $ s_0 \\equal{} 2$, $ h_{max} \\equal{} \\boxed{\\frac 53}$.",
  "Solution_2": "[quote=\"Yongyi781\"]First, calculate the velocity of the 6kg box just before it hits the ground. We have four forces: $ m_1g,m_2g,F_{T1} \\equal{} m_2g,F_{T2} \\equal{} m_1g$. Since $ m_1 \\equal{} 6$ and $ m_2 \\equal{} 2$, the net force on the 6kg box will be $ 4g$ down, and thus the 6kg box will accelerate at $ \\frac 23 g\\ \\mathrm{m/s^2}$.\n\nThe velocity of the 6kg box just before it hits the ground is $ \\sqrt {\\frac 43 g} \\equal{} \\frac 23 \\sqrt {3g}$ down because $ v^2 \\equal{} 2ax,a \\equal{} g,x \\equal{} 1$. Therefore, the velocity of the 2kg box will also be $ \\frac 23 \\sqrt {3g}\\ \\mathrm{m/s}$ up. At this point, there will be no more tension in the string, so the only force acting on the 2kg box will be gravity, $ 2g\\ \\mathrm{m/s^2}$ down. (Also, the 2kg box is 2 meters in the air.) Now, we just use $ s \\equal{} \\minus{} \\frac 12 gt^2 \\plus{} v_0t \\plus{} s_0$, plug and chug, we get the maximum height is $ s_0 \\plus{} \\frac 23$, and since $ s_0 \\equal{} 2$, $ h_{max} \\equal{} \\boxed{\\frac 53}$.[/quote]\r\nI think your solution is wrong since we have the acceleration of the system, as well as of these two boxes, is 5(from these equations: T-m_1g=m1.a and m2.g-T=m2a. Adding them we get: a=(m2-m1).g/(m1+m2)=5). I also got the part of velocity of the 6kg box as well as the 2kg box but I can't manage to find the [b]2.5 above the ground[/b] as the answer given by the book. I just get from the equation: s=v0.t+1/2.at^2=1. I think it seems to contradict since when the 2kg box moves, it still moves even when the 6kg box touches the ground because of its inertial frame, so the distance it moves must be greater than that of 6kg box moves.",
  "Solution_3": "Nobody minds helping me with this problem? :(",
  "Solution_4": "as the 6 kg object has more masss ,  it will go downward\r\n\r\nwriting the equations for both the objects\r\n\r\n(T-2g)=2a\r\nand 6g-T=6a\r\nwe get 4g=8a\r\na=g/2\r\n\r\nwhen the 6 kg block hits the groundit will stop but since the 2 kg block will have some velocity it will rise and tension will be zero\r\nthe 6 kg block stops when it travels a distance of 1 m\r\nhence the 2 kg block also travels a distance of 1 m\r\n\r\nhence final velocity of the 2kg block=sqrt(2*(g/2)*1)=sqrt(g)\r\n\r\nnote that it is now 2 m above the ground\r\nnow the 2kg block will rise without tension \r\nlet the maximum height attained be H\r\nthe sqrt(2gH)=sqrt(g)\r\nH=1/2\r\n\r\nhence total height =2+1/2=2.5\r\nor H=",
  "Solution_5": "[hide=\"My Solution:\"]\n[quote=\"ghjk\"]Two objects with masses 6.00 kg and 2.00 kg, respectively, hang 1.00m above the floor from the ends of a cord 4.00 m long passing over a frictionless pulley.Both objects start from rest. Find the maximum height reached by 2.00 kg-object[/quote]\nDraw the FBD for each mass:\nFor $ 6kg:$\n$ 60 \\minus{} T \\equal{} 6a$\n$ (T \\equal{} tension, \\; a \\equal{} acceleration \\; of \\; masses)$\nFor $ 2kg:$\n$ 3T \\minus{} 60 \\equal{} 6a$\nSolve linearly, get $ a \\equal{} 5m/s^2$.\nNow acceleration is applied till $ 6kg$ hits floor.\nThen,\n$ t \\equal{} \\sqrt {\\frac {2}{5}}$\nMax. height, $ H \\equal{} \\frac {v^2}{2g}$\n$ v \\equal{} at \\equal{} \\sqrt {10}$\nSubstituting, we get $ H \\equal{} 0.5$\nTherefore, max. height=$ \\boxed {1m \\plus{} 1m \\plus{} 0.5m \\equal{} 2.5m}$\n[/hide]",
  "Solution_6": "as the cord is massless the tension is uniform throught the cord .hence both the masses move with equal acceleration.on applying newtons second law we have 2 equations and  2 unknowns,tension and acceleration.\r\nusing kinematic equations we can find out the distance moved",
  "Solution_7": "[quote=\"aadil\"]as the cord is massless the tension is uniform throught the cord .hence both the masses move with equal acceleration.on applying newtons second law we have 2 equations and  2 unknowns,tension and acceleration.\nusing kinematic equations we can find out the distance moved[/quote]\r\nAND FINALLY, YOU GET $ 2.5m$!"
}
{
  "Tag": [
    "calculus",
    "calculus computations"
  ],
  "Problem": "$ x_n$ ($ n$ positive integer) is the solution of \r\n$ x^{\\frac{1}{n}}\\equal{}\\sqrt{1\\minus{}x}$,\r\nascertain the nature of the series $ \\sum_1^{\\plus{}\\infty}x_n$.",
  "Solution_1": "[hide=\"Hint\"]Show that $ x_n > \\frac {1}{n}$. It may be easier to do this using the inverse functions.[/hide]",
  "Solution_2": "[quote=\"Spasty\"]$ x_n$ ($ n$ positive integer) is the solution of \n$ x^{\\frac {1}{n}} \\equal{} \\sqrt {1 \\minus{} x}$,\nascertain the nature of the series $ \\sum_1^{ \\plus{} \\infty}x_n$.[/quote]\r\neasy:\r\n$ x^{1/n}$~$ 1 \\minus{} x/2$\r\nthen\r\n$ x$~$ 1 \\minus{} nx/2$\r\nor\r\n$ x_{n}$~$ \\frac {2}{1 \\plus{} n}$\r\ntherefore serie divergent"
}
{
  "Tag": [
    "ratio",
    "geometric series"
  ],
  "Problem": "Can someone remind me what the formula is for the sum of an infinite geometric series????\r\n\r\n\r\nSome other useful formulas would be great... :D",
  "Solution_1": "$\\frac{a}{1-r}$ where $|r|<1$ and $a$ is the first term of the sequence",
  "Solution_2": "Well, just so you know, it is quite simple to find the formula if you forgot it.  Just do teh following.\r\n$S_{\\infty}=a+ar+ar^2+...+ar^{\\infty}\\\\\\implies S_{\\infty}(r-1)=ar+ar^2+...+ar^{\\infty}-a-ar-ar^2-...-ar^{\\infty}=-a\\\\\\implies S_{\\infty}=\\frac{a}{1-r}$\r\n\r\nYou coudl do the exact same to find the formula for $S_n=\\frac{a(r^n-1)}{r-1}$\r\nMasoud Zargar",
  "Solution_3": "boxedexe, be careful with your notation, infinity cannot be used like a number, i agree that is the correct idea, but the proof is more rigirous to say that the limit of the sum is $\\frac{a_0}{1-r}$",
  "Solution_4": "Clearly, you are correct with the limit idea. :)",
  "Solution_5": "its a/ 1-r where a is the first term and r is the common ratio \r\n\r\n :alien:  :thumbup:"
}
{
  "Tag": [
    "calculus",
    "integration",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Prove that \r\n\r\n$\\int_{0}^{\\frac{\\pi}{2}}\\arccos(\\frac{ \\cos(x)}{1+2 \\cos(x)})dx=\\frac{5\\pi^2}{24}$\r\n\r\n----------------------------------------------------------------------------------\r\n\r\nHere a proof, it use only material from MPSI/MP* \r\n http://www.dma.ens.fr/culturemath/contenu/dossiers.html#CoxeterInt\r\n\r\n\r\nIf you have another proof post it\r\nthx",
  "Solution_1": "[url]https://math.stackexchange.com/questions/4037652/solve-int-0-pi-2-arccos-left-dfrac-cosx12-cosx-right-dx[/url] , [url]https://math.stackexchange.com/questions/3956051/help-with-int-0-pi-2-arccos-frac-cosx12-cosxdx?noredirect=1[/url]"
}
{
  "Tag": [
    "greatest common divisor",
    "prime divisor",
    "number theory",
    "inequalities"
  ],
  "Problem": "For any positive integer $ n>1$, let $ P\\left(n\\right)$ denote the largest prime divisor of $ n$. Prove that there exist infinitely many positive integers $ n$ for which\n\\[ P\\left(n\\right)<P\\left(n\\plus{}1\\right)<P\\left(n\\plus{}2\\right).\\]",
  "Solution_1": "In fact, I had seen it for the first time in a romanian selection test in 1984 I think. But I don't know to solve this extremely beautiful problem.",
  "Solution_2": "It is wonderful :) I adore it :)\r\nI heard first of it last year, from one of my IMO team mates.\r\nBut he didn't have any idea of a solution; neither do I.\r\nSo ... a solution would be really really really appreciated.",
  "Solution_3": "It just so happens that i have a proof of this right in front of me. I'd love this to be mine, but it's copied word by word from a book:\r\n\r\n[quote=\"problem\"]For any positive integer $ n > 1$, let $ P\\left(n\\right)$ denote the largest prime which divides $ n$.\nProve that there exist infinitely many positive integers $ n$ for which\n\n$ P\\left(n\\right) < P\\left(n \\plus{} 1\\right) < P\\left(n \\plus{} 2\\right)$.[/quote]\r\n\r\nLet $ p>2$ be a prime, and let $ n_k\\equal{}p^{2^k}\\minus{}1$. We have $ P\\left(n_k\\plus{}1\\right)\\equal{}p$. We look at the sequence $ \\left(n_k\\plus{}2\\right)\\ \\left\\{k\\geq 1\\right\\}$. We can see that any 2 distinct terms have 2 as their greatest common divisor, because for $ i<j$ we have $ \\left(p^{2^j}\\plus{}1\\right)\\minus{}2 \\equal{} \\left(p\\minus{}1\\right)\\left(p\\plus{}1\\right)\\left(p^2\\plus{}1\\right) ...\\left(p^{2^i}\\plus{}1\\right) ...\\left(p^{2^{j\\minus{}1}}\\plus{}1\\right)$.\r\n\r\nSince none of the terms of the sequence considered is divisible by 4, it means that the sequence $ P\\left(n_k\\plus{}2\\right)$ contains arbitrarily large primes. Let $ k$ be the smallest integer for which $ P\\left(n_k\\plus{}2\\right) > p$. Since $ n_k\\equal{}\\left(p\\minus{}1\\right)\\cdot\\left(p\\plus{}1\\right)\\cdot n_1\\cdot n_2\\cdot ...\\cdot n_{k\\minus{}1}$ and all prime divisors of the number on the right are $ <p$, it means that $ P\\left(n_k\\right)<p\\equal{}P\\left(n_k\\plus{}1\\right)<P\\left(n_k\\plus{}2\\right)$, and such a number $ n_k$ can be found for any prime $ p$.",
  "Solution_4": ":? I'm a bit confused ...\r\n\r\nWhen you type n_k + 1 for example, do you mean n<sub>k</sub> + 1 or n<sub>k + 1</sub>?\r\n...",
  "Solution_5": "[quote=\"grobber\"]Let p>2 be a prime, and let n_k=p^(2^k)+1. We have P(n_k+1)=p.[/quote]\r\n\r\n:?:?:? Why that? Am I missing something?",
  "Solution_6": "When I say a_k+1 I mean (a_k) + 1, not the (k+1)'th term of the sequence. \r\n\r\nAbout that other thing, it's simple: I wrote + instead of -. I've edited it, so see if it makes sense now. Sorry! :D",
  "Solution_7": "Ok, thanks ... :)",
  "Solution_8": "Another idea...\r\nLet p<sub>1</sub>, p<sub>2</sub>, ... be the sequence of prime numbers.\r\nLet f<sub>n</sub> be the product of the first n primes: f<sub>n</sub> = p<sub>1</sub>p<sub>2</sub>...p<sub>n</sub>.\r\nTake a positive integer k with...\r\n*[i]P[/i](k) \\leq p<sub>n</sub>, and\r\n*kp<sub>1</sub>p<sub>2</sub>...p<sub>n</sub> + 1 = p<sub>n + 1</sub><sup>m</sup> for some m.\r\nDefine m = kf<sub>n</sub>. Then we have\r\n*[i]P[/i](m) = p<sub>n</sub>\r\n*[i]P[/i](m + 1) = p<sub>n + 1</sub> > p<sub>n</sub>\r\n*[i]P[/i](m + 2) > p<sub>n + 1</sub>, since it's easy to check that none of p<sub>1</sub>, ..., p<sub>n + 1</sub> divides m + 2.\r\nSo we should prove the existence of such a number k.\r\nIt's easy to prove that there is a \"k\" such that kp<sub>1</sub>p<sub>2</sub>...p<sub>n</sub> + 1 = p<sub>n + 1</sub><sup>m</sup> for some m,\r\nbut how to prove that there exists such a k for which [i]P[/i](k) \\leq p<sub>n</sub>?\r\nMaybe I'm overlooking something obvious :D or maybe it's not tht obvious.",
  "Solution_9": "i tried this problem  for a long time. i  was kinda close in a way because i knew that the exponent  had to  be   power of two . but  choosing these numbers is very  clever,  how  can  that  natural to someone?",
  "Solution_10": "[quote=\"Arne\"]Another idea...\nLet $ p_1, p_2, \\ldots$ be the sequence of prime numbers.\nLet $ f_n$ be the product of the first $ n$ primes: $ f_n \\equal{} p_1p_2...p_n$.\nTake a positive integer k with...\n$ P(k) \\leq p_n,$ and\n$ kp_1p_2...p_n \\plus{} 1 \\equal{} p_{n \\plus{} 1}m$ for some $ m.$\nDefine $ m \\equal{} kf_n.$ Then we have\n$ P(m) \\equal{} p_n$\n$ P(m \\plus{} 1) \\equal{} p_{n \\plus{} 1} > p_n$\n$ P(m \\plus{} 2) > p_{n \\plus{} 1},$ since it's easy to check that none of $ p_1, ..., p_{n\\plus{}1}$ divides $ m \\plus{} 2.$\n[/quote]\r\nI think, this is wrong! Why do you think that between \r\n$ p_{n}$ and $ p_{n \\plus{} 1}$  is not another prime number?",
  "Solution_11": "I give full solution in Russian forum.",
  "Solution_12": "can one say that there are infinite primes of the form $ 3^n \\plus{} 2$ ?\r\nif so , then \r\nP($ 3^n$) = 3\r\n$ 3^n \\plus{} 1$ > P($ 3^n \\plus{} 1$) > 3 , because $ 3^n \\plus{} 1 \\equal{} 2^s$ has only trivial solutions and $ 3^n \\plus{} 1$ is not divisible by 3 and because its one less than a prime , it must have a divisor\r\nP($ 3^n \\plus{} 2$) = $ 3^n \\plus{} 2$",
  "Solution_13": "to Grobber : you have to write : $ n_k\\equal{}(p\\minus{}1)(p\\plus{}1)(n_1\\plus{}2)(n_2\\plus{}2)..(n_{k\\minus{}1}\\plus{}2)$ instead of : $ n_k\\equal{}\\left(p\\minus{}1\\right)\\cdot\\left(p\\plus{}1\\right)\\cdot n_1\\cdot n_2\\cdot ...\\cdot n_{k\\minus{}1}$",
  "Solution_14": "Thats a very nice solution grobber. i love it",
  "Solution_15": "Had to throw in my 2 cents:\n\nIt is easy to see if n + 1 is the square of an odd prime, n will behave as needed.  We only need to show EITHER n + 1 is a square of an odd prime and n + 2 is forced in certain circumstances to follow the rest OR n + 2 be a square of an odd prime and show that P(n) < P(n + 1) for some specific situations.  I will try to unsuccessfully show the former.\n\nBut I am using Dirichlet's.  And from Dirichlet's we know there are infinitely many primes with the ones digit being 1 (i.e. the sequence 10k + 1 will hit infinitely many primes).  Call this prime p.  Note that P(p^2) = p.  Note further that $P(p^2 -1) = P(n)$ is less than HALF p + 1 as p + 1 and p - 1 are both even.\n\nEh, figured it might go somewhere.  Anyone with the rest of this proof?",
  "Solution_16": "[quote=grobber]It just so happens that i have a proof of this right in front of me. I'd love this to be mine, but it's copied word by word from a book:\n\n[quote=\"problem\"]For any positive integer $ n > 1$, let $ P\\left(n\\right)$ denote the largest prime which divides $ n$.\nProve that there exist infinitely many positive integers $ n$ for which\n\n$ P\\left(n\\right) < P\\left(n \\plus{} 1\\right) < P\\left(n \\plus{} 2\\right)$.[/quote]\n\nLet $ p>2$ be a prime, and let $ n_k\\equal{}p^{2^k}\\minus{}1$. We have $ P\\left(n_k\\plus{}1\\right)\\equal{}p$. We look at the sequence $ \\left(n_k\\plus{}2\\right)\\ \\left\\{k\\geq 1\\right\\}$. We can see that any 2 distinct terms have 2 as their greatest common divisor, because for $ i<j$ we have $ \\left(p^{2^j}\\plus{}1\\right)\\minus{}2 \\equal{} \\left(p\\minus{}1\\right)\\left(p\\plus{}1\\right)\\left(p^2\\plus{}1\\right) ...\\left(p^{2^i}\\plus{}1\\right) ...\\left(p^{2^{j\\minus{}1}}\\plus{}1\\right)$.\n\nSince none of the terms of the sequence considered is divisible by 4, it means that the sequence $ P\\left(n_k\\plus{}2\\right)$ contains arbitrarily large primes. Let $ k$ be the smallest integer for which $ P\\left(n_k\\plus{}2\\right) > p$. Since $ n_k\\equal{}\\left(p\\minus{}1\\right)\\cdot\\left(p\\plus{}1\\right)\\cdot n_1\\cdot n_2\\cdot ...\\cdot n_{k\\minus{}1}$ and all prime divisors of the number on the right are $ <p$, it means that $ P\\left(n_k\\right)<p\\equal{}P\\left(n_k\\plus{}1\\right)<P\\left(n_k\\plus{}2\\right)$, and such a number $ n_k$ can be found for any prime $ p$.[/quote]\n\nWhy $p(n_k)<p$?",
  "Solution_17": "[quote=nmd27082001]\nWhy $p(n_k)<p$?[/quote]\nActually, the equation there is not quite correct and it should be $n_k=(p-1)(p+1)(n_1+1)(n_2+1)\\dotsc (n_{k-1}+2)$ as the factors are of the form $p^{2^j}+1$, not $p^{2^j}-1$.\nClearly, all prime divisors of $p-1$ and $p+1$ are less than $p$. And by assumption $P(n_j+2) \\le p$ for all $j<k$, hence all primes dividing one of the factors on the RHS are $\\le p$, so $P(n_k) \\le p$ but clearly $p$ itself does not divide $n_k$ either, so $P(n_k)<p$.\n\n"
}
{
  "Tag": [
    "invariant",
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "Grandson of a Baron Minhauzen states that in his country there exist $n\\geq 3$ major cities , each two of which are connected with the airline in one direction. He also claims that from each city it is possible to arrive two any other with at most one change of the planes. For which values of $n\\geq 3$ is it possible that the grandson of Barone Minhauzen claims the truth?",
  "Solution_1": "I can only prove that it works for odd $n$ up until now:\r\n\r\nLabel the cities $0,1,2,\\ldots,n-1$, and draw an arc from $x$ to $y$ iff the smallest non-negative residue of $y-x$ modulo $n$ is odd. It's easy to prove that it works.",
  "Solution_2": "i think it works for all $n\\geq 5$ and for $n=3$.(hopefully i haven't goofed up anywhere..)\r\n\r\nbasically if it works for $n$, then it is easy to see that it works for $n+2$ as well. just add $2$ new vertices to the vertex set $V$, call them $x,y$ and have directed edges from each vertex of $V$ to x and from $y$ to each vertex of $V$.finally add an edge from $x$ to $y$.now for  $x$ and $y$ clearly both trips are possible by paths of length $\\leq 2$.for any $v\\in V$, $v\\rightarrow x$ is possible by definition.also, $x\\rightarrow y\\rightarrow v$ so the reverse is also possible. finally for any $v\\in V, y\\rightarrow v$ by definition and $v\\rightarrow x\\rightarrow y$ and so the claim holds.\r\n\r\nsince for $n=3$ the directed triangle works so for all odd n the above construction works.\r\nfor $n=6$ a grid is possible,so for all even $\\geq 6$ it works again. \r\nthe grid for $n=6$ is:\r\n  $(1,2),(1,3),(4,1),(5,1),(6,1),(3,2),(2,4),(2,5),(2,6),(4,3),(3,5),(6,3),(5,4),(4,6),(6,5)$.",
  "Solution_3": "Ahh, I see. I actually thought it didn't work for even $n$, and was too lazy to look for a construction for $6$ :(.",
  "Solution_4": "actually so did i! but i couldn't find any invariants,so tried to see why it wouldn't work for 6 and somehow fooling around  got that configuration."
}
{
  "Tag": [
    "symmetry",
    "probability",
    "articles",
    "Euler",
    "calculus",
    "geometry",
    "number theory"
  ],
  "Problem": "I didnt really want to make a new topic..but i couldnt really find a topic similiar to this one...and sorry if there is one where in this site....but I have been reading some wonderful ...truly wonderful books on te subject of mathematics. I have read The Milliennium Problems by Keith Devlin (who is a really good author) and I am currently reading Godel, Escher, Back by Douglas Hofstadler and although everytime I read it I only get 1/4 of what I am reading. so I end up reading each page 4 times to understand what he is trying to say.....i still love it...its really really good and I'm not even done with the book yet....I am acutally planning on reading more boks by Keith Devlin becuase his kind of books are what I am interested in....Logic...thoeries...and well I cant really explain it...I hope that reading these books might....but I am more interested in reading about thoreies than tryin to do themm....although godel, escher, bach's MU puzzle was actually really interesting for me to do it...and I am reading a logic book...Its really hard for me to explian exactly what i mean in words...I was wondering if anyone can recommend me to some other great and wonderful books that I might like and enjoy reading...THANKS",
  "Solution_1": "I agree Keith Devlin books are quite good. So are Ian Stewart books. I'll add more books to this reply through an edit tomorrow or the next day.",
  "Solution_2": "Have you read Music of the Primes?\r\n\r\nAlso, the various Simon Singh books are good for the mathematical layperson.",
  "Solution_3": "I like reading math biographies. Two that stand our are [i]The Man Who Loved Only Numbers[/i] and [i]The Man Who Knew Infinity[/i].",
  "Solution_4": "these books seems pretty interested..but can u provide a little summary and the author of the book so that i know a bit more about the book?",
  "Solution_5": "If you're looking for casual mathematics reading, I would recommend [i]The Colossal Book of Mathematics[/i] by Martin Gardner.  It has some very interesting chapters on topics as varied as Diophantine equations, symmetry, higher dimentions, probability, fractals, and paradoxes.  It even has a chapter about the book [i]Gdel, Escher, Bach[/i] that SLeon mentioned earlier.",
  "Solution_6": "[quote=\"dts\"]If you're looking for casual mathematics reading, I would recommend [i]The Colossal Book of Mathematics[/i] by Martin Gardner.  It has some very interesting chapters on topics as varied as Diophantine equations, symmetry, higher dimentions, probability, fractals, and paradoxes.  It even has a chapter about the book [i]Gdel, Escher, Bach[/i] that SLeon mentioned earlier.[/quote]\r\n\r\nThe other collections of Martin Gardner's articles are also good for casual reading.",
  "Solution_7": "[i]Countdown, the Art of Problem Solving, the Art and Craft of Problem Solving. [/i] and [i]Of Men and Numbers[/i] are good books...",
  "Solution_8": "\"Prime Obssesion\", by John Derbyshire is definately the most detailed math book for a lay-person that I've seen (In my opinion, it's much better than Music of the Primes and the Millenium Problems). It's also about the Riemman Hypothesis, and actually discusses and shows part of the derivation for the formula for primes less than a given number, the paths and efforts of those who are trying to prove the hypothesis, plus a healthy does of 19th century math history. \r\n\r\n\"Mathematical Mysteries\", by Calvin C. Clawson is also a good all around cool and inspirational mainly number theory math book, with topics such as formulas for pi, the euler-mascheroni constant, ect.. Paraphrasing Amazon.com, \"He delves into the heart of number theory to see and understand the exquisite relationships among natural numbers.\"",
  "Solution_9": "I think an interesting math book is Lapses in Mathematical Reasoning by Bradis, MInkovskii, and Kharchea  :)",
  "Solution_10": "is that book going ot be soemthing that needs like a calcalus background to understand?",
  "Solution_11": "[quote=\"rrusczyk\"]Also, the various Simon Singh books are good for the mathematical layperson.[/quote]\r\n\r\nHe speaks the truth.  Reading Fermat's Enigma is what really got me into math.",
  "Solution_12": "[quote=\"elston\"]I think an interesting math book is Lapses in Mathematical Reasoning by Bradis, MInkovskii, and Kharchea  :)[/quote]\n\n[quote=\"SLeon\"]is that book going ot be soemthing that needs like a calcalus background to understand?[/quote]\r\n\r\nNo, that book has no calculus in it (as far as I can tell).  The most complicated things it has are geometry, logs, and trigonometry.  Some of the stuff in the earlier chapters just involve arithmetic...",
  "Solution_13": "THANKS FOR RECOMMANDING SIMON SINGH'S FERMAT'S ENGIMA...IT WAS REALLY REALLY GOOD... IM GLAD THAT I HAD THE CHANCE TO READ IT BECUASE IT [i]WAS [/i]BY FAR THE BEST BOOK I READ",
  "Solution_14": "My favorites are  books by George Gamow - Eg One, two three infinity and Mr. Tomkins in Wonderland.\r\n\r\nMost of his other books are physics related but are fun."
}
{
  "Tag": [],
  "Problem": "Let L be a six digit (base ten) integer whose last digit is not 0, and let M be the number formed by moving this last digit to the front. For example, if L = 354268, then M = 835426. Find all possible L such that L = 3M. The sum of all the digits of all the possible L is:\r\n\r\na) 36\r\nb) 72\r\nc) 54\r\nd) 48\r\ne) None of these\r\n\r\n[hide=\"answer\"]\nc) 54\n[/hide]\r\n\r\n\r\nHow do you do this? Please explain to me like a absolute beginner because I forgot some of AoPS 1.  :blush:",
  "Solution_1": "We know the sum must be a multiple of 3, but this doesnt help us!\r\nI would use case work for the first digits of L,M",
  "Solution_2": "OK I think I've got the answer.\r\n\r\n[hide=\"Click to view hint\"]\nSuppose L = ABCDEF. Then M = FABCDE. Now let x = ABCDE and y = F.\nWork on from there.\n\n[hide=\"Click to view solution\"]\nL = 10x + y and M = 100000y + x.\nSince L = 3M, we have the equation: 10x + y = 300000y + 3x.\nSo 7x = 299999y and hence x = 42857y.\nNote that x is a one-digit number and y is a five-digit number.\nSuppose x is 1. Then y is 42857.\nSuppose x is 2. Then y is 85714.\nSuppose x is 3 or bigger. Then y is 128571 or bigger, and isn't a five digit number anymore.\nSo L = 142857 or 285714\nAdding the digits up gives $ \\boxed{54}$.\n[/hide]\n\n[/hide]",
  "Solution_3": "Brilliant! Thank you so much smileyboi!\r\n\r\n :D",
  "Solution_4": "Nice:\r\nTry multiplying 142857 by 1,2,3,4,5,6,7. What do you get?",
  "Solution_5": "Well, the number is basically 1/7, without the repeating part and multiplied by 1000000. So, you can figure out what it is by examining 1000000*1/7*n, where n=1,2,3,4,5,6,7 except that for n=7, the answer won't be 10000000, it'll be 1 less.",
  "Solution_6": "142857*2=285714\r\n142857*3=428571\r\n142857*4=571428\r\n142857*5=714285\r\n142857*6=857142\r\n142857*7=999999\r\nDo you see the magic?",
  "Solution_7": "That's basically what I was saying since 285714 is the repeating part after the decimal in the expansion of $ 2*\\frac{1}{7}$. It can be extended to each of the other $ n*\\frac{1}{7}$, for n=1,2,3,4,5,6. The n=7 is an interesting case though.",
  "Solution_8": "Interesting..."
}
{
  "Tag": [
    "linear algebra",
    "matrix"
  ],
  "Problem": "In sistemul cartezian xOy se considera punctele coliniare $A_{i}(x_{i},y_{i}),i=1,4$,astfel incat sa existe matricile inversabile M din $M_{4}(C)$ in care primele doua linii sunt $(x_{1},x_{2},x_{3},x_{4})$ si $(y_{1},y_{2},y_{3},y_{4})$.Aratati ca suma elementelor matrici $M^{-1}$ nu depinde de matricea $M$ data.",
  "Solution_1": "Moubinool asked me to translate this in English.\r\n\r\nIn the plane $xOy$ consider the collinear points $A_i(x_i,y_i),\\ i=\\overline{1,4}$ s.t. there is at least an invertible matrix $M\\in\\mathcal M_4(\\mathbb C)$ for which the first two lines are $(x_1,x_2,x_3,x_4)$ and $(y_1,y_2,y_3,y_4)$. Show that for all such invertible matrices $M$, the sum of the elements of $M^{-1}$ is the same.",
  "Solution_2": "Thank you grobber"
}
{
  "Tag": [],
  "Problem": "The numbers 3, $ a$, $ b$, $ c$, $ d$, 38, $ \\ldots$ form an arithmetic sequence. What is the value the sum $ b\\plus{}c$?",
  "Solution_1": "b+c=a+d because you subtract the common difference from b and add the common difference to c. Doing this again, you find that b+c=3+38=[b]41[/b].",
  "Solution_2": "Another way to do this problem would be to find $ b$ and $ c$. We can figure out the common difference between consecutive terms, which is $ 7$. The sequence is $ 3, 10, 17, 24, 31, 38$ and $ 17 \\plus{} 24 \\equal{} \\boxed{41}$",
  "Solution_3": "[quote=\"UberYoda\"]Another way to do this problem would be to find $ b$ and $ c$. We can figure out the common difference between consecutive terms, which is $ 7$. The sequence is $ 3, 10, 17, 24, 31, 38$ and $ 17 \\plus{} 24 \\equal{} \\boxed{41}$[/quote]\nAleph0's way is a more favorable approach considering the times pressure in a $45$ second limit--especially since you need to solve the problem before your opponent does."
}
{
  "Tag": [
    "inequalities"
  ],
  "Problem": "Suppose $ a,b,c$ are three sides of triangle.\r\n\r\nProve that\r\n\r\n$ \\sqrt{a\\plus{}b\\minus{}c} \\plus{}\\sqrt{b\\plus{}c\\minus{}a} \\le 2\\sqrt{b}$\r\n\r\n :wallbash:",
  "Solution_1": "[hide]Let a = x+y, b = y+z, c = z+x. Then we get that this is equivalent to $ \\sqrt{2y} \\plus{} \\sqrt{2z} \\le 2\\sqrt{y\\plus{}z},$ which is true by QM-AM.[/hide]",
  "Solution_2": "Hi vinksman since both sides are positive first square them\r\n\r\nThen cancelling commom terms and squaring again what u want to prove is \r\n$ (a\\plus{}b\\minus{}c)(b\\plus{}c\\minus{}a)$<=$ b^2$\r\n\r\njust direct expansion yields what u want to prove is \r\n\r\n$ \\minus{}(a\\minus{}c)^2$ <=0\r\nWhich is obviously true",
  "Solution_3": "b = (b+a-c + b-a+c)/2.\r\n\r\nso, setting b+a-c = x\r\n\r\n                b-a+c = y,\r\n\r\nthe inequality is equivalent to\r\n\r\nsqrt(x) + sqrt(y) <= 2sqrt((x+y)/2)\r\n\r\n<=> x + y + 2sqrt(xy) <= 2(x+y) = x + y + (x+y)\r\n\r\n<=> x + y => 2sqrt(xy)\r\n\r\nwhich is true by AM-GM.\r\n\r\ntherefore, the original inequality is true.  :)"
}
{
  "Tag": [
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "let $a,b,c,d \\in [1,2]$\r\nfind min and max $A=\\sum_{cyc}\\frac{a}{b+cd}$",
  "Solution_1": "[hide=\"min\"]We have: $c+d \\ge cd$ \nso that $b+cd \\le b+c+d$\n$\\Rightarrow \\frac{a}{b+cd}\\ge \\frac{a}{b+c+d}$\n$\\Rightarrow LHS \\ge \\sum_{cyc}\\frac{a}{b+c+d}\\ge \\frac{4}{3}$[/hide]\n\n[hide]I think that max=2[/hide]",
  "Solution_2": "[quote=\"tunganh\"]I think that max=2[/quote]\r\nYou are right since, $A(a),$ $A(b),$ $A(c)$ and $A(d)$ are convex functions. Hence, $\\max A=$\r\n$=\\max\\{A(1,1,1,1),A(1,1,1,2),A(1,1,2,2),A(1,2,2,2),A(2,2,2,2)\\}=2.$"
}
{
  "Tag": [],
  "Problem": "Find two consecutive integers with the property that the sums of their digits are each divisible by $2006$.",
  "Solution_1": "This number x is $x=y*10^k-1$, were $y\\not=9(mod \\ 10)$, and the sum digits y+1 divisible 2006, $2006|(9k-1)\\Longrightarrow k=223+2006*m$. For example minimal number is $y=10^{223}-3, x=10^{446}-2*10^{223}-1$.",
  "Solution_2": "Well this is not from the german olympiad but from the german national contest (Bundeswettbewerb Mathematik)."
}
{
  "Tag": [
    "function",
    "calculus",
    "integration",
    "induction",
    "real analysis",
    "real analysis solved"
  ],
  "Problem": "Let be given a possitive integer n . Consider a continous function $ f(x):[0;n]-->R $ satisfying $ f(0)=f(n) $ . Prove that : there exist n couples of numbers $ a_i , b_i (i=1,2,..,n) $ belonging to $ [0;n] $ such that $ b_i-a_i $ are possitive integers and $ f(a_i) = f(b_i) $ for all $ i=1,2,..,n $",
  "Solution_1": "Infact we can find infinitely many such couples\r\n Let $c$ be a point in $(0,n)$ so that $f(c)$ is different than $f(0)$ and therefore $f(n)$\r\n ( If such $c$ does not exist then $f$ is constant and the conclusion holds trivially)\r\n Apply the Intermediate value property, we find $a_1 \\in (0,c)$ ,$b_1\\in (c,n)$ so that\r\n $f(a_1) = f(b_1) = \\frac {f(c) + f(0)}{2}$.\r\n Now consider $(a_1,b_1)$ alone and follow the same procedure above we have $a_2,b_2$, then $a_3,b_3$....",
  "Solution_2": "dickchimney, I don't see that you have shown that $b_1-a_1$ is an integer. In fact, with that integral-difference condition, there are in general only finitely many such pairs of ponts.\r\n\r\nIn fact, to be assured of having $n$ such pairs of points, we must include $(0,n)$ as one of the pairs.",
  "Solution_3": "If $(a_i,b_i)$ is such a couple, can we also take $(b_i,a_i)$ as being one?",
  "Solution_4": "[quote=\"Kent Merryfield\"]dickchimney, I don't see that you have shown that $b_1-a_1$ is an integer. In fact, with that integral-difference condition, there are in general only finitely many such pairs of ponts.\n\n[/quote]\r\n\r\n Yep!! Thank you for that!! :blush:  \r\n But it must be true that there exist $x$ so that $f(x)=f(x+1)$ where $x \\in [0,n-1]$\r\n Otherwise we will have $g(x)=f(x)-f(x+1) > 0$ or $<0$ everwhere which leads to\r\n $f(0) > f(1) >...>f(n)$ or $f(0) < f(1) <... <f(n)$, contradiction.\r\n Once we have $f(x)=f(x+1)$ we can glue two points $x, x+1$ ( even if $x=0$ or $n-1$) together without affecting the hypothesis, especially the interger-diffrerence property!! Then we can process by induction, right?",
  "Solution_5": "I do not think it's that easy, man. I think there can be some integers $m<n$ for which there's no $x$ s.t. $f(x)=f(x+m)$. We can use your method to find such $x$'s for the case $m|n$ with no problems, but it doesn't look good if we want to find $n$ pairs.. :?\r\n\r\n[Edit: I think I see what you mean now; I hadn't read it properly :); sorry about that  :blush: ]",
  "Solution_6": "[quote=\"grobber\"] I think there can be some integers $m<n$ for which there's no $x$ s.t. $f(x)=f(x+m)$. [/quote]\r\n\r\n You're right!! But I did not prove that!! \r\n I process by induction on the length of the interval, (after the glueing thing, the length decreases by 1)\r\n Please check it again!! :D",
  "Solution_7": "[quote=\"grobber\"] I hadn't read it properly :blush: ][/quote]\r\n\r\n No problem, that is the case of my first \"proof\" :blush:",
  "Solution_8": "Here's another approach.\r\n\r\nFor $k\\in\\{1,2,\\dots\\}$, let $g_k(x)=f(x+k)-f(x).$ We note that $g_n(0)=0.$ We count that as one of our solutions. What we now need is for there to be a total of at least $n-1$ roots of the collection of equations $g_k(x)=0$ for $1\\le k\\le n-1.$\r\n\r\nConsider $f$ to be continuous and periodic of period $n.$ (That's what $f(0)=f(n)$ is good for.) Then define each $g_k$ to be continous and periodic. Each $g_k$ has average value zero.  Thus either $g_k$ is indentically zero (in which case we have infinitely many solutions) or it has at least one positive and at least one negative value in each period. Thus (by the Intermediate Value Theorem) for each $k,$, $1\\le k \\le n-1,$ there are at least two roots of $g_k(x)=0$ in $[0,n).$\r\n\r\nThis apparently gives us $2(n-1)$ solutions, but can we count them all?\r\n\r\nIf $g_k(x)=0$ for $0\\le x \\le n-k,$ then this is a solution on our list. But if $n-k<x<n$ then we can rewrite this as $g_{n-k}(x-n+k)=0$ and it counts as a root of $g_{n-k}(x)=0.$ Thus each solution we found does count as one of our solutions, but we may be counting the same solution twice (but no more than twice). Allowing for the double counting, we still have $(n-1)$ solutions, plus the one solution of $g_n(x)=0$ which we handle separately.",
  "Solution_9": "What Kent just proved is somewhat stronger than the desired result but I cannot state it clearly",
  "Solution_10": "dear Kent Merryfield , why can you write $ g_{n-k}(x-n+k)=0 $ where $ n-k\\leq x\\leq n $ ?",
  "Solution_11": "[quote=\"romano\"]dear Kent Merryfield , why can you write $ g_{n-k}(x-n+k)=0 $ where $ n-k\\leq x\\leq n $ ?[/quote]\r\n\r\n Let's see what is $ g_{n-k}(x-n+k) $ \r\n By definition It it just $f(x)-f(x-n+k)$. We have\r\n $g_{n-k}(x-n+k)= f(x)-f(x-n+k)=f(x)-f(x+k)=-(f(x+k)-f(x)) =-(g_k(x))$\r\n So the former is zero iff the latter is zero !! That's all! :D",
  "Solution_12": "Thanks for covering that, dickchimney. By the way, the double counting always happens. There is a one-to-one correspondence between a root of $g_k$ in the \"wrong\" part of the interval and a root of $g_{n-k}$ in the \"right\" part of the interval, and vice versa. The only way to have more than the specified minimum number of solutions is to have some $g_k$ have more than two roots.\r\n\r\nIt took me a while to understand your solution in #5, but now that I do, I see that it is quite elegant. You are inducting on $n$, and your inductive hypothesis is that the whole theorem is true for $n-1.$",
  "Solution_13": "Thank u Kent! ;) \r\n What is your opinion Romano ? Is it solved ?",
  "Solution_14": "Yeah , nice solutions , Kent , Dickchimney !"
}
{
  "Tag": [
    "probability",
    "number theory",
    "relatively prime"
  ],
  "Problem": "What is the probability that two integers picked at random are relatively prime?",
  "Solution_1": "The probability for two random numbers to be relatively prime is proven to be $P[\\gcd(a,b)=1]=[\\zeta (2)]^{-1}=\\frac{6}{\\pi^2}$, where $\\zeta (s)=\\sum_{k=1}^{\\infty}\\frac{1}{k^s}$.\r\n\r\nMasoud Zargar",
  "Solution_2": "Yes, but I thought people might be interested in proving it for themselves.  From the proof should follow quickly (by generalization) that the probability that $n$ random numbers are relatively prime is $[\\zeta(n)]^{-1}$."
}
{
  "Tag": [
    "arithmetic sequence"
  ],
  "Problem": "In the arithmetic sequence $ 17, a, b, c, 41$, what is the value\nof $ b$?",
  "Solution_1": "Since b is right in the middle, b is the average of 17 and 41, which is 29.",
  "Solution_2": "Since it is an arithmetic sequence, there is some common difference. Let's call it $ x$\r\n\r\n17+2x=b\r\n41-2x=b\r\n\r\nAdding the two equations,\r\n\r\n58=2b\r\n\r\n29=b"
}
{
  "Tag": [
    "probability",
    "AMC",
    "AIME",
    "linear algebra",
    "matrix"
  ],
  "Problem": "Hi, this is my first post on this forum!\r\n\r\nThis is the question:\r\n\"13.  A bug moves around a wire triangle. At each vertex it has 1/2 chance of moving towards each of the other two vertices. What is the probability that after crawling along 10 edges it reaches its starting point?\"\r\n\r\nI tried (unsuccessfully) to do it like this:\r\n\r\nI label the three vertices S, A, and B with S being the starting point. I then let S(n), A(n), and B(n) be the number of ways to get back to S starting from S, A, and B respectively using n moves. From this we get the following equations:\r\n\r\nS(n)=A(n-1)+B(n-1)\r\nA(n)=B(n-1)+S(n-1)\r\nB(n)=A(n-1)+S(n-1)\r\n\r\nWhen you sub the last two equations into the first, you get one equation in terms of one variable.\r\n\r\nS(n)=2S(n-2)+S(n-1)\r\n\r\nThen I proceeded to solve for S(3) and S(4) to get the chain started. I got S(3)=2 and S(4)=4. However, this gave me a different answer for S(10) than the answer I found on this site:\r\nhttp://www.kalva.demon.co.uk/aime/soln/answ03b.html\r\n\r\nDid I mess something up in my work? I tried to find an error in my logic but I couldn't find one. Can someone help out? \r\n\r\nThanks for your time.",
  "Solution_1": "${S(n)=2 S(n-2)+S(n-1)}$\r\n\r\n${S(x)=\\alpha \\beta^{x}}$\r\n\r\n${S(n)=2 S(n-2)+S(n-1)}$\r\n${\\alpha \\beta^{n}=2 \\alpha \\beta^{n-2}+\\alpha \\beta^{n-1}}$\r\n${\\alpha \\beta^{n-2}\\left(\\beta^{2}-\\beta-2\\right)=0}$\r\n\r\n${\\alpha \\neq 0\\land \\beta^{n-2}\\neq 0}$\r\n\r\n${\\beta^{2}-\\beta-2=0}$\r\n${\\beta =-1\\lor \\beta =2}$\r\n\r\n${S(n)=(-1)^{n}\\alpha_{1}+2^{n}\\alpha_{2}}$\r\n\r\n${S(3)=2\\land S(4)=4}$\r\n\r\n${\\alpha_{1}=0\\land \\alpha_{2}=\\frac{1}{4}}$\r\n\r\n${S(n)=\\frac{2^{n}}{4}}$\r\n\r\n${S(10)=256}$\r\n\r\n\r\nI think it. but i didnt understand your system....",
  "Solution_2": "[hide=\"I do not like the Kalva solutions\"]\nlet's set up a recursion\n\nafter 1 move, it's impossible to be back at S\n\nafter 2 moves, the probability is 1/2, because, after 1 move, the bug has a 1/2 chance of going back to the start \n\nafter 3 moves - the probability now that the bug is at the start is 1/2 (the probability that the bug was not at the start after 2 moves) times 1/2 (the probability that the bug goes to the start rather than the other vertex) - 1/4\n\nafter 4 - the probability now is (1-(1/4))(1/2)=3/8\n\nyou will find that if $p_{n}$ is the probability that the bug is at the start after $n$ moves, then $p_{n}=(1-p_{n-1})(\\frac{1}{2})$\n\nyou do the math - I'm too tired.\n[/hide]\r\n\r\nclicking on the \"Resources\" icon in the top right corner of the forum index and then clicking \"USA\" under \"National and Regional Competitions\" takes you to a link where you can see all of the AIME problems, a majority of which have very nice-looking solutions.",
  "Solution_3": "sapphyre571, I just used your recursive formula to find the tenth move but it doesn't match the answer on the kalva site. Your equation gives 169/512.\r\n\r\nBut my question is to find the problem in my solution because I want to apply this method to future problems of this type (it seems versatile).",
  "Solution_4": "the adjacency matrix is:\r\n\r\n0 1 1\r\n1 0 1\r\n1 1 0\r\n\r\ntaking this to the 10th power gives\r\n\r\n342 341 341\r\n341 342 341\r\n341 341 342\r\n\r\nthe desired probability is 342/1024=171/512, so 683",
  "Solution_5": "Sorry I haven't learned matrices yet. But what I'm looking for is not an alternative solution. I'm looking for a problem with mine.",
  "Solution_6": "Is there an easy way to raise the adjacency matrix to large powers? (ie not multiplying it together lots of times)?",
  "Solution_7": "If you want to raise a matrix to a large power, diagonalization is the way to go. That matrix has eigenvalues $2,-1,-1$; the eigenvector for $2$ is $(1,1,1)^{T}$ and the eigenvectors for $-1$ are the orthogonal complement.\r\nThis is college-level math, and the AIME problem certainly has other solutions.",
  "Solution_8": "Ok sorry, but I took LinAlg last summer and I've forgotten most of it. What do I do once I know the eigenvector?",
  "Solution_9": "[hide]\nUm...how about:\n\nProbability of returning to point A after 1 step = 0\nAfter 2 steps = 1/2 * (1-0) = 1/2 \nAfter 3 steps = 1/2 * (1-1/2) = 1/4\nAfter 4 steps = 1/2 * (1-1/4) = 3/8\nAfter 5 steps = 1/2 * (1-3/8) = 5/16\netc\nAfter 10 steps = 1/2 * (1-85/256) = 171/512\nedit: Sorry, someone already posted this method lol[/hide]"
}
{
  "Tag": [
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Why is the last statement on the attachment true? Shouldn't the intersection of f^-1 [B] be larger than the inverse image of the intersection, and why can we conclude that they are equal?",
  "Solution_1": "Let's see: \r\n\r\n$ x\\in f^{\\minus{}1}\\left[\\bigcap_{\\lambda} B_{\\lambda}\\right]$ iff $ f(x)\\in \\bigcap_{\\lambda} B_{\\lambda}$ iff $ f(x)\\in B_{\\lambda}$ for all $ \\lambda.$\r\n\r\n$ x\\in \\bigcap_{\\lambda}f^{\\minus{}1}[B_\\lambda]$ iff $ x\\in f^{\\minus{}1}[B_\\lambda]$ for all $ \\lambda$ iff $ f(x)\\in B_{\\lambda}$ for all $ \\lambda.$\r\n\r\nThey're the same.\r\n\r\nCan you provide an example showing why it's different for forward images?",
  "Solution_2": "Suppose $ \\exists a \\in A_1$ and $ b \\in A_\\lambda \\forall \\lambda \\neq 1$ s.t. $ f(a) \\equal{} y, f(b) \\equal{} y$.\r\n\r\nThen, $ y \\in \\bigcap f[A_\\lambda]$ but $ y \\notin f[\\bigcap A_\\lambda]$"
}
{
  "Tag": [
    "Gauss",
    "logarithms"
  ],
  "Problem": "\u0391\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03c4\u03b5 \u03cc\u03c4\u03b9 $ \\sum^{\\infty}_{n \\equal{} 1}\\frac {ln2 \\minus{} (\\sum^{n}_{i \\equal{} 1}\\frac {1}{n \\plus{} i})}{n} \\equal{} \\sum^{\\infty}_{n \\equal{} 1}\\frac {\\sum^{n}_{i \\equal{} 1}\\frac {1}{i}}{(2n \\plus{} 1)(2n \\plus{} 2)}$",
  "Solution_1": "[quote=\"\u0393\u03b9\u03ce\u03c1\u03b3\u03bf\u03c2\"]\u0391\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03c4\u03b5 \u03cc\u03c4\u03b9 $ \\sum^{\\infty}_{n \\equal{} 1}\\frac {ln2 \\minus{} (\\sum^{n}_{i \\equal{} 1}\\frac {1}{n \\plus{} i})}{n} \\equal{} \\sum^{\\infty}_{n \\equal{} 1}\\frac {\\sum^{n}_{i \\equal{} 1}\\frac {1}{i}}{(2n \\plus{} 1)(2n \\plus{} 2)}$[/quote]\r\n\r\n \u0393\u03b9\u03ce\u03c1\u03b3\u03bf , \r\n \u03c4\u03b5\u03bb\u03b5\u03af\u03c9\u03c2 \u03c6\u03b9\u03bb\u03b9\u03ba\u03ac \u03bd\u03b1 \u03c0\u03c9 \u03c4\u03bf\u03cd\u03c4\u03bf : \r\n \u0391\u03c6\u03bf\u03cd \u03c3\u03b5 \u03b5\u03bd\u03b4\u03b9\u03b1\u03c6\u03ad\u03c1\u03bf\u03c5\u03bd \u03bf\u03b9 \u03bf\u03bb\u03c5\u03bc\u03c0\u03b9\u03ac\u03b4\u03b5\u03c2 , \u03bc\u03b7\u03bd \u03b1\u03bd\u03bf\u03af\u03b3\u03b5\u03c3\u03b1\u03b9 \u03c3\u03b5 \u03c0\u03bf\u03bb\u03bb\u03ac \u03bc\u03ad\u03c4\u03c9\u03c0\u03b1. \u0391\u03c5\u03c4\u03ad\u03c2 \u03bf\u03b9 \u03b1\u03c3\u03ba\u03ae\u03c3\u03b5\u03b9\u03c2 \u03b4\u03b5\u03bd \u03b1\u03c0\u03bf\u03c4\u03b5\u03bb\u03bf\u03cd\u03bd \u03b8\u03ad\u03bc\u03b1\u03c4\u03b1 \u03b3\u03b9\u03b1 \u03bf\u03bb\u03c5\u03bc\u03c0\u03b9\u03ac\u03b4\u03b5\u03c2.\u0394\u03b5\u03bd \u03ba\u03ac\u03bd\u03bf\u03c5\u03bd \u03cc\u03bb\u03b5\u03c2 \u03bf\u03b9 \u03b1\u03c3\u03ba\u03ae\u03c3\u03b5\u03b9\u03c2 \u03c4\u03bf\u03c5 CRUX \u03b3\u03b9\u03b1 \u03bf\u03bb\u03c5\u03bc\u03c0\u03b9\u03ac\u03b4\u03b5\u03c2.\u0388\u03c4\u03c3\u03b9 \u03b8\u03b1 \u03ba\u03b5\u03c1\u03b4\u03af\u03c3\u03b5\u03b9\u03c2 \u03c7\u03c1\u03cc\u03bd\u03bf \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03b1\u03c3\u03c7\u03bf\u03bb\u03b7\u03b8\u03b5\u03af\u03c2 \u03bc\u03b5 \u03c4\u03b9\u03c2 shortlist \u03ba\u03b1\u03b9 \u03ac\u03bb\u03bb\u03b1 \u03b2\u03b1\u03c3\u03b9\u03ba\u03ac \u03bf\u03bb\u03c5\u03bc\u03c0\u03b9\u03b1\u03ba\u03ac \u03c0\u03b5\u03b4\u03af\u03b1.\u0397 \u03af\u03b4\u03b9\u03b1 \u03c0\u03c1\u03bf\u03c4\u03c1\u03bf\u03c0\u03ae \u03b1\u03c0\u03b5\u03c5\u03b8\u03cd\u03bd\u03b5\u03c4\u03b1\u03b9 \u03ba\u03b1\u03b9 \u03c3\u03b5 \u03cc\u03bb\u03bf\u03c5\u03c2 , \u03cc\u03c3\u03bf\u03b9 \u03c3\u03c4\u03bf\u03c7\u03b5\u03cd\u03b5\u03c4\u03b5 \u03c3\u03c4\u03b9\u03c2 \u0399\u039c\u039f.\r\n\u03a4\u03bf \u03c3\u03cd\u03bd\u03b8\u03b7\u03bc\u03b1 \u03c4\u03bf\u03c5 Gauss \u03ae\u03c4\u03b1\u03bd : Pauca sed matura = \u03bb\u03af\u03b3\u03b1 \u03b1\u03bb\u03bb\u03ac \u03ce\u03c1\u03b9\u03bc\u03b1 ! \u0395\u03c0\u03b9\u03bb\u03ad\u03be\u03c4\u03b5 \u03bb\u03bf\u03b9\u03c0\u03cc\u03bd \u03c4\u03b1 \u03ba\u03b1\u03c4\u03ac\u03bb\u03bb\u03b7\u03bb\u03b1 \u03b8\u03ad\u03bc\u03b1\u03c4\u03b1. \r\n \u0391\u03c5\u03c4\u03ad\u03c2 \u03c4\u03b9\u03c2 \u03b1\u03c3\u03ba\u03ae\u03c3\u03b5\u03b9\u03c2 \u03b2\u03ac\u03bb\u03c4\u03b5\u03c2 \u03ba\u03b1\u03bb\u03cd\u03c4\u03b5\u03c1\u03b1 \u03c3\u03c4\u03bf \u03c3\u03c4\u03b9\u03ba\u03c5 \u03bc\u03b5  \u03c4\u03bf\u03c5\u03c2 \u03c6\u03bf\u03b9\u03c4\u03b7\u03c4\u03ad\u03c2.\u038c\u03c7\u03b9 \u03b3\u03b9\u03b1\u03c4\u03af \u03b5\u03b4\u03ce \u03b2\u03bb\u03ac\u03c0\u03c4\u03b5\u03b9 , \u03b1\u03bb\u03bb\u03ac \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03b2\u03bb\u03ad\u03c0\u03b5\u03c4\u03b5 \u03c0\u03b9\u03bf \u03c0\u03bf\u03bb\u03cd \u03b1\u03c3\u03ba\u03ae\u03c3\u03b5\u03b9\u03c2 \u03b3\u03b9\u03b1 \u03bf\u03bb\u03c5\u03bc\u03c0\u03b9\u03ac\u03b4\u03b5\u03c2 \u03ba\u03b1\u03b9 \u03cc\u03c7\u03b9  \u03b1\u03c3\u03ba\u03ae\u03c3\u03b5\u03b9\u03c2 \u03b1\u03c0\u03cc \u03c4\u03bf\u03c5\u03c2 \u03ac\u03bb\u03bb\u03bf\u03c5\u03c2 \u03ba\u03bb\u03ac\u03b4\u03bf\u03c5\u03c2 , \u03cc\u03c3\u03bf \u03c9\u03c1\u03b1\u03af\u03b5\u03c2 \u03ba\u03b1\u03b9 \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9. \r\n \u039c\u03b9\u03b1 \u03b6\u03b5\u03c3\u03c4\u03ae \u03b3\u03bd\u03ce\u03bc\u03b7 \u03b5\u03be\u03ad\u03c6\u03c1\u03b1\u03c3\u03b1 \u03ba\u03b1\u03b9 \u03b4\u03b5 \u03b8\u03ad\u03bb\u03c9 \u03bd\u03b1 \u03c4\u03b7\u03bd \u03c0\u03b1\u03c1\u03b5\u03be\u03b7\u03b3\u03ae\u03c3\u03b5\u03b9\u03c2. \u0395\u03af\u03bc\u03b1\u03b9 \u03c3\u03af\u03b3\u03bf\u03c5\u03c1\u03bf\u03c2 \u03cc\u03c4\u03b9 \u03b8\u03b1 \u03c4\u03b7\u03bd \u03b1\u03be\u03b9\u03bf\u03c0\u03bf\u03b9\u03ae\u03c3\u03b5\u03b9\u03c2 \u03ba\u03b1\u03c4\u03ac\u03bb\u03bb\u03b7\u03bb\u03b1 \u03ba\u03b1\u03b9 \u03c3\u03c4\u03bf \u03c4\u03ad\u03bb\u03bf\u03c2 \u03b8\u03b1 \u03bf\u03c6\u03b5\u03bb\u03b7\u03b8\u03b5\u03af\u03c2. \u0391\u03bb\u03bb\u03ac \u03b1\u03c2  \u03c0\u03bf\u03c5\u03bd \u03ba\u03b1\u03b9 \u03c4\u03b1 \u03ac\u03bb\u03bb\u03b1 \u03c0\u03b1\u03b9\u03b4\u03b9\u03ac \u03c4\u03b7 \u03b3\u03bd\u03ce\u03bc\u03b7 \u03c4\u03bf\u03c5\u03c2. \u03a0\u03b9\u03b8\u03b1\u03bd\u03cc\u03bd \u03bd\u03b1 \u03ba\u03ac\u03bd\u03c9 \u03bb\u03ac\u03b8\u03bf\u03c2.\r\n  \u039c\u03c0\u03ac\u03bc\u03c0\u03b7\u03c2",
  "Solution_2": "\u03a4\u03c1\u03b9\u03b1 hints:\r\n- \u0397 \u03c3\u03b5\u03b9\u03c1\u03ac $ \\ln 2$.\r\n- Catalan's identity.\r\n- \u0395\u03bd\u03b1\u03bb\u03bb\u03b1\u03b3\u03ae \u03c4\u03c9\u03bd \u03c3\u03c5\u03bc\u03b2\u03cc\u03bb\u03c9\u03bd \u03ac\u03b8\u03c1\u03bf\u03b9\u03c3\u03b7\u03c2 (\u03bb\u03cc\u03b3\u03c9 \u03b1\u03c0\u03cc\u03bb\u03c5\u03c4\u03b7\u03c2 \u03c3\u03cd\u03b3\u03ba\u03bb\u03b9\u03c3\u03b7\u03c2 - \u03c3\u03b5 \u03c0\u03bf\u03b9\u03bf \u03bc\u03ad\u03bb\u03bf\u03c2, \u03c4\u03bf \u03b1\u03c6\u03ae\u03bd\u03c9 \u03c3\u03b5 \u03c3\u03ad\u03bd\u03b1).\r\n\r\n\u0391\u03c0\u03cc \u03b5\u03b4\u03ce \u03ba\u03b1\u03b9 \u03c0\u03ad\u03c1\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b5\u03cd\u03ba\u03bf\u03bb\u03bf."
}
{
  "Tag": [
    "calculus",
    "derivative",
    "calculus computations"
  ],
  "Problem": "Find the derivative of\r\n\\[ y = (x^3 + 1)^{ - 4/3}.\r\n\\]\r\n[hide=\"What I Did\"]\\begin{align*} y' & = - \\frac {4}{3}(x^3 + 1)^{ - 7/3} \\cdot (3x^2) \\\\\n& = \\frac { - 4x^2}{(x^3 + 1)^{7/3}} \\end{align*}\n\nIs that correct?  Is there a way to simplify the denominator, $ (x^3 + 1)^{7/3}$?[/hide]\r\n\r\nThanks in advance!",
  "Solution_1": "hello, your derivative is ok, you can simplify to $ y^' = \\frac { - 4x^2}{(x^3 + 1)^2\\sqrt [3]{x^3 + 1}}$.\r\nSonnhard.",
  "Solution_2": "[quote=\"Dr Sonnhard Graubner\"]hello, your derivative is ok, you can simplify to $ y^' = \\frac { - 4x^2}{(x^2 + 1)\\sqrt [3]{x^2 + 1}}$.\nSonnhard.[/quote]\r\n\r\nOk, thanks!  (I'm going to assume that you meant $ y^' = \\frac { - 4x^2}{(x^3 + 1)\\sqrt [3]{x^3 + 1}}$.)",
  "Solution_3": "hello yes, this was a typo, it is $ y'(x)\\equal{}\\minus{}\\frac{4x^2}{(x^3\\plus{}1)^2\\sqrt[3]{x^3\\plus{}1}}$.\r\nSorry,Sonnhard.",
  "Solution_4": "[quote=\"Dr Sonnhard Graubner\"]hello yes, this was a typo, it is $ y'(x) \\equal{} \\minus{} \\frac {4x^2}{(x^3 \\plus{} 1)^2\\sqrt [3]{x^3 \\plus{} 1}}$.\nSorry,Sonnhard.[/quote]\r\nyou could also solve it using logarithm."
}
{
  "Tag": [
    "inequalities",
    "function",
    "inequalities solved"
  ],
  "Problem": "Let k and n be \\in N with 1 < k \\leq n; x_1, x_2, x_3,... , x_k are k positive integers, whose sums equals its product.\r\n\r\na.) Prove:\r\n\r\nx^n-1_1 + x^n-1_2 + ... + x^n-1_k \\geq kn .\r\n\r\nb.) Which additional sufficient and necessary properties must be satisfied for k,n and x_1, x_2, x_3,... , x_k in order to get equality:\r\n\r\nx^n-1_1 + x^n-1_2 + ... + x^n-1_k = kn .",
  "Solution_1": "Suppose x1+...+xk=x1*x2*...*xk.\r\nWe apply Cochy's inequality:\r\nx1+...+xk=x1*x2*...*xk\\leq ((x1+...+xk)/k)^k ==>\r\nx1*x2*...*xk=x1+...+xk\\geq k^(k/k-1).\r\n\r\nx1^(n-1)+...+xk^(n-1)\\geq k(x1*x2*...*xk)^((n-1)/k)\\geq k*k^((n-1)/(k-1))\\geq kn since n\\geq k and x^(1/(x-1)) is decreasing function for x\\geq 2.\r\n\r\nEqulity raises if \r\n1) k=n -- from last inequality\r\n2) x1=x2=...=xk -- from Cochy's enequality\r\nBut in such case x1*k=x1^k ==> k=x1^(k-1) ==> k=2, n=2, x1=x2=2\r\n\r\nI feel some imperfection with this problem",
  "Solution_2": "myth try to use Cauchy instead of Cochy! - in rest everything looks fine :)"
}
{
  "Tag": [
    "calculus",
    "calculus computations"
  ],
  "Problem": "Solve the Clairaut equation:\r\n\r\n$ x\\equal{}tx'\\minus{}f(x')$\r\n\r\nAnd consider the equation with $ f(x)\\equal{}\\frac{x^2}{2}$. For which initial conditions will the equation have a solution?",
  "Solution_1": "The general solution of the Clairaut equation $ x \\equal{} tx' \\minus{} f(x')$ is $ x \\equal{} Ct \\minus{} f(C)$. There can also be a singular solution by excluding p from the two equations $ x \\equal{} tp \\minus{} f(p)$ and $ t \\minus{} f'(p) \\equal{} 0$. For $ f(x) \\equal{} \\frac{x^2}{2}$, and so $ f(x') \\equal{} \\frac{(x')^2}{2}$ we have as the general solution\r\n\r\n$ x \\equal{} Ct \\minus{} \\frac{C^2}{2}$.\r\n\r\nTo find a possible singular solution we eliminate p from $ x \\equal{} tp \\minus{} \\frac{p^2}{2}$ and $ t \\minus{} p \\equal{} 0$, from which we obtain $ x \\equal{} \\frac{t^2}{2}$. This is indeed a singular solution of the original equation, as can be readly shown by substitution."
}
{
  "Tag": [],
  "Problem": "Let $n$ be a fixed positive integer. To any choice of real numbers satisfying \\[0\\le x_{i}\\le 1,\\quad i=1,2,\\ldots, n,\\] there corresponds the sum \\[\\sum_{1\\le i<j\\le n}|x_{i}-x_{j}|.\\] Let $S(n)$ denote the largest possible value of this sum. Find $S(n)$.",
  "Solution_1": "[hide] Since the summation is symmetric, assume WLOG that $x_{1}\\le x_{2}\\le \\cdots \\le x_{n}$. Then we can rewrite the sum as\n\n$\\sum_{k=1}^{n-1}(n-k)|x_{k+1}-x_{k}| = \\sum_{k=1}^{n-1}(n-k)a_{k}$\n\nwhere $a_{k}$ are positive reals summing to $1$. Clearly, it makes the most sense for $a_{1}= 1$, from which we get $S(n) = n-1$.[/hide]",
  "Solution_2": "WLOG $x_1 \\leq x_2 \\leq x_3 \\leq \\cdots \\leq x_n$. The sum is actually\n$$ \\sum_{k=1}^{n-1} k(n-k) (x_{k+1} - x_k) = \\sum_{k=1}^n (2k-n-1)x_{k}$$\nHence we need $x_k=1$ for $2k-n-1 \\geq 0$ and $x_k = 0$ otherwise. This gives the total as $\\boxed{S(n) = \\left \\lfloor \\frac{n}{2} \\right \\rfloor \\left \\lceil \\frac{n}{2} \\right \\rceil}$."
}
{
  "Tag": [
    "arithmetic sequence",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "consider n is a natural number .prove that there exist a prime number like p that every one of these \r\nnumbers p-1,p+1,p+2 have atleast n different prime divisor.",
  "Solution_1": "Let $ \\{ q_n \\}$ be an enumeration of the primes with $ q_1 \\equal{} 2$.  The system of congruences\r\n\r\n$ x \\equiv 1 \\bmod q_1 q_2 ... q_n$\r\n$ x \\equiv \\minus{} 1 \\bmod q_{n \\plus{} 1} q_{n \\plus{} 2} ... q_{2n}$\r\n$ x \\equiv \\minus{} 2 \\bmod q_{2n \\plus{} 1} q_{2n \\plus{} 2} ... q_{3n}$\r\n\r\nhas a solution set $ x \\equiv k \\bmod q_1 q_2 ... q_{3n}$ with $ \\gcd(k, q_1 q_2 ... q_{3n}) \\equal{} 1$ by CRT.  This solution set is an arithmetic progression that contains a prime by Dirichlet's theorem.\r\n\r\nThis problem is in a class of similar problems where it is very difficult to explicitly describe the solution set but where, thanks to CRT, it is very easy to show that it exists.  A related question:  show that for any $ n$ there exist $ n$ consecutive positive integers, none of which can be written as the sum of two squares.  (Consider a system of congruences modulo primes congruent to $ 3 \\bmod 4$.)"
}
{
  "Tag": [
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "Let a;b;c;d be real numbers satisfying: $\\large \\frac{1}{2} \\leq  a^2+b^2+c^2+d^2 \\leq  1$.\r\nFind the maximum and minimum of:\r\n$\\large Q = (a-2b+c)^2+(b-2c+d)^2+(b-2a)^2+(c-2d)^2$",
  "Solution_1": "hello, i think we have\r\n$ \\frac {1}{4}(7\\minus{}3\\sqrt{5})\\le Q\\le \\frac{5}{2}(3\\plus{}\\sqrt{5})$.\r\nSonnhard."
}
{
  "Tag": [
    "inequalities proposed",
    "inequalities"
  ],
  "Problem": "Let a,b,c,d >0 such that $ a^3\\plus{}b^3\\equal{}c^3\\plus{}d^3$ prove that    if $ ab \\geq cd$ then $ a\\plus{}b \\geq c\\plus{}d$",
  "Solution_1": "First condition $ \\Rightarrow (a \\plus{} b)(a^2 \\minus{} ab \\plus{} b^2) \\equal{} (c \\plus{} d)(c^2 \\minus{} cd \\plus{} d^2)$ let $ a \\plus{} b \\equal{} t$ and $ c \\plus{} d \\equal{} q$ if $ ab\\geq cd$ then if $ t < q$ this implies that $ t(t^2 \\minus{} 2ab) < q(q^2 \\minus{} 2cd)$ contradiction then $ t\\geq q$",
  "Solution_2": "Let a,b,c>0 such that a+b+c=1 Prove that\r\n$ \\sum\\sqrt[3]{a\\minus{}b\\plus{}c^3} \\leq 1$"
}
{
  "Tag": [
    "number theory",
    "modular arithmetic"
  ],
  "Problem": "A collection S of integers is defined by the following three rules (I) 2 is in S; (II) for every x in S, 3x and x+7 are also in S; (III) no integers except those defined by rules (I) and (II) are in S. What is the smallest integer greater than 2004 which is NOT in S?\r\n\r\nThis question prevented me from getting a 46 on the 2004 National Test, I still don't know how to do it...",
  "Solution_1": "[quote=\"13375P34K43V312\"]A collection S of integers is defined by the following three rules (I) 2 is in S; (II) for every x in S, 3x and x+7 are also in S; (III) no integers except those defined by rules (I) and (II) are in S. What is the smallest integer greater than 2004 which is NOT in S?\n\nThis question prevented me from getting a 46 on the 2004 National Test, I still don't know how to do it...[/quote]\r\n[hide=\"I love this question, really I do\"]\nAny number that =2/6/4/5/1/3 mod 7 is in the set. So, we need a number divisible by 7. 2009 is the smallest number that fits this description. So, our answer is 2009[/hide]\r\nIs this good enough?",
  "Solution_2": "I think the answer is[hide]2009\n\nI know it involves using mod 7, in which you can never get a multiple of 7, therefore 2009 is the first multiple of 7 greater than 2004.[/hide]\r\n\r\n-jorian",
  "Solution_3": "[quote=\"jhredsox\"]I think the answer is[hide]2009\n\nI know it involves using mod 7, in which you can never get a multiple of 7, therefore 2009 is the first multiple of 7 greater than 2004.[/hide]\n\n-jorian[/quote]\r\nDid you read my solution?\r\nHide yours.",
  "Solution_4": "I did hide mine. No, I did not read your solution, hence i posted 30 seconds afterwards :wink: \r\n-jorian",
  "Solution_5": "[quote=\"jhredsox\"]I did hide mine. No, I did not read your solution, hence i posted 30 seconds afterwards :wink: \n-jorian[/quote]\r\nSorry, I'm stupid. Anyways, this took me about 2.5 minutes, that isn't too bad :) . The reason I asked did you read my solution is because I show why it works in more detail.",
  "Solution_6": "ya, c, u shouldn't feel stupid as i gave up, and cheated (don't ask how :rotfl: )\r\n-jorian",
  "Solution_7": "This problem is interesting.  [hide=\"The key\"] Every number of the form $2 \\cdot 3^{k}$ is in this set for $k = 0, 1, 2, ...$.  Now,\n\n$2 \\equiv 2 \\bmod 7$\n$2 \\cdot 3 \\equiv 6 \\bmod 7$\n$2 \\cdot 3^{2}\\equiv 4 \\bmod 7$\n$2 \\cdot 3^{3}\\equiv 5 \\bmod 7$\n$2 \\cdot 3^{4}\\equiv 1 \\bmod 7$\n$2 \\cdot 3^{5}\\equiv 3 \\bmod 7$\n$2 \\cdot 3^{6}\\equiv 2 \\bmod 7$\n\nAnd so forth.  (This is a very interesting pattern!  Why does it give every non-zero remainder upon division by $7$?  Under what conditions does a similar pattern exist for, say, multiplication by $b \\bmod p$?)\n\nSince we can add $7$ to any element of the set to get more elements, after \"some point\" every number that is not divisible by $7$ is in the set.  Specifically, after $2 \\cdot 3^{5}= 486 < 2004$.  [/hide]\r\nEdit:  By the way, [b]bpms[/b], your original statement is slightly incorrect.  Since rule (II) only states $x \\in S \\implies x+7 \\in S$ and not the reverse, we are missing many numbers $< 486$.",
  "Solution_8": "[quote=\"t0rajir0u\"]\nEdit:  By the way, [b]bpms[/b], your original statement is slightly incorrect.  Since rule (II) only states $x \\in S \\implies x+7 \\in S$ and not the reverse[/quote]\r\nWait how is the reverse not implied? If x-7 is in the set, so is x.",
  "Solution_9": "$x-7 \\implies x \\neq x \\implies x-7$. The standard falsity of converse.\r\n\r\nt0rji0u's proposed extension is a pretty well known result related to Euler's totient theorem.",
  "Solution_10": "All I remember is, when I actually took the test in practice, I found a nifty way to do it quickly and got it right.\r\n\r\nHere's my order of skill level:\r\n\r\nworst: a random problem\r\nmiddle: practice tests\r\nbest: real competition\r\n\r\nEdit: Someone else already gave a really good explanation of how to do it, so I won't repeat. However, if you saw this problem and had no clue how to do it, I would recommend.\r\n\r\n[hide]starting with two and adding 7 and multiplying by three systematically. For example, multiply $2$ by $3$ until multiplying it by $3$ would make it greater than 2009. Then add $7$ from then on to find a number to \"rule out\" e.g. $2*3^{6}= 1458$. 1458 is 2 mod 7, so obviously another number that is 2 mod 7, 2004, can be obtained. So you can rule out 2004. Then repeat by adding 7 and following the same operations you did above. Find numbers between 2002 and 2010 to see what numbers you can get. Pretty soon you'll have ruled out enough different numbers to realize that 2009 is the right answer. Although realizing the x is congruent to a (mod 7) thing is much better[/hide]",
  "Solution_11": "You could explicitly calculate ways to get the first few numbers greater than $2004$ until you can't get one anymore, but that seems harder to me, not easier.  \r\n\r\nThe problem is essentially a modular problem.  Rather than avoiding the topic, why not try to learn some modular arithmetic?",
  "Solution_12": "[quote=\"asianhottie8293\"]DOES ANYBODY HAVE A WAY WITHOUT MOD[/quote]\r\nThis question was a hard National-level question.  Very few students solved this question in the alotted time.\r\n\r\nI think there are two approaches:\r\n\r\n[hide=\"1.  Use logical reasoning\"]\nMy intuition tells me the problem probably has something to do with multiples of 3 or with multiples of 7.\n\nThink about the small numbers in the set:  2, 6, 18, 9, 16, 23, 13, 20, 27, 25.  We have multiples of 3.  We have multiples of 3, plus 1 (16, 13, 25).  And we have multiples of 3 minus 1 (23, 20).  Maybe it doesn't have anything to do with multiples of 3.\n\nWe also have lots of non-multiples of 7 in the set.  But in the first few items above, I don't see any multiples of 7.\n\nSo how would we get a multiple of 7 in the set?  If there's not already a multiple of 7 in the set, then adding 7 won't get us a multiple of 7.  And if there's not already a multiple of 7 in the set, then multiplying by 3 won't get us a multiple of 7.\n\nSo, there's no multiple of 7.  The smallest multiple of 7 greater than 2004 is 2009.[/hide]\n\n[hide=\"2.  Take advantage of your calculator\"]\nWork fast and systematically:\n\n2 is in the set.  So is 9, 16, ..., 2004, 2011, ...\n2*3=6 is in the set.  So is 13, 20, ..., 2008, 2015, ....\n6*3=18 is in the set.  So is 25, 32, ..., 2006, 2013, ...\n18*3=54 is in the set.  So is 61, 68, ..., 2007, 2014, ...\n54*3=162 is in the set.  So is 169, 176, ..., 2010, 2017, ...\n162*3=486 is in the set.  So is 493, 500, ..., 2005, 2012, ...\n\nSo far, every number greater than 2004 is in the set, except 2009, 2016, ...  \n\nIf you happen to try others, such as $9\\times 3=27$:\n9*3=27 is in the set.  So is 34, 41, ..., 2008, 2015, ...\nThen, you'll find you are duplicating those you have already found to be in the set.  Hopefully, your intuition will get you back on track with the sequence shown above, before your time runs out.[/hide]",
  "Solution_13": "Basically, when you don't know what to do on a counting problem like this, try to look for a pattern (if you didn't finsih the other problem(s), do those instead then). You could have figured there was no 7, 14, etc. (it's using mods, but in a middle school fashion).\r\n\r\nJorian Hoover"
}
{
  "Tag": [
    "search"
  ],
  "Problem": "How many different ways can change for 1 dollar be made using nickels, dimes, and/or quarters?\r\n\r\nI've seen these problems a lot. Is there any quick way to do them?",
  "Solution_1": "Related, but uses half-dollars as well...[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=2143529262&t=161837[/url]",
  "Solution_2": "There was a similar problem on the Maryland Math League last week or so. The only way I did these problems would be to systematically brute force; that is, make a careful list of every possible combination. Starting with say, 10 dimes, work down to 1 dime, and see how many combinations this would lead with other coins. This ensures you won't miss the \"stray\" combination when you're done."
}
{
  "Tag": [
    "Functional Analysis",
    "topology",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Does there exist an infinite dimensional Banach space X with a subset S such that each element of X is a finite linear combination of elements of S?",
  "Solution_1": "What did you really mean? :? You can just take $S=X$.",
  "Solution_2": "I meant with S countable, sorry.",
  "Solution_3": "Suppose $X$ is an infinite dimesional Banach space.\r\n\r\nSince $X$ is infinite dimensional, there exists an infinite linearly independent set. WLOG, we can choose this set to be $S=\\{v_{k}: k\\in \\mathbb{N}\\}$ in such a way that $\\|v_{k}\\|=1$ for each $k.$\r\n\r\nNow consider the set of every sum $\\sum_{k=1}^{\\infty}a_{k}v_{k}$ for scalars such that $\\sum_{k=1}^{\\infty}|a_{k}|<\\infty.$\r\n\r\nEvery such series converges, because its partial sums form a Cauchy sequence in $X$ and $X$ was presumed to be complete.\r\n\r\nEvery such series gives a point in the closure of the span of $S.$ What we need to show is that not every such series can itself be in the span of $S.$\r\n\r\nHmm... if we were working with a Schauder basis for $X,$ this would be easy, but I'm not quite sure how to finish it in general.",
  "Solution_4": "In other words, you'd want to know whether an infinite dimensional Banach space can have a countable Hamel basis (over its scalar field). No, it cannot: \r\n\r\nLet $\\{e_{n}\\}_{n\\ge 0}$ be such a basis for our Banach space $X$. Then our space is equal to $\\bigcup_{n\\ge 0}\\langle e_{0},\\ldots,e_{n}\\rangle$, where $\\langle\\ldots\\rangle$ denotes the linear span of whatever is between the brackets. Each $\\langle e_{0},\\ldots,e_{n}\\rangle$, however, is a proper closed linear subspace of $X$, and hence a closed nowhere dense subset. By [url=http://mathworld.wolfram.com/BaireCategoryTheorem.html]Baire's theorem[/url], a complete metric space is not a countable union of such sets, and we have the desired contradiction.",
  "Solution_5": "I should have thought of Baire category. Thanks, grobber.",
  "Solution_6": "in this case your set is closure(span(S)) and this is infact equal to X. \r\n\r\nthink of L2[T]. it is certainly true that, because of fourier series (stone weierstrass etc) that if S={e^inx:n in N} then L2[T]=closure(span(S))."
}
{
  "Tag": [
    "topology",
    "advanced fields",
    "advanced fields unsolved"
  ],
  "Problem": "Let X and Y be normed spaces. How would you show that a linear mapping T: X-->Y\r\nis continuous wrt the usual normed topologies iff it is continuous wrt the weak\r\ntopologies on X and Y.",
  "Solution_1": "Let $ f: X\\to Y$ be linear.\r\n\r\nOne direction is easy:\r\nLet $ f: X\\to Y$ be continuous with respect to the norm topology. The sets of the form $ V\\equal{}y_1'^{\\minus{}1}(U_1)\\cap\\ldots\\cap y_n'^{\\minus{}1}(U_n)$ with open $ U_n$ and functionals $ y'\\in Y'$ are a subbase for the weak topology on $ Y$. Since $ y'\\circ f\\in X'$ for each $ y'\\in Y'$ the preimage $ f^{\\minus{}1}(V)\\equal{}(y_1'\\circ f)^{\\minus{}1}(U_1)\\cap\\ldots\\cap(y_n'\\circ f)^{\\minus{}1}(U_n)$ are open in the weak topology on $ X$. So $ f$ is continuous with respect to the weak topologies on $ X$ and $ Y$.\r\n\r\nNow consider the other case. $ f$ is linear it is continouos for the norm topologies iff it is bounded. So let $ A\\subseteq X$ be a bounded set. We prove that $ f(A)$ is bounded. In normed spaces boundedness is equivalent to weak boundedness. So let $ y'\\in Y'$ be arbitrary. $ y'\\circ f$ is continouos with respect to the weak topology on $ X$ and so it is for the norm topology too (the weak topology is coarser than the norm topology). Hence $ y'\\circ f\\in X'$ and so $ y'(f(A))$ is bounded. Hence $ f(A)$ is weakly bounded in $ Y$ and therefore bounded. So $ f$ is bounded and continuous for the norm topologies.",
  "Solution_2": "Could you explain why the sets like $ V$ form a subbase for the weak topology? Also what do you mean by weakly bounded? \r\n\r\nMany Thanks",
  "Solution_3": "These set are a base because of the definition of the weak topology. The weak topology is usually defined as coarsest topology on $ X$ s.t. all $ x'\\in X'$ are still continuous. This means nothing else as that the weak topology is the initial topologie with respect to the maps $ x': X\\to\\mathbb{K}$. And the initial topology has this base.\r\n\r\n\"$ A\\subseteq X$ is weakly bounded\" means that $ x'(A)$ is bounded for each $ x'\\in X'$."
}
{
  "Tag": [
    "induction"
  ],
  "Problem": "How do I show that if $ a$ and $ b$ are positive integers, then there is a smallest positive integer of the form $ a\\minus{}bk$ where $ k$ is $ Z$",
  "Solution_1": "If you have the well-ordering principle (can be proven by induction for $ \\mathbb{Z}_ \\plus{}$), then $ \\{a \\minus{} bk, k \\in \\mathbb{Z}\\} \\cap \\mathbb{Z}_ \\plus{}$ is a nonempty subset of $ \\mathbb{Z}_ \\plus{}$ and thus has a least element."
}
{
  "Tag": [
    "geometry",
    "geometric transformation",
    "reflection"
  ],
  "Problem": "i just want to know if Georgia Tech requires SAT subject tests?  i couldn't find it online.\r\n\r\nand online it said \"jan. sat scores may not be considered for admission\" what exactly does this mean?\r\n\r\nthanks",
  "Solution_1": "Thread retitled by moderator to reflect that \"GT\" is Georgia Tech."
}
{
  "Tag": [
    "puzzles"
  ],
  "Problem": "ok, think about this...it is really not that bad, it is a very old puzzle...\r\nWell, there are 2 roses infront of a rose, 2 roses in the back of a rose, one in front, one in the back...how many roses are there in all?  :?:",
  "Solution_1": "To clarify this, as I see it:\r\n\r\nThere are 2 roses in front of a rose and 2 roses behind a rose.\r\nHow many roses are there in all, if the fewest number of roses\r\nare allowed?",
  "Solution_2": "Way too easy, I think.\r\nThree?  :)",
  "Solution_3": "ive said it before ill say it again\r\n[hide]3?[/hide]",
  "Solution_4": "[quote=\"Cesario\"]Way too easy, I think.\nThree?  :)[/quote]\r\n\r\n :10:  wow, that was exactly what i thought, but it took so many people like a million guesses to get it right...lol. :rotfl:",
  "Solution_5": "klutzycherry14,\r\n\r\nno, you take the hit on this one.  Your statement of this OLD \r\nproblem was mucked up, and I gave you the chance to clarify\r\nit after my post.  You never acknowledged that you hadn't presented it exactly right.\r\n\r\nYour statement \"...it took so many people like a million guesses\r\nto get it right...\" is beyond silly.",
  "Solution_6": "[quote=\"box of rocks\"]klutzycherry14,\n\nno, you take the hit on this one.  Your statement of this OLD \nproblem was mucked up, and I gave you the chance to clarify\nit after my post.  You never acknowledged that you hadn't presented it exactly right.\n\nYour statement \"...it took so many people like a million guesses\nto get it right...\" is beyond silly.[/quote]\r\n\r\nim sorry, i should have realized that before...i just assumed that that since u said that and i had not contradicted it, everyone would assume that it was correct. and i admit that u were right in your assumption, so i apologize for that, and i was not referring to the people on here when i said the million people thing, i meant when i asked people like in person, like in my family/friends and stuff. so, i apologize for that...hope that is clear now... :oops:"
}
{
  "Tag": [
    "MATHCOUNTS"
  ],
  "Problem": "for those who took the Mathcounts competition this year, can u please tell me ur\r\n1. grade \r\n2. ur scores for chapter, state (if availiable) , and national (if avaliable) tests.\r\n3. ur ranking for chapter, state, and national competitions\r\n4. state\r\n\r\nthanx\r\n(just doing this for fun)",
  "Solution_1": "mine is \r\n1. 8th grade\r\n2. 44 for chapter and 43 for state\r\n3. 2nd in chapter and 7th in state\r\n4. California",
  "Solution_2": "1. 8th grade\r\n2. 30th at chapter and 7th at states\r\n3. Chapter: 36 State: 37\r\n4. Massachusetts",
  "Solution_3": "1. 8th grade\r\n2. 9th Chapter\r\n3. 2nd Chapter, forgot what in state..but not top ten :-x IT WAS SOOO EASY!!\r\n4. Pencil vania hehehhe",
  "Solution_4": "scores                        rank\r\nChapter      43                                1st...2nd last year\r\n \r\nState          40                                4th...38th last year\r\n\r\nNationals    34                                59th (really really sucky by my standards)",
  "Solution_5": "1. 7th\r\n2. 44, 40, 30(extremely bad)\r\n3. 2nd, 2nd, 118th(extremely bad)\r\n4. Oklahoma",
  "Solution_6": "Wow, this was one of the first posts in the Mathcounts forum. :omighty:",
  "Solution_7": "[quote=\"Nexmus\"]Wow, this was one of the first posts in the Mathcounts forum. :omighty:[/quote]\r\n\r\n\r\nKID, DONT REVIVE OLD TOPICS!\r\nITS GOING TO GET DELETED ANYWAYS>>>>>",
  "Solution_8": "dude, I was just joking. LOL :rotfl:",
  "Solution_9": "[quote=\"Nexmus\"]dude, I was just joking. LOL :rotfl:[/quote]\r\n\r\nDude, get some friends.  And Im not joking.",
  "Solution_10": "No flaming.\r\n\r\nNo reviving old topics.\r\n\r\nLocked  :harhar:"
}
{
  "Tag": [
    "function",
    "number theory proposed",
    "number theory"
  ],
  "Problem": "For every positive integer $n$ let $\\varphi(n)$ be the Euler\u2019s function defined as follows: $\\varphi(n) =n(1-\\frac{1}{p_1})(1-\\frac{1}{p_2})\\ldots(1-\\frac{1}{p_m}),$ where $p_1, p_2, \\ldots, p_m$ are all pairwise distinct prime divisors of $n.$ For example, $\\varphi(1)=\\varphi(2)=1,$ $\\varphi(3)=\\varphi(4)=2.$ Find all $n \\le 1000$ such that $\\varphi(n) = \\varphi(n + 1).$",
  "Solution_1": "[quote=rogue]For every positive integer $n$ let $\\varphi(n)$ be the Euler\u2019s function defined as follows: $\\varphi(n) =n(1-\\frac{1}{p_1})(1-\\frac{1}{p_2})\\ldots(1-\\frac{1}{p_m}),$ where $p_1, p_2, \\ldots, p_m$ are all pairwise distinct prime divisors of $n.$ For example, $\\varphi(1)=\\varphi(2)=1,$ $\\varphi(3)=\\varphi(4)=2.$ Find all $n \\le 1000$ such that $\\varphi(n) = \\varphi(n + 1).$[/quote]\nSee [url=https://artofproblemsolving.com/community/c6h160894]here [/url]for a solution.\n",
  "Solution_2": "[quote=Tintarn][quote=rogue]For every positive integer $n$ let $\\varphi(n)$ be the Euler\u2019s function defined as follows: $\\varphi(n) =n(1-\\frac{1}{p_1})(1-\\frac{1}{p_2})\\ldots(1-\\frac{1}{p_m}),$ where $p_1, p_2, \\ldots, p_m$ are all pairwise distinct prime divisors of $n.$ For example, $\\varphi(1)=\\varphi(2)=1,$ $\\varphi(3)=\\varphi(4)=2.$ Find all $n \\le 1000$ such that $\\varphi(n) = \\varphi(n + 1).$[/quote]\nSee [url=https://artofproblemsolving.com/community/c6h160894]here [/url]for a solution.[/quote]\n\nBut.. there is not a solution? It's said that it's the only case when n is even, he didn't look at n-1 e.g."
}
{
  "Tag": [
    "AMC",
    "AIME",
    "Hi",
    "geometry",
    "inequalities",
    "area of a triangle",
    "Heron\\u0027s formula"
  ],
  "Problem": "Given that $x$, $y$, and $z$ are real numbers that satisfy:\r\n\r\n\\[ x=\\sqrt{y^2-\\frac{1}{16}}+\\sqrt{z^2-\\frac{1}{16}} \\]\r\n\\[ y=\\sqrt{z^2-\\frac{1}{25}}+\\sqrt{x^2-\\frac{1}{25}} \\]\r\n\\[ z=\\sqrt{x^2-\\frac{1}{36}}+\\sqrt{y^2-\\frac{1}{36}} \\]\r\n\r\nand that $x+y+z=\\frac{m}{\\sqrt{n}}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime, find $m+n$.",
  "Solution_1": "I don't have latex on this computer, so I won't bother trying to type it up. To start off, all you have to do is go through a standard squaring procedure to get rid of the radicals (move one radical to one side, square, then isolate the radical again and square). This should give you some fairly workable equations to solve.",
  "Solution_2": "[hide]Each is in the general form \n$a=\\sqrt{b^2-d^2}+\\sqrt{c^2-d^2}$ Isolate a radcal and square.\n\n$b^2-d^2=a^2-2a\\sqrt{c^2-d^2}+c^2-d^2$ Cancel, isolate the radical, and square\n\n$a^4+b^4+c^4+2a^2c^2-2a^2b^2-2b^2c^2=4a^2c^2-4a^2d^2$\n\n$a^4+b^4+c^4=2a^2b^2+2a^2c^2+2b^2c^2-4a^2d^2$\nEverything is cyclic but the last term, so \n\n$-4x^2\\cdot\\frac1{16}=-4y^2\\cdot\\frac1{25}=-4z^2\\cdot\\frac1{36}$ so\n\n$x: y: z=4: 5: 6$ so $x=\\frac{4y}5$ and $z=\\frac{6y}5$ Plug it into the middle equation\n\n$\\frac{256y^4+625y^4+1296y^4}{625}=\\frac{800y^4}{625}+\\frac{1800y^4}{625}+\\frac{1152y^4}{625}-\\frac{100y^2}{625}$\n\n$1575y^4=100y^2$ $y$ is not $0$\n\n$y^2=\\frac4{63}$ $y=\\frac2{3\\sqrt7}$\n\n$x+y+z=3y=\\frac{2}{\\sqrt7}$ The answer is $\\boxed{009}$[/hide]",
  "Solution_3": "I was talking to my friend about his.  I think we found an amazing solution but I never actually worked out the numbers.\r\n\r\n[hide]\n\nConsider a triangle with sides x,y,z and altitudes of 1/4, 1/5, 1/6 to each side, respectively.  It is then clear that a triangle with these side lengths would satisfy the three given equations.  Then you can write x * 1/4 = y *1/5 = z * 1/6 = 2 *AREA = 2*Heron's Formula.  Solve for y and z in terms of x, then plug this into Heron's formula and solve for x (the algebra isn't too hard)  Then easily solve for y and z.\n\n[/hide]",
  "Solution_4": "[quote=\"mkkool64\"]I was talking to my friend about his.  I think we found an amazing solution but I never actually worked out the numbers.\n\n[hide]\n\nConsider a triangle with sides x,y,z and altitudes of 1/4, 1/5, 1/6 to each side, respectively.  It is then clear that a triangle with these side lengths would satisfy the three given equations.  Then you can write x * 1/4 = y *1/5 = z * 1/6 = 2 *AREA = 2*Heron's Formula.  Solve for y and z in terms of x, then plug this into Heron's formula and solve for x (the algebra isn't too hard)  Then easily solve for y and z.\n\n[/hide][/quote]\r\nwow, that is so simple.  There's barely any algebra.",
  "Solution_5": "I found the same solution as mkkool64. Here's the complete thing:\r\n\r\n[hide]Since the area is constant, you get that $y=\\frac54 x$ and $z=\\frac64 x$. So $s=\\frac{x+y+z}2=\\frac{15}8 x$, and $\\displaystyle\\left(\\frac{x}8\\right)^2=s(s-x)(s-y)(s-z)=\\frac{15}8x\\left(\\frac78x\\right)\\left(\\frac38x\\right)\\left(\\frac58x\\right)$. Cancelling, we get $15\\cdot7\\cdot3\\cdot5x^2=64\\implies x=\\frac8{15\\sqrt7}$. Now, since $x+y+z=\\frac{15}4x=\\frac{15}4\\frac8{15\\sqrt7}=\\frac2{\\sqrt7}$, so our answer is $2+7=\\fbox{009}$.[/hide] If this is the solution that goes in the contests section, I could edit it to be more comprehensive.",
  "Solution_6": "Try looking at the transcript in the AIME ii math jam.",
  "Solution_7": "How does a solution get chosen to be a link next to the amc/aime questions?",
  "Solution_8": "yes, that is exactly what I want to know, do we have to join the math jam to have our solution up there?",
  "Solution_9": "I think it's a mod (joml88 maybe) that chooses them for each contest. Not that it really matters, just as long as it gets done.",
  "Solution_10": "Generally, either 4everwise or I put the questions/solutions in the database for the AIME.  It has nothing to do with the mathjam.  We just go through the particular thread and find the post with the best solution and then input that one into the database.  If another proof uses an alternate method, we'll input that one as well.  We highly prefer solutions to be as polished as possible and for the post to have nothing but the solution.\r\n\r\nSince my spring break just started, I'll try and find some time to input the AIME II 2006 solutions into the database...if they're not in after a few days, someone remind me :P",
  "Solution_11": "I've noticed that 4everwise has a tendency of choosing his own solutions...  :P",
  "Solution_12": "[quote=\"matt276eagles\"]I've noticed that 4everwise has a tendency of choosing his own solutions...  :P[/quote]\r\n\r\nOf course :D  But in all fairness, his solutions usually are very well written, clear, concise, and correct :)",
  "Solution_13": "Yes, I agree. :lol:",
  "Solution_14": "4everwise also has the tendency to post about 10 problems in seperate topics all at once in the GS forum and post solutions if noone posts the answer in 15 mins. \r\n\r\nThis is very educational, so I don't mind.",
  "Solution_15": "[quote=\"mkkool64\"]I was talking to my friend about his.  I think we found an amazing solution but I never actually worked out the numbers.\n\n[hide]\n\nConsider a triangle with sides x,y,z and altitudes of 1/4, 1/5, 1/6 to each side, respectively.  It is then clear that a triangle with these side lengths would satisfy the three given equations.  Then you can write x * 1/4 = y *1/5 = z * 1/6 = 2 *AREA = 2*Heron's Formula.  Solve for y and z in terms of x, then plug this into Heron's formula and solve for x (the algebra isn't too hard)  Then easily solve for y and z.\n\n[/hide][/quote]\r\n\r\n*brings up a dead topic* [ but in all fairness, the given solution is not complete, so i should revive discussion]\r\n\r\ni found the same solution, except there is one thing i cannot get, how do you know that x,y,z satisfy the triangle inequality, i.e. $x+y\\ge z$, and symmetric?",
  "Solution_16": "In response to Altheman: Taking the first equation, we have $x=\\sqrt{y^{2}-\\frac1{16}}+\\sqrt{z^{2}-\\frac1{16}}<\\sqrt{y^{2}}+\\sqrt{z^{2}}=y+z$, which is the triangle inequality. Similarly, $y<x+z$ and $z<x+y$. Thus, a triangle can be constructed with sides of lengths $x,y,z$.",
  "Solution_17": "[quote=\"mkkool64\"]I was talking to my friend about his.  I think we found an amazing solution but I never actually worked out the numbers.\n\n[hide]\n\nConsider a triangle with sides x,y,z and altitudes of 1/4, 1/5, 1/6 to each side, respectively.  It is then clear that a triangle with these side lengths would satisfy the three given equations.  Then you can write x * 1/4 = y *1/5 = z * 1/6 = 2 *AREA = 2*Heron's Formula.  Solve for y and z in terms of x, then plug this into Heron's formula and solve for x (the algebra isn't too hard)  Then easily solve for y and z.\n\n[/hide][/quote]\r\n\r\nat my internship i gave this prob to one of the engineer's wives who was also an engineer. She came up with that solution.",
  "Solution_18": "[quote=nat mc][hide]Each is in the general form \n$a=\\sqrt{b^2-d^2}+\\sqrt{c^2-d^2}$ Isolate a radcal and square.\n\n$b^2-d^2=a^2-2a\\sqrt{c^2-d^2}+c^2-d^2$ Cancel, isolate the radical, and square\n\n$a^4+b^4+c^4+2a^2c^2-2a^2b^2-2b^2c^2=4a^2c^2-4a^2d^2$\n\n$a^4+b^4+c^4=2a^2b^2+2a^2c^2+2b^2c^2-4a^2d^2$\nEverything is cyclic but the last term, so \n\n$-4x^2\\cdot\\frac1{16}=-4y^2\\cdot\\frac1{25}=-4z^2\\cdot\\frac1{36}$ so\n\n$x: y: z=4: 5: 6$ so $x=\\frac{4y}5$ and $z=\\frac{6y}5$ Plug it into the middle equation\n\n$\\frac{256y^4+625y^4+1296y^4}{625}=\\frac{800y^4}{625}+\\frac{1800y^4}{625}+\\frac{1152y^4}{625}-\\frac{100y^2}{625}$\n\n$1575y^4=100y^2$ $y$ is not $0$\n\n$y^2=\\frac4{63}$ $y=\\frac2{3\\sqrt7}$\n\n$x+y+z=3y=\\frac{2}{\\sqrt7}$ The answer is $\\boxed{009}$[/hide][/quote]\n\nSorry for the bump but what being cyclic mean?",
  "Solution_19": "It means the terms are symmetric and only the variables are switched out.\nI think it can be represented with $\\sum_{cyclic}$ or something similar",
  "Solution_20": "[quote=mkkool64]I was talking to my friend about his.  I think we found an amazing solution but I never actually worked out the numbers.\n\n[hide]\n\nConsider a triangle with sides x,y,z and altitudes of 1/4, 1/5, 1/6 to each side, respectively.  It is then clear that a triangle with these side lengths would satisfy the three given equations.  Then you can write x * 1/4 = y *1/5 = z * 1/6 = 2 *AREA = 2*Heron's Formula.  Solve for y and z in terms of x, then plug this into Heron's formula and solve for x (the algebra isn't too hard)  Then easily solve for y and z.\n\n[/hide][/quote]\nHow come the equations would be satisfied\n\n",
  "Solution_21": "[quote=gary2003][quote=mkkool64]I was talking to my friend about his.  I think we found an amazing solution but I never actually worked out the numbers.\n\n[hide]\n\nConsider a triangle with sides x,y,z and altitudes of 1/4, 1/5, 1/6 to each side, respectively.  It is then clear that a triangle with these side lengths would satisfy the three given equations.  Then you can write x * 1/4 = y *1/5 = z * 1/6 = 2 *AREA = 2*Heron's Formula.  Solve for y and z in terms of x, then plug this into Heron's formula and solve for x (the algebra isn't too hard)  Then easily solve for y and z.\n\n[/hide][/quote]\nHow come the equations would be satisfied[/quote]\nFor the first equation, if you split the side with length $x$ into two segments split by the altitude called $m$ and $n$ with $m$ being adjacent the side with length $y$ and $n$ being adjacent to the side with length $z$, it follows that $m^2+\\frac{1}{16}=z^2$ and $n^2+\\frac{1}{16}=y^2$ because of the Pythagorean theorem. Rearranging, you get $m=\\sqrt{z^2-\\frac{1}{16}}$ and $n=\\sqrt{y^2-\\frac{1}{16}}$.\n Because $m+n=x$, $x=\\sqrt{z^2-\\frac{1}{16}}+\\sqrt{y^2-\\frac{1}{16}}$.\nYou can use this method on the other sides, and you get the other two equations.",
  "Solution_22": "Wow, what was the motivation behind the triangle solution?",
  "Solution_23": "[quote=NewBeginning]Wow, what was the motivation behind the triangle solution?[/quote]\n\nThere are a lot of \"Algebra\" problems that contain these Pythagorean terms; generally, if you can find a geometric interpretation, the problem will solve itself after that. Here, the main goal is to recognize that the terms are square roots of differences of squares, which are legs of a right triangle.\n\nAnother problem with this approach is [url=http://artofproblemsolving.com/wiki/index.php?title=1991_AIME_Problems/Problem_15]1991 AIME Problem 15[/url]\n",
  "Solution_24": "Problem\nGiven that $x, y,$ and $z$ are real numbers that satisfy:\n\n$$x = \\sqrt{y^2-\\frac{1}{16}}+\\sqrt{z^2-\\frac{1}{16}}$$$$y = \\sqrt{z^2-\\frac{1}{25}}+\\sqrt{x^2-\\frac{1}{25}}$$$$z = \\sqrt{x^2 - \\frac 1{36}}+\\sqrt{y^2-\\frac 1{36}}$$and that $x+y+z = \\frac{m}{\\sqrt{n}},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime, find $m+n.$\n\nSolution\nLet $XY = z$, $XZ=y$, and $YZ = x$ in $\\Delta XYZ$ and $h_x$ be the altitude from $X$ to $YZ$, $h_y$ be the altitude from $Y$ to $XZ$, and $h_z$ be the altitude from $Z$ to $XY$. Then, by Pythagoras' theorem, $x = \\sqrt{y^2 - h_x^2} + \\sqrt{z^2 - h_x^2}$. We find that $h_x=\\frac{1}{4}$, $h_y=5$, and $h_z=6$. Because the area of a triangle is constant, $x\\cdot h_x = y\\cdot h_y = z \\cdot h_z \\Longrightarrow \\frac{x}{4} = \\frac{y}{5} = \\frac{z}{6}=2[ABC]$. Alternatively, we can find the area of $\\Delta XYZ$ by using Heron's formula: $s=\\frac{x+y+z}{2}=\\frac{8A+10A+12A}{2}=15A$ and $A=[ABC]=\\sqrt{s(s-x)(s-y)(s-z)}=$ $\\sqrt{(15A)(15A-x)(15A-y)(15A-z)}$=$\\sqrt{15A(15A-8A)(15A-10A)(15A-12A)} =$ $\\sqrt{15A \\cdot 7A \\cdot 5A \\cdot 3A} = $ $15A^2\\sqrt{7}$. Hence, since $A=15A^2\\sqrt{7}$, $A=\\frac{1}{15\\sqrt{7}}$ and $x+y+z=2s=15A=\\frac{1}{\\sqrt{7}}$. $m+n=1+7=\\boxed{8}$",
  "Solution_25": "@above,  I don't think that $x+y+z$ is equal to $s$, maybe $2s$? ;)",
  "Solution_26": "Consider triangle $ABC$ with $AB=x,BC=y,CA=z$, and let $h_A,h_B,h_C$ be the altitudes from $A,B,C$, respectively. The problem statement implies $h_A=\\tfrac14,h_B=\\tfrac15,h_C=\\tfrac16$. If $K$ denotes the area of $\\triangle ABC$, then $xh_A=yh_B=zh_C=2K$, so $x:y:z=4:5:6$.\n\nLet $x=4t,y=5t,z=6t$. The area of a triangle with sides of length $4,5,6$ is $\\tfrac{15}4\\sqrt7$ by Heron's formula, which implies $K=\\tfrac{15}4\\sqrt7t^2$. Then \\[\\frac15=h_B=\\frac{2K}y=\\frac{3\\sqrt7}2t\\;\\Longrightarrow\\; t=\\frac2{15\\sqrt7},\\]\nso the perimeter of $\\triangle ABC$ is $15t=\\tfrac2{\\sqrt7}\\;\\Longrightarrow\\;2+7=\\boxed{009}$.",
  "Solution_27": "Manipulate merely the first equation:\n\\begin{align*}\nx &= \\sqrt{y^2-\\frac{1}{16}}+\\sqrt{z^2-\\frac{1}{16}} \\\\\nx^2 &= y^2+z^2 - \\frac 18 + \\sqrt{4y^2z^2 - \\frac 14 (y^2+z^2) + \\frac 1{64}} \\\\\n\\left(x^2-y^2-z^2+\\frac 18\\right)^2& = 4y^2z^2 - \\frac 14 (y^2+z^2) + \\frac 1{64} \\\\\nx^4 - 2 x^2 y^2 - 2 x^2 z^2 + \\frac 14 x^2+y^4 & \\\\ +2y^2z^2-\\frac 14 y^2+z^4 - \\frac 14 z^2 + \\frac 1{64} &= 4y^2z^2-\\frac 14(y^2+z^2) + \\frac 1{64} \\\\\n(x^4+y^4+z^4) - 2(x^2y^2+x^2z^2+y^2z^2) + \\frac 14 x^2 &=0.\n\\end{align*}\nSimilarly, \n\\begin{align*}\n(x^4+y^4+z^4) - 2(x^2y^2+x^2z^2+y^2z^2) + \\frac 14 x^2 &=0 \\\\\n(x^4+y^4+z^4) - 2(x^2y^2+x^2z^2+y^2z^2) + \\frac 4{25} y^2 &=0 \\\\\n(x^4+y^4+z^4) - 2(x^2y^2+x^2z^2+y^2z^2) + \\frac 19 z^2 &=0.\n\\end{align*}\nHenceforth $$\\frac 14 x^2 = \\frac 4{25} y^2 = \\frac 19 z^2 \\implies \\frac 12 x = \\frac 25 y = \\frac 13 z.$$ This means that $$y = \\frac 54 x, z = \\frac 32 x.$$ Substituting back in, $$x = \\sqrt{\\left(\\frac 54 x\\right)^2 - \\frac 1{16}} + \\sqrt{\\left(\\frac 32 x\\right)^2 - \\frac 1{16}},$$ so $x = \\frac 8{15 \\sqrt 7}$. Therefore $$(x, y, z) = \\left(\\frac 8{15 \\sqrt 7}, \\frac 2{3 \\sqrt 7}, \\frac 4{5 \\sqrt 7}\\right).$$ This extracts an answer of $\\boxed{009}$.",
  "Solution_28": "We can geometrically interpret this as a triangle with altitudes $\\frac{1}{4}, \\frac{1}{5}, \\frac{1}{6}$. Let the sides be $8p, 10p, 12p$. Then the area is $\\sqrt{15p \\cdot 7p \\cdot 5p \\cdot 3p}$ from Heron, so $\\sqrt{15p \\cdot 7p \\cdot 5p \\cdot 3p} = p$. This means $p=\\frac{1}{15\\sqrt7}$, for an answer of $30 \\cdot \\frac{1}{15\\sqrt7} = \\frac{2}{\\sqrt7} \\implies \\boxed{9}$.",
  "Solution_29": "ugh why does there always have to be an annoying quick solution after you're done bashing",
  "Solution_30": "Redacted",
  "Solution_31": "These looks really like pythagorean theorems, which is clear that $x, y, z$ are the three sides of a triangle with height $\\frac{1}{4}, \\frac{1}{5}, \\frac{1}{6}$. This gives $x:y:z=4:5:6$. This is basically a similar triangle from a triangle with sides $4, 5, 6$ which has an area of $\\frac{15\\sqrt{7}}{4}$. Now, look at one height of the triangle. One height is $\\frac{3\\sqrt{7}}{2}$. Therefore, the answer is $(4+5+6)\\frac{2}{3\\sqrt{7}\\cdot 5}=\\frac{2}{\\sqrt{7}}\\to \\boxed{009}$",
  "Solution_32": "[hide=solution]Note that $x,y,z$ form the side lengths of a triangle with heights of length $\\frac{1}{4},\\frac{1}{5},\\frac{1}{6}.$ This means that $x:y:z=4:5:6.$ We can easily find the lengths of the heights of a triangle with side lengths $4,5,6$ and scale appropriately to find $x,y,z.$ The final answer is $(x,y,z)=\\left(\\frac{8\\sqrt{7}}{105},\\frac{10\\sqrt{7}}{105},\\frac{12\\sqrt{7}}{105}\\right).$[/hide]",
  "Solution_33": "not too bad :P there's some symmetry so we really only need to worry about equation 1\n[hide=writeup]It's evident that $x$, $y$, and $z$ are all positive. Let's focus on the first equation for now, and eliminate the square roots. \n\\[x-\\sqrt{y^2-\\frac{1}{16}}=\\sqrt{z^2-\\frac{1}{16}}\\]\nsquare both sides and cancel the $\\frac{1}{16}$ term that results\n\\[x^2-2\\sqrt{y^2-\\frac{1}{16}}+y^2=z^2\\]\nnow move the terms without radicals to one side and the term with a radical to the other, and square yet again, then move the terms all to one side:\n\\[x^4+y^4+z^4-2x^2y^2-2y^2z^2-2z^2x^2+\\frac{1}{4}x^2=0.\\]\nSimilarly 'simplify' the other two equations to get:\n\\begin{align*}x^4+y^4+z^4-2x^2y^2-2y^2z^2-2z^2x^2+\\frac{4}{25}y^2&=0\n\\\\ x^4+y^4+z^4-2x^2y^2-2y^2z^2-2z^2x^2+\\frac{1}{9}z^2&=0\\end{align*} \nSo we obtain that $\\frac{1}{4}x^2=\\frac{4}{25}y^2=\\frac{1}{9}z^2$, that is, $\\frac{1}{2}x=\\frac{2}{5}y=\\frac{1}{3}z$. Hence $y=\\frac{5}{4}x$ and $z=\\frac{3}{2}x$. Plugging this into the first equation (or actually any equation) results in\n\\[x=\\sqrt{\\frac{25}{16}x^2-\\frac{1}{16}}+\\sqrt{\\frac{9}{4}x^2-\\frac{1}{16}}.\\]\nSolving this, $x=\\frac{8}{15\\sqrt{7}}$. So $(x, y, z)=\\left(\\frac{8}{15\\sqrt{7}}, \\frac{2}{3\\sqrt{7}}, \\frac{4}{5\\sqrt{7}}\\right)$. \n$x+y+z=\\frac{2}{\\sqrt{7}}$. Finally, $m+n=\\boxed{009}$. $\\blacksquare$[/hide]",
  "Solution_34": "[hide=solution]\nLook at the equations, and how closely they resemble the Pythagorean Theorem. Note that $x$, $y$, and $z$ could represent the sides of a triangle with altitudes of $\\frac{1}{4}$, $\\frac{1}{5}$, and $\\frac{1}{6}$ on the $x$, $y$, and $z$ sides. This satisfies the equation.\nThe formula for the area of a triangle gives $\\frac{1}{4}x=\\frac{1}{5}y=\\frac{1}{6}z$, so $y=\\frac{5}{4}x$ and $z=\\frac{3}{2}x$.\nPlugging this back into the equation for $x$ gives\n\\[x=\\sqrt{\\frac{25}{16}x^2-\\frac{1}{16}}+\\sqrt{\\frac{36}{16}x^2-\\frac{1}{16}},\\]\nso\n\\[x=\\frac{1}{4}\\sqrt{25x^2-1}+\\frac{1}{4}\\sqrt{36x^2-1}.\\]\nMultiplying both sides by $4$ and moving $\\sqrt{25x^2-1}$ to the other side gives\n\\[4x-\\sqrt{25x^2-1}=\\sqrt{36x^2-1}.\\]\nSquaring both sides gives\n\\[(4x-\\sqrt{25x^2-1})^2=16x^2-8x\\sqrt{25x^2-1}+25x^2-1=36x^2-1,\\]\nso simplifying gives $5x^2=8x\\sqrt{25x^2-1}$. Dividing both sides by $x$ gives $5x=8\\sqrt{25x^2-1}$, so squaring both sides again gives\n\\[25x^2=64(25x^2-1),\\]\nso $x^2=\\frac{64}{63\\cdot 25}$.\nHence, $x=\\frac{8}{3\\cdot 5\\sqrt{7}}=\\frac{8}{15\\sqrt{7}}$. This gives $y=\\frac{2}{3\\sqrt{7}}$ and $z=\\frac{4}{5\\sqrt{7}}$, so $x+y+z=\\frac{2}{\\sqrt{7}}$, which gives an answer of $\\boxed{009}$.\n[/hide]",
  "Solution_35": "Alg mani FTW\nInterpret as triangle ABC side lengths x=BC,y=CA,z=AB and altitudes 1/4,1/5,1/6, respectively. Then $x/4=y/5=z/6$ and simple substitution solves it. ",
  "Solution_36": "Consider an acute triangle with sides $x$, $y$, and $z$, with corresponding altitudes $\\frac 14$, $\\frac 15$, and $\\frac 16$. This cleanly satisfies the equations in the problem.\n\nFrom the height ratios, we have $x : y : z = 4 : 5 : 6$. Note the height to the side with length 4 in a 4-5-6 triangle has length $\\frac{15\\sqrt{7}}{8}$, so the scale ratio is $\\frac{2\\sqrt{7}}{105}$, and our solution is\n\\[(x,y,z)=\\left(\\frac{8\\sqrt{7}}{105},\\frac{2\\sqrt{7}}{21},\\frac{4\\sqrt{7}}{35}\\right) \\implies x+y+z = \\frac{2}{\\sqrt{7}} \\implies \\boxed{009}.~\\blacksquare\\]",
  "Solution_37": "Notice that $x,y,z$ are the sides of a triangle with heights $\\frac{1}{4}, \\frac{1}{5}, \\frac{1}{6}$, where the side lengths and height correspond. Then, we have\n\n\\[\\frac{x}{4}=\\frac{y}{5}=\\frac{z}{6}\\]\n\nso $y=\\frac{5}{4}x$ and $z=\\frac{3}{2}x$. This means $s=\\frac{x+y+z}{2}=\\frac{15}{8}x$. By Heron's formula we have\n\n\\[2\\sqrt{s(s-x)(s-y)(s-z)}=\\frac{x}{4}\\]\n\\[\\implies \\frac{30x^2\\sqrt{7}}{64}=\\frac{x}{4}\\]\n\nThus, $x=\\frac{8}{15\\sqrt{7}}$, so $y=\\frac{5}{4} \\cdot \\frac{8}{15\\sqrt{7}}$ and $z=\\frac{3}{2} \\cdot \\frac{8}{15\\sqrt{7}}$. Simplifying, we get\n\n\\[(x,y,z)=\\left(\\frac{8\\sqrt{7}}{105}, \\frac{2\\sqrt{7}}{21}, \\frac{4\\sqrt{7}}{35} \\right)\\]"
}
{
  "Tag": [
    "advanced fields",
    "advanced fields unsolved"
  ],
  "Problem": "Let $f \\in X^{*}-\\{0\\}$. Put $\\alpha = inf\\{\\parallel x\\parallel : f(x) = 1\\}$. Prove: $\\parallel f\\parallel = 1/ \\alpha$",
  "Solution_1": "Maybe telling us the definition of $X^{*}$ would help? :huh:",
  "Solution_2": "I guess that $X$ is a Banach space and $X^{*}$ is its dual ..",
  "Solution_3": "Set $L : = f^{-1}(0)$, then\r\n\\[\\|f\\| = \\sup_{x \\in X}\\frac{\\left\\langle x,f \\right\\rangle}{\\|x\\|}= \\sup_{x \\in L}\\frac{\\left\\langle x,f \\right\\rangle}{\\|x\\|}= \\frac 1{\\inf_{x \\in L}\\|x\\|}.\\]"
}
{
  "Tag": [],
  "Problem": "Estimate the following to a nearest ten:\n\\[ \\frac {0.5 \\times 0.12 \\times 2500} {31 \\times 0.09}\\]",
  "Solution_1": "[quote=\"GameBot\"]Estimate the following to a nearest ten:\n\\[ \\frac {0.5 \\times 0.12 \\times 2500} {31 \\times 0.09}\n\\]\n[/quote]\r\n\r\n$ \\frac{.5 \\cdot .12 \\cdot 2500}{31 \\cdot .09}\\equal{}\r\n\\frac{.5 \\cdot 12 \\cdot 2500}{31 \\cdot 9}\\equal{}\r\n\\frac{6 \\cdot 2500}{31 \\cdot 9}\\equal{}\r\n\\frac{5000}{93}$\r\n$ 50 < \\frac{5000}{93} < 55$\r\n50.",
  "Solution_2": "[quote=\"BOGTRO\"][quote=\"GameBot\"]Estimate the following to a nearest ten:\n\\[ \\frac {0.5 \\times 0.12 \\times 2500} {31 \\times 0.09}\n\\]\n[/quote]\n\n$ \\frac{.5 \\cdot .12 \\cdot 2500}{31 \\cdot .09}\\equal{}\n\\frac{.5 \\cdot 12 \\cdot 2500}{31 \\cdot 9}\\equal{}\n\\frac{6 \\cdot 2500}{31 \\cdot 9}\\equal{}\n\\frac{5000}{93}$\n$ 50 < \\frac{5000}{93} < 55$\n50.[/quote]\n\nTo make things a little more clear :D :\n\nXD $\\frac{0.5\\times 0.12\\times 2500}{31\\times 0.09}$\n\n $= \\frac{0.5\\times 12\\times 2500}{31\\times 9}$\n\n $= \\frac{6\\times 2500}{31\\times 9} = \\frac{5000}{93}$\n\n $= \\frac{50}{0.93} < \\frac{50}{0.90} $ and we know that\n\n $= \\frac{50}{0.90} = 55.\\overline{555...}$ so\n\n $= \\frac{50}{0.93} \\approx 54$ $\\text{or}$ $53$, which are still rounded to $50$\n\n $= \\frac{50}{0.93} \\approx \\boxed{50}$."
}
{
  "Tag": [
    "superior algebra",
    "superior algebra theorems"
  ],
  "Problem": "Someone is expert of the theory of group?\r\nPlease help-me\r\nthanks\r\nsastra",
  "Solution_1": "You can ask as many questions as you wish on the forum. As long as you don't do it like you did a lot of times before:\r\n- do not post any question more than once\r\n- do not write words in capitals, especially \"HOW TO DEMONSTRATE\r\n\r\nMore than half of capitals, and see my comment there:\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?p=733515#733515\r\nDoubleThread of the first one:\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?p=733516#733516\r\nJust a impudence:\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?p=734532#734532\r\n\r\nAll three locked now, and be sure that locking and deleting of any multithread will be done from now, for all threads."
}
{
  "Tag": [
    "geometry unsolved",
    "geometry"
  ],
  "Problem": ":(  :(  :(",
  "Solution_1": "Let $A',B',C'$ denote the apices of the equilateral triangles $\\triangle A'BC,$ $\\triangle B'CA,$ $\\triangle C'AB$ erected outwardly. Their centers are $X,Y,Z,$ respectively, thus $Np \\equiv AX \\cap BY \\cap CZ$ is the 1st Napoleon point of $\\triangle ABC.$\n\nThe point $Z$ is constructible and $C$ is then on $ZNp,$ such that $\\tfrac{BX}{BC}=\\tfrac{\\sqrt{3}}{3}$ and $\\angle XBC=30^{\\circ}.$ Hence, $C$ is found as the intersection of $ZNp$ with the line image of $ANp$ under the spiral similarity with center $B,$ coefficient $\\tfrac{\\sqrt{3}}{3}$ and  rotational angle $30^{\\circ}$ counterclockwise. "
}
{
  "Tag": [
    "logarithms",
    "function",
    "calculus",
    "derivative",
    "calculus computations"
  ],
  "Problem": "Prove that :\r\n\r\n$e^{y}>\\frac{16(1-y)}{(2-y)^{4}},\\forall y\\in (0,1)$",
  "Solution_1": "${e^{y}>\\frac{16 (1-y)}{(2-y)^{4}}}$\r\n\r\n${(0<y<1)\\land \\left(e^{y}>1\\right)}$\r\n\r\n${y>\\log \\left(\\frac{16 (1-y)}{(2-y)^{4}}\\right)}$\r\n${y>\\log (16)+\\log \\left(\\frac{1-y}{(2-y)^{4}}\\right)}$\r\n${y-\\log \\left(\\frac{1-y}{(2-y)^{4}}\\right)>\\log (16)}$\r\n\r\n${\\frac{\\partial \\left(y-\\log \\left(\\frac{1-y}{(2-y)^{4}}\\right)\\right)}{\\partial y}=0}$\r\n${\\frac{y}{y^{2}-3 y+2}=0}$\r\n${y=0}$\r\n\r\n${\\left(y-\\log \\left(\\frac{1-y}{(2-y)^{4}}\\right)\\text{/.}y\\to 0\\right)=\\log (16)}$\r\n\r\nSince $y \\neq 0$ and this the minimum in this interval, and the function is crescent. so it's true",
  "Solution_2": "Correction: $\\frac{d}{dy}\\left(y-\\ln\\left(\\frac{1-y}{(2-y)^{4}}\\right)\\right)= \\frac{y^{2}}{(1-y)(2-y)}.$\r\n\r\nIn other words, $y^{2}$ in the numerator, not $y.$ \r\n\r\nThe conclusion still holds.\r\n\r\nTwo bits of notation: there's no need to use the partial derivative symbol $\\partial$; this is an ordinary derivative, not a partial derivative. And \"crescent\" in English would be \"increasing.\"",
  "Solution_3": "That for $x\\in \\left(0,\\frac12\\right)$ is sharpen than $e^{x}>x+1$. It's nice isn't it?",
  "Solution_4": "i use partial because im typing on mathematica\r\n\r\ni use D[f[x],x]\r\n\r\n\r\nif i use Dt[f[x],x]   mathematica will differentiate all constants too\r\n\r\nDt[a]  + Dt[c]   etc...\r\n\r\n\r\nand this sucks...\r\n\r\nand sorry dor the y^2"
}
{
  "Tag": [
    "geometry",
    "geometric transformation",
    "reflection",
    "trapezoid",
    "perpendicular bisector",
    "angle bisector",
    "geometry unsolved"
  ],
  "Problem": "let $ABC$ be triangle  where $\\angle A=80^o$ and $\\angle B=\\angle C$ .\r\n let $M$ be a point in triangle $ABC$  satisfying $\\angle MAC=20^o$and $\\angle MCA=30^o$\r\n find value of$\\angle MBC$",
  "Solution_1": "Are you sure about this problem? The value of $\\angle{MBC}$ is NOT an integer...",
  "Solution_2": "i'm sory  .\r\n i have a mistake  in my post \r\n     i  have just complete it",
  "Solution_3": "Reflect $M$ through $AC$ to $M'$, then $\\angle{M'CA}=\\angle{MCA}=30^o$. Since $\\angle{ACB}=50^o$, so $\\angle{BCM'}=80^o$. \r\n\r\nBecause $\\angle{MAC}=\\angle{M'AC}=20^o$, and $\\angle{BAC}=80^o$, so $\\angle{BAM'}=100^o$. Now we see that $\\angle{BAM'}+\\angle{BCM'}=180^o$, so $A,B,C,M'$ are cyclic points. Therefore $\\angle{M'BA}=30^o$ and $\\angle{M'BC}=20^o$.\r\n\r\nSince: $\\angle{AM'M}=\\frac{180^o-\\angle{M'AM}}{2}=\\frac{180^o-40^o}{2}=70^o$, and $\\angle{AM'B}=\\angle{ACB}=50^o$, so $\\angle{BM'M}=20^o=\\angle{BCM}$, thus $BM'=BC$. From here we can get $\\triangle{BM'M} \\cong \\triangle{BCM}$, hence $\\angle{M'BM}=\\angle{CBM}$. But because $\\angle{M'BM}+\\angle{CBM}=20^o$, so finally we have $\\angle{MBC}=10^o$.",
  "Solution_4": "it,s very nice .  i can't  solve this problem  but you are very mart \r\n thanks veru much ,shobber!!!",
  "Solution_5": "Reflect $M$ in perpendicular bisector of $BC$ in $M'$, so triangle $\\triangle AM'M$ is $40^\\circ-70^\\circ-70^\\circ$; construct equilateral triangle $\\triangle MM'P$ ($P$ inside $\\triangle AMM'$, on perpendicular bisector of $BC$, hence $\\angle PAM = \\angle CAM$ and $APMC$ is cyclic ($\\angle APM=150^\\circ$, so $\\angle APM + \\angle ACM=180^\\circ$), $AM$ angle bisector of $\\angle PAC$, so $PM = MC$; consequently $CM = MM' = M'B$, $BCMM'$ is isosceles trapezoid (i.e. cyclic), and $BM$ is angle bisector of $\\angle CBM'=20^\\circ$, with $\\angle CBM=10^\\circ$.\n\nBest regards,\nsunken rock"
}
{
  "Tag": [],
  "Problem": "William's father was older than his grandfather. How did that happen ?",
  "Solution_1": "[hide]William's father was really old. William's grandfather is not his father's father, he is his mother's father. He was younger than his son-in-law.[/hide]\r\n\r\n...",
  "Solution_2": "The grandfather is on William's mother's side, so William's mother is really young compared to his father.    :blush:",
  "Solution_3": "The grandfather is on William's mother's side, so William's mother is really young compared to his father. \r\n\r\nOr...\r\n\r\nWilliam's father was really old. William's grandfather is not his father's father, he is his mother's father. He was younger than his son-in-law.",
  "Solution_4": "Or maybe he was born on a leap day.",
  "Solution_5": "[quote=\"Stank\"]Or maybe he was born on a leap day.[/quote] :rotfl:  :rotfl:  :rotfl:",
  "Solution_6": "a couple has a childat around 20. that child marries someone around 40 when the child was around 20.\r\nor it could be a stepgrandfather"
}
{
  "Tag": [
    "calculus",
    "integration",
    "calculus computations"
  ],
  "Problem": "[b](1)  Integrate\n  e^x*(x^3-x+2)/(1+x^2)^2 with respect to x\nhere e^x is multiplied by(x^3-x+2).[/b]",
  "Solution_1": "Actually,the answer is $ \\frac{e^{x}(1\\plus{}x)}{1\\plus{}x^2}$",
  "Solution_2": "As there have been no further replies, I will offer my naive approach.\r\n\r\nWith the quotient rule in mind, I assume (admiting it's quite an assumption and perhaps unwarranted) that the answer must be of the form \r\n\r\n$ \\frac{e^{x}f(x)}{1\\plus{}x^{2}}$.\r\n\r\nDifferentiating, $ f(x)$ must be linear: $ ax\\plus{}b$.\r\n.\r\nSubstituting and comparing coefficients, $ a\\equal{}b\\equal{}1$ and so $ I\\equal{} \\frac{e^{x}(1\\plus{}x)}{1\\plus{}x^{2}}$.\r\n\r\nHope this is of some use.",
  "Solution_3": "[quote=\"AndrewTom\"]admiting it's quite an assumption and perhaps unwarranted[/quote]\r\nWell, there are pretty much two ways it could go: either the assumption is right, or it's not and there's no elementary antiderivative at all. If we just wrote something similar at random, we would most likely be in the second case.",
  "Solution_4": "hmmmm.. a bit confuse of the equations.  :(",
  "Solution_5": "Thanks jmerry.\r\n\r\nNot sure what you mean, emjhay, so I'll try to write out all the steps:\r\n\r\n$ I = \\int {\\frac {e^{x}(x^{3} - x + 2)}{(1 + x^{2})^{2}} dx}$.\r\n\r\nBecause the denominator is $ (1 + x^{2})^{2}$, it seems that the integrand could be derived (no pun intended)  by applying the quotient rule to $ y = \\frac {e^{x}f(x)}{(1 + x^{2})}$, giving \r\n\r\n$ \\frac {(1 + x^{2}(e^{x}f'(x) + e^{x}f(x)) - 2xe^{x}f(x)}{(1 + x^{2})^{2}}$ \r\n\r\n= $ \\frac {e^{x}((1 - 2x + x^{2})f(x) + (1 + x^{2})f'(x))}{(1 + x^{2})^{2}}$.\r\n\r\nBy comparison with the given integrand, $ f(x)$ must be linear, that is, $ f(x) = ax + b$ and so $ f'(x) = a$.\r\n\r\nSubstituting, we get\r\n\r\n$ ax + b - 2ax^{2} - 2bx + ax^{3} + bx^{2} + a + ax^{2} = x^{3} - x + 2$.\r\n\r\nTherefore $ a = 1$ and $ b = a = 1$, so that $ f(x) = 1 + x$, and\r\n\r\n$ I = \\frac {e^{x}(1 + x)}{1 + x^{2}}$.\r\n\r\nI hope that this is of some help to you, emjhay; I tried rearranging $ x^{3} - x + 2$ (without differentiating the answer given by centry57 and using this to get a rearrangement), various substitutions and integration by parts but got nowhere."
}
{
  "Tag": [],
  "Problem": "[color=green]Does anybody have a fast way (less than 2 minutes time) to do this problem?\n\nSuppose a coin purse contains 30 coins which are either nickels, dimes, and/or quarters. How many different combinations of these coins are there whose value is $5 ?[/color]",
  "Solution_1": "if you can do mental math quickly and are organized and work fairly fast it is possible to do it within the 2 min time limit :?",
  "Solution_2": "This is took me about 3 minutes to work out . . . [hide]\n\nLet $N$ = number of nickels,\n$D$ = number of dimes,\n$Q$ = number of quarters.\n\nThere are 30 coins:\n$.\\qquad[1]\\;\\;N + D + Q = 30$\n\nTheir total value is 500 cents:\n$.\\qquad\\;5N + 10D + 25Q = 500\\;\\;\\Rightarrow\\;\\;[2]\\;\\;N + 2D + 5Q = 100$\n\nSubtract [1] from [2]: $D + 4Q = 70$\n\nIf $Q = 13,\\; D = 18\\;\\;$ . . .  and there are more than 30 coins.\n\nWe have the following solutions:\nQ = 14, D = 14, N = 2\nQ = 15, D = 10, N = 5\nQ = 16, D = 6, N = 8\nQ = 17, D = 2, N = 11[/hide]",
  "Solution_3": "There is a fairly obvious minimum on the number of quarters. After that, the problem simplifies significantly, at least in terms of the size of the numbers. The problem itself, of course, is simply tedious and not complicated.",
  "Solution_4": "There is a fairly obvious minimum on all of the three coins. I think you mean maximum :P .",
  "Solution_5": "Soroban's solution is very nice.  Thanks.\r\n\r\nThe difficulty is to understand that Q<18, then to start from Q=17 to get the desired solutions."
}
{
  "Tag": [
    "induction",
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "Given a sequence$ \\{a_n\\}$. such that $ a_1\\equal{}1,  a_{n\\plus{}1}\\equal{}(1\\plus{}\\frac{1}{n^2\\plus{}n})a_n\\plus{}\\frac{1}{2^n},  n\\in N^*$  \r\n(1).prove that  $ a_n\\ge 2$       $ (n\\ge 2)$  \r\n(2).Prove that$ a_n<e^2$         $ (n\\ge 1)$  if $ ln(1\\plus{}x)<x$ for any $ x>0$, where $ e\\equal{}2.71828...$",
  "Solution_1": "hello, to 1) we prove by induction that $ a_n\\geq 2$ for all $ n\\ge2$.\r\nAt first let $ n \\equal{} 2$, then $ a _2 \\equal{} (1 \\plus{} 1/2)a_1 \\plus{} 1/2 \\equal{} 2\\geq2$. Now we assume that $ a_n\\geq2$, then we get\r\n$ a_{n \\plus{} 1} \\equal{} \\left(1 \\plus{} \\frac {1}{n^2 \\plus{} n}\\right)a_n \\plus{} \\frac {1}{2^n}\\geq\\left(1 \\plus{} \\frac {1}{n^2 \\plus{} n}\\right)2 \\plus{} \\frac {1}{2^n}\\geq2 \\Leftrightarrow \\frac {2}{n^2 \\plus{} n} \\plus{} \\frac {1}{2^n}\\geq0$ and this is true.\r\nSonnhard.",
  "Solution_2": "[quote=\"sxgfly\"]Given a sequence$ \\{a_n\\}$. such that $ a_1 \\equal{} 1, a_{n \\plus{} 1} \\equal{} (1 \\plus{} \\frac {1}{n^2 \\plus{} n})a_n \\plus{} \\frac {1}{2^n}, n\\in N^*$  \n(1).prove that  $ a_n\\ge 2$       $ (n\\ge 2)$  \n(2).Prove that$ a_n < e^2$         $ (n\\ge 1)$  if $ ln(1 \\plus{} x) < x$ for any $ x > 0$, where $ e \\equal{} 2.71828...$[/quote]\r\n1)$ a_n\\geq a_{n \\minus{} 1} \\geq \\cdots a_2 \\equal{} 2$\r\n2)we prove $ a_n < e^{2 \\minus{} \\frac {2}{n}}$\r\nfor $ n \\equal{} 1$ right\r\nif for $ n \\equal{} k$ is right\r\nwhen $ n \\equal{} k \\plus{} 1$\r\n$ a_{k \\plus{} 1} \\equal{} (1 \\plus{} \\frac {1}{k^2 \\plus{} k})a_k \\plus{} \\frac {1}{2^k} < (1 \\plus{} \\frac {1}{k^2 \\plus{} k})e^{2 \\minus{} \\frac {2}{k}} \\plus{} \\frac {1}{2^k}$\r\n$ < e^{2 \\minus{} \\frac {2}{k}}(1 \\plus{} \\frac {1}{k^2 \\plus{} k} \\plus{} \\frac {1}{2^ke^{2 \\minus{} \\frac {2}{k}}})$\r\n$ < e^{2 \\minus{} \\frac {2}{k}}e^{\\frac {2}{k(k \\plus{} 1)}} \\equal{} e^{2 \\minus{} \\frac {2}{k \\plus{} 1}}$",
  "Solution_3": "[quote=\"ye109\"][quote=\"sxgfly\"]Given a sequence$ \\{a_n\\}$. such that $ a_1 \\equal{} 1, a_{n \\plus{} 1} \\equal{} (1 \\plus{} \\frac {1}{n^2 \\plus{} n})a_n \\plus{} \\frac {1}{2^n}, n\\in N^*$  \n(1).prove that  $ a_n\\ge 2$       $ (n\\ge 2)$  \n(2).Prove that$ a_n < e^2$         $ (n\\ge 1)$  if $ ln(1 \\plus{} x) < x$ for any $ x > 0$, where $ e \\equal{} 2.71828...$[/quote]\n1)$ a_n\\geq a_{n \\minus{} 1} \\geq \\cdots a_2 \\equal{} 2$\n2)we prove $ a_n < e^{2 \\minus{} \\frac {2}{n}}$\nfor $ n \\equal{} 1$ right\nif for $ n \\equal{} k$ is right\nwhen $ n \\equal{} k \\plus{} 1$\n$ a_{k \\plus{} 1} \\equal{} (1 \\plus{} \\frac {1}{k^2 \\plus{} k})a_k \\plus{} \\frac {1}{2^k} < (1 \\plus{} \\frac {1}{k^2 \\plus{} k})e^{2 \\minus{} \\frac {2}{k}} \\plus{} \\frac {1}{2^k}$\n$ < e^{2 \\minus{} \\frac {2}{k}}(1 \\plus{} \\frac {1}{k^2 \\plus{} k} \\plus{} \\frac {1}{2^ke^{2 \\minus{} \\frac {2}{k}}})$\n$ < e^{2 \\minus{} \\frac {2}{k}}e^{\\frac {2}{k(k \\plus{} 1)}} \\equal{} e^{2 \\minus{} \\frac {2}{k \\plus{} 1}}$[/quote]\r\n\r\nHow to obtain the last inequality!  can you tell me?",
  "Solution_4": "[quote=\"sxgfly\"][quote=\"ye109\"][quote=\"sxgfly\"]Given a sequence$ \\{a_n\\}$. such that $ a_1 = 1, a_{n + 1} = (1 + \\frac {1}{n^2 + n})a_n + \\frac {1}{2^n}, n\\in N^*$  \n(1).prove that  $ a_n\\ge 2$       $ (n\\ge 2)$  \n(2).Prove that$ a_n < e^2$         $ (n\\ge 1)$  if $ ln(1 + x) < x$ for any $ x > 0$, where $ e = 2.71828...$[/quote]\n1)$ a_n\\geq a_{n - 1} \\geq \\cdots a_2 = 2$\n2)we prove $ a_n < e^{2 - \\frac {2}{n}}$\nfor $ n = 1$ right\nif for $ n = k$ is right\nwhen $ n = k + 1$\n$ a_{k + 1} = (1 + \\frac {1}{k^2 + k})a_k + \\frac {1}{2^k} < (1 + \\frac {1}{k^2 + k})e^{2 - \\frac {2}{k}} + \\frac {1}{2^k}$\n$ < e^{2 - \\frac {2}{k}}(1 + \\frac {1}{k^2 + k} + \\frac {1}{2^ke^{2 - \\frac {2}{k}}})$\n$ < e^{2 - \\frac {2}{k}}e^{\\frac {2}{k(k + 1)}} = e^{2 - \\frac {2}{k + 1}}$[/quote]\n\nHow to obtain the last inequality!  can you tell me?[/quote]\r\n$ < e^{2 - \\frac {2}{k}}(1 + \\frac {1}{k^2 + k} + \\frac {1}{2^ke^{2 - \\frac {2}{k}}})$\r\n${ \\leq  e^{2 - \\frac {2}{k}}(1 + \\frac {1}{k^2 + k} + \\frac {1}{2^k(1+2 - \\frac {2}{k})}})$\r\n${ \\leq  e^{2 - \\frac {2}{k}}(1 + \\frac {1}{k^2 + k} + \\frac {1}{(1+k+\\frac{k(k-1)}{2})(1+2 - \\frac {2}{k})}})$\r\n$ \\leq e^{2 - \\frac {2}{k}}(1 + \\frac {2}{k^2 + k})$\r\n$ < e^{2 - \\frac {2}{k}}e^{\\frac {2}{k(k + 1)}} = e^{2 - \\frac {2}{k + 1}}$"
}
{
  "Tag": [
    "inequalities"
  ],
  "Problem": "let $ a,b$and $ c$ be positive reals numbers , prove that :\r\n$ Min((b\\minus{}c)^2,(c\\minus{}a)^2,(a\\minus{}b)^2))\\le\\frac{a^2\\plus{}b^2\\plus{}c^2}{5}$",
  "Solution_1": "Wolog $ a\\le b\\le c$, and let $ b \\equal{} a \\plus{} x, c \\equal{} b \\plus{} y$ then as $ a,b,c$ are positive reals, $ b \\equal{} a \\plus{} x > x$.\r\nThen:\r\n$ \\min((b \\minus{} c)^2,(c \\minus{} a)^2,(a \\minus{} b)^2))\\le\\frac {a^2 \\plus{} b^2 \\plus{} c^2}{5}$\r\n$ \\Leftrightarrow \\min (x^2,y^2)\\le \\frac {(b \\minus{} x)^2 \\plus{} b^2 \\plus{} (b \\plus{} y)^2}{5}$ (as $ x^2,y^2\\le(x \\plus{} y)^2$)\r\n$ \\Leftrightarrow \\min (x^2,y^2)\\le \\frac {b^2 \\plus{} x^2 \\plus{} y^2 \\plus{} 2by \\plus{} 2b^2 \\minus{} 2bx}{5}$\r\n$ \\Leftrightarrow \\min (x^2,y^2)\\le \\frac {b^2 \\plus{} x^2 \\plus{} y^2 \\plus{} 2by \\plus{} 2b(b \\minus{} x)}{5}$  (*)\r\nNow:\r\n$ b^2 \\plus{} x^2 \\plus{} y^2 \\plus{} 2by \\plus{} 2b(b \\minus{} x) > b^2 \\plus{} x^2 \\plus{} y^2 \\plus{} 2by$ (as $ b > x$)\r\n$ > 2x^2 \\plus{} y^2 \\plus{} 2xy > 5\\min (x^2,y^2)$ and so (*) is true, and the original inequality follows.",
  "Solution_2": "[quote=\"tchebytchev\"]let $ a,b$and $ c$ be positive reals numbers , prove that :\n$ Min((b \\minus{} c)^2,(c \\minus{} a)^2,(a \\minus{} b)^2))\\le\\frac {a^2 \\plus{} b^2 \\plus{} c^2}{5}$[/quote]\r\nWhen you reduce the same quantity c=min(a,b,c)>=0 then we have:LHS is the same, but the RHS will decrease.Therefore, we  need to prove when c=0! Then it becomes so easy! :)",
  "Solution_3": "[img]http://s2.sinaimg.cn/middle/0018zOAxgy724JAxHk521&690[/img]"
}
{
  "Tag": [
    "combinatorics proposed",
    "combinatorics"
  ],
  "Problem": "Ten teams participated in a football competition where each team played against every other team exactly one .When the comprtition was over ,it turned out that for every three team $A;B;C$,if $A$ defeated $B$  and $B$  defeated $C$ then $A$  defeated $C$. \r\nprove that there were 4 teams $A: B: C$ such that $A$ defeated $B$ ;$B$  defeated $C$ ;$C$  defeated $D$ or such that each match between them was a draw",
  "Solution_1": "If a team won no games, we call it \"rank 0.\"  If a team won games only against rank 0 teams, we call it rank 1.  If a team won games only against rank 0 and rank 1 teams, we call it rank 2.  And so on.  If there are no chains of length 3 (that is, involving 4 teams) then every team is of rank 0, 1 or 2.  But there are 10 teams, so at least one of these ranks must have 4 teams.  The 4 teams at this rank must have all draws between them, since no team beats other teams of its rank."
}
{
  "Tag": [
    "email",
    "AMC"
  ],
  "Problem": "My school registered for the AMC A test in December, we got a notification that we would receive our AMC stuff by early January. However, that hasn't arrived yet (as of 1/21) so I was wondering if any schools have received their AMC A stuff?",
  "Solution_1": "i'm taking AMC 10b and i just got my packet on friday",
  "Solution_2": "We are caught up on A-date shipping, having now sent approximately 3000 packages since January 2.\r\nIf your contest manager provided an email  address, the contest manager will receive an automatically generated email with a tracking number.  Use the tracking number to find the current location of your package.  If you are within the contiguous 48 states, shipping should take 5 days or less.  If you are in Alaska or Hawaii, it will take about 5-7 days.  Outside the US, it is variable, about 5-7 days depending on distance and location with the country.\r\n\r\nIf you believe your order or your package has been lost, contact the AMC office by any of the means printed on all of our materials.\r\n\r\nSteve Dunbar\r\nMAA Director, American Mathematics Competitions"
}
{
  "Tag": [
    "quadratics",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "How many primes are the Pisano period?",
  "Solution_1": "I think you should better define what is the Pisano period... ;) \r\n\r\nPierre.",
  "Solution_2": "The m pisano Period ($\\pi (m)$) are the period of Fibonacci sequance mod m\r\nfor example: $\\pi(4) = 6$\r\nsince the residue of Fibonacci mod 4 are [1,2,3,1,0,1],1,2,......  \r\nfrom what I know, there is only one prime {3} \r\nfreom $\\pi(2)=[1,0,1],1,0...........$ =3\r\nbecause pisano Period are even if m>2",
  "Solution_3": "Is it trival Pisano Period are even if m>2?",
  "Solution_4": "[quote=\"wonbert\"]Is it trival Pisano Period are even if m>2?[/quote]\r\n\r\nI don't know if you'll consider it trivial, but it certainly isn't hard.\r\n\r\nLet $u=\\frac{-1+\\sqrt 5}2$ and $\\bar u=\\frac{-1-\\sqrt 5}2$. It sufices to prove it for prime $m$'s. If $p=5$, check it by hand. If $p\\ne 5$, we work in the field formed by adjoining $\\sqrt 5$ to $\\mathbb Z_p$ (this field either has $p^2$ elements if $5$ is not a quadratic residue modulo $p$, or $p$ elements if $5$ is a quadratic residue modulo $p$). \r\n\r\nWe want to show that a $t$ satisfying $(u^{n+t}-u^n)-(\\bar u^{n+t}-\\bar u^n)=0,\\ \\forall n\\ (*)$ must be even (remember, $(*)$ must hold in the mentioned field). $(*)$ translates to $u^n(u^t-1)-\\bar u^n(\\bar u^t-1)=0,\\ \\forall n$. By considering two consecutive values of $n$, we get a system of two equations with the unknowns $u^t-1$ and $\\bar u^t-1$, and we conclude that these must both be $0$. Since $\\bar u=\\frac {-1}u$, the conclusion that $t$ must be even follows quickly."
}
{
  "Tag": [
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "Let$0\\leq p_i \\leq 1,\\forall i = 1.,2,...,n$.Prove that:$\\exists x \\in [0,1]$ such that:\r\n$\\sum_{i=1}^n \\displaystyle\\frac {1}{|x-p_i|} \\leq 8n(1+\\frac{1}{3} +.....+\\frac{1}{2n-1})$",
  "Solution_1": "Hi pigfly!\r\n\r\nI think, there are more sharp estimate.\r\nDivide [0,1] into $2n$ equal segments. From Pigeon-hole principle we obtain that at least n of these segments don't contain $p_i$ for all $i=1..n$. Let $x_i$ be midpoints of these segments.\r\nWe have \r\n\\[\\sum_{j=1}^n\\sum_{i=1}^n\\frac{1}{|x_j-p_i|}=\\sum_{i=1}^n\\sum_{j=1}^n\\frac{1}{|x_j-p_i|}\\]. \r\nIt is easy to show that \r\n\\[\\sum_{j=1}^n\\frac{1}{|x_j-p_i|} < 2\\cdot\\left(\\frac{1}{1/2n}+\\frac{1}{3/2n}+\\dots+\\frac{1}{(2[n/2]+1)/2n}\\right)=4n\\left(\\frac{1}{1}+\\frac{1}{3}+\\dots+\\frac{1}{2[n/2]+1}\\right).\\]\r\nHence we conclude that there exists $x_j$ s.t.\r\n\\[\\sum_{i=1}^n\\frac{1}{|x_j-p_i|} < 2\\cdot\\left(\\frac{1}{1/2n}+\\frac{1}{3/2n}+\\dots+\\frac{1}{(2[n/2]+1)/2n}\\right)=4n\\left(\\frac{1}{1}+\\frac{1}{3}+\\dots+\\frac{1}{2[n/2]+1}\\right).\\]"
}
{
  "Tag": [
    "function",
    "logarithms",
    "geometry",
    "geometric transformation",
    "rotation",
    "homothety",
    "advanced fields"
  ],
  "Problem": "Function f(z) is regular in the strip $|Rez|<\\frac{\\pi}{4}$ and satisfy |f(z)|<1 and f(0)=0. Prove that |f(z)|<|tg(z)|",
  "Solution_1": "the trick here is to transform the unit disc via a homography into the upper halfplane ,later by taking the squre to transform the halfplane into the whole C but a halfline and then by taking a logarithm to transform it into the strip  -pi<im z< pi then transform this strip by rotation and homotety into the strip from the problem. by combining these functions and applying f to what's derived you receive a function from the unit disc into the unit disc (since |f(z)|<1) take care the zero to be transformed into zero. now just apply shwarz lemma",
  "Solution_2": "2qw:Could you please help me with understanding, how i can get |tgz| using you solution(Shwarz lemma says that |f(z)|<|z| and as far as i understand function tgz doesn't include in our transformation functions)"
}
{
  "Tag": [
    "function",
    "linear algebra",
    "matrix",
    "pigeonhole principle",
    "combinatorics proposed",
    "combinatorics"
  ],
  "Problem": "A problem I came up with a short while ago. I don't know the answer yet, but then I haven't worked on it much yet. Here it is.\r\n\r\nThere are $ n$ players at a Go tournament (as you may know, there are no draws in modern Go tournaments). The tournament is in a round-robin format, meaning that each player in the tournament plays a game against each other player in the tournament, receiving 1 point for a win and 0 points for a loss, for a total of $ n \\minus{} 1$ games per player. The tournament ends with Victor as the decisive winner (he had a higher score than anyone else, no SOS involved). What is the largest number of games that Victor could conceivably have lost, as a function of $ n$?\r\n\r\nFor instance, for $ n \\equal{} 5$, it's possible that he lost up to one games, using the following scoresheet (Victor on first row):\r\n\\[ \\begin{bmatrix} X & 1 & 1 & 0 & 1 \\\\\r\n0 & X & 1 & 1 & 0 \\\\\r\n0 & 0 & X & 1 & 1 \\\\\r\n1 & 0 & 0 & X & 0 \\\\\r\n0 & 1 & 0 & 1 & X \\end{bmatrix}\r\n\\]\r\n\r\nI'm not sure how hard this is, but I suspect not very.\r\n\r\n[b]Edit:[/b] gave a more compelling example.",
  "Solution_1": "it looks easy :) \r\n\r\nVictor won n-k-1 games and lost k games; then, he earned n-k-1 points.\r\n\r\nat all other games there were (n-2)+(n-3)+...+1= (n-1)(n-2)/2 points for n-1 people... but all them plated against Victor, then there are  (n-1)(n-2)/2 +k points for n-1 people.\r\n\r\nVictor won more than the half of the games (if not, at least one rival earnt more points than Victor or tied with him, because dividing (n-1)(n-2)/2 +k points to n-1 people gives one with at least [(n-2)/2]* +1 points, following the Pigeonhole principle, and it is more than [(n-1)/2]*), then k < (n-1)/2\r\n\r\nNOTE: when I write [x] i mean \"the smallest integer higher than or equal to x\"\r\n\r\nThen, the max number of points for n-1 people is (n-1)(n-2)/2 +[(n-3)/2]. This number is smaller than the number\r\n(n-1)(n-2)/2 +[(n-2)/2]. And this is:\r\n\r\nn(n-2)/2, for n even (case 1)\r\nor\r\n((n-1)^2)/2, for n odd. (case 2)\r\n\r\nthen, if we divide for n-1:\r\nfor case 2, we have n-2, but this number is bigger than the original maximum, and we have the demostration!\r\n\r\nand, for case 1, we have, for AM-GM:\r\n\r\nfor case 1 we have that AM1=n-1, but we have not equality, then GM1 is smaller than it.\r\n\r\nAM1>GM1\r\n\r\nThen, (n-1)^2>n(n-2), and dividing for n-1, we have it again!\r\n\r\nThen, the minimum value of k is\r\n\r\n                                                                      k=[(n-3)/2] \r\n\r\n :lol:"
}
{
  "Tag": [
    "geometry",
    "rhombus",
    "congruent triangles"
  ],
  "Problem": "Prove that all rhombuses (rhombi  :D ) are parallelograms.",
  "Solution_1": "[quote=\"236factorial\"]Must they be?[/quote]\r\n\r\n[hide=\"Yes.\"]\nBut what is your definition of rhombus? If you include in the definition the fact that the sides are parallel, then we're done. Definition please. :D\n[/hide]",
  "Solution_2": "Rhombus-a quadrilateral with all four sides equal  :D",
  "Solution_3": "No proofs in getting started.",
  "Solution_4": "Prove that.  :roll:",
  "Solution_5": "[hide=\"my proof\"]\nDraw a diagonal acroll the rhombus. This creates two triangles. Since all the sides of the rhombus are equal, and the diagonal is congruent to itself, the two triangles are congruent by SSS. Then, using CPCTC (Corresponding Parts of Congruent Triangles are Congruent) and the base angles theorem, combined with the transitive property of congruence, we know that the four angles that have the diagonal as a side are congruent. Then, by the converse of the alternate interior angles theorem, we have that the opposite sides are parallel.\n[/hide]",
  "Solution_6": "Good proof  :) \r\n\r\nI failed to notice because that the diagonal was congruent to itself, maybe next I'll work on cleaner paper  :roll:  :blush:"
}
{
  "Tag": [
    "probability",
    "ratio",
    "floor function",
    "factorial"
  ],
  "Problem": "1. How many terminal zeroes are there in the combination of 383 taken 144?\r\n\r\n2. A, B, and C are playing a game with a coin by flipping it several times. If A flips heads, then the coin goes to B, if he flips tails, it goes to C. If B flips heads, then the coin goes to C, if he flips tails, it goes to A. If C flips heads, then the coin goes to A, if he flips tails, it goes to B.After each flip, the person now holding the flips it again, and so on. Suppose A makes the first flip of the coin, what is the probability that after the sixth flip, the coin goes to B? (Is there any other way other than using the tree method??)  \r\n\r\n3. In a large party, the ratio of boys to girls is 5:4 . Each boy shook hands with all the girls, and also half of the boys. Each girl shook hands with all the boys, and also half of the girls. If there were 5929 handshakes made, and no pair of person shook hands more than once, how many people were there?\r\n\r\n4. Jonah is inside a whale at sea. At the dawn of each day, there is a 25% chance that he will write in his diary, and at noon, there is a 1/3 chance that the whale will swallow him. If he remains stuck indefinitely, what is the probability that Jonah will be swallowed without writing anything in his diary?\r\n\r\n5. In how many ways can you arrange the letters of NUMERO such that no two vowels are together?",
  "Solution_1": "1. Do you mean $ \\binom{383}{144}$? If yes: Let $ f(x)\\equal{}$number of zeroes in which $ x$ ends.\r\n$ f(383)\\equal{}\\lfloor \\frac{383}{5}\\rfloor \\plus{}\\lfloor \\frac{383}{25}\\rfloor \\plus{}\\lfloor \\frac{383}{125}\\rfloor\\equal{}94$\r\nDo the same for $ f(144)$ and $ f(383\\minus{}144)$.\r\nAnswer: $ 3$",
  "Solution_2": "Yes, but the correct answer is 1.. :(",
  "Solution_3": "Do you know how to do it? I think that it is $ 3$, but correct me!",
  "Solution_4": "I dont know. These problems are the ones I was not able to answer in a competition. The quizmaster said that it is not as easy as what i think which i answered as 3.",
  "Solution_5": "[quote=\"Bugi\"]Let $ f(x) \\equal{}$number of zeroes in which $ x$ ends.\n$ f(383) \\equal{} \\lfloor \\frac {383}{5}\\rfloor \\plus{} \\lfloor \\frac {383}{25}\\rfloor \\plus{} \\lfloor \\frac {383}{125}\\rfloor \\equal{} 94$\nDo the same for $ f(144)$ and $ f(383 \\minus{} 144)$.\nAnswer: $ 3$[/quote]\r\n\r\nFirst of all, what you wrote as the definition of $ f$ is not what you meant.  Second of all, $ 10/2 \\equal{} 5$ is the ratio of a number that ends in 1 zero over a number that ends in 0 zeroes, yet the quotient has 0 (not 1 - 0 = 1) zeroes -- yet according to your solution, this is not possible!  So, perhaps time to reconsider some of your (implicit) assumptions!  :)",
  "Solution_6": "[quote=\"orangefronted\"]4. Jonah is inside a whale at sea. At the dawn of each day, there is a 25% chance that he will write in his diary, and at noon, there is a 1/3 chance that the whale will swallow him. If he remains stuck indefinitely, what is the probability that Jonah will be swallowed without writing anything in his diary?[/quote]\r\n\r\n\r\n4) $ \\frac14$ chance that he writes. $ \\frac34\\cdot\\frac13 \\equal{} \\frac14$ chance that he gets eaten without writing anything. There's a $ \\frac{\\frac14}{\\frac14\\plus{}\\frac14}\\equal{}\\frac12$ chance that he gets swallowed without writing anything.",
  "Solution_7": "1) 10 = 2(5)\r\nfloor(383/2) = 191\r\nor 383! = k(2^191)\r\nfloor(383/5) = 76\r\nso there are 76 terminal zeros in (383!) . \r\n\r\nfloor(144/2) = 72   ,  floor(144/5) = 28\r\nfloor(239/2) = 119 ,   floor(239/5) = 47\r\n\r\n76 - (28 + 47) = 1\r\nThere is 1 terminal zero in 383C144 or 383C239",
  "Solution_8": "2) (i don't know any other way to do it) 10/32\r\n3) 2(5929) = 11858 = b(g + b/2) + g(b+ g/2)\r\n    4b = 5g\r\n\r\n    11858 = 2bg + (b^2 + g^2) / (2)\r\nsubstitute\r\n             = (10g^2)/4 + (41g^2)/32\r\n             = (121g^2)/32\r\n           g = 56  so   b = 70\r\n           number of people = 126\r\n4) 0 \r\n5) 72\r\nv - vowel   c - consonant\r\ncvcvcv and vcvcvc ,  so (3!)(3!)2 =72",
  "Solution_9": "[quote=\"ManuelS\"]1) 10 = 2(5)\nfloor(383/2) = 191\nor 383! = k(2^191)\nfloor(383/5) = 76\nso there are 76 terminal zeros in (383!) . \n\nfloor(144/2) = 72   ,  floor(144/5) = 28\nfloor(239/2) = 119 ,   floor(239/5) = 47\n\n76 - (28 + 47) = 1\nThere is 1 terminal zero in 383C144 or 383C239[/quote]\r\n\r\nYou fail to consider that repeated powers in the factorials (such as $ 125 \\equal{} 5^3$) should be counted multiple times.\r\n\r\nas has been shown before by Bugi, $ 5^3 \\parallel{} \\binom{383}{144}$\r\n\r\nNow we must find $ k$ such that $ 2^k \\parallel{} \\binom{383}{144}$, and our answer will be $ min(3,k)$.\r\n\r\n$ \\lfloor \\frac{383}{2} \\rfloor \\equal{} 191$, $ \\lfloor \\frac{191}{2} \\rfloor \\equal{} 95$, $ \\lfloor \\frac{95}{2} \\rfloor \\equal{} 47$, $ \\lfloor \\frac{47}{2} \\rfloor \\equal{} 23$, $ \\lfloor \\frac{23}{2} \\rfloor \\equal{} 11$, $ \\lfloor \\frac{11}{2} \\rfloor \\equal{} 5$, $ \\lfloor \\frac{5}{2} \\rfloor \\equal{} 2$, $ \\lfloor \\frac{2}{2} \\rfloor \\equal{} 1$\r\nso $ 383! \\equal{} 2^{375}a$ where $ a$ is odd\r\n\r\n$ \\lfloor \\frac{144}{2} \\rfloor \\equal{} 72$, $ \\lfloor \\frac{72}{2} \\rfloor \\equal{} 36$, $ \\lfloor \\frac{36}{2} \\rfloor \\equal{} 18$, $ \\lfloor \\frac{18}{2} \\rfloor \\equal{} 9$, $ \\lfloor \\frac{9}{2} \\rfloor \\equal{} 4$, $ \\lfloor \\frac{4}{2} \\rfloor \\equal{} 2$, $ \\lfloor \\frac{2}{2} \\rfloor \\equal{} 1$\r\nso $ 144! \\equal{} 2^{142}b$ where $ b$ is odd\r\n\r\n$ \\lfloor \\frac{239}{2} \\rfloor \\equal{} 119$, $ \\lfloor \\frac{119}{2} \\rfloor \\equal{} 59$, $ \\lfloor \\frac{59}{2} \\rfloor \\equal{} 29$, $ \\lfloor \\frac{29}{2} \\rfloor \\equal{} 14$, $ \\lfloor \\frac{14}{2} \\rfloor \\equal{} 7$, $ \\lfloor \\frac{7}{2} \\rfloor \\equal{} 3$, $ \\lfloor \\frac{3}{2} \\rfloor \\equal{} 1$\r\nso $ 239! \\equal{} 2^{232}c$ where $ c$ is odd\r\n\r\nThus we have $ k\\equal{}375\\minus{}142\\minus{}232\\equal{}1$, so $ min(1,3) \\equal{} 1$.",
  "Solution_10": "Wow, i didn't noticed :blush: \r\nI guess i though it was simpler.",
  "Solution_11": "I've worked simpler problems such as how many trailing zeros a factorial has, in which the 2's don't matter.  If you start thinking only a trailing zero on bottom can get rid of a trailing zero on top, then it is quite easy to forget about the 2's.  In short: be careful with your assumptions.\r\n\r\n@manuel: well, I guess you were just lucky\r\n\r\nNow #5:\r\n\r\n[hide=\"Hint\"]Use a ball and divider argument[/hide]\n\n[hide]\nLet the balls be consonants and the dividers be balls.  So we have $ 3$ balls and thus $ 4$ places to put the dividers.  Since we have $ 3$ dividers, we have $ \\binom{4}{3} \\equal{} 4$ ways to choose where to place the dividers.  But the dividers (vowels) and the balls (consonants) are not indistinguishable.  We have $ 3! \\equal{} 6$ possible permutations of the vowels, and $ 3! \\equal{} 6$ possible permutations of the consonants, giving a total of $ 6*6*4 \\equal{} 144$ possible arrangements.\n[/hide]",
  "Solution_12": "The correct answers are as follows:\r\n1. 1\r\n2. 21/64\r\n3. 126\r\n4. 1/2\r\n5. 144"
}
{
  "Tag": [
    "trigonometry",
    "geometry",
    "circumcircle",
    "geometry proposed"
  ],
  "Problem": "$\\triangle ABC,\\ A>90^{\\circ}: \\ P,N\\in (BC),\\ PA=PB,\\ NA=NC;$\r\n\r\n$[AM]\\cap BC\\ne\\emptyset,\\ MB=MC,\\ m(\\widehat {BCM})=A-90^{\\circ}.$\r\n\r\n[u]Ascertain the measures of the MNP's angles and prove that[/u] $\\frac{MN}{\\cos B}=\\frac{NP}{\\sin A}=\\frac{PM}{\\cos C}.$",
  "Solution_1": "M isnt fixed.then by moving it,the angle BCM wont be fixed.\r\n\r\nplz get information about it ;)",
  "Solution_2": "$MB=MC,\\ m(\\widehat {CBM})=m(\\widehat {BCM})=90^{\\circ}-A\\Longrightarrow$the point $M$ is fixed !.\r\n\r\n.The line $BC$ separates the points $A,M$. The all points are fixed !.\r\n\r\n$A>90^{\\circ}\\Longrightarrow P\\in (BN)$. Indeed,\r\n\r\n$P\\in (BN)\\Longleftrightarrow a>\\frac{c}{2\\cos B}+\\frac{b}{2\\cos C}\\Longleftrightarrow$\r\n\r\n$2a\\cos B\\cos C>c\\cos C+b\\cos B\\Longleftrightarrow$\r\n\r\n$2\\sin A\\cos B\\cos C>\\sin C\\cos C+\\sin B\\cos B \\Longrightarrow$\r\n\r\n$4\\sin A\\cos B\\cos C>\\sin 2C+\\sin 2B\\Longleftrightarrow$\r\n\r\n$4\\sin A\\cos B\\cos C>2\\sin (B+C)\\cos (B-C)\\Longleftrightarrow$\r\n\r\n$2\\cos B\\cos C>\\cos (B-C)\\Longleftrightarrow$\r\n\r\n$2\\cos B\\cos C>\\cos B\\cos C+\\sin B\\sin C\\Longleftrightarrow$\r\n\r\n$\\cos (B+C)>0\\Longleftrightarrow \\cos A<0.$",
  "Solution_3": "The three interior angles of $\\triangle MNP$ can be found by angle chasing. Then the desired equality can be obtained by applying sine law to this triangle.",
  "Solution_4": "The point $M$ is the circumcenter of the triangle $ABC$ !"
}
{
  "Tag": [
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "we have $102$ coins $100$ regular and $2$ unregular\r\nfind that the $2$ coins are heavier or lighter than regulars by 3 weightings :?:",
  "Solution_1": "Hit me if I'm wrong, but:\r\n\r\n1)Weigh a regular coin\r\n2)Weigh first unregular coin\r\n3)Weigh second unregular coin.\r\n\r\nHannes",
  "Solution_2": "I think we don't know which coins are the irregular ones...",
  "Solution_3": "Is it posible?",
  "Solution_4": "[quote=\"M4RI0\"]Is it posible?[/quote]\r\nSince we just want to find out if they are heavier or not, and not to find out which ones, I think yes (I haven't thought about it till now).",
  "Solution_5": "Divide the coins into 3 groups of 34. Of these three groups, weigh every combination of two (first and second, first and third, second and third).\r\n\r\nIf the two unusual coins are in two different groups, these will balance. Then consider whether the other group, consisting of only regular coins, are heavier or lighter than these two.\r\n\r\nIf the two unusual coins are in the same group, the other two will balance. Then consider whether they are heavier or lighter than this one.",
  "Solution_6": "First, you only need 2 weighings to carry that out.  Second, that doesn't work because it presupposed I know the answer.  For instance: suppose I do what you say, and I find that two piles are heavier than the third.  Are the coins heavy or light?",
  "Solution_7": "anyone has an answer :?:",
  "Solution_8": "let it be known. \r\n\r\ndivide the 102 coins into 2 piles, A and B, each of 51 coins (duh). compare these two piles. \r\n\r\n1) if they're different, this means the irregular coins are either both in A or both in B. choose either one of the two piles--say A!!--and divide it into 3 piles, X, Y, Z, each of 17 coins. compare X to Y and X to Z. there are two possible results: \r\n\r\n(a) X equals Y and X equals Z, in which case the irregular coins must have been in pile B; since A differed from B, then the result we got when we compared A to B tells us if the irregular coins are lighter or heavier; \r\n(b) if X differs from either Y or Z, then the irregulars must both be in A and again, since A differed from B, we know. (and we also know (and we also know))\r\n\r\n2) if A = B, this means there's one irregular coin in each of A and B. choose either pile, say A!! divide it into 3 piles, X, Y, Z, each of 17 coins. compare X and Y and X and Z. keeping in mind that there's exactly one irregular coin somewhere among X, Y, Z, we have 3 possible results:\r\n\r\n(a) X differs from Y and X differs from Z. in that case, the irregular is clearly in X, and the comparison of X and Y tells us if it's light or heavy.\r\n(b) X differs from Y and X equals Z. in that case, the irregular's in Y and again we know.\r\n(c) X equals Y and X differs from Z. in that case, the irregular's in Z and again, ya know.\r\n\r\n-letting it be, Holografio Insinuendo"
}
{
  "Tag": [
    "function",
    "algebra",
    "domain",
    "calculus",
    "derivative",
    "calculus computations"
  ],
  "Problem": "If a function haas a local max or a local min value at an interior point c of its domain, and if the first derivative, f', exists at c, then f'(c)=\r\n\r\nc\r\nf\r\n-c\r\n1\r\nnone of these",
  "Solution_1": "$ f'(c)\\equal{}0$.  In order for a function $ f$ to have a critical point, its derivative must equal 0 at that point.",
  "Solution_2": "[quote=\"JRav\"]In order for a function $ f$ to have a critical point, its derivative must equal 0 at that point.[/quote]\r\n\r\n[u]Or[/u] the derivative be undefined at that point! (in this case, of course, he specified that the derivative exists - but in general it is something to be considered).\r\n\r\nAlso note that being a critical point of f does not gaurantee that f has a max or a min at that point. It is a necessary, but not sufficient, condition.",
  "Solution_3": "Obviously, but I'm going based on the info given."
}
{
  "Tag": [
    "modular arithmetic"
  ],
  "Problem": "Two brothers sold a herd of sheep which they owned. For each sheep, they received as many dollars are the number of sheeps there are originally in the herd. The money was taken and divided in the following manner. The older brother takes 10 dollars, then the younger takes 10 dollars, after the elder takes again, and the younger and so on. At the end of the division, the younger brother, whose turn it was found there were less than 10 dollars left, so he took what remained. To make the division fair, the older brother then gave the younger his penknife. How much is the penknife worth????\r\nHave fun  :D Wont let me use the dollar sign...",
  "Solution_1": "[hide] If the brothers had $ x$ sheep in their flock, then they made $ x^2$ dollars. Now, whenever the older brother and younger brother make a division, there are dividing twenty dollar holdings. Thus, on the last division, the older brother made $ 10$ dollars and the younger brother made $ y$ dollars, where $ y < 10$. However, if considered $ \\bmod20$, it suffices to find the units digit, given that the tens digit is odd. We see that $ 4^2$ and $ 6^2\\equiv16\\pmod{20}$, and these are the only digits (we need only cosider units digits in our test). In either case, the last digit is $ 6$, and the penknife is thus worth $ 10 \\minus{} 6 \\equal{} \\boxed{4}$ dollars.[/hide]"
}
{
  "Tag": [
    "calculus",
    "integration",
    "number theory",
    "prime factorization"
  ],
  "Problem": "Can you say -2 is a factor of 4? Can there be negative factors of positive numbers?\r\n\r\nAlso when a problem asks you to find the zeros of an equation which of the following way to write it is correct? (or multiple ones):\r\na.) x=a,b,c,...\r\nb.) x={a,b,c,...}\r\nc.) a,b,c,...\r\nd.) {a,b,c,...}\r\ne.) (given equation)=0 when x=a,b,c,...",
  "Solution_1": "[quote=\"Tare\"]Can you say -2 is a factor of 4? Can there be negative factors of positive numbers?[/quote]\r\n\r\nYes, you can.",
  "Solution_2": "So 4 has 6 factors?",
  "Solution_3": "It depends on your definition of factors.  It is meaningful to define numbers with negative factors, but it is more useful to only work with the positive ones.  Usually, people stick to a definition of positive factors.",
  "Solution_4": "Oh OK, so there's 3 factors for 4.",
  "Solution_5": "[quote]It depends on your definition of factors. It is meaningful to define numbers with negative factors, but it is more useful to only work with the positive ones. Usually, people stick to a definition of positive factors.[/quote]\r\n\r\nOk, yeah, this is often true, but you can still say that -2 divides 4.",
  "Solution_6": "I would say that 4 has 6 factors: 3 positive factors and 3 negative factors.",
  "Solution_7": "In that case why not say it has an infinite number of factors.\r\n\r\nIn number theory you only consider positive integral factors, because the negatives are considered associates. Basically meaning they obviously correspond to the positive ones. This also makes the statement of the fundamental theorem of arithmatic - that each number has a unique prime factorization - meaningful."
}
{
  "Tag": [
    "geometry",
    "AMC",
    "AIME",
    "circumcircle",
    "trapezoid",
    "trigonometry",
    "symmetry"
  ],
  "Problem": "(This is the edited version of problem #10 on Mock AIME 2.) ABCDE is a cyclic pentagon with BC = CD = DE. The diagonals AC and BE intersect at M. N is the foot of the altitude from M to side AB.  We have MA = 25, [b]MD = 113[/b], and MN = 15.  The area of triangle ABE can be expressed as m/n where m and n are relatively prime positive integers. Determine the remainder obtained when m+n is divided by 1000.\r\n\r\n(It is elsewhere on forums, but nobody has yet solved it.)\r\n\r\nEdit- Changed MD = 39 to MD = 113.",
  "Solution_1": "Source of the problem: [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=cyclic+pentagon&t=14899]Mock AIME 2 Contest   \r\n[/url]\r\n\r\nLet $(O)$ be the circumcircle of the pentagon $ABCDE$ with center $O$ and radius $R = OA = OB = ...$. Since $BC = CD = DE$, the quadrilateral $BCDE$ is an isosceles trapezoid and $BE \\parallel CD$. Assume that this isosceles trapezoid is fixed and the point $A$ moves on the minor arc $BE$. The angle $\\alpha = \\measuredangle BAC \\equiv \\measuredangle BAM$ then spans the major arc $BC$ of the circumcircle $(O)$ and it is constant. The area of the triangle $\\triangle ABM$ is\r\n\r\n$|\\triangle ABM| = \\frac 1 2 AB \\cdot MN = \\frac 1 2 AB \\cdot AM \\sin \\alpha$\r\n\r\n$\\sin\\alpha = \\frac{MN}{AM} = \\frac {15}{25} = \\frac 3 5$\r\n\r\n$\\cos\\alpha = \\sqrt{1 - \\sin^2{\\alpha}} = \\frac 4 5$\r\n\r\nSince the triangle $\\triangle CDE$ is isosceles with the base $CE$ and the lines $BE \\parallel CD$, we have\r\n\r\n$\\measuredangle DEC = \\measuredangle DCE = \\alpha$\r\n\r\n$\\measuredangle CBE = \\measuredangle DEB = \\measuredangle DEC + \\measuredangle DCE = 2 \\alpha$\r\n\r\n$\\cos{\\measuredangle CBE}  = \\cos{2 \\alpha} = \\cos^2{\\alpha} - \\sin^2{\\alpha} = \\frac{7}{25}$\r\n\r\n$\\sin{\\measuredangle CBE}  = \\sin{2 \\alpha} = 2 \\sin \\alpha \\cos \\alpha = \\frac{24}{25}$\r\n\r\n$BE = BC \\cos{2 \\alpha} + CD + DE \\cos{2 \\alpha} = BC (1 + 2 \\cos{2 \\alpha}) =$\r\n\r\n$= BC \\left( 1 + \\frac{14}{25} \\right) = \\frac{39}{25} BC$\r\n\r\nSince the circle $(O)$ is also the circumcircle of the triangles $\\triangle BCD, \\triangle BCE$, \r\n\r\n$BC = 2 R \\sin \\alpha = \\frac 6 5 R$\r\n\r\n$BD = 2 R \\sin {(180^o - 2\\alpha)} = 2 R \\sin {2\\alpha} = \\frac{48}{25} R = \\frac 8 5 BC$\r\n\r\nLet $P, Q$ be the feet of normals from the vertices $C, D$ to the diagonal $BE$. Then\r\n\r\n$CP = DQ = BC \\sin{2 \\alpha} = \\frac{24}{25} BC$\r\n\r\n$BP = EQ = BC \\cos{2 \\alpha} = \\frac{7}{25} BC$\r\n\r\n$BQ = EP = BD \\cos{\\alpha} = \\frac{32}{25} BC$\r\n\r\nUsing Pythagorean theorem for the right angle triangles $\\triangle CPB, \\triangle CPM$,\r\n \r\n$MC^2 = CP^2 + MP^2 = CP^2 + (BM - BP)^2 = BC^2 - \\frac{14}{25} BC \\cdot BM + BM^2$\r\n\r\nUsing Pythagorean theorem for the right angle triangles $\\triangle DQB, \\triangle DQM$,\r\n\r\n$MD^2 = DQ^2 + MQ^2 = DQ^2 + (BQ - BM)^2 = \\frac{64}{25} BC^2 - \\frac{64}{25} BC \\cdot BM + BM^2$\r\n\r\nBy the intersecting chords theorem, we have\r\n\r\n$MA \\cdot MC = MB \\cdot ME = BM \\cdot (BE - BM)$\r\n\r\n$\\frac{MA}{MD} = \\frac{BM \\cdot (BE - BM)}{MC \\cdot MD} = \\frac {25}{39}$\r\n\r\nDenoting $x = \\frac{BM}{BE}$ and substituting $BC = \\frac{25}{39}BE$ into the equations for $MC^2, MD^2$, \r\n\r\n$\\left(\\frac{MC}{BE}\\right)^2 = \\left(\\frac{25}{39}\\right)^2 - \\frac{14}{39}\\ x + x^2$\r\n\r\n$\\left(\\frac{MD}{BE}\\right)^2 = \\left(\\frac{40}{39}\\right)^2 - \\frac{64}{39}\\ x + x^2$\r\n\r\nand we get the following equation for $x$:\r\n\r\n$x^2 \\left(1 - x\\right)^2 = \\frac {25^2}{39^2} \\left[\\left(\\frac{25}{39}\\right)^2 - \\frac{14}{39}\\ x + x^2\\right] \\left[\\left(\\frac{40}{39}\\right)^2 - \\frac{64}{39}\\ x + x^2\\right]$\r\n\r\nChanging the variable to $x = y + \\frac 1 2$, in order to achieve some symmetry, and using $25^2 + 25 \\cdot 39 = 40^2$,\r\n\r\n$\\left(y^2 - \\frac 1 4\\right)^2 = \\frac {25^2}{39^2} \\left[\\left(\\frac{25}{39}\\right)^2 + \\frac 1 4 - \\frac{7}{39} + \\frac{25}{39}\\ y + y^2\\right] \\left[\\left(\\frac{25}{39}\\right)^2 + \\frac 1 4 + \\frac{25}{39} - \\frac{32}{39} - \\frac{25}{39}\\ y + y^2\\right]$\r\n\r\n$\\left(y^2 - \\frac 1 4\\right)^2 = \\frac {25^2}{39^2} \\left[\\left(\\left(\\frac{25}{39}\\right)^2 + \\frac 1 4 - \\frac{7}{39} + y^2\\right)^2 - \\left(\\frac{25}{39}\\right)^2 y^2\\right]$\r\n\r\nThus the odd powers of $y$ disappeared and we obtained a quadratic equation for $y^2$.\r\n\r\n$16 \\cdot 39^4 (39^2 - 25^2)\\ y^4 - 39^2 [8(39^4 + 2929 \\cdot 25^2) - 16 \\cdot 25^4]\\ y^2 + 39^6 - 25^2 \\cdot 2929^2 = 0$\r\n\r\n$2^4 \\cdot 7 \\cdot 39^4 y^4 - 2^8 \\cdot 39^2 \\cdot 821\\ y^2 - 17 \\cdot 19 \\cdot 109 \\cdot 409 = 0$\r\n\r\nThe roots of this quadratic equation are\r\n\r\n$y^2 = \\frac{2^5 \\cdot 821 \\pm \\sqrt{2^{10} \\cdot 821^2 + 7 \\cdot 17 \\cdot 19 \\cdot 109 \\cdot 409}}{2^2 \\cdot 7\\cdot 39^2} = \\frac{2^5 \\cdot 821 \\pm \\sqrt{3^4 \\cdot 5^{10}}}{2^2 \\cdot 7 \\cdot 39^2} =$\r\n\r\n$=\\frac{2^5 \\cdot 821 \\pm 3^2 \\cdot 5^5}{2^2 \\cdot 7 \\cdot 39^2} = +\\frac{19 \\cdot 409}{2^2 \\cdot 39^2}$ [color=white].[/color] or [color=white].[/color] $-\\frac{17 \\cdot 109}{2^2 \\cdot 7 \\cdot 39^2}$\r\n\r\nSince $y^2 > 0$, the negative root is not acceptable and $y = \\pm \\frac{\\sqrt{19 \\cdot 409}}{2 \\cdot 39} \\doteq \\pm 1.13017$. Returning to the variable $x$,\r\n\r\n$x = \\frac{BM}{BE} = \\frac 1 2\\left(1 \\pm \\frac{\\sqrt{19 \\cdot 409}}{39}\\right) \\doteq +1.63017$ [color=white].[/color] or [color=white].[/color] $-0.63017$\r\n\r\nConsequently, the point $M$ cannot lie within the segment $BE$ and the defined pentagon $ABCDE$ does not exist. Rest assured that the results of all equations, including the roots of the final quadratic equation, check with the Sketchpad. :D",
  "Solution_2": "You are right, Yetti. I apologize to anyone who agonized over this fallacious problem. I don't know if I'll ever  trust myself when writing another geometry problem.\r\n\r\nFortunately, this one is not over-constrained. Rather, it's a matter of bounds. $MA \\le \\frac{81}{125}R\\cdot\\sec{\\frac{\\alpha}{2}} = .6147R$, but $MD \\ge \\frac{6}{5}R\\cos{2\\alpha} = 1.1520R$. In fact, these limiting cases do not correspond, so $\\frac{MA}{MD} < \\frac{6.147R}{1.1520R} = .5336$, but $\\frac{25}{39} = .6410$, impossible. It is easily seen, however, that $\\frac{MA}{MD}$ can become arbitrarily small; simply translate $A$ towards $B$. The trapezoid $BCDE$ remains fixed w.r.t. its circumcircle, but $M$ also approaches $B$ as $MD$ approaches $BD$. A quick lower estimate: $MD < 2R$, so it is possible to obtain $\\frac{MA}{MD} = \\frac{.6147R}{2R} = .3074$ and any lower, positive ratios.\r\n\r\nIn my sloppy effort to find nice numbers, I chose $MD = 39$, which fails to be defined.  I invite you to try this problem with $MD = 113$. If the area of $ABE$ is expressed as $\\frac{m}{n}$ with two relatively prime positive positive integers $m$ and $n$, then what remainder is obtained when $m+n$ is divided by 1000?",
  "Solution_3": "[quote=\"Mildorf\"]Fortunately, this one is not over-constrained. Rather, it's a matter of bounds...[/quote]\nUsing the cosine theorem for the triangles $\\triangle BCM, \\triangle EDM$ \n\n$MC^2 = BC^2 - 2 BC \\cdot BM \\cos{2 \\alpha} + BM^2$\n\n$MD^2 = BC^2 - 2 BC \\cdot (BE - BM) \\cos{2 \\alpha} + (BE - BM)^2$\n\nDividing these equations by $BE$, denoting  $x = \\frac{BM}{BE}$ and substituting $\\frac{BC}{BE} = \\frac{1}{1 + 2 \\cos{2 \\alpha}}$, we get\n\n$\\left( \\frac {MC}{BE} \\right)^2 = \\frac{1}{(1 + 2 \\cos{2 \\alpha})^2} - \\frac{2 \\cos{2 \\alpha}}{1 + 2 \\cos{2 \\alpha}}\\ x  + x^2$\n\n$\\left( \\frac {MD}{BE} \\right)^2 = \\frac{1}{(1 + 2 \\cos{2 \\alpha})^2} - \\frac{2 \\cos{2 \\alpha}}{1 + 2 \\cos{2 \\alpha}}\\ (1 - x) + (1 - x)^2$\n\nBy the intersecting chords theorem, we have\n \n$MA \\cdot MC = MB \\cdot ME = BM \\cdot (BE - BM)$\n\n$\\frac{MA}{MD} = \\frac{BM \\cdot (BE - BM)}{MC \\cdot MD}$\n\nSquaring this equation, reducing the right fraction by $BE$ and substituting $x = \\frac{BM}{BE}$ and the above expressions for $\\frac{MC}{BE}, \\frac{MD}{BE}$ in terms of $x$, we get the following quartic equation for $x$, where we denoted $p = \\frac {MA}{MD}$:\n\n$x^2 (1 - x)^2 = p^2 \\left[ \\frac{1}{(1 + 2 \\cos{2 \\alpha})^2} - \\frac{2 \\cos{2 \\alpha}}{1 + 2 \\cos{2 \\alpha}}\\ x + x^2\\right] \\times$\n\n$\\times \\left[ \\frac{1}{(1 + 2 \\cos{2 \\alpha})^2} - \\frac{2 \\cos{2 \\alpha}}{1 + 2 \\cos{2 \\alpha}}\\ (1 - x)  + (1 - x)^2\\right]$\n\nBecause of the symmetry of the isosceles trapezoid $BCDE$, this quartic equation can always be symmetrized into a quadratic equation for $y^2$ by the substitution $y = x - \\frac 1 2$:\n\n$\\left( \\frac 1 4 - y^2 \\right)^2 = p^2 \\left[ \\frac{1}{(1 + 2 \\cos{2 \\alpha})^2} -  \\frac{2 \\cos{2 \\alpha}}{1 + 2 \\cos{2 \\alpha}} \\left( y + \\frac 1 2 \\right ) + \\left(y + \\frac 1 2 \\right)^2 \\right] \\times$\n\n$\\times \\left[ \\frac{1}{(1 + 2 \\cos{2 \\alpha})^2} + \\frac{2 \\cos{2 \\alpha}}{1 + 2 \\cos{2 \\alpha}} \\left( y - \\frac 1 2 \\right) + \\left(y - \\frac 1 2 \\right)^2 \\right] = $\n\n$= p^2 \\left[ \\left( \\frac{5 - 4 \\cos^2{2 \\alpha}}{4(1 + 2 \\cos{2 \\alpha})^2} + y^2 \\right)^2 - \\frac{y^2}{(1 + 2 \\cos{2 \\alpha})^2} \\right]$\n\nUsing $\\tan{\\alpha} = \\frac{MN}{MA} = \\frac 3 4$ and\n\n$\\cos{2 \\alpha} = cos^2{\\alpha} - sin^2{\\alpha} = \\frac{1 - \\tan^2{\\alpha}}{1 + \\tan^2{\\alpha}} = \\frac{MA^2 - MN^2}{MA^2 + MN^2} = \\frac{4^2 - 3^2}{4^2 + 3^2} = \\frac{7}{25}$\n\nwe can calculate the coefficients of the quadratic equation for $y^2$:\n\n$q = \\frac{5 - 4 \\cos^2{2 \\alpha}}{(1 + 2 \\cos{2 \\alpha})^2} = \\frac{5 \\cdot 25^2 - 4 \\cdot 7^2}{(25 + 2 \\cdot 7)^2} = \\frac{2929}{4 \\cdot 39^2} = \\frac{29 \\cdot 101}{2^2 \\cdot 3^2 \\cdot 13^2}$\n\n$r = \\frac{1}{1 + 2 \\cos{2 \\alpha}} = \\frac{25}{39} = \\frac{5^2}{3 \\cdot 13}$\n\nThe quartic polynomial\n \n$P(y) = \\left( \\frac 1 4 - y^2 \\right)^2 - p^2 \\left[ (q + y^2)^2 - r^2 y^2 \\right] =$\n\n$= (1 - p^2) y^4 - \\left( 2p^2 q - p^2 r^2 + \\frac 1 2 \\right) y^2 + \\frac{1}{16} - p^2 q^2$\n\nis an even function of $y$, its graph having the shape of a bottle bottom. Since the derivative is equal to zero at $y = 0$, $P(-1/2) = P(1/2) < 0$ and $P(y) \\rightarrow +\\infty$ for $y \\rightarrow \\pm \\infty$, it has a maximum at $y = 0$. Hence, it always has 2 real roots symmetrical WRT zero outside of the interval $(-1/2, +1/2)$. Depending on the ratio $p = \\frac{MA}{MD}$, it also can have 2 real roots inside of this interval. The upper limit of this ratio can be obtained when $P(y)$ has a double root equal to zero, i.e., when\n\n$P(0) = \\frac{1}{16} - p^2 q^2 = 0$\n\n$p = \\frac{MA}{MD} = \\frac{4}{q} = \\frac{39^2}{2929} \\doteq 0.51929$\n\nIf $MA = 25$, then the segment $MD$ has to have the length at least $MD = \\frac {MA}{p} = \\frac{25 \\cdot 2929}{39^2} = \\frac{5^2 \\cdot 29 \\cdot 101}{3^2 \\cdot 13^2} \\doteq 48.143$.   \n\n[quote=\"Mildorf\"]In my sloppy effort to find nice numbers, I chose $MD = 39$, which fails to be defined.  I invite you to try this problem with $MD = 113$.[/quote]\nThe nice integral lengths of the segment $MD$ are those, for which the discriminant of $P(y)$, considered as a quadratic polynomial of $y^2$, is a full square. It is a full square for $MD = 39$, but this value is below the lower limit for $MD$. The next (and the last!) nice integral length is $MD = 113$. Then\n\n$a = (1 - p^2) = \\frac{2^4 \\cdot 3 \\cdot 11 \\cdot 23}{113^2}$ \n\n$b = -\\left( 2p^2 q - p^2 r^2 + \\frac 1 2 \\right) = -\\frac{2^3 \\cdot 7^2 \\cdot 26111}{3^2 \\cdot 13^2 \\cdot 113^2}$\n\n$c = \\frac{1}{16} - p^2 q^2 = \\frac{7^2 \\cdot 11 \\cdot 19 \\cdot 41 \\cdot 59 \\cdot 61}{3^4 \\cdot 13^4 \\cdot 113^2}$\n\nThe discriminant is then\n\n$D = b^2 - 4ac = \\left( 2p^2 q - p^2 r^2 + \\frac 1 2 \\right)^2 - 4(1 - p^2) \\left( \\frac{1}{16} - p^2q^2 \\right) = \\left( \\frac{2^8 \\cdot 5^5 \\cdot 7}{3^2 \\cdot 13^2 \\cdot 113^2} \\right)^2$\n\nand the roots of the quadratic equation for $y^2$ are\n\n$y^2 = \\frac{-b \\pm \\sqrt D}{2a} = \\frac{2^3 \\cdot 7^2 \\cdot 26111 \\pm 2^8 \\cdot 5^5 \\cdot 7}{2^5 \\cdot 3^3 \\cdot 11 \\cdot 13^2 \\cdot 23} = \\frac{7^2 \\cdot 26111 \\pm 2^5 \\cdot 5^5 \\cdot 7}{2^2 \\cdot 3^3 \\cdot 11 \\cdot 13^2 \\cdot 23} =$\n\n$ = \\frac{7 \\cdot 11 \\cdot 19 \\cdot 41}{2^2 \\cdot 3^2 \\cdot 13^2 \\cdot 23}$ [color=white].[/color] or [color=white].[/color] $\\frac{7 \\cdot 59 \\cdot 61}{2^2 \\cdot 3^3 \\cdot 11 \\cdot 13^2}$\n\nAs a check,\n\n$y_+^2 \\cdot y_-^2 = \\frac{7 \\cdot 11 \\cdot 19 \\cdot 41}{2^2 \\cdot 3^2 \\cdot 13^2 \\cdot 23} \\times \\frac{7 \\cdot 59 \\cdot 61}{2^2 \\cdot 3^3 \\cdot 11 \\cdot 13^2} = \\frac{7^2 \\cdot 19 \\cdot 41 \\cdot 59 \\cdot 61}{2^4 \\cdot 3^5 \\cdot 13^4 \\cdot 23}$\n\n$\\frac c a = \\frac{7^2 \\cdot 11 \\cdot 19 \\cdot 41 \\cdot 59 \\cdot 61}{3^4 \\cdot 13^4 \\cdot 113^2} \\times \\frac{113^2}{2^4 \\cdot 3 \\cdot 11 \\cdot 23} = \\frac{7^2 \\cdot 19 \\cdot 41 \\cdot 59 \\cdot 61}{2^4 \\cdot 3^5 \\cdot 13^4 \\cdot 23}$\n\nWe need only the smaller root (the one with the minus sign), because the larger one leads to $y$ outside of the interval $(-1/2, 1/2)$, i.e., to the positions of the point $M$ outside of the segment $BE$;\n\n$y_-^2 = \\frac{7 \\cdot 59 \\cdot 61}{2^2 \\cdot 3^3 \\cdot 11 \\cdot 13^2} \\doteq 0.12548 < \\frac 1 4$\n\nas required (checks with the Sketchpad). Returning to the variable $x$,\n\n$x(1 - x) = \\frac 1 4 - y_-^2 = \\frac{3^3 \\cdot 11 \\cdot 13^2 - 7 \\cdot 59 \\cdot 61}{2^2 \\cdot 3^3 \\cdot 11 \\cdot 13^2} = \\frac{2 \\cdot 5^5}{3^3 \\cdot 11 \\cdot 13^2}$\n\nThus we found the value of\n\n$x(1 - x) = \\frac{BM \\cdot ME}{BE^2} = \\frac{MC \\cdot MA}{BE^2}$\n\n[quote=\"Mildorf\"]If the area of $ABE$ is expressed as $\\frac{m}{n}$ with two relatively prime positive positive integers $m$ and $n$, then what remainder is obtained when $m+n$ is divided by 1000?[/quote]\r\n\r\nThe area of this triangle has to be expressed with the help of this ratio and some of the known lengths $MA, MN, MD$.\r\n\r\n$|\\triangle ABE| = |\\triangle ABM| + |\\triangle EBM| = \\left(\\frac{MB}{MC}\\right)^2 |\\triangle CEM| +  \\left(\\frac{MA}{MB}\\right)^2 |\\triangle CBM|$\r\n\r\nbecause the quadrilateral $ABCE$ is cyclic and the triangle pairs $\\triangle ABM \\sim \\triangle CEM$ and $\\triangle EBM \\sim \\triangle CBM$ are similar. The triangles $\\triangle CEM, \\triangle CBM$ have the same altitudes from the common vertex $C$ to the line $BE$ and consequently, their areas are in the ratio of their bases $ME, BM$. Hence\r\n\r\n$|\\triangle ABE| = \\left( \\frac{BM \\cdot ME}{MC^2} + \\frac{MA^2}{BM^2} \\right) |\\triangle CBM| =  \\left( \\frac{MA}{MC} + \\frac{MA^2}{BM^2} \\right) |\\triangle CBM|$\r\n\r\nwhere we used $BM \\cdot ME = MA \\cdot MC$. Again, the triangles $\\triangle CBM$ and $\\triangle CBE$ have the same altitudes from the common vertex $B$ to the line $BE$ and consequently, their areas are in the ratio of their bases $BM, BE$. Hence\r\n\r\n$|\\triangle ABE| = \\left( \\frac{MA}{MC} + \\frac{MA^2}{BM^2} \\right) \\frac{BM}{BE} |\\triangle CBE| = \\left( \\frac{MA \\cdot BM}{MC \\cdot BE} + \\frac{MA^2}{BM \\cdot BE} \\right) |\\triangle CBE|$\r\n\r\nThe area of the triangle $\\triangle CBE$ is equal to\r\n\r\n$|\\triangle CBE| = \\frac 1 2 BE \\cdot BC \\sin{2 \\alpha} = \\frac{\\sin{2 \\alpha}}{2(1 + 2 \\cos{2 \\alpha})} BE^2$\r\n\r\n$|\\triangle ABE| = \\frac{\\sin{2 \\alpha}}{2(1 + 2 \\cos{2 \\alpha})} \\left( \\frac{MA \\cdot BM}{MC \\cdot BE} + \\frac{MA^2}{BM \\cdot BE} \\right) BE^2 = \\frac{\\sin{2 \\alpha}}{2(1 + 2 \\cos{2 \\alpha})} \\left( \\frac{BE \\cdot BM}{MC \\cdot MA} + \\frac{BE}{BM} \\right) MA^2$\r\n\r\nUsing $BM \\cdot ME = MA \\cdot MC$ again,\r\n\r\n$|\\triangle ABE| = \\frac{\\sin{2 \\alpha}}{2(1 + 2 \\cos{2 \\alpha})} \\left( \\frac{BE}{ME} + \\frac{BE}{BM} \\right)\\ MA^2 = \\frac{\\sin{2 \\alpha}}{2(1 + 2 \\cos{2 \\alpha})} \\frac{BE \\cdot BM + BE \\cdot ME}{BM \\cdot ME}\\ MA^2 =$\r\n\r\n$= \\frac{\\sin{2 \\alpha}}{2(1 + 2 \\cos{2 \\alpha})} \\frac{BE^2}{BM \\cdot ME}\\ MA^2 = \\frac{\\sin{2 \\alpha}}{2(1 + 2 \\cos{2 \\alpha})} \\frac{BE^2}{MC \\cdot MA}\\ MA^2$\r\n\r\nwhich is the desired equation for the area of the triangle $\\triangle ABE$. Substituting the values:\r\n\r\n$\\frac{\\sin{2 \\alpha}}{2(1 + 2 \\cos{2 \\alpha})} = \\frac{24}{2(25 + 14)} = \\frac{4}{13}$\r\n\r\n$|\\triangle ABE| = \\frac{4}{13} \\cdot \\frac{3^3 \\cdot 11 \\cdot 13^2}{2 \\cdot 5^5} \\cdot 25^2 = \\frac{2 \\cdot 3^3 \\cdot 11 \\cdot 13}{5} = \\frac{7722}{5} = \\frac m n$\r\n\r\n$m + n \\mod 1000 = 7727 \\mod 1000 = 727$\r\n\r\nAll prime factorizations check all right and the result checks with the Sketchpad. :lol: It is a very nice problem but it seems too difficult for the (Mock) AIME contest.\r\n\r\nRegards, yetti.",
  "Solution_4": "[quote=\"yetti\"]$|\\triangle ABE| = \\frac{12}{39} \\cdot \\frac{3^3 \\cdot 11 \\cdot 13^2}{2 \\cdot 5^5} \\cdot 25^2 = \\frac{2 \\cdot 3^4 \\cdot 11 \\cdot 13}{5} = \\frac{23166}{5}$\n\nI still have to check all those prime factorizations, because the result does not check with the Sketchpad. It is a nice problem but to difficult for the (Mock) AIME contest.\n\nyetti[/quote]\r\n\r\nI'll save you the time: you added a factor of 3 right there.\r\n\r\nMy solution:\r\n[hide]Pythagoras gives $AN = 20$.  We draw $BD$ and $AD$, and construct the altitute $MP$ to $AD$, with $P$ on $AD$, and altitude $MM'$ to $AE$, with $M'$ on $AE$. Because $BC = CD = DE$, angles $BAC$, $CAD$, and $DAE$ are congruent.  Because $P$ is on $AD$, triangles $MNA$ and $MPA$ are congruent by AAS, so $MP = 15$ and $PA = 20$, from which Pythagoras gives $PD = 112$, implying $AD = 132$.\n\nLet $\\alpha = m\\angle{BAC}$, so $m\\angle{MAE} = 2\\alpha$, and $m\\angle{NAE} = 3\\alpha$.  Because we have $\\sin{\\alpha} = \\frac{3}{5}$ and $\\cos{\\alpha} = \\frac{4}{5}$, we compute $\\sin(2\\alpha) = \\frac{24}{25}$, and $\\sin(3\\alpha) = \\frac{117}{125}$.  We find that $MM' = 24$ using $\\sin(2\\alpha) = \\frac{24}{25}$.  By a simple Law of Sines argument $DE$ : $EB$ : $BD = 25$ : $39$ : $40$.\n\nLet $[ABE]$ = the area of $ABE$.\nWe have $[ABE] = 1/2(15 \\cdot AB + 24 \\cdot AE)$.\n\nPtolemy on $ABDE$ yields $AB \\cdot DE + AE \\cdot BD = AD \\cdot BE$.  Using the abundance of facts that we have ascertained previously, this gives:\n\n$AB \\cdot 25x + AE \\cdot 40x  =  132 \\cdot 39x$\n$25AB + 40AE  =  39 \\cdot 132$\n$15AB + 24AE  =  \\frac{39 \\cdot 132 \\cdot 3}{5}$\n\nFinally, $[ABE] = \\frac{1}{2} \\cdot (15AB + 24AE) = \\frac{1}{2}\\frac{39 \\cdot 132 \\cdot 3}{5} = \\frac{7722}{5}$. Therefore, the answer is $722 + 5 = 727$.[/hide]",
  "Solution_5": "[quote=\"Mildorf\"][quote=\"yetti\"]$|\\triangle ABE| = \\frac{12}{39} \\cdot \\frac{3^3 \\cdot 11 \\cdot 13^2}{2 \\cdot 5^5} \\cdot 25^2 = \\frac{2 \\cdot 3^4 \\cdot 11 \\cdot 13}{5} = \\frac{23166}{5}$\n\nI still have to check all those prime factorizations, because the result does not check with the Sketchpad. It is a nice problem but too difficult for the (Mock) AIME contest.\n\nyetti[/quote]\n\nI'll save you the time: you added a factor of 3 right there.[/quote]\nHey, that was just a few seconds before I hit the <Submit> button on my final edit.\n \n[quote=\"Mildorf\"]My solution:\n<snip>\n[/quote]\r\nI suspected that there would be a simpler solution."
}
{
  "Tag": [
    "IMC",
    "college contests"
  ],
  "Problem": "Do not forget to post the problems of IMC 2008.\r\n\r\nThank you",
  "Solution_1": "But where are the problems 4-6 of the first day?",
  "Solution_2": "I typed problems 1-3 yesterday, but I didn't have the paper with me, so I only typed those problems of which I was sure I remembered accurately.",
  "Solution_3": "And the results."
}
{
  "Tag": [
    "geometry",
    "geometric transformation",
    "reflection",
    "inequalities",
    "triangle inequality"
  ],
  "Problem": "Please help me in solving the following two problems which appear to interrelated.\r\n\r\n1. C and D are two points on the same side of a straight line AB.Find a point X on AB such that the angles CXA & DXB are equal.\r\n\r\n2. C and D are two points on the same side of a straight line AB and P is any point on AB. Show that PC + PD is least when angles CPA and DPB are equal.\r\nRegards \r\nPinaki Mishra",
  "Solution_1": "[hide=\"answer to 1\"]\nConnect C to B and D to A. The intersection of these two lines is the point X. Then CXA and DXB are vertical angles, and therefore are equal.\n[/hide]",
  "Solution_2": "[hide=\"2\"]\nLet $D'$ be the point obtained by reflecting $D$ over $AB$.\n$PD=PD'$, so $PC+PD=PC+PD'$.\nClearly, $PC+PD'$ is minimized when $C$, $P$, $D'$ are collinear (which can be proven by triangle inequality).\nIf $P$ is on $CD'$, then $\\angle CPA=\\angle D'PB=\\angle DPB$.\nTherefore, $PC+PD$ is minimized when $\\angle CPA=\\angle DPB$.\n[/hide]",
  "Solution_3": "[hide=\"1\"]Reflect D over AB to D'. X lies on the intersection of CD' and AB, since then $\\angle DXB=\\angle D'XB$, since they are relflections, and $\\angle D'XB=\\angle CXA$ from vertical angles, implying $\\angle DXB=\\angle CXA$.[/hide]\n\n[hide=\"2\"]Reflect D over AB to D'. PC+PD=PC+PD', and PC+PD' is minimum when P lies on CD', since,if P does not lie on PC,  from triangle inequality, PC+PD'>PC. $\\angle CPA=\\angle D'PB$ from vertical angles, and $\\angle D'PB=\\angle DPB$ since they are reflections $\\implies \\angle CPA=\\angle DPB$.[/hide]"
}
{
  "Tag": [
    "LaTeX"
  ],
  "Problem": "Hi all,\r\n\r\nI want to create a small text at the bottom of the first page of a document, like the attachment below. Could you please tell me how to do that? Someone suggested me to use \\footnote{} but it will automatically create a number above the text, and I don't like that, I just wanted to create exactly the same as shown in the attachment.\r\n\r\nThanks for your time :-)",
  "Solution_1": "[code]\\makeatletter\n\\newcommand\\unmarkfootnote[1]{% \n  \\begingroup \n    \\let\\@makefntext\\noindent \n    \\footnotetext{#1}% \n  \\endgroup} \n\\makeatother[/code]Then use \\unmarkfootnote{text} for your unnumbered footnote.",
  "Solution_2": "Incredible!!! This is exactly what I want to do!!! Thanks a lot, dear [b]stevem[/b], you are the  :first:"
}
{
  "Tag": [
    "inequalities",
    "trigonometry",
    "inequalities unsolved"
  ],
  "Problem": "Given $a_i, b_i\\geq 0$ and and $p,q$ reals with $\\frac 1p+ \\frac 1q=1$.\r\nProve that $\\sum_{i=1}^k \\sum_{j=1}^k \\frac {a_i b_j}{i+j} \\leq \\frac {\\pi} {\\sin {\\frac {\\pi}p}} (\\sum_{i=1}^k a_i^p)^{\\frac 1p} (\\sum_{i=1}^k b_i^q) ^{\\frac 1q}$",
  "Solution_1": "i think it has already been posted ..",
  "Solution_2": "http://www.mathlinks.ro/Forum/viewtopic.php?highlight=hilbert+inequality&t=20996"
}
{
  "Tag": [
    "percent"
  ],
  "Problem": "Please help me with these two questions...I've tried everything, but am clueless. I have the chart, but can't figure out the answers. Any help?\r\n\r\n1. Assume that the heights of 18 year old males are normally distributed with a mean of 69 inches and a standard deviation of 6 inches. \r\n\r\na) What % of 18 year old males are between 69 and 75 inches tall? \r\nb) If 1000 18 year old males are selected at random how many will be less than 72 inches tall? \r\n\r\n\r\n2. A placement exam is given to all entering students at a certain College. The scores are normally distributed with a mean of 80 and a standard deviation of 8. \r\na) What percent of the students scored between 74 and 86\r\nb) What percent of the students scored above 92?",
  "Solution_1": "I assume that when you say 'chart' you are referring to a Normal Distribution Table.\r\n\r\nFor the first problem ,we have to convert the info given into standard-normal form, i.e., calculate the z-scores.\r\nWe are given that $\\mu = 69 $ , $\\sigma = 6$. The z-score for some value of a random variable $X$ representing height is given by $z= \\frac{x- \\mu}{ \\sigma}$. for $x=69$ we get a z-score of $z= \\frac{69-69}{6} =0$, and for our transformed random variable $Z$, $P(Z<0)=.5000$ (read this off of your table).\r\nFor $x=75$, we get a z-score of $z= \\frac{75-69}{6} = \\frac{6}{6} =1$, and $P(Z<1)=.8413$. Now we want to find;\r\n$P(69<X<75)=P(0<Z<1)=P(Z<1)-P(Z<0)=.3413$. I.e., $34.13\\%$.\r\n\r\nPart(b) is an easy application of expectation it seems.\r\n\r\nProblem 2 is the same idea as 1(a)-- transform $X$ to $Z$, and calculate the z-scores."
}
{
  "Tag": [
    "email"
  ],
  "Problem": "Someone in my club would like to know about the commitment needed for AYS. If you or someone you know has participated, I'd greatly appreciate being able to pass on the contact information. Use a PM to share emails or phone numbers.\r\n\r\nKiwi!\r\n\r\nBilly",
  "Solution_1": "Do you mean ASYO (Atlanta Symphony Youth Orchestra)?",
  "Solution_2": "You mean Atlanta Youth Symphony Orchestra, right?  If so, the usual acronym is ASYO.  If you use this acronym, everybody who does these things will know exactly what you are talking about.  If you use a different acronym, most people probably won't.\r\n\r\nI go to school with several participants this year.  It is too late to audition for this year, unless your friend has already made it, in which case (s)he should probably do it.  There's a website at [url]http://www.atlantasymphony.org/communityandeducation/youthorchestra/default.aspx[/url];\r\nit has a [url=http://www.atlantasymphony.org/communityandeducation/youthorchestra/asyorehearsalsasyo.aspx]rehearsal schedule[/url].  If your friend is wondering about outside-of-rehearsal commitment, his/her private teacher should be able to give a good answer.  Anyway, making ASYO is a non-trivial feat---there are people I know who didn't get in and probably . . . should have?  (I kind of felt I needed to be able to do math team stuff on Saturdays; I'm doing the [url=http://www.music.emory.edu/eyso]Emory Youth Orchestra[/url].  It is also too late to audition for Emory this year, unfortunately, along with most other youth orchestras in the area.)\r\n\r\n. . . Or did you by chance mean the [url=http://www.music.emory.edu/AYWS/]Atlanta Youth [i]Wind[/i] Symphony[/url] at Emory?  They rehearse every Monday night at Emory for two hours.  Most of what I said about where to find information for ASYO holds here."
}
{
  "Tag": [
    "search"
  ],
  "Problem": "I know I saw a proof on here a while back (probably 4 or 5 months ago now) that proved that 1+1=3. I was just wondering if someone knew where I could find it. I googled it, and also searched through the art of problem solving forum search, but got now results. Thanks\r\n\r\nJosh",
  "Solution_1": "hmm....\r\n\r\nmy solution would be ,,,\r\n\r\n1 drop of water + 1 drop of water = 1 drop of water\r\n\r\napplying this rule\r\n\r\n1 drop of water / 3 = 3 drops of water\r\n\r\n :D",
  "Solution_2": "Do a search for 1=2.  There have been numerous threads on this.  Like [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=1%3D2&t=33143]this[/url]  one for example.",
  "Solution_3": "well, you can prove it by that negative thingy...\r\n\r\nassume: x /= 3\r\nmultiply by 0: 0 /= 0\r\n\r\nbut 0 does = 0\r\n\r\nso x=3, where x is anything (or everything, since everything is a thing)"
}
{
  "Tag": [
    "group theory",
    "abstract algebra",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "Let $ G$ be a group. $ H$ is called a characteristic subgroup of $ G$ if $ f(H)\\equal{}H$ for any automorphism $ f$ on $ G$. Show that if $ H$ is a characteristics subgroup of $ N$ and $ N$ is normal subgroup of $ G$, then prove or disprove that $ H$ is a normal subgroup of $ G$. Thanks in advance.",
  "Solution_1": "For every $ a\\in G$ consider automorphism $ f_a(x): \\equal{}axa^{\\minus{}1}$. Since $ N$ is normal so $ f|_{N}$ is an automorphism of $ N$. So $ H$ is normal subgroup of $ G$."
}
{
  "Tag": [
    "trigonometry",
    "geometry",
    "parallelogram",
    "AoPSwiki",
    "AMC",
    "AIME",
    "search"
  ],
  "Problem": "having trouble with these:\r\n\r\n1. find arccotan(3)\r\n2. find arcsc(1)\r\n\r\ncan someone explain these well?\r\n\r\nalso, can someone please post problems in which you need to use stewart's theorem in order to solve? I need to get better at applying stewart's theorem.",
  "Solution_1": "For the first, memorize your common angles. The second one is $ \\csc^{\\minus{}1}1\\equal{}\\frac{\\pi}{2}.$ You can use a calculator to get the answer to the first.\r\n\r\nDon't really know what you want me to explain...:maybe: I guess you find an angle $ \\theta$ such that $ \\csc\\theta\\equal{}1.$\r\n\r\nFirst of all, do you know what Stewart's Theorem is?\r\n\r\n[hide=\"Problems with Stewarts Theorem\"]\n[hide=\"1)\"]\nShow that if AD is an angle bisector of $ \\triangle ABC$ then \\[ AD^2\\plus{}(BD)(CD)\\equal{}(AB)(AC).\\][/hide]\n\n[hide=\"2)\"]\nSegment $ CX$ divides $ AB$ such that $ AX\\equal{}2BX\\equal{}4.$ If $ AC\\equal{}7$ and $ BC\\equal{}5,$ find $ CX.$[/hide][/hide]",
  "Solution_2": "maybe I should specify... I mean problems involving Stewart's Theorem that are NOT in AoPS volume 2. :P",
  "Solution_3": "It was the only resource I had at the moment :P\r\n\r\n[hide=\"1)\"]\nIn an isosceles triangle with two sides of length $ 17,$ a line with length $ 16$ is drawn from the vertex to the base. If one segment of the base, as cut by the line, exceeds the other by $ 8,$ find the lengths of the base.[/hide]\n\n[hide=\"2)\"]\nProve that the sum of the squares of the distances from the vertex of the right angle, in a right triangle, to the trisection points along the hypotenuse, is equal to $ \\frac{5}{9}$ the square of the length of the hypotenuse.[/hide]\n\n[hide=\"3)\"]\nProve that a triangle with two equal angle bisectors must be isosceles.[/hide]\n\n[hide=\"4)\"]\nProve that the sum of the squares of the sides of a parallelogram is equal to the sum of the squares of the diagonals.[/hide]\n\n[hide=\"5)\"]\nTwo shorter sides of a triangle measure $ 9$ and $ 18.$ If the internal angle bisector drawn to the longest side has length $ 8,$ find the length of the longest side.[/hide]\r\n\r\nI don't believe these are in Volume 2, but I never bothered to check.",
  "Solution_4": "[url=http://www.mathlinks.ro/Wiki/index.php/Special:WhatLinksHere/Stewart's_Theorem]AoPSWiki: Links to Stewart's Theorem[/url]\r\n[url]http://www.google.com/search?hl=en&q=site:mathlinks.ro+stewart+theorem+aime[/url]\r\n[url]http://www.mathlinks.ro/search.php?search_keywords=stewart%20AND%20theorem&return_chars=-1[/url]",
  "Solution_5": "[quote=\"RoFlLoLcOpT\"]It was the only resource I had at the moment :P\n\n[hide=\"1)\"]\nIn an isosceles triangle with two sides of length $ 17,$ a line with length $ 16$ is drawn from the vertex to the base. If one segment of the base, as cut by the line, exceeds the other by $ 8,$ find the lengths of the base.[/hide]\n\n[hide=\"2)\"]\nProve that the sum of the squares of the distances from the vertex of the right angle, in a right triangle, to the trisection points along the hypotenuse, is equal to $ \\frac {5}{9}$ the square of the length of the hypotenuse.[/hide]\n\n[hide=\"3)\"]\nProve that a triangle with two equal angle bisectors must be isosceles.[/hide]\n\n[hide=\"4)\"]\nProve that the sum of the squares of the sides of a parallelogram is equal to the sum of the squares of the diagonals.[/hide]\n\n[hide=\"5)\"]\nTwo shorter sides of a triangle measure $ 9$ and $ 18.$ If the internal angle bisector drawn to the longest side has length $ 8,$ find the length of the longest side.[/hide]\n\nI don't believe these are in Volume 2, but I never bothered to check.[/quote]\r\n\r\nsome were, but these are great! Thanks! Where did you get these anyways?",
  "Solution_6": "Does anyone besides me have an unreasonable hatred for Stewart's Theorem and think that it does not deserve to be called a theorem?",
  "Solution_7": "Well it did have to be proven, did it not?"
}
{
  "Tag": [
    "MATHCOUNTS",
    "AMC",
    "AIME",
    "USA(J)MO",
    "USAMO",
    "AMC 8",
    "AMC 10"
  ],
  "Problem": "Vote above, talk below.\r\n\r\nI got a 41 on chapter (27 sprint, 7 target) I am sad, I goofed up on four problems with silly mistakes :mad:  :(",
  "Solution_1": "Since no one's posting, I'll post. I got 29/16=45... I forgot what y=x^3 looked like :P",
  "Solution_2": "I feel like invading your forum.\r\n\r\nI got a 38(22/16), I know really bad. Placed like I forgot. Then in states I did so much better. :lol:",
  "Solution_3": "LOL stargirl it seemed nice you're posting :lol:",
  "Solution_4": "I will post..... I got 36 and tied for 2nd :rotfl: but after tiebreakers got 5th :(",
  "Solution_5": "29/16",
  "Solution_6": "[quote=\"cognos599\"]29/16[/quote]\r\nshouldn't you be hanging out with yo little indiana homies",
  "Solution_7": "shouldn't you be hanging out with your little california homies? you do realize this is the illinois forum......",
  "Solution_8": "dang isnt this really late to keep this active huh\r\n30/16",
  "Solution_9": "[cough cough] i got a better score on my chapter then on my school mc but i still made state because my team is amazing (river forest roosevelt) :blush:",
  "Solution_10": "wut did u actually get :?:",
  "Solution_11": "don't laugh... :blush: \r\n\r\nthese are my sprint round scores i cant remember my other ones :blush: \r\n\r\nSchool  - 13\r\nChapter- 17\r\nState   - 13 \r\n :blush:",
  "Solution_12": ":rotfl:  :rotfl:  :rotfl: \r\n\r\n(lol, jk)\r\n\r\nHere are my FULL scores:\r\n\r\nChapter: 41\r\nState: 30",
  "Solution_13": "This is the first year that I have been doing any math competitions. Even though there was not a large break between school and chapter competitions I learned a huge amount. \r\nAfter I had the thrill of going to MC state, I decided to join AoPS. It's pretty cool.",
  "Solution_14": "Do AMC 8, and AMC 10/12 A/B if possible.\r\nMake sure you don't overdevote yourself to MC, or in high school you might fry.",
  "Solution_15": "Yes, indeed. :lol:",
  "Solution_16": "[quote=\"dynamo729\"]The year was 1991 (The only year AL won woohoo). I would venture a guess and say bobby or darryl would probably get a perfect if they were to take the same test as MCrawford.[/quote]\r\nDo you know what the person in second got in 1991?",
  "Solution_17": "no.\r\n\r\nbtw did you notice one person on the poll voted 9 or below as chapter score? who the heck was that? :o",
  "Solution_18": "Probably a joke...",
  "Solution_19": "might have been me, i don't remember.",
  "Solution_20": "Aryth did YOU do that as a joke? :o",
  "Solution_21": "[quote=\"gauss1181\"]Aryth did YOU do that as a joke? :o[/quote]\r\nYah remember for the state forum he voted 40-46.   :rotfl:",
  "Solution_22": "lol did he really score that high? :huh:",
  "Solution_23": "No i mean he was joking.",
  "Solution_24": "what did he really get?",
  "Solution_25": "between 35 and 46  :wink:",
  "Solution_26": "Was that in 2007?",
  "Solution_27": "yeah, for eighth grade",
  "Solution_28": "My teacher didn't show us our score!!!!",
  "Solution_29": "Lol for reference this was 9 years ago :)"
}
{
  "Tag": [
    "inequalities proposed",
    "inequalities"
  ],
  "Problem": "let $ (x,y,z) \\geq 0$ , such : $ xyz\\equal{}1$\r\nprove that $ (xy\\plus{}yz\\plus{}xz)^5 \\geq ( (x\\plus{}y)^2\\plus{}(y\\plus{}z)^2\\plus{}(x\\plus{}z)^2)^2 \\frac{27}{16}$",
  "Solution_1": "[quote=\"anas\"]let $ (x,y,z) \\geq 0$ , such : $ xyz \\equal{} 1$\nprove that $ (xy \\plus{} yz \\plus{} xz)^5 \\geq ( (x \\plus{} y)^2 \\plus{} (y \\plus{} z)^2 \\plus{} (x \\plus{} z)^2)^2 \\frac {27}{16}$[/quote]\r\nTry $ a\\equal{}4$ and $ b\\equal{}c\\equal{}\\frac{1}{2}.$ :wink:",
  "Solution_2": "u are right  :blush:  , I made a mistake in my calculation..., but with the same conditions the following is true  :D :\r\n\r\n$ 16(xy\\plus{}yz\\plus{}xz)^4(x^2\\plus{}y^2\\plus{}z^2) \\geq  27 ((x\\plus{}y)^2\\plus{}(x\\plus{}z)^2\\plus{}(y\\plus{}z)^2)^2$ , I hope that u give a solution with SOS  :)",
  "Solution_3": "[quote=\"anas\"]u are right  :blush:  , I made a mistake in my calculation..., but with the same conditions the following is true  :D :\n\n$ 16(xy \\plus{} yz \\plus{} xz)^4(x^2 \\plus{} y^2 \\plus{} z^2) \\geq 27 ((x \\plus{} y)^2 \\plus{} (x \\plus{} z)^2 \\plus{} (y \\plus{} z)^2)^2$ , I hope that u give a solution with SOS  :)[/quote]\r\nIt easy follows from $ (xy \\plus{} xz \\plus{} yz)^2\\geq3(x \\plus{} y \\plus{} z)xyz.$ :wink:",
  "Solution_4": "yes u are right , but I hope to see a solution with SOS :)"
}
{
  "Tag": [
    "geometry",
    "angle bisector",
    "geometry proposed"
  ],
  "Problem": "Problem 2. \r\nLet $ABC$ the triangle, $D$ the point in $BC$ that $BD*AB=DC*AC$.\r\n$M$ and $N$ is midpoints of segments $AD$ and $BC$ respectively .We know that \r\n$\\angle{MBC}+\\angle{MCA}=\\angle{MBA}+\\angle{MCB}$.Prove that circumcentre  of triangle $BMC$ lie on $MN$.",
  "Solution_1": "[quote=\"Tiks\"].Prove that circumcentre  of triangle $BMC$ lie on $MN$.[/quote]\r\nThis big statement is equivalent $\\angle{BMC}=90^o$.",
  "Solution_2": "[quote=\"pavel kozlov\"][quote=\"Tiks\"].Prove that circumcentre  of triangle $BMC$ lie on $MN$.[/quote]\nThis big statement is equivalent $\\angle{BMC}=90^o$.[/quote]\r\nI am not sure !:) ,also can be $BM=CM$ :P  :rotfl: .",
  "Solution_3": "Doesn't the statement BD. AB = DC . AC, \r\nimply that AD is the angular bisector.\r\n\r\nSanjith.",
  "Solution_4": "[quote=\"Sathej\"]Doesn't the statement BD. AB = DC . AC, \nimply that AD is the angular bisector.\n\nSanjith.[/quote]\r\nNo!for angle bisector need this statement $BD*AC=DC*AB$ ;) .\r\nGuys this easy problem still unsloved!:P  :D .",
  "Solution_5": "Tiks can you post the solution for this problem \r\n\r\nDavron",
  "Solution_6": "It's impossible...................... :o \r\nI created this problem one month ago .Is it possible to find this in Armenia :o \r\nI have no comments...........What can I say ?Maybe two people in the world had the same idea..... \r\n\r\nOk. A hint is to draw the angle bisector $AK$ . Then  $BK=DC$ and we can see that \r\n$AK//MN$  and  after some angle chasing or using isogonal point we find that  $\\widehat{ BMC}=90$  and we are done",
  "Solution_7": "Thank you very much! But could you write your full soluiton explaining each point please ...\r\nBecause i am preaparing for the IMO this way...but i could understand everything my self but i need them exlained to learn methods of proof \r\n\r\nThanks\r\n\r\nDavron",
  "Solution_8": "Ara, Tiko inch es erexeqi vra kayfavat linum. Meghqen :( . Chnayac vor kardaci ceric cer es postd, mtqov es artahaytutyun@ ancav: vax hors arev. Ara, es prchot@ ova :huh:  karoxa hesa tam oberkot fra. :mad: Vechni tuftum a u bltcnum vonc vor debil-degenrat lini. Gagiki asac super buta  :rotfl:"
}
{
  "Tag": [
    "inequalities",
    "continued fraction",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "Prove that  there is a positive integer  $n$  such that \r\n\r\n$sin\\ n<10^{-k}$  for any positive integer  $k$  and  $sin\\ n>0$",
  "Solution_1": ":P [b]Observe that: sin(4) = -0.7568 < 0 < 10^(-k) for any k. Thus n = 4 did the job and is the smallest positive integer you get.[/b]",
  "Solution_2": "Sorry I missed a condition .It should be  $sin\\ n>0$",
  "Solution_3": "Silouan:\r\n[b]\nThe TI-83 graphing calculator did the finding and no computer involved and that is a proof for your proposal.\n\nTruthfully it is not worth to use heavy machine to find a much longer proof. You just need to realize that if the\n\ninequality holds for all k > 0 and by letting k tends to infinity it must hold as well. Thus the natural n you seek must\n\nbe such that sin(n) <= 0. One easy way out to see n = 4 is the first of such n's is to draw a unit circle and mark on\n\nthis circle 1, 2, 3, 4, 5,... radians and find the mark that gives you sin negative. This shouldn't be a difficult\n\ntask.\n\n :) [/b]",
  "Solution_4": "See my correction of the problem above .",
  "Solution_5": "According to Hurwitz's theorem, there are infinitely many pairs of integers $(m,n)$ such that $\\epsilon: =|\\pi-n/m|<1/(\\sqrt{5}m^2)$. Because $\\pi>0$, we may assume that $m,n>0$. Now $n=m\\pi-m\\epsilon$, so\r\n\\begin{eqnarray*} |n\\text{ sin }n|&=&|n\\text{ sin }(m\\pi-m\\epsilon)|\\\\ &=&n|\\text{sin }m\\epsilon|\\leq nm|\\epsilon|<\\frac{n}{\\sqrt{5}m}\\\\ &=&\\frac{m\\pi-m\\epsilon}{\\sqrt{5}m}\\\\ &=&\\frac{\\pi-\\epsilon}{\\sqrt{5}}<\\frac{\\pi}{\\sqrt{5}}+\\frac{1}{5m^2}. \\end{eqnarray*}\r\nThus, for any $c>\\pi/\\sqrt{5}$ there are infinitely many positive integers $n$ such that $|n\\text{ sin }n|<c$. Now by choosing $c=10$ you can find infinitely many integers such that $|\\text{sin }n|<10^{-k}$ as $n>10^{k+1}$.\r\n\r\nActually this is not a valid proof. We don't know if $\\text{sin }n>0$. At least I gave some ideas which might lead to a solution.",
  "Solution_6": "All that needs to been proven is that there are positive integers $m$ and $n$ with\r\n$0 < n-m \\pi < 10^{-k}.$\r\nAny all but finitely many of the convergents $n/m$ to the continued\r\nfraction of $\\pi$ satisfy $0 < |n-m \\pi| < 10^{-k}$. These convergents\r\nalternate between being greater and less than $\\pi$."
}
{
  "Tag": [
    "group theory",
    "abstract algebra",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "$K$ a field. If $-1$ is a sum of squares in $K$, let $f(K)$ be the minimal number $n$ such that there exists $(x_1,..,x_n) \\in K^n$ satisfying $x_1^2+x_2^2+..+x_n^2=-1$.\r\nProve that $f(K)$ is a power of $2$.",
  "Solution_1": "nobody ? I found it nice ...",
  "Solution_2": "Your problem is so beautiful.It's true for many fields such that:C , Z/p (p is a prime ) ,....But I didn't find the solution base on the solution for Z/p.If you have the solution can you post it ?",
  "Solution_3": "[quote=\"alekk\"]$K$ a field. If $-1$ is a sum of squares in $K$, let $f(K)$ be the minimal number $n$ such that there exists $(x_{1},..,x_{n}) \\in K^{n}$ satisfying $x_{1}^{2}+x_{2}^{2}+..+x_{n}^{2}=-1$.\nProve that $f(K)$ is a power of $2$.[/quote]\r\n\r\nThis is a consequence of the problem posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=70349]here[/url]. I'll restate it: if $m$ is a power of $2$, then the set of non-zero elements of $K$ which are sums of $m$ squares is a group under multiplication.\r\n\r\nDenote $f(K)$ by $n$, and let $k$ be the unique non-negative integer for which $2^{k}\\le n<2^{k+1}$. Notice that $n+1$ is the smallest $t$ for which there exist $t$ non-zero squares in $K$ which add up to $0$. Let $x_{1}^{2}+\\ldots+x_{2^{k+1}}^{2}=0$ be a non-trivial solution (i.e. not all $x_{i}$ are zero). Since $2^{k}<n+1$, we have $a=x_{1}^{2}+\\ldots+x_{2^{k}}^{2}\\ne 0$. But $x_{2^{k}+1}^{2}+\\ldots+x_{2^{k+1}}^{2}=-a$, so both $a$ and $-a$ are sums of $2^{k}$ squares. According to the problem cited above, $-1$ must then be a sum of $2^{k}$ squares as well (because it's in the multiplicative subgroup of $K^{*}$ generated by $\\pm a$), so $n\\le 2^{k}$. Combined with $2^{k}\\le n$, this gives the desired result: $n=2^{k}$.\r\n\r\n\r\nP.S.\r\n\r\nThat other problem is stated for fields of characteristic different from $2$, but that's all we need anyway. In fact, fields of finite characteristic are not interesting here: in such a field, $-1$ is either a square or a sum of two squares, so there's nothing to prove.",
  "Solution_4": "[quote=\"alekk\"]$K$ a field. If $-1$ is a sum of squares in $K$, let $f(K)$ be the minimal number $n$ such that there exists $(x_{1},..,x_{n}) \\in K^{n}$ satisfying $x_{1}^{2}+x_{2}^{2}+..+x_{n}^{2}=-1$.\nProve that $f(K)$ is a power of $2$.[/quote]\r\n\r\nThis was considered very surprising when Pfister discovered it.  The proof went through several simplifications before reaching the textbooks.  So it can be quite difficult as an exercise if proposed without context or hints."
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "AMC",
    "USA(J)MO",
    "USAMO",
    "ARML"
  ],
  "Problem": "Let $f$ be a polynomial of degree 98, such that $f(k)=\\frac 1k$ for $k=1,2,3\\ldots99.$  Determine $f(100).$",
  "Solution_1": "Ah, a classic... where does this problem come from?",
  "Solution_2": "[quote=\"zabelman\"]Ah, a classic... where does this problem come from?[/quote]\r\n\r\nI'm pretty sure this problem has been posted on this board at LEAST 3 times, in one form or another :).",
  "Solution_3": "This one is from USAMTS(round 4), but I have seen a similar problem from USAMO and BMO.",
  "Solution_4": "And ARML team round.",
  "Solution_5": "..and IMO.\r\n\r\n...and IMO shortlist."
}
{
  "Tag": [
    "geometry"
  ],
  "Problem": "The surface area and volume of a prisim r given find the surfase area and volume of a prisim with (a) double ,(b) triple,(c) half the dimensions of the given prisim? \r\n\r\n1a)                    S.A= 23 ft sq\r\n1b)                    Volume= 12 ft cubed\r\n1c)\r\n\r\n\r\n2a)                     S.A = 61.7 metrs sq\r\n\r\n2b)                     volume+ 28.2 meters cubed\r\n2c)",
  "Solution_1": "http://www.artofproblemsolving.com/Forum/viewtopic.php?t=223308"
}
{
  "Tag": [
    "search",
    "function",
    "MATHCOUNTS",
    "\\/closed"
  ],
  "Problem": "Especially when viewing marathon threads or threads with many hide tags, it is sometimes difficult to find the instances of the searched term. For instance, searching \"balls and urns\" returns a top result of the MATHCOUNTS Marathon thread. Good luck finding where balls and urns is explained in the 230+ pages. :P \r\n\r\nI'm wondering if there could be a feature that allows the viewer to \"jump\" to the first instance of the searched term, opening hide tags if necessary and such. That would be pretty helpful, especially when searching in longer threads.",
  "Solution_1": "You can search by posts instead of threads, which almost always does the job."
}
{
  "Tag": [
    "logarithms"
  ],
  "Problem": "\u039d\u03b1 \u03b4\u03b5\u03af\u03be\u03b5\u03c4\u03b5 \u03cc\u03c4\u03b9:\r\n\r\n$\\frac{lnx}{x}<\\frac{lny}{x}+\\frac{1}{y}$ \u03bc\u03b5 $0<y<x$.",
  "Solution_1": "BOUZOUKMAN,sigura ine etsi i ekfonisi? :maybe: \r\ngrafete isodinama  $\\frac{In\\frac{x}{y}}{x}<\\frac{1}{y}$\r\n diladi $In\\frac{x}{y}<\\frac{x}{y}$\r\n  $\\frac{x}{y}<e^{\\frac{x}{y}}$ pu isxii apo tin gnosti $e^{a}\\geq a+1>a$\r\nali8ia to $y<x$ pu xriazete?\r\n\r\n :D",
  "Solution_2": "\u0391\u03c0\u03cc \u03c5\u03c0\u03cc\u03b8\u03b5\u03c3\u03b7: $\\ln x>\\ln y$. \u0388\u03c7\u03bf\u03c5\u03bc\u03b5:\r\n\r\n$\\frac{\\ln x}{x}<\\frac{\\ln y}{x}+\\frac{1}{y}\\implies\\frac{\\ln x}{x}<\\frac{y\\ln y+x}{xy}$\r\n\r\n\u0391\u03c6\u03bf\u03cd $x>0$, \u03b4\u03b9\u03b1\u03b9\u03c1\u03ce \u03bc\u03b5 $x$ \u03ba\u03b1\u03b9 \u03ad\u03c7\u03c9:\r\n\r\n$\\ln x<\\frac{y\\ln y+x}{y}\\implies y\\ln x<y\\ln y+x\\implies y\\ln\\frac{x}{y}<x\\implies \\ln\\frac{x}{y}<\\frac{x}{y}$ \r\n\r\n\u03c0\u03bf\u03c5 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9. \r\n\r\nOupssssssss... \u03bc\u03b5 \u03c0\u03c1\u03cc\u03bb\u03b1\u03b2\u03b5 \u03bf \u03a1\u03ac\u03bb\u03bb\u03b7\u03c2 !  :P"
}
{
  "Tag": [
    "analytic geometry",
    "number theory",
    "least common multiple",
    "geometry",
    "geometric transformation",
    "vector",
    "homothety"
  ],
  "Problem": "prove that there dont exist a polygon like $A_1A_2A_3A_4...A_{2n+1}$ such that $|A_iA_{i+1}|=1$ ,$A_{2n+2}=A_1$ such that the coordinates of $A_i$be rational.",
  "Solution_1": "I don't think it's true...What about the square with vertices $(0,0),(1,0),(1,1),(0,1)$?\r\n\r\nPierre.",
  "Solution_2": "Oops my bad ,It has odd  vetrices .",
  "Solution_3": "I am not sure but I *think* this might be some sort of descent?",
  "Solution_4": "Ok. So, assume that such a polygon does exist. With an appropriate change of scale (just multiply by the lcm of all the denominators of the coordinates of the vertices), we deduce that all the vertices may be assumed to be integer points, and its side is a positive integer.\r\nThus, there exists a $k$-gon with equal sides with integer length and integer points as vertices.\r\nAmong all of these polygons, let's choose one with the smallest side length, say $d>0$ (recall that $d$ is an integer).\r\nWlog (use translation), we may assume that $A_1 = O$, the origin (I keep the notation $A_1, A_2, \\cdots$).\r\nLet the vector $A_iA_{i+1}$ has coordinates $(a_i,b_i).$\r\nThen $a_i^2 + b_i^2 = d^2$ for each $i$. (1)\r\nMoreover $\\sum_{i=1}^k a_i = \\sum_{i=1}^k b_i = 0.$\r\nIt follows that $(\\sum_{i=1}^k a_i )^2 + ( \\sum_{i=1}^k b_i )^2 = 0$, which leads to $2\\sum_{i \\neq j} (a_ia_j + b_ib_j) = -kd^2.$\r\nSince $k$ is odd, it forces $d$ to be even. Thus $d^2 = 0 \\mod [4]$ and, using (1), we deduce that both $a_i$ and $b_i$ are even, for each $i$. Thus, we may use an homothety with ratio $\\frac 1 2 $ to construct another $k$-gon with equal side and integer points as vertices, but with side length $\\frac d 2$, which contradicts the minimality of $d$. Hence result.\r\n\r\nPierre."
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "calculus",
    "calculus open"
  ],
  "Problem": "x' = f_a(x)\r\nno e.q. pt if a<0\r\n1 e.q. pt if a=0\r\n4 e.q. pts if a>0\r\n\r\nThanks in advance.....",
  "Solution_1": "What are we assumed to do? To find an example of $f_a$ smoothly depending on $a$ with this property?  :?",
  "Solution_2": "yes....",
  "Solution_3": "Well, here is a polynomial: $f_a(x)=(6a^2+1)(a-x^2)+a x^4$.",
  "Solution_4": "thanks fedja....\r\n\r\n[quote=\"fedja\"]Well, here is a polynomial: $f_a(x)=(6a^2+1)(a-x^2)+a x^4$.[/quote]"
}
{
  "Tag": [
    "ratio",
    "complex analysis",
    "complex analysis unsolved"
  ],
  "Problem": "For what values of $ z$ is $ \\sum_{k\\equal{}1}^\\infty(\\frac{z}{1\\plus{}z})^n$ convergent?",
  "Solution_1": "That's a geometric series. Where is the ratio less than $ 1$ in absolute value?"
}
{
  "Tag": [
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "Hi\r\nWhen do 2 elt's in SO(3) commute? Intuitively, I think two elements should commute iff the eigenvectors with eigenvalue 1 of both matrices are the same. I can prove that if this is the case then they commute, but I can't prove it the other way round.\r\nThanks",
  "Solution_1": "[hide=\"Hint\"]The matrices are simultaneously diagonalizable.[/hide]",
  "Solution_2": "1) The matrices are simultaneously diagonalizable by an unitary matrix.\r\n2) If $ r_1\\equal{}Rot(D_1,\\pi),r_2\\equal{}Rot(D_2,\\pi)$ and $ D_1,D_2$ are orthogonal, then $ r_1r_2\\equal{}r_2r_1$."
}
{
  "Tag": [
    "inequalities",
    "calculus",
    "integration",
    "trigonometry",
    "function",
    "calculus computations"
  ],
  "Problem": "Let $I_n=\\int^{\\pi/4}_0\\tan^n\\theta d\\theta$ ,prove that\r\n\r\n$\\frac{1}{2(n+1)}<I_n<\\frac{1}{2(n-1)}$ for natural number , $n>1$\r\n\r\nIm stuck at this one , anyone help?",
  "Solution_1": "1. I(n+2) <  I(n)  < I(n-2) \r\n\r\n2. I(n) = 1/(n-1) - I(n-2)\r\n\r\n3. I(n+2) = 1/(n+1) - I(n)\r\n\r\nUse 2. and 3. to find I(n-2) and I(n+2) in terms of I(n) and write these expressions into inequality 1 which you can then adjust to the required form.\r\n\r\nEquations 2 and 3  come from the  standard reduction formula for the tangent function raised to a positive integral power \r\n.\r\nInequality 1 is true because (TanX)^(n+2)  < (TanX)^n  < (TanX)^(n-2) in the given range of integration."
}
{
  "Tag": [
    "geometry",
    "perpendicular bisector"
  ],
  "Problem": "how many circles do two points determine?",
  "Solution_1": "let's assume that these two points lie on a line.\r\nTherefore, all possible centers of the circles lie on the perpendicular bisector of the line, because the radius to the two points have to be equal (equidistant to the two points)\r\n\r\nHowever, the point has to lie within the circle created when the diameter is the distance between the two points.\r\n\r\nTherefore, there is an infinite amount of circles, but the center has to lie on the perpendicular bisector, and the center has to be within the semicircle made when those points diametrically determine a circle.",
  "Solution_2": "As the $ 15$th definition in the $ 1$st book of Euclid's [u][i]Element[/i][/u], he writes:\r\n[i]\"A circle is a plane figure contained by one line such that all the straight lines falling upon it from one point among those lying within the figure equal one another.\"[/i]\r\nTwo points determine only $ 1$ line.  Therefore, to determine this \"$ 1$ point\" that all of these straight lines fall upon, you must need at least 3 points.  In other words, there are $ \\boxed {infinite}$ circles that go through 2 points."
}
{
  "Tag": [],
  "Problem": "If I already sent in the resume for my application (my school makes us send in a resume, which they send along with the counselor recommendation) and I want to add something, how do I do that? Do I e-mail them or send them a letter? Also, what should I say? The main reason I'm doing this is that I found out that I was a semi-finalists in Siemens and I definately want to let my college know.",
  "Solution_1": "I would just lay out the fact that you were just notified that you are a Westinghouse-Siemens semifinalist, and send a postal mail letter with a photocopy of the notification to the school. Ask your school to be sure what kosher procedure is. I rather suspect that all of the schools to which you are applying receive a copy of the national list of semifinalists, but it can't hurt to be sure that the admission office knows of your selection. \r\n\r\nCongratulations. Good luck on your applications.",
  "Solution_2": "You could also call up the admissions office of whatever colleges you've applied to and politely ask them how you should tell them.  Also, can't you just include it in the portion of your application that you send out personally?",
  "Solution_3": "Thanks! Also, do you know if we should send the entire research report to them along with our application? Do they even read it?",
  "Solution_4": "You could, but I doubt it's likely to make much difference.  I'd think they'll probably take Siemen's word that it's a good paper.  No harm in sending a copy along, I guess.",
  "Solution_5": "Having worked in college offices years ago, my tendency is to minimize the amount of extra papers included with an application. Send what they ask for. Noting that you are a Siemens-Westinghouse semifinalist is likely to be enough; if you knew a faculty member in the school you are applying to who would really be interested in reading the paper, and flagging your application for the admission office's attention, that would be one thing, but the admission office can simply take your work about being a semifinalist, again because there is a verifiable [url=http://www.siemens-foundation.org/docs/semi_finalist_list_2004-05.pdf]national list of semifinalists[/url]. \r\n\r\nThe [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=16376]books I recommend about college admission[/url] explain more about what large piles of paper admissions offices have to deal with."
}
{
  "Tag": [],
  "Problem": "Start at M in the diagram and form a path by moving to an adjacent letter to the right, left, up or down. How many paths spell the word MATH?\n\n[asy]for (int i=0; i&lt;3; ++i) {\nlabel(\"H\",(-3+i, i),W);\nlabel(\"H\",(i,3-i),W);\nlabel(\"H\",(3-i,-i),W);\nlabel(\"H\",(-i,-3+i),W);\n}\nfor (int i=0; i&lt;2; ++i) {\nlabel(\"T\",(-2+i, i),W);\nlabel(\"T\",(i,2-i),W);\nlabel(\"T\",(2-i,-i),W);\nlabel(\"T\",(-i,-2+i),W);\n}\nlabel(\"A\",(0,1),W);\nlabel(\"A\",(0,-1),W);\nlabel(\"A\",(1,0),W);\nlabel(\"A\",(-1,0),W);\nlabel(\"M\",(0,0),W);[/asy]",
  "Solution_1": "WLOG pick the top A (otherwise, rotate). Then there are two \"side T\"s that have 2 H's, and one \"corner T\" with 3 H's. 2*2+3=7, 7*4=28.",
  "Solution_2": "Note that there is only one path from the central M to the H's on the exact top, right, left, and bottom. Also, there are 3 paths to each of the remaining H's. Then the sum is 4*1 + 8*3 = 28.",
  "Solution_3": "... sorry dont understand. i in 5th grade, gifted",
  "Solution_4": "hint: separate it into 4 quadrants and then count the number of ways in each quadrant",
  "Solution_5": "[quote=sap2018]... sorry dont understand. i in 5th grade, gifted[/quote]\n\nwhat don't you understand? It seems simple enough to me.\n\n",
  "Solution_6": "[quote=cowcow][quote=sap2018]... sorry dont understand. i in 5th grade, gifted[/quote]\n\nwhat don't you understand? It seems simple enough to me.[/quote]\n\nDo not discourage other learners.\n\n[hide=Solution]The strategy we will use for this problem is multiplying individual cases of counting to find a group case. We start at $M$. Starting at $M$, there are $4$ possibilities of $A$'s we can move too. From each of those $A$'s, there are $3$ possibilities of $T$'s we can move to. From each of the $T$'s, there are $3$ possibilities of $T$ we can move to. Therefore, the solution is $4\\times3\\times3=48$ possible paths.\n\nTry a harder problem [hide=here]look below[/hide][/hide]"
}
{
  "Tag": [],
  "Problem": "The Sprint Round competition consists of 30 problems with a time limit of 40 minutes. You completed 20 Sprint Round problems in 25 minutes. On average, how many times longer will you be able to spend on each remaining problem than you did on each of the first 20? Express your answer as a common fraction.",
  "Solution_1": "[hide]My first 20 were done at 5/4 of a minute each.\n\nNow I have 10 left, and 15 minutes.\nSo Now I can spend 3/2 of a minute each. \n3/2 divided by 5/4 = 6/5[/hide]",
  "Solution_2": "[hide=\" :nhl: \"]\nYou spent 25 min on 20 problems\n\nYou have 15 min left for 10 problems.\nThat is 90 seconds per problem.\n\nFor the first 20, you spent 25 min=1500 seconds.\nPer problem, that is 75 seconds.\n\n$90/75=\\frac 65$[/hide]",
  "Solution_3": "[hide]there's 15 minutes left\nthere are 10 problems left\n15min/10problems= 3/2 minute per problem\n\noriginally you did 25min/20problems= 5/4 minutes per problem\n\n3/2=5/4*x\n3/2*4/5=x\n\nx=6/5[/hide]"
}
{
  "Tag": [
    "AMC",
    "AIME"
  ],
  "Problem": "What/who do you guys think impacts math motivation the most? I mean like, do you think most people get motivated into math because of their parents, teachers, friends, or themselves?\r\n\r\nAlso, what kinds of factors might be the leading cause for losing that motivation? I'm wondering about this because I have a friend who used to always improve herself and incline herself to study for math contests when she was in her early middle school years, she was constantly doing very well. But now she's sorta having a burnout-losing her grip on practicing, and she doesn't really seem to be doing much math practice.\r\n\r\nAny ideas of how to help someone like this?",
  "Solution_1": "Why do you think not being motivated to seriously invest oneself in math contests is something that needs fixing?\r\n\r\nThis is a very common and natural phenomenon, and I'll attribute it to two reasons:\r\n\r\n[b](1) Increasing opportunity cost:[/b]\r\nThe direct cost of mathematics is paper and ink, but the opportunity cost is whatever you could be doing with the time you spend. In early middle school, this opportunity cost is very small -- there's not as much else you can be doing with your time. As you grow into high school, and then into college, the opportunity cost goes higher and higher. You could be working, interning, doing research/personal projects, exploring fields of potential interest, volunteering, building friendships, taking care of siblings, etc.\r\n\r\n[b](2) Decreasing value:[/b]\r\nIn middle school, competition mathematics is basically right in line with any interests in math, science, and engineering. As you mature, your interests become more specific, and the fields themselves broaden at a rapid pace. By the time college comes around, a chemistry major won't find the same appeal in pure mathematics anymore. But they might've been in Kumon back in middle school.",
  "Solution_2": "I personally think that a really good/nice/awesome math teacher helps create a liking for math.  And yes, if you're reasonably good at math and all your friends are participating in competitive math, then that would also motivate you to do it as well.  \r\n\r\nHmm.  There are (see previous post) quite a few reasons that some one would stop doing competitive math.  Another reason might be that the person lost interest in it.  I'm in middle school still, so I don't know that much about high school competitions.  But I'm getting the impression that they're EXTREMELY hard.  Failing stuff demotivates people a lot.",
  "Solution_3": "The thing that is motivating me to do math right now is the hunger to be able to participate in the International Mathematical Olympiad, and that is all I am looking forward to. (I have quite a way to go, I'm not even in the AIME level yet, but the feeling for the Olympiad has cornered me.) :)",
  "Solution_4": "I think one important motivation is the knowledge. As much as you know more, you like it more. You make associations between different topics, you understand what lies behind, you know where you stand and so on.\r\nThis is valid not only for mathematics, but for other disciplines.\r\n Another factor is the medium. In a medium that is predisposed to defeatism, or where knowledge/science/math is downplayed, the average person will be less likely to grow interested on these subjects.\r\nThese two factors are example of internal versus external motivations.\r\nI find TZF's post quite interesting.",
  "Solution_5": "For me, I feel that having a peer is extremely important. When I run a race in track, if someone is in front of me, then I feel that I have to work harder to catch up, but when I can't see anyone else, I feel more lax. In a similar way, I believe having a person that you compete with will motivate you. This doesn't just apply to math, I think this applies to everything you do."
}
{
  "Tag": [],
  "Problem": "these are some exercises that my teacher has given me. however it's quite difficult and i want to see good solutions from you! ^^\r\n1. Given $ f: R\\rightarrow R$ that\r\n$ f(u \\plus{} v) \\plus{} f(u \\minus{} v) \\equal{} 2f(u)f(v);$with any $ u,v\\in R$\r\na) prove that f is \r\nb) Quality $ f(0)$ to prove that $ f(x) \\geq \\minus{} 1;$ with any $ x\\in R$\r\n2. Given $ f: R\\rightarrow R$ that\r\n$ f(u \\plus{} v) \\plus{} f(u \\minus{} v) \\equal{} 2f(u)g(v);$with any $ u,v\\in R$\r\n$ f\\not \\equal{} 0; \\left|f(x) \\right|\\leq 1$\r\nprove that $ \\left|g(x) \\right|\\leq 1$",
  "Solution_1": "This belongs in the Olympiad section. Maybe Pre-Olympiad, but certainly not High School Basics.",
  "Solution_2": "Hello!\r\n1/a) there is no  a question  :?: \r\n\r\nb/ $ (u,v) \\equal{} (0,0) \\Rightarrow f(0) \\equal{} f(0)^2 \\Rightarrow f(0) \\equal{} 0$ or $ f(0) \\equal{} 1$ \r\n\r\ncase 1 : $ f(0) \\equal{} 0$\r\n\r\n$ (u,v) \\equal{} (\\frac {x}{2} ,\\frac {x}{2}) \\Rightarrow f(x) \\equal{} 0 > \\minus{} 1$\r\n\r\ncase 2 : f(0)=1\r\n\r\n$ (u,v) \\equal{} (\\frac {x}{2} ,\\frac {x}{2}) \\Rightarrow f(x) \\equal{} 2f^2(\\frac {x}{2}) \\minus{} 1 \\geq    \\minus{} 1$",
  "Solution_3": "Oh! I'm sorry. because there is a word in the first question that i can't translate into English.I am Vietnamese. I don't know exactly that word but it means bounded or round function. Do you know it?\r\n :?:  :!:"
}
{
  "Tag": [
    "logarithms",
    "function",
    "calculus",
    "calculus computations"
  ],
  "Problem": "find the antiderivative of e^(2ln u) du",
  "Solution_1": "$ e^{2 \\ln u} \\equal{} u^2$",
  "Solution_2": "$ e^{2\\ln u} \\equal{} u^{2}$, $ u > 0$\r\n(Just to be picky)",
  "Solution_3": "$ 2\\ln u \\equal{} \\ln u^2$ so it makes no difference if $ u > 0$ or $ u < 0$ so long as $ u\\neq0$.\r\n\r\n(Just to be more picky) :wink:",
  "Solution_4": "[quote=\"Rajiv\"]$ 2\\ln u \\equal{} \\ln u^2$[/quote]\r\nYes, but the LHS is only well-defined as a real function (and therefore this property is only true) when $ u > 0$.  ;) ;)",
  "Solution_5": "Okay!\r\n\r\n[hide=\"And...\"]$ 2\\ln x$ exists only in first and fourth quadrant while $ \\ln u^2$ exists in all four quadrants and is the mirror image of $ 2\\ln x$ about the Y-Axis.[/hide]",
  "Solution_6": "$ \\ln x^2\\equal{}2\\ln |x|\\ (x\\neq 0)$."
}
{
  "Tag": [
    "email"
  ],
  "Problem": "This is the first year that my school is doing Mandelbrot (I got them into it  :D ), so we ordered a contest book. It says in the beginning that w/o expressed permission from the writers, we arent allowed to make photocopies, etc. Well, considering that you were one of hte writers, can we (pretty please  8-) ) make photocopies so that we dont have to use the book 1 at a time? Also, am I allowed to post select problems that I enjoy from the book onto this site?",
  "Solution_1": "You can post selected problems on this site (I doubt even copyright law could stop that even if problem authors wanted to).  As for photocopying, I'm not comfortable answering that question as that book is nearly entirely the work of Sam Vandervelde (Sandor and I only contributed to the first year or two of that spiral book).  You can email him at info@mandelbrot.org.  I suspect his policy will be much like the one we explicitly state in AoPS - photocopying of bits and pieces for educational purposes (as I'm guessing you're planning) is ok, but reproducing the entire book is not (and probably not economic, at any rate)."
}
{
  "Tag": [
    "function",
    "complex numbers",
    "complex analysis"
  ],
  "Problem": "having trouble with this question:\r\n\r\ngiven sinhz, show that u & v are harmonic conjugates.\r\n\r\nu = real part of sinhz          = sinhxcosy\r\nv = imaginary part of sinhz = icoshxsiny\r\n\r\nhow do i show that these are harmonic conjugates?",
  "Solution_1": "Harmonic conjugates get me going every time.\r\n\r\nI can feel your pain!\r\n\r\n :o"
}
{
  "Tag": [
    "Euler",
    "algorithm",
    "modular arithmetic",
    "number theory",
    "Euclidean algorithm",
    "number theory proposed"
  ],
  "Problem": "What's a smart way to calculate $ 3^{15^{62944331} \\plus{} 2} \\mod 37$\r\n\r\n[hide]\nI set $ e \\equal{} 15^{62944331} \\plus{} 2$, then wanted to do $ e \\mod \\phi(37) \\equal{} e \\mod 36$, but then $ 15^{62944331} \\plus{} 2 \\mod 36$ is nasty again, so I looked for $ 15^k$ where $ k \\equal{} 62944331 \\mod \\phi(36) \\equal{} 62944331 \\mod 12$, which is nasty as well (I can only use euclidean algorithm here)\n\n[/hide]",
  "Solution_1": "[color=darkblue]I think the most important idea is $ 27 \\cdot 37 \\equal{} 999$\n$ 3^{15^{62944331}\\plus{}2} \\equal{} 3^{9k \\plus{} 2}$. Notice that $ k$ is odd.\nBut $ 3^{9k} \\equal{} 27^{3k}\\equiv ( \\minus{} 10)^{3k} \\equal{} ( \\minus{} 1000)^k \\equal{} \\minus{} 1000^k\\equiv \\minus{} 1^k \\pmod{37}$. So $ \\boxed{3^{9k \\plus{} 2}\\equiv \\minus{} 9 \\equiv 28 \\pmod{37}}$.[/color]"
}
{
  "Tag": [
    "probability",
    "linear algebra",
    "matrix",
    "symmetry",
    "function",
    "induction"
  ],
  "Problem": "The elements of a determinant are arbitrary integers. Determine the probability that the value of the determinant is odd.",
  "Solution_1": "You mean the elements of a matrix (not a determinant) are integers...\r\n\r\nanyway, the probability must be one half by symmetry; assume it isn't. The function of calculating a determinant is as symmetric as possible, except for one thing: the choice between positive and negative (i.e. letting the top element on the left be first, so positive when calculating the determinant of its mini matrix... as opposed to negative)\r\nbut negative/positive is symmetric with respect towards even/odd (except zero maybe-but that particular case has zero probability...)\r\nso it must be symmetric\r\n\r\nthat was a really bad explanation, sorry, tell me if I'm wrong",
  "Solution_2": "can't you start with a 2 by 2 and prove it by induction?",
  "Solution_3": "[quote=\"me@home\"]You mean the elements of a matrix (not a determinant) are integers...\n\nanyway, the probability must be one half by symmetry; assume it isn't. The function of calculating a determinant is as symmetric as possible, except for one thing: the choice between positive and negative (i.e. letting the top element on the left be first, so positive when calculating the determinant of its mini matrix... as opposed to negative)\nbut negative/positive is symmetric with respect towards even/odd (except zero maybe-but that particular case has zero probability...)\nso it must be symmetric\n\nthat was a really bad explanation, sorry, tell me if I'm wrong[/quote]\r\n\r\nEr, you may be right, but I don't think that \"the formula for a determinant is rather symmetric\" is a sufficient solution for the problem. Anyhow, I just worded it as it is worded in the book.\r\n\r\nEDIT: it's actually not true. The probability is $1/3$ for a 2 by 2 matrix.",
  "Solution_4": "[quote=\"Phelpedo\"]EDIT: it's actually not true. The probability is $1/3$ for a 2 by 2 matrix.[/quote]\r\n\r\nI got $\\frac{3}{8}$ for a 2 by 2 :maybe:",
  "Solution_5": "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=determinant+odd&t=43828]look here[/url]\r\n\r\ne: Yes, the answer is $3/8$ for a matrix in $\\mathcal{M}_{2}(\\mathbb{Z}/2\\mathbb{Z})$",
  "Solution_6": "Darn, I must have miscounted one case  :blush:"
}
{
  "Tag": [
    "USAMTS",
    "group theory"
  ],
  "Problem": "I've just begun studying a little bit of group theory and abstract algebra.  I was wondering if there are any math contest questions (at the high school level) where group theory would provide a simple, or at very least elegant, solution to the problem.",
  "Solution_1": "Not really. Good contest-writers want to avoid giving an advantage to people who have studied more advanced math, so if a problem can be solved very easily with group theory but not easily with other methods, then it isn't a very good problem for a high school contest.\r\n\r\nHaving said that, there's a problem from USAMTS 2001-2002 (the one about BAABAABAA) that I solved using, basically, group theory. I went overboard on that problem, using ninth roots of unity I think. I knew it could be done modulo 3, but I liked my solution more. That didn't stop the grader from saying something along the lines of ``You know, you could have done that by looking at it modulo 3.''"
}
{
  "Tag": [],
  "Problem": "Many Math Counts and/or AMC problems I get wrong are problems of this type...\r\n\r\n1. The median and the mean of a set of 5 integers is 8. What is the largest/smallest number that can be in this set?\r\n\r\n2. The median of a set of 7 integers is 10. The mean is 13. What is the largest/smallest number that can be in this set?\r\n\r\nYup....",
  "Solution_1": "OK, don't worry, these problems can be really easily done if you know the technique. The thing u must understand is average multiplied by the number of integers in the set equals the sum of all the numbers in the set. I'll give you an example from your own problem. So for #1 you said the average is $ 8$ and the number of integers in the set is $ 5$. This must mean that the sum of all the integers in the set is $ 8*5$ or $ 40$. AFTER we know that, we can see what your problem is asking. First of all your problem looks like this {a,b,c,d,e} where $ a\\plus{}b\\plus{}c\\plus{}d\\plus{}e\\equal{}40$ and so we can assume a and b being $ 1$ and $ c\\equal{}8$ because it said the median is 8 as well. This must mean $ d\\equal{}9$ because you are trying to get \"e\" to be highest as possible, so u must make all the other variables to be small as possible. Sooo we have the equation $ 1\\plus{}1\\plus{}8\\plus{}9\\plus{}e\\equal{}40$ and you solve for \"e\" to get $ e\\equal{}21$. Thus, the greatest possible number that can be in such a set is 21!! Ask me if your unclear with anything. Using this technique, solve the next problem.",
  "Solution_2": "On the contrary, as the numbers need not be distinct, the maximum is 22 (1,1,8,8,22 achieves this).\r\n\r\n\r\nIf however they all had to be different, it would be (1,2,8,9,20)\r\n\r\n\r\nOf course, this assumes the problem meant positive integers only."
}
{
  "Tag": [
    "geometry open",
    "geometry"
  ],
  "Problem": "I have a question, is the following problem true?\r\nProblem: If the hexgon ABCDEF has AD,BE,CF are concurrent, then six segments of that hexagon all touch a conic.\r\nHope to receive your replies soon.",
  "Solution_1": "yes it's true.",
  "Solution_2": "Yes???\n\nIs it true????"
}
{
  "Tag": [
    "geometry",
    "trigonometry",
    "parallelogram",
    "algebra",
    "polynomial",
    "perpendicular bisector",
    "Pythagorean Theorem"
  ],
  "Problem": "Let $\\triangle{ABC}$ be a right triangle with $\\angle{C}=90^{\\cir}$ and $D$ be the midpoint of $BC$. If $AC=BC=6$ and $E$ , $F$ are two points on sides $AC$ and $AB$,respectively, so that $\\triangle{AEF}$ is congruent to $\\triangle{DEF}$, find the area of quadrilateral $AEDF$. (without using trigonometrics )\r\n[img]http://i5.photobucket.com/albums/y194/jelly119/001-1.jpg[/img]",
  "Solution_1": "See the attached diagram.\r\n\r\n$AFDE$ can be either a parallelogram or a kite. If it's a parallelogram, then the problem is quite trivial and the answer is $[AFDE]={a^{2}\\over 4}=9$\r\n\r\nIf it's a kite, then $EF$ is the perpendicular bisector of $AD$. Let $M$ be the midpoint of $AD$. Draw $DN\\perp AB$. Then $EM: AM=CD: CA\\implies EM={1\\over 2}AM={1\\over 4}AD$, and $FM: MA=DN: NA$, where $DN=\\frac{DB}{\\sqrt{2}}=\\frac{a}{2\\sqrt{2}}$ and $NA=AB-BN=a\\sqrt{2}-\\frac{a}{2\\sqrt{2}}=\\frac{3a}{2\\sqrt{2}}$, hence $FM: MA=1: 3\\implies FM={1\\over 3}MA={1\\over 6}AD$. Therefore $EF=EM+FM={5\\over 12}AD$. On the other hand, Pythagorean theorem on $\\triangle ACD$ gives $AD={a\\sqrt{5}\\over 2}\\implies EF=\\frac{5a\\sqrt{5}}{24}$. Now $[AFDE]={1\\over 2}AD\\cdot EF={25a^{2}\\over 96}={75\\over 8}$",
  "Solution_2": "And can you guys believe that this is a problem in our midterm for 8 grades... :wacko:",
  "Solution_3": "Well, essentially it's not THAT complicated... Some basic similarity, Pythagorean Theorem... You just have to spot a thing or two - I would expect this to be solved by an 8-grader who knows his maths.",
  "Solution_4": "a teenie tiny mistake: $[AEDF]=\\frac{75}{4}$.",
  "Solution_5": "$\\frac{25\\cdot 36}{96}=\\frac{25\\cdot 6}{16}=\\frac{25\\cdot 3}{8}$\r\n\r\nBesides that, ${75\\over 4}$ exceeds the area of the triangle itself, which is $18$.",
  "Solution_6": "i'm dumb  :blush:",
  "Solution_7": "[quote=\"Farenhajt\"]Well, essentially it's not THAT complicated... Some basic similarity, Pythagorean Theorem... You just have to spot a thing or two - I would expect this to be solved by an 8-grader who knows his maths.[/quote]\r\n\r\nBut our 8-graders haven't learned the concepts and skills of similarity yet!  :(   Would it be possible that  there is any trick about this problem??",
  "Solution_8": "[quote=\"jelly119\"][quote=\"Farenhajt\"]Well, essentially it's not THAT complicated... Some basic similarity, Pythagorean Theorem... You just have to spot a thing or two - I would expect this to be solved by an 8-grader who knows his maths.[/quote]\n\nBut our 8-graders haven't learned the concepts and skills of similarity yet!  :(   Would it be possible that  there is any trick about this problem??[/quote]\r\n\r\nall you do is use the pythagorean theorem a bunch of times...what is the big deal? FE is the perpendicular bisector of AD",
  "Solution_9": "[quote=\"Altheman\"]all you do is use the pythagorean theorem a bunch of times...what is the big deal? FE is the perpendicular bisector of AD[/quote]\r\n\r\nI would be grateful to know the details about your no-big-deal solutions.",
  "Solution_10": "There's another approach which circumvents similarity but relies on equations and basic polynomial skills.\r\n\r\nDenote $x=AE$. Then $CE=a-x, ED=x$, hence by Pythagorean we have\r\n\r\n\\begin{eqnarray*}(a-x)^{2}+\\left({a\\over 2}\\right)^{2}=x^{2}&\\iff& a^{2}-2ax+x^{2}+{a^{2}\\over 4}=x^{2}\\\\ &\\iff&{5a^{2}\\over 4}=2ax\\\\ &\\iff& x={5a\\over 8}\\end{eqnarray*}\r\n\r\nThen find $EM$ by applying Pythagorean on $\\triangle AME$ - the answer is $\\frac{a\\sqrt{5}}{8}$ (keep in mind, of course, that $AM={AD\\over 2}$ and $AD$ is calculated from $\\triangle ACD$)\r\n\r\nSimilarly, denote $y=AF$. In the same manner as in my previous solution, find that $AN=\\frac{3a}{2\\sqrt{2}}, DN=\\frac{a}{2\\sqrt{2}}$, which gives\r\n\r\n\\begin{eqnarray*}\\left(\\frac{3a}{2\\sqrt{2}}-y\\right)^{2}+\\left(\\frac{a}{2\\sqrt{2}}\\right)^{2}=y^{2}&\\iff&{9a^{2}\\over 8}-{3a\\over\\sqrt{2}}y+y^{2}+{a^{2}\\over 8}=y^{2}\\\\ &\\iff&{5a^{2}\\over 4}={3a\\over\\sqrt{2}}y\\\\ &\\iff& y=\\frac{5a\\sqrt{2}}{12}\\end{eqnarray*}\r\n\r\nThen by Pythagorean on $\\triangle AFM$ find $FM=\\frac{a\\sqrt{5}}{12}$.\r\n\r\nFrom there proceed to finding $EF=EM+FM$, as above.",
  "Solution_11": "[quote=\"Farenhajt\"]There's another approach which circumvents similarity but relies on equations and basic polynomial skills.\n\nDenote $x=AE$. Then $CE=a-x, ED=x$, hence by Pythagorean we have\n\n\\begin{eqnarray*}(a-x)^{2}+\\left({a\\over 2}\\right)^{2}=x^{2}&\\iff& a^{2}-2ax+x^{2}+{a^{2}\\over 4}=x^{2}\\\\ &\\iff&{5a^{2}\\over 4}=2ax\\\\ &\\iff& x={5a\\over 8}\\end{eqnarray*}\n\nThen find $EM$ by applying Pythagorean on $\\triangle AME$ - the answer is $\\frac{a\\sqrt{5}}{8}$ (keep in mind, of course, that $AM={AD\\over 2}$ and $AD$ is calculated from $\\triangle ACD$)\n\nSimilarly, denote $y=AF$. In the same manner as in my previous solution, find that $AN=\\frac{3a}{2\\sqrt{2}}, DN=\\frac{a}{2\\sqrt{2}}$, which gives\n\n\\begin{eqnarray*}\\left(\\frac{3a}{2\\sqrt{2}}-y\\right)^{2}+\\left(\\frac{a}{2\\sqrt{2}}\\right)^{2}=y^{2}&\\iff&{9a^{2}\\over 8}-{3a\\over\\sqrt{2}}y+y^{2}+{a^{2}\\over 8}=y^{2}\\\\ &\\iff&{5a^{2}\\over 4}={3a\\over\\sqrt{2}}y\\\\ &\\iff& y=\\frac{5a\\sqrt{2}}{12}\\end{eqnarray*}\n\nThen by Pythagorean on $\\triangle AFM$ find $FM=\\frac{a\\sqrt{5}}{12}$.\n\nFrom there proceed to finding $EF=EM+FM$, as above.[/quote]\r\n\r\nYep...that's the way I taught my 8-grades brother! But the equations really look terrible for an ordinary student...With the concepts of similariy in your previous solutions would be much better!!",
  "Solution_12": "Hey, are you sure that the creators of the problem didn't have a PARALLELOGRAM on their minds? :)",
  "Solution_13": "[quote=\"Farenhajt\"]Hey, are you sure that the creators of the problem didn't have a PARALLELOGRAM on their minds? :)[/quote]\r\n\r\nThey gave the correct answer $\\frac{75}8$. So I don't think that's an \"accident\". :wink: As I know, no one got the right answer in that test!! After all, there are many other complicated problems -for an 8-grader- to be done in 45 minutes!!"
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "Prove that: nonnegative real numbers $ x1 x2 ... xn$ satisfy the following inequality:\r\n\r\n$ (x1x2...xn)(1/x1 \\plus{} 1/x2 \\plus{} ... \\plus{} 1/xn)<\\equal{}(x1)^(n\\minus{}1)\\plus{}(x2)^(n\\minus{}1)\\plus{}... \\plus{}(xn)^(n\\minus{}1)$",
  "Solution_1": "You mean:\r\n$ (x_1\\cdot x_2\\cdot \\ldots \\cdot x_n)(\\frac{1}{x_1}\\plus{}\\frac{1}{x_2}\\plus{}\\ldots \\plus{}\\frac{1}{x_n})\\le (n\\minus{}1)(x_1\\plus{}x_2\\plus{}\\ldots \\plus{}x_n)$?",
  "Solution_2": "[quote=\"Bugi\"]You mean:\n$ (x_1\\cdot x_2\\cdot \\ldots \\cdot x_n)(\\frac {1}{x_1} \\plus{} \\frac {1}{x_2} \\plus{} \\ldots \\plus{} \\frac {1}{x_n})\\le (n \\minus{} 1)(x_1 \\plus{} x_2 \\plus{} \\ldots \\plus{} x_n)$?[/quote]\r\nThat is wrong, so I think it is: \r\n\r\n$ (x_1\\cdot x_2\\cdot \\ldots \\cdot x_n)(\\frac {1}{x_1} \\plus{} \\frac {1}{x_2} \\plus{} \\ldots \\plus{} \\frac {1}{x_n})\\le x_1^{n \\minus{} 1} \\plus{} x_2^{n \\minus{} 1} \\plus{} \\ldots \\plus{} x_n^{n \\minus{} 1}$\r\n\r\nWhich is just rearrangement, muirhead or AM-GM :)",
  "Solution_3": "Yes, (x_1)^(n-1) appears as $ (x_1)^(n \\minus{} 1)$",
  "Solution_4": "Could anybody show how to prove this inequality directly from AM-GM inequality?",
  "Solution_5": "[quote=\"matex (L)(L)(L)\"]Could anybody show how to prove this inequality directly from AM-GM inequality?[/quote]\r\n$ \\frac{x_1^{n\\minus{}1} \\plus{} .... \\plus{} x_{n\\minus{}1}^{n\\minus{}1}}{n} \\ge x_1x_2...x_{n\\minus{}1}$. Just do this $ n$ times and add up!"
}
{
  "Tag": [
    "inequalities",
    "email",
    "calculus"
  ],
  "Problem": "prove that:\r\n\r\na^4+b^4+c^4+d^4>=(a^2)bc+(b^2)cd+(c^2)da+(d^2)ab",
  "Solution_1": "[hide=\"solution\"]with am-gm, i get :\n$a^{4}+a^{4}+b^{4}+c^{4}\\geq 4a^{2}bc$. then i sum up cyclically this inequality, and the problem is solved.[/hide]",
  "Solution_2": "[quote=\"maky\"][hide=\"solution\"]with am-gm, i get :\n$a^{4}+a^{4}+b^{4}+c^{4}\\geq 4a^{2}bc$. then i sum up cyclically this inequality, and the problem is solved.[/hide][/quote]thank you for your answer you are so smart :wink:  :P \r\nplease email me abook about calcules if you have.\r\nif you have. :wink:"
}
{
  "Tag": [
    "\\/closed"
  ],
  "Problem": "Recently, my post rating has decreased dramatically, but I don't know which of my posts are being rated as spam. I have never intentionally spammed and always keep my posts as high-quality as possible. Thus, if I really am spamming, may somebody please be kind enough to mention to me what is wrong with my posts? I have never intended to do anything harmful; I love this site.  :)  But, really, I wish to know if I'm doing something wrong. Sorry for any inconveniences I may have caused if I have really spammed. Thanks in advance.",
  "Solution_1": "[b] @r15s11z55y89w21 [/b] : Hello, What is post rating? How can it decrease/increase? What was your post rating before, and what is it now? \r\n\r\nI'm new to this site, so I want to know about posts and post rating.\r\n\r\nThank you.",
  "Solution_2": "Post rating information can be found [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=146659]here[/url]. Remember to read the stickies for the forum before posting on any forum.",
  "Solution_3": "OK, post ratings are somekind of important, but if you scroll down in this forum, you'll find several topics related to post ratings. Why are people so obsessed? Of course, r15... is right, when he's intentionally falling because of someone, that should be reported... But there are dozens of topics on this subject, and just PM an admin...\r\n\r\nAnd it would be nice (I have no skills to tell the simplicity of that) that the poster can see who rated him and how...",
  "Solution_4": "We will enable the ability to see which of your posts were rated in the future. However it is very unlikely we will give you the ability to [b]see who[/b] rated your posts and how he rated them (it would be a flame war ignition sequence)",
  "Solution_5": "[b] @ Valentine Vornicu [/b] :  :D \r\n\r\nYes, I think that should be fine. I also have the same problem as the [b] OP [/b], many of my good (I mean-- posts pertaining to providing solutions to abstruse puzzles, technical posts etc) posts aren't rated at all or worse they are rated as being \"spam\". I think a high percentage of my posts ( I mean posts that were rated) were intentionally being rated as spam.\r\n\r\nI had observed that most (almost 95%) of my posts weren't rated. (I do acknowledge some of my earliest/earlier posts were silly); I think life would be much better for me and the [b] OP [/b] (and people like us) if people found some time to dispassionately 'rate' posts otherwise we have to bear the burden of being labeled as \"worthless/spam posters\" ([b] which in fact, we're not [/b]) and receive brickbats from petty moderator(s) and other pesky posters.\r\n\r\n[b] P.S.: [/b] By petty moderator(s) I meant, As far as I'm concerned there was only one petty modertor whom I know,who has  indulged in some gross over-simplification regarding my post rating. \r\n\r\nThanks!",
  "Solution_6": "@Gen8:\r\n\r\nhttp://www.mathlinks.ro/viewtopic.php?p=1516485#1516485\r\n\r\nhttp://www.mathlinks.ro/viewtopic.php?p=1515172#1515172\r\n\r\nThese aren't so early posts...\r\n\r\n@Valentin:\r\n\r\nWell, someone rated r15... a 1 again... And I don't see why his post is so bad.",
  "Solution_7": "My rating has also dropped from 3.8 to 3.1 in the past few days. (a lot considering i've been here for 3 years) And I don't think I've spammed.",
  "Solution_8": "The problem in general for low ratings is that most people rate spam, but not good posts. So someone may have 95 good posts and 5 spam, and have all the spam rated 1's and have 3 good post rated 5's. This brings the average to 2.5. BTW, I have a rating of 2.2, an improvement from 1.9. I have only been here a month and a half, though.\r\n\r\nAnother quesiton: Why can only the user see their own rating?",
  "Solution_9": "Mods can see all the ratings... And I think it's better to keep ratings to yourself (not allow others to see)"
}
{
  "Tag": [
    "calculus",
    "calculus computations"
  ],
  "Problem": "Let u''(w) < 0, show that E(u(X)) <= u(E(X)), by expressing u(w) as a series around the point w = E(X) and terminate the expansion with an error term involving the second derivative.",
  "Solution_1": "Let $ w_0\\equal{}E[X]$. For any $ w\\in \\mathbb R$ we have $ u(w)\\equal{}u(w_0)\\plus{}u'(w_0)(w\\minus{}w_0)\\plus{}\\frac{u''(s)}{2}(w\\minus{}w_0)^2$ where $ s$ is between $ w$ and $ w_0$. The last term is nonpositive, therefore...",
  "Solution_2": "Since the last term is non-positive, therefore\r\n\r\nu(w) <= u(w0) + u'(w0)(w - w0)\r\nu(X) <= u(E(X)) + u'(E(X))(X- E(X))\r\nE(u(X)) <= u(E(X))   Taking expectation on both sides\r\n\r\nHowever, from the Taylor series, why the first equation is true?\r\nu(w) = u(w0) + u'(w0)(w-w0) + (u''(s)/2)(w-w0)^2 ?",
  "Solution_3": "ok, it is the Taylor's series with Lagrange remainder"
}
{
  "Tag": [
    "Alcumus",
    "Support"
  ],
  "Problem": "How many problems are supposed to be done before you master one of the tasks ex: counting independent events. I did 160 problems on alcumus and never mastered any  :(",
  "Solution_1": "[quote=\"tcjy\"]How many problems are supposed to be done before you master one of the tasks ex: counting independent events. I did 160 problems on alcumus and never mastered any  :([/quote]\r\n\r\nYou can master and unmaster it in one question, it's all about the percentile, I think it has to be at lest 87.5% on the subject, and have done at least 10 questions (maybe more, like 15) to master it.",
  "Solution_2": "i don't think it's based solely on the percentile\r\nit's based on how others do at the topic as well\r\n\r\nsay you have 70% questions right\r\nif the bulk of AoPS have on 60% right, alcumus will consider you to have \"master\" the topic\r\n\r\nhowever, if you have 70% right while the bulk of AoPS have 80% right, you will not \"master\" the topic"
}
{
  "Tag": [],
  "Problem": "This is a poll for the people that are competing in the Problem Solving Tournament. How far do you think you are going to make it?",
  "Solution_1": "I voted top 6.... but I am being optimistic today! :P\r\n\r\nI wonder what BOGTRO will vote......",
  "Solution_2": "Where's BOGTRO...??? xD",
  "Solution_3": "#1!! (too short  :P )",
  "Solution_4": "mewto55555 isn't even in the tournament",
  "Solution_5": "I predict 5th.",
  "Solution_6": "I predict last place.",
  "Solution_7": "Is it too late to join the tournament????",
  "Solution_8": "Well, we are currently near the end of the 2nd round...",
  "Solution_9": "it is too late to join the tournament, but I have started another one called \"Another Tournament\"",
  "Solution_10": "Guess what I voted!!!",
  "Solution_11": "1st????????????????",
  "Solution_12": "When the one going on currently is over, I will be starting another one. There are currently fourteen people left, and on August 11th, I will be narrowing it down to the top 12."
}
{
  "Tag": [
    "rotation",
    "geometry",
    "3D geometry",
    "prism",
    "analytic geometry",
    "ratio"
  ],
  "Problem": "Let A(0,0), B(18,0), and C(10,6) be the vertices of a triangular region that is rotated about the x-axis.  What is the volume, in cubic units, of the solid that is formed.  Express your answer in terms of pi.\r\nI know how to it but I was wondering why find the are of the triangle and mutiplying it by the circumfirenz with radius 6.\r\nCan someone please tell me why?",
  "Solution_1": "I don't quite understand what you're asking  ;)",
  "Solution_2": "Find the area of the triangle and take the furthest point, 10,6, and find the circumfrence with radius 6.  Multiply those.  Why does it not come out to the right answer.",
  "Solution_3": "Well I don't know why that would work anyway. The way I'd do it is take the volume of a frustrum with radii 18 and 6 and then subtract the upside-down cone. \r\n\r\n$\\frac 13 \\cdot \\pi \\cdot 10 \\cdot (324 + 36 + 288) - 120 \\pi = 2040 \\pi$",
  "Solution_4": "your answer is wrong . it is 216pi, which I know how to get. I just won't use my way from now on.",
  "Solution_5": "[quote=\"kchande\"]Find the area of the triangle and take the furthest point, 10,6, and find the circumfrence with radius 6.  Multiply those.  Why does it not come out to the right answer.[/quote]\r\n\r\nVisualize the solid that is created when you rotate the triangle about the x-axis.  You get two cones with a circular base of radius 6 joined at the bases.  What you are finding using the method above is the volume of a triangular prism with base having area of the triangle and height equal to the circumference of a circle with radius 6.  \r\n\r\nNaga",
  "Solution_6": "I got my x coordinates mixed up with my y-coordinates :(\r\nI thought 18,0 was graphed like 0, 18",
  "Solution_7": "I no that way with the two cones, but thanks anyway",
  "Solution_8": "6 is not the height of the center of mass of the triangle.  What you're looking for is the height of the intersection of medians.  Since medians cut other medians into 2:1 ratios, you can easiliy see that the height of the center of mass is actually 2, since the total height is 6.  Then you get 2*2*pi*area = 4pi*54 = 216pi as required.",
  "Solution_9": "What you would do, kchande, is find the area of the triangle, multiply by the circumference of the circle and then divide by three to get the volume of the two cones which is ${216}\\pi$.",
  "Solution_10": "By the way, this concept is called Pappus' Centroid Theorem or some variant of that name.",
  "Solution_11": "when I did it, I thought the problem was easy, but now it seems so hard.",
  "Solution_12": "Actually, you don't multiply the area of the tirangle by the circumference of a circle with radius six to get the answer. You actually need to find the centroid of the triangle, find its distance x from the x-axis, and multiply the area of the triangle by the circumference of a circle with radius x. This is actually a theorem proved by some ancient mathematician about 2000 years ago.",
  "Solution_13": "Well, whenever you draw the heighth of a triangle to the base and then the centroid to the base, the distance from the centroid to the base is always one third of the altitude to the base because the ratio of the distance of the midpoint to the centroid to the distance of the centroid to the vertex is always one to two. Also if you have a triangle with coordinates $(x_1,y_1), (x_2,y_2),$ and $(x_3,y_3)$, the center of mass is at point $(\\displaystyle \\frac{x_1+x_2+x_3}{3},\\displaystyle \\frac{y_1+y_2+y_3}{3})$",
  "Solution_14": "well, finding the areas of the cones is a bit easier",
  "Solution_15": "I found it very simple o_O ... add the volumes of the two cones together, and voila...",
  "Solution_16": "[quote=\"kchande\"]well, finding the areas of the cones is a bit easier[/quote]\r\n\r\nHow do find you the area of a cone?"
}
{
  "Tag": [
    "LaTeX",
    "geometry",
    "3D geometry",
    "prism",
    "Support",
    "Asymptote",
    "ratio"
  ],
  "Problem": "[size=200][color=green]Introduction[/color][/size]\r\n\r\nAoPSia is certainly an interesting planet. It is known for being an incredible outlier on the positive end of the plot of average intelligence of populations of selected planets. Of course, AoPSians do not view their status as \u201coutlier\u201d at all, and rather as quite a normal state, and often exclaim a \u201cphailure\u201d when they make a stupid mistake that would definitely be noted a brilliant statement in another world. The planet and its surrounding moons are a very peaceful, gorgeous place, abundant in mathematical exhibits and monuments and thousands of mathematical tournaments taking place. Everything always goes well, as the president of AoPSia, rrusczyk, as well as all subordinates, and heck, just about all of the citizens, are always able to use the awesome art of problem solving to pretty much solve all problems that may arise. Quite a while ago, the population of AoPSia decided for a revolution for the good, remodeling all mathematical structures to the next degree of technology, practically proclaiming to all of the worlds of life that AoPSia is to be the new superpower, above all others. \r\n\r\nOf course, the other words out there are often jealous of AoPSian intelligence, and thus, tease the inhabitants of AoPSia as \u201cnerds.\u201d The people of AoPSia are perfectly fine with that label; in fact, the majority is proud of it, and, to cap stuff off, have something that actually hurts the other parties to shoot back at them: calling them \u201c\u00f70 brains.\u201d Most other worlds\u2019 inhabitants are devastated by such a comeback, and it is probably of the better of them to just give up right there. Of course, most people of AoPSia know of once when an Minicephleran challenged an AoPSian to prove the brainless status of those people. The AoPSian, being a good AoPSian, took out a piece of paper and neatly made a convincing two-column proof. \r\n\r\nGiven: Diagram of Minicephleran Brain as Labeled -->\r\nProve: Minicephleran$ \\cong$brainless\r\n\r\nStatements ------------------- Reasons ------------------\r\n1. Minicehpleran$ \\cong$brainless | 1. Reflexive Property\r\n\r\nThe poor guy couldn\u2019t believe his eyes. The AoPSian has proved his intelligence level right in front of his eyes. He advised others of his type to try not to bother the \u201camazingly smart\u201d population of AoPS. \r\n\r\nAoPSia is certainly a wonderful prosperous world. Recently, however, much attention has shifted to the rapidly-advancing moon of Forthewinia, the site of most math tournaments. Colonization of that moon just started a while ago by levans, who became, as he should, the president of Forthewinia. Many loved the idea, and quickly followed, causing the population of Forthewinia to grow practically exponentially. Before long, Forthewinia was the brightest spot of AoPSia\u2019s night sky. Lights are abundant, and there are often celebrations of great Forthewinian legends, like AIME15, Mewto55555, Jongao, stevenmeow, and BOGTRO, as well as newly-rising stars, like UberYoda and miller4math. It was pretty much, the moon that never sleeps, the moon of joyful competition and paradise. After all, what type of AoPSian won\u2019t like to show off their elite math skills? \r\n\r\nForthewinia is certainly unique, and the members of Forthewinia highlight this uniqueness with a slightly unusual system. For example, they commonly write in LaTeX, and some solve equations in base 42. Forthewinia, as well as the rest of AoPSia, often taunts at foreign units of measurement and methods of communication because many don\u2019t make logical sense. \r\n\r\nOne foreign unit most laughed at is the hit-point, or HP. It is often jokingly referred to as \u201cheadless pigs\u201d and others of the sort that HP [i]might[/i] stand for. The Forthewinian equivalent of the HP, the MRU, stands for mathpower resistance unit, and has no funny alternative acronym comprehension. One benefit of using this unit is that Forthewinians, being AoPSians, can gauge the ability of resistance of an outsider in a flash, given their HP, but those outsiders fry their brains just trying to comprehend the unbelievably complicated formula for converting MRU to HP:\r\n\r\n1 MRU = 1 HP. \r\n\r\nOf course, this unit of MRU is hardly ever used, as Forthewinia is usually very peaceful; only at very special times, usually in contact with other worlds, is the unit noticed; usually, it\u2019s not even worried about, as mathematical attacks would most likely overpower all other attacks.  But, yes, it is not a surprise that AoPSians often laugh uncontrollably when in discourse about other worlds. \r\n\r\nAlbeit the great ascending development of Forthewinia, the competitive nature of the planet has triggered a few heated arguments. One such was revolved around a decision levans made near the genesis of Forthewinia. levans selected advisors, popularly known as moderators or just \u201cmods,\u201d to aid him in the protection of the greatness and wealth of Forthewinia. So far, only four have been appointed to hold the special title and honor, and one retired. Many Forthewinians, being quite complacent in character from the experience of many tournaments, have questioned exactly what quality of them marks them as inferior to the ones that were selected as \u201cmods,\u201d and indeed, many have petitioned the king to gain moderator status and have been rigidly denied. Mods lived in the highest of possible luxury in Forthewinia, each with their own huge mansion, and a fourth building where they actually spend most of their time discussing organization in Forthewinia and loquaciously rambling about math. This sparked much jealousy, and in some instances, hate. In one worst case, an all-out war nearly commenced between one of the great Forthewinian legends and one of levans\u2019s advisors. Usually, peace is retained, and most Forthewinians realize the importance of that. But just once in a while, some unrest arises. \r\n\r\nAs anyone can see, Forthewinia is certainly an interesting moon. It is now by far the most prosperous moon existent. However, the guys from the other moons of other planets...are still jealous, and one, that has actually prepared for an invasion for many years, now makes an unexpected strike...Forthewinia has been invaded. All heptagonal prism will break loose. Thus, we commence with Chapter 1, and on a dark note.",
  "Solution_1": "[size=200]Chapter 1[/size]\r\n\r\n\u201cHmmmmm. It must be night in Forthewinia,\u201d remarked james4l to himself. (A fair amount of light is radiated from AoPSia, but only for a certain sector of the surrounding space, giving moons day when it enters the radiation zone and night when it is away.) \u201cThis is the very moon that is becoming the apex of wealth of all of the worlds of space. Well, I\u2019ll be there soon.\u201d \r\n\r\nA distinctively intelligent man, even in the AoPSian world, james4l is probably one of the most welcomed tourists in Forthewinia, highly famous for his good judgment in decisions, mathematical genius, and displays of nerdiness. Previously, james4l lived in Forthewinia himself, but a while ago moved back to the main world and planet of AoPSia. He occasionally comes to Forthewinia to visit, one such instance being now, as he travels in his private spaceship, the 4l-42 (pronounced \u201cfour-ell-forty-two,\u201d not \u201cnegative one\u201d). Hilariously, he came to visit just in time for a very big occurrence, which of course, he doesn\u2019t know of yet. \r\n\r\n--\r\n\r\nForthewinia boasts its strong 24/7 protection against outside invasion. From a very well-built guard tower, anyone can see anything coming. r15s11z55y89w21 runs the night shift every 10th day. However, lazy as he is, r15s11z55y89w21 usually dozes off halfway through his shift and pays his assistant, dragon96, to do the rest of the shift for him. Even though he was still skeptical about it, levans decided to approve what is going on, as the guarding work does get done correctly, and so, half of the time dragon96 is looking through the specially-manufactured Forthewinian telescopes that spot distant approaching spaceships even though r15s11z55y89w21 should be doing it. \r\n\r\n\u201cHey r15s11z55y89w21,\u201d says dragon96, \u201ccome here!\u201d \r\nHesitant from laziness, r15s11z55y89w21 comes over. \r\n\u201cLook! Someone\u2019s coming! And look closer! It says 4l-42!\u201d \r\n\u201cIt\u2019s our old friend, james4l!\u201d \r\n\u201cWell he\u2019s coming for a visit. How nice. We\u2019ll give him a warm welcome to the new Forthewinia that is so much different from what it was the last time he\u2019s been here.\u201d \r\n\u201cAllrighty! He\u2019s almost landing.\u201d \r\n\r\n--\r\n\r\n*DING*\r\nThe elevator door to the third floor opens, and james4l enters, but to r15s11z55y89w21\u2019s and dragon96\u2019s astonishment, bearing a melancholy expression on his face. r15s11z55y89w21 and dragon96 suddenly realize that james4l is holding in his hand a weird sheet of foreign-looking paper. Simple staring was all that happened for a short period of silence, and dragon96 finally enunciated, \u201cUh...what\u2019s...wrong?\u201d \r\n\u201cThis,\u201d said james4l, handing the obviously non-AoPSian-manufactured paper over. \r\n\u201cOh gosh!\u201d exclaimed r15s11z55y89w21, peering over dragon96 as he reads the paper. It was a poem. Short, but definitely not sweet. \r\n\r\n[i]You evil nerds of Forthewinia, \nSo arrogantly bright for the universe to see, \nTo force you to realize your wrongness, \nWe will deprive a group of three. [/i]\r\n\r\n\u201cGosh, so threatening. Geez, at this guard station, we could easily spot any intruders and tell everyone of the invasion to defend,\u201d said dragon96. r15s11z55y89w21 reinforced dragon96\u2019s statement, \u201cExactly, and who could possibly succeed against us?\u201d \r\nDeep in thought, james4l finally commented, \u201cExactly who is this group of three? Us?\u201d\r\nUsing his elite logic skills, dragon96 explained, \u201cObviously not. They didn\u2019t even know you were coming. What other groups of three are there?\u201d\r\nr15s11z55y89w21\u2019s face turned slightly red. \u201c...mods.\u201d\r\ndragon96, unaware of that possibility, exclaimed, \u201choly calculator.\u201d\r\njames4l advised, \u201cWell, in that case  we better go and check the mods\u2019 mansions.\u201d \r\n\r\n--\r\n\r\njames4l came back from the mansions. \u201cIt happened. No one is answering the door; they always do.\u201d r15s11z55y89w21 and dragon96 looked at each other. r15s11z55y89w21 questioned, \u201cWhat puzzles me is exactly how they did it. Tonight we only found two spaceships coming, yours and levans\u2019s, as he returns from his business trip to main-planet AoPSia.\u201d \r\n\r\ndragon96 reminded r15s11z55y89w21, \u201cWe don\u2019t really have time to think. We got to plan something to do for this case we\u2019ve got.\u201d \r\n\r\n--\r\n\r\nImmediately the next morning all were informed of the mods\u2019 disappearance. r15s11z55y89w21, dragon96, and james4l discovered that intruders hijacked levans\u2019s plane and came sneakily via that. levans escaped, but the hijackers went on for more. They captured the three mods and then used an inflatable spaceship they had carried with them (by far the most advanced invention outside of AoPSia) to leave the planet. \r\n\r\nTo levans\u2019s dismay, the lack of mods wreaked havoc in Forthewinia. Discourse immediately took a big turn for the bad, and fights and arguments arose, many concerning who the new mods to be replaced are; in such the heat, some citizens were so enveloped in jealousy as to think of this as a priority before saving the mods. \r\n\r\nOne day, r15s11z55y89w21 discussed with levans that he would go on a mission to try to find where the mods were, and levans approved. r15s11z55y89w21 deduced that the paper the horrid poem was written on was paper of Eipia (Ee-igh-PIE-ya), a planet that is becoming a rival to Forthewinia. After informing his trusted friend, james4l, of where the mods are being held and what he plans to do, r15s11z55y89w21 plans a trip to Eipia. He decided that dragon96 should go with him. \r\n\r\n\u201cBring lots of any helpful stuff, we might have a lot to do, especially food,\u201d he said to dragon96. \r\n\u201cOkay. I\u2019ll help with the stuff.\u201d\r\n\r\nr15s11z55y89w21 brought 4200 LaTeX-dwollars (Forthewinian currency), and dragon96 brought some weapons, a lot a food, and a bunch of other miscellaneous items. And so they left for Eipia, and the rest of Forthewinia awaited their return with the mods, some mischievous citizens taking the last opportunity to commit nefarious actions before the policing system comes back. The only problem is, r15s11z55y89w21 and dragon96 did not come back. \r\n\r\njames4l now became quite popular on Forthewinia, and people often asked him what was really going on. james4l just said that what is going on is really going on, r15s11z55y89w21 and dragon96 left and for some reason aren\u2019t coming back. \r\n\r\nUnrest has now became a huge crisis on Forthewinia, making the moon a terror of a place. Evenually, levans jumped on his spaceship and fled to nearby Alcumia, noting that even though he is king, he only has 196 MRU, and thus isn\u2019t really in a very good position. \r\n\r\nIt took a long time for Forthewinia to realize what problem this fighting could become if it continues. At last, the current greatest citizen in accomplishment, AIME15, decided that he will support the cause to go get isabella2296, jjx1, vallon22 (, r15s11z55y89w21, and dragon96) back. james4l, now much more than a tourist, is also going. Sixteen other Forthewinians agreed to help the cause. AIME15 designated lotsofmath to take half of the crew, thus making a split in case an occasion calls for two tasks to be done. He still decided however, to have one group bigger than the other so that more can be done with this group. The whole crew of eighteen thus boarded the ASYMPTOTE XLII, and headed for Eipia. \r\n\r\n\"Surely with this many people we can conquer those Eipians,\" proclaimed AIME15. \"However, note that they are getting smarter as well. We\u2019re going to be going against people of our own kind. We should supposedly, be more powerful than them, but they\u2019ve got their whole planet there and we have eighteen, possibly twenty if we find r15s11z55y89w21 and dragon96 somewhere and they weren't also captured. \"\r\n\r\nThe crew: \r\nAIME15, leader\r\nBOGTRO, funtwo, Meta_Knight, pythag011, AIME15's top advisors\r\nmz94, mathblitz, vahalla, eggylv999, rest of crew\r\nlotsofmath, leader of secondary group\r\ncoolts, Joe10112, 1=2, Dojo, Garyzx, secondary group crew\r\n\r\nTwo days into their journey, AIME15 realized that he probably should have brought some money. \r\n\r\n\"Uh-oh, we might need to buy stuff out there, considering we don't have that many weapons, plus other stuff.\" \r\n\"Hmmmmmmm, I have 3500 LaTeX-dwollars with me,\" commented BOGTRO. \r\n\r\nOthers also presents bits of money they found in their pocket or packages they brought. Collecting all of the money they had, AIME15 instantly summed in his head that in total they had 4725 LaTeX-dwollars to spend. \"We're sort of short on supplies. So we'll take a stop at the AoPSian system-boundary store and try to see what we can buy there. Hopefully something useful.\" Thus, they will stop at the store on their trip.",
  "Solution_2": "Hmm... so what do we do?\r\n\r\n[b]Buy Combinatorial Nullstellensatz from the store.[/b]",
  "Solution_3": "The first thing that you would do, before embarking on the journey to Eipia, is to decide what you will buy at the AoPSian system-boundary shop. That post will come soon.",
  "Solution_4": "This is too long to read. A summary please?",
  "Solution_5": "Blah blah blah you are on a planet with a store, you have 4725 dollars, what do you want to buy.",
  "Solution_6": "Realize how excellent that summary was.\r\n\r\n[b]I buy the AoPS website[/b]",
  "Solution_7": "one question\r\n\r\nhow did you pick the groups and leaders and such?\r\n\r\nby mru?\r\n\r\nanyway, where is the store list?\r\n\r\ndid he pm it to you?",
  "Solution_8": "the store list will be up soon. be patient. He's recovering from Carpal Tunnel.",
  "Solution_9": "sorry\r\n\r\nwhats the definition of \"carpal tunnel?\"\r\n\r\ni know, im a n00b. o well\r\n\r\nand i assume n00b is a very inexperianced player.",
  "Solution_10": "[quote=\"Joe10112\"]sorry\n\nwhats the definition of \"carpal tunnel?\"\n\ni know, im a n00b. o well\n\nand i assume n00b is a very inexperianced player.[/quote]\r\n\r\nGoogle is your friend.",
  "Solution_11": "but, i cannot go on google!\r\n\r\ni have blocked internet  :(",
  "Solution_12": "r15 is an excellent writer.  How much must I pay you to write my essay for me?\r\n\r\nMeh, so the store inventory list/prices is coming soon?",
  "Solution_13": "[quote=\"wiki\"]Carpal tunnel syndrome (CTS), or median neuropathy at the wrist, is a medical condition in which the median nerve is compressed at the wrist, leading to paresthesias, numbness and muscle weakness in the hand. The diagnosis of CTS is often misapplied to patients who have activity-related arm pain.\n\nMost cases of CTS are idiopathic (without known cause), genetic factors determine most of the risk, and the role of arm use and other environmental factors is disputed.\n\nNight symptoms and waking at night--the hallmark of this illness--can be managed effectively with night-time wrist splinting in most patients. The role of medications, including corticosteroid injection into the carpal canal, is unclear. Surgery to cut the transverse carpal ligament is effective at relieving symptoms and preventing ongoing nerve damage, but established nerve dysfunction in the form of static (constant) numbness, atrophy, or weakness are usually permanent and do not respond predictably to surgery.\n\n[/quote]",
  "Solution_14": "ohhhhhhh\r\n\r\nthanks dojo\r\n\r\ni hope this post isnt too short",
  "Solution_15": "Are we going to continue playing?",
  "Solution_16": "I don't know. Announce if you're still here. If at least half of the people respond within 24 hours, the game shall continue. \r\n\r\nHeh, I must be pro at indirectly reviving games.",
  "Solution_17": "lurker is here",
  "Solution_18": "I am here.\r\n\r\nWhile I'm here, might as well advertise:\r\n\r\nGo to http://cyneer.com, host of the first National Internet Math Olympiad!",
  "Solution_19": "bool(mehere)= 1\r\n\r\n/me chuckles at C++ ref.\r\n\r\nYes, I'm here.",
  "Solution_20": "hi[b][/b]",
  "Solution_21": "i am here but i dont come on very much so youll have to count me out",
  "Solution_22": "noooooooooo................... anyone want to send prods?",
  "Solution_23": "I'm still here. AIME15 decide what we do.",
  "Solution_24": "I will be more active monday.",
  "Solution_25": "nooooo!!!! :(",
  "Solution_26": "It appears that LTM has failed. \r\n\r\n=(\r\n\r\nAnyway, if you like LTM-ish stuff, go [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=288353]here[/url]. \r\n\r\n[hide=\" \"]360-->3G0\n363-->3G3\n384-->3B4\nNGG: 14:42-->coordinate strategy\nE6--><6\nDoctor, Silver, Oxygen, Nitrogen-->Dr., Ag, O, N\n512 was 11th-->15n\nr15's transmitter code-->785-IY-452\nEipian003: 110\nEipian004: 110\nEipian013: 125\nEipian014: 125\nEipian015: 145\nEipian016: 145\nEipian017: 90\nEipian018: 90\nEipian019: 100\nEipian020: 150\nEipian021: 150\nEipian022: 150\nEipian023: 175\nEipian024: 430\nEipian025: 300\nEipian026: 125\nEipian027: 125\nEipian028: 125\nEipian029: 200\nEipian030: 200\npispeakerfreaker-->202-->382-->411\n(x,y)-->xth letter in yth line<--5\n13-->lotsofmath\n15-->BOGTRO\n22-->AIME15[/hide]",
  "Solution_27": "You mean this topic is officially done with??? :?:",
  "Solution_28": "Do not post on this thread if you're not in the game. This rule still applies, even though at least this part of the game ended. Additionally, do not post unless you have something meaningful to say.",
  "Solution_29": "[quote=\"r15s11z55y89w21\"]It appears that LTM has failed. \n\n=(\n\nAnyway, if you like LTM-ish stuff, go [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=288353]here[/url]. \n\n360-->3G0\n363-->3G3\n384-->3B4\nNGG: 14:42-->coordinate strategy\nE6--><6\nDoctor, Silver, Oxygen, Nitrogen-->Dr., Ag, O, N\n512 was 11th-->15n\nr15's transmitter code-->785-IY-452\nEipian003: 110\nEipian004: 110\nEipian013: 125\nEipian014: 125\nEipian015: 145\nEipian016: 145\nEipian017: 90\nEipian018: 90\nEipian019: 100\nEipian020: 150\nEipian021: 150\nEipian022: 150\nEipian023: 175\nEipian024: 430\nEipian025: 300\nEipian026: 125\nEipian027: 125\nEipian028: 125\nEipian029: 200\nEipian030: 200\npispeakerfreaker-->202-->382-->411\n(x,y)-->xth letter in yth line<--5\n13-->lotsofmath\n15-->BOGTRO\n22-->AIME15[/quote]\r\n\r\nHow were we supposed to figure this out?"
}
{
  "Tag": [
    "AMC",
    "AIME",
    "geometry",
    "inequalities",
    "triangle inequality"
  ],
  "Problem": "A set of positive numbers has the triangle property if it has three distinct elements that are the lengths of the sides of a triangle whose area is positive. Consider sets $ \\{4, 5, 6, \\ldots, n\\}$ of consecutive positive integers, all of whose ten-element subsets have the triangle property. What is the largest possible value of n? \r\n\r\nHints only, please. Thanks in advance!",
  "Solution_1": "I haven't tried this yet, but I would work backwards. Also, look for the smallest value of n that does not have the property rather than the largest value that does.",
  "Solution_2": "[hide=\"Hint\"]Try to find the 10-element subset that meets this property, making the largest element in that subset as small as possible.\n[hide=\"Little More\"]Use 4 and 5, they are the smallest elements to start out with.\n[hide=\"Complete Spoiler\"] Triangle Inequality to find the next smallest-as-possible element that is viable.\n[/hide][/hide][/hide]",
  "Solution_3": "[hide=\"mah hintz\"]Consider the two smallest elements of one of the ten-element sets.\n[hide=\"More\"]What must all of the other elements be bigger than?[/hide][/hide]",
  "Solution_4": "Thanks for the hints! When I get a chance, I'll post the answer I get (because I don't have the key), and hopefully then someone who does have the answers can please let me know if it's right.",
  "Solution_5": "Is the answer 254?",
  "Solution_6": "Nope, sorry.\r\n\r\nEdit: You basically got it. Reread the question and see if you answered it correctly. You found the first set that does not have the property. That's not what the question is asking.",
  "Solution_7": "Oh, I think I see what I did! It's 253, right?",
  "Solution_8": "Yeah, it is."
}
{
  "Tag": [
    "induction",
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "How many ways can we select n-2 subset A_1...A_{n-2} of S={1;2...n} satisfying for i<j $|A_i|<|A_j|$ and $A_i\\notin A_j$",
  "Solution_1": "Can anyone solve it?I found n!",
  "Solution_2": "There cannot be a set of size 1 and n-1 at the same time. Let $a_n$ be the number of set systems which contain sets of sizes 2 to n-1 and $b_n$  the number of set systems which contain sets of sizes 1 to n-2. Let $c_n$ be the total number of the systems.\r\nIf we wanna construct one of the sets in $b_n$, we can choose the set $A_1$ of size 1 in n ways. The other sets are of sizes 2 to n-2 and are subsets of $\\{1,2,...,n\\}-A_1$, so for each $A_1$ there are $a_{n-1}$ such subsets, so\r\n(1) $b_n=n.a_{n-1}$.\r\nIf we wanna construct one of the sets in $a_n$, we can choose $A_1$ of size n-1 in n ways. Denote x the number not being in $A_1$. The other sets consist of the number x and a subset of size 1 to n-1. There are $b_{n-1}$ such subsets for each $A_1$, so\r\n(2) $a_n=n.b_{n-1}$\r\nFrom the equations (1) and (2) follows\r\n(3)$c_n=n.c_{n-1}$. \r\nAs $c_3=6$, from (3) by induction  $c_n=n!$."
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "ratio"
  ],
  "Problem": "A cone and a cylinder have the same height and a base with the\nsame diameter. What is the ratio of the volume of the cone to\nthe volume of the cylinder? Express your answer as a common\nfraction.",
  "Solution_1": "Let r be the radius of the cone and cylinder, and h the height. Since the diameters are the same, the radii are the same.\r\n\r\nVolume of cone= (1/3)$ \\pi$*$ r^2$*$ h$.\r\n\r\nVolume of cylinder= $ \\pi$*$ r^2$*$ h$.\r\n\r\nRatio is $ \\boxed{\\frac {1}{3}}$"
}
{
  "Tag": [
    "geometry",
    "geometric transformation",
    "rotation"
  ],
  "Problem": "there are 16 books on a shelf.in how many ways can these be arrangedif 12 of them are volumes on history and must be kept together in order.( i will like to see yours)my anwser was 1 less than the real anwser",
  "Solution_1": "I am definitely NOT even close to gifted, I estimate my iq somewhere around 60-70\r\n    [hide=\"bogus solution\"] since the 12 history books have to always be together and in the same order then just imagine that all 12 are one block so then there would be 5 blocks which can be rearranged 120 ways. [/hide]",
  "Solution_2": "5!=120\r\nso 120 is the anss",
  "Solution_3": "Just imagine that the group of 12 is only 1 book.\r\n\r\nThen there are 5 books, and 5 spaces, leaving 5! as the correct solution.\r\n\r\nIf some items must be kept together in the same order, you might as well treat 'em as one object.",
  "Solution_4": "The 12 books are one \"book\" \r\nThere are 5 spaces.\r\nSo:\r\n$ 5\\cdot4\\cdot3\\cdot2\\cdot1\\equal{}120$\r\nor\r\n$ 5!\\equal{}120$",
  "Solution_5": "This line of thinking may be useful in contest problems when you have a situation similar to the following:\r\n\r\n[b]Problem:[/b] [i]Seven cheap cars and seven expensive cars are parked in a parking lot with 32 spaces, arranged in one row. Noone wants to risk scratching an expensive car while opening his/her door, so there can't be any cars in the two spaces adjacent to an expensive car. In how many ways can this be done?\n[/i]\r\n[b]Solution:[/b] [i]There are four cases: no expensive cars in the leftmost or rightmost slot, expensive car in only the leftmost slot, expensive car in only the rightmost slot, and expensive car in the leftmost and rightmost slots.\n\nIn the first case, we group each expensive car with the 2 adjacent spaces to form \"units\", burning up 14 of the 32 spaces. So now, out of effectively 18 spaces, we pick 7 for cheap cars, and 7 for expensive car units. This gives 18!/(7! 7! 4!) combos.\n\nIn both the second and third cases, the expensive car on the edge is grouped with the single adjacent slot, reducing the number of slots to 31. Now we want to fit 7 regular cars, and 6 expensive car units (for which we burn 12 slots). Now, we have 18!/(7!6!5!) ways of doing this for each of these two cases (we multiply that number by 2).\n\nIn the fourth case, place an expensive car on each end, leaving 28 slots. Now, group each of the 5 remaining expensive cars with the two adjacent spaces, burning another 10 spaces. There are 18!/(7!5!6!) ways of arranging these.\n\nThe total number of arrangements is 42007680[/i]\r\n\r\n\r\nObviously you wouldn't see those exact numbers on a contest since the numbers are huge. But it's an example of a slightly less obvious situation where grouping might help.",
  "Solution_6": "I don't see where the IQ of 129 comes in...",
  "Solution_7": "Perhaps his IQ was estimated to be 129.",
  "Solution_8": "[url]http://en.wikipedia.org/wiki/IQ_reference_chart[/url]\r\nI'm sorry, but above 129 is only moderately gifted. :P\r\n\r\nAnd by Zuton Force's poke-hint, you should deduce that the answer is [i]not[/i] 120.",
  "Solution_9": "well, above 129 is where they start calling people gifted, moderately or not, i guess",
  "Solution_10": "If it's on a shelf, why can't it be 2*5!\r\nThe volumes can be counting up or counting down right? They just have to be in numerical order?\r\n\r\nedit: They're unmarked so I'm sure the answer is just 5!",
  "Solution_11": "Oh, well I was assuming the books were marked and distinct from each other.",
  "Solution_12": "[quote=\"kysuke\"]If it's on a shelf, why can't it be 2*5!\nThe volumes can be counting up or counting down right? They just have to be in numerical order?\n\nedit: They're unmarked so I'm sure the answer is just 5![/quote]\r\n\r\nWell then you have to think about flipping books upside down and stuff. Well I don't suppose flipping a book upside down is a different arrangement",
  "Solution_13": "I don't understand The Zuton Force's post.  Specifically, if each expensive car is grouped with a space on each of its sides, then how is it possible for there to be an expensive car followed by a space followed by another expensive car?\r\nThe arrangement \"expensive car, space, expensive car\" is not possible.  Instead, the arrangement would have to be \"expensive car, space, space, expensive car.\"  So isn't this undercounting the true count by not including arrangements in which there is only one space between expensive cars?",
  "Solution_14": "ALL OF YOU AER WRONG!!!!!! THe history books are not distinct, so its $ 5!\\cdot12!$ because yo can rearrange the history books!",
  "Solution_15": "[quote=\"Euler221\"]I don't understand The Zuton Force's post.  Specifically, if each expensive car is grouped with a space on each of its sides, then how is it possible for there to be an expensive car followed by a space followed by another expensive car?\nThe arrangement \"expensive car, space, expensive car\" is not possible.  Instead, the arrangement would have to be \"expensive car, space, space, expensive car.\"  So isn't this undercounting the true count by not including arrangements in which there is only one space between expensive cars?[/quote]\r\n\r\nYeah, I was smoking... and then they wouldn't let me delete it. But the point was, that grouping is a useful technique :P",
  "Solution_16": "Why do so many people not read the original post?\r\n\r\nIt clearly says that the 16 history books must be kept in order. (So they are distinct, mewto.)\r\n\r\nI am almost certain that rotations do not count in these types of problems, but then again, the problem should have specified.",
  "Solution_17": "[quote=\"Euler221\"]I don't understand The Zuton Force's post.  Specifically, if each expensive car is grouped with a space on each of its sides, then how is it possible for there to be an expensive car followed by a space followed by another expensive car?\nThe arrangement \"expensive car, space, expensive car\" is not possible.  Instead, the arrangement would have to be \"expensive car, space, space, expensive car.\"[/quote]\r\n\r\nActually, no: the problem only requires that the spaces [i]adjacent[/i] to the expensive car (i.e., the space to the right of the car and the space to the left of the car) be left empty, so the arrangement \"expensive car, space, expensive car\" is indeed possible (as is the arrangement you suggested). Perhaps a clearer problem statement would be as follows:\r\n\r\n[i]An employee of the Legendre Hotel has been ordered to park seven cheap cars and seven expensive cars in a parking lot. The parking lot consists of a single row of 32 parking spaces. The manager of the hotel doesn't want to risk damaging any of the expensive cars and, therefore, has told the employee that there must be at least one empty parking space between an expensive car, and any other car. If each car must be parked in a parking space, in how many ways can the cars be arranged? [/i]",
  "Solution_18": "[quote=\"metafor\"][quote=\"Euler221\"]I don't understand The Zuton Force's post.  Specifically, if each expensive car is grouped with a space on each of its sides, then how is it possible for there to be an expensive car followed by a space followed by another expensive car?\nThe arrangement \"expensive car, space, expensive car\" is not possible.  Instead, the arrangement would have to be \"expensive car, space, space, expensive car.\"[/quote]\n\nActually, no: the problem only requires that the spaces [i]adjacent[/i] to the expensive car (i.e., the space to the right of the car and the space to the left of the car) be left empty, so the arrangement \"expensive car, space, expensive car\" is indeed possible (as is the arrangement you suggested). Perhaps a clearer problem statement would be as follows:\n\n[i]An employee of the Legendre Hotel has been ordered to park seven cheap cars and seven expensive cars in a parking lot. The parking lot consists of a single row of 32 parking spaces. The manager of the hotel doesn't want to risk damaging any of the expensive cars and, therefore, has told the employee that there must be at least one empty parking space between an expensive car, and any other car. If each car must be parked in a parking space, in how many ways can the cars be arranged? [/i][/quote]\r\n\r\nI am aware that the arrangement \"expensive car, space, expensive car\" is possible, as you stated.  My complaint is that the Zuton Force's solution does not allow for this arrangement, and therefore undercounts the total number of possible arrangements.  In his solution, the Zuton Force groups each expensive car with a space on both of its sides, forming what he calls \"units.\"  Let's say that two of these \"units\" are placed next to one another.  Then the arrangement would be \"space, expensive car, space, space, expensive car, space.\"  As you can see, there are two spaces between the expensive cars.  Furthermore, there is no way to get just one space between two expensive cars if they are grouped into the \"units.\"  Therefore, the Zuton Force's solution is incorrect.",
  "Solution_19": "Ah \u2014 apologize for the misunderstanding, then. -_-;",
  "Solution_20": "I think that this isn't an IQ problem. Its just counting here is the solution : consider the 12 history books as just one book \r\nthen at first you have 4 choices then 3 choices... so the are $ 4!$ solutions read Intro C and P you will never have a question again  :D",
  "Solution_21": "When he says \"in order\" maybe he means that you can also reverse the order of history books. The answer would then be $ 2\\cdot 5!$ . You may say: \"If so, you can find an opportune definition of order and arrange the books as you wish\". That's correct. I assumed that order means something like alphabetical order. And another thing... the history books are usually big: you can put other books on them and also put them horizontally. That's another question I guess, if not I just say I'm not >129.\r\nBye, Gil."
}
{
  "Tag": [],
  "Problem": "Hey guys..could you help me with this question?\r\n\r\n8^(3x-3)-8^3x=448\r\n\r\nThanks.",
  "Solution_1": "Take a closer look.\r\n[hide]$ 8^{3x\\minus{}3}\\minus{}8^{3x} \\equal{} 8^{3x\\minus{}3}(1\\minus{}8^3) \\equal{} 448$, so $ (8^3)^{x\\minus{}1} \\equal{} \\minus{}\\frac {448} {8^3 \\minus{} 1}$, but all this is useless, since LHS is negative while RHS is positive, so never any real solution[/hide]",
  "Solution_2": "$ 8^{3x \\minus{} 3} \\minus{} 8^{3x} \\equal{} 448$\r\n\r\nlet $ 8^{3x}\\equal{}t$\r\n\r\n$ t/8^{3} \\minus{} t \\equal{}448$\r\n\r\n$ t\\equal{}\\minus{}32768/73$\r\n\r\n$ 8^{3x}\\equal{}\\minus{}32768/73$\r\n\r\nso there is no real solution to the equation"
}
{
  "Tag": [
    "calculus",
    "integration",
    "geometry",
    "algebra",
    "binomial theorem",
    "calculus computations"
  ],
  "Problem": "Let $ A$ be the region : $ \\left\\{(x,\\ y)|x\\geq 0,\\ y\\geq 0, \\left(\\frac{x}{a}\\right)^{\\frac{1}{2}}\\plus{}\\left(\\frac{y}{1\\minus{}a}\\right)^{\\frac{1}{2}}\\leq 1\\right\\}$ for $ 0<a<1.$\r\n\r\n(1) Find the area of $ A.$\r\n\r\n(2) Find the maximum volume of the solid obtained by revolving $ A$ around the $ x$ axis.",
  "Solution_1": "[hide]Solve for y to obtain\n\n$ y \\equal{} (1\\minus{}a)\\left(1 \\minus{} 2(x/a)^{1/2} \\plus{} (x/a)\\right), \\quad 0 \\le x \\le a \\le 1$.\n\nWith the change of variable\n\n$ u \\equal{} x/a, \\quad du \\equal{} 1/a \\, dx$, we obtain\n\n$ A(a) \\equal{} a(1\\minus{}a) \\int_{u\\equal{}0}^1 (1 \\minus{} 2 \\sqrt{u} \\plus{} u) \\, du \\equal{} \\frac{a(1\\minus{}a)}{6}$.\n\nThe volume V obtained by revolving the region A about the x-axis is given by\n\n$ V(a) \\equal{} \\int_{x\\equal{}0}^a \\pi y(x)^2 \\, dx \\equal{} \\pi a (1\\minus{}a)^2 \\int_{u\\equal{}0}^1 (1 \\minus{} \\sqrt{u})^4 \\, du$.\n\nSince the integral is merely a constant, we can immediately differentiate the above with respect to a to obtain the value that maximizes V:\n\n$ V'(a) \\sim (1\\minus{}a)(1\\minus{}3a) \\equal{} 0$\n\nimplies V attains its maximal volume for a = 1/3.  (The only other critical points are a = 0 and a = 1, both of which give V(a) = 0.)  Now to compute the integral, we employ another substitution (to avoid expanding via the binomial theorem):\n\n$ t \\equal{} 1 \\minus{} \\sqrt{u}$,\n$ u \\equal{} (1 \\minus{} t)^2$,\n$ du \\equal{} 2(t\\minus{}1) \\, dt$;\n\n$ \\int_{u\\equal{}0}^1 (1\\minus{}\\sqrt{u})^4 \\, du \\equal{} \\int_{t\\equal{}1}^0 2(t\\minus{}1)t^4 \\, dt \\equal{} 2 \\Bigl[\\frac{t^6}{6} \\minus{} \\frac{t^5}{5}\\Bigr]_{t\\equal{}1}^0 \\equal{} \\frac{1}{15}$,\n\nand therefore\n\n$ V\\left(\\frac{1}{3}\\right) \\equal{} \\frac{4\\pi}{405}$.[/hide]",
  "Solution_2": "In view of double integral, we see that \\[ v\\left( A \\right) \\equal{} \\int_0^{1 \\minus{} a} {\\int_0^{a\\left( {1 \\minus{} \\sqrt {\\frac{y}\r\n{{1 \\minus{} a}}} } \\right)^2 } {dx} dy}  \\equal{} a\\int_0^{1 \\minus{} a} {\\left( {1 \\minus{} \\sqrt {\\frac{y}\r\n{{1 \\minus{} a}}} } \\right)^2 dy}  \\equal{} \\cdots\r\n\\]",
  "Solution_3": "[quote=\"atomicwedgie\"][hide]Solve for y to obtain\n\n$ y \\equal{} (1 \\minus{} a)\\left(1 \\minus{} 2(x/a)^{1/2} \\plus{} (x/a)\\right), \\quad 0 \\le x \\le a \\le 1$.\n\nWith the change of variable\n\n$ u \\equal{} x/a, \\quad du \\equal{} 1/a \\, dx$, we obtain\n\n$ A(a) \\equal{} a(1 \\minus{} a) \\int_{u \\equal{} 0}^1 (1 \\minus{} 2 \\sqrt {u} \\plus{} u) \\, du \\equal{} \\frac {a(1 \\minus{} a)}{6}$.\n\nThe volume V obtained by revolving the region A about the x-axis is given by\n\n$ V(a) \\equal{} \\int_{x \\equal{} 0}^a \\pi y(x)^2 \\, dx \\equal{} \\pi a (1 \\minus{} a)^2 \\int_{u \\equal{} 0}^1 (1 \\minus{} \\sqrt {u})^4 \\, du$.\n\nSince the integral is merely a constant, we can immediately differentiate the above with respect to a to obtain the value that maximizes V:\n\n$ V'(a) \\sim (1 \\minus{} a)(1 \\minus{} 3a) \\equal{} 0$\n\nimplies V attains its maximal volume for a = 1/3.  (The only other critical points are a = 0 and a = 1, both of which give V(a) = 0.)  Now to compute the integral, we employ another substitution (to avoid expanding via the binomial theorem):\n\n$ t \\equal{} 1 \\minus{} \\sqrt {u}$,\n$ u \\equal{} (1 \\minus{} t)^2$,\n$ du \\equal{} 2(t \\minus{} 1) \\, dt$;\n\n$ \\int_{u \\equal{} 0}^1 (1 \\minus{} \\sqrt {u})^4 \\, du \\equal{} \\int_{t \\equal{} 1}^0 2(t \\minus{} 1)t^4 \\, dt \\equal{} 2 \\Bigl[\\frac {t^6}{6} \\minus{} \\frac {t^5}{5}\\Bigr]_{t \\equal{} 1}^0 \\equal{} \\frac {1}{15}$,\n\nand therefore\n\n$ V\\left(\\frac {1}{3}\\right) \\equal{} \\frac {4\\pi}{405}$.[/hide][/quote]\r\n\r\nThat's the correct answer. :)\r\n\r\nIn case of $ \\sqrt {\\frac {x}{a}} \\plus{} \\sqrt {\\frac {y}{b}} \\equal{} 1\\ (a > 0,\\ b > 0),$\r\n\r\n$ S \\equal{} \\frac {1}{6}ab,\\ V_x \\equal{} \\frac {\\pi}{15}ab^2,\\ V_y\\equal{}\\frac{\\pi}{15}a^2b.$"
}
{
  "Tag": [
    "trigonometry",
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "prove or disprove:\r\n$ ( \\sin \\frac A 2 \\plus{} \\sin \\frac B 2 \\plus{} \\sin \\frac C 2)(\\frac 1 {\\sin \\frac A 2} \\plus{} \\frac 1 {\\sin \\frac B 2} \\plus{} \\frac 1 {\\sin \\frac C 2}) \\ge \\frac 9{\\sqrt {8\\sin \\frac A 2\\sin \\frac B 2\\sin \\frac C 2}}$\r\nwhere $ A,B,C$ are  three angles of triangle.",
  "Solution_1": "[quote=\"zhaobin\"]prove or disprove:\n$( \\sin \\frac A 2+ \\sin \\frac B 2+ \\sin \\frac C 2)(\\frac 1 { \\sin \\frac A 2}+\\frac 1 { \\sin \\frac B 2}+\\frac 1 {\\sin \\frac C 2}) \\ge \\frac 9{\\sqrt{8\\sin \\frac A 2\\sin \\frac B 2\\sin \\frac C 2}}$\nwhere $A,B,C$ are  three angles of triangle.[/quote]\r\n\r\nAlas, this one isn't true. There are cases where it holds and there are cases where it doesn't, and both can be arbitrary close to A = B = C.\r\n\r\n  Darij",
  "Solution_2": "[quote=\"darij grinberg\"][quote=\"zhaobin\"]prove or disprove:\n$( \\sin \\frac A 2+ \\sin \\frac B 2+ \\sin \\frac C 2)(\\frac 1 { \\sin \\frac A 2}+\\frac 1 { \\sin \\frac B 2}+\\frac 1 {\\sin \\frac C 2}) \\ge \\frac 9{\\sqrt{8\\sin \\frac A 2\\sin \\frac B 2\\sin \\frac C 2}}$\nwhere $A,B,C$ are  three angles of triangle.[/quote]\n\nAlas, this one isn't true. There are cases where it holds and there are cases where it doesn't, and both can be arbitrary close to A = B = C.\n\n  Darij[/quote]\r\ncan you expain more?I can't understand :(",
  "Solution_3": "[quote=\"zhaobin\"]can you expain more?I can't understand :([/quote]\r\n\r\nI mean that\r\n[b]1.[/b] it is wrong\r\nand\r\n[b]2.[/b] (apparently - I have just checked with numeric data,) for every $\\epsilon>0^{\\circ}$, there exists a triangle ABC whose angles are all in the interval $\\left[60^{\\circ}-\\epsilon;\\;60^{\\circ}+\\epsilon\\right]$ for which the inequality is true, and there exists a triangle ABC whose angles are all in the interval $\\left[60^{\\circ}-\\epsilon;\\;60^{\\circ}+\\epsilon\\right]$ for which the inequality is wrong. This means that we can't make the inequality correct by demanding that triangle ABC should be acute-angled or something like that.\r\n\r\n  Darij",
  "Solution_4": "[quote=\"darij grinberg\"][quote=\"zhaobin\"]can you expain more?I can't understand :([/quote]\n\nand there exists a triangle ABC whose angles are all in the interval $\\left[60^{\\circ}-\\epsilon;\\;60^{\\circ}+\\epsilon\\right]$ for which the inequality is wrong.   Darij[/quote]\r\nthank you, :)"
}
{
  "Tag": [
    "integration",
    "function",
    "calculus",
    "calculus computations"
  ],
  "Problem": "\\[ Evaluate : \\int_{0}^{1}\\sqrt[3]{2x^3 \\minus{} 3x^2 \\minus{} x \\plus{} 1 }   dx\\]",
  "Solution_1": "[quote=\"tariq1440\"]\n\\[ Evaluate : \\int_{0}^{1}\\sqrt [3]{2x^3 \\minus{} 3x^2 \\minus{} x \\plus{} 1 } dx\n\\]\n[/quote]\r\n\r\nHi tariq1440 :) !!!\r\n\r\nwe have: $ \\sqrt [3]{2x^3 \\minus{} 3x^2 \\minus{} x \\plus{} 1} \\equal{} \\sqrt [3]{2}\\sqrt [3]{(x \\minus{} \\frac {1}{2})( x \\minus{} \\phi) (x \\minus{} 1 \\plus{} \\phi)}$\r\n\r\nwith $ \\phi \\equal{} \\frac {1 \\plus{} \\sqrt {5}}{2}$  \r\n\r\nby changing the variable $ x \\equal{} t \\plus{} \\frac {1}{2}$\r\n\r\nwe find that: \r\n\r\n$ \\int_{0}^{1}\\sqrt [3]{2x^3 \\minus{} 3x^2 \\minus{} x \\plus{} 1 } dx \\equal{} \\sqrt [3]{2}\\int_{ \\minus{} \\frac {1}{2}}^{\\frac {1}{2}}\\sqrt [3]{t(t^2 \\minus{} \\frac {5}{4}) } dx\\equal{}0$\r\n\r\nbecause \r\n\r\n$ t \\rightarrow t(t^2 \\minus{} \\frac{5}{4})$ are odded (impair) functions \r\n\r\n$ .......$\r\n\r\nand thanks",
  "Solution_2": "Let $ u\\equal{}1\\minus{}x$\r\n\t\r\n$ \\int_{0}^{1}\\root{3}\\of{2x^{3}\\minus{}3x^{2}\\minus{}x\\plus{}1}dx\\equal{}\\allowbreak \\minus{}\\int_{0}^{1}\\root{3}\\of{2u^{3}\\minus{}3u^{2}\\minus{}u\\plus{}1}\\,du$\r\n\t\r\n$ \\int_{0}^{1}\\root{3}\\of{2x^{3}\\minus{}3x^{2}\\minus{}x\\plus{}1}dx\\equal{}0$",
  "Solution_3": "yes :) Indochina you're right :) !!\r\n\r\nif : $ f(x) \\equal{} \\sqrt[3]{2x^3 \\minus{} 3x^2 \\minus{}x \\plus{}1}$\r\n\r\nwe find that $ \\Omega ( \\frac{1}{2} ; 0)$ are a center of symetry\r\n\r\nthe same of my integral \r\n.....",
  "Solution_4": "WOW! :rotfl:",
  "Solution_5": "[[/quote]\r\n\r\nHi tariq1440 :) !!!\r\n\r\nwe have: $ \\sqrt [3]{2x^3 \\minus{} 3x^2 \\minus{} x \\plus{} 1} \\equal{} \\sqrt [3]{2}\\sqrt [3]{(x \\minus{} \\frac {1}{2})( x \\minus{} \\phi) (x \\minus{} 1 \\plus{} \\phi)}$\r\n\r\n[/quote]\r\n\r\n\r\nhow do you know this , please ?\r\n\r\nthanks\r\n\r\n :lol:  :)",
  "Solution_6": "[quote=\"IndoChina\"]Let $ u \\equal{} 1 \\minus{} x$\n\t\n$ \\int_{0}^{1}\\root{3}\\of{2x^{3} \\minus{} 3x^{2} \\minus{} x \\plus{} 1}dx \\equal{} \\allowbreak \\minus{} \\int_{0}^{1}\\root{3}\\of{2u^{3} \\minus{} 3u^{2} \\minus{} u \\plus{} 1}\\,du$\n\t\n$ \\int_{0}^{1}\\root{3}\\of{2x^{3} \\minus{} 3x^{2} \\minus{} x \\plus{} 1}dx \\equal{} 0$[/quote]\r\n\r\n\r\nvery nice , thanks",
  "Solution_7": "[quote=\"tariq1440\"]\n\n[quote=\"mathema*\"]\n\nHi tariq1440 :) !!!\n\nwe have: $ \\sqrt [3]{2x^3 \\minus{} 3x^2 \\minus{} x \\plus{} 1} \\equal{} \\sqrt [3]{2}\\sqrt [3]{(x \\minus{} \\frac {1}{2})( x \\minus{} \\phi) (x \\minus{} 1 \\plus{} \\phi)}$\n\n[/quote]\n\n\nhow do you know this , please ?\n\nthanks\n\n :lol:  :)[/quote]\r\n\r\nOk tariq !\r\n\r\nequation $ 2x^3 \\minus{} 3x^2 \\minus{} x \\plus{} 1 \\equal{} 0$ have as oubvious solution $ x \\equal{} \\frac {1}{2}$ than\r\n\r\n$ (2x^3 \\minus{} 3x^2 \\minus{} x \\plus{} 1) \\equal{} 2(x \\minus{} \\frac {1}{2})(x^2 \\minus{} x \\minus{} 1) \\equal{} ....$\r\n\r\n:)",
  "Solution_8": "[color=green]Moderator edit: redundant quote removed.[/color]\r\n\r\n\r\nthanks , my friend"
}
{
  "Tag": [
    "induction",
    "quadratics",
    "Gauss",
    "modular arithmetic",
    "number theory",
    "prime numbers",
    "number theory proposed"
  ],
  "Problem": "For all integers $n\\geq 1$ we define $x_{n+1}=x_1^2+x_2^2+\\cdots +x_n^2$, where $x_1$ is a positive integer. Find the least $x_1$ such that 2006 divides $x_{2006}$.",
  "Solution_1": "We have that $x_{n+1}=x_n+x_n^2$ hence all $x_n$ are even. We will prove by a backward induction that $x_2$ must be divisible by $1003$. \r\n\r\n$x_{n}\\equiv 0 (\\mod 1003)\\Leftrightarrow x_{n-1}^2+x_{n-1}\\equiv 0 (\\mod 1003)\\Leftrightarrow x_{n-1}\\equiv -1$ or $x_{n-1}\\equiv 0 (\\mod 1003)$. \r\n\r\nWe will prove that $x_{n-1}\\equiv -1$ cannot hold for $n-1\\ge 2$: \r\n\r\n$x_{n-1}=x_{n-2}^2+x_{n-2}\\equiv -1 (\\mod 1003)\\Leftrightarrow (2x_{n-2}+1)^2\\equiv -3 (\\mod 1003)$.\r\n\r\nBut from quadratic reciprocity law we get that $3$ is quadratic residue $\\mod 1003$, but since $1003=4\\cdot 250+3$ we have that $-1$ is not a quadratic residue $(\\mod 1003)$. Hence contradiction.\r\n\r\nSince $1003\\mid x_2$, we have that $x_1\\equiv 0,-1 (\\mod 1003)$. Hence the least value for $x_1$ is 1002, which works.",
  "Solution_2": "I think $x_1=1.$ But i am wrong. \r\nLet y=x(x+1)(mod 1003). If y=0, then x=0 or x=-1 (mod 1003). If y=-1=x(x+1) we have $(2x+1)^2=3 (mod 1003)$. It give contradition, because $(\\frac{3}{1003})=-(\\frac{1003}{3})=-1$. Therefore if $x_1\\not= 0,-1$, then $1003\\not | x_n$ for all n>1.",
  "Solution_3": "I proved that it is not 1. :P",
  "Solution_4": "You prove: if $x_1=0,-1(mod 1003)$, then  $2006|x_n$ for n>1. But you don't prove, that $x_1\\not =0,-1$ don't work.",
  "Solution_5": "I agree that $x_{n+1}=x_n^2+x_n$ but only for $n\\geq 2$ . Since from the question we can see that\r\n\r\n$x_2=x_1^2\\not =x_1+x_1^2$ \r\n\r\n@Rust : The question does say \" Positive Integer \" , so I think we doesnt have to prove the case for $x_1=0,-1$ .",
  "Solution_6": "There is some part not complete since I cant prove it , but i think it is true .\r\n\r\nI claim that $x_1=472$\r\n\r\nFrom question , $x_2=x_1^2$ and $x_n=x_{n-1}(1+x_{n-1})$ for $n\\geq 3$ \r\n\r\nOr $x_n=x_1^2(x_2+1)(x_3+1)\\cdots (x_{n-1}+1)$ ,$n\\geq 3$ .....(i)\r\n\r\nSince any $(x_i+1,x_j+1)=1$ and $(x_i+1,x_1)=1$ , the terms are all pairwise coprime .\r\n\r\nWe also notice that $x_n$ is even for all $n\\geq 3$ and $x_1^2(x_2+1)$ is even .\r\n\r\nIf $x_{2006}$ is divisible by $2006=2\\cdot 17\\cdot 59$ , then we know that there must be ONLY one $x_i+1$ or $x_1$ who has the factor of $17$ or $59$ .\r\n\r\nIf it is $x_k+1$ ($k\\geq 3)$ who has the factor of $17$ or $59$  , then \r\n(i) $x_k+1\\equiv 0\\mod 59$ $\\implies x_{k-1}(x_{k-1}+1)\\equiv -1\\mod 59$ \r\nHere is the part I dont know how to prove but i think there is no solution for $x_{k-1}$ which satisfy this ?\r\n\r\nif that is true , then we can say that it is either $x_2+1$ or $x_1$ who has the factor of $59$ . But $x_2+1\\equiv 0\\mod 59$ $\\implies x_1^2\\equiv -1\\mod 59$ (anyway to prove no solution for this other that brute force all cases?) . Hence we conclude $x_1\\equiv 0\\mod 59$......(i)\r\n\r\n(ii)As for the factor of $17$ , we do the same way and found that it must also be either $x_2+1$ or $x_1$ who has the factor of $17$ . But $x_2+1\\equiv 0\\mod 17$ $\\implies x_1\\equiv \\pm 4\\mod 17$ .....(ii) (another case is $x_1\\equiv 0\\mod 17$ but that will give us $x_1\\equiv 0\\mod 1003$ when combine with (ii) )\r\n\r\nHence by (i),(ii) , we use Chinese Remainder Theorem and conclude that either $x_1\\equiv 472\\mod 1003$ or $\\equiv 531\\mod 1003$ . So the smallest such $x_1$ is $\\boxed{472}$\r\n\r\n\r\nSomeone plz complete the \"incomplete\" part for me to make this solution valid :P",
  "Solution_7": "[quote=\"shyong\"] \n$\\implies x_{k-1}(x_{k-1}+1)\\equiv -1\\mod 59$ \nHere is the part I dont know how to prove but i think there is no solution for $x_{k-1}$ which satisfy this ?\n\nSomeone plz complete the \"incomplete\" part for me to make this solution valid :P[/quote]\r\n\r\nYou need to show that the equation\r\n$x^2 +x +1\\equiv 0 (\\mod 59 )$\r\ndoes not has solution in $Z_{59}$\r\nThis is equivalent to \r\n$(2x +1)^2 = -3 (\\mod 59)$\r\n\r\nNow $(\\frac{56}{59})= (\\frac{7}{59}) (\\frac{2}{59})^3= (\\frac{59}{7})=(\\frac 37)=-1$ , because $(\\frac{2}{prime(8k+3)} )=-1.$\r\n\r\nHence -3 is not quadratic residue modulo 59 !",
  "Solution_8": "which solution is true?",
  "Solution_9": "[quote=\"xirti\"]We have that $x_{n+1}=x_n+x_n^2$ hence all $x_n$ are even. We will prove by a backward induction that $x_2$ must be divisible by $1003$. \n\n$x_{n}\\equiv 0 (\\mod 1003)\\Leftrightarrow x_{n-1}^2+x_{n-1}\\equiv 0 (\\mod 1003)\\Leftrightarrow x_{n-1}\\equiv -1$ or $x_{n-1}\\equiv 0 (\\mod 1003)$. \n\nWe will prove that $x_{n-1}\\equiv -1$ cannot hold for $n-1\\ge 2$: \n\n$x_{n-1}=x_{n-2}^2+x_{n-2}\\equiv -1 (\\mod 1003)\\Leftrightarrow (2x_{n-2}+1)^2\\equiv -3 (\\mod 1003)$.\n\nBut from quadratic reciprocity law we get that $3$ is quadratic residue $\\mod 1003$, but since $1003=4\\cdot 250+3$ we have that $-1$ is not a quadratic residue $(\\mod 1003)$. Hence contradiction.\n\nSince $1003\\mid x_2$, we have that $x_1\\equiv 0,-1 (\\mod 1003)$. Hence the least value for $x_1$ is 1002, which works.[/quote]\n\nHy I can't get something 1003 is not a prime number so how can you apply Gauss lemma in your first line ?",
  "Solution_10": "[quote=\"shyong\"]There is some part not complete since I cant prove it , but i think it is true .\n\nI claim that $x_1=472$\n\nFrom question , $x_2=x_1^2$ and $x_n=x_{n-1}(1+x_{n-1})$ for $n\\geq 3$ \n\nOr $x_n=x_1^2(x_2+1)(x_3+1)\\cdots (x_{n-1}+1)$ ,$n\\geq 3$ .....(i)\n\nSince any $(x_i+1,x_j+1)=1$ and $(x_i+1,x_1)=1$ , the terms are all pairwise coprime .\n\nWe also notice that $x_n$ is even for all $n\\geq 3$ and $x_1^2(x_2+1)$ is even .\n\nIf $x_{2006}$ is divisible by $2006=2\\cdot 17\\cdot 59$ , then we know that there must be ONLY one $x_i+1$ or $x_1$ who has the factor of $17$ or $59$ .\n\nIf it is $x_k+1$ ($k\\geq 3)$ who has the factor of $17$ or $59$  , then \n(i) $x_k+1\\equiv 0\\mod 59$ $\\implies x_{k-1}(x_{k-1}+1)\\equiv -1\\mod 59$ \nHere is the part I dont know how to prove but i think there is no solution for $x_{k-1}$ which satisfy this ?\n\nif that is true , then we can say that it is either $x_2+1$ or $x_1$ who has the factor of $59$ . But $x_2+1\\equiv 0\\mod 59$ $\\implies x_1^2\\equiv -1\\mod 59$ (anyway to prove no solution for this other that brute force all cases?) . Hence we conclude $x_1\\equiv 0\\mod 59$......(i)\n\n(ii)As for the factor of $17$ , we do the same way and found that it must also be either $x_2+1$ or $x_1$ who has the factor of $17$ . But $x_2+1\\equiv 0\\mod 17$ $\\implies x_1\\equiv \\pm 4\\mod 17$ .....(ii) (another case is $x_1\\equiv 0\\mod 17$ but that will give us $x_1\\equiv 0\\mod 1003$ when combine with (ii) )\n\nHence by (i),(ii) , we use Chinese Remainder Theorem and conclude that either $x_1\\equiv 472\\mod 1003$ or $\\equiv 531\\mod 1003$ . So the smallest such $x_1$ is $\\boxed{472}$\n\n\nSomeone plz complete the \"incomplete\" part for me to make this solution valid :P[/quote]\nYou're right, that is the right value. My solution is similar to xirti one, but the problem in his solution is that he forget some Ideas, this is a correct form.\n[hide=\" Solution\"]\nFirst, not that the proprety $x_{n+1}=x_{n}+x_{n}^2$ is true only for $n\\geq 2$ and so for all $n\\geq 3$, $x_n$ is even. Now let $p\\in \\{17,59\\}$, it's then sufficient to find the smallest $x_1$ for which $p|x_{2006}$. Let $n\\geq 3$ if $p|x_{n+1}$ the $p|x_n$ or $p|x_n+1$. If $p|x_{n}$ then $p|x_{n-1}^2+x_{n-1}+1$ or $p|(2x_{n-1}+1)^2+3$ which is wrong because, $-3$ is not a quadratic residue modulo $17$ or $59$ (that's a simplificated way to use reciprocity law on prime numbers), and then $p|x_{n}$, so what we have shown is that, for every integer $n\\geq 3$, if $p|x_{n+1}$ consequently: $p|x_{n}$, so $p|x_3=x_1^4+x_1^2$\ntherefore $p|x_1^2(x_1^2+1)$. In order to minimise $x_1$, we have to find the smallest that verify one of the following cases:\n[b]case 1.[/b] $17|x_1^2$ and $59|x_1^2+1$\n[b]case 2.[/b] $59|x_1^2$ and $17|x_1^2+1$\nThe first case is obviously impossible since $-1$ isn't a quadratic residu modulo $59$.\nOn the second case, we put $x_1=59k$ remark that $59^2\\equiv 13\\pmod{17}$ so, $17|13k^2+1$ and so, $k^2\\equiv 13 \\pmod{17}$ which is equivalent with $k\\equiv 8\\pmod{17}$ or $k\\equiv 9 \\pmod{17}$. So $k_{min}=8$ and hence the smallest $x_1$ verifying the problem is $8\\times 59$ or, $x_1=472$.\nThe reciproc is obvious.\n[/hide]",
  "Solution_11": "I find that[b] x_1=118 [/b]satisfies the problem, (since x-2=118*119=7*2006) thus all x_i greater than 1 is divisible by 2006. for this reason consider the first index d such that 17 or 59 ( no matter because they have [b]reminder -1 while dividing by 6[/b])divide  x_d . thus we must have x_d =-1 (mod 17 or 59) and (x_d-1)^2 +x_d-1+1=0 (mod 17 or 59) but all of primes dividing x^2 +x+1 is of the form 6k+1. contradiction. thus we must have this cases:[b] x_1 =-1(mod 59,17) x_1=0 (mod 17) and -1 (mod 59) and x_1=-1(mod 17) and x_1=0 (mod 59)[/b]. after some calculations we can see that 118 obtained from the third cases.",
  "Solution_12": "I think you made a mistake  :)  $ x_{2}=x_{1}^{2} $  :wink:"
}
{
  "Tag": [
    "\\/closed"
  ],
  "Problem": "Is there going to be Intermediate Num Theo or something of that nature?",
  "Solution_1": "We expect to offer that next year some time."
}
{
  "Tag": [],
  "Problem": "We were asked to measure the density of water. however, when the water was transfered to the graduated cylinder (to obtain the volume) we were asked to compute the density of water using the mass and weight excluding the 1mL...why is this so? thanks",
  "Solution_1": "Excluding which 1 mL?  And density is based on mass, not weight.",
  "Solution_2": "ok. let me explain to you this. We were given an unknown. We were asked to compute the density of the liquid. The instructions are as follows: obtain 1 mL of that liquid. record the mass of the liquid. Now, add 1mL to the 1mL liquid (in other words, now, there are 2 mL of the liquid). Record the new mass of the two mL liquid. Compute the density of the liquid by excluding the initial 1mL's mass and volume.\r\n\r\nSo like, if this is the data:\r\n                  Weight of graduated cylinder + 1mL           Weight of graduated cylinder + 2mL   Final Volume\r\n                             30.67                                                                 31.57                        2\r\n\r\nSo, the density of the liquid (Excluding the initial 1mL's volume and mass) is: (31.57-30.67)/(2-1)=0.9\r\n\r\nin other words, why couldnt we just have weighed the initial 1mL and use the mass and volume directly to get the density? that's my questions...",
  "Solution_3": "I'm not sure if I completely understand it yet.  In order to measure the density, essentially all you have to do is mass a graduated cylinder, pour out 1 mL, and mass the combination.  That should be it.",
  "Solution_4": "oh never mind. someone told me that the base of the qraduated cylinders were not standardized. that's why"
}
{
  "Tag": [
    "function",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "Find n0 \\in N* such that for all n \\in N this ineq holds:\r\n\r\n284^n>n^24\r\n\r\nI tried using the function 284^x - x^24 and to aproximate it's solutions using Newton's methods.. but i'm not sure it is good...",
  "Solution_1": "i'm sorry for deleting Kuba's messege, but this problem is in a ROmanian Gazzette and it's too soon to answer it!\r\n\r\nI also locked the topic!",
  "Solution_2": "I guess we can unlock it now?"
}
{
  "Tag": [],
  "Problem": "i realy want to see your proof for these:\r\n$a,b,c$ are real numbers prove that:\r\n1)$(-a)(-b)=ab$\r\n2)$(-a)(b)=-ab$\r\n3)$\\dfrac{a}{a}=1$\r\n4)$a \\times \\b=0 \\rightarrow a=0 \\bigvee b=0$\r\n5)a(b+c)=a.b+ac\r\nthere is more but i think it is enough for now.please dont say these are obvious ,please prove these even they are obvious and send me your proof. :D",
  "Solution_1": "come on boys,you can do it and you will like it,so dint waist the time and prove these.i promisse you will lose nothing.",
  "Solution_2": "Good morning,\r\n\r\nhttp://smard.cqu.edu.au/Database/Junior/Number/Integers/jn_in002.txt\r\n\r\nsums it up nicely.\r\n\r\n-- Owlbert",
  "Solution_3": "Please pay greater attention to the level of each forum.  This isn't the first time I've explained that you are posting inappropriately difficult material in the middle school forums.  If you continue, I'll have to take away your ability to post in the middle school area.",
  "Solution_4": "i though this problems are easy and so i possed it here if i knew they are not i won't post it here.i am realy sorry.and thanks you.\r\n :D"
}
{
  "Tag": [
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "If the last three digits of $1978^{m}$ & $1978^{n}$ are the same with $1\\leq m<n$ , then find the least value of $m+n$.",
  "Solution_1": "This was one of the contest problems on IMO '78. Thus I'm sure this was posted before :)",
  "Solution_2": "http://www.mathlinks.ro/Forum/viewtopic.php?p=367368#p367368"
}
{
  "Tag": [
    "vector",
    "trigonometry",
    "linear algebra",
    "linear algebra unsolved"
  ],
  "Problem": "Let $ A \\in M_2(\\mathbb{Z})$ such that there exists $ n \\in \\mathbb{N}^{*}$, $ (n,6)\\equal{}1$, and $ A^n\\equal{}I_2$. Prove that $ A\\equal{}I_2$.",
  "Solution_1": "From $ A^n\\equal{}I$ follows that if $ \\lambda , \\overline{\\lambda} \\in\\mathbb{C}$ are eigenvalues of $ A$ then they must be roots of the equation $ z^n\\equal{}1$.Suppose that $ \\lambda\\equal{} \\cos \\frac{2k\\pi}{n} \\plus{}i \\sin \\frac{2k\\pi}{n}$ and since $ tr(A)\\equal{}\\lambda \\plus{}\\overline{\\lambda} \\equal{}2\\cos \\frac{2k\\pi}{n} \\in \\mathbb{Z}$ we must have $ 2\\cos \\frac{2k\\pi}{n} \\in \\{\\pm 1,\\pm 2, 0\\}$.And since $ (n,6)\\equal{}1$ it follows that $ \\cos \\frac{2k\\pi}{n}$ can only be equal to $ 1$. So both eigenvalues of $ A$ are equal to $ 1$.Therefore, from Cayley-Hamilton theorem we get that $ A^2\\equal{}2A\\minus{}I$ , and so on, inductively we get that $ A^n\\equal{}nA\\minus{}(n\\minus{}1)I$ which from the hypothesis we know that it is equal to $ I$. So $ A^n\\equal{}nA\\minus{}(n\\minus{}1)I\\equal{}I$ $ \\Rightarrow$ $ A\\equal{}I$."
}
{
  "Tag": [
    "trigonometry",
    "geometry",
    "rectangle",
    "circumcircle",
    "geometry unsolved"
  ],
  "Problem": "Let ABC be a right triangle with right angle at A has altitude AH > Let N,M be the midpoints AB,AC, respectively. The two lines BM and AH intersect at D and the two lines CN and AH intersect at E .Prove that $ \\hat{ACD}\\equal{}\\hat{ABE}$",
  "Solution_1": "[size=134][color=darkred][b]Nice problem ![/b][/color][/size]\r\n\r\n[quote=\"tranquoc\"] Let $ ABC$ be a $ A$ - right triangle . Denote $ \\left\\|\\begin{array}{c} H\\in BC\\ ,\\ AH\\perp BC \\\\\n \\\\\nN\\in [AB]\\ ,\\ NA \\equal{} NB \\\\\n \\\\\nM\\in [AC]\\ ,\\ MA \\equal{} MC\\end{array}\\right\\|$ and $ \\left\\|\\begin{array}{c} E\\in CN\\cap AH \\\\\n \\\\\nD\\in BM\\cap AH\\end{array}\\right\\|$ . Prove that $ \\widehat{ACD}\\equiv\\widehat{ABE}$ . [/quote]\r\nProve easily that $ \\boxed {\\ \\tan\\widehat {ACD} \\equal{} \\tan\\widehat {ABE} \\equal{} \\frac {bc}{a^2}\\le \\frac 12\\ }$ .",
  "Solution_2": "We denote the points $ K\\equiv BE\\cap ON$ and $ L\\equiv BC\\cap AK,$ where $ O$ is the midpoint of the side-segment $ BC$ of the given triangle $ \\bigtriangleup ABC.$\r\n\r\nBecause of the line segment $ ON,$ is the midperpendicular of $ AB,$ we have that $ \\angle ABE \\equal{} \\angle BAL$ and so, it is enough to prove that $ \\frac {LH}{LB} \\equal{} \\frac {DH}{DA},$ \r\nbecause of the similar right triangles $ \\bigtriangleup HAB$ and $ \\bigtriangleup HCA,$ with $ \\angle HAB \\equal{} \\angle HCA.$\r\n\r\nApplying the [b][size=100]Menelaus theorem[/size][/b], we can say that:\r\n\r\nFrom the right triangle $ \\bigtriangleup HBE,$ with transversal $ AKL$ $ \\Longrightarrow$ $ \\frac {LH}{LB}\\cdot \\frac {KB}{KE}\\cdot \\frac {AE}{AH} \\equal{} 1$ $ \\Longrightarrow$ $ \\frac {LH}{LB} \\equal{} \\frac {KE}{KB}\\cdot \\frac {AH}{AE}$ $ ,(1)$\r\n\r\nFrom the triangle $ \\bigtriangleup BEC,$ with transversal $ OKN$ $ \\Longrightarrow$ $ \\frac {KE}{KB}\\cdot \\frac {OB}{OC}\\cdot \\frac {NC}{NE} \\equal{} 1$ $ \\Longrightarrow$ $ \\frac {KE}{KB} \\equal{} \\frac {NE}{NC}$ $ ,(2)$\r\n\r\nFrom the right triangle $ \\bigtriangleup HEC,$ with transversal $ ANB$ $ \\Longrightarrow$ $ \\frac {AH}{AE}\\cdot \\frac {NE}{NC}\\cdot \\frac {BC}{BH} \\equal{} 1$ $ \\Longrightarrow$ $ \\frac {AH}{AE} \\equal{} \\frac {NC}{NE}\\cdot \\frac {BH}{BC}$ $ ,(3)$\r\n\r\nFrom $ (1),$ $ (2),$ $ (3)$ $ \\Longrightarrow$ $ \\frac {LH}{LB} \\equal{} \\frac {BH}{BC}$ $ ,(4)$\r\n\r\nFrom the right triangle $ \\bigtriangleup HAC,$ with transversal $ BDM$ $ \\Longrightarrow$ $ \\frac {DH}{DA}\\cdot \\frac {MA}{MC}\\cdot \\frac {BC}{BH} \\equal{} 1$ $ \\Longrightarrow$ $ \\frac {DH}{DA} \\equal{} \\frac {BH}{BC}$ $ ,(5)$\r\n\r\nFrom $ (4),$ $ (5)$ $ \\Longrightarrow$ $ \\frac {LH}{LB} \\equal{} \\frac {DH}{DA}$ $ ,(6)$ and the proof is completed.\r\n\r\nKostas Vittas.",
  "Solution_3": "[quote=\"tranquoc\"] [color=darkred]Let $ ABC$ be a $ A$ - right triangle . Denote $ \\left\\|\\begin{array}{c} H\\in BC\\ ,\\ AH\\perp BC \\\\\n \\\\\nN\\in [AB]\\ ,\\ NA \\equal{} NB \\\\\n \\\\\nM\\in [AC]\\ ,\\ MA \\equal{} MC\\end{array}\\right\\|$ and $ \\left\\|\\begin{array}{c} E\\in CN\\cap AH \\\\\n \\\\\nD\\in BM\\cap AH\\end{array}\\right\\|$ . Prove that $ \\widehat{ACD}\\equiv\\widehat{ABE}$ .[/color] [/quote]\r\n\r\n[color=darkblue][b][u]Proof I (synthetic).[/u][/b] Let $ \\left\\|\\begin{array}{ccc} P\\in AB\\cap CD \\\\\n \\\\\nR\\in AC\\cap BE\\end{array}\\right\\|$ and $ \\left\\|\\begin{array}{c} AH \\equal{} h \\\\\n \\\\\n\\phi \\equal{} m(\\widehat {ACD}) \\\\\n \\\\\n\\psi \\equal{} m(\\widehat {ABE})\\end{array}\\right\\|$ . Since $ D$ , $ E$ belong to the medians $ BM$ , $ CN$ respectively obtain :\n\n$ \\boxed {\\ \\left\\|\\begin{array}{c} HP\\parallel AC \\implies\\ HP\\perp AB\\implies PHA\\sim ABC\\implies\\frac {AP}{CA} \\equal{} \\frac {AH}{CB}\\implies\\tan\\phi \\equal{} \\frac {h}{a} \\\\\n \\\\\nHR\\parallel AB \\implies HR\\perp AC\\implies RHA\\sim ACB\\implies\\frac {AR}{BA} \\equal{} \\frac {AH}{BC}\\implies\\tan\\psi \\equal{} \\frac {h}{a}\\end{array}\\right\\|\\implies\\widehat{ACD}\\equiv\\widehat {ABE}\\ }$ .\n\n[b][u]Proof II (metric).[/u][/b] Apply the [b]Menelaus' theorem[/b] to the transversal $ \\overline {BDM}$ and the triangle $ ACH$ :\n\n$ \\frac {BH}{BC}\\cdot\\frac {MC}{MA}\\cdot\\frac {DA}{DH} \\equal{} 1\\implies$ $ \\boxed {\\ \\frac {DA}{DH} \\equal{} \\frac {a^2}{c^2}\\ }$ .  Apply an well-known relation to the cevian $ [CD$ in the triangle $ ACH$ : \n\n$ \\frac {DA}{DH} \\equal{} \\frac {CA}{CH}\\cdot\\frac {\\sin\\widehat {DCA}}{\\sin\\widehat {DCH}} \\equal{}$ $ \\frac {a\\sin\\phi}{b\\sin (C \\minus{} \\phi)}$ $ \\implies$ $ \\boxed {\\ \\frac {DA}{DH} \\equal{} \\frac {a^2\\tan\\phi}{b(c \\minus{} b\\tan\\phi )}\\ }$ . Thus, $ \\frac {1}{c^2} \\equal{} \\frac {\\tan\\phi}{b(c \\minus{} b\\tan\\phi )}$ $ \\implies$ $ \\tan\\phi \\equal{} \\frac {bc}{b^2 \\plus{} c^2}$ .\n\nAnalogously obtain $ \\tan\\psi \\equal{} \\frac {bc}{b^2 \\plus{} c^2}$ . In conclusion, $ \\widehat{ACD}\\equiv\\widehat{ABE}$ . Shortly, $ \\tan\\phi \\equal{} \\tan \\psi \\equal{} \\frac {bc}{a^2} \\equal{} \\frac {h_a}{a}$ .[/color]\r\n \r\n\r\n[color=darkred][b][u]Remark.[/u][/b] Construct, outside of $ \\triangle ABC$ , the rectangle $ BCST$ , where $ BT \\equal{} CS \\equal{} h$ . Denote the circumcircle $ \\Gamma$ of this rectangle. \n\nProve easily that $ S\\in PD$ , $ T\\in RD$ , $ \\widehat {CBS}\\equiv\\widehat {ACD}\\equiv\\widehat {ABE}$ , i.e. the points $ P$ , $ R$ belong to the circle $ \\Gamma$ . \n\nIf $ \\{X,Y\\} \\equal{} AH\\cap \\Gamma$ , then $ XY \\equal{} h\\sqrt 3$ and $ AX \\equal{} \\frac {\\sqrt 7 \\minus{} \\sqrt 3}{2}\\cdot h$ .[/color]",
  "Solution_4": "Here is a solution similar to [b]Virgil Nicula's[/b], but differs at the end:\r\n[hide=\"Solution\"]\nExtend $ CD$ and $ BE$ to meet $ AB$ and $ AC$ at $ Y$ and $ X$ respectively.  Then, Ceva's Theorem gives that $ \\frac {YB}{AY}\\cdot \\frac {AM}{MC}\\cdot \\frac {CH}{BH} \\equal{} 1$.  Since $ AM\\equal{}MC$, we have that $ \\frac {YB}{AY} \\equal{} \\frac {BH}{CH}$ so $ YH\\parallel AC$.  Similarly, $ HX\\parallel AB$.  Now, this means that since $ AC\\perp AB$, we have that $ HY\\perp AB$.  Similarly, $ HX\\perp AC$.  Since $ \\angle AHB\\equal{}90$ and $ HY\\perp AB$, we have that $ \\angle YHA \\equal{} \\angle ABC$.  Yet, $ \\angle HYA \\plus{} \\angle HXA \\equal{} 90 \\plus{} 90 \\equal{} 180$, so $ YHXA$ is cyclic, which implies that $ \\angle YHA \\equal{} \\angle YXA$.  This means that $ \\angle YBC \\equal{} \\angle YXA$, so $ BYXC$ is cyclic, so $ \\angle ABE \\equal{} \\angle YBX \\equal{} \\angle YCX \\equal{} \\angle ACD$. [/hide]",
  "Solution_5": "See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=365571][color=red][b]here[/b][/color][/url]"
}
{
  "Tag": [
    "function",
    "LaTeX"
  ],
  "Problem": "When creating user-defined functions, which should be used - \\newcommand or \\def?\r\n\r\nWhat are the differences between these two commands (apart from \\newcommand having the restriction that the name of your function can't begin with \"end\")? Are there specific cases where one should be used over the other?\r\n\r\nThanks.",
  "Solution_1": "\\def is a TeX primitive whilst \\newcommand is a \\LaTeX macro which, amongst other things, checks for clashes of names with other commands and will allow commands like \\par in their definition. Thus \\newcommand calls \\def at some point but does other things, so is preferred if possible.",
  "Solution_2": "Alright. Thanks stevem!"
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "analytic geometry",
    "geometric transformation",
    "rotation",
    "reflection"
  ],
  "Problem": "Have you ever seen those problems about folding a group of squares into a cube?  For example, in the diagram below, the first group of squares can be folded into a cube, whereas the second group of squares cannot.  (Each O represents a square.  Sorry for the crude picture.) \r\n\r\n[code]\nOO\n OO\n  OO\n\n O\nOO\n OOO\n[/code]\r\n\r\nAnyway, I think the usual way of solving these problems is to try to visualize the folding in 3 dimensions.  Because I have trouble visualizing in 3D, and for curiousity's sake, I was wondering whether there is a method to figure out the answer by sticking to 2 dimensions.  Is there some geometric or graph-theoretic principle that would allow us to determine when a group of squares can be folded into a cube?",
  "Solution_1": "[hide]\n\nI cannot find any set formula for unfolded cubes, but I have noticed two different trends for acceptable arrangements:\n\n\n\nThe first is based off the traditional 'cross' arrangement; given an arrangement of 5 squares that form a 'T' (three high, a cross bar three wide, with a shared square at the top), the 6th square can be placed anywhere on the top or bottom of the 'T'.\n\n\n\nFor example:\n\n0\n\n000\n\n-0\n\n-0\n\n\n\n(the hyphens are place holders)\n\n\n\nThe second comes from the idea that each square is involved in forming a corner.  If each square is part of a three-square right angle (a square can be part of more than one), provided two adjacent groups don't mirror each other to form a 'u', then a square can be formed.\n\n\n\nFor example:\n\n-0\n\n000      works\n\n--00\n\n\n\nbut\n\n\n\n-0\n\n000      doesn't\n\n0  0 \n\n[/hide]\n\n\n\nI am definitely open for someone to come up with a more generalized explanation.",
  "Solution_2": "[quote=\"Ravi B\"]Have you ever seen those problems about folding a group of squares into a cube?  For example, in the diagram below, the first group of squares can be folded into a cube, whereas the second group of squares cannot.  (Each O represents a square.  Sorry for the crude picture.) \n\n[code]\nOO\n OO\n  OO\n\n O\nOO\n OOO\n[/code]\n\nAnyway, I think the usual way of solving these problems is to try to visualize the folding in 3 dimensions.  Because I have trouble visualizing in 3D, and for curiousity's sake, I was wondering whether there is a method to figure out the answer by sticking to 2 dimensions.  Is there some geometric or graph-theoretic principle that would allow us to determine when a group of squares can be folded into a cube?[/quote]\r\n\r\nWe mathematicians regularly work in four and more dimensions and most of us cannot visualize in 4D. We maneuver by coordinates instead.\r\n\r\nThe following solution to your question is longer and more tedious than required. There are simpler ways to tell if a pattern of squares can be folded into a cube. But this is a chance to teach the power of coordinate systems.\r\n\r\nIf a flat pattern of squares were folded into a cube, one of those squares would be the bottom face of the cube. So imagine that the folded cube is in Cartesion space with its corners at (0,0,0), (0,0,1), (0,1,0), (0,1,1), (1,0,0), (1,0,1), (1,1,0), and (1,1,1). The bottom face has the corners (0,0,0), (0,1,0), (1,0,0), and (1,1,0), which I will abbreviate as 000, 010, 100, and 110. The edge between 000 and 100 is the set {([i]x[/i],0,0) | 0 :le: [i]x[/i]  :le: 1 }, which I will abbreviate as x00. The edge between 100 and 110 is the set {(1,[i]y[/i],0) | 0 :le: [i]y[/i]  :le: 1 }, which I will abbreviate as 1y0. And so on. The square itself I will call xy0.\r\n\r\n[tex]\\begin{picture}(100,80)\r\n\\put(20,10){\\framebox(60,60){}}\r\n\\put(0,0){$000$}\r\n\\put(81,0){$100$}\r\n\\put(81,72){$110$}\r\n\\put(0,72){$010$}\r\n\\put(40,37){$xy0$}\r\n\\put(40,0){$x00$}\r\n\\put(81,37){$1y0$}\r\n\\put(40,72){$x10$}\r\n\\put(0,37){$0y0$}\r\n\\end{picture}[/tex]\r\n\r\nNow suppose we add another square to the pattern.\r\n\r\n[tex]\\begin{picture}(160,80)\r\n\\put(20,10){\\framebox(60,60){}}\r\n\\put(0,0){$000$}\r\n\\put(71,0){$100$}\r\n\\put(71,72){$110$}\r\n\\put(0,72){$010$}\r\n\\put(40,37){$xy0$}\r\n\\put(40,0){$x00$}\r\n\\put(81,37){$1y0$}\r\n\\put(40,72){$x10$}\r\n\\put(0,37){$0y0$}\r\n\\put(80,10){\\framebox(60,60){}}\r\n\\end{picture}[/tex]\r\n\r\nThe edge in the new square that touches vertex 100 will be perpendicular to the original square xy0 when the pattern is folded. So it must go in the z direction. Thus, the third digit in its label is z. The other two digits we can take from the point 100, so the edge is 10z.\r\n\r\nLikewise the top edge of the new square starts at 110 and points in the z direction after folding, so it is 11z.\r\n\r\nThe other end of 10z must be 101 and the other end of 11z must be 111. The edge between 101 and 111 changes in the y coordinate, so it is 1y1. This makes it parallel to edge 1y0, which makes sense because opposite sides of a square are parallel.\r\n\r\nFinally, the whole square has edges in the y direction and the z direction, so it must be in the yz-plane. The x coordinate of all its vertices is 1, so the square is 1yz.\r\n\r\n[tex]\\begin{picture}(160,80)\r\n\\put(20,10){\\framebox(60,60){}}\r\n\\put(0,0){$000$}\r\n\\put(71,0){$100$}\r\n\\put(71,72){$110$}\r\n\\put(0,72){$010$}\r\n\\put(40,37){$xy0$}\r\n\\put(40,0){$x00$}\r\n\\put(81,37){$1y0$}\r\n\\put(40,72){$x10$}\r\n\\put(0,37){$0y0$}\r\n\\put(80,10){\\framebox(60,60){}}\r\n\\put(141,0){$101$}\r\n\\put(141,72){$111$}\r\n\\put(105,0){$10z$}\r\n\\put(141,37){$1y1$}\r\n\\put(105,72){$11z$}\r\n\\put(105,37){$1yz$}\r\n\\end{picture}[/tex]\r\n\r\nLet us add another squares above the second square. I already labeled it in the diagram below. I had to leave out some old labels, because apparently the TeX simulator in this forum has a limit in how many picture elements it can handle.\r\n\r\n[tex]\\begin{picture}(160,140)\r\n\\put(20,10){\\framebox(60,60){}}\r\n\\put(62,72){$110$}\r\n\\put(0,72){$010$}\r\n\\put(40,37){$xy0$}\r\n\\put(40,72){$x10$}\r\n\\put(81,37){$1y0$}\r\n\\put(80,10){\\framebox(60,60){}}\r\n\\put(141,0){$101$}\r\n\\put(141,72){$111$}\r\n\\put(105,0){$10z$}\r\n\\put(141,37){$1y1$}\r\n\\put(105,72){$11z$}\r\n\\put(105,37){$1yz$}\r\n\\put(80,70){\\framebox(60,60){}}\r\n\\put(141,132){$011$}\r\n\\put(141,97){$x11$}\r\n\\put(105,132){$01z$}\r\n\\put(105,97){$x1z$}\r\n\\put(62,132){$010$}\r\n\\put(61,97){$x10$}\r\n\\end{picture}[/tex]\r\n\r\nIn this diagram, two edges and two vertices have the same label. But that is fine, because when the sides are folded, those edges and vertices are going to be glued together into a single edge and vertex.\r\n\r\nWe can see that it is impossible to now add a new square above the original square xy0. That square would have two perpendicular edges both labeled x10. Since x-direction edges are parallel, they cannot be perpendicular. Also, that new square would end up labeled x1z (if we labeled via its bottom edge) or xy0 (if we labeled via its right edge), and we already have squares labeled x1z and xy0. Repeating labels on edges is fine, but repeating labels on squares means that those squares will end up on the same face when the pattern is folded into a cube.\r\n\r\nIf we added three squares to the left of the second square, for a total of six squares, we would have enough squares to make a cube. But does that pattern fold into a cube? The diagram below shows the labels on the squares. No square folds over onto face x0z, so this cannot form a cube.\r\n\r\n[tex]\\begin{picture}(125,50)\r\n\\put(0,0){\\framebox(25,25){}}\r\n\\put(25,0){\\framebox(25,25){}}\r\n\\put(50,0){\\framebox(25,25){}}\r\n\\put(75,0){\\framebox(25,25){}}\r\n\\put(100,0){\\framebox(25,25){}}\r\n\\put(25,25){\\framebox(25,25){}}\r\n\\put(3,8){$xy0$}\r\n\\put(28,8){$1yz$}\r\n\\put(53,8){$xy1$}\r\n\\put(78,8){$0yz$}\r\n\\put(103,8){$xy0$}\r\n\\put(28,33){$x1z$}\r\n\\end{picture}[/tex]\r\n\r\nOn the other hand, if the extra xy0 square on the left is moved to beneath the square it touches, it becomes labeled as the x0z square that we need. The pattern below will fold into a cube.\r\n\r\n[tex]\\begin{picture}(100,75)\r\n\\put(0,25){\\framebox(25,25){}}\r\n\\put(25,25){\\framebox(25,25){}}\r\n\\put(50,25){\\framebox(25,25){}}\r\n\\put(75,25){\\framebox(25,25){}}\r\n\\put(25,50){\\framebox(25,25){}}\r\n\\put(75,0){\\framebox(25,25){}}\r\n\\put(3,33){$xy0$}\r\n\\put(28,33){$1yz$}\r\n\\put(53,33){$xy1$}\r\n\\put(78,33){$0yz$}\r\n\\put(28,58){$x1z$}\r\n\\put(78,8){$x0z$}\r\n\\end{picture}[/tex]\r\n\r\nThe fastest method of determining whether a pattern of six squares will fold into a cube is to memorize the ones that do. There are only eleven of them, not counting rotations and reflections (I checked rigorously). I leave it as a challenge to the rest of you to find them all.\r\n\r\nErin Schram",
  "Solution_3": "Thanks for the explanation and the pictures!  I will think more about your coordinate method."
}
{
  "Tag": [
    "AMC",
    "AIME",
    "analytic geometry",
    "AMC 10"
  ],
  "Problem": "So what did you get?",
  "Solution_1": "129. fail. :(",
  "Solution_2": "I got an okay score. At least I beat my goal of qualifying for AIME by a positive multiple of 10.",
  "Solution_3": "Hmm Yongyi I am sure your score was \"okay.\"\r\n\r\nI got 144. Yay for failing similar triangles on #17.",
  "Solution_4": "I don't deserve to talk about my score.\r\n\r\nSo I won't :(.",
  "Solution_5": "Well, either do I^\r\n\r\nI feel so disappointed...now I need like a 9 on the AIME inorder to pass because I guessed away 7.5 points !!!!!!!!!!!!!!!!!!  \r\n\r\nSad thing was, if I had done 22 in the time limit (I did it in lunch in my head), I wouldve not guessed on the 4 remaining and wouldve gotten 132.\r\n\r\nWhich woulda beaten Lyndon!!  Too bad I didnt and only got a 120...127.5 would sure look nice.",
  "Solution_6": "Apparently Lyndon epic phailed compared to last year?",
  "Solution_7": "xpmath, I coordinate bashed number 17 :D\r\n\r\n138 for me (trivialized 22 even though it was epicly trivial, put 1/6 and failed number 25).  My brother got number 25 right because he thought 2^7=64  :| \r\n\r\nI hope 7 on AIME will qualify for USAMO.",
  "Solution_8": "Coordinate bashing on #17 should be automatic disqualification from AIME. :P\r\n\r\nSeriously, I solved it in like 10 seconds with similar triangles.",
  "Solution_9": "YES WELL SOME OF US AREN'T AS GOOD AS YOU ARE.\r\n\r\nAhem. So apparently this AMC 10 was relatively easy, according to everybody? Then again I thought last year's 10A was around the same difficulty level, so...\r\n\r\nDarn I still haven't gotten rid of my spamming urge. I'll go back to my own forum now.",
  "Solution_10": "i took AMC 12A. and phailed. i got like a 130.5",
  "Solution_11": "114. I hope to do better.  Actually, I will do better!!"
}
{
  "Tag": [
    "function",
    "complex analysis",
    "complex analysis unsolved"
  ],
  "Problem": "$ K\\equal{}\\{z\\in\\mathbb{C}: |z|\\le 1\\},\\quad T\\equal{}\\{z\\in\\mathbb{C}: |z|\\equal{}1\\},\\quad |a|<1$\r\n\r\ndefine $ h: K\\to C$ by $ h(z)\\equal{}\\frac{z\\minus{}a}{1\\minus{}\\overline{a}z}$\r\n\r\nfind $ \\min_{z\\in T}|h(z)|$",
  "Solution_1": "Note that $ h$ is a Mobious Transformation, which maps circles to circles and quick calculation shows\r\n\\[ |h(e^{i\\theta})| \\equal{} \\left|\\frac {e^{i\\theta} \\minus{} a}{1 \\minus{} \\overline{a}e^{i\\theta}}\\right| \\equal{} \\left|\\frac {e^{i\\theta} \\minus{} a}{e^{ \\minus{} i\\theta} \\minus{} \\overline{a}}\\right| \\equal{} 1.\r\n\\]\r\nHence $ h(T) \\equal{} T$. It can be further shown that the maps $ \\phi_a(z) : \\equal{} \\frac {z \\minus{} a}{1 \\minus{} \\overline{a}z}$ for $ |a| < 1$ are all of the conformal maps of the open unit disk onto itself (up to rotation).",
  "Solution_2": "thanks Jimmy !! :)"
}
{
  "Tag": [
    "function",
    "quadratics",
    "calculus",
    "algebra",
    "quadratic formula",
    "algebra unsolved"
  ],
  "Problem": "1. Maria did a study of babysitting tares in her neighborhood over the last three years and discovered that the function b(t)=6.25(1.2)^t represents the change in pricing, where t is the number of years since 2000. What was the babysitting cost per hour in Maria's neighborhood in 2000? \r\n[color=darkred]My answer was: first, b=6.25(1.2)^3=10.8 --> pricing change. But i don't know how to find her cost per hour?[/color]\r\n\r\n2. An archer shoots an arrow into the air such that its hieght at any time, t, is given by the function h(t)=-16t^2+kt+3. If the maximum height of the arrow occurs at time t=4, what is the value of k? [color=darkred] I totally have no idea.[/color]\r\n\r\nThanks a lot for answering!",
  "Solution_1": "This question should be posted in a different forum, but I will try to answer them anyway.\r\n\r\nFor your first part, I might not understand correctly, but I would believe that the question is just asking for $6.25\\cdot (1.2)^{0}=6.25$ as the answer.  I don't think there is enough info for the question to be asking anything else....\r\n\r\nThe second part is easy: the apex of a quadratic occurs at $\\frac{-b}{2a}$ (just look at the quadratic formula, or use some differential calculus), so we must have $\\frac{-k}{2\\cdot (-16)}=4$ which means $k=128$."
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "tetrahedron",
    "function",
    "Putnam"
  ],
  "Problem": "Is there any generalization of Pick's Theorem to 3+ dimensions?",
  "Solution_1": "Consider the tetrahedron with vertices at (0,0,0), (1,0,0), (0,1,0), and (1,1,k) for any integer k. Each of these has four vertex lattice points, no edge or face or interior lattice points, but they all have different volume. So volume can't be a function of number of lattice points.",
  "Solution_2": "Maybe there's one that has to do with faces passing through \"lattice squares\" (I just made that term up; I mean squares of side 1 that have all four vertices as lattice points) as well as lattice points?",
  "Solution_3": "It would probably be more like lattice lines.  Lattice points are 0 dimensional, so you'd want something 1 dimensional.",
  "Solution_4": "http://mathworld.wolfram.com/EhrhartPolynomial.html\r\n\r\nThough I doubt that will be useful for any math contests...",
  "Solution_5": "Cool.  I was just curious if such a formula existed.  I wasn't worried about contests.",
  "Solution_6": "There is a preprint of a book on this subject at [url]http://math.sfsu.edu/beck/papers/ccd.pdf[/url].  It should be of interest to members of this forum, and even refers to a problem from a past Putnam contest"
}
{
  "Tag": [
    "MATHCOUNTS",
    "Euler",
    "AMC",
    "USA(J)MO",
    "USAMO",
    "geometry",
    "trigonometry"
  ],
  "Problem": "Kevin Lei, Lewis Chen, some other guy from hopkins, me.\r\n\r\nYay I made nats. \r\n\r\nSo... now what?",
  "Solution_1": "johnny ho didn't make it?? that is...",
  "Solution_2": "which one's james4l? and the other ones some random yury dude.",
  "Solution_3": "[quote=\"nikeballa96\"]which one's james4l? and the other ones some random yury dude.[/quote]\r\n\r\nConfusing ca with tx?",
  "Solution_4": "ca=/=tx\r\n\r\nHowever, just adding a letter makes it true:\r\n\r\nca=tax\r\n\r\nOkay, I just made another lame \"joke\"...\r\n\r\nAnyway, where's Kevin Lei from? (just wondering)",
  "Solution_5": "[quote=\"r15s11z55y89w21\"]ca=/=tx\n\nHowever, just adding a letter makes it true:\n\nca=tax\n\nOkay, I just made another lame \"joke\"...\n\nAnyway, where's Kevin Lei from? (just wondering)[/quote]\r\n\r\nKevin Lei is Euler_Apprentice, and he's from miller.\r\n\r\nSo are there any training sessions or something?",
  "Solution_6": "james4l=Lewis Chen\r\n\r\n5th place :mad:\r\n\r\nIf any one of you four get cancer, can I take your spot?\r\n\r\nOr, if pythag011 wants those extra 3-4 days to study for MOP, he can give up his place to me :D",
  "Solution_7": "good job!!\r\nI didn;t even make states....ruined the history of my school\r\nwe were tied with another school and they used some tiebreaker which they won so they went to states :mad: \r\no well even if i went to states I wouldn't have gotten top 4  :|",
  "Solution_8": "hm, i wouldn't have made nats there.\r\ni wuold've gotten top seven or so, but not nats.",
  "Solution_9": "This is the frickin stupid thing about MATHCOUNTS.\r\n\r\nI can have a score that's double the score of say, someone in Guam, yet they win their state and I don't get a thing except a 1st place Countdown trophyand a 4th place trophy, yet I still don't get to go to nats.\r\n\r\nWhy don't they just take the top 228 scores throughout the nation!?!?!?!?!?!\r\n\r\nAnd, [b]one arithmetic error[/b] cost me nats, as well as first place, as well as <insert something here after we can discuss scores>.\r\n\r\n[size=67]I keep making excuses.  Sigh.[/size]",
  "Solution_10": "that's a good point\r\n\r\ni think it'd be better if every state was balanced between good and bad.\r\nbut it wouldn't be good to take top 228 scores because other people don't get a chance",
  "Solution_11": "If you had grown up in Guam, I'd imagine you would have had far less opportunity to get interested in and learn about math. There are huge advantages to living in a competitive state (or at least the regions of a competitive state that do the competing). It means a school-wide and/or regional network of people with the same interest and the same or higher ability. I think this is probably the motivation behind state USAMO qualifiers as well.",
  "Solution_12": "Of course, I wouldn't mind if they split California along the same lines as the splitting of the state championship meet, and had two teams. Note that pythag011 is the first Southern Californian to make the team in the last three or four years.",
  "Solution_13": "...well, in that case, good job pythag011...and wow, three from the Bay Area?",
  "Solution_14": "[quote=\"Kent Merryfield\"]Of course, I wouldn't mind if they split California along the same lines as the splitting of the state championship meet, and had two teams. Note that pythag011 is the first Southern Californian to make the team in the last three or four years.[/quote]\r\n\r\nLast two years actually; someone from southern California made it in 2007.",
  "Solution_15": "[quote=\"r15s11z55y89w21\"]...and wow, three from the Bay Area?[/quote]\r\nI don't know this for sure, but I'd guess that the last non-Bay Area northern Californian to make the state team was Austin Shapiro in - what would that be - 1997? Or did David Stolp ever make the state team?",
  "Solution_16": "hi guys\r\n\r\nNC has at least one person who would be guaranteed to make CA nats\r\n\r\nbecause he got a[color=indigo]n AHEM[/color]?",
  "Solution_17": "not_trig, no scores. please edit that out.",
  "Solution_18": "Rankings - Norcal\r\n\r\n(if there is a space I think they are not in AOPS)\r\n\r\n1. Eulers_Apprentice\r\n2. james4l\r\n3. Michael2Choi\r\n4. AIME15\r\n5. EcstaticPotter\r\n6. Matthew Chow\r\n7. miller4math\r\n8. Spencer Yee\r\n9. Charles Liu\r\n10. Nikhil Buduma\r\n\r\nOk, here is where things get a bit more unpredictable, seeing as they only called up the top 10. (I used the fact that the countdown was a top-bottom style)\r\n\r\n11. Matthew Lin\r\n12. Lexing Tong\r\n13. Kevin Chen\r\n14. Julia Huang\r\n15. Grace Lin\r\n16. some person who beat Kevin Lei\r\n\r\npythag011 would have been around first to third, anyways making the team.",
  "Solution_19": "BTW, does anyone know anyone about the coach?\r\n\r\nAccording to Eulers_Apprentice on ftw, I got 1st in ca... but he's not very sure...",
  "Solution_20": "Coach = Miller's coach",
  "Solution_21": "Oh darn Jordan only had Grace Lin in the top 16?\r\nAnd oh man, Calvin from JLS did not place, nor did anyone from Terman :O",
  "Solution_22": "woah is Grace korean?\r\n\r\nshe sounds hot",
  "Solution_23": "Err....Lin=\u6797.\r\nAs for your other interests...I don't know who she is, other than that she is the little sister of some person who was in my math class a while ago.",
  "Solution_24": "[quote=\"serialk11r\"]Err....Lin=\u6797.\nAs for your other interests...I don't know who she is, other than that she is the little sister of some person who was in my math class a while ago.[/quote]\r\n\r\nlol sister of Albert Lin? I know her, family friends. She seriously did not know any math until recently. She's not really hot as least of what I can remember of her but I haven't seen her in years -- if I thought she was hot though, I would be a pedophile then as she's in like 6th/7th grade.\r\n\r\nNo JLS people? I'm disappointed.",
  "Solution_25": "Actually today the 7th grade kid who won Peninsula Chapter told me he placed in top 16, so I'm assuming he's the 16th place person. (from JLS)",
  "Solution_26": "wow... i just discovered this topic.  :o \r\nwhy are you guys talking about grace lin? i dont even know who she is...\r\nand ur not really a pedophile because u have to be like 50 to be a pedophile. wow this is random.\r\nanyways\r\n\r\ni forgot, but either\r\nmiller4math got 7th and EcstaticPotter got 6th\r\nor\r\nmiller4math got 6th and EcstaticPotter got 5th\r\n\r\ni think its the second one. i take that back.\r\ni know its the second one. im not gonna erase what i just typed cause im too lazy  :) \r\nwow this post is getting really long so im just going to stop.",
  "Solution_27": "[quote=\"Eulers_Apprentice\"]wow... i just discovered this topic.  :o \nwhy are you guys talking about grace lin? i dont even know who she is...\nand ur not really a pedophile because u have to be like 50 to be a pedophile. wow this is random.\nanyways\n\ni forgot, but either\nmiller4math got 7th and EcstaticPotter got 6th\nor\nmiller4math got 6th and EcstaticPotter got 5th\n\ni think its the second one. i take that back.\ni know its the second one. im not gonna erase what i just typed cause im too lazy  :) \nwow this post is getting really long so im just going to stop.[/quote]\r\n\r\nno AIME15 was 5th.\r\n\r\nAlso, I get free plane tickets to bay area, yay.",
  "Solution_28": "errr pythag actually\r\nAIME15 got 4th in [b]norcal[/b]\r\nwhich is what were talking about. not CA in total",
  "Solution_29": "So the countdown was not top-bottom entirely? I remember Johnny beat Nikhil in cd.\r\n\r\nOh well - results are out. http://www.mathcounts-ca.org/uoptop25and402009.pdf",
  "Solution_30": "[quote=\"bowser\"]good job!!\nI didn;t even make states....ruined the history of my school\nwe were tied with another school and they used some tiebreaker which they won so they went to states :mad: \no well even if i went to states I wouldn't have gotten top 4  :|[/quote]\r\n\r\n :ninja: its okay, i was so stupid i didnt take math counts",
  "Solution_31": "Just to clarify, in 2007 someone from SoCal called Daichi something (he's Japanese, and went back to Japan after nats) made the nats team.\r\n\r\nalso these rankings are totally different from last year, things fluctuate FAST as people graduate middle school.",
  "Solution_32": "[quote=\"bookaholic\"]If you had grown up in Guam, I'd imagine you would have had far less opportunity to get interested in and learn about math. There are huge advantages to living in a competitive state (or at least the regions of a competitive state that do the competing). It means a school-wide and/or regional network of people with the same interest and the same or higher ability. I think this is probably the motivation behind state USAMO qualifiers as well.[/quote]\r\nMeh\r\n\r\nWhat sucks the most, then, is living just outside of a competitive region?",
  "Solution_33": "Yeah, fo sho that's definitely the worst =("
}
{
  "Tag": [
    "number theory",
    "relatively prime",
    "number theory unsolved"
  ],
  "Problem": "Prove that n is prime iff $\\sigma(n)+\\phi(n)=nd(n)$.",
  "Solution_1": "can you explain the functioons",
  "Solution_2": "$\\phi (n)$ is the number of positive integers less than $n$ which are relatively prime with $n$.\r\n$d(n)$ is the number of positive divisors of $n$ (including 1 and $n$).\r\n$\\sigma (n)$ is the sum of the positive divisors of $n$.",
  "Solution_3": "I guess it must be moved to number theory section.",
  "Solution_4": "Clearly, if $n$ is prime then $\\sigma (n) = n+1, \\phi (n) = n-1$ and $d(n)=2$ so that the equality holds.\r\n\r\nNow, assume that $n$ is a composite number.\r\nNote that $\\phi (n) \\leq n-1.$\r\nLet $1=d_1 < d_2 < \\cdots < d_k = n$ be the positive divisors of $n$. Then $k \\geq 3.$\r\nTherefore :\r\n$nd(n) - \\sigma (n) = kn - (d_1 + \\cdots + d_k) = (n-d_1) + \\cdots + (n-d_k) \\geq (n-1) + (n-d_2) > n-1 > \\phi (n).$\r\nThus $n$ does not satisfy the desired equality, and we are done.\r\n\r\nPierre.",
  "Solution_5": "Thanks Pierre. :)"
}
{
  "Tag": [
    "geometry",
    "geometric transformation",
    "reflection",
    "function",
    "analytic geometry"
  ],
  "Problem": "the inverse of the relation $ {(2,\\minus{}3),(5,\\minus{}4),(8,\\minus{}2)}$ is?\r\n\r\n\r\n(the answer to this is just $ {(\\minus{}3,2),(\\minus{}4,5),(\\minus{}2,8)}$ right? or am i missing something :huh: )",
  "Solution_1": "Sorry, I can't understand the problem? What do you mean \"inverse\" in this case?",
  "Solution_2": "that's exactly what i was stuck on. \r\n\r\n\r\nVerbatim, the problem reads:\r\n\r\n\"The inverse of the relation $ {(2,\\minus{}3),(5,\\minus{}4),(8,\\minus{}2)}$ is\r\n\r\nA. $ {(\\minus{}2,3),(\\minus{}5,4),(\\minus{}8,2)}$\r\n\r\nB. $ {(2,3),(5,4),(8,2)}$\r\n\r\nC. $ {(\\minus{}2,\\minus{}3),(\\minus{}5,\\minus{}4),(\\minus{}8,\\minus{}2)}$\r\n\r\nD. $ {(\\minus{}3,2),(\\minus{}4,5),(\\minus{}2,8)}$\r\n\r\nE. None of the above",
  "Solution_3": "Oh, maybe the reflection of a point across the line $ y\\equal{}x$, $ inverse$ comes from the inverse fuction?",
  "Solution_4": "I'm pretty sure it's that.",
  "Solution_5": "i mean, i guess, but this is supposed to be a $ very$ easy test, so thinking is not supposed to be involved in the process....\r\n\r\nso would B be right?",
  "Solution_6": "You mean D?",
  "Solution_7": "an inverse of a function $ f$ is the function $ g$ such that $ g(f(x))\\equal{}x$.for instance, the inverse of\r\n$ f(x)\\equal{}2x\\plus{}3$ would be $ g(x)\\equal{}\\frac{x\\minus{}3}{2}$",
  "Solution_8": "so, to put it all in a nutshell...  \r\n\r\nthe inverse of given coordinates is  just the x and y's flipped?(as in: switch the x's and the y's)",
  "Solution_9": "That's right."
}
{
  "Tag": [
    "modular arithmetic",
    "function"
  ],
  "Problem": "question is-- find last two digits of $ 7^{7^{1000}}$\r\n[hide=\"approach 1\"]$ \\phi 100 \\equal{} 40$. By euler's theorem, $ a^{\\phi(n)}\\equiv1\\pmod{n}$. So, here if we divide the exponent $ 7^{1000}$ by 40, we get an equivalent power less than \"n\". Again, we hav to solve $ 7^{1000}$ in mod(40). taking $ \\phi$ again , $ \\phi 40 \\equal{} 16$ .dividing 1000 by 16 , remainder=8. We are left with $ 7^8$ .But, $ 7^2\\equiv1\\pmod{16}$. so, $ 7^8$ is 1(mod 16). Expression left is$ 7^1 \\pmod{100}$. Hence, [b]answer is 07.[/b][/hide]\n[hide=\"approach 2\"]\nfirst take $ 7^7$, break into power of 4 and 3. bt, $ 7^4 \\equiv 1 \\pmod{100}$.so,we are left with $ 7^{3^{1000}} \\equal{} 7^{9^{500}}\\equiv7^{500}\\equiv7^{20}\\equiv1\\pmod{100}$ so, [b]answer is 01[/b][/hide]\n[hide=\"approach 3\"]break mod inot 4 and 25. Afetr calc. we obtain $ x\\equiv1\\pmod{4} and x\\equiv1\\pmod{25}$, where x is given expression. solving here, i got by chinese remainder [b]01 as answer.[/b] [/hide]\r\nwhere is the error?",
  "Solution_1": "In approach 1 you applied $ \\bmod 16$ to the wrong level of exponentiation.  Write your work done more clearly.\r\n\r\nApproach 2 is completely invalid.  First of all, $ 7^{7^{1000}} \\neq (7^7)^{1000}$, and second of all it is not at all valid to \"break into powers of $ 4$ and $ 3$\" except by the binomial theorem.\r\n\r\nYou haven't shown any work in approach 3, but $ 7^{7^{1000}} \\equiv \\minus{}1 \\bmod 4$, so something is wrong.\r\n\r\nThe fastest check here is [hide=\"fixing approach 1\"] $ \\lambda(100) \\equal{} 20$ where $ \\lambda(n)$ is the [url=http://en.wikipedia.org/wiki/Carmichael_function]Carmichael function[/url] and $ \\lambda(20) \\equal{} 4$, so $ 7^{1000} \\equiv 1 \\bmod 20$ hence $ 7^{7^{1000}} \\equiv \\boxed{07} \\bmod 100$.  [/hide]",
  "Solution_2": "notice that the last 2 digits in the powers of 7 repeat in groups of 4\r\n(01,07,49,43...)\r\nand since $ 7^2 \\equiv 1 \\pmod{4}, (7^2)^{500} \\equiv 1 \\pmod 4$\r\nso since the exponent $ 7^{{1000}}$ is 1 mod 4, $ 7^{7^{1000}} \\equiv 7 \\pmod{100}$\r\n$ \\boxed{07}$ is the answer",
  "Solution_3": "In approach 1, wut i tried to do was that in $ a^p\\pmod{n}$, we can split 'p' into $ (k\\cdot%Error. \"phin\" is a bad command.\n+r)$. And as $ a^{\\phi(n)}\\equiv{1}\\pmod{n}$, r is the resultant exponent power which is less than n and also equivalent to the initial expression in $ Z_n$. Hence, to obtain this we have to divide the exponent by $ \\phi(n)$, so as to obtain 'r'. This means same as finding equivalent power in $ Z_{\\phi(n)}$. So, i solved this and got 1 as resultant exponent.\r\nI think nothing shud be wrong with this approach.",
  "Solution_4": "[quote=\"analytic\"]we hav to solve $ 7^{1000}$ in mod(40). taking $ \\phi$ again , $ \\phi 40 \\equal{} 16$ .dividing 1000 by 16 , remainder=8. We are left with $ 7^8$ .But, $ 7^2\\equiv1\\pmod{16}$. so, $ 7^8$ is 1(mod 16). [/quote]\r\nThis step is incorrect.  You are evaluating $ 7^8 \\bmod 40$, not $ \\bmod 16$."
}
{
  "Tag": [],
  "Problem": "Dr. Alan visits 9 patients on his ward on Monday. On Tuesday, and each day thereafter, one patient is discharged and two more are admitted to the ward, except on Saturday and Sunday, when no one is discharged. How many patients are on the ward at the end of the following Tuesday?",
  "Solution_1": "9+8+2=19.The message is too small. Please make the message longer before submitting.\r\n\r\n[color=blue]Please read the sticky on writing solutions well. -Isabella2296[/color]",
  "Solution_2": "[quote=\"aleph0\"]9+8+2=19.The message is too small. Please make the message longer before submitting.\n\n[color=blue]Please read the sticky on writing solutions well. -Isabella2296[/color][/quote]\n...",
  "Solution_3": "Well, the solution is, add one patient to the group every day, except Saturday and Sunday, when there are two patients added. Therefore, we have two weekend days and six weekday days... (sounded weird). Therefore, $2\\cdot2+6\\cdot1=4+6=10$, but remember, this isn't the answer. This is the number of patients increased from the first Monday to the next (week) Tuesday. Since there were nine patients, we add: $9+10=\\boxed{19}$."
}
{
  "Tag": [
    "geometry",
    "geometric transformation",
    "reflection",
    "homothety",
    "AMC",
    "USA(J)MO",
    "USAMO"
  ],
  "Problem": "Let ABCD be a convex quadrilateral, the diagonals of which intersect orthogonally at O.\r\nProve that the reflections of O with respect to the sides of the quadrilateral are concyclic.",
  "Solution_1": "(I use the code window in order to make the ASCII graphics below.)\r\n\r\n[code]\n        B       Y     C \n        /-------|-----\\_ \n      X/__'-._  | __-'__\\_Z \n      /   --__'-O---''    \\_ \n     /     __--'/'-_        \\_ \n    /   __-    |    '-_       \\_ \n   /__--      /        '-_      \\_ \n A/-___      |            '-_     \\_ \n       ----_/__              '-._   \\_ \n            W  ----____          '-_  \\_ \n                       ----____     '-_ \\_ \n                               ----____'-_\\_ \n                                       ----_\\D \n\nThe reflections of the point O in the sidelines AB, BC, CD, DA are the images of the orthogonal projections X, Y, Z, W of the point O on these sidelines AB, BC, CD, DA under the homothety with center O and factor 2. Hence, it is enough to show that the points X, Y, Z, W are concyclic (since the images of concylic points under a homothety are concyclic points again).\n\nSince < AWO = 90\u00b0 and < AXO = 90\u00b0, the points W and X lie on the circle with diameter AO; hence, < XWO = < XAO, or, equivalently, < XWO = < BAO. Similarly, < ZWO = < CDO, thus\n\n  < XWZ = < XWO + < ZWO = < BAO + < CDO,\n\nand similarly < XYZ = < ABO + < DCO.\n\nNow, in order to establish the concyclicity of the points X, Y, Z, W, it is enough to show < XWZ + < XYZ = 180\u00b0. In other words, we have to prove\n\n  (< BAO + < CDO) + (< ABO + < DCO) = 180\u00b0.\n\nHowever, since triangles AOB and COD are right-angled, we have < BAO + < ABO = 90\u00b0 and < CDO + < DCO = 90\u00b0, and thus\n\n  (< BAO + < CDO) + (< ABO + < DCO)\n  = (< BAO + < ABO) + (< CDO + < DCO)\n  = 90\u00b0 + 90\u00b0 = 180\u00b0,\n\nqed.[/code]\r\n\r\n  darij",
  "Solution_2": "Again, this problem was intended for the beginners Darij :)\r\n\r\nThere is also a solution using inversion :) Is anyone interested?",
  "Solution_3": "Yes, I would like to see that solution.",
  "Solution_4": "By the way, here's my solution before I looked at the above solution.\r\n\r\nLet cyclic quadrilateral $ABCD$ have projections of $O$ onto $AB$,$BC$,$CD$, and $AD$ as $Y$,$W$,$Z$, and $X$, respectively.\r\nIf $\\angle XZW+\\angle XYW=180^{\\circ}$, quadrilateral $XYZW$ is cyclic (if we have a qudrilateral $PQRS$, then $\\angle P+\\angle R=\\frac{arcQRS+arcQPS}{2}=\\pi rad=180^{\\circ}$ implies that if two opposite angles of a quadrilateral sum up to $180^{\\circ}$, then the quadrilateral is cyclic).\r\n\r\nQuadrilaterals $OXDZ$,$OZCW$,$OWBY$, and $OXAY$ are cyclic since they all have 2 opposite 90 degree angles.\r\n\r\n$\\angle ODA=\\angle OZX$, $\\angle OCD=\\angle OZW$, $\\angle OYW=\\angle OBC$, $\\angle OYX=\\angle OAD$, $\\angle OAD+\\angle ODA=90^{\\circ}$, and $\\angle OCB+\\angle OBC=90^{\\circ}$.\r\n\r\nHence,\r\n\r\n$\\\\\\angle XZW+\\angle XYW=\\angle OZX+\\angle OZW+\\angle OYX+\\angle OYW\\\\ =\\angle ODA+\\angle OCB+\\angle OBC+\\angle OAD\\\\ =(\\angle ODA+\\angle OAD)+(\\angle OCB+\\angle OBC)\\\\ =90^{\\circ}+90^{\\circ}=180^{\\circ}$\r\n\r\nAnd the conclusion follows.\r\n\r\nMasoud Zargar",
  "Solution_5": "Actually, this was on an old USAMO, but it's not hard. What about that inversion solution? I have one--invert twice about O... :D",
  "Solution_6": "How to use inversion?",
  "Solution_7": "http://www.amazon.com/Geometry-Revisited-Mathematical-Association-Textbooks/dp/0883856190\r\n\r\nIf you don't have it, you should get it.",
  "Solution_8": "I suppose the inversion maps the points to a line and turns it into a collinearity problem.",
  "Solution_9": "Seeing that this thread has been revived, for those of you who are interested in a solution via inversion, look for Inversion in Mathematical Excalibur (google!).",
  "Solution_10": "[quote=\"Differ\"]http://www.amazon.com/Geometry-Revisited-Mathematical-Association-Textbooks/dp/0883856190\n\nIf you don't have it, you should get it.[/quote]\r\nhmm, is there a site where I can read that book for free (or website with similar articles on inversions), as I don't feel like buying it (too expensive...I'm not rich at all... moreover, even if i have that book, spending time finishing that book would not give me time to do other problems posted on this site and learn other strategies - right now i don't even think i have seen at most 1% of the problems posted on this site)? But the book is pretty cool, though (only gotta read 6 pages... so stingy, lol)\r\nThanks for any info...  :D",
  "Solution_11": "[quote=\"mathwizarddude\"][quote=\"Differ\"]http://www.amazon.com/Geometry-Revisited-Mathematical-Association-Textbooks/dp/0883856190\n\nIf you don't have it, you should get it.[/quote]\nhmm, is there a site where I can read that book for free (or website with similar articles on inversions), as I don't feel like buying it (too expensive...I'm not rich at all... moreover, even if i have that book, spending time finishing that book would not give me time to do other problems posted on this site and learn other strategies - right now i don't even think i have seen at most 1% of the problems posted on this site)? But the book is pretty cool, though (only gotta read 6 pages... so stingy, lol)\nThanks for any info...  :D[/quote]\r\n\r\nBorrow it from your friends! One of them is bound to have it. (everyone from Southern California has a copy (or two))",
  "Solution_12": "Oh really? Perhaps not for the place I'm living here (I know CA has a lot of Asians, but not for UT)... i bet less than 1% of the people even *like* math here...",
  "Solution_13": "[quote=\"boxedexe\"]Seeing that this thread has been revived, for those of you who are interested in a solution via inversion, look for Inversion in Mathematical Excalibur (google!).[/quote]\n\nI can't be bothered reading the solution in Excalibur, but it is different to this:\n\nAs noted previously, by a clear homothecy it suffices to prove $X,Y,Z,W$ is cyclic where $X,Y,Z,W$ are the orthogonal projections from $O$ to the sides $AB,BC,CA,AD$. Invert with centre $O$. Let the line passing through $A,X,B$ be mapped to the circle $\\Gamma$. Since $OX$ is the shortest distance from $O$ to the line $AB$, it follows $OX'$ is the longest distance from $O$ to $\\Gamma$ again. Then $X'$ is diametrically opposite $O$ in $\\Gamma$ and then $\\angle X'B'O=\\angle X'A'O'=90^{\\circ}$. But clearly also $\\angle A'OB'=90^{\\circ}$ since $A'B'$ lie on the fixed rays $OA,OB$. So $OA'X'B'$ is a rectangle. Repeat for $Y',Z',W'$ to conclude $X'Y'Z'W'$ is a rectangle which implies $XYZW$ is cyclic."
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "Given $ a,b,c \\geq0$.Prove that\r\n$ \\frac1{5a^2\\plus{}bc}\\plus{}\\frac1{5b^2\\plus{}ca}\\plus{}\\frac1{5c^2\\plus{}ab} \\geq \\frac3{a^2\\plus{}b^2\\plus{}c^2\\plus{}ab\\plus{}bc\\plus{}ca}$",
  "Solution_1": "[quote=\"dduclam\"]Given $ a,b,c \\geq0$.Prove that\n$ \\frac1{5a^2 \\plus{} bc} \\plus{} \\frac1{5b^2 \\plus{} ca} \\plus{} \\frac1{5c^2 \\plus{} ab} \\geq \\frac3{a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} ab \\plus{} bc \\plus{} ca}$[/quote]\r\n$ \\sum_{cyc}\\frac {1}{5a^2 \\plus{} bc} \\equal{} \\sum_{cyc}\\frac {b^2c^2}{5a^2b^2c^2 \\plus{} b^3c^3}\\geq\\frac {(ab \\plus{} ac \\plus{} bc)^2}{15a^2b^2c^2 \\plus{} a^3b^3 \\plus{} a^3c^3 \\plus{} b^3c^3}.$\r\nId est, it remains to prove that $ (ab \\plus{} ac \\plus{} bc)^2(a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} ab \\plus{} ac \\plus{} bc)\\geq3(a^3b^3 \\plus{} a^3c^3 \\plus{} b^3c^3) \\plus{} 45a^2b^2c^2,$\r\n which obviously true.\r\nI think, the following similar inequality nicer:\r\nFor psitive $ a,$ $ b$ and $ c$ prove that\r\n\\[ \\frac {1}{a^2 \\plus{} bc} \\plus{} \\frac {1}{b^2 \\plus{} ac} \\plus{} \\frac {1}{c^2 \\plus{} ab}\\geq\\frac {9}{a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} ab \\plus{} c \\plus{} bc}\r\n\\]",
  "Solution_2": "by Cauchy we have :\r\n$ \\frac{1}{a^{2}\\plus{}bc} \\plus{} \\frac{1}{b^{2}\\plus{}ac} \\plus{} \\frac {1}{c^{2}\\plus{}ab} \\geq  \\frac{(1\\plus{}1\\plus{}1)^2}{a^{2}\\plus{}b^{2}\\plus{}c^{2}\\plus{}ab\\plus{}bc\\plus{}ca}$",
  "Solution_3": "OK,khashi70!  What about the following Vasc's inequality:\r\n\\[ \\frac {1}{a^2 \\plus{} bc} \\plus{} \\frac {1}{b^2 \\plus{} ac} \\plus{} \\frac {1}{c^2 \\plus{} ab}\\geq\\frac {3}{ ab \\plus{} ac \\plus{} bc}\r\n\\]\r\n:wink:",
  "Solution_4": "[quote=\"arqady\"]\\[ \\frac {1}{a^2 \\plus{} bc} \\plus{} \\frac {1}{b^2 \\plus{} ac} \\plus{} \\frac {1}{c^2 \\plus{} ab}\\geq\\frac {3}{ ab \\plus{} ac \\plus{} bc}\n\\]\n:wink:[/quote]\r\n\r\nNice inequality ... Here is what I did ... The inequality is equivalent to \r\n\r\n$ (a\\minus{}b)^2(b\\minus{}c)^2(c\\minus{}a)^2\\plus{}abc(a^3\\plus{}a^2b\\plus{}b^2a\\plus{}b^3\\plus{}c^2a\\plus{}a^2c\\plus{}c^3\\plus{}b^2c\\plus{}c^2a\\plus{}3abc)\\geq 0$ which is true  :) \r\n\r\n\r\nThe following more general inequality holds\r\nIf $ \\frac{2}{3}\\leq r\\leq 3\\plus{}\\sqrt{7}$  then \r\n\r\n$ \\frac {1}{ra^2 \\plus{} bc} \\plus{} \\frac {1}{rb^2 \\plus{} ac} \\plus{} \\frac {1}{rc^2 \\plus{} ab}\\geq\\frac {r\\plus{}2}{r( ab \\plus{} ac \\plus{} bc)}$\r\n\r\nthe proof is similar  :wink:",
  "Solution_5": "Given $ a,b,c\\geq0$.Prove that\r\n$ \\frac a{b^3\\plus{}8c^2a}\\plus{}\\frac b{c^3\\plus{}8a^2b}\\plus{}\\frac c{a^3\\plus{}8b^2c}\\geq\\frac2{a^2\\plus{}b^2\\plus{}c^2\\plus{}ab\\plus{}bc\\plus{}ca}$"
}
{
  "Tag": [
    "calculus",
    "integration",
    "trigonometry",
    "inequalities",
    "calculus computations"
  ],
  "Problem": "For $ f(x)\\equal{}1\\minus{}\\sin x$, let $ g(x)\\equal{}\\int_0^x (x\\minus{}t)f(t)\\ dt.$\r\n\r\nShow that $ g(x\\plus{}y)\\plus{}g(x\\minus{}y)\\geq 2g(x)$ for any real numbers $ x,\\ y.$",
  "Solution_1": "hi kunny-kun  :D \r\n\r\n$ g(x) \\equal{} \\int_0^x (x \\minus{} t) f(t) \\, dt$\r\n\r\nIt's easy to see $ g(z) \\equal{} g( \\minus{} z)$\r\n\r\nmaking a small substituition $ x \\equal{} \\frac {a \\minus{} b}{2}\\land y \\equal{} \\frac {a \\plus{} b}{2}$\r\n\r\nwe get the inequality $ \\frac {g(a) \\plus{} g(b)}{2}\\geq g\\left(\\frac {a \\minus{} b}{2}\\right)$\r\n\r\nmaking $ b \\to \\minus{} b$ ,  $ \\frac {g(a) \\plus{} g(b)}{2}\\geq g\\left(\\frac {a \\plus{} b}{2}\\right)$\r\n\r\nI think there is a Famouse Inequality like above, but i cannot remember it.\r\n\r\n:( could someone help me?",
  "Solution_2": "Perhaps Jensen's Inequality.",
  "Solution_3": "thanks Carcul-kun\r\n\r\nSo $ g$ is convex, according to the inequality\r\n\r\nso $ g''(z)  \\geq 0$\r\n\r\nUsing Cauchy Integration formula to differentiate this in a easy way\r\n\r\n$ g''(x)\\equal{}f(x)\\equal{}1\\minus{}\\sin (x)\\geq 0$\r\n\r\nSo $ g$ is convex and Jesen is true and we are donne"
}
{
  "Tag": [
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "Prove that for all real numbers $a_1,a_2,...,a_n$, we have:\r\n$\\displaystyle a_1^2+(\\frac {a_1+a_2}{2})^2+...+(\\frac {a_1+a_2+...+a_n}{n})^2 \\leq 4(a_1^2+a_2^2+...+a_n^2)$.",
  "Solution_1": "[quote=\"hxtung\"]Prove that for all real numbers $a_{1},a_{2},...,a_{n}$, we have:\n$ a_{1}^{2}+(\\frac{a_{1}+a_{2}}{2})^{2}+...+(\\frac{a_{1}+a_{2}+...+a_{n}}{n})^{2}\\leq 4(a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2})$.[/quote]",
  "Solution_2": "it's aspecial cas of hardy inequality just see:\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?highlight=inequality+hardy&t=29756"
}
{
  "Tag": [
    "number theory",
    "prime factorization"
  ],
  "Problem": "Q- Prove that when three consecutive even numbers are squared and the results are added, that the sum always has at least 3 divisors.\r\n\r\nThank you for your help.",
  "Solution_1": "[hide]\n\n:P I didn't notice the \"squared\"  :D \n\n$(2n)^2 + (2n+2)^2 + (2n+4)^2 = 4 \\cdot (3n^2+6n+5)$\n[/hide]",
  "Solution_2": "[quote=\"t\u00b5t\u00b5\"][hide]$2n + (2n+2) + (2n+4) = 2 \\cdot 3 \\cdot (n+1)$[/hide][/quote]\r\n\r\nHow does that show when they're squared? \r\n\r\nConsidering the numbers to be 2n, 2n+2, and 2n+4, squaring and adding them gives us $4(3n^2+2n+5)$, which I think has three factors (2, 4, and $3n^2+2n+5$). Don't forget 1 either  :D",
  "Solution_3": "Let the numbers be x-2, x, and x+2.\r\n\r\nWhen the numbers are squared and added, we get (3x^2 + 8), where x mod 2 == 0.\r\n\r\nSo, we can replace x with 2x, giving us 3(2x)^2 + 8, yielding 12x^2 + 8, which is only divisible by certain numbers. When factored, this gives us 4(3x^2+2), which is obviously divisible by all factors of 4, which are 1, 2, and 4. Thus, all consecutive even numbers when squared and added are divisible automatically by 1, 2, and 4. And finally, for the last divisor, the number is divisible by itself, yielding four divisors total (including 1 and the number itself). This also suggests that when the result is obtained, there have to be [b]at least [/b] four divisors for the number given.[/b]",
  "Solution_4": "[quote=\"Karth\"]Let the numbers be x-2, x, and x+2.\n[/quote]\r\nHere you make a mistake, because x could be an odd number ;)\r\n\r\nThe correct notation (used by t\u00b5t\u00b5 and 236!) would be : 2n, 2n+2, 2n+4",
  "Solution_5": "Let the numbers be $2(n-1), 2n, 2(n+1)$ where $n$ is any integer.  Squaring and adding we get $4(n^2 -2n + 1 + n^2 + n^2 +2n + 1) = 2^2 (3n^2 + 2)$.  To find the divisors of this, we do the prime factorization of number, add one to the exponents and multiply.  No matter what the factorization of the second part is, it will always be multiplied by the 3 from the $2^2$, therefore the minimum number of divisors is 3.",
  "Solution_6": "@mathmanman:\r\n\r\nThat's why on the last step, I replaced x with 2x, cuz no matter what x is (when it is an integer), 2x is always even.  :D",
  "Solution_7": "Yeah, I saw that :)\r\nBut, I was just talking about the notation ;) \r\nI mean, your proof is good, but.. you do not begin very well :P\r\nLike, for example, an integer is usually denoted by \"n\""
}
{
  "Tag": [
    "search",
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "for x,y,z>0  \r\nx,y,z<=1 and x+y+z<=3/2;\r\nprove that:\r\n\r\n$\\sqrt{\\frac{288}{3}x^5+\\frac{1}{y^3}} +\\sqrt{\\frac{288}{3}y^5+\\frac{1}{z^3}}+\\sqrt{\\frac{288}{3}z^5+\\frac{1}{x^3}}\\geq4\\sqrt{6}$\r\n\r\nin my opion it 's $3\\sqrt{11}$",
  "Solution_1": "Where do you use z in your problem, and what exactly are you summing?",
  "Solution_2": "It's symmetric sum notation, do a search for the Inequalities thread, and it should become clear what is intended.",
  "Solution_3": "no answer for this ,agree or disagree :(",
  "Solution_4": "What is $3\\sqrt{11}$?",
  "Solution_5": "yes, it comes from a books\r\n\r\ni mean that it must be $3\\sqrt{11}$ except for $4\\sqrt{6}$",
  "Solution_6": "you mean the right hand side of the inequality should be $3 \\sqrt{11}$ instead of $4 \\sqrt{6}$?",
  "Solution_7": "Hello, Cuong.\r\nYour problem smells good. Here's my solution:\r\n  By AM-GM:  $ 96x^5+\\frac{1}{y^3} = 3.(32x^5)+ 8.(\\frac{1}{8y^3}) \\geq 11\\sqrt[11]{\\frac{32^3.x^{15}}{8^8.y^{24}}}$ (*)\r\n  By AM-GM+ (*) : $ LHS \\geq 3.\\sqrt{11}.\\sqrt[22]{\\frac{32^3}{8^8}}.\\sqrt[66]{\\frac{1}{(xyz)^9}}$\r\n  AM-GM again $ (xyz)^9 \\leq (\\frac{x+y+z}{3})^{27} \\leq (\\frac12)^{27}$;\r\n  So $ LHS \\geq 3.\\sqrt{11}$\r\n Equality occurs if and only if x=y=z=1/2; :) \r\nGood luck!",
  "Solution_8": "many thanks: :?"
}
{
  "Tag": [
    "trigonometry",
    "vector",
    "trig identities",
    "Law of Cosines"
  ],
  "Problem": "If O is the origin, A is the point (4,-2) and B is the point (7,-5), find m<AOB to the nearest tenth of a degree.\r\n\r\n\r\nI get about 8.9 degrees\r\n\r\nCould someone check it for me? \r\n\r\nTHanks!!",
  "Solution_1": "[hide=\"solution\"]We graph the points and use the Law of Cosines to get\n\nAOB\u22488.972626615\n\nSo your estimation is very close but I would say about 9.[/hide]",
  "Solution_2": "Using a vector approach:\r\n[hide]We need to find the angle between $<4,-2>$ and $<7,-5>$. The cosine of the angle between two vectors is equal to the dot product divided by the product of the magnitudes of the vectors; thus calling the angle $x$, $cosx = \\frac{38}{\\sqrt{20 \\times74}}$ and it follows that $x\\approx 8.973$.[/hide]"
}
{
  "Tag": [
    "Olimpiada de matematicas"
  ],
  "Problem": "Equipe brasile\u00f1a:\r\n\r\nBRA1: Gustavo Lisb\u00f4a Empinotti\r\nBRA2: Marcelo Tadeu de S\u00e1 Oliveira Sales\r\nBRA3: Matheus Ara\u00fajo Marins\r\nBRA4: Matheus Secco Torres da Silva\r\n\r\n\r\nDigam suas equipes!!",
  "Solution_1": "PERUVIAN TEAM:\r\nPER1.- CESAR CUENCA .\r\nPER2.- FERNANDO MANRIQUE.\r\nPER3.- IVAN ALEXANDER.\r\nPER4.- SALAS HUAMAN.\r\n\r\n                                            ESTE ES EL PODEROSISIMO!!! EQUIPO PERUANO...\r\n\r\nPD:\r\n Yo ya toy viejo pa esto ..:P..",
  "Solution_2": "Quem es tu en la equipe?",
  "Solution_3": "[color=blue][b]Paraguay:[/b]\n\n1. Ariel Schvartzman.\n2. Marcos Mart\u00ednez.\n3. Edgar \u00c1valos.\n4. Edgar Elizeche.\n\n\nSaludos de parte de un nuevo ex-ol\u00edmpico.[/color]",
  "Solution_4": "Chle\r\n\r\n1 Anibal Veloso\r\n2 Benjamin Baeza\r\n3 Daniel Contreras\r\n4 Matias Garrido"
}
{
  "Tag": [
    "number theory",
    "greatest common divisor",
    "Gauss",
    "Euler",
    "number theory unsolved"
  ],
  "Problem": "Probably this is an easy one but so far i havent solved any gcd problem so its hard for me...  :( \r\n$1.$ Prove that if $ab+cd$ is devisible by $a-c$ then $ad+bc$ is also devisible by $a-c$\r\n$2.$ Prove that if $ad-bc$ and $a-b$ are devisible by $k$ and $(a,k)=1$ then $c-d$ is also devisible by $k$",
  "Solution_1": "Notice that \\[ (ab + cd) - (ad + bc) = (a - c)(b - d) \\] so, if $a - c$ divides $ab + cd$, then $a - c$ must divide $ad + bc$ as well.\r\n\r\nDo you see why?\r\n\r\nAnd do you see how to do the second part?",
  "Solution_2": "[quote=\"Gohan\"]\n$2.$ Prove that if $ad-bc$ and $a-b$ are devisible by $k$ and $(a,k)=1$ then $c-d$ is also devisible by $k$[/quote]\r\n$k|(ad-bc)$  and  $k|(a-b)$  then $k|(ac-bc)-(ad-bc)$\r\n$k|a(c-d)$\r\n$k|(c-d)$ (gauss)\r\n:)",
  "Solution_3": "10x Arne and Fermat-Euler...  :D",
  "Solution_4": "[quote=\"Gohan\"]10x Arne and Fermat-Euler...  :D[/quote]\r\nthanx\r\nArne  :winner_first:\r\nFermat-Euler  :winner_second:"
}
{
  "Tag": [
    "induction",
    "extremal principle"
  ],
  "Problem": "Consider a row of $2n$ squares colored alternately black and white. A legal move consists of choosing any contiguous set of squares and inverting their colors. What is the miniimum number of moves required to make the entire row be one color? \r\n\r\nThe answer is pretty obvious, but be careful with your proofs!",
  "Solution_1": "so can we pick one square only..like lets say u change the one square from black to white..or do u have to pick more then one square in a row",
  "Solution_2": "[hide=\"Well...\"] Say we want to turn the squares black.  If we invert an even number of squares, we have done nothing.  If we invert an odd number of squares, we can increase the number of black squares by at most $1$, so it will take at [b]least[/b] $n$ moves.\n\nIf we are allowed to flip one square at a time, we can flip the $n$ white squares, so it takes at [b]most[/b] $n$ moves.\n\nErgo it takes [b]exactly[/b] $n$ moves. QED. \n\n(We must be allowed to flip one square at a time, or else the case $n = 1$ is insoluble.) [/hide]",
  "Solution_3": "ya thats true...and i guess that solution is also valid..nicely done",
  "Solution_4": "[quote=\"t0rajir0u\"][hide=\"Well...\"] Say we want to turn the squares black.  If we invert an even number of squares, we have done nothing.  If we invert an odd number of squares, we can increase the number of black squares by at most $1$, so it will take at [b]least[/b] $n$ moves.\n\nIf we are allowed to flip one square at a time, we can flip the $n$ white squares, so it takes at [b]most[/b] $n$ moves.\n\nErgo it takes [b]exactly[/b] $n$ moves. QED. \n\n(We must be allowed to flip one square at a time, or else the case $n = 1$ is insoluble.) [/hide][/quote]\r\n\r\nThe argument is a little trickier, I think. It's not always true that you can only increase the number of black squares by at most $1$. For example, say you invert one black square. Now you have 3 white squares; you can invert all of those, thus increasing the number of black squares by $3$. Obviously, this isn't \"efficient,\" but it makes the proof a little trickier.",
  "Solution_5": "Hmm, true.  I hadn't considered that.   :|",
  "Solution_6": "[hide=\"induction? :maybe: \"]I think the minimum is $n$.\n\nBase case.\nFor $n=1$, the figure looks like $\\text{BW}$, and clearly it takes only 1 move to change to all black or all white. \n\nInductive step.\nAssume that it takes only $n$ moves turn $2n$ squares of BW's into one color. For $2(n+1)$ squares, the figure looks like $BW\\cdots BW(BW)$. By assumption, it takes $n$ moves to turn it into $BB\\cdots BB(BW)$, and $1$ more move to make the entire $2(n+1)$ row into W's.\n\nSo it takes a minimum of n moves.(?)[/hide]",
  "Solution_7": "[quote=\"pianoforte\"][hide=\"induction? :maybe: \"]I think the minimum is $n$.\n\nBase case.\nFor $n=1$, the figure looks like $\\text{BW}$, and clearly it takes only 1 move to change to all black or all white. \n\nInductive step.\nAssume that it takes only $n$ moves turn $2n$ squares of BW's into one color. For $2(n+1)$ squares, the figure looks like $BW\\cdots BW(BW)$. By assumption, it takes $n$ moves to turn it into $BB\\cdots BB(BW)$, and $1$ more move to make the entire $2(n+1)$ row into W's.\n\nSo it takes a minimum of n moves.(?)[/hide][/quote]\r\n\r\nThat makes the assumption that if you have $2(n+1)$ squares the optimal solution is to change the first $2n$ squares before the last $2$. Which is nontrivial and not even completely true since there exist many strategies for reaching the optimal solution of $n+1$ moves.\r\n\r\nAnother inductive \"solution\": After applying a single move to $2(n+1)$ squares (assuming you don't affect the end squares), combine all contiguous white and black squares. In one move, we have reduced it to the $2n$ case. The problem is that we don't know that the optimal solution doesn't affect only some of the squares we combined (e.g. only the left square in a pair of white squares). This induction seems a little more promising, though.",
  "Solution_8": "I am under the impression that there should exist a \"slick\" and simple solution- the result is, after all, intuitively obvious.",
  "Solution_9": "How about the extremal principle?\r\n\r\n[hide=\"Hint\"]Define the [i]weight[/i] of a single move to be the number of squares flipped by it.  Define the [i]weight[/i] of a sequence of moves to be the sum of the weights of all of its moves.  Among all the sequence of moves that result in a unicolor row, consider those that use the minimum number of moves.  Among all these minimum-move sequences, consider one that has the minimum weight.  In such a \"minimum minimum\" sequence, show that its moves are disjoint.[/hide]",
  "Solution_10": "[quote=\"Ravi B\"]How about the extremal principle?\n\n[hide=\"Hint\"]Define the [i]weight[/i] of a single move to be the number of squares flipped by it.  Define the [i]weight[/i] of a sequence of moves to be the sum of the weights of all of its moves.  Among all the sequence of moves that result in a unicolor row, consider those that use the minimum number of moves.  Among all these minimum-move sequences, consider one that has the minimum weight.  In such a \"minimum minimum\" sequence, show that its moves are disjoint.[/hide][/quote]\n\n[hide=\"Neat Idea!\"]Well, by contradiction, suppose two moves overlap. If instead of using these two moves we invert each of the parts affected by only one of these two moves, we achieve the same result, since the squares affected by both end up unchanged anyway, and have a corresponding weight loss (haha) of $2 \\cdot (\\text{number of overlapped squares})$, a contradiction to the minimum weight assumption.\n\nThen, we may apply the originally \"flawed\" arguments to achieve the result.[/hide]"
}
{
  "Tag": [],
  "Problem": "Hi,\r\n\r\nIs there an easy way to evaluate $ \\sum_{i\\equal{}1}^{\\infty}\\frac{1}{i^2}$?\r\n\r\nSorry if this is too hard/easy for this forum. :maybe: \r\n\r\nThanks,\r\nThinkFlow",
  "Solution_1": "I don't think it can get easier than [url=http://en.wikipedia.org/wiki/Basel_problem]this[/url].\r\n\r\nAnd [url=http://www.secamlocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf]this[/url] is also useful."
}
{
  "Tag": [
    "algorithm",
    "AMC"
  ],
  "Problem": "I have gotten worse at math because I have gotten much more careless since last year. Is there any way to practice avoiding them?",
  "Solution_1": "basically, just check over your work. double check your arithmatic as you go along, and later during the test, perhaps at the two-hour mark or so, check your answers again. read the questions several times, especially lengthy ones. make sure you give what they asked for (for example, if you solve for x and y, and the problem asks for x, don't give y)",
  "Solution_2": "Yeah, check.\r\nI rather went overboard with checking this year, and ran out of time 7/8 of the way through doing #11. And I STILL didn't catch my mistake on #2 which plagued half the population.  :D",
  "Solution_3": "[quote=\"davidyko\"]Yeah, check.\nI rather went overboard with checking this year, and ran out of time 7/8 of the way through doing #11. And I STILL didn't catch my mistake on #2 which plagued half the population.  :D[/quote]\r\n\r\ni had a bit of time to briefly check 1 through 5. I did catch my mistake in #2 because from my interpretation there could be two answers.",
  "Solution_4": "Be [b]really[/b] careful the first time you do the problem. That ensures that you'll spend enough time on each problem and sometimes even eliminates the need to check back later. I usually like to run through my work right after doing a problem, but this year it seemed like checking was taking a really long time and I needed to make sure I got to all the problems. So instead of the usual procedure, I decided to mark a few answers I wasn't confident about and to check on them again later.",
  "Solution_5": "Chess 64's method of using mod 9 to check arithmetic works pretty well",
  "Solution_6": "besides what has already been said, no.",
  "Solution_7": "[quote=\"bpms\"]Chess 64's method of using mod 9 to check arithmetic works pretty well[/quote]\r\nwhat's Chess64's method?\r\n\r\nfor multiplication, i do standard multiplication and then check with this algorithm that i discovered 4 years ago in pre-algebra class :P\r\nthe product of numbers  equals the square of their arithmetic mean subtracted by the square of half their difference.\r\n$xy = (\\frac{x+y}{2})^{2}-(\\frac{x-y}{2})^{2}$\r\nit's useful for checking products of two integers that have an even sum.\r\nto check a square:\r\n$x^{2}= (x+y)(x-y)+y^{2}$\r\nuseful when you're trying to check $48^{2}$\r\n$48^{2}= 46*50+2^{2}$",
  "Solution_8": "[quote=\"fractalz\"][quote=\"bpms\"]Chess 64's method of using mod 9 to check arithmetic works pretty well[/quote]\nwhat's Chess64's method?[/quote]\r\nMod 9",
  "Solution_9": "I heard that people should just do work very fast in practice, even if you make more mistakes. Then, when you are used to working so fast, then working just a little bit slowly  will greatly reduce mistakes under time constraint.",
  "Solution_10": "Write NEATLY on your scrap paper. Put only one question per sheet of scrap paper. Number each scrap paper with the question number.\r\n\r\nTry to do things a different way when you check over your work. If you have a sequence of numbers to add, try adding them in a different order and see if you get the same answer. Plug in your answers to the original problem and make sure they work (this is how I caught the planets question -- when I was moving the planets to the 210 position I realized they would be collinear at 105  too). Underline the question. Put a box around key phrases of the question that aren't italicized, but should be."
}
{
  "Tag": [
    "geometry",
    "geometric transformation",
    "homothety",
    "geometry open"
  ],
  "Problem": "Construction Trapezoidal know two diagonal measure of length be m and n , $ \\widehat{BAI} \\equal{} \\frac{1}{2}  \\widehat{BIA}$ (I be intersection two diagonal)",
  "Solution_1": "here's a solution to your nice problem \r\nconstruct a line $ (L)$ and put on it the lenght $ m\\plus{}n \\equal{} A'B'$ then draw a circle with radius $ m$ and center $ R$ \r\nnow draw a lenght egal to $ \\sqrt{(m\\plus{}n)n}$ it's a well known technic  \r\ndraw a circle with center $ B'$ and radius $ \\sqrt{(m\\plus{}n)n}$ so we get $ I$ \r\nyou can see now that  $ ABI$ is deduced from $ A'B'I$ by homothety"
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "let $ a,b,c \\ge0$ such that $ \\frac {1}{{{a^2} \\plus{} {b^2} \\plus{} 1}} \\plus{} \\frac {1}{{{b^2} \\plus{} {c^2} \\plus{} 1}} \\plus{} \\frac {1}{{{c^2} \\plus{} {a^2} \\plus{} 1}} \\ge 1$.\r\nProve that $ ab \\plus{} bc \\plus{} ca\\le3$",
  "Solution_1": "It is a double- application of Cauchy-Schwartz! :)",
  "Solution_2": "[quote=\"great math\"]It is a double- application of Cauchy-Schwartz! :)[/quote]\r\n\r\ni don't think so.( also use cauchy Schwart).Will you show me your proof. I think i's not really easy",
  "Solution_3": "Well, if you want to know the answer, here is my solution\r\n\r\nFrom the given condition, we have\r\n\r\n$ 2 \\ge \\sum_{cyc}\\frac{a^2\\plus{}b^2}{a^2\\plus{}b^2\\plus{}1}$\r\n\r\nApplying Cauchy- Schwartz as follows\r\n\r\n$ 2\\left(2(a^2\\plus{}b^2\\plus{}c^2)\\plus{}3\\right) \\ge \\left(2(a^2\\plus{}b^2\\plus{}c^2)\\plus{}3\\right)\\left(\\sum_{cyc}\\frac{a^2\\plus{}b^2}{a^2\\plus{}b^2\\plus{}1}\\right) \\ge (\\sum_{cyc}\\sqrt{a^2\\plus{}b^2})^2$\r\n\r\nIt implies us to the following evaluation\r\n\r\n$ 2\\left(2(a^2\\plus{}b^2\\plus{}c^2)\\plus{}3\\right) \\ge (\\sum_{cyc}\\sqrt{a^2\\plus{}b^2})^2$ \r\n\r\nAmazingly, by long expansion, we get\r\n\r\n$ a^2\\plus{}b^2\\plus{}c^2 \\plus{}3 \\ge \\sum_{cyc}\\sqrt{(a^2\\plus{}b^2)(a^2\\plus{}c^2)} \\ge \\sum_{cyc} (a^2\\plus{}bc)$\r\n\r\nIn conclusion, we obtain\r\n\r\n$ 3 \\ge ab\\plus{}bc\\plus{}ca$            \r\n as desired.\r\n\r\nWe complete our proof."
}
{
  "Tag": [
    "function"
  ],
  "Problem": "Develop a mathematical procedure by which you might judge when one function fits a data set better than another.\r\n\r\n??? How would you do that?",
  "Solution_1": "Well, think about it.  Given a set of data and some possible functions, what makes some of them closer approximations than others?",
  "Solution_2": "they have to go through or almost go through most points in the data. but is the question looking for something more?  :maybe:",
  "Solution_3": "A more concrete example:  Suppose I have the data points\r\n\r\n$(0, 0.1), (1, 1.1), (2, 4.1), (3, 9.1), (4, 16.1), (5, 25.1)$\r\n\r\nClearly the function $f(x) = x+0.1$ goes through two of these points, whereas the function $f(x) = x^{2}$ goes through none of them - yet which one is the better fit?",
  "Solution_4": "correlation coefficient",
  "Solution_5": "[quote=\"t0rajir0u\"]A more concrete example:  Suppose I have the data points\n\n$(0, 0.1), (1, 1.1), (2, 4.1), (3, 9.1), (4, 16.1), (5, 25.1)$\n\nClearly the function $f(x) = x+0.1$ goes through two of these points, whereas the function $f(x) = x^{2}$ goes through none of them - yet which one is the better fit?[/quote]\r\n\r\n\r\nThe $f(x) = x^{2}$ one because it follows the trend of the data points.",
  "Solution_6": "What do you mean by that?\r\n\r\nYou probably mean that if you use the two rules to generate data points, the first rule produces\r\n\r\n$(0, 0.1), (1, 1.1), (2, 2.1), (3, 3.1), (4, 4.1), (5, 5.1)$\r\n\r\nWhereas the second rule produces\r\n\r\n$(0, 0), (1, 1), (2, 4), (3, 9), (4, 16), (5, 25)$\r\n\r\nAnd clearly that second set of points is much \"closer\" than that first set of points.  So can you encapsulate this \"closeness\" mathematically?"
}
{
  "Tag": [],
  "Problem": "This is a problem that I saw on a Japanese TV show and I'm pretty stumped\r\n(I'll use \"times\" for the multiplication sign):\r\n\r\nabc times cba = 2002 times x\r\n\r\nFind x if a, b, c, and x are different integers and abc and cba are two 3-digit integers which have reversed digits.",
  "Solution_1": "i don't know if this helps, but 2002=2*7*11*13.\r\n\r\nthen we know either a or c is divisible by 2 without loss of generality let it be a, and let a=2d, 2d-b+c is divisible by 11. actually, 2d-b+c is either 11 or 0.\r\n\r\nanother thing you could do is say that abc=100a+10b+c, cba=100c+10b+a.\r\n\r\nabc x cba=10001ac+1010bc+1010 ab+100a^2+100b^2\r\n+100c^2. all of this we know is equal to 0 mod 2, 7, 11, 13, which give us the equations:\r\n\r\nac=0 mod 2.\r\n5ac+2bc+2ab+2a^2+2b^2+2c^2=0 mod 7\r\n2ac+9bc+9ab+a^2+b^2+c^2=0 mod 11\r\netc. you get the point. not sure if this is helpful tho...",
  "Solution_2": "this becomes quite trivial if you know the divisibility rules.\r\n\r\nIt's well explained [url=http://www.m-a.org.uk/xplusyfiles/home/issue_11/multiple_tricks/index.htm]here[/url]:\r\n\r\nand use the fact that a,b,c are integers and try and set a bound for x.\r\nThen we can also find x as with the remainder rules we can determine the l.c.f. of x"
}
{
  "Tag": [],
  "Problem": "$ \\frac{3 \\times 5}{9 \\times 11} \\times \\frac{7 \\times 9 \\times 11}{3 \\times 5 \\times 7}\\equal{}$\r\n\r\n\\[ \\textbf{(A)}\\ 1 \\qquad\r\n\\textbf{(B)}\\ 0 \\qquad\r\n\\textbf{(C)}\\ 49 \\qquad\r\n\\textbf{(D)}\\ \\frac{1}{49} \\qquad\r\n\\textbf{(E)}\\ 50\r\n\\]",
  "Solution_1": "[quote=\"AIME15\"]$ \\frac {3 \\times 5}{9 \\times 11} \\times \\frac {7 \\times 9 \\times 11}{3 \\times 5 \\times 7} \\equal{}$\n\\[ \\textbf{(A)}\\ 1 \\qquad \\textbf{(B)}\\ 0 \\qquad \\textbf{(C)}\\ 49 \\qquad \\textbf{(D)}\\ \\frac {1}{49} \\qquad \\textbf{(E)}\\ 50\n\\]\n[/quote]\r\n\r\n[hide]Cancel, cancel, cancel, cancel, and then cancel.\n\n$ \\boxed{1}$[/hide]",
  "Solution_2": "1.Everything will get canceled",
  "Solution_3": "[hide=\"Solution\"]\nNotice that all terms cancel and we are left with $\\boxed{\\text{(A)} 1}$\n[/hide]",
  "Solution_4": "...why did you revive a 5.5 year old thread that had already been resolved?",
  "Solution_5": "Lol, since the threads were already bumped by [b]@competitivecoder[/b] ( :o ), I decided to give nice additional solutions that were $\\LaTeX$ified for future thread viewers :)",
  "Solution_6": "Here is a solution with a motivation, to help future viewers  :coolspeak: \n\nWe firstly must understand the meaning of $ \\times $. This problem is on the [b]AMC 8[/b], so this is meant for [i]8th[/i] graders to solve. Thus, we may hypothesize that $ \\times $ means a [i]variable[/i], specifically the variable $X$. \n\n[color=#f00]Lemma 1 : 7 < 8[/color] \n[i]Proof:[/i] \nNote that if we subtract $7$ from both sides, we get that: \n$$ 0 < 1 $$\nNow, we use binomial theorem to solve: note that $0^2 + 1^2 = 1^2$, so we have a right triangle with side lengths $0,1,$ and $1$. Since $1$ is the hypothenuse, and $0$ is a leg, and since a leg is always smaller than the hypothenuse, we find that $0 < 1$. $\\Box$\n[color=#f00]Lemma 2 : The $\\times$ is a variable[/color] \n[i]Proof:[/i] \nWe refer to this postulate:\n[quote = my 7th grade teacher]\nFrom now on, we use dots for multiplication and the \"x\" is a variable, not the multiplication sign.\n[/quote]\nUsing Lemma 1, we get that $7<8$, and since this is the AMC 8 (proved earlier), which is meant for 8th graders, and because we must use dots for multiplication, we can conclude that the $x's$ are variables. There is an alternative proof at the end. $\\Box$\n\nNow, since the $\\times$ are variables, we can make the expression:\n$$ \\frac{3x5}{9x11}x\\frac{7x9x11}{3x5x7}$$\nBy communative property and algebra, this equal\n$$\\frac{15x^2}{99x}\\frac{693x^2}{105x^2} =   \\frac{10395x^4}{10395x^3} = x$$\nTo find $x$, we guess that it equals $0$, so the answer is $\\boxed{A}$. \n\n[hide=alternative proof]\nNote that the main confusion is the $x's$. Since my ex is a [i]pig[/i] and is annoying, and since multiplication is also pretty annoying, the $\\times$ is multiplication. Therefore, everything cancels out, and we get the answer of $\\boxed{1}$.\n[/hide]",
  "Solution_7": "Love your solution but I don't think we should bump over 10 year old threads  :-D ",
  "Solution_8": "[quote=AIME15]$ \\frac{3 \\times 5}{9 \\times 11} \\times \\frac{7 \\times 9 \\times 11}{3 \\times 5 \\times 7}\\equal{}$\n\n\\[ \\textbf{(A)}\\ 1 \\qquad\n\\textbf{(B)}\\ 0 \\qquad\n\\textbf{(C)}\\ 49 \\qquad\n\\textbf{(D)}\\ \\frac{1}{49} \\qquad\n\\textbf{(E)}\\ 50\n\\][/quote]\n\nIt cancels out",
  "Solution_9": "[quote=Posymk25]Love your solution but I don't think we should bump over 10 year old threads  :-D[/quote]\n\nit was only 5 years since the last post before him",
  "Solution_10": "Wasn't this the first problem from the first AMC 8 ever?",
  "Solution_11": "Imagine some poor kid actually multiplying all that out :noo:",
  "Solution_12": "Link to original problem: [url]https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_1[/url]",
  "Solution_13": "[quote=franzliszt]Imagine some poor kid actually multiplying all that out :noo:[/quote]\n\nI almost did xD",
  "Solution_14": "The people who did this problem on the real test are probably in their 40's. That's crazy.... :huh: ",
  "Solution_15": "[hide]A is the answer",
  "Solution_16": "I am trying to imagine a smart kid tricking everyone in his class to multiply all that out...",
  "Solution_17": "[hide]1 of course. Because they all cancel out[/hide]",
  "Solution_18": "[hide]everything eliminates: 3/3, 5/5, 7/7. 9/9, 11/11. the answer is A. [/hide]\nalso, why are we bumping threads that are older than some of you?",
  "Solution_19": "[quote=pubulous][hide]everything eliminates: 3/3, 5/5, 7/7. 9/9, 11/11. the answer is A. [/hide]\nalso, why are we bumping threads that are older than some of you?[/quote]\n\nlol. interesting bump tho",
  "Solution_20": "The answer is $1$, or a, everything is cancelled",
  "Solution_21": "[quote=AIME15]$ \\frac{3 \\times 5}{9 \\times 11} \\times \\frac{7 \\times 9 \\times 11}{3 \\times 5 \\times 7}\\equal{}$\n\n\\[ \\textbf{(A)}\\ 1 \\qquad\n\\textbf{(B)}\\ 0 \\qquad\n\\textbf{(C)}\\ 49 \\qquad\n\\textbf{(D)}\\ \\frac{1}{49} \\qquad\n\\textbf{(E)}\\ 50\n\\][/quote]\n[quote=Quentissentia[l][hide=:D, sol]A[/hide][/quote]\nthats my answer",
  "Solution_22": "[quote=Quentissential][quote=AIME15]$ \\frac{3 \\times 5}{9 \\times 11} \\times \\frac{7 \\times 9 \\times 11}{3 \\times 5 \\times 7}\\equal{}$\n\n\\[ \\textbf{(A)}\\ 1 \\qquad\n\\textbf{(B)}\\ 0 \\qquad\n\\textbf{(C)}\\ 49 \\qquad\n\\textbf{(D)}\\ \\frac{1}{49} \\qquad\n\\textbf{(E)}\\ 50\n\\][/quote]\n[quote=Quentissentia[l][hide=:D, sol]A[/hide][/quote]\nthats my answer[/quote]\n\nwhy revive a 11 yr old forum",
  "Solution_23": "You mean thread :)",
  "Solution_24": "[hide]A is the answer[/hide]",
  "Solution_25": "[quote=mathapple101]Here is a solution with a motivation, to help future viewers  :coolspeak: \n\nWe firstly must understand the meaning of $ \\times $. This problem is on the [b]AMC 8[/b], so this is meant for [i]8th[/i] graders to solve. Thus, we may hypothesize that $ \\times $ means a [i]variable[/i], specifically the variable $X$. \n\n[color=#f00]Lemma 1 : 7 < 8[/color] \n[i]Proof:[/i] \nNote that if we subtract $7$ from both sides, we get that: \n$$ 0 < 1 $$\nNow, we use binomial theorem to solve: note that $0^2 + 1^2 = 1^2$, so we have a right triangle with side lengths $0,1,$ and $1$. Since $1$ is the hypothenuse, and $0$ is a leg, and since a leg is always smaller than the hypothenuse, we find that $0 < 1$. $\\Box$\n[color=#f00]Lemma 2 : The $\\times$ is a variable[/color] \n[i]Proof:[/i] \nWe refer to this postulate:\n[quote = my 7th grade teacher]\nFrom now on, we use dots for multiplication and the \"x\" is a variable, not the multiplication sign.\n[/quote]\nUsing Lemma 1, we get that $7<8$, and since this is the AMC 8 (proved earlier), which is meant for 8th graders, and because we must use dots for multiplication, we can conclude that the $x's$ are variables. There is an alternative proof at the end. $\\Box$\n\nNow, since the $\\times$ are variables, we can make the expression:\n$$ \\frac{3x5}{9x11}x\\frac{7x9x11}{3x5x7}$$\nBy communative property and algebra, this equal\n$$\\frac{15x^2}{99x}\\frac{693x^2}{105x^2} =   \\frac{10395x^4}{10395x^3} = x$$\nTo find $x$, we guess that it equals $0$, so the answer is $\\boxed{A}$. \n\n[hide=alternative proof]\nNote that the main confusion is the $x's$. Since my ex is a [i]pig[/i] and is annoying, and since multiplication is also pretty annoying, the $\\times$ is multiplication. Therefore, everything cancels out, and we get the answer of $\\boxed{1}$.\n[/hide][/quote]\n\nwow you're so cool and funny! bumping a 5 year old thread to make a joke?!?! i never would become so funny in my wildest dreams! ",
  "Solution_26": "cross multiplication",
  "Solution_27": "[quote=motorfinn][quote=mathapple101]Here is a solution with a motivation, to help future viewers  :coolspeak: \n\nWe firstly must understand the meaning of $ \\times $. This problem is on the [b]AMC 8[/b], so this is meant for [i]8th[/i] graders to solve. Thus, we may hypothesize that $ \\times $ means a [i]variable[/i], specifically the variable $X$. \n\n[color=#f00]Lemma 1 : 7 < 8[/color] \n[i]Proof:[/i] \nNote that if we subtract $7$ from both sides, we get that: \n$$ 0 < 1 $$\nNow, we use binomial theorem to solve: note that $0^2 + 1^2 = 1^2$, so we have a right triangle with side lengths $0,1,$ and $1$. Since $1$ is the hypothenuse, and $0$ is a leg, and since a leg is always smaller than the hypothenuse, we find that $0 < 1$. $\\Box$\n[color=#f00]Lemma 2 : The $\\times$ is a variable[/color] \n[i]Proof:[/i] \nWe refer to this postulate:\n[quote = my 7th grade teacher]\nFrom now on, we use dots for multiplication and the \"x\" is a variable, not the multiplication sign.\n[/quote]\nUsing Lemma 1, we get that $7<8$, and since this is the AMC 8 (proved earlier), which is meant for 8th graders, and because we must use dots for multiplication, we can conclude that the $x's$ are variables. There is an alternative proof at the end. $\\Box$\n\nNow, since the $\\times$ are variables, we can make the expression:\n$$ \\frac{3x5}{9x11}x\\frac{7x9x11}{3x5x7}$$\nBy communative property and algebra, this equal\n$$\\frac{15x^2}{99x}\\frac{693x^2}{105x^2} =   \\frac{10395x^4}{10395x^3} = x$$\nTo find $x$, we guess that it equals $0$, so the answer is $\\boxed{A}$. \n\n[hide=alternative proof]\nNote that the main confusion is the $x's$. Since my ex is a [i]pig[/i] and is annoying, and since multiplication is also pretty annoying, the $\\times$ is multiplication. Therefore, everything cancels out, and we get the answer of $\\boxed{1}$.\n[/hide][/quote]\n\nwow you're so cool and funny! bumping a 5 year old thread to make a joke?!?! i never would become so funny in my wildest dreams![/quote]\n\nactually, you are being much funny right now ! you so sarcastic it make me laugh ! you must be in wildest dreams ! ",
  "Solution_28": "[quote=PerfectPanther13][quote=franzliszt]Imagine some poor kid actually multiplying all that out :noo:[/quote]\n\nI almost did xD[/quote]\n\nMe too then i realized whats the point you can just cancel out everything.",
  "Solution_29": "everything cancels out so we ge[t $1$",
  "Solution_30": "CANCELLATION CONSTELLATION",
  "Solution_31": "[quote=mathapple101]Here is a solution with a motivation, to help future viewers  :coolspeak: \n\nWe firstly must understand the meaning of $ \\times $. This problem is on the [b]AMC 8[/b], so this is meant for [i]8th[/i] graders to solve. Thus, we may hypothesize that $ \\times $ means a [i]variable[/i], specifically the variable $X$. \n\n[color=#f00]Lemma 1 : 7 < 8[/color] \n[i]Proof:[/i] \nNote that if we subtract $7$ from both sides, we get that: \n$$ 0 < 1 $$\nNow, we use binomial theorem to solve: note that $0^2 + 1^2 = 1^2$, so we have a right triangle with side lengths $0,1,$ and $1$. Since $1$ is the hypothenuse, and $0$ is a leg, and since a leg is always smaller than the hypothenuse, we find that $0 < 1$. $\\Box$\n[color=#f00]Lemma 2 : The $\\times$ is a variable[/color] \n[i]Proof:[/i] \nWe refer to this postulate:\n[quote = my 7th grade teacher]\nFrom now on, we use dots for multiplication and the \"x\" is a variable, not the multiplication sign.\n[/quote]\nUsing Lemma 1, we get that $7<8$, and since this is the AMC 8 (proved earlier), which is meant for 8th graders, and because we must use dots for multiplication, we can conclude that the $x's$ are variables. There is an alternative proof at the end. $\\Box$\n\nNow, since the $\\times$ are variables, we can make the expression:\n$$ \\frac{3x5}{9x11}x\\frac{7x9x11}{3x5x7}$$\nBy communative property and algebra, this equal\n$$\\frac{15x^2}{99x}\\frac{693x^2}{105x^2} =   \\frac{10395x^4}{10395x^3} = x$$\nTo find $x$, we guess that it equals $0$, so the answer is $\\boxed{A}$. \n\n[hide=alternative proof]\nNote that the main confusion is the $x's$. Since my ex is a [i]pig[/i] and is annoying, and since multiplication is also pretty annoying, the $\\times$ is multiplication. Therefore, everything cancels out, and we get the answer of $\\boxed{1}$.\n[/hide][/quote]\n\nbtw, 6th graders can also take this...",
  "Solution_32": "Why is it so overcomplicated?",
  "Solution_33": "[quote=royliang]Why is it so overcomplicated?[/quote]\n\nno u :love:",
  "Solution_34": "[quote=eevee_12][quote=royliang]Why is it so overcomplicated?[/quote]\n\nno u :love:[/quote]\n\nwhat a thing to say to someone thats just asking a question! ",
  "Solution_35": "[hide=bruh.... :dry: ]its $1$ cause everything cancels out![/hide]",
  "Solution_36": "Ok no one needs the answer anymore. The answer is one. We all now know. :roll: ",
  "Solution_37": "[quote=NInjas25]Ok no one needs the answer anymore. The answer is one. We all now know. :roll:[/quote]\n\nhaha yea"
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "quadratics"
  ],
  "Problem": "simplify \\[ \\sqrt [3]{2\\plus{}\\sqrt{5}}\\plus{}\\sqrt [3]{2\\minus{}\\sqrt{5}}\\].",
  "Solution_1": "Look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=245005]here[/url].",
  "Solution_2": "[quote=\"DaReaper\"]simplify\n\\[ \\sqrt [3]{2 \\plus{} \\sqrt {5}} \\plus{} \\sqrt [3]{2 \\minus{} \\sqrt {5}}\n\\]\n.[/quote]\r\n\r\nIf, $ \\sqrt [3]{\\sqrt {5} \\plus{} 2} \\plus{} \\sqrt [3]{\\sqrt {5} \\minus{} 2}$,\r\n$ \\sqrt [3]{\\left(\\frac{\\sqrt {5}}{2} \\plus{} \\frac{1}{2}\\right)^3} \\plus{} \\sqrt [3]{\\left(\\frac{\\sqrt {5}}{2} \\minus{} \\frac{1}{2}\\right)^3} \\equal{} \\left(\\frac{\\sqrt {5}}{2} \\plus{} \\frac{1}{2}\\right) \\plus{} \\left(\\frac{\\sqrt {5}}{2} \\minus{} \\frac{1}{2}\\right) \\equal{} \\sqrt{5}$.\r\n\r\nBut, $ \\sqrt [3]{2 \\minus{} \\sqrt {5}} \\not \\in \\mathbb{R}$.\r\nCf. [url=http://www.mathlinks.ro/viewtopic.php?t=243230]a nice eqution 1![/url]",
  "Solution_3": "[hide=\"identity\"]If a+b+c=0, then $ a^3\\plus{}b^3\\plus{}c^3\\equal{}3abc$.[/hide]\n[hide=\"proof\"]Consider the cubic with roots a,b, and c. It is $ x^3\\minus{}(a\\plus{}b\\plus{}c)x^2\\plus{}(ab\\plus{}ac\\plus{}bc)x\\minus{}abc$. Substituting x=a,b, and c, and then adding the resulting equations gives $ a^3\\plus{}b^3\\plus{}c^3\\minus{}(a\\plus{}b\\plus{}c)(a^2\\plus{}b^2\\plus{}c^2)\\plus{}(a\\plus{}b\\plus{}c)(ab\\plus{}ac\\plus{}bc)\\minus{}3abc\\equal{}0$, or $ a^3\\plus{}b^3\\plus{}c^3\\equal{}(a\\plus{}b\\plus{}c)(a^2\\plus{}b^2\\plus{}c^2\\minus{}ab\\minus{}ac\\minus{}bc)\\plus{}3abc$. Since a+b+c=0, we have $ a^3\\plus{}b^3\\plus{}c^3\\equal{}3abc$, as desired.[/hide]\n\n[hide=\"solution to problem\"] Let $ x\\equal{}\\sqrt[3]{2\\plus{}\\sqrt{5}}\\plus{}\\sqrt[3]{2\\minus{}\\sqrt{5}}$, so $ x\\minus{}\\sqrt[3]{2\\plus{}\\sqrt{5}}\\minus{}\\sqrt[3]{2\\minus{}\\sqrt{5}}\\equal{}0$. Using our identity, we now have $ x^3\\minus{}4\\equal{}3x\\sqrt[3]{\\minus{}1}$. If we're looking for only real solutions, which I assume we are, as otherwise it'll get messy, we have $ x^3\\plus{}3x\\minus{}4\\equal{}0$. Noticing that the coefficients add to 0, we have x=1 as a root. Dividing (x-1) out, we have $ (x\\minus{}1)(x^2\\plus{}x\\plus{}4)$. The quadratic clearly has no real roots, so we find that the only real solution is x=1.[/hide]",
  "Solution_4": "easy  [hide]\\[ \\[\n\\begin{array}{l}\n x = \\sqrt[3]{{2 + \\sqrt 5 }} + \\sqrt[3]{{2 - \\sqrt 5 }} \\\\ \n  \\Rightarrow x^3  = 4 + 3x\\sqrt[3]{{4 - 5}} \\\\ \n  \\Rightarrow x^3  + 3x - 4 = 0 \\\\ \n  \\Rightarrow (x - 1)(x^2  + x + 4) = 0 \\\\ \n  \\Rightarrow x = 1 \\\\ \n \\end{array}\n\\]\n\n [/hide]"
}
{
  "Tag": [
    "trigonometry",
    "function",
    "algebra",
    "polynomial",
    "algebra unsolved"
  ],
  "Problem": "Solve the eq\r\n$(\\sin 3x+4\\sin 2x\\cos 3x)^{2}=11\\cos^{2}3x$",
  "Solution_1": "hey guys...\r\n\r\ndont make theses impossibles problems..\r\n\r\nis a lost of time\r\n\r\n\r\nthe solution is a arc-function of a polynomial of degree 5\r\n\r\nthis dont have a explicit formula.\r\n\r\n\r\nso  the solution is only approximations...\r\n\r\n\r\nto do this, use newton method or other thing\r\n\r\n\r\nor just make a good problem that have a good answer...\r\n\r\n\r\nhumpt"
}
{
  "Tag": [
    "floor function",
    "function",
    "modular arithmetic",
    "inequalities",
    "number theory",
    "relatively prime"
  ],
  "Problem": "Let $n$ be a natural number and $f(n) = 2n - 1995 \\lfloor \\frac{n}{1000} \\rfloor$($\\lfloor$ $\\rfloor$ denotes the floor function).\r\n\r\n1. Show that if for some integer $r$: $f(f(f...f(n)...))=1995$ (where the function $f$ is applied $r$ times), then $n$ is multiple of $1995$.\r\n\r\n2. Show that if $n$ is multiple of 1995, then there exists r such that:$f(f(f...f(n)...))=1995$ (where the function $f$ is applied $r$ times). Determine $r$ if $n=1995.500=997500$",
  "Solution_1": "[hide=\"1\"]\nLet $f^r$ denote $\\underbrace{f\\circ f\\circ f\\circ \\cdots \\circ f}_r$. Since $f(n)\\equiv 2n\\pmod{1995}$, $f^r(n)\\equiv 2^rn\\pmod{1995}$. Then if $f^r(n)$ is divisible by 1995, so is $2^rn$. But 1995 and 2 are relatively prime, so 1995 must divide $n$.\n[/hide]\n[hide=\"2\"]\nIf $n$ is a multiple of 1995, then so is $f^r(n)$ for any $r$. Also, $\\frac{a}{1000}-1< \\left\\lfloor \\frac{a}{1000} \\right\\rfloor \\leq \\frac{a}{1000}$, so $\\frac{a}{200}\\leq f(a)< \\frac{a}{200}+1995$ for any positive integer $a$.\n\nSuppose there exists a positive integer $n$ which is a multiple of 1995 but $f^r(n)\\neq 1995$ for any $n$. Let $m$ be the least positive value of $f^r(n)$. Then $m\\geq 1995\\cdot 2$ since $m$ is a multiple of 1995. But the inequality gives \\[ 0<\\frac{m}{200}\\leq f(m)<\\frac{m}{200}+1995\\leq \\frac{101m}{200}<m \\] so $0<f(m)<m$ which is a contradiction. Therefore the sequence must eventually reach 1995.\n\nWhen $n=1997\\cdot 500$, $\\frac 52 \\cdot 1995\\leq f(n)<\\frac 72 \\cdot 1995$ so $f(n)$ must equal $3\\cdot 1995$. Since $\\frac{3\\cdot 1995}{200}$ is clearly less than 1995, $f^2(n)=1995$, so $r=2$ in this case.\n[/hide]"
}
{
  "Tag": [
    "vector"
  ],
  "Problem": "When you add vectors together, is it necessary to draw a 'sum' vector as well? \r\n\r\n[geogebra]38c9fff7c5195921e033ddc64b18a52f57cb6b60[/geogebra]",
  "Solution_1": "I'm not sure what you mean; if we add $ <a,b>\\plus{}<c,d>\\equal{}<a\\plus{}c,b\\plus{}d>$, we don't need to draw it to figure out the result, if that is what you mean."
}
{
  "Tag": [
    "modular arithmetic",
    "linear algebra",
    "matrix",
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "we have $2n+1$ real numbers $a_{1},a_{2},a_{3}...a_{2n+1}$ \r\nfor all $1\\le i\\le 2n+1$ \r\nwe can divide numbers $a_{1},a_{2},...a_{i-1},a_{i+1},...a_{2n+1}$\r\ninto two groups (each with $n$ elements) with equal sums\r\nprove that $a_{1}=a_{2}=...=a_{2n+1}$",
  "Solution_1": "[b]Solution.[/b]  Assume to the contrary that there exists a set $\\{a_{i}\\}_{i=1}^{2n+1}$ with all numbers not equal and with minimal sum of values.  Let $S=a_{1}+\\hdots+a_{2n+1}$.  Since when excluding $a_{i}$, $S-a_{i}$ can be divided into two sets with equal sums, $a_{i}\\equiv S\\pmod 2$, that is, $a_{i}$ have equal parity since $S\\mod 2$ remains invariant.\r\n\r\nIf $a_{i}\\equiv 0\\pmod 2\\ \\forall i\\in [1,2n+1]\\cap\\mathbb{Z}$, then we obtain a set $\\left\\{ \\frac{a_{i}}{2}\\right\\}_{i=1}^{2n+1}$ satisfying the properties of the problem with an even smaller sum of values. Contradiction.\r\n\r\nSimilarly, if all the $a_{i}$ are odd, then we have a contradiction since we can obtain a set $\\left\\{ \\frac{a_{i}+1}{2}\\right\\}_{i=1}^{2n+1}$ with a smaller total sum, not all elements equal. Contradiction.\r\n\r\nAnd the conclusion follows.$\\blacksquare$",
  "Solution_2": "note that $a_{1},...a_{2n+1}$ are $\\in R$  \r\ni solve it if $a_{1},...a_{2n+1}$ are $\\in Q$",
  "Solution_3": "You get the result for $\\mathbb R$ out of that for $\\mathbb Q$ by choosing a (Hamel)basis of $\\mathbb R$ over $\\mathbb Q$ and applying the result for rationals.",
  "Solution_4": "Of course, we may avoid Hamel basis (and hence axiom of choice) by using uniform approximation theorem: consider N such that $Na_{i}$ differ from some integers by less then 0.0001. Then Na_i satisfy the same condition.\r\n\r\nOr, we may say that linear system have only trivial solution over Q, hence it have only trivial solution over R (bacuse rank of a matrix does not depend on a field, over whiich it is considered. Of course, if the characteristics of field is 0)"
}
{
  "Tag": [
    "number theory",
    "prime factorization",
    "prime numbers",
    "number theory unsolved"
  ],
  "Problem": "Prove: For each positive integer is the number of divisors whose decimal representations ends with a 1 or 9 not less than the number of divisors whose decimal representations ends with 3 or 7.",
  "Solution_1": "Let $ n$ be the positive integer.\r\nLet set $ A$ contain all the divisors of $ n$ whose decimal representations end with $ 1$ or $ 9$.\r\nLet set $ B$ contain all the divisors of $ n$ whose decimal representations end with $ 3$ or $ 7$.\r\nWe will show that the cardinality of set $ A$ is greater than or equal to that of set $ B$.\r\nThe numbers in $ A$ and the numbers in $ B$ are not divisible by $ 2$ or $ 5$, so we only need to consider divisors of $ n$ that do not have $ 2$'s or $ 5$'s in their factorizations. The size of $ A$ and the size of $ B$ do not change if the exponent of 2 or 5 in $ n$'s prime factorization is changed to $ 0$. Thus we can let $ n$ not be divisible by $ 2$ or $ 5$.\r\n$ n$ can be expressed as \r\n$ n \\equal{} \\displaystyle\\prod_{w \\equal{} 1}^a p_w \\cdot \\displaystyle\\prod_{x \\equal{} 1}^b q_x \\cdot \\displaystyle\\prod_{y \\equal{} 1}^c r_y \\cdot \\displaystyle\\prod_{z \\equal{} 1}^d s_z$\r\nwhere $ p_w$ is a prime ending with $ 1$, $ q_x$ is a prime ending with $ 3$, $ r_y$ is a prime ending with $ 7$, $ s_z$ is a prime ending with $ 9$, and $ a, b, c$ and $ d$ are the sums of exponents on prime numbers ending with $ 1$, $ 3$, $ 7$, and $ 9$ respectively.\r\nNow consider a divisor of $ n$. Let $ m$ be that divisor. Then we can express $ m$ as\r\n$ m \\equal{} \\displaystyle\\prod_{w \\equal{} 1}^e p_w \\cdot \\displaystyle\\prod_{x \\equal{} 1}^f q_x \\cdot \\displaystyle\\prod_{y \\equal{} 1}^g r_y \\cdot \\displaystyle\\prod_{z \\equal{} 1}^h s_z$\r\nwhere $ 0 \\le e \\le a$, $ 0 \\le f \\le b$, $ 0 \\le g \\le c$, and $ 0 \\le h \\le d$.\r\nTaking $ m$ mod $ 10$, we get\r\n$ m \\equiv 1^e \\cdot 3^f \\cdot 7^g \\cdot 9^h \\equiv 3^f \\cdot ( \\minus{} 3)^g \\cdot ( \\minus{} 1)^h \\equiv 3^{f \\plus{} g} \\cdot ( \\minus{} 1)^{g \\plus{} h} \\mod 10$.\r\nBy Euler's totient theorem $ 3^4 \\equiv 1 \\mod 10$, so if $ f \\plus{} g \\equiv 0 \\mod 4$, $ 3^{f \\plus{} g} \\equiv 1$ so $ m \\equiv \\pm 1$ and $ m$ ends with a 1 or a 9. \r\nIf $ f \\plus{} g \\equiv 1 \\mod 4$, $ 3^{f \\plus{} g} \\equiv 3$ so $ m \\equiv \\pm 3$ and $ m$ ends with 3 or 7. \r\nIf $ f \\plus{} g \\equiv 2 \\mod 4$, $ 3^{f \\plus{} g} \\equiv 9$ so $ m \\equiv \\pm 9$ and $ m$ ends with a 1 or a 9. \r\nIf $ f \\plus{} g \\equiv 3 \\mod 4$, $ 3^{f \\plus{} g} \\equiv 7$ so $ m \\equiv \\pm 7$ and $ m$ ends with a 3 or a 7. \r\nIf $ f \\plus{} g$ is even, $ m$ is in set $ A$. If $ f \\plus{} g$ is odd, $ m$ is in set $ B$.\r\nFor each $ e, h$ such that $ 0 \\le e \\le a$ and $ 0 \\le h \\le d$,\r\nThere are $ i$ values of $ f$ that are even and $ j$ of them that are odd, such that $ i \\ge j$. There are $ k$ values of $ g$ that are even and $ l$ of them that are odd, such that $ k \\ge l$. $ ik \\plus{} jl \\ge il \\plus{} jk$ by rearrangement. Since $ ik \\plus{} jl$ values of $ f \\plus{} g$ are even and $ il \\plus{} jk$ values of $ f \\plus{} g$ are odd, the number of values of $ f \\plus{} g$ that are even is greater than or equal to the number of values of $ f \\plus{} g$ that are odd. Therefore, the cardinality of set $ A$ is greater than that of set $ B$, and we are finished."
}
{
  "Tag": [],
  "Problem": "Give a counting explanation for why (2n)!(2m)! / [m!n!(m+n)!] is an integer.",
  "Solution_1": "heya there!\r\n\r\nwe know that the exponent of a prime factor for the canonic form of  (2n)!(2m)! if \r\n\r\nsum { [2n/p_i^k]+[2m/p_i^k]} , the sum if after k\r\n\r\nand for m!n!(m+n)! is\r\n\r\nsum {[n/p_i^k]+[m/p_i^k]+[(m+n)/p_i^k]} again the sum is after k.\r\n\r\nwe know that [2a]+[2b]>=[a]+[b]+[a+b] for any a and b from R+.\r\nso the number is integer!!! :D\r\n\r\ncheers!",
  "Solution_2": "is it a counting explanation ?",
  "Solution_3": "I don't know!!! I just gave a solution.. didn't even read what he needed(sorry)! :D what do you mean by a counting explanation... I think that this is a counting explanation... :D\r\n\r\ncheers!",
  "Solution_4": "[quote=\"Lagrangia\"]I don't know!!! I just gave a solution.. didn't even read what he needed(sorry)! :D what do you mean by a counting explanation... I think that this is a counting explanation... :D\n\ncheers![/quote]\r\n\r\nI think he means a combinatorial argument... like counting something in two ways.",
  "Solution_5": "I think that such an explanation would be hard to find.. guess that the solution I have posted is good enough! :D"
}
{
  "Tag": [],
  "Problem": "Two people take turns cutting up a rectangular $6\\times 8$ chocolate bar. You are allowed to cut only along a division between the squares and the cut can only be in a straight line. The last player who can break the chocolate wins. Is there a winning strategy for either player? Generalize.",
  "Solution_1": "[hide=\"Seems too easy to be right . . .\"]\nEach time the chocolate is broken, the number of pieces increases by one.  Initially there is 1 piece and at the end there are 48 pieces.  After the first player breaks the chocolate , there are 2 pieces, after the second player goes there are 3, after the first player goes again there are 4, etc.  The first player always leaves an even number of pieces and the second player always leaves an odd number of pieces.  Since 48 is even, Player 1 wins.  For an $m \\times n$ bar, Player 1 wins iff either $m$ or $n$ is even; Player 2 wins iff both are odd.\n[/hide]",
  "Solution_2": "[quote=\"E^(pi*i)=-1\"][hide=\"Seems too easy to be right . . .\"]\nEach time the chocolate is broken, the number of pieces increases by one.  Initially there is 1 piece and at the end there are 48 pieces.  After the first player breaks the chocolate , there are 2 pieces, after the second player goes there are 3, after the first player goes again there are 4, etc.  The first player always leaves an even number of pieces and the second player always leaves an odd number of pieces.  Since 48 is even, Player 1 wins.  For an $m \\times n$ bar, Player 1 wins iff either $m$ or $n$ is even; Player 2 wins iff both are odd.\n[/hide][/quote]\r\n\r\nI agree that that sounds too easy to be right, but it seems right.\r\n\r\nOh yes that's right. The version I was thinking about is the normal cutcakes;\r\n\r\nYou have a 6x8 chocolate bar. The same rules apply but player 1 can only cut horizontally and player 2 can only cut vertically. Pieces always maintain their orientation. Who has a winning strategy?",
  "Solution_3": "Which direction is the $6$ in?\r\n\r\nSo this problem is basically a matter of survivor: outlast.",
  "Solution_4": "Hey, I don't understand the problem. For example, player 1 breaks the 6x8 into a 6x2 and a 6X6. If player 2 breaks the 6x6 into a 5x6 and a 1x6. This means we have 3 pieces of bars. What next? could player 1 choose to break any of them until what's left are 48 1x1 chocolates bars?",
  "Solution_5": "I think a player should break every remaining block (that's not $ 1\\times 1$) once during a single turn, otherwise the problem is trivial.",
  "Solution_6": "hm, so, correct me if i'm wrong. once we have 48 1x1 bars, the game ends?"
}
{
  "Tag": [
    "group theory",
    "abstract algebra",
    "linear algebra",
    "matrix",
    "algebra",
    "polynomial",
    "inequalities"
  ],
  "Problem": "Is a finite subgroup of $ \\mathrm{GL}_n(\\mathbb{C})$ necessarily nipotent?",
  "Solution_1": "Any finite group can be embedded as a finite subgroup of $ GL_n(K)$, for any field $ K$ and large enough $ n$.",
  "Solution_2": "I think you mean \"unipotent\". If so then we can say that there are no non-trivial finite unipotent subgroups $ G$ of $ Gl_n(\\mathbb C)$. To see this let $ A\\in Gl_n(\\mathbb C)$ a unipotent matrix of finite order: $ A^m \\equal{} E$. Then $ A$ is diagonalizable, since the minimal polynomial $ f_A(X)$ divides the polynomial $ X^m \\minus{} 1$, thus the zeros of $ f_A$ are all simple. If $ A$ is unipotent, then $ A \\equal{} E \\plus{} N$ with some nilpotent matrix $ N$ which must be 0 since $ A$ is diagonalizable. Thus $ A \\equal{} E$.\r\n\r\nNB: This remains true if $ \\mathbb C$ is replaced by a field of characteristic 0.",
  "Solution_3": "There's another meaning here: a nilpotent group is one in which the chain of central groups ($ Z_{n\\plus{}1}/Z_n$ is the center of $ G/Z_n$) eventually reaches the whole thing.\r\n\r\nFor this meaning, well, finite groups that aren't nilpotent exist. $ S_3$ is the first, and we can embed that in $ GL_2(\\mathbb{C})$, or $ GL_3(K)$ for any $ K$.",
  "Solution_4": "[quote=\"jmerry\"]$ S_3$ is the first, and we can embed that in $ GL_2(\\mathbb{C})$, or $ GL_3(K)$ for any $ K$.[/quote]\r\n\r\nIn case you are curious, S3 embeds in GL(2,K) for any (associative, unital) ring K.  This is because it has a 2-dimensional representation over the integers.  Representations of Sn are triply nice: (1) the trace of every matrix in a representation over a field of characteristic 0 is rational, (2) up to conjugation every representation over a field of characteristic 0 is actually over the field of rationals, (3) up to conjugation every matrix representation over the rationals is actually over the integers.  Once you have a representation over the integers, you get a representation over any associative unital ring.  Sn only has a few normal subgroups, so it is usually easy to check the representation is faithful.\r\n\r\n#1 is studied in the papers mentioned in the \"all elements of the order are conjugate\".  #2 is related to the Schur index, and #3 is why combinatorics is so interested in the representation theory of Sn.\r\n\r\nThe representation over an arbitrary ring takes a specified element of order 2 to [0,1;1,0] and a specified element of order 3 to [0,1;-1,-1].  [0,1;-1,-1] is never the identity, so the representation is always faithful."
}
{
  "Tag": [
    "calculus",
    "integration",
    "logarithms",
    "LaTeX",
    "derivative",
    "calculus computations"
  ],
  "Problem": "Hello everyone:\r\n\r\nI have questions regarding the following two integrals. \r\n\r\nThank you very much.\r\n\r\n---\r\n\r\n1. Differentiate: $ \\frac {d}{dx}\\left[\\int_{e^{2x}}^3 (y^2 \\minus{} 5 \\ln y) dy \\right]$\r\n\r\nI did not have much trouble calculating it, but I got an answer of $ \\minus{} 2e^{6x} \\plus{} 20xe^{2x}$ instead of the given [hide] $ 2e^{6x} \\minus{} 20xe^{2x}$[/hide].\r\n\r\nWhen I looked at the answer, my mistake took place here:\r\n\r\n$ \\frac {d}{dx}\\left[\\int_{e^{2x}}^3 (y^2 \\minus{} 5 \\ln y) dy \\right] \\equal{} \\minus{} \\frac {d}{dx}\\left[\\int_3^{e^{2x}} (y^2 \\minus{} 5 \\ln y) dy \\right]$\r\n\r\n= $ [e^{4x} \\minus{} 5(2x)][2e^{2x}]$\r\n\r\nOriginally, I thought that the negative sign was to be carried over after the integration part. Why does it go away after integration?\r\n\r\n---\r\n\r\n2. Differentiate: $ \\frac {d}{dx}\\left(\\int_{x \\minus{} 2}^5 9b^2 db \\right)$\r\n\r\nAfter some steps, I arrived at:\r\n\r\n$ 375x^3 \\minus{} 3(x \\minus{} 2)^3$.\r\n\r\nI started to differentiate here, and I applied the Chain Rule to $ \\frac {d}{dx} 3(x \\minus{} 2)^3$.\r\n\r\nHowever, this resulted in an incorrect answer as the given answer simplified $ 3(x \\minus{} 2)^3$ before differentiation. Why is my method erroneous?",
  "Solution_1": "[quote=\"catalan\"]-\n\n1. Differentiate: $ \\frac {d}{dx}\\left[\\int_{e^{2x}}^3 (y^2 \\minus{} 5 \\ln y) dy \\right]$\n\nI did not have much trouble calculating it, but I got an answer of $ \\minus{} 2e^{6x} \\plus{} 20xe^{2x}$[/quote]\r\n\r\nThat's the correct answer.",
  "Solution_2": "I assume you meant this: [quote=\"catalan\"]$ \\frac {d}{dx}\\left(\\int_{x \\minus{} 2}^{5x} 9b^2 db \\right)$[/quote]  (It's always a good idea when using LaTeX to preview your posts, or at least read over the final product, to make sure you haven't misplaced any {}s, etc.)\n\n[quote=\"catalan\"]I applied the Chain Rule to $ \\frac {d}{dx} 3(x \\minus{} 2)^3$.\n\nHowever, this resulted in an incorrect answer as the given answer simplified $ 3(x \\minus{} 2)^3$ before differentiation. Why is my method erroneous?[/quote]  Well, either you or the text made a computational error, and if you don't show your computations or your answer then it's impossible to tell who it actually is.  Any valid differentiation rule correctly applied will give the correct final answer, and the chain rule leads to the same thing as the product rule or as expanding everything out and applying the power rule; all three methods give something equal to $ 9(x \\minus{} 2)^2$ (which will appear with a minus sign in the final answer).",
  "Solution_3": "$ Q\\equal{}\\int_{e^{2x}}^3 (y^2 \\minus{} 5 \\ln y) dy$\r\n\r\nby parts (i know this is the hard way, instead of FTC)\r\n\r\n$ u\\equal{}y^2\\minus{}5\\ln y$, $ dv\\equal{}dy$\r\n$ du\\equal{}2y\\minus{}\\frac{5}{y}dy$, $ v\\equal{}y$\r\n\r\n$ Q\\equal{}(y^3\\minus{}5y \\ln y) |_{e^{2x}}^3 \\minus{} \\int_{e^{2x}}^3 2y^2\\minus{}5 dy$\r\n$ \\equal{} 27\\minus{}15\\ln 3 \\minus{}e^{6x}\\plus{}10xe^{2x} \\minus{} (\\frac{2y^3}{3}\\plus{}5y )|_{e^{2x}}^3$\r\n$ \\equal{} 27\\minus{}15\\ln 3 \\minus{}e^{6x}\\plus{}10xe^{2x} \\minus{} 18\\plus{}15\\plus{}\\frac{2e^{6x}}{3}\\minus{}5e^{2x}$\r\n$ \\equal{}\\frac{\\minus{}e^{6x}}{3}\\plus{}(2x\\minus{}1)5e^{2x}\\plus{}C$\r\n\r\nso you are rightwhen you diff",
  "Solution_4": "In the well-known conditions, we have  $ \\boxed {\\ \\int_{a(x)}^{b(x)}f(t)\\ \\mathrm {dt}\\equal{}f[b(x)]\\cdot b'(x)\\minus{}f[a(x)]\\cdot a'(x)\\ }$ .",
  "Solution_5": "Thank you very much for your replies!",
  "Solution_6": "[quote=\"Virgil Nicula\"]In the well-known conditions, we have  $ \\boxed {\\ \\int_{a(x)}^{b(x)}f(t)\\ \\mathrm {dt} \\equal{} f[b(x)]\\cdot b'(x) \\minus{} f[a(x)]\\cdot a'(x)\\ }$ .[/quote]\r\n\r\nI think you forgot to put a $ \\frac{d}{dx}$ in front.",
  "Solution_7": "You are right. :lol:"
}
{
  "Tag": [
    "LaTeX"
  ],
  "Problem": "anybody knows how to bold stuff in Latex?",
  "Solution_1": "Wrong forum but OK, I know how to bold letters\r\n\r\n[code]{\\bf Hello}[/code]\r\n\r\n$\\text{Hello}~{\\bf Hello}$",
  "Solution_2": "but what does $\\text{bf}$ stand for?",
  "Solution_3": "Bold font? I don't know..."
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "inequalities",
    "algorithm",
    "inequalities proposed"
  ],
  "Problem": "Let $ a,b,c$ be positive real numbers such that $ a\\plus{}b\\plus{}c\\equal{}1.$ Prove that for any $ 0<k \\le 2\\sqrt{6} \\minus{}4,$ we have\r\n$ \\frac{a^3}{a\\plus{}k(a\\minus{}b)^2}\\plus{}\\frac{b^3}{b\\plus{}k(b\\minus{}c)^2}\\plus{}\\frac{c^3}{c\\plus{}k(c\\minus{}a)^2} \\ge \\frac{1}{3}.$\r\n:)",
  "Solution_1": "Of course, $ 2\\sqrt{6}\\minus{}4\\equal{}0.898979485566...$ is not the best possible, for example, $ \\frac{59}{65}\\equal{}0.907692307692...$ is a better one.\r\n\r\nIn fact, the best possible numer is one of the 5 real roots of the following polynomial,\r\n\r\n$ 603979776\\,k^{29}\\minus{}18924699648\\,k^{28}\\plus{}289134870528\\,k^{27}\\minus{}2848483016704\\,k^{26}\\plus{}20225656193024\\,k^{25}\\minus{}109773054031872\\,k^{24}\\plus{}471820389211136\\,k^{23}\\minus{}1643287718855936\\,k^{22}\\plus{}4710311338973696\\,k^{21}\\minus{}11232540105181952\\,k^{20}\\plus{}22456479447729024\\,k^{19}\\minus{}37850088513976064\\,k^{18}\\plus{}54001862026633568\\,k^{17}\\minus{}65395647141882224\\,k^{16}\\plus{}67313415153512784\\,k^{15}\\minus{}58897753746129680\\,k^{14}\\plus{}43751305461719036\\,k^{13}\\minus{}27520743653021705\\,k^{12}\\plus{}14602143321125408\\,k^{11}\\minus{}6500926905187368\\,k^{10}\\plus{}2412231014113856\\,k^9\\minus{}739886997661168\\,k^8\\plus{}185756734585344\\,k^7\\minus{}37735657172736\\,k^6\\plus{}6118618383360\\,k^5\\minus{}778206852864\\,k^4\\plus{}75643158528\\,k^3\\minus{}5354809344\\,k^2\\plus{}248168448\\,k\\minus{}5640192,$\r\n\r\nnamely, $ k_{max}\\equal{}0.9077537983....$",
  "Solution_2": "Thank you very much for your result, but I think it is very hard for us to prove the inequality for the result you find if we dont have computer's helping. However, for $ k\\equal{}2\\sqrt{6} \\minus{}4,$ we can prove it in a very simple way.",
  "Solution_3": "We have a similar result is:\r\nIf $ a,b,c >0,a\\plus{}b\\plus{}c\\equal{}1,$ then for any $ 0 \\le k \\le \\frac{3\\plus{}\\sqrt{6}}{6},$ the following inequality holds\r\n$ \\sum \\frac{a^3}{a\\plus{}k(b\\minus{}c)^2} \\ge \\frac{1}{3}.$\r\nIn here, $ \\frac{3\\plus{}\\sqrt{6}}{6}$ is the best constant. :)",
  "Solution_4": "Dear can_hang2007,\r\nI don't know whether it can be proven by your simple way that for $ k\\equal{}\\frac{59}{65}\\,$ and even $ k\\equal{}\\frac{9}{10}$ which both are stronger than $ 2\\sqrt{6}\\minus{}4$.",
  "Solution_5": "[quote=\"Guest\"]Of course, $ 2\\sqrt {6} \\minus{} 4 \\equal{} 0.898979485566...$ is not the best possible, for example, $ \\frac {59}{65} \\equal{} 0.907692307692...$ is a better one.\n\nIn fact, the best possible numer is one of the 5 real roots of the following polynomial,\n\n$ 603979776\\,k^{29} \\minus{} 18924699648\\,k^{28} \\plus{} 289134870528\\,k^{27} \\minus{} 2848483016704\\,k^{26} \\plus{} 20225656193024\\,k^{25} \\minus{} 109773054031872\\,k^{24} \\plus{} 471820389211136\\,k^{23} \\minus{} 1643287718855936\\,k^{22} \\plus{} 4710311338973696\\,k^{21} \\minus{} 11232540105181952\\,k^{20} \\plus{} 22456479447729024\\,k^{19} \\minus{} 37850088513976064\\,k^{18} \\plus{} 54001862026633568\\,k^{17} \\minus{} 65395647141882224\\,k^{16} \\plus{} 67313415153512784\\,k^{15} \\minus{} 58897753746129680\\,k^{14} \\plus{} 43751305461719036\\,k^{13} \\minus{} 27520743653021705\\,k^{12} \\plus{} 14602143321125408\\,k^{11} \\minus{} 6500926905187368\\,k^{10} \\plus{} 2412231014113856\\,k^9 \\minus{} 739886997661168\\,k^8 \\plus{} 185756734585344\\,k^7 \\minus{} 37735657172736\\,k^6 \\plus{} 6118618383360\\,k^5 \\minus{} 778206852864\\,k^4 \\plus{} 75643158528\\,k^3 \\minus{} 5354809344\\,k^2 \\plus{} 248168448\\,k \\minus{} 5640192,$\n\nnamely, $ k_{max} \\equal{} 0.9077537983....$[/quote]\r\n\r\nHmm...how did you got that evil polynomial? :D",
  "Solution_6": "[quote=\"Stephen\"][quote=\"Guest\"]Of course, $ 2\\sqrt {6} \\minus{} 4 \\equal{} 0.898979485566...$ is not the best possible, for example, $ \\frac {59}{65} \\equal{} 0.907692307692...$ is a better one.\n\nIn fact, the best possible numer is one of the 5 real roots of the following polynomial,\n\n$ 603979776\\,k^{29} \\minus{} 18924699648\\,k^{28} \\plus{} 289134870528\\,k^{27} \\minus{} 2848483016704\\,k^{26} \\plus{} 20225656193024\\,k^{25} \\minus{} 109773054031872\\,k^{24} \\plus{} 471820389211136\\,k^{23} \\minus{} 1643287718855936\\,k^{22} \\plus{} 4710311338973696\\,k^{21} \\minus{} 11232540105181952\\,k^{20} \\plus{} 22456479447729024\\,k^{19} \\minus{} 37850088513976064\\,k^{18} \\plus{} 54001862026633568\\,k^{17} \\minus{} 65395647141882224\\,k^{16} \\plus{} 67313415153512784\\,k^{15} \\minus{} 58897753746129680\\,k^{14} \\plus{} 43751305461719036\\,k^{13} \\minus{} 27520743653021705\\,k^{12} \\plus{} 14602143321125408\\,k^{11} \\minus{} 6500926905187368\\,k^{10} \\plus{} 2412231014113856\\,k^9 \\minus{} 739886997661168\\,k^8 \\plus{} 185756734585344\\,k^7 \\minus{} 37735657172736\\,k^6 \\plus{} 6118618383360\\,k^5 \\minus{} 778206852864\\,k^4 \\plus{} 75643158528\\,k^3 \\minus{} 5354809344\\,k^2 \\plus{} 248168448\\,k \\minus{} 5640192,$\n\nnamely, $ k_{max} \\equal{} 0.9077537983....$[/quote]\n\nHmm...how did you got that evil polynomial? :D[/quote]\r\nI am sorry she is not very beautiful indeed, but it is really an algebraic number of degree 29. I got her by an algorithm from a Chinese book titled \"Automated Proving and Discovering on Inequalities\" by L. Yang and B. C. Xia, published in 2008, with a computer of course."
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "quadratics",
    "3D geometry",
    "tetrahedron",
    "Pythagorean Theorem",
    "geometry proposed"
  ],
  "Problem": "Let three circles (1),(2),(3) have the same center O and their radius are $ R_1,R_2,R_3$ respectively. Three points A,B,C move on (1),(2),(3) respectively. Known that $ R_1 \\equal{} 1,R_2 \\equal{} \\sqrt {2},R_3 \\equal{} \\sqrt {5}$, find the greatest value of the area of triangle ABC.",
  "Solution_1": "$ O$ is point in the $ \\triangle ABC$ with distances from the vertices $ AO : BO : CO \\equal{} 1 : \\sqrt 2 : \\sqrt 5.$ Suppose $ A, O, B$ are given, then for max area, $ CO \\perp AB,$ etc. $\\Longrightarrow$ $ O$ is orthocenter of the $ \\triangle ABC.$ Rename it $ H$ and let $ (O,R)$ be the circumcircle of this triangle. Let $ AH$ cut $ (O)$ again at $ D.$ The isosceles $ \\triangle BDO \\sim \\triangle DHC$ are similar, having equal angles $\\Longrightarrow$ $ BH \\cdot CH = BD \\cdot CH \\equal{} HD \\cdot OD \\equal{} HD \\cdot R.$ Power of $ H$ to $ (O)$ is $ R^2 \\minus{} OH^2 \\equal{} AH \\cdot HD \\equal{} \\frac{AH \\cdot BH \\cdot CH}{R}.$ On the other hand, \n\n$ AH^2 \\plus{} BH^2 \\plus{} CH^2 \\equal{} 12R^2 \\minus{} (a^2 \\plus{}  b^2 \\plus{}  c^2) \\equal{} 3R^2 \\plus{} OH^2 \\equal{} 4R^2 \\minus{} \\frac{AH \\cdot BH \\cdot CH}{R}$\n\nSubstituting $ AH, BH, CH,$ we get a cubic for $ R$: $ 4R^3 \\minus{} 8R \\minus{} \\sqrt{10} \\equal{} 0.$\nSubstituting $ R \\equal{} \\varrho \\sqrt{10},$ this becomes $ 40 \\varrho^3 \\minus{} 8 \\varrho \\minus{} 1 \\equal{} 0.$ Looking for a rational root, $ \\varrho \\equal{} \\frac{1}{2}.$ Factoring it out, the remaining quadratic $ 20 \\varrho^2 \\plus{} 10 \\varrho \\plus{} 1 \\equal{} 0$ has 2 more real roots, both negative, not acceptable. With $ R \\equal{} \\varrho \\sqrt{10} \\equal{} \\frac{\\sqrt{10}}{2},$ the triangle sides are\n\n$ a \\equal{} \\sqrt{4R^2 \\minus{} AH^2} \\equal{} 3,\\ \\ b \\equal{} \\sqrt{4R^2 \\minus{} BH^2} \\equal{} \\sqrt{8},\\ \\ c \\equal{} \\sqrt{4R^2 \\minus{} CH^2} \\equal{} \\sqrt{5}.$ \n\n$ S \\equal{} \\frac{abc}{4R} \\equal{} 3.$",
  "Solution_2": "Another solution:\r\nP is the point satisfying PO is the perpendicular to the plane (ABC) and OP=1. \r\nApply Pythagorean theorem, $ PA\\equal{}\\sqrt{2},PB\\equal{}\\sqrt{3},PC\\equal{}\\sqrt{6}$\r\nThe volume of the tetrahedron PABC is denoted by V.\r\n$ \\frac{1}{3}OP.S(ABC)\\equal{}V\\leq \\frac{1}{6}PA.PB.PC\\equal{}\\frac{1}{6}.\\sqrt{2}.\\sqrt{3}.\\sqrt{6}\\equal{}1$\r\n$ \\Rightarrow S(ABC)\\leq 3$\r\nThe equality occurs when $ \\angle APB\\equal{}\\angle BPC\\equal{}\\angle CPA\\equal{}90^o$, then $ AB\\equal{}\\sqrt{5},BC\\equal{}3,CA\\equal{}\\sqrt{8}$.\r\nWe can choose O(0,0), A(0,1), B(1,-1), C(-2,-1)."
}
{
  "Tag": [
    "calculus",
    "integration",
    "function",
    "calculus computations"
  ],
  "Problem": "Can this integral be integrated? $ \\int{\\frac{e^{x}}{x}\\left(1-\\frac{1}{x}\\right)dx}$",
  "Solution_1": "[quote=\"emjay285\"]Can this integral be integrated? $ \\int{\\frac{e^{x}}{x}\\left(1-\\frac{1}{x}\\right)dx}$[/quote]\r\n\r\nSure, you can integrate it as follow:\r\n$ \\int{\\frac{e^{x}}{x}\\left(1-\\frac{1}{x}\\right)dx}$\r\n$ \\Rightarrow\\int{\\frac{e^{x}}{x}dx-\\int{\\frac{e^{x}}{x^{2}}dx}}$\r\nIntegration by parts:\r\n$ \\Rightarrow\\int{\\frac{e^{x}}{x}dx-(-\\frac{e^{x}}{x}+\\int{\\frac{e^{x}}{x}dx)}}$\r\n$ \\Rightarrow\\frac{e^{x}}{x}$",
  "Solution_2": "ah thank you",
  "Solution_3": "it's a particular case of a well known theorem\r\n\r\n[color=darkred][b][u]Theorem[/b][/u][/color]\r\nLet $ f$ be a differentiable function then\r\n$ \\boxed{\\int e^{x}(f(x)\\plus{}f'(x))\\text{dx}\\equal{}e^{x}f(x)\\plus{}c}$\r\nthe result is obvious just differentiate $ e^{x}f(x)$"
}
{
  "Tag": [
    "geometry",
    "incenter"
  ],
  "Problem": "Let $ABC$ be an acute angled triangle with $\\angle{BAC}=60^\\circ$ and $AB > AC$. Let $I$ be the incenter, and $H$ the orthocenter of the triangle $ABC$ . Prove that $2\\angle{AHI}= 3\\angle{ABC}$.",
  "Solution_1": "[hide]only see that B,C,I,H are concyclic..  :)[/hide]",
  "Solution_2": "in fact $B,C,H,O,I$ are concyclic and that leads to an easy solution.\r\nby the way i was not a contestent at apmo .",
  "Solution_3": "Here is my solution:\n\nEasy angle chasing yields $\\angle BIC=120^\\circ$, while more angle chasing yields $\\angle BHC=120^\\circ$. Hence $BIHC$ is cyclic, and if we denote the foot of the perpendicular from $C$ to $AB$ as $D$, then $\\angle IHD=\\angle IBC$. Now, let $\\angle ABC=2\\beta$. It is easy to find $\\angle DCB=90-2\\beta$, hence $\\angle AHD=2\\beta$. Therefore, $3\\angle ABC=6\\beta=2\\angle AHI$.",
  "Solution_4": "we first prove a lemma\nlemma = let $ABC$ be a triangle with $D$ as midpoint of minor arc $BC$ of circumcircle of $ABC$\nalso let $I$ be incentre of triangle $ABC$ than we prove that $D$ is centre of circumcircle of $BIC$.\n\nproof = it is evident that $A$ excentre of triangle $ABC$ lie on circumcircle of $BIC$\nalso $BD=DC$ and thus $\\angle CBD=A/2 = \\angle CAD=\\angle BAI$\nthus $A,I,D$ are collinear.\nnow $\\angle IBD = 90-C/2=\\angle JCB=\\angle BID$ where $J$ is $A$ excentre.\nand thus $BD=DI$\nthus $D$ is circumcentre of $BIC$. \n\nmain proof = by this lemma and $\\angle A = 60$ we get that $\\angle BOC=\\angle BIC=\\angle BHC = 120$\nand hence $B,H,O,I,C$ are concyclic.\nby concyclicity of $HIJB$ we have $\\angle HBJ = \\angle AIH = 120-C-B/2=60+B/2$\nin triangle $AHI$ we thus get $\\angle AHI=3B/2$ since $\\angle HAI=60-B$\nand thus $2\\angle AHI = 3\\angle B$.\n\nwe are done :D\n\n\n\n",
  "Solution_5": "A nice problem .\nWe will show that points B,I, H and C are concyclic wich is too easy since angle<BIC = 90\u00b0+A/2 = 120\u00b0=180\u00b0-A=<BHC.\nLet (w) the cercle ( BIHC). Let Y be the intersection point of line AH with (w);then angle <YBC = <ABC .\n[u]Prof[/u]\nBy simple angle chasing <BAY = <HCB =<HYB = <AYB  and (AY) perpendicular to (BC) so triangle BAY is isoceles then the equality follows.\nNow denote by X the center of (w) ,then :\n2<AHI=360\u00b0-2<IHY = <IXY = 2<IBY =<ABC+ 2<CBY = <ABC + 2<ABC = 3<ABC , QED\n\n\n",
  "Solution_6": "[hide=Solution]\nFirst, it is well known that $\\angle BIC=90+\\frac{1}{2}A=120=180-A=\\angle BHC$, so $BIHC$ is cyclic with vertices in this order. Let the midpoint of minor arc $\\overarc{BC}$ be $M$. Observe $\\odot(BIHC)$ is centered at $M$ by the Incenter-Excenter Lemma.\n\nNow, $\\angle AHI + \\angle IHC + \\angle BHC = 360$.\n$\\angle AHC = 180-B$ is obvious. \nWe chase $\\angle IHC = \\angle BHC + \\angle IHB = 120+\\frac{C}{2}$ since $\\angle IHB = \\frac{1}{2} \\angle IMB = \\frac{1}{2}C$.\nThis means that $\\angle AHI = 60+B-\\frac{C}{2} = A+B-\\frac{C}{2} = 180-\\frac{3C}{2}$.\n\nTo finish, $B+C=120 \\implies 180=\\frac{3}{2}(B+C) \\implies 180-\\frac{3C}{2}=\\frac{3B}{2}$. Hence, $\\angle AHI = \\frac{3B}{2}$ as desired.\n[/hide]",
  "Solution_7": " This problem is easy.\nFirst we should know $B,I,H,C$ are concyclic.\n$<BHC=<CIB=120$\n",
  "Solution_8": "Let altitudes intersect sides $a,b,c$ in points $F,D,E$. Now, $\\angle IBH=\\angle ABD-\\angle ABI=30^\\circ-\\frac{\\beta}{2}$.\n$\\angle ICH=\\angle ACI-\\angle ACH=30^\\circ-\\frac{\\beta}{2} \\implies \\angle IBH=\\angle ICH$. Using that quadrilateral $BCHI$ is cyclic we obtain:\n$\\angle BCI=\\angle BHI=\\frac{\\gamma}{2}$ and $\\angle FHB=90^\\circ-\\angle FBH=90^\\circ-\\beta+\\angle ABH=120^\\circ-\\beta$\nFinally, $\\angle AHI=180^\\circ-\\angle BHI-\\angle FHB=60^\\circ-\\frac{\\gamma}{2}+\\beta=60^\\circ-\\frac{120^\\circ-\\beta}{2}+\\beta=\\frac{3\\beta}{2} \\implies 2\\angle AHI=3\\beta$ :)",
  "Solution_9": "Note $\\angle BHC=\\angle BIC=\\angle BOC=120^{\\circ},$ so $H,O$ lie on $(BIC).$\n\nWe claim that $AHO$ is isosceles with $AH=AO.$ We use complex numbers. Put $ABC$ on the unit circle and note that $H=A+B+C=A+D,$ so $AH=OD=AO.$ Now note that $\\angle CAH=\\angle BAO=90^{\\circ}-\\angle C,$ so $\\angle HAO=\\angle A-(180^{\\circ}-\\angle 2C)=\\angle C-\\angle B=2\\angle C-120^{\\circ}.$ Thus $\\angle AHO=150^{\\circ}-\\angle C=\\angle B+30^{\\circ}.$\n\nAlso note $\\angle HIO=180^{\\circ}-\\angle CHO=180^{\\circ}-\\angle A+\\angle B=120^{\\circ}+\\angle B,$ so $\\angle IHO = 30^{\\circ}-\\frac{1}{2}\\angle B.$ Thus $\\angle AHI=\\frac{3}{2}\\angle B,$ as desired.",
  "Solution_10": "[hide=solution sketch]Key Observation B, H, I, C are concyclic. Note that $\\angle AHI = =180- \\angle IHB-\\angle BHK$, where K is the foot of altitude to BC from A. Now angle chase with the cyclic quadrilateral, and we get the desired result[/hide]",
  "Solution_11": "Let $M$ be the midpoint of minor arc $BC$ of $(ABC)$ and $O$ be its center. We have that $\\triangle COM$ is equilateral and hence $OM=AH$. Also evidently $AH || OM$, hence $AOMH$ is a parallelogram. But since $OM=OA$ it's a rhombus, whence $OM=MH$. Now from the excenter/incenter lemma we have that $CHIOB$ is cyclic. We have $\\angle AHI=\\angle AOI=\\angle AOC-\\angle COH=\\angle AOC-\\angle IBC=\\angle AOC-\\frac{1}{2}\\angle ABC=\\frac{3}{2}\\angle ABC$, as desired.",
  "Solution_12": "Note that we have $$\\angle BIC = 180^{\\circ}-\\angle IBC - \\angle ICB = 180^{\\circ} - \\frac{\\angle ABC + \\angle ACB}{2} = 180^{\\circ} - \\frac{180^{\\circ}-\\angle A}{2}$$ and since $\\angle A = 60^{\\circ}$, then $\\angle BIC = 120^{\\circ}$. Now, it is well-known that $$\\angle BHC = 180^{\\circ}-\\angle A = 180^{\\circ}-60^{\\circ}=120^{\\circ}$$ so as a result we have $\\angle BIC = \\angle BHC$ so $BIHC$ is cyclic. Now, reflecting $A$ over $BC$ and letting that point be $A'$ we know $A'$ is on the circumcircle of $BIHC$, so $$\\angle AHI = 180^{\\circ}- \\angle A'HI = \\angle A'BI = \\angle B + \\angle CBI = \\frac32\\angle B$$ so $2\\angle AHI = 3 \\angle B$.",
  "Solution_13": "First note that $BIC$ are concyclic. I claim that $H$ lies on this circle. Let $\\Gamma_1$ denote the circumcircle of $\\triangle ABC$ and $\\Gamma_2$ denote $(BIC)$. Then, arc $\\overarc {BAC}$ has measure $240^{\\circ}$. Then, since $\\angle BIC=90^{\\circ}+\\frac{\\angle A}{2}=120^{\\circ}$, looking at $\\Gamma_2$, $\\overarc{BC}$ not containing $I$ has measure $240^{\\circ}$ as well. Hence, $\\Gamma_1$ and $\\Gamma_2$ are reflections of each other across $BC$. Since the reflection of $H$ across $BC$ lies on $\\Gamma_1$, $H$ must lie on $\\Gamma_2$. \n\nNow, \n\\[3\\angle ABC=6\\angle IBC=6(180-\\angle IHC)=1080-6(180-\\angle AHI+\\angle ABC)=6\\angle AHI-6\\angle ABC,\\]\nwhich implies the desired result",
  "Solution_14": "Points $I, H, O, B, C$ Will lie on same circle.. \nAnd then by some angle chase we will get the desired results",
  "Solution_15": "We start with a lemma.\n\n[color=red][b]Lemma:[/b][/color] In a triangle $ABC$ with orthocenter $H$, $AH=2R\\cos A$.\n\n[i]Proof.[/i] Consult problem Exercise 4.4.7 in the AoPS Precalculus Book. $\\square$\n\nNow note that $AH=2R\\cos A=2R\\cdot \\frac12=R=AO$. Also, $AI=AI$ and $\\measuredangle HAI=\\measuredangle IAO$ since $H$ and $O$ are isogonal conjugates. Thus, $\\triangle AHI\\cong \\triangle AOI$. \n\nNow observe the following angle equalities: \\begin{align*} \\angle BHC&=180^\\circ-\\angle A=120^\\circ\\\\ \\angle BIC&=90^\\circ+\\frac{\\angle A}2=120^\\circ\\\\ \\angle BOC&=2\\angle A=120^\\circ \\end{align*}\n\nso hexagon $CHIOBI_A$ is cyclic. \n\nThe rest is just angle chasing: $$\\measuredangle AHI=\\measuredangle IOA=\\measuredangle IOC+\\measuredangle COA=\\measuredangle IBC+2\\measuredangle CBA=4\\measuredangle CBI- \\measuredangle CBI=3\\measuredangle CBI$$ and multiplying both sides by $2$ gives the desired $2\\measuredangle AHI=6\\measuredangle CBI=3\\measuredangle CBA$. $\\blacksquare$\n\n\n[hide=Remarks]This problem was quite brutal for directed angles...one must write the condition as $2\\measuredangle AHI=3\\measuredangle CBA$ and NOT $2\\measuredangle AHI=3\\measuredangle ABC$.\n\nSome cool properties about a triangle $ABC$ with $\\angle A=60^\\circ$ and $\\overline{AI}\\cap(ABC)=M$ are\n\n[b]Property 1:[/b] $(ABC)$ and $(IBC)$ (which we will call $\\omega_1$ and $\\omega_2$ respectively) are reflections over $BC$.\n[i]Proof.[/i] Clearly $H\\in \\omega_2$ and $H'\\in\\omega_1$ where $H'$ is the refllection of $H$ over $BC$. This is true by the Reflecting the Orthocenter Lemma. Now, for a point $P$ on $\\omega_1$, we have $\\measuredangle BPC=\\measuredangle BAC=60^\\circ$ and $\\measuredangle BP'C=-\\measuredangle BPC=120^\\circ=\\measuredangle BIC$ as needed.\n\n[b]Property 2:[/b] $\\triangle COM$ is equilateral.\n[i]Proof.[/i] Clearly, $IM$ bisects $\\angle BIC=120^\\circ$, so $\\triangle MIC$ is isosceles with $\\angle MIC=60^\\circ$, implying the result.\n\n[b]Property 3:[/b] $CHIOBI_A$ is cyclic.\n[i]Proof.[/i] Proven in the solution to the main problem.\n\n[b]Property 4:[/b] $\\triangle AHI\\cong \\triangle AOI$.\n[i]Proof.[/i] Proven in the solution to the main problem.\n\n[b]Property 5:[/b] $\\triangle IMH\\cong \\triangle IMO$.\n[i]Proof.[/i] We have $MH=MO$ and $MI=MI$. Also note that $\\measuredangle HMI=\\measuredangle IMO$ since $H$ and $O$ are isogonal conjugates and thus reflections across the line $\\overline{AIM}$. Thus $\\triangle IMH\\cong \\triangle IMO$. [url]https://artofproblemsolving.com/community/c6h1408296p7899381[/url]\n\n[b]Property 6:[/b] $AOMH$ is a rhombus.\n[i]Proof.[/i] Observe that $AH\\parallel OM$. From the above, we have $MH=MO=AO=AH$, finishing. (Actually, the parallel condition was not even needed.)\n\n[b]Property 7:[/b] Line $AIM$ is the perpendicular bisector of $\\overline{OH}$. \n[i]Proof.[/i] This follows from $AO=AH$ (or $MH=MO$) combined with the $O$ and $H$ isogonality.\n\n[b]Property 8:[/b] If $N_9$ is the center of the nine-point circle of $\\triangle ABC$, then it lies on the line $\\overline{AIM}$.\n[i]Proof.[/i] $N_9$ is the midpoint of $\\overline{OH}$ which does the trick.\n\nProblems to try: [url=2007 APMO Problem 2]https://artofproblemsolving.com/community/c6h141391p799589[/url]\n[/hide]",
  "Solution_16": "[quote=APMO 2007 P2]Let $ABC$ be an acute angled triangle with $\\angle{BAC}=60^\\circ$ and $AB > AC$. Let $I$ be the incenter, and $H$ the orthocenter of the triangle $ABC$ . Prove that $2\\angle{AHI}= 3\\angle{ABC}$.[/quote]\n[b]Solution.[/b] Since $\\angle A=60^{\\circ}$, we have that $BIHC$ is cyclic. Now just by angle chasing, we get that:\n$$\\angle AHI=180^{\\circ}-\\angle IHD=180^{\\circ}-(\\angle IHB+\\angle BHD)=180^{\\circ}-(\\frac{\\angle C}{2}+\\angle C)=180^{\\circ}-\\frac{3\\angle C}{2}$$\n$$2\\angle AHI=360^{\\circ}-3\\angle C=3(120^\\circ-\\angle C)=3\\angle B.\\blacksquare$$",
  "Solution_17": "[b][color=lightblue]Claim:[/color][/b] $BIHC$ is cyclic.\n\n[i]Proof.[/i] Notice that \\begin{align*}\\angle HCI&=\\tfrac{1}{2}\\angle C-(90-60)\\\\&=\\tfrac{1}{2}\\angle C-30+\\tfrac{1}{2}(\\angle B+\\angle C-120)\\\\&=\\tfrac{1}{2}\\angle B-(90-\\angle C)\\\\&=\\angle HBI.\\end{align*} $\\blacksquare$\n\nHence, $$\\angle AHI=(180-\\angle IHC)+(180-AHC)=\\tfrac{1}{2}\\angle B+180-\\angle A-\\angle C=\\tfrac{3}{2}\\angle B.$$ $\\square$",
  "Solution_18": "Let AD,BE and CF be altitudes of triangle. \u2220EHF = 180 - 60 = 120 = \u2220BHC and \u2220BIC = \u2220A + \u2220B/2 + \u2220C/2 = 120 so BIHC is cyclic.\n\u2220AHI = \u2220IHF + \u2220FHA = \u2220IBC + \u2220DHC = \u2220IBC + \u2220ABC ---> \u2220AHI = 3/2 \u2220ABC.\nwe're Done.",
  "Solution_19": "$BHIC$ is cyclic since $\\angle BHC = 180^{\\circ} - \\angle A = 120^{\\circ} = 90^{\\circ} + \\frac{\\angle A}{2}$. $D$ be the foot from $A$ to $BC$. Consider $DHIC$ where $\\angle HDC = 90^{\\circ}$, $\\angle IBC = \\frac{\\angle B}{2}$, $\\angle HIC = 180^{\\circ} - 90^{\\circ} + \\angle B = 90^{\\circ} + \\angle B$. Now, trivial calculation gives, $\\angle AHI = \\frac{3}{2}\\angle B$ and check that, this is the result we wanted. ",
  "Solution_20": "[asy]\n/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */\nimport graph; size(42cm); \nreal labelscalefactor = 0.5; /* changes label-to-point distance */\npen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ \npen dotstyle = black; /* point style */ \nreal xmin = -10.982013289036544, xmax = 16.091971207087482, ymin = -7.555732004429678, ymax = 8.00945071982281;  /* image dimensions */\n\n /* draw figures */\ndraw((-0.2750542635658916,5.839758582502767)--(-9.51982945736434,-4.371944629014396), linewidth(2)); \ndraw((-0.2750542635658916,5.839758582502767)--(2.932316722037652,-0.5513997785160583), linewidth(2)); \ndraw((-9.51982945736434,-4.371944629014396)--(2.932316722037652,-0.5513997785160583), linewidth(2)); \ndraw(circle((-0.9208974895600552,1.0958517060424493), 2.7050286774217995), linewidth(2)); \ndraw((xmin, -3.25925925925926*xmin + 4.943285427176898)--(xmax, -3.25925925925926*xmax + 4.943285427176898), linewidth(2)); /* line */\ndraw((xmin, 0.5018450184501846*xmin + 0.405534360659222)--(xmax, 0.5018450184501846*xmax + 0.405534360659222), linewidth(2)); /* line */\ndraw((xmin, -0.9053117782909931*xmin + 2.103261087624264)--(xmax, -0.9053117782909931*xmax + 2.103261087624264), linewidth(2)); /* line */\ndraw((-0.2750542635658916,5.839758582502767)--(-0.9208974895600541,1.0958517060424489), linewidth(2)); \ndraw((-0.9208974895600541,1.0958517060424489)--(1.2064943515156186,1.0110075407556212), linewidth(2)); \ndraw((-0.9208974895600541,1.0958517060424489)--(-9.51982945736434,-4.371944629014396), linewidth(2)); \ndraw((-0.9208974895600541,1.0958517060424489)--(2.932316722037652,-0.5513997785160583), linewidth(2)); \n /* dots and labels */\ndot((-0.2750542635658916,5.839758582502767),dotstyle); \nlabel(\"$A$\", (-0.1807198228128475,6.07559468438538), NE * labelscalefactor); \ndot((-9.51982945736434,-4.371944629014396),dotstyle); \nlabel(\"$B$\", (-9.425495016611295,-4.136108527131783), NE * labelscalefactor); \ndot((2.932316722037652,-0.5513997785160583),dotstyle); \nlabel(\"$C$\", (3.0266511627906953,-0.3155636766334448), NE * labelscalefactor); \ndot((-0.9208974895600541,1.0958517060424489),dotstyle); \nlabel(\"$I$\", (-0.8174772978959038,1.3352890365448493), NE * labelscalefactor); \ndot((-0.12745211158699393,-1.490192488832711),dotstyle); \nlabel(\"$D$\", (-0.03921816168327944,-1.2589080841638987), NE * labelscalefactor); \ndot((-2.9262243575179765,2.9112977389281434),dotstyle); \nlabel(\"$E$\", (-2.822084163898118,3.151227021040973), NE * labelscalefactor); \ndot((1.4967668262111673,2.3091444991970103),dotstyle); \nlabel(\"$F$\", (1.5880509413067534,2.538053156146178), NE * labelscalefactor); \ndot((1.2064943515156186,1.0110075407556212),dotstyle); \nlabel(\"$H$\", (1.3050476190476172,1.240954595791804), NE * labelscalefactor); \nclip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); \n /* end of picture */\n[/asy]\n\n\n[color=#0ff][b]Claim:-[/b][/color]points $B,I,H,C$ are concylic\n\n[color=#f0f][b]Proof:-[/b][/color] we notice that $\\angle{BHC}=180^{\\circ}-\\angle{A}=120^{\\circ}$ and $\\angle{BIC}=90^{\\circ}+\\frac{A}{2}=120^{\\circ}$ so we get $\\angle{BHC}=\\angle{BIC}$ hence points $B,I,H,C$ are concylic $\\blacksquare$\\\\\n\nnow we have $\\angle{AHI}+\\angle{AHC}+\\angle{CHI}=360^{\\circ} \\implies \\angle{AHI}=360^{\\circ}-\\left(180^{\\circ}-\\frac{\\angle{B}}{2}\\right)-(180^{\\circ}-\\angle{B}) \\implies \\angle{AHI}=\\frac{3\\angle{B}}{2} \\implies 2\\angle{AHI}=3\\angle{ABC}$ $\\square$",
  "Solution_21": "Let $\\angle ABC=2\\theta$ and $\\angle ACB=120-2\\theta$. We wish to show that $\\angle AHI=3\\theta$.\\\\\n\nClaim: $HIBC$ is cyclic. This is because $$\\angle BHC=180-\\angle A=120$$ and $$\\angle BIC=90+(\\angle A/2)=120.$$\n\nThen, $$\\angle IHB=\\angle ICB=60-\\theta.$$ Let $E$ be the foot from $B$ to $AC$. Then, $$\\angle AHE=\\angle C=120-2\\theta,$$ so $$\\angle AHI=180-\\angle AHE-\\angle IHB=180-(120-2\\theta)-(60-\\theta)=3\\theta,$$ hence done.",
  "Solution_22": "Observe that $B, C, H, I$ are concyclic, as $\\angle BHC = 120^\\circ = \\angle BIC$ from some simple angle chasing.\nNow we have $\\angle AHC = 180^\\circ - B, \\angle CHI = 180^\\circ - \\frac{B}{2} \\implies \\angle AHI = \\frac{3B}{2}$, and we are done. $\\square$"
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "Prove that \r\n\r\n$\\sqrt{\\frac{m_a}{a}}+\\sqrt{\\frac{m_b}{b}}+\\sqrt{\\frac{m_c}{c}}\\geq\\sqrt{6}$",
  "Solution_1": "Nice inequality! It's really yours? :?  I saw this i think some time ago, but....i don't know if this yours?\r\n Anyway, inequalities connected to this one can be found in my book for inequalities. :D  ;)",
  "Solution_2": "Sorry, I discovered  that in 1976!\r\n\r\nIn Russian montly \"Kvant\" 1976, number 7, page 16 :\r\nprove $\\frac{m_a}{a}+\\frac{m_b}{b}+\\frac{m_c}{c}\\geq\\frac{3\\sqrt{3}}{2}$\r\nauthor: V. Matizen.\r\n\r\nMaybe what you saw is this?\r\n\r\nWhen I proved this inequality, I thought of mine.",
  "Solution_3": "You are V.Matizen? :lol:",
  "Solution_4": "No.",
  "Solution_5": "I think that the enunciation is wrong. See you for $a=b=c=1\\ (m_a=m_b=m_c=\\frac{\\sqrt 3}{2})$.",
  "Solution_6": "[quote=\"Ph-An\"]I think that the enunciation is wrong. See you for $a=b=c=1\\ (m_a=m_b=m_c=\\frac{\\sqrt 3}{2})$.[/quote]\r\n \r\nno! a=b=1, c-->2",
  "Solution_7": "Indeed, but then must  $\\sum \\sqrt \\frac{m_a}{a}>\\sqrt 6$ (strict !). You are right, thus you misleaded me.",
  "Solution_8": "Cezar, can you solve this?  :roll:",
  "Solution_9": "Which inequality? :?  The one that Ph-An posted? :)",
  "Solution_10": "[quote=\"cezar lupu\"]Which inequality? :?  [/quote]\r\nprove that:\r\n$\\sqrt{\\frac{m_a}{a}}+\\sqrt{\\frac{m_b}{b}}+\\sqrt{\\frac{m_c}{c}}\\geq\\sqrt{6}$\r\nCezar, can you solve this? :)",
  "Solution_11": "[quote=\"Ph-An\"]Indeed, but then must  $\\sum \\sqrt \\frac{m_a}{a}>\\sqrt 6$ (strict !). [/quote]\r\nYou are right. If  $\\sum \\sqrt \\frac{m_a}{a}>\\sqrt 6$ then  $\\sum \\sqrt \\frac{m_a}{a}\\geq\\sqrt 6.$ :)",
  "Solution_12": "Okay, this is my solution:\r\n\r\nFirst, let's write out the problem completely:\r\n\r\n[color=blue][b]Problem.[/b] Let ABC be a triangle with medians $m_a$, $m_b$, $m_c$. Prove the inequality\n\n$\\sqrt{\\frac{m_a}{a}}+\\sqrt{\\frac{m_b}{b}}+\\sqrt{\\frac{m_c}{c}}>\\sqrt6$.[/color]\r\n\r\nNow, the [i]solution[/i]: First, we will prove the inequality\r\n\r\n$\\sqrt{\\frac{a}{m_a}}+\\sqrt{\\frac{b}{m_b}}+\\sqrt{\\frac{c}{m_c}}>2\\sqrt2$.\r\n\r\nIn fact, we have $m_a\\leq\\frac12\\left(b+c\\right)$, $m_b\\leq\\frac12\\left(c+a\\right)$ and $m_c\\leq\\frac12\\left(a+b\\right)$ (these are well-known and easy inequalities; they are even true with an < instead of a $\\leq$; a short proof of them was given in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=252755#p252755]http://www.mathlinks.ro/Forum/viewtopic.php?t=21462 post #16[/url]). Also, we have $\\sqrt{\\frac{a}{b+c}}+\\sqrt{\\frac{b}{c+a}}+\\sqrt{\\frac{c}{a+b}}>2$ (this is also an old inequality; see http://www.mathlinks.ro/Forum/viewtopic.php?t=48825 , http://www.mathlinks.ro/Forum/viewtopic.php?t=50390 , http://www.mathlinks.ro/Forum/viewtopic.php?t=48311 , http://www.mathlinks.ro/Forum/viewtopic.php?t=46428 , http://www.mathlinks.ro/Forum/viewtopic.php?t=27481 ). Thus,\r\n\r\n$\\sqrt{\\frac{a}{m_a}}+\\sqrt{\\frac{b}{m_b}}+\\sqrt{\\frac{c}{m_c}}\\geq\\sqrt{\\frac{a}{\\frac12\\left(b+c\\right)}}+\\sqrt{\\frac{b}{\\frac12\\left(c+a\\right)}}+\\sqrt{\\frac{c}{\\frac12\\left(a+b\\right)}}$\r\n$=\\sqrt2\\cdot\\left(\\sqrt{\\frac{a}{b+c}}+\\sqrt{\\frac{b}{c+a}}+\\sqrt{\\frac{c}{a+b}}\\right)>\\sqrt2\\cdot 2=2\\sqrt2$.\r\n\r\nThus, the inequality\r\n\r\n$\\sqrt{\\frac{a}{m_a}}+\\sqrt{\\frac{b}{m_b}}+\\sqrt{\\frac{c}{m_c}}>2\\sqrt2$\r\n\r\nis proven. Now, applying this inequality to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=59415]the medians triangle[/url] of triangle ABC, whose sidelengths are $m_a$, $m_b$, $m_c$ and whose medians are $\\frac34a$, $\\frac34b$, $\\frac34c$, we get\r\n\r\n$\\sqrt{\\frac{m_a}{\\frac34a}}+\\sqrt{\\frac{m_b}{\\frac34b}}+\\sqrt{\\frac{m_c}{\\frac34c}}>2\\sqrt2$.\r\n\r\nAfter multiplication by $\\frac{\\sqrt3 }{2}$, this becomes $\\sqrt{\\frac{m_a}{a}}+\\sqrt{\\frac{m_b}{b}}+\\sqrt{\\frac{m_c}{c}}>\\sqrt6$, and the problem is solved.\r\n\r\nBy the way, this inequality:\r\n\r\n[quote=\"arqady\"]In Russian montly \"Kvant\" 1976, number 7, page 16 :\nprove $\\frac{m_a}{a}+\\frac{m_b}{b}+\\frac{m_c}{c}\\geq\\frac{3\\sqrt{3}}{2}$\nauthor: V. Matizen.[/quote]\r\n\r\nis a particular case of http://www.mathlinks.ro/Forum/viewtopic.php?t=21016 .\r\n\r\n  Darij",
  "Solution_13": "[quote=\"darij grinberg\"]\nNow, the [i]solution[/i]: First, we will prove the inequality\n\n$\\sqrt{\\frac{a}{m_a}}+\\sqrt{\\frac{b}{m_b}}+\\sqrt{\\frac{c}{m_c}}>2\\sqrt2$.\n[/quote]\nMy proof:\n$\\sqrt\\frac{a}{m_a}=\\sqrt2\\cdot\\sqrt[4]{\\frac{1}{\\frac{2a^2+2b^2-a^2}{a^2}\\cdot1\\cdot1\\cdot1}}\\geq\\frac{4\\sqrt2}{3+\\frac{2b^2+2c^2-a^2}{a^2}}=2\\sqrt2\\cdot\\frac{a^2}{a^2+b^2+c^2}.$\n[quote=\"darij grinberg\"]\nBy the way, this inequality:\n[quote=\"arqady\"]In Russian montly \"Kvant\" 1976, number 7, page 16 :\nprove $\\frac{m_a}{a}+\\frac{m_b}{b}+\\frac{m_c}{c}\\geq\\frac{3\\sqrt{3}}{2}$\nauthor: V. Matizen.[/quote]\nis a particular case of http://www.mathlinks.ro/Forum/viewtopic.php?t=21016 .\n  Darij[/quote]\r\nMy proof:\r\n${\\frac{m_a}{a}=\\frac{\\sqrt3}{2}\\sqrt{\\frac{2b^2+2c^2-a^2}{3a^2}}=\\frac{\\sqrt3}{2}\\cdot\\sqrt{\\frac{1}{\\frac{3a^2}{2b^2+2c^2-a^2}\\cdot1}}\\geq\\frac{\\sqrt3}{2}\\cdot\\frac{2}{\\frac{3a^2}{2b^2+2c^2-a^2}+1}}=$\r\n$=\\frac{\\sqrt3}{2}\\cdot\\frac{2b^2+2c^2-a^2}{a^2+b^2+c^2}.$ \r\nRozenberg Michael (arqady). :)"
}
{
  "Tag": [
    "trigonometry"
  ],
  "Problem": "Solve for the equation $ \\sin (\\sin x) \\equal{} \\cos (\\cos x)$\r\n\r\nHere's my nearly done solution:\r\n[hide]\nI first substituted $ x \\equal{} \\frac {\\pi}{2} \\minus{} t$.\nSo, the equation would now be: $ \\sin (\\cos t) \\equal{} \\cos (\\sin t)$\nFrom here, $ \\sin t \\equal{} \\arccos (\\sin (\\cos t))$.\nIf we draw some triangles, we would arrive at $ \\sin t \\equal{} \\frac {\\pi}{2} \\minus{} \\cos t$\n\neventually, $ \\sin x \\plus{} \\cos x \\equal{} \\frac {\\pi}{2}$\n\nIs this right?\n[/hide]\r\nthanks!:)",
  "Solution_1": "Seems right...\r\nOf course, since $ \\sin x \\plus{} \\cos x \\leq \\frac{\\sqrt{2}}{2}<\\frac{\\pi}{2}$, this implies there are no solutions...",
  "Solution_2": "You should probably be more careful. \r\n\r\n$ \\sin x\\equal{}\\sin y$ doesn't imply that $ x\\equal{}y$. In particular $ x\\minus{}y$ is a multiple of $ 2\\pi$ so you may want to include this.",
  "Solution_3": "[quote=\"Brut3Forc3\"]Seems right...\nOf course, since $ \\sin x \\plus{} \\cos x \\leq \\frac {\\sqrt {2}}{2} < \\frac {\\pi}{2}$, this implies there are no solutions...[/quote]\r\nBut isn't $ \\frac{\\pi}{2} <2$? This means that there could be solutions because the maximum value of $ \\sin x \\plus{} \\cos x \\equal{}2$?",
  "Solution_4": "A [i]slightly [/i]more straightforward approach would be to solve $ \\sin A \\equal{} \\cos B$, and then substitute in $ \\sin x, \\; \\cos x$ for $ A, \\; B$, respectively.",
  "Solution_5": "Is correctly \"for any real $ x$ we have $ \\sin x\\plus{}\\cos x\\le\\sqrt 2$ \" ...",
  "Solution_6": "Yes. $ |\\sin x \\plus{} \\cos x| \\equal{} \\sqrt{\\sin 2x \\plus{} 1}\\leq \\sqrt 2$.",
  "Solution_7": "So basically, this has no solutions? i'm actually considering this as an approach to solve russia 1995 trig equation.",
  "Solution_8": "[quote=\"Brut3Forc3\"]Seems right...\nOf course, since $ \\sin x \\plus{} \\cos x \\leq \\frac {\\sqrt {2}}{2} < \\frac {\\pi}{2}$, this implies there are no solutions...[/quote]\r\n\r\nNo, $ \\sin x \\plus{} \\cos x \\leq \\sqrt 2$.",
  "Solution_9": "hello, we have\r\n$ \\sin(\\sin(x))\\minus{}\\cos(\\cos(x))\\equal{}\\minus{}2\\sin\\left(\\frac{\\pi}{4}\\minus{}\\frac{\\cos(x)}{2}\\minus{}\\frac{\\sin(x)}{2}\\right)\\sin\\left(\\frac{\\pi}{4}\\plus{}\\frac{\\cos(x)}{2}\\minus{}\\frac{\\sin(x)}{2}\\right)$\r\nSonnhard."
}
{
  "Tag": [
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "let $a,b,c>0$,$a+b+c=1$, prove\r\n\r\n$\\frac{1}{(a^{2}+b)}+\\frac{1}{(b^{2}+c)}+\\frac{1}{(c^{2}+a)}\\geq \\frac{13}{2}$.",
  "Solution_1": "The inequality inversely holds!\r\n\r\n(a+b+c)^2/(a^2+b^2+c^2+b*(a+b+c))+(a+b+c)^2/(b^2+c^2+a^2+c*(a+b+c))\r\n\r\n+(a+b+c)^2/(c^2+a^2+b^2+a*(a+b+c))<=13/2\r\n\r\n<==>16*cg[3,6,1]+25*cg[3,6,2]+75*cg[3,6,3]+160*cg[3,6,4]+210*cg[3,6,5]+232*cg[3,6,6]\r\n\r\n+864*cg[3,6,10]>=0,\r\n\r\nin which\r\n\r\n cg[3,6,1] = c^4*(c-b)*(c-a)+b^4*(b-c)*(b-a)+a^4*(a-c)*(a-b)>=0;\r\n\r\ncg[3,6,2] = c^3*(a+b)*(c-b)*(c-a)+b^3*(c+a)*(b-c)*(b-a)+a^3*(b+c)*(a-c)*(a-b)>=0;\r\n\r\ncg[3,6,3] = c^2*(b-a)^2*(c-b)*(c-a)+b^2*(c-a)^2*(b-c)*(b-a)+a^2*(c-b)^2*(a-c)*(a-b)\r\n\r\n>=0;\r\n\r\ncg[3,6,4] = b^2*a^2*(c-b)*(c-a)+c^2*a^2*(b-c)*(b-a)+c^2*b^2*(a-c)*(a-b)>=0;\r\n\r\ncg[3,6,5] = c^2*b*a*(c-b)*(c-a)+b^2*c*a*(b-c)*(b-a)+a^2*c*b*(a-c)*(a-b)>=0;\r\n\r\ncg[3,6,6] = c*b*a*(a+b)*(c-b)*(c-a)+b*c*a*(c+a)*(b-c)*(b-a)+a*c*b*(b+c)*(a-c)*(a-b)\r\n\r\n>=0;\r\n\r\ncg[3,6,10] = a^2*b^2*c^2>=0.",
  "Solution_2": "Mr. fjwxcsl, you would  show that \r\n\r\n$(a+b+c)^{2}(\\frac{1}{a^{2}+b(a+b+c)}+\\frac{1}{b^{2}+c(a+b+c)}+\\frac{1}{c^{2}+a(a+b+c)})\\geq 13/2$\r\n\r\nnot \r\n$(a+b+c)^{2}(\\frac{1}{a^{2}+b^{2}+c^{2}+b(a+b+c)}+\\frac{1}{b^{2}+a^{2}+c^{2}+c(a+b+c)}+\\frac{1}{c^{2}+a^{2}+b^{2}+a(a+b+c)})\\geq 13/2$.",
  "Solution_3": "Sorry. \r\n\r\nWe have\r\n\r\n(a+b+c)^2/(a^2+b*(a+b+c))+(a+b+c)^2/(b^2+c*(a+b+c)) +(a+b+c)^2/(c^2+a*(a+b+c))>=13/2        (*)\r\n\r\n<==>\r\n0<=F(a,b,c)= 8*a*c^5+2*b^6-3*b*c^5+8*b^5*c+5*b^4*c^2-4*b^2*c^4+2*c^6+2*a^6+5*a^2*c^4+a^2*b*c^3\r\n-4*a^4*c^2+8*a^5*b-3*c*a^5+5*a^4*b^2-4*a^2*b^4-3*a^4*b*c+a^3*b^2*c+3*a^3*b*c^2+3*a^2*b^3*c\r\n-27*a^2*b^2*c^2-3*a*b^5-3*a*b^4*c+a*b^3*c^2+3*a*b^2*c^3-3*a*b*c^4,\r\n\r\nlet a=x+y+z,b=x+y,c=x, or a=x+y+z,b=x+y,c=x,x,y,z>=0,then\r\n\r\nF(x+y+z,x+y,x)\r\n379*y^2*z^3*x+335*z^2*x^3*y+325*z^3*x^2*y+502*y^3*z^2*x+648*y^2*z^2*x^2+466*x^2*y^3*z+309*x^3*y^2*z+146*x^3*y^3+86*z^3*x^3+284*z*x*y^4+140*y^3*z^3+132*y*z^4*x+58*y*z*x^4+61*z*y^5+143*x^2*y^4+17*x*z^5+63*x*y^5+20*y*z^5+53*z^4*x^2+58*x^4*y^2+75*y^2*z^4+136*y^4*z^2+10*y^6+2*z^6+58*z^2*x^4 >=0;\r\n\r\nF(x+y+z,x,x+y)\r\n=(58*z^2+58*y*z+58*y^2)*x^4+(86*z^3+155*z^2*y+129*y^2*z+146*y^3)*x^3+(53*z^4+145*z^3*y+108*y^2*z^2\r\n\r\n+106*y^3*z+143*y^4)*x^2+(17*z^5+59*y*z^4+53*z^3*y^2-4*y^3*z^2+31*z*y^4+63*y^5)*x+2*z^6+9*y*z^5\r\n\r\n+11*y^2*z^4-6*y^3*z^3-19*y^4*z^2-z*y^5+10*y^6>=0\r\n\r\n( in which  \r\n\r\n   17*z^5+59*y*z^4+53*z^3*y^2-4*y^3*z^2+31*z*y^4+63*y^5>=0;\r\n\r\n   2*z^6+9*y*z^5+11*y^2*z^4-6*y^3*z^3-19*y^4*z^2-z*y^5+10*y^6>=0).\r\n\r\nThrefore the inequality  (*) holds.",
  "Solution_4": "Let   f1(y,z)=17*z^5+59*y*z^4+53*z^3*y^2-4*y^3*z^2+31*z*y^4+63*y^5, \r\n\r\nf2(y,z)=2*z^6+9*y*z^5+11*y^2*z^4-6*y^3*z^3-19*y^4*z^2-z*y^5+10*y^6,\r\n\r\nwe have\r\n\r\nf1(y,z)+f1(z,y)=80*(y+z)*(y-z)^4+330*(y+z)*y*z*(y-z)^2+219*(y+z)*y^2*z^2>=0;\r\n\r\nf1(y,z)*f1*(z,y)=1071*(y-z)^10+14954*y*z*(y-z)^8+76537*y^2*z^2*(y-z)^6+171086*y^3*z^3*(y-z)^4\r\n\r\n+154550*y^4*z^4*(y-z)^2+47961*y^5*z^5>=0;\r\n\r\n\r\nf2(y,z)+f2(z,y)=12*(y-z)^6+80*y*z*(y-z)^4+132*y^2*z^2*(y-z)^2+12*y^3*z^3>=0;\r\n\r\nf2(y,z)*f2(z,y)=20*(y-z)^12+328*y*z*(y-z)^10+2023*y^2*z^2*(y-z)^8+5570*y^3*z^3*(y-z)^6\r\n\r\n+5811*y^4*z^4*(y-z)^4+116*y^5*z^5*(y-z)^2+36*y^6*z^6>=0,\r\n\r\nthen  f1(y,z)>=0, f2(y,z)>=0.",
  "Solution_5": "[quote=\"nanren\"]Let $ a,b,c > 0$ and $ a \\plus{} b \\plus{} c \\equal{} 1,$ prove that\n\n$ \\frac {1}{a^2 \\plus{} b} \\plus{} \\frac {1}{b^2 \\plus{} c} \\plus{} \\frac {1}{c^2 \\plus{} a} > \\frac {13}{2}.$[/quote]Using the arithmetic-mean \u2014 geometric-mean inequality,\r\n\r\n$ \\sum_{cyc}{\\frac {1}{a^2 \\plus{} b}} \\plus{} \\sum_{cyc}{\\frac {(3 \\minus{} b \\plus{} 3c)^2(a^2 \\plus{} b)}{(3bc \\plus{} 3ca \\plus{} 3ab \\plus{} 1)^2}} > 2\\sum_{cyc}{\\frac {3 \\minus{} b \\plus{} 3c}{3bc \\plus{} 3ca \\plus{} 3ab \\plus{} 1}}$\r\n\r\n$ \\equal{} \\frac {22}{3bc \\plus{} 3ca \\plus{} 3ab \\plus{} 1},$ \r\n\r\nit remains to prove that\r\n\r\n$ 44(3bc \\plus{} 3ca \\plus{} 3ab \\plus{} 1) > 13(3bc \\plus{} 3ca \\plus{} 3ab \\plus{} 1)^2 \\plus{} 2\\sum_{cyc}{(3 \\minus{} b \\plus{} 3c)^2\\left(a^2 \\plus{} b\\right)}.$\r\n\r\nThis comes down to the following quartic homogeneous inequality:\r\n\r\n$ 44(a \\plus{} b \\plus{} c)^2\\left(a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} 5bc \\plus{} 5ca \\plus{} 5ab\\right)$\r\n\r\n$ \\minus{} 13\\left(a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} 5bc \\plus{} 5ca \\plus{} 5ab\\right)^2 \\minus{} 2\\sum_{cyc}{(3a \\plus{} 2b \\plus{} 6c)^2\\left[a^2 \\plus{} b(a \\plus{} b \\plus{} c)\\right]}$\r\n\r\n$ \\equiv F(a,b,c) \\equal{} F(a,a \\plus{} s,a \\plus{} t)$\r\n\r\n$ \\equal{} 12a^4 \\plus{} 16(s \\plus{} t)a^3 \\plus{} 8(16s^2 \\minus{} 13st \\plus{} 16t^2)a^2 \\plus{} 2(51s^3 \\plus{} 46s^2t \\minus{} 71st^2 \\plus{} 51t^3)a$\r\n\r\n$ \\plus{} \\frac {(2s \\minus{} t)^2(4s^2 \\plus{} 84st \\plus{} 19t^2) \\plus{} (2s^2 \\minus{} t^2)^2}{4} > 0,$\r\n\r\nwhich is clearly true for $ a \\equal{} \\min\\{a,b,c\\}.$\r\n\r\nSee also here : http://www.mathlinks.ro/viewtopic.php?t=120780"
}
{
  "Tag": [
    "articles",
    "inequalities",
    "quadratics",
    "algebra"
  ],
  "Problem": "Some article names contain misprints (like Chebyshevs inequality, which should actually be Chebyshev's inequality, etc.) Is there any easy way to edit them? The standard edit option doesn't seem to allow you to edit the article title  :?",
  "Solution_1": "That's because the article title is an intrinsic and immutable property of the page. I assume you'd have to actually move the contents to the uncreated page titled Chebyshev's Inequality (http://www.artofproblemsolving.com/Wiki/index.php?title=Chebyshev%27s_inequality&action=edit).  I'd also suggest making the original page a redirect after that.",
  "Solution_2": "I've been gradually cleaning up article names.  Two common mistakes that people should avoid are as follows:\r\n\r\n1. Don't just capitalize the first letter in every word.  Only do so when the article name is a proper noun.  Otherwise, users will only be able to link with the same incorrect capitalization.  I've changed about 10 of these already and don't want to change 100 more.\r\n\r\n2. Don't pluralize entries without good reason.  I just changed \"quadratic equations\" to quadratic equation.  With internal links, you can still write \"quadratic equations\" and have it link to the appropriate articles by using this syntax:\r\n[[quadratic equation | quadratic equations]]\r\nwhere the first term is the article name, and the second shows up in the text.",
  "Solution_3": "You can even write [[quadratic equation]]s and it will still be a link with \"quadratic equations\". ;)",
  "Solution_4": "But how can we redirect the pages? For example Rearrangement inequality to Rearrangement Inequality.",
  "Solution_5": "See the topic called \"Two names for the same topic\". It says: \r\n\r\n\"#REDIRECT [[American Regions Mathematics League]] \"\r\n\r\nUsing that as the content of any article that would redirect to the article titled American Regions Mathematics League\r\n\r\nEdit (Gah, need to think out my posts): You might also want to wait for the ongoing capitalization issue to settle before you set it one way or another."
}
{
  "Tag": [
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Prove, that equation $x^x=y^3+z^3$ has infinitely many solutions in positive integers",
  "Solution_1": "Set $y=2^k , z= 2^k$, then we need $x^x=y^3+z^3 = 2^{3k} \\cdot (1^3+1^3) = 2^{3k+1}$.\r\nAssume $x=2^m$, then we need $2^{2^m \\cdot m } = x^x = 2^{3k+1}$, thus $3k+1 = 2^m \\cdot m$.\r\nNow any $m \\equiv 4 \\mod 6$ (and also $m \\equiv 5 \\mod 6$) fulfills $2^m \\cdot m \\equiv 1 \\mod 3$, thus for these $m$ such $k$ exists, giving us infinetely many solutions.\r\nThis procedure should work for most other starting terms, too, as long as $3 \\nmid y+z$ and best when $y^3+z^3$ is a prime power.",
  "Solution_2": "Ah,beautiful proof."
}
{
  "Tag": [
    "function",
    "calculus",
    "derivative",
    "limit",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Hi everyone,\r\n\r\nIs it possible to find a sequence of differentiable function $ \\{f_n\\}$ over $ (0,1)$ which converges pointwise\r\nto $ f$, but such that the sequence of derivatives $ \\{f'_n\\}$ over $ (0,1)$ converges pointwise to a function\r\n$ g$, with $ g\\not \\equal{} f'$? \r\n\r\nMoreover, what is the simplest example of a series $ \\sum a_n$ that converges but such that $ \\sum a_n^2$ diverges?\r\n\r\nThank you.",
  "Solution_1": "It's easy enough to construct examples of differentiable $ f_n\\to f$ for which perhaps the sequence $ f_n'$ diverges at some point, or for which $ f'$ doesn't exist somewhere. To do exactly what you asked is going to be a bit trickier. I'll have to think about that.\r\n\r\nAs for the series: you can't have that for $ a_n>0$ or absolute convergence, so the example must involve cancellation and conditional convergence. I'd go with $ a_n\\equal{}\\frac{(\\minus{}1)^n}{\\sqrt{n}}$ as my example.",
  "Solution_2": "Thank you for your answer. For the first problem, if the interval were $ [0,1]$ instead of $ (0,1)$  then the sequence is \r\n\r\n\\[ f_n(x)\\equal{}\\frac{x^n}{n}\\]\r\n\r\nwhich converges to $ f\\equal{}0$ on $ [0,1]$. Now $ \\{f'_n\\} \\equal{}\\{x^{n\\minus{}1}\\}$ converges to $ g\\equal{}0$ for $ x\\not\\equal{}1$ and \r\nto $ g\\equal{}1$ for $ x\\equal{}1$. Obviously $ g\\not\\equal{}f'$.\r\n\r\nBut, what about the case in which the interval is $ (0,1)$? \r\n\r\nFor the second problem, what does your series converge to?",
  "Solution_3": "[quote=\"vnca1\"]For the second problem, what does your series converge to?[/quote]  No reason to think the value it converges to is particularly nice, but it's easy to see it converges by the alternating series test, and easy to see that the square series diverges.",
  "Solution_4": "The first question is not hard as stated: just take some differentiable function $ f$ that has derivative $ 1$ at $ 0$ and vanishes outside $ [\\minus{}1,1]$ and put $ f_n(x)\\equal{}\\frac 1nf(nx)$. Then $ f_n\\to 0$ on the entire real line but $ f_n'(0)\\to 1$ (though $ f_n'(x)\\to 0$ for all $ x\\ne 1$). \r\n\r\nLet's make the question more interesting. Does there exist a sequence of continuously differentiable functions $ f_n$ such that $ f_n\\to 0$ pointwise on the entire real line and $ f'_n$ converge pointwise to a non-vanishing function $ g$ on the entire real line. ;)",
  "Solution_5": "I'm sorry but I don't understand how everything you talked about is related to the first question.",
  "Solution_6": "I'll be a little more formula-driven than fedja, although this is essentially his example. (Just without the compact support.)\r\n\r\nLet $ f_n(x)\\equal{}\\frac{x}{1\\plus{}n^2x^2}$ on $ \\mathbb{R}.$ Then $ \\lim_{n\\to\\infty}f_n(x)\\equal{}0$ for all $ x\\in \\mathbb{R}.$\r\n\r\nIn fact, we even have that $ f_n\\to0$ uniformly, since $ |f_n(x)|\\le\\frac1{2n}.$\r\n\r\nSo $ f_n\\to f,$ where $ f(x)\\equal{}0$ and thus $ f'(x)\\equal{}0$ everywhere.\r\n\r\nIn the meantime, $ f_n'(x)\\equal{}\\frac{1\\minus{}n^2x^2}{(1\\plus{}n^2x^2)^2}$ Taking pointwise limits,\r\n\r\n$ \\lim_{n\\to\\infty}f_n'(x)\\equal{}\\begin{cases}1&x\\equal{}0\\\\0&x\\ne 0\\end{cases}.$\r\n\r\nSo that is almost what you asked for - a sequence of functions whose derivatives converge, but to something other than the derivative of the limit.\r\n\r\nThe only objection is that you wanted this on the interval $ (0,1).$ All right then, just let $ g_n(x)\\equal{}f_n\\left(x\\minus{}\\frac12\\right).$ Then the problematical point lies at $ \\frac12,$ in the middle of your interval.",
  "Solution_7": "Thank you, I see it now. This is very clever."
}
{
  "Tag": [
    "quadratics"
  ],
  "Problem": "Prove that there don't exist rational numbers $ x,\\ y$ such that $ x^2\\plus{}y^2\\equal{}3$.",
  "Solution_1": "Then we'd have $ a^2 \\plus{} b^2 \\equal{} 3c^2$ for integers. So a and b must both be divisible by 3. So a=3n, b=3m, then $ 3n^2\\plus{}3m^2\\equal{}c^2$. Then c must be divisible by 3 and we have a contradiction by infinite descent.",
  "Solution_2": "Sorry, I can't understand your solution. :(",
  "Solution_3": "$ a^2 \\equiv 0, 1 \\bmod 3 \\implies a^2 \\plus{} b^2 \\equiv 0, 1, 2 \\bmod 3$ and $ 0 \\bmod 3$ is possible if and only if $ a^2, b^2 \\equiv 0 \\bmod 3$, so $ a$ and $ b$ are both divisible by $ 3$.  Equivalently, $ \\minus{}1$ is not a quadratic residue $ \\bmod 3$."
}
{
  "Tag": [
    "trigonometry",
    "analytic geometry",
    "complex numbers"
  ],
  "Problem": "[b]For how many real values of $ K$ will $ (2 \\plus{} Ki)^3$ be a real number?[/b]\r\n[hide=\"my solution\"]the trigonometric representation of $ 2 \\plus{} Ki$ is $ r\\cdot (\\cos \\theta \\plus{} i\\sin \\theta)$ for now we can ignore the $ r$ \n$ \\cos 3\\theta \\plus{} i\\sin \\theta \\equal{} (\\cos \\theta \\plus{} i\\sin \\theta)^3 \\equal{} \\cos^3\\theta \\plus{} 3i\\cos^2\\theta \\sin \\theta \\minus{} 3\\cos \\theta \\sin^2\\theta \\minus{} i\\sin^3\\theta$ \nsince we want the solution to be all real,\n$ 3i\\cos^2\\theta\\sin\\theta \\minus{} i\\sin^3\\theta \\equal{} 0$\nwe can solve for theta to get $ \\cos\\theta \\equal{} \\frac {1}{2} \\Longrightarrow \\theta \\equal{} 60$ deg. $ z \\equal{} 2 \\plus{} Ki \\equal{} (r,60)$ in the complex plane there is only one coordinate with those polar coordinates that will result in a completely real result. thus, there is only one real value of K.\ndoes that work? i know i have the right answer, but is my solution valid?[/hide]\r\n\r\n*sorry  :oops:",
  "Solution_1": "Your question is incomplete.  Is it \"for how many real values of $ k$ is $ (2 \\plus{} ki)^3$ real?  You don't need de Moivre's; just expand.",
  "Solution_2": "You don't really need a trigonometric representation for such small powers.. expanding would do just fine.",
  "Solution_3": "[quote=\"t0rajir0u\"]Your question is incomplete.  Is it \"for how many real values of $ k$ is $ (2 \\plus{} ki)^3$ real?  You don't need de Moivre's; just expand.[/quote]\r\n\r\n$ (2 \\plus{} ki)^3 \\equal{} 8 \\plus{} 12ki \\minus{} 6k^2 \\minus{} k^3i \\equal{} (8 \\minus{} 6k^2) \\plus{} k(12 \\minus{} k^2)i$\r\n\r\nSo we simply find the values of $ k$ at which the imaginary part vanishes, which are $ k \\equal{} 0, k \\equal{} 2\\sqrt {3}, k \\equal{} \\minus{} 2\\sqrt {3}$."
}
{
  "Tag": [],
  "Problem": "Determine (with proof) the largest integer $ n$ such that $ n$ is divisible by all positive integers less than $ \\sqrt[3]{n}$.",
  "Solution_1": "Posted before.\r\nhttp://www.mathlinks.ro/viewtopic.php?p=456076#456076"
}
{
  "Tag": [
    "function",
    "floor function",
    "induction"
  ],
  "Problem": "An integer sequence is defined by \\[a_{n}= 2a_{n-1}+a_{n-2}\\]\r\nand $a_{0}=0, a_{1}=1$\r\n\r\nProve that $2^{k}| a_{n}\\iff 2^{k}| n$",
  "Solution_1": "does this help at all..the fact that:\r\n$a_{n}=2^{n}-1$",
  "Solution_2": "That my friend, is completely wrong! Please don't post if it is just a wild guess.",
  "Solution_3": ":blush:  that generalization is wrong i didnt know that\r\n$a_{0}= 0$\r\n\r\n..my bad\r\n\r\nEDIT: Ya i clearly misread the sequence formula..it doesnt make sense how i got that....my apologies",
  "Solution_4": "[hide=\"This might be of some help...\"]\nI'm not sure how to proceed next, but the recurrence has the explicit formula of $a_{n}= \\frac{ (1+\\sqrt 2)^{n}-(1-\\sqrt 2)^{n}}{2\\sqrt 2}$.[/hide]",
  "Solution_5": "By means of generating functions, e.g., we obtain \\[a_{n}= \\frac{(1+\\sqrt{2})^{n}-(1-\\sqrt{2})^{n}}{2\\sqrt{2}}\\]By Newtons binomial, we've got \\[\\begin{aligned}(1+\\sqrt{2})^{n}-(1-\\sqrt{2})^{n}& = \\sum_{i=0}^{n}(\\binom{n}{i}(\\sqrt{2})^{i}-\\binom{n}{i}(-\\sqrt{2})^{i}\\\\ & = \\sum_{\\begin{array}{c}i=0 \\\\ i \\textrm{ even}\\end{array}}^{n}(\\binom{n}{i}(\\sqrt{2})^{i}-\\binom{n}{i}(-\\sqrt{2})^{i})+\\sum_{\\begin{array}{c}i=0 \\\\ i \\textrm{ odd}\\end{array}}^{n}(\\binom{n}{i}(\\sqrt{2})^{i}-\\binom{n}{i}(-\\sqrt{2})^{i}) \\\\ & = 0+\\sum_{\\begin{array}{c}i=0 \\\\ i \\textrm{ odd}\\end{array}}^{n}2 \\binom{n}{i}(\\sqrt{2})^{i}\\\\ & = 2\\sqrt{2}\\sum_{j=0}^{\\lfloor\\frac{n-1}{2}\\rfloor}\\binom{n}{2j+1}2^{j}\\end{aligned}\\]So \\[a_{n}= \\sum_{j=0}^{\\lfloor\\frac{n-1}{2}\\rfloor}\\binom{n}{2j+1}2^{j}\\]How to proceed...  :P I'm not even sure if this is the right way to tackle it  :maybe:",
  "Solution_6": "Well, I have to go to bed soon, I think it can be done by induction:\r\n\r\n$a_{n}-a_{n-2}=2a_{n-1}$, so terms that are 2 terms apart differ by a multiple of 2.\r\n\r\n$a_{n}=2a_{n-1}+a_{n-2}=4a_{n-2}+2a_{n-3}+2a_{n-3}+a_{n-4}$\r\n$\\implies a_{n}-a_{n-4}= 4(a_{n-2}+a_{n-3})$\r\n, so terms that are 4 terms appart differ by a multiple of 4.\r\n\r\nIf you can induct this and show that terms that are $2^{k}$ apart differ by a multiple of $2^{k}$, and you know that $0$ is a multiple of all $2^{k}$, so you can have $2k|n\\implies2k|a_{n}$.",
  "Solution_7": "Kurt G\u00f6del: Come on, you are very close for 1 part of the question.  :wink: \r\n\r\nThe Zuton Force: Can you please elaborate your solution and write the general form? (You only treated the cases where the difference is $2$ or $4$) Splendid idea though  :)",
  "Solution_8": "Zuton Force has revealed a glimpse of the idea... Why doesn't anyone notice how funny the statement is? \r\nTake a close look at that recursion... do you see why $a_{n}= a_{k}a_{n-k+1}+a_{k-1}a_{n-k}$? :)\r\n\r\nIt is given for $k=2$; but substituting the formula in itself gives the next $k$. Now the induction step is in hand reach.  :wink:"
}
{
  "Tag": [
    "geometry",
    "quadratics",
    "Pythagorean Theorem",
    "algebra",
    "system of equations"
  ],
  "Problem": "Let D= a^2+b^2+c^2, where a and b are consecutive integers and c=ab. Then Sqrt(D) is\r\n(A) Always an even integer\r\n(B) Sometimes an odd integer, sometimes not\r\n(C) Always an odd integer\r\n(D) Sometimes rational, sometimes not\r\n(E) Always irrational\r\n\r\nI will try to show my work. I have three equations in the beginning. The two listed in the question and a+1=b. I can use the two equations above to substitute into the third equation to get a^2+(a+1)^2+(ab)^2=D. I don't see how factoring helps. I think I got to the fact that D always is odd but can not figure out what the sqrt(D) must always be.    \r\n\r\nAnother problem that I don't even know how to attack is:\r\nQuadrilateral ABCD is inscribed in a circle with side AD, a diameter of length 4. If sides AB and BC each have lenght 1, then CD has length\r\n(A)7/2 (B) 5 Sqrt(2)/2 (C)Sqrt(11) (D) Sqrt(13) (E)2sqrt(3)",
  "Solution_1": "Let's write b = a+1 and c = a^2+a. Then a^2+b^2+c^2 = a^2+(a+1)^2+(a^2+a)^2 = a^4+2a^3+3a^2+2a+1 = (a^2+a+1)^2, so sqrt(D) = a^2+a+1. Now if a is even, then a^2+a+1 is odd. But if a is odd, then a^2+a+1 is also odd, so sqrt(D) is always an odd integer (C).",
  "Solution_2": "Thanks except I don't know how you spotted the factorization of a^4+2a^3+3a^2+2a+1 to get\r\n(a^2+a+1)^2. Also, I hope the geometry problem is solvable without the picture.",
  "Solution_3": "You don't really need that.\r\n\r\nLeave b as it is for a:^2:b:^2:\r\n\r\nSo you get\r\n\r\na:^2:+(a+1):^2:+a:^2:b:^2: = 2a:^2:+2a+1+a:^2:b:^2: \r\n= 2(a:^2:+a) + a:^2:b:^2: + 1\r\n\r\nLook at that. It's trivial that the first part is even. The second part is even, because either a or b is even, thus the square is. Adding one to an even number is always odd.",
  "Solution_4": "Here is a hint for the geometry problem:\r\n\r\nAll you need to use is the Pythagorean Theorem and Ptolemy's Theorem. Try to somehow use them to get a system of equations that you can solve.",
  "Solution_5": "Can you give me a bigger hint then saying Ptolemy's theorem and pythagorean theorem because I still can't find the solution.",
  "Solution_6": "Well, there is a right angle at ABD since it is inscribed in a circle and AD is the diameter, so we can apply the Pythagorean Theorem to find BD = Sqrt(15). Now using Ptolemy's Theorem would result in an equation with two variables AC and CD, both of which we do not know right now. Is there another way we can get an equation including AC and CD so that we can solve both of them simultaneously?",
  "Solution_7": "Sorry, I know I am giving away how bad I am in math but I still don't get what equations I can set up.",
  "Solution_8": "Well, ACD is also a right angle since AD is the diameter and the angle is inscribed in the triangle. Therefore, we can establish the Pythagorean Theorem to obtain\r\n\r\nAC^2 + CD^2 = AD^2 = 4^2 = 16\r\n\r\nApplying Ptolemy's Theorem to ABCD, we have\r\n\r\nAB*CD + BC*AD = BD*AC\r\n1*CD + 1*4 = Sqrt(15) * AC\r\nCD = Sqrt(15) * AC - 4\r\n\r\nSubstituting, we have\r\n\r\nAC^2 + (Sqrt(15) * AC - 4)^2 = 16\r\n\r\nSimplifying and solving the quadratic, we obtain AC = 0 or AC = Sqrt(15)/2. It is clearly the latter since AC must be greater than 0. Substituting again, we have\r\n\r\nCD = Sqrt(15) * (Sqrt(15)/2) - 4 = 15/2 - 4 = 7/2\r\n\r\nSo the answer is (A).",
  "Solution_9": "Thanks, know I get it. It was hard for me to realize that both angle ABD and BCD were right angles."
}
{
  "Tag": [
    "linear algebra",
    "matrix",
    "algorithm",
    "vector",
    "algebra",
    "polynomial",
    "invariant"
  ],
  "Problem": "I wonder if there is an algorithm to transform any matrix such as \r\n\r\n$A = \\left[\\begin{array}{ccc} A_{1,1} & A_{1,2} & A_{1,3} \\\\\r\nA_{2,1} & A_{1,2} & A_{2,3} \\\\ A_{3,1} & A_{3,2} & A_{3,3}\r\n\\end{array}\\right]$\r\n\r\ninto its companion form \r\n\r\n$B = \\left[\\begin{array}{ccc} 0 & 1 & 0 \\\\\r\n0 & 0 & 1 \\\\ b_1 & b_2 & b_3\r\n\\end{array}\\right]$\r\n\r\nCould this be done effectively for matrices of any order?  :oops:",
  "Solution_1": "The naive approach is to start with a random row vector $v$, and take basis $v,vA,vA^2,\\dots$, terminating when we reach a vector which is a linear combination of the previous. This basis is then used to transform the matrix; if $X$ has rows $v,vA,\\dots$, $XAX^{-1}=B$ This almost always works; if the minimal polynomial is the characteristic polynomial, it works if the initial vector is not in some finite collection of proper subspaces. If the field is infinite, this is almost certain. If the minimal polynomial is of smaller degree than the characteristic polynomial, there is no chance; in that case, the form must include several diagonal blocks.\r\n\r\nThis method has no guarantee of being numerically decent.",
  "Solution_2": "A problem is:  \"What sort of matrices can be transformed into companion form\"?\r\nThere is an explicit algorithm for matrices to be transformed into Jordan canonoical form. Are there explicit algorithm to transform matrices into thier companion form as well?  :?:",
  "Solution_3": "[quote=\"bchui\"]\"What sort of matrices can be transformed into companion form\"? [/quote]\r\nA $n\\times n$ matrix $A$ is similar to one in companion form if and only if its minimal polynomial $p$ has degree $n$\r\n\r\nIf you're using a standard of \"explicit\" that allows the standard arguments for Jordan form, I'll just add to my previous algorithm that we choose $v$ outside the invariant subspaces of $q(A)$ for each $q|p$. This is always possible, so we do it.\r\n\r\nThe whole thing is really very easy, as long as you start in the right place."
}
{
  "Tag": [
    "vector",
    "inequalities",
    "trigonometry",
    "geometry proposed",
    "geometry"
  ],
  "Problem": "Let ABC be a triangle having I is the incenter.\r\n  Prove that (IA+IB+IC)^2 >= 4r(4R+r).",
  "Solution_1": "Here is my solution \r\n    We have 4r(4R+r)= 2(ab+bc+ca)-a^2-b^2-c^2=4[(p-a)(p-b)+(p-b)(p-c)+(p-c)(p-a)]=4[IA.IBcos(A/2)cos(B/2)+...]<=(IA+IB+IC)^2. :cool:",
  "Solution_2": "[quote]We have 4r(4R+r)= 2(ab+bc+ca)-a^2-b^2-c^2[/quote]\r\n\r\nwhy? Can you explain it.\r\n\r\nCan you have a solution without sin cos tan",
  "Solution_3": "You have ab+bc+ca=p^2+r^2+4Rr.",
  "Solution_4": "[quote=\"treegoner\"]4r(4R+r)= 2(ab+bc+ca)-a^2-b^2-c^2=4[(p-a)(p-b)+(p-b)(p-c)+(p-c)(p-a)][/quote]\n\nThis can be simplified a bit: if we involve the exradii $r_{a},$ $r_{b},$ $r_{c}$, then a well-known result of Steiner tells us that $4R=r_{a}+r_{b}+r_{c}-r$, and we also know $rr_{a}=\\left( p-b\\right) \\left( p-c\\right) $, $rr_{b}=\\left( p-c\\right) \\left( p-a\\right) $, $rr_{c}=\\left( p-a\\right) \\left( p-b\\right) $, so that we have $4r\\left( 4R+r\\right) =4r\\left( r_{a}+r_{b}+r_{c}\\right) =4\\left(\nrr_{a}+rr_{b}+rr_{c}\\right) =4\\left( \\left( p-b\\right) \\left( p-c\\right)\n+\\left( p-c\\right) \\left( p-a\\right) +\\left( p-a\\right) \\left( p-b\\right)\n\\right) $. But what I don't understand is the following observation:\n\n[quote=\"treegoner\"]4[IA.IBcos(A/2)cos(B/2)+...]<=(IA+IB+IC)^2[/quote]\r\n\r\nCan you explain it to me? Thanks!\r\n\r\n  Darij",
  "Solution_5": "We have cos(A/2)cos(B/2)=sin((B+C)/2)sin((C+A)/2)<=sin^2((A+B+2C)/4)=sin^2(X).\r\n   Similarly we have XYZ is a triangle.\r\n   but we have sum 4xysin^2Z<=(x+y+z)^2 \r\n   I think it is a very cool inequality.  :cool:",
  "Solution_6": "Can we put the vector on IA;IB;IC to have your last inequality?\r\nWe have:\r\n$\\sum 4\\cdot IA\\cdot IB\\cdot \\cos \\frac{A}{2}\\cos \\frac{B}{2} =\\sum 2\\cdot\r\nIA\\cdot IB\\cdot \\left( \\cos \\frac{A+B}{2}+\\cos \\frac{A-B}{2}\\right)  \\\\\r\n\\leq -\\sum 2\\cdot IA\\cdot IB\\cdot \\cos \\frac{\\pi +C}{2}+2\\sum IA\\cdot IB$.\r\nWe have $\\left( \\overrightarrow{IA}+\\overrightarrow{IB}+\\overrightarrow{IC}\\right)^{2}\\geq 0$\r\n$\\Longleftrightarrow \\sum IA^{2}\\geq -\\sum 2\\cdot IA\\cdot IB\\cos \\frac{\\pi +C}{2}$\r\nSo $\\sum 4\\cdot IA\\cdot IB\\cdot \\cos \\frac{A}{2}\\cos \\frac{B}{2}\\leq \\sum\r\nIA^{2}+2\\sum IA\\cdot IB=\\left( IA+IB+IC\\right) ^{2}$ :)",
  "Solution_7": "Thanks, actually both of your proofs are cool :-)\r\n\r\n  Darij"
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "trigonometry"
  ],
  "Problem": "On hypotenuse $ AB$ of right triangle $ ABC$ are taken points $ K$ and $ L$ such that $ AK \\equal{} KL \\equal{} LB$. Find angles of triangle $ ABC$, if $ CK \\equal{} \\sqrt {2}CL$",
  "Solution_1": "Use the appolonius theorem",
  "Solution_2": "Midpoint $ O$ of $ AB$ is the circumcenter, $ R \\equal{} OA \\equal{} OB \\equal{} OC$ is the circumradius, $ \\phi \\equal{} \\angle COB < 90^\\circ.$ $ O \\equal{} (0, 0),$ $ A \\equal{} ( \\minus{} R, 0),$ $ B \\equal{} ( \\plus{} R, 0),$ $ K \\equal{} ( \\minus{} R/3, 0),$ $ L \\equal{} ( \\plus{} R/3, 0),$ $ C \\equal{} (R \\cos \\phi, R \\sin \\phi).$\r\n\r\n$ R^2[(\\cos \\phi \\plus{} 1/3)^2 \\plus{} \\sin^2 \\phi] \\equal{} CK^2 \\equal{} 2\\ CL^2 \\equal{} 2R^2[(\\cos \\phi \\minus{} 1/3)^2 \\plus{} \\sin^2 \\phi]$\r\n \r\n$ \\cos \\phi \\equal{} \\frac {5}{9},\\ \\ \\sin \\widehat{ABC} \\equal{} \\cos \\widehat{CAB} \\equal{} \\cos \\frac {\\phi}{2} \\equal{} \\sqrt {\\frac {1 \\plus{} \\cos \\phi}{2}} \\equal{} \\frac {\\sqrt 7}{3}$",
  "Solution_3": "I do not remember the name of the following theorem, so whoever knows it, please, communicate, thanks:\n\nFor a point $\\{D\\}\\in (AB)$, the hypothenuse of a right-angled triangle, there exists following relation:\n$BC^2\\cdot AD^2+AC^2\\cdot BD^2=CD^2\\cdot AB^2$; apply it for $K$ and $L$, getting: $9CK^2=a^2+4b^2, \\ 9CL^2= 4a^2+b^2$, where $a=BC, b=AC$; now easily get $\\tan(\\widehat{BAC})=\\sqrt{\\frac{3}{7}}$, a.s.o.\n\nBest regards,\nsunken rock",
  "Solution_4": "Let $AK=KL=LB=x, CL=y, CK=y\\sqrt{2},BC=a$.\n\nThen in $\\triangle BCK$ segment $CK$ is a median, hence $CL^2={BC^2+CK^2\\over 2}-{BK^2\\over 4}\\iff y^2={a^2+2y^2\\over 2}-{4x^2\\over 4}\\iff a=x\\sqrt{2}$.\n\nHence $b=\\sqrt{(3x)^2-a^2}=x\\sqrt{7}$\n\nNow $\\angle BAC=\\arctan\\sqrt{3\\over 7}$"
}
{
  "Tag": [
    "geometry"
  ],
  "Problem": "I'm confused as to when resonance is necessary to describe a Lewis structure. I'll use an example from a recent test:\r\n\r\nThe question asked which structure needed resonance to accurately describe it and two of the choices were $ \\text{NO}_{2}^{\\minus{}}$ and $ \\text{ClO}_{4}^{\\minus{}}$. When I drew the structures I found that both of them exhibit resonance but why is it that the first one NEEDS resonance to accurately describe it?\r\n\r\nAlso, on the subject of resonance, my textbook doesn't do the greatest of jobs describing it. Does it mean that when a structure has resonance, what does that say about the bond length of each specific bond? Does it mean that each one can vary or just that all of the lengths are the same?",
  "Solution_1": "A common mistake people make when thinking about resonance is they assume that the structure of say $ \\ce{NO2\\minus{}}$ rapidly interconverts between its two forms.\r\n\r\nTHIS IS WRONG!\r\n\r\nThe correct way to picture molecular bonding is as a superpositioning of all the possible resonance structures but each structure does not necessarily contribute as much to the overal form. By the way don't both structures need resonance to be described accurately?",
  "Solution_2": "The ressonance concept arises as a result of the natural limitation of Lewis structures: in Lewis notation, electron pairs are represented by small lines, between atoms if they are bonding, or on the atoms if they are non-bonding. However, this representation makes the electrons' position artificially static, localized in a given region of space. However, there are plenty of situations where we know that the electrons are free to move between three or more nuclei. But how can we continue to represent that situation with Lewis structures? The answer is ressonance and ressonance structures. Each of those structures represent a possible localization for electrons, but none is the true picture of the molecule.",
  "Solution_3": "Remember, science is not fact. Chemistry is not 100% correct. Resonance was established as a result of some molecules that could be drawn in multiple Lewis structures. In reality, the actual geometry of the molecule is the [i]average[/i] of the total amount of Lewis diagrams drawn. For example, in $ \\ce{SO2}$, the bond lengths between the sulfur and the oxygen are the average of the lengths of two Lewis diagrams. Experiments have been carried out that prove that the bond lengths between the sulfur and oxygen are all the same, and that neither is shorter or longer than the other."
}
{
  "Tag": [
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "let a,b,c be the length of the sides of triangle ABC\r\nprove that:\r\n$ \\frac{1}{(a\\plus{}b\\plus{}c)^2}\\plus{}\\frac{1}{(ab\\plus{}bc\\plus{}ca)^2}\\geq \\frac{1}{3(a^2b\\plus{}b^2c\\plus{}c^2a)}\\plus{}\\frac{1}{3(ab^2\\plus{}bc^2\\plus{}ca^2)}$",
  "Solution_1": "any idea????",
  "Solution_2": "It's an old and nice problem. Can_hang had a nice proof for it. But I can't find his link. Sorry  :wink:",
  "Solution_3": "[quote=\"nguoivn\"]It's an old and nice problem. Can_hang had a nice proof for it. But I can't find his link. Sorry  :wink:[/quote]\r\ncan you ask him??thanks\r\nI really want to know the solution of this inequality.",
  "Solution_4": "[quote=\"tuandokim\"]let a,b,c be the length of the sides of triangle ABC\nprove that:\n$ \\frac {1}{(a \\plus{} b \\plus{} c)^2} \\plus{} \\frac {1}{(ab \\plus{} bc \\plus{} ca)^2}\\geq \\frac {1}{3(a^2b \\plus{} b^2c \\plus{} c^2a)} \\plus{} \\frac {1}{3(ab^2 \\plus{} bc^2 \\plus{} ca^2)}$[/quote]\r\n$ \\frac {1}{(a \\plus{} b \\plus{} c)^2} \\plus{} \\frac {1}{(ab \\plus{} bc \\plus{} ca)^2}\\geq \\frac{2}{(a\\plus{}b\\plus{}c)(ab\\plus{}ac\\plus{}bc)}$ and easy SOS. :wink:",
  "Solution_5": "[quote=\"arqady\"][quote=\"tuandokim\"]let a,b,c be the length of the sides of triangle ABC\nprove that:\n$ \\frac {1}{(a \\plus{} b \\plus{} c)^2} \\plus{} \\frac {1}{(ab \\plus{} bc \\plus{} ca)^2}\\geq \\frac {1}{3(a^2b \\plus{} b^2c \\plus{} c^2a)} \\plus{} \\frac {1}{3(ab^2 \\plus{} bc^2 \\plus{} ca^2)}$[/quote]\n$ \\frac {1}{(a \\plus{} b \\plus{} c)^2} \\plus{} \\frac {1}{(ab \\plus{} bc \\plus{} ca)^2}\\geq \\frac {2}{(a \\plus{} b \\plus{} c)(ab \\plus{} ac \\plus{} bc)}$ and easy SOS. :wink:[/quote]\r\nCan you tell me more about your solution???thanks :blush:",
  "Solution_6": "[quote=\"tuandokim\"][quote=\"arqady\"][quote=\"tuandokim\"]let a,b,c be the length of the sides of triangle ABC\nprove that:\n$ \\frac {1}{(a \\plus{} b \\plus{} c)^2} \\plus{} \\frac {1}{(ab \\plus{} bc \\plus{} ca)^2}\\geq \\frac {1}{3(a^2b \\plus{} b^2c \\plus{} c^2a)} \\plus{} \\frac {1}{3(ab^2 \\plus{} bc^2 \\plus{} ca^2)}$[/quote]\n$ \\frac {1}{(a \\plus{} b \\plus{} c)^2} \\plus{} \\frac {1}{(ab \\plus{} bc \\plus{} ca)^2}\\geq \\frac {2}{(a \\plus{} b \\plus{} c)(ab \\plus{} ac \\plus{} bc)}$ and easy SOS. :wink:[/quote]\nCan you tell me more about your solution???[/quote]\r\nOf cause!\r\nWe need to prove that\r\n$ \\frac {2}{(a \\plus{} b \\plus{} c)(ab \\plus{} ac \\plus{} bc)}\\geq\\frac {1}{3(a^2b \\plus{} b^2c \\plus{} c^2a)} \\plus{} \\frac {1}{3(ab^2 \\plus{} bc^2 \\plus{} ca^2)}.$\r\nBut $ \\frac {2}{(a \\plus{} b \\plus{} c)(ab \\plus{} ac \\plus{} bc)}\\geq\\frac {1}{3(a^2b \\plus{} b^2c \\plus{} c^2a)} \\plus{} \\frac {1}{3(ab^2 \\plus{} bc^2 \\plus{} ca^2)}\\Leftrightarrow$\r\n$ \\Leftrightarrow6\\sum_{cyc}(a^4bc \\plus{} a^3b^3 \\plus{} a^2b^2c^2)\\geq\\sum_{cyc}(a^2b \\plus{} a^2c)\\sum_{cyc}(a^2b \\plus{} a^2c \\plus{} abc)\\Leftrightarrow$\r\n$ \\Leftrightarrow\\sum_{cyc}( \\minus{} a^4b^2 \\minus{} a^4c^2 \\plus{} 4a^3b^3 \\plus{} 4a^4bc \\minus{} 5a^3b^2c \\minus{} 5a^3c^2b \\plus{} 4a^2b^2c^2)\\geq0\\Leftrightarrow$\r\n$ \\Leftrightarrow \\minus{} \\sum_{cyc}(a^4b^2 \\minus{} 2a^3b^3 \\plus{} a^2b^4) \\plus{} \\sum_{cyc}(c^3a^3 \\minus{} c^3a^2b \\minus{} c^3ab^2 \\plus{} c^3b^3) \\plus{}$\r\n$ \\plus{} 2\\sum_{cyc}(a^4bc \\minus{} a^3b^2c \\minus{} a^2b^3c \\plus{} b^4ac) \\minus{} 2\\sum_{cyc}(a^3bc^2 \\minus{} 2a^2b^2c^2 \\plus{} ab^3c^2)\\geq0\\Leftrightarrow$\r\n$ \\Leftrightarrow\\sum_{cyc}(a \\minus{} b)^2( \\minus{} a^2b^2 \\plus{} c^3(a \\plus{} b) \\plus{} 2abc(a \\plus{} b) \\minus{} 2abc^2)\\geq0.$\r\nLet $ a\\geq b\\geq c$ and $ S_c \\equal{} \\minus{} a^2b^2 \\plus{} c^3(a \\plus{} b) \\plus{} 2abc(a \\plus{} b) \\minus{} 2abc^2.$\r\nHence, $ S_a\\geq0,$ $ S_b\\geq0$ and $ (a \\minus{} c)^2\\geq(a \\minus{} b)^2.$\r\nId est, $ \\sum_{cyc}(a \\minus{} b)^2S_c\\geq(a \\minus{} c)^2S_b \\plus{} (a \\minus{} b)^2S_c\\geq(a \\minus{} b)^2\\left(S_b \\plus{} S_c\\right).$\r\nThus, we need to prove that $ S_b \\plus{} S_c\\geq0.$\r\nBut $ S_b \\plus{} S_c \\equal{} \\minus{} a^2c^2 \\plus{} b^3(a \\plus{} c) \\plus{} 2abc(a \\plus{} c) \\minus{} 2ab^2c \\minus{}$\r\n$ \\minus{} a^2b^2 \\plus{} c^3(a \\plus{} b) \\plus{} 2abc(a \\plus{} b) \\minus{} 2abc^2 \\equal{}$\r\n$ \\equal{} a(b^3 \\plus{} c^3) \\plus{} b^3c \\plus{} c^3b \\plus{} 4a^2bc \\minus{} a^2b^2 \\minus{} a^2c^2\\geq$\r\n$ \\geq a^2(b^2 \\minus{} bc \\plus{} c^2) \\plus{} b^3c \\plus{} c^3b \\plus{} 4a^2bc \\minus{} a^2b^2 \\minus{} a^2c^2 \\equal{}$\r\n$ \\equal{} b^3c \\plus{} c^3b \\plus{} 3a^2bc\\geq0.$ Done! :)",
  "Solution_7": "nice solution arqady.thanks :D"
}
{
  "Tag": [
    "integration",
    "trigonometry",
    "calculus",
    "vector",
    "function",
    "derivative",
    "analytic geometry"
  ],
  "Problem": "Let $ \\bold{F} \\equal{} \\minus{} 150 \\bold{j}$\r\n\r\n(a) If $ C$ is the straight line segment $ y \\equal{} x$ from $ P(100,100)$ to $ Q(0,0)$ find the work.\r\n\r\nSo $ x \\equal{} x, \\ y \\equal{} x, \\ 0 \\leq x \\leq 100$\r\n\r\n$ \\minus{} \\int_{0}^{!00} ( \\minus{} 150 \\bold{j}) \\cdot (\\bold{i} \\plus{} \\bold{j}) \\equal{} 15000$\r\n\r\n\r\n\r\n(b) $ C$ is the circular arc $ x \\equal{} 100 \\sin t, \\ y \\equal{} 100 \\cos t$, $ P(0,100)$ to $ Q(0,0)$. I think this is a typo and it should be $ Q(100,0)$. Find the work.\r\n\r\nSo $ \\int_{0}^{\\pi/2} ( \\minus{} 150 \\bold{j}) \\cdot (100 \\sin t \\bold{i} \\plus{} 100 \\cos t \\bold{j}) \\equal{} \\minus{} 15,000$\r\n\r\n\r\n\r\n(c) $ C$ is the parabolic arc $ y \\equal{} \\frac {x^2}{100}$ from $ P(100,100)$ to $ Q(0,0)$. Find the work.\r\n\r\n$ x \\equal{} x, \\ y \\equal{} \\frac {x^2}{100}$\r\n\r\n$ \\int_{0}^{100} ( \\minus{} 150 \\bold{j}) \\cdot (x \\bold{i} \\plus{} \\frac {x^2}{100} \\bold{j}) \\equal{} \\minus{} 66666.6$\r\n\r\n\r\nAre these correct?",
  "Solution_1": "Well first $ W \\equal{} \\int_a^b \\bold{F} \\cdot \\bold{r'(t)} ~dt$.\r\n\r\n$ \\bold{(a)}$ So thus your first integral is correct if that $ !$ is a typo for $ 1$.\r\n\r\n$ \\bold{(b)}$ This one I believe you got a little mixed up on.  We'll get:\r\n\r\n$ \\bold{r}(t) \\equal{} \\bold{<} 100 \\sin t, 100 \\cos t  \\bold{>}$\r\n$ \\implies \\bold{r}'(t) \\equal{} \\bold{<} 100 \\cos t, \\minus{} 100 \\sin t  \\bold{>}$\r\n\r\nThus: $ W \\equal{} \\int_0^{\\frac{\\pi}{2}} \\bold{<} 0, \\minus{}150 \\bold{>} \\cdot \\bold{<} 100 \\cos t, \\minus{} 100 \\sin t  \\bold{>} ~dt$\r\n\r\n$ \\bold{(c)}$ The whole $ \\bold{r}'(t)$ thing also needs fixing in this one as well.  Also, we are going from $ x\\equal{}100$ to $ x\\equal{}0$ so flip your bounds.",
  "Solution_2": "Thanks.\r\n\r\nI get $ 15000$ for all of them (number 2 its negative).",
  "Solution_3": "This is a conservative force. The work should be path-independent. That is, we only need to know the endpoints of the path, not the intermediate details, in order to calculate the work.",
  "Solution_4": "Can you use a differential form to evaluate this?\r\n\r\nMaybe a 1-form: $ A \\ dx \\plus{} B \\ dy \\plus{} C \\ dz$?",
  "Solution_5": "The distinction between a vector field and a 1-form is one of notation, not substance. Changing the notation won't change anything of any importance about the computations. I called it a conservative field; I could just as easily have called it an exact differential form. The meaning is the same.",
  "Solution_6": "(Technical definitions: A vector field $ v$ is an operator from scalar-valued functions to scalar-valued functions which behaves like a derivative (linear at each point, product rule). It naturally corresponds to a vector-valued function, with the vectors tangent to the underlying space. A $ 1$-form is an operator from vector fields to scalar-valued functions, which naturally corresponds to a \"covector\"-valued function.)\r\n\r\nActually, there is a difference between vector fields and 1-forms: the spaces they live in are \"dual\" to each other. They are (locally) isomorphic, but that isomorphism depends on a choice of coordinates. This distinction between vectors and covectors gets blurred in the standard first treatments of multivariable calculus, but it shows up if you try to apply coordinate transformations- it's covariant for one and contravariant for the other.\r\n\r\nThat said, this $ F$ is a 1-form, not a vector field. The things that we integrate, that have the Stokes family of theorems for them, are forms.\r\n\r\nAn example of what I was talking about in the first paragraph: What happens if we scale our coordinate system- the point which was represented by $ (x,y,z)$ is now represented by $ (2x,2y,2z)$? The vector field is doubled (it should give the same result when applied as a derivative to a function, now stretched out over twice the coordinate range) and the 1-form is halved (it should give the same result when integrated over a curve, now covering twice the coordinate range).\r\n\r\nAnd now- I'm not really contradicting Kent. We have to do our computations in coordinates anyway, and on $ \\mathbb{R}^n$ we have a standard basis to relate everything to. If we want to write these things down, we need to use some sort of vector notation, and duplicating that of vector fields works pretty well. You're not even using \"real\" vector fields much anyway."
}
{
  "Tag": [],
  "Problem": "hey the NSO results are out but we dont remember the roll no how do we check last time we used to enter the school code???now what could we do??? :blush:",
  "Solution_1": "[schoolcode]2M[yourcode]\r\n\r\nAnyways I got AIR222.... Not a big deal but I guess I'll get a minimum of 250Rs. :D",
  "Solution_2": "Was too bored to go to school.. So din write nso.. :D",
  "Solution_3": "i got air 55 :) \r\nbut i heard theres gonna be a 2nd round also  :| \r\nis that true",
  "Solution_4": "yeah ur rite shreyas. 250 bucks indeed :D\r\n\r\ni got 'AIR' 304  :wink: \r\n\r\n@anitashu: yes, there is gonna be a 2nd round.. and jus 4 the record- the AIR rank one in the second round gets 51 grand!!   \r\n:w00tb:",
  "Solution_5": "hey wats d eligibility for 2nd round...\r\nim AIR 500...\r\nn when is it likely to be??",
  "Solution_6": "@abishek: Top 500 is eligibility. But I'm not sure if you've either missed the bus or just hopped onto it. ie: Dunno if AIR 500 is included. :(\r\n\r\nAnyways, anyone here writing NCO 2nd Level. Cause it clashes with Fiitjee open test?",
  "Solution_7": "Got AIR 562 . Am shocked  :o \r\nDid the paper terribly for words ( got 35 or 36  :maybe: )\r\nExpected rank less than 2000.\r\nNow am sad that if i had got 2 more questions would have got thro.  :( \r\nAnyways wanna write cyber and then paper 2 of fiitjee ( mostly paper 2 will be tougher )\r\n\r\n@pardesi : Did you write nco and nso?",
  "Solution_8": "yes and the [b]most[/b] miserable performance 1000+ :rotfl: \r\nnone from our school qualifies this time and last time we had got the best school medal :o"
}
{
  "Tag": [
    "induction",
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "If $n\\geq 3$ and $a_{1}\\geq a_{2}\\geq .... \\geq a_{n}\\geq 0$, prove\r\n$a_{1}a_{2}a_{3}+a_{2}a_{3}a_{4}+..+a_{n-2}a_{n-1}a_{n}\\leq \\frac{a_{1}+a_{2}+...+a_{n}}{27}^{3}$\r\n\r\n :wink:",
  "Solution_1": "[quote=\"Fitim\"]If $n\\geq 3$ and $a_{1}\\geq a_{2}\\geq .... \\geq a_{n}\\geq 0$, prove\n$a_{1}a_{2}a_{3}+a_{2}a_{3}a_{4}+..+a_{n-2}a_{n-1}a_{n}\\leq \\frac{a_{1}+a_{2}+...+a_{n}}{27}^{3}$\n\n :wink:[/quote]\r\n\r\n\r\nI think that in $RHS$ is $\\frac{(a_{1}+a_{2}+...+a_{n})^{3}}{27}$?",
  "Solution_2": "[quote=\"pohoatza\"]\nI think that in $RHS$ is $\\frac{(a_{1}+a_{2}+...+a_{n})^{3}}{27}$?[/quote]\r\n\r\nYes! I have mistakes. :wink:",
  "Solution_3": "[quote=\"Fitim\"]If $n\\geq 3$ and $a_{1}\\geq a_{2}\\geq .... \\geq a_{n}\\geq 0$, prove\n$a_{1}a_{2}a_{3}+a_{2}a_{3}a_{4}+..+a_{n-2}a_{n-1}a_{n}\\leq \\frac{a_{1}+a_{2}+...+a_{n}}{27}^{3}$\n\n :wink:[/quote]\r\nnot hard.by indution .n=3 obvious\r\nsoppose n=k we have $a_{1}a_{2}a_{3}+a_{2}a_{3}a_{4}+..+a_{k-2}a_{k-1}a_{k}\\leq \\frac{(a_{1}+a_{2}+...+a_{k})^{3}}{27}$\r\nthen $a_{1}a_{2}a_{3}+a_{2}a_{3}a_{4}+..+a_{k-1}a_{k}a_{k+1}\\leq \\frac{(a_{1}+a_{2}+...+a_{k})^{3}}{27}+a_{k-1}a_{k}a_{k+1}\\leq \\frac{(a_{1}+a_{2}+...+a_{k})^{3}}{27}+\\frac19 a_{k+1}(a_{1}+\\dots+a_{k})^{2}\\leq \\frac{(a_{1}+a_{2}+...+a_{k+1})^{3}}{27}$\r\nanother a little bit harder problem:\r\nIf $n\\geq 4$ and $a_{1}\\geq a_{2}\\geq .... \\geq a_{n}\\geq 0$, prove\r\n$a_{1}a_{2}a_{3}+a_{2}a_{3}a_{4}+..+a_{n-2}a_{n-1}a_{n}+a_{n-1}a_{n}a_{1}+a_{n}a_{1}a_{2}\\leq \\frac{(a_{1}+a_{2}+...+a_{n})^{3}}{16}$"
}
{
  "Tag": [
    "integration",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "Find all values of $ a, b, x$ and $ y$ that satisfy the simultaneous equations:\r\n\r\n$ a\\plus{}b\\equal{}1$\r\n\r\n$ ax\\plus{}by\\equal{}\\frac{1}{3}$\r\n\r\n$ ax^{2} \\plus{} by^{2} \\equal{}\\frac{1}{5}$\r\n\r\n$ ax^{3}\\plus{}by^{3}\\equal{}\\frac{1}{7}$.",
  "Solution_1": "Do you calculate Gaus's formula for $ \\int_0^1f(x)dx?$.\r\nYou can calculate as \\[ by^2(y\\minus{}x)\\equal{}\\frac 17 \\minus{}\\frac x5,\\]\r\n\\[ by(y\\minus{}x)\\equal{}\\frac 15 \\minus{}\\frac x3,\\]\r\n\\[ b(y\\minus{}x)\\equal{}\\frac 13 \\minus{}x.\\]\r\nThen \\[ y\\equal{}\\frac{3\\minus{}5x}{5(1\\minus{}3x)}\\equal{}\\frac{15\\minus{}21x}{21\\minus{}35x}\\to 35x^2\\minus{}30x\\plus{}3\\equal{}0,...\\]",
  "Solution_2": "Hi rust. Could you explain in more detail, please: I don't know what Gauss's formula for $ \\int_{0}^{1} f(x)\\ dx$ is, but your method looks like it might be quite interesting?\r\n\r\nI used \r\n\r\n$ (ax^{n}\\plus{}by^{n})(x\\plus{}y) \\equal{} (ax^{n\\plus{}1}\\plus{}by^{n\\plus{}1}) \\plus{}xy(ax^{n\\minus{}1}\\plus{} by^{n\\minus{}1})$\r\n\r\n(which may also have a name).",
  "Solution_3": "Are you out there Rust?\r\n\r\nOr could someone else explain?",
  "Solution_4": "hello, solving this system i have got\r\n$ a_1\\equal{}\\frac{1}{2}\\plus{}\\frac{1}{36}\\sqrt{30}$\r\n$ b_1\\equal{}\\frac{1}{2}\\minus{}\\frac{1}{36}\\sqrt{30}$\r\n$ x_1\\equal{}\\frac{3}{7}\\minus{}\\frac{2}{35}\\sqrt{30}$\r\n$ y_1\\equal{}\\frac{3}{7}\\plus{}\\frac{2}{35}\\sqrt{30}$\r\nor\r\n$ a_2\\equal{}\\frac{1}{2}\\minus{}\\frac{1}{36}\\sqrt{30}$\r\n$ b_2\\equal{}\\frac{1}{2}\\plus{}\\frac{1}{36}\\sqrt{30}$\r\n$ x_2\\equal{}\\frac{3}{7}\\plus{}\\frac{2}{35}\\sqrt{30}$\r\n$ y_2\\equal{}\\frac{3}{7}\\minus{}\\frac{2}{35}\\sqrt{30}$\r\ni don't want to type all my calculations from my sheet of paper in this editor.\r\nSonnhard.",
  "Solution_5": "Thanks Sonnhard. They're the same as my answers. I was looking for some other ways to solve the problem, and I'm still confused by Rust's mention of Gauss's formula. I like Rust's method of solution but I can't see what Gauss's formula has to do with it; perhaps because I don't know what Gauss's formula is. Can anyone shed some light on this?",
  "Solution_6": "Has anyone figured out what Rust is alluding to with $ \\int_{0}^{1} f(x)\\ dx$?"
}
{
  "Tag": [
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "I was reasearching a little bit of [url=http://mathworld.wolfram.com/FigurateNumber.html]Figurate Numbers[/url], such as square numbers, pentagonal numbers, etc.  It is possible to find the formulas and the first few numbers of each of these types fairly easily. However, I was just wondering, what exactly is an r-gonal number? For example, what would the first few r-gonal numbers be? Is it possible to write a formula to determine the nth r-gonal number?",
  "Solution_1": "the n-th r-gonal number is $ n \\cdot \\left( \\frac {(r-2)(n-1)}{2}+1 \\right) $",
  "Solution_2": "[quote=\"ZetaX\"]the n-th r-gonal number is $ n \\cdot \\left( \\frac {(r-2)(n-1)}{2}+1 \\right) $[/quote]\r\n\r\nAnd what exactly is a r-gonal number? Is that formula the same thing as:\r\n\r\n[hide]$\\frac{1}{2}r[(n-2)r-(n-4)]$[/hide]\r\n\r\nWhere $r$ represents the number of sides, and $n$ is the $n$th r-gonal number?",
  "Solution_3": "You can define a m-th r-gone by dividing every side of an r-gone into m segments, giving their endpoints as the 'm-th r-gone points'. Now if you set the first n 'm-th r-gones' (m from 0 to n) 'inside' each other, having two edges and one vertice in common, you get the n-th r-gonal number as the number of different 'r-gone-points'.\r\n(sorry for my bad explanation  :( )"
}
{
  "Tag": [
    "limit"
  ],
  "Problem": "hmm\r\nI dunno how to make polls\r\nbut vote how many posts will get in this topic before it gets locked?\r\n\r\nit would be really sad if a moderator was the first to look at this and locked it immediately...\r\n\r\nand that's why I don't think that this is a very good idea...",
  "Solution_1": "Well...on the top next to \"message body\", it should say \"poll\", which would allow you to insert a poll.\r\n\r\nYay! $ \\ge2$",
  "Solution_2": "yayx0rz thanks Carbon",
  "Solution_3": "If this thread is to survive, the moderator must have a secondary goal besides locking the thread. Perhaps he wants to make our poll as inaccurate as possible. If the moderator has just the one goal, we are doomed no matter what, so we may as well assume that this is not the case. We must therefore be extremely careful in voting so that the moderator will not lock the thread for fear of us winning.\r\n\r\nThis is very important! Think before you vote!",
  "Solution_4": "YESH INDEED",
  "Solution_5": "here is a question: can mods look at the results of a poll BEFORE they vote? After all they are mods.",
  "Solution_6": "everyone can\r\nand this is just spam\r\ni think ill lock it",
  "Solution_7": "this is pretty tricky\r\n\r\nits like game theory omg",
  "Solution_8": "that's what she said",
  "Solution_9": "I think the amount of posts needed to [b]arouse[/b] the attention of a mod is on the order of $ \\lim_{x\\to\\infty} x$\r\n\r\nTry making a TWSS joke out of that!",
  "Solution_10": "-wonders who unlocked this- just out of curiosity and such.",
  "Solution_11": "Wasn't me. *Re-locks*"
}
{
  "Tag": [
    "induction"
  ],
  "Problem": "Somebody told me recently that in a \"show that\" problem, you could use the result and work backwards as your reasoning, but in a proof problem you had to come to your conclusion without what was asked to prove in the question. This seems wrong to me as styles of proof such as contradiction and mathematical induction [b]require[/b] you to use what was to prove as part of the proof. But that lead me to thinking; is there a difference between these types of problems? Is it just in the level of rigour?",
  "Solution_1": "In a proof, you [i]can[/i] work backwards from the requested statement, but you then have to show that all of your steps are reversible. If something is necessary for something else (that is, $p\\implies q$) it may not be true that it's sufficient ($q\\implies p$). Showing the reversibility of your steps is -very- important if you plan on working backwards."
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "Let $x,y$ two real numbers which are not both $0$. Prove that\r\n\r\n$\\dfrac{x+y}{x^2-xy+y^2}\\leq\\dfrac{2\\sqrt 2}{\\sqrt{x^2+y^2}}$.\r\n\r\nPlease I want from the moderators to send all the problems to Olympiad recousrces.\r\nI will put all the for problems from 2004.",
  "Solution_1": "Hi silouan:\r\n  I think it is an easy inequality.  :D \r\n   1>If $x+y<0$ then $LHS<0<RHS$\r\n   2> If $x+y>0$ then square all of them we have to solve the new inequality:\r\n$V(x,y)=8(x^2-xy+y^2)^2-(x^2+y^2)(x^2+2xy+y^2)\\geq0$\r\n$V(x,y)= (x-y)^2(7x^2-4xy+7y^2)\\geq0$\r\n  Done! :!:",
  "Solution_2": "Siouan, your inequality follows immediately from this easy one: \r\n$2(x^{3}+y^{3})\\geq (x+y)(x^{2}+y^{2})$ which is a direct consequence of Chebyshef. ;) , but it can be also proved by calculations. very easy. ;)"
}
{
  "Tag": [],
  "Problem": "Given a natural number $n.$ Generate new numbers by the following mapping $\\Phi.$ Erase the last digit and add to this number 4 times the last digit, e.g. $\\sigma = 1995 \\mapsto 199 + 4 \\cdot 5 = 219 = \\tau.$ \r\n\r\na.) Determine starting numbers $\\sigma$ for which after a finite number $k$ of compositions of $\\Phi,$ $(\\Phi^{(k)}),$ a number $\\tau$ occurs several times. Does there $\\exists$ a starting numbers $\\sigma$ for which after a finite number applications of $\\Phi,$ $(\\Phi^{(k)}),$ only the same number $\\tau$ occurs (called one-cycle).\r\n\r\nb.) Let $1001 = \\tau$ for a certain $\\Phi^{(k)}, k \\geq 1.$ Prove that there does not exist any $\\tau$ for a certain $\\Phi^{(k)}$ so that $\\tau$ is a prime.",
  "Solution_1": "Edited for error:\n\n[hide]a) For the first part, any number, after repeated applications of Phi, will eventually lead to a cycle.  Consider any four digit number.  When you erase the last digit of the number, it decreases by at least 100, and when you multiply the former last digit by 5 and add it to the new number, you add at most 45.  Therefore, any application of Phi to a four digit number will result in a smaller number (the same argument holds for numbers with more digits; for an n digit number, an application of Phi will decrease the number by at least (10)^(n-2) and increase it by at most 45).  Thus, if we have a number of four or more digits, we will eventually get a number with three or fewer digits.  When we have a number with three or fewer digits, two things can happen.  Either an application of Phi will lead to another number with three or fewer digits, or we get a number with four or more digits (I know, this second one is impossible, but I'm not sure how to prove it quickly), either of which will lead to another number with three or fewer digits.  Either way, we'll eventually run out of such small numbers, and we'll have to repeat some.  For the second part, application of Phi to 26 will result in 26, thus there is a starting number (49) that will result in a one-cycye.[/hide]\n\n\n\nI'm not sure about part b)...",
  "Solution_2": "b) goes as follows: \r\n\r\n1001 -> 104 -> 26 -> 26 -> ... furher the row always evolves 26-26-26-... so what is the problem? :?\r\n\r\nOr do you mean we have to run back and check that 1001 cannot come from any primes, regardless of how far you return?",
  "Solution_3": "[quote=\"orl\"] \nb.) Let $1001 = \\tau$ for a certain $\\Phi^{(k)}, k \\geq 1.$ Prove that there does not exist any $\\tau$ for a certain $\\Phi^{(k)}$ so that $\\tau$ is a prime.[/quote]\r\n\r\nAt first I have to say that you cannot start the sequence with $1001$ as you have to look for $ \\tau$ and not $ \\sigma.$ $ \\tau$ cannot occur before the second element in the sequence (first element is $ \\sigma$) of your numbers, e.g. you can choose $(9971, 1001, 104, 26, \\ldots).$ And this case would mean that you chose $k =1$ and I said $k \\geq 1.$ So you have to consider all previous numbers in the sequence too.  :)",
  "Solution_4": "Somebody can tackle part b.)  :)",
  "Solution_5": "part b:\n\n\n\n[hide]The transformation 10a+b -> a+4b is equivalent to multiplication by 4 in modulo 39, i.e. 4(10a+b)=40a+4b=a+4b mod 39. Since 1001=26 mod 39, which is divisible by 13, any number in a sequence containing 1001 will be divisible by 13. The only prime in such a sequence is 13. But 1001 and 13 cannot be in the same sequence, for the following reason:\n\n\n\n13 can't come after 1001, since 1001->104->26->26..., which does not contain 13.\n\n\n\n1001 cant come after 13, since 13->13->..., which doesnt contain 1001.[/hide]",
  "Solution_6": "Btw, this also happens to be a divisibility test for 13; it retains divisibility by 13, though any remainders will get distorted by the x4 part.\r\n\r\nFor example, we all know 1001=7x11x13\r\n1001-->100+4x1=104-->10+4x4=26, which is a multiple of 13, hence 1001 is a multiple of 13"
}
{
  "Tag": [
    "geometry",
    "AMC",
    "AIME"
  ],
  "Problem": "A square is drawn inside a circle so that one vertex lies at the center of the circle and another vertex lies on the circumference. The are of the circle is $ 72\\pi$ square inches. How many square inches are in the area of the square?\n\n[asy]size(50);\ndraw(Circle((0,0),5));\ndraw((-5,0)--(5,0));\ndraw((0,5)--(0,-5));\ndraw((3.5,0)--(3.5,-3.5)--(0,-3.5));[/asy]",
  "Solution_1": "The radius of the circle is $ 6\\sqrt{2}$, so the diagonal is $ 6\\sqrt{2}$, so the side of the square is $ 6$ and its area is $ \\boxed{36}$.",
  "Solution_2": "[quote=\"AIME15\"]The radius of the circle is $ 6\\sqrt{2}$, so the diagonal is $ 6\\sqrt{2}$, so the side of the square is $ 6$ and its area is $ \\boxed{36}$.[/quote]\n\n\nJust to add a little to what AIME said:\n\nNow that you know what the radius of the circle is, you can use it to find the area of the square, so (6sqrt2)^2 divided by 2 is 72/2 which is 36!!! The formula to find the area of the square is diagonal^2 divided by two.",
  "Solution_3": "You can just divide by 2pi. It works every time for a problem like this",
  "Solution_4": "[quote=GameBot]A square is drawn inside a circle so that one vertex lies at the center of the circle and another vertex lies on the circumference. The are of the circle is $ 72\\pi$ square inches. How many square inches are in the area of the square?\n\n[asy]size(50);\ndraw(Circle((0,0),5));\ndraw((-5,0)--(5,0));\ndraw((0,5)--(0,-5));\ndraw((3.5,0)--(3.5,-3.5)--(0,-3.5));[/asy][/quote]\n\nyou made a typo"
}
{
  "Tag": [
    "trigonometry",
    "function",
    "inequalities",
    "algebra proposed",
    "algebra"
  ],
  "Problem": "A calculator treats angles as radians. It initially displays 1. What is the largest value that can be achieved by pressing the buttons cos or sin a total of 2001 times? (So you might press cos five times, then sin six times and so on with a total of 2001 presses.)",
  "Solution_1": "It is easy to show maximum is $max=cos(sin(sin \\dots 1)\\dots )$ It give approch $cos(\\sqrt{\\frac 3n})=1-\\frac{3}{2n}+O(n^{-2}), \\ n=2000.$",
  "Solution_2": "No,since $\\tan1>1,\\sin1>\\cos1$,so $\\cos(\\sin(\\sin\\cdots(\\sin1)))<\\cos(\\sin(\\sin\\cdots(\\cos1)))$.",
  "Solution_3": "Yes. But formula $max=1-\\frac{3}{2n}+O(n^{-2})$ works.",
  "Solution_4": "Yes you are right.But I think for this problem we needn't to estimate the value.",
  "Solution_5": "After some years thinking and computing :D , I will post a solution:\r\n\r\nThinking in Dynamic Programming, we will define the sequences $ M_k$ and $ m_k$ as the maximum and the minimum value of the values of pressing sequence, with $ k$ presses.\r\n\r\nWe can assume all values are in the first quadrant (between 0 and $ \\pi/2$).\r\n\r\nAs $ sin$ and $ cos$ are monotonic (increasing and decreasing, respectively) functions, we can say that $ \\{M_{k\\plus{}1},m_{k\\plus{}1}\\} \\in \\{sin(M_k),sin(m_k),cos(M_k),cos(m_k)\\}$.\r\n\r\nThe idea is to put restrictions in this fact.\r\n\r\nThese are easily verifiable facts about $ sin$ and $ cos$:\r\n\r\n1 - $ sin(x) > cos(y) \\Leftrightarrow x\\plus{}y > \\pi/2$\r\n2 - $ sin(x) < cos(y) \\Leftrightarrow x\\plus{}y < \\pi/2$\r\n3 - $ sin(x) > sin(y) \\Leftrightarrow x>y$\r\n4 - $ cos(x) > cos(y) \\Leftrightarrow x<y$\r\n\r\nIf $ M_k\\plus{}m_k > \\pi/2$ then $ M_{k\\plus{}1}\\equal{}sin(M_k)$ and $ m_{k\\plus{}1}\\equal{}cos(M_k)$\r\nIf $ M_k\\plus{}m_k < \\pi/2$ then $ M_{k\\plus{}1}\\equal{}cos(m_k)$ and $ m_{k\\plus{}1}\\equal{}sin(m_k)$\r\n\r\nSo, we conclude that $ M_k^2\\plus{}m_k^2\\equal{}1$ for $ k \\geq 1$.\r\n\r\nBut using Power Mean Inequality, we cn assure that $ M_k\\plus{}m_k \\leq \\sqrt{2} \\leq \\pi/2$, and so we can use the formula $ M_{k\\plus{}1}\\equal{}cos(m_k)$ and $ m_{k\\plus{}1}\\equal{}sin(m_k)$ for any $ k \\geq 1$.\r\n\r\nNow the problem is almost obvious :)"
}
{
  "Tag": [],
  "Problem": "Let the two base angles of a triangle be $A$ and $B$, with $B$ larger than $A$.  The altitude to the base divides the vertex angle $C$ into two parts, $C_1$ and $C_2$, with $C_2$ adjacent to side $a$.  Then:\r\n(A) $C_1 + C_2 = A + B$\r\n(B) $C_1 - C_2 = B - A$\r\n(C) $C_1 - C_2 = A - B$\r\n(D) $C_1 + C_2 = B - A$\r\n(E) $C_1 - C_2 = A + B$",
  "Solution_1": "[hide]The altitude to the base makes a 90 degree angle, making the other two angles in the triangles inside ABC euqal each other: \n\nC1 + A = C2 + B. \n\n\n\nWe need to simplify it, since that isn't one of the answer choices.\n\n\n\nC1-C2 = B-A, or B.[/hide]"
}
{
  "Tag": [
    "probability"
  ],
  "Problem": "Six cards with numbers 1, 1, 3, 4, 4, 5 are given. We are drawing 3 cards from 6 given cards one by one and are forming a three-digit number with the numbers over the cards drawn according to the drawing order. Find the probability that this three-digit number is a multiple of 3. (The card drawn is not put back)\n\n$\\textbf{(A)}\\  \\frac {1}{5} \\qquad\\textbf{(B)}\\  \\frac {2}{5} \\qquad\\textbf{(C)}\\  \\frac {3}{7} \\qquad\\textbf{(D)}\\  \\frac {1}{2} \\qquad\\textbf{(E)}\\ \\text{None}$",
  "Solution_1": "The number of all outcomes is $ \\frac {6!}{2!2!} \\equal{} 180$\r\n\r\nThe number of favorable outcomes is:\r\n\r\n1. Digits $ 114$ - either one of the two fours can be drawn, and the drawn digits can be ordered in $ 3$ ways, hence $ 2\\cdot 3 \\equal{} 6$\r\n\r\n2. Digits $ 135$ - either one of the two one's can be drawn, and the drawn digits can be ordered in $ 6$ ways, hence $ 2\\cdot 6 \\equal{} 12$\r\n\r\n3. Digits $ 144$ - either one of the two one's can be drawn, and the drawn digits can be ordered in $ 3$ ways, hence $ 2\\cdot 3 \\equal{} 6$\r\n\r\n4. Digits $ 345$ - either one of the two fours can be drawn, and the drawn digits can be ordered in $ 6$ ways, hence $ 2\\cdot 6 \\equal{} 12$\r\n\r\nTotal: $ 6 \\plus{} 12 \\plus{} 6 \\plus{} 12 \\equal{} 36$\r\n\r\nProbability: $ {36\\over 180} \\equal{} {1\\over 5}$",
  "Solution_2": "Farenhajt, are you sure the total number of outcomes is 180? That seems a little high to me. I think you calculated the number of ways to form 6 digit numbers instead of 3 digit numbers.",
  "Solution_3": "You're right, I should really go to sleep :roll:\r\n\r\nDesignate the digits thus: $ 1, 1', 3, 4, 4', 5$. Then drawing a 3-digit number is non-repetitive variation, so there's $ 6\\cdot 5\\cdot 4 \\equal{} 120$ outcomes.\r\n\r\nNothing will change in the count of favorable cases, they will only split, like $ 11'4 \\plus{} 11'4'$ etc.\r\n\r\nBut then the result is $ {36\\over 120} \\equal{} {3\\over 10}$, so the answer is $ E$...?",
  "Solution_4": "Okay, now I agree with your denominator but I disagree with your numerator (according to the way you are counting). For example, 114 could be either 11'4, 1'14, 11'4', or 1'14' (plus all the permutations of those).",
  "Solution_5": "Grrrrr!\r\n\r\nNow all the digits are distinct, hence each possibility yields $ 6$ orders, and there are $ 8$ possibilities, hence $ P\\equal{}{48\\over 180}\\equal{}{4\\over 15}$\r\n\r\nAnd if this is also wrong, please just give the solution (it's 4:50 am here).",
  "Solution_6": "I get $ {48 \\over 120} \\equal{} {2 \\over 5}$, which is answer choice $ B$",
  "Solution_7": "You see what I'm doing? Writing $ 180$ instead of $ 120$ in the denominator... Tragic.",
  "Solution_8": "i think the answer is $ \\frac{2}{5}$. \r\n114, 135, 144, 345 are the multiples of 3.\r\n113,115,134,145,344,445 are not. \r\n\r\nFor 1,1,4 ; we have $ \\frac{3!}{2!}\\frac{2}{6}\\frac{1}{5}\\frac{2}{4} \\equal{} \\frac{1}{10}$. \r\n\r\nFor 1,3,5 ; we have $ 3!\\frac{2}{6}\\frac{1}{5}\\frac{1}{4} \\equal{} \\frac{1}{10}$. \r\n\r\nFor 1,4,4 ; we have $ \\frac{3!}{2!}\\frac{2}{6}\\frac{1}{5}\\frac{2}{4} \\equal{} \\frac{1}{10}$. \r\n\r\nFor 3,4,5 ; we have $ 3!\\frac{2}{6}\\frac{1}{5}\\frac{1}{4} \\equal{} \\frac{1}{10}$. \r\n\r\nSum of four is $ \\frac{2}{5}$. \r\n\r\nAlso to check the result, let's calculate others.\r\n\r\nFor (1,1,3) , (1,1,5) , (3,4,4) , (4,4,5) ; we have $ \\frac{3!}{2!}\\frac{2}{6}\\frac{1}{5}\\frac{1}{4} \\equal{} \\frac{1}{20}$. Sum of them is $ \\frac{1}{5}$.\r\n\r\nFor (1,3,4) , (1,4,5) ; we have $ 3!\\frac{2}{6}\\frac{2}{5}\\frac{1}{4} \\equal{} \\frac{1}{5}$. Sum of them is $ \\frac{2}{5}$.\r\n\r\nLast check is $ \\frac{2}{5} \\plus{} \\frac{1}{5} \\plus{} \\frac{2}{5} \\equal{} 1$."
}
{
  "Tag": [
    "trigonometry",
    "function"
  ],
  "Problem": "This isn't a typical AMC type problem but the idea is quite classic and useful if you haven't seen it before.\r\n\r\n(Me) \r\n\r\n$ \\displaystyle \\frac{1}{\\sqrt{2} \\plus{} \\sqrt{3}} \\plus{} \\frac{1}{\\sqrt{3} \\plus{} \\sqrt{4}} \\plus{} \\frac{1}{\\sqrt{4} \\plus{} \\sqrt{5}} \\plus{} \\frac{1}{\\sqrt{5} \\plus{} \\sqrt{6}} \\equal{} 2 \\sin k (\\tan m \\minus{} \\tan k)$\r\n\r\nIf $ 0 < k,m < \\frac{\\pi}{2},$ then find $ k\\plus{}m$.\r\n\r\n$ \\displaystyle \\text{(A)} \\ \\frac{5\\pi}{12} \\qquad \\ \\text{(B)} \\ \\frac{\\pi}{2} \\qquad \\text{(C)} \\ \\frac{7\\pi}{12} \\qquad \\text{(D)} \\ \\frac{2\\pi}{3} \\qquad \\text{(E)} \\ \\frac{5\\pi}{6}$",
  "Solution_1": "[hide]Answer: $ (\\text{C}) \\frac{7\\pi}{12}$\n\nMultiply the numerator and denominator of each fraction by the conjugate (?) of the denominator ($ \\sqrt {2} \\minus{} \\sqrt {3}$, $ \\sqrt {3} \\minus{} \\sqrt {4}$, etc.)\n\nThis will result in denominator of $ \\minus{} 1$ for all of the fractions, which we may now add together. \nWe now find that $ \\frac {\\sqrt {6}}{2} \\minus{} \\frac {\\sqrt {2}}{2} \\equal{} \\sin{k}(\\tan{m} \\minus{} \\tan{k}) \\equal{} \\sin{k}\\tan{m} \\minus{} \\sin{k}\\tan{k}$\nFrom here I assumed that $ \\sin{k}\\tan{k} \\equal{} \\frac {\\sqrt {2}}{2}$ which gave me $ k \\equal{} \\frac {\\pi}{4}$. Substituting this value for $ k$ in $ \\sin{k}\\tan{m} \\equal{} \\frac {\\sqrt {6}}{2}$ gave me $ m \\equal{} \\frac {\\pi}{3}$\n\n$ k\\plus{}m\\equal{}\\frac{7\\pi}{12}$[/hide]",
  "Solution_2": "There is obviously not a unique answer to this...",
  "Solution_3": "[quote=\"Altheman\"]There is obviously not a unique answer to this...[/quote]\r\n\r\nI know, I realized it after I wrote the problem so I decided not to put on the contest but like, the way I did it was:\r\n\r\n[hide=\"Essentially same\"]\nAfter simplifying the fractions, it came down as $ \\sqrt{6} \\minus{} \\sqrt{2} \\equal{} \\sqrt{2} (\\sqrt{3} \\minus{} 1) \\equal{} 2 \\frac{\\sqrt{2}}{2} (\\sqrt{3} \\minus{} 1) \\equal{} 2 \\sin \\frac{\\pi}{4} (\\tan \\frac{\\pi}{3} \\minus{} \\tan \\frac{\\pi}{4})$ [/hide]\r\n\r\nHmm, I wonder if it's possible to find ALL the $ (k,m)$ pairs that satisfy this value at all? I mean, the numbers will have to be rounded but would there be a way to do this? Multiplying $ \\sin k$ only gives $ \\frac{\\sqrt{6} \\minus{} \\sqrt{2}}{2} \\equal{} \\frac{\\sin m \\sin k}{\\cos m} \\minus{} \\frac{\\sin^2 k}{\\cos k}$ which doesn't really help anything though...\r\n\r\nOf course, that was just a question though.",
  "Solution_4": "[quote=\"Silverfalcon\"]$ \\frac {\\sqrt {6} \\minus{} \\sqrt {2}}{2} \\equal{} \\frac {\\sin m \\sin k}{\\cos m} \\minus{} \\frac {\\sin^2 k}{\\cos k}$[/quote]\r\n\\[ \\equal{} 2\\sin{15^{\\circ}}\r\n\\]\r\n\r\n\r\nDunno if that's any help...",
  "Solution_5": "Wait, I don't get it.. but then again I'm not exactly creative...\r\n\r\nSo i simplified it down to $ \\frac{\\sqrt{6} \\minus{} \\sqrt{2}}{2} \\equal{} sin(tan(k)\\minus{}tan(m))$ \r\n\r\nbut... how do I make this...work?",
  "Solution_6": "You probably misread the question. The the argument of the sine function does not call the tangent function -- you have sin(stuff) times tan(stuff)"
}
{
  "Tag": [
    "group theory",
    "abstract algebra",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "In simple group A_5, what is number of subgroups of order 12?",
  "Solution_1": "There are 5 - namely, the copies of $ A_4$ given by the permutations that fix $ k$, for $ 1\\leq k\\leq 5$.\r\n\r\nTo show there are no more...\r\n\r\n[hide]suppose $ H$ is a subgroup of order 12. $ H$ contains an element of order 3 (a 3-cycle). But $ H$ cannot contain two 3-cycles which are independent (that is, one not generating the other). For if $ H$ contained two such cycles, one of the following must hold about the sets of elements moved by the cycles:\n\n- They would intersect in 1 element. WLOG let the cycles be $ (123),(145)$. Then $ (123)(145)\\equal{}(14523)$, a 5-cycle, which could not be in $ H$.\n\n- They would intersect in 2 elements. WLOG let the cycles be $ (123),(124)$. These cycles generate all even permutations fixing 5, contrary to the assumption that $ H$ was not a copy of $ A_4$.\n\nSo $ H$ contains exactly one subgroup of 3-cycles. WLOG let this subgroup be $ <(123)>$. If $ K$ is the subgroup of $ A_5$ fixing 1, then we have\n\\[ |H\\cap K|\\equal{}\\frac{|H\\parallel{}K|}{|HK|}\\equal{}\\frac{12^2}{|HK|}\\geq \\frac{12^2}{60}\\equal{}\\frac{12}{5}>2\\]\nBut also $ H\\cap K$ is a subgroup of both $ H$ and $ K$, so cannot contain a 3-cycle. Our only option is that $ H\\cap K$ is the subgroup $ <(23)(45),(24)(35)>$ of $ K$. In particular, $ (24)(35)\\in H$. We find that $ H$ contains $ (123)(24)(35)\\equal{}(12435)$ which is a 5-cycle, a contradiction. So $ H$ must be one of the 5 copies of $ A_4$.[/hide]\r\n\r\nThere's probably a shorter/easier way to do this..."
}
{
  "Tag": [],
  "Problem": "What is the product of the numbers $ 1,2,3,4,5,6,7,8,9,$ and 0?",
  "Solution_1": "This is $ 0$ because $ 0$ times anything is $ \\boxed{0}$."
}
{
  "Tag": [
    "logarithms"
  ],
  "Problem": "$\\log_{3x+4}(4x^{2}+4x+1)+\\log_{2x+1}(6x^{2}+11x+4)=4$",
  "Solution_1": "Let $\\log_{3x+4}(2x+1) = a$ and $\\log_{2x+1}(3x+4) = b$. We then have $ab = 1$ and $4 = \\log_{3x+4}((2x+1)^{2})+\\log_{2x+1}((2x+1)(3x+4)) = 2a+1+b$, so $3 = 2a+b$, i.e. $b^{2}-3b+2 = 0$. So we have $b = 1$ or $b = 2$.\r\n\r\nWhen $b = 1$, we obtain $\\log_{2x+1}(3x+4) = 1$, i.e. $3x+4 = 2x+1$. But then $x =-3$, which means that $3x+4$ and $2x+1$ or both negative, which makes both logarithms undefined. Therefore, no solution when $b = 1$.\r\n\r\nWhen $b = 2$, that means $\\log_{2x+1}(3x+4) = 2$, i.e. $(2x+1)^{2}= 3x+4$, so $0 = 4x^{2}+x-3 = (x+1)(4x-3)$. For $x =-1$ we have the same undefined logarithms, so the only solution is $x = \\frac{3}{4}$.",
  "Solution_2": "[quote=\"Kurt G\u00f6del\"] = , i.e. $ b^{2} \\minus{} 3b \\plus{} 2 \\equal{} 0$. [/quote]\r\n\r\nHow did you derive  that  :huh:",
  "Solution_3": "[quote=\"ashrafmod\"]$ \\log_{3x \\plus{} 4}(4x^{2} \\plus{} 4x \\plus{} 1) \\plus{} \\log_{2x \\plus{} 1}(6x^{2} \\plus{} 11x \\plus{} 4) \\equal{} 4$[/quote] \r\n$ \\Leftrightarrow \\log_{3x \\plus{} 4}(2x \\plus{} 1)^2 \\plus{} \\log_{2x \\plus{} 1}((3x \\plus{} 4)(2x \\plus{} 1)) \\equal{} 4$\r\n$ \\Leftrightarrow 2\\log_{3x \\plus{} 4}(2x \\plus{} 1) \\plus{} \\log_{2x \\plus{} 1}(3x \\plus{} 4) \\plus{} \\log_{2x \\plus{} 1}(2x \\plus{} 1) \\equal{} 4$\r\n$ \\Leftrightarrow 2\\log_{3x \\plus{} 4}(2x \\plus{} 1) \\plus{} \\log_{2x \\plus{} 1}(3x \\plus{} 4) \\equal{} 3$\r\nSet $ \\log_{3x \\plus{} 4}(2x \\plus{} 1) \\equal{} t$\r\nWe have:\r\n$ 2t \\plus{} \\frac {1}{t} \\equal{} 3$\r\nOr: $ (2t \\minus{} 1)(t \\minus{} 1) \\equal{} 0 \\Leftrightarrow t \\equal{} 1 V t \\equal{} \\frac {1}{2}$\r\n$ t \\equal{} 1: \\log_{3x \\plus{} 4}(2x \\plus{} 1) \\equal{} 1 \\Leftrightarrow x \\equal{} \\minus{} 3$,impossible\r\n$ t \\equal{} \\frac {1}{2}: \\log_{3x \\plus{} 4}(2x \\plus{} 1) \\equal{} \\frac {1}{2} \\Leftrightarrow (2x \\plus{} 1)^2 \\equal{} (3x \\plus{} 4)$\r\n$ \\Leftrightarrow 4x^2 \\plus{}x\\minus{}3 \\equal{} 0 \\Leftrightarrow (4x\\minus{}3)(x\\plus{}1)\\equal{}0 \\Leftrightarrow x\\equal{}\\frac{3}{4}$ \r\nThen equatiom has one root: $ x\\equal{}\\frac{3}{4}$\r\n\r\n__________\r\nEditted!",
  "Solution_4": "[quote=\"NguyenDungTN\"]$ t \\equal{} \\frac {1}{2}: \\log_{3x \\plus{} 4}(2x \\plus{} 1) \\equal{} \\frac {1}{2} \\Leftrightarrow 2x \\plus{} 1 \\equal{} (3x \\plus{} 4)^2$[/quote]\r\n\r\nNice work, but this step is wrong.  You also could have established the incorrectness of your solution by checking to see that the value K.G. gave is in fact a root.",
  "Solution_5": "[quote=\"queen\"][quote=\"Kurt G\u00f6del\"] = , i.e. $ b^{2} \\minus{} 3b \\plus{} 2 \\equal{} 0$. [/quote]\n\nHow did you derive  that  :huh:[/quote]\r\nFrom $ ab\\equal{}1$, we have $ a\\equal{}\\frac{1}{b}$.  Substituting this into $ 3\\equal{}2a\\plus{}b$ gives us $ b^2\\minus{}3b\\plus{}2\\equal{}0$ after multiplication by $ b$.",
  "Solution_6": "[quote=\"JBL\"][quote=\"NguyenDungTN\"]$ t \\equal{} \\frac {1}{2}: \\log_{3x \\plus{} 4}(2x \\plus{} 1) \\equal{} \\frac {1}{2} \\Leftrightarrow 2x \\plus{} 1 \\equal{} (3x \\plus{} 4)^2$[/quote]\n\nNice work, but this step is wrong.  You also could have established the incorrectness of your solution by checking to see that the value K.G. gave is in fact a root.[/quote]\r\nI'm sorry! I've editted my solutions!"
}
{
  "Tag": [
    "real analysis",
    "limit",
    "real analysis unsolved"
  ],
  "Problem": "Let $ limsup|a_n|^{\\frac{1}{n}}\\equal{}R$ where $ R\\in[0,\\infty]$.\r\n\r\nWhy $ limsup|na_n|^{\\frac{1}{n\\minus{}1}}\\equal{}limsup|na_n|^{\\frac{1}{n}}$??",
  "Solution_1": "$ \\lim \\frac{n}{n\\minus{}1} \\equal{} 1$, does that help?"
}
{
  "Tag": [
    "inequalities",
    "number theory",
    "prime numbers"
  ],
  "Problem": "Find all pairs of natural numbers $(n,k)$ for which \r\n\r\n$(n+1)^{k}-1 = n!$.",
  "Solution_1": "if $n>3$ and $n+1$ was composite then we would have $n+1 \\mid n!$ so we get that $n+1 \\mid 1$...\r\nso we have $n\\leq 3$ or $n+1$ is a prime number...\r\nso assume that $n+1$ is prime,then we have $n!=(n+1)^{k}-1=n[(n+1)^{k-1}+...+1]$\r\nso we get that $(n-1)!=(n+1)^{k-1}+...+1$ now if $n>4$ then its a composite number,thus we would have:\r\n$(n+1)^{k-1}+...+1 \\equiv k (\\bmod n)$ and also $n \\mid (n-1)!$ so we must have $n \\mid k$ so $k\\geq n$ but one can see that if $n\\geq 2$:\r\n$(n+1)^{k}\\geq (n+1)^{n}>(n-1)!+1$ so in this case we wouldnt have any solutions...\r\nso the remain case is $n\\leq 4$ which would give us $n=4,k=2$ and $n=2,k=1$...",
  "Solution_2": "[quote=\"BaBaK Ghalebi\"]if $n>3$ and $n+1$ was composite then we would have $n+1 \\mid n!$ so we get that $n+1 \\mid 1$.[/quote]\r\n\r\nCan you explain this further?  :)",
  "Solution_3": "maybe the folowing explanation would make it clear...\r\nwell if $n+1$ was composite then we would have $n+1=pq$ s.t $\\gcd (p,q)=1$ and $p,q>1$ so we get that $p,q<n+1$ so $p,q\\leq n$ so we get that:\r\n$i)p \\mid n!$\r\n$ii)q \\mid n!$\r\nnow from $i),ii)$ and $\\gcd (p,q)=1$ we get that $pq \\mid n!$ so $n+1 \\mid n!$...\r\nnow in the equation $(n+1)^{k}-1=n!$ we have $n+1 \\mid RHS$ so we get that $n+1 \\mid LHS$ so $n+1 \\mid 1$...",
  "Solution_4": "[quote=\"BaBaK Ghalebi\"]well if $n+1$ was composite then assume that $p^{\\alpha}\\mid n+1$ s.t $\\alpha>1$ [/quote]\r\n\r\nAlright; everything made sense except for why you can assume that $\\alpha >1$. It doesn't seem like this treats the case of something like, $n+1=6=2 \\cdot 3$. Then, you wouldn't have $2p \\leq p^{\\alpha-1}<n \\iff 4 \\leq 1 <5$. Unless I'm missing something, it doesn't look like you treat the cases where $n+1 = \\prod p_{i}$; only cases where $n+1$ is prime or is composed of primes with higher powers than $1$.",
  "Solution_5": "[quote=\"cincodemayo5590\"][quote=\"BaBaK Ghalebi\"]well if $n+1$ was composite then assume that $p^{\\alpha}\\mid n+1$ s.t $\\alpha>1$ [/quote]\n\nAlright; everything made sense except for why you can assume that $\\alpha >1$. It doesn't seem like this treats the case of something like, $n+1=6=2 \\cdot 3$. Then, you wouldn't have $2p \\leq p^{\\alpha-1}<n \\iff 4 \\leq 1 <5$. Unless I'm missing something, it doesn't look like you treat the cases where $n+1 = \\prod p_{i}$; only cases where $n+1$ is prime or is composed of primes with higher powers than $1$.[/quote]\r\nsorry it was a typo...\r\nI edited my post,I think its correct now...",
  "Solution_6": "[quote=\"BaBaK Ghalebi\"][quote=\"cincodemayo5590\"][quote=\"BaBaK Ghalebi\"]well if $n+1$ was composite then assume that $p^{\\alpha}\\mid n+1$ s.t $\\alpha>1$ [/quote]\n\nAlright; everything made sense except for why you can assume that $\\alpha >1$. It doesn't seem like this treats the case of something like, $n+1=6=2 \\cdot 3$. Then, you wouldn't have $2p \\leq p^{\\alpha-1}<n \\iff 4 \\leq 1 <5$. Unless I'm missing something, it doesn't look like you treat the cases where $n+1 = \\prod p_{i}$; only cases where $n+1$ is prime or is composed of primes with higher powers than $1$.[/quote]\nsorry it was a typo...\nI edited my post,I think its correct now...[/quote]\r\n\r\nOK, that makes more sense now. Thanks. :)",
  "Solution_7": "Everything is fine until here\r\n[quote=\"BaBaK Ghalebi\"]now if $n>4$ then its a composite number,thus we would have:\n$(n+1)^{k-1}+...+1 \\equiv k (\\bmod n)$ and also $n \\mid (n-1)!$ so we must have $n \\mid k$ so $k\\geq n$ but one can see that if $n\\geq 2$:\n$(n+1)^{k}\\geq (n+1)^{n}>(n-1)!+1$ so in this case we wouldnt have any solutions...\nso the remain case is $n\\leq 4$ which would give us $n=4,k=2$ and $n=2,k=1$...[/quote]\r\nThen I think your english gets in the way of the rest of the proof, at least for me... :blush:",
  "Solution_8": "Well yeah, my proof is pretty much the same.\r\n\r\nFor $n>4$ it is clear that $n$ muc be even so let $n=2m$ and so $n!= (2m)! = (2m).(2m-1)!$ so it is clear that $n^{2}| n!$. So $n^{2}| (n+1)^{k}-1$. Using binomial expansion we see that if $n^{2}| (n+1)^{k}-1$ then $n|k$ and so $k \\geq n$.\r\n\r\nTherefore $(n+1)^{k}\\geq (n+1)^{n}> (n-1)!+1$. Now all we have to do is check for $n=1,2,3,4$.",
  "Solution_9": "[quote=\"me@home\"]Everything is fine until here\n[quote=\"BaBaK Ghalebi\"]now if $n>4$ then its a composite number,thus we would have:\n$(n+1)^{k-1}+...+1 \\equiv k (\\bmod n)$ and also $n \\mid (n-1)!$ so we must have $n \\mid k$ so $k\\geq n$ but one can see that if $n\\geq 2$:\n$(n+1)^{k}\\geq (n+1)^{n}>(n-1)!+1$ so in this case we wouldnt have any solutions...\nso the remain case is $n\\leq 4$ which would give us $n=4,k=2$ and $n=2,k=1$...[/quote]\nThen I think your english gets in the way of the rest of the proof, at least for me... :blush:[/quote]\r\nwell we assumed that $n+1$ is prime and also if $n>4$ then $n$ must be composite (because there doesnt exist two consecutive prime numbers greater than $4$) so $n$ is a composite number so $n \\mid (n-1)!$ now from the equation $(n-1)!=(n+1)^{k-1}+...+1$ we get that $n \\mid LHS$ so $n \\mid RHS$ but $RHS \\equiv 1+1+...+1 \\equiv k (\\bmod n)$ so we get that $n \\mid k$ so $n\\leq k$,but we know that the folowing inequality is true for $n\\geq 2$:\r\n$(n+1)^{k}\\geq (n+1)^{n}>(n-1)!+1$ contradicting the equality of the problem...\r\nso we have $n\\leq 4$ so we only need to check these cases...\r\nI hope its clear now...  :maybe:"
}
{
  "Tag": [
    "search",
    "advanced fields",
    "advanced fields unsolved"
  ],
  "Problem": "Can anyone versed in Physics help me with this problem?.  \r\n\r\n\"Two slabs, each one inch thick, each have one surface at 0 degrees and the other at 100 degrees. At t=0, they are stacked with their 100 degree faces together and then the outside surfaces are held at 100 degrees. Find u(x,t) for t>0\".\r\n\r\nCan someone help?.  Should this problem contain u_0 and u_f\r\n\r\nThanks,\r\nCody",
  "Solution_1": "i think you need $u(x,0)$. Then you just have to find a solution to the heat diffusion equation, which can be done with fourier expansion. Try to search solution of the form $u(x,t)=f(x)g(t)$, i think (not sure) it will help.",
  "Solution_2": "I think the initial condition is that the temperature varies linearly within each block from 0 at the outside edge to 100 at the inside edge- then we introduce a discontinuity by holding the outside edges at 100. The system will of course decay toward a uniform temperature of 100.",
  "Solution_3": "another pb: find the evolution of the heat on a ring .",
  "Solution_4": "Because there are two slabs, u_0 is piecewise.  One for x in (0,1) where the temperature goes from 0 to 100 and one for x in (1,2) where the temperature gores from 100 to 0.  This gives u_0-u_f=100x-100 and 100-100x."
}
{
  "Tag": [
    "logarithms",
    "function"
  ],
  "Problem": "Determinati solutiile sistemului:\r\n\r\n \r\n\r\n   2006^x=2005+y\r\n   2006^y=2005+z\r\n   2006^z=2005+x",
  "Solution_1": "$a>0\\ ,\\ x\\in R\\ : \\ f(x)=(a+1)^x\\ \\ (\\uparrow),\\ g(x)=x+a\\ \\ (\\uparrow)\\Longrightarrow$ \\[ \\boxed {\\ \\{ \\begin{array}{ccc} f(x) & = & g(y)\\\\ f(y) & = & g(z)\\\\ f(z) & = & g(x) \\end{array} \\ }  \\] [u]The first case[/u] : $\\boxed {\\ x\\le y\\ }\\Longrightarrow f(x)\\le f(y)\\Longrightarrow g(y)\\le g(z)\\Longrightarrow$\r\n\r\n$\\boxed {\\ y\\le z\\ }\\Longrightarrow f(y)\\le f(z)\\Longrightarrow g(z)\\le g(x)\\Longrightarrow \\boxed {\\ z\\le x\\ }\\Longrightarrow x=y=z\\ .$\r\n\r\n[u]The second case[/u] : $x>y\\Longrightarrow \\ldots \\Longrightarrow x>y>z>x\\Longrightarrow x>x$, what is false.\r\n\r\nTherefore, $\\boxed {\\ x=y=z\\ \\ ;\\ \\ (a+1)^x=x+a\\ }.$ Denote $b=a+1>1$. \r\n\r\n$b\\ln b<1\\Longrightarrow 0<x_1<x_2=1\\ ;\\ b\\ln b=1$$\\Longrightarrow x_0=1\\ ;\\ b\\ln b>1\\Longrightarrow 1=x_1<x_2\\ .$",
  "Solution_2": "Thank you for this solution. \r\n\r\nHere is my solution: If 2006^x - 2005=f(x), we have:\r\n\r\nf(x)=y\r\nf(y)=z\r\nf(z)=x.   \r\n So f(f(f(x)))=x.  But f is monotone, so f(x)=x (a known problem), and we have x=y=z.  We obtain that only (1,1,1) is solution.",
  "Solution_3": "In my solution, the function $f(x)=(a+1)^x-x-a$ isn't always monotonely; this point depends of the basis $a\\ !$"
}
{
  "Tag": [
    "calculus",
    "derivative",
    "function",
    "trigonometry",
    "topology",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Prove or (give a counterexample ) that for a derivative function there exists an interval $(a,b)$ such that $f$ is bounded on $(a,b)$.",
  "Solution_1": "[quote=\"xirti\"]Prove or (give a counterexample ) that for a derivative function there exists an interval $(a,b)$ such that $f$ is bounded on $(a,b)$.[/quote]\r\n$f(x)=\\sqrt{x}$\r\n$a=0,b=1$",
  "Solution_2": "$e^x$",
  "Solution_3": "Give a counterexample. not an example of function bounded on an interval. every continuous function is bounded on some intervals. I asked to prove that for a function there exists [b]an[/b] interval such that f is bounded on it. Or, if this statment is false you may give a [b]counter[/b]example (an example of derivative function unbounded an every interval).",
  "Solution_4": "I assume that the question is whether the derivative $f=g'$ of an [i]everywhere[/i] differentiable function $g$ can be unbounded on every open interval. Then the answer is \"no\". Indeed, $f$ is a pointwise limit of a sequence of continuous functions $f_n(x)=n\\left(g\\left(x+\\frac 1n\\right)-g(x)\\right)$. Let $F_N=\\{x\\in\\mathbb R\\,: \\, |f_n(x)-f_m(x)|\\le 1\\text{ for all }n,m>N\\}$. Each $F_N$ is closed and their union is the entire real line. Hence, by the Baire theorem, one of $F_N$ contains an open interval. The rest should be clear. Actually, a slight modification of this reasoning shows that $f$ must even have a lot of points of continuity.",
  "Solution_5": "Note carefully what fedja said: there must be [b]some[/b] interval on which $g'$ is bounded (and some interval on which $g'$ is continuous.) That certainly allows $g'$ to be discontinuous at some points and unbounded in some neighborhoods.\r\n\r\nWith that said, the example of $g(x) =\\sqrt{x}$ on $(0,1)$ is not really what we're looking for. For every $a\\in(0,1),$ there is a neighborhood of $a$ on which $g'$ is bounded. And it won't do to try to put this example on $[0,1)$ (or, for that matter, to use $\\sqrt[3]{x}$ on the whole line) because then the function will not be differentiable at zero.\r\n\r\nBut here's a member of the classic family of examples:\r\n\r\nLet $g(x)=\\begin{cases}x^2\\sin\\left(\\frac1{x^{10}}\\right)&x\\ne0\\\\0&x=0\\end{cases}.$\r\n\r\nThis function is differentiable everywhere (even at zero) but the derivative is not bounded in any neighborhood of zero. In fact, $g'$ is not (absolutely) integrable in any neighborhood of zero.",
  "Solution_6": "Great solution fedja. I was even trying to find an counterexample :D  I have a question: It is wrong if I work with \r\n$G_N=\\{x\\in \\mathbb{R}: |f_n(x)-f(x)|\\le 1 ,\\forall n\\ge N\\}$ instead $F_N$ on the same method? It seems more natural to me.",
  "Solution_7": "The problem with that approach: we show that $F_N$ is closed by using the fact that both $f_m$ and $f_n$ are continuous for all pairs $m,n$; $F_N$ is an intersection of closed sets.\r\n\r\nYou don't know that $f$ is continous, so your $G_N$ sets are not obviously closed.\r\n\r\nRandom notational fact: $F$ usually stands for a closed set, $G$ for an open set."
}
{
  "Tag": [
    "calculus",
    "integration",
    "calculus computations"
  ],
  "Problem": "For $ a>0$ evaluate\r\n\\[ \\int_{0}^{1}\\frac{1}{\\sqrt{1\\minus{}x^a}}\\;dx\\]",
  "Solution_1": "[hide]Performing the substitution $ x^a \\equal{} u$ we have \n\n$ I(a) \\equal{} \\int_0^1 \\frac{1}{\\sqrt{1 \\minus{} x^a}}\\,dx \\equal{} \\frac{1}{a} \\int_0^1 u^{\\frac{1}{a} \\minus{} 1} (1\\minus{}u)^{\\frac12 \\minus{}1}\\,du$\n\n$ \\equal{} \\frac{1}{a} B\\left(\\frac{1}{a}, \\frac12\\right) \\equal{} \\frac{1}{a} \\frac{\\Gamma \\left(\\frac{1}{a}\\right) \\Gamma \\left(\\frac12\\right)}{\\Gamma \\left(\\frac{1}{a} \\plus{} \\frac12\\right)} \\equal{} \\frac{\\sqrt{\\pi}}{a } \\frac{\\Gamma \\left(\\frac{1}{a}\\right)}{\\Gamma \\left(\\frac{1}{a} \\plus{} \\frac12\\right)}.$[/hide]"
}
{
  "Tag": [],
  "Problem": "Guess the age based on the discussion below...\r\n\r\n\r\nA: \"Write down my date of birth in the DDMMYYYY format.  My age this year multiplied by DD and MM gives you YYYY.\"\r\n \r\nB: \"That's not good enough.\"\r\n \r\nA: either DD or MM or both could be single-digit numbers or double-digit numbers.\r\n \r\nB: \"I'm still clueless.\"\r\n \r\nA: \"My son will join my birthday celebration this year.\"\r\n \r\nB: \"I'm still clueless.\"\r\n \r\nA: \"Both DD and MM are prime numbers.\" \r\n \r\nB: \"I got it!\"\r\n \r\nWhat is A's date of birth based on above dialogue?",
  "Solution_1": "$x\\implies\\textrm{year}$\r\n\r\n$y\\implies\\textrm{month}$\r\n\r\n$z\\implies\\textrm{day}$\r\n\r\nThen we have\r\n\r\n$\\left(2006-x\\right)yz=x\\implies x=\\frac{2006yz}{yz+1}$\r\n\r\nand $y\\in\\left\\{2,3,5,7,11\\right\\}$, $z\\in\\left\\{2,3,5,7,11,13,17,19,23,29,31\\right\\}$.  So now we have $2006=2\\cdot 17\\cdot 59$ so $yz=1,16,58$.  The only one that works here is $58=2\\cdot 29$, so the date of birth is $\\boxed{29/02/1972}$.",
  "Solution_2": "[quote=\"amcavoy\"]$x\\implies\\textrm{year}$\n\n$y\\implies\\textrm{month}$\n\n$z\\implies\\textrm{day}$\n\nThen we have\n\n$\\left(2006-x\\right)yz=x\\implies x=\\frac{2006yz}{yz+1}$\n\nand $y\\in\\left\\{2,3,5,7,11\\right\\}$, $z\\in\\left\\{2,3,5,7,11,13,17,19,23,29,31\\right\\}$.  So now we have $2006=2\\cdot 17\\cdot 59$ so $yz=1,16,58$.  The only one that works here is $58=2\\cdot 29$, so the date of birth is $\\boxed{29/02/1972}$.[/quote]\r\n\r\nhmm.... does this mean that the date of birth in this year is 29th Feb??\r\n\r\nbut 2006 is not a leap year?\r\n\r\nright? what are we missing...\r\n\r\nanyone...",
  "Solution_3": "February 29th, 1972 existed.  In fact, it was a Tuesday.  Remember that we aren't talking about 2006 here."
}
{
  "Tag": [],
  "Problem": "If the equation  $ ax^2\\plus{}bx\\plus{}c \\equal{} 0$ ,with $ a , b , c$ non zero digits, has only one integer root, find all triples $ (a,b,c)$",
  "Solution_1": "[hide=\"An attempt...\"]\nWe have $ a,b,c\\in \\{1,2,3,\\ldots,9\\}$, $ b^2 - 4ac = 0$, and $ \\dfrac{ - b}{2a}\\in\\mathbb{Z}$. To make $ \\frac { - b}{2a}$ an integer, we have $ b = 2na$, where $ n\\in\\mathbb{Z}$.\nNow we have $ (2na)^2 - 4ac = 0 \\iff 4n^2a^2 - 4ac = 0\\iff n^2a - c = 0\\iff n^2a = c$.\nSo $ (a,b,c) = (a,2na,n^2a).$\nIt is obvious from this that $ n$ is $ 1,2,$ or $ 3$.\nSo $ (a,b,c\\) = (a,2a,a), (a,4a,4a),$ or $ (a,6a,9a)$,\n$ (a,b,c) = (1,2,1), (2,4,2), (3,6,3), (4,8,4), (1,4,4), (2,2,8), (1,6,9)$.\n[/hide]\r\nWas that right?\r\nEDIT: Never mind... it wasn't.",
  "Solution_2": "what about like (2x+1)(x+3)?",
  "Solution_3": "Oops, I read it to mean only one root, which happened to be an integer T_T. It seemed a bit easy...",
  "Solution_4": "[quote=\"Brut3Forc3\"][hide=\"An attempt...\"]\nWe have $ a,b,c\\in \\{1,2,3,\\ldots,9\\}$, $ b^2 - 4ac = 0$, and $ \\dfrac{ - b}{2a}\\in\\mathbb{Z}$. To make $ \\frac { - b}{2a}$ an integer, we have $ b = 2na$, where $ n\\in\\mathbb{Z}$.\nNow we have $ (2na)^2 - 4ac = 0 \\iff 4n^2a^2 - 4ac = 0\\iff n^2a - c = 0\\iff n^2a = c$.\nSo $ (a,b,c) = (a,2na,n^2a).$\nIt is obvious from this that $ n$ is $ 1,2,$ or $ 3$.\nSo $ (a,b,c\\) = (a,2a,a), (a,4a,4a),$ or $ (a,6a,9a)$,\n$ (a,b,c) = (1,2,1), (2,4,2), (3,6,3), (4,8,4), (1,4,4), (2,2,8), (1,6,9)$.\n[/hide]\nWas that right?\nEDIT: Never mind... it wasn't.[/quote]\r\n\r\n[color=red][i]There was a typo in the set of solutions :[/i][/color]\r\n\r\n\r\n$ (a,b,c) = (1,2,1), (2,4,2), (3,6,3), (4,8,4), (1,4,4), (2,8,8), (1,6,9)$."
}
{
  "Tag": [
    "induction"
  ],
  "Problem": "Prove there there exists positive integers $ a_1, a_2,..., a_ n$ such that $ |a_i\\minus{}a_j|$ divides $ a_i\\plus{}a_j$ for all distinct $ i$ and $ j$.",
  "Solution_1": "so you mean for all n?\r\nit is easy to prove if say n=2 or 3",
  "Solution_2": "[quote=\"stevenmeow\"]so you mean for all n?\nit is easy to prove if say n=2 or 3[/quote]\r\n\r\nyes, using induction"
}
{
  "Tag": [
    "integration",
    "algebra",
    "function",
    "domain",
    "calculus",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Suppose that $ f,g \\in L^p([0,1])$ for $ 1<p$ finite. Assume that for all $ y \\in (0,\\infty)$\r\n\r\n$ |\\left\\{g(x) > y\\right\\}| \\leq \\frac{1}{y} \\int_{g(x) > y}f(x)dx$\r\n\r\nShow that $ |g|_p \\leq q |f|_p$, where $ q$ is the conjugate of p.",
  "Solution_1": "There seems to be a tacit assumption here that both $ f$ and $ g$ are nonnegative functions. I'll keep that assumption; without it, we'd have a number of absolute values in the statement but it would otherwise work the same way. It also doesn't matter at all that the domain is $ [0,1].$ The domain could be any measurable $ E,$ of finite or infinite measure. \r\n\r\nA piece of exponent arithmetic that we'll use later: $ \\frac{p}{p\\minus{}1}\\equal{}q.$\r\n\r\n$ \\|g\\|_p^p\\equal{}\\int_0^{\\infty}py^{p\\minus{}1}|\\{g(x)>y\\}|\\,dy \\le\\int_0^{\\infty}py^{p\\minus{}1}\\int_{\\{g(x)>y\\}}f(x)\\,dx\\,dy$\r\n\r\nOf course we use Tonelli. Why else write a double integral?\r\n\r\n$ \\equal{}\\int_Ef(x)\\int_0^{g(x)}py^{p\\minus{}2}\\,dy\\,dx\\equal{}\\frac{p}{p\\minus{}1}\\int_Ef(x)g(x)^{p\\minus{}1}\\,dx$\r\n\r\nNow we use H\u00f6lder. Note that $ (g(x)^{p\\minus{}1})^q\\equal{}g(x)^p.$\r\n\r\n$ \\le \\frac{p}{p\\minus{}1}\\|f\\|_p\\left(\\int_Eg^p\\right)^{\\frac{p\\minus{}1}p}\\equal{}q\\|f\\|_p\\|g\\|_p^{p\\minus{}1}.$\r\n\r\nIf $ \\|g\\|_p<\\infty$ (which we assumed), we can then divide $ \\|g\\|_p^{p\\minus{}1}$ to get $ \\|g\\|_p\\le q\\|f\\|_p.$"
}
{
  "Tag": [],
  "Problem": "What's a radical axis/radical center?",
  "Solution_1": "[url]http://mathworld.wolfram.com/RadicalLine.html[/url]\r\n\r\n[url]http://mathworld.wolfram.com/RadicalCenter.html[/url]"
}
{
  "Tag": [
    "linear algebra",
    "linear algebra unsolved"
  ],
  "Problem": "Let $A,B$ - real matrices $n \\times n$. Suppose that there exist pairwise distinct $t_1, ..., t_{n+1} \\in \\mathbb{R}$ such, that $A+t_iB$ are nilpotent matrices. Show that $A$ and $B$ are nilpotent.",
  "Solution_1": "I remember this problem was discussed many times on Mathlinks. Also it was on IMC 1995, you can find it from the http://www.imc-math.org.",
  "Solution_2": "Oh sorry, I know the solution, I was hoping there are more than one solution to this problem."
}
{
  "Tag": [
    "geometry",
    "incenter",
    "circumcircle",
    "geometry unsolved"
  ],
  "Problem": "Let $ I$ and $ O$ denote respectively the incenter and the circumcenter of a triangle $ ABC$. Given that $ \\hat{AIO} \\equal{} 90$. Prove that the area of the triangle $ ABC$ is less than $ \\frac {3\\sqrt {3}}{4}AI^2$",
  "Solution_1": "[url]http://www.mathlinks.ro/viewtopic.php?t=168836&search_id=79333260&start=20[/url]\r\nYou should know you was posting many problems of M&Y.\r\nLock this topic!"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "trigonometry",
    "Vieta",
    "trig identities"
  ],
  "Problem": "Let $ a < b < c$ and $ a,$ $ b$ and $ c$ are roots of the equation: $ x^3 \\minus{} 3x \\plus{} 1 \\equal{} 0.$ Prove that\r\n\\[ \\frac {a}{b \\plus{} 1} \\plus{} \\frac {b}{c \\plus{} 1} \\plus{} \\frac {c}{a \\plus{} 1} \\equal{} \\minus{} 3\r\n\\]",
  "Solution_1": "[hide]Note from the coefficient of $ x^{2}$ that $ a \\plus{} b \\plus{} c \\equal{} 0$. Then $ a\\plus{}c \\equal{} \\minus{}b$, $ b\\plus{}c\\equal{}\\minus{}a$, $ a\\plus{}b\\equal{}\\minus{}c$, so it suffices to show that:\n\n\\[ \\frac{1}{a\\plus{}1} \\plus{} \\frac{1}{b\\plus{}1} \\plus{} \\frac{1}{c\\plus{}1} \\equal{} 0\\]\n\nNote that $ P(x\\minus{}1) \\equal{} (x\\minus{}1)^{3} \\minus{} 3(x\\minus{}1) \\plus{} 1$ has roots $ a\\plus{}1, b\\plus{}1, c\\plus{}1$. So the polynomial with reverse coefficients has the roots $ 1/(a\\plus{}1), 1/(b\\plus{}1), 1/(c\\plus{}1)$. So we must show that the $ x^{2}$ coefficient of the polynomial with reverse coefficents is $ 0$; the $ x$ coefficient of $ P(x\\minus{}1)$ is $ 0$. This is easy: The $ x$ coefficient is simply $ 3\\minus{}3\\equal{}0$ (from binomial expansion), as desired.[/hide]",
  "Solution_2": "[quote=\"vishalarul\"][hide]Note from the coefficient of $ x^{2}$ that $ a \\plus{} b \\plus{} c \\equal{} 0$. Then $ a \\plus{} c \\equal{} \\minus{} b$, $ b \\plus{} c \\equal{} \\minus{} a$, $ a \\plus{} b \\equal{} \\minus{} c$, so it suffices to show that:\n\\[ \\frac {1}{a \\plus{} 1} \\plus{} \\frac {1}{b \\plus{} 1} \\plus{} \\frac {1}{c \\plus{} 1} \\equal{} 0\n\\][/hide][/quote]\r\n\r\nHow do you get that?  I can't figure out how to get from the given equation to this equation.",
  "Solution_3": "Yeah, i was wondering that too.\r\n\r\nAll I get is (2-c)/(b+1)+(2-b)/(a+1)+(2-a)/(b+1)=0.",
  "Solution_4": "Sorry, I messed up (and I can't edit that post anymore...)\r\n\r\n[hide=\"Ideas\"]It turns out that indeed $ \\frac {1}{a \\plus{} 1} \\plus{} \\frac {1}{b \\plus{} 1} \\plus{} \\frac {1}{c \\plus{} 1} \\equal{} 0$. I thought we had to prove that\n\n$ \\frac {a \\plus{} c}{b \\plus{} 1} \\plus{} \\frac {a \\plus{} b}{c \\plus{} 1} \\plus{} \\frac {b \\plus{} c}{a \\plus{} 1} \\equal{} \\minus{} 3$\n\n(which should be true, since $ a \\plus{} b \\plus{} c \\equal{} 0$)\n\nHowever, this is indirectly helpful. It remains to show that\n\n$ \\frac {c}{b \\plus{} 1} \\plus{} \\frac {a}{c \\plus{} 1} \\plus{} \\frac {b}{a \\plus{} 1} \\equal{} 0$\n\nBut I bet that's just as hard as the original problem. Well, if we expand this, we need to show that \n\n$ a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} ab \\plus{} bc \\plus{} ca \\plus{} a^2 b \\plus{} b^2 c \\plus{} c^2 a \\equal{} 0$\n\nSince \n\n$ 0 \\equal{} (a \\plus{} b \\plus{} c)^2 \\equal{} a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} ab \\plus{} bc \\plus{} ca \\plus{} (ab \\plus{} bc \\plus{} ca) \\equal{} a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} ab \\plus{} bc \\plus{} ca \\minus{} 3$\n\nwe need to show that $ a^2 b \\plus{} b^2 c \\plus{} c^2 a \\equal{} \\minus{} 3$. If we look at $ (a \\plus{} b \\plus{} c)^3 \\equal{} 0$ and expand,\n\n$ a^3 \\plus{} b^3 \\plus{} c^3 \\plus{} 3(a^2 b \\plus{} b^2 c \\plus{} c^2 a \\plus{} a b^2 \\plus{} b c^2 \\plus{} c a^2) \\plus{} 6abc \\equal{} 0$\n\nFrom $ a^3 \\plus{} b^3 \\plus{} c^3 \\minus{} 3abc \\equal{} (a \\plus{} b \\plus{} c)(STUFF) \\equal{} 0$, we get $ a^3 \\plus{} b^3 \\plus{} c^3 \\equal{} 3abc \\equal{} \\minus{} 3$,  so we need to show that:\n\n$ a b^2 \\plus{} b c^2 \\plus{} c a^2 \\equal{} 6$\n\nLet $ X \\equal{} a^2 b \\plus{} b^2 c \\plus{} c^2 a$ and $ Y \\equal{} a b^2 \\plus{} b c^2 \\plus{} c a^2$. Notice that:\n\n$ X \\plus{} Y \\equal{} ab(a \\plus{} b) \\plus{} bc(b \\plus{} c) \\plus{} ca(c \\plus{} a) \\equal{} \\minus{} 3abc \\equal{} 3$\n\nAnd now, I'm stuck... I'm not sure how to incorporate $ a < b < c$ into this...\n\nEDIT: $ Y \\minus{} X \\equal{} (b \\minus{} a)(c \\minus{} b)(c \\minus{} a)$, but that doesn't really help...\n\nEDIT2: Here's some actual progress. $ \\minus{}X\\equal{}a/b\\plus{}b/c\\plus{}c/a$ and $ \\minus{}Y\\equal{}b/a\\plus{}c/b\\plus{}a/c$. Mutiplying these two together and expanding,\n\n$ XY \\equal{} 3\\plus{}(ac/b^2\\plus{}ab/c^2\\plus{}bc/a^2) \\plus{} (b^2/ac\\plus{}c^2/ab\\plus{}a^2/bc)$\n\nThe second sum, $ (b^2/ac\\plus{}c^2/ab\\plus{}a^2/bc)$ is easy: put a common denominator $ abc\\equal{}\\minus{}1$:\n\n$ (b^2/ac\\plus{}c^2/ab\\plus{}a^2/bc) \\equal{} \\minus{}(a^3\\plus{}b^3\\plus{}c^3) \\equal{} 3$\n\nThe first sum is more annoying. We put a common denominator, and we wish to find $ a^3b^3\\plus{}b^3c^3\\plus{}c^3a^3$. Noting that this equals $ [(a^3\\plus{}b^3\\plus{}c^3)^2\\minus{}(a^6\\plus{}b^6\\plus{}c^6)]/2 \\equal{} \\minus{}24$ (well, at least it should: a formal proof requires newton's sums, which aren't that hard to compute). So FINALLY, $ XY \\equal{} 3 \\plus{} 3 \\minus{} 24 \\equal{} \\minus{}18$\n\nSo $ (X,Y)\\equal{}(\\minus{}3,6)$ or $ (6,\\minus{}3)$. To figure out which one we use $ Y\\minus{}X \\equal{} (b\\minus{}a)(c\\minus{}b)(c\\minus{}a)>0$ (it's not that hard to prove; just plug in and factor) since $ a<b<c$.\n\nSo $ X<Y$, implying that $ X \\equal{} \\minus{}3$. Which means we're done.[/hide]\r\n\r\nThis problem is a lot harder than I thought. :blush: (I really, really, hope I didn't make any mistakes...)",
  "Solution_5": "Hints\r\n\r\nUse Trig Substitutions to explicitly calculate a,b, and c.\r\n\r\nThen simplify that expression to prove to prove something else, and then use trig identities and Vieta's sums to get to the core.",
  "Solution_6": "$ a\\plus{}b\\plus{}c\\equal{}0$ is due to the fact that a cubic polynomial $ a_3x^3\\plus{}a_2x^2\\plus{}a_1x\\plus{}a_0$ is equal to $ a_3[x^3\\minus{}(r_1\\plus{}r_2\\plus{}r_3)x^2\\plus{}(r_1r_2\\plus{}r_1r_3\\plus{}r_2r_3)x\\minus{}r_1r_2r_3]$, where $ r_1$, $ r_2$, and $ r_3$ are the roots of the polynomial. In this case $ a$, $ b$, and $ c$ are the roots of the polynomial. Therefore, vishalarul is correct.",
  "Solution_7": "[hide=\"Vieta + Newton, almost there\"]First define $ P(x) = x^3 - 3x + 1$. $ Q(x) = P(x - 1) = x^3 - 3x^2 + 3$ has roots $ a + 1$, $ b + 1$, $ c + 1$. $ R(x) = x^3Q(1/x) = 3x^3 - 3x + 1$ has roots $ \\frac {1}{a + 1}$, $ \\frac {1}{b + 1}$, $ \\frac {1}{c + 1}$. Since its second coefficient is $ 0$,\n\\[ \\frac {1}{a + 1} + \\frac {1}{b + 1} + \\frac {1}{c + 1} = 0\n\\]\nDefine\n\\[ X = \\frac {c}{a + 1} + \\frac {a}{b + 1} + \\frac {b}{c + 1} \\\\\nY = \\frac {b}{a + 1} + \\frac {c}{b + 1} + \\frac {a}{c + 1}\n\\]\nWe have:\n\\begin{eqnarray*} X + Y & = & \\frac {b + c}{a + 1} + \\frac {a + c}{b + 1} + \\frac {a + b}{c + 1} \\\\\n& = & \\frac { - a - 1}{a + 1} + \\frac { - b - 1}{b + 1} + \\frac { - c - 1}{c + 1} \\\\\n& = & - 3 \\end{eqnarray*}\nDefine $ q_n$ to be the sum of the $ n$th powers of the roots of $ Q(x)$ and $ r_n$ to be the same for $ R(x)$. By Newton's sums:\n\\[ q_2 = - 3q_1 = 9 \\\\\nq_3 - 3q_2 + 9 = 0 \\\\\nq_3 = 18 \\\\\n3r_3 - 3r_1 + 3 = 0 \\\\\nr_3 = - 1\n\\]\nMultiplying $ X$ and $ Y$:\n\\begin{eqnarray*} XY & = & 3 + \\frac {(a + 1)^2}{(b + 1)(c + 1)} + \\frac {(b + 1)^2}{(a + 1)(c + 1)} + \\frac {(c + 1)^2}{(a + 1)(b + 1)} \\\\\n& & + \\frac {(a + 1)(b + 1)}{(c + 1)^2} + \\frac {(a + 1)(c + 1)}{(b + 1)^2} + \\frac {(b + 1)(c + 1)}{(a + 1)^2} \\\\\n& = & 3 + \\frac {q_3}{ - 3} - 3r_3 \\\\\n& = & 3 - 6 + 3 \\\\\n& = & 0 \\end{eqnarray*}\nNow either $ X = 0$ and $ Y = - 3$ or $ X = - 3$ and $ Y = 0$. We just have to prove now that $ X < Y$.\n[/hide]",
  "Solution_8": "[hide=\"Trig Method\"]Expanding,\n\n$ \\sum_{cyc} a(a\\plus{}1)(c\\plus{}1) \\equal{} \\minus{}3(a\\plus{}1)(b\\plus{}1)(c\\plus{}1)$\n$ \\iff$\n$ \\sum_{cyc} a^{2}c \\plus{} ac \\plus{} a^{2} \\plus{} a \\equal{} \\minus{}3(abc \\plus{} ab \\plus{} bc \\plus{} ca \\plus{} a \\plus{} b \\plus{} c \\plus{} 1)$\n$ \\iff$\n$ \\sum_{cyc} a^{2} c \\plus{} (a\\plus{}b\\plus{}c)^{2} \\minus{} (ab\\plus{}bc\\plus{}ca) \\plus{} (a\\plus{}b\\plus{}c) \\equal{} \\minus{}3 (abc\\plus{}ab \\plus{} bc \\plus{} ca \\plus{} a \\plus{} b \\plus{} c \\plus{} 1)$\n$ \\iff \\text{Vieta}$\n$ \\sum_{cyc} a^{2} c \\equal{} \\minus{}3 \\minus{} 3(\\minus{}1\\minus{}3\\plus{}1) \\equal{} 6$\n$ \\iff$\n$ \\sum_{cyc} a^{2} c/abc \\equal{} \\minus{}6$\n$ \\iff$\n$ \\sum_{cyc} a/b \\equal{} \\minus{}6$\n\nSince the roots of $ x^{3} \\minus{} 3x \\plus{} 1 \\equal{} 0$ are $ a \\equal{} 2\\cos 160^{\\circ}$, $ b\\equal{}2\\cos 80^{\\circ}$, $ c\\equal{}2\\cos 40^{\\circ}$, we need to show:\n$ \\cos 160^{\\circ}/\\cos 80^{\\circ} \\plus{} \\cos 80^{\\circ}/\\cos 40^{\\circ} \\plus{} \\cos 40^{\\circ}/\\cos 160^{\\circ} \\equal{} \\minus{}6$\n\nWe use double angle formula for cosine:\n\n$ 2\\cos 80^{\\circ} \\minus{} 1/\\cos 80^{\\circ} \\plus{} 2\\cos 40^{\\circ} \\minus{} 1/\\cos 40^{\\circ} \\minus{} 2\\cos 20^{\\circ} \\plus{} 1/\\cos 20^{\\circ} \\equal{} \\minus{}6$\n$ \\iff$\n$ 2\\cos 80^{\\circ} \\minus{} 1/\\cos 80^{\\circ} \\plus{} 2\\cos 40^{\\circ} \\minus{} 1/\\cos 40^{\\circ} \\plus{} 2\\cos 160^{\\circ} \\minus{} 1/\\cos 160^{\\circ} \\equal{} \\minus{}6$\nThe sum of the roots is $ 0$, so we need to show that twice the sum of the reciprocal of the roots is $ 6$. ($ 1/\\cos 40^{\\circ} \\plus{} 1/\\cos 80^{\\circ} \\plus{} 1/\\cos 160^{\\circ} \\equal{} 6$). But this is clearly true: the polynomial with negative roots is the polynomial with reverse coefficients: $ x^{3} \\minus{} 3x^{2} \\plus{} 1$. The sum of the roots of this polynomial is $ 3$, so twice the sum of the roots (i.e. $ 1/\\cos 40^{\\circ} \\plus{} 1/\\cos 80^{\\circ} \\plus{} 1/\\cos 160^{\\circ}$) is indeed $ 6$.[/hide]\nThat was pretty long. I wonder if there's a shorter way to do it.\n[hide=\"And is there a way to prove that...\"]the roots of $ x^{3} \\minus{} 3x \\plus{} 1$ are $ \\cos 40^{\\circ}$, $ \\cos 80^{\\circ}$, and $ \\cos 160^{\\circ}$?[/hide]\r\nwhich was the key fact used in the proof?",
  "Solution_9": "[quote=\"vishalarul\"]the roots of $ x^{3} \\minus{} 3x \\plus{} 1$ are $ \\cos 40^{\\circ}$, $ \\cos 80^{\\circ}$, and $ \\cos 160^{\\circ}$?\nwhich was the key fact used in the proof?[/quote]\r\nNo! With them this is nicer. For example:\r\nLet $ a < b < c$ and  $ a,$ $ b$ and $ c$ are roots of the equation: $ x^3 \\minus{} 6x \\plus{} 1 \\equal{} 0.$  Prove that\r\n\\[ \\frac {a}{b \\plus{} 1} \\plus{} \\frac {b}{c \\plus{} 1} \\plus{} \\frac {c}{a \\plus{} 1} \\equal{} \\minus{} \\frac {5 \\plus{} \\sqrt {93}}{4}\r\n\\]",
  "Solution_10": "That is so random, but there must be some nice way of getting it."
}
{
  "Tag": [
    "probability"
  ],
  "Problem": "[size=150]Next Question[/size]\r\n\r\nSuppose that we are to be presented with $n$ distinct prizes in sequence.  After being presented with a prize we must immediately decide whether to accept it or reject it and consider the next prize.  The only information we are given when deciding whether to accept a prize is the relative rank of that prize compared to ones already seen.  Suppose that once a prize is rejected it is lost, and that our objective is to maximize the probability of obtaining the best prize.  Assuming that all $n!$ orderings of the prizes are equally likely, what is the probability that this strategy will select the best prize?",
  "Solution_1": "Why did you post the same question twice?",
  "Solution_2": "sry, you can delete this one....I had clicked the wrong link :oops:"
}
{
  "Tag": [
    "AMC",
    "AIME",
    "USA(J)MO",
    "USAMO"
  ],
  "Problem": "I was wondering if we could null the AIME A score and intead count the AIME B score to be official? Also, is it legit to have skipped AIME A to take AIME B (@ another testing location)?",
  "Solution_1": "I know that you cannot take the AIME A and then say I want to not count that and take the AIME B for sure... I don't know about the skipping thing, but it sounds reasonable.",
  "Solution_2": "AIME & USAMO Registration\r\n\r\nThere is no additional registration fee for the American Invitational Mathematics Examination, unless you choose to take the second sitting of the exam. An additional administrative/shipping fee is charged, with a minimum of $25.00 for the first 10 students and $50.00 for over 10 students. This is explained more thoroughly in the AMC 10/12 and AIME Teacher's Manual.\r\n\r\nShouldn't this apply, such that I could register for the B?",
  "Solution_3": "[quote=\"spang\"]Shouldn't this apply, such that I could register for the B?[/quote]\r\n\r\nNot if you already took the first AIME this year, as explained in the [url=http://www.unl.edu/amc/d-publication/d1-pubarchive/2003-4pub/04AIME/html/04-04aime-text.html]AIME manual[/url]: [b]Under no circumstances can a student take both AIME's.[/b] So, no, no one can annull an AIME score from today's test just to take the test on the second date.",
  "Solution_4": "[quote=\"spang\"]I was wondering if we could null the AIME A score and intead count the AIME B score to be official? Also, is it legit to have skipped AIME A to take AIME B (@ another testing location)?[/quote]\r\n\r\num, you can skip the AIME a to take the AIME b, but in general the B is harder and it's not considered very legit.",
  "Solution_5": "Well umm...this isn't very uplifting to someone here who's taking the AIME II.  LoL., wish me luck.",
  "Solution_6": "It doesn't matter which is harder, because for the USAMO they take a number of people from each test proportional to the number who took each one. So, if the B is a lot harder, they'll still take the top 1% on the B.",
  "Solution_7": "[quote=\"WarpedKlown1335\"]Well umm...this isn't very uplifting to someone here who's taking the AIME II.  LoL., wish me luck.[/quote]\r\nheh, sorry.  but you didn't skip the first one on purpose, right?  good luck!  :)",
  "Solution_8": "THANK YOU!  :-D :-D",
  "Solution_9": "[quote=\"z9R4C3\"][quote=\"WarpedKlown1335\"]Well umm...this isn't very uplifting to someone here who's taking the AIME II.  LoL., wish me luck.[/quote]\nheh, sorry.  but you didn't skip the first one on purpose, right?  good luck!  :)[/quote]\r\n\r\nheck...there's a lot of us that'll need good luck for this test!!",
  "Solution_10": "The first AIME would have been during my spring break so I couldn't take it.",
  "Solution_11": "[quote]It doesn't matter which is harder, because for the USAMO they take a number of people from each test proportional to the number who took each one. So, if the B is a lot harder, they'll still take the top 1% on the B.[/quote]\r\n\r\nCould someone knowledgable comment more on this, because as far as I know, the USAMO cutoff index has remained constant for both AIME 1 and AIME 2 takers in the past few years (ie the cutoff was 226 for everyone last year).",
  "Solution_12": "The proportional thing doesn't happen.",
  "Solution_13": "Really? So what do they do when one test is obviously easier than another?",
  "Solution_14": "Nothing at all.",
  "Solution_15": "I presume that if a serious discrepancy between the tests should present itself, they would do something about it.  However, that seems rather improbable, especially given that the average score still hovers consistently at a 3 or below, right?"
}
{
  "Tag": [
    "geometry",
    "geometric transformation",
    "reflection"
  ],
  "Problem": "Suppose we have an equilateral triangle with vertices at points $(0, 0), \\ (10, 0), \\text{ and }(5, 5\\sqrt{3})$. If a point $(a,b)$ is chosen within the triangle, what is the shortest distance that the point can travel so that it touches all three sides? (Touching a vertex is the same as touching 2 of the sides.) Is there a \"best\" point that minimizes this distance? What if the point is outside of the triangle?",
  "Solution_1": "[hide=\" :maybe:.. \"]\nJust a guess..but can you start at the vertex and make your way down to the opposite side thus touching all three sides and its $5\\sqrt{3}$\n[/hide]",
  "Solution_2": "[hide]Something tells me it is just the altitude of $5\\sqrt{3}$[/hide]",
  "Solution_3": "That's what I'm thinking also. Can anyone prove it? \r\n\r\nWhat about the general case?",
  "Solution_4": "I think the general is that you take the vertex opposite the largest side and draw the altitude and that is the answer.",
  "Solution_5": "heres an attempted proof that the altitude is the minimal distance, but its not the only path with that length\r\n\r\nClearly we must start on a side, otherwise would just be stupid.  Simply reflect the triangle over the edge other than the one you are startng at, particularly the one your starting point is closer to, so then go from whatever point you start on to the opposite edge of the reflected triangle, the minimal length of this is the perpendicular of 5sqrt(3).  Then relfect the part of the path on the reflected triangle back on to the original, and it has the same distance."
}
{
  "Tag": [
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "im really confused over this problem.\r\nCan anybody please help or advise please?\r\n\r\n=========================\r\nMark each statement true or false\r\n1. { x | x = A } = {A}\r\n2. { x | x $\\in$ A } = A\r\n3. { x | {x} $\\subset$ A } = {A}\r\n4. { x | {x} $\\subset$ $\\varnothing$ } = $\\varnothing$\r\n5. { x | {x} = A } = {A}\r\n6. { x | {x} $\\in$ A } = A\r\n=========================\r\n\r\nThank you very much for your advise ^^!",
  "Solution_1": "true: 1,2,4\r\nfalse : 3,5,6 \r\n\r\n:D",
  "Solution_2": "how do you know???\r\ncould anyone give me some suggestions????",
  "Solution_3": "First, we don't solve homework problems here (at least, not immediate homework problems), secondly, try posting it in some high-school or pre-olympiad forum. As you can see, this is the advanced forum, so probably nobody will bother to answer you.",
  "Solution_4": "[quote=\"Ointer\"]how do you know???\ncould anyone give me some suggestions????[/quote]\r\n\r\nLet sets $A$ and $B$. Do you know defined of $A=B$ ?  :huh:"
}
{
  "Tag": [
    "function",
    "algebra proposed",
    "algebra"
  ],
  "Problem": "Find all functions f:$ R \\to R$ such that\r\n$ f(x^3\\minus{}y)\\plus{}2y(3f^2(x)\\plus{}y^2)\\equal{}f(y\\plus{}f(x))$ for all $ x;y \\in R$",
  "Solution_1": "$ P(x,0): f(x^3)\\equal{}f(f(x))$----------------(1)\r\nNow we prove $ f(x)$ is injective. Suppose it is not true and $ f(a)\\equal{}f(b)$ for $ a\\neq b$.\r\n$ P(a,y)\\minus{}P(b,y): f(a^3\\minus{}y)\\equal{}f(b^3\\minus{}y)$. So $ f(x)$ is periodic with period $ a\\minus{}b$.\r\n$ P(x,a\\minus{}b): f(x^3)\\plus{}6(a\\minus{}b)f(x)^2\\plus{}2(a\\minus{}b)^3\\equal{}f(f(x))\\equal{}f(x^3)$. This implies\r\n$ 3f(x)^2\\plus{}(a\\minus{}b)^2\\equal{}0$, a contradiction with $ f(x)\\in R$.\r\nNow formula (1) implies $ f(x)\\equal{}x^3$.",
  "Solution_2": "$ y: \\equal{} \\frac {x^3 \\minus{} f(x)}{2}$\r\n\r\n[b]Done[/b]\r\n\r\n :!:  :w00t:  :P\r\n\r\n[hide=\"BTW\"]I Hope you meant $ f^2(x)\\equal{}(f(x))^2$ and not $ f(f(x))$. [/hide]",
  "Solution_3": "Nice solution."
}
{
  "Tag": [
    "analytic geometry",
    "geometry",
    "parallelogram",
    "trigonometry",
    "absolute value",
    "trig identities",
    "Law of Cosines"
  ],
  "Problem": "Given a triangle $ ABC$, let $ M$ be mid point of $ BC$ such that $ AB\\equal{}4,\\ AM\\equal{}1$. Find the possible least measure of angle $ \\angle{BAC}$.",
  "Solution_1": "[hide=\"my solution...\"]\nLet $ A$ be the center of a circle of raidus $ 1$. Now $ M$ is a point on that circle and the midpoint of $ BC$. \nNow we can see that $ \\angle BAC$ is minimised then $ \\overline{BC}$ is tangent to the cirlce\n\n$ \\Rightarrow Min: \\angle BAC \\equal{} 2\\arccos \\left(\\frac{1}{4}\\right) \\sim 151.045...$[/hide]",
  "Solution_2": "Regrettably your answer is incorrect.",
  "Solution_3": "[hide]\nLet's place triangle $ ABC$ in coordinate system such that $ A (0,0)$, $ B (4,0)$ and $ C$ lies in II. quadrant. Let $ M$ lie on circle $ x^2 + y^2 = 1 \\implies M (sin \\phi,cos \\phi)$. Let $ x_0$ and $ y_0$ be the coordinates of point $ C$. Since $ M$ is midpoint of $ BC$, $ sin \\phi = \\frac {x_0 + 4}{2}$ and $ cos\\phi = \\frac {y_0}{2} \\implies x_0 = 2(sin\\phi - 2), y_0 = 2cos\\phi$. Straight line $ AC$ has equation $ y = \\frac {y_0}{x_0}x \\implies y = \\frac {cos\\phi}{sin\\phi - 2}x$. The angle we are looking for is the angle which this line closes with positive x-axis, so $ tg\\alpha = k = \\frac {cos\\phi}{sin\\phi - 2}$. This angle will achieve minimum for $ |k|_{max}$ (because $ C$ is in II. quadrant). Let $ f(\\phi) = \\frac {cos\\phi}{sin\\phi - 2}$. Now let's find $ f'(\\phi)$ in order to find extremes of the function. Since $ f'(\\phi) = \\frac {2sin\\phi - 1}{(sin\\phi - 2)^2}$ which equals $ 0$ for $ sin\\phi = \\frac {1}{2} \\implies f(\\phi)_{min} = - \\frac {1}{\\sqrt {3}}$ (notice that maximum has same absolute value since $ f(\\phi)$ is a trigonometric function). Using $ f(\\phi) = tg\\alpha \\implies \\boxed{\\alpha_{min} = 150^{\\circ}}$.[/hide]",
  "Solution_4": "Good. :)",
  "Solution_5": "[hide=\"or\"]\nBy parallelogram theorem,we have\n$ AB^2\\plus{}AC^2\\equal{}2(AM^2\\plus{}BM^2)\\Longrightarrow 16\\plus{}b^2\\equal{}2(1\\plus{}\\frac{a^2}{4})\\Longrightarrow a^2\\equal{}2b^2\\plus{}28$.\nand by law of cosines and AM-GM we have\n$ \\cos\\angle BAC\\equal{}\\frac{16\\plus{}b^2\\minus{}a^2}{8b}\\equal{}\\frac{\\minus{}b^2\\minus{}12}{8b}\\equal{}\\minus{}\\frac{1}{8}\\left(b\\plus{}\\frac{12}{b}\\right)\\le \\minus{}\\frac{2}{8}\\sqrt{b\\cdot \\frac{12}{b}}\\equal{}\\minus{}\\frac{\\sqrt{3}}{2}$,\nequality occurs when $ b\\equal{}2\\sqrt{3}$.\n\nHence the least is $ \\boxed{150^\\circ}$ \n[/hide]"
}
{
  "Tag": [
    "calculus",
    "calculus computations"
  ],
  "Problem": "Find the condition of $ f(x)$to this equation is true\r\n$ lim_{x\\to 0}(\\frac {1}{x}(1 - \\frac {1}{f(x)}) = lim_{x\\to 0}(\\frac {1-\\frac {1}{L}}}{x}$\r\nwhere $ L = lim_{x\\to 0} f(x)$\r\nAnd can you show your genaral one",
  "Solution_1": "Do you mean:\r\n\r\n$ lim_{x \\to 0} \\left ( \\frac{1\\minus{}\\frac{1}{f(x)}}{x} \\right ) \\equal{} lim_{x \\to 0} \\left( \\frac{1 \\minus{} \\frac{1}{L}}{x} \\right )$, where $ L : \\equal{} lim_{x \\to 0} f(x)$?\r\n\r\nFor each $ L \\neq 1$ the RHS is 'equal' to negative or positive infinity I would say. Writing such an equation down is kind of meaningless in those cases.\r\nSo you would have to have $ lim_{x \\to 0} f(x) \\equal{} 1$. We then get the inderminate form $ \\frac{0}{0}$ on the LHS, so we can try to apply l'Hopital to get:\r\n\r\n$ lim_{x \\to 0} \\left ( \\frac{1\\minus{}\\frac{1}{f(x)}}{x} \\right ) \\equal{} lim_{x \\to 0} \\left ( \\frac{\\frac{1}{(f(x))^2}f^`(x)}{1} \\right )\r\n\\equal{} lim_{x \\to 0} \\frac{f^`(x)}{(f(x))^2},$ which would have to be $ 0$ to fit the RHS.\r\n\r\nI thus come to the following two conditions:\r\n\r\n- $ lim_{x \\to 0} f(x) \\equal{} 1$\r\n\r\n- $ lim_{x \\to 0} \\frac{f^`(x)}{(f(x))^2} \\equal{} 0$\r\n\r\n\r\nDid I make any sense in the above?",
  "Solution_2": "[quote=\"Tyl\"]\n\n\n\n- $ lim_{x \\to 0} f(x) \\equal{} 1$\n\n- $ lim_{x \\to 0} \\frac {f^`(x)}{(f(x))^2} \\equal{} 0$\n\n\n[/quote]\r\nThank you,I am too"
}
{
  "Tag": [],
  "Problem": "How many integers satisfy $ \\minus{}4 \\le 2x\\plus{}1 \\le 6$?",
  "Solution_1": "Subtracting 1 from all,\r\n\r\n$ \\minus{}5 \\le 2x \\le 5$\r\n\r\nDividing by 2\r\n\r\n$ \\minus{}2.5 \\le x \\le 2.5$\r\n\r\nso x=-2,-1,0,1,2 or 5 vlues."
}
{
  "Tag": [
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "Solve in $ \\mathbb{R}$ the equation $ \\left| x \\minus{}\\frac{\\pi}{6} \\right|\\plus{}\\left| x\\plus{}\\frac{\\pi}{3} \\right|\\equal{}\\arcsin \\frac{x^3\\minus{}x\\plus{}2}{2}.$",
  "Solution_1": "Let $ f(x)=|x-\\frac{\\pi}{6}|+|x+\\frac{\\pi}{3}|,g(x)=arcsin\\frac{x^3-x+2}{2}.$\r\n$ g(x)$ dtfined, when  $ -2\\le x^3-x+2\\le 2$. If $ x>\\frac{\\pi}{6}$, then $ f(x)>\\frac{\\pi}{2}\\ge g(x)$ - no solution.\r\nIf $ -\\frac{\\pi}{3}\\le x\\le \\frac{\\pi}{6}$, then ${ f(x)=\\pi}{2}$, so $ x^3-x=0\\to x=0 \\ or \\ x=-1$.\r\nIf $ x<\\frac{-\\pi}{3}$, then $ f(x)=-\\frac{\\pi}{6}-2x>\\frac{\\pi}{2}$ - no solution.\r\nOnly $ x=0,x=-1$ are solutions."
}
{
  "Tag": [
    "induction",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Define a sequence by $ n(n\\minus{}1)a_{n} \\equal{} (n\\minus{}1)(n\\minus{}2)a_{n\\minus{}1}\\minus{}(n\\minus{}3)a_{n\\minus{}2}$, $ n \\ge 2$, where $ a_0 \\equal{} a_1$ are any real numbers. Prove that $ \\displaystyle\\sum_{n\\equal{}0}^{\\infty}a_{n}$ converges to $ a_{0}e$.",
  "Solution_1": "[hide=\"answer\"] Note that $ \\frac {1}{(n \\minus{} 2)!} \\equal{} \\frac {1}{(n \\minus{} 3)!} \\minus{} \\frac {n \\minus{} 3}{(n \\minus{} 2)!}$, so by induction it's easy to show $ a_n \\equal{} \\frac {a_0}{n!}$, and the sum is $ a_0\\sum_{n \\equal{} 0}^\\infty{\\frac {1}{n!}} \\equal{} a_0 e$ [/hide]"
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "probability",
    "expected value",
    "probability and stats"
  ],
  "Problem": "a funny one I was asked yesterday during a job interview:\r\nthere is a ant walking on the edges of a cube. Every time she arrives at a vertex of the cube, she chooses randomly another edge, and continues to walk on this edge. Suppose that it takes $ 1$ minute to travel one edge: what is the average amount of time it takes for the ant to go from one vextex to the opposite one ?",
  "Solution_1": "I'm surprised I wasn't able to find this already on the forum anywhere.\r\n\r\n[hide]Let $ A$ be your starting vertex, call all three of the vertices adjacent to $ A$ $ B$, call all three of the vertices adjacent to the $ B$ $ C$.  Let $ a$ be the expected number of steps from $ A$, $ b_A$ be the expected number of steps from one of the $ B$ if you've just arrived from $ A$, $ b_C$ be the expected number of steps from one of the $ B$ if you've just arrived from $ C$, and $ c$ the expected number of steps from one of the $ C$.  Then it's easy to see that\n\\begin{align*} a & = 1 + b_A \\\\\nb_A & = 1 + c \\\\\nb_C & = 1 + a/2 + c/2 \\\\\nc & = 1 + b_C/2. \\end{align*}\nSolving this system gives $ a = 6, b_A = 5, b_C = 6$ and $ c = 4$.\n\nThat's pretty good savings compared with a \"dithering\" ant that chooses from among the three edges randomly, regardless of where it came from.[/hide]"
}
{
  "Tag": [
    "counting",
    "distinguishability"
  ],
  "Problem": "Steal the avatar!\r\nhttp://www.artofproblemsolving.com/Forum/images/avatars/1678273634461505c0c3a30.gif\r\n[img]http://www.artofproblemsolving.com/Forum/images/avatars/1678273634461505c0c3a30.gif[/img]\r\nSupporters:\r\nfootballrocks41237",
  "Solution_1": "dude why magixter and not me",
  "Solution_2": "dud why magister and not ME??? :P",
  "Solution_3": "Please don't, this is retarded.\r\nThe point of avatars are to make the posts more distinguishable or whatever, if a bunch of people have same avatars, I actually have to check their username before reading their posts, which gets annoying",
  "Solution_4": "Dude, everyone should just steal my avatar. It's so creative.  :D",
  "Solution_5": "I use avatars as a way to express my general mood. But no one can ever figure out what each one means.",
  "Solution_6": "yeah, and least of all you yourself. :P",
  "Solution_7": "w00t yeah i stole it too cause i was bored XD",
  "Solution_8": "you poor unoriginal things"
}
{
  "Tag": [],
  "Problem": "How many positive integers $ n$ satisfy $ 200 &lt; n^2 &lt; 900$?",
  "Solution_1": "The smallest number that can be $ n$ is $ 15$ because $ 15^2\\equal{}225$. The largest number that can be $ n$ is $ 29$ because $ 29^2\\equal{}841$. Therefore, the numbers that can be $ n$ are $ 15\\minus{}29$, which is a total of $ 15$ numbers"
}
{
  "Tag": [],
  "Problem": "Hello everyone,\r\n\r\nI am having some trouble with the following two problems pertaining to exponents, and I would appreciate any help or hints!\r\n\r\nThank you!\r\n\r\n---\r\n\r\n1. Let $ a$ and $ b$ be real numbers with $ a > 1$ and $ b > 0$. If $ ab = a^b$ and $ \\frac{a}{b} = b^3$, find and b. (Hint: Use substitution or multiply the equations side by side)\r\n\r\nFirst off, could anyone please explain what is meant by multiplying the equations side by side?\r\n\r\n[u]My Work[/u]:\r\n\r\nSince $ \\frac{a}{b} = b^3$, $ a = b^4$.\r\n\r\nI substituted this into the other equation:\r\n\r\n$ b^{4}(b) = (b{}^4){}^b$\r\n\r\n$ b^5 = b^{4b}$\r\n\r\n$ 5 = 4b$\r\n\r\n$ b = \\frac{4}{5}$. \r\n\r\nHowever, the provided answer for b is 0.5 (I wanted to get the right answer for $ b$ before continuing to find $ a$.Where might I have erred?)\r\n\r\n\r\n2. Let $ f(x) = 2^{kx} + 9$ where $ k$ is a real number. If $ f(3): f(6) = 1: 3$, find the value of $ 2^{3k}$.\r\n\r\nSince $ f(6) = f(3) x 3$:\r\n\r\n$ \\frac{2^{6k} + 9}{2^{3k} + 9} = 3$\r\n\r\n$ \\frac{2{}^{3k}{}^2 + 9}{2^{3k} + 9} = 3$\r\n\r\nLet $ y = 2^{3k}$:\r\n\r\n$ \\frac{y^2 + 9}{y + 9} = 3$\r\n\r\n$ y^2 + 9 = 3y + 27$\r\n\r\n$ y^2 - 3y - 18 = 0$\r\n\r\n$ (y - 6)(y + 3) = 0$\r\n\r\n$ y = -3, 6$\r\n\r\nThe provided answer gives $ y = 2^{3k} = 6$. Here, I am unsure of how I can rule out -3 as a solution for y. \r\n\r\n\r\n\r\n[/i]",
  "Solution_1": "This place: \r\n5 = 4b\r\nb is 5/4, NOT 4/5   :)\r\n\r\nAnd about second. It's impossible to get negative value, when you are involuting positive number!"
}
{
  "Tag": [
    "quadratics",
    "algebra",
    "algebra proposed"
  ],
  "Problem": "Determine all real numbers $ r$ such that there is precisely one pair $ (x,y)$ of real numbers satisfying the conditions :\r\n\r\n$ 1. y\\minus{}x\\equal{} r$\r\n$ 2.  x^2 \\plus{}y^2 \\plus{} 2x \\le 1$.",
  "Solution_1": "The values are $ r \\in \\{ \\minus{}1, 3 \\}$; for $ r \\in (\\minus{}1, 3)$ there are two solutions, while for $ r \\in (\\minus{}\\infty, \\minus{}1) \\cup (3, \\plus{}\\infty)$ there are none.\r\n$ x^2 \\plus{} y^2 \\plus{} 2x \\equal{} (x\\plus{}1)^2 \\plus{} y^2 \\minus{}1$, so the second relation describes the points inside the closed disk $ D$ of center $ (\\minus{}1,0)$ and radius $ \\sqrt{2}$, while the first describes the line $ \\ell$ given by $ y \\equal{} x\\plus{}r$, parallel to the first bisector. The unique solution is obtained when $ \\ell$ is tangent to $ D$.",
  "Solution_2": "hello, the discriminant of the quadratic equation\r\n$ x^2\\plus{}x(r\\plus{}1)\\plus{}\\frac{r^2\\minus{}1}{2}\\equal{}0$ must be zero, this is the case if\r\n$ \\left(\\frac{r\\plus{}1}{2}\\right)^2\\equal{}\\frac{r^2\\minus{}1}{2}$\r\n$ \\minus{}r^2\\plus{}2r\\plus{}3\\equal{}0$ with the solutions\r\n$ r\\equal{}3$ or $ r\\equal{}\\minus{}1$\r\nSonnhard."
}
{
  "Tag": [
    "geometry",
    "AMC",
    "AIME",
    "search"
  ],
  "Problem": "Dear Math Enthusiast, \r\n\r\nThe 2006 Four-by-Four Competition took place on Thursday, February 9th. This competition consists of ten rounds in which teams of four are challenged to solve four problems in three minutes. The first-place winners in each division are: \r\n\r\nSchool: Vestavia Hills High School, Vestavia Hills, AL\r\n\r\nMiddle School Team:\r\n     Jack Dickerson\r\n     Kevin Shutzberg\r\n     Max LaVictoire\r\n     Sam Kallman\r\n     from The Davis Academy, Atlanta, GA\r\n\r\n\"Rock\" Division Team:\r\n     Alan Chou\r\n     Peter Cheng\r\n     Kathryn Brewer\r\n     Rebecca Mai\r\n     from Grissom High School, Huntsville, AL\r\n\r\n\"Paper\" Division Team:\r\n     Linda Liu\r\n     William Dobny\r\n     Stephanie Wang\r\n     Zane Blanton\r\n     from Parkview High School, Lilburn, GA\r\n \r\n\"Scissors\" Division Team:\r\n     Amir Talebi\r\n     Eric Chatterjee\r\n     Elliot Lee\r\n     Dianna Wu\r\n     from Eastlake High School, Sammamish, WA\r\n\r\n\r\nThe recognized scores in each division were: \r\n\r\nSchools: 83-210\r\nMiddle School Teams: 19-40\r\n\"Rock\" Division Teams: 60-64\r\n\"Paper\" Division Teams: 61-72\r\n\"Scissors\" Division Teams: 68-86\r\n\r\nComplete results for the competition (the top 25 school and the top 5 teams in each division) are available on our website; take a look to see who in your area you can congratulate!",
  "Solution_1": "[quote=\"tom_clymer\"]\n\"Scissors\" Division Team:\n     Amir Talebi\n     Eric Chatterjee\n     Elliot Lee\n     Dianna Wu\n     from Eastlake High School, Sammamish, WA\n[/quote]\r\n\r\nyay :) \r\n\r\na nice energy boost before the AIME :)",
  "Solution_2": "[quote=\"tom_clymer\"]Dear Math Enthusiast, \n\n\"Scissors\" Division Team:\n     Amir Talebi\n     Eric Chatterjee\n     Elliot Lee\n     Dianna Wu\n     from Eastlake High School, Sammamish, WA\n[/quote]\r\n\r\nGood to know I won!\r\nWait, I never did this...\r\n\r\n-Eric Chatterjee",
  "Solution_3": "Hello.\r\n\r\nIn case there is any confusion arising from the previous post, I just wanted to let everyone know that I (Eric Chatterjee, Eastlake High School Class of 2007, Sammamish, WA), am the one who took part in the team that won the Scissors Division of Four by Four. \r\n\r\nHope that this is now clear to everyone viewing this topic. :)",
  "Solution_4": "According to my research, there are actually 3 of us. You in WA, me in OH and another \"Eric Chatterjee\" somewhere in Germany. Know any others? :)",
  "Solution_5": "Based on my search, I believe there is a director/producer by this name.  ;)"
}
{
  "Tag": [
    "advanced fields",
    "advanced fields unsolved"
  ],
  "Problem": "$ E$ is a fibration over $ B$ with fiber $ F$, and $ F\\minus{}>E\\minus{}>B$ is an exact sequence,\r\nif $ F$ is simply connected , then $ \\pi(E)\\equal{}\\pi(B)$. Is there such a thm? in which book can \r\nI find the proof?",
  "Solution_1": "[url]http://en.wikipedia.org/wiki/Homotopy_group#Long_exact_sequence_of_a_fibration[/url]\r\n\r\ntake that exact sequence and apply it. they probably also give references."
}
{
  "Tag": [
    "geometry"
  ],
  "Problem": "I have a question about midpoints.  I am sure it is a simple solution but I am having difficulty solving.  P Q R S ant T are on line k such that Q is the midpoint of PT, R is the  midpoint of QT and S is the midpoint of RT.  If PS=9, then what is PT?\r\n\r\nThis is how far I have gotten\r\nP________Q____R__S__T\r\nQT=PT/2\r\nRT=QT/2\r\nSt=RT/2\r\nI know that PQ=QT or PQ=PT/2\r\nNow I am stuck.\r\n\r\nAny help would be appreciated and thanks :D",
  "Solution_1": "[quote]This is how far I have gotten \nP________Q____R__S__T \nQT=PT/2 \nRT=QT/2 \nSt=RT/2 \nI know that PQ=QT or PQ=PT/2 \n[/quote]\r\n\r\nNow make use of these statements to form an expression for PS:\r\n\r\nQT = PQ = 2QR = 4RS.\r\n\r\nPS = PQ + QR + RS \r\n     = 7RS\r\n\r\nSo, RS = PT / 8 = 9 / 7\r\n\r\nSo, PT = 72 / 7\r\n\r\nAlternatively,\r\n\r\nPS = PQ + QR + QS\r\n     = PT / 2 + PT / 4 + PT /8\r\n     = 7PT / 8\r\n\r\nAgain, 7PT / 8  = 9\r\n                7PT = 72\r\n                  PT = 72 / 7\r\n\r\nUse your equality statements to form an expression for a portion of PT and solve for that portion.  Then use the relationship of that portion to PT to solve for PT.",
  "Solution_2": "Easy!\r\n[hide] As you know: \n  ST = 1/2 RT =1/4 QT = 1/8 PT\nor  ST = 1/7 PS = 1/7 * 1/9 = 1/63\n=> PT = 9 + 1/63 = 568/63[/hide]\r\n Maybe I am wrong. I am not sure! :(",
  "Solution_3": "[hide]\nPQ= 1/2 PT\n\nQR= RT= 1/2 PQ= 1/4 PT\n\nRS= ST= 1/2 RT= 1/2 QR= 1/4 PQ= 1/8 PT\n\nPS= PT- ST\n\nST= 1/8PT\n\nPS= 7/8PT\n\n7/8PT= 9\n\nPT= 9*7/8= 63/8\n\n[/hide]",
  "Solution_4": "[hide=\"Answer to dholstein\"]\nAlright $PQ=\\frac{1}{2}PT, QR=\\frac{1}{4}PT, RS=ST=\\frac{1}{8}PT$. Let's make the value of $PT=x$ so all the named segments are the named fraction of $x$. Now $PS=PQ+QR+RS=\\frac{x}{2}+\\frac{x}{4}+\\frac{x}{8}=\\frac{7x}{8}$. So we set $\\frac{7x}{8}=9=\\frac{7x}{8}=\\frac{72}{8}$ which we can make $7x=72$, so $x=\\frac{72}{7}=PT$[/hide]\n\n[hide=\"Correction for Chocolate milk\"]\nIn the second line of your work you have $PS=(\\frac{1}{7})(\\frac{1}{9})$. However, you are supposed to instead have $9(\\frac{1}{7})$ because $9$ is 7 times the amount of $ST$, which is $\\frac{9}{7}$ then add that to $9$ to obtain $\\frac{72}{7}$[/hide]\n\n[hide=\"Correction for mustafa\"]\nIn the last step you are supposed to divide $\\frac{9}{(\\frac{7}{8})}$ which is basically $(\\frac{8}{7})(9)=\\frac{72}{7}$[/hide]",
  "Solution_5": "Whoops, lol I think i'm too tired to do math right now :oops:"
}
{
  "Tag": [
    "geometry proposed",
    "geometry"
  ],
  "Problem": "Let $ ABCD$ be a quadrilateral , $ X$ and $ Y$ be the mid points of $ AC$ and $ BD$ respectively. The lines through $ X$ and $ Y$ respectively parallel to $ BD$ and $ AC$ meet at $ o$. Let $ P,Q,R,S$ be the mid points of $ AC,BC,CD,DA$ respectively. Prove that $ [APoS]\\equal{}\\frac{1}{4}[ABCD]$.",
  "Solution_1": "Because of $ (APS) \\plus{} (CQR) \\equal{} \\frac {(ABD)}{4} \\plus{} \\frac {(CBD)}{4} \\equal{} \\frac {(ABCD)}{4},$ it is enough to prove that $ (OPS) \\equal{} (CQR)$ $ ,(1)$\r\n\r\nBecause of now, the triangles $ \\bigtriangleup OPS,\\ \\bigtriangleup CQR,$ have $ PS \\equal{} \\frac {BD}{2} \\equal{} QR,$ it is enough to prove that their altitudes through $ O,\\ C$ respectively, are congruent.\r\n\r\nBut it is true, because of $ XM \\equal{} KN$ $ ($ easy to prove from $ MC \\equal{} MK$ and $ XA \\equal{} XC$ $ ),$\r\nwhere $ K\\equiv AC\\cap BD,\\ M\\equiv AC\\cap QR,\\ N\\equiv AC\\cap PS$ and the proof is completed. \r\n\r\nKostas Vittas."
}
{
  "Tag": [
    "trigonometry",
    "limit",
    "function",
    "integration",
    "real analysis",
    "real analysis solved"
  ],
  "Problem": "I still hope somebody will solve the nested root problem. Meanwhile, let me post one more. It is not really hard but I like it anyway...\r\n\r\nFind $\\sum_{n\\geq 1}\\frac1{2^n}\\tan \\frac1{2^n}$.",
  "Solution_1": "I think that if we also add the term corresponding to $n=0$, which is more natural anyway :), we get $1-2\\cot 2$, but I'm not that sure. I might have made some computational errors.\r\n\r\nFirst of all, we develop $x\\tan x$ in a Taylor series around the origin: $x\\tan x=\\sum_{n\\ge 1}\\frac{(-1)^{n-1}\\cdot 2^{2n}\\cdot (2^{2n}-1)\\cdot B_{2n}}{(2n)!}\\cdot x^{2n}$. After this, we fix $n$, and sum up all the terms corresponding to $n$ as $x$ runs through the powers of $\\frac 12$ from $0$ to $\\infty$. That ugly $2^{2n}-1$ cancels out, and a $2^{2n}$ appears, so the series looks like $\\sum_{n\\ge 0}\\frac{(-1)^{n-1}\\cdot 4^{2n}\\cdot B_{2n}}{(2n)!}\\ (*)$.\r\n\r\nNow, since $x\\cot x=\\sum_{n\\ge 1}\\frac{(-1)^n\\cdot B_{2n}}{(2n)!}\\cdot (2x)^{2n}$ (for $x\\in[-\\pi,\\pi]$), this matches $(*)$, if we take $x=2$ and multiply it by $-1$.\r\n\r\nIs it Ok?\r\n\r\n[Edited]",
  "Solution_2": "Very nice, grobber! You lost $+1$ somewhere (I leave it to you to find the exact spot) because the answer for summation including $0$ is $1-2\\cot 2$ (and without $n=0$ it is $1-\\cot 1$). Otherwise it's right. I had a slightly different solution in mind but congratulations anyway :lol:. Did you try the nested root problem?",
  "Solution_3": "Nope, I didn't try it, but I don't think it's the sort of problem I like to try really hard to solve :). As for this one, it doesn't look like I'll be finding that lost $1$ any time soon :D.",
  "Solution_4": "In general\r\n\\[\r\n\\displaystyle\\sum_{n=0}^{+\\infty}\\frac{1}{2^n}\\tan\\frac{x}{2^n}=\\frac{1}{x}-2\\cot 2x\r\n\\]\r\nfollowing another way like this:\r\nApplying the identity $\\cot x =2\\cot 2x+\\tan x \\quad \\textrm{for}\\quad x\\not=k\\frac{\\pi}{2}\\quad k\\in\\mathbb{Z}$  and the fact that if $\\displaystyle\\lim_{n\\to+\\infty}a^{-n}f(b^{-n}x)=l(x)$\r\nthen \r\n\\[\r\n\\displaystyle\\sum_{n=0}^{+\\infty}a^{-n}g(b^{-n}x)=\\frac{l(x)-af(bx)}{c}\r\n\\]\r\nwhere a,b,c are non zero costant and the functions f and g satisfy the condition\r\n$f(x)=af(bx)+cg(x).$\r\nIn our case $f(x)=\\cot x,\\quad g(x)=\\tan x,\\quad a=2,\\quad b=2\\quad\\textrm{and}\\quad c=1.$\r\nRemember, also, that\r\n\\[\r\n\\displaystyle\\lim_{n\\to+\\infty}\\frac{1}{2^n}\\cot \\frac{1}{2^n}=\\frac{1}{x}\r\n\\]",
  "Solution_5": ":)  :)  :)",
  "Solution_6": "O.K Myth\r\nThe smile is the salt of life.",
  "Solution_7": "I meant that I don't like grobber's approach, since it is too ugly, though it leads to the answer ;) (without any offences). Your solutions is very nice. I was expecting it, but I haven't in mind that identity.",
  "Solution_8": "I figured out where I lost the $1$ :). The series for $x\\tan x$ starts from $n=1$, so in the end, we \"almost\" get the series for $x\\cot x$, without the initial term, which is $1$. When we multiply it by $-1$, the $1$ we were missing becomes an extra $1$. I'll edit it.\r\n\r\nEdit: no offence taken, Myth :)",
  "Solution_9": "Find this one $\\sum_{n\\geq 0}\\frac{1}{2^{2k}}\\tan^2(\\frac{x}{2^k})$",
  "Solution_10": "Some words about first sum: $S(x)=\\sum_{n\\geq0}\\frac{1}{2^n}tan(\\frac{x}{2^n}) $. Integrating by parts we may obtain that $\\int S(x)dx=\\sum_{n\\geq0}(-ln(cos(\\frac{x}{2^n})))=-ln(\\frac{sin(2x)}{x})=ln(x)-ln(sin(2x))$, so we obtain $S(x)$ \r\n(sorry for not being very precise)",
  "Solution_11": "[quote=\"Moubinool\"]Find this one $\\sum_{n\\geq 0}\\frac{1}{2^{2k}}\\tan^2(\\frac{x}{2^k})$[/quote]\r\nBy integrating we get $\\sum_{k\\geq 0}\\frac{1}{2^k}\\left(\\tan(x/2^k)-x/2^k\\right)=\\sum_{k\\geq 0}\\frac{1}{2^k}\\tan\\frac{x}{2^k}-\\frac{4x}{3}$."
}
{
  "Tag": [
    "modular arithmetic",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Find all positive integer solutions $ (a,b,c,n)$ of the equation: $ 2^n\\equal{}a!\\plus{}b!\\plus{}c!$.",
  "Solution_1": "[quote=\"moldovan\"]Find all positive integer solutions $ (a,b,c,n)$ of the equation: $ 2^n \\equal{} a! \\plus{} b! \\plus{} c!$.[/quote]\r\n\r\nWlog say $ a\\geq b \\geq c$\r\n\r\nIf $ c\\geq 3$, $ 3$ divides RHS but not LHS. So $ c<3$\r\n1) $ c\\equal{}1$\r\nThen $ b\\equal{}1$ and $ a>1$ else RHS would be odd while LHS even. So $ 2^n\\equal{}a!\\plus{}2$ \r\nIf $ n\\equal{}1$, this equation has no solution\r\nIf $ n\\equal{}2$, we get the solution $ 2^2\\equal{}2!\\plus{}1!\\plus{}1!$\r\nIf $ n>2$, $ a<4$ else $ LHS\\equal{}0\\pmod 4$ while $ RHS\\equal{}2\\pmod 4$. And we get the solution $ 2^3\\equal{}3!\\plus{}1!\\plus{}1!$\r\n\r\n2) $ c\\equal{}2$\r\nThen $ 2^n\\equal{}a!\\plus{}b!\\plus{}2\\geq 6$ and so $ n\\geq 3$\r\nIf $ b\\geq 4$, then $ LHS\\equal{}0\\pmod 4$ while $ RHS\\equal{}2\\pmod 4$. So $ b<4$\r\n\r\n2.1) $ c\\equal{}2$ and $ b\\equal{}2$\r\nThen $ 2^n\\equal{}a!\\plus{}4$. \r\nIf $ a\\geq 4$, $ RHS\\equal{}4\\pmod 8$ and so $ n<3$ and so no solution\r\nWe just have to test $ a\\in[2,3]$ and no solution.\r\n\r\n2.2) $ c\\equal{}2$ and $ b\\equal{}3$\r\nThen $ 2^n\\equal{}a!\\plus{}8$\r\nIf $ a\\geq 8$, $ RHS\\equal{}8\\pmod 16$ and so $ n<4$ and so no solution\r\nWe just have to test $ a\\in[3,7]$ and two solutions :\r\n$ 2^5\\equal{}4!\\plus{}3!\\plus{}2!$\r\n$ 2^7\\equal{}5!\\plus{}3!\\plus{}2!$\r\n\r\n\r\nSo 4 solutions :\r\n$ 2^2\\equal{}2!\\plus{}1!\\plus{}1!$\r\n$ 2^3\\equal{}3!\\plus{}1!\\plus{}1!$\r\n$ 2^5\\equal{}4!\\plus{}3!\\plus{}2!$\r\n$ 2^7\\equal{}5!\\plus{}3!\\plus{}2!$"
}
{
  "Tag": [
    "number theory",
    "relatively prime",
    "Diophantine equation",
    "IMO Shortlist"
  ],
  "Problem": "(a) Let $ n$ be a positive integer. Prove that there exist distinct positive integers $ x, y, z$ such that\r\n\r\n\\[ x^{n\\minus{}1} \\plus{} y^n \\equal{} z^{n\\plus{}1}.\\]\r\n\r\n(b) Let $ a, b, c$ be positive integers such that $ a$ and $ b$ are relatively prime and $ c$ is relatively prime either to $ a$ or to $ b.$ Prove that there exist infinitely many triples $ (x, y, z)$ of distinct positive integers $ x, y, z$ such that\r\n\r\n\\[ x^a \\plus{} y^b \\equal{} z^c.\\]",
  "Solution_1": "[quote=\"orl\\]\r\n(b) Let $ a, b, c$ be positive integers such that $ a$ and $ b$ are relatively prime and $ c$ is relatively prime either to $ a$ or to $ b.$ Prove that there exist infinitely many triples $ (x, y, z)$ of distinct positive integers $ x, y, z$ such that\r\n\\[ x^a \\plus{} y^b \\equal{} z^c.\r\n\\][/quote]\r\n\r\nSee it here  http://www.mathlinks.ro/viewtopic.php?t=150638",
  "Solution_2": "My solution.\r\nExist u, v, w such that: $ u \\plus{} v \\equal{} w$ and:\r\n$ u \\equal{} \\prod_i^n p_i^{\\alpha_i}$;\r\n$ v \\equal{} \\prod_i^n p_i^{\\beta_i}$;\r\n$ w \\equal{} \\prod_i^n p_i^{\\gamma_i}$;\r\nBy chinese remainder theorem, exist $ \\delta_i, i \\equal{} 1,..n$  such that:\r\n\r\n$ a\\mid\\delta_i \\plus{} \\alpha_i$\r\n$ b\\mid\\delta_i \\plus{} \\beta_i$\r\n$ c\\mid\\delta_i \\plus{} \\gamma_i$\r\n\r\nChoose $ z \\equal{} \\prod_i^n p_i^{\\delta_i}$, This problem proven"
}
{
  "Tag": [
    "AMC",
    "AIME",
    "MATHCOUNTS",
    "search",
    "geometry",
    "USA(J)MO",
    "USAMO"
  ],
  "Problem": "I'd like to repost something that rrusczyk posted about this time last year:\r\n\r\n[quote=\"rrusczyk\"]This is in response to the recent wave of students who do really well on the AMC exams then complain about their performance.\n\nPlease stop and think for a minute before complaining about what are for most people excellent results. Imagine you are a student who is just getting started and seeing someone complaining about a performance that is far, far better than what you are able to do. How do you feel? What do you think about the person who is complaining?\n\nI'm guessing most of us don't like to see outright bragging, and we don't see much bragging on these boards. However, most people read complaints about high scores as essentially equivalent to bragging (if not worse).\n\nWe'd like our community to be inclusive and attractive to beginning students. The prevailing attitude we've seen recently of very strong students complaining about their excellent scores doesn't advance this goal (or, I think, reflect what the students complaining hope to reflect).\n\nI understand these high scores may not be up to your standards, but complaining about doesn't help that!\n\nI also realize that some of you may be tounge-in-cheek joking in your complaints, but it's not always obvious when that is the case.[/quote]\r\nPlease keep this advice in mind as you discuss this year's contest.  :)",
  "Solution_1": "By good scores do you mean good scores for aops, or good scores in the world?\r\n\r\nFor example I got a 132 on the 10 and am really disappointed. By AoPS standards that's pretty low, but by the world's standards that's good.\r\n\r\nCan I complain about it?",
  "Solution_2": "I think you know the answer to that question.\r\n\r\nThere is never any point in complaining about something you cannot change. Always work toward the future.",
  "Solution_3": "[quote=\"white_horse_king88\"]I think you know the answer to that question.\n\nThere is never any point in complaining about something you cannot change. Always work toward the future.[/quote]\r\nI think I know it, but I just wanted to clarify :oops:\r\n\r\nYou're right, complaining isn't good, though I'm not that upset about it, I'm just disappointed because I could've done better.",
  "Solution_4": "I'm sure you'll do fine on the AIME! And you have a month still.  :)",
  "Solution_5": "You say \"by AoPS standards\".  My (and rrusczyk's) point was that we get new students every day here at AoPS, some who are just starting out with problem solving.  So the phrase \"AoPS standards\" doesn't hold a lot of meaning to me.\r\n\r\nThe initial post was not a rule, it was [i]advice[/i].  It's not that you aren't \"allowed\" to complain about your score.  It's that you should use your judgment as to how others might perceive such complaining.  That's all. :)",
  "Solution_6": "well i'm complaining about a really low score. \r\n\r\ni got like 80 on the AMC 10  :|",
  "Solution_7": "well some people may complain if they got a 144 and they missed question one. is that different?  :lol:",
  "Solution_8": "you can be annoyed!\r\n\r\nlike i was with the amc 8!\r\n\r\n-jorian",
  "Solution_9": "Don't complain if you qualified for the AIME.",
  "Solution_10": "How about no complaing at all.  It does nothing at all at this point.  The time you spend complaining could be better spent doing something productive so you don't have something to complain about next time.",
  "Solution_11": "[quote=\"mna851\"]How about no complaing at all.  It does nothing at all at this point.  The time you spend complaining could be better spent doing something productive so you don't have something to complain about next time.[/quote]\r\n\r\n\r\nexactly. that's very good advice.",
  "Solution_12": "I'm a bit disappointed, I guess, with the time limit. An extra um 30 seconds would've sufficed  :D",
  "Solution_13": "I did really crappy this year and ended up with an 80 or so. Does that warrant a wave of complaints? -_-",
  "Solution_14": "well, i completely agree with not complaining about good scores.  Think to yourself, how good were you when you started AoPS?  \r\n\r\nLast year as a sixth grader, i started posting on here but i saw that peple were complaining about 39's and 40's at MATHCOUNTS States. I got really discouraged and i was like, \"woah :o , i feel stupid- i got a 27\".  \r\n\r\nSo it's a great idea to keep your complaints to yourself!! :)",
  "Solution_15": "To akkaren:\r\n\r\nI apologize for my harsh and inapproriate language in my previous posting. Apparently I wasn't in the right state of mind.",
  "Solution_16": "Complaining doesn't help anyone. It doesn't help you, people reading it, our your ability. Saying something like \"I expected about 10 points higher, I usually got much higher scores when I practice,\" is all right, but saying something like \"I sucked so much, I can't believe I was so retarded to get X,\" doesn't help anyone. What if I read that and I got X-10? That would mean that person thinks I suck alot and I'm really retarded. Who benefits from that?",
  "Solution_17": "Well, when you read people's AMC scores, you should take them in context. For example, say Arnav gets a 138 on the AMC, and then starts complaining like the whiner he is. If you got like a 100, Arnav's complaining shouldn't make you feel bad.\r\n\r\nIf you're a n00b who's never heard of Arnav, well, all the high scores you see should motivate you to study harder. I'm not going to stop discussing AMC scores because of some n00bs ok.",
  "Solution_18": "[quote=\"chess64\"]Well, when you read people's AMC scores, you should take them in context. For example, say Arnav gets a 138 on the AMC, and then starts complaining like the whiner he is. If you got like a 100, Arnav's complaining shouldn't make you feel bad.\n\nIf you're a n00b who's never heard of Arnav, well, all the high scores you see should motivate you to study harder. I'm not going to stop discussing AMC scores because of some n00bs ok.[/quote]\r\nSpam Deleted...",
  "Solution_19": "I'm sorry if my previous post sounded harsh, I wasn't intending it to be so.  I'm not saying that flaming is ok, even if it technically is against yourself, but I still think that people should be able to post their scores, and what they think of them (assuming the latter part is done in a civilized manner).  I personally would find it more deterring to have a bunch of people being all secretive about their scores than to hear about high ones.  If they aren't even willing to mention their scores around me, that would make me feel like I'm so crappy, they don't even think I'm worthy to hear about what they do.  Hearing about scores a lot higher than my own, on the other hand, motivates me to do better.  I've already heard about people 2 grades lower than me doing better than me on the AMC, and I'm not offended by it.  I think it's really cool that they can do that.  Also, since I live in a pretty rural area, there's really no competition for me out here, so with these discussions, I can put things in perspective, and not get complacent about my own performance.",
  "Solution_20": "I'm only complaining about my 144 because I made a really stupid mistake on #25 that, had I not made that mistake, would have given me perfect. I would not be disappointed with, say, 132, had I not really had a good shot at a perfect.",
  "Solution_21": "I have very mixed feelings on this subject, having been on both sides of it. I think that if someone is disappointed with their score, they have every right to express it in terms \"I didn't do as well as I could have\". I don't think people have a right to say \"that's a bad score\" in absolute terms. For comparison, think about olympic level athletes. I think that if we see an speedskater get emotional after slipping and falling in the finals of an event that they could've won, we empathize with them. They have taken a combination of hard work and natural talent to get to a place where they can really do something special, and they didn't get the opportunity to do as well as they could've more or less due to random bad luck. On the other hand, when a football player complains about their salary being cut from $10^{8}to a mere$8*10^7, nobody really feels bad for them, because that is so far into the 99th percentile of income that it doesn't really matter. So, to complete the analogy, if you say \"I got a 138 and I'm disappointed because I've worked hard and I feel like I could've gotten a perfect score\", you can get a little sympathy, but if you say \"I got a 144 and I can't believe it b/c now I won't get a cool plaque and someone beat me and now I can't put it on my college resume....\" you're going to get no sympathy from me",
  "Solution_22": "[quote=\"PenguinIntegral\"][quote=\"chess64\"]Well, when you read people's AMC scores, you should take them in context. For example, say Arnav gets a 138 on the AMC, and then starts complaining like the whiner he is. If you got like a 100, Arnav's complaining shouldn't make you feel bad.\n\nIf you're a n00b who's never heard of Arnav, well, all the high scores you see should motivate you to study harder. I'm not going to stop discussing AMC scores because of some n00bs ok.[/quote]\nNever hearing of Arnav is a necessary and sufficient condition for being a n00b.[/quote]\r\n\r\nIndeed? That sounds pretty slavish  :| \r\n\r\nTo make this post not get deleted:\r\n\r\nDo you guys think the AMC B is going to be harder than the A?",
  "Solution_23": "[quote=\"Karth\"][quote=\"PenguinIntegral\"][quote=\"chess64\"]Well, when you read people's AMC scores, you should take them in context. For example, say Arnav gets a 138 on the AMC, and then starts complaining like the whiner he is. If you got like a 100, Arnav's complaining shouldn't make you feel bad.\n\nIf you're a n00b who's never heard of Arnav, well, all the high scores you see should motivate you to study harder. I'm not going to stop discussing AMC scores because of some n00bs ok.[/quote]\nNever hearing of Arnav is a necessary and sufficient condition for being a n00b.[/quote]\n\nIndeed? That sounds pretty slavish  :| \n\nTo make this post not get deleted:\n\nDo you guys think the AMC B is going to be harder than the A?[/quote]\r\nWell it depends on what you mean.\r\nThe AMC 10 A, as someone said, was easier to get a perfect score (no killer) than usual, but harder to make AIME.\r\nSo what perspective?",
  "Solution_24": "I think the difficulty depends on your strengths and weaknesses; if your weaknesses don't really show up much on the B test, but did on the A test, then your opinion would be somewhat different.",
  "Solution_25": "[quote=\"chess64\"]The AMC is not a standardized test.\n\nAlso, akkaren, it sounds like you would consider the following complaining:\n\n[quote=\"hypothetical person\"]\nI got a 144, because I misread #7 :( :( :( I can't believe myself! :mad:\n[/quote]\n\nIn my opinion, this person would definitely have the right to make this statement. What would you say if that happened to you? Nothing?[/quote]\r\n\r\nYes. I would keep my mouth shut. It's a shame that a person so good at math can't read a sentence in English correctly. But, why not? Complain. Let's see if tomorrow you'll be magically transported back to the exact instant you were working on seven. Maybe things will fix themselves then. Or maybe. \"You'd better believe yourself.\"\r\n\r\nIt's a shame how elitist this forum has become.\r\n\r\nWhen there are two options, go with the one less likely to discourage someone. If some USAMOer who made MOP and I were talking about a problem and he was just ranting about how he or she missed a point or two on some technicality while I sit here with a miserably low score, well I'd rather not hear it.\r\n\r\nThen again, if you say that, \"If they don't want to hear it stay away from the thread,\" then you're clearly missing the point of a productive and encouraging forum engaged at higher education and promoting learning.\r\n\r\nThis is not your thread. Thus, be as respectful as possible. You may not think complaining about a 144 is disrespectful, but many do. It's not, \"Oh, I don't think I mean to be disrespectful, so it's ok.\" A parallel would be Senator Biden's remarks on Obama as the first 'clean and articulate' African-American. Did he mean to be disrespectful? Of course not. Yet, it's not his call whether or not his remarks were injurious. See?",
  "Solution_26": "ok....we've had like 20 people say the same thing now...",
  "Solution_27": "[quote=\"indianamath\"]ok....we've had like 20 people say the same thing now...[/quote]\r\nand yet randomdragoon still made a good point...",
  "Solution_28": "I didn't even read randomdragoon's point, but if I interpreted the last two posts correctly, randomdragoon made a good point. That's already been made 20 times.\r\n\r\n/pickiness  :rotfl: \r\n\r\nOverall, I think the blame falls in both courts. It is your duty not to complain about high scores. However, it is also the duty of readers not to get all emo about it if their score is lower and the person said something like \"blech, I missed #19 because I made a arithmetic mistake, darnit, I would've had perfect\". Since you, as the poster with the 144, you can only control what YOU do. So just keep it minimum.\r\n\r\nBut also, as a person that's been in both situations, I'd have to say when someone beats me by a lot, and I know that person is better than me, and he/she complains about it, I always think to myself \"that's by HIS (or her) standards, not mine\".",
  "Solution_29": "[quote=\"Fanatic\"]I didn't even read randomdragoon's point, but if I interpreted the last two posts correctly, randomdragoon made a good point. That's already been made 20 times.[/quote]\r\n:D well I was saying that randoomdragoon stated things in a more obvious way."
}
{
  "Tag": [
    "analytic geometry",
    "graphing lines",
    "slope"
  ],
  "Problem": "I am the equation y= 8/3\r\nWhat is the equation of a line that intersects perpendicularly at (-3,8/3)?\r\n\r\nI'm really stuck on this problem.. if I pick any point for that equation, for example 0 i get my points (0,8/3) and (-3,8/3). So the slope would be 0/-3 which would be 0.\r\n\r\nThe inverse (since its perpendicular) of the slope is -3/0.. but you cant divide by 0. So what do I do?",
  "Solution_1": "[hide=\"Hint\"]When graphed, what does y=8/3 look like?  What type of line is perpendicular to this line?[/hide]",
  "Solution_2": "Since the slope is $ \\frac30$, the answer is simply $ \\boxed{\\text{undefined}}$.",
  "Solution_3": "ernie, you misread the question.  The OP wanted to find the equation of the line, not its slope.",
  "Solution_4": "You're running into problems because you're forcing yourself to think in terms of slope-intercept form.\r\n\r\nTry to think in terms of the standard form of a line:\r\n\r\n$ \\boxed{Ax \\plus{} By \\equal{} C}$, where $ A,B,C$ are constants, and $ A$ and $ B$ are [i]not both zero[/i].\r\n\r\n[hide=\"Hint\"]Two lines $ A x \\plus{} B y \\equal{} C$ and $ D x \\plus{} E y \\equal{} F$ are perpendicular if and only if $ AD \\plus{} BE \\equal{} 0$.[/hide]",
  "Solution_5": "Then isnt it\r\n[hide]x=-3, since it is perpendicular to y=n for a number n and it intersects the equation? [/hide]",
  "Solution_6": "Yes it is JOngao. AN equation for a line doesn't have to have a real slope or have a y term in it (in the case of vertical lines)"
}
{
  "Tag": [
    "AMC 8",
    "AMC 10",
    "AMC"
  ],
  "Problem": "Umm, just wondering, cause I've heard this going around my school, but do people who score really high on AMC 8 get invited to take the AMC 10 or something? :huh:  Just wondering...cause I only missed one problem on AMC 8... so yeah...if you can clear that up, thanks.  :)",
  "Solution_1": "I think (and I could be completely wrong about this) when they say \"invited\" they mean \"recommended\". They want you to take the AMC 10, but they are not giving a formal recommendation. You should ask whoever is in charge of the AMC 10 at your school if you can take it, because a 24 on the AMC 8 proves that you are.",
  "Solution_2": "If they don't let you take it, go to another school and take it :P",
  "Solution_3": "Schools that have limited spots may make you qualify (unofficially) to take it, but it is an open contest.",
  "Solution_4": "Often if it isn't offered at your middle school, you can ask around to see if you can take it at your local high school.",
  "Solution_5": "[quote]but do people who score really high on AMC 8 get invited to take the AMC 10 [/quote]\r\n\r\nFrequently Asked Question AMC 8, #1\r\nhttp://www.unl.edu/amc/f-miscellaneous/faq.shtml\r\n\r\nSteve Dunbar\r\nMAA Director, American Mathematics Competitions"
}
{
  "Tag": [
    "modular arithmetic",
    "inequalities",
    "algebra",
    "polynomial",
    "number theory proposed",
    "number theory"
  ],
  "Problem": "prove that if $ n \\in \\mathbb{N}$ then $ n^2\\plus{}1 | n!$ if and only if these two statements aren't true:\r\n1-there exists a prime $ p$ such that $ n^2 \\equiv \\minus{}1 \\pmod {p},0<n<p$.\r\n2-there exists a prime $ p$ such that $ 2p^2\\equal{}n^2\\plus{}1$.",
  "Solution_1": "what? no taker... :(",
  "Solution_2": "IMHO the result is quite surprising, however you can see my proof here:\r\n\r\n[url=http://bboyjordan.wordpress.com/2009/10/31/every-number-m2m-nm-has-all-digits/]Problem 9-When n^2+1 divides n![/url]\r\n\r\n\r\nShoki, do you know the source of this problem?",
  "Solution_3": "these two problems are well-known:\r\n(1)Prove that there exists infinitely many $ n \\in N$ such that $ n^2\\plus{}1 \\not | n!$ ,and also (2) Prove that there exists infinitely many $ n \\in N$ such that $ n^2\\plus{}1 | n!$.\r\nand they are both easy.\r\ni wondered what can be the condition(s) for which $ n^2\\plus{}1 \\not |n!$ and i've found those conditions but i didn't see sth talking about these conditions before...\r\nmy solution :\r\nobviously if one of those conditions is satified by an integer $ n$ then we'll have $ n^2\\plus{}1 \\not | n!$.\r\nnow suppose that we have $ n^2\\plus{}1 \\not | n!$\r\nin this case there must exist a prime $ p$ for which $ v_p(n!) < v_p(n^2\\plus{}1)$.note that this prime can't be $ 2,3$. since $ 4,3 \\not| n^2\\plus{}1$ and so $ p \\ge 5$.\r\nwe consider three cases:\r\n1- $ v_p(n^2\\plus{}1) \\equal{} t \\ge 3$ which implies $ v_p(n!) \\le t\\minus{}1$\r\nsuppose that $ t$ is even in this case we can't have $ p^t\\equal{}n^2\\plus{}1$ and so we'll have $ p^{\\frac{t}{2}} < \\sqrt{n^2\\plus{}1}$ but since $ p^{\\frac{t}{2}}$ is an integer and we have $ [\\sqrt{n^2\\plus{}1}]\\equal{}n$ so we get $ p^{\\frac{t}{2}} \\le n$ so it follows that :\r\n$ v_p(n!) \\ge [\\frac{n}{p}]\\plus{}...\\plus{} [\\frac{n}{p^{\\frac{t}{2}}}] \\ge p^{\\frac{t}{2}\\minus{}1}\\plus{}...\\plus{}p\\plus{}1$\r\nbut we must have $ v_p(n!)<t$ or in other words $ \\frac{p^{\\frac{t}{2}}\\minus{}1}{p\\minus{}1}<t$ and the rest is trivial by inequality:\r\n$ p^{\\frac{t}{2}}\\minus{}1<t(p\\minus{}1)<tp\\minus{}1 \\implies p^{\\frac{t}{2}\\minus{}1}<t \\ , \\ p \\ge 5 \\implies 5^{\\frac{t}{2}\\minus{}1}<t$ which is clearly false for all $ t \\ge 4$.\r\nnow let's check the case for which $ t$ is odd.note that obviously $ tp >n$ (1) because $ v_p(n!)<t$.  we get $ p^{\\frac{t}{2}} \\le \\sqrt{n^2\\plus{}1}$ but in this case we'll have $ p^{\\frac{t\\minus{}1}{2}} \\le n$ and so we get similarly \r\n$ v_p(n!) \\ge \\frac{p^{\\frac{t\\minus{}1}{2}}\\minus{}1}{p\\minus{}1} \\ , \\ v_p(n!)<t \\implies p^{\\frac{t\\minus{}3}{2}}<t \\ , \\ p\\ge 5 \\implies 5^{\\frac{t\\minus{}3}{2}}<t$ which can be true for only  $ t\\equal{}3$ and also $ p\\equal{}5$.and also in this case by (1) we get $ 15 >n$.by checking the remaining cases we are done!  \r\n2- $ v_p(n^2\\plus{}1) \\equal{} 2$ which implies $ p^2 \\parallel{} n^2\\plus{}1$ but clearly $ p^2$ can't be equal to $ n^2\\plus{}1$.so we must have $ 2p^2 \\le n^2\\plus{}1$.if $ n^2\\plus{}1\\equal{}2p^2$ then we obtain the condition 2.also $ 3p^2,(2p)^2 \\neq n^2\\plus{}1$ so we have to check the case $ 5p^2 \\le n^2\\plus{}1$.now we get $ 2p \\le 2\\sqrt{\\frac{n^2\\plus{}1}{5}}$ but we must have $ 2p >n$ which implies $ 2\\sqrt{\\frac{n^2\\plus{}1}{5}} > n$ and the rest is trivial.\r\n3-$ v_p(n^2\\plus{}1) \\equal{} 1$ which is the first condition.\r\nwe r done!",
  "Solution_4": "I didn't check carefully the computational part of your solution, but I see that the resolutive idea is exactly the same, and that's enough for me (altough you didn't answer me about the source  :P ). Anyway:\r\n[quote=\"shoki\"][b](1)[/b]Prove that there exists infinitely many $ n \\in N_0$ such that $ n^2 \\plus{} 1 \\not | \\text{ }n!$[/quote]\nIt is a result of Nagel that there \"[i]For all fixed $ m \\in \\mathbb{R} \\cap (0,\\infty)$ there exist infinitely many $ n \\in \\mathbb{N}_0$ such that $ \\text{gpf}(n^2 \\plus{} 1) > mn$\"[/i], and it is generalized in a non linear form (by Hooley) in \"[i]There exist infinitely many $ n \\in \\mathbb{N}_0$ such that $ \\text{gpf}(n^2 \\plus{} 1) > n^{\\frac {11}{10}}$[/i], but the conjecture about $ |p(\\mathbb{N}_0) \\cap \\mathbb{P}| \\equal{} \\plus{} \\infty$ where $ p(n): \\equal{} n^2 \\plus{} 1$ or other non linear irreducible polynomial in $ \\mathbb{Z}[x]$ remains still unsolved. But obviusly for this problem is more than enough having in mind the easier [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=1190546#1190546]IMO 2008/3[/url].\n[quote=\"shoki\"][b](2)[/b] Prove that there exists infinitely many $ n \\in N$ such that $ n^2 \\plus{} 1 \\mid n!$.[/quote]\r\nCan the previous problem help us?Sure: taking $ 2 \\mid n$ we have that condition (2) is fulfilled, and it is enough to check that $ \\text{gpf}(n^2 \\plus{} 1) \\le n$, but $ n^2 \\plus{} 1 \\not \\in \\mathbb{P}$ when $ n \\equal{} 2k^2$ for some $ k \\in \\mathbb{N}_0$ (see [url=http://www.artofproblemsolving.com/Wiki/index.php/Sophie_Germain_Identity]Sophie-German identity[/url]): the two factors are coprime so it enough to check that $ \\text{gpf}(2k^2 \\plus{} 2k \\plus{} 1) \\le 2k^2$, and it is trivial since it holds when $ 2k^2 \\plus{} 2k \\plus{} 1 \\not \\in \\mathbb{P}$ (e.g. when $ \\frac {k \\minus{} 1}{5} \\in \\mathbb{Z}$..).[]",
  "Solution_5": "well, i'm not familiar with $ Z[i]$ and things like that , so if my solution is the same sorry for that.\r\nfor the problem 1:\r\ni (first) saw it [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=1358045#1358045]here[/url] and for the second one i don't remember the source but it has been posted many times in this forum. :)"
}
{
  "Tag": [],
  "Problem": "the chord CD of a circle with center O is prependicular to its diameter AB,and the chord AE bisects the radius \r\n\r\nOC.prove that the chord DE,bisects the chord BC.",
  "Solution_1": "what the ... are u doing?(shokhi bood)",
  "Solution_2": "Do you think this is funny? :mad: Topic locked, user banned!"
}
{
  "Tag": [
    "linear algebra",
    "matrix",
    "logarithms",
    "linear algebra unsolved"
  ],
  "Problem": "here is a nice one:\r\nprove that $E=exp(Mn(R))=f(Gl_n(R))$ where $f(M)=M^2$.",
  "Solution_1": "Sorry, I am not familiar with all high algebra notions. What is $Gl_n(\\mathbb R)$  :blush: ?",
  "Solution_2": "The multiplicative group of $n\\times n$ invertible linear matrices.",
  "Solution_3": "Paraphrasing: an invertible real matrix has a logarithm iff it has a square root. Or $\\exists\\,M$ such that $A=e^M$ iff $\\det(A)\\ne0$ and $\\exists\\,B$ such that $A=B^2.$\r\n\r\nOne direction is cheap: If $A=\\exp(M)$, then $A=\\left[\\exp\\left(\\frac12M\\right)\\right]^2.$\r\n\r\nNote that this shows that $\\det(A)>0$. Of course, we knew that anyway, from $\\det(e^M)=e^{\\text{tr}(M)}.$\r\n\r\nI don't have the main result; I'll have to think about it.",
  "Solution_4": "Yes, one direction is clear...\r\nIs it possible that the result is true for an arbitrary banach space?",
  "Solution_5": "Anybody knows how to prove the other direction?",
  "Solution_6": "it has been proved on the forum on an other thread (started by Harazi I think ..)"
}
{
  "Tag": [
    "geometry",
    "IMO",
    "circumcircle",
    "perpendicular bisector",
    "angle bisector",
    "IMO 1959"
  ],
  "Problem": "An arbitrary point $M$ is selected in the interior of the segment $AB$. The square $AMCD$ and $MBEF$ are constructed on the same side of $AB$, with segments $AM$ and $MB$ as their respective bases. The circles circumscribed about these squares, with centers $P$ and $Q$, intersect at $M$ and also at another point $N$. Let $N'$ denote the point of intersection of the straight lines $AF$ and $BC$.\r\n\r\na) Prove that $N$ and $N'$ coincide;\r\n\r\nb) Prove that the straight lines $MN$ pass through a fixed point $S$ independent of the choice of $M$;\r\n\r\nc) Find the locus of the midpoints of the segments $PQ$ as $M$ varies between $A$ and $B$.",
  "Solution_1": "I shall use analytic geometry to solve this problem.\r\n\r\n(1): Define coordinate system: Let $M$ be the origin and $AB$ be the x-axis. \r\nLet $A: (-a,0)$ and $B: (b,0)$. It is well-known that the circumcentre of a square is the intersection of its diagonals. Hence: $P: (-\\dfrac{a}{2}, \\dfrac{a}{2})$ and $Q: (\\dfrac{b}{2}, \\dfrac{b}{2})$.\r\nLet the circumcircles of $AMCD$ and $MBEF$ be $C_1$ and $C_2$, thus we can write the equations of the circle: \r\n\\[ C_1: (x+\\dfrac{a}{2})^2+(y-\\dfrac{a}{2})^2=R_1^2= \\dfrac{a^2}{2} \\]\r\n\\[ C_2: (x-\\dfrac{b}{2})^2+(y-\\dfrac{b}{2})^2=R_2^2= \\dfrac{b^2}{2} \\]\r\nExpand and notice that we can cancel out $\\dfrac{a^2}{2}$ and $\\dfrac{b^2}{2}$, the equations become:\r\n\\[ C_1: x^2+ax+y^2-ay=0 \\cdots (1) \\]\r\n\\[ C_2: x^2-bx+y^2-by=0 \\cdots (2) \\]\r\nNow comes the computation of finding the intersection of the two circles. First we substract $(2)$ from $(1)$, which gives:\r\n\\[ (a+b)x+(b-a)y=0 \\]\r\n\\[ \\Longleftrightarrow (a+b)x=(a-b)y \\]\r\n\\[ \\Longleftrightarrow x=\\dfrac{a-b}{a+b}\\cdot y\\cdots (3) \\]\r\nPlug $(3)$ into $(1)$, we get:\r\n\\[ y^2 \\cdot (\\dfrac{a-b}{a+b})^2 + \\dfrac{a(a-b)}{a+b}\\cdot y +y^2-ay=0 \\]\r\nBecause by the definition of the coordinate system, we know that one intersection of $C_1$ and $C_2$ is at the origin. But we want the other one. Hence we can divide the two sides of the equation by $y$:\r\n\\[ y \\cdot (\\dfrac{a-b}{a+b})^2+\\dfrac{a(a-b)}{a+b}+y-a=0 \\]\r\nFactoring: \r\n\\[ y\\cdot (\\dfrac{a^2-2ab+b^2}{(a+b)^2}+1)=a-\\dfrac{a(a-b)}{a+b} \\]\r\n\\[ y\\cdot (\\dfrac{a^2-2ab+b^2+a^2+2ab+b^2}{(a+b)^2})=a\\cdot (\\dfrac{a+b+(b-a)}{a+b}) \\]\r\n\\[ y\\cdot (\\dfrac{2a^2+2b^2}{(a+b)^2})=\\dfrac{2ab}{a+b} \\]\r\n\\[ y\\cdot \\dfrac{a^2+b^2}{a+b}=ab \\]\r\n\\[ y=\\dfrac{ab(a+b)}{a^2+b^2} \\]\r\nPlug $y$ into $(3)$ we get:\r\n\\[ x=\\dfrac{ab(a-b)}{a^2+b^2} \\]\r\nTherefore we get:\r\n\\[ N: (\\dfrac{ab(a-b)}{a^2+b^2},\\dfrac{ab(a+b)}{a^2+b^2}) \\]\r\n\r\nNow we've done most of the work. Since we are trying to prove that $A$, $F$ and $N$ are collinear, so:\r\n\\[ \\dfrac{\\dfrac{ab(a+b)}{a^2+b^2}-0}{\\dfrac{b(a-b)}{a^2+b^2}+a}=\\dfrac{b(a+b)}{b(a-b)+a^2+b^2}=\\dfrac{b(a+b)}{a(a+b)}=\\dfrac{b}{a} \\]\r\nAnd: \r\n\\[ \\dfrac{b-\\dfrac{ab(a+b)}{a^2+b^2}}{0-\\dfrac{ab(a-b)}{a^2+b^2}}=\\dfrac{a^2+b^2-(a^2+ab)}{a(b-a)}=\\dfrac{b(b-a)}{a(b-a)}=\\dfrac{b}{a} \\]\r\nThus by analytic geometry, $N$, $A$, $F$ are collinear. \r\nDo the same for $N$, $C$, $B$, we can also prove these points are collinear. Therefore question (1) is solved.\r\n\r\n\r\n(2): The formula of the perpendicular bisector of $AB$ is $x=\\dfrac{b-a}{2}$. The equation of line $NM$ is $y=\\dfrac{a+b}{a-b}\\cdot x$. Thus the y-ordinate of the intersection of the two lines it:\r\n\\[ y=-\\dfrac{a+b}{2} \\]\r\nBecause $a+b=AB$, $AB$ is a contant value, hence the $MN$ passes through a fixed point. The point is the third vertex of the isoscele right triangle with side $AB$.\r\n\r\n(3): $P: (-\\dfrac{a}{2},\\dfrac{a}{2})$, $Q: (\\dfrac{b}{2}, \\dfrac{b}{2})$. Let the mid-point of $PQ$ be $X: (x_0, y_0)$. Hence:\r\n\\[ x_0=\\dfrac{x_1+x_2}{2}=\\dfrac{b-a}{4} \\]\r\n\\[ y_0=\\dfrac{y_1+y_2}{2}=\\dfrac{a+b}{4} \\]\r\nOnce again, $a+b$ is constant. Hence the locus of $X$ is a segment: $y=\\dfrac{a+b}{4}$, $x \\in [-\\dfrac{a+b}{2}, \\dfrac{a+b}{2}]$.\r\n\r\nThe problem is solved.\r\n\r\nP.S I am also interested in a geometrical proof.  :)",
  "Solution_2": "Geometrical proof is quite easy. \r\n\r\na) We have $\\angle CBM = \\angle AFM$ so $\\angle AN'B = \\frac{\\pi}{2}$. It means, that $N'$ lay on the circle circumscribed about squares $AMCD$ and $MBEF$ so $N = N'$. \r\n\r\nb) Triangles $CMB$ and $ANB$ are similar, hence $\\frac{CM}{MB} = \\frac{AN}{NB}$ and $\\frac{AM}{MB} = \\frac{AN}{NB}$. It menas that $NM$ is bisector of $\\angle ANB$. $N$ lay on a circle with center in midpoint of $AB$ and radius $\\frac{AB}{2}$ so $NM$ always pass through center of arc $AB$.",
  "Solution_3": "I am sorry, Shobber, the genuine and shortest solution is the Michael Marcinkovski's !",
  "Solution_4": "a) $ \\angle ANM \\equal{} \\angle MNC \\equal{} \\angle FNE \\equal{} \\angle MNB \\equal{} \\angle BNE \\equal{} 45^\\circ$. So $ A,N,F$ are collinear, $ B,C,N$ are collinear.\r\n\r\nb) from a, we know $ NM$ is angle bisector of $ \\angle ANB$. So $ NM$ always pass through center of arc $ AB$.\r\n\r\nc) Draw perpendicular lines from $ P$, $ Q$, and the midpoint ($ R$) to the segment $ AB$ (Call them $ X, Y, Z$). $ PX \\equal{} AM / 2$, $ QZ \\equal{} MB / 2$. So $ RY \\equal{} \\frac {AM \\plus{} MB} {4} \\equal{} AB/4$ which means a parallel line segment to $ AB$.",
  "Solution_5": "Dear Mathlinkers,\nan article concerning the Vecten's figure and its developpement can be seeing on my site\nhttp://perso.orange.fr/jl.ayme , la figure de Vecten  vol. 5 p. 25\nSincerely\nJean-Louis",
  "Solution_6": "[hide=a]We use coordinates. Let $A$ be the origin, $M$ be a point $(2a,0)$ and $B$ be the point $(2b,0)$. It is not hard to see that $D$ has coordinates $(0,2a)$, $C$ has coordinates $(2a,2a)$, $E$ has coordinates $(2b,2b-2a)$, and $F$ has coordinates $(2a,2b-2a)$. We have that the equation of $AF$ is $y = \\left( \\frac{b-a}{a}\\right)x$ and the equation of $BC$ is $y = \\left( \\frac{a}{a-b} \\right) x - 2b \\left( \\frac{a}{a-b}\\right)$. Their intersection, or in other words, point $N'$  has coordinates $\\left( \\frac{2a^2}{2a-b}, \\frac{2ab-2a^2}{2a-b}\\right)$. It is easy to prove that $PQ \\perp N$. Therefore, $N'$ lies on the radical axis of $(DCAM)$ and $(FEBM)$. Now, all that we need to prove is that $N'$ lies on $(DCMA)$. This is the same as proving $DN \\perp NM$, which we can confirm is indeed true. Therefore, $N$ is point $N\u2019$.[/hide]\n[hide=b] Since $DNCA$ is cyclic, $\\angle ANB = 90$. Therefore, $$\\triangle CBM \\sim \\triangle ABN \\implies \\frac{CM}{AN} = \\frac{BM}{BN} \\implies \\frac{AM}{AN} = \\frac{BM}{CN}.$$ Thus, $MN$ bisects $\\angle ANB$. As a result, $MN$ passes through the bisector of arc $AB$. [/hide]\n[hide=c]We use the same coordinates as we did in part a. Note that since $P$ has coordinates $(a+b,b-a)$ and $Q$ has coordinates $(a,a)$, we know that the midpoint of $PQ$ is $\\left( \\frac{2a+b}{2},\\frac{b}{2}\\right)$. Since $b$ is a constant, the locus of the midpoints of $PQ$ is a line.[/hide]",
  "Solution_7": "Solved with [b]franchester[/b], [b]ingenio[/b], [b]tworigami[/b], [b]GameMaster402[/b], [b]anish9876[/b], [b]peeyushmaths[/b], [b]AIME12345[/b], [b]SD2014[/b], [b]AOPS12142015[/b], [b]huangyi_99[/b], [b]sriraamster[/b], [b]kothasuhas[/b], [b]budu[/b], and [b]mathfun5[/b].\n\nThe central motivation of the following is that since we are cognizant of the placement at a measly #5 of a trivial contest, the use of genuine advanced techniques is not imperative.\n\n(a) First, we show that $N'$ lies on the circle with diameter $AC$. We proceed with moving points, varying $M$ on $AB$ in a linear fashion. We desire the demonstration of the following:\n\n-----\n[color=#f00][b]Lemma:[/b][/color] The map $$M \\rightarrow C \\rightarrow BC \\cap (AB)$$ tethered to $$AB \\rightarrow \\ell \\rightarrow (AB)$$\nwhere $\\ell$ is the fixed angle bisector of $\\angle BAD$, as well as the corresponding map for $A$ is projective, and these two maps are identical.\n\n[i]Proof:[/i] To show that it is projective, note that it is a composition of two perspectivities centered at $\\infty_{AD}$ and $B$, respectively. The proof is similar for the corresponding map for $A$. Now, it suffices to check that they coincide in $3$ cases. $M \\rightarrow A$, $M \\rightarrow B$, and $MA=MB$ are simple, so the maps are the same, as desired.\n-----\n\nNow, we know that $N'$ lies on the circle with diameter $AB$, so $\\angle AN'C = 90^{\\circ} = \\angle ADC$ and $\\angle BNF = 90^{\\circ} = \\angle BMF$,  so $N'$ lies on the same two circles as $N$, and since $N' \\neq M$, implying $N' = N$, as desired. $\\blacksquare$\n\n(b) The simple lemma preceding the clearly intended solution can be stated as follows:\n-----\n[color=#f00][b]Lemma:[/b][/color] Let $\\alpha\\beta\\gamma$ be a triangle with $\\angle \\beta = 90^{\\circ}$. The bisector of $\\angle \\beta$ meets $AC$ at $\\delta$, the midpoint of $\\beta \\delta$ is $\\mu$, and point $\\tau$ on the opposite side of $\\beta \\gamma$ satisfies $\\tau \\beta = \\tau \\gamma$ and $\\angle \\beta \\tau \\gamma = 90^{\\circ}$. Then, $\\alpha$, $\\mu$, $\\tau$ are collinear.\n\n(I genuinely recommend trying to prove that, it's pretty cool)\n\n[i]Proof:[/i] First, project $(\\beta,\\delta;\\mu,\\infty)=-1$ through $\\alpha$ to get that if $\\kappa$ lies on $BC$ with $\\angle \\beta \\kappa \\alpha = 45^{\\circ}$ and $\\chi$ is the intersection of $\\alpha \\mu$ with $\\beta \\gamma$, $(\\beta,\\gamma;\\chi,\\kappa)=-1$.\n\nNow, let $\\tau'$ be the reflection of $\\tau$ over $\\beta \\gamma$ and let $\\tau'\\kappa$ meet the circle with diameter $BC$ at $\\phi$. Projecting $(\\beta,\\gamma;\\chi,\\kappa)=-1$ through $\\phi$ gives $\\phi$, $\\kappa$, $\\tau$ collinear.\n\nFinally, by Pascal's theorem, $\\kappa = \\tau'\\phi \\cap \\beta \\gamma$, $\\infty_{\\kappa \\alpha} = \\beta\\tau' \\cap \\tau \\gamma$ and $\\beta \\beta \\cap \\tau\\phi$ are collinear. Since $\\kappa\\infty_{\\kappa \\alpha}$ and $\\beta\\beta$ meet at $\\alpha$, $\\tau\\phi$ also goes through $\\alpha$.\n\nTherefore, $\\tau$, $\\kappa$, $\\phi$, $\\mu$ $\\alpha$ are collinear, as desired.\n-----\nNow, we claim that $MN$ passes through the the point through the point $T$ on the opposite side of $BC$ satisfying $TB=TC$ and $\\angle BTC = 90^{\\circ}$.\n\nBy Pascal's Theorem on points on the circle with diameter $AC$, $AA \\cap MN$, $AD \\cap NC$, and $AC \\cap DM = P$ are collinear. Applying the lemma to the triangle formed by $AD \\cap BC$, $A$, and $C$ gives $J,P,T$ collinear (since $PA=PC$ and $\\angle DAC = \\angle MAC = 45^{\\circ}$). It is easy to check that $AA$ passes through $T$ as well, so $N,M,T$ collinear, as desired. $\\blacksquare$\n\n(c) The heart of this part (which I actually really love, the wedding is in May) is in the following lemma (which I also recommend trying)\n\n-----\n[color=#f00][b]Lemma:[/b][/color] Let $ABC$ be a triangle, let $D$ be a point on $BC$, pick $E$ on $AC$ and $F$ on $AB$ with $DE || AB$ and $DF || BC$. Then, the circumcenter of $AEF$ lies on the perpendicular bisector of $A$ and the midpoint of the $A$-symmedian chord.\n\n[i]Proof:[/i] Trivially, it suffices to show that the midpoint (call it $K$) of the $A$-symmedian chord lies on $(AEF)$. It is well known (and you should feel bad if you don't know this /s), $K$ is the center of the spiral similarity mapping $AC$ to $BA$.\n\nWe have $BF:FA=BD:DC=AE:EC$, so this spiral similarity also maps $E$ to $F$. Thus, $\\angle EKF = \\angle BKA = 180^{\\circ} - \\angle A = 180^{\\circ} - \\angle EAF$, where $\\angle BKA = 180^{\\circ} - \\angle A$ is also well-known. Thus, we have $AEFK$ cyclic, as desired.\n-----\n\nBack to this trivial problem, let $Z = AP \\cap BQ$. We can see that $ZAB$ is a 45-45-90 triangle with right angle at $Z$, and $MP || BZ$ and $MQ || AZ$. Clearly, $\\angle PAQ = 90^{\\circ}$, so the circumcenter of $APQ$ is the midpoint of $PQ$.\n\nBy the lemma, since the midpoint of $BC$ is the midpoint of the $Z$-symmedian chord of $(ZBC)$, the midpoint of $PQ$ varies on the perpendicular bisector of $Z$ and the midpoint of $BC$, as desired. $\\blacksquare$",
  "Solution_8": "[quote=mira74]The heart of this part (which I actually really love, the wedding is in May) [/quote]\n\nAm I invited?\n",
  "Solution_9": "[color=#960000][b](a)[/b][/color] $M$ - center spiral similarity that sends $AC$ to $FB$. In other words it's a Miquel point of $ACBF$ and intersection of sides $AF$, $BC$ is common point of $\\odot MAC$ and $\\odot MFB$.\n\nEDIT: looking on diagram again and, oh my, it's just obvious: $F$ orthocenter of $ABC$ and conclusion follows."
}
{
  "Tag": [
    "function",
    "trigonometry",
    "parameterization",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "A particle of mass m moves in a horizontal straight line undera force equal to m(n^2) times the displacement from a point O on the line and directed towards O;in addition the motion of the particle is resisited by a force equal to mk times the square of the speed.\r\n\r\nThe particle is projected  with speed V from O along the line\r\n\r\nBy considering the equation of motion in dimensionless form show that the displacement  x is of the form  x =(V/n). f( u/V ,kV/n)  where u is the non zero speed of the particle when at x.\r\nShow that the particle first comes to rest  at a distannce z  of the form z=V/n.g(kV/n)\r\n\r\nIf kV/n is small show that g(kV/n) is approximately 1-(2kV)/(3n)",
  "Solution_1": "The equation of movement is $m\\frac{dv}{dt}=-mn^2 x+mkV^2$,right?",
  "Solution_2": "Kunny\r\n\r\nV is the initial speed at O .Let u be the speed when particle is distant x from O the equation of motion is\r\n\r\n\r\nm.dv/dt = -m(n^2)x - mk(u^2)   AS both forces are resisitances we must take negative signs \r\n\r\n\r\n\r\n\r\nMichael",
  "Solution_3": "Thank you, Michael.\r\nSorry for asking,again.\r\n\r\nThe equation of the motion is $m\\frac{d^2 x}{dt^2}=-mn^2x-mk\\left(\\frac{dx}{dt}\\right)^2$\r\n\r\nInitial condition At $t=0$ $x'(t)=V,x(t)=0$ Is this right?\r\n\r\nkunny",
  "Solution_4": "We can write the equation of the motion so:\r\n\\[\r\nm\\frac{d^2x}{dt^2}+mk\\left|\\frac{dx}{dt}\\right| \\frac{dx}{dt}+mn^2x=0\r\n\\]\r\nThe damping force has to be written as proportional to $|\\dot{x}|\\dot{x}$ rather than to $\\dot{x}^2$ to ensure that its direction is always such as to oppose the  motion.\r\n$\\dot{x}=\\frac{dx}{dt}$",
  "Solution_5": "Kunny\r\n\r\n\r\nConcerning your second post:\r\n\r\nThe equation of motion is correct and the initial conditions are correct-at time t=0 the particle is projected from the originO (in the positive direction) with a speed V.Now you must solve the question.Good luck!\r\n\r\n\r\n\r\nMichael",
  "Solution_6": "Thank you for your reply, Michael. :) \r\n\r\nI will try it.\r\n\r\nkunny",
  "Solution_7": "The differential equation is not exactly soluble by elementary methods. We can use a trial function (TF) of the form:\r\n\\[\r\nx*=A_0(1-Kt/T)\\cos \\omega t\r\n\\]\r\nwhere $  K=\\frac {\\textrm {decay of amplitude in one cycle}}{\\textrm{amplitude at start of cycle}},\\quad \\omega_0=n,\\quad T=2\\pi/\\omega$ and assumed that the damping parameter mk is small.\r\nFind the residual equation $\\mathfrak {R} $ by substitution of TF in the differential equation. Then let's use \"harmonic balancing\" to find $\\omega$ and K.",
  "Solution_8": "[quote=\"Nen\u00e9\"]We can use a trial function (TF) of the form:\n\\[\nx*=A_0(1-Kt/T)\\cos \\omega t\n\\]\n[/quote]\r\nI'm not sure this will work (at any rate I couldn't make anything rigorous out of this idea). On the other hand, the first part of the problem is a simple exercise in scaling. The second part ($g(t)\\approx 1-\\frac23t$) is also doable but requires more careful analysis of the equation. I'd like to wait until kunny comes up with his solution though :)",
  "Solution_9": "I have just solved the differential equation.\r\n\r\n$m\\frac{d^2 x}{dt^2}=-mn^2 x-mk(\\frac{dx}{dt})^2$\r\n\r\n\r\n\\[u=|\\frac{dx}{dt}|=\\sqrt{(V^2-\\frac{n^2}{2k^2}})e^{-2kx}-\\frac{n^2}{k}x+\\frac{n^2}{2k^2}\\]\r\n\r\nI will post the detail later.\r\n\r\nkunny"
}
{
  "Tag": [
    "MATHCOUNTS"
  ],
  "Problem": "Greetings! I am hosting a mock MATHCOUNTS Masters' Round Contest on AoPS. For more information, see my thread in the MATHCOUNTS forum at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=205407.",
  "Solution_1": "Umm, why is this in Classroom math?\r\n :huh:",
  "Solution_2": "This is here so that more people see and sign up for the contest. The contest is MATHCOUNTS-based and part of middle-school math, so I posted it here as well as the highschool beginner area.",
  "Solution_3": "[quote=\"12markkram34\"]This is here so that more people see and sign up for the contest.[/quote]\r\n\r\nWhich doesn't mean posting the same thread in multiple forums  :wink: . Someone lock this."
}
{
  "Tag": [
    "search",
    "national olympiad"
  ],
  "Problem": "[color=red][update: see the partial results at the bottom of the page][/color]\r\n[color=red][update: all the tests problems have links here][/color]\r\n\r\nHi. These days the Romanian Team Selection tests for the Balkan MO and Junior Balkan MO are taking place in Bucharest. These tests also count for the selection of the IMO team. \r\n\r\n[u]You can check the problems in the download section here http://www.mathlinks.ro/index.php?f=121[/u]\r\nAlso I will put the links where the problems are discussed on MathLinks, in case you are too lazy to search (for source only ;) ) for them. \r\n\r\nSENIORS (for Balkan MO, IMO and Bulgarian MO, Yacutz Intl. Contest)\r\n\r\n1st Test\r\n\r\nproblem 01 - http://www.mathlinks.ro/viewtopic.php?t=5448\r\nproblem 02 - http://www.mathlinks.ro/viewtopic.php?t=5446\r\nproblem 03 - http://www.mathlinks.ro/viewtopic.php?t=5449\r\nproblem 04 - http://www.mathlinks.ro/viewtopic.php?t=5450\r\n\r\n2nd Test \r\n\r\nproblem 05 - http://www.mathlinks.ro/viewtopic.php?t=5451\r\nproblem 06 - http://www.mathlinks.ro/viewtopic.php?t=5453\r\nproblem 07 - http://www.mathlinks.ro/viewtopic.php?t=5452\r\nproblem 08 - http://www.mathlinks.ro/viewtopic.php?t=5447\r\n\r\n3rd Test\r\n\r\nproblem 09 - http://www.mathlinks.ro/viewtopic.php?t=5488\r\nproblem 10 - http://www.mathlinks.ro/viewtopic.php?t=5487\r\nproblem 11 - http://www.mathlinks.ro/viewtopic.php?t=5485\r\nproblem 12 - http://www.mathlinks.ro/viewtopic.php?t=5486\r\n\r\n4th Test\r\n\r\nproblem 13 - http://www.mathlinks.ro/viewtopic.php?t=5859\r\nproblem 14 - http://www.mathlinks.ro/viewtopic.php?t=5860\r\nproblem 15 - http://www.mathlinks.ro/viewtopic.php?t=5861\r\n\r\n5th Test\r\n\r\nproblem 16 - http://www.mathlinks.ro/viewtopic.php?t=5889\r\nproblem 17 - http://www.mathlinks.ro/viewtopic.php?t=5888\r\nproblem 18 - http://www.mathlinks.ro/viewtopic.php?t=5887\r\n\r\nJUNIORS (for Junior Balkan MO)\r\n\r\n1st Test\r\n\r\nproblem 01 - http://www.mathlinks.ro/viewtopic.php?t=5457\r\nproblem 02 - http://www.mathlinks.ro/viewtopic.php?t=5455\r\nproblem 03 - http://www.mathlinks.ro/viewtopic.php?t=5459\r\nproblem 04 - http://www.mathlinks.ro/viewtopic.php?t=5458\r\n\r\n2nd Test\r\n\r\nproblem 05 - http://www.mathlinks.ro/viewtopic.php?t=5460\r\nproblem 06 - http://www.mathlinks.ro/viewtopic.php?t=5461\r\nproblem 07 - http://www.mathlinks.ro/viewtopic.php?t=5456\r\nproblem 08 - http://www.mathlinks.ro/viewtopic.php?t=5454\r\n\r\n3rd Test \r\n\r\nproblem 09 - http://www.mathlinks.ro/viewtopic.php?t=5489\r\nproblem 10 - http://www.mathlinks.ro/viewtopic.php?t=5491\r\nproblem 11 - http://www.mathlinks.ro/viewtopic.php?t=5490\r\nproblem 12 - http://www.mathlinks.ro/viewtopic.php?t=5492\r\n\r\n4th Test \r\n\r\nproblem 13 - http://www.mathlinks.ro/viewtopic.php?t=5824\r\nproblem 14 - http://www.mathlinks.ro/viewtopic.php?t=5825\r\nproblem 15 - http://www.mathlinks.ro/viewtopic.php?t=5827\r\nproblem 16 - http://www.mathlinks.ro/viewtopic.php?t=5826\r\n\r\n5th Test \r\n\r\nproblem 17 - http://www.mathlinks.ro/viewtopic.php?t=5890\r\nproblem 18 - http://www.mathlinks.ro/viewtopic.php?t=5893\r\nproblem 19 - http://www.mathlinks.ro/viewtopic.php?t=5892\r\nproblem 20 - http://www.mathlinks.ro/viewtopic.php?t=5891",
  "Solution_1": "Hi folks.\r\n\r\nThe results are as follows:\r\n\r\nThe Romanian Balkan MO Team 2004 is \r\n\r\nNegut Andrei (aka Andrei Negut :D)\r\nKreindler Gabriel (aka Kappa)\r\nUngureanu Bogdan\r\nPachitariu Marius\r\nDinu Razvan (aka mZero)\r\nRosoiu Alexandru (aka ral) \r\n\r\nGOOD LUCK guys!\r\n\r\nRunner ups (and also the Romanian Team going to Bulgarian National MO) are\r\n\r\nZahariuc Adrian (aka dzeta)\r\nStefanescu Andrei (aka andreis)\r\nIonescu Dragos Ciprian \r\nBazavan Eduard\r\nTalau Cristian\r\nTurea Lucian  \r\n\r\nThe last 6, except Turea Lucian, and also \r\n\r\nGratie Cristian (aka blondu)\r\nIsmail Andrei (aka iandrei) \r\n\r\nare runner-ups for the IMO team, for the two remaining TSTs at the end of May. \r\n\r\nThe scores are attached in the xls file.",
  "Solution_2": "Maybe you know the fantastic math olympiad [url=http://140.247.141.165/~ajorza/math/]site[/url] by Andrei Jorza. [url=http://140.247.141.165/~ajorza/math/mathfiles/selection/]Here[/url] you can find a collection Romanian TSTs. But it seems that grobber has transformed to Gorbber...  :D  BTW how did you come up with your user name ?\r\n\r\nThe same for harazi. I read it has some connection with reverting his name. Maybe harazi wants to tell me.  :)",
  "Solution_3": "No, harazi wants to keep secret!  ;) Friends know why. By the way, I like very much the very false and idiot solution that appear on one of these materials from the cited site. Man, how could they write something like this? Now, guess what's the problem I am talking about.  ;)",
  "Solution_4": "[quote=\"orl\"]Maybe you know the fantastic math olympiad [url=http://140.247.141.165/~ajorza/math/]site[/url] by Andrei Jorza. [url=http://140.247.141.165/~ajorza/math/mathfiles/selection/]Here[/url] you can find a collection Romanian TSTs. But seems to have transformed to Gorbber...  :D  BTW how did you come up with your user name ?\n\nThe same for harazi. I read it has some connection with reverting his name. Maybe harazi wants to tell me.  :)[/quote]Yes I know the site, I also gaved him some materials, but unfortunately, as Harazi pointed out, there are many flaws in his papers.",
  "Solution_5": "DEAR VALENTIN VORNICU!WHERE COULD I GET ROMANIAN 2003,2002 OLYMPIADS AND SELECTION TESTS?",
  "Solution_6": "[quote=\"spider_boy\"]DEAR VALENTIN VORNICU!WHERE COULD I GET ROMANIAN 2003,2002 OLYMPIADS AND SELECTION TESTS?[/quote]Please [b]do not[/b] use CapsLock on (only capital letters) to write your messages. \r\n\r\nYou can get the most of the Romanian 2003 and 2002 selection tests in the Forums. Just search for them.",
  "Solution_7": "[quote=\"orl\"]Maybe you know the fantastic math olympiad [url=http://140.247.141.165/~ajorza/math/]site[/url] by Andrei Jorza. [url=http://140.247.141.165/~ajorza/math/mathfiles/selection/]Here[/url] you can find a collection Romanian TSTs. But it seems that grobber has transformed to Gorbber...  :D  BTW how did you come up with your user name ?\n\nThe same for harazi. I read it has some connection with reverting his name. Maybe harazi wants to tell me.  :)[/quote]\r\n\r\nI'm sorry,but I can't open sites this.Why?"
}
{
  "Tag": [],
  "Problem": "An exciting new ride in a theme park is described as follows: \"The vehicle plus riders takes only a few seconds to accelerate to a spine-tingling 53m/s. Then (with motor off) it shoots vertically upwards to the height of a forty-two storey building and pauses.\" \r\n\r\nCalculate the time for the vehicle plus riders to make the vertical climb to 126m. State an assumption that you make.",
  "Solution_1": "The applicable formula here is the following:\r\n$ \\Delta x \\equal{} x_o \\plus{} V_0*t \\plus{} (\\frac {1}{2})(a)(t^2)$\r\n\r\nSo we have $ 126m \\equal{} 0 \\plus{} (53 \\frac {m}{s})(t) \\minus{} (4.9 \\frac {m}{s^2})(t^2)$\r\nSolve for $ t$, and you have $ t \\equal{} 3.528 \\text{ or } 7.288$\r\nObviously, it will reach 126m twice (once up, once down) so we take the lesser of the two values.\r\n\r\nThe implicit assumption we are making is that acceleration due to gravity is constant."
}
{
  "Tag": [
    "AMC"
  ],
  "Problem": "answers for AMC 10B 2008 anyone...\r\n\r\nThanks!!!",
  "Solution_1": "http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2008-12a/07-2-1012-Answers.shtml",
  "Solution_2": "ahhhh thank you soo much!!!"
}
{
  "Tag": [
    "geometry",
    "3D geometry"
  ],
  "Problem": "In how many ways can 8 people be seated in a row of chairs if three of the people, John, Wilma and Paul, refuse to sit in consecutive seats?",
  "Solution_1": "Deal with restriction first so first we have an arrangement of 3 people-3!=6\r\nThen to deal with the restriction we place people like this *|*|*so thats 6*5*4 so far. Then from there, there are 6*7*8 ways for the other people.\r\n6*5*4*6*7*8 but we have to divide by 5! because the other people can be seated in any order that are the same but we counted it twice-this is probably really unclear. So we have 336 as the answer.",
  "Solution_2": "Shentang, I don't think your solution is correct.\r\n\r\nBasically, consider the three people as a superperson.  Now, the number of ways we can arrage them when they sit together is clearly $ 6! \\cdot 3!$.  Now we subtract this from $ 8!$ to get $ 40320 \\minus{} 4320 \\equal{} \\boxed{36000}$.",
  "Solution_3": "I don't get the part with 6!*3!.\n:( Help?",
  "Solution_4": "I don't like this problem...are we restricting the permutations so that:\n1)  The three people listed can't sit together in that specified order?\n2)  The three people listed can't sit in three consecutive seats, but putting two of them next to each other is okay?\n3)  Any two of the three people listed can't sit next to each other?\n\nI'm guessing that 2 is the actual problem, but a reminder to anyone tinkering around writing combinatorics problems:  be sure to state what's distinct and what isn't!  Is a vanilla on top-chocolate on bottom ice cream cone the same as a chocolate-vanilla ice cream cone?  Not to teenager, but just try and convince a 4 year-old that it tastes the same.\n\nSo, that said, I'll try and crack the problem with option 2 being the restriction.\n\n[hide]If we bunch the three people together as one person, sitting them together, and have the other five people sit individually, there are now six \"groups\" of people to seat, so there are $6!$ ways to seat them.  But within that one group of three, there are $3!$ ways to seat them.  Thus, there are $6!3! =  4320$ ways to seat everyone so that the trio are in three consecutive seats.  All of these permutations are illegal, and all illegal permutations have been accounted.\n\nThere are $8! = 40320$ ways to seat everyone without restrictions.  Thus, there are $40320 - 4320 = 36000$ ways to seat everyone so that the three people are not in three consecutive seats.[/hide]",
  "Solution_5": "It's 2. This is just how problems are written, @bove. There is no need to doubt yourself.\n\n[quote=elevate]I don't get the part with 6!*3!.\n:( Help?[/quote]\n\n[hide=answer/basically sol]If we want everyone to be together, we can \"group\" John, Wilma, and Paul in one group, so there are as if $6$ people, so $6!$ ways. There are also $3!$ ways to order our people in the group inside the group. So our answer is $6!\\cdot3!$[/hide]\n\nIf anyone tells me I bumped, I will rage.",
  "Solution_6": "[hide]\nConsider 8 slots\n\n_ _ _ _ _ _ _ _\n\nNow, JWP is a person and there are 6 ways to insert them into the slots, together, and there are 3! ways to arrange them once they are adjacent, so 6*6, and the rest of the 5 people can be arranged in 5! ways\nAns: 8!-6*6*5!=36000\n[/hide]\n",
  "Solution_7": "Use complementary counting: \n[hide]\n\n1. Think of John Wilma and Paul stuck together (remember you are finding the opposite and then subtracting)\n2. If JWP is 1 person, then there are 6 people in total\n3. How many ways can you arrage 6 people? --> 6! = 720\n4. There can be 6 ways to arrange JWP so dp 720 * 6 = 4320\n5. Subtract :) --> there are 8! ways to arrange everyone so just do:\n8! - 4320 = [b]36000[/b]\n[/hide]",
  "Solution_8": "ughgghg\ni forgot the fact that the 3 people could be rearranged"
}
{
  "Tag": [
    "invariant",
    "calculus",
    "calculus computations"
  ],
  "Problem": "Find the volume of the shape formed by rotating the graph  \r\n\\[ x^{2}\\plus{}12(y\\minus{}10e^{2009})^{2}\\minus{}1728\\equal{}0\\] about the y axis.",
  "Solution_1": "The $ 10e^{2009}$ doesn't matter, as it only translates the region of interest along the y-axis, and the volume is invariant under this translation.  After this observation it should become immediately obvious what the shape is and how to calculate its volume.",
  "Solution_2": "and that is......"
}
{
  "Tag": [
    "linear algebra",
    "matrix"
  ],
  "Problem": "Consider the set G of all 2\u00d72 non-zero matrices of the\r\nform (a b )\r\n.......(b a + b)\r\nwhere a, b belong to Z3, the set of congruence classes modulo 3.\r\n\r\n(1) Find the number of elements in the set G.\r\n\r\n(2) Prove that G is a group under matrix multiplication modulo 3.\r\n\r\n(3) Complete the multiplication table for G.\r\n\r\n(4) Is the group G isomorphic to the group of symmetries of a square?\r\n\r\n(two groups are isomorphic if there exists a bijection\r\nbetween them that preserves the group operations, in other words,\r\nafter an appropriate permutation and relabeling of the elements,\r\ntheir multiplication tables are identical.",
  "Solution_1": "This is not the place for group theory problems. Go to the Superior Algebra forum in the College Playground.",
  "Solution_2": "Or the linear algebra forum.  But stop posting college level math here!"
}
{
  "Tag": [
    "group theory",
    "abstract algebra",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "Show that the group which is freely generated by $ n$ elements contains a normal subgroup of index 2.",
  "Solution_1": "If F(S) is the free group over the (non empty) set S and $ s_0\\in S$ then the normal closure of $ \\{s_0^2\\}\\cup\\{s\\neq s_0\\}$ in F(S) is such a normal subgroup.",
  "Solution_2": "I prefer the group of all elements represented by a string of even length.",
  "Solution_3": "I was thinking, are you sure that the strings that have even length form a normal subgroup of the free group? Because I was thinking, let's say that our group is freely generated by $ r, s$ and $ t$. So $ rsrt$ is in our normal subgroup. So is $ trsr$. But then $ rsrttrsr\\equal{}rsrt^{2}rsr$ which doesn't have even length? Or does it? I seem to be confused on what you mean by even length.",
  "Solution_4": "$ t^2\\equal{}tt$ has even length if we're going to make any sense of the idea.\r\n\r\nIf you prefer, we're defining a homomorphism $ \\sigma: F\\to \\{\\pm1\\}$ by letting $ \\sigma$ be $ \\minus{}1$ on each generator.  Such a homomorphism is guaranteed to exist by the universal property of free groups.  Call a string \"even\" if its image is $ \\plus{}1$.  There you go."
}
{
  "Tag": [
    "function",
    "combinatorics proposed",
    "combinatorics"
  ],
  "Problem": "Let $ x_1,x_2,\\dots,x_n$ be real numbers so that $ |x_i|\\geq 1$ for each $ i$, and let $ \\mathcal{J}$ be any interval of length $ 2$ open at at least one end. Prove that the number of sums $ \\sum_{i\\equal{}1}^n \\varepsilon_i x_i$ (where $ \\varepsilon_i\\equal{}\\pm 1$) lying in $ \\mathcal{J}$ is at most $ \\binom{n}{\\left[\\frac{n}{2}\\right]}$\r\n\r\nAs a corollary we have \r\n\r\nLet $ x_1,x_2,\\dots,x_n$ be real numbers so that $ |x_i|\\geq 1$ for each $ i$, and let $ \\mathcal{J}$ be any interval of length $ 2r$ open at at least one end, where $ r$ is a natural number. Then the number of sums $ \\sum_{i\\equal{}1}^n \\varepsilon_i x_i$ (where $ \\varepsilon_i\\equal{}\\pm 1$) lying in $ \\mathcal{J}$ is at most $ r\\binom{n}{\\left[\\frac{n}{2}\\right]}$\r\n\r\nBut as a challenge you can try proving the stronger bound of $ \\sum_{i\\equal{}1}^r \\binom{n}{\\left[\\frac{n\\plus{}i}{2}\\right]}$",
  "Solution_1": "The first part is essentially the same as http://www.artofproblemsolving.com/Forum/viewtopic.php?t=279781",
  "Solution_2": "I was about to tell the story of this problem, but realized it had probably appeared here before.\r\n\r\nIndeed, with the key names [url=http://en.wikipedia.org/wiki/Littlewood-Offord_problem]Littlewood and Offord[/url] one quickly finds, e.g., the following threads : \r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=23686\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=103843\r\n\r\nThe results in [b]Albanian Eagle[/b]'s post come from the article by Paul Erd\u0151s about this problem : \r\n[url=http://www.ams.org/bull/1945-51-12/S0002-9904-1945-08454-7/S0002-9904-1945-08454-7.pdf]On a lemma of Littlewood and Offord[/url],\r\nwhere the generalization to higher dimensions is conjectured.\r\nAs [b]pbornsztein[/b] writes in the first of the above two threads, that conjecture was proved by Daniel Kleitman, at first in the complex case in [url=http://math.ucsd.edu/~dwildstr/vu/kleitman.pdf]On a lemma of Littlewood and Offord on the distribution of certain sums[/url]\r\n\r\nFinally, as mentioned by [b]kevinatcausa[/b] in the second thread, such results are still fashionable, with results like the one by Terence Tao and Van Vu about an [url=http://terrytao.wordpress.com/2009/02/16/a-sharp-inverse-littlewood-offord-theorem/]inverse Littlewood-Offord theorem.[/url]",
  "Solution_3": "Thanks guys for the links, I was not aware the problems were already in the forum.\r\n\r\nFor the sake of completeness I am going to write proofs for them here in case I need to reference them in other posts:\r\n\r\n[b]Proof of the first statement:[/b]\r\n\r\nLike JBL, we can assume all $ x_i$'s are positive otherwise we exchange $ x_i\\to \\minus{}x_i$.\r\nWe associate to each sum $ \\sum_{i\\equal{}1}^n\\varepsilon_ix_i$ the set $ A\\equal{}\\{i: \\varepsilon_i\\equal{}1\\}$.\r\nNow if $ A_1\\subset A_2$ were such sets, the corresponding sums differ at least by 2 and so they can not both be in $ \\mathcal{J}$. We conclude that the sets corresponding to sums in $ \\mathcal{J}$ must form an antichain, and the result follows from Sperner's theorem.\r\nThe corollary follows immediately from considering $ \\mathcal{J}$ as a union of $ r$ disjoint intervals of length $ 2$ open at at least one end.\r\n\r\n[b]Proof of the second statement:[/b]\r\n\r\nLike in the proof above we assume all $ x_i$'s are positive and associate $ A\\equal{}\\{i: \\varepsilon_i\\equal{}1\\}$ to the sum $ \\sum_{i\\equal{}1}^n\\varepsilon_ix_i$. Now if $ A_1\\subset A_2$ are two such sets and $ |A_2/A_1|\\geq r$ then the corresponding sums will differ by at least $ 2r$ so at most one of them can be in $ \\mathcal{J}$.\r\nNow the result follows from the lemma:\r\n(Erdos) If $ \\mathcal{A}$ is a collection of subsets of an n-set such that no $ k\\plus{}1$ members of $ \\mathcal{A}$ form a chain then\r\n\\[ |\\mathcal{A}|\\le \\sum_{i\\equal{}0}^{k\\minus{}1}\\binom{n}{\\left[\\frac{n\\plus{}i}{2}\\right]}\\]\r\n\r\n\r\nThis last lemma I think is proved in Erdos's article. It's interesting to note it is in fact a special case of a result by [b]Kleitman (1974):[/b]\r\n\r\nLet $ P$ be a poset with rank function $ r$ and with the LYM property, and let $ \\mathcal{C}$ be any regular covering of $ P$ by chains. Then, if $ \\lambda$ is a real valued function defined on the elements of $ P$, and $ R\\subset P$,\r\n\\[ \\sum_{x\\in R}\\frac{\\lambda(x)}{N_{r(x)}}\\le \\max_{C\\in \\mathcal{C}}\\sum_{x\\in C\\cap R}\\lambda(x)\\]\r\n\r\nSo I hope somebody will post a proof for the above result. Here's another one:\r\n\r\nLet $ g_d(n,r)$ denote the maximum number of sums $ \\sum_{i\\equal{}1}^n \\varepsilon_ix_i$ where $ \\varepsilon_i\\equal{}0$ or $ 1$, lying inside a ball of diameter $ r$, where the $ x_i$ are in $ \\mathbb{R}^d$ with norm at least $ 1$\r\n\r\nSo far we have Erdos's result that\r\n\\[ g_1(n,r)\\equal{}\\sum_{i\\equal{}1}^r \\binom{n}{\\left[\\frac{n\\plus{}i}{2}\\right]}\\]\r\nAnd Kleitman's \r\n\\[ g_d(n,1)\\equal{}\\binom{n}{\\left[\\frac{n}{2}\\right]}\\]\r\n\r\nProve the bound\r\n\\[ g_d(n,r)\\le 2^{d\\minus{}1}[rd^{\\frac{1}{2}}]\\binom{n}{\\left[\\frac{n}{2}\\right]}\\]"
}
{
  "Tag": [
    "quadratics"
  ],
  "Problem": "Find all ordered pairs $(x,y)$ of real numbers for which $x^2+xy+x=14$ and $y^2+xy+y=28$.",
  "Solution_1": "[hide]\nAdd the equations together.\n$x^2 + 2xy + y^2 + (x + y) = 42$\n$(x+y)(x+y+1) = 42$\nLet a = x+y\n$a(a+1) = 42$\n$a = 6 \\textbf{ or } a = -7$\nSo all solutions are also solutions of the equation x + y = 6, x + y = - 7\n[/hide]",
  "Solution_2": "[quote=\"JavaMan\"][hide]\nAdd the equations together.\n$x^2 + 2xy + y^2 + (x + y) = 42$\n$(x+y)(x+y+1) = 42$\nLet a = x+y\n$a(a+1) = 42$\n$a = 6 \\textbf{ or } a = -7$\nSo all solutions are also solutions of the equation x + y = 6, x + y = - 7\n[/hide][/quote]\r\n\r\nSee thats as far as i got but i dont know what to do after that because you are supposed to actually find the ORDERED PAIRS.",
  "Solution_3": "There's an infinite number of ordered pairs, I don't think you can specify them all :)",
  "Solution_4": "lkryptonicl is right- There are only finite number of solutions.\r\n\r\nFrom $x+y=6$, for instance, we can substitute $y=6-x$ into one of the original equations... ;) \r\n\r\n\r\nThere's also another cool approach:\r\n[hide=\"Hint\"]\nFactoring out $x$ from the first equation, we get $x(x+y+1)=14$.\nSimilarly, from the second equation, we get $y(x+y+1)=28$.\n\nWhat does this imply?\n[/hide]",
  "Solution_5": "Does the hint imply [hide]$\\frac{x}{y} = \\frac{1}{2}$[/hide]?",
  "Solution_6": "[quote=\"aidan\"]Does the hint imply [hide]$\\frac{x}{y} = \\frac{1}{2}$[/hide]?[/quote]\r\n\r\nor $x+y=-1$",
  "Solution_7": "[quote=\"shadysaysurspammed\"][quote=\"aidan\"]Does the hint imply [hide]$\\frac{x}{y} = \\frac{1}{2}$[/hide]?[/quote]\n\nor $x+y=-1$[/quote]\r\n\r\nif x+y = -1, then doesnt that make the equation 0 = 14, 0 = 28?",
  "Solution_8": "Of courese,I was just saying that it does imply that $x+y=-1$ but that's not possible.",
  "Solution_9": "So, how are we supposed to write the solutions then?",
  "Solution_10": "Simple when you find $y=2x$.Just substitute this in any of the equations and solve the quadratic to get 2 values of $x$ or 2 values of $y$ and then find the ordered pairs...\r\n\r\nSubstituting $y=2x$ in first equation,we get\r\n\r\n$x=-7/3$ and $x=2$.So the ordered pairs are (-7/3,-14/3),(2,4)",
  "Solution_11": "Using frt's hint I got the same thing as shadysaysurspammed  :)"
}
{
  "Tag": [
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "Prove that if a,b,c are sides of a triangle snd \r\n\r\n2(ab^2+bc^2+ca^2)=a^2b+b^2c+c^2a+3abc, then the triangle is equilateral.\r\n\r\nBomb",
  "Solution_1": "See http://www.mathlinks.ro/Forum/viewtopic.php?t=41715 . Also, the problem should be from an old short/longlist, since it appears in one of the IMO problem books (of the MAA series) as one of the \"40 additional problems\".\r\n\r\n  darij"
}
{
  "Tag": [
    "advanced fields",
    "advanced fields unsolved"
  ],
  "Problem": "Having trouble with this one.\r\n\r\nFind an explicit formula for a solution of the problem below. \r\n\r\nSorry it is an attachment.  I'm new and didn't know how to do it inline.\r\n\r\nCarson",
  "Solution_1": "That's really blurry. I'm going to attempt to copy it. Is this what it says? Tell me if it's not.\r\n\r\n$u_t-\\Delta u+cu=f(t,x),\\,t>0,\\,x\\in\\mathbb{R}^n$\r\n\r\n$u(0,x)=g(x),\\,x\\in\\mathbb{R}^n$\r\n\r\nwhere $c\\in\\mathbb{R}.$ Assume that $f,g$ are smooth and have compact support.",
  "Solution_2": "That is correct.\r\n\r\nCarson",
  "Solution_3": "Substitution $u\\rightarrow v e^{-ct}$ can bring it to the original form. It is really a tough problem if your c is not a constant .\r\nThe real heat equation is $u_t-\\vec{\\nabla}\\cdot\\left(k\\left(x\\right)\\vec{\\nabla}u\\right)=f$, also impossible to solve exactly."
}
{
  "Tag": [
    "calculus",
    "integration",
    "trigonometry",
    "calculus computations"
  ],
  "Problem": "$ \\int \\sqrt{\\frac{1\\minus{}\\cos x}{\\cos \\alpha \\minus{} \\cos x}} \\; dx$\r\n\r\nwhere $ 0 < \\alpha < x < \\pi$\r\n\r\n\r\n  [hide=\"Ans\"] $ \\minus{}2 \\sin ^{\\minus{}1} \\frac{\\cos \\frac{x}{2}}{\\cos \\frac{ \\alpha}{2}} \\plus{}c$ [/hide]",
  "Solution_1": "Use half-angle formula to obtain $ \\int \\sqrt{\\frac{1\\minus{}\\cos x}{\\cos \\alpha \\minus{} \\cos x}} \\, dx \\equal{} 2 \\int \\frac{\\sin t}{\\sqrt{\\cos^2 \\beta \\minus{} \\cos^2 t}} \\, dt$, where $ t \\equal{} \\tfrac{x}{2}$ and $ \\beta \\equal{} \\tfrac{\\alpha}{2}$.\r\n\r\nThen use the substitution $ \\cos t \\equal{} (\\cos \\beta) u$. This directly gives the answer."
}
{
  "Tag": [
    "modular arithmetic"
  ],
  "Problem": "Find all natural numbers $ x$ that makes from  $ 2^x \\minus{} 1$ a complete square.\r\n\r\nI was thinking its infinity. :read:",
  "Solution_1": "No it isn't. \r\nFor a natural n, $ n^2\\equiv0,1\\pmod4$, but $ 2^x\\minus{}1\\equiv1,3\\pmod4$. The only case when $ 2^x\\minus{}1\\equiv1\\pmod4$ is x=1,n=1.",
  "Solution_2": "[hide]We can easily find trivial solution $ x\\equal{}1$. Let's prove that it is only solution.\n\nLet $ x\\equal{}2k$ (where $ k$ is positive integer)\n\n$ 2^{2k}\\minus{}1\\equal{}(2^k\\plus{}1)(2^k\\minus{}1)$\n\n$ 2^k\\minus{}1\\equal{}a \\\\\n2^k\\plus{}1\\equal{}a\\plus{}2$\n\n$ a(a\\plus{}2)\\equal{}b^2$\n\n$ a^2\\plus{}2a\\plus{}1\\equal{}b^2\\plus{}1$\n\n$ (a\\plus{}1)^2\\equal{}b^2\\plus{}1$\n\nbut this is not possible\n\nNow let $ x\\equal{}2k\\plus{}1$\n\n$ 2^{2k\\plus{}1}\\minus{}1\\equal{}a^2$             ($ a$ must be odd)\n\n$ 2^{2k\\plus{}1}\\minus{}2\\equal{}a^2\\minus{}1$\n\n$ 2(2^{2k}\\minus{}1)\\equal{}(a\\plus{}1)(a\\minus{}1)$\n\n$ 2(2^k\\plus{}1)(2^k\\minus{}1)\\equal{}(a\\plus{}1)(a\\minus{}1)$\n\n$ RHS$ is divisible by $ 4$ but $ LHS$ isn't. Contradiction.[/hide]"
}
{
  "Tag": [
    "geometry",
    "incenter",
    "induction",
    "cyclic quadrilateral",
    "geometry proposed"
  ],
  "Problem": "In a circle with center $O$ is inscribed a polygon, which is triangulated. Show that the sum of the squares of the distances from $O$ to the incenters of the formed triangles is independent of the triangulation.",
  "Solution_1": "[img]9232[/img]\r\n\r\nFirst, we prove it for a cyclic quadrilateral.\r\nSince $\\angle AI_{A}B = 90+\\frac{1}{2}\\angle ADB = 90+\\frac{1}{2}\\angle ACB = \\angle AI_{B}B$ and so on, we get that the quadrilaterals $AI_{A}I_{B}B, BI_{B}I_{C}C, CI_{C}I_{D}D, DI_{D}I_{A}A$ are all cyclic. Then we can calculate $\\angle I_{D}I_{A}I_{B}= 360-\\angle AI_{A}I_{B}-\\angle AI_{A}I_{D}=$ $\\angle ABI_{B}+\\angle ADI_{D}=$ $\\frac{1}{2}\\angle ADC+\\frac{1}{2}\\angle ABC = 90$.\r\n\r\nWe conclude that $I_{A}I_{B}I_{C}I_{D}$ is a rectangle.\r\nTherefore, for each point P in the plane, $PI_{A}^{2}-PI_{D}^{2}= PI_{B}^{2}-PI_{C}^{2}$: you just have to draw the projection of P on a side and apply Pithagora's theorem. Take P=O and the case \"cyclic quadrilateral\" is done.\r\n\r\nRemark: from Euler's formula, $OI^{2}= R^{2}-2rR$, we get the very interesting fact that the sum of the inradii of $ABC,ADC$ equals the sum of the inradii of $BCD,BAD$.\r\n\r\nNow proving the general case is just a combinatorial problem.\r\n\r\nIn fact it'easy to see that you can transform any triangulation to an other by doing this operation:\r\n- take two adiacent triangles in the triangulation that form the cyclic quadrilateral ABCD, and swap its diagonal.\r\nWe have proved that this operation does not change the sum of the distances from O to the incenters.\r\n\r\nWe prove that for any triangulations and any vertex X of the polygon, using that operation you can obtain the triangulation \"all vertex linked to X\".\r\nThis can be done by induction: of no vertex is linked to X, then the vertices Y,Z adjacent to X are linked, take the quadtilateral XYZT where YZT forms another triangle of the triangulation, and change the diagonal from YZ to ZT.\r\nOnce you have done this, the diagonal XT divides the polygon in two polygons, and continue doing this on each of the two polygons, you will end up with the desired configuration.\r\n\r\nNice problem  :)",
  "Solution_2": "as for quadrilateral,it's a question appeared in China TST."
}
{
  "Tag": [
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Prove that :\r\n If $ a,b \\in$ {$ 1;3;5;...;2^n \\minus{} 1$} and  $ ab \\equiv 1 (mod 2^n) , a \\equiv b \\equiv 1$ or $ a \\equiv b \\equiv \\minus{} 1 (mod 2^n)$",
  "Solution_1": "It is not true . \r\nYou can prove that it is true if and only if $ n\\leq 3$ :lol:",
  "Solution_2": "[quote=\"TTsphn\"]It is not true . \nYou can prove that it is true if and only if $ n\\leq 3$ :lol:[/quote]\r\nExactly $ n\\le 2$, because $ 3*3\\equal{}1\\mod 2^3$."
}
{
  "Tag": [
    "inequalities",
    "function",
    "number theory",
    "number theory unsolved"
  ],
  "Problem": "Solve in positive integers the equation\r\n\\[ n \\equal{} \\varphi(n) \\plus{} 402 ,\r\n\\]\r\nwhere $ \\varphi(n)$ is the number of positive integers less than $ n$ having no common prime factors with $ n$.",
  "Solution_1": "IMPORTANT: This applies to the wrong version of the problem, where it was \"402\" instead of \"420\".\r\n\r\n[hide]If $n=p_1^{\\alpha_1}p_2^{\\alpha_2}\\dots p_k^{\\alpha_k}$ where $p_i$ is prime, then $\\phi(n)=p_1^{\\alpha_1-1}(p_1-1)p_2^{\\alpha_2-1}(p_2-1)\\dots p_k^{\\alpha_k-1}(p_k-1).$ Hence\n\n$p_1^{\\alpha_1}p_2^{\\alpha_2}\\dots p_k^{\\alpha_k}-p_1^{\\alpha_1-1}(p_1-1)p_2^{\\alpha_2-1}(p_2-1)\\dots p_k^{\\alpha_k-1}(p_k-1)=2\\cdot 3\\cdot 67.$\n\nIt's now obvious that $\\alpha_i\\ngtr 2$, since otherwise LHS wouldn't be squarefree.\n\nDenote $\\{u_i\\}_{i=1}^{r}=\\{p_i|\\alpha_i=1\\}$ and $\\{v_i\\}_{i=1}^{s}=\\{p_i|\\alpha_i=2\\}.$\n\n\\begin{eqnarray*}u_1u_2\\dots u_rv_1^2v_2^2\\dots v_s^2-(u_1-1)(u_2-1)...(u_r-1)v_1(v_1-1)v_2(v_2-1)\\dots v_s(v_s-1)=2\\cdot 3\\cdot 67.\\end{eqnarray*}\n\nSince LHS is divisible by $v_i$, it follows that $s=3$ and $v_1=2, v_2=3, v_3=67.$ Hence\n\n\\begin{eqnarray*}u_1u_2\\dots u_r2^23^267^2-(u_1-1)(u_2-1)...(u_r-1)\\cdot 2\\cdot 1\\cdot 3\\cdot 2\\cdot 67\\cdot 66=2\\cdot 3\\cdot 67\\end{eqnarray*}, or\n\n\\begin{eqnarray*}402u_1u_2\\dots u_r-132(u_1-1)(u_2-1)...(u_r-1)=1.\\end{eqnarray*}\n\nBut in that case LHS is even, hence $n$ is squarefree.\n\nTherefore,\n\n$p_1p_2\\dots p_k-(p_1-1)(p_2-1)\\dots (p_k-1)=402$\n\nIf all $p_i$ are odd, then LHS is odd, hence $p_1=2$:\n\n$2p_2\\dots p_k-(p_2-1)\\dots(p_k-1)=402$\n\n$(p_2-1)\\dots(p_k-1)$ is divisible by $2^{k-1}$, hence $2^{k-2}|p_2p_3\\dots p_k-201$, and I'm stuck :)[/hide]",
  "Solution_2": "i can't manage to solve this problem. can you help me?",
  "Solution_3": "This is a very nasty problem. I wonder if there is a nice solution \r\n(my solution is a dirty brute-force which took me something like 3 hours :()",
  "Solution_4": "Could you write a sketch of the solution? I mean main results and such.",
  "Solution_5": "[quote=\"Farenhajt\"]\nIf $n=p_1^{\\alpha_1}p_2^{\\alpha_2}\\dots p_k^{\\alpha_k}$ where $p_i$ is prime, then $\\phi(n)=p_1^{\\alpha_1-1}(p_1-1)p_2^{\\alpha_2-1}(p_2-1)\\dots p_k^{\\alpha_k-1}(p_k-1).$ Hence\n\n$p_1^{\\alpha_1}p_2^{\\alpha_2}\\dots p_k^{\\alpha_k}-p_1^{\\alpha_1-1}(p_1-1)p_2^{\\alpha_2-1}(p_2-1)\\dots p_k^{\\alpha_k-1}(p_k-1)=2\\cdot 3\\cdot 67.$\n\nIt's now obvious that $\\alpha_i\\ngtr 2$, since otherwise LHS wouldn't be squarefree.\n\nDenote $\\{u_i\\}_{i=1}^{r}=\\{p_i|\\alpha_i=1\\}$ and $\\{v_i\\}_{i=1}^{s}=\\{p_i|\\alpha_i=2\\}.$\n\n\\begin{eqnarray*}u_1u_2\\dots u_rv_1^2v_2^2\\dots v_s^2-(u_1-1)(u_2-1)...(u_r-1)v_1(v_1-1)v_2(v_2-1)\\dots v_s(v_s-1)=2\\cdot 3\\cdot 67.\\end{eqnarray*}\n\nSince LHS is divisible by $v_i$, it follows that $s=3$ and $v_1=2, v_2=3, v_3=67.$ Hence\n\n\\begin{eqnarray*}u_1u_2\\dots u_r2^23^267^2-(u_1-1)(u_2-1)...(u_r-1)\\cdot 2\\cdot 1\\cdot 3\\cdot 2\\cdot 67\\cdot 66=2\\cdot 3\\cdot 67\\end{eqnarray*}, or\n\n\\begin{eqnarray*}402u_1u_2\\dots u_r-132(u_1-1)(u_2-1)...(u_r-1)=1.\\end{eqnarray*}\n\nBut in that case LHS is even, hence $n$ is squarefree.\n\nTherefore,\n\n$p_1p_2\\dots p_k-(p_1-1)(p_2-1)\\dots (p_k-1)=402$\n\nIf all $p_i$ are odd, then LHS is odd, hence $p_1=2$:\n\n$2p_2\\dots p_k-(p_2-1)\\dots(p_k-1)=402$\n\n$(p_2-1)\\dots(p_k-1)$ is divisible by $2^{k-1}$, hence $2^{k-2}|p_2p_3\\dots p_k-201$, and I'm stuck :)[/quote]\r\nI think Farenhajt's solution is good.Let me try to complete it.While\r\n$2p_2\\dots p_k-(p_2-1)\\dots(p_k-1)=402$\r\nthen we can know $p_2\\dots p_k<402$,so$k<5$for$3*5*7*11>402$\r\nthen the problem turn an easy equation.\r\nwe only need to make out three equation.\r\n2a-(a-1)=402,ab-(a-1)(b-1)=402,abc-(a-1)(b-1)(c-1)=402(a,b,c are all prime)\r\nthe first and the second is quite easy,so I only do someting about the third.\r\nwe could know $abc<402$,so $min{a,b,c}<8$.we could try a=3,5,7,then the third can turn the second,so we solve the problem.finally we know n=802,546.",
  "Solution_6": "[quote=\"Farenhajt\"]\\begin{eqnarray*}u_1u_2\\dots u_rv_1^2v_2^2\\dots v_s^2-(u_1-1)(u_2-1)...(u_r-1)v_1(v_1-1)v_2(v_2-1)\\dots v_s(v_s-1)=2\\cdot 3\\cdot 67.\\end{eqnarray*}\n\nSince LHS is divisible by $v_i$, it follows that $s=3$ and $v_1=2, v_2=3, v_3=67.$[/quote]\r\nI do not see why it follows. I think it follows that $s\\leq 3$ and $v_i\\in\\{2,3,67\\}$ for $i=1..s$, nothing more.",
  "Solution_7": "It is very esay question.. I have solved that",
  "Solution_8": "Ditdo: Please post your solution.",
  "Solution_9": "ok! :)  ;)",
  "Solution_10": "You can see my solution by one click! ;) \r\n[hide]$\\[ n = \\varphi(n) + 402$. \nIf n is odd then,  $\\varphi(n)$ -even and $\\[ n = \\varphi(n) + 402$ is even.Contradiction,so $n=2m$ and $\\varphi(2m) = \\varphi(m)$, then ${{\\ 2m = \\varphi(m) + 402\\ <} \\ m + 402\\implies\\ m \\ <} \\ 402$ and ${{2m >} 402\\implies\\ m >} 201$\nIf m is divisible by 4, then $2m - \\varphi(m) = 402$ is divisible by 4. Contaradiction.\nIf $m=2p\\implies\\varphi(2p)=p-1\\implies\\4p=p-1+402\\implies\\3p=401\\ \\phi$. So m is odd.\nIf m is divisible by 67, then $\\varphi(m)$ is divisible by 66 ${\\implies\\ 402+\\varphi(m)=2m\\vdots{3}\\implies m\\vdots {3}\\implies m\\vdots{3}*67, m<}402$ and also $m\\vdots{201}\\implies m=201\\ \\phi$\nIf $m\\vdots {p^3}\\implies\\varphi(p)\\vdots {p^2}\\implies\\ 402\\vdots {p^2} \\ \\phi$\nIf $m\\vdots {p^2}\\implies\\varphi(p)\\vdots {p}\\implies\\ 402\\vdots {p}\\implies\\ p=3$\nIf m have 4 distinct prime divisors, then ${m\\ge\\ 3*5*7*11>}402$. So m have at most 3 distinct prime divisors.\n$I.$  If $m=p^{\\alpha}$ then $2p^{\\alpha\\ }=p^{\\alpha-1\\)(p -1)+402}\\implies\\ p^{\\alpha} -p^{\\alpha-1}=402\\implies\\ p=401,{\\alpha}=1, n=802$.\n$II.$  If $m=p^{\\alpha}q^{\\beta}$ then $2p^{\\alpha}q^{\\beta}=p^{\\alpha -1}q^{\\beta -1} {(p -1)(q -1)}+402$\n $\\implies\\ p^{\\alpha -1}q^{\\beta -1}={(pq+p+q -1)}=402=2*3*67$\n$i)$ ${\\beta}=1$ ; ${\\alpha}=1\\implies\\ {(p+1)(q+1)=402}\\ \\phi$.\n$ii)$ ${\\beta}=1$ ; ${\\alpha}=2\\implies\\ p=3\\ {(p+1)(q+1)=138}\\ \\phi$\n$III.$ If $m=p^{\\alpha}q^{\\beta}r^{\\gamma}$ then $2p^{\\alpha}q^{\\beta}r^{\\gamma}=p^{\\alpha -1}q^{\\beta -1}r^{\\gamma -1} {(p -1)(q -1)(r -1)}+402$\n$\\implies\\ p^{\\alpha -1}q^{\\beta -1}r^{\\gamma -1}={(2pqr -(p -1)(q -1)(r -1)}=402$\nFrom here we get $n=546=2*3*7*13$[/hide]",
  "Solution_11": "[quote=\"Ditdo\"]so $n=2m$ and $\\varphi(2m) = \\varphi(m)$[/quote]\r\nOnly if $m$ is odd that may not be the case.",
  "Solution_12": "Maxal,$m$ is odd,because if $m$ is even then $n\\vdots4$\r\nLet us return to the equation $n=\\phi(n)+402$\r\nBecause $n\\vdots4$, and $\\phi(n)\\vdots4$ it follows that $402\\vdots4$.Contradiction!!!!!\r\nAlso $n\\not=2^{k}$.If so then we find that\r\n$2^k=2^{k}-2^{k-1} +402$,more clearly\r\n$2^k=2^{k-1}+402$,clearly $k\\le2$,because 402 is not divisible by $2^k$ for any $k\\ge2$.By checking we get no answer.",
  "Solution_13": "[quote=\"spider_boy\"]Maxal,$m$ is odd,because if $m$ is even then $n\\vdots4$\nLet us return to the equation $n=\\phi(n)+402$\nBecause $n\\vdots4$, and $\\phi(n)\\vdots4$[/quote]\r\nWhy $\\phi(n)\\vdots4$ ?",
  "Solution_14": "[quote=\"maxal\"][quote=\"spider_boy\"]Maxal,$m$ is odd,because if $m$ is even then $n\\vdots4$\nLet us return to the equation $n=\\phi(n)+402$\nBecause $n\\vdots4$, and $\\phi(n)\\vdots4$[/quote]\nWhy $\\phi(n)\\vdots4$ ?[/quote]\r\nLook, maxal ,let $n=p_1^{q_1}p_2^{q_2}\\ldots{p_k^{q_k}}$,then \r\n$\\phi(n)=p_1^{q_1-1}\\ldots{p_k^{q_k-1}}(p_1-1)(p_2-1)\\ldots{(p_k-1)}$\r\nNow,if $n\\not=2^{k}$ then n has another prime divisor other than $2$.\r\nnow,if m is even,then $LHS$ of the equation is divisible by $4$.And as we said\r\n$\\phi(n)=p_1^{q_1-1}\\ldots{p_k^{q_k-1}}(p_1-1)(p_2-1)\\ldots{(p_k-1)}$\r\nHere $p_1^{q_1-1}$ is divisible by $2$.and one of $(p_2-1),\\ldots,{(p_k-1)}$ is divisible by at least $2$,because all of $p_2,p_3,.....,p_k$ are odd.There is at least one such expression,beacuse n has another prime divisor that$2$..But $402$ is not divisible by $4$.Contradiction.",
  "Solution_15": "As with any problem, its usually easier to work with a smaller case before the general case. In this problem. Lets first assume that  $n$ is a semiprime. Let $n=pq$ For primes $p,q$. Then $\\phi(n)=\\phi(pq)=(p-1)(q-1)=pq-(p+q)+1$.\r\n$n=\\phi(n)+402$\r\n$pq=pq-(p+q)+1+402$\r\n$p+q=403$\r\nThe only solution for $(p,q)$ is:\r\n(401,2), so $n=802$\r\n\r\nJust writing this off the cuff. Maybe this will help??\r\nThere's a nice little theorem in [u]Number Theory[/u] by George E. Andrews. It says that:\r\n$\\displaystyle\\frac{\\pi(x)}{x}\\leq\\displaystyle\\frac{\\phi(k)}{k}+\\displaystyle\\frac{2k}{x}$\r\nRearrange the terms:\r\n$\\displaystyle\\frac{\\phi(k)}{k}\\geq\\displaystyle\\frac{\\pi(x)}{x}-\\displaystyle\\frac{2k}{x}$\r\nBut since $\\phi(k)=k-402$,\r\n$\\displaystyle\\frac{k-402}{k}\\geq\\displaystyle\\frac{\\pi(x)}{x}-\\displaystyle\\frac{2k}{x}$\r\n$1-\\displaystyle\\frac{402}{k}\\geq\\displaystyle\\frac{\\pi(x)}{x}-\\displaystyle\\frac{2k}{x}$\r\n$\\displaystyle\\frac{402}{k}\\leq1-\\displaystyle\\frac{\\pi(x)}{x}+\\displaystyle\\frac{2k}{x}$\r\n$402x\\leq k(x+2-\\pi(x))$\r\nPlugging in the formula given (but setting $x=k$), we get...\r\n$402x\\leq k(x-\\phi(x))$\r\nApparently it doesn't help. Oh well.",
  "Solution_16": "can anyone tell me what is $\\{u_i\\}_{i=1}^{r}=\\{p_i|\\alpha_i=1\\}$ and $\\{v_i\\}_{i=1}^{s}=\\{p_i|\\alpha_i=2\\}.$  and theorem of George E.Andrews?",
  "Solution_17": "because $\\varphi(n) \\leq n - \\sqrt(n)$ we find that $n\\leq 20$\r\nnow one can actually do stuff by hand tghrough elimination\r\nthat inequality was taken from here  http://en.wikipedia.org/wiki/Euler's_phi_function\r\n :lol:\r\nthis is for n composite as in the link above is mentioned.\r\nfor n prime we have $n=n-1+402$ wich is very unlikly for any n.",
  "Solution_18": "[quote=\"spx2\"]because $\\varphi(n) \\leq n - \\sqrt(n)$ we find that $n\\leq 20$\n.[/quote]\r\n\r\nYou are wrong here .From $\\varphi(n) \\leq n - \\sqrt{n}$  we take that  $n\\leq 1604$  \r\n\r\nwhich means nothing .",
  "Solution_19": "1).$\\varphi(n) \\leq n- \\sqrt{n}$\r\n2).$\\varphi(n)=n-402$\r\n\r\n1,2 =>  $n-402\\leq n- \\sqrt{n}$ <=> $\\sqrt{n}\\leq 402$\r\nso $n \\leq 402^2$ so $n\\leq 161604$.\r\nyes i am wrong",
  "Solution_20": "i thought again about the problem,hopefully this time i'm right.\r\n$n-\\varphi(n)=402$\r\n$\\prod p_i ^{\\alpha_i-1} (\\prod p_i - \\prod (p_i -1) ) = 402$\r\nnow 402=2*3*67\r\nnow one can easily take combinations of these as beeing the numbers from\r\n$\\prod p_i ^{\\alpha_i-1}$\r\nis this right ?",
  "Solution_21": "this isn't good because there may exist any prime at power 1 in n wich will not appear\r\nin that product altough it appears in the paranthesis",
  "Solution_22": ":)  One more answer $n=802$\r\n\r\nDavron",
  "Solution_23": "$n=\\sum_{d|n} \\varphi(d)$\r\nso if $n=\\prod_{i=1}^{k} p_i^{\\alpha_i}$ *\r\n$n-\\varphi(n)=\\sum_{i=1}^{k} \\sum_{d|\\frac{n}{p_i}}\\varphi(d)$\r\nbut $\\sum_{d|\\frac{n}{p_i}}\\varphi(d) = \\frac{n}{p_i}$ so\r\n$n-\\varphi(n)=\\sum_{i=1}^{k} \\frac{n}{p_i}$\r\nbut if we consider n in the form * we have\r\n$\\sum_{i=1}^{k} \\frac{n}{p_i}=\\prod_{i=1}^{k} p_i^{\\alpha_i -1} (\\sum_{i=1}^{k} p_i )$\r\n\r\nis it simpler now ?\r\n\r\ndavron are you sure what you wrote is a solution because $802=2 \\cdot 401$\r\nboth $2,401$ are prime both appear at power 1 in 802.\r\nthis means that $\\prod_{i=1}^{k} p_i^{\\alpha_i -1} (\\sum_{i=1}^{k} p_i )=403$\r\nwich is a little weird ...\r\ncould someone please help ?",
  "Solution_24": "please anyone does this help in any way?\r\nhow can we solve this problem ?",
  "Solution_25": "spx2, please stop posting wrong solutions in every topic! Post only full and complete solutions. It's  :spam:",
  "Solution_26": "Substitute $ n$ by $ P_{1}^{\\alpha_{1}} \\cdot P_{2}^{\\alpha_{2}} \\cdots P_{x}^{\\alpha_{x}}$. As we know, $ \\varphi(n) = P_{1}^{\\alpha_{1} - 1}\\cdot P_{2}^{\\alpha_{2} - 1}\\cdots P_{x}^{\\alpha_{x} - 1}$. So the givven equation rewrites as the following: ${ P_{1}^{\\alpha_{1} - 1}\\cdot P_{2}^{\\alpha_{2} - 1}\\cdots P_{x}^{\\alpha_{x} - 1}(P_{1}P_{2}\\cdots P_{x} - (P_{1} - 1)(P_{2} - 1)\\cdots(P_{x} - 1}) = 402$. $ (1)$.\r\nIf none of the numbers $ P_{1},P_{2},...,P_{x}$ is $ 2$, than in $ (1)$ equality $ LHS$ is odd, while $ 402$ is even-contradiction. So, wlog $ P_{1} = 2$. If $ \\alpha_{1} > 1$, than in $ (1)$ equality $ LHS$ is divisible by $ 4$, while $ 402$ isn't divisible by $ 4$. So, $ \\alpha_{1} = 1$. Assume there exists the number $ \\alpha_{i}$ such that $ \\alpha_{i} > 1$. As $ P_{1}^{\\alpha_{1} - 1}\\cdot P_{2}^{\\alpha_{2} - 1}\\cdots P_{x}^{\\alpha_{x} - 1}|402 = 2\\cdot 3\\cdot 67$, $ P_{i}$ is either $ P_{i} = 3$ or $ P_{i} = 67$. $ n$ is even. So, $ \\varphi(n) < = \\frac {n}{2}$, giving $ 402 < = n < = 804$. If wlog $ P_{2} = 67$, than the last inequality gives: $ 3 < = P_{3}\\cdots P_{x} < = 6$ which follows $ P_{3 } = P_{x} = 5$. So,$ n = P_{1}P_{2}P_{3} = 2\\cdot 67\\cdot 5 = 670$. $ \\varphi(n) = 132$, but $ 670$ isn't equal to $ 132 + 402$-contradiction. If $ P_{2} = 3$,than the same argument leads us to that: $ 67 < = P_{3} \\cdot P_{x} < = 134$. By checking all possible numbers instead of $ P_{3},P_{4},...,P_{x}$, using the fact that if $ i$ doesn't equal to $ j$,where $ 2 < = i,j < = x$, than $ P_{i}$ doesn't equal to $ P_{j}$, we make sure there aren't the solutions to the given equation.\r\nNow it remains to take a look to the case $ \\alpha_{1} = \\alpha_{2} = ... = \\alpha{x} = 1$.\r\nOur equation rewrites so: $ 2P_{2}P_{3}\\cdots P_{x} = (P - {1} - 1)(P_{2} - 1)\\cdots(P_{x} - 1) + 402$. $ (2)$. Now we see the two possible cases: $ (a)$ $ x$ is odd; $ (b)$ $ x$ is even;\r\n$ (a)$ $ x = 2m + 1$. ($ m$ is positive integer, $ m > = 2$(If $ m = 1$,we have no solutions. If $ m = 0$, we have no solutions).\r\nWlog assume $ P_{1} < P_{2} < ... < P_{x}$. If $ P_{m + 1} > = 19$, than $ 804 > = n > = 2P_{m + 1}P_{m + 2} > = 2\\cdot 19\\cdot 23 > 804$-contradiction. So, $ P_{m + 1} < = 17$.\r\nIf $ P_{m + 1} = 17$, than $ 804 > = n > = 2\\cdot 3 \\cdot 17 \\cdot 19 > 804$-contradiction. If $ P_{m + 1} = 13$, than $ 804 > = 2 \\cdot 13 \\cdot 17 > 804$-contradiction. If $ P_{m + 1} = 11$, than $ 804 > = n > = 2 \\cdot 3 \\cdot 11 \\cdot 13 > 804$-contradiction. If $ P_{m + 1} = 7$, $ 804 > = n > = P_{m - 1}P_{m}P_{m + 1}P_{2m + 1} > = 2 \\cdot 3 \\cdot 7 \\cdot 23 > 804$-contradiction, if $ P_{2m + 1} > = 23$. If not, checking all the cases we get the solution $ n = 546$. If $ p_{m + 1} = 5$, than $ P_{m} = 3$ and $ P_{m - 1} = 2 = P_{1}$.So, $ m = 2$. $ 804 > = P_{1}P_{2}P_{3}P_{4}P_{5} > = 2 \\cdot 3 \\cdot 5 \\cdot7 \\cdot 11 > 804$-contradiction.\r\nIf \r\n$ P_{m + 1} = 3$, $ P_{m} = P_{1}$ giving $ m = 1$, which we already checked.\r\n(b) $ x = 2m$. If $ m = 1$, $ n = 802$ is the solution. Assume $ m > = 2$.\r\nBy the same argument as in case $ (a)$ we get there is no more solution.\r\nFinally, we have two possible solutions:$ 802$ and $ 546$.",
  "Solution_27": "[quote=\"Ditdo\"]You can see my solution by one click! ;) \n[hide]$ \\[ n = \\varphi(n) + 402$. \nIf n is odd then,  $ \\varphi(n)$ -even and $ \\[ n = \\varphi(n) + 402$ is even.Contradiction,so $ n = 2m$ and $ \\varphi(2m) = \\varphi(m)$, then ${{ \\ 2m = \\varphi(m) + 402\\ < } \\ m + 402\\implies\\ m \\ < } \\ 402$ and ${{ 2m > } 402\\implies\\ m > } 201$\nIf m is divisible by 4, then $ 2m - \\varphi(m) = 402$ is divisible by 4. Contaradiction.\nIf $ m = 2p\\implies\\varphi(2p) = p - 1\\implies\\4p = p - 1 + 402\\implies\\3p = 401\\ \\phi$. So m is odd.\nIf m is divisible by 67, then $ \\varphi(m)$ is divisible by 66 ${ \\implies\\ 402 + \\varphi(m) = 2m\\vdots{3}\\implies m\\vdots {3}\\implies m\\vdots{3}*67, m < }402$ and also $ m\\vdots{201}\\implies m = 201\\ \\phi$\nIf $ m\\vdots {p^3}\\implies\\varphi(p)\\vdots {p^2}\\implies\\ 402\\vdots {p^2} \\ \\phi$\nIf $ m\\vdots {p^2}\\implies\\varphi(p)\\vdots {p}\\implies\\ 402\\vdots {p}\\implies\\ p = 3$\nIf m have 4 distinct prime divisors, then ${ m\\ge\\ 3*5*7*11 > }402$. So m have at most 3 distinct prime divisors.\n$ I.$  If $ m = p^{\\alpha}$ then $ 2p^{\\alpha\\ } = p^{\\alpha - 1\\)(p - 1) + 402}\\implies\\ p^{\\alpha} - p^{\\alpha - 1} = 402\\implies\\ p = 401,{\\alpha} = 1, n = 802$.\n$ II.$  If $ m = p^{\\alpha}q^{\\beta}$ then $ 2p^{\\alpha}q^{\\beta} = p^{\\alpha - 1}q^{\\beta - 1} {(p - 1)(q - 1)} + 402$\n $ \\implies\\ p^{\\alpha - 1}q^{\\beta - 1} = {(pq + p + q - 1)} = 402 = 2*3*67$\n$ i)$ $ {\\beta} = 1$ ; $ {\\alpha} = 1\\implies\\ {(p + 1)(q + 1) = 402}\\ \\phi$.\n$ ii)$ $ {\\beta} = 1$ ; $ {\\alpha} = 2\\implies\\ p = 3\\ {(p + 1)(q + 1) = 138}\\ \\phi$\n$ III.$ If $ m = p^{\\alpha}q^{\\beta}r^{\\gamma}$ then $ 2p^{\\alpha}q^{\\beta}r^{\\gamma} = p^{\\alpha - 1}q^{\\beta - 1}r^{\\gamma - 1} {(p - 1)(q - 1)(r - 1)} + 402$\n$ \\implies\\ p^{\\alpha - 1}q^{\\beta - 1}r^{\\gamma - 1} = {(2pqr - (p - 1)(q - 1)(r - 1)} = 402$\nFrom here we get $ n = 546 = 2*3*7*13$[/hide][/quote]\r\n$ \\phi(2m)=\\phi(m)$ only for odd values of m, otherwise it doesn't hold (since function multiplicity doesn't apply)",
  "Solution_28": "Obviosly $ n\\equal{}2m$ and m is odd.\r\nTherefore $ 2m\\minus{}\\phi (m)\\equal{}402$. It give $ 201<m<402$.\r\nObviosly $ m\\equal{}rad(m)$ or $ m\\equal{}3rad (m)$ (because $ 402\\equal{}2*3*67$ and $ 67^2>402$).\r\n1. If $ 3\\not |m$, then $ m\\not \\equal{}p_1p_2p_3$, because else $ 3\\not p_i\\minus{}1$ and $ 5*11*17>402$.\r\nIf $ m\\equal{}p_1p_2$, then $ 2m\\minus{}\\phi(m)\\plus{}1\\equal{}(p_1\\plus{}1)(p_2\\plus{}1)\\equal{}404\\equal{}4*101$ give contradition.\r\nIt prove, that $ m\\equal{}3k$ and $ k$ is not prime. If $ k\\equal{}3l$ we get $ 3l\\minus{}\\phi(l)\\equal{}67, 23\\le l\\le 41$ - not solution.\r\nTherefore $ m\\equal{}3p_1p_2,3<p_1<p_2$. It give $ 2p_1p_2\\plus{}p_1\\plus{}p_2\\equal{}202$ and $ 67<p_1p_2<134$.\r\nOr $ (2p_1\\plus{}1)(2p_2\\plus{}1)\\equal{}405\\equal{}3^4*5\\equal{}15*27$, or $ p_1\\equal{}7,p_2\\equal{}13$.\r\nOnly solution $ n\\equal{}2*3*7*13$."
}
{
  "Tag": [],
  "Problem": "On a graph, a lattice point is an ordered pair (x,y) with integers x and y. Exactly 15 lattice points lie strictly in the interior of the triangular region with vertices (0,0), (n,0), and (n,n). What is n?",
  "Solution_1": "[hide]\nregion is isosceles right triangle.\n$15=1+2+...+x$\n$n=x+2$\n$x=5$\n$n=7$\n[/hide]",
  "Solution_2": "[hide]The figure is an isosceles right triangle.  On the bottom row, there are n-1 lattice points.  On the 2nd row, there are n-2 points.  That pattern continues until there is just 1 point left.  That sum will be $\\displaystyle\\frac{(n-2)(n-1)}{2}=15$ Simplify this to get ...\n$(n-2)(n-1)=30$\nn=7\nThat makes n equal to $\\fbox{7}$.[/hide]"
}
{
  "Tag": [
    "algebra",
    "polynomial"
  ],
  "Problem": "how can one come up with the factorisation: $ \\minus{}x^2y\\minus{}x^2z\\plus{}xy^2\\plus{}xz^2\\plus{}y^2z\\minus{}yz^2\\equal{}(x\\minus{}y)(x\\minus{}z)(y\\minus{}z)$\r\nwithout any prior knowledge/hints?",
  "Solution_1": "You may see that if you set $ y\\equal{}x$, the whole thing vanishes. Thus it is (as a polynomial) divisible by $ (x\\minus{}y)$. The same argument gives that it is divisible by $ (y\\minus{}z)$ and $ (z\\minus{}x)$. In total, it is divisible by $ (x\\minus{}y)(y\\minus{}z)(z\\minus{}x)$. Now the original term and this product have the same degree $ 3$, thus they can only differe by a constant $ c$. But to find this constant is easy (either factor it out completely or compare the coefficient of $ x^2y$).",
  "Solution_2": "Alternately, treat as a polynomial in $ x$ and treat $ y, z$ as constants.  You seem to have written something wrong in the expression; as the RHS is written, the coefficient of $ x^2 z$ in the LHS should be $ 1$."
}
{
  "Tag": [
    "geometry",
    "inradius",
    "number theory",
    "greatest common divisor"
  ],
  "Problem": "Prove that the area of a right angled triangle is never a perfect square.( TST 2007 )",
  "Solution_1": "Are you sure this is a tst problem?\r\n\r\nYou must surely mean when the sides of the triangle are integers.\r\n[hide]Recall that $ x^2 \\plus{} y^2 \\equal{} z^2$, after suitable cancelling out the common factors has the solution $ x \\equal{} m^2 \\minus{} n^2, y \\equal{} 2mn,z \\equal{} m^2 \\plus{} n^2$\nSo, it is equivalent to proving that there does not exist an integer k, such that $ k^2 \\equal{} mn(m^2 \\minus{} n^2)$ for integers m and n such that $ gcd(m,n) \\equal{} 1$\n\nSay such an integer exists and let m>n.\nI denote $ gcd(m,n) \\equal{} (m,n)$\nObserve that $ (m,n) \\equal{} (m,m \\minus{} n) \\equal{} (m,m \\plus{} n) \\equal{} (n,m \\plus{} n) \\equal{} (n,m \\minus{} n) \\equal{} 1$\nSo, $ (mn,m^2 \\minus{} n^2) \\equal{} 1$\nSo lets say $ mn \\equal{} u^2, m^2 \\minus{} n^2 \\equal{} v^2$\nThis implies m and n must themeselves be squares.$ m^2 \\minus{} n^2 \\equal{} v^2$, So $ m^2\\equal{}k_1^2\\plus{}k_2^2$\nSo $ n^2\\equal{}k_1^2\\minus{}k_2^2$ or $ n^2\\equal{}2k_1k_2$\nBoth these cases yield nonsense when plugged in the original equation.\nAm i right??[/hide]",
  "Solution_2": "r u sure? about the question",
  "Solution_3": "The condition you actually want is $ 2k^2 \\equal{} mn(m^2 \\minus{} n^2)$, which is more than a little harder to work with.",
  "Solution_4": "Probably I am not understanding the question,\r\n\r\nbut if you take triangle with sides $ 3,4,5$,it is right triangle $ (3^2\\plus{}4^2\\equal{}5^2)$\r\nand the area is $ \\frac{1}{2}\\cdot 3 \\cdot 4 \\equal{}6$\r\n\r\nBut I remember solving a question like inradius of a right triangle is an integer and like that\r\n\r\nPlease clarify",
  "Solution_5": "Here's my solution for the problem.\r\nYou can reduce the problem to a primitive pythagorean triplet.Then the legs are 2mn and m^2-n^2.Then the area is mn(m^2-n^2).Now gcd of m and n is 1 by the definition of primitive pythagorean triplet.Now gcd of mn and m+n and mn and m-n is 1 by utilizing the gcd properties.This implies that mn and m^2-n^2 are squares.Now we can deduce that m and n are squares because gcd of m and n is 1. This leads to a contradiction of the fact that both m ,n and m^2-n^2 are squares. Actually hell_ever your solution is very good and you are right this is not a TST problem however it was discussed in the IMOTC of 2007."
}
{
  "Tag": [
    "combinatorics proposed",
    "combinatorics"
  ],
  "Problem": "Let $ A_1,A_2,\\dots, A_n$ be any $ n$ sets. Show that we can choose $ \\geq \\sqrt{n}$ of them such that the union of no two is a third.",
  "Solution_1": "It is sufficient to prove that when $ n \\equal{} k^2 \\plus{} 1$, we can choose $ k \\plus{} 1$ sets such that the union of no two is a third.",
  "Solution_2": "Ok.. I think you only need one or two more sentences  :D",
  "Solution_3": "Only my thought(not a complete solution)\r\n\r\nTake $ F$ such that $ F \\subset \\{A_1,A_2,...A_n\\}$ the union of no two in $ F$ is a third.And $ |F|$ is [b]maximal[/b]\r\n\r\nWLOG $ F\\equal{}\\{A_1,...A_k\\}$\r\n\r\n$ A_i$ cannot be addded into $ F$ where $ i>k$\r\n\r\nso $ A_i\\equal{}A_p \\cup A_q$ or $ A_p\\equal{}A_i\\cup A_q$ where $ p,q\\le k$\r\n\r\nI think after some simple argument we can establish $ 2\\binom{k}{2}\\ge n\\minus{}k$ which implies $ k\\ge \\sqrt{n}$\r\n\r\nBut what confuse me is that if $ A_i\\equal{}A_p \\cup A_q$ then we can map $ i$ to $ (p \\cup q)$ and $ A_p \\cup A_q$ is unique,however,there may exist more than \r\n\r\none $ A_i$ so that $ A_p\\equal{}A_i\\cup A_q$ ,so we are not allowed to map $ i$ to $ (p \\cap q)$\r\n\r\nDid I make what I said clear?",
  "Solution_4": "I understand what you wrote, and in fact you do have a problem as you mentioned that there might be more than one $ A_i$ so that $ A_p\\equal{}A_i\\cup A_q$  :wink: \r\nThe problem becomes trivial if one thinks of the possible partial orders one can associate to $ \\{A_1,A_2,\\dots,A_n\\}$",
  "Solution_5": "[quote=\"Albanian Eagle\"]I understand what you wrote, and in fact you do have a problem as you mentioned that there might be more than one $ A_i$ so that $ A_p \\equal{} A_i\\cup A_q$  :wink: \nThe problem becomes trivial if one thinks of the possible partial orders one can associate to $ \\{A_1,A_2,\\dots,A_n\\}$[/quote]\r\nThank you very much for your hint,you mean Dilworth theorem kills it :D \r\nBut I am eager to find an elementary solution :)",
  "Solution_6": "For given n , I proved that we can find sqrt(2n)-1 of them,but we cannot always find  2sqrt(n)+1 of them",
  "Solution_7": "Why don't you post your proof, it seems like a nice result."
}