{ "Tag": [ "modular arithmetic", "number theory proposed", "number theory" ], "Problem": "Prove that for each prime number $ p>3$ and for each positive integer $ k$ there exists a positive integer $ n$ such that $ p^{k}\\parallel{} n^{p\\minus{}2}\\plus{}2,$ i.e. $ p^{k}| n^{p\\minus{}2}\\plus{}2$ but $ p^{k\\plus{}1}\\not | n^{p\\minus{}2}\\plus{}2.$", "Solution_1": "Choose a number $ s$ such that $ 2s\\equiv 1(\\mod p^k)$ so $ s^{p\\minus{}2}\\equiv 2(\\mod p^k)$ \r\n\r\nNow choose the numer $ n$ such that $ p^k\\parallel{}n\\plus{}s$ \r\nNow we know that if $ p^k\\parallel{}a\\minus{}b$ and $ p^l\\parallel{}n$ then $ p^{k\\plus{}l}\\parallel{}a^n\\minus{}b^n$ \r\n\r\nUsing this fact we have that $ p^k\\parallel{}n^{p\\minus{}2}\\plus{}s^{p\\minus{}2}\\equiv n^{p\\minus{}2}\\plus{}2$ and we are done", "Solution_2": "[quote=\"silouan\"]Choose a number $ s$ such that $ 2s\\equiv 1(\\mod p^k)$ so $ s^{p \\minus{} 2}\\equiv 2(\\mod p^k)$[/quote]\r\nWhy is that? But it's clear that such $ s$ exists anyway, since the map from $ \\mathbb{Z}^*_{p^k}$ to $ \\mathbb{Z}^*_{p^k}$: $ x \\minus{}> x^{p\\minus{}2}$ is bijective.", "Solution_3": "I believe it's because we know that $ s^{p - 1} \\equiv 1 \\pmod {p^k}$, so:\r\n\r\n\\begin{eqnarray*} 2*s^{p - 1} & \\equiv & 2 \\pmod {p^k} \\\\\r\n2*s*s^{p - 2} & \\equiv & 2 \\pmod {p^k} \\\\\r\ns^{p - 2} & \\equiv & 2 \\pmod {p^k}\\end{eqnarray*}\r\n\r\n\r\nOn the other hand...\r\n[quote]Now we know that if $ p^k\\parallel{}a - b$ and $ p^l\\parallel{}n$ then $ p^{k + l}\\parallel{}a^n - b^n$ [/quote]\r\nWhat result tells us this? Can someone name it please?\r\n\r\nEDIT: nevermind, that top part is wrong... we can't actually say that $ 2s \\equiv 1 \\pmod {p^k} \\Rightarrow s^{p - 2} \\equiv 2 \\pmod {p^k}$, but your point about bijective mapping still holds so that's ok.", "Solution_4": "[quote=\"kops723\"]\nEDIT: nevermind, that top part is wrong...[/quote]\r\n\r\nWhy ? :huh:", "Solution_5": "Because $ s^{p\\minus{}1} \\equiv 1 \\mod{p^k}$ does not necessarily hold.", "Solution_6": "In fact I thought that the system of congruences $ 2s\\equiv 1(\\mod p^k)$ and $ s^{p\\minus{}1}\\equiv 1(\\mod p^k)$ \r\n\r\nhas a solution ... But in fact this is not trivial (and maybe wrong .Is it? ) \r\n\r\nHopefully the bijection saves my proof :)", "Solution_7": "[quote=\"silouan\"]But in fact this is not trivial (and maybe wrong .Is it? ) [/quote]\r\n\r\nYes. Note that existence of such solution implies that $ 2^{p-1} \\equiv 1 \\mod{p^k}$ which usually doesn't hold even for $ k=2$ (primes for which congruence hold for $ k=2$ are called Wieferich primes and they occur very rarely and we don't know if there are infinitely many of them). But since $ (p-2, \\varphi(p^k))=1$ map $ x^{p-2}$ is an automorphism of multiplicative group of ${ \\mathbb{Z}_p^k}$ so it's not a big deal.", "Solution_8": "[quote]On the other hand...\n[quote]Now we know that if $ p^k\\parallel{}a \\minus{} b$ and $ p^l\\parallel{}n$ then $ p^{k \\plus{} l}\\parallel{}a^n \\minus{} b^n$ [/quote]\nWhat result tells us this? Can someone name it please?\n[/quote]\r\nSee for example here : http://reflections.awesomemath.org/2007_3/Lifting_the_exponent.pdf" } { "Tag": [ "number theory open", "number theory" ], "Problem": "Has this problem been discussed/solved on the forum?\r\n\r\nFind the sum:\r\n\r\n$ \\sum\\limits_{k\\equal{}0}^{n}\\biggm|\\binom{n}{k}\\minus{}2^{k}\\biggm|\\equal{}?$\r\n\r\nIn other words when is $ 2^{k}>\\binom{n}{k}$?", "Solution_1": "when 2^k>nCk\r\nin such cases the sum is\r\n| 2^n - (2^n+1 - 1)|\r\n[/code][/hide][/list]" } { "Tag": [ "AMC", "AIME" ], "Problem": "Just curious, but how much do you think one could raise their score by working through all the past Aime tests?", "Solution_1": "I'd say about 2-6 pts. For me, I raised my AIME score from a 6 in 2004 to a 12 in 2005, mostly by just doing old AIME tests. In my opinion, doing old AIMEs is BY FAR the most effective way to practice for the AIME.", "Solution_2": "Hmm.. \r\n\r\nI only took one AIME in my life ( :D ) so far and that was 2002 AIMEA. I made two to three careless mistakes on that exam and got 5. If I practice AIME for next two months, will there be any chance for me to reach the floor value?\r\n\r\nI mean, you don't know what that will be but let's just say around 7-8. :) In another word, do you think I can raise my score by 3-4 points at least in next two months by practicing old AIMEs?\r\n\r\nThanks a lot!", "Solution_3": "Well, in around 4 months (May-September), I slowly but steadily raised my scores around 5 points by taking a [b]lot [/b]of AIMEs. See:\r\n[url=http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=352]My AIME Scores[/url]", "Solution_4": "Well...\r\n\r\nThe first time I ever saw an AIME was when I took the 2005 AIME 1. I got a 3. After joining AoPS at the end of April, after about six months I could solve over 10 problems on the same AIME. I think it varies between people, but you should expect probably at least 4 points improvement over the course of a few months.", "Solution_5": "[quote=\"1234567890\"]Well, in around 4 months (May-September), I slowly but steadily raised my scores around 5 points by taking a [b]lot [/b]of AIMEs. See:\n[url=http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=352]My AIME Scores[/url][/quote]\r\n\r\nLol, I'm at where you were in August right now :blush: I need to get my score up at least 5 points if I want to get a 10. I'm modifying my schedule to include an AIME every week :!: :D", "Solution_6": "Hmm... I'm doing an AIME a day during breaks. :) :lol:", "Solution_7": "I'm just doing any AIME problems that are posted on AoPS... :roll:", "Solution_8": "I do an AIME a week... I get about a 10 on them. But that's with the pressure off, with the pressure on I'll probably get like a 6 or something." } { "Tag": [ "probability", "complex analysis", "real analysis" ], "Problem": "I will be graduating next quarter with a degree in mathematics (God willing). Problem is: during my first and second years in UCSD(after transferring from WLAC), I really messed up my gpa and I was not really focused. I've been trying to clean up my act ever since. \r\nI started researching for jobs and stuff and I stumbled across this place and I learnt more about careers in math than I ever knew.\r\n\r\nI'm interested in Actuaries since that's a path that I can still take (due to the fact that I'm graduating soon)... \r\n\r\nCan one of you please point me in the right direction?", "Solution_1": "If you look through this forum you'll find a fair amount of discussion about actuarial work.\r\n\r\nOtherwise, you might do better to ask specific questions.", "Solution_2": "[quote=\"MCrawford\"]If you look through this forum you'll find a fair amount of discussion about actuarial work.\n\nOtherwise, you might do better to ask specific questions.[/quote]\r\n\r\nThanks for your reply. I'll be more specific.\r\nAfter I get my B.A. in Applied Mathematics, will I be able to become an actuary? \r\nI haven't taken any classes in finance, but I have probability and statistics in my bag along with some programming and a bunch of mathematics (complex analysis, real analysis, abstract algebra, numerical analysis, numerical pde's).", "Solution_3": "I don't know enough about the actuarial world to have an opinion on if you're screwed or not, but I strongly suggest you just proceed at this point as if you are not screwed. Start researching jobs. Check out the major actuarial societies for guidance.", "Solution_4": "I also recommend taking the first exam. If you're solid with your prob/stats, then you'll pass it easily with a high score. It should be easier than a 200 level math course exam at UCSD, but you'll want to make sure you get a high score, so looking at a couple of practice tests is probably a good idea.", "Solution_5": "[quote=\"MCrawford\"]I also recommend taking the first exam. If you're solid with your prob/stats, then you'll pass it easily with a high score. It should be easier than a 200 level math course exam at UCSD, but you'll want to make sure you get a high score, so looking at a couple of practice tests is probably a good idea.[/quote]\r\n\r\nThank you. I've already looked at some of the practice tests and they don't seem that bad. I'll try to take the exam.", "Solution_6": "A good place that answers a lot of questions about actuarial careers is the Actuarial Advice page:\r\n\r\nhttp://users.aol.com/fcas/advice.html" } { "Tag": [ "advanced fields", "advanced fields theorems" ], "Problem": "Hi, I am just interested in a simple formula. I just need it to do other stuff.\r\n\r\nSuppose you are in 3-dimension Euclidean with f(x,y,z)=k where k>0.\r\nSuppose that I have two points (a1,b1,c1) and (a2,b2,c2) such that ai,bi,ci>0 and \r\nf(ai,bi,ci)=o for i=1,2. Then what is the exact equation of the geodesics (smallest curve on the plane given by f) that go from one point to the other on f? Thanks.", "Solution_1": "It's hard in general to find the exact solution of the geodesic-problem, even in this simple case. The geodetic that connect two points on a surface is the solution of a second order sistem of linear differential equations that involves coefficients of the metric tensor of the surface. You can find it in any text of Riemannian geometry..." } { "Tag": [ "geometry", "trapezoid", "geometric transformation", "reflection", "Gauss", "cyclic quadrilateral", "perpendicular bisector" ], "Problem": "Given a cyclic quadrilateral ABCD, and the points of intersection $F=AB\\cap CD$ and $G=BC\\cap DA$. Let M be the midpoint of the segment FG. Denote by A', B', C' and D' the nine-point centers of triangles AFG, BFG, CFG and DFG.\r\n\r\nProve that these points A', B', C', D' and M lie on one circle. More precisely, the quadrilateral A'D'B'C' is an isosceles trapezium, and the perpendicular bisectors of the segments A'C' and B'D' coincide and pass through the point M.\r\n\r\nThis was indicated as a conjecture in\r\n\r\nChristopher J. Bradley, [i]The Diagonal Points of a Cyclic Quadrangle[/i], Crux Mathematicorum 3 / 31 (2005), pages 168-172.\r\n\r\nHowever, I must say the result is true and rather easy to prove. Enjoy!\r\n\r\n Darij", "Solution_1": "If you reflect $B$ in $FG$ to get $B''$, the quadrilateral $B''GDF$ will be cyclic, so the circumradii of $BFG,DFG$ are equal, which means that so are the radii of their nine-point circles. We thus get $B'M=D'M$. In the exact same way we find $C'M=A'M$.\r\n\r\nLet $X,Y,Z$ be the midpoints of $BG,BF,BD$ respectively. We have $\\angle XMY=\\angle GBF$ and $\\angle XZY=\\angle GDF=\\pi-\\angle GBF=\\pi-\\angle XMY$, which means that $X,Y,Z,M$ are concyclic, i.e. the midpoint of $BD$ lies on the nine-point circle of $BGF$. It must also lie on the nine-point circle of $DGF$ then, which means that $B'D'$ is perpendicular to the line connecting the midpoints of $FG,BD$, i.e. the Gauss line of $ABCD$. Of course, $A'C'$ must also be perpendicular to this line, and we have $A'C'\\|B'D'$. Together with what we have from the paragraph above, we get the stronger conclusion written by Darij.", "Solution_2": "You forgot to show that the points A', B', C', D' and M lie on one circle. Funnily enough, so did I when I first saw the problem.\r\n\r\nAnyway, now I have a [i]complete[/i] solution of the problem:\r\n\r\nWe will work with directed angles modulo 180\u00b0.\r\n\r\nLet U and V be the midpoints of the segments BD and AC. Then, the Gauss line g of the complete quadrilateral formed by the lines AB, BC, CD, DA passes through the midpoints U, V and M of its diagonals BD, AC and FG.\r\n\r\nLet X, Y, R, W be the midpoints of the segments BG, BF, DG, DF, respectively.\r\n\r\nSince the points M and Y are the midpoints of the sides FG and BF of triangle BFG, we have MY || GB. Similarly, UY || DF, MX || FB and UX || DG. Thus, < (MY; UY) = < (GB; DF) and < (MX; UX) = < (FB; DG). Since the quadrilateral ABCD is cyclic, < BCD = < BAD. Hence,\r\n\r\n < MYU = < (MY; UY) = < (GB; DF) = < BCD = < BAD = < (FB; DG) = < (MX; UX) = < MXU.\r\n\r\nThus, the points M, X, Y and U lie on one circle; call this circle $k_B$. This circle $k_B$, passing through the midpoints M, X and Y of the sides FG, BG and BF of the triangle BFG, must be the nine-point circle of triangle BFG; hence, its center is the nine-point center B' of triangle BFG. Similarly, the points M, R, W and U lie on one circle $k_D$, which is the nine-point circle of triangle DFG and has center D'. Now, as we saw above, the chordal angle of the chord MU in the circle $k_B$ equals < MYU = < BAD; similarly, we can see that the chordal angle of the chord MU in the circle $k_D$ equals < DCB. Since < DCB = - < BCD = - < BAD, it follows that the chordal angles of the chord MU in the circles $k_B$ and $k_D$ are opposite. Thus, these circles are symmetric to each other with respect to the line MU. Hence, their centers B' and D' are also symmetric to each other with respect to the line MU. Since the line MU is the Gauss line g of the complete quadrilateral ABCD, it follows that the points B' and D' are symmetric to each other with respect to the Gauss line g. Thus, the Gauss line g is the perpendicular bisector of the segment B'D'. Similarly, the Gauss line g is the perpendicular bisector of the segment A'C'. Hence, we have shown that the perpendicular bisectors of the segments A'C' and B'D' coincide - namely, they both coincide with the Gauss line g -, and pass through the point M (since g passes through M).\r\n\r\nSince the perpendicular bisectors of the segments A'C' and B'D' coincide, the quadrilateral A'D'B'C' is an isosceles trapezium.\r\n\r\nNow, it remains to show that the points A', B', C', D' and M lie on one circle. Prepare for another angle chase:\r\n\r\nFirst, since the points M, U and Y lie on the circle $k_B$, while the center of this circle is B', the central angle theorem yields < UMB' = 90\u00b0 - < MYU. Since < MYU = < BCD, this becomes < UMB' = 90\u00b0 - < BCD. Similarly, < UMD' = 90\u00b0 - < DAB. Thus,\r\n\r\n < B'MD' = < UMD' - < UMB' = (90\u00b0 - < DAB) - (90\u00b0 - < BCD) = < BCD - < DAB = < BCD + < BAD.\r\n\r\nSince < BCD = < BAD, this becomes < B'MD' = 2 < BCD.\r\n\r\nLet S and T be the feet of the perpendiculars from the points F and G to the lines BC and CD. Then, < FSG = 90\u00b0 and < FTG = 90\u00b0; thus, the points S and T lie on the circle with diameter FG. The center of this circle is, of course, the midpoint M of the segment FG. Hence, by the central angle theorem, < SMT = 2 < GST. But < GST = < (BC; GT) = < (BC; CD) + < (CD; GT), and $GT\\perp CD$ yields < (CD; GT) = 90\u00b0, so that we get < GST = < (BC; CD) + 90\u00b0 = < BCD + 90\u00b0, and thus\r\n\r\n < SMT = 2 < GST = 2 (< BCD + 90\u00b0) = 2 < BCD + 180\u00b0 = 2 < BCD (since 180\u00b0 = 0\u00b0).\r\n\r\nSince < B'MD' = 2 < BCD, this becomes < SMT = < B'MD'.\r\n\r\nThe nine-point circles of triangles BFG and CFG have two points in common: the point M, being the midpoint of the common side FG of the two triangles, and the point S, being the foot of the altitude from the vertex F to the side BG aka CG. Hence, the line MS is the common chord of these two nine-point circles. But the centers of these two nine-point circles are B' and C', respectively. Since the common chord of two circles is perpendicular to the line joining their centers, we thus have $MS\\perp B^{\\prime}C^{\\prime}$, so that < (MS; B'C') = 90\u00b0. Similarly, < (MT; C'D') = 90\u00b0. Hence,\r\n\r\n < B'C'D' = < (B'C'; C'D') = - < (MS; B'C') + < (MS; MT) + < (MT; C'D')\r\n = - 90\u00b0 + < SMT + 90\u00b0 = < SMT = < B'MD'.\r\n\r\nHence, the point C' lies on the circle through the points B', D' and M. Similarly, the point A' lies on the circle through the points B', D' and M. Hence, the points A', B', C', D' and M lie on one circle. And the problem is solved.\r\n\r\n darij" } { "Tag": [ "linear algebra", "matrix", "induction", "linear algebra unsolved" ], "Problem": "Consider $ A, B \\in M_n (\\mathbb{R})$ such that $ B\\minus{}A\\equal{}I_n$. What conditions must the matrix $ A$ have such that for any $ k \\in \\mathbb{N}^*$, the equality $ B\\plus{}B^2\\plus{}...\\plus{}B^k\\equal{}kI_n\\plus{}\\frac{k(k\\plus{}1)}{2}\\cdot A$ holds?", "Solution_1": "for $ k\\equal{}2$ we find $ A^2\\equal{}0$.\r\nnow suppose that $ A^2\\equal{}0$ then $ B^2\\equal{}2B\\minus{}I$ ,and by induction $ B^k\\equal{}kB\\plus{}(1\\minus{}k)I$ for every natural number $ k$.\r\nand in this case we have : $ B\\plus{}B^2\\plus{}...\\plus{}B^k\\equal{}kI_n\\plus{}\\frac{k(k\\plus{}1)}{2}\\cdot A$." } { "Tag": [], "Problem": "Provide reasonable synthesis of each of the following amine drugs (click to see structure and relevant information):\r\n\r\n1) [url=http://en.wikipedia.org/wiki/Pargyline]Pargyline[/url]\r\n\r\n2) [url=http://en.wikipedia.org/wiki/Tranylcypromine]Tranylcypromine[/url]\r\n\r\n3) [url=http://en.wikipedia.org/wiki/Rasagiline]Rasagiline[/url]", "Solution_1": "from what reagents? :maybe:", "Solution_2": "From readily available reagents, like benzene or toluene, alcohols, etc.", "Solution_3": "for the second one, is it possible to use a simmons-smith reaction without affecting the benzne ring?", "Solution_4": "i just realized that you wud probably get tropilidene as well.\r\nneways how abt this for the frist one?\r\nn,nmethyl,benzylamine + formaldehyde + ethyne(one end protected by TMS) in presence of CuBr \r\n and foll by deprtoection of the TMS group?\r\nsomething like this?\r\nhttp://www.orgsyn.org/orgsyn/prep.asp?prep=v84p0001", "Solution_5": "And what is the purpose of that?" } { "Tag": [ "limit", "linear algebra", "matrix", "real analysis", "real analysis unsolved" ], "Problem": "Let $ (F_{n})_{n\\geq 1}$ be the Fibonacci number, defined as $ F_{1}=F_{2}=1$ and $ F_{n+2}=F_{n+1}+F_{n}$. Prove that:\r\n\r\n a) $ \\lim_{n\\to\\infty}\\sum_{k=1}^{n}\\frac{F_{k}}{2^{k}}=2$;\r\n\r\n b) $ \\sum_{n=1}^{\\infty}\\arctan{F_{2n-1}}=\\frac{\\pi}{2}$.", "Solution_1": "With a) we can use Binet's theorem and $ 1+x+...+x^{n}=\\frac{x^{n+1}-1}{x-1}$. :D", "Solution_2": "Hmm, Kuba told me a solution for a). I would be interrested if you can post a full solution. :)", "Solution_3": "Let $ A = \\begin{pmatrix}1 & 1 \\\\ 1 & 0 \\end{pmatrix}$. For appropriate* $ x$ we have that\r\n\\[ \\sum_{n=0}^\\infty \\left( xA \\right)^{n}= \\left( I-xA \\right)^{-1}. \\]\r\nNow let's look at the upper-right element of both sides. In the left, we have $ \\sum_{n=1}^\\infty x^{n}F_{n}$. In the right, it's $ \\frac{x}{1-x-x^{2}}$.\r\n[I assumed that we all know that $ A^{n}= \\begin{pmatrix}F_{n+1}& F_{n}\\\\ F_{n}& F_{n-1}\\end{pmatrix}$.]\r\n\r\nPlug in $ x = \\frac12$ and there you go :)\r\n\r\n\r\n*We do have to find out what the range of $ x$ can be. The condition is $ (xA)^{n}\\to 0$, i.e. $ |x| < \\frac{\\sqrt 5-1}2$.", "Solution_4": "Here is the solution for number 1)\r\n\r\n$ \\sum_{n=1}^{\\infty}\\frac{F_{n}}{2^{n}}=\\frac{1}{\\sqrt{5}}\\sum_{n=1}^{\\infty}\\left(\\left(\\frac{\\alpha}{2}\\right)^{n}-\\left(\\frac{\\beta}{2}\\right)^{n}\\right)$\r\n\r\n$ =\\frac{1}{\\sqrt{5}}\\left(\\frac{\\alpha}{2-\\alpha}-\\frac{\\beta}{2-\\beta}\\right)=2$,\r\n\r\nwhere $ \\alpha=\\frac{1+\\sqrt{5}}{2}$ and $ \\beta=\\frac{1-\\sqrt{5}}{2}$.", "Solution_5": "The simple trick without Binet's formula for number 1 can be found, for example, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=160771]here[/url].", "Solution_6": "I think you meant \r\n\r\n$ \\sum_{n=0}^{\\infty}\\arctan\\frac{1}{F_{2n+1}}=\\frac{\\pi}{2}$.\r\n\r\nWe have that\r\n\r\n$ \\arctan F_{2k+2}-\\arctan F_{2k}=\\arctan\\frac{F_{2k+1}}{1+F_{2k+2}F_{2k}}=\\arctan\\frac{1}{F_{2k+1}}$, since \r\n\r\n$ 1+F_{2k+1}F_{2k}=F_{2k+1}^{2}$.\r\n\r\nIterate and the result would follow." } { "Tag": [], "Problem": "Provide the shortest way of making the following functional group interconversion:\r\n\r\n$ \\ce{R\\minus{}NH2} \\longrightarrow \\ce{R\\minus{}CHO}$", "Solution_1": "how about Hofmann elimination (excess methyl iodide; silver oxide, water, heat), addition of HBr, conversion to a Grignard (oxidative insertion of magnesium metal), attack on formaldehyde and quench, and oxidation via PCC ?", "Solution_2": "how about converting $ \\ce{RNH2}$ to $ \\ce{RCN}$ and then convert into $ \\ce{RCN}$ and then hydrolyse to $ \\ce{RCOOH}$ and then reduce to $ \\ce{RCHO}$ :maybe:", "Solution_3": "[b]MysticTerminator[/b], the conversion should be independent of the existence or not of a beta hydrogen. And how about if R is aryl?\r\n\r\n\r\n[b]Madness[/b], your synthesis envolves at least 6 steps. Anything shorter?", "Solution_4": "1. $ \\ce{R \\minus{} NH_2}$ + $ \\ce{CHCl_3}$ + $ \\ce{KOH \\longrightarrow R \\minus{} NC}$\r\n\r\n2. $ \\ce{R \\minus{} NC}$ +$ \\ce{\\Delta}$ $ \\longrightarrow$ $ \\ce{R \\minus{} CN}$\r\n\r\n3. $ \\ce{R \\minus{} CN}$ + $ \\ce{SnCl_2}$ /$ \\ce{HCl \\longrightarrow R \\minus{} CHO}$", "Solution_5": "yes ofc Stephen's reduction :fool:", "Solution_6": "That method of reduction is almost never used nowadays. A more modern one is to simply use LiAlH(O-tBu)3, for example. Also, isonitriles are seldom made that way. But the idea is ok." } { "Tag": [ "LaTeX", "\\/closed" ], "Problem": "http://www.artofproblemsolving.com/NewClassroom/ClientDoc.php#latex\r\n\r\nNote that we do in fact need dollar signs around math and it is *not* different from the forum procedure (except the semicolon).", "Solution_1": "I have just tried it without the dollar signs and it works properly. Can you provide the formula you are using which is not working?", "Solution_2": "Ok, nvm. I was confused by \r\n[quote]This differs from the forum in that you should not use dollar signs around the LaTeX code.[/quote]\r\n\r\nbut then having to use \\text{}. Apologies :)." } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Let n>4, and N a normal subgroup of An (a group of even permutations of Sn), and N $ \\neq$ {(1)}.\r\nProve that there is at least one three-cycle in N.", "Solution_1": "$ A_n$ is simple for $ n > 4$; so $ N$ is necessarily either $ 0$ or $ A_n$.\r\n\r\nThough I suppose you're not supposed to be using that fact...?" } { "Tag": [ "inequalities theorems", "inequalities" ], "Problem": "help me :\r\n\r\nhttp://www.mathlinks.ro/Forum/topic-28678.html\r\n\r\ni can't understand why we can assume a+b+c=1 in this problem,\r\n\r\nsory if i post in the wrong sub forum :D ;)", "Solution_1": "Hi , \r\n This is a homogeneous ineq ! \r\n Bat dang thuc nay thuan nhat ( co nghia la f(a,b,c) = f(ka,kb,kc) )nen ta co the gan cho a+b+c 1 gia tri nao do (khac 0 ) (them gia thiet de bai)" } { "Tag": [], "Problem": "It's probably related to the server move, but all images on the wiki (with the exception of those uploaded extremely recently) seem to have disappeared. :( Thumbnails have the error message \"Error creating thumbnail: sh: /usr/bin/convert: No such file or directory\".\r\n\r\n[I wasn't sure which forum it would be better to post this]", "Solution_1": "Ooops. We will copy the images over. Sorry about that :D", "Solution_2": "[quote=\"azjps\"]It's probably related to the server move, but all images on the wiki (with the exception of those uploaded extremely recently) seem to have disappeared. :( Thumbnails have the error message \"Error creating thumbnail: sh: /usr/bin/convert: No such file or directory\".\n\n[I wasn't sure which forum it would be better to post this][/quote]\r\n\r\nI copied over the image, but there were two that were posted after the server switch but before I moved the images directory that were lost. Azjps, I think you posted both; can you repost (glance through the recent entries, you'll see 'em).", "Solution_3": "Okay, I have re-posted them. Thanks :)" } { "Tag": [ "abstract algebra", "superior algebra", "superior algebra solved" ], "Problem": "Let $G$ be a non-abelian group of order $p^3$ for some prime $p$. Prove that $Z(G)=G'$.", "Solution_1": "I can prove that $Z(G)$ has order $p$, but I guess that is not sufficient?", "Solution_2": "Since $G/Z(G)$ has order $p^2$, it's abelian, which means that $G'\\le Z(G)\\ (*)$. Now, $Z(G)$ has order $p$ and $G'$ cannot be trivial, so $(*)$ implies the desired result.", "Solution_3": "What is $G'$ ?!", "Solution_4": "The commutator group of $G$." } { "Tag": [], "Problem": "cho minh hoi muon xem de thi cac nuoc cac nam cach day nhieu nam thi vao cho nao \r\nminh nghe noi trong mathlinks.ro co de chon doi tuyen vong 2 tu nam 64 minh khong biet cho nao? giup minh nhe\r\n ai co trang web toan gi hay noi cho minh , minh cam on nhieu . \r\n de thi cac nuoc ban nao co the tot bung cung cap cho minh khong \r\n\r\n WTO_tientethegioi@yahoo.com \r\n lien he voi minh nhe", "Solution_1": "[quote=\"tranthanhnam\"]cho minh hoi muon xem de thi cac nuoc cac nam cach day nhieu nam thi vao cho nao \nminh nghe noi trong mathlinks.ro co de chon doi tuyen vong 2 tu nam 64 minh khong biet cho nao? giup minh nhe\n ai co trang web toan gi hay noi cho minh , minh cam on nhieu . \n de thi cac nuoc ban nao co the tot bung cung cap cho minh khong \n\n WTO_tientethegioi@yahoo.com \n lien he voi minh nhe[/quote]\r\nVao cho High shool contest" } { "Tag": [], "Problem": "Did you hear the one about how Arnav's dad got ran over by a truck? Now, Delong has to take his place.", "Solution_1": "what a mean thing to say because I love jesus and god and dont like saying people die in real life when they ddint.", "Solution_2": "[quote=\"rebirth\"]what a mean thing to say because I love jesus and god and dont like saying people die in real life when they ddint.[/quote]\r\n\r\nLike Jesus?\r\n\r\noh wait, i interpreted your statement backwards", "Solution_3": "[quote=\"rebirth\"]what a mean thing to say because I love jesus and god and dont like saying people die in real life when they ddint.[/quote]\r\n\r\nHe never mentioned death. He only mentioned the accident.", "Solution_4": "[quote=\"Jesus sucks\"]Did you hear the one about how Arnav's dad got ran over by a truck? Now, Delong has to take his place.[/quote]\r\n\r\n Mystic Terminator hater, you should be ashamed of your self." } { "Tag": [ "calculus", "integration", "trigonometry", "calculus computations" ], "Problem": "Please find:\r\n\\[{ I=\\int^{ \\pi}_{-\\pi}\\frac{\\sin(nx)dx}{(1+2^{x})\\sin(x)}}\\]", "Solution_1": "Using the substitution $ x =-t$,\r\n\r\n$ \\int_{-\\pi}^{\\pi}\\frac{\\cos nx}{1+2^{x}}\\; dx = \\int_{-\\pi}^{\\pi}\\frac{2^{t}\\cos nt}{1+2^{t}}\\; dt$\r\n\r\nSo $ \\int_{-\\pi}^{\\pi}\\frac{\\cos nx}{1+2^{x}}\\; dx = 0$ for all nonzero integer $ n$.\r\n\r\nNow let ${ I_{n}= \\int_{-\\pi}^{\\pi}\\frac{\\sin nx}{(1+2^{x})\\sin x}}\\; dx$. Then $ I_{n+2}-I_{n}= \\int_{-\\pi}^{\\pi}\\frac{2\\cos(n+1)x}{1+2^{x}}\\; dx = 0$ for all $ n \\geq 0$, with initial values $ I_{0}= 0$ and $ I_{1}= \\pi$." } { "Tag": [ "search" ], "Problem": "Thanks a lot", "Solution_1": "Resistor network:\r\nIf the resistors are in series, current is the same and voltages add. Resistance values add.\r\nIf the resistors are in parallel, currents add and voltage is the same. Reciprocal values of resistances add. \r\n\r\nCapacitor network:\r\nIf the capacitors are in series, charge is the same and voltages add. Reciprocal values of capacitances add.\r\nIf the capacitors are in parallel, charges add and voltage is the same. Capacitance values add. \r\n\r\n(1) C = 4.4 uF, V = 9.8 V.\r\n\r\nInitial charge $Q_i = VC_i = V \\frac{C \\cdot C}{C + C} = \\frac 1 2 VC$\r\n\r\nFinal charge $Q_f = VC_f = V \\frac{2C \\cdot C}{2C + C} = \\frac 2 3 VC$\r\n\r\nAdded charge $\\triangle Q = Q_f - Q_i = \\frac 1 6 VC$\r\n\r\n\r\n(2) V = 88 V. $C_1, C_2$ are in parallel, $C_{12} = C_1 + C_2$\r\n\r\n$C_3, C_{12}$ are in series, the total capacitance is $C = \\frac{C_{12}C_3}{C_{12} + C_3} = \\frac{(C_1 + C_2)C_3}{C_1 + C_2 + C_3}$\r\n\r\nTotal charge is $Q = CV$. Charges on $C_{12}, C_3$ are the same, $Q_{12} = Q_3 = Q$. Voltages add, $V_{12} + V_3 = V$.\r\n\r\n$V_3 = \\frac{Q_3}{C_3} = \\frac{Q}{C_3} = V \\frac{C}{C_3} = V \\frac{C_1 + C_2}{C_1 + C_2 + C_3}$\r\n\r\n$V_{12} = \\frac{Q_{12}}{C_{12}} = \\frac{Q}{C_{12}} = V \\frac{C_3}{C_1 + C_2 + C_3}$\r\n\r\nCharges on $C_1, C_2$ add, $Q_1 + Q_2 = Q_{12} = Q$. Voltages are the same, $V_1 = V_2 = V_{12}$.\r\n\r\n$Q_1 = C_1 V_1 = C_1 V_{12} = V \\frac{C_1 C_3}{C_1 + C_2 + C_3}$\r\n\r\n$Q_2 = C_2 V_2 = C_2 V_{12} = V \\frac{C_2 C_3}{C_1 + C_2 + C_3}$\r\n\r\n\r\n(3) C = 9.88 uF, V = 10 V. Label $C_3, C_4, C_5$ the remaining capacitors. $C_2, C_3$ are in series, $C_{23} = \\frac{C^2}{2C} = \\frac 1 2 C$. $C_{23}, C_4$ are parallel, $C_{234} = C_{23} + C_4 = \\frac 3 2 C$. $C_{234}, C_5$ are in series, $C_{2345} = \\frac{\\frac 3 2 C^2}{\\frac 3 2 C + C} = \\frac 3 5 C$. $C_1, C_{2345}$ are parallel, $C_{12345} = C_2 + C_{2345} = \\frac 8 5 C$.\r\n\r\nTotal charge is $Q = C_{12345}V = \\frac 8 5 CV$. $C_1, C_{2345}$ are parallel, charges add and voltages are the same, $Q_1 + Q_{2345} = Q, V_1 = V_{2345} = V$.\r\n\r\n$Q_1 = C_1 V = CV$\r\n\r\n$Q_{2345} = C_{2345} V = \\frac 3 5 CV$\r\n\r\n$C_{234}, C_5$ are in series, charges are the same and voltages add, $Q_{234} = Q_5 = Q_{2345}, V_{234} + V_5 = V$.\r\n\r\n$V_{234} = \\frac{Q_{234}}{C_{234}} = \\frac{Q_{2345}}{C_{234}} = \\frac{\\frac 3 5 CV}{\\frac 3 2 C} = \\frac 2 5 V$\r\n\r\n$V_5 = \\frac{Q_5}{C_5} = \\frac{Q_{2345}}{C_5} = \\frac{\\frac 3 5 CV}{C} = \\frac 3 5 V$\r\n\r\n$C_{23}, C_4$ are parallel, charges add and voltages are the same, $Q_{23} + Q_4 = Q_{234}, V_{23} = V_4 = V_{234}$.\r\n\r\n$Q_{23} = C_{23}V_{23} = C_{23}V_{234} = \\frac 1 2 C \\cdot \\frac 2 5 V = \\frac 1 5 CV$\r\n \r\n$Q_4 = C_4V_4 = C_4V_{234} = C \\cdot \\frac 2 5 V = \\frac 2 5 CV$\r\n\r\n$C_2, C_3$ are in series, charges are the same and voltages add, $Q_2 = Q_3 = Q_{23}, V_2 + V_3 = V_{23}$.\r\n\r\n$V_2 = \\frac{Q_2}{C_2} = \\frac{Q_{23}}{C_2} = \\frac{\\frac 1 5 CV}{C} = \\frac 1 5 V$\r\n \r\n$V_3 = \\frac{Q_3}{C_3} = \\frac{Q_{23}}{C_3} = \\frac{\\frac 1 5 CV}{C} = \\frac 1 5 V$\r\n\r\nThe answer is $Q_1 = CV$ and $Q_2 = \\frac 1 5 CV$.", "Solution_2": "Thank you so much for your help. I think I understand it now. However, I was still unable to answer the second part of the first question which read:\r\n\r\n(b) what is the increase in the sum of the charges: the charge on the positive plate of one capacitor plus the charge on the positive plate of the other capacitor? \r\n\r\nWhat am i doing wrong? Thanks", "Solution_3": "Initially, the positive charge $+Q_i = \\frac 1 2 CV$ is on the positive plate of the capacitor $C_1$ and the negative charge $-Q_i = -\\frac 1 2 CV$ of equal magnitude is on its negative plate. The same is true for the capacitor $C_2$, because the 2 capacitors are in series and their charges are equal. The initial sum of charges on the positive plates of the 2 capacitors is $+2Q_i = CV$. Afterwards, the positive charge $+Q_f = \\frac 2 3 CV$ is on the positive plate of the capacitor $C_1$, the negative charge $-Q_f = \\frac 2 3 CV$ on its negative plate and the same is true for the other capacitor. The final sum of charges on the positive plates of the 2 capacitors is $+2Q_f = \\frac 4 3 CV$. The increase in the sum of the positive charges is $2Q_f - 2Q_i = \\frac 4 3 CV - CV = \\frac 1 3 CV$.", "Solution_4": "Thanks for your help.", "Solution_5": "Since [b]yetti[/b] has provided sufficiently detailed solutions of the past problems, I will present only the main steps of the work here:\r\n\r\n1) In short: \\[ V_3 = V, \\qquad q_3 = V_3C_3 = VC_3, \\qquad U_3 = \\frac{V^2C_3}{2}, \\]\r\n\\[ q_1 = q_2 = V\\frac{C_1C_2}{C_1+C_2}, \\qquad V_1 = \\frac{q_1}{C_1} = V\\frac{C_2}{C_1+C_2}, \\qquad V_2 = \\frac{q_2}{C_2} = V\\frac{C_1}{C_1+C_2}, \\]\r\n\\[ U_1 = \\frac{V_1^2 C_1}{2} = \\frac{V^2C_2^2C_1}{2(C_1+C_2)^2}, \\qquad U_2 = \\frac{V_2^2 C_2}{2} = \\frac{V^2C_1^2C_2}{2(C_1+C_2)^2}. \\]\r\n\r\n2) When $C_3$ breaks down, the voltage on both capacitors left will be $V$, so that is the answer for b). As for a), the total charge on the system $C_1||C_2$ ($C_1$ parallely connected to $C_2$) will be $q_{12} = V\\frac{C_3(C_1+C_2)}{C_1+C_2+C_3}$ which is further equal to $q_3$. The voltage on this system is $V_{12} = V\\frac{C_3}{C_1+C_2+C_3} \\Rightarrow q_1 = V_1C_1 = V\\frac{C_1C_3}{C_1+C_2+C_3}$. After the breakdown the charge on $C_1$ will be $q'_1 = VC_1$, which means that the change of the charge is $\\Delta q = q'_1 - q_1 = V\\frac{C_1(C_1+C_2)}{C_1+C_2+C_3}$.", "Solution_6": "Thank you so much. I believe you were slightly incorrect with the algebra in determining U1 and U2 (when you squared V1 and V2, respectively) but I figured it out nonetheless. Thank you very much.", "Solution_7": "Oh, yes, I forgot to write down the two-s in the final expressions... :oops:", "Solution_8": "its ok i figured it out anyway..\r\n\r\nthank you very much. i really appreciate your help", "Solution_9": "[quote=\"jeter123\"]Thanks a lot[/quote]\r\n\r\nIt is not a good idea to erase the initial post of a thread and especially not the central question.! :mad: Try to restore these.\r\n Others have to search the whole topic to find out what it is all about, so please\r\ndont edit without consideration." } { "Tag": [ "function", "number theory", "greatest common divisor", "combinatorics proposed", "combinatorics" ], "Problem": "There are $ n$ runners waiting at distinct points of a circular race track. At the starting signal each runner takes off. Each person arbitrarily chooses which of the two possible directions to go. Each has the same constant speed and whenever two of them meet they both change direction (instantaneously). This continues indefinitely. Show that there is a moment (after the start) at which each runner is back at his/her starting point.\r\n\r\nGiven the values of $ n$ and the time $ t$ it takes a runner to run a full lap (unobstructed), is there an upper bound on the amount of time it can take for the above to happen, and if there is, what is it (as a function of $ n$ and $ t$)?", "Solution_1": "We can think of the situation in a different way: when two runners encounter each other, rather than each changing directions, the two continue as they were, but trade identities. Considering the situation like this, it is clear that after every period of time $ t$, there is exactly one runner in each of the starting locations. However, after time $ t$, the runners in those initial positions have possibly changed identities. Yet, it is clear that the identities of the runners in each starting position cycle the same way every time $ t$. Since this is represented by some finite set of finite cycles, after the GCD of the sizes of these cycles times $ t$, all the runners will be in their initial positions, as desired.\r\n\r\nTo find an upper bound on the amount of time it can take for all the runners to return to their initial positions, we make one more observation: the order of the runners never changes. If we look around the track, the runners will always appear in the same order. Thus, our set of cycles from before is actually a single cycle containing all $ n$ runners, and it follows that the maximum amount of time it can take for all the runners to return to their initial positions is $ nt$, which can be obtained, for instance, when all the runners are running the same way except for one." } { "Tag": [ "algebra", "polynomial", "modular arithmetic", "search", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $ f$ be an irreducible polynomial in $ \\mathbb{Z}[x]$. Let $ \\mathcal{P}$ be the family of prime ideals in $ \\mathbb{Z}[x]$ that contain $ f$. Is $ \\#\\mathcal{P}$ finite?", "Solution_1": "Not in general.\r\nPut $ f: \\equal{} x^{2} \\plus{} 1.$\r\nThen for every $ p\\in \\mathbb{Z}$ prime st $ p \\equiv \\minus{} 1\\pmod{4} \\ f$ is irreducible in $ \\mathbb{Z}_{p}$, thus ideal $ I_{p}: \\equal{} (f, p) \\subset \\mathbb{Z}[x]$ is prime\r\n(it's even maximal, see: http://www.mathlinks.ro/viewtopic.php?search_id=1643396135&t=72993).\r\nSince we have infinitely many primes $ p\\in \\mathbb{Z}$ satisfying $ p\\equiv \\minus{} 1\\pmod{4},$ and $ I_{p_{1}}\\neq I_{p_{2}}$ for such primes $ p_{1} \\neq p_{2},$ then the set $ \\mathcal{P}$ is infinite." } { "Tag": [ "calculus", "algebra", "partial fractions", "real analysis", "real analysis solved" ], "Problem": "a(n)=3/(1*2*3)+5/(2*3*4)+...+(2n-1)/((n-1)n(n+1)) for n \\geq 2\r\n\r\ni) Compute a(n) \r\nii) Show that a(n) is monotone and bounded\r\niii) lim a(n)\r\niv) lim ((4-a(n))/5)^n \r\n---------------------------------------------------------------------------\r\nIf you have another solution post it, here I see this nice one\r\n\r\nWatching how a(n) is defined we see f(x)=(2x-1)/((x-1)x(x+1)) partial fractions gives f(x)=1/2.1/(x-1)+1/x -3/2.1/(x+1)\r\na(n)= \\sum (k=2 to n)f(k)=1/2(H(n)-1/n) +(H(n)-1) -3/2(H(n)-1/(n+1)-1/2-1) where H(n)= \\sum (k=1 to n)1/k \r\na(n)=5/4 - 1/(2n) - 3/2.1/(n+1) with this expression we have the expression for i)\r\nii) a(n) is increasing, 0 \\leq a(n) \\leq 5/4\r\niii) lim a(n)=5/4\r\niv) a(n)= 5/4- 1/(2n) - 3/2.1/n + o(1/n) = 5/4 - 2/n +o(1/n) \r\nI am not sure for the question iv) maybe\r\nI think it is more nicer to compute lim (-1/4 + a(n))^n, indeed\r\n(-1/4 + a(n))^n = (1 -1/(2n) + o(1/n))^n =\r\nexp(n.ln(1- 1/(2n) + o(1/n)))=exp( -1/2 + o(1)) \r\nlim (-1/4 + a(n))^n = 1/ \\sqrt e", "Solution_1": "[quote=\"Moubinool\"]\niv) a(n)= 5/4- 1/(2n) - 3/2.1/n + o(1/n) = 5/4 - 2/n +o(1/n) \n lim (-1/4 + a(n))^n[/quote]\r\n\r\nJust en error of calculus \r\n(-1/4 +a(n))^n = ( 1 - 2/n+o(1/n))^n, its is exp(-2)", "Solution_2": "As Lagrangia said compute lim (4/5.a(n))^n\r\n(4/5.a(n))^n=((4/5(5/4-2/n+o(1/n)))^n=(1-8/5.1/n+o(1/n))^n \r\nits limit is exp(-8/5)" } { "Tag": [ "absolute value" ], "Problem": "What is $ |x^{3}|\\div{x^{2}}$? Is the answer $ x$ or $ |x|$?\r\n\r\nThen what is $ |x^{3}|^{3}$?\r\n\r\nHow about $ 5|x|y^{2}\\minus{}4xy^{2}$? Is it $ |x|y^{2}$, $ xy^{2}$, or something completely different?", "Solution_1": "For the first one, we have \\begin{align*}\r\n|x^3|\\div x^2 &= |x^3|\\div |x^2| \\\\\r\n&= |x^3\\div x^2| \\\\\r\n&= |x| \\\\\r\n\\end{align*}\r\nThe second one is just \\begin{align*}\r\n|x^3|^3 &= |x^3|\\cdot |x^3|\\cdot |x^3| \\\\\r\n&= |x^3\\cdot x^3\\cdot x^3| \\\\\r\n&= |x^9| \\\\\r\n\\end{align*}\r\nFor that last one, if $ x\\geq 0$, then we just get $ 5xy^2-4xy^2=xy^2=|x|y^2$. However, if $ x\\leq 0$, we end up with $ -9xy^2$", "Solution_2": "Why $ \\minus{} 9xy^{2}$?\r\n\r\nAnd multiplying absolute value quantities is like multiplying radicals with similar indexes?\r\n\r\n$ |x|*|y| \\equal{} |xy|$?\r\n\r\nBut:\r\n\r\n$ |x^{3}|*|x^{2}| \\equal{} x^{6}$?\r\n\r\nSince the power in the product is even?", "Solution_3": "okay, think about it. how can $ |a^2|\\cdot|a^3|$ equal $ |a^6|$? suppose they are all positive:\r\n\r\n$ a\\cdot{a}$ is $ a^2$ and $ a\\cdot{a\\cdot{a}}$ is $ a^3$. therefore, $ |a^2|\\cdot|a^3|\\equal{}|a^5|$\r\n\r\ni hope that made sense?", "Solution_4": "I made a typo. I meant:\r\n\r\n$ |x^{3}|*|x^{3}|\\equal{}x^{6}$\r\n\r\nSince even powers are always positive?" } { "Tag": [ "MATHCOUNTS" ], "Problem": "I need the old mathcounts tests from about the first one to the the year 2000. does anybody know where I can get them?", "Solution_1": "look on the stickie \" MATHCOUNTS Resources Here\"" } { "Tag": [ "algebra", "polynomial", "Gauss", "algebra proposed" ], "Problem": "Greetings. It's been quite a long time since I last entered the forum,\r\nso I decided to share the following original irreducibility criterion, which\r\nhas a nice (btw classical) corollary.\r\n\r\nTheorem. (Original) Let $p$ be a prime and $l$ be an integer not divisible by\r\n$p$. Assume that a polynomial\r\n\\[\r\nf(x) = a_n x^n + a_{n-1}x^{n-1} + \\cdots + a_0 \\, \\in \\mathbb{Z} [x]\r\n\\]\r\nhas the following properties.\r\n\r\n(i) $p \\, \\nmid \\, a_0$ and $p \\, \\nmid \\, a_{n-1}$;\r\n(ii) The leading coefficient $a_n$ is of the form $lp^k$.\r\n(iii) All complex roots of $f(x)$ lie in the annulus $|z| <\r\n l^{-1}$. \r\n\r\nThen, $f$ is irreducible over $\\mathbb{Z}$.\r\n\r\nWith $l=1$, $n=p^k$ and $f(x) = (1 + x + \\cdots + x^n)' = 1 + 2x + 3x^2 + \\cdots + nx^{n-1}$, we obtain the following cute corollary.\r\n\r\nCorollary. If $n$ is a prime power, then the polynomial\r\n\\[\r\n1 + 2x + 3x^2 + \\cdots + nx^{n-1}\r\n\\]\r\nis irreducible.\r\n\r\nIt has been conjectured that this polynomial is irreducible for all values of $n$, and, to the best of my knowledge, this conjecture is still open. (Although it has been shown that the above polynomial is irreducuble for \"almost all\" $n$.)\r\n\r\nI would welcome other nice corollaries of my criterion.\r\n\r\nHave a great day!\r\n\r\n--Vesselin", "Solution_1": "What's the proof for this criterion?", "Solution_2": "Welcome back, Vess, we missed you. Please, do not post the proof and let us the pleasure to try. Of course, I will fail, but let us think a little bit. I also knew that this is an open question and that is solved for almost all n, but I forgot where I wrote this.", "Solution_3": "Here is a little ammendment. In condition (iii), it actually suffices to have $l^{-1}$ instead of $l^{1-n}$, sorry for the misjudgment. This is actually much better, and now condition (iii) does not depend on $n$.\r\n\r\n\r\n\r\nTo answer iura's question, here is an outline. Assume that $f=gh$ with $g,h \\in \\mathbb{Z}[x]$ --- non-constant.\r\n\r\n1. Using (i) (only), show that either $g$ or $h$ has leading coefficient which is not divisible by $p$.\r\n2. Assume that the leading coefficient of $g$ is not divisible by $p$. By (ii), this coefficient is at most $l$.\r\n3. Since, by (iii), all complex roots of $f$ are in the annulus $|z| n$ for some $ n$.\r\nSo we ruled out $ f(x) \\equal{} c \\cdot x^{2m} \\plus{} r(x), c \\neq 0$ where $ r(x)$ is an odd polynomial, but every polynomial which can not be written on that form is a solution.", "Solution_2": "[quote=\"Mathias_DK\"][quote=\"Stephen\"]Find all polynomial $ f$ that all coefficients are all intergers which satisfies the term below.\n\nThere are infinitly many natural numbers $ a, b$ that satisfies $ gcd(a,b) \\equal{} 1$ and $ a \\plus{} b|f(a) \\plus{} f(b)$[/quote]\nIt is obvious that $ a \\plus{} b \\mid a^n \\plus{} b^n$ when $ n$ is odd. So all odd polynomials satisifies the condition.\nLet $ f(x) \\equal{} x r(x^2) \\plus{} q(x^2)$ (where $ xr(x^2)$ is the odd part of $ f$ and $ q(x^2)$ is even part of $ f$)\n$ a \\plus{} b \\mid f(a) \\plus{} f(b) \\equal{} (ar(a^2) \\plus{} br(b^2)) \\plus{} q(a^2) \\plus{} q(b^2) \\iff$\n$ a \\plus{} b \\mid q(a^2) \\plus{} q(b^2)$\n$ q(b^2) \\equiv q(( \\minus{} a)^2) \\equal{} q(a^2) \\bmod a \\plus{} b$\nSo $ a \\plus{} b \\mid 2q(a^2),2q(b^2)$\nAssume that $ q(x) \\equal{} c \\cdot x^m, c \\neq 0$ for some $ m$. Then\n$ a \\plus{} b \\mid 2c \\cdot a^{2m} \\iff a \\plus{} b \\mid 2c$. But there are not infinitly many pairs $ (a,b)$ such that $ a \\plus{} b \\le 2c$ so this is impossible.\nHence we rule out $ q(x) \\equal{} c \\cdot x^m, c \\neq 0$.\nNow let $ q(x) \\equal{} x^t \\cdot r(x)$ s.t. $ r(0) \\neq 0$.\nNow $ a \\plus{} b \\mid 2q(a^2) \\iff a \\plus{} b \\mid 2a^2t r(a^2) \\iff a \\plus{} b \\mid 2r(a^2)$. Now we can find infinitly many pairs by choosing $ b \\equal{} |2r(a^2)| \\minus{} a$. Fore then $ |2r(a^2)| \\minus{} a \\equiv |2r(0)| \\bmod a$. And hence we choose $ a \\equal{} k|2r(0)| \\plus{} 1$ for some $ k$. Then $ \\gcd(a,b) \\equal{} 1$. Since $ r(x)$ isn't constant, because then $ q(x)$ would have been on the form $ c \\cdot x^m$, we see that $ |2r(a^2)|$ grows faster than $ a$, and hence $ |2r(a^2)| \\minus{} a$ must be positive for all $ a > n$ for some $ n$.\nSo we ruled out $ f(x) \\equal{} c \\cdot x^{2m} \\plus{} r(x), c \\neq 0$ where $ r(x)$ is an odd polynomial, but every polynomial which can not be written on that form is a solution.[/quote]\r\n\r\nThank you so much! :lol:" } { "Tag": [ "geometry", "parallelogram" ], "Problem": "I've gone through all of AOPS vol. 1 twice now. Looking back, I can solve all questions requiring a simple answer consisting of a number or a few variables. \r\n\r\nHowever, I have skipped quite a few proofs, even the fundamental ones in the exercises throughout the chapter (I've gone through some of the thought processes needed to prove a statement, but did not formulate anything onto paper).\r\n\r\nNow I am going back to the proofs found in the problem section \r\nat the end of each chapter, beginning with 11 (triangles). I think I can remember the proofs that were part of the exercises, and I was wondering if most of their results are considered well-known?\r\n\r\nFor instance, \"median to opposite side of a triangle is half that side, then the triangle is right\" or \"sum of the squares of the sides of a parallelogram is equal to the sum of the squares of its diagnols\" are examples. Are some of these considered well-known theorems? In other words, when working on the proofs from selected contests such as M&IQ or AHSME (these were MC questions turned into proofs by authors?), can I state most of the results from the exercises found throughout the first part of the chapter?", "Solution_1": "Yes, though you should know how to derive them in case you run into something you can't cite. (Like in Aops 2).\r\n\r\nBut you should REALLY REALLY REALLY know the proofs by heart. People won't always be there to tell you what is trivial and what isn't, and you should err on the side of caution.", "Solution_2": "[quote=\"PenguinIntegral\"]Yes, though you should know how to derive them in case you run into something you can't cite. (Like in Aops 2).\n\nBut you should REALLY REALLY REALLY know the proofs by heart. People won't always be there to tell you what is trivial and what isn't, and you should err on the side of caution.[/quote]\r\nI agree, and once you do the proofs, you don't forget the stuff you learned. I had to learn the hard way. :blush: (not the whole book, only one chapter)\r\nbtw, nice quote snipez, Harlan Ellison should have meet Einstein. :D" } { "Tag": [ "geometry solved", "geometry" ], "Problem": "What is \"inner\" point ? Can someone explain it in Russian, please?\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=60810", "Solution_1": "Inner point of a figure = \u0432\u043d\u0443\u0442\u0440\u0435\u043d\u043d\u044f\u044f \u0442\u043e\u0447\u043a\u0430 \u0444\u0438\u0433\u0443\u0440\u044b.\r\n\r\n\u0415\u0441\u043b\u0438 \u0432\u043e\u043f\u0440\u043e\u0441 \u043e \u0441\u0442\u0440\u043e\u0433\u043e\u043c \u043e\u043f\u0440\u0435\u0434\u0435\u043b\u0435\u043d\u0438\u0438 \u044d\u0442\u043e\u0433\u043e \u043f\u043e\u043d\u044f\u0442\u0438\u044f - \u0432\u043e\u0442 \u043e\u043f\u0440\u0435\u0434\u0435\u043b\u0435\u043d\u0438\u0435 \u0434\u043b\u044f \u043c\u043d\u043e\u0433\u043e\u0443\u0433\u043e\u043b\u044c\u043d\u0438\u043a\u043e\u0432:\r\n\r\n\u041d\u0435\u0441\u0430\u043c\u043e\u043f\u0435\u0440\u0435\u0441\u0435\u043a\u0430\u044e\u0449\u0438\u0439\u0441\u044f \u043c\u043d\u043e\u0433\u043e\u0443\u0433\u043e\u043b\u044c\u043d\u0438\u043a \u0440\u0430\u0437\u0434\u0435\u043b\u044f\u0435\u0442 \u043f\u043b\u043e\u0441\u043a\u043e\u0441\u0442\u044c \u043d\u0430 \u0434\u0432\u0435 \u0447\u0430\u0441\u0442\u0438 (\u0442\u0430\u043a\u0438\u0435, \u0447\u0442\u043e \u043b\u044e\u0431\u043e\u0439 \u043e\u0442\u0440\u0435\u0437\u043e\u043a, \u0441\u043e\u0435\u0434\u0438\u043d\u044f\u044e\u0449\u0438\u0439 \u0442\u043e\u0447\u043a\u0443 \u0432 \u043e\u0434\u043d\u043e\u0439 \u0447\u0430\u0441\u0442\u0438 \u0441 \u0442\u043e\u0447\u043a\u043e\u0439 \u0432 \u0434\u0440\u0443\u0433\u043e\u0439 \u0447\u0430\u0441\u0442\u0438 \u043e\u0431\u044f\u0437\u0430\u0442\u0435\u043b\u044c\u043d\u043e \u043f\u0435\u0440\u0435\u0441\u0435\u043a\u0430\u0435\u0442 \u043d\u0435\u043a\u043e\u0442\u043e\u0440\u0443\u044e \u0441\u0442\u043e\u0440\u043e\u043d\u0443 \u043c\u043d\u043e\u0433\u043e\u0443\u0433\u043e\u043b\u044c\u043d\u0438\u043a\u0430). \u0421\u0440\u0435\u0434\u0438 \u044d\u0442\u0438\u0445 \u0434\u0432\u0443\u0445 \u0447\u0430\u0441\u0442\u0435\u0439 \u043e\u0434\u043d\u0430 - \u043e\u0433\u0440\u0430\u043d\u0438\u0447\u0435\u043d\u043d\u0430\u044f, \u0434\u0440\u0443\u0433\u0430\u044f - \u043d\u0435\u043e\u0433\u0440\u0430\u043d\u0438\u0447\u0435\u043d\u043d\u0430\u044f. \u041e\u0433\u0440\u0430\u043d\u0438\u0447\u0435\u043d\u043d\u0430\u044f \u0447\u0430\u0441\u0442\u044c \u043d\u0430\u0437\u044b\u0432\u0430\u0435\u0442\u0441\u044f [i]\u0432\u043d\u0443\u0442\u0440\u0435\u043d\u043d\u043e\u0441\u0442\u044c\u044e[/i] \u043c\u043d\u043e\u0433\u043e\u0443\u0433\u043e\u043b\u044c\u043d\u0438\u043a\u0430. \u0412\u043d\u0443\u0442\u0440\u0435\u043d\u043d\u044f\u044f \u0442\u043e\u0447\u043a\u0430 \u043c\u043d\u043e\u0433\u043e\u0443\u0433\u043e\u043b\u044c\u043d\u0438\u043a\u0430 - \u044d\u0442\u043e \u0442\u043e\u0447\u043a\u0430, \u043b\u0435\u0436\u0430\u0449\u0430\u044f \u0432\u043e \u0432\u043d\u0443\u0442\u0440\u0435\u043d\u043d\u043e\u0441\u0442\u0438 \u043c\u043d\u043e\u0433\u043e\u0443\u0433\u043e\u043b\u044c\u043d\u0438\u043a\u0430.\r\n\r\n[quote=\"Mamat\"] http://www.mathlinks.ro/Forum/viewtopic.php?t=60810 [/quote]\r\n\r\n\u0410 \u0442\u0443\u0442 \u0440\u0435\u0447\u044c \u0438\u0434\u0451\u0442 \u043f\u0440\u043e\u0441\u0442\u043e \u043e \u0432\u043d\u0443\u0442\u0440\u0435\u043d\u043d\u0435\u0439 \u0442\u043e\u0447\u043a\u0435 \u043e\u0442\u0440\u0435\u0437\u043a\u0430...\r\n\r\n Darij" } { "Tag": [ "induction", "superior algebra", "superior algebra unsolved" ], "Problem": "The following problem is very famous and easy to solve. But sth is not clear for me.So I decided to discuss about it.\r\n\r\n:Let $a, b$ be elements of a group such that $a^{-1}ba=b^i$. Prove that $a^{-n}ba^n=b^{i^n}$for all $n \\in N \\ and\\ i\\ge 0$.\r\nIn all my sources the condition on $i$ is stated, whereas in my opinion this is unnecessary. What do you think?", "Solution_1": "I prove the theorem with induction.\r\nIt is obvious to see when $n=1$ is ture.\r\nAnd if we have $a^{-(n-1)}ba^{n-1}=b^{i^{n-1}}$\r\nthen we have:$a^{-n}ba^{n}=a^{-1}b^{i^{n-1}}a$\r\nbut it is easily to know:\r\n$a^{-1}b^{i^{n-1}}a=\\underbrace{b^i b^i \\cdots b^i}_{i^{n-1} times}=b^{i^n}$\r\n\r\nFor your problem:\r\nI don't what do you mean by \"stated\",it is mean $i \\ge 0$\r\nif it is I also think it is unnecessary.\r\nPS:It is my first post in this section. :)", "Solution_2": "[quote=\"The Captain\"]The following problem is very famous and easy to solve. But sth is not clear for me.So I decided to discuss about it.\n\n:Let $a, b$ be elements of a group such that $a^{-1}ba=b^i$. Prove that $a^{-n}ba^n=b^{i^n}$for all $n \\in N \\ and\\ i\\ge 0$.\nIn all my sources the condition on $i$ is stated, whereas in my opinion this is unnecessary. What do you think?[/quote]\r\n\r\nI can prove it for $i=-1$. For $i=-1$, the condition of the problem becomes $a^{-1}ba=b^{-1}$. \r\n\r\nWe can invert both sides of the original premise to give $a^{-1}b^{-1}a=b$. We'll refer to this as the derived premise.\r\n\r\nWe proceed by induction. The result is trivially true for $n=1$. We assume the result is true for $n=k$ and consider the $n=k+1$ case. \r\n\r\nWe consider two subcases. If $k$ is even then:\r\n\r\n$a^{-k-1}ba^{k+1}=a^{-1}b^{i^{k}}a=a^{-1}ba=b^{-1}$ \r\n\r\nusing the original premise. \r\n\r\nAnd if $k$ is odd then \r\n\r\n$a^{-k-1}ba^{k+1}=a^{-1}b^{i^{k}}a=a^{-1}b^{-1}a=b$ \r\n\r\nusing the derived premise. \r\n\r\nSo in either case the result is true for $n=k+1$ so by mathematical induction it is true for all positive integers $n$.\r\n\r\nI suspect a similar argument works for $i<-1$ but I haven't worked it out yet." } { "Tag": [ "advanced fields", "advanced fields unsolved" ], "Problem": "How is it possible that the infinite series:\r\n\r\n1/(x+3) + 1/[(x+3)(x+4)] + 2/[(x+3)(x+4)(x+5)] + 6/[(x+3)(x+4)(x+5)(x+6) + ....\r\n\r\nis simply 1/(x+2)? \r\n\r\nHowever, I am applying the methods in numerical analysis. \r\n\r\nIt should be the summation (from n = 0 to +oo) n! / [(x+3)(x+4)...(x+3 + n)]\r\n\r\nFrom here, I am clueless.\r\n[/list]", "Solution_1": "We are looking at $f(y)=\\sum\\limits_{n=0}^{\\infty}\\dfrac{n!}{y\\cdot(y+1)\\cdots(y+n)}$\r\n(I substituted $y=x+3$). By comparison to $\\sum\\dfrac1{n^y}$, this series diveges for $x\\le1$ and converges for $y>1$.\r\n\r\nWe have ${\\dfrac{n!}{y\\cdot(y+1)\\cdots(y+n)}=\\dfrac{(n-1)!}{y\\cdot(y+1)\\cdots(y+n-1)}- \\dfrac{(n-1)!}{(y+1)\\cdot(y+2)\\cdots(y+n)}}$ for $n\\ge1$, so $f(y)=\\dfrac1y+f(y)-f(y+1)$, and $f(y+1)=\\dfrac1y$, at least for $y>1$ where everything converges absolutely and this rearrangement works. This result should hold in general, but we need a slightly different argument to get at $f(y)$ for $1 sqrt(2) on a number line.[/b]\r\n\r\nMy teacher said that the answer is:\r\n\r\n[b]<======!===O[/b]--I-------|-------I--[b]O===!=======>[/b]\r\n -2 -1 0 1 2\r\n\r\nbut I think that the answer is:\r\n\r\n<-------!---[b]O==I======|======I======!=======>[/b]\r\n -2 -1 0 1 2\r\n\r\nbecause x can be any number greater than -1.4142 ...\r\n\r\n(=====) means shaded in,\r\n(------) means unshaded,\r\n( O ) means greater than, but not including\r\n( ! ) means 2 and -2\r\n( I ) means 1 and -1\r\n( | ) means 0.\r\n\r\n\r\nThanks,\r\nKevin K.", "Solution_1": "Wait, is the problem asking you to graph $x > \\sqrt2$ or $x >-\\sqrt2$? Either way, your teacher is wrong. If the first is the problem, you're both wrong. It should be all points to the right of $\\sqrt2$. If it's the second, then you're right.\r\n\r\nYour teacher graphed $|x| > \\sqrt2$.", "Solution_2": "I meant to graph x > sqrt(2) for both the negative and positive values, -1.414 and 1.414.\r\n\r\nAlso, I thought a radical sign without anything meant both roots, positive and negative. One with a negative in front would denote the negative square root.", "Solution_3": "the radical sign usually denotes the positive.\r\ni see why you think that your way does both the positive and negative values, but think about it this way: 0 is on your graph, but $0<\\sqrt{2}$. Anything bigger than $\\sqrt{2}$ is also bigger than the negative value. also, your teacher is wrong unless he/she is talking about $|x|$", "Solution_4": "the problem was probably supposed to be: graph x^2>2" } { "Tag": [ "geometry", "perimeter", "similar triangles" ], "Problem": "I need help on several 2000 National Sprint questions...\r\n\r\n1. In triangle ABC, angle C is a right angle. Point M is the midpoint of segment AB, point N is the midpoint of segment AC, and point O is the midpoint of segment AM. The perimeter of triangle ABC is 112 cm, and ON = 12.5 cm. What is the number of square centimeters in the area of quadrilateral MNCB?\r\n[hide]I've gotten to the point where I've figured out it's a 7-24-25 triangle, so the side lengths are 14, 48, and 50. But I still don't know how to find the quadrilateral area. :( [/hide]\r\n\r\n2. Alyssa rode her bike for 7 hours. She started on level ground and rode at a rate of 8 mph. She came to a hill that slowed her down to a rate of 5 mph. Upon reaching the top of the hill, she turned around and descended at a rate of 20 mph. Finally, she returned home on the flat portion at 8 mph. What is the total number of miles she rode?\r\n\r\n3. For positive integers a, b and c, what is the value of the product abc?\r\n$ \\frac{1}{a\\plus{}\\frac{1}{b\\plus{}\\frac{1}{c}}} \\equal{} \\frac{3}{8}$", "Solution_1": "2: Hint. On the hill find the average speed.\r\n[hide]It's 8mph as well. Therefore her average speed is 8mph, *7hr=56 miles[/hide]\r\n\r\n3: continued fractions.\r\n\r\na+(1/(b+1/c))=8/3, so a=2\r\n1/(b+1/c)=2/3\r\nb+1/c=3/2 so b=1, c=1", "Solution_2": "[hide=\"Hint for 1\"]What kind of a quadrilateral is MNCB?[/hide]\n[hide=\"Bigger hint for 1\"]Show via similar triangles that MNCB is a trapezoid.[/hide]\r\n[quote]a+(1/(b+1/c))=8/3, so a=2 \n1/(b+1/c)=2/3 \nb+1/c=3/2 so b=1, c=1[/quote]\r\nYou mean c=2, and the product is $ 2\\cdot 1\\cdot 2\\cdot\\equal{}\\boxed{4}$, right? :wink:" } { "Tag": [], "Problem": "Hi all of my dears in spain community, :) \r\n\r\nsorry for interferring , but i had to ask this question. :wink: \r\n\r\ncould any body plz show me , how to buy all of the salvatore dali's\r\n\r\npictures?\r\n\r\nor even show me any links about it?\r\n\r\nthanks alot my dears.", "Solution_1": "This will maybe help you: http://ar.search.yahoo.com/search/images?p=salvador+dal%ED&sm=Buscar&fr=FP-tab-img-t&toggle=1&ei=ISO-8859-1&meta=all%3D1", "Solution_2": "oh tanks dear jose,\r\n\r\ni searched alot ,but\r\n\r\ninfact i want to buy a book from salvatore dali, if it exists.\r\n\r\ndeos in spain exist any book from him?\r\n\r\nor if it is , how can i buy? :wink:" } { "Tag": [ "\\/closed" ], "Problem": "I found something about a calendar and setting options for it in my profile. I haven't seen anything like a calendar on AoPS, so could anyone tell me what it is?", "Solution_1": "The settings reqard the dating system on the board. Maybe Calendar wasn't the best naming but that's about all there is to it." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Prove that\r\n$\\sum_{A,B,C}\\sqrt{2Sin^{2}A+2Sin^{2}B-Sin^{2}C}\\geq 2 \\sum_{A,B,C}Sin^{2}A$ \r\nWhere $A,B,C$ are angles of triangle $ABC$", "Solution_1": "Any trouble? Your ineq is not homogenized.", "Solution_2": "The inequality is equivalent to\r\n$\\sum_{a,b,c}m_{a}\\geq \\frac{\\sum_{a,b,c}a^{2}}{2R}$\r\nwhere $m_{a}, m_{b}, m_{c}$ are the lengths of the medians of triangle and $R$ is the circumradius.($a,b,c$ are the sides) :idea: \r\n\r\nor\r\n\r\n$\\sum_{a,b,c}\\sqrt{2a^{2}+2b^{2}-c^{2}}\\geq 2 \\frac{\\sum_{a,b,c}a^{2}}{abc}\\sqrt{s(s-a)(s-b)(s-c)}$ :idea: \r\n\r\nNow substituting,\r\n\r\n$a= x+y$,$b = y+z$ ,$c = z+x$,\r\nwe get a homogenous inequality in $x,y and z$\r\n\r\n$Sin A$ etc are dimensionless and do not have any relationship with homogeneity of any inequality :wink: ....." } { "Tag": [ "probability", "percent" ], "Problem": "I have two problems I can't solve!!! D:\r\n\r\nThe 1st term of a sequence is 2. Each term after the 1st is obtained by multiplying the preceding term by 5 and then subtracting 6. What is the sum of the first five terms of the sequence?\r\nHere I got 398... It should be right... but the book says 458...\r\n\r\nRonnie, Ben, and Jevon are in the same math class. There are 24 students in the class, and the teacher has announced that she will be dividing the class into groups of 3. What is the probability that Ronnie, Ben, and Jevon will be in the same group? Express as a common fraction.\r\n\r\nI'm still a little shaky on THESE types of probablilty questions, so I don't know where to start.", "Solution_1": "[hide=\"Number One\"]\nLet's list out the terms.\n\n1st Term: 1\n2nd Term: $ (1 \\cdot 5) \\minus{} 6 \\equal{} \\minus{}1$\n3rd Term: $ (\\minus{}1 \\cdot 5) \\minus{} 6 \\equal{} \\minus{}11$\n4th Term: $ (\\minus{}11 \\cdot 5) \\minus{} 6 \\equal{} \\minus{}61$\n5th Term: $ (\\minus{}61 \\cdot 5) \\minus{} 6 \\equal{} \\minus{}311$\n\n$ 1 \\minus{} 1 \\minus{} 11 \\minus{} 61 \\minus{} 311 \\equal{} \\boxed{\\minus{}383}$\n\nHmm, I'm getting an answer completely different from both your and the book's answer.\n[/hide]\n\n[hide=\"Number Two\"]\nThere are $ 24 \\div 3 \\equal{} 8$ total groups, so there are $ 8 \\times 3! \\equal{} 48$ total ways to divide the class into groups of 3. There's only 1 way for Ronnie, Ben, and Jevon to be in the same group, so our answer is $ \\boxed{\\frac{1}{48}}$. \n[/hide]", "Solution_2": "Nooo... wait a minute...\r\n\r\n1) The first term of the sequence is two.\r\n\r\n2) Aren't there three ways for Ronnie, Ben, and Jevon to be in the same group? They could be in the first group together, the second group together, and the third group together... right? :huh:", "Solution_3": "1.\r\n1st-2\r\n2nd-2*5-6=4\r\n3rd-4*5-6=14\r\n4th-14*5-6=64\r\n5th-64*5-6=314\r\n314+64+14+4+2=398\r\ni believe you are right.\r\n#2 Suppose your were those three were picked right away then that would be 3/24x2/23x1/222=1/2024\r\nSuppose that they weren't picked first and picked second\r\n21/24x20/23x19/22x3/21x2/20x1/19=1/2024\r\nso order they are picked in they have a 1/2024 chance. There are eight different places the group can be picked 1st, 2nd, 3rd, 4th, 5ht, 6th, 7th or 8th. So 1/2024x8=1/253.", "Solution_4": "But for problem number two, there are only three groups. How did you get eight? :huh: \r\n\r\n\r\n\r\nOh yeah. There is another problem I don't get... :blush:\r\n\r\nKeith and Chi-An are playing a game. Keith chooses a poistive integer less than 100. Chi-An then tries to guess Keith's number by asking questions to which Keith answers yes or no. What is the number of questions Chi-An must ask so that, even if she is unlucky, she will always be able to identify the number?", "Solution_5": "eh this is like common sense\r\n\r\nso keith chooses a number among 1~99.\r\n\r\nif the number was 99\r\n\r\nchi-an can ask 1, 2, 3, ..., 98, which is asking 98 times.\r\n\r\nanswer : 98", "Solution_6": "Nah... She could ask, \"is the number greater than x\", etc. So, rounding up, you start with 50 possible, than 25, 13, 7, 4, 2, 1. Thus, you have to ask at the very most $ 7$ questions.\r\n\r\nFor number 2, one of the three has to first be in that group to \"pick out\" that group. Then there are 23 people left, so there's a $ \\frac{2}{23}$ chance of one of them being in that group. Finally, there are 22 people left, so there's a $ \\frac{1}{22}$ chance of the group containing Robbie, Ben, and Jevon. Multiplying gives the answer $ \\frac{1}{253}$. \r\n\r\nWait, I think I messed up somewhere...", "Solution_7": "Well, on number three it can't be seven, because If it is greater than fifty, she'll ask six questions...", "Solution_8": "Yes, we are [b]always[/b] looking at the worse-case scenario to find out the number of times that we must do something to assure that it will work every time, not just when you're lucky. All of the numbers thus must be taken into account. \r\n\r\nLol nick :D.\r\n\r\nCan anyone solve #2 with a solid solution?", "Solution_9": "blah i suck at counting but\r\n\r\nthere are total $ \\binom{24}{3}\\cdot\\binom{21}{3}\\cdots\\binom{6}{3}\\cdot\\binom{3}{3}$ ways to divide 24 students to 8 groups of 3.\r\n\r\nand if ronnie, ben and jevon are in the same group, there would be total $ \\binom{21}{3}\\cdot\\binom{18}{3}\\cdots\\binom{6}{3}\\cdot\\binom{3}{3}$\r\n\r\nso the probability would be $ \\frac{\\binom{21}{3}\\cdot\\binom{18}{3}\\cdots\\binom{6}{3}\\cdot\\binom{3}{3}}{\\binom{24}{3}\\cdot\\binom{21}{3}\\cdots\\binom{6}{3}\\cdot\\binom{3}{3}}\\equal{}\\frac{1}{\\binom{24}{3}}$\r\n\r\nhmm i got $ \\frac{1}{2024}$.", "Solution_10": "uh, that's my answer from before.", "Solution_11": "[quote=\"shentang\"]#2 Suppose your were those three were picked right away then that would be 3/24x2/23x1/222=1/2024\nSuppose that they weren't picked first and picked second\n21/24x20/23x19/22x3/21x2/20x1/19=1/2024\nso order they are picked in they have a 1/2024 chance. There are eight different places the group can be picked 1st, 2nd, 3rd, 4th, 5ht, 6th, 7th or 8th. So 1/2024x8=1/253.[/quote]\r\n\r\nblah, didnt see that\r\n...you said the answer was 1/253.\r\nbesides, i solved it differently anyways", "Solution_12": "Wow. That makes sense! :)\r\n\r\nThe [i]transient Triangle[/i] is a region with three cities: A-ville, B-burg andC-town. The ppopulation of each of these cities was the same at the end of 1996 as it was at the end of 1995, but the residents had moved between cities. What percent of the residents of the Transient Triangle were residents of A-ville at the end of 1996 if:\r\n\r\n- 10% of the 1995 residents of A-ville moved to B-burg in 1996;\r\n- 10% of the 1995 residents of A-ville moved to C-town in 1996;\r\n- 20% of the 1995 residents of B-burg moved to A-ville in 1996;\r\n- 10% of the 1995 residents of B-burg moved to C-town in 1996;\r\n- 10% of the 1995 residents of C-town moved to A-ville in 1996; and\r\n- 10% of the 1995 residents of C-town moved to B-burg in 1996?", "Solution_13": "[quote=\"FantasyLover\"][quote=\"shentang\"]#2 Suppose your were those three were picked right away then that would be 3/24x2/23x1/222=1/2024\nSuppose that they weren't picked first and picked second\n21/24x20/23x19/22x3/21x2/20x1/19=1/2024\nso order they are picked in they have a 1/2024 chance. There are eight different places the group can be picked 1st, 2nd, 3rd, 4th, 5ht, 6th, 7th or 8th. So 1/2024x8=1/253.[/quote]\n\nblah, didnt see that\n...you said the answer was 1/253.\nbesides, i solved it differently anyways[/quote]\r\n\r\nOh, you just missed that the trio could be picked 1st 2nd 3rd....8th you assumed they were picked 1st. so 1/2024*8=1/253", "Solution_14": "actually, it doesn't matter.\r\n\r\nyou have to pick 3 out of 21, out of 18, etc. because you must not pick one of 3 guys.\r\nso eventually it becomes JUST 21C3*18C3*...*3C3.", "Solution_15": "Successful: pick a group, order the three people. 8*3! = 48\r\nTotal: place the three people in 24 slots. 24P3 = 24*23*22 = 48*23*11\r\n48/24P3 = 1/253\r\n\r\nI thought of previous solutions, but here's a different one for variety.", "Solution_16": "Sooo... is\r\n\r\n[u]2+SQRT(2)[/u]\r\n2+2SQRT(2)\r\n\r\nin simplest radical form as a common fraction?\r\n\r\n(Irrelevant question)", "Solution_17": "No, because you can't have radicals in the denominator. Multiply both the numerator and the denominator by the conjugate(in this case, it's $ 2 \\minus{} 2\\sqrt {2}$), and the final answer would be $ \\frac {\\sqrt {2}}{2}$.", "Solution_18": "This is called rationalizing the denominator.", "Solution_19": "Alternatively, in this case, you can do something sneaky with this special form of fraction. If you have $ \\frac {n \\plus{} \\sqrt n}{n \\plus{} n\\sqrt n}$, factor our $ \\sqrt n$ from the numerator and $ n$ from the denominator, we get \\[ \\frac {\\sqrt n(\\sqrt n \\plus{} 1)}{n(1 \\plus{} \\sqrt n)} \\equal{} \\frac {\\sqrt n}n \\left(\\frac {1 \\plus{} \\sqrt n}{1 \\plus{} \\sqrt n}\\right) \\equal{} \\frac {\\sqrt n}n\\]" } { "Tag": [ "modular arithmetic", "combinatorics proposed", "combinatorics" ], "Problem": "$ (a)$ Prove that $ \\mathbb{N}$ can be partitioned into three (mutually disjoint) sets such that, if $ m,n \\in \\mathbb{N}$ and $ |m\\minus{}n|$ is $ 2$ or $ 5$, then $ m$ and $ n$ are in different sets. \r\n$ (b)$ Prove that $ \\mathbb{N}$ can be partitioned into four sets such that, if $ m,n \\in \\mathbb{N}$ and $ |m\\minus{}n|$ is $ 2,3,$ or $ 5$, then $ m$ and $ n$ are in different sets. Show, however, that $ \\mathbb{N}$ cannot be partitioned into three sets with this property.", "Solution_1": "[quote=\"moldovan\"]$ (a)$ Prove that $ \\mathbb{N}$ can be partitioned into three (mutually disjoint) sets such that, if $ m,n \\in \\mathbb{N}$ and $ |m - n|$ is $ 2$ or $ 5$, then $ m$ and $ n$ are in different sets. \n$ (b)$ Prove that $ \\mathbb{N}$ can be partitioned into four sets such that, if $ m,n \\in \\mathbb{N}$ and $ |m - n|$ is $ 2,3,$ or $ 5$, then $ m$ and $ n$ are in different sets. Show, however, that $ \\mathbb{N}$ cannot be partitioned into three sets with this property.[/quote]\r\n\r\n[b](a)[/b] $ n \\equiv 1 \\pmod{3} \\Rightarrow n \\in A_1$\r\n$ n \\equiv 2 \\pmod{3} \\Rightarrow n \\in A_2$\r\n$ n \\equiv 0 \\pmod{3} \\Rightarrow n \\in A_3$\r\n\r\nNow clearly $ A_1,A_2,A_3$ are partions of $ \\mathbb{N}$ satisfying the condition as difference of any two numbers in the same set must be divisible by $ 3$.\r\n\r\n[b](b)[/b] Similarly to (a) we take $ n \\equiv 1 \\pmod{4} \\Rightarrow n \\in B_1$\r\n$ n \\equiv 2 \\pmod{4} \\Rightarrow n \\in B_2$\r\n$ n \\equiv 3 \\pmod{4} \\Rightarrow n \\in B_3$\r\n$ n \\equiv 0 \\pmod{4} \\Rightarrow n \\in B_4$\r\n\r\nand $ B_1,B_2,B_3,B_4$ are partitions of $ \\mathbb{N}$ satisfies the condition.\r\n\r\n\r\nFor second part of (b), assume that the $ 3$ partions $ C_1,C_2,C_3$ exist and let $ 1 \\in C_1$, then $ 3 \\nin$ subset containing $ 1 \\Rightarrow 3 \\in C_2$ WLOG.\r\n\r\nSimilarly $ 6 \\notin$ subsets containing $ 1,3 \\Rightarrow 6 \\in C_3$\r\n\r\n$ 4 \\notin$ subsets containing $ 1,6 \\Rightarrow 4 \\in C_2$\r\n\r\n$ 8 \\notin$ subsets containing $ 3,6 \\Rightarrow 8 \\in C_1$\r\n\r\n$ 5 \\notin$ subsets containing $ 3,8 \\Rightarrow 5 \\in C_3$\r\n\r\n$ 7 \\notin$ subsets containing $ 4,5 \\Rightarrow 7 \\in C_1$\r\n\r\n$ 2 \\notin$ subsets containing $ 4,5,7 \\Rightarrow 2 \\notin C_1,C_2,C_3$ , contradiction, done." } { "Tag": [ "calculus", "derivative", "real analysis", "real analysis unsolved" ], "Problem": "Consider f(x) = 4/(1+x^2)\r\n\r\nBy finding a bound on f(4) show that \r\n\r\n| pi - SIMPn | <= 8/15n^4.", "Solution_1": "[quote=\"numero1\"]Consider f(x) = 4/(1+x^2)\n\nBy finding a bound on f(4) show that \n\n| pi - SIMPn | <= 8/15n^4.[/quote]\r\nDo you mean a bound on $f(4)$ or on $f^{(4)}$, the fourth derivative? That would be the right kind of bound for this problem.\r\n\r\nBtw, this is a standard homework problem.", "Solution_2": "Yes the fourth derivative.\r\nI am having difficulty taking the fourth derivative..LOL.\r\nwhen I use the quotoent rule the denominator just blowa up too big." } { "Tag": [ "geometry", "rhombus", "3D geometry", "AMC 10" ], "Problem": "The AMC 10 has 25 questions. If there are an equal number of correct answers being (A), (B), (C), (D), and (E), what is the number of questions will I get correct that if I select (C) for each question?\r\n\r\n$ \\textbf{(A)} 4\\qquad\\qquad\\qquad\\textbf{(B)} 5\\qquad\\qquad\\qquad\\textbf{(C)} 6\\qquad\\qquad\\qquad\\textbf{(D)} 7\\qquad\\qquad\\qquad\\textbf{(E)} 8$\r\n\r\nA rhombus is given with diagonals of which one is double the length of the other. If the length of the longer diagonal is x, what is the area of the rhombus?\r\n\r\n$ \\textbf{(A)} {x^2}\\over {2} \\qquad\\qquad\\qquad\\textbf{(B)} x^2\\qquad\\qquad\\qquad\\textbf{(C)} x^2-4\\qquad\\qquad\\qquad\\textbf{(D)} {x^2}\\over 4\\qquad\\qquad\\qquad\\textbf{(E)} x$\r\n\r\n\r\nJoe uses a ridiculous phone provider which charges an amount x cents per minute for which he talks in a month. The phone company also has a flat rate for which it charges all customers regardless of how much they talk. In January, the phone company charges Joe $ \\$$14.20. The following month, Joe does a lot of business conferences and talks on the phone for four times as long as he did in January. His monthly bill is $ \\$$37.30. What is the fixed rate that the phone company charges to the nearest cent?\r\n\r\n$ \\textbf{(A)} \\$3.55\\qquad\\qquad\\qquad\\textbf{(B)} \\$6.50\\qquad\\qquad\\textbf{(C)} \\$9.33\\qquad\\qquad\\qquad\\textbf{(D)} \\$5.78\\qquad\\qquad\\textbf{(E)} \\$23.10$\r\n\r\n\r\nWhat is the maximum possible number of intersection points between two circles and two lines?\r\n\r\n$ \\textbf{(A)} 8\\qquad\\qquad\\qquad\\textbf{(B)} 9\\qquad\\qquad\\qquad\\textbf{(C)} 11\\qquad\\qquad\\qquad\\textbf{(D)} 12\\qquad\\qquad\\qquad\\textbf{(E)} 15$\r\n\r\n\r\n\\item In a standard cube, an ant is set at a corner. What is the maximum possible lengths an ant can walk if it cannot visit any point twice or overlap any of the lengths, if the ant must end up in the point it started at?\r\n\r\n$ \\textbf{(A)} 5\\qquad\\qquad\\qquad\\textbf{(B)}\r\n6\\qquad\\qquad\\qquad\\textbf{(C)} \r\n7\\qquad\\qquad\\qquad\\textbf{(D)} \r\n8\\qquad\\qquad\\qquad\\textbf{(E)} \r\n9$", "Solution_1": "[hide=\"Problem 1\"]We know that there are 25 questions. When we select (C) for each question, and there are an equal number of (A)'s,(B)'s,(C)'s,(D)'s, and (E)'s, there are 5 (C)'s, which implies that the answer is $ \\boxed{\\textbf{(C)}}$.[/hide]\n\n[hide=\"Problem 2\"]If the length of the longer diagonal is $ x$, then the length of the shorter diagonal is $ x/2$. The area of a rhombus is $ \\dfrac{d_1d_2}{2}$. Applying this gets us $ x^2/4\\implies \\boxed{\\textbf{D}}$.[/hide]\n\n[hide=\"Problem 3\"]We translate these statements into equations. It now becomes quite clear that the correct answer is $ \\boxed{\\textbf{(B)}}$.[/hide]\n\n[hide=\"Problem 4\"]We can simply draw this out and verify that the correct answer is $ \\boxed{\\textbf{(C)}}$. For example, we can see such a situation in the below image:\n\n[img]http://myfiles.zapto.org/AMC10Images/CirclesandLine.jpg[/img]\n[/hide]\n\n[hide=\"Problem 5\"]One method of solving this problem is drawing a cube, starting the ant at the foremost top left hand corner, travel around the top of the cube to the point at the foremost top right hand corner, then travelling downwards, making a similar travelling, and then heading back to the point it started at. The answer is $ \\boxed{\\textbf{(D)}}$.[/hide]", "Solution_2": "I think you have a typo for problem 2.", "Solution_3": "Wasn't there also a typo for problem 1 or its answer?\r\n[quote]The AMC 10 has 25 questions. If there are an equal number of correct answers being (A), (B), (C), (D), and (E), what is the number of questions will I get correct that if I select (C) for each question?\n(A)4 (B)5 (C)6 (D)7 (E)8[/quote]\n[quote]...there are 5 (C)'s, which implies that the answer is (C). \n[/quote]\r\nAnyway, those were enjoyable problems with or without the typos.", "Solution_4": "Oh yeah answer is (B)", "Solution_5": "yea i liked the telephone problem.. Solution:\r\nlet m be the minutes used in January, and f be the fixed rate:\r\n$ xm + f = 14.20$\r\n $ 4xm + f = 37.30$ \r\nmultiplying the first equation by 4 and subtracting we get $ 3f = 19.50 \\implies f = 6.50 \\implies \\boxed {B}$\r\n\r\nhowever, these were more difficult than the average 1-5 on the AMC 10" } { "Tag": [ "calculus", "calculus computations" ], "Problem": "you have a stick of length $x$, and a paper with parallel lines $y$ distance apart. suppose $x>y$. \r\nwhen you throw the stick on the paper $n$ times, you find, out of those $n$ times, that the stick crosses \r\n[u]at least[/u] one of those parallel lines, $m$ times.\r\n\r\nshow:\r\n\r\n $\\frac{n}{m}\\left[ \\Omega\\;+\\;\\sec^{-1}{\\Omega}\\;-\\sqrt{\\Omega^{2}-1}\\;\\;\\right] \\;=\\;\\boxed{\\frac{\\pi}{2}}\\;\\;\\;\\quad\\;\\;\\;\\quad where\\;\\;\\;\\;\\;\\Omega\\;=\\;\\frac{x}{y}$", "Solution_1": "This is called Buffons's Needle Problem.\r\n\r\nMore is to read on this on say [url=http://mathworld.wolfram.com/BuffonsNeedleProblem.html]mathworld[/url] and http://en.wikipedia.org/wiki/Buffon's_needle\r\n\r\n(sorry for the wiki link, couldn't fix it )" } { "Tag": [ "geometry", "3D geometry", "prism", "MATHCOUNTS" ], "Problem": "i am looking for a way to do te volume of a rectangular prisim knowin only side areas. i know how to work the problems, but am looking for a less trial and error method that would work with large numbers and no calculator\r\n\r\nheres what i have so far\r\n\r\nfactor all three numbers \r\nEX. side areas = 70, 30, 21.\r\n 70=2*5*7\r\n 30=2*3*5\r\n 21=3*7\r\nnotice 70 and 21 both have a seven in common, so the edge they meet at=7\r\ncontinue this with the other numbers and you can ultimantly find the volume\r\nBUT, ive noticed that doesnt always work\r\n\r\ni was wondering if any aopsers could think up some cool methods to solve these kinds of problems without guess and check.\r\n\r\nthanks", "Solution_1": "here you go.\r\n\r\nIf the dimensions are x, y, and z then your face areas would be xy, yz, and xz. Those would be equal to the 70, 30, 21.\r\n\r\nIf you take the product of the face areas both with the variables and the numbers you get $x^{2}y^{2}z^{2}= 70\\cdot30\\cdot21$\r\n\r\nIf you take the square root of both sides you get \r\n\r\n$\\sqrt{x^{2}y^{2}z^{2}}= \\sqrt{70\\cdot30\\cdot21}$ = \r\n$xyz = \\sqrt{70\\cdot30\\cdot21}$\r\n$xyz = \\sqrt{7\\cdot10\\cdot10\\cdot3\\cdot3\\cdot7}$\r\n$xyz = \\sqrt{7^{2}\\cdot10^{2}\\cdot3^{2}}$\r\n$xyz = 7\\cdot10\\cdot3$\r\n$xyz = 210$\r\n\r\nand since $xyz$ is the volume of the prism, there you go. You could probably simplify the radical faster than I did, I just wanted to show all the steps out for you.", "Solution_2": "so \r\n volume= sqrt(area1*area2*area3)", "Solution_3": "The side lengths are $a$, $b$, and $c$.\r\n\r\nYou are given $ab=70$, $bc=30$, and $ac=21$, and you know $V=abc$.\r\n\r\n$V=abc=\\sqrt{(abc)^{2}}=\\sqrt{ab\\cdot bc\\cdot ac}$\r\n\r\nUsing this formula, you can find the volume.", "Solution_4": "Wow, do those two solutions look identical... I wonder why...\r\n(*cough*i_like_pie*cough*)", "Solution_5": "They do look similar don't they...", "Solution_6": "Yeah, but it's still worth noting that this solution works in any case, not just the surface area/volume ones...\r\n\r\nBut the s. area and volume is by far the most common in mathcounts" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Noah has 8 species of animals to fit into 4 cages of the ark. He plans to put the species in each cage.\r\n\r\nIt turns out that, for each spieces, there are at most 3 other species with which it cannot share the accomodation.\r\n\r\nProve that there is a way to assign the animals to their cages so that each species shares with compatible species.", "Solution_1": "total possible pairs = 8C2=28\r\nmaximum incompatible pairs has to be less than 24 (8*3,assuming incompatibility relation is symmetric)\r\nThus compatible pairs has to be greater than 4", "Solution_2": "Consider the graph whose vertices are the species and the edges join any two species which cannot share accomodation.\r\nThus, each vertex has degree at most $ 3$.\r\nIt is a basic result to prove that any simple graph with maximal degree $ \\Delta$ can be properly colored with at most $ \\Delta \\plus{}1$ colors.\r\nHere, one color is one cage.\r\n\r\nPierre.", "Solution_3": "In the \"sledgehammer\" category:\r\n\r\nConsider the graph whose vertices are the species, with edges joining species that can share accommodation (the complement of pbornsztein's graph). It has 8 vertices and minimum degree 4, so by Dirac's theorem it has a Hamiltonian circuit. But the 8-cycle contains two natural perfect matchings, so we're done." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Does there exist 10 integers $ a_1,a_2,\\cdots,a_{10}$ satisfying $ a_i\\ne a_j$ for $ i\\ne j$ and $ S\\minus{}a_i$ is perfect square for every $ i\\equal{}1,\\cdots,10$ with $ S\\equal{}\\sum_{i\\equal{}1}^{10} a_i$", "Solution_1": "Let $ b_1,b_2\\dots,b_{10}$ be integers such that $ \\sum_{i\\equal{}1}^{10}b_i^2\\equal{}9S$, for some $ S$. Then $ a_i\\equal{}S\\minus{}b_i^2$ for $ i\\equal{}1,2,\\dots,10$ work.\r\n\r\nFor example: 376, 349, 304, 241, 160, 61, -56, -191, -344, -515 is such a sequence. Here $ a_i\\equal{}385\\minus{}9i^2$ and $ S\\minus{}a_i\\equal{}(3i)^2\\,\\forall i\\in\\{1,2,\\dots,10\\}$." } { "Tag": [ "probability", "factorial", "AoPSwiki", "MATHCOUNTS", "geometry", "trigonometry", "calculus" ], "Problem": "I have a problem . How can I improve on counting and probability problems besides Alcumus. I can never et those.", "Solution_1": "I suggest reading Introduction to Counting and Probability.", "Solution_2": "Tip: Just get used to as many different types of those problems. Counting and probability problems are generally very limited in their types so just practice getting used to them. (Similar to stoichiometry problems). For example, identify which are independent events, which are dependent, which need combinations, which are permutations, when to use factorials, when to use combinatorics, etc.", "Solution_3": "izzy how can i geet this book. How much does it cost?", "Solution_4": "[url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=202]This[/url] is Introduction to Counting and Probability. You can look at the excerpts/tests to see if it suits you. If it's too expensive, there are many places to find problems, including:\r\n[url=http://www.artofproblemsolving.com/Wiki/index.php/Category:Introductory_Combinatorics_Problems]AoPSWiki[/url]\r\n[url=http://mathcounts.saab.org/mc.cgi]Elias Saab's MC website[/url]\r\n\r\nIf you look hard enough, you can find many more. :wink: The advantage of the text is that it gives thorough examples and problems. However, it is possible to just look at the table of contents and learn the topics by yourself.", "Solution_5": "If you can find a cheaper, used copy at a local bookstore, that's even better. I know someone who has all the books, so I can borrow them. There are lots of ways to get a copy.", "Solution_6": "Not that many bookstores have AoPS books. :wink: But if you're lucky, you can borrow them from somewhere.", "Solution_7": "If you got a gift card or something, you could specially order an AoPS book and pay less than the normal price. That's what I'm doing for All-Time Greatest MathCounts Problems.", "Solution_8": "Lots of people will give you gift cards for AoPS. :wink:", "Solution_9": "[quote=\"math154\"]Not that many bookstores have AoPS books. :wink: But if you're lucky, you can borrow them from somewhere.[/quote]\r\nI believe some Barnes and Nobles have them, and Bookmans (depending on where you live)...but don't try eBay =]", "Solution_10": "Eh...My Barnes & Nobles has a single math section...like one shelf...and it's filled with...Pre-Algebra for Dummies, Algebra for Dummies, Geometry for Dummies, Algebra II for dummies, Trigonometry for dummies, Precalculus for dummies, Calculus for dummies, ...", "Solution_11": "And there's always the library", "Solution_12": "Since when could you find AoPS books in typical libraries...", "Solution_13": "jjx never specifically asked for aops books.\r\n\r\nedit: oh wait he did haha", "Solution_14": "I suggest looking at different counting problems on the web, and analyzing there solutions. That is how I learned, and I got pretty good at it.", "Solution_15": "[quote=\"math154\"]Eh...My Barnes & Nobles has a single math section...like one shelf...and it's filled with...Pre-Algebra for Dummies, Algebra for Dummies, Geometry for Dummies, Algebra II for dummies, Trigonometry for dummies, Precalculus for dummies, Calculus for dummies, ...[/quote]\r\nAs I said, it depends on where you live. Some have bigger sections. (and both of the B&Ns in my city have ______ for Dummies too =P plus more!!!)", "Solution_16": "[quote=\"math154\"]Eh...My Barnes & Nobles has a single math section...like one shelf...and it's filled with...Pre-Algebra for Dummies, Algebra for Dummies, Geometry for Dummies, Algebra II for dummies, Trigonometry for dummies, Precalculus for dummies, Calculus for dummies, ...[/quote]\r\n\r\nLol. The exact same thing happened to me, I wasn't able to find a single helpful book at my nearest Barnes and Noble book store.", "Solution_17": "While AoPS books are very good, keep in mind that they're not the only good books in the world. I've found some pretty good math books at B&N that show proofs and such.", "Solution_18": "[quote=\"isabella2296\"]While AoPS books are very good, keep in mind that they're not the only good books in the world. I've found some pretty good math books at B&N that show proofs and such.[/quote]\r\nEven the Dummies series is okay. (but it's great for beginners)", "Solution_19": "Actually, I find that Dummies books and The Complete Idiot's Guide to ______ books are pretty good if you're doing something totally new/difficult for you. Especially if there's no AoPS book on it yet, like Calculus.", "Solution_20": "i like the ____for dummies books. they're easy to understand but they cover the information. the aops books are very brief, though they give the info too.", "Solution_21": "[quote=\"isabella2296\"]Actually, I find that Dummies books and The Complete Idiot's Guide to ______ books are pretty good if you're doing something totally new/difficult for you. Especially if there's no AoPS book on it yet, like Calculus.[/quote]\r\nBut sometimes it just breezes over the concept.\r\nWhich isn't always a good thing", "Solution_22": "Really? That's never happened in my experience.", "Solution_23": "Okay, Thank you for your help everyone! :D I went to Barnes and Nobles and got some really good AOPS books. I hope they help me because currently I am horrible at counting and probabilty. I can never get them(seriously) unless I do guess and check which doesn;t work", "Solution_24": "try Counting and Probability for Smarties", "Solution_25": "[quote=\"isabella2296\"]Really? That's never happened in my experience.[/quote]\r\nIt doesn't happen that often though.", "Solution_26": "How do you guys practice geometry?", "Solution_27": "There are tons of resources on the web - a quick Google search will yield a bunch. Elias Saab's website has some pretty good geometry problems, and there's a lot of books out there too. :)", "Solution_28": "[quote=\"isabella2296\"]There are tons of resources on the web - a quick Google search will yield a bunch. Elias Saab's website has some pretty good geometry problems, and there's a lot of books out there too. :)[/quote]\r\n\r\nThanks. I'll check it out. :lol: :)", "Solution_29": "i usually just go to AoPSWiki and check out the AIME Problems and Solutions. Those are pretty helpful", "Solution_30": "[quote=\"Wickedestjr\"]How do you guys practice geometry?[/quote]\r\n\r\nAoPS geometry class.\r\n\r\njjx - how'd you find aops books at barnes and noble!?!?!?!? i could never find them at my local bookstore...", "Solution_31": "[quote=\"AIME15\"][quote=\"Wickedestjr\"]How do you guys practice geometry?[/quote]\n\nAoPS geometry class.\n\njjx - how'd you find aops books at barnes and noble!?!?!?!? i could never find them at my local bookstore...[/quote]\r\n\r\nyeah, read the rest of the thread some people just have them and some people don't. i personally have never seen them at my bookstore.", "Solution_32": "I know you can purchase them from \"Bookstore\" on the top of the sidebar.", "Solution_33": "[quote=\"AIME15\"]I know you can purchase them from \"Bookstore\" on the top of the sidebar.[/quote]\r\n\r\nof course, they are AoPS books, if they didn't sell them here, where else would they sell them?", "Solution_34": "go take the intro class", "Solution_35": "Duh hongsquare, i already said htat.", "Solution_36": "I think that book stores would get a lot of money by selling AoPS books.", "Solution_37": "[quote=\"Wickedestjr\"][quote=\"isabella2296\"]There are tons of resources on the web - a quick Google search will yield a bunch. Elias Saab's website has some pretty good geometry problems, and there's a lot of books out there too. :)[/quote]\n\nThanks. I'll check it out. :lol: :)[/quote]\r\nif you have intro to geo DO THE CHALLENGE PROBLEMS!!! they are gonna make you beast in geometry.\r\nAlso, search Math books in AoPs wiki and you can get some pretty satisfying results. I have found that for counting and probability (especially the latter) it helps to do A LOT of problems even the easy ones.", "Solution_38": "[quote=\"Poincare\"][quote=\"Wickedestjr\"][quote=\"isabella2296\"]There are tons of resources on the web - a quick Google search will yield a bunch. Elias Saab's website has some pretty good geometry problems, and there's a lot of books out there too. :)[/quote]\n\nThanks. I'll check it out. :lol: :)[/quote]\nif you have intro to geo DO THE CHALLENGE PROBLEMS!!! they are gonna make you beast in geometry.\nAlso, search Math books in AoPs wiki and you can get some pretty satisfying results. I have found that for counting and probability (especially the latter) it helps to do A LOT of problems even the easy ones.[/quote]\r\n\r\nI really wanted to get the geometry intro book, and I was planning on buying it soon. My MC coach let me borrow a copy of Intro to C & P, and it was really helpful. The challenge problems in that book were good too.", "Solution_39": "If you've already taken the geometry class, or any aops class for that matter, over a year ago, best to review everything by doing the challenge probs in the book.\r\n\r\nWicked, buy the books. It'll be worth the mony. Why do you use the past tense \"I really WANNTED\", \"I was PLANNING\"?", "Solution_40": "I don't know, it was a long time ago, and it sort of slipped past my mind. I was waiting til Christmas." } { "Tag": [], "Problem": "Suppose that 7 boys and 13 girls line up in a row. Let S be the number of places in the row where a boy and a girl are standing next to each other. For example, for the row GBBGGGBGBGGGBGBGGBGG we have S = 12. If all possible orders of these 20 people are considered, what is the average value of S?\r\n\r\nI forget if we ever solved this, but it was one of the harder ones there.\r\n\r\nAlso, if you have a problem to post, post it here, so we can revive the SC Forum!! :P", "Solution_1": "We have $7$ boys and $13$ girls, however, if we find out the ways the boys can be positioned, this will determine the position of girls as well (since there are no extra spaces in the row).\r\n\r\nBecause BG is the same as GB, we can select two side by side spaces to put these. For instance, if we select the first two spaces, we have:\r\n\r\nBG_ _ _ _ .... (18 blank spaces)\r\n\r\nWe now have $18$ spaces and $6$ remaining boys, so we can determine the number of ways of determining the sequence of the remaining spaces as $\\binom{18}{6}$. \r\n\r\nFor the second case, we have\r\n\r\n_BG_ _ _ _ (18 blank spaces)\r\n\r\nSimilarily, we get $\\binom{18}{6}$\r\n\r\nContinuing this way, we find the total number of arrangements with BG are $19 \\cdot \\binom{18}{6}$ and the total number of ways we can arrange the sequence (regardless of having BG) is $\\binom{20}{7}$, thus our average would be $\\boxed{\\frac{91}{20}}$" } { "Tag": [ "calculus", "integration", "limit", "algebra", "binomial theorem", "real analysis", "real analysis unsolved" ], "Problem": "Find\r\n\r\n$\\lim_{n\\rightarrow \\infty}\\frac{n}{2^{n}}\\int_{0}^{1}x^{2n}(x^{2}+1)^{n}dx$.\r\n\r\nDidi", "Solution_1": "Well expanding using the binomial theorem we can get\r\n\r\n$\\int_{0}^{1}\\sum_{k=0}^{n}\\binom{n}{k}x^{2k+2n}dx = \\sum_{k=0}^{n}\\binom{n}{k}\\frac{1}{2n+2k+1}$\r\n\r\nso\r\n\r\n$\\frac{n}{2^{n}}\\sum_{k=0}^{n}\\binom{n}{k}\\frac{1}{2n+2k+1}= \\frac{1}{2^{n}}\\sum_{k=0}^{n}\\binom{n}{k}\\frac{1}{2+\\frac{2k+1}{n}}$.\r\n\r\nSince the sum of the coefficients is $2^{n}$, it's a weighted average of all those $\\frac{1}{2+\\frac{2k+1}{n}}$, which range from $\\frac{1}{2}$ to $\\frac{1}{4}$ so it's probably around $\\frac{1}{3}$, but I don't know how to evaluate it.", "Solution_2": "I think it's not hard to prove that\r\n\\[\\sum_{|k-\\frac{n}{2}|>n^{\\tfrac23}}\\binom nk=o(2^{n})\\]\r\nThat's why the lim is equal to $\\frac13$", "Solution_3": "Let's try it without expanding out the binomial. We'll start with a sequence of substitutions:\r\n\r\n$\\frac{n}{2^{n}}\\int_{0}^{1}x^{2n}(1+x^{2})^{n}\\,dx$ Let $y=x^{2}.$\r\n\r\n$=\\frac{n}{2^{n+1}}\\int_{0}^{1}y^{n-1/2}(1+y)^{n}\\,dt$ Let $u=1-y.$\r\n\r\n$=\\frac{n}{2}\\int_{0}^{1}(1-u)^{n-1/2}\\left(1-\\frac{u}2\\right)^{n}\\,du$ Let $v=nu.$\r\n\r\n$=\\frac12\\int_{0}^{n}\\left(1-\\frac{v}{n}\\right)^{n-1/2}\\left(1-\\frac{v}{2n}\\right)^{n}\\,dv$\r\n\r\nNow, take the limit as $n\\to\\infty,$ interchanging limit and integral. We can find a Dominated Convergence Theorem justification for the exchange. The limit will be\r\n\r\n$\\frac12\\int_{0}^{\\infty}e^{-v}e^{-v/2}\\,dv= \\frac12\\int_{0}^{\\infty}e^{-3v/2}\\,dv=\\frac12\\cdot\\frac23= \\frac13.$" } { "Tag": [ "topology" ], "Problem": "Prove: If every infinite subset of X always have limit point $ \\Rightarrow$ X is compact space", "Solution_1": "This is false. If $ \\omega$ denotes the first uncountable ordinal, then $ [0, \\omega)$ in the order topology is limit point compact but not compact (it's not even Lindelof). Did you intend for $ X$ to be a metric space?", "Solution_2": "[quote=\"sondhtn\"]Prove: If every infinite subset of X always have limit point $ \\Rightarrow$ X is compact space[/quote]\r\n\r\nYes, X is a metric space", "Solution_3": "Could somebody help me ??", "Solution_4": "Do you know that a metric space $ X$ is compact if and only if it is complete and for every $ \\varepsilon>0$, one can find a finite $ \\varepsilon$-net in $ X$?", "Solution_5": "I can proof $ X$ is complete but I can't find a finite $ \\varepsilon$-net in $ X$. Can you help me !", "Solution_6": "Just take any maximal $ \\varepsilon$-separated set. It is a $ \\varepsilon$-net and has no limit points." } { "Tag": [ "complex analysis", "graph theory" ], "Problem": "\u5feb\u958b\u5b78\u4e86, \u5148\u795d\u9858\u5404\u4f4d\u5b78\u696d\u66f4\u9032\u5e7e\u6b65, \r\n\u7562\u696d\u73ed\u7684\u80fd\u8003\u9032\u7406\u60f3\u7684\u5927\u5b78, \u53c3\u52a0\u6578\u5b78\u7af6\u8cfd\u7684\u80fd\u596a\u5f97\u512a\u7570\u6210\u7e3e...\r\n\u4e0d\u77e5\u5927\u5bb6\u6709\u4f55\"\u65b0\u5b78\u5e74\"\u9858\u671b?\r\n\u5c0d\u65bc\u6211\u4f86\u8aaa, \u80fd\u53d6\u5f97\u4fdd\u9001\u8cc7\u683c\u5df2\u7d93\u662f\u4e0d\u932f\u7684\u4e86 :P", "Solution_1": "[quote=\"ifai\"]\u53d6\u5f97\u4fdd\u9001\u8cc7\u683c :P[/quote]\r\nWhich university?", "Solution_2": "\u6d59\u5927\u5427! \u4e0d\u904e\u6210\u529f\u6a5f\u6703\u4e0d\u5927...\r\n\u64da\u6211\u6240\u77e5, \u6211\u6821\u66fe\u6210\u529f\u4fdd\u9001(\u7372\u9304\u53d6)\u6700\u597d\u7684\u5927\u5b78\u53ea\u662f\u4e2d\u5c71\u5927\u5b78. \u800c\u5317\u5927, \u6e05\u83ef\u7b49\u7684\u540d\u984d\u90fd\u88ab\u5225\u7684\u5b78\u6821\u4f54\u4e86. \u6211\u6821\u6709\u4e00\u4f4d\u66fe\u4ee3\u8868\u6fb3\u9580\u53c3\u52a0cmo, \u4e26\u7372\u5168\u6fb3\u6821\u969b\u6578\u5b78\u6bd4\u8cfd\u51a0\u8ecd, imo \u9285\u734e, \u800c\u4e14\u591a\u624d\u591a\u85dd\u7684\u6821\u53cb, \u4ed6\u7576\u6642\u4fdd\u9001\u53bb\u6e05\u83ef, \u4f46\u6c92\u6709\u88ab\u9304\u53d6. \u6700\u5f8c\u4ed6\u9084\u662f\u81ea\u5df1\u8003\u4e86\u4e0a\u53bb. \u73fe\u5728\u4ed6\u5347\u5927\u56db, \u7269\u7406\u7cfb(\u57fa\u790e\u79d1\u5b78\u73ed).", "Solution_3": "\u6211\u4eec\u8fd9\u91cc\u60c5\u51b5\u4e0d\u4e00\u6837. \u7531\u4e8e\u6e05\u534e\u5317\u5927\u90fd\u662f\u5317\u4eac\u7684\u5b66\u6821, \u6240\u4ee5\u53ea\u8981\u5728\u5317\u4eac\u8d5b\u533a\u62ff\u4e00\u7b49\u5956\u5c31\u884c\u4e86.", "Solution_4": "[quote=\"shobber\"]\u6211\u4eec\u8fd9\u91cc\u60c5\u51b5\u4e0d\u4e00\u6837. \u7531\u4e8e\u6e05\u534e\u5317\u5927\u90fd\u662f\u5317\u4eac\u7684\u5b66\u6821, \u6240\u4ee5\u53ea\u8981\u5728\u5317\u4eac\u8d5b\u533a\u62ff\u4e00\u7b49\u5956\u5c31\u884c\u4e86.[/quote]\r\n\u7fa8\u6155...\u4e0d\u904e\u6211\u5011\u7684\u9ad8\u8003\u6bd4\u4f60\u5011\u7c21\u55ae\u5f97\u591a", "Solution_5": "Shobber,\u4f60\u592a\u723d\u4e86,\u6211\u4eec\u8fd9\u8fb9\u8fdb\u4e86CMO\u90fd\u4e0d\u4e00\u5b9a\u80fd\u4e0a\u5317\u5927\u554a.......", "Solution_6": "dingdongdog, which county do u live?", "Solution_7": "\u70ba\u4f55\u9019\u500b Chinese Communities \u7e3d\u662f\u90a3\u9ebc\u51b7\u6e05?\r\n\u9019\u88e1view\u7684\u4eba\u4e0d\u5c11, \u4f46post\u7684\u4eba(\u6578)\u5be6\u5728...", "Solution_8": "Of course I'm in China.\r\nMy goal in this term is to be in CMO.", "Solution_9": "[quote=\"dingdongdog\"]Of course I'm in China.\nMy goal in this term is to be in CMO.[/quote]\r\n\u4f60\u5728\u5ee3\u6771\u4f55\u5e02? \u4eca\u5e74\u5e7e\u5e74\u7d1a?\r\n(\u4e0d\u4fbf\u56de\u7b54\u7684\u8a71\u53ef\u4ee5\u4e0d\u7b54)", "Solution_10": "\u6211\u4f4f\u5728\u6df1\u5733,\u6211\u5bb6\u7a97\u524d\u662f\u6df1\u5733\u6cb3,\u6cb3\u5bf9\u9762\u662f\u65b0\u754c.\u73b0\u5728\u4e0a\u9ad8\u4e09.", "Solution_11": "[quote=\"dingdongdog\"]\u6211\u4f4f\u5728\u6df1\u5733,\u6211\u5bb6\u7a97\u524d\u662f\u6df1\u5733\u6cb3,\u6cb3\u5bf9\u9762\u662f\u65b0\u754c.\u73b0\u5728\u4e0a\u9ad8\u4e09.[/quote]\r\n\u6df1\u5733? \u4e0d\u932f\u554a. \u6211\u4e5f\u662f\u4e0a\u9ad8\u4e09, \u4e0d\u904e\u5967\u6578\u4e00\u5b9a\u8ddf\u4f60\u76f8\u5dee\u5f88\u9060. \u4f60(\u5011)\u61c2\u5fae\u7a4d\u5206\u55ce? \u570b\u5167\u5728\u9ad8\u4e2d\u968e\u6bb5\u6703\u5426\u6559\u521d\u7b49\u5fae\u7a4d\u5206?", "Solution_12": "\u5fae\u79ef\u5206\u662f\u81ea\u5b66\u7684.", "Solution_13": "[quote=\"dingdongdog\"]\u5fae\u79ef\u5206\u662f\u81ea\u5b66\u7684.[/quote]\r\n\u73fe\u8655\u65bc\u54ea\u500b\u7a0b\u5ea6? \u6b63\u5728\u5b78\u751a\u9ebc\u5167\u5bb9?", "Solution_14": "\u9ad8\u4e09\u8bfe\u672c\u5e94\u8be5\u6709\u6781\u9650\u548c\u521d\u6b65\u79ef\u5206\u5427? \u4e0d\u8fc7\u9ad8\u8003\u6d89\u53ca\u5f88\u5c11, \u6240\u4ee5\u6ca1\u4eba\u628a\u5b83\u4eec\u5f53\u56de\u4e8b.", "Solution_15": "\u53e6\u4e00\u4ef6\u60f3\u63d0\u7684\u662f, \u6fb3\u9580\u548c\u53bb\u5e74\u4e00\u6a23, \u90fd\u4e0d\u641eCMO\u9810\u8cfd, \u76f4\u63a5\u7531\u4e4b\u524d\"\u5168\u6fb3\u6578\u5b78\u6bd4\u8cfd\"\u7684\u512a\u52dd\u8005\u53c3\u8cfd. \u96d6\u7136\u641e\u9810\u8cfd\u6211\u4e5f\u4e00\u5b9a\u4e0d\u6703\u5165\u9078, \u4f46\u7e3d\u7b97\u53ef\u4ee5\u5b78\u9ede\u6771\u897f. \u4eca\u5e74\u53c8\u4e0d\u641e, \u771f\u662f\u6709\u9ede\u5931\u671b...", "Solution_16": "Would you write it in English, please!", "Solution_17": "[quote=\"Lilian\"]Would you write it in English, please![/quote]\r\nThis is a forum for CHINESE, don't you know that!", "Solution_18": "[quote=\"Lilian\"]Would you write it in English, please![/quote]\r\nJust like asking you to post in Chinese.", "Solution_19": "hello, everyone, \u6211\u662f\u4f86\u81ea\u6fb3\u9580\u7684 :) \u5c31\u8b80\u6fe0\u6c5f\u4e2d\u5b78, \u8acb\u554fifai\u662f\u5c31\u8b80\u90a3\u9593\u5b78\u6821\u7684? \u672c\u4eba\u5c31\u8b80\u9ad8\u4e8c, \u90fd\u4fc2\u7814\u7a76\u6578\u5b78\u7684, \u751a\u9ebc\u5167\u5bb9\u4e5f\u53ef(Real analysis, Complex analysis, Graph theory, Combination...), \u591a\u591a\u6307\u6559 :)", "Solution_20": "[quote=\"Mathsniper\"]hello, everyone, \u6211\u662f\u4f86\u81ea\u6fb3\u9580\u7684 :) \u5c31\u8b80\u6fe0\u6c5f\u4e2d\u5b78, \u8acb\u554fifai\u662f\u5c31\u8b80\u90a3\u9593\u5b78\u6821\u7684? \u672c\u4eba\u5c31\u8b80\u9ad8\u4e8c, \u90fd\u4fc2\u7814\u7a76\u6578\u5b78\u7684, \u751a\u9ebc\u5167\u5bb9\u4e5f\u53ef(Real analysis, Complex analysis, Graph theory, Combination...), \u591a\u591a\u6307\u6559 :)[/quote]\r\n\r\nsniper, do you remember me?\r\n\u9084\u6709\uff0c\u4f60\u8b80\u5f97\u5481\u5feb\uff1fcomplex analysis??", "Solution_21": "[quote=\"Soarer\"][quote=\"Mathsniper\"]hello, everyone, \u6211\u662f\u4f86\u81ea\u6fb3\u9580\u7684 :) \u5c31\u8b80\u6fe0\u6c5f\u4e2d\u5b78, \u8acb\u554fifai\u662f\u5c31\u8b80\u90a3\u9593\u5b78\u6821\u7684? \u672c\u4eba\u5c31\u8b80\u9ad8\u4e8c, \u90fd\u4fc2\u7814\u7a76\u6578\u5b78\u7684, \u751a\u9ebc\u5167\u5bb9\u4e5f\u53ef(Real analysis, Complex analysis, Graph theory, Combination...), \u591a\u591a\u6307\u6559 :)[/quote]\n\nsniper, do you remember me?\n\u9084\u6709\uff0c\u4f60\u8b80\u5f97\u5481\u5feb\uff1fcomplex analysis??[/quote]\r\nyaaaa, \u6211\u7576\u7136\u8a18\u5f97\u4f60 :P \u6211\u73fe\u5728\u5617\u8a66\u4e86\u89e3\u5c07\u89c0\u9ede\u653e\u9ad8\u4e00\u9ede\u53bb\u770b\u6578\u5b78, \u5230\u4e00\u5b9a\u6642\u9593\u6642\u5728\u4e2d\u9014\u6df1\u5165\u4e86\u89e3\u6240\u5b78\u7684\u77e5\u8b58, \u597d\u4f3c\u662fReal and Complex anaylsis, \u7576\u7136, \u73fe\u5728\u6211\u53ea\u662f\u61c2, \u4e00\u822c\u7684\u9084\u80fd\u505a, \u96e3\u7684\u9084\u6709\u5f85\u7814\u7a76, \u56e0\u70ba\u5169\u500b\u7406\u8ad6\u5df2\u662f\u591a\u5e74\u4f86\u901a\u904e\u591a\u4f4d\u6578\u5b78\u5bb6\u5960\u7acb\u7684, \u4ee5\u6211\u89d2\u5ea6, \u8981\u7d30\u7a76\u5247\u5fc5\u7136\u8981\u4e86\u89e3\u5176\u4e00\u5207(\u5305\u62ec\u7406\u8ad6\u5275\u7acb\u7684\u80cc\u666f\u53ca\u5176\u610f\u7fa9). \u6240\u4ee5\u8981\u6295\u653e\u66f4\u591a\u7684\u6642\u9593\u4e0b\u53bb, \u4f46\u73fe\u5728\u6211\u9084\u4e0d\u5f97\u4e0d\u5feb\u7814\u7a76\u5716\u8ad6... \u56e0\u70ba\u5176\u61c9\u7528\u5ee3\u6cdb, \u672c\u4eba\u662f\u6709\u5b78\u7fd2\u7a0b\u5e8f\u8a2d\u8a08\u7684\u80cc\u666f(\u5f9e\u521d\u4e00, \u73fe\u5728\u9ad8\u4e8c), \u5728\u6bd4\u8cfd\u4e2d\u4f7f\u7528\u7387\u9ad8(\u4f46\u6211\u8981\u6307\u660e... \u6211\u5c0d\u6bd4\u8cfd\u4e26\u4e0d\u770b\u91cd, \u6211\u53ea\u5728\u4e4e, \u5b78\u61c2, \u7cbe), \u800c\u4e14Graph theory\u4ea6\u662f\u91cd\u8981\u7406\u8ad6!!! \u6240\u4ee5\u6211\u4ea6\u90fd\u60f3\u6b64\u70ba\u9996\u8981\u5b78\u7fd2\u5167\u5bb9.", "Solution_22": "[quote=\"Mathsniper\"][quote=\"Soarer\"][quote=\"Mathsniper\"]hello, everyone, \u6211\u662f\u4f86\u81ea\u6fb3\u9580\u7684 :) \u5c31\u8b80\u6fe0\u6c5f\u4e2d\u5b78, \u8acb\u554fifai\u662f\u5c31\u8b80\u90a3\u9593\u5b78\u6821\u7684? \u672c\u4eba\u5c31\u8b80\u9ad8\u4e8c, \u90fd\u4fc2\u7814\u7a76\u6578\u5b78\u7684, \u751a\u9ebc\u5167\u5bb9\u4e5f\u53ef(Real analysis, Complex analysis, Graph theory, Combination...), \u591a\u591a\u6307\u6559 :)[/quote]\n\nsniper, do you remember me?\n\u9084\u6709\uff0c\u4f60\u8b80\u5f97\u5481\u5feb\uff1fcomplex analysis??[/quote]\nyaaaa, \u6211\u7576\u7136\u8a18\u5f97\u4f60 :P \u6211\u73fe\u5728\u5617\u8a66\u4e86\u89e3\u5c07\u89c0\u9ede\u653e\u9ad8\u4e00\u9ede\u53bb\u770b\u6578\u5b78, \u5230\u4e00\u5b9a\u6642\u9593\u6642\u5728\u4e2d\u9014\u6df1\u5165\u4e86\u89e3\u6240\u5b78\u7684\u77e5\u8b58, \u597d\u4f3c\u662fReal and Complex anaylsis, \u7576\u7136, \u73fe\u5728\u6211\u53ea\u662f\u61c2, \u4e00\u822c\u7684\u9084\u80fd\u505a, \u96e3\u7684\u9084\u6709\u5f85\u7814\u7a76, \u56e0\u70ba\u5169\u500b\u7406\u8ad6\u5df2\u662f\u591a\u5e74\u4f86\u901a\u904e\u591a\u4f4d\u6578\u5b78\u5bb6\u5960\u7acb\u7684, \u4ee5\u6211\u89d2\u5ea6, \u8981\u7d30\u7a76\u5247\u5fc5\u7136\u8981\u4e86\u89e3\u5176\u4e00\u5207(\u5305\u62ec\u7406\u8ad6\u5275\u7acb\u7684\u80cc\u666f\u53ca\u5176\u610f\u7fa9). \u6240\u4ee5\u8981\u6295\u653e\u66f4\u591a\u7684\u6642\u9593\u4e0b\u53bb, \u4f46\u73fe\u5728\u6211\u9084\u4e0d\u5f97\u4e0d\u5feb\u7814\u7a76\u5716\u8ad6... \u56e0\u70ba\u5176\u61c9\u7528\u5ee3\u6cdb, \u672c\u4eba\u662f\u6709\u5b78\u7fd2\u7a0b\u5e8f\u8a2d\u8a08\u7684\u80cc\u666f(\u5f9e\u521d\u4e00, \u73fe\u5728\u9ad8\u4e8c), \u5728\u6bd4\u8cfd\u4e2d\u4f7f\u7528\u7387\u9ad8(\u4f46\u6211\u8981\u6307\u660e... \u6211\u5c0d\u6bd4\u8cfd\u4e26\u4e0d\u770b\u91cd, \u6211\u53ea\u5728\u4e4e, \u5b78\u61c2, \u7cbe), \u800c\u4e14Graph theory\u4ea6\u662f\u91cd\u8981\u7406\u8ad6!!! \u6240\u4ee5\u6211\u4ea6\u90fd\u60f3\u6b64\u70ba\u9996\u8981\u5b78\u7fd2\u5167\u5bb9.[/quote]\r\n\r\n\u592a\u5feb\u4e86... complex analysis =.=\"", "Solution_23": "[quote=\"Soarer\"][/quote][quote=\"Mathsniper\"][quote=\"Soarer\"][quote=\"Mathsniper\"]hello, everyone, \u6211\u662f\u4f86\u81ea\u6fb3\u9580\u7684 :) \u5c31\u8b80\u6fe0\u6c5f\u4e2d\u5b78, \u8acb\u554fifai\u662f\u5c31\u8b80\u90a3\u9593\u5b78\u6821\u7684? \u672c\u4eba\u5c31\u8b80\u9ad8\u4e8c, \u90fd\u4fc2\u7814\u7a76\u6578\u5b78\u7684, \u751a\u9ebc\u5167\u5bb9\u4e5f\u53ef(Real analysis, Complex analysis, Graph theory, Combination...), \u591a\u591a\u6307\u6559 :)[/quote]\n\nsniper, do you remember me?\n\u9084\u6709\uff0c\u4f60\u8b80\u5f97\u5481\u5feb\uff1fcomplex analysis??[/quote]\nyaaaa, \u6211\u7576\u7136\u8a18\u5f97\u4f60 :P \u6211\u73fe\u5728\u5617\u8a66\u4e86\u89e3\u5c07\u89c0\u9ede\u653e\u9ad8\u4e00\u9ede\u53bb\u770b\u6578\u5b78, \u5230\u4e00\u5b9a\u6642\u9593\u6642\u5728\u4e2d\u9014\u6df1\u5165\u4e86\u89e3\u6240\u5b78\u7684\u77e5\u8b58, \u597d\u4f3c\u662fReal and Complex anaylsis, \u7576\u7136, \u73fe\u5728\u6211\u53ea\u662f\u61c2, \u4e00\u822c\u7684\u9084\u80fd\u505a, \u96e3\u7684\u9084\u6709\u5f85\u7814\u7a76, \u56e0\u70ba\u5169\u500b\u7406\u8ad6\u5df2\u662f\u591a\u5e74\u4f86\u901a\u904e\u591a\u4f4d\u6578\u5b78\u5bb6\u5960\u7acb\u7684, \u4ee5\u6211\u89d2\u5ea6, \u8981\u7d30\u7a76\u5247\u5fc5\u7136\u8981\u4e86\u89e3\u5176\u4e00\u5207(\u5305\u62ec\u7406\u8ad6\u5275\u7acb\u7684\u80cc\u666f\u53ca\u5176\u610f\u7fa9). \u6240\u4ee5\u8981\u6295\u653e\u66f4\u591a\u7684\u6642\u9593\u4e0b\u53bb, \u4f46\u73fe\u5728\u6211\u9084\u4e0d\u5f97\u4e0d\u5feb\u7814\u7a76\u5716\u8ad6... \u56e0\u70ba\u5176\u61c9\u7528\u5ee3\u6cdb, \u672c\u4eba\u662f\u6709\u5b78\u7fd2\u7a0b\u5e8f\u8a2d\u8a08\u7684\u80cc\u666f(\u5f9e\u521d\u4e00, \u73fe\u5728\u9ad8\u4e8c), \u5728\u6bd4\u8cfd\u4e2d\u4f7f\u7528\u7387\u9ad8(\u4f46\u6211\u8981\u6307\u660e... \u6211\u5c0d\u6bd4\u8cfd\u4e26\u4e0d\u770b\u91cd, \u6211\u53ea\u5728\u4e4e, \u5b78\u61c2, \u7cbe), \u800c\u4e14Graph theory\u4ea6\u662f\u91cd\u8981\u7406\u8ad6!!! \u6240\u4ee5\u6211\u4ea6\u90fd\u60f3\u6b64\u70ba\u9996\u8981\u5b78\u7fd2\u5167\u5bb9.[/quote][quote=\"Soarer\"]\n\n\u592a\u5feb\u4e86... complex analysis =.=\"[/quote]\r\nkeke~ \u4f60\u7576\u4f62\u8208\u8da3\u66f8\u5481\u7747\u5427 :P Complex analysis\u6211\u53ea\u662f\u6620\u5c04\u90a3part\u6c92\u6709\u770b, \u4f46\u554f\u904e\u963fsir\u8a71\u6620\u5c04\u500b\u5ea6\u597d\u9b3c\u591a\u7406\u8ad6, \u6211\u963fsir\u4ee5\u524d\u4fc2\u5b78\u51fd\u6578\u8ad6, so\u5f97\u9592\u90fd\u6435\u4f62\u6559\u4e0b\u6211 :D", "Solution_24": "Happy New Year\uff01\r\n \r\nEveryone\u3002 :|", "Solution_25": "[quote=\"Mathsniper\"]keke~ \u4f60\u7576\u4f62\u8208\u8da3\u66f8\u5481\u7747\u5427 $\\cdots$\u5f97\u9592\u90fd\u6435\u4f62\u6559\u4e0b\u6211 :D[/quote]\r\n\u770b\u4e0d\u61c2...", "Solution_26": "[quote=\"shobber\"][quote=\"Mathsniper\"]keke~ \u4f60\u7576\u4f62\u8208\u8da3\u66f8\u5481\u7747\u5427 $\\cdots$\u5f97\u9592\u90fd\u6435\u4f62\u6559\u4e0b\u6211 :D[/quote]\n\u770b\u4e0d\u61c2...[/quote]\r\n\u4f60\u5f53\u5b83\u5174\u8da3\u4e66\u90a3\u6837\u770b\u5427 \r\n\u6709\u65f6\u95f4\u90fd\u627e\u4ed6\u6559\u4e00\u4e0b\u6211", "Solution_27": "\u90a3\u4e9b\u90fd\u662f\u5e7f\u4e1c\u90a3\u8fb9\u7684\u81ea\u521b\u8bcd\u6c47\u4e48? \u8fd8\u6709, \"\u5f97\u95f2\" \u4e8c\u5b57, \u5927\u6709\u53e4\u98ce...", "Solution_28": "\u662f\u5e7f\u4e1c\u8bdd\uff08\u6216\u9999\u6e2f\u8bdd\uff09\u7684\u53e3\u8bed\u3002", "Solution_29": "[quote=\"shobber\"]\u90a3\u4e9b\u90fd\u662f\u5e7f\u4e1c\u90a3\u8fb9\u7684\u81ea\u521b\u8bcd\u6c47\u4e48? \u8fd8\u6709, \"\u5f97\u95f2\" \u4e8c\u5b57, \u5927\u6709\u53e4\u98ce...[/quote]\r\nNot exactly \u81ea\u521b\u8bcd, just the difference between Cantonese and Putonghua." } { "Tag": [], "Problem": "Mike paid $ \\$1.25$ for a stamp three years ago. He was just offered double that amount for the stamp. Assuming the stamp's offer price doubles every three years, how many dollars will he be offered in 12 more years?", "Solution_1": "He will be offered $ \\$1.25\\cdot2^{1\\plus{}\\frac{12}3}\\equal{}\\$\\frac54\\cdot2^5\\equal{}\\$5\\cdot2^3\\equal{}\\boxed{\\$40}$." } { "Tag": [ "inequalities", "geometry", "triangle inequality" ], "Problem": "Find smallest real number $p$ for which is this statement true: \r\n\r\nFor every positive integer $n$ this inequality holds:\r\n$\\sqrt{1^2+1}+\\sqrt{2^2+1}+\\ldots\\sqrt{n^2+1}\\leq \\frac{1}{2}n(n+p)$", "Solution_1": "I think $p=2\\sqrt{2}-1$.I'll post my solution if answer is right.", "Solution_2": "It would seem to be, theres just one part of my proof i cant do! Because im to lazy to multiply out two hugh brackets.... :)", "Solution_3": "Yup your right, a completely brief proof follows... its not even a proof, it just outlines the method, ask if you want more!\r\n\r\n [hide]You can easily prove that when you add an extra term to the sequence that the RHS grows more than the LHS so the min value is indeed 2 (:sqrt:2)-1, corresponding to the first term in the sequence.[/hide]", "Solution_4": "Yes, that was clear . ;)", "Solution_5": "Yes, it's right.\r\nAlso you can find an geometric interpretation (after you find $p$)...", "Solution_6": "[quote=\"ondrob\"]Yes, it's right.\nAlso you can find an geometric interpretation (after you find $p$)...[/quote]\r\nYes, I've just found very nice geometric solution for it. I'll post it soon. :)", "Solution_7": "Ok, here is my solution. We have that: $p\\geq{\\frac{2\\sum_{i=1}^{n}{\\sqrt{i^{2}+1}}}{n}}-n.$Now I'll prove (using geometry) that RHS is decreasing ,since $p=2\\sqrt{2}-1$ when $n=1$.So we have to prove that $\\sum_{i=1}^{n}{\\sqrt{i^{2}+1}}+\\frac{n(n+1)}{2}\\geq{n\\sqrt{(n+1)^{2}+1}}.$\r\nLet $P=(0;1)$ and $A_{i}=(i;0)$ $(i=1,2,..,n)$.Using triangle inequality we obtain $\\sqrt{i^{2}+1}+(n-i+1)>\\sqrt{(n+1)^{2}+1}$ $(i=1,2,..,n)$, adding these up we get the result , since $\\frac{n(n+1)}{2}=\\sum_{i=1}^{n}{i}.$\r\nVery nice!! :)", "Solution_8": "Yes, congratulations Armo, you've found very nice geometrical solution. I'll post another inequality that can be done with geometrical approach soon..." } { "Tag": [], "Problem": "hi :lol: \r\n\r\nIm Rachel...Just joined this forum out of my interest in educational forums.\r\n\r\nI like to say hi to all of ya n would like to give a small intro abt me..\r\n\r\nwell...im just a high school student..like browsing the net...i do chat all the time.(in the forums usually)\r\n\r\nso thought to register in this wonderful interesting forum talking much abt ma interests\r\n\r\nok guys...so can i know all of ya :roll: :roll: :roll:", "Solution_1": "Hi there.\r\nI'm *points to my username* nutz_for2.718281828.\r\nI'm officially in middle school, but I sometimes surf around the highschool forums for something to do, instead of the usual beginners stuff.\r\n\r\nYou can take classes here:\r\n\r\nIntroductory classes, intermediate classes and olympiad classes. I reccomend introductory if you domn't have a firm grip on it. I then suggest intermediate. After you have mastered that, then the olympiad courses would be just right for you.\r\n\r\nAlso, you can browse around the forums for questions out of the blue. like Billy Bob questions. I'm just entering intermediate, but the problems are too hard for me.\r\n\r\nWell, for your entertainment, there is also a games and fun factory forum.\r\n\r\nI hope you enjoy AoPS, I do.", "Solution_2": "Hello,\r\n\r\nwelcome to Mathlinks. I'm fredbel6, university maths student.\r\n\r\nIn \"Games and Fun Factory\" there is a \"sticky\" topic where people can introduce themselves (I have done so , and maybe you want to do it too). Click on this link to go there :\r\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=15747[/url]", "Solution_3": "Hi :lol: ,\r\n\r\nThanx to all the members for ur warm heart welcome.\r\n\r\nBye :)", "Solution_4": "Hello rachel!\r\n\r\nWelcome to AoPS! It is a great place to be if you are interested in maths - both serious (olympiad type) and for fun (bubka type :D ).\r\n\r\nExplore games and fun factory - you will find lots of interesting things there. Enjoy and keep writing about your experiences.\r\n\r\nCheers,\r\nbubka", "Solution_5": "Thanx for u all fun loving guyz. \r\n\r\nbyeeeeeeeeee!" } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Calculate: $\\sum_{k=1}^{n}k.2^{k}$", "Solution_1": "Let $P(x)=x+x^{2}+...+x^{n}=x.\\frac{x^{n}-1}{x-1}(x\\not = 1)$.\r\n\r\nWe have $P'(x)=1+2x+...+kx^{k-1}+...+nx^{n-1}$ therefore sum is $2.P'(2)$.", "Solution_2": "[quote=\"N.T.TUAN\"]Let $P(x)=x+x^{2}+...+x^{n}=x.\\frac{x^{n}-1}{x-1}(x\\not = 1)$.\n\nWe have $P'(x)=1+2x+...+kx^{k-1}+...+nx^{n-1}$ therefore sum is $P'(2)$.[/quote]\r\nis there any problem if the sum is a not continous function?", "Solution_3": "is there sample? :wink:", "Solution_4": "[quote=\"santosguzella\"]Calculate: $\\sum_{k=1}^{n}k.2^{k}$[/quote]\r\nif $U_{n}=\\sum_{k=1}^{n}k.2^{k}$\r\n$U_{n+1}=\\sum_{k=1}^{n+1}k.2^{k}=U_{n}+(n+1)2^{n+1}=2U_{n}+\\sum_{k=1}^{n+2}=2U_{n}+2^{n+2}-2$\r\nso , $U_{n}=(n+1-2)2^{n+1}+2=(n-1)2^{n+1}+2$" } { "Tag": [ "limit", "logarithms", "calculus", "calculus computations" ], "Problem": "Find the limit:\r\n\r\n$\\lim_{n\\rightarrow \\infty}\\frac{1}{n}\\sum_{i=1}^{n}\\frac{1}{k}$", "Solution_1": "It is equal to $0$ because $\\sum_{i=1}^{n}\\frac{1}{i}-\\ln{n}$ is convegers.", "Solution_2": "Or by Cesaro-Stoltz $\\lim_{n\\to\\infty}\\frac{\\sum_{k=1}^{n}\\frac{1}{k}}{n}=\\lim_{n\\to\\infty}\\frac{\\sum_{k=1}^{n+1}\\frac{1}{k}-\\sum_{k=1}^{n}\\frac{1}{k}}{n+1-n}=\\lim_{n\\to\\infty}\\frac{1}{n+1}=0$" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Happy New Year !\r\n\r\n\r\nMircea Lascu", "Solution_1": "HAPPY NEW YEAR!!!! \r\n\r\n\r\nI'm SURE 2007 will be quite a special year! :lol:" } { "Tag": [ "FTW" ], "Problem": "How many times does the digit 9 appear in the list of all integers from 1 to 500?", "Solution_1": "$ **9 : 5 \\times 10 \\equal{} 50$\r\n$ *9* : 5 \\times 10 \\equal{} 50$\r\n$ *99 : 5$\r\n$ 50 \\plus{} 50 \\minus{} 5 \\equal{} \\boxed{95}$.", "Solution_2": "FTW says the answer is 100... however I checked by computer and the solution is indeed 95.", "Solution_3": "Actually, there are TWENTY 9's for each set of 100 numbers starting at _01 and going to _99, (except for the 900's, which are not included here), so 5*20=100.", "Solution_4": "Maybe the confusion comes from the fact that 99 has 2 9's in it.", "Solution_5": "yeah. i messed up on that the first time.", "Solution_6": "an alternate way of explaining the \"there are 20 9's from 00 to 99\" is as follows\nthere are 100 total numbers, each number is two-digit.\nEach digit has the same frequency in this interval, and there are $100\\cdot2 = 200$ digits.\nSo there are $\\frac{200}{10}=20$ 9s.\nand since 9 doesn't appear in the hundreds digit from 0 to 500, we can multiply $20$ by $\\frac{500-0}{100}=5$ to obtain our answer, 100." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "inequalities", "geometry", "function", "number theory" ], "Problem": "What does everyone think about the topics of the USAMO questions are gonna be? I looked online, and there seems to be no general pattern that they follow.\r\n\r\n\r\nAlso, what do you think will be the cutoffs for MO(S)P?", "Solution_1": "There's no way to tell what the USAMO problems will be; topic distribution is pretty random, except that usually the first question each day is not very technical, so as to give more people a chance.\r\n\r\nThe name of the residential math program to which you referred is MOP. The cutoff for black (the top twelve or so) is usually in the 20s; blue (next 15-18 non-seniors) is usually in the high teens or low 20s, and red (9th grade) is more variable.", "Solution_2": "Last year the cutoffs for blue and red were 17 and 9. I'm not sure what the cutoff for black was.\r\n\r\nI think generally you should need 2.5 problems to make blue and 1.5 to make red. 3.5-4 is probably good for black.", "Solution_3": "All I'm going to say is, I think pure inequalities have gone the way of the dodo.\r\n\r\nDon't take my word for it though, but I would bet money on this assertion.", "Solution_4": "[quote=\"randomdragoon\"]All I'm going to say is, I think pure inequalities have gone the way of the dodo.\n\nDon't take my word for it though, but I would bet money on this assertion.[/quote]\r\n\r\nThey come up when you least expect them: STUDY INEQUALITIES!!! :D\r\n\r\n(of course, I'm telling you this to throw you off so you study inequalities while I prepare for the 6 geometry problems on the 2008 USAMO :P )", "Solution_5": "Hmmm one thing I noticed was there's a good few problems like \"find with justification, the largest number of...\". Maybe I just haven't seen all that many problems.\r\nWait what's .5 of a question? Usually they give very little credit for BS, right, so you need to get a ton of good BS and collect a few points.", "Solution_6": "No...the way to get partial credit is if you have significant parts of a solution. \r\n\r\nMost olympiad problems at least have some sort of bounding aspect to them. I doubt that you will see a three variable symmetric inequality or something...those are pretty much 'solved'. But you will most likely need to use inequalities as parts of solutions.", "Solution_7": "Actually usually MOSP is somewhere around 3-3.5 problems for blue and 4 for black. Last year was really weird, because the test was... bad. I hated that test.\r\n\r\nWhat will be the fate of this new era of combo problem favoring rigor over creativity (last year's #2, #4)? As long as those don't show up en masse, I will be happy.\r\n\r\n[quote=\"randomdragoon\"]All I'm going to say is, I think pure inequalities have gone the way of the dodo.\n\nDon't take my word for it though, but I would bet money on this assertion.[/quote]\r\n\r\nThere were pure inequalities on the IMO just two and three years ago. And last year's USAMO #6 may not have been pure, but it definitely used the same ideas as your standard inequality. They might be afraid of putting them on the USAMO because the level of inequalities these days that would challenge really good people might be impossibly difficult for the neophytes (?).", "Solution_8": "Yeah, actually I've noticed a general decline in quality in AMC/AIME/USAMO problems of late. Maybe I need to submit some problems after I graduate :o", "Solution_9": "[quote=\"jb05\"]There were pure inequalities on the IMO just two and three years ago. And last year's USAMO #6 may not have been pure, but it definitely used the same ideas as your standard inequality. They might be afraid of putting them on the USAMO because the level of inequalities these days that would challenge really good people might be impossibly difficult for the neophytes (?).[/quote]\r\n\r\nYeah, I was referring to the USAMO about this claim, for the reasons you outlined above.", "Solution_10": "Last year, with the question on Tyrannosaurus Rexes (dinosaur problem), if you had simply found the answer with little justification how many points should you get?", "Solution_11": "1, possibly 0 or 2. 0 corresponds to them deciding that the answer is easy enough to see and not enough work was done. 2 corresponds to the answer being slightly nontrivial and your making nontrivial progress towards the solution.", "Solution_12": "[quote]actually I've noticed a general decline in quality in AMC/AIME/USAMO problems of late[/quote]\r\n\r\nCan you quantify that, or make your statement precise? Exactly how, and in what degree is there a general decline on quality?\r\nCan you compare to the quality over time of the CMO, of the BMO, the Australian MO and the corresponding preliminary contests?\r\n\r\nI just finished looking through 60 years of the AMC contests in preparation for posters and brochures for our 2008-2009 contest year. My perception is that the problems now have greater variety (more number theory, more interesting counting, more sequences, more meaningful questions about functions) and less routine computation. I think the problem statements are now more precise. \r\n\r\nYou probably have too great a familiarity with the problems, you have studied them too much and so they seem routine to you now. What once was fresh and new, now is ordinary for you. But objectively, I believe the quality is bettr, and certainly no worse.\r\n\r\nSteve Dunbar\r\nMAA Director, American Mathematics Competitions", "Solution_13": "[quote=\"Hamster1800\"]1, possibly 0 or 2. 0 corresponds to them deciding that the answer is easy enough to see and not enough work was done. 2 corresponds to the answer being slightly nontrivial and your making nontrivial progress towards the solution.[/quote]\r\nSo would \"nontrivial progress\" be equivalent to writing down the first half of a full solution, or writing down a \"full\" solution but with nontrivial holes?", "Solution_14": "Nontrivial progress is any nontrivial work that will lead to a correct solution, or a solution with a major flaw in it that could (with difficulty) be fixed. (Depending on how big the flaw is, the latter may receive points comparable to the former, so for this reason many people overestimate their scores on the USAMO.)\r\n\r\n\r\n\r\nNote the \"nontrivial\" qualifier on the work; you can't write down something mildly obvious and expect to get points from it. Nor will you get points for an interesting observation that doesn't get you anywhere.", "Solution_15": "As an example of nontrivial progress, on #1 last year I had the correct idea that when you get a term equal to the average of all the terms before it, the sequence becomes constant (this is pretty much trivial, actually, if you ask me), but my proof for the fact that there must exist such a term was complete BS, basically i did a whole bunch of random stuff and ended up where i started with completely different notation (pathetically, I had no idea that it was wrong until like the next day).\r\n\r\nI believe I managed to get a 3 for that, and thinking back, I probably should have gotten a 1. But I mean, I did get a 6 total, and I had the standard BS on #4 and #5 and that's it, so I probably didn't get any lower than a 3.\r\n\r\nI guess that counts as nontrivial, hmmm", "Solution_16": "Hmmm... I got 11 last year, which included #1 and \"standard BS\" on #4, #5. I am inclined to believe that I got 2 on each of those. (7,2,2)\r\n\r\nThis would make your distribution 2,2,2, which I think is more feasible than anything involving a 3 or a 4 on one problem, which is always rare.\r\n\r\nStill, if you got points, then that's a good example of \"nontrivial\" nonetheless." } { "Tag": [], "Problem": "How many shells of elecrons orbit the nucleus of an atom?", "Solution_1": "Er.. it depends on the atom (how many electrons it has)?\r\nH has one shell.. while Na has 3 (energy levels)", "Solution_2": "Yeah...but are you (ln(dx/dy)) referring to orbitals of a certain sublevel?", "Solution_3": "look up in the periodic table.", "Solution_4": "that is a completely vague question. it can definitely range depending on principal quantum number or energy level, excited states, etc. and theyre not all really shells per se, the s orbitals resemble a shell of sorts but the p-orbitals are dumbbell shaped with x, y, or z orientation. the d and f-orbitals are still more complexly contructed because they can contain more electrons.", "Solution_5": "maximum seven!", "Solution_6": "it can be much more!!!\r\nwith new elements being discovered everyday,nothing is impossible :rotfl: :D :rotfl:", "Solution_7": "my post edited: :rotfl: [size=150]uptil now seven[/size]" } { "Tag": [ "AwesomeMath", "summer program" ], "Problem": "Questions for people who've been to AMSP.\r\n\r\n Is it possible to take an Intermediate Aops class during AMP without skipping classes? Would it be a good idea to bring an instrument to practice? Does the university wash your laundry for you? Sorry I haven't been to AMSP before.", "Solution_1": "The AMSP classes are earlier in the day. I think that somebody tried to do some AoPS midterm assignment or something but then forgot about it (I'll ask him). I think that you'll be pretty busy (especially if want to work on extra problems after class is over). \r\n\r\nThere some free time after dinner that you can do assorted activities. Some people did sports like frisbee or soccer. Some did more math, some just hung out, talked, went swimming, played things like cards, etc. There was also ping-pong/DDR. I know there were some people who practiced violin daily so you can definitely do that if you want. \r\n\r\nThe way that laundry worked last year was that you sort of live in an apartment that has a washer and a dryer (and soap). So you do have to wash your own clothes, but it is pretty easy.", "Solution_2": "whewww\r\nits good that we can bring our instrumentss!\r\nyayy ;]\r\n\r\nso it's pretty much self laundry...and soap's provided so we dont have to bring any or buy any there?\r\nthankss", "Solution_3": "There is a place where you can play piano: there are two practice rooms with upright pianos that aren't too bad, as well as a concert hall that may or may not be locked and may or may not have no lights on and may or may not have the lighting room locked.\r\n\r\ndiophantient was taking an AoPS class that ran during AMP, I forget what he did though. Maybe that's the same person that Altheman is referring to.", "Solution_4": "[quote=\"CatalystOfNostalgia\"]There is a place where you can play piano: there are two practice rooms with upright pianos that aren't too bad, as well as a concert hall that may or may not be locked and may or may not have no lights on and may or may not have the lighting room locked.\n\ndiophantient was taking an AoPS class that ran during AMP, I forget what he did though. Maybe that's the same person that Altheman is referring to.[/quote]\r\nYeah, I ended up just missing 3 classes and screwing the midterm.", "Solution_5": "Midterm? AoPS classes have midterms? Does that mean finals too? Are they hard?", "Solution_6": "some i guess. the ones i did didnt have mdtrms, but i guess some do...\r\ngood luck on acceptance!\r\ni feel stupid saying this as i havent been accepted yet either.\r\nwish me luck 2!", "Solution_7": "DAI! did you get your plane tickets yet?\r\n\r\nor even any other peoplee?\r\n\r\noh and quickk question,\r\nwhen we arrive at the airport, how do we get to the campuss?", "Solution_8": "[quote=\"sooozyy\"]DAI! did you get your plane tickets yet?\n\nor even any other peoplee?\n\noh and quickk question,\nwhen we arrive at the airport, how do we get to the campuss?[/quote]\r\nThere are AwesomeMath people there who have AwesomeMath shirts and AwesomeMath signs who lead you to an AwesomeMath shuttle which takes you to AwesomeMath.", "Solution_9": "[quote=\"diophantient\"][quote=\"sooozyy\"]DAI! did you get your plane tickets yet?\n\nor even any other peoplee?\n\noh and quickk question,\nwhen we arrive at the airport, how do we get to the campuss?[/quote]\nThere are AwesomeMath people there who have AwesomeMath shirts and AwesomeMath signs who lead you to an AwesomeMath shuttle which takes you to AwesomeMath.[/quote]\r\n\r\nAirport? Plane tickets? Where do you find the instructions to go onto AwesomeMath??", "Solution_10": "TPBM is the (in-)famous David, Suzy.\r\nAnyways, I'm probably considering WOOT next year instead. And no, I didn't buy my tickets yet. My parents will probably wait til later. Did you?", "Solution_11": "next year..i'm not doing any academic camps.\r\nprobably focus on my early college apps and intership or whatnot.\r\nnope didn't buy my ticket yet.\r\ni should though.", "Solution_12": "Speaking of planes,\r\n\r\n\r\nI already set up the flight to Dallas in Flight Simulator X and I'm gonna fly it tomorrow. I'll try an ILS appraoch, which I hate to do. Anybody else like flying or flight simulators?\r\n\r\n:pilot:", "Solution_13": "DAI\r\nhopefully you got my pm.\r\n\r\ni reserved my flight todayy.\r\ninfact like ten minutes ago.\r\ncheckk your inbox!" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Let $(F_{n})$ be a decreasing sequence (i.e. $F_{1}\\supseteq F_{2}\\supseteq...$) of closed bounded nonempty sets in $R^{k}$. Prove that $F=\\bigcap_{n=1}^{\\infty}F_{n}$ is also closed, bounded, and nonempty.\r\n\r\nproving that $F$ is closed and bounded it is easy but i having trouble in proving that $F$ is non empty.", "Solution_1": "Note that while the first and second property can be proved by themselves, the third is wrong if you omit the closed and bounded hypotheses (aka compactness). owk", "Solution_2": "Choose a $f_{j}\\in F_{j}$ for each $j$. The sequence $(f_{j})_{j=1}^\\infty$ is contained in $F_{1}$. $F_{1}$ is compact, so according to Bolzano-Weiestrass there is a convergent subsequence with limit in $f$ in $F_{1}$. But this subsequence, except possibly the first element, is also in $F_{2}$. So since $F_{2}$ is closed, $f$ is also in\u00a0$F_{2}$. Similarly we have $f \\in F_{j}$ for all $j$." } { "Tag": [ "geometry", "perpendicular bisector" ], "Problem": "Let $ABC$ be a triangle\r\n$M$ is a variable point on the side $BC$ and $D$ is a point on the line $AB$ such that $MA=MD$.\r\nThe parallel line from $B$ to $MD$ intersects the line $AM$ at $P$.\r\nFind the locus of $P$", "Solution_1": "oh, of course", "Solution_2": "\"The parallel line from B to MD intersects the line AM at P.\"\r\n\r\nI assume you mean perpendicular line. I can't figure out what it would be parallel to.", "Solution_3": "[quote=\"MatrixFrog\"]\"The parallel line from B to MD intersects the line AM at P.\"\n\nI assume you mean perpendicular line. I can't figure out what it would be parallel to.[/quote]\r\n\r\nMy guess is \"parallel to MD and passing through B\".", "Solution_4": "pontios, can you please explain that part please????.", "Solution_5": "And while he's doing that, can someone refresh my memory of what a locus is. I've used them before, but it's been quite a while...", "Solution_6": "[quote=\"Farenhajt\"][quote=\"MatrixFrog\"]\"The parallel line from B to MD intersects the line AM at P.\"\n\nI assume you mean perpendicular line. I can't figure out what it would be parallel to.[/quote]\n\nMy guess is \"parallel to MD and passing through B\".[/quote]\n\nYes. $BP \\parallel MD, \\ P \\in AM$\nI'm sorry\n\n[quote=\"DanK\"]And while he's doing that, can someone refresh my memory of what a locus is. I've used them before, but it's been quite a while...[/quote]\r\n\r\nYou have to find where the point $P$ is moving on (a line or a curve).\r\n\r\nIn other words the locus is the set of all the positions of $P$\r\n\r\nFor example, a circle $(O,R)$ is the locus of the points $M$ with the property $MO=R$", "Solution_7": "Hmmm, all I can figure out so far is that it will be an expontential equation...", "Solution_8": "[hide]\nSince $DM\\parallel BP$, we have $\\angle ADM=ABP$.\nBut we also know that $\\angle ADM=\\angle MAD$ because $AM=MD$, so $\\angle MAD=\\angle ABP$ (or $\\angle PAB=\\angle PBA$), which implies $PA=PB$.\nHence the locus of $P$ is simply the perpendicular bisector of $AB$, but only to the direction of $C$.\n\n[hide=\"Strictly speaking\"]If $P$ is on the other side of $C$ with respect to $AB$, then $M$ will not be on $\\overline{BC}$, so the perpendicular bisector of $AB$ to the opposite direction from $C$ cannot be included.\n\nIf $P$ is on the same side as $C$ but inside the triangle, then $M$ will be still on $\\overline{BC}$, while $D$ will not be on $\\overline{AB}$. However, the problem states that $D$ could be anywhere on $\\overleftrightarrow{AB}$, so this case is still fine.\n\nFinally, if $P$ is on the same side as $C$ and outside the triangle, the property obviously holds, so this case is fine as well.[/hide]\n[/hide]", "Solution_9": "Oops. I forgot to accomodate for the fact that $MA=MD$. :blush:" } { "Tag": [ "geometry", "3D geometry", "octahedron", "real analysis", "real analysis unsolved" ], "Problem": "Let $H$ be a hexagone symmetrical with respect to the origin and having non void interiour. There exists a plane $P$ in $\\mathbb{R}^{3}$ and a cube $C$ centered at the origin such that the intersection between $C$ and $P$ is $H$.", "Solution_1": "In dual situation we must proof that arbitrary symmetrical hexagone is a projection of octahedron, but it's obvious." } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "If we have $v_0$, and $\\forall n \\;v_{n+1}=\\frac{1}{a+bv_n}$. Can we evaluate $\\prod_{k=0}^{n}v_k$?", "Solution_1": "Define $P_{n}\\equiv \\prod_{k=0}^{n} v_{k}$, $Q_{n} \\equiv 1/P_{n}$ \r\n$v_{n+1}=\\frac{1}{a+bv_n}$\r\n$\\iff a v_{n+1}+b v_{n+1}v_{n}=1$\r\n$\\iff a \\frac{Q_{n}}{Q_{n+1}}+b \\frac{Q_{n}}{Q_{n+1}}\\frac{Q_{n-1}}{Q_{n}}=1$\r\n$\\iff Q_{n+1}-a Q_{n}-b Q_{n-1}=0$", "Solution_2": "thanks :) :)" } { "Tag": [ "number theory solved", "number theory" ], "Problem": "Determine all positive integers n with the following property: Every natural number which is written in base 10 has n digits, in particular (n-1) Ones and one prime !", "Solution_1": "I can't exactly understand what you're asking. Could you explain?", "Solution_2": "Determine all positive integers n with the following property: Every natural number whose decimal expansion contains n digits, one digit is 7 and the other (n-1) digits are 1s, is a prime !\r\n\r\nI hope the question is clear now ! :)", "Solution_3": "I hope I interpreted the qn correctly.\r\n\r\nFirstly, the qn is equivalent to finding all n satisfying the condition that (10^n - 1)/9 + 6*10^k is prime for all k \\in {0, 1, ..., n-1}. Obviously n is \\neq 0 (mod 3), since otherwise the sum of digits of the number is = 0 (mod 3), implying the number = 0 mod 3 and so cannot be prime.\r\n\r\nConsider n \\geq 6. Note that 6*10^k are all distinct (mod 7) for k = 0, 1, ..., 5, corresponding to 1, 2, ..., 6 (mod 7) in some order, so if (10^n - 1)/9 \\neq 0 (mod 7), then there exist some k \\in {0, 1, ..., 5} s.t. (10^n - 1)/9 + 6*10^k = 0 (mod 7) and therefore not prime.\r\n\r\nThus, for all n \\geq 6 satisfying the above condition, we must have (10^n - 1)/9 = 0 (mod 7), implying n = 0 (mod 6), which is a contradiction since we have shown earlier that n \\neq 0 (mod 3).\r\n\r\nIt then suffices to check the cases for n < 6. Note that 3 divides 711, 13 divides 7111, & 17 divides 71111. Note that 7, 17 & 71 are all prime. Therefore the only possible values for n are 1 & 2." } { "Tag": [], "Problem": "I'm a beginner in C, and I got some troubles to understand this code, after running it, it shows some values of N P Q, but I don't understand how can we guess them from the code, for example, how can the first part of the code give us this ==> C : N=6 P=2 Q=1 ? normally if N=5, then after executing the program, we should have N=6 and Q=5 (since we have ++ after the N) !! I am confused, can someone explain at least 1 part of the program :help: \r\nThis is what u see after running the program :\r\nC : N=6 P=2 Q=1\r\nD : N=6 P=3 Q=1\r\nE : N=6 P=2 Q=0\r\nF : N=6 P=3 Q=1\r\nG : S S\r\nH : 83 83\r\nI : 53 53\r\n\r\n[code]#include \n#include \nmain()\n{\n int N=10, P=5, Q=10, R;\n char C='S';\n\n N = 5; P = 2;\n Q = N++ > P || P++ != 3;\n printf (\"C : N=%d P=%d Q=%d\\n\", N, P, Q);\n\n N = 5; P = 2;\n Q = N++ < P || P++ != 3;\n printf (\"D : N=%d P=%d Q=%d\\n\", N, P, Q);\n\n N = 5; P = 2;\n Q = ++N == 3 && ++P == 3;\n printf (\"E : N=%d P=%d Q=%d\\n\", N, P, Q);\n\n N=5; P=2;\n Q = ++N == 6 && ++P == 3;\n printf (\"F : N=%d P=%d Q=%d\\n\", N, P, Q);\n\n N=C;\n printf (\"G : %c %c\\n\", C, N);\n printf (\"H : %d %d\\n\", C, N);\n printf (\"I : %x %x\\n\", C, N);\n getchar();\n return 0;\n}[/code]\r\n[/hide]", "Solution_1": "[code]Q = N++ > P || P++ != 3;[/code]\nLet me add brackets to show you how the compiler understands it:\n[code]Q = (N++ > P) || (P++ != 3)[/code]\n\nIn other words, its not setting Q = N++; its setting Q to be the result of OR-ing those two bits together. First it compares N>P which is true, then it does N++ and sets N to 11. Since it already knows true || anything is true, it doesn't need to look at the second bracket, and therefore never does P++. \nSo we get N=6, P is still 2, and Q is true, ie Q=1.\n\n[code]Q = N++ < P || P++ != 3;[/code]\n\nThis time N1$ and $<\\sqrt n$. That's because if $n$ is prime we take $2$ and if $n$ isn't prime we take $p-1$, where $p$ is the smallest prime factor of $n$.\r\n\r\nEdit: \r\n\r\nLooks like I keep missing cases :): if $2|n$ then this might not work; Anyway, we know that $\\phi(n)$ must be a power of $2$, which tells us that $n$ has form $2^tp_1p_2\\ldots p_k$, with $p_i$ odd primes of the form $2^a+1$. Morevoer, we know that there are no numbers $a$ s.t. $(a,n)=1$ and $125$ that we have to check is $30$, and it doesn't work because we take $a=7$ again.\r\n\r\nThe solutions can now be found manually.", "Solution_7": "I think Harazi means that one :\r\nhttp://www.kalva.demon.co.uk/short/soln/sh00n1.html\r\n\r\nPierre.", "Solution_8": "Yes, I rushed a little bit, but it's practically the same problem. Anyway, what's with these complicated solutions? The solution given by Sasha to my problem for the TST is just perfect: smart and very short.", "Solution_9": "ab+1 divides n isn't the same as ab = 1 mod n... :D", "Solution_10": "Really? Well, Peter, when I said they are practically the same I wasn't wrong. Think a little bit: they both ask the numbers n for which any relatively prime a with n has order 2 in Z_n? Isn't this the same?", "Solution_11": "[quote=\"harazi\"]they both ask the numbers n for which any relatively prime a with n has order 2 in Z_n? Isn't this the same?[/quote]\r\n\r\nSorry hazari, I just learnt the euler phi function yesterday, orders are planned for in a couple of days, currently I got even no clue what they are :D", "Solution_12": "[quote=\"Peter VDD\"]ab+1 divides n isn't the same as ab = 1 mod n... :D[/quote]\r\n\r\nHe replaced b by -b. Then, of course, the condition \"ab + 1 is divisible by n\" becomes \"ab = 1 mod n\", and the condition \"a + b is divisible by n\" becomes \"a = b mod n\".\r\n\r\n Darij", "Solution_13": "Yes, sorry\r\n\r\ndidn't think of that since both problems state natural numbers :blush:", "Solution_14": "Sorry, of course I meant: replace b by any number equivalent to -b modulo n. This number needs not to be negative.\r\n\r\n Darij", "Solution_15": "[color=#00f]Hi, this is my first post! :welcomeani: \nI usually try to be very formal in my proofs, I hope you find it useful.\nAny flaws or misstypes that you find you can tell me, I'd like to know. Any kind of constructive feedback is welcomed![/color]\n\n[color=#f00][size=150][b][u]Solution:[/u][/b][/size][/color]\n\nLet $a \\equiv b \\pmod{n}$ and write $a = ka' + r$ and $b = kb' + r$, so $gcd(a,n)=gcd(b,n)=gcd(r,n)$. So we simplify the problem to be that if $r$ is coprime with $n$ then $r^2 \\equiv 1 \\pmod{n}$.\n\nNow we are going to bound the prime factors of $n$. So, take $p$ to be the greatest prime that divides $n$, suppose that $p \\geq 5 \\implies \\phi (p) = p-1 \\geq 4$. It means there exists three distinct positive integers $x, y, z < p$ such that $gcd(x,p)=gcd(y,p)=gcd(z,p)=1$. Now, consider $x+px'$, we know that $gcd(x+px',p)=gcd(x,p)=1$ and let $n_x = \\dfrac{n}{p^{V_p(n)}}$. By definition we have that $gcd(n_x,p)=1$ so there exists a multiplicative inverse of $p$ modulo $n_x$, call that $p^{-1}$. Now take $x' \\equiv p^{-1}(1-x) \\pmod{n_x} \\implies x+px' \\equiv 1 \\pmod{n_x}$ so $gcd(x+px',n_x)=1$ and because we have $gcd(x+px',p)=1 \\implies gcd(x+px', p^{V_p(n)})=1 \\implies gcd(x+px',n)=1$ so we have that $(x+px')^2 \\equiv 1 \\pmod{n} \\implies (x+px')^2 \\equiv x^2 \\equiv 1 \\pmod{p}$. Analogously we get that $y^2 \\equiv z^2 \\equiv 1 \\pmod{p}$, but if we look at $k^2 \\equiv 1 \\pmod{p} \\iff p|k^2-1=(k-1)(k+1)$ so $p|k-1$ or $p|k+1$, resulting in only two solutions modulo $p$, so by the [i]Pigeonhole Principle[/i] we obtain that two of $x, y, z$ are congruent modulo $p$. Let them be $x$ and $y$, then $x \\equiv y \\pmod{p} \\implies p|x-y \\implies x=y$ since $0 1$. And in fact all of those work! :D", "Solution_16": "We claim that the answer is $n = 2,4,8,3,6,12,24.$ These clearly work. Now we will show that these are the only ones that work.\n\nThe given condition is equivalent to $a^2 \\equiv 1 \\pmod{n}$ for all $a$ such that $\\gcd(a,n) = 1.$ We will split things up on which primes do not divide $n$ or not.\n\n[b]Case 1:[/b] $2 \\nmid n.$ Plugging in $a = 2$ yields $4 \\equiv 1 \\pmod{n},$ yielding $n = 3.$\n\n[b]Case 2:[/b] $2 \\mid n$ but $3 \\nmid n.$ Plugging in $a = 3$ yields $9 \\equiv 1 \\pmod{n},$ implying that $n = 2,4,8.$\n\n[b]Case 3:[/b] $2 \\mid n$ and $3 \\mid n,$ but $5 \\nmid n.$ Plugging in $a = 5$ yields $25 \\equiv 1 \\pmod{n},$ implying that $n = 6,12,24.$\n\n[b]Case 4:[/b] $2,3,5 \\mid n$ but $7 \\nmid n.$ Plugging in $a = 7$ yields $49 \\equiv 1 \\pmod{n},$ implying that $5 \\nmid n,$ contradiction.\n\n[b]Case 5:[/b] $2,3,5,7 \\mid n,$ but $11 \\nmid n.$ Plugging in $a = 11$ yields $121 \\equiv 1 \\pmod{n},$ implying that $7 \\nmid n,$ contradiction.\n\nIn general, if $p_1, p_2, \\dots, p_k$ are the first $k$ primes in order, then\n\n[b]Case $n$:[/b] $p_1, p_2, \\dots, p_{k-1} | n$ but $p_k \\nmid n.$ Plugging in $a = p_k$ yields $p_k^2 \\equiv 1 \\pmod{n},$ or $(p_k+1)(p_k-1) \\equiv 0 \\pmod{n}.$ The second-largest prime factor of $p_k - 1$ is $\\frac{p_k - 1}{2},$ and this is less than $p_{k-1}$ by Bertrand's Postulate. Hence $p_{k-1} \\nmid p_k - 1.$ Again by Bertrand's Postulate, we can deduce that $p_{k-1} \\nmid p_k + 1$. This is a contradiction.\n\nTherefore, the only $n$ that work are $n = 2,3,4,6,8,12,24.$", "Solution_17": "Not sure what I was doing during June that gave me such a weird solution.. came across this again in otis/otis excerpts somewhere\n\n\n Let $n=\\prod_{i=1}^{k}p_i^{e_i}:a^2\\equiv1\\pmod{p_i^{e_i}}\\implies a\\equiv\\{1,-1\\}\\pmod{p_i^{e_i}}$, which implies p is 2 or 3. Now, setting a=5 implies $25\\equiv1\\pmod n\\implies n\\mid24$, so all factors of 24 (except 1) are our answers. It's easy to check that these work because the square of any integer relatively prime to 8 or 3 is 1 mod 3,8. $\\blacksquare$ \n\nfixed latex", "Solution_18": "[hide=Sol]\n\nIf $a \\equiv b \\pmod{n}$, and $ab \\equiv 1 \\pmod{n}$, for all $a$, $b$ relatively prime to $n$ then we must have for any residue $r$ relatively prime to $n$, $r^2 \\equiv 1 \\pmod{n}$, implying the order is at most 2. Then the only primes dividing $n$ can be 2 or 3. \n\nClaim: Any divisor of 24 is a valid $n$. \nProof. Let $\\nu_3(n) = k$, and $\\nu_2(n) = \\ell$. It is well known that the maximal order modulo $n$, is the lcm of the maximal orders modulo its prime factors. Thus $k \\leq 3$, and $\\ell \\leq 1$, as desired. \n[/hide]", "Solution_19": "Note that the condition is equivalent to saying that for all mods $k$ relatively prime to $n$, $k^2$ is $1$ mod $n$. \\\\\\\\\nI claim that no prime $p\\ge5$ can divide into $n$. FTSOC, assume that such a $p$ divides into $n$. Now note that for any $p\\geq 5$, $2^2=4$ is not $1$ mod $p$, but rather $4$ mod $p$ since $p>4$. So if we find a number $x$ mod $n$ that is $1$ mod all other prime divisors and $2$ mod $p$ (this is always possible by CRT), then $x^2$ cannot be $1$ mod $n$ since $x^2$ is not $1$ mod $p$. \\\\\\\\\nFrom this, we have that $n$ is in the form of $2^a3^b$ for some nonnegative integers $a$ and $b$. Now I also claim that $9$ also cannot divide into $n$. This is because from a similar argument used before for $p\\ge 5$, we have that $2^2$ is not $1$ mod $9$. \\\\\\\\\nFinally, I also claim that $16$ cannot divide $n$. This is because from an argument similar to the argument used before, $3^2$ is not $1$ mod $16$, meaning that $16$ cannot divide $n$. \\\\\\\\\nTherefore this gives us that the only possible values for $n$ are $2$, $4$, $8$, $3$, $6$, $12$, and $24$. Through trial and error, we find that all of these work, therefore our solutions for $n$ are $2$, $4$, $8$, $3$, $6$, $12$, and $24$, finishing the problem.", "Solution_20": "This is the same as finding all $n\\ge 2$ such that $a^2\\equiv 1\\pmod{n}$ for all $a$ such that $\\gcd(a, n)=1$. We only need to focus on when $n$ is a power of a prime because of CRT. Plugging in $a=2$ means that all odd $x>4$ can't divide $n$. Plugging in $a=3$ gives that all $3\\not|x>9$ can't divide $n$. This means the solutions for $n$ are $2, 3, 4, 6, 8, 12, 24$.", "Solution_21": "This equivalent to showing that $a^2 \\equiv 1 \\pmod n$ for all $a$ relatively prime to $n$. By CRT, we can split this into a system of congruences of the form $a^2-1 \\equiv 0 \\pmod{p^k}$ for $a$ relaitvely prime to $p$.\n\nIf $p \\ne 2$, we need $2^2-1 \\equiv 0 \\pmod{p^k} \\implies p=3, k=0,1$.\n\nIf $p=2$, we need $3^2-1 \\equiv 0 \\pmod{2^k} \\implies k=0,1,2,3$.\n\nThus our solutions for $n$ are $\\boxed{2,3,4,6,8,12,24}$, which we can test to ensure they all work. $\\blacksquare$", "Solution_22": "Sketch:\n\nFundamentally, this is stating that when you square a relatively prime residue of $n$, we get $1 \\pmod{n}$. This reminds of us the very well-known fact that $a^2 - 1 \\equiv 0 \\pmod{24}$ if $\\gcd(a, 24) = 1$. Thus, we can easily deduce that all factors of $24$ suffice (besides $1$ as $n \\geq 2$). Now if a prime $p > 3$ existed such that $p | n$, then we know that any relatively prime to $n$ residue that is $2 \\pmod {p}$ will become $4 \\pmod {p}$ when squared which is not $1$. A similar argument holds if $9 | n$. Moreover, if $16 | n$, then we choose some residue relatively prime to $n$ that is $3 \\pmod{16}$. Squaring this would yield $9 \\pmod{16}$ which is not $1 \\pmod{16}$. ", "Solution_23": "This question is essentially finding all positive integers $n$ such that all $a$ satisfying $\\gcd(n,a) = 1$ also satisfies $a^2 \\equiv 1 \\pmod{n}$. Because of CRT, only need to find the possible prime factors of $n$. Plugging in $a=2$ shows that every odd prime greater than $3$ fails, so $n$ consists of only factors of $2$ and $3$. Then, plugging in $a=5$ shows that all factors of $24$ work except $1$. \n\nThe answer is $n = \\boxed{2,3,4,6,8,12,24}$.", "Solution_24": "This problem is equivalent to finding all $n$ so that all $a$ relatively prime to $n$ satisfy $a^2 \\equiv 1\\pmod{n}$.\nBy CRT we can just look at the separate prime divisors $p^m$ of $n$, so if $a^2 \\equiv 1 \\pmod{n}$, then $a^2 \\equiv 1 \\pmod{p^m}$ also holds.\n\nIf $p > 3$, then $a = 2$ immediately fails, so the prime divisors of $n$ are $2$ and $3$.\n$\\newline$\n\\[a^2 \\equiv 1 \\pmod{2^m}\\]\nIt's easy to check that all $m \\leq 3$ work, as plugging in $a = 3$ for $m > 3$ fails.\n$\\newline$\n\\[a^2 \\equiv 1 \\pmod{3^m}\\]\nAll $m \\leq 1$ work, as plugging in $a = 2$ for $m > 1$ fails.\n$\\newline$\nSo our answers are $n = 2, 3, 6, 8, 12, 24$(as $1$ is not included due to the problem statement.)", "Solution_25": "[hide=Sketch]\nLet $p$ be a prime such that it divides n. Then ,$a^2 \\equiv 1 (mod p), GCD(a,p ) =1$. From this it is clear that $p$ can take the values of $2,3$ only. Let $n=2^x3^y$, now we just have to check the numbers of the form of $n$ upto $25$. Therefore we get $1,2,3,4,6,8,12,24.$ :D ", "Solution_26": "[b][color=#f00][size=150]A nice and standard one.[/size][/color][/b]\nNotice that suppose we have $a \\equiv b (mod \\ n)$ and $ab \\equiv 1 (mod \\ n)$, it is very tempting to consider $a^2 \\equiv 1 (mod \\ n)$ [W.L.O.G]. Infact, reverse engineering we get that having this condition imposes both the conditions starting from either, given that $gcd(p,a)=gcd(p,b)=1$.\nWe take cases, either n is odd or it is even.\n[color=#0f0][b]Case 1: n is odd[/b][/color], $gcd(n,2)=1$ thus $2^2 \\equiv 1 (mod \\ n)$ $\\implies$ $3 \\equiv 0 (mod \\ n)$. We get $\\boxed{\\text n=1,3}$ which satisfy our conditions.\n[color=#0f0][b]Case 2: n is even[/b][/color], now we take subcase where $gcd(n,3)=1$ then $3^2-1 \\equiv 8 \\equiv 0 (mod \\ n)$ $\\implies$ $\\boxed{\\text n=2,4,8}$\nElse, $3 \\mid n$ and we take further cases $gcd(n,5)=1$ wherein we achieve, $5^2-1 \\equiv 24 \\equiv 0 (mod \\ n)$ $\\implies$ $\\boxed{\\text n=6,12,24.}$\nAgain we can assume $5|n$ which would lead to further cases for the prime being $7$ but manual checking gives no further solution.\n[b][size=150][color=#f00]Claim: The maximum prime dividing n must either be 2 or 3.[/color][/size][/b]\n[b]Proof:[/b] Consider $n=p_1^{z_1}p_2^{z_2}\\ldots p_k^{z_k}$ wherein $k \\in \\mathbb{N}$. \n[color=#f00]By FTA (Fundamental Theorem Of Arithmetic)[/color], we know that the form is finite and representable. Combined with euler's proof of existence of infinite primes. Let $P$ be the minimum prime not occuring in n greater than the maximal prime occuring in it. Thus we get $P \\nmid n$ and $P>p_k$ assuming that $p_k>p_{k-1}>\\cdots>p_1$. We consider the equation, $P^2 \\equiv 1 (mod \\ p_k)$ $\\implies$ $P \\equiv 1,(-1) (mod \\ p_k)$. Notice that we can extend the argument to $P_k$ representing the $k-th$ smallest prime greater than the maximal prime dividing n such that $P_k \\nmid n$. \nThis would imply that every prime not dividing n must be in this sequence but more importantly, all primes $> p_k$ are of the form $P_k=p_k*t \\pm 1$. Now, since the form of $n$ is finite it cannot just contain something like [color=#0f0][b]\"all primes of a dirichlet form\"[/b][/color]. We use the general statement of [b]Dirichlet's Theorem[/b] which tells us that there are infinite primes in the arithmetic progression (let us consider a particular form) say, $a_n=p_k*(n-1)+2$ and obviously all such primes cannot be in n. The only other possibility is when atleast this form corresponds to $P_k=p_k*t \\pm 1$ (we can extend to $a_n=p_k*(n-1)+3$ and so on) thus we get $2 \\equiv \\pm 1 (mod \\ p_k)$ $\\implies$ $3,1 \\equiv 0 (mod \\ p_k)$ $\\implies$ $p_k=1,3$.Notice that, $p_k=1$ doesn't make any sense whereas $p_k=3$ is our original maximum through manual work. Infact, the reason for $p_k=2$ not occuring is because we need to also take into account that $p_k*t+2 \\equiv 0 (mod \\ p_k)$ which then forces $p_k=2$ and is our edge caese. This completes the proof of claim.\n[b]Now, we use this claim combined with our original cases thus limiting us to the fact that primes starting from 5 and so on can never divide n and thus the answers we found are the only ones.[/b]\n[size=100][b]Q.E.D[/b][/size] $\\blacksquare$", "Solution_27": "Everything is trivial by CRT :) [INMOTC Karnataka insider joke]\nConsider the case where n is prime. Clearly as $a^2 \\equiv 1$ mod $n$, $a \\equiv 1, -1$ mod $n$ $\\implies n = 2$ or $3$.\nNow by CRT, every such number has to be of the form $2^x3^y$. \nWe look for the highest exponent of 2, which by trial and error is $2^3 = 8$. Similarly, the highest power of 3 is $3^1 = 3$. Therefore, the only possibilities are all factors of $2^3 \\cdot 3^1 = 24$ (except $1$ as it is excluded). Clearly all these numbers work, so we are done. $\\square$\n\nNote: for higher powers of $2$, consider the case $a = 3$, and for higher powers of $3$, consider the case $a = 2$ for contradictions.", "Solution_28": "Why is the problem equivalent to $(A)$ $x^2 \\equiv 1 \\pmod n$ for all $\\gcd(x,n) = 1$?\nWe are given that $a \\equiv b \\pmod n \\Leftrightarrow ab \\equiv 1 \\pmod n$ for $a$ and $b$ relatively prime to $n$.\nBut we used $a \\equiv b \\pmod n \\Rightarrow ab \\equiv 1 \\pmod n$ to get $(A)$?", "Solution_29": "Only if case:\nFrom the first $12$ values, get that $2,3,4,6,8$, and $12$ work.\nThen, if $n$ does not divide $2$, then $a=b=2$ results in $4 \\equiv 1 \\pmod n$, which is false.\nSimilarly, if $n$ does not divide $3$, then $a=b=3$ results in $9 \\equiv 1 \\pmod n$, which is false.\nSo $n$ has to divide $6$\nGet that $24$ also works but $18$ does not.\nIf $n \\ge 25$, then if $n$ does not divide $5$, then $a=b=5$ results in $25 \\equiv 1 \\pmod n$, which is false.\nSo $n$ has to divide $30$.\nGet that $30$ does not work and for $n \\ge 60$, $a=7$ results in it not working\nSo $n$ has to divide $420$, but then $a=11$ does not work, so it has to divide $420*11$, and then it repeats, so it does not work since $p_1*p_2*...*p_n > (p_{n+1})^2$ for large $n$ where $p_n$ is the nth prime. This can be proved by induction because it is true for $n=4$ and if $n$ works, then the LHS is multiplied by $p_{n+1}$, but the RHS is multiplied by $(\\frac{p_{n+2}}{p_{n+1}})^2 \\ge 2^2 = 4$, so the LHS will remain larger. So the only numbers that work are $\\boxed{2,3,4,6,8,12,24}$\nTesting, all of those satisfy the if case." } { "Tag": [ "trigonometry", "conics", "parabola", "quadratics" ], "Problem": "What's the greatest possible value of $ a^3b\\minus{}ab^3$ if $ a^2\\plus{}b^2\\equal{}1$?", "Solution_1": "You can set $ a \\equal{} \\cos \\theta ,\\ b \\equal{} \\sin \\theta$, so $ a^3b\\minus{}ab^3 \\equal{} ab(a^2 \\minus{} b^2) \\equal{} \\frac {1}{4}\\sin 2\\theta \\leq \\frac{1}{4}$.", "Solution_2": "[quote=\"kunny\"]You can set $ a \\equal{} \\cos \\theta ,\\ b \\equal{} \\sin \\theta$, so $ a^3b \\minus{} ab^3 \\equal{} ab(a^2 \\minus{} b^2) \\equal{} \\frac {1}{4}\\sin 2\\theta \\leq \\frac {1}{4}$.[/quote]\r\n\r\nIs any other method without using trigonometry? :maybe:", "Solution_3": "[hide=\"Solution Without Trig\"]$ S = ab(a^2 - b^2) \\implies S = ab(a + b)(a - b)$\n\n$ S^2 = a^2 b^2 (a + b)^2 (a - b)^2$\n[size=59][color=white]-[/color][/size]\n$ = a^2 b^2 (a^2 + b^2 + 2ab)(a^2 + b^2 - 2ab)$\n[size=59][color=white]-[/color][/size]\n$ = a^2 b^2 (1 + 2ab)(1 - 2ab)$\n[size=59][color=white]-[/color][/size]\n$ = a^2 b^2 (1 - 4a^2b^2)$\n[size=59][color=white]-[/color][/size]\n$ = \\frac {1}{4} (4a^2b^2)(1 - 4a^2b^2)$\n\nLet $ x = 4 a^2 b^2$:\n[size=59][color=white]-[/color][/size]\n$ \\implies 4 S^2 = x(1 - x)$\n\nThis is a parabola, and achieves a maximum value of $ 1/4$ when $ x = 1/2$.\n\nThus, $ 4S^2 \\leq \\frac {1}{4} \\implies S^2 \\leq \\frac {1}{16} \\implies S \\leq \\frac {1}{4}$\n\nNow we have to show that this is attainable. By GM-QM, we have that:\n[size=59][color=white]-[/color][/size]\n$ \\sqrt{ab} \\leq \\sqrt{\\frac{a^2+b^2}{2}} = \\sqrt{\\frac{1}{2}}$\n[size=59][color=white]-[/color][/size]\n$ \\implies ab \\leq \\frac{1}{2}$\n[size=59][color=white]-[/color][/size]\n$ \\implies 4 a^2 b^2 \\leq 1$\n[size=59][color=white]-[/color][/size]\n$ \\implies x \\leq 1$\n\nTherefore, $ x=\\frac{1}{2}$ is attainable, the maximum of $ S$ is indeed $ \\frac{1}{4}$\n[/hide]\r\n\r\n[b]Notice:[/b] Kunny had a typo in his solution; the last two steps should read $ = \\frac{1}{4} \\sin 4 \\theta \\leq \\frac{1}{4}$", "Solution_4": "[hide=\"Explanation without parabolas\"]\n\nI used TZF's solution, just if you don't understand parabolas:\n\n$ x(1\\minus{}x)\\equal{}x\\minus{}x^2\\equal{}\\minus{}(x^2\\minus{}x\\plus{}\\frac{1}{4})\\plus{}\\frac{1}{4}\\equal{}\\minus{}(x\\minus{}\\frac{1}{2})^2\\plus{}\\frac{1}{4}$\nIt's maximum for $ \\minus{}(x\\minus{}\\frac{1}{2})^2\\equal{}0$ or $ x\\minus{}\\frac{1}{2}$[/hide]", "Solution_5": "[quote=\"Bugi\"][hide=\"Explanation without parabolas\"]\n\nI used TZF's solution, just if you don't understand parabolas:\n\n$ x(1 \\minus{} x) \\equal{} x \\minus{} x^2 \\equal{} \\minus{} (x^2 \\minus{} x \\plus{} \\frac {1}{4}) \\plus{} \\frac {1}{4} \\equal{} \\minus{} (x \\minus{} \\frac {1}{2})^2 \\plus{} \\frac {1}{4}$\nIt's maximum for $ \\minus{} (x \\minus{} \\frac {1}{2})^2 \\equal{} 0$ or $ x \\minus{} \\frac {1}{2}$[/hide][/quote]\r\n\r\nI understand parabolas and quadratic equations :)\r\n\r\nThanks for all solution you've posted :)" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "solve in Z x^3 +y^3 +z^3 +t^3=3 :)", "Solution_1": "its not that hard try in mod 7 :wink:", "Solution_2": "What does $\\mod 7$ help\u00bf\r\nIt's an open problem to solve $x^{3}+y^{3}+z^{3}=3$. Note that $x=4=y, z=-5$ is a solution.", "Solution_3": "sorry zetax we can prove that x^3 +y^3 +z^3 +t^3=3 has ifinitly solutions in Z :ninja: YOU CAN TRY AGAIN", "Solution_4": "[quote=\"fermat3\"]sorry zetax we can prove that x^3 +y^3 +z^3 +t^3=3 has ifinitly solutions in Z :ninja: YOU CAN TRY AGAIN[/quote]\r\nzetax didn't say the opposite :ninja: .", "Solution_5": "hi mathlinkers i waited a another solution for my problem then i ll poste mine let x^3 +y^3 +z^3 +t^3=3 . we find infinitly many solutions in Z let a=4-24n^3 b=4+24n^3 c=-24n^2 and t=-5 and done .azbi :ninja:" } { "Tag": [ "geometry", "LaTeX", "3D geometry", "sphere" ], "Problem": "An air chamber of volume V has a neck area of cross section a into which a ball of mass m just fits and can move up and down without any friction. For figure: http://www.flickr.com/photos/30014811@N06/3249114189/\r\n\r\nShow that when the ball is pressed down a little and released, it executes SHM. Obtain an expression for the time period of oscillations assuming the pressure-volume variations of air to be isothermal.", "Solution_1": "Interesting question. I am pretty new, so I can't type in Latex (yet). So please bare with my lack of professionalism :blush: \r\n\r\nF{net} = rho g V - mg= ma\r\n\r\nv = 4/3 pi r^3\r\n\r\nso \r\n\r\nrho g 4/3 pi r^3 - mg = m r\" (??)\r\n(then solve the differential equation, and I think you can go from here..I think)\r\n\r\nI think that's how I would do it. Not sure if I am right. :(", "Solution_2": "Seems the ball holds the air inside the bottle (no air in the neck) in order to make sens the \"pressure-volume variations of air to be isothermal\".\r\nIn this case $ \\rho$ varies.", "Solution_3": "[quote=\"Immanuel Bonfils\"]Seems the ball holds the air inside the bottle (no air in the neck) in order to make sens the \"pressure-volume variations of air to be isothermal\".\nIn this case $ \\rho$ varies.[/quote]\r\n\r\nInteresting observation. I completely miss that line. :surrender: \r\n\r\nI don't see how this would contribute to the problem solving though.", "Solution_4": "You have to fix your differential equation quite a bit!\r\n\r\nThe way you put it's not linear ($ r^3$ !)\r\n\r\nAlso r, the sphere radius, is a constant; so r\" =0 !", "Solution_5": "[quote=\"MNrule\"]An air chamber of volume V has a neck area of cross section a into which a ball of mass m just fits and can move up and down without any friction. For figure: http://www.flickr.com/photos/30014811@N06/3249114189/\n\nShow that when the ball is pressed down a little and released, it executes SHM. Obtain an expression for the time period of oscillations assuming the pressure-volume variations of air to be isothermal.[/quote]\r\n[hide=\"My thoughts\"]\n\nThe differential equation describing SHM is well known:\n\\[ \\ddot{z} \\plus{}\\omega^2z \\equal{} 0\\]\n\nwhere, in this case, $ z$ is the vertical coordinate.\n\nFrom the Lagrangian for this system:\n\\[ m\\ddot{z} \\plus{} mg \\equal{} F_z\\]\n\nOne now has to determine $ F_z$.\n\\[ F_z \\equal{} F_{\\text{buoyant}} \\plus{} F_{\\text{press}} \\]\n\nIf the ball is neutrally buoyant then $ F_{\\text{buoyant}} \\equal{} mg$.\n\nUnder this assumption, $ m\\ddot{z} \\equal{} F_{\\text{press}}$.\n\nHowever there is no indication in the problem statement as to how $ F_{press}$ ought to be modeled.\n\nBut if one further lets $ F_{\\text{press}} \\equal{} p_{acoustic}A \\equal{}\\rho_{air}A^2v^2$, where $ A$ is the cross section of the ball, and $ v$ is the speed of sound in the air column. In order to solve the problem, one needs to set this equal to some spring force, i.e. $ F_{\\text{press}} \\equal{} \\minus{}kz$, and then solve for $ k$.\n\nThen $ \\omega \\equal{}\\sqrt{\\frac{k}{m}} \\equal{}\\frac{2\\pi}{T}$, where $ T$ is the oscillation period to be found.\n\nAt this point, however, I am uncertain how to make the expression $ \\rho_{air}A^2v^2\\propto z$.[/hide]", "Solution_6": "[hide=\"With the assumptions I posted earlier,\"]\nthe body's form doesn't matter; it could be a cilinder (piston) as well..\n\\[ T \\equal{} {{2\\pi}\\over a} \\sqrt {{{mV} \\over p_0}}\\]\n\nwhere $ p_0$ is the equilibrium pressure, when the volume of air in the bottle is V. \n [/hide]" } { "Tag": [ "geometry", "LaTeX" ], "Problem": "A 12 in. by 16 in. painting is surrounded by a frame of uniform width. The area of the frame equals the area of the painting. Find the width of the frame.", "Solution_1": "[hide]Let the width be $w$. $(12+2w)(16+2w)=2\\cdot 12\\cdot 16=384$. So $w=\\boxed{\\sqrt{97}-7}$[/hide]", "Solution_2": "[quote=\"linzk3\"]A 12 in. by 16 in. painting is surrounded by a frame of uniform width. The area of the frame equals the area of the painting. Find the width of the frame.[/quote]\r\n[hide]$(12+2w)(16+2w)=2(12)(16)$\n\n$4w^{2}+56w+192=384$\n\n$w^{2}+14w-48=0$\n\n$w=\\frac{-14\\pm\\sqrt{196+192}}2$\n\n$w=\\frac{-14\\pm2\\sqrt{97}}2$\n\n$w=-7+\\sqrt{97}\\approx\\boxed{2.85\\text{ inches}}$[/hide]", "Solution_3": "[quote=\"linzk3\"]A 12 in. by 16 in. painting is surrounded by a frame of uniform width. The area of the frame equals the area of the painting. Find the width of the frame.[/quote]\r\n\r\n[hide]\n$(12+x)(16+x)=12(16)(2)$\n$(12+x)(16+x)=384$\n$x^{2}+28x+192=384$\n$x^{2}+28x-192=384$\n$x=\\frac{-28+\\sqrt{1552}}{2}$\n$x=5.6977156035922094434924228298352$\n$\\frac{x}{2}=2.8488578017961047217462114149175$\ni think i did something wrong\n[/hide]", "Solution_4": "[quote=\"ProtestanT\"]$(12+x)(16+x)=12(16)(2)$[/quote]\nSould be $(12+2x)(16+2x)=2\\cdot12\\cdot16$ because the frame is on both sides.\n\n[quote=\"ProtestanT\"]$x^{2}+28x+192=384$\n$x^{2}+28x-192=384$[/quote]\r\nUmm...", "Solution_5": "[quote=\"i_like_pie\"][quote=\"ProtestanT\"]$(12+x)(16+x)=12(16)(2)$[/quote]\nSould be $(12+2x)(16+2x)=2\\cdot12\\cdot16$ because the frame is on both sides.\n\n[quote=\"ProtestanT\"]$x^{2}+28x+192=384$\n$x^{2}+28x-192=384$[/quote]\nUmm...[/quote]\r\n\r\nIt doesnt matter if its x or 2x because you can just divide by 2 in the end, which is what he did.\r\n\r\nhow did ProtestanT get two equations from 1?\r\n\r\nthe one with +192 is the right one.", "Solution_6": "[hide]A=12*16\n =192\n\n(12+2x)(16+2x)=2A\n192+56x+4x^2=384\n48+14x+x^2=96\n0=x^2+14x-48\n\nx=(-14(+-)sqrt(196+192))/2\n= 2.849 in[/hide]\r\n\r\nI'll try to learn how to use Latex soon.", "Solution_7": "[quote=\"Castor496\"]I'll try to learn how to use Latex soon.[/quote]\r\n\r\nWelcome to AoPS! Here are some sites that might help with $\\LaTeX$:\r\n\r\nhttp://tobi.oetiker.ch/lshort/lshort.pdf\r\n\r\nhttp://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php", "Solution_8": "[quote=\"i_like_pie\"][quote=\"Castor496\"]I'll try to learn how to use Latex soon.[/quote]\n\nWelcome to AoPS! Here are some sites that might help with $\\LaTeX$:\n\nhttp://tobi.oetiker.ch/lshort/lshort.pdf\n\nhttp://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php[/quote]\r\n\r\n$Thanks,$ $those$ $links$ $were/are$ $helpful.$ :)", "Solution_9": "[quote=\"Castor496\"][quote=\"i_like_pie\"][quote=\"Castor496\"]I'll try to learn how to use Latex soon.[/quote]\n\nWelcome to AoPS! Here are some sites that might help with $\\LaTeX$:\n\nhttp://tobi.oetiker.ch/lshort/lshort.pdf\n\nhttp://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php[/quote]\n\n$Thanks,$ $those$ $links$ $were/are$ $helpful.$ :)[/quote]\r\ndon't put text in latex, some people have slow computers/internet", "Solution_10": "[quote=\"i_like_pie\"][quote=\"ProtestanT\"]$(12+x)(16+x)=12(16)(2)$[/quote]\nSould be $(12+2x)(16+2x)=2\\cdot12\\cdot16$ because the frame is on both sides.\n\n[quote=\"ProtestanT\"]$x^{2}+28x+192=384$\n$x^{2}+28x-192=384$[/quote]\nUmm...[/quote]\r\nactually, the second one is supposed to be $x^{2}+28x-192=0$\r\n\r\ni subtracted 384 from both sides\r\n\r\nsry bout the typo" } { "Tag": [], "Problem": "http://www.youtube.com/watch?v=UTby_e4-Rhg", "Solution_1": "That actually was pretty good...but the people needed to stop going \"heh eheh heheh\" every five seconds.", "Solution_2": "[quote=\"asdf2345\"]That actually was pretty good...but the people needed to stop going \"heh eheh heheh\" every five seconds.[/quote]\r\n\r\n\r\nyeah, if they didn't keep saying it, it would be a lot better\r\n\r\n\r\n\r\nI thought the guy on the left was really freaky." } { "Tag": [ "integration", "limit", "real analysis", "real analysis unsolved" ], "Problem": "$\\displaystyle\\int_{R_+}{f(x) dx}<\\infty$, $\\displaystyle f(x)\\searrow 0$ as $x\\to\\infty$. Prove that $\\displaystyle \\lim_{x\\to\\infty} xf(x)=0$.", "Solution_1": "It is really well known:\r\n\r\nWe have that $lim _{x \\to \\infty} \\int_x^{2x}f(t)dt=0$, but $xf(2x)\\leq \\int_x^{2x}f(t)dt$, so $\\lim_{x \\to \\infty}xf(x)=0$" } { "Tag": [ "algebra", "function", "domain", "number theory proposed", "number theory" ], "Problem": "Let $f(z) : = \\sum_{n \\geq 0}z^{2^{n}}$ on $|z| < 1$. Prove that $f(2/3)$ is irrational. \r\n\r\n[The general result, due to Mahler, is that $f(\\alpha)$ is transcendental for every algebraic number $\\alpha \\neq 0$ in the domain $|z| < 1$ of $f(z)$; this is not much harder to prove than the special case presented above. ]", "Solution_1": "This is my approching method.\r\n\r\nIf f(2/3) is rational, Then there exist coprime integer a, b such that $f(\\frac{2}{3})=a/b$. so $a f(\\frac{2}{3})=b$.\r\nThen there exist integer n such that $2^{n}>a$\r\nBut in base-3, $f(\\frac{2}{3})=0.22020002000000020 ...._{(3)}$\r\nand $2^{n+1}>2a=a_{k}a_{k-1}...a_{0}_{(3)}$\r\nTherefore, $b=a f(\\frac{2}{3})=0. blahblahblah... a_{k}a_{k-1}...a_{0}0a_{k}...a_{0}00a_{k}..._{(3)}$ can't be integer! contradiction!\r\n\r\nSo $f(\\frac{2}{3})$ is irrational.\r\n\r\nI think that we can apply to z is rational, than f(z) is transcendental by this method. Is it true?\r\n\r\nI'm sorry, my English is poor.", "Solution_2": "It's not this simple: of course the problem is equivalent to showing that the the base-3 expansion of $f(2/3)$ is not periodic, but this expansion is not what you claim it to be! Notice that, in order to obtain the expansion, you must write each $2^{2^{n}}$ in base $3$, and then proceed to add all terms $2^{2^{n}}/3^{2^{n}}$ and make all base-$3$ carries --- and you'll see it's not at all that simple. In fact, you don't have any obvious pattern for the base-3 expansion of $f(2/3)$, so don't try to look for that expansion directly; try some other method. \r\n\r\nIf the problem were, instead, to show that $f(1/3)$ is irrational, then this would be trivial just by looking at the base-2 expansion of that number. However, for $f(2/3)$, things are not as straightforward. This is why I gave the problem with $2/3$ --- it is the simplest non-trivial case.", "Solution_3": ":oops: \r\nsorry." } { "Tag": [ "USAMTS" ], "Problem": "The Round 3 scores are now available on the USAMTS web site http://www.usamts.org.", "Solution_1": "I'm just curious, do you have an estimate as to approximately how long it will take for them to grade the Round 4 submissions?", "Solution_2": "We're hoping second week of April. Shouldn't be too much later than that." } { "Tag": [ "analytic geometry", "graphing lines", "slope" ], "Problem": "when the slope is given?\r\n\r\nAs in \"Find y if the slope of a line passing through points A (5,y) and B (2,3) is equal to 1.\"\r\n\r\nHow do you do it? Any help would be appreciated.", "Solution_1": "First the equation for slope (denoted $m$) is:\r\n\r\n\\[m=\\frac{y_2-y_1}{x_2-x_1}\\]\r\n\r\nif you are given two points $(x_1,y_1)$ and $(x_2,y_2)$. So in your example $(x_1,y_1)$ is $(5,y)$ and $(x_2,y_2)$ is $(2,3)$. \r\n\r\nCan you solve it from here?", "Solution_2": "An alternative approach.. though joml88's is faster.\r\n\r\nFor the equation of a line, y = mx +b.\r\nsubstitute m=1, and (2, 3) to find b.\r\nThen substitute (5, y) in the equation.", "Solution_3": "[quote=\"joml88\"]First the equation for slope (denoted $m$) is:\n\n\\[m=\\frac{y_2-y_1}{x_2-x_1}\\]\n\nif you are given two points $(x_1,y_1)$ and $(x_2,y_2)$. So in your example $(x_1,y_1)$ is $(5,y)$ and $(x_2,y_2)$ is $(2,3)$. \n\nCan you solve it from here?[/quote]\r\n\r\nNope. This is the other problem of the same type that I'm having problems with.\r\n\r\nFind x if the slope of a line passing through points A (-2,x) and B (4,1) is equal to -2/3.\r\n\r\nThe book I'm using doesn't explain how to do it, it just gives a solution to it.", "Solution_4": "[quote=\"cheezymoogle\"][quote=\"joml88\"]First the equation for slope (denoted $m$) is:\n\n\\[m=\\frac{y_2-y_1}{x_2-x_1}\\]\n\nif you are given two points $(x_1,y_1)$ and $(x_2,y_2)$. So in your example $(x_1,y_1)$ is $(5,y)$ and $(x_2,y_2)$ is $(2,3)$. \n\nCan you solve it from here?[/quote]\n\nNope. This is the other problem of the same type that I'm having problems with.\n\nFind x if the slope of a line passing through points A (-2,x) and B (4,1) is equal to -2/3.\n\nThe book I'm using doesn't explain how to do it, it just gives a solution to it.[/quote]\r\n\r\nWe can plug the numbers into the formula: \\[\\frac{(x)-(1)}{(-2)-4}=-\\frac{2}{3}\\]\r\n\r\n\\[\\frac{x-1}{-6}=-\\frac{2}{3}=-\\frac{4}{6}\\]\r\n\r\n\\[x-1=-4\\]\r\n\r\n\\[x=-3\\]", "Solution_5": "[quote=\"cheezymoogle\"]when the slope is given?\n\nAs in \"Find y if the slope of a line passing through points A (5,y) and B (2,3) is equal to 1.\"\n\nHow do you do it? Any help would be appreciated.[/quote]\r\n\r\nwith the definition of slope, you know that the slope 1=y/x; meaning that for every change in y, the x changes the exact same amount, and vice versa. in this problem, the x goes from 5->2, a change of -3. so y->3 must be a change in -3. \r\nsetting up the equation, you get:\r\ny-3=3\r\ny=6", "Solution_6": "Or we can simply write $y=mx+b$", "Solution_7": "[quote=\"cheezymoogle\"]when the slope is given?\n\nAs in \"Find y if the slope of a line passing through points A (5,y) and B (2,3) is equal to 1.\"\n\nHow do you do it? Any help would be appreciated.[/quote]\r\n\r\nif the slope is 1, y=x+b\r\n\r\n3=2+b\r\n\r\nb=1\r\n\r\ny=x+1\r\n\r\ny=5+1=[b]6[/b]\r\n\r\ny=mx+b(standard form), where m is the slope\r\nb is the yintercept" } { "Tag": [ "ARML", "quadratics", "conics", "ellipse", "geometry", "trapezoid", "AMC" ], "Problem": "I got a 4. (1,2,3,4). Should have gotten 123456 and maybe 7, but oh well I have four more years.", "Solution_1": "I made so many careless mistakes. Like for number 7, I set an equation and got sin(75+x)=-1, so 75+x=270, but then I forgot to subtract the 75 and get 195.", "Solution_2": "[quote=\"nat mc\"]I got a 4. (1,2,3,4). Should have gotten 123456 and maybe 7, but oh well I have four more years.[/quote]\r\n\r\nI got a 4 also. (1, 2, 5, 7) Definitely should have had 3 and maybe 4(i was using weighted averages backwards), but I don't blame myself for missing 6 or 8. I started to work on 6 after taking about 7 minutes to convince myself of my answer to number 5 (a fairly guessable question in my opinion). After looking at the solution, I saw how far I was away from the answer. I had just barely subtracted to cancel y from the equations when time was up and put down a guess. Then number 8, well, I can't say that I personally woiuld have guessed at a root in that equation, but I also didn't think to rewrite as quadratic in a which is a nice solution. Anyway, I'm not too unsatisfied and I've got a couple more years. Congrats to all. :)", "Solution_3": "I also got 4, (1,3,4, and 7).\r\nI was a little miffed cuz for 2 I had the right answer, then I was using pythagorean theorem and checking and I messed it up and got a idfferent answer and put it down and it was wrong, of course. then later I saw how you didn't even need to pythag it, oh well. And #5 too :( but whats past is past, I guess.", "Solution_4": "yeah, i got a 4 also. i realized i put the amount of positive numbers in that 5 member set (3) and so i got that wrong, and i just had no clue how to do that ellipse problem. i realized the similar triangles in the trapezoid problem with only about 30 seconds left and i put a<0 for that last problem because of descarte's rule of signs. ohh welllll! it's all good i suppose", "Solution_5": "I got a 3 (1,3,5), but I really should have had at least a five, since I made extremely dumb mistakes on #2 and #7, and #4 and #8 were actually extremely easy also...", "Solution_6": "I got 4 (1,2,5,7). Really stupid; misread #3 and spent the whole 10 minutes on it, but could've got #4 pretty easily too... *sigh* well maybe next year... I should've spent more time doing #8, too, but I kept checking #7 to make sure I was right.", "Solution_7": "almost identical situation to paladin8. 1,2,5,7. misread 3, spent most of the time on 7", "Solution_8": "I got a 4 also - 1,3,4,5. I had absolutely no idea on the circle and ellipse one. Can someone explain how to get the solution to that?", "Solution_9": "hmm I wonder who that 1 vote of 7 is... maybe Zachary Abel?\r\n\r\nAnyhow, I got a 5 (1,2,3,4,5) yes,yes, #7 was easy.... And I realized that by using the equation that I had, I could have pulled out #8, had I not stopped and just guessed :mad: which makes me very mad. Could have been up there for Site competition had I put 195 instead of 255, and colud have been up there for National competition had I put $\\frac{3}{4}$ instead of $\\frac{1}{2}$....\r\n\r\nbut I have 2 more yrs... and I tied a USAMO HM(hehe).", "Solution_10": "I could have been up for site if I hadn't made any stupid mistakes and had gotten the incredibly easy #4, and I could have been up for national - in 9th grade - if I had gotten the easy #8. But I didn't. So, don't worry about this year, just practice for next year, and get it then. ;)", "Solution_11": "Yeah, 7 was a very interesting problem. A lot of people got 255 instead because when you look at the coordinates, you realize that a line going from the two points would pass through the origin, so it would seem like you could just add 180 to 75, but they switched the sin and the cos for one of the coordinates, so instead of calculating 180+75 it would be 180+15.\r\n\r\nI used the distance formula instead (surprisingly it was quite easy to simplify). I didn't have to worry about the trap, but I ended up making an arithmetic error. Sheesh.", "Solution_12": "I got 5 (don't remember which ones). For two of them I ran out of time and correctly guessed them :) . Anyway, 5 is good and I still have four more years so I'm happy. And nat mc: 4 is great for an 8th grader. You said you should have gotten 6 or 7, but very, very few middle schoolers are at the top of the [b]nation[/b] in high school competitions. Perhaps Gabriel, Reed, or Tiankai, but they were in middle school too long ago so I wouldn't know.", "Solution_13": "3 here. (1,3,5) 7 tricked me cuz i fell into the sin-cos trap cuz i didn't read the problem closely. and suprisingly enough, i got #1 right by guessing a random prime. i think that used up all my luck for the year. :lol:", "Solution_14": "I got 1. One problem out of the whole set. And it wasn't 1 - 4 either. One problem. All your scores trump mine. Oh well, I still have a few more years... :(", "Solution_15": "[quote=\"Elemennop\"]I didn't see what was so hard about #7, either, so long as you stayed in the algebraic mode instead of going geometrically.[/quote]\r\n\r\nWhat do you mean? Almost instantly after I decided to go geometric I saw the solution...", "Solution_16": "[quote=\"joml88\"][quote=\"Elemennop\"]I didn't see what was so hard about #7, either, so long as you stayed in the algebraic mode instead of going geometrically.[/quote]\n\nWhat do you mean? Almost instantly after I decided to go geometric I saw the solution...[/quote]\r\n\r\nReally? I went algebraic and didn't hit a single snag, but later tried to do it geometrically and couldn't get it.\r\n\r\nMaybe that's just cause I'm really bad at geometry.", "Solution_17": "I think joml might mean in terms of a circle of radius 1 and a circle of radius 2, which, if they are at opposite degree measurements, are obviously 3 units apart. That's what I did.", "Solution_18": "My score is too embarrasing to talk about... but I must say the problems often do appear harder than they turned out to be. Any tips for ARML?\r\n\r\nI mean, I'm just weak at ARML. That's weird. I always never end up finishing some problem or other because I split my time wrong.", "Solution_19": "[quote=\"white_horse_king88\"]My score is too embarrasing to talk about... but I must say the problems often do appear harder than they turned out to be. Any tips for ARML?\n\nI mean, I'm just weak at ARML. That's weird. I always never end up finishing some problem or other because I split my time wrong.[/quote]\r\n\r\nSame. This year I missed both 3 and 4 because I misread 3 and spent the whole 10 minutes on it. Blah.", "Solution_20": "Does anyone know who was the highest scoring freshmen at ARML (and the score)?", "Solution_21": "Well, I got a 3. :P\r\n\r\nI think one on the Southern California A team got a 5.", "Solution_22": "We (South Carolina) had a freshman with a 5 and also an 8th grader with a 5.", "Solution_23": "[quote=\"JesusFreak197\"]I think one on the Southern California A team got a 5.[/quote]\r\nWe had two freshmen with 5's - Macky and Yolanda. I'm sure if you look nationally, you'll turn up a several 6's - there are a lot of teams, after all.", "Solution_24": "We (LV!!!) had a freshman with a 5.\r\n\r\nMe, I only managed to get 3 (1,2,5); messed up on 3 and 7 and took too long trying to do 6. (If only I could have seen the cheap calculus trick then...)", "Solution_25": "I (Central Jersey) got 4 {1, 2, 3, 5}. Could you tell me how to do number 6 w/ calculus?", "Solution_26": "6 works like this: [hide]u write ${\\bf f}(t)=<5\\cos t,4\\sin t,0>$ where $$ are the components. Then use the formula for curvature:\n$\\frac{||{\\bf f}^\\prime (t) \\times {\\bf f}^{\\prime \\prime} (t) ||}{||{\\bf f}^\\prime (t)||^3}=\\frac{20}{(16+9\\sin^2 t)^\\frac{3}{2}}$\n\nAt $t=0$ this is $\\frac{5}{16}$. Take the reciprocal to get $\\boxed{\\frac{16}{5}}$. (The circle fits here because curvature is at a maximum at $t=0$; it wouldn't work for $t=\\frac{\\pi}{2}$.)[/hide]", "Solution_27": "i love how the score distribution (for the mathlinkers/AoPSers who voted) is sort of like a bell curve!", "Solution_28": "Did anyone get both #6 and #8 correct?", "Solution_29": "I'm pretty sure a few did, but whoever did probably made a dumb mistake on another problem, as those were definitely the two hardest." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Find all positive integers that cannot be written as the sum of two composite numbers.\r\nCan anyone give me the solution of this problem? Thanks a lot. :lol:", "Solution_1": "[quote=\"Nam Luu\"]Find all positive integers that cannot be written as the sum of two composite numbers.\nCan anyone give me the solution of this problem? Thanks a lot. :lol:[/quote]\r\n\r\nI suppose we are speaking about sum of [u][b]positive [/b][/u]composite numbers.\r\n\r\nLet $ n\\ge 12$ such a number. This means that $ n \\minus{} 4$, $ n \\minus{} 6$ and $ n \\minus{} 8$ are primes(else $ n \\equal{} 4 \\plus{} (n \\minus{} 4)$, or $ n \\equal{} 6 \\plus{} (n \\minus{} 6)$ or $ n \\equal{} 8 \\plus{} (n \\minus{} 8)$ would be a decomposition in sum of two positive composite numbers)\r\nBut exactly one of the three numbers $ n \\minus{} 4$, $ n \\minus{} 6$ or $ n \\minus{} 8$ is divisible by $ 3$ and greater than $ 3$, so non prime.\r\n\r\nSo $ n < 12$ and then :\r\n$ 8 \\equal{} 4 \\plus{} 4$\r\n$ 10 \\equal{} 6 \\plus{} 4$\r\n\r\nHence the result : $ \\{1,2,3,4,5,6,7,9,11\\}$", "Solution_2": "[quote=\"pco\"]\nI suppose we are speaking about sum of [u][b]positive [/b][/u]composite numbers.\n\nLet $ n\\ge 12$ such a number. This means that $ n \\minus{} 4$, $ n \\minus{} 6$ and $ n \\minus{} 8$ are primes(else $ n \\equal{} 4 \\plus{} (n \\minus{} 4)$, or $ n \\equal{} 6 \\plus{} (n \\minus{} 6)$ or $ n \\equal{} 8 \\plus{} (n \\minus{} 8)$ would be a decomposition in sum of two positive composite numbers)\nBut exactly one of the three numbers $ n \\minus{} 4$, $ n \\minus{} 6$ or $ n \\minus{} 8$ is divisible by $ 3$ and greater than $ 3$, so non prime.\n\nSo $ n < 12$ and then :\n$ 8 \\equal{} 4 \\plus{} 4$\n$ 10 \\equal{} 6 \\plus{} 4$\n\nHence the result : $ \\{1,2,3,4,5,6,7,9,11\\}$[/quote]\r\nOf course we are talking about positive number.\r\nThanks again for the solution.\r\nThis problem can't be called \"very difficult\" :D :D :D", "Solution_3": "thanks :)" } { "Tag": [ "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "Consider a matrix $A\\in\\mathcal M_{n}(R)$ such that $A^{n}=\\alpha\\cdot A$ with $\\alpha\\neq\\{\\pm 1\\}$. Proove that $B=A+I_{n}$ is inversable.", "Solution_1": "The condition $A+I_n$ is invertible is equivalent to \"$A$ hasn't eigenvalue -1\". But from the condition $A^n=\\alpha A$ we get $\\lambda^n=\\alpha \\lambda$ which means that $\\lambda \\neq -1$ under the hypothesis.", "Solution_2": "Is this all :D ?", "Solution_3": "yes beacause $\\det(X)=\\prod x_i$ where $x_i$ are the eigenvalues of $X$." } { "Tag": [ "induction", "combinatorics proposed", "combinatorics" ], "Problem": "A nice third problem: :D\r\n\r\nThe edges of a regular 2k-gon are coloured in red and blue. At each step every edge whose adjacent edges have the same colour, turns into red and the other edges turn into blue. Prove that after 2k - 1 steps all edges will be red, and that 2k - 1 is the best possible.", "Solution_1": "[quote=\"Arne\"]A nice third problem: :D\n\nThe edges of a regular 2k-gon are coloured in red and blue. At each step every edge whose adjacent edges have the same colour, turns into red and the other edges turn into blue. Prove that after 2k - 1 steps all edges will be red, and that 2k - 1 is the best possible.[/quote]\r\n \r\n Your problem remind me one that I solved last week( a bit easier).\r\n Let A1,..An be a regular n-gon. At each of the vertices of the n-gon we have a number(real) a_1,a_2,..a_n.at each step we replace the number at Ai by a_i*a_(i+1)\r\n 1)Prove that if n=2 k then after some steps all the number\r\nat each vertice are positive.\r\n 2)prove that the statement is not always true if n<>2 k\r\nAnswer:\r\n 1)by induction we can prove that\r\n after m step the number at Ai will be\r\n a_i^(0,m)*a_(i+1)^(1,m)*...*(a_i+m)^(m,m)\r\n with (a,b)=b!/a!/(b-a)!\r\n and consider a_(n+1)=a1...\r\n if n=2 k then after 2k steps the number at Ai will be \r\n ai^2*a_(i+1)*(1,n)+...\r\n If n=2 k it's easy to prove that (1,n),(2,n)...(n-1,n) are even so the number at Ai is positive.\r\n I strongly believe that this technique can also be used to prove your problem. :D :D :D \r\n I also think that my question2 is also right for your problem\r\n :D :D :D", "Solution_2": "Something is wrong here.\r\nConsider k=2 and a square with 2^k=4 edges BBBR.\r\n\r\nAfter one step, this becomes RRBB.\r\nAfter two steps, this becomes RBRB.\r\nAfter three steps, this becomes BBBB.\r\nAfter four steps, this becomes RRRR.\r\n\r\nFour steps are not 2^(k-1) steps.", "Solution_3": "$BBBR\\rightarrow BRBR\\rightarrow RRRR$ and that is two steps." } { "Tag": [ "ARML", "LaTeX" ], "Problem": "Check it out: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=81064", "Solution_1": "That's a great idea. The only problem in Latex in AIM. I know you can use latex in GAIM, but it's hard to set up, or so I am told.", "Solution_2": "that is a good idea, here's the website that allows you to download it, after you have downloaded GAIM http://linux.softpedia.com/get/Communications/Chat/Gaim-LaTeX-3544.shtml\r\n\r\nAlthough if this doesn't work, I still think it's worthwhile doing the AIM practices anyway since not all questions require latex anyway.", "Solution_3": "Yes, but it's not easy to compile latex in GAIM. You also need to download other stuff. Scroll down on that website for details." } { "Tag": [ "limit", "real analysis", "real analysis unsolved" ], "Problem": "Please try the following problem and post the solution please.\r\n\r\na(n)=a(n)+[1+{a(n)}^2]^(1/2),a(0)=0.Find Lim a(n) as n---> infinity.\r\nThanks in advance", "Solution_1": "I think you wanted to write that $ x_{n \\plus{} 1} \\equal{} x_n \\plus{} \\sqrt {1 \\plus{} x_n^2}$ , right ? If so then we have that $ x_{n \\plus{} 1} \\minus{} x_n > 0$ so $ x_n$ is strictly increasing therefore $ \\exists \\lim_{n\\to \\infty} x_n \\equal{} l$ ($ l > 0$) .If $ l \\in \\mathbb{R}$ taking the initial relation to limit we obtain $ l \\equal{} l \\plus{} \\sqrt {1 \\plus{} l^2}$ contradiction.So the only posibility is that $ \\lim_{n \\to \\infty} x_n \\equal{} \\infty$", "Solution_2": "Alternately, $ x_1\\equal{}1$ and each term is at least twice the previous term.", "Solution_3": "An other one : $ x_{n\\plus{}1} \\geq x_n\\plus{}1 \\geq x_{n\\minus{}1}\\plus{}2 \\geq ... \\geq x_1\\plus{}n$ . Manifestly $ x_n \\rightarrow \\plus{}\\infty$\r\n :cool:" } { "Tag": [ "logarithms", "algebra unsolved", "algebra" ], "Problem": "Let $x*y$=$\\frac{x+y}{1+xy}$\r\n Calculate (((2*3)*4).....)*1995)", "Solution_1": "the given is \r\n\r\n tan(tan\u25182 + tan\u25183 + .... tan\u25181995)", "Solution_2": "the given is\r\n\\[ th(arcth(2)+arcth(3)+...+arcth(1995))=th(\\frac 12 \\ln (\\prod_{k=2}^{1995} \\frac{k+1}{k-1}))=\\frac{998*1995-1}{998*1995+1}=\\frac{1991009}{1991011} \\]", "Solution_3": "We define $X_1=0 , X_{n}= \\frac{X_{n-1}+n}{1+nX_{n-1}}$. Then :\r\n$X_{2m}=\\frac{2m^2+m+1}{2m^2+m-1} ,\\ X_{2m+1}=\\frac{2m^2+3m}{2m^2+3m-2}$ \r\n:cool:", "Solution_4": "Sorry, I made a signe mistake..:rotfl: .\r\n$X_{2m}=\\frac{2m^2+m+1}{2m^2+m-1} ,\\ X_{2m+1}=\\frac{2m^2+3m}{2m^2+3m+2}$ \r\n:cool:" } { "Tag": [ "geometry", "incenter", "circumcircle", "vector", "geometry unsolved" ], "Problem": "In triangle ABC , 100\n9(10)<100\nSo we see that 90$. By Holder,\r\n\r\n\\[\\left(\\sum_{\\text{cyc}} \\frac{1}{p^4}\\right)^{1/5} \\left(\\sum_{\\text{cyc}} p\\right)^{4/5} \\geq 4,\\]\r\n\r\nso we have\r\n\r\n\\[\\sum_{\\text{cyc}} \\frac{1}{p^4} \\geq \\frac{4^5}{8^4} = \\frac{1}{4}.\\]", "Solution_2": "HM-GM yields:\r\n\r\n\\[\r\n\\frac{4}{\\frac{1}{a^4}+\\frac{1}{b^4}+\\frac{1}{c^4}+\\frac{1}{d^4}} \\leq \\sqrt[4]{a^4b^4c^4d^4} = |a||b||c||d|\r\n\\]\r\n\\[\r\n\\frac{1}{a^4}+\\frac{1}{b^4}+\\frac{1}{c^4}+\\frac{1}{d^4} \\geq \\frac{4}{|a||b||c||d|}\r\n\\]\r\n\r\nSo we still have to prove $\\frac{4}{|a||b||c||d|} \\geq \\frac{1}{4}$ . We substitute $w:=|a|;x:=|b|;y:=|c|;z:=|d|$ with $w,x,y,z>0$.\r\nNow, AM-GM gives $ (\\frac{w+x+y+z}{4})^4 \\geq wxyz $ , but $w+x+y+z=8$, so $wxyz \\leq 16$, but resubstituting, this proves $\\frac{4}{|a||b||c||d|} \\geq \\frac{1}{4}$, so we're done.", "Solution_3": "Thank you for your replies, ThAzN1,Urs Schoenenberger.\r\n\r\nHere is the back ground of making this problem.\r\n\r\n$\\frac{1}{x^4}\\geqq -\\frac{1}{2}|x|+\\frac{17}{16}\\ (x\\neq 0)$" } { "Tag": [ "complex numbers", "complex analysis", "complex analysis unsolved" ], "Problem": "Let z_1, z_2, z_3 be complex numbers such that abs(z_1)=abs(z_2)=abs(z_3). Prove that z_1, z_2, z_3 are vertices of an equilateral triangle if and only if z_1 + z_2 + z_3 = 0.", "Solution_1": "The points $z_{i}$ divide the unit circle into three arcs. The triangle is equilateral iff all three arcs are equal (to $2\\pi/3$)." } { "Tag": [ "FTW" ], "Problem": "Who do you think and why.", "Solution_1": "Rafael Nadal FTW!!!\r\n\r\nThat's why. Cuz I said so.", "Solution_2": "I I voted voted for for Rafael nadal Rafael Nadal. I have trouble with polls too. :D", "Solution_3": "I'm always rooting for Roddick, just because he's fun to watch. And Blake, too.\r\n\r\nAs for who will win, I really don't know. Federer seems to have lost it - I mean he got eliminated from the Olympics AND Wimbledon, but he was always pretty good on the solid court." } { "Tag": [ "vector", "geometry", "3D geometry", "integration", "calculus", "calculus computations" ], "Problem": "Consider the vector field $ \\vec{F} \\equal{} 10\\vec{i} \\plus{} y\\vec{j} \\plus{} 5\\vec{k}$ .\r\n\r\nFind the flux of the right face of the cube in the first octant of the given vector field. The cube edge length is c.", "Solution_1": "The flux is\r\n$ \\Phi_f \\equal{} \\int_S \\mathbf{F} \\cdot \\mathbf{dA}$ where $ S$ is the surface you're integrating over and $ dA$ is the surface normal.\r\n\r\nIn our case, I think the surface normal is c^2 in the positive x direction.\r\n\r\nThe cube face we're looking at has $ x \\equal{} c$, $ 0 anyhow, we have a subgroup $ FT$ of $ G$ of order $ 15$, and since clearly $ P \\cap FT \\equal{} 1$, we get $ G \\equal{} P*FT$ ==> $ G$ is the semidirect product of $ P$ and $ FT$ ==> there exists a homomorphism $ FT \\to Aut(P)$, but this homomorphism has to be the trivial one (why?) ==> the above product is actually direct, and since both $ P$ and $ FT$ are abelian you're done.", "Solution_4": "[quote=\"puuhikki\"]This was posted in Ask an algebraist.\n\nLet H be the unique Sylow 17-subgroup of G; then N_G(H)/C_G(H) is isomorphic to a subgroup of Aut(H), so |N_G(H)/C_G(H)| divides both 255 and 16. Hence N_G(H) = C_(H) = G, so H is central, in which case |Z(G)| has order a multiple of 17 and is a divisor of 225. It follows |G/Z(G)| = 1, 3, 5, 15, hence G must abelian.\n\nFinally, since the order of G is a product of distinct primes, it follows from the fundamental theorem of finite abelian groups \n\nOr also: since there's a unique Sylow 17-sbgp P it is normal. There can't be both 51 Sylow 5-sbgps. and 85 Sylow 3-sbgps. (why?), so either there's unique and thus normal Sylow 5-sbgp F or 3-sbgp T ==> anyhow, we have a sbgp. FT of G of order 15, and since clearly P /\\ FT = 1, we get G = P*FT ==> G is the semidirect product of P and FT ==> there exists a homom. FT --> Aut(P), but this homom. HAS to be the trivial one (why?) ==> the above product is actually direct, and since both P and FT are abelian you're done.[/quote]\r\n\r\nThe 1st I have it .\r\n\r\nlet me check if I got the second one \r\n\r\n\"There can't be both 51 Sylow 5-sbgps. and 85 Sylow 3-sbgps. (why?)\" we will exceed the elements of $ G$.\r\n\r\n\"HAS to be the trivial one (why?) ==> the above product is actually direct\"\r\n\r\nsince |Aut(P)| is coprime with |FT| .\r\n\r\nIt seem similar to my thinking for second solution .\r\n\r\nAre my explanation true ???\r\n\r\n :)" } { "Tag": [], "Problem": "How many integers from 100 through 999, inclusive, do not\ncontain any of the digits 2, 3, 4 or 5?", "Solution_1": "There are $ 5$ choices for the first digit and $ 6$ for each of the other two. Thus, the answer is $ 5 \\times 6 \\times 6\\equal{}\\boxed{180}$." } { "Tag": [], "Problem": "A set of circles in the plane with the sum of diameters equal to 41 can be completely covered with a circle of diameter 10. Prove that there exists a line which intersects more than 4 of the circles from the given set.\r\n\r\n[i]Croatia 1993, 11th grade problem[/i]", "Solution_1": "This is a well-known idea (I would have posted this earlier, but this is the first time I see your post).\r\n\r\nWe project everything on a line, so the problem can be reduced to this: a set of collinear segments with the sum of their lengths 41 can be covered by a segment of length 10. Show that there is a point which belongs to more than 4 of the segments. This is obviously true: \r\n\r\nLet's assume every pt on the segm of length 10 belongs to at most 4 segms. We add up all the lengths of the segments, so every point appears at most 4 times, so the length can't be more than 4*10=40, but the length we get is 41, so we have a contradiction." } { "Tag": [ "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Hello,\r\n \r\nI have to proof that:\r\n\r\n$ \\begin{gathered}\r\n {\\bigwedge ^m}\\left( \\mathbb{Q} \\right) \\equal{} 0{\\text{ (with }}m > 1) \\hfill \\\\\r\n \\mathbb{Q} {\\text{ should be regarded as a }}\\mathbb{Z}{\\text{ \\minus{} module}} \\hfill \\\\ \r\n\\end{gathered}$\r\n\r\nHas anybody a hint for me?", "Solution_1": "Well, you have to show that any element $ \\frac{p_1}{q_1} \\wedge \\frac{p_2}{q_2}$ for $ \\frac{p_1}{q_1}, \\frac{p_2}{q_2} \\in \\mathbb{Q}$ is actually zero. How can you show that this is zero?\r\n\r\n[hide=\"Hint\"] Remember that any element of the form $ r \\wedge r$ is zero. Use the multilinearity over $ \\mathbb{Z}$ of the wedge product.[/hide]", "Solution_2": "Hi,\r\n\r\nI know this calculation rules of the wedge prodcut:\r\n\r\n$ \\frac{{{p_1}}}\r\n{{{q_1}}},\\frac{{{p_2}}}\r\n{{{q_2}}} \\in \\mathbb{Q} \\hfill \\\\\r\n \\frac{{{p_1}}}\r\n{{{q_1}}} \\wedge \\frac{{{p_2}}}\r\n{{{q_2}}} \\equal{} \\left( {\\frac{{{p_1}}}\r\n{{{q_1}}} \\plus{} \\frac{{{p_2}}}\r\n{{{q_2}}}} \\right) \\wedge \\frac{{{p_2}}}\r\n{{{q_2}}} \\equal{} \\frac{{{p_1}}}\r\n{{{q_1}}} \\wedge \\left( {\\frac{{{p_2}}}\r\n{{{q_2}}} \\plus{} \\frac{{{p_1}}}\r\n{{{q_1}}}} \\right) \\hfill \\\\\r\n r \\in \\mathbb{Z} \\hfill \\\\\r\n r\\left( {\\frac{{{p_1}}}\r\n{{{q_1}}} \\wedge \\frac{{{p_2}}}\r\n{{{q_2}}}} \\right) \\equal{} \\frac{{r \\cdot {p_1}}}\r\n{{{q_1}}} \\wedge \\frac{{{p_2}}}\r\n{{{q_2}}} \\equal{} \\frac{{{p_1}}}\r\n{{{q_1}}} \\wedge \\frac{{r \\cdot {p_2}}}\r\n{{{q_2}}} \\hfill $\r\n\r\nBut unfortunately I still couldn't show that\r\n$ {\\bigwedge ^m}(\\mathbb{Q}) \\equal{} 0$\r\n\r\nIt would be nice, if anybody would help me again :) \r\n\r\nBye\r\nEdgar", "Solution_3": "[quote=\"edgar01\"]Hi,\n\nI know this calculation rules of the wedge prodcut:\n\n$ \\frac {{{p_1}}} {{{q_1}}},\\frac {{{p_2}}} {{{q_2}}} \\in \\mathbb{Q} \\hfill \\\\\n\\frac {{{p_1}}} {{{q_1}}} \\wedge \\frac {{{p_2}}} {{{q_2}}} \\equal{} \\left( {\\frac {{{p_1}}} {{{q_1}}} \\plus{} \\frac {{{p_2}}} {{{q_2}}}} \\right) \\wedge \\frac {{{p_2}}} {{{q_2}}} \\equal{} \\frac {{{p_1}}} {{{q_1}}} \\wedge \\left( {\\frac {{{p_2}}} {{{q_2}}} \\plus{} \\frac {{{p_1}}} {{{q_1}}}} \\right) \\hfill \\\\\nr \\in \\mathbb{Z} \\hfill \\\\\nr\\left( {\\frac {{{p_1}}} {{{q_1}}} \\wedge \\frac {{{p_2}}} {{{q_2}}}} \\right) \\equal{} \\frac {{r \\cdot {p_1}}} {{{q_1}}} \\wedge \\frac {{{p_2}}} {{{q_2}}} \\equal{} \\frac {{{p_1}}} {{{q_1}}} \\wedge \\frac {{r \\cdot {p_2}}} {{{q_2}}} \\hfill$\n\nBut unfortunately I still couldn't show that\n$ {\\bigwedge ^m}(\\mathbb{Q}) \\equal{} 0$\n\nIt would be nice, if anybody would help me again :) \n\nBye\nEdgar[/quote]\r\n\r\n$ \\frac{p_1}{q_1} \\wedge \\frac{p_2}{q_2} \\equal{} p_1p_2q_1q_2 \\left(\\frac{1}{q_1q_2} \\wedge \\frac{1}{q_1q_2} \\right).$", "Solution_4": "Oh, thank you for your great help, ysharifi." } { "Tag": [ "symmetry", "AMC", "AIME" ], "Problem": "Let $n=2^{31}3^{19}.$ How many positive integer divisors of $n^2$ are less than $n$ but do not divide $n$?", "Solution_1": "[hide]\nour final answer is the total number of divisors of $n^2$ which are less than $n$, minues the factors of $n$(which are obviously factors of $n^2$).\n$n^2=(2^{31}3^{19})^2=2^{62}3^{38}$ thus $n^2$ has 63*39 factors, since we have a perfect square it will have an odd number of divisors, thus the product of pairs of factors taken from the from the first and last elements of factors, will result in $n^2$, looking at symmetry since each pair of factors we see that there are $\\frac{63*39-1}{2}=1228$ factors of $n^2$ less than $n$.\n$n$ itself has 639 factors (other than itself) so the desired number is: $1228-639=\\boxed{589}$ [/hide]", "Solution_2": "Sorry to revive, but I'm trying to study for AIMEs and I'm wondering: how did you find out how many factors of $n^2$ are less than $n$? I didn't get the explanation.", "Solution_3": "If $d$ is a divisor of $n^2$, then so is $\\frac{n^2}{d}$. That means that after we subtract $1$ (to account for $n$), half of the remaining divisors are less than $n$.", "Solution_4": "31 x 19 = 589", "Solution_5": "First we will find the amount of factors of $n^2$ that are less than $n$. $n^2$ will have $63 \\cdot 39$ total factors so out of those $\\frac{63 \\cdot 39-1}{2}$ are less than $n$, which is equal to $1228$. Out of these we subtract $32 \\cdot 20 = 640$ to get $588$, but we add back one for $n$ itself to give $\\boxed{589}$.", "Solution_6": "n^2 has $63\\cdot39=2457$, now since we have an odd number of factors, $\\frac{2457-1}{2}$ are less than n so we have $1228$, now we subtract the number of factors of $n$ which is 640, so 1228-640+1=$\\boxed{589}$", "Solution_7": "$n^2$ has $63 \\cdot 39$ total factors. Out of those, $\\frac{63 \\cdot 39-1}{2}$ are less than $n$ and after subtracting the number of factors of $n,$ which is $32 \\cdot 20$ and adding back one for $n,$ we get our answer of $\\boxed{589}.$" } { "Tag": [ "geometry", "rectangle", "ratio", "3D geometry", "calculus", "integration", "inequalities" ], "Problem": "1. A while ago in a laboratory in an african country, a dangerous bacteria named \"Ishangololo\" was made by an accident. After maturity, this bacteria splits into two bacterias like itself. The age of maturity might not be the same in all bacterias. We say that the bacteria A is a child of the bacteria B if A has came into existence by the division of the bacteria B of by the division of one of B's children. Right now, we have 3k bacterias in the lab. Prove that there is a bacteria whose number of children among the current bacterias is a number greater or equal than k and smaller or equal than 2k. \r\n\r\n2. Around a circle, we have written n numbers each either 0 or 1 and one of these n numbers is underlined. In each round, we can do one of the below : \r\ni) Remove the line below the underlined number and underline one of the numbers next to it. \r\nii) Change the underlined number from 0 to 1 and vice versa. \r\nFind the least number of rounds needed so that we can change any first condition to another. \r\n\r\n3. A 8x8 square on the plane divided into 64 squares 1x1must be covered by 64 black and 64 white isosceles triangles (each square must be covered by two triangles). A covering is \"fine\" if any two neighboring (having a common side) triangles are of different colors. How many different \"fine\" colorings are there? \r\n\r\n4. A journalist is looking for a person named Z at a meeting of N persons. He has been told that Z knows all the other people at the meeting but none of them knows Z. The journalist may ask any person X : 'Do you know that man?' about any Y. \r\na) Can he be certain to find Z by asking less than N questions? \r\nb) What minimal number of questions are needed to find Z? (Prove that a lesser number won't work.) \r\n\r\n5. A rectangle axb (a>b) is cut into right triangles so that any common side of two triangles is a leg (cathetus) for one of them and a hypotenuse for the other. (Every two triangles have a common side or a common vertex or no common points.) Prove that the ratio a/b is not less than 2. \r\n\r\n6. 8 students were asked to solve 8 problems. (the same ones) \r\na) Each problem was solved by 5 students. Prove that one can find two students so that each problem was solved by at least one of them. \r\nb) If each problem was solved by 4 students, then it is possible that no such a pair of students exists. Prove this. \r\n\r\n7. The integers 1,2, ... , n each appear n times in an nxn matrix. Show that there is a row or column containing at least n1/2 distinct integers. \r\n\r\n8. There is a race car on a circular racetrack of n miles. There are fuel stations around the track, such that the total amount of fuel at all the stations is just enough to travel n miles. Show that there is a point on the track at which the race car may start, such that it can complete a circuit by picking up fuel along the way. \r\n\r\n9. Define a sequence of strings as follows: Let R0 = 1 , and let Rn be the string obtained by replacing each instance of 'i' in Rn-1 with '123..i' and finally adding 'n+1' at the end. So for instance, R1 = 12 , R2 = 1123 , R3 = 11121234 . Show that for n>0 , if we write Rn down and Rn backwards under that, then each column will contain exactly one '1' . \r\n\r\n10. Let m and n be positive integers. Let x1 , x2, ... , xn be non negative integers, whose average is less than or equal to m+1, and let y1 , y2, ... , yn be non negative integers, whose average is less than or equal to n+1. Show that some non empty subset of the xi's has sum equal to some non empty subset of the yj's. \r\n\r\n11. 8 singers take part in an art festival. The organizer wants to plan m concerts. 4 singers go on stage in every concert. Restrict that the times of which every two singers go on stage in a concert are all the same. What is the minimal of m? (Make a design.) \r\n\r\n12. Between 2n-1 positive integers, prove that we can choose n of them in which the sum of these n integers is divisible by n. \r\n\r\n13. On the island of Camelot, live 13 gray, 15 brown and 17 crimson chameleons. If two chameleons of different colors meet, they both simultaneously change color to the third color. Is it possible that they will eventually all be the same color? \r\n\r\n14. The numbers x1 , x2 , ... , xn are such that x1 + x2 + ... + xn = 0 and x12 + x22 + ... + xn2 = 1 . Prove that there are two numbers among them whose product is no greater than -1/n. \r\n\r\n15. We say that some positive integer m \"covers\" the number 1998, if 1, 9, 9, 8 appear in this order as digits of m. (For instance 1998 is covered by 215993698 but not by 213326798.) Let k(n) be the number of positive integers that cover 1998 and have exactly n digits (n>4), all different from 0. What is the remainder of k(n) in division by 8? \r\n\r\n16. The cube S with edge 2 consists of 8 unit cubes. We will call the cube S with one unit cube removed, a \"piece\". The cube T with edge length 2n consists of (2n)3 unit cubes. Prove that if one unit cube is removed from T, then the remaining solid can be built with pieces. \r\n\r\n17. Peter has three accounts in a bank, each with an integral number of dollars. He is only allowed to transfer money from one account to another so that the amount of money in the latter is doubled. \r\na) Prove that Peter can always transfer his money into two accounts. \r\nb) Can Peter always transfer his money into one account? \r\n\r\n18. On a straight line, n red points and n blue points (not necessarily distinct) are marked arbitrarily. Prove that the sum of the distances between every two points of different colors is not less than the sum of the distances between every two points of the same color. \r\n\r\n19. A 7x7 square is made up of 16 '1x3' tiles and 1 '1x1' tile. Prove that the 1x1 tile lies either at the center of the square or adjoins one of its boundaries. \r\n\r\n20. At a round table are 1994 girls playing a deck of n cards. Initially, one girl holds all of the cards. In each turn, if at least one girl holds at least two cards, one of these girls must pass a card to each of her two neighbors. The game ends when and only when each girl is holding at most one card. \r\na) Prove that if n > 1993, then the game cannot end. \r\nb) Prove that if n < 1994, then the game must end.", "Solution_1": "[hide=\"1\"] \nThe problem statement is equivalent to the following: Given a full binary tree with $ 3n$ leaves, show there exists a full binary subtree with at most $ 2n$ leaves and at least $ n$ leaves.\n\nThe proof of this is simple. Take the largest full proper subtree of our tree. Clearly it has at least half of the leaves. If the number of leaves is too large, take the largest full proper subtree of that. Keep doing this until we are in the desired interval. Because each subsequent subtree has at least half of the leaves of the tree before it, we cannot skip through the interval. By the Well Ordering Principle, we must be in the interval at some point.[/hide]\n\n[hide=\"2\"] Uh...$ 2n\\minus{}1$? Assuming the places of the numbers matters.[/hide]\n\n[hide=\"Idea on 3\"]\nDepending on the orientation of the diagonal separating the two isoceles triangles in a 1 x 1 square, assign that square either 0 or 1. The sum of the numbers in a 2 x 2 square must be even. So we can calculate the number of ways to number the squares of a chessboard with 0s and 1s so that the sum of the numbers in any 2 x 2 square is even, and then multiply by two (because black and white are interchangeable).[/hide]\n\n[hide=\"19\"]\nNumber the square in two ways as follows:\n\n1 2 3 1 2 3 1 -1-2-3-1-2-3-1\n2 3 1 2 3 1 2 -2-3-1-2-3-1-2\n3 1 2 3 1 2 3 -3-1-2-3-1-2-3\n1 2 3 1 2 3 1 -1-2-3-1-2-3-1\n2 3 1 2 3 1 2 -2-3-1-2-3-1-2\n3 1 2 3 1 2 3 -3-1-2-3-1-2-3\n1 2 3 1 2 3 1 -1-2-3-1-2-3-1\n\nEvery 1 x 3 tile covers one of each kind of number. But there are 16 2's, 16 3's, and 17 1's. Therefore, the 1 x 1 tile is on a 1. Similarly, the 1 x 1 tile is on a -1. This yields the desired result.\n\nThis double numbering trick is very useful when you have a symmetric shape.[/hide]", "Solution_2": "[b]11.[/b] Every singer sings with 3k other singers in total. But that number is divisible by 7 because he sings with each of the remaining 7 singers equal number of times. Thus there are minimal 7 concerts in which one singer sings. That gives 56 concerts in total, but every concert is sung by 4 of them so the real number of concerts is minimal 14. example for 14: take singers a,b,c,d,e,f,g,h and concerts as follows: abcd, abef, abgh, aceg, acfh, adeh, adfg, bcef, bcgh, bdeg, bdfh, cdeh, cdfg, efgh.\r\n\r\n[b]13.[/b] Let a, b, c be numbers of gray, brown and crimson chameleons. a, b, c will always make the complete system of residues modulo 3 (easy to check) so the statement is impossible.", "Solution_3": "What do you mean in Problem 7?\r\n\r\nI claim that you mean:\r\n\r\n7. The integers $ 1,2,\\dots, n$ each appear $ n$ times in an $ n\\times{n}$ matrix. Show that there is a row or column containing at least $ \\sqrt{n}$ distinct integers.\r\n\r\nYou can give counterexamples for $ \\frac{n}{2}$, $ n\\plus{}\\frac{1}{2}$, and $ \\frac{n\\plus{}1}{2}$. So I believe you mistyped it and it should be $ n^{1/2}$.\r\n\r\n[hide=\"Solution\"]\nLet $ r_{i}$ and $ c_{i}$ be the $ i$-th row and the $ i$-th column for each $ i \\equal{} 1,2,\\dots,n$. Denoting by $ R$ and $ C$ are the sets $ R: \\equal{}\\{r_{1},r_{2},\\dots,r_{n}\\}$ and $ C: \\equal{}\\{c_{1},c_{2},\\dots,c_{n}\\}$. Let $ G$ be a bipartite graph with two sets of vertices $ V_{1}\\equal{} R\\cup{C}$ and $ V_{2}\\equal{}\\{1,2,\\dots,n\\}$. A vertex $ v\\in V_{1}$ is connected to a vertex $ k$ in $ V_{2}$ iff $ k$ belongs to the row or the column for which $ v$ is the representative. (You actually don't need to use Graph Theory. It is essentially the Double Counting Technique.)\n\nSuppose for each $ k$, the number of the neighbors of $ k$ in $ R$ is $ x_{k}$ and the number of the neighbors of $ k$ in $ C$ is $ y_{k}$. Therefore, $ x_{k}y_{k}\\geq n$ (because each integer occurs $ n$ times). Using the AM-GM Inequality, one obtains $ \\deg(k) \\equal{} x_{k}\\plus{}y_{k}\\geq 2\\sqrt{x_{k}y_{k}}\\geq 2\\sqrt{n}$. Thus,\n\\[ \\sum_{v\\in{V_{1}}}\\deg{v}\\equal{}\\sum_{k\\in{V_{2}}}\\deg(k)\\geq 2n\\sqrt{n}\\,.\\]\nHence, $ \\frac{1}{2n}\\sum_{v\\in{V_{1}}}\\deg{v}\\geq\\sqrt{n}$ and according to the Pigeonhole Principle, there must exist a row or a column with more than $ \\sqrt{n}$ distinct numbers.\n\n[/hide]", "Solution_4": "19. A $ 7\\times7$ square is made up of $ 16$ '$ 1\\times3$' tiles and $ 1$ '$ 1\\times1$' tile. Prove that the $ 1\\times{1}$ tile lies either at the center of the square or adjoins one of its boundaries.\r\n\r\n[hide=\"Solution Using Generating Function\"]\nLet $ P(x,y) \\equal{} (1\\plus{}x\\plus{}x^{2}\\plus{}\\cdots\\plus{}x^{6})(1\\plus{}y\\plus{}y^{2}\\plus{}\\cdots\\plus{}y^{6})$. The term $ x^{i}y^{j}$ represents the square at the intersection of the $ (i\\plus{}1)$-st row and $ (j\\plus{}1)$-st column. Suppose that we can cover the $ 7\\times7$ board with sixteen $ 1\\times3$ tiles and one $ 1\\times1$ tile. If the $ 1\\times1$ tile is situated at the $ (a\\plus{}1)$-st row and the $ (b\\plus{}1)$-st column, then,\n\\[ P(x,y) \\equal{} x^{a}y^{b}\\plus{}(1\\plus{}x\\plus{}x^{2})Q(x,y)\\plus{}(1\\plus{}y\\plus{}y^{2})R(x,y)\\,,\\]\nfor some polynomial $ Q$ and $ R$.\n\nLet $ \\omega \\equal{}\\frac{\\minus{}1\\plus{}\\sqrt{\\minus{}3}}{2}$ be the primitive cube root of unity. Thus,\n\\[ P(\\omega,\\omega) \\equal{}\\omega^{a\\plus{}b}\\,.\\]\nOn the other hand,\n\\[ P(\\omega,\\omega) \\equal{}\\left(1\\plus{}(\\omega\\plus{}\\omega^{4})(1\\plus{}\\omega\\plus{}\\omega^{2})\\right)\\left(1\\plus{}(\\omega\\plus{}\\omega^{4})(1\\plus{}\\omega\\plus{}\\omega^{2})\\right)\\equal{}1\\,.\\]\nHence, $ \\omega^{a\\plus{}b}\\equal{} 1$ or $ a\\plus{}b$ is a multiple of $ 3$.\n\nSimilarly, $ \\omega^{a\\minus{}b}\\equal{} P\\left(\\omega,\\frac{1}{\\omega}\\right) \\equal{} 1$. Thus, $ 3$ divides $ a\\minus{}b$. Consequently, $ a$ and $ b$ are multiples of $ 3$ and hence, the $ 1\\times1$ tile is at the square $ (1,1)$, $ (1,4)$, $ (1,7)$, $ (4,1)$, $ (4,4)$, $ (4,7)$, $ (7,1)$, $ (7,4)$, or $ (7,7)$. Hence, it is either on the boundary or at the center of the board.\n[/hide]", "Solution_5": "Are you sure about the 16th problem?\r\n\r\n\r\n16. The cube $ S$ with edge $ 2$ consists of $ 8$ unit cubes. We will call the cube $ S$ with one unit cube removed, a \"piece\". The cube $ T$ with edge length $ 2n$ consists of $ (2n)^{3}$ unit cubes. Prove that if one unit cube is removed from $ T$, then the remaining solid can be built with pieces.\r\n\r\nIt is obviously not true when $ n\\equiv 0,3,5,6 (\\mathrm{mod}\\,7)$. (For $ (2n)^{3}\\equiv\\minus{}1 (\\mathrm{mod}\\,7)$ for these numbers.)\r\n\r\nI think the problem is:\r\n16. The cube $ S$ with edge $ 2$ consists of $ 8$ unit cubes. We will call the cube $ S$ with one unit cube removed, a \"piece\". The cube $ T$ with edge length $ 2^{n}$ consists of $ (2^{n})^{3}$ unit cubes. Prove that if one unit cube is removed from $ T$, then the remaining solid can be built with pieces.\r\n\r\nIf this is the case, then the solution is quite simple. Just induct on $ n$.", "Solution_6": "[quote=\"Batominovski\"]Are you sure about the 16th problem?\n\n\n16. The cube $ S$ with edge $ 2$ consists of $ 8$ unit cubes. We will call the cube $ S$ with one unit cube removed, a \"piece\". The cube $ T$ with edge length $ 2n$ consists of $ (2n)^{3}$ unit cubes. Prove that if one unit cube is removed from $ T$, then the remaining solid can be built with pieces.\n\nIt is obviously not true when $ n\\equiv 0,3,5,6 (\\mathrm{mod}\\,7)$. (For $ (2n)^{3}\\equiv\\minus{}1 (\\mathrm{mod}\\,7)$ for these numbers.)\n\nI think the problem is:\n16. The cube $ S$ with edge $ 2$ consists of $ 8$ unit cubes. We will call the cube $ S$ with one unit cube removed, a \"piece\". The cube $ T$ with edge length $ 2^{n}$ consists of $ (2^{n})^{3}$ unit cubes. Prove that if one unit cube is removed from $ T$, then the remaining solid can be built with pieces.\n\nIf this is the case, then the solution is quite simple. Just induct on $ n$.[/quote]\r\n\r\nI assumed the question was \"Find all $ n$ for which this is possible\"", "Solution_7": "12. is the Erdos-Grinburg-Ziv theorem :wink:", "Solution_8": "12.\r\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=159531[/url]", "Solution_9": "18 is leningrad some years ago,I'll send the exact year and the solution later..." } { "Tag": [ "function", "calculus", "calculus computations" ], "Problem": "Let $ f$ and $ g$ be functions that are differentiable on the interval $ (a,b)$. Let $ h(x) \\equal{} \\begin{cases} f(x) ;& a < x < c \\\\ g(x);& c \\leq x< b \\end{cases}$. Show that if $ f(c) \\equal{} g(c)$ and $ f'(c) \\equal{} g'(c)$ then $ h$ is differentiable at $ c$.", "Solution_1": "Use definition." } { "Tag": [ "function", "calculus", "trigonometry", "algebra", "domain", "absolute value", "inequalities unsolved" ], "Problem": "$a^{2}+b^{2}= 1, |a+\\frac{a}{b}+b+\\frac{b}{a}| >/= 2-\\sqrt{2}$\r\n\r\nI tried letting a=sin x and b= cos x, and tried to find the global minimum of the function, but the absolute value signs make things harder...Anyway, there is meant to be a purely algebraic method with no calculus involved. Help please. The >/= is meant to be more or equal to btw.", "Solution_1": "A hint : Put $y=\\sin x+\\cos x$, we have $\\frac{\\cos{x}}{\\sin{x}}+\\frac{\\sin{x}}{\\cos{x}}=\\frac{1}{\\sin{x}\\cos{x}}=\\frac{2}{y^{2}-1}$.", "Solution_2": "$| y+\\frac{2}{y^{2}-1}|$ So what should I do now? The Absolute value signs are still annoying when finding the global minimum >.<", "Solution_3": "Now you can use real analysis with $f(y)=y+\\frac{2}{y^{2}-1}$ on $[-\\sqrt{2},\\sqrt{2}]\\setminus \\{-1,1\\}$. :wink:", "Solution_4": "I have no idea what you mean by real analysis, or why you choose that domain or range. Sorry im new at this", "Solution_5": "[quote=\"Gib Z\"]I have no idea what you mean by real analysis[/quote]\n$f'(y)=\\frac{y^{4}-2y^{2}-4y+1}{(y^{2}-1)^{2}},...$,\n\n[quote] or why you choose that domain or range.[/quote]\r\nBecause $y=\\sqrt{2}\\cdot\\sin (x+\\frac{\\pi}{4})$ and $y^{2}-1\\not = 0$.\r\nNote: I see that the equation $f'(y)=0$ isn't easy! Sorry :blush:", "Solution_6": "Ahh perhaps I should start over, this is meant to be a precalc question anyway, it an elementary solution.." } { "Tag": [], "Problem": "Another math question for whoever wants to answer it for me. Thank you.\r\n\r\n\r\n The speed of light is 299 793 km/s. If it takes 8 mins. for light to reach the earth from the sun, which of the following is the best estimate of the distance (in kilometres) between the earth and the sun?\r\n\r\nA. 300 000 x 8 x 60\r\n\r\nB. 300 000 x 8\r\n\r\nC. [u]300 000[/u]\r\n x 60\r\n 8\r\n\r\n\r\nD. [u]300 000[/u]\r\n 8\r\n\r\n\r\nE. [u]300 000[/u]\r\n x 8\r\n 60", "Solution_1": "so approximately light goes 300000 km in a second.\r\n\r\n8 minutes = 8*60 seconds\r\n\r\nso the answer would be A (300000*8*60)" } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "Find $n$ such that the sum of the first $n$ positive integers is a three digit number having equal digits.\r\n\r\nNo solution required! The problem was proposed by Professor Tiberiu Roman, who had the idea of an international math competition. He was the one who invented the IMO. The first two IMO's, as everybody knows, were organized by Romania in 1959 and 1960.\r\nI felt that prof. Roman's name has to be mentioned somewhere in this site.", "Solution_1": "[quote=\"enescu\"]\nI felt that prof. Roman's name has to be mentioned somewhere in this site.[/quote]\r\n\r\n :) \r\n\r\nPierre.", "Solution_2": "Is Professor Tiberiu Roman still alive ? Tell us more about his life, connections to mathematical olympiads and how the idea of an first international mathematical olympiad evolved in Romania. :)", "Solution_3": "Well, unfortunately I don't have detailed information about this. As far as I know, prof. Roman was involved in math olympiads until late 60's. After that he retired. Nevertheless, in 1999, when the IMO was held in Romania, prof. Roman was invited to the opening ceremony and got a big hand from all the participants. I was there and I reckon I was deeply touched. I will look for further information and keep you posted." } { "Tag": [ "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "Let $ A\\in M_{n\\times m}(F),B\\in M_{m\\times n}(F)$,where $ F$ is a field.If $ AB$ is Idempotent.(Which means $ (AB)^{2}\\equal{} AB$),and $ rank(AB) \\equal{} rank(BA)$.\r\nProve:$ BA$ is Idempotent.", "Solution_1": "$ spectrum(AB)\\equal{}\\{0_{1},\\cdots,0_{p},1_{1},\\cdots,1_{q}\\}$ with $ dim(ker(AB))\\equal{}p,dim(ker(AB\\minus{}I))\\equal{}q.$\r\n$ det(BA\\minus{}\\lambda{I})\\equal{}\\pm{det(AB\\minus{}\\lambda{I})\\lambda^{m\\minus{}n}}$, thus $ spectrum(BA)\\equal{}\\{0_{1},\\cdots,0_{p\\plus{}m\\minus{}n},1_{1},\\cdots,1_{q}\\}.$\r\n$ q\\equal{}rank(AB)\\equal{}rank(BA)$ thus $ dim(ker(BA))\\equal{}m\\minus{}q\\equal{}m\\minus{}n\\plus{}p$.\r\nMoreover: $ dim(ker(BA\\minus{}I))\\equal{}dim(ker(AB\\minus{}I))\\equal{}q$ (cf:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?p=894873#894873 ).\r\nConclusion: $ BA$ is diagonalizable; thus it's a projector. QED.", "Solution_2": "Thank you,loup blanc.\r\nMy proof is completely different from you.\r\nBy $ (AB)^{2}\\equal{}AB$,we can easily get $ rank(BAB)\\equal{}rank(AB)\\equal{}rank(BA)$\r\nthus there exist a matrix $ C$ such that $ BABC\\equal{}BA$\r\nMultiply $ A$ both side,we have: $ ABABC\\equal{}ABA\\equal{}ABC$\r\nThus $ BA\\equal{}B(ABC)\\equal{}B(ABA)$" } { "Tag": [ "blogs", "\\/closed" ], "Problem": "not allowing me to change my blog background. =(", "Solution_1": "What happens when you try?", "Solution_2": "Sometimes, it doesn't come right away, so try hitting CTRL+F5 when viewing your blog. I had the same problem and ctrl+F5 usually fixes it." } { "Tag": [ "geometry", "rectangle", "analytic geometry", "trapezoid" ], "Problem": "Let $ABCD$ be a rectangle with center $O.$ Let $N\\in (AO),$ let $M$ be the midpoint of $[AD]$ and let $P= MN \\cap CD,\\, E= OP\\cap BC.$ Prove that $NE\\perp BC.$", "Solution_1": "[hide]Easy with coordinate geometry... though I'm sure there has to be a better solution. If nobody comes up with something I'll post my solution.[/hide]", "Solution_2": "[hide]Let $PM\\cap AB=Q$, $PO\\cap AB=F$.\nSince $AB\\parallel PD$, $\\triangle AQM \\sim\\triangle DPM$, and $AM = MD\\Longrightarrow AQ = PD$. $\\triangle POD\\sim\\triangle FOB$, and $DO = OB\\Longrightarrow PD=BF$. Then $AQ = BF$.\n$\\triangle AQN \\sim\\triangle CPN$ and $\\triangle BFE \\sim\\triangle CPE$. $PC = PC$ and $AQ=BF$, so $\\frac{AN}{NC}=\\frac{BE}{EC}$, which means that $NE\\parallel AB\\Longrightarrow NE\\perp BC$.[/hide]", "Solution_3": "Wrong message !", "Solution_4": "[quote][color=darkred][b][u]Generalization.[/u][/b] Let $ABCD$ be a trapezoid ($AD\\parallel BC$). Denote the middlepoints $O$, $M$ of the segments $[BD]$, $[AD]$ respectively.\nFor a point $N\\in AC$ define the intersections $P\\in NM\\cap CD$ and $R\\in PO\\cap BC$. Prove that $NR\\parallel AB.$[/color][/quote]\r\n[color=darkblue][b][u]Proof.[/u][/b] Denote the intersections $S\\in NP\\cap BC$ and $T\\in OP\\cap AD$. Observe that $TD=BR$ and\n\n$\\left\\{\\begin{array}{c}\\frac{PD}{PC}=\\frac{TD}{RC}=\\frac{RB}{RC}\\\\\\\\ \\frac{PD}{PC}=\\frac{MD}{SC}=\\frac{MA}{SC}=\\frac{NA}{NC}\\end{array}\\right\\|$ $\\Longrightarrow$ $\\frac{RB}{RC}=\\frac{NA}{NC}$ $\\Longrightarrow$ $NR\\parallel AB$.\n\n[b][u]Commentary.[/u][/b] From the [u]hypothesis of the April's problem[/u] I used in the my proof only the following essential deeds :\n\n$1\\blacktriangleright AD\\parallel BC$, i.e. the quadrilateral $ABCD$ is a trapezoid ;\n$2\\blacktriangleright$ the point $O$ is the middlepoint of the diagonal $[BD]$ ;\n$3\\blacktriangleright$ the point $M$ is the middlepoint of the side $[AD]$ ;\n$4\\blacktriangleright$ the point $N$ belongs to the line $AC$.\n\nThus, you can extend this problem to a trapezoid ! You must learn to \"clean\" the hypothesis, to remove the useless things. [/color]", "Solution_5": "[quote][color=darkred][b][u]Generalization II.[/u][/b] Let $ABCD$ be a trapezoid ($AD\\parallel BC$). Denote two points $O\\in BD$, $M\\in AD$ for which $OM\\parallel AB$.\nFor a point $N\\in AC$ define the intersections $P\\in NM\\cap CD$ and $R\\in PO\\cap BC$. Prove that $NR\\parallel AB.$[/color][/quote][hide=\"Proof.\"]$S\\in NP\\cap BC$ and $T\\in OP\\cap AD\\Longrightarrow\\left\\{\\begin{array}{c}OM\\parallel AB\\Longleftrightarrow \\frac{MD}{MA}=\\frac{OD}{OB}\\\\\\\\\\frac{PD}{PC}=\\frac{TD}{RC}=\\frac{TD}{RB}\\cdot \\frac{RB}{RC}=\\frac{OD}{OB}\\cdot\\frac{RB}{RC}\\\\\\\\ \\frac{PD}{PC}=\\frac{MD}{SC}=\\frac{MD}{MA}\\cdot\\frac{MA}{SC}=\\frac{MD}{MA}\\cdot\\frac{NA}{NC}\\end{array}\\right\\|$ $\\Longrightarrow$ $\\frac{RB}{RC}=\\frac{NA}{NC}$ $\\Longrightarrow$ $NR\\parallel AB$.[/hide]", "Solution_6": "Thank you for two interesting problems and useful advice, [color=indigo][b]Mr. Virgil Nicula[/b][/color] :lol:", "Solution_7": "[hide=\"A small present (bonus) for April.\"][quote=\"Virgil Nicula\"][color=darkred][b][u]An equivalent enunciation of the second generalization.[/u][/b]\n\nGiven are in space two triangles $ABC$, $A'BC$ and the points $\\{\\begin{array}{c}M\\in AB\\ ,\\ N\\in AC\\\\\\ M'\\in A'B\\ ,\\ N'\\in A'C\\end{array}$ so that $MN\\parallel M'N'$.\n\nDefine the intersection $I\\in MM'\\cap NN'$. Prove that $I\\in AA'\\Longleftrightarrow MN\\parallel BC$.[/color][/quote]\n\n[color=darkblue][b]The latest commentary.[/b] This problem is easily, but it is a very good example for a development of a lesson in which the students will understand\nthat one problem isn't abandoned when they found already at least a proof. Only from here the creation (innovation) is starting and has the last word.[/color][/hide]", "Solution_8": "[quote=\"Virgil Nicula\"][quote][color=darkred][b][u]Generalization II.[/u][/b] Let $ABCD$ be a trapezoid ($AD\\parallel BC$). Denote two points $O\\in BD$, $M\\in AD$ for which $OM\\parallel AB$.\nFor a point $N\\in AC$ define the intersections $P\\in NM\\cap CD$ and $R\\in PO\\cap BC$. Prove that $NR\\parallel AB.$[/color][/quote][/quote] Generalization:\r\nLet $ABCD$ be [color=darkred]any quadrilateral[/color]. Denote two points $O\\in BD$, $M\\in AD$ for which $OM\\parallel AB$.\r\nFor a point $N\\in AC$ define the intersections $P\\in NM\\cap CD$ and $R\\in PO\\cap BC$. Prove that $NR\\parallel AB$.[hide=\"Proof\"]\nApply Menelaus for $\\triangle ADC$ and transversal $NMP$, and for $\\triangle BDC$ and transversal $ROP$. Hope i am right :) [/hide]", "Solution_9": "Yes, that right. [color=red]Very nice problem and nice solution too![/color]\r\n\r\nThanks. :P", "Solution_10": "For the initial problem: Denote $Q=MN\\cap AB, R=PE\\cap AB$; as $OM$ is midline in triangle $\\Delta PQR \\implies QR=2\\cdot OM=AB\\iff AQ=RB$. Next, $\\frac{QN}{NP}=\\frac{AQ}{CP}=\\frac{RB}{CP}=\\frac{RE}{PE}$, so $\\frac{RE}{PE}=\\frac{QN}{NP}\\iff NE\\parallel AB\\iff NE\\bot BC$.\n\nBest regards,\nsunken rock" } { "Tag": [], "Problem": "Enrollment is open for kids who will be in 6th or 7th grade in September 2005.\r\nThe full time school is free for the families of students.\r\nTeachers are also welcome to apply.\r\n\r\nhttp://www.amsacs.org", "Solution_1": "I would suggest that you also post this in the Massachusetts forum." } { "Tag": [ "calculus", "integration", "abstract algebra", "function", "real analysis", "real analysis unsolved" ], "Problem": "can anyone find it please\r\n\r\ni've got a result from this page:\r\n\r\nhttp://wims.unice.fr/wims/wims.cgi?module=tool/analysis/function.en\r\n\r\nbut i doubt if it's right", "Solution_1": "It's not an elementary function. There might be expressions for it if we add enough additional functions to the system.", "Solution_2": "There are some ways to estimate definite integrals of $x^{x}$. See for example\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=34003[/url].\r\n\r\nBuffalo's method was the nicest and most accurate." } { "Tag": [ "inequalities open", "inequalities" ], "Problem": "a,b,c sides of a triangle with a^2 + b^2 + c^2 = s\r\n\r\nprove \r\n\r\n(root(s-2ab)+root(s-2bc)+root(s-2ca))/root(s) < j\r\n\r\nfor j = 2root(2), or improve on it [smaller j]", "Solution_1": "\\[ \\begin{array}{l} a = x + y/////////b = x + z//////c = y + z \\\\ \\\\ \\sqrt{2(x^2 + y^2 )} + \\sqrt{2(x^2 + z^2 )} + \\sqrt{2(y^2 + z^2 )} < \\\\ \\\\ 2\\sqrt{2((x + y)^2 + (x + z)^2 + (y + z)^2 )} \\\\ \\\\ \\Rightarrow (\\sqrt{x^2 + y^2 } + \\sqrt{x^2 + z^2 } + \\sqrt{y^2 + z^2 } )^2 < \\\\ \\\\ 6(x^2 + y^2 + z^2 ) < 4((x + y)^2 + (x + z)^2 + (y + z)^2 ) \\\\ \\end{array} \\]\r\ni sink my solution is true ,isn't it?", "Solution_2": "yes,\r\n\r\nalthough, from your proof, j= root(6) is a better bound since,\r\n\r\n6(x^2 + y^2 + z^2) <= 3((x+y)^2 + (y+z)^2 + (z+x)^2)\r\n\r\nI think we could get better though?" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "let $a,b,c$ be positive real such that : \r\n$(a+b-c)(\\frac{1}{a}+\\frac{1}{b}-\\frac{1}{c})=4$\r\n find minimal of expression P\r\n $P=(a^4+b^4+c^4)(\\frac{1}{a^4}+\\frac{1}{b^4}+\\frac{1}{c^4})$\r\n I want to have a solution complete", "Solution_1": "Square both side etc." } { "Tag": [ "algebra", "polynomial", "Vieta", "geometry", "3D geometry", "AMC", "AIME" ], "Problem": "If a cubic polynomial like $ 8x^3\\plus{}1001x\\plus{}2008\\equal{}0$ has no x-squared term, then the sum of the roots is 0. Why is this the case? Can this rule be generalized to other polynomials?\r\n\r\nThanks!", "Solution_1": "It can be applied to all $ d$-degree polynomials $ P(x)$ that do not have a $ d \\minus{} 1$-degree term.\r\n\r\nTherefore, for example, in a fourth-degree polynomial of the form \r\n$ P(x) \\equal{} ax^{4} \\plus{} bx^{2} \\plus{} cx \\plus{} d$ that doesn't have a cubic term (of the form $ kx^{3}$)\r\nthe sum of its roots is $ 0$. \r\nThis is simply an application of [url=http://mathworld.wolfram.com/VietasFormulas.html]Vieta's Formulas[/url]", "Solution_2": "In direct response to your question, we can consider a polynomial with roots $ r,s,t$\r\n\r\n$ P(x) \\equal{} a(x \\minus{} r)(x \\minus{} s)(x \\minus{} t) \\equal{} a(x^3 \\minus{} (r \\plus{} s \\plus{} t)x^2 \\plus{} (rs \\plus{} st \\plus{} tr)x \\minus{} rst)$\r\n\r\nSo if there is no $ x^2$ term then we must have $ r \\plus{} s \\plus{} t \\equal{} 0$ \r\n\r\nAnd this then generalises to Vieta's Formula", "Solution_3": "Learning a lot about Vieta's is really useful (and pretty simple too). You can do a lot with it like find the sum of the squares of the roots, cubes of the roots, etc. and the sum of the reciprocals of the roots, sum of the squares of the reciprocals, etc.\r\n\r\nI think that's the polynomial from an AIME II 2008 question." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "$ a,b,c$natural $ a!b!\\equal{}a!\\plus{}b!\\plus{}c!$ find $ a,b,c$", "Solution_1": "See here : http://www.mathlinks.ro/viewtopic.php?t=169922" } { "Tag": [ "geometry", "3D geometry" ], "Problem": "I'm stuck with this problem and i can't solve it right now, while i did in the past. :maybe: :mad: \r\nSimplify the: $ \\sqrt[3]{{26 \\plus{} 15\\sqrt 3 }} \\plus{} \\sqrt[3]{{26 \\minus{} 15\\sqrt 3 }}$\r\n\r\nThe answer as i have in my old papers is the number 4, but now i can't prove it.", "Solution_1": "[hide]\nLet $ a \\equal{} \\sqrt [3]{{26 \\plus{} 15\\sqrt 3 }} \\plus{} \\sqrt [3]{{26 \\minus{} 15\\sqrt 3 }}$, $ b \\equal{} \\minus{} \\sqrt [3]{{26 \\plus{} 15\\sqrt 3 }}$ and $ c \\equal{} \\minus{} \\sqrt [3]{{26 \\minus{} 15\\sqrt 3 }}$.\n\nClearly, $ a \\plus{} b \\plus{} c \\equal{} 0$, and so, $ a^3 \\plus{} b^3 \\plus{} c^3 \\minus{} 3abc \\equal{} (a \\plus{} b \\plus{} c)(a^2 \\plus{} b^2 \\plus{} c^2 \\minus{} ab \\minus{} bc \\minus{} ca) \\equal{} 0$.\n\n$ a^3 \\plus{} b^3 \\plus{} c^3 \\minus{} 3abc \\equal{} 0$\n$ a^3 \\plus{} b^3 \\plus{} c^3 \\equal{} 3abc$\n$ a^3 \\minus{} (26 \\plus{} 15\\sqrt 3) \\minus{} (26 \\minus{} 15\\sqrt 3) \\equal{} 3a\\sqrt [3]{(26 \\plus{} 15\\sqrt 3)(26 \\minus{} 15\\sqrt 3)}$\n$ a^3 \\minus{} 52 \\equal{} 3a\\sqrt [3]{676 \\minus{} 675}$\n$ a^3 \\minus{} 52 \\equal{} 3a$\n$ a^3 \\minus{} 3a \\minus{} 52 \\equal{} 0$\n\nWe're looking to factor $ a^3 \\minus{} 3a \\minus{} 52$. From the form of that expression, it's probably $ (a \\minus{} r_1)(a^2 \\plus{} ar_1 \\plus{} r_2)$. We have that\n\n$ r_1r_2 \\equal{} 52$\n$ r_1^2 \\minus{} r_2 \\equal{} 3$\n\nFrom which we conclude that $ r_1 \\equal{} 4$, $ r_2 \\equal{} 13$.\n\n$ (a \\minus{} 4)(a^2 \\plus{} 4a \\plus{} 13) \\equal{} 0$\n\nClearly, $ a \\in \\mathbb{R}$, so $ a \\equal{} 4$, the only real solution to that equation.\n[/hide]", "Solution_2": "[quote=\"ffao\"][hide]\nLet $ a \\equal{} \\sqrt [3]{{26 \\plus{} 15\\sqrt 3 }} \\plus{} \\sqrt [3]{{26 \\minus{} 15\\sqrt 3 }}$, $ b \\equal{} \\minus{} \\sqrt [3]{{26 \\plus{} 15\\sqrt 3 }}$ and $ c \\equal{} \\minus{} \\sqrt [3]{{26 \\minus{} 15\\sqrt 3 }}$.\n\nClearly, $ a \\plus{} b \\plus{} c \\equal{} 0$, and so, $ a^3 \\plus{} b^3 \\plus{} c^3 \\minus{} 3abc \\equal{} (a \\plus{} b \\plus{} c)(a^2 \\plus{} b^2 \\plus{} c^2 \\minus{} ab \\minus{} bc \\minus{} ca) \\equal{} 0$.\n\n$ a^3 \\plus{} b^3 \\plus{} c^3 \\minus{} 3abc \\equal{} 0$\n$ a^3 \\plus{} b^3 \\plus{} c^3 \\equal{} 3abc$\n$ a^3 \\minus{} (26 \\plus{} 15\\sqrt 3) \\minus{} (26 \\minus{} 15\\sqrt 3) \\equal{} 3a\\sqrt [3]{(26 \\plus{} 15\\sqrt 3)(26 \\minus{} 15\\sqrt 3)}$\n$ a^3 \\minus{} 52 \\equal{} 3a\\sqrt [3]{676 \\minus{} 675}$\n$ a^3 \\minus{} 52 \\equal{} 3a$\n$ a^3 \\minus{} 3a \\minus{} 52 \\equal{} 0$\n\nWe're looking to factor $ a^3 \\minus{} 3a \\minus{} 52$. From the form of that expression, it's probably $ (a \\minus{} r_1)(a^2 \\plus{} ar_1 \\plus{} r_2)$. We have that\n\n$ r_1r_2 \\equal{} 52$\n$ r_1^2 \\minus{} r_2 \\equal{} 3$\n\nFrom which we conclude that $ r_1 \\equal{} 4$, $ r_2 \\equal{} 13$.\n\n$ (a \\minus{} 4)(a^2 \\plus{} 4a \\plus{} 13) \\equal{} 0$\n\nClearly, $ a \\in \\mathbb{R}$, so $ a \\equal{} 4$, the only real solution to that equation.\n[/hide][/quote]\r\nIngenious solution! Many thanks.\r\n \r\n I don't think that i had solved it that way back then, so if you or anyone else have different solutions please post them too. :oops:", "Solution_3": "If you don't see the previous solution you can always expand the cube:\r\n\r\n[hide=\"Solution #2\"]\nLet $ a \\equal{} \\sqrt[3]{{26 \\plus{} 15\\sqrt 3 }} \\plus{} \\sqrt[3]{{26 \\minus{} 15\\sqrt 3 }}$, as previously.\n\n$ a^3 \\equal{} 26 \\plus{} 15\\sqrt 3 \\plus{} 3\\sqrt[3]{(26 \\plus{} 15\\sqrt 3)^2(26 \\minus{} 15\\sqrt 3)} \\plus{} 3\\sqrt[3]{(26 \\plus{} 15\\sqrt 3)(26 \\minus{} 15\\sqrt 3)^2} \\plus{} 26 \\minus{} 15\\sqrt 3$\n\n$ a^3 \\equal{} 52 \\plus{} 3\\sqrt[3]{{26 \\plus{} 15\\sqrt 3 }} \\plus{} 3\\sqrt[3]{{26 \\minus{} 15\\sqrt 3 }}$\n$ a^3 \\equal{} 52 \\plus{} 3(\\sqrt[3]{{26 \\plus{} 15\\sqrt 3 }} \\plus{} \\sqrt[3]{{26 \\minus{} 15\\sqrt 3 }})$\n$ a^3 \\equal{} 52 \\plus{} 3a$\n[/hide]", "Solution_4": "$ \\sqrt[3]{{26 \\plus{} 15\\sqrt 3 }}\\equal{}\\sqrt[3]{{8\\plus{}12\\sqrt 3 \\plus{}18\\plus{}3\\sqrt3}}\\equal{}\\sqrt[3]{{(2\\plus{}\\sqrt 3)^3}}\\equal{}2\\plus{}\\sqrt 3$\r\n$ \\sqrt[3]{{26 \\minus{} 15\\sqrt 3 }}\\equal{}\\sqrt[3]{{8\\minus{}12\\sqrt 3 \\plus{}18\\minus{}3\\sqrt3}}\\equal{}\\sqrt[3]{{(2\\minus{}\\sqrt 3)^3}}\\equal{}2\\minus{}\\sqrt 3$\r\nSo $ \\sqrt[3]{{26 \\plus{} 15\\sqrt 3 }} \\plus{} \\sqrt[3]{{26 \\minus{} 15\\sqrt 3 }}\\equal{}2\\plus{}\\sqrt 3 \\plus{} 2\\minus{}\\sqrt 3\\equal{}4$" } { "Tag": [ "geometry", "perimeter", "real analysis", "real analysis unsolved" ], "Problem": "How do you divide a circular cake into n pieces of equal area while minimizing the total length of all the cuts? I can't figure out how you go about doing it.", "Solution_1": "Does anyone know how to do this?", "Solution_2": "How about a slightly simpler beginning question: assuming a radius of $ 1,$ for what values of $ n$ is it possible to cut the cake into equal area pieces with total length of cut strictly less than $ n$? (You can get $ n$ from equally spaced radial cuts.)", "Solution_3": "[asy]size(100);\nint n=10;\nD(CR((0,0),10));\nD(CR((0,0),3),red);\nfor (int k=0;kR s.t. \r\n\r\n(i) f(xy)=f(x)f(y)+f(-x)f(-y) for all x,y in R\r\n(ii) f(x+y)<=f(x)+f(y) for all x,y in R\r\n(iii) f(1999)=1000", "Solution_1": "ok time for a partial solution :D \r\nnote that f can not be a constant function.putting x=-1,y=1 we get\r\nf(-1)=2f(1)f(-1)\r\ncase 1) f(-1)=0 and f(0)=1/2 or 0 (i'll consider f(0)=1/2 )\r\nnote that putting x=x and y=0 we get f(x)+f(-x)=1==>f(1)=1\r\nnow consider the function g(x) =2f(x)-x.after plugging it into the functional equation we get g(xy)+xy=[g(x)+x-1][g(y)+y-1] + 1\r\nnow putting x=x,y=1 we get g(x)=1-x (not possible by initial conditions) or g(y)=1 ==> f(x) =(x+1)/2 and this can be verified\r\n f(0) = 0 about this i'm still thinking .mayb i'm overlooking some obvious fact.anyways something is better than nothing\r\n :)" } { "Tag": [ "function", "analytic geometry", "real analysis", "real analysis unsolved" ], "Problem": "Let X be a finite set and Y a countable set. Let F(X, Y) the set of functions defined in X with values in Y. Prove that F(X, Y) is countable.\r\n \r\nI need some help, please!", "Solution_1": "[quote=\"Duncreggan\"]Let X be a finite set and Y a countable set. Let F(X, Y) the set of functions defined in X with values in Y. Prove that F(X, Y) is countable.\n \nI need some help, please![/quote]\r\n\r\nThere's a very simple identification of $ F(X,Y)$ with $ N^k$.", "Solution_2": "Do you mean F(X,Y) to be the set of functions from X to Y? (the word 'in' confused me, \"defined in X\"). Such a function is a special kind of relation from A to B, i.e. a subset of the cartesian product AxB (namely such that every first coordinate has a unique second coordinate). Write $ Y\\equal{}\\{y_1,y_2,...\\}$.\r\n\r\nTo get the idea, first assume |X|=1, say X={x}. Then all functions from X to Y are of the form $ \\{(x,y_i)\\}$, i.e. $ F(X,Y)\\equal{}\\bigcup_{i\\in\\mathbb{N}}\\{(x,y_i)\\}$. We know that the countable union of countable (in this case even finite) sets is again countable, so the conclusion follows.\r\n\r\nNow assume $ |X|\\equal{}n\\in\\mathbb{N}$, write $ X\\equal{}\\{x_1,...,x_n\\}$. Then all functions from X to Y are of the form $ \\{(x_1,y_{i_1}),...,(x_1,y_{i_n})\\}$ with $ i_1,..,i_n\\in\\mathbb{N}$. So $ F(X,Y)\\equal{}\\bigcup_{(i_1,...,i_n)\\in\\mathbb{N}^n}\\{(x_1,y_{i_1}),...,(x_1,y_{i_n})\\}$. Since $ \\mathbb{N}^n\\equal{}\\mathbb{N}\\times...\\times\\mathbb{N}$ ($ n$ times) is again countable (assuming you know this), this again is the countable union of countable (finite) sets, so the conclusion follows.", "Solution_3": "I would like to know if my solution is correct.\r\n\r\nSince X is finite we can write X = $ \\{x_1,...,\r\nx_n\\}$. Then, define the map $ \\phi : \\mathcal{F}(X, Y) \\rightarrow Y^n$ by\r\n\\[ \\phi(f) \\equal{} (f(x_1),..., f(x_n)).\\]\r\nNote that $ \\phi$ is a one-to-one function. Since $ Y^n \\equal{} Y\\times ... \\times Y$ is\r\ncountable because Y is countable, we have $ \\mathcal{F}(X, Y)$ is\r\ncountable. (The result follows by the injection of $ \\phi$).", "Solution_4": "Nice solution, yes it's correct :)" } { "Tag": [ "algebra", "polynomial", "Galois Theory", "superior algebra", "superior algebra unsolved" ], "Problem": "How can i show that the polynomial x^5 + 20x +16 in Q[X]\r\n>is not solvable by radicals by showing it is the Galois group A5\r\n>\r\n>I have tried but I can only show that the splitting field is of degree 5\r\n>so it contains a 5-cycle and since it has 1 real root and 4 complex ones\r\n>it also contains a 2-cycle but this leads me to S5 and that is apparently wrong I should be getting A5\r\nCan anyone help me see how there are no 2-cycle but a product of two 2-cycle\r\n\r\nThanks", "Solution_1": "The conjugation switches two pairs of roots. Thus it is of type $ (23)(45)$ where $ 2,3,4,5$ stand for complex roots (and $ 1$ being the real root).", "Solution_2": "For a full proof that the group is contained in $ A_5$, you would need to show that the discriminant $ D$ is a perfect square. Even permutations of the roots fix $ \\sqrt{D}$, while odd permutations send it to $ \\minus{}\\sqrt{D}$.\r\n\r\nOf course, $ S_5$ isn't solvable either; that would have been an easier exercise." } { "Tag": [ "geometry", "trapezoid", "perimeter" ], "Problem": "Point E lies on the base AD of the trapezoid ABCD. The triangles ABE, BCE and CDE have equal perimeters.\r\nProve that $|BC| = |\\frac{AD}{2}|$.", "Solution_1": "my work seems kind of wrong\r\n\r\n[hide]Let altitude from E of triangle BEC meet BC at F\n\nLet altitude from B of triangle ABE meet AE at G\n \nLet altitude from C of triangle CDE meet ED at H\n\nby AAS triangle BGE is congruent to triangle BFE\n\nby AAS triangle EFC is congruent to triangle ECH\n\nlet the length of EF by a.\n\nlet the length of AG be b.\n\nlet the length of FC by c.\n\nbecause the perimeters of the three triangles are equal, b+sqrt(b^2+a^2)=c+sqrt(c^2+a^2) and c and b must be equal. same goes for the lengths of HE and BF. Since all the triangles are congruent and AD is two bases while BC is only one base of the same triangle, AD/2 is equal to BC[/hide]", "Solution_2": "I think you are on the right track, and there seems to be nothing wrong, just a little more explanation I think. Here it is: since AB=FC by what you have done with the right triangles, and when you drew the altitudes of the triangles, you have BF=GE and FC=EH, which means (since AB=FC), AB=EH, and when you proved once again that EH=BF, you also have EH=BF=GE, so therefore all six length are equal and there are two on top and four on bottom so therefore BC=AD/2." } { "Tag": [], "Problem": "It was aked simplfy for $k=0,1,2,3$\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=19463\r\n-----------------------------------------------------------------------------\r\n\r\nSimplify for $k=4,5,7$\r\n\r\n$\\frac{a^k}{(a-b)(a-c)}+\\frac{b^k}{(b-c)(b-a)}+\\frac{c^k}{(c-a)(c-b)}$\r\n\r\nfor a,b,c in $K$ commutative field $a \\neq b$, $b \\neq c$, $a\\neq c$", "Solution_1": "Indeed I can do it for generic k ;) .\r\n\r\nAgain, see my post #8 at http://mathlinks.ro/Forum/viewtopic.php?t=16360 .\r\n\r\n Darij" } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "Harvard", "college", "MIT", "Princeton" ], "Problem": "Why is it that the people who are always winning math competitions always come from one of these three schools?\r\n\r\nDoes someone who makes it to USAMO and goes to one of these schools really deserve it? After all, they have MUCH more resources than most other students. The disparity between what students at these schools and regular run-of-the-mill public schools is enormous. Then, not only are these people rewarded for their \"genius\" etc, etc, they go onto Harvard, MIT, Princeton, etc, etc.\r\n\r\nI am sure that most people here will disagree with this, but this is the impression that I have been getting ever since I started participating in all this math competition and stuff. It is really frustrating to know that you have no next to no chance because you go to a lousy school and started late because there was no math team at your school.\r\n\r\nOn this note, I would like to say how greatly appreciative I am of Art of Problem Solving. While many people have stated how much they owe to AoPS, this praise is definately not undeserved.\r\n\r\nI curious as to what people think about what I have said because this has been running through my head a lot for the past year or so.", "Solution_1": "there was discussion on this before... maybe called 'usamo and college applications' or something like that. check the college forum. \r\n\r\nBut i think most people actually share your beliefs...", "Solution_2": "First, I am pleased that you appreciate the Art of Problem Solving. I personally think that there is enough material on this site to make up for a lot of the extra resources that students at the top school have access to. You may find other sources for mathematics education that help you in your pursuit as well.\r\n\r\nSecond, we are currently running a Math Jam (right now) on the AMC B problems from this year. Going over the solutions might be a good way to supplement your education. We run these free Math Jams 2-3 times per month usually.\r\n\r\nThird, do not despair. You can find resources outside of your school. You can also work to help your own school understand how best to build a program. Many of us had this experience as students. The desire to move forward is much of the inspiration for the Art of Problem Solving as it is today.\r\n\r\nMathematics is first and foremost about curiosity. \"How do I figure this out? How do I solve this problem? Why does this work the way it works?\" If you stay on the inquisitive course and keep working then in the end you will find yourself amongst the serious problem solvers.", "Solution_3": "Hrmph ...\r\n\r\nI'm upset that Stuy isn't on that list ...\r\n\r\n\r\nanyhow, with respect to what you actually said:\r\n\r\nI hope this doesn't sound mean, but that's life -- there is no place on earth and never has been where everyone has equal opportunities. I don't think that's a good thing (and I find the disparity in opportunity in mathematics among different groups of people to be quite upsetting), but on the large scale there isn't really much to be done about it, at least on a short-term basis. The closest thing to a solution for the short term is to do your best to make up for it. I'm all for people starting math teams in their own schools -- if you find a teacher willing to meet with 5-6 students once or twice a week, you have a math team. There are lots of resources on line, including AoPS and many others that I won't try to name.\r\n\r\n\r\nOne final point about the schools you named -- all the people who go to TJ, AAST, Exeter, or Stuyvesant are chosen to go there. Now, there are distinct limitations on that choosing, whether it be geographical or financial or whatever. But to say that they (we) are \"rewarded for our 'genius'\" is silly -- the schools pick groups of students who are highly likely to succeed already. The fact that Stuy is a good school has much more to do with the student body than the faculty.* So again, I think the best solution is to try and find the people who are (or might be) interested and band together with them in doing things like starting a math team or whatever else.\r\n\r\n\r\n*Note that this is coming from someone who thinks very highly of the Stuyvesant faculty (particularly in math).", "Solution_4": "Yeah, your personal motivation is much more important that the resources that your school can happen to provide. Even if a student's school does not have nearly as many resources as Exeter or TJ or, in Joel's case, Stuy, if that student is truly motivated and eager to succeed in mathematics and math competions during their high school years, they will not be deterred at all. I am sure that any of the students on past US IMO teams who went to very good high schools would still have accomplished what they did had they had the same motivation, but had gone to a less encouraging school.", "Solution_5": "I don't know. I went to a normal public high school that didn't offer all the resources that a math magnet school would although we were much better off than most other public schools, but I had the good fortune to be the second-best mathematician at my school when I was a freshman. I think that was my primary motivation. I have never been a big fan of being the second-best, and so I worked hard to try to change it around. Had I been the best, I doubt I would be nearly as good at math as I am now.", "Solution_6": "[quote] I had the good fortune to be the second-best mathematician at my school when I was a freshman. I think that was my primary motivation. I have never been a big fan of being the second-best, and so I worked hard to try to change it around. Had I been the best, I doubt I would be nearly as good at math as I am now.[/quote]\r\n\r\nReally? But even if you had been the best at your school, you would certainly not be the best in the US, or the world. There are always higher goals to reach for, right?", "Solution_7": "I go to AAST, and I have worked at least an hour a day on math since about 8th grade. We have all the past AMC's, AIMEs, and tons of national and international olympiads in storage so we can practice. I think our successes come from the fact that we draw the best math students from the county, and that we have a lot of study and practice resources available to us. On top of that, we devote a lot of time to it.", "Solution_8": "i forgot i think mr. president bush went to either Philips andover or Philips Exeter. i received a 100 page book from them the SCHOOL COSTS A LOT. but that doesn't make him smart. his SAT is about 200 lower than Clinton's", "Solution_9": "[quote]but that doesn't make him smart. his SAT is about 200 lower than Clinton's[/quote]\r\n\r\ntrue..but having a higher SAT doesn't make Clinton smarter either :wink:", "Solution_10": "Cliton provided millions of jobs and he reduced the debt. Then look at Bush reversing everything good Clinton did.", "Solution_11": "churchhill, that avatar fooled me into thinking you were mr. crawford until i noticed the poor grammar", "Solution_12": "We have different signatures so there isn't really any confusion.", "Solution_13": "[quote=\"churchilljrhigh\"]We have different signatures so there isn't really any confusion.[/quote]\r\n\r\nAre you trying to pose as one of the AoPS instructors? Why not make an avatar based on Jiang Zemin's photograph?", "Solution_14": "Well, fubu on AOPS told me to change my old one.", "Solution_15": "what are your top 3 scores in AMC 12A and 12B?", "Solution_16": "12B : 136, 132.5, 132. \r\n12A: 144, 138, 136,", "Solution_17": "[quote]Why is it that the people who are always winning math competitions always come from one of these three schools [AAST, Exeter, TJ]? . . . I curious as to what people think about what I have said because this has been running through my head a lot for the past year or so.[/quote]\n\nIt's unfair to say that its solely the schools that are responsible for the success. There is a lot of talent in the selection pool for the schools (as an AAST junior, i would know).\n\nStudents at AAST spend their weekends in school, practicing math. Mr Holbrook, the math team coach for the school, effectively has more than one copy of AoPS 1 and 2 per student. He has a massive treasure trove of mathematical problems and books.\n\nWhile the availability of problems is an advantage, it would not do justice to my fellow students to say that this is the sole reason for success. There is a lot of dedication on the part of the students. As mentioned above, many kids spend an extra [b] 8 - 16 hours a week [/b] in school doing math [b] [i] (on the weekends). [/i] [/b] It is this type of dedication that brings victories. Mathematics is not something that is learned haphazardly. It takes hours upon hours of studying and math work in order to reach the level we are currently at.\n\nIt is unfair for those who go to local schools and don't have the resources that a public magnet school could provide. However, there is a plethora of resources available for those whose schools do not provide these resources. For instance, an amazing web site, [url]http://www.kalva.demon.co.uk[/url] has a whole slew of olympiad problems, including all the IMO's, USAMO's, AIME's, and other math competitions. MAA writes great books that you can order. ART OF PROBLEM SOLVING is the bible of mathematics :)\n\nAs for eighth graders competing, half the time they are better than freshmen. For motivated eighth-graders, there are always programs available. One must search and find the programs. \n\n[quote]But I agree with the fact that the magnet schools do have more of an advantage over a regular public school. Our school selects all of the best students from northern NJ, and the ones dedicated to math spend four to eight hours every weekend at math team practice, not to mention the many hours during the day and at home.[/quote]\r\n\r\nIgnoring variables like sickness, performance is directly related to time spent working on mathematics. Students like Lun Mu devote a large part of their free time to mathematics. Thus, it is unfair to whine about the performance of Exeter or TJ or AAST without appreciating the amount of time devoted to mathematics. I do not have personal experience in TJ or Exeter, but as I am told the TJ math team is mostly a student-run club, spearheaded by incredibly devoted and talented students with little assistance from teachers.\r\n\r\n(If you got bored reading the long paragraphs, just jump down here) No one can carp about someone performing better than another unless the niggler recognizes and appreciates the dedication and effort put into mathematics (or in any other field).", "Solution_18": "Does AAST have 8th graders? I was under the impression that it goes from 9th to 12th grade.", "Solution_19": "God are you a student or a teacher?", "Solution_20": "he said he was a junior in HS.", "Solution_21": "AAST does [b] not [/b] have eighth-graders. AAST is a high school. There is a program called \"Academy After Hours\" where the math team coach teachers night classes to motivated youngsters. Our coach allows these youngsters, if they are good enough, to compete in math olympiads. \r\n\r\nAs for ARML, there are 15 people to a team. Just like how some home-schooled kids form a team, just like how entire states send teams, we allow interested eighth-graders an opportunity to compete by bringing them together. This can be compared to the NYC teams, a combination of high-school students from different public schools that are together for the purpose of ARML.\r\n\r\n[quote]the AAST 8th graders do not really compare with teams that have seniors on them.[/quote]\r\n\r\nIt is unfair to compare our 8th graders to division A or division B teams at ARML. You are trying to compare kids with a gap of 4 years, so it is expected that they do not compare. However, their relative performance is impressive, noting that this is the first time for these kids, whereas seniors have competed for four years. Furthermore, you are trying to compare first-string teams from regions like Eastern Mass to eighth-graders from the small Bergen county. If you wish to rank the AAST 8th graders, do not compare them to first-string seniors with years and awards under their belts. You are trying to compare fledglings and aces.", "Solution_22": "True, although if you'd continued reading what I wrote, you would note that I went on to say\r\n\r\n[quote=\"JBL\"]They tend to do pretty terribly overall, although not so badly considering that most people there are a couple or three years older than them.[/quote]\r\n\r\n\r\nAlso, it's worth pointing out that \"little Bergen County\" has a population of roughly 900,000 people and that it is one of the richest counties in one of the richest states in the country.", "Solution_23": "I don't know if they still do, but in the early 90s Maryland brought a team of 8th graders that did very well at ARML. I know they came in ahead of several states A Teams. And in 2000 an 8th grader came in first place at ARML - Tiankai Liu.", "Solution_24": "In all honesty, the only advantage we have over other schools is that we pull good students out of a large pool of people. But we only get people with potential; we still put in a lot of hours (just as much, if not more, than other schools).", "Solution_25": "Being able to put in many hours is a serious advantage -- many schools can't find (or pay for) teachers who will do that for their kids. The NYC team can't do it because we aren't chosen soon enough and because it's just hard to organize when there's no good structure in place to contact everyone, etc., and no one being paid to do it.", "Solution_26": "[quote]\nBeing able to put in many hours is a serious advantage -- many schools can't find (or pay for) teachers who will do that for their kids. [/quote]\r\n\r\nThe ability to put in extra hours is dependent on the student, not the teacher. One can have a teacher willing to devote every waking moment to mathematics, but it means little if the student is not willing to devote time to mathematics. It might be nice to have a teacher, but many of the greatest IMO team members (e.g. Reid Barton) are homeschooled, so it is unfair to attack the school.\r\n\r\nThe presence of a math coach (or lack thereof) does not determine success. It is all about how much time you are willing to put into mathematics outside of school (and no teacher can change how much effort you devote ouside of the school)", "Solution_27": "[quote=\"God\"]The presence of a math coach (or lack thereof) does not determine success. It is all about how much time you are willing to put into mathematics outside of school (and no teacher can change how much effort you devote ouside of the school)[/quote]\r\n\r\nI respectfully disagree, particularly with the second point. Thousands of students every year have the love of mathematics (and plenty of other subjects) squashed out of them by teachers who don't care about the subject and don't allow it to demonstrate its beauty. The quality of a teacher has very much to do with invigorating students -- particularly in the non-magnet school situation where the vast majority of the student body isn't already interested on their own. (Remember, you're coming from an atypical school in many respects.)\r\n\r\nI do agree with you, of course, that the will of the student is important in all of this, but it's not the only important thing. \r\n\r\n\r\nAnd, just to be clear, I'm not trying in any way to disparage AAST -- I think your performance is quite impressive, and I commend you for it. But it's not replicable everywhere by any old random group of kids who like math. There are a lot of factors involved that other places simply cannot replicate.", "Solution_28": "I completely agree with JBL's opinion.", "Solution_29": "[quote=\"JBL\"]Being able to put in many hours is a serious advantage -- many schools can't find (or pay for) teachers who will do that for their kids. The NYC team can't do it because we aren't chosen soon enough and because it's just hard to organize when there's no good structure in place to contact everyone, etc., and no one being paid to do it.[/quote]\r\n\r\nThey can't do it.. yet.. Wait till you see the totally automated system I\"m about to put up...." } { "Tag": [ "algebra", "polynomial", "inequalities proposed", "inequalities" ], "Problem": "Given $ a,b,c \\ge 0$.Prove that:\r\n(a) $ (a \\plus{} b \\plus{} kc)(b \\plus{} c \\plus{} ka)(c \\plus{} a \\plus{} kb) \\ge \\frac {(k \\plus{} 2)^3}{9\\sqrt {3}}(a \\plus{} b \\plus{} c)^2\\sqrt {ab \\plus{} bc \\plus{} ca}$\r\nfor k such that:$ 54k \\ge (k \\plus{} 2)^3$\r\n(b) $ (a \\plus{} b \\plus{} kc)(a \\plus{} c \\plus{} kb)(b \\plus{} c \\plus{} ka) \\ge 2(k \\minus{} 1)\\sqrt {k}(a \\plus{} b \\plus{} c)^2\\sqrt {ab \\plus{} bc \\plus{} ca}$\r\nfor all k>1.Equality occur if and only $ k \\ge 3\\plus{}\\sqrt{8}$.For $ k \\equal{} 3 \\plus{} \\sqrt {8}$,equality occur if and only a=b,c=0 or any cyclic permution.\r\n(c)Prove that:,we don't have k>0 such that:\r\n$ (a \\plus{} b \\plus{} kc)(b \\plus{} c \\plus{} ka)(c \\plus{} a \\plus{} kb) \\le \\frac {(k \\plus{} 2)^3}{9\\sqrt {3}}(a \\plus{} b \\plus{} c)^2\\sqrt {ab \\plus{} bc \\plus{} ca}$\r\n(d)Prove that for a,b,c is side length in triangle,\r\n$ (a \\plus{} b \\plus{} kc)(b \\plus{} c \\plus{} ka)(c \\plus{} a \\plus{} kb) \\le \\frac {(k \\plus{} 2)^3}{9\\sqrt {3}}(a \\plus{} b \\plus{} c)^2\\sqrt {ab \\plus{} bc \\plus{} ca}$\r\nfor k such that:$ 9k(\\sqrt {12} \\plus{} 3) \\le (k \\plus{} 2)^3$", "Solution_1": "If $ {k}\\ge0,$ then\r\n\r\n$ (a \\plus{} b \\plus{} kc)(b \\plus{} c \\plus{} ka)(c \\plus{} a \\plus{} kb)\\ge\\frac {(k \\plus{} 2)^{3}}{9\\sqrt {3}}(a \\plus{} b \\plus{} c)^{2}\\sqrt {ab \\plus{} bc \\plus{} ca}$\r\nholds for all$ {a,b,c}\\ge{0}$ iff $ {k_1}\\le{k}\\le{k_2},$ where\r\n $ k_1 \\equal{} 0.148097507... ,k_2 \\equal{} 2.971699821...$\r\n is roots of the following polynomial\r\n$ 1664 \\minus{} 6456k \\minus{} 29346k^2 \\minus{} 20896k^3 \\plus{} 7419k^4 \\plus{} 906k^5 \\plus{} 53k^6.$" } { "Tag": [], "Problem": "Find the coefficient of $ x^2$ in the expansion of \\[ \\left(x\\plus{}\\frac{1}{x}\\right)^6\\].", "Solution_1": "hello, the coefficient is $ \\binom{6}{2}x^4\\left(\\frac{1}{x}\\right)^2\\equal{}15x^2.$\r\nSonnhard." } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "Let $ (u_{n})$ be a sequence given by $ u_{1}$ and the formula $ u_{n\\plus{}1}\\equal{}\\frac{k\\plus{}u_{n}}{1\\minus{}u_{n}}$ ,where $ k\\geq 0;n\\equal{}1;2;...$ \r\nGiven that $ u_{1}\\equal{}u_{13}$.Find value of $ k$", "Solution_1": "Posted 3-4 days before.", "Solution_2": "[quote=\"Rust\"]Posted 3-4 days before.[/quote]\r\n\r\nAnd killed because it was a live topic in some Vietnam math review.\r\nDon't answer this problem." } { "Tag": [ "logarithms" ], "Problem": "Find $ (\\log_2{x})^2$ if $ \\log_2{(\\log_8{x})} \\equal{} \\log_8{(\\log_2{x})}$. Please clearly show all algebraic manipulation.\r\n\r\nThanks in advance!", "Solution_1": "hello, we have $ \\log_8 x \\equal{} \\frac {\\ln(x)}{\\ln(8)} \\equal{} \\frac {\\ln(x)}{3\\ln(2)}$, so we have $ \\log_2\\left(\\frac {\\ln(x)}{3\\ln(2)}\\right) \\equal{} \\log_8\\left(\\frac {\\ln(x)}{\\ln(2)}\\right)$, setting $ \\frac {\\ln(x)}{\\ln(2)} \\equal{} u$ so we get\r\n$ \\log_2 \\left({\\frac {u}{3}}\\right) \\equal{} \\log_8 u$, this is equivalent with $ \\frac {\\ln(\\frac {u}{3})}{\\ln(2)} \\equal{} \\frac {\\ln(u)}{3\\ln(2)}$, simplifying we get $ \\ln(\\frac {u}{3}) \\equal{} \\frac {\\ln(u)}{3} \\Leftrightarrow 3\\ln(u) \\minus{} 3\\ln(3) \\equal{} \\ln(u)$ or $ \\ln(u) \\equal{} \\ln(3^{\\frac {3}{2}})$ and so we have $ u \\equal{} 3^{\\frac {3}{2}}$, our $ u \\equal{} \\log_2 x$ so we have $ \\log_2 x \\equal{} 3^{\\frac {3}{2}}$ and from here follows $ \\left(\\log_2 x\\right)^2 \\equal{} 3^3 \\equal{} 27$.\r\nSonnhard." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Prove by combinatorial reasoning that\r\n\r\n$ \\sum_{k\\equal{}1}^{n}k\\binom{n\\plus{}1}{k\\plus{}1}\\equal{}(n\\minus{}1)2^{n}\\plus{}1$.", "Solution_1": "You've already posted this problem here:\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=168173[/url]\r\n\r\nDo not double post." } { "Tag": [ "function" ], "Problem": "If the lengths of two sides of a right triangle are 3 and 4, what is the least possible length of the third side?", "Solution_1": "make four the hypotenuse\r\nso it's 7^.5", "Solution_2": "Yeah, that's what I got.", "Solution_3": "[quote=\"colts18\"]Yeah, that's what I got.[/quote]\r\nhow did you get that? YOu don't even know what a hypotenuse is. Or even how to say it. :P", "Solution_4": "[quote=\"jimli\"]make four the hypotenuse\nso it's 7^.5[/quote]\r\nWhy do you raise 7 by .5??", "Solution_5": "They probably can't do the sqrt function on latex. \r\n\r\nBasically that's $\\sqrt{16-9}$ or $\\sqrt{7}$", "Solution_6": "that makes more sense to me", "Solution_7": "yup if u make 4 the hypotenuse, c squared is 16, and 3 squared is 9, so its the root of (c squared minus 9 (sqrt 16-9) which is 7... making the other side 2.6457513110645905905016157536393 (calculator..)", "Solution_8": "sqrt(16-9)=sqrt(7)", "Solution_9": "To clarify the definition of hypotenuse, it is the longest side of a right triangle, or the side opposite the right angle. The other two sides are known as legs.", "Solution_10": "[quote=\"gr8sk8r\"][quote=\"jimli\"]make four the hypotenuse\nso it's 7^.5[/quote]\nWhy do you raise 7 by .5??[/quote]\r\n\r\nthe radical of 7 (sqrt(7)) is the same thing as 7^1/2=7^0.5 because, squareroot is the same thing as the 2nd root of x-variable. Some example will clarify this: \r\n\r\n7^1/2=squareroot of 7\r\n7^1/3=cubedroot of 7\r\n7^2/2=squareroot of 7^2\r\n7^3/2=squareroot of 7^3\r\n\r\nTherefore, sqrt(4^2-3^2)=sqrt(16-9)=sqrt(7)=7^1/2 or 7^0.5\r\n\r\n-Masoud Zargar", "Solution_11": "If it is writen in the form $b^{\\frac{x}{y}}$ it is the same as \r\n$\\sqrt[y]{b^x}$", "Solution_12": "exactly.", "Solution_13": "[quote=\"math92\"]If the lengths of two sides of a right triangle are 3 and 4, what is the least possible length of the third side?[/quote]\r\n\r\n[hide]If $4$ was the hypotenuse, then the other leg would be\n\n$\\sqrt{4^2-3^2}$\n\n$\\sqrt{(4-3)(4+3)}$\n\n$\\sqrt{1*7}$\n\n$\\sqrt{7}$[/hide]", "Solution_14": "[hide] square root of 7[/hide]", "Solution_15": "[hide]\n$x=\\sqrt{c^2-a^2}$, $x=\\sqrt{4^2-3^2}$, $x=\\sqrt{7}$[/hide]" } { "Tag": [ "geometry", "3D geometry", "probability", "circumcircle", "similar triangles", "angle bisector" ], "Problem": "20. In acute triangle ABC, points D and E lie on BC and AC, respectively. If AD and BE intersect at T so that AT/DT=3 and BT/ET=4, what is CD/BD?\r\n\r\n23. Each face of a cube is painted either red or blue, each with probability 1/2. The color of each face is determined independently. What is the probability that the painted cube can be placed on a horizontal surface so that the four vertical faces are all the same color?\r\n\r\n24. In triangle ABC, we have AB=7, AC=8, and BC=9. Point D is on the circumcircle of the triangle so that AD pisects angle BAC. What is the value of AD/CD?", "Solution_1": "[hide=\"20\"] Draw TX parrallel to AC so that X is on BC. Thus, we obtain two pairs of similar triangles: BTX and BEC and DTX and DAC. Now, call DX=a. So, by the second pair of similar triangles, we have that\n\\[ \\frac {DX}{XC} \\equal{} \\frac {TD}{AT}\n\\]\n\n\\[ XC \\equal{} 3a\n\\]\nSo, by the first pair,\n\\[ \\frac {BX}{XC} \\equal{} \\frac {BT}{TE}\n\\]\n\n\\[ \\frac {BX}{3a} \\equal{} 4\n\\]\n\n\\[ BX \\equal{} 12a\n\\]\n\n\\[ BD \\equal{} 11a\n\\]\n\n\\[ DC \\equal{} 4a\n\\]\n\n\\[ \\frac {CD}{BD} \\equal{} \\boxed{\\frac {4}{11}}\n\\]\n[/hide]\n\n[hide=\"24\"] Say that AD and BC intersect at P. So, by the Angle Bisector Theorem,\n\\[ \\frac {BP}{CP} \\equal{} \\frac {AB}{AC}\n\\]\n\n\\[ \\frac {BP}{CP} \\equal{} \\frac {7}{8}\n\\]\n\n\\[ BP \\plus{} CP \\equal{} BC \\equal{} 9\n\\]\n\n\\[ BP \\equal{} \\frac {21}{5}\n\\]\n\n\\[ CP \\equal{} \\frac {24}{5}\n\\]\nNow, ACDB is cyclic, so $ \\angle DAC \\equal{} \\angle BCD$ and $ \\angle ADC \\equal{} \\angle PDC$, so triangle ADC is similar to triangle CDP. Thus,\n\\[ \\frac {AC}{PC} \\equal{} \\frac {AD}{CD}\n\\]\n\n\\[ \\frac {AD}{CD} \\equal{} \\frac {8}{\\frac {24}{5}} \\equal{} \\frac {40}{24} \\equal{} \\boxed{\\frac {5}{3}}\n\\]\n[/hide]", "Solution_2": "23. \r\n\r\n[hide]The total permutation of painting is $ 2^6\\equal{}64$\n\nCase 1: all 6 faces are the same color, which always satisfies the condition; permutation: $ 1*2\\equal{}2$\n\nCase 2: 5 faces are the same color, which also always satisfies the condition; permutation: $ 6*2\\equal{}12$\n\nCase 3: 4 faces are the same color, which satisfies the condition if and only if the other 2 faces are opposite; so it has a permuation $ 3*2\\equal{}6$\n\nHence, the probability is: $ (2\\plus{}12\\plus{}6)/64\\equal{}5/16$[/hide]\r\n\r\nP.S.: At last I know how to use LaTeX...thanks @Icy and the creators...it looks much clearer now :)", "Solution_3": "[hide=\"23\"]Count the complement:\nIf 5 of one color, then, 0 ways such that there are no 4 square \"ring\" of some same color.\nIf 4, then one for every edge, 12.\nIf 3, then either one for every vertex, or one for every pair of edges that are on opposite sides of a face square, 12.\n\nThen, $ \\frac{12\\plus{}20\\plus{}12}{2^6}\\equal{}5/16$.[/hide]\n\n[hide=\"24\"]Let AD intersect BC at E. Since they bisect the same arc, angle ADC = angle ABC. By AA, $ \\triangle ABE ~ \\triangle ADC$. From angle-bisector, BE=7*9/15. AD/DC=AB/BE=5/3.[/hide]" } { "Tag": [ "trigonometry", "algebra unsolved", "algebra" ], "Problem": "$ x\\plus{}tnx\\equal{}0$", "Solution_1": "Lol can someone offer a explanation please.", "Solution_2": "hello, do you mean $ x\\plus{}\\tan(x)\\equal{}0$?\r\nSonnhard.", "Solution_3": "[quote=\"Dr Sonnhard Graubner\"]hello, do you mean $ x \\plus{} \\tan(x) \\equal{} 0$?\nSonnhard.[/quote]\r\nHa that makes more sense.", "Solution_4": "yes)) \r\n\r\n$ x\\plus{}tan(x)\\equal{}0$\r\n\r\nsorry)", "Solution_5": "hello, what do you want to know about it, the roots of the equation?\r\nSonnhard.", "Solution_6": "yes. if it has roots.. and it's given that $ 0 < \\equal{} x < \\equal{} \\pi/2$\r\n\r\nor just find the maximum of\r\n\r\n$ \\sin (2 \\alpha) \\left(800 \\minus{} \\frac{500}{\\pi}\\alpha \\right)$\r\n\r\nif $ 0<\\equal{}\\alpha<\\equal{} \\pi/2$", "Solution_7": "First of all $ \\tan(\\frac {\\pi}{2})$ is undefined, so we have $ 0 \\le x < \\frac {\\pi}{2}$. \r\n\r\nOver this interval, we have $ \\tan(x) \\ge 0$. \r\n\r\nSo, the only possible zero occurs when $ x \\equal{} \\tan(x) \\equal{} 0$, which is true for $ x \\equal{} 0$. Thus, $ x \\equal{} 0$ is the only solution.", "Solution_8": "[b]Xantos C. Guin[/b]\r\n\r\nI got it :) \r\n\r\ncan you help me to find the max $ f(\\alpha)$ if $ 0\\le \\alpha \\le \\frac {\\pi}{2}$\r\n\r\n$ f(\\alpha) = \\sin (2 \\alpha) \\left(800 - \\frac {500}{\\pi}\\alpha \\right)$\r\n\r\nI tried by this way but got another problem...\r\n$ f(\\alpha)'=0$\r\n\r\n$ 1600cos2\\alpha-frac{500}{\\pi}sin2\\alpha+2\\alpha cos2\\alpha=0$\r\n\r\n$ (1600-\\frac{1000\\alpha}{\\pi})cos2\\alpha=\\frac{500}{\\pi}sin2\\alpha$\r\n\r\n$ tan2\\alpha=\\frac{500}{1600\\pi-1000\\alpha}$\r\n\r\nor $ tan2\\alpha=\\frac{5}{16\\pi-10\\alpha}$", "Solution_9": "hello, differentiate your $ f(\\alpha)$ with respect to $ \\alpha$ we get\r\n$ f'(\\alpha)\\equal{}2\\cos(2\\alpha)\\left(800\\minus{}\\frac{500}{\\pi}x\\right)\\plus{}\\sin(2\\alpha)\\left(\\minus{}\\frac{500}{\\pi}\\right)$.\r\nNow you must solve the equation $ f'(\\alpha)\\equal{}0$.\r\nSonnhard.", "Solution_10": "hello, i have got $ 2\\cos(2\\alpha)\\left(800\\minus{}\\frac{500\\alpha}{\\pi}\\right)\\equal{}\\frac{500}{\\pi}\\sin(2\\alpha)$\r\n$ \\frac{2\\pi}{500}\\left(800\\minus{}\\frac{500\\alpha}{\\pi}\\right)\\equal{}\\tan(2\\alpha)$\r\n$ \\frac{16\\pi}{5}\\minus{}2\\alpha\\equal{}\\tan(2\\alpha)$\r\nSonnhard.", "Solution_11": "[b]Dr Sonnhard Graubner[/b]\r\n\r\nyour one is true) i made a mistake there))) there must be 1/tan(2$ \\alpha)$\r\n\r\nand how to find the roots now?", "Solution_12": "hello, i think it can only be solved by a numerical way.\r\nSonnhard." } { "Tag": [ "videos", "HMMT" ], "Problem": "So who's seen Napoleon Dynamite? I haven't, but when I do I probably won't laugh because everyone in my school quotes something from it every day, and it kind of gets annoying after a while. But what I don't understand is why the movie is getting so much attention after it came out on DVD, and not while it was in theaters (At least I think that's what happenned).", "Solution_1": "Eh, I saw it.\r\n\r\nIt was fun, but the stupidist movie I ever watched EVER.\r\n\r\nI dont' see the point in making such a pointless movie...", "Solution_2": "That was greatly fuuny and dumb!!", "Solution_3": "that movie has no right to exist\r\n\r\nit's like the ultimate feel-bad movie (hey! look at all these dweebs we found! what do you think we can make them do??????)", "Solution_4": "HAHA.... That was such a funny movie but really pointless and stupid.", "Solution_5": "The people at my school are like the people at hs_potter's. There's alwyas a bunch of people saying stuff like \"Tina you fat lard\" or something like that, I've never seen that movie.", "Solution_6": "we had to watch it in our school.\r\nit was really messed up (funny, but messed up)", "Solution_7": "i found it to be the dumbest most pointless movie i have ever seen. i just sat there the whole time and waited for them to snap out of it! and it didn't happen! i was like...oh my word", "Solution_8": "dont be hatin'. this movie was awesome. especially since:\r\na). i was reminded of several people i knew.\r\nb). i cracked up laughing at all the understated funnies.\r\nc). it was an independent film that made it big.", "Solution_9": "i would have to agree with everyone else: it was really dumb but still highly amusing. it's kinda nice to see something original even if it is that pointless. Also, I think it'll die off pretty soon.", "Solution_10": "Yes it was stupid, and was very one-dimensional, but it was still funny and I agree with bleumoose.\r\n\r\nAlso it will die off once another stupid funny movie takes its place.", "Solution_11": "[quote=\"mcalderbank\"]Yes it was stupid, and was very one-dimensional, but it was still funny and I agree with bleumoose.\n\nAlso it will die off once another stupid funny movie takes its place.[/quote]\r\n\r\ni honestly don't think that any movie can be dumber than napoleon dynamite. :)", "Solution_12": "can some1 actually tell me exactly is this movie about?\r\nLike some scenes or something", "Solution_13": "there is no plot so theres really nothing much to say", "Solution_14": "funny/absurd/gross scene:\r\n\r\nGuy shoots a cow while a schoolbus of children passes by.\r\n\r\nSound effects: Kids screaming, 'meeuh\"", "Solution_15": "It was so dumb it was funny", "Solution_16": "[quote=\"jli\"]It was so dumb it was funny[/quote]\r\n\r\nLol... You got that right! People are always asking why it was so funny. I'm like because it was stupid.", "Solution_17": "Someone brought it along to a XC-ski meet and we tried to watch it on the way back, but the bus's DVD player was screwed up, so we saw it in 3-second segments separated by 10 seconds of black. It was still funny.\r\n\r\nLast Wednesday, I met a kid they could have based the movie on.", "Solution_18": "[quote=\"Danbert\"]Someone brought it along to a XC-ski meet and we tried to watch it on the way back, but the bus's DVD player was screwed up, so we saw it in 3-second segments separated by 10 seconds of black. It was still funny.\n\nLast Wednesday, I met a kid they could have based the movie on.[/quote]\r\n\r\nOh you know thats not actually the DVD players fault. It was a promotion video and to prevent illegal coping they make it that way. I saw in that version also. Same way I saw Anchorman.", "Solution_19": "On our way back home from HMMT, we watched Napoleon Dynamite. I heard from people that this is a great movie. But you know what? Napoelon Dynamite is the WORST MOVIE EVER! There is no plot, it's just retarded....... I don't know why people came up to me saying it's the funniest movie in the world.... Maybe I missed something?", "Solution_20": "[quote=\"Jhyon8149\"]On our way back home from HMMT, we watched Napoleon Dynamite. I heard from people that this is a great movie. But you know what? Napoelon Dynamite is the WORST MOVIE EVER! There is no plot, it's just retarded....... I don't know why people came up to me saying it's the funniest movie in the world.... Maybe I missed something?[/quote]\r\n\r\nboy did u hit the nail squarely on the head! i couldn't agree more. i kno i kno i already posted my opinions here but this was so right on i had to say something. :)", "Solution_21": "[quote=\"Tokyo\"]can some1 actually tell me exactly is this movie about?\nLike some scenes or something[/quote]\r\n\r\nIt's um about Napoleon. And really about nothing and everything all at once.", "Solution_22": "[quote=\"Tokyo\"]can some1 actually tell me exactly is this movie about?\nLike some scenes or something[/quote]\r\nIt consists of a fair amount of scenes of someone just randomly staring at a wall. And then some Mexican dude gets elected school president. Thats basically the whole plot.", "Solution_23": "haha :D come on, you guys, theres a bit more to the movie than that.\r\n\r\nI thought it was funny because I could relate to a lot of the people in the movie, and the scene where Napoleon dances is pretty cool" } { "Tag": [ "geometry", "circumcircle", "incenter", "geometry open" ], "Problem": "Triangles ABC and A'B'C' lie on the same circle and have the same orthocenter. AA', BB',CC' intersect at points X,Y,Z. Prove that the circumcenter of XYZ lie on OH. ( H is a common orthocenter and O is circumcenter)", "Solution_1": "I have nice solution for this problem\r\nbut it so long and my picture not same to your picture\r\nIn my figure A,A',B,B',C,C' are on the circle wiht its order\r\nI use some lemma :\r\n 1) triangle ABC ,I,O be incenter and circumcenter of ABC ,I touch side of ABC at D,E,F then orthocenter of DEF ,O,I are conlinear\r\n 2) Let M,N,P be intersection of tangent of(O) at A,B,C\r\n M',N',P' be intersection of tangent of(O) at A',B',C'\r\n then M.N,P,M',N',P' are concyclic\r\n 3) AA',BB',CC' met at X,Y,Z on MM',NN',PP' and O'X perpendicular to MM'...etc...\r\n 4) two circle with ccenter O,O' .line $\\delta$ met (O) and (O') at A,B,C,D( a,b on (O) and C,D on (O')) let M be intersection of tangent at A,C of (O) and (O') ...similar ly we have N,P,Q then M,N,P,Q are concyclic wiht center O'' such O,O'and O'' are conlinear .\r\nSo you have proof", "Solution_2": "Someone replied :). I'll check your solution, I hope it's correct. Because mine is awful and too long. Thank you very much.", "Solution_3": "[quote=\"Mamat\"]Someone replied :). I'll check your solution, I hope it's correct. Because mine is awful and too long. Thank you very much.[/quote]\r\nI and two my friend was proved this problem and it correct -I think ( we take 6 day to prove it\r\nare you have solution for your problem? :)", "Solution_4": "Nice :). Now I have a good solution for this problem.", "Solution_5": "if you have time you may post so lution for this pro MAMAT :)", "Solution_6": "Ok, thank you hunter_cvp_ic for nice solution", "Solution_7": "Finally I have some time to check your solution. What is Q? Which order do you choose when you find M,N,P?", "Solution_8": "[quote=\"Mamat\"]Finally I have some time to check your solution. What is Q? Which order do you choose when you find M,N,P?[/quote]\r\n well ! You want to ask point Q on lemma 4 ,is't it?\r\nIf i'm correct then\r\n M be intersecttion of tangents from A and C of circle (O) and (O') N be intersection of tangents from A and D of (O) and (O') ,P be be intersection of tangents from B and C of (O) and (O') ,Q be be intersection of tangents from B and D of (O) and (O') . :lol:\r\nand you can choose M,N,P on any order for which they be intersection of one tangent of (O) and one of (O')", "Solution_9": "Nice work hunter_cvp_ic :)", "Solution_10": "How did you prove that circumcircle of XYZ is tangent to MM', NN', PP'?", "Solution_11": "Sorry ! it so long \r\nthen may be i can't post it now ..I will post it latter( some days)", "Solution_12": "It's ok, because I think that I have the proof, but it's really very long (if it's true).", "Solution_13": "This problem is identical to 2022 G7" } { "Tag": [ "probability", "ratio" ], "Problem": "Phillip flips an unfair coin eight times. This coin is twice as likely to come up heads as tails. How many times as likely is Phillip to get exactly three heads than exactly two heads?", "Solution_1": "The probability that he gets heads on any flip is $ \\frac23$ and $ \\frac13$ for tails. Hence, the probability that he gets exactly three heads is $ \\binom83\\left(\\frac23\\right)^3\\left(\\frac13\\right)^5$ and the probability that he gets exactly two heads is $ \\binom82\\left(\\frac23\\right)^2\\left(\\frac13\\right)^6$, so the ratio is $ \\frac {\\binom83\\left(\\frac23\\right)^3\\left(\\frac13\\right)^5}{\\binom82\\left(\\frac23\\right)^2\\left(\\frac13\\right)^6} \\equal{} \\frac {\\binom83\\cdot\\frac23}{\\binom82\\cdot\\frac13} \\equal{} \\frac {56\\cdot2}{28} \\equal{} 2\\cdot2 \\equal{} \\boxed4$.", "Solution_2": "[hide=Alternative solution]\nSo it's twice as likely to be heads than tails, as given in the problem.\nC(8,3)=8*7*6/3*2*1\nC(8,2)=8*7/2*1\n\nC(8,3)/C(8,2)=6/3=2\n\nThus there are twice as many ways and each of them is twice as likely as tails.\n2*2=4", "Solution_3": "Wait, why is this in Intermediate Algebra?\n\n[hide=Solution]The probability he gets heads is $\\frac{2}{3}$ and the probability he gets tails is $\\frac{1}{3}$. The answer is hence $\\frac{\\binom{8}{3}\\left(\\frac{2}{3}\\right)^3\\left(\\frac{1}{3}\\right)^5}{\\binom{8}{2}\\left(\\frac{2}{3}\\right)^2\\left(\\frac{1}{3}\\right)^6} = \\boxed{4}$.[/hide]", "Solution_4": "[hide=Solution.]We use generating functions. Denote by $1$ flipping tails and by $x$ flipping heads. Since the coin is twice as likely to come up heads, we need $2x,$ not $x.$ Thus, the generating function for one flip is $(1+2x).$ For eight flips, it is $(1+2x)^8.$ When expanded, the coefficient of $x^3$ is $\\binom83(2)^3(1)^5$ and the coefficient of $x^2$ is $\\binom82(2)^2(1)^6.$ This means our ratio is $\\frac{\\binom83(2)^3(1)^5}{\\binom82(2)^2(1)^6}=\\boxed4.$", "Solution_5": "[hide=long solution]There are $\\dbinom{8}{3}=56$ ways to get exactly $3$ heads. There is a $\\dfrac{2}{3}$ chance of Phillip getting a heads and a $\\dfrac{1}{3}$ chance of tails. For each case there is a $\\left(\\dfrac{2}{3}\\right)^3\\cdot\\left(\\dfrac{1}{3}\\right)^5=\\dfrac{8}{6561}$ chance of it occuring. So, there is a $56\\cdot\\dfrac{8}{6561}=\\dfrac{448}{6561}$ of getting exactly $3$ heads. We can do the same for $2$ heads: $\\dbinom{8}{2}=28$ ways, $\\left(\\dfrac{2}{3}\\right)^2\\cdot\\left(\\dfrac{1}{3}\\right)^6=\\dfrac{4}{6561}$. $28\\cdot\\dfrac{4}{6561}=\\dfrac{112}{6561}$. $\\dfrac{448}{6561}\\div\\dfrac{112}{6561}=\\boxed{4}$.[/hide]", "Solution_6": "[quote=Blue_banana4][hide=long solution]There are $\\dbinom{8}{3}=56$ ways to get exactly $3$ heads. There is a $\\dfrac{2}{3}$ chance of Phillip getting a heads and a $\\dfrac{1}{3}$ chance of tails. For each case there is a $\\left(\\dfrac{2}{3}\\right)^3\\cdot\\left(\\dfrac{1}{3}\\right)^5=\\dfrac{8}{6561}$ chance of it occuring. So, there is a $56\\cdot\\dfrac{8}{6561}=\\dfrac{448}{6561}$ of getting exactly $3$ heads. We can do the same for $2$ heads: $\\dbinom{8}{2}=28$ ways, $\\left(\\dfrac{2}{3}\\right)^2\\cdot\\left(\\dfrac{1}{3}\\right)^6=\\dfrac{4}{6561}$. $28\\cdot\\dfrac{4}{6561}=\\dfrac{112}{6561}$. $\\dfrac{448}{6561}\\div\\dfrac{112}{6561}=\\boxed{4}$.[/hide][/quote]\n\nMe: gets same problem again, goes to 'Page Feed, ' looks at my own solution and inputs correct answer." } { "Tag": [ "function" ], "Problem": "Find the values for each function below after plugging f(x + 1).\r\n\r\n(1) f(x) = sqrt{x^2 + 2}\r\n\r\n(2) f(x) = 1 - (1/(x + 2)^2", "Solution_1": "do u mean this..\r\n\r\n[hide=\"1\"]$f(x+1)=\\sqrt{(x+1)^{2}+2}=\\sqrt{x^{2}+2x+3}$[/hide]\n[hide=\"2\"]$f(x+1)=1-\\frac{1}{(x+1+2)^{2}}=1-\\frac{1}{(x+3)^{2}}= \\frac{(x+3)^{2}-1}{(x+3)^{2}}=\\frac{(x+2)(x+4)}{(x+3)^{2}}$[/hide]", "Solution_2": "Yes, I needed to learn HOW to do the questions and your reply shows me how.\r\n\r\nThanks!", "Solution_3": "well basically if it says find f(x+1), you just put x+1 in all the places where you see x\r\n\r\nexample:\r\n$f(x)=x^{2}$\r\nfind $f(x+1)$\r\nthen you plug $(x+1)$ into $X$\r\nso you get\r\n$(x+1)^{2}$", "Solution_4": "Yes, I got it.\r\n\r\nThanks!", "Solution_5": "do you have a book or something? all these problems are trivial and are covered in a algebra book", "Solution_6": "To altheman:\r\n\r\nYes, I have many books but none of the books are easy to follow, which is very common in terms of math, physics and astronomy." } { "Tag": [], "Problem": "Show that there do not exist positive integers $ x$,$ y$,$ u$, and $ v$ such that $ (x, y) \\equal{} 1$,$ (u, v) \\equal{} 1$, $ xy \\equal{} uv$, and $ x^2 \\plus{} y^2 \\equal{} u^2 \\minus{} v^2$.", "Solution_1": "Just set $ v\\equal{}xy/u$ and get $ (xy)^2$+$ (yu)^2$+$ (xu)^2$=$ u^4$.\r\n\r\n In page 3 there is an exercise I' ve posted, according to which the previous equation doesn't have solutions in positive integers.", "Solution_2": "[hide=\"panos_lo\"]\n\u03a3\u03a5\u0393\u03a7\u0391\u03a1\u0397\u03a4\u0397\u03a1\u0399\u0391 \u0393\u0399\u0391 \u03a4\u0397\u039d \u0395\u03a0\u0399\u03a4\u03a5\u03a7\u0399\u0391 \u03a3\u039f\u03a5 \u03a3\u03a4\u0397\u039d J\u0392\u039c\u039f :!: :!: \n[/hide]", "Solution_3": "Page 3? \r\n\r\nAnd Dimitris X, what are you saying?", "Solution_4": "[quote=\"Zhero\"]Page 3?[/quote]\n\nYep, page 3: http://www.mathlinks.ro/viewtopic.php?t=284204\n\n[quote=\"Zhero\"]And Dimitris X, what are you saying?[/quote]\r\n\r\nFrom what I gather, he's wishing panos success in the JBMO (should things like this really be said in a problem thread?)" } { "Tag": [ "FTW" ], "Problem": "There is no pass but please tell highest rating. If you did not start playing, go to http://www.artofproblemsolving.com/Edutainment/g1/index.php\r\nI will probably play from 9:00-9:30 Eastern Time.", "Solution_1": "Oops. I forgot to ask for your current rating.", "Solution_2": "Maybe you wanna post this in the FTWOS forum?", "Solution_3": "But I'm asking people in Algebra 1 class.", "Solution_4": "[quote=\"Dillinger\"]But I'm asking people in Algebra 1 class.[/quote]\r\n\r\nBut you're talking about FTW, right? ;)", "Solution_5": "Why don't you just pm them? Agreed that it's in the wrong forum though." } { "Tag": [ "geometry", "3D geometry", "number theory solved", "number theory" ], "Problem": "Find all natural number $n$ such that $3^n-71$ is perfect cube.", "Solution_1": "[hide]$3^{n} - 71 = x^3$.\n\nWe try $\\mod 8$.\n\n$3^{n} \\equiv 1, 3$.\n\n$x^3 \\equiv 0, 1, 3, 5, 7$.\n\n$71 \\equiv 7$.\n\nWe have only two cases.\n1. $x^3 \\equiv 1 - 7 \\equiv 2$.\n\n2. $x^3 \\equiv 3 - 7 \\equiv 4$.\n\nBut $2$ and $4$ are not cubic residues $\\mod 8$. No solutions.[/hide]" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Let n be a natural number such that n= a^2 + b^2 + c^2, for some natural numbers a,b,c. Prove that 9n = (p1a + q1b + r1c)^2 + (p2a + q2b + r2c)^2 + (p3a + q3b + r3c)^2, where pi, qi, ri are all non zero integers. Further,if 3 does not divide at least one of a,b,c, prove that 9n can be expressed in the form x^2 + y^2 +z^2, where x,y,z are natural numbers none of which is divisble by 3.", "Solution_1": "Let $n$ be a natural number such that $n= a^{2}+b^{2}+c^{2}$, for some natural numbers $a,b,c$ . Prove that $9n = (p_{1}a+q_{1}b+r_{1}c)^{2}+(p_{2}a+q_{2}b+r_{2}c)^{2}+(p_{3}a+q_{3}b+r_{3}c)^{2}$, where $p_{i}, q_{i}, r_{i}$ are all non zero integers. Further,if $3$ does not divide at least one of $a,b,c,$ prove that $9n$ can be expressed in the form $x^{2}+y^{2}+z^{2}$, where $x,y,z$ are natural numbers none of which is divisble by $3$.\r\n\r\n\r\nNow, I only see $9n=(a+2b+2c)^{2}+(2a-2b+c)^{2}+(2a+b-2c)^{2}$ :)", "Solution_2": "[quote=\"N.T.TUAN\"]Let $n$ be a natural number such that $n= a^{2}+b^{2}+c^{2}$, for some natural numbers $a,b,c$ . Prove that $9n = (p_{1}a+q_{1}b+r_{1}c)^{2}+(p_{2}a+q_{2}b+r_{2}c)^{2}+(p_{3}a+q_{3}b+r_{3}c)^{2}$, where $p_{i}, q_{i}, r_{i}$ are all non zero integers. Further,if $3$ does not divide at least one of $a,b,c,$ prove that $9n$ can be expressed in the form $x^{2}+y^{2}+z^{2}$, where $x,y,z$ are natural numbers none of which is divisble by $3$.\n\n\nNow, I only see $9n=(a+2b+2c)^{2}+(2a-2b+c)^{2}+(2a+b-2c)^{2}$ :)[/quote]\r\nHow can you say that each term is not divisible by 3???????\r\n\r\nWell The correct solution must be \r\nif $3$ does not divide $(a+b+c)$ then $9n$ can be expressed as\r\n$9n=(-a+2b+2c)^{2}+(2a-b+2c)^{2}+(2a+2b-c)^{2}=x^{2}+y^{2}+z^{2}$where $x,y,z$ are natural numbers none of which is divisble by $3$.\r\nbut if $3|(a+b+c)$ then this method fails as each of $x,y,z$ are divisible by $3$\r\nWLOG assume 3 does not divide $c$ (it's given).\r\nlet $a+b+c=3k$ for some k being a natural number , substituting \r\n$c=3k-a-b$ we get that, $9n$ can be expressed as\r\n$9n=(a+b+3k)^{2}+(a+4b-6k)^{2}+(4a+b-6k)^{2}=x^{2}+y^{2}+z^{2}$where $x,y,z$ are natural numbers none of which is divisble by $3$.\r\nbecause each term is congruent to $(a+b)$i.e $-c$ i.e non zero modulo $3$." } { "Tag": [ "trigonometry" ], "Problem": "cos^-1(x(sqrt3)) + cosx = pi/2\r\n\r\nsolve for x", "Solution_1": "hello, do you mean $ \\cos^{\\minus{}1}(x\\sqrt{3})\\plus{}\\cos(x)\\equal{}\\frac{\\pi}{2}$?\r\nSonnhard." } { "Tag": [ "algebra", "polynomial", "calculus", "integration", "function", "domain", "linear algebra" ], "Problem": "Let $ A$ and $ B$ be two commutative rings with unity such that $ A\\subseteq B$. Let $ u\\in B$ and $ v\\in B^{\\times}$. Assume that $ u\\in A\\left[v\\right]\\cap A\\left[v^{ \\minus{} 1}\\right]$. \r\n\r\n[b](a)[/b] Prove that $ u$ is integral over $ A$.\r\n\r\n(This is [url=http://jmilne.org/math/CourseNotes/math676.html]J. S. Milne, [i]Algebraic Number Theory[/i], version 3.02[/url], exercise 2-5, without the useless \"integral domains\" condition.)\r\n\r\n[b](b)[/b] If $ P\\in A\\left[X\\right]$ and $ Q\\in A\\left[X\\right]$ are polynomials satisfying $ u \\equal{} P\\left(v\\right) \\equal{} Q\\left(v^{ \\minus{} 1}\\right)$, then show that $ u$ is the root of a monic polynomial of degree $ \\deg P \\plus{} \\deg Q$ over $ A$. (Of course, this is a stronger version of part [b](a)[/b].)\r\n\r\n[b](c)[/b] Modify the above statements to replace the condition $ v\\in B^{\\times}$ by $ v\\in B$.\r\n\r\n([i]Remark.[/i] While [b](c)[/b] is easy, [b](b)[/b] seems to be harder than [b](a)[/b]. I know two solutions for [b](a)[/b], but only one of them still shows [b](b)[/b].)", "Solution_1": "Let the degrees of the polynomials be $ n$ and $ m$ respectively. I think (b) can be proven reasonably easily by simply writing the two equations for $ u$ (clearing denominators in the second equation) then multiplying the first equation by $ v^{i}$ for all $ i m\\end{array}\\right.$ for all $ i\\in\\left\\{0,1,...,n \\plus{} m\\right\\}$.\r\n\r\nThen, $ \\sum_{i \\equal{} 0}^{n \\plus{} m}a_iv^i \\equal{} 0$ and $ \\sum_{i \\equal{} 0}^n a_{i \\plus{} m}v^i \\equal{} u$. So it is enough to prove the following fact:\r\n\r\n[color=blue][b]Theorem 2.[/b] Let $ A$ and $ B$ be two rings (commutative, with unity) such that $ A\\subseteq B$. Let $ N\\in\\mathbb{N}$. Let $ a_0$, $ a_1$, ..., $ a_N$ be $ N \\plus{} 1$ elements of $ A$, and let $ v\\in B$ be such that $ \\sum_{i \\equal{} 0}^N a_iv^i \\equal{} 0$. Then, $ \\sum_{i \\equal{} 0}^{N \\minus{} k} a_{i \\plus{} k}v^i$ is integral over $ A$ for every $ k\\in\\left\\{0,1,...,N\\right\\}$.[/color]\r\n\r\nAnd this can be proven by backward induction over $ k$:\r\n\r\nFor $ k \\equal{} N$, we have $ \\sum_{i \\equal{} 0}^{N \\minus{} k} a_{i \\plus{} k}v^i \\equal{} a_N$, which is obviously integral over $ A$.\r\n\r\nSet $ \\rho_k \\equal{} \\sum_{i \\equal{} 0}^{N \\minus{} k} a_{i \\plus{} k}v^i$ for every $ k\\in\\left\\{0,1,...,N\\right\\}$. Then, $ \\rho_{k \\minus{} 1} \\equal{} \\sum_{i \\equal{} 0}^{N \\minus{} k} a_{i \\plus{} k \\minus{} 1}v^i$ satisfies $ \\rho_{k \\minus{} 1} \\equal{} \\rho_k v \\plus{} a_{k \\minus{} 1}$. Now, $ \\sum_{i \\equal{} 0}^N a_iv^i \\equal{} 0$ yields\r\n\r\n$ \\sum_{i \\equal{} 0}^{k \\minus{} 1}a_i\\rho_k^{k \\minus{} 1 \\minus{} i}\\left(\\rho_k v\\right)^i \\plus{} \\left(\\rho_k v\\right)^k \\equal{} \\rho_k^{k \\minus{} 1}\\left(\\sum_{i \\equal{} 0}^{k \\minus{} 1}a_iv^i \\plus{} \\rho_k v^k\\right)$\r\n$ \\equal{} \\rho_k^{k \\minus{} 1}\\left(\\sum_{i \\equal{} 0}^{k \\minus{} 1}a_iv^i \\plus{} \\sum_{i \\equal{} 0}^{N \\minus{} k} a_{i \\plus{} k}v^i v^k\\right) \\equal{} \\rho_k^{k \\minus{} 1}\\cdot\\underbrace{\\sum_{i \\equal{} 0}^N a_iv^i}_{ \\equal{} 0} \\equal{} 0$,\r\n\r\nso that $ \\rho_k v$ is integral over $ A\\left[\\rho_k\\right]$. Thus, $ \\rho_{k \\minus{} 1} \\equal{} \\rho_k v \\plus{} a_{k \\minus{} 1}$ is integral over $ A\\left[\\rho_k\\right]$ as well. Hence, if $ \\rho_k$ is integral over $ A$, then so is $ \\rho_{k \\minus{} 1}$, and we can proceed by induction.\r\n\r\nIf you try to get an estimate on the degree of $ \\rho_k$ from this proof, you get something like $ \\left(n \\minus{} k\\right)!$...\r\n\r\n darij", "Solution_3": "Nice notes, I learned a lot from them actually - thanks. (I like $ a\\in A$ implies $ a$ is 1-integral in particular :P). Seriously though, thm 2 is rather nice.", "Solution_4": "Well, I wrote them up just to have the proofs written out with every detail, so I can think about how they generalize to the $ \\left(A,I\\right)$-integral case. This is the case when $ A\\subseteq B$ is a ring extension, and $ I$ is an ideal of $ A$, and we call some $ u\\in B$ integral over $ \\left(A,I\\right)$ if and only if there exists some $ n\\in\\mathbb{N}$ and elements $ i_k\\in I^k$ for every $ k\\in\\left\\{1,2,...,n\\right\\}$ such that $ u^n \\plus{} \\sum_{k \\equal{} 1}^ni_ku^{n \\minus{} k} \\equal{} 0$. Note that this generalizes two well-known things: integrality over a ring (when $ I \\equal{} A$) and integrality over an ideal within the same ring (when $ B \\equal{} A$). If I have not made mistakes (what is quite a strong assumption), we have the following facts:\r\n\r\n[b]Analogue of Theorem 4.[/b] Let $ A$ and $ B$ be two rings such that $ A\\subseteq B$. Let $ v\\in B$ and $ u\\in B$. Let $ m\\in\\mathbb{N}$ and $ n\\in\\mathbb{N}$. Assume that $ v$ is $ m$-integral over $ A$ (not necessarily over $ \\left(A,I\\right)$ !), and that $ u$ is $ n$-integral over $ \\left(A\\left[v\\right],IA\\left[v\\right]\\right)$. Then, $ u$ is $ nm$-integral over $ \\left(A,I\\right)$.\r\n\r\n[b]Analogue of Theorem 5.[/b] Let $ A$ and $ B$ be two rings such that $ A\\subseteq B$.\r\n[b](a)[/b] Let $ a\\in I$. Then, $ a$ is $ 1$-integral over $ \\left(A,I\\right)$.\r\n[b](b)[/b] Let $ x\\in B$ and $ y\\in B$. Let $ m\\in \\mathbb{N}$ and $ n\\in \\mathbb{N}$. Assume that $ x$ is $ m$-integral over $ \\left(A,I\\right)$, and that $ y$ is $ n$-integral over $ \\left(A,I\\right)$. Then, $ x \\plus{} y$ is integral (and the degree is $ nm$) over $ \\left(A,I\\right)$.\r\n[b](c)[/b] Let $ x\\in B$ and $ y\\in B$. Let $ m\\in \\mathbb{N}$ and $ n\\in \\mathbb{N}$. Assume that $ x$ is $ m$-integral over $ \\left(A,I\\right)$, and that $ y$ is $ n$-integral over $ A$ (not necessarily over $ \\left(A,I\\right)$ !). Then, $ xy$ is integral (and the degree is $ nm$) over $ \\left(A,I\\right)$.\r\n\r\nNote that Theorem 4 can be proven using resultants, but the analogue of Theorem 5 doesn't follow from it anymore as nicely as Theorem 5 followed from Theorem 4. But at least, the analogue of Theorem 4 allows you to reduce the general case to the case $ A \\equal{} I$ and the case $ B \\equal{} A$. The first case is known, and the second one is thoroughly studied in [url=http://people.reed.edu/~iswanson/book/index.html]Irena Swanson and Craig Huneke, [i] Integral Closure of Ideals, Rings, and Modules[/i], 2006-2009[/url]. Now what somewhat frightens me is that I don't see any literature about the general case. Anyway, if you are still reading this, I could as well tell you the most general case: Let $ A\\subseteq B$ be a ring extension, and let $ \\left(I_k\\right)_{k\\in\\mathbb{N}}$ be a filtration of ideals in $ A$ (that is, $ I_1\\subseteq I_2\\subseteq I_3\\subseteq ...$ are ideals of $ A$ satisfying $ I_aI_b\\subseteq I_{a \\plus{} b}$ for all $ a$ and $ b$). An element $ u\\in B$ will be called integral over $ \\left(A,\\left(I_k\\right)_{k\\in\\mathbb{N}}\\right)$ if and only if there exists some $ n\\in\\mathbb{N}$ and elements $ i_k\\in I_k$ for every $ k\\in\\left\\{1,2,...,n\\right\\}$ such that $ u^n \\plus{} \\sum_{k \\equal{} 1}^ni_ku^{n \\minus{} k} \\equal{} 0$. How do elements of integral over $ \\left(A,\\left(I_k\\right)_{k\\in\\mathbb{N}}\\right)$ behave with respect to addition, multiplication and integrality over integrality? See the new version of [url=http://www.cip.ifi.lmu.de/~grinberg/Integrality.pdf]my integrality notes[/url] for some answers.\r\n\r\n darij" } { "Tag": [ "ratio" ], "Problem": "The ratio of losses to wins for Kyle's team is 3 to 2. If the team had played the same number of games, but had won twice as many of its games, what would the ratio of losses to wins have been? Express your answer as a common fraction.", "Solution_1": "This may be represented as 3 losses for every 2 wins. If the number of games is to stay the same but the number of wins is to double, then it would be 1 loss for every 4 wins. This is written in ratio as 1:4 or 1/4" } { "Tag": [ "algebra", "polynomial", "quadratics", "number theory unsolved", "number theory" ], "Problem": "Let [tex]p>4[/tex] be a prime number. Show that [tex]p^2 | \\sum_{j = 1}^{p - 1} \\frac{( p - 1 ) !}{j}[/tex].", "Solution_1": "The RHS is equal to $(2)(3)(4)...(p-1)+(1)(3)(4)...(p-1)+(1)(2)(4)...(p-1)+...+(1)(2)(3)...(p-2)$\r\n\r\nThe first term plus the last term is divisible by p. The second term plus the second to last term is divisible by p, etc. Rewriting and dividing by p, we have that the RHS divided by p is equal to:\r\n\r\n$(2)(3)(4)...(p-2)+(1)(2)(3)...(p-4)(p-3)(p-1)+...+(1)(2)...(.5(p-3))(.5(p+3))(.5(p+5))...(p-1)$\r\n\r\nWe want this too be congruent to 0 mod p. Dividing by (p-1)! mod p, we get:\r\n\r\n$1^{-1}(p-1)^{-1}+2^{-1}(p-2)^{-1}+...+(.5(p-1))^{-1}(.5(p+1))^{-1}$\r\n\r\nEquivalent to (mod p again):\r\n$(-1)(1^{2}+2^{2}+...(.5(p-1))^{2})$\r\nWhich is, by the well known formula for the sum of the first n squares, divisible by p... anyone got an easier way?", "Solution_2": "I guess I could have made the first couple steps more clear... the point is that:\r\n\r\n$(p-1)!/j + (p-1)!/(p-j) = (p-j + j) (p-1)!/(j(p-j)) = (p) (p-1)!j^{-1}(p-j)^{-1}$\r\nAnd then too, the sum for j=1...(p-1)/2 of $(p-1)!j^{-1}(p-j)^{-1}$ is divisible by p, so the original bit was divisible by pp.\r\n\r\nAnyhow, the reason it only works for p>4 is that the sum of the first (3-1)/2 squares is not divisible by three, while for all other odd primes, the sum of the first (p-1)/2 squares is divisible by p.", "Solution_3": "The RHS is the coefficient of $x$ in the polynomial $Q(x)=(x-1)(x-2)\\ldots(x-(p-1))$ (well, give or take a sign change, which doesn't matter anyway). $Q(p)=(p-1)!$, so $p^3|Q(p)-Q(0)$. \r\n\r\nNow, all the coefficients of $Q$ (except for the first one and the last one, of course) are divisible by $p$ (this is easy to prove by reducing $Q$ modulo $p$), so all the monomials of $Q$ of degree at least $2$ are divisible by $p^3$ (when the argument is $p$). This means that the monomial of $Q$ of degree $1$ is also divisible by $p^3$, so the coefficient of $x$ must be divisible by $p^2$.\r\n\r\n\r\n\r\nP.S.:\r\n\r\nIt's also easy to see why the above doesn't work for $p=3$: in that case, the degree of $Q$ is $2$, so it's not true that all the monomials of $Q$ of degree $\\ge 2$ are divisible by $p^3$.", "Solution_4": "That would be an easier way :)", "Solution_5": "let's note that :\r\n$\\sum_{j=1}^{p-1}\\frac{(p-1)!}{j}=\\sum_{j=1}^{[\\frac{p}{2}]}(p-1)!(j^{-1}+(p-j)^{-1})=\r\n\\sum_{j=1}^{[\\frac{p}{2}]}(p-1)!p(j(p-j)^{-1}=\r\np(p-1)!\\sum_{j=1}^{[\\frac{p}{2}]}(j(p-j))^{-1}=$\r\n$p(p-1)!\\sum_{j=1}^{[\\frac{p}{2}]}(j(p-j))^{-1}(\\mod p^2)$.\r\n\r\nSo let's proove that $\\sum_{j=1}^{[\\frac{p}{2}]}(j(p-j))^{-1}=0$ $(\\mod p)$.We know that \r\n$(j(p-j))^{-1}$ $(\\mod p)$ $=$ ${-j^2}^{-1}$ $(\\mod p)$,so\r\n${\\sum_{j=1}^{[\\frac{p}{2}]}(j(p-j))^{-1}=\\sum_{j=1}^{[\\frac{p}{2}]}{-j^2}^{-1}=}=-\\sum_{j=1}^{[\\frac{p}{2}]}{j^2}^{-1}=\\sum_{j=1}^{[\\frac{p}{2}]}{j^2}=0$ $(\\mod p)$,\r\ncause the sum of quadratic residuums of $p$ is $0$ $(\\mod p)$ if $p>3$,q.e.d.\r\n\r\nps. I haven't seen other posts while writing this text, maybe other solutions are similar in some way." } { "Tag": [], "Problem": "what did your team place at county? we got 1st :)", "Solution_1": "We got 7th, and yeah, we really stink. cuz everyone made ME do ALL of the problems on the team round alone, and so we probably failed it", "Solution_2": "We came 2nd;stupid mistake on team round.", "Solution_3": "well, at states, we got 6 and placed 9th cuz we changed 3 right answers to wrong answer at the last sec. literaly", "Solution_4": "He asked for county :wink: \r\nBut we came 3rd at states.", "Solution_5": "oh...we got 1st like him", "Solution_6": "I'm pretty sure we have the hardest chapter in MI...", "Solution_7": "your location says ohio. did you just move?", "Solution_8": "Ya, I moved in June.", "Solution_9": "so you live in michigan. not ohio....or vise versa?", "Solution_10": "No, I live in Ohio, and moved from Michigan. \r\nWe're getting off topic.", "Solution_11": "oh you got me confused when you said \"have\". ok", "Solution_12": "This discussion is getting pretty off-topic, and we have plenty of threads about results / scores already, so I'm going to lock this topic." } { "Tag": [ "induction", "real analysis", "real analysis unsolved" ], "Problem": "Can one find continuum many disjoint subsets of $\\mathbb{R}$ all of positive measure ?", "Solution_1": "If by \"measure\" you mean \"outer measure\", then I believe so. I think I even started a topic on the subject, in which fedja answered. \r\n\r\nThe proof I have in mind involves well-ordering the $2^{\\aleph_{0}}$ open sets of measure $1$, say, and building our $2^{\\aleph_{0}}$ sets by transfinite induction, making sure that they are disjoint and not contained in any of the open sets of measure $1$. That way you get sets of outer measure $>1$. I don't have the time to go through this in detail though." } { "Tag": [ "geometry", "calculus", "integration" ], "Problem": "find the area contained by the points of intersection of the graphs of $ 12x^2\\minus{}20xy\\plus{}7y2\\equal{}0$ and $ 2x\\minus{}3y\\plus{}4\\equal{}0$. all answers are in square units\r\n\r\n*btw, this is a comp. test, so no calculus should be needed*", "Solution_1": "[quote=\"DaReaper\"]find the area contained by the points of intersection of the graphs of $ 12x^2 \\minus{} 20xy \\plus{} 7y2 \\equal{} 0$ and $ 2x \\minus{} 3y \\plus{} 4 \\equal{} 0$. all answers are in square units\n\n*btw, this is a comp. test, so no calculus should be needed*[/quote]\r\nActually DaReaper...I wouldn't be surprised if the solution contained calculus...this year the problems had some calculus in it...especially ciphering...a possible suggestion would be to find the intersection points and take the integral of one subtracted from the other from the intersection points.\r\nI couldn't find a non-calculus method when I was taking the test :|", "Solution_2": "so there is no way to do this w/o calc??", "Solution_3": "Using calculus is by far the simplest method. The region is very irregular, so it would be difficult to calculate." } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "The $ 2n$ move chess game has the same rules as the regular one,with only one exception each player has to make $ 2n$ consecutive moves at a time. Prove that white (who plays first) always has a non losing strategy.\r\n\r\nHere $ n$ is a natural number!", "Solution_1": "Let's suppose second player has a winning strategy.\r\n\r\nWhite moves knight $ 2n$ times, back and forth, back and forth, to give black the same starting position.\r\n\r\nThen white must have at this point a non-losing strategy because the option for a consecutive move draw has been left open, and second player (now white) from absolute start has a winning strategy.", "Solution_2": "For $ n\\ge 2$ prove that white has a winning strategy :D :P", "Solution_3": ":lol: \r\n\r\ne4 Bc4 Qf1 Qf7#", "Solution_4": "Ok lets say I make a condition that we are not to repeat two moves at a time!Then?", "Solution_5": "[quote=\"manjil\"]Ok lets say I make a condition that we are not to repeat two moves at a time!Then?[/quote]\r\nThen what? I can't find a solution then. Is there any?", "Solution_6": "[quote=\"ChessTal\"][quote=\"manjil\"]Ok lets say I make a condition that we are not to repeat two moves at a time!Then?[/quote]\nThen what? I can't find a solution then. Is there any?[/quote]\r\n\r\nWell that's my question right!", "Solution_7": "[quote=\"manjil\"][quote=\"ChessTal\"][quote=\"manjil\"]Ok lets say I make a condition that we are not to repeat two moves at a time!Then?[/quote]\nThen what? I can't find a solution then. Is there any?[/quote]\n\nWell that's my question right![/quote]\r\nOK then:\r\n\r\nSo you are asking to prove that:\r\n[i] The $ 2n$ move chess game has the same rules as the regular one,with only one exception each player has to make $ 2n$ consecutive moves at a time. Prove that white (who plays first) always has a non losing strategy but with not repeating two moves at a time*.[/i]\r\n\r\n*I guess by \"at a time\" you mean \"per turn\".\r\n There is also one ambiguity about what we would do with checks. I mean if in a 6(n=3) Chess game, white for example has in a move a 6-moves move, then what happens if in the 4th move of the series for e.g, he gives a check? Does his turn ends(in the 4th move stopping the 6 move series) and it's the opponent's turn to play his 6 moves or he would capture the King in his 5th move and win? The last case obviously is not a valid Chess game since in Chess there is no King capture.\r\n\r\n[u] Having the last thing in mind and by adopting the second non-Chess case of a [b]\"captured King or checkmated King=win condition\"[/b] we have that:[/u]\r\n\r\n\u2022For $ n\\ge2$ both sides have $ 2n\\ge4$ moves per turn to play. So white just plays at the start as it was mentioned before 1.e4, Bc4, Qf3, Qxf7 mate!\r\nSo for $ n\\ge2$ this game guarantees always white a win.\r\n\r\n\u2022For $ n \\equal{} 1$ both sides have 2 moves per turn to play. So white starts with 1.Nc3, e3\r\nIn his second move and always depending on black's answer, he plays 2.Qf3, Bc4\r\nNow threats for black to be mated or captured(his King) are too strong and i haven't found saving moves for him, but there might be. And obviously this in not a complete solution to the $ n \\equal{} 1$ case.\r\n\r\nAnd i'm not even sure if you mean this kind of game. So you need to clarify the ambiguity....", "Solution_8": "I doubt there's a winning strategy for White in the $ n \\equal{} 1$ variant (Alessandro Castelli wouldn't agree: he'd say 1. e4/Nf3 or 1. d4/Nf3 are theoretically 1:0 situations :) ).\r\n\r\nThe reason is: both [url=http://www.chessvariants.org/multimove.dir/marseill.html]Marseillais[/url]and [url=http://www.chessvariants.org/multimove.dir/doublemove.html]Doublemove[/url]chess are still played for fun, even by GMs. If there was such a strategy, the games would be abandoned (although-White has a considerable advantage in the game).\r\n\r\nMaybe I'll return to this some day,when I'm not too busy and analyze it further. ;)" } { "Tag": [ "MATHCOUNTS" ], "Problem": "I am Shi-Ke Xue; I used to go to Takoma Park Middle School before I moved. I just thought I might ask/find out how many people posting in the Maryland AoPS forum goes/went there.", "Solution_1": "I'm not from Takoma, and unfortunately I have a great dislike (no offense to you, or any AoPSer that goes there) for them. Mostly jealousy, since they destroy us in math and chess :P but they were pretty arrogant about it. \r\n\r\nFortunately for us, we got back at them in band?", "Solution_2": "I think everyone in Maryland who doesn't go to Takoma Park hates them. Their perpetual domination of Mathcounts gets sort of annoying.\r\n\r\nAs you might guess, I didn't go there, although I met a few of their students when I trained with their coach as the alternate for National MC in 8th grade. They were less arrogant than I had been led to believe they were.\r\n\r\nmchoi815 is a Takoma Park alumni, I believe.", "Solution_3": "alumnus? or is \"alumni\" acceptable?\r\n\r\nand yes I am.", "Solution_4": "Yes, yes, I forget that \"alumni\" is plural . . .", "Solution_5": "I know this hasn't been posted on in a month, so this should be considered spam, but yeah i currently go to Takoma Park.", "Solution_6": "I go to TPMS too. Unfortunately, i did not make any competition teams. The basic idea is that a lot of the good people get picked to go to TPMS, they pick out the good people and they come (it is a magnet school). \r\n\r\nWe also have lots of alumni, and i think that a lot of good, Blair students came from TPMS.", "Solution_7": "Wow - Takoma's a magnet school? For what subject? I thought it was a public school. Is Montgomery Blair also a magnet school?", "Solution_8": "Magnet schools, including Takoma and Blair, are public schools. I think they both have math magnets.\r\n\r\nNote that there are actually a [i]lot[/i] of magnet schools around, most of which aren't any good. Takoma and Blair are unusual because they have such a strong reputation that they get almost everyone in the county who is good at math.", "Solution_9": "[quote=\"E^(pi*i)=-1\"]Note that there are actually a [i]lot[/i] of magnet schools around, most of which aren't any good.[/quote]\r\n\r\nNorth County is one of them. Their STEM program is supposedly really good, but they perform dismally at math and programming competitions. Ironically, my school has placed 1st at our county competitions for the past two years and managed to attract some of the best (math) students in the county with a non-math program (that I myself succumbed to). (And, on that note, our programming team is made up solely of students that are in that exact same program) :)", "Solution_10": "[quote=\"E^(pi*i)=-1\"] I think they both have math magnets.\n[/quote]\r\n\r\nyeah. it has the math, science, adn computer science magnet program", "Solution_11": "Can Montgomery Blair \"magnet\" students from outside of Montgomery County?", "Solution_12": "Sadly, no. I would go there if they could.", "Solution_13": "oh that's too bad.\r\ni'm sure they already have a handful with moco though. :)", "Solution_14": "i can haz chat on ftw if i post?", "Solution_15": "who is everyone\nidk any aops usernames", "Solution_16": "I might know someone on here cuz I go to a school not named Takoma Park or Montgomery Blair where I go to Montgomery Blair HS from my school so I can ride the bus to go home.", "Solution_17": "who went to the math competition on the Jan. 11th?", "Solution_18": "like\nthe one at takoma park?", "Solution_19": "[quote=presto502]Mathalon was trash.\n\nAlso yes we (Takoma) got first (some random seventh grade team did).\nThis was due to the fact that they screwed up the team round scoring. I doubt they will fix it.\n\nA certain person who does not go to Takoma got first in individual.[/quote]\n\nxd that was me\nsorry if this is old or something", "Solution_20": "Gj :P\nAlso this post was originally from 2008 so you're good", "Solution_21": "[quote=pie314159265]like\nthe one at takoma park?[/quote]\n\nyes", "Solution_22": "Who is ohsomething? I currently go to Eastern and I am on the math team going to MathCounts state round. Are you an alumni or a current student?", "Solution_23": "current. I will pm you for privacy reasons.", "Solution_24": "btw mathalon was troll this year..... i got 29 and was like 4th LOL (also on 3rd place team)", "Solution_25": "i go to takoma\ni can list some other aops users who go if u like", "Solution_26": "[quote=statman]btw mathalon was troll this year..... i got 29 and was like 4th LOL (also on 3rd place team)[/quote]\n\nThere were three perfect scores, All CMP students, Daniel Yuan (RFMS), Matthew Liu (BMMS), and Brandon Du (EMMS). So you were in good company. On the tie-breakers, Matthew came in first, Daniel came in second, and Brandon came in third because he forgot to write his tie breaker questions on his score sheet instead of the test sheet. Had he put them in the right place, he might have come in first.", "Solution_27": "[quote=CatherineAsaro][quote=statman]btw mathalon was troll this year..... i got 29 and was like 4th LOL (also on 3rd place team)[/quote]\n\nThere were three perfect scores, All CMP students, Daniel Yuan (RFMS), Matthew Liu (BMMS), and Brandon Du (EMMS). So you were in good company. On the tie-breakers, Matthew came in first, Daniel came in second, and Brandon came in third because he forgot to write his tie breaker questions on his score sheet instead of the test sheet. Had he put them in the right place, he might have come in first.[/quote]\n\nsorry, that comment i made was from 2016 :D", "Solution_28": "Oh, yeah, sorry, that was from the previous year. The three perfect scores were this year.", "Solution_29": "clemente fell off severely" } { "Tag": [ "algebra", "polynomial", "inequalities", "function", "maximization", "IMO Shortlist" ], "Problem": "Let $ P(x)$ be the real polynomial function, $ P(x) \\equal{} ax^3 \\plus{} bx^2 \\plus{} cx \\plus{} d.$ Prove that if $ |P(x)| \\leq 1$ for all $ x$ such that $ |x| \\leq 1,$ then\r\n\r\n\\[ |a| \\plus{} |b| \\plus{} |c| \\plus{} |d| \\leq 7.\\]", "Solution_1": "Let $ P( \\minus{} 1) \\equal{} y_1,\\ P( \\minus{} 1/2) \\equal{} y_2,\\ P(1/2) \\equal{} y_3,\\ P(1) \\equal{} y_4$. Then from Lagrange's interpolation formula (or from a linear system of equations) we obtain:\r\n$ P(x) \\equal{} \\frac 1 3 (( \\minus{} 2y_1 \\plus{} 4y_2 \\minus{} 4y_3 \\plus{} 2y_4)x^3 \\plus{} (2y_1 \\minus{} 2y_2 \\minus{} 2y_3 \\plus{} 2y_4)x^2 \\plus{}$\r\n$ (\\frac 1 2 y_1 \\minus{} 4y_2 \\plus{} 4y_3 \\minus{} \\frac 1 2 y_4)x \\plus{} ( \\minus{} \\frac 1 2y_1 \\plus{} 2y_2 \\plus{} 2y_3 \\minus{} \\frac 1 2 y_4)$.\r\nIt is easy to check, that if $ y_i \\in [ \\minus{} 1,1]$, then $ \\pm a \\pm b \\pm c \\pm d \\le 7$ for any combination of pluses and minuses. So, $ |a| \\plus{} |b| \\plus{} |c| \\plus{} |d| \\le 7$.\r\n\r\n[b]Remark.[/b] This solution applies the technics, which is common for polynomials, bounded on [-1, 1]. Actually we consider Chebyshev's polynomial $ T_n(x)\\equal{}\\cos(n\\arccos(x))$ of the same degree and compare it with our polynomial in points, where $ T_n(x)\\equal{} \\pm 1$. In our case it will be a polynomial $ 4x^3\\minus{}3x$ and the corresponding points are $ \\minus{}1,\\ \\minus{}1/2,\\ 1/2,\\ 1.$", "Solution_2": "[quote=\"Sasha Rybak\"]It is easy to check, that if $ y_i \\in [ \\minus{} 1,1]$, then $ \\pm a \\pm b \\pm c \\pm d \\le 7$ for any combination of pluses and minuses. So, $ |a| \\plus{} |b| \\plus{} |c| \\plus{} |d| \\le 7$.[/quote]\r\nCan you explain this further? Why is it \"easy to check,\" and how do you check it?", "Solution_3": "Let $a \\le 0, b \\le 0$.Then set $Q(x)=-P(x)$\nLet $a \\le 0,b \\ge 0$.Then set $Q(x)=P(-x)$\nLet $a \\ge 0,b \\le 0$.Then set $Q(x)=-P(-x)$\n\nIn each of these cases we see that $Q(x)$ satisfies all the conditions of the problem and the coefficient of $x^3$ and $x^2$ are nonnegative.So wlog let $a \\ge 0,b \\ge 0$.\n\n$c \\ge 0,d \\ge 0 \\Rightarrow |a|+|b|+|c|+|d|=a+b+c+d=P(1) \\le 1$\n$c \\ge 0,d < 0 \\Rightarrow |a|+|b|+|c|+|d|=a+b+c-d=P(1)-2P(0) \\le 3$\n$c < 0,d \\ge 0 \\Rightarrow |a|+|b|+|c|+|d|=a+b-c+d=\\frac{4}{3}P(1)-\\frac{1}{3}P(-1)-\\frac{8}{3}P(\\frac{1}{2})+\\frac{8}{3}P(-\\frac{1}{2}) \\le 7$\n$c < 0,d < 0 \\Rightarrow |a|+|b|+|c|+|d|=a+b-c-d=\\frac{5}{3}P(1)-4P(\\frac{1}{2})+\\frac{4}{3}P(-\\frac{1}{2}) \\le 7$\n\nSo we are done!!", "Solution_4": "This problem is very reminiscent of Chebyshev bounds on polynomials. What I have so far is that $|f(x)| \\le |C_3(x)|$ for $|x| \\ge 1$, where $C_3(x)$ is the 3rd degree Chebyshev polynomial. However, I cannot seem to isolate into $|a| + |b| + |c| + |d|$? Does anyone have a completion?", "Solution_5": "[quote=orl]Let $ P(x)$ be the real polynomial function, $ P(x) \\equal{} ax^3 \\plus{} bx^2 \\plus{} cx \\plus{} d.$ Prove that if $ |P(x)| \\leq 1$ for all $ x$ such that $ |x| \\leq 1,$ then\n\n\\[ |a| \\plus{} |b| \\plus{} |c| \\plus{} |d| \\leq 7.\\][/quote]\nLet $f(x)=a+bx+cx^2+dx^3$ and $|f(0)|\\leq 1,|f'(0)|\\leq 1,|f''(0)|\\leq 1,|f(1)|\\leq \\frac{1}{2}.$\nThen for all $x\\in [0,1]$ [url=https://artofproblemsolving.com/community/c6h1425490p8030153] we have [/url]$|f(x)|\\leq \\frac{11}{8}.$\n", "Solution_6": "[quote=sayantanchakraborty]Let $a \\le 0, b \\le 0$.Then set $Q(x)=-P(x)$\nLet $a \\le 0,b \\ge 0$.Then set $Q(x)=P(-x)$\nLet $a \\ge 0,b \\le 0$.Then set $Q(x)=-P(-x)$\n\nIn each of these cases we see that $Q(x)$ satisfies all the conditions of the problem and the coefficient of $x^3$ and $x^2$ are nonnegative.So wlog let $a \\ge 0,b \\ge 0$.\n\n$c \\ge 0,d \\ge 0 \\Rightarrow |a|+|b|+|c|+|d|=a+b+c+d=P(1) \\le 1$\n$c \\ge 0,d < 0 \\Rightarrow |a|+|b|+|c|+|d|=a+b+c-d=P(1)-2P(0) \\le 3$\n$c < 0,d \\ge 0 \\Rightarrow |a|+|b|+|c|+|d|=a+b-c+d=\\frac{4}{3}P(1)-\\frac{1}{3}P(-1)-\\frac{8}{3}P(\\frac{1}{2})+\\frac{8}{3}P(-\\frac{1}{2}) \\le 7$\n$c < 0,d < 0 \\Rightarrow |a|+|b|+|c|+|d|=a+b-c-d=\\frac{5}{3}P(1)-4P(\\frac{1}{2})+\\frac{4}{3}P(-\\frac{1}{2}) \\le 7$\n\nSo we are done!![/quote]\n\nWhat is Q(x) here\n\n", "Solution_7": "Also if |X| \u2264 1 then isn't X only 1and -1", "Solution_8": "Let $x_1=-1,x_2=-\\frac 12,x_3=\\frac 12,x_4=1.$ From Lagrange's interpolation formula$,$\n$$\\begin{aligned}P(x)&=\\sum\\limits_{k=1}^4P(x_k)\\prod\\limits_{j\\neq k}\\frac {x-x_j}{x_k-x_j}\\\\&=-\\frac 23P(-1)\\left(x^3-x^2-\\frac x4+\\frac 14\\right)+\\frac 43P\\left(-\\frac 12\\right)\\left(x^3-x^2-\\frac x4+\\frac 14\\right)\\\\&\\quad\\text{ }-\\frac 43P\\left(\\frac 12\\right)\\left(x^3+\\frac {x^2}2-x-\\frac 12\\right)+\\frac 23P(1)\\left(x^3+x^2-\\frac x4-\\frac 14\\right).\\end{aligned}$$\nFor $p,q,r,s\\in\\mathbb R,$ define $(p,q,r,s)=pP(x_1)+qP(x_2)+rP(x_3)+sP(x_4).$\nThen $a=\\left(-\\frac 23,\\frac 43,-\\frac 43,\\frac 23\\right),b=\\left(\\frac 23,-\\frac 23,-\\frac 23,\\frac 23\\right),c=\\left(\\frac 16,-\\frac 43,\\frac 43,-\\frac 16\\right),d=\\left(-\\frac 16,\\frac 23,\\frac 23,-\\frac 16\\right).$\nTherefore\n$a+b=\\left(0,\\frac 23,-2,\\frac 43\\right),a-b=\\left(-\\frac 43,2,-\\frac 23,0\\right)\\Rightarrow |a|+|b|=\\max\\{|a+b|,|a-b|\\}\\leqslant 4.$\n$c+d=\\left(0,-\\frac 23,2,-\\frac 13\\right),c-d=\\left(\\frac 13,-2,\\frac 23,0\\right)\\Rightarrow |c|+|d|=\\max\\{ |c+d|,|c-d|\\}\\leqslant 3.$\n$\\therefore |a| + |b| + |c| + |d|\\leqslant 4+3=7.\\blacksquare$", "Solution_9": "[img]https://i.ibb.co/x6SRRwK/1996-ISL-A5.png[/img]\n[hide=P(x) is a real polynomial function, why do you subtitute i?]In [url=https://mathworld.wolfram.com/RealPolynomial.html]Wolfram Mathworld[/url], I saw this definition of [b]Real Polynomial[/b]:\n[img]https://i.ibb.co/wQ7rSNz/def.png[/img]Also, it's given that $ |P(x)| \\le 1$ $\\boxed{\\text{for all} \\; x\\; \\text{such that}\\; |x| \\le 1},$ not $\\text{for all real}\\; x\\; \\text{such that}\\; |x|\\le 1.$[/hide]" } { "Tag": [ "geometry", "power of a point", "radical axis", "geometry unsolved" ], "Problem": "For a circle [b]O[/b] with its chord [b]AB[/b] , let two circles [b]P[/b] , [b]Q[/b] intouch the circle [b]O[/b] at [b]R[/b] , [b]S[/b]\r\n[b]AB[/b] = external common tangent of [b]P[/b] and [b]Q[/b]\r\nIntersection point of internal common tangents of [b]P[/b] and [b]Q[/b] = [b]L[/b] \r\nmidpoint of arc [b]AB[/b] = [b]M[/b] , midpoint of arc [b]RS[/b] = [b]N[/b]\r\n\r\n[b]L[/b] , [b]M[/b] , [b]N[/b] are collinear \r\n\r\nIs this right ?\r\nIf so , May I ask you a favor ? ( as synthetic ones as possible )\r\nThank you in advance :) \r\nHorrible chores :(", "Solution_1": "I am sorry , but this problem maybe wrong :(", "Solution_2": "Thanks !\r\nI feel that MN is the radical axis of two circles P and Q from your nice figure.\r\nMay I ask you the certification or the proof ? :blush: sorry to bother you", "Solution_3": "I don't think so , because PQ and MN are not perpendicularity" } { "Tag": [ "logarithms", "inequalities unsolved", "inequalities" ], "Problem": "Let $ n \\in Z, n \\geq 2$ and $ x,y \\geq 0$. Prove that:\r\n\r\n $ \\sqrt[n]{x^n\\plus{}y^n} \\ge \\sqrt[n\\plus{}1]{x^{n\\plus{}1}\\plus{}y^{n\\plus{}1}}$", "Solution_1": "[quote=\"ngoclam1088\"]Let $ n \\in Z, n \\geq 2$ and $ x,y \\geq 0$. Prove that:\n\n $ \\sqrt [n]{x^n \\plus{} y^n} \\ge \\sqrt [n \\plus{} 1]{x^{n \\plus{} 1} \\plus{} y^{n \\plus{} 1}}$[/quote]\r\nIf $ x \\equal{} 0$ or $ y \\equal{} 0$, we obtain the equality. Hence, suppose that $ x > 0$ and $ y > 0$. Then equivalently\r\n$ \\sqrt [n]{t^n \\plus{} 1}\\ge\\sqrt [n \\plus{} 1]{t^{n \\plus{} 1} \\plus{} 1},\\qquad\\text{where}\\quad t: \\equal{} \\frac {x}{y}.$\r\nNow, you can take logarithms to get a more elegant form of this inequality. The rest is easy." } { "Tag": [ "trigonometry", "function", "AMC" ], "Problem": "If\r\n\r\n\\[ f\\left(\\frac{x}{x-1}\\right) = \\frac{1}{x}, \\ \\text{for all} \\ x \\neq 0,1 \\]\r\n\r\nand $0 < \\theta < \\pi/2$, then $f(\\sec^2 \\theta)=$\r\n\r\n$\\text{(A)} \\ \\sin^2 \\theta \\qquad \\text{(B)} \\ \\cos^2 \\theta \\qquad \\text{(C)} \\ \\tan^2 \\theta \\qquad \\text{(D)} \\ \\cot^2 \\theta \\qquad \\text{(E)} \\ \\csc^2 \\theta$", "Solution_1": "[hide]\n$f\\left(\\frac{1}{\\cos^2{\\theta}}\\right) = f\\left(\\frac{1}{1-\\sin^2{\\theta}}\\right) = f\\left(\\frac{\\csc^2{\\theta}}{\\csc^2{\\theta}-1}\\right) = \\frac{1}{\\csc^2{\\theta}} = \\sin^2{\\theta}$. So A. [/hide]" } { "Tag": [ "algebra", "function", "domain", "topology", "real analysis", "real analysis unsolved" ], "Problem": "Let f:U -> R^m be continuously differentiable where m>1 and U is an open set in R^m. Suppose {x in U: det [f'(x)]=0 } only contains isolated points. Prove f is an open map. Use this result to prove the Fundamental Theorem of Algebra.", "Solution_1": "I have seen this old problem but never manage to solve it. (not even the first part). it is only necessary to prove that if\r\n$ f: \\mathbb{R}^n\\rightarrow\\mathbb{R}^n, f \\in C^1, n>1$\r\n$ det(f'(x)) \\equal{} 0$ iff $ x \\equal{} 0$,\r\n\r\nf is injective for some neighborhood around 0.\r\n\r\nthe theorem is false in $ \\mathbb{R}$", "Solution_2": "sorry I misread the problem... it says prove that it is an open map, injectivity is a stronger result by invariance of domain (i wonder if this stronger result is true though)." } { "Tag": [], "Problem": "There is this ant of mass $m$ that is on a rotating compact disk. Given that the coefficient of friction between the ant and the disk is $\\mu$. What is the greatest frequency the disk can be rotating at such that the ant is able to stay on? (The radius of the disk is $R$, the gravitational acceleration is $g$).", "Solution_1": "If the ant rests in the center of the disk it can stay there arbitrarily long...", "Solution_2": "Djole is right!you must specify the position of the ant OR else the solution to your problem is quite simple I suppose \r\n \r\n$mxw^{2}={\\mu}mg$\r\nthis implies$w^{2}= \\frac{{\\mu}g}{x}$\r\n\r\nthus $w$ is maximised for the position $x = 0$ i.e when thwe ant is situated at the centre of the circle." } { "Tag": [ "function", "trigonometry", "calculus", "derivative" ], "Problem": "I don't know if this belongs to this forum but oh well...\r\n\r\nIf $f(x)=\\sin x$, find $f^{51}(x)$.", "Solution_1": "Do you mean the $51$-th deriviative? If so, $f^{(4)}=f^{(0)}$, thus $f^{(51)}= f^{(3)}=-\\cos x$.", "Solution_2": "When $f(x)=\\sin x$ $f^{(n)}=\\sin (x+\\frac{n\\pi}{2})$", "Solution_3": "Is $f^{51}(x)$ a common way of writing the $51$-th deriviative of $f$?", "Solution_4": "I've seen it meaning the derivative a number of times, I've also seen it as the number of function applications, $f^{3}(x)$ being $f(f(f(x)))$.", "Solution_5": "It's more common to write $f^{(n)}(x)$ for the n-th derivative." } { "Tag": [ "number theory open", "number theory" ], "Problem": "Let x 2\r\np Prime\r\n\r\nfind all Solutions so that\r\n\r\n $ y\\minus{}px | y^2\\minus{}xy\\minus{}2x$", "Solution_1": "I guess there are infinitely many.\r\n\r\n$ (y \\minus{} px)m \\equal{} y^2 \\minus{} xy \\minus{} 2x \\\\\r\n\\implies (x \\plus{} m)^2 \\minus{} 4x(pm \\minus{} 2) \\equal{} a^2$ where all symbols represent naturals.\r\n\r\nFrom which we get\r\n$ x^2 \\plus{} x(10 \\minus{} 4pm) \\plus{} (m^2 \\minus{} a^2) \\equal{} 0$ \r\n\r\nIf \r\n$ x^2 \\minus{} x(10 \\minus{} 4pm) \\equal{} 0 \\\\\r\n\\implies x \\equal{} (4pm \\minus{} 10)$, then obvious, we get infinitely many solutions.", "Solution_2": "I think there is an error in your transformation" } { "Tag": [ "Support", "LaTeX" ], "Problem": "I'm looking for information on how to combine computers and the internet and handwritten math notes.\r\nMy basic idea is to use a graphics tablet and an online canvas that supports collaborative works.\r\n\r\nI have few relevant experiences myself, but a number of advantages of combining handwritten notes and computers are obvious:\r\n- Handwritten notes are natural for doing math.\r\n- They are much quicker than TeX etc. and avoid disruptions from coding formulae.\r\n- It's easy to add simple diagrams etc. in handwritten notes.\r\n- Via the internet, collaborative work with all its benefits becomes possible.\r\n- The material can be made available for a large audience.\r\n- Adding audio streams etc. makes the written material easier to understand.\r\n- An electronic canvas or blackboard can be made much larger than a physical one.\r\n- It's easy to manipulate elements on the canvas: remove parts, move them, highlight them, change their size, zoom in and zoom out, add comments (written or audio) ...\r\n- It's easy to keep track of different versions.\r\n(I know there are a number of severe shortcomings - messy handwriting being a major one.)\r\n\r\nSo - any ideas on this?", "Solution_1": "[quote=\"thomasteepe\"](I know there are a number of severe shortcomings - messy handwriting being a major one.)[/quote] Large file sizes is the other: LaTeX documents can be distributed as plaintext, and many people live in places without ready access to broadband internet.\r\n\r\nI've seen a few people use tablet computers to very good effect. They also work very well for presentations, as you can project slides but still have much of the freedom of a chalk talk." } { "Tag": [ "parameterization", "geometry", "3D geometry", "sphere", "vector", "function", "conics" ], "Problem": "Hey guys,\r\n\r\nI'm reading a course in 2nd year calculus. I have some questions. If this isn't the right place to post them let me know.\r\n\r\nFind a vector-valued function g(t): R map to R3, to describe the curve of intersection between x^2 + y^2 + z^2=6, and x+y+z=0.", "Solution_1": "We want to express $x$, $y$, and $z$ in terms of one variable (say, $t$) called the parameter. First, note that $z=-x-y$, so if we express $x$ and $y$ in terms of $t$, then $z$ will be easy to handle. The first equation tells us that $x^{2}+y^{2}+(-x-y)^{2}=6$. This is an equation of an ellipse, which can be brought into standard form and parametrized. \r\n\r\nA more geometric approach is to realize that you have a circle of radius $\\sqrt{6}$ in the plane $x+y+z=0$. Find two perpendicular unit vectors in the plane, say $\\vec u=\\langle 1,-1,0\\rangle/\\sqrt{2}$ and $\\vec v=\\langle 1,1,-2\\rangle/\\sqrt{6}$. Then the equation is $\\vec r(t)=\\sqrt{6}(\\vec u\\cos t+\\vec v\\sin t)$.", "Solution_2": "Thanks, the geometric approach makes sense, but I'm still not sure how to parametrize 2x^2+2xy+2y^2=6 in the \"standard\" way.", "Solution_3": "The axes of this ellipse make the angle of $\\pi/4$ with coordinate axes. So we need a new coordinate system $(u,v)$ in which the axes are rotated by $\\pi/4$. For example, $x=u+v$ and $y=u-v$ work. In this system the equation becomes $6u^{2}+2v^{2}=6$, which can be parametrized by $u=\\cos t$ and $v=\\sqrt{3}\\sin t$. Finally, return to $x$ and $y$.", "Solution_4": "Great, thanks." } { "Tag": [ "function", "calculus", "integration", "real analysis", "real analysis unsolved" ], "Problem": "Let h:[0,1]->R \r\n h(x)= 0 if x is irrational\r\n 1/n if x is rational, i.e. x=m/n , where m and n are\r\n natural numbers and coprime.\r\n\r\n Let [a,b] be any interval contained in { x: x>0 }\r\n\r\n Show that h is integrable on [a,b] in Riemann's sense.", "Solution_1": "Don't assume that you know the characterization of integrable functions in terms of almost-everywhere continuity. This problem would occur in a course just after the raw definition of the Riemann integral, and long before you would be in position to prove that characterization.\r\n\r\nA key to the proof: pick an arbitrary $\\epsilon>0.$ For how many points in the interval of integration is it true that $h(x)>\\epsilon?$", "Solution_2": "Direct proof using Kent's hint is tricky but informative ($\\left\\{ f(x) > \\epsilon \\right\\}$ is finite then $N = |\\left\\{f(x) > \\epsilon \\right\\}|$ then cut with intervals of length at most $\\epsilon / N$)\r\n\r\nThere is a killer argument : $f$ is continuous on irrationals (classical exercice, sure it was solved in this forum) so $f$ is continuous almost-everywhere (= except on a set of measure $0$) so it is Riemann-integrable (Lebesgue's :D theorem), and its integral is $0$" } { "Tag": [ "geometry", "geometry proposed" ], "Problem": "In triangle $ ABC$, $ M$ is the midpoint of side $ BC$ and $ G$ is the centroid of triangle $ ABC$. A line $ l$ passes through $ G$, intersecting line $ AB$ at $ P$ and line $ AC$ at $ Q$, where $ P\\ne B$ and $ Q\\ne C$. If $ [XYZ]$ denotes the area of triangle $ XYZ$, show that $ \\frac{[BGM]}{[PAG]}\\plus{}\\frac{[CMG]}{[QGA]}\\equal{}\\frac32$.", "Solution_1": "[quote=\"Johan Gunardi\"]In triangle $ ABC$, $ M$ is the midpoint of side $ BC$ and $ G$ is the centroid of triangle $ ABC$. A line $ l$ passes through $ G$, intersecting line $ AB$ at $ P$ and line $ AC$ at $ Q$, where $ P\\ne B$ and $ Q\\ne C$. If $ [XYZ]$ denotes the area of triangle $ XYZ$, show that $ \\frac {[BGM]}{[PAG]} \\plus{} \\frac {[CMG]}{[QGA]} \\equal{} \\frac{3}{2}$.[/quote]\r\nNice problem :) \r\nHere,\r\n$ [BGM]\\equal{}[CMG]\\equal{}\\frac{1}{6}\\cdot[ABC]$\r\nSo, We need to prove that,\r\n$ \\frac{[ABC]}{9[PAG]}\\plus{}\\frac{[ABC]}{9[QGA]}\\equal{}1......(1)$\r\n\r\nHere,\r\n$ \\frac{[PBG]}{[PGA]}\\equal{}\\frac{BP}{PA} \\Longrightarrow \\frac{[ABC]}{3[PGA]}\\equal{}\\frac{c}{PA}.....(2)$\r\n[as, $ [ABG]\\equal{}\\frac{[ABC]}{3}$\r\nAnalogously,$ \\frac{[QGC]}{[AGQ]}\\equal{}\\frac{CQ}{QA} \\Longrightarrow \\frac{[ABC]}{3[QGA]}\\equal{}\\frac{b}{QA}.....(3)$\r\n\r\nSo, $ (1)$ becomes,\r\n$ \\frac{c}{3PA}\\plus{}\\frac{b}{3QA}\\equal{}1$\r\n\r\nWe draw $ A'B' \\parallel AB$ through $ G$\r\n\r\n[Note: Here we shall use directed length]\r\nUsing Menelau's Theorem on $ \\triangle AB'C'$ and transvarsal $ PQ$ we get,\r\n$ \\frac{AP}{PB'}\\equal{}\\minus{}\\frac{QA}{C'Q} \r\n\\\\\r\n\\Longrightarrow \\frac{PB'}{AP}\\plus{}\\frac{C'Q}{QA}\\equal{}0\r\n\\\\\r\n\\Longrightarrow \\frac{PB'}{AP}\\plus{}1\\plus{}\\frac{C'Q}{QA}\\plus{}1\\equal{}2\r\n\\\\ \r\n\\Longrightarrow \\frac{AB'}{AP}\\plus{}\\frac{C'A}{QA}\\equal{}2\r\n\\\\ \r\n\\Longrightarrow \\frac{\\frac{2AB}{3}}{AP}\\plus{}\\frac{\\frac{2CA}{3}}{QA}\\equal{}2\r\n\\\\\r\n\\Longrightarrow \\frac{AB}{3AP}\\plus{}\\frac{CA}{3QA}\\equal{}1$\r\n\r\nSo,\r\n$ \\frac{c}{3PA}\\plus{}\\frac{b}{3QA}\\equal{}1$\r\nwhich implies,\r\n$ \\frac {[BGM]}{[PAG]} \\plus{} \\frac {[CMG]}{[QGA]} \\equal{} \\frac{3}{2}$\r\n$ Q.E.D$", "Solution_2": "[quote=\"Johan Gunardi\"][color=darkred] In triangle $ ABC$ , $ M$ is the midpoint of side $ BC$ and $ G$ is the centroid of triangle $ ABC$ . A line $ d$ passes through $ G$\n\nintersecting sideline $ AB$ at $ P$ and sideline $ AC$ at $ Q$ , where $ P\\ne B$ and $ Q\\ne C$ . Show that $ \\frac {[BGM]}{[PAG]} \\plus{} \\frac {[CMG]}{[QGA]} \\equal{} \\frac32$ .[/color] [/quote]\n\n[quote][color=darkred][b][u]Lemma (well-known or you can prove it easily !).[/u][/b] $ P\\in AB\\ ,\\ Q\\in AC\\ ,\\ G\\in PQ\\ \\Longrightarrow\\ \\frac {\\overline {AB}}{\\overline {AP}} \\plus{} \\frac {\\overline {AC}}{\\overline {AQ}} \\equal{} 3$ .[/color][/quote]\r\n\r\n[color=darkblue][b][u]Proof of the proposed problem.[/u][/b] Denote the distance $ \\delta (X)$ of the point $ X$ to the line $ AM$ . \n\nTherefore, $ \\left\\|\\begin{array}{c} \\frac {[BGM]}{[AGP]} \\equal{} \\frac {GM\\cdot\\delta (B)}{AG\\cdot \\delta (P)} \\equal{} \\frac 12\\cdot\\frac {\\overline {AB}}{\\overline {AP}} \\\\\n \\\\\n\\frac {[CGM]}{[AGQ]} \\equal{} \\frac {GM\\cdot\\delta (C)}{AG\\cdot\\delta (Q)} \\equal{} \\frac 12\\cdot\\frac {\\overline {AC}}{\\overline {AQ}} \\end{array}\\right\\|$ $ \\stackrel{LEMMA}{\\ \\ \\ \\ \\implies\\ \\ \\ \\ }$ $ \\boxed {\\ \\frac {[BGM]}{[AGP]} \\plus{} \\frac {[CMG]}{[AGQ]} \\equal{} \\frac32\\ }$ .[/color]", "Solution_3": "[quote=\"Virgil Nicula\"][color=darkred][b]Lemma (well-known or you can prove it easily !). $ P\\in AB\\ ,\\ Q\\in AC\\ ,\\ G\\in PQ\\ \\Longrightarrow\\ \\frac {\\overline {AB}}{\\overline {AP}} \\plus{} \\frac {\\overline {AC}}{\\overline {AQ}} \\equal{} 3$ .[/b][/color][/quote]\r\nI have proved it in my sollution :wink: (But it was not known to me....)", "Solution_4": "[quote=\"Moonmathpi496\"]... So $ \\frac {c}{3PA} \\plus{} \\frac {b}{3QA} \\equal{} 1$ which implies ... \nI have proved it in my sollution :wink: (But it was not known to me)[/quote]\r\n[color=darkblue][b]Very well ![/b] You passed through the proof of this well-known (remarkable !) property of a line $ d$ , $ G\\in d$ . Generally,\n\n$ \\left\\|\\begin{array}{c} M\\in (BC)\\ ,\\ P\\in (AM) \\\\\n\\ X\\in (AB)\\ ,\\ Y\\in (AC)\\ ,\\ P\\in XY\\end{array}\\right\\|$ $ \\implies$ $ \\boxed {\\ \\frac {AB}{AX}\\cdot MC \\plus{} \\frac {AC}{AY}\\cdot MB \\equal{} \\frac {AM}{AP}\\cdot BC\\ }$ .[/color]" } { "Tag": [ "inequalities", "induction", "inequalities proposed" ], "Problem": "Prove that for every $n\\in \\mathbb{N}$, we have $(n!)^{2}\\geq n^{n}$.", "Solution_1": "[quote=\"mathematica\"]Prove that for every $n\\in \\mathbb{N}$, we have $(n!)^{2}\\geq n^{n}$.[/quote]\r\nMaybe is $(n!)^{2}\\leq n^{n}$", "Solution_2": "[quote=\"Zamfirmihai\"][quote=\"mathematica\"]Prove that for every $n\\in \\mathbb{N}$, we have $(n!)^{2}\\geq n^{n}$.[/quote]\nMaybe is $(n!)^{2}\\leq n^{n}$[/quote]\r\nI think the initial problem is ok.Here's a proof using induction:\r\nWe only have to prove that:\r\n$((n+1)!)^{2}\\geq (n+1)^{n+1}$ $\\Leftrightarrow$ $(n!)^{2}(n+1)^{2}\\geq (n+1)^{n+1}$,but by the induction hypothesys we have:\r\n$(n!)^{2}(n+1)^{2}\\geq n^{n}(n+1)^{2}\\geq (n+1)^{n+1}$\r\nAnd the last one is equivalent to:\r\n$n^{n}(n+1)\\geq (n+1)^{n}=\\sum_{k=0}^{n}n^{n-k}\\cdot C_{n}^{k}$ which is true ,because $n^{n}\\geq C_{n}^{k}\\cdot n^{n-k},\\Leftrightarrow T_{1}\\geq T_{k+1}\\forall k=\\overline{0,n-1}$ ,which is proved by the fact that $\\frac{T_{k+2}}{T_{k+1}}<1$ ,where $T_{k+1}=C_{n}^{k}\\cdot n^{n-k}$", "Solution_3": "$(n!)^{2}=\\prod_{i=1}^{i=n}((n+1-i)*i)$ so you need to prove that $(n+1-i)*i\\geq n$ if $1\\leq i\\leq n$ which is equivalent to $(n-i)(i-1)\\geq 0$ which is true.\r\nthis is a very well known proof for this problem." } { "Tag": [ "function", "limit", "real analysis", "real analysis unsolved" ], "Problem": "Show the Riesz Marcell theorem in case $ p\\equal{}2$, with Fourier Transform.\r\nGive a countraexample in $ p\\equal{}\\infty$.", "Solution_1": "I know two closely related theorems of Marcel Riesz, and each theorem could be stated either for $ \\mathbb{T}$ or for $ \\mathbb{R}.$\r\n\r\nHere it is on $ \\mathbb{T}:$\r\n\r\n1. Suppose $ f$ is a \"sufficiently nice\" function on $ \\mathbb{T}.$ Define $ \\widetilde{f}(x)\\equal{}\\sum_{k\\equal{}\\minus{}\\infty}^{\\infty}\\minus{}i\\text{sgn}\\,(k)\\widehat{f}(k)e^{ikx}.$ (Note: this can also be obtained by convolving $ f$ with a constant times the principal value of $ \\cot(x/2).$) Then for $ 1 1$. But $ |\\mathbb{F}_q^*|\\equal{}q\\minus{}1$, so it is not true in this case.\r\n\r\nWhen $ n\\equal{}1$, however, $ Aut(G) \\cong Aut(Z_p) \\cong Z_{p\\minus{}1} \\cong \\mathbb{F}_p^*$.", "Solution_2": "Oh, it was not that hard efter all. I should have had tried harder. Thanks LydianRain!" } { "Tag": [ "AMC", "AIME" ], "Problem": "i know this has been done on the regular amc forum, but who needs all those extra non-hoosier(or whose ear) posts.", "Solution_1": "who really cares? everyone basically knows everone else's scores, but whatever...\r\n\r\n135 on 10a\r\n25 on 8\r\nAIME: i'll edit it in later\r\n\r\ntoo lazy to list seventh grade..." } { "Tag": [], "Problem": "Find the value of x\r\n\r\nr and s are parallel.\r\n\r\n[img]http://img193.imageshack.us/img193/7902/melhorq.jpg[/img]", "Solution_1": "From the arcs, you can see that $ 4x \\plus{} 30 \\equal{} 2x \\plus{} 10$, so $ x \\equal{} \\minus{}10$, unless I'm missing something, which I think I must be... :huh:", "Solution_2": "If I interpret that correctly, it's just that the arcs aren't distinguishing noncongruent angles.\r\n\r\nDraw a third line parallel to r and s through that point in the middle and then you end up with \\[ 2x\\plus{}10\\plus{}180\\minus{}(5x\\plus{}20)\\equal{}4x\\plus{}30\\Rightarrow x\\equal{}\\boxed{20}\\]", "Solution_3": "I don't understand your logic ,5849206328x.", "Solution_4": "cf.\r\nhttp://www.mathlinks.ro/viewtopic.php?t=293598\r\nhttp://www.mathlinks.ro/viewtopic.php?t=293696", "Solution_5": "Ok, but What do you do in this example ?\r\nDo you have a site to learn ?", "Solution_6": "Ah that's ok. Now I understand. Thanks for your help ;)" } { "Tag": [], "Problem": "You are given a string of digits: $718923$. You are allowed to perform the following operation any number of times:\r\n\r\n1) Take any two digits that are next to each other.\r\n\r\n2) Place their positive difference anywhere in the string.\r\n\r\n3) Place their sum anywhere in the string. If their sum is greater than 10, only place its unit digit in the string. After a series of operations, is it possible to transform the original string into $505731218798236$? If not, prove it.", "Solution_1": "[hide=\"answer\"]\nNO because seeing the $1$st $2$ digits $70...$ $0$ can never be formed because $7$ is next to it and $70$ can never be formed by adding $2$ digits\nIf it is then there would be more digits in the front.\n[/hide]", "Solution_2": "Are you saying that that 70 is too large to be formed by adding two digits? That's true, but 0 can be a result of taking the difference of two equal digits, and 7 can be a result of adding, say, 3 and 4. (remember, after computing the sum and difference, you can place them anywhere you want)", "Solution_3": "yes I get your point then I think it is Yes It can be formed.", "Solution_4": "Step 3 should have been fixed (fixed now). Also, the final string is shorter." } { "Tag": [ "induction", "inequalities", "least common multiple", "More Sequences" ], "Problem": "Does there exist positive integers $a_{1} lcm(a_n^d a_{n \\minus{} 1}^{d \\plus{} 1},a_n^{d \\plus{} 1} a_{n \\minus{} 1}^d)\r\n\\]\r\nor equivalently\r\n\\[ \\ell^d a_{n \\minus{} 1}^{d \\plus{} 1} > a_n^d a_{n \\minus{} 1}^d \\ell\r\n\\]\r\nwhere $ \\ell \\equal{} lcm(a_{n \\minus{} 1},a_n).$ The last inequality is equivalent to\r\n\\[ \\left(\\frac {\\ell}{a_n}\\right)^d > \\frac {\\ell}{a_{n \\minus{} 1}}\r\n\\]\r\nwhich has a solution (w.r.t. $ d$) since $ \\frac {\\ell}{a_n} > 1.$ \r\nTherefore, we can select $ d$ such that the sequence (*) of $ n \\plus{} 1$ elements satisfies the property 1).\r\nIt is also easy to see that the property 2) for the original sequence (i.e., $ a_n < lcm(a_{n \\minus{} 1},a_n)$) implies the property 2) for the sequence (*):\r\n\\[ a_n^{d \\plus{} 1} a_{n \\minus{} 1}^d < a_n^d a_{n \\minus{} 1}^d lcm(a_{n \\minus{} 1},a_n) \\equal{} lcm(a_n^d a_{n \\minus{} 1}^{d \\plus{} 1},a_n^{d \\plus{} 1} a_{n \\minus{} 1}^d).\r\n\\]\r\nQ.E.D." } { "Tag": [ "Mafia" ], "Problem": "Shock! is a game is a Mafia-esque game where a group of people stand in a circle. One of the people is a \"shocker\" and the aim of the game is to discover who. The game is played as follows:\r\n\r\n- I will PM the shocker a number.\r\n- The shocker will PM one of the people \"standing\" next to him one less than that number.\r\n- That person will PM me the number they recieved and PM one of the people next to him/her one less than the number.\r\n- Whoever recieves one shock, will be eliminated from the game and I will post in this thread who it was.\r\n- The players converse on the thread. After the conversation is completed, a vote takes place. Whoever has the most votes drops dead. Whether or not they were the shocker is revealed. The next round begins.\r\n\r\nRules and specifics:\r\n- We need at least 20 people to play. As soon as I recieve 20 PM's, I will PM one of the 20 people with a message saying that they are the shocker.\r\n- During the \"shock period\", NO POSTS MAY BE MADE ON THIS THREAD.\r\n- During the conversation period, you may tell people what number you got, but people can lie about this.\r\n- The conversation period is over once everybody has said in BLUE TYPE who they vote for.\r\n\r\nNotice that you will know who the shocker is if you are standing next to him/her. You can vote for this person, but you may not tell people your true reasoning for doing so.\r\n\r\nTo play, you must be an acitve AoPSer, in other words, you will be able to participate whenever need be.\r\n\r\nPlease submit joining PM's as soon as possible.", "Solution_1": "Sign-up list:\r\n\r\nperfect628\r\npianoforte\r\nmath92", "Solution_2": "May I sign up?", "Solution_3": "You did not indicate how you shock someone.\r\n\r\nThis rule makes it stupid:\r\n[quote]Notice that you will know who the shocker is if you are standing next to him/her. You can vote for this person, but you may not tell people your true reasoning for doing so. [/quote]\r\nas we can just say: VOTE FOR THE SHOCKER ONLY IF YOU ARE STANDING NEXT TO HIM\r\nand when two people vote for one person, the game is over.\r\nOr the persons standing next to the shocker might be just the only one who voted without reason.\r\nOr the persons standing next to the shocker could vote for the same person over a period of rounds.\r\nThe thing is, you can never actually enforce this rule.\r\n\r\nAnd depending how you shock, it may be too obvious as well considering you can just post you received a number and we know the person is at most 2 spaces from you and if you post who you sent it to or who you received it from then we've narrowed it down to 2 people and the game is quickly over.", "Solution_4": "Wait a minute.\r\n\r\nSo we basically give each other shockers.\r\n\r\nDo you know how incredibly wrong this sounds?", "Solution_5": "[quote=\"Ignite168\"]You did not indicate how you shock someone.\n\nThis rule makes it stupid:\n[quote]Notice that you will know who the shocker is if you are standing next to him/her. You can vote for this person, but you may not tell people your true reasoning for doing so. [/quote]\nas we can just say: VOTE FOR THE SHOCKER ONLY IF YOU ARE STANDING NEXT TO HIM\n[b]and when two people vote for one person, the game is over.[/b]\n[i]Or the persons standing next to the shocker might be just the only one who voted without reason.[/i]\nOr the persons standing next to the shocker could vote for the same person over a period of rounds.\nThe thing is, you can never actually enforce this rule.\n\nAnd depending how you shock, it may be too obvious as well considering you can just post you received a number and we know the person is at most 2 spaces from you and if you post who you sent it to or who you received it from then we've narrowed it down to 2 people and the game is quickly over.[/quote]\r\n\r\nI don't see your reasoning for the bold sentence.\r\nFor the italics sentence, you can make up a different reason as to why you are voting for the person. \r\nIf it makes the game better in your opinion, I can delete the rule in which I say the number aloud to everybody.\r\n\"Shocking\" is done by PMing somebody next to you with a number.", "Solution_6": "You could change the game up slightly so that the people next to the shocker don't know who it is. Pick a random positive integer to send to the shocker, and let anyone who is shocked shock anyone else. Furthermore, don't do the shocking by direct PM--have the people PM you with who they want to shock, then PM the shocked person with the number. That would be a lot harder, though.", "Solution_7": "Ok then the shocker is obviously the one that's next to the spot in which people keep dying.", "Solution_8": "Ignite, you misunderstood the rules. If you are the shocker, let's say you get the number \"6\". Then you pass \"5\" to somebody. That person passes \"4\" to somebody. That person passes \"3\" to somebody. And so on until somebody recieves \"1\", who dies.", "Solution_9": "Ok then in that case, people claim what number they got and then the shocker has to either claim he never received a number, or he received a number from the person next to him. Then, one of the people next to him will counter whatever he claims and the game is quickly over after throwing two people out the group.", "Solution_10": "I wrote this incorrectly on the original rules, but you are NOT allowed to claim what number you got.", "Solution_11": "Sorry to sound like a broken record, but does anybody want to join? I only have 3 players and I'm going away in a month for a month with no internet, so....", "Solution_12": "[quote=\"miyomiyo\"]Sorry to sound like a broken record, but does anybody want to join? I only have 3 players and I'm going away in a month for a month with no internet, so....[/quote]\r\n\r\nIn two months, you should get plenty of players I guess ..." } { "Tag": [ "Asymptote", "limit", "calculus", "calculus computations" ], "Problem": "Let $ p,q \\in \\mathbb{Q}$ such that $ p\\plus{}q\\equal{}1$. Prove that $ y\\equal{}x\\plus{}ap\\plus{}bq$ is a slant asymptote of the curve $ y\\equal{}(a\\plus{}x)^p(b\\plus{}x)^q$.", "Solution_1": "This is a basic problem about limits. We are trying to find m and c satisfying\r\n\r\n$ \\lim_{x\\rightarrow \\infty }\\left( (a\\plus{}x)^{p}(b\\plus{}x)^{q}\\minus{}(mx\\plus{}c)\\right) \\equal{}0$\r\n\r\nYou will find that $ m\\equal{}1,c\\equal{}pa\\plus{}bq$ will satisfy the criteria.", "Solution_2": "[quote=\"IndoChina\"]This is a basic problem about limits. We are trying to find m and c satisfying\n\n$ \\lim_{x\\rightarrow \\infty }\\left( (a \\plus{} x)^{p}(b \\plus{} x)^{q} \\minus{} (mx \\plus{} c)\\right) \\equal{} 0$\n\nYou will find that $ m \\equal{} 1,c \\equal{} pa \\plus{} bq$ will satisfy the criteria.[/quote]How exactly do we prove that?" } { "Tag": [ "FTW" ], "Problem": "What is the remainder when the product of the first 25 prime numbers is divided by 4?", "Solution_1": "Its a multiple of 2 (because of 2) so its an odd number divided by 2 so the answer is 1.", "Solution_2": "To be a little more clear:\r\n\r\nThere is only one even prime number ($ 2$). Then the product of the first $ 25$ numbers is $ 2(\\text{some odd junk})$. Divide by $ 4$, and we get $ \\frac{\\text{some odd junk}}{2}$, so it has remainder $ \\boxed{1}$.", "Solution_3": "FTW says the correct answer is 2...I dont get it.", "Solution_4": "AIME15 made a typo at the last part. Since 2(some odd junk) = 2(2n+1) = 4n+2, the answer is 2.", "Solution_5": "wait, shouldn't these solutions be hidden?", "Solution_6": "[quote=\"Mrdavid445\"]wait, shouldn't these solutions be hidden?[/quote]\n\nThere's a great reason to revive a 3-year old thread.", "Solution_7": "The answer is 2. The product of the first three primes is 2*3*5=30. 30/4=7 R2. You can see that 30*any odd number will result with a remainder of 2 when divided by 4. For example, 30*7=210. 210/4=52 R2." } { "Tag": [], "Problem": "What is the sum of all the digits in the sequence 1, 2, 3, 4, 5, 6, 7,...99, 100?", "Solution_1": "hehe :)\r\n$\\frac {n(n+1)}{2}$ over and over again.. ;)\r\n\r\n[hide]100*101/2 = 50*101=5050[/hide]", "Solution_2": "that's incorrect.", "Solution_3": "Huh? :huh: \r\nThen, I think I misunderstood the question..", "Solution_4": "It's asking for the digit's sum.\r\n[hide]\nS=1+2+3+...+9=45\nSince there are 10 different ten's, 10, 20, 30...,\nyou do 45*10=450\nSince there are the ten's digits, you add 45 more.\nPlus 1 for 100=\n496[/hide]", "Solution_5": "[quote=\"mathmanman\"]Huh? :huh: \nThen, I think I misunderstood the question..[/quote]\r\nyou did sum of numbers and the question asked for sum of [u]digits[/u]", "Solution_6": "Oh! :rotfl: \r\nYeah, so sorry...\r\nWell, pascal_1623 has solved it :)" } { "Tag": [ "geometry", "circumcircle", "Euler", "inequalities", "geometry proposed" ], "Problem": "$O$ is an interior point of a triangle $ABC$. The feet of the perpendiculars from $O$ to $BC$, $CA$ and $AB$ are denoted by $A_{1}$, $B_{1}$, $C_{1}$ respectively. The bisector of $\\widehat{BOC}$ intersects $BC$ at point $A_{2}$, The bisector of $\\widehat{COA}$ intersects $CA$ at point $B_{2}$, The bisector of $\\widehat{AOB}$ intersects $AB$ at point $C_{2}$. Prove that $area(A_{1}B_{1}C_{1}) \\leq area(A_{2}B_{2}C_{2})$", "Solution_1": "Yes, it is nice, I change the your notation $O$ by $M$ and circumcenter is $O$\r\nlet $MA=x,MB=y,MC=z$ note that $AA_{2},BB_{2},CC_{2}$ are concurrent at $I(MA,MB,MC)$ therefore\r\n$S_{A_{2}B_{2}C_{2}}=\\frac{2xyz}{(x+y)(y+z)(z+x)}S_{ABC}$ and by Euler formula\r\n$S_{A_{1}B_{1}C_{1}}=\\frac{R^{2}-OM^{2}}{4R^{2}}S_{ABC}$\r\nnow we need prove that\r\n$\\frac{8xyz}{(x+y)(y+z)(z+x)}\\ge(1-\\frac{MO^{2}}{R^{2}})$\r\neasily seen the condition $R-OM0$ and $u-vv>0$ we will get\r\n$\\frac{8xyz}{(x+y)(y+z)(z+x)}\\ge 1-\\frac{v^{2}}{u^{2}}$\r\nI have solution of the last inequality by analysis, it isn't nice therefore I want to find a nice solution from anybody!:D :D", "Solution_2": "[quote=\"gemath\"]\nIf $x,y,z>0$ and $u-vv$.I edited it.", "Solution_6": "Ok, I must misunderstand something all the time, so I just leave it as it is with only small comment. If $u>v$ as you claim problem is obvious as LHS is greater than zero whereas RHS is not :D", "Solution_7": "no when $u>v\\Rightarrow 1-\\frac{v^{2}}{u^{2}}>0$ it in't obvious, actually I want to find nice solution!" } { "Tag": [], "Problem": "Cac ban cho hoi lam the nao de dang nhap vao trang nay?\r\nTruoc kia lam rat don gian.", "Solution_1": "[quote=\"N.T.TUAN\"]Cac ban cho hoi lam the nao de dang nhap vao trang nay?\nTruoc kia lam rat don gian.[/quote]\r\n\r\nDe dang nhap vao trang nay`. chi can` vao` trang web nay` http://www.artproblemsolving.com. Xong roi` Login , sau do' vao` NATIONAL &COMMUNITIES/ASIA/VIETNAM thui !! ma` khong biet toi tra loi co dung' muc. dich' cua cau hoi khong nua~!! :lol:", "Solution_2": "Cam on ban,minh lam duoc roi,nhung khong phai bang cach ban noi.", "Solution_3": "It's alright. :D" } { "Tag": [], "Problem": "We need more Emoticons!", "Solution_1": "Did you try the \"More emoticon\" link? What more could you want?", "Solution_2": "Maybe you could make some.", "Solution_3": "There are many emoticons:\r\n :moose: :ddr: :spam:", "Solution_4": "How did you do that spiderboy? Oops. nvm.. Found out!!!\r\n\r\n\r\n :spider: :starwars: :gathering: \r\n\r\n\r\nI'm just having fun with a new discovery." } { "Tag": [ "algebra", "polynomial", "superior algebra", "superior algebra unsolved" ], "Problem": "if $ f(x)\\equal{} a_{0}\\plus{}a_{1}x\\plus{}a_{2}x^{2}\\plus{}....\\plus{}a_{n}x^{n} \\in R[x]$ with all roots in $ R$(real number),\r\nshow that the roots of $ g(x)\\equal{}\\binom{n}{0}a_{0}\\plus{}\\binom{n}{1}a_{1}x\\plus{}\\binom{n}{2}a_{2}x^{2}\\plus{}....\\plus{}\\binom{n}{n}a_{n}x^{n}$ are also all in $ R$", "Solution_1": "is there any theorem about real roots of polynomial ?", "Solution_2": "[quote=\"mathxxxx\"]is there any theorem about real roots of polynomial ?[/quote]\r\nMay be [url=http://en.wikipedia.org/wiki/Sturm's_theorem]Sturm's theorem[/url] can help, but I'm not sure.\r\nWhence you took this problem?" } { "Tag": [ "algebra", "function", "domain", "polynomial", "rational function", "superior algebra", "superior algebra unsolved" ], "Problem": "Hello, I would have a hint for this problem:\r\n \r\nLet $ K$ be a field. I note $ K[X]$ the domain of polynomials over $ K$\r\nand $ K' \\equal{} K(X)$ the field of fractions of $ K[X].$\r\n $ v : K(X) \\rightarrow \\mathbb{Z}\\cup \\plus{} \\infty$ defined by $ v(\\frac{f}{g}) \\equal{} v(f) \\minus{} v(g)$ with $ v(f) \\equal{} \\minus{}deg(f)$ if $ f \\neq 0$ and $ v(0) \\equal{} \\plus{}\\infty$.\r\n\r\nShow that $ K(X)$ is not complete for the valuation $ N$ such that $ N(\\frac{f}{g})\\equal{} \\rho^{v(\\frac{f}{g})}$ with $ 0 < \\rho < 1.$", "Solution_1": "$ 1, 1\\plus{}X, 1\\plus{}X\\plus{}X^2, \\ldots$ is a Cauchy sequence without limit.", "Solution_2": "[quote=\"berti\"]$ 1, 1 \\plus{} X, 1 \\plus{} X \\plus{} X^2, \\ldots$ is a Cauchy sequence without limit.[/quote]\r\n\r\nAre you sure this is a Cauchy sequence?\r\n\r\n darij", "Solution_3": "it's a Cauchy sequence but has a limit $ L\\equal{}\\frac{1}{1\\minus{}X} \\in K'.$ So it does not prove that $ K'$ is not complete!", "Solution_4": "Why is it a Cauchy sequence? For $ n s$ we have $ u \\minus{} 2^i < 0$ and thus $ q_{u \\minus{} 2^i} \\equal{} 0$, and for $ i < s$ we have $ i\\leq s \\minus{} 1$ and thus $ u \\minus{} 2^i \\equal{} 2^s \\minus{} 2^i\\geq 2^s \\minus{} 2^{s \\minus{} 1} \\equal{} 2^{s \\minus{} 1}$, so that $ u \\minus{} 2^i$ is bounded from below by $ 2^{s \\minus{} 1}$, and if $ s$ is big enough, this will mean that $ q_{u \\minus{} 2^i} \\equal{} 0$). Hence, for every integer $ s$ big enough, we can take $ u \\equal{} 2^s$ and then $ p_u \\equal{} \\sum_{i \\equal{} 0}^{\\infty}q_{u \\minus{} 2^i}\\neq 0$ (because a sum with one and only one nonzero term must be $ \\neq 0$). Thus, we have found infinitely many $ u$ such that $ p_u\\neq 0$, but this contradicts to $ p_u$ being $ 0$ for all sufficiently large $ u$.\r\n\r\nThe contradiction shows that $ \\sum_{i \\equal{} 0}^{\\infty}X^{2^i}$ does not represent a rational function, and we are done.\r\n\r\nA slightly different way to prove this would be to show that, if $ \\sum_{u \\equal{} 0}^{\\infty} p_uX^u$ and $ \\sum_{v \\equal{} 0}^{\\infty} q_vX^v$ are two polynomials, then the coefficients of the formal series $ \\frac {\\sum_{u \\equal{} 0}^{\\infty} p_uX^u}{\\sum_{v \\equal{} 0}^{\\infty} q_vX^v}$ (when written as a sum of powers of $ X$) satisfy a linear recurrence, while the coefficients of the formal series $ \\sum_{i \\equal{} 0}^{\\infty}X^{2^i}$ cannot satisfy a linear recurrence (because a sequence satisfying an $ n$-term linear recurrence is zero if it has $ n$ consequent zero entries, while the series of the coefficients of $ \\sum_{i \\equal{} 0}^{\\infty}X^{2^i}$ has arbitrary large blocks of consequent zero entries but are not all zero).\r\n\r\n darij", "Solution_6": "is there a nice description of the completion?", "Solution_7": "As Darij already noticed, once you substitute $ X\\to X^{\\minus{}1}$, $ v(f)$ is just the $ X$-adic valuation. So, the completion (for the original $ v$) is $ K((X^{\\minus{}1}))$. owk" } { "Tag": [ "trigonometry", "inequalities", "rearrangement inequality" ], "Problem": "Let ABC is a triangle, AB=c,AC=b, BC=a.Prove that\r\n$ \\frac{p}{{q \\plus{} r}}a^2 \\plus{} \\frac{q}{{p \\plus{} r}}b^2 \\plus{} \\frac{r}{{p \\plus{} q}}c^2 \\ge 2\\sqrt 3 S$\r\np,q,r is the three real positive numbers :) .", "Solution_1": "[quote=\"TRAN THAI HUNG\"]Let ABC is a triangle, AB=c,AC=b, BC=a.Prove that\n$ \\frac {p}{{q \\plus{} r}}a^2 \\plus{} \\frac {q}{{p \\plus{} r}}b^2 \\plus{} \\frac {r}{{p \\plus{} q}}c^2 \\ge 2\\sqrt 3 S$\np,q,r is the three real positive numbers :) .[/quote]\r\nUse\r\nhttp://reflections.awesomemath.org/2008_1/unexpected_ineq.pdf", "Solution_2": "[quote=\"arqady\"][quote=\"TRAN THAI HUNG\"]Let ABC is a triangle, AB=c,AC=b, BC=a.Prove that\n$ \\frac {p}{{q \\plus{} r}}a^2 \\plus{} \\frac {q}{{p \\plus{} r}}b^2 \\plus{} \\frac {r}{{p \\plus{} q}}c^2 \\ge 2\\sqrt 3 S$\np,q,r is the three real positive numbers :) .[/quote]\nUse\nhttp://reflections.awesomemath.org/2008_1/unexpected_ineq.pdf[/quote]How can we use it for this case? :)", "Solution_3": "[quote=\"TRAN THAI HUNG\"][quote=\"arqady\"][quote=\"TRAN THAI HUNG\"]Let ABC is a triangle, AB=c,AC=b, BC=a.Prove that\n$ \\frac {p}{{q \\plus{} r}}a^2 \\plus{} \\frac {q}{{p \\plus{} r}}b^2 \\plus{} \\frac {r}{{p \\plus{} q}}c^2 \\ge 2\\sqrt 3 S$\np,q,r is the three real positive numbers :) .[/quote]\nUse\nhttp://reflections.awesomemath.org/2008_1/unexpected_ineq.pdf[/quote]How can we use it for this case? :)[/quote]\r\nWe know that $ \\frac {p}{{q \\plus{} r}}(y \\plus{} z) \\plus{} \\frac {q}{{p \\plus{} r}}(x \\plus{} z) \\plus{} \\frac {r}{{p \\plus{} q}}(x \\plus{} y) \\ge \\sqrt {3(xy \\plus{} xz \\plus{} yz)},$ \r\nwhere $ p,$ $ q,$ $ r,$ $ x,$ $ y$ and $ z$ are positive numbers.\r\nLet $ y \\plus{} z \\equal{} a^2,$ $ x \\plus{} z \\equal{} b^2$ and $ x \\plus{} y \\equal{} c^2,$ where $ a,$ $ b$ and $ c$ are positive numbers. \r\nHence, $ a \\plus{} b \\minus{} c > 0,$ $ a \\plus{} c \\minus{} b > 0$ and $ b \\plus{} c \\minus{} a > 0$ and we obtain your inequality for acute-angle triangle. \r\nFor other triangles it's obvious:\r\nLet $ c\\equal{}max\\{a,b,c\\}.$ Hence, $ \\frac {p}{{q \\plus{} r}}a^2 \\plus{} \\frac {q}{{p \\plus{} r}}b^2 \\plus{} \\frac {r}{{p \\plus{} q}}c^2 \\ge$\r\n$ \\geq\\frac {p}{{q \\plus{} r}}a^2 \\plus{} \\frac {q}{{p \\plus{} r}}b^2 \\plus{} \\frac {r}{{p \\plus{} q}}(a^2\\plus{}b^2) \\ge \\sqrt 3 ab\\geq2\\sqrt 3 S.$ :wink:", "Solution_4": "WoW! I did not see it even after reading it.\r\nThanks for showing us.", "Solution_5": "see Topics in Inequalities by Hoojoo Lee. :P", "Solution_6": "By Rearrangement Inequality, $ LHS \\ge \\frac {a^2\\plus{}b^2\\plus{}c^2}{2}$\r\n\r\nFrom Weitzenbock inequality, $ a^2\\plus{}b^2\\plus{}c^2 \\ge 4 \\sqrt 3 S$\r\n\r\nThe given inequality follows from the above two", "Solution_7": "[quote=\"hsbhatt\"]By Rearrangement Inequality, $ LHS \\ge \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{2}$\n[/quote]\r\nExplain it please. Thank you!", "Solution_8": "[quote=TRAN THAI HUNG]Let ABC is a triangle, AB=c,AC=b, BC=a.Prove that\n$ \\frac{p}{{q \\plus{} r}}a^2 \\plus{} \\frac{q}{{p \\plus{} r}}b^2 \\plus{} \\frac{r}{{p \\plus{} q}}c^2 \\ge 2\\sqrt 3 S$\np,q,r is the three real positive numbers :) .[/quote]\nBy C-S $\\frac{p}{{q \\plus{} r}}a^2 \\plus{} \\frac{q}{{p \\plus{} r}}b^2 \\plus{} \\frac{r}{{p \\plus{} q}}c^2=(p+q+r)\\left(\\frac{a^2}{q+r}+\\frac{b^2}{p+r}+\\frac{c^2}{p+q}\\right)-a^2-b^2-c^2\\geq$\n$\\geq(p+q+r)\\cdot\\frac{(a+b+c)^2}{\\sum\\limits_{cyc}(p+q)}-a^2-b^2-c^2=\\frac{1}{2}\\sum_{cyc}(2ab-a^2)\\geq2S\\sqrt3$.\n\n" } { "Tag": [ "calculus", "integration", "function", "calculus computations" ], "Problem": "Let $ g: (\\alpha,\\beta)\\rightarrow\\Re$ be a differentiable function, with $ g'(x)$ being a continuous function, and let $ f(x)$ be a continuous function for which $ (f\\circ g)(x)$ is well-fined as a function on $ (\\alpha,\\beta)$. Then $ \\int^{b}_{a}f(g(x))g'(x)dx \\equal{}\\int^{g(b)}_{g(a)}f(u)du$ for any $ a,b\\in(\\alpha,\\beta)$.\r\n\r\nVerify the substitution method by the Fundamental Theorem of Calculus and the Mean Value Theorems (for Differential Calculus, Integral Calculus).\r\n\r\nThank you for helping. :P", "Solution_1": "I was told to let $ H(x)\\equal{}\\int^{b}_{a}f(g(x))g'(x)dx\\minus{}\\int^{g(b)}_{g(a)}f(u)du$, then prove that $ H'(x)\\equiv 0$, and $ H(a)\\equal{}0$ for some $ a\\in(\\alpha,\\beta)$.\r\n\r\nI don't quite get how to prove $ H'(x)\\equiv 0$, can someone please help.\r\nThank you. :P", "Solution_2": "[quote]I was told to let $ H(x) \\equal{}\\int^{b}_{a}f(g(x))g'(x)du\\minus{}\\int^{g(b)}_{g(a)}f(u)du$ , then prove that $ H(x)\\equiv 0$[/quote]\r\nNote that $ H(x) \\equal{}\\int^{x}_{a}f(g(u))g'(u)du\\minus{}\\int^{g(x)}_{g(a)}f(u)du$\r\nLet $ F'(x): \\equal{} f(x)$ and $ G'(x): \\equal{} f(g(x))g'(x)$.So we have\r\n$ H(x) \\equal{} G(x)\\plus{}F(g(a))\\minus{}F(g(x))\\minus{}G(a)$.It's easy to see $ H(a) \\equal{} 0$ on the other hand:\r\n$ H'(x) \\equal{} G'(x)\\minus{}(F(g(x)))'\\equal{}0$\r\n$ \\iff H(x) \\equal{} G'(x)\\minus{}g'(x)F'(g(x)) \\equal{} f(g(x))g'(x)\\minus{}g'(x)f(g(x)) \\equal{} 0$\r\nSo $ H(x)\\equiv 0$" } { "Tag": [ "real analysis", "complex analysis", "real analysis unsolved" ], "Problem": "Hey guys, I have Problem that I can't solve. I would appreciate if someone would help me out with suggestions. It can be found in Rudin's Real and Complex Analysis after the second chapter. It goes like this:\r\n\r\nFind a measurable set E, such that for each interval I, you have:\r\n\r\n0 < m( E n I) < m(I)\r\n\r\nWhere m is the Lebesgue measure, and E n I means the intersection of I with E.", "Solution_1": "More precisely ,the problem is about a borle set E and an open interval I ." } { "Tag": [ "calculus", "integration", "limit", "complex analysis" ], "Problem": "Hi \r\nI have tried to evaluate the integral on the unit circle\r\nint {(e^z - 2)/(e^z -2z-1)}\r\nbut I end up with ln(0) ie undefined.\r\nAny suggestion?\r\n\r\nThx", "Solution_1": "[quote=\"romy\"]Hi \nI have tried to evaluate the integral on the unit circle\nint {(e^z - 2)/(e^z -2z-1)}\nbut I end up with ln(0) ie undefined.\nAny suggestion?\n\nThx[/quote]\r\nU mean $\\int_{|z|=1}\\frac{e^{z}-2}{e^{z}-2z-1}=2 \\pi iRes(\\frac{e^{z}-2}{e^{z}-2z-1},0)$\r\n\r\n$Res(\\frac{e^{z}-2}{e^{z}-2z-1},0)=\\lim_{z\\to 0}{\\frac{e^{z}-2}{e^{z}-2z-1}}=\\frac{e^{z}-2}{(e^{z}-2z-1)'}|_{z=0}=1$\r\nSo ur integral is $2 \\pi i$" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let $ a_1, a_2, ..., a_n$ be positive reals and integer $ n>2$. Define\r\n$ S \\equal{} \\sum a_i$\r\n$ b_i \\equal{} S \\minus{} a_i$\r\n$ T \\equal{} \\sum b_i$\r\nShow that: $ \\frac {\\prod a_i}{\\prod (S\\minus{}a_i)} \\leq \\frac {\\prod b_i}{\\prod (T\\minus{}b_i)}$\r\n(All summations and products range from $ 1$ to $ n$).\r\n\r\nNote: Searched and didn't find it. This problem appeared in Amer. Math Monthly about 1 year back. I'm pretty sure the deadline has passed.", "Solution_1": "this problem is hard, but not nice. oh, can you post a solution for it?\r\n :D" } { "Tag": [], "Problem": "Tom has six U.S. coins each of a different denomination (penny, nickel, dime, quarter, fifty cent piece, and silver dollar). How many different combinations of money can he formed using at least one coin?\r\n :lol: Have fun :D\r\n\r\nFor people who try to trial and error this, It'll take much too long :? not that I would know about that :lol:", "Solution_1": "Weeelllllllllll...\r\n[hide]\nThis is equivalent to the subset problem, except that the null subset is not allowed. Thus, we have $2^6-1=63$ ways.\n[/hide]", "Solution_2": "is it one of each coin, or 6 of each coin?", "Solution_3": "I think it's one of each coin. If so, it would be\r\n[hide]$\\binom61+\\binom62+\\binom63+\\binom64+\\binom65+\\binom66=63$ ways.[/hide]" } { "Tag": [ "geometry", "incenter" ], "Problem": "[url=http://i38.photobucket.com/albums/e146/imdabest112/GeometryProblem.jpg]Picture[/url]\r\n\r\nI need to prove that angle D = 90 + .5 (angle A)\r\n\r\nHelp?\r\n\r\nPS. What is this rating under my name? It constantly fluctuates, and I don't know what it's rating. Thanks.", "Solution_1": "[quote=\"Icy\"]What is this rating under my name? It constantly fluctuates, and I don't know what it's rating. Thanks.[/quote]\r\n\r\nThis is the average of the ratings that your posts get, on a scale of 0 to 6.", "Solution_2": "[quote=\"Icy\"][url=http://i38.photobucket.com/albums/e146/imdabest112/GeometryProblem.jpg]Picture[/url]\n\nI need to prove that angle D = 90 + .5 (angle A)\n\n[/quote]\r\nWell, you forgot to say that $ D$ is incenter.\r\n\r\nDefine some angles with $ \\alpha,\\beta$, play with them.\r\n\r\nIs not a hard problem.", "Solution_3": "Sometimes, there are some problems that are incredibly easy. I hit my head against them for an hour or two, and right when I'm about to give up, the answer comes to me and I feel like the dumbest person on the planet. Does this ever go away with geometry proofs? Or is it just for me?", "Solution_4": "Draw perpediculars from incenter $ D$ to $ AB$ and $ AC$ call the foot of the perpediculars $ R$ and $ S$ respectively. Since we can easily prove by HL that $ \\triangle{ADS}\\cong\\triangle{ADR}$ we can call $ \\angle{ADS}\\cong\\angle{ADR}\\equal{}\\beta$ so $ \\angle{A}\\equal{}180\\minus{}2\\beta$. Now draw lines from $ D$ to $ B$ and $ C$ and a perpendicular from $ D$ to $ BC$ calling it $ T$. Similarly we can prove that $ \\angle{RDB}\\equal{}\\angle{TDB}\\equal{}\\alpha$ and $ \\angle{SDC}\\equal{}\\angle{TDC}\\equal{}\\mu$. Now we can show that $ \\angle{BDC}\\equal{}\\alpha\\plus{}\\mu$ and that $ 2\\alpha\\plus{}2\\mu\\equal{}360\\minus{}2\\beta$ so $ \\angle{BDC}\\equal{}180\\minus{}\\beta$. Now we can easily show that $ \\frac{1}{2}(180\\minus{}2\\beta)\\plus{}90\\equal{}180\\minus{}\\beta$", "Solution_5": "[quote=\"Icy\"][img]http://i38.photobucket.com/albums/e146/imdabest112/GeometryProblem.jpg[/img]\n\nI need to prove that angle D = 90 + .5 (angle A)[/quote]\r\nCall $ A\\equal{}2\\alpha,\\,B\\equal{}2\\beta,\\,C\\equal{}2\\gamma$\r\n\r\nSince $ \\measuredangle\\,D\\equal{}2\\alpha\\plus{}\\beta\\plus{}\\gamma$ (very easy to prove), yields $ \\measuredangle\\,D\\equal{}90^{\\circ}\\plus{}\\alpha$\r\n\r\nwhich the asked it's direct $ \\blacksquare$", "Solution_6": "[quote=\"Hetidemek\"][quote=\"Icy\"][img]http://i38.photobucket.com/albums/e146/imdabest112/GeometryProblem.jpg[/img]\n\nI need to prove that angle D = 90 + .5 (angle A)[/quote]\nCall $ A \\equal{} 2\\alpha,\\,B \\equal{} 2\\beta,\\,C \\equal{} 2\\gamma$\n\nSince $ \\measuredangle\\,D \\equal{} 2\\alpha\\plus{}\\beta\\plus{}\\gamma$ (very easy to prove), yields $ \\measuredangle\\,D \\equal{} 90^{\\circ}\\plus{}\\alpha$\n\nwhich the asked it's direct $ \\blacksquare$[/quote]\r\n\r\nYes, I found this solution about 2 minutes after I read your post. It's what I was referring to in the other post...the solution was so obvious I was ashamed that I asked. But thanks to both of you anyway :D" } { "Tag": [ "geometry", "perimeter", "geometry unsolved" ], "Problem": "XY = 10cm, YZ = 14cm, XZ = 16cm. P bisects YZ into equal halves. Many lines can be drawn thru P, meeting XY (possibly extended) at A and XZ at B (possibly extended). Describe how to find the minimum perimeter of triangle XAB.", "Solution_1": "Use Menelaus.", "Solution_2": "still cant figure this out. need some help here.", "Solution_3": "I will sketch my approach:\r\nConsider any point P inside the angle ZXY and draw a line through it. Then draw the excircle to that line. One will notice that the length of the line is equal to the two tangents to this excircle, so the perimeter of the triangle is equal to the two long (equal!) tangents from X to the excircle. Since the angle is constant, it becomes clear that to minimize the perimeter, we have to minimize the radius of the excircle. This radius is minimized when the excircle passes through P (anything smaller and P will be inside it, which is not possible). So to find the min. perimeter, construct the circle tangent to XY and XZ that passes through P and construct the tangent to that circle through P, and this will be the desired line AB. Hopefully I didn't miss anything. :maybe:", "Solution_4": "I will sketch my approach:\r\nConsider any point P inside the angle ZXY and draw a line through it. Then draw the excircle to that line. One will notice that the length of the line is equal to the two tangents to this excircle, so the perimeter of the triangle is equal to the two long (equal!) tangents from X to the excircle. Since the angle is constant, it becomes clear that to minimize the perimeter, we have to minimize the radius of the excircle. This radius is minimized when the excircle passes through P (anything smaller and P will be inside it, which is not possible). So to find the min. perimeter, construct the circle tangent to XY and XZ that passes through P and construct the tangent to that circle through P, and this will be the desired line AB. Hopefully I didn't miss anything. :maybe:" } { "Tag": [ "videos" ], "Problem": "Click any of the images below! Please have speakers on!\r\n\r\n[url=http://www.leekspin.com][img]http://upload.wikimedia.org/wikipedia/en/9/9d/Loituma.gif[/img][/url]\r\n\r\n[url=http://www.youtube.com/watch?v=vjvVBCNcL_A][img]http://upload.wikimedia.org/wikipedia/en/9/9d/Loituma.gif[/img][/url]\r\n\r\n[url=http://www.leekspin.com][img]http://upload.wikimedia.org/wikipedia/en/9/9d/Loituma.gif[/img][/url]\r\n\r\n[url=http://www.leekspin.com][img]http://upload.wikimedia.org/wikipedia/en/9/9d/Loituma.gif[/img][/url]\r\n\r\n[url=http://www.netnebulo.hu/loituma_clock.swf][img]http://msnbcmedia.msn.com/j/msnbc/Components/Video/060513/tdy_brown_mathcompete_060513.300w.jpg[/img][/url]\r\n\r\n[url=http://www.leekspin.com][img]http://upload.wikimedia.org/wikipedia/en/9/9d/Loituma.gif[/img][/url]\r\n\r\n[url=http://www.leekspin.com][img]http://upload.wikimedia.org/wikipedia/en/9/9d/Loituma.gif[/img][/url]\r\n\r\n[url=http://swgalaxies.datastart.hu/modules/Videos/files/vader-loituma.swf][img]http://upload.wikimedia.org/wikipedia/en/9/9d/Loituma.gif[/img][/url]\r\n\r\n[url=http://www.leekspin.com][img]http://upload.wikimedia.org/wikipedia/en/9/9d/Loituma.gif[/img][/url]\r\n\r\n[url=http://www.leekspin.com][img]http://upload.wikimedia.org/wikipedia/en/9/9d/Loituma.gif[/img][/url]", "Solution_1": "Links to a stupid song. One piece of advice - don't wait for the song to stop, because it won't. Enjoy!", "Solution_2": "err, ok. That was kind of random...", "Solution_3": "errr isn't this spam....I mean with the pictures(links) repeated 200 times and stuff...well not spam but....uhhh", "Solution_4": "[quote=\"junggi\"]errr isn't this spam....I mean with the pictures(links) repeated 200 times and stuff...well not spam but....uhhh[/quote]\r\n\r\nFirst of all, no picture is repeated 200 times.\r\n\r\nSecond of all, I asked Klebian about it first and he said as long as there were 2 different pictures, 4 different links, and the song wasn't in French it was ok.", "Solution_5": "Not worth the effort it took to click the link. Seriously. The animation is meaningless, as is the gibberish being sung. Do geeks spread things like these just so they have a sort of in-joke? Or just because it's the \"popular\" thing to do? Or do you really think it's funny?", "Solution_6": "[quote=\"Ignite168\"][quote=\"junggi\"]errr isn't this spam....I mean with the pictures(links) repeated 200 times and stuff...well not spam but....uhhh[/quote]\n\nFirst of all, no picture is repeated 200 times.\n\nSecond of all, I asked Klebian about it first and he said as long as there were 2 different pictures, 4 different links, and the song wasn't in French it was ok.[/quote]\r\n\r\nHm, sure, if you say so.\r\n\r\nDon't worry, jmadsen, most people think the song sucks. Ignite thinks it's \"awesome\" though...", "Solution_7": "[quote=\"Klebian\"][quote=\"Ignite168\"][quote=\"junggi\"]errr isn't this spam....I mean with the pictures(links) repeated 200 times and stuff...well not spam but....uhhh[/quote]\n\nFirst of all, no picture is repeated 200 times.\n\nSecond of all, I asked Klebian about it first and he said as long as there were 2 different pictures, 4 different links, and the song wasn't in French it was ok.[/quote]\n\nHm, sure, if you say so.\n\nDon't worry, jmadsen, most people think the song sucks. Ignite thinks it's \"awesome\" though...[/quote]\r\n\r\nPlease find sources for both your statements. Wikipedia seems to disagree at least.", "Solution_8": "well i find the song quite interesting. :blush: \r\nit has something unusual. and that language is very nice. what language is it, ignite?", "Solution_9": "He says it is in suomi. I won't go as far as actually saying it sucks, but I must admit, as a song, it leaves much to be desired.", "Solution_10": "The second link has the video of the song. The 5th link is a clock. The 7th link is Star Wars.\r\n\r\nYou guys didn't even bother looking at all 4 of them did you?\r\n\r\nSo jmadsen, shut your mouth if you don't know all the facts :)", "Solution_11": "WRONG!!! 8th is Star Wars.", "Solution_12": "i believe jmadsen wins", "Solution_13": "I found them amusing...and they took over my mind...*hums eerie song* :blush:", "Solution_14": "what language *keeps singing*", "Solution_15": "Finnish\r\nMore about loituma you have [url=http://en.wikipedia.org/wiki/Loituma]here[/url].", "Solution_16": "THANKS *continues*" } { "Tag": [ "symmetry", "geometry", "angle bisector" ], "Problem": "Prove that in any triangle $ ABC$ with angle bisectors $ AP$, $ BQ$, and $ CR$, $ \\frac{1}{AR}\\plus{}\\frac{1}{BP}\\plus{}\\frac{1}{CQ}\\equal{}\\frac{1}{AQ}\\plus{}\\frac{1}{BR}\\plus{}\\frac{1}{CP}$\r\n\r\nMaybe not the coolest proof, but I like the result.", "Solution_1": "Are $ P,Q,$ and $ R$ on segments $ BC, AC,$ and $ AB$, respectively?", "Solution_2": "Yes; angle bisectors must be on the side opposite the angle they bisect.", "Solution_3": "[hide]I will rename $ P, Q, R$ as $ A', B', C'$ if you don't mind, and I will call the lengths of the triangles $ a,b,c$ in the conventional way. Thus the identity is, in my notation,\n\\[ \\frac{1}{AC'} \\plus{} \\frac{1}{BA'} \\plus{} \\frac{1}{CB'} \\equal{} \\frac{1}{AB'} \\plus{} \\frac{1}{BC'} \\plus{} \\frac{1}{CA'} .\\]\n\nNotice that by the angle bisector theorem, $ AB' \\equal{} b \\cdot \\frac{c}{a\\plus{}c}$, and similarly for the other lengths. Then we have\n\\[ \\frac{1}{AC'} \\plus{} \\frac{1}{BA'} \\plus{} \\frac{1}{CB'} \\equal{} \\frac{a\\plus{}b}{cb} \\plus{} \\frac{b\\plus{}c}{ac} \\plus{} \\frac{c \\plus{} a}{ba} \\equal{} \\left( \\frac{a}{cb} \\plus{} \\frac{b}{ac} \\plus{} \\frac{c}{ba} \\right) \\plus{} \\left( \\frac{1}{c} \\plus{} \\frac{1}{a} \\plus{} \\frac{1}{c} \\right) . \\]\nBy symmetry of $ a$, $ b$, and $ c$, the other expression is also equal to this quantity, so the identity is proven. $ \\blacksquare$[/hide]\r\nNice relation." } { "Tag": [ "Support" ], "Problem": "A carpenter is preparing to put a roof on a garage that is 30 feet by 50 feet by by 30 feet. A steel support beam 56 feet in length is positioned in the center of the garage. To support the roof, another beam will be attached to the top of the center beam. At what angle of elevation is the new beam? In other words, what is the pitch of the roof?", "Solution_1": "I can't visulaize this. Please describe it better." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Prove that in any triangle we have $\\frac{R}{2r}\\geq\\frac{m_a}{h_a}$.", "Solution_1": "the hour is late so i'm sorry if i'm speaking nonsense in here but if you just use wellknown substitutions like R=abc/4Sand r=S/p and m_a^2=2b^2+2c^2-a^2/4 and a=(x+y)/2},b=y+z/2,c=x+z/2 the ineq turns out to be very weak.equality takes place only when the triangle is degenerate doesn't it?anyway i shall look over it again in the morning cuz now i'm just tooooo sleepy", "Solution_2": "@Bodom: Of course, an equilateral triangle ABC is an equality case!\r\nSee http://www.mathlinks.ro/Forum/viewtopic.php?t=2558 for this problem...\r\n\r\n darij" } { "Tag": [ "algebra", "polynomial", "function", "geometry", "complex numbers", "AoPS Books" ], "Problem": "hello, does anyone know a website or book or something that can help me with:\r\npolynomial functions (the more indepth stuff)\r\n\r\narguments\r\n\r\nde moivres theorem\r\n\r\ncomplex numbers\r\n\r\n\r\n\r\nthanks a bunch!", "Solution_1": "Mathworld's always a good place to look, and so are the AoPS books.", "Solution_2": "Yes the [url=http://www.artofproblemsolving.com/Books/AoPS_B_About.php]AoPS Books[/url] might help you, or this particular [url=http://www.artofproblemsolving.com/Classes/AoPS_C_ClassesS.php#inttrig]AoPS Class[/url] might help you even more.", "Solution_3": "Barbeau's book about Polynomials (Springer-Verlag) is quite good. Hahn's book Complex Numbers and Geometry is a good book too.", "Solution_4": "okay, thanks" } { "Tag": [ "calculus", "integration" ], "Problem": "Let n be a five digit number and let m be the four digit number obtained by removing\r\nthe middle digit of n. Find all such m,n such that n/m is an integer.", "Solution_1": "[hide=\"A Start\"]\nLet N = $ 10000a+1000b+100c+10d+e$\nM is then $ 1000a+100b+10d+e$\n\nWe need to divide N into M and see what the remainder is.\nThen we need to show for what values the remainder is integral, and that's the solution.\n[/hide]" } { "Tag": [ "logarithms", "algebra unsolved", "algebra" ], "Problem": "10^(x-4)= 72^(x+2)\r\nsolve for x\r\nthanks", "Solution_1": "$x - 4 = (x + 2) \\log 72$\r\n$(1 - \\log 72)x = 4 + 2 \\log 72$\r\n$x = \\frac{4 + 2 \\log 72}{1 - \\log 72} \\approx -8.99845...$\r\n\r\nDoes not really have a rational solution... (although, with calculator or given the value of $\\log 3$ and $\\log 2$, this problem is too easy and should not belong in this forum.)" } { "Tag": [ "conics", "puzzles" ], "Problem": "Fill in the number so that this sudoku satisfies the properties:\r\n1. The sum of numbers in the adjacent boxes with the same color must equal to the top left figure.\r\n2. The numbers in the adjacent boxes with same color must be all different.", "Solution_1": "And it has to be a regular sudoku too!??!?!", "Solution_2": "Yes!!! :wink:", "Solution_3": "Oh man. There was a...fun sudoku in the comics section. Jason is doing it, and instead of numbers, it has \"hard\" math problems, such as $3\\sqrt{9}$ instead of 6. Here it is:\r\n\r\nhttp://www.gocomics.com/foxtrot/2006/10/15/\r\n\r\n[hide=\"Spoiler - ANSWERS\"]\nhttp://adventure.joystiq.com/2006/10/15/foxtrots-sodoku-ometry/\n[/hide]", "Solution_4": "[quote=\"mysmartmouth\"]Oh man. There was a...fun sudoku in the comics section. Jason is doing it, and instead of numbers, it has \"hard\" math problems, such as $3\\sqrt{9}$ instead of 6. Here it is:\n\nhttp://www.gocomics.com/foxtrot/2006/10/15/\n\n[hide=\"Spoiler - ANSWERS\"]\nhttp://adventure.joystiq.com/2006/10/15/foxtrots-sodoku-ometry/\n[/hide][/quote]\r\n\r\nUnless I'm mistaken $-(2^{-2}) =-1/4$ (square upper-right of lower-middle), so it isn't possible", "Solution_5": "[quote=\"besttate\"]\n\nUnless I'm mistaken $-(2^{-2}) =-1/4$ (square upper-right of lower-middle), so it isn't possible[/quote]\r\n\r\nThat says $-(i^{2})$, which would be 1.", "Solution_6": "[quote=\"mysmartmouth\"]\nThat says $-(i^{2})$, which would be 1.[/quote]\r\n\r\nAh, the $i$ was just super squiggly so that it looked like a 2 and its \"dot\" was a negative sign. heh", "Solution_7": "I thought it was $2$ too.", "Solution_8": "Yeah, at first my math teacher who showed me this thought it was a 2 as well." } { "Tag": [ "geometry", "combinatorics proposed", "combinatorics" ], "Problem": "This one is a very nice one. I have a beautiful proof and hope you do too.\r\n Let ABCD be a quadrilateral and n be a positive interger > 1. We divide the sides AB, BC, CD, DA each into n equal segments. We draw lines through 2 corresponding points lying on AB and CD, AD and BC and we obtain $n^2$ quadrilaterals. Prove that the sum of areas of n different small quadrilaterals which pairwisely don't lie on the same row or column is equal to $\\frac{1}{n}$ the area of ABCD.\r\n I note that a corresponding pairs of points are points lying on two opposite sides of ABCD and divide their sides with the same ratio. :lol:", "Solution_1": "You might want to look at [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16137&highlight=]this[/url] :)." } { "Tag": [ "real analysis", "integration", "function", "Support", "geometry", "geometric transformation", "real analysis unsolved" ], "Problem": "Let $ f\\in L^{p}$ and $ g\\in L^{q}$, where $ p$ and $ q$ are Holder conjugates. Prove that $ f*g$ is bounded and continuous.\r\n\r\nObviously, $ \\left|(f*g)(x)\\right|=\\left|\\int_{\\mathbb{R}}f(y)g(x-y)\\,dy\\right|\\leq\\left(\\int_{\\mathbb{R}}\\left|f(y)\\right|^{p}\\right)^{1/p}\\left(\\int_{\\mathbb{R}}\\left|g(x-y)\\right|^{q}\\right)^{1/q}$\r\n$ =\\left\\|f\\right\\|_{p}\\left\\|g\\right\\|_{q}<\\infty$, so we get the bounded claim. Now for continuity, we get that\r\n$ \\left|(f*g)(x+h)-(f*g)(x)\\right|\\leq\\left\\|f\\right\\|_{p}\\left\\|g(x+h)-g(x)\\right\\|_{q}$ from above. However, I don't see why the $ g$ term necessarily goes to 0 without the assumption that $ g$ is continuous (which I never assumed).\r\n\r\nIn one solution I have seen, the the notation $ \\tau_{\\xi}(g): =g(\\xi-x)\\in L^{q}$ is used (not a problem) and it ultimately arrives at the conclusion\r\n$ \\cdots\\leq\\left\\|f\\right\\|_{p}\\left\\|\\tau_{\\xi}(g)-\\tau_{\\xi_{0}}(g)\\right\\|_{q}=\\left\\|f\\right\\|_{p}\\left\\|\\tau_{\\xi-\\xi_{0}}(\\tau_{\\xi_{0}}(g))-\\tau_{\\xi_{0}(g)\\right\\|_{q}}$, which goes to 0 as $ \\xi\\rightarrow\\xi_{0}$. Can someone justify this last step and/or my work?", "Solution_1": "I would start over. Write $ f\\equal{}f_1\\plus{}f_2,g\\equal{}g_1\\plus{}g_2,$ with $ f_1,f_2\\in C_c$ and $ \\|f_2\\|_p,\\|g_2\\|_q<\\epsilon.$\r\n\r\nThen $ f*g\\equal{}f_1*g_1\\plus{}f_1*g_2\\plus{}f_2*g*1\\plus{}f_2*g_2.$\r\n\r\n$ f_1*g_1\\in C_c.$ As for the other terms, we have $ \\|f_1*g_2\\|_{\\infty}\\le \\|f_1\\|_p\\|g_2\\|_q<\\epsilon\\|f_1\\|_p,$ and so on.\r\n\r\nThe fact that $ f*g$ can by uniformly approximated by continuous compactly supported functions means that it is itself continuous and goes to zero at infinity.", "Solution_2": "I'm reading through Wheeden and Zygmund's book right now and I can't seem to find it, but is there a theorem that guarantees that functions in $ L^{p}$ can be uniformly approximated by compactly supported functions?\r\n\r\nOr, does my proof follow from Theorem 8.19, which is continuity in $ L^{p}$? See [url=http://books.google.com/books?id=YDkDmQ_hdmcC&dq=wheeden+and+zygmund&printsec=frontcover&source=bl&ots=Md_KarLMOl&sig=OcPp2HgnQxW81e2TGfXIquq90JU&hl=en&ei=FLqESqmkI4umMZuahfEE&sa=X&oi=book_result&ct=result&resnum=6#v=onepage&q=&f=false]here[/url] for the statement, bottom of page 134.", "Solution_3": "Not uniformly approximated, of course, but approximated in the $ L^p$ sense.", "Solution_4": "Okay, that's just from the fact that continuous functions with compact support are dense in $ L^{p}$, right?\r\n\r\nAnd does Theorem 8.19 justify the final step of my proof?", "Solution_5": "Yes, theorem 8.19, which is the continuity of translation in $ L^p(\\mathbb{R}^n)$ for $ 1\\le p<\\infty,$ does justify your last step. Of course the proof of that theorem also rests on the density of $ C_c$ functions in $ L^p,$ and the failure of that theorem for $ L^{\\infty}$ comes from the failure of that density." } { "Tag": [ "analytic geometry", "geometry", "3D geometry", "prism" ], "Problem": "Consider the 4x4x4 array of lattice points whose coordinates are either 0, 1, 2, or 3. How many right rectangular solids have eight of these points as vertices and have edges of integer length?", "Solution_1": "[hide=\"Solution\"]You need to choose two distinct coordinates in each dimension to define a rectangular prism. In a $ 4\\times4\\times4$ space, there are $ \\tbinom{4}{2}$ ways to choose two distinct points in each dimension, so there are $ \\tbinom{4}{2}^3\\equal{}\\boxed{216}$ such rectangular prisms.\n\nIn general, there are $ \\tbinom{n}{2}^3$ such rectangular prisms in an $ n\\times n\\times n$ space.[/hide]" } { "Tag": [ "geometry", "similar triangles", "geometry proposed" ], "Problem": "Two circles with centres C and D intersect at A and B.A st.line through the point A intersects the circle with centre C at P and that with centre D at Q.prove $ \\angle PBQ \\equal{} \\angle CAD$ and $ \\angle BPC \\equal{} \\angle BQD$. Where am I going wrong?\r\n\r\nI have never faced a problem like this before.Why am I unable to these sums on circles?.In fact,these sums are supposed to be easy.Please suggest ways to overcome this .E-books can be suggested.( which start right from the basics)\r\n\r\n[b]I am actually looking for the reason why I am getting stuck.[/b]\r\n\r\nThank you very much.", "Solution_1": "Can someone solve it?", "Solution_2": "Draw in AB, then $ \\angle PBQ \\equal{} \\angle PBA \\plus{} \\angle ABQ \\equal{} 1/2 \\angle PCA \\plus{} 1/2 \\angle ADQ \\equal{} 90 \\minus{} \\angle PAC \\plus{} 90 \\minus{} \\angle QAD$ (since PAC and QAD are isocles)$ \\equal{} \\angle CAD$.\r\nThis forum, in my opinion, is the best way to get better. Learn the basics about cyclic quads and similar triangles and everything (which you seem to know) and then just practice a lot and read others' posts." } { "Tag": [], "Problem": "Let $ x_1\\equal{}1, x_2\\equal{}16, x_{n\\plus{}2}\\equal{}7x_{n\\plus{}1}\\minus{}x_n\\plus{}10$. Prove that $ x_n$ is a perfect square for every positive integer $ n$.", "Solution_1": "[hide]Let $ P(n)$ be the proposition $ x_{n + 2} = (3\\sqrt {x_{n + 1}} - \\sqrt {x_{n}})^2$ and $ x_{n + 1} + x_{n} - 3{\\sqrt {x_{n + 1}x_{n}} - 5 = 0}$ \n\n$ P(1)$,\n\n$ x_3 = 7\\times16 - 1 + 10 = (4\\times3 - 1)^2$\n\nand\n\n$ x_2 + x_1 - 3{\\sqrt {x_2\\times x_1} - 5 = 16 + 1 - 3\\times4 - 5 = 0}$\n\nTherefore P(1) is true.\n\n$ P(2)$,\n\n$ x_4 = 7\\times121 - 16 + 10 = (3\\times11 - 4)^2$\n\n$ x_3 + x_2 - 3{\\sqrt {x_3 x_2}} - 5 = 121 + 16 - 3(11\\times 4) - 5 = 0$\n\nAssume $ P(n)$ is true for $ 1\\le n\\le k$ .\n\n$ P(k + 1)$,\n\n$ x_{k + 3} = 7x_{k + 2} - x_k + 10$\n\n$ x_{k + 3} = 3(\\sqrt {x_{k + 2}} - \\sqrt {x_{k + 1}})^2 - 2({x_{k + 2} + x_{k + 1} - 3\\sqrt {x_{k + 2}x_{k + 1}} - 5)}$\n\n$ x_{k + 2} + x_{k + 1} - 3\\sqrt {x_{k + 2}x_{k + 1}} - 5$\n\n$ = (3\\sqrt {x_{k + 1}} - \\sqrt {x_{k}})^2 + x_{k + 1} - 3(3\\sqrt {x_{k + 1}} - \\sqrt {x_k})\\sqrt {x_{k + 1}} - 5$\n\n$ = x_{k + 1} + x_{k} - 3(\\sqrt {x_{k}x_{k + 1}}) - 5 = 0$\n\n$ P(k+ 1)$ is true.\n\n$ P(n)$ is true for all $ n\\ge2$.\n\nSince $ x_1$ and $ x_2$ are perfect sqaures, $ x_n$ is a perfect square for all positive intergers $ n$.[/hide]" } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $ |G| \\equal{} 3^n$ and $ p \\equal{} 2\\cdot 3^t \\plus{} 1$ be prime, where $ t\\leq n$. If\r\nthe number of cyclic subgroups for every possible order of $ G$ is a\r\ndivisor of $ 3^n\\cdot p^l$, then is $ G$ cyclic?", "Solution_1": "what means \"for every possible order of $ G$\"?", "Solution_2": "It means that \"the number of cyclic subgroups of order $ 3^k$ for every possible $ k$\", certainly $ k\\leq n$. thanks again." } { "Tag": [ "algebra", "polynomial" ], "Problem": "I was trying out a problem but it didn't have the answers so I am wondering if these are the roots to the Polynomial : $ x^8\\plus{}x^4\\plus{}1$.\r\n\r\nRoots:\r\n$ x\\equal{}\\dfrac{\\sqrt{3}\\pm i}{2}$\r\n$ x\\equal{}\\dfrac{\\minus{}\\sqrt{3}\\pm i}{2}$\r\n$ x\\equal{}\\dfrac{1\\pm i\\sqrt{3}}{2}$\r\n$ x\\equal{}\\dfrac{\\minus{}1\\pm i\\sqrt{3}}{2}$", "Solution_1": "yes, they are the answers\r\nbased on the format of your answers, it looks like (i just said looks like) you factored the polynomial, but you could also multiply the original polynomial by $ x^4\\minus{}1$ (with roots 1, -1, i, -i)\r\nand get $ x^12\\minus{}1$ whose roots are the 12th roots of unity excluding 1, -1, i, and -i results in your answers", "Solution_2": "Put braces around the exponent, if it's more than one character, stevenmeow.\r\n\r\ne.g.\r\nx^12-1 yields $ x^12\\minus{}1$\r\n\r\nbut\r\n\r\nx^{12}-1 gives $ x^{12}\\minus{}1$\r\n :wink:" } { "Tag": [], "Problem": "Source: AMC 12 2000 #14\r\n\r\nWhen the mean, median, and mode of the list: $10,2,5,2,4,2,x$ are arranged in increasing order, they form a non-constant arithemetic progression. What is the sum of all real possible values of $x$?", "Solution_1": "[hide]The mode has to be 2 since there are already 3 of them. The median is $x$ if $2 \\le x \\le 4$. But x must be greater than 2 because otherwise, both the mode and median would be 2, violating the condition of it being a non-constant arithemetic progression. When x is greater than 4, the median is always 4. The mean is: $\\frac{25 + x}{7}$. First, looking at the case where $x>4$, the last term must be 6 because the first two terms of the progression are fixed: 2 and 4. With 6 as the mean:\n\n$\\frac{25 + x}{7} = 6 \\rightarrow x = 42 - 25 = 17$\n\nThat's the only value for x when x is greater than 4. The only value that works for x between 2 and 4 is 3. So the sum is 17 + 3 = 20.[/hide]" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "The vertices and midpoints of the sides of a regular $2007$-gon are to be numbered with numbers $1,2,...,4014$, so that the sum of three numbers on every side is the same. Show that such a numbering is possible", "Solution_1": "Anybody solved it?", "Solution_2": "Can nobody? :(" } { "Tag": [ "geometry", "trigonometry", "circumcircle", "trig identities", "Law of Cosines", "perpendicular bisector" ], "Problem": "In the triangle ABC, the angle B has 15 degrees, and the angle C has 30 degrees. Let AD=AB with B,A,D colineary.Find the angle ADC.", "Solution_1": "hint please?\r\n\r\ndoes it require any high-level theorems?", "Solution_2": "[hide=\"trig bash\"] Drop a perpendicular from $ A$ to $ BC$, and let the length of the perpendicular be 1. We find that $ AC\\equal{}2,AB\\equal{}\\sqrt{6}\\plus{}\\sqrt{2}\\equal{}AD$. By angle chase, $ \\angle{DAC}\\equal{}45^{\\circ}$, so by the law of cosines, $ CD\\equal{}2\\sqrt{2}$, and by the law of cosines again, $ \\angle{CAD}\\equal{}30^{\\circ}$.[/hide]", "Solution_3": "$ \\triangle XBC$ is equilateral, $ X$ on the same side of $ BC$ as $ A.$ $ \\angle BCA \\equal{} 30^\\circ,$ $ A$ is on bisector of the $ \\angle BCX \\equal{} 60^\\circ,$ $ AX \\equal{} AB.$ Circle $ (A)$ with radius $ AB$ passes through $ X$ and cuts $ AB$ again at $ D.$ Perpendicular bisector of $ BC$ cuts the circle $ (A)$ at $ X$ and again at $ Y.$ $ \\triangle BYC$ is isosceles with $ YB \\equal{} YC.$ $ \\angle BDY \\equal{} \\angle BXY \\equal{} 30^\\circ.$ $ BD$ is a diameter of $ (A)$ $ \\Longrightarrow$ $ \\angle BYD \\equal{} 90^\\circ.$ Then $ \\angle ABY \\equiv \\angle DBY \\equal{} 60^\\circ$ and $ \\angle CBY \\equal{} \\angle ABY \\minus{} \\angle ABC \\equal{} 45^\\circ.$ Since $ \\triangle BYC$ is isosceles, $ \\angle CBY \\equal{} 45^\\circ$ and $ \\angle BYD \\equal{} 90^\\circ,$ $ YD$ passes through $ C$ and $ \\angle ADC \\equal{} \\angle BDY \\equal{} 30^\\circ.$", "Solution_4": "nice solution :D", "Solution_5": "Let $X$ like in above solution; clearly $AX\\bot AB$, hence $DX=BX=CX=BC$, so $X$ is the circumcenter of $\\triangle BCD$ and $m(\\widehat{BCD})=\\frac{m(\\widehat{BXC})}{2}=30^\\circ$.\n\nBest regards,\nsunken rock" } { "Tag": [ "linear algebra", "matrix", "calculus", "limit", "trigonometry", "calculus computations" ], "Problem": "Hello.\r\n\r\nIm reading this book about numerical analysis, Absolute and relative error and there's a part I do not understand. It says as follows:\r\n\"\r\nA given size of error is usually more serious when the magnitude of the true value is small. \r\nFor example:\r\n$ 1036.62\\pm0.010$ is accurate to five significant digits and may be adequate to precision, wheres\r\n$ 0.005\\pm0.010$ is a clear disaster.\r\n\"\r\n\r\nCan someone explain how $ 0.005\\pm0.010$ is a clear disaster? not so clear for me. I also didnt understand the first part because 1036.62 itself has 6 significant digits.\r\n\r\nAny help would be appreciated.\r\nThanks a lot.", "Solution_1": "You're not paying attention to the $ \\pm.$\r\n\r\n$ 1036.62\\pm0.01$ could be $ 1036.61$ or $ 1036.63$ - so while we at least have a hint about that sixth digit, we don't know it with certainty.\r\n\r\nAs for $ 0.005\\pm0.010:$\r\n\r\nLet's say I show you an $ n\\times n$ matrix $ A$ and ask you whether $ A$ is invertible. You perform Gaussian elimination, doing the arithmetic in a floating point system so that there is roundoff error. At the end of that Gaussian elimination process, the element in the lower right corner is $ 0.005\\pm0.010.$ So tell me - is the matrix invertible? Even if it is invertible, is the determinant positive or negative? The point is that your arithmetic is now so screwed up that you can't even answer those questions.\r\n\r\nHere's an exercise for you. You know from calculus that $ \\lim_{x\\to0}\\frac {1 \\minus{} \\cos x}{x^2} \\equal{} \\frac12.$\r\n\r\nNow try to convince yourself of that with a floating point device. Take a calculator (or a spreadsheet on your computer - and make sure you're set to \"radians\"). Put in various values of $ x$ and record your answer. Then keep pushing - to smaller, and smaller and smaller values of $ x.$ Is smaller always better? If you were doing arithmetic in the real numbers, smaller would be better, but you're not doing arithmetic in the real numbers, you're doing floating point arithmetic.\r\n\r\nThe error in calculating $ \\cos x$ should be roughly constant for any value of $ x,$ and $ 1$ should be an exact value. The problem is that that makes the error in $ 1 \\minus{} \\cos x$ roughly constant in size, but as $ x\\to 0,$ that number (the numerator of your fraction) goes to zero, creating exactly the problem your textbook is talking about.\r\n\r\nAdded in edit: Using Excel and its standard set of numbers, I seem to get optimum results in that fraction for $ x$ somewhere around $ 0.001$ to $ 0.003.$ For $ x$ smaller than about $ 10^{\\minus{}8},$ I just get zero." } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Prove that there exists a bijection between [0,1)x[0,1) and [0,1) using properties of R. Please do not use schroder-bernstein theorem or other deep set theoretic facts. My instructor says it is possible to show a bijection exists using only binary expansion and similar stuff.", "Solution_1": "Let $ x$ have decimal expansion $ 0.x_1x_2x_3\\dots.$ Let $ y$ have decimal expansion $ 0.y_1y_2y_3.dots.$\r\n\r\nSince decimal expansions aren't quite in a one-to-one correspondence to real numbers, we must give a rule to resolve the ambiguity. Let it be this rule: if $ x$ or $ y$ is a rational number that can be written with a power of $ 10$ as the denominator, choose the expansion that ends in all zeros. That way, we do not have any expansions that end in all $ 9$s.\r\n\r\nNow let $ z\\equal{}0.x_1y_1x_2y_2x_3y_3\\dots$ Once we note that this cannot end in all $ 9$s, we see that the mapping that sends $ (x,y)$ to $ z$ is one-to-one and maps $ [0,1)\\times[0,1)$ onto $ [0,1).$ In particular, it should be apparent what the inverse of this mapping is.\r\n\r\nQuestion: is the mapping I described both measurable and measure-preserving?", "Solution_2": "Actually, that isn't onto- it's missing numbers like $ 0.9090909090\\dots$. The ambiguity still causes trouble, and you need to handle some countable subset separately.", "Solution_3": "Oops. OK, pull out the entire set of numbers of the form $ \\frac{j}{10^m}.$ That countable set can be placed into a one-to-one correspondence with its cartesian product with itself. And then do the decimal expansions splice for all of the other real numbers.", "Solution_4": "[quote]Oops. OK, pull out the entire set of numbers of the form \\frac{j}{10^m}. That countable set can be placed into a one-to-one correspondence with its cartesian product with itself. And then do the decimal expansions splice for all of the other real numbers.[/quote]\r\n\r\nWhen you remove a countable 'A' set from [0,1), you are actually removing Ax[0,1) from the cartesian product so Kent Merryfield's argument is wrong." } { "Tag": [], "Problem": "For $ x,y \\in N$ (x,y>0 )\r\nFind the smallest value of x+y such that $ x^3\\plus{}7xy\\plus{}y^3$ is a perfect squa :D re", "Solution_1": "first assume that one of the numbers is one to make the sum as low as possible.\r\nSo we get 1^3+7y+y^3=k^2\r\nPerfect squares are either 1mod3 or 0mod3. So lets first look at the case where 0 mod3=k^2 so this means that 1+7y+y^3=0mod3. 7y+y^2=2mod3. y(y^2+7)=2mod3. This splits into two more cases where y^2=0 mod 3 or 1mod3. It does not work when it is 0 mod 3 but y^2 can be 1 mod 3. If k^2 is 1 mod 3. Then y^3+7y would have to be 0 mod 3. Which works. Now we see using inspection to find if y would be greater if it were 0 mod 3 or 1 mod 3. We see that when y is also 1(1 mod 3) it is a perfect square which happens to be 9. So the minimum value of x+y is 2.", "Solution_2": "Sorry guy what I meant is $ x^{3} \\plus{} 6xy \\plus{} y^{3}$ :blush: [/quote]" } { "Tag": [ "group theory", "abstract algebra", "inequalities", "function", "number theory", "relatively prime", "superior algebra" ], "Problem": ":?: I have some difficulty with the following problem of Herstein!!\r\n If Sylow-p subgroup P of a finite group G lies in the center of G, then prove that there exists a normal subgroup N of G such that, P & N intersects at unity & G=PN.\r\n\r\nNormality is not a problem,it can also be shown that it suffices to prove for o(G)=pq (q!=0 mod p).It is now required to prove that x->x^q is a homomorphism. Here I got stuck.", "Solution_1": "This follows from Schur-Zassenhaus.", "Solution_2": "Can someone show how you proved the existence of normal subgroup N with PN=G ?", "Solution_3": "[quote=\"Subhajit\"]it suffices to prove for o(G)=pq (q!=0 mod p).It is now required to prove that x->x^q is a homomorphism. Here I got stuck.[/quote]\r\nThis is an odd approach. In this case you know there is a Sylow $ q$-subgroup $ Q$, which is clearly the required subgroup.\r\n\r\nIn fact, I don't think you can show $ x\\mapsto x^q$ is a homomorphism without also showing $ G$ is abelian, except for the case $ q\\equiv1\\mod p$: since $ G'\\leq P\\leq\\mathcal{Z}(G)$, $ x^q y^q (xy)^{\\minus{}q} \\equal{} [x,y]^{\\frac{q(q\\minus{}1)}2}$. If $ q\\neq 1\\mod p$, this is 1 only when $ [x,y]\\equal{}1$.\r\n\r\n\r\nMore generally, we have the following two theorems, either of which proves the result in question:\r\n\r\nBurnside: Whenever $ C_G(P)\\equal{}N_G(P)$, then $ P$ has a normal complement.\r\nSchur-Zassenhaus: Any normal Hall subgroup has a complement.", "Solution_4": "Is there a more elementary proof without using Schur-Zassenhaus or Burnside's normal complement theorem ?", "Solution_5": "The proof of Schur-Zassenhaus in the case when a Hall subgroup $ H$ is central goes like this: pick a section $ \\sigma: K\\to G$ of the SES\r\n\\[ 1\\to H \\to G \\to K \\to 1,\r\n\\]\r\nwhere $ K \\equal{} G/H$, which has order relatively prime to $ H$. Define a function $ f: K\\times K\\to H$ by $ \\sigma(ab) \\equal{} f(a,b)\\sigma(a)\\sigma(b)$. Computing $ \\sigma(abc)$ in two different ways shows that\r\n\\[ (df)(a,b,c) \\stackrel{def} \\equal{} f(b,c) \\minus{} f(ab, c) \\plus{} f(a,bc) \\minus{} f(a,b) \\equal{} 0,\r\n\\]\r\nusing additive notation for $ H$. So $ f$ defines an element of $ H^2(K,H)$, which is by definition\r\n\\[ H^2(K,H) \\equal{} \\frac {\\{f: K^2\\to H | df \\equal{} 0\\}}{\\{f : K^2\\to H\\,|\\,\\exists \\varphi: K\\to H,\\,f(a,b) \\equal{} \\varphi(b) \\minus{} \\varphi(ab) \\plus{} \\varphi(a)\\}}\r\n\\]\r\nIt's easy to see that $ H^2(K,H)$ has exponent dividing $ |H|$. The only non-trivial bit is that since $ (|H|,|K|) \\equal{} 1$, we must have $ H^2(K,H) \\equal{} 0$. So by definition, there exists a function $ \\varphi: K\\to H$ such that $ f(a,b) \\equal{} \\varphi(b) \\minus{} \\varphi(ab) \\plus{} \\varphi(a)$. Now it is easy to see that this implies that the section $ s(a) \\equal{} \\varphi(a)\\sigma(a)$ is a homomorphism, splitting the above SES, and that $ G \\equal{} H\\rtimes s(K)$.\r\n\r\n(Similarly, the fact that $ H^1(K,H) \\equal{} 0$ shows that any two complements of $ H$ are conjugate.)", "Solution_6": "[quote=\"Subhajit\"]It is now required to prove that x->x^q is a homomorphism. Here I got stuck.[/quote]\r\nOkay, now I see where that's coming from. Suppose $ P\\leq \\mathcal{Z}(G)$ is a Sylow $ p$-subgroup of $ G$, with index $ m$, $ p\\nmid m$. We show that $ V(g)\\equal{}g^m$ is the transfer homomorphism $ V: G\\to P$.\r\n\r\nLet $ \\{x_1,\\ldots,x_m\\}$ be a set of coset representatives of $ P$, i.e., $ G\\equal{}\\coprod x_i P$. Let $ g\\in G$. For each $ i$, there is a unique $ j\\equal{}\\sigma_g(i)$ such that $ x_{\\sigma_g(i)}^{\\minus{}1}gx_i \\in P$. The transfer from $ G$ to $ P$ is defined as\r\n\\[ V(g) \\equal{} \\prod x_{\\sigma_g(i)}^{\\minus{}1}gx_i \\in P.\\]\r\nSince\r\n\\[ x_{\\sigma_{gh}(i)}^{\\minus{}1} gh x_i \\equal{} \\left(x_{\\sigma_g(\\sigma_h(i))}^{\\minus{}1}g x_{\\sigma_h(i)}\\right)\\left(x_{\\sigma_h(i)}^{\\minus{}1}hx_i\\right),\\]\r\nand $ P$ is abelian, it is easy to see that $ V$ is a homomorphism. (If $ P$ were not abelian, we would only have a homomorphism $ G\\to P/P'$.)\r\n\r\nNow consider an orbit of size $ k_i$, $ \\{x_i,gx_i,\\ldots,g^{k_i\\minus{}1}x_i\\}$. Since $ P$ is abelian, the product of the factors corresponding to this orbit can be rearranged to give $ x_i^{\\minus{}1}g^{k_i}x_i \\equal{} g^{k_i}$, since $ g^{k_i}\\in P$ is central. Taking the product over each orbit, we just get $ V(g)\\equal{}g^m$.\r\n\r\nNow if $ N\\equal{}\\ker V$, then because $ (m,|P|)\\equal{}1$, we have $ N\\cap P\\equal{}1$. But also $ G/N \\leq P$ is a $ p$-group, so in fact $ G\\equal{}NP$." } { "Tag": [ "MIT", "college" ], "Problem": "Hallo!\r\nDamit es auf der Zugfahrt nach und von Rostock nicht so langweilig ist, f\u00e4nde ich es cool, wenn m\u00f6glichst viele Teilnehmer in einem Zug fahren.\r\nSo gibt es z.B. am 10.2.05 um 7:44 ab M\u00fcnchen eine Verbindung, mit der man um 17:20 in L\u00fctten-Klein ankommt.\r\n\r\nM\u00fcnchen Hbf \t10.02.05 \tab \t07:44 \t18 \t\tICE 882 \tInterCityExpress\r\nBordRestaurant\r\nHamburg Hbf \t10.02.05 \tan \t13:54 \t12a/b\r\nHamburg Hbf \t10.02.05 \tab \t14:19 \t6a \t\tRE 33013 \tRegionalExpress\r\nFahrradmitnahme begrenzt m\u00f6glich, Fahrzeuggebundene Einstiegshilfe: Anmeldung 01805-512512 (12 ct/Min.)\r\nRostock Hbf \t10.02.05 \tan \t16:54 \t8\r\nRostock Hbf \t10.02.05 \tab \t17:07 \t1 \t\tS \tS-Bahn\r\nFahrradmitnahme begrenzt m\u00f6glich\r\n\r\nAls R\u00fcckfahrt bietet sich bieten sich beispielsweise folgende Z\u00fcge an:\r\n\r\nL\u00fctten Klein \t17.02.05 \tab \t08:46 \t2 \t\tS \tS-Bahn\r\nFahrradmitnahme begrenzt m\u00f6glich\r\nRostock Hbf \t17.02.05 \tan \t08:59 \t1\r\nRostock Hbf \t17.02.05 \tab \t09:06 \t9 \t\tRE 33006 \tRegionalExpress\r\nFahrradmitnahme begrenzt m\u00f6glich, Fahrzeuggebundene Einstiegshilfe: Anmeldung 01805-512512 (12 ct/Min.)\r\nHamburg Hbf \t17.02.05 \tan \t11:38 \t6b\r\nHamburg Hbf \t17.02.05 \tab \t12:05 \t14a/b \t\tICE 789 \tInterCityExpress\r\nBordRestaurant\r\nM\u00fcnchen Hbf \t17.02.05 \tan \t18:13 \t18\r\nDauer: 9:27; f\u00e4hrt t\u00e4glich, nicht 21. Feb bis 8. M\u00e4r 2005, 30. M\u00e4r bis 20. Apr 2005\r\n\r\nL\u00fctten Klein \t17.02.05 \tab \t09:46 \t2 \t\tS \tS-Bahn\r\nFahrradmitnahme begrenzt m\u00f6glich\r\nRostock Hbf \t17.02.05 \tan \t09:59 \t1\r\nRostock Hbf \t17.02.05 \tab \t10:07 \t3 \t\tIC 2373 \tInterCity\r\nFahrradmitnahme reservierungspflichtig, Fahrradmitnahme begrenzt m\u00f6glich, BordBistro\r\nHamburg-Harburg \t17.02.05 \tan \t12:37 \t5\r\nHamburg-Harburg \t17.02.05 \tab \t13:12 \t4 \t\tICE 589 \tInterCityExpress\r\nBordRestaurant\r\nM\u00fcnchen Hbf \t17.02.05 \tan \t19:01 \t13\r\nDauer: 9:15; f\u00e4hrt t\u00e4glich, nicht 25. Feb, 4. M\u00e4r\r\n\r\n\r\nIch sch\u00e4tze mal, die Leute aus S\u00fcddeutschland fahren alle entlang dieser Linie. Was meint ihr zu den Vorschl\u00e4gen?", "Solution_1": "Ich bin dabei!\r\n\r\nDer erste Zug ist gut. Bei den R\u00fcckfahrtsz\u00fcgen w\u00fcrde ich eher mit dem zweiten rechnen, ich glaube nicht, dass wir so wahnsinnig schnell fr\u00fchst\u00fccken und dann das durchaus nicht kurze St\u00fcck zur S-Bahn rennen wollen.", "Solution_2": "ich wollte nach dem bwm-kolloquium (7.-8.2.) mehr oder weniger direkt nach rostock weiterreisen...\r\nbei der r\u00fcckfahrt kann ich dann aber auf jeden fall mitfahren. :)", "Solution_3": "Komme nat\u00fcrlich (wie bereits gsagt) auch mit! :)", "Solution_4": "Bin auch dabei! :)" } { "Tag": [ "calculus", "calculus computations" ], "Problem": "How do you find out on what intervals $ y \\equal{} x^5 \\minus{} 5x$ is both increasing and concave up?", "Solution_1": "$ \\frac{dy}{dx}\\geq 0 \\Rightarrow$ increasing.\r\n$ \\frac{d^{2}y}{dx^{2}} \\geq 0\\Rightarrow$ concave up.", "Solution_2": "[hide]$ {y}'\\equal{}5x^4\\minus{}5$\n$ {y}''\\equal{}20x^3$\nTherefore $ \\boxed{x\\geq1}$.[/hide]" } { "Tag": [ "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "Ahoj :)\r\n\r\nLet A be the n x n matrix, the [tex](k,l)^{th}[/tex] complex entry is [tex](-1)^{\\frac{2 k l}{n+1}}[/tex] for k,l=1,...,n . Find and prove the formula for [tex]det(A^2)[/tex] .", "Solution_1": "Do you mean for the entries to be $e^{\\frac{2\\pi ikl}{n+1}}$ ?", "Solution_2": "Yes, that what it is, sorry if it isn't clear.", "Solution_3": "This is a Vandermonde.", "Solution_4": "Hello\r\n\r\n\"Vandermonde\" is not any \"closed formed\" solution for me :roll:\r\n\r\np.s. some german journal states that he was [b]van der Monde[/b] ;)" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Let $ X$ be a subset of $ R^n$, such that the distance between any pair of points $ x$ and $ y$ is in given set $ \\{ d_1, d_2, \\ldots, d_s \\}$, i.e. $ \\parallel{}x \\minus{} y\\parallel{} \\equal{} \\sqrt { (x_1 \\minus{} y_1)^2 \\plus{} (x_2 \\minus{} y_2)^2 \\plus{} \\ldots \\plus{} (x_n \\minus{} y_n)^2} \\equal{} d_k$, where $ 1 \\leqslant k \\leqslant s$. \r\nProve that \r\n\\[ |X| \\leqslant \\binom{n \\plus{} s \\plus{} 1}{s}.\\]", "Solution_1": "The current best bound is $ {n\\plus{}s} \\choose{s}$ by Bannai, Bannai and Stanton.\r\n\r\nThe paper is free here :\r\nhttp://www.springerlink.com/content/9th7t0l322644p94/\r\n\r\nPierre." } { "Tag": [], "Problem": "I need to prove n^4+64 is composite (non-prime) for all positive integer values of n. \r\n\r\nIt is straightforward enough except in the cases where n is of the form 5(2p+1). Can anyone sort this out?", "Solution_1": "Too easy and too well known for Pre-Olympiad.\r\n\r\n$n^{4}+64 = n^{4}+4(2)^{4}= (n^{2}-4n+8)(n^{2}+4n+8)=((n-2)^{2}+4)((n+2)^{2}+4)$\r\n\r\nClearly none of the factors can be $1$ so it is composite.\r\nThis is called the Sophie-Germain identity" } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "Let $f(z)\\equiv az^2+bz+c=0$ be a ecuation with complex coeficients and $f(z_1)=f(z_2)=0$. Ascertain $|z_1|+|z_2|.$", "Solution_1": "You square and get it????..\r\n\r\n[hide]I'll post my soln later!!![/hide]" } { "Tag": [ "ratio", "probability", "expected value", "real analysis", "real analysis unsolved" ], "Problem": "I don't know if this post is in the right thread or not but I'll ask anyway.\r\n\r\nThis is the problem i need to solve:\r\n\r\nTeam A is playing Team B. Team A offense averages 29.5 points against teams that on average give up 25.76 points on defense. Team B gives up on defense an average of 32.1 points against teams that average 25.48 points on offense. What is the best estimate of how many points Team A will score???\r\n\r\n\r\n\r\nI have 2 ways to solve this problem but I'm getting different answers. I always set these up as fractions. Like 29.5/25.76 and 32.1/25.48 as actual on top and opponents average on bottom.\r\n\r\nMethod 1. Find out the differences. Team A outscores their oppontents averages by (29.5 - 25.76) = 3.74 and then Team B defense the same thing (32.1 - 25.48) = 6.62 Add these two together 3.74 + 6.62 = 10.36 and then add 10.36 to the average of the denominators which is (average of 25.76 and 25.48=25.62) which gives you 35.98\r\n\r\n\r\nMethod 2. Cross Multiply and take averages. Team A averages 29.5 against teams that give up 25.76 But Team B doesn't give up 25.76 they give up 32.1 so Team A has 36.76 But then Team B gives up 32.1 against teams that put up 25.48 but Team A doesn't put up 25.48 they put up 29.5 so Team B defense has 37.17 Take the average of the two (36.76 and 37.17) and you get 36.96\r\n\r\n\r\n35.98 and 36.96 are too far apart for me. Maybe there is another more accurate way to solve this than what i'm doing. there shouldn't be two different answers for 1 problem. I understand it's averages, so maybe a range of numbers are possible. Please help me out i need the most accurate way to find out the solution whether it's an actual number or a range.", "Solution_1": "The problem with both methods is that you have two sets of statistical data to work with: one that tells you how Team A performs on average and one that tells you how Team B performs on average. You can't assume that both averages will hold in a game between the two, so you have to take some kind of average of the two averages. \r\n\r\nFor example, in Method 1 you assume that the spread of the game will be the sum of the spread of an average game that A plays and the spread of an average game that B plays, which is obviously absurd (consider what happens if Team A and Team B are the only teams in the league and always play each other, and Team A always scores $ 30$ points and Team B zero. Your method will then predict a $ 60$-point victory). You need to average $ 3.74$ and $ 6.62$, not add them, if you're going to do it this way. \r\n\r\nIn Method 2 you appear to be assuming that the ratio between points scored and points given up (rather than the difference) is constant. I'm not sure what you're doing here, but this is a different statistical assumption than the first method.\r\n\r\nYou shouldn't expect there to be one answer to this; the question isn't well-defined enough yet. It should be phrased as an expected-value question, but we are given two separate expected values." } { "Tag": [ "pigeonhole principle" ], "Problem": "On the chess board we want to put bishops such that no bishop threatens another one but each of the unoccupied field is at least threatened by one bishop.\r\n\r\na.) Find such a configuration of 8 bishops.\r\n\r\nb.) Determine the maximum number of bishops that can be put on the chess board satisfying above rules.\r\n\r\nc.) Analyse for which number $n$ there is such a configuration with $8 < n < m$.", "Solution_1": "a) The red squares are the ones occupied by bishops.", "Solution_2": "b) Probably more than this, but this is what I got.\r\n\r\nSorry about the double post, but it won't let you attach more than one document.", "Solution_3": "i figure that there is a maximum of 14 bishops on the board. The 8 on the first rank and then 6 on the 8th rank but none in either of the corners", "Solution_4": "[hide=\"a start\"]well when a bishop is placed on any of the sides, it eliminates the position for 8 squares and another bishop can be placed anywhere and they can only have one intersecting \"line of fire\" and every other square gives us a solvable equation so we just occupy all of the squares available to us then[/hide]", "Solution_5": "i think that that is the max. Its kinda like putting the 8 queens on the board in such a way that none is attacking the other. In that case you put them a knight's move away from each other... Trust me i solved it at chess camp this summer", "Solution_6": "[quote=\"Treething\"]a) The red squares are the ones occupied by bishops.[/quote]\r\n\r\nWell, that's not correct.\r\n\r\nLook at some of the top squares not occupied by bishops. They aren't threatened by bishops either, and the problem asks for an arrangement where other squares are threatened also. So, that doesn't work.\r\n\r\nLet's say the chessboard is facing you, so that the bottom left square is a1, and the top right square is h8.\r\n\r\nPlace the bishops on a1 through h1, and then place bishops on b8 through g8. You will find that this is the maximum arrangement as demji has mentioned already.\r\n\r\nThere are 14 total, and every unoccupied square is being attacked by a bishop.[/img]", "Solution_7": "Try pigeonhole principle (the floor variant)." } { "Tag": [ "probability" ], "Problem": "A jar contains 6 red balls, 5 green balls and 9 yellow balls. Danny draws a ball at random from the jar and does not return it. Celine draws another ball at random from the jar. What is the probability that they drew a ball of the same color? Give your answer as a common fraction in lowest terms.", "Solution_1": "[quote=\"Quevvy\"]A jar contains 6 red balls, 5 green balls and 9 yellow balls. Danny draws a ball at random from the jar and does not return it. Celine draws another ball at random from the jar. What is the probability that they drew a ball of the same color? Give your answer as a common fraction in lowest terms.[/quote]\r\n\r\n[hide=\"solution\"]The number of ways to draw 2 red balls is ${6\\choose 2}=15$\n\nThe number of ways to draw 2 green balls is ${5\\choose 2}=10$\n\nThe number of ways to draw 2 yellow balls is ${9\\choose 2}=36$\n\nThe total number of ways to draw 2 balls is ${6+5+9\\choose 2}=190$\n\n$\\frac{15+10+36}{190}=\\boxed{\\frac{61}{190}}$[/hide]" } { "Tag": [ "topology", "geometry", "3D geometry", "sphere", "combinatorial geometry", "advanced fields", "advanced fields unsolved" ], "Problem": "Let $ X\\equal{}l^2(\\mathbb{N})$. Prove that the closure of the unit sphere in the weak topology is the unit ball.\r\n\r\n* I think one can prove this using Krein-Milman: the closed(in the norm topology) unit ball is the closure of the convex hull of it's extreme points. Here, since we are dealing with the convex set (unit ball in particular), there is no difference between weak closure and norm closure(Hahn-Banach theorem consequence). And it is only left to notice that extreme points of closed unit ball of $ l^2(\\mathbb{N}$ is the unit sphere.\r\n\r\nI'm interested whether there is a straight way, without invoking lots of theorems.", "Solution_1": "$ a\\plus{}cf_n\\to a$ weakly, where $ (f_n)$ is an ONB for $ a^{\\perp}$ and $ c\\equal{}\\sqrt{1\\minus{}\\|a\\|^2}$." } { "Tag": [ "trigonometry" ], "Problem": "A man walks $x$ miles due west, turns $150^\\circ$ to his left and walks 3 miles in the new direction. If he finishes a a point $\\sqrt{3}$ from his starting point, then $x$ is\r\n\r\n$\\text{(A)} \\ \\sqrt 3 \\qquad \\text{(B)} \\ 2\\sqrt{5} \\qquad \\text{(C)} \\ \\frac 32 \\qquad \\text{(D)} \\ 3 \\qquad \\text{(E)} \\ \\text{not uniquely determined}$", "Solution_1": "[hide]A triangle is formed with side lengths $x$, $3$, and $\\sqrt{3}$. Angle opposite side of length $3$ is uniquely determined. Law of Sine shows this is not an ambiguous case as the sin of the angle is $\\frac{\\sqrt{3}}{2}$, which denotes either $60^\\circ$ or $150^\\circ$, and the latter's sum with $30^\\circ$ is over $180^\\circ$. The angle is $60^\\circ$ and thus the it is a $30-60-90$ triangle, $x$ is thus $2\\sqrt{3}$\n\n$\\boxed{E}$[/hide]", "Solution_2": "[hide=\"Solution\"]$\\text{Answer: (E)}$\n\nThis problem is easy to figure out with a drawing. Let's say he started on some line $l$. The man walks for some distance x until he reaches a point $A$ and turns 150 degrees, forming a 30 degree angle. Next, he walks 3 miles to a point $B$.\n\nNow, there are two such points located on line $l$ that are of a distance $\\sqrt{3}$ from point $B$, so the answer is not uniquely determined.[/hide]", "Solution_3": "[hide]The triangle formed is not uniquely determined (SSA), as two triangles can be formed using the Door-Swinging Theorem. So the answer is $\\boxed{E}$[/hide]" } { "Tag": [ "trigonometry", "geometry", "circumcircle", "incenter", "trig identities", "Law of Cosines", "geometry unsolved" ], "Problem": "In a triangle $ ABC$,$ BC$ \u00ed smallest edge.Choose $ M$ and $ N$ on $ AB$ and $ AC$ respectively such that $ BM\\equal{}BC$ and $ CN\\equal{}CB$\r\nProve that\r\n$ \\frac{MN}{BC}\\equal{}\\sqrt{3\\minus{}2(cosA\\plus{}cosB\\plus{}cosC)}$", "Solution_1": "[color=darkblue]$ \\sqrt {3 - 2\\sum \\cos A} = \\sqrt {3 - 2\\bigg(1 + \\dfrac{r}{R}\\bigg)} = \\sqrt {1 - \\dfrac{2r}{R}}$\n\n$ BN = 2BC\\sin \\dfrac{C}{2}$\n\n$ \\small{MN = \\sqrt {BN^2 + BM^2 - 2BN\\cdot BM \\cos \\angle MBN} =}$ \n\n${ BC\\sqrt {1 + 4\\sin^2 \\dfrac{C}{2} - 4\\sin \\dfrac{C}{2} \\sin\\bigg( B + \\dfrac{C}{2}\\bigg)}}$\n\nSo, we need to prove that\n\n$ 1 + 4\\sin^2 \\dfrac{C}{2} - 4\\sin \\dfrac{C}{2} \\sin\\bigg( B + \\dfrac{C}{2}\\bigg) = 1 - \\dfrac{2r}{R}$\n\n$ 4\\sin \\dfrac{C}{2} \\sin\\bigg( B + \\dfrac{C}{2}\\bigg) - 4\\sin^2 \\dfrac{C}{2} = \\dfrac{2r}{R}$\n\n$ 4\\sin \\dfrac{C}{2} \\bigg( \\sin B \\cos\\dfrac{C}{2} + \\cos B \\sin \\dfrac{C}{2}\\bigg) - 4\\sin^2 \\dfrac{C}{2} = \\dfrac{2r}{R}$\n\n\n$ 2\\sin C \\sin B + 4 \\cos B \\sin ^2 \\dfrac{C}{2} - 4\\sin^2 \\dfrac{C}{2} = \\dfrac{2r}{R}$\n\n$ \\sin C \\sin B + 2\\sin ^2 \\dfrac{C}{2}( \\cos B - 1) = \\dfrac{r}{R}$\n\n\n$ \\sin C \\sin B - 4 \\sin ^2 \\dfrac{C}{2} \\sin ^2 \\dfrac{B}{2} = \\dfrac{r}{R}$\n\n$ \\dfrac{bc}{4R^2} - 4\\dfrac{(p - a)^2 (p - c)(p - b)}{bca^2} = \\dfrac{r}{R}$\n\n$ \\dfrac{a^2b^2c^2 - 16R^2(p - a)^2(p - c)(p - b)}{4R} = \\dfrac{bca^2 S}{p} = \\dfrac{b^2c^2a^3}{4Rp}$\n\n$ a^2b^2c^2p - 16R^2p(p - a)^2(p - b)(p - c) = b^2c^2a^3$\n\n$ a^2b^2c^2p - 16R^2 S^2(p - a) = b^2c^2a^3$\n\n$ a^2b^2c^2p - 16R^2 \\dfrac{a^2b^2c^2}{16R^2}(p - a) = b^2c^2a^3$\n\n$ p - (p - a) = a$\n\nThat's all.... :!: [/color]", "Solution_2": "From the isosceles triangle $ BCM$ we get $ \\angle BCM = \\frac {\\pi - B}{2}$ and $ CM = 2BC\\cos\\frac {\\pi - B}{2}$. We also have $ CN = BC$ and $ \\angle MCN = C - \\frac {\\pi - B}{2} = \\frac {C - A}{2}$. Thus we deduce from the law of cosines,\r\n\\begin{align*}MN^2 & = CM^2 + CN^2 - 2CM\\cdot CN\\cos\\angle MCN \\\\\r\n& = 4BC^2\\sin^2\\frac B2 + BC^2 - 4BC^2\\sin\\frac B2\\cos\\frac {C - A}{2} \\\\\r\n& = BC^2\\left(4\\sin^2\\frac B2 + 1 - 2\\sin\\frac {B + C - A}{2} - 2\\sin\\frac {A + B - C}{2}\\right) \\\\\r\n& = BC^2\\left(3 - 2 + 4\\sin^2\\frac B2 - 2\\sin\\frac {\\pi - 2A}{2} - 2\\sin\\frac {\\pi - 2C}{2}\\right) \\\\\r\n& = BC^2\\left(3 - 2\\cos B - 2\\cos A - 2\\cos C\\right).\\end{align*}", "Solution_3": "[b]General problem.[/b] Let $\\triangle ABC$ be a scalene triangle and $ BC$ is its shortest side. $ M,N$ are two points on $AC, AB$ such that $ BN \\equal{} CM \\equal{} BC.$ Then $ MN$ is perpendicular to $OI,$ where $ O , I$ are the circumcenter and incenter of $ \\triangle ABC,$ and $\\frac {_{MN}}{^{BC}} \\equal{} \\frac {_{OI}}{^{R}}.$\n\n[b]Proof:[/b] In the isosceles triangles $ \\triangle OAC$ and $ \\triangle OAB,$ we have\n\n$ R^2 \\minus{} OM^2 \\equal{} AM \\cdot MC , \\ R^2 \\minus{} ON^2 \\equal{} AN \\cdot BN$\n\n$\\Longrightarrow ON^2 \\minus{} OM^2 \\equal{} AM \\cdot MC \\minus{} AN \\cdot BN$ \n\n$ \\Longrightarrow ON^2 \\minus{} OM^2\\equal{} (CA \\minus{} BC)BC \\minus{} (AB \\minus{} BC)BC \\equal{} BC(CA \\minus{} AB)$\n\nSince $\\triangle MIC$ and $\\triangle MIB$ are isosceles, we have $ IM \\equal{} IB,$ $IN \\equal{} IC.$ Then\n\n$ IN^2 \\minus{} IM^2 \\equal{} IC^2 \\minus{} IB^2 \\equal{} BC(CA \\minus{} AB)$ \n\n$ \\Longrightarrow ON^2 \\minus{} OM^2 \\equal{} IN^2 \\minus{} IM^2 \\Longrightarrow \\ IO \\perp MN$\n\nOn the other hand, $ \\angle NBM \\equal{} \\angle OCI \\equal{} 90^{\\circ} \\minus{} \\frac {_1}{^2}\\angle C \\minus{} \\angle A.$ If $ L \\equiv{} CI \\cap BM,$ then it follows that $ \\angle NMB$ and $\\angle OIC$ are supplementary, which implies that $ \\triangle IOC$ and $ \\triangle MNB$ are pseudo-similar, hence \n\n$ \\frac {MN}{OI} \\equal{} \\frac {BN}{OC} \\Longrightarrow \\frac {MN}{BC} \\equal{} \\frac {OI}{R}.$", "Solution_4": "[quote=\"dragon001\"]In a triangle $ ABC$,$ BC$ \u00ed smallest edge.Choose $ M$ and $ N$ on $ AB$ and $ AC$ respectively such that $ BM \\equal{} BC$ and $ CN \\equal{} CB$\nProve that\n$ \\frac {MN}{BC} \\equal{} \\sqrt {3 \\minus{} 2(cosA \\plus{} cosB \\plus{} cosC)}$[/quote]\r\n\r\nThis problem is on Mathematical and Youth magazine and it can be solved easily by using vector. Don't need trigonometry.\r\nTo dragon001: You are bad :mad:", "Solution_5": "[b]Livetolove[/b] [b]is right[/b], this topic must be deleted, the problem that has been proposed in this topic is the problem in [b]Mathematics and Youth magazine[/b] in [b]Vietnam[/b], the latest number [b](May/2009)[/b]." } { "Tag": [ "conics", "parabola", "integration", "calculus", "calculus computations" ], "Problem": "use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the x-axis. sketch the region and a typical shell. \r\n\r\nx=1+y^2, x=0, y=1, y=2", "Solution_1": "Solution\r\n[hide]Well I'm not going to sketch it but its a parabola opening to the right with a vertex at $(1,0)$.\n\n$x=1+y^{2}$, $x=0$, $y=1$, $y=2$.\n\nSo we're revolving around the x-axis:\n\n$V= 2 \\pi \\int_{a}^{b}p(x) \\cdot h(x) dx$\n\nWhere $p(x)$ is the distance from the axis of revolution and $h(x)$ is your representative rectangle. Now:\n\n$V=2 \\pi \\int_{1}^{2}y(1+y^{2}) dy$\n$V=2 \\pi \\int_{1}^{2}(y+y^{3}) dy$\n$V=2 \\pi \\left[ \\frac{y^{2}}{2}+\\frac{y^{4}}{4}\\right]_{1}^{2}$\n$V=2 \\pi \\left[ \\left( \\frac{2^{2}}{2}+\\frac{2^{4}}{4}\\right)-\\left( \\frac{1^{2}}{2}+\\frac{1^{4}}{4}\\right) \\right]$\n$V=2 \\pi \\left( \\frac{3}{2}+\\frac{15}{4}\\right)$\n$\\boxed{V=\\frac{21 \\pi}{2}}$[/hide]" } { "Tag": [], "Problem": "Let a, b, c be real numbers such that $ \\frac{\\left(a\\minus{}b\\right)\\left(b\\minus{}c\\right)\\left(c\\minus{}a\\right)}{\\left(a\\plus{}b\\right)\\left(b\\plus{}c\\right)\\left(c\\plus{}a\\right)}\\equal{}\\frac{1}{11}$. What is the value of\r\n\\[ \\frac{a}{a\\plus{}b}\\plus{}\\frac{b}{b\\plus{}c}\\plus{}\\frac{c}{c\\plus{}a}\r\n\\]", "Solution_1": "[quote=\"STARS\"]Let a, b, c be real numbers such that $ \\frac {\\left(a - b\\right)\\left(b - c\\right)\\left(c - a\\right)}{\\left(a + b\\right)\\left(b + c\\right)\\left(c + a\\right)} = \\frac {1}{11}$. What is the value of\n\\[ \\frac {a}{a + b} + \\frac {b}{b + c} + \\frac {c}{c + a}\n\\]\n[/quote]\r\n\r\n\r\n$ \\frac{a}{b}=x;\\frac{b}{c}=y;\\frac{c}{a}=z$---->$ xyz=1$------>$ \\frac {\\left(a - b\\right)\\left(b - c\\right)\\left(c - a\\right)}{\\left(a+b\\right)\\left(b+c\\right)\\left(c+a\\right)}=\\frac{(x-1)(y-1)(z-1)}{(x+1)(y+1)(z+1)}=\\frac{1}{11}$------>$ \\frac{xyz-xy-xz-yx+x+y+z-1}{xyz+xy+xz+yz+x+y+z+1}=\\frac{1}{11}$------->$ 10(x+y+z)=2+12(xy+xz+yz)$------> $ 5(x+y+z)=1+6(xy+xz+yz)$-----> $ A=xy+xz+yz$-----> $ x+y+z=\\frac{1+6A}{5}$\r\n\r\n$ \\frac{a}{a+b} + \\frac{b}{b+c} + \\frac{c}{c+a}=\\frac{x}{x+1}+\\frac{y}{y+1}+\\frac{z}{z+1}=\n\\frac{x(y+1)(z+1)+y(x+1)(z+1)+z(x+1)(y+1)}{(x+1)(y+1)(z+1)}=\\frac{3xyz+2(xy+xz+yz)+x+y+z}{xyz+xy+xz+yz+x+y+z+1}=\n\\frac{3+2(xy+xz+yz)+x+y+z}{2+xy+xz+yx+x+y+z}=\\frac{3+2A+\\frac{1+6A}{5}}{2+A+\\frac{1+6A}{5}}=\n\\frac{16+16A}{11+11A}=\\frac{16}{11}$" } { "Tag": [], "Problem": "There are 7 people on a quidditch team. If there are 7 people who tried out and will be on the team and,2 of them have the same chances to be seeker, how many ways can the quidditch team be formed?\r\n\r\n[hide] is there a way to overcount this?[/hide]", "Solution_1": "what about the beaters, and all the other jobs that end with eepers. are there restrictions on the other positions?", "Solution_2": "no not at all. they all are very capable of every position but seeker and then only two of them have the ability to catch the snitch.", "Solution_3": "so why isn't it $2\\times 6!$?", "Solution_4": "i really was just wondering if there was a way to overcount it...\r\n[hide] this is actually about a question i myself have i thought it'd be 2*2*6 :oops: i'm only sure for it if i had 4 people 2 only seekers..2*2*3=12[/hide]", "Solution_5": "[hide]Choose the person to be seeker first. There are two choices for the seeker. Are the other 6 positions distinct? If they are, then there are $6!$ ways to choose the others, and if not, only $\\binom66=1$ way. Thus, there are either $2\\times6!=1440$ or $2\\times1=2$ ways.[/hide]", "Solution_6": "yeah actually, are the positions idstinct? the beaters are the same i guess and so are the three offensive people, forgot what their name is.", "Solution_7": "yes 7 people tried out for the team 2 can seek and 5 can't but all 7 can chase,beat,and keep.", "Solution_8": "just to let you guys know your answers are correct! now i only need my hidden question answered.", "Solution_9": "I have a different answer. first, there are 2 ways to chose the seeker. then, there are $\\frac{6!}{3!\\cdot2!}=60$ways to scramble the other players. so isn't it $\\boxed{120}$? :maybe:", "Solution_10": "I am not sure..the only one of these types of problems i have memorized is if i had 4 people and 2 could be seeker...the answer is 12 which is $2*3!$ so I can't tell you where you went wrong..or for that case if I went wrong but I have more faith in theirs...as that is how I did the problem above. And I know that is the right answer. :read: Hmmm...I'll be thinking about it. Maybe I'll get AoPS 1 out.", "Solution_11": "ahha! i meant if flint and malfoy could be any player and bletchley,pucey,montague,crabbe and goyle could be any of the remaining ones! so they are distinct people.." } { "Tag": [ "geometry", "analytic geometry", "trigonometry" ], "Problem": "Consider the following system:\r\n\r\n$x^{2}y=xy^{2}$\r\n$x^{2}+y^{2}=4$\r\nThe real solutions make a convex regular polygon when graphed on the Cartesian plane. Find the area of the polygon.", "Solution_1": "[hide]Well the first one is true iff $x=y$ or $x=0$ or $y=0$...\nAnd the second one just makes a circle with radius 2... \nI don't know what kind of polygon that can possibly make... it's just a kind of plus sign with a diagnol line through it.l [/hide]\r\n\r\nI think I may be missing something here...", "Solution_2": "Yeah, what you said is correct.\r\n\r\nIt does make a polygon. Substitute the first equation that you wrote into $x^{2}+y^{2}=4$ and see what you get.", "Solution_3": "are you sure its regular? seems like a convex hexagon but not regular.", "Solution_4": "[hide=\"Interesting\"] $x^{2}y = xy^{2}\\Leftrightarrow (x-y) xy = 0$\n\nHence $x = y, x = 0, y = 0$ make up our solutions. We also want points on a circle of radius $2$. This gives us six points\n\n$(0, \\pm 2)$\n$(\\pm 2, 0)$\n$(\\pm \\sqrt{2}, \\pm \\sqrt{2})$\n\nWhich want to, but cannot, form either a regular hexagon or a regular octagon. Odd. [/hide]", "Solution_5": "[quote=\"t0rajir0u\"][hide=\"Interesting\"] $x^{2}y = xy^{2}\\Leftrightarrow (x-y) xy = 0$\n\nHence $x = y, x = 0, y = 0$ make up our solutions. We also want points on a circle of radius $2$. This gives us six points\n\n$(0, \\pm 2)$\n$(\\pm 2, 0)$\n$(\\pm \\sqrt{2}, \\pm \\sqrt{2})$\n\nWhich want to, but cannot, form either a regular hexagon or a regular octagon. Odd. [/hide][/quote]\r\n\r\nTo be completely clear: the last two points are $\\pm(\\sqrt{2},\\sqrt{2})$ (since $(\\pm\\sqrt{2},\\pm\\sqrt{2})$ could be interpreted as $4$ points).", "Solution_6": "[quote=\"t0rajir0u\"]\nWhich want to, but cannot, form either a regular hexagon or a regular octagon. Odd. [/quote]\r\n\r\nWhat do you mean \"want to\" make a regular hexagon?", "Solution_7": "I just mean that they are six of the eight points of a regular octagon. They do not, by themselves, form either a regular hexagon or a regular octagon. :P Sorry if I was confusing.", "Solution_8": "Doesn't that make 8 points?", "Solution_9": "[quote=\"13375P34K43V312\"]Consider the following system:\n\n$x^{2}y=xy^{2}$\n$x^{2}+y^{2}=4$\nThe real solutions make a convex regular polygon when graphed on the Cartesian plane. Find the area of the polygon.[/quote]\r\n\r\n\r\n[hide]The first equation can be rewritten as $x^{2}y-xy^{2}=xy(x-y)$.\n\nSo, the first graph is the line $y=x$ and the coordinate axes.\n\nThe second graph is just a circle with radius 2.\n\nWe see that all of the points of intersection of the circle are $(2,0)$, $(0,2)$, $(-2,0)$, $(0,-2)$, $(\\sqrt{2},\\sqrt{2})$, and $(-\\sqrt{2},-\\sqrt{2})$. This isn't a regular polygon\n\nOne can check that the area is $4*\\frac{1}{2}*\\sin 45^\\circ*4+2*\\frac{1}{2}*4=4+4\\sqrt{2}$.\n\n$4+4\\sqrt{2}$[/hide]", "Solution_10": "[quote=\"13375P34K43V312\"]Doesn't that make 8 points?[/quote]\r\n\r\nNo. We are missing $(\\sqrt{2},-\\sqrt{2})$ and $(-\\sqrt{2}, \\sqrt{2})$." } { "Tag": [ "number theory", "greatest common divisor", "relatively prime" ], "Problem": "Suppose $a, b, n \\in \\mathbb{Z} $ with $n > 0$. Suppose $ab \\equiv 1 \\bmod {n}$. Prove that $a$ and $b$ are relatively prime to $n$.\r\n\r\nHow can I prove this?", "Solution_1": "Well then let ab=xn+1. Therefore, ab and xn are relatively prime. Thus, ab and n are relatively prime, and the condition follows.", "Solution_2": "The largest number that divides the LHS of ab - xn = 1 is 1 so the GCD of ab and xn is 1, and GCD(ab,n)=1.", "Solution_3": "He wants $a$ and $b$ seperately. How do we prove that?", "Solution_4": "$ab = 1 + nq$\r\n\r\n$ab - nq = 1$\r\n\r\nThe $GCD(a,n)|1$ and $GCD(b,n)|1$ so they both equal one.", "Solution_5": "OK Now I get it. Just a difference of perspective!" } { "Tag": [], "Problem": "I'm going to post 38 Fantastic problems for you to solve from the MOEMS problem of the day. :D Please give comments.\r\n\r\nSomeone can make up solutions please...", "Solution_1": "Are there any comments?", "Solution_2": "Why did you take the time to compile those problems if they're all posted in another topic?", "Solution_3": "where are they???", "Solution_4": "[quote=\"anirudh\"]where are they???[/quote]\r\nMOEMS Problem of the Day.", "Solution_5": "its all in the document, you can print it and answer it anywhere..." } { "Tag": [], "Problem": "Find the sum of all digits of $2^{60}$", "Solution_1": "$2^{60}=1152921504606846976,$\r\n$1+1+5+2+9+2+1+5+4+6+6+8+4+6+9+7+6=\\boxed{82}$", "Solution_2": "That would've been ridiculously easy. Yay, Google calculator! :lol:", "Solution_3": "i think he meant without a calculator but whatever...it's still trivial, just a bunch of arithmetic to do:\r\n\r\n$((1000+24)^2)^3$", "Solution_4": "Yes it has to be done without calculaors.How did you get that iversonfan?\r\n\r\nAny more easier methods anyone?", "Solution_5": "Iversonfan just simplified the $2^{60}$ to get $((1000+24)^2)^3$ because powering 1000's gives you a digital sum of 1.", "Solution_6": "How about if the expression is written in binary - what is the digital sum? :)", "Solution_7": "[hide=\"Gyans Extension\"]In binary it's just 100000000000... with 59 0's. So the digital sum is 1.[/hide]", "Solution_8": "What is digital sum and how does it help to get the answer?", "Solution_9": "It doesn't really. It's more a joke I guess.", "Solution_10": "I still don't see how to do this problem.", "Solution_11": "[quote=\"mna851\"]I still don't see how to do this problem.[/quote]\r\n\r\nHere is basically how do attack this problem.\r\n\r\n$2^{60} = (2^{10})^{6} = (1024)^6 = (1000+24)^6 = ((1000+24)^2)^3$\r\n\r\nNow, from here, computing $(1000+24)(1000+24) = 1000^2 + 48000+576$ is not too bad. Write this number as $1000^2+48576$. Now, apply the Binomial Theorem.\r\n\r\n$(1000^2+48576)^3 = \\binom{3}{0} (1000^2)^3 + \\binom{3}{1} (1000^2)^2 \\cdot 48576 + \\binom{3}{2} (1000^2) \\cdot 48576^2 + \\binom{3}{3} 48576^3$\r\n\r\nJust multiply numbers out. It's not \"too\" bad.\r\n\r\nWhat Gyan said is another question and actually more of joke because finding $2^{60}$ in base 2 is fairly trivial. That's my Centry said \"It's a joke.\"" } { "Tag": [], "Problem": "The obtuse angle of an isosceles triangle is bisected, and each resulting angle is $ 54$ degrees larger than a base angle. How many degrees are in the measure of the obtuse angle.", "Solution_1": "Let the base angle be $ x$.\r\n\r\n$ \\frac{180\\minus{}2x}{2}\\equal{}90\\minus{}x$.\r\n\r\n$ 90\\minus{}x\\minus{}x\\equal{}54$\r\n$ 2x\\equal{}36$\r\n$ x\\equal{}18$.\r\n$ 180\\minus{}2x\\equal{}144$." } { "Tag": [ "analytic geometry", "geometry", "circumcircle", "radical axis", "geometry unsolved" ], "Problem": "Let $ABC$ be a triangle, $l$ a line and $L,M,N$ the feet of perpendiculars to $BC,CA,AB$ through $L,M,N$ respectively, are concurrent. Their intersection is called the orthopole of the line $l$ and the triangle $ABC$.", "Solution_1": "You can find a proof of the [url=http://www.cut-the-knot.org/Curriculum/Geometry/Orthopole.shtml]Orthopole[/url]. This have a litle generalization, the [i]x-isopole[/i] (I haven't found any result of it, I have a solution with barycentric coordinates but it's quite long).\r\n\r\nThey are very interesting results about orthopole:\r\n\r\n$\\blacksquare$ The orthopole, for a given triangle of a circundiameter of that triangle lies on the nine-point circle of the triangle.\r\n$\\blacksquare$ If a line meets the circumcircle of a triangle, the Simson lines of the points of intersection with the circle meet in the orthopole of the line for a triangle.\r\n$\\blacksquare$ The [url=http://mathworld.wolfram.com/FeuerbachPoint.html]Feuerbach points[/url] of a triangle are the orthopolres, for that triangle, of the circumdiameters passing throught the tritangent centers (respectly).\r\n$\\blacksquare$The orthopole of a line for a triangle is the radical center of the three circles tangente to the given line and havian for centers the vertices of the anticomplementary triangle of the given triangle." } { "Tag": [ "AMC", "AIME", "algorithm", "USA(J)MO", "USAMO", "AMC 10", "AMC 12" ], "Problem": "[quote]The 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. [/quote]\r\n\r\nThis is taken from the FAQ on the AMC website. Is it true that you must take the AMC 10 to take advantage of the floor score cutoff. What if you are still in tenth grade or less, but choose to take the AMC 12 can you still take advantage of the floor score?", "Solution_1": "This has always confused me too. :|", "Solution_2": "> Is it true that you must take the AMC 10 to take advantage of the floor score cutoff. \r\n\r\nNo.\r\n\r\n> What if you are still in tenth grade or less, but choose to take the AMC 12 can you still take > advantage of the floor score?\r\n\r\nYes.\r\n\r\nIn practice, at least the last two years, I have selected approximately 80 students in 10th grade and below who scored at or above the \"floor AIME score\" without regard to whether teh students took the AMC 10 or the AMC 12. (Note that this was broader than the written algorithm, and worked to no one's disadvantage.) \r\n\r\nThis rule will be omitted in the 2006 AIME/USAMO Teacher's Manual, which is just about ready to go to the printer. When the final copy goes to the printer, the amended version with the omitted \"rule\" will also be posted on our website.\r\n\r\nAMC Director\r\nSteve Dunbar", "Solution_3": "I was watching at what amounted to the birth of this rule in 2002 - its intent was clearly to protect the chances of younger students who took that AMC-12 and not the AMC-10. Among students I knew at the time, the original announcement of USAMO qualification included two. One was a 12th grader who qualified by index; the other was a 10th grader with a 126 AMC-10 and a 6 AIME. (Remember that scores are not constant year-to-year; in particular, AIME scores ran lower in 2002 than in some more recent years.)\r\n\r\nBut another student I knew, a 9th grader, had a 117 AMC-12 and a 6 AIME. The folks in Lincoln must have been thinking about cases like his, and how there was just no fair way to compare an AMC-12 score to an AMC-10 score. The 9th grader did get a belated invitation to take the USAMO, and, it being the summer of greatly expanded MOP, he got the opportunity to attend MOP.\r\n\r\nSo the \"floor\" was invented specifically to avoid comparing the AMC-10 to the AMC-12, and to leave opportunities open for students taking the AMC-12.", "Solution_4": "Why do you select only 80, shouldn't it be all students in 10th grade and below who score at or above the AIME floor score? Isn't that the rule?", "Solution_5": "Ok, perhaps I should rephrase my question. I am currently a sohpmore. If I take the AMC-12, and qualify for the AIME, and score the floor score or above on the AIME, but it's still not enough to qualify by index will I be guranteed USAMO qualification? Similarly, if I take the AMC-10, qualify for the AIME, and score the floor score or above, but once again still not enough to qualify by index, will I be guranteed USAMO qualification?", "Solution_6": "[quote=\"mna851\"]Ok, perhaps I should rephrase my question. I am currently a sohpmore. If I take the AMC-12, and qualify for the AIME, and score the floor score or above on the AIME, but it's still not enough to qualify by index will I be guranteed USAMO qualification? Similarly, if I take the AMC-10, qualify for the AIME, and score the floor score or above, but once again still not enough to qualify by index, will I be guranteed USAMO qualification?[/quote]\r\n\r\nYes. Wait but for MOP index(USAMO index + 9*USAMO score) don't we have an advantage of taking AMC-10(not that the few extra points will make much of a difference)?", "Solution_7": "I don't think index even matters for 10th or younger. While I could be completely off, my understanding is that we just need to score either $\\ge 120$ on the AMC10 or $\\ge 100$ on the AMC12, and then whatever the floor is on the AIME; there's no index taken into account until 11th and 12th.", "Solution_8": "[quote=\"JesusFreak197\"]I don't think index even matters for 10th or younger. While I could be completely off, my understanding is that we just need to score either $\\ge 120$ on the AMC10 or $\\ge 100$ on the AMC12, and then whatever the floor is on the AIME; there's no index taken into account until 11th and 12th.[/quote]\r\n\r\nHmm yes there is an index for 9th and 10th graders, at least for people who take AMC12 (I am not sure how index for people who took AMC10 is calculated). Otherwise we are saying that people who are in 10th grade or below cannot place in top 160.", "Solution_9": "JesusFreak thats what I thought too until I read the quote posted in my first post on this thread.", "Solution_10": "The only differences if you take the AMC12 instead of the 10 is that you have the ability to set the AIME floor value and your score counts towards your school team's score.\r\n\r\nOne other question though. If say the index cutoff for USAMO is 210 and the floor value is a 7. what if some 10th grader took the AMC12 and gets 130/7? Is the index lowered to 200 or does it stay at 210? I mean, the cutoff should stay at 210, right?", "Solution_11": "I will attempt to answer the questions in order.\r\n\r\n> Why do you select only 80, shouldn't it be all students in 10th grade and below who score at > or above the AIME floor score? Isn't that the rule?\r\n\r\nThe answer is that we currently can accept approximately 250 for the USAMO. If we take approximately 160 based on index and then approximately 80 young students based on AIME score above the floor, and then approximately 10 from states not otherwise represented, that makes a total of approximately 250. When the rules were revised in 2002, those proportions were observed to hold in previous years. Based on a combination of those observations and a design that attempted to give first consideration to the AMC 12 and the index, the Committee on the American Mathematics Competitions approved what are now the existing rules. \r\n\r\nIt is my observation from selection in 2002, 2003, 2004, and 2005 that after the approximately 160 are chosen by index, then there are approximately 80 young students above the floor AIME value. The proportions have worked well. But it is conceviable that some year, I would observe significantly more than 80 young students above the floor value. In such a case, I have to reserve an objective criterion that allows me to select approximately 80 young students to preserve the other design goal of approximately 250 students being invited to the USAMO. In practice for the last four years, it has worked out that ALL the 10th grade and below students who scored at or above the floor AIME score have been invited. it's been around 80 each year. But it could be that some year, either many more or many fewer than 80 would meet that criterion. \r\n\r\n> I am currently a sohpmore. If I take the AMC-12, and qualify for the AIME, and score the > floor score or above on the AIME, but it's still not enough to qualify by index will I be \r\n> guranteed USAMO qualification?\r\n\r\nYes, provided there are not more than 80 10th graders and below in the same situation. That has been the case in 2002, 2003, 2004, and 2005, but of course, I can't guarantee it for the future.\r\n\r\n> Similarly, if I take the AMC-10, qualify for the AIME, and score the floor score or above,\r\n> but once again still not enough to qualify by index, will I be guranteed USAMO >qualification?\r\n\r\nSame answer. \r\n\r\n> Yes. Wait but for MOP index(USAMO index + 9*USAMO score) don't we have an advantage > of taking AMC-10(not that the few extra points will make much of a difference)?\r\n\r\nMy observation from those 10th graders and below who take both the AMC 10 and the AMC 12 in a given year and reach that decision point, the difference between the two indices makes no difference. The weight attached to the AIME and the USAMO are so great that in actual practice the difference between the AMC 10 and the AMC 12 gets washed out. In short, the few extra points do not make much difference. Although in some mythical \"principle\" a 10th grader or below \"should\" score higher on the AMC 10 than on the AMC 12, in practice I don't see it among those students above the floor value. The variations I see at this level appear to be due as much to chance, or \"test-taking mood\", or \"luck\" or call it what you will. \r\n\r\n\r\n\r\n> I don't think index even matters for 10th or younger. While I could be completely off, my > understanding is that we just need to score either \\ge 120 on the AMC10 or \\ge 100 on \r\n> the AMC12, and then whatever the floor is on the AIME; there's no index taken into >account until 11th and 12th.\r\n\r\nYes, that's been the case the last 4 years but of course, I can't guarantee it for the future. Some day, I might have to rely on an index to select the approximately 80. But based on 4 year's experience, I don't expect to.\r\n\r\n> The only differences if you take the AMC12 instead of the 10 is that you have the ability to > set the AIME floor value and your score counts towards your school team's score.\r\n\r\nThat's right, that's a good observation.\r\n\r\n> One other question though. If say the index cutoff for USAMO is 210 and the floor value is > a 7. what if some 10th grader took the AMC12 and gets 130/7? Is the index lowered to \r\n> 200 \r\n\r\nNo.\r\n\r\n> or does it stay at 210? I mean, the cutoff should stay at 210, right?\r\n\r\nYes, it does.\r\n\r\nSteve Dunbar\r\nAMC Director", "Solution_12": "Then why does it say on the website that students choosing between the AMC 10 and AMC 12 should be advised to take the AMC 12 since it will give them an advantage in USAMO selection?\r\n\r\nDoes this have something to do with the process if there are more than 80 students above the floor value?", "Solution_13": "[quote=\"dasherm\"]Then why does it say on the website that students choosing between the AMC 10 and AMC 12 should be advised to take the AMC 12 since it will give them an advantage in USAMO selection?[/quote]I don't think so.\r\n\r\nIt is because they can qualify through two ways, either a high index, or scoring above the Aime floor.", "Solution_14": "> Then why does it say on the website that students choosing \r\n> between the AMC 10 and AMC 12 should be advised to take \r\n> the AMC 12 since it will give them an advantage in USAMO \r\n> selection? \r\n\r\nWe (the AMC and I) say that on our website and every time we are asked on the basis of observed data. I just checked again, and for the 2005 USAMO, there were 115 students in 10th grade and below who qualified for, and were invited to, the USAMO. Of those 115, 68 (=58%) qualified after taking the AMC 12, and the other 47 qualified after they took the AMC 10. This further indicates that about 35 10th graders and below were selected for the USAMO on the basis of a high index using their AMC 12 score. That is consistent with my notes from the selection process.\r\n\r\nFurthermore, for 2004 which is the most recent year for which I have handy data, 1565 students in 9th and 10th grade qualified for AIME from the AMC 12. On the other hand 1309 qualified for the AIME from the AMC 10. That is, 54% of the 9th and 10th grade students who qualified for AIME did so by means of the AMC 12. Since you have to take the AIME to qualify for USAMO, the observed data appears to be roughly consistent from year to year, and across criteria.\r\n\r\nIf you are in 10th grade or below, you should choose which contest, AMC 10 or AMC 12, you elect to take by using this information and your own self-knowledge of where you believe you will do your best work. Since we do not have the advantage of having that knowledge about you, the AMC and I must rely on the observed facts to suggest decision-making. The observed facts suggest taking the AMC 12.\r\n\r\nSteve Dunbar\r\nAMC Director" } { "Tag": [ "trigonometry", "geometry", "angle bisector", "cyclic quadrilateral", "geometry solved" ], "Problem": "Let $ABC$ be a tringle with Fermat point $F$ and isodynamic point $S$ (isogonal conjugate of $F$) denote $d_a$ is distance from $S$ to internal bisector of $A$ similar define $d_b,d_c$ prove that \r\n$SA+SB+SC=FA+FB+FC+ 2(\\frac{d_a^2}{SA}+\\frac{d_b^2}{SB}+\\frac{d_c^2}{SC})$", "Solution_1": "More generally:\r\n\r\n[color=blue][b]Theorem 1.[/b] Let P and Q be two points inside a triangle ABC. Let x, y, z be the distances from the point Q to the angle bisectors of the angles PAQ, PBQ, PCQ, respectively. Then,\n\n$AQ\\cdot\\sin\\measuredangle BPC+BQ\\cdot\\sin\\measuredangle CPA+CQ\\cdot\\sin\\measuredangle APB$\n$=AP\\cdot\\sin\\measuredangle BPC+BP\\cdot\\sin\\measuredangle CPA+CP\\cdot\\sin\\measuredangle APB+2\\left(\\frac{x^2}{AQ}\\cdot\\sin\\measuredangle BPC+\\frac{y^2}{BQ}\\cdot\\sin\\measuredangle CPA+\\frac{z^2}{CQ}\\cdot\\sin\\measuredangle APB\\right)$.[/color]\r\n\r\nWhy is this more general than your problem? Just let P = F and Q = S. As we know, the first Fermat point P = F and the first isodynamic point Q = S of triangle ABC are isogonal conjugates; thus, the lines AP and AQ are symmetric to each other with respect to the angle bisector of the angle CAB. Thus, the angle bisector of the angle CAB is also the angle bisector of the angle PAQ. Combining this with the fact that Q = S, we see that the distance x from the point Q to the angle bisector of the angle PAQ is the distance $d_a$ from the point S to the angle bisector of the angle CAB. So we have $x=d_a$, and similarly $y=d_b$ and $z=d_c$.\r\n\r\nThis, together with P = F and Q = S, transforms the above equation into\r\n\r\n$AS\\cdot\\sin\\measuredangle BFC+BS\\cdot\\sin\\measuredangle CFA+CS\\cdot\\sin\\measuredangle AFB$\r\n$=AF\\cdot\\sin\\measuredangle BFC+BF\\cdot\\sin\\measuredangle CFA+CF\\cdot\\sin\\measuredangle AFB+2\\left(\\frac{d_a^2}{AS}\\cdot\\sin\\measuredangle BFC+\\frac{d_b^2}{BS}\\cdot\\sin\\measuredangle CFA+\\frac{d_c^2}{CS}\\cdot\\sin\\measuredangle AFB\\right)$.\r\n\r\nSince F is the first Fermat point of triangle ABC, we have < BFC = < CFA = < AFB = 120\u00b0, so this equation becomes\r\n\r\n$AS\\cdot\\sin 120^{\\circ}+BS\\cdot\\sin 120^{\\circ}+CS\\cdot\\sin 120^{\\circ}$\r\n$=AF\\cdot\\sin 120^{\\circ}+BF\\cdot\\sin 120^{\\circ}+CF\\cdot\\sin 120^{\\circ}+2\\left(\\frac{d_a^2}{AS}\\cdot\\sin 120^{\\circ}+\\frac{d_b^2}{BS}\\cdot\\sin 120^{\\circ}+\\frac{d_c^2}{CS}\\cdot\\sin 120^{\\circ}\\right)$, thus\r\n$AS+BS+CS=AF+BF+CF+2\\left(\\frac{d_a^2}{AS}+\\frac{d_b^2}{BS}+\\frac{d_c^2}{CS}\\right)$,\r\n\r\nand the problem is solved. Remains to prove Theorem 1.\r\n\r\n[i]Proof of Theorem 1.[/i] Let X, Y, Z be the orthogonal projections of the point Q on the lines AP, BP, CP. Then, since < PXQ = 90\u00b0, < PYQ = 90\u00b0, < PZQ = 90\u00b0, the points X, Y, Z lie on the circle with diameter PQ. Thus, < ZXY = < ZPY, so that $\\sin\\measuredangle ZXY=\\sin\\measuredangle ZPY=\\sin\\left(180^{\\circ}-\\measuredangle BPC\\right)=\\sin\\measuredangle BPC$. Similarly, $\\sin\\measuredangle XYZ=\\sin\\measuredangle CPA$ and $\\sin\\measuredangle YZX=\\sin\\measuredangle APB$. Now, by the sine law in triangle XYZ, we have $YZ: ZX: XY=\\sin\\measuredangle ZXY: \\sin\\measuredangle XYZ: \\sin\\measuredangle YZX$, what therefore becomes $YZ: ZX: XY=\\sin\\measuredangle BPC: \\sin\\measuredangle CPA: \\sin\\measuredangle APB$.\r\n\r\nNow, the Ptolemy theorem for the cyclic quadrilateral PXZY yields\r\n\r\n$PX\\cdot YZ+PY\\cdot ZX=PZ\\cdot XY$.\r\n\r\nSince $YZ: ZX: XY=\\sin\\measuredangle BPC: \\sin\\measuredangle CPA: \\sin\\measuredangle APB$, this transforms into\r\n\r\n$PX\\cdot\\sin\\measuredangle BPC+PY\\cdot\\sin\\measuredangle CPA=PZ\\cdot\\sin\\measuredangle APB$.\r\n\r\nNow, this equation depends on the arrangement of the points X, Y, Z with respect to the segments AP, BP, CP. Using [i]directed[/i] segments on the lines AP, BP, CP, where these lines are directed in such a way that the segments AP, BP, CP are positive, this equation becomes\r\n\r\n$PX\\cdot\\sin\\measuredangle BPC+PY\\cdot\\sin\\measuredangle CPA+PZ\\cdot\\sin\\measuredangle APB=0$.\r\n\r\nNow, if U is the orthogonal projection of the point Q on the angle bisector of the angle PAQ, then $\\measuredangle UAQ=\\frac{\\measuredangle PAQ}{2}$, and the distance x from the point Q to the angle bisector of the angle PAQ is simply the length QU. In the right-angled triangle AUQ, we have $QU=AQ\\cdot\\sin\\measuredangle UAQ$, what therefore becomes $x=AQ\\cdot\\sin\\frac{\\measuredangle PAQ}{2}$. Thus,\r\n\r\n$2\\frac{x^2}{AQ}=2\\frac{\\left(AQ\\cdot\\sin\\frac{\\measuredangle PAQ}{2}\\right)^2}{AQ}=2\\sin^2\\frac{\\measuredangle PAQ}{2}\\cdot AQ$\r\n$=\\left(1-\\cos\\measuredangle PAQ\\right)\\cdot AQ=AQ-AQ\\cdot\\cos\\measuredangle PAQ$.\r\n\r\nNow, in the right-angled triangle AXQ, we have $AX=AQ\\cdot\\cos\\measuredangle XAQ=AQ\\cdot\\cos\\measuredangle PAQ$, so this becomes\r\n\r\n$2\\frac{x^2}{AQ}=AQ-AQ\\cdot\\cos\\measuredangle PAQ=AQ-AX=AQ-\\left(AP+PX\\right)=AQ-AP-PX$,\r\n\r\nso that\r\n\r\n$AP+2\\frac{x^2}{AQ}=AQ-PX$.\r\n\r\nSimilarly, $BP+2\\frac{y^2}{BQ}=BQ-PY$ and $CP+2\\frac{z^2}{CQ}=CQ-PZ$. Hence,\r\n\r\n$AP\\cdot\\sin\\measuredangle BPC+BP\\cdot\\sin\\measuredangle CPA+CP\\cdot\\sin\\measuredangle APB+2\\left(\\frac{x^2}{AQ}\\cdot\\sin\\measuredangle BPC+\\frac{y^2}{BQ}\\cdot\\sin\\measuredangle CPA+\\frac{z^2}{CQ}\\cdot\\sin\\measuredangle APB\\right)$\r\n$=\\left(AP+2\\frac{x^2}{AQ}\\right)\\cdot\\sin\\measuredangle BPC+\\left(BP+2\\frac{y^2}{BQ}\\right)\\cdot\\sin\\measuredangle CPA$\r\n$+\\left(CP+2\\frac{z^2}{CQ}\\right)\\cdot\\sin\\measuredangle APB$\r\n$=\\left(AQ-PX\\right)\\cdot\\sin\\measuredangle BPC+\\left(BQ-PY\\right)\\cdot\\sin\\measuredangle CPA+\\left(CQ-PZ\\right)\\cdot\\sin\\measuredangle APB$\r\n$=\\left(AQ\\cdot\\sin\\measuredangle BPC+BQ\\cdot\\sin\\measuredangle CPA+CQ\\cdot\\sin\\measuredangle APB\\right)-\\left(\\underbrace{PX\\cdot\\sin\\measuredangle BPC+PY\\cdot\\sin\\measuredangle CPA+PZ\\cdot\\sin\\measuredangle APB}_{=0}\\right)$\r\n$=AQ\\cdot\\sin\\measuredangle BPC+BQ\\cdot\\sin\\measuredangle CPA+CQ\\cdot\\sin\\measuredangle APB$,\r\n\r\nand Theorem 1 is proven.\r\n\r\n Darij", "Solution_2": "[quote=\"darij grinberg\"]More generally:\n\n[color=blue][b]Theorem 1.[/b] Let P and Q be two points inside a triangle ABC. Let x, y, z be the distances from the point Q to the angle bisectors of the angles PAQ, PBQ, PCQ, respectively. Then,\n\n........[/quote]\r\n\r\nwow darij, I must say, you truly impress me sometimes :).\r\n\r\nDaniel", "Solution_3": "Thank you Darij, Your proof is very nice!!! :) :)" } { "Tag": [], "Problem": "[b]If x is positive integer greater or equal to 8,\r\nf:x->x,\r\nf(x+y)=f(xy)\r\nfor all x>=4\r\ny>=4\r\nif f(8)=9\r\nthen what is value of f(9)?[/quote]", "Solution_1": "[hide=\"Solution\"]\nUse:\n\n$ f(9) \\equal{} f(4 \\plus{} 5) \\equal{} f(4 \\cdot 5) \\equal{} f(20)$\n$ f(20) \\equal{} f(4 \\plus{} 16) \\equal{} f(4 \\cdot 16) \\equal{} f(64)$\n$ f(64) \\equal{} f(8 \\cdot 8) \\equal{} f(8 \\plus{} 8) \\equal{} f(16)$\n$ f(16) \\equal{} f(4 \\cdot 4) \\equal{} f(4 \\plus{} 4) \\equal{} f(8)$\n\nHence, $ \\boxed{f(9) \\equal{} f(8) \\equal{} 9}$ is a solution.\n[/hide]\r\n\r\n[b]Follow-up:[/b] Can we assert that $ f(m) \\equal{} f(n) \\; \\forall m,n \\in \\mathbb{N}^{\\geq 8}$? If not, can we assert that $ f$ is indeed a function." } { "Tag": [ "algebra", "polynomial" ], "Problem": "Prove that $ a$ is an upper bound of the roots of $ P(x)$ if the synthetic division of $ P(x)$ by $ a$ yields all positives. Prove that $ a$ is a lower bound if synthetic division yields alternating positives and negatives.", "Solution_1": "If a synthetic division by a yields a polynomial in which all coefficients are positive. Then the roots of this polynomial must all be negative. We couldn't have a positive root since that would yield a positive number since every single term would be positive. But I don't see how a needs to necessarily be an upper bound... a could be negative just like the other roots. It works for a > 0 though.\r\n\r\nFor the part I'm going to assume that the term with the highest degree is positive and the next one is negative and so on. The other way around is essentially the same thing. If the highest degree of the polynomial is even then the roots most be positive otherwise the whole polynomial would be positive and thus could not equal zero. If the highest degree of the polynomial is odd then the roots must be positive otherwise the whole thing would be negative and would not be able to equal zero. But like the previous one a will only be a lower bound for sure if it has a restriction on it. Specifically a < 0.", "Solution_2": "When you say \"synthetic division by a,\" do you mean $ \\frac{P(x)}{a}$ or $ \\frac{P(x)}{x \\minus{} a}$?", "Solution_3": "oh, sorry, i meant $ \\frac{P(x)}{x\\minus{}a}$" } { "Tag": [ "geometry", "3D geometry", "search", "number theory unsolved", "number theory" ], "Problem": "If in an arithmetic progression of positive integers a square and a cube occur then a sixth power also occur.", "Solution_1": "It is a problem from shortlist - Russian proposition - don't remember the year but it was posted before - try to search for it", "Solution_2": "Yes, it's ISL 1997 : http://www.mathlinks.ro/Forum/viewtopic.php?t=658 . :)" } { "Tag": [], "Problem": "What is the greatest integer $ x$ such that $ 2^x$ is a factor of $ 1^1 \\cdot 2^2 \\cdot 3^3 \\cdot 4^4 \\cdots 13^{13}$?", "Solution_1": "Since odd numbers won't contribute factors of $ 2$, we only need to count the number of 2s in $ 2^2\\cdot 4^4\\cdot 6^6\\cdots 12^{12}$.\r\n\r\nFrom here we just count the number of 2s in each factor:\r\n$ 2^2$ clearly has $ 2$\r\n$ 4^4\\equal{}2^8$, so $ 8$\r\n$ 6^6\\equal{}2^6\\cdot 3^6$, so $ 6$\r\n$ 8^8\\equal{}2^{24}$ which gives us $ 24$\r\n$ 10^{10}\\equal{}2^{10}\\cdot 5^{10}$, so $ 10$\r\n$ 12^{12}\\equal{}2^{24}\\cdot 3^{12}$, so $ 24$\r\n\\[ 2\\plus{}8\\plus{}6\\plus{}24\\plus{}10\\plus{}24\\equal{}\\boxed{74}\\]" } { "Tag": [ "probability", "expected value", "absolute value", "AMC" ], "Problem": "1. I have 25 pennies, 10 nickels, 5 dimes, and 1 quarter in my pocket. If I remove a coin from my pocket, and each is equally likely to be chosen, what is the expected value of the coin I take, in cents?\r\n\r\n2. Compute the sum of all 5-digit integers whose digits are 1, 3, 5, 7 and 9 in some order.\r\n\r\n3. Three boxes are set out on the table. One contains two black balls, one contains two white balls, and one contains a black and a white ball. Matt removes a ball at random from box A, and it is black. If he then removes a ball from box B, compute the probability it is white.\r\n\r\n4. If |x| + x + y = 10 and |y| + x - y = 12, compute x + y.", "Solution_1": "[hide]For #2, there are 5! = 120 different numbers that can be made using the digits 13579. In 24 cases, the ones digit is 1, in 24 cases, it is 3, in 24 cases it is 5, etc. Thus the sum of all the ones digits is 24(1) + 24(3) + ... 24(9) = 24(1+3+5+7+9) = 24*25 = 600. Similarly, the sum of all the tens digits is the same as that of the ones digits, only it's worth 10 times as much, so 6000. So the hundreds digit is worth 60,000, the thousands digit is worth 600,000, and the ten-thousands digit is worth 6 million. Adding it all together, we get 6,666,600. Which is a big number. Don't quote me on that.[/hide]\n\n\n\nFor #3, I assume the chance Matt picks a particular box to be Box A or Box B is 1/3? And can box B be the same box as box A?", "Solution_2": "For number 2, good job. For number 3, boxes A and B are distinct, and he picks boxes at random, and thus with equal probability at each point.", "Solution_3": "[hide]\n\n1. The total amount of money in there is 150 cents. The total number of coins is 41. So the expected value is 150/41\n\n\n\n3. There are two cases in which this can happen. He can pick the first one from the one with two black balls, and the second from the one with one white and one black, or vice verca. The probability of the first is 1/3*1/2*1/2 = 1/12. The second is also 1/12, so the answer is 1/6.\n\n\n\n[/hide]", "Solution_4": "X is positive, since if x is negative, the |x| and the x in equation 1 cancel and y=10. But in equation 2, 10+x-10=12, x=12, and 12 is positive. So x is positive.\r\nY is negative, since if y were positive, the |y| and -y in equation 2 would cancel, leaving x= to 12. But equation 1, using this and the fact that x is positive and |x|=x, simplifies to 24+y=10. This means that y=-14, a negative. So |y|=-y.\r\nUsing this, we can remove absolute value and make a system.\r\n\r\n 2(2x+y=10) 2x+y=10\r\n x-2y=12 -2(x- 2y=12)\r\n\r\n 4x+2y=20 2x+y=10\r\n x-2y=12 -2x+4y=-24\r\n\r\n 5x=32 3y=-14\r\n x=32/5 y=-14/3", "Solution_5": "SouthPaw, your number 1 is good, but your number 3 needs a bunch of work. First of all, you answered a different question that I asked. Second of all, even if you were answering the same problem, your method is not sound. (1) Think about what finite probability means, and (2) check what the givens of the problem are -- what you know for sure happens.\r\n\r\nTopper -- Good job, although you technically did not answer the question asked.", "Solution_6": "Ok, i think i got #3 [hide]Matt has a 1/2 chance to choose Box A to be the box with 2 blacks and 1/2 to choose box A with 1 black. This means his probability to get a white will be, half the time 3/4 and half the time 1/2. overall probability is 5/8 (1/2*3/4+1/2*1/2)[/hide]", "Solution_7": "Your first bit is wrong. It's not [hide]1/2 and 1/2 like you said.[/hide]", "Solution_8": "JBL wrote:3. Three boxes are set out on the table. One contains two black balls, one contains two white balls, and one contains a black and a white ball. Matt removes a ball at random from box A, and it is black. If he then removes a ball from box B, compute the probability it is white.\n\n\n\n[hide]For simplicity's sake, let us call the box with the two black balls \"box1\", the box with the two white balls \"box2\", and the box with one of each \"box3\".\n\nConsider the probability of getting a black ball from each box. The probabilities are\n\nblack out of box1 = 1\n\nblack box2 = 0\n\nblack box3 = 1/2.\n\n\n\nNow, we take the case where he may have picked any of the boxes to be box A. The chance of getting each of box1, 2 and 3 is 1/3, so the chance of getting a black is 1/3*1 + 1/3*1/2 = 1/2. But we know that Matt picked a black, so we need to multiply by two. Now, the probability that he picked a black from box 1 is 1/3 * 2 = 2/3 and the probability that he picked a black from box3 is 1/6*2 = 1/3.\n\n\n\nBut this is the probability that he picked box1 versus box 3 to be box A. Hence, the chance that box A = box1 is 2/3; the chance that box A = box3 is 1/3.\n\n\n\nNow, a ball will be white all the time if we pick box2 as box B. It will be white half the time if we pick box3 as box B.\n\n\n\n2/3 of the time, we will pick box1 as box A. This leaves box2 and box3 open to be picked as boxB. In this case, the probability that the ball is white is then (1/2)*(1) + (1/2)(1/2) = 3/4. Since this only happens 2/3 of the time, we multiply by 2/3 to get 1/2.\n\n\n\n1/3 of the time, we will pick box3 as box A. This leaves box1 and box3 to be picked as boxB. In this case, the probability that the ball is white is then (1/2)(0) + (1/2)(1) = 1/2. But this only happens 1/3 of the time, so we multiply by 1/3 to get 1/6.\n\n\n\nSumming these two possibilities together, we obtain 2/3. Not bad.[/hide]\n\n\n\nNow for the moment of truth: am I right?", "Solution_9": "To the best of my knowledge. I do feel that there are simpler ways to do this, however. Draw three boxes, with two blacks in one, a black and a white in another, and two whites in a third. Now, we know he picked a black ball and there are 3 of them, so the probability he picked any one of the three is 1/3. If he picked one of the two in the BB box, the probability of picking a white is 3/4. If he picked the B in the BW, the probability of getting a white is 1/2. Thus, his total probability is 2/3*3/4 + 1/3*1/2 = 2/3.\r\n\r\nThe reason we can leave out the box-choosing step is that it's symmetric. When we're in the case where we've picked a B from BB, there are 4 marbles left and we'll get each of them with equal probability, so that probability is 1/4. This wouldn't be true if there were 3 marbles in one box and 1 in another -- but in our situation, it lets us simplify somewhat.\r\n\r\nGood job, mathfanatic.", "Solution_10": "The answer is 26/15. Sorry! :oops:", "Solution_11": "Actually, pardon me. You made a slight arithmetic (reading) mistake near the end -- just look at your last 3 lines, see where you went wrong." } { "Tag": [ "function", "inequalities", "algebra unsolved", "algebra" ], "Problem": "1)$ a,b,c,d\\in \\Re$ \r\n2) $ a\\neq b\\neq c\\neq d$ \r\n3)$ ac\\equal{}bd$\r\n4)$ \\frac {a} {b} \\plus{}\\frac {b} {c} \\plus{}\\frac {c} {d} \\plus{}\\frac {d} {a} \\equal{}4$\r\n\r\nFind the maximum value of \r\n$ \\frac {a} {c} \\plus{}\\frac {b} {d} \\plus{}\\frac {c} {a} \\plus{}\\frac {d} {b}$\r\n\r\nDaniel", "Solution_1": "$ \\frac{a}{b}+\\frac{b}{c}+\\frac{c}{d}+\\frac{d}{a}=4\\implies a^2cd+ab^2d+abc^2+bcd^2=4abcd$\r\n$ \\implies bd(a+c)(b+d)=4abcd\\implies (a+c)^2(b+d)^2=16abcd$\r\n\\begin{eqnarray*}\\frac{a}{c}+\\frac{b}{d}+\\frac{c}{a}+\\frac{d}{b}&=& \\frac{1}{abcd}(da^2b+ab^2c+bc^2d+cd^2a)\\\\\r\n&=&\\frac{1}{abcd}(cd+ab)(da+bc)\\\\\r\n&\\leq& \\frac{1}{4abcd}(ab+bc+cd+da)^2\\\\\r\n&=&\\frac{1}{4abcd}(a+c)^2(b+d)^2\\\\\r\n&=&4\\end{eqnarray*}", "Solution_2": "On squaring (4) and using (3) we obtain\r\n\r\n\\frac {a} {c} +\\frac {b} {d} +\\frac {c} {a} +\\frac {d} {b}=6-(1/2)((a/b)^2+(b/c)^2+(c/d)^2+(d/a)^2) <=4.(A.M>=G.M)[/quote]", "Solution_3": "I understand what you have done, but you have to prove that it is indeed the maximum...if you don't the problem is not finished\r\n( and I think is not the maximum yet I haven't solved it)\r\nDaniel", "Solution_4": "Well I will write what I have reached...I can't finish the problem. \r\nObviously the numbers are nonzero.\r\nSuposse $ a$, $ b$, $ c$ and $ d$ are positive ( if all are negative is the same). Then by $ AM \\minus{} GM$ on $ 4)$ we get that $ \\frac {a}{b} \\equal{} \\frac {b}{c}$. Because of $ 3)$ we get that $ b \\equal{} d$ a contradiction. So there is at least 1 negative number ( but there is at least 1 positive number) Because of $ 3)$ we obtain that excatly two of them are positive and two of them negative. If $ a$ and $ c$ are negative then all terms in $ 4)$ are negatives and their sum can't be $ 4$. Then, WLOG let $ a$ and $ b$ negative and $ c$ and $ d$ are positive. So in the expression we're trying to maximize all terms are negative and it is easy to see that by $ AM \\minus{} GM$ the expression is $ \\le \\minus{} 4$. Here I'm completly lost. I don't know if I can attain the value $ \\minus{} 4$ or if I have to improve the maximum\r\nThank's for your help\r\n\r\nDaniel\r\n\r\nPS: Maybe you have misunderstood $ 1)$ It should have said that the four numbers $ a,b,c,d$ are distinct", "Solution_5": "[color=green]Let\u00b4s see...\n\n$ \\frac{d}{a}\\equal{}\\frac{c}{b}$\n\n$ \\frac{c}{d}\\equal{}\\frac{b}{a}$\n\nso\n\n$ \\frac{a}{b}\\plus{}\\frac{b}{c}\\plus{}\\frac{c}{d}\\plus{}\\frac{d}{a}\\equal{}\\frac{a}{b}\\plus{}\\frac{b}{c}\\plus{}\\frac{b}{a}\\plus{}\\frac{c}{b}$\n\ni.e. $ \\frac{a}{b}\\plus{}\\frac{b}{c}\\plus{}\\frac{c}{d}\\plus{}\\frac{d}{a}\\equal{}(\\frac{a}{b}\\plus{}\\frac{b}{a})\\plus{}(\\frac{b}{c}\\plus{}\\frac{c}{b})$\n\nbut... you want \n\n$ (\\frac{a}{b}\\plus{}\\frac{b}{a})\\plus{}(\\frac{b}{c}\\plus{}\\frac{c}{b})\\equal{}4$\n\nbut if $ x$ is not $ 1$\n\n$ (\\sqrt{x}\\minus{}\\frac{1}{\\sqrt{x}})^2>0$\n\nwich implies $ x\\plus{}\\frac{1}{x}>2$\n\nfor $ x\\equal{}\\frac{b}{c}$ and $ x\\equal{}\\frac{a}{b}$ adding we get\n\n$ (\\frac{a}{b}\\plus{}\\frac{b}{a})\\plus{}(\\frac{b}{c}\\plus{}\\frac{c}{b})>4$\n\nI must be missing something here :blush: can you help me?\n\n\n\n\nLife is GREEN! :D \n\n\n\nTaUrUs*[/color]", "Solution_6": "[color=green]uhmm sorry i have just see... \n\n\n$ x$ must be negative...\n\n\n\n\nTaUrUs*\n[/color]", "Solution_7": "[color=green]well we may supose without lose of generality that $ x\\equal{}\\frac{a}{b}<0$ and $ \\frac{b}{c}>0$ \n\nthis is because $ \\frac{a}{b}\\plus{}\\frac{b}{a}\\plus{}\\frac{b}{c}\\plus{}\\frac{c}{b}\\equal{}4$\n\nso... again without lose of generality $ a<0,d<0,b>0,c>0$\n\nand $ \\frac{a}{c}\\plus{}\\frac{b}{d}\\plus{}\\frac{c}{a}\\plus{}\\frac{d}{b}<0$\n\nso the maximun can not be $ 4$ :wink: \n\n\nTaUrUs*[/color]", "Solution_8": "[color=green]\n\nwe have $ |x\\plus{}\\frac{1}{x}|>2$ so since all terms in \n\n$ \\frac{a}{c}\\plus{}\\frac{b}{d}\\plus{}\\frac{c}{a}\\plus{}\\frac{d}{b}$ must be negative... we have the maximun when $ \\frac{a}{c}\\plus{}\\frac{c}{a}\\equal{}\\minus{}2$ and $ \\frac{b}{d}\\plus{}\\frac{d}{b}\\equal{}\\minus{}2$\nthat is when $ a\\equal{}\\minus{}c$ and $ b\\equal{}\\minus{}d$ then the maximun is $ \\minus{}4$.\n\n\n\nLife is GREEN! :D \n\n\n\n\nTaUrUs*[/color]", "Solution_9": "If $ a\\equal{}\\minus{}c$ and $ b\\equal{}\\minus{}d$ then $ \\minus{}a^2\\equal{}\\minus{}b^2$ then $ a\\equal{}b$ or $ a\\equal{}\\minus{}b\\equal{}d$ Since the numbers are distinct you reach a contradiction\r\nIt is driving me mad :o \r\n\r\n :D Daniel", "Solution_10": "hello, try it with\r\n$ a\\equal{}\\minus{}\\frac{3\\minus{}2\\sqrt{2}}{\\minus{}3\\plus{}2\\sqrt{2}}$\r\n$ b\\equal{}\\minus{}1$\r\n$ c\\equal{}\\minus{}3\\plus{}2\\sqrt{2}$\r\n$ d\\equal{}3\\minus{}2\\sqrt{2}$\r\nSonnhard.", "Solution_11": "I believe the maximum value doesn't exist.\r\n\r\n$ ac \\equal{} bd\\Longleftrightarrow \\frac {a}{b} \\equal{} \\frac {d}{c} \\equal{} k$, yielding $ a \\equal{} bk,\\ d \\equal{} ck$.\r\n\r\n$ \\frac {a}{b} \\plus{} \\frac {b}{c} \\plus{} \\frac {c}{d} \\plus{} \\frac {d}{a} \\equal{} \\frac {c}{b} \\plus{} \\frac {b}{c} \\plus{} k \\plus{} \\frac {1}{k} \\equal{} 4$.\r\n\r\n$ \\therefore \\frac {a}{c} \\plus{} \\frac {b}{d} \\plus{} \\frac {c}{a} \\plus{} \\frac {d}{b} \\equal{} \\left(k \\plus{} \\frac {1}{k}\\right)\\left(\\frac {c}{b} \\plus{} \\frac {b}{c}\\right) \\equal{} \\left(k \\plus{} \\frac {1}{k}\\right)\\left\\{4 \\minus{} \\left(k \\plus{} \\frac {1}{k}\\right)\\right\\}$.\r\n\r\nSince $ \\left|k \\plus{} \\frac {1}{k}\\right| \\equal{} |k| \\plus{} \\left|\\frac {1}{k}\\right|\\geq 2\\sqrt {|k|\\cdot \\left|\\frac {1}{k}\\right|} \\equal{} 2$, we have $ \\frac {a}{c} \\plus{} \\frac {b}{d} \\plus{} \\frac {c}{a} \\plus{} \\frac {d}{b}\\leq 4$, \r\n\r\nthe equality holds when $ k \\plus{} \\frac {1}{k} \\equal{} 2\\Longleftrightarrow k \\equal{} 1$, i.e. $ a \\equal{} b$ and $ d \\equal{} c$, which contradicts the assumption. $ \\therefore \\frac {a}{c} \\plus{} \\frac {b}{d} \\plus{} \\frac {c}{a} \\plus{} \\frac {d}{b} < 4$.", "Solution_12": "Indeed I have just read a very nice solution and the maximum is $ \\minus{}12$\r\nSo how did you get that result Sonnhard? It will be nice if you post the complete solution not only the answer :) \r\nThank you\r\n\r\nDaniel", "Solution_13": "hello Daniel, here is my solution. From $ ac=bd$ we get $ d=\\frac{ac}{d}$ for\r\n$ d\\not=0$ and so we have $ \\frac{a}{b}+\\frac{b}{a}+\\frac{c}{c}+\\frac{c}{b}=4$. Let $ f(a,b,c)=\\frac{a}{c}+\\frac{c}{a}+\\frac{b^2}{ac}+\\frac{ac}{b^2}$. From the equation above we have\r\n$ c_1=\\frac{-a^2+4ab-b^2-(a-b)\\sqrt{a^2-6ab+b^2}}{2a}$\r\nor\r\n$ c_2=\\frac{-a^2+4ab-b^2+(a-b)\\sqrt{a^2-6ab+b^2}}{2a}$\r\nBy inserting $ c_1$ in $ f(a,b,c)$ we have\r\n$ f_1(a,b)=-\\frac{(a^2+b^2)(a^2-4ab+b^2)}{a^2b^2}$.\r\nThis function has a maximum for $ a=-b$ it is $ -12$. Analogously for the case\r\nwhen we inserting $ c_2$ in $ f(a,b,c)$ and we get $ f_2(a,b)=-\\frac{(a^2+b^2\r\n)(a^2-4ab+b^2}{a^2b^2}$.\r\nSonnhard.", "Solution_14": "Thank you Sonnhard! :) \r\n\r\nDaniel", "Solution_15": "hello Daniel, a little remark. We can prove the inequality only by simplifying\r\n$ \\frac{(a^2\\plus{}b^2)(a^2\\minus{}4ab\\plus{}b^2)}{a^2b^2} \\le \\minus{}12$ , this is equivalent with\r\n$ \\frac{(a\\plus{}b)^2(a^2\\minus{}6ab\\plus{}b^2)}{a^2b^2} \\geq0$. This is true because for \r\nreal $ c$ must be the inequality $ a^2\\minus{}6ab\\plus{}b^2 \\geq0$ fulfilled.\r\nSonnhard." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let $ x,y,z\\in\\mathbb{R}$ such that $ \\minus{}1\\leq x\\plus{}y\\plus{}z\\leq 1$, $ \\minus{}1\\leq x\\minus{}y\\plus{}z\\leq 1$ and $ \\minus{}8\\leq 4x\\plus{}2y\\plus{}z\\leq 8$, $ \\minus{}8\\leq 4x\\minus{}2y\\plus{}z\\leq 8$\r\n\r\nProve or disprove $ |x|\\plus{}4|y|\\leq 7$", "Solution_1": "hello, one solution of your system is $ \\minus{}37$.\r\nSonnhard." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "If $ a,b,c$ are positive real numbers such that $ abc\\equal{}1$, then\r\n\r\n$ \\frac 1{(a\\plus{}1)(a\\plus{}2)}\\plus{}\\frac 1{(b\\plus{}1)(b\\plus{}2)}\\plus{}\\frac 1{(c\\plus{}1)(c\\plus{}2)} \\ge \\frac 1{2}$.", "Solution_1": "let $ a\\equal{}\\frac{yz}{x^2}$ and $ b\\equal{}\\frac{xz}{y^2}$ and $ c\\equal{}\\frac{xy}{z^2}$\r\nso we most prove that \r\n\r\n \\[ \\sum\\frac{x^4}{(yz\\plus{}x^2)(yz\\plus{}2x^2)}\\geq\\frac12\\]\r\n after using Titu inequality we have\r\n\\[ 2(\\sum x^2)^2\\geq 2\\sum x^4 \\plus{}3\\sum x^2yz \\plus{}\\sum y^2 z^2\\]\r\n or\r\n\\[ \\sum x^2y^2\\geq\\sum x^yz\\]", "Solution_2": "[quote=\"Vasc\"]If $ a,b,c$ are positive real numbers such that $ abc \\equal{} 1$, then\n\n$ \\frac 1{(a \\plus{} 1)(a \\plus{} 2)} \\plus{} \\frac 1{(b \\plus{} 1)(b \\plus{} 2)} \\plus{} \\frac 1{(c \\plus{} 1)(c \\plus{} 2)} \\ge \\frac 1{2}$.[/quote]\r\n\r\n It's hungkhtn'problem which he posted on VIF :wink: Here my proof:\r\n\r\n$ \\frac 1{(a \\plus{} 1)(a \\plus{} 2)} \\plus{} \\frac 1{(b \\plus{} 1)(b \\plus{} 2)} \\plus{} \\frac 1{(c \\plus{} 1)(c \\plus{} 2)} \\ge \\frac 1{2}$\r\n\r\n<=> $ \\sum2(a\\plus{}1)(a\\plus{}2)(b\\plus{}1)(b\\plus{}2)\\ge (a\\plus{}1)(b\\plus{}1)(c\\plus{}1)(a\\plus{}2)(b\\plus{}2)(c\\plus{}2)$\r\n\r\n<=> $ 4\\sum a^2\\plus{}3\\sum a\\ge 3\\sum ab\\plus{}12$ which true by AM-GM :)", "Solution_3": "We have some nother results: For $ a,b,c>0$ such that $ abc\\equal{}1$ then\r\n\r\n1/ $ \\frac 1{(a \\plus{} 1)(b \\plus{} 2)} \\plus{} \\frac 1{(b \\plus{} 1)(c \\plus{} 2)} \\plus{} \\frac 1{(c \\plus{} 1)(a \\plus{} 2)} \\le \\frac 1{2}$\r\n\r\n2/ $ \\frac 1{\\sqrt{(a \\plus{} 1)(b \\plus{} 1)}} \\plus{} \\frac 1{\\sqrt{(b \\plus{} 1)(c \\plus{} 1)}} \\plus{} \\frac 1{\\sqrt{(c \\plus{} 1)(a \\plus{} 1)}} \\le \\frac 3{2}$", "Solution_4": "This statement is also true:\r\n\r\nIf $ a,b,c,d$ are positive real numbers such that $ abcd \\equal{} 1$, then\r\n\r\n$ \\frac 1{(a \\plus{} 3)(2a \\plus{} 1)} \\plus{} \\frac 1{(b \\plus{} 3)(2b \\plus{} 1)} \\plus{} \\frac 1{(c \\plus{} 3)(2c \\plus{} 1)}\\plus{} \\frac 1{(d \\plus{} 3)(2d \\plus{} 1)} \\ge \\frac 1{3}$.", "Solution_5": "More general:\r\n\r\nIf $ a_1,a_2,...,a_n$ are positive real numbers such that $ a_1a_2...a_n \\equal{} 1$, then\r\n\r\n$ \\sum\\frac 1{(a_1 \\plus{} n\\minus{}1)((n\\minus{}2)a_1 \\plus{} 1)} \\ge \\frac 1{n\\minus{}1}$.", "Solution_6": "[quote=\"Vasc\"]More general:\n\nIf $ a_1,a_2,...,a_n$ are positive real numbers such that $ a_1a_2...a_n \\equal{} 1$, then\n\n$ \\sum\\frac 1{(a_1 \\plus{} n \\minus{} 1)((n \\minus{} 2)a_1 \\plus{} 1)} \\ge \\frac 1{n \\minus{} 1}$.[/quote]\r\n\r\nI think we can set $ a_i\\equal{}\\frac{\\prod_{k\\equal{}1}^{n}{x_k}}{x_{i}^{n}}$ and use Cauchy - Schwarz inequality.", "Solution_7": "[quote=\"NguyenDungTN\"][quote=\"Vasc\"]More general:\n\nIf $ a_1,a_2,...,a_n$ are positive real numbers such that $ a_1a_2...a_n \\equal{} 1$, then\n\n$ \\sum\\frac 1{(a_1 \\plus{} n \\minus{} 1)((n \\minus{} 2)a_1 \\plus{} 1)} \\ge \\frac 1{n \\minus{} 1}$.[/quote]\n\nI think we can set $ a_i \\equal{} \\frac {\\prod_{k \\equal{} 1}^{n}{x_k}}{x_{i}^{n}}$ and use Cauchy - Schwarz inequality.[/quote]\r\n\r\nTry this method for $ n\\equal{}4$ and $ n\\equal{}5$. :?:", "Solution_8": "More general:\r\n\r\nLet $ a_1,a_2,...,a_n$ be positive real numbers such that $ a_1a_2...a_n = 1$. If $ p,q$ are nonnegative real numbers such that ${ p+q=n-1}$, then\r\n\r\n$ \\frac 1{1+pa_1+qa_1^2}+\\frac 1{1+pa_2+qa_2^2}+...+\\frac 1{1+pa_n+qa_n^2} \\ge 1$.\r\n\r\n[hide]Nice particular cases: $ p=n-1$ , $ q=0$ and $ p=2r$, $ q=r^2$ with $ r=\\sqrt{n}-1$.[/hide]", "Solution_9": "let a,b,c>0 such that abc=1 Prove that\r\n$ (ab\\plus{}bc\\plus{}ca)^2[(a\\plus{}b\\plus{}c)^2\\plus{}ab\\plus{}bc\\plus{}ca] \\geq 4(a\\plus{}b\\plus{}c)^3$" } { "Tag": [ "vector", "geometry", "parallelogram", "probability", "expected value", "combinatorics proposed", "combinatorics" ], "Problem": "we have vectors $v_1,..,v_n \\in \\mathbb{R}^n$ such that $|v_n|=1$. Prove that there exist $\\epsilon_i \\in \\{-1,1\\}$ where $i=1,..,n$ such that $|\\epsilon_1v_1+..+\\epsilon_nv_n|\\leq \\sqrt{n}$", "Solution_1": "We apply the parallelogram formula repeatedly ($\\|u+t\\|^2+\\|u-t\\|^2=2\\|u\\|^2+2\\|t\\|^2$, which holds in all Hilbert spaces), we get $\\sum\\|\\varepsilon_1v_1+\\ldots+\\varepsilon_nv_n\\|^2=2^nn$, where the sum is taken over all $n$-uples $(\\varepsilon_1,\\ldots,\\varepsilon_n)\\in\\{\\pm 1\\}^n$. Since the sum has $2^n$ terms, the conclusion follows immediately.", "Solution_2": "thanks for your solution, this idea is quite similar to the one which i know\r\nlet $\\epsilon_i$ be $-1,1$ with the same probabilities, then the expected value of $X=|\\sum \\epsilon_iv_i|^2$ is \r\n$E(X)=E((\\sum \\epsilon_iv_i)(\\sum \\epsilon_iv_i))=\\sum_{i,j}(v_i \\cdot v_j)E(\\epsilon_i \\epsilon_j)$\r\nbut if $i \\neq j$ then $E(\\epsilon_i \\cdot \\epsilon_j )=E(\\epsilon_i)E(\\epsilon_j)=0$ and if $i=j$ then $E(\\epsilon_i^2)=1$ so $E(X)=\\sum v_i\\cdot v_i=n$\r\nso there must be sequence of $\\epsilon_i$ which gives what we want" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let n be a positive interger and a,b,c are positive numbers. Prove that\r\n$ (a^n\\plus{}b^n)^{\\frac 1 n}\\plus{}(c^n\\plus{}d^n)^{\\frac 1 n}\\ge ((a\\plus{}c)^n\\plus{}(b\\plus{}d)^n)^{\\frac 1 n}$", "Solution_1": "This is the http://en.wikipedia.org/wiki/Minkowski_inequality" } { "Tag": [], "Problem": "a,b,c reals\r\n1(a+b+c)/(a+b+c)=1. The same is true for S2=b/(a+b)+c/(b+c)+a/(a+c). But S1+S2=3, so if S2>1 it means that S1<2, so we proved 1 \\sigma(j)$\r\n $ \\pi(\\sigma)$ denote the number of if $ i < j$ then $ \\sigma(i) < \\sigma(j)$\r\n For $ (i,j) \\in \\{1,2,...,n\\}^2$\r\nCalculate :\r\n1) $ \\sum_{\\sigma \\in \\sigma_n}{\\frac {( \\minus{} 1)^{I(\\sigma)}}{I(\\sigma) \\plus{} \\pi(\\sigma) \\plus{} 1}} \\equal{} ?$\r\n[hide=\"Answer of me\"] $ \\frac {2}{n \\plus{} 1}$[/hide]\n2) $ \\sum_{\\sigma \\in \\sigma_n}{\\frac {( \\minus{} 1)^{I(\\sigma)}}{(v(\\sigma) \\plus{} 1)(I(\\sigma) \\plus{} \\pi(\\sigma) \\plus{} 1)}}$\n[hide=\"Answer of me\"] :roll: [/hide]", "Solution_1": "I'm reminded of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=64449]a Putnam problem[/url] although it isn't exactly the same.", "Solution_2": "Putnam trick works on #1:\n\nLet $ A_x = Jx + (1 - x)I$ where $ J$ is the all-ones matrix.\n\nThen it is easy to see that $ \\int_0^1 \\det(A_x) \\mathrm{d}x$ is the desired summation (for part (a)), \nthis follows by the same logic as Putnam 2005 B6. We also have that \n\n$\\det A_x = \\det (Jx + (1 - x)I) = (1 - x)^n \\det(I + J(x/(1 - x))) = (1 - x)^{n - 1}(1 - x + nx)$.\n\nIntegrating the above from 0 to 1 gives $ \\frac {2}{n + 1}$.\n\n\nPart (b) should follow by computing the determinant of $ Jx + (y-x)I$ and integrating both $ x$ and $ y$ from 0 to 1 -- should be a simple computation.", "Solution_3": "Yes ,It is correct !.... It a is nice solution in matrix", "Solution_4": "For Answer (2) you can read with $ m\\equal{}n\\minus{}1$ http://www.mathlinks.ro/viewtopic.php?p=1498321#1498321", "Solution_5": "Calculate \r\nEdit mistake of me : :blush: For (1),(2) solution same as [b]Nukular[/b]\r\n Call $ G(\\sigma)$ are set fixed point of $ \\sigma$ and $ v(\\sigma)$ denote the number of $ G(\\sigma)$\r\n$ \\pi(\\sigma) \\equal{} \\{\\sigma(1),\\sigma(2),...,\\sigma(n)\\} \\minus{} G(\\sigma)$ are set point remain not $ \\subset$ $ G(\\sigma)$ of $ \\sigma$ and $ I^{'}(\\sigma)$ denote the number of $ \\pi(\\sigma)$\r\n$ I(\\sigma)$ denote the number of if $ i < j$ then $ \\sigma(i) > \\sigma(j)$ ,$ (i,j) \\in \\{1,2,..,n\\}^2$ \r\n1)$ \\sum_{\\sigma\\in\\sigma_{n}}{\\frac {( \\minus{} 1)^{I(\\sigma)}}{I^{'}(\\sigma) \\plus{} 1}} \\equal{} ?$ \r\n2)$ \\sum_{\\sigma\\in\\sigma_{n}}{\\frac {( \\minus{} 1)^{I(\\sigma)}}{(v(\\sigma) \\plus{} 1)(I^{'}(\\sigma) \\plus{} 1)}}$\r\n3)\r\na)$ \\sum_{\\sigma\\in\\sigma_{n}}{\\frac {( \\minus{} 1)^{I(\\sigma)}}{2^{v(\\sigma)}}} \\equal{} ?$\r\n[hide=\"Answer of me\"]$ \\frac {2n \\minus{} 1}{2}(\\frac { \\minus{} 1}{2})^{n \\minus{} 1}$[/hide]\nb)$ \\sum_{\\sigma\\in\\sigma_{n}}{( \\minus{} 1)^{I(\\sigma)}(e\\minus{}1)^{v(\\sigma)}} \\equal{} ?$\n[hide=\"Answer of me\"][/hide]\n4)\n$ \\sum_{\\sigma\\in\\sigma_{n}}{\\frac {( \\minus{} 1)^{I(\\sigma)}}{2^{v(\\sigma)}(I^{'}(\\sigma) \\plus{} 1)}}$\n[hide=\"Answer\"] :play_ball: [/hide]", "Solution_6": "Sorry, but can anyone translate the problem into English for me? I am not even able to reverse-engineer it from Nukular's solution... What exactly prevents $ I\\left(\\sigma\\right) \\plus{} \\pi\\left(\\sigma\\right)$ from being $ \\frac {n\\left(n \\minus{} 1\\right)}{2}$ all the time? Or $ 1$, if $ i$ and $ j$ are fixed and $ i \\sigma(j)$\r\nThen :\r\n1) $ \\sum_{\\sigma\\in\\sigma_{n}}{\\frac {( \\minus{} 1)^{I(\\sigma)}}{n \\plus{} 1 \\minus{} v(\\sigma)}} \\equal{} \\frac {2}{n \\plus{} 1}$\r\n2)$ \\sum_{\\sigma\\in\\sigma_{n}}{\\frac {( \\minus{} 1)^{I(\\sigma)}}{(v(\\sigma) \\plus{} 1)(n \\plus{} 1 \\minus{} v(\\sigma))}} \\equal{}$\r\n [hide]$ \\frac {2 \\plus{} ( \\minus{} 1)^{n \\plus{} 1}n}{(n \\plus{} 1)(n \\plus{} 2)}$[/hide]\n solution same as Nukular \n3) $ \\sum_{\\sigma\\in\\sigma_{n}}{\\frac {( \\minus{} 1)^{I(\\sigma)}}{v(\\sigma) \\plus{} m}} \\equal{} \\int_{0}^{1}{y^{m \\minus{} 1}(y \\plus{} n \\minus{} 1)(y \\minus{} 1)^{n \\minus{} 1}dy} \\equal{} \\frac {( \\minus{} 1)^{n \\plus{} 1}n!(m \\plus{} n \\minus{} 1)}{m(m \\plus{} 1)...(m \\plus{} n)} ,\\forall m \\in \\mathbb{N^{*}}$ \nUse function $ B$ to calculate or read [b]putnam 2005[/b] have a diferent solution !\nOr use [hide=\"Hint\"]$ x^{m \\minus{} 1} \\det(A_{x}) \\equal{} x^{m \\minus{} 1}\\det( Jx \\plus{} (1 \\minus{} x)I) \\equal{} \\sum_{\\sigma \\in \\sigma_n}{( \\minus{} 1)^{I(\\sigma)}x^{v(\\sigma) \\plus{} m \\minus{} 1}}$[/hide]\n3)$ \\sum_{\\sigma \\in \\sigma_n}{\\frac {(\\minus{}1)^{I(\\sigma)}2^{v(\\sigma)}}{(v(\\sigma) \\plus{} 1))(v(\\sigma) \\plus{} 2)}} \\equal{} ?$\n[hide=\"Hint\"]$ A_{x \\plus{} y} \\equal{} J_{x \\plus{} y} \\plus{} (1 \\minus{} (x \\plus{} y))I$ ,$ \\det(A_{x \\plus{} y}) \\equal{} \\sum_{\\sigma \\in \\sigma_n}{( \\minus{} 1)^{I(\\sigma)}(x \\plus{} y)^{v(\\sigma)}}$[/hide]\r\n............. ...\r\n Sorry everyone :blush:", "Solution_11": ":blush: Note $ A_x \\equal{} \\begin{bmatrix}x & 1 & 1 & 1 & ... & 1 & 1 \\\\\r\n1 & x & 1 & 1 & ... & 1 & 1 \\\\\r\n1 & 1 & x & 1 & ... & 1 & 1 \\\\\r\n1 & 1 & 1 & x & ... & 1 & 1 \\\\\r\n... & ... & ... & ... & ... & ... & ... \\\\\r\n1 & 1 & 1 & 1 & ... & x & 1 \\\\\r\n1 & 1 & 1 & 1 & ... & 1 & x\\end{bmatrix}$ \r\n For $ \\sum_{\\sigma\\in\\sigma_{n}}{\\frac {( \\minus{} 1)^{I(\\sigma)}}{2^{v(\\sigma)}(n \\minus{} v(\\sigma) \\plus{} 1)}}$\r\nNote $ A \\equal{} \\begin{bmatrix}\\frac {1}{2} & y & y & y & ... & y & y \\\\\r\ny & \\frac {1}{2} & y & y & ... & y & y \\\\\r\ny & y & \\frac {1}{2} & y & ... & y & y \\\\\r\ny & y & y & \\frac {1}{2} & ... & y & y \\\\\r\n... & ... & ... & ... & ... & ... & ... \\\\\r\ny & y & y & y & ... & \\frac {1}{2} & y \\\\\r\ny & y & y & y & ... & y & \\frac {1}{2}\\end{bmatrix}$ !\r\n For 3) implies for $ y\\equal{}1$ don't cacilate $ \\int$ :|" } { "Tag": [ "probability", "geometry", "3D geometry" ], "Problem": "If 3 fair dice are tossed, what is the probability that the sum of the faces tossed is greater than 12?\r\n\r\nA) $ \\frac{7}{72}$\r\n\r\nB) $ \\frac{1}{6}$\r\n\r\nC) $ \\frac{1}{3}$\r\n\r\nD) $ \\frac{7}{27}$\r\n\r\nE) $ \\frac{3}{8}$", "Solution_1": "Just use casework to solve it, it doesn't take a lot of time.\r\n\r\n-rst2007", "Solution_2": "Or...consider $ a\\plus{}b\\plus{}c\\ge13\\implies(6\\minus{}a)\\plus{}(6\\minus{}b)\\plus{}(6\\minus{}c)\\le5$.", "Solution_3": "Think there is a cube with the edge length 6\r\nA(0,0,0)B(0,6,0)C(6,0,0)D(6,6,0)A'(0,0,6)B'(0,6,6)C'(6,0,6)D'(6,6,6)\r\nThink there is a plane which satisfy the equation x+y+z=12\r\nThe number of points above the plane and in the cube is 56\r\n(the integer point in a isosceles right-angle equals to n(n+1)/2 so the sum of them is 6*7/2+5*6/2+4*5/2+3*4/2+2*3/2+1*2/2)\r\nand the number of whole points is 216\r\nso the answer is 56/216=7/27\r\n\r\n\r\n[/img]", "Solution_4": "you roll the first dice = 6 ways\r\nroll 2nd dice = 6 ways\r\nroll 3rd dice = 6 ways\r\n\r\ntherefore total ways = 6x6x6\r\n\r\nthen you need to find the sum of the 3 dices\r\ndiagram may help\r\n\r\n[img]http://i43.tinypic.com/212ajqx.jpg[/img]\r\n\r\nGo clock wise direction with blue dice rolling first. \r\neg. roll a 6, roll a 6 with black die and roll 6 with red die = 18\r\nso shade the top right square\r\n\r\nroll 6,5,6 so shade in next square etc etc\r\nkeep proceeding like this!\r\n\r\nbut there are only 8 possibilities because rolling 5,5,4 is same as rolling 4,5,5", "Solution_5": "Assume they're all 6.\r\nNow taking them away, x+y+z=5 You have 3 dividers and 5 of a number, so 8C3=56.\r\n\r\n56/216=7/27" } { "Tag": [ "modular arithmetic", "number theory unsolved", "number theory" ], "Problem": "Show that there are no integer solutions to $ 2^n \\plus{} 3^n \\equal{} x^p$ with $ p\\geq 2$.\r\n\r\n(I have shown $ p>1000$ and $ n>10^{200000}$, but I'm a bit stuck.)", "Solution_1": "[quote=\"Eigenray\"]I have shown $ p > 1000$ and $ n > 10^{200000}$.[/quote]\r\nPlease show us how you did it.", "Solution_2": "I will assume here that $ n$ is odd. The case $ n$ even is not too difficult.\r\n\r\nFirst note that $ 5|x$, and $ p\\geq 2$, so $ 5^2|2^n\\plus{}3^n$, which implies $ 5|n$. Now suppose $ d|n$, and that $ q^\\alpha \\parallel{} 2^d\\plus{}3^d$ with $ q\\neq 5$, and $ \\alpha$ not divisible by $ p$, say $ \\alpha\\equal{}kp\\minus{}r$, $ 02 is equal to 1024, there are 31 perfect squares less than 1000.", "Solution_2": "[hide]100, because 100 2 is 10,000.[/hide]\n\n[hide=\"4everwise\"] you did all the squares up to 1,000. the question asks for all the squares to 10,000,[/hide]", "Solution_3": "actaully both of you are wrong. [hide]The question asks for how many squares exist smaller than 10,000. However, 100^2 equals exactly 10,000 and there are 99 squares smaller than 10,000[/hide]", "Solution_4": "Yeah, the ultimate tricks.\r\n[hide] :sqrt:10000 = 100, so ones whose squares are [b]less[/b] than 10000 only number 99. [/hide]" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "for a,c,b>0 and $a^2+b^2+c^2\\leq3$\r\nprove\r\n$a^2+b^2+c^2+3abc\\leq 2(a+b+c)$ :D", "Solution_1": "woops... long live $x^2\\le 0$", "Solution_2": "I'm sorry to tell you, but there is a fatal mistake in your proof.", "Solution_3": "I found that for ${p \\ge \\sqrt{3}-1}$, the inequality holds\r\n$a^2+b^2+c^2+3pabc \\le (p+1)(a+b+c)$.\r\n\r\nThus, the best inequality of this type is\r\n$a^2+b^2+c^2+3(\\sqrt3-1)abc \\le \\sqrt3(a+b+c)$.", "Solution_4": "The ugliest solution:\r\nHomogenalise the inquality(we may do so because the degree of RHS is smaller than the degree of every term of LHS).\\[(\\sum a^2)\\sqrt{\\sum a^2}+3(3-\\sqrt{3})abc\\le\\sum a^2\\sum a\\]Since $(\\sum a^2)(\\sum a-\\sqrt{\\sum a^2})=(\\sum a^2)\\frac{2\\sum ab}{\\sum a+\\sqrt{\\sum a^2}}$, it is enough to prove $2\\sum a^2\\sum ab\\ge3(3-\\sqrt{3})abc(\\sum a+\\sqrt{\\sum a^2})$. Note that $\\sum a^3(b+c)\\ge2abc\\sum a$, thus, it remains to prove\\[\\sum a^3(b+c)\\ge2abc\\sqrt{3\\sum a^2}\\]Expanding,\\[(6,2,0)+(4,4,0)+(6,1,1)+(3,3,2)+2(4,3,1)\\ge6(4,2,2)\\]which is obvious.", "Solution_5": "Your solution is not ugly if in inequality\r\n$2\\sum a^2\\sum ab\\ge3(3-\\sqrt{3})abc(\\sum a+\\sqrt{\\sum a^2})$\r\nreplace $\\sum a$ with $\\sqrt{3\\sum a^2}$, then use AM-GM Inequality. ;)", "Solution_6": "I'm too keen on expansion... :blush:", "Solution_7": "[quote=\"mecrazywong\"]Expanding,\\[(6,2,0)+(4,4,0)+(6,1,1)+(3,3,2)+2(4,3,1)\\ge6(4,2,2)\\]which is obvious.[/quote]\r\nWhy is this obvious?", "Solution_8": "From Muirhead all are true.", "Solution_9": "Isn't that $(3, 3, 2)\\leq (4, 2, 2)$?", "Solution_10": "[quote=\"leepakhin\"]Isn't that $(3, 3, 2)\\leq (4, 2, 2)$?[/quote]\r\nyou have right. I will make the computations because it is possible that mecrazywong \r\nmake a typo.I will see this.", "Solution_11": "[quote=\"leepakhin\"]Isn't that $(3, 3, 2)\\leq (4, 2, 2)$?[/quote]\r\n$(3,3,2)+(6,2,0)\\ge2(4.5,2.5,1)\\ge2(3,3,2)$...", "Solution_12": "Thank you very much, Sam." } { "Tag": [], "Problem": "An algebraic expression of the form a + bx has the value of 15 when x = 2 and the value of 3 when x = 5.\r\nCalculate a + b.", "Solution_1": "Hopefully 19 !!!", "Solution_2": "That's what I get.[hide]\nFrom the given, we have a sytem of equations.\n\n$ a\\plus{}2b\\equal{}15$\n$ a\\plus{}5b\\equal{}3$\n\nSubtracting, we find that $ \\minus{}3b\\equal{}12 \\implies b\\equal{}\\minus{}4$.\n\nPlugging this into either of our expressions, we find that $ a\\equal{}23$.\n\nSo, $ 23\\minus{}4\\equal{}\\boxed{19}$.[/hide]", "Solution_3": "Great ! By the way which standard are you in ?", "Solution_4": "You could set up the equation a+2b=15 and a+5b=3. Put the first equation in the variable As form which would be A=15-2b. Plug that in for the second equation to get 15-2b+5b=1. THen get 15+3b=3. 3b=-12 b=-4 plug that in for B and solve.\r\n\r\nYou should get 19." } { "Tag": [ "geometry" ], "Problem": "The quadrilateral shown below has its vertices at the points $ (0,1)$, $ (3,4)$, $ (4,3)$, and $ (3,0)$. How many square units are in its area?\n[asy]size(75);\ndraw((-.5,0)--(4.5,0));\ndraw((0,-.5)--(0,4.5));\ndraw((0,1)--(3,4)--(4,3)--(3,0)--cycle);[/asy]", "Solution_1": "This is 2 triangles with base 4 (from $ (3,0)$ to $ (3,4)$).\r\nThe left triangle has a height of 3, and the right has a height of 1.\r\nTherefore, your area is:\r\n$ \\frac{4*3}{2} \\plus{} \\frac{4*1}{2} \\equal{} \\boxed{8}$" } { "Tag": [ "inequalities", "geometry", "circumcircle", "inradius", "perimeter", "ratio", "trigonometry" ], "Problem": " Let ABC be a right triangle.\r\nprove that R $ \\ge$ (1+ $ \\sqrt2$)r", "Solution_1": "Sometimes it's useful to explain the variables, but i guess that $ R$ is the circumradius and $ r$ the inradius of triangle $ ABC$. We have $ R \\equal{} \\frac {c}{2}$ and $ r \\equal{} \\frac {A}{s}$, where $ s$ is the semi-perimeter and $ A$ is the area of triangle $ ABC$, repsectively. Moreover, $ c \\equal{} \\sqrt {a^2 \\plus{} b^2}$. We have to show that the ratio $ \\frac {R}{r}$ is greater or equal to $ 1 \\plus{} \\sqrt {2}$. Expressing $ R/r$ with the above formulas, one gets\r\n\\[ \\frac {R}{r} \\equal{} \\frac {\\sqrt {a^2 \\plus{} b^2}(a \\plus{} b \\plus{} \\sqrt {a^2 \\plus{} b^2})}{2ab} \\equal{} \\frac {1}{2} \\sqrt {\\frac {a}{b} \\plus{} \\frac {b}{a}} \\left( \\sqrt {\\frac {a}{b}} \\plus{} \\sqrt {\\frac {b}{a}} \\plus{} \\sqrt {\\frac {a}{b} \\plus{} \\frac {b}{a}} \\right) \\ge \\frac {1}{2} \\sqrt {2} ( 2 \\plus{} \\sqrt {2}) \\equal{} 1 \\plus{} \\sqrt {2},\\]\r\nwhere the last inequality follows from AmGm.", "Solution_2": "hello, just another solution, using the formula $ r\\equal{}\\frac{a\\plus{}b\\minus{}c}{2}$ so our inequality ist equivalent to\r\n$ c\\geq(1\\plus{}\\sqrt{2})(a\\plus{}b\\minus{}c)$\r\n$ c(2\\plus{}\\sqrt{2})\\geq(1\\plus{}\\sqrt{2})(a\\plus{}b)$\r\n$ \\sqrt{a^2\\plus{}b^2}(2\\plus{}\\sqrt{2})\\geq(1\\plus{}\\sqrt{2}(a\\plus{}b)$\r\nsquaring both sides we get\r\n$ 2(a^2\\plus{}b^2)(3\\plus{}2\\sqrt{2})\\geq(3\\plus{}2\\sqrt{2})(a^2\\plus{}b^2\\plus{}2ab)$\r\n$ (a\\minus{}b)^2\\geq0$ which is true.\r\nSonnhard.", "Solution_3": "[quote=\"zool007\"][color=darkred]Let $ ABC$ be a right triangle. Prove that $ R\\ge \\left(1\\plus{} \\sqrt2\\right)r$ .[/color] [/quote]\n\n[color=darkblue][b][u]Proof 1[/u].[/b] If suppose w.l.o.g. that $ A \\equal{} 90^{\\circ}$ , then $ \\boxed {2R \\equal{} a}$ and $ \\boxed {2r \\equal{} b \\plus{} c \\minus{} a}$ . Therefore $ p \\equal{} r \\plus{} 2R$ and $ 4R^2 \\equal{} a^2 \\equal{}$\n\n$ \\boxed {b^2 \\plus{} c^2\\ge 2bc} \\equal{} 4S \\equal{} 4pr$ $ \\Longleftrightarrow$ $ R^2\\ge pr$ $ \\Longleftrightarrow$ $ R^2\\ge (2R \\plus{} r)r$ $ \\Longleftrightarrow$ $ R^2 \\minus{} 2Rr \\minus{} r^2\\ge 0$ $ \\Longleftrightarrow$ $ \\boxed {R\\ \\ge (1 \\plus{} \\sqrt 2)r\\ }$ .\n\n[b][u]Proof 2[/u].[/b] If suppose w.l.o.g. that $ A \\equal{} 90^{\\circ}$ , then $ \\boxed {2R \\equal{} a}$ and $ \\boxed {2r \\equal{} b \\plus{} c \\minus{} a}$ . Therefore\n\n $ \\boxed {\\ R\\ge \\left(1 \\plus{} \\sqrt 2\\right)r\\ }\\Longleftrightarrow$ $ 2R\\ge \\left(1 \\plus{} \\sqrt 2\\right)\\cdot 2r$ $ \\Longleftrightarrow$ $ a\\ge \\left(1 \\plus{} \\sqrt 2\\right)(b \\plus{} c \\minus{} a)$ $ \\Longleftrightarrow$\n\n$ a\\left(2 \\plus{} \\sqrt 2\\right)\\ge \\left(1 \\plus{} \\sqrt 2\\right)(b \\plus{} c)\\Longleftrightarrow$ $ a\\sqrt 2\\ge b \\plus{} c\\Longleftrightarrow \\boxed {\\ 2\\left(b^2 \\plus{} c^2\\right)\\ge (b \\plus{} c)^2\\ }$ .\n\n[b][u]Proof 3[/u].[/b] Suppose w.l.o.g. that $ A \\equal{} 90^{\\circ}$ . I\"ll use the well-known relation $ \\boxed {\\cos A \\plus{} \\cos B \\plus{} \\cos C \\equal{} 1 \\plus{} \\frac rR}$ , \n\ni.e. in this case $ \\cos B \\plus{} \\cos C \\equal{} 1 \\plus{} \\frac rR$ . Thus $ \\boxed {\\ R\\ge \\left(1 \\plus{} \\sqrt 2\\right)r\\ }\\Longleftrightarrow$ $ \\frac {1}{\\sqrt 2 \\plus{} 1}\\ge \\frac rR$ $ \\Longleftrightarrow \\sqrt 2\\ge 1 \\plus{} \\frac rR$ $ \\Longleftrightarrow$\n\n$ \\sqrt 2\\ge \\cos B \\plus{} \\cos C\\Longleftrightarrow \\sqrt 2\\ge 2\\cos\\frac {B \\plus{} C}{2}\\cos\\frac {B \\minus{} C}{2}$ $ \\Longleftrightarrow \\boxed {\\ 1\\ge\\cos\\frac {B \\minus{} C}{2}\\ }$ because $ B \\plus{} C \\equal{} 90^{\\circ}$ .\n\n[b][u]Proof 4[/u].[/b] Suppose w.l.o.g. that $ A \\equal{} 90^{\\circ}$ . I\"ll use the well-known relation $ \\boxed {4\\sin\\frac A2\\sin\\frac B2\\sin\\frac C2 \\equal{} \\frac rR}$ , \n\ni.e. in this case $ \\frac rR \\equal{} 2\\sqrt 2\\sin\\frac B2\\sin\\frac C2$ . Thus $ \\boxed {\\ R\\ge \\left(1 \\plus{} \\sqrt 2\\right)r\\ }\\Longleftrightarrow$ $ \\frac {1}{\\sqrt 2 \\plus{} 1}\\ge2\\sqrt 2\\sin\\frac B2\\sin\\frac C2$ $ \\Longleftrightarrow$\n\n$ \\sqrt 2 \\minus{} 1\\ge \\sqrt 2\\left(\\cos\\frac {B \\minus{} C}{2} \\minus{} \\cos\\frac {B \\plus{} C}{2}\\right)\\Longleftrightarrow$ $ \\sqrt 2 \\minus{} 1\\ge \\sqrt 2\\cos\\frac {B \\minus{} C}{2} \\minus{} 1$ $ \\Longleftrightarrow \\boxed {\\ 1\\ge \\cos\\frac {B \\minus{} C}{2}\\ }$ .\n\n[b][u]Proof 5 [/u].[/b] Denote the incentre $ I$ and the second intersection $ D$ of $ AI$ with the circumcircle $ C(O,R)$ . Observe that $ ID \\equal{} DB \\equal{} R\\sqrt 2$ ,\n\n $ OI^2 \\equal{} R^2 \\minus{} 2Rr$ and $ \\boxed {ID\\le OI \\plus{} OD}$ $ \\Longleftrightarrow$ $ R\\sqrt 2\\le \\sqrt {R^2 \\minus{} 2Rr} \\plus{} R$ $ \\Longleftrightarrow$ $ R\\left(\\sqrt 2 \\minus{} 1\\right)\\ge r$ $ \\Longleftrightarrow$ $ \\boxed {\\ R\\ge\\left(1 \\plus{} \\sqrt 2\\right)r\\ }$ .\n\n[b][u]Remark[/u].[/b] We have equality if and only if $ B \\equal{} C \\equal{} 45^{\\circ}$ , i.e. $ \\frac b1 \\equal{} \\frac c1 \\equal{} \\frac {a}{\\sqrt 2}$ .[/color]\n\n[quote][color=darkred][b][u]An easy extension[/u].[/b] Let $ ABC$ be a triangle with the inradius $ r$ and the circumradius $ R$ . Prove that $ \\boxed{\\ R\\ \\ge\\ \\frac {1 \\plus{} \\sin\\frac A2}{\\sin A\\cos\\frac A2}\\cdot r\\ }$ .[/color][/quote]\r\n\r\n[color=darkblue][b][u]Hint[/u].[/b] You can prove it as with the same idea from the last proof ![/color]" } { "Tag": [ "analytic geometry" ], "Problem": "How fast can you add? How many coordinates are you plot in 60 seconds?\r\n\r\n[b]So, you think you are the best!?! Prove it![/b]\r\n\r\nCompete against other math students in a collection of [u]free[/u] [b]math games[/b] at:\r\n\r\nhttp://www.xpmath.com/\r\n\r\nHave fun!\r\n\r\n-Hui", "Solution_1": "like, what exactly is it?", "Solution_2": "It's a series of free math games that tests your speed and accuracy in various math skills.\r\n\r\nYou can compete with others over the internet for [b]prizes[/b] such as [i]Amazon gift certificates[/i]. :roll: \r\n\r\nTake a look and you'll see.[/i][/b]" } { "Tag": [ "geometry", "3D geometry", "tetrahedron", "geometric transformation", "reflection", "vector", "geometry proposed" ], "Problem": "Prove that if the bimedians of a tetrahedron are the common perpendiculars to respective sides then its opposite sides are equal.", "Solution_1": "What is bimedian?", "Solution_2": "The segment connecting the midpoints of two opposite sides of a tetrahedron.\r\n\r\nAs for a solution,\r\n\r\nLet $M_{XY}$ be the midpoint of $XY$, where $X,Y$ are vertices from the set $\\{A,B,C,D\\}$. We get $M_{XY}U=M_{XY}V$ for all permutations $X,Y,U,V$ of $A,B,C,D$ (call this observation $(*)$). From here the conclusion follows easily. We can see that if the tetrahedron satisfies the conclusion, then $(*)$ holds. On the other hand, given $BCD$, $(*)$ determines the tetrahedron uniquely, so it must satisfy the conclusion.\r\n\r\nThere are other beautiful things to say about this tetrahedron. For example, it's the only tetrahedron with all faces equivalent (i.e. having the same area), or the only tetrahedron for which the angles around each vertex sum up to $\\pi$.", "Solution_3": "Ok!\r\nKunny posted this problem some time ago (and I solved it). But I don't know whether this thread disappeared or not?", "Solution_4": "May be, I've missed something, but it seems obvious :\r\ntwo opposite sides are reflected of each other with respect to the corresponding bimedian hence they have the same length.\r\nKind regards. Jean-Pierre", "Solution_5": "Maybe I am missing something, but I don't understand your idea. :?", "Solution_6": "The reflection wrt the line joining the midpoints of AB and CD maps A to B and C to D, hence AD = BC and AC = BD\r\nKind regards. Jean-Pierre", "Solution_7": "I have an other approach:\r\nLet vectorAB=x, vector AC=y, vector AD=z then vector MN=(y+z-x)/2, where M and N are the midponts of AB and CD. Hence\r\n(y+z-x)/2*x=(y+z-x)/2*(z-y)=0\r\nThen xy+xz-x 2 =0 and\r\n z 2 -y 2 -xz+xy=0. hence \r\nz 2 =(y-x) 2 and y 2 =(z-x) 2 .Finally AD=BC and AC=BD and the conlusion follows." } { "Tag": [ "IMO Shortlist", "3D geometry", "packing", "combinatorics", "Coloring" ], "Problem": "A staircase-brick with 3 steps of width 2 is made of 12 unit cubes. Determine all integers $ n$ for which it is possible to build a cube of side $ n$ using such bricks.", "Solution_1": "can someone please find a solution or at least a hint please?", "Solution_2": "There are 12 unit cubes in each brick, hence $ n$ must be divisible by 2 and 3, i.e., $ 6|n$. Let $ n\\equal{}6k$. Two staircase-bricks can be used to build a $ 2\\times3\\times4$ box. Hence it is easy to construct a cube of side $ 12$. Thus $ n\\equal{}6k$ is possible for even $ k$. We claim that if $ k$ is odd, we cannot build a cube of side $ n$, i.e., the number of bricks must be even.\r\n\r\nAssume there are $ m$ bricks. Assign them to the Cartesian coordinates such that the center of a corner cube is at (0,0,0). Consider the cubes in $ (x,y,z)$ such that $ x,y,z$ are even. We call such cubes \"good\". In a brick, there can be either 1 or 3 good cubes. Assume that there are $ p$ bricks that contain 3 good cubes. Hence the total number of good cubes is $ 3p\\plus{}m\\minus{}p\\equal{}2p\\plus{}m$. But the number of good cubes is $ \\frac{12m}{8}$. Hence $ 2p\\plus{}m\\equal{}\\frac{12m}{8}$, or $ m\\equal{}4p$. So $ m$ is even. Q.E.D.", "Solution_3": "Actually you need $ 4|m$ ($ m$ even does not suffice, since there are $ n^3 \\equal{} 12m$ would just imply $ 8|n^3$, which gives nothing).\r\n\r\nOr did I miss something?", "Solution_4": "@above yeah we need 4|m\nFor this we will use comlex combinatorics Let $\\epsilon^4=1$ and $\\epsilon \\neq 1$ and we will assign to each square with coordinates $(x,y,z)$ with $\\epsilon^{x+y+z}$\nThen for each staircase-brick with starting unit cube $(x,y,z)$ which $x+y+z$ is minimum Then our sum will be 0(it is easy to calculate).And we know that The general sum is \n$(\\epsilon^0+......+\\epsilon^{m-1})^3$ and it is 0.Then 4|m and then we are done", "Solution_5": "@elcinmusazade what is the reasoning for $4|m?$ As I understand it you proved that the sum of the staircase brick is $0$ and the total sum is $0$ and any number of staircase bricks should sum to $0$ so there should be no restriction?\n\njgnr has a good idea and I built on it.\n\nNote that $12|n^3$ so $6|n.$ Let $n=6k.$ Note that two staircase bricks can be put together to form a $2\\times3\\times4$ prism like in the following picture. \n[img width=50]https://media.discordapp.net/attachments/925784397469331477/950060134556663839/Screen_Shot_2022-03-06_at_8.59.17_AM.png?width=1108&height=1170[/img]\n\nWe can place a three-dimensional array of $6\\times4\\times3$ of these $2\\times3\\times4$ prisms to form a $12\\times12\\times12$ cube. These cubes can then be made into a big cube of side length divisible by $12,$ i.e. when $k$ is divisible by $2.$\n\nNow, we split the $6k\\times 6k\\times 6k$ into a $3k\\times 3k\\times 3k$ grid of $2\\times 2\\times 2$ cubes and in each such cube we call the bottom-left-back corner unit cube [i]special.[/i] Note that in the two layers of the staircase brick, one of them cannot have any special cubes.\n\n[img width=50]https://media.discordapp.net/attachments/925784397469331477/950073693189586984/Screen_Shot_2022-03-06_at_9.53.22_AM.png?width=1332&height=1169[/img]\n\nIn the above picture, observe that differently-colored cubes cannot be both special, and one of the four colors must be a special cube, so the number of special cubes per staircase brick is $1$ or $3.$ Note that the number of staircase bricks is $18k^3$ which is even. Thus, the number of special cubes is even and so the number of total $2\\times 2\\times 2$ cubes is even. This implies that $27k^3$ is even, so $k$ is also even. Therefore the only solutions are $n=12k$ for all positive (nonnegative if you happen to count $0$) integers $k$.", "Solution_6": "Notice that the volume of such a cube must be a multiple of $12$ and thus $n$ should be a multiple of $6$. We claim that $n$ can be $\\boxed{\\text{all multiples of }12}$. Notice that two staircases can be combined to create a $2\\times 3\\times 4$ box, which can clearly be stacked to fill any cube with side length a multiple of $12$. \n\nIt suffices to show that $n=12k-6$ for positive integer $k$ is impossible. Suppose we can fill such a cube completely with staircases. Break up such a cube into unit subcubes and let their centers be $(a, b, c)$ for $1\\le a,b,c\\le n$. Give each subcube a [i]label[/i] from $0$ to $3$ equivalent to the sum of it's coordinates modulo $4$. Now it is easy to check that all staircases placed inside such a cube must have an odd number of unit cubes of every type. Since there are an even number of staircases in filling up a cube, the cube must have an even number of unit cubes of each type.\n\nBreak up the side length $12k-6$ cube into one $12k-4$ cube, three $(12k-4)\\times (12k-4) \\times 2$ prisms, three $(12k-4)\\times 2\\times 2$ prisms, and one $2\\times 2\\times 2$ cube. Everything except the $2\\times 2\\times 2$ cube has an even number of unit cubes of each type, and the $2\\times 2\\times 2$ cube has an odd number of unit cubes of each type. This means that the big cube has an odd number of unit cubes of each type, contradiction.", "Solution_7": "We claim the answer is all positive multiples of $12$. Evidently, if $n$ is an integer that works, we have $12\\mid n^3$, so $6\\mid n$. Hence, $n=6k$. Now, note that we may place two staircase-bricks together to form a $2\\times3\\times4$ prism. Since $2,3,4\\mid12$, it readily follows that the cube can be built when $k$ is even. Now we prove the cube cannot be built if $k$ is odd. Suppose it can and consider a tiling of a $6k\\times6k\\times6k$ cube with staircase-bricks. We separate the brick into six $6\\times6$ layers. In each layer, the staircase-brick fills either a $2\\times1,2\\times2$, or $2\\times3$. If the staircase-brick fills a $2\\times2$, then it must fill a $2\\times1$ and $2\\times3$ in the above and below layers. Hence, the bottom-most layer has no $2\\times2$ filled by staircase-bricks. It follows the layer right on top is filled only by $2\\times2$. It is straightforward to see that the only way to tile a $6k\\times6k$ layer using $2\\times2$ pieces is to lay out the pieces in a $3k\\times3k$ grid. Therefore, we use $9k^2$ pieces. Now, look at the bottom layer. It is filled by $x$ $2\\times3$ and $y$ $2\\times1$. Since it is the bottom layer, these must be part of one of the $9k^2$ pieces above. Hence, $x+y=9k^2$. On the other hand, summing areas we see $6x+2y=36k^2$, or $3x+y=18k^2$. Thus, $2x=9k^2$. However, $k$ is odd, so this is impossible. Hence, the cube cannot be built when $k$ is odd as desired. $\\blacksquare$", "Solution_8": "The answer is all multiples of $12$. Evidently these work, because we can form $2\\times 3\\times 4$ bricks by combining two staircase-bricks.\n\nNow, for odd multiples of $6$, label each cube with a number from $0$ to $3$, representing its 3-D taxicab distance to some predetermined corner. Note that the number of cubes labeled with each number is odd, since we can \"reduce\" the dimensions $\\bmod\\text{ 4}$ to obtain a $2\\times 2\\times 2$ square, where the counts of all numbers are all odd.\n\nSuppose FTSOC that we can form such a cube with staircase-bricks. Then we must clearly use an even number of staircase-bricks. I claim that the counts of each of the labels is odd; this will give a contradiction, since it implies that the counts of each of the numbers in the cube is even. There are two cases to consider. The first case is where the closest cube to the predetermined corner is a cube on the bottom step of the staircase-brick. In this case, each of the labels from $0$ to $3$ appears exactly three times. The second case is where the closest cube is one of the two \"back\" corners of the staircase-brick. This case gives one label which appears five times, two which appear three times, and one which appears once. \n\nThen, we have our desired contradiction, so we are done. $\\blacksquare$", "Solution_9": "We claim that the answer is $12|n$.\n\nWe can cap the staircase on top of itself to form a 2x3x4 block, which can be combined to build a 12x12x12 block, so all multiples of 12 are possible.\n\nObviously, $n$ is divisible by 6 by volume. Let $n=6m$. Assign coordinates $(i,j,k)$ with $1\\leq i,j,k\\leq 6m$ to each unit cube. Denote a unit cube as orz if $i,j,k$ are all even.\n\nClaim: If a staircase-brick is placed anywhere, it will occupy an odd number of orz unit cubes. Note that the staircase-brick can be decomposed into a 2x2x2 cube and two small 1x1x2 blocks. Note that exactly one unit cube in the 2x2x2 is orz. Now, the four remaining unit cubes can be paired into two pairs such that the two unit cubes in each pair are either both orz or both not orz (since each coordinate differs by either 0 or 2), so there are an odd number of orz unit cubes.\n\nNote that the total number of staircase-bricks is $\\frac{(6m)^3}{12}=18m^3$, which is even, so the total number of orz cubes must be even since each staircase-brick has an odd number of orz cubes. Therefore, $m$ must be even since there are $27m^3$ total orz cubes, so $n$ must be divisible by 12." } { "Tag": [], "Problem": "Let a and b be positive integers such that (1 + ab)|(a^2 + b^2). Show that the integer (a^2 + b^2)/(1 + ab) must be a perfect square.\r\n\r\n[i]p. 20 An Introduction to the Theory of Numbers, Ivan Niven et al[/i]", "Solution_1": "Wasn't this on the IMO?", "Solution_2": "Didn't know." } { "Tag": [ "arithmetic sequence", "algebra unsolved", "algebra" ], "Problem": "let $N$ denote the set of all natural numbers\r\nlet it be divided into $k$ disjoint subsets namely $S_{1},S_{2}\\dots S_{k}$ each of which has it's elements in $arithmetic$ $progression$ .\r\nalso let the common differeences be $\\alpha_{1},\\alpha_{2}\\dots\\alpha_{k}$\r\nthen prove that \r\n\r\n$\\sum_{i=1}^{k}\\frac{1}{\\alpha_{i}}$ $=$ $1$", "Solution_1": "See here, problem 22, page 19, good luck! :wink:", "Solution_2": "thank you :)" } { "Tag": [ "function", "vector", "topology", "group theory", "abstract algebra", "algebra", "polynomial" ], "Problem": "Here's a truly beautiful and I think very very difficult problem:\r\n Consider $f$ be a $C^1$ class function from $R^n$ into itself such that it vanishes only in 0, it tends to infinity (in norm, of course) when the argument tends to infinity and $df(x)$ is invertible at any point. Then $f$ is a difeomorphisme.", "Solution_1": "[quote=\"harazi\"]Here's a truly beautiful and I think very very difficult problem:\n Consider $f$ be a $C^1$ class function from $R^n$ into itself such that it vanishes only in 0, it tends to infinity (in norm, of course) when the argument tends to infinity and $df(x)$ is invertible at any point. Then $f$ is a difeomorphisme.[/quote]\r\nYou don't need the requirement that $f(x) = 0 \\, \\implies \\, x = 0$.\r\nFirst we show that $f$ is onto. Let $K \\subset \\mathbb{R}^n$ be compact, then $f^{-1}(K) = \\{x \\, | \\, f(x) \\in K\\}$ is closed (since $f$ is continuous) and bounded (since otherwise we can find $x_n \\to \\infty$, yet $f(x_n) \\in K$). In short, [i]$f$ is proper.[/i]\r\n\r\nSo let $a \\in \\mathbb{R}^n$, and consider $A = f^{-1}(B_{|a|+1}(0))$ (pre-image of closed ball of radius $|a|+1$). This is a compact set, hence the minimum of $g(x) = |f(x)-a|$ is attained at some $x^* \\in A$. \r\n\r\nSuppose $f(x^*) = b \\neq a$, then there are an open neighborhood $U(x^*)$ and an open ball $B_{\\epsilon}(b)$ such that $f(U(x^*)) = B_{\\epsilon}(b)$, by the inverse function theorem. Thus there is a point $z \\in U(x^*)$ such that $f(z) = b + \\delta(a-b)$, and therefore $z \\in A$ and also $|f(z) - a| < |b-a|$, contradiction. Hence $f(x^*) = a$, and $f$ is onto.\r\n\r\nTo show that $f$ is 1-1, assume that $f(a) = f(b)$ yet $a \\ne b$. Consider $g(t) = f(a + t(b-a))$. Then $g(b) = g(a)$ and hence $0 = g'(c) = Df(a+c(b-a))(b-a)$ for some $0 < c < 1$. Thus $Df(a+c(b-a))$ has a non-trivial null space, contradicting the assumption that $Df(x)$ is invertible everywhere. ;)", "Solution_2": "Oh, no! You do not have Rolle's theorem in dimensions greater than $1$! The surjectivity is simple, indeed, but the injectivity is much harder ;)", "Solution_3": "Ok, this might look like some nonsense....Because I did not make it rigurous...But...can we do something on the lines of :\r\nTake a point $\\alpha \\in \\mathbb{R}^n$. Then locally, by the Inverse function theorem, $f$ will have an inverse $f_{\\alpha}$ in some neighbourhood $U_{\\alpha}$. Now, take a compact ball $B_r$...And by the Heine-Borel (I think) theorem there will be an open cover of it with sets $U_{\\alpha}$, finite in number. Then, it is easy to see that they will be coherent in their intersections, as inverses of the same functions, so we would have an inverse in the ball...Can we now extend it to $\\mathbb{R}^n$ ?\r\nSorry for this nonsense... :blush:", "Solution_4": "Continuing the nonsense....We would need the condition of increasing norms for the fact that $f(B_r)$ would be unbounded and thus cover all of $\\mathbb{R}^n$....And the $f(0)=0$...is it necessay? I mean, if we take $f(x)=x+c$, $c$ a constant vector, nothing changes, right?\r\nSorry for wasting your time :blush:", "Solution_5": "[quote=\"fedja\"]Oh, no! You do not have Rolle's theorem in dimensions greater than $1$! The surjectivity is simple, indeed, but the injectivity is much harder ;)[/quote]Right, I noticed it an hour after posting this nonsense :blush:", "Solution_6": "[quote=\"DusT\"]Ok, this might look like some nonsense....Because I did not make it rigurous...But...can we do something on the lines of :\nTake a point $\\alpha \\in \\mathbb{R}^n$. Then locally, by the Inverse function theorem, $f$ will have an inverse $f_{\\alpha}$ in some neighbourhood $U_{\\alpha}$. Now, take a compact ball $B_r$...And by the Heine-Borel (I think) theorem there will be an open cover of it with sets $U_{\\alpha}$, finite in number. Then, it is easy to see that they will be coherent in their intersections, as inverses of the same functions, so we would have an inverse in the ball...Can we now extend it to $\\mathbb{R}^n$ ?\nSorry for this nonsense... :blush:[/quote]\r\nHere's how this can be made precise: \r\n\r\nFor each $x \\in %Error. \"mathnn\" is a bad command.\n{R}^n$, let $k(x) = \\#\\{z \\, | \\, f(x)=f(z)\\}$, i.e. the number of points whose image agrees with $f(x)$. Since $f$ is onto, $k(x) \\ge 1$ for all $x$, and $k(0) = 1$ by one of the original assumptions. Note that $k(x)$ is finite for all $x$, since pre-images of single points cannot have accumulation points (because $f$ is locally invertible) and are compact (because $f$ is proper). By local invertibility of $f$, $C_m = \\{x \\, | \\, k(x) = m\\}$ is open for each $m$. Thus $C_1$ is open, non-empty, and since the complement of $C_1$ is $\\cup_{m>1}C_m$ and hence open, $C_1$ is also closed. Thus $C_1 = \\mathbb{R}^n$, and $f$ is 1-1.", "Solution_7": "Actually, we don't need the assumption that $f^{-1}(0)=\\{0\\}$. It follows from the other hypotheses that $f$ is one-to-one, but we need a little more work. Let's take it from the top:\r\n\r\nFirst of all, $f$ is a local homeomorphism, so it's an open map. Secondly, it's proper, and any proper map from a metric space to itself is also closed (easy to prove: if $f(x_{n})\\to y$, the preimage of $\\{f(x_{n})\\}\\cup\\{y\\}$ is compact, so if $x$ is the limit of a subsequence of $(x_{n})_{n}$, then $f(x)$ is the limit of a subsequence of $(f(x_{n}))_{n}$, i.e. $f(x)=y$). From these two we find that $f(\\mathbb R^{n})$ is a non-empty clopen subset of the connected space $\\mathbb R^{n}$, so it must be the entire space, i.e. $f$ is onto.\r\n\r\n$f: \\mathbb R^{n}\\to\\mathbb R^{n}$ is onto, and our hypotheses imply that it's a covering space. Since $\\mathbb R^{n}$ is connected, the preimage of every point has the same (finite, because $f$ is proper) number of elements (this is a basic fact on covering spaces; it's also just a restatement of what hpe did, and I'm sorry I didn't see it before he wrote it down :)). Knowing that $|f^{-1}(0)|=1$ immediately tells us that this common value of all $|f^{-1}(x)|$ is $1$, but as I said, we can reach this conclusion without this hypothesis: \r\n\r\nAnother basic fact about covering spaces is that if $p: \\widetilde X\\to X$ is a covering map (for well-behaved spaces $\\widetilde X,X$, i.e. which have all the connectedness properties we might need: they're connected, locally connected, path-connected, and locally path-connected), then the common value $|p^{-1}(x)|$ is precisely the index of the subgroup $p_{*}(\\pi_{1}(\\widetilde X))$ in $\\pi_{1}(X)$, where $\\pi_{1}(\\cdot)$ denotes the fundamental group, and $p_{*}$ is the group homomorphism indiced by $p$. In our case, both $\\widetilde X$ and $X$ are simply connected, i.e. this index can only be $1$, and we're done.\r\n\r\nIn conclusion, we now know that:\r\n\r\nA $C^{1}$ diffeomorphism which also happens to be a proper map is actually a diffeomorphism.\r\n\r\nNote that we can replace the hypothesis of being proper with that of being a closed map. We saw above that proper implies closed. Conversely, for local homeomorphisms, closed implies proper.", "Solution_8": "S.I Pinchuk, in \"A counterexample to the Real Jacobian conjecture\" (Math. Z. 217, 1994), gives an example of a non-injective polynomial mapping from $R^{2}$ into itself of degree(p,q)=(10,25) and whose jacobian determinant is everywhere positive in $R^{2}$.\r\n\r\nWhere is the mistake?", "Solution_9": "That mapping is probably not proper (i.e. the preimage of a compact set is not always compact). That can happen with polynomials of more than one variable." } { "Tag": [ "algebra", "polynomial", "LaTeX", "AoPSwiki" ], "Problem": "$ 17)$ If ${ \\frac{x^3+9x^2+11x+5}{x(x+1)^3}=\\frac{A}x}+\\frac{B}{(x+1)}+\\frac{C}{(x+1)^2}+\\frac{D}{(x+1)^3}$, then find $ A+B+C+D$.\r\n\r\n\r\n\r\n$ 19)$ Solve for $ x$: \r\n\r\n$ (2x^2+12x+19)^2-4(2x^2+12x+19)+3=0$", "Solution_1": "Hi\r\nFor 19:\r\nX=2x\u00b2+12x+19\r\nthe equation becomes:\r\nX\u00b2-4X+3=0\r\n...\r\nHow we can use the latex languege here?thx", "Solution_2": "First one is a partial fraction decompostion:\r\n[hide]\nFixify the denominators.\n$ x^3\\plus{}9x^2\\plus{}11x\\plus{}5\\equal{}A(x\\plus{}1)^3\\plus{}Bx(x\\plus{}1)^2\\plus{}Cx(x\\plus{}1)\\plus{}Dx$\nRearrange. Warning: do not try this at home\n$ x^3\\plus{}9x^2\\plus{}11x\\plus{}5\\equal{}(A\\plus{}B)x^3\\plus{}(3A\\plus{}2B\\plus{}C)x^2\\plus{}(3A\\plus{}B\\plus{}C\\plus{}D)x\\plus{}A$\nEquationify. | is just so you don't confuse the equations together.\n$ A\\plus{}B\\equal{}1$ | $ 3A\\plus{}2B\\plus{}C\\equal{}9$ | $ 3A\\plus{}B\\plus{}C\\plus{}D\\equal{}11$ | $ A\\equal{}5$\nUsually you would have to solvify for these, but I'm going to take a shortcut for this one.\nSubtract 2 times the 4th equation from the 3rd equation. $ A\\plus{}B\\plus{}C\\plus{}D\\equal{}1$\n[/hide]", "Solution_3": "[quote=\"sami.d\"]Hi\nFor 19:\nX=2x\u00b2+12x+19\nthe equation becomes:\nX\u00b2-4X+3=0\n...\nHow we can use the latex languege here?thx[/quote]\r\n\r\nyep its the same as $ x^4\\plus{}bx^2 \\plus{} c$ where you let $ \\alpha\\equal{}x^2$\r\n\r\n--\r\n\"@sami.d\" - to use latex scroll over the $ x^4\\plus{}bx^2 \\plus{} c$ and you will see the code. But you need to use the dollar sign to activate it.. so dollarsign1234dollarsign and it will become like latex\r\n\r\nfor more info\r\n\r\nhttp://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideSym.php\r\nhttp://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideSym.php", "Solution_4": "I got the 2nd one, but for 17, I still am not sure how to approach the problem the correct way. If someone could explain that in further detail, that would be awesome =D", "Solution_5": "Like I said, it's a partial fraction decomposition. Look it up or see my post above.", "Solution_6": "I thought partial fraction decomposition would make it ${ \\frac {x^3 + 9x^2 + 11x + 5}{x(x + 1)^3} = \\frac {A}x} + \\frac {B}{(x + 1)} + \\frac {C}{(x + 1)} + \\frac {D}{(x + 1)}$\r\ninstead of \r\n${ \\frac {x^3 + 9x^2 + 11x + 5}{x(x + 1)^3} = \\frac {A}x} + \\frac {B}{(x + 1)} + \\frac {C}{(x + 1)^2} + \\frac {D}{(x + 1)^3}$\r\n???", "Solution_7": "[quote=\"erdogankerem123\"]I thought partial fraction decomposition would make it ${ \\frac {x^3 + 9x^2 + 11x + 5}{x(x + 1)^3} = \\frac {A}x} + \\frac {B}{(x + 1)} + \\frac {C}{(x + 1)} + \\frac {D}{(x + 1)}$\ninstead of \n${ \\frac {x^3 + 9x^2 + 11x + 5}{x(x + 1)^3} = \\frac {A}x} + \\frac {B}{(x + 1)} + \\frac {C}{(x + 1)^2} + \\frac {D}{(x + 1)^3}$\n???[/quote]\r\n :?: \r\nWhy would you want to have three fractions with the same denominator? I don't think that would help very much.", "Solution_8": "The format $ \\frac {x^3 \\plus{} 9x^2 \\plus{} 11x \\plus{} 5}{x(x \\plus{} 1)^3} \\equal{} \\frac {A}{x} \\plus{} \\frac {B}{(x \\plus{} 1)} \\plus{} \\frac {C}{(x \\plus{} 1)^2} \\plus{} \\frac {D}{(x \\plus{} 1)^3}$ is correct.\r\n[color=red]It is essential that the format is correct, or else the methods used below will not work at all.[/color]\r\n\r\n[hide=\"Solution\"]First, multiply through $ x (x \\plus{} 1)^3$:\n\n$ x^3 \\plus{} 9x^2 \\plus{} 11x \\plus{} 5 \\equal{} A(x \\plus{} 1)^3 \\plus{} Bx(x \\plus{} 1)^2 \\plus{} Cx(x \\plus{} 1) \\plus{} Dx$\n\n$ x : \\equal{} 0 \\implies \\boxed{A \\equal{} 5}$\n$ x : \\equal{} \\minus{} 1 \\implies 2 \\equal{} \\minus{} D \\implies \\boxed{D \\equal{} \\minus{} 2}$\n\nWe now subtract $ 5(x \\plus{} 1)^3 \\minus{} 2x$ from each side:\n\n$ \\minus{} 4x^3 \\minus{} 6x^2 \\minus{} 2x \\equal{} Bx(x \\plus{} 1)^2 \\plus{} Cx(x \\plus{} 1)$\n$ \\minus{} 2x (x \\plus{} 1)(2x \\plus{} 1) \\equal{} Bx(x \\plus{} 1)^2 \\plus{} Cx(x \\plus{} 1)$\n$ \\minus{} 2(2x \\plus{} 1) \\equal{} B(x \\plus{} 1) \\plus{} C$\n$ \\minus{} 4(x \\plus{} 1) \\plus{} 2 \\equal{} B(x \\plus{} 1) \\plus{} C$\n\n$ \\boxed{B \\equal{} \\minus{} 4}$ and $ \\boxed{C \\equal{} 2}$[/hide]", "Solution_9": "[quote=\"archimedes1\"][quote=\"erdogankerem123\"]I thought partial fraction decomposition would make it ${ \\frac {x^3 + 9x^2 + 11x + 5}{x(x + 1)^3} = \\frac {A}x} + \\frac {B}{(x + 1)} + \\frac {C}{(x + 1)} + \\frac {D}{(x + 1)}$\ninstead of \n${ \\frac {x^3 + 9x^2 + 11x + 5}{x(x + 1)^3} = \\frac {A}x} + \\frac {B}{(x + 1)} + \\frac {C}{(x + 1)^2} + \\frac {D}{(x + 1)^3}$\n???[/quote]\n :?: \nWhy would you want to have three fractions with the same denominator? I don't think that would help very much.[/quote]\r\nI don't know; I'm pretty new to this whole partial fraction decomposition thingy. I just read the AopsWiki on it, and all the notation was somewhat confusing, but that's how I interpreted it. If I'm wrong could someone explain exactly what partial fraction decomposition is to me and how and when to use it?" } { "Tag": [ "geometry", "inequalities", "complex numbers", "national olympiad" ], "Problem": "Problem 1\r\nExamine if we can place 9 convex 6-angled polygons (the one next to the other) to construct a convex 39-angled.\r\n\r\nProblem 2\r\nProve that: For each x,y,z in the R have power that\r\n(x^2-y^2)/ (2x^2+1) + (y^2-z^2)/(2y^2+1)+(z^2-x^2)/ (2z^2+1)<=0\r\n\r\nProblem 3\r\nLet the midpoint M of the side AB of an 4sided, inscribed in circle,ABCD.Let P the point of section of MC with BD. Let the parallel from the point C to the AP which intersects the BD at S. If CAD angle=PAB angle= BMC/2 angle, prove that BP=SD.\r\n\r\nProblem 4\r\nFind all the positive integers n , n>=3 such that n/ (n-2)! \r\n\r\nTry to find the solutions of the problems. If you find it difficult i have the solutions.", "Solution_1": "What do you mean by Greek TST 2005 ? :? maybe a TST for the JBMO, but surely not for the IMO. Be more specific when you compose such posts. ;) \r\nI don't think that these problems are difficult.", "Solution_2": "[quote=\"silouan\"]Problem 1\nExamine if we can place 9 convex 6-angled polygons (the one next to the other) to construct a convex 39-angled.\n\nProblem 2\nProve that: For each x,y,z in the R have power that\n(x^2-y^2)/ (2x^2+1) + (y^2-z^2)/(2y^2+1)+(z^2-x^2)/ (2z^2+1)<=0\n\nProblem 3\nLet the midpoint M of the side AB of an 4sided, inscribed in circle,ABCD.Let P the point of section of MC with BD. Let the parallel from the point C to the AP which intersects the BD at S. If CAD angle=PAB angle= BMC/2 angle, prove that BP=SD.\n\nProblem 4\nFind all the positive integers n , n>=3 such that n/ (n-2)! \n\nTry to find the solutions of the problems. If you find it difficult i have the solutions.[/quote]\r\n\r\n1. Let us try to dissect a 39gon into 9 6-gons. Each 6 gon can only occupy at most 5 sides of the 39 gon. If we have 8 6gons occupying, then there are 8 sides in the middle. If we have sone 6gons occupying less than 5 sides, we use all 6 gons covering the sides, and there are none left to cover the middle, which is concave.\r\n\r\n2. We prove for positive reals $x,y,z$, that $\\sum \\frac{x}{2x+1} \\leq \\sum \\frac{y}{2x+1}$. Obviously, $(x,y,z)$ and $(\\frac{1}{2x+1},\\frac{1}{2y+1}, \\frac{1}{2z+1})$ are oppositely sorted. Suppose $x \\ge y \\ge z$. By rearrangement, we have the result immediately. Suppose $x \\ge z \\ge y$. We also have the result by rearrangement.\r\n\r\n3. no idea\r\n\r\n4. I dont understand the question", "Solution_3": "sorry it is for JBMO. For the IMO i also give the problems", "Solution_4": "hi, silouan!\r\nnow, i look over the problems.\r\ndid you have only 4 problems?\r\n\r\nif i understood everything at the 4th, then n can be any positive integer that is not a prime except the number 4.\r\nindeed if n is not a prime then \r\nI. n = a*b, with 2<=a,b<=n div 2< n-1 and a<>b so a|(n-2)! and b|(n-2)! so n=ab|(n-2)!\r\nII. n=p*p, where p is a prime, p>2 then p|(n-2)! and p|2p< n-1 so 2p|(n-2)!\r\nso n=p*p | p*2p | (n-2)!.\r\nnow if n= 4then (n-2)! = 2 and 4 isn't a divisor of 2.\r\nand if n is a prime then (n,t)=1 any 1<=t<=n-2\r\ni think this is a easy problem\r\nsomething is un clear at the geometry problem(maybe i did a mistake).\r\ni reformulte the condition and u tell me if it is ok\"\r\nABCD - inscribed[u] CONVEX [/u]quadrilateral, thus A,B,C,D- concylclic\r\nm is on (AB), such that BM=MA. \r\nP is the intersection point of MC and BD. S is on BD, such that CS||AP.\r\n(angle CAD) = (angle PAB) = 1/2*(angle BMC) then BP=SD. \r\n\r\nthe inequality is rather simple and indeed it's obvious to use re-arrangements inequality, although you may prove it in a different way. here is my solution.\r\nset a=x^2 b=y^2 z=c^2.\r\n\r\nweh have ab+bc+ca<=a^2+b^2+c^2 <==>\r\n4(ab+bc+ca)<=2(a^2+b^2+c^2)+2(ab+bc+ca) <==>\r\n12abc+(a+b+c)+4(ab+bc+ca)<=2(a^2+b^2+c^2)+2(ab+bc+ca)+12abc+(a+b+c)\r\n\r\nnow note that.\r\n12abc+a+b+c+4(ab+bc+ca) = a(2b+1)(2c+1) + \r\n b(2a+1)(2c+1) + \r\n c(2a+1)(2b+1).\r\n\r\nand 2(a^2+b^2+c^2)+2(ab+bc+ca)+12abc+a+b+c = b(2b+1)(2c+1)+ \r\n c(2c+1)(2a+1)+\r\n a(2a+1)(2b+1).\r\n\r\ndividing by (2a+1)(2b+1)(2c+1) the obtained inequality and subtracting the right side from the left side we get what is asked.\r\n\r\n\r\nthe first problem it is beautiful and suitablke for the for the first problem in a JBMO tst . i have an other solution.\r\nwell, the sum of the angles of a 39convex n gon is 37*180.\r\nbut the sum of the angles of a 6 gon is 4*180, so the sum of the angles of 9 6gons is 9*4*180 = 36*180<37*180.\r\nthat is a contradiction, because, if a convex Ngon is divided into convex Kgons, thatn the sum of the angles of all the Kgons must be greater or equal than the sum of the angles of the bin Ngon.\r\nthus the answer is no.", "Solution_5": "Freemind you solve great the 1,2,4 problem. The geometry problem is quite difficult.I will post his solution the night. I hope that you will have your computer open. I also will post the 4 solutions of Moldova TST. Noow try to solve the TST Greek for the IMO. IT has very clever problems. See you!", "Solution_6": "thx silouan.\r\ni don't think i'll be online this night.\r\ntomorrow i think i'll enter on mathlinks just for a little.i'll think more at the geometry problem and i hope i'll enter on moldova's tst.\r\nthey were kinda easy, except some problems.\r\nand i will also look this night (but not at the computer over the tst for imo problems).\r\nhope i'll solve something.\r\ndo you really have only one TST?\r\nbye\r\n keep in touch.", "Solution_7": "I will post my solutions for Greek TST JBMO problems.\r\nThe first is exactly similar with yours freemind.\r\nProblem 2\r\nWe post a=2*(x^2)+1, b=2(y^2)+1 , c= 2(z^2)+1 , so a,b,c>0\r\nand (x^2)-(y^2)=(a-b)/2 , (y^2)-(z^2)=(b-c)/2 , (z^2)-(x^2)=(c-a)/2\r\nsowe shpold prove that , (a-b)/(2a)+(b-c)/(2b)+(c-a)/ (2c)<=0 <=>\r\n(a-b)/(a)+(b-c)/(b)+(c-a)/ (c)<=0 <=> (b/a)+(c/b)+(a/c)>=3 <=>\r\n(a^2)b +(b^2)c + (c^2)a>=3abc <=> [(a^2)b +(b^2)c + (c^2)a] / 3 >= 3rd root of (abc)^3 which is the inequality of AGM.\r\n\r\nProblem 3\r\nIt is a very good problem!\r\nAt first, at the triangle MAP the angle PMB is exterior, so angle PMB=angle MAP+angle MPA. If angle MAP=x , the angle PMB=2x from the hypothesis. So 2x=x+angle MPA. So angle MPA=x. Triangle MPA isosceles with MA=MP.But MA=MB because M is midpoint. Now at the triangle PAB the MP which is medium =MA=MP. So from the theory the triangle PAB is a right-anlged triangle with angle BPA=90 . But from hypothesis we have that AP// CF(F is the point where the CS intersects the AD). So angle PSF=90.\r\nLet MPB=m.So x+m=90.The triangle MBP is isosceles because MB=MP so angle MBP=m. Angle DBC=x because it is going to the same arc of the circle with angle DAC. So angle B=x+m=90. The ABCD is insribed angle B+angle D=180 so angle D=90. Because angle B=angle D=90 the diagonal AC pass through the O which is the center of the circle.So AC=2R, R is the ray of the circle.Now we make a trick. We extend the CS until CS intersects the circle in N. We also form the lines AN and ND. The angle SNA =90 because it is going to the middle of the circle. The APSN has all of his angles=90 so it is an right-angled parallelogram.So NS=AP. The triangles NSD and APB are equals because there are right-angled and they have: AP=NS and angle DNS=angle PAB=x(angle DNS=x decause it is going to hte same arc of the circle with angle DAC). Finally because NSD and APB are equals they have SD=BP. Good?\r\n\r\nProblem 4\r\nI will use the theorem of Wilson.This theorem says: The positive integer p>2 is a prime number if only if: (p-1)! \u2261 -1(modp). So n is not a prime number.Now if n=ab the a,b will appear at the (n-2)! so if n=ab n/ (n-2)!\r\nFinally if n=(q^2) ( q>2) we have that q=sqrt(n). In addition 2q 2*sqrt(n) (n^2)-6n+1>0 which have power for n>=6 . So 2(q^2)/ (n-2)! and (q^2)/(n-2)!\r\nAfter all these we conclude that n/(n-2)! when n complex number>=6\r\n\r\nFreemind what is your opinion for the geometry?", "Solution_8": "Here freemind i put all the solutions. I wait for your comments. See you!", "Solution_9": "hi!\r\nthe solution at geometry is nice indeed, and it seems to be correct :))\r\nthe inequality is also nice. \r\nbut b/a +c/b + a/c >=3 results immediatelly from the AM-GM inequality because a/b*b/c*c/a=1. \r\nand the 4th is correct.\r\njust make a correction. don't use \"complex numbers\" but nonprime or smth like that.\r\nbecause complex numbers is something totally different.", "Solution_10": "Freemind do you believe that the geometry problem was difficult? Also I want to tell you that in Greece we write only 4 problems. In two days i will sent you the problems of my national olympiad in your e-mail. See you.", "Solution_11": "the geometry was kind difficult. but it was beautiful as it ha an elementary non-algebraical solution :)\r\nthx for the problems you sent me, tonight i'll try to solve them and i hope i will post tomorrow the solutions (if i can solve them :D )\r\nkeep in touch", "Solution_12": "Ok freemind i wait your solutions in my e-mail. See you" } { "Tag": [ "videos", "articles" ], "Problem": "I wish you a happy new year!\r\n\r\nBut let us remember the many victims of the desaster in Asia!\r\n :( :( :(", "Solution_1": "We will remember it for many many years...", "Solution_2": "Over 100000 victims, may they'll receive a lot of help from rich countries quickly.", "Solution_3": "[quote=\"Moubinool\"]Over 100000 victims, may they'll receive a lot of help from rich countries quickly.[/quote]\r\nFrom France, for example...", "Solution_4": "CNN reports the death toll at 135,000 and it should breach the 150,000 mark as remote regions report deaths.", "Solution_5": "A German organization which helps there even estimates more than 200,000 victims. :( :( :( \r\nI hate to see these videos from tourists in which you can see how the people try to save themselves, but you don't know if the did or not and you sit in front of the TV and can't do anything. The most horrible thing is that you know:\r\nThat wasn't an action film: THAT WAS TRUE. And maybe these people are dead now.", "Solution_6": "Pray for the souls of the dead.\r\n\r\nkunny", "Solution_7": "speaking to all german speaking people here, i would like to point your attention to this: http://matheplanet.com/matheplanet/nuke/html/viewtopic.php?topic=29686 , what is mainly a discussion of the article at http://derstandard.at/?id=1905744 . having no clue about physics and being not the best in geography either, i cannot comment, but what would you say about it?\r\n\r\n darij", "Solution_8": "i read something about a \"bidding war\" on who can pay more between US and some other countries in google news", "Solution_9": "WHO is also trying to stop the epidemics that are caused because of the tidal wave which could double the death toll.", "Solution_10": "[quote=\"Scrambled\"]i read something about a \"bidding war\" on who can pay more between US and some other countries in google news[/quote]\r\n\r\nUS raised it's 35 million figure to 350 million and I believe that's all that's going to happen now. UN's already got about 1.1 billion secured with more to come soon obviously as more countries get permission from federal gov't and etc. to donate.", "Solution_11": "i believe that one of those \"other countries\" is japan, not sure though", "Solution_12": "it's so depressing watching the ppl run for their lives and then just get swept away so helplessly........ :(", "Solution_13": "Yes, I was thinking of organizing the ASIA club at my school to actually do fundraisers." } { "Tag": [], "Problem": "[China 1996] In a soccer tournament, each team plays another team exactly once and receives 3 points for a win, 1 point for a draw, and 0 points for a loss. After the tournament, it is observed that there is a team which has earned both the most total points and won the [i]fewest[/i] games. Find the smallest number of teams in the tournament for which this is possible.\r\n\r\nEnjoy! :)", "Solution_1": "[quote=\"Silverfalcon\"][China 1996] In a soccer tournament, each team plays another team exactly once and receives 3 points for a win, 1 point for a draw, and 0 points for a loss. After the tournament, it is observed that there is a team which has earned both the most total points and won the [i]fewest[/i] games. Find the smallest number of teams in the tournament for which this is possible.\n\nEnjoy! :)[/quote]\r\n\r\n\r\n[hide]\nyou never said it was unique most and unique fewest\nso you could have 3 teams\neach 1 beat one of the other teams and lost to one of the other teams.\neach has 3 points, each has 1 win.\n\n[/hide]", "Solution_2": "I typed up exactly what it said on the book.\r\n\r\nAnd no, the answer is not 3.", "Solution_3": "[hide]I found a solution with 8 teams.\n\nName the 8 teams A,B,C,D,E,F,G,and H.\n\nA defeats B and C, and draws against D,E,F,G,and H for 2 wins and 11 points.\nB defeats C, E, and F for 3 wins and 9 points.\nC defeats D, E, and G for 3 wins and 9 points.\nD draws against A and defeats B, F, and H for 3 wins and 10 points.\nE draws against A and defeats D, F, and H for 3 wins and 10 points.\nF draws against A and defeats C, D, and G for 3 wins and 10 points. \nG draws against A and defeats B, E, and H for 3 wins and 10 points.\nH draws against A and defeats B, C, and F for 3 wins and 10 points.\n\nTherefore, A has the fewest wins and the most points.\n\nI believe 8 teams is the fewest for which this is possible.[/hide]", "Solution_4": "yes cf249 you're right,the answer is 8,I can wirte the official solution here if anybody is interested.", "Solution_5": "It'd be nice, since I don't quite know how to prove that it's the minimum.", "Solution_6": "ok here it goes:\r\n[hide=\"the official solution\"]\nlet $ W$ be the desired team,now let $ n$ denote the number of soccer teams which participated in this tournament,so there are $ \\binom{n}{2} \\equal{} \\frac {n(n \\minus{} 1)}2$ matches played among these teams,now note that each game gives at least $ 2$ points in total,so sum of the points of all teams is greater than $ n(n \\minus{} 1)$ so the average is at least $ (n \\minus{} 1)$.\nnow note that $ W$ has played $ (n \\minus{} 1)$ games,and also we know that $ W$ has the most total points,so the total points of $ W$ is more than $ (n \\minus{} 1)$ so $ W$ must have won at least one game...\nnow as the problem said $ W$ has won the fewest number of games,so every other team must have won at least $ 2$ games.thus every other team has at least $ 6$ points.hence $ W$ must draw at least $ 4$ games in order to have the most total points,but note that if team $ A$ had a draw with team $ W$ then the total points of $ A$ would be at least $ 7$,hence $ W$ must have drawn at least $ 5$ games.which means $ n\\geq 7$.\nnow we show that $ n\\not \\equal{} 7$:\nif $ n \\equal{} 7$ then let $ S$ denote the set of all teams,now team $ W$ has won a single match and has drew $ 5$ matches,so the total points of $ W$ is equal to $ 8$,now note that every other team has won exactly two games,and drew at most one game.thus every other team has lost at least $ 3$ games,so there are at least $ 6\\times 3 \\equal{} 18$ looses in $ S\\setminus W$,so there must be at least $ 18$ wins in $ S$,but we know that the number of wins in $ S$ is equal to $ 1 \\plus{} 6\\times 2 \\equal{} 13$,which is a contradiction...\nso $ n\\geq 8$,now we will give an example for $ n \\equal{} 8$:\n\nlet $ S \\equal{} \\{W,A_1,A_2\\ldots , A_7\\}$ be the teams which participated in the tournament,now by $ X\\to Y$ we mean $ X$ won $ Y$ and by $ X\\iff Y$ we mean $ X,Y$ drew in their match:\n\n$ W\\to A_1$\n$ W\\to A_2$\n$ W\\iff A_i$ for $ i\\geq 3$\n\nand for every $ 1\\leq i\\leq 7$:\n\n$ A_i\\to A_{i \\plus{} 1}$\n$ A_i\\to A_{i \\plus{} 2}$\n$ A_i\\to A_{i \\plus{} 3}$\n$ A_{i \\plus{} 4}\\to A_i$\n$ A_{i \\plus{} 5}\\to A_i$\n$ A_{i \\plus{} 6}\\to A_i \\left(A_{i \\plus{} 7} \\equal{} A_i\\right)$.\n\nso $ W$ would have $ 11$ points in total,$ A_1,A_2$ would have $ 9$ points in total (with $ 3$ wins),and $ A_3,A_4,\\ldots ,A_7$ have $ 10$ points in total (with 3 wins)...[/hide]", "Solution_7": "Yup, basically what he wrote (although he paraphrased it bit). :)", "Solution_8": "[quote=\"Silverfalcon\"]although he paraphrased it bit[/quote]\r\n\r\nyeah because I have the translated version,so I had to translate it back to english again... :)", "Solution_9": "[quote=\"BaBaK Ghalebi\"][quote=\"Silverfalcon\"]although he paraphrased it bit[/quote]\n\nyeah because I have the translated version,so I had to translate it back to english again... :)[/quote]\r\n\r\nOh. \r\n\r\nI have English version here so I could've typed it up and saved you time. :D" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Given a convex 2n polygon. (n is a positive integer ) \r\nProve that exists at least one diagonal that is not parallel to any of the edges.", "Solution_1": "This has already been posted:\r\n\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=polygon&t=85912[/url]" } { "Tag": [ "geometry", "rhombus", "number theory", "least common multiple" ], "Problem": "On Saturday we had the inter-provincial in south africa and unfortunately Gauteng won :( I say unfortunately cos im western province. We usually win but we lost our two best members to the computer olympiad so we ended up coming 4th. Well done Boland for coming 3rd!(they also lost their two best members to the CO).\r\n\r\nLast 3 questions from individual paper:\r\n\r\n13. Find the area of the rhombus ABCD given that the radii of the circles circumscribed around triangles ABD and ACD are 12.5 and 25 respectively.\r\n\r\nA) 50 B)100 C) 200 D) 400 E)800\r\n\r\n14. In a sequence consisting of 2006 digits, each number formed by two successive digits is divisible by 19 or 23. If the first digit of the sequence is 1, the last possible digits are...\r\n\r\nA) 2 and 5 B) 2 and 7 C) 3 and 5 D) 3 and 7 E) 9\r\n\r\n15. Which of the following is equal to 1/(sqrt(2006+sqrt((2006^2)-1))?\r\n \r\n\r\n (sqrt = square root ^2 =squared.... think its pretty obvious but just in case...)\r\n\r\nA) (sqrt(2007)-sqrt(2005))/sqrt(2)\r\n \r\nB) (sqrt(2007)+sqrt(2005))/ sqrt(2)\r\n\r\nC) (sqrt(2006)-sqrt(2004))/sqrt(2)\r\n\r\nD) (sqrt(2006)+sqrt(2004))/ sqrt(2)\r\n\r\nE) 1/2(sqrt(2006^2-1))\r\n\r\nI know they're probably simple for most of you (especially since they're just multiple choice)... but normal school students who dont have any other training in olympiads take part so they dont have any written solution type questions. The team paper has 10 questions and these require numerical or algebraic answers but no explanations.\r\n\r\nThe last 3 of this paper were:\r\n\r\n8. Given triangle ABC with AB=5=AC and BC=6. P is a point inside the triangle such that x\\}$\r\n1) Are variables $ X$ and $ Y$ independent? \r\n2) Does $ X$ or $ Y$ belong to $ L^{1}$?\r\n3)Compute $ E(X|Y)$\r\n4)Are variables $ \\frac {Y}{X}$ and $ X$ independent?\r\n\r\nI don't quite see how I am supposed to compute marginal densities (densities of X and Y, to check whether $ g_{x}(x)g_{y}(y)\\equal{}g(x,y)$. I mean how we compute density of $ X$, when D is $ D \\equal{} \\{(x,y) : 0 < x < e \\minus{} 2 ; y > x\\}$", "Solution_1": "$ D$ is just some region. Note that $ e\\minus{}2$ is a constant, so $ 00$ such that $\\prod U_{i}$ contains the open ball $B(x,r)$. \r\n2) if $x\\in Q$ and $r>0$, then there is a product $\\prod U_{i}$ of the above form such that $a\\in \\prod U_{i}\\subset B(a,r)$.", "Solution_6": "Schwarz inequality shows that:\r\n$|Tf(y)-Tf(x)|=|\\int_{[x,y]}f(t) dt| \\leq \\parallel f\\parallel _{2}\\sqrt{|y-x|}$\r\nHence one immediately sees that the image of the closed unit ball\r\nof $L^{2}$ is a uniformy continuous subset of continuous functions on the compact $[0;1]$. Ascoli theorem \r\nsays that this set is relatively compact for the uniform norm $\\parallel .\\parallel _{\\infty;[0;1]}$.\r\nHence it is relatively compact in $L^{2}([0;1])$.\r\n\r\nTry to prove it is also true for:\r\n$Tf(x)=\\int_{[0;1]}K(x,y) f(y) dy$\r\nwhere $K(.,.)$ is any function in $L^{2}([0;1] \\times [0;1])$", "Solution_7": "Just wondering, is this way right:\n\nI will use next: Let $A:\\mathcal{H}\\to \\mathcal{H}$ is linear operator. Then $A$ is compact iff for every $x_n \\overset{w}{\\to} 0$ implies $Ax_n \\overset{s}{\\to} 0.$ ($\\mathcal{H}$ is notation for Hilbert space)\n\nWe want to show that $T:L^2[0,1] \\to L^2[0,1]$ is compact operator, where $\\ Tf(x) =\\int_{0}^{x}f(t)\\, dt $. Let $f_n$ be sequence which weakly converges to $0$. That means that for every function $g \\in L^2[0,1]$ is true that $\\lim_{n \\to +\\infty} \\langle f_n, g \\rangle=0$, or $\\lim_{n \\to +\\infty} \\int_0^1 f_n(t) \\overline{g(t)}\\, dt= 0. \\qquad (1)$\n\n Let $g_x(t)=\\chi_{[0,x]}(t)$. Obviously, $g_x \\in L^2[0,1]$. Then $(1)$ is giving us $\\lim_{n \\to +\\infty} \\int_0^x f_n(t)dt=0$, that is $(Tf_n)(x) \\to 0$ when $n \\to +\\infty$.\n\nWe want to prove that $\\| Tf_n - 0 \\| \\to 0$ as $n \\to +\\infty$. This is same as proving that $\\| Tf_n \\|^2=\\int_0^1 |(Tf_n)(x)|^2\\, dx \\to 0$ as $n \\to +\\infty$. \n\nSo actually we want to swap $\\lim$ and $\\int$ and then, because $(Tf_n)(x) \\to 0$ as $n \\to +\\infty$, it is obvious that $\\| Tf_n \\|^2 \\to 0$, or $Tf_n \\overset{s}{\\to} 0$, and then $T$ is compact operator.\n\nWe will use dominated convergence theorem: It's not hard to show that $|Tf(x)| \\leqslant \\| f \\|_2 \\sqrt{x}$ for every $f \\in L^2[0,1]$ (use Cauchy-Schwarz inequality). From that we see that $|Tf_n(x)|^2 \\leqslant \\| f_n \\|^2_2 x.$ We are almost here, we just don't want that $n$ in index. But, we can use fact that every weakly convergent sequence is bounded, so there is some constant $M$ such that $\\| f_n \\|_2 \\leqslant M$ (for all $n$). So $|Tf_n(x)|^2 \\leqslant M^2 x$ and our dominant is $M^2 x$, which of course belong to $L^2[0,1]$.\n\nIs this correct?" } { "Tag": [ "calculus", "integration", "probability", "expected value", "calculus computations" ], "Problem": "Evaluate $\\int_{0}^{a}(\\frac{x^{2}}{2n}-2 \\pi floor(\\frac{x^{2}}{4n \\pi}) ) * (\\frac{1}{\\sqrt{2 \\pi}\\sigma}e^{-\\frac{1}{2}(\\frac{x-\\mu}{\\sigma})^{2}}) dx$.\r\n\r\n$n, \\mu, \\sigma \\in \\mathbb{R}^{+}$.\r\n\r\nEither as it is, or limit of that as $a \\to \\infty$, whichever is do-able (if not both).", "Solution_1": "No one?\r\n\r\nAnyone at least have any ideas?", "Solution_2": "Well, i do have some ideas:\r\n\r\n- notice that one of the factors of the integrand is equal to $N(\\mu, \\sigma^{2})$, so I suggest it has something to do with propability.\r\n- Moreover, the integral is the expected value of $\\frac{x^{2}}{2n}-2\\pi floor \\frac{x^{2}}{4n\\pi}$ when $x$ is a stochastic variable which is normally distributed with mean $\\mu$ and variance $\\sigma^{2}$ for $x$ between $0$ and $a$, and 0 elsewhere.", "Solution_3": "Well, yeah.\r\n\r\nThat's where the integral CAME FROM.\r\n\r\nI'm trying to find the expected value of $\\theta$ (but just the part between $0$ and $2 \\pi$)." } { "Tag": [ "absolute value" ], "Problem": "On a circle, there are $ n$ $ (n>3)$ numbers with a total sum of 94, such that each number is equal to the absolute value of the difference between the two numbers which follow it in clockwise order. What is the possible value of $ n$? :)", "Solution_1": "The numbers are non-negative, as they take absolute value.\r\nLet them be $ a_1, a_2, ..., a_k$ arranged in clockwise direction.\r\nThen $ |a_{k\\minus{}2}|\\equal{}|a_k\\minus{}a_{k\\minus{}1}| \\le max(a_k, a_{k\\minus{}1})$ and equality holds only when one of the $ a_k, a_{k\\minus{}1}\\equal{}1$.\r\nNow choose $ a_i\\equal{}k$ such that $ a_i$ is the largest of the numbers.Then it's easy to see that $ \\begin{cases} a_{i\\plus{}1}\\equal{}0 \\\\ a_{i\\plus{}2}\\equal{}k \\end{cases}$ or $ \\begin{cases} a_{i\\plus{}1}\\equal{}k \\\\ a_{i\\plus{}2}\\equal{}0 \\end{cases}$ are the only possible values of $ a_{i\\plus{}1}$ and $ a_{i\\plus{}2}$. Both of them gives patterns like $ k,k,0,k,k,0,k,k,0,...,k,k,0$\r\n\r\n$ 94\\equal{}47 \\times 2\\equal{} 94 \\times 1$, so $ 1,1,0,1,1,0,...,1,1,0$ is the only possible answer , which gives $ n\\equal{}141$", "Solution_2": "[quote=\"stephencheng\"]The numbers are non-negative, as they take absolute value.\nLet them be $ a_1, a_2, ..., a_k$ arranged in clockwise direction.\nThen $ |a_{k \\minus{} 2}| \\equal{} |a_k \\minus{} a_{k \\minus{} 1}| \\le max(a_k, a_{k \\minus{} 1})$ and equality holds only when one of the $ a_k, a_{k \\minus{} 1} \\equal{} 1$.\nNow choose $ a_i \\equal{} k$ such that $ a_i$ is the largest of the numbers.Then it's easy to see that $ \\begin{cases} a_{i \\plus{} 1} \\equal{} 0 \\\\\na_{i \\plus{} 2} \\equal{} k \\end{cases}$ or $ \\begin{cases} a_{i \\plus{} 1} \\equal{} k \\\\\na_{i \\plus{} 2} \\equal{} 0 \\end{cases}$ are the only possible values of $ a_{i \\plus{} 1}$ and $ a_{i \\plus{} 2}$. Both of them gives patterns like $ k,k,0,k,k,0,k,k,0,...,k,k,0$\n\n$ 94 \\equal{} 47 \\times 2 \\equal{} 94 \\times 1$, so $ 1,1,0,1,1,0,...,1,1,0$ is the only possible answer , which gives $ n \\equal{} 141$[/quote]\r\n\r\nThanks Very Much :)" } { "Tag": [], "Problem": "$ABC$ and $A'B'C$are regular triangles with the same orientation.let $P,Q,R$be the midpoints of the segments\r\n\r\n$AB',BC,A'C$\r\n\r\nrespectively.show that: triangle$PQR$is regular", "Solution_1": "[quote=\"ashegh\"]$ABC$ and $A'B'C$are regular triangles with the same orientation.let $P,Q,R$be the midpoints of the segments\n$AB',BC,A'C$\nrespectively.show that: triangle$PQR$is regular[/quote]\r\nIn space is not correct. :?" } { "Tag": [ "geometry", "3D geometry", "MATHCOUNTS", "percent" ], "Problem": "If you know them than answer these questions:\r\nCubes are just like exponents.\r\n1=\r\n2=\r\n3=\r\n4=\r\n5=\r\n6=\r\n7=\r\n8=\r\n9=\r\n10=\r\n11=\r\n12=\r\n13=\r\n14=\r\n15=\r\n16=\r\n17=\r\n18=\r\n19=\r\n20=", "Solution_1": "1\r\n8\r\n27\r\n64\r\n125\r\n216\r\n343\r\n512\r\n729\r\n1000\r\n1331\r\n1728\r\n\r\n\r\nThat's all I know by memory....", "Solution_2": "i know that $ 13^3\\equal{}2197$ and $ 15^3\\equal{}3375$.\r\n\r\nforgot 14... think it might be 2744 or something but im not sure\r\n\r\nmeh too lazy to get my calc", "Solution_3": "1\r\n8\r\n27\r\n64\r\n125\r\n216\r\n343\r\n512\r\n729\r\n1000\r\n1331\r\n1728\r\n2197\r\n2744\r\n3375\r\n4096\r\n4913\r\n5832\r\n6859\r\n8000", "Solution_4": "ya i posted it cuz my math counts teacher made me memorize squares and cubes.", "Solution_5": "I've found that for MathCounts, knowing them up to 12 is sufficient. You usually don't need to know 19^3 or something.", "Solution_6": "meh i memorized squares to 41...\r\n\r\ni need to memorize more...", "Solution_7": "I learned a neat trick from \"How to Calculate Quickly\" by Henry Sticker, and it involves learning how to square numbers ending in 5... with hardly any work at all.\r\n\r\nSuppose you have the number A5 where A is some string of digits.\r\n\r\nThe trick to squaring A5 has just two steps:\r\n1: multiply A(A+1)\r\n2: Concatenate (attach) 25 to the right end of the resultant number\r\nViola :)\r\n\r\nBelow, I use [] to mean concatenation... can't remember if there's standard notation for it!\r\n\r\n$ 15^2\\equal{}1\\cdot(1\\plus{}1)[]25\\equal{}225$\r\n$ 25^2\\equal{}2\\cdot(2\\plus{}1)[]25\\equal{}625$\r\n$ 35^2\\equal{}3\\cdot(3\\plus{}1)[]25\\equal{}1225$\r\n$ 125^2\\equal{}12\\cdot(12\\plus{}1)[]25\\equal{}15625$ etc\r\n\r\nGood exercise to try and prove this little factoid! Good luck!", "Solution_8": "Factorization is the key to why it works. Let A5 be 10n+5. So,\r\n\r\n$ (10n\\plus{}5)^2\\equal{}100n^2\\plus{}100n\\plus{}25\\equal{}100(n^2\\plus{}n)\\plus{}25\\equal{}100n(n\\plus{}1)\\plus{}25$\r\n\r\nNow, it's pretty clear.", "Solution_9": "How about using $ c^3\\equal{}(a\\plus{}b)^3$\r\n\r\n$ c^3\\equal{}a^3\\plus{}b^3\\plus{}3ab(a\\plus{}b)$?\r\n\r\nFor example take 53\r\n\r\n$ 53^3\\equal{}(50\\plus{}3)^3$\r\n$ 53^3\\equal{}125000\\plus{}27\\plus{}450(53)$\r\n$ 53^3\\equal{}125000\\plus{}27\\plus{}23850$\r\n$ 53^3\\equal{}148877$\r\n\r\nSimple....... :lol:", "Solution_10": "To me squares are harder after 30 but i still have to memorize them cubes is much easier to me", "Solution_11": "RightAnnihilator Said: I learned a neat trick from \"How to Calculate Quickly\" by Henry Sticker, and it involves learning how to square numbers ending in 5... with hardly any work at all. \r\n\r\nSuppose you have the number A5 where A is some string of digits. \r\n\r\nThe trick to squaring A5 has just two steps: \r\n1: multiply A(A+1) \r\n2: Concatenate (attach) 25 to the right end of the resultant number \r\nViola \r\n\r\nBelow, I use [] to mean concatenation... can't remember if there's standard notation for it! \r\n\r\n \r\n \r\n \r\n etc \r\n\r\nGood exercise to try and prove this little factoid! Good luck!\r\n\r\nmyruggy89 (myself) says: That is a Factoid Indeed. Good Job on Figuring this Out", "Solution_12": "Another one, which I think is in Henry Sticker's book also, which generalizes the previous one I posted:\r\n\r\nIf AC and AD represent two numbers where C and D are single digits, and C+D=10, then we have:\r\n\r\n$ (AC)\\cdot (AD) \\equal{} A\\cdot (A \\plus{} 1)[]C\\cdot D$\r\n\r\nWhere [] is again the concatenation (attaching) symbol.\r\n\r\nAgain, examples:\r\n24(26)=[u]6[/u]24\r\n92(98)=[u]90[/u]16\r\n123(127)=[u]156[/u]21\r\n\r\nIf you proved the last trick, this one will be nearly as easy! Of course, 5+5 is 10, so this proof would cover the previous one as well...", "Solution_13": "How to square AB quicker:\r\n\r\n$ AB\\equal{}10A\\plus{}B$\r\n\r\n$ AB^2\\equal{}(10A\\plus{}B)^2\\equal{}100A^2\\plus{}20AB\\plus{}B^2$\r\n\r\nAnd that should simplify squares after 41.", "Solution_14": "Well, that's just the expansion of a binomial squared. So technically, you could do this with any number, like\r\n\r\n$ 53^2\\equal{}(40\\plus{}13)^2\\equal{}40^2\\plus{}2(40)(13)\\plus{}13^2$.\r\n\r\nAlthough, separating it the first way would be much easier.", "Solution_15": "Awesome Work Everyone, I will eventually post similar stuf like this to elaborate on this maybe perfect squares but only numbers between 40-100 than harder to see how you do", "Solution_16": "for squaring and multiplying in general, putting some algebra into it makes it easier, for eample\r\n\r\n$ 41^2\\equal{}(40\\plus{}1)^2\\equal{}40^2\\plus{}2(40)(1)\\plus{}1^2\\equal{}1600\\plus{}80\\plus{}1\\equal{}1681.$\r\n\r\nits' not as hard as it looks. another example, using algebra slightly differently is\r\n\r\n$ 102*106\\equal{}(100\\plus{}2)(100\\plus{}6)\\equal{}100^2\\plus{}(2\\plus{}6)(100)\\plus{}12\\equal{}10000\\plus{}800\\plus{}12\\equal{}10812$\r\n$ 92*108\\equal{}(100\\minus{}8)(100\\plus{}8)\\equal{}100^2\\minus{}8^2\\equal{}10000\\minus{}64\\equal{}99936$\r\n$ 52*89\\equal{}(50\\plus{}2)(90\\minus{}1)\\equal{}4500\\plus{}180\\minus{}50\\minus{}2\\equal{}4680\\minus{}52\\equal{}4628$ \r\n\r\ntheres also a concept called the vinculum(i think?), where the digits 6 thru 9 are not really used. for example, 48 would be written as 52, with a line over the 2 to represent 50 - 2. 9801 would be 10000-199, so would be written as 10199 with a line over the 199. it takes some practice to get used to the new arithmetic properties, but it is an interesting alternative to conventional arithmetic.", "Solution_17": "I get what the concept of what you're saying is, but I looked up \"vinculum\" on Wikipedia, and it said that a vinculum is just a bar placed over a mathematical expression (as in repeating decimals), generally meaning that it is a group.\r\n\r\nhttp://en.wikipedia.org/wiki/Vinculum_(symbol)\r\n\r\nA couple of examples of using algebra have been given in some of the previous posts.\r\n\r\nAnd this is not just for squaring, as your examples show. The difference of 2 squares is useful in particular.", "Solution_18": "I got on my first test a 31 out of 60 in 5 minutes. Not bad Right? :blush:", "Solution_19": "On a test of cubes?\r\n\r\nWell, in general, I wouldn't be happy with a 50% in any case (or whatever your percent is).\r\nBut I guess cubes are kinda hard to do in 5 minutes.", "Solution_20": "we did squares and cubes in 5 minutes" } { "Tag": [ "function", "number theory proposed", "number theory" ], "Problem": "For the $t(n)$ function defined for non-negative whole numbers we have:$t(0)=t(1)=0$, $t(2)=1$.\r\nIf $n \\gneqq 2$, then $t(n)$ is the smallest positive whole number that doesnt divide $n$.\r\nLets have :\r\n$T(n)=t(t(t(n)))$.\r\nFind the value of $S$ if :\r\n$S=T(1)+T(2)+T(3)+...+T(2005)+T(2006)$", "Solution_1": "if $2 |n$ then $t(n)=p$ odd prime so $t(t(n))=2$ so $t(t(t(n))=1$.\r\nif $2 \\not |n$ then $t(n)=2$, $t(t(t(n)))=0$. Hence $S=1002$ (as $t(1)=0$)", "Solution_2": "nope thats not right :(" } { "Tag": [ "ARML", "conics", "geometry", "probability", "Science Olympiad", "number theory" ], "Problem": "What have you acccomplished so far this summer? What do you have left to accomplish? (It doesn't only have to be in math.) My summer vacation is more than half over and I want to do more before school starts. \r\n\r\nFor me, \r\n\r\nMy accomplishments: \r\n 1. Reviewed old math competitions\r\n 2. Organized old school papers \r\n 3. Went on vacation and relaxed :) \r\n\r\nugh not much else \r\n\r\nLeft to do:\r\n 1. Start up a sport\r\n 2. Finish summer homework\r\n 3. More math\r\n 3. Practice piano more efficiently\r\n\r\nWhat about you?", "Solution_1": "1) Math\r\n2) Computer class\r\n3) Slept all day\r\n\r\n...actually, not really. I hogged the computer all day because there is nothing better at our house.", "Solution_2": "did arml, slept, did math not really, surfed the internet, read some web comics (life on forbez, girl genius, questionable content), practiced piano and cello kinda, studied earth science, made the usa international earth science olympiad team (stop laughing), got a superior rating slash trophy at M.A.M.A in A division piano solo, went to the Omaha Zoo after doing that, and that was this weekend. have not started on summer reading.", "Solution_3": "ARMLed, pianoed, violined, mathed, and summer homeworked\r\n\r\nUSA International Earth Science Olympiad TEam? :huh: Congratulations!", "Solution_4": "1. Call of Duty 4 and Halo 3. Lots of it. \r\n\r\n2. Summer reading\r\n\r\n3. [Currently on a] Vacation! :pilot: \r\n\r\n4. [Currently doing] CTY online chem course.", "Solution_5": "1. Survived the boyfriend's family on multiple occassions (including the scary aunt)\r\n2. Found, and will move into a nice apartment\r\n3. Getting a kitten on Saturday!\r\n4. Joined a choir as a piano player\r\n5. Started and memorized two small piano pieces to perform for my family when I finally go home again.\r\n6. Surviving work...can't wait till school starts in the fall.\r\n7. Finally went up the CN tower =P\r\n8. Consistently played on the dodgeball team, 2nd/19 teams! Yay!", "Solution_6": "[quote=\"Sunny\"]1. Survived the boyfriend's family on multiple occassions (including the scary aunt)\n\n7. Finally went up the CN tower =P\n[/quote]\r\n\r\n1.) Well that must be fun :P . It's worse though when you're the guy and you have to visit your girlfriend's father :| .\r\n\r\n7.) I went up to the CN tower and sat on the floor glass - it was creepy. Although I did go to Tokyo Tower two days ago and did pretty much the same thing.", "Solution_7": "[quote=\"7h3.D3m0n.117\"][quote=\"Sunny\"]1. Survived the boyfriend's family on multiple occassions (including the scary aunt)\n\n7. Finally went up the CN tower =P\n[/quote]\n\n1.) Well that must be fun :P . It's worse though when you're the guy and you have to visit your girlfriend's father :| .\n\n7.) I went up to the CN tower and sat on the floor glass - it was creepy. Although I did go to Tokyo Tower two days ago and did pretty much the same thing.[/quote]\r\n\r\nHahahaha, yeah...Dad's are scary. =) Not mine though, my boyfriend has it ridiculously easy in that department hahaha. With my family, he suffers the risk of being stuffed to death with food if anything!\r\n\r\nYeah, the glass floor is really disconcerting, especially with everyone jumping on it! I actually thought the elevator ride up would be scary...but it wasn't at all. \r\nOh Tokyo Tower eh? That's pretty cool!", "Solution_8": "umm...\r\n\r\nStuff for fun:\r\n1. went crazy in New York\r\n2. watched Rent on Broadway (that leaves on September 7)\r\n3. ate ginormo Philly cheesesteaks at Reading terminal\r\n4. CTY LANCASTER 08.1!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1\r\n5. made a facebook and got addicted (ohnoes!)\r\n\r\nStuff for academics/preparation for school:\r\n1. CTY LANCASTER 08.1!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1\r\n2. practiced violin nearly 2 hrs a day (thats a lot for me cuz i haven't practiced in a long time)\r\n3. read about 10 uber boring books (Great Expectations, Jane Eyre, etc.)\r\n4. Learned AP Statistics\r\n5. Learned Spanish II (my sis was taking a course, and i was just using her books)\r\n\r\nwoww.. i thought my list of fun would be much longer, but never mind! :D\r\n\r\ni'm proud of myself.", "Solution_9": "went through aops intro to geometry and intro to counting and probability almost done with number theory and i feel really smart although it probably woud have ehlped to study alittle more and wut i have left is to do intro to algebra to enhance my algebra skills and get a start on my java book and if anyone here makes it to thomas jefferson let me know ill probably be there 2 :alien: :starwars: :spam: is actually a type of food", "Solution_10": "went to ARML, GIV Math Camp, volunteered at church summer camp, taking Inter. Alg, doing other math, piano, guzheng, facebook, computer, aim, tv, the usual summer time wasting...", "Solution_11": "1. Picked up golf, can't hit it past 150 yards with the longest club. :( \r\n2. Joined AoPS\r\n3. Practiced flute\r\n4. Finally wrote a 4-page story\r\n5. practiced writing C# a little.. can't get past console applications without my dad's help :( \r\n6. taught my sister how to speak english :lol: \r\n7. nothing in particular", "Solution_12": "i haven't done much...\r\n\r\n1. slept (yay! who hasn't?)\r\n2. went to summer skool (world history is boring)\r\nat summer skool i...\r\n~got eaten, stabbed, and shot at by a crazy guy that likes hitting me on the head with history books\r\n3. ran after a bunch of hyperactive kids at a summer camp (arg i'm tired)\r\n4. taken an english class at sabio (very fun)\r\n5. partied with my friends!!!!! (yay!!)\r\n\r\nhmm i think i did more than that...ahh well can't think :D" } { "Tag": [ "inequalities", "function", "inequalities proposed" ], "Problem": "$A \\ge B$\r\n\r\nhow can I proof that?\r\n\r\nI can proof that finding $C$ such as $A \\ge C$ so i proof that $C >B$?", "Solution_1": "$A-B\\geq 0$. If $A\\geq B>0$, then $\\frac{A}{B}\\geq 1$.", "Solution_2": "err...\r\n\r\nyou dont understand\r\n\r\nA,B are functions\r\n\r\ni just want to know the ways I can proof an inequality.", "Solution_3": "Even $A,\\ B$ are functions, my opinion holds. :wink:" } { "Tag": [ "geometry", "IMO Shortlist", "reflection", "circumcircle" ], "Problem": "Let $\\Gamma$ be a circle and let $d$ be a line such that $\\Gamma$ and $d$ have no common points. Further, let $AB$ be a diameter of the circle $\\Gamma$; assume that this diameter $AB$ is perpendicular to the line $d$, and the point $B$ is nearer to the line $d$ than the point $A$. Let $C$ be an arbitrary point on the circle $\\Gamma$, different from the points $A$ and $B$. Let $D$ be the point of intersection of the lines $AC$ and $d$. One of the two tangents from the point $D$ to the circle $\\Gamma$ touches this circle $\\Gamma$ at a point $E$; hereby, we assume that the points $B$ and $E$ lie in the same halfplane with respect to the line $AC$. Denote by $F$ the point of intersection of the lines $BE$ and $d$. Let the line $AF$ intersect the circle $\\Gamma$ at a point $G$, different from $A$.\r\n\r\nProve that the reflection of the point $G$ in the line $AB$ lies on the line $CF$.", "Solution_1": "If K = AE^GB one can see that B is the orthocenter of triangle AKF. Then KF == d is the polar of P = EG^AB wrt $\\Gamma$ and EG is the polar of D wrt $\\Gamma$. Therefore DG is tangent to $\\Gamma$ at G. It is easy to see that GCDF is cyclic, then 0 & \\text{ in }\\Omega \\end{aligned}\\right.\\]\r\nor, equivalently, of the equation\r\n\\[\\left\\{ \\begin{aligned}-\\Delta u & = (u^{+})^{p}& \\text{ in }\\Omega, \\\\ u & = 0 & \\text{ on }\\partial\\Omega, \\end{aligned}\\right.\\]\r\nbacause of the strong maximum principle.\r\n\r\nHere one should make use of the Mountain-Pass Lemma:\r\n\r\n[b]Mountain-Pass Lemma.[/b] Let $X$ be a Banach space, $I \\in C^{1}(X,\\mathbf R)$. Suppose there exist $x_{1}\\in X \\setminus \\{0\\}$ and $0 < R < \\|x_{1}\\|$ such that\r\n\\[\\max (I(0),I(x_{1})) < \\inf_{\\|x\\|=R}I(x).\\]\r\nNow denote\r\n\\[c : = \\inf_{\\gamma\\in\\Gamma}\\max_{0 \\le t \\le 1}I(\\gamma(t)),\\]\r\nwhere $\\Gamma : = \\{\\gamma \\in C([0,1],X) \\mid \\gamma(0)=0 \\text{ and }\\gamma(1)=x_{1}\\}$.\r\n\r\nThen there exists a sequence $(x_{n})$ (\"Palais-Smale sequence\") in $X$ such that\r\n\\[\\lim I(x_{n}) = c \\text{ and }\\lim \\|I'(x_{n})\\|_{*}= 0.\\]\r\n\r\nThus if $I$ satisfies the conditions of Mountain-Pass Lemma and satisfies (PS)-condition, then $I$ has always a critical point.\r\n\r\nIn our case, $I$ satisfies (PS)-condition for $p$ defined above and satisfies all the conditions of Mountain-Pass Lemma. Hence the system of PDE defined above has a least one solution.", "Solution_3": "my fault, it was of course a mistake: $|DI(u_{n}).u_{n}| \\leq \\|DI(u_{n})\\|_{H^{-1}}.\\|u_{n}\\|_{H^{1}_{0}}$. And one can conclude in my 'proof' the boundedness of $u_{n}$ because $\\|DI(u_{n})\\|_{H^{-1}}$ is bounded (tends towards $0$)\r\n\r\nPS: anyone with a proof of this Mountain pass lemma ? very intuitive in finite dimension, but ?!? I find it beautiful to conclude the problem :)" } { "Tag": [ "AMC", "AIME", "conics", "hyperbola" ], "Problem": "Could someone help me with this problem?\r\nHow many lattice points lie on the hyperbola x^2-y^2=2000^2?\r\n\r\nIt doesn't seem to be a hard problem but I just can't get the right answer.", "Solution_1": "[hide]Try factoring the left side and then finding factors of the right side.[/hide]", "Solution_2": "[hide]I got that (x+y)(x-y)=2^8*5^6 but i cant seem to do anything else. This means that 2000^2 has 63 factors but I am still missing what to do next.[/hide]", "Solution_3": "[hide]It means that one of the factors has to be x+y and one has to be x-y.[/hide]", "Solution_4": "[hide]\n\n(x+y)(x-y)=2^8*5^6\n\n(x+y) and (x-y) could be both positive or both negative.\n\n\n\n\n\n\n\nBoth positive - \n\n(x+y), (x-y) must be of same parity or else x and y would not be integers, so they're both even.\n\n\n\n\n\n\n\n\n\nLet (x+y)=2(a), (x-y)=2(b).\n\n\n\n2(a)*2(b)=2^8*5^6. So ab=2^6*5^6. \n\n\n\nSo (x+y) could be (6+1)(6+1)=49 numbers.\n\n\n\nEach value of (x+y) corresponds to a distinct lattice point, so 49 lattice points.\n\n\n\n\n\n\n\n\n\nBoth negative-\n\nSame as above.\n\n\n\nSo there's 49*2=98 lattice points.\n\n\n\n[/hide]", "Solution_5": "Ahh, I see. Thanks for the explanation. I understand it now." } { "Tag": [ "algebra", "system of equations" ], "Problem": "How many ordered triples $ (x,y,z)$ of integers satisfy the system of equations below?\r\n\r\n\\[ x^2 \\minus{} 3xy \\plus{} 2yz \\minus{} z^2 \\equal{} 31\\]\r\n\\[ \\minus{}x^2\\plus{}6yz\\plus{}2z^2 \\equal{} 44\\]\r\n\\[ x^2 \\plus{} xy \\plus{} 8z \\equal{} 100\\]\r\n\r\n$ \\textbf{(A) } 0 \\qquad \\textbf{(B) } 1 \\qquad \\textbf{(C) } 2 \\qquad \\textbf{(D) } \\text{a finite number greater than } 2 \\qquad \\textbf{(E) } \\text{infinitely many}$", "Solution_1": "Already shown here: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=401136#401136[/url]", "Solution_2": "Sorry, searched through resources for this problem, but couldn't find it. This is solely for that." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "i have found this inequality: consider a,b: integer and 3<=a<=b then we have : a^b>=b^a. I know 2 solutions which don't use logarith.How many solutions you have? :D :P", "Solution_1": "I think I've seen this topic somewhere before :?", "Solution_2": "We can do it by induction.\r\nSuppose $a^b\\geq b^a$ and $b\\geq a$. We need to show that $a^{b+1}\\geq (b+1)^a$, but $a^{b+1}=a\\cdot a^b\\geq ab^{a}$. Thus we obtained new inequlity to prove: $ab^a\\geq (b+1)^a$, but\r\n\\[(b+1)^a=\\sum_{k=0}^{a}C_a^kb^{a-k}<1+\\sum_{k=0}^{a-1}a^kb^{a-k}\\leq 1+\\sum_{k=0}^{a-1} b^a=1+ab^a,\\]\r\nhence result :D\r\nI suspect that there is a nice short proof.", "Solution_3": "we can prove series inequalities :\r\n 3^(1/3)>=4^(1/4)>=...>=n^(1/n) by this simple inequality:\r\n for all a>=3 then a^(a+1)>=(a+1)^a by induction. \r\n from this series inequalities, we get the solution\r\n\r\nI'm bolzano-ykl, I have forgotten my password. :D :lol: is it a nice short solution? Can you give another one?" } { "Tag": [ "calculus", "real analysis", "real analysis unsolved" ], "Problem": "Hello,\r\n I read somewhere that a bounded sequence is convergent iff it is monotonic.Is it correct?But I have a counter-example for this statement.Consider the sequence (-1/2)^n i.e., -1/2, 1/4, -1/8, 1/16....Clearly this is bounded and convergent but not monotonic.", "Solution_1": "No, the statement is not correct, as you have yourself demonstrated.\r\n\r\nMaybe you are thinking of one way to formulate the completeness axiom for real numbers: A sequence of real numbers is convergent if it is monotone and bounded.", "Solution_2": "I like to formulate that particular equivalent to the completeness of the real numbers as the \"monotone sequence alternative\": a nondecreasing sequence either converges or tends to $+\\infty.$ (Of course, you can also say that a nonincreasing sequence either converges or tends to $-\\infty.$)\r\n\r\nThe partial sums of a series with positive terms falls under this alternative - hence the power of comparison tests.\r\n\r\nBut as thealchemist has realized, this is not an \"iff.\" With series, it is certainly possible to have convergent series whose terms vary in sign.", "Solution_3": "Thanks for clarifying.And Kalle,is that the completeness 'axiom' or can it be proved?If I am not wrong,axioms cannot be proved.", "Solution_4": "It is common in calculus or analysis courses to state (as axioms) the properties of a complete ordered field, and then to call those the axioms for the real numbers. There's a set of algebraic axioms which are the axioms for a field (and there are many, many other fields, explored in detail in an algebra course). \r\n\r\nThe use of axioms is a way to generality: any theorem you can prove using only the field axioms would then be true for any field whatsoever. And that's the way the word \"axiom\" is most often used in mathematics - not to mean \"cannot be proved\" but to give a starting place for a class of mathematical objects.\r\n\r\nThen there's a set of order axioms; put that together with the field axioms and you have an ordered field. This is more restrictive, but there are a number of ordered fields, including $\\mathbb{Q}$ (every ordered field must contain a copy of $\\mathbb{Q}$) and some more exotic examples, including non-archimedean ordered fields.\r\n\r\nFinally, there's the axiom of completeness. Most of the textbooks I know state it as \"every nonempty set that is bounded above has a least upper bound.\" Start there, and you can prove as a theorem that \"monotone bounded sequences convege.\" But it's also possible to start with \"monotone bounded sequences converge\" and get back to the least upper bound property as a theorem based on that. I can think of several other places to start.\r\n\r\nFinally: can any of this be proved? That question turns out to be equivalent to asking if the real numbers can be constructed. They can. There are several possible constructions, the most popular being \"Dedekind cuts.\" One fallout from the construction is that one can also prove this: given any two complete ordered fields, then there is an order-preserving isomorphism from one field onto the other. In effect, there is (up to isomorphism) only one complete orderd field. That is, there is only one object that satisfies that list of properties you called the axioms of $\\mathbb{R}.$" } { "Tag": [ "geometry", "trigonometry", "complex numbers", "geometry proposed" ], "Problem": "Let $ ABCD$ be a quadrilateral which has an incircle. The incircle touches sides $ AB, BC, CD, DA$ at $ E,F,G,H$ respectively. Prove that $ AC, BD, EG, FH$ concur.", "Solution_1": "The well-known Newton's theorem. Posted many times before :wink:\n\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=50&t=19549[/url]\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=147923[/url]\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=208680[/url]", "Solution_2": "can you show some proof?", "Solution_3": "Let $ S \\equiv{} EG \\cap HF.$ $ X \\equiv{} BC \\cap AD$ is the pole of line $ HF$ WRT incircle $(I)$ and $ Y \\equiv{} AB \\cap DC$ is the pole of $ EG$ WRT $(I)$ $\\Longrightarrow$ $ XY$ is the polar of $S$ WRT $(I).$ Thus, points $ W \\equiv EF \\cap HG$ and $V \\equiv{} EH \\cap FG$ lie on $XY.$ $ W$ and $V$ are the poles of $DB$ and $AC$ WRT $(I).$ If $ X,Y,V,W$ are collinear, then $ EG,HF,AC,DB$ concur.", "Solution_4": "Actually the proof is quite simple: let's call P the intersection of AC and EG. See that AE = AH ( 1 ), CG = CF ( 2 ), \r\nangle AEP + angle CGP = 180 degs ( 3 ). Applying the sine theorem to the triangles AEP and PCG, then to the triangles AHP and CFP we get that both GE and FH intersect AC at the same point. Similarly we can prove they are concurrent on BD.\r\n\r\nBest regards,\r\nsunken rock", "Solution_5": "[quote=\"Johan Gunardi\"]Let $ ABCD$ be a quadrilateral which has an incircle. The incircle touches sides $ AB, BC, CD, DA$ at $ E,F,G,H$ respectively. Prove that $ AC, BD, EG, FH$ concur.[/quote]\r\nLet $ a,b,c,d,e,f,g,h$ be the corresponding complex numbers to $ A,B,C,D,E,F,G,H$, and consider the complex plane with the same center as the incircle. (Assume wlog $ |e|\\equal{}|f|\\equal{}|g|\\equal{}|h|\\equal{}1$)\r\nThen we easily get:\r\n$ a \\equal{} \\frac{2he}{h\\plus{}e}, b\\equal{} \\frac{2ef}{e\\plus{}f}, c \\equal{} \\frac{2fg}{f\\plus{}g}, d \\equal{} \\frac{2gh}{g\\plus{}h}$.\r\n$ AC,EG,FH$ are concurrent iff:\r\n$ ((a\\minus{}c) \\minus{} (e\\minus{}g))(\\overline{(e\\minus{}g)\\minus{}(f\\minus{}h)}) \\equal{} (\\overline{(a\\minus{}c) \\minus{} (e\\minus{}g)})((e\\minus{}g)\\minus{}(f\\minus{}h))$, which should be true :)\r\n(I don't feel like doing it by hand right now, but notice that $ \\overline a \\equal{} \\frac{2}{h\\plus{}e}, \\overline c \\equal{} \\frac{2}{f\\plus{}g}$... It should be pretty straightforward.. )" } { "Tag": [], "Problem": "The system of equations\r\n\r\n$2x+4y=7$\r\n$5x+cy=\\pi^{\\pi}\\cdot67^{34}\\sqrt{3}$\r\n\r\nhas no solutions. Find the value of $c$.", "Solution_1": "[hide]we need to get y by itself and have x with the same coefficients so that they are parallel. x here is 2/4=5/c, 2c=20, so c=10[/hide]", "Solution_2": "[quote=\"13375P34K43V312\"]The system of equations\n\n$2x+4y=7$\n$5x+cy=\\pi^{\\pi}\\cdot67^{34}\\sqrt{3}$\n\nhas no solutions. Find the value of $c$.[/quote]\r\n[hide=\"a comment\"]If you wanted this to be trickier you shouldn't have made it obvious that there is a smooth solution. YOu should have made the rhs of the second equation something like $c^{3}$ so that the person might hope that they can brute force it...[/hide]\r\nJust a comment...\r\nThe same thing goes for the bonus in you last round of SD III, if you had said an average american male more people would have instantly turned to wikipedia and bam gotten the wrong answer. Just a comment..." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Prove that there exist infinitely many $n\\in N$ such that the largest prime divisor of $n^4+1$ is greater than $2n$.", "Solution_1": "http://www.mathlinks.ro/Forum/viewtopic.php?t=28606\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=75957" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "$ a,b,c\\geq 0$\r\n$ \\sum_{cyc} a\\sqrt{a^2\\plus{}3bc}\\geq 2(ab\\plus{}bc\\plus{}ca)$", "Solution_1": "[quote=\"Bloody-Joker\"]$ a,b,c\\geq 0$\n$ \\sum_{cyc} a\\sqrt {a^2 \\plus{} 3bc}\\geq 2(ab \\plus{} bc \\plus{} ca)$[/quote]\r\nSee here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=33515", "Solution_2": "Thanks a lot arqady :lol:" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $n\\geq 3$ be natural numbers, and let $a_1,\\ a_2,\\ \\cdots,\\ a_n,\\ \\ b_1,\\ b_2,\\ \\cdots,\\ b_n$ be positive numbers such that $a_1+a_2+\\cdots +a_n=1,\\ b_1^2+b_2^2+\\cdots +b_n^2=1.$ Prove that $a_1(b_1+a_2)+a_2(b_2+a_3)+\\cdots +a_n(b_n+a_1)<1.$", "Solution_1": "see http://www.mathlinks.ro/Forum/viewtopic.php?t=39364 .\r\n\r\n dg", "Solution_2": "solution\n$S=\\sum_{i=1}^n a_ib_i+\\sum_{i=1}^{n-1} a_ia_{i+1}+a_na_1$\nobjective is to prove S<1\n\nConsider,$(\\sum_{i=1}^n a_i)^2=1 $\n$\\implies \\sum_{i=1}^n a_i^2+2\\sum_{i\\neq j}^n a_ia_j=1$\n$\\implies \\sum_{i=1}^n a_ia_{i+1} +a_na_1<1/2-\\sum_{i=1}^n a_i^2/2$\nTherefore, $S<\\sum_{i=1}^n a_ib_i+1/2-\\sum_{i=1}^n a_i^2/2$\nNow $\\sum_{i=1}^n (b_i-a_i)^2>0$\n$\\implies \\sum_{i=1}^n a_i^2+\\sum_{i=1}^n b_i^2-2\\sum_{i=1}^n a_ib_i>0$\n$\\sum_{i=1}^n a_ib_i<1/2+\\sum_{i=1}^n a_i^2/2 (because \\sum_{i=1}^n b_i^2=1)$\nplugging the inequality in the inequality derived before\n$S<1/2+\\sum_{i=1}^n a_i^2/2+1/2-\\sum_{i=1}^n a_i^2/2$\n$\\implies S<1$", "Solution_3": "We have\n$\\sum{a_ib_i}+\\sum{a_ia_{i+1}} \\le \\sum{{a_i}^2}+\\sqrt{\\sum{a_i}^2}$....(1)\n\nBy power mean inequality\n$\\sum{{a_i}^2} \\ge \\frac{1}{n}$...(2)\n\n$n \\ge 3$\n$\\Rightarrow n^2-3n+1 >0$\n$\\Rightarrow 1-\\frac{3}{n}+\\frac{1}{n^2} >0$\n$\\Rightarrow (\\sum{{a_i}^2})^2-3\\sum{{a_i}^2}+1 >0$ (Using (2))\n$\\Rightarrow (1-\\sum{{a_i}^2})^2 >\\sum{{a_i}^2}$\n$\\Rightarrow \\sum{{a_i}^2}+\\sqrt{\\sum{{a_i}^2}}<1$....", "Solution_4": "By Cauchy we get\n\n$$\\sum a_i b_i \\le \\sqrt{\\sum a_i^2}.$$\n\nHence it suffices to prove that $\\sqrt{\\sum a_i^2} + \\sum a_ia_{i+1} \\le 1.$\n\nWe have that $\\sum a_i a_{i+1} \\le \\frac12 ((\\sum a_i)^2 - (\\sum a_i^2)) \\le \\frac{(\\sum a_i)^2 - (\\sum a_i^2)}{\\sqrt{(\\sum a_i)^2} + \\sqrt{\\sum a_i^2)}}.$\n\nThis is just $\\sum a_i - \\sqrt{\\sum a_i^2},$ and so we're done.\n\n$\\square$", "Solution_5": "[quote=Kunihiko_Chikaya]Let $n\\geq 3$ be natural numbers, and let $a_1,\\ a_2,\\ \\cdots,\\ a_n,\\ \\ b_1,\\ b_2,\\ \\cdots,\\ b_n$ be positive numbers such that $a_1+a_2+\\cdots +a_n=1,\\ b_1^2+b_2^2+\\cdots +b_n^2=1.$ Prove that $a_1(b_1+a_2)+a_2(b_2+a_3)+\\cdots +a_n(b_n+a_1)<1.$[/quote]\n\nUsing [b]A.M.-G.M. inequality[/b] we have\n\n$$a_1b_1 \\leq \\frac{a_1^2+b_1^2}{2}, \\quad \\quad \\quad a_1b_1 \\leq \\frac{a_1^2+b_1^2}{2}, \\quad \\quad \\quad \\cdots \\quad \\quad \\quad a_nb_n \\leq \\frac{a_n^2+b_n^2}{2} $$\n\nTherefore, \n\n$$a_1(b_1+a_2)+a_2(b_2+a_3)+\\cdots +a_n(b_n+a_1) = \\left(a_1a_2+a_2a_3+a_3a_4 +\\cdots +a_na_1\\right) + \\left(a_1b_1+ a_2b_2+\\cdots +a_nb_n\\right) \\leq \\left(a_1a_2+a_2a_3+a_3a_4 +\\cdots +a_na_1\\right) + \\frac{1}{2}\\left(a_1^2+b_1^2+a_2^2+b_2^2+\\cdots +a_n^2+b_n^2\\right) $$\n\nNow using the fact $ \\quad b_1^2+b_2^2+\\cdots +b_n^2=1$,\n\n$$\\left(a_1a_2+a_2a_3+a_3a_4 +\\cdots +a_na_1\\right) + \\frac{1}{2}\\left(a_1^2+b_1^2+a_2^2+b_2^2+\\cdots +a_n^2+b_n^2\\right) = \\left(a_1a_2+a_2a_3+a_3a_4 +\\cdots +a_na_1\\right)+\\frac{1}{2}\\left(a_1^2+a_2^2+\\cdots +a_n^2\\right) +\\frac{1}{2}$$\n\nSo, it suffices to show that\n\n $$\\left(a_1a_2+a_2a_3+a_3a_4 +\\cdots +a_na_1\\right)+\\frac{1}{2}\\left(a_1^2+a_2^2+\\cdots +a_n^2\\right) +\\frac{1}{2} < 1$$\n\n$$\\Leftrightarrow 2a_1a_2+2a_2a_3+2a_3a_4 +\\cdots +2a_na_1+a_1^2+a_2^2+\\cdots +a_n^2 \\leq 1$$\n\nusing the fact $\\quad a_1+a_2+\\cdots +a_n=1 \\quad $, the last inequality is equivalent to\n\n$$\\Leftrightarrow 2a_1a_2+2a_2a_3+2a_3a_4 +\\cdots +2a_na_1+a_1^2+a_2^2+\\cdots +a_n^2 <(a_1+a_2+a_3+\\cdots+a_n)^2$$\n\n$$\\Leftrightarrow 2a_1a_2+2a_2a_3+2a_3a_4 +\\cdots +2a_na_1+a_1^2+a_2^2+\\cdots +a_n^2 $$\n\n$$< (2a_1a_2+2a_2a_3+2a_3a_4+\\cdots +2a_na_1) + (a_1a_3+a_1a_4 + \\cdots+a_1a_{n-1}) + ( a_2a_4+a_2a_5+a_2a_6+\\cdots+a_2a_n) + (a_3a_5+a_3a_6+\\cdots a_3a_1) + \\cdots + (a_na_2 + a_na_3+a_na_4 +\\cdots+a_na_{n-2}) + a_1^2+a_2^2+\\cdots +a_n^2$$\n\n$$\\Leftrightarrow 0 < (a_1a_3+a_1a_4 + \\cdots+a_1a_{n-1}) + ( a_2a_4+a_2a_5+a_2a_6+\\cdots+a_2a_n) + (a_3a_5+a_3a_6+\\cdots a_3a_1) + \\cdots + (a_na_2 + a_na_3+a_na_4 +\\cdots+a_na_{n-2}) $$\n\nwhich is obviously true." } { "Tag": [ "percent" ], "Problem": "What percent of one hour is five seconds? Round your answer to the nearest tenth.", "Solution_1": "[hide] In $ 1$ hour, there are $ 3600$ seconds. Therefore, $ \\frac{5}{3600}\\equal{}\\frac{1}{720}\\approx\\boxed{.1\\%}$.[/hide]" } { "Tag": [ "linear algebra", "matrix", "algebra", "polynomial" ], "Problem": "If $x$ is $n \\times 1$-matrix on a field, prove that \\[\\det (I_{n}+xx^{t}) = 1+x^{t}x\\]", "Solution_1": "I would say calculate the eigenvalues, that would give you the characteristic polynomial and hence the result :\r\n$1+x^{t}x$", "Solution_2": "A nice solution:\r\n\r\nLet $A = \\left({\\begin{array}{*{20}c}{I_{n}}& x \\\\ 0 & 1 \\\\ \\end{array}}\\right)$ and $B = \\left({\\begin{array}{*{20}c}{I_{n}}&{-x}\\\\{x^{t}}& 1 \\\\ \\end{array}}\\right).$ So $AB = \\left({\\begin{array}{*{20}c}{I_{n}+xx^{t}}& 0 \\\\{x^{t}}& 1 \\\\ \\end{array}}\\right)$ and $BA = \\left({\\begin{array}{*{20}c}{I_{n}}& 0 \\\\{x^{t}}&{xx^{t}+1}\\\\ \\end{array}}\\right)$. \r\nThen $\\det (I_{n}+xx^{t}) = \\det (AB) = \\det (BA) = 1+xx^{t}.$", "Solution_3": "Because of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=87254]this topic[/url] you see that $\\det (I_{n}+xx^{t})=ch_{xx^{t}}(1)=ch_{x^{t}x}(1)=\\det(I_{n}+x^{t}x)=1+x^{t}x$\r\n\r\nP.S.\r\nOf Course the solution you wrote is the hint I gave at that topic", "Solution_4": "[quote=\"Omid Hatami\"]Because of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=87254]this topic[/url] you see that $\\det (I_{n}+xx^{t})=ch_{xx^{t}}(1)=ch_{x^{t}x}(1)=\\det(I_{n}+x^{t}x)=1+x^{t}x$\n\nP.S.\nOf Course the solution you wrote is the hint I gave at that topic[/quote]\r\n\r\nI'm sorry. I do not have any information about your topic.", "Solution_5": "[quote=\"mofidy\"]I'm sorry. I do not have any information about your topic.[/quote]\r\nJust click on the link I have given ([url=http://www.mathlinks.ro/Forum/viewtopic.php?t=87254]this topic[/url])" } { "Tag": [], "Problem": "Firstly, how do you think the difficulty of 2002 and 2003 quarterfinals are compared to 2009 f=ma?\r\n\r\nSecondly, do you think it would be a good idea to go through the quarterfinals to help prepare for the 2010 F=ma? Or do they stress different stuff?", "Solution_1": "well it definitely wouldn't hurt if you have nothing else to do to prepare...I think it would probably be more helpful to just review all the concepts from previous f=ma tests and ingrain in your head the harder to derive and useful equations...since there are only a couple of days left to take the 2010 prelim exam." } { "Tag": [ "geometry", "3D geometry", "sphere", "Hi" ], "Problem": "Chords $AA^{\\prime}$, $BB^{\\prime}$, $CC^{\\prime}$ of a sphere meet at an interior point $P$ but are not contained in a plane. The sphere through $A$, $B$, $C$, $P$ is tangent to the sphere through $A^{\\prime}$, $B^{\\prime}$, $C^{\\prime}$, $P$. Prove that $\\, AA' = BB' = CC'$.", "Solution_1": "Ok!\r\nConstruct plane $ABPA'B'$, then intersection of this plane and big sphere is a circle, so we conclude $AP\\cdot PA'=BP\\cdot PB'$ [b](*)[/b]. Moreover, intersection with small spheres is a two tangent circles. On this figure it is easy to show that $AB||A'B'$. It follows that $AP:BP=A'P:B'P$ [b](**)[/b].\r\nFrom [b](*)[/b] and [b](**)[/b] we conclude $AA'=BB'$ :)", "Solution_2": "Or, you could invert wrt $P$ with power $PA\\cdot PA'$. The two tangent spheres turn into two parallel planes, and it's clear from here.", "Solution_3": "[quote=\"grobber\"]Or, you could invert wrt $P$ with power $PA\\cdot PA'$. The two tangent spheres turn into two parallel planes, and it's clear from here.[/quote]\r\nHmm... In my opinion you use the same arguments...", "Solution_4": "I see that now, but I hadn't read your message when I posted mine. :)", "Solution_5": "A detailed solution:\r\n\r\nConsider the inversion with pole $P$ fixing the sphere. It sends $X$ to $X'$ and vice-versa for all $X\\in\\{A,B,C\\}$. Since the spheres $(PABC),(PA'B'C')$ are tangent, the planes $(A'B'C'),(ABC)$, which are the images of those spheres respectively, must be parallel, and we're done: on the one hand, we have $PA\\cdot PA'=PB\\cdot PB'$, and on the other hand we have $\\frac{PA}{PA'}=\\frac{PB}{PB'}$, so $\\{PA,PA'\\}=\\{PB,PB'\\}\\Rightarrow AA'=PA+PA'=PB+PB'=BB'$, and we do the same to prove $AA'=CC'$.", "Solution_6": "Invert at $P$ with power $\\sqrt{PA \\cdot PA'}$ followed by reflection in $P$. The spheres $\\mathcal{S}$ through $A, B, C, P$ and $\\mathcal{S}'$ through $A', B', C', P$ are mapped to planes parallel to each other and tangent to the original sphere. It follows that $PA \\cdot PA'=PB \\cdot PB'$ and $$\\frac{PA}{PA'}=\\frac{PB}{PB'},$$ so $AA'=BB'$. Similarly, $AA'=CC'$ and we may conclude.", "Solution_7": "Solved with the help of 17 people (mostly with their cameras off and no participation)\n\n[asy]\ndraw(circle((-2,0),1));\ndraw(circle((1,0),2));\npair A,B,C,D,P;\nA=(-2,1);\nB=(-2.455111059789108,-0.8904346821608);\nC=(1,-2);\nD=(1.910222119578216,1.780869364392161);\nP=(-1,0);\ndot(A^^B^^C^^D^^P);\nlabel(\"$A$\",A,N);\nlabel(\"$B$\",B,S);\nlabel(\"$A'$\",C,S);\nlabel(\"$B'$\",(1.910222119578216,1.921869364392161),NW);\nlabel(\"$P$\",(-0.7,0),E);\ndraw(A--B--C--D--B--C--A--B--C--D--A--B--D);\n[/asy]\n\nConsider the cross-section going through the plane defined by the two intersecting lines $AA'$ and $BB'.$ By Power of a Point, we have $$PA~ \\cdot ~PA' = PB~ \\cdot ~PB' ~~(\\clubsuit)~~ \\implies \\frac{PA}{PB}=\\frac{PB'}{PA'} ~~(\\alpha)~~\\implies \\triangle PAB \\sim \\triangle PB'A'.$$ From here, we present two solutions:\nFirst, note that $$\\triangle PAB \\sim \\triangle PB'A' \\implies \\angle PAB = \\angle PA'B' \\qquad (\\heartsuit).$$ Since $A,B,A',B'$ were on the circumference of a sphere and the cross-section of a sphere is a circle, we have that $ABA'B'$ is a cyclic quadrilateral and that $$\\angle PAB = \\angle PB'A' \\qquad (\\spadesuit).$$ Together $\\heartsuit$ and $\\spadesuit$ imply that $$\\angle PA'B' = \\angle PB'A' \\implies PB'=PA' \\stackrel{\\alpha}\\implies PA=PB.$$\n\nOr, we can proceed without angles:\n$$\\triangle PAB \\sim \\triangle PB'A' \\implies AP ~ \\cdot ~ B'P=BP ~\\cdot ~A'P ~~(\\diamondsuit).$$ Adding $\\clubsuit$ and $\\diamondsuit$ and rearranging yields $$PA~ \\cdot ~PA' ~+~AP ~ \\cdot ~ B'P= PB~ \\cdot ~PB' ~+~BP ~\\cdot ~A'P \\implies PA(PA'~+~PB') = BP(PA'~+~PB') \\implies PA = BP \\stackrel{\\alpha}\\implies PB'=PA'.$$\n\nEither way, we obtain $PA'=PB'$ and $PA=PB,$ therefore $$PA+PA'=PB+PB' \\implies AA'=BB'.$$ We can derive $BB'=CC'$ in the exact same manner, concluding our proof.\n\n$\\mathbb{Q.E.D.}$", "Solution_8": "By PoP $|PA|\\cdot |PA'|=|PB|\\cdot |PB'|$ and by homothety $\\frac{|PA|}{|PA'|}=\\frac{|PB|}{|PB'|}.$ Hence $|PA|=|PB|\\implies |AA'|=|BB'|.$ \nAnalogously $|BB'|=|CC'|.$\n", "Solution_9": "[quote=JAnatolGT_00]By PoP $|PA|\\cdot |PA'|=|PB|\\cdot |PB'|$ and by homothety $\\frac{|PA|}{|PA'|}=\\frac{|PB|}{|PB'|}.$ Hence $|PA|=|PB|\\implies |AA'|=|BB'|.$ \nAnalogously $|BB'|=|CC'|.$[/quote]\n\nSeems to be what I did except you just made me realize I could've multiplied the first two equations you wrote down and square-rooted. :roll:", "Solution_10": "[hide=Another solution]Note that inversion $A\\leftrightarrow A'$ wrt $P$ maps sphere $ABCP$ onto plane $A'B'C',$ and sphere $A'B'C'P$ onto $ABC.$ \nHence these two planes are parallel, and conclusion follows easily.[/hide]", "Solution_11": "Do we really need to use inversion?! I guess not. Here another simple approach.\n\n[asy]\ndraw(circle((-2,0),1));\ndraw(circle((1,0),2));\npair A,B,C,D,P;\nA=(-2,1);\nB=(-2.455111059789108,-0.8904346821608);\nC=(1,-2);\nD=(1.910222119578216,1.780869364392161);\nP=(-1,0);\ndot(A^^B^^C^^D^^P);\nlabel(\"$A$\",A,N);\nlabel(\"$B$\",B,S);\nlabel(\"$A'$\",C,S);\nlabel(\"$B'$\",(1.910222119578216,1.921869364392161),NW);\nlabel(\"$P$\",(-0.7,0),E);\ndraw(A--B--C--D--B--C--A--B--C--D--A--B--D);\n[/asy]\n\nDraw a common tangent passing from point $P$ and suppose that the tangent line cuts segments $BA'$ and $AB'$, respectively, at point $M$ and $N$. Then, $\\angle{PB'A'}=\\angle{BAP}=\\angle{BPM}=\\angle{NPB'}=\\angle{PA'B'}=\\angle{ABP}$ from which we get that triangles $ABP$ and $A'B'P$ are isosceles, and have the same base angles. From here we get that $AA'=AP+A'P=BP+B'P=BB'$." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "$ a,b,c \\ge 0$. Prove that :\r\n$ \\sum_{cyclic} \\sqrt []{\\frac {a(b \\plus{} c)}{a^2 \\plus{} bc}} \\ge 2$", "Solution_1": "[quote=\"bigbang195\"]$ a,b,c \\ge 0$. Prove that :\n$ \\sum_{cyclic} \\sqrt []{\\frac {a(b \\plus{} c)}{a^2 \\plus{} bc}} \\ge 2$[/quote]\r\n\r\nIt's old problem of Vasc :)", "Solution_2": "[quote=\"caubetoanhoc94\"][quote=\"bigbang195\"]$ a,b,c \\ge 0$. Prove that :\n$ \\sum_{cyclic} \\sqrt []{\\frac {a(b \\plus{} c)}{a^2 \\plus{} bc}} \\ge 2$[/quote]\n\nIt's old problem of Vasc :)[/quote]\r\n\r\nwhy !!. I don't understand . (em ko hi\u1ec3u, anh l\u00e0m r\u00f5 \u0111i \u1ea1 :wink: )", "Solution_3": "[quote=\"bigbang195\"]$ a,b,c \\ge 0$. Prove that :\n$ \\sum_{cyclic} \\sqrt []{\\frac {a(b \\plus{} c)}{a^2 \\plus{} bc}} \\ge 2$[/quote]\r\nSee here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=87751" } { "Tag": [ "ARML" ], "Problem": "I'm not sure if there might be restrictions or something, since it's so soon after the competition. I wanted to see how many people can solve certain problems, and how much time it takes them.", "Solution_1": "Probably. Although I think ARML will post them up, like they did with Last year's problem, on their site at arml.com.", "Solution_2": "I was thinking of having more discussion here, though, unless arml.com has a forum.", "Solution_3": "I dunno. But if it is allowed, then you should post it on Other US Contests and Programs Forum because i think that is where it is mostly suitable.", "Solution_4": "2005 ARML problems are posted on the ARML website, so you can post any questions you have. Attached are the 2005 Questions as well as the Answers. For a complete index of all the rounds of the 2005 ARML, click [url=http://www.arml.com/2005contest.htm]here.[/url]" } { "Tag": [ "function", "real analysis", "calculus", "calculus computations" ], "Problem": "I'm new to delta-epsilon proofs, and I'm struggling to prove that this statement is true:\r\n$ \\exists \\epsilon > 0, \\forall \\delta > 0$ such that $ \\exists x \\in (0,1), \\exists y \\in (0,1), |x\\minus{}y| < \\delta$ \t implies $ |\\frac{1}{x} \\minus{} \\frac{1}{y}| < \\epsilon$.\r\n\r\nThanks for the help.", "Solution_1": "For good reason. That's not true.\r\n\r\nIn writing it down formally, you mixed up the quantifiers; whatever your $ \\epsilon$, I can choose $ \\delta\\equal{}1$, make $ y$ close to zero and have $ \\left|\\frac1x\\minus{}\\frac1y\\right|$ arbitrarily large.\r\n\r\nOK, we fix that obvious mistake and match the definition of a limit: for any $ \\epsilon>0$, there exists a $ \\delta>0$ such that...\r\nThen this is the definition of uniform continuity, and it's still not true; $ \\frac1x$ is not uniformly continuous on $ (0,1)$. For any $ \\delta$ and any $ \\epsilon$, we can choose $ x$ and $ y$ close to zero so that $ \\left|\\frac1x\\minus{}\\frac1y\\right|>\\epsilon$; $ x\\equal{}\\frac1N$ and $ y\\equal{}\\frac2N$ work for large enough $ N$.", "Solution_2": "[quote=\"perlypython\"]I'm new to delta-epsilon proofs, and I'm struggling to prove that this statement is true:\n$ \\exists \\epsilon > 0, \\forall \\delta > 0$ such that $ \\exists x \\in (0,1), \\exists y \\in (0,1), |x \\minus{} y| < \\delta$ \t implies $ |\\frac {1}{x} \\minus{} \\frac {1}{y}| \\geq \\epsilon$.[/quote]\r\n\r\nI think it should be >= rather than <.", "Solution_3": "Then it's the negation of uniform continuity, and I just gave you a construction.\r\n\r\n[Added in edit]\r\nThe part after \"such that\" needs cleaning up too; it's ungrammatical with both \"exists\" and \"implies\" there. For what you're asking, the full statement should be:\r\nThere exists an $ \\epsilon>0$ such that for any $ \\delta>0$, there exist $ x,y\\in (0,1)$ with $ |x\\minus{}y|<\\delta$ and $ \\left|\\frac1x\\minus{}\\frac1y\\right|\\ge\\epsilon$.\r\n\r\nUniform continuity of a function $ f$ on an interval $ I$ would be:\r\nFor any $ \\epsilon>0$, there exists a $ \\delta>0$ such that for all $ x,y\\in I$ with $ |x\\minus{}y|<\\delta$, $ |f(x)\\minus{}f(y)|<\\epsilon$.\r\n\r\nI strongly recommend using words rather than symbolic quantifiers if you're not clear. It's much easier to see what statements make sense.", "Solution_4": "If we didn't have to prove uniform continuity and just had to show continuity, could we do this if we set $ \\delta \\equal{} xy\\epsilon$:\r\n\r\n$ |\\frac{1}{x} \\minus{}\\frac{1}{y}| \\equal{} |\\frac{y\\minus{}x}{xy}| \\le |\\frac{x\\minus{}y}{xy}| < \\frac{\\delta}{xy} \\equal{} \\epsilon$ ?", "Solution_5": "You don't get to choose both variables before choosing $ \\delta$- that just doesn't make sense.", "Solution_6": "How am I suppose to make it make sense then?", "Solution_7": "What limit are you looking for exactly?" } { "Tag": [], "Problem": "Harry Potter has n mixtures in front of him, arranged in a row. Each mixture has one of 100 different colors (colors have numbers from 0 to 99).\r\n\r\nHe wants to mix all these mixtures together. At each step, he is going to take two mixtures that stand next to each other and mix them together, and put the resulting mixture in their place.\r\n\r\nWhen mixing two mixtures of colors a and b, the resulting mixture will have the color (a+b) mod 100.\r\n\r\nAlso, there will be some smoke in the process. The amount of smoke generated when mixing two mixtures of colors a and b is a*b.\r\n\r\nFind out what is the minimum amount of smoke that Harry can get when mixing all the mixtures together.\r\nInput\r\n\r\nThere will be a number of test cases in the input.\r\n\r\nThe first line of each test case will contain n, the number of mixtures, 1 <= n <= 100.\r\n\r\nThe second line will contain n integers between 0 and 99 - the initial colors of the mixtures.\r\nOutput\r\n\r\nFor each test case, output the minimum amount of smoke.\r\nExample\r\n\r\nInput:\r\n2\r\n18 19\r\n3\r\n40 60 20\r\n\r\nOutput:\r\n342\r\n2400\r\n\r\nIn the second test case, there are two possibilities:\r\n\r\n * first mix 40 and 60 (smoke: 2400), getting 0, then mix 0 and 20 (smoke: 0); total amount of smoke is 2400\r\n * first mix 60 and 20 (smoke: 1200), getting 80, then mix 40 and 80 (smoke: 3200); total amount of smoke is 4400 \r\n\r\nThe first scenario is a much better way to proceed. \r\n\r\nDoes any one have a idea?", "Solution_1": "the source is spoj.. I posted it earlier and the idea is Dynami c Programming..check tripleM solution", "Solution_2": "For the DP, consider dp[a][b], which holds the minimum smoke for mixing the mixtures between index a and index b. Then consider states in order of length, i.e. (b-a)." } { "Tag": [ "calculus", "integration", "trigonometry", "calculus computations" ], "Problem": "Hi, could you please help me on this problem.\r\nIntegrate $ \\int F(x)$\r\nwhere $ F(x) \\equal{} \\sqrt {4 \\minus{} x^2}$", "Solution_1": "Let $ x \\equal{} 2\\sin t$" } { "Tag": [], "Problem": "What is the sum of the first 80 positive odd integers subtracted from the sum of the first 80 positive even integers?", "Solution_1": "[hide] We know that the sum of the first $ n$ positive odd integers is $ n^2$, and that for even integers it is $ n(n\\plus{}1)$. Solving this general case, we have $ n(n\\plus{}1)\\minus{}n^2\\equal{}n^2\\plus{}n\\minus{}n^2\\equal{}n$. Applying this to our specific case, we see that $ n\\equal{}\\boxed{80}$.[/hide]" } { "Tag": [ "integration", "calculus", "calculus computations" ], "Problem": "If $ f(x)$ is continuous and $ |f(x)| \\le 1/(|x|^3\\plus{}1)$ for all $ x$ show that,\r\n$ \\int \\int \\int \\bigtriangledown \\cdot f \\,dx \\equal{} 0$", "Solution_1": "There is some truth to this statement, but strictly speaking, it's false. First, $ \\nabla \\cdot \\vec f$ does not necessarily exist. Second, the equality is true only if $ \\iiint_{\\mathbb R^3}$ is interpreted in a particular way." } { "Tag": [], "Problem": "Found this neat game, currently stuck on lvl. 22.\r\n\r\nhttp://www.newgrounds.com/portal/view/408447", "Solution_1": "OMGOMGOMGOMGOMG I luv this game!!!!!!!\r\n\r\n\r\nMinor note: stuck on lvl 13.", "Solution_2": "yayyy thx i rly love this game :D i didnt know there a sequel.\r\n\r\n[img]http://img66.imageshack.us/img66/1216/logicdc1.jpg[/img]\r\n\r\np.s. lvl 22 and 23 seem to be a bit dificult but once u pass them u storm away in minutes :D" } { "Tag": [ "geometry", "inradius", "area of a triangle", "Heron\\u0027s formula", "geometry unsolved" ], "Problem": "in a triangle ABC, $\\mid AB \\mid$ =2cm, $\\mid AC \\mid$ =4cm and the inscribe circle in ABC has a radius 1cm, find $\\mid BC \\mid$.", "Solution_1": "[quote=\"yrhc@eht\"]in a triangle ABC, $\\mid AB \\mid$ =2cm, $\\mid AC \\mid$ =4cm and the inscribe circle in ABC has a radius 1cm, find $\\mid BC \\mid$.[/quote]\r\n\r\nNational olympiad???\r\nApply Heron's formula for the area of a triangle and the fact that the area equals inradius multiplied by semiperimeter. The rest is computation...\r\n\r\n darij" } { "Tag": [ "AoPS Books" ], "Problem": "AOPS should make a book just full of problems..( and a solutions book) with nothing else.....", "Solution_1": "Well, that's not the real point of a book. What kind of problems should there be, what level, etc, are all questions you can ask.\r\nThere are a lot of questions available on the internet, so you can always find questions.", "Solution_2": "And I'm sure there are already tons of books that are full of just problems and solutions. The great thing about the AoPS books is that they [b]teach[/b] the concept very efficiently and then have problems at the end of each chapter that test if you truly grasped/understood the chapter's concepts.", "Solution_3": "Well said." } { "Tag": [ "LaTeX" ], "Problem": "how do you draw lines in latex? like, as in line AB, the little line above AB... thanks. :D\r\n\r\nEDIT never mind.", "Solution_1": "Use\r\n[code]\\bar{x}[/code]\nor\n[code]\\overline{AB}[/code]\r\nto get $\\bar{x}$ and $\\overline{AB}$." } { "Tag": [ "induction", "inequalities open", "inequalities" ], "Problem": "For any natural numbers $ n\\ge 2$,\r\n\r\nprove that $ (n\\plus{}1)^{n\\minus{}1}(n\\plus{}2)^n > 3^n (n!)^2$", "Solution_1": "Any body has ieda?", "Solution_2": "By induction\n1 step $ n=2 $ it's true\n2 step let for $ n=k $ it's true\n3 step $ n=k+1 $\nwe need to prove $ (k+2)^{k}(k+3)^{k+1}>3^{k+1}((n+1)!)^{2} $\nwe know $ (1+\\frac{2}{k+1})^{k+1}>3 $ by Bernoulli\nmultiply it by $ (k+2)^{k} $\nthen we get $ (k+2)^{k}(k+3)^{k+1}>3\\cdot (k+1)^{k+1}\\cdot (k+2)^{k} $\n$ >3^{k+1}((n+1)!)^{2} $. that's all" } { "Tag": [ "geometry", "AMC", "USA(J)MO", "USAMO", "AMC 10", "AMC 8" ], "Problem": "I'd like to hear some recommendation on how appropriate the AMC 10 is for middle schoolers based on their AMC 8 scores. For example, should student who scored say 15 on the AMC 8 be encouraged to take the AMC 10? Or only those with scores of ~20? \r\n\r\nMy reasoning: The AMC 10 and 12 exams have not been offered ever at any of the high schools in our town (according to the helpful AMC 10 registration coordinator). I managed to get my 8th grade son's middle school to offer the AMC 8 by paying the registration fee, and a number of students took the test. Two scored highly enough (22) that I know they should take the AMC 10, but I'm not sure about the rest. I'm trying to decide whether to lobby the middle school to offer the AMC 10 or to approach one of the high schools instead. The middle school will be more inclined to agree if they have several students who should take it. And though a high school offering might be more broadly appropriate, I have much better contacts at the middle school.\r\n\r\nThanks!", "Solution_1": "I would encourage you to take the AMC 10 as a middle schooler just for the experience. It's different from AMC 8, but it isn't too difficult. I wouldn't know the exact conversions from AMC 8 to AMC 10. Take the AMC 10 because there are also more opportunities from doing well on it.", "Solution_2": "I think letting them take the AMC 10 is an excellent idea. As long as the middle schoolers have a sound knowledge of basic Algebra 1 or even just Pre-algebra, they wouldn't do too badly on the AMC 10. At my town's middle school, we had everybody who received greater than a 20 on the AMC 8 take the AMC 10, but that restriction was only due to the fact there were over 1000 kids at the school and many got 10-20. Theoretically, I think a person who got 15 on the AMC 8 can figure out at least 7 or 8 problems on the AMC 10. Those with more than 20 can range from 6 right to 15+ right.", "Solution_3": "i would have to say it would be a great thing to them get a feel of taking the AMC-10 for the future. Just make sure that they have a solid base on math topics like geometry, algebra. Also just to mention, AMC-8 scores definately don't correlate to AMC-10 scores. For example, i knew 8th grades who got a 24 on AMC-8 and an eight-something on AMC-10, and an 8th grader who got a 22 on AMC-8 and 144 on AMC-10 (actually they got to the USAMO too, but that's a completely different case). Also make sure that the middle schoolers just can't get down on themselves if they don't get it some problems, especially towards the end with the geometry question as I have experienced before as a 7th grader. I wish them luck on the AMC-10!" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let $ a_{1},a_{2},...,a_{n}>0$ such that $ a_{1}a_{2}...a_{n}\\equal{}\\frac{1}{2}$ then find the minimum value of\r\n\\[ \\sum_{cyc}\\sqrt{a_{1}\\plus{}\\frac{1}{a_{1}}}\\]", "Solution_1": "[quote=\"M.A\"]Let $ a_{1},a_{2},...,a_{n} > 0$ such that $ a_{1}a_{2}...a_{n} \\equal{} \\frac {1}{2}$ then find the minimum value of\n\\[ \\sum_{cyc}\\sqrt {a_{1} \\plus{} \\frac {1}{a_{1}}}\n\\]\n[/quote]\r\nIt 'll be solve by Minkopski", "Solution_2": "[quote=\"onlylove_math\"][quote=\"M.A\"]Let $ a_{1},a_{2},...,a_{n} > 0$ such that $ a_{1}a_{2}...a_{n} \\equal{} \\frac {1}{2}$ then find the minimum value of\n\\[ \\sum_{cyc}\\sqrt {a_{1} \\plus{} \\frac {1}{a_{1}}}\n\\]\n[/quote]\nIt 'll be solve by Minkopski[/quote]\r\nDetail it my friend,I only solve it by AMGM or C_S.Help me", "Solution_3": "Follows Minkowski Inequality\\[{ |z_1|+|z_2|+....+|z_n|}\\geq|z_1+z_2+....+z_n|\\] we have\r\n\r\n${{ LHS\\geq\\sqrt{(\\sum{\\sqrt{a_i})^2+(\\sum{\\frac{1}{\\sqrt{a_i}})^2}}} \\geq\\sqrt{n^2\\sqrt[n]{a_1a_2..a_n}+n^2\\frac{1}{\\sqrt[n]{a_1a_2...a_n}}}}=n\\sqrt{\\frac{1}{\\sqrt[n]{2}}+\\sqrt[n]{2}}}$", "Solution_4": "[quote=\"Allnames\"][quote=\"onlylove_math\"][quote=\"M.A\"]Let $ a_{1},a_{2},...,a_{n} > 0$ such that $ a_{1}a_{2}...a_{n} \\equal{} \\frac {1}{2}$ then find the minimum value of\n\\[ P \\equal{} \\sum_{cyc}\\sqrt {a_{1} \\plus{} \\frac {1}{a_{1}}}\n\\]\n[/quote]\nIt 'll be solve by Minkopski[/quote]\nDetail it my friend,I only solve it by AMGM or C_S.Help me[/quote]\r\n$ P \\equal{}\\sum_{cyc}\\sqrt {a_{1} \\plus{} \\frac {1}{a_{1}}} \\ge\\ \\sqrt {(\\sqrt {a_{1}}\\plus{}\\sqrt {a_{2}}\\plus{}...\\plus{}\\sqrt {a_{n}})^2 \\plus{} (\\frac {1}{sqrt {a_{1}}}\\plus{}\\frac {1}{\\sqrt {a_{2}}}\\plus{}...\\plus{}\\frac {1}{\\sqrt {a_{n}}})^2}$\r\n$ P \\ge\\ \\sqrt {n^2\\sqrt[n]{a_{1}a_{2}...a_{n}} \\plus{} \\frac {n^2}{\\sqrt[n]{a_{1}a_{2}...a_{n}}}}$\r\n$ P \\ge\\ n\\sqrt {\\frac {1}{\\sqrt [n]{2}}\\plus{}\\sqrt [n]{2}}$" } { "Tag": [ "calculus", "rotation", "geometry", "geometric transformation", "reflection" ], "Problem": "Consider the region bounded by f(x)=x^2 and g(x)=5x. Find the volume of the solid formed by rotating the region about the line x=2.", "Solution_1": "You know how to do general rotational volumes, right? The problem for this one is that it's self-intersecting . . . when you rotate it, you end up filling in volume you've already filled in. Thus, you have to break it into two seperate pieces and rotate them seperately. I recommend that you reflect the two graphs about the line x = 2 in order to find a nice way of splitting.", "Solution_2": "Actually, on further thought, you might want to take the volume through normal means, and then compute the volume of the overlap seperately.\r\n\r\nMake sure that you consider all the area on one side of the line of rotation -- that is, reflect part (either the bottom or the top). Otherwise its volume will count for negative, which will be bad.\r\n\r\nBy the way, I get a very ugly answer.", "Solution_3": "I understand that it is self-intersecting. I did it, with some decimal approximations (my calc teacher assigned this, but I doubt that she meant it), and I wanted to know if I got it right.", "Solution_4": "Decimal approximations :(\r\n\r\nwell, what did you get?", "Solution_5": "129.949 or so. I lobbed decimals at 4 places, and lobbed the answer at 3" } { "Tag": [ "LaTeX" ], "Problem": "I can't get the \\boxed{whatever i want boxed} to work. The error window pops up and says \"undefined control sequence.\" Is there a package I need to include?", "Solution_1": "\\boxed is an amsmath command so have you got \\usepackage{amsmath} in your document's preamble?", "Solution_2": "That's what I was missing. Thanks." } { "Tag": [ "linear algebra", "matrix", "vector", "linear algebra unsolved" ], "Problem": "Let $A$ an hyperplan of $M_n(K),($where $K$ is $\\Re$ or $C)$,which is stable by product.Show that $I_n \\in A$.", "Solution_1": "It is known that the hyperplane will contain an invertible matrix, thus all the subalgebra generated by this matrix, thus also its inverse (which is in the subalgebra), thus their product, $I_n$.", "Solution_2": "In fact, I think I've managed to prove that for $n>2$ there aren't such hyper- planes. ;)", "Solution_3": "What I said in the first message is total non-sense. But what I said in the second remains true.", "Solution_4": "Could someone please delete this? :blush: :)", "Solution_5": "I have another solution.\r\nSuppose that $ R $ is a hyperplane in $ M_{n}(K) $, which is\r\nstable by product. Then the linear relation which defines $ R $\r\ncan be written as $ tr(XA) = 0 $ , $ X \\in M_{n}(K) $, for some\r\nmatrix $ A $. Then if we take some $ Y \\in R $ then for any $ X\r\n\\in M_{n}(K) $ we have $ XY \\in M_{n}(K) $, so $ tr(XYA) = 0 $.\r\nSo every $ X \\in M_{n}(K) $ satisfies $ tr(X(YA)) = 0 $,\r\ntherefore because $ R $ is a hyperplane, we have that $ XA =\r\n\\lambda A $ for some $ \\lambda \\in K $, for every $ X \\in M_{n}(K)\r\n$. Now if we denote by $ T \\subseteq M_{n}(K) $ the set of all\r\nmatrices $ X $, such that $ XA $ is proportional to $ A $, then it\r\nis a linear subspace of $ M_{n}(K) $. Also it is proper, because $\r\nA \\neq 0 $, so there exists some vector $ v $, such that $ w = Av\r\n\\neq 0 $, therefore if we take some linear operator $ X \\in\r\nM_{n}(K) $, such that $ Xw $ is not proportional to $ w $, then $\r\nXA $ will not be proportional to $ A $. Therefore $ T $ is a\r\nproper linear subspace of $ M_{n}(K) $, which contains a\r\nhyperplane $ R $, therefore it is equal to $ R $. Now, if we take\r\n$ X = I_{n} \\in M_{n}(K) $, then of course $ XA = A $ is\r\nproportional to $ A $, so $ I_{n} \\in T = R $.\r\n\r\nI hope it is correct.", "Solution_6": "Killer PM-ed me to tell me what was wrong with my solution, and I felt like an idiot :). Anyway, I wanted to post the solution along the same lines as my first post, which will hopefully be deleted by a moderator :).\r\n\r\nAssume our hyperplane $\\mathcal H$ does not contain $I_n$. Then every $n\\times n$ matrix $A$ can be written uniquely as $M_A+\\lambda_AI_n$ for a matrix $M_A\\in\\mathcal H$ and $\\lambda_A\\in K$. This means that for all $i\\ne j$, our hyperplane will contain a matrix $A_{ij}$ with some $\\lambda_{ij}$ on the main diagonal and a $1$ in position $(i,j)$. Since $A_{ij}^k\\in\\mathcal H,\\ \\forall k\\in\\mathbb N^*$, we immediately find that for all $i\\ne j$ our hyperplane contains $M_{ij}$, the matrix with $0$ everywhere except for the position $(i,j)$, where it has a $1$, and from here it's easy: we add such matrices to get two different permutation matrices which are inverse to one another, and their product, which is $I_n$, will also belong to $\\mathcal H$, contradiction.\r\n\r\nHopefully, it's Ok this time." } { "Tag": [ "geometry", "trigonometry", "geometric transformation", "homothety", "ratio", "reflection", "geometry unsolved" ], "Problem": "[color=darkblue]Let $ ABC$ is a triangle. Let $ (O;R)$ is a change circle but $ (O;R)$ always pass point $ A$ and center $ O$ of circle lie on $ BC$. Let $ (O;R)$ meet $ AB, AC$ at $ M$ and $ N$ respectively. Let $ H$ is orthocenter of $ AMN$ triangle. Find locus of point $ H$.[/color]", "Solution_1": "Hi is it necessary circle $ (O)$ ? Let $ \\angle BAC\\equal{}\\alpha,d$ be bisector of $ \\angle A$ and $ f_1\\equal{}H_{A}^{2|\\cos\\alpha|}$ is homothecy center $ A$ ratio $ 2|\\cos\\alpha|$ and $ f_2\\equal{}R_d$ is reflection axis $ d$ then $ H\\equal{}f_1\\circ f_2(O)$ so $ \\{H\\}\\equal{}f_1\\circ f_2(BC)$." } { "Tag": [ "geometry", "circumcircle", "cyclic quadrilateral", "geometry unsolved" ], "Problem": "consider ABC is an triangle with acute angles, and o is circumcenter of ABC triangle.\r\nS is the circumcircle of ABO.lines CA and CB hit S in P and Q.prove that CO and PQ are perpendicular.", "Solution_1": "[quote=\"behdad.math.math\"]consider ABC is an triangle with acute angles, and o is circumcenter of ABC triangle.\nS is the circumcircle of ABO.lines CA and CB hit S in P and Q.prove that CO and PQ are perpendicular.[/quote]\r\nIt is true, for every circle with cord the side-segment $ AB$ of $ \\bigtriangleup ABC,$ which intersects its sidelines $ AC,\\ BC,$ at points $ P,\\ Q,$ respectively.\r\n\r\nThis is the well known property of the [b][size=100]Antiparallel lines[/size][/b]. The line segment $ PQ,$ is always parallel to the tangent line of the circumcircle $ (O)$ of $ \\bigtriangleup ABC,$ at vertex $ C$ ( easy to prove from the cyclic quadrilateral $ APQB$ ) and so, we have $ CO\\perp PQ.$\r\n\r\nThe line segment $ PQ$ is called, the antiparallel line of the line segment $ AB,$ with respect to the angle $ \\angle C$ ( = with respect to the rays of the angle $ \\angle C$ ).\r\n\r\nKostas vittas." } { "Tag": [ "function" ], "Problem": "Starting with n = 0\r\n\r\n-1, 0, 1/2, 1/6, ...\r\n\r\nFind the function in terms of n, and then, prove:\r\n\r\nlim f(n) = 0\r\nn-->infinity\r\n\r\nHave fun! Easy once you find the pattern!", "Solution_1": "I can't prove things :( But I think I have the pattern!!!1 \r\n[hide] are you adding 1/2 times the last thing you added to any number n starting with -1 as a(when a is the first term) and 1 being the first add-on(i know there has to be a better way to say this...\nso that is is\n-1,0,1/2,1/6,1/14 and so on...?(i'd put more but it's soon bed time...)[/hide]", "Solution_2": "hmm... I think there is a problem with this, I mean all you did was give us the first four terms, so like I don't really know what kind of pattern your asking for. For instance, those four points define a cubic, so if I felt so inclined, I could just figure out what that was. Or it could possibly be some sort of recursion that we could find a closed formula for I mean it kind of looks like [hide] A_n = A_(n-1) - A_(n-2) x 1/n[/hide] but again I don't really know what your asking for.", "Solution_3": "f(n) = (n-1)/n!\r\n\r\nAs n approaches infinity, f(n) approaches 0.", "Solution_4": "[quote=\"neelnal\"]f(n) = (n-1)/n!\n\nAs n approaches infinity, f(n) approaches 0.[/quote]\r\n\r\nThe problem with all these \"find the pattern\" problems is that there are lots of different solutions.\r\n\r\nI have a pattern that correctly generates the given terms:\r\n\r\nCriterion 1: The first term will be -1.\r\nCriterion 2: The second term will be 0.\r\nCriterion 3: The third term will be 1/2.\r\nCriterion 4: The fourth term will be 1/6.\r\nCriteiron 5: The subsequent terms will all be 42.", "Solution_5": "I know that there are ways to get around problems like this, but you really just have to go for the most sensible solution.\r\n\r\nIs it...\r\n[hide]$(n-1)/n!$[/hide]" } { "Tag": [ "analytic geometry" ], "Problem": "The number of points with positive rational coordinates selected from the set of points in the xy-plane such that $x+y\\leq 5$, is:\r\n\r\n$\\text{(A)} \\ 9 \\qquad \\text{(B)} \\ 10 \\qquad \\text{(C)} \\ 14 \\qquad \\text{(D)} \\ 15 \\qquad \\text{(E)} \\ \\text{infinite}$", "Solution_1": "$E$, as we can use negative coordinates. :P\r\n\r\n*Edit* I can't read :blush:\r\n\r\n...Though still $E$", "Solution_2": "[hide]\n(B) 10\n\nBased on the conditions: \n$x>0$\n$y>0$\n$x+y\\le15$\n\nThis makes a triangle with 10 lattice points in it. \n\nSo the answer is 10. \n[/hide]", "Solution_3": "[quote=\"Xantos C. Guin\"](B) 10\n\nBased on the conditions: \n$x>0$\n$y>0$\n$x+y\\le15$\n\nThis makes a triangle with 10 lattice points in it. \n\nSo the answer is 10.[/quote]Careful. It said positive rational coordinates. ;)", "Solution_4": "Oh duh I was thinking integers. \r\n\r\n[hide]\n(E) Infinite since there are an infinite number of rational coordinates. \n[/hide][/hide]" } { "Tag": [ "topology", "advanced fields", "advanced fields unsolved" ], "Problem": "Let $X,Y$ be two topological spaces such that $X$ is homeomorphic to a subspace of $Y$, and $Y$ is homeomorphic to a subspace of $X$. Prove that $X$ can be partitioned into two parts $X_1,X_2$, and $Y$ can be partitioned into $Y_1,Y_2$ such that $X_i$ is homeomorphic to $Y_i$ for $i=1,2$.", "Solution_1": "Could you show me an example of the partition about X=R, Y=[0,1]? I found it at least not very trivial, if possible." } { "Tag": [], "Problem": "What is the greatest value of $ x$ such that $ 10x$ is a divisor of the sum of the first twenty natural numbers?", "Solution_1": "The sum of the first twenty natural numbers is $ \\frac{20\\cdot21}2\\equal{}210$. Since $ 10x|210\\implies x|21$, it is easy to see that $ x\\leq\\boxed{21}$." } { "Tag": [ "videos", "probability", "expected value", "Support" ], "Problem": "I just finished the 60 diagnostic problems a couple of hours ago and have been watching the videos. I have watched three so far (Venn Diagrams Part II, Counting part II, and Expected Value part I), but my karma is at 990 (while one would expect it to be at 985). Also, I rated all the videos I watched, and commented on Venn Diagrams II. Interestingly, the ratings \"disappear\" when I refresh the page, my comment doesn't show up, and my karma only goes down :huh:\r\n\r\n$ E(\\text{My karma every time I watch a video})\\equal{}\\minus{}5$ :rotfl:\r\n\r\nOn a serious note: Y'know, fix it.\r\n\r\nA little more background information: Firefox 3.0.4 on Mac OS X 10.5.5. No weird extensions or anything, cookies+javascript enabled, latest flash player, etc, etc.", "Solution_1": "Have you rated any questions?", "Solution_2": "Thanks for the report. We'll look into the video ratings not storing next week.", "Solution_3": "No, I don't believe I've rated any questions. I'll try that and post back here if I see any positive/negative result.\r\n\r\nThanks.\r\n\r\nEdit:\r\nThis is getting weird...\r\nMy karma has dropped to 945, which is odd, because all I've done so far is a few problems and watched a couple of videos...\r\nI left a comment on a video and rated a couple of videos but, as I mentioned before, the ratings/comments disappear when I refresh the page, and my karma only goes down.", "Solution_4": "Define \"a few\". Because your karma goes down by $ 5$ for getting a question.", "Solution_5": "It goes down by 5?", "Solution_6": "after you watch a video, do you click the back button?\r\nif you do, maybe the rating doesn't save if you go back\r\ni usually just click the get problem button and it seems to work fine", "Solution_7": "Yes, I do click the back button, but I can't imagine why that would affect the rating (once I submit a rating, it should be passed to the server-side script that inserts it into their db, so me going back should not change that).\r\n\r\nI think I did 9 problems (my problem count is 69 I think). The thing is, I rated every solution and every video that I watched, so my karma should still be 1000.", "Solution_8": "To repeat:\r\n\r\n[quote=\"levans\"]Thanks for the report. We'll look into the video ratings not storing next week.[/quote]" } { "Tag": [ "floor function", "limit", "real analysis", "real analysis unsolved" ], "Problem": "For $n \\in N$ let $f(n)=\\sum_{i=1}^{n}\\frac{1}{\\lfloor 10^{i}\\pi \\rfloor }$\r\n\r\nClearly $f$ is bounded.\r\nFind \\[\\lfloor \\lim_{n \\rightarrow \\infty}f(n) \\rfloor\\]", "Solution_1": "Er... is there some mistake?\r\n\r\n$\\lfloor 10^{i}\\pi \\rfloor > 3\\cdot 10^{i}$ and hence $\\frac{1}{\\lfloor 10^{i}\\pi\\rfloor}<\\frac{1}{3\\cdot 10^{i}}$.\r\n\r\nso, $\\lim f(n) < \\sum_{i=1}^{\\infty}\\frac{1}{3\\cdot 10^{i}}<1$. Therefore the answer is $0$ since the sum is obviously positive." } { "Tag": [ "linear algebra", "matrix", "geometry", "geometric transformation", "homothety", "linear algebra unsolved" ], "Problem": "Find all $n \\in \\mathbb{N}$ for which there exist square $n \\times n$ matrices $A$ and $B$ such that\r\n$\\mathrm{rank}\\, A+\\mathrm{rank}\\, B\\leq n$ and every square real matrix $X$ which commutes with $A$ and $B$ is of the form $X=\\lambda I,\\,\\lambda \\in \\mathbb{R}.$", "Solution_1": "Each integer n is appropriate:\r\ntake A=diag(0,1,...,n-1) and B=[Bij] with Bij=1 for all i,j.\r\nRank(A)=n-1,rank(B)=1,UA=AU iff U is a diagonal matrix.\r\nIf U is a diagonal matrix then BU=UB implies that U is an homothety (cf. the first row of UB-BU)." } { "Tag": [], "Problem": "There are 3 hats with numbers 1,2, and 3 on them, owned by Mr. Ratteh Relwob. He also owns 3 bowling balls numbered 1,2,3. If Mr. Relwob insists on storing his bowling balls in his hats and does not wish to put a ball in a hat with the same number, how many ways can he do it? (He makes large enough hats to put 3 bowling balls in)\r\n\r\nBilly", "Solution_1": "Can one hat go waste in a condition Billy?", "Solution_2": "Yes, you don't have to use all the hats.", "Solution_3": "Im getting $2\\cdot3 + 2 = 8$ combinations... is that right?", "Solution_4": "8 is indeed the correct answer.", "Solution_5": "[hide=\"Answer\"]Well, the hats may be large enough to put 3 bowling balls in, but he can't put all three in one hat, or else the number will overlap. ;) Let's make a chart:\n\n[code] 1 2 3 \n---------\n B C A\n B A/C\n C A B\n A/C B\n C A/B\n C A/B\nB/C A\nB/C A[/code]\n\nAnd that covers everything. There are three combinations of two, and each combination of two can only go in one hat; but, while that combination of two is in one hat, there are two places for the other ball to go, so we have $6$. Then, there are two places where each ball can go individually, and two combinations will cover that, we our answer is $\\boxed{8}$.[/hide]" } { "Tag": [ "trigonometry", "modular arithmetic", "LaTeX", "calculus", "function" ], "Problem": "hey guys i was just wondering how this is true;\t\r\n\r\n(i^(n-k)-(-i)^(n-k))/(2i)\t\r\n\t=\tsin[1/2(n-k)pi].", "Solution_1": "Let me be sure I got this right...\r\n\r\n$\\frac{i^{n-k}-(-i)^{n-k}}{2i} = \\sin{\\frac12(n-k)\\pi}$\r\n\r\nHmm... I'll need to work on that one", "Solution_2": "yeh thats what it is\r\n\r\nexcept\r\neverything after sin is in brackets\r\n\r\nthe second part should be sin(1/2(n-k)pi)", "Solution_3": "[hide]\nLet $p=n-k$ and try mod 4.\nThis should be helpful because you can simplify $i^p$ based on what $p$ is equivalent in mod 4.\nFor instance, if you know $p=1\\pmod{4}$, $i^p=i$, etc...\n[/hide]", "Solution_4": "[hide=\"Hint\"]Remember that $i^n$ is cyclic - $i^1=i,i^2=-1,i^3=-i,i^4=1$, then $i^5=i$ again.[/hide]", "Solution_5": "[hide]\n\nIt isn't true for all real numbers, but it is true for integers. It's very easy to find a conterexample using your calculator. Try $n-k=.2$ and you'll see for yourself.\n\nHowever I will show it is true for all integers.\n\nFor ease of computations, let $x = n - k$, since $n-k$ is never actually manipulated.\n\n$\\frac{i^x - (-i)^x}{2i} = \\frac{i^x - (i^3)^x}{2i} = \\frac{i^{x+1} - i(i^3)^x}{-2} = \\frac{i^{3x + 1} - i^{x+1}}2 = \\frac{i^{x+1}(i^{2x} - 1)}2 = \\frac{i^{x+1}(i^x-1)(i^x+1)}2$\n\nNow we examine this casewise. \n\nIf $x \\in \\{...,-4,0,4,8,...\\}$ then $i^x = 1$, and the numerator (which includes $i^x - 1$) becomes zero. $\\sin{\\frac12x\\pi} = \\sin{2k\\pi} = 0$ and it is equivilant.\n\nIf $x \\in \\{...,-3,1,5,9,...\\}$ then $i^x = i$, and the numerator becomes $-(i-1)(i+1) = -i^2 + 1 = 2$, and since the denominator equals 2, the whole thing equals 1. $\\sin{\\frac12x\\pi} = 1$ and it is equivilant.\n\nYou cans work out the last two yourself; they're very similar.\n\n[/hide]\r\n\r\nEdit: I should have hid this the first time.", "Solution_6": "Let $x=n-k$\r\n$\\\\\\frac{i^{n-k}-(-i)^{n-k}}{2i}=\\frac{(\\cos\\frac{\\pi}{2}+i\\sin\\frac{\\pi}{2})^x-(\\cos\\frac{\\pi}{2}-i\\sin\\frac{\\pi}{2})^x}{2i}\\\\=\\frac{e^{ix\\frac{\\pi}{2}}-e^{-ix\\frac{\\pi}{2}}}{2i}=\\frac{\\sinh(ix\\frac{\\pi}{2})}{i}=\\sin(x\\frac{\\pi}{2})=\\sin(\\frac{1}{2}\\pi(n-k))$\r\n\r\nQ.E.D\r\n\r\nEDIT: I had several problems with latex, so calculus lover beated me. :(\r\nMasoud Zargar", "Solution_7": "[quote=\"boxedexe\"]Let $x=n-k$\n$\\\\\\frac{i^{n-k}-(-i)^{n-k}}{2i}=\\frac{(\\cos\\frac{\\pi}{2}+i\\sin\\frac{\\pi}{2})^x-(\\cos\\frac{\\pi}{2}-i\\sin\\frac{\\pi}{2})^x}{2i}\\\\=\\frac{e^{ix\\frac{\\pi}{2}}-e^{-ix\\frac{\\pi}{2}}}{2i}=\\frac{\\sinh(ix\\frac{\\pi}{2})}{i}=\\sin(x\\frac{\\pi}{2})=\\sin(\\frac{1}{2}\\pi(n-k))$\n\nQ.E.D\n\nEDIT: I had several problems with latex, so calculus lover beated me. :(\nMasoud Zargar[/quote]\r\n\r\nI'm not sure that hyperbolic functions generally, let alone of a complex argument, are for an intermediate topic.", "Solution_8": "I though someone would say that. :P", "Solution_9": "[quote=\"boxedexe\"]beated[/quote]\r\n\r\n:D", "Solution_10": "ok i understand now :lol: \r\nthank you everybody :)", "Solution_11": "[quote=\"chess64\"][quote=\"boxedexe\"]beated[/quote]\n\n:D[/quote]\r\n\r\nYou're right. beat would be the correct one according to the illogical language of english. (illogical$\\rightarrow$ :()", "Solution_12": "English is illogical*, so English=illogical.\r\n\r\nIllogical = impractical by the fundamental theorem of illogicalness and impracticalness, so English = impractical by transivity.\r\n\r\nImpractical things should not exist, so English should not exist, by transivity again. $\\boxed {}$\r\n\r\n*[size=75]English is illogical because ${\\small \\text {``beated''} \\ne \\text { past participle of ``beat''.}}$ For other examples see the proof that English is illogical in reference [1].[/size]\r\n\r\n[u]References[/u]\r\n1. English is illogical by chess64, published 2007.", "Solution_13": "[i]Dear Sen[/i], here is a particular case of a more general question:\r\n\r\n$z=\\cos x+i\\sin x\\Longrightarrow z ^p=\\cos px+i\\sin px,\\ \\overline z ^p=\\cos px-\\sin px\\Longrightarrow$\r\n$\\sin px=\\frac{z^p-\\overline z^p}{2i}$. For $z: =i\\ (\\ x=\\frac{\\pi}{2},\\ \\overline z =-i\\ )$ and $p: =n-k$ we get:\r\n\r\n$\\boxed {\\ \\sin \\frac{(n-k)\\pi }{2}=\\frac{i^{n-k}-(-i)^{n-k}}{2i}\\ }$" } { "Tag": [ "inequalities", "inequalities solved" ], "Problem": "Let a,b,c are positive real satisfied (ab) 2 +(bc) 2 +(ca) 2 \\geq (abc) 2 .\r\nProve that:\r\n 1/(c 3*(a 2 +b 2 )+1/(b 3*(c 2 +a 2)+1/(a 3*(b 2 +c 2 ) \\geq \\sqrt 3/2", "Solution_1": "Sorry for wrong:\"Prove that: \r\n1/(c 3*(a 2 +b 2 )+1/(b 3*(c 2 +a 2)+1/(a 3*(b 2 +c 2 ) 3/2:\"\r\nProve that:\r\n(ab) 2 /c 3 *(a 2 +b 2 )+(cb) 2 /a 3 *(c 2 +b 2 )+ (ac) 2 /b 3 *(a 2 +c 2) \\geq \\sqrt 3/2", "Solution_2": "From what I have seen, this magazine has beautiful problems. This is one of them. Let 1/a^2=x and so on. Then the inequality becomes sum xsqrt(x)/(y+z)>=sqrt(3)/2. But sum xsqrt(x)/(y+z)=sum x^2/[sqrt(x)(y+z)]. But sum sqrt(x)(y+z)=sum sqrt(y+z)*sqrt(x(y+z))<=sqrt(sum (y+z))*sqrt(sum x(y+z))=2sqrt[(sum x)(sum xy)]<=2sqrt[(sum x)*(sum x)^2/3]<=2sqrt[(sum x)^4/3]=2/sqrt(3)*(sumx)^2 and the solution is finished. I have used that sum x>=1." } { "Tag": [], "Problem": "Can the Ti-89 calculator be used on the PSAT and SAT tests?", "Solution_1": "yes", "Solution_2": "rly? i just took my SAT's this morning. u don't need it. i didn't take a calculator at all.", "Solution_3": "Well, you don't need it. That's not what he was asking though... for the PSAT, SAT Math, and the SAT Math Level 2, you can use a TI-89 Calculator." } { "Tag": [ "geometry", "incenter", "geometric transformation", "homothety", "trapezoid", "analytic geometry", "reflection" ], "Problem": "Let $\\omega$ be incircle of $ABC$. $P$ and $Q$ are on $AB$ and $AC$, such that $PQ$ is parallel to $BC$ and is tangent to $\\omega$. $AB,AC$ touch $\\omega$ at $F,E$. Prove that if $M$ is midpoint of $PQ$, and $T$ is intersection point of $EF$ and $BC$, then $TM$ is tangent to $\\omega$.\r\n[i]By Ali Khezeli[/i]", "Solution_1": "Let $I$ be the incenter and $L$ is the tangency point of $\\omega$ with $BC.$ Now let $AL\\cap \\omega =S.$ Now it is easily seen that $FLES$ is harmonic quadrangle and so $TS$ is tangent to $\\omega.$ (It follows easily from polus and polar) Let $AL\\cap PQ=R.$ Now it is easily seen that a homothety $h$ with center $A$ which maps $\\omega$ to $\\omega_{a}$ also maps $PQ$ to $BC.$ So now we conclude that if $PQ\\cap\\omega =W$ we have $PR=WQ.$ Because of the fact that $PQ\\parallel BC$ it follows that $WL$ is diameter of the incurcle and therefore $\\angle WSR=\\pi/2.$ But if $TS\\cap PQ=M_{1}$ in the right-angle triangle $RSQ$ we have $M_{1}W=M_{1}S$ (because $SM_{1}$ and $M_{1}W$ are tangents to $\\omega$) and from well-known problem we conclude that $M_{1}$ is the midpoint of $RW.$ Thus $M_{1}P=M_{1}R+RP=M_{1}W+WQ=M_{1}Q$ and so $M_{1}=M.$ QED", "Solution_2": "I have a solution with complex numbers!\r\nIf the origin of complex plane is the incenter, and the incircle is the unit circle, we will have:\r\n\r\n$m = \\frac{x(xz+xy-2yz)}{(x-z)(x-y)}$ and $t = \\frac{x(xy+xz-2yz)}{x^{2}-yz}$, \r\n\r\nwhere y = E, z = F and x = K (K is the intersection of the incircle with BC)!\r\n\r\nTo prove that TM is tangent to the incircle is easy!", "Solution_3": "Another \"projective\" solution, a little simpler from the one bilarev gave, should be by considering $N$ the midpoint of $(BC)$, and denote $V \\in BQ \\cap CP$.\r\n\r\nBecause $BCQP$ is a trapezium, we have that $A,M,V,N$ are collinear and $(A,M,V,N)$ form a harmonical division.\r\n\r\nNow because $BCQP$ is a circumscribe quadrilateral from the Newton theorem, we have that $V \\in EF$, thus $T(A,M,V,N)$ is a harmonical division, and now by intersecting it with $AD$ we have that $(AX_{1}YD)$ is a h.div, where $X \\in TM \\cap AD$ and $Y \\in TV \\cap AD$.\r\n\r\nBut because $EF$ is the polar of $A$, we have that $(AX_{2}YD)$ is a h.div, where $X_{2}\\in AD \\cap w$. Thus $X_{1}\\equiv X_{2}$, but now $TX_{2}$ is trivially tangent to $w$, because $AD$ is the polar of $T$, thus the point $X_{2}\\in AD \\cap w$, is the place where the second tangent cuts the circle, thus $TM$ is tangent to $w$.", "Solution_4": "Actually I improved my solution and now I think this is the shortest one:\r\n\r\nJust consider $ X \\in AD \\cap w$ and $ Y \\in AD \\cap EF$, because that $ AD$ is the polar of $ T$ we have that $ TX$ is tangent to $ w$, but we have that $ (AXYD) \\equal{} \\minus{} 1$, so $ B(AXYD) \\equal{} \\minus{} 1$, therefore by intersecting it with $ PQ$, we have that $ (P,M,Q, \\infty) \\equal{} \\minus{} 1$, thus $ M$ is the midpoint of $ (PQ)$., and thus the problem is solved. (I used that $ BY$ cuts $ PQ$ in $ Q$ because of the Newton theorem).", "Solution_5": "[size=130][color=darkred][b]PP ([u]Iran TST 2007[/u]).[/b] Let $\\omega$ be the incircle of the triangle $ABC$. Let $P\\in AB$ and $Q\\in AC$ be two points such that\n\n$PQ\\parallel BC$ and $PQ$ is tangent to the circle $\\omega$. The lines $AB$, $AC$ touch the circle $\\omega$ at $F$, $E$ respectively. Denote \n\nthe middlepoint $M$ of $[PQ]$ and $T\\in EF\\cap BC$. Prove that the line $TM$ is tangent to the circle $\\omega$ ([u]Ali Khezeli[/u])[/color][/size]\n\n[size=115][color=darkblue][b][u]Proof[/u].[/b] Denote $D\\in BC\\cap w$ , $N\\in IM\\cap BC$ and the midpoints $S$ , $V$ of the segments $[AD]$ , $[BC]$ respectively. Observe that \n\n$M\\in AV$ , $IM=IN$ . From the well-known properties $\\boxed{\\ S\\in IV\\ }\\ (1)$ and $\\boxed{\\ TI\\perp AD\\ }\\ (2)$ obtain immediatelly that $MN\\parallel AD$ ,\n\ni.e. $TI\\perp MN$ and the line $TM$ is tangent to the circle $w$ because the lines $TM$ , $TN$ are symmetrically w.r.t the line $TI$ .\n\n$(1)\\blacktriangleright$ Suppose w.l.o.g. $b\\ne c$ . Denote the intersection $L\\in AI\\cap BC$ . Prove easily that $VL=\\frac{a|b-c|}{2(b+c)}$ , $VD=\\frac{|b-c|}{2}$ , \n\n$\\frac{IA}{IL}=\\frac{b+c}{a}$ . Apply the [u][b]Menelaus[/b]' theorem[/u] to the transversal $\\overline{VIS}$ and the triangle $ADL$ , i.e. $\\frac{VD}{VL}\\cdot\\frac{IL}{IA}\\cdot\\frac{SA}{SD}=1$ $\\Longleftrightarrow$ $S\\in IV$ .\n\n$(2)\\blacktriangleright$ The line $EF$ is the polar $\\alpha$ of the point $A$ w.r.t. the circle $w$ and the point $T\\in\\alpha$ $\\implies A$ belongs to the polar $\\tau$ of the point $T$ \n\nw.r.t. the circle $w$ and $D\\in\\tau$ $\\implies$ the polar of $T$ is $\\tau=AD$ $\\implies$ $TI\\perp AD$ . [u][b]Otherwise[/b][/u] prove easily that $TA^{2}-TD^{2}=IA^{2}-ID^{2}$ .\n\n[u]See [b]PP14[/b][/u] from[/color] [url=http://www.artofproblemsolving.com/blog/101045][color=red][b]here[/b][/color][/url][/size]", "Solution_6": "let $AD$ intersect incircle at $S$ and $PQ$ at $D'$ and consider incircle is tangent to $PQ$ at $R$ , we know \r\n\r\nthat $MD'=MR$ and $\\angle RSD=90^{\\circ}$ hence $MD'=MR=MS$ also we have $\\angle ADB=\\angle MD'S=\\angle D'SM$ \r\n\r\nhence $MS$ is tangent to incircle , let $MS\\cap BC=T'$ , polar of $T'$ pass through $A$ so polar of $A$ (EF) \r\n\r\npass through $T'$ hence $T'\\equiv T$", "Solution_7": "E.lopes, how do you prove that the line TM is tangent to the unit circle?", "Solution_8": "[quote=\"Joao Guerreiro\"]E.lopes, how do you prove that the line TM is tangent to the unit circle?[/quote]\r\n\r\nif $z+t.z' = c$ is the equation of TM (where $z'$ is the conjugate of $z$), the foot of the perpendicular, trought the incenter, to TM, will have cordinates $\\frac{c}{2}$. So, we can prove that $|\\frac{c}{2}| = 1$!", "Solution_9": "hoch ghaste tohin be aghaye khezeli ro nadaram valiiiiiiii\r\nin soal akhe chie ke dadan bara TS....\r\nmmidadan marhale 2 laaghal mishod ye karish kard:d", "Solution_10": "Another approach:\n\nBy direct calculation we get $RM\\cdot TD=IR^2$, i.e. $\\triangle RIM \\sim \\triangle DTI$, done.\n\nBest regards,\nsunken rock", "Solution_11": "What is R? \nWhat is D? \nWhat is I?\nYour solution, sunken rock, is very unclear! :(", "Solution_12": "@CDP100: \nSorry, $\\{R\\} \\equiv \\omega \\cap PQ$, $I$ - the center of $\\omega$, $\\{D\\} \\equiv \\omega \\cap BC$.\nFurther, after having seen, by compiling $TD$ and $RM$ that $IR^2=RM\\cdot TD$, the reflection of $D$ in $IT$ coincides with the reflection of $R$ in $IM$, which solves the problem.\n\n[b]Note[/b]: It has been a [b][u]typo[/u][/b], I edited it. Sorry for any inconveniences caused!\n\nBest regards,\nsunken rock", "Solution_13": "Dear Mathlinkers,\n1. let I the center of the incircle, D be the point of contact of the incircle and BC, D' the antipode of D, D\" the second point of intersection of AD with the incircle and D+ the point of intersection of AD and PQ.\n2. It is known that TD\" is tangent to the incircle at D\"\n3. M is the midpoint of D'D+\n4. IM is parallel to AD and perpendicular to D'D\"\n5. By symmetry wrt MI, MD\" is tangent to the incircle at D\"\u2026\nand we are done \nSincerely\nJean-Louis", "Solution_14": "[quote=\"sunken rock\"]@CDP100: \nSorry, $\\{R\\} \\equiv \\omega \\cap PQ$, $I$ - the center of $\\omega$, $\\{D\\} \\equiv \\omega \\cap BC$.\nFurther, after having seen, by compiling $TD$ and $RM$ that $IR^2=RM\\cdot TD$, the reflection of $D$ in $ID$ coincides with the reflection of $R$ in $IM$, which solves the problem.\n\nBest regards,\nsunken rock[/quote]\n\nThank you very much!", "Solution_15": "The four lines $YX$, $YD$, $YE$, $YF$.", "Solution_16": "Let $BQ \\cap CP = X$.\n$X \\in EF$ (by Brianchon on $PQECBF$).\nSince $X \\in AM$ we have $TM$ is tangent to $\\omega$ (by converse of Brianchon on $TBAEQM$).", "Solution_17": "Let $AD$ intersect $TM$ at $U$, $D$ be the $A$-intouch point, and $DG$, where $G$ is the antipode of $D$ on $\\omega$, intersect the line parallel to $BC$ at $Y$. Let the concurrency point of $EF, DG$, and $AN$ (where $N$ is the midpoint of $BC$) be $X$, and let $FE$ intersect $AD$ be $V$.\n\nThen observe $$(A, V; U, D) = (A, X; M, N) = (Y, X; G, D) = (P_{\\infty}, N; AG \\cap BC, D) = -1.$$ As $V$ lies on the polar of $A$, $U \\in \\omega$, and as $AD$ is the polar of $T$, this is equivalent to $U$ being the other tangency point of $T$.", "Solution_18": "Harmonic cookie :) (also probably the most complicated solution with a million intersections)\n\nLet $K=\\overline{AD} \\cap \\omega$, $(K,D;F,E) = -1 = (\\overline{TT} \\cap \\omega,D;F,E)$ so $TK$ is tangent to $\\omega$. Let $D'$ be the andipode of $D$ in $\\omega$ which is the touchpoint of $\\overline{PQ}$ with $\\omega$, also let $N$ be the midpoint of $\\overline{BC}$, $D'' = \\overline{AD'} \\cap \\overline{BC}$, $R=\\overline{DD'} \\cap \\overline{EF}$ and $S=\\overline{KD} \\cap \\overline{EF}$. By incircle concurrency lemma $R$ lies on $\\overline{AN}$.\n[b]Key Claim: [/b] $(A,R;M,N) = -1$.\n[i]Proof: [/i]\n$$(A,R;M,N) \\stackrel{D'}{=} (D'' , D;\\infty , N) = -1$$\nwhere $N$ is midpoint of $\\overline{D' D''}$ because of a well known config. Then note that \n$$(A, R ; \\overline{TK} \\cap \\overline{AN}, N) \\stackrel{T}{=} (A,S; K,D) = -1 = (A,R;M,N)$$\nSo $M \\equiv \\overline{TK} \\cap \\overline{AN} \\implies T-K-M$ and thus $TM$ is tangent to $\\omega$ at $K$. \n\nThis is such a bruh solution", "Solution_19": "Let $N,N'$ be touch point of $\\omega$ with $AB,PQ.$ Let $TR$ tangent to $\\omega$ at $R\\neq N,$ and redefine $M$ as $RT\\cap PQ.$ \nObserve that $NFRE$ is harmonic, so by duality wrt $\\omega$ $$(PQM\\infty )=N'(EFRN)=-1,$$ i.e. $RT$ bisects $PQ$ as desired.", "Solution_20": "[asy]\n/* TAKEN FROM V_ENHANCE AND EDITED */\nsize(10cm);\ndefaultpen(fontsize(9pt));\npair A = dir(130);\npair B = dir(210);\npair C = dir(330);\ndraw(A--B--C--cycle);\n\ndraw(incircle(A, B, C), blue);\npair I = incenter(A, B, C);\n\npair F = foot(I, A, B);\npair E = foot(I, A, C);\npair T = extension(E, F, B, C);\n\ndraw(E--T--B);\n\npair P = extension(A, B, I, I+rotate(-90)*(I-B));\npair Q = extension(A, C, I, I+rotate(90)*(I-C));\ndraw(P--Q);\npair M = midpoint(P--Q);\ndraw(T--M, red);\n\npair D = foot(I, B, C);\ndraw(A--D, dotted);\n\npair X = extension(A, D, M, T);\npair Y = 2*I-D;\npair Sp = extension(A,D,P,Q);\n\ndot(\"$A$\", A, dir(A));\ndot(\"$B$\", B, dir(B));\ndot(\"$C$\", C, dir(C));\ndot(\"$I$\", I, dir(I));\ndot(\"$F$\", F, dir(150));\ndot(\"$E$\", E, dir(45));\ndot(\"$T$\", T, dir(T));\ndot(\"$P$\", P, dir(P));\ndot(\"$Q$\", Q, dir(20));\ndot(\"$M$\", M, dir(90));\ndot(\"$D$\", D, dir(D));\ndot(\"$S$\", X, dir(170));\ndot(\"$R$\", Y, dir(90));\ndot(\"$S'$\", Sp, dir(110));\n\n/* Source generated by TSQ */\n[/asy]\nDefine $D$ the tangency of $\\omega$ with $BC.$ Let line $AD$ intersect $\\omega$ at $S$ and $PQ$ at $S'.$ Let $R$ be the tangency of $PQ$ with $\\omega.$\n\nSince $AF$ and $AE$ are tangents of $\\omega$ at $E$ and $F,$ it follows $FDES$ is a harmonic quadrilateral. By well known properties, this implies $TS$ is tangent to $\\omega$ at $S.$ It suffices to show $MS$ is tangent to $\\omega.$ First, consider the homothety $\\mathcal{H}$ centered at $A$ bringing $\\omega$ to the $A$-excircle. It follows $\\mathcal{H}(S')=D$ and $\\mathcal{H}(R)=R',$ where $R'$ is the tangency point of $BC$ and the $A$-excircle.\n\nBy excircle properties, it follows $BD=CR',$ which implies the midpoint of $DR'$ is also the midpoint of $BC,$ and therefore by the homothety the midpoint of $S'R$ is also the midpoint of $PQ,$ which happens to be $M.$ Thus, $MS'=MR.$ Now notice since $D,I,R$ are collinear it follows $R$ and $D$ are antipodes. Thus $\\angle DSR=90^\\circ.$\n\nNote then $\\angle RSS'=90^\\circ,$ so the circumcircle of $SS'R$ is centered at $M,$ and thus $MS'=MR=MS,$ so $MS$ is tangent to $\\omega,$ and therefore $TM$ is tangent to $\\omega,$ as required.", "Solution_21": "Let $N$ the midpoint of $BC$, and let the tangents from $N$ to $\\omega$ be $D,K$ where $D$ lies on $BC$, let $G$ the point where $PQ$ is tangent to $\\omega$ (so $I,D,G$ are colinear), let $AM$ intersect $\\omega$ at $X,Y$ and let $AD \\cap \\omega=L$. Now since its known that $A,G,K$ are colinear we project and take polars w.r.t. $\\omega$\n$$-1=(X,Y; D, K) \\overset{A}{=} (Y, X; L, G) \\implies \\mathcal P_M=LG \\implies M \\in \\mathcal P_L$$\nBut since $\\mathcal P_L=TL$ we have that $T,L,M$ are colinear hence $TM$ is tangent to $\\omega$ as desired.", "Solution_22": "Let $D$ be the contact point of $\\omega$ with $BC$, and let $R$ be the antipode of $D$ in $\\omega$. Let $S\\neq R$ in $\\omega$ such that $MS$ tangent to $\\omega$. Finally, let $D'=AD\\cap PQ$. \nBy homothety $D'$ is the contact point of the incircle of $\\triangle APQ$ with side $PQ$, and so $MD'=MR=MS$, so $RS\\perp SD'$.\nAlso, $DR$ diameter in $\\omega$, so $RS\\perp SD$, and so $A-D'-S-D$ collinear. Finally, \n$T\\in \\mathcal{P}(A) \\Longrightarrow AD=\\mathcal{P}(T) \\Longrightarrow S\\in \\mathcal{P}(T) \\Longrightarrow T\\in \\mathcal{P}(S)$, \nand so $TM$ is tangent to $\\omega$ at $S$.", "Solution_23": "Let $\\omega$ touch $BC$ at $D$, $AD$ meet $\\omega$ again at $R$ and intersect $PQ$ at $Y$, $PQ$ touch $\\omega$ at $X$, and $AX \\cap BC = X_1$. \n\nSince $DERF$ is harmonic, we know $TR$ is tangent to $\\omega$. Hence, it suffices to show $MR$ is tangent to $\\omega$.\n\nThe Diameter of the Incircle Lemma implies $BD = CX_1$. Now, since $APQ$ and $ABC$ are homothetic, it follows that $MX = MY$.\n\nBecause $DX$ is a diameter of $\\omega$, we have $$\\angle XRY = 180^{\\circ} - \\angle XRD = 90^{\\circ}$$ which means $M$ is the circumcenter of $XRY$. This yields $MR = MX$, as required. $\\blacksquare$\n\n\n[b]Remark:[/b] The solution is quite simple. At first, however, I sort of overcomplicated this problem by adding $AM \\cap DX \\cap EF$ and attempting to use harmonic bundles.", "Solution_24": "Let the incircle be the unit circle, and set $d = -i$. Then, we have\n$$p = \\frac{2fi}{f+i} \\text{ and } q = \\frac{2ei}{e+i},$$so we can compute$$m = \\frac{2ief-e-f}{(e+i)(f+i)}.$$Next, using a Cartesian coordinate system, we can compute$$t = \\frac{f+\\overline f}2 - \\frac{i(e+\\overline e - f - \\overline f)}{e-\\overline e - f + \\overline f}\\left(\\frac{f-\\overline f}{2i} + 1\\right) - i = \\frac{-i(e+i)(f+i)}{ef+1} - i.$$Then, the reflection $d'$ of $D$ over $\\overline{TO}$ is$$d' = \\frac{t \\cdot i}{\\overline t} = \\frac{2ef+i(e+f)}{e+f+2i}$$by the complex reflection formula; we have $\\overline{TD\"}$ is tangent to the unit circle. Then, it suffices to show that $M$ is the tangent intersection of the tangent at $i$ and the tangent at $d'$. In other words, we can simply check that$$\\frac{2i\\left(\\frac{2ef+i(e+f)}{e+f+2i}\\right)}{\\frac{2ef+i(e+f)}{e+f+2i} + i} = \\frac{2i ef-e-f}{(e+i)(f+i)} = m,$$so it follows that $\\overline{TD'}$ passes through $M$, and thus $\\overline{TM}$ is tangent to the incircle.", "Solution_25": "Let $D$ be the $A$-contact point, $D'$ be the tangency point of $\\overline{PQ}$ with $\\omega$, and $D'',X$ respectively be the intersections of $\\overline{AD}$ with $\\overline{PQ}$ and $\\omega$. Then $D'$ and $D''$ are reflections of each other over $M$ and $D'$ and $D$ are reflections of each other over the incenter $I$.\n\nBecause $T$ lies on the polar of $A$ wrt $\\omega$, $A$ lies on the polar of $T$, which passes through $D$ as well. Hence it passes through $X$, so $\\overline{TX}$ is tangent to $\\omega$. It then suffices to show that $\\overline{MX}$ is also tangent to $\\omega$, i.e. $MX=MD'$. Since $\\angle D''XD'=\\angle DXD'=90^\\circ$ and $M$ is the midpoint of $\\overline{D'D''}$, it follows that it is the circumcenter of right triangle $D'XD''$, hence $MX=MD'=MD''$. $\\blacksquare$", "Solution_26": "Take the circle the unit circle. Let the circle touch $BC$ at $D$ .We denote the complex co-ordinate of a point $X$ by $x$ .\nWe set $d=1$ it is clear $PQ$ toiuches the unit circle at $-1$ .By standard formulae $b=\\frac {2f}{1+f}$ and $c=\\frac{2e}{1+e}$ and $p=\\frac{-2f}{f-1}$ and $q=\\frac{-2e}{e-1}$. Note that $ \\Re{(t)} = 1$ and the equation of chord $EF$ is $z=e+f-ef\\overline{z}$ and $t+\\overline{t} = 2 \\implies \\overline t = 2-t$ and putting this in the equation of $EF$ we get $t = \\frac{e+f-2ef}{1-ef}$ and $ m= \\frac{e+f-2ef}{(e-1)(f-1)}$.let $o'$ be the complex co-ordinate of the reflection of origin wrt the line $MT$ $$\\frac{\\overline{m}}{\\overline t - \\overline m } = \\frac{o' - m}{t-m}$$. let the mid-point of $O$ and $O'$ is $P$. then $p=\\frac{o'+0}{2} = -\\frac{e+f-2ef}{e+f-2}$(Stepped the calculation part ) Now note that $ \\overline {p} = -\\frac{e+f-2}{e+f-2ef}$.So , $p\\overline p = 1$ and so $| p| = 1$ and so $P$ lies on the unit circle and so $OP\\perp TM$ and hence we are done .", "Solution_27": "dkaoskoas\n\nhi because\n\n$I$ is the incenter\n$N=AD\\cap \\omega$\n$Y$ is midpoint of $BC$\n$D'=DI\\cap \\omega$\n$X=DI\\cap AM\\cap EF$ (well-known to exist)\n$Z=AD\\cap EF$\n\nSince $T$ lies on the polar of $A$ so $A$ lies on the polar of $T$ hence $AD$ is the polar hence $TN$ is tangent to $\\omega$ hence it suffices to show that $M,N,T$ are collinear. If we can show that $(A,X;M,Y)=-1$ we will be done since we also have $(A,Z;N,D)=-1$ and we could finish by Prism Lemma (aka phantom points).\n\nNow let $\\infty_{BC}$ be the point at infinity along line $BC$ and project onto $\\overline{DID'}$, suppose $A\\to A'$. It suffices to show that $(A',X;D',D)=-1$ which is trivial: realize that $A'\\in (AEIF)$ hence by Shooting Lemma\n\\[IX\\cdot IA'=IE^2=IF^2=ID^2=ID'^2\\]\ndone. Yay!", "Solution_28": "Let $\\omega$ intersect $\\overline{BC}$ and $\\overline{PQ}$ at $D$ and $G$, and define $K = \\overline{AD} \\cap \\omega, L = \\overline{AD} \\cap \\overline{PQ}$. First, we note that $KDFE$ is harmonic, so $\\overline{TK}$ is tangent to $\\omega$. Thus, it suffices to show that $\\overline{MK}$ is tangent to $\\omega$ too.\n\nHowever, since a homothety at $A$ takes $\\overline{PQ}$ to $\\overline{BC}$ and $G$ to the $A$-extouch point, we have $LM = MG$. But $\\overline{GD}$ is a diameter of $\\omega$ because $\\overline{PQ} \\parallel \\overline{BC}$, so $\\overline{GK} \\perp \\overline{LKD}$, and $M$ is the circumcenter of right $\\triangle LKG$. Therefore $MK = MG$, and since $MG$ is tangent to $\\omega$, $MK$ is as well.", "Solution_29": "Let $D$ and $D'$ be the intersections of $\\overline{BC}$ and $\\overline{PQ}$ with $\\omega$, respectively \u2013 let $D''$ be the intersection between $\\overline{AD}$ and $\\overline{PQ}$, and lastly, let $U$ be the second intersection of $\\overline{AD}$ with $\\omega$. Because $DD'$ is a diameter of $\\omega$, $U$ is the foot from $D'$ to $\\overline{AD}$.\n\nWe claim that $\\overline{MU}$ and $\\overline{TU}$ are both tangent to $\\omega$, which implies the problem.\n[list=1]\n[*]To prove that $\\overline{MU}$ is tangent to $\\omega$: a homothety of factor $\\tfrac{1}{2}$ centered at $D'$ sends $D$ to $I$ and sends $D''$ to $M$. Thus, $MI$ is the perpendicular bisector of $\\overline{UD'}$. Since $MD'$ is obviously tangent to $\\omega$, it follows that $MU$ is tangent to $\\omega$ as well.\n[*]To prove that $\\overline{TU}$ is tangent to $\\omega$: let $G$ be the foot from $I$ to $\\overline{AD}$. Note that $AFGIE$ is cyclic with diameter $AI$. Now, radical axis theorem on $(AFGIE)$, $(GID)$ and $\\omega$ means that $T$, $G$ and $I$ are collinear. So, $TI$ is the perpendicular bisector of $\\overline{UD}$, so $\\overline{TU}$ is tangent to $\\omega$.\n[/list]" } { "Tag": [ "modular arithmetic", "number theory", "divisibility tests", "algebra unsolved", "algebra" ], "Problem": "Let $ M$ and $ N$ be two nine-digit positive integers with the property that if ANY one digit of $ M$ is replaced by the digit of $ N$ in the corresponding place (e.g. the \"tens\" digit of $ M$ replaced by the \"tens\" digit of $ N$) then the resulting integer is a multiple of $ 7$.\r\n\r\nProve that any number obtained by replacing a digit of $ N$ by the corresponding digit of $ M$ is also a multiple of $ 7$.\r\n\r\nFind an integer $ d > 9$ such that the above result concerning divisibility by $ 7$ remains true when $ M$ and $ N$ are two $ d\\minus{}$digit positive integers.", "Solution_1": "[quote=\"AndrewTom\"]Let $ M$ and $ N$ be two nine-digit positive integers with the property that if ANY one digit of $ M$ is replaced by the digit of $ N$ in the corresponding place (e.g. the \"tens\" digit of $ M$ replaced by the \"tens\" digit of $ N$) then the resulting integer is a multiple of $ 7$.\n\nProve that any number obtained by replacing a digit of $ N$ by the corresponding digit of $ M$ is also a multiple of $ 7$.\n\nFind an integer $ d > 9$ such that the above result concerning divisibility by $ 7$ remains true when $ M$ and $ N$ are two $ d \\minus{}$digit positive integers.[/quote]\r\n\r\nLet $ d_p(x)$, for $ p\\ge 0$ and $ x\\in\\mathbb N$ be the digit in $ (p\\plus{}1)^{th}$ position from the right in decimal representation of $ x$\r\nLet $ \\Delta_p\\equal{}d_p(N)\\minus{}d_p(M)$\r\n\r\nThe property may be written $ M\\plus{}\\Delta_p10^p\\equal{}0\\pmod 7$ $ \\forall p\\in\\{0,1,...,8\\}$\r\n\r\nFrom this we deduce $ \\Delta_p10^p\\equal{}\\minus{}M\\pmod 7$ and so $ N\\equal{}M\\plus{}\\sum_p\\Delta_p10^p\\equal{}M\\minus{}9M\\pmod 7$ and so $ M\\plus{}N\\equal{}0\\pmod 7$\r\n\r\nSo $ N\\minus{}\\Delta_p10^p\\equal{}0\\pmod 7$ and the property true for $ (M,N)$ is symetrical and true for $ (N,M)$\r\n\r\nAnd this property is obviously true for any $ d\\equal{}2\\pmod 7$ (in order to have $ M\\minus{}dM\\equal{}\\minus{}M\\pmod 7$) and so, for example, for $ d\\equal{}16$" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "$ a\\plus{}b\\plus{}c\\equal{}1$, $ a,b,c$ Real positive. Proof :\r\n$ \\frac{ab\\plus{}c}{a\\plus{}b}\\plus{}\\frac{ac\\plus{}b}{a\\plus{}c}\\plus{}\\frac{bc\\plus{}a}{b\\plus{}c}\\geq 2$", "Solution_1": "The inequality is equivalent to \\[ \\sum\\frac {ab \\plus{} 1\\minus{}a\\minus{}b}{a \\plus{} b}\\ge 2\\iff\\sum\\frac {(1 \\minus{} a)(1 \\minus{} b)}{a \\plus{} b}\\ge 2\\]\r\nSo we need only prove that\r\n\\[ \\sum\\frac {(b \\plus{} c)(c \\plus{} a)}{a \\plus{} b}\\ge 2\r\n\\]\r\nLet $ a \\plus{} b \\equal{} x,b \\plus{} c \\equal{} y,c \\plus{} a \\equal{} z$. Then we have $ x \\plus{} y \\plus{} z \\equal{} 2$ and we need to prove that \r\n\r\n$ \\frac {xy}{z} \\plus{} \\frac {yz}{x} \\plus{} \\frac {zx}{y}\\ge 2\\iff x^2y^2 \\plus{} y^2z^2 \\plus{} z^2x^2\\ge xyz(x \\plus{} y \\plus{} z)$\r\n\r\n$ \\iff \\sum_{cyc}(xy \\minus{} yz)^2\\ge 0$, obviously true.", "Solution_2": "The following stronger one is also true for $ a,b,c > 0$ and $ a \\plus{} b \\plus{} c \\equal{} 1$\r\n\\[ \\frac {ab \\plus{} c}{b \\plus{} c} \\plus{} \\frac {bc \\plus{} a}{c \\plus{} a} \\plus{} \\frac {ca \\plus{} b}{a \\plus{} b}\\ge 2\r\n\\]\r\n\r\nEDIT: This must be an equality, sorry for not noticing it earlier. Then we get another proof of the original inequality by rearrangement.", "Solution_3": "[quote=\"nayel\"]The following stronger one is also true for $ a,b,c > 0$ and $ a \\plus{} b \\plus{} c \\equal{} 1$\n\\[ \\frac {ab \\plus{} c}{b \\plus{} c} \\plus{} \\frac {bc \\plus{} a}{c \\plus{} a} \\plus{} \\frac {ca \\plus{} b}{a \\plus{} b}\\ge 2\n\\]\n[/quote]\r\n\r\n\r\n$ b\\plus{}c\\equal{}x , c\\plus{}a\\equal{}y , a\\plus{}b\\equal{}z$ ----> $ a\\equal{}1\\minus{}x,b\\equal{}1\\minus{}y,c\\equal{}1\\minus{}z$ ----> $ \\frac {ab \\plus{} c}{b \\plus{} c} \\plus{} \\frac {bc \\plus{} a}{c \\plus{} a} \\plus{} \\frac {ca \\plus{} b}{a \\plus{} b}\\equal{}x\\plus{}y\\plus{}z\\equal{}2$", "Solution_4": "Any other solution?", "Solution_5": "There is an other solution ,too:\r\n$ ab \\plus{} c \\equal{} ab \\plus{} c(a \\plus{} b \\plus{} c) \\equal{} (b \\plus{} c)(c \\plus{} a)$\r\nThen continue by the same way of solution no.1 :spam:\r\nPut $ b \\plus{} c \\equal{} x;c \\plus{} a \\equal{} b;a \\plus{} b \\equal{} z$\r\nWe have:\r\n$ \\sum{ab \\plus{} c}{a \\plus{} b} \\equal{} \\frac {xy}{z} \\plus{} \\frac {yz}{x} \\plus{} \\frac {zx}{y}$\r\nWe will prove that: $ \\sum{xy}{z}\\geq x \\plus{} y \\plus{} z(1)$\r\nThat right!\r\nBy AM-GM: \r\n$ \\frac {xy}{z} \\plus{} \\frac {yz}{x}\\geq 2y$\r\n$ \\frac {yz}{x} \\plus{} \\frac {zx}{y}\\geq 2z$\r\n$ \\frac {xy}{z} \\plus{} \\frac {zx}{y}\\geq 2x$\r\n$ \\Rightarrow (1) right$ :D\r\n$ \\Rightarrow \\sum{ab \\plus{} c}{a \\plus{} b}\\geq x \\plus{} y \\plus{} z \\equal{} 2(a \\plus{} b \\plus{} c) \\equal{} 2$\r\nThat inequalities solved" } { "Tag": [ "geometry", "topology" ], "Problem": "What is the main difference between mathematics and physics?\r\n\r\nI would like an answer in simple terms.\r\n\r\nThanks", "Solution_1": "Physics is an attempt to understand the real world using mathematical models.\r\n\r\nThe central difference, then, is that the axioms of physics are assumptions about the real world that might change based on new experimental evidence, whereas the axioms of mathematics are arbitrary and change whenever it is useful for them to change. (Well, perhaps \"change\" is the wrong word. I mean that we can introduce new axioms that produce new systems without regard for the real world - non-Euclidean geometry, for example.)", "Solution_2": "So, a mathematician would not be able to solve physics problems and vice-versa?\r\n\r\nWhy do colleges have a physics department and a math department?\r\n\r\nIf BOTH teach numbers, why not include BOTH into one department?\r\n\r\nI am looking for many views.", "Solution_3": "[quote=\"Interval\"]So, a mathematician would not be able to solve physics problems and vice-versa?[/quote]\n\nI'm not sure you understood what I meant.\n\n[b]Physics is an application of mathematics.[/b] Thinking of them as two separate fields on equal ground with each other ignores the universality of mathematics.\n\nAnything which studies patterns and structure in any kind of logical environment can be called mathematics. All theoretical work in physics is mathematics.\n\nIn fact, one of the most fascinating things I ever learned about the world was that some mathematical systems which appeared to have no corollary in the real world were found later to have usefulness as physical models.\n\nMuch of modern physics - quantum mechanics and so forth - is, in its technical details, heavily mathematical. A mathematician would excel at such theoretical work, and a physicist would have to be a good mathematician.\n\n[quote=\"Interval\"]Why do colleges have a physics department and a math department?[/quote]\n\nThe studies have different focuses (again, I'm not sure you understood what I meant). Physics has the additional focus of [b]experiment.[/b] Physicists have to be scientists, and the main goal of science is to both postdict and predict observable phenomena.\n\nPure mathematics is free of such constraints, and is free to work in any system that mathematics deem interesting and worthy of study.\n\nIn practice, much of the mathematics you will learn at a college level is a prerequisite for much of the physics you will be studying - just because the departments are distinct doesn't mean they have nothing to do with each other, yet again, they are distinct fields of study. \n\n[quote=\"Interval\"]If BOTH teach numbers, why not include BOTH into one department?[/quote]\r\n\r\nYou might as well throw the Business department in there too, since business is about numbers.\r\n\r\nFirst off, as I have said repeatedly, [b]mathematics is not just about numbers.[/b] There are entire fields of mathematics in which concrete numbers hardly ever appear, and mathematicians are instead studying things more bizarre - topology, for example, or set theory. \r\n\r\nSecond, [b]physics only employs numbers as a tool.[/b] Quantifying physical phenomena make it easier to suggest a particular model which produces quantifiable and testable predictions.", "Solution_4": "Your reply is well-understood.\r\n\r\nGood job!", "Solution_5": "I think theres another way that u can look at it.\n\nphysics is a system in which in reality, there are an infinite amount of variables(air resistance, pressure,etc.) therefore, in physics, each problem can only focus on a single principle or law. For example, if trying to calculate the amount of pressure on a hydraulic system through pascals principle(easy physics) then there are a million variables, etc. fluid, area, air pressure, fluid pressure, etc. In mathematics you can create a world that is completely arbitrary. in mathematics, you control everything, including the formulas, equation, etc. though there are limitations, like basic axioms, etc. but youn stil can do anything you want. In physics you are limited my gravity, compressibility of fluids, gas laws, etc.\n\nhope you like my view :surf:", "Solution_6": "mathematics is a tool in which to further the work of physics, (as well as every other science).\n", "Solution_7": "The main difference is that mathematics is fixed and independent of all sciences. That is, if a statement in mathematics is a theorem for us, it will also be a theorem for any alien, even if the alien lives in a universe where the laws of physics are completely different than those of our own universe.", "Solution_8": "Physics makes calculation to answer the question: What and why is happening?\nMathematics learns the question what the calculation itself is and how the calculations are happening. As a result it provides a good calculational systems for the various sites of the reality.\nSo, mathematics is a tool for physics. And that is why there were times when physics has caused the development of mathematics putting up new challenges for it.", "Solution_9": "Physics necessarily uses maths but maths does not requires physics, that's the main difference\n\nthank you ", "Solution_10": "Math is love, physics is life.", "Solution_11": "If math and physics are so related, why does AoPS have ChemWOOT instead of something akin to 'PhysWOOT'?\n", "Solution_12": "They should!!!!!", "Solution_13": "[quote=QuantumMech]If math and physics are so related, why does AoPS have ChemWOOT instead of something akin to 'PhysWOOT'?[/quote]\n\nI think levans said something about where they didn't have they correct people/it takes a long time to develop the initial curriculum.", "Solution_14": "It would be awesome if AOPS had physics classes. Proposed curriculum:\n\nWeek 1: SI Units and Vectors\nWeek 2: Particle and collision mechanics\nWeek 3: Rotational motion\nWeek 4: Fluid mechanics\nWeek 5: Kinetic gas theory\nWeek 6: First and Second Law of thermodynamics\nWeek 7: Electrostatic and Gravitational fields (I grouped them together, since they are so similar.)\nWeek 8: Circuit analysis \nWeek 9: Magnetostatics\nWeek 10: Induction, AC circuits and Faraday`s Law\nWeek 11: Maxwell`s Equations, EM waves, and Optics\nWeek 12: Special Relativity\nWeek 13: Quantum Phenomena\nWeek 14: Special techniques like Lagrangian and Hamiltonians", "Solution_15": "I'm still waiting for PhysWOOT", "Solution_16": "me too :)", "Solution_17": "It would be kinda cool if we wrote the curriculum, kinda like just posting big LaTeX documents on the Physics Forum.", "Solution_18": "physics=awesome\nmath=not as awesome(notice i didn't say math isnt awesome)", "Solution_19": "Someone may say the reverse, you cannot say straightforward that maths is not as awesome....", "Solution_20": "physics=awesome,\nmath=awesome as well!", "Solution_21": "[quote=RPFeynman110]It would be awesome if AOPS had physics classes. Proposed curriculum:\n\nWeek 1: SI Units and Vectors\nWeek 2: Particle and collision mechanics\nWeek 3: Rotational motion\nWeek 4: Fluid mechanics\nWeek 5: Kinetic gas theory\nWeek 6: First and Second Law of thermodynamics\nWeek 7: Electrostatic and Gravitational fields (I grouped them together, since they are so similar.)\nWeek 8: Circuit analysis \nWeek 9: Magnetostatics\nWeek 10: Induction, AC circuits and Faraday`s Law\nWeek 11: Maxwell`s Equations, EM waves, and Optics\nWeek 12: Special Relativity\nWeek 13: Quantum Phenomena\nWeek 14: Special techniques like Lagrangian and Hamiltonians[/quote]\n\nHold on.... You mean to say you will cover all of physics in 14 weeks?! I was also thinking about this and there should be at least 3 courses ~18 weeks each. Mechanics, E&M, Misc topics (thermo, optics, modern, etc.).\n\nProposed Curriculum for mech:\n\nweek 1: Math review (calculus, vectors, dimensions, SI units, etc.)\nweek 2: 1D kinematics\nweek 3: 2D and 3D kinematics\nweek 4: Introduction to Newton's laws\nweek 5: Intermediate newton's laws, friction\nweek 6: Hard problems in Newton's Laws\nweek 6: Work and energy\nweek 7: Problem solving with the conservation of energy\nweek 8: Momentum\nweek 9: Collisions\nweek 10: Rotational kinematics, angular KE\nweek 11: Torque and Angular momentum\nweek 12: Hard problems in rotation\nweek 13: Introduction to gravitation\nweek 14: More gravitation, Kepler's laws\nweek 15: Fluid statics\nweek 16: Fluid Dynamics\nweek 17: Challenging Problems I\nweek 18: Challenging Problems II" } { "Tag": [ "trigonometry" ], "Problem": "Find the value of:\r\n$ A\\equal{}sin\\frac{\\pi}{2^n}$\r\n\r\n$ B\\equal{}cos\\frac{\\pi}{2^n}$\r\n$ n\\in N$", "Solution_1": "Kyoshiro, I can't quite understand your question - whose value we have to find?\r\nCould you explain it to me a little better?", "Solution_2": "He wants closed-form expressions for $ \\sin \\frac{\\pi}{2^n}$ and $ \\cos \\frac{\\pi}{2^n}$\r\n\r\nIt's cleaner for cosine, because you have $ \\cos 2x \\equal{} 2 \\cos ^2 x \\minus{} 1 \\implies \\cos x \\equal{} \\frac{1}{2} \\sqrt{2x\\plus{}2}$, so using that recursively its just:\r\n\r\n$ \\cos \\frac{\\pi}{2^n} \\equal{} \\frac{1}{2} \\sqrt{2 \\plus{} \\sqrt{2\\plus{} \\sqrt{2\\plus{} \\ldots \\sqrt{2 \\plus{} \\sqrt{2}}}}}$, where the total number of surd signs is equal to $ n\\minus{}1$ for $ n\\geq 2$\r\n\r\nThen you can just use pythagorean on that to get the formula for sine..." } { "Tag": [ "AMC", "AIME", "ratio", "geometry" ], "Problem": "Let cevians $AD, BE$, and $CF$ of $\\triangle ABC$ meet at an interior point $P$. If $AB = 3\\cdot FB$ and $AC = 4\\cdot AE$, then compute the value of $\\frac{PA\\cdot PB\\cdot PC}{PD\\cdot PE\\cdot PF}$.", "Solution_1": "I think there are two things that jump out: Ceva's Theorem and calculating ratios by calculating area ratios. No time to solve it right now though.", "Solution_2": "\r\nWhat's a cevian?", "Solution_3": "[hide]By Ceva $\\frac{BD}{DC}=\\frac {1}{6}\n\nNow use Menelaus to finish. (e.g, start by going down along a Cevian)[/hide]", "Solution_4": "dear thazn1,\r\nmay I put this question in the training notes for lower form students in my school?\r\nThanks! :D", "Solution_5": "If they use my solution, make them prove Ceva and Menelaus :D", "Solution_6": "[quote=\"blahblahblah\"]If they use my solution, make them prove Ceva and Menelaus :D[/quote]\r\n\r\nmost junior students in my school don't know these two theorems yet, so they must prove Ceva and menelaus :D", "Solution_7": "That's a lot of green smilies.\r\n\r\n\r\n :D :D :D", "Solution_8": "[quote=\"siuhochung\"]dear thazn1,\nmay I put this question in the training notes for lower form students in my school?\nThanks! :D[/quote]\r\n\r\nHeh, sure. :D My solution actually uses just Ceva, and a lot of area ratios with it (which ends up to be basically the same thing). :)", "Solution_9": "[quote=\"ASquare\"]\nWhat's a cevian?[/quote]\r\n\r\nA segment connecting a vertex of a triangle to a point on the opposite side.", "Solution_10": "[quote=\"ThAzN1\"][quote=\"siuhochung\"]dear thazn1,\nmay I put this question in the training notes for lower form students in my school?\nThanks! :D[/quote]\n\nHeh, sure. :D My solution actually uses just Ceva, and a lot of area ratios with it (which ends up to be basically the same thing). :)[/quote]\r\n\r\nThank you!! :D (green smiles again)" } { "Tag": [ "function", "integration", "calculus", "calculus computations" ], "Problem": "Let $ 0\\le k\\le1.$ Find all continuous functions $ f: [0,1] \\to \\mathbb{R}_ + \\cup \\{ 0\\}$ that satisfy:\r\n\r\n\\begin{eqnarray*}\\int_0^1 {f(u)\\,du} & =& 1\\\\\\int_0^1 {uf(u)\\,du} &=& k\\\\\\int_0^1 {u^2 f(u)\\,du} &=& k^2 .\\end{eqnarray*}", "Solution_1": "The equations imply $ \\int_0^1 (u\\minus{}k)^2f(u)\\ du\\equal{}0$. Since $ f$ continuous and nonnegative, so is $ (u\\minus{}k)^2f(u)$. This requires $ (u\\minus{}k)^2f(u)\\equal{}0$ for all $ u\\in [0,1]$, which means $ f(u)\\equal{}0$ for all $ u\\in [0,1]$ (except possibly at $ k$, but then continuity shows that $ f(k)\\equal{}0$ as well). But that would make $ \\int_0^1 f(u)\\ du\\equal{}0$. So there is no solution." } { "Tag": [ "integration", "calculus", "logarithms", "limit", "real analysis", "real analysis unsolved" ], "Problem": "$a(n)=\\int_{0}^{1}\\frac{1}{\\sqrt{1+(nx^{n-1})^2}}dx$ \r\n\r\nasymptotic expansion with two terms when n tends +oo ?", "Solution_1": "Observe, first of all, that $\\int_1^\\infty \\frac 1{\\sqrt{1+(nx^{n-1})^2}}\\,dx\\le \\int_1^{\\infty}\\frac 1{nx^{n-1}}\\,dx=\\frac 1{n(n-2)}=O(n^{-2})$. So, we can expand the interval of integration to $[0,+\\infty)$ if we can neglect terms of order $\\frac 1{n^2}$. Let $x_n=n^{-\\frac 1{n-1}}=1-\\frac{\\ln n}{n}+O(n^{-2}\\ln^2 n)$. Making change of variable $x=x_ny$, we can write the integral as $x_n\\int_{0}^{\\infty}\\frac 1{\\sqrt{1+y^{2n-2}}}\\,dy$. Now substitute $y=e^{\\frac t{n-1}}$. We get $\\int_{0}^{\\infty}\\frac 1{\\sqrt{1+y^{2n-2}}}\\,dy=\\frac 1{n-1} \\int_{-\\infty}^\\infty \\frac{e^{t/(n-1)}}{\\sqrt{1+e^{2t}}}\\,dt=1+\\frac1{n-1}\\int_{-\\infty}^\\infty e^{t/(n-1)}g(t)\\,dt$ where $g(t)=\\left\\{\\begin{aligned} \\frac 1{\\sqrt{1+e^{2t}}},&\\quad t\\ge 0\\\\ \\frac 1{\\sqrt{1+e^{2t}}}-1,&\\quad t<0\\end{aligned}\\right.$. Since $g(t)=O(e^{-|t|})$, we see that the last integral can be written as a sum of inverse powers of $n-1$ with some coefficients (just decompose $e^{t/(n-1)}$ into its Taylor series). We are interested in the free term $c$ of that series, which is $c=\\int_{-\\infty}^{\\infty}g(t)\\,dt$. If we find it, we can conclude that the original integral is $\\left(1-\\frac{\\ln n}{n}+O(n^{-2}\\ln^2n\\right)\\left(1+\\frac cn+O(n^{-2})\\right)=1-\\frac{\\ln n}n+\\frac cn +O(n^{-2}\\ln^2 n)$. Now, making change of variable $y=e^t$, we see that the integral expressing $c$ can be written as $\\lim_{\\delta\\to 0+}\\int_\\delta^\\infty \\frac 1{y\\sqrt{1+y^2}}\\,dy-\\ln \\frac 1\\delta$. The first integral can be computed using the substitution $z=\\sqrt{y^2+1}$ and we get $c=\\ln 2$. Thus $\\int_0^1 \\frac 1{\\sqrt{1+(nx^{n-1})^2}}\\,dx=1-\\frac{\\ln n}n+\\frac{\\ln 2}n +O(n^{-2}\\ln^2 n)$. (I hope I haven't made a stupid mistake somewhere... :) )" } { "Tag": [ "ratio", "number theory", "relatively prime" ], "Problem": "In a telephone directory for a certain city in South Dakota, there are 213 people named Smith and 327 people named Johnson. What is the ratio of the number of people named Smith to the number of people named Johnson? Express your answer as a common fraction.", "Solution_1": "Factor out a 3 so you get\r\n$ \\frac{71}{109}$\r\nand both these numbers are prime.", "Solution_2": "[quote=\"AceOfDiamonds\"]Factor out a 3 so you get\n$ \\frac{71}{109}$\nand both these numbers are prime.[/quote]\nThey don't have to be both prime, they just have to be relatively prime meaning they have no common factors other than 1." } { "Tag": [ "AMC" ], "Problem": "if:\r\n2a+b+c+d+e=5\r\na+2b+c+d+e=15\r\na+b+2c+d+e=30\r\na+b+c+2d+e=20\r\na+b+c+d+2e=50\r\n\r\nFind abcde.", "Solution_1": "darkquantum wrote:if:\n2a+b+c+d+e=5\na+2b+c+d+e=15\na+b+2c+d+e=30\na+b+c+2d+e=20\na+b+c+d+2e=50\n\nFind abcde.\n\n[hide]\n\n6(a+b+c+d+e) = 120\n\n(a+b+c+d+e) = 20\n\n\n\nso... \n\nsubtrcting that from each of the equations we find:\n\na = -15\n\nb = -5\n\nc = 10\n\nd = 0\n\nfrom here you know the product is going to be ... 0[/hide]", "Solution_2": "yea\r\nactaully\r\nu don't even need to solve for a,b,c,e\r\n\r\nall u need to see is that a+b+c+d+e=a+b+c+2d+e.\r\nThis means d=0.\r\nSo ur done", "Solution_3": "oh. didn't see that. thanks." } { "Tag": [ "search", "combinatorics unsolved", "combinatorics" ], "Problem": "Find a closed form of the following sum\r\n\\[ f(i, j) \\equal{} \\sum_{k \\equal{} 0}^i \\binom{2 i \\minus{} 2k}{i \\minus{} k} \\binom{k \\plus{} j}{k}.\r\n\\]", "Solution_1": "i dont think it has a good close form but we know that:\r\n$ f(i,j)$ = cofficients of $ x^i$ in $ ( \\frac {1}{1\\minus{}x})^{1 \\plus{} j}\\frac {1}{\\sqrt{1 \\minus{} 4x}}$\r\n\r\ndo we know $ f(i,1)$ in close form?", "Solution_2": "Similar problems were posted by me here:\r\nhttp://www.mathlinks.ro/viewtopic.php?search_id=1368030362&t=171763\r\nProbably that approach works here either.", "Solution_3": "Even $ f(n,0) \\equal{} \\sum_{k \\equal{} 0}^n \\binom{2k}{k}$ doesn't have a nice closed form, at least as far as the [url=http://www.research.att.com/~njas/sequences]OEIS[/url] knows (it's sequence [url=http://www.research.att.com/~njas/sequences/A006134]A006134[/url]). $ f(n,1)$ isn't listed in the OEIS." } { "Tag": [ "linear algebra", "matrix", "superior algebra" ], "Problem": "Let $A,B \\in M_n \\left( \\mathbb{C} \\right)$ matrixs, such that $A^3 = A^2 $ and $A + B = I_n $\r\n\r\nProve that $\\exists X \\in M_n \\left( \\mathbb{C} \\right)$ such that: $X \\cdot \\left( {AB + I_n } \\right) = \\left( {AB + I_n } \\right) \\cdot X = I_n $", "Solution_1": "We have to show that $I_n+A-A^2$ is invertible, and this follows from the fact that $(A-A^2)^2=O_n$, so $(I_n+A-A^2)(I_n-A+A^2)=I_n-(A-A^2)^2=I_n$.", "Solution_2": "Metru, this post should be in the [url=http://www.mathlinks.ro/Forum/index.php?f=8]Superior Algebra[/url] forums. I've moved it there, and please post all college-related problems there. Thanks!" } { "Tag": [ "geometry", "Euler", "Olimpiada de matematicas" ], "Problem": ":D ya que mi novia no entra al msn colgare problemas, que tal?\r\n\r\n[color=blue][b]Problema 1. [/b]Dado el triangulo $ABC$ recto en $C,$ sea $D$ un punto en $AB$ tal que los triangulo $ACD$ y $DCB$ tienen el mismo inradio, demostrar que $[ABC]=CD^2$ ($[ABC]$ significa area dle triangulo $ABC$)\n\n[b]Problema 2.[/b] Dado un triangulo, demostrar que sus puntos de Fermat, el centro de su circunferencia de Euler y el circuncentro estan en una circunferencia. [/color]\r\n\r\n(vean en:\r\n$\\square$ [url]http://http://centros5.pntic.mec.es/~marque12/matem/fermat.htm[/url], \r\n$\\square$ http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=fermat+%5C%5Ctriangle&t=47955, \r\n $\\square$ [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=triangles+constructed++the+outside+%28on+the+inside&t=26139]www.artofproblemsolving.com/Forum/viewtopic.php?highlight=triangles+constructed++the+outside+%28on+the+inside&t=26139[/url])\r\n\r\n\r\n[color=blue][b]Problema 3. [/b]En un trapecio se conoce que la suma de ambas diagonales es 4 y que el \u00e1rea es 2. Seg\u00fan lo anterior se pide calcular la altura.\n\n[b]Problema 4.[/b] Sea $ABCDE$ un pent\u00e1gono convexo de \u00e1rea $1$ (las diagonales quedan dentro del pent\u00e1gono). Sean $P,$ $Q,$ $R$ y $S$ los baricentros de los tri\u00e1ngulos $ABE,$ $BCE,$ $CDE$ y $DAE,$ respectivamente. Indicar la naturaleza del cuadril\u00e1tero $PQRS$ y dar su \u00e1rea.\n\n[b]Problema 5.[/b] La circunferencia inscrita en $\\triangle ABC$ tiene centro $O$ y es tangente a los lados $BC,$ $AC$ y $AB$ en los puntos $X,$ $Y$ y $Z,$ respectivamente. Las rectas $BO$ y $CO$ intersecan a la recta $YZ$ en los puntos $P$ y $Q,$ respectivamente. Si $XP$ y $XQ$ tienen la misma longitud, Indicar la naturaleza del triangulo $ABC.$ \n\n[b]Problema 6.[/b] Los v\u00e9rtices de un triangulo $ABC$ pertenecen a la hip\u00e9rbola $xy=1.$ Demostrar que el ortocentro tambi\u00e9n pertenece a la hip\u00e9rbola.[/color]\r\n\r\n(Sug. Usar tangentes)\r\n\r\n[color=blue][b]Problema 7.[/b] En $ABC,$ sean $AD,BE,CF$ cevianas concurrentes en $M.$ Si los tri\u00e1ngulos $BDM,$ $CME$ y $AMF$ tienen la misma \u00e1rea y el mismo per\u00edmetro, entonces $ABC$ es equil\u00e1tero.[/color]\r\n\r\n\r\n[color=red][b]Problema 8.[/b] (Dedicado a Tipe) Los n\u00fameros enteros del 1 al 2002, ambos inclusive, se escriben en una pizarra en orden creciente 1, 2, . . . , 2001, 2002. Luego, se borran los que ocupan el primer lugar, cuarto lugar, s\u00e9ptimo lugar, etc., es decir, los que ocupan los lugares de la forma 3k + 1. En la nueva lista se borran los n\u00fameros que est\u00e1n en los lugares de la forma 3k +1. Se repite este proceso hasta que se borran todos los n\u00fameros de la lista. \u00bfCu\u00e1l fue el \u00faltimo n\u00famero que se borr\u00f3? :dry: $(detesto$ $la$ $aritmetica)$\nEspero manden sus soluciones prontito ;) [/color]\r\n\r\nCuidense\r\n\r\n$Jose^/ Carlos$", "Solution_1": "[color=red][b]Problema 3. [/b]En un trapecio se conoce que la suma de ambas diagonales es $4$ y que el \u00e1rea es $2$. Seg\u00fan lo anterior se pide calcular la altura.[/color]\r\n\r\n[color=blue][b]Lema.[/b] Dado un trapecio, sean $d_1$ y $d_2$ las longitudes de las diagonales y sea $x$ el angulo que forman, entonces el area del trapecio es $(d_1.d_2.senx)/2$.[/color]\r\n\r\nEntonces tenemos que $(d_1.d_2.senx)/2=2$, de ahi que $d_1.d_2.senx=4$, es decir $d_1.d_2\\geq 4$, por la desigualdad de las medias tenemos que $\\frac{d_1+d_2} {2}\\geq \\sqrt{d_1.d_2}\\geq 2$, asi se tiene que $2\\geq \\sqrt{d_1.d_2}\\geq 2$, de ahi que $d_1=d_2=2$.\r\n\r\nAhora bien, como su area es $2$ entonces si $b_1, b_2,h$ son su base mayor, base menor y altura respectivamente, $2=\\frac{h(b_1+b_2)} {2}$ y tambien, tenemos que $d_1^2=h^2+(b_1+b_2)^2$ y de ahi que $h^4+4=4h^2$ y se sigue que $h=\\sqrt{2}$. $\\square$", "Solution_2": "jeje, veo que tu novia volvio a irse... :D", "Solution_3": "[quote=\"Pascual2005\"]jeje, veo que tu novia volvio a irse... :D[/quote]\r\n\r\nalgo asi, jejeje... me gustaria que $CUELGUEN$ $SOLUCIONES!!!$" } { "Tag": [ "limit", "calculus", "calculus computations" ], "Problem": "For each of the series below, decide which of the descriptions: \r\n\r\n(i) Does not converge; \r\n(ii) Converges, but not absolutely;\r\n(iii) Converges absolutely\r\n\r\napplies, and prove your answer.\r\n\r\n1. [sum] from n=1 to oo sin(n*[pi] + (1/n))\r\n\r\n2. [sum] from n=1 to oo (e - (1 + (1/n^{2})^{n^{2}})\r\n\r\n3. [sum] from n=1 to oo (e - (1 + ((1/n))^{n} / (1 - (1/2n))))", "Solution_1": "$\\sum_{n=1}^{\\infty}\\left(\\sin(n\\pi)+\\frac{1}{n}\\right)$ diverges. note that $\\sin(n\\pi)$ is 0 whenever n is an integer so this reduces to just\r\n\r\n$\\sum_{n=1}^{\\infty}\\frac{1}{n}$ which is the harmonic series and is known to diverge.", "Solution_2": "I think that series is $\\sum_{n=1}^{\\infty}\\sin(n\\pi+1/n)$.\r\nNow, $\\sin(n\\pi+1/n)=\\sin(n\\pi)\\cos(1/n)+\\cos(n\\pi)\\sin(1/n)=(-1)^{n}\\sin(1/n)$ for $n=1,2,...$, hence the answer is (ii).", "Solution_3": "[quote=\"Spasty\"]I think that series is $\\sum_{n=1}^{\\infty}\\sin(n\\pi+1/n)$.\nNow, $\\sin(n\\pi+1/n)=\\sin(n\\pi)\\cos(1/n)+\\cos(n\\pi)\\sin(1/n)=(-1)^{n}\\sin(1/n)$ for $n=1,2,...$, hence the answer is (ii).[/quote]\r\no woops guess i missed that set of parentheses. :blush: \r\nquestion, what do u mean by \"hence the answer is (ii)\"?", "Solution_4": "The series converges for Leibniz, but not absolutely.", "Solution_5": "Any ideas about the other two series?", "Solution_6": "2) Using Taylor's formula, we can show that\r\n$\\lim_{x\\to+\\infty}x(e-(1+\\frac{1}{x})^{x})=\\frac{e}{2}$.\r\nHence, put $x=n^{2}$, we get\r\n$\\lim_{n\\to+\\infty}n^{2}(e-(1+\\frac{1}{n^{2}}){}^{n}{}^{2})=\\lim_{n\\to+\\infty}\\frac{(e-(1+\\frac{1}{n^{2}}){}^{n}{}^{2})}{1/n^{2}}=\\frac{e}{2}$.\r\nThis means that the series has the same character of $\\frac{1}{n^{2}}$ which converges." } { "Tag": [ "geometry", "rectangle", "MATHCOUNTS" ], "Problem": "How many squares having all 4 vertices on a dot are possible in a 5 by 5 geoboard, like:\r\n\r\n. . . . . \r\n. . . . . \r\n. . . . .\r\n. . . . .\r\n. . . . .\r\n\r\n\r\n\r\n--question: Is there an extremely efficient formula or way to count squares rather than drawing them all out?--", "Solution_1": "This problem is much more fun for rectangles...\r\n\r\nAnyway, yes, there is an extremely efficient way to do this. Count all the possible squares that are $k \\times k$ for $k = 1, 2, 3, 4$, and add all that up. \r\n\r\n[hide=\"Generalization\"] The number of $k \\times k$ squares in an $n \\times n$ lattice is the number of possible placements of, say, the top left corner of that square. Take a moment to see why the possible positions of this corner are a $(n-k) \\times (n-k)$ square subset of the lattice. Hence our total number of squares is\n\n$\\sum_{k=1}^{n}(n-k)^{2}= \\sum_{k=1}^{n-1}k^{2}= \\frac{(n-1)(n)(2n-1)}{6}$ \n\nFor $n = 5$, we have\n\n$\\frac{4(5)(9)}{6}= \\boxed{ 30 }$ squares. \n\n(Edited. Then we need to add in the diagonal squares.) [/hide]", "Solution_2": "[quote=\"t0rajir0u\"]This problem is much more fun for rectangles...[/quote]\r\nHere are two ways to find all rectangles in an $m\\times n$ grid:\r\n\r\n[hide=\"First way\"]$x=\\binom{m}{2}\\binom{n}{2}$[/hide]\n[hide=\"Second way\"]$x=\\frac{mn(m+1)(n+1)}{4}$[/hide]", "Solution_3": "The second one is written wrong, and both formulas (in the correct form) are equivalent, so why did you write them separately?", "Solution_4": "[quote=\"t0rajir0u\"]The second one is written wrong, and both formulas (in the correct form) are equivalent, so why did you write them separately?[/quote]\r\nHow is the second one wrong?\r\n\r\nThey are separate because although they are equivalent, they look completely different.", "Solution_5": "The question does not exclude squares with diagonal sides.\r\n\r\n[hide=\"Casework\"]There are the $\\frac{n(n+1)(2n+1)}{6}=30$ squares already mentioned.\n\n$9$ with sidelength $\\sqrt{2}$.\n\n$8$ with sidelength $\\sqrt{5}$.\n\n$2$ with sidelength $\\sqrt{10}$.\n\n$1$ with sidelength $2\\sqrt{2}$.\n\nSo I think it is $30+9+8+2+1=50$, I think :maybe: .[/hide]", "Solution_6": "[quote=\"i_like_pie\"]How is the second one wrong?[/quote]\n\nIt should be ${m \\choose 2}{n \\choose 2}= \\frac{mn(m-1)(n-1)}{4}$.\n\n[quote=\"pianoforte\"]The question does not exclude squares with diagonal sides.[/quote]\r\n\r\nHmm, fair enough. We'll have to take those into consideration too. (It's less generalizable, though :( ) [hide=\"A little trickier...\"] We'll have to count a few different kinds. First, the $\\sqrt{2}\\times \\sqrt{2}$ squares. The set of possible locations of their topmost corners is a $3 \\times 3$ square, so there are $9$ of these.\n\nThen the $2 \\sqrt{2}\\times 2 \\sqrt{2}$ squares. We can only fit $1$ of these in.\n\nNext, the $\\sqrt{5}\\times \\sqrt{5}$ squares. These can come in two orientations, and there are $4$ positions of the topmost corner for both, so there are $8$ of these.\n\nNext, the $\\sqrt{10}\\times \\sqrt{10}$ squares. Two orientations, and we can only fit $1$ of each, so $2$ of these.\n\nIn total, we have added $9+1+8+2 = 20$ squares to our nice generalizable answer, making our final total... $\\boxed{ 50 }$? [/hide]", "Solution_7": "[quote=\"t0rajir0u\"][quote=\"i_like_pie\"]How is the second one wrong?[/quote]\nIt should be ${m \\choose 2}{n \\choose 2}= \\frac{mn(m-1)(n-1)}{2}$.[/quote]\r\nTry $m=2,n=3$.\r\n\r\n$\\frac{mn(m-1)(n-1)}{2}$ yields $6$. However, the correct answer, obtained with $\\frac{mn(m+1)(n+1)}{4}$, is $18$.", "Solution_8": "I think t0rajir0u meant $\\frac{mn(m-1)(n-1)}{4}$, while thinking of a grid of m horizontal lines, and n vertical lines, while you were thinking of a grid with m rows and n columns.", "Solution_9": "[quote=\"i_like_pie\"]$\\frac{mn(m-1)(n-1)}{2}$ yields $6$. However, the correct answer, obtained with $\\frac{mn(m+1)(n+1)}{4}$, is $18$.[/quote]\n\nIn either case either your first expression is wrong or your second one is.\n\n[quote=\"pianoforte\"]I think t0rajir0u meant $\\frac{mn(m-1)(n-1)}{4}$, while thinking of a grid of m horizontal lines, and n vertical lines, while you were thinking of a grid with m rows and n columns.[/quote]\r\n\r\nYes. Usually, when I speak of an $m \\times n$ lattice grid, I expect there to be $m \\times n$ lattice points in it.", "Solution_10": "[quote=\"t0rajir0u\"][quote=\"i_like_pie\"]$\\frac{mn(m-1)(n-1)}{2}$ yields $6$. However, the correct answer, obtained with $\\frac{mn(m+1)(n+1)}{4}$, is $18$.[/quote]\n\nIn either case either your first expression is wrong or your second one is.\n\n[quote=\"pianoforte\"]I think t0rajir0u meant $\\frac{mn(m-1)(n-1)}{4}$, while thinking of a grid of m horizontal lines, and n vertical lines, while you were thinking of a grid with m rows and n columns.[/quote]\n\nYes. Usually, when I speak of an $m \\times n$ lattice grid, I expect there to be $m \\times n$ lattice points in it.[/quote]\r\nI, too, was talking about a grid with $m\\times n$ points. I still don't see how mine is wrong as it gives results that correspond with reality. Try several cases and compare the results that $\\frac{mn(m+1)(n+1)}{4}$ and $\\frac{mn(m-1)(n-1)}{2}$ give with the real answer.", "Solution_11": "Did I write that over two? Sorry, it should be\r\n\r\n${m \\choose 2}{n \\choose 2}= \\frac{mn(m-1)(n-1)}{4}$\r\n\r\nConsider the degenerate case $m = n = 1$. This is a point. A point is not a rectangle (are you counting degenerate rectangles? I'm not).", "Solution_12": "This problem is from 2003 Mathcounts, the 2nd problem on the Team round, to be precise. The answer key says that it is [hide] 50[/hide]", "Solution_13": "Aha. I was counting the degenerate case before, so subtract $5^{2}= 25$ from my answer and it's correct. I see.\r\n\r\nEdit: In any case, i_like_pie, I don't see how you can't see that your expressions aren't equivalent. If you accept that\r\n\r\n${n \\choose 2}= \\frac{n(n-1)}{2}$\r\n\r\nThen clearly \r\n\r\n${n \\choose 2}{m \\choose 2}= \\frac{mn(m-1)(n-1)}{4}$\r\n\r\nWhich is not your second expression.", "Solution_14": "can you explain why you subtract 5^2 and what is \"degenerate?\" It would help me understand. Thanks.", "Solution_15": "I was counting points as squares without realizing it. So I counted all $25$ points already in the square in my original argument...\r\n\r\n(A point is a \"degenerate\" square because it's like a square of side length $0$.)", "Solution_16": "oh. thanks. it makes much more sense now. i thought that you meant degenerate, as in not a square originally, and did not take it as points." } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "Do there exist a circle and an infinite set of points on it such that the distance between any two of the points is rational ?", "Solution_1": "http://www.mathlinks.ro/Forum/viewtopic.php?t=44864\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=112824" } { "Tag": [], "Problem": "What is the molar mass of barium nitrate, Ba(NO3)2?\r\n \r\n1)261 g/mole\r\n \r\n2)199 g/mole\r\n \r\n3)213 g/mole\r\n \r\n4)167 g/mole\r\n \r\n5)none of these", "Solution_1": "$137.33+2\\cdot(14.01+3\\cdot16.00)=137.33+2\\cdot(14.01+48.00)=137.33+2\\cdot62.01=137.33+124.02=261.35\\approx261=\\boxed{1)261}$" } { "Tag": [], "Problem": "What is the tens digit of $12+3^4\\times 67^5+89$?", "Solution_1": "[quote=\"rcv\"]What is the tens digit of $12+3^4\\times 67^5+89$?[/quote]\r\n[hide=\"solution\"]\nAll you need are the last two digits of each number.\n12 + 81 + 89 + 67:^5:\n182 + 67:^5:\n82 + 07\n89\n[b]8[/b]\n[/hide]", "Solution_2": "Treething: Your method should work, but your order of operations are a little messed up. Carefully compare your formula with the original problem.", "Solution_3": "[hide]S=12+3^4*67^5+89=101+(3*67)^4*67=101+(201)^4*67\nSince 201^4 :equiv: 1mod100, S=101+1*67 :equiv: 168 :equiv: 68mod100. SO the tens digit is 6.\n\n$S=12+3^4 \\times 67^5+89=101+(3 \\times 67)^4 \\times 67=101+(201)^4 \\times 67$\nSince $201^4 \\equiv 1 \\mod 100, S=101+1 \\times 67 \\equiv 168 \\equiv 68 \\mod 100$. SO the tens digit is 8.[/hide]\n\n[size=75][color=darkred][Edit: Hidden and LaTex'ed by moderator rcv][/color][/size]" } { "Tag": [ "AMC", "AIME", "USA(J)MO", "USAMO", "geometry", "3D geometry", "tetrahedron" ], "Problem": "I have something I was wondering if people related to/ could give advice about.\r\n\r\nI took practice AMC, AIME, and USAMO tests this summer.\r\n\r\nOn the AMCs, I tended to score in the 120-135 range.\r\n\r\nOn the AIMEs, I scored in the 8-12 range.\r\n\r\nOn the USAMO, I would get about 1 completely right. Sometimes 2.\r\n\r\nI am just worried because for me, the AMC seems so much tougher to improve in than the AIME or USAMO. My score does not budge and I feel too pressed for time and make so many silly mistakes. Does anyone else find the AIME an easier test taking experience than the AMCs? Is there anything I can do other than practice to help me in timing during the AMC? What do all of you perfect scorers out there do to get a perfect score?", "Solution_1": "this is irrelevant, but I think that scores take a LOT of work to go up, especially some kinds. Geometry scores on Vestavia practices go up really slowly. I've been practicing forever, as in 3 weeks lol. But once, it did go up a lot. \r\n\r\nI am just a green geometer, but I know that practice works and it's the only way as well as studying. \r\n\r\nHaha that probably didn't help just like that tetrahedron thing I posted formulas for a long tiem ago. I know.... :blush:", "Solution_2": "Don't worry about it; if you qualify for USAMO, who cares what your index was, and if you can qualify comfortably, why obsess over slightly better scores when you can be doing olympiad stuff? Work harder problems and the easier ones take care of themselves.\r\n\r\n(By the way, just so you know, don't be too sure of how well you're really doing on the USAMO\u2014look at some of the threads [url=http://www.artofproblemsolving.com/Forum/index.php?f=133&ppage=50&sort=lastpost&order=DESC&start=185]here[/url], especially [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=85142]this one[/url]. :wink: )", "Solution_3": "I had my teacher at school grade it. He is pretty good. Ph.D from Cornell.\r\n\r\nI mean there are those questions where they ask \"find all possible solutions\" and I do that and I show that there are no other possible solutions, or the geometry problems where it says to prove something, etc." } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "I'm reading paper at http://www.emis.de/journals/EJC/Volume_10/PDF/v10i1r8.pdf . In figure 2, there is that the best known packing has side length 3.8772 and tilt angle 40.182 degrees. How can I verify those values, as another source (Which Way did the Bicycle Go) says that the side length is about 3.877083?", "Solution_1": "Anyone? This irritates me as I don't know which source I can trust." } { "Tag": [ "LaTeX" ], "Problem": "How many ways can you color the squares of a $ 2 \\times 2008$ grid in 3 colors such that no two squares of the same color share an edge?", "Solution_1": "[hide=\"A Beginning...\"]Let the colors be red, blue, and green.\n\nThe top row can be colored in $ 3\\cdot2^{2007}$ ways.\n\nIt remains to determine in how many ways the bottom row can be colored. :ninja:[/hide]", "Solution_2": "[hide=\"You want to think the other way, since 2008 is arbitrary :wink: \"]\nThere are $ 3 \\cdot 2 = 6$ choices for the leftmost two squares, without loss of generality say $ \\textcolor{red}{R}\\textcolor{green}{G}$. For the next column, there are $ 3$ possibilities, $ \\textcolor{green}{G}\\textcolor{red}{R}, \\textcolor{blue}{B}\\textcolor{red}{R}, \\textcolor{green}{G}\\textcolor{blue}{B}$, and so forth for the rest of the columns. \n\\begin{tabular}{|r\\parallel{}r|r|}\n\\hline\n\\cdots &\\textcolor{red}{R}&\\textcolor{green}{G}\\\\\n\\hline\n\\cdots &\\textcolor{green}{G}&\\textcolor{red}{R}\\\\\n\\hline\n\\end{tabular}, \\quad\n\\begin{tabular}{|r\\parallel{}r|r|}\n\\hline\n\\cdots &\\textcolor{red}{R}&\\textcolor{blue}{B}\\\\\n\\hline\n\\cdots &\\textcolor{green}{G}&\\textcolor{red}{R}\\\\\n\\hline\n\\end{tabular}, \\quad\n\\begin{tabular}{|r\\parallel{}r|r|}\n\\hline\n\\cdots &\\textcolor{red}{R}&\\textcolor{green}{G}\\\\\n\\hline\n\\cdots &\\textcolor{green}{G}&\\textcolor{blue}{B}\\\\\n\\hline\n\\end{tabular}\nThus, the answer is $ 6 \\cdot 3^{2007} = \\boxed{2 \\cdot 3^{2008}}$. \n[/hide]", "Solution_3": "Nice solution; I was obviously thinking differently. :D\r\n\r\n[hide=\"LaTeX Tip\"]You can use [b]\\color{}[/b] instead of [b]\\textcolor{}{}[/b].\n\nTo be able to continue in black afterwards without explicitly saying [b]\\color{black}[/b], use [b]{\\color{}}[/b].\n\nAlso note that [b]\\textcolor{}{}[/b] affects all parts of [b],[/b] not just actual text. Example: $ \\textcolor{blue}{\\begin{tabular}{|r|r|}\\hline A & B \\\\\n\\hline C & D \\\\\n\\hline \\end{tabular}}$[/hide]" } { "Tag": [ "\\/closed" ], "Problem": "See attached.", "Solution_1": "Doesn't this have to do with that Guest viewing profile thing, about how the thing isn't updated in real-time so it delays it a bit?? (idk). lol...I'm looking at the guest at the bottom who is apparently in the Missouri forum... :rotfl: :rotfl: :rotfl:" } { "Tag": [ "algorithm", "geometry", "algebra", "polynomial", "search", "floor function", "induction" ], "Problem": "Prison Warden Willy is in charge of guarding a not-necessarily convex polygonal prison. He is incredibly lazy, and does not want to have to manually watch the prisoners. However, he feels for safety reasons that he needs to be able to remotely see into every nook and cranny of each and every cell.\r\n\r\nThen he goes to Home Depot and notices that they spell special \"vertex-mounted\" swivel cameras that have a 360 degree field of view, blocked only when their line of sight is interrupted.\r\n\r\nCan Warden Willy devise an algorithm that will let him place these cameras on particular vertices so as to be guaranteed two things?\r\n\r\n1) Complete visual coverage of the prison\r\n2) Minimum number of cameras purchased\r\n\r\nFor any given prison, do there exist multiple camera configurations which use the same (optimal) number of cameras?\r\nExample of such a set-up:\r\n [geogebra]9828ce4deb89c177de5f05debccbe8d371973610[/geogebra]", "Solution_1": "This is not at all a graph theory problem. Anyway, I will go ahead and conjecture that one needs at least as many cameras as concave angles in the prison.", "Solution_2": "I think you're right. This is more of a geometry problem, in retrospect.", "Solution_3": "If you take a star shape, like take a regular $ n$-gon and join sides $ A_i$ and $ A_{i\\plus{}k}$ (indices mod $ n$) for some $ k$ not equivalent to $ \\minus{}1,0,1$ mod $ n$, then make the region the interior area, a camera in the center can easily cover it, but you have as many concave angles as you want.\r\n\r\nI would guess this problem is NP-hard or something. You can probably easily come up with a suboptimal algorithm involving some sort of triangulation and then placing a camera where it covers the most triangles, but I feel like the challenge of coming up with an algorithm that guarantees the least amount of cameras without brute force is similar to the challenge for doing it for TSP and the like.", "Solution_4": "Can we replace t0r's \"at least\" with \"at most\" just by placing a camera at each angle of measure more than 180 degrees?\r\n\r\nEdit (Re: Ravi's post that follows): I knew this problem sounded familiar!", "Solution_5": "I didn't read the problem carefully, but it sounds like the [url=http://en.wikipedia.org/wiki/Art_gallery_theorem]Art Gallery Problem[/url].", "Solution_6": "[quote=\"Ravi B\"]I didn't read the problem carefully, but it sounds like the [url=http://en.wikipedia.org/wiki/Art_gallery_theorem]Art Gallery Problem[/url].[/quote]\r\n\r\n(I'm not sure if my reduction works; I took a course in algorithms over the summer and we brushed over reductions, so this might not be correct.)\r\n\r\n[hide]It is the Art Gallery Problem, which is NP-hard as MellowMelon claimed. If you can solve the dominating set problem, then you can solve this problem as follows: determine if a dominating set of size one (one camera) suffices for the polygon, if not, then check if a dominating set of size two, and iterate up until you obtain a valid number of cameras. This takes linear time in the number of cameras, which runs in polynomial time. (This could be potentially improved for by using binary search, although it is unnecessary.) Therefore, the Art Gallery Problem is NP-hard.[/hide]", "Solution_7": "Wow! I didn't know this sort of thing was studied so intensely.\r\n\r\nWell, I think that I have a solution for the special-case I gave here, which IS indeed a graph-theoretical solution.\r\nIf people are interested, I can post it soon.", "Solution_8": "Can these irregular n-gons intersect themselves?\r\n\r\nHere are some of my idea's\r\n\r\nFor any irregular, non-self-intersecting n-gon, there are atmost $ \\left\\lfloor \\frac{n}{2} \\right\\rfloor$ concave angles. Hence we need at most that many cameras.\r\n\r\nIf there are only $ m$ concave angles in the n-gon, and they are all next to each other then we have to have $ m$ cameras", "Solution_9": "No intersection. I think you're on the right track.", "Solution_10": "I think you guys are liking this concave angle idea a bit too much. Both your $ \\lfloor n/2 \\rfloor$ claim and your claim that $ m$ consecutive concave angles means $ m$ cameras are false. Consider a right triangle, but push the hypotenuse in towards the vertex a tiny bit to make an arc. You can pick an arbitrary amount of vertices on this arc, all with concave angles. We can get arbitrarily many concave angles with 3 convex angles. Also, one camera on the vertex of the triangle will suffice to cover everything.", "Solution_11": "I suppose it's been sufficiently long. I'll post what I think is one solution:\r\n[hide]\nFirst, take the graph and triangulate it. Then construct a three-coloring of the vertices. This will label approximately $ \\frac {n}{3}$ of the total vertices with each color. (In reality, it will be slightly more for some, is the graph does not have a nodes in a multiple of three (for example, the one given). )\n\nDetermine which of the vertex colors is the least represented on the graph, and place one camera at each of these vertices. This will always be $ \\lfloor\\frac{n}{3}\\rfloor$, though there are sometimes multiple ways to cover the prison with this many.\n\nShort explanation of correctness:\nWhen you triangulate the graph, you divide it into regions in which every single camera on the vertices of that triangle can see the entire interior. Therefore, for each triangle you choose, you only need to place at MOST one camera on the vertices. It would be suboptimal to have any triangle with two or more cameras, from a heuristic point of view. By three-coloring the graph, we isolate which triangles have the vertices with the fewest camera-overlaps, and from here we can generate one solution.\n\nHere's an image of two possible solutions:\nThe three-coloring has labeled the vertices 1, 2, or 3. The YELLOW vertices (choosing 1's) are one optimal covering, and the large BLUE vertices (choosing 3's) are another.\n\n\n\n[img]19146[/img][/hide]", "Solution_12": "[quote=\"DiscreetFourierTransform\"]First, take the graph and triangulate it. Then construct a three-coloring of the vertices.[/quote] Why is it obvious that a 3-coloring exists?", "Solution_13": "[hide=\"Proof a 3 coloring exists\"]You can prove easily by induction that a triangulation of an $ n$-gon always has $ n\\minus{}2$ triangles. If every triangle shared at most one edge with the $ n$-gon, the $ n$-gon could only have $ n\\minus{}2$ edges, which is a contradiction. So at least one triangle has to share two edges with the $ n$-gon.\n\nTriangles are clearly $ 3$-colorable. Suppose we can color all $ k$-gons. In a $ k\\plus{}1$-gon, find a vertex $ V$ such that both of the edges linking to it are contained in the same triangle (exists by the above paragraph). Temporarily ignore $ V$ and use the inductive hypothesis to color everything else. $ V$ is connected to only two other vertices. Pick the color neither one uses.[/hide]", "Solution_14": "What's the complexity class of generating first a triangulation, then a three-coloring?\r\nIt can't be THAT hard. Or is it still NP?\r\n\r\nI'm fairly sure triangulating would be linear, right?", "Solution_15": "Graph coloring is [url=http://en.wikipedia.org/wiki/Graph_coloring]pretty hard[/url], although the specific problem of finding a $ 3$-coloring isn't quite the same as the decision problems listed in the Wikipedia article. See, for example, [url=http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6WH3-4D1V78P-3&_user=501045&_rdoc=1&_fmt=&_orig=search&_sort=d&view=c&_acct=C000022659&_version=1&_urlVersion=0&_userid=501045&md5=1597f94db1962e290407227f449534c4]this paper[/url].", "Solution_16": "$ \\lfloor n/3 \\rfloor$ is not necessarily the minimum; take either of the two examples I've posted in this thread. The problem is constructing the actual minimum. What you have is a suboptimal algorithm. For most NP-hard problems these are pretty easy to come up with.", "Solution_17": "Feel free to correct me anywhere you find mistake.\r\n\r\nIn given set-up minimum number of cameras would be 2 so your solution isn't correct.\r\n\r\nMy knowledge is too poor to rigorously prove it but consider this algorithm:\r\n\r\n$ 1)$ Find all concave angles and consider placing camera on any of them.\r\n\r\n$ 2)$ Determine \"field of vision\" for every possible place of camera (field of vision is polygon for which every point inside the polygon can be connected with vertex of the angle with straight line without crossing sides of the original polygon).\r\n\r\n$ 3)$ Count the number of polygons which are not included in the \"field of vision\". Let those be the \"other\" polygons.\r\n\r\n$ 4)$ Determine angle which \"field of vision\" splits original polygon in least possible \"other\" polygons. Place a camera there. In case that some angles produce same number of the \"other\" polygons, there might be multiple solutions to the problem.\r\n\r\n$ 5)$ Treat any remaining \"other\" polygon same way as the original one (in other words repeat the algorithm).", "Solution_18": "I'm sorry, could you demonstrate how it's possible to do the example I posted with two cameras?", "Solution_19": "DFT: Pick the two #3 vertices on concave angles in your last diagram. That covers everything.\r\n\r\nFlame: I am 99.9% sure that with enough thought you could construct a room for which that algorithm did not come up with the [i]optimal[/i] configuration. There are some fast and very good algorithms for TSP, for example, but none of them can guarantee the minimum.", "Solution_20": "The wikipedia article has a number of good references (and I think the reference DiscreetFourierTransform describes is actually covered therein).", "Solution_21": "I agree, in the way it was, prison can be constructed so the algorithm wouldn't produce optimal solution but I think I corrected the flaws. If you still think it's wrong, please prove or demonstrate.\r\n\r\n[hide=\"algorithm\"]\n$ 1)$ Check if there is any trivial solution (solution with only 1 camera)\n\n$ 2)$ Find all concave angles and consider placing camera on any of them.\n\n$ 3)$ Determine \"field of vision\" for every possible place of camera (field of vision is polygon for which every point inside the polygon can be connected with vertex of the angle with straight line without crossing sides of the original polygon).\n\n$ 4)$ Count the number of polygons which are not included in the \"field of vision\". Let those be the \"other\" polygons.\n\n$ 5)$ If there are 2, or more \"other\" polygons that can be seen from a single vertex, treat it as 1 \"other\" polygon.\n\n$ 6)$ Determine angle which \"field of vision\" splits original polygon in least possible \"other\" polygons. Place a camera there. In case that some angles produce same number of the \"other\" polygons, there might be multiple solutions to the problem.\n\n$ 7)$ Treat any remaining \"other\" polygon same way as the original one (in other words repeat the algorithm).[/hide]\n\nThis is my solution (one of) for the proposed set-up:\n\n[hide]\n[img]http://i217.photobucket.com/albums/cc7/Rakia69/200902211348_609.jpg[/img][/hide]\r\n\r\nI still have a feeling I'm missing something big as this seems to easy to be right." } { "Tag": [ "function", "number theory", "Diophantine equation" ], "Problem": "I don't see much of a difference between infinite descent and the well-ordering-principle. I assume it's only because of the way I first saw it/used it, but I would like to know more. Could someone explain the WOP to me?\r\n\r\nThanks.", "Solution_1": "The phrase \"well-ordering principle\" is used to mean two distinct but related concepts. The one you are thinking of is that the natural numbers are well-ordered; in other words, any subset of the natural numbers has a minimum.\r\n\r\nInfinite descent refers to a proof technique for showing that a (usually Diophantine) equation has no solutions by showing that if that equation has solutions, then it is possible to construct a subset of the natural numbers without a minimum, which violates the well-ordering principle. The two proofs, for any given problem, are exactly equivalent - that is, a proof phrased using infinite descent can just as well be phrased using well-ordering. It is usually considered a little nicer to use well-ordering.\r\n\r\n(The well-ordering principle in a more general context states that it is possible to define a well-ordering on any set. This statement is equivalent to the axiom of choice (the least element of a set is a choice function), and in this form is quite unintuitive; what exactly does a well-ordering of the reals look like?)", "Solution_2": "Ah, thanks, the middle paragraph was exactly what I was looking for :)" } { "Tag": [], "Problem": "Prove the following for triangular numbers $ t_{n}$:\r\n\r\n$ t_{1}\\plus{}t_{2}\\plus{} ...\\plus{}t_{n} \\equal{} n(n\\plus{}1)(n\\plus{}2)/6$.", "Solution_1": "$ t_{n}\\equal{}\\frac{n(n\\plus{}1)}{2}$\r\n\r\n$ \\sum_{1}^{n} t_{n}\\equal{}\\frac{1}{2}\\sum_{1}^{n}n^{2}\\plus{}\\frac{1}{2}\\sum _{1}^{n}n$\r\n\r\n$ \\sum_{1}^{n} t_{n}\\equal{}\\frac{n(n\\plus{}1)(2n\\plus{}1)}{12}\\plus{}\\frac{n(n\\plus{}1)}{4}$\r\n\r\n$ \\sum_{1}^{n} t_{n}\\equal{}\\frac{n(n\\plus{}1)(n\\plus{}2)}{6}$", "Solution_2": "A similar proof can be found [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=306743]here[/url]." } { "Tag": [ "AMC", "AIME", "ratio", "Pascal\\u0027s Triangle" ], "Problem": "Can someone help me with this problem?\r\nIn Pascal's Triangle, each entry is the sum of the two entries above it. In which row of Pascal's Triangle do three consecutive entries occur that are in the ratio 3:4:5?", "Solution_1": "[hide=\"Hint\"]\nThe $ k$th element of the $ n$th row of Pascal's Triangle is $ \\binom{n}{k}$. Thus, $ \\binom{n}{k\\minus{}1}: \\binom{n}{k}: \\binom{n}{k\\plus{}1}\\equal{}3: 4: 5$. Simplify these.\n[/hide]", "Solution_2": "Copy of a Mock AiME question.", "Solution_3": "On the AoPS wiki it says n=62.\r\nI also got 62 for the top number, but doesnt that mean that it is the 63rd row?", "Solution_4": "[quote=\"Jongao\"]On the AoPS wiki it says n=62.\nI also got 62 for the top number, but doesnt that mean that it is the 63rd row?[/quote]\r\n\r\nThe top \"row\" is considered the 0th row.", "Solution_5": "Heres how I did it. Let entries be \n$\\binom{x}{m-1}$, $\\binom{x}{m}$ and $\\binom{x}{m+1}$\n\nSo write that out.\n\n$\\frac{(x)(x-1)...(x-(m-1)+1)}{(3)(m-1)(m-2)...1}$=\n$\\frac{(x)(x-1)...(x-m+1)}{(4)(m)(m-1)(m-2)...1}$=\n$\\frac{(x)(x-1)...(x-(m+1)+1)}{(5)(m+1)(m)(m-1)(m-2)...1}$\n\nNow let $(x)(x-1)...(x-(m-1)+1)$ which simplifies to $(x)(x-1)...(x-m+2)$ be A.\n\nAnd let $(m-1)(m-2)...1$ be B. Substitute.\n\n$\\frac{A}{(3)B}$=\n$\\frac{A(x-m+1)}{(4)mB}$=\n$\\frac{A(x-m+1)(x-m)}{(5)m(m+1)B}$. \n\nIt is much easier to pair equation 1&2 and 2&3 instead of 1&3 since we only need two equation anyway. Get.\n\n$\\frac{A}{(3)B}$=$\\frac{A(x-m+1)}{(4)mB}$. Simplify. Get.\n$\\frac{1}{(3)}$=$\\frac{(x-m+1)}{(4)m}$ As you can see A and B are nice place holders simplify tricky/uglyness. Cross-multiply.\n\n$4m=3x-3m+3$ which is $7m=3x+3.$ Next equation!\n\n$\\frac{A(x-m+1)}{(4)mB}$=$\\frac{A(x-m+1)(x-m)}{(5)m(m+1)B}$. Simplfiy.\n$\\frac{1}{(4)}$=$\\frac{(x-m)}{(5)(m+1)}$. Cross-multiply.\n\n$5m+5=4x-4m$ So $9m=4x-5$ Now you have two equations.\n\n$7m=3x+3.$ and $9m=4x-5$. Solve.\n$63m=27x+27$ and $63m=28x-35$. Equal.\n$27x+27=28x-35$ which you get x=62. Remember what x was in beginning. Since \"$x$ choose anything\" is part of \"row$ x+1$\", answer is 63.\n(1st row is row 0, as AIME15 says)." } { "Tag": [ "limit", "integration", "real analysis", "real analysis unsolved" ], "Problem": "Let $f(x)=2x(1-x), x\\in\\mathbb{R}$ and denote $f_n=f\\circ f\\circ ... \\circ f$, $n$ times.\r\n(a) Find $\\lim_{n\\rightarrow\\infty} \\int^1_0 f_n(x)dx$.\r\n(b) Now compute $\\int^1_0 f_n(x)dx$.", "Solution_1": "ok,\r\n$f_n(x)=\\frac 1 2 -2^{2^n-1}(x-\\frac 1 2)^{2^n}$\r\nby induction. :)", "Solution_2": "[quote=zhaobin]ok,\n$f_n(x)=\\frac 1 2 -2^{2^n-1}(x-\\frac 1 2)^{2^n}$\nby induction. :)[/quote]\n\nelaborate plz! ", "Solution_3": "Checking inductively that zhaobin's formula works is pretty trivial. However, I can try to explain how zhaobin may have arrived at his formula.Well, start with the obvious fact that $f_{n+1}=-2f_n^2+2f_n$. Now, we complete the square:\n\n$f_{n+1}-\\frac{1}{2}=-2f_n^2+2f_n-\\frac{1}{2}= -2(f_n -\\frac{1}{2})^2$\nWhich leaves us with:\n$f_{n+1}= \\frac{1}{2}-2(f_n -\\frac{1}{2})^2$\nHowever,\n$f_1= \\frac{1}{2}-2(x-1/2)^2$, which leads pretty quickly to zhaobin's formula." } { "Tag": [], "Problem": "When you have a fraction in the exponent, you're supposed to evaluate the numerator before the denominator, right?", "Solution_1": "I do not believe it matters which order you evaluate but by convention, you evaluate the numerator before the denominator" } { "Tag": [ "linear algebra", "matrix", "quadratics", "vector", "calculus", "derivative", "function" ], "Problem": "Let $A=(a_{ij})_{n{\\times}n}$ be a symmetric matrix with real entries.Further let's consider the quadratic form $W(\\overline{x})=<\\overline{x},A\\overline{x}>$,for $\\overline{x}{\\in}L$,where $L$ is a $n$ dimensional\r\nlinear space.It is well known that all proper values of $A$(sorry if the word is not right) are real and it has a diagonal canonical(Jordanian) form.So if all proper values of $A$ are positive then $W(\\overline{x})$ will be a positively defined quadratic form.\r\nNow I came upto an interesting question.What if $A$ is not a symmetric one.I have a conjecture which seems to be right:\r\n[i]If all real parts of proper values of[/i] $A$ [i]are positive then $W(\\overline{x})$ is a positive defined.[/i]\r\nWhat do you think about it?Is it right?", "Solution_1": "Does \"proper values\" mean \"eigenvalues\"?", "Solution_2": "I'm sure that's what it means (that's sort of what it sounds like in Romanian too :)).\r\n\r\nAnyway, I don't think it's true. Think about the matrix $A=\\left(\\begin{array}{cc}0&1\\\\-2&2\\end{array}\\right)$. If ${\\bf v}$ is the column vector $\\left(\\begin{array}{c}x&y\\end{array}\\right)$, then you get ${\\bf v}^t\\cdot A\\cdot {\\bf v}=-xy+2y^2$. Maybe I'm missing something? :?", "Solution_3": "Indeed [b]grobber[/b] the conjecture fails even when n=2.Strange that I didn't notice that :blush: .Anyway thanks for the example.Now I must think on a new way to prove the lemma in the paper. :)", "Solution_4": "What was the lemma? Could you share it with us? :) Or maybe the location of the paper, if it can be found online?", "Solution_5": "Ok,[b]grobber[/b] I'll post the lemma later.The paper is on [i]Lyapunov[/i]'s theorem for steady equilibrium of autonom differential equations.", "Solution_6": "This conjecture is, however, true for normal matrices, i.e. matrices which satisfy $AA^t=A^tA$ (a similar matrix is a special type of normal matrix).\r\n\r\nA matrix is normal iff it's unitarily diagonalizable (we're working in $\\mathbb C$ here). Given the vector ${\\bf x}=(x_1\\ x_2\\ \\ldots\\ x_n)^t$, we have ${\\bf x}^*A{\\bf x}={\\bf\\tilde x}^*U^*AU{\\bf \\tilde x}$, where $U$ is a unitary matrix s.t. $U^*AU$ is diagonal, and ${\\bf\\tilde x}$ is the unique solution to $U{\\bf\\tilde x}={\\bf x}$.\r\n\r\nNow, if $\\lambda_1,\\ldots,\\lambda_n$ are the eigenvalues of $A$, and ${\\bf\\tilde x}=(\\tilde x_1\\ \\ldots\\ \\tilde x_n)^t$, then ${\\bf\\tilde x}^*U^*AU{\\bf \\tilde x}=\\lambda_1|\\tilde x_1|^2+\\ldots+\\lambda_n|\\tilde x_n|^2\\ (*)$. We've been working with complex numbers up until now, but we had started off with a real matrix $A$, and real ${\\bf x}$, so $(*)$ is real, meaning that its equal to its real part, which is $\\sum\\Re\\lambda_i\\cdot|\\tilde x_i|^2$, and this is always $>0$, unless ${\\bf\\tilde x}=0$, which is equivalent to ${\\bf x}=0$.\r\n\r\nI hope it's Ok.", "Solution_7": "Sorry that I type the lemma so late.As I told we consider the autonom system of differential equations:\r\n$\\dot{x_{1}}=f_{1}(x_{1},x_{2},\\ldots,x_{n}) \\\\\r\n\\dot{x_{2}}=f_{2}(x_{1},x_{2},\\ldots,x_{n}) \\\\\r\n\\vdots \\\\\r\n\\dot{x_{n}}=f_{n}(x_{1},x_{2},\\ldots,x_{n})$,\r\nwhere derivatives are taken over time $t$ and where all functions are defined in some region $\\Delta{\\subset}\\mathbb{R}^{n}$ and have continious derivatives up to the second order.We will say that position $\\vec{a}{\\in}\\mathbb{R}^{n}$ is an equibrilium position for this system if $\\vec{x}(t)=\\vec{a}$ is a solution for this system or equivalently $\\vec{f}(\\vec{a})=\\vec{0}$.Also let's denote by $\\vec{\\varphi}(t,\\vec{\\xi})$ the solution of that system with starting parametres $0,\\vec{\\xi}$.\r\nDenote $a_{ij}=\\frac{\\partial{f_{i}(\\vec{a})}}{\\partial{x_{j}}}$ and $A=(a_{ij})_{n{\\times}n}$.\r\nThe position $\\vec{a}$ is called steady by Lyapunov if\r\na)For sufficently small $\\delta>0$ the solutions $\\vec{\\varphi}(t,\\vec{\\xi})$ where $|\\vec{\\xi}-\\vec{a}|<\\delta$ are defined for all $t{\\geq}0$.\r\nb)For every $\\epsilon>0$ there exists a $\\sigma>0(\\sigma{\\leq}\\delta)$ s.t. $|\\vec{\\varphi}(t,\\vec{\\xi})-\\vec{a}|<\\epsilon$ for all $t{\\geq}0$,as soon as $|\\vec{\\xi}-\\vec{a}|<\\sigma$.\r\nIf we add up third condition \r\nc)There exists a $\\sigma>0(\\sigma{\\leq}\\delta)$ s.t. $\\lim_{t{\\rightarrow}\\infty}{|\\vec{\\varphi}(t,\\vec{\\xi})-\\vec{a}|}=0$,as soon as $|\\vec{\\xi}-\\vec{a}|<\\sigma$.\r\nwe call the position $\\vec{a}$ asymptotically steady equbrilium.\r\nLyapunov's theorem states that if all eigenvalues of matrix $A$ have negative real parts then \r\n$\\vec{a}$ is an asymptotically steady equibrilium.Moreover there exists an $\\sigma>0$ s.t.\r\n$|\\vec{\\varphi}(t,\\vec{\\xi})-\\vec{a}|{\\leq}r|\\vec{\\xi}-\\vec{a}|e^{-{\\alpha}t}$ for $t{\\geq}0$,as soon as $|\\vec{\\xi}-\\vec{a}|<\\sigma$,where $r,\\alpha>0$ are independent of $\\vec{\\xi}$.\r\nThe main idea of the proof in the paper is to find some positively defined quadratic form $W$ s.t. the distance of the solution from equbrilium in terms of that quadratic form:$w(t)=W(\\vec{\\varphi}(t,\\vec{\\xi})-\\vec{a})$,have a tendence to decraese i.e. $\\dot{w}(t){\\leq}0$.Moreover $\\dot{w}(t){\\leq}-2{\\alpha}w(t)$ for some $\\alpha>0$.If we take standart norm we come up to the lemma that I posted." } { "Tag": [ "\\/closed" ], "Problem": "Someone has been messing with my profile and i haven't told anyone my password!", "Solution_1": "Make sure you log out whenever you are finished on the site (and if you have a computer set to automatically log in, you are giving others the opportunity to edit your profile)." } { "Tag": [ "geometry", "geometry unsolved" ], "Problem": "Inside triangle ABC, P is a point such that angle(PBA)=angle(PCA). Q and R are the feet of the perpendiculars on AB and AC respectively from P. M is the midpoint of BC. Determine whether QM=RM or not.\r\n\r\nTanvir", "Solution_1": "Call $ D, E$ be the midpoints of $ BP, CQ$ respectively\r\n\r\n$ QDM$ and $ MER$ are equal triangles\r\n\r\nSo $ MQ=MR$", "Solution_2": "Thanks. Too easy. stupid of me, not to see this.\r\n\r\nTanvir", "Solution_3": "[quote=\"tanvirabs\"]Inside triangle ABC, P is a point such that angle(PBA)=angle(PCA). Q and R are the feet of the perpendiculars on AB and AC respectively from P. M is the midpoint of BC. Determine whether QM=RM or not.\n\nTanvir[/quote][/quote]\r\n\r\n [b]History ![/b]\r\n\r\n I think that this nice problem was proposed by Great Britain at an IMO ( Short List) , but I can not remember the year." } { "Tag": [ "logarithms", "trigonometry" ], "Problem": "[size=200]$ find$ $ i^{i^{i}}$[/size]", "Solution_1": "Just another great question.... :roll:", "Solution_2": "someone got an answer???", "Solution_3": "[quote=\"asintota\"][size=200]$ find$ $ i^{i^{i}}$[/size][/quote]\r\n\r\nThe conventiol definition of exponents is used here ... \r\n\r\n$ i^{i}= e^{i\\log i}$ now $ \\log i =\\ln |i|+i \\arg (i) = \\frac{\\pi}{2}i$\r\n\r\nAnd so, $ i^{i}= e^{-\\pi/2}= z$\r\n\r\nSo, $ i^{z}= e^{z\\log (i)}$ but $ z\\log (i) =e^{-\\pi/2}\\cdot \\frac{\\pi}{2}i$\r\n\r\nSo, $ i^{z}= i^{i^{i}}= \\cos \\left( \\frac{\\pi}{2}e^{-\\pi/2}\\right)+i\\sin \\left( \\frac{\\pi}{2}e^{-\\pi/2}\\right)$" } { "Tag": [ "AMC", "AIME", "probability", "USA(J)MO", "USAMO", "AIME II" ], "Problem": "Another tumbleweed :maybe: \r\nso how did everyone do on the AIME I?", "Solution_1": "took it for practice and got an 8.\r\n\r\ndid some stupid stuff, but could've messed up even more.", "Solution_2": "AIME II: 6\r\nAnswered 1-7, Missed #1\r\n\r\nGot to work hard for next year... No more of this missing the cutoff stuff and stress about large probability of missing the cutoff... \r\n\r\nAnd... Nice job, Kongster and good luck on the USAMO!", "Solution_3": "i officially got a 1. \r\n\r\nbut i put 7 correct answers\r\n\r\ni didn't bubble my ansewrs to 1-8 :|", "Solution_4": "Lol thanks, im totally going to fail due to lack of studying though :P\r\nwait unimpossible, you just didnt bubble in the answers?", "Solution_5": "[quote=\"Kongster\"]Lol thanks, im totally going to fail due to lack of studying though :P\nwait unimpossible, you just didnt bubble in the answers?[/quote]\r\n\r\nyeah i didn't bubble ansewrs for the first line of questions (#1-8) XD", "Solution_6": "I haven't posted anything in ages.\r\nDo I still count as a student of Virginia? \r\nDo my roots go down deep enough for me to \r\nhold this view hundreds of miles away?\r\nAnyways, as the year flies past and I turn\r\ninto a depressing lower, I wish all of you guys\r\nluck as freshmen and 8th graders, and there's\r\ndefinitely something to look forward to in high school...\r\nthat is, graduation and senior slackoff.\r\nSurvive for one more month~", "Solution_7": "In Young! Gosh! You're still a student of Virginia in our hearts here at TJ for those who remember you from longfellow. I would guess your AIME score this year was double digit but that's just my speculation.\r\n\r\nNice poem and have fun at MOP!", "Solution_8": "i got a 7 this year.\r\n\r\ni came out of the test pretty relieved that i didn't make more than a couple mistakes. I thought a 7 would be safe for USAMO with a 138 amc 10.\r\n\r\nturns out 7 was pretty average. Looks like USAMO will evade me for at least another year.", "Solution_9": "unimpossible, what grade are u in?", "Solution_10": "unimpossible is in 9th grade\r\n\r\n@unimpossible, the average for AIME is between 2 and 3. I'm sure that the USAMO index will be around 204 this year, cuzz it was like a medium hard AIME.\r\n\r\nI really really really sucked at this year's AIME,", "Solution_11": "yeah i really didn't improve much from last year.\r\n\r\nif you want some advice, if you want to do well in extracurricular academic stuff, don't do sports.\r\n\r\ni got pretty pathetic at math during basketball...", "Solution_12": "What extracurricular academy stuff do you do, unimpossible?" } { "Tag": [ "floor function", "ceiling function", "inequalities", "algebra proposed", "algebra" ], "Problem": "Suppose $x_{1},x_{2},\\dots,x_{n}\\in\\mathbb R$ are different numbers. Prove that there exist $f(x_{1}),f(x_{2}),\\dots,f(x_{n})$ such that $f(x_{i})\\in\\{\\lfloor x_{i}\\rfloor,\\lceil x_{i}\\rceil\\}$ and for each $S\\subset \\{1,2,\\dots,n\\}$: \\[|\\sum_{\\sigma\\in S}f(x_\\sigma)-\\sum_{\\sigma\\in S}x_\\sigma|\\leq \\frac{n+1}4\\]", "Solution_1": "I don't undestand, what is x? $f(x_{i})$ integer?\r\nIf $f(x_{i})=x_{i}$ the sum is 0.", "Solution_2": "It had a little typo. I corrected it. You see $f(x_{i})\\in\\mathbb Z$. You must choose $f(x_{i})$ such that the inequality holds.", "Solution_3": "$f(x_{i})$ can be choose following way: \r\n1. $f(x_{1})$ is nearest integer. It give $|f(x_{1})-x_{1}|\\le \\frac{1}{2}$.\r\n2. Let $f(x_{1}),..,f(x_{k})$ chose, suth that $|S_{k}|\\le \\frac{1}{2}$, were \\[S_{k}=\\sum_{i=1}^{k}(f(x_{i})-x_{i})\\] .\r\nLet ${q=[x_{k+1}}-x_{k+1},-1\\frac{n+1}4$. Thus $x_{k+1}>\\frac{n+1}{4(k+1)}$. Then $x_{k+1}+\\dots+x_{n}\\geq \\frac{(n+1)(n-k)}{4(k+1)}\\geq (n-k)-\\frac{n+1}4$. And the last inequality holds with simple AM-GM." } { "Tag": [ "AMC", "USA(J)MO", "USAMO" ], "Problem": "How much do you have to score on CMO (Canadian Mathematics Olympiads) to represent Canada at IMO. \r\nAnd to say, lowest required score.\r\nI know it's based upon a percentile but from your experiences what should be the minimum score?\r\nThanks", "Solution_1": "To represent at the IMO, your USAMO, APMO and CMO scores are taken into consideration", "Solution_2": "So it is not only your CMO score? I thought that it was the top six who were selected annually from the CMO. Also, how do you participate in the APMO?" } { "Tag": [ "calculus", "derivative", "limit", "induction", "real analysis", "real analysis unsolved" ], "Problem": "Say f:R->R has n derivatives at zero, and they all have value zero. Then is it true that f is little-oh of x^n at zero?\r\n\r\nIf we have that f is n+1 times continuously differentiable in a neighborhood of zero, then by Taylor's theorem, we could show this to be true. However, we only assume f to have an nth derivative at the point x=0 (note that for this to be true, f must have n-1 derivatives in a neighborhood of zero).\r\n\r\nBy definition of the derivative, this is true for n=1.", "Solution_1": "You sould keep disctinct Taylor theorem and Peano theorem (in Italy we use this terminology). According to the latter \r\n$ f(x) \\minus{} P_n(x) \\equal{} o((x \\minus{} x_0)^n)$ as soon as $ f$ admits n derivatives at $ x \\equal{} x_0.$ The proof makes use of the l'Hopital's rule $ n \\minus{} 1$ times on the quantity \r\n$ {f(x) \\minus{} P_n(x)\\over (x \\minus{} x_0)^n}$ until you obtain $ {1\\over n!}\\lim_{x\\to x_0}\\left({f^{(n \\minus{} 1)}(x) \\minus{} f^{(n \\minus{} 1)}(x_0)\\over x \\minus{} x_0} \\minus{} f^{(n)}(x_0)\\right) \\equal{} \\lim_{x\\to x_0} (f^{(n)}(x_0) \\minus{} f^{(n)}(x_0)) \\equal{} 0$ (actually an induction to make the proof rigorous would be necessary). By the Peano theorem you just know that the remainder is $ o((x \\minus{} x_0)^n)$ and nothing more. The Taylor theorem gives you more information about the remainder but at least the $ (n \\plus{} 1)$-th derivative except at $ x_0$ is necessary", "Solution_2": "Thanks, that was very helpful!", "Solution_3": "Without L'Hospital, you can do the following. Let me just assume $ f(0) \\equal{} f'(0) \\equal{} f''(0) \\equal{} 0$ for simplicity. The mean value theorem gives $ f(x) \\equal{} f(x) \\minus{} f(0) \\equal{} f'(c_x)x \\equal{} [(f'(c_x)\\minus{}f'(0))/c_x]c_xx$. As $ x\\rightarrow 0$, the term in brackets $ \\rightarrow f''(0) \\equal{} 0$, and $ |c_xx| < x^2$ , so $ f(x) \\equal{} o(x^2)$.", "Solution_4": "Nice. \r\n\r\n[quote=\"WWW\"]Without L'Hospital, you... [/quote]" } { "Tag": [ "limit", "logarithms", "calculus", "integration", "calculus computations" ], "Problem": "Evaluate\r\n$\\displaystyle\\lim_{n\\to\\infty}\\left(\\prod_{k=1}^n\\left(1+\\frac{k}{n}\\right)\\right)^{\\frac{1}{n}}$\r\n\r\ni have been able to simplify it to $\\displaystyle\\lim_{n\\to\\infty}\\left(\\frac{1}{n}\\right)((n+1)(n+2)\\dots(2n))^{\\frac1n}$\r\nbut from here i have no clue.\r\ncould someone please point me in the right direction.\r\nthanks.", "Solution_1": "[quote=\"maokid7\"]\ncould someone please point me in the right direction.\nthanks.[/quote]\r\nTake the logarithm of the product and think of the resulting expression as of a Riemann sum.", "Solution_2": "ok so i got an answer. i just want to make sure it is correct\r\n[hide=\"my solution\"]\nso as suggested i first take the natural log.\n$f(n)=\\displaystyle\\lim_{n\\to\\infty}\\left(\\prod_{k=1}^n\\left(1+\\frac{k}{n}\\right)\\right)^{\\frac{1}{n}}$\nso\n$ln(f(n))=\\displaystyle\\lim_{n\\to\\infty}\\frac{1}{n}\\sum_{k=1}^nln\\left(1+\\frac{k}{n}\\right)$\nso turning this into an integral we have\n$ln(f(n))=\\displaystyle\\int_{1}^2ln(x)dx$ (if i made a mistake it is most like that it was made here :? )\nso evaluating this we find that $ln(f(n))=2ln2-1$.\nto find $f(n)$ we just undo the $ln$ by raising $e$ to both sides.\nso $f(n)=e^{2ln2-1}=\\frac{4}{e}$.\n[/hide]", "Solution_3": "You did it correctly.\r\n\r\n$\\frac1{n} = dx$\r\n\r\nThus the limits of integration vary are from $0$ to $1$, this makes $x = k dx$.\r\n\r\n$\\frac{k}{n} = x \\rightarrow \\ln (f(n)) = \\int_0^1 \\ln (1+x) \\, dx = \\int_1^2 \\ln x \\, dx$\r\n\r\n$\\int \\ln x \\, dx= x\\ln x - x$\r\n\r\n$\\int_1^2 \\ln x \\, dx = 2\\ln 2 - 2 - \\ln 1 + 1 = 2\\ln 2 - 1$\r\n\r\nI just wrote in the extra steps for clarification.\r\n\r\nGood job." } { "Tag": [ "geometry", "3D geometry", "frustum", "symmetry", "LaTeX" ], "Problem": "20. A fuel tank is in the shape of a right frustum (truncated cone) as shown. Its parallel circular bases have area square feet and square feet respectively. When the fuel tank is half filled as shown (up to the axis of symmetry), the area of the surface of the fuel (cross section perpendicular to the bases, including the axis of symmetry) is 108 sq. ft. What is the volume, in cubic feet, of the fuel in the tank?", "Solution_1": "I believe that in the line \"area __ square feet and __ square feet respectively\", the blanks need to be filled in in order to do this problem.", "Solution_2": "b-flat: I don't believe you can copy text that has some latex and expect the latex to show up when pasted unless you copy the code itself - same with the other threads that besttate said he couldn't answer.", "Solution_3": "20. A fuel tank is in the shape of a right frustum (truncated cone) as shown. Its parallel circular bases have area square feet and square feet respectively. When the fuel tank is half filled as shown (up to the axis of symmetry), the area of the surface of the fuel (cross section perpendicular to the bases, including the axis of symmetry) is 108 sq. ft. What is the volume, in cubic feet, of the fuel in the tank? that is the question.", "Solution_4": "The problem still has not been fixed...", "Solution_5": "A fuel tank is in the shape of a right frustum (truncated cone) as shown. Its parallel circular bases have area 36 pie square feet and 9 pie square feet respectively. When the fuel tank is half filled as shown (up to the axis of symmetry), the area of the surface of the fuel (cross section perpendicular to the bases, including the axis of symmetry) is 108 sq. ft. What is the volume, in cubic feet, of the fuel in the tank?\r\n\r\nnow it has been fixed." } { "Tag": [ "modular arithmetic", "number theory", "Divisibility Theory" ], "Problem": "Determine all pairs $(n,p)$ of nonnegative integers such that [list] [*] $p$ is a prime, [*] $n<2p$, [*] $(p-1)^{n} + 1$ is divisible by $n^{p-1}$. [/list]", "Solution_1": "If $ p \\equal{} 2$ then $ (p \\minus{} 1)^{n} \\plus{} 1 \\equal{} 2$ so $ n \\equal{} 2$.\r\nIf $ n \\equal{} 1$ then $ (p \\minus{} 1)^{n} \\plus{} 1$ is divisible by $ n^{p \\minus{} 1}$ for any prime $ p$. If $ n \\equal{} 2$, then $ (p \\minus{} 1)^{n} \\plus{} 1$ must be even so $ (p \\minus{} 1)^{n}$ is odd so $ p$ is even but it is prime therefore $ p \\equal{} 2$.\r\nThis gives the solutions $ (1,p)$ and $ 2,2$ when $ n\\leq 3$ or $ p\\leq 3$.\r\n\r\nNow look at the case when $ p,n\\geq 3$. Because then $ p$ is odd, $ (p \\minus{} 1)^{n} \\plus{} 1$ is odd so $ n$ is odd. Let $ x$ be the greatest prime divisor of $ n$, then we have $ x|n|n^{p \\minus{} 1}|((p \\minus{} 1)^{n} \\plus{} 1)$ so $ (p \\minus{} 1)^{n}\\equiv \\minus{} 1\\pmod x$. Now because $ x|n$ then $ x \\minus{} 1$ and $ n$ are coprime so there will exist positive integers $ a,b$ such that $ an \\equal{} b(x \\minus{} 1) \\plus{} 1$. Because $ n$ is odd, $ x$ is odd so $ b(x \\minus{} 1)$ is odd so $ b(x \\minus{} 1) \\plus{} 1$ is even hence $ a$ is odd. But then, if $ (p \\minus{} 1)^{n}\\equiv \\minus{} 1\\pmod x$, $ (p \\minus{} 1)^{an}\\equiv \\minus{} 1\\pmod x$ therefore $ (p \\minus{} 1)^{b(x \\minus{} 1)}*(p \\minus{} 1)\\equiv \\minus{} 1\\pmod x$. But by Fermat's Little Theorem $ (p \\minus{} 1)^{x \\minus{} 1}\\equiv 1\\pmod x$ because $ x$ and $ p \\minus{} 1$ are coprime (otherwise, it would be impossible to have $ n^{p \\minus{} 1}|((p \\minus{} 1)^{n} \\plus{} 1)$ as $ x|n$. Then, $ (p \\minus{} 1)^{b(x \\minus{} 1)}*(p \\minus{} 1)\\equiv \\minus{} 1\\pmod x$ gives $ (1)^{b}*(p \\minus{} 1)\\equiv \\minus{} 1\\pmod x$ so $ p \\minus{} 1\\equiv \\minus{} 1\\pmod x$ so $ x|p$ so $ x \\equal{} p$. But $ x|n$ and $ n < 2p$ so $ n \\equal{} p$.\r\nThen $ n^{p \\minus{} 1}|((p \\minus{} 1)^{n} \\plus{} 1)$ is equivalent to $ p^{p \\minus{} 1 }|((p \\minus{} 1)^{p} \\plus{} 1)$ but if we expand $ (p \\minus{} 1)^{p} \\plus{} 1$ we get:\r\n$ p^{p \\minus{} 1}| p^{p} \\minus{} p^{p \\minus{} 1}*\\binom{p}{1} \\plus{} p^{p \\minus{} 2}*\\binom{p}{2} \\plus{} ... \\plus{} p*\\binom{p}{1} \\minus{} 1 \\plus{} 1$\r\nor:\r\n$ p^{p \\minus{} 1}|p^{2}(p^{p \\minus{} 2} \\minus{} p^{p \\minus{} 3}*\\binom{p}{1} \\plus{} ... \\plus{} \\binom{p}{p \\minus{} 2} \\plus{} 1)$\r\nand therefore $ p^{p \\minus{} 1}|p^{2}$ because the expression in brackets is not divisible by $ p$ hence the only solution possible is if $ p \\minus{} 1\\leq 2$ so $ p\\leq 3$. But we assumed $ p\\geq 3$ so $ p \\equal{} 3$ so $ n \\equal{} 3$.\r\n\r\nSo the solutions are $ (1,p), (2,2), (3,3)$", "Solution_2": "If not have n < 2p , can you prove it ?", "Solution_3": "It is an IMO problem and it also relate with an IMO problem\nFind all positive integer n such that :$\\frac {2^n+1}{n^2}$ is an integer $(*)$\n(put p=3 in A72)...\nDR.TITU ANDREESCU wrote a comment in his book(Number theory:structures,examples and problems),the condition $n<2p$ is not necessary.\nVery lucky that i found a solution of $(*)$ difference from the official solution by Lifting the exponent lemma in this link \nhttp://reflections.awesomemath.org/2007_3.html\nWork similarly,i think we can prove $A72$ without condition $n<2p$", "Solution_4": "[quote=\"taylorcute56\"]If not have n < 2p , can you prove it ?[/quote]\n\nSuppose $ p>2 $ and $ n>1 $. (otherwise, the case is trivial.)\nSince $p$ is odd prime, $n>1$ is odd integer.\nSuppose $q$ is the samllest prime divisor of $n$ s.t. $ q|n|n^{p-1}|(p-1)^{n}+1 $.\n$ (p-1)^{n}\\equiv-1\\pmod q $.................................. [b][i](1)[/i][/b]\n$ (p-1)^{q-1}\\equiv 1\\pmod q $\n$ (p-1)^{gcd(2n, q-1)}\\equiv 1\\pmod q $\nSo, $ (p-1)^{2}\\equiv 1\\pmod q $\nSo, $ p\\equiv 2\\pmod q $ or $ q|p $\n\nIf $ p\\equiv 2\\pmod q $ , by [i][b](1)[/b][/i] and $q>2$, contraduction.\nSo, $ q|p $\nSo, $q=p$\nSo, $ p|n $\n\n$n=p^{s}u$, $ s\\geq 1 $, $ p>2 $, $(u,2)=(u,p)=1$\n$ n^{p-1}|(p-1)^{p^{s}u}+1 $\n$ (p-1)^{p^{s}u}+1 \\equiv p^{s+1} ( \\mod p^{s+2})$\n\n$(p-1)s =v_{p}( n^{p-1}) \\leq v_{p}( (p-1)^{p^{s}u}+1) = s+1 $\n$(p-1)s \\leq s+1$\n$ 2s+1 \\geq sp $\n$ 3 \\geq 2+ \\frac{1}{s} \\geq p \\geq 3 $\nso, $s=1$ and $p=3$.\nso, $n=3u$, $ u\\geq 1 $, $(u,2)=(u,3)=1$.\n\nSuppose $u>1$.\nSuppose $r$ is the samllest prime divisor of $u$ s.t. $ r|u|9u^{2}|8^{u}+1 $.\nSo, $r>3$.\n$ 8^{u}\\equiv-1\\pmod r $\n$ 8^{r-1}\\equiv 1\\pmod q $\n$ 8^{gcd(2u, r-1)}\\equiv 1\\pmod r $\n$ 8^{2}\\equiv 1\\pmod r $\n$ r|63 $\nSo, $r=7$\nSo, $ 7 | 8^{u}+1 \\equiv 2\\pmod 7$, contradiction.\nSo, $u=1$\nSo, $n=3$\nSo, $(n,p)=(3,3)$.\n\n\n\n.", "Solution_5": "[quote=\"Thanh.Minh2009\"]Find all positive integer n such that :$\\frac {2^n+1}{n^2}$ is an integer $(*)$ [/quote]\n\nSuppose $n>1$\nFirst, $ n|2^{n}+1 $\nSuppose $p>2$ is the smallset prime divisor of$n$.\n$ 2^{n}\\equiv-1\\pmod p $\n$ 2^{p-1}\\equiv 1\\pmod p $\n$ 2^{gcd(2n,p-1)}\\equiv 1\\pmod p $\nSo, $p=3$\n$n=3^{s}u$, $s\\geq 1$, $(u,2)=(u,3)=1$\n\n$ 3^{2s}u^{2}|2^{3^{s}u}+1 $\n$2^{u}=x\\equiv-1\\pmod 3$\n$ x^{3^{s}}+1 =(x+1)\\prod_{i=0}^{s-1}(x^{3^{i}2}-x^{3^{i}}+1) $\nFor $i=0, 1, ..., s-1$\n$x^{3^{i}2}-x^{3^{i}}+1=(x^{3^{i}}+1)(x^{3^{i}}-2)+3 \\equiv 3\\pmod 9 $\nand since $(u,2)=(u,3)=1$, \n$x+1=2^{u}+1\\equiv -3, 3\\pmod 9 $\nSo, $ x^{3^{s}}+1$ have exactly $(s+1)$ $3$'s as prime divisor.\nSo, $2s\\leq s+1$\nSo, $s\\leq1$\nSo, $s=1$.\nSo, $n=3u$, $ u\\geq 1 $, $(u,2)=(u,3)=1$.\nSo, $ 9u^{2}|8^{u}+1 $.\n\nAs we have said in previous post, \n[quote=\"luapsedree\"] Suppose $u>1$.\nSuppose $r$ is the samllest prime divisor of $u$ s.t. $ r|u|9u^{2}|8^{u}+1 $.\nSo, $r>3$.\n$ 8^{u}\\equiv-1\\pmod r $\n$ 8^{r-1}\\equiv 1\\pmod q $\n$ 8^{gcd(2u, r-1)}\\equiv 1\\pmod r $\n$ 8^{2}\\equiv 1\\pmod r $\n$ r|63 $\nSo, $r=7$\nSo, $ 7 | 8^{u}+1 \\equiv 2\\pmod 7$, contradiction.\nSo, $u=1$\nSo, $n=3$\n [/quote]\nSo, $n=3$.\n\n.", "Solution_6": "Let $q$ be the smallest prime divisor of $n$. Hence $q|(p-1)^{n}-1$. Again from F.L.T we get $q|(p-1)^{q-1}-1$. Hence we get that $q|(p-1)^{gcd(q-1,2n)}-1$. Since q is the smallest prime divisor of $n$ so $gcd(2n,q-1)=2$. Hence $q|(p-1)^{2}-1$ i.e $q|p$ or $q|p-2$. Since $gcd((p-1)^{n}-1,p-2)=1$ so we conclude that $q|p$. and so $q=p$. Hence $p|n$. Now $n<2p$.So $n=p$. Now the problem is to find all primes such that $p^{p-1}|(p-1)^{p}-1$. Now let $p>4$. Then $p^{3}|(p-1)^{p}-1$ which forces that $p^{3}|p$. a contradiction. so $p<4$ and hence $p=2,3$.Now we check that these two values satisfies the required relations and hence $(n,p)=(2,2),(3,3)$.", "Solution_7": "[quote=rem] Now because $ x|n$ then $ x \\minus{} 1$ and $ n$ are coprime[/quote]\n\nIs this always true?? What about $n=21$, and $x=7$, for example??" } { "Tag": [], "Problem": "Hi everybody!\r\nthe BMO is near(27 april)\r\nWho is coming! \r\n:)", "Solution_1": "Albanian team\r\n\r\n1- Erion Dula\r\n2- \u0130lirian Aliaj\r\n3- Ersiljo Tushaj\r\n4- Gjergji Zaimi (Albanian Eagle)\r\n5- Keler Marku (me)\r\n6- Klevis Ymeri (unexpected)", "Solution_2": "common isnt anyone interested on this topic.\r\nArent there any students partecipating to BMO this year!", "Solution_3": "is BMO = Balkan Mathematical Olympiad ?", "Solution_4": "It's the Balkan Olympiad indeed.", "Solution_5": "My friend, the Bogy Man and I are coming", "Solution_6": "Are you partecipating for the first time this year?", "Solution_7": "The UK (who for a second year running are an invited country) team consists of:\r\n\r\nJos Gibbons\r\nTim Hennock\r\nJonathan Lee\r\nDaniel Lightwing\t\r\nJulia Robson\r\nDominic Rowland\r\n\r\nNo Mathlinkers in the group unfortunately.", "Solution_8": "The Bulgarian team is also ready:\r\n1.Boris Stranjev\r\n2.Anton Bikov\r\n3.Emil Baronov\r\n4.Tzvetelina Tzeneva\r\n5.Deyan Simeonov\r\n6.Dimitar Simeonov(me)\r\n\r\nThe last names of BUL5 and BUL6 are same because of coincidence ;)\r\nHope we have fun and present ourselves well.", "Solution_9": "The Romanian team is as follows: \r\n\r\n(11) Adrian Zahariuc (dzeta) - bronze IMO 2004, gold IMO 2005\r\n(12) Pachitariu Marius (ikap) - silver BMO 2004, 2005, bronze IMO 2005\r\n(10) Dumitrescu Mihai - gold JBMO 2004\r\n(12) Bazavan Eduard\r\n(11) Turea Lucian - gold JBMO 2003\r\n(12) Micnea Dragos - gold JBMO 2003, IMO 2005", "Solution_10": "who do you think will take the first place this year! :D", "Solution_11": "[quote=\"Albanian Eagle\"]who do you think will take the first place this year! :D[/quote]Romania of course :D" } { "Tag": [ "linear algebra", "linear algebra unsolved" ], "Problem": "Here are some interesting questions that I am not quite certain how to do. Any help would be appreciated.\r\n\r\n\r\nhttp://www.math.tamu.edu/~rojas/hw3.pdf", "Solution_1": "Of all people I didn't expect you to try to cheat on your homework :) Anyway, next time attach the PDF in the topic and/or write the problem in a separate post (which is the preferred way).", "Solution_2": "Haha, it's not really hw. I'm doing research at Texas A&M and these are suggested problems to get us to try and think about a research subject. I'm not fishing for answers here. Collaboration is definitely allowed and I posted the link so people know I'm not trying to be sneaky about it. Sorry if it came off that way! :wink:", "Solution_3": "This is for an REU?! The U in REU stands for undergraduate, but this seems to be algebraic geometry. I wish I went there!" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let $ a,b,c\\geq0$ such that $ a\\plus{}b\\plus{}c\\equal{}5$. Prove that\r\n\r\n$ \\frac{(a\\minus{}1)(b\\minus{}1)}{c^2\\minus{}2c\\plus{}2}\\plus{}\\frac{(b\\minus{}1)(c\\minus{}1)}{a^2\\minus{}2a\\plus{}2}\\plus{}\\frac{(c\\minus{}1)(a\\minus{}1)}{b^2\\minus{}2b\\plus{}2}\\leq1$", "Solution_1": "It's can be write into: Given $ a, b, c \\geq\\ 0$ and $ a\\plus{}b\\plus{}c\\equal{}2$. Prove that: $ \\sum\\ \\frac{ab}{c^2\\plus{}1} \\leq\\ 1$\r\nIt's well-known ineq :maybe:", "Solution_2": "[quote=\"nguoivn\"]It's can be write into: Given $ a, b, c \\geq\\ 0$ and $ a \\plus{} b \\plus{} c \\equal{} 2$. Prove that: $ \\sum\\ \\frac {ab}{c^2 \\plus{} 1} \\leq\\ 1$\nIt's well-known ineq :maybe:[/quote]\r\n\r\nThank you very much", "Solution_3": "[quote=\"nguoivn\"]It's can be write into: Given $ a, b, c \\geq\\ 0$ and $ a \\plus{} b \\plus{} c \\equal{} 2$. Prove that: $ \\sum\\ \\frac {ab}{c^2 \\plus{} 1} \\leq\\ 1$\nIt's well-known ineq :maybe:[/quote]\r\n\r\nIt must be $ a,b,c\\ge \\minus{}1$" } { "Tag": [ "inequalities", "geometry", "incenter", "vector", "inequalities proposed" ], "Problem": "In a triangle ABC ,G denotes the centroid and I the incentre\r\n\r\nProve 3(AG^2 +BG^2 +CG^2)/2 > AI^2 + BI^2 +CI^2 >=AG^2 +BG^2+CG^2\r\n\r\nThis is based on well known results so probably it cant be claimed to be original.", "Solution_1": "Let $s=(a+b+c)/2$, $x=s-a$, $y=s-b$ and $z=s-c$, $r$ be the inradius. Then \r\n\r\n$AI^2 + BI^2 +CI^2 = (r^2+x^2) + (r^2+y^2) + (r^2+z^2) = (x^2+y^2+z^2) + 3r^2$.\r\n\r\n$AG^2 +BG^2+CG^2 = (2m_a/3)^2 + (2m_b/3)^2 + (2m_c/3)^2 = (4/9)(m_a^2+m_b^2+m_c^2)$\r\n\r\n$ = (1/3)(a^2+b^2+c^2) = (1/3) ((x+y)^2+(y+z)^2+(z+x)^2)$\r\n\r\n$ = (2/3)(x^2+y^2+z^2+xy+yz+zx) $.\r\n\r\n$r^2= S^2/s^2 = (s-a)(s-b)(s-c)/s =(xyz)/(x+y+z)$.\r\n\r\nSo AI^2 + BI^2 +CI^2 >=AG^2 +BG^2+CG^2 is equivalent to:\r\n\r\n$(x^2+y^2+z^2) + (3xyz)/(x+y+z) \\geq (2/3)(x^2+y^2+z^2+xy+yz+zx) $.\r\n\r\nOr equivalent to: $(x^2+y^2+z^2)(x+y+z) + 9xyz \\geq 2(xy+yz+zx)(x+y+z)$.\r\n\r\nThe last one is equivalent to Schur's inequality.\r\n\r\n\r\nThe other side 3(AG^2 +BG^2 +CG^2)/2 > AI^2 + BI^2 +CI^2 is equivalent to:\r\n\r\n$(1/2)(a^2+b^2+c^2) > AI^2 + BI^2 +CI^2$.\r\n\r\nSince $\\angle{BIC} = \\pi - B/2 - C/2 > \\pi/2$, $a^2 > BI^2 +CI^2$. The rest is trivial.", "Solution_2": "For all M we have : sum(MA^2)>=sum(GA^2)", "Solution_3": "[quote=\"tranminhhoang\"]For all M we have : sum(MA^2)>=sum(GA^2)[/quote]\r\n\r\nYou are right!\r\n\r\nIn general, for any point M, and any reals x,y,z, using vectors, we can prove:\r\n\r\n$(x+y+z)(xMA^2 + yMB^2 + z MC^2) \\geq yza^2 + zxb^2 + xyc^2$.\r\n\r\nTaking x=y=z=1, gives: $MA^2 + MB^2 + MC^2 \\geq \\frac{1}{3}(a^2 + b^2 + c^2) = GA^2+GB^2+GC^2$." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "search", "probability", "geometry", "trigonometry" ], "Problem": "What is usually needed on USAMO to make MOP and how many people are taken on average each year? Also, if anyone knows what the grade level split is it would be greatly appreciated! :lol:", "Solution_1": "Ahhh..\r\n\r\nThis topic has been asked many many times. Why don't you go to:\r\n\r\nhttp://www.unl.edu/amc\r\n\r\nor search this forum? I keep seeing the same question over and over and if people just try to find it, I know they'll find them.\r\n\r\nFor topics, ahhhh, again. All stuffs are in there. Number theory, counting, probability, geometry, trigonometry, sequences and series, and all stuffs plus your knowledge of proof writing. That's the most important part.\r\n\r\n ;)" } { "Tag": [ "LaTeX", "function", "combinatorics unsolved", "combinatorics" ], "Problem": "Does anybody have a nice solution for this problem ( like Double Counting ) ?\r\n\r\nProve that :\r\n\r\n ${\\sum_{k=0}^n } {2k \\choose k} . {2n-2k \\choose n-k} = 4^n$", "Solution_1": "a small hint of LaTeX : use [code]{n \\choose k}[/code] to implement the binomial coefficient ${n \\choose k}$. :) [/list]", "Solution_2": "assume that $a_0=1 ,a_1,a_2,...$ is sequence st$ \\sum a_k.a_{n-k}=1$ now assume g(x) is generating function of {$a_n$} so$ g(x)^2 $ is generating function of {$b_n$} st$ \\sum a_ka_{n-k}=b_n=\\frac{1}{1-x} ,g(x)=\\frac{1}{(1-x)^\\frac{1}{2}} $we can easily see that the coefficient of $x^n$ in$ (1-x)^\\frac{-1}{2} $equals$ {n\\choose k}\\div{4^n}$\r\nso we have$ 4^n=\\sum{2k\\choose k}.{2n-2k\\choose n-k}$ and then we are done also in concrete mathematics at page 186 u can find another solution.", "Solution_3": "A small correction.\r\nSam-n wanted to say that the coefficient of $x^n$ in $(1-x)^{-1/2}$ is $\\frac{(-1)^n}{4^n}{2n \\choose n}$.\r\n\r\nP.S. Russian edition of Concrete Mathematics contains this example at page 214." } { "Tag": [ "algebra", "polynomial", "number theory", "prime numbers", "superior algebra", "superior algebra unsolved" ], "Problem": "For which prime numbers is the quotient ring\r\n\r\n$\\mathbb{F}_{p}[x]/(x^{2}+1)$ a field??\r\n\r\n\r\nWe r given $(x^{2}+1)$ is reducible in $\\mathbb{F}_{p}$ \r\nand in the first part of this question I have\r\nProved that the multiplicative\r\ngroup of Fp contains an element of order 4 and Deduced that p is congruent to 1 modulo 4.\r\n\r\n\r\nI know that its a field iff its irreducible , any one knows how to apply this to the question, or wat idea i must focus on .. Thanks again", "Solution_1": "[quote=\"ueamath\"]For which prime numbers is the quotient ring\n\n$\\mathbb{F}_{p}[x]/(x^{2}+1)$ a field??\n[/quote]\r\n\r\nWe need that $x^{2}+1$ to be irreducible in $\\mathbb{F}_{p}[x]$\r\n\r\nSince this polynomial has degree 2 it is necessary and sufficient that $x^{2}+1$ have no zeros in $\\mathbb{F}_{p}$\r\n\r\nThat is, $x^{2}+1 = 0$ has no solutions for all $x\\in \\mathbb{F}_{p}$. That means, $x^{2}\\not =-1$. Thus, $-1$ is a quadradic [b]non[/b]-residue modulo prime $p$. \r\n\r\nBut it is a know fact that the Legendre symbol $(-1/p)=-1$ if and only if $p \\equiv 3 \\mod 4$ for odd primes. And it is trivial to see that for $p=2$ (even primes) it cannot be irreducible. Thus, $p=4k+3$ are all the primes." } { "Tag": [ "floor function", "search", "combinatorics unsolved", "combinatorics" ], "Problem": "The maximum number of pairwise incomparable subsets of a given set $X$ of $n$ elements is $\\displaystyle \\binom{n}{\\left\\lfloor \\frac{n}{2} \\right\\rfloor}$.", "Solution_1": "You are right this is Sperner's theorem about antichains.\r\nUse the search button with keywords 'Sperner' and/or 'antichain'...\r\nFor exemple :\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?highlight=sperner&t=1162\r\n\r\nPierre." } { "Tag": [ "vector" ], "Problem": "1. The line which contains all points $(x,y,z)$ which are of the form $(x,y,z)=(2,-2,5)+k(1,-3,2)$ intersects the plane $2x-3y+4z=163$ at $P$ and intersects the $yz$-plane at $Q$. Find the distance $PQ$. Express your answer as a simplified radical expression.\r\n\r\n2. The intersection of the planes $2x-y-3z=8$ and $x+2y-4z=14$ is the line $L$. Find the value $A$ so that $L$ is perpendicular to the line through $(A,2,2)$ and $(6,11,-1)$.", "Solution_1": "Anyone? :(", "Solution_2": "[hide=\"#1\"]Points of the line are $(2+k,-2-3k,5+2k)$. Plug them into the plane equation to find $k$:\n\n$2(2+k)+3(2+3k)+4(5+2k)=163$\n\n$19k=133$\n\n$k=7$\n\nHence $P(9,-23,19)$\n\nThe equation of $yz$-plane is $x=0$, therefore $2+k=0\\iff k=-2$ gives $Q(0,4,1)$\n\nThe distance is $PQ=\\sqrt{9^{2}+27^{2}+18^{2}}=9\\sqrt{1+3^{2}+2^{2}}=9\\sqrt{14}$\n[/hide]\n\n[hide=\"#2\"]Rewrite the equations of the planes thus:\n\n$2x-y=8+3z$\n$x+2y=14+4z$\n\nMultiply the first by $2$ and add it to the second to obtain\n\n$5x=30+10z\\implies x=6+2z$\n\nMultiply the second by $-2$ and add it to the first to obtain\n\n$-5y=-20-5z\\implies y=4+z$\n\nTherefore, the general point of the line is $(6+2z,4+z,z)$. Its vector is $\\langle 2,1,1\\rangle$. The vector of the other line is $\\langle 6-A, 9,-3\\rangle$. The dot product of the two must be zero:\n\n$2(6-A)+9-3=0\\implies 18-2A=0\\implies A=9$[/hide]" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "The computers in a computer room are in a network such that every computer is connected by a cable to exactly three others. The computers can exchange data directly or indirectly (via other computers). Let $ k$ be the smallest number of computers that need to be removed so that two of the remaining computers can no longer exchange data or there is only\r\none computer left. Let $ l$ be the smallest number of cables that need to be removed so that some two of the remaining computers can no longer exchange data. Show that $ k \\equal{} l$.", "Solution_1": "http://www.artofproblemsolving.com/Forum/viewtopic.php?t=40251" } { "Tag": [ "Columbia", "Princeton", "college", "articles", "geometry" ], "Problem": "Hey ya'll,\r\n\r\nI just came back from a few college tours and I've been thinking about what I want to do in college. One of the things that has been in the front of my mind is really learning a language. I'm currently taking spanish in highschool and will have completed 4 years by the time I graduate, but I'm not quite sure if that's what I want to do in college.\r\n\r\nAfter hearing that Columbia offers classes that teach 40+ different languages, I was trying to figure out which language I might take when I attend college.\r\n\r\nWhat I want is a language that I will be able to use quite frequently, that I will get some mileage out of. I want to master this language (probably spending a semester or two doing a 'study abroad' program in the country of the language and maybe obtaining a minor in it) and if I don't put it to any use, I'm sure that I won't recall it. \r\n\r\nSo, any suggestions on what the best language to learn nowadays for an aspiring mathematician are quite welcome. Reasoning behind the suggestion is also encouraged!!\r\n\r\nThanks :)", "Solution_1": "russian.", "Solution_2": "[quote=\"manuel\"]russian.[/quote]\r\n\r\nWhy russian?", "Solution_3": "Strongly encourage you to learn Chinese, or other Far East Languages like Korean or Japanese, if you want to do well in math. Not only try to speak those languages, but to memorize the characters and to read and to write them!!! Because of this:\r\n\r\nhttp://www.msnbc.msn.com/id/13560741/", "Solution_4": "The linked study is probably baloney (I say this as a parent of children who are native speakers of both Mandarin and English). Chinese is a fine language to try to learn, and Columbia would be a good place to try to learn it, because it is only a subway ride away from neighborhoods in New York City where you can hear Mandarin conversations all over the place. Princeton requires its math graduate students to show proficiency in either French, German, or Russian--presumably because a lot of important articles about mathematics have been published in those languages. See what your graduate school requirements might be, but other than that, take whatever language you feel like, or even two or three.", "Solution_5": "To me, the three most useful languages to know are: English, Chinese, and Spanish. \r\n\r\nEnglish because you live in the U.S., and its spoken widely around the world.\r\n\r\nChinese because China is developing very quickly and becoming very important in the economics of the world, plus there are a lot of mathematicians in China.\r\n\r\nSpanish because of all the Hispanics living in the U.S. I guess I'm fairly biased, though, living in Texas, where in many communities Spanish is the primary language used.\r\n\r\nEdit: I agree with tokenadult on the study.", "Solution_6": "I would say russian because of the beauty of the russian style of doing mathematics and of researching.\r\nIf you are interested in applications of math into other areas or in the technical part of mathematics, well, yes chinese.\r\nBut i would say, that if you want to read a lof of creative stuff in math you should mainly learn an eastern european language. \r\nAlthough, it won't be \"useful\" in the \"real world\", but it will allow you to learn a lot of beautiful stuff in math.", "Solution_7": "Yeah i always was under the assumption that German is the language used quite frequently in the sciences (including mathematics, obviously).", "Solution_8": "Many graduate schools in math require you to pass (apparently very easy) exams in two of French, German, and Russian, so you may consider one of those. Also, if you want to go to Budapest Semesters in Mathematics (which is apparently an excellent program; JBL can tell you more since he attended), you may consider learning Hungarian (which is probably a more difficult language to learn than most)." } { "Tag": [ "function", "algebra solved", "algebra" ], "Problem": "f is a real-valued function on the reals. It satisfies \r\nf(x^3+y^3)=(x+y)(f 2 (x)+f(x)f(y)+f 2 (y)) for all x,y. Prove that f(1996x)=1996f(x) for all x.", "Solution_1": "Chinese 1996.\r\nSolution in the 1996 - 1997 book from Andreescu and Feng. See \r\n\r\nhttp://www.unl.edu/amc/a-activities/a4-for-students/mc96-97-01feb.pdf ." } { "Tag": [], "Problem": "A yo-yo has an outer \"wheel\" with a diameter of 7 cm and an inner \"wheel\" with a diameter of 1 cm. The inner wheel is wound with 30 cm of string and there is 40 cm of string left out. If the yo-yo is placed on a table and the string is pulled, then how far and in what direction will the yo-yo move before the string is either stretched completely out or used up.", "Solution_1": "I still can't prove the direction the yo-yo's going to go...plus there's the problem of inertia and friction, so I'm guessing all these non-mathematical/physical variables changes the answer..." } { "Tag": [ "function", "inequalities unsolved", "inequalities" ], "Problem": "Let $a,b,c$ be three positive real numbers. Prove that:\r\n$\\sqrt{\\frac{a+b}{c}+\\frac{b+c}{a}+\\frac{c+a}{b}}+2\\sqrt{\\frac{ab+bc+ca}{a^2+b^2+c^2}} \\geq \\sqrt{6}+2$ :)", "Solution_1": "Oh no, it is dropping to the bottom :spider: , help me out of it.", "Solution_2": ":D First I think that you had solved it :P", "Solution_3": "Yes, I had a solution, use ABC :D .", "Solution_4": "[quote=\"Anh Cuong\"]Yes, I had a solution, use ABC :D .[/quote]\r\n\r\nwhat is ABC?", "Solution_5": "Oh, apology for my lateness. It was long time since the last time i came here.\r\n$ABC$ is my new method in proving inequality. The solution using $ABC$ for this problem is quite short, you can download this method in http://www.mathnfriend.net :), but of course it is Vietnamese :P .", "Solution_6": "please write method ABC", "Solution_7": "please write method ABC \r\n\r\n Alisherstudios@inbox.ru", "Solution_8": "Hey Anh Chuong can you write this method for all of us?", "Solution_9": "See here : http://www.mathlinks.ro/Forum/viewtopic.php?p=440463#p440463 :)", "Solution_10": "[quote=\"mathmanman\"]See here : http://www.mathlinks.ro/Forum/viewtopic.php?p=440463#p440463 :)[/quote]\r\n\r\nThank you very much !!! :)", "Solution_11": "[quote=\"Anh Cuong\"]Oh, apology for my lateness. It was long time since the last time i came here.\n$ABC$ is my new method in proving inequality. The solution using $ABC$ for this problem is quite short, you can download this method in http://www.mathnfriend.net :), but of course it is Vietnamese :P .[/quote]\r\n\r\nCan you complete this problem?", "Solution_12": "solution by ABC method :\r\nThe function in abc is sunken so it attains minimum when b=a or b=0 . Then we have to find the minimum of \r\n$\\sqrt{2x+2+\\frac{2}{x}}+2\\sqrt{\\frac{x^{2}+2x}{2x^{2}+1}}$ \r\n\r\nwhere $x=\\frac{a}{c}$ . The first is root has minium $\\sqrt{6}$ which can be found easy by $x+\\frac{1}{x}\\geq 2$ and the second has minimum $2$ and it easy to find this by derivation .\r\n\r\nWhen one variable is tend to $0$ it is easy .\r\nQED" } { "Tag": [], "Problem": "A balloon made up of rubber permeable to all isotopic forms of hydrogen is filled with pure deuterium ($ \\ce{D2}$) and is placed inside a box filled with pure hydrogen ($ \\ce{H2}$). What will happen?", "Solution_1": "Since H2 has a lower molar mass than D2, it has a faster effusion rate. Eventually, all the D2 will be replaced by H2, assuming there is enough H2 in the box to replace the D2 in the balloon." } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "Let $ f(G)$ be the minimal number of paths (repetition of vertices is allowed, but repetition of edges isn't) such that together they contain all the edges in the graph, each exactly one time (meaning, the paths are disjoint of one another). The paths may be open or close.\r\n\r\nNow, let $ G$ be a graph which has at least two vertices in any connected component of it ([u]there are no isolated vertices[/u]).\r\n\r\nShow that $ f(G) = \\max (\\frac{r}{2},1)$ for a [b]connected[/b] graph, where $ r$ is the number of vertices with an odd degree.\r\n\r\nCan you give a formula for $ f(G)$ if $ G$ is not a connected graph?", "Solution_1": "I am correcting something in the statement which may not seem clear. In the formula $ f(G)=\\max (r/2,1)$, $ r$ is the number of vertices with an odd degree but is not necessarily the number of all the vertices in the graph (meaning, there could be some vertices with even degree)." } { "Tag": [ "induction", "Additive Number Theory" ], "Problem": "Let $a_{1}=1$, $a_{2}=2$, $a_{3}$, $a_{4}$, $\\cdots$ be the sequence of positive integers of the form $2^{\\alpha}3^{\\beta}$, where $\\alpha$ and $\\beta$ are nonnegative integers. Prove that every positive integer is expressible in the form \\[a_{i_{1}}+a_{i_{2}}+\\cdots+a_{i_{n}},\\] where no summand is a multiple of any other.", "Solution_1": "We had this in the QEDMO (an olympiad organized by Darij and me in the QED association), so I just copy the solution I proposed there:\r\n\r\nWe use induction:\r\nFor $n=1=2^0 3^0$, it's true.\r\nIf $n>1$ is even, then $n=2m$ with positive integer $m$. Now $m$ is by induction hypothesis of the desired form, $m=2^{a_1} 3^{b_1} + 2^{a_2} 3^{b_2} + ... + 2^{a_k} 3^{b_k}$. But then $n=2^{a_1+1} 3^{b_1} + 2^{a_2+1} 3^{b_2} + ... + 2^{a_k+1} 3^{b_k}$ is also of the desired form.\r\nIf $n>1$ is odd, then let $n=2^0 3^p$ if $n$ is a power of $3$ and we are done then.\r\nOtherwise $n-3^p$ is even and $m=\\frac{n-3^p}{2}$ is of desired type, $m=2^{a_1} 3^{b_1} + 2^{a_2} 3^{b_2} + ... + 2^{a_k} 3^{b_k}$. This gives $n=2^{a_1+1} 3^{b_1} + 2^{a_2+1} 3^{b_2} + ... + 2^{a_k+1} 3^{b_k} + 2^0 3^p$. This fulfills all requirements, because $0< a_i+1$ and $p>b_i$ fr alle $i$, such that the new summand does neither divide one of the old ones nor is divided by one of them." } { "Tag": [ "geometry", "circumcircle", "trigonometry", "perpendicular bisector", "angle bisector", "geometry solved" ], "Problem": "Points P, Q lie inside angle BAC such that BP = CP and BQ = CQ. Suppose that ABP + ACQ = 180. Prove that BAP = CAQ.", "Solution_1": "Here's a hint: Draw the circumcircle of triangle ABC. Pick an arbitrary point R along minor arc BC (excluding points B and C). Then the perpendicular bisector of BC meet lines BR and CR at P and Q, respectively. Now after using sine law, all that is left is to prove:\r\n\r\n|AP|*|BQ|=|AQ|*|BP|, where || denotes the length of the line segment. :)", "Solution_2": "Here are two solutions:\r\n\r\nThe first one:\r\n\r\nLet $D,E$ be the points where the internal, and, respectively, external bisector of $\\angle BAC$ cuts the circumcircle of $ABC$. We have $\\angle ABP+\\angle ACQ=\\pi=\\angle ABD+\\angle ACD=(\\angle ABP+\\angle PBD)+(\\angle ACQ-\\angle DCQ)$, and from ehre we deduce $\\angle PBD=\\angle DCQ$, which is, in turn, equal to $\\angle DBQ$, so $BD$ is the angle bisector of $\\angle PBQ$. In the same way we prove that $BE$ is the external angle bisector of $\\angle PBQ$, so the circumcircle of $ABC$ is, in fact, an apollonain circle of the triangle $PBQ$. The conclusion follows immediately from here, since the apollonian circle of $PBQ$ corresponding to vertex $B$ is the locus of those points $A$ for which $AD$ is the bisector of $\\angle PAQ$, and $AD$ is also the angle bisector of $\\angle BAC$.\r\n\r\nThe second one:\r\n\r\nWe move $P$ on the perpendicular bisector $\\ell$ of $BC$ and move $Q$ accordingly, so as to preserve the condition given in the hypothesis. The map $P\\mapsto Q$ is an involution on $\\ell$, so it induces an involution $AP\\mapsto AQ$ on the pencil of lines through $A$. An involution is determined by the images of two lines, so it suffices to prove the conclusion for two particular positions of $P$, chosen so that it's easier to prove there. To this end, take $P=D$ ($D$ is defined above), and then $P=O$, the circumcenter of $ABC$. \r\n\r\nIn the first case we'll have $P=Q$, so the conclusion holds automatically, while in the second case, $Q$ will go to infinity, and $AQ$ will be the $A$-altitude of $ABC$, and it's well-known and easy to prove (very simple angle chase) that the $A$-altitude is isogonally conjugate to $AO$.", "Solution_3": "Just wanted to tell that I have updated my note \"Isogonal conjugation with respect to a triangle\" (avaliable at http://www.cip.ifi.lmu.de/~grinberg/geometry2.html and http://www.artofproblemsolving.com/Forum/viewtopic.php?t=18472 ), and it contains a result (Theorem 19) that slightly extends this problem:\n\n[color=blue][b]Theorem.[/b] Let $ABC$ be a triangle, and let $P$ and $Q$ be two points on the perpendicular bisector of the segment $BC$. We use directed angles modulo $180^{\\circ}$.\n\n[b](a)[/b] The following five assertions $\\mathcal{F}_{1}$, $\\mathcal{F}_{2}$, $\\mathcal{F}_{3}$, $\\mathcal{F}_{4}$ and $\\mathcal{F}_{5}$ are pairwisely equivalent:\n\n[b]Assertion $\\mathcal{F}_{1}$:[/b] We have $\\measuredangle ABP=\\measuredangle ACQ$.\n[b]Assertion $\\mathcal{F}_{2}$:[/b] We have $\\measuredangle ACP=\\measuredangle ABQ$.\n[b]Assertion $\\mathcal{F}_{3}$:[/b] We have $\\measuredangle BCP+\\measuredangle BCQ=\\measuredangle BAC$.\n[b]Assertion $\\mathcal{F}_{4}$:[/b] We have $\\measuredangle PBC+\\measuredangle QBC=\\measuredangle BAC$.\n[b]Assertion $\\mathcal{F}_{5}$:[/b] The points $P$ and $Q$ are inverse to each other wrt the circumcircle of triangle $ABC$.\n\n[b]b)[/b] If one of the five assertions $\\mathcal{F}_{1}$, $\\mathcal{F}_{2}$, $\\mathcal{F}_{3}$, $\\mathcal{F}_{4}$ and $\\mathcal{F}_{5}$ holds, then the lines $AP$ and $AQ$ are isogonal to each other wrt the angle $CAB$.[/color]\n\nYour problem is a different way to state that, if Assertion $\\mathcal{F}_{1}$ holds, then the lines AP and AQ are isogonal to each other wrt the angle CAB. In fact, your conditions BP = CP and BQ = CQ mean that the points P and Q lie on the perpendicular bisector of the segment BC; your condition < ABP + < ACQ = 180\u00b0 becomes Assertion $\\mathcal{F}_{1}$ when it is rewritten using directed angles modulo 180\u00b0; finally, your assertion < BAP = < CAQ is equivalent to saying that the lines AP and AQ are isogonal to each other wrt the angle CAB.\n\n darij" } { "Tag": [ "AMC", "AIME" ], "Problem": "What is the alternate AIME?", "Solution_1": "It's another version of the AIME for schools that are on vacation during the week of the regular AIME.", "Solution_2": "Can you take both?", "Solution_3": "No, you can only take one.", "Solution_4": "Is it your choice as to when you want to take it? I'd like some extra time to study since its my first time.", "Solution_5": "It's your school's choice more than yours, I think. You also have to pay extra to take the alternate one.", "Solution_6": "Everyone has to take the normal date, unless you have a conflict (such as your school being closed). The alternate AIME is not like the AMC B exams -- it isn't just open to normal registration.", "Solution_7": "An extreme method will be to get out of school due to either \"a dire need of a doctor\" or \"religious ceremony\" though that might not be accepted (well the first one; the second is ensured by the Bill of Rights I guess)", "Solution_8": "I'm pretty sure schools actually do have the option of which AIME to take, but then they have to pay more to take the alternate AIME, so it doesn't seem to make much sense to do that unless there is a school holiday. I think it's even possible to split it up so that some people at a school take the normal AIME and some people take the alternate. (My teacher wanted to order one copy of the alternate AIME just so that he would have the extra problems, but then I convinced him that it would be simple enough to get the test without doing that.)", "Solution_9": "There's also the problem that, in general, the AIMEB is harder than the AIMEA, even though the AMC committee tries to make them about the same difficulty. If I were you, I'd take the A exam.", "Solution_10": "It is possible for schools to split, but I'm fairly certain the expectation is \"only with a conflict.\" I took the alternate AIME one year and thought it was pretty similar to the first AIME. I don't think you'll get a whole lot of advantage putting it off for two weeks, anyhow." } { "Tag": [ "calculus", "trigonometry", "geometry", "rectangle", "perimeter", "inequalities" ], "Problem": "This is a problem I have been working on but just haven't quite got. Maximize the sum: $S=\\sin(a)+\\sin(b)+\\sin(c)$, where $a + b + c = \\pi$ and a,b, and c are the angles of a triangle. All angles are in radians. P.S. the sum may not have a maximum, but a value it approaches, so, yeah.", "Solution_1": "Seems to me that Lagrange multipliers (good ole cheating with calculus) gives a value of pi/3 for each of the angles.", "Solution_2": "Ok, Thanks\r\nHere's what I did:\r\n$S=\\displaystyle \\sin a + \\sin b + \\sin c\\\\\r\n=\\displaystyle \\sin a + \\sin b + \\sin (\\pi-a-b)\\\\\r\n=\\displaystyle \\sin a + \\sin b + \\sin (a+b)\\\\\r\n=\\displaystyle \\sin a + \\sin b + \\sin a \\cos b + \\sin b \\cos a\\\\\r\n=\\displaystyle \\sin a\\left(1+\\cos b\\right) + \\sin b\\left(1+\\cos a\\right)\\\\\r\n=\\displaystyle \\sqrt{1-\\cos^{2} a}\\left(1+\\cos b\\right) + \\sqrt{1-\\cos^{2} b}\\left(1+\\cos a\\right)\\\\\r\n=\\displaystyle \\sqrt{1-v^{2}}\\left(1+w\\right) + \\sqrt{1-w^{2}}\\left(1+v\\right)\\Rightarrow\\cos a = v,\\cos b = w\\\\\r\n=\\displaystyle \\sqrt{(1+v)(1+w)}\\left(\\sqrt{(1-v)(1+w)}+\\sqrt{(1+v)(1-w)}\\right)\\\\\r\n=\\sqrt{hg}\\left(\\sqrt{(2-h)(g)}+\\sqrt{(h)(2-g)}\\right)\\Rightarrow v+1=h,w+1=g\\\\\r\n=g\\sqrt{(2-h)(h)}+h\\sqrt{(g)(2-g)}\\Rightarrow0 1, y > 1, z > 1\n[/hide]", "Solution_26": "[hide=Solution] We can convert to exponential form to get three equations, which are \\begin{align*} x^{24} &= w \\\\ y^{40} &= w \\\\ (xyz)^{12} &=w .\\end{align*} Now because these all have a LCM of $120$ we can make all these exponents equal that and work from their. So raising our first equation to the $5th$, our second to the $3rd$ and our last equation to the $10th$ we get three equations. Our equations are \\begin{align*} x^{120} &= w^5 \\\\ y^{120} &= w^3 \\\\ (xyz)^{120} &= w^{10}. \\end{align*} Now we can divide the third equation by the first and second equations to get $$z^{120} = w^2.$$ Now taking the square root we get $$z^{60} = w.$$ Taking the logarithm we get $$\\boxed{\\log_{z} w = 60}.$$ $\\square$", "Solution_27": "[hide=my sol]\nAfter converting into exponents, we get the following equations:\n\n$x^{24}=w$ \n\n$y^{40}=w$ \n\n$x^{12}*y^{12}*z^{12}=w$\n\nAfter switching out the $x^{12}$ term in terms of $w$ from the first equation and doing the same thing with $y$, we are left with:\n$w^{1/2}*w^{3/10}*z^{12}=w$\n\nBecause,\n$1-\\frac{1}{2}-\\frac{3}{10}=\\frac{1}{5}$\n\nSo,\n $z^{12}=w^{\\frac{1}{5}}$\n\nWhich means $\\log_z w=\\boxed{060}$\n\n[/hide]", "Solution_28": "[hide=mysol]\nI did the same thing a1001 did", "Solution_29": "The given equations rewrite as $x^{24}=w$, $y^{40}=w$, and $x^{12}y^{12}z^{12}=w$, respectively. Raising the third of these equations to the $10$th power, we have that $x^{120}y^{120}z^{120}=w^{10}$, and substituting $x^{24}=w\\implies x^{120}=w^5$ and $y^{40}=w\\implies y^{120}=w^3$ yields $w^8z^{120}=w^{10}$ or $z^{120}=w^2$. Hence, $z^{60}=w\\implies \\log_zw=\\boxed{60}$, as requested.", "Solution_30": "[hide=sol]\n$$\\log_x w = 24 \\implies x^{24} = w \\implies x = w^{1/24},$$\n$$\\log_y w = 40 \\implies y^{40} = w \\implies y = w^{1/40},$$\n$$\\log_{xyz} w = 12 \\implies (xyz)^{12} = w.$$\nSince $x = w^{1/24}$ and $y = w^{1/40},$\n$$(w^{1/24} w^{1/40}z)^{12} = w \\implies z^{60} = w.$$\nHence $\\log_z w = \\boxed{60}.$\n[/hide]", "Solution_31": "We have that $x^{24}=y^{40}=w$, and $x^{12} \\cdot y^{12} \\cdot z^{12}=w$, so $z^{60}=w$, by some boring computations, which I do not want to write. So we have $\\boxed{60}.$", "Solution_32": "[hide]With logarithms, it is much easier to work with logs of the same base than logs of the same argument based on the properties we have. With this in mind, we take the reciprocals of all three equations to get $w$ as the base. We now have $\\[\\log_w x = \\frac{1}{24},\\quad \\log_w y = \\frac{1}{40} \\quad\\text{and}\\quad \\log_{w} xyz = \\frac{1}{12}.\\]$ Now that we have multiple logs of the same base, we can do operations like adding the first two equations, which gives us $\\log_{w} xy = \\frac{1}{15}$. Finally, we can subtract this from the last equation to get $\\log_{w} z = \\frac{1}{60}$, so our answer is $60$. Note that we didn't have to do any hard application of logs after the initial realization to take the reciprocal.[/hide].", "Solution_33": "Let $w=a^{120}$. Then $x=a^5$, $y=a^3$ and $za^8=a^{10}$ so $z=a^2$ so our answer is $60$.", "Solution_34": "These equations are equivalent to $w=x^{24}$, $w=y^{40}$, and $w=(xyz)^{12}$. Then, we get that \n\n\\[w^{1/2} w^{3/10} w^{k} = x^{12}y^{12}z^{12}=w,\\]\n\nso $k=\\frac{1}{5}$, making our answer $\\frac{12}{1/5}=\\boxed{60}$." } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $ G$ be a finite group of order $ n$, $ k|n$ s.t. $ \\left(k,\\frac{n}{k}\\right) \\equal{} 1$.\r\n(a) Suppose there exists a normal subgroup $ N$ of $ G$ of order $ k$. Prove that $ N$ is the only subgroup of order $ k$.\r\n(b) Prove or disprove: all subgroups of order $ k$ are conjugate.", "Solution_1": "(a) suppose $ U$ is a subgroup of order $ k$. we want to show $ U \\equal{} N$; it's sufficient to prove $ U \\subseteq N$. if $ a \\in U$, then $ a^k \\equal{} 1$. now take integers $ u,v$ such that $ uk \\plus{} v(n/k)\\equal{}1$. in the group $ G/N$ of order $ n/k$ we compute $ a \\equal{} (a^k)^u (a^v)^{n/k} \\equiv 1$ mod $ N$, so $ a \\in N$.", "Solution_2": "You might like to read [url=http://en.wikipedia.org/wiki/Hall_subgroup]here[/url] for (b). There is much to say.\r\nBtw $ PSL(2,7)$ is the simple group of order $ 168$.", "Solution_3": "(a) can be generalized to the following: let $ G$ be a group, $ N,H$ subgroups, s.t. $ N$ is finite, $ H$ has finite index and $ (|N|,[G: H]) \\equal{} 1$. Suppose further that either $ N$ or $ H$ is normal. Prove that $ N < H$.\r\nThanks for the link, olorin, I didn't know about that. Problem (a) just led me to this question.", "Solution_4": "This has been discussed [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=215646]here[/url].\r\nAnd we only need $ NH\\le G$ ($ \\Leftrightarrow NH\\equal{}HN$) instead of either group being normal.", "Solution_5": "Thanks - I have exactly the same solution, but I didn't want to assume that $ HN$ is a subgroup because then it's easier to get the right idea." } { "Tag": [ "MIT", "college", "summer program", "Mathcamp", "AMC", "USA(J)MO", "USAMO", "\\/closed" ], "Problem": "Will there be some sort of Programming Math Jam or class in the future? since programming also requires a lot of creativity and problem solving?", "Solution_1": "I agree that would be awesome", "Solution_2": "It would be even better to have a programming class :)\r\n\r\nAlthough this is AoPS...\r\n\r\nbut it is more related to math then some of the other Math Jams", "Solution_3": "well, it is art of problem solving, not art of math problem solving so...\r\n\r\nBut, didn't they have USACO math jam once? I think they did...", "Solution_4": "What's USACO? Also, I think it would be possible, as long as the connection to problem solving is there and there has to be someone that volunteers to host it. I think that last part and just the time it consumes are the reasons there aren't many math jams like these. There have been non-AoPS math jams, like linguistics, MIT, AMC 10/12, and Mathcamp math jams.", "Solution_5": "[quote=\"PowerOfPi\"]What's USACO? Also, I think it would be possible, as long as the connection to problem solving is there and there has to be someone that volunteers to host it. I think that last part and just the time it consumes are the reasons there aren't many math jams like these. There have been non-AoPS math jams, like linguistics, MIT, AMC 10/12, and Mathcamp math jams.[/quote]\r\n\r\nUSA Computing Olympiad.\r\n\r\nBasically, usaco problems are not programming problems. (Though you do need decent programming skills code them. Most of the difficulty comes from finding the algorithms. (Unless you happen to be bad at programming... being good at programming helps a lot.)\r\n\r\nAnd yes, there has been a usaco math jam. There wasn't exactly a lot of programming in the math jam though. \r\n\r\nAlso: There is a high correlation between people good at usamo and people good at usaco:\r\n\r\nIOI USA team this year:\r\n\r\nBrian Hamrick\r\nNeal Wu\r\nTravis Hance\r\nWenyu Cao\r\n\r\nAll of these are either moppers or formerly moppers, and Hamrick and Wenyu are usamo winners.", "Solution_6": "Well, the level of programmers, from \"what's programming\" to \"I created a web 2.0 browser\" is too variable for this to come true." } { "Tag": [ "probability" ], "Problem": "What is the least number of persons required if the probability that two or more of them have the same birthday exceeds $ \\frac 12$", "Solution_1": "[quote=\"#H34N1\"]What is the least number of persons required if the probability that two or more of them have the same birthday exceeds $ \\frac 12$[/quote]\r\n\r\n[hide]\n$ \\frac{366}{366}\\cdot\\frac{365}{366}$ .... and on and on\n\n\n[/hide]", "Solution_2": "[hide=\"hint\"] Count the complement [/hide]\n[hide=\"complete solution\"] This problem is easy if you count the complement. For nobody to share a birthday, there are 365 possibilities for the first person (forgetting about leap years), then 364 for the next, then 363, and so on. Multiplying these together yields $ \\displaystyle \\frac{365!}{(365\\minus{}n)!}$, where $ n$ is the number of people. The total possible combinations of birthdays is $ 365^n$, so we need to find when $ \\displaystyle \\frac{365!}{365^n(365\\minus{}n)!}$ first is less than 1/2, which by trial and error, occurs at $ \\boxed{n\\equal{}23}$[/hide]" } { "Tag": [ "trigonometry", "inequalities proposed", "inequalities" ], "Problem": "$Prove:$ $\\frac{cosx}{sin^{2}x(cosx-sinx)}>8$ $for$ $x\\in (0,\\frac{\\pi}{4})$", "Solution_1": "[quote=\"new_login\"]$Prove:$ $\\frac{cosx}{sin^{2}x(cosx-sinx)}>8$ $for$ $x\\in (0,\\frac{\\pi}{4})$[/quote]\r\n\r\n$\\frac{\\cos x}{\\sin^{2}x(\\cos x-\\sin x)}>8$\r\n\r\n$\\frac{\\sin^{2}x(\\cos x-\\sin x)}{\\cos x}<\\frac{1}{8}$\r\n\r\n$\\sin^{2}x\\cos x-\\sin^{3}x<\\frac{\\cos x}{8}$\r\n\r\nDivide both sides by $\\cos^{3}x$ and we get:\r\n\r\n$\\tan^{2}x-\\tan^{3}x<\\frac{1}{8\\cos^{2}x}=\\frac{1+\\tan^{2}x}{8}$\r\n\r\n$8\\tan^{3}x-7\\tan^{2}x+1>0$\r\n\r\nLet $y=\\tan x$, where $y\\in (0;1)$\r\n\r\n$8y^{3}-7y^{2}+1=(y-1)(8y^{2}+y+1)+2>0$\r\n\r\nBy AM-GM: $8y^{2}+y+1\\ge6y$\r\n\r\n$(y-1)(8y^{2}+y+1)+2\\ge (y-1)6y+2=6y^{2}-6y+2>0$\r\n\r\nIt is true because $D=36-48<0$" } { "Tag": [], "Problem": "The 39th. Rudolf Ortvay international theoretical physics competition for university students (phd. students included) starts tommorow.\r\nThe competition is strictly INDIVIDUAL. NO groups alowed.\r\nThe problems are not textbook style, but lots of thinking required ones.\r\n\r\nThe previous problems and further info can be seen [url=http://ortvay.mafihe.hu/main.html]here[/url].\r\n\r\nThe new problems will also appear on this website.", "Solution_1": "nice questions Thaakisfox are u writing it?", "Solution_2": "well yes I will try to solve as many as I can. :D \r\nWhat about you? :)", "Solution_3": "i will try to solve but i am in my last year of high school :(" } { "Tag": [], "Problem": "In my book it says this: $({A\\cap}{B}\\cap{X})\\cup({A}\\cap{B}\\cap{C}\\cap{X}\\cap{Y})\\cup({A}\\cap{X}\\cap\\bar{A})$ can be reduced to this: $({A\\cap}{B}\\cap{X})$ can some one show me the steps to do this (really how to get the $Y$ and $C$ out)?\r\n\r\n\r\n\r\n\r\nThank You", "Solution_1": "$A\\cap B\\cap C\\cap X\\cap Y$ is contained in $A\\cap B\\cap X$, and $A\\cap X\\cap \\bar{A}$ is contained in $A\\cap \\bar{A}$ which is clearly the empty set. The union of these two with $A\\cap B\\cap X$ must then be $A\\cap B\\cap X$." } { "Tag": [ "algebra", "polynomial", "algebra unsolved" ], "Problem": "For fixed $ \\lambda \\in (0,1)$ and $ n,N \\in \\mathbb{N}$ consider the polynomial\r\n$ P(\\lambda) \\equal{} \\sum_{i\\equal{}0}^n a_i \\lambda^i$\r\nwhere $ a_i \\in \\mathbb{Z}$ and $ \\minus{}N \\leq a_i \\leq N$ $ \\forall$ $ i \\equal{}0,1,...,n$\r\n\r\nGiven that $ P(\\lambda) > 0$ :\r\n(i) Prove that if $ \\lambda$ is a rational number $ \\lambda \\equal{} \\frac{p}{q}$ then $ P(\\lambda) \\geq \\frac{1}{q^n}$\r\n(ii) Find a positive lower bound on $ P(\\lambda)$ for irrational $ \\lambda$.", "Solution_1": "Its really easy for rational $ \\lambda \\equal{} \\frac{p}{q}$ , since $ P(\\lambda)q^n$ is a positive integer it must be atleast $ 1$. \r\n\r\nWhich gives $ P(\\lambda)\\geq\\frac{1}{q^{n}}$\r\n\r\nBut this result cannot be trivially extended to irrationals. \r\n\r\nThis problem will be solved if one can find a rational approximation $ \\frac{p}{q}$ for $ \\lambda$ such that\r\n\r\n$ |\\lambda \\minus{} \\frac{p}{q}| < \\frac{C}{q^n}$\r\n\r\nwhere $ C \\equal{} \\frac{1}{2n^2N}$ is a constant.\r\n\r\nI am not sure if one can always do this. Need Help.", "Solution_2": "This depends on exactly what $ \\lambda$ is. For any real $ n>2$, for almost every $ \\lambda$, there is some $ C$ so that $ \\left|\\lambda\\minus{}\\frac pq\\right|>\\frac{C}{q^n}$ for all integers $ p,q$. At $ n\\equal{}3,C\\equal{}\\frac1{18}$, the set of values in $ [0,1]$ with even one approximation better than that has measure less than $ \\frac1{10}$.\r\n\r\nSo no; you can't find those approximations most of the time. Even for $ n\\equal{}2$, you can only get within $ \\frac{1}{kq^2}$ if there is an entry of at least $ k$ in the continued fraction." } { "Tag": [ "group theory", "abstract algebra", "number theory unsolved", "number theory" ], "Problem": "Hi,\r\n\r\nHow can we show that for all integers n and m we have \r\n\r\nn!m!(n+m)! divide (2n)!(2m)!\r\n\r\n\r\nfor example, if n=m we have n!^2 (2n)! divide (2n)!^2, it's clear because (2n)!/(n!)^2 is a binomial number\r\n\r\n\r\nthanks", "Solution_1": "$ [2x] \\plus{} [2y] \\ge [x] \\plus{} [y] \\plus{} [x \\plus{} y]$ where $ [x]$ is the integer part of $ x$.", "Solution_2": "I know this proof. And I thank you very much.\r\n\r\nI'm looking for a proof that enables to see (2n)!(2m)! / (n!m!(m+n)!) as a cardinal of a finite set.\r\n\r\nFor example, (2n)!/ (n!^2) is exactly the cardinal of sets of cadinal n in a big set of 2n elements.\r\n\r\nOr maybe, for another proof, does there exists a subgroup of S_(2n) X S_(2m) which is isomorph to S_n x S_m x S_(n+m), where S_n is the symmetric group ?\r\n\r\nThanks", "Solution_3": "$ \\binom{2n \\plus{} 2m}{m,n,m \\plus{} n}$ is a multinomial, which is clearly integer.", "Solution_4": "If I recall correctly, $ \\binom{2n \\plus{} 2m}{m, n, m \\plus{} n}$ is not the same thing as $ \\frac{(2n)! (2m)!}{m! n! (m \\plus{} n)!}$." } { "Tag": [ "complex analysis", "complex analysis unsolved" ], "Problem": "For which sets of points on the boundary of a circle can one find a power series that converges exactly on those boundary points (and in the interior)?", "Solution_1": "I posted it some time ago and Fedja told me it is still an open problem.", "Solution_2": "Is it known for finite and cofinite sets?", "Solution_3": "Assume the disk is the unit disk. Then the Maclauren series of $\\sum_{k=1}^{N}\\log(1-z/z_{k})$, with distinct $z_{k}$ on the unit circle $\\mathbb{T}$, converges precisely on $\\mathbb T\\setminus\\{z_{k}\\colon k=1,\\dots,N\\}$. So the answer is yes for cofinite sets.\r\n\r\nAn attempt at generalization: $f(z)=\\sum_{k=1}^{\\infty}c_{k}\\log(1-z/z_{k})$, where $z_{k}\\in\\mathbb T$ and $\\sum_{k}|c_{k}|<\\infty$. Now what? :huh:" } { "Tag": [ "AMC", "AIME" ], "Problem": "We have rescheduled the AIME Math Jam for [b]Monday, March 24[/b] at [b]7 PM Eastern / 4 PM Pacific[/b]. During the Math Jam we will discuss the solutions to all 15 problems on the AIME.", "Solution_1": "The [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=220]transcript of the Math Jam[/url] is now available." } { "Tag": [], "Problem": "For every integer $ a>1$ an infinite list of integers is constructed $ L(a)$, as follows:\r\n\r\n[list] $ a$ is the first number in the list $ L(a)$.[/list]\n[list] Given a number $ b$ in $ L(a)$, the next number in the list is $ b\\plus{}c$, where $ c$ is the largest integer that divides $ b$ and is smaller than $ b$.[/list]\r\nFind all the integers $ a>1$ such that $ 2002$ is in the list $ L(a)$.", "Solution_1": "[hide]If $ b$ is a number in the sequence, and $ p$ is the smallest prime dividing $ b$, then the next number is $ b\\plus{}\\frac{b}{p}\\equal{}b\\cdot\\frac{p\\plus{}1}{p}$. From this, we immediately see that the list is increasing, so we any other $ a$ must be less than 2002.\n\nSay that we had an even number in the sequence. Let this number be $ 2^i \\cdot m$, where $ m$ is odd. Clearly, for the next $ i$ steps in the list, 2 will be the smallest prime, so we will multiply by $ \\frac{3}{2}$ each time. After these $ i$ steps, we will be left with $ 3^i \\cdot m$. Then, 3 will be the smallest prime, so the next number will be $ 2^2\\cdot3^{i\\minus{}1} \\cdot m$. Then, the next numbers will be $ 2\\cdot 3^i \\cdot m, 3^{i\\plus{}1}\\cdot m$, and from here, a pattern of the smallest prime being 3,2,2 will repeat forever. Each number in this sequence is divisible by $ 4$ or $ 3$, and since $ 2002\\equal{}2\\cdot7\\cdot11\\cdot13$ is not divisible by $ 4$ nor $ 3$, it follows that no even number can precede 2002.\n\nIf the first number is even, then it must be $ 2002$, giving us one valid value of $ a$.\n\nOtherwise, if $ a$ is odd, then the next number in the list must be even, and thus must be 2002. In this case, we must have $ 2002\\equal{}a\\cdot\\frac{p\\plus{}1}{p}$, or $ 2002p\\equal{}a(p\\plus{}1)$. Since $ a$ is odd and $ \\gcd(p,p\\plus{}1)\\equal{}1$, we see that $ p\\plus{}1$ must be an even factor of 2002. Testing out all 8 even factors of 2002, we see that the only such factors that are 1 greater than a prime are $ 2\\cdot7\\equal{}14$ and $ 2\\cdot7\\cdot13\\equal{}182$. The first case corresponds to $ a\\equal{}11\\cdot13^2\\equal{}1859$, and the second corresponds to $ a\\equal{}11\\cdot 181 \\equal{} 1991$. But in both these cases, $ 11$ is the smallest prime, and the next value should be $ \\frac{12a}{11}$, which will not give a value of 2002.\n\nThus, our only valid value of $ a$ is $ 2002$.[/hide]", "Solution_2": "Let $a_i$ be the number on the $i-th$ position on $L(a)$, so $a=a_0$. Do $a_{t}=2012$ and $a_{t-1}=bp$ with $b$ the largest proper divisor of $a_{t-1}$ and $p$ prime so is obvious $b>p$. From this we have;\n\n$2002=b(p+1)$, the divisors of $2002$ are $1,2,7,11,13,14,22,26,77,91,143,154,182,286,1001,2002$ the posible pairs of integers $(b,p+1)$ that satisfy $b>p$ are $(77,26), (91,22), (143,14), (154,13), (182,11), (286,7), (1001,2), (2002,1)$ and the only one that satisfy that $p$ is prime is $(143,14)$, so this leaves us with $bp=1859$ whose largest proper divisor is $169$ which is a contradiction, what means that there can\u00b4t be any number before $2002$ on $L(a)$, so $a=a_{0}=2002$" } { "Tag": [ "algebra", "polynomial" ], "Problem": "A polynomial $ F(x)$ of degree $ n$ satisfies $ F(k) \\equal{} F_k, k \\equal{} 0, 1, 2, ... n$, where $ F_k$ is the Fibonacci sequence with $ F_0 \\equal{} 0, F_1 \\equal{} 1$. Compute $ F(n\\plus{}1)$.\r\n\r\n[hide=\"Big hint\"] See the technique described [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=211560]here[/url]. [/hide]", "Solution_1": "It is easily verified that $ \\Delta^k(p(x))_{x\\equal{}0}\\equal{}F_{\\minus{}k}$\r\n\r\nThen since $ p(x)\\equal{}\\sum_{k\\equal{}0}^n\\frac{ x_{(k)} \\Delta^k(p(x))_{x\\equal{}0}}{k!}$, \r\n\r\nThe problem can be solved by the method that I gave in your other thread:\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=212474\r\n\r\n(we have to write $ a_k\\equal{}F_{\\minus{}k}$ as an linear recurrence with increasing indices but thats easy enough...", "Solution_2": "[hide=\"more generally\"] one can establish\n\\[ f_n(x) = \\begin{cases} \\sum_{k = 1}^{n}{( - 1)^{k + 1}F_k\\frac {\\Gamma(x + 1)}{\\Gamma(x + 1 - k)k!} & x\\notin \\mathbb{Z}\\cap \\lbrace x0$ \u03c0\u03bf\u03c5 \u03bf\u03b9 \u03c0\u03b1\u03c1\u03b1\u03bd\u03bf\u03bc\u03b1\u03c3\u03c4\u03b5\u03c2 \u03c4\u03b7\u03c2 \u03c0\u03b1\u03c1\u03b1\u03ba\u03b1\u03c4\u03c9 \u03b1\u03bd\u03b9\u03c3\u03c9\u03c3\u03b7\u03c2 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b8\u03b5\u03c4\u03b9\u03ba\u03bf\u03b9 \u03b9\u03c3\u03c7\u03c5\u03b5\u03b9\r\n\r\n$ \\frac{(a\\plus{}b\\plus{}c)^2}{a^2\\plus{}b^2\\plus{}c^2}\\leq\\frac{1}{ac\\plus{}3ab\\minus{}b^2}\\plus{}\\frac{1}{ab\\plus{}3cb\\minus{}c^2}\\plus{}\\frac{1}{bc\\plus{}3ac\\minus{}a^2}$", "Solution_1": "\u0397 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03b3\u03c1\u03ac\u03c6\u03b5\u03c4\u03b1\u03b9 \u03b9\u03c3\u03bf\u03b4\u03cd\u03bd\u03b1\u03bc\u03b1\r\n\r\n$ \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ac \\plus{} 3ab \\minus{} b^2} \\plus{} \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} 3cb \\minus{} c^2} \\plus{} \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{bc \\plus{} 3ac \\minus{} a^2}\\geq (a \\plus{} b \\plus{} c)^2$ \u03c3\u03c4\u03b7\u03bd \u03bf\u03c0\u03bf\u03af\u03b1 \u03bb\u03cc\u03b3\u03c9 \u03bf\u03bc\u03bf\u03b9\u03bf\u03b3\u03ad\u03bd\u03b5\u03b9\u03b1\u03c2 \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bd\u03b1 \u03c5\u03c0\u03bf\u03b8\u03ad\u03c3\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 $ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} 1$. \r\n\r\n\u0388\u03c4\u03c3\u03b9 \u03b7 \u03c0\u03c1\u03bf\u03c2 \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03b3\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \r\n\r\n$ \\frac {1}{ac \\plus{} 3ab \\minus{} b^2} \\plus{} \\frac {1}{ab \\plus{} 3cb \\minus{} c^2} \\plus{} \\frac {1}{bc \\plus{} 3ac \\minus{} a^2}\\geq 1 \\plus{} 2(ab \\plus{} bc \\plus{} ac)$\r\n\r\n\r\n\r\n\u038c\u03bc\u03c9\u03c2 \u03bb\u03cc\u03b3\u03c9 \u03c4\u03b7\u03c2 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1\u03c2 Andreescu \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \r\n\r\n$ LHS \\geq \\frac {(1 \\plus{} 1 \\plus{} 1)^2}{4(ab \\plus{} bc \\plus{} ac) \\minus{} 1} \\equal{} \\frac {9}{4(ab \\plus{} bc \\plus{} ac) \\minus{} 1} \\stackrel{(\\star)}{\\geq} \\frac {9}{4\\cdot 1 \\minus{} 1} \\equal{} 3 \\stackrel{(\\star)}{\\geq} 1 \\plus{} 2(ab \\plus{} bc \\plus{} ac)$,\r\n\r\n$ (\\star)$ \u0394\u03b9\u03cc\u03c4\u03b9 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 $ ab \\plus{} bc \\plus{} ac \\leq a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} 1$.\r\n\r\n\u0395\u03bb\u03c0\u03af\u03b6\u03c9 \u03bd\u03b1 \u03bc\u03b7 \u03bc\u03bf\u03c5 \u03ad\u03c7\u03b5\u03b9 \u03be\u03b5\u03c6\u03cd\u03b3\u03b5\u03b9 \u03ba\u03ac\u03c4\u03b9.\r\n\r\n\u0391\u03bb\u03ad\u03be\u03b1\u03bd\u03b4\u03c1\u03bf\u03c2", "Solution_2": "\u039f.\u039a. \u0391\u03bb\u03b5\u03be\u03b1\u03bd\u03b4\u03c1\u03b5\r\n\u03a0\u03c1\u03bf\u03c6\u03b1\u03bd\u03c9\u03c2 \u03b5\u03bd\u03bf\u03b5\u03b9\u03c2 \u03bc\u03b9\u03b1 \u03b9\u03c3\u03bf\u03b4\u03c5\u03bd\u03b1\u03bc\u03b7 \u03b1\u03c5\u03c4\u03b7\u03c2 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03bf\u03bc\u03bf\u03b3\u03b5\u03bd\u03b7\u03c2..\r\n\u0397 \u03b4\u03b9\u03ba\u03b9\u03b1 \u03bc\u03bf\u03c5 \u03bb\u03c5\u03c3\u03b7 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c0\u03b5\u03c1\u03b9\u03b3\u03c1\u03b1\u03c6\u03b9\u03ba\u03b1 \u03b5\u03bb\u03bb\u03b1\u03c4\u03c9\u03c3\u03b7 \u03c4\u03bf\u03c5 \u0392 \u03bc\u03b5\u03bb\u03bf\u03c5\u03c2 \u03bc\u03b5 \u03ba\u03b1\u03c4\u03b1\u03bb\u03bb\u03b7\u03bb\u03b7 \u03b1\u03c5\u03be\u03b7\u03c3\u03b7 \u03c4\u03c9\u03bd \u03c0\u03b1\u03c1\u03b1\u03bd\u03bf\u03bc\u03b1\u03c3\u03c4\u03c9\u03bd \u03ba\u03b1\u03b9 \u03bc\u03b5\u03c4\u03b1 \u03bc\u03b5 \u0391ndreescu \u03c4\u03b1 \u03b9\u03b4\u03b9\u03b1...", "Solution_3": "\u03c1\u03b5 \u03c0\u03b1\u03b9\u03b4\u03b9\u03b1 \u03c0\u03bf\u03b9\u03b1 \u03b9\u03c3\u03bf\u03b4\u03c5\u03bd\u03b1\u03bc\u03b7 \u03b1\u03bd\u03b9\u03c3\u03bf\u03c4\u03b7\u03c4\u03b1 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03bf\u03bc\u03bf\u03b3\u03b5\u03bd\u03b7\u03c2 \u03c3\u03c4\u03b7\u03bd \u03bb\u03c5\u03c3\u03b7 \u03c4\u03bf\u03c5 \u03b1\u03bb\u03b5\u03be\u03b1\u03bd\u03b4\u03c1\u03bf\u03c5; :|", "Solution_4": "\u03b7 \u03b1\u03bd\u03b9\u03c3\u03bf\u03c4\u03b7\u03c4\u03b1 \u03b3\u03c1\u03b1\u03c6\u03b5\u03c4\u03b1\u03b9 \r\n$ \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ac \\plus{} 3ab \\minus{} b^2} \\plus{} \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} 3cb \\minus{} c^2} \\plus{} \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{bc \\plus{} 3ac \\minus{} a^2}\\geq (a \\plus{} b \\plus{} c)^2$\r\n\u03b1\u03c0\u03bf \u03c4\u03b7\u03bd \u0391\u039c-\u0393\u039c \u03c0\u03b1\u03b9\u03c1\u03bd\u03bf\u03c5\u03bc\u03b5\r\n$ 3ab\\leq a^2\\plus{}b^2\\plus{}ab$etc\r\n\u03b1\u03c1\u03b1 \u03c4\u03bf LHS \u03b5\u03b9\u03bd\u03b1\u03b9(ANDREESCU and $ 3(a^2\\plus{}b^2\\plus{}c^2)\\geq (a\\plus{}b\\plus{}c)^2$\r\n$ LHS\\geq \\sum \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ac \\plus{}a^2 \\plus{}ab}\\geq \\frac {(3(a^2\\plus{}b^2\\plus{}c^2))^2}{(a\\plus{}b\\plus{}c)^2}\\geq (a\\plus{}b\\plus{}c)^2$\r\nQED :D", "Solution_5": "\u0395\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03c9 \u03b3\u03b9\u03b1 \u03c4\u03bf ''\u03c0\u03b1\u03b9\u03b4\u03b9\" \u03b5\u03b9\u03c7\u03b1 \u03c7\u03c1\u03bf\u03bd\u03b9\u03b1 \u03bd\u03b1 \u03c4\u03bf \u03b1\u03ba\u03bf\u03c5\u03c3\u03c9 \r\n\u0395\u03c7\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b7\u03bd \u03b9\u03b4\u03b9\u03b1 \u03bb\u03c5\u03c3\u03b7\r\n\u0393\u03b9\u03b1 \u03c4\u03b7\u03bd \u03bf\u03bc\u03bf\u03b3\u03b5\u03bd\u03b5\u03b9\u03b1 \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03b5\u03b9\u03bc\u03b1\u03b9 \u03b5\u03b9\u03bb\u03b9\u03ba\u03c1\u03b9\u03bd\u03b7\u03c2 \u03c5\u03c0\u03b5\u03b8\u03b5\u03c3\u03b1 \u03bf\u03c4\u03b9 \u03b8\u03b1 \u03c5\u03c0\u03b1\u03c1\u03c7\u03b5\u03b9 \u03ba\u03b1\u03c0\u03bf\u03b9\u03b1 \u03b9\u03c3\u03bf\u03b4\u03c5\u03bd\u03b1\u03bc\u03b7 \u03b3\u03b9\u03b1\u03c4\u03b9 \u03b5\u03ba\u03b5\u03b9\u03bd\u03b7 \u03c0\u03bf\u03c5 \u03b1\u03bd\u03b1\u03c6\u03b5\u03c1\u03b5\u03c4\u03b1\u03b9 \u03bf \u0391lex \u03b5\u03ba \u03c0\u03c1\u03c9\u03c4\u03b7\u03c2 \u03bf\u03c8\u03b5\u03c9\u03c2 .....\r\n\u0398\u03b1 \u03bc\u03b1\u03c2 \u03c0\u03b5\u03b9 \u03bf \u0391\u03bb\u03b5\u03be\u03b1\u03bd\u03b4\u03c1\u03bf\u03c2 ...", "Solution_6": "\u0388\u03c7\u03b5\u03c4\u03b5 \u03b1\u03c0\u03cc\u03bb\u03c5\u03c4\u03bf \u03b4\u03af\u03ba\u03b9\u03bf. \u0397 \u03ba\u03b1\u03c4\u03ac\u03c3\u03c4\u03b1\u03c3\u03b7 \u03b4\u03b5 \u03c3\u03ce\u03b6\u03b5\u03c4\u03b1\u03b9 \u03bb\u03cc\u03b3\u03c9 \u03c4\u03bf\u03c5 \u03b4\u03b5\u03be\u03b9\u03bf\u03cd (\u03c3\u03c4\u03b7 \u03b4\u03b9\u03ba\u03ae \u03bc\u03bf\u03c5 \"\u03bb\u03cd\u03c3\u03b7\") \u03bc\u03ad\u03bb\u03bf\u03c5\u03c2 \u03c0\u03bf\u03c5 \u03ba\u03ac\u03bd\u03b5\u03b9 \u03c4\u03b7\u03bd \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf\u03bc\u03bf\u03b3\u03b5\u03bd\u03ae\u03c2! \u0389\u03c4\u03b1\u03bd \u03ba\u03b5\u03ba\u03c4\u03b7\u03bc\u03ad\u03bd\u03b7 \u03c4\u03b1\u03c7\u03cd\u03c4\u03b7\u03c4\u03b1! \u0396\u03b7\u03c4\u03ce \u03c3\u03c5\u03b3\u03bd\u03ce\u03bc\u03b7 \u03c0\u03bf\u03c5 \u03c3\u03b1\u03c2 \u03c4\u03b1\u03bb\u03b1\u03b9\u03c0\u03ce\u03c1\u03b7\u03c3\u03b1. \u0397 \u03bb\u03cd\u03c3\u03b7 \u03c0\u03bf\u03c5 \u03ad\u03b3\u03c1\u03b1\u03c8\u03b5 \u03bf Athinaios \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03ac\u03c1\u03b1 \u03c0\u03bf\u03bb\u03cd \u03ba\u03b1\u03bb\u03ae!\r\n\r\n\u03a6\u03b9\u03bb\u03b9\u03ba\u03ac,\r\n\u0391\u03bb\u03ad\u03be\u03b1\u03bd\u03b4\u03c1\u03bf\u03c2" } { "Tag": [], "Problem": "Ok so I need 3 people who are usual to the games forum.\r\n\r\npost here if you fill this requirement and you want to play.\r\n\r\nIf you post also post one of these character types:\r\n\r\nAxeman\r\nSwordsman\r\nBowman\r\nSpearman\r\nMage\r\n\r\nwhen you have chosen I will post participants and all others are not to post in this thread. Pms will be sent out with more detailed information about character and plotline\r\n\r\nEDIT: your team of 3 may have the same profession\r\n\r\nplotline and more info coming up in next post", "Solution_1": "All starting people will have a base hp of 10 and a base energy of 10\r\nSpecial attacks are an attack the mod (me) chooses to give you. Special attack take up large amounts of energy but do major damage (can be used to remove physical obstacles including annoying merchants)\r\nKeep in mind you can die in this game and while the angel will restore your life you will be lose 20% Hp and 10% energy and thus easier to kill. If the angel dies and then you die then you won't come back to life.\r\n\r\n[hide=\"profession info\"]Swordsman:\n\nCan become Pirate or Knight. \nIf Pirate gains use of double sword wielding and may upgrade to mercenary. +lots of attack\nIf Knight gains use of wielding sword and shield and can become a Lord. +lots of defense\n\nbase attack/defense/speed and special will come in pm.\n\nBowman:\n\nCan become Daggermaster or Sharpshooter. \nIf you become a daggermaster you gain the ability to use daggers, dirks, throwing knives, etc and can become a assasin. +speed\nIf you become a Sharpshooter you gain use of a crossbow and can become a sniper. +attack & special.\n\nbase attack/defense/speed and special will come in pm.\n\nAxeman:\n\nCan become Bandit or Master Axeman.\nIf you become bandit you gain the use of throwing axe and can become a Axethrower. +gain long range attack and speed\nIf you become a MA you can use a battle axe and become a General. + uber amounts of attack - a little speed\n\nSpearman:\n\nCan become a javelin thrower or Lanceman.\nIf you become a javelin thrower gain use of a javelin and can become a Javelin Master. +defense +speed\nIf you become a Lanceman you gain use of a lance and can become a paladin. +defense +attack\n\nMage:\n\nbecomes a wizard\nCan become a warlock or a Sage\nIf you become a Sage you specialize in fire magic. Can become a pyromancer. +attack +defense\n*If you become a Warlock you specialize in dark magic. Can become a sorcerer and then a necromancer. + special +attack +speed +defense\n\n*Necromancers is the hardest rank to achieve but if you become it you can overcome almost any opponent by yourself.\n\nTo level up you must have a crest.[/hide]\n\n[b]Angel/Demons/elemental assistants are too hard to manage and thus will not be a choice in this version and instead 1 angel will be used instead[/b]\n\n[hide=\"Plotline\"]\nThe Angel of weaponsmaster Bob came back \"mortally\" wounded. The pair had tried to single handedly take out the Archdevil in the volcano of doom. They failed and Weaponsmaster Bob died there. His final words were to take these 3 crests to the 3 best warriors in the kingdom to finish the work he started. After saying these words the Angel \"died.\"\nThe 3 best warriors were sought out and given the crests but they did not know how to use them. The language on the crests were only known to the Weaponmaster bob, his angel, and a its creator a seer in the city of Thor. The three banded together and went on the path of the forest to this city to seek out the seer. On the way there an Angel who was almost \"killed\" by the Archdevil joined their band to defeat the ArchDevil. This Angel is on his way to the seer too to have his crest translated.[/hide]\r\n\r\nThe next level you choose after the profession you choose is what your crest will make you. also decide whether or not you wish to make your angel guide a heavenly sniper, an archmage or a Divine lord.\r\n\r\n[b]THERE IS A SUPER CREST IN THE GAME THAT WILL CHANGE ANY PLAYER THAT IS AT A TOP LEVEL INTO A WEAPONSMASTER EXCEPT FOR A MAGE.\nTHERE IS ALSO ONE FOR THE ANGEL TO BECOME AN ARCHANGEL CALLED THE HOLY CREST.[/b] :ninja: \r\n\r\nYou will occasionally find weapons in this game but the best ones come from the super bosses in this game. when you switch a weapon or shield or scroll or armor... you will sell and have to buy it back after you move from where you stand.\r\n\r\nEditted to make shorter\r\n\r\nEDIT: After a while more players will be allowed to join The band after the monster, Archdevil is destroyed for inter user war. If this becomes a success we might even make a computer game for it! Someone else will have to though because I cant program.", "Solution_2": "i'll be an axeman", "Solution_3": "i think i will be a mage\r\n\r\nif i can still join (?)", "Solution_4": "i want to play- mage", "Solution_5": "all right we have three people. sign ups are still up for replacements and to join later. pm me because posting here will clutter it up. Pms with more info will be sent and I will start the story when I get back from orchestra", "Solution_6": "pms sent.\r\n\r\nWelcome to the test rpg. in your pms note that anything I sent you may or may not be told to team mates. Also from now on I will not be sending you your stats constantly but as you gain new weapons I will tell you your attack. You can level up and it will always do something positive and sometimes a little negative.\r\n\r\n2 of you your regular attack will always hit. if I do not specifically say you have a chance of missing then you are one of the 2. Only the captain has any rejuvinators. (cant speln :( )\r\n\r\nIn this game you may kill your team mates if they hinder you, but to do so will require a huge advantage on you part because if you try and you do not have over 300% better stats then you are most likely to die (this is to discourage it).\r\n\r\nAs this is a test I will constantly change things to see what is best.\r\n[b]\nPlotline time![/b]\r\n\r\nIn the barracks 2 of the best mages assemble. Joe who was currently studying wind was given a tome of wind (forgot to add that in pm) and eggy who was a priest at that time was given a tomb of light. (forgot that too :blush: ) Funtwo who was headed as captain was given a hatchet because he was a specialist with axes. Each of you were given the crest and then the Angel gave his advice (the one thats still \"alive\"):\r\n\r\nThe seer who is the last to be able to translate that to you resides in the city of Thor. To get there we must travel through the forest of bandits. hopefully your captain will be of use here. I myself will help guide you and fight with you to find the meaning of my crest. And so you headed out into the forest.\r\n\r\nAbout half-way you were attacked by 5 bandits! And since the captain was not yet a bandit you had to fight them!\r\n\r\n[b]Each Bandit:\nHP:5\nenergy:4\n[/b]\r\nThings to do list:\r\nDiscuss what to do\r\nAttack\r\n\r\nNote: I have decide to be the angel I will help fight but will not be very useful. I will drop in hints every now and then to help guide you.\r\n\r\nHint: These bandits will be easy to kill. This is training.\r\n\r\nYou have 3 days to attack (once each) then the bandits attack", "Solution_7": "speed-5\r\nspecial-8\r\ndefense-1\r\nattack-3\r\n\r\nyou are currently a light mage your attacks do more damage but may miss the opponent.\r\nspecial: lightning- x energy 20x% chance of hit does 2x damage\r\nregular: 2 energy 40% chance of hit 5 damage\r\n\r\n\r\nhit regular", "Solution_8": "HIT!\r\n\r\n1 bandit lies dead on the floor. You pick up his weapon, a wooden mallet and discard it because funtwo's hatchet is better.\r\n\r\nBandits are getting restless at seeing one of their comrades dead. They will attack sooner then they planned...\r\n1 bandit steps forward to attack and the angel replies by pulling out a scepter, draining his energy and restoring eggy's energy. This kills the bandit.\r\n\r\n3 Bandits left. 1 raises a horn and blows in it. I wonder what it means...", "Solution_9": "speed-5 \r\nspecial-8 \r\ndefense-1 \r\nattack-3 \r\n\r\nyou are currently a wind mage your attacks do less damage but can paralyze the opponent for some time. \r\nspecial: whirlwind- x energy 10x% chance of paralysis does x damage \r\nregular: 2 energy 40% chance of paralysis 1 damage \r\n\r\n\r\n\r\nregular atak on the one blowing the horn", "Solution_10": "regular attack to teh one with the horn", "Solution_11": "Joe: Hit! slight paralysis. -1 damage\r\nFuntwo: Hit! 2 damage.\r\n\r\nThe bandit is so damaged that he does not have time to counter attack and funtwo manages a SECOND SLICE without costing any energy and kills the bandit.\r\n\r\nThe remaining 2 bandits charge towards us waving their wooden mallets over their heads striking Joe for 2 damage. I pull out my scepter of hp/energy stealing and restores everyone's health and energy.\r\n\r\nThe last bandit throws a book at us and it breaks my scepter (drat) and runs. I pick up my scepter and start taking supplies from the bandits.\r\n\r\nItem gains:\r\n\r\n4 bags (we can carry stuff now!!!)\r\n1 AMP energy drink\r\n3 bandages\r\nTome of darkness \r\n3 wooden mallets\r\n2 shards of a broken scepter\r\n\r\nI take my scepter and remove the orb at the top. I then take the silver shards and in a flash of light I The black orb is covered in silver. This is my regeneration orb. It can bring you back to life but cant steal energy or life anymore.\r\n\r\nThings to do:\r\n\r\ndistribute spoils-captains job\r\ndecide what to do with tome of darkness and wooden mallets.-everyone participates captain gives me final decision.\r\ndecide whether or not to follow the rogue bandit to the camp or to stay here and sleep the night.-up to you not me.", "Solution_12": "follow the guys, give mathblitz the scepter pieces, give tome of darkness to a mage (me or joe10112)", "Solution_13": "hmmm \r\n\r\nmaybe we should follow the bandits.\r\n\r\nis there currency in this game?\r\n\r\nbandits have bountys.......\r\n\r\nand we get muny!!\r\n\r\nfirst, funtwo, distribute stuff!", "Solution_14": "wait\r\n\r\nwhat is our hp??\r\n\r\nand somehow, this seems sort of like dragonfable...........", "Solution_15": "10/10 all. I restored it all before my wand broke.\r\n\r\nI dont know what dragon fable is...\r\n\r\nYes there is currency, but it wont be as useful as you think it is. All the things that you need cant be bought. They must be earned in this game. but you can buy some useful stuff. Like.. nevermind you'll see for yourselves when we get to Thor.", "Solution_16": "4 bags (we can carry stuff now!!!)\r\n1 AMP energy drink\r\n3 bandages\r\nTome of darkness\r\n3 wooden mallets\r\n2 shards of a broken scepter\r\ngive mallets to everyone else\r\nwe share the bags\r\nif i can use bandages on other ppl, ill keep them and be the doctor person\r\ntomb of darkness to eggylv\r\nshards and AMP to joe", "Solution_17": "Keep in mind that I used the shards to make my silver orb thingy that will bring you back to life if you die. so you cant really distribute it....\r\n\r\nAlso I assume that you will pursue the bandits? yes or no and i will continue to the story funtwo or I will accept if the other 2 say yes first.", "Solution_18": "oh ok dont distribute shards\r\nand pursue bandits", "Solution_19": "You follow the bandit quietly and he leads you to their camp... in a cave. The problem is that the bandit got there first and now there are 2 sentries. Both are alert and since there is only one way in.... what to do.... OH NOES! a bandit spots us and runs inside to inform the leader. looks like it time to invade.\r\n\r\nWe burst out of our hiding place and quickly slaughter the remaining sentry. Time to invade. \r\n\r\nTo speed the plot up I will just tell the invasion. NOTE: all things done in the story line cost no energy. that means everyone still has full health and energy bars.\r\n\r\nJoe enters first and unleashes a deadly whirlwind. This it turns out is a general attack which hits all enemies. Eggy now very hyper slaughters 2 bandits with a wave of his hand. Funtwo jumps up and slashes at the leader which kills him on impact. The others throw down their weapons and run and being merciful we spare them.\r\n\r\nOnly now do you realize the amount of stuff in this cave. There are 3 sacks of gold, but some of it turns out to be fake gold.\r\nyou gain a 3 hatchets. Unfortunately in the slaughter everything else was ruined. In the corner is a captive who all the \"gold\" belonged too. should you free him? he seems scared to death by you.", "Solution_20": "a cornered enemy is the most dangerous. does everyone agree?\r\nwait for others' opinions", "Solution_21": "well here are the options:\r\n\r\nundo his gag (leave the restraints on) and see what he has to say.\r\nkill him.\r\nfree him and talk.\r\n\r\nkeep in mind if hes a friend and you treat him like a suspect he wont help you very much, but if he is a enemy and you remove the restraints well...", "Solution_22": "i vote that egglv and i hold his arms, while funtwo undoes the gag\r\n\r\nthen, we chat.\r\n\r\nand we tell him that we are nicey nicey guys, and just want to talk. we are holding your arms, just in case you atak us.\r\n\r\nsorry, but it is better then killing you, dont you agree?\r\n\r\nwe say to the captive.\r\n\r\nif he turns out to be a good guy, i vote we invite him into our group\r\n\r\n:)", "Solution_23": "pretty good idea. waiting for eggylv's opinion on this", "Solution_24": "make sure hes still tied, but okay.", "Solution_25": "ok, we do as we planned\r\n\r\nnow what?", "Solution_26": "wats a tome?\r\n\r\nand plus, what is a AMP drink?\r\n\r\nwhat does it do?\r\n\r\nand what happens after we do as we planned??", "Solution_27": "A tome is a big book. In this case its a big magic book. oh you cant equip it until you become a dark mage thingy.\r\n\r\nAMP is an energy drink. It will give you +5 energy I believe.\r\n\r\nAs for what happens next is this:\r\n\r\nYou undo his gag first and tell him that you are friends and wish to help him, but are afraid that if you untie him he will attack you. Understandingly he nods and tells his story.\r\n\r\n\"I am a master blacksmith from Thor. I was on my way to the barracks to be employed as a master weapons creator. Thats when the bandits got me. One loosed an arrow into my leg then bound me up. When I got here they tried to find out where my money was. I had only taken some fools gold and a little regular gold to fool petty robbers who would take me to have body guards, but they attacked and here I am now. If you will set me free and give me my mallets to work with I will help you on whatever journey you have before I go on my way to Thor.\r\n\r\nYou untie his bonds and give him the mallets. He tries to stand up but falls down. \"Im so tired.\" he says. \"The bandits starve me as a torture to find the truth. Do you have anything that will help me?\"\r\n\r\nYes you say and give him the AMP drink. Feeling better but not so much better he proposes to stay the night. Being tired you agree.\r\n\r\nIn the morning you all feel refreshed when you ask him...\r\n\r\nDo you know the seer of Thor?\r\nCan you make us any weapons right now?\r\nDo you know about any crests and how to use them?\r\n\r\nYou may only choose one right now.", "Solution_28": "you can decide", "Solution_29": "[quote=\"eggylv999\"]you can decide[/quote]\r\n\r\nwho me?", "Solution_30": "the rest of the people", "Solution_31": "right now, i vote weapons question.", "Solution_32": "i vote crests, they can point us in the right direction", "Solution_33": "but only the seer knows about the crest.", "Solution_34": "so, funtwo, what question are you gonna ask??\r\n\r\ni vote on the weapons!", "Solution_35": "i am currently in favor of asking about the seer because its a lot easier to come across a blacksmith than a person from thor. this is some valuable information", "Solution_36": "yah sure, u guys win.......\r\n\r\no well.\r\n\r\nlets get on with the game :lol: \r\n\r\nps what does :D mean?\r\n\r\noen-mouthed happy?? and why is he green???\r\n\r\nand why are ppl obsessed w/ him???", "Solution_37": "(there was a problem with the story the blacksmith was supposed to say on my way to the barracks) sry\r\n\r\nYou ask the blacksmith what he knows about the seer and he responds by telling you the way to find the seer from the entrance to the city then leaves on his way to the barracks. Upon arriving you go directly to the house and talk to the seer. You got lucky. Many people were tracking him down with questions about the crests he made that they had and he was about to go into hiding. But he sees you are good hearted people. He asks however to be given your things and money except 40 gold and your equiped weapons/tomes so that he may survive in trade for how to use the crests. You agree and he shows you that you need only to press the :D on the back of all his crests. All three of you do this and...\r\n\r\nThe story from here decides on your choice. Oh I dont know how long this test will last because I will be too busy with projects/fairs/finals. Chances are I will ask another to continue this. Probably funtwo will take the story line and be the angel while someone else takes his place in the game.\r\n\r\nFurther instructions: Go to post 2 of this thread. Choose the level up you wish. After I post the next part I will probably have to give this game up." } { "Tag": [ "inequalities" ], "Problem": "What is the greatest integer that satisfies $ 6 \\minus{} x > \\minus{}9$ ?", "Solution_1": "Solving the inequality gives $ x<15$. We can't have $ x$ equal to 15, so the largest integer that satisfies the inequality is $ \\boxed{14}$." } { "Tag": [ "trigonometry", "function", "logarithms" ], "Problem": "1 Find the max and min of $\\displaystyle f(x)=2\\sin^2{\\pi x}-1$ are:\r\n\r\n2The function:$\\displaystyle f(x)=\\frac{\\sin{x}+\\cos{x}}{\\sin{x}-\\cos{x}}$ is a:\r\nA Odd fuchtion. B Even function. C Even and Odd. D Neither even nor odd.\r\n\r\n3 The number of the roots of: $\\displaystyle\\log_4{2x}=2\\cos{x}$ is:\r\nA 1. B 2. C.3. D. more than 3\r\n\r\n4 Find the range of $\\displaystyle y=\\sin{x}+ \\sqrt{2}\\cos{x}$ where $x \\in [0, 90^{o}]$\r\n\r\n5 Find the range of $f(x)=4\\sin^2{x}+6\\cos{x}-6$\r\n\r\n6 Find the monotonically decreasing interval of the function $\\displaystyle f(x)=1-\\sin{\\frac{(\\pi}{3}-2x)}$\r\n\r\n7 Find the monotonically increasing interval of the function $\\displaystyle f(x)=\\log_{\\frac12}{\\frac12 \\sin2x}$\r\n\r\n8 If $\\sin x=\\frac13$, $x\\in (0, 2\\pi)$, find $x$.", "Solution_1": "Aren't they basic trig exercises?", "Solution_2": "[quote=\"liyi\"]Aren't they basic trig exercises?[/quote]\r\nthere was a debate on whether we can post trig at the getting started forum. The result is: Basic trigs belongs at least to the intermediat forum......but ok, i shall change the name." } { "Tag": [ "geometry", "ratio", "similar triangles" ], "Problem": "Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the trangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is $m$ times the area of the square. The ratio of the area of the other small right triangle to the area of the square is\r\n\r\n$\\text{(A)}\\ \\frac1{2m+1}\\qquad\\text{(B)}\\ m \\qquad\\text{(C)}\\ 1-m\\qquad\\text{(D)}\\ \\frac1{4m} \\qquad\\text{(E)}\\ \\frac1{8m^2}$", "Solution_1": "[hide]Let $s=$ the side length of the square. The triangle whose area is $m$ times the area of the square has an area of $ms^2$, and one of its legs has length $s$. Therefore, the other leg must have length $2ms$.\n\nOne of the legs of the other triangle also has length $s$. Let $x=$ the length of the other leg. By similar triangles, $\\frac{2ms}{s}=\\frac{s}{x}$, or $2m=\\frac{s}{x}$. Solving for $x$ gives $x=\\frac{s}{2m}$. The triangle has an area of $\\frac{1}{2}sx=\\frac{s^2}{4m}$. Therefore, it has an area $\\boxed{\\frac{1}{4m}}$ times the area of the square.[/hide]", "Solution_2": "I got it this way, but is there any faster way to do this. :idea:", "Solution_3": "Sorry for reviving, but I didn't feel like reposting for a quicker solution.\n\n[hide=\"Super Quick Solution\"]Let the side length be $x$. So then, we know that one of the small triangle has leg $2mx$. So using [url=http://www.artofproblemsolving.com/blog/82154]Tyler's Favorite Area Trick[/url], we have \n$(\\text{Height of Other Triangle})\\cdot2mx = x^2$ which gives $x/2m$. So the area is $\\frac{1}{2}x/2m\\cdot x = x^2/4m$ which gives $\\frac{1}{4m}$[/hide]", "Solution_4": "[quote=\"thkim1011\"]Sorry for reviving, but I didn't feel like reposting for a quicker solution.\n\n[hide=\"Super Quick Solution\"]Let the side length be $x$. So then, we know that one of the small triangle has leg $2mx$. So using [url=http://www.artofproblemsolving.com/blog/82154]Tyler's Favorite Area Trick[/url], we have \n$(\\text{Height of Other Triangle})\\cdot2mx = x^2$ which gives $x/2m$. So the area is $\\frac{1}{2}x/2m\\cdot x = x^2/4m$ which gives $\\frac{1}{4m}$[/hide][/quote]\nyou made my day", "Solution_5": "Sorry for reviving, but what is Tyler's Favorite Area Trick? I can't seem to find it anywhere.\n", "Solution_6": "I think it was in one of the AoPS books like Simon's Favorite Factoring Trick", "Solution_7": "Do you know which one? I can't find it on google or in Intro to Geometry.", "Solution_8": "oops sorry i made that up (i was in middle school back in those days heh...). Btw tyler is an alias of mine.\n\nIt merely derives the formula $\\frac 1s = \\frac 1b + \\frac 1h$, where $s$ is the length of the inscribed square, $b$ is the base in which the square lies on, and $h$ is the height wrt the base. The method is to double the square in the following way\n[img]https://qph.is.quoracdn.net/main-qimg-de1c14dbd02964062a886e5d936cbf68?convert_to_webp=true[/img]\nThen you should see that the area of the square is equal to the sum of the area of the two rectangles, which you could combine. The only problem with this \"trick\" is that this is the only thing in which this is applicable.", "Solution_9": "[hide=Solution]Let the other leg length of the one $m$ times the area be $a$. Then $ab=x^2$ if $x$ is the side length of the square and $b$ is the other leg length of the other small right triangle. Also, the area is $mx^2$, so $a=2mx$. This means $2mx \\cdot b = x^2$, or $2m \\cdot b = x$. We hence have $b=\\frac{x}{2m}$, so the area is $\\frac{x^2}{4m}$. This makes the ratio $\\frac{1}{4m} \\implies \\boxed{\\text{(D)}}$.[/hide]" } { "Tag": [ "FTW" ], "Problem": "Ok I know I am late and it is winter already but never mind that.\r\nHere is the previous one:\r\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=82168&start=20[/url]\r\n\r\nSame rules - autocracy prevails.", "Solution_1": "mine pwns... I think. :P", "Solution_2": "yeah ur 1 PWNS JK", "Solution_3": "dang no one beats my avatar", "Solution_4": "who doesn't love walking around the river", "Solution_5": "Who doesn't like a 64x64 black square with a single photoshop effect put on it?", "Solution_6": "Captain Obvious is my favorite super hero and it should be your favorite too....I mean, he comes to the rescue in every other thread in this forum.", "Solution_7": "Koda is too cute to ignore.", "Solution_8": "Koda is disqualified for obvious reasons.\r\nThe rest I am counting as signups.", "Solution_9": "what about mine? :lol:", "Solution_10": "[quote=\"deimos\"]what about mine? :lol:[/quote]\r\n\r\nlol :D It's nicer than mine, for sure... :D", "Solution_11": "What about the hypercube? Who says we just need 3D?", "Solution_12": "Pirates FTW.", "Solution_13": "Home made fractals ftw.", "Solution_14": "Copyright infringements ftw.", "Solution_15": "[img]http://www.venganza.org/piratesarecool4.jpg[/img]\r\n\r\nPirates FTW.", "Solution_16": "I dont even know wtf my avatar is", "Solution_17": "[quote=\"Elemennop\"]I dont even know wtf my avatar is[/quote]\r\nIt's like a yellow rat on a chair, WTF is that?\r\n\r\nYou need a better avatar...", "Solution_18": "I think that's pickachu", "Solution_19": "[quote=\"ragnarok23\"]I think that's pickachu[/quote]\r\noh really i didnt notice\r\n\r\nanyway i vote for lmnop's avatar", "Solution_20": "Elemennop's avatar is pretty awesome.\r\nI vote for it too.", "Solution_21": "Mine is a low resolution pic of me with a caption. \r\n\r\nPenguin FTW.", "Solution_22": "[quote=\"junggi\"]\nanyway i vote for lmnop's avatar[/quote]\r\n\r\n\r\nwho is Imnop and can we still vote?", "Solution_23": "lmnop is elemennop, and your votes don't count officially. However I can keep a most popular award. :)", "Solution_24": "Come on you all have to agree my Cheburashka pwns", "Solution_25": "Excuse me? Albert Hitchcock FTW!", "Solution_26": "[url=http://www.lunchlady.org/sadhump.gif]No.[/url]", "Solution_27": "[quote=\"kyyuanmathcount\"]Captain Obvious is my favorite super hero and it should be your favorite too....I mean, he comes to the rescue in every other thread in this forum.[/quote]\r\n\r\nsorry, but i believe it's every 1.43943974937479374 topics (this is a math foum) :P :oops: \r\n-jorian", "Solution_28": "[quote=\"Treething\"][url=http://www.lunchlady.org/sadhump.gif]No.[/url][/quote]\r\n :rotfl: :rotfl: :rotfl:", "Solution_29": "on one other forum one of my friends has this one:\r\n\r\n\r\n\r\n[hide][url]http://img123.imageshack.us/img123/7656/brokenimagefightersat1ow7.gif[/url][/hide]\r\n\r\n :rotfl:", "Solution_30": "[quote=\"Treething\"][url=http://www.lunchlady.org/sadhump.gif]No.[/url][/quote]\r\n\r\n :rotfl: :rotfl: :rotfl: \r\nyou should definitely make this your avatar.", "Solution_31": "My avatar is a photoshop thing too.", "Solution_32": "ok, since everyone's saying it, FTW!!!!\r\n\r\nP.S. Are contestants allowed to vote?", "Solution_33": "Mine is the blandestone ever", "Solution_34": "yay for Drizzt!\r\n\r\ni like treethings animation and the broken sign thing\r\nand elenmops. go pikachu!", "Solution_35": "I think my av is the dizzle sauce\r\nGO BIG RED", "Solution_36": "Should I close submissions or is anyone else interested?", "Solution_37": "uhh... I don't understand the showcase method. Do random people vote, or do chosen people vote ???", "Solution_38": "GOOOOOOOOOOO MARS ROVER\r\ni mean that thing can last\r\nits like a year after its supposed to have died", "Solution_39": "[quote=\"Thu7\"]GOOOOOOOOOOO MARS ROVER\ni mean that thing can last\nits like a year after its supposed to have died[/quote]\r\n\r\nits like a year after jason's goldfish supposed to have died so GOOOOOOOOOOO JASON'S GOLDFISH", "Solution_40": "Anybody else or should I declare the results?", "Solution_41": "Consult captain obvious for the results.", "Solution_42": "obviously not.", "Solution_43": "(see first post on second page of Debate)" } { "Tag": [], "Problem": "Is the fundamental counting principle an axiom or theorem?", "Solution_1": "This is not a Getting Started question, it is more of a philosophical/foundations of math type of question. Mods, please move this topic.", "Solution_2": "Sorry about posting in the wrong forum...\r\n\r\nAny answers?" } { "Tag": [ "logarithms", "combinatorics unsolved", "combinatorics" ], "Problem": "Let $ n$ be a positive integer, and $ S$ an $ N$-element set of positive integers. Show that there is a subset $ A$ of $ S$ such that $ A$ has at least $ \\log_3n$ elements and for every pair a,b of distinct elements in $ A$, $ a\\plus{}b$ is not in $ S$.", "Solution_1": "http://www.mathlinks.ro/viewtopic.php?t=227025 was posted by you, so I guess you're just trying to revive this? Are you assuming there's an elementary solution to the case of $ \\log_3(n)$?", "Solution_2": "Well, I thought so. Are there not an elementary solution?", "Solution_3": "I've no idea. Is there reason to believe that there is one? Was this a contest problem?" } { "Tag": [ "function", "Euler", "number theory unsolved", "number theory" ], "Problem": "Prove that\r\n\\[ \\sum_{n = 1}^{\\infty} \\frac{\\varphi ( n )}{n!} \\]\r\nis irrational. :)\r\n\r\nRemark: $\\varphi$ is the Euler totient function.", "Solution_1": "It is obviosly. If this means is $\\frac pq$ then n=q give contradition.", "Solution_2": "Sorrry, I don't know what $p,q,n$ is. :roll:\r\nTherefore I cannot understand your solution/hint.\r\nBe more talkative, please ! :P", "Solution_3": "Let $\\frac pq = \\sum_{n = 1}^{\\infty} \\frac{\\varphi ( n )}{n!}$. Multiple n!=q!. Left expression is integer, right not.", "Solution_4": "@rust: hear to {x} and give more explicit answers!\r\nThere are a lot of people here not understanding what you write because of your to high compression.\r\nThus please be (a lot\u00bf) more explicit (I don't talk only about this special problem but in general)." } { "Tag": [], "Problem": "\u0391\u03c5\u03c4\u03ae\u03bd \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03c4\u03b7\u03bd \u03b2\u03c1\u03ae\u03ba\u03b1 \u03c3\u03b5 \u03c0\u03b1\u03bb\u03b9\u03ac \u03b8\u03ad\u03bc\u03b1\u03c4\u03b1 \u03c3\u03c4\u03b7 \u03c3\u03c7\u03bf\u03bb\u03ae \u03bc\u03bf\u03c5 \u03c3\u03c4\u03bf \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03bf\u03b9\u03c7\u03bf \u03bc\u03ac\u03b8\u03b7\u03bc\u03b1 \u03ba\u03b1\u03b9 \u03b3\u03b9' \u03b1\u03c5\u03c4\u03cc \u03c4\u03b7\u03bd \u03b2\u03ac\u03b6\u03c9 \u03b5\u03b4\u03ce.\u0394\u03c5\u03c3\u03c4\u03c5\u03c7\u03ce\u03c2 \u03b4\u03b5\u03bd \u03ad\u03c7\u03c9 \u03ba\u03b1\u03c4\u03b1\u03c6\u03ad\u03c1\u03b5\u03b9 \u03bd\u03b1 \u03c4\u03b7\u03bd \u03bb\u03cd\u03c3\u03c9 \u03b5\u03b4\u03ce \u03ba\u03b1\u03b9 \u03bc\u03b5\u03c1\u03b9\u03ba\u03ad\u03c2 \u03bc\u03ad\u03c1\u03b5\u03c2 \u03ba\u03b1\u03b9 \u03b5\u03bb\u03c0\u03af\u03b6\u03c9 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03bc\u03bf\u03c5 \u03b4\u03b9\u03b1\u03c6\u03b5\u03cd\u03b3\u03b5\u03b9 \u03c4\u03b7\u03c2 \u03c0\u03c1\u03bf\u03c3\u03bf\u03c7\u03ae\u03c2 \u03ba\u03ac\u03c4\u03b9 \u03c0\u03bf\u03bb\u03cd \u03b2\u03b1\u03c3\u03b9\u03ba\u03cc \u03b3\u03b9\u03b1\u03c4\u03af \u03c4\u03bf \u03b4\u03af\u03bd\u03c9 \u03c3\u03c4\u03b1 \u03c4\u03ad\u03bb\u03b7 \u03c4\u03bf\u03c5 \u03bc\u03b7\u03bd\u03cc\u03c2 \u03c4\u03bf \u03bc\u03ac\u03b8\u03b7\u03bc\u03b1.... :lol: \r\n\r\n\u0388\u03c3\u03c4\u03c9 \u03bc\u03af\u03b1 \u03b1\u03ba\u03bf\u03bb\u03bf\u03c5\u03b8\u03af\u03b1 $ n$ \u03b1\u03ba\u03b5\u03c1\u03b1\u03af\u03c9\u03bd \u03b1\u03c1\u03b9\u03b8\u03bc\u03ce\u03bd $ a_{1},a_{2},...,a_{n}$.\u039d\u03b1 \u03b4\u03b5\u03b9\u03c7\u03b8\u03b5\u03af \u03cc\u03c4\u03b9 \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03c0\u03ac\u03bd\u03c4\u03b1 \u03ad\u03bd\u03b1 \u03c5\u03c0\u03bf\u03c3\u03cd\u03bd\u03bf\u03bb\u03bf \u03b1\u03c0\u03cc \u03b1\u03c5\u03c4\u03ce\u03bd \u03c4\u03c9\u03bd \u03b1\u03c1\u03b9\u03b8\u03bc\u03ce\u03bd \u03c0\u03bf\u03c5 \u03c4\u03bf \u03ac\u03b8\u03c1\u03bf\u03b9\u03c3\u03bc\u03ac \u03c4\u03bf\u03c5\u03c2 $ \\sum a_{i}$ \u03b3\u03b9\u03b1 \u03b4\u03b9\u03b1\u03b4\u03bf\u03c7\u03b9\u03ba\u03bf\u03cd\u03c2 \u03b4\u03b5\u03af\u03ba\u03c4\u03b5\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03bf\u03bb\u03bb\u03b1\u03c0\u03bb\u03ac\u03c3\u03b9\u03bf \u03c4\u03bf\u03c5 $ n$.\r\n\r\n\u03a5.\u0393:\u0388\u03ba\u03b1\u03bd\u03b1 \u03bc\u03b9\u03b1 \u03bc\u03b9\u03ba\u03c1\u03ae \u03b1\u03bb\u03bb\u03b1\u03b3\u03ae \u03c3\u03c4\u03b7\u03bd \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03b4\u03b9\u03cc\u03c4\u03b9 \u03b5\u03af\u03c7\u03b1 \u03be\u03b5\u03c7\u03ac\u03c3\u03b5\u03b9 \u03b1\u03ba\u03c1\u03b9\u03b2\u03ce\u03c2 \u03c4\u03b7\u03bd \u03b5\u03ba\u03c6\u03ce\u03bd\u03b7\u03c3\u03ae \u03c4\u03b7\u03c2....", "Solution_1": "Just a hint: \r\n\r\n[hide]\u0398\u03b5\u03ce\u03c1\u03b7\u03c3\u03b5 \u03c4\u03b1 $ n$ \u03c3\u03c4\u03bf\u03b9\u03c7\u03b5\u03af\u03b1 $ A_1,A_2,..., A_n$ \u03c4\u03ad\u03c4\u03bf\u03b9\u03b1 \u03ce\u03c3\u03c4\u03b5 $ A_i\\equal{}a_1\\plus{}a_2\\plus{}...\\plus{}a_i$ \u03ba\u03b1\u03b9 \u03c0\u03ac\u03c1\u03b5 \u03c4\u03b1 \u03c5\u03c0\u03cc\u03bb\u03bf\u03b9\u03c0\u03b1 \u03c0\u03bf\u03c5 \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bd \u03bd\u03b1 \u03b1\u03c6\u03ae\u03bd\u03bf\u03c5\u03bd \u03b4\u03b9\u03b1\u03b9\u03c1\u03bf\u03cd\u03bc\u03b5\u03bd\u03b1 \u03bc\u03b5 \u03c4\u03bf $ n$. \n[/hide]", "Solution_2": "\u039c\u03b9\u03b1 \u03bb\u03cd\u03c3\u03b7 \u03c0\u03bf\u03c5 \u03c3\u03ba\u03ad\u03c6\u03c4\u03b7\u03ba\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03b5\u03be\u03ae\u03c2.\r\n\r\n\u03b8\u03b5\u03c9\u03c1\u03bf\u03cd\u03bc\u03b5 \u03c4\u03b1 \u03b1\u03b8\u03c1\u03bf\u03af\u03c3\u03bc\u03b1\u03c4\u03b1 $ A_{i,j}\\equal{}a_{i}\\plus{}a_{i\\plus{}1}\\plus{}...\\plus{}a_{j}$ \u03bc\u03b5 $ i\\leq j$.\u0388\u03c3\u03c4\u03c9 \u03cc\u03c4\u03b9 \u03ba\u03b1\u03bd\u03ad\u03bd\u03b1 \u03b1\u03c0\u03cc \u03c4\u03b1 $ A_{i,j}$ \u03b4\u03b5\u03bd \u03ad\u03c7\u03b5\u03b9 \u03c5\u03c0\u03cc\u03bb\u03bf\u03b9\u03c0\u03bf \u03bc\u03b7\u03b4\u03ad\u03bd.\u03a3\u03cd\u03bc\u03c6\u03c9\u03bd\u03b1 \u03bc\u03b5 \u03c4\u03b7\u03bd \u03c0\u03b1\u03c1\u03b1\u03c0\u03ac\u03bd\u03c9 \u03c5\u03c0\u03cc\u03b8\u03b5\u03c3\u03b7 \u03bf\u03cd\u03c4\u03b5 \u03ba\u03b1\u03b9 \u03c4\u03b1 $ A_{1,j}$ \u03ad\u03c7\u03bf\u03c5\u03bd \u03c5\u03c0\u03cc\u03bb\u03bf\u03b9\u03c0\u03bf \u03bc\u03b7\u03b4\u03ad\u03bd \u03ac\u03bc\u03b1 \u03c4\u03b1 \u03b4\u03b9\u03b1\u03b9\u03c1\u03ad\u03c3\u03bf\u03c5\u03bc\u03b5 \u03bc\u03b5 \u03c4\u03bf\u03bd $ n$.\u0386\u03c1\u03b1 \u03b1\u03bd\u03b1\u03b3\u03ba\u03b1\u03c3\u03c4\u03b9\u03ba\u03ac \u03c4\u03bf\u03c5\u03bb\u03ac\u03c7\u03b9\u03c3\u03c4\u03bf\u03bd \u03b4\u03cd\u03bf \u03ad\u03c7\u03bf\u03c5\u03bd \u03c4\u03bf \u03af\u03b4\u03b9\u03bf \u03c5\u03c0\u03cc\u03bb\u03bf\u03b9\u03c0\u03bf $ m$.\u0386\u03c1\u03b1 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03cc\u03c4\u03b9 \u03c5\u03c0\u03ac\u03c1\u03c7\u03bf\u03c5\u03bd $ A_{1,k}$ \u03ba\u03b1\u03b9 $ A_{1,l}$ \u03c4\u03ad\u03c4\u03bf\u03b9\u03b1 \u03ce\u03c3\u03c4\u03b5 $ A_{1,k}\\equal{}bn\\plus{}m$ \u03ba\u03b1\u03b9 $ A_{1,l}\\equal{}cn\\plus{}m$ \u03c0\u03c1\u03ac\u03b3\u03bc\u03b1 \u03c0\u03bf\u03c5 \u03c3\u03b7\u03bc\u03ac\u03b9\u03bd\u03b5\u03b9 \u03cc\u03c4\u03b9 $ A_{k\\plus{}1,l}\\equal{}A_{1,l}\\minus{}A_{1,k}\\equal{}(c\\minus{}b)n$.\u0394\u03b7\u03bb\u03b1\u03b4\u03ae \u03c4\u03bf $ A_{k\\plus{}1,l}$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03bf\u03bb\u03bb\u03b1\u03c0\u03bb\u03ac\u03c3\u03b9\u03bf \u03c4\u03bf\u03c5 $ n$ \u03c0\u03c1\u03ac\u03b3\u03bc\u03b1 \u03ac\u03c4\u03bf\u03c0\u03bf.", "Solution_3": "\u0395\u03c0\u03b5\u03b9\u03b4\u03ae \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c4\u03bf \u03af\u03b4\u03b9\u03bf \u03b1\u03bd\u03c4\u03b9\u03ba\u03b5\u03af\u03bc\u03b5\u03bd\u03bf \u03b5\u03af\u03c0\u03b1 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03c6\u03c4\u03b9\u03ac\u03be\u03c9 \u03ac\u03bb\u03bb\u03bf post...\r\n\r\n\u039d\u03b1 \u03b2\u03c1\u03b5\u03b8\u03b5\u03af \u03b1\u03bd\u03b1\u03bb\u03c5\u03c4\u03b9\u03ba\u03cc\u03c2 \u03c4\u03cd\u03c0\u03bf\u03c2 \u03c4\u03b7\u03c2 \u03b1\u03ba\u03bf\u03bb\u03bf\u03c5\u03b8\u03af\u03b1\u03c2\r\n\r\n$ a_{n}\\plus{}(n\\minus{}2)a_{n\\minus{}1}\\equal{}(n\\minus{}1)!$ \r\n\r\n\u03bc\u03b5 $ a_{0}\\equal{}4$.", "Solution_4": "\u03a0\u03bf\u03bb\u03bb\u03b1\u03c0\u03bb\u03b1\u03c3\u03af\u03b1\u03c3\u03b5 \u03ba\u03b1\u03b9 \u03c4\u03b1 \u03b4\u03c5\u03bf \u03bc\u03ad\u03bb\u03b7 \u03c4\u03bf\u03c5 \u03b1\u03bd\u03b1\u03b4\u03c1\u03bf\u03bc\u03b9\u03ba\u03bf\u03cd \u03c4\u03cd\u03c0\u03bf\u03c5 \u03bc\u03b5\r\n$ \\frac{(\\minus{}1)^{n\\minus{}2}}{(n\\minus{}2)!}$\r\n\u039c\u03b5\u03c4\u03b1 \u03b1\u03bd \u03b8\u03ad\u03c3\u03b5\u03b9\u03c2\r\n$ b_n\\equal{}\\frac{(\\minus{}1)^{n\\minus{}2}}{(n\\minus{}2)!}.a_n$" } { "Tag": [ "algebra", "polynomial", "calculus", "derivative", "linear algebra", "matrix", "number theory" ], "Problem": "Suppose $P$ and $Q$ are two monic polynomials. It is known that $\\{z\\mid P(z)=0\\}=\\{z\\mid Q(z) =0\\}$ and $\\{z\\mid P'(z)=0\\}=\\{z\\mid Q'(z)=0\\}$. Is it true that $P^n=Q^m$ for some naturals $m$ and $n$?", "Solution_1": "From the first condition we have:\r\n $P(x)=(x-a_1)^{m_1}...(x-a_k)^{m_k},Q(x)=(x-a_1)^{n_1}...(x-a_k)^{n_k}$\r\n where $a_1<...0$.\r\nWe'll prove that $m_i=n_i\\ell ,i=1,...,k$.\r\nDenote by $s_k(x_1,...,x_n)$ the $k$-th symmetric polynomial of $x_1,...,x_n$\r\nFrom (1) we easily get the following matrix-equality:$AB=0$ where $A$ is the following $k\\times k$matrix:\r\n$(row 1)1 1 ...1$\r\n$(row 2)s_1(x_2,...,x_k) s_1(x_3,...,x_k,x_1) ...s_1(x_1,x_2,...,x_{k_1})$\r\n . . ... .\r\n$(row k)s_{k-1}(x_2,...,x_k) s_{k-1}(x_3,...,x_1)...s_{k-1}(x_1,...,x_{k-1})$\r\nand $B$ is the $k\\times 1$matrix:\r\n$(m_1-\\ell n_1,...,m_k-\\ell n_k)^T$\r\nNote that $\\det(A)=\\pro_{ik$", "Solution_1": "hello, i think there is no maximum, $ (2x\\plus{}y)(2x\\plus{}z)$ can be infinity.\r\nSonnhard.", "Solution_2": "[quote=\"Dr Sonnhard Graubner\"]hello, i think there is no maximum, $ (2x \\plus{} y)(2x \\plus{} z)$ can be infinity.\nSonnhard.[/quote]\r\nWhy?The example?Find $ k$ max Equivalent find min $ (2x\\plus{}y)(2x\\plus{}z)$", "Solution_3": "hello, oh, you have right, this was an error of mine, the searched minimum is $ 2\\sqrt{2}$\r\nSonnhard.", "Solution_4": "[quote=\"shenyixin\"]Let $ a,b,c$ be positive real numbers such that $ xyz (x \\plus{} y \\plus{} z) \\equal{} 1$. Find maxium $ k$ such that :$ (2x \\plus{} y)(2x \\plus{} z) > k$[/quote]\r\nhere is my solution ,and when in the original problem,it can approch with x->0,y->0,z->infinity", "Solution_5": "Yes,you are right\uff01Thank you\uff01" } { "Tag": [ "calculus", "integration", "limit", "calculus computations" ], "Problem": "2 problems inspired by the integral in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=229813]this thread[/url].\r\n\r\n[b]Definitions:[/b] [i]Define a sequence as follows:[/i] $ a_n \\equal{} \\int_0 ^1 x^n (1 \\minus{} x)^n dx$\r\n\r\n\r\n[b]Problem 1:[/b] Prove or disprove the existence of $ z \\in \\mathbb{R}^{ \\plus{} }$ such that $ \\lim_{n \\to \\infty} \\left( z^n a_n \\right) \\in \\mathbb{R}^{ \\plus{} }$\r\n\r\n[b]Problem 2:[/b] Show that $ a_{n \\plus{} 1} \\equal{} \\frac {n \\plus{} 1}{4n \\plus{} 6} a_n$", "Solution_1": "Might want to clarify your use of 'x's in problem 1", "Solution_2": "Fair enough --\r\ndone :)", "Solution_3": "There is a closed form for this integral. Those who have computed the Jordan content of an n-ball will be familiar with it. :)\r\n\r\nIf the problems are not taken care of when I get back, I will try to answer them.", "Solution_4": "[b]Problem 3.[/b] Compute $ \\lim_{n\\to\\infty}\\sqrt{n}\\cdot 4^na_n.$ Do not quote Stirling's approximation - use the integral in a more direct way than that.\r\n\r\n[hide=\"Hint:\"]The substitution $ 2x\\minus{}1\\equal{}\\frac{u}{\\sqrt{n}}$ has its uses.[/hide]" } { "Tag": [ "function", "limit", "real analysis", "real analysis unsolved" ], "Problem": "Proove or disprove:\r\n\r\nIf $g$ is continuous in $[a,b]$ and $f$ is continuous in $[g(a),g(b)]$,then for every $c$ in $[a,b], \\lim_{x \\to c} f(g(x))=\\lim_{g(x) \\to g(c)} f(g(x))$\r\n\r\nIf it's wrong what extra properties should $f$ and $g$ have so it's correct?", "Solution_1": "--------------------------", "Solution_2": "What do you mean? You mean it can not be proved or disproved?", "Solution_3": "If g is monotone on $[a,b]$ , then everything must be OK. Otherwise g(c) does not probably belong to [g(a) , g(b)] and then we have great problems.For examle the limit may have not sense undre some special conditions. \r\n\r\n[u]Babis[/u]", "Solution_4": "yes, i agree" } { "Tag": [ "pigeonhole principle", "induction", "graph theory", "combinatorics unsolved", "combinatorics" ], "Problem": "[b]Problem:[/b] Let $ G$ be a graph with $ n$ vertices and $ m$ edges. If $ m \\ge n$ prove that $ G$ has a cycle.\r\n\r\nNOTE: This is a homework problem. As such, I am not asking for a solution, but more along the lines of feedback or hints towards a method that would lead to a solution.\r\n\r\nThoughts: Obviously, we assume $ G$ is simple, because if it has a loop or parallel edge, it has a cycle. I feel in my bones that I have to use the pigeonhole for this, but I am not sure if a) that is really what this problem needs, or b) what to do to get it there. This problem frustrates me because it is so obvious when you look at an example. Obviously if you draw n vertices and put n-1 edges between them, if you add an edge anywhere, it makes a cycle. However, I am new to graph theory, and the proof seems frustratingly difficult.\r\n\r\n[b]Idea 1:[/b] If $ G$ is acyclic, then it must simply be a collection of paths and it may/or may not contain isolated vertices. The most amount of edges any path of length $ i$ can have is $ i\\minus{}1$. At this point, I want to show that if we sum the number of vertices and edges, there will always be more vertices than edges. This would produce a contradiction (since we assume $ m \\ge n$. Unfortunately, I am not sure how to show this (partly because paths can share vertices), so I am stuck here.\r\n\r\n[b]Idea 2:[/b] I was thinking along the lines of letting $ G$ be an acyclic graph with $ n\\minus{}1$ edges, and showing that if I add an edge, then it creates a cycle. However, I am not sure how to show that either.\r\n\r\n[b]Idea 3:[/b] Pigeonholing it somehow. But how!?!? :[.\r\n\r\n[b]Idea 4:[/b] Consider the longest path like in the proof for if the minimum degree is $ \\ge 2$. This does not seem to take me in the write direction for this problem though.\r\n\r\nWoe is me for I have been working this problem for 3 hours now. Any comments are appreciated.", "Solution_1": "An acyclic graph is also known as a [url=http://mathworld.wolfram.com/Tree.html]tree[/url], so equivalently you want to prove that a tree on $ n$ vertices has at most $ n\\minus{}1$ edges. \r\n\r\nLet me throw out some ideas:\r\n\r\n- Induction?\r\n- Make all the edges directed. Where do they point?\r\n- [url=http://mathworld.wolfram.com/RootedTree.html]Root[/url] the tree. (This is one way to think about the previous idea.)" } { "Tag": [ "Pascal\\u0027s Triangle" ], "Problem": "$\\sum_{i=1}^{\\(n}\\(i}+\\sum_{i=1}^{\\(n-1}\\(i}+\\sum_{i=1}^{\\(n-2}\\(i}+....+\\sum_{i=1}^{\\(1}\\(i}$\r\n\r\nKeep in mind that $\\sum_{i=1}^{\\(n}\\(i}= \\frac{n(n+1)}{2}$ and $\\sum_{i=1}^{\\(n}\\(i^{2}}= \\frac{n(n+1)(2n+1)}{6}$\r\n\r\n\r\nSo far, I have...\r\n\r\n\r\n$\\sum_{i=1}^{\\(n}\\(i}+\\sum_{i=1}^{\\(n-1}\\(i}+\\sum_{i=1}^{\\(n-2}\\(i}+....+\\sum_{i=1}^{\\(1}\\(i}$\r\n\r\n\r\n= $\\frac{(n)(n+1)}{2}+\\frac{(n-1)(n)}{2}+\\frac{(n-2)(n-1)}{2}+....+\\frac{2}{2}$\r\n\r\n= $\\frac{1}{2}[ (n)(n+1)+(n-1)(n)+(n-2)(n-1)+....+2]$\r\n\r\nDoes anybody know how to simplyfy this further?", "Solution_1": "Very good! :D \r\n\r\nBut the stuff in parentheses is equal to\r\n\\[\\sum_{i=1}^{n}{i(i+1)}= \\sum_{i=1}^{n}{(i^{2}+i)}= \\sum_{i=1}^{n}{i^{2}}+\\sum_{i=1}^{n}{i}\\]\r\n\r\nand you can plug in your formulas to get\r\n\r\n\\[\\frac{n^{2}+n}{2}+\\frac{2n^{3}+3n^{2}+n}{6}=\\frac{2n^{3}+6n^{2}+4n}{6}=\\frac{n^{3}+3n^{2}+2n}{3}\\]\r\n\r\nso the final sum is half of that, or\r\n\\[\\frac{n^{3}+3n^{2}+2n}{6}\\]", "Solution_2": "[quote=\"not_trig\"]Very good! :D \n\nBut the stuff in parentheses is equal to\n\\[\\sum_{i=1}^{n}{i(i+1)}= \\sum_{i=1}^{n}{(i^{2}+i)}= \\sum_{i=1}^{n}{i^{2}}+\\sum_{i=1}^{n}{i}\\]\nand you can plug in your formulas to get\n\\[\\frac{n^{2}+n}{2}+\\frac{2n^{3}+3n^{2}+n}{6}=\\frac{2n^{3}+6n^{2}+4n}{6}=\\frac{n^{3}+3n^{2}+2n}{3}\\]\nso the final sum is half of that, or\n\\[\\frac{n^{3}+3n^{2}+2n}{6}\\]\n[/quote]\r\n\r\nThat is awesome, it actually works!\r\n\r\nHowever, I am sorry, but I lost you from the start,\r\n\r\nhow did you get this:\r\n\r\n\\[\\sum_{i=1}^{n}{i(i+1)}= \\sum_{i=1}^{n}{(i^{2}+i)}= \\sum_{i=1}^{n}{i^{2}}+\\sum_{i=1}^{n}{i}\\]\r\n\r\n?", "Solution_3": "[quote=\"problem_solver\"]= $\\frac{1}{2}[ (n)(n+1)+(n-1)(n)+(n-2)(n-1)+....+2]$[/quote]\n[quote=\"not_trig\"]But the stuff in parentheses is equal to \n$\\sum_{i=1}^{n}{i(i+1)}= \\sum_{i=1}^{n}{(i^{2}+i)}= \\sum_{i=1}^{n}{i^{2}}+\\sum_{i=1}^{n}{i}$[/quote]\r\n$(n)(n+1)+(n-1)(n)+(n-2)(n-1)+....+2$ is equivalent to $\\sum_{i=1}^{n}{i(i+1)}$, because:\r\n$(n)(n+1)$ is the value of $i(i+1)$ when $i=n$,\r\n$(n-1)(n)$ is the value of $i(i+1)$ when $i=n-1$,\r\n$(n-2)(n-1)$ is the value of $i(i+1)$ when $i=n-2$, \r\n...\r\n$(1)(2)$ is the value of $i(i+1)$ when $i=1$.", "Solution_4": "[hide=\"hint\"]the hockeystick identity or w/e is helpful here[/hide]", "Solution_5": "[quote=\"minsoens\"][quote=\"problem_solver\"]= $\\frac{1}{2}[ (n)(n+1)+(n-1)(n)+(n-2)(n-1)+....+2]$[/quote]\n[quote=\"not_trig\"]But the stuff in parentheses is equal to \n$\\sum_{i=1}^{n}{i(i+1)}= \\sum_{i=1}^{n}{(i^{2}+i)}= \\sum_{i=1}^{n}{i^{2}}+\\sum_{i=1}^{n}{i}$[/quote]\n$(n)(n+1)+(n-1)(n)+(n-2)(n-1)+....+2$ is equivalent to $\\sum_{i=1}^{n}{i(i+1)}$, because:\n$(n)(n+1)$ is the value of $i(i+1)$ when $i=n$,\n$(n-1)(n)$ is the value of $i(i+1)$ when $i=n-1$,\n$(n-2)(n-1)$ is the value of $i(i+1)$ when $i=n-2$, \n...\n$(1)(2)$ is the value of $i(i+1)$ when $i=1$.[/quote]\r\n\r\nNOW THAT IS ARTISTIC! :wink:", "Solution_6": "It's the same as\r\n\\[\\binom{n+1}{2}+\\binom{n}{2}+...+\\binom{2}{2}\\]\r\n\r\nUsing hockeystick gives,\r\n\\[\\binom{n+2}{3}= \\frac{n(n+1)(n+2)}{6}\\]", "Solution_7": "[quote=\"JavaMan\"]It's the same as\n\\[\\binom{n+1}{2}+\\binom{n}{2}+...+\\binom{2}{2}\\]\nUsing hockeystick gives,\n\\[\\binom{n+2}{3}= \\frac{n(n+1)(n+2)}{6}\\]\n[/quote]\r\n\r\nSometimes i simply get flattered by mathematics :)", "Solution_8": "Why do you call that identity \"Hockey Stick\"? :) As far I know it's called Chu-Shih-Chieh's identity, or something like that!", "Solution_9": "[quote=\"Jutaro\"]Why do you call that identity \"Hockey Stick\"? :) As far I know it's called Chu-Shih-Chieh's identity, or something like that![/quote]\r\n\r\nif you look at it's representation on a pascal's triangle, it looks like a hockey stick" } { "Tag": [], "Problem": "I turned in my responses to Admission test A before the deadline (feb 12) but have yet to turn in the $\\$20$grading fee or an application or the camp tuition.\r\n\r\nWhat do you think is my best bet? Turn in the application now with the fee, and expect them to grade the test A that I turned in a month ago?\r\n\r\n\r\nAlso, do we still need personal and teacher references if we attended camp last year?", "Solution_1": "sup jwill :P \r\n\r\ni'm pretty sure that you can still turn in the application and fee and they will grade the test. i'm not sure whether you'll get 10% off the price though. and you don't need personal reference and teacher reference if you went last year. the only thing you need to fill out is the personal information form i think.", "Solution_2": "New questions:\r\n\r\nIf I live in Dallas, do I have to live on site - this camp is like uberexpensive and saving the thousand dollars on room/board would be nice.\r\n\r\nAnd if I take admission test B, do I have to send in my teacher recs and everything with the test, or can I send those in later (I will send in the grading fee with the test)?", "Solution_3": "So, is the fee that you have to send (in USD) $\\$20$ or $\\$50$??? Also, could we send in cash??\r\n\r\n- Werdna Nawahd", "Solution_4": "also, how much is good proof writing or general appeal of the qualifying quiz worth in an application?" } { "Tag": [], "Problem": "Supposse $a_1,...,a_{2002}$ is a sequence of 2002 numbers, each of which equals $1$ or $-1$. What is the smallest possible value of the sum $\\sum_{iAHb (Hc in AB : .<", "Solution_8": "There are 14 numbers, not including 100.\r\n\r\nPlus, the answer sheet says 14. :D", "Solution_9": "[quote=\"neutrus\"]How many positive integers less than 100 have reciprocals with terminating decimal representations?[/quote]\n\nQuestion says \"less than 100\" so that's why 100 doesn't count.\n\n[quote=\"Textangle\"]$ \\frac {1}{a}$ terminates iff $ a$ has no distinct prime factors other than $ 2$ and $ 5$. So, we find the number of positive integers under 100 that is divisible by 2 and/or 5.\n\n\n$ 1 \\equal{} 1$\n$ 2 \\equal{} 2$\n$ 2^2 \\equal{} 4$\n$ 2^3 \\equal{} 8$\n$ 2^4 \\equal{} 16$\n$ 2^5 \\equal{} 32$\n$ 2^6 \\equal{} 64$\n$ 5 \\equal{} 5$\n$ 5^2 \\equal{} 25$\n$ 2\\times5 \\equal{} 10$\n$ 2^2\\times5 \\equal{} 20$ \n$ 2^3\\times5 \\equal{} 40$\n$ 2^4\\times5 \\equal{} 80$\n$ 2\\times5^2 \\equal{} 50$\n\nSo $ \\boxed{13}$ numbers (not including 100 itself).\n\nDoes anyone know how to do this without listing?\n\nDarn. Beaten.[/quote]\r\n\r\nThere are 14 in the list. That's how the answer is 14 without including 100. No other way than listing." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Sorry for my stupid mistake...\r\na_1,a_2,...,a_n positive real numbers satisfy a_1+a_2+...+a_n=1\r\nShow that sum of (a_i*a_j)/(a_i+a_j) numbers at most (n-1)/4", "Solution_1": "\\[\\frac{a_i a_j}{a_i +a_j} = \\frac{1}{\\frac{1}{a_i}+\\frac{1}{a_j}} \\le \\frac{a_i+a_j}{4}\\]\r\n\\[\\sum_{1 \\le i < j \\le n}{\\frac{a_i a_j}{a_i + a_j}} \\le \\sum_{1 \\le i < j \\le n}{\\frac{a_i+a_j}{4}} = \\frac{1}{4}\\cdot(n-1)(a_1+a_2+\\cdots+a_n) = \\frac{n-1}{4}\\]", "Solution_2": "Maybe, you're talking about [b]positive[/b] numbers?..", "Solution_3": "Do you think it is true for all real numbers?", "Solution_4": "[quote=\"Myth\"]Do you think it is true for all real numbers?[/quote]\r\nNo, of course.\r\nIt was just remark to the text. \r\nIt's fixed, as I see :)" } { "Tag": [ "probability" ], "Problem": "A bowl contains fewer than 50 marbles, and each is either red, green or blue. The probability of drawing a red marble is $ \\frac{2}{5}$ and the probability of drawing a green marble is $ \\frac{3}{7}$. If two marbles are drawn without replacement, what is the probability that both are blue? Express your answer as a common fraction.", "Solution_1": "The probability of drawing a blue marble is $ 1\\minus{}\\frac{2}{5}\\minus{}\\frac{3}{7}\\equal{}\\frac{35}{35}\\minus{}\\frac{14}{35}\\minus{}\\frac{15}{35}\\equal{}\\frac{6}{35}$.\r\n\r\n$ \\frac{6}{35} \\cdot \\frac{5}{34}\\equal{}\\frac{6}{238}\\equal{}\\frac{3}{119}$. This might be wrong though.", "Solution_2": "BOGTRO, your solution is correct." } { "Tag": [], "Problem": "How many triples of positive integers a, b, and c are there with a < b < c and a + b + c = 401?", "Solution_1": "What I posted in the GS topic:\n\n\n\n[hide]\n\n\n\nThe equation in positive integers is equal to the equation in non-negative integers (the substitution and so on), so we can use the formula for the last one to find the # of solutions to be . However, we want , which is one of six permutations on the order of the three variables, so we divide our answer by yielding . However, we overcounted times for solutions where and , so our final answer is .\n\n\n\nI did that a bit hastily, so it might be wrong.[/hide]", "Solution_2": "My answer varies slightly..\r\n\r\n[hide]Consider when a=1, 2, 3...132. Through casework, we see that when a=1, there are 198 solutions which satisfy the equation and inequality. When a=2, there are 197. We also see that a=3 gives us 195 solutions. Through more casework, we see that numbers congruent to 1mod3 (numbers when divided by three which leave remainder 1) are not included. Thus, the total number of solutions is (198+197)+(195+194)+...+(3+2). This reduces to a summation of 6n-1 from 1 to 66.\n\n[b]13200[/b][/hide]", "Solution_3": "Solution\r\n[hide]\nUrns a, b, and c will contain 401 balls, representing the sum of 401. The problem specifies positive integers, so we can place 1 ball in each of a, b, and c, leaving 398. The possibilities for 398 balls in 3 urns is 400C398. This leaves positive a, b, and c, but they may be equal and/or out of order. We cannot have a=b=c because 401 isn't divisible by 3. \n\nIf we have a=b, then there's the diophantine 2a + c = 401, which has 200 positive integer solutions. a, b, and c can be assigned each solution in 3!/2! ways. So we are left with 400C398 - 200*3!/2! unequal a, b, c that may be out of order.\n\nBecause a, b, and c are unequal, each triplet solution can be arranged 3! ways. Only one of those is a0$ . Show that :\r\n\r\n\\[ \\frac{3(a^3\\plus{}b^3\\plus{}c^3)}{(a\\plus{}b\\plus{}c)(a^2\\plus{}b^2\\plus{}c^2)}\\plus{}\\frac{3(a\\plus{}b)(b\\plus{}c)(c\\plus{}a)}{a^3\\plus{}b^3\\plus{}c^3}\\ge 9\\]", "Solution_1": "[quote=\"Pain rinnegan\"]Let $ a,b,c > 0$ . Show that :\n\\[ \\frac {3(a^3 \\plus{} b^3 \\plus{} c^3)}{(a \\plus{} b \\plus{} c)(a^2 \\plus{} b^2 \\plus{} c^2)} \\plus{} \\frac {3(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)}{a^3 \\plus{} b^3 \\plus{} c^3}\\ge 9\\]\n[/quote]\r\n\r\nI've proved that $ \\text{LHS}\\le 9$\r\n\r\nwe have \r\n\r\n$ 3\\sum_{cyc}a^2.\\sum_{cyc}a.\\sum_{cyc}\\left((a \\plus{} b)(a \\minus{} b)^2\\right) \\ge \\sum_{cyc}a^3.\\sum_{cyc}\\left((a \\plus{} b)(a \\minus{} b)^2\\right)$ because $ 3\\sum_{cyc}a^3 \\plus{} ... \\ge \\sum_{cyc}a^3$\r\n$ \\implies 2\\sum_{cyc}a^2.\\sum_{cyc}a(\\sum_{cyc}a^3 \\minus{} 3abc) \\plus{} 3\\sum_{cyc}a^2.\\sum_{cyc}a.\\sum_{cyc}\\left((a \\plus{} b)(a \\minus{} b)^2\\right)$$ \\ge \\sum_{cyc}a^3.\\sum_{cyc}\\left(a \\plus{} b)(a \\minus{} b)^2\\right)$\r\n\r\nand Notice that:\r\n\r\n$ \\frac {3(a^3 \\plus{} b^3 \\plus{} c^3)}{(a \\plus{} b \\plus{} c)(a^2 \\plus{} b^2 \\plus{} c^2)} \\minus{} 1 \\plus{} \\frac {3(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)}{a^3 \\plus{} b^3 \\plus{} c^3} \\minus{} 8$\r\n$ \\equal{} \\frac {\\sum(a \\plus{} b)(a \\minus{} b)^2}{\\sum a.\\sum a^2} \\minus{} \\frac {3(\\sum(a \\plus{} b)(a \\minus{} b)^2 \\plus{} 2(\\sum a^3 \\minus{} 3abc)}{\\sum a^3} \\le 0$", "Solution_2": "[quote=\"Pain rinnegan\"]Let $ a,b,c > 0$ . Show that :\n\\[ \\frac {3(a^3 \\plus{} b^3 \\plus{} c^3)}{(a \\plus{} b \\plus{} c)(a^2 \\plus{} b^2 \\plus{} c^2)} \\plus{} \\frac {3(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)}{a^3 \\plus{} b^3 \\plus{} c^3}\\ge 9\\]\n[/quote]\r\n\\[ 9 \\minus{} \\left(\\frac {3(a^3 \\plus{} b^3 \\plus{} c^3)}{(a \\plus{} b \\plus{} c)(a^2 \\plus{} b^2 \\plus{} c^2)} \\plus{} \\frac {3(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)}{a^3 \\plus{} b^3 \\plus{} c^3}\\right)\\]\r\n\\[ \\equal{} \\sum {\\left( {\\frac {{ {a \\plus{} b} }}{{\\left( {a \\plus{} b \\plus{} c} \\right)\\left( {a^2 \\plus{} b^2 \\plus{} c^2 } \\right)}} \\minus{} \\frac {{4a \\plus{} 4b \\plus{} c}}{{a^3 \\plus{} b^3 \\plus{} c^3 }}} \\right)\\left( {a \\minus{} b} \\right)^2 }\\]\r\nand\r\n\\[ {\\frac {{ {a \\plus{} b}}}{{\\left( {a \\plus{} b \\plus{} c} \\right)\\left( {a^2 \\plus{} b^2 \\plus{} c^2 } \\right)}} \\minus{} \\frac {{4a \\plus{} 4b \\plus{} c}}{{a^3 \\plus{} b^3 \\plus{} c^3 }}}\\]\r\n\\[ \\equal{} \\frac {{c^2 \\left( {5a^2 \\plus{} 8ab \\plus{} 5b^2 } \\right) \\plus{} 5c(a \\plus{} b)\\left( {a^2 \\plus{} b^2 } \\right) \\plus{} (a \\plus{} b)^2 \\left( {3a^2 \\plus{} ab \\plus{} 3b^2 } \\right) \\plus{} 4c^3 (a \\plus{} b) \\plus{} c^4 }}{{\\left( {a \\plus{} b \\plus{} c} \\right)\\left( {a^2 \\plus{} b^2 \\plus{} c^2 } \\right)\\left( {a^3 \\plus{} b^3 \\plus{} c^3 } \\right)}}\\]\r\n$ > 0$\r\nso\r\n\\[ \\frac {3(a^3 \\plus{} b^3 \\plus{} c^3)}{(a \\plus{} b \\plus{} c)(a^2 \\plus{} b^2 \\plus{} c^2)} \\plus{} \\frac {3(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)}{a^3 \\plus{} b^3 \\plus{} c^3}\\le 9\\]", "Solution_3": "[quote=\"Evariste-Galois\"][quote=\"Pain rinnegan\"]Let $ a,b,c > 0$ . Show that :\n\\[ \\frac {3(a^3 \\plus{} b^3 \\plus{} c^3)}{(a \\plus{} b \\plus{} c)(a^2 \\plus{} b^2 \\plus{} c^2)} \\plus{} \\frac {3(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)}{a^3 \\plus{} b^3 \\plus{} c^3}\\ge 9\\]\n[/quote]\n\nI've proved that $ \\text{LHS}\\le 9$\n[/quote]\r\n\r\nYes , i had a typo . It is $ LHS\\le 9$", "Solution_4": "Let's say that $ a^3\\plus{}b^3\\plus{}c^3\\equal{}x$ , $ a^2\\plus{}b^2\\plus{}c^2\\equal{}y$ and $ a\\plus{}b\\plus{}c\\equal{}z$\r\n\r\nInequality equivalents to $ f(x)\\equal{}3x^2\\minus{}10xyz\\plus{}yz^4 \\leq0$\r\n\r\nBecause of $ xn+p$, $\\left(m+n\\right)+\\left(n+p\\right)>p+m$ and $\\left(n+p\\right)+\\left(p+m\\right)>m+n$, it is clear that the numbers n + p, p + m and m + n are sidelengths of a triangle. The semiperimeter $s=\\frac{\\left(n+p\\right)+\\left(p+m\\right)+\\left(m+n\\right)}{2}$ of this triangle is simply $s=m+n+p$, and thus we have $s-\\left(n+p\\right)=m$, $s-\\left(p+m\\right)=n$ and $s-\\left(m+n\\right)=p$. Hence, after the Heron formula, the area of this triangle is\r\n\r\n$\\sqrt{s\\left(s-\\left(n+p\\right)\\right)\\left(s-\\left(p+m\\right)\\right)\\left(s-\\left(m+n\\right)\\right)}=\\sqrt{\\left(m+n+p\\right)mnp}$.\r\n\r\nHence, the Corollary in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=204393#204393]http://www.mathlinks.ro/Forum/viewtopic.php?t=2958 post #3[/url], applied to the triangle with sidelengths n + p, p + m, m + n and area $\\sqrt{\\left(m+n+p\\right)mnp}$ and to the three nonnegative reals x, y, z, yields\r\n\r\n$x\\left(n+p\\right)^{2}+y\\left(p+m\\right)^{2}+z\\left(m+n\\right)^{2}\\geq 4\\sqrt{yz+zx+xy}\\cdot\\sqrt{\\left(m+n+p\\right)mnp}$.\r\n\r\nEquivalently,\r\n\r\n$x\\left(n+p\\right)^{2}+y\\left(p+m\\right)^{2}+z\\left(m+n\\right)^{2}\\geq 4\\sqrt{\\left(yz+zx+xy\\right)mnp\\left(m+n+p\\right)}$.\r\n\r\nThis is exactly your inequality after a change of variable names (you used quite an asymmetric way of naming variables).\r\n\r\n Darij", "Solution_3": "Yes! I solved it the same way even I hadn't seen that corollary before.\r\nThis was just a more \"algebraic\" form of Pedoe's inequality...", "Solution_4": "[quote=\"hieuchuoi@\"]Prove the inequality below with $m,n,p,x,y,z$ are positive numbers:\n$x(m+n)^{2}+y(n+p)^{2}+z(m+p)^{2}\\geq 4\\sqrt{(xy+yz+zx)mnp(m+n+p)}$[/quote]\r\nSetting $a=m+n,b=n+p,c=p+m$ then $a,b,c$ are three sides of an triangle ABC with area $S=\\sqrt{mnp(m+n+p)}$. The inequality becomes\r\n\\[xa^{2}+yb^{2}+zc^{2}\\ge 4\\sqrt{xy+yz+zx}S \\]\r\n\\[\\Leftrightarrow xa^{2}+yb^{2}+z(a^{2}+b^{2}-2ab\\cos{C}) \\ge 2\\sqrt{xy+yz+zx}ab\\sin{C}\\]\r\n\\[\\Leftrightarrow (x+z)a^{2}+(y+z)b^{2}\\ge 2ab\\left(z\\sin{C}+\\sqrt{xy+yz+zx}\\cos{C}\\right) \\]\r\nApplying AM - GM inequality, we have\r\n\\[(x+z)a^{2}+(y+z)b^{2}\\ge 2ab\\sqrt{(x+z)(y+z)}\\]\r\nOn the other hand, using Cauchy inequality, we have\r\n\\[\\left(z\\sin{C}+\\sqrt{xy+yz+zx}\\cos{C}\\right) \\le \\sqrt{(\\sin^{2}{C}+\\cos^{2}{C})(z^{2}+xy+yz+zx)}=2ab\\sqrt{(x+z)(y+z)}\\]\r\n\r\nWe are done. :)" } { "Tag": [ "number theory", "relatively prime", "prime factorization" ], "Problem": "Solve in N:\r\n$a(a+1)=d^n$", "Solution_1": "Wow this is a classroom topic. It is really hard......or perhaps I am just not seeing something", "Solution_2": "are you sure that this is middle school classroom material? seems a bit too hard...unless there's some sort of formula that i haven't learned yet? the best i can get is $\\ a^2+a=d^n$... yea...i know that not much...", "Solution_3": "[quote=\"Beat\"]Solve in N:\n$a(a+1)=d^n$[/quote]\r\n\r\na is relatively prime to a+1, so\r\nif n is not equal to 1,\r\nthen \r\n$a=e^n$\r\n$a+1=f^n$\r\nfor some e , f in N.\r\nBut it has no positive integer solution.\r\n\r\nso it's only \r\n$n=1,a=m,d=m(m+1).$", "Solution_4": "[quote=\"Beat\"]Solve in N:\n$a(a+1)=d^n$[/quote]\r\n\r\nI think that $a=1$ or $a+1=1$ from where $a=0$, so $a=1$, then $d^n=2$, so $d=2$ and $n=1$... but 0% sure, I only think because $a$ and $a+1$ are relatively primes...", "Solution_5": "I can prove the case where d is prime, (I accidentally interpreted the problem wrong). I'll give the other case (d is composite) a shot later.\r\n\r\nLemma: For any natural number $a$, $a+1$ and $a$ are relatively prime.\r\n\r\nProof: Notice that the smallest positive linear combination of $a$ and $a+1$ is 1: $(a+1)-(a)=1$. Thus $(a+1,a)=1$, and the two numbers are relatively prime.\r\n\r\nNow to the problem\r\nI will prove that the only solution in $N$ is $(a,n,d)=(1,1,2)$.\r\n\r\nProof: First note that one of $a$ or $a+1$ must equal 1. To prove this, suppose that neither $a$ nor $a+1$ equal 1. Both numbers contain atleast one prime greater than 1. Further, the primes in their respective prime factorizations must be distinct, since we know that $(a,a+1)=1$. Thus the prime factorization of $a(a+1)$ will contain atleast two distinct primes, contradicting the equation we are given.\r\nThus one of $a$ or $a+1$ equals 1. If $a+1=1$, then $a=0$, and we would no longer be working in $N$. Thus $a=1$, from which it follows that $a+1=2, d=2,$ and $n=1$.\r\n\r\nCheck out the sweet music! [url]http://www.djperan.com/webmp3/Sandro-PeranRemix.mp3[/url]" } { "Tag": [ "algebra solved", "algebra" ], "Problem": "Let $ n$ be a positive integer.\r\nLet $ \\left\\{a_1,a_2,...,a_n\\right\\}$ and $ \\left\\{b_1,b_2,...,b_n\\right\\}$ be two disjoint $ n$-element subsets of the set $ \\left\\{1,2,3,...,2n\\right\\}$. (Then, of course, $ \\left\\{a_1,a_2,...,a_n\\right\\}\\cup\\left\\{b_1,b_2,...,b_n\\right\\} \\equal{} \\left\\{1,2,3,...,2n\\right\\}$.)\r\nAssume that $ a_1 > a_2 > a_3>... > a_n$ and $ b_1 < b_2 < b_3 < ... < b_n$.\r\nProve that\r\n$ \\sum_{i \\equal{} 1}^{n} \\left|a_i \\minus{} b_i\\right| \\equal{} n^2$.", "Solution_1": "[url=http://www.cut-the-knot.org/Curriculum/Games/ProizvolovGame.shtml]Here[/url]'s a Java applet which might help you figure it out. I remember trying to prove it a long time ago and failing. After checking the box \"clue\" in that applet I figured it out in about a second :)." } { "Tag": [ "\\/closed" ], "Problem": "Is there a problem with the Classroom? Whenever I click on the link, I'm directed not to the classroom but to the artofproblemsolving website. Previously, I clicked the embedded classroom to get in--but now it says \"No access\". Is this going to be fixed?? Just wondering...cause there's WOOT class today.", "Solution_1": "Yes, it's going to be fixed." } { "Tag": [ "geometry", "inequalities", "\\/closed" ], "Problem": "I haven't noticed any geometry classes on the \"Online Classes\" page. I have only checked recently, so it might have been offered before (and now it's over). Will there be any geometry classes any time over the summer? I would be interested if there will.", "Solution_1": "The olympiad geometry class just ended a few weeks ago. Im sure they'll offer it again soon, because it was truly an amazing class. Take it.", "Solution_2": "Also of great interest would be an inequalities class. I believe there was one, but unfortunately, I was too busy to take it.", "Solution_3": "Ya, the 11th class is tonight. It's also a very good class. Take it too :)", "Solution_4": "Ugh. Now I [i]really[/i] wish I hadn't been so busy :-). I think the classes should have subject stuff over the summer and MathCounts/AMC stuff during the school year. That way, people who might want to study a particular subject (for school next year, for example) can do it over the summer, and those who need to prepare for MathCounts/AMC can study right before it before they forget/get rusty like I did.", "Solution_5": "We will soon be adding a second Geometry course that will not be as advanced as the Olympiad Geometry course we offer.\r\n\r\nBoth the Olympiad Geometry and Inequalities courses will be offered during the school year.\r\n\r\nOur current plan is to eventually offer three Geometry and two Inequalities classes to cover the spectrum of those subjects.", "Solution_6": "That'd be good if there were 3 Geometry courses.\r\nI just finished Geometry and understood everything pretty well. I can usually do the first 20 questions involving Geometry the AMC 10. I would think the intro class would be too easy for me (I'll have to take the Do You Need This test in a week or two to find out for sure), and yet the Olympiad class would most likely be too difficult.", "Solution_7": "[quote=\"NightFlarer\"]That'd be good if there were 3 Geometry courses.\nI just finished Geometry and understood everything pretty well. I can usually do the first 20 questions involving Geometry the AMC 10. I would think the intro class would be too easy for me (I'll have to take the Do You Need This test in a week or two to find out for sure), and yet the Olympiad class would most likely be too difficult.[/quote]\r\n\r\nWe do plan to eventually offer three levels of Geometry, though the first Geometry course will be valuable for students who can already work most AMC geometry problems.", "Solution_8": "I would be really interested in a AMC level Geometry class. I still have some problems at some areas in geometry.", "Solution_9": "Unfortunately, it's pretty unlikely it will be ready by then, and having seen your work in other areas, I'd agree that the next level is more suitable for you than intro.\r\n\r\nYou'll be getting some boot-camp geometry training in the Indep Study, though, and we can talk more about geometry approaches in office hours when you show up.", "Solution_10": "The independent study assignments will start off with the harder parts of geometry in volume 1." } { "Tag": [], "Problem": "Well think about it: if clock parts move at x speed and y distance, a certain amount of time has passed and the clock shows the correct time. However, if they move at t speed, it shows the wrong time because they've moved v distance. Got it, or have I confused you completely?", "Solution_1": "Are you just saying that the clock hands have to move y distance? Or else it wouldn't be right? ISn't that a kinda contradiction, in the sense that anything else would be wrong? I don't know what he was asking, but I think he wanted something specific.", "Solution_2": "Well the clock's speed tells us how fast a second or a minute or whatever it is you're measuring is and the distance tells us the actual time it takes for an event to occur. I'm not sure if that's the answer or not though...", "Solution_3": "nevermind. I had an elxplination but after further thought on the problem I found that it was irrelevant to what u were looking for. Although last period I found that if u connect the hour hand and the minute hand then u can find the time opposite of the angle of the hypotenuse. I was bored so I did math or science or whatever u want to call it something." } { "Tag": [ "geometry", "analytic geometry", "algorithm" ], "Problem": "Find the number of square units in the area of the shaded triangle in the figure. The coordinates of the three vertices are labeled.\n[asy]size(150);\npair A = (0,0), B = (7,0), C = (0,7), D=(4,0), E=(4,4),F=(0,6);\ndot(A);\ndraw(A--B,EndArrow);\ndraw(A--C,EndArrow);\ndraw(D--E--F--cycle);\nfill(D--E--F--cycle,gray(0.5));\nlabel(\"$(0,6)$\",F,W);\nlabel(\"$(4,4)$\",E,E);\nlabel(\"$(4,0)$\",D,NE);[/asy]", "Solution_1": "Perfect time to use the shoelace algorithm!\r\n\r\nThe shoelace algorithm (for a triangle) is:\r\n\r\n$ \\frac{1}{2} |x1y2 \\plus{} x2y3 \\plus{} x3y1 \\minus{} x1y3 \\minus{} x2y1 \\minus{} x3y2|$\r\n\r\nThus, simply plug it in.\r\n\r\n$ \\frac{1}{2} |16 \\plus{} 24 \\plus{} 0 \\minus{} 24 \\minus{} 24 \\minus{} 0|$\r\n\r\nThus, the answer is $ \\boxed{4}$.", "Solution_2": "are you sure?\r\ni got 8\r\n\r\nhere's the pic:\r\n[asy]size(150);\npair A = (0,0), B = (7,0), C = (0,7), D=(4,0), E=(4,4),F=(0,6);\ndot(A);\ndraw(A--B,EndArrow);\ndraw(A--C,EndArrow);\ndraw(D--E--F--cycle);\nfill(D--E--F--cycle,gray(0.5));\nlabel(\"$(0,6)$\",F,W);\nlabel(\"$(4,4)$\",E,E);\nlabel(\"$(4,0)$\",D,NE);[/asy]\r\n\r\nif we extend the right side of the triangle up until (4,6), then we have a right triangle and it'll be easy to find the area of that\r\n\r\n[asy]size(150);\npair A = (0,0), B = (7,0), C = (0,7), D=(4,0), E=(4,6),F=(0,6);\ndot(A);\ndraw(A--B,EndArrow);\ndraw(A--C,EndArrow);\ndraw(D--E--F--cycle);\nfill(D--E--F--cycle,gray(0.5));\nlabel(\"$(0,6)$\",F,W);\nlabel(\"$(4,4)$\",E,E);\nlabel(\"$(4,0)$\",D,NE);[/asy]\r\n\r\nthe area of that shaded region would be $ 4\\times6\\div2 \\equal{} 12$\r\n\r\nbut then we'd have to subtract the part that we put in\r\nthe area of that part would be $ 2\\times4\\div2\\equal{}4$ (i would make a pic, but i'm not sure how to draw this exact diagram)\r\n\r\nanyway, we subtract 4 from 12 to get 8\r\ndid i do something wrong?", "Solution_3": "I made a mistake.\r\n\r\nIt should be this:\r\n\r\n$ \\frac{1}{2} |16 \\plus{} 24 \\plus{} 0 \\minus{} 24 \\minus{} 0 \\minus{} 0|$\r\n\r\nThis equals $ \\boxed{8}$.\r\n\r\nThanks for catching my error, vallon." } { "Tag": [ "inequalities solved", "inequalities" ], "Problem": "Let $a,b,c \\in R$ so that $b+c=a>0$ . If $x,y \\in R $ and $ \\sqrt{a-bx-cy}+ \\sqrt{a+by+cx}=a $ prove that $|x+y| \\leq a$.\r\n\r\ncheers! :D :D :cool:", "Solution_1": "Nice:\r\n Put a-bx-cy=u^2, a+by+cx=v^2, with u,v>=0. Then u+v=b+c and x=(u^2b+v^2c-a^2)/(c^2-b^2) and y=(a^2-v^2b-u^2c)/(c^2-b^2) and thus |x+y|=|u^2-v^2|/(b+c)=|u-v|<=u+v=a." } { "Tag": [ "vector", "combinatorics unsolved", "combinatorics" ], "Problem": "there are $ n$ Space Vectors in a three dimensional space.The angles between any two of the vectors are obtuse angles. Find the maximum value of $ n$.\r\n\r\n\r\n\r\n :)", "Solution_1": "i think the answer is $ 4$", "Solution_2": "Lets start from vector $ v_1$. Let $ S$ be a plane which contais the origin and $ S\\bot v.$ $ S$ is divides $ R^{3}$ into two half-space. Call them $ E_1$ and $ E_2$ (they are open). If $ v_1\\in E_1$ then $ v_2,...,v_n \\in E_2$. If $ n>4$ then consider $ v_2,..,v_5$. They are form (in some order. we suppose that in order that was specified ) some angle. it follows from the conditions of the problem that these angle is convex. So $ \\angle (v_2,v_3) \\plus{} ... \\angle(v_5,v_2) <360^{\\circ}$. So one of them is low than $ 90^{\\circ}$. Contradiction.\r\nIt is easy to buid an example for $ n\\equal{}4$..\r\nsorry for my english )", "Solution_3": "your solution may be the best one. :)" } { "Tag": [ "Gmaas" ], "Problem": "what's ur age??", "Solution_1": "dont know why it really matters, but i'm 14", "Solution_2": "15 1/2 (exactly as of today)", "Solution_3": "15 1/6 as of tomorrow!", "Solution_4": "13", "Solution_5": "almost 14 1/2.", "Solution_6": "13 here", "Solution_7": "15 in 6 days", "Solution_8": "slightly greater than 15 1/6", "Solution_9": "16...yes...the oldest", "Solution_10": "I'm 16 also.", "Solution_11": "[quote=\"KBabe\"]How old [b]r u[/b]?[/quote]\r\n\r\nI'm old enough to not use \"r\" and \"u\" in that context... which puts me just over 14.5", "Solution_12": "umm, 14 1/3? give or take some days :D", "Solution_13": "Another old geezer here: approx 16.5", "Solution_14": "12 years old.", "Solution_15": "[quote=ThriftyPiano]The first few posters are like 35+ years old by now.\nThe oldest is either 37 or 38.[/quote]\n\nNah the oldest is like 34", "Solution_16": "1 8 7 $ $", "Solution_17": "[quote=stringbean]i'm 9 [hide=important]pls don't tease me that will make me very sad[/hide][/quote]\n\n[hide=dude your like, way younger than me ] and probably smarter than me[/hide]\n\nEdit: I'm 13 (just turned 13 on the 23 so yaaaaa)", "Solution_18": "hMmmmMmMmMMmM like 18769665420 years old ", "Solution_19": "[quote=captcha000]dont know why it really matters, but i'm 14[/quote]\n\nWoahhh you're like 30 or something now! *Mind Blown*", "Solution_20": "[quote=ArtyA6][quote=stringbean]i'm 9 [hide=important]pls don't tease me that will make me very sad[/hide][/quote]\n\n[hide=dude your like, way younger than me ] and probably smarter than me[/hide]\n\nEdit: I'm 13 (just turned 13 on the 23 so yaaaaa)[/quote]\n\nlol my last map reading score was an 202[hide=but like]my math was a 268 so it kinda cancels each other out kinda idk i'm bad at reading [/hide]", "Solution_21": "$~~~~~~~~~~~~$", "Solution_22": "i am 13 currently, but i is uneducated\n\nnever knew pith0n was younger than me..", "Solution_23": "\\bump 14 now", "Solution_24": "9...$ $", "Solution_25": "12 give or take...", "Solution_26": "These people are both old and young :0", "Solution_27": "10.... (not kidding, I'm actually 10)", "Solution_28": "[quote=tedy]16 or 17[/quote]\n\n????", "Solution_29": "13 $ $" } { "Tag": [ "quadratics", "number theory unsolved", "number theory" ], "Problem": "Let $m ,n \\in N , a \\in Z$ and $gcd(m,n)=1$ . Prove that : \r\n If there exist $x,y\\in N$ such that : $x^2 \\equiv a (\\mod m)$ and $y^2\\equiv a (\\mod n)$ , then there exist $z \\in N$ such that : $z^2 \\equiv a (\\mod mn)$", "Solution_1": "Just look up the chineese remainder theroem. (it is a useful and simple lemma ;) \r\n-Ali", "Solution_2": "Could you please give me a link to Chinese remainder ?", "Solution_3": "The asked link is there:\r\nhttp://mathworld.wolfram.com/ChineseRemainderTheorem.html\r\nthen we can find z by solving the system\r\n$x\\equiv z (mod m)$,\r\n$y\\equiv z (mod n)$,\r\nby the above cited theorem we know the solution exists." } { "Tag": [ "vector", "combinatorics unsolved", "combinatorics" ], "Problem": "We say that a set $\\displaystyle A$ of non-zero vectors from the plane has the property $\\displaystyle \\left( \\mathcal S \\right)$ iff it has at least three elements and for all $\\displaystyle \\overrightarrow u \\in A$ there are $\\displaystyle \\overrightarrow v, \\overrightarrow w \\in A$ such that $\\displaystyle \\overrightarrow v \\neq \\overrightarrow w$ and $\\displaystyle \\overrightarrow u = \\overrightarrow v + \\overrightarrow w$.\r\n\r\n(a) Prove that for all $\\displaystyle n \\geq 6$ there is a set of $\\displaystyle n$ non-zero vectors, which has the property $\\displaystyle \\left( \\mathcal S \\right)$.\r\n\r\n(b) Prove that every finite set of non-zero vectors, which has the property $\\displaystyle \\left( \\mathcal S \\right)$, has at least $\\displaystyle 6$ elements.\r\n\r\n[i]Mihai Baluna[/i]", "Solution_1": "Nobody wants to solve it?", "Solution_2": "(a) If $n = 2m$, then we can choose $(-m, 0); (-m+1, 0); \\cdots ; (-1,0); (1,0); \\cdots; (m-1, 0); (m, 0)$.\r\n\r\nIf $n = 2m+1$, we can just add $(m+1, 0)$.\r\n\r\nNot sure about (b) yet..." } { "Tag": [ "number theory", "prime factorization" ], "Problem": "How many positive integers less than 1000 are multiples of 11 and\nare also perfect squares?", "Solution_1": "A perfect square that is a multiple of $ 11$ must be the square of a multiple of $ 11$, since $ 11$ has no repeating factors in its prime factorization (It is not divisible by any perfect squares other than 1). So we do a simple check.\r\n\r\n$ 11^2\\equal{}121$\r\n$ 22^2\\equal{}484$\r\n\r\n$ 33^3\\equal{}1089$\r\n\r\nOops! The problem only asks for numbers under $ 1000$! Thus, there are $ \\boxed{2}$ perfect squares under $ 1000$ that are multiples of 11, $ 121$ and $ 484$." } { "Tag": [ "function", "calculus", "derivative", "calculus computations" ], "Problem": "Show that $e^{x^{2}y+z}-(1+x^{2}+y^{2})$ defines a function $z=z(x,y)$ near $(0,0)$. And show\r\nthat $(0,0)$ i a critical point of this function. Also what type of critical point is it.\r\n\r\n[hide=\"my try\"]\nWe use implicit function thm.\nNeed som help with the other questions.\n[/hide]\r\n\r\n[color=red]EDIT I mean :[/color] $e^{x^{2}y+z}-(1+x^{2}+y^{2})=0$", "Solution_1": "I suppose that you meant \"$e^{x^{2}y+z}-(1+x^{2}+y^{2})=0$\"\r\n\r\nImplicit Function theorem also tells you what $\\partial z/\\partial x$ and $\\partial z/\\partial y$ are. They both are equal to zero at $(0,0)$, which means that you have a critical point. You can apply the IFT again to find all second-order partials and run the second-derivative test... but $z=\\ln(1+x^{2}+y^{2})-x^{2}y$, so why bother? it's a local minimum, because the 3rd-order term $x^{2}y$ does not matter.", "Solution_2": "Could you develop the part where one examines the crtitical point, please?", "Solution_3": "Never mind my sketchy explanation about the 3rd order. Just find the second derivatives of $z$ using the explicit formula for it, and apply the 2nd der.test." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Find $max c$, where $c=const$ such that the sequence {$a_n$}$_1^\\infty$ $a_n=\\frac{1}{1^2}+\\frac{1}{2^2}+\\frac{1}{3^2}+..+\\frac{1}{n^2}+\\frac{1}{n+c}$ is monotonely decreasing.\r\n\r\n[hide]I don`t see the sense of that problem, but.. [/hide]", "Solution_1": "Monotonically decresaing means:\r\n\r\n$n < m \\rightarrow a_n > a_m$\r\n\r\nSo set $m=n+1$, do the maths, and i got that:\r\n\r\n$c < \\sqrt{2}-1$\r\n\r\nIm not sure, but this means a max dosnt exist... you can put c as close as you want to it, but never equal to it, so you can always get a higher c.\r\n\r\n\r\n(PS not to sure about mods liken homework posted!!!)" } { "Tag": [ "geometry", "3D geometry", "prism", "sphere", "AMC", "AIME", "rectangle" ], "Problem": "Consider the set of points that are inside or within one unit of a rectangular parallelepiped (box) that measures 3 by 4 by 5 units. Given that the volume of this set is $(m + n \\pi)/p$, where $m$, $n$, and $p$ are positive integers, and $n$ and $p$ are relatively prime, find $m + n + p$.", "Solution_1": "[hide=\"Answer\"]The box's volume is 60.\n\nThe points external to the box within one unit can be broken up into three types of pieces: rectangular prisms, quarter cylinders, and eighth-spheres. The prisms can be found by 2(3*4 + 4*5 + 5*3) = 94. There will be a total of 12 quarter cylinders, each with radius 1. Four will have a height of 3, four will have a height of 4, and four will have a height of 5. We can thus combine them, and we have 3 cylinders of radius 1 and heights 3, 4, and 5. Their volume is $\\pi(3 + 4 + 5) = 12\\pi$. For the eighth-spheres, we have a total of 8, so we can combine them to form a regular sphere with radius 1. Its area is $\\frac{4}{3} \\pi$. Adding these volumes, we have\n\n$60 + 94 + 12\\pi + \\frac{4}{3}\\pi=\\frac{462 + 40\\pi}{3}$\n\nTherefore the answer is 505.[/hide]\r\n\r\nI haven't done an AIME problem in a while, so I might have messed up somewhere.", "Solution_2": "I think you forgot to subtract 6. 487 sounds more correct.", "Solution_3": "Uh, I think 505 sounds more correct.", "Solution_4": "[quote=\"Treething\"]I think you forgot to subtract 6. 487 sounds more correct.[/quote]Hmm.. i think 487 sounds correct. Treething is reminding you to subtract the volume inside the rectangular parallelepiped that is not within one unit of the sides. You can visualize it in 2-d by thinking about the area in a 3 x 4 rectangle that is at least 1 unit from the sides. What he did was $(3-1-1)(4-1-1)(5-1-1)=6$", "Solution_5": "Note that the question asks for the volume \"[b]inside[/b] or within one unit of a rectangular parallelepiped\" (emphasis mine) :wink: \r\n\r\n505 is the correct answer (checked on AMC website).", "Solution_6": "Even if you did have to subtract 6, since when did 505 - 6 = 487?", "Solution_7": "[quote=\"E^(pi*i)=-1\"]Note that the question asks for the volume \"[b]inside[/b] or within one unit of a rectangular parallelepiped\"[/quote]oh oops.. :blush: \n[quote=\"Ubemaya\"]Even if you did have to subtract 6, since when did 505 - 6 = 487?[/quote]$60+94+12\\pi+\\frac{4}{3}\\pi-6=\\frac{462-18+40\\pi}{3}=\\frac{444+40\\pi}{3}$ which adds up to 487.", "Solution_8": "problems like this are impossible to visualize for me :( any suggestions?", "Solution_9": "[quote=\"Ubemaya\"]Even if you did have to subtract 6, since when did 505 - 6 = 487?[/quote]\r\nsubtracting six is the same as subtracting 18 from the answer, because there's a denominator of 3. :wink:", "Solution_10": "[quote=\"BarbieRocks\"][quote=\"Ubemaya\"]Even if you did have to subtract 6, since when did 505 - 6 = 487?[/quote]\nsubtracting six is the same as subtracting 18 from the answer, because there's a denominator of 3. :wink:[/quote]This has been responded by minsoens 3 years ago...making your comment [u]superfluous[/u].\n\n[quote=\"mathemonster\"]problems like this are impossible to visualize for me :( any suggestions?[/quote]It isn't impossible but it takes practice. Draw more 3D figures, study them, rotate them, do more problems related to solid geometry. The point is to develop your mind to \"visualize\" objects from familiar ones and make more complex shapes. Even watch more 3D movies (if that helps :P ).", "Solution_11": "[quote=chesspro][hide=\"Answer\"]The box's volume is 60.\n\nThe points external to the box within one unit can be broken up into three types of pieces: rectangular prisms, quarter cylinders, and eighth-spheres. The prisms can be found by 2(3*4 + 4*5 + 5*3) = 94. There will be a total of 12 quarter cylinders, each with radius 1. Four will have a height of 3, four will have a height of 4, and four will have a height of 5. We can thus combine them, and we have 3 cylinders of radius 1 and heights 3, 4, and 5. Their volume is $\\pi(3 + 4 + 5) = 12\\pi$. For the eighth-spheres, we have a total of 8, so we can combine them to form a regular sphere with radius 1. Its area is $\\frac{4}{3} \\pi$. Adding these volumes, we have\n\n$60 + 94 + 12\\pi + \\frac{4}{3}\\pi=\\frac{462 + 40\\pi}{3}$\n\nTherefore the answer is 505.[/hide]\n\nI haven't done an AIME problem in a while, so I might have messed up somewhere.[/quote]\n\nHello, in this solution does anyone know where he gets the quarter cylinders from? (I'm not great at visualization and trying to understand a good way to do this problem). Thank you in advance.", "Solution_12": "I assume its from the points that are on the edge but are not vertices, as the vertices are the eighth spheres and not part of the cylinder" } { "Tag": [ "vector" ], "Problem": "Could anyone tell me how exactly one derives (rigorously, if possible) from Maxwell's equations (any subset of the latter) Coloumb's law - or point me to an appropriate source. Also, in the case of e.g. electrostatics, does the the priniciple of superposition of electric fields as caused by different charges (i.e. the elctric field at a point in space is the vector sum of the electric fields caused by individual charges) follow from Maxwell's equations (any subset of them)? And to finish it off (O.K. I'm obviously asking too many questions, I hope you lot can answer at least some of them :roll: ), what exactly does it mean that light is an electromagnetic wave - that is, (precisely) which conditions must the E and H vectors in space satify for us to be in a position to claim that \"there is\" light in this space, or to be more illustrative, when does an electromagnetic field have the \"god-given\" right to say: \"Let there be light!\"? \r\nThanks for any answers and thoughts on the subject. :lol:", "Solution_1": "I don't know about Coulomb's law, but the principle of superposition of electric fields follow from the same principle of forces. I.e. (I hope you can follow):\r\n\r\n$F=\\sum_i F_i = \\sum_i K\\frac{qq_i}{r_i^2} = qK\\sum_i \\frac{q_i}{r_i^2}=q\\sum_i E_i$\r\n\r\n$F$ being the net force on a charge, $F_i,q_i,r_i$ being the appropriate things for each individual charge that affects the \"test\" charge. Since we desire $F=qE$ it follows that\r\n\r\n$E=\\sum_i E_i$\r\n\r\nAnd that's superposition for ya. (Replace everything with vector equivalences for a stringent proof.)\r\n\r\n--\r\n\r\nElectromagnetic waves [b]in vacuum[/b] satisfy the [i]wave equations[/i]:\r\n\r\n$\\nabla^2 \\vec{E} = \\frac{1}{c^2} \\frac{\\partial \\vec{E}}{\\partial t}$\r\n$\\nabla^2 \\vec{B} = \\frac{1}{c^2} \\frac{\\partial \\vec{B}}{\\partial t}$\r\n\r\nThis follows from Maxwell's equations ($c$ being the speed of light). I don't know of any such equations for electromagnetic radiation through matter." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "i searched but couldn't find this problem in the forum. so i posted it.\r\n\r\nFind all primes $p,q$ such that $p+q=(p-q)^{3}$.", "Solution_1": "[quote=\"nayel\"]i searched but couldn't find this problem in the forum. so i posted it.\n\nFind all primes $p,q$ such that $p+q=(p-q)^{3}$.[/quote]\r\n\r\nmaybe i'm wrong...i'm not so good :blush: but this is what i've found:\r\n\r\nif $p+q=(p-q)^{3}$ then $(p-q) | (p+q)$\r\n$\\frac{p+q}{p-q}= \\frac{p-q+q+q}{p-q}=1+\\frac{2q}{p-q}$\r\nthen $(p-q)| 2q$ => $2q=np-nq \\Rightarrow p= q+\\frac{2q}{n}$\r\neither $n=1 \\Rightarrow p=3q$ but both $p$ and $q$ are primes\r\nor $n=2 \\Rightarrow p=2q$ but both $p$ and $q$ are primes\r\nor $n=q \\Rightarrow p=q+2 (1)$ \r\nor $n=2q \\Rightarrow p=q+1 (2)$\r\nif $(2)$then $2q+1=1$ but $q$ is a prime\r\nif $(1)$then $2q+2= (q+2-q)^{3}$\r\n $2q+2=8$\r\n $q=3 ; p=5$\r\n \r\nThe only solution is $(5,3)$\r\n\r\nEDIT: I added the case n=2q", "Solution_2": "nice solution! :D \r\n\r\ni have the following one:\r\nif $p,q\\ge 5$, we check the four cases: (congruences are taken $\\bmod 6$ if otherwise not stated)\r\n1. $p\\equiv q\\equiv 1$, then $LHS\\equiv 2$ while $RHS\\equiv 0$\r\n2. $p\\equiv 1, q\\equiv-1$ then $LHS\\equiv 0, RHS\\equiv 2$\r\n3. $p\\equiv-1, q\\equiv 1$ then $LHS\\equiv-2, RHS\\equiv0$\r\n4. $p\\equiv-1, q\\equiv-1$ then $LHS\\equiv 0, RHS\\equiv-2$\r\n\r\nso both $p$ and $q$ cannot be >3.\r\n\r\nif $p=3$ we get $q<3$ so nothing. similarly $p=2$ implies nothing.\r\nif $q=3$ we have after expanding, $p(p^{2}-9p+26)=30$ so $p|30$ and from here $p=5$ because $p=3$ isn't a solution. \r\nif $q=2$, $p(p^{2}-6p+11)=10\\Rightarrow p|10\\Rightarrow$ no solution. :wink:", "Solution_3": "Could you please explain what do LHS and RHS mean...i haven't found it :blush: \r\nthanks", "Solution_4": "[quote=\"EUCLA\"]Could you please explain what do LHS and RHS mean...i haven't found it :blush: \nthanks[/quote]\r\n\r\nLHS=Left Hand Side (of the equation) \r\nRHS=Right Hand Side (of the equation) :)", "Solution_5": "[quote=\"nayel\"][quote=\"EUCLA\"]Could you please explain what do LHS and RHS mean...i haven't found it :blush: \nthanks[/quote]\n\nLHS=Left Hand Side (of the equation) \nRHS=Right Hand Side (of the equation) :)[/quote]\r\nthanks, now i've understood your solution :lol:" } { "Tag": [ "logarithms", "quadratics", "algebra", "quadratic formula" ], "Problem": "Solve each log equation given.\r\n\r\n(1) log_5 (x^2 + x + 4) = 2\r\n\r\n(2) log_3 (3^x) = -1", "Solution_1": "[quote=\"sharkman\"]Solve each log equation given.\n\n(1) log_5 (x^2 + x + 4) = 2\n\n(2) log_3 (3^x) = -1[/quote] Use the log rule that $ \\log_x (a) \\equal{} b$ can be written as $ x^b \\equal{} a$, to get a nice expression. (Eg. $ \\log_5 (x) \\equal{}2$ can be written as $ x\\equal{} 5^2$.)\r\n\r\n Simplify then solve for x.", "Solution_2": "1)\r\n$ \\log_5(x^2 + x + 4) = 2\\implies5^2 = x^2 + x + 4\\implies x^2 + x - 21 = 0$\r\nusing the quadratic formula $ x_1 = 4.11$ and $ x_2 = - 5.11$ is the solutions\r\n\r\n2)\r\n$ \\log_3(3^x) = - 1\\implies3^{ - 1} = 3^x\\implies\\frac {1}{3} = 3^x$\r\n$ \\ln{\\frac {1}{3} = \\ln{3^x}\\implies x = \\frac {\\ln{\\frac {1}{3}}}{\\ln{3^x}} = - 1\\implies x = - 1}$", "Solution_3": "In particular, the exact solutions to number 1 are $ \\frac{\\minus{}1 \\pm \\sqrt{85}}{2}$.\r\n\r\nIn number 2, if you use the same binom did in his 3rd step on his first step, you immediately get x=-1", "Solution_4": "[quote=\"binomial_4eva\"]1)\n$ \\log_5(x^2 + x + 4) = 2\\implies5^2 = x^2 + x + 4\\implies x^2 + x - 21 = 0$\nusing the quadratic formula $ x_1 = 4.11$ and $ x_2 = - 5.11$ is the solutions\n\n2)\n$ \\log_3(3^x) = - 1\\implies3^{ - 1} = 3^x\\implies\\frac {1}{3} = 3^x$\n$ \\ln{\\frac {1}{3} = \\ln{3^x}\\implies x = \\frac {\\ln{\\frac {1}{3}}}{\\ln{3^x}} = - 1\\implies x = - 1}$[/quote] Alternatively recall that $ \\log_a a = 1$. \r\n\r\n[hide=\"Hence,\"] $ \\log_3(3^x) = - 1 \\Leftrightarrow x \\log_3 3 = - 1.$ \n\nI.e. $ x = - 1$.[/hide]" } { "Tag": [ "AMC", "AIME", "ratio", "geometry", "angle bisector" ], "Problem": "Does anyone have any non-proof AIME-like probs which I can try for these? The AoPS book is kinda short on them.\r\nAlso, has Menalaus/Ceva appeared in previous AIMEs?\r\n\r\nThanks.", "Solution_1": "Menelaus/Ceva [i]very[/i] rarely comes up on non-Olympiad, non-proof type problems. It wont show up nor ever has shown up on AIME (unless there's an alternate way using Ceva/Menelaus).\r\n\r\nNonetheless, here is one: \r\n\r\nGiven triangle ABC, with AB = 12 and CA = 15. The median from B, the altitude from C, and the angle bisector from A are all concurrent inside the triangle. Compute the length of BC.", "Solution_2": "[quote=\"joml88\"]Menelaus/Ceva [i]very[/i] rarely comes up on non-Olympiad, non-proof type problems. It wont show up nor ever has shown up on AIME (unless there's an alternate way using Ceva/Menelaus).\n\nNonetheless, here is one: \n\nGiven triangle ABC, with AB = 12 and CA = 15. The median from B, the altitude from C, and the angle bisector from A are all concurrent inside the triangle. Compute the length of BC.[/quote]\r\n\r\nI thought that was an old aime problem?\r\n\r\nUsing angle bisector, we know that the parts of $BC$ are $12x$ and $15x$ which are separated by the angle bisector, for some common ratio $x$, so using Ceva, we have $12x\\cdot\\frac{15}{2}\\cdot a=15x\\cdot\\frac{15}{2}\\cdot b \\Longrightarrow a=\\frac{5}{4}b$, where $a$ and $b$ are the parts of $AB$ which are seperated by the altitude.\r\n\r\nFrom using Pythagorus, we have $15^2-a^2=(27x)^2-b^2$, but since $a+b=12$, so $b=\\frac{16}{3}$, we can simplify this to $(27x)^2=15^2-(\\frac{20}{3})^2+(\\frac{16}{3})^2$, and someone can solve this..." } { "Tag": [], "Problem": "\u0420\u0435\u0431\u044f\u0442\u0430 \u043f\u043e\u043c\u043e\u0433\u0438\u0442\u0435 \u043c\u043d\u0435 \u0434\u0432\u0435 \u043f\u043e\u0441\u043b\u0435\u0434\u043d\u0438\u0435 \u0437\u0430\u0434\u0430\u0447\u0438!\r\n \u0420\u0410\u041d\u041d\u0415\u0415 \u0421\u041f\u0410\u0421\u0418\u0411\u041e!!!\r\nhttp://rapidshare.com/files/145515024/IMG_3961.JPG.html", "Solution_1": "Out of curiosity,where do you study?\r\nquite clear that in Russia,but where exactly?due to the problems it is your first year?", "Solution_2": "[quote=\"Erken\"]Out of curiosity,where do you study?\nquite clear that in Russia,but where exactly?due to the problems it is your first year?[/quote]\r\n\u0414\u0430. \u042d\u0442\u043e \u0434.\u0437 \r\n\u042f \u0432 \u0422\u043e\u043c\u0441\u043a\u0435 \u0438 \u044d\u0442\u043e \u0437\u0430\u0434\u0430\u0447\u0438 \u0434\u043b\u044f \u043f\u0435\u0440\u0432\u043e\u0433\u043e \u043a\u0443\u0440\u0441\u0430.\r\n \u042f \u0443\u0436\u0435 \u0440\u0435\u0448\u0438\u043b \u0432\u0441\u0435 \u043d\u043e \u043d\u0435 \u043e\u0447\u0435\u043d\u044c \u043a\u0440\u0430\u0441\u0438\u0432\u043e.\r\n\u0418 \u0445\u043e\u0447\u0443 \u0432\u0441\u0435 \u0432\u043c\u0435\u0441\u0442\u0435 \u0440\u0435\u0448\u0430\u0442." } { "Tag": [ "geometry" ], "Problem": "I've been a TA for 5 schools in the Westchester Area (30 miles north from NYC). Each year some way or the other I'm stuck dealing with the graduation gown ordering. Each of the schools use a rep that comes in an measures all of the students. I've had many problems with this as for some odd reason when it comes graduation time we are left with a few students that don't have a graduation gown.\r\n\r\nHas anyone thought of ordering directly from a company and placing the order yourself? \r\n\r\n[url=http://www.graduationsource.com]Graduation Source[/url]", "Solution_1": "The most appropriate forum for this question is probably [url=http://www.artofproblemsolving.com/Forum/index.php?f=138]the Round Table[/url]. \r\n\r\nI'm not sure how it works in your area, but at my high school the seniors are all given order forms on a certain day after school and told to turn them in a week or so afterward. The students without graduation gowns are just the students that didn't order them, which is more or less their problem." } { "Tag": [ "geometry", "search", "videos", "puzzles" ], "Problem": "A ball is elevated off the ground. A person is trying to get the ball knocked down onto the ground. After two minutes of failed attempts, the ball is finally on the ground. However, once the ball hits the floor, he regrets having tried to knock it down in the first place. What is going on?\r\n\r\nFeel free to ask questions.", "Solution_1": "Erm... The ball is a bowling ball. He is trying to get it down from a high shelf to bowl. But, once he gets it down, it hits his toe, making him regret getting the ball down.\r\n\r\nSecond guess: The ball is an egg of a bird. The guy is hungry and tries to get it from the nest. But, he accidentally hits the egg over without catching it and it splatters everywhere, leaving him with no food.", "Solution_2": "No to both of those. There was virtually no chance of him getting hurt in the process.\r\n\r\nAgain, feel free to ask questions.", "Solution_3": "did knocking the ball down do any damage to anyone or the environment in which he was standing (ex. break a floor)?", "Solution_4": "Did knocking the ball down do any damage to anyone or the environment in which he was standing? [b]No[/b]", "Solution_5": "[hide=\"Hint\"]The person was nowhere near the ball. But he was trying to get it knocked down... how could this be?[/hide]", "Solution_6": "[hide]The ball was stuck on top of something, and the person threw something at the ball to knock it down. When he finally knocked the ball down, the other thing got stuck up there?[/hide]", "Solution_7": "The ball was stuck on top of something, and the person threw something at the ball to knock it down. When he finally knocked the ball down, the other thing got stuck up there? [b]No[/b]", "Solution_8": "How was the ball elevated?\r\n\r\n...Or are we only allowed yes or no questions?\r\nIn that event, what the ball elevated with a string?", "Solution_9": "Is the ball spherical?", "Solution_10": "Please ask yes/no questions only. However, The_Scintillator, you could find out how the ball was elevated by asking more questions.\r\n\r\nIn that event, what the ball elevated with a string? [b]No[/b]\r\nIs the ball spherical? [b]Yes[/b]", "Solution_11": "Question time! :D\r\n\r\nWas he playing Cricket? Or some odd subsort? With a ball instead of stumps....\r\n\r\nWas he playing Bocce? (I don't know much about that game, though.)\r\n\r\nWas the ball elevated on a solid surface?\r\n\r\nWas the ball elevated on a liquid surface?\r\n\r\nWas the ball elevated onto a shelf?\r\n\r\nWas the ball elevated onto a pole?\r\n\r\nWas the ball involved in some sort of sport?\r\n\r\nWas the ball and scenario insolved in some sort of sport?\r\n\r\nWas the ball setting off a chain reaction?\r\n\r\nDoes the person now suffer from psychological/emotion trauma due to the effects of the ball? (Sounds kind of silly, but would rule out some stuff.)\r\n\r\nDoes it matter who elevated the ball?\r\n\r\nDoes it matter what sort of space/dimension/universe he was in?", "Solution_12": "Was the diameter of the ball $ \\ge 5cm$ ? $ \\ge 10cm$ ? $ \\ge 20cm$ ? $ \\ge 50cm$ ? $ \\ge 1m$ ? $ \\ge 10m$ ? $ \\ge 1km$ ?\r\nWas the ball red? green? blue? yellow? cyan? magenta? white? black? transparent?\r\nWas the ball shattered after it fell down?\r\nWas something destroyed after it fell down?\r\nDid the incident cause any loss to the person?", "Solution_13": "Was he playing Cricket? Or some odd subsort? With a ball instead of stumps.... [b]No[/b]\r\n\r\nWas he playing Bocce? (I don't know much about that game, though.) [b]No[/b]\r\n\r\nWas the ball elevated on a solid surface? [b]Yes, if by that you mean it was supported by something solid[/b]\r\n\r\nWas the ball elevated on a liquid surface? [b]No[/b]\r\n\r\nWas the ball elevated onto a shelf? [b]Yes[/b]\r\n\r\nWas the ball elevated onto a pole? [b]Yes, but not an upright pole[/b]\r\n\r\nWas the ball involved in some sort of sport? [b]Outside this scenario, probably not[/b]\r\n\r\nWas the ball and scenario insolved in some sort of sport? [b]Some consider it a sport, others don't[/b]\r\n\r\nWas the ball setting off a chain reaction? [b]No[/b]\r\n\r\nDoes the person now suffer from psychological/emotion trauma due to the effects of the ball? (Sounds kind of silly, but would rule out some stuff.) [b]No[/b]\r\n\r\nDoes it matter who elevated the ball? [b]No[/b]\r\n\r\nDoes it matter what sort of space/dimension/universe he was in? [b]No[/b]\r\n\r\nWas the diameter of the ball...? [b]Based on official measurements, the diameter is about 1 meter (officially 40 inches)[/b]\r\n\r\nWas the ball red? Green? Blue? ... [b]It was either red or blue. Theoretically, the ball could've been another color. However, in the scenario given, the balls are either red or blue.[/b]\r\n\r\nWas the ball shattered after it fell down? [b]No[/b]\r\n\r\nWas something destroyed after it fell down? [b]No[/b]\r\n\r\nDid the incident cause any loss to the person? [b]Yes, but not physically[/b]", "Solution_14": "Did the incident cause an emotional loss? Any mental loss...?\r\n\r\nWas the object on the pole AND the shelf? At the same time? \r\n\r\nWas the object elevated onto a sideways pole? \r\n\r\nIs the sport known in the U.S.?\r\n\r\nDoes the sport merit a page in Wikipedia? \r\n\r\nWhat sport is this, anyways? (Not an official question.)\r\nGrarg. I don't know.", "Solution_15": "[hide=\"Is the first letter of the first word of the name of the game:\"]\na?\nb?\nc?\nd?\ne?\nf?\ng?\nh?\ni?\nj?\nk?\nl?\nm?\nn?\no?\np?\nq?\nr?\ns?\nt?\nu?\nv?\nw?\nx?\ny?\nz?\n[/hide]\n\n[hide=\"Is the second letter of the first word of the name game:\"]\na?\nb?\nc?\nd?\ne?\nf?\ng?\nh?\ni?\nj?\nk?\nl?\nm?\nn?\no?\np?\nq?\nr?\ns?\nt?\nu?\nv?\nw?\nx?\ny?\nz?\n[/hide]\n\n[hide=\"Is the third letter of the first word of the name of the game:\"]\na?\nb?\nc?\nd?\ne?\nf?\ng?\nh?\ni?\nj?\nk?\nl?\nm?\nn?\no?\np?\nq?\nr?\ns?\nt?\nu?\nv?\nw?\nx?\ny?\nz?\n[/hide]\r\n\r\nHow many words are in the name of the game?", "Solution_16": "[quote]How many words are in the name of the game?[/quote]\r\n\r\nPSST!! Not a yes or no question..\r\n\r\nnice approach, though.", "Solution_17": "Cheap shot. But whatever. :P \r\n\r\nThe first three letters of the name of the game: [b]FIR[/b]", "Solution_18": "Is the game directed towards...\r\nadults (20+)?\r\nseniors (65+)?\r\nteenagers (13-19)?\r\nchildren (12 and under)?\r\n\r\nTo win, do you try to score as many points as possible?\r\n\r\nIs the ball put into motion by anything other than a human? (such as a club or whatnot)", "Solution_19": "Is the game directed towards... \r\nadults (20+)? [b]No[/b]\r\nseniors (65+)? [b]No[/b]\r\nteenagers (13-19)? [b]Yes[/b]\r\nchildren (12 and under)? [b]No[/b]\r\nTo win, do you try to score as many points as possible? [b]Yes[/b]\r\nIs the ball put into motion by anything other than a human? [b]Yes[/b]", "Solution_20": "Does the name contain \"ball\"?\r\nIs it called \"fireball\"?\r\nDoes the sport involve the hand?", "Solution_21": "Does the name contain \"ball\"? [b]No[/b]\r\nIs it called \"fireball\"? [b]No[/b]\r\nDoes the sport involve the hand? [b]Yes, but they don't touch the ball[/b]", "Solution_22": "[hide=\"Hint\"]The organization that created this game also created another game called \"Triple Play.\" This game also has its own Wikipedia article.[/hide]", "Solution_23": "Now it seems too easy\r\n\r\nIs it FIRST Overdrive?", "Solution_24": "Is it FIRST Overdrive? [b]Yes[/b]\r\n\r\nI just gave it away, didn't I?", "Solution_25": "are there usually mmore than 5 participants in this game?\r\n\r\n\r\ndoes playing the game involve more than the ball, the pole, and the participant?\r\n\r\n\r\nare there usually more than 10 balls?", "Solution_26": "I'm officially calling this solved.\r\n\r\n[hide=\"Explanation\"][url=http://www.youtube.com/watch?v=yViRWaSODIc]General overview of game[/url]\nHe was trying to knock the ball down in order to move it around the track, earning points. However, he got it down before the end of the game, so he lost an opportunity for a 12-point bonus in the end. In addition, he lost the opportunity to go around the track, which would still earn points.[/hide]", "Solution_27": "I tried to offer a hint with the question I asked.\r\n[quote=\"krsattack\"]Does this sport require [b]human[/b] (physical) strength?[/quote]\n\nOf course, then lingomaniac88 changed my question (BTW, did you do that on purpose?):\n[quote=\"lingomaniac88\"]Does this sport require physical strength? [b]No[/b][/quote]\r\n\r\nI was hoping that someone would catch my wording...", "Solution_28": "Well, by that, krsattack, I assumed that you explicitly meant physical strength.\r\n\r\nOf course, it does require mental strength, i.e., strategy. \"Should I try to grab and hurdle (throw over the overpass) the ball, or should I just herd it under the finish line?\" Efficiency vs. quantity of points. For those who have no idea what I'm talking about, watch this:\r\n\r\n[youtube]yViRWaSODIc[/youtube]", "Solution_29": "Wow... the \"40 inches to be exact\" and the \"red or blue. in this situation\" should have immediately tipped me off..." } { "Tag": [ "LaTeX", "modular arithmetic" ], "Problem": "Consider a string of $n$ 7's, $7777\\cdots77$, into which $+$ signs are inserted to produce an arithmetic expression. For example, $7+77+777+7+7=875$ could be obtained from eight 7's in this way. For how many values of $n$ is it possible to insert $+$ signs so that the resulting expression has value 7000?", "Solution_1": "(Sorry, haven't installed latex yet)\r\n\r\n\r\n[hide]Looking at the number 7000, we obviously see the maximum number of sevens: a string of one thousand sevens. Then, we see that the minimum is 28 sevens: 777 X 9 + 1 =7000. The next step is to see by what interval the value of N increases. Since 777 is 3 sevens, and 77*10+7 is 21 sevens, we can convert a 777 into 77's and 7's and add 18 to the value of N. Since we have 9 777's to work with, this gives us 28,46,64,82,100,118,136,154,172,190 as values for N. Since 77 can be converted into 7*11, we can add 9 to N by converting 77 into 7's. Our N=190, which has 0 777's 90 77's and 10 7's. We therefore can add 9 to N 90 times by doing this. All values of N not covered by this can be dealt with with the N=46 (8 777's 10 77's 2 7's) up to 190. We also have our 28 term of 9 777's and 1 7. (1000-46)/9=106, but this does not cover N=46 or N=28. So we have a grand total of 108 possible values. [/hide]", "Solution_2": "[quote=\"musicheck\"](Sorry, haven't installed latex yet)\n\n\n[hide]Looking at the number 7000, we obviously see the maximum number of sevens: a string of one thousand sevens. Then, we see that the minimum is 28 sevens: 777 X 9 + 1 =7000. The next step is to see by what interval the value of N increases. Since 777 is 3 sevens, and 77*10+7 is 21 sevens, we can convert a 777 into 77's and 7's and add 18 to the value of N. Since we have 9 777's to work with, this gives us 28,46,64,82,100,118,136,154,172,190 as values for N. Since 77 can be converted into 7*11, we can add 9 to N by converting 77 into 7's. Our N=190, which has 0 777's 90 77's and 10 7's. We therefore can add 9 to N 90 times by doing this. All values of N not covered by this can be dealt with with the N=46 (8 777's 10 77's 2 7's) up to 190. We also have our 28 term of 9 777's and 1 7. (1000-46)/9=106, but this does not cover N=46 or N=28. So we have a grand total of 108 possible values. [/hide][/quote]\r\n\r\nYou DONT have to have [i]anything[/i] installed to use $\\LaTeX$ on the forums. Where does this myth come from :?", "Solution_3": "To use Latex you only need to write dollar signs $. \r\nIf you want to write x^2 just write $x^2$.", "Solution_4": "wait...dondigo, how did you post those dollar signs without having them turn on LaTeX?", "Solution_5": "Check the box that says \"Disable BBCode in this post\" when you are posting (it's below the message body in the options section).", "Solution_6": "[quote=\"musicheck\"]\n[hide=\"fixed up latex for him...\"]Looking at the number $7000$, we obviously see the maximum number of $7's$: a string of $1000 \\ 7's$. Then, we see that the minimum is $28 \\ 7's: \\ 777*9 + 1 =7000$. The next step is to see by what interval the value of $N$ increases. Since $777$ is $3 \\ 7's, \\ 77*10+7 is 21 \\ 7's$, we can convert a $777$ into $77's$ and $7's$ and add $18$ to the value of $N$. Since we have $9 \\ 777's$ to work with, this gives us $28,46,64,82,100,118,136,154,172,190 (=28+18n | 1\\leq n\\leq 9)$ as values for $N$. Since $77$ can be converted into $7*11$, we can add $9$ to $N$ by converting $77$ into $7's$. Our $N=190$, which has $0 \\ 777's \\ 90 \\ 77's \\ 10 7's$. We therefore can add $9$ to $N \\ 90$ times by doing this. All values of $N$ not covered by this can be dealt with with the $N=46 \\ (8 \\ 777's \\ 10 \\ 77's \\ 2 \\ 7's)$ up to $190$. We also have our $28$ term of $9 \\ 777's$ and $1 \\ 7$. $(1000-46)/9=106$, but this does not cover $N=46, N=28$. So we have a grand total of $\\boxed{108}$ possible values. [/hide][/quote]\n\nBy the way, are you sure that [hide=\"answer\"]108[/hide] is the answer?\nHere is how I get my answer:\n[hide=\"??\"]Divide through by $7$ in the question to rewrite it as a string of $1's$ adding to $1000$. So we have $a+11b+111c=1000$. Notice we can rewrite variables to get $a+d+e=1000$, which normally has $\\binom{1002}{2}$ solution in positive integers. However, we can only rewrite when $d$ and $e$ are divisible by $11$ and $111$ respectively. This happens $\\frac{1}{11*111}$ of the time, so we find our answer by $\\lfloor \\frac{\\binom{1002}{2}}{11*111} \\rfloor = \\boxed{136}$. Since the numbers are so big, even using a ballpark figure still gives 136...[/hide]", "Solution_7": "[hide]Yes, $108$ is the answer. (I checked the official answers on the AMC page.) Basically, $n$ can be any number equivalent to $1\\pmod{9}$ between $28$ and $1000$ inclusive, except $37$. I got $109$ because I didn't get rid of $37$. :lol:[/hide]", "Solution_8": "hm aren't there a lot more than $ 108$ solutions to $ a\\plus{}11b\\plus{}111c\\equal{}1000$, or am I overlooking something in the problem?", "Solution_9": "Yes, but many of them give the same total number of $ 7$'s.\r\n\r\nHere is a fairly quick method for the beginning:\r\n[hide]\nWe have \n\n$ a\\plus{}11b\\plus{}111c\\equal{}1000$\n\nand\n\n$ a\\plus{}2b\\plus{}3c\\equal{}n$.\n\nSo subtract the second from the first to get\n\n$ 9b\\plus{}108c\\equal{}1000\\minus{}n$.\n\nAnd now it's obvious that $ 9|1000\\minus{}n\\Rightarrow n\\equiv 1\\pmod 9$.\n[/hide]", "Solution_10": "108 is the right answer" } { "Tag": [ "function", "geometry", "3D geometry", "inequalities" ], "Problem": "Determine all the functions $f : Z \\rightarrow Z$ such that\r\n$f(x + 1959) \\leq f(x) + 1959$ and $f(x + 2002) \\geq f(x) + 2002$.\r\n\r\nDetermine the smallest base $b > 1$ for which there exist a perfect cube $xyxy$ in base $b$", "Solution_1": "Problem 2: [hide]Let's rewrite this algebriacally. xb^3+xb+yb^2+y=z^3 where x and y are less than b. Now, let's factor this equation. xb(b^2+1)+y(b^2+1) or (xb+y)(b^2+1). This means our cubic has to be divisible by the base squared plus one. Now, we test our possibilities. 2 yields 5 which fails because the smallest cube that's divisible by 5 is 125 which is too big. 3 yields 10 which fails because the smallest cube that's divisible by 10 is 1000 which is too big. 4 yields 17 which fails since the smallest cube that's divisible by 17 is 17^3, too big. Similar results come in for 5(26) and 6(37). But when we try 7, we get 50 which is divides 1000 or 10^3. Dividing 50 into 1000 yields 20 so we have 7x+y=20 where x and y are under 7. (2,6) is a solution. Just to verify, 2(343)+6(49)+2(7)+6=686+294+14+6=980+20=1000. So seven is the smallest mod that works.[/hide]", "Solution_2": "Let's assume f(x) is linear for now. We'll expand the case later. y=a(x+1959+b) greaterthanorequalto y=a(x+b)+1959. We cancel out the x and b terms to yeild 1959a islessthanorequalto 1959, so a islessthanorequalto 1. But when we apply this function to 2002, after simplifing, we get 2002a isgreaterthanorequalto 2002, so a isgreaterthanorequalto 1. This pretty much shows that a=1. So all linear equations that come in the form y=x+b will work. Now, we expand to quadratics. a(x+1959)^2+b(x+1959)+c islessthanorequalto ax^2+bx+c+1959 so we cancel out the bx, c, ax^2 terms. We get 1959(2)(ax)+(1959)^2(a)+1959(b) islessthanorequalto 1959 so we divide by 1959. We get 2ax+1959a+b islessthanorequalto 1. Applying the same process to the second inequality, we get 2ax+2002a+b isgreaterthanorequalto 1. Obviously, a isn't negative. But now I'm stuck. Any ideas?" } { "Tag": [ "Support" ], "Problem": "I was just curious how many Indian Mathlinkers are from the North East and Assam in particular!\r\n\r\nSo if anyone is from the NE then please give a brief introduction here!", "Solution_1": "[quote=\"manjil\"]I was just curious how many Indian Mathlinkers are from the North East and Assam in particular!\n\nSo if anyone is from the NE then please give a brief introduction here![/quote]\r\n\r\nSurprising it seems I am the only one!", "Solution_2": "Probably u can spread it that side.", "Solution_3": "[quote=\"chemrock\"]Probably u can spread it that side.[/quote]\r\n\r\nYes why not? I am trying, but the people here are very lazy! :rotfl: :P", "Solution_4": "You've just got another mathlinker from the north- east.....................................hi this is supranta here.", "Solution_5": "why are you guys dividing india into parts\r\n? :huh: :rotfl:", "Solution_6": "Hey man, I am not dividing India into parts......I was just curious!", "Solution_7": "Lol, he was joking! One of the purposes this site performs is to bring the best minds from all over India together. You'll meet people from all parts of the country here. Why look for people in your region when you can meet them everyday in any case? :rotfl:", "Solution_8": "dont take my post seriously (i dont know how you do it :wink: :rotfl: )", "Solution_9": "[quote=\"nuclear_alchemist\"]Lol, he was joking! One of the purposes this site performs is to bring the best minds from all over India together. You'll meet people from all parts of the country here. Why look for people in your region when you can meet them everyday in any case? :rotfl:[/quote]\r\n\r\n(and in the US too) :P", "Solution_10": "[quote=\"manjil\"]I was just curious how many Indian Mathlinkers are from the North East and Assam in particular!\n\nSo if anyone is from the NE then please give a brief introduction here![/quote]\r\nmathlinkers..........!?\r\nThe site artofproblemsolving too supports the same forums.", "Solution_11": "[quote=\"Rajiv\"][quote=\"manjil\"]I was just curious how many Indian Mathlinkers are from the North East and Assam in particular!\n\nSo if anyone is from the NE then please give a brief introduction here![/quote]\nmathlinkers..........!?\nThe site artofproblemsolving too supports the same forums.[/quote]\r\n\r\nSure that is implied!", "Solution_12": "Anyone.I know one or 2 from tripura who visit here regularly", "Solution_13": "well thats too distant\r\n\r\nnever knew Aops is so widespread :)", "Solution_14": "fbaggins!!!", "Solution_15": "he is frm madhya pradesh :D", "Solution_16": "oh....i thought west bengal...hehe :D", "Solution_17": "o..many from WB r here anyways!! :D", "Solution_18": "There's me ,murgi,muks,akashnil,bengal(NRI)\r\nThat's all I can remember! :D", "Solution_19": "and Keshav. Also, I lived in Kolkata for 6 years :P", "Solution_20": "i am from tripura \r\n\r\n[hide]but i live in delhi now, in Vmc reg batch[/hide]", "Solution_21": "which group\r\nand what's your name", "Solution_22": "I am in for jee 2011,\r\n in reg-4.\r\n\r\nMy name is Shankhadip Majumder." } { "Tag": [ "geometry", "3D geometry", "tetrahedron", "sphere", "circumcircle", "calculus", "integration" ], "Problem": "What is the 4d volume of a 4d tetrahedron with side length one? Show work...\r\n\r\nThe tessaract is the wrong name for it, but I can't change that\r\n\r\n(and if there are two solids here then list them both... I dont know the answer im just curious)", "Solution_1": "I always thought a tesseract was the 4d equivalent of a 3d cube. (Which would imply that it has volume 1 unit4)", "Solution_2": "Yes, I also thought that a tesseract was a hyper-cube.\r\n\r\nAlso, there are 6 platonic solids in 4 dimensional space -- I was under the impression that there were [i]two[/i] simplexes, and thus you're asking a question with a reference to two different shapes. (I do know there are 6 platonic solids in 4-D . . . I don't know if the second bit is correct.)", "Solution_3": "im sorry i accendately wrote the wrong thing... Ill change it", "Solution_4": "Okay, so I looked it up on Mathworld and it didn't tell me anything drastically interesting, except that there is only 1 real equivalent of the tetrahedron in 4-space.\r\n\r\nNow, I'm going to outline how you would find its volume, and maybe do it some other time.\r\n\r\nSolution by analogy --\r\nto get from a line segment (a 1-simplex) to an equilateral triangle (a 2-simplex), we go to the centroid and draw a perpendicular and take a point on it such that all three segments we end up with are equal.\r\nTo get from the equilateral triangle to the tetrahedron (3-simplex), we similarly go to the centroid, draw a perpendicular to the plane, and pick the point on it such that all 6 segments are equal.\r\nTo get from there to the 4-simplex, we have to draw a perpendicular to the space we are in, and go a distance such that all the segments connecting it to the vertices of the tetrahedron are equal in length to the side of the tetrahedron.\r\n\r\nNow, I can do this -- but I'm not going to do it now. See if you can figure it out. (Note that finding the length you need to do doesn't solve the problem entirely, but it's about 1/2 of the way there. You're also going to need to solve for the area of the tetrahedron and its height in order to get anywhere.)", "Solution_5": "There is one unique 4-simplex (or hyper-tetrahedron, or pentatope as they are called) in four dimensions. As somebody already pointed out, the tessaract is actually another name for the hypercube.\r\n\r\nThis is a very cool problem though. There are lots of interesting ways to solve it. I worked on this quite a bit when I was in high school, as well as finding the volumes of hyper-spheres. They are fun problems to try. It leads to an interesting study of the sphere packing problem in higher dimensions.", "Solution_6": "Has anyone made any progress with this problem yet?", "Solution_7": "This was very much fun. I claim that the volume of an n-dimensional simplex of side s is given by \r\nVn = sn:sqrt:(n + 1)/(n! * 2n/2).\r\n\r\nTo do this, I need the following fact: the circumradius of an n-dimensional simplex is (n - 1)/n times the height of the same simplex. I'm sure that this can be verified with an integral or a volume argument identical to that which can be used to prove that the medians trisect each other at the centroid. Anyhow, let me take it as a given. Then we have that if hn is the height of an n-dimensional simplex,\r\nhn = s2 - ((n - 1)/n * hn - 1)2\r\nwhich is just an application of the Pythagorean Theorem.\r\n\r\nThen, inductively we can show that in general,\r\nhn = s:sqrt:(n*(n + 1)/2)/n.\r\n\r\nNow, Vn = Vn - 1*hn/n. This definitely can be verified by integral, I think :). Anyhow, if we continuously apply that identity to itself, and then make the substitution for the hi's, you get exactly the formula I wrote above.", "Solution_8": "Yep that's right. Now a further project you could try, is to find the volume of an n-sphere. Then figure out what is the ratio of the volume of an n-sphere inscribed in an n-simplex to the volume of the simplex. And what is the ratio of the total volume of \"packed\" n-spheres inscribed in an n-simplex.\r\n\r\nAnd finally what happens to this ratio as the number of inscribed spheres tends to infinity? The ratio turns out to be very unexpected as n varies.", "Solution_9": "What do you think about this interesting fact? The ``volume'' of an n-sphere of radius 1 is maximized when n=5.", "Solution_10": "[quote=\"ComplexZeta\"]What do you think about this interesting fact? The ``volume'' of an n-sphere of radius 1 is maximized when n=5.[/quote]\r\n\r\nDo you know how to show that? And in what dimension is the \"n-tetrahedron\" maximized?", "Solution_11": "Well, for s = 1, we can see that in will start getting smaller pretty quickly (since both of the pieces of the denominator get big quite fast, while the numerator doesn't) and quickly checking the first few cases shows that the volume is minimized at n = 1 (or 0, I suppose). However, if we take bigger values of s, we should be able to make it be just about any dimension we want. Letting s = :rt2: makes it uniquely n = 1. Letting s be bigger should definitely make it become higher dimensions, until the n! and :rt2:n catch up. I don't really want to calculate higher values, though, because :rt2: was just so nice and I'm very lazy. Perhaps when I'm back at school with Mathematica . . ." } { "Tag": [ "geometry", "MATHCOUNTS", "function", "trigonometry", "AMC", "USA(J)MO", "USAMO" ], "Problem": "Sorry if this is going to violate the \"no whining\" rule. I just had to get it out.\r\n\r\nRecently, my math teacher had criticized me again. For what seemed to be the \"1 followed by an excessive amount of zeros\" times. My math--(middle school Algebra and Geometry) teacher had criticized me for thinking that I was a braggart, an egomaniac, and a show-off. (My math teacher plays favorites, they're the MC coach--and only favors those who are good at MATHCOUNTS, and I'm not good at MC at all.) Putting that aside, though, we got the AMC 10 and 12 scores. (Yes, this is a bit dated, and I apologize.) I scored dismally. I don't mean 144 and didn't get a perfect score--I mean outright failed. I was unable to function at all that day, and scored, well, well below 80 on both. \r\n\r\nWhy did this matter? Well, at my school, like many other public schools throughout the nation, many kids pretended to be stupid. I, myself, and not a victim and hope to never succumb to this [i]disease[/i]. I didn't hide the fact that math was something interesting for me. I never hid the fact that it is truly an art and is breathtakingly beautiful. However, I [i]did[/i] hide the fact that I was more advanced than others. I didn't feel I needed to rub it in other peoples' faces, but I did want them to appreciate the beauty of it. This is where the problem began.\r\n\r\nOne day, I let it accidentally slip when we were deriving the Law of Cosines that \"oblique triangles were frustrating...*mumble**mumble**mumble*...\". Right triangles can be so much easier to work with, on occasion. I didn't think anyone had heard, except for my friend sitting next to me, who then promptly asked, \"What are oblique triangles?\". And me being me, I answered that \"They, in short, are triangles that are....erm....not right.\" That's probably not the exact definition, as I was frazzled by the fact that I had lost control and let a little bit of vocabulary slip. That usually didn't happen. However, I shook it off and proceeded.\r\n\r\nI let little bits of vocabulary slip-but didn't feel this was a huge deal. So I knew a little bit of advanced vocabulary. Whoop-e-dee-doo-dah-day. My teacher, however, did not quite feel the same way. My teacher remembered all these incidents and strung them together in her mind. My teacher then proceeded to nail me about everything. Not only did she target me for one little thing I'd say, for example, \"Could you please keep it down over there?\", and she'd get all mad and have her tirade about how they were discussing mathematics and how it was perfectly fine--even if I had just voiced everyone else's opinion. I knew I wasn't the only one who was about to say it. I just happened to be the only that did. She would always jump on me about things like that, and really chastise me in front of everyone else. Most people just looked the other way. This is middle school, remember?\r\n\r\nNow let's bring this information all back into the (somewhat) present. The day the scores came. My teacher was as shameless as usual, and let me the scores--but not before \"inconspicuously\" flaunting everyone else's for a few moments before pointing me to my scores. Needless to say, they were at the very bottom. That's when the lecture started. She managed to impact me in such a brutal way in such few sentences that I was partially traumatized. She talked about how, and this is a direct quote--\"...you really don't know all those things...\" I kept on my serious veneer the entire time I was in there. Just because she wanted the satisfaction of seeing me break doesn't mean I was going to give it to her. Another quote was:\r\nMe: \"But I never said I knew all those things.\"\r\nTeacher: \"But you always used advanced vocabulary and other things....\"\r\nMe: (Trying not to break....)\r\n\r\nAnd in case you were wondering, I did break but managed to keep myself together. The way I've grown up--I have excellent self-control over my emotions. I shed a few tears, and no one even noticed. I felt much better after my friends tried to cheer me up when i traumatized and wasn't talking or really, just showing any emotion whatsoever.\r\n\r\n\r\nI'm sorry if depresses anyone. But I really felt I needed to get it out--just to type it. I'll probably take it off after a little while and I've cooled off. Thanks. :)\r\n\r\nBut, um, out of curiosity (not to meantion blatent disregard for other's feelings), has anyone had someone do this to them? I know this has happened to me a lot, but has anyone else ever been subject to anything like this--at all? Just a simple yes or or not or maybe will suffice, I don't need stories that will make me use up an entire box of tissues--but I'd like to hear them anyways, even if we run out of tissues. :P", "Solution_1": "[quote=\"The_Scintillator\"]I'm sorry if depresses anyone. But I really felt I needed to get it out--just to type it. I'll probably take it off after a little while and I've cooled off. Thanks. :)[/quote]\r\n\r\nThis is why we have blogs. :) \r\n\r\nAnd to your problem, the advanced vocabulary is simply a better way to get your point across. The next time that incident happened where your teacher criticizes you for, simply explain that it's a better way to communicate (not to show off), and the other possibility is to have you speak in redudancy (using words like \"thingamajig, whachacallit, blah\"). Granted you might get detention for being smart with the teacher, but this is middle school (like it matters?), and the teacher needs to be taught that gifted math kids should not be limited. Using the term \"oblique\" triangle is not a crime. \r\n\r\nHowever, I do think that you should tone your vocab down a bit. If I can picture myself sitting in a regular math class with no \"AoPS\" math in my mind, I do find that people using advanced terminology is quite self-centering. I think it's best if you're modest, and trudge along math class like everyone else. I do this frequently and as I'm getting by in math, my peers still see me as a relatively advanced math student. The respect is there, but the effort in math class diminishes. \r\n\r\nDon't worry about it. You're going to end up with bad [math] teachers in high school. It's important that you pass the class (with an A) and not make too many enemies. Being a good student with a \"secret AoPS math life\" is the modest way to get by.", "Solution_2": "I truly feel sorry for you. I do not know what grade you are in, possibly middle school, and I never thought that the teachers would be involved in your life as much as you described. I already feel this in elementary school, when less-gifted students in another class get a higher grade than me when I score [i]slightly[/i] less on a harder, more advanced assessment. As you said, they \"nailed\" me and I haven't emotionally developed a rock-hard emotionless feeling to last for quite a while.\r\n\r\nOf course, there isn't a lot of fancy vocabulary that is difficult to master in elementary school. Nonetheless, I have a slight ability to have a good sense on what things should be. I'm involved with every club after school, which are difficult (especially Black Saga!), there are many \"Bees\" (a.k.a. Spelling Bee, Geography Bee), classwork assignments, homework, [b]science fair*[/b] and socializing with friends and classmates.\r\n\r\n*About science fair. Last year, it was tragic. I wasn't really counting on winning; after all, you don't win anything in fourth grade. But to see such...such insolent people winning and gloating here and there, I was completely shattered. My emotions got carried away, and the bathroom became my permanent home (for two minutes). So, I thought, \"Why not get help from my brother and his friends? That'll help me win!\" and indeed, I got that help from them. I was so sure, because after attending science fair night, the competition was blown away with my board.\r\n\r\nDuring the announcements to see who would receive Top 10 spots, I was pretty happy for my friends, but still waiting for my name to be called. After every person, my hands were shaking and I was growing nervous. Why wasn't my name being called for the Top 10? I wasn't [i]too[/i] arrogant, was I? The judges can't hold anything against you, right? I'm pretty sure they can't, but they did give some of the kids who needed help learning English and can't get along very well with others the Top 10 spot. Of course, I was outraged and shocked.\r\n\r\nFrom there on out I was emotionless, stony, hard, cold, like a robot. No feelings. No heart. I had no choice anymore. If life was that unfair, think of the [b] real world [/b]. But after consulting with friends, resuming normal life on the cybernetwork and just chilling around made me feel better.\r\n\r\nThere are things you want, but things you can't get. Teachers will try to challenge you and think of you so highly that their standards are practically in the sky and you have to die to reach them. It's sad to think of the changes I've seen come and gone in the same school for my whole life, me starting at the same school in Pre-K and my brother starting in the same school since Kindergarten (he's now in ninth grade).", "Solution_3": "[quote=\"7h3.D3m0n.117\"]Being a good student with a \"secret AoPS math life\" is the modest way to get by.[/quote]\r\n\r\nThis is good advice. Letting the world (or at least the school) know of your AoPS life may not be the best decision you make... but regarding the vocab, I think it's more of the teacher's fault than yours. Sure, advanced vocab may make other students feel bad but from your post, it doesn't seem like you are getting carried away with it.", "Solution_4": "@D3m0n Shad0w: You probably lost points for using extensive vocabulary in your research report (you used my college biology textbook, remember?) so they probably thought you cheated.\r\n\r\n@Buzzer11: Yeah telling your high school you're a math nerd won't help you :P .", "Solution_5": "Aye, getting carried away [b]will and always[/b] lead to arrogance. You could slip up, because arrogance is the banana peel or the bar of soap just sitting there. You could avoid it by previewing the other chapters, brushing up on some old units, or even moving on to another subject. Teachers do socialize, don't they? They're bound to know you're doing excellently in say, math, but failing in Biology.\r\n\r\n@7h3.D3m0n.117: I did not cheat, just used you as a \"resource\" that others wouldn't think or have the heart to use. One girl who got 3rd place used her mom, who was a real scientist.", "Solution_6": "I :( feel :| comforted :)!\r\nHe. Yeah, I modified that from someone's sig...\r\n\r\nI guess I can just live my double life happily and avoid arrogance. \r\nThanks for all the advice. :D\r\n\r\n:dry: And I'm not failing any of my classes...\r\n\r\nI think I need a blog. \r\n\r\nAnd *small chuckle*....a kid in elementary school? using a college textbook for a science fair, by then, i think they would think that you've been plaigerizing. I know that happened to me before. Apparently, when you don't do a baking soda volcano for a science project, questions arise. :ninja: :lol:", "Solution_7": "aaaaaaaaaaaaaa", "Solution_8": "@illusiat\r\nHmm. I wonder how the \"1st smartest person\" feels at your school. Then I wonder how all the other, i.e., \"3rd smartest person\" and below feel. Ouch. :o \r\n\r\nHow would you be able to judge, anyways? :huh:", "Solution_9": "...\r\n\r\nThis will take time, I was much the same in middle school...but you have to be comfortable with your abilities and love what you're doing for the sake of doing it, not for the sake of getting any kind of recognition (this isn't a direct response to the OP, but rather to some other things that were said). I remember in my freshman year, I made it a big goal to qualify for the USAMO, and almost judged my worth by my competition scores. By the time I had gotten to junior year, on the other hand, I was more comfortable with my mathematical abilities, and took competitions more as an opportunity for fun rather than a way to measure myself.\r\n\r\nNowadays, although it's generally accepted that I'm far away the best (if you can define \"best\") math student at my school, I don't try to measure mathematical ability. At that point, it's not so much about competing with the next person and trying to score the highest. Instead of being \"best\" or \"second best\" or whatever, you all love the subject and pursue it together. These fields become even more team-oriented as you move up into college and beyond (I don't know if you're looking that far). You'll probably encounter some sort of camp or research program in your high school years: I got some experience with the team nature of math at PROMYS.\r\n\r\nSo although it's great to pursue things you love, you don't necessarily have to be the best in every competition, and you shouldn't feel too bad if competitions don't go your way. If you do badly, the outcome is that your abilities haven't depreciated, you've learned something interesting, and all you've lost is a chance to add another entry to your list of achievements you use to qualify yourself. Also I should say some of the kids at my school don't do very well on competitions, but are extremely smart. I share math problems with them all the time, and they know me not just as a person who loves and is good at math, but also as a helpful person who is open-minded and willing to share knowledge in mathematics.\r\n\r\nI'm not saying competition is bad: it's fun, and drives you to do well. But don't take it so seriously that you alienate your competition: remember, they're your peers, and it becomes much less pressured and more fun when it's a friendly competition rather than a fight to the death.\r\n\r\n\r\n\r\nAs far as dealing with teachers goes...I can guess a lot of your peers say that you are smarter than the teacher. The teacher probably [b]does not[/b] like this. Instead, take a position as the open-minded student, only seeking to learn more. I'm not saying to suck up to the teacher, but just swallow your ego and take the stance of \"I'm just trying to learn from you\"...from the position you're in now, it may help to approach the teacher, apologize (even though you aren't at fault in any way, the teacher will feel better and will be more forgiving), and try to find out what the she wants. If you're in class throwing terms around (even if it's unintentional), it disrupts the teacher's plan and makes it appear to the rest of the class that you know more than the teacher. When you're talking, students will tend to focus their attention on you rather than the teacher because you are known to be smart, and the teacher knows this. I'm sure it's just a misunderstanding and the teacher feels the class is unstable with someone like you around (this is bound to happen in a structured sort of class in middle school level): it's not like she randomly wants you to suffer.\r\n\r\nWhen you get to high school (and even more in college, I'm sure), you'll get teachers who are more involved in their subjects.", "Solution_10": "[quote=\"The_Scintillator\"]\nPutting that aside, though, we got the AMC 10 and 12 scores. (Yes, this is a bit dated, and I apologize.) I scored dismally. I don't mean 144 and didn't get a perfect score--I mean outright failed. I was unable to function at all that day, and scored, well, well below 80 on both. \n[/quote]\r\n\r\nI'm not trying to be harsh or anything. But are you sure you're not rationalizing?", "Solution_11": "[quote=\"The_Scintillator\"]And *small chuckle*....a kid in elementary school? using a college textbook for a science fair, by then, i think they would think that you've been plaigerizing. I know that happened to me before. Apparently, when you don't do a baking soda volcano for a science project, questions arise. :ninja: :lol:[/quote]\r\n\r\nWell, the college textbook was more reliable to use. And, a college textbook wouldn't be as straightforward as what I put in my research report. Even my teacher said it was ok, but I guess it wasn't.", "Solution_12": "One more thing: Don't worry about showing your teachers how smart you are. I have found that a student's brightness shines through without the need for help from the student him or herself. AMC scores, for one. Even if your scores this year weren't that great, in your opinion, you'll get more chances. These are the things that show your teachers, as well as your peers, how intellegent you are, not the vocabulary you use in class or how much you talk about higher levels of mathematics.", "Solution_13": "Still feeling comforted. :)\r\nBut yeah, I guess I've only got a quarter left with that teacher, and she can only continue to influence me for a few more years....she knows people at the high school I'll be attending.\r\nI'm hopeful that she won't know people at the college I'm attending. Hopefully.\r\nI'm also not particularly worried about showing my teachers how smart I am. I don't ask for that kind of attention--it drives me insane. I don't want recognition. I mostly just want teachers to leave me alone. I just want to do what I do. And if failure comes along with it, if recognition comes along with it, so be it. \r\nEh. I'm rambling. Forgive me.\r\n\r\n@SimonM\r\nI might be rationalizing about my scores--I knew most of the things on there, but applying them was more difficult. [i]I[/i] think it was due to the fact that I was hyper about my birthday, the four gifts I had just received, and the amount of candy I had ingested--all without regard for the fact I was taking the AMC 10 and 12 that day, since I had forgotten. I almost didn't go. Ah, well. I've pretty much gotten over my scores by now. :D\r\n\r\n@K81o7\r\nI've tried the approach with the whole \"I just want to learn from you\". I honestly do. I want to learn from them, anything and everything they can teach me. It doesn't work. :(\r\nBut the advice is much appreciated. Thank you. \r\n\r\n@buzzer11\r\nI should hope I get more chances. Thanks. :)\r\n\r\n@D3m0n Shad0w\r\nHeh. Enough said. :lol:", "Solution_14": "Confront the teacher. Tell her that she made you feel bad, and that you feel she purposely tried to break your spirit. Tell her how immature it is of a teacher to feel the need to bring down a student with sarcasm and malicious comments. And if she feels there is something wrong with your attitude, she has handled it very poorly if she almost made you cry. She should and probably will feel ashamed.\r\n\r\nWell, I don't know how old you are, so I guess it can be really hard to confront her since you're not an adult, but at least it's what I'd like to do", "Solution_15": "[quote=\"The_Scintillator\"]I'm also not particularly worried about showing my teachers how smart I am. I don't ask for that kind of attention--it drives me insane. I don't want recognition. I mostly just want teachers to leave me alone. I just want to do what I do. And if failure comes along with it, if recognition comes along with it, so be it. [/quote]\r\n\r\nThere could be full potential underneath your skin. Having teachers recognize a talent (I don't know your age, so if your in high school this could very well not work) that you don't usually show will allow them to involve you with clubs, extra classes, and private tutors. It could work well with college.", "Solution_16": "A bit on advanced vocabulary, though: math isn't about vocabulary, and sometimes people can get carried away with using very specific terms that most other people don't understand. I wouldn't call it \"your fault\" if you do that, but you should be aware that using such vocabulary can sometimes be more of a hindrance than an aid to communication, even when talking to other people who are good at math. I for one didn't know what an oblique triangle was.", "Solution_17": "That doesn't like a very nice teacher. I'm in middle school too, and my geometry teacher also scolded me when I overly expressed my knowledge of math, but in private, and acknowledged that I was indeed a \"bright student\", so I didn't feel depressed afterwards. A teacher should never try to humiliate a student, especially when he or she does not fully understand why a student acts that way (I think your teacher hasn't interacted with too many advanced learners). \r\n I think it's a bit sad that mathematically gifted students are often overlooked, whereas those best in other subjects, especially sports, can easily show off their talent without being criticized. I have also been thought as a braggart before, but mostly by other students rather than teachers (most of my teachers were pretty easygoing). If I really needed to show off something, I would scribble it on paper, show it to the person and quickly cross it away.", "Solution_18": "In life, you have to dumb down your vocabulary, mannerisms, logic, everything, when you're around people who aren't as interested in pointlessly technical things as you are. I learned this in, what, 9th grade? Now I do this on a subconscious level, and life is much easier.", "Solution_19": "[quote=\"Lazarus\"]In life, you have to dumb down your vocabulary, mannerisms, logic, everything, when you're around people who aren't as interested in pointlessly technical things as you are.[/quote]\r\n\r\nAt least until you get a job that let's you have an advanced vocabulary. Once you get a really good job, you can say words like rapport.", "Solution_20": "[quote=\"1=2\"][quote=\"Lazarus\"]In life, you have to dumb down your vocabulary, mannerisms, logic, everything, when you're around people who aren't as interested in pointlessly technical things as you are.[/quote]\n\nAt least until you get a job that let's you have an advanced vocabulary. Once you get a really good job, you can say words like rapport.[/quote]\r\nDude wtf that appears in english class...", "Solution_21": "[quote=\"serialk11r\"][quote=\"1=2\"][quote=\"Lazarus\"]In life, you have to dumb down your vocabulary, mannerisms, logic, everything, when you're around people who aren't as interested in pointlessly technical things as you are.[/quote]\n\nAt least until you get a job that let's you have an advanced vocabulary. Once you get a really good job, you can say words like rapport.[/quote]\nDude wtf that appears in english class...[/quote]\r\n\r\nI mean on a daily basis.", "Solution_22": "I don't think this should be a question of whether advanced vocabulary is allowed or not. In my opinion, high level vocabulary can be used, but with a limit in mind. If you start stringing your sentences with mathematics jargon, you probably went too far. :)", "Solution_23": "wow.. i really need to get something out right now...\r\nit relates to the topic..\r\n\r\nOne day in Geometry, we were talking about right triangles and solving for angles. I raised my hand because I knew a much simpler way that leads for less error. My teacher, who knew that I was studying PreCalc at home said, \"Now you're not relating to the Law of Sines or anything like that. right?\"\r\nThat just blew me off. All because I knew some more than the class should know, she jumped to the conclusion of thinking that I was going to show off again. I maintained my composure and said, \"No.\" and the stupid thing is that the problem didn't even need to use the freakin law of sines.\r\n\r\nAlso!\r\n\r\nmy teacher is one of the most unprepared teacher of all times. She loses her transparencies, she grades papers so slowly (and her excuse is that she gives out partial credit, but the partial credit she gives out is like half a point off for every single question).\r\nAlso, we were doing standardized test preparation. she knew she didn't have to go over it, so we just did really quick problems that the whole class solved in 2 minutes. Guess who gets all three questions wrong. my teacher. She claims that she has an off day. By now, i'm really sick and tired of her..\r\nOh. and I taught point symmetry to the class. She had NO FREAKING IDEA WHAT THE HECK POINT SYMMETRY WAS!!\r\n\r\nMy math teacher claims to like me. So far only two math teachers i've had at school liked me. the rest hated me for being accelerated.\r\nWhat's the big deal?\r\nshouldn't they be happy that they have bright students in their class?\r\n\r\nSure, with genius comes arrogance, but it's at least controlled.\r\nI could doze off in class if I wanted to, but the kid behind me keeps asking me questions, and i'm right next to the overhead cart, which means that she could catch me doing anything. \r\n\r\n@Scintillator, I am very sorry if I am in a better situation than you are, but yes, We all feel that way...\r\n\r\nMy favorite subject right now is math.\r\nMy least favorite subject right now is history.\r\nMy favorite teacher is my US History teacher.\r\nMy least favorite teacher is my math teacher.\r\n\r\nI'm so glad that I have AoPS to resort to.", "Solution_24": "[quote=\"1=2\"][quote=\"Lazarus\"]In life, you have to dumb down your vocabulary, mannerisms, logic, everything, when you're around people who aren't as interested in pointlessly technical things as you are.[/quote]\n\nAt least until you get a job that let's you have an advanced vocabulary. Once you get a really good job, you can say words like rapport.[/quote]\r\n\r\n\r\n...\r\n\r\n>_>", "Solution_25": "[quote=\"1=2\"][quote=\"Lazarus\"]In life, you have to dumb down your vocabulary, mannerisms, logic, everything, when you're around people who aren't as interested in pointlessly technical things as you are.[/quote]\n\nAt least until you get a job that let's you have an advanced vocabulary. Once you get a really good job, you can say words like rapport.[/quote]\r\nQFT at lazarus's reply.\r\nSeriously, ?????????", "Solution_26": "Shouldn't this be in a blog though?\r\nMeh, the worst thing a teacher who hated me did was give me a C as a grade." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Determine the smallest natural number $ n$ such that among any $ n$ integers one can choose $ 18$ integers whose sum is divisible by $ 18$.", "Solution_1": "*)Lemma. Among any $ 2n \\minus{} 1$ integers one choose $ n$ integers whose sum is divisible by $ n$.\r\nProof. If lemma holds for $ m,n$ then lemma holds for $ mn$.\r\nSo we have prove for $ n$ is prime. Let $ 2n \\minus{} 1$ integers are $ a_1,a_2, . . . ,a_{2n \\minus{} 1}$.\r\nIndeed, assume any $ n$ integers whose sum isn't divisible by $ n$, then apply theorem Fermat:\r\n$ \\sum_{1\\leq i_1 < i_2 < . . . < i_n\\leq 2n \\minus{} 1}(a_{i_1} \\plus{} a_{i_2} \\plus{} . . . \\plus{} a_{i_n})^{n \\minus{} 1}\\equiv (^{2n \\minus{} 1}_n)$(mod n).\r\nImpossible, hand-left is divisible by $ n$, hand-right isn't divisible by $ n$.\r\n*)We have $ 2n \\minus{} 2$ integer:\r\n$ (0,0, . . . ,0,1,1, . . . ,1)$ satisfy no $ n$ integer whose sum is divisible by $ n$.\r\n*)So $ n$ smallest in problem is $ 35$.", "Solution_2": "Hello, I have two questions about your proof...\r\nWhy is the following true?[quote=\"Thjch Ph4 Trjnh\"]If lemma holds for $ m,n$ then lemma holds for $ mn$.[/quote]\nAlso why is the left hand side divisible by $ n$ if $ n$ is prime?[quote=\"Thjch Ph4 Trjnh\"]$ \\sum_{1\\leq i_1 < i_2 < . . . < i_n\\leq 2n \\minus{} 1}(a_{i_1} \\plus{} a_{i_2} \\plus{} . . . \\plus{} a_{i_n})^{n \\minus{} 1}\\equiv \\binom{2n \\minus{} 1}{n}\\pmod{n}$[/quote]\r\nThank you,", "Solution_3": "http://www.mathlinks.ro/viewtopic.php?p=389030", "Solution_4": "[quote=\"moldovan\"]Determine the smallest natural number $ n$ such that among any $ n$ integers one can choose $ 18$ integers whose sum is divisible by $ 18$.[/quote]\r\n\r\nsorry, but i could not understand your question.\r\naccording to the statement of the question, i am allowed to choose any $ n$ naturals out of which i choose some $ 18$ numbers whose sum is divisible $ 18$.\r\nif so then $ n$ can be $ 18$ itself as i can choose $ 18$ multiples of $ 18$ and their sum is obviously divisible by $ 18$.\r\n\r\nthere is something missing..(either in the question or my comprehension)..please clear it", "Solution_5": "\"Any\" means \"for all\".", "Solution_6": "[quote]Hello, I have two questions about your proof...\nWhy is the following true?[quote=\"Thjch Ph4 Trjnh\"]If lemma holds for $ m,n$ then lemma holds for $ mn$.[/quote]\n[/quote]\r\nAssume lemma holds for $ m,n$. Let $ 2mn\\minus{}1$ integers are:$ a_1,a_2, . . .,a_{2mn\\minus{}1}$.\r\nAmong $ a_1,a_2, . . .,a_{2n\\minus{}1}$ has $ n$ number whose sum is divisible by $ n$, assume is $ a_1,a_2, . . . ,a_n$.\r\nAmong $ a_{n\\plus{}1},a_{n\\plus{}2},. . .,a_{3n\\minus{}1}$ ...\r\n....\r\nSo have $ (2m\\minus{}1)$ $ n\\minus{}$tuples $ (a_{i_1},a_{i_2},. . .,a_{i_n})$.\r\nDonote $ b_i\\equal{}\\frac{a_{i_1}\\plus{}a_{i_2}\\plus{}. . .\\plus{}a_{i_n}}{n}$ so $ b_i\\in Z$.\r\nWe obtain $ (2m\\minus{}1)$ $ n\\minus{}$tuples $ b_i$.\r\nSo have $ m$ numbers of these whose sum is divisible by m.\r\nCompletes.", "Solution_7": "Hi, can someone post a clearer proof of the generalization? Thanks :)", "Solution_8": "Hi, the generalization actually cannot be proven with elementary means I believe; see [url=http://en.wikipedia.org/wiki/Zero-sum_problem]here[/url].", "Solution_9": "As mavropnevma sir says\n[quote=\"mavropnevma\"]Erd\u00f6s-Ginsburg-Ziv theorem: [color=#0000FF]Out of $2n-1$ integers there exist $n$ with sum divisible by $n$. [/color] \nthe general proof is elementary, but kind of complicated;[/quote]\n\nin this thread British MO 2000 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=42&t=592776\n\nThis is a very similar but much simpler question", "Solution_10": "[quote=Thjch Ph4 Trjnh]\n$ \\sum_{1\\leq i_1 < i_2 < . . . < i_n\\leq 2n \\minus{} 1}(a_{i_1} \\plus{} a_{i_2} \\plus{} . . . \\plus{} a_{i_n})^{n \\minus{} 1}\\equiv (^{2n \\minus{} 1}_n)$(mod n).\n.[/quote]\nHow do your prove that the LHS is divisible by n?\n\n", "Solution_11": "Actually it has been a long time since the discussion, but anyway.\nSo the problem can be generalized for every k positive integer and here is why:\n[b][i][size=150][size=110]Claim1[/size][/size][/i]:[/b] It works for any $p \\in P$. \nIndeed, it is straightforward that the answer is at least $2p-1$, otherwise there is a tuple $(0,0,0,0...0,1,1,1...1)$, where $0$s and $1$s appear exactly $p-1$ times, the sum is never divisible by $p$.\nSo it is left to prove that it is sufficient. For the sake of contrary let $a_1, a_2, ..., a_{2p-1}$ be the numbers, so that the $(a_1+a_2+...+a_p)$ is not divisible by $p$ for all permutations of $a_i$. There are exactly $\\binom{2p-1}{p}$ ways to pick $a_i,$ it is clear that $\\binom{2p-1}{p} \\not\\vdots p$.\nLet's count the coefficients of $a_1a_2...a_j$ in $\\sum{(a_1+a_2+...+a_p)}^{p-1}$ . From the one hand, by Fermat's, $LHS$ is just $1+1+...+1$ $\\binom{2p-1}{p}$ times, meaning that $LHS \\not\\vdots p$. From the other, each term $a_1^{e_1}*...*a_j^{e_j}$ for $e_1+...+e_j=p-1$ appears once for $(a_1+...+a_j+...)^p$, so there are $\\binom{2p-1-j}{p-j}$ appearances of $a_1^{e_1}*...*a_j^{e_j}$, which is divisible by $p$, the contradiction.\n[b][i][size=150][size=110]Claim2[/size][/size][/i]:[/b] if it works for $k_1, k_2$, then it works for $k_1k_2$\nLet's pick $2k_1-1$ among $2k_1k_2-1$, by hypothesis, there are $k_1$ numbers, whose sum $\\vdots k_1$. Return the rest $k_1-1$ numbers and do this operation again and again until we get $2k_2-1$ groups where each group's sum $\\vdots k_1$. let $b_i=s_i/k_1$, then $ b_i \\in Z$. For $b_1, b_2, ..., b_{2k_2-1}$ we have $k_2$ numbers whose sum $\\vdots k_2$, so then the union of all these consists of $k1*k2$ terms, the sum is divisible by $k_1*k_2$ and we are done\nAnswer: $2*18-1=35$" } { "Tag": [ "ratio" ], "Problem": "A substance contains $40 \\% C$, $53.33 \\% O$, $6.66 \\% H$. $13.39mg$ of this substance has volume $10mL$ in normal conditions.\r\n\r\nFind the formula of the substance.\r\n\r\n\r\nI get some \"leftovers\" when computing. Am I doing something wrong?", "Solution_1": "I'm rly not sure but I got $CHO_2$ just assuming it's out of 100 g......or is it suppose to be out of 13.39 mg?.....I dont' know how the mg and vol. fit in....though", "Solution_2": "I'd focus on the weight ratios. Since we are looking for a formula, we don't care how much is really there. Just divide the wt% by mole weight and end up with 40% C/ 12 = 3.33; 53.33% O/16 = 3.33; 6.66% H/1 = 6.66.\r\n\r\nThis leads me to the mole ratio of C:O:H of 1:1:2, which leads me to the formula of CH2O.\r\n\r\nNow, why would we get volume and mass information? They must want to give us an ability to calculate the molecular weight (from mass and number of moles from PV=nRT).\r\n\r\nI'll leave the rest of this to you.", "Solution_3": "PV=nRT? We're probably not working with gases, as the stuff is more dense than water.\r\n\r\nI'm not sure where we can get the molecular mass.\r\n\r\nIt's probably some sort of monosaccharide, as they have the $CH_2O$ empirical formula. I don't know which one in particular; it could be glucose, galactose, fructose, ribose...\r\n\r\nI guess I could look for a table of densities.", "Solution_4": "I was also thinking something like glucose or fructose. I don't know if you can really find the molecular formula from the density alone (without a table of densities) for solids and liquids.", "Solution_5": "If water has a density of 1 g / 1 mL, and the lowest density liquid has a density of 0.5 g/1 mL, then the compound must be a gas if 0.013 g occupies 10 mL. Formaldehyde is the only possibility." } { "Tag": [ "calculus", "derivative", "function" ], "Problem": "Find, step by step, second derivative of function:\r\n\r\n$ f(x)\\equal{}x^{4}(1\\minus{}x)$", "Solution_1": "Please refer to the second announcement in this sub-forum.\r\n\r\n[hide]$ f(x)\\equal{}x^4\\minus{}x^5$\n\n$ f'(x)\\equal{}4x^3\\minus{}5x^4$\n\n$ f''(x)\\equal{}12x^2\\minus{}20x^3$[/hide]" } { "Tag": [ "trigonometry" ], "Problem": "A half-cylinder of glass has radius $R$ and index of refraction $n$. A ray of light lies in\r\nthe plane of the figure as shown, and it is incident on the plane face at an angle $\\theta$. It goes\r\non to hit the curved surface; for what range of incident positions on the flat surface will the\r\nlight be able to leave the cylinder when it first hits the curved surface? Let the position\r\nthat it hits the flat surface be $x$ as measured from the centerline of the flat. For what\r\ncombination of $n$ and $\\theta$ will this range extend over the whole flat surface $(-R6\\cdot3$. However, one can prove\r\nthat for $n>4$ the statement almost holds, then we have \r\n\r\n\\[\\min[a_i,a_j]\\leq6(\\lfloor n/2\\rfloor+1),\\]\r\n (I replaced $<$ with $\\leq$) and the equality occurs for numbers $n+1,n+2,\\dots,2n$ (it is not so hard to prove).\r\nOn the site 20000 problems under the see I found the same problem, with the\r\nsame formulation. I wonder if I am misunderstanding something here. Does\r\nanyone have access to AMM? This was problem AMM3834 by Paul Erdos.", "Solution_1": "As you notice, the strict inequality is incorrect, and has to replaced by a large one, and only holds for $n \\neq 4$.\r\nSince $n=1,2,3$ can be checked by hand, let's assume that $n \\geq 5$.\r\n\r\nNow, here is the published answer from C. R. Phelps :\r\n\r\nFor any $n$, consider first the sequence $a_i = n+i$.\r\nThe first even number in this sequenec is $2( [ \\frac n 2 ] + 1)$ and the number $3( [ \\frac n 2 ] + 1)$ is also in the sequence, so min$ lcm(a_i , a_j) \\leq 6( [ \\frac n 2 ] + 1)$ for this special sequence.\r\n\r\nFor any other sequence $a_1 < a_2 < \\cdots < a_{2n} \\leq 2n$, for any $a_i$ there exists an integer $k_i \\geq 1$ such that $n < k_ia_i \\leq 2n$. If $k_ia_i = k_ja_j$ for some distinct $i,j$ then $lcm (a_i,a_j)$ divides $k_ia_i$ and therefore is not greater than $2n$.\r\nOtherwise the $k_ia_i$ are all distinct ($i=1, \\cdots, n$) and form exactly the special sequence in some order, so that the above inequality holds.\r\n\r\nMoreover, the bound $6( [ \\frac n 2 ] + 1)$ is best possible since it is the minimum for the special sequence for each $n$ : Suppose that for some $n$, for the special sequence $lcm(a_i,a_j) =M < 6( [ \\frac n 2 ] + 1)$ where $a_i < a_j$. Then $a_j \\geq n+3$ so that $a_j > 2( [ \\frac n 2 ] + 1)$ with $M = ma_j$ for some integer $m$. Hence $m=2$, and $M=ka_i = 2a_j$, so that $k \\geq 3$.\r\nIf $k=3$ then $a_i < 2( [ \\frac n 2 ] + 1)$ with $a_i$ even, a contradiction.\r\nIf $k \\geq 4$ thn $a_i < \\frac 3 2( [ \\frac n 2 ] + 1)$ so that $a_i < n+1$ also impossible. Hence $M < 6( [ \\frac n 2 ] + 1)$ is impossible.\r\n\r\nPierre.", "Solution_2": "sbos, does that site with 20000 problems work? I've been trying to browse it for months, getting only stuff like \"layout not found\" (whatever that means). It's a great resource, but I thought it had too many problems to be useful. Could you give me the address? (maybe ou're talking about some other site?)", "Solution_3": "Now I have troubles with that site too. I found the above problem long time ago when the site worked normally.\r\nThank you pbornsztein!" } { "Tag": [ "calculus", "integration", "number theory", "Diophantine equation", "superior algebra", "superior algebra unsolved" ], "Problem": "Let d be a square free integer.\r\n\r\nProve that if (x^2)-d(y^2)=-1 has an integral solution, \r\nthen (x^2)-d(y^2)=1 also has an integral solution.", "Solution_1": "Think about $ (x \\plus{} y\\sqrt {d})^2$.", "Solution_2": "sorry. I'm still kinda stumped.", "Solution_3": "$ (x\\plus{}y\\sqrt{d})(x\\minus{}y\\sqrt{d})\\equal{}\\minus{}1$. Square both sides. See what you get.", "Solution_4": "Ok. I squared it and it simplified to the second equation!\r\n\r\nSo does this simply mean that any integral solution to equation 1 is also an integral solution to equation 2?", "Solution_5": "Not quite.\r\n\r\n[b]Definition:[/b] Given a number $ z \\equal{} x \\plus{} y \\sqrt {d}$ where $ x, y$ are integers, let $ \\bar{z} \\equal{} x \\minus{} y \\sqrt {d}$ be its [i]conjugate.[/i]\r\n\r\n[b]Definition:[/b] Given $ z \\equal{} x \\plus{} y \\sqrt {d}$, let $ N(z) \\equal{} z \\cdot \\bar{z} \\equal{} x^2 \\minus{} dy^2$ be its [i]norm.[/i]\r\n\r\n[b]Lemma:[/b] $ N(zw) \\equal{} N(z) N(w)$.\r\n\r\nThe first Diophantine equation means $ z \\equal{} x \\plus{} y \\sqrt {d}$ has norm $ \\minus{} 1$. That means $ z^2 \\equal{} (x^2 \\plus{} dy^2) \\plus{} (2xy) \\sqrt {d}$ has norm $ 1$, so $ (x^2 \\plus{} dy^2, 2xy)$ is a solution to the second Diophantine equation." } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "The integer sequence $ a_i$ is defined by $ a_0 \\equal{} m, a_1 \\equal{} n, a_2 \\equal{} 2n\\minus{}m\\plus{}2, a_{i\\plus{}3} \\equal{}3(a_{i\\plus{}2} \\minus{} a_{i\\plus{}1})\\plus{} a_i$. It contains arbitrarily long sequences consecutive terms which are squares. Show that every term is a square.", "Solution_1": "It is easy to show, that $ a_i\\equal{}i^2\\plus{}(n\\minus{}m\\minus{}1)i\\plus{}m$." } { "Tag": [ "MATHCOUNTS", "AMC", "AMC 8", "AMC 10", "AMC 12", "USA(J)MO", "USAMO" ], "Problem": "Greeting,\r\n\r\nMaybe in next week or before, I'll post thirty questions for beginners to do. Answers must be send to me by P.M. with explanation, meaning if you only put the answer, you won't get a credit at all. Questions will cover from Mathcounts to maximum of AHSME. No questions will be exactly same with any of past years' competition so, there is no way you're going to cheat. Problems, however, will be similar.\r\n\r\nFor 25 questions, it's like this:\r\n\r\n5 MC\r\n5 AMC8\r\n5 AMC10\r\n5 AMC12\r\n5 AHSME\r\n\r\n\r\nTotal point is 100, meanig correct question worth 4 points each. Unanswered question is given 1.5 points while incorrect answers are given no point at all. Plus, any answes with no expalnation is absolutely disqualified so, make you sure you do right.\r\nTest will be on for exactly 3 days, and all answers must be submitted before I draw up the questions. After two or three days, there will be rank of people from first to last. Although it's for beginners, intermediater and advancer may join too. Unlike Rep123max's getting started contest, it's actually a competition in here except there isn't going to be prize (sorry!). \r\n\r\nSilverfalcon\r\n\r\n[i]Edited: I erased last part about next level competition[/i]", "Solution_1": "Are calculators allowed? And, I assume that it goes without saying that work needs to be shown?", "Solution_2": "Calculators are perfectly allowed. For showing work, calculations doesn't have to be shown like this:\r\n\r\n[tex]5x = 20[/tex]\r\n[tex]x = 4[/tex]\r\n\r\nBut what I meant by explanation is that whenever somebody sees your answer, they must say, \"oh, that's the way you do it\"\r\n\r\nSo, try to put some words that match with your answers to get full credit.", "Solution_3": "What is the \"CMO?\"", "Solution_4": "According to what I know, CMO is Canadian Mathematics Olympiad. I think it's in level of AMC10.", "Solution_5": "CMO is far harder than AMC10. It's the Canadian equivalent of the USAMO, but the questions are easier.", "Solution_6": "Thank you.", "Solution_7": "I changed little bit.", "Solution_8": "Unlike having all questions at once, I'll post 5 questions from each level.\r\n\r\nNumber one from MC level:\r\n\r\n1. 6372 has 24 factors. What is the sum of the largest and smallest factor?\r\n\r\nNumber two from AMC8 level:\r\n\r\n2. Chord with length of 140cm is drawn inside the circle, and it's 48cm away from the center. Then what is the area of circle? Write your answer in terms of [size=150]:pi:.[/size]\r\n\r\nNumber three from AMC10 level:\r\n\r\n3. [tex](x^2+z^3)^4)^5)^6)^7[/tex] has degree of what?\r\n\r\n\r\nNumber four from AMC12 level:\r\n\r\n4. Prove that [tex]x^7-x^6-3x^5+3x^4+3x^3-3x^2-x+1[/tex] has two solutions 1, and -1 by factoring it out.\r\n\r\nNumber five from AHSME level:\r\n\r\n5. [tex]10!9!8!7!6!5!4!3!2!1![/tex] has how many factors?\r\n\r\nPlease don't discuss them under here. If you think you have the answer, please p.m. to me with them, and they are only accepted by Wednesday, 5/19.", "Solution_9": "My questions aren't totally for intermediaters.\r\nIf you understand the questions and pretty much terms of math, you are able to do most of the questions.\r\nBut it's true that most of them are pretty hard for beginners...", "Solution_10": "anytime on the 19th? So like, could I PM u in the evening, or do you want it before then?", "Solution_11": "Rep, in my opinion these are easier than your questions. =)", "Solution_12": "In fact, ones I just posted are the ones that are pretty \"easy\"", "Solution_13": "This message has been deleted.", "Solution_14": "NVM I take that back =)", "Solution_15": "Thats confusing silver falcon. No one send me answers so everyone turn them in??? Do you want them yet or not? You are contridicting yourself or else I am very tired and just confusing myself.", "Solution_16": "[quote=\"BHorseMath\"]Thats confusing silver falcon. No one send me answers so everyone turn them in??? Do you want them yet or not? You are contridicting yourself or else I am very tired and just confusing myself.[/quote]\r\n\r\nPretty sure he meant \"no one sen[b]t[/b] me answers yet.\" Give him a break ;-)", "Solution_17": "sorry I was cranky and confused.", "Solution_18": "[quote=\"BHorseMath\"]Thats confusing silver falcon. No one send me answers so everyone turn them in??? Do you want them yet or not? You are contridicting yourself or else I am very tired and just confusing myself.[/quote]\r\n\r\nSorry about my bad wording. But I hope you understand that you hadn't p.m.ed me with your rest of solution in round two since like last two weeks. If you don't p.m.ed me soon, your score would be 4, point for only question you got right.\r\n\r\nSincerely,\r\n\r\nP.S. Thanks probability 1.01", "Solution_19": "Did I send you a part solution silverfalcon? if you can remind me what problem I will send the rest.", "Solution_20": "Finally!!! Last five questions are made:\r\n\r\nNumber twenty-one from MC level:\r\n\r\nIn this arithmetic sequence: 1,4,7..... What's the 26th term?\r\n\r\nNumber twenty-two from AMC8 level:\r\n\r\nWord Latex has 120 different combinations. Word Telax is nth number. Define n.\r\n\r\nNumber twenty-three from AMC10 level:\r\n\r\nHow many different values of b's are possible in expression [tex]x^2+bx-15[/tex]to factor? (Trust me, it can be factored).\r\n\r\nNumber twenty-four from AMC12 level:\r\n\r\nFor this question, listen very carefully. Segment AB and AC are tangent to circle Z, and they are same length. Square of the length of the AB is equal to product of segement AD, which is secant to circle Z and pass point E, whch is the one of outer point of the circle Z. Chord ED is bisected perpendicularly by diameter KP at J. If segment AC has length of 8, AE has length of 4, and KJ has length of 16, find the length of JP.\r\n\r\nNumber twenty-five from AHSME level:\r\n\r\nRationalize the numerator in following expression:\r\n\r\n[tex]\\displaystyle \\frac {\\sqrt {x+h} - \\sqrt x}{h}[/tex]\r\n\r\nGood luck! These five questions are considered hardest five questions of all and I hope you all do pretty well. Especial luck to our two finales, Chinaboy and Ragingg.....", "Solution_21": "Silverfalcon, can you check your PM box, cuz i sent you my answers 3 days ago!", "Solution_22": "[quote=\"Silverfalcon\"]Number twenty-two from AMC8 level:\n\nWord Latex has 120 different combinations. Word Telax is nth number. Define n.\n[/quote]\nPlease clarify what you mean by the nth number and the \"Word Telax\"\n\n\n[quote=\"Silverfalcon\"]\nNumber twenty-three from AMC10 level:\n\nHow many different values of b's are possible in expression [tex]x^2+bx-15[/tex]to factor? (Trust me, it can be factored).\n[/quote]\r\nDo you mean integer values of b? Cuz there are infinite values of b by which you are phrasing it.", "Solution_23": "For the first question, it's just word \"Telax\" like \"Latex,\" \"Texla,\" etc... For the second question, it meant integer value to factor the trinomial into two binomials. Here's hint:\n\n[hide]There isn't really infinite number of solutions because remember, in FOIL process, the middle term is sum of in and out like this:\n\n[tex](a+2)(a+3) --> 2a+3a --> 5a[/tex]So, think on how many factors are in the last term..[/hide]", "Solution_24": "I assume that you mean they are ordered in alphabetical order. That is probably the part that confused some people.", "Solution_25": "Yes, my bad on that.", "Solution_26": "[quote=\"Silverfalcon\"]Number twenty-four from AMC12 level:\n\nFor this question, listen very carefully. Segment AB and AC are tangent to circle Z, and they are same length. [b]Square of the length of the AB is equal to product of segement AD, which is secant to circle Z and pass point E, whch is the one of outer point of the circle Z.[/b] Chord ED is bisected perpendicularly by diameter KP at J. If segment AC has length of 8, AE has length of 4, and KJ has length of 16, find the length of JP.\n[/quote]\r\n\r\nYou totally lost me there. How can there be the product of a segment and a point.", "Solution_27": "I know my word is confusing but try this way:\n\n\n\n[hide]What are the two other ways of power of a point theorem other than two chords intercepting? Now think about it. K? It should be a big help if you see which one to use[/hide]", "Solution_28": "Contest has been ended.\r\n\r\nWinner: theone833\r\n\r\nAnswers will be soon posted.\r\n\r\nSincerely,\r\n\r\nSilverfalcon", "Solution_29": "Solutions: Odds first\r\n\r\n1. 6372's largest factor is 6372 and smallest is 1. Therefore, it's 6372+1 or 6373.\r\n\r\n3. Simply 3*4*5*6*7 = 2520\r\n\r\nMore will be posted." } { "Tag": [ "algebra", "polynomial", "inequalities", "function", "inequalities proposed" ], "Problem": "If $ a,b,c$ are nonnegative real numbers, then\r\n\r\n$ (a) \\ \\ \\ \\ 10\\sum a^4\\plus{}64\\sum a^2b^2 \\ge 33\\sum ab(a^2\\plus{}b^2)$;\r\n\r\n$ (b) \\ \\ \\ \\ 29\\sum a^4\\plus{}118\\sum a^2b^2 \\ge 72\\sum ab(a^2\\plus{}b^2)$.", "Solution_1": "[quote=\"Vasc\"]If $ a,b,c$ are nonnegative real numbers, then\n\n$ (a) \\ \\ \\ \\ 10\\sum a^4 \\plus{} 64\\sum a^2b^2 \\ge 33\\sum ab(a^2 \\plus{} b^2)$;\n\n[/quote]\r\nLet $ a\\plus{}b\\plus{}c\\equal{}3u,$ $ ab\\plus{}ac\\plus{}bc\\equal{}3v^2,$ $ abc\\equal{}w^3$ and $ u^2\\equal{}tv^2.$ \r\nHence, $ \\sum_{cyc}(10 a^4 \\minus{} 33a^3b\\minus{}33a^3c\\plus{}64 a^2b^2 )\\geq0\\Leftrightarrow$\r\n$ \\Leftrightarrow270u^4\\minus{}657u^2v^2\\plus{}450v^4\\geq55uw^3.$\r\nBut $ (a\\minus{}b)^2(a\\minus{}c)^2(b\\minus{}c)^2\\geq0$ gives $ w^3\\leq3uv^2\\minus{}2u^3\\plus{}2\\sqrt{(u^2\\minus{}v^2)^3}.$\r\nId est, it remains to prove that\r\n$ 270u^4\\minus{}657u^2v^2\\plus{}450v^4\\geq55u\\left(3uv^2\\minus{}2u^3\\plus{}2\\sqrt{(u^2\\minus{}v^2)^3}\\right).$\r\nBut $ 270u^4\\minus{}657u^2v^2\\plus{}450v^4\\geq55u\\left(3uv^2\\minus{}2u^3\\plus{}2\\sqrt{(u^2\\minus{}v^2)^3}\\right)\\Leftrightarrow$\r\n$ \\Leftrightarrow190u^4\\minus{}411u^2v^2\\plus{}225v^4\\geq55u\\sqrt{(u^2\\minus{}v^2)^3}\\Leftrightarrow$\r\n$ \\Leftrightarrow190t^2\\minus{}411t\\plus{}225\\geq55\\sqrt{t(t\\minus{}1)^3}\\Leftrightarrow$\r\n$ \\Leftrightarrow(75t^2\\minus{}155t\\plus{}81)(21t\\minus{}25)^2\\geq0,$ which is obviously true.", "Solution_2": "Let f(a,b,c) = L.H.S - R.H.S\r\nFix S = a+b+c and T = ab+bc+ca then f(a,b,c) is linear function on P = abc. So, f(a, b, c) achieves its minimum when P max or min. In these cases, two variables are equal. So, it is sufficient to show that \r\n $ f(a, b, b) \\ge 0$\r\nIn firts case this leads to $ (5a^2 \\minus{} 3ab \\plus{} b^2)(a \\minus{} 3b)^2 \\ge 0$\r\nIn second case we come to $ (29a^2 \\minus{} 28ab \\plus{} 8b^2)(a \\minus{} 2b)^2 \\ge 0$\r\n\r\nBy this method, we can prove the generalization:\r\n\r\nIf $ (2A \\plus{} B \\minus{} 2C)x^4 \\minus{} 2Cx^3 \\plus{} 2Bx^2 \\minus{} 2Cx \\plus{} A \\ge 0$ for all x\r\n\r\nthen $ A(a^4 \\plus{} b^4 \\plus{} c^4) \\plus{} B(a^2b^2 \\plus{} b^2c^2 \\plus{} c^2a^2) \\ge C(ab(a^2 \\plus{} b^2) \\plus{} bc(c^2 \\plus{} a^2) \\plus{} ca(c^2 \\plus{} a^2))$ for all a,b,c.", "Solution_3": "Nice, Arqady.\r\nNice, Namdung.\r\nFor the second method, we need to verify also that $ f(0,b,c) \\ge 0$ for any nonnegative $ b$ and $ c$. We need to show that $ f(0,b,c)\\ge 0$ and $ f(a,1,1) \\ge 0$ for all $ a,b,c$.\r\nFor instant, the inequality\r\n$ \\sum a^4 \\plus{} 8\\sum a^2b^2 \\plus{} 8abc\\sum a \\ge 5\\sum ab(a^2 \\plus{} b^2)$\r\nis not true, but the condition $ f(a,b,b)\\ge 0$ (or $ f(a,1,1)\\ge 0$) is verified.", "Solution_4": "Thank you Vasc! I forgot the case P = 0!\r\n\r\nNote that inequalities of these type are independent in such meaning: no one can be obtained from another one. In other words, we can not say 1) is stronger that 2) or versi versa.", "Solution_5": "Namdung, your reasoning works for all symmetric homogeneous polynomial of degree 5 with 3 non-negative variables.", "Solution_6": "Yes. Indeed. \r\n\r\nIt is interesting to characterize a simple criterium for A, B, C to have\r\n $ A(\\Sigma a^4) \\plus{} B(\\Sigma {a^2b^2}) \\ge C(\\Sigma {ab(a^2 \\plus{} b^2})$" } { "Tag": [], "Problem": "Does anyone know how many students there are per room and per apartment?", "Solution_1": "2 people per room, 4 rooms per apartment.", "Solution_2": "Then why are there only three lines for rooming requests? If we write in more students below the lines, will those requests be honored?", "Solution_3": "apparently the group size in each apartment is becoming smaller. Last year I think there were 8 students, plus a counselor, per apartment, which meant that everyone slept in a double except the counselor, who slept in the common room.\r\n\r\nThis year, there are six students, so the counselor gets a room too. Which is good news for me, because I'll be a counselor. =)\r\n\r\nI will see some of you there. I look forward to getting to know all of you! And if you're unlucky, maybe you'll even be in my apartment/class (I think I'm assisting in computational geo).", "Solution_4": "[quote=\"roadnottaken\"]Then why are there only three lines for rooming requests? If we write in more students below the lines, will those requests be honored?[/quote]\r\nso can we write more?", "Solution_5": "well i think they already confirmed my room but i wrote down all five people on the three lines, it should be fine", "Solution_6": "[quote=\"13375P34K43V312\"]well i think they already confirmed my room but i wrote down all five people on the three lines, it should be fine[/quote]\r\n:rotfl: :rotfl:", "Solution_7": "[quote=\"bpms\"][quote=\"13375P34K43V312\"]well i think they already confirmed my room but i wrote down all five people on the three lines, it should be fine[/quote]\n:rotfl: :rotfl:[/quote]\r\ni dont get it", "Solution_8": "I haven't gotten something asking for roommate preferences...", "Solution_9": "it came with all the information and forms and such.", "Solution_10": "oh, gotcha.\r\nthanks." } { "Tag": [ "trigonometry", "limit", "integration", "logarithms" ], "Problem": "fie f:[0;1] -> [b]R[/b] , definita astfel:\r\nf(x) = arctg(1/x) , daca x apartine (0;1] ; pi/2 , daca x=0\r\n\r\na) Sa se arate ca f e continua (evident)\r\nb) Sa se arate ca f e derivabila si sa se calculeze f'(x) , pentru orice x din [0;1]\r\nc) sa se calculeze integrala de la 0 la 1 din f(x)dx", "Solution_1": "[quote=\"abedhu\"]Fie$\\ f: [0;1]\\rightarrow\\mathbb{R}$ , definita astfel:\n$\\ f(x) =\\arctan\\left(\\frac{1}{x}\\right)$ , daca $x\\in\\ (0,1]$ si $\\frac{\\pi}{2}$ daca $\\ x=0$.\n\na) Sa se arate ca $\\ f$ e continua.\nb) Sa se arate ca $\\ f$ e derivabila si sa se calculeze $\\ f'(x)$ , pentru orice $\\ x$ din$\\ [0;1]$.\nc) Sa se calculeze $\\int_{0}^{1}f(x)dx$[/quote]\r\nMai bine pui problemele pe Avansati Olimpiada, ca aici nu prea te baga nimeni in seama.", "Solution_2": "Ba eu cred ca aici e locul acestei probleme !\r\n\r\npentru b) aplicam regula lui H\u00f4pital ca sa gasim derivata (daca exista!) in 0\r\nObtinem f'(0)= -1.\r\npentru c) ma mai gindesc...sa gasesc o primitiva ... :mad:", "Solution_3": "Un anume [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=17823]stancioiu sorin[/url] a postat punctul (c) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=107358]aici[/url].", "Solution_4": "$2.\\blacktriangleright\\ R_{0}(x)\\equiv\\frac{f(x)-f(0)}{x-0}=-\\frac{\\frac{\\pi}{2}-\\arctan\\frac{1}{x}}{\\tan (\\frac{\\pi}{2}-\\arctan\\frac{1}{x})}$ $\\Longrightarrow f'(0)\\equiv f'_{d}(0)=\\lim_{x\\searrow 0}R_{0}(x)=-1\\Longrightarrow f'(0)=-1\\ .$ \r\n$f'(x)=\\{\\begin{array}{ccc}-1 & \\mathrm{daca}& x=0\\\\-\\frac{1}{x^{2}+1}& \\mathrm{daca}& 0. The trisection points split the whole distance into thirds, so they will be found by moving (9/3)=3 units right and (6/3)= 2 units up twice from the point (-6, 2). \n\nThe first trisection point will be (-6[color=red][b]+3[/b][/color], 2[color=red][b]+2[/b][/color]) or (-3, 4)\nThe second trisection point will be (-6[color=red][b]+3+3[/b][/color], 2[color=red][b]+2+2[/b][/color]) or (0,6)\n[/hide]" } { "Tag": [ "modular arithmetic", "number theory unsolved", "number theory" ], "Problem": "Let $n$ be an integer greater than $6$. Prove that if $n-1$ and $n+1$ are both prime, then $n^{2}(n^{2}+16)$ is divisible by $720$. Is the converse true?", "Solution_1": "If $n-1$ and $n+1$ are both prime, then $n$ is obviously a multiple of $6$.\r\nLet $n=6p$\r\n$n^2(n^2+16)=(6p)^2[(6p)^2+16]=144p^2(9p^2+4)$.\r\nIf $p=5q$, then $n^2(n^2+16)$ is a multiple of $144\\times 5=720$.\r\nIf $p=5q\\pm 1$, then $n=30q\\pm 6$. Is is easy to see that either one of $n-1$ or $n+1$ is divisible by $5$.\r\nIf $p=5q\\pm 2$, then $9p^2+4\\equiv 9(4)+4\\equiv 0\\pmod{5}$, i.e. $n^2(n^2+16)$ is a multiple of $144\\times 5=720$.\r\nThe converse is not true. For instance, when $n=720$, $n^2(n^2+16)$ is a multiple of $720$. But $721=7\\times 103$ is not a prime.", "Solution_2": "Prove that $n$ is a multiple of 6.", "Solution_3": "[quote=\"kennylcc\"]Prove that $n$ is a multiple of 6.[/quote]\r\n\r\n$n-1\\equiv 1(mod 2)$ $\\implies n\\equiv 0(mod 2)$\r\n\r\nAny three consecutive integer has at least one of them is a multiple of three . $n-1$ and $n+1$ are prime , so $n$ must be the one with multiple of three :D", "Solution_4": "At Olympiad level would you have to prove that $n \\equiv{0} \\mod{6}$ or would they take it as \"obvious\"? I proved it, but my entire solution was two whole sides of A4 (and considering I only started this problem with 20 mins to go, that was bad :)), so it could have used some cutting down.", "Solution_5": "As a special case, $n=4$. Contradiction", "Solution_6": "There is no contradiction $n=4$ since $n>6$...\r\n\r\nTo Ilthigore's question: you can prove it the following way:\r\nAll primes except $2$ and $3$ are $\\equiv \\pm 1 \\mod 6$. Prove: For $\\equiv 0,2,4$ they are divisible by $2$ and for $\\equiv 3$ they are divisible by $3$.\r\nSo when we have two primes ($\\neq 2;3$) with difference $2$, then the smaller one ($n-1$) is either $\\equiv -1 \\mod 6$ and it follows $n \\equiv 0 \\mod 6$ or it is $\\equiv 1 \\mod 6$, but then the greater one would be $\\equiv 3 \\mod 6$ which is impossible.", "Solution_7": ":blush: :blush: \r\n\r\nIll read it next time!" } { "Tag": [], "Problem": "In a big city, all the streets have double-hand, but they have only two possible directions (north-south or east-west). During a tour in a car, the same place wasn\u00b4t visited more than 1 time, and the tour finished in the start-point. 100 left turns were done. How many right-turns were done, then???? (all posibilities are asked).", "Solution_1": "I'm guessing the asnswer is either all even integers or all integers divisible by four, I couldn't really understand the question...", "Solution_2": "[quote=\"me@home\"]I couldn't really understand the question...[/quote]\r\n\r\nIf it helps for anyone that can translate it better, look for \"Nivel Juvenil\", problem 4...\r\n\r\nhttp://www.oma.org.ar/enunciados/ptic052.htm", "Solution_3": "did you get my asnwer to your other question?", "Solution_4": "I saw it, but I don't know the answer, so I can't tell you if your answer is OK or not...", "Solution_5": "Does the proof seem okay conceptually?\r\nI was guessing from your other post with that 1997 problem that the two were similar, and from the same test possibly. Is this the case? Also, how come this problem is so easy to crack compared to the 1997 problem?" } { "Tag": [ "inequalities", "algebra", "polynomial", "inequalities unsolved" ], "Problem": "For $a,b,c$ positive reals such that $a+b+c=1$ prove that\r\n\r\n$ 5(a^2+b^2+c^2) + 18abc \\geq \\frac{7}{3}$", "Solution_1": "This is quite routine.\r\nFirst we homogenize, making the inequality\r\n\\[\r\n5(a^2+b^2+c^2)(a+b+c)+18abc \\geq \\frac{7}{3}(a+b+c)^3\r\n\\]\r\nafter expanding it, the ineq becomes\r\n\\[\r\n4(a^3+b^3+c^3)+6abc \\geq 3(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b)\r\n\\]\r\nand this is just the sum of Schurs and AM-GM:\r\n\\begin{eqnarray*}\r\n3(a^3+b^3+c^3)+9abc &\\geq& 3(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b) \\\\\r\na^3+b^3+c^3 &\\geq& 3abc\r\n\\end{eqnarray*}", "Solution_2": "hi, may i know what is \"homogenize\"? do you mean making each term into same power?\r\nif so, on the right hand side, why did you choose (a+b+c)^3 instead of (a^3+b^3+c^3) or any other possible choice?\r\nFurthermore, if there's no a+b+c=1 or similar condition, how will you do the homogenization?\r\nThanks!", "Solution_3": "Homogenize means to make all terms of the same power. If all terms are not of the same power, well, then you don't have too many options. This is useful because AM-GMing, Cauchy, Schur's, etc, applied to terms of degree t will give you more terms of degree t.\r\n\r\na+b+c=1 essentially de-homogenizes the equation, so bill is just reversing that.\r\n\r\nOn the RHS, he picked (a+b+c)^3 because it is equal to 7, which is convienient, and because it transforms the LHS into a polynomial of degree 3. After this, all he has left is a big symmetric polynomial of degree 3, and we all know that such inequalities are usually very easy with Schur and weighted AM-GM (and Muirhead if necessary)", "Solution_4": "[quote=\"blahblahblah\"]Homogenize means to make all terms of the same power. If all terms are not of the same power, well, then you don't have too many options. This is useful because AM-GMing, Cauchy, Schur's, etc, applied to terms of degree t will give you more terms of degree t.\n\na+b+c=1 essentially de-homogenizes the equation, so bill is just reversing that.\n\nOn the RHS, he picked (a+b+c)^3 because it is equal to 7, which is convienient, and because it transforms the LHS into a polynomial of degree 3. After this, all he has left is a big symmetric polynomial of degree 3, and we all know that such inequalities are usually very easy with Schur and weighted AM-GM (and Muirhead if necessary)[/quote]\r\n\r\nThank you!" } { "Tag": [ "geometry", "circumcircle", "trigonometry", "angle bisector" ], "Problem": "In a triangle $ ABC$ with $ AB\\equal{}1,\\ BC\\equal{}\\sqrt{7},\\ AC\\equal{}2$ the interior bisector of $ \\angle{CAB}$ intersects $ BC$ at $ D$ and intersects the circumcircle of $ ABC$ again at $ E$. Let $ O$ be the circumcenter of $ BED$.\r\n\r\n(1) Find the length of $ DE$.\r\n\r\n(2) Find the value of $ \\tan \\angle{EBO}$.", "Solution_1": "[hide=\"number 1\"]\nLet $ \\overline{AD} \\equal{} h, \\angle CAD\\equal{}\\angle BAD \\equal{} \\alpha$\n\nFrom the cosine law\n$ c_1^2 \\equal{} a_1 \\plus{} h^2 \\minus{} 2a_1h\\cos{\\alpha} \\qquad(1)$\n$ c_2^2 \\equal{} a_1 \\plus{} h^2 \\minus{} 2a_2h\\cos{\\alpha}\\qquad(2)$\n\nSub (1) and (2)\n$ a_2c_1^2 \\minus{} a_2a_1^2 \\minus{} a_2h^2 \\equal{} a_1c_2^2 \\minus{} a_1a_2^2 \\minus{}a_2h^2$\n$ h^2(a_1\\minus{}a_2) \\equal{} (a_1c_2^2\\minus{}a_2c_1^2) \\plus{}a_1a_2(a_2\\minus{}a_1)$\n\nUsing the angle bisector theorem this simplifies to\n$ h^2 \\equal{} \\minus{}\\frac{a_1a_2}{(a_1\\plus{}a_1)^2} \\plus{} a_1a_2$\n\nWith $ a_1 \\equal{} 2, a_2 \\equal{} \\sqrt{7}$\n$ h^2 \\equal{} \\minus{}\\frac{2\\sqrt{7}}{(2\\plus{}\\sqrt{7})^2}\\plus{}2\\sqrt{7} \\equal{} \\frac{56\\minus{}4\\sqrt{7}}{9}$\n\n$ \\Rightarrow h \\equal{} \\frac{2\\sqrt{14\\minus{}\\sqrt{7}}}{3}$...\n\nSo now we apply the formula $ \\overline{AD}\\cdot\\overline{DE}\\equal{}\\overline{BD}\\cdot\\overline{DC}$\n\n$ \\overline{DE}\\cdot\\frac{2\\sqrt{14\\minus{}\\sqrt{7}}}{3}\\equal{}\\frac{2\\sqrt{7}}{(2\\plus{}\\sqrt{7})^2}$\n\n$ \\overline{DE} \\equal{} \\frac{11\\sqrt{7}\\minus{}28}{3\\sqrt{14\\minus{}\\sqrt{7}}}$ :) \n\n[/hide]", "Solution_2": "Regrettably your answer is incorrect.", "Solution_3": "Bah, I always make silly mistakes with your questions.\r\nHere I have put $ \\overline{AB} \\equal{} 2$ and $ \\overline{BC} \\equal{} 1$ instead of the other way arround\r\n\r\nIs the answer; $ \\frac {7}{3}$\r\n\r\nEDIT: :rotfl: I made a mistkae in my correction as well. IT'S A CURSE!\r\n\r\nNote to self; just work more thoroughly before posting!", "Solution_4": "Incorrect.", "Solution_5": "Yet another mistake, how about now??????", "Solution_6": "[quote=\"ocha\"]Bah, I always make silly mistakes with your questions.\nHere I have put $ \\overline{AB} \\equal{} 2$ and $ \\overline{BC} \\equal{} 1$ instead of the other way arround\n\nIs the answer; $ \\frac {7}{3}$\n\nEDIT: :rotfl: I made a mistkae in my correction as well. IT'S A CURSE!\n\nNote to self; just work more thoroughly before posting![/quote]\r\n\r\nNow You get the answer of Q1.", "Solution_7": "[hide]Part1. $ \\frac {CD}{DB} = \\frac {AC}{AB} \\Rightarrow CD = \\frac {2\\sqrt {7}}{3}, \\; BD = \\frac {\\sqrt {7}}{3}$\n$ AD^2 = AC.AB - CD.BD = 2 - \\frac {14}{9}\\Rightarrow AD = \\frac {2}{3}$\n$ CD\\cdot DB = AD.DE \\Rightarrow DE = \\frac {7}{3}$[/hide]\n[hide]2.\n$ \\cos \\angle CAB = \\frac {AC^2 + AB^2 - BC^2}{2AC\\cdot AB} = - \\frac {1}{2} \\Rightarrow \\angle CAB = 120^\\circ \\Rightarrow EBC = 60^\\circ \\; \\Rightarrow \\Delta EBC$is equilateral \n$ \\Rightarrow EB = \\sqrt {7}$\n$ \\cos \\angle EDB = \\frac {ED^2 + BD^2 - EB^2}{2ED\\cdot DB} = - \\frac {1}{2\\sqrt {7}} \\;\\;\\sin \\angle EDB = \\frac {3\\sqrt {3}}{2\\sqrt {7}}$\n$ EB = 2R\\sin \\angle D \\\\\nRightarrow R = \\frac {7}{3\\sqrt {3}}$ skipping some obvious steps $ \\tan \\angle EBO = \\frac {1}{3\\sqrt {3}} = \\frac {\\sqrt {3}}{9}$[/hide]", "Solution_8": "That's correct. :)" } { "Tag": [ "function", "geometry", "geometric transformation", "rotation", "parameterization", "real analysis", "calculus" ], "Problem": "Let $f: \\mathbb{R}\\to \\mathbb{R}$ be a $C^{\\infty}$ function. Suppose that $u: \\mathbb{R}^{n}\\to \\mathbb{R}$ is a $C^{2}$ function that satisfies\r\n$\\Delta u+f(u)=0$\r\nin the ball $B=B(0;R)$ and $u=0$ on the boundary $\\partial B$. Prove that $u$ is radially symetric !", "Solution_1": "So what was the answer?", "Solution_2": "if the solution u is unique, it must be radial, because rotations of u also satisfy the equation.\r\n\r\nIs solution unique?", "Solution_3": "Even for the linear function $f(t)=ct$ there may be a one-parameter family of solutions (eigenfunctions of the Laplacian). \r\n\r\nI tried to think this way: if $u$ is a solution, then it is equal to the convolution of $-f(u)$ with Green's function for $B$ (which can be written explicitly, and does have certain symmetry). In other words, $u$ is a fixed point of an explicit nonlinear integral operator. However, this did not lead me anywhere. :(" } { "Tag": [ "calculus", "logarithms", "function" ], "Problem": "3^x +4^x=5^x.\r\nobviously,x=2,but HOW???", "Solution_1": "If $ x$ has to be an integer, well Fermat's Last Theorem, that's how.", "Solution_2": "Alternatively, show that $ f(x) \\equal{} \\left ( \\frac {3}{5} \\right )^x \\plus{} \\left ( \\frac {4}{5} \\right )^x$ is strictly decreasing", "Solution_3": "[quote=\"williams\"]3^x +4^x=5^x.\nobviously,x=2,but HOW???[/quote]\r\n[hide=\"Hint\"]\nI think calculus will be useful to prove that there is one solution\n[/hide]", "Solution_4": "If we want to use calculus to show strictly decreasing, then\r\n$ f(x)\\equal{}e^{x\\ln{.6}}\\plus{}e^{x\\ln{.8}}$\r\nso\r\n$ f'(x)\\equal{}\\ln{.6}e^{x\\ln{.6}}\\plus{}\\ln{.8}e^{x\\ln{.8}}\\equal{}\\ln\\frac{3}{5}\\left(\\frac{3}{5}\\right)^{x}\\plus{}\\ln\\frac{4}{5}\\left(\\frac{4}{5}\\right)^{x}$\r\nNow you can proceed from here.", "Solution_5": "Why? It is a basic property of exponential functions that $ b^x$ is strictly decreasing for $ 0 < b < 1$ and strictly increasing for $ b > 1$; see, for example, Rudin's discussion of the construction of the exponential.", "Solution_6": "$ 5^n > 4^n \\plus{} 4^n > 4^n \\plus{} 3^n$ (for $ n\\geq4$ )\r\nso $ n < 4$ $ \\implies$ $ \\boxed{n \\equal{} 2}$", "Solution_7": "[hide]\n\nTaking modulo 4, $ (\\minus{}1)^x\\equiv 1 \\mod4$. So, $ x$ must be even. Let $ x\\equal{}2a$. So, now:\n\n$ (3^a)^2\\plus{}(4^a)^2\\equal{}(5^a)^2$\n\nSince $ (3^a, 4^a, 5^a)$ is a primitive Pythagorean triple, \n\n$ 3^a\\equal{}m^2\\minus{}n^2$ (1)\n\n$ 4^a\\equal{}2mn$ (2)\n\n$ 5^a\\equal{}m^2\\plus{}n^2$ (3)\n\nFrom (2), m and n must be powers of 2. Also, one is even and one is odd. The only power of 2 that is odd is 1, and since $ m>n$ due to (1), $ n\\equal{}1$. So, $ 3^a\\equal{}m^2\\minus{}1\\equal{}(m\\plus{}1)(m\\minus{}1)$. Hence, $ m\\equal{}2$. $ a\\equal{}1$ and thus, $ x\\equal{}2$\n\n[/hide]", "Solution_8": "[quote=\"lifeisacircle\"] $ (3^a, 4^a, 5^a)$ is a primitive Pythagorean triple, \n\n$ 3^a \\equal{} m^2 \\minus{} n^2$ (1)\n\n$ 4^a \\equal{} 2mn$ (2)\n\n$ 5^a \\equal{} m^2 \\plus{} n^2$ (3)\n[/quote]\r\n\r\n$ (3^a, 4^a, 5^a)$ is not a primitive pythagorean triple; $ (3a, 4a, 5a)$ is.", "Solution_9": "I thought that (3,4,5) was the only primitive triple of the form (3a,4a,5a)...", "Solution_10": "yeah, but i was just pointing out that $ (3^a,4^a,5^a)$ can't be a pythagorean triple unless a=1", "Solution_11": "Pythagorean triples have nothing to do with this problem except that they inform the only solution; in particular, $ x$ is not required to be an integer." } { "Tag": [ "probability", "integration", "probability and stats" ], "Problem": "Two components of a minicomputer have the following joint pdf for their useful lifetimes $ X$ and $ Y$: $ f(x,y) \\equal{} \\begin{cases} xe^{ \\minus{} x(1 \\plus{} y)} \\ x \\geq 0 \\ \\text{and} \\ y \\geq 0 \\\\\r\n0 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\text{otherwise} \\end{cases}$\r\n\r\n(a) What is the probability that the lifetime $ X$ of the first component exceeds $ 3$?\r\n\r\n(b) What are the marginal pdf's of $ X$ and $ Y$? Are the two lifetimes independent? \r\n\r\n(c) What is the probability that the lifetime of at least one component exceeds $ 3$?", "Solution_1": "$ \\Pr[X > 3] \\equal{} \\int_{x \\equal{} 3}^\\infty \\int_{y \\equal{} 0}^\\infty f_{X,Y}(x,y) \\, dy \\, dx$.\r\n\r\n$ f_X(x) \\equal{} \\int_{y \\equal{} 0}^\\infty f_{X,Y}(x,y) \\, dy$.\r\n\r\n$ f_Y(y) \\equal{} \\int_{x \\equal{} 0}^\\infty f_{X,Y}(x,y) \\, dx$.\r\n\r\nX and Y are independent iff $ f_{X,Y}(x,y) \\equal{} f_X(x) f_Y(y)$.\r\n\r\n$ \\Pr[X > 3 \\cup Y > 3] \\equal{} 1 \\minus{} \\Pr[X \\le 3 \\cap Y \\le 3] \\equal{} 1 \\minus{} \\int_{x \\equal{} 0}^3 \\int_{y \\equal{} 0}^3 f_{X,Y}(x,y) \\, dy \\, dx$;\r\n\r\nif X and Y are independent, then the above simplifies to\r\n\r\n$ \\Pr[X > 3 \\cup Y > 3] \\equal{} 1 \\minus{} \\Pr[X \\le 3] \\Pr[Y \\le 3] \\equal{} 1 \\minus{} F_X(3)F_Y(3)$.", "Solution_2": "(a) and (c) are just integrals. Find the appropriate bounds and compute (Using the complement makes one simpler).\r\n\r\nThe joint density of a pair of independent random variables is of the form $ f(x)g(y)$. There's no way you're going to get an $ e^{\\minus{}xy}$ term out of that." } { "Tag": [ "MATHCOUNTS", "modular arithmetic", "LaTeX", "floor function", "number theory", "prime factorization" ], "Problem": "I have some questions I don't understand. Please help.\r\n\r\n\r\n1.Find the value of x such that the four digit number x15x is divisible by 18.\r\n \r\n2.How many 9 digit zip codes are possible if the first digit cannot be zero.\r\n\r\n3.The walls of standard houses are constructed with 2 by 4 studs placed 16\" apart, center to center. How many studs are needed for a wall 48 feet wide?\r\n\r\n4.What is the largest perfect square that divides 64,800?\r\n \r\n\r\nPlease help me. Thanks, if you do.", "Solution_1": "[hide=\"1\"]18=2*9\ndivisible by both\n\na15b\nb=even\na+b=3 or 12\na+b=3\n\nb=2, a=1 or\nb=0, a=3\n\na+b=12\nb=4, a=8\n[b]b=6, a=6[/b]\n\nthat fills in for \n\nx15x\nif x=x\n\nso its 6156[/hide]\n[hide=\"2\"]1st cant be 0. 9 other #s possible. all others: 10 possible #s\n9x10x10x10x10x10x10x10x10=900000000 codes[/hide]\n\n[hide=\"4\"]prime factorization\n\n64800=2^5*5^2*3^4\n\nbiggest square: (2^2*5*3^2)^2=2^4*5^2*3^4=64800/2=[b]32400[/b]\n[/hide]\r\nim thinking about #3", "Solution_2": "[quote=\"mathirons\"]1.Find the value of x such that the four digit number x15x is divisible by 18.[/quote]\n[hide]$ x$ must be even because the number is divisible by 2. Also, $ 2x\\plus{}6\\equal{}9k$ because the number is a multiple of 9. Quite clearly, $ \\boxed{x\\equal{}6}$ since no other even digits satisfy $ 2x\\plus{}6\\equal{}9k$.[/hide]\n\n[quote=\"mathirons\"]2.How many 9 digit zip codes are possible if the first digit cannot be zero.[/quote]\n[hide]The first digit can have 9 possible values, and the other 8 can have 10. The answer is $ 9\\cdot10^{8}$, or $ \\boxed{900000000}$.[/hide]\n\n[quote=\"mathirons\"]4.What is the largest perfect square that divides 64,800?[/quote]\n[hide]$ 64800\\equal{}2^{5}\\cdot3^{4}\\cdot5^{2}$\n\nThe largest square that divides $ 64800$ is $ 2^{4}\\cdot3^{4}\\cdot5^{2}\\equal{}\\boxed{32400}$.[/hide]", "Solution_3": "[quote=\"mz94\"][hide=\"1\"]18=2*9\ndivisible by both\n\na15b\nb=even\na+b=3 or 12\na+b=3\n\nb=2, a=1 or\nb=0, a=3\n\na+b=12\nb=4, a=8\n[b]b=6, a=6[/b]\n\nthat fills in for \n\nx15x\nif x=x\n\nso its 6156[/hide]\n[/quote]\r\nWTH", "Solution_4": "hes saying the sum of all the digits is divisible by 9 and x is even...i think...atleast...thats what i think\r\n\r\nso $ 2x\\plus{}6$[u]=[/u]$ 0\\pmod9$ and $ x$ [u]=[/u] $ 0\\pmod2$", "Solution_5": "I'm not sure why he introduced two new variables, but what abacadaea said is correct.\r\n\r\n@ abacadaea - The congruent symbol is written \\equiv in $ \\text{\\LaTeX}$.", "Solution_6": "$ \\equiv$ ok thanks", "Solution_7": "for 3, would it be rounded down(48*12/20). sorry i dodnt know how to do that cool boldy thingy.", "Solution_8": "\"That cool boldy thingy\" is called [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]LaTeX[/url]. Simply put dollar signs around mathematical expressions, and it turns into a nice-looking image. Do not, however, type [i]everything[/i] in $ \\text{\\LaTeX}$ as this causes the site to be much slower for those without high speed internet connections.\r\n\r\nAs for #3, is the width of the studs 2\" or 4\"?", "Solution_9": "so, i only do it on the math part like 2x+3 making it look like $ 2x\\plus{}3$ but not $ like this?$", "Solution_10": "Yes, unless you have a good reason (e.g. math symbols like $ \\mathbb{N}$ and such, if they count as text at all).", "Solution_11": "what are those $ \\mathbb{N}$'s used for?", "Solution_12": "N's i think are the set of all counting numbers?\r\n\r\ni know Z is for integers\r\nQ is for rationals\r\nW i think is for whole numbers. ectr.", "Solution_13": "$ \\mathbb{N}$ represents all natural numbers, in other words, all positive non-zero integers. Sometimes zero is included so that must be specified.", "Solution_14": "$ \\mathbb{C}$ stands for complex. But we're getting of topic.", "Solution_15": "well arent all the problems solved?\r\n\r\ni guess this topic should be locked", "Solution_16": "[/offtopic]\r\n\r\nMy question (Are the studs 2\" or 4\" wide?) hasn't been answered yet, so I can't answer #3.", "Solution_17": "usuallyits length by witdth so i think its 2 wide", "Solution_18": "Actually you don't need to know the width. [hide]*Center* to *center* means you would require $ \\lfloor 48*12/16\\rfloor\\plus{}1$ studs, because it doesn't matter how wide the studs are you must have one at the beginning and end. Also, studs are always placed so the smaller dimension is facing the outside of the wall, since most stress on the studs will be up-down or into-out of the house, so you want to have as little flex in these dimensions as possible. You don't usually push parallel to the wall... :)[/hide] Having some construction experience helps with this question ;).", "Solution_19": "oh.....\r\n\r\nThe message is too small. Please make the message longer before submitting.The message is too small. Please make the message longer before submitting.The message is too small. Please make the message longer before submitting.The message is too small. Please make the message longer before submitting.The message is too small. Please make the message longer before submitting.The message is too small. Please make the message longer before submitting.The message is too small. Please make the message longer before submitting.The message is too small. Please make the message longer before submitting.The message is too small. Please make the message longer before submitting." } { "Tag": [ "videos", "geometry" ], "Problem": "We got so wrapped up in this on the Intro to Geometry forum that I am moving it here.", "Solution_1": "HALO PWNS!!!" } { "Tag": [ "integration", "trigonometry", "limit", "function", "abstract algebra", "real analysis", "real analysis unsolved" ], "Problem": "Set\r\n\r\nf(x)= 0 if x<=0 \r\n sin(pi/x) if x>0\r\n\r\nand \r\n\r\ng(x)= 0 if x<=0\r\n 1 if x>0\r\n\r\n\r\nProve that f has no antiderivative but g does not.\r\n\r\nThanks", "Solution_1": "$ f$ admits an antiderivative but $ g$ does not. \r\n\r\n\r\nLetting $ F(x) \\equal{} \\int_0^x \\sin {\\pi\\over t} dt,$ and $ F(0) \\equal{} 0,$ $ F'(x) \\equal{} f(x)$ by the\r\nfundamental theorem at the points of continuity of $ f$ so $ x\\ne0.$\r\nIf $ x \\equal{} 0,$ $ F'(0) \\equal{} 0$ holds. A proof runs as follows\r\n\r\n$ F^\\prime(0) \\equal{} \\lim_{h\\to0} {1\\over h}\\int_0^h \\sin{\\pi\\over z} dz.$\r\nChanging variable $ \\pi/z \\equal{} y$ we have $ \\lim_{h\\to0} {\\pi\\over h} \\int_{\\pi\\over h}^{ \\plus{} \\infty} dy {\\sin y\\over y^2} \\equal{} \\lim_{h\\to0} {\\pi\\over h} \\Bigl[ \\minus{} {\\cos z\\over z^2}\\Bigl\\vert_{\\pi\\over h}^{ \\plus{} \\infty} \\plus{} \\int_{\\pi\\over h}^{ \\plus{} \\infty} dz { \\minus{} 2\\sin z\\over z^3}\\Bigr].$ The limit of the first part is clearly zero. As for the second contribution we can bound\r\n$ \\vert\\int_{\\pi\\over h}^{ \\plus{} \\infty} dz { \\minus{} 2\\sin z\\over z^3}\\vert\\le 2\\int_{\\pi\\over h}^{ \\plus{} \\infty} dz {1\\over z^3} \\equal{} h^2/\\pi^2$ and of course $ \\lim_{h\\to0} {\\pi \\over \\vert h\\vert} \\vert\\int_{\\pi\\over h}^{ \\plus{} \\infty} dz { \\minus{} 2\\sin z\\over z^3}\\vert\\le \\lim_{h\\to0} {1\\over \\pi}{h^2\\over \\vert h\\vert} \\equal{} 0$ concluding the proof.\r\n\r\n\r\nAs for $ g(x),$ an easy and standard application of the Lagrange's theorem forbids\r\nthe functions having a jump or removable\r\ndiscontinuity from being derivatives.\r\n\r\nThe application of Lagrange's theorem is the following\r\n\r\n[i]\nLet $ h\\colon[a,b]\\to {\\bf R}$ be a differentiable function such that $ \\lim_{x\\to x_0^{\\pm}} h'(x) \\equal{} l^{\\pm}\\in {\\bf R}\\cup( \\plus{} \\infty)\\cup( \\minus{} \\infty)$ and let $ x_0\\in [a,b].$ Then $ \\lim_{h\\to0^{\\pm}} {1\\over h}(h(x_0 \\plus{} h) \\minus{} h(x_0)) \\equal{} l^{\\pm}$\n(of course if $ x_0 \\equal{} a$ we consider only right limits and left limits if $ x_0 \\equal{} b$).\n[/i]\r\n\r\nUse Lagrange to prove the result above. The conclusions about jump or removable discontinuities are easy consequences.", "Solution_2": "Thank you Arrigo-Sacchi,\r\n\r\nVery Thorough and helpful.\r\n\r\nI appreciate the effort you put in helping me.", "Solution_3": "your welcome\r\n\r\n\r\n[quote=\"AlwaysMathLover\"]Thank you Arrigo-Sacchi,\n\nVery Thorough and helpful.\n\nI appreciate the effort you put in helping me.[/quote]" } { "Tag": [ "videos" ], "Problem": "I just discovered the MLG Gaming website today and became a user. \r\n\r\nHow do I join a video game competition there? (if anyone has experience)", "Solution_1": "What particular game are you trying to get into? \r\n\r\nThe most popular MLG game that exists is Halo 3, but there have been small factions of Call of Duty 4 or Gears of War, of even SSBB. To be identified as an MLG Professional player, you would have to dedicate your time to tournaments (paid entries) and winning them." } { "Tag": [ "modular arithmetic", "number theory unsolved", "number theory" ], "Problem": "How many perfect squares are there (mod $ 2^n$) ?", "Solution_1": "For $ n$ even: The perfect squares are only $ 0,1,4,\\ldots,(2^{n/2}\\minus{}1)^2$, call this set $ S\\equal{}\\{0,1,4,\\ldots,2^{n/2}\\minus{}1\\}$. Because if $ k\\equiv r\\pmod{2^n}$ ($ \\minus{}2^{n/2}\\le r\\le2^{n/2}\\minus{}1$), then $ k^2\\equiv r^2\\pmod{2^n}$. Since $ r^2\\in S$, then $ k^2\\in S$ too. We claim that these numbers are pairwise distinct. Assume $ a^2\\equiv b^2\\pmod{2^n}$ ($ 0\\le ao does not necessarily mean that the shape is a square.", "Solution_3": "When he said AB=CD, it does sound more if though the parallelogram is a rectangle instead. Otherwise, he would probably ask to prove that all the sides are equal.", "Solution_4": "[quote=\"monotonic\"]Let ABCD be a parallelogram and $\\measuredangle A=90^0$.\n 1) Prove that $\\measuredangle B=C=D=90^0$\n 2) Prove that AC=BD[/quote]\r\n\r\nIf DAB = 90, DA and AB are perpendicular. As the shape, by defenition, has 4 sides with opposing sides parallel, the only possible positions of BC and CD are parallel to DA and AB, and perpendicular to AB and DC respectively...\r\n\r\nthis is so easy it's hard. I'm no good at geometry :(", "Solution_5": "[hide]1. ABCD is a parallelogram with ang A = 90.\n\n\n\n That means ang C = 90.\n\n As interior angle measure = 360, (360 - 180)/2 = 90\n\n So, ang B = ang C = ang D\n\n \n\n2. 1 proves that each ang = 90\n\n\n\n As ABCD is a parallelogram AB = CD and BC = AD\n\n Again, as each angle is 90, AB and CD are perpendicular to AD and BC.\n\n \n\n Pythagorean theorem (a^2 + b^2 = c^2) applies (Right triangles ABC and ACD)\n\n AB = CD = a; BC = AD = b; AC = BD = c\n\n Therefore, AC and BD are equal.\n\n \n\n\n\nNot very direct, but everything is stated/implied.[/hide]", "Solution_6": "I'm going to promote this to Intermediate, because as usual problems involving geometry content and requiring the problem-solver to \"prove . . . \" are not at the getting-started level. \r\n\r\nThanks for posting the problem.", "Solution_7": "\"Prove ... \" is just a synonym of \"Show that ... \", tokenadult.\r\n\r\nAs for intermediate, there's really no use for moving this to intermediate. Then it's just a waste of forum space to post it. This is really the basics of geometry, which you learn in the very first geometry lessons. These things we learnt in primary school (4th-5th grade) so if things for primary school are too hard for beginner's section, then what the heck do we have that section? Be honest, how many members from primary school do we have here?? :?" } { "Tag": [ "algebra open", "algebra" ], "Problem": "Does $ a^2 \\plus{} b^2 \\equal{} c^2 \\plus{} d^2$ have solution???If yes please explain it?", "Solution_1": "[quote=\"Flakky\"]Does $ a^2 \\plus{} b^2 \\equal{} c^2 \\plus{} d^2$ have solution???If yes please explain it?[/quote]\r\n\r\nIf you are talking about positive integer, I am sure that it has a lot...\r\n\r\nfor example $ 11^2\\plus{}10^2\\equal{}14^2\\plus{}5^2\\equal{}221$\r\n\r\nYou can consider $ (a^2\\plus{}b^2)(c^2\\plus{}d^2)\\equal{}(ac\\plus{}bd)^2\\plus{}(ad\\minus{}bc)^2\\equal{}(ad\\plus{}bc)^2\\plus{}(ac\\minus{}bd)^2$ :roll:" } { "Tag": [ "geometry", "trigonometry", "LaTeX" ], "Problem": "$\\text{Find the largest area of the triangle whose three sides } a,b,c \\text{ satisfy the conditions: }$\r\n\r\n$0 < a \\leq 2 \\leq b \\leq 4\\leq c \\leq 6$\r\n\r\nNote that there are no specifications on the type of numbers a,b, and c must be, except that they must satisfy the above inequality.", "Solution_1": "i think the largest area will be a right triangle of 2,4, 2sqrt5 with an area of 4", "Solution_2": "how did you get that answer though? brute force or what?", "Solution_3": "One way the area of the triangle may be expressed is $\\frac{1}{2}ab\\sin{C}$.\r\nMaximizing $a,b,\\sin{C}$, the area is $\\frac{1}{2}(2)(4)(1)=4$.\r\n\r\nSince $\\sin{C}=1$ implies $\\angle ACB=90^\\circ$, then $c=\\sqrt{a^2+b^2}=2\\sqrt{5}<6$, and this triangle does satisfy all requirements.", "Solution_4": "The smallest side must be maximum, $a = BC = 2$. The remaining vertex $A$ of the triangle $\\triangle ABC$ must be within light yellow annulus $((B))$ centered at the vertex $B$ with radii $c_1 = 4, c_2 = 6$ and within the light green annulus $((C))$ centered at the vertex $C$ with radii $b_1 = 2, b_2 = 4$. For the maximum triangle area, the altitude from the vertex $A$ must be maximum, hence, $b = 4$ and the angle $\\angle C = 90^o$ is right. Then $c^2 = a^2 + b^2 = 20,\\ 16 < c^2 < 36,\\ c = 2\\sqrt 5$ and the triangle area is $|\\triangle ABC| = 4$.", "Solution_5": "no...brutal forcing this problem would be....impossible....\r\n\r\ni did it like Samson but as u should know i am Latex handicapped and lazy so i didn't give any work....", "Solution_6": "yes, you want it to be a right triangle, obviously, because this makes a maximum height of the other side, since $\\sin{90}=1$, and if the angle is less or greater than 90, then $\\sin{\\theta}<1$ making the area and the height decrease" } { "Tag": [ "geometry", "parallelogram" ], "Problem": "The sides of this parallelogram measure 7,9, $ 8y\\minus{}1$ and $ 2x\\plus{}3$ units, consecutively. What is the value of $ x\\plus{}y$?\n\n[asy]draw((0,0)--(21,0)--(30,25)--(9,25)--cycle);\nlabel(\"$8y-1$\",(10,0),S);\nlabel(\"9\",(25.5,12.5),E);\nlabel(\"7\",(19.5,25),N);\nlabel(\"$2x+3$\",(4.5,12.5),W);[/asy]", "Solution_1": "$ 2x\\plus{}3\\equal{}9\\Rightarrow x\\equal{}3$\r\n$ 8y\\minus{}1\\equal{}7\\Rightarrow y\\equal{}1$\r\n\r\nThen $ x\\plus{}y\\equal{}4$." } { "Tag": [ "videos" ], "Problem": "I wear glasses since the finish of 7th grade(now I'm 11th) :agent: . I could get rid of them next year if I make an surgical operation but I have to be 18 years old. How about you :P?", "Solution_1": "I've worn glasses since 3rd grade (I'm currently in 8th)\r\n\r\nI heard that laser eye surgery wears off after a year or so... :?", "Solution_2": "Wel, I had glasses(since around 3 rd grade), but i got contacts a month ago. I like contacts much better. So should I vote yes or no?", "Solution_3": "I have glasses, but I dont wear them much, in fact, I havn't wore them ever since the start of summer break.", "Solution_4": "Since 5th grade. And I don't plan to make any surgical operations.", "Solution_5": "Glasses since 1st grade. Contacts since 8th.\r\n\r\n(Oh, and I'm in 12th now)", "Solution_6": "Since end of 7th, and I'm going into 8th. But, I barely wear them (only during a period in school where i need to see the board for a while, and thats it), and I'm gettign contacts soon.", "Solution_7": "technically, ive had glasses since 10th grade (now going into 12th), but i still dont really wear them that often.", "Solution_8": "So many have glasses.. O.o\r\n\r\nI have 20/25 vision or something like that. My left eye is just a tiny bit below 20/20 alone, but my right eye makes up for the difference when I have both eyes open. Glasses look so inconvenient, I would go for the surgery or at least contacts.", "Solution_9": "i wore glasses since fouth grade.", "Solution_10": "I wear contact lenses, about -375 something like that.", "Solution_11": "I have glasses. I got them just a bit ago but I don't wear tham much...", "Solution_12": "glasses since 6th! w00t!\r\nyay for 600/650 vision!", "Solution_13": "I don't know about measure system form US. or other countries. So I can't tell about my vision size. :huh: \r\nLeft eye: -4.00 \r\nRight eye: -6.00\r\n\r\nBoth eyes are blue :rotfl: !!\r\n\r\nIn american measure system might be -600 and -400... but I'm not sure of that.", "Solution_14": "[quote=\"236factorial\"]I wear contact lenses, about -375 something like that.[/quote]\r\nDon't you mean -3.75?\r\nBoth of my eyes are -5.25", "Solution_15": "I've worn glasses since 4th grade (I'm in 9th now)\r\nI want to get contacts and when I turn 21, I want to get the eye surgery", "Solution_16": "Glasses for the last, oh, 10 years (since 1st grade)\r\nGot contacts two years ago for sports, but they are uncomfortable and don't properly correct for my astigmatism. Oh well. Even the toric lenses they told me to use don't help much more (but it is a bit better)", "Solution_17": "I wear black framed rectangular glasses! They are so cool though, i'm proud of them! lol (I haven't been on in a while, school) Yeah, when I went to the place the lady at the counter looked at my prescription and said, \"Do you wear contacts?\" \r\n\r\nMe: \"No\"\r\nlady: \"Do you have another pair of glasses?\"\r\nMe: \"yes, but they give me a headache because their too small, and the words are still a little blurry that I can't read them.\"\r\nlady: \"You'll probably want contacts, their better to drive with. What do you wear when driving.\"\r\nMe: \"I'm 11, I don't drive!\"\r\nlady: \"YOU'RE 11?! Oh, well still how do you get around with this prescription?!\"\r\nMe: \"i bump into every wall, it's not that bad.\"\r\n1 week later\r\nlady: \"Here's your new glasses!\"\r\n*I put them on*\r\nMe: \"OH MY GOD!!!!!! I was blind! Those are bannas? I thought they were boxes of cereal!!!\" \r\nI don't know my prescription, but it was that bad... I heard a 7.75 or something with a 7, I have no idea... I don't even know where it is. For the longest time people's faces were little pale (tan, dark, whatever the case may be) ovals that had two black dots and a pink/red line. Yep, my sight was bad... Still is! But my glasses give me personality, and my eyes look so small without them, even though their not. Their not dork/geek glasses, they have an edge to them. Or at least I make them that way... :lol:", "Solution_18": "wow glasses since 2nd grade, contacts since 3rd because of soccer... and im worse than -7.00 in both eyes. :D", "Solution_19": "im 20-20, so i dont have glasses :)", "Solution_20": "i'm 25-20 and the doctor says if i get to 30-20 i have to wear glasses :(", "Solution_21": "I wear glasses, I've worn them since 6th grade (9th now). My right and left eyes are both -3, my middle eye is -3.5.", "Solution_22": "[quote=\"captcha000\"]im a sophomore, worn glasses since about 1st grade. i wore contacts for two or three years but my parents have heard that they make your eyesight worse (?) so im back to glasses[/quote]\r\n\r\nthat's SO not true. i actually had to wear contacts for three years as part of the study to keep my eyesight from GETTING worse. and for the person with astigmitism: there are contacts for that now. i have it too, i used to have to wear hard contacts but now they make soft ones. they are the kind you keep for 3 months and then throw away. (much better than hard, BELIEVE ME).", "Solution_23": "[quote=\"sponge008\"]I wear glasses, I've worn them since 6th grade (9th now). My right and left eyes are both -3, my middle eye is -3.5.[/quote]\r\n\r\nmiddle-eye????? :?\r\nI have soft contacts too. I switch mine every 2 weeks.", "Solution_24": "Glasses from 1st to 6th.\r\n\r\nNo glasses for me anymore.", "Solution_25": "[quote=\"Treething\"]Glasses from 1st to 6th.\n\nNo glasses for me anymore.[/quote]\r\n\r\nerrrr contacts then? or do you just walk around blindly lol?\r\nand why does everybody use the -5.00 or w/e system...what happend to like 600/20 and 550/20???", "Solution_26": "I just got contacts today. But I voted for glasses, since voted a week ago I think.", "Solution_27": "how many people have lasik", "Solution_28": "[quote=\"noneoftheabove\"]how many people have lasik[/quote]\r\n\r\nwhat's that?", "Solution_29": "Lasik is eye surgery." } { "Tag": [ "limit" ], "Problem": "Se da $f:[0,\\infty ) \\to R$,continua cu $\\mathop {\\lim }\\limits_{x \\to \\infty } f(x) = 0$.\r\nSa se arate ca f este marginita.", "Solution_1": "Weierstrass", "Solution_2": "Totusi, asta nu este o solutie. Ce se intelege dintr-un simplu \"Weierstrass\"? :?", "Solution_3": "Ca se aplica teorema lui Weierstrass, dupa ce ai \"taiat jos\" un interval catre $\\infty$.", "Solution_4": "Si pun pariu ca aproape nimeni n-a inteles mai nimic cu \"taiatul asta jos\"? :P", "Solution_5": "Ideea este sa nu uitam totusi nivelul acestui sub-forum (bacalaureat) ;)", "Solution_6": "Asa ziceam si eu, Valentin. ;)", "Solution_7": "Fie. In primul rand nu este necesar ca limita sa fie $0$, ci doar sa fie finita.\r\n\r\nFie $\\lim_{x\\rightarrow\\infty}f(x)=l\\in{\\bf R}$. Exista atunci $a>0$ astfel incat(apropo, cui ii place prescurtarea a.i.?) $f(x)\\in[l-1,l+1],\\,(\\forall)x\\geq a$. Deci $f$ este marginita pe $[a,\\infty)$. Cum $f$ este continua, ea este marginita pe intervalul compact $[0,a]$. Prin urmare $f$ este marginita pe $[0,a]\\cup[a,\\infty)=[0,\\infty)$.\r\n\r\nSper ca toata lumea este acum multumita.", "Solution_8": "[i]Valentin. V.[/i], aici nu sunt de acord cu tine. Aceasta problema era in manualul vechi de analiza (XI), daca nu ma insel. Solutia era \"cu taiatul jos\" cum foarte bine s-a exprimat Mchis68, dar care a avut rabdarea (mai bine spus \"darul\") de a detalia ulterior riguros solutia problemei. De aici se poate trage o concluzie: daca postezi un reply cu indicatii, atunci ai bunavointa sa oferi cat de curand (daca solicitantul nu a inteles \"indicatiile\") si o solutie detaliata. \r\n\r\nImi permit sa mai fac in comentariu legat de cele mentionate mai sus. Un bun manual (sau culegere de probleme) trebuie sa contina solutiile tuturor problemelor si aceste solutii sa fie prezentate pe trei nivele: [u]solutie completa[/u]; [u]indicatii[/u]; [u]raspuns final[/u] (care corespund celor trei nivele ale dialogului profesor-elev:[u]meditatie[/u]; [u]consultatie[/u]; [u]cerc mate[/u]).\r\n\r\nIata un alt exemplu (problema de manual) la nivel [u]bac[/u] sau [u]admitere[/u]: \r\n\"o functie $f: R\\rightarrow R$ continua si marginita are cel putin un punct fix.\"\r\n[u]Indicatie:[/u] se considera functia $g: R\\rightarrow R,\\ g(x)=f(x)-x$ si se aplica o consecinta a proprietatii Darboux (extinsa la capete deschise) a unei functii continue: $\\lim_{x\\to -\\infty} g(x)=\\infty$ si $\\lim_{x\\to\\infty} g(x)=-\\infty$ etc." } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "Let M, N, P, Q be the midpoints of the sides AB, BC, CD and DA of the quadrilateral\r\n ABCD respectively. Assume that X=(AX)\u2229(BQ), Y=(BQ)\u2229(CM), Z=(CM)\u2229(DN), T=(DN)\u2229(AP).\r\n Prove that S_{XYZT}=S_{AQX}+S_{BMY}+S_{CNZ}+S_{DPT}.", "Solution_1": "[quote=\"Sergey Ginsburg\"]X=(AX)\u2229(BQ)[/quote]\r\n\r\ndo you mean X=(AP)\u2229(BQ)? if it's this the solution is simple:\r\n\r\n$ [XYZT] \\equal{} [ABCD] \\minus{} [MBC]\\minus{}[NCD]\\minus{}[PDA]\\minus{}[QAB]\\plus{}[AQX]\\plus{}[BMY]\\plus{}[CNZ]\\plus{}[DPT] \\equal{} [ABCD] \\minus{} \\frac{[ACB]}{2} \\minus{} \\frac{[BCD]}{2} \\minus{} \\frac{[CDA]}{2} \\minus{} \\frac{[DAB]}{2} \\plus{} [AQX]\\plus{}[BMY]\\plus{}[CNZ]\\plus{}[DPT] \\equal{} [ABCD] \\minus{} \\frac{[ABCD]}{2} \\minus{} \\frac{[ABCD]}{2} \\plus{} [AQX]\\plus{}[BMY]\\plus{}[CNZ]\\plus{}[DPT] \\equal{} [AQX]\\plus{}[BMY]\\plus{}[CNZ]\\plus{}[DPT]$" } { "Tag": [ "probability" ], "Problem": "There are 3 groups.The 1st group contains 3 girls & and 1 boy.The 2nd group contains 2girls and 2 boys whereas the 3rd contain 1girl and 3 boys.Another group is formed by picking one student from the existing groups.What is the probability that the group formed will contain 1 boy & 2 girls.(in lowest form)", "Solution_1": "Group #1 - 3G, B\r\nGroup #2 - 2G, 2B\r\nGroup #3 - G, 3B\r\n\r\n1) Boy comes from #1, and the girls come from #2 and #3.\r\n\r\n$ \\frac{1}{4}\\times \\frac{1}{2} \\times \\frac{1}{4}\\equal{}\\frac{1}{32}$\r\n\r\n2) Boy comes from #2, and the girls come from #1 and #3.\r\n\r\n$ \\frac{3}{4}\\times\\frac{1}{2}\\times\\frac{1}{4}\\equal{}\\frac{3}{32}$\r\n\r\n3) Boy comes from #3, and the girls come from #1 and #2.\r\n\r\n$ \\frac{3}{4}\\times\\frac{1}{2}\\times\\frac{3}{4}\\equal{}\\frac{9}{32}$\r\n\r\n$ \\frac{1}{32}\\plus{}\\frac{3}{32}\\plus{}\\frac{9}{32}\\equal{}\\frac{13}{32}$\r\n\r\nanswer : $ \\frac{13}{32}$" } { "Tag": [ "symmetry", "geometry", "IMO Shortlist" ], "Problem": "Let $AB$ be a diameter of a circle; let $t_1$ and $t_2$ be the tangents at $A$ and $B$, respectively; let $C$ be any point other than $A$ on $t_1$; and let $D_1D_2. E_1E_2$ be arcs on the circle determined by two lines through $C$. Prove that the lines $AD_1$ and $AD_2$ determine a segment on $t_2$ equal in length to that of the segment on $t_2$ determined by $AE_1$ and $AE_2.$", "Solution_1": "The lines $AD_1$ and $AE_1$ determine an involution on the tangent from $C$ to the circle (the tangent different from $CA$). This means that they also determina an ivnolution on $t_2$, and since it's obvious that the infinity point of $t_2$ corresponds to itself in this involution, it means that the involution is a symmetry wrt a point (a well-known rersult), and from here the conclusion follows (we find the midpoint of $D_1'E_1'$ to be the same as that of $D_2'E_2'$, where, in general, $X'=AX\\cap t_2$).", "Solution_2": "[hide=\"Solution\"]Let $X_1$, $X_2$, $Y_1$, $Y_2$ be the intersections of lines $AD_1$, $AD_2$, $AE_1$, $AE_2$ with $t_2$. Furthermore, let $CT$ be the other tangent from $C$ to the circle. Finally, let $M=AT\\cap t_2$. Since quadrilateral $AE_1TE_1$ is harmonic, the perspectivity at $A$ onto $t_2$ yields $-1=(A, D_1;T, E_1)=(P_{\\infty}, X_1;M, Y_1)$, where $P_{\\infty}$ is the point at infinity on $t_2$. It follows that $M$ is the midpoint of segment $X_1Y_1$. Similarly, $M$ is the midpoint of segment $X_2Y_2$, so $X_1X_2=Y_1Y_2$ and the result follows. $\\blacksquare$[/hide]" } { "Tag": [ "function", "quadratics", "inequalities", "limit", "algebra", "real analysis", "real analysis unsolved" ], "Problem": "say that $ f(n) \\equal{} 12n^2 \\plus{} 6n$ and $ g(n) \\equal{} n^3$\r\n\r\nand $ f(n) < c * g(n)$, for all c > 0 and n > n0\r\n\r\nthere has to be exist an n0 that satisfies this statement\r\n\r\nhow do we find an no that is valid for this statement??", "Solution_1": "Clearly $ n_0$ must depend on $ c$. So all you have to do is put in those two functions and check. \r\n\r\nWe want $ 12n_0^2 \\plus{} 6n_0 < cn_0^3$, or $ cn_0^2 \\minus{} 12n_0 \\minus{} 6 > 0$. The quadratic equation says that the larger root of this is \r\n$ n_0 \\equal{} \\frac{6 \\plus{} \\sqrt{36 \\plus{} 6c}}{c}$\r\nand therefore if $ n$ is greater than that $ n_0$ then your inequality will hold.\r\n\r\nIf you don't want $ n_0$ to depend on $ c$, then you're going to have a hard time finding a single value that always works.", "Solution_2": "[quote=\"Xevarion\"]Clearly $ n_0$ must depend on $ c$. So all you have to do is put in those two functions and check. \n\nWe want $ 12n_0^2 \\plus{} 6n_0 < cn_0^3$, or $ cn_0^2 \\minus{} 12n_0 \\minus{} 6 > 0$. The quadratic equation says that the larger root of this is \n$ n_0 \\equal{} \\frac {6 \\plus{} \\sqrt {36 \\plus{} 6c}}{c}$\nand therefore if $ n$ is greater than that $ n_0$ then your inequality will hold.\n\nIf you don't want $ n_0$ to depend on $ c$, then you're going to have a hard time finding a single value that always works.[/quote]\r\n\r\non the book it says that no = (12+6)/c\r\n\r\nI am just confused where they find this equation?", "Solution_3": "$ 12n^2 \\plus{} 6n < 12n^2 \\plus{} 6n^2 < cn^3$\r\n\r\nif $ n > n_0$ where $ n_0 \\equal{} 18/c$. \r\n\r\nThe idea is that in order to find a simpler expression for $ n_0$, you replace $ f$ by something that's pretty similar, but simpler and larger.\r\n\r\nedit: obviously the $ n_0$ you get this way will be worse, but generally in this kind of situation you don't care too much.", "Solution_4": "[quote=\"Xevarion\"]$ 12n^2 \\plus{} 6n < 12n^2 \\plus{} 6n^2 < cn^3$\n\nif $ n > n_0$ where $ n_0 \\equal{} 18/c$. \n\nThe idea is that in order to find a simpler expression for $ n_0$, you replace $ f$ by something that's pretty similar, but simpler and larger.\n\nedit: obviously the $ n_0$ you get this way will be worse, but generally in this kind of situation you don't care too much.[/quote]\r\n\r\nokay the book also quotes that too, but how do I get from \r\n\r\n$ 12n^2 \\plus{} 6n < 12n^2 \\plus{} 6n^2$\r\n\r\nsorry I am a little bit long at understanding something", "Solution_5": "Well, it should be obvious that $ 12n^2 \\plus{} 6n < 12n^2 \\plus{} 6n^2$ if $ n > 1$. So, if you have some inequality like \r\n$ 12n^2 \\plus{} 6n^2 < cn^3$ (1)\r\nthen you put that one and the previous one together and you get \r\n$ 12n^2 \\plus{} 6n < cn^3$ (2)\r\nSo, the objective was to find $ n_0$ such that (2) is true for all $ n > n_0$, right? The idea here is that all we need to do is find $ n_0$ such that (1) is true for all $ n > n_0$, and then that one will work for (2) as well! \r\n\r\nBut (1) is easy: \r\n$ 18n^2 < cn^3$\r\n$ 18 < cn$\r\n$ 18/c < n$\r\n\r\nSo you just say $ n_0 \\equal{} 18/c$, and the inequality holds for all $ n > n_0$. Since you've found $ n_0$ for (1), this also works for (2), and you are done!", "Solution_6": "In context, this is precisely about showing that $ \\lim_{x\\to\\infty}\\frac{f(x)}{g(x)}\\equal{}0$, by the definition.\r\n\r\nWe have $ \\frac{f(x)}{g(x)}\\equal{}\\frac{12}{x}\\plus{}\\frac{6}{x^2}$; each term clearly goes to zero, and they're easy to estimate separately. Now, how do we prove the addition part of the arithmetic theorem? The usual way is to estimate each piece to $ \\frac{\\epsilon}{2}$, and combine by the triangle inequality.\r\n\r\nWhen is $ \\frac{12}{n}<\\frac c2$? We need $ n> \\frac{24}{c}$.\r\nWhen is $ \\frac{6}{n^2}<\\frac c2$? We need $ n> \\sqrt{\\frac{12}{c}}$\r\n\r\nWe can therefore choose $ n_0\\equal{}\\max\\{\\frac{24}{c},\\sqrt{\\frac{12}{c}}\\}$. Since $ \\frac{24}{c}\\ge \\sqrt{\\frac{12}{c}}$ when $ c\\ge \\frac1{48}$, we could use the simpler $ n_0\\equal{}\\max\\{\\frac{24}{c},\\frac12\\}$.\r\n\r\nThere is no obligation to have the smallest $ n_0$ possible, or even close to it. Crude estimates of the error are perfectly fine." } { "Tag": [ "trigonometry", "geometry", "geometric transformation", "reflection", "angle bisector", "geometry proposed" ], "Problem": "In an isosceles triangle $ ABC , (AB \\equal{}AC)$ we have $ \\angle A \\equal{} 100$. In the interior ot the triangle we take \r\n [b]a)[/b] a point $ P$ such that $ \\angle PAC\\equal{} 20 , \\angle PCA \\equal{} 10$ \r\n [b]b)[/b]a point $ M$ such that $ \\angle MAB \\equal{} 20 , \\angle MCB \\equal{} 10$ .\r\n Prove that : \r\n [b]a)[/b] $ BA\\equal{}BP$ [b]b)[/b] $ MA\\equal{}MB$\r\n \r\n[color=red][b]NOTE[/b][/color]\r\n\r\n The trigonometric form of Ceva's theorem gives simple solutions , to both questions, since it is sufficient to prove that $ P,M$ are on the bisector of angle $ B$. \r\n[color=red][b] I'd like to see a synthetic solution![/b][/color]", "Solution_1": "but why serching for a syntetic solution when with ceva on 1 staps we conclude?\r\n$ \\frac {\\sin \\angle PBC}{\\sin \\angle PBA} \\equal{} \\frac {\\sin20}{\\sin80}\\frac {\\sin 30}{\\sin{10}} \\equal{} \\frac {\\sin 20}{2 \\sin10 \\cos10} \\equal{} 1$ so P stay on the angle bisector of B and the same for M because P an M are isogonal cojugate.", "Solution_2": "$ AP$ meets the bisetor of $ \\angle B$ at $ P'.$ $ \\triangle ABP'$ is isosceles with $ BA \\equal{} BP'$ base $ \\angle P'AB \\equal{} 80^\\circ$ and vertex $ \\angle ABP' \\equal{} 40^\\circ.$ Bisector of the $ \\angle ABP'$ cuts bisector of the $ \\angle CAB$ at $ Q.$ $ QA \\equal{} QP',$ $ \\triangle AQP'$ is isosceles with base $ \\angle P'AQ \\equal{} 30^\\circ,$ vertex $ \\angle AQP' \\equal{} 120^\\circ.$ $ QP'$ cuts $ AC$ at $ C'.$ $ \\angle AC'Q \\equal{} 180^\\circ \\minus{} (\\angle CAQ \\plus{} \\angle AQP') \\equal{} 10^\\circ$ $ \\Longrightarrow$ line $ QP'C'$ is reflection of line $ QB$ in $ AQ$ and $ C' \\equiv C, P' \\equiv P$ are identical.", "Solution_3": "For reading easiness, it is a good idea to combine the [b][size=100]yetti\u2019s[/size][/b] solution, with the configuration of a regular [b][size=100]9-gon[/size][/b].\r\n\r\nI think, after the result $ \\angle AQP' \\equal{} 120^{o},$ we can also say that the line segment $ QP',$ as the reflection of $ QB$ with respect to $ AQ,$ passes through the point $ C,$ as the reflection of $ B,$ wrt $ AQ.$\r\n\r\nSo, because of $ \\angle ACQ \\equal{} 10^{o},$ from the triangle $ \\bigtriangleup AQC,$ where $ \\angle AQC \\equal{} 120^{o}$ and $ \\angle CAQ \\equal{} 50^{o},$ we conclude that $ CP'\\equiv CP$ $ \\Longrightarrow$ $ P'\\equiv P$ and we are done.\r\n\r\nKostas vittas.", "Solution_4": "Construct equilateral triangle DPC, D and A are on different sides of BC\r\nWe have PD \u22a5 BC , \u25b3PBC and \u25b3DBC are congruence => PB=BD\r\n\u2220CPQ=\u2220PAC + \u2220PCA =20+10=30 => \u2220DPQ=60-\u2220CPQ=30 \u25b3APD and \u25b3APC are congruence => AD=AC \u2220BAD=100-2*20=60 => \u25b3ABD is equilateral triangle => AB=BP", "Solution_5": "We can construct P' is the intersection of the bisector of $ \\hat{B}$ with the ray CP and prove P=P' by proving $ \\hat{P'AC}$ =20.\r\nConstruct MBC is the equilateral triangle, so $ \\bigtriangleup$BMA= $ \\bigtriangleup$BCP' then BA=BP'. So we have $ \\hat{BAP'}$ =80 then $ \\hat{P'AC}$=20, result is P'=P so BA=BP.", "Solution_6": "[quote=\"plane geometry\"]Construct equilateral triangle DPC, D and A are on different sides of BC\nWe have PD \u22a5 BC , \u25b3PBC and \u25b3DBC are congruence => PB=BD\n\u2220CPQ=\u2220PAC + \u2220PCA =20+10=30 => \u2220DPQ=60-\u2220CPQ=30 \u25b3APD and \u25b3APC are congruence => AD=AC \u2220BAD=100-2*20=60 => \u25b3ABD is equilateral triangle => AB=BP[/quote]\r\n\r\n[color=red][b] YES ! [/b][/color]\r\n\r\n I was seeking excactly such a solution ! Could someone try for a similar solution to question b) ?\r\n\r\n[b] Note[/b]\r\n\r\n Thanks to all solvers for their nice solutions.\r\n\r\n Babis", "Solution_7": "[quote=\"stergiu\"][quote=\"plane geometry\"]Construct equilateral triangle DPC, D and A are on different sides of BC\nWe have PD \u22a5 BC , \u25b3PBC and \u25b3DBC are congruence => PB=BD\n\u2220CPQ=\u2220PAC + \u2220PCA =20+10=30 => \u2220DPQ=60-\u2220CPQ=30 \u25b3APD and \u25b3APC are congruence => AD=AC \u2220BAD=100-2*20=60 => \u25b3ABD is equilateral triangle => AB=BP[/quote]\n\n[color=red][b] YES ! [/b][/color]\n\n I was seeking excactly such a solution ! Could someone try for a similar solution to question b) ?\n\n[b] Note[/b]\n\n Thanks to all solvers for their nice solutions.\n\n Babis[/quote]\r\nExactly the same way", "Solution_8": "For the result $ MA \\equal{} MB,$ we can also say that we have $ M'A \\equal{} M'B$ $ ($ $ \\angle ABM' \\equal{} \\angle BAM' \\equal{} 20^{o}$ $ ),$ where $ M'$ is the intersection point of the line segment $ AM,$ from the line segment $ BP,$ as the angle bisector of $ \\angle B.$\r\n\r\nThen, because of the point $ M'$ is the isogonal conjugate of $ P,$ with respect to the given triangle $ \\bigtriangleup ABC$ ( easy to prove ), we conclude that $ \\angle BCM' \\equal{} \\angle ACP \\equal{} 10^{o}$ and so, we have $ CM'\\equiv CM.$\r\n\r\nHence, we conclude that $ M'\\equiv M$ and we are done. \r\n\r\nKostas Vittas.", "Solution_9": "There are many problem, which are similarity with this problem. The main idea often be used is construct a suitable equilateral triangle." } { "Tag": [], "Problem": "Try this problem on your own. (Note: This is not a trick question of any sort like the turtle or the circle problem I posted earlier)\r\n\r\nProve that there exist two positive prime integers with at least 10,000 composite integers in between them, and give two possible numbers.", "Solution_1": "I think this may be slightly misworded.. as I think 2 and any huge prime would satisfy this condition :).. perhaps you mean to say prove there exists 10000 consecutive composite numbers?", "Solution_2": "Arne posted this problem earlier. I think it was on the Advanced board although it probably should have been Intermediate.", "Solution_3": "sorry, i mean consecutive composites.", "Solution_4": "o, ok, i didn't know somebody posted that already. i saw it at a talk at presentation high.", "Solution_5": "Oops. I was wrong. Arne's problem was different. Sorry. It's just that I used the same argument to solve that problem as I would to solve this one. That's why I was confused.\r\n\r\nSPIKEzZ, try not to use too much SFBA terminology. I know what you mean when you say you saw it at a talk at Presentation High because I am from SFBA and went to those talks, but it's just gibberish to most others.", "Solution_6": "Is this the argument?\r\n\r\nWe take 2+10000!, 3+10000!, ..., 10000+10000!. These are all composite.\r\n\r\nPretty cute. :D", "Solution_7": "Well, that's only 9999, but it's the right idea." } { "Tag": [ "Euler", "number theory proposed", "number theory" ], "Problem": "Find $n$ positive integer so that\r\n\r\n $n = 2 \\phi(n)+150$\r\n\r\nwhere $\\phi$ is Euler number.", "Solution_1": "[quote=\"RDeepMath91\"]Find $n$ positive integer so that\n\n $n = 2 \\phi(n)+150$\n\nwhere $\\phi$ is Euler number.[/quote]\r\nWe have $2|n$, and $4\\not |n$ (because $\\phi(n)$ is even, when n>2).\r\nIf prime p|n and p>5 then $p^{2}\\not |n$. Let $n=2m_{1}m_{2},m_{1}=3^{k_{1}}5^{k_{2}},m_{2}=p_{1}p_{2}...p_{s}$. It is easy to chek, that $k_{1}\\le 2,k_{2}\\le 3, s\\le2$.\r\nIt give $m_{1}m_{2}-\\phi(m_{1})\\phi(m_{2})=75$\r\nIf $m_{1}=1$ we have s=2 and $p_{1}+p_{2}=76$. There are 3 solutions\r\n$m_{2}=29*47,23*53,17*59,n=2*29*47,2*23*53,2*17*59$.\r\nLet $m_{1}>1$. If s=0 by chek $m_{1}=3^{2}*5^{3}$ we know, there are no solution. If $m_{1}\\ge 75$ no solution, because $m_{1}m_{2}-\\phi(m_{1})\\phi(m_{2})>m_{1}(m_{2}-\\phi(m_{2}))>75$. All variants are \r\n$m_{1}=3,\\phi(3)=2,m_{1}=5,\\phi(5)=4,m_{1}=9,\\phi(9)=6,m_{1}=15,\\phi(15)=8,m_{1}=25,\\phi(25)=20,m_{1}=45,\\phi(45)=24$\r\nIf s=1 $m_{1}p-\\phi(m_{1})p=75-\\phi(m_{1})$. It give $m_{1}=3,p=73,m_{1}=5,p=71,m_{1}=9,p=23,m_{1}=25,p=11$ 4 new solutions\r\n$n=2*3*73,2*5*71,2*9*23,2*25*11$.\r\nIf $s=2$ we have $(m_{1}-\\phi(m_{1}))p_{1}p_{2}+\\phi(m_{1})(p_{1}+p_{2}-1)=75$\r\nhave not solutions.", "Solution_2": "Thx...\r\nThat's very good.... :D" } { "Tag": [ "Support", "geometry", "calculus", "integration", "Yale", "college", "articles" ], "Problem": "GOOD teachers might make a whole lot more money if parents, the clients of the school system, could shop readily for teachers. But the schoolteacher unions oppose voucher plans and most other policy proposals that help parents shop for teachers because many union members have a rational fear that their salaries would drop if clients could choose teachers on the basis of quality of teaching.", "Solution_1": "I know this is getting way off topic and what I will say might seem really crazy but I think that the reason that some people get more then others and not necessarily that, like how can you tell how good is something? you probably cant exactly, so there is a problem in quantification. I think that the reason that most problems exist and are very difficult to solve is becuase it is impossible to determine their exact value, I call this my theory of quantification problems. LOL, I am just kidding", "Solution_2": "[quote=\"hello\"]I know this is getting way off topic and what I will say might seem really crazy but I think that the reason that some people get more then others and not necessarily that, like how can you tell how good is something? you probably cant exactly, so there is a problem in quantification. I think that the reason that most problems exist and are very difficult to solve is becuase it is impossible to determine their exact value, I call this my theory of quantification problems. LOL, I am just kidding[/quote]\r\n\r\nYou make a good point. This is part of the reason we need more problem solvers to go around in fields outside of mathematics. Usually when we quantify problems correctly, the solution either becomes apparent to the public or is accepted by the public over time due to confidence from the academic or expert communities.\r\n\r\nThere are many problems that have not been quantified or quantified well that society will battle with for years to come. Teacher productivity is one of them.", "Solution_3": "[quote=\"tokenadult\"]GOOD teachers might make a whole lot more money if parents, the clients of the school system, could shop readily for teachers. But the schoolteacher unions oppose voucher plans and most other policy proposals that help parents shop for teachers because many union members have a rational fear that their salaries would drop if clients could choose teachers on the basis of quality of teaching.[/quote]\r\n\r\nTokenadult, there are two assumptions embedded in your statement which I'd like to call into question. \r\n(1) School systems have clients.\r\nThe use of the word \"client\" assumes a corporate model for school systems. A corporate model is inappropriate here; school systems are a public service, not a corporation. Similarly, public universities provide a public service, and both public and private universities provide educational opportunities (not guarantees of diplomas for class attendance).\r\n(2) Parents are able to tell which teachers are \"good.\"\r\nI totally believe that you personally can assess the quality of a school teacher, but I don't think that a very high percentage of the general parent population has the experience or insight to judge this. There are gross misperceptions as to what teachers' workloads and responsibilities and training and support actually are...\r\n\r\n\r\nAlso, I doubt that knowing which teachers are \"good\" would make a difference in salaries or in overall educational quality. We don't have enough public funding to change the former or enough well-trained teachers to change the latter.", "Solution_4": "I don't see how the points you made affect tokenadult's argument. It doesn't matter whether or not the corporate analogy is literally accurate or not, and as long as you can judge the teacher for your own needs, the whole workload/training and such thing would evolve to meet that. It's selfish, but that's how America works in general.", "Solution_5": "[quote=\"smbelcas\"]Tokenadult, there are two assumptions embedded in your statement which I'd like to call into question. \n(1) School systems have clients.\nThe use of the word \"client\" assumes a corporate model for school systems. A corporate model is inappropriate here; school systems are a public service, not a corporation.[/quote]\r\n\r\nCould you lay out the additional major premises you need here to tighten up your argument? I have a feeling I disagree with them, but let's be sure before I further discuss the point. \r\n\r\nJust as one example, no one disputes that lawyers have clients, and a few lawyers even think that the practice of law is a public service ;) but hardly any law firm is organized as a corporation. \r\n\r\nAnd even if we agree about what is, couldn't we also agree that what should be is different from what now is?", "Solution_6": "[quote=\"smbelcas\"]\nTokenadult, there are two assumptions embedded in your statement which I'd like to call into question. \n(1) School systems have clients.\nThe use of the word \"client\" assumes a corporate model for school systems. A corporate model is inappropriate here; school systems are a public service, not a corporation. Similarly, public universities provide a public service, and both public and private universities provide educational opportunities (not guarantees of diplomas for class attendance).[/quote]\n\nYou assert by fiat that schools are a public service, but that's not only untrue in many observable cases, but it is false by definition. You cannot have a public service of a scarce resource and if you've ever walked down the halls of a public school that isn't top-notch, you would know that education is still a scarce resource. In fact, there is no such thing as a company that profits from any activity if there are no scarce resources. \n\nEconomics is the study of scarce resources. To me this means that private education models are extremely appropriate -- and even necessary. Besides, where does the \"magic line\" begin and end that allows for private university but not private elementary or secondary education? Why the double-standard?\n\nI know that many people have a knee-jerk reaction to corporate [insert anything here], but unless we're specifically critiquing poor management of individual leaders, all that getting corporations involved in schooling does is creates a fiduciary duty for educators to educate (hence the term \"client\"). As it is, teachers in most school systems have little incentive to do their job. Two summers ago I played ultimate frisbee with a math teacher from Chula Vista who stated quite frankly that she hates math and works just to pick up the paycheck because she can't imagine she can work a better job. Unions are strong and it's difficult to filter those kinds of teachers out of the system.\n\nAdditionally, every major study indicates that even as we strip out socioeconomic variables, [url=http://news.yahoo.com/news?tmpl=story&u=/ap/20050525/ap_on_re_us/charter_school_study_1]charter schools[/url] and [url=http://aspe.os.dhhs.gov/hsp/97trends/ea2-3.htm]private schools[/url] are doing much better jobs than public schools with secular private schools leading the way. Public schools even outspend their private counterparts (something like 7k to 4.5k per student as of just a few years ago).\n\nHaving been through a public school system that is often ranked very high nationally among public schools in a variety of statistics, I didn't feel like a took a single course that challenged me or interested most of my peers. Many of the top students whose 1500+ SAT scores and 8ish AP exams push up national rankings are taking summer college classes, attending corporate academic programs, or have private tutors. \n\nEven at the top public schools in California, motivated students are bored and frustrated as evidenced by what we see on AoPS blogs. It saddens me the number of times I read something along the lines of \"I am so bored in school.\"\n\nIt is very very hard for me to imagine a pretty picture of public education. Even our leaders of industry such as Bill Gates are telling us very frankly that our kids are not being well enough educated to hire.\n\nBeyond that, I think that the assertion that corporations should stay out of education is offensive. What is AoPS? I would gladly contrast our model for private math education with that of most any public school.\n\nOne last question: If public schools are doing an adequate job, why do [url=http://www.faegre.com/articles/article_1601.aspx]H-1B visas for foreign workers without post-graduate degrees run out on the very first day of the fiscal year[/url]?\n\nIf you want a technology job in this country, you might still be better off with an education from a struggling, resource-poor, financially strapped former Soviet satellite than from an ordinary, well-funded American public school.\n\nMy answer is that the status quo system of education has dropped the ball in preparing students for the technology age. Schools botched the job so badly, that I cannot even fathom how we could call it a service, much less a public service.\n\n(note: that's not to say I don't consider the great teachers at all schools, public and private, among the heroes of our culture. I hold them in high esteem as they struggle through a difficult system.)\n\n\n[quote=\"smbelcas\"]\n(2) Parents are able to tell which teachers are \"good.\"\nI totally believe that you personally can assess the quality of a school teacher, but I don't think that a very high percentage of the general parent population has the experience or insight to judge this. There are gross misperceptions as to what teachers' workloads and responsibilities and training and support actually are...\n\nAlso, I doubt that knowing which teachers are \"good\" would make a difference in salaries or in overall educational quality. We don't have enough public funding to change the former or enough well-trained teachers to change the latter.[/quote]\r\n\r\nSo long as we stick to a public system, salaries will more more about politics than merit. But I disagree that it doesn't affect salaries. It will as education becomes more of an enterprise. Besides that, giving up on the idea of a meritocracy in public education seems far more like an indictment of the system than anything else.\r\n\r\nYou're right that most parents are not involved and knowledgeable enough to make judgements about good teachers. This seems like a good reason to judge based on what the students are capable of accomplishing.\r\n\r\nDoesn't that at least beg us to experiment more to see if non-public forms of education get the job done better?", "Solution_7": "[quote=\"probability1.01\"]It's selfish, but that's how America works in general.[/quote]\r\n\r\nFortunately, unless a situation is a prisoners dilemma, self interest results in meeting mutual needs. This drives all cooperative economics. Certainly education can and should be a win-win game.\r\n\r\nI hope the two general economic \"camps\" (corporate and anti-corporate) can come to grips with the fact that their differences are mostly a matter of viewing only one side or the other of game theoretic situations.\r\n\r\nCooperation can consistently take place to the mutual benefit of large groups of people so long as people are only working with others that they trust. \r\n\r\nOn the other hand, the existence of scarce resources in a world in which some people do not act in trustworthy ways means that prisoners dilemmas necessarily exist. \r\n\r\nWhen I stand back and evaluate what kind of system would allow for Nash equilibriums of cooperative scenerios and promote iterated cooperation in regards to scarce resources, I cannot imagine how this could be achieved without the option of families to choose those with whom they trust and cooperate.\r\n\r\nForced cooperation does not promote tit-for-tat, an evolutionarily robust strategy. It's a quick fix that makes the voicing of self-interest look greedy when it's exactly what makes our species strong.\r\n\r\nWe just need to get down to the job of educating people so that they understand that building a [i]better world is a matter of self-interest[/i].", "Solution_8": "I recently read a few books by Ayn Rand such as Atlas Shrugged and The Fountainhead. MCrawford, you seem to have very similar views. Do you know anything about Objectivism? I agree with you entirely, by the way.", "Solution_9": "We're pretty off topic now but I'd like to comment on the charter school issue. Since 1993, over 100 charter schools have opened in Colorado. The results have been mixed. The worst schools suffer from poor management and lack of focus. The best charter schools outperform the neighborhood schools by a wide margin.\r\n\r\nMy kids have benefited from attending one of these fine charter schools, which is not only one of the best performing middle schools in the state but has repeatedly won national recognition as one of the finest schools in the country. Does our community applaud this school and use it as a shining example for other schools in the district? On the contrary, our charter school is repeatedly attacked for elitism in spite of a lottery-based admissions policy. The local school board (and teachers union) is very antagonistic toward our school and would love to see it disappear. Fortunately the state of Colorado currently supports charter schools and encourages their continuation. Unfortunately our local school board has almost complete control over charter school funding! This means our school receives 70% of the per-pupil funding awarded to non-charter schools. And it still outperforms year after year, as do other charter schools in our district.\r\n\r\nSince the introduction of charter schools to our community, the overall performance of all the schools has improved because they're now competing for students. Some of this competition has had negative consequences but in general I think our community and our children have benefited from the increased number of choices available. We no longer accept the old \"one size fits all\" policy. There are now Montessori schools, Core Knowledge schools, arts schools, etc., to choose from. Parents now look at standardized test results and other performance data before selecting a school for their children.\r\n\r\nOne of the biggest problems with the introduction of charter schools has been an increase in ethnic and economic segregation. The district could handle this problem by providing free busing for low-income students, making it easier for them to attend charter schools. Instead they're proposing a takeover of the charter schools admissions process in order to discourage families from choosing non-neighborhood schools. It's been very frustrating watching the local school board repeatedly place handcuffs on our charter schools, hoping they will fail.\r\n\r\nBecause of the lower per-pupil funding, charter schools in our district cannot match the teacher salaries offered by the regular public schools. Unfortunately this has resulted in high teacher turnover. Fortunately at our school we've had a loyal group of excellent teachers who have stayed for years because they believe in the mission of our school (to challenge students with high academic standards) and because they appreciate having more autonomy in their classrooms. It's a struggle to manage costs on such small budgets but our local charter schools have succeeded, mostly by reducing administrative overhead. If charter schools were able to match regular teacher salaries, I think they would be able to attract more of the best teachers.\r\n\r\nSome charter schools in other communities have suffered from management problems but in our community I think charter schools have been very successful in providing alternatives for families dissatisfied with their neighborhood schools and raising academic standards across the entire school district.", "Solution_10": "Thank you for posting your experiences.\r\n\r\n[quote=\"wild octagon\"]\nOne of the biggest problems with the introduction of charter schools has been an increase in ethnic and economic segregation.[/quote]\r\n\r\nIt seems to me that this is a problem in schools in most areas of the country no matter what system is used (unless integration is forced). I remember the bus students at my own high school being lost in the process of fitting in. For the most part it was like there were two distinct cultures in the same school -- the larger, wealthier local culture, and the less affluent bus-student culture. While I'm sure that helped some of the students, there was no cultural integration.\r\n\r\nI think cultural integration occurs more naturally when better schools simply elevate more students of different race backgrounds. Perhaps the first step is on the local level where standards would be raised for low-income schools leading to great economic development. While the benefits of cultural integration would not be immediate, there may be no good immediate solution (even if many well-meaning people want there to be one).", "Solution_11": "I would love to see more charter schools established near low income neighborhoods, which would reduce white flight and segregation. The problem is the parents who take the initiative to start new schools tend to live in wealthier, more educated, neighborhoods. Maybe some of these successful charter school founders should share their experiences with parents in low income neighborhoods.\r\n\r\nIn Colorado, one simple way to improve the performance of low-performing schools would be to eliminate ESL and bilingual programs or offer them as afterschool options only. As the child of immigrant parents who did not speak English, I sympathize with the desire to raise bilingual and bicultural children but I firmly believe that bilingual education should take place outside of school. All young Americans deserve to learn English in school and to learn it well. How else will they be able to compete for jobs when they're adults? The last time I read a debate about bilingual education in the local paper, the Hispanic parents were against it while the English-speaking parents were in favor because they wanted their children to be exposed to multiculturalism!", "Solution_12": "[quote=\"wild octagon\"]I would love to see more charter schools established near low income neighborhoods, which would reduce white flight and segregation. The problem is the parents who take the initiative to start new schools tend to live in wealthier, more educated, neighborhoods. Maybe some of these successful charter school founders should share their experiences with parents in low income neighborhoods.\n[/quote]\r\n\r\nUnfortunately, lower income parents are not nearly as likely to have the time and resources (including know-how and social connections) to initiate such action.\r\n\r\nI often wonder if a company might develop that specializes in providing the know-how and a plan for such schools. States could hire them to help clean up the problem. This might solve the problem of states spending money that isn't used well in some schools due to the level of poverty and chaos.", "Solution_13": "[quote]I often wonder if a company might develop that specializes in providing the know-how and a plan for such schools. States could hire them to help clean up the problem. [/quote]\r\n\r\nSuch companies exist already---notably Edison Schools, which made the flamboyant move of hiring the president of Yale, Benno Schmidt, to head the company over a decade ago. The company currently runs charter schools serving a quarter million kids in 20 states. http://www.edisonschools.com/overview/ov0.html \r\n\r\nThere's a number of other companies doing the same sort of thing on various scales around the country.\r\n\r\nHowever, such businesses face lots of obstacles--to begin with, big poitical opposotion from teachers unions and government employee unions generally.\r\n\r\nIn NY, teacher unions and allies are the single biggest political donor block--in terms of their contributions to state legislative campaigns. Most of the very large troubled urban school systems are located in states that happen to have heavy union organization. And privately operated charter schools rarely save the district money in the short run, because the district is typically still stuck with heavy fixed costs for bond issues, teacher pensions, etc So when some kids go off to privately operated charters (followed by their share of the tax dollars), the total school district budget often rises and the local taxpayers often end up paying more, at least in the short run.\r\n\r\nIn states where teacher & government employee unions are not so strongly organized, it's a bit easier, but there are stilll difficulties. How to measure value-added is controversial, especially if there is a lot of student/family turnover (which often happens in low-income areas.)\r\n\r\nThere was a lot of excitement and sizzle back when the Edison Schools project was launched in the early 90s. It was being sold as \"the next big hot thing\" to investors. I can imagine investors must be far more jaded now.\r\n\r\nIt's hard to do a great job of educating kids who come from very troubled and chaotic family backgrounds, who may have a lot of health problems, who move around a lot, who have parents with alcohol and drug problems, domestic violence, etc. Just keeping some semblance of order in the classroom isn't easy. \r\n\r\n And if your company's schools manage to educate some kids, it's often hard to prove or establish the value-added to a skeptical public constituency--you may have had a real impact on kids who have moved out of your school by the time the test comes around. And high-stakes standardized tests present enormous incentives and opportunities for cheating/integrity issues, especially if corporate profits are at stake. \r\n\r\nDifferent scholars have looked at the same data from charter schools and tuition voucher plans and come up with diametrically opposite conclusions as to whether those schools do in fact do a better job of educating kids than the regular public schools. After all, it's necessary to adjust for the fact that the sample of students whose parents choose to enroll them may not be representative. Reasonable people can differ about how to do the statistical adjustments necessary--and the public winds up confused in the end.\r\n\r\n(I'm struck by the parallels to the drug tests run by the big pharmaceutical companies to support the introduction of the new medicines they've invented. There seems to be a lot of concern about the integrity of the research protocols, given that the drug companies fund and sponsor the testing and have an incentive to make sure that the studies come out in a way that will present their new drug in the most favorable possible light. And yet, drug testing can be a lot more straightforward than educational testing. You can do double-blind placebo-controlled studies for drug tests. You can't really do random-assignment double-blind placebo-controlled studies for educational interventions. Whatever the flaws in drug-testing methodology, I suspect the flaws in educational-program--testing methodology are far greater. ) \r\n\r\nIn the end, I'm pessimistic about the idea that investor-backed charter schools are going to be wildly successful on a large scale in turning out incredibly well-educated kids. \r\n\r\nThe discipline problems in school are huge. And school administrators have a limited amount of ability to control those discipline problems if they have to take \"all comers,\" as charter schools are generally required to do. The liability exposure if bad stuff happens to a kid in a school is big too. \r\n\r\nI'm not against market solutions in general, but as a a potential investor looking for a place to put my money, I'd cast a jaundiced eye on private companies that think they can make a quick buck delivering high quality education in troubled neighborhoods.\r\n\r\nThe big success story, in my eyes, are the struggling Catholic schools that manage to operate on a shoestring, with dedicated and caring teachers who work for a pittance (lower salaries and MUCH lower benefits), but who manage to create an orderly environment in which kids who want to learn can do so--because they have the power to expel disruptive kids. A big part of the reason for their success is the fact that the kids' parents are committed to the enterprise---they may not be able to pay a lot of tuition, but whatever they do pay is a real sacrifice and means that they have a commitment and stake in their kids' education. They are going to make sure their kid makes the most of the opportunity. If the kid gets in trouble at school, the kid knows he's going to be in trouble at home too. Increasingly, a majority of the kids attending urban Catholic schools are not Catholic---but parents of the enrolled kids see the school as a haven of safety and order and caring in which their kids can learn. But the impetus for those schools doesn't come from a profit motive--it comes from a sense of a larger mission of service and dedication to the community. And the success of those schools comes from the fact that the sense of mission is shared by the parents as well as the school staff.\r\n\r\nI'm no longer Catholic, but I'm awed by the personal charisma and dedication of some of the teachers I know who work in inner city schools (both public schools and Catholic schools.) It is possible for dedicated and caring public school teachers to make a huge difference in urban schools--but too many of the best burn out. The daunting and silly educational bureaucracy frustrates a lot of the best public school teachers. Susan Ohanian, a former Teacher of the Year from a troubled urban school district in NY, writes eloquently of the idiocy of the bureaucratic constraints she faced as a teacher.", "Solution_14": "sophia,\r\n\r\nI realized right after I posted that I had heard of some companies like that, but had never been in an area where they operated (at least to the extent that it was noticeable).\r\n\r\nI agree with what you say. Dedicated staff are the key, not profit motive. Unfortunately I think Catholic schools are not easily replicated in creating a missionary environment, so I keep looking for somebody who will devise a solution that scales. It seems difficult to create a large scale solution without profit motive. At the very least, an economically robust system is likely to stand the test of time to provide worthy experimentation in education, particularly when the obstacles in experimentation as so great as you mention.\r\n\r\nIn addition to the problem of selecting an appropriate vehicle for action, you hint at the problem of a nonexistent incentive structure. Unions and therefore government have lower incentive for supporting the creation of alternative environments, particularly those that create competition. That's a shame given that evolution in virtually any environment is dictated by competition for resources (and I think that students are often implicitly viewed as \"resources\" to teachers unions and therefore to politicians).\r\n\r\nAdditionally, the incentives for parents to enroll students in experimental or new schools may be skewed for many of the reasons you mention.\r\n\r\nA problem solving approach might be a matter of providing a new, more sane incentive structure, but it's hard for me to imagine that happening within the status quo system. And it's a mind-bender to formulate a solution from outside the system which often views itself as both a paternal entity and with great defensiveness.", "Solution_15": "Wildoctagon, what schools in NYC did you attend?", "Solution_16": "[quote=\"JBL\"]Wildoctagon, what schools in NYC did you attend?[/quote]Uh ... I'd rather not say, if you don't mind. Few have ever heard of my elementary school and junior high. My magnet HS ([i]not[/i] Stuyvesant, Bronx Science, Brooklyn Tech, or Hunter HS) was a very good school with an excellent math dept. (btw - this was eons ago, probably before you were born.)", "Solution_17": "Fair enough. Just as an aside, I don't think your suggestion for parental involvement in the poorest city areas is realistic unless we implement a far superior social safety net. And until we do that, schools will continue getting stuck with issues that they really have no business or expertice dealing with and that undoubtedly take away from the academic programs.", "Solution_18": "[quote=\"JBL\"]Fair enough. Just as an aside, I don't think your suggestion for parental involvement in the poorest city areas is realistic unless we implement a far superior social safety net. And until we do that, schools will continue getting stuck with issues that they really have no business or expertice dealing with and that undoubtedly take away from the academic programs.[/quote]\r\n\r\nI think admitting that we don't have a system for dealing with those locales means admitting that we can't school them.\r\n\r\nI'm not saying I disagree. Unfortunately I agree with you.\r\n\r\nMy thought is that continued economic improvement will hopefully generate more jobs to get such locales more integrated into the social fabric.", "Solution_19": "[quote=\"JBL\"]Fair enough. Just as an aside, I don't think your suggestion for parental involvement in the poorest city areas is realistic unless we implement a far superior social safety net. And until we do that, schools will continue getting stuck with issues that they really have no business or expertice dealing with and that undoubtedly take away from the academic programs.[/quote]Unfortunately inner city schools have to make it their business to deal with social issues as well as academic ones. I realize there are no easy solutions to inner city education problems but there have been a few success stories like East Harlem's [url=http://www.thenotebook.org/editions/2003/fall/harlem.htm][u]District 4[/u][/url] that have demonstrated the progress that can be made when bureaucratic micromanagement is eliminated and school choice is offered to families, teachers and principals.", "Solution_20": "I think we should also keep in mind that inner city schools aren't the only ones with major problems. When I hear discussions about bad schools I always hear people bring up inner city schools as examples. But it may be that, as a whole, poor rural schools are even worse off academically.\r\n\r\nSouth Carolina for instance has the lowest average Math SAT score in the country (I think around 475), and a high school graduation rate just barely over 50%.\r\n\r\nSchool choice can only do so much for people in rural areas because there may only be one or two high schools within a gazzillion miles of them. Plus, its harder to get good teachers to come to rural areas than to come to schools in urban areas.\r\n\r\nWhen you look nation wide I bet there are just as many kids who live in these type of areas than who in poor urban environments - but no one ever sees or talks about them. This is just an example of the scale and diversity of educational situations. So even in reform tactics, one size doesn't fit all.", "Solution_21": "The fact that school choice only works when there are choices seems like a great reason to continue to develop better internet resources. At the very least the internet is likely to begin supplementing the education of more and more of those students.", "Solution_22": "For students with high speed connections, the Internet is a great resource. I recently discovered MIT's OpenCourseWare [url=http://ocw.mit.edu/OcwWeb/Global/OCWHelp/avocw.htm][u]video lectures[/u][/url] in physics, linear algebra, and a few other subjects. We can now audit, for free, courses taught by top MIT professors. Imagine if all our top universities followed MIT's example and offered free video lectures on the Internet.", "Solution_23": "Yeah, I love MIT Open Sourceware. I've probably spent 100 hours reading from their lecture notes over the past couple of years when I've had the itch to learn about something from their huge online course selection.", "Solution_24": "The Intro to Bio course includes lectures by Eric Lander, former US Math Olympiad team member.", "Solution_25": "[quote=\"gauss202\"]I think we should also keep in mind that inner city schools aren't the only ones with major problems. When I hear discussions about bad schools I always hear people bring up inner city schools as examples. But it may be that, as a whole, poor rural schools are even worse off academically.[/quote]As a native New Yorker, I don't know much about rural schools but I'm interested in learning more. I assume they don't have to deal with the crime problems prevalent in urban ghettoes. Are there problems specific to rural schools that are not found in urban and suburban settings?", "Solution_26": "[quote=\"wild octagon\"]Are there problems specific to rural schools that are not found in urban and suburban settings?[/quote]\r\n\r\nIt's damn hard to attract teachers. Students must travel over great distances. Funding can be very hard to come by. Uneducated parents. Students needed to work at home (on farm, for example). \r\nNot an exhaustive list, of course :) -- urban and rural schools have a lot of overlapping problems and a lot of great differences (especially racial, at least in the northeast).", "Solution_27": "[quote=\"JBL\"]It's - hard to attract teachers. Students must travel over great distances. Funding can be very hard to come by. Uneducated parents. Students needed to work at home (on farm, for example). \nNot an exhaustive list, of course :) -- urban and rural schools have a lot of overlapping problems and a lot of great differences (especially racial, at least in the northeast).[/quote]The working on the farm issue is the only one that seems unique to rural schools. If kids have to spend hours each day on farm chores then it would be difficult for them to get homework done. On the other hand, they're learning about hard work and discipline instead of watching TV or playing video games so that should help them in the long run.\r\n\r\nUrban schools also have difficulty hiring teachers, for different reasons obviously. (I remember one NYC substitute teacher who clutched her purse tightly the whole time she was writing on the blackboard. I guess she thought we would steal her money if she put her purse down. :lol: My school must have neglected to inform her that our class was one of the good ones.) Commute time is also an issue in urban areas. In NYC, for example, it's not uncommon for middle and high school students to commute an hour or more each way on buses and trains to reach magnet schools. I did. I guess the difference with rural districts is the little kids also have to commute long distances. And of course the issue of uneducated parents is common in many areas of the country.\r\n\r\nI'm not sure what funding problem you're referring to. Are rural districts allocated less money than urban districts? In New York State the opposite has been true. For many years the state government siphoned off funds that should have gone to NYC and allocated them to wealthier districts instead.", "Solution_28": "Many of the problems rural districts face are similar but take on a different form. For example, in New York City, those students (and I too was among them) who travel long distances to school have a good transit network to travel in and could have chosen a nearer school, had they so desired. In rural districts, there is very often one single school, with only small numbers (30-100) of students per class, but covering very large geographic regions. Also, rural districts suffer brain drain to a much, much worse degree than urban districts do.\r\n\r\n[quote=\"wild octagon\"]I'm not sure what funding problem you're referring to. Are rural districts allocated less money than urban districts? In New York State the opposite has been true. For many years the state government siphoned off funds that should have gone to NYC and allocated them to wealthier districts instead.[/quote]\r\n\r\nThat's not the opposite effect, it's just a different effect. Rural districts aren't wealthy, just as urban districts aren't. Property values are low, as is the ability of residents to pay property taxes. It's the suburban districts that have constantly been funded at higher levels than urban districts in New York, but rural districts get screwed by the system as well. I don't know that they are allocated less money on a per-pupil basis than NYC and New York's other urban areas, but like the inner city, they require more state funding. You can't generate such convenient numbers (like the 5 or 6 billion dollars the state should be paying the city but isn't) for rural districts because they are much more spread out and serve many fewer students and have different needs, but it's clear in New York that the suburbs have a plum situation while much of the rest of the state, city and rural, does not.", "Solution_29": "[quote=\"qvc122\"]If you go back 100 years, most people were illiterate.[/quote]\r\n\r\nThat statement is not factually true about the United States. It surely is true about, for example, China. Horace Mann, when he promoted compulsory school attendance laws, published literacy figures for Massachusetts in his Common School Journal. Those literacy figures, from the pen of an advocate of compulsory school attendance, were above 90 percent. It's rather doubtful that literacy is much higher today than it was when Mann was writing. Mann's desire was not to bring an illiterate society into literacy, but rather to bring an increasingly diverse society into uniformity. He wanted to standardize religion and political attitudes at a time when immigration from various countries (particularly Ireland) was populating Massachusetts with people who didn't agree with Horace Mann on those issues. \r\n\r\nMuch more about the history of compulsory school attendance can be found in an [url=http://learninfreedom.org/school_state.html]online bibliography about school and state issues[/url] that I refer to from time to time." } { "Tag": [], "Problem": "[b]ROUND 1 HAS BEGUN[/b]\r\n\r\nPlayers:\r\nKlebian\r\npianoforte\r\nAnonAmoz\r\nazhang\r\nMustafa\r\n\r\nHere are the current powers:\r\n[list]\n[b]Cloned Sidekick[/b]\n\tCloned Sidekick has 50 Energy and a Power of your choice from those in the current game.\n\n[b]Curse[/b]\n The lowest bid(s) is stuck with this curse. This curse cannot be discarded. The highest bidder pays for it. You can only use one activated power per turn. You can only attack one person per turn. All normal attacks you make deal 10 less damage than normal. (THIS IS BAD, IN CASE YOU DIDN'T REALIZE IT. IF YOU DON'T BID, YOU BID 0 ON THIS, MEANING YOU AND ANY OTHER PEOPLE WHO BID 0 WIN THIS CURSE)\n\n[b]Sniper[/b]\n Global Effect (This Power may not be bid on. It affects all Heroes): Each round, Sniper makes a 10 Point Attack against the opponent with the lowest total Defense , after all other Powers are applied. \n\n[b]Assassin[/b]\n Choose one player in the game. Choice is not revealed to others until end of round. That player's attack and defense is lowered to zero for one turn. Max damage to that player's energy that turn is 50. Can only be used once.\n\n[b]Army of Noobs[/b]\nSummons army of 10 noobs who can each do 2 damage. They are controlled by the owner of this power. Once they attack, they are removed from the game. Their damage is unblockable, but no more than 5 noobs can be used on the same person. If you are attacked by them, you cannot attack next turn. \n\n[b]High-Powered Energy Shield[/b]\n\tBurn X: Add 10X to your Defense.\n\n[b]Masochist[/b]\nEach time you take Damage, gain a permanent +1 to your regular Attacks. \n\n\n[b]I think I can...[/b]\nYou can use one other hero's power (of your choice) each turn. (You can pick each turn) \n\n[b]Regeneration[/b]\nGain 3 Energy at the beginning of each round.\n\n[b]Glass Swords[/b]\nAdd 40 to each of your Attacks. Lose this power the first time you sustain any damage. \n[/list]", "Solution_1": "Please nobody take Pseudonite, that one is just cruel as a cucumber.", "Solution_2": "Wait what the heck, you chose the powers??????", "Solution_3": "Yes, I wanted to choose just to see how it would work out, since some people may have not played before.", "Solution_4": "Just because someone hasn't played doesn't mean that they are helpless, plus Klebian already made a game of this a while ago.", "Solution_5": "Ok... so... should I get rid of the list and allow everyone to send in two?\r\n\r\nEDIT: I will, since you two seem unhappy.", "Solution_6": "[quote=\"mustafa\"]Just because someone hasn't played doesn't mean that they are helpless, plus Klebian already made a game of this a while ago.[/quote]\r\nOh, huh, I actually did run it. I forgot I had.\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=113443 is the link.", "Solution_7": "Some people have submitted one power, one has submitted none, please submit ASAP. There will be 10 powers at the end. If I have less than 7 at the deadline(6:35 PM PST), it will be extended another day. If I have 7 or more, the remaining powers will be selected by me.", "Solution_8": "So for Army of Noobs, at the beginning of the game, a player gets 10 noobs, and can have them attack when he wants to. Max damage allowed using noobs the entire game is 10. Can noobs be attacked?", "Solution_9": "I'd have to say no. I never really thought of that, because this power was made by a player, but since they don't really have energy, they just do damage, you cannot attack them.", "Solution_10": "Yeah I didn't think so, wasn't sure.", "Solution_11": "One person has not submitted.\r\n\r\nEDIT: The deadline has passed, I will make two powers.\r\n\r\nRegeneration\r\n\tGain 3 Energy at the beginning of each round.\r\n\r\nGlass Swords\r\n\tAdd 40 to each of your Attacks. Lose this power the first time you sustain any damage.\r\n\r\n\r\nKlebian, could you please add these to the list and change the bold text to say \"Send in your bids within 48 hours\"?\r\n\r\nEDIT: I don't think you'll see this, so I'll just PM you.", "Solution_12": "Bump", "Solution_13": "Oooohh... bold text. :lol: It looks better like that, thanks!", "Solution_14": "How many people still haven't sent in bids?", "Solution_15": "Deadline is in like 10 minutes, but I'll extend it another day, if the people who have submitted agree. If not, all the people who didn't submit use half their energy on defense(the other half is not used, this is your \"punishment\" for not submitting).", "Solution_16": "It's ok with me if you extend the deadline, this game is pretty pointless if only half the people play. You might want to send out reminder pms, in case the missing people forgot about this", "Solution_17": "AnonAmoz can't play anymore, and I think it's kinda late for a replacement, so his energy will all be allocated to defense every turn, unless that makes it impossible for the game to end.\r\n\r\nMan, I think this game is failing. :o", "Solution_18": "It's a fun game, we just need more committed people.", "Solution_19": "Wow, this sucks. azhang has forfeited. :( \r\n\r\nDo you guys still wanna continue, or should I try to start a new game with more people? :o", "Solution_20": "Let's continue, the way the game was going, it was basically going to turn into a showdown between me and Pulak anyways", "Solution_21": "Alright. Round 2 results coming soon in this post. azhang's hero will become a time bomb. Every turn he doesn't take damage(his defense will be set at the highest possible, and the curse will be removed.), the counter goes up by 1. Every turn he does take damage, the counter goes up by 2. Once the counter goes past 5, he will make an all out attack on one surviving person the next turn, determined by http://www.random.org, and will continue doing so until he dies.\r\n\r\nKlebian\r\nPowers: Assassin, I think I can...(used for army of noobs)\r\nStarting Energy: 61\r\nAttacks: AnonAmoz-20, pianoforte- 6\r\nDefense: 25\r\nAttacked by: pianoforte- 4\r\nEnding Energy: 61\r\n\r\nKlebian's Sidekick\r\nPowers: I think I can...(army of noobs)\r\nStarting Energy: 40\r\nAttacks: AnonAmoz- 30 + 2 noobs(4), azhang- 2 noobs(4)\r\nDefense: 10\r\nAttacked by: nothing\r\nEnding Energy: 40\r\n\r\npianoforte\r\nPowers: Army of Noobs\r\nStarting Energy: 4\r\nAttacks: Klebian-4\r\nDefense: 0\r\nAttacked by: Klebian- 6\r\nEnding Energy: 0 DEAD(also hit by sniper)\r\n\r\nazhang- TIME BOMB\r\nPowers: none\r\nStarting Energy: 50\r\nAttacks: nothing\r\nDefense: 50\r\nAttacked by: Klebian- 2 noobs(4)\r\nEnding Energy: 46\r\nOther: Counter - 2\r\n\r\nAnonAmoz\r\nPowers: Masochist, Regeneration\r\nStarting Energy: 36(regeneration)\r\nAttacks: nothing\r\nDefense: 36\r\nAttacked by: Klebian- 20, Klebian's sidekick- 30 + 2 noobs(4)\r\nEnding Energy: 18\r\nOther: Masochist bonus total- 2\r\n\r\nMustafa\r\nPowers: Cloned Sidekick, High-Powered Energy Shield, \r\nStarting Energy: 44\r\nAttacks: nothing\r\nDefense: 43 + BURN 1 to use Energy Shield(10)= 53\r\nAttacked by: nobody\r\nEnding Energy: 43\r\n\r\n\r\nMustafa's Sidekick\r\nPowers: I think I can...(Energy Shield)\r\nStarting Energy: 50\r\nAttacks: azhang- 30\r\nDefense: 18 + BURN 2(20)= 38\r\nAttacked by: nobody\r\nEnding Energy: 48\r\n\r\n\r\n\r\n\r\nCheck to make sure it's right. I was too lazy to bold it and stuff. Send in round 3 moves ASAP.", "Solution_22": "This thing sure seems dead, I've only got moves from one person.... But then again, there only are 2 active ones left in the game.\r\n\r\nShould I end this and start possibly start a new game?", "Solution_23": "If so, I'd like to join. (As long as everybody is active.)", "Solution_24": "MUSTAFA! Where are you?", "Solution_25": "[quote=\"CheeseIsGood\"]MUSTAFA! Where are you?[/quote]\r\n\r\nME?!?!?! I sent in my stuff a long time ago...", "Solution_26": "Really? :huh: \r\n\r\n[url=http://img207.imageshack.us/my.php?image=inboxan7.jpg][img]http://img207.imageshack.us/img207/7163/inboxan7.th.jpg[/img][/url]\r\nI don't see it.", "Solution_27": "Whatever, I'll just send stuff in again", "Solution_28": ":o \r\nResults will be up tomorrow... I'm kinda busy, and the interest in this game seems to have dropped to 0.", "Solution_29": "Argh... I don't think I can do this anymore, it takes about 5 minutes, but the interest in the game has died. Closed. :(" } { "Tag": [], "Problem": "otaghi darim ba iek lamp,va kharej az an 3 kelid .vali nemidanim ke kodam kelid baraye lamp ast,chegoone mitavan faghat ba 2 bar estefade az kelid ha va iek bar varede otagh shodan kelide lamp ra yaft?", "Solution_1": "Aval iek kelid ra be tasadof roshan mikonim , [i]bad az gozashte chand daghighe[/i] yeki az kelid haye baghimande ra roshan mikonim[u] va haman lahze varede otagh mishavim [/u], agar lamp roshan bood anvaght be lamp dast mizanim ,agar lamp garm bood mifamim ke kelid lamp kelid aval boode vali agar sard bood mifahmim ke kelid lamp kelide dovom boode (chon ke taze roshanash kardim) vali agar roshan nabood maloom ast ke kelid sevom kelid lamp boode.", "Solution_2": "[quote=\"ehsan2004\"]Aval iek kelid ra be tasadof roshan mikonim , [i]bad az gozashte chand daghighe[/i] yeki az kelid haye baghimande ra roshan mikonim[u] va haman lahze varede otagh mishavim [/u], agar lamp roshan bood anvaght be lamp dast mizanim ,agar lamp garm bood mifamim ke kelid lamp kelid aval boode vali agar sard bood mifahmim ke kelid lamp kelide dovom boode (chon ke taze roshanash kardim) vali agar roshan nabood maloom ast ke kelid sevom kelid lamp boode.[/quote]\r\n\r\nghabule azadeh? :maybe: \r\nmitunim be lamp dast bezanim? :huh:", "Solution_3": "are,javabeshoon doroste", "Solution_4": ":mad: :huh:" } { "Tag": [ "ratio" ], "Problem": "there is a competition here in the philippines that asked this question:\r\n\r\n\"A = 16^(0.249999999....)\r\n\r\nWrite A as a fraction in simplest form.\"\r\n\r\nSome of the contestants answered 2, while others answered 2/1. \r\n\r\ninitially, the correct answer that flashed on the screen was 2. but of course, those students with 2/1 as an answer protested. in the end, the judges considered both answers as correct. \r\n\r\nany opinions on this?", "Solution_1": "Every integer is also a rational number, and as 2/1=2 and 2/1 is reduced, both answered should be considered correct.", "Solution_2": "If we were to really seek for an argument, 2 is 'wrong' (notice that I've put it in single quotes). Why? Literally, assuming that we take word for word and following all instructions, 2 is not 'written' as a fraction; on the other hand, 2/1 is written as a fraction in simplest form (as a ratio of two integers). Thus, for the sake of winning an argument and biting at the problem committee, they were 'wrong', literally :wink:", "Solution_3": "I have no idea if there is a formal definition of fraction that takes the case of integers into account. Just compute A, then say \"Saying 'Express 2 as a common fraction' makes no sense. Hence my answer will be A=2\".\r\n\r\n\r\nActually in retrospect I think this competition is multiple choice.", "Solution_4": "[quote=\"leprosy\"]\"A = 16^(0.249999999....)\n\nWrite A as a fraction in simplest form.\"[/quote]\r\n\r\nI would have to say that both 2 and 2/1 are correct as well. One of them is simplest and the other is a \"fraction\". Take your pick.\r\n\r\nTo avoid confusion the next time around, I'd say \"write A as a rational number in the simplest form\". That way you don't have people protesting that it was supposed to be a fraction.", "Solution_5": "If it says A in a fraction, 2/1 would be correct. If its simplest form, 2 would be correct.", "Solution_6": "Stop the discussion, everything has been said, and you are just repeating." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "[i][color=brown]This comes out from another problem [/color][/i]\r\n\r\nFind $ m$ such that $ 2223-m^{2}\\mid 14553$\r\n\r\n[hide=\"I dont know if this is a hint\"] $ 14553 = 21^{2}\\cdot 33$\n\n$ 2223 = 33^{2}+21^{2}+21\\cdot 33$[/hide]", "Solution_1": "[quote=\"pontios\"][i][color=brown]This comes out from another problem [/color][/i]\n\nFind $ m$ such that $ 2223-m^{2}\\mid 14553$\n[/quote]\r\n\r\n$ 14553 = 3^{3}7^{2}11$ and we have just $ 24$ divisors for $ 14553$ : 1, 3, 7, 9, 11, 21, 27, 33, 49, 63, 77, 99, 147, 189, 231, 297, 441, 539, 693, 1323, 1617, 2079, 4851 and 14553 which give $ 24$ possibilities for $ m^{2}$ : 2222, 2220, 2216, 2214, 2212, 2202, 2196, 2190, 2174, 2160, 2146, 2124, 2076, 2034, 1992, 1926, 1782, 1684, 1530, 900, 606, 144, -2628 and -12330 and only two are perfect squares : $ 900$ and $ 144$.\r\n\r\nSo the only solutions are $ m=12$ and $ m=30$", "Solution_2": "Why not just use that $ 2223=3^{2}\\cdot 13 \\cdot 19$? Then, you are solving\r\n\\[ 3^{2}\\cdot 13 \\cdot 19-m^{2}\\mid 3^{3}\\cdot 7^{2}\\cdot 11\\] i.e. solve \\[ 3^{3}\\cdot 7^{2}\\cdot 11=y(3^{2}\\cdot 13 \\cdot 19-m^{2}).\\] Notice that if $ 3$ does not divide $ y$, then there is no solution for $ m$ because at most $ 3^{2}\\mid (3^{2}\\cdot 13 \\cdot 19-m^{2})$. So, one only needs to check the cases where $ 3\\mid y$. Altogether, you just need to check:\r\n$ (3^{2}\\cdot 13 \\cdot 19-m^{2}) = 3^{a}7^{b}11^{c}$ where $ a=0,1,$ or $ 2$, $ b=0,1,$ or $ 2$, and $ c=0$ or $ 1$ (18 equations). For example, the case where $ a=b=c=0$: $ 2223-m^{2}= 3^{2}\\cdot 13 \\cdot 19-m^{2}= 1$ implies $ 2224=m^{2}$ which is not true.\r\n\r\nThis may not be the most elegant, but it gets the job done. :( (For the moment I don't see anything easier).", "Solution_3": "[quote=\"pco\"]So the only solutions are $ m=12$ and $ m=30$[/quote]\r\nWhat about $ m=54$? :) I think you might want to check those again. But, the ideas above should work (its just the computation that is a little annoying).", "Solution_4": "[quote=\"baz\"] What about $ m=54$? :) [/quote]\r\n\r\nI considered everything was in $ \\mathbb N$, including $ 2223-m^{2}$ :blush:", "Solution_5": "[quote=\"pco\"][quote=\"pontios\"][i][color=brown]This comes out from another problem [/color][/i]\n\nFind $ m$ such that $ 2223\\minus{}m^{2}\\mid 14553$\n[/quote]\n\n$ 14553 \\equal{} 3^{3}7^{2}11$ and we have just $ 24$ divisors for $ 14553$ : 1, 3, 7, 9, 11, 21, 27, 33, 49, 63, 77, 99, 147, 189, 231, 297, 441, 539, 693, 1323, 1617, 2079, 4851 and 14553 which give $ 24$ possibilities for $ m^{2}$ : 2222, 2220, 2216, 2214, 2212, 2202, 2196, 2190, 2174, 2160, 2146, 2124, 2076, 2034, 1992, 1926, 1782, 1684, 1530, 900, 606, 144, -2628 and -12330 and only two are perfect squares : $ 900$ and $ 144$.\n\nSo the only solutions are $ m \\equal{} 12$ and $ m \\equal{} 30$[/quote]\r\n\r\nI use the same method.\r\nAlso, I consider the negative divisors for 14553\r\nThe solutions are $ m\\equal{}12,30,54$" } { "Tag": [ "geometry", "perimeter", "trigonometry", "LaTeX", "area of a triangle", "trig identities", "Law of Cosines" ], "Problem": "This was a problem from a team round that I wasn't able to figure out (source in the subject description):\r\n\r\nA triangle has an area of 2, a perimeter of 8, and an angle of 30. What is the length of the side opposite the 30 degree angle?", "Solution_1": "[hide=\"icky way\"]\nArea of a triangle is $ \\frac{ab \\sin C}{2}$, in this case, $ C\\equal{}30$, so $ \\frac{ab}{4} \\equal{} 2$.\nObviously, $ ab\\equal{}8$. Express all the sides in terms of $ a$. The sides of the triangle are $ a$, $ \\frac{8}{a}$, $ 8\\minus{}(a\\plus{}\\frac{8}{a})$\nPlug these values into herons formula and set it equal to 2 and solve the equation.\nThere has to be an easier way though.\n[/hide]", "Solution_2": "[quote=\"Lulze\"]This was a problem from a team round that I wasn't able to figure out (source in the subject description):\n\nA triangle has an area of 2, a perimeter of 8, and an angle of 30. What is the length of the side opposite the 30 degree angle?[/quote]\r\n\r\n[hide=\"probably a tad advanced and long\"]\ncall the sides a, b, and c\nc is the side opposite the 30 degree angle\nwe know area is $ \\frac {1}{2}ab\\sin{C}$\nwhich in this case leads to $ 8 \\equal{} ab$\nby law of cosines $ a^2 \\plus{} b^2 \\minus{} ab\\sqrt {3} \\equal{} c^2$\nnow add $ ab(2 \\plus{} \\sqrt {3})$\n$ a^2 \\plus{} b^2 \\plus{} 2ab \\equal{} c^2 \\plus{} 16 \\plus{} 8\\sqrt {3}$\n$ (8 \\minus{} c)^2 \\equal{} c^2 \\plus{} 16 \\plus{} 8\\sqrt {3}$\n$ 16c \\equal{} 48 \\minus{} 8\\sqrt {3}$\n$ c \\equal{} 3 \\minus{} \\frac{\\sqrt {3}}{2}$\n\ni think thats wrong\nlooks too ugly\n\n\n[/hide]\r\n\r\nand it looks like mathcountsfreak had the same starting idea as me, ehe.", "Solution_3": "shouldn't the final answer be $ 3\\minus{}\\frac{\\sqrt{3}}{2}$?\r\n\r\nMight've been a typo.", "Solution_4": "[quote=\"mathcountsfreak\"]shouldn't the final answer be $ 3 \\minus{} \\frac {\\sqrt {3}}{2}$?\n\nMight've been a typo.[/quote]\r\n\r\noh thanks. that was probably a latex error on my part", "Solution_5": "[quote=\"Lulze\"]This was a problem from a team round that I wasn't able to figure out (source in the subject description):\n\nA triangle has an area of 2, a perimeter of 8, and an angle of 30. What is the length of the side opposite the 30 degree angle?[/quote]\r\n[hide]Let $ c$ be the side opposite to angle 30. So $ \\frac{1}{2} ab \\sin 30 \\equal{} 2$ or $ ab \\equal{} 8$. \n\nUsing cosine rule, we have $ c^2 \\equal{} a^2 \\plus{} b^2 \\minus{} 2ab \\cos 30$ or $ a^2 \\plus{} b^2 \\minus{} c^2 \\equal{} 8\\sqrt{3}$.\n\nSince $ a\\plus{}b\\plus{}c \\equal{} 8$, we have $ (8\\minus{}c)^2 \\equal{} a^2 \\plus{} b^2 \\plus{} 2ab$ or $ (8\\minus{}c)^2 \\equal{} a^2 \\plus{} b^2 \\plus{} 16$. Subtracting this from the previous eqn, and solving for $ c$, we obtain $ c \\equal{} 3 \\minus{} \\frac{\\sqrt{3}}{2}$.[/hide]", "Solution_6": "Thank you very much! But I'm confused as to how you went from \r\n\r\n$ a^2 \\plus{} b^2 \\plus{} 2ab \\equal{} c^2 \\plus{} 16 \\plus{} 8\\sqrt {3}$ \r\nto \r\n$ (8 \\minus{} c)^2 \\equal{} c^2 \\plus{} 16 \\plus{} 8\\sqrt {3}$", "Solution_7": "[quote=\"Lulze\"]Thank you very much! But I'm confused as to how you went from \n\n$ a^2 \\plus{} b^2 \\plus{} 2ab \\equal{} c^2 \\plus{} 16 \\plus{} 8\\sqrt {3}$ \nto \n$ (8 \\minus{} c)^2 \\equal{} c^2 \\plus{} 16 \\plus{} 8\\sqrt {3}$[/quote]\r\n\r\nperimeter is 8\r\nand perimeter is a+b+c\r\nso a+b+c=8\r\na+b=8-c\r\non the LHS on the top equation\r\nyou see (a+b)^2\r\nso its (8-c)^2\r\n\r\ntoo lazy to latexify", "Solution_8": "[quote=\"CircleSquared\"][quote=\"Lulze\"]Thank you very much! But I'm confused as to how you went from \n\n$ a^2 \\plus{} b^2 \\plus{} 2ab \\equal{} c^2 \\plus{} 16 \\plus{} 8\\sqrt {3}$ \nto \n$ (8 \\minus{} c)^2 \\equal{} c^2 \\plus{} 16 \\plus{} 8\\sqrt {3}$[/quote]\n\nperimeter is 8\nand perimeter is a+b+c\nso a+b+c=8\na+b=8-c\non the LHS on the top equation\nyou see (a+b)^2\nso its (8-c)^2\n\ntoo lazy to latexify[/quote]\r\n\r\nAhh, thank you!" } { "Tag": [ "algorithm" ], "Problem": "I'm interested in how many schools offer computer science, and how many students take it.\r\n\r\nIf it is offered at your school, is it some sort of requirement for the students there? What language is it taught in? Does your school offer AP computer science?", "Solution_1": "It's an elective at my school. There is one teacher. There are between 10 and 20 kids in all of his classes. He offers regular classes (in basic computer literacy and BASIC), an honors class in BASIC and C++, and an AP class (in Java). The honors class is the prerequisite for the AP class.", "Solution_2": "My school only has AP computer science (Java) which is an elective.", "Solution_3": "nope.\r\nI don't have one, that's probably cuz I'm only in middle school..", "Solution_4": "My school has\r\n\r\nProgramming for the Internet (half yr)\r\nwhich is a recommended prerequisite for\r\nAP Comp Sci (full yr)\r\n\r\nI think there are some other courses as well that at least dabble in programming\r\n\r\nI'm planning on taking both, first next yr, then AP the yr after. We don't always have enough people sign up for the AP though...", "Solution_5": "We've got Basic, C++, and AP Java. I'll probably take advantage of this, but not in my schedule for next year.", "Solution_6": "My school used to offer one, but they closed it about 3 years ago.", "Solution_7": "at my school there were courses in grades 10, 11 and 12. VB6, C++ and Java respectively. no AP stuff in my high school at all. really simplistic material. no algorithms. cool final projects though. i prepared for the Canadian Computing Contest a lot by myself.", "Solution_8": "The school I used to go to offers AP Computer Science, but from what I hear, it bairly even succeeds in getting students to AP level.\r\n\r\nMy current school offers ~1 year introduction of C++ programming which is self-paced, so you can finish the course in 3 weeks if you have the time. After that, it offers an introdcution to Java, which essentially means reading a big book on Java. The third course in the sequence it offers is AP Computer Science, which is fairly trivial if you've taken the first two (Intro to C++ and intro to Java).\r\n\r\nReally, AP Computer Science is a nice introduction to ideas in programming, but that course alone won't get you far in programming. (...although, the same could be said for all other AP courses...)", "Solution_9": "AP CS used to be not too bad.\r\nThen they cut the AB, bc not many people took it.\r\nARGH.\r\n\r\nI was gonna take the AB too, this year. Too bad for me.\r\nSo I'm taking CS A, but as the actual \"main course,\" I'm reading Intro to Algorithms and doing usaco :D", "Solution_10": "I know that my school has Computer Science, and AP Computer Science. I've only seen Java textbooks lying around. Also, I accidentally walked in on a class ( :blush: ), not sure if it was AP or not, but it only had 6 people or so...", "Solution_11": "high school = AP comp sci. It is an elective, but I basically HAVE to take it because we are required to take 2 math courses in High school. By the end of the year, I'll have completed half of calc BC, so I was hoping that during the summer, I would study the rest of Calc BC and take the AP test. But now I have to actually take the CLASS calc BC at the high school :(", "Solution_12": "We have two Computer Programming classes. We have Computer Programming I (the class I am in now), which teaches QBasic, and Computer Programming II (the class I want to take next year) which teaches C++ or Java. The teacher for that class goes super slow, and I want to be a computer programmer when I grow up, so :dry: .", "Solution_13": "Fortunately, my school has a club, but no course. The courses on programming are only offered at the High School", "Solution_14": "We only have Java courses + general software/tech courses. The only classes that are geared specifically toward programming are Computer sciences I, II, III, IV and Advanced Object-Oriented Design.", "Solution_15": "My school(middle school) mandates \"Computer Keyboarding\" for all 6th graders. There are also electives: Computer Graphics, Digital Tools 1, Digital Tools 2, and Microsoft Office. But there is no computer programming for the Munster School District.", "Solution_16": "Our computer science program has slowly dwindled to nonexistence. We used to offer one section of AP Computer Science A and at least two sections of Computer Science Programming Java (an introduction to coding in Java). This year, we're only offering the one section of AP Computer Science A, and we're having two seniors and one sophomore take an online class in Data Structures (which covers the now defunct AP Computer Science AB)." } { "Tag": [ "function", "LaTeX", "calculus", "integration", "real analysis", "calculus computations" ], "Problem": "Find the laplace of the following function:\r\n\r\n[color=white].........[/color]$ \\{t$ $ 0{\\le}t<1$\r\n$ f(t)$=$ \\{$\r\n[color=white].........[/color]$ \\{0$ $ t{\\ge}1$\r\n\r\nf(t) = t-U(t-1)?\r\nis the equation of the above function?\r\n\r\n\r\nsimilar qn\r\n\r\n[color=white].........[/color]$ \\{1\\minus{}e^t$ $ 01$\r\n\r\nfind $ L{f(t)}$ using the unit step function", "Solution_1": "1. Isn't it $ t(U(t) - U(t - 1))$ ? The term in brackets is 0 for $ t < 0$, one for $ 0 < t < 1$ and zero again for $ t > 1$.\r\n\r\nApplying similar logic, you can obtain an equation for the second.\r\n\r\n$ \\text{\\LaTeX}$ help:\r\n\r\n$ f(t) = \\begin{cases}t & t\\in[0,1) \\\\\r\n0 & t\\in[1,\\infty)\\end{cases}$\r\n\r\n$ f(t) = \\begin{cases}1 - e^t & t\\in(0,1) \\\\\r\n0 & t\\in(1,\\infty)\\end{cases}$\r\n\r\nPS: what's the value of $ f(1)$ in 2nd problem? :)", "Solution_2": "[quote=\"hsiljak\"]PS: what's the value of $ f(1)$ in 2nd problem? :)[/quote]\r\nIrrelevant. Changing the function on a set of measure zero does not change its Laplace transform. In practice, I tend to care about the value at a point of continuity, but not at a point of discontinuity.\r\n\r\nOne approach is simply to brute-force the appropriate integrals:\r\n\r\nFor the first example\r\n\r\n$ F(s)\\equal{}\\int_0^1te^{\\minus{}st}\\,dt\\equal{} \\left.\\minus{}\\frac{t}{s}e^{\\minus{}st}\\right|_0^1\\plus{} \\frac1s\\int_0^1e^{\\minus{}st}\\,dt$\r\n\r\n$ \\equal{}\\minus{}\\frac{e^{\\minus{}s}}{s}\\minus{}\\frac{e^{\\minus{}s}}{s^2}\\plus{}\\frac1{s^2}.$\r\n\r\nIf we now try to reverse-engineer a formula for $ f$ by running that through our table of inverse Laplace transforms, we get\r\n\r\n$ f(t)\\equal{}U(t\\minus{}1)(\\minus{}1\\minus{}(t\\minus{}1))\\plus{}t\\equal{}t\\minus{}U(t\\minus{}1)t.$\r\n\r\n(I'm treating $ U(t)$ as understood in context. This is consistent with what hsiljak wrote.)\r\n\r\nWe could similarly get the second example just by doing the integral." } { "Tag": [ "geometry", "vector", "trigonometry", "geometric transformation", "reflection", "geometry proposed" ], "Problem": "Two circles $S_{1}$ and $S_{2}$ with centers $O_{1}$ and $O_{2}$, respectively, intersect at two points $A$ and $B$. The tangents to the circles $S_{1}$ and $S_{2}$ at the point $A$ intersect the segments $BO_{2}$ and $BO_{1}$ at the points $K$ and $L$, respectively. Prove that $KL \\parallel O_{1}O_{2}$.\r\n\r\n[i]Russian olympiad/2003/proposed by S.Berlov[/i]\r\n\r\nNot really that hard, but I felt like posting something :D", "Solution_1": "well after 10 mins of trying the problem with my lucky pen (a blue parker vector :D ) i think i've got the problem.\r\nfirst,let P_1= BO_1 \\cap AK and P_2=BO_2 \\cap AL.\r\n0$,\r\n $(x+r_{1})(x+r_{2})......(x+r_{n})\\geq(x+({r_{1}\\cdot r_{2}\\cdot.....r_{n}})^{\\frac{1}{n}})^{n}$\r\n\r\n2.\r\nIf $a,b,c$ are positive and $a+b+c=6$, show that\r\n$(a+\\frac{1}{b})^{2}+(b+\\frac{1}{c})^{2}+(c+\\frac{1}{a})^{2}\\geq \\frac{75}{4}$\r\n\r\nJBL-yeah I fgot! thanks", "Solution_1": "The first one fails for $n = 1$ if $r_{1}< 1$. And it fails for $n \\geq 3$ if I make the $r_{i}$ very large, for instance. You probably want it to be the $n$th root.", "Solution_2": "Changed it now! :blush:", "Solution_3": "[hide]The first one is just a special case of the generalized Holder inequality.\n\nThe second we can say\n\n$\\sum_{cyc}\\left(a+\\frac{1}{b}\\right)^{2}\\ge \\frac{1}{3}\\left(\\sum \\left(a+\\frac{1}{a}\\right)\\right)^{2}$\n\nby AM-GM and\n\n$\\sum \\left(a+\\frac{1}{a}\\right) = 6+\\sum \\frac{1}{a}\\ge \\frac{15}{2}$\n\nby AM-HM/Cauchy. Then just plug in and we have\n\n$\\sum_{cyc}\\left(a+\\frac{1}{b}\\right)^{2}\\ge \\frac{1}{3}\\left(\\frac{15}{2}\\right)^{2}= \\frac{75}{4}$.[/hide]\r\n\r\nI hope my inequalities still go the right way...", "Solution_4": "[quote=\"paladin8\"][hide]The first one is just a special case of the generalized Holder inequality.\n\nThe second we can say\n\n$\\sum_{cyc}\\left(a+\\frac{1}{b}\\right)^{2}\\ge \\frac{1}{3}\\left(\\sum \\left(a+\\frac{1}{a}\\right)\\right)^{2}$\n\nby AM-GM and\n\n$\\sum \\left(a+\\frac{1}{a}\\right) = 6+\\sum \\frac{1}{a}\\ge \\frac{15}{2}$\n\nby AM-HM/Cauchy. Then just plug in and we have\n\n$\\sum_{cyc}\\left(a+\\frac{1}{b}\\right)^{2}\\ge \\frac{1}{3}\\left(\\frac{15}{2}\\right)^{2}= \\frac{75}{4}$.[/hide]\n\nI hope my inequalities still go the right way...[/quote]\r\nOkay, thanks for posting but I dont understand how you got the 1st line of your solution. :maybe:", "Solution_5": "[quote=\"riddler\"][quote=\"paladin8\"][hide]The first one is just a special case of the generalized Holder inequality.\n\nThe second we can say\n\n$\\sum_{cyc}\\left(a+\\frac{1}{b}\\right)^{2}\\ge \\frac{1}{3}\\left(\\sum \\left(a+\\frac{1}{a}\\right)\\right)^{2}$\n\nby AM-GM and\n\n$\\sum \\left(a+\\frac{1}{a}\\right) = 6+\\sum \\frac{1}{a}\\ge \\frac{15}{2}$\n\nby AM-HM/Cauchy. Then just plug in and we have\n\n$\\sum_{cyc}\\left(a+\\frac{1}{b}\\right)^{2}\\ge \\frac{1}{3}\\left(\\frac{15}{2}\\right)^{2}= \\frac{75}{4}$.[/hide]\n\nI hope my inequalities still go the right way...[/quote]\nOkay, thanks for posting but I dont understand how you got the 1st line of your solution. :maybe:[/quote]\r\n\r\nFrom $x^{2}+y^{2}+z^{2}\\ge \\frac{(x+y+z)^{2}}{3}$, where $x = a+\\frac{1}{b}$, $y = b+\\frac{1}{c}$, $z = c+\\frac{1}{a}$?", "Solution_6": "[quote=\"riddler\"]\n2.\nIf $a,b,c$ are positive and $a+b+c=6$, show that\n$(a+\\frac{1}{b})^{2}+(b+\\frac{1}{c})^{2}+(c+\\frac{1}{a})^{2}\\geq \\frac{75}{4}$\n\nJBL-yeah I fgot! thanks[/quote]\r\n\r\ni am just learning inequalities so bear with me...\r\n\r\n[hide=\"solution\"]By am-hm \\[\\frac{ \\sum a}{3}\\ge \\frac{3}{ \\sum \\frac{1}{a}}\\Longleftrightarrow \\sum\\frac{1}{a}\\ge 1.5 \\Longleftrightarrow\\] \\[\\sum a+\\sum\\frac{1}{a}\\ge 1.5+6=7.5\\] By cauchy, \\[\\left[ (a+\\frac{1}{b})^{2}+(b+\\frac{1}{c})^{2}+(c+\\frac{1}{a})^{2}\\right]\\cdot (1^{2}+1^{2}+1^{2}) \\ge \\left( \\sum a+\\sum\\frac{1}{a}\\right)^{2}\\ge (7.5)^{2}\\] \\[\\Longleftrightarrow (a+\\frac{1}{b})^{2}+(b+\\frac{1}{c})^{2}+(c+\\frac{1}{a})^{2}\\ge \\frac{75}{4}\\] [/hide]", "Solution_7": "Thanks for your solutions. Can anyone post the proof of the 1st one?", "Solution_8": "[hide=\"Inelegant proof of #1\"]Expand the LHS. The general term is of the form\n\n$\\sum_{1\\leqslant i_{1} 0$ \r\nChoose $\\delta = \\frac{\\epsilon}{2}$\r\nthen for \r\nfor $0 <|x-1|< \\delta$\r\n$|f(x)-L| = |2x+1-3|$\r\n$= |2x-2|$\r\n$= 2|x-1|$\r\n$< 2 \\delta = 2(\\frac{\\epsilon}{2})= \\epsilon$", "Solution_1": "Actually, x-1 is not equal to epsilon! This is just a preliminary analysis (as you state) so that you can get x - 1 after working with f(x) - L.\r\nThe reason you need x-1 is due to the condition you have :\r\nthe absolute value of x - 1 is less than epsilon.", "Solution_2": "could you also do\r\n\r\n$|x_{1}-x_{2}|<\\delta$ and \r\n$|f(x_{1})-f(x_{2})|=|2x_{1}+1-2x_{2}-1|=2|x_{1}-x_{2}|<\\epsilon$\r\nso\r\njust choose $x$ values that are withing $\\frac{\\epsilon}{2}$", "Solution_3": "[quote=\"maokid7\"]could you also do\n\n$|x_{1}-x_{2}|<\\delta$ and \n$|f(x_{1})-f(x_{2})|=|2x_{1}+1-2x_{2}-1|=2|x_{1}-x_{2}|<\\epsilon$\nso\njust choose $x$ values that are withing $\\frac{\\epsilon}{2}$[/quote]\r\n\r\nDo you know any good site with this kind of stuff?", "Solution_4": "I got that from ACoPS and I know it is in an apendix of my calculus book.\r\n\r\nbasically all it says is that for a [b]uniformly continuous function[/b] there is a $\\delta$ that satisfies\r\n\r\n$|x_{1}-x_{2}|<\\delta$ and $|f(x_{1})-f(x_{2})|<\\epsilon$\r\n\r\nand then you just have to play around w/ $|f(x_{1})-f(x_{2})|$ to obtain a $|x_{1}-x_{2}|$.\r\n\r\nand the value of $\\delta$ depends only on $\\epsilon$ and not on the values of $x_{1}$ or $x_{2}$." } { "Tag": [ "geometry", "3D geometry" ], "Problem": "Person A fills up a sugar cone with sprinkles and is about to devour it. However, he suddenly has to use the bathroom. Person B comes along and sees the cone sitting upright in a glass and is very tempted. He eats some, then some more, and by the time Person A exits the bathroom the top 3/4 of the cone has been devoured, and there are only 75 sprinkles left in it.\n\nPerson A loves his sprinkles, and of course will be very unhappy upon seeing the tiny bit left. He will yell at Person B in a loudness of one decibal for every 25 sprinkles Person B has eaten. How many decibals is the yelling that Person B receives?", "Solution_1": "wow, wat a random problem\r\n\r\n[hide]$25\\times25=\\boxed{625}\\text{earsplitting decibels}$[/hide]", "Solution_2": "intresting problem.......", "Solution_3": "thats the fun of making up problems. You can come up with anything, for example, the person C could come in and go shoot everyone and eat a third of the remaining sprinkles and give the rest of them to person D. How many sprinkles does person D get?\r\n\r\nsee!", "Solution_4": "intresting problem... did you make it up....", "Solution_5": "[hide=\"person D eats...\"]50 sprinkles[/hide]\n\nI'm sorry, I just had to... :D " } { "Tag": [ "calculus", "integration", "logarithms", "calculus computations" ], "Problem": "How do I integrate exp(-2*sqrt(t))\r\n\r\nIt's been a while I don't remember how to do these. Thanks", "Solution_1": "Try integration by parts.", "Solution_2": "Step 1: make the substitution $ x\\equal{}2\\sqrt{t},$ which you can write as $ t\\equal{}\\frac{x^2}4.$ Differentiate that to get $ dt\\equal{}\\frac{x}{2}\\,dx.$\r\n\r\nNow we have $ \\int e^{\\minus{}2\\sqrt{t}}\\,dt\\equal{}\\frac12\\int xe^{\\minus{}x}\\,dx.$\r\n\r\nStep 2 will be an integration by parts. Can you see how to proceed from here?", "Solution_3": "[color=darkblue]Denote $ I\\equiv\\int e^{ - 2\\sqrt t}\\ \\mathrm {dt}\\ .$ Apply the [b]substitution[/b] $ \\{\\begin{array}{c} x = \\phi (t) = e^{ - 2\\sqrt t} \\\\\n \\\\\n(\\ \\ln x = - 2\\sqrt t\\ ) \\\\\n \\\\\n\\phi '(t) = \\frac { - 1}{\\sqrt t}\\cdot e^{ - 2\\sqrt t}\\end{array}$ \n\nThus, $ I = \\int e^{ - 2\\sqrt t}\\cdot\\frac { - \\sqrt t}{e^{ - 2\\sqrt t}}\\cdot \\underline {|\\ \\phi '(t)\\ \\mathrm {dt}}$ $ \\Longrightarrow$ $ \\boxed {I = \\frac 12\\cdot F(e^{ - 2\\sqrt t}) + \\mathcal C}\\ ,$\n\nwhere $ F(x) = \\int \\ln x\\ \\mathrm {dx}\\ .$ Apply [b]the integration by parts[/b] :\n\n$ \\{\\begin{array}{ccc} u(x) = \\ln x & \\Longrightarrow & u'(x) = \\frac 1x \\\\\n \\\\\nv'(x) = 1 & \\Longrightarrow & v(x) = x\\end{array}\\|$ $ \\implies$ $ \\boxed {F(x) = x\\ln x - x + \\mathcal C}\\ .$\n\nIn conclusion, $ \\boxed {\\boxed {I = - (\\sqrt t + \\frac 12)\\cdot e^{ - 2\\sqrt t} + \\mathcal C}}\\ .$[/color]", "Solution_4": "[quote=\"Kent Merryfield\"]Step 1: make the substitution $ x \\equal{} 2\\sqrt {t},$ which you can write as $ t \\equal{} \\frac {x^2}4.$ Differentiate that to get $ dt \\equal{} \\frac {x}{2}\\,dx.$\n\nNow we have $ \\int e^{ \\minus{} 2\\sqrt {t}}\\,dt \\equal{} \\frac12\\int xe^{ \\minus{} x}\\,dx.$\n\nStep 2 will be an integration by parts. Can you see how to proceed from here?[/quote]\r\n\r\nyes thank you", "Solution_5": "$ \\int e^{\\minus{}2 \\sqrt{t}}\\,dt \\equal{} \\minus{} \\int \\sqrt{t} \\cdot \\left(\\minus{} \\frac{1}{\\sqrt{t}}\\right) e^{\\minus{}2 \\sqrt{t}}\\,dt \\equal{} \\minus{}\\left(e^{\\minus{}2 \\sqrt{t}} \\cdot \\sqrt{t} \\minus{} \\int e^{\\minus{}2 \\sqrt{t}} \\cdot \\frac{1}{2 \\sqrt{t}}\\,dt\\right)$\r\n\r\n$ \\equal{} \\minus{} \\sqrt{t} \\cdot e^{\\minus{}2 \\sqrt{t}} \\minus{} \\frac{1}{2} \\int \\left(\\minus{} \\frac{1}{\\sqrt{t}}\\right) e^{\\minus{}2 \\sqrt{t}}\\,dt \\equal{} \\minus{}e^{\\minus{}2 \\sqrt{t}} \\left(\\sqrt{t} \\plus{} \\frac{1}{2} \\right) \\plus{} C$.", "Solution_6": "I've got one more\r\n\r\nIntegrate:\r\nexp((.02*t)/(1+.05t))" } { "Tag": [ "geometry", "quadratics", "inequalities", "algebra", "quadratic formula" ], "Problem": "This problem is really cool!\r\n\r\nABCD is a convex quadrilateral with AC and BD intersecting at M. If the areas of AMB and CMD are 4 and 9 respectively, what is the minimum area of ABCD?", "Solution_1": "[hide]A(MBC)/A(AMB)=A(DMC)/A(MAD) because of sharing altitude. (A(AMD) means area of triangle AMD)\n\n\n\nsub 4 and 9 in \n\nwe have A(MBC)/4=9/A(MDA). Let A(MBC) be x,A(MDA) be y.\n\nxy=36. y=36/x\n\n\n\nLet x+y=k\n\nx+36/x=k\n\nx :^2: +36=kx\n\nx :^2: -kx+36=0\n\nQuadratic Formula\n\n(k+ :pm: :sqrt: (k :^2: -144))/2\n\n so k :ge: 12. therefore 12 is the minimum value of x+y. 12+4+9=25= minimum value of the quad.[/hide]", "Solution_2": "good job!\n\n\n\nNote: [hide]an easier way to minimize x+36/x is with AM-GM inequality: x+36/x >= 2 sqrt(x*36/x) = 12.[/hide]" } { "Tag": [ "superior algebra", "superior algebra unsolved" ], "Problem": "(Galois closure of $ \\mathbb{Q}(\\sqrt[4]{2},\\zeta_6)$ over $ \\mathbb{Q}$)$ \\equal{} \\mathbb{Q}(\\sqrt[12]{2},\\zeta_{12})$?\r\n\r\nIs it right?", "Solution_1": "Why the $ \\sqrt[12]2$\u00bf It should be a $ \\sqrt[4]2$." } { "Tag": [ "Gauss", "number theory proposed", "number theory" ], "Problem": "Let $a,b$ be integers, and $p,q\\ (q\\neq0)$ be rational numbers.\r\nProve that if $p+qi$ is the root of $x^2+ax+b=0$,then show that both $p$ and $q$ are integers.", "Solution_1": "Since second root is $p-qi$, from Viet theorem (am I correctly wrote it?) we obtain $2p=-a$ and $p^2+q^2=b$. Suppose $a$ is odd, then $4b=a^2+(2q)^2$, so $2q$ is a rational and $(2q)^2$ is an integer. It follows that $2q$ is integer. But in this case $a^2+(2q)^2=1$ or $2$ modulo 4 -- contradiction. Therefore, $a$ is even, $p$ is an integer, and $q$ must e integer too.", "Solution_2": "Naturally you are right as usual.\r\n\r\nkunny", "Solution_3": ":D :D :D" } { "Tag": [], "Problem": "Aici sunt subiectele de la etapa judeteana http://scoala18.xhost.ro/. Au fost aspru criticate. Ce parere aveti ??? Va multumesc.", "Solution_1": "In ce sens au fost criticate?" } { "Tag": [ "linear algebra", "matrix" ], "Problem": "This is just a technicality to which i am unsure.\r\n\r\n If $ A \\vee B \\Rightarrow C$\r\n\r\nThen is it correct to say\r\n\r\n$ A \\Rightarrow C$ or $ B \\Rightarrow C$", "Solution_1": "If you're unsure if a logic formula is a tautology, you can always write its truth table:\r\n\r\n$ \\begin{matrix}A & B & C & A\\lor B & (A\\lor B)\\implies C & (A\\implies C) & (B\\implies C) & (A\\implies C)\\lor (B\\implies C) \\\\\r\nT & T & T & T & T & T & T & T \\\\\r\nT & T & F & T & F & F & F & F \\\\\r\nT & F & T & T & T & T & T & T \\\\\r\nT & F & F & T & \\boxed{F} & F & T & \\boxed{T} \\\\\r\nF & T & T & T & T & T & T & T \\\\\r\nF & T & F & T & \\boxed{F} & T & F & \\boxed{T} \\\\\r\nF & F & T & F & T & T & T & T \\\\\r\nF & F & F & F & T & T & T & T\\end{matrix}$\r\n\r\nAs you can see, the boxed cases disprove the equivalence between the two formulas.\r\n\r\nYou can use elementary transformations of basic logic operations to get to the correct tautology. Having in mind that $ (p\\implies q)\\iff(\\lnot p\\lor q)$, and that $ \\land$ and $ \\lor$ are distributive to one another, we get\r\n\r\n$ [(A\\lor B)\\implies C]\\iff[\\lnot(A\\lor B)\\lor C]\\iff[(\\lnot A\\land\\lnot B)\\lor C]\\iff [(\\lnot A\\lor C)\\land(\\lnot B\\lor C)]\\iff [(A\\implies C)\\boxed{\\boxed{\\land}}(B\\implies C)]$", "Solution_2": "From Farenhajt's truth table, if $ A \\lor B \\implies C$, then $ A \\implies C$ or $ B \\implies C$. However, the converse is not true. In symbolic notation:\r\n\r\n$ (A \\lor B \\implies C) \\implies ((A \\implies C) \\lor (B \\implies C))$\r\n\r\nBut:\r\n\r\n$ \\lnot ((A \\lor B \\implies C) \\iff ((A \\implies C) \\lor (B \\implies C)))$" } { "Tag": [ "calculus", "derivative", "articles", "geometry", "number theory" ], "Problem": "Hey folks,\r\n\r\nI have a series of second-round quant interviews for a summer internship position coming up later this week and am wondering whether any of you who have gone through such a process have any advice on how to prepare and do well in the interviews. Any input would be greatly appreciated. Thanks!", "Solution_1": "i just had my second interview for jane street capital this thursday. i think i did pretty well on the first interview, and then i bombed the second by completely missing the first brainteaser question. :oops_sign: \r\n\r\nat the end he said I would hear back next week.\r\n\r\nhow do you think you did?", "Solution_2": "Since it\u2019s a not a full time role it will not be quite as tough, but I\u2019d still recommend preparing the same way. You might want to pick up \u201cHeard On the Street\u201d, \u201cC++ Design Patterns and Derivatives Pricing\u201d, and \u201cOptions, Futures, and Other Derivatives\u201d (Hull). It\u2019s probably a bit late for the second two, but definitely read through the first.", "Solution_3": "I just got a job as a quantitative analyst, and it surprised me a bit, because I did pure maths, so am not as well qualified as a lot of people. \r\n\r\nI think it helped that I read the annual report of the company, and google-searched them and read every news article about them that I could find, and then carefully prepared answers to the common questions like \"why do want to work here?\" and \"what are your skills?\" and so forth. So I would recommend doing that. I think it helps to show enthusiasm about the specific company you're applying to.", "Solution_4": "As a headhunter, I am struck by the number of people who get the \"why do you want do do this\" question wrong.", "Solution_5": "[quote=\"Dominic Connor\"]As a headhunter, I am struck by the number of people who get the \"why do you want do do this\" question wrong.[/quote]\r\n\r\nWhat is the correct/best answer?", "Solution_6": "The answer should be honest...\r\n\r\nIt should show that you've thought about this kind of work, and this particular job.\r\n\r\nMoney, is not a great motivation, though it does not hurt to say something like\r\n\"obviously the money is a factor, but I'm enthusiastic about...\"\r\n\r\nSend the message that your enthusiasm is for this job.\r\nIt is an opportunity for some naked bragging of the form \"this lets me apply my abilities in XX\"\r\nThat's more important than you might think.\r\n\r\nZeta4g2 was surprised to get past, and I can see the issues he faced. He had deep skills, but not the ones the interviewers might have asked about. Thus if you feel part of your light has been shaded, it's a chance to bring it in, as useful. \r\nThus everything your hear about the job has value, even if it's the standard BS about \"being a good team player\".\r\n\r\nWith all due respect to Zeta, he didn't know what they wanted, few entry level people do with precision. I recall one very well known quant, famous in his area who understands classical quant maths at a deep level. Sadly in his first job he found that he needed number theory, which is of course pure maths. \r\n\r\nSome wrong answers we hear are quite tragically bad.\r\nShocked silence, expressing an interest in a totally different sort of work, and mumbling random things.", "Solution_7": "mb1551- i am doing an internship with a Quant Team. I got it by just sending them a resume, and with no financial skills. Its pretty good- at N A B, but what skills do I need to know to get an internship at a bigger bank? how did u guys find out about these? i just did b/c my family member worked at the particular bank." } { "Tag": [ "geometry", "3D geometry", "Pythagorean Theorem" ], "Problem": "In the picture attached, there is a cube, having three vertices called A, B, and C. Find the measure of angle ABC that is formed by the dashed segments. Anyone got a solution?", "Solution_1": "[hide=\"Hint\"]\nA and C are just begging to be connected.[/hide]", "Solution_2": "Hmm, looking at the post below, it is a special triangle...but what kind of triangle is it?\r\n\r\nIs it not an equilateral triangle?", "Solution_3": "[hide=\"Solution\"]Once we connect A and C, what do we discover? A special triangle![/hide]", "Solution_4": "So am I right or no?", "Solution_5": "[quote=\"fanaticsm\"]So am I right or no?[/quote]\r\n\r\n[hide=\"Answer\"]You're right. Everyone of the sides is a diagonal of a face of the cube, so unless it's not a perfect cube,you can be pretty confident in your answer.[/hide]", "Solution_6": "[hide]\nconnect A and C, which makes an equilateral triangle. If you don't get this, assume that this is a unit cube. Using pythagorean theorem or the formula for a 45-45-90 triangle, you wil find that each side is equal to sqrt(2). So it is an equilateral triangle which has each angle equal to 60 degrees.[/hide]" } { "Tag": [ "HCSSiM", "Columbia", "AMC", "AMC 8", "AMC 10", "AMC 10 B" ], "Problem": "I had the highest score on the AMC 10B in my state, but someone had a higher score on the 10 A. will i get a state winners plaque, or is it just who had the highest score on either test. The scores on the A were much higher than on the b so i think it would be a little unfair if I didnt.", "Solution_1": "No, you won't. I know from being in the exact position that you are in.", "Solution_2": "hm I had the highest score on the amc 10 B in my state last year and the director at the site I took at it told me she got a plaque for me. However, I haven't received it yet so I'm not sure :huh:", "Solution_3": "[quote=\"Tenoreoz\"]hm I had the highest score on the amc 10 B in my state last year and the director at the site I took at it told me she got a plaque for me. However, I haven't received it yet so I'm not sure :huh:[/quote]In one year? lol", "Solution_4": "hrm... i think it's just your score that matters, and that you don't really need awards.\r\ni mean.. what use are they??\r\nmaybe i just feel bad cuz im in CA :( , but still, awards don't really matter.", "Solution_5": "I had the highest 10B score last year, but someone else had a higher score on the 10A. I still got a plaque for the highest 10B scorer.", "Solution_6": "Really? In that case I earned 2 plaques that I never got . . . :mad:", "Solution_7": "[quote=\"E^(pi*i)=-1\"]Really? In that case I earned 2 plaques that I never got . . . :mad:[/quote]\r\n\r\nnow you know how I feel, except worse lol.", "Solution_8": "You know, sometimes it really sucks to live in CA... :(", "Solution_9": "According to the AMC website, intramural awards are given for each test, but regional and national awards are for the score in a year. Hence, if one is beaten by someone on the other test, one should not get a plaque.", "Solution_10": "I hope that seperate plaques are awarded, b/c otherwise i won't get one :(\\\r\n\r\nIs it possible that this varies by state?", "Solution_11": "[quote=\"xscapezaer\"]You know, sometimes it really sucks to live in CA... :([/quote]\r\nyou can look at it in a more positive manner: If someone does achieve very well in CA, he'd probably feel better than if he did in some less populated state.", "Solution_12": "Do ties get plaques? (ie. we have 3 141's on the 12A and 1 141 on the 12B in MA, do we all get plaques)", "Solution_13": "[quote=\"ProtestanT\"][quote=\"xscapezaer\"]You know, sometimes it really sucks to live in CA... :([/quote]\nyou can look at it in a more positive manner: If someone does achieve very well in CA, he'd probably feel better than if he did in some less populated state.[/quote]\r\nhehe yeah same way I felt when I got first in Oregon with a 24 AMC 8.\r\nEDIT: err I meant I didn't feel as accomplished as I would've for getting first in a more populated state..\r\n\r\nand The Zuton Force - I got a plaque for the AMC 8 mentioned above, even though it was a three-way tie.", "Solution_14": "Cant we just ask the amc director? there seems 2 b alot of confusion", "Solution_15": "a) why the triple post \r\nb) i have an AMC 8 top scorer plaque :10:", "Solution_16": "[quote=\"Quickster94\"]i'm happy, i did end up getting a plaque for top AMC-10 A scorer :lol:[/quote]\r\n\r\ni only got a pin :(", "Solution_17": "[quote=\"Quickster94\"]i'm happy, i did end up getting a plaque for top AMC-10 A scorer :lol:[/quote]\r\n\r\ndid you get a 120? i got a 121.5 on the 10A on i was like 10th in my state, we had 2 perfects. :(", "Solution_18": "hmmm, yes in fact i did get a 120, how did u know?...", "Solution_19": "I got a perfect on the 10 but i haven't gotten anything yet...", "Solution_20": "[quote=\"Quickster94\"]hmmm, yes in fact i did get a 120, how did u know?...[/quote]\r\n\r\nyou are from vermont and you said you got a 10A plaque so i looked at the stats.\r\n\r\nEDIT: 800th POST!!!!!!!!!", "Solution_21": "It's wierd. People said that you get a plaque if get first in the state. But at our school the state champion got a trophy.", "Solution_22": "i won AMC 10A and 10B at my school and i only got 1 pin. :(", "Solution_23": "Maybe they give different awards in different states or counties. :maybe:", "Solution_24": "Maybe they give different awards in different states or counties. :maybe:", "Solution_25": "i got my plaque today!!!\r\nAMC 10B winner :lol: finally", "Solution_26": "Congratulations!! They r very nice :lol:", "Solution_27": "thanks :) agreed", "Solution_28": "Strange... I'm the highest scorer in GA on the 10A, but I never even got so much as a \"school winner's pin\" for it...", "Solution_29": "oh, ur school probably has them somewhere... I didn't get my plaque until june, idk when they got it though" } { "Tag": [ "algebra", "polynomial", "number theory unsolved", "number theory" ], "Problem": "for positive integers n and k, show that\\[(n^{4}-1)(n^{3}-n^{2}+n-1)^{k}+(n+1)n^{4k-1}\\]\r\nis divisible by $n^{5}+1$", "Solution_1": "[hide=\"hint\"]Put $A=(n^{4}-1)(n^{3}-n^{2}+n-1)^{k}+(n+1)n^{4k-1}$. Because $(n,n^{5}+1)=1$, we only need prove $n^{5}+1|n^{k+1}A$.[/hide]", "Solution_2": "[quote=\"ice_age\"]for positive integers n and k, show that\n\\[(n^{4}-1)(n^{3}-n^{2}+n-1)^{k}+(n+1)n^{4k-1}\\]\nis divisible by $n^{5}+1$[/quote]\r\n$n^{5}+1=(n+1)g(n), \\ g(n)=n^{4}-n^{3}+n^{2}-n+1$. It is obviosly $(n+1)|f(n)=(n^{4}-1)(n^{3}-n^{2}+n-1)^{k}+(n+1)n^{4k-1}$. \r\nBecause $(n,g(n))=1$ we can consider $g(n)|n^{k+1}f(n)$.\r\n We have $n^{5}-n=(n+1)(g(n)-1)$, therefore \r\n$n^{k+1}f(n)=(n^{5}-n)(g(n)-1)^{k}+(n+1)n^{5k}=-(n+1)(g(n)-1)^{k+1}+(n+1)[(n+1)g(n)-1]^{k}$ \r\nor $n^{k+1}f(n)=(n+1)[(g(n)-1)^{k+1}+((n+1)g(n)-1)^{k}]=0(mod \\ g(n))$.", "Solution_3": "[quote=\"Rust\"][quote=\"ice_age\"]for positive integers n and k, show that\n\\[(n^{4}-1)(n^{3}-n^{2}+n-1)^{k}+(n+1)n^{4k-1}\\]\nis divisible by $n^{5}+1$[/quote]\n$n^{5}+1=(n+1)g(n), \\ g(n)=n^{4}-n^{3}+n^{2}-n+1$. It is obviosly $(n+1)|f(n)=(n^{4}-1)(n^{3}-n^{2}+n-1)^{k}+(n+1)n^{4k-1}$. \nBecause $(n,g(n))=1$ we can consider $g(n)|n^{k+1}f(n)$.\n We have $n^{5}-n=(n+1)(g(n)-1)$, therefore \n$n^{k+1}f(n)=(n^{5}-n)(g(n)-1)^{k}+(n+1)n^{5k}=-(n+1)(g(n)-1)^{k+1}+(n+1)[(n+1)g(n)-1]^{k}$ \nor $n^{k+1}f(n)=(n+1)[(g(n)-1)^{k+1}+((n+1)g(n)-1)^{k}]=0(mod \\ g(n))$.[/quote]\r\n\r\nthanks.", "Solution_4": "if it's \"legal\" then it follows very easy when using roots of unity... \r\n\r\nnow i ask: is it legal to use them? i mean, since we're working with polynomials with integer coefficients, does their use affect in something?", "Solution_5": "A polynomial divides another iff the dividend contains all of the roots of the divisor; this is over their splitting field, regardless of the coefficients.", "Solution_6": "Counted with multiplicity." } { "Tag": [], "Problem": "What is the greatest possible number of points of intersection for eight distinct lines in a plane?", "Solution_1": "[hide]1 line = 0 points\n2 lines = 1 point\n3 lines = 3 points\n4 lines = 6 points\n5 lines = 10 points\n6 lines = 15 points\n7 lines = 21 points\n8 lines = 28 points\n\nSee a pattern? Because to maximize intersection points, the nth line must cross through all (n-1) other lines on the plane. So each 'term' increases by n-1. Therefore, A(n) [# intersection points where n is the # of distinct lines in the plane] = 1 + 2 + 3... (n-1) = (n)(n-1)/2. 8(7)/2 = [b]28.[/b]\n[/hide]\r\nlol I must be really bored.", "Solution_2": "I did this one in my math club. Fundamentally, you start off with pictures, slowly changing to patterns at a destined time. Just like, Superwizard188, you have to use patterns, and also pictures, which he probably did as well (just not included, obviously).", "Solution_3": "[hide]Whenever we choose a pair of lines, they should meet. So $\\binom{8}{2}=28$.[/hide]" } { "Tag": [ "ratio", "limit", "inequalities", "algebra", "polynomial", "arithmetic sequence", "geometric sequence" ], "Problem": "Is it possible to partition N, the set of all positive integers, into two disjoint sets A and B such that:\r\n\r\ni) There exists no infinite (non-constant) arithmetic sequence in A, and\r\nii) There exists no infinite (non-constant) geometric sequence in B?\r\n\r\n :D", "Solution_1": "Yes it is possible. And, if I'm not wrong the following partition excludes any infinite (non constant) arithmetical or geometrical progression in each set.\r\n\r\nConsider the sequence $(U_n)$ defined by $U_1 = 1$ and $U_{n+1} = n(U_1 + \\cdots + U_n).$\r\n\r\nNow, color the positive integers alternatively red and blue by groups of size $U_1$ (red), $U_2$ (blue), $U_3$ (red), and so on...\r\nLet $A$ be the set of red integers and $B$ be the set of blue integers.\r\n\r\nLet $a,d>0$ be the first term and the ratio, respectively, of an arithmetical sequence. Since $\\lim U_n = +\\infty$, it is clear that there are arbitrary large strings of consecutive red (resp. blue) integers after $a$. Thus at least one term of the sequence must be red and at least one must be blue.\r\nTherefore there is no monochrome arithmetical sequence.\r\n\r\nLet $a,q$ be the first term and the ratio, respectively, of a geometrical sequence. Now, let $n>q$ be even such that $aq^k $ belongs to the string of $U_n$ for some integer $k$. Wlog, we may assume that $k$ is maximal, that is $aq^{k+1}$ does not belong to this string.\r\nBut, since $aq^k$ belongs to the string, we clearly have $aq^k \\leq U_1+U_2 + \\cdots + U_n$. Thus $aq^{k+1} \\leq q(U_1 + \\cdots + U_n) < U_{n+1}$ so that $aq^{k+1}$ must be red.\r\nWe have a similar reasoning for to prove that any geometrical sequence cannot be all red.\r\nTherefore, there is no monochrome geometrical sequence.\r\n\r\nPierre.", "Solution_2": "Pierre, please correct me if I am wrong, but from your argument, it doesn't seem like A :cup: B = N, the entire set of positive integers because {U_n} = {1, 1, 4, 18, ... }.... :(", "Solution_3": "No, you don't color the numbers $U_n$. You color the positive integers by groups of consecutive integers, and the size of the $n^{th}$ group is $U_n$.\r\n\r\nPierre.", "Solution_4": "[quote]No, you don't color the numbers U_n. You color the positive integers by groups of consecutive integers, and the size of the nth group is U_n. \n\nPierre[/quote]\r\n\r\nOh! Oops, then it's an ingenius solution! Thanks a lot! :)", "Solution_5": "Actually, Pierre, if you are interested to know, the original problem is still open on CRUX:\r\n\r\nIs it possible to partition N, the set of all positive integers, into two disjoint sets A, B such that:\r\n\r\n1) No three numbers in A make an (non-constant) arithmetic sequence.\r\n2) There is no infinite (non-constant) arithmetic sequence in B.\r\n\r\n\r\nI don't know, did I change it to an easier one? :lol:", "Solution_6": "I know solution for this one, it can't be open.", "Solution_7": "So, it is your turn to answer Harazi! :P \r\n\r\nPierre.", "Solution_8": "You will have a solution soon. :P", "Solution_9": "Let $ x_1, x_2,...$ be an enumeration of all the infinite arithmetic progressions of natural numbers. Take any number $a_1$ from $x_1$, then take $ a_2>2a_1$ from $ x_2$ then $ a_3>2a_2$ from $x_3$ and so on. Put all numbers $ a_k$ in a class A and the rest in B. Then B clearly does not have infinite arithmetic progressions. And since $ a_k+a_i>2a_{k-1}$, no three elements in A are in arithmetic progression.", "Solution_10": "Using the same idea, here is another partition that avoids in both classes infinite arithmetic or geometric progressions. Let $ a_i$ be all infinite arithmetic progressions and $g_i$ all infinite geometric progressions. Take $ x_1$ in $a_1$ and $ z_1>x_1$ in $g_1$. Take $ y_1>z_1$ in $ a_1$ and $ t_1>z_1$ in $ g_1$. Then take $ x_2>t_1$ in $ a_2$, $ z_2>x_2$ in $ g_2$, $ y_2>z_2$ in $ a_2$, $t_2>y_2$ in $ g_2$ and continue the process. Then, put all numbers $ x_i, z_i$ in B and all the other numbers in A. Any infinite arithmetic or geometric progression will have elements both from A and B.", "Solution_11": "Yes, I think this kind of tools can be used for quite any countable set of things which has to be avoided.\r\n\r\nPierre.", "Solution_12": "Amazing solutions, harazi! Many thanks to you and Pierre! :lol: \r\n\r\nbtw, I didn't renew my CRUX subscription, so it was written on the September of 2002 issue that the problem was open...(bet it isn't now!) :lol:", "Solution_13": "Actually, now that I think about this again, I am a bit confused :? ...why is there no arithmetic sequence in B? First time I thought it was because one term has been taken out of every infinite sequence, but that doesn't mean it can't be replaced because there are other sequences containing that term without it being removed. So suppose we took out a_1 from x_1, there are surely other sequences containing a_1. But we take out a_i (different from a_1) from each of them, so when putting them altogether in B, a_1 exists as well as every other term in x_1. So we get an infinite sequence, don't we? I can see I'm really confused now... :? :lol: :?", "Solution_14": "Any infinite arithmetical sequence is defined by its first term and its ratio, which both are positive integers. Thus, the set of infinite arithmetical sequence is countable, so that you can enumerate it with the positive integers, and that is what Harazi said.\r\nThen, following this ordering, we choose (at least) one term in each of this sequences and put it in $A$, therefore none of these sequences is completely contained in $B$.\r\nThe choices made have to be done carefully to avoid that a three-term arithmetical sequence appears in $A$. That's why Harazi add the inequality condition in his choices. Note that he can do that because the initial arithmetical sequences are supposed to be infinite.\r\n\r\nThe key is that they form a countable set. That's why I said that this method can be extended to other such sets. For example, instead of arithmetical sequences, which are degree 1 case, you may forbid any sequence of the form $P(0),P(1), \\cdots$ where $P$ ranges over the set non-constant polynomials with integer coefficients. You may also add the geometrical sequences (since an union of two countable sets is countable) and so on...\r\n\r\nI hope it is clearer.\r\n\r\nPierre.", "Solution_15": "[quote]Any infinite arithmetical sequence is defined by its first term and its ratio, which both are positive integers. Thus, the set of infinite arithmetical sequence is countable, so that you can enumerate it with the positive integers, and that is what Harazi said. \nThen, following this ordering, we choose (at least) one term in each of this sequences and put it in A, therefore none of these sequences is completely contained in B.[/quote]\r\n\r\nI understand the enumeration part, and that we take term(s) out of each these arithmetic sequences. But as long as we take out finite number of terms (in fact, I think Harazi indicated that we take one term out of the sequence), the problem still remains. Consider the following:\r\n\r\n{1, 3, 5, 7, 9, 11, 13, 15, ... } is an infinite arithmetic sequence. So we take term(s) out of it (say 3, 7) and put them in A, then the rest in B. But we STILL have the infinite arithmetic sequence (9, 11, 13, 15, ... ) in B, don't we? Recall that a sequence, though might not be \"complete\", can still be infinite.\r\n\r\nSorry for my extreme slowness. :blush:", "Solution_16": "No, because the infinite remaining sequence is counted as another one in the enumeration : The first one is associated to $(1,2)$ and the remaining one to $(9,2)$ (first term and ratio). Thus, the remaining sequence will appear somewhere in the enumeration, so that at least one of its term will be put into $A$, and therefore the remaining sequence will not be entirely contained in $B$.\r\n\r\nPierre.", "Solution_17": "I got it. Thanks Pierre, for your vast patience. :)" } { "Tag": [ "ratio", "limit", "Euler", "geometric series", "calculus", "calculus computations" ], "Problem": "Prove that $ \\frac{2}{1} \\times \\frac{3}{2} \\times \\frac{5}{4} \\times \\frac{7}{6} \\times \\cdots \\times \\frac{p}{p \\minus{} 1} \\times \\cdots$ diverges.", "Solution_1": "Well we can tell that each term is greater than 1 and with infinite terms slightly greater then 1 multiplied by each other the result will be infinite and not 1. Everthing diverges unless the numbers form a geometric series that have fractional common ratio like 1+1/2+1/4... which we know equals 2. As p tends to infinite the value of the fraction is merely one because infinite/infinite +1 still equals 1 which we learn in limits.", "Solution_2": "[quote=\"varunrocks\"]Well we can tell that each term is greater than 1 and with infinite terms slightly greater then 1 multiplied by each other the result will be infinite and not 1.[/quote]\r\nNope. For example, $ \\lim_{n \\to \\infty} \\left( 1 \\plus{} \\frac {1}{n} \\right)^n \\equal{} e$. If you don't count that as a legitimate product, $ \\prod_{i \\equal{} 1}^{\\infty} \\left( 1 \\plus{} \\frac {1}{r^i} \\right)$ converges for all $ r > 1$.\r\n\r\nOne way to do this problem is to write the product as $ \\prod_{i \\equal{} 1}^{\\infty} \\left( \\frac {1}{1 \\minus{} \\frac {1}{p_i}} \\right)$, which \"converges\" to the harmonic series. This is the Euler product." } { "Tag": [ "parameterization", "function", "advanced fields", "advanced fields unsolved" ], "Problem": "I have the parametetrization $t \\mapsto (t^2 - 1, t(t^2-1))$ of the curve $y^2 = x^2(x+1) = x^3+x^2$. How does this help me to find an isomorphismn $K[t] \\to K[x,y]/\\langle{y^2 -x^2(x+1)}\\rangle$?", "Solution_1": "They are not isomorphic - Indeed, $K[t]$ is integrally closed, the other ring not. Another possibility was to see that the local ring of $K[x,y]/(y^2-x^2(x+1))$ at $(0,0)$ is not regular, whereas all local rings of $K[t]$ are regular. But you DO get a morphism $K[x,y]/(y^2-x^2(x+1)) \\to K[t]$ given by $x\\to t^2-1, y\\to t(t^2-1)$. The fact that this is not an isomorphism corresponds to the fact that the preimage of the point $(0,0)$ contains two points.", "Solution_2": "[quote=\"Peter Scholze\"]They are not isomorphic - Indeed, $K[t]$ is integrally closed, the other ring not. [/quote]How can I show the latter?\r\n\r\nBTW, I have $K = \\mathbf{R}$.", "Solution_3": "The field you work on does not really matter, but let's take $K=\\mathbb{R}$. $R=\\mathbb{R}[x,y]/(y^2-x^2(x+1))$ is not integrally closed, since $\\frac{y}{x}$ is an algebraic integer in the field of fractions of $R$: $(\\frac{y}{x})^2 - (x+1) = 0$. But clearly $\\frac{y}{x}$ is not contained in $R$." } { "Tag": [ "search" ], "Problem": "http://www.lmgtfy.com/?q=how+to+find+chuck+norris&l=1", "Solution_1": "Oh. My gosh.\r\n :rotfl: \r\nGoogle won't search for Chuck Norris because it knows you don't find Chuck Norris, he finds you.", "Solution_2": "Ah... 'let me google that for you.' I use that site all the time. :)", "Solution_3": "lol$ $" } { "Tag": [], "Problem": "[size=150]You have received a $2,000 invoice from the Acme Swamp Water Company for raw materials to use in your beer manufacturing plant. The terms are 3/10, net 40. If yor fail to take advantage of this discount. You are paying what interest rate?\n\n1.22.7%\n2.36,735%\n3.37.11%\n4.55.67%\n5.None of the above\n\nThanks[/size]", "Solution_1": "Where's the problem from?\r\n\r\nI should know this, but what is meant by 3/10 and net 4?", "Solution_2": "[quote=\"mathnerd314\"]Where's the problem from?\n\nI should know this, but what is meant by 3/10 and net 4?[/quote]\r\n\r\n\r\nThanks.\r\n \r\n\r\nLook at this Problem.\r\n\r\nIf credit terms of \"2/10, net 40\" are offered, the approximate cost of not taking the discount and paying at the end of the credit period would be closest to which of the following? (Assume a 365-day year.)\r\n\r\n1.18.6%\r\n\r\n2.4.3%\r\n\r\n3.24.8%\r\n\r\n4.30.0%\r\n\r\n\r\n[2/(100-2)]*[365/(40-10)]=.248", "Solution_3": "well can you tell us more about it? like can you explain what is terms 2/10 and net 40? what do they mean? and what subject is this? finance? :?", "Solution_4": "[quote=\"Palytoxin\"]well can you tell us more about it? like can you explain what is terms 2/10 and net 40? what do they mean? and what subject is this? finance? :?[/quote]\r\n\r\n\r\n Yes, finance." } { "Tag": [], "Problem": "It isn't.\r\n\r\nSome people are born in rich families, some aren't. Some people are born in family in with their parents has Uni degree, some aren't.\r\n\r\nThat could be also some reason why whole IMO isn't fair... You have there equal conditions, that is right, but because the life before the IMO fair isn't, whole IMO isn't fair eighter.\r\n\r\nLife isn't fair!!\r\n\r\nWhat do u think about it?", "Solution_1": "No one said it was. We can only do the best we can with what we have.\r\nYour comments about IMO can be applied to anything else in life. There is no way to make things perfectly equal for everyone because we can't measure quantitatively how much one person has that another doesn't.", "Solution_2": "I don't think we share the same deffenition for the word 'fair'. :wink:", "Solution_3": "Maybe I should begin lobbying again for the creation of a Self Pity forum.", "Solution_4": "[quote=\"JBL\"]Maybe I should begin lobbying again for the creation of a Self Pity forum.[/quote]\r\n\r\nIt would pretty much instantly become the most popular fourm. Math is hard, and peoples egos are fragile. :P :roll:", "Solution_5": "Life is ruled by the rules of nature. They are what they are, we can not judge them, though we are all subjects to it. This applies to absolutely all segments of our lives. Which are just a small part of ubiquitous nature :) \r\n\r\nAnd of course, whatever you do, you can't fool nature :D", "Solution_6": "[quote=\"setout\"]It isn't.\n\nSome people are born in rich families, some aren't. Some people are born in family in with their parents has Uni degree, some aren't.\n\nThat could be also some reason why whole IMO isn't fair... You have there equal conditions, that is right, but because the life before the IMO fair isn't, whole IMO isn't fair eighter.\n\nLife isn't fair!!\n\nWhat do u think about it?[/quote]\r\nWas it ever assumed life [i]was[/i] fair? Are you delusional?", "Solution_7": "Wait a minute, setout, isn't he the same user who started that topic about Peter Scholze's clicking of pens and who made at least three separate topics about acquiring romantic partners? :|", "Solution_8": "[quote=\"fredbel6\"]Wait a minute, setout, isn't he the same user who started that topic about Peter Scholze's clicking of pens and who made at least three separate topics about acquiring romantic partners? :|[/quote]\r\nyeah, he is.\r\n\r\nLife was never fair, never is fair, and never will be fair. There is completely no doubt about that. :!: \r\nThe thing that Life is not fair is pretty much a fact accepted by nearly everyone in the world.", "Solution_9": "I hope you understand what I mean when I say \"The greatest inequity is complete and total equality\".", "Solution_10": "Maybe a bit unrelated. But there was a user with the signature something like \"Life is unfair. Otherwise I could not get what I do not deserve.\". :D", "Solution_11": "First of all we are differend; clever or not beautifull or not, etc", "Solution_12": "[quote=\"orl\"]Maybe a bit unrelated. But there was a user with the signature something like \"Life is unfair. Otherwise I could not get what I do not deserve.\". :D[/quote]\r\n\r\nhttp://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=11966", "Solution_13": "[quote=\"setout\"]It isn't.\n\nSome people are born in rich families, some aren't. Some people are born in family in with their parents has Uni degree, some aren't.\n\nThat could be also some reason why whole IMO isn't fair... You have there equal conditions, that is right, but because the life before the IMO fair isn't, whole IMO isn't fair eighter.\n\nLife isn't fair!!\n\nWhat do u think about it?[/quote]\r\n\r\nHello setout. How was IMO? And did you find the girl you were looking for? (I actually ask these because you seem upset)", "Solution_14": "IMO tests how well people can solve math problems. well, some people will be better than others, and that's competition. it's unfair because some people are naturally more effective with math problems than others and were born with more math aptitude. but society does not celebrate mediocrity. if life were completely fair, competition is obsolete. read the book Anthem by Ayn Rand. very intense book.", "Solution_15": "that's why you should sell all your things and become a hobo and eventually end up in an asylum. life isn't fair and it never will be...sometimes you have the \"unfair\" advantage and sometimes not.\r\n\r\nedit: please don't think i meant the hobo part...", "Solution_16": "[quote=\"wobster109\"] read the book Anthem by Ayn Rand. very intense book.[/quote]\r\n\r\nbarf\r\n\r\nthe last thing we need is more randroids", "Solution_17": "[quote=\"wobster109\"] \nsociety does not celebrate mediocrity. [/quote]\r\n\r\nI completely disagree with you there. Unfortunately, in most cases, society DOES celebrate mediocrity :!:", "Solution_18": "[quote=\"Cesario\"][quote=\"wobster109\"] \nsociety does not celebrate mediocrity. [/quote]\n\nI completely disagree with you there. Unfortunately, in most cases, society DOES celebrate mediocrity :!:[/quote]\r\n\r\nPerhaps you can offer some proof or example? I haven't heard of anyone being rewarded for being normal or for doing something ordinary.", "Solution_19": "[quote=\"Elemennop\"]\nbarf\n\nthe last thing we need is more randroids[/quote]\r\n\r\nat least it's better than watching so much tv that you hallucinate pokemon watching tv. :P", "Solution_20": "All men are created equal. All men shall rightly share in the pursuit of life, liberty, and happiness. All goods and services shall be subject to the prices assigned to them by the government and, thus, no man shall stand richer nor any man poorer in the wake of his neighbour. In the last few sentences I have adressed the problem proposed by the originator of this thread and, at the same time, created a form of government free of imperfection.", "Solution_21": "[quote=\"Bictor717\"][quote=\"Cesario\"][quote=\"wobster109\"] \nsociety does not celebrate mediocrity. [/quote]\n\nI completely disagree with you there. Unfortunately, in most cases, society DOES celebrate mediocrity :!:[/quote]\n\nPerhaps you can offer some proof or example? I haven't heard of anyone being rewarded for being normal or for doing something ordinary.[/quote]\r\n\r\nExamples are all around you, Bictor. I did not mean there were actually prizes given for mediocrity, but generally, average people are always better accepted in and by the society than geniuses are. Don't you agree?", "Solution_22": "[quote=\"G-UNIT\"]All men are created equal. All men shall rightly share in the pursuit of life, liberty, and happiness. All goods and services shall be subject to the prices assigned to them by the government and, thus, no man shall stand richer nor any man poorer in the wake of his neighbour. In the last few sentences I have adressed the problem proposed by the originator of this thread and, at the same time, created a form of government free of imperfection.[/quote]\r\n\r\nin short, the ideal communist government. but is that really fair? suppose you work much harder than your neighbor. should you two really be at absolute economic equality?", "Solution_23": "[quote=\"wobster109\"][quote=\"G-UNIT\"]All men are created equal. All men shall rightly share in the pursuit of life, liberty, and happiness. All goods and services shall be subject to the prices assigned to them by the government and, thus, no man shall stand richer nor any man poorer in the wake of his neighbour. In the last few sentences I have adressed the problem proposed by the originator of this thread and, at the same time, created a form of government free of imperfection.[/quote]\n\nin short, the ideal communist government. but is that really fair? suppose you work much harder than your neighbor. should you two really be at absolute economic equality?[/quote]\r\n\r\nYes! Really no...communism is actually wrong however appealing so when you become a power hungry dictator don't do it. Communism can't affect brains... doctor and bagger at grocery store..same?", "Solution_24": "[quote=\"Bictor717\"][quote=\"Cesario\"][quote=\"wobster109\"] \nsociety does not celebrate mediocrity. [/quote]\n\nI completely disagree with you there. Unfortunately, in most cases, society DOES celebrate mediocrity :!:[/quote]\n\nPerhaps you can offer some proof or example? I haven't heard of anyone being rewarded for being normal or for doing something ordinary.[/quote]\r\n\r\nWhat about awards for perfect attendance? They reward you for consistantly showing up every day." } { "Tag": [ "function", "trigonometry", "probability and stats" ], "Problem": "suppose $ \\{\\xi_{n} \\}_{n \\geq 1}$ be sequence of independent random variables such that :\r\n$ \\mathbb{P} \\{\\xi_{n} \\equal{} \\minus{} n \\} \\equal{} \\mathbb{P} \\{\\xi_{n} \\equal{} n \\} \\equal{} \\frac {1}{n}$\r\nand \r\n$ \\mathbb{P} \\{\\xi_{n} \\equal{} 0 \\} \\equal{} 1 \\minus{} \\frac {2}{n}$ \r\nShow that such sequence of independent random variables don't satisfy the strong law of large numbers .\r\nI know it's simple question for the peoples who are familiar with Probabilities and Statistics.\r\nThis is what I have done.\r\nAt first let's check that it satisfay to the law of large numbers.\r\nSince $ \\mathbb{E}\\xi_{n} \\equal{} 0$ hence by P.Levy's theorem it's enough to show that \r\n$ f_{\\eta_{n}}(t) \\to f(t) \\equal{} 1$ \r\nwhere $ f_{n}(t)$ and $ f(t)$ is correspondent characteristic function of $ \\eta_{n} \\equal{} \\frac {1}{n}\\sum_{1\\leq j\\leq n} \\xi_{j}$ and $ 0$.\r\nnote that $ f_{\\xi_{j}} (t) \\equal{} 1 \\minus{} \\frac {2}{j} \\plus{} \\frac {2\\cos jt}{j}$ thus we have\r\n\\[ f_{\\eta_{n}}(t) \\equal{} \\prod_{j \\equal{} 1}^{n}f_{\\xi_{j}/n}(t) \\equal{} \\prod_{j \\equal{} 1}^{n}\\left( 1 \\minus{} \\frac {2}{j} \\plus{} \\frac {2\\cos jt/n}{j}\\right) \\approx \\exp \\left( \\sum_{j \\equal{} 1}^{n} \\minus{} \\frac {2}{j} \\plus{} \\frac {2 \\minus{} (tj)^{2}/n^{2}}{j} \\right) \\approx \\exp \\left( \\minus{} \\frac {t^{2}}{2} \\right)\r\n\\]\r\nthus it don't satisfy the (weak ) law of large numbers and hence strong.\r\n\r\nI'd like to see any other solutions . (simple solutions )", "Solution_1": "or you can use the (second) Borel-Cantelli lemma: \r\nhttp://en.wikipedia.org/wiki/Borel-Cantelli_lemma", "Solution_2": "I am doubting that it will help. :wink: \r\nCan you write a part of your variant of this theorem proof ?", "Solution_3": "the events $ A_n \\equal{} \\{ \\xi_n\\equal{}n \\}$ occur infinitely many times, almost surely, since $ \\sum P(\\xi_n\\equal{}n)\\equal{}\\plus{}\\infty$ and these events are independant (second Borel Cantelli lemma). Hence $ S_n \\equal{} \\frac{\\xi_1 \\plus{} \\ldots \\plus{} \\xi_n}{n}$ is not a Cauchy sequence since $ S_{n\\plus{}1}\\minus{}S_n$ does not go to $ 0$." } { "Tag": [ "videos", "MATHCOUNTS" ], "Problem": "Will someone give me problems including Venn Diagrams and then explain how to use them?\r\nThanks.", "Solution_1": "hard or easy", "Solution_2": "Both would be nice.", "Solution_3": "Look on alcumus. There is a whole section of problems and even videos explaining the concept.", "Solution_4": "I don't think I have unlocked that part yet...\r\nI just need one or two example problems and solutions.", "Solution_5": "[hide=\"Easier Problem\"]Suppose every student in the school takes French, Spanish, both, or neither. There are 33 French, 33 Spanish, and 12 neither. With 66 people in the school, how many students take French and Spanish?[/hide]\r\nTry this.\r\n\r\nBy the way, most problems (like mine) won't include a direct instruction to use a venn diagram. As you get more advanced you won't even need one (PIE)", "Solution_6": "If set $ A$ has $ n$ elements, compute the total number of pairs of sets $ (X, Y)$ such that $ \\emptyset \\subseteq X \\subseteq Y \\subseteq A$.\r\n[hide=\"Solution\"]The Venn diagram is made of three regions: $ X$, $ Y \\setminus X$ and $ A\\setminus Y$. Any of the $ n$ elements may belong to any of these regions, so the answer is $ 3^n$.[/hide]", "Solution_7": "Most problems in Mathcounts will not use that notation.\r\n\r\nActually, most high school contests will not either.", "Solution_8": "Easy problem; \r\n\r\nA school teaches the languages Spanish and French. There are 40 students that go to the school. Five students don't learn any languages. There are 6 people who decided to learn both languages, and the spanish class is seven students larger than the French class. How many people are taking French? \r\n\r\n\r\nHard problem; \r\n\r\nAt Mallard High School there are three intermural sports leagues: football,\r\nbasketball, and baseball. There are 427 students participating in these sports: 128\r\nplay on football teams, 291 play on basketball teams, and 318 play on baseball\r\nteams. If exactly 36 students participate in all three of the sports, how many\r\nstudents participate in exactly two of the sports?\r\n\r\nI got that one from a competition. Enjoy! :)", "Solution_9": "[hide=\" Easy problem\"]\n\n$ 40 \\minus{} 5 \\equal{} 35$ study at least one language $ x$ study Spanish but not French, so $ x\\minus{}7$ study French but not Spanish.\n\n$ x\\plus{}6 \\plus{}x\\minus{}7 \\equal{} 35$\n\n$ x\\equal{}18$\n\n$ 17$ study French.\n\n[/hide]\n\n[hide=\" Hard problem\"]\n\n\n$ IEP$ gives $ 427 \\equal{} 128 \\plus{} 291 \\plus{}318 \\minus{} x \\plus{}36$\n\n$ x\\equal{}346$.\n\n[/hide]", "Solution_10": "Hard problem:\r\n\r\n427 = 128 + 291 +318 - x - 2 * 36\r\n\r\nx=238.", "Solution_11": "Hi bull98. You're right. I forgot to subtract the $ 36$'s." } { "Tag": [ "search", "function", "number theory unsolved", "number theory" ], "Problem": "Prove that the number$2^{2^{n}}+1$ has a prime factor greater than $2^{n+2}(n+1)$", "Solution_1": "This problem posted (and somebody solved) by harazi. Please search or PM to he :wink:", "Solution_2": "http://www.mathlinks.ro/Forum/viewtopic.php?highlight=factor+prime&t=98980\r\ni don't think this.LINK? Do you know? :lol:", "Solution_3": "n.t.tuan: could you post the link? i don't have nice results with search function\r\n\r\nthe topic http://www.mathlinks.ro/Forum/viewtopic.php?highlight=factor+prime&t=98980 don't have solutions :blush:", "Solution_4": "see http://www.mathlinks.ro/Forum/viewtopic.php?t=46125 :wink:" } { "Tag": [ "algebra", "polynomial" ], "Problem": "Find all positive integer solutions to $abc - 2 = a + b + c$.", "Solution_1": "Here's how I'd try approaching this problem. \r\n\r\n[hide=\"Possible Method\"]$abc=a+b+c+2$, so we're looking for a cubic polynomial with the sum of the roots 2 larger than the product.[/hide]", "Solution_2": "4everwise, i do not see how relating them to the roots of a polynomial would help. could you please explain a little more.", "Solution_3": "[hide=\"Solution\"]\n$abc = a + b + c + 2$\n\nWe will do a little casework before proving the nonexistence of large solutions. First, assume WLOG that $a \\ge b \\ge c$. \n\nCase: $c = 1$. Our equation becomes $ab = a + b + 3 \\implies (a-1)(b-1) = 4$, which has solutions $\\boxed{a = 5, b = 2, a = 3, b = 3}$.\n\nCase: $c = 2$. We have $2ab = a + b + 4$ which has solution $\\boxed{a = 2, b = 2}$. Now, suppose we increment either $a$ or $b$. The LHS increases by at least two while the RHS increases by only one. Hence there cannot be any larger solutions.\n\nCase: $c > 2$. Suppose we begin with $a = b = c = 3$ (we have already verified that no smaller solutions exist) and attempt to find a larger solution by incrementing any of $a, b, c$. The LHS increases by at least $9$ (from its initial value of $27$) while the RHS increases by one (from its initial value of $11$). Hence there cannot be any larger solutions.\n\nQED.\n[/hide]" } { "Tag": [ "ratio", "geometry" ], "Problem": "The sides of $\\triangle ABC$ are $AB=8$, $BC=12$, and $AC=9$. if $\\overline r{GF}$, $\\overline{IE}$, and $\\overline{HD}$ are respectively perpendicular to the sides as shown, and $GH=4$, find the ratio of the area of $\\triangle GHI$ to that of $\\triangle ABC$.", "Solution_1": "Okay this is what Elemmnop said. And I get it now :D \r\n\r\n\r\nSince AB= 8 and GH= 4, and since the two triangles are similar, the area ratio from the small to large triangle is 4 squared to 8 squared. Therefore the ratio is 16:64 which simplifies to 1:4.\r\n\r\n1:4 is then the answer.", "Solution_2": "How do you know the triangles are similar?", "Solution_3": "triangles in problems liket this are similar most of the times hehe :P", "Solution_4": "Think angles with perpendicular sides.", "Solution_5": "Ok, I figured out why they are similar, but how do we know $GH$ is the smallest side?", "Solution_6": "By establishing which angle is equal to which, determine the pairs of the corresponding vertices.", "Solution_7": "[quote=\"Farenhajt\"]By establishing which angle is equal to which, determine the pairs of the corresponding vertices.[/quote]\r\nOh, of course! I should have realized that. Thank you all for your help. :)", "Solution_8": "yeah, farenhajt is awesome at explaining :D \r\n\r\nI asked like several people about a similar problem from 1991 Vestavia Tournament it was like number 21. except the measures were different. But exact same picture, exact same segments, just different measure for GH AB and BC" } { "Tag": [ "limit", "real analysis", "real analysis theorems" ], "Problem": "Let $ X,Y$ be metric spaces and $ f: X\\rightarrow Y$. How to prove following equivalence:\r\n $ x\\in cl(A) \\Rightarrow f(x)\\in cl(f(A)) \\Leftrightarrow x\\equal{}\\lim x_n \\Rightarrow f(x)\\equal{}\\lim f(x_n)$ ?\r\n\r\ncl(A) is closure of A", "Solution_1": "The $ \\Leftarrow$ direction - If $ x \\in cl(A)$, then there is $ x_n \\in A$ with $ x_n \\to x$ so by assumption $ f(x_n) \\to f(x)$, but $ f(x_n) \\in A$ so $ f(x) \\in cl(A)$\r\n\r\nI dont think the other direction is true. For example let $ f: \\mathbb{R}\\to \\mathbb{R}$ given by $ f(x) \\equal{} x$ except $ f(0) \\equal{} 1$ and $ f(1) \\equal{} 0$. Then take $ A\\equal{} (0,1)$ and notice that $ x \\in cl(0,1)\\equal{}[0,1] \\Rightarrow f(x) \\in cl(f(0,1)) \\equal{} [0,1]$, but, for example $ x_n \\equal{} 1/n \\to 0$, but $ f(1/n)\\to 0 \\not \\equal{} f(0) \\equal{} 1$" } { "Tag": [ "trigonometry" ], "Problem": "I need some help with solving a trig problem using difference formulas.\r\n\r\ncosxcosy(tanx+tany)\r\n\r\nI know it equals sin(x+y) cause its in the back of my book but I need to see the steps to actually find that out.", "Solution_1": "\\[\\begin{aligned}\\cos x\\cos y\\left(\\tan x+\\tan y\\right) & = \\cos x\\cos y\\left(\\frac{\\sin x}{\\cos x}+\\frac{\\sin y}{\\cos y}\\right) \\\\ & = \\cos y\\sin x+\\cos x\\sin y = \\sin(x+y).\\end{aligned}\\]", "Solution_2": "Yeah...I dont get how that works out at all. Is there any more work?" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "In R, show that all intervals are Borel sets. \r\n\r\nI'm a bit confused here. If I understand correctly, a Borel set is the smallest sigma-algebra of set S that contains all the elements of set S. As a sigma-algebra, it must contain the empty set, it must contain the complement of any element of the set and it must contain the union of any elements of the set. \r\n\r\nI just don't see how to prove this statement.\r\n\r\nThanks for any help", "Solution_1": "Hint: $[a,b]=\\bigcap_{n=1}^{\\infty}(a-1/n,b+1/n)$." } { "Tag": [ "inequalities", "topology", "triangle inequality", "real analysis", "real analysis solved" ], "Problem": "This is very easy, but it's sort of interesting, I think:\r\n\r\nFind the necessary and sufficient conditions which a real set must satisfy in order to be the set of distances between pairs of points in some metric space.", "Solution_1": "Necessary and sufficient condition: the set contains only positive numbers. (we ignore trivial segments and exclude zero as a distance)\r\n\r\nFor a set $A$ of positive numbers, let our metric space be $A\\cup0$, with metric $d(x,y)=\\max(x,y)$. Since every triangle is isosceles, the triangle inequality is satisfied. Obviously, the possible distances are exactly the elements of $A$." } { "Tag": [ "inequalities", "geometry", "Putnam" ], "Problem": "If 3x+2y+5z=1 then find the minumum value of (1/x)+(1/y)+(1/z)", "Solution_1": "By Cauchy, we have:\r\n\r\n$\\displaystyle \\left( 3x+2y+5z \\right)\\left(\\frac {1}{x}+\\frac {1}{y}+\\frac {1}{z}\\right)\\geq (\\sqrt 3 + \\sqrt 2 +\\sqrt 5)^2$\r\n\r\nSince $\\left( 3x+2y+5z \\right)=1$, we see that $(\\sqrt 3 + \\sqrt 2 + \\sqrt 5)^2$ is the minimum possible value (it shouldn't be hard to show that it's attainable.)\r\n\r\nHowever, this is only valid for positive numbers. In fact, if we allow negative numbers, it's not hard to show that we can make the quantity go to negative infinity.", "Solution_2": "mmr:\r\n\r\nThe Cauchy-Schwarz (Buniakowski) inequality states that, for any $x_i,y_i\\in\\mathbb{R}$\r\n\r\n$\\displaystyle\\left(\\sum^{n}_{i=0} x_i^2\\right)\\left(\\sum^{n}_{i=0} y_i^2\\right)\\geq\\left(\\sum^{n}_{i=0} x_iy_i\\right)^2$ with equality iff $\\frac {a_i}{b_i}=\\frac{a_j}{b_j}$ for all $i,j$ such that $0\\leq\\ i,j,\\leq n$.", "Solution_3": "actually, I used Cauchy to finish off an ugly geometry problem (which I coordinatized) on Putnam 96 A2 the other day." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "geometry", "circumcircle", "analytic geometry" ], "Problem": "Let $A_1A_2A_3$ be a triangle and let $\\omega_1$ be a circle in its plane passing through $A_1$ and $A_2.$ Suppose there exist circles $\\omega_2, \\omega_3, \\dots, \\omega_7$ such that for $k = 2, 3, \\dots, 7,$ $\\omega_k$ is externally tangent to $\\omega_{k-1}$ and passes through $A_k$ and $A_{k+1},$ where $A_{n+3} = A_{n}$ for all $n \\ge 1$. Prove that $\\omega_7 = \\omega_1.$", "Solution_1": "When replying to the problem, I ask that you make posts for solutions and submit comments, jokes, smilies, etc. separately. Furthermore, please do not \"hide\" any portion of the solution. Please use LaTeX for posting solutions. Thanks.", "Solution_2": "Clearly, $\\omega_i$ is determined by its center $O_i$, since we know that it passes through $A_i,A_{i+1}$. We also know that for all $i,\\ O_i,A_{i+1},O_{i+1}$ are collinear. We must prove that in these conditions, $O_7=O_1$.\r\n\r\nSuppose we prove that for all $i$, the angles $\\angle OA_iO_i$ and $\\angle OA_iO_{i+2}$ are equal (where $O$ is the circumcenter of $A_iA_{i+1}A_{i+2}$). The result would then follow, since $O_i\\mapsto O_{i+3}$ would be an involution on the perpendicular bisector $d_i$ of $A_iA_{i+1}$, mapping a point on this line to its harmonic conjugate wrt $O$ and the intersection of the tangents in $A_i,A_{i+2}$ to the circumcircle of $A_iA_{i+1}A_{i+2}$. Since $O_1\\mapsto O_4\\mapsto O_7$, we must have $O_7=O_1$, as desired.\r\n\r\nAll that\u2019s left to prove is:\r\n\r\nLet $O_1$ be any point on $d_1$, and put $O_2=O_1A_2\\cap d_2,\\ O_3=O_2A_3\\cap d_3$. Then, $A_1O_1,\\ A_1O_3$ make equal angles with $OA_1$.\r\n\r\nThis, however, is a simple angle chase :).", "Solution_3": "[quote=\"MithsApprentice\"]Let $ A_1A_2A_3$ be a triangle and let $ \\omega_1$ be a circle in its plane passing through $ A_1$ and $ A_2.$ Suppose there exist circles $ \\omega_2, \\omega_3, \\dots, \\omega_7$ such that for $ k \\equal{} 2, 3, \\dots, 7,$ $ \\omega_k$ is externally tangent to $ \\omega_{k \\minus{} 1}$ and passes through $ A_k$ and $ A_{k \\plus{} 1},$ where $ A_{n \\plus{} 3} \\equal{} A_{n}$ for all $ n \\ge 1$. Prove that $ \\omega_7 \\equal{} \\omega_1.$[/quote]\r\nSorry to revive an old topic, but I have a rather different solution than the one posted previously:\r\n[hide=\"Solution\"]\nInvert with arbitrary radius about $ A_1$. Let the image of point $ P$ map to $ P'$. Now, notice that $ \\omega_3$ and $ \\omega_4$ are both tangent and pass through $ A_1$, so they are tangent at $ A_1$. The same applies for $ \\omega_6$ and $ \\omega_7$. Now, since $ \\omega_3$ passes through $ A_3$, $ \\omega_3'$ is a line through $ A_3'$ and similarly, $ \\omega_4'$ passes through $ A_2'$. Furthermore, $ \\omega_3'\\parallel \\omega_4'$. Now, $ \\omega_5$ passes through both $ A_2$ and $ A_3$ and is tangent to $ \\omega_4$, so $ \\omega_5'$ passes through $ A_2'$ and $ A_3'$ and is tangent to $ \\omega_4'$ (at $ A_2')$. $ \\omega_6$ is tangent to $ \\omega_ 5$ and passes through $ A_3$ and $ A_1$, so $ \\omega_6'$ is a line passing through $ A_3'$ and tangent to $ \\omega_5'$. $ \\omega_7$ passes through $ A_1$ and $ A_2$ and was tangent to $ \\omega_6$ at $ A_1$, so $ \\omega_7'$ is a line parallel to $ \\omega_6'$ passing through $ A_2'$. $ \\omega_2$ passes through $ A_2$ and $ A_3$ and is tangent to $ \\omega_3$ at $ A_3$, so $ \\omega_2'$ is a circle tangent to $ \\omega_3'$ at $ A_3'$. $ \\omega_1$ is tangent to $ \\omega_2$ at $ A_2$ and passes through $ A_1$, so $ \\omega_1'$ is a line through $ A_2'$, tangent to $ \\omega_2'$ and $ A_2'$. Let $ \\omega_4'$ and $ \\omega_6'$ meet at $ P$ and let $ \\omega_1'$ and $ \\omega_3'$ meet at $ Q$. Furthermore, let $ \\omega_5'$ meet $ \\omega_1'$ and $ \\omega_3'$ at $ B$ and $ A$ respectively and let $ \\omega_2'$ meet $ \\omega_6'$ and $ \\omega_4'$ at $ C$ and $ D$ respectively. Since $ QA_3'$ and $ QA_2'$ are tangents to $ \\omega_5'$, we have that $ \\angle QA_3'A_2' \\equal{} \\angle QA_2'A_3'$. Since $ \\omega_4'\\parallel \\omega_3'$, $ \\angle QA_3'A_2' \\equal{} \\angle PA_2'A_3'$. For a similar reason as before,\n\\[ \\angle PA_2'A_3' \\equal{} \\angle PA_3'A_2'\\implies \\angle PA_3'A_2' \\equal{} \\angle QA_2'A_3'\\implies QA_2'\\parallel PA_3'\\implies \\omega_1'\\parallel \\omega_6'\\implies \\omega_1' \\equal{} \\omega_7'\\]\nsince $ \\omega_7'\\parallel \\omega_6'$ and $ \\omega_7'$ passes through $ A_2'$. Finally, we get the result that $ \\omega_1 \\equal{} \\omega_7$, as desired. [/hide]", "Solution_4": "[hide=\"Easy Coordinates\"]\nLet the centers of $ \\omega_1, \\omega_2, \\ldots, \\omega_7$ be $ O_1, O_2, \\ldots, O_7$. Since $ \\omega_k$ and $ \\omega_{k\\plus{}1}$ are externally tangent at $ A_{k\\plus{}1}$, we see that $ A_{k\\plus{}1}$ is the midpoint of $ O_k$ and $ O_{k\\plus{}1}$. Let $ A_1\\equal{}(0,a)$, $ A_2\\equal{}(b,0)$, and $ A_3\\equal{}(c,0)$, and let $ O_1\\equal{}(x,y)$. Then it is easy to calculate that\n\n$ O_2\\equal{}(2b\\minus{}x,\\minus{}y)$\n\n$ O_3\\equal{}(2c\\minus{}2b\\plus{}x,y)$\n\n$ O_4\\equal{}(2b\\minus{}2c\\minus{}x, 2a\\minus{}y)$\n\n$ O_5\\equal{}(2c\\plus{}x,y\\minus{}2a)$\n\n$ O_6\\equal{}(\\minus{}x,2a\\minus{}y)$\n\n$ O_7\\equal{}(x,y)$\n\nas desired.\n[/hide]", "Solution_5": "[quote=\"E^(pi*i)=-1\"]Since $ \\omega_k$ and $ \\omega_{k \\plus{} 1}$ are externally tangent at $ A_{k \\plus{} 1}$, we see that $ A_{k \\plus{} 1}$ is the midpoint of $ O_k$ and $ O_{k \\plus{} 1}$.[/quote]\r\n\r\nWhy?\r\n\r\n darij", "Solution_6": "Oops, you're right, it's wrong. :oops: \r\n\r\nIt seems like there should be a constant ratio between radii of circles through $ A_1$ and $ A_2$ and those through $ A_2$ and $ A_3$, though . . . maybe there is a way to multiply the coordinates by these factors such that it works out?", "Solution_7": "Hmm I think you can do this with just trivial angle chasing: $ A=\\angle{A_{3}A_{1}A_{2}},B=\\angle{A_{1}A_{2}A_{3}},C=\\angle{A_{2}A_{3}A_{1}},X=\\angle{O_{1}A_{1}A_{2}}$. Then, measures of the angles $ \\angle{O_{i}A_{i}A_{i+1}}$ are easy to get, we find that $ \\angle{O_{7}A_{1}A_{2}}=X$, and we also need $ O_{1},O_{7}$ on the same side of $ A_{1}A_{2}$ (which follows from the fact that we need $ A_{i}$ between $ O_{i+1}$ and $ O_{i+2}$, blah.", "Solution_8": "We let the radius of the circle $\\omega_i$ by $r_i$.As usual denote the side opposite to vertex $A_i$ by $a_i$ for $i=1,2,3$.\nLet $O_1A_2A_1=\\theta$,where $O_1$ is the centre of $\\omega_1$.\nBy applications of sine rule we may easily deduce the following relations:\n\n$\\frac{r_1}{r_2}=-\\frac{a_1}{a_3}\\frac{\\cos\\theta}{\\cos(A_2+\\theta)}$\n\n$\\frac{r_2}{r_3}=-\\frac{a_2}{a_1}\\frac{\\cos(A_2+\\theta)}{\\cos(A_2-A_3+\\theta)}$\n\n$\\frac{r_3}{r_4}=-\\frac{a_3}{a_2}\\frac{\\cos(A_2-A_3+\\theta)}{\\cos(A_1-A_2+A_3-\\theta)}$\n\nMultiplying these out we get\n\n$\\frac{r_1}{r_4}=-\\frac{\\cos\\theta}{\\cos(A_1-A_2+A_3-\\theta)}$.\n\nSimilarly it is easy to observe that\n\n$\\frac{r_4}{r_7}=-\\frac{\\cos(A_1-A_2+A_3-\\theta)}{\\cos(A_1-A_2+A_3-(A_1-A_2+A_3-\\theta))}=-\\frac{\\cos(A_1-A_2+A_3-\\theta)}{\\cos\\theta}$.\n\nMultiplying these equalities we obatain $r_7=r_1$.Also because $7 \\equiv 1\\pmod{3}$ we note that $\\omega_7$ passes through $A_1$ and $A_2$.We thus obtain $\\omega_7=\\omega_1$.", "Solution_9": "[hide=Hint]You only care about the circle centers, which happen to be uniquely determined by the previous center ($O_2=\\overline{O_1A}\\cap \\ell$, where $\\ell$ is the perpendicular bisector of the next side).", "Solution_10": "The idea is to keep track of the subtended arc $\\widehat{A_i A_{i+1}}$ of $\\omega_i$ for each $i$. To this end, let $\\beta = \\measuredangle A_1 A_2 A_3$, $\\gamma = \\measuredangle A_2 A_3 A_1$ and $\\alpha = \\measuredangle A_1 A_2 A_3$.\n\n\n\n[asy] size(8cm); defaultpen(fontsize(9pt)); pair A2 = (0,0); pair O1 = (-8,0); pair O2 = (5,0); pair A1 = abs(A2-O1)*dir(285)+O1; pair A3 = abs(A2-O2)*dir(245)+O2; pair O3 = extension(O2, A3, midpoint(A1--A3), midpoint(A1--A3)+dir(A1-A3)*dir(90));\n\nfilldraw(circle(O1,abs(A1-O1)), invisible, deepgreen); filldraw(circle(O2,abs(A2-O2)), invisible, deepgreen); filldraw(circle(O3,abs(A3-O3)), invisible, deepgreen); filldraw(A1--A2--A3--cycle, invisible, red);\n\npair J = (0,-3.2); draw(O1--O2--O3, deepcyan+dotted);\n\ndot(\"$O_1$\", O1, dir(90), deepcyan); dot(\"$O_2$\", O2, dir(90), deepcyan); dot(\"$O_3$\", O3, dir(-90), deepcyan); dot(\"$A_1$\", A1, dir(A1-J), red); dot(\"$A_2$\", A2, dir(A2-J), red); dot(\"$A_3$\", A3, dir(A3-J), red);\n\nlabel(\"$\\alpha$\", A1, 2.4*dir(J-A1), red); label(\"$\\beta$\", A2, 2*dir(J-A2), red); label(\"$\\gamma$\", A3, 2*dir(J-A3), red); [/asy]\n\n\n\nInitially, we set $\\theta = \\measuredangle O_1 A_2 A_1$. Then we compute \\begin{align*} \t\\measuredangle O_1 A_2 A_1 &= \\theta \\\\ \t\\measuredangle O_2 A_3 A_2 &= -\\beta-\\theta \\\\ \t\\measuredangle O_3 A_1 A_3 &= \\beta-\\gamma+\\theta \\\\ \t\\measuredangle O_4 A_2 A_1 &= (\\gamma-\\beta-\\alpha)-\\theta \\\\ \\end{align*} and repeating the same calculation another round gives \\[ \\measuredangle O_7 A_2 A_1 = k-(k-\\theta) = \\theta \\] with $k = \\gamma-\\beta-\\alpha$. This implies $O_7 = O_1$, so $\\omega_7 = \\omega_1$.\n", "Solution_11": "I think you can do this with moving points, but here is the angle chase approach.\n\nLet $O_i$ be the center of $\\omega_i$ for $i=1,2,\\dots,7$. \n\nObserve\n\\[\\measuredangle O_7A_1A_2=-\\measuredangle A_2A_1A_3-\\measuredangle A_3A_1O_6=-\\measuredangle A_2A_1A_3-\\measuredangle O_6A_3A_1=-\\measuredangle A_2A_1A_3+\\measuredangle A_1A_3A_2+\\measuredangle A_2A_3O_5=\\]\n\\[-\\measuredangle A_2A_1A_3+\\measuredangle A_1A_3A_2+\\measuredangle O_5A_2A_3=-\\measuredangle A_2A_1A_3+\\measuredangle A_1A_3A_2-\\measuredangle A_3A_2A_1-\\measuredangle A_1A_2O_4.\\]\nA similar angle chase finishes.", "Solution_12": "Let $O_i$ be the center of $\\omega_i$. Also let $\\theta_1=\\measuredangle A_1A_2A_3,\\theta_2=\\measuredangle A_2A_3A_1,\\theta_3=\\measuredangle A_3A_1A_2$. Observe that $\\omega_1$ is uniquely determined by $\\measuredangle O_1A_2A_1$, and $\\omega_7$ is uniquely determined by $\\measuredangle O_7A_2A_1$, so if we can show those are equal we're done. Now, let $\\measuredangle O_1A_2A_1=\\theta$. By basic angle chasing (since $O_i,A_{i+1},O_{i+1}$ are collinear), we can find:\n\\begin{align*}\n\\measuredangle O_1A_2A_1&=\\theta\\\\\n\\measuredangle O_2A_3A_2&=-\\theta_1-\\theta\\\\\n\\measuredangle O_3A_1A_2&=\\theta_1-\\theta_2+\\theta\\\\\n\\measuredangle O_4A_2A_1&=-\\theta_1+\\theta_2-\\theta_3-\\theta\\\\\n\\measuredangle O_5A_3A_2&=\\theta_1-\\theta_2+\\theta_3-\\theta_1+\\theta\\\\\n\\measuredangle O_6A_1A_2&=-\\theta_1+\\theta_2-\\theta_3+\\theta_1-\\theta_2-\\theta\\\\\n\\measuredangle O_7A_2A_1&=\\theta_1-\\theta_2+\\theta_3-\\theta_1+\\theta_2-\\theta_3+\\theta=\\theta.\n\\end{align*}\nhence $\\omega_1=\\omega_7$ as desired. $\\blacksquare$", "Solution_13": "Interesting problem. We will aply an inversion on $\\Gamma$ with center $A_1$ and ratio $R$ (it's not important). Define $Inv _{\\Gamma} (\\omega _i)=\\Omega _i$ and $Inv _{\\Gamma} (A_i)=B_i$. Note that $\\Omega _i$ is a line for $i=1, 3, 4, 6, 7$ because $\\omega _i$ pass through $A_1$. Also note that $\\Omega _i$ and $\\Omega _{i+1}$ are parallels for $i=3, 6$ because $\\Omega _i$ and $\\Omega _{i+1}$ are tangents in $A_1$. We want that $\\Omega _1=\\Omega _7$, it's sufficient to prove that $\\Omega _1 \\parallel \\Omega _6$. Now we will use directed angles. Let $\\ell=B_2B_3$:\n$$\\measuredangle (\\Omega _1,\\ell)=\\measuredangle(\\ell,\\Omega _3)=\\measuredangle(\\ell,\\Omega _4)=\\measuredangle (\\Omega _6,\\ell).$$\n\nSo we have that $\\Omega _1$ and $\\Omega _6$ are parallels as desired.", "Solution_14": "Not sure if this works.\n\nThrough some manipulations we find that it suffices to show $\\angle O_1A_2O_4=\\angle O_7A_1O_4$, but this is true because $$\\angle O_1A_2A_4=\\angle O_5A_2O_2=\\angle O_5A_3O_2=\\angle O_3A_3O_6=\\angle O_3A_1O_6=\\angle O_7A_1O_4$$", "Solution_15": "wait i think this works only when using directed angles (because $A_1O_4$ bisecting $\\angle O_1A_1O_7$ is possible)", "Solution_16": "Let $O_i$ be the center of $\\omega_i$. Define $\\angle A_3A_1A_2=\\alpha,\\angle A_1A_2A_3=\\beta,\\angle A_2A_3A_1=\\gamma.$ Define $\\angle A_iA_{i+1}O_i=\\angle A_{i+1}A_iO_i=\\theta_i.$ It suffices to show $\\theta_1=\\theta_7.$ In fact, note that \n\\begin{align*}\n\\theta_2 &= 180^\\circ-\\beta-\\theta_1 \\\\\n\\theta_3 &= 180^\\circ-\\gamma-\\theta_2 \\\\\n&= \\beta-\\gamma +\\theta_1 \\\\\n\\theta_4 &= 180^\\circ - \\alpha -\\theta_3 \\\\\n&= 180^\\circ - \\alpha - \\beta + \\gamma - \\theta_1 \\\\\n&= 2\\gamma - \\theta_1\n\\end{align*}\nNote that this process is the exact same for $\\theta_4\\to \\theta_7$, and since function $f(k)=2\\gamma -k$ is cyclic with order $2$, we are done.", "Solution_17": "Let $\\theta=\\angle A_1O_1A_2$, where $O_i$ is the center of $\\omega_i$. Furthermore, let $\\alpha, \\beta, \\gamma$ be the angles of the triangle. Then, check that we have the sequence $$\\theta \\to 2\\beta - \\theta \\to \\theta+2\\gamma-2\\beta \\to 2\\alpha-2\\gamma+2\\beta-\\theta=360^\\circ-4\\beta-\\theta.$$ Upon applying another sequence of three transformations, this returns the identity $\\theta$, so the centers of $\\omega_1$ and $\\omega_7$ coincide, which suffices.", "Solution_18": "Define $O_{k}$ as the centers of each respective circle. Let $\\alpha = \\angle{O_{1}A_{2}A_{1}}$. \n\\begin{align*}\n \\angle{A_{2}O_{4}A_{1}} &= 180-\\angle{A_{1}}-\\angle{O_{3}A_{1}A_{3}} \\\\\n &= 180-\\angle{A_{1}}-(180-\\angle{A_{3}-\\angle{A_{2}A_{3}O_{2}}}) \\\\\n &= \\angle{A_{3}}+\\angle{A_{2}A_{3}O_{2}}-\\angle{A_{1}} \\\\\n &= \\angle{A_{3}}+180-\\angle{A_{2}}-\\alpha-\\angle{A_{1}}\n\\end{align*}\nWe can apply this transformation twice to get that, \n\\[\\angle{A_{2}A_{1}O_{7}} = \\alpha\\implies \\omega_{1} = \\omega_{7}\\]", "Solution_19": "Consider if we extend the internal tangent of all of the pairs of circles $\\omega_{k-1}$ and $\\omega_k$ for $k = 2, 3, \\dots, 8$. Name these $\\ell_1, \\ell_2, \\dots \\ell_7$. Since $\\ell_x$ is just $\\ell_{x-1}$ reflected over the perpendicular bisecter of some side, and because the internal tangent is not dependent on the circle (it is in fact what makes the next circle), we just have to prove that $\\ell_1 = \\ell_7$. \n\nCall $\\angle A_2A_1A_3$ $a$, $\\angle A_3A_2A_1$ $b$, and $\\angle A_1A_3A_2$ $c$ \nWe know that the angle formed by $A_1$, $A_2$, and $\\ell_1$ is less than $a$ (otherwise $\\omega_2$ does not exist). Thus we call this $x$ and we know that the angle formed by $\\ell_1$, $A_1$, and $A_3$ is $a-x$ we can see that this is equal to the angle formed by $A_1$, $A_3$, and $\\ell_2$, so we know that the angle formed by $\\ell_2$, $A_3$, $A_2$ is $b-a+x$. Continuing this pattern for subsequent angles we find that the next angles are $180-2b-x$, $a+2b+x - 180$, $180 - a - b - x$, and $x$. Thus we find that $\\ell_7 = \\ell_1$ and we are done. $\\blacksquare$\n", "Solution_20": "For $1 \\le i \\le 7$, let $O_i$ be the center of $\\omega_i$ and $\\theta_i = \\measuredangle O_iA_iA_{i + 1} = \\measuredangle A_iA_{i + 1}O_i$. Then for $2 \\le i \\le 7$, the tangency condition tells us that $O_{i - 1}, A_i$, and $O_i$ are collinear, so\n$$\\theta_i = \\measuredangle O_iA_iA_{i + 1} = \\measuredangle O_{i - 1} A_iA_{i + 1} = \\measuredangle O_{i - 1}A_iA_{i - 1} + \\measuredangle A_{i - 1}A_iA_{i + 1} = -\\theta_{i - 1} + \\measuredangle A_{i - 1}A_iA_{i + 1}.$$\nTherefore, applying this identity in succession, we have\n$$\\theta_7 = \\theta_1 + \\sum_{i = 2}^7 (-1)^{7 - i} \\measuredangle A_{i - 1}A_iA_{i + 1} = \\theta_1 + \\sum_{i = 2}^4 ((-1)^{7 - i} + (-1)^{4 - i})(\\measuredangle A_{i -1}A_iA_{i + 1}) = \\theta_1.$$\nThus, $\\measuredangle O_1A_1A_2 = \\measuredangle O_7A_1A_2$ and $\\measuredangle A_1A_2O_1 = \\measuredangle A_1A_2O_7$. This is enough to imply that $O_1 = O_7$ since we would otherwise have $\\angle O_1A_1A_2 + \\angle O_7A_1A_2 = 180^{\\circ}$ (using non-directed angles), which is impossible since $\\angle O_1A_1A_2, \\angle O_7A_1A_2 < 90^{\\circ}$. So, we are done." } { "Tag": [ "MATHCOUNTS" ], "Problem": "does anyone know the mathcounts practice site, that you choose which kind of problems you want (target, countdown, sprint, etc) and how many problems you want to do? and the site scrambles the numbers in the problems and you submit your answers...? sorry, I know this is kind of vague. I had this site last year but I haven't visited it in a year so I don't know where it went?\r\n\r\nany help would be appreciated :D", "Solution_1": ":D Heeh, Im pretty sure this is it: http://mathcounts.saab.org/mc.cgi\r\n\r\n Just type \"mathcounts drills\" on google." } { "Tag": [], "Problem": "1. Vijay takes 5 hours to mow the lawn. His sister Deepali takes 3 hours to mow the same lawn. Working together, how long will it take them to mow the lawn?\r\n\r\n2. Wei drives from Chicago to New York, averaging 45 miles per hour. He then drives back the same route, but averages 60 mph on the return trip. What is his average speed?\r\n\r\n3. Yan takes 90 seconds to walk up an up escalator. Walking at the same pace down the up escalator, it takes Yan 150 seconds to reach the bottom. Walking at the same pace, how long would it take Yan to climb the same escalator if the escalator was turned off? How long would it take Yan to ride up the escalator (if she wasn't walking, but the escalator was on?)\r\n\r\n4. Alex and his two brothers, Rob and Paul, are going to paint the 4 walls of Alex's cubical house. The decide that Alex and Rob will paint the first side, Rob and Paul will paint the second, Paul and Alex will paint the third, and all three together will paint the 4th. If Alex paints 1 wall in 6 hours, Rob paints one wall in 2 hours, and Paul paints 1 wall in 3 hours, what total fraction of the four walls will each of the brothers paint?\r\n\r\n5. Jooho is two thirds of the way from one end of a railway tunnel when he sees a train 100 feet away approaching the tunnel. Calculating quickly, he realizes that no matter which way he runs, he will escape from the tunnel at exactly the same moment he would be plowed down by the train. How long is the tunnel? How many times faster does the train travel than Jooho?\r\n\r\n6. David travels from A to B and then from B to C. He realizes that if he makes the first segment of the trip at 4 mph and the second at 4 mph, the trip will take him 3 hours. If he travels the first at 2 mph and the second at 6 mph, the trip will take him 3 hours. If A, B, and C are in the same plane, what is the shortest possible distance between A and C? The largest?\r\n\r\n7. Runners A and B race along a hexagonal track, always running at constant speeds. B begins to fall behind, and when he is exactly half done with the race realizes he can't possibly win. So he cuts directly across the center of the racetrack, catching A just as she crosses the finish line. If A runs at 18 kilometers per hour, how fast does B run?\r\n\r\n8. Dongbo always runs 5 miles every morning at a constant speed of 5 miles per hour before leaving for work. One day, he decided that he would start out at 7 miles per hour, and after every mile he ran, he would decrease his speed by 1 mile per hour. (So he ran his first mile at 7 mph, the second at 6 mph, and the last at 3 mph.) He was surprised to find that under this system, he ended up being late for work. How much more time did his run take under this system?\r\n\r\n9. Jim drove from Lawrence, KS, to Amherst, MA taking an out-of-the-way route in order to travel at 60 mph. Returning home to Lawrence, he took a more direct route but was only able to travel 40 mph. He discovered to his surprise that the two routes had taken the same amount of time. What was his average speed for the round trip?", "Solution_1": "Nice choice of problems Joel :twisted:\r\n\r\nBy the way if the number's a fraction/decimal is the unit in singular form or plural? What if it's 0 or negative? imaginary?", "Solution_2": "Hmm ... if it's rational fractional, you say \"one third of a mile per hour,\" or some such. But if it's irrational and still between 0 and 1, you say \"square root of two minus 1 mile[b]s[/b] per hour.\" For the first case, you could also use decimals to say \"point three-repeating mile[b]s[/b] per hour.\" I hope none of the answers are zero, irrational, or imaginary, although I did just make the numbers up off of the top of my head, so it's theoretically possible.\r\n\r\n\r\nPlease put your solutions in spoiler!", "Solution_3": "1[hide]15/8[/hide]\n\n2[hide]360/7[/hide]\n\n3[hide]225/2, and 450[/hide]\n\n4[hide]Alex = 1/4(1/4 + 1/3 + 1/6) = 3/16\n\nRob = 1/4(3/4 + 3/5 + 1/2) = 37/80\n\nPaul= 1/4(2/5 + 2/3 + 1/3) = 7/20[/hide]", "Solution_4": "Don't be lazy -- write up your solutions!", "Solution_5": "Well. It is generally the same solution for 1,2, and 4. \n\n\n\n1)[hide]Vijay takes 5 h and Deepali takes 3h. Each hour, Vijay mows 1/5 of the lawn and Deepali mows 1.3 of the lawn. Together in one hour they mow 1/3+1/5 of the entire lawn. Since the lawn is one whole, their time together is 1/ ( 1/3 + 1/5) = 15/8.[/hide]\n\n2)[hide] Set the distance from the cities to be x. Then it takes Wei x/45 hours for one way and x/60 on the way back. His entire time is x/45 + x/60. The distance traveled is 2x. 2x / (x/45+x/60) = 360/7[/hide]\n\n3) [hide]x is the person's speed, and y is the escalator speed. Z is the escalator length. z/(x+y) = 90 and z/ (x-y) = 150 . Do some simultaneous equations to end up with what you want : z/x and z/y. They turn out to be 225/2 and 450.[/hide]\n\n4)[hide]This one takes some time. You want the fractions of the 4 walls each person paints. Alex and Rob together take 1/(1/6 + 1/2) hours, Alex and Paul together take 1/(1/6+1/3) h, Paul and rob together take 1/(1/2+1/3)h, and all three brothers take 1/(1/6+1/3+1/2)h. You solve and for A and R 3/2 hours, for A and P 2 hours, for R and P 6/5 hours, and for A, P, and R 1 hour. Since rate*time = total distance or job done, you solve for each person.\n\nAlex did. 3/2 * 1/6 + 2 * 1/6 + 1* 1/6 for the three walls he did altogether. Simplified, Rob did 3/4 + 3/5 + 1/2 of all the walls, and Paul did 2/3 + 2/5 + 1/3 of all the walls. Since there are 4 walls, the sum of A + R +P = 4, so you divide the portion of the 4 walls they did by 4 to find the fraction of the total they painted.[/hide]", "Solution_6": "Haha... I feel sorry for the guy in problem 5. Why did he waste time calculating? If he ran absentmindedly in one direction before calculating, he might have made it out.", "Solution_7": "5) Maybe he was a mathematician and calculated really quickly \n\n[hide]\n\nOk so let y = length of tunnel, x = speed of Jooho, and kx = speed of train (k times faster than Jooho)\n\n\n\nWe can make two equations with these:\n\n\n\n(y/3)/x = (100-y/3)/kx time it takes to get to closer end of tunnel\n\n\n\n(2y/3)/x = (100+2y/3)/kx time it takes to get to further end\n\n\n\nDo a bit of simplifying and cross-multiplying to get:\n\n\n\n3kyx = 900x-3xy \n\nand\n\n6kyx = 900x+6xy\n\n\n\nSubtract the first from the second to get 3kyx = 9xy wihch means k=3. Plug it back into the first equation and solve for y because the x cancels out. You get 75.\n\n\n\nSo the tunnel is 75 feet long and the train is 3 times faster.[/hide]", "Solution_8": "7)[hide] so assume the hexagon is regular to calculate it more easily. and call the side lengths 3. So at 18kph it will take 1 hour. now, half of it is obviously 9, and if you draw equilateral triangles you can see that cutting across will be 6. Therefore, it is 15 km in one hour. So the speed is 15kph.[/hide]\n\n\n\n8)[hide] well for the first mile, it takes 1/7 of an hour, then 1/6, 1/5, 1/4, and 1/3. Add them up and you get 459/420, which is 13/140 hours later, approximately 5.5 minutes.[/hide]" } { "Tag": [ "inequalities", "geometry", "trigonometry", "symmetry", "geometry proposed" ], "Problem": "Given are $\\triangle ABC$ area $S$ with any $M$ on plane prove that\r\n$\\cos\\frac{A}{2}MA^{2}+\\cos\\frac{B}{2}MB^{2}+\\cos\\frac{C}{2}MC^{2}\\ge 2S$\r\n$\\cos\\frac{B}{2}MA^{2}+\\cos\\frac{C}{2}MB^{2}+\\cos\\frac{A}{2}MC^{2}\\ge 2S$", "Solution_1": "[color=darkred][b]A classical and remarkable geometrical inequality.[/b] \n\nFor any real numbers $x\\ ,\\ y\\ ,\\ z$ for which $x+y+z> 0$ and for any point $M$ from the plane of the triangle $ABC$ there is the following inequality :\n\n$\\boxed{\\ x\\cdot MA^{2}+y\\cdot MB^{2}+z\\cdot MC^{2}\\ge \\frac{yza^{2}+zxb^{2}+xyc^{2}}{x+y+z}\\ }$ with equality if and only if $M: =P\\ (\\ x\\ ,\\ y\\ ,\\ z\\ )\\ .$[/color]", "Solution_2": "Can you show show clearly Virgil, actually I tried but I can't solve by this way! :)", "Solution_3": "[quote=\"gemath\"][color=darkred]Can you [b][u]show , show clearly[/u][/b] Virgil? Actually [u]I tried but I can't solve by this way ![/u][/color][/quote]\r\n[color=darkblue][b][u]All done ![/u][/b] I found a proof using the well-known inequality $\\boxed{\\ x+y+z>0\\Longrightarrow\\sum xMA^{2}\\ge\\frac{\\sum xya^{2}}{\\sum x}\\ \\ (*)\\ }$ and the strong inequality\n$\\boxed{\\ \\sum a\\sin\\frac{A}{2}\\ge p\\ \\ (**)\\ }$ ( [u]see the topic[/u] http://www.mathlinks.ro/Forum/viewtopic.php?t=80834 [u]where appears and a nice, easy and geometrical proof of it[/u] ).\n\n[b]Proof.[/b] $\\left\\{\\begin{array}{c}p\\ge 3r\\sqrt 3\\Longrightarrow\\boxed{\\ p^{2}\\ge 3S\\sqrt 3}\\\\\\\\ \\boxed{DOWN\\equiv\\sum\\cos\\frac{A}{2}\\le \\frac{3\\sqrt 3}{2}}\\mathrm{\\ (\\cap-Jensen\\ for\\ Cos)}\\\\\\\\ \\boxed{UP\\equiv\\sum a^{2}\\cos\\frac{B}{2}\\cos\\frac{C}{2}=p\\cdot\\sum a\\sin\\frac{A}{2}\\ge^{(**)}p^{2}}\\end{array}\\right\\|$ $\\Longrightarrow$ $\\sum MA^{2}\\cos\\frac{A}{2}\\ge^{(*)}\\frac{UP}{DOWN}$ $\\ge$ ${\\frac{2p^{2}}{3\\sqrt 3}}\\ge 2S$ $\\Longrightarrow$ $\\boxed{\\boxed{\\ \\sum MA^{2}\\cos\\frac{A}{2}\\ge 2S\\ }}\\ .$\n\n[b]Remark.[/b] $\\boxed{\\sum MA^{2}\\cos\\frac{A}{2}\\ge \\frac{2\\sqrt 3}{9}\\cdot p^{2}}\\ge 2S\\ .$ If you please, I \"pester\" you a bit : mine is [u]stronger[/u] than yours ![/color]", "Solution_4": "ok thank Virgil for your stronger ineq, it gives me an interesting idea! :)", "Solution_5": "Congratulations Virgil ! The beautiful solutions one after the other :lol: \r\n(This was really surprising)\r\n@gemath: Your inequalities are nice but how come you never post your solutions? For example this one has been solved now , why don't you post yours, too? :P :wink:", "Solution_6": "I agree with you dear Albanian Eagle. \r\n\r\nIn my opinion, when a different solution of an our proposed problem, is presented in the forum, then the solver directly (at first, by previous message) and later, by post message all the other mathlinkers, must know the solution we have in mind (if we have), as a minimum intication of honorific.\r\n\r\nKostas Vittas.", "Solution_7": "yes vittasko and Albanian Eagle I know I need to explain more than only say thank you :D .\r\nReally I found the second before it base on [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=114248]my inequality[/url], my solution of that like of msceok,\r\nand by simple (but most important) inequality\r\n$aMA^{2}+bMB^{2}+cMC^{2}\\ge abc$ add them and use law of sin we have:\r\n$(\\sin A+\\sin C)MA^{2}+(\\sin B+\\sin A)MB^{2}+(\\sin C+\\sin B)MC^{2}\\ge\\frac{2abc}{2R}=4S$\r\nand note that\r\n$\\sin A+\\sin C=2\\sin\\frac{A+C}{2}\\cos\\frac{A-C}{2}\\le 2\\cos\\frac{B}{2}$\r\nsimilar we will have the second mine.\r\nI think that method can't solve the non-symmetry problem as the second!\r\nwhen I found the second I think the first is true and post both of them and actually it is true ! :P" } { "Tag": [ "LaTeX", "inequalities proposed", "inequalities" ], "Problem": "a,b,c >0 and \r\n[latex]a^2 \u00a0+ b^2 \u00a0+ c^2 \u00a0\\le 12[/latex]\r\nmin ? :s=\r\n\r\n[latex]\\frac{{a^6 }}{{ab + 2\\sqrt {1 + c^3 } }} + \\frac{{b^6 }}{{bc + 2\\sqrt {1 + a^3 } }} + \\frac{{c^6 }}{{ca + 2\\sqrt {1 + b^3 } }}[/latex]\r\n[/code][/hide]", "Solution_1": "Let $ a,b,c$ be three real numbers so that $ a,b,c>0$ and $ a^2\\plus{}b^2\\plus{}c^2 \\le 12$. \r\n\r\nFind $ \\min s\\equal{}\\frac{{a^6 }}{{ab \\plus{} 2\\sqrt {1 \\plus{} c^3 } }} \\plus{} \\frac{{b^6 }}{{bc \\plus{} 2\\sqrt {1 \\plus{} a^3 } }} \\plus{} \\frac{{c^6 }}{{ca \\plus{} 2\\sqrt {1 \\plus{} b^3 } }}$.\r\n :D :D :D" } { "Tag": [ "floor function", "Irrational numbers" ], "Problem": "Find the smallest positive integer $n$ such that \\[0< \\sqrt[4]{n}-\\lfloor \\sqrt[4]{n}\\rfloor < 0.00001.\\]", "Solution_1": "$\\sqrt[4]n$ has to be slightly bigger than an integer. Thus $n=a^4+1$ (replacing $1$ by a bigger number always results in a greater difference and bigger n, thus is contraproductive).\r\n\r\nWe want $\\sqrt[4]{a^4+1} - a < \\epsilon$ (here $\\epsilon=0.00001$). This means $a^4+1 < (a+\\epsilon)^4=a^4+\\epsilon(4a^3+6a^2\\epsilon+4a\\epsilon^2+\\epsilon^3)$.\r\nNow this requires that $\\frac 1{4\\epsilon} < a^3$, giving (using my PC :D) that $a \\geq 30$. A small calculation then shows that $a=30$ works (we don't need to plug it in the first, but rather in the third inequality)." } { "Tag": [ "limit", "real analysis", "real analysis unsolved" ], "Problem": "Let $ a_{k}$, $ k \\equal{} 0, 1, 2, \\dots$, be a sequence of non-negative real numbers such that \r\n\r\n(a) $ a_{m \\plus{} n} \\le a_{m}\\cdot a_{n}$ for all $ m$, $ n \\ge 0$;\r\n(b) there exists an $ m_0$ such that $ a_{m_{0}} < 1$.\r\n\r\nShow that $ \\displaystyle\\sum_{k \\equal{} 0}^{\\infty}a_{k} < \\infty$, i.e. the $ \\displaystyle\\lim_{N\\rightarrow\\infty}\\displaystyle\\sum_{k \\equal{} 0}^{\\infty}a_{k}$ exists and is finite.", "Solution_1": "$ a_{p \\plus{} km_0} \\leq a_pa_{km_0}\\leq a_pa_{m_0}^k$ , so $ \\displaystyle\\sum_{n \\equal{} 0}^{\\infty}a_{n} \\equal{} \\sum_{p \\equal{} 0}^{m_0 \\minus{} 1}\\sum_{k \\equal{} 0}^{\\infty}a_ {p \\plus{} km_0} \\leq \\sum_{p \\equal{} 0}^{m_0 \\minus{} 1}a_p( \\displaystyle\\sum_{k \\equal{} 0}^{\\infty}a_{m_0}^k) \\equal{} \\displaystyle (\\sum_{p \\equal{} 0}^{m_0 \\minus{} 1}a_p)/(1 \\minus{} a_{m_0})$ :cool:" } { "Tag": [ "geometry", "probability", "analytic geometry" ], "Problem": "$S$ is the set of all points with coordinates $(m,n)$ such that $m$ and $n$ are positive integers with $m < 4$ and $n < 4$ . Two points from $S$ are chosen at random. What is the probability that the midpoint of the segment of the two points is also in the $S$ ? Give your answer as a common fraction in lowest term.", "Solution_1": "[quote=\"#H34N1\"]$S$ is the set of all points with coordinates $(m,n)$ such that $m$ and $n$ are positive integers with $m < 4$ and $n < 4$ . Two points from $S$ are chosen at random. What is the probability that the midpoint of the segment of the two points is also in the $S$ ? Give your answer as a common fraction in lowest term.[/quote]\r\nWould this be 1/1 because the midpoint will always be in S?\r\n\r\n500th post!", "Solution_2": "No. Read the problem.", "Solution_3": "[hide]That would be 9 points total. There are 36 ways to pick two out of 9. 8 pairs have their midpoints also in S so it's 8/36, 2/9.[/hide]", "Solution_4": "[hide] Your choices are 1,2 or 3 for both M and N as the person above said there are 36 different pairs then think of it like tic-tac toe, there 8 different lines. 8/36= 2/9[/hide]" } { "Tag": [], "Problem": "My parents won't let me buy it so I am looking for an ebook but I can't find one. Anyone have it? Thanks.", "Solution_1": "Post full data (full author name, exact title, edition and publication year (if you know them), etc.). Also, specify what languages you can read. Then, maybe, someone will be able to help you :).", "Solution_2": "Apparently it is a very famous book, but here are the details. \r\n\r\nAptitude test problems in physics\r\nSS Krotov\r\nMIR Publication\r\n\r\nThat is all I know about it. I know English." } { "Tag": [ "linear algebra", "matrix", "algebra", "polynomial" ], "Problem": "Let M be a square matrix of order n.\r\n\r\nThen show that there exists an integer k>0 such that for all n>k,\r\n$M-\\lambda\\textbf{I}_{n}$ is invertible.", "Solution_1": "how do u add an $n \\times n$ and $m \\times m$ matrix? :?", "Solution_2": "Should that be $\\lambda>k$? The question makes no sense as written.", "Solution_3": "Alright then. I had the same doubt. But here is the solution given.\r\nCan someone clearly explain this?\r\n\r\nSoln: Since |$M-\\lambda \\textbf I_{n}$| is a polynomial in $\\lambda$, it has finitely many roots. \r\nThus as soon as k is an integer which is greater than the absolute values of the roots, we have n > k $\\Longrightarrow$ |$M-\\lambda \\textbf I_{n}$| $\\not=0$.\r\n\r\nTherefore $M-\\lambda \\textbf I_{n}$ is invertible.", "Solution_4": "That solution would be correct if we replaced \"$n>k$\" with \"$\\lambda>k$\"." } { "Tag": [], "Problem": "What would happen if not all of your team is at States for the Team Round (even if you have 2 team, 1 ind or something)???\r\n\r\nCan you pair up with a person from another school???\r\n\r\nWhat if your whole team is at states, but you are the only individual? What happens then?\r\n\r\n-jorian", "Solution_1": "[quote=\"jhredsox\"]What would happen if not all of your team is at States for the Team Round (even if you have 2 team, 1 ind or something)???\n\nCan you pair up with a person from another school???\n\nWhat if your whole team is at states, but you are the only individual? What happens then?\n\n-jorian[/quote]\r\nIf your team has three members they don't take the average, they divide by four. Otherwise, my team last year probably would have had three people. :|", "Solution_2": "Individuals pair up with three other individuals to make a team...but they don't usually turn out well.", "Solution_3": "(unless you purposely throw off getting team to get indiv's so u plan who ur with)\r\n\r\n-jorian", "Solution_4": "They're not an official team.", "Solution_5": "I was ind. last year at states, and we got 7. they weren't that bad...", "Solution_6": "it'd be cool if u got together with the best people around the state and did individuals...and they counted it :P\r\n\r\n-jorian", "Solution_7": "[quote=\"perfect628\"]I was ind. last year at states, and we got 7. they weren't that bad...[/quote]\r\n\r\nI thought you said they were bad and you did the whole thing...oh well that's better than what we got as a team last year, bleh", "Solution_8": "[quote=\"easyas3.14159...\"]Individuals pair up with three other individuals to make a team...but they don't usually turn out well.[/quote]\r\nHow would you know?\r\n\r\nI don't think that they count the scores for teams made up of individuals...", "Solution_9": "It's up to the states, I know CT did different things in '05 than '06", "Solution_10": "Oh, I ended up pretty much doing the whole team round by myself. I made my usual dumb mistakes and got a 7. Yeah, they divide by 4 no matter how many members you have.", "Solution_11": "[quote=\"Ubemaya\"][quote=\"easyas3.14159...\"]Individuals pair up with three other individuals to make a team...but they don't usually turn out well.[/quote]\nHow would you know?\n\nI don't think that they count the scores for teams made up of individuals...[/quote]\r\n\r\nYou don't have to go to nationals to figure that 4 strangers probably wont work as good as 4 teammates.", "Solution_12": "unless last year\r\n\r\nyou had\r\n\r\nandrew\r\nneal\r\nkevin\r\ndaesun\r\n\r\nthey would own (even though they prob weren't top 4 written...but still)\r\n\r\ndidn't daesun get 5th, neal 1st (im sure about neal)\r\n\r\n-jorian", "Solution_13": "I'm pretty sure they all missed number 10 on the team round.\r\n\r\n1st: Neal\r\n2nd: Daniel Li\r\n3rd: Kevin\r\n4th: Nathan Benjamin", "Solution_14": "How'd you forget Daniel? He's like the biggest to forget.", "Solution_15": "I think he had the top 4 from countdown.", "Solution_16": "[quote=\"13375P34K43V312\"][quote=\"perfect628\"]I was ind. last year at states, and we got 7. they weren't that bad...[/quote]\n\nI thought you said they were bad and you did the whole thing...oh well that's better than what we got as a team last year, bleh[/quote]\r\n\r\nThat was my school team. This year they're not TOO bad.", "Solution_17": "[quote=\"Klebian\"]I'm pretty sure they all missed number 10 on the team round.\n\n1st: Neal\n2nd: Daniel Li\n3rd: Kevin\n4th: Nathan Benjamin[/quote]\r\n\r\nSo did basically everyone else.", "Solution_18": "[quote=\"perfect628\"][quote=\"Klebian\"]I'm pretty sure they all missed number 10 on the team round.\n\n1st: Neal\n2nd: Daniel Li\n3rd: Kevin\n4th: Nathan Benjamin[/quote]\n\nSo did basically everyone else.[/quote]\r\nWe still don't know who got number 10 right...\r\nBut a team of those four could've had the time to spare to go through 10.", "Solution_19": "I thought a bunch of people established that the Virgin Islands guessed #10 right?", "Solution_20": "I seriously doubt that...", "Solution_21": "[quote=\"pianoforte\"]I seriously doubt that...[/quote]\n*cough* *cough*\n[quote=\"Ignite168\"]Virgin Islands got #10 right, this is how it went with 1 minute left\n\"Hey, man, I have no idea what the answer is\"\n\"How about we put that answer to everything from that one book\"\n\"Yeah, that's a great idea, what's that number man?\"\n\"Uhh.. I forgot\"\n\"Hmm.. I don't know either\"\n\"Beats me\"\n\"Wait isn't it like 52 or something?\"\n\"Yeah that's it!\"\n\"Let's write it\"\n\nAnd they got it right :D[/quote]" } { "Tag": [ "inequalities", "induction", "algebra", "polynomial", "quadratics", "difference of squares", "special factorizations" ], "Problem": "If [tex]a_0=1[/tex] and [tex] a_i < a_{i+1} \\leq a_i + 1[/tex] for [tex] 0 \\leq i \\leq n[/tex] , show that\r\n\r\n[tex]\\displaystyle \\left(\\sum_{i=0}^n a_i \\right)^2 \\geq \\sum_{i=0}^n a_i^3[/tex]\r\n\r\nIdeas?", "Solution_1": "I think induction works. Figure out what inequality you need to make the induction go through. That inequality in turn might need its own inductive proof.", "Solution_2": "A solution by indirect induction:\n\nAlso, for convenience, I set a_0 = 0. This includes the original as a special case a_1 = 1.\n\n[hide]\n\nLet S(m) = :Sigma: (a_k) : 0 :le: k :le: m, and C(n) = :Sigma: (a_k :^3:) : 0 :le: k :le: m\n\nThen C(m+1) - C(m) = a_(m+1):^3: and \n\nS(m+1):^2: - S(m):^2: = a_(m+1)*(a_(m+1) + 2*S(m)) = a_(m+1)*(-a_(m+1) + 2S(m+1))\n\nWe use induction on the hypothesis that a_(m+1):^3: :le: S(m+1):^2: - S(m):^2:.\n\nThis is obvious for m=0, as a:^3: :le: a:^2: for a :le: 1\n\nWe can write the hypothesis for m-1 as \n\na_m :^3: + a_m :^2: - 2*S(m)*a_m :le: 0, or equivalently a_m :^2: + a_m - 2S(m) :le: 0.\n\nLet the roots of x:^2: + x - 2*S(m) = 0 be r and -1-r, with r :ge: 0. The inequality a_m :^2: + a_m - 2*S(m) :le: 0 implies a_m :le: r.\n\nBy definition of a_m, a_(m+1) :le: r+1\n\nThen the roots of x:^2: - x - 2*S(m) = 0 are 1+r and -r, so\n\n a_(m+1):^2: - a_(m+1) - 2*S(m) :le: 0 and \n\na_(m+1):^3: - a_(m+1):^2: - 2S(m)*a_(m+1) :le: 0.\n\nThis gives a_(m+1):^3: :le: S(m+1):^2: - S(m):^2: as desired, so the inequality is true for all m.\n\nFinally, S(n):^2: - C(n) = :Sigma: (S(m+1):^2: - S(m):^2: - a_(m+1):^3:) : 0 :le: m :le: n-1, and we are done.\n\nEquality is only possible if a_(m+1) = a_m + 1 for all m < n.[/hide]", "Solution_3": "There is a simple and elegant solution using a variation on the formula that alison gave here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=12175\r\nThe formula is\r\n[tex]n^3 = (a_1 + a_2 + ... + a_n)^2 - (a_1 + a_2 + ... + a_{n-1})^2[/tex]\r\nFor the conditions of this particular inequality, we can use the fact (easily proven by induction) that\r\n[tex](a_1 + a_2 + ... + a_n)^2 - (a_1 + a_2 + ... + a_{n-1}) \\ge a_n^3[/tex] (the only equality condition being when [tex]a_{i+1} = a_i + 1[/tex])\r\nWe can add the series of inequalities having [tex]a_i^3[/tex] as the right side, and the left side will telescope and become the desired expression.", "Solution_4": "Thanks for the help folks.\r\n\r\nYeah, I tried induction on it, but I didn't realize to perform the necessary 'induction within the induction'.\r\n\r\nBy the way, jmerry, when you wrote\r\n\r\n[quote=\"jmerry\"]\nThen the roots of x:^2: - x - 2*S(m) = 0 are 1+r and -r, so\n a_(m+1):^2: - a_(m+1) - 2*S(m) :le: 0 and \na_(m+1):^3: - a_(m+1):^2: - 2S(m)*a_(m+1) :le: 0.\nThis gives a_(m+1):^3: :le: S(m+1):^2: - S(m):^2: as desired, so the inequality is true for all m.\n[/quote]\n\ndid you actually mean to write the following?\n\n[quote]\nThen the roots of x:^2: + x - 2*S(m+1) = 0 are 1+r and -r, so\n a_(m+1):^2: + a_(m+1) - 2*S(m+1) :le: 0 and \na_(m+1):^3: + a_(m+1):^2: - 2S(m+1)*a_(m+1) :le: 0.\nThis gives a_(m+1):^3: :le: S(m+1):^2: - S(m):^2: as desired, so the inequality is true for all m.\n[/quote]\r\n\r\nHmm, the solution still seems to be difficult to discover though. How long did it take you to solve the problem?", "Solution_5": "I meant what I wrote. Since I hadn't calculated anything about S(m+1), I wasn't going to use it in an inequality. Your second quote is actually wrong- the roots of that polynomial are between -r-1 and -r and between r+1 and r+2. It's not enough to prove anything.\r\nI also don't believe schulmannerism about the ease- the induction he refers to is most of my proof. \r\n\r\nI originally worked backwards to find that -a_m and a_(m+1) were between the two roots of x:^2: - x - 2*S(m) = 0; this could be written formally as the induction I needed. The key was that I could write the factor S(m) + S(m+1) as either 2*S(m) + a_(m+1) or 2*S(m+1) - a_(m+1)\r\n\r\nThe solution didn't take me long to find, but I'm not sure anyone else would get it quickly. The step that you questioned is one of my favorite tricks- \"flipping the quadratic\" by switching focus from one root to the other. I'd never applied it to an inequality before, but it was obvious to me.", "Solution_6": "[quote=\"jmerry\"]\nI also don't believe schulmannerism about the ease- the induction he refers to is most of my proof. [/quote]\r\n\r\nHere's how the induction goes.\r\n[tex](a_1 + a_2 + ... + a_n)^2 - (a_1 + a_2 + ... + a_{n-1})^2 \\ge a_n^3[/tex]\r\nIs manipulated (using difference of squares) to become\r\n[tex]2(a_1 + a_2 + ... + a_{n-1}) \\ge a_n^2 - a_n[/tex]\r\nThis is our induction hypothesis. Add [tex]2a_n[/tex] to both sides\r\n[tex]2(a_1 + a_2 + ... + a_{n-1} + a_n) \\ge a_n^2 + a_n[/tex]\r\nNow we just need to prove that [tex]a_n^2 + a_n \\ge a_{n+1}^2 - a_n[/tex]\r\n[tex]a_n^2 + a_n = a_n(a_n + 1) \\ge (a_{n+1}-1)(a_{n+1}) \\ge (a_{n+1}-1)(a_{n+1})[/tex] as desired.\r\n\r\n\r\nBTW, making your posts nice with TeX is easy. Just putting the begin and end tags around a_1 + 2s^2 turns it into [tex]a_1 + 2S^2[/tex] Also, if you quote someone, you can see their TeX.", "Solution_7": "Another way to prove that inner inequality is via telescoping sum:\r\n[tex]\\begin{eqnarray*}\r\na_n^2 & = & \\sum_{i=0}^{n-1} (a_{i+1}^2 - a_i^2) \\\\\r\n & = & \\sum_{i=0}^{n-1} (a_{i+1} - a_i)(a_{i+1} + a_i) \\\\\r\n & \\le & \\sum_{i=0}^{n-1} (a_{i+1} + a_i) \\\\\r\n & = & a_n + 2 \\sum_{i=0}^{n-1} a_i \\, .\r\n\\end{eqnarray*}\r\n[/tex]", "Solution_8": "I get it now. Thanks guys." } { "Tag": [ "modular arithmetic", "algorithm", "number theory", "Diophantine equation", "Euclidean algorithm" ], "Problem": "I tried this problem but my solution was very \"painful\". Here is my solution:\r\n\r\n\r\nClearly $ 13^{\\varphi(1000)} \\equiv 1 \\pmod{1000} \\equal{}> 13^{400} \\equiv 1 \\pmod{1000}$ $ \\equal{}> 13^{398}*169 \\equiv 1 \\pmod{1000} (1)$\r\n\r\nWhat I did now was find a solution to the congruence $ 169x \\equiv 1 \\pmod{1000}$ to get rid of the 169 on (1). I had a hard time finding the solution because I dont know how to solve this type of congruence. Finally I got x= 929, so $ 13^{398} \\equiv 929 \\pmod{1000}$\r\n\r\nId like to see other solutions or how do you solve the congruence I mentioned.", "Solution_1": "The congruence you were trying to solve amounts to finding the multiplicative inverse of 169 (mod 1000). This can be done using the Euclidean Algorithm.\r\n\r\nBy definition, $ 169x\\equiv1 (mod1000)$ means that $ 169x\\minus{}1000y\\equal{}1$ for integers $ x$ and $ y$. Now we have a diophantine equation and we can use the Euclidean Algorithm:\r\n\r\n$ 1000\\equal{}5\\cdot169\\plus{}155$\r\n$ 169\\equal{}1\\cdot155\\plus{}14$\r\n$ 155\\equal{}11\\cdot14\\plus{}1$\r\n$ 14\\equal{}14\\cdot1\\plus{}0$\r\n\r\nRunning this in reverse, we get\r\n\r\n$ 1\\equal{}155\\minus{}11\\cdot14$\r\n$ 1\\equal{}155\\minus{}11(169\\minus{}155)\\equal{}12\\cdot155\\minus{}11\\cdot169$\r\n$ 1\\equal{}12(1000\\minus{}5\\cdot169)\\minus{}11\\cdot169\\equal{}12\\cdot1000\\minus{}71\\cdot169$\r\n\r\nThis gives the solution $ x\\equal{}\\minus{}71$, $ y\\equal{}12$, but we want a positive integer solution for $ x$. To do this, add 1000 to $ x$ and subtract 169 from the (meaningless) value of $ y$ to get $ x\\equal{}929$, the solution you found. There might have been a more direct way to arrive at the answer, but that's the quickest I know off the top of my head.", "Solution_2": "Since $ 1001 \\equal{} 7 \\cdot 11 \\cdot 13$, we have $ 13\\cdot77 \\equiv 1\\pmod{1000} \\Rightarrow 13^{\\minus{}1} \\equiv 77 \\pmod{1000}$. \r\n\r\nSo, $ 169^{\\minus{}1} \\equiv (13^{\\minus{}1})^2 \\equiv 77^2 \\equal{} 5929 \\equiv 929 \\pmod{1000}$.", "Solution_3": "Thanks a lot abnormal distribution!!", "Solution_4": "[hide]When congruencies don't work, this is how u can ponder with it....\n$ 13^{398}\\equal{}(170\\minus{}1)^{199}$\nNow try....[/hide]" } { "Tag": [ "calculus", "algebra unsolved", "algebra" ], "Problem": "Calculate:\r\n\r\n$ \\sum_{n\\equal{}1}^{2009}\\frac{n}{n^4\\plus{}n^2\\plus{}1}$", "Solution_1": "Factorize $ n^4\\plus{}n^2\\plus{}1$ then use Telescoping Method.", "Solution_2": "I've tried that, I'd like a solution if possible.", "Solution_3": "$ n^4\\plus{}n^2\\plus{}1\\equal{}(n^2\\plus{}1)^2\\minus{}n^2\\equal{}(n^2\\plus{}1\\plus{}n)(n^2\\plus{}1\\minus{}n)$\r\n\r\n$ \\equal{}(n(n\\plus{}1)\\plus{}1)((n\\minus{}1)n\\plus{}1)$", "Solution_4": "I have managed to factor it, i have transformed it to sum of terms twice, and the result still doesn't cancel out. This is what I ended up with (or something like that, I don't remember fully, the test was yesterday:\r\n\r\n$ \\frac{1^2}{1\\cdot3}\\plus{}\\frac{2^2}{3\\cdot7}\\plus{}\\frac{3^2}{7\\cdot 13}\\plus{}\\cdots$\r\n\r\nAgain, I need a [i]solution[/i] , thanks in advance for whoever takes some time to post it.", "Solution_5": "Calculate $ \\frac{1}{(k\\minus{}1)k\\plus{}1}\\minus{}\\frac{1}{k(k\\plus{}1)\\plus{}1}$.", "Solution_6": "What is the answer?", "Solution_7": "[quote=\"triplebig\"]What is the answer?[/quote]\r\n\r\n$ \\frac{2019045}{4038091}$", "Solution_8": "[quote=\"triplebig\"]What is the answer?[/quote]\r\n\r\nAre you unable to do the calculus yourself ?\r\n\r\nKenny suggested you could calculate $ \\frac {1}{(k \\minus{} 1)k \\plus{} 1} \\minus{} \\frac {1}{k(k \\plus{} 1) \\plus{} 1}$\r\n\r\nThis calculus gives $ \\frac{2k}{k^4\\plus{}k^2\\plus{}1}$\r\n\r\nSo $ \\sum_{n \\equal{} 1}^{2009}\\frac {n}{n^4 \\plus{} n^2 \\plus{} 1}$ $ \\equal{}\\frac{1}{2}\\sum_{n \\equal{} 1}^{2009}(\\frac {1}{(n \\minus{} 1)n \\plus{} 1} \\minus{} \\frac {1}{n(n \\plus{} 1) \\plus{} 1})$\r\n\r\nSo $ \\sum_{n \\equal{} 1}^{2009}\\frac {n}{n^4 \\plus{} n^2 \\plus{} 1}$ $ \\equal{}\\frac{1}{2}\\sum_{n \\equal{} 1}^{2009}\\frac {1}{(n \\minus{} 1)n \\plus{} 1} \\minus{} \\frac{1}{2}\\sum_{n \\equal{} 1}^{2009}\\frac {1}{n(n \\plus{} 1) \\plus{} 1}$\r\n\r\nSo $ \\sum_{n \\equal{} 1}^{2009}\\frac {n}{n^4 \\plus{} n^2 \\plus{} 1}$ $ \\equal{}\\frac{1}{2} \\sum_{n \\equal{} 1}^{2009}\\frac {1}{(n \\minus{} 1)n \\plus{} 1} \\minus{} \\frac{1}{2}\\sum_{n \\equal{} 2}^{2010}\\frac {1}{(n \\minus{}1)n \\plus{} 1}$\r\n\r\nSo $ \\sum_{n \\equal{} 1}^{2009}\\frac {n}{n^4 \\plus{} n^2 \\plus{} 1}$ $ \\equal{}\\frac{1}{2} \\minus{} \\frac{1}{2}\\frac {1}{2009*2010 \\plus{} 1}$\r\n\r\nSo $ \\sum_{n \\equal{} 1}^{2009}\\frac {n}{n^4 \\plus{} n^2 \\plus{} 1}$ $ \\equal{}\\frac {2009*1005}{2009*2010 \\plus{} 1}$ $ \\equal{}\\frac{2019045}{4038091}$, as Stephen said.", "Solution_9": "Could you express it in terms of the general sum formula?", "Solution_10": "[quote=\"triplebig\"]Could you express it in terms of the general sum formula?[/quote]\r\n\r\nIf you turn 2009 into k, the answer will be:\r\n\r\n$ \\frac{k(k\\plus{}1)}{2k(k\\plus{}1)\\plus{}2}$" } { "Tag": [ "blogs", "\\/closed" ], "Problem": "I'm sorry if this has already been discussed but there r a number of problems in my blog like:-\r\n\r\n1)Edit/delete an entry\r\n2)Contributors can't do anything\r\n\r\nare these common problems or just with my blog?", "Solution_1": "Yeah, it would be really helpful to be able to edit/delete blog entries :P", "Solution_2": "You can edit entries, just it takes some work. :D\r\n\r\n" } { "Tag": [ "function" ], "Problem": "i have a major doubt...\r\n\r\nlet $ \\omega_r$ be the resonant frequency\r\n\r\nand $ \\omega_1$ and $ \\omega_2$ be the frequencies when power is half the max. power\r\n\r\nnow.....\r\n[b]1.[/b] acc. to NCERT,the graph is symmetrical\r\nso $ \\omega_1 , \\omega_2 \\equal{} \\omega_r \\pm \\delta \\omega$\r\n\r\ni.e. $ \\omega_r \\equal{} \\frac{\\omega_1 \\plus{} \\omega_2}{2}$\r\n\r\n[b]2.[/b] graph is actually not symmetrical\r\n\r\nan alternate way to find $ \\omega_r$\r\nat resonance $ L\\omega_1 \\minus{} \\frac{1}{C\\omega_1} \\equal{}\\frac{1}{C\\omega_2} \\minus{} L\\omega_2$\r\n\r\nsolving,we get $ \\omega_r \\equal{} \\sqrt{\\omega_1 \\omega_2}$\r\n\r\nwhich one is true???\r\nif 1st is wrong ,it means whole derivation of NCERT is wrong (the most possible case)\r\n\r\nif both formulae are applicable ,solve the 2 equations\r\nit will give $ \\omega_1 \\equal{}\\omega_2 \\equal{} \\omega_r$\r\n :rotfl: not possible....", "Solution_1": "I dont think graph is symmetrical. The function 1/wC - wL doesnt look symmetric at all. f(wo+w') =/ f(wo - w')", "Solution_2": "then NCERT is wrong...\r\n\r\nand so is the formula $ Q \\equal{} \\frac{\\omega_r L}{R}$???\r\n\r\n\r\nmeanwhile,\r\nthe problem is 2 days,i had two tests\r\nboth hav the same questions\r\nbut diff formula had to be applied to get the answer :rotfl: now what??", "Solution_3": "fk the derivation. remember the result and its significance.", "Solution_4": "actually for a medical streamer ppl say prefer wat is given in NCERT\r\n :rotfl:" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Prove that if $a,b,c\\geq 0$ then the following inequality is true \r\n\r\n$a^{3}+b^{3}+c^{3}+2a^{2}b+2b^{2}c+2c^{2}a+6abc\\geq 5ab^{2}+5bc^{2}+5ca^{2}$", "Solution_1": "If (a,b,c)=(1,1,0), then sum(a^3+2*a^2*b)+6*a*b*c-5*sum(a*b^2)=-1,the inequality does not hold! We have\r\n\r\n sum(a^3+6*a^2*b)>=sum(5*a*b^2)+6*a*b*c.", "Solution_2": "[quote=\"fjwxcsl\"]sum(a^3+6*a^2*b)>=sum(5*a*b^2)+6*a*b*c.[/quote]\r\n\r\nCould you post your solution thank you ,", "Solution_3": "fjwxcsl just wants to say that your inequality is wrong...take $a=1,b=1,c=0$" } { "Tag": [ "limit", "logarithms", "calculus", "derivative", "function", "algebra", "binomial theorem" ], "Problem": "Evaluate $ \\lim_{x\\to\\infty} \\frac{x\\plus{}1}{x\\minus{}1}$.", "Solution_1": "[hide]$ \\lim_{x\\to\\infty}\\frac {x \\plus{} 1}{x \\minus{} 1} \\equal{} \\lim_{x\\to\\infty} 1 \\plus{} \\frac {2}{x \\minus{} 1} \\equal{} 1 \\plus{} 0 \\equal{} 1$[/hide]\n\n[hide]Since $ \\lim_{x\\to\\infty}x\\plus{}1 \\equal{} \\lim_{x\\to\\infty}x\\plus{}1 \\equal{} \\infty$, we can use L'Hopital's Rule: \n$ \\lim_{x\\to\\infty}\\frac{x\\plus{}1}{x\\minus{}1} \\equal{} \\lim_{x\\to\\infty} \\frac{1}{1} \\equal{} 1$[/hide]", "Solution_2": "HEre is a way . whenever we have these forms like $ \\lim_{x\\to\\infty}$ if possible always take the highest powers of the variable $ \\rightarrow \\infty$ outside . then it becomes easy . \r\n\r\nFor example in this case take $ x$ outside and write it as $ \\lim_{x\\to\\infty}\\frac { x(1\\plus{}1/ x ) }{x( 1 \\minus{}1/ x ) }$ \r\n\r\nAs $ x \\rightarrow \\infty$ therfore $ 1/x \\rightarrow 0$. hence the limit is $ 1$ .", "Solution_3": "Just don't take \"highest powers\" too literally; think about what's big and what's small, and what's bigger than what else.\r\n\r\n$ \\lim_{x\\to\\infty}\\frac{x^2\\plus{}x\\plus{}1\\plus{}2^x}{x^2\\minus{}x\\plus{}1\\minus{}2^x}\\equal{}?$", "Solution_4": "[quote=\"ThetaPi\"]Evaluate $ \\lim_{x\\to\\infty} \\frac {x \\plus{} 1}{x \\minus{} 1}$.[/quote]\r\n\r\nOops, the question is to evaluate $ \\lim_{x\\to\\infty} \\left(\\frac {x \\plus{} 1}{x \\minus{} 1}\\right)^x.$", "Solution_5": "The interesting challenge then is to do that in your head without writing anything down.\r\n\r\n[hide=\"Short answer.\"]$ e^2$[/hide]", "Solution_6": "I should have realized that the problem was too simple. \r\n[hide]\n$ \\lim_{x\\to\\infty}\\left(\\frac {x \\plus{} 1}{x \\minus{} 1}\\right)^{x} \\equal{} \\lim_{x\\to\\infty}\\left(1 \\plus{} \\frac {2}{x \\minus{} 1}\\right)^{x}$ \n\nLet $ y \\equal{} \\frac {x \\minus{} 1}{2}$, then $ x \\equal{} 2y \\plus{} 1$, and as $ x \\to \\infty$, $ y \\to \\infty$. \n\nTherefore: \n$ \\lim_{x\\to\\infty}\\left(\\frac {x \\plus{} 1}{x \\minus{} 1}\\right)^{x} \\equal{} \\lim_{x\\to\\infty}\\left(1 \\plus{} \\frac {2}{x \\minus{} 1}\\right)^{x}$ \n\n$ \\lim_{x\\to\\infty}\\left(\\frac {x \\plus{} 1}{x \\minus{} 1}\\right)^{x} \\equal{} \\lim_{y\\to\\infty}\\left(1 \\plus{} \\frac {1}{y}\\right)^{2y \\plus{} 1}$ \n\n$ \\lim_{x\\to\\infty}\\left(\\frac {x \\plus{} 1}{x \\minus{} 1}\\right)^{x} \\equal{} \\lim_{y\\to\\infty}\\left(1 \\plus{} \\frac {1}{y}\\right)\\left[\\left(1 \\plus{} \\frac {1}{y}\\right)^y\\right]^2$ \n\n$ \\lim_{x\\to\\infty}\\left(\\frac {x \\plus{} 1}{x \\minus{} 1}\\right)^{x} \\equal{} \\left[\\lim_{y\\to\\infty}\\left(1 \\plus{} \\frac {1}{y}\\right)\\right]\\left[\\lim_{y\\to\\infty}\\left(1 \\plus{} \\frac {1}{y}\\right)^y\\right]^2$\n\n$ \\lim_{x\\to\\infty}\\left(\\frac {x \\plus{} 1}{x \\minus{} 1}\\right)^{x} \\equal{} 1 \\cdot e^2 \\equal{} e^2$[/hide]", "Solution_7": "However, the problem I face is that I [i]do[/i] know that the value is $ e^2$, but I have no idea how to tackle the exponent. Using the binomial theorem does not seem to work either.", "Solution_8": "The logarithm of your limit is the limit of $ x\\ln\\left(\\frac{x\\plus{}1}{x\\minus{}1}\\right)\\equal{}x\\ln\\left(1\\plus{}\\frac2{x\\minus{}1}\\right).$\r\n\r\nWe can expand out the Taylor series of that logarithm:\r\n\r\n$ \\ln\\left(1\\plus{}\\frac2{x\\minus{}1}\\right)\\equal{}\\frac2{x\\minus{}1}\\plus{}O(x^{\\minus{}2})$\r\n\r\n$ x\\ln\\left(1\\plus{}\\frac2{x\\minus{}1}\\right)\\equal{}\\frac{2x}{x\\minus{}1}\\plus{}O(x^{\\minus{}1})\\equal{}2\\plus{}O(x^{\\minus{}1})$\r\n\r\nLet $ x\\to\\infty$ and that tends to $ 2;$ hence our answer.", "Solution_9": "I am sorry, but as you can see, I have only just started calculus at school; I am still in the midst of understanding product rule, chain rule, intermediate value theorem, continuity and differentiability etc.\r\n\r\nA simple solution would be good. I do not understand why such a question has appeared in my homework.", "Solution_10": "All right, here's the elementary version of that:\r\n\r\nAs before, we take the logarithm $ x\\ln\\frac{x\\plus{}1}{x\\minus{}1}$. We don't like limits at $ \\infty$ very much, so we substitute $ t\\equal{}\\frac1x$; it becomes $ \\frac1t\\cdot\\ln\\frac{\\tfrac1t\\plus{}1}{\\tfrac1t\\minus{}1}\\equal{}\\frac{\\ln\\frac{1\\plus{}t}{1\\minus{}t}}{t}$\r\n\r\nOK, now it's $ \\lim_{t\\to 0}\\frac{\\ln\\frac{1\\plus{}t}{1\\minus{}t}}{t}$. That's a derivative; if we define $ f(t)\\equal{}\\ln\\frac{1\\plus{}t}{1\\minus{}t}$, it's $ \\lim_{t\\to 0}\\frac{f(t)\\minus{}f(0)}{t}\\equal{}f'(0)$. If the logarithm is on the list of functions you know how to differentiate, you now have the tools.", "Solution_11": "I don't understand why people take the logarithm.\r\n\r\nThe limit\r\n\r\n$ \\lim_{x \\to \\infty} (1 \\plus{} \\frac {a}{x})^x$ is one of the fundamental ones and equals $ e^a$.\r\n\r\nIn fact, some texts use this to define the constant $ e$...", "Solution_12": "I agree with you. Neither L'Hopital nor Landau sign do need for the limit. :wink:" } { "Tag": [ "probability", "combinatorics unsolved", "combinatorics" ], "Problem": "Let $ F$ be the family of the subsets of the set $ A\\equal{}\\{1,2,...,3n\\}$ which have the sum of their elements divisible by $ 3$\r\nFor each element of $ F$,compute the square of sum of its element.What is the value of the sum of all the obtained numbers?", "Solution_1": "We will prove that every element of $ A$ belongs to exactly\r\n\\[ f(n) \\equal{} \\frac {2^n (2^{2n \\minus{} 1} \\plus{} 1)}{3}\r\n\\]\r\nmembers of $ F,$ after which computation is immediate and the total sum becomes\r\n\\[ S \\equal{} \\frac {2^n (2^{2n \\minus{} 1} \\plus{} 1) n (3n \\plus{} 1)(6n \\plus{} 1)}{6}\r\n\\]\r\nThe number of members of $ F$ that contain $ i \\in A$ is the same for all $ i \\equiv_3 j,$ so we denote it by $ f(n, i \\bmod 3).$ Then we can assume $ i \\in \\{1, 2, 3\\}$. If it is $ 3,$ then it can be or not be in a member of $ F$ with equal probability, so $ f(n, 3) \\equal{} \\tfrac{|F|}2.$ Now denote by $ \\{b_1, \\ldots, b_{3n}\\}$ a set in $ F$, with $ b_i \\equal{} 1$ if $ i \\in \\text{the set}$ and $ b_i \\equal{} 0$ otherwise. Then $ f(n, 1) \\equal{} f(n, 2) \\equal{} \\tfrac{|F|}{2},$ because the map $ F \\mapsto F$ that sends\r\n\\[ \\{b_1, b_2, b_3, b_4, \\ldots, b_{3n}\\} \\mapsto \\{1 \\minus{} b_1, 1 \\minus{} b_2, b_3, b_4, \\ldots, b_{3n}\\}\r\n\\]\r\nbijectively maps members of $ F$ to which $ 1$ belongs to members to which $ 1$ doesn't (and similarly for $ 2$). We have reduced our assertion, then, to finding $ |F|.$ We know that\r\n\\[ |F| \\equal{} 2^n \\times \\text{\\# of } \\subset \\{1,2,4,5,\\ldots,3n \\minus{} 2,3n \\minus{} 1\\} \\text{ whose sum of elements } \\equiv_3 0\r\n\\]\r\nIt is enough to prove that this number above equals $ g(n) \\equal{} (2^{2n} \\plus{} 2) / 3.$ The subsets of $ \\{1,2,4,5 ,\\ldots, 3n \\minus{} 2, 3n \\minus{} 1\\}$ whose sum of elements is divisible by $ 3$ can be split in two classes, with one consisting of supersets of $ \\{3n \\minus{} 2,3n \\minus{} 1\\}.$ This class has $ g(n \\minus{} 1)$ members, while the other has $ 2^{2n \\minus{} 2}$ elements (choose among the first $ 2n \\minus{} 2$ elements arbitrarily, and then you have a single choice for the last one.) The relation $ g(n) \\equal{} g(n \\minus{} 1) \\plus{} 2^{2n \\minus{} 2}$ and $ g(1) \\equal{} 2$ ($ F$ contains the empty set) gives $ g(n) \\equal{} 2 \\plus{} 2^2 \\plus{} 2^4 \\plus{} \\cdots \\plus{} 2^{2n \\minus{} 2} \\equal{} (4^n \\plus{} 2) / 3$ and we're done.", "Solution_2": "bodan, it asks for the sum of the squares of the sums, not the sum of the squares of the elements ;)", "Solution_3": "So then we also need to find, for a set $ X \\equal{} \\{a, b\\} \\subset A,$ how many members of $ F$ contain $ X.$ \r\n\r\nThis time it's $ \\tfrac{|F|}{4},$ because $ a$ and $ b$ can be taken to be in $ a \\in \\{1,2,3\\}$ and $ b \\in \\{4,5,6\\},$ (this works only for $ n > 1,$ we have to check $ n \\equal{} 1$ separately) and again the map\r\n\\[ \\{b_1, b_2, b_3, b_4, \\ldots, b_{3n}\\} \\mapsto \\{1 \\minus{} b_1, 1 \\minus{} b_2, b_3, b_4, \\ldots, b_{3n}\\}\r\n\\]\r\ntakes bijectively members of $ F$ that contain $ a$ and contain $ b$ to members that do not contain $ a$ but contain $ b,$ and since the total number of members that contain $ b$ was proven to be $ \\tfrac{|F|}{2},$ the result follows. \r\n\r\nSo, for $ n \\geq 2,$\r\n\\[ \\sum_{3 \\,|\\, a \\plus{} b \\plus{} \\cdots} (a \\plus{} b \\plus{} \\cdots)^2 \\equal{} \\sum_{3 \\,|\\, a \\plus{} b \\plus{} \\cdots} \\sum_{a,b} ab \\equal{} \\frac {|F|}{2} \\sum_{1 \\leq i \\leq j \\leq 3n} ij\r\n\\]\r\nand since we know $ |F|,$ we can simplify this.", "Solution_4": "We can generalize this as follows: denote, for a subset $ \\{a_1, \\ldots, a_k, a_{k\\plus{}1}, \\ldots\\} \\equal{} X \\subset A$ with $ k \\leq n,$ the $ k\\minus{}$th symmetric sum of $ X$ by \\[ \\sigma_k (X) \\equal{} a_1 \\cdots a_k \\plus{} a_1 \\cdots a_{k\\minus{}1}a_{k\\plus{}1} \\plus{} \\cdots\\] Let $ s(X)$ be the sum of elements of $ X,$ and put \\[ S_k (A) \\equal{} \\sum_{3 \\,|\\, s(X)} \\sigma_k (X)\\] Then each $ k\\minus{}$element subset of $ A$ appears in the sum $ S_k$ exactly $ \\tfrac{|F|}{2^{k}}$ times (because $ k \\leq n,$ we can put the elements of this $ k\\minus{}$element subset each in $ \\{3i \\plus{} 1, 3i\\plus{}2,3i\\plus{}3\\},$ for $ 0 \\leq i < k;$ then partition the members of $ F$ in $ 2^k$ classes of the same size, only one of which contains the entire $ k\\minus{}$element subset.) So\r\n\\[ S_k (A) \\equal{} \\frac{2^{n\\minus{}k\\plus{}1}(2^{2n\\minus{}1} \\plus{} 1)}{3} \\sum_{1 \\leq i_1 < \\cdots < i_k \\leq mn} i_1 \\cdots i_k\\]", "Solution_5": "See also http://www.mathlinks.ro/viewtopic.php?t=290343 :)" } { "Tag": [ "geometry", "analytic geometry", "trigonometry", "function", "inequalities", "HCSSiM", "probability" ], "Problem": "Post away :lol:", "Solution_1": "How can the area of the quadrilateral in problem 3 be $8i$?", "Solution_2": "[quote=\"stupidkid\"]How can the area of the quadrilateral in problem 3 be $8i$?[/quote]\r\nI think on the Math Jam he realized he made a mistake on that. Here are my solutions. Feel free to criticize, especially problem 2 (I used a solution that, while workable, isn't particularly elegant):", "Solution_3": "Yeah, that's wrong, as I found out in the Math Jam. I thought that coordinates => area", "Solution_4": "I'm trying to figure out how we could have gotten the same answers yet my solutions are 2-3 times as long. :huh:", "Solution_5": "I wasn't at the math jam, but from what I've gathered on the forums, my solutions for 2, 4, and 5 are definitely correct. I'm fairly confident that my solutions for 1 and 3 are also correct, and these are all brief, with pictures too! The only question is... were they too brief... we'll see...\r\n\r\nEdit: Bah, #3 wrong, I simplified root(4r^6) into 4r^3. Hopefully I'll only lose a point for that. Everything else is correct.", "Solution_6": "[quote=\"sonar\"]Post away :lol:[/quote]\r\nI loved your solution to #2 - so simple and elegant (especially the way you generalized it).", "Solution_7": "Wow, your solutions are very detailed...I kinda said, \"Similarly it can be shown....\" quite a lot. Haha, probably will get some points deducted for that.", "Solution_8": "i wasn't in the math jam and yes i knew that i got a lot of them wrong, but here are my solutions. Criticize away :)", "Solution_9": "I can't post mine because I handwrote it. Oh well. :( \r\n\r\nMy solutions sounded really good. What they didn't accomplish in equations, they made up for in language, hehe. More points for me! :D", "Solution_10": "here are mine. i think i was kind of too undetailed for some stuff.", "Solution_11": "1/lnx, your solutions weren't that bad. :) \r\n\r\nAll the trig on number 5 made me dizzy for a while though. :D", "Solution_12": "I'll follow suit and post my solutions. If you feel like it, you can read or skim over my solution to #1, which was probably way overdone. I defined a function [i]f[/i] which is basically equivalent to modulus, except I wasn't so sure about modular arithmetic so I didn't want to risk making a mistake.\r\n\r\nI like the way I did #5. I haven't seen any other people do it that way. It was kinda long, though.\r\n\r\nMy solution to #2 was also pretty good. It was actually just a page, making it my shortest one.", "Solution_13": "[quote=\"1/(ln x)\"]i wasn't in the math jam and yes i knew that i got a lot of them wrong, but here are my solutions. Criticize away :)[/quote]\r\nI think your problem with #4 is you assumed it was sufficient for $a + b$ to be expressable in more than one way as a sum. In fact, you find that $a + b$ cannot be expressable as a sum of squares. You also say at one point that $a \\neq a'_1$ and $b \\neq b'_1$, and then reverse that inequality in the next paragraph - so, I'm not sure why you ended up with the solution you did.\r\n\r\nAs a general note, did anybody else provide a motivation for their factorization in #3? I tried to, because just showing the factorization seemed a bit arbitrary.", "Solution_14": "[quote=\"zanttrang\"]Feel free to criticize, especially problem 2 (I used a solution that, while workable, isn't particularly elegant):[/quote]\n\nBut your solution for 3 makes up for it :)\n\n[quote=\"zanttrang\"]I loved your solution to #2 - so simple and elegant (especially the way you generalized it).[/quote]\n\nThanks, HCSSiM Probability Maxi represent :P\n\n[quote=\"stupidkid\"]Wow, your solutions are very detailed...I kinda said, \"Similarly it can be shown....\" quite a lot. Haha, probably will get some points deducted for that.[/quote]\r\n\r\nNot necessarily. Dr. Schram noted in last year's posts that if the method is exactly the same, saying \"similarly ...\" is acceptable. I'll try and find the exact post to give a bit of context.", "Solution_15": "Some of you guys should learn how solutions are properly written. ;) I thought that a lot of them were too informal. Look through the published solutions for year 16, for example. :)\r\n\r\nWell here are my solutions. I didn't get #5 and #2 is wrong :stink: but I think the rest are good.\r\n\r\nFeel free to laugh though! :D", "Solution_16": "Here they are: I had to hunt up my tex compiler again to recompile. I don't really mind you guys getting my ID number - that's public information - and I trust you guys not to use my name. ;)", "Solution_17": "I think your sum notation is wrong on #2, and also the answer is wrong... but so is mine :( Nice job on #1 and #5 though! :)", "Solution_18": "Yes, I had the wrong sum. :( I couldn't figure out how to include an image, so I didn't have a diagram included with the last one, and I forgot to *prove* that the segments connecting the midpoints of 2 sides of a triangle has half the length of the third side and also that if a quadrilateral has an inscribed circle tangent to all 4 sides the sum of any 2 opposite sides is equal to any other sum. Hopefully I won't lose a point for that.", "Solution_19": "[quote=\"jb05\"]All right, I don't have PDF writers, and all my solutions are handwritten, so I'm just going to choose my favorite (and one of my shortest), #3:\n\n[hide]Let $w=rz$ and $q=r^4$. Then the expression reduces to $w^4+(10q-2)w^2-16qw+(q^2+10q+1)=0$ or $9q^2+(10w^2-16+10)q+(w^4-2w^2+1)=0$. We use the quadratic formula on the above (in my proof I go through it, but here I will just give the result) and get $q=-(9w^2-18w+9)/9$ or $w=-(w^2+2w+1)/9$. Therefore, $-q=(w-1)^2$ or $-9q=(w+1)^2$. So $w=1\\pm i\\sqrt q$ or $w=-1\\pm 3i\\sqrt q$. Since w=r*z and $q=r^4$, $z=1/r\\pm ir$ or $z=-1/r\\pm 3ir$ (again, sorry but I'm leaving out intermediate steps to save space). Since this figure has 2 sets of 2 points each having identical x-coordinates, it is a trapezoid. The average of the bases is (6r+2r)/2=4r and the height is 2/r. Multiplying, the area is always 8.[/hide][/quote]\r\nNice job -- that's a really clever substitution. :)", "Solution_20": "[quote=\"solafidefarms\"]Yes, I had the wrong sum. :( I couldn't figure out how to include an image, so I didn't have a diagram included with the last one, and I forgot to *prove* that the segments connecting the midpoints of 2 sides of a triangle has half the length of the third side and also that if a quadrilateral has an inscribed circle tangent to all 4 sides the sum of any 2 opposite sides is equal to any other sum. Hopefully I won't lose a point for that.[/quote]\r\nYou don't have to prove that the midsegment of a triangle is half the length of the corresponding side. That's already a well-known fact. ;)", "Solution_21": "[quote=\"chess64\"]Some of you guys should learn how solutions are properly written. ;) I thought that a lot of them were too informal. Look through the published solutions for year 16, for example. :)\n\nWell here are my solutions. I didn't get #5 and #2 is wrong :stink: but I think the rest are good.\n\nFeel free to laugh though! :D[/quote]\r\nNice diagram on #3. :)", "Solution_22": "I somehow kept thinking that the $10r^4$ term was $4r^4$ for some reason and so I missed that factoring thing and didn't get the answer. :wallbash_red: :wallbash_red: :wallbash_red:", "Solution_23": "I'll post the pdf version of my solutions. Maybe I shouldn't have sent in a dvi...\r\n\r\nEDIT: I posted something completely different, something from my math club. My bad. :D \r\n\r\nHere's the zip I sent in, anyways...\r\n\r\nEDIT2: Sorry i had to delete it...was taking up space...but ask me if you want it...", "Solution_24": "[quote=\"K81o7\"]I'll post the pdf version of my solutions. Maybe I shouldn't have sent in a dvi...[/quote]\n\n[quote=\"USAMTS\"]We will only accept the following file types: DVI, PDF, RTF, TEX, TXT, TIF, GIF, BMP, JPG, EPS. [/quote]\r\nYou should be fine.\r\n\r\np.s. What problem is your solution for? Doesn't look like something from the USAMTS. :huh: \r\nEither something's messed up or I'm just crazy . . .", "Solution_25": "[quote=\"ZennyK\"]\np.s. What problem is your solution for? Doesn't look like something from the USAMTS. :huh: \nEither something's messed up or I'm just crazy . . .[/quote]\r\nNo, you're not crazy - I'm not sure what his solution is for either, but it doesn't look like USAMTS stuff.", "Solution_26": "Wayyyyyy too informal, but it *might* do the job.", "Solution_27": ":o Whoops.... :D \r\nThis is the solution i wrote for a different problem, but i converted at the same time. I'll put my USAMTS solutions in when I get home...\r\nI'm such an idiot...\r\nJust pretend you didn't see it. :?", "Solution_28": "[quote=\"solafidefarms\"]Wayyyyyy too informal, but it *might* do the job.[/quote]\r\n\r\nWhose thing are you talking about?", "Solution_29": "Well here's the pdf wtih my solutions...and the diagram i drew for #5..." } { "Tag": [ "trigonometry", "function", "real analysis", "calculus", "integration", "derivative", "calculus computations" ], "Problem": "Inverse Laplase transformation of 1 / (s^2 + 1)^3? That s^2 is what baffles me. Someone help please? Much thanks.", "Solution_1": "Are you sure this isn't an inverse Lapla[b]c[/b]e transform problem? Is this possibly what the problem really said?\r\n\r\nFind $f(t)$ such that $\\mathcal{L}(f)(s)=\\frac1{(s^{2}+1)^{3}}.$", "Solution_2": "Oops, sorry. It's inverse. :blush:", "Solution_3": "One possible approach (and a messy one):\r\n\r\nLet $y(t)=\\sin t.$ Then $\\mathcal{L}(y)(s)=\\frac1{s^{2}+1}.$\r\n\r\nThe function we want is $f(t)=y*y*y(t).$ \r\n\r\nAdmittedly, computing those convolution integrals isn't nice.", "Solution_4": "The transform of $tf(t)$ is $-F'(s)$. So the transform of $t^{2}\\sin t$ should be $\\frac{2(3s^{2}-1)}{(s^{2}+1)^{3}}$. Well, this is not quite what you want, but it may give you an idea.\r\n\r\nTry [url=http://www.eecircle.com/applets/007/ILaplace.html]this applet[/url], but don't trust it blindly.", "Solution_5": "And where the sine is, so also should be the cosine. The Laplace transform of $\\cos t$ is $\\frac{s}{s^{2}+1}.$\r\n\r\nUsing the derivative relationship mlok gave, try making yourself a little table with the Laplace transforms of $\\sin t,\\ t\\sin t,\\ t^{2}\\sin t,\\ \\cos t,\\ t\\cos t,$ and $t^{2}\\cos t.$", "Solution_6": "One last perspective:\r\n\r\nThe function we want is the solution of this differential equation:\r\n\r\n$y^{[6]}+3y^{[4]}+3y''+y=0;\\ y(0)=y'(0)=y''(0)=y'''(0)=y^{[4]}(0),\\ y^{[5]}(0)=1$\r\n\r\nThis solution has the form\r\n\r\n$y(t)=(At^{2}+Bt+C)\\cos t+(Dt^{2}+Et+F)\\sin t$\r\n\r\nThere's an implied $6\\times 6$ system of linear equations to solve for the coefficients of that linear combination; you might want to seek electronic help with that linear system." } { "Tag": [ "videos" ], "Problem": "Hmmm? I can do 70 wpm without punctuation. (Real sentences, reasonably common words; I read off a sentence and then type it from memory, so if it's not grammatically complete or spelled correctly :D I am slooow.)", "Solution_1": "70-100+, depending on what I'm typing (the words, whether there are numbers/symbols, capitalization, etc.)", "Solution_2": "According to the typing program at school I can get up to 720 :D :D \r\nReally I am usually at about 100-150 wpm", "Solution_3": "biff world record was 360 so stop lying ;) \r\n\r\nI think you meant words by the way biff, he meant gwam (5 characters a word), not numbers ;)", "Solution_4": "around 75-85\r\n\r\non micro type pro, the fastest i got was 117, but it was easy words like \"fast hide good lukc to you he she it\" :D", "Solution_5": "Approximately 110...\r\nBut yeah, that's bad grammar and stuff.", "Solution_6": "I can consistently type around 130. [for extended periods of time]\r\n\r\nReach peak speeds of 150.\r\n\r\nIf you don't believe me you haven't played trivia against me under the old AOPSTI format.", "Solution_7": "If I type with complete accuracy, and am going pretty quickly, maybe 70 wpm. I generally don't just speed type making grammar mistakes, but just now I did a sample, typing the last paragraph of my english assignment sheet, with a few words mispelled at the end, and got around 90-100 wpm.", "Solution_8": "I was like 35 GWAM..and I might have gotten like 40 a couple times", "Solution_9": "Somewhere over 100, with good grammar and punctuation, etc.\r\n\r\nWho said over 200? I don't believe you. :P", "Solution_10": "I want to see a video of typing that fast.", "Solution_11": "[quote=\"la mafioso\"]biff world record was 360 so stop lying ;) \n\nI think you meant words by the way biff, he meant gwam (5 characters a word), not numbers ;)[/quote]\r\nI can type six words in one second...\r\nbut they're all \"a\"", "Solution_12": "[quote=\"biffanddoc\"][quote=\"la mafioso\"]biff world record was 360 so stop lying ;) \n\nI think you meant words by the way biff, he meant gwam (5 characters a word), not numbers ;)[/quote]\nI can type six words in one second...\nbut they're all \"a\"[/quote]\r\n\r\ni want see u do that ;)", "Solution_13": "On typing programs, i get 160. Consistantly, i get about 70 if im concentrating (usually not).", "Solution_14": "I type between 100-150 and I use proper grammar and stuff. It's really not that hard unless you don't type very often. By the way, I believe Lucky707. He killed us in Trivia. :D", "Solution_15": "Depends on the Language. In English I can type around 120-150 wpm. In Romanian I can easily do 150-180 wpm (without special characters, ie replacing \u00e2 with a, \u008d\u008d\u0090\u00ee with i and so on). With diacritics, it goes way down to 50-80 wpm.", "Solution_16": "I usually get 60-70 wpm, and the highest I got was like 110 or something.", "Solution_17": "WOW you know what....all of you people are like KRAZY and must have 11 brains one on each finger+ one in ur head. I only have one. I average 50 and 60 when i am feeling good 80 when I am typing really really fast", "Solution_18": "I did a 2 minute online typing test just now, with gross speed of 82 wpm and a net speed of 73 wpm. That's about where I figured I was at.", "Solution_19": "am i just slow???? my words per minute is at 45 right now...", "Solution_20": "According to an online test, I type at around 80 wpm at 98+ accuracy (an insertion/deletion throws everything off, and you can't backspace). But the passage I typed was pretty complex, and it lasted 3 minutes. When I'm typing off the top of my head instead of reading, I'm guessing my speed goes over 100 wpm.", "Solution_21": "Around 45-50 WPM with numbers, punctuation and high accuracy. I go ALOT faster with no numbers and symbols.", "Solution_22": "[quote=\"probability1.01\"]According to an online test, I type at around 80 wpm at 98+ accuracy (an insertion/deletion throws everything off, and you can't backspace). But the passage I typed was pretty complex, and it lasted 3 minutes. When I'm typing off the top of my head instead of reading, I'm guessing my speed goes over 100 wpm.[/quote]\r\n\r\nI usually type faster when I'm reading something, because otherwise I don't always know what I want to say. However, I generally get lower accuracy when I do that, because I'm typing so fast that I don't always think about what I'm doing, and of course it doesn't let you backspace. >_<", "Solution_23": "I'm at 52 words per minutes, but it's probably higher on a normal day, cause I'm really really tired right now.", "Solution_24": "INSANELY FAST. if I really really have to. usually though i'd guess... 100-150 rpm? lol, wpm, not rpm.", "Solution_25": "Err.. well the last time i measured was 3 years ago and then i did 55. But now, on average of just typing casually (no rush) i do 70. When im in a rush i can do over 100 easily. And its cool to type while your reading something (ie copying it onto computer) - i duno its just a cool feeling." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,\\ b,\\ c$ be real numbers.\r\n\r\nProve: $ (a^2 \\minus{} ab \\plus{} b^2)|a \\minus{} b| \\plus{} (b^2 \\minus{} bc \\plus{} c^2)|b \\minus{} c| \\plus{} (c^2 \\minus{} ca \\plus{} a^2)|c \\minus{} a|$\r\n\r\n$ \\plus{}2(\\sqrt{1\\minus{}a^2}\\plus{}\\sqrt{1\\minus{}b^2}\\plus{}\\sqrt{1\\minus{}c^2})\\leq 12.$\r\n\r\n[color=red]Last edited[/color]", "Solution_1": "[quote=\"kunny\"]Let $ a,\\ b,\\ c$ be real numbers.\n\nProve: $ (a^2 \\minus{} ab \\plus{} b^2)|a \\minus{} b| \\plus{} (b^2 \\minus{} bc \\plus{} c^2)|b \\minus{} c| \\plus{} (c^2 \\minus{} ca \\plus{} a^2)|c \\minus{} a|$\n\n$ \\geq \\frac {4}{27}(a \\plus{} b \\plus{} c)^4 \\plus{} \\frac 13 (ab \\plus{} bc \\plus{} ca)^2$[/quote]\r\n\r\nWas the inequality supposed to be reversed from the above? Because $ a\\equal{}b\\equal{}c\\equal{}1$ immediately yields a contradiction.\r\n\r\nCheers,\r\n\r\nRofler", "Solution_2": "Thank you for pointing out it, but I have changed the problem a little bit.", "Solution_3": "[quote=\"kunny\"]Let $ a,\\ b,\\ c$ be real numbers.\n\nProve: $ (a^2 \\minus{} ab \\plus{} b^2)|a \\minus{} b| \\plus{} (b^2 \\minus{} bc \\plus{} c^2)|b \\minus{} c| \\plus{} (c^2 \\minus{} ca \\plus{} a^2)|c \\minus{} a|$\n\n$ \\plus{} 2(\\sqrt {1 \\minus{} a^2} \\plus{} \\sqrt {1 \\minus{} b^2} \\plus{} \\sqrt {1 \\minus{} c^2})\\leq 12.$\n\n[color=red]Last edited[/color][/quote]\r\nHow about if $ a>1,\\, b>1\\,\\,$ or $ \\,\\,c>1$ ?", "Solution_4": "We are to suppose that $ a,\\ b,\\ c\\in{[\\minus{}1,\\ 1]}$, which is common sense, isn't it? :wink:", "Solution_5": "[quote=\"kunny\"]We are to suppose that $ a,\\ b,\\ c\\in{[ \\minus{} 1,\\ 1]}$, which is common sense, isn't it? :wink:[/quote]\r\nCommon sense? I do not think so.", "Solution_6": "Why? Can you say $ \\sqrt{1\\minus{}a^2}<0?$ :P" } { "Tag": [ "AMC" ], "Problem": "when is the IMO going to be?\r\n\r\nGO USA!!!! :winner_first: :yoda: :thumbup:", "Solution_1": "The contest itself is on July 25 and 26. See [url=http://www.imo2007.edu.vn/]the official site[/url] for more details." } { "Tag": [ "geometry", "calculus", "probability", "number theory" ], "Problem": "My math team participate in the City of Chicago Math League and its format is as follows:\r\n\r\nEach contest consists of a 20 question exam which must be completed in 50 minutes. Contests are administered at three sites; schools are assigned to sites based on their locations. Competition is divided into four grade level/topic areas: Algebra 1 (9th grade), Geometry (10th grade), Algebra 2 (11th grade), and Pre-Calculus (12th grade). An advanced student who has already completed the course for her grade level (for example, a sophomore who has already completed geometry) can take an \"Accelerated\" version of that contest, or can take a higher-level contest. As a result, there are eight levels of competition.\r\n\r\nRather than having people practice independently at practice (which is really boring for them and for the group), I tried to do different things with the old practice tests. In the past couple of months I did:\r\n\r\n1) Relays (of another contest)\r\n-> Our team captain did a Ti-89 workshop.\r\n2) Speed rounds (one on one and category vs. category)\r\n3) Counting and Probability Seminar (I go through all the old tests and extract all the problems related to the topic. I explained the basics, then each team solve their own probability problems then present the hard ones).\r\n4) Musical chairs (where the ones left standing up had to write their memorized topic formulas- that is given on an index card- on the board)\r\n\r\nI ran out of ideas so fast that I am freaking out now. What are other FUN and productive ways to practice for such contests on this format? It is really hard because the team is divided among FOUR topic areas (algebra-precalc) and FOUR subs (in-progress and accelerated). \r\n\r\nI am a co-captain and I really want the math team practices to be \"better than math class\" . \r\n\r\nAny ideas for me? Thanks!", "Solution_1": "Well, since you had a C/P seminar, you might as well have a basic number theory seminar :D \r\nsorry, that wasnt very creative\r\n\r\nactually, i would say occasionally take a break and do some challenge problems. Although it isn't from the contest, it would indirectly help them by improving their problem solving skills. they like challenge problems, right? :maybe:" } { "Tag": [ "algorithm", "number theory open", "number theory" ], "Problem": "Hi , i'm not god on mathematics sow please help me.I'm working on software who must calculate an algorithm :\r\n i have 16 rows with 3 values and i have to calculate the maximum of options for this algo.\r\n ex: 1. a1 a2 a3 \r\n 2. b1 b2 b3\r\n 3. c1 c2 c3\r\n ....................\r\n ....................\r\n 16. q1 q2 q3\r\n the algo is like the game for socker .you must choice 1 value for 1 string : a1 \r\n or a2 or a3 for first string so on until the end.So , my problem is what algo to use ?Please help me bigs..!!!", "Solution_1": "use recursion to exhaust the choices" } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Let $ f: \\mathbb{R}\\rightarrow\\mathbb{R}$ be such that \r\n\r\n$ f(\\frac{x\\plus{}y}{2})\\plus{}f(\\frac{2xy}{x\\plus{}y})\\equal{}f(x)\\plus{}f(y)$ for all positive $ x,y$. Prove that\r\n\r\n$ 2f(\\sqrt{xy})\\equal{}f(x)\\plus{}f(y)$ for all positive $ x,y$.", "Solution_1": "Not sure if the following link gives the answer.\r\nhttp://www.mathlinks.ro/viewtopic.php?t=241518", "Solution_2": "Here the continuity is not given..." } { "Tag": [ "number theory open", "number theory" ], "Problem": "is it true or not:if k=/=1 there is n>1 such that 2^n =k(modn) are there infinitely many such n?", "Solution_1": "I think that this problem is an open problem, i dont think anybody has found a counterexample, but i dont think the proof is known.", "Solution_2": "This is F10 in UPINT3." } { "Tag": [ "modular arithmetic", "algebra open", "algebra" ], "Problem": "Is is true that: ${{ktp^{l}}\\choose{k}}\\equiv{0}\\pmod{p}$ where $p$ is a prime and the rest are positive integers :?:", "Solution_1": "[quote=\"dondigo\"]Is is true that: ${{ktp^{l}}\\choose{k}}\\equiv{0}\\pmod{p}$ where $p$ is a prime and the rest are positive integers :?:[/quote]\r\n\r\nIs it true. It is also true that $\\binom{ap^{b}}{c}\\equiv 0 \\pmod{p}$ for $p | \\frac{ap^{b}}{c}$, because $\\binom{ap^{b}}{c}= \\frac{ap^{b}}{c}\\times \\prod_{i=1}^{b-1}{\\frac{ap^{b}-i}{i}}$, and since $p^{x}| i \\Rightarrow p^{x}| (ap^{b}-i)$, thus if $p | \\frac{ap^{b}}{c}$ then $p | \\binom{ap^{b}}{c}$. The original is a trivial consequence of this." } { "Tag": [ "floor function" ], "Problem": "Solve in the equation. \r\n\r\n$ [\\frac{3x^2}{x^2\\plus{}x\\plus{}1}]\\equal{}x$,$ x\\in R$ :blush: \r\n\r\n\r\n\r\n\r\n__________________________________\r\n$ \\textbf{Azerbaijan Land of Fire }$", "Solution_1": "Let $ f(x)\\equal{}3x^2/(x^2\\plus{}x\\plus{}1)$. First note that $ 3x^2\\ge0$ and $ x^2\\plus{}x\\plus{}1\\equal{}(x\\plus{}1/2)^2\\plus{}3/4>0$, so $ \\min{f(x)}\\equal{}0$ when $ x\\equal{}0$. Furthermore,\r\n\\[ f(x)\\le4\\Longleftrightarrow(x\\plus{}2)^2\\ge0,\\]so $ \\max{f(x)}\\equal{}4$ when $ x\\equal{}\\minus{}2$. Thus $ \\lfloor{f(x)}\\rfloor\\equal{}x\\equal{}\\{0,1,2,3,4\\}$.\r\n\r\n[list][*]$ x\\equal{}0\\implies\\lfloor{f(x)}\\rfloor\\equal{}0$, which works.\n[*]$ x\\equal{}1\\implies\\lfloor{f(x)}\\rfloor\\equal{}1$, which works.\n[*]$ x\\equal{}2\\implies\\lfloor{f(x)}\\rfloor\\equal{}1$, which does not work.\n[*]$ x\\equal{}3\\implies\\lfloor{f(x)}\\rfloor\\equal{}2$, which does not work.\n[*]$ x\\equal{}4\\implies\\lfloor{f(x)}\\rfloor\\equal{}2$, which does not work.[/list]\r\nHence $ x\\equal{}\\boxed{0,1}$." } { "Tag": [ "geometry", "calculus", "integration", "rectangle", "analytic geometry", "conics", "parabola" ], "Problem": "Is shoelace theorem the only way to solve this problem other than graphing and counting squares:\r\n\r\nWhat is the area of a triangle with vertices at the points (-8,5) (3,1) and (5,-4)", "Solution_1": "There are various other ways, but they're definitely more complicated for this kind of problem.", "Solution_2": "oh okay ty, i guess i will just learn the shoelace theorem :(", "Solution_3": "The other ways involve using the distance formula and heron's to get really ugly numbers which give a clean answer. I'm surprised that the shoelace never gives ugly numbers i.e. radicals.", "Solution_4": "Heron's formula simplifies fairly nicely in almost all cases with integral coordinates.\r\n\r\n\r\nShoelace gives nice answers because these polygons can be inscribed in a rectangle with integral sidelengths, and you only need to chip off rectangles with integral areas and triangles with area ending in .0 or .5.\r\n\r\nBasically, the area of a polygon with only integral coordinates can be expressed in the form $\\frac{C}{2}$, where C is an integer.", "Solution_5": "hmm... I know herons. which is the sq rt of semiperimeter times semi minus first side, semi minus 2nd, and semi minus 3rd.\r\n\r\nIt is kinda hard sometimes though cuz of ugly numbers yeah...\r\n\r\nShoelace? Is it hard to put in your head? Because everytime I see it I can't draw it myself with those diagonals. IF it gives nice numbers, maybe it's worth memorizing? im thinking about it :roll:", "Solution_6": "[quote=\"bpms\"]The other ways involve using the distance formula and heron's to get really ugly numbers which give a clean answer. I'm surprised that the shoelace never gives ugly numbers i.e. radicals.[/quote]\r\n\r\nAnd you'll also be pretty surprised when you find that the area between two parabolas will usually turn out nice, like 17/3 or something.\r\n\r\nShoelace is worth learning, graphing and counting points would work if you wanted to use Pick's, but that takes a long time.", "Solution_7": "Shoelace is easiest unless you're dealing with three points such that 2 of them have either the same x or y coordinate.", "Solution_8": "oh okay. I will learn it this weekend and get it stuck in my head. \r\n\r\nALSO, WHEN given 4 coordinates for a tetrahedron, does shoelace work? or does a similar formula work?", "Solution_9": "I doubt it, how are you suggesting you go about finding the volume?\r\n\r\nIt works for all polygons, though, as long as all the coordinates are in the correct order.", "Solution_10": "tetrahedron surface area \r\n\r\nnot volume", "Solution_11": "When would you ever have the coordinates for the vertices of a tetrahedron and be asked to find the surface area?\r\n\r\n@bpms: You shouldn't be surprised that it almost never gives radicals (unless you start with radical coefficients) because if the coordinates are all integers, then the worst thing that could happen would be a fraction :o", "Solution_12": "on my fact sheet i got from school it says you can find the volume of a tetrahedron with vertices x1 y1 z1, x2 y2 z3, and x4 y4 and z 4 \r\n\r\n\r\nis volume equals 1/6 times the absolute value of x1 y1 z1 1\r\n x2 y2 z2 1\r\n x3 y3 z3 1\r\n x4 y4 z4 1", "Solution_13": "You can also use matrices to find the area of a triangle.", "Solution_14": "Pick's Theorem is much cooler :D", "Solution_15": "what is pick's theorem?", "Solution_16": "$I+\\frac{B}{2}-1$ is equal to the area of any polygon in the coordinate system, where $I$ is the amount of integer coordinates contained inside the polygon and $B$ is the amount of integer coordinates contained on the perimeter of the polygon.", "Solution_17": "Shoelace is much more powerful than pick's", "Solution_18": "Is there a proof for the shoelace theorem?", "Solution_19": "I think you can prove the Shoelace's theorem by thinking the sides of polygon as vectors, calculate ther crossproducts, and summing up.", "Solution_20": "[quote=\"Totally Zealous\"]Is there a proof for the shoelace theorem?[/quote]\r\n\r\nWell obviously, as it's a theorem. :wink: \r\n\r\nGoogle it, you'll find results.", "Solution_21": "hmm i've never heard of this theorem before. What is it?", "Solution_22": "[quote=\"13375P34K43V312\"]Google it, you'll find results.[/quote]", "Solution_23": "[quote=\"bpms\"]Shoelace is much more powerful than pick's[/quote]\r\n\r\nI agree.", "Solution_24": "Hey just wondering is it also called shoelace [i]algorithm[/i]?\r\nThe person that taught it to me said it was the shoelace algorithm.", "Solution_25": "Lots of questions, here's some answers:\r\n\r\nhttp://staff.imsa.edu/math/journal/volume2/articles/Shoelace.pdf", "Solution_26": "Call it whatever you want.\r\n\r\nFor some odd reason bpms calls the Fermat-Euler theorem Euler's totient. You can call things whatever you want, just know how to use it and (hopefully) why it works.", "Solution_27": "[quote=\"now a ranger\"]on my fact sheet i got from school it says you can find the volume of a tetrahedron with vertices x1 y1 z1, x2 y2 z3, and x4 y4 and z 4 \n\n\nis volume equals 1/6 times the absolute value of x1 y1 z1 1\n x2 y2 z2 1\n x3 y3 z3 1\n x4 y4 z4 1[/quote]\r\n\r\nHave you learned vectors? This is derived directly from the vector product of the three defining vectors of a tetrahedron.", "Solution_28": "[quote=\"13375P34K43V312\"]Call it whatever you want.\n\nFor some odd reason bpms calls the Fermat-Euler theorem Euler's totient. You can call things whatever you want, just know how to use it and (hopefully) why it works.[/quote]\r\nThat is how it was referred to when I first saw it. I have seen it called Fermat-Euler and Euler's generalization.", "Solution_29": "i know a little bit about vectors, just vaguely\r\n\r\nyeah, just like for example in algebra. MY teacher last year called the point slope form the \"equation writer\" \r\n\r\nOr like slope intercept the slopin but that's just kinda the south i guess", "Solution_30": "[quote=\"now a ranger\"]\n\nyeah, just like for example in algebra. MY teacher last year called the point slope form the \"equation writer\" \n\nOr like slope intercept the slopin but that's just kinda the south i guess[/quote]\r\n\r\nLOL.", "Solution_31": "[quote=\"goldendomer\"][quote=\"bpms\"]Shoelace is much more powerful than pick's[/quote]\n\nI agree.[/quote]\r\n\r\nI disagree. Pick's Theorem can be used to prove a surprising number of very powerful results in some very unexpected areas. For example, it makes for some very elegant proofs in the theory of continued fractions. \r\n\r\nHere's an example of what I'm talking about:\r\n\r\nGiven positive integers $a, b, c, d$ such that $ad-bc = 1$, show that there does not exist a rational $\\frac{b}{a}< r < \\frac{d}{c}$ with a denominator (in lowest terms) less than $a+c$.", "Solution_32": "Duke number I8 was easily done with an application of Shoelace followed up by Pick's:\r\n\r\nThe hexagon defined by the 6 points $(i, 2^{i}-1)$, where i takes the integer values 0 - 5 has how many lattice points in its interior?", "Solution_33": "Can you post the solution with Picks & shoelace? That's not the way I did it...", "Solution_34": "[hide=\"A sketch\"] Find the area by Shoelace. The only possible non-vertex border points are on the line between $(0, 0)$ and $(5, 31)$ (there aren't any), so by Pick's we can calculate the number of interior lattice points. [/hide]", "Solution_35": "Yeah, what he said. Coordinates will give you area by shoelace. Then use the area (K) in Pick's:\r\n\r\nK = I +B/2 - 1\r\n\r\n(I = interior points, B=boundary points)\r\n\r\nObviously there are only 6 boundary points, the given ones: Since the line between two consecutive points only goes over by 1, while it goes up a lot, it won't hit any lattice points until it goes fully over 1 (to the next given point).", "Solution_36": "pick's theorem is pretty cool, but I messed up counting the dots under time control once. :( \r\n\r\n\r\ncan someone please post with latex the shoelace theorem. I have to make sure I don't have it wrong.", "Solution_37": "It really isn't LaTeX.\r\n\r\nYou go around the polygon and you write all the coordinates down in order, in two neat columns, starting and ending at the same point.\r\n\r\nThen you multiply diagonally. You take the first x-coordinate and you multiply it by the second y-coordinate; the second x by the third y; and so on.\r\n\r\nAdd all these numbers up.\r\n\r\nDo the same thing the other way: the first y-coordinate with the second x-coordinate, etc.\r\n\r\nAdd all these numbers up too.\r\n\r\n(If you've been drawing lines between the numbers you're trying to multiply, you should see a row of X shapes that look like shoelaces.)\r\n\r\nSubtract the two totals and divide by two.", "Solution_38": "yeah i knew that part lol\r\n\r\nthat's why its calle shoelace, and u multiply by 1/2 when ur done adding and subtracting.", "Solution_39": "[quote=\"13375P34K43V312\"]And you'll also be pretty surprised when you find that the area between two parabolas will usually turn out nice, like 17/3 or something.[/quote]\r\n\r\nYes, and the same goes pretty much for all polynomials of even degree (and odd degree for that matter, but then you have to define horizontal boundaries...), as long as the exponents are all integers. But that's all Calculus anyway.\r\n\r\nBtw, I personally think that the Shoelace Theorem has more applications, but I can see where Pick's would be useful too..." } { "Tag": [ "geometry", "circumcircle", "incenter", "geometric transformation", "reflection", "power of a point", "radical axis" ], "Problem": "Let $ (I_a), (I_b), (I_c)$ be the excircles of triangle $ ABC. (I_a)$ touches $ AB, AC$ at $ C'_a, B'_a$, similar to $ B'_c, A'_c, C'_b, A'_b$. Let $ O_a, O_b, O_c$ be circumcenters of triangle $ AB'_cC'_b, BC'_aA'_c, CA'_bB'_a$. $ l_a$ be a line which passes through $ I_a$ and perpendicular to $ O_bO_c$. Simiar to $ l_b, l_c$. Prove that $ l_a, l_b, l_c$ are concurrent.", "Solution_1": "Let $ (O)$ to the circumcircle and $ (I)$ incircle of the $ \\triangle ABC.$ Incircle $ (I)$ touches $ BC, CA, AB$ at $ D, E, F.$ Excircles $ (I_a), (I_b), (I_c)$ touch $ BC, CA, AB$ at $ D_a, F_b, F_c.$ Internal angle bisectors $ AI, BI, CI$ cut $ (O)$ again at $ X, Y,Z.$ Perpendicular bisectors $ OX, OY, OZ$ of $ BC, CA, AB$ cut these at their midpoints $ A', B', C'$ and the circumcircle $ (O)$ again at $ X', Y', Z'.$ \r\n\r\n[color=red]Lemma 1:[/color] Perpendicular bisectors of $ AD, BE, CF$ cut segments $ A'X', B'Y', C'Z'$ at their midpoints $ P_a, P_b, P_c.$ Similarly, perpendicular bisectors of $ AB_c', BC_a', CA_b'$ cut sgments $ A'X, B'Y, C'Z$ at their midpoints $ O_a, O_b, O_c.$\r\n\r\nParallel of $ BC$ through $ A$ cuts $ (O)$ again at $ A''.$ By Ptolemy for $ AA''BC,$ $ \\overline {AA''} \\equal{} \\frac {c^2 \\minus{} b^2}{a}.$ Let $ ID$ cut $ AA''$ at $ D''$ and tangent of $ (O)$ at $ X'$ at $ X''.$ $ \\overline {X'X''} \\equal{} \\overline {A'D} \\equal{} \\frac {_1}{^2} \\overline {D_aD} \\equal{} \\frac {_1}{^2}(c \\minus{} b).$ \r\n\r\n$ \\minus{} \\overline {D''A} \\cdot \\overline {D''A''} \\equal{} \\frac {_1}{^4} (\\overline {A''A} \\minus{} \\overline {D_aD}) \\cdot (\\overline {A''A} \\minus{} \\overline {D_aD}) \\equal{}$\r\n$ \\equal{} \\frac {_1}{^4}\\left(\\frac {(c^2 \\minus{} b^2)^2}{a^2} \\minus{} (c \\minus{} b)^2\\right) \\equal{} \\frac {p(p \\minus{} a)(c \\minus{} b)^2}{a^2}$\r\n\r\n$ \\minus{} \\overline {D''D} \\cdot \\overline {D''X''} \\equal{} \\overline {DD''} \\cdot (\\overline {A'X'} \\minus{} \\overline {DD''}) \\equal{} h_a \\cdot \\left(\\frac {r_b \\plus{} r_c}{2} \\minus{} h_a\\right) \\equal{}$\r\n$ \\equal{} \\frac {2pr}{a} \\left(\\frac {pr}{2(p \\minus{} b)} \\plus{} \\frac {pr}{2(p \\minus{} c)} \\minus{} \\frac {2pr}{a}\\right) \\equal{} \\frac {p^2r^2}{a^2} \\cdot \\frac {(c \\minus{} b)^2}{(p \\minus{} b)(p \\minus{} c)}$\r\n\r\n$ \\Longrightarrow$ $ \\overline {D''A} \\cdot \\overline {D''A''} \\equal{} \\overline {D''D} \\cdot \\overline {D''X''}$ $ \\Longrightarrow$ $ AX''A''D$ is cyclic and $ D_a \\in \\odot (AX''A''D)$ by symmetry. Its circumcenter is the common midpoint $ P_a$ of $ D_aX''$ and $ A'X'.$ Therefore, $ P_a$ is circumcenter of $ \\triangle ADD_a$ and similarly, $ O_a$ is circumcenter of $ \\triangle AB_c'C_b'.$ $ \\square$\r\n\r\n[color=red]Lemma 2:[/color] Nagel lines $ AD_a, BE_b, CF_c$ of $ \\triangle ABC$ are perpendicular to $ P_bP_c, P_cP_a, P_aP_b.$ Similarly, Gergonne lines $ AD, BE, CF$ of $ \\triangle ABC$ are perpendicular to $ O_bO_c, O_cO_a, O_aO_b.$\r\n\r\nLet $ (P_b)$ be circle with center $ P_b$ passing through $ A, C$ and let $ (P_c)$ be circle with center $ P_c$ passing through $ A, B.$ Let $ (P_b), (P_c)$ cut $ BC$ again at $ D_b, D_c,$ respectively. Powers of $ B, E$ to $ (P_b)$ are both equal to $ \\minus{} (p \\minus{} a)(p \\minus{} c)$ and likewise, powers of $ C, F$ to $ (P_c)$ are both equal to $ \\minus{} (p \\minus{} a)(p \\minus{} b)$ $ \\Longrightarrow$\r\n\r\n$ \\overline{D_aC} \\cdot \\overline{D_aD_b} \\equal{} \\overline{D_aC} \\cdot (\\overline{CB} \\plus{} \\overline{BD_b} \\minus{} \\overline{CD_a}) \\equal{}$ $ \\minus{} (p \\minus{} b) \\left(a \\plus{} \\frac {(p \\minus{} a)(p \\minus{} c)}{a} \\minus{} (p \\minus{} b)\\right) \\equal{} \\minus{} \\frac {(p \\minus{} b)(p \\minus{} c)}{a}$ \r\n$ \\overline{D_aB} \\cdot \\overline{D_aD_c} \\equal{} \\overline{D_aB} \\cdot (\\overline{BC} \\plus{} \\overline{CD_c} \\minus{} \\overline{BD_a}) \\equal{}$ $ \\minus{} (p \\minus{} c) \\left(a \\plus{} \\frac {(p \\minus{} a)(p \\minus{} b)}{a} \\minus{} (p \\minus{} c)\\right) \\equal{} \\minus{} \\frac {(p \\minus{} b)(p \\minus{} c)}{a}$\r\n\r\n$ \\Longrightarrow$ $ AD_a$ is radical axis of $ (P_b), (P_c)$ $ \\Longrightarrow$ $ AD_a \\perp P_bP_c.$ $ \\square$\r\n\r\n[color=red]Lemma 3:[/color] I-symmedians of the in/excentral triangles $ \\triangle II_bI_c,\\triangle II_bI_c,\\triangle II_bI_c$ are parallel to Nagel lines $ AD_a, BE_b, CF_c$ of their common orthic triangle $ \\triangle ABC.$ Similarly, $ I_a, I_b, I_c$-symmedians of the excentral triangle $ \\triangle I_aI_bI_c$ are parallel to Gergonne lines $ AD, BE, CF$ of its orthic triangle $ \\triangle ABC.$\r\n\r\nLet $ D'$ be midpoint of $ AD.$ Newton line $ A'D'$ of degenerated tangential quadrilateral $ ABDC$ with diagonals $ AD, BC$ goes through its incenter $ I.$. Therefore $ (A'I \\equiv AD') \\parallel AD_a.$ is midline of $ \\triangle AD_aD.$ $ BC$ is antiparallel of $ I_bI_c$ WRT $ \\angle I_cII_b.$ The I-symmedian of $ \\triangle II_bI_c$ has to cut this antiparallel in half $ \\Longrightarrow$ $ A'I$ is the I-symmedian of this triangle.\r\n________________________________________________________________\r\n\r\nMidpoints $ O_a, O_b, O_c$ of $ A'X, B'Y, C'Z$ from lemma 1 are circumcenters of the $ \\triangle AB_c'C_b', \\triangle BC_a'A_c', \\triangle CA_b'B_a',$ respectively. By lemmas 2, 3, perpendiculars $ l_a, l_b, l_c$ to $ O_bO_c, O_cO_a, O_aO_b$ through $ I_a, I_b, I_c$ are symmedians of the excentral triangle $ \\triangle I_aI_bI_c,$ concurrent at its symmedian point. $ \\square$", "Solution_2": "Dear Yetti, \r\nThanks you for your solution, but I think that using radical center can be shorter :)", "Solution_3": "Let $Y_a,Y_b,Y_c$ be the tangency points of $ (I_a),(I_b),(I_c)$ with $ BC,CA,AB.$ Note that circles $ (O_a),(O_b)$ and $(O_c)$ are the reflections of the circles $ \\odot(AY_cY_b),$ $\\odot(BY_aY_c)$ and $\\odot(CY_aY_b)$ about the external bisectors of $ A,B,C.$ Circles $\\odot (BY_aY_c), $ $\\odot(CY_aY_b)$ obviously pass through the Bevan point $ B_e$ of $\\triangle ABC,$ namely the concurrency point of the perpendiculars from $ I_a,I_b,I_c$ to $BC,CA,AB.$ Therefore, $ I_a$ has equal power WRT $(O_b),(O_c)$ $\\Longrightarrow$ $ \\ell_a$ is radical axis of $ (O_b),(O_c).$ Mutatis mutandis, $ \\ell_b,\\ell_c$ are radical axes of the pairs $ (O_a),(O_c)$ and $ (O_a),(O_b)$ respectively $\\Longrightarrow$ $ \\ell_a,\\ell_b,\\ell_c$ meet at the radical center of $(O_a),(O_b),(O_c)$ (Mittenpunkt of ABC)." } { "Tag": [ "calculus", "integration", "calculus computations" ], "Problem": "Evaluate $\\int \\int_R (x-1) dA$, where $R$ is the region in the first quadrant enclosed between $y = x$ and $y = x^3$.\r\n\r\nSo I did\r\n\r\n[hide]\n$\\int_0^1 \\int_{x^3}^x (x-1) dy dx$ but I got the wrong answer (according to the back of the book, anyway)... is there something wrong with this? [/hide]", "Solution_1": "I can try:\r\n\r\n$\\int^1_0\\int^x_{x^3}(x-1)dA=\\int^1_0[(x-1)y]^x_{x^3}dx=\\int^1_0((x-1)x)-((x-1)x^3)=$\r\n\r\n$\\int^1_0x^2-x-x^4+x^3=[\\frac{1}{3}x^3-\\frac{1}{2}x^2-\\frac{1}{5}x^5+\\frac{1}{4}x^4]^1_0=$\r\n\r\n$\\frac{1}{3}-\\frac{1}{2}-\\frac{1}{5}+\\frac{1}{4}=-\\frac{7}{60}$\r\n\r\nStrainge that I find A negatief answer is this ok? What is the solution given by your book?\r\n\r\nGreets.", "Solution_2": "The answer better be negative -- $x - 1$ is negative over that entire region.", "Solution_3": "ok\u00e9 is my solution also ok\u00e9? Greets", "Solution_4": "Yes your answer is right. :)", "Solution_5": "[quote=\"Bert\"]I can try:\n\n$\\int^1_0\\int^x_{x^3}(x-1)dA=\\int^1_0[(x-1)y]^x_{x^3}dx=\\int^1_0((x-1)x)-((x-1)x^3)=$\n\n$\\int^1_0x^2-x-x^4+x^3=[\\frac{1}{3}x^3-\\frac{1}{2}x^2-\\frac{1}{5}x^5+\\frac{1}{4}x^4]^1_0=$\n\n$\\frac{1}{3}-\\frac{1}{2}-\\frac{1}{5}+\\frac{1}{4}=-\\frac{7}{60}$\n\nStrainge that I find A negatief answer is this ok? What is the solution given by your book?\n\nGreets.[/quote]\r\n\r\nYes, that's what I got, too. The book says $-\\frac{1}{2}$, which I guess is wrong; it confused me very much." } { "Tag": [ "LaTeX", "linear algebra", "matrix" ], "Problem": "I'm looking for a list with more new and interesting tricks, more specifically on things that can be done with the forum-latex configuration on MathLinks. \r\n\r\nSome tricks I recently discovered include [list][*]\\to can be used instead of \\rightarrow[*]$\\text{sgn}(t)=\\begin{cases}1&\\text{ voor }t>0\\\\0&\\text{ voor }t=0\\\\-1&\\text{ voor }t<0\\end{cases}$ using \\begin{cases} instead of working with \\left\\{\\begin{array}[*]\\begin{matrix} instead of also with array[*]\\buildrel & \\stackrel commands[*]the use of \\limits for better sub&super-scripting.[*]...[/list]\r\nNo world-shocking new things, but I keep discovering new things over and over and I wondered: there has to be a finite list of these, isn't it? Where can I find such a list?\r\n\r\nAgain, only taking the packages from this forum into account, else the list would of course not be finite as an infinity of packages can be made.", "Solution_1": "Most of things you mention either don't use packages or use the amsmath package which is implemented here (you can read what it can do in its [url=http://tug.ctan.org/tex-archive/macros/latex/required/amslatex/math/amsldoc.pdf]User Guide[/url]). Also I strongly recommend getting hold of a copy of [url=http://www.amazon.co.uk/-Guide-LATEX/dp/0321173856/sr=8-1/qid=1158007648/ref=sr_1_1/202-0137577-3200634?ie=UTF8&s=gateway]Kopka & Daly's A Guide to Latex[/url]. Yes there's far more than can be used on a forum, but it is an excellent guide to typing LaTeX mathematics.\r\n\r\nFun exercise: Without looking at the code, can you see how this is done with just one command: $\\genfrac{)}{[}{4pt}{2}{1}{2}$\r\nThe answer is in both publications above." } { "Tag": [ "Support", "blogs" ], "Problem": "The never beginning story. We start at the end of the story and the next person writes a setence or two that should occur immediately before the last post.\r\n\r\n\r\n\r\nAnd so, Alice and Bob learned the secret of unbreakable encryption.", "Solution_1": "The great lord of the universe threw them into Nottingham where Robin Hood was.", "Solution_2": "... which would enable them to escape. Unfortunately, this scheme did not turn out feasible because shortly after that ...", "Solution_3": "Carmen tried to help them by trying to break the encryption but died trying to do so, thus distracting the great Lord of the Universe...\r\n\r\n(By the way, let's keep a rule like in the other story, no less than 5 words per \"sentence\" and no more than 25 words)", "Solution_4": "Things were looking rather bleak for them at the moment, what with the reinforcements cut off and the Great Lord returning victorious.", "Solution_5": "Iversonfan2007 commited suicide.", "Solution_6": "Iversonfan2007 got owned by Max300 and felt bad so...", "Solution_7": "but then he started watching the match between max300 and iversonfan 2007 \r\n[color=green]\n[size=42](edited by mod, your post is supposed to come [i]before[/i] the previous one)[/size][/color]", "Solution_8": "Things were going well for Dino,", "Solution_9": "...Dino, Alice's brother, invested in google two years ago, so", "Solution_10": "Alice, Bob, and Carmen had invested in Enron, and were now bankrupt and in a cage under the control of the great lord.... However,", "Solution_11": "...Alice, Bob, and Carmen each wanted to invest in an energy company, so", "Solution_12": "and now that Alice, Bob and Carmen were brought back to life, they were poor and", "Solution_13": "...Alice and Bob, along with their daughter Carmen, unwisely decided to bid their 30 million dollars they won in the lottery on a six-pack of Gatorade on EBAY.", "Solution_14": "...but fortune must have felt sorry for Bob, Alice, and Carmen because shortly after, they won the lottery number 667! They received 30 million dollars, and It was the oddest coincidence ever, but", "Solution_15": "A long time ago, there lived the first humans - Alice and Bob - and there daughter Carmen. That was much before Adam and Eve, who were only the first in the universe, which was yet to be formed. One day they sat in their drawing room, discussing that at this rate the universe would never be formed.", "Solution_16": "so they quickly went to the computer and called dell tech support, but then the universe blew up and everything had to start at the [b]very[/b] beginning again.", "Solution_17": "The dell computer went bezerk,", "Solution_18": "so they called chuck norris to", "Solution_19": "They went to his computer to compute the square root of 1681, but", "Solution_20": "a picture of chuck norris appeared", "Solution_21": "Carmn was tinkering on the picture machine with Alice and Bob, and the knew that if the saw Chuck, they will have to find the square root of 1681. Then,", "Solution_22": "and then..........", "Solution_23": "alkara1, this is the never [b]beginning[/b] story. Read the first post of this thread and understand the rules before posting...\r\n\r\nso we have...\r\n\r\n[quote]and then... Carmn was tinkering on the picture machine with Alice and Bob, and the knew that if the saw Chuck, they will have to find the square root of 1681. Then, a picture of chuck norris appeared. They went to his computer to compute the square root of 1681, but The dell computer went bezerk, etc etc etc...[/quote]\r\n\r\n\r\nCarmen said to herself, \"Oh my, I may have to go look for a picture of Chuck Norris!\"", "Solution_24": "SRRY\r\n\r\nalright then,\r\n\r\nCarmen loved chuck norris. she even had a chuck norris life-size house somewhere up in Canada. Then", "Solution_25": "Alice then learned a terrible secret.", "Solution_26": "Alice found herself seeing Carmen stare at a Chuck Norris picture in her sparetime. She also heard Carmen making long-distance phone calls to Canada, saying \"He is a bit taller, make the house a bit taller.\"", "Solution_27": "alice is a stalker\r\nshe stalked carmen in her free time\r\nalice was astounded at where carmen went: she once followed carmen all the way to canada, and she didn't know why she was there", "Solution_28": "[quote=\"akalra1\"]alice is a stalker\nshe stalked carmen in her free time\nalice was astounded at where carmen went: she once followed carmen all the way to canada, and she didn't know why she was there[/quote]\r\n\r\nu do know that alice is carmen's mom, right?", "Solution_29": "Carmen wanted to know why Alice followed her around all the time. One day, she was browsing Myspaces when she saw, written on Bob's blog:\r\n\r\n\r\n(you know, posts are supposed to be between 5-25 words :wink: )" } { "Tag": [ "search", "AMC" ], "Problem": "I was just looking at that one over the past couple days. What a coincidence. Unfortunately I don't have the answers with me. I think my teacher has them so I can get them for you tomorrow if you still need them by then.", "Solution_1": "here they are:\r\n\r\nE\r\nE\r\nA\r\nC\r\nE\r\n\r\nE\r\nB\r\nC\r\nD\r\nB\r\n\r\nB\r\nD\r\nD\r\nC\r\nB\r\n\r\nD\r\nD\r\nD\r\nB\r\nE\r\n\r\nA\r\nB\r\nC\r\nD\r\nD\r\n\r\nC\r\nC\r\nB\r\nB\r\nB", "Solution_2": "Does anyone know a website that has 1991 ASHME problems?", "Solution_3": "i bet kalva does but anyways you can search \"1991 AHSME problems\" and you're bound to find something.", "Solution_4": "I thought that the easiest stuff that Kalva had was Aime(hence no ASHME/AMC)." } { "Tag": [ "limit", "logarithms", "floor function", "real analysis", "real analysis unsolved" ], "Problem": "Is it true that:\r\n\\[ \\lim_{n \\rightarrow \\infty} ( \\{ H_n \\} - \\{ \\ln n \\} ) = \\gamma \\]\r\nWhere $H_n$ is the n-th harmonic number $\\gamma$ is Euler's constant and $\\{x\\}$ denotes the fractional part of a real number $x$.", "Solution_1": "No, it won't be true. Look at the integer parts:\r\n\r\n$\\lim_{n\\to\\infty}\\lfloor H_n \\rfloor=\\lim_{n\\to\\infty}\\lfloor \\ln n\\rfloor=\\infty$\r\n\r\nso each sequence crosses over every positive integer. But they won't cross over the integers at the same time.\r\n\r\nThere are infinitely many $n$ for which $\\lfloor H_n \\rfloor=\\lfloor \\ln n\\rfloor$ and there are infinitely many $n$ for which $\\lfloor H_n \\rfloor\\ne\\lfloor \\ln n\\rfloor.$\r\n\r\nThus the sequence of the differences of the fractional parts should have two cluster points.", "Solution_2": "In order to prove that, do you just use the asymptotics for harmonic numbers (in the sense that you prove that {ln n}>1-y>0.4 infinitely often), or do you have a more ingenious proof ?", "Solution_3": "Just the asymptotics - nothing fancy.", "Solution_4": "Thanks :)" } { "Tag": [ "ARML", "probability", "geometry", "trigonometry", "analytic geometry", "algebra", "polynomial" ], "Problem": "1.\r\nWhen the sum of the first $k$ terms of the series\r\n\r\n\\[ 1^2 + 2^2 + 3^2 + \\cdots + n^2 + \\cdots \\]\r\nis subtracted from the sum of the first $k$ terms of the series\r\n\r\n\\[ 1 \\cdot 2 + 2 \\cdot 3 + 3 \\cdot 4 + \\cdots + n(n+1) + \\cdots, \\]\r\nthe result is 210. Find the numerical value of $k$.\r\n\r\n\r\n2.\r\nThree coplanar circles are concentric, and the lengths of their respective radii are 10, 20, and 30. A square is circumscribed about the smallest circle, and squares are inscribed in the other two circles. Find the length of the shortest line segment that could contain a point on each square.\r\n\r\n\r\n3.\r\nFind the numerical value of $k$ for which\r\n\r\n\\[ \\frac{7}{x+y} = \\frac{k}{x+z} = \\frac{11}{z-y}. \\]\r\n\r\n\r\n4.\r\nSpender Sam always carried only dollars and pennies, but he never carried more than 99 pennies. When leaving home one day, Sam noted that the number of dollars he had was 150% of the number of pennies he had. After spending 10 dollars and 78 cents, Sam discovered that the number of dollars he had remaining was 1 less than the number of pennies he then had. With how much money did Sam leave home that day?\r\n\r\n\r\n5.\r\nTwo cards are randomly selected from a standard 52 card deck of playing cards, without replacement. Find the probability that at least on king and at least one heart are selected (though not necessarily on different cards).\r\n\r\n\r\n6.\r\nFind the ordered pair of positive integers $(a, b)$, with $a Art of Owna\r\n\r\n$factor(24x^{4}+100x^{3}+48x^{2}-100x+24) = 4(3x-1)(2x-1)(x+3)(x+2)$", "Solution_12": "Maple $>$ TI-89 $\\geq$ Art of Owna\r\n\r\n :D", "Solution_13": "[quote=\"undefined117\"]Maple $>$ TI-89 $\\geq$ Art of Owna\n\n :D[/quote]\r\n\r\nWhats with the $\\ge$? Equality holds iff doomsday is upon us.\r\n\r\nDude Art of Owna, how come you weren't at the top 25 banquet? There were a lot of freshman n00bs.", "Solution_14": "[quote=\"stupidkid\"][quote=\"undefined117\"]Maple $>$ TI-89 $\\geq$ Art of Owna\n\n :D[/quote]\n\nWhats with the $\\ge$? Equality holds iff doomsday is upon us.[/quote]\r\nlol\r\n\r\nwell according to my math teacher, \"your calculator is only as smart as you are\"\r\n\r\nor something along those lines.", "Solution_15": "[quote=\"stupidkid\"][quote=\"undefined117\"]Maple $>$ TI-89 $\\geq$ Art of Owna\n\n :D[/quote]\n\nWhats with the $\\ge$? Equality holds iff doomsday is upon us.\n\nDude Art of Owna, how come you weren't at the top 25 banquet? There were a lot of freshman n00bs.[/quote]\r\nwtf i got all As and an A- in honors bio (I already posted somewhat like this in the \"what about other subjects besides math\") because I got a B+ on a grading period and she doesnt do percent so there was no way I could get an A in that class after getting the B+. So my GPA's screwed (3.974 including this year and some middle school stuff they added), meaning I wont be getting a scholorship to some prestigious college...\r\n\r\nBut there were only 5 asians i think (in top 25), only 2 male asians (one is just specialized at school stuff so he'll get into like northwestern or michigan.., jimmy's actually pretty good at other stuff-he was 1 point away on amc 12 and 1.5 on amc 10 from passing, and he acutally does sports-lol he was my tennis doubles partner on team..most annying kid ever though)", "Solution_16": "[quote=\"undefined117\"]heh but it's pretty sad how i have NO CLUE as to how you got to where you were on number 8\n\nif i ever get on the ARML team.. i'm hopeless.[/quote]\r\nbut you passed AIME..so you must have to be good at math (i dont know a single person that passes AIME and sucks at math)", "Solution_17": "[quote=\"undefined117\"]heh but it's pretty sad how i have NO CLUE as to how you got to where you were on number 8\n\nif i ever get on the ARML team.. i'm hopeless.[/quote]\r\n\r\nThese are the problems from the individual round of the 1979 ARML event. I don't know how the results were that year, but I-8 is always meant to be a back-breaker of a problem. \r\n\r\nSometimes they're hard because they're tedious and have a load of boring calculations (like this one, or like the one from last year). Sometimes, though, they require a brilliant insight and things work out beautifully if approached from the proper point of view (look at the 1997 and 1998 tests).", "Solution_18": "[quote=\"generating\"][quote=\"undefined117\"]heh but it's pretty sad how i have NO CLUE as to how you got to where you were on number 8\n\nif i ever get on the ARML team.. i'm hopeless.[/quote]\n\nThese are the problems from the individual round of the 1979 ARML event. I don't know how the results were that year, but I-8 is always meant to be a back-breaker of a problem. \n\nSometimes they're hard because they're tedious and have a load of boring calculations (like this one, or like the one from last year). Sometimes, though, they require a brilliant insight and things work out beautifully if approached from the proper point of view (look at the 1997 and 1998 tests).[/quote]\r\n\r\nbut there must be an easier way for number 8? I mean im pretty sure they wont be that mean and have you use the ratioan root theorm and guess the roots in less than 10 mins.", "Solution_19": "[quote=\"Art of Owna\"][quote=\"undefined117\"]heh but it's pretty sad how i have NO CLUE as to how you got to where you were on number 8\n\nif i ever get on the ARML team.. i'm hopeless.[/quote]\nbut you passed AIME..so you must have to be good at math (i dont know a single person that passes AIME and sucks at math)[/quote]\r\nno really. i suck at mathcounts, especially at countdown. i REALLY should start reading up on the AoPS books :blush: \r\n\r\nbut otherwise, i usually need some time to think about my strategies and stuff.", "Solution_20": "[quote=\"Art of Owna\"]wtf i got all As and an A- in honors bio (I already posted somewhat like this in the \"what about other subjects besides math\") because I got a B+ on a grading period and she doesnt do percent so there was no way I could get an A in that class after getting the B+. So my GPA's screwed (3.974 including this year and some middle school stuff they added), meaning I wont be getting a scholorship to some prestigious college...\n\nBut there were only 5 asians i think (in top 25), only 2 male asians (one is just specialized at school stuff so he'll get into like northwestern or michigan.., jimmy's actually pretty good at other stuff-he was 1 point away on amc 12 and 1.5 on amc 10 from passing, and he acutally does sports-lol he was my tennis doubles partner on team..most annying kid ever though)[/quote]\r\nDude chill out, GPA means nothing, don't be like those Asians who cry over a non A. I have 2 A- and 1 B+, although I'm ranked 24th, so barely in top 25. Btw, top colleges, e.g. Harvard, MIT, Stanford, etc, [i]do not[/i] give out merit based scholarships, only financial aid. You get scholarships from other organizations, such as Siemens and they don't give about your GPA (well some probably do, but those are worthless scholarships anyways).", "Solution_21": "[quote=\"stupidkid\"][quote=\"Art of Owna\"]wtf i got all As and an A- in honors bio (I already posted somewhat like this in the \"what about other subjects besides math\") because I got a B+ on a grading period and she doesnt do percent so there was no way I could get an A in that class after getting the B+. So my GPA's screwed (3.974 including this year and some middle school stuff they added), meaning I wont be getting a scholorship to some prestigious college...\n\nBut there were only 5 asians i think (in top 25), only 2 male asians (one is just specialized at school stuff so he'll get into like northwestern or michigan.., jimmy's actually pretty good at other stuff-he was 1 point away on amc 12 and 1.5 on amc 10 from passing, and he acutally does sports-lol he was my tennis doubles partner on team..most annying kid ever though)[/quote]\nDude chill out, GPA means nothing, don't be like those Asians who cry over a non A. I have 2 A- and 1 B+, although I'm ranked 24th, so barely in top 25. Btw, top colleges, e.g. Harvard, MIT, Stanford, etc, [i]do not[/i] give out merit based scholarships, only financial aid. You get scholarships from other organizations, such as Siemens and they don't give about your GPA (well some probably do, but those are worthless scholarships anyways).[/quote]\r\n\r\nI was being sarcastic about the scholorship, I dont cry over not getting a non-A, and Im not like all those other Asians in general (sterotype..)...I actually told my mom when I got that GPA that I dont really care about my GPA and Ill just make it up with some award (like qualification for USAMO or like Intel and stuff), but she just said GPA is the easiest thing to get, along with National Finalist. Yea and like I said, if you just get 4.0s, you'll go to a respectable school (Northwestern is actually a good example), but you wont go to any of the top colleges because those colleges value abilities/talent more than your GPA. For example, Harvard isnt looking for future students that will become ordinary doctors or lawyers that make 200,000 a year (which is way above average still), but they are looking for students that will become future millionaires and billionarires that will be come famous that will give Harvard perstige (becuase no one is going to hear about a ordinary doctor, but if that person becomes famous, then Harvard will be mentioned even more). 4.0=ordinary hardworking, dedicated doctors, lawyers,etc...not bad)\r\nNational Award or special talent=better chance at becoming famous and that is what top colleges are looking for out of their alumni (pres. bush couldnt go to yale with his grades..but his leadership qualities and money made yale want him because of what he can and already was)\r\nI mean harvard knows how to find talent-for better or worse (Its got some of the most famous and infamous alumni..but their still well-known)\r\n\r\nIf I sounded like I was mad or something or wining..that wasnt really inteneded (I mean to be honest I was disappointed but I really dont care about it that much-definenlty not as much as all those other Asians in my grade-or even americans-who begged my orchestra teacher for a higher participation grade..-hmm*cough* udayan? atealst you wenret the only one)", "Solution_22": "Meh, dunno about the billionaires part. Those people are just born with the ability to make money. If you look at the world's richest 20 people (forbes.com), I think half of them never even finished college! o.O One dude didn't even [i]go[/i] to college.\r\n\r\nOh btw, USAMO alone is definitely not enough these days to get you into a top school. The competition is crazy now. Go ask Mohit, he'd know. \r\n\r\nYo, Art of Owna, wanna join the CHS webteam? We need some underclassman who either knows Linux or is willing to learn (so that means you'll need to install it and spend however long it takes you to get fluent with it, it took me probably 50 hrs in 2 weeks), and it's Gentoo Linux, one of the hardest to learn. If no one knows Linux, carmelhighschool.net is screwed after I graduate.\r\n\r\nOh also, (damn its the 4th edit), you can only participate in the Intel Talent Search when you're a Senior, so it's too late for college apps anyways. Just fyi.", "Solution_23": "[quote=\"stupidkid\"]Meh, dunno about the billionaires part. Those people are just born with the ability to make money. If you look at the world's richest 20 people (forbes.com), I think half of them never even finished college! o.O One dude didn't even [i]go[/i] to college.\n\nOh btw, USAMO alone is definitely not enough these days to get you into a top school. The competition is crazy now. Go ask Mohit, he'd know. \n\nYo, Art of Owna, wanna join the CHS webteam? We need some underclassman who either knows Linux or is willing to learn (so that means you'll need to install it and spend however long it takes you to get fluent with it, it took me probably 50 hrs in 2 weeks), and it's Gentoo Linux, one of the hardest to learn. If no one knows Linux, carmelhighschool.net is screwed after I graduate.\n\nOh also, (- its the 4th edit), you can only participate in the Intel Talent Search when you're a Senior, so it's too late for college apps anyways. Just fyi.[/quote]\r\nUh..I dont know Linux..you can go ask udayan (that new indian kid in that sunday class)-he apparently thinks he knows alot of computers..uh I can probalby learn it, but Im pretty busy these days...but Im taking comp sci next year, if that has anything to do with it..", "Solution_24": "...\r\n\r\nHow are you busy as a frosh? Man, I didn't do anything freshmen year, just kinda slacked off and played some starcraft.", "Solution_25": "[quote=\"stupidkid\"]...\n\nHow are you busy as a frosh? Man, I didn't do anything freshmen year, just kinda slacked off and played some starcraft.[/quote]\r\n\r\nI need to be more busy now..I havent done my homework at home (SRT plus orchestra=free SRT) for the past 6 weeks..and it almost cost me another A-..\r\nbut i mean on weekends and stuff I have lots of things to do. and now I have to volunteer at the library sometime (apparently you need that for college :o )\r\nbut i dont play games any more..havent played my XBOX in like 2 months.." } { "Tag": [ "modular arithmetic", "quadratics", "number theory open", "number theory" ], "Problem": "Let call:\r\n$M_q=2^q-1$\r\n$\\eta(q) = \\ \\text{the number of distinct numbers } 3^n+1/3^n \\pmod{M_q}$.\r\n\r\nFor q=3,5,7,17,19, we have: $order(3,M_q) = M_q-1 = 2 \\eta(q)$.\r\n\r\nFor q=13,31, we have: $order(3,M_q) = 2 \\eta(q) - 2$.\r\nand: $order(3,M_{13}) = \\frac{M_q - 1}{9} \\ \\ ; \\ \\ order(3,M_{31}) = \\frac{M_q - 1}{3} \\ .$\r\n\r\nCan you explain why ? and generalize the result ?\r\n\r\nT.\r\n\r\nValues:\r\n[code]q - order(3,Mq) - eta(q)\n--------------------------\n3 6 3\n5 30 15\n7 126 63\n13 910 456\n17 131070 65535\n19 524286 262143\n31 715827882 357913942[/code]", "Solution_1": "Lets forget $3$ and write $k$ instead, and lets also forget that we have Mersenne primes and thus let $p$ be any odd prime.\r\nLet $f(x) : = x + \\frac{1}{x} \\mod p$, then we want the size (lets call it $\\eta_p(k)$) of the set $\\{ f(k^n) | n \\in \\mathbb{N}\\}$.\r\n\r\nFirst lets find out how often $f(x) \\equiv f(y) \\mod p$ with $x,y \\not\\equiv 0 \\mod p$ happens:\r\nThis means $x + \\frac{1}{x} \\equiv y + \\frac{1}{y} \\mod p$\r\n$\\iff x^2y+y \\equiv xy^2+x \\mod p$\r\n$\\iff (xy-1)(x-y) \\equiv 0 \\mod p$.\r\nThis means that either $x \\equiv y \\mod p$, the trivial case, or $xy \\equiv 1 \\mod p$.\r\nBut: when $x \\equiv \\pm 1 \\mod p$, then only the case $x \\equiv y \\mod p$ can occure.\r\n\r\nLook at the set $\\mathrm{Pow}(k) : = \\{ k^n \\mod p | n \\in \\mathbb{Z}\\}$ (we can use $\\mathbb{Z}$ instead of $\\mathbb{N}$ because of Fermat's Little Theorem).\r\nIt has size $|\\mathrm{Pow}(k)| = ord_p(k)$. Additionally, we can pair up the elements $k^n \\mod p$ and $k^{-n} \\mod p$ for each $n$, since they give the same value $f(k^n) \\equiv f(k^{-n}) \\mod p$, and only those are equal (note that $1, -1 \\mod p$ will be left allone, but each noted as \"pair\" with one element). Since different pairs give different values, we have that $\\eta_p(k)= \\text{\\\"number of such pairs\\\"}$.\r\nThus when $-1 \\in \\mathrm{Pow}(k)$ ($1$ is always in the set), there will be $\\frac{ord_p(k)-2}{2}+2 = \\frac{ord_p(k)+2}{2}$ pairs, thus by the above $\\eta_p(k) = \\frac{ord_p(k)+2}{2} \\iff 2 \\eta_p(k) = ord_p(k)+2$.\r\nSimilar when $-1$ is not in the set: $2 \\eta_p(k) = ord_p(k)+1$\r\n\r\nThis for example gives $\\eta_7(3) = 4$, but you stated $=3$ above :? \r\n\r\nTo fnd out if $-1$ is in the set, we need to know if the order of $k \\mod p$ is even or odd (this suffices to know: when $ord_p(k)$ would be odd, we couldn't have $2 \\eta_p(k) = ord_p(k)+2 \\mod 2$, and analogous for the other case).\r\nWhen $s$ is the biggest integer with $2^s | p-1$, we could calculate $k^{\\frac{p-1}{2^s}} \\mod p$ (since $\\frac{p-1}{2^s}$ is the biggest odd divisor of $p-1$) and look if it is $1 \\mod p$ or not (the order is odd iff it is $1 \\mod p$).\r\nWhen $4 \\nmid p-1$, we just ask whether $k$ is a quadratic residue $\\mod p$ or not, which can be checked by Jacobi symbols.\r\n\r\nSpecial case $k=3, p=2^q-1$:\r\nThen $4 \\nmid p-1$, thus we use Legendre symbols (Jacobi is not needed since both numbers are prime) and the law of quadratic reciprocity:\r\n$\\left( \\frac{3}{2^q-1} \\right) = - \\left( \\frac{2^q-1}{3} \\right) = -1$.\r\nThis shows that the order of $3 \\mod p$ is even.\r\nThus for Mersenne primes, it should always be $2 \\eta_p(3) = ord_p(3)+2$, but the above says something different...", "Solution_2": "[quote=\"ZetaX\"]Thus for Mersenne primes, it should always be $2 \\eta_p(3) = ord_p(3)+2$, but the above says something different...[/quote]Okay. Seems PARI/gp does not say the truth or I mis-use it.\r\nI used the program (given by maxal in GIMPS forum):\r\n[code]{ tt(q,b) = local(M);\nM=2^q-1;\nlength(Set(vector(2^(q-1),n,lift(Mod(b,M)^n+Mod(b,M)^(-n)))))\n}[/code]The results given in post#1 have been computed with tt(q,3) for q =3..31 .\r\nBut the main question is not this mistake about 1 missing element for q=3,5,7,17,19 .\r\n\r\nWhile computing $\\eta_p(3)$ for q= 13, I've found a different result than the one stated by Shallit and Vasiga in their paper:\r\nThe DiGraph under $x^2-2$ modulo $M_{13}$ has 9 cycles of length 6.\r\nOut of these 9 cycles, I've found that only 5 of these cycles have their elements generated by $g(n)=(3^n+1/3^n) \\pmod{M_q}$. Here is a list of one element of each of these 5 cycles: 288, 101, 532, 920, 2627 . The 4 other 6-cycles starting with : 23, 722, 1812, 2255 have no element generated by $g(n)$, even when 0 <= n <= 8191.\r\nShallit and Vasiga say that all Cycles are generated by $g(n)$ ; computation shows that this is true for q=3,5,7,17,19 but not for q=13 and 31. Discussing with \"maxal\" in GIMPS forum, we came to compute $\\eta(q)$ and $order(3,M_q)$.\r\n\r\nFor q=3,5,7,17,19, we have: $order(3,M_q) = M_q-1$ .\r\n\r\nFor q=13,31, we have: $order(3,M_{13}) = \\frac{M_q - 1}{9} \\ \\ ; \\ \\ order(3,M_{31}) = \\frac{M_q - 1}{3}$.\r\n\r\nWhy these formulae are different ?\r\n\r\nTony", "Solution_3": "Well, I don't speak PARI :blush: \r\n\r\nI haven't read the paper till now, but it is somewhat natural that not all are of the type above.\r\nIt's an open conjecture by Artin that every nonsquare (here $3$) is a primitive root $\\mod$ infinetely many primes $p$.\r\nAs order of $3 \\mod p=2^q-1$, we can expect some big divisors of $p-1$, but not that they are $=p-1$.\r\nAnd I would expect that the Denominator of that fraction can get as big as they want (I would expect every odd integer to occure, or at least a multiple of every odd integer), but also get $1$ often.", "Solution_4": "[quote=\"ZetaX\"]Well, I don't speak PARI :blush: [/quote]The basic of this tool is easy.\nI do not like the syntax (why did they not used a C-like syntax ?!), but it is really useful.\n\n[quote]I haven't read the paper till now, but it is somewhat natural that not all are of the type above.\nIt's an open conjecture by Artin that every nonsquare (here $3$) is a primitive root $\\mod$ infinetely many primes $p$.\nAs order of $3 \\mod p=2^q-1$, we can expect some big divisors of $p-1$, but not that they are $=p-1$.\nAnd I would expect that the Denominator of that fraction can get as big as they want (I would expect every odd integer to occure, or at least a multiple of every odd integer), but also get $1$ often.[/quote]Mersenne numbers are very special simple and unic numbers. I think we can expect to find simple properties.\r\nI've computed $\\frac{M_q-1}{order(3,M_q)}$ for more values of $q$ and here are the results, showing that we have the conjecture:\r\n\\[ \\frac{M_q-1}{order(3,M_q)}=3^n \\ \\ \\text{with} \\ \\ n=0,1,2 \\]\r\nT.\r\n\r\n[code]q (Mq-1)/order(3,Mq)\n-----------------\n3 1\n5 1\n7 1\n13 9\n17 1\n19 1\n31 3\n61 9\n89 1\n107 1\n127 3\n521 1\n607 3\n1279 3[/code]" } { "Tag": [ "calculus", "calculus computations" ], "Problem": "On $ x\\minus{}y$ plane, find the maximum length of the line segment involved the region (including the boarder) bounded by the curve $ y\\equal{}x^2\\minus{}4$ and the $ x$ axis.", "Solution_1": "may i ask, how come you like min and max problems sooooo much ? :maybe: these days you seem to post questions on those things only...", "Solution_2": "That's by mere chance, misan. :lol:", "Solution_3": "so fix the point on the x axis at $ x\\equal{}\\alpha \\ge 0$\r\n\r\nwe can draw a line from any point on the curve..\r\n$ (x,x^2\\minus{}4)$ to $ (\\alpha,0)$\r\n\r\nand maximize distance squared\r\n$ d^2\\equal{}(x\\minus{}\\alpha)^2\\plus{}(x^2\\minus{}4)^2\\equal{}x^2\\minus{}2\\alpha x \\plus{}\\alpha^2\\plus{}x^4\\minus{}8x^2\\plus{}16$\r\n\r\n$ d_\\alpha \\equal{} 2(\\alpha \\minus{}x)\\equal{}0$ then $ \\alpha_c \\equal{}0,x,2$ \r\n$ d_x \\equal{}2x \\minus{}2\\alpha_c\\plus{}4x^3\\minus{}16x\\equal{}4x^3\\minus{}14x\\minus{}2\\alpha_c\\equal{}2(2x^3\\minus{}7x\\minus{}\\alpha_c)\\equal{}0$\r\n\r\nso critical points at $ 2x^3\\minus{}7x\\minus{}\\alpha_c\\equal{}0$\r\n\r\ncase 1: $ \\alpha_c \\equal{}0$\r\n\r\n$ \\to 2x^3\\minus{}7x\\equal{}0 \\to 2x(x^2\\minus{}\\frac{7}{2})\\equal{}0 \\to x_c\\equal{}0, \\pm \\sqrt{\\frac{7}{2}}$\r\n\r\n$ d^2(x_c)\\equal{}x_c^2\\plus{}x_c^4\\minus{}8x_c^2\\plus{}16\\equal{}x_c^2(x_c^2\\minus{}7)\\plus{}16\\equal{} 16, \\frac{15}{4}$\r\n\r\ncase 2: $ \\alpha_c\\equal{}x$\r\n\r\n$ \\to 2x^3\\minus{}8x\\equal{}0 \\to 2x(x^2\\minus{}4)\\equal{}0 \\to x_c\\equal{}0, \\pm 2$\r\n\r\n$ d^2(x_c)\\equal{}(x_c^2\\minus{}4)^2\\equal{}16,0$\r\n\r\ncase 3: $ \\alpha_c\\equal{}2$\r\n\r\n$ \\to 2x^3\\minus{}7x\\minus{}2\\equal{}0 \\to 2(x\\minus{}2)(x^2\\plus{}2x\\plus{}\\frac{1}{2} )\\equal{}0 \\to x_c\\equal{} 2, \\minus{}1 \\pm \\frac{1}{\\sqrt{2}}$\r\n\r\n$ d^2(x_c)\\equal{}(x_c\\minus{}2)^2\\plus{}(x^2\\minus{}4)^2\\equal{}0,$ or $ (\\minus{}3\\pm \\frac{1}{\\sqrt{2}} )^2\\plus{}((\\minus{}1\\pm \\frac{1}{\\sqrt{2}})^2\\minus{}4)^2\\equal{}$\r\n\r\n$ 9 \\mp 3\\sqrt 2\\plus{} \\frac{1}{2}\\plus{} (1\\mp \\sqrt 2 \\plus{} \\frac{1}{2} \\minus{}4)^2\\equal{}\\frac{19}{2} \\mp 3\\sqrt 2 \\plus{}(\\frac{5}{2}\\pm \\sqrt 2)^2\\equal{}$\r\n\r\n$ \\frac{19}{2} \\mp 3\\sqrt 2 \\plus{}\\frac{25}{4} \\pm 5\\sqrt 2\\plus{} 2\\equal{}\\frac{71}{8} \\pm 2 \\sqrt 2 < 16$ either way..\r\n\r\nso max $ d^2$ is 16, meaning max distance is 4. either on the boundary or at x=0 straight up", "Solution_4": "Your answer is incorrect." } { "Tag": [ "topology" ], "Problem": "Is it possible for the fundamental group to be finite but not trivial?", "Solution_1": "Sure; for example, the fundamental group of $ \\mathbb{R} P^2$ is $ \\mathbb{Z}/2\\mathbb{Z}$. In fact we can say much more: [b]every group[/b] is the fundamental group of [url=http://en.wikipedia.org/wiki/Eilenberg%E2%88%92MacLane_space]some space[/url]." } { "Tag": [ "geometry", "rectangle", "combinatorics proposed", "combinatorics" ], "Problem": "\"Knight-rook\" can make horizontal or vertical move like rook,but after each horizontal move he must do vertical or vice versa.We arrange $ n$ traps on the chessboard $ 7*7$,such that \"knight-rook\" has only finite numbers of moves.\"Knight rook\" gets in to trap if it stands on it ,or goes through the trap(Starting position of \"knight-rook\" is arbitrarily).Find the minimum value of $ n$.", "Solution_1": "[hide=\"Answer:\"]13[/hide]", "Solution_2": "Does anybody want to see a solution?I can post it, if someone wants. I know two solutions,one of them is mine, and the official one.", "Solution_3": "I would like so...", "Solution_4": "Here is my solution,\r\nAssume that we have put $ k$ traps and the table satisfies the condition. Now turn the table into a graph such that its vertices are the squares which haven't got traps and put edges to the adjacent ones. \r\nLemma: If the number of edges of a simple graph is greater than or equal to its number of vertices, then it contains a cycle. It is a well-known fact.\r\nAt first we got $ 2.7.6$ edges and we have deleted some of them after the trapping operation. A trap can delete at most $ 4$ edges. Another point is we must put at least one trap to the edges of the table and this can delete at most $ 3$ edges. So we have at least $ 84\\minus{}4(k\\minus{}1)\\minus{}3$ edges and exactly $ 49\\minus{}k$ vertices. Due to the condition this graph can't contain a cycle. Hence from the lemma $ 84\\minus{}4(k\\minus{}1)\\minus{}3<49\\minus{}k$, then $ k>12$.\r\nIf we put traps to $ (2,2), (4,2), (6,2), (3,3), (5,3), (7,3), (2,4), (3,5), (5,5), (7,5), (2,6), (4,6), (6,6)$ the knight-rook can only have finitely many moves.", "Solution_5": "[quote=\"Umut Varolgunes\"]Here is my solution,\nAssume that we have put $ k$ traps and the table satisfies the condition. Now turn the table into a graph such that its vertices are the squares which haven't got traps and put edges to the adjacent ones. \nLemma: If the number of edges of a simple graph is greater than or equal to its number of vertices, then it contains a cycle. It is a well-known fact.\nAt first we got $ 2.7.6$ edges and we have deleted some of them after the trapping operation. A trap can delete at most $ 4$ edges. Another point is we must put at least one trap to the edges of the table and this can delete at most $ 3$ edges. So we have at least $ 84 \\minus{} 4(k \\minus{} 1) \\minus{} 3$ edges and exactly $ 49 \\minus{} k$ vertices. Due to the condition this graph can't contain a cycle. Hence from the lemma $ 84 \\minus{} 4(k \\minus{} 1) \\minus{} 3 < 49 \\minus{} k$, then $ k > 12$.\nIf we put traps to $ (2,2), (4,2), (6,2), (3,3), (5,3), (7,3), (2,4), (3,5), (5,5), (7,5), (2,6), (4,6), (6,6)$ the knight-rook can only have finitely many moves.[/quote]\r\nYour solution is totally the same as the official one,but nobody at olympiad didn't solve the problem in this way.In fact,there was only one solution,it was mine :) .I'll post it later...", "Solution_6": "I loved the question and i am wondering the solution of yours, Erken.", "Solution_7": "The first idea that came to my mind was to use tilings(in fact it is standart idea :P ).But the question was how to use tilings in that kind of problems and after thinking for an hour,I came to conclusion that I must find some \"almost fixed\" arrangement of these traps and after that find some \"cycle\" to obtain a contradiction.\r\n[b]Solution: [/b]\r\nWe will call a cell(path)[i] \"safety\" [/i]if there is no trap(s) on cell(path). \r\nSuppose that we have some arrangement that satisfies to our conditions and contains less than $ 13$ traps.\r\nWe will prove that there must be at least $ 12$ traps.\r\nIt is clear that in each square $ 2\\times 2$ there must be at least $ 1$ trap and at least $ 2$ traps must be in square $ 3\\times 3$.\r\nAlso it is not hard to show that there must be at least $ 3$ traps in rectangle $ 4\\times 3$.\r\nWe will cut our board on a small squares and rectangles as it showed on a picture:\r\n[img]http://img185.imageshack.us/img185/7490/adzb0.png[/img]\r\nTherefore,there must be at least $ 12$ traps,and each tagged squares and rectangles there must be minimal number of traps,i.e there must be exactly $ 4$ traps in tagged $ 4\\times 4$, exactly $ 2$ traps in tagged $ 3\\times 3$ square and so on...\r\nAnd after that we will turn our picture four times to $ 90^{\\circ}$ around the center of a board.From where we conclude that in each green tagged square $ 3\\times 3$ there are exactly $ 2$ traps and in each blue tagged rectangly $ 1\\times 3$ there are exactly $ 1$ trap(see a picture below :roll: ):\r\n[img]http://img362.imageshack.us/img362/5819/adyr8.png[/img]\r\nAnd obviously in the center of each tagged $ 3\\times 3$ square there is a trap,otherwise there exist a [i]\"safety\"[/i] cycle.\r\n[b]Here is the main idea:[/b]\r\nIt is easy to show that if we have some $ 3\\times 3$ square with $ 2$ traps(one of the in a center) and two [i]\"safety\"[/i] cells on it, then there exist a path between them.\r\nSo if we place our \"knight-rook\" one some cell inside square $ 3\\times 3$ then we can move in a clockwise direction and come to the border of rectangle $ 1\\times 3$,and considering several cases ,we finally obtain that \"knight-rook\" can move to the next square $ 3\\times 3$,and doing it four times we will return to the first square $ 3\\times 3$,and as it was noticed,we can return to the initial cell,therefore there is a cycle.\r\n[b]Here is an example[/b] of how can traps be placed:\r\n[img]http://img145.imageshack.us/img145/9243/adbl1.png[/img]\r\n,where red squares are traps...\r\nQ.E.D\r\nP.S:I can explain my solution in details,if someone wants ... and sorry for bad pictures... :blush:" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Given a sequence of eight integers $x_{1},x_{2},...,x_{8}$ in a single operation one replaces these numbers with $|x_{1}-x_{2}|,|x_{2}-x_{3}|,...,|x_{8}-x_{1}|$. Find all the eight-term sequences of integers which reduce to a sequence with all the terms equal after finitely many single operations.", "Solution_1": "It's Ducci Sequence [url]http://en.wikipedia.org/wiki/Ducci_sequence[/url]", "Solution_2": "Since the wikipedia page on Ducci sequence doesn't seem to give an explicit solution, I will provide one.\n\nA few initial observations:\n\n[list=1]\n[*]All entries are always integers and after the first operation, all entries are nonnegative.\n[*]From $(1)$, from the second 8-tuple onward, the maximum entry is non-increasing over successive operations.\n[*]The $8$-tuples ends up with all the same entries at some point iff it eventually ends up as an $8$-tuple of all $0$s.\n[*]If after some number of operations we have a sequence of $(a_1, a_2, \\ldots, a_8)$ of the form $(kb_1, kb_2, \\ldots, kb_8)$ for integers $k, b_1, b_2, \\ldots, b_8,$ then the $8$-tuple can be reduced to $(b_1, b_2, \\ldots, b_8)$ and this new $8$-tuple eventually ends up as all $0$s iff the old one does.\n[*]When viewed $\\pmod 2$, the operation is equivalent to applying $(x_1, x_2, \\ldots, x_8) \\pmod 2 \\to (x_1+x_2, x_2+x_3, \\ldots, x_8+x_1) \\pmod 2$.\n[/list]\n\nFrom $(3)$ it suffices to show that the entries all eventually end up as $0$. Proceed by induction on the maximum entry $m$ in the $8$-tuple resulting from a single operation. This method is legal due to $(1)$ and $(2)$. As a base case, with $m = 0$ we are already done. Consider $m = k > 0$ and assume the claim follows for all nonnegative integers less than $k$. From $(5)$, we observe the following for successive operations $\\pmod 2$:\n\\begin{align*}\n(x_1, x_2, \\ldots, x_8) &\\to (x_1+x_2, x_2+x_3, \\ldots, x_8+x_1)\\\\\n&\\to (x_1+x_3, x_2+x_4, \\ldots, x_8+x_2)\\\\\n&\\to (x_1+x_2+x_3+x_4, x_2+x_3+x_4+x_5, \\ldots, x_8+x_1+x_2+x_3)\\\\\n&\\to (x_1+x_5, x_2+x_6, \\ldots, x_8+x_4)\\\\\n&\\to (x_1+x_2+x_5+x_6, x_2+x_3+x_6+x_7, \\ldots, x_8+x_1+x_4+x_5)\\\\\n&\\to (x_1+x_3+x_5+x_7, x_2+x_4+x_6+x_8, \\ldots, x_8+x_2+x_4+x_6)\\\\\n&\\to (x_1+x_2+\\cdots+x_8, x_1+x_2+\\cdots+x_8, \\ldots, x_1+x_2+\\cdots+x_8)\\\\\n&\\to (0, 0, \\ldots, 0) \\pmod 2\n\\end{align*}\n\nThus after a finite number of steps, the $8$-tuple is in the form $(2a_1, 2a_2, \\ldots, 2a_8)$ for integers $a_1, a_2, \\ldots, a_8$ and from $(4)$, the $8$-tuple can be reduced so that all entries are divided by a factor of $2$. Since the maximum entry is non-increasing and we began with $m = k$, we now have $m \\le \\lfloor k / 2 \\rfloor < k$ (for $k > 0$). The $8$-tuple now reaches the form $(0, 0, \\ldots, 0)$ in a finite number of steps by the IH. $\\blacksquare$\\\\\n\n[hide=\"Remark\"]\nIn the list of properties of the Ducci sequence given on its wikipedia page, it says that for $n$-tuples where $n$ is not a power of $2$, the sequence either reaches $(0, 0, \\ldots, 0)$ or enters a cycle where all entries are either $0$ or $M$ for a fixed constant $M$. The fact that it at least enters a cycle is clear from $(1)$ and $(2)$ above since the number of $n$-tuples with entries all nonnegative integers at most some fixed maximum is finite, so repeats must eventually occur.\n\nTo see that the $n$-tuples must all eventually contain only $0$ and $M$ for a fixed $M$ (if not eventually reaching $(0, 0, \\ldots, 0)$), note that this requires the non-increasing maximum entry to stop decreasing and remain stationary, in this case no more than $M$, and in fact, exactly $M$. Indeed, call the $i$-th tuple $x_i$ and its $j$-th entry $x_i^{(j)}$. If the the maximum persists as $M$ for $n$ $n$-tuples $x_i, x_{i+1}, \\ldots, x_{i+n-1}$, we can prove that all entries of $x_i$ must be either $0$ or $M$: let $x_{i+n-1}^{(j)} = M$.\n\nThen $\\{x_{i+n-2}^{(j)}, x_{i+n-2}^{(j+1)}\\} = \\{0, M\\}$ (with superscripts taken $\\pmod n$), as any other pair of integers in $\\{0, 1, \\ldots, M\\}$ could not have achieved this difference. From similar reasoning, each of $x_{i+n-3}^{(j)}, x_{i+n-3}^{(j+1)}, x_{i+n-3}^{(j+2)} \\in \\{0, M\\}$ with at least one being $M$. Repeating this reasoning down to $x_i$, all of its entries must be in $\\{0, M\\}$ and it is straightforward that this then must always be the case for subsequent $n$-tuples in the iterations.\n[/hide]", "Solution_3": "@above What is the proof of observation 4?" } { "Tag": [ "algebra", "function", "domain", "trigonometry" ], "Problem": "define domain of function $f$ bescribed by the following equation:\r\n$f(x)-f^{2}(x)+f^{3}(x)-\\ldots = \\sin x$", "Solution_1": "[hide=\"Answer\"]$\\mathcal{D}_{f}=\\bigcup_{k\\in\\mathbb{Z}}\\left(-{7\\pi\\over 6}+2k\\pi,{\\pi\\over 6}+2k\\pi\\right)$[/hide]" } { "Tag": [ "algebra", "polynomial", "superior algebra", "superior algebra unsolved" ], "Problem": "Given a field $K$ and an irreducible polynomial $f(x)\\in K[x]$, degree of $f$ is at least 3. Is it true that there always exist irreducible polynomial $g(x)\\in K[x]$ such that degree(g) is more than degree(f)? \r\n\r\nIn other words, if degrees of irreducible polynomials are bounded, must they be bounded by 2?", "Solution_1": "See http://www.mathlinks.ro/Forum/viewtopic.php?t=30853 where I asked this already :)" } { "Tag": [ "AMC", "AMC 12", "probability", "AMC 10", "AMC 10 A", "AMC 10 B", "geometry" ], "Problem": "So... what's everyone's AMC12 goals and fears?(by fears, I mean things like lots of NT, lots of geo, etc.)\r\n\r\nMy goal : 132+\r\nfear : lots of probability", "Solution_1": "I'm going to get a perfect score.", "Solution_2": "well I didn't study enough... at all.\r\n\r\nSo I can't say I'm expecting that much. 120+ would make me happy.\r\n\r\nI'd say my main fear is trig. Or any problems that take a lot of time...", "Solution_3": "Goal would be like 120-125, but not taking it- AMC10A and AMC10B this year.", "Solution_4": "My fear is that there'll be a lot of math on it.", "Solution_5": "Goal: 126+\r\nFear: Lots of NT", "Solution_6": "Goal: 150. But realistically, 120.\r\n\r\nFear: lots of geo in problems 20-25 :o", "Solution_7": "Goal: 150.\r\nRealistically: 141.\r\n\r\nFear: Brain freeze.", "Solution_8": "Goal: 140+ 150 would be awesome\r\n\r\nfear- multiple geometry problems in the higher numbers", "Solution_9": "goal: 130\r\n\r\nfear: lots of geo and trig, pretty much problems that take a long time.", "Solution_10": "Never taken it before, but I am planning to do lots of practices.\r\nGoal: 125.\r\nFear: Trigonometry.", "Solution_11": "Goal: 120.\r\nFear: A little combinatorics, but mainly geo.", "Solution_12": "Goal: Finding somewhere to take it (my high school is proving to be very stubborn). Otherwise, I'm hoping for a 145.5.\r\nFear: Geometry.", "Solution_13": "Goal: >140\r\nFear: <140\r\n :huh:", "Solution_14": "What's with the fear of geometry?", "Solution_15": "Goal: 135\r\nFears: lots of hard geo and trig, or <110", "Solution_16": "Goal: 132+\r\nHope: 139.5+\r\n\r\nFear: Any hard trig questions, or any involving the complex plane.", "Solution_17": "Goal: 100+\r\nFear: Geo problems, because they suck up a lot of time.", "Solution_18": "Goal: >135.\r\nFear: Computation.", "Solution_19": "Goal: 150, obviously. I haven't gotten a perfect score on any AMC's for real, and haven't taken any practice ones.\r\n\r\nFear: Problems that require a deep insight for which I don't have time.", "Solution_20": "[quote=\"math154\"]Goal: >135.\nFear: Computation.[/quote]\r\nditto", "Solution_21": "Goal: 144+\r\n\r\nFear: Bashy Math >.<", "Solution_22": "Goal: 110\r\nFear: Stupid mistakes and things I don't know", "Solution_23": "[quote=\"1=2\"]What's with the fear of geometry?[/quote]\r\n\r\ngeometry problems can be quite time consuming on the AMCs... Especially if they're in the 20-25 range.", "Solution_24": "Goal: 130+\r\nFear: Spending too much time on the easy problems and running out of time.", "Solution_25": "Goal: 120-125 (just want to make AIME and hopefully USAMO)\r\n\r\nFear: Combinatorics and problems that exceed my problem solving capability", "Solution_26": "Goal: 132+\r\nFear: Not being able to achieve it and/or brain cramp/nervousness or just failing in general.", "Solution_27": "haha, my goal of 135 is very unrealistic at the moment. I guess my 'real' goal is 120+.", "Solution_28": "Goal: 150\r\nFear: Overconfidence... :P", "Solution_29": "Goal: Make AIME\r\nFears: Careless mistakes, inability to read properly on the day of.", "Solution_30": "\"there is nothing to fear but fear itself\" - FDR", "Solution_31": "To beat my last year's score of 136.5, though I probably won't since I haven't really studied this year.", "Solution_32": "[quote=\"Smartguy\"]\"there is nothing to fear but fear itself\" - FDR[/quote] This is actually the abridged version. He then said, \"and hard questions that happen on the AMC 12.\"", "Solution_33": "Goal: 144+ (BUT NOT A 145.5 SINCE THAT WOULD MEAN I IGNORED ALL OF MY PRINCIPLES!)\r\n\r\nFear: Not enough BASHYNESS", "Solution_34": "[quote=\"Mewto55555\"]Goal: 144+ (BUT NOT A 145.5 SINCE THAT WOULD MEAN I IGNORED ALL OF MY PRINCIPLES!)\n[/quote]\r\n\r\nhuh?", "Solution_35": "The only way to get a 145.5 is to get 24 right and leave one blank. I think he is implying that his principles are to answer all the questions.", "Solution_36": "Goal: 130+\r\n\r\nFears: That I get less than a 130, do the test too slowly, get sleep deprived the night before (I have to take the 12A this year, and possibly the 12B as well if I fail the 12A...and also you know kids at my school get sleep deprived all the time...), don't beat certain beastly people (e.g. xpmath, abacadaea, laughinghead505, whitesoxfan), or if there's too much NT and/or combo :huh: \r\n\r\nBiggest fear ever: I have too many fears. :( Plus homework is eating away at my studying for math contests :(", "Solution_37": "Goal: 115+\r\nFears: Modular arithmetic, trigonometry" } { "Tag": [], "Problem": "The mean of a collection of five positive integers is 6. What is the greatest possible median of the collection?", "Solution_1": "sum of 5 integers are 30.\r\n\r\nthe problem is about finding greatest median, so let's make two integers 1.\r\n\r\nso it's now 1, 1, _, _, _.\r\n\r\nand other 3 integers could be x, x, and x+1.\r\n\r\nso it's total 3+3x=30.\r\n\r\n3x=27, and x=9.\r\n\r\nthe answer is 9." } { "Tag": [ "modular arithmetic", "number theory unsolved", "number theory" ], "Problem": "Find all $ n,m\\in N$ such that\r\n\r\n$ n^5\\plus{}n^4\\equal{}7^m\\minus{}1$", "Solution_1": "I think I found something ...Nice problem \r\n\r\nThe condition is equivalent to $ n^5\\plus{}n^4\\plus{}1\\equal{}7^m$ or $ (n^2\\plus{}n\\plus{}1)(n^3\\minus{}n\\plus{}1)\\equal{}7^m$ so \r\n$ n^2\\plus{}n\\plus{}1\\equal{}7^p$ (1) and $ n^3\\minus{}n\\plus{}1\\equal{}7^q$ (2) with $ p\\plus{}q\\equal{}m$ \r\nWith eaasy divisibilities we can find that \r\n$ n\\equiv 2(\\mod\\ 7^p)$ . Subtituting in the relation (1) gives $ 7^p|7$ so p=0 or 1 from which we can find the solutions\r\n(m,n)=(0,0) or (m,n)=(2,2)", "Solution_2": "[quote=\"silouan\"]\nWith eaasy divisibilities we can find that \n$ n\\equiv 2(\\mod\\ 7^p)$.[/quote]\r\nDo you mind explaining this? Thanks.", "Solution_3": "[quote=\"baz\"][quote=\"silouan\"]\nWith eaasy divisibilities we can find that \n$ n\\equiv 2(\\mod\\ 7^p)$.[/quote]\nDo you mind explaining this? Thanks.[/quote]\r\n\r\nIn fact, Silouan wrote : \"$ n^2 \\plus{} n \\plus{} 1 \\equal{} 7^p$ (1) and $ n^3 \\minus{} n \\plus{} 1 \\equal{} 7^q$ (2) with $ p \\plus{} q \\equal{} m$\"\r\n\r\nAnd, since $ n^3\\minus{}n\\plus{}1\\equal{}(n\\minus{}1)(n^2\\plus{}n\\plus{}1)\\minus{}(n\\minus{}2)$, we have $ 7^q\\equal{}(n\\minus{}1)7^p\\minus{}(n\\minus{}2)$ and $ n\\equal{}2\\pmod{\\min(p,q)}$\r\nThen, if $ p\\leq q$, $ n\\equal{}2\\pmod{7^p}$ and, with (1), we have $ 7\\equal{}0\\pmod{7^p}$ and so $ p\\equal{}0$ or $ p\\equal{}1$\r\nElse, if $ q\\leq p$, $ n\\equal{}2\\pmod{7^q}$ and, with (2), we have $ 7\\equal{}0\\pmod{7^q}$ and so $ q\\equal{}0$ or $ q\\equal{}1$\r\n\r\nAnd hence the solution $ 2^5\\plus{}2^4\\equal{}7^2\\minus{}1$", "Solution_4": "cancer How come u r posting all the indan MO Postal problems ??", "Solution_5": "[quote=\"pardesi\"]cancer How come u r posting all the indan MO Postal problems ??[/quote]\r\n\r\nI didn't mean to, pardesi, I don't even know anything about it. I got the problems from here: http://www.imomath.com/othercomp/lirymnbiueb/BlrMO07.pdf\r\n\r\nI hope you understand me :) \r\n\r\nSo feel free to lock any problems which coincide with your Indian problems :)", "Solution_6": "And actually, as a side note, I don't think it doesn't look like a good idea that the problems which students can do at home are copied from such sources... :wink:", "Solution_7": "Consider equation\r\n$ x^{3m \\plus{} 2} \\plus{} x^{3m \\plus{} 1} \\plus{} 1 \\equal{} p^a$\r\na)Prove that if $ p\\equiv 5(\\mod 6)$ then the equation has no solution.\r\nb)Prove that the equation has only finite solution (we can find out it for each m).\r\nMore $ v_p(3m^2 \\plus{} 3m \\plus{} 1) \\equal{} k$ then\r\n$ x^2 \\plus{} x \\plus{} 1 \\equal{} p^c$ where $ c\\leq k$", "Solution_8": "sorry cancer but it does turn out that the people setting the questions are copying from there :)", "Solution_9": "anyways the last date for submitting the solutions was sept 12... :P \r\n\r\nno harm done.. :D", "Solution_10": "actually this problem has been posted before [url=http://www.mathlinks.ro/viewtopic.php?t=164126]here[/url] and those who wanted to cheat have already done so :D \r\nactually my question to cancer was more about the coincidence that two posts of his had postal problems \r\n[url=http://www.mathlinks.ro/viewtopic.php?t=168857]here[/url] is the second one." } { "Tag": [ "Euler", "modular arithmetic", "number theory proposed", "number theory" ], "Problem": "Prove that if $ (a,n)\\equal{}1$ and $ k$ is the number of odd prime factors of $ n$, then $ a^{\\frac{\\phi(n)}{2^{k\\minus{}1}}}$ is congruent to $ 1$ mod $ n$.", "Solution_1": "let $ n \\equal{} 2^rp_1^{r_1}\\cdots p_k^{r_k}$. by Euler's Theorem, $ a^{\\phi(p_i^{r_i})\\prod_{j\\neq i}\\phi(p_j^{r_j})/2^{k \\minus{} 1}}\\equiv 1\\pmod{p_i^{r_i}}$ because $ \\prod_{j\\neq i}\\phi(p_j^{r_j})/2^{k \\minus{} 1}$ is an integer. also $ a^{\\phi(2^r)\\prod_{j \\equal{} 1}^k\\phi(p_j^{r_j})/2^{k \\minus{} 1}}\\equiv 1\\pmod{2^r}$ because $ \\prod_{j \\equal{} 1}^k\\phi(p_j^{r_j})/2^{k \\minus{} 1}$ is an integer. the result follows." } { "Tag": [ "calculus", "derivative", "function" ], "Problem": "Im unsure of where this goes (considering its calculus), but a question of the question's wording came up in regard to well, heh, this question:\r\n\r\nIf $f(x) = x^2 + 6x$, $x > 0$, find the derivative of the inverse of $f(x)$ in terms of $x$.", "Solution_1": "The function is injective over $x\\in[0,\\infty)$, so it has an inverse.\r\n\r\nYou can solve for $f^{-1}$, by using the identity $f(x)=y \\Rightarrow f^{-1}(y)=x$\r\n\r\nIt's straightforward to find that $f^{-1}(x)=\\sqrt {x+9}-3$, it follows immediately that\r\n\r\n$f^{-1}^{'}(x)=\\frac {1}{2\\sqrt{x+9}}$, $x\\in[0,\\infty)$", "Solution_2": "Thanks. This was just to settle a dispute that a couple people had in my school." } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "Each side of a convex quadrilateral $ ABCD$ is tangent to a circle of radius $ r$ whose centre is denoted by $ O$. Show that at least one of the circle,circumscribed about triangles $ OAB$ and $ OCD$, respectively, has a radius greater tham or equal to $ r$.", "Solution_1": "It is easy to prove that $ \\angle DOC\\plus{}\\angle AOB\\equal{}180^{\\circ} \\Rightarrow$ max$ (\\angle AOB,\\angle DOC) \\ge 90^{\\circ}$\r\n\r\n$ W.L.O.G.$ suppose that $ \\angle AOB \\ge 90^{\\circ} \\Rightarrow R_{\\triangle AOB} \\ge r$" } { "Tag": [ "vector", "geometry", "circumcircle", "inequalities unsolved", "inequalities" ], "Problem": "Let $ ABC$ is a triangle, let $ x, y, z \\in R$. Prove that: $ \\boxed{x^2 \\plus{} y^2 \\plus{} z^2\\ge \\minus{}2xycos2C \\minus{} 2yzcos2A \\minus{} 2zxcos2B}$", "Solution_1": "[quote=\"thanhnam2902\"]Let $ ABC$ is a triangle, let $ x, y, z \\in R$. Prove that: $ \\boxed{x^2 \\plus{} y^2 \\plus{} z^2\\ge 2xycos2C \\plus{} 2yzcos2A \\plus{} 2zxcos2B}$[/quote]\r\nIt is wrong . Must be : $ {x^2 \\plus{} y^2 \\plus{} z^2 \\plus{} 2xycos(2C) \\plus{} 2yzcos(2A) \\plus{} 2zxcos(2B) \\ge 0}$ ;\r\nLet $ \\overline{a}$ , $ \\overline{b}$ , $ \\overline{c}$ - vectors , then : $ (\\overline{a}\\minus{}\\overline{b})^2\\plus{}(\\overline{b}\\minus{}\\overline{c})^2\\plus{}(\\overline{c}\\minus{}\\overline{a})^2 \\le 3(\\overline{a}^2\\plus{}\\overline{b}^2\\plus{}\\overline{c}^2)$ ;", "Solution_2": "Yes. I'm sory.", "Solution_3": "Let $ (O;R)$ is circumscircle of $ \\triangle ABC$.\r\n\r\nShow that: $ (x.\\overrightarrow{OA} \\plus{} y.\\overrightarrow{OB} \\plus{} z.\\overrightarrow{OC})^2 \\ge 0$\r\n\r\nOK!", "Solution_4": "[color=darkblue]Let $ (O;R)$ is circumscircle of $ \\triangle ABC$.[/color]\r\n\r\n[color=darkblue]note: [/color]$ \\| \\begin {array} {c}OA=OB=OC=R\\\\\\overrightarrow{OA}.\\overrightarrow{OB} = OA.OB.cos2C = R^2cos2C \\\\\r\n\\overrightarrow{OB}.\\overrightarrow{OC} = OB.OC.cos2A = R^2cos2A \\\\\r\n\\overrightarrow{OC}.\\overrightarrow{OA} = OC.OA.cos2B = R^2cos2B \\end {array}$\r\n$ \\forall x, y, z \\in R$ [color=darkblue]we have [/color]$ (x.\\overrightarrow{OA} + y.\\overrightarrow{OB} + z.\\overrightarrow{OC})^2 \\ge 0$ \r\n\r\n$ \\Longleftrightarrow x^2R^2 + y^2R^2 + z^2R^2+ 2xyR^2cos2C + 2yzR^2cos2A + 2zxR^2cos2B \\ge 0$\r\n\r\n$ \\Longleftrightarrow \\boxed{x^2 + y^2 + z^2\\ge - 2xycos2C - 2yzcos2A - 2zxcos2B}$\r\n\r\n[color=darkblue]OK![/color]" } { "Tag": [ "search" ], "Problem": "Can someone please explain this problem:\r\n\r\nWhat is the smallest number of disks radius \u00bd that can cover a disk radius 1?\r\n\r\n\r\nSolution:\r\nhttp://web.archive.org/web/20040814192240/www.kalva.demon.co.uk/eotvos/esoln/sol1947c.html\r\n\r\nthanks!", "Solution_1": "The smaller disks need to cover the boundary of the larger disk. Consider the amount of the boundary that a single disk of radius $ \\frac {1}{2}$ can cover. ($ P$ and $ Q$ in the solution you've linked are the intersections of the boundaries of the two disks.)\r\n\r\nHere are some threads on the forum that deal with this or similar problems:\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=254821974&t=2047\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=254821974&t=169465\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=254821974&t=56774\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=254821974&t=56482", "Solution_2": "thanks for the insight :D :D" } { "Tag": [ "logarithms", "number theory unsolved", "number theory" ], "Problem": "how many digits does the following number have in the base of $ 10$ ?\r\n\r\n$ {3}^{40}$\r\n\r\nhmmm...i'm confused...is there Really an elementary way to calculate the digits of a number without using Logarithms or whatever???\r\n\r\n( actually,the question has given 5 choices to choose from (Iraninan Level 1 contest is a 5-choice-to-choose-from-like exam) : 1) 17 , 2) 18 , 3) 19 , 4)20 , 5)21 )", "Solution_1": "$ 3^{40}\\equal{} (10\\minus{}1)^{20}\\equal{} 10^{20}(1\\minus{}\\frac{1}{10})^{20}\\equal{}\\frac{10^{20}}{e^{2}\\plus{}\\epsilon}$. Therefore it had 20 digits (>$ 10^{19}$) and first and last digits are 1." } { "Tag": [ "trigonometry", "geometry", "trig identities", "Law of Sines" ], "Problem": "Solve the following system:\r\n\r\n$ 3(x \\plus{} \\frac {1}{x}) \\equal{} 4(y \\plus{} \\frac {1}{y}) \\equal{} 5(z \\plus{} \\frac {1}{z});\\ xy \\plus{} yz \\plus{} zx \\equal{} 1$", "Solution_1": "This is a well-known trick, we can let x=tan(A/2), y=tan(B/2), z=tan(C/2), where A,B, and C are angles of a triangle\r\nProof here: [url]http://www.goiit.com/posts/show/174404.htm[/url]\r\nSubbing in with half-angle formulas, we get\r\n$ \\frac {sin A}{3} \\equal{} \\frac {sin B}{4} \\equal{} \\frac {sin C}{5}$.\r\nBy the law of sines they are the angles of a 3-4-5 right triangle.\r\nSo $ x \\equal{} \\frac {1}{3}$, $ y \\equal{} \\frac {1}{2}$, $ z \\equal{} 1$ :D\r\nBy the way, does anyone have another proof of the substitution?", "Solution_2": "[quote=\"thomaskurian89\"]Solve the following system:\n\n$ 3(x + \\frac {1}{x}) = 4(y + \\frac {1}{y}) = 5(z + \\frac {1}{z});\\ xy + yz + zx = 1$[/quote]\r\n\r\n[color=darkred]Another solution:[/color]\r\n[hide]\nLet $ xy = a,yz = b,xz = c,a + b + c = 1$\n\n$ 3(x + 1/x) = 4(y + 1/y) = 5(z + 1/z)$ times $ xyz$ for each term, we get $ 3(ac + b) = 4(ab + c) = 5(bc + a)$\n\n$ a + b + c = 1 \\implies b = 1 - a - c ,c = 1 - a - b, a = 1 - b - c$\n\nThen \n\n$ 3(ac + b) = 4(ab + c) = 5(bc + a)$\n\n$ \\implies 3(ac + 1 - a - c) = 4(ab + 1 - a - b) = 5(bc + 1 - b - c)$\n\n$ \\implies3(a - 1)(c - 1) = 4(a - 1)(b - 1) = 5(b - 1)(c - 1)$\n\n$ \\because x\\not = 0,y\\not = 0,z\\not = 0$\n$ \\therefore a\\not = 1,b\\not = 1,c\\not = 1$\n\n$ 3(a - 1)(c - 1) = 4(a - 1)(b - 1) = 5(b - 1)(c - 1)\\implies \\frac 3 {b - 1} = \\frac 4 {c - 1} = \\frac 5 {a - 1}$\n\nSuppose $ \\frac {b - 1} 3 = \\frac {c - 1} 4 = \\frac {a - 1} 5 = t$\n\n$ \\{\\begin{array}{c}{1} a = 5t + 1 \\\nb = 3t + 1 \\\nc = 4t + 1 \\end{array}$\nand $ a + b + c = 1 \\implies (5t + 1) + (3t + 1) + (4t + 1) = 1\\implies t = \\frac { - 1}{6}$\n\n$ \\therefore a = \\frac 1 6 ,b = \\frac 1 2 , c = \\frac 1 3$\n\n$ \\implies x =\\pm \\sqrt {\\frac {ac} b} =\\pm \\frac 1 3 , y = \\pm \\sqrt {\\frac {ab} c} =\\pm \\frac 1 2,z = \\pm \\sqrt {\\frac {bc} a} =\\pm 1$\n\nSolution $ (x,y,z) = (\\frac 1 3,\\frac 1 2,1)$ or $ (x,y,z) = (-\\frac 1 3,-\\frac 1 2,-1)$\n[/hide]", "Solution_3": "In general, how would someone figure out that a trigonometric substitution would be in order?", "Solution_4": "The 3-4-5 reminded me of a 3-4-5 right triangle, and I had seen a somewhat similar problem before." } { "Tag": [ "probability" ], "Problem": "If the tens digit $ d$ is replaced with one of the digits 0-9, what is the probability that the four-digit positive integer $ 12d4$ is divisible by 12? Express your answer as a common fraction.", "Solution_1": "If 12d4 is divisible by 12, d4 must be divisible by 12 since 1200 is divisible by 12.\r\nThus, d can equal 2 or 8. Since there are 10 possible values, the answer is 2/10=1/5.", "Solution_2": "In order for a number to be divisible by 12, the number must be divisible by 3 and 4.\n\n3 - Sum of digits needs to be divisible by 3 ( 48: 4+8=12, 12/3=4 YES )\n\n4 - Last two digits form a number divisible by 4 ( 124: 24/4=6 YES )\n\nWe have the digits 12d4.\n\nTo make the sum of the digits divisible by 3, we have: 1+2+d+4=mult. of 3\n 7+d=mult. of 3\nSo we have d= 2, 5, and 8\n\nNow we check if 24, 54, and 84 are divisible by 4.\n24/4=6 YES, 54/4=13.5 NO, and 84/4=21 YES.\n\nSo, the probability of the number being divisible by 12 is: 2/10 = [size=150]1/5[/size]", "Solution_3": "Please do not revive threads this old when you have no contribution." } { "Tag": [ "algebra", "polynomial", "calculus", "integration", "algebra unsolved" ], "Problem": "find all polynoms in $R$\r\nsuch that \r\nfor all rational $r$ there is a rational $x$ such $P(x)=r$\r\n\r\n[color=red]Edited by Megus: Don't use dollar sign before every word, also you shouldn't name a topic like \"?????\"[/color]", "Solution_1": "ehh... What do you mean by \"T.O.T\" ?? :?", "Solution_2": "By using the Lagrange interpolation formula, we find that $P$ (our polynomial) must actually have rational coefficients. By scaling the coefficients, we may assume it has integral coefficients, so let $P(x)=a_nx^n+\\ldots+a_1x+a_0\\in\\mathbb Z[X]$.\r\n\r\nFor any integer $m$, we can find a rational $\\frac{p_m}{q_m}$ s.t. $P\\left(\\frac{p_m}{q_m}\\right)=m$. After writing this expression, we see that we have $q_m|a_n,\\ \\forall m\\in\\mathbb Z$, so all the rational solutions to equations of the form $P(x)=m,m\\in\\mathbb Z$ belong to $\\frac 1{a_n}\\cdot\\mathbb Z$. \r\n\r\nIf we assume the degree of $P$ is $\\ge 2$, then, as $m\\to\\infty$, we have $P^{-1}(m+1)-P^{-1}(m)\\to 0$ (if $m$ is large enough, then $P^{-1}$ makes sense, since $P$ is injective on a neighbourhood of infinity), so we will find pairs of rationals belonging to $\\frac 1{a_n}\\cdot\\mathbb Z$ as close to each other as we want, but this is a contradiction (two such points must have distance at least $\\frac 1{a_n}$ between them).", "Solution_3": "thank you very much \r\nbut how we assum that $p(x)$ has in tegral coefficient", "Solution_4": "This can be done since you can substitute $x=c\\cdot y$ where $c$ is some common multiple of the denominator of high enough powers of the coefficients" } { "Tag": [ "geometry solved", "geometry" ], "Problem": "AB=CD=EF in cyclic hexagon ABCDEF. \r\n prove mdpts of BC,DE,FA form equilatrel triangle.", "Solution_1": "See http://www.mathlinks.ro/Forum/viewtopic.php?t=45874 .\r\n\r\n darij" } { "Tag": [ "calculus", "integration", "function", "calculus computations" ], "Problem": "[color=darkblue]For any $a\\in R$, $\\int_{0}^{1}|f(x)-a|\\ dx\\ge\\int_{0}^{1}\\left|f(x)-f\\left(\\frac{1}{2}\\right)\\right|\\ dx$, where the continue function $f: [0,1]\\rightarrow R$ is monotonously.\n\n[b]Remark.[/b] See the my message from the Kunny's topic http://www.mathlinks.ro/Forum/viewtopic.php?t=106178 [/color]", "Solution_1": "Let $t$ such that $f(t)-a=0\\ (02,\\; x,y\\in\\mathbb{R}^{k},\\; |x-y|=d>0,$ and $r>0$. Prove \r\na)If $2r>d$, there are infinitely many $z\\in\\mathbb{R}^{k}$ such that $|z-x|=|z-y|=r$.\r\nb)If $2r=d$, there is exactly one such $z$.\r\nc)If $2rB$ we have $|f_{n}(x)-f(x)|<\\frac{\\varepsilon}{2}+2^{-n}$.\r\n\r\nTake now $n$ also sufficiently large to have $2^{-n}<\\frac{\\varepsilon}{2}$, and you're done. :)" } { "Tag": [ "algebra", "polynomial", "calculus", "calculus computations" ], "Problem": "How do I solve by hand the equation\r\n\r\n$ x^3\\minus{}7 x^2\\plus{}14 x\\minus{}7\\equal{}0$\r\n\r\nthanks guys", "Solution_1": "You don't.\r\n\r\n[url=http://en.wikipedia.org/wiki/Eisenstein%27s_criterion]Eisenstein's[/url] tells us that this polynomial is irreducible, so all of its roots are irrational. The only way to find the exact solutions, therefore, is the [url=http://mathworld.wolfram.com/CubicFormula.html]cubic formula[/url]." } { "Tag": [ "AMC", "AIME", "number theory", "prime numbers" ], "Problem": "Ok, i have a $50 bet riding on this, so i need some help. The problem on the test that said n=a^bc, and a,b,c are all prime, how many factors of n are there whose divisors are greater than 2 and less then the number. Thats the problem as best as i can recall. As is, what is the answer?", "Solution_1": "I'm fairly certain that if the College Board finds out that you posted this, they'll send their College Board SWAT Team to hunt you down and kill you.", "Solution_2": "yes, but i live dangerously, come on, help me out here.", "Solution_3": "Any one here take the sat and remember this problem?", "Solution_4": "Well, that, and the way you worded the question, I'm fairly certain that it makes no sense.", "Solution_5": "yea, what about a = 2? Doesn't seem to make much sense", "Solution_6": "[quote]are there whose divisors are greater than 2 and less then the number. [/quote]\r\n\r\nAre you sure the problem was greater than 2 (and not greater than 1 (or greater or equat to two ) ?\r\n\r\nIf so the answer would be (bc-1)\r\n\r\nYhe way you posted the problem, the answer is one less when a=2", "Solution_7": "If you go to the college confidential board, there's a huge post about SAT problems.... I think the question and answer was on there.", "Solution_8": "This problem should get moved down, no way a SAT question is Pre-Olympiad level.....(be pretty scary if they are)", "Solution_9": "I think the answer is bc+1, if a is greater than 2. If a = 2, then the answer is 0.\r\n\r\nAnd like beta said, this definetly isn't a Pre-Olympiad problem. If it were, I'd be a gold medalist in the IMO.", "Solution_10": "Wait. Is it even legal to make a bet on confidential problems?\r\n\r\nThey are not allowed to be distributed until a certain date, like this example.\r\n\r\nIt's like owning old money.\r\n\r\nIf it's five years old, it, crud\r\nIf it's fifteen years old, it's illegal because of forgery problems\r\nIf it's fifty years old it's an antique...\r\n\r\nSomeone explain how old bills exist!\r\n\r\nBy the way, explain the problem better; is it just $\\&a^{bc}\\&$?\r\n\r\nwhere a,b, and c are prime?\r\n\r\nI didn't get the rest of the problem...\r\n\r\nIt doesn't look like a SAT problem; what section did you get this from?", "Solution_11": "[quote=\"pakagawa\"]Wait. Is it even legal to make a bet on confidential problems?[/quote]\r\n\r\nI think underage betting in general is illegal.", "Solution_12": "I think all betting that isn't reported to the government is illegal. Including home poker games etc. Am I right?", "Solution_13": "Googled!\r\n[url]http://www.homepokergames.com/homepokerlaw.php[/url]", "Solution_14": "[quote=\"Andretti\"]The problem on the test that said n=a^bc, and a,b,c are all prime, how many factors of n are there whose divisors are greater than 2 and less then the number.[/quote]\r\nI think the original problem was more like this: If $\\textstyle n = a^2 b c$, where $a$, $b$, and $c$ are different prime numbers greater than 2, how many positive divisors does $n$ have besides 1 and $n$?\r\n\r\nThe answer is 10.", "Solution_15": "The answer would be 12 then, actually.", "Solution_16": "Did you see the word \"besides\"? The original wording might have been more emphatic, like \"not including 1 and n\".", "Solution_17": "Yea.. .it was 10, but i'm sure they stressed that it was NOT 1 and the number", "Solution_18": "this is an example of a Pre-Olympiad Problem\r\n\r\nAIME 2004A #15\r\nDefine f(1) = 1, f(n) = n/10 if n is a multiple of 10 and f(n) = n+1 otherwise. For each positive integer m define the sequence a_1, a_2, a_3, ... by a_1 = m, a_n+1 = f(a_n). Let g(m) be the smallest n such that an = 1. For example, g(100) = 3, g(87) = 7. Let N be the number of positive integers m such that g(m) = 20. How many distinct prime factors does N have? \r\n\r\nThat's a lot harder than SAT problem", "Solution_19": "[quote=\"Ravi B\"]Did you see the word \"besides\"? The original wording might have been more emphatic, like \"not including 1 and n\".[/quote]\r\n\r\nFine. See if I care. :)" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "How many pairs of positive integers $ (x,y)$ with $ x \\le y$ are there such that $ \\gcd (x,y)=5!$ and $ \\text{lcm} (x,y)=50!?$", "Solution_1": "[quote=\"moldovan\"]How many pairs of positive integers $ (x,y)$ with $ x \\le y$ are there such that $ \\gcd (x,y) = 5!$ and $ \\text{lcm} (x,y) = 50!?$[/quote]\r\n\r\nLet $ x=(5!)x_1, y=(5!)y_1$, then $ \\gcd(x_1,y_1)=1$\r\n\r\n$ x_1y_1= \\text{lcm} (x_1,y_1)=\\frac{50!}{5!}$ has $ 15$ prime factor.\r\n\r\nWrite $ \\frac{50!}{5!}=p_1^{a_1}p_2^{a_2}p_3^{a_3}...p_{15}^{a_{15}}$\r\nThen each of $ p_i^{a_i}$ divides either $ x_1$ or $ y_1$.\r\n\r\nSo number of pairs $ =2^{15}$" } { "Tag": [], "Problem": "Find all numbers of four digits that are perfect squares, and that perfect square is the result of adding a two digit number made by the first two digits of that square, and another two digit number made by the last two digits of that square. In example 9801 because $ 9801\\equal{}(98\\plus{} 01) ^2$\r\n\r\nPlease explain it clear, and step-by-step!", "Solution_1": "Probably not much help but $ 0001 \\equal{} (00 \\plus{} 01)^{2} \\equal{} 01^{2}$. Putting this with your example $ 99^{2}$, we have $ 01 \\plus{} 99 \\equal{} 100 \\equal{} 10^{2}$. Looks interesting: I'll try it later if no-one else answers.\r\n\r\nP.S. Reversing, $ (10 \\plus{} 89)^{2} \\equal{} 9801$.\r\n\r\nP.P.S. I've found two more: $ 2025 \\equal{} (20 \\plus{} 25)^{2}$ and $ 30 \\plus{} 25 \\equal{} (30 \\plus{} 25)^{2}$. Still have nothing general.\r\n\r\nAnd $ 81 \\equal{} (8 \\plus{} 1)^{2} \\equal{} 9^{2}$.", "Solution_2": "Just looked on the internet and found this: http://wapedia.mobi/en/D._R._Kaprekar\r\n\r\nHope this is useful.", "Solution_3": "[quote=\"AndrewTom\"]Probably not much help but $ 0001 \\equal{} (00 \\plus{} 01)^{2} \\equal{} 01^{2}$. Putting this with your example $ 99^{2}$, we have $ 01 \\plus{} 99 \\equal{} 100 \\equal{} 10^{2}$. Looks interesting: I'll try it later if no-one else answers.\n\nP.S. Reversing, $ (10 \\plus{} 89)^{2} \\equal{} 9801$.\n\nP.P.S. I've found two more: $ 2025 \\equal{} (20 \\plus{} 25)^{2}$ and $ 30 \\plus{} 25 \\equal{} (30 \\plus{} 25)^{2}$. Still have nothing general.\n\nAnd $ 0801 \\equal{} (08 \\plus{} 01)^{2} \\equal{} 9^{2}$.[/quote]\r\n\r\n$ 0801\\equal{}9^2$?", "Solution_4": "$ 81 \\equal{} (8 \\plus{} 1)^{2} \\equal{} 9^{2}$.\r\n\r\nEdited." } { "Tag": [ "AoPSwiki", "\\/closed" ], "Problem": "On the solution on http://www.artofproblemsolving.com/Wiki/index.php/2006_IMO_Problems/Problem_5 shouldn't the last line after Lemma 1 have just $ a_i\\plus{}b_i$ instead of $ {a_i\\plus{}b_i}/2$", "Solution_1": "Yes, that must have been a typo, I have now fixed it. But strangely, this is the first time I hear about an IMO solution being posted in the Wiki; I thought the discussion topics were enough.\r\n\r\nCan we harmonize the Wiki with the forum? For instance, every solution will first be posted in the forum, then receive some feedback (e. g. typo corrections, improvements) until it becomes \"publishable\", and then be posted on the Wiki. And I think the Resources Section (e. g. [url=http://www.mathlinks.ro/Forum/resources.php?c=1&cid=16&year=1989]here[/url]) should link to the Wiki as well, not just to the discussion topic in the forum.\r\n\r\n darij", "Solution_2": "Indeed in the [url=http://www.mathlinks.ro/resources.php]resources section[/url] you can add up to three solutions for each problem. Surely not everybody can edit the forum posts except mods/admins but it also a form of protection against vandalism. So my question is: How often do regular non-mod/admin user edit the problems sections of the wiki by adding problems/solutions, correcting errors/typos etc. in the wiki?\r\n\r\nAs Darij I feel it is kind of duplication but I also see the advantages that everybody can edit it. So what about the following idea: When we post problems and/or solutions to the resources section it will automatically also be posted to the wiki? But then we need to think how to synchronize it. Wiki entries should not be overwritten by an update of the resources section but merged. If a certain competition existed in the wiki before the resources section there is also the issue to find this mapping. If you consecutively update a competition in the resources section with problems $ P \\equal{} \\{P_1, \\ldots, P_6\\}$ then the merging process should not turn into adding multiple instance copies of $ P$ to the wiki. But it may be a good idea for a subset of solutions corresponding to problems of $ P$ have changed a lot. Then those items could be combined manually. To simplify synchronization a solution should always first posted to the forum.", "Solution_3": "I agree with orl completely. I spend a huge amount of time doing stuff in the Resources section, so it would be *really* helpful if we didn't have the duplication - there are contests which are in the Wiki but not the Resources section and vice versa. Obviously, it would be difficult to resolve conflicts." } { "Tag": [ "Ross Mathematics Program" ], "Problem": "Where can anybody but you sit? \r\n\r\n\r\nWhy would a cleaning person prefer to wash a mirror \r\nthan a window of the same size, assuming both are easily \r\naccessible? \r\n\r\n\r\nHow can you throw a ball so it will reverse direction and return \r\nto you without the ball bouncing against or touching any \r\nsolid object? \r\n\r\n\r\nYour bed and light switch are 15 feet apart. Without using \r\nany object or mechanical device, no pole, no remote \r\nswitch, how can you turn off the light and get into bed \r\nbefore the room is dark? \r\n\r\n\r\nRusty's Montana cabin is almost buried in snow, and the \r\ntemperature is 25 below zero. Rusty's eyes move from the \r\nsingle match he has in his hand to a candle, an oil lamp and \r\na fireplace with kindling all ready to be lit. Which does he light \r\nfirst? \r\n\r\n\r\nWhat common mechanized transports in New York City \r\n(120,000 of them) carry 3 times as many passengers daily \r\nas all the cities subways, trains, buses, cars, and taxis \r\ncombined? \r\n\r\n\r\nWhy didn't William Howard Taft run for a third term as \r\npresident? \r\n\r\n\r\nWhat is cowhide chiefly used for? \r\n\r\n\r\nCan you arrange these letters into one long word: \r\nd o o r n o n e g w l \r\n\r\n\r\nIf Betsy Ross were living today, what would she be most \r\nnoted for? \r\n\r\nI got these from a Reader's Digest. \r\nPlease hide your answers so other people will have a chance to figure them out on their own. :)", "Solution_1": "Why would a cleaning person prefer to wash a mirror\r\nthan a window of the same size, assuming both are easily\r\naccessible? \r\n[hide]A mirror has only one side, whereas a window has two[/hide]\n\nYour bed and light switch are 15 feet apart. Without using\nany object or mechanical device, no pole, no remote\nswitch, how can you turn off the light and get into bed\nbefore the room is dark? \n[hide]It's daytime[/hide]\n\nRusty's Montana cabin is almost buried in snow, and the\ntemperature is 25 below zero. Rusty's eyes move from the\nsingle match he has in his hand to a candle, an oil lamp and\na fireplace with kindling all ready to be lit. Which does he light\nfirst?\n[hide]The match[/hide]\n\nWhat common mechanized transports in New York City\n(120,000 of them) carry 3 times as many passengers daily\nas all the cities subways, trains, buses, cars, and taxis\ncombined? \n[hide]Is it like pigeons or some bird like that?[/hide]\n\nWhat is cowhide chiefly used for?\n[hide]To be worn on cows of course!!![/hide]", "Solution_2": "Where can anybody but you sit?\r\n[hide]On top of my stomach. Or under my legs. Something like that.[/hide]\nHow can you throw a ball so it will reverse direction and return \nto you without the ball bouncing against or touching any \nsolid object? \n[hide]Throw straight up.[/hide]\nWhy didn't William Howard Taft run for a third term as \npresident?\n[hide]Because the amendment to ban more than two terms was already in effect by then? Just guessing.[/hide]\nCan you arrange these letters into one long word: \nd o o r n o n e g w l \n[hide]one long word.[/hide]\nIf Betsy Ross were living today, what would she be most \nnoted for? \n[hide]Defying the usual limits on the length of human lives?[/hide]", "Solution_3": "[quote=\"LynnelleYe\"]\nWhy didn't William Howard Taft run for a third term as \npresident?\n[hide]Because the amendment to ban more than two terms was already in effect by then? Just guessing.[/hide][/quote]\n\n[hide]I don't think he ever made a second term.[/hide]", "Solution_4": "Where can anybody but you sit? \n\n[hide]Next to you, perhaps?[/hide]\n\n\n\nWhat common mechanized transports in New York City\n\n(120,000 of them) carry 3 times as many passengers daily\n\nas all the cities subways, trains, buses, cars, and taxis\n\ncombined?\n\n[hide]Elevators, presumably.[/hide]\n\n\n\nWhy didn't William Howard Taft run for a third term as\n\npresident?\n\n[hide]Usually one has to become president for two terms before running for a third.[/hide]", "Solution_5": "Where can anybody but you sit? [hide] on my lap[/hide]\n\nHow can you throw a ball so it will reverse direction and return \nto you without the ball bouncing against or touching any \nsolid object? [hide] throw the ball straight up [/hide]\n\n\nYour bed and light switch are 15 feet apart. Without using \nany object or mechanical device, no pole, no remote \nswitch, how can you turn off the light and get into bed \nbefore the room is dark? [hide] just do it during the day when there is light [/hide]\n\n\nRusty's Montana cabin is almost buried in snow, and the \ntemperature is 25 below zero. Rusty's eyes move from the \nsingle match he has in his hand to a candle, an oil lamp and \na fireplace with kindling all ready to be lit. Which does he light \nfirst? [hide] the match [/hide]\n\n\nWhat common mechanized transports in New York City \n(120,000 of them) carry 3 times as many passengers daily \nas all the cities subways, trains, buses, cars, and taxis \ncombined? [hide] elevators? [/hide]\n\nCan you arrange these letters into one long word: \nd o o r n o n e g w l \n[hide] one long word [/hide]", "Solution_6": "Why would a barber want to cut two out-of-towners than one resident of the town?\r\n[hide]\nBecause he would get twice the money[/hide]", "Solution_7": "why did I get a 19-year bump?\n[hide]Because I necropost[/hide]" } { "Tag": [], "Problem": "A group of $ N$ students, where $ N < 50$, is on a field trip. If their teacher puts them in groups of 8, the last group has 5 students. If their teacher instead puts them in groups of 6, the last group has 3 students. What is the sum of all possible values of $ N$?", "Solution_1": "N mod 8 = 5\r\nN mod 6 = 3\r\n\r\nSo, N+3 = 5+3 = 8 = 0 (mod 8)\r\nand N+3 = 3+3 = 6 = 0 (mod 6)\r\n\r\nSo, N+3 is divisible by both 6,8. There are two possible ways: 24 and 48.\r\n\r\nSo, N=21 or 45, so our answer is 66." } { "Tag": [], "Problem": "A new sports stadium has a seating capacity of 24,000 people. One-fifth of the seats are box seats, and $ \\frac58$ of the remaining seats are reserved seats; all other seats are general admission. How many seats are general admission?", "Solution_1": "After the $ \\frac{1}{5}$ of box seats are taken away, $ 1\\minus{}\\frac{1}{5}\\equal{}\\frac{4}{5}$ of the original seats are left. When $ \\frac{5}{8}$ of the remaining are reserved, only $ 1\\minus{}\\frac{5}{8}\\equal{}\\frac{3}{8}$ of the seats that aren't box seats are also not reserved seats. The number of remaining seats, therefore, is $ \\frac{3}{8}(\\frac{4}{5})(24000)\\equal{}\\fbox{7200}$." } { "Tag": [], "Problem": "\u0397 \u03c0\u03b1\u03c1\u03b1\u03ba\u03ac\u03c4\u03c9 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03bf\u03bb\u03cd \u03ba\u03b1\u03bb\u03ae \u03b3\u03b9\u03b1 \u03b4\u03b9\u03b1\u03b3\u03c9\u03bd\u03b9\u03c3\u03bc\u03bf\u03cd\u03c2. \u03a7\u03b1\u03c1\u03b5\u03af\u03c4\u03b5 \u03c4\u03b7\u03bd!", "Solution_1": "\u0395\u03bd\u03b4\u03b9\u03b1\u03c6\u03ad\u03c1\u03bf\u03bd \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03ba\u03b1\u03b9 \u03bc\u03c0\u03bf\u03c1\u03ce \u03bd\u03b1 \u03c0\u03c9 \u03cc\u03c7\u03b9 \u03b5\u03cd\u03ba\u03bf\u03bb\u03bf . \u039d\u03b1 \u03b7 \u03bb\u03cd\u03c3\u03b7 \u03bc\u03bf\u03c5 \r\n\r\n\u0388\u03c3\u03c4\u03c9 $ F,R$ \u03bf\u03b9 \u03c0\u03c1\u03bf\u03b2\u03bf\u03bb\u03ad\u03c2 \u03c4\u03c9\u03bd \u03ba\u03ad\u03bd\u03c4\u03c1\u03c9\u03bd $ O_1,\\ O_2$ \u03c3\u03c4\u03b7\u03bd $ AB$ . \u03a0\u03c1\u03bf\u03c6\u03b1\u03bd\u03ce\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 $ \\angle{DMC}\\equal{}90$ \u03bf\u03c0\u03cc\u03c4\u03b5 \u03b1\u03c1\u03ba\u03b5\u03af \u03bd\u03b1 \u03b4\u03b5\u03af\u03be\u03bf\u03c5\u03bc\u03b5 \r\n\u03cc\u03c4\u03b9 \u03b1\u03bd $ O$ \u03c4\u03bf \u03bc\u03ad\u03c3\u03bf \u03c4\u03bf\u03c5 $ DC$ \u03ba\u03b1\u03b9 $ K$ to \u03bc\u03ad\u03c3\u03bf \u03c4\u03bf\u03c5 $ AB$ \u03c4\u03cc\u03c4\u03b5 $ OK\\equal{}OM$ .\r\n\u0398\u03b5\u03c9\u03c1\u03ce $ P$ \u03c4\u03bf \u03bc\u03ad\u03c3\u03bf \u03c4\u03bf\u03c5 $ O_1O_2$ \u03ba\u03b1\u03b9 $ S_1,\\ S_2$ \u03bf\u03b9 \u03c4\u03bf\u03bc\u03ad\u03c2 \u03c4\u03c9\u03bd $ O_1O_2,\\ DM$ \u03ba\u03b1\u03b9 $ O_1F,\\ DM$ \r\n\r\n\u0398\u03b1 \u03b4\u03b5\u03af\u03be\u03c9 \u03b1\u03c1\u03c7\u03b9\u03ba\u03ac \u03cc\u03c4\u03b9 \u03b7 $ OP$ \u03c0\u03b5\u03c1\u03bd\u03ac \u03b1\u03c0\u03cc \u03c4\u03bf \u03bc\u03ad\u03c3\u03bf $ L$ \u03c4\u03bf\u03c5 FR \u03b4\u03b5\u03af\u03c7\u03bd\u03bf\u03bd\u03c4\u03b1\u03c2 \u03cc\u03c4\u03b9 $ OP//O_1F$ . \r\n\u0388\u03c7\u03bf\u03c5\u03bc\u03b5 \u03c4\u03ce\u03c1\u03b1 \u03cc\u03c4\u03b9 \u03b7 \u03ba\u03ac\u03b8\u03b5\u03c4\u03b7 \u03b1\u03c0\u03cc \u03c4\u03bf $ O$ \u03c3\u03c4\u03b7\u03bd $ DM$ \u03c0\u03b5\u03c1\u03bd\u03ac \u03b1\u03c0\u03cc \u03c4\u03bf \u03bc\u03ad\u03c3\u03bf \u03c4\u03bf\u03c5 $ DM$ \u03ac\u03c1\u03b1 \u03ba\u03b1\u03b9 \u03b1\u03c0\u03cc \u03c4\u03bf $ O_1$ \u03ba\u03b1\u03b9 \u03cc\u03bc\u03bf\u03b9\u03b1 \r\n\u03b7 $ OO_2$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03ba\u03ac\u03b8\u03b5\u03c4\u03b7 \u03c3\u03c4\u03b7\u03bd $ MC$ \u03ac\u03c1\u03b1 \u03b8\u03b1 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 $ \\angle{O_1OO_2}\\equal{}90$ \r\n\r\n\u03a4\u03ce\u03c1\u03b1 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 $ \\angle{O_1S_1S_2}\\equal{}90\\minus{}\\angle{S_1MN}\\equal{}90\\minus{}\\angle{AMD}\\equal{}\\angle{FS_2M}$ \u03ac\u03c1\u03b1 $ O_1S_1S_2$ \u03b9\u03c3\u03bf\u03c3\u03ba\u03b5\u03bb\u03ad\u03c2 \r\n\u03ba\u03b1\u03b9 $ O_1O$ \u03cd\u03c8\u03bf\u03c2 \u03ac\u03c1\u03b1 \u03ba\u03b1\u03b9 \u03b4\u03b9\u03c7\u03bf\u03c4\u03cc\u03bc\u03bf\u03c2 \u03ac\u03c1\u03b1 $ \\angle{S_2O_1O}\\equal{}\\angle{OO_1S_1}\\equal{}\\angle{O_1OP}$ \u03ac\u03c1\u03b1 $ OP//O_1F$ \r\n\u03ba\u03b1\u03b9 \u03c3\u03c5\u03bd\u03b5\u03c0\u03ce\u03c2 $ OP\\bot FR$ \u03ac\u03c1\u03b1 $ OP$ \u03c0\u03b5\u03c1\u03bd\u03ac \u03b1\u03c0\u03cc \u03c4\u03bf \u03bc\u03ad\u03c3\u03bf $ L$ \u03c4\u03bf\u03c5 FR \r\n\r\n\u038c\u03bc\u03c9\u03c2 \u03c4\u03bf $ L$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03ba\u03b1\u03b9 \u03bc\u03ad\u03c3\u03bf \u03c4\u03bf\u03c5 \u039a\u039c \u03bf\u03c0\u03cc\u03c4\u03b5 $ OL$ \u03cd\u03c8\u03bf\u03c2 \u03ba\u03b1\u03b9 \u03b4\u03b9\u03ac\u03bc\u03b5\u03c3\u03bf\u03c2 \u03c3\u03c4\u03bf \u039a\u039f\u039c \u03ac\u03c1\u03b1 \u039a\u039f=\u039a\u039c , \u03c4\u03bf \u03b6\u03b7\u03c4\u03bf\u03cd\u03bc\u03b5\u03bd\u03bf", "Solution_2": "\u03a3\u03b9\u03bb\u03bf\u03c5\u03b1\u03bd\u03ad , \r\n \u03c0\u03bf\u03bb\u03cd \u03ba\u03b1\u03bb\u03ae \u03c3\u03ba\u03ad\u03c8\u03b7 \u03ba\u03b1\u03b9 \u03c9\u03c1\u03b1\u03af\u03b1 \u03bb\u03cd\u03c3\u03b7 :lol: . \u0398\u03b1 \u03c3\u03c4\u03b5\u03af\u03bb\u03bb\u03c9 \u03bc\u03b9\u03b1 \u03c0\u03b1\u03c1\u03b1\u03bb\u03bb\u03b1\u03b3\u03ae \u03c3\u03c4\u03bf \u03b1\u03b3\u03b3\u03bb\u03b9\u03ba\u03cc \u03c6\u03cc\u03c1\u03bf\u03c5\u03bc , \u03b1\u03bb\u03bb\u03ac \u03c3\u03b5 \u03c0\u03b1\u03c1\u03b1\u03ba\u03b1\u03bb\u03ce \u03bd\u03b1 \u03bc\u03b7 \u03c3\u03c4\u03b5\u03af\u03bb\u03bb\u03b5\u03b9\u03c2 \u03bb\u03cd\u03c3\u03b7 , \u03bc\u03ad\u03c7\u03c1\u03b9 \u03bd\u03b1 \u03bc\u03b1\u03b6\u03ad\u03c8\u03bf\u03c5\u03bc\u03b5 \u03ba\u03b1\u03b9 \u03ba\u03b1\u03bc\u03b9\u03ac \u03ac\u03bb\u03bb\u03b7 \u03b9\u03b4\u03ad\u03b1.\r\n \u039a\u03ac\u03c0\u03bf\u03b9\u03b1 \u03c3\u03c4\u03b9\u03b3\u03bc\u03ae , \u03b1\u03bd \u03b8\u03b5\u03bb\u03ae\u03c3\u03b5\u03c4\u03b5 , \u03b8\u03b1 \u03c3\u03c4\u03b5\u03af\u03bb\u03bb\u03c9 \u03ba\u03b1\u03b9 \u03bc\u03b9\u03b1 \u03ac\u03bb\u03bb\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03c0\u03bf\u03c5 \u03ad\u03c3\u03c4\u03b5\u03b9\u03bb\u03b1 \u03c3\u03c4\u03bf CRUX \u03b3\u03b9\u03b1 \u03c4\u03bf \u03b8\u03ad\u03bc\u03b1 \u03b1\u03c5\u03c4\u03cc. \u0391\u039d \u03b8\u03ad\u03bb\u03b5\u03b9\u03c2 \u03b3\u03c1\u03ac\u03c8\u03b5 \u03c4\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03c3\u03bf\u03c5 \u03c3\u03c4\u03b7\u03bd \u03b1\u03b3\u03b3\u03bb\u03b9\u03ba\u03ae \u03bc\u03b5 \u03c4\u03bf \u03cc\u03bd\u03bf\u03bc\u03ac \u03c3\u03bf\u03c5 \u03b1\u03c0\u03cc \u03ba\u03ac\u03c4\u03c9(\u03ba\u03b1\u03b9 \u03c4\u03bf \u03c3\u03c7\u03ae\u03bc\u03b1) \u03ba\u03b1\u03b9 \u03c3\u03c4\u03b5\u03af\u03bb\u03c4\u03bf (\u03ae \u03c3\u03c4\u03b5\u03af\u03bb\u03c4\u03bf \u03c3\u03b5 \u03bc\u03ad\u03bd\u03b1 \u03bd\u03b1 \u03c4\u03bf \u03ba\u03ac\u03bd\u03c9 \u03b5\u03b3\u03c9) \u03c3\u03c4\u03bf\u03bd R.Woodrow \u03c4\u03bf\u03c5 CRUX [2007:411], moldavian olympiad\r\n\r\n \u039c\u03c0\u03ac\u03bc\u03c0\u03b7\u03c2", "Solution_3": "\u0388\u03be\u03c5\u03c0\u03bd\u03b7 \u03bb\u03cd\u03c3\u03b7, \u03a3\u03b9\u03bb\u03bf\u03c5\u03b1\u03bd\u03ad!\r\n\u039a\u03b1\u03bb\u03ae \u03c7\u03c1\u03bf\u03bd\u03b9\u03ac \u03c3\u03b5 \u03cc\u03bb\u03bf\u03c5\u03c2! :) \r\n\r\n\r\n\u0397 \u03b2\u03b1\u03c3\u03b9\u03ba\u03ae \u03b9\u03b4\u03ad\u03b1 \u03c4\u03b7\u03c2 \u03bb\u03cd\u03c3\u03b7\u03c2 \u03c0\u03bf\u03c5 \u03c0\u03c1\u03bf\u03c4\u03b5\u03af\u03bd\u03c9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03c4\u03bf $ A$ \u03ad\u03c7\u03b5\u03b9 \u03b4\u03b9\u03c0\u03bb\u03ac\u03c3\u03b9\u03b1 \u03b4\u03cd\u03bd\u03b1\u03bc\u03b7 \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 \u03c4\u03bf\u03bd $ \\Gamma_2$ \u03c3\u03b5 \u03c3\u03c7\u03ad\u03c3\u03b7 \u03bc\u03b5 \u03c4\u03b7 \u03b4\u03cd\u03bd\u03b1\u03bc\u03b7 \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 \u03c4\u03bf\u03bd \u03ba\u03cd\u03ba\u03bb\u03bf \u03bc\u03b5 \u03b4\u03b9\u03ac\u03bc\u03b5\u03c4\u03c1\u03bf $ DC$. \r\n\r\n\r\n\u0388\u03c3\u03c4\u03c9 $ \\phi \\equal{} \\widehat {NMC} \\equal{} \\widehat {CMB}$. \u0391\u03c0\u03cc \u03c4\u03bf \u03b5\u03b3\u03b3\u03b5\u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03bd\u03bf \u03c3\u03c4\u03bf\u03bd $ \\Gamma_1$ \u03c4\u03b5\u03c4\u03c1\u03ac\u03c0\u03bb\u03b5\u03c5\u03c1\u03bf $ MADN$ \u03c0\u03c1\u03bf\u03ba\u03cd\u03c0\u03c4\u03b5\u03b9 $ \\widehat {DAN} \\equal{} \\widehat {MNB} \\equal{} 2\\phi \\ (1)$\r\n\r\n\r\n\u0397 $ AC$ \u03c4\u03ad\u03bc\u03bd\u03b5\u03b9 \u03c4\u03bf\u03bd \u03ba\u03cd\u03ba\u03bb\u03bf $ \\Gamma_2$ \u03c3\u03c4\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf $ L$\r\n\r\n\u0395\u03af\u03bd\u03b1\u03b9 $ \\widehat {NLC} \\equal{} \\phi \\equal{} > \\widehat {ALN} \\equal{} \\pi \\minus{} \\phi \\ \\ (2)$\r\n\u0391\u03c6\u03bf\u03cd \u03c4\u03bf $ D$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03bc\u03ad\u03c3\u03bf\u03bd \u03c4\u03bf\u03c5 \u03c4\u03cc\u03be\u03bf\u03c5 $ AN$, \u03b8\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 $ DA \\equal{} DN$. \u0388\u03c3\u03c4\u03c9 $ d$ \u03bf \u03ba\u03cd\u03ba\u03bb\u03bf\u03c2 \u03bc\u03b5 \u03ba\u03ad\u03bd\u03c4\u03c1\u03bf \u03c4\u03bf $ D$ \u03ba\u03b1\u03b9 \u03b1\u03ba\u03c4\u03af\u03bd\u03b1 $ DA$.\r\n\u0391\u03c0\u03cc \u03c4\u03b9\u03c2 $ (1),(2)$ \u03c0\u03c1\u03bf\u03ba\u03cd\u03c0\u03c4\u03b5\u03b9 \u03cc\u03c4\u03b9 \u03c4\u03bf $ L$ \u03b1\u03bd\u03ae\u03ba\u03b5\u03b9 \u03c3\u03c4\u03bf\u03bd \u03ba\u03cd\u03ba\u03bb\u03bf $ d$\r\n\r\n[img]11902[/img]\r\n\u0388\u03c3\u03c4\u03c9 \u03c4\u03ce\u03c1\u03b1 $ K$ \u03b7 \u03c0\u03c1\u03bf\u03b2\u03bf\u03bb\u03ae \u03c4\u03bf\u03c5 $ D$ \u03c0\u03ac\u03bd\u03c9 \u03c3\u03c4\u03b7\u03bd $ AL$. \u03a3\u03cd\u03bc\u03c6\u03c9\u03bd\u03b1 \u03bc\u03b5 \u03c4\u03b1 \u03c0\u03c1\u03bf\u03b7\u03b3\u03bf\u03cd\u03bc\u03b5\u03bd\u03b1, \u03c4\u03bf $ K$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03bc\u03ad\u03c3\u03bf\u03bd \u03c4\u03bf\u03c5 $ AL$.\r\n\r\n\u039f \u03ba\u03cd\u03ba\u03bb\u03bf\u03c2 $ k$ \u03bc\u03b5 \u03b4\u03b9\u03ac\u03bc\u03b5\u03c4\u03c1\u03bf $ DC$ \u03b4\u03b9\u03ad\u03c1\u03c7\u03b5\u03c4\u03b1\u03b9 \u03c0\u03c1\u03bf\u03c6\u03b1\u03bd\u03ce\u03c2 \u03b1\u03c0\u03cc \u03c4\u03bf $ K$ \u03ba\u03b1\u03b9 \u03c4\u03bf $ M$. \u0388\u03c3\u03c4\u03c9 \u03cc\u03c4\u03b9 \u03c4\u03ad\u03bc\u03bd\u03b5\u03b9 \u03c4\u03b7\u03bd \u03b5\u03c5\u03b8\u03b5\u03af\u03b1 $ AB$ \u03c3\u03b5 \u03ad\u03bd\u03b1 \u03b1\u03ba\u03cc\u03bc\u03b7 \u03c3\u03b7\u03bc\u03b5\u03af\u03bf $ P$\r\n\r\n\r\n\u0394\u03cd\u03bd\u03b1\u03bc\u03b7 \u03c4\u03bf\u03c5 $ A$ \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 \u03c4\u03bf\u03bd \u03ba\u03cd\u03ba\u03bb\u03bf $ \\Gamma_2: \\ AM \\cdot AB \\equal{} AL \\cdot AC$\r\n\r\n\u0394\u03cd\u03bd\u03b1\u03bc\u03b7 \u03c4\u03bf\u03c5 $ A$ \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 \u03c4\u03bf\u03bd \u03ba\u03cd\u03ba\u03bb\u03bf $ k: \\ AM \\cdot AP \\equal{} AK \\cdot AC$\r\n\r\n\u0394\u03b9\u03b1\u03b9\u03c1\u03ce\u03bd\u03c4\u03b1\u03c2 \u03ba\u03b1\u03c4\u03ac \u03bc\u03ad\u03bb\u03b7 \u03c0\u03b1\u03af\u03c1\u03bd\u03bf\u03c5\u03bc\u03b5 $ \\frac {AB}{AP} \\equal{} 2$\r\n\r\n\r\n-----------------------------------------------------------------------------------------\r\n[b]\u03a0\u03b1\u03c1\u03b1\u03c4\u03ae\u03c1\u03b7\u03c3\u03b7![/b]\r\n\r\n\u0391\u03bd \u03c4\u03bf $ M$ \u03b2\u03c1\u03af\u03c3\u03ba\u03b5\u03c4\u03b1\u03b9 \u03b5\u03ba\u03c4\u03cc\u03c2 \u03c4\u03bf\u03c5 \u03b5\u03c5\u03b8\u03c5\u03b3\u03c1\u03ac\u03bc\u03bc\u03bf\u03c5 \u03c4\u03bc\u03ae\u03bc\u03b1\u03c4\u03bf\u03c2 $ AB$, \u03c6\u03b1\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03b4\u03b5\u03bd \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9. \u0394\u03b5\u03bd \u03c4\u03bf \u03ad\u03c8\u03b1\u03be\u03b1 \u03b1\u03bb\u03bb\u03b1 \u03bf\u03c1\u03af\u03c3\u03c4\u03b5 \u03ad\u03bd\u03b1 \u03b1\u03bd\u03c4\u03b9\u03c0\u03b1\u03c1\u03ac\u03b4\u03b5\u03b9\u03b3\u03bc\u03b1 (S \u03bc\u03ad\u03c3\u03bf\u03bd \u03c4\u03bf\u03c5 AB):\r\n\r\n[img]11901[/img]\r\n[i]\u0395\u03be\u03ac\u03bb\u03bb\u03bf\u03c5, \u03c3\u03c4\u03b7\u03bd \u03c0\u03b5\u03c1\u03af\u03c0\u03c4\u03c9\u03c3\u03b7 \u03b1\u03c5\u03c4\u03ae \u03bf \u03ba\u03cd\u03ba\u03bb\u03bf\u03c2 d \u03b4\u03b5\u03bd \u03c0\u03b5\u03c1\u03bd\u03ac\u03b5\u03b9 \u03bf\u03cd\u03c4\u03b5 \u03ba\u03b1\u03bd \u03b1\u03c0\u03cc \u03c4\u03bf M\n\u038c\u03c0\u03c9\u03c2 \u03c4\u03bf \u03c0\u03b1\u03c1\u03b1\u03c4\u03b7\u03c1\u03ce \u03cc\u03bc\u03c9\u03c2 \u03ad\u03c7\u03b5\u03b9 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf \u03b5\u03bd\u03b4\u03b9\u03b1\u03c6\u03ad\u03c1\u03bf\u03bd \u03ba\u03b1\u03b9 \u03af\u03c3\u03c9\u03c2 \u03b8\u03b1 \u03ae\u03c4\u03b1\u03bd \u03ba\u03b1\u03bb\u03cc \u03bd\u03b1 \u03c4\u03bf \u03b4\u03bf\u03cd\u03bc\u03b5 \u03c3\u03b1\u03bd \u03be\u03b5\u03c7\u03c9\u03c1\u03b9\u03c3\u03c4\u03ae \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7[/i]\r\n\r\n\u039b\u03bf\u03c5\u03ba\u03ac\u03c2" } { "Tag": [ "search", "number theory unsolved", "number theory" ], "Problem": "Let $ A\\equal{}\\{2^{2^n}\\plus{}1|n\\in\\mathbb{N}\\}, B\\equal{}\\{6^{2^n}\\plus{}1|n\\in\\mathbb{N}\\}$. Prove that $ A\\cup B$ contains infinitely many composite numbers.", "Solution_1": "[url]http://www.mathlinks.ro/viewtopic.php?search_id=1921896294&t=143722[/url]" } { "Tag": [ "group theory", "abstract algebra", "modular arithmetic", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $ G$ be a finite group, en assume that every maximal subgroup $ M \\leq G$ has the property that $ [G : M]$ is prime. \r\nLet $ p$ be the biggest prime divisor of $ \\left|G\\right|$, and let $ P$ be a Sylow $ p$-subgroup of $ G$.\r\n\r\nProve that $ P \\trianglelefteq G$.", "Solution_1": "[hide=\"Some Progress\"]Suppose $ P \\ntrianglelefteq G$. Then $ N(P) \\neq G$, so let $ M \\ge N(P)$ be a maximal subgroup. Then we have that $ [G: M] \\equal{} q$, where $ q < p$ is prime.\n\nLetting $ G$ act on the left cosets of $ M$, we get a homomorphism from $ G$ into $ S_q$ with kernel $ K$ equal to the largest normal subgroup of $ G$ contained in $ M$. Since $ p \\nmid q!$, we must have $ P \\le K$. Now, since $ K$ is normal and contains $ P$, it must contain all $ p$-Sylow subgroups of $ G$, of which there are \n$ [G: N(P)]$. This means that $ [K: K \\cap N(P)] \\equal{} [G: N(P)]$.[/hide]", "Solution_2": "Ok, you have very good ideas! I don't know if you can complete them to a full solution though, mine was different.\r\nHere's something you might try and which will work: What can you say about the Sylow $ p$-subgroups of $ M$ ? :)", "Solution_3": "Let $ N$ be the normaliser of $ P$ in $ G$. Suppose that $ N\\neq G$ and let $ M$ be a maximal subgroup of $ G$ containing $ N$. Then the number of Sylow $ p$-subgroups in $ G$ is equal to $ \\frac{|G|}{|N|} \\equal{} \\frac{ q\\cdot |M|}{|N|} \\equiv 1 \\pmod p$ for some prime number $ q$ dividing $ |G|$. Clearly $ q\\neq p$. \r\nBut $ M$ contains $ N$, so the normaliser of $ P$ in $ M$ is equal to $ N$. It follows that the number of Sylow $ p$-subgroups in $ M$ is equal to $ \\frac{|M|}{|N|} \\equiv 1\\pmod p$.\r\nIt follows that $ q\\equiv 1\\pmod p$, clearly contradicting $ p$ being the biggest prime divisor of $ |G|$.", "Solution_4": "Exactly what I answered Yimin, so I think and hope this is correct. \r\nYou wrote it down short and elegant!" } { "Tag": [ "function" ], "Problem": "Prove that there are no integers $ x>1, y>1, z>1$ such that $ x!+y!=z!$", "Solution_1": "Assume WLOG $ x \\geq y$ so $ x!+y! \\leq 2x! <(x+1)!$ and at the same time $ x!+y! > x!$ so we have $ x!1,y>1,z>1$.\r\n\r\nEdit: Hope my reasoning is correct. Also is the integer restriction necessary? Because don't $ !$ only apply for non-negative integers anyway?\r\n\r\nThis reminded me of this problem: Find all solutions to $ x!+y!+z! = t!$.", "Solution_2": "[quote=\"BanishedTraitor\"]Also is the integer restriction necessary? Because don't $ !$ only apply for non-negative integers anyway?[/quote]\r\n\r\n[url]http://en.wikipedia.org/wiki/Factorial#The_gamma_function[/url]" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "for all positive real $a,b,c$ such that $abc=1$ prove that\r\n\r\n$\\frac{a^{2}-4a+3}{a^{2}+4a+3}+\\frac{b^{2}-4b+3}{b^{2}+4b+3}+\\frac{c^{2}-4c+3}{c^{2}+4c+3}\\geq 0$", "Solution_1": "[quote=\"shyong\"]for all positive real $a,b,c$ such that $abc=1$ prove that\n\n$\\frac{a^{2}-4a+3}{a^{2}+4a+3}+\\frac{b^{2}-4b+3}{b^{2}+4b+3}+\\frac{c^{2}-4c+3}{c^{2}+4c+3}\\geq 0$[/quote]\r\n$\\frac{a^{2}-4a+3}{a^{2}+4a+3}+\\frac{b^{2}-4b+3}{b^{2}+4b+3}+\\frac{c^{2}-4c+3}{c^{2}+4c+3}\\geq 0\\Leftrightarrow$\r\n$\\Leftrightarrow\\sum_{cyc}(9a^{2}b^{2}+12a^{2}b+12a^{2}c+27a^{2}-44ab+20a-36)\\geq0\\Leftrightarrow$\r\n$\\Leftrightarrow\\sum_{cyc}(9a^{2}b^{2}+a^{2}b+b^{2}a+6.5a^{2}+6.5b^{2}+10a+10b-44ab)+$\r\n$+\\sum_{cyc}(11a^{2}b+11a^{2}c+14a^{2}-36)\\geq0,$ which obvious.", "Solution_2": "[quote=\"arqady\"][quote=\"shyong\"]for all positive real $a,b,c$ such that $abc=1$ prove that\n\n$\\frac{a^{2}-4a+3}{a^{2}+4a+3}+\\frac{b^{2}-4b+3}{b^{2}+4b+3}+\\frac{c^{2}-4c+3}{c^{2}+4c+3}\\geq 0$[/quote]\n$\\frac{a^{2}-4a+3}{a^{2}+4a+3}+\\frac{b^{2}-4b+3}{b^{2}+4b+3}+\\frac{c^{2}-4c+3}{c^{2}+4c+3}\\geq 0\\Leftrightarrow$\n$\\Leftrightarrow\\sum_{cyc}(9a^{2}b^{2}+12a^{2}b+12a^{2}c+27a^{2}-44ab+20a-36)\\geq0\\Leftrightarrow$\n$\\Leftrightarrow\\sum_{cyc}(9a^{2}b^{2}+a^{2}b+b^{2}a+6.5a^{2}+6.5b^{2}+10a+10b-44ab)+$\n$+\\sum_{cyc}(11a^{2}b+11a^{2}c+14a^{2}-36)\\geq0,$ which obvious.[/quote]\r\n\r\nWow :o . you are really an expanding expert :D\r\n\r\nSo what do you think with 4 variables $abcd=1$ and \r\n\r\n$E(a,b,c,d)\\geq 0$ where $E(a)=\\frac{a^{2}-4a+3}{a^{2}+4a+3}$\r\n\r\nI dont have a proof by I conjecture that it is true ." } { "Tag": [ "algebra", "polynomial", "induction", "superior algebra", "superior algebra solved", "first college math post" ], "Problem": "If A,B are invertible and the set {Ak - Bk | k is a natural number} is finite , then there exists a natural number m such that Am = Bm.", "Solution_1": "seems to me that you fogot to mention what A and B are? I suppose matrixes with real entries ... ?", "Solution_2": "Yes , A and B are square matrices of size n with complex entries. Sorry , I was a bit in a rush last night :)", "Solution_3": "So, let's denote Ak-Bk by Ck. Thus, we have C0=0 and we must find a Cr, with r>0, that equals 0.\r\n\r\nLet us consider the sequences (Ci,Ci+1,...,Ci+n), for all i \\geq 0. Because the Ck's take on finitely many values, therefore there are only finitely many values that the sequences may take on, because n is fixed (being the order of the matrices). Therefore we will be able to find m \\geq 0 and r>0 such that:\r\n(Cm,Cm+1,...,Cm+n)= (Cm+r,Cm+r+1,...,Cm+r+n).\r\nThus we have Cm=Cm+r, Cm+1=Cm+r+1, ..., Cm+n=cm+n+r.\r\n\r\nNow comes the interesting part.\r\nLet Xn+an-1Xn-1+...+a0 and Let Xn+bn-1Xn-1+...+b0 be the characteristic polynomials of the matrices A and B, respectively. Because A and B are invertible, and (-1)na0 is the determinant of A, and (-1)nb0 that of B, a0 and b0 will be different from 0. According to the Cayley-Hamilton theorem, we have:\r\n\r\nAn+an-1An-1+...+a0=0 and\r\n\r\nBn+bn-1Bn-1+...+b0=0, so for any k \\geq 0 we have\r\n\r\nAn+k+an-1An+k-1+...+a0Ak=0 (1) and\r\nBn+k+bn-1Bn+k-1+...+b0Bk=0, which leads to\r\n\r\nBn+k+an-1Bn+k-1+...+a0Bk = (an-1-bn-1)Bn+k-1+...+(a0-b0)Bk=Bk(-X) (2), \r\nwhere -X= (an-1-bn-1)Bn-1+...+(a0-b0).\r\n\r\nBy substracting (2) from (1) we obtain, for any k \\geq 0, the following:\r\n\r\nCn+k+an-1Cn+k-1+...+a0Ck=BkX.\r\n\r\nNow let's return to the m and r we found at the beginning of the solution. We have\r\nCn+m+an-1Cn+m-1+...+a0Cm=BmX and\r\nCn+m+r+an-1Cn+m+r-1+...+a0Cm+r=Bm+rX\r\n\r\nBut for any i with 0 \\leq i \\leq n we have Cm+i=Cm+r+i. \r\nTherefore by substracting the two equalities above we have Bm(Br-1)X=0.\r\nBecause B is invertible we have (Br-1)X=0.\r\n\r\nNow, we know that \r\nCn+m-1+an-1Cn+m-2+...+a0Cm-1=Bm-1X and \r\n\r\nCn+m+r-1+an-1Cn+m+r-2+...+a0Cm+r-1=Bm+r-1X. \r\n\r\nBy substracting we have\r\na0(Cm+r-1-Cm-1)=Bm-1(Br-1)X=0, therefore Cm+r-1=Cm-1. \r\nIn the same manner we obtain Cm+r-2=Cm-2,..., and by induction, in the end we will have Cr=C0=0, which completes the proof.\r\n\r\n______________________________________________________", "Solution_4": "Funny thing ! It seems like this is the only solution , in the book (the dreaded Caragea:D) there is a solution in the same manner , slightly simplified.", "Solution_5": "nevertheless andrei negut's solution look nice to me ... i would have probably cooked up the same solution :) well done andrei", "Solution_6": "[quote=Valentin Vornicu]nevertheless andrei negut's solution look nice to me ... i would have probably cooked up the same solution :) well done andrei[/quote]\n\nyeah nice job :)", "Solution_7": "[quote=iandrei]If A,B are invertible and the set {Ak - Bk | k is a natural number} is finite , then there exists a natural number m such that Am = Bm.[/quote]\n\nwhat does mean????", "Solution_8": "[quote=mathpro12345][quote=iandrei]If A,B are invertible and the set {Ak - Bk | k is a natural number} is finite , then there exists a natural number m such that Am = Bm.[/quote]\n\nwhat does mean????[/quote]\n\nits actually [code][sup][/sup][/code] but it is in AoPS wiki formatting :)" } { "Tag": [ "geometry", "geometric transformation", "rotation", "real analysis", "real analysis unsolved" ], "Problem": "Can someone please help me with this question? I can do nothing more for this than draw pictures at the moment.\r\n\r\nLet $ x, y \\in \\mathbb{R}^n$ ($ n \\geq 3$) and $ \\left\\Vert x\\minus{}y \\right\\Vert\\equal{}d>0$ according to the standard euclidean (\"dot product\") norm. Set $ r>0$ and let $ Z\\equal{}\\{z: \\left\\Vert z\\minus{}x\\right\\Vert\\equal{}\\left\\Vert z\\minus{}y\\right\\Vert\\equal{}r \\}$. Then,\r\n$ 2r>d \\Rightarrow Z$ is infinite,\r\n$ 2r\\equal{}d \\Rightarrow |Z|\\equal{}1$.\r\n\r\nThanks for any help.", "Solution_1": "By a rotation and translation, you can assume $ x\\equal{}(\\minus{}d/2,0,0,\\cdots)$, $ y\\equal{}(d/2,0,0,\\cdots)$.", "Solution_2": "Ok so then we have (nothing changes for n>3)\r\n\r\n$ \\left( \\left(z_1\\plus{}\\frac{d}{2} \\right) ^2 \\plus{} z_2^2 \\plus{} z_3^2 \\right) ^{1/2} \\equal{} \\left( \\left(z_1\\minus{}\\frac{d}{2} \\right) ^2 \\plus{} z_2^2 \\plus{} z_3^2 \\right) ^{1/2}$\r\n$ \\Rightarrow \\left|z_1\\plus{}\\frac{d}{2} \\right| \\equal{} \\left|z_1 \\minus{}\\frac{d}{2} \\right| \\Rightarrow z_1\\equal{}0$\r\n\r\nThen $ 2r\\equal{}d \\Rightarrow 4\\left( \\left(\\frac{d}{2} \\right)^2 \\plus{} z_2^2 \\plus{} z_3^2 \\right) \\equal{} d^2 \\Rightarrow z_2\\equal{}z_3\\equal{}0$\r\nand $ 2r > d \\Rightarrow 4\\left( \\left(\\frac{d}{2} \\right)^2 \\plus{} z_2^2 \\plus{} z_3^2 \\right) > d^2 \\Rightarrow z_2\\plus{}z_3>0$.\r\n\r\nAnd that seems to be all... Thanks for the tip." } { "Tag": [ "function", "calculus", "derivative", "probability", "Functional Analysis", "topology", "real analysis" ], "Problem": "give example of function which is continuous at all real pts. but differentiable nowhere", "Solution_1": "E.g. $f(x)=\\sum_{n=0}^{\\infty}\\frac{\\cos(8^{n}x)}{2^{n}}$", "Solution_2": "am interested in knowing other example your example is well known.can we in general say something of such class of func.?", "Solution_3": "I don't know of any compelling reason to be interested in this class, but here is a curious theorem: \"Every separable Banach space is isometric to a space of continuous nowhere differentiable functions\". This is actually the title of a 1995 paper by L. Rodr\u00edguez-Piazza. It is a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=111284]classical theorem[/url] of Banach and Mazur that any separable Banach space is isometric to a linear subspace of $C[0,1]$. Rodr\u00edguez-Piazza shows that the linear subspace can be chosen so that all of its elements, except for zero function, are nowhere differentiable.", "Solution_4": "There are \"more\" nowhere differentiable continuous functions than there are continuous functions with a derivative at some point.\r\n\r\nYou can't use cardinality for this - the sets you're interested in have the same cardinality. But there are at least two other senses of \"more\": Baire category and measure.\r\n\r\nBaire category: the set of all contionuous functons on $[a,b]$ that have a derivative at any point can be contained is a countable union of nowhere dense subsets of the metric space $C[a,b].$\r\n\r\nMeasure (probability): Construct a probability measure on the set of continuous functions with a certain fixed value at $a$ by constructing Brownian motion. The probability that such a continuous function has a derivative at any point is zero." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,b,c$ be positive real numbers such that\r\n\r\n$ abc \\ge 1$\r\n\r\nProve 2 following inequalities\r\n\r\n\r\n[quote]$ \\dfrac{1}{{3{a^4} \\plus{} 4{b^4} \\plus{} 5}} \\plus{} \\dfrac{1}{{3{b^4} \\plus{} 4{c^4} \\plus{} 5}} \\plus{} \\dfrac{1}{{3{c^4} \\plus{} 4{a^4} \\plus{} 5}} \\le \\dfrac{1}{4}$[/quote]\n\n \n [quote]$ \\dfrac{1}{{{a^7} \\plus{} {b^5} \\plus{} 1}} \\plus{} \\dfrac{1}{{{b^7} \\plus{} {c^5} \\plus{} 1}} \\plus{} \\dfrac{1}{{{c^7} \\plus{} {a^5} \\plus{} 1}} \\le 1$[/quote]", "Solution_1": "[quote=\"leviethai\"]Let $ a,b,c$ be positive real numbers such that\n\n$ abc \\ge 1$\n\nProve 2 following inequalities\n\n\n[quote]$ \\dfrac{1}{{3{a^4} \\plus{} 4{b^4} \\plus{} 5}} \\plus{} \\dfrac{1}{{3{b^4} \\plus{} 4{c^4} \\plus{} 5}} \\plus{} \\dfrac{1}{{3{c^4} \\plus{} 4{a^4} \\plus{} 5}} \\le \\dfrac{1}{4}$[/quote]\n\n \n [quote]$ \\dfrac{1}{{{a^7} \\plus{} {b^5} \\plus{} 1}} \\plus{} \\dfrac{1}{{{b^7} \\plus{} {c^5} \\plus{} 1}} \\plus{} \\dfrac{1}{{{c^7} \\plus{} {a^5} \\plus{} 1}} \\le 1$[/quote][/quote]\r\nThe first inequality follows from the two known inequalities:\r\n$ \\frac{1}{a^4\\plus{}b^4\\plus{}1}\\plus{}\\frac{1}{b^4\\plus{}c^4\\plus{}1}\\plus{}\\frac{1}{c^4\\plus{}a^4\\plus{}1} \\le 1,$\r\n$ \\frac{1}{a^4\\plus{}2}\\plus{}\\frac{1}{b^4\\plus{}2}\\plus{}\\frac{1}{c^4\\plus{}2} \\le 1.$\r\n\r\nFor the second one, see my proof here: http://can-hang2007.blogspot.com/2010/01/inequality-38-l-v-hai.html", "Solution_2": "[quote=\"leviethai\"]Let $ a,b,c$ be positive real numbers such that\n\n$ abc \\ge 1$\n\nProve 2 following inequalities\n\n$ \\dfrac{1}{{3{a^4} \\plus{} 4{b^4} \\plus{} 5}} \\plus{} \\dfrac{1}{{3{b^4} \\plus{} 4{c^4} \\plus{} 5}} \\plus{} \\dfrac{1}{{3{c^4} \\plus{} 4{a^4} \\plus{} 5}} \\le \\dfrac{1}{4}$\n \n[/quote]\r\n\r\nSee the general problem here (#28). :) \r\nhttp://www.maths.vn/forums/showthread.php?t=16412&page=3\r\nYour problem is the case with $ k \\equal{} \\frac{1}{3}$." } { "Tag": [ "ratio", "geometric sequence", "AMC" ], "Problem": "Positive integers $ a$, $ b$, and $ 2009$, with $ a $ 2^p\\plus{}p^2\\equal{}3$ !!!!!!!\r\n\r\nso $ p\\equal{}3$", "Solution_3": "[quote=\"rogerlee\"]\n5, Find all prime numbers p for which $ 2^p \\plus{} p^2$ is prime.\n[/quote]\nIf $ p > 3$, then p is odd and $ 2^p \\plus{} p^2$ is divisible by 3, but $ 2^p \\plus{} p^2 > 3$.\nIt easy to check, that p=2 not solution, and p=3 - solution.\n[quote=\"rogerlee\"]\n7, If $ p$ is any prime other than $ 2$ or $ 5$, prove that $ p$ divides infinitely many of the integers $ 9, 99, 999, 9999,$ \u2026.[/quote]\r\nIn sequence 9, 99, 999, ... there are exist pair (x, y) that x=y (mod p), because the set {0, 1, ..., p-1} is finite.\r\nSo, $ (x \\minus{} y)/10^k$ in this sequence and divisible by $ p$ (here $ k$ - the number of zeros in the end of $ x \\minus{} y$).\r\n\r\nSo, we can see, there exist $ z \\equal{} 999...9 \\equal{} 10^t \\minus{} 1$, which is divisible by $ p$.\r\n\r\nThus, every element in sequence $ z_1 \\equal{} z \\equal{} 10^t \\minus{} 1$, $ z_n \\equal{} 10^{nt} \\minus{} 1$ is divisible by $ p$.", "Solution_4": "[quote=\"rogerlee\"]\n6, Prove if a, b, c, d are integers satisfying\n\\[ (3a \\plus{} 5b)(7b \\plus{} 11c)(13c \\plus{} 17d)(19d \\plus{} 23a) \\equal{} 2001^{2001}\n\\]\nthen a must be even.\n[/quote]\r\nIf $ xyzt \\equal{} 2001^{2001}$, then $ xyzt\\equal{}1$ $ (mod 4)$\r\nSo, $ x \\plus{} y \\plus{} z \\plus{} t$ is divisible by 4.\r\nIn this problem we have $ x \\plus{} y \\plus{} z \\plus{} t \\equal{} 26a \\plus{} 12b \\plus{} 24c \\plus{} 36d$, so $ a$ is even.", "Solution_5": "[quote=\"rogerlee\"]\n3, Determine the largest of all integers n with the property that n is divisible by all positive integers that are less than $ ^3\\sqrt n$.\n[/quote]\r\nSuppose $ n > 512$.\r\n$ n$ divisible by $ 8*7*5*3 = 840$, so $ n\\geq840$.\r\n$ ^3\\sqrt 840 > 9$.\r\nSo, we see, that $ n$ divisible by $ 8*7*5*9 = 2520$, and $ n$ must be divisible by $ 11$.\r\n\r\nLet $ p$ - the largest prime number, less then $ ^3\\sqrt n$. \r\nIt's well-known, that $ \\frac {^3\\sqrt n}{2}\\leq p < ^3\\sqrt n$.\r\nWe have $ n\\leq 8p^3 (1)$.\r\nLet $ p_1 = 5, p_2 = 7, ..., p_k, p$ - prime numbers, and $ n$ must be divisible by $ p_1, p_2, ..., p_k, p$.\r\n\r\nSo, $ n \\geq 9*(2p_k)*(4p_{k - 1})*p \\geq 9*p*p*p > 8p^3$ - if $ p \\geq 11$ then $ p_k$ and $ p_{k - 1}$ exists both. \r\nContradiction with (1).\r\nSo, we have $ n\\leq 512$.\r\n\r\nThe answer is $ n = 420$.\r\nIf $ n > 420$, then $ n$ divisible by 3*4*5*7=420, which impossible in $ 420 < n\\leq512$.", "Solution_6": "4c) By Bertrand Postulate, we have that $ \\forall n\\in\\mathbb{N}\\ \\exists i: \\ n\\le p_i\\le 2n$.\r\nIn $ [2,7]$ there are $ 4$ primes, and for every $ n\\in\\mathbb{N}$ we have \\[ 2^3\\le p_{a_1}\\le 2^4\\] \\[{ 2^4\\le p_{a_2}}\\le 2^5\\] \\[ \\ldots\\] \\[{ 2^{n-1}\\le p_{a_{n-3}}\\le 2^{n}}\\] So there are at least $ n-3+4=n+1$ primes less than $ 2^n<2^{2^n}$", "Solution_7": "[quote=\"Ibn Rochde\"][1] http://www.mathlinks.ro/viewtopic.php?t=212837\n\n[2] $ S(n) \\equal{} \\sum \\frac {1} {k}$ and $ P(n) \\equal{} \\sum \\frac {1} {2k \\minus{} 1}$ \n\n$ S(2n) \\equal{} P(n) \\plus{} \\frac {1} {2}S(n)$ so $ P(n) \\equal{} S(2n) \\minus{} \\frac {1} {2} S(n)$\n\nand we know $ S(k)$ is not an integer that mean $ P(k)$ is not integer \n\n\n[/hide][/quote][b][/b]and we know $ S(k)$ is not an integer that mean $ P(k)$ is not integer\uff1f\uff1f\uff1f", "Solution_8": "[quote=\"rogerlee\"]\n2, Show that $ 1 \\plus{} \\frac {1}{3} \\plus{} \\frac {1}{5} \\plus{} ... \\plus{} \\frac {1}{2n \\minus{} 1}$ is not an integer\n[/quote]\r\n\r\nLet $ S_n \\equal{} 1 \\plus{} \\frac {1}{3} \\plus{} \\frac {1}{5} \\plus{} ... \\plus{} \\frac {1}{2n \\minus{} 1} \\equal{} \\frac {\\alpha_n}{\\beta_n}$, where denominator is $ \\beta_n \\equal{} 1*3*5*...*(2n \\minus{} 1)$.\r\n\r\nLet $ p$ - the largest prime number, less or equal $ 2n \\minus{} 1$.\r\nThere is only one number in the set {1, 3, 5, ..., 2n-1}, which is divisible by $ p$ (we can use Bertrand Postulate to prove this fact: $ n < p < 2n$, so $ 2k \\minus{} 1$ not divisible by $ p$, if $ p\\neq 2k \\minus{} 1$).\r\n\r\nSo, the nominator $ \\alpha_n$ is a summ of $ n$ elements and $ n \\minus{} 1$ of them are divisible by $ p$, and one of them isn't divisible by $ p$. \r\nWe see that nominator isn't divisible by $ p$, but denominator is divisible by $ p$, and $ \\frac {\\alpha_n}{\\beta_n}$ is not an integer.", "Solution_9": "As we can see, some of these problems (1-5) can be solved by using Bertrand Postulate very easy.", "Solution_10": "[quote=\"rogerlee\"][quote=\"Ibn Rochde\"][1] http://www.mathlinks.ro/viewtopic.php?t=212837\n\n[2] $ S(n) \\equal{} \\sum \\frac {1} {k}$ and $ P(n) \\equal{} \\sum \\frac {1} {2k \\minus{} 1}$ \n\n$ S(2n) \\equal{} P(n) \\plus{} \\frac {1} {2}S(n)$ so $ P(n) \\equal{} S(2n) \\minus{} \\frac {1} {2} S(n)$\n\nand we know $ S(k)$ is not an integer that mean $ P(k)$ is not integer \n\n\n[/hide][/quote][b][/b]and we know $ S(k)$ is not an integer that mean $ P(k)$ is not integer\uff1f\uff1f\uff1f[/quote]", "Solution_11": "Ibn Rochde, what means your post?", "Solution_12": "[quote=\"Kirill\"]Ibn Rochde, what means your post?[/quote]\r\nsorry i had some PC problems .i want to say that his right and we can solve it by using the same method of the first question" } { "Tag": [ "AMC", "AIME", "geometry" ], "Problem": "Here's a challenge...\r\n\r\nABC is a triangle, D lies on the side BC and E lies on the side AC. AE = 3, EC = 1, CD = 2, DB = 5, AB = 8. AD and BE meet at P. The line parallel to AC through P meets AB at Q, and the line parallel to BC through P meets AB at R. Find area PQR/area ABC.", "Solution_1": "Ragingg wrote:Here's a challenge...\n\nABC is a triangle, D lies on the side BC and E lies on the side AC. AE = 3, EC = 1, CD = 2, DB = 5, AB = 8. AD and BE meet at P. The line parallel to AC through P meets AB at Q, and the line parallel to BC through P meets AB at R. Find area PQR/area ABC.\n\nSpoiler:\n\n[hide]\n\nLet x=[APB]/[ABC] then [APC]/[ABC]=2x/5 and [CPB]/[ABC]=x/3. So\n\nx + 2x/5 + x/3 =1. Solve x = 15/26\n\n\n\nLet F be the intersection of CP and AB, then \n\n[PQR]/[ABC] = (PR/CB)^2 = (PF/CF)^2 = ([APB]/[ABC])^2 = x^2 = (15/26)^2 = 225/676\n\n[/hide]" } { "Tag": [ "email", "search", "floor function", "geometry", "3D geometry", "algebra", "factorization" ], "Problem": "I found some numbers that are really interesting!\r\n\r\nn:number of digits\r\n\r\nn=3,\r\n1^3+5^3+3^3=153\r\n3^3+7^3+0^3=370\r\n3^3+7^3+1^3=371\r\n4^3+0^3+7^3=407\r\n\r\nn=4,\r\n8^4+2^4+0^4+8^4=8208\r\n9^4+4^4+7^4+4^4=9474\r\n1^4+6^4+3^4+4^4=1634\r\n\r\nn=5, \r\n92727\r\n54748\r\n93084\r\n\r\nn=6,\r\n548834\r\n\r\nn=7,\r\n9926315\r\n\r\nif you can find any other such numbers, please email me at wuchengyuan@gmail.com.", "Solution_1": "wow", "Solution_2": "I'm quite curious - how did you actually find the numbers that you have?\r\n\r\nI did a search for solutions by brute force (using Excel), which I'm sure is not the most efficient thing to do. I managed to check all the numbers up to 20 million without too much difficulty even at that though, so I'm sure a more refined approach could go a lot further.\r\n\r\nYour set of solutions for $n=3,4,5,6$ is complete (as far as I can see), but there are three other solutions for $n=7$, giving a full set:\r\n\r\n$1741725=1^7+7^7+4^7+1^7+7^7+2^7+5^7$\r\n$4210818=4^7+2^7+1^7+0^7+8^7+1^7+8^7$\r\n$9800817=9^7+8^7+0^7+0^7+8^7+1^7+7^7$\r\n$9926315=9^7+9^7+2^7+6^7+3^7+1^7+5^7$\r\n\r\nI haven't been able to find any solution with $8$ digits.\r\n\r\nIncidentally, there are also (trivial) solutions for $n=1$ (i.e. all numbers from $1$ to $9$), but interestingly enough there are no solutions for $n=2$.\r\n\r\nIt's interesting that the problem (which is intuitively very clear) has an extremely messy formal statement. If I've formulated it correctly, we're actually looking for solutions to the equation\r\n\r\n\\[n=\\displaystyle\\sum_{i=1}^{\\lfloor log_{10}n\\rfloor +1}\\left( \\lfloor 10^{-i+1}n \\rfloor-10\\lfloor 10^{-i}n\\rfloor \\right)^{\\lfloor log_{10}n\\rfloor +1}.\\]", "Solution_3": "Or of course, you could just go and look it up... :blush: \r\n\r\nApparently they're called Armstrong numbers...\r\n\r\n[url]http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eismum.cgi[/url]", "Solution_4": "[quote=\"Diarmuid\"]Or of course, you could just go and look it up... :blush: \n\nApparently they're called Armstrong numbers...\n\n[url]http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eismum.cgi[/url][/quote]\r\n\r\nThat link doesn't work.. But I think Armstrong nubmers are nubmers that are equal to sum of cubes of their digits, so it's not this..\r\n\r\nAnd to topic starter.. There are more of these numbers.. Starting with all one-digit numbers.. And there are those I e-mailed you.. Here are they:\r\n\r\n1741725\r\n4210818\r\n9926315\r\n24678050\r\n24678051\r\n88593477\r\n\r\nI don't find this fascinating as others probably do.. And you didn't write any question concerning those numbers.. Are there infinitely many of them, are there infinitely many of pairs $(n, n+1)$ of those numbers..", "Solution_5": "It can be proved if k is a \"perfect-digit-number\" then k has less than 61 digits. Thus, they must be finite.\r\n\r\n\r\n(115132219018763992565095597973971522400 and 115132219018763992565095597973971522401, n=39)", "Solution_6": "[quote=\"kayo\"]It can be proved if k is a \"perfect-digit-number\" then k has less than 61 digits. Thus, they must be finite.\n\n\n(115132219018763992565095597973971522400 and 115132219018763992565095597973971522401, n=39)[/quote]\r\n\r\nYeah, I realised what stupid thing I wrote.. Then.. How many of them are there?\r\n\r\nAnd about those examples with 39 digits.. How can we check it? And more important, how did you find them?", "Solution_7": "[quote=\"kayo\"]It can be proved if k is a \"perfect-digit-number\" then k has less than 61 digits.[/quote]\r\n\r\nHow?", "Solution_8": "Sorry if the link doesn't work - but just type the first few into the integer sequence search engine and you'll get it.\r\n\r\nI'm not sure if this would prove the bound of $61$, but it's easy to see that this sequence can only have finitely many terms, since the sum of the $n^th$ powers of $n$ digits is at most $n9^n$, whereas an $n$ digit number is at least $10^{n-1}$, which is larger for sufficiently large $n$.\r\n\r\nAccording to the Sloane website, the last Armstrong number is the one kayo posted, with only $39$ digits.", "Solution_9": "[quote=\"idioteque\"]\nAnd about those examples with 39 digits.. How can we check it? And more important, how did you find them?[/quote]\r\n\r\nSloane, right ;)", "Solution_10": "The sequence mentioned by [b]Diarmuid[/b] is here : http://www.research.att.com/~njas/sequences/A005188 . ;)" } { "Tag": [ "algebra", "polynomial", "Rational Root Theorem", "algebra unsolved" ], "Problem": "Let n >= 3 be an integer and x be a real number such that the number x, x 2, x^n have the same fractional parts. Prove that x is an integer.", "Solution_1": "Have a look here :\r\n\r\nhttp://www.animath.fr/\r\n\r\nThis is pb #83 (solutions are at the end of the book)\r\n\r\nPierre.", "Solution_2": "sorry,I can read English only. I cannot find the solution.\r\nWould you mind showing the solution?", "Solution_3": "I didn't read it, but I have a solution.\r\nSuppose that $x^2-x=k$ for some positive integer $k$ and also that $x$ is not an integer. By the Rational Root theorem, $x$ is irrational.\r\nConsider the sequence $a_n$ defined by $a_0=1,a_1=x,a_n=a_{n-1}+ka_{n-2}$. Then $a_n=x^n$. We also have $a_n=b_n+xc_n$, where $b_n$ and $c_n$ are integer sequences defined by the same recursion with $b_0=1,b_1=0,c_0=0,c_1=1.$ Therefore $x^n\\equiv xc_n\\mod 1.$ We have $c_1=c_2=1$, and $c_n$ increases; $c_n\\ge c_3=k+1\\ge2$ for all $n\\ge3$. Since $x$ is irrational, we can only have $xc_m\\equiv xc_n\\mod 1$ if $c_m=c_n$. Since this never happens for $n\\ge3$, we are done.", "Solution_4": "([tex]\\Longrightarrow )[/tex] We have [tex]x = \\frac{a}{b}[/tex], [tex]x^2 = \\frac{c}{b}, x^n =\r\n\\frac{d}{b}[/tex] for [tex]n \\geqslant 3[/tex] and [tex]a, c, d \\in N[/tex]\r\n\r\nThus [tex]v_p ( x^2 ) = 2 v_p ( a ) - 2 v_p ( b ) = v_p ( c ) - v_p ( b )\r\n\\Leftrightarrow v_p ( c ) = 2 v_p ( a ) - v_p ( b )[/tex]\r\n\r\nBut because [tex]c \\in N[/tex] we have [tex]v_p ( c ) \\geqslant 0[/tex] which means that [tex]2 v_p\r\n( a ) \\geqslant v_p ( b )[/tex] (i)\r\n\r\nAnd [tex]v_p ( x^n ) = n ( v_p ( a ) - v_p ( b ) ) = v_p ( d ) - v_p ( b )\r\n\\Leftrightarrow v_p ( d ) = nv_p ( a ) - ( n - 1 ) v_p ( b )[/tex]\r\n\r\nBut because [tex]d \\in N[/tex] we have [tex]v_p ( d ) \\geqslant 0[/tex] which means that [tex]nv_p (\r\na ) - ( n - 1 ) v_p ( b )[/tex] (ii)\r\n\r\nThen substracting (ii)-(i) gives us [tex]( n - 2 ) v_p ( a ) \\geqslant ( n - 2 )\r\nv_p ( b )[/tex] thus [tex]v_p ( a ) \\geqslant v_p ( b )[/tex] and [tex]b|a[/tex]\r\n\r\nAnd [tex]x[/tex] must be an integer.\r\n\r\n[tex]( \\Longleftarrow )[/tex] If [tex]x[/tex] is an integer then [tex]x = \\frac{x}{1}[/tex] and thus [tex]x^2\r\n= \\frac{x^2}{1}[/tex], and [tex]x^n = \\frac{x^n}{1}[/tex] have the same fractional part.", "Solution_5": "but who has told you that $x$ is a rational number??", "Solution_6": "Right, when he said fractional part , I forgot x is after all a real number" } { "Tag": [ "geometry", "incenter", "trigonometry", "Asymptote", "geometry proposed" ], "Problem": "Through the incenter $ I$ of $ \\Delta ABC$ we draw the perpendicular to $ AC$ that cuts the sidelines $ BC$ and $ AB$ at $ M,N$ \r\nrespectively. Show that if $ \\frac {1}{IM^2} \\plus{} \\frac {1}{IN^2} \\equal{} \\frac {1}{r^2}$ then $ \\angle A$ is right\r\n\r\n$ r$ is the inradii\r\n\r\n[b]Are there any other triangles that satisfy the condition besides the right-angled triangle?[/b]\r\n\r\n[hide]The answer is yes.figure it out.[/hide]", "Solution_1": "[quote=\"Virgil Nicula\"] [color=darkred]Let $ ABC$ be a triangle with the incircle $ w \\equal{} C(I,r)$ . Denote $ M\\in BC$ , $ N\\in AB$ so that $ I\\in MN$ \n\nand $ \\overline{IMN}\\perp AC$ . Prove that $ \\frac {1}{IM^2} \\plus{} \\frac {1}{IN^2} \\equal{} \\frac {1}{r^2}\\ \\Longleftrightarrow\\ B \\equal{} 90^{\\circ}\\ \\ \\vee\\ \\ |A \\minus{} C| \\equal{} 90^{\\circ}$ .[/color] [/quote]\r\n\r\n[color=darkblue][b][u]Proof[/u].[/b] $ r \\equal{} MI\\cdot\\cos C \\equal{}$ $ NI\\cdot\\cos A$ . Therefore, $ \\frac {1}{IM^2} \\plus{} \\frac {1}{IN^2} \\equal{} \\frac {1}{r^2}\\ \\Longleftrightarrow\\ \\cos^2C \\plus{}$ $ \\cos^2A \\equal{}$ $ 1\\ \\Longleftrightarrow$\n\n$ \\cos 2A \\plus{} \\cos 2C \\equal{}$ $ 0\\ \\Longleftrightarrow\\ \\cos (A \\plus{} C)\\cos (A \\minus{} C) \\equal{} 0\\ \\Longleftrightarrow\\ \\ldots\\ \\Longleftrightarrow\\ B$ $ \\equal{} 90^{\\circ}\\ \\ \\vee\\ \\ |A \\minus{} C| \\equal{} 90^{\\circ}$ .[/color]", "Solution_2": "Thanks for your fast post[b] Virgil[/b].Your a master at trigonometry, as i can see in all your posts,What do you think of the question?.I found actually that if we fixe the side BC, the locus of A that satisfies the condition is a [b]Descartes Folium[/b].\r\nWell i used cabri to visualize the curve and it seems like a Desartes Folium,Im performing the computations to see if this curve is really a Descartes folium", "Solution_3": "after simplifications, the hypotesis is equivalent to this following equation\r\n\r\n$ b^2 \\equal{} \\frac {(a^2 \\minus{} c^2)^2}{a^2 \\plus{} c^2}$\r\n\r\nIf we fix $ BC \\equal{} a$ $ B(0,0)$, $ C(a,0)$, then $ A(x,y)$, the locus of $ A$ is the curve $ y^2 \\equal{} \\frac {x(x \\minus{} a)^2}{2a \\minus{} x}$\r\n\r\nthis is the locus of all the points A of the other triangles besides the right angle that satisfy the above condition.\r\n\r\nthis curve has asymptote $ x \\equal{} 2a$ and a loop around $ BC$. It really looks like a Descartes folium, but i'm not sure if that \r\nformula represents a Descartes Folium." } { "Tag": [ "calculus", "calculus computations" ], "Problem": "Find two convergent series $\\Sigma_{k=1}^{\\infty}x_{k}$ and $\\Sigma_{k=1}^{\\infty}y_{k}$ s.t. $\\Sigma_{k=1}^{\\infty}x_{k}y_{k}$ diverges. Can this happen if one of the series is aboslutely convergent?", "Solution_1": "1. One of typical example is:\r\n\r\n$\\sum_{n=1}^{\\infty}\\frac{(-1)^{n}}{\\sqrt{n}}$ converges but $\\sum_{n=1}^{\\infty}\\left( \\frac{(-1)^{n}}{\\sqrt{n}}\\right)^{2}= \\sum_{n=1}^{\\infty}\\frac{1}{n}$ diverges to $\\infty$.\r\n\r\n\r\n2. Suppose $\\sum_{n=1}^{\\infty}x_{n}$ absolutely converges and $\\sum_{n=1}^{\\infty}y_{n}$ converges. Then $y_{n}$ is bounded by some positive real number $M$, so $|x_{n}y_{n}| \\leq M|x_{n}|$ and by comparison test, $\\sum_{n=1}^{\\infty}x_{n}y_{n}$ absolutely converges." } { "Tag": [ "probability", "number theory", "least common multiple" ], "Problem": "A four-digit number is chosen at random from all four-digit numbers. Express as a common fraction the probability that the number is divisible by 2, 3, 4, and 5.", "Solution_1": "Is this just $ \\frac1{60}$? Can someone explain?", "Solution_2": "To be divisible by 2,3,4, and 5, it must be divisible by their LCM, which is 60. The number of 4-digit numbers divisible by 60 is as follows:\r\n\r\n$ 9999 \\div 60 \\approx 166$\r\n\r\n$ 1000 \\div 60 \\approx 17$\r\n\r\n$ 166 \\minus{} 17 \\equal{} 149 \\plus{} 1 \\equal{} 150$\r\n\r\nThere are $ 9999 \\minus{} 1000 \\plus{} 1 \\equal{} 9000$ 4-digit numbers in total, so the probability is indeed $ \\boxed{\\frac{1}{60}}$." } { "Tag": [ "inequalities", "integration", "calculus", "function", "geometry", "rectangle", "factorial" ], "Problem": "This question may not be too clear:\r\nCan anyone give an upper bound for this \r\n\r\nsum of square roots of 1 ... n, i.e. \r\nsqrt1 + sqrt 2 + ... + sqrt n <= x.\r\nFind suitable x.\r\n\r\nOr, can anyone find a lower bound?", "Solution_1": "By Cauchy-Schwarz,\r\n${(\\sqrt{1}\\cdot 1+\\sqrt{2}\\cdot 1+\\sqrt{3}\\cdot 1+\\cdots+\\sqrt{n}\\cdot 1)}^{2}< (1+2+3+\\cdots+n)(1^{2}+\\cdots+1^{2})$\r\n$\\Rightarrow \\sqrt{1}+\\sqrt{2}+\\cdots+\\sqrt{n}< n\\sqrt{\\frac{n+1}{2}}$\r\n\r\nHowever, it is certain there are stricter upper bounds.", "Solution_2": "We can get a nice bound with calculus.\r\n\r\n$\\int_{1}^{n+1}\\sqrt{x}\\, dx > \\sum_{k=1}^{n}\\sqrt{k}> \\int_{0}^{n}\\sqrt{x}\\, dx$\r\n\r\nWhere $\\int \\sqrt{x}\\, dx = \\frac{2}{3}x^{ \\frac{3}{2}}$. Hence our final bound is given by\r\n\r\n$\\frac{2}{3}(n+1)^{ \\frac{3}{2}}-\\frac{2}{3}> \\sum_{k=1}^{n}\\sqrt{k}> \\frac{2}{3}n^{ \\frac{3}{2}}$\r\n\r\nThis is better than the one given by Cauchy since the equality case of Cauchy doesn't really lend itself to giving a good approximation of a sum where the terms go to infinity.", "Solution_3": "[quote=\"t0rajir0u\"]We can get a nice bound with calculus.\n\n$\\int_{1}^{n+1}\\sqrt{x}\\, dx > \\sum_{k=1}^{n}\\sqrt{k}> \\int_{0}^{n}\\sqrt{x}\\, dx$\n\nWhere $\\int \\sqrt{x}\\, dx = \\frac{2}{3}x^{ \\frac{3}{2}}$. Hence our final bound is given by\n\n$\\frac{2}{3}(n+1)^{ \\frac{3}{2}}-\\frac{2}{3}> \\sum_{k=1}^{n}\\sqrt{k}> \\frac{2}{3}n^{ \\frac{3}{2}}$\n\nThis is better than the one given by Cauchy since the equality case of Cauchy doesn't really lend itself to giving a good approximation of a sum where the terms go to infinity.[/quote]\r\n\r\nhow do u know that $\\sum_{k=1}^{n}\\sqrt{k}>\\int_{0}^{n}\\sqrt{x}dx$?", "Solution_4": "because the sum is spaced out farther", "Solution_5": "The sum in question is a right-hand approximation of the right integral, and a left-hand approximation of the left integral...and both are increasing, so...", "Solution_6": "[quote=\"maokid7\"]how do u know that $\\sum_{k=1}^{n}\\sqrt{k}>\\int_{0}^{n}\\sqrt{x}dx$?[/quote]\r\n\r\n[img]http://www.ecr.unimelb.edu.au/~xlr8tr/techs/calculus/integration2.png[/img]\r\n\r\nThis may not be the correct function, but I hope it helps.\r\n\r\nEssentially, the sum is the area of the rectangles and the integral is the area under the curve. We can pick the sum so that we get rectangles that are exactly within the curve, or so that the curve is exactly within the rectangles.", "Solution_7": "[quote=\"thealchemist\"]This question may not be too clear:\nCan anyone give an upper bound for this \n\nsum of square roots of 1 ... n, i.e. \n$\\sqrt1+\\sqrt 2+...+\\sqrt n \\leq x$.\nFind suitable x.\n\nOr, can anyone find a lower bound?[/quote]\r\n\r\nFor a lower bound, you can use AM-GM (i'm not sure of a way to get a more strict bound) to get\r\n\r\n$\\sum_{k=1}^{n}{\\sqrt{k}}\\geq n\\cdot \\sqrt[2n]{n!}$. (for integral $n$.)", "Solution_8": "[quote=\"cincodemayo5590\"]For a lower bound, you can use AM-GM (i'm not sure of a way to get a more strict bound) to get\n\n$\\sum_{k=1}^{n}{\\sqrt{k}}\\leq n\\cdot \\sqrt[2n]{n!}$. (for integral $n$.)[/quote]\r\n\r\nI think you mean $\\geq$.\r\n\r\nIn any case, the bound I gave is both sharper and easier to calculate :wink:\r\n\r\nEdit: As with Cauchy, the problem with using AM-GM is that the terms tend to infinity, which means that the result will not be close to the equality case (where the terms are equal).", "Solution_9": "Using Stirling's approximation for factorial, that lower bound simplifies to $\\frac{n^{3/2}}{\\sqrt{e}}$ (I think.. maybe I made a mistake :P). That will be more accurate for larger $n$. \r\n\r\nOf course, an integral will give the sharpest approximation again. And if you want to get an even better estimate, you can approximate the difference between the integral and the sum by another integral, and so on. It just gets more tedious and complicated as you go on.", "Solution_10": "[quote=\"t0rajir0u\"][quote=\"cincodemayo5590\"]For a lower bound, you can use AM-GM (i'm not sure of a way to get a more strict bound) to get\n\n$\\sum_{k=1}^{n}{\\sqrt{k}}\\leq n\\cdot \\sqrt[2n]{n!}$. (for integral $n$.)[/quote]\n\nI think you mean $\\geq$.\n\nIn any case, the bound I gave is both sharper and easier to calculate :wink:\n\nEdit: As with Cauchy, the problem with using AM-GM is that the terms tend to infinity, which means that the result will not be close to the equality case (where the terms are equal).[/quote]\r\n\r\nRight :oops: (It should be better now)" } { "Tag": [ "real analysis", "real analysis solved" ], "Problem": "Here we go :)\r\n\r\nShow that $(0,1]$ and $[0,1]$ have the same cardinality", "Solution_1": "Let $S=\\{1,\\frac{1}{2},\\frac{1}{3},\\dots,\\frac{1}{n},\\dots\\}$\r\nThen we define\r\n$f(x) = \\left\\{\r\n\\begin{array}{cc}\r\n0, & x =1\\\\\r\n\\frac{1}{n-1}, & x=\\frac{1}{n}\\\\\r\nx, & x\\notin S\r\n\\end{array}\r\n\\right.\r\n$\r\nthen $f:(0,1]\\to[0,1]$ is a bijection", "Solution_2": "The cardinality of the sets is c." } { "Tag": [ "linear algebra", "matrix" ], "Problem": "How many squares are pictured?\n\n[asy]path p = (0,0)--(2,0)--(2,2)--(0,2)--(0,0);\nfor(int i = 0; i < 4; ++i){\nfor(int k = 0; k < 4; ++k){\ndraw(shift((2*i,2*k))*p);}}\ndraw(shift((1,3))*p);\ndraw(shift((5,3))*p);[/asy]", "Solution_1": "Let's ignore the two \"decorative\" squares near the middle for a moment. In the $ 4$-by-$ 4$ matrix of squares, there is the $ 1$ largest square, $ 4$ of the $ 3$-by-$ 3$ squares, $ 9$ of the $ 2$-by-$ 2$s, and $ 16$ small squares. Adding these four quantities up gives $ 16\\plus{}9\\plus{}4\\plus{}1\\equal{}30$ squares. \r\n\r\nNow for the extra ones we've ignored previously: there are $ 8$ tiny squares and $ 2$ larger squares, making $ 10$. Our grand total is $ 30\\plus{}10\\equal{}\\fbox{40}$ squares." } { "Tag": [], "Problem": "Varfurile unui poligon regulat cu $\\ n$ laturi se coloreaza in culori diferite, astfel incat punctele cu aceeasi culoare sunt varfurile unor poligoane regulate.Sa se arate ca printre aceste poligoane se gasesc doua egale.", "Solution_1": "any ideas?", "Solution_2": "Observam ca pentru un poligon regulat cu $n$ laturi, orice sub-poligon regulat cu $k$ laturi verifica $k | n$.\r\n\r\nDeci $n$ nu este prim. Fie $\\frac{n}{k}$ numarul laturilor pe care le are un sub-poligon al sau in configuratia data. Daca $k$ este numar prim si $k \\| n$ (adica $k|n$ si $k^{2}\\nmid n$), atunci orice alt divizor al lui $n$ verifica $(d,k) = 1$, deci vom avea $k$ sub-poligoane regulate, toate egale intre ele.\r\n\r\n[i]Observatie.[/i] In pasul precedent, daca exista $k$ a.i. $k$ e prim si indeplineste conditiile, atunci il alegem neaparat. Altfel facem pasul 2 de mai jos.\r\n\r\nDaca $k$ nu-i prim sau $k^{2}| n$, atunci fie $p$ prim, $p|k$. Consideram apoi sub-poligonul $A_{1}\\ldots A_{\\frac{n}{p}}$, luat din $p$ in $p$ varfuri. Facem chestia asta pentru ca daca un sub-poligon al celui initial are un varf in $A_{1}\\ldots$, atunci acolo are toate varfurile (sper sa fie corect ce-am zis eu in ultima fraza). Acest sub-poligon n-are cum sa fie monocromatic, deci putem aplica aceasta procedura pentru $A_{1}\\ldots$ (de la \"Fie $\\frac{n}{k}$\" incolo).\r\n\r\nDaca notam $\\prod n_{i}^{\\alpha_{1,i}}$ factorizarea lui $n$ in pasul 1, $\\prod n_{i}^{\\alpha_{2,i}}$ factorizarea noului $n$ (adica $\\frac{n}{p}$) in pasul 2 etc. observam ca $\\sum \\alpha_{1,i}> \\sum \\alpha_{2,i}> \\ldots$, deci chestia asta se va termina pana la urma (mai exact, atunci cand vom da de un $k$ prim & $k | n$). \r\n\r\n\r\nNu stiu cat de bine e, dar macar am incercat :D" } { "Tag": [ "linear algebra", "matrix" ], "Problem": "[2 5]*[x]=[4] how would you solve for x and y??\r\n[3 1] [y] [7]\r\n\r\n\r\n\"[\"\r\n [ is supposed to be 1 bracket...sorry i didnt know how to do that![/img]", "Solution_1": "[quote=\"linzk3\"][2 5]*[x]=[4] how would you solve for x and y??\n[3 1] [y] [7]\n\n\n\"[\"\n [ is supposed to be 1 bracket...sorry i didnt know how to do that![/img][/quote]\r\n\r\nI have no experience with matricies, but...[hide] You could just multiply out the matricies, to get \n$2x+5y=4$\n$3x+y=7$\nAnd then solve. [/hide]", "Solution_2": "Another way would be to change the augmented matrix to reduced row-echelon form. (or find the inverse of the matrix, use cramer's rule, etc.)" } { "Tag": [ "geometry", "projective geometry", "geometry proposed" ], "Problem": "Projective geometry: \r\n\r\nlet $\\Delta ABC$ be a triangle in the projective plane, let $D$ be a fixed point on $AB$ and let $H$ and $K$ be fixed points which do not lie on one of the sides of the triangle. Let $P$ be a variable point on $CD$. Let $Q = AP \\cap DH$ and $R = BP \\cap DK$. Show that $QR$ intersects $AB$ in a fixed point. .", "Solution_1": "Let lines $AC=a$, $BC=b$, $DK=k$, $DH=h$, $DC=d$. \r\nConsider a central proiection which put line $AB$ to infty.Then $h' \\parallel d' \\parallel k'$ ,angels $(b',d')$ and $(k',d')$ are fixed. All triangels $P'Q'R'$ which could be get when $P'$ goes over $d'$ are equal and making equal angels with $d'$.So all $P'Q'$ are parallel $\\Longleftrightarrow$ they passing throug a fixed point on $A'B'$.", "Solution_2": "Why my posts are rated like a spam?! :mad: \r\nSomeone don't know what is 'central projection' or how I use it in solution or why there are angels in it?\r\nOr you know but think there is mistake.\r\nPlease ask questions." } { "Tag": [ "geometry", "circumcircle", "incenter", "geometry unsolved" ], "Problem": "Let two circles $(O1)$ and $(O2)$ meet at $A,B$ ( $A$ $\\neq$ $B$) . Let circumcircle of triangle $O1AO2$ meet $(O1) ,(O2)$,$AB$ at $M1,M2$,$K$ ,respectively. Prove that $KM1$ = $KM2$", "Solution_1": "Let $N_{1}, N_{2}$ be points where $KM_{1}$, $KM_{2}$ meets O_1, O_2 respectively. It's enough to prove that $KN_{1}=KN_{2}$. We have:\r\n\r\n$2 \\angle M_{1}N_{1}A=\\angle M_{1}O_{1}A=\\pi-\\angle M_{1}KA= \\angle M_{1}N_{1}+\\angle N_{1}KA$ so $KN_{1}=AK$. Analogically $KN_{2}=AK$ so $KN_{1}=KN_{2}$ as wanted. QED", "Solution_2": "[quote=\"Megus\"]Let $N_{1}, N_{2}$ be points where $KM_{1}$, $KM_{2}$ meets O_1, O_2 respectively. It's enough to prove that $KN_{1}=KN_{2}$. We have:\n\n$2 \\angle M_{1}N_{1}A=\\angle M_{1}O_{1}A=\\pi-\\angle M_{1}KA= \\angle M_{1}N_{1}+\\angle N_{1}KA$ so $KN_{1}=AK$. Analogically $KN_{2}=AK$ so $KN_{1}=KN_{2}$ as wanted. QED[/quote]\r\n\r\nthere's something wrong with your angle calculations... :>", "Solution_3": "It 's obvious that $B$ is the intersection of $M_{2}O_{1}$ and $M_{1}O_{2}$. Since these are the bisectors of angles $AM_{2}M_{1}$ and $AM_{1}M_{2}$, we obtain $B$ is the incenter of $O_{1}AO_{2}$. Therefore $KM_{1}=KM_{2}$.", "Solution_4": "[quote=\"gollywog\"]there's something wrong with your angle calculations... :>[/quote]\r\n\r\nwhere exactly? :huh:" } { "Tag": [ "trigonometry", "algebra proposed", "algebra" ], "Problem": "Solve:\r\n\\[ \\frac {1}{x^2} \\plus{} \\frac {1}{(4 \\minus{} \\sqrt {3}x)^2} \\equal{} 1\r\n\\]\r\n[hide]hint:\n\\[ x \\equal{} \\frac {1}{\\cos{70^o}},x \\equal{} \\frac {1}{\\cos{30^o}},x \\equal{} \\frac {1}{\\cos{50^o}},x \\equal{} \\frac {1}{\\cos{170^o}}\n\\]\n[/hide]", "Solution_1": "Let $ x\\equal{}\\frac{1}{cosy}$, then $ \\frac{cosy}{4cosy\\minus{}\\sqrt 3}\\equal{}siny$. It equavalent to $ sin(y\\plus{}\\frac{\\pi}{6})\\equal{}sin2y$.\r\nIt give solution $ x\\equal{}\\frac{1}{cosy}$, were $ y\\equal{}\\frac{\\pi}{6}$ or $ y\\equal{}\\frac{5\\pi}{18}$.", "Solution_2": "But there are 4 roots :D", "Solution_3": "Yes. Consider $ \\pi\\minus{}\\pi/6\\minus{}y\\plus{}2k\\pi\\equal{}2y$. It give $ y\\equal{}\\frac{5\\pi}{18}\\plus{}\\frac{2k\\pi}{3},k\\equal{}0,1,2.$" } { "Tag": [ "limit", "LaTeX", "calculus", "calculus computations" ], "Problem": "$ \\lim_{x\\to 0}\\left( \\frac{1}{x}\\minus{}\\frac{1}{\\ln(x\\plus{}1)} \\right)$", "Solution_1": "[hide=\"Solution.\"]\n\nEdit: oops. Apparently, fraction bars can't magically be transformed into minus signs.\n\n[/hide]", "Solution_2": "[hide=\"Solution\"][img]http://imgcash5.imageshack.us/img213/9093/lospital1as9.jpg[/img]\n\nUsing L'ospital twice you can find -1/2.[/hide]\r\n\r\nIf you want more information ask me ( i didn't put more because of my dificult in Latex )", "Solution_3": "There's something novel going on with exponents in the first solution", "Solution_4": "Starting where others left off\r\n\r\n$ \\lim_{x\\to{0}}\\frac{\\ln(1+x)-x}{x\\ln(1+x)}$\r\n\r\nNow inserting the Maclaurin series gives\r\n\r\n${ \\lim_[x\\to{0}}\\frac{x-\\frac{x^2}{2}+\\cdots-x}{x\\left(x-\\frac{x^2}{2}+\\cdots\\right)}=\\lim_{x\\to{0}}\\frac{\\frac{-x^2}{2}+\\cdots}{x^2-\\frac{x^3}{2}+\\cdots}=\\frac{-1}{2}$", "Solution_5": "Essentially the same as Mathstud28's solution but organized slightly differently (avoiding the common denominator):\r\n\r\n$ \\frac1x\\minus{}\\frac1{\\ln(1\\plus{}x)}\\equal{}\\frac1x\\minus{}\\frac1{x\\minus{}\\frac12x^2\\plus{}O(x^3)}$\r\n\r\n$ \\equal{}\\frac1x\\minus{}\\frac1x\\cdot\\frac1{1\\minus{}\\frac12x\\plus{}O(x^2)}$\r\n\r\n$ \\equal{}\\frac1x\\minus{}\\frac1x\\left(1\\plus{}\\frac12x\\plus{}O(x^2)\\right)$\r\n\r\n$ \\equal{}\\frac1x\\minus{}\\frac1x\\minus{}\\frac12\\plus{}O(x)\\equal{}\\minus{}\\frac12\\plus{}O(x).$\r\n\r\nAnd the limit is $ \\minus{}\\frac12.$" } { "Tag": [ "modular arithmetic" ], "Problem": "The number $ 2^{1993}\\plus{}3^{1993}$ is a multiple of 5. What is the units digit of the quotient $ \\frac{2^{1993}\\plus{}3^{1993}}{5}$?", "Solution_1": "The units digit of $ 2^{1993}$ is 2 and the units digit of $ 3^{1993}$ is 3. The the units digit of the sum is 5, so when divided by 5, it has a units digit of 1.", "Solution_2": "Not necessarily...15/5=3", "Solution_3": "how do you solve this?\r\ni tried\r\n$ {(5 - 3)^{1993} + (5 - 2)^{1993} \\pmod{50}}$ with binom. theorem, didn't work :(", "Solution_4": "If you don't like ugly format, don't read this, or else write it up nicely for someone else (if you confirm the approach).\r\n\r\nFactoring 2^1993 + 3^1993 we get\r\n(2 + 3)(2^1992 - 2^1991*3^1 + 2^1990*3^2 - 2^1989*3^3 + . . . + 3^1992)\r\n\r\nDividing by 5 leaves\r\n(2^1992 - 2^1991*3^1 + 2^1990*3^2 - 2^1989*3^3 + . . . + 3^1992)\r\n\r\nPatterns for powers of 2 and powers of 3 give us the following numbers with only the units digits showing.\r\n[(_6) \u2013 (_8)(_3) + (_4)(_9) \u2013 (_2)(_7) + (_6)(_1). . . + (_2)(_7) + (_1)]\r\n\r\nWhich simplifies to the repeating pattern:\r\n[_6 \u2013 _4 + _6 \u2013 _4 \u2013 . . . + _6 \u2013 _4 + _1]\r\n\r\nReducing to:\r\n[ _2 * 996 + _1] = _2 + _1 = _3\r\n[/hide]", "Solution_5": "I need better explanation. ", "Solution_6": "$2^1+3^1=5$. This is a multiple of $5$, $5\\cdot1$.\n$2^3+3^3=35$. This is $5\\cdot 7$.\n$2^5+3^5=275$. This is $5\\cdot 55$ (units digit 5)\n$2^7+3^7=128+2187=2315$ This is $5\\cdot463$ (units digit 3)\n\nsince the numbers are gonna get big now, it's gonna be easier to reduce everything to $\\pmod{50}$\n\n$2^9\\equiv 12\\pmod{50}$\n$3^9\\equiv3^7*9\\equiv87*9\\equiv33\\pmod{50}$\n\nso\n\n$2^9+3^9\\equiv 12+33\\equiv 45 \\pmod{50}$ and dividing by $5$ gives us $9$.\n\nif you keep doing this, you'll see that the sequence goes $1, 7, 5, 3, 9, 9, 3, 5, 7, 1$, repeating every 10 odd numbers. since $1993+1/2 = 997$th odd number, which means its the $7$th term, $\\boxed{3}$\n\n[hide=\"birdmath's solution\"]\n\n$x^a+y^a$ can be factored into\n$(x+y)(x^{a-1}-x^{a-2}y^1+x^{a-3}y^2...-x^1y^{a-2}+y^{a-1})$\n\nwhich is where birdmath got\n$(2+3)(2^{1992}-2^{1991}\\cdot3...-2\\cdot3^{1991}+3^{1992})$\n\nand dividing by five, we're only left with the polynomial.\n\nnow we're only looking for the units digit.\n\nthis gives us\n\n$(6-8(3)+4(9)...-4+1)$\n\nfinding the units digit we get $2$ $\\frac{1992}{2}=996$ times with an extra $1$ at the end.\n\nso we have $996*2$ (units digit of 2) plus $1$ so $2+1=3$.\n\ngood bump." } { "Tag": [ "real analysis", "advanced fields", "advanced fields solved" ], "Problem": "Show that a countable number of lines on [b]R[/b]2 cannot fill the whole plane.\r\n(Please give an elementary proof without using Baire category)\r\n\r\n[i]Chinese undergraduate students camp, year 1992[/i]", "Solution_1": "Let's give a countable number of lines of the plane.\r\nThere are an uncountable number of different directions for the lines in the plane. Thus, we may find a line D which is not parallell to any of the given lines. Each of the given lines meet D in exactly one points, thus they cover only a at most countable part of D, and thus do not cover D entirely.\r\n\r\nPierre.", "Solution_2": "Quite nice. This is exactly the first proof in official solutions.. it's the easiest one to understand..", "Solution_3": "To be honest, I did not even try to think about any Baire's theorem. :D \r\nPierre.", "Solution_4": "another simple proof is to consider a circle on the plane.\r\nDoes anyone have other proofs?", "Solution_5": "The Lebesgue measure of countable lines is zero where the Lebesegue mesure of $\\mathbb{R^2}$ is not zero." } { "Tag": [ "number theory", "prime numbers" ], "Problem": "A pair of $ \\it{emirps}$ consists of two prime numbers such that reversing the digits of one number gives the other. How many pairs of two-digit emirps exist such that each number in the pair is greater than 11?", "Solution_1": "Zomg, emirp is prime backwards! How clever.\r\n\r\nSo, basically, you test all primes greater than 11 that could be emirps. This means that primes starting with an even number are out (when reversed, they'll be divisible by 2), and primes starting with 5 are out (when reversed, they'll be divisible by 5, of course). Test all others and you get:\r\n\r\n(13,31)\r\n(17,71)\r\n(37,73)\r\n(79,97)\r\n\r\nWhich is 4." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,b,c>0$.Prove that:\r\n$ 3\\sqrt{3}(\\sum \\frac{a^3}{2a^2\\plus{}b^2}) \\geq \\sum \\sqrt{a^2\\plus{}ab\\plus{}b^2}$", "Solution_1": "It seems that the following inequality is true too.\r\nLet $ a,$ $ b$ and $ c$ are positive numbers. Prove that\r\n\\[ \\sum_{cyc}\\frac {18a^3}{2a^2 \\plus{} b^2}\\geq\\sum_{cyc}\\sqrt {17a^2 \\plus{} 2ab \\plus{} 17b^2}\r\n\\]\r\nFor $ k>8.5$ the inequality \r\n$ \\sum_{cyc}\\frac {a^3}{2a^2 \\plus{} b^2}\\geq\\frac{1}{3\\sqrt{2k\\plus{}1}}\\sum_{cyc}\\sqrt {ka^2 \\plus{} ab \\plus{} kb^2}$ is wrong already.", "Solution_2": "[quote=\"No Reason\"]Let $ a,b,c > 0$.Prove that:\n$ 3\\sqrt {3}(\\sum \\frac {a^3}{2a^2 \\plus{} b^2}) \\geq \\sum \\sqrt {a^2 \\plus{} ab \\plus{} b^2}$[/quote]\r\n\r\nSee also : http://www.mathlinks.ro/viewtopic.php?t=228984" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Each of $ n > 2$ lines on a plane is cut by the others into $ 2$ rays and $ n \\minus{} 2$ equal segments.\r\nProve that $ n \\equal{} 3$.", "Solution_1": "http://www.mathlinks.ro/viewtopic.php?t=21906" } { "Tag": [ "floor function", "number theory proposed", "number theory" ], "Problem": "Prove that if $ b^{n}-1$ divides $ a$ then the sum of digits of $ a$ in base $ b$ representation is not less than $ n(b-1).$", "Solution_1": "Some definitions:\r\n\r\ns(n) is the sum of the digits of n in base b.\r\nf(n) is the greatest integer k such that $ b^{k}|n$\r\n$ g(n)=\\sum_{i=1}^{n}f(n)$ (and g(0)=0)\r\n\r\nNow (by simple double counting): $ g(x)=\\sum_{i=1}^{\\infty}\\left\\lfloor\\frac{n}{b^{i}}\\right\\rfloor=\\frac{x-s(x)}{b-1}$\r\n\r\nMoreover $ g(x)+g(y)\\le g(x+y)$ (by the first equality, because $ \\left\\lfloor\\frac{x}{b^{i}}\\right\\rfloor+\\left\\lfloor\\frac{y}{b^{i}}\\right\\rfloor\\le\\left\\lfloor\\frac{x+y}{b^{i}}\\right\\rfloor$)\r\n\r\nLet $ a=k(b^{n}-1)$ where k is a positive integer\r\n\r\nSo we have that:\r\n\r\n$ g(a+k-1)-g(a)\\ge g(k-1)$\r\n\r\n$ g(a+k-1)+f(a+k)-g(a)\\ge g(k-1)+f(a+k)$\r\n\r\nBut $ f(a+k)=f(kb^{n})=f(k)+n$, so:\r\n\r\n$ g(a+k)-g(a)\\ge g(k-1)+f(k)+n$\r\n\r\n$ \\frac{a+k-s(a+k)}{b-1}-\\frac{a-s(a)}{b-1}\\ge g(k)+n$\r\n\r\n$ \\frac{k-s(a+k)}{b-1}+\\frac{s(a)}{b-1}\\ge \\frac{k-s(k)}{b-1}+n$\r\n\r\nBut $ s(a+k)=s(kb^{n})=s(k)$, so: $ \\frac{s(a)}{b-1}\\ge n$, that is to say $ s(a)\\ge n(b-1)$ as we wished to prove." } { "Tag": [ "inequalities" ], "Problem": "how many solutions $(a,b,c)$ to \r\n\r\n$a+b+c\\le d$", "Solution_1": "infinite\r\nunless you specify $d$ and restrict $a,b,c$", "Solution_2": "sorry a, b, c positive integres, d is positive integer", "Solution_3": "", "Solution_4": "[hide=\"shazam\"]$a-1+b-1+c-1=d-3$\nlet $(a-1,b-1,c-1)=(u,v,w)$\nimagine $d-3$ dots in a line\nwe need to partition these dots into three sections\nthis requires two partitions\nso we have $d-1$ total objects, so we can place the partitions in ${d-1}\\choose2$ ways. now we know $u,v,w$ are nonnegative integers, so $a,b,c$ are positive integers[/hide]\nI think\nEDIT: no wait\nthat's for equality\nto satisfy the inequality, you get\n[hide=\"this\"]\n${{d-1}\\choose{2}}+{{d-2}\\choose{2}}+{{d-3}\\choose{2}}+...+{2\\choose{2}}$\nwhich, by the hockeystick theorem, equals\n${d-1}\\choose3$[/hide]", "Solution_5": "are you sure it is not [b]dC3[/b]?\r\n\r\nTry a case: a+b+c <=5, wehre a,b,c are positive integers.\r\n\r\nso a+b+c can equal 2,1,or 0\r\n\r\nwhen a+b+c=2, there are 6 solutions\r\nwhen a+b+c=1, ther are 3 solutions\r\nwhen a+b+c=0, ther is 1 solution\r\nso 6+3+1=10, which is percisely 5C3,\r\n\r\nyou can think about it like this: a+1+b+1+c+1+x=d, where x is nonnegative, so get rid of the less than sign,\r\n\r\nso a+b+c+x=d-3, now thats just dC3..", "Solution_6": "Sorry folks, I had thought it was an $\\emph{equality}$ as well.", "Solution_7": "[quote=\"yeppyyep\"]How many solutions $(a,b,c)$ to \n$a+b+c\\le d$\nexist for $(a,b,c) \\in \\mathbb{N}^{3}$ and $d \\in \\mathbb{N}$?[/quote]\r\n\r\n[hide=\"Solution\"]\nFirst, let's change the restriction on $a,b,c$: put $a=x+1, \\ b=y+1, \\ c=z-1$ so that $(x,y,z) \\in \\left( \\mathbb{N}^{*}\\right)^{3}$.\n\nNow, we have $x+y+z \\leq d-3$. Imagine a line of $d-3$ dots, each one representing one unit of value. We can place three dividers either directly before, directly after, or within the line of dots. The number of dots to the left of all dividers will be the value of $x$; the number between the first two dividers will be $y$; the number between the second and third will become the value of $z$; and whatever is left (past the last divider on the right) will be left-over. We allow the left-over since the value of $x+y+z$ will not always be equal to $d-3$.\n\nConsequently, the number of distinct ways to place the dividers, and the number of ordered solutions, is:\n\\[\\dbinom{ (d-3)+3 }{3}= \\boxed{ \\dbinom{d}{3}}\\]\n[/hide]", "Solution_8": "you can always make it an equality by adding a fourth variable thats nonnegative.", "Solution_9": "Good call. The so-called \"pivot\" variable.", "Solution_10": "[quote=\"Art of Owna\"]you can always make it an equality by adding a fourth variable thats nonnegative.[/quote]\r\nvery nice art of owna!", "Solution_11": "my bad - hockey stick says the answer is $d\\choose3$" } { "Tag": [ "function", "inequalities unsolved", "inequalities" ], "Problem": "Prove that:\r\n$\\frac{1}{a^2} + \\frac{1}{b^2} + \\frac{1}{c^2} + 3 \\geq 2(a+b+c)$\r\nWhere a,b,c are positives numbers satisfy condition: abc=1", "Solution_1": "[quote=\"thienthan89\"]Prove that:\n$\\frac{1}{a^2} + \\frac{1}{b^2} + \\frac{1}{c^2} + 3 \\geq 2(a+b+c)$\nWhere a,b,c are positives numbers satisfy condition: abc=1[/quote]\r\nFirst, substitute $a=e^x,b=e^y,c=e^z$. We now have $x+y+z=0$. WLOG $x\\geq y\\geq z$.\r\nNow there are two cases: $x\\geq y\\geq 0\\geq z$ and $x\\geq 0\\geq y\\geq z$. As I'm lazy, I will only show the first case.\r\nWe have $-2x\\leq -2y\\leq 0\\leq 0\\leq 0\\leq -2z$, $z\\leq z\\leq y\\leq y\\leq x\\leq x$ and\r\n\\[ -2x\\leq -x-y=z \\\\-2x-2y=2z\\leq z+z \\\\-2x-2y-0\\leq z+z+y \\\\-2x-2y-0-0\\leq z+z+y+y \\\\-2x-2y-0-0-0\\leq z+z+y+y+x \\\\-2x-2y-0-0-0-2z=z+z+y+y+x+x \\]\r\nThus we can use Karamata on the convex function $e^x$ to show \\[ e^{-2x}+e^{-2y}+e^{-2z}+e^{0}+e^{0}+e^{0}\\geq e^{x}+e^{x}+e^{y}+e^{y}+e^{z}+e^{z} \\] :D :D :D" } { "Tag": [], "Problem": "Not sure if you guys can help this guy, but I run the Q&A site http://www.ExpertBee.com and someone offered to pay to find out \" how do you seperate the oxygen and hydrogen molecules in water,\" and no one has bid to answer.\r\nIf you know, help the guy out and make a little money. here is the posting link:\r\nhttp://www.expertbee.com/auction-query_i12899.html\r\nThanks\r\nExpertBee", "Solution_1": "I don't know anything about paying for answers.\r\n\r\nThe question as asked is meaningless -- the question assume a fact that is incorrect: there is no molecules of hydrogen and oxygen to seperate. If what was meant was how to break down the water into oxygen atoms and hydrogen atoms, which will seperately combine into oxygen molecules and hydrogen molecules that can be seperately collected. The answer is an extremely easy electrochem. question. Run a direct current through an electrolyte solution of water, be careful to collect the oxygen and hydrogen before they can recombine in a fire or explosion.\r\n\r\nIf the question is how to do this comercially, there are questions of the amount of electrolyte (perhaps table salt) voltage and gas seperation techniques that I am not qualified to address, having never done this reaction commercially. The real question woudl be why would one want to develop hydrogen and oxygen this way. Normally, hydrogen is evolved as a by-product of a chemical reaction conducted for other purposes (sometimes evolution of chlorine from brine, or salt water). Normally, oxygen (the by-product waste gas of plants living and making sugars) is made by cooling air and fractionally distilling the resulting liquid into the separate gases as needed (nitrogen, oxygen, etc.).\r\n\r\nI hope this helps.", "Solution_2": "Thanks for your help!", "Solution_3": "be careful guys, expertbee.com is a well documented scam.", "Solution_4": "why would somone want to develop hydrogen from water....dah...lots of water around, which could be used to make valuable hydrogen for running hydrogen cars and more oxygen for human consumption, they sell it in japan dont they....there are wonderings if there are oceans of oil underground, to run machines, we already know there are oceans of water above and below ground which could be used minimally to run clean burning machines..that could use both hydrogen and oxygen separately...doing this with plants from gasses is just way too slow and expensive sort of like the current cosly alternative of making ethanol from corn, eventually we will probably learn to make ethanol cheaply from other bio sources, but for now we need people who can do the math...........before asking silly questions..." } { "Tag": [ "rotation", "combinatorics unsolved", "combinatorics" ], "Problem": "Can we put positive integers $1,2,3, \\cdots 64$ into $8 \\times 8$ grids such that the sum of the numbers in any $4$ grids that have the form like $T$ ( $3$ on top and $1$ under the middle one on the top, this can be rotate to any direction) can be divided by $5$?", "Solution_1": "The answer is no. \r\nSuposse it is possible:\r\nLet's enumerate de grids in the following way: in the first row with 1,2,3,4,5,6,7,8; in the second row with 9,10,11,12,13,14,15,16, etc. \r\nLet $A$ the number in the square 2, taking the $T$'s (9,10,11,2) and (9,10,11,18) it follows that the number in the square 18 is congruent to $A$ mod 5, analogous the numbers in the squares 34,50,52,36,20,4,22,6,38,54,56,40,24 are congruent to $A$ mod 5, the trouble is in 8, since the numbers in 7 and 23 are congruent, and taking the $T$'s (6,7,8,15) and (22,23,24,15) and using that the numbers in 6,22,24 are congruent mod 5, it follows that the number in 8 is also congruent the number in 24, i.e. is congruent to $A$. \r\nAnalogous the numbers in:\r\n 1,3,5,7,17,19,21,23,33,35,37,39,49,51,53,55 are conguent mod 5, say to $B$.\r\n9,11,13,15,25,27,29,31,41,43,45,47,57,59,61,63 are congruent mod 5, say to $C$.\r\n10,12,14,16,26,28,30,32,42,44,46,48,58,60,62,64 are congruent mod 5, say to $D$.\r\nIt's obvious that $A;B;C;D$ are pairwise distinct mod 5 (if not we have 32 numbers in 1,2,3,...,64 mod 5) , so we have only 4 possible residues mod 5, it is a contradiction because for example 50,51,52,53,54 are pairwise distinct mod 5.\r\n\r\n$Tipe$" } { "Tag": [ "LaTeX" ], "Problem": "Find an integer $n$ with two digits which satisfies \\[\\sqrt{n + \\sqrt{n + 7}} \\in \\mathbb{N}.\\]", "Solution_1": "[hide]\n\n\n\n is a square. We are given that n is two digits. Because is an integer, it follows is.\n\n\n\nThis gives us only a few possibilities for n+7: .\n\n\n\nSo the expression could be 18+5, 29+6, 42+7, 57+8, 74+9, 93+10. Only one of these is a square, namely n = 42.\n\n\n\n[/hide]", "Solution_2": ":10:", "Solution_3": "[quote=\"Fermat -Euler\"]:10:[/quote]\r\n\r\nPlease don't do that again, it's a useless post. And I suggest you learn LaTeX (and stop posting attachments) since I have to spend a LOT of time on editing your posts." } { "Tag": [ "limit", "calculus", "calculus computations" ], "Problem": "For $ 0 \\leq x \\leq 1$, evaluate $ \\lim_{n\\to\\infty} \\sum_{k\\equal{}0}^{n} \\binom{n\\plus{}k}{n} x^n (1\\minus{}x)^k$.", "Solution_1": "hello, it is $ \\lim_{n\\to\\infty} \\sum_{k\\equal{}0}^{n} \\binom{n\\plus{}k}{n} x^n (1\\minus{}x)^k\\equal{}\\frac{1}{x}$.\r\nSonnhard." } { "Tag": [ "geometry", "Support", "AMC", "USA(J)MO", "USAMO", "\\/closed" ], "Problem": "I know that rrusczyk was suggesting to xxreddevilzxx to get a sponsorship to help him (or her) to pay for WOOT.\r\n\r\nBut I was just wondering if anyone has ever tried this, for WOOT, or for anything else, and how you went about doing it, because I'd like to get sponsored, but I have no idea of who to ask, or how to ask them.", "Solution_1": "I thought you get sponsored if you make it to MOSP.", "Solution_2": "[quote=\"h_s_potter2002\"]I know that rrusczyk was suggesting to xxreddevilzxx to get a sponsorship to help him (or her) to pay for WOOT.\n\nBut I was just wondering if anyone has ever tried this, for WOOT, or for anything else, and how you went about doing it, because I'd like to get sponsored, but I have no idea of who to ask, or how to ask them.[/quote]\r\n\r\nxxreddevilzxx actually succeeded with this, getting wealthy individuals in his area to sponsor him; you may want to hunt him down to ask how.", "Solution_3": "Actually it was just one individual who had interest in where I was going with the AoPS education. I actually met with him after writing the letter and we talked about what AoPS is (types of things done here). Then we got into financial matters, which didn't take long. He paid through most (I paid minimal amount) of the cost for WOOT and I was really happy I got to do it. I am not sure if I could do WOOT again (I am not sure, is it more expensive this year?) but good luck with the sponsorship for those who are interested. I would suggest writing a letter around all the businesses, or a few if time is of essence. Just write about a few things you have done, a little bit about AoPS and WOOT, and then what you would get by participating. \r\n\r\nHope this helped.", "Solution_4": "Would one be more likely to get a team sponsored or an individual sponsored?", "Solution_5": "I think it would be more likely to get an individual sponsored, because it wouldn't cost as much money.\r\n\r\nBasically you're trying to convince someone that your enrollment in WOOT is worth their money. If you talk to someone you want to sponsor your team, you can tell them that their money is going to support the (your school here) math team. That may be a good way to gain support, but remember that you need a few thousand dollars, and I doubt a single person would be willing to fork over that much money. You may need to talk to a lot of different people. xxreddevilzxx only needed a few hundred dollars, because he is only one person, and because the cost of WOOT was less last year. Good luck to you and your team! :)", "Solution_6": "Er, if xxredddevilzxx is still here, how do you speak to the people such that they get persuaded? Here are my main reasons for taking WOOT:\r\n\r\n- My interest in math\r\n- My weakness in proofs\r\n- Making USAMO next year\r\n- Having fun with other top people in the world\r\n\r\nHow would I phrase it in such a way that it is convincing?" } { "Tag": [], "Problem": "Find $ x, y \\in R$ of this system of this equation: $ \\begin {cases} 1\\plus{}x^3y^3\\equal{}19x^3 \\\\ y\\plus{}xy^2\\equal{}\\minus{}6x^2\\end {cases}$", "Solution_1": "Ok I solved it a bit long, but in a nice way, I hope :D\r\n\r\n$ y(xy \\plus{} 1) \\equal{} \\minus{} 6x^2 \\implies$ from the second equation,\r\n\r\n$ (xy \\plus{} 1) \\equal{} \\frac { \\minus{} 6x^2}{y}$\r\n\r\nThen I have taken the 3rd power of both sides, which gave me,\r\n\r\n$ x^3y^3 \\plus{} 3x^2y^2 \\plus{} 3xy \\plus{} 1 \\equal{} \\frac { \\minus{} 216x^6}{y^3}$\r\n\r\nI already know the value for $ x^3y^3 \\plus{} 1$.. which is $ 19x^3$\r\n\r\n$ 19x^3 \\plus{} 3xy(xy \\plus{} 1) \\equal{} \\frac { \\minus{} 216x^6}{y^3}$ I also know the value for $ xy \\plus{} 1$\r\n\r\n$ 19x^3 \\minus{} 18x^3 \\equal{} \\frac { \\minus{} 216x^6}{y^3}$\r\n\r\n$ x^3 \\equal{} \\frac { \\minus{} 216x^6}{y^3}$\r\n\r\n$ y^3 \\equal{} \\minus{} 216x^3 \\implies y \\equal{} \\minus{} 6x$ --- I used this value at the second equation, it looks easier for me.\r\n\r\n$ \\minus{} 6x \\plus{} x(36x^2) \\equal{} \\minus{} 6x^2 \\implies 36x^3 \\equal{} 6x \\minus{} 6x^2$\r\n\r\n$ 6x^2 \\equal{} 1 \\minus{} x \\implies 6x^2 \\plus{} x \\minus{} 1 \\equal{} 0$\r\n\r\nNow from that, I have the roots $ \\minus{} 1/2,1/3$ Since I now, know the value of x, why shouldn't I put that in the equation I found $ y \\equal{} \\minus{} 6x$\r\n\r\nI get for $ x \\equal{} \\minus{} 1/2$ , $ y \\equal{} 3$ and for $ x \\equal{} 1/3$ , $ y \\equal{} \\minus{} 2$" } { "Tag": [ "trigonometry", "ratio", "geometry", "circumcircle", "angle bisector" ], "Problem": "how do you prove that the interior angles of a triangle are concurrent using ceva's theorem?", "Solution_1": "You mean the interior angle [b]bisectors[/b]?\r\n\r\nWell, if you use the trig version of Ceva, it's very easy. Using the other form of Ceva, you can use the angle bisector theorem.\r\n\r\n[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=422[/img]\r\n\r\n[hide=\"Using Ceva\"]\nBy Ceva, we wish to show $\\frac{BF}{CF}\\cdot\\frac{AE}{BE}\\cdot\\frac{CD}{AD}=1.$ By the angle bisector theorem:\n\n\\begin{eqnarray*} \\frac{BF}{CF} &=& \\frac{AB}{AC}\\\\ \\frac{AE}{BE} &=& \\frac{AC}{BC}\\\\ \\frac{CD}{AD} &=& \\frac{BC}{AC} \\end{eqnarray*}\n\nMultiplying these produces the desired result.[/hide]\n\n[hide=\"Trig Ceva\"]\nWell, $\\sin \\alpha\\sin \\beta\\sin \\gamma=\\sin \\alpha\\sin \\beta\\sin \\gamma$ obviously holds...so we're done![/hide]", "Solution_2": "Question: when you do the trig Ceva, in which direction does one take the ratios of the sin's? More specifically, what is the general form of the trigonometric version of Ceva?", "Solution_3": "Triangle ABC with Cevians AD, BE, CF. The three cevians are concurrent if and only if:\r\n\\[ \\frac{\\sin{\\angle{DAB}}}{\\sin{\\angle{EBA}}} \\cdot \\frac{\\sin{\\angle{EBC}}}{\\sin{\\angle{FCB}}} \\cdot \\frac{\\sin{\\angle{FCA}}}{\\sin{\\angle{DAC}}} = 1 \\]", "Solution_4": "Edit: see below", "Solution_5": "Joml88, doesn't that assume each of those triangles in which you take the sin of one of its angles has the same circumcircle? I am somewhat confused.", "Solution_6": "Oops, what I meant to say was use \\[ AE=\\frac{CE\\sin\\gamma}{\\sin \\angle CAB} \\] and similar relations. Then they will cancel.", "Solution_7": "[quote=\"joml88\"]Oops, what I meant to say was use \\[ AE=\\frac{CE\\sin\\gamma}{\\sin \\angle CAB} \\] and similar relations. Then they will cancel.[/quote]\r\n\r\nCool stuff, thanks for the clarification. It took me some time to realize you were using the diagram up above, heh. :blush:" } { "Tag": [], "Problem": "isn't there a solutions guide at the end?", "Solution_1": "First, you do 6 c 2, which is 15. Then you subtract two. THe reason is that there are 3 ways to make a circle with points across from eachother. However, there are four units between each thing across from eachother (the ones where you choose a unit farthest from it lies on only one circle. Not three). you subtract two because it must be distinct." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Given:\r\n\r\n$-\\frac{1}{\\sqrt{a}}\\leq x_{i}\\leq \\sqrt{a}$\r\n\r\nand\r\n\r\n$x_{1}+x_{2}+\\cdots x_{m}= b*\\sqrt{a}$\r\n\r\nwhat is the maximum value of:\r\n\r\n$x_{1}^{p}+x_{2}^{p}+\\cdots x_{m}^{p}$", "Solution_1": "what about the constrait of $p$?$p\\geq 1$?", "Solution_2": "k is an even positive integer." } { "Tag": [ "number theory", "prime numbers" ], "Problem": "Use each of the digits 1,2,3,4,5,6,7,8,9 exactly twice to form distinct prime numbers whose sum is as small as possible. What must this minimal sum be? (Note: The five smallest primes are 2,3,5,7, and 11)", "Solution_1": "Hello,\r\n\r\n23,29,43,47,619,857 and 6581. The sum is 8199. It existe a better solution ?", "Solution_2": "We can't use 2 or 5 as units digits unless we include 2 and 5 themselves. So our first two numbers are 2 and 5. However, we still need to use the digits 2 and 5 once more, and the digits 4, 6, and 8 twice more. None of these can be units digits, so to minimize the sum we can try to use them as tens digits. To show that this is possible,\r\n\r\n$2,5,23,41,47,59,61,67,83,89$\r\n\r\nThe sum is $477$." } { "Tag": [], "Problem": "[url=http://usinfo.state.gov/ei/Archive/2005/Apr/19-681391.html]Madagascar[/url]\r\n\r\nThis program is one of the many programs instituted by major world powers designed to promote the economic growth of LDC, or Third world countries. Just out of curiousity, what are your opinions on the role of the US and other developed and industrialized nations in aiding those that are not? What limits, if any, should be placed on such activity, if allowed at all? Post your thoughts.", "Solution_1": "It seems to me that LDCs do better as they trade more and that artificial external programs cause at least us much long term harm as good. When wealth is imported into a nation before the education required to manage and sustain that wealth, it seems to create corruption or even warlordism depending on local stability.\r\n\r\nI'm sure there are counterexamples to my point and I would be interested in hearing them for the sake of diagnosing from exactly where the failures of various economic programs stem.\r\n\r\nI also take some issue with the word \"responsibility\" which seems to me highly mutable and easily politicized. I think it often steers attention away from real economic analysis. Questions of responsibility can be more meaningfully phrased in terms of the utility of an individual." } { "Tag": [], "Problem": "[size=150]Find two numbers y so that the distance from point \nS(-3, 2) to point T(5, y) = 10.[/size]", "Solution_1": "[hide]\n\n$D^{2}=(5+3)^{2}+(y-2)^{2}$\n$100=64+(y-2)^{2}$\n$36=(y+2)^{2}$\n$\\pm6=y-2$\n$y=-4,8$[/hide]", "Solution_2": "Great reply but what does D^2 mean here?", "Solution_3": "I just used the distance formula which is\r\n\r\n$D=\\sqrt{(x-x_{1})^{2}+(y-y_{1})^{2}}$, so then\r\n$D^{2}=(x-x_{1})^{2}+(y-y_{1})^{2}$" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Let $f: \\{1,2, ... \\}\\to \\{1,2,...\\}$ such that: \r\ni. $f(n+1)= f(n)-1$ or $4f(n)-1$\r\nii. $\\forall m \\geq 1 , \\exists n \\geq 1$ such that $f(n)=m$\r\n\r\nFind $f(2002)$", "Solution_1": "[quote=\"Bazooka\"]Let $f: Z^{+}\\to Z^{+}$ such that: \ni. $f(n+1)= f(n)-1$ or $4f(n)-1$\nii. $\\forall m \\in Z^{+}, \\exists n \\in Z^{+}$ such that $f(n)=m$\n\nFind $f(2002)$[/quote]\r\nMay be $f: N\\to N$.\r\nIf $f(n)=0$, then $f(n+1)=-1$ :lol:", "Solution_2": "$f(2002)=1069$ or $2093$.", "Solution_3": "$f(2002)$ can be any number. Because \r\n1) and 2) don't define function.\r\nIf $f$ is bijective, then $f(2^{n}+k)=2^{n+1}-k-1,0\\le k<2^{n}$, therefore\r\n$f(2002)=f(2^{10}+978)=2^{11}-979=1069$.", "Solution_4": "$\\forall m\\geq 1$ then $\\exists !n$ such that \r\n$f(n)=m$" } { "Tag": [ "videos", "email" ], "Problem": "Here's a really funny video - [url]http://www.bobricci.com/flash/blockedme.htm[/url]", "Solution_1": "that's hilarious!!! my mom was sitting behind me and was like \"what does 'wtf' mean?\" i'm just like um....lol. but it was great!", "Solution_2": "hahahaha yea - but um... at the end he subscribes to um... susn... p***, don't wanna say pg-18 word here, so edit out, so i dunno - rather funny yes - but don't wanna ur parents behind u like Mathfiend... lol... my momdunno what wtf means either", "Solution_3": "[quote=\"white_horse_king88\"]my momdunno what wtf means either[/quote]this is funnier than the movie itself :D", "Solution_4": "lol lol lol. funny. it actually has a plot and stuff not lyke those songs that spend five minutes whining about life and not really getting anywhere. good one for chat ppl.", "Solution_5": "that was HILARIOUS!!!!!!!!!!!! Wonder who bothered to make the movie.....\r\nbut it was alsome", "Solution_6": "That was great. At a summer camp that I went to one year (by the way, not math related), there was a kid who brought a stereo and played the song that the movie parodies (and the word \"frickin'\" wasn't used) consecutively time after time for a whole day at full volume.", "Solution_7": "[quote=\"themonster\"]That was great. At a summer camp that I went to one year (by the way, not math related), there was a kid who brought a stereo and played the song that the movie parodies (and the word \"frickin'\" wasn't used) consecutively time after time for a whole day at full volume.[/quote]You should email him the topic :)", "Solution_8": "You know that in this generation, our minds have been dirtied (or at least mine has been) when you know what deals with the internet and rhymes with \"torn\"", "Solution_9": "[quote=\"theone853\"]You know that in this generation, our minds have been dirtied (or at least mine has been) when you know what deals with the internet and rhymes with \"torn\"[/quote]\r\n\r\nHey, what do you have against corn? Corn's good, especially corn on the cob.", "Solution_10": ":ninja: \r\n\r\nit's so yellow, that's why i'm prejudice against corn\r\n\r\n ;) :alien: ;)", "Solution_11": "o_O\r\n\r\n...\r\n\r\nInteresting o_O ...", "Solution_12": "Off topic - GASP!!! the holiday smilies are gone.", "Solution_13": "o yes........did u listen to the \"mrs. claus\" one? its so funny. just click on the \"other songs by bob ricci\" link and choose mrs. claus. but then you would need to be familiar with \"stacy's mom\" by fountains of wayne\" but not really, cause theyre both equally funny but.....still innapropriate. id rate it a little less than pg-13", "Solution_14": "[quote=\"Athena\"]o yes........did u listen to the \"mrs. claus\" one? its so funny. just click on the \"other songs by bob ricci\" link and choose mrs. claus. but then you would need to be familiar with \"stacy's mom\" by fountains of wayne\" but not really, cause theyre both equally funny but.....still innapropriate. id rate it a little less than pg-13[/quote]\r\n\r\nLOL, that's awesome.", "Solution_15": "[quote=\"peter\"]Off topic - GASP!!! the holiday smilies are gone.[/quote]It took you some time to notice that :D :D" } { "Tag": [ "algebra", "system of equations" ], "Problem": "Solve the system of equations \\[ \\{\\begin{array}{lll} y+xy^2&=&6x^2\\\\ 1+x^2y^2&=&5x^2\\end{array}\\]", "Solution_1": "[quote=\"April\"]Solve the system of equations\n\\[ \\{\\begin{array}{lll} y + xy^2 & = & 6x^2 \\\\\n1 + x^2y^2 & = & 5x^2\\end{array}\n\\]\n[/quote]\r\ndivide both sides of both equation by x^2 we get the system :\r\n\\[ \\{\\begin{array}{lll} \\frac{y}{x^2} + \\frac{y^2}{x} & = & 6 \\\\\r\n\\frac{1}{x^2} + y^2 & = & 5\\end{array}\r\n\\]\r\nnow let $ \\frac{1}{x}=a ; y=b$ the system is equivalent to\r\n\\[ \\{\\begin{array}{lll} a^2b + b^a & = & 6 \\\\\r\na^2 + b^2 & = & 5\\end{array}\r\n\\]\r\nnow let S=a+b and P=ab we have \r\n\\[ \\{\\begin{array}{lll} SP & = & 6 \\\\\r\n\\S^2-2P& = & 5\\end{array}\r\n\\]\r\nso we have $ S^3-5S-12=0$ which means $ (S-3)(S^2+3S+4)=0$ gives $ S=3$", "Solution_2": "[quote=\"HTA -- typos fixed\"]divide both sides of both equation by x^2 we get the system :\n\\[ \\{\\begin{array}{lll} \\frac {y}{x^2} + \\frac {y^2}{x} & = & 6 \\\\\n\\frac {1}{x^2} + y^2 & = & 5\\end{array}\n\\]\nnow let $ \\frac {1}{x} = a ; y = b$ the system is equivalent to\n\\[ \\{\\begin{array}{lll} a^2b + \\mathbf{\\color{red}{b^2 a}} & = & 6 \\\\\na^2 + b^2 & = & 5\\end{array}\n\\]\nnow let S=a+b and P=ab we have\n\\[ \\{\\begin{array}{lll} SP & = & 6 \\\\\n\\mathbf{\\color{red}{S^2}} - 2P & = & 5\\end{array}\n\\]\nso we have $ S^3 - 5S - 12 = 0$ which means $ (S - 3)(S^2 + 3S + 4) = 0$ gives $ S = 3$[/quote]\r\n\r\nUsing $ SP = 6$, we get $ P = 2$. Thus, $ a$ and $ b$ are the roots of $ x^2 - 3x + 2$, or $ 1$ and $ 2$.\r\n\r\n$ (a,b) = (1,2) \\implies \\boxed{(x,y) = (1,2)}$\r\n$ (a,b) = (2,1) \\implies \\boxed{(x,y) = ( \\frac {1}{2}, 1 )}$\r\n\r\nI believe there are also four very ugly complex solutions, which result from using the two complex values of $ S$ from that cubic in $ S$." } { "Tag": [ "geometry", "trigonometry", "calculus", "integration", "analytic geometry", "function", "trig identities" ], "Problem": "I would like to find the area under a certain polar curve. However, I can not solve for $r$. Is there any general way this can be done? If not, is there a way to get a calculator or Mathematica to give me a good estimate?\r\n\r\nI'm trying to integrate $r^{2}$ from $0$ to somewhere around $\\pi/4$ (wouldn't want to spend time finding the exact upper bound if I won't be able to complete it anyway), where $r=\\sqrt{r^{2}+1-2r\\cos{\\theta}}+\\sqrt{r^{2}+2-2\\sqrt{2}r\\cos{(\\pi/4-\\theta)}}$.\r\n\r\nAny input appreciated.", "Solution_1": "There is a TeX error in the formula: it should contain $\\cos(\\pi/4-\\theta)$. You may be able to do this in Mathematica by applying [url=http://documents.wolfram.com/mathematica/functions/Solve]Solve[/url] to your equation and then using [url=http://documents.wolfram.com/mathematica/functions/NIntegrate]NIntegrate[/url]. \r\nBut are you sure that the problem is set up correctly? It's not often that one integrates $r(\\theta)$ with $r$ representing a polar curve.", "Solution_2": "heh good point, $r^{2}$ then I suppose. Problem editted, I'll give it a little more thought with the adjustment and try out Mathematica, but I'll probably need a bit more assistance.", "Solution_3": "Yikes, here is how Mathematica solved for $r$, which after seeing, I had some doubts. Sure enough when I tried to integrate it says: \r\nNIntegrate::slwcon: Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration being 0, oscillatory integrand, or insufficient WorkingPrecision. If your integrand is oscillatory try using the option Method->Oscillatory in NIntegrate.\r\n\r\nI don't know how to do that, but I doubt it will help, right?\r\n\r\n[quote]\\!\\(NIntegrate[\\((1\\/12\\ \\((4\\ \\@2\\ Cos[\u03c0\\/4 - t] + 4\\ \n Cos[t])\\) - 1\\/2\\ \u221a\\((\\(-\\(4\\/3\\)\\) + 8\\/3\\ Cos[\n \u03c0\\/4 - t]\\^2 - 8\\/9\\ \\@2\\ Cos[\n \u03c0\\/4 - t]\\ Cos[t] + \\(4\\ Cos[t]\\^2\\)\\/3 - \\((64\\ \n 2\\^\\(1/3\\)\\ Cos[\u03c0\\/4 - t]\\^4)\\)/\\((\\(-1259712\\) + \n 746496\\ Cos[\u03c0\\/4 - t]\\^6 + 1679616\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ \\\nCos[t] - 2239488\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^5\\ Cos[t] + 1259712\\ \n Cos[t]\\^2 - 5038848\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^2 + 5598720\\ \\\nCos[\u03c0\\/4 - t]\\^4\\ \n Cos[t]\\^2 + 2519424\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^3 - \\\n3732480\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 - 839808\\ Cos[t]\\^4 + 2799360\\ \\\nCos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 - 559872\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^5 + 93312\\ \\\nCos[t]\\^6 + \u221a\\((1586874322944 - 1880739938304\\ Cos[\u03c0\\/4 - t]\\^6 - \\\n4231664861184\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t] + 5642219814912\\ \\@2\\ Cos[\u03c0\\/4 - \\\nt]\\^5\\ Cos[t] - 3173748645888\\ Cos[t]\\^2 + 12694994583552\\ Cos[\u03c0\\/4 - t]\\^2\\ \\\nCos[t]\\^2 - 14105549537280\\ Cos[\u03c0\\/4 - t]\\^4\\ Cos[t]\\^2 - 6347497291776\\ \\@2\\ \\\nCos[\u03c0\\/4 - t]\\ Cos[t]\\^3 + 9873884676096\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 + \\\n3702706753536\\ Cos[t]\\^4 - 8463329722368\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 + \\\n2115832430592\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^5 - 470184984576\\ \\\nCos[t]\\^6)\\))\\)\\^\\(1/3\\) - \\((96\\ 2\\^\\(5/6\\)\\ Cos[\u03c0\\/4 - t]\\ \\\nCos[t])\\)/\\((\\(-1259712\\) + 746496\\ Cos[\u03c0\\/4 - t]\\^6 + 1679616\\ \\@2\\ Cos[\u03c0\\/4 \\\n- t]\\^3\\ Cos[t] - 2239488\\ \\@2\\ \n Cos[\u03c0\\/4 - t]\\^5\\ Cos[t] + 1259712\\ Cos[t]\\^2 - 5038848\\\n\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^2 + \n 5598720\\ Cos[\u03c0\\/4 - t]\\^4\\ Cos[t]\\^2 + 2519424\\ \\@2\\ \\\nCos[\u03c0\\/4 - t]\\ Cos[t]\\^3 - 3732480\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 - 839808\\\n\\ Cos[t]\\^4 + 2799360\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 - 559872\\ \\@2\\ Cos[\u03c0\\/4 - \\\nt]\\ Cos[t]\\^5 + 93312\\ Cos[t]\\^6 + \u221a\\((1586874322944 - 1880739938304\\ \\\nCos[\u03c0\\/4 - t]\\^6 - 4231664861184\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t] + \\\n5642219814912\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^5\\ Cos[t] - 3173748645888\\ Cos[t]\\^2 + \\\n12694994583552\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^2 - 14105549537280\\ Cos[\u03c0\\/4 - \\\nt]\\^4\\ Cos[t]\\^2 - 6347497291776\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^3 + \\\n9873884676096\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 + 3702706753536\\ Cos[t]\\^4 - \\\n8463329722368\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 + 2115832430592\\ \\@2\\ Cos[\u03c0\\/4 - \\\nt]\\ Cos[t]\\^5 - 470184984576\\ Cos[t]\\^6)\\))\\)\\^\\(1/3\\) + \\((128\\ 2\\^\\(5/6\\)\\ \\\nCos[\u03c0\\/4 - t]\\^3\\ Cos[t])\\)/\\((\\(-1259712\\) + 746496\\ Cos[\u03c0\\/4 - t]\\^6 + \\\n1679616\\ \\@2\\ Cos[\n \u03c0\\/4 - t]\\^3\\ Cos[t] - 2239488\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^5\\ \\\nCos[t] + 1259712\\ Cos[t]\\^2 - 5038848\\ Cos[\n \u03c0\\/4 - t]\\^2\\ Cos[t]\\^2 + 5598720\\ Cos[\u03c0\\/4 - t]\\^4\\ \\\nCos[t]\\^2 + 2519424\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^3 - 3732480\\ \\@2\\ Cos[\u03c0\\/4 - \\\nt]\\^3\\ Cos[t]\\^3 - 839808\\ Cos[t]\\^4 + 2799360\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 - \\\n559872\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^5 + 93312\\ Cos[t]\\^6 + \u221a\\((1586874322944 \\\n- 1880739938304\\ Cos[\u03c0\\/4 - t]\\^6 - 4231664861184\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ \\\nCos[t] + 5642219814912\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^5\\ Cos[t] - 3173748645888\\ Cos[t]\\\n\\^2 + 12694994583552\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^2 - 14105549537280\\ Cos[\u03c0\\/4 \\\n- t]\\^4\\ Cos[t]\\^2 - 6347497291776\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^3 + \\\n9873884676096\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 + 3702706753536\\ Cos[t]\\^4 - \\\n8463329722368\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 + 2115832430592\\ \\@2\\ Cos[\u03c0\\/4 - \\\nt]\\ Cos[t]\\^5 - 470184984576\\ Cos[t]\\^6)\\))\\)\\^\\(1/3\\) + \\((96\\ 2\\^\\(1/3\\)\\ \\\nCos[t]\\^2)\\)/\\((\\(-1259712\\) + 746496\\ Cos[\u03c0\\/4 - t]\\^6 + 1679616\\ \\@2\\ Cos[\n \u03c0\\/4 - t]\\^3\\ Cos[t] - 2239488\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^5\\ \\\nCos[t] + 1259712\\ Cos[t]\\^2 - 5038848\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^2 + 5598720\\ \\\nCos[\u03c0\\/4 - t]\\^4\\ Cos[t]\\^2 + 2519424\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^3 - \\\n3732480\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 - 839808\\ Cos[t]\\^4 + 2799360\\ \\\nCos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 - 559872\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^5 + 93312\\ \\\nCos[t]\\^6 + \u221a\\((1586874322944 - 1880739938304\\ Cos[\u03c0\\/4 - t]\\^6 - \\\n4231664861184\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t] + 5642219814912\\ \\@2\\ Cos[\u03c0\\/4 - \\\nt]\\^5\\ Cos[t] - 3173748645888\\ Cos[t]\\^2 + 12694994583552\\ Cos[\u03c0\\/4 - t]\\^2\\ \\\nCos[t]\\^2 - 14105549537280\\ Cos[\u03c0\\/4 - t]\\^4\\ Cos[t]\\^2 - 6347497291776\\ \\@2\\ \\\nCos[\u03c0\\/4 - t]\\ Cos[t]\\^3 + 9873884676096\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 + \\\n3702706753536\\ Cos[t]\\^4 - 8463329722368\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 + \\\n2115832430592\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^5 - 470184984576\\ \\\nCos[t]\\^6)\\))\\)\\^\\(1/3\\) - \\((192\\ 2\\^\\(1/3\\)\\ Cos[\u03c0\\/4 - t]\\^2\\ \\\nCos[t]\\^2)\\)/\\((\\(-1259712\\) + 746496\\ Cos[\u03c0\\/4 - t]\\^6 + 1679616\\ \\@2\\ Cos[\u03c0\\\n\\/4 - t]\\^3\\ Cos[t] - 2239488\\ \\@2\\ Cos[\n \u03c0\\/4 - t]\\^5\\ Cos[t] + 1259712\\ Cos[t]\\^2 - 5038848\\ \\\nCos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^2 + 5598720\\ Cos[\u03c0\\/4 - t]\\^4\\ Cos[t]\\^2 + 2519424\\ \\\n\\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^3 - 3732480\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 - \\\n839808\\ Cos[t]\\^4 + 2799360\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 - 559872\\ \\@2\\ Cos[\u03c0\\\n\\/4 - t]\\ Cos[t]\\^5 + 93312\\ Cos[t]\\^6 + \u221a\\((1586874322944 - 1880739938304\\ \\\nCos[\u03c0\\/4 - t]\\^6 - 4231664861184\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t] + \\\n5642219814912\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^5\\ Cos[t] - 3173748645888\\ Cos[t]\\^2 + \\\n12694994583552\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^2 - 14105549537280\\ Cos[\u03c0\\/4 - \\\nt]\\^4\\ Cos[t]\\^2 - 6347497291776\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^3 + \\\n9873884676096\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 + 3702706753536\\ Cos[t]\\^4 - \\\n8463329722368\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 + 2115832430592\\ \\@2\\ Cos[\u03c0\\/4 - \\\nt]\\ Cos[t]\\^5 - 470184984576\\ Cos[t]\\^6)\\))\\)\\^\\(1/3\\) + \\((64\\ 2\\^\\(5/6\\)\\ \\\nCos[\u03c0\\/4 - t]\\ \n Cos[t]\\^3)\\)/\\((\\(-1259712\\) + 746496\\ Cos[\u03c0\\/4 - t]\\^6 + \\\n1679616\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[\n t] - 2239488\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^5\\ Cos[t] + 1259712\\ \\\nCos[t]\\^2 - 5038848\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^2 + 5598720\\ Cos[\u03c0\\/4 - t]\\^4\\ \\\nCos[t]\\^2 + 2519424\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^3 - 3732480\\ \\@2\\ Cos[\u03c0\\/4 - \\\nt]\\^3\\ Cos[t]\\^3 - 839808\\ Cos[t]\\^4 + 2799360\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 - \\\n559872\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^5 + 93312\\ Cos[t]\\^6 + \u221a\\((1586874322944 \\\n- 1880739938304\\ Cos[\u03c0\\/4 - t]\\^6 - 4231664861184\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ \\\nCos[t] + 5642219814912\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^5\\ Cos[t] - 3173748645888\\ Cos[t]\\\n\\^2 + 12694994583552\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^2 - 14105549537280\\ Cos[\u03c0\\/4 \\\n- t]\\^4\\ Cos[t]\\^2 - 6347497291776\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^3 + \\\n9873884676096\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 + 3702706753536\\ Cos[t]\\^4 - \\\n8463329722368\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 + 2115832430592\\ \\@2\\ Cos[\u03c0\\/4 - \\\nt]\\ Cos[t]\\^5 - 470184984576\\ Cos[t]\\^6)\\))\\)\\^\\(1/3\\) - \\((16\\ \n 2\\^\\(1/3\\)\\ Cos[t]\\^4)\\)/\\((\\(-1259712\\) + 746496\\ Cos[\u03c0\\/4 \\\n- t]\\^6 + 1679616\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t] - 2239488\\ \\@2\\ Cos[\u03c0\\/4 - \\\nt]\\^5\\ Cos[t] + 1259712\\ Cos[t]\\^2 - 5038848\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^2 + \\\n5598720\\ Cos[\u03c0\\/4 - t]\\^4\\ Cos[t]\\^2 + 2519424\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^3 \\\n- 3732480\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 - 839808\\ Cos[t]\\^4 + 2799360\\ \\\nCos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 - 559872\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^5 + 93312\\ \\\nCos[t]\\^6 + \u221a\\((1586874322944 - 1880739938304\\ Cos[\u03c0\\/4 - t]\\^6 - \\\n4231664861184\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t] + 5642219814912\\ \\@2\\ Cos[\u03c0\\/4 - \\\nt]\\^5\\ Cos[t] - 3173748645888\\ Cos[t]\\^2 + 12694994583552\\ Cos[\u03c0\\/4 - t]\\^2\\ \\\nCos[t]\\^2 - 14105549537280\\ Cos[\u03c0\\/4 - t]\\^4\\ Cos[t]\\^2 - 6347497291776\\ \\@2\\ \\\nCos[\u03c0\\/4 - t]\\ Cos[t]\\^3 + 9873884676096\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 + \\\n3702706753536\\ Cos[t]\\^4 - 8463329722368\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 + \\\n2115832430592\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^5 - 470184984576\\ \\\nCos[t]\\^6)\\))\\)\\^\\(1/3\\) - \\(\\(1\\/\\(81\\ 2\\^\\(1/3\\)\\)\\)\\((\\((\\(-1259712\\) + \\\n746496\\ Cos[\u03c0\\/4 - t]\\^6 + 1679616\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t] - 2239488\\ \\\n\\@2\\ Cos[\u03c0\\/4 - t]\\^5\\ Cos[t] + 1259712\\ Cos[t]\\^2 - 5038848\\ Cos[\u03c0\\/4 - \\\nt]\\^2\\ Cos[t]\\^2 + 5598720\\ Cos[\n \u03c0\\/4 - t]\\^4\\ \n Cos[t]\\^2 + 2519424\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^3 \\\n- 3732480\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 - 839808\\ Cos[t]\\^4 + 2799360\\ \\\nCos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 - 559872\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^5 + 93312\\ \\\nCos[t]\\^6 + \u221a\\((1586874322944 - 1880739938304\\ Cos[\u03c0\\/4 - t]\\^6 - \\\n4231664861184\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t] + 5642219814912\\ \\@2\\ Cos[\u03c0\\/4 - \\\nt]\\^5\\ Cos[t] - 3173748645888\\ Cos[t]\\^2 + 12694994583552\\ Cos[\u03c0\\/4 - t]\\^2\\ \\\nCos[t]\\^2 - 14105549537280\\ Cos[\u03c0\\/4 - t]\\^4\\ Cos[t]\\^2 - 6347497291776\\ \\@2\\ \\\nCos[\u03c0\\/4 - t]\\ Cos[t]\\^3 + 9873884676096\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 + \\\n3702706753536\\ Cos[t]\\^4 - 8463329722368\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 + \\\n2115832430592\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^5 - 470184984576\\ \\\nCos[t]\\^6)\\))\\)\\^\\(1/3\\))\\)\\))\\) - 1\\/2\\ \u221a\\((4\\/3 - 8\\/3\\ Cos[\u03c0\\/4 - t]\\^2 + \\\n1\\/12\\ \\((\\(-4\\)\\ \\@2\\ Cos[\u03c0\\/4 - t] - 4\\ Cos[t])\\)\\^2 + 8\\/9\\ \\@2\\ Cos[\u03c0\\/4 \\\n- t]\\ Cos[t] - \\(4\\ Cos[t]\\^2\\)\\/3 - 2\\/3\\ \\((\n 6 - 8\\ Cos[\n \u03c0\\/4 - t]\\^2 + 8\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t] - 4\\ \\\nCos[t]\\^2)\\) + \\((64\\ 2\\^\\(1/3\\)\\ Cos[\n \u03c0\\/4 - t]\\^4)\\)/\\((\\(-1259712\\) + 746496\\ Cos[\u03c0\\/4 -\n t]\\^6 + 1679616\\ \\@2\\ Cos[\n \u03c0\\/4 - t]\\^3\\ Cos[t] - 2239488\\ \\@2\\ Cos[\n \u03c0\\/4 - t]\\^5\\ Cos[t] + 1259712\\ Cos[\n t]\\^2 - 5038848\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^2 + 5598720\\ \\\nCos[\u03c0\\/4 - t]\\^4\\ Cos[\n t]\\^2 + 2519424\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^3 - 3732480\\ \\\n\\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 - 839808\\ Cos[t]\\^4 + 2799360\\ Cos[\u03c0\\/4 - t]\\\n\\^2\\ Cos[t]\\^4 - 559872\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^5 + 93312\\ Cos[t]\\^6 + \u221a\\\n\\((1586874322944 - 1880739938304\\ Cos[\u03c0\\/4 - t]\\^6 - 4231664861184\\ \\@2\\ \\\nCos[\u03c0\\/4 - t]\\^3\\ Cos[t] + 5642219814912\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^5\\ Cos[t] - \\\n3173748645888\\ Cos[t]\\^2 + 12694994583552\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^2 - \\\n14105549537280\\ Cos[\u03c0\\/4 - t]\\^4\\ Cos[t]\\^2 - 6347497291776\\ \\@2\\ Cos[\u03c0\\/4 - \\\nt]\\ Cos[t]\\^3 + 9873884676096\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 + \\\n3702706753536\\ Cos[t]\\^4 - 8463329722368\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 + \\\n2115832430592\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^5 - 470184984576\\ \\\nCos[t]\\^6)\\))\\)\\^\\(1/3\\) + \\((96\\ 2\\^\\(5/6\\)\\ Cos[\u03c0\\/4 - t]\\ \\\nCos[t])\\)/\\((\\(-1259712\\) + 746496\\ Cos[\u03c0\\/4 - t]\\^6 + 1679616\\ \\@2\\ Cos[\u03c0\\/4 \\\n- t]\\^3\\ Cos[t] - 2239488\\ \\@2\\ \n Cos[\u03c0\\/4 - t]\\^5\\ Cos[t] + 1259712\\ Cos[t]\\^2 - \n 5038848\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^2 + \n 5598720\\ Cos[\u03c0\\/4 - t]\\^4\\ Cos[t]\\^2 + 2519424\\ \\@2\\ \\\nCos[\u03c0\\/4 - t]\\ Cos[t]\\^3 - 3732480\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 - 839808\\\n\\ Cos[t]\\^4 + 2799360\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 - 559872\\ \\@2\\ Cos[\u03c0\\/4 - \\\nt]\\ Cos[t]\\^5 + 93312\\ Cos[t]\\^6 + \u221a\\((1586874322944 - 1880739938304\\ \\\nCos[\u03c0\\/4 - t]\\^6 - 4231664861184\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t] + \\\n5642219814912\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^5\\ Cos[t] - 3173748645888\\ Cos[t]\\^2 + \\\n12694994583552\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^2 - 14105549537280\\ Cos[\u03c0\\/4 - \\\nt]\\^4\\ Cos[t]\\^2 - 6347497291776\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^3 + \\\n9873884676096\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 + 3702706753536\\ Cos[t]\\^4 - \\\n8463329722368\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 + 2115832430592\\ \\@2\\ Cos[\u03c0\\/4 - \\\nt]\\ Cos[t]\\^5 - 470184984576\\ Cos[t]\\^6)\\))\\)\\^\\(1/3\\) - \\((128\\ 2\\^\\(5/6\\)\\ \\\nCos[\u03c0\\/4 - t]\\^3\\ Cos[t])\\)/\\((\\(-1259712\\) + 746496\\ Cos[\u03c0\\/4 - t]\\^6 + \\\n1679616\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ \n Cos[t] - 2239488\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^5\\ Cos[t] + \\\n1259712\\ Cos[t]\\^2 - 5038848\\ Cos[\u03c0\\/4 - \n t]\\^2\\ Cos[t]\\^2 + 5598720\\ Cos[\u03c0\\/4 - t]\\^4\\ Cos[t]\\^2 + \\\n2519424\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^3 - 3732480\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ \\\nCos[t]\\^3 - 839808\\ Cos[t]\\^4 + 2799360\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 - 559872\\\n\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^5 + 93312\\ Cos[t]\\^6 + \u221a\\((1586874322944 - \\\n1880739938304\\ Cos[\u03c0\\/4 - t]\\^6 - 4231664861184\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ \\\nCos[t] + 5642219814912\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^5\\ Cos[t] - 3173748645888\\ Cos[t]\\\n\\^2 + 12694994583552\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^2 - 14105549537280\\ Cos[\u03c0\\/4 \\\n- t]\\^4\\ Cos[t]\\^2 - 6347497291776\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^3 + \\\n9873884676096\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 + 3702706753536\\ Cos[t]\\^4 - \\\n8463329722368\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 + 2115832430592\\ \\@2\\ Cos[\u03c0\\/4 - \\\nt]\\ Cos[t]\\^5 - 470184984576\\ Cos[t]\\^6)\\))\\)\\^\\(1/3\\) - \\((96\\ 2\\^\\(1/3\\)\\ \\\nCos[t]\\^2)\\)/\\((\\(-1259712\\) + 746496\\ Cos[\u03c0\\/4 - t]\\^6 + 1679616\\ \\@2\\ Cos[\u03c0\\\n\\/4 - t]\\^3\\ Cos[t] - 2239488\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^5\\ Cos[t] + \n 1259712\\ Cos[t]\\^2 - 5038848\\ Cos[\u03c0\\/4 - t]\\^2\\ \\\nCos[t]\\^2 + 5598720\\ Cos[\u03c0\\/4 - t]\\^4\\ Cos[t]\\^2 + 2519424\\ \\@2\\ Cos[\u03c0\\/4 - \\\nt]\\ Cos[t]\\^3 - 3732480\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 - 839808\\ Cos[t]\\^4 \\\n+ 2799360\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 - 559872\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ \\\nCos[t]\\^5 + 93312\\ Cos[t]\\^6 + \u221a\\((1586874322944 - 1880739938304\\ Cos[\u03c0\\/4 - \\\nt]\\^6 - 4231664861184\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t] + 5642219814912\\ \\@2\\ \\\nCos[\u03c0\\/4 - t]\\^5\\ Cos[t] - 3173748645888\\ Cos[t]\\^2 + 12694994583552\\ \\\nCos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^2 - 14105549537280\\ Cos[\u03c0\\/4 - t]\\^4\\ Cos[t]\\^2 - \\\n6347497291776\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^3 + 9873884676096\\ \\@2\\ Cos[\u03c0\\/4 - \\\nt]\\^3\\ Cos[t]\\^3 + 3702706753536\\ Cos[t]\\^4 - 8463329722368\\ Cos[\u03c0\\/4 - t]\\^2\\\n\\ Cos[t]\\^4 + 2115832430592\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^5 - 470184984576\\ \\\nCos[t]\\^6)\\))\\)\\^\\(1/3\\) + \\((192\\ 2\\^\\(1/3\\)\\ Cos[\u03c0\\/4 - t]\\^2\\ \\\nCos[t]\\^2)\\)/\\((\\(-1259712\\) + 746496\\ Cos[\u03c0\\/4 - t]\\^6 + 1679616\\ \\@2\\ Cos[\u03c0\\\n\\/4 - t]\\^3\\ Cos[t] - 2239488\\ \\@2\\ Cos[\n \u03c0\\/4 - t]\\^5\\ Cos[t] + 1259712\\ Cos[t]\\^2 - 5038848\\ \\\nCos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^2 + 5598720\\ Cos[\u03c0\\/4 - t]\\^4\\ Cos[t]\\^2 + 2519424\\ \\\n\\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^3 - 3732480\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 - \\\n839808\\ Cos[t]\\^4 + 2799360\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 - 559872\\ \\@2\\ Cos[\u03c0\\\n\\/4 - t]\\ Cos[t]\\^5 + 93312\\ Cos[t]\\^6 + \u221a\\((1586874322944 - 1880739938304\\ \\\nCos[\u03c0\\/4 - t]\\^6 - 4231664861184\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t] + \\\n5642219814912\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^5\\ Cos[t] - 3173748645888\\ Cos[t]\\^2 + \\\n12694994583552\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^2 - 14105549537280\\ Cos[\u03c0\\/4 - \\\nt]\\^4\\ Cos[t]\\^2 - 6347497291776\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^3 + \\\n9873884676096\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 + 3702706753536\\ Cos[t]\\^4 - \\\n8463329722368\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 + 2115832430592\\ \\@2\\ Cos[\u03c0\\/4 - \\\nt]\\ Cos[t]\\^5 - 470184984576\\ Cos[t]\\^6)\\))\\)\\^\\(1/3\\) - \\((64\\ 2\\^\\(5/6\\)\\ \\\nCos[\u03c0\\/4 - t]\\ Cos[\n t]\\^3)\\)/\\((\\(-1259712\\) + 746496\\ Cos[\u03c0\\/4 - \n t]\\^6 + 1679616\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[\n t] - 2239488\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^5\\ Cos[t] + 1259712\\ \\\nCos[t]\\^2 - 5038848\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^2 + 5598720\\ Cos[\u03c0\\/4 - t]\\^4\\ \\\nCos[t]\\^2 + 2519424\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^3 - 3732480\\ \\@2\\ Cos[\u03c0\\/4 - \\\nt]\\^3\\ Cos[t]\\^3 - 839808\\ Cos[t]\\^4 + 2799360\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 - \\\n559872\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^5 + 93312\\ Cos[t]\\^6 + \u221a\\((1586874322944 \\\n- 1880739938304\\ Cos[\u03c0\\/4 - t]\\^6 - 4231664861184\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ \\\nCos[t] + 5642219814912\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^5\\ Cos[t] - 3173748645888\\ Cos[t]\\\n\\^2 + 12694994583552\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^2 - 14105549537280\\ Cos[\u03c0\\/4 \\\n- t]\\^4\\ Cos[t]\\^2 - 6347497291776\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^3 + \\\n9873884676096\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 + 3702706753536\\ Cos[t]\\^4 - \\\n8463329722368\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 + 2115832430592\\ \\@2\\ Cos[\u03c0\\/4 - \\\nt]\\ Cos[t]\\^5 - 470184984576\\ Cos[t]\\^6)\\))\\)\\^\\(1/3\\) + \\((16\\ 2\\^\\(1/\n 3\\)\\ Cos[t]\\^4)\\)/\\((\\(-1259712\\) + 746496\\ Cos[\u03c0\\/4 - \\\nt]\\^6 + 1679616\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t] - 2239488\\ \\@2\\ Cos[\u03c0\\/4 - \\\nt]\\^5\\ Cos[t] + 1259712\\ Cos[t]\\^2 - 5038848\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^2 + \\\n5598720\\ Cos[\u03c0\\/4 - t]\\^4\\ Cos[t]\\^2 + 2519424\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^3 \\\n- 3732480\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 - 839808\\ Cos[t]\\^4 + 2799360\\ \\\nCos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 - 559872\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^5 + 93312\\ \\\nCos[t]\\^6 + \u221a\\((1586874322944 - 1880739938304\\ Cos[\u03c0\\/4 - t]\\^6 - \\\n4231664861184\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t] + 5642219814912\\ \\@2\\ Cos[\u03c0\\/4 - \\\nt]\\^5\\ Cos[t] - 3173748645888\\ Cos[t]\\^2 + 12694994583552\\ Cos[\u03c0\\/4 - t]\\^2\\ \\\nCos[t]\\^2 - 14105549537280\\ Cos[\u03c0\\/4 - t]\\^4\\ Cos[t]\\^2 - 6347497291776\\ \\@2\\ \\\nCos[\u03c0\\/4 - t]\\ Cos[t]\\^3 + 9873884676096\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 + \\\n3702706753536\\ Cos[t]\\^4 - 8463329722368\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 + \\\n2115832430592\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^5 - 470184984576\\ \\\nCos[t]\\^6)\\))\\)\\^\\(1/3\\) + \\(\\(1\\/\\(81\\ 2\\^\\(1/3\\)\\)\\)\\((\\((\\(-1259712\\) + \\\n746496\\ Cos[\u03c0\\/4 - t]\\^6 + 1679616\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t] - 2239488\\ \\\n\\@2\\ Cos[\u03c0\\/4 - t]\\^5\\ Cos[t] + 1259712\\ Cos[t]\\^2 - 5038848\\ Cos[\u03c0\\/4 - \\\nt]\\^2\\ Cos[t]\\^2 + 5598720\\ Cos[\u03c0\\/4 - t]\\^4\\ Cos[t]\\^2 + 2519424\\ \\@2\\ Cos[\u03c0\\\n\\/4 - t]\\ Cos[t]\\^3 - 3732480\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[\n t]\\^3 - 839808\\ Cos[t]\\^4 + 2799360\\ \\\nCos[\u03c0\\/4 - t]\\^2\\ \n Cos[t]\\^4 - 559872\\ \\@2\\ Cos[\u03c0\\/4 - \\\nt]\\ Cos[t]\\^5 + 93312\\ Cos[t]\\^6 + \u221a\\((1586874322944 - 1880739938304\\ \\\nCos[\u03c0\\/4 - t]\\^6 - 4231664861184\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t] + \\\n5642219814912\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^5\\ Cos[t] - 3173748645888\\ Cos[t]\\^2 + \\\n12694994583552\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^2 - 14105549537280\\ Cos[\u03c0\\/4 - \\\nt]\\^4\\ Cos[t]\\^2 - 6347497291776\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^3 + \\\n9873884676096\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 + 3702706753536\\ Cos[t]\\^4 - \\\n8463329722368\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 + 2115832430592\\ \\@2\\ Cos[\u03c0\\/4 - \\\nt]\\ Cos[t]\\^5 - 470184984576\\ Cos[t]\\^6)\\))\\)\\^\\(1/3\\))\\)\\) - \\\n\\((\\(-\\(8\\/3\\)\\)\\ \\((4\\ \\@2\\ Cos[\u03c0\\/4 - t] - 4\\ Cos[t])\\) + 1\\/3\\ \\((\\(-4\\)\\ \\\n\\@2\\ Cos[\u03c0\\/4 - t] - 4\\ Cos[t])\\)\\ \\((\\(-\\(1\\/9\\)\\)\\ \\((\\(-4\\)\\ \\@2\\ Cos[\u03c0\\/4 \\\n- t] - 4\\ Cos[t])\\)\\^2 + 4\\/3\\ \\((6 - 8\\ \n Cos[\u03c0\\/4 - t]\\^2 + 8\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ \\\nCos[t] - 4\\ Cos[t]\\^2)\\))\\))\\)/\\((4\\ \u221a\\((\\(-\\(4\\/3\\)\\) + \n 8\\/3\\ Cos[\u03c0\\/4 - \n t]\\^2 - \n 8\\/9\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t] + \\(4\\ \\\nCos[t]\\^2\\)\\/3 - \\((64\\ 2\\^\\(1/3\\)\\ Cos[\u03c0\\/4 - \n t]\\^4)\\)/\\((\\(-1259712\\) + 746496\\ \\\nCos[\u03c0\\/4 - t]\\^6 + 1679616\\ \\@2\\ Cos[\n \u03c0\\/4 - t]\\^3\\ \n Cos[t] - 2239488\\ \\@2\\ Cos[\n \u03c0\\/4 - t]\\^5\\ Cos[t] + 1259712\\ Cos[t]\\^2 - \\\n5038848\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^2 + 5598720\\ Cos[\u03c0\\/4 - \n t]\\^4\\ Cos[t]\\^2 + 2519424\\ \\@2\\ Cos[\u03c0\\/4 \\\n- t]\\ Cos[t]\\^3 - 3732480\\ \\@2\\ \n Cos[\u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 - 839808\\ \\\nCos[t]\\^4 + 2799360\\ Cos[\u03c0\\/4 - \n t]\\^2\\ Cos[t]\\^4 - 559872\\ \\@2\\ Cos[\u03c0\\/4 - \\\nt]\\ Cos[t]\\^5 + 93312\\ Cos[t]\\^6 + \u221a\\((1586874322944 - 1880739938304\\ \\\nCos[\u03c0\\/4 - t]\\^6 - 4231664861184\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t] + \\\n5642219814912\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^5\\ Cos[t] - 3173748645888\\ Cos[t]\\^2 + \\\n12694994583552\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^2 - 14105549537280\\ Cos[\u03c0\\/4 - \\\nt]\\^4\\ Cos[t]\\^2 - 6347497291776\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^3 + \\\n9873884676096\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 + 3702706753536\\ Cos[t]\\^4 - \\\n8463329722368\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 + 2115832430592\\ \\@2\\ Cos[\u03c0\\/4 - \\\nt]\\ Cos[t]\\^5 - 470184984576\\ Cos[t]\\^6)\\))\\)\\^\\(1/3\\) - \\((96\\ 2\\^\\(5/6\\)\\ \\\nCos[\u03c0\\/4 - t]\\ Cos[t])\\)/\\((\\(-1259712\\) + 746496\\ Cos[\u03c0\\/4 - t]\\^6 + 1679616\\\n\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t] - 2239488\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^5\\ Cos[t] + \\\n1259712\\ Cos[t]\\^2 - 5038848\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^2 + 5598720\\ Cos[\u03c0\\/4 \\\n- t]\\^4\\ Cos[t]\\^2 + 2519424\\ \\@2\\ \n Cos[\u03c0\\/4 - t]\\ Cos[t]\\^3 - 3732480\\ \\@2\\ Cos[\n \u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 - 839808\\ Cos[\n t]\\^4 + 2799360\\ Cos[\u03c0\\/4 - t]\\^2\\ \\\nCos[t]\\^4 - 559872\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^5 + 93312\\ Cos[t]\\^6 + \\\n\u221a\\((1586874322944 - 1880739938304\\ Cos[\u03c0\\/4 - t]\\^6 - 4231664861184\\ \\@2\\ \\\nCos[\u03c0\\/4 - t]\\^3\\ Cos[t] + 5642219814912\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^5\\ Cos[t] - \\\n3173748645888\\ Cos[t]\\^2 + 12694994583552\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^2 - \\\n14105549537280\\ Cos[\u03c0\\/4 - t]\\^4\\ Cos[t]\\^2 - 6347497291776\\ \\@2\\ Cos[\u03c0\\/4 - \\\nt]\\ Cos[t]\\^3 + 9873884676096\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 + \\\n3702706753536\\ Cos[t]\\^4 - 8463329722368\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 + \\\n2115832430592\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^5 - 470184984576\\ \\\nCos[t]\\^6)\\))\\)\\^\\(1/3\\) + \\((128\\ 2\\^\\(5/6\\)\\ Cos[\u03c0\\/4 - t]\\^3\\ \\\nCos[t])\\)/\\((\\(-1259712\\) + 746496\\ Cos[\u03c0\\/4 - t]\\^6 + 1679616\\ \\@2\\ Cos[\u03c0\\/4 \\\n- t]\\^3\\ Cos[t] - 2239488\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^5\\ Cos[t] + 1259712\\ Cos[t]\\^2 \\\n- 5038848\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^2 + 5598720\\ Cos[\u03c0\\/4 - t]\\^4\\ \n Cos[t]\\^2 + 2519424\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ \\\nCos[t]\\^3 - 3732480\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[\n t]\\^3 - 839808\\ Cos[t]\\^4 + 2799360\\ \\\nCos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 - 559872\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^5 + 93312\\ \\\nCos[t]\\^6 + \u221a\\((1586874322944 - 1880739938304\\ Cos[\u03c0\\/4 - t]\\^6 - \\\n4231664861184\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t] + 5642219814912\\ \\@2\\ Cos[\u03c0\\/4 - \\\nt]\\^5\\ Cos[t] - 3173748645888\\ Cos[t]\\^2 + 12694994583552\\ Cos[\u03c0\\/4 - t]\\^2\\ \\\nCos[t]\\^2 - 14105549537280\\ Cos[\u03c0\\/4 - t]\\^4\\ Cos[t]\\^2 - 6347497291776\\ \\@2\\ \\\nCos[\u03c0\\/4 - t]\\ Cos[t]\\^3 + 9873884676096\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 + \\\n3702706753536\\ Cos[t]\\^4 - 8463329722368\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 + \\\n2115832430592\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^5 - 470184984576\\ \\\nCos[t]\\^6)\\))\\)\\^\\(1/3\\) + \\((96\\ 2\\^\\(1/3\\)\\ Cos[t]\\^2)\\)/\\((\\(-1259712\\) + \\\n746496\\ Cos[\u03c0\\/4 - t]\\^6 + 1679616\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t] - 2239488\\ \\\n\\@2\\ Cos[\u03c0\\/4 - t]\\^5\\ Cos[t] + 1259712\\ Cos[t]\\^2 - 5038848\\ Cos[\u03c0\\/4 - \\\nt]\\^2\\ Cos[t]\\^2 + 5598720\\ Cos[\u03c0\\/4 -\n t]\\^4\\ Cos[t]\\^2 + 2519424\\ \\@2\\ Cos[\u03c0\\/4 \\\n- t]\\ Cos[t]\\^3 - 3732480\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 - 839808\\ \\\nCos[t]\\^4 + 2799360\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 - 559872\\ \\@2\\ Cos[\u03c0\\/4 - t]\\\n\\ Cos[t]\\^5 + 93312\\ Cos[t]\\^6 + \u221a\\((1586874322944 - 1880739938304\\ Cos[\u03c0\\/4 \\\n- t]\\^6 - 4231664861184\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t] + 5642219814912\\ \\@2\\ \\\nCos[\u03c0\\/4 - t]\\^5\\ Cos[t] - 3173748645888\\ Cos[t]\\^2 + 12694994583552\\ \\\nCos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^2 - 14105549537280\\ Cos[\u03c0\\/4 - t]\\^4\\ Cos[t]\\^2 - \\\n6347497291776\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^3 + 9873884676096\\ \\@2\\ Cos[\u03c0\\/4 - \\\nt]\\^3\\ Cos[t]\\^3 + 3702706753536\\ Cos[t]\\^4 - 8463329722368\\ Cos[\u03c0\\/4 - t]\\^2\\\n\\ Cos[t]\\^4 + 2115832430592\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^5 - 470184984576\\ \\\nCos[t]\\^6)\\))\\)\\^\\(1/3\\) - \\((192\\ 2\\^\\(1/3\\)\\ Cos[\u03c0\\/4 - t]\\^2\\ \\\nCos[t]\\^2)\\)/\\((\\(-1259712\\) + 746496\\ Cos[\u03c0\\/4 - t]\\^6 + 1679616\\ \\@2\\ Cos[\u03c0\\\n\\/4 - t]\\^3\\ Cos[t] - 2239488\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^5\\ Cos[t] + 1259712\\ Cos[\n t]\\^2 - \n 5038848\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^2 + \\\n5598720\\ Cos[\u03c0\\/4 - t]\\^4\\ Cos[t]\\^2 + 2519424\\ \\@2\\ Cos[\n \u03c0\\/4 - t]\\ Cos[t]\\^3 - 3732480\\ \\@2\\ \\\nCos[\u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 - 839808\\ Cos[t]\\^4 + 2799360\\ Cos[\u03c0\\/4 - t]\\^2\\ \\\nCos[t]\\^4 - 559872\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^5 + 93312\\ Cos[t]\\^6 + \\\n\u221a\\((1586874322944 - 1880739938304\\ Cos[\u03c0\\/4 - t]\\^6 - 4231664861184\\ \\@2\\ \\\nCos[\u03c0\\/4 - t]\\^3\\ Cos[t] + 5642219814912\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^5\\ Cos[t] - \\\n3173748645888\\ Cos[t]\\^2 + 12694994583552\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^2 - \\\n14105549537280\\ Cos[\u03c0\\/4 - t]\\^4\\ Cos[t]\\^2 - 6347497291776\\ \\@2\\ Cos[\u03c0\\/4 - \\\nt]\\ Cos[t]\\^3 + 9873884676096\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 + \\\n3702706753536\\ Cos[t]\\^4 - 8463329722368\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 + \\\n2115832430592\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^5 - 470184984576\\ \\\nCos[t]\\^6)\\))\\)\\^\\(1/3\\) + \\((64\\ 2\\^\\(5/6\\)\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^3)\\)/\\((\\\n\\(-1259712\\) + 746496\\ Cos[\u03c0\\/4 - t]\\^6 + 1679616\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ \\\nCos[t] - 2239488\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^5\\ Cos[t] + \n 1259712\\ Cos[t]\\^2 - 5038848\\ Cos[\u03c0\\/4 - t]\\^2\\ \n Cos[t]\\^2 + \n 5598720\\ Cos[\u03c0\\/4 - t]\\^4\\ Cos[\n t]\\^2 + 2519424\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\\n\\^3 - 3732480\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 - 839808\\ Cos[t]\\^4 + 2799360\\\n\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 - 559872\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^5 + 93312\\\n\\ Cos[t]\\^6 + \u221a\\((1586874322944 - 1880739938304\\ Cos[\u03c0\\/4 - t]\\^6 - \\\n4231664861184\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t] + 5642219814912\\ \\@2\\ Cos[\u03c0\\/4 - \\\nt]\\^5\\ Cos[t] - 3173748645888\\ Cos[t]\\^2 + 12694994583552\\ Cos[\u03c0\\/4 - t]\\^2\\ \\\nCos[t]\\^2 - 14105549537280\\ Cos[\u03c0\\/4 - t]\\^4\\ Cos[t]\\^2 - 6347497291776\\ \\@2\\ \\\nCos[\u03c0\\/4 - t]\\ Cos[t]\\^3 + 9873884676096\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 + \\\n3702706753536\\ Cos[t]\\^4 - 8463329722368\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 + \\\n2115832430592\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^5 - 470184984576\\ \\\nCos[t]\\^6)\\))\\)\\^\\(1/3\\) - \\((16\\ 2\\^\\(1/3\\)\\ Cos[t]\\^4)\\)/\\((\\(-1259712\\) + \\\n746496\\ Cos[\u03c0\\/4 - t]\\^6 + 1679616\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t] - 2239488\\ \\\n\\@2\\ Cos[\u03c0\\/4 - t]\\^5\\ Cos[\n t] + 1259712\\ Cos[t]\\^2 - 5038848\\ Cos[\u03c0\\/4 \\\n- t]\\^2\\ Cos[t]\\^2 + 5598720\\ Cos[\u03c0\\/4 - t]\\^4\\ Cos[t]\\^2 + 2519424\\ \\@2\\ \\\nCos[\u03c0\\/4 - t]\\ Cos[t]\\^3 - 3732480\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 - 839808\\\n\\ Cos[t]\\^4 + 2799360\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 - 559872\\ \\@2\\ Cos[\u03c0\\/4 - \\\nt]\\ Cos[t]\\^5 + 93312\\ Cos[t]\\^6 + \u221a\\((1586874322944 - 1880739938304\\ \\\nCos[\u03c0\\/4 - t]\\^6 - 4231664861184\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t] + \\\n5642219814912\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^5\\ Cos[t] - 3173748645888\\ Cos[t]\\^2 + \\\n12694994583552\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^2 - 14105549537280\\ Cos[\u03c0\\/4 - \\\nt]\\^4\\ Cos[t]\\^2 - 6347497291776\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^3 + \\\n9873884676096\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 + 3702706753536\\ Cos[t]\\^4 - \\\n8463329722368\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 + 2115832430592\\ \\@2\\ Cos[\u03c0\\/4 - \\\nt]\\ Cos[t]\\^5 - 470184984576\\ Cos[t]\\^6)\\))\\)\\^\\(1/3\\) - \\(\\(1\\/\\(81\\ \\\n2\\^\\(1/3\\)\\)\\)\\((\\((\\(-1259712\\) + 746496\\ Cos[\u03c0\\/4 - t]\\^6 + 1679616\\ \\@2\\ \\\nCos[\u03c0\\/4 - t]\\^3\\ Cos[t] - 2239488\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^5\\ Cos[t] + 1259712\\ \n Cos[\n t]\\^2 - 5038848\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^2 \\\n+ 5598720\\ Cos[\u03c0\\/4 - t]\\^4\\ Cos[t]\\^2 + 2519424\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ \\\nCos[t]\\^3 - 3732480\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 - 839808\\ Cos[t]\\^4 + \\\n2799360\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 - 559872\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^5 \\\n+ 93312\\ Cos[t]\\^6 + \u221a\\((1586874322944 - 1880739938304\\ Cos[\u03c0\\/4 - t]\\^6 - \\\n4231664861184\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t] + 5642219814912\\ \\@2\\ Cos[\u03c0\\/4 - \\\nt]\\^5\\ Cos[t] - 3173748645888\\ Cos[t]\\^2 + 12694994583552\\ Cos[\u03c0\\/4 - t]\\^2\\ \\\nCos[t]\\^2 - 14105549537280\\ Cos[\u03c0\\/4 - t]\\^4\\ Cos[t]\\^2 - 6347497291776\\ \\@2\\ \\\nCos[\u03c0\\/4 - t]\\ Cos[t]\\^3 + 9873884676096\\ \\@2\\ Cos[\u03c0\\/4 - t]\\^3\\ Cos[t]\\^3 + \\\n3702706753536\\ Cos[t]\\^4 - 8463329722368\\ Cos[\u03c0\\/4 - t]\\^2\\ Cos[t]\\^4 + \\\n2115832430592\\ \\@2\\ Cos[\u03c0\\/4 - t]\\ Cos[t]\\^5 - 470184984576\\ \\\nCos[t]\\^6)\\))\\)\\^\\(1/3\\))\\)\\))\\))\\))\\))\\)^2, t, 0, .916670976316]\\)[/quote]", "Solution_4": "Maybe someone would help if you told us how this problem came about. :maybe:", "Solution_5": "http://www.artofproblemsolving.com/Forum/viewtopic.php?t=29997\r\n\r\nI was trying a different approach using D as the pole because then it would for sure only have one $r$ for each $\\theta$ so it seemed nicer. Because when making my sketch of the curve I used a ruler pivoting at D. Law of cosines lead to my aforementioned equation.", "Solution_6": "I think the conversion to polar coordinates does not help much. You could try to define a function u(x,y) = 1 if AP+BP < DP and 0 otherwise. (Here P has coordinates (x,y). Mathematica has a function [url=http://documents.wolfram.com/v5/TheMathematicaBook/PrinciplesOfMathematica/EvaluationOfExpressions/2.6.8.html]If[/url]). Then integrate $u(x,y)$ numerically via NIntegrate.\r\n\r\nEdit: after much wailing and gnashing of teeth, Mathematica spat put 0.113882, confirming an earlier computation of [b]JBL[/b].\r\n\r\n[size=75]I doubt that the problem has any nice answer or solution.[/size]" } { "Tag": [ "geometry", "LaTeX", "trigonometry" ], "Problem": "I was reading the book AoPS volume II, and in chapter 4 the last question (Page 50) wasn't solved clearly in the solution book. The author said \"after lots of algebra we get the disired expression.\" :D \r\nCan someone help me with this? \r\n(OMG i tried to write the problem in latex code but i can't, can someone write the question in latex code with ABCD and s in italic?)\r\nProve that if in AMCD we let a=AB, b=BC, c=CD, and d=DA, we have [ABCD]^2=(s-a)(s-b)(s-c)(s-d)=abcd cos^2((B+D)/2) where s is the semiperimeter of the quadrilateral. What does this expression, called Brahmagupta's formula, yield for a cyclick quadrilateral?", "Solution_1": "Prove that if in $ABCD$ we let $a=AB$, $b=BC$, $c=CD$, and $d=DA$, we have $[ABCD]^2=(s-a)(s-b)(s-c)(s-d)-abcd\\ cos^2((B+D)/2)$ where $s$ is the semiperimeter of the quadrilateral. (its $ABCD$ not $AMCD$, i guess!)", "Solution_2": "the same (similar) thing is true for a triangle. that how you can easilly find the area of the trianlge, if you have the three sides of the triangle.", "Solution_3": "[quote=\"amirhtlusa\"]Prove that if in $ABCD$ we let $a=AB$, $b=BC$, $c=CD$, and $d=DA$, we have $[ABCD]^2=(s-a)(s-b)(s-c)(s-d)=abcd cos^2((B+D)/2)$ where $s$ is the semiperimeter of the quadrilateral. What does this expression, called Brahmagupta's formula, yield for a cyclick quadrilateral?\n(its $ABCD$ not $AMCD$, i guess!)[/quote]\r\n\r\nactually your version of Brahmagupta is not correct. it should be $[ABCD]^2=(s-a)(s-b)(s-c)(s-d) - abcd cos^2((B+D)/2)$", "Solution_4": "oops. i fixed it :D \r\nthanks", "Solution_5": "use \\cos to make the cos in $\\LaTeX$ look nicer: $\\cos ABC$ as opposed to $cos ABC$ ;)", "Solution_6": "http://www.artofproblemsolving.com/Forum/viewtopic.php?p=112006\r\n\r\nProof by darij.", "Solution_7": "What is B And D in cos2((B+D)/2)", "Solution_8": "sorry guys I was abit lost I know whats A abd B", "Solution_9": "Hey I got Heron fromula for this one \r\nBut i havnt been able to yet figure out the other one\r\nIs there anyother way apart from Brahmaguptas generallised formula to prove it \r\nSicne I teach in class 8 and 9 We dont yet teach trigonometry so using that method would be abstract" } { "Tag": [ "inequalities", "function", "real analysis", "real analysis unsolved" ], "Problem": "Prove that :\r\n\r\n$ \\sum\\limits_{i\\equal{}1}^{k}{\\sum\\limits_{j\\equal{}1}^{k}}{|x_i \\minus{}x_j|} \\leq k^2$,\r\n\r\nwhere $ x_t \\in [0,2]$ , $ \\forall t\\equal{}1,2,\\cdots k$.\r\n\r\nDoes the equality hold ? if so when?\r\n\r\n :D :D :D", "Solution_1": "The left-hand side is a piecewise convex function of any given $ x_i$, so it takes on its maximum values at the endpoints, namely $ x_i\\equal{}0$, $ 2$, or $ x_j$ for some $ j$. Since setting some of the variables equal to each other [i]still[/i] yields a piecewise convex function, we can assume without loss of generality that all of the $ x_i$ are equal to $ 0$ or $ 2$.\r\n\r\nAssume that $ n$ of the $ x_i$ are equal to $ 0$, and $ k\\minus{}n$ are equal to $ 2$. Then we compute that the sum is equal to $ 4n(k\\minus{}n)\\leq k^2$, with equality if and only if $ k\\equal{}2n$.\r\n\r\nNote that if $ k$ is odd then the sum's maximum is $ k^2\\minus{}1$, achieved when either $ \\frac{k\\minus{}1}{2}$ or $ \\frac{k\\plus{}1}{2}$ of the variables are $ 0$.", "Solution_2": ":first: :D" } { "Tag": [ "inequalities" ], "Problem": "[img]http://www.mathlinks.ro/Forum/files/abc_107.gif[/img]", "Solution_1": "An ugly but straitforward solution may be to use Cauchy-Schwarz ($(\\sum a_i/b_i) \\cdot (\\sum a_i b_i ) \\geq (\\sum a_i)^2$), which ends proving $8 (\\sum a^3)^2 \\geq 3 (\\sum a^3(a+b)^3)$ which is true by rearrangement after expansion.", "Solution_2": "See also http://www.mathlinks.ro/Forum/viewtopic.php?t=45819 .\r\n\r\n darij" } { "Tag": [ "geometry", "geometric transformation", "reflection", "rotation", "search", "geometry proposed" ], "Problem": "Let ABC be a triangle.\r\nReflect the vertex A in the sideline BC; you get a new point A'.\r\nConstruct an equilateral triangle CAB' on the side CA of triangle ABC inwardly (i. e. such that the points B and B' lie in the same halfplane with respect to the line CA).\r\nConstruct an equilateral triangle ABC' on the side AB of triangle ABC outwardly (i. e. such that the points C and C' lie in different halfplanes with respect to the line CA).\r\nShow that the points A', B', C' are collinear.\r\n\r\nMentioned in\r\n\r\n[url=http://forumgeom.fau.edu/FG2003volume3/FG200304index.html]Lawrence S. Evans, [i]Some Configurations of Triangle Centers[/i], Forum Geometricorum 3 (2003) pp. 49-56.[/url]\r\n\r\n Darij", "Solution_1": "Take a point $A^*$ on $BC$ s.t. $\\angle A'AA^*=\\frac {\\pi}3$ (a directed angle). It's clear that $AA^*=AA'$, and all we have to do now is rotate the line $BCA^*$ around $A$ with an angle of $-\\frac {\\pi}3$, and we get the line $B'C'A'$.", "Solution_2": "Cool proof, I didn't think it was so simple! (Actually, when I posted the problem, I thought: it should not be very hard, but I have no time to search for a synthetic proof now, at least I have one using barycentrics...)\r\n\r\n Darij" } { "Tag": [ "inequalities" ], "Problem": "Prove that $AM \\geq HM$ for $\\{a_0,a_1,\\dots,a_n\\}$ where all $a_i > 0$ for $0\\leq {i} \\leq {n}$. Using either AM-GM or Cauchy and show that there is equality iff all $a_i=1$.\r\n\r\nnotes:\r\n$AM=\\frac{a_0+a_1+a_2+\\dots+a_n}{n}$\r\nand\r\n$HM=\\frac{n}{\\frac{1}{a_0}+\\frac{1}{a_1}+\\frac{1}{a_2}+\\dots+\\frac{1}{a_n}}$", "Solution_1": "[hide=\"Cauchy-Schwarz inequality Solution\"]By the Cauchy-Schwarz inequality, we have\n\n$(a_1 + a_2 + \\dots + a_n)(\\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_n}) \\geq (1 + 1 + \\dots + 1)^2 = n^2$\n\nRearranging, we get\n$\\mbox{AM} = \\frac{a_1 + a_2 + \\dots + a_n}{n} \\geq \\frac{n}{ \\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_n}} = \\mbox{HM}$\nwith equality iff\n$a_1^2 = a_2^2 = \\dots = a_n^2$\nor since all $a_i > 0$,\n$a_1 = a_2 = \\dots = a_n$\n(instead of all $a_i = 1$)[/hide]\r\n\r\nHopefully I didn't mess anything up... :P\r\nEDIT: I did :P \r\nEDIT 2: I did again :P :rotfl:", "Solution_2": "nebula beat me to it. :P\r\n\r\nIncidentally, this is one of several inequalities that can be proven very simply with a slight rewriting of Jensen's.", "Solution_3": "o yeah woops idk what i was thinking when i wrote the $a_i=1$ sorry :blush: :oops:", "Solution_4": "[quote=\"nebula42\"]with equality iff\n$\\frac{a_0}{\\frac{1}{a_0}} = \\frac{a_1}{\\frac{1}{a_1}} = \\frac{a_2}{\\frac{1}{a_2}} = ... = \\frac{a_n}{\\frac{1}{a_n}}$\nor\n$a_0^2 = a_1^2 = a_2^2 = ... = a_n^2$\n(instead of all $a_i = 1$)\n\nHopefully I didn't mess anything up... :P[/quote]\nHmm are you sure about this last part? Let $a_1=3,a_2=3,a_3=-3$, then $a_1^2=a_2^2=a_3^2$ But $\\frac{3+3-3}3\\neq\\frac3{\\frac13+\\frac13-\\frac13}$\nI think equality holds iff $a_1=a_2=\\cdots=a_n$\n\n[quote=\"maokid7\"]$AM=\\frac{a_0+a_1+a_2+\\dots+a_n}{n}$\nand\n$HM=\\frac{n}{\\frac{1}{a_0}+\\frac{1}{a_1}+\\frac{1}{a_2}+\\dots+\\frac{1}{a_n}}$[/quote]\r\nI think it's\r\n$AM=\\frac{a_1+a_2+\\dots+a_n}{n}$\r\nand\r\n$HM=\\frac{n}{\\frac{1}{a_1}+\\frac{1}{a_2}+\\dots+\\frac{1}{a_n}}$\r\n\r\nBecause by yours the AM of $a$ and $b$ would be $a+b$", "Solution_5": "[quote=\"nat mc\"][quote=\"nebula42\"]with equality iff\n$\\frac{a_0}{\\frac{1}{a_0}} = \\frac{a_1}{\\frac{1}{a_1}} = \\frac{a_2}{\\frac{1}{a_2}} = ... = \\frac{a_n}{\\frac{1}{a_n}}$\nor\n$a_0^2 = a_1^2 = a_2^2 = ... = a_n^2$\n(instead of all $a_i = 1$)\n\nHopefully I didn't mess anything up... :P[/quote]\nHmm are you sure about this last part? Let $a_1=3,a_2=3,a_3=-3$, then $a_1^2=a_2^2=a_3^2$ But $\\frac{3+3-3}3\\neq\\frac3{\\frac13+\\frac13-\\frac13}$\nI think equality holds iff $a_1=a_2=\\cdots=a_n$\n\n[quote=\"maokid7\"]$AM=\\frac{a_0+a_1+a_2+\\dots+a_n}{n}$\nand\n$HM=\\frac{n}{\\frac{1}{a_0}+\\frac{1}{a_1}+\\frac{1}{a_2}+\\dots+\\frac{1}{a_n}}$[/quote]\nI think it's\n$AM=\\frac{a_1+a_2+\\dots+a_n}{n}$\nand\n$HM=\\frac{n}{\\frac{1}{a_1}+\\frac{1}{a_2}+\\dots+\\frac{1}{a_n}}$\n\nBecause by yours the AM of $a$ and $b$ would be $a+b$[/quote]\r\n\r\nOh yeah, I can't believe I didn't notice that :oops: ... oh well. Is there a stronger inequality that can prove that $a_1 = a_2 = ... = a_n$ is necessary?", "Solution_6": "yes i know it should be $a_1,a_2,\\dots,a_n$ and not $a_0,a_1,\\dots,a_n$\r\n\r\nand if you noticed it said $a_i>0$ so $a_1^2=a_2^2,\\dots,a_n^2$ works.", "Solution_7": "[quote=\"maokid7\"]yes i know it should be $a_1,a_2,\\dots,a_n$ and not $a_0,a_1,\\dots,a_n$\n\nand if you noticed it said $a_i>0$ so $a_1^2=a_2^2,\\dots,a_n^2$ works.[/quote]\r\n\r\nOh yeah... I can't believe I didn't see that either... :blush: :P", "Solution_8": "[quote=\"maokid7\"]yes i know it should be $a_1,a_2,\\dots,a_n$ and not $a_0,a_1,\\dots,a_n$\n\nand if you noticed it said $a_i>0$ so $a_1^2=a_2^2,\\dots,a_n^2$ works.[/quote]\r\noh ok, I didn't see that. Then it's fine :)" } { "Tag": [ "inequalities", "logarithms", "calculus", "derivative", "function", "search", "inequalities proposed" ], "Problem": "If a,b,c,d,e are positive real numbers such that $ abcde\\equal{}1$, prove that\r\n\\[ \\frac{a\\minus{}1}{a^2\\plus{}a\\plus{}1}\\plus{}\\frac{b\\minus{}1}{b^2\\plus{}b\\plus{}1}\\plus{}\\frac{c\\minus{}1}{c^2\\plus{}c\\plus{}1}\\plus{}\\frac{d\\minus{}1}{d^2\\plus{}d\\plus{}1}\\plus{}\\frac{e\\minus{}1}{e^2\\plus{}e\\plus{}1}\\leq0.\\]", "Solution_1": "by AM-GM we know that:\r\n$ a^2\\plus{}a\\plus{}1\\geq 3a$\r\nthus we get that:\r\n$ \\frac{a\\minus{}1}{a^2\\plus{}a\\plus{}1}\\leq\\frac{a\\minus{}1}{3a}$\r\nthus its sufficient to show that:\r\n$ \\frac{a\\minus{}1}{3a}\\plus{}\\frac{b\\minus{}1}{3b}\\plus{}\\frac{c\\minus{}1}{3c}\\plus{}\\frac{d\\minus{}1}{3d}\\plus{}\\frac{e\\minus{}1}{3e}\\leq 0$\r\n$ \\Rightarrow \\frac{a\\minus{}1}{a}\\plus{}\\frac{b\\minus{}1}{b}\\plus{}\\frac{c\\minus{}1}{c}\\plus{}\\frac{d\\minus{}1}{d}\\plus{}\\frac{e\\minus{}1}{e}\\leq 0$\r\nwhich is equivalent to:\r\n$ 5\\leq \\frac{1}{a}\\plus{}\\frac{1}{b}\\plus{}\\frac{1}{c}\\plus{}\\frac{1}{d}\\plus{}\\frac{1}{e}$\r\nwhich can be obtained directly from AM-GM...", "Solution_2": "[quote=\"BaBaK Ghalebi\"]thus we get that:\n$ \\frac {a \\minus{} 1}{a^2 \\plus{} a \\plus{} 1}\\leq\\frac {a \\minus{} 1}{3a}$\n[/quote]\r\n\r\nI that here you are wrong because you don't know the sign of $ a\\minus{}1$ , you you can't multiply ...", "Solution_3": "This is actually a problem in Secrets in Inequalities, and I (or zaizai) did post this problem in this forum, too. Try to search.\r\n\r\n@ hitek. I think using a fake nickname in Mathlinks at this time is not a good idea.", "Solution_4": "[quote=\"hitek\"]If a,b,c,d,e are positive real numbers such that $ abcde \\equal{} 1$, prove that\n\\[ \\frac {a \\minus{} 1}{a^2 \\plus{} a \\plus{} 1} \\plus{} \\frac {b \\minus{} 1}{b^2 \\plus{} b \\plus{} 1} \\plus{} \\frac {c \\minus{} 1}{c^2 \\plus{} c \\plus{} 1} \\plus{} \\frac {d \\minus{} 1}{d^2 \\plus{} d \\plus{} 1} \\plus{} \\frac {e \\minus{} 1}{e^2 \\plus{} e \\plus{} 1}\\leq0.\n\\]\n[/quote]\r\n$ \\sum_{cyc}\\frac{a\\minus{}1}{a^2\\plus{}a\\plus{}1}\\leq0\\Leftrightarrow\\sum_{cyc}\\left(\\frac{a\\minus{}1}{a^2\\plus{}a\\plus{}1}\\minus{}\\frac{\\ln a}{3}\\right)\\leq0.$\r\nLet $ f(a)\\equal{}\\frac{a\\minus{}1}{a^2\\plus{}a\\plus{}1}\\minus{}\\frac{\\ln a}{3}.$ Then $ f'(a)\\equal{}\\frac{(1\\minus{}a)(a^3\\plus{}3a^2\\plus{}9a\\plus{}1)}{2(a^2\\plus{}a\\plus{}1)^2}.$\r\nThus, $ a_{max}\\equal{}1$ and it's done.", "Solution_5": "[quote=\"arqady\"]\nLet $ f(a) \\equal{} \\frac {a \\minus{} 1}{a^2 \\plus{} a \\plus{} 1} \\minus{} \\frac {\\ln a}{3}.$ Then $ f'(a) \\equal{} \\frac {(1 \\minus{} a)(a^3 \\plus{} 3a^2 \\plus{} 9a \\plus{} 1)}{2(a^2 \\plus{} a \\plus{} 1)^2}.$\nThus, $ a_{max} \\equal{} 1$ and it's done.[/quote]\r\nwhy is $ a_{max}\\equal{}1$\r\nfor example $ (a,b,c,d,e)\\equal{}(100,200,1,\\frac{1}{100},\\frac{1}{200})$", "Solution_6": "[quote=\"pardesi\"][quote=\"arqady\"]\nLet $ f(a) \\equal{} \\frac {a \\minus{} 1}{a^2 \\plus{} a \\plus{} 1} \\minus{} \\frac {\\ln a}{3}.$ Then $ f'(a) \\equal{} \\frac {(1 \\minus{} a)(a^3 \\plus{} 3a^2 \\plus{} 9a \\plus{} 1)}{2(a^2 \\plus{} a \\plus{} 1)^2}.$\nThus, $ a_{max} \\equal{} 1$ and it's done.[/quote]\nwhy is $ a_{max} \\equal{} 1$\nfor example $ (a,b,c,d,e) \\equal{} (100,200,1,\\frac {1}{100},\\frac {1}{200})$[/quote]\r\n$ f(100) \\equal{} \\frac {99}{10101} \\minus{} \\frac {\\ln100}{3} \\equal{} \\minus{} 1.525255719... < 0.$ \r\nBy the way, is proved that $ f(a)\\leq f(1) \\equal{} 0$ for all $ a > 0.$\r\nHence, $ f(200) < 0,$ $ f\\left(\\frac {1}{100}\\right) < 0$ and $ f\\left(\\frac {1}{200}\\right) < 0.$\r\nThus, $ \\sum_{cyc}f(a)\\leq0.$ :wink:\r\nMaybe you have meant to $ \\max a$ $ ?$\r\nBut I saw about $ a_{max}.$ :wink:", "Solution_7": "[quote=\"arqady\"]Let $ f(a) \\equal{} \\frac {a \\minus{} 1}{a^2 \\plus{} a \\plus{} 1} \\minus{} \\frac {\\ln a}{3}$. Then $ f'(a) \\equal{} \\frac {(1 \\minus{} a)(a^3 \\plus{} 3a^2 \\plus{} 9a \\plus{} 1)}{2(a^2 \\plus{} a \\plus{} 1)^2}.$\nThus, $ a_{max} \\equal{} 1$ and it's done.\n[/quote]\r\nu said $ a_{max} \\equal{} 1$ which is wrong as a can be any real no.\r\nactually is suppose u meant $ f_{max}$ is at $ a\\equal{}1$", "Solution_8": "[quote=\"arqady\"]\n$ \\sum_{cyc}\\frac {a \\minus{} 1}{a^2 \\plus{} a \\plus{} 1}\\leq0\\Leftrightarrow\\sum_{cyc}\\left(\\frac {a \\minus{} 1}{a^2 \\plus{} a \\plus{} 1} \\minus{} \\frac {\\ln a}{3}\\right)\\leq0.$\n\n[/quote]\r\nIs it $ \\sum_{cyc}\\left(\\frac {a \\minus{} 1}{a^2 \\plus{} a \\plus{} 1} \\minus{} \\frac {\\ln a}{3}\\right)\\leq0.$ instead of \r\n\r\n$ \\sum_{cyc}\\left(\\frac {a \\minus{} 1}{a^2 \\plus{} a \\plus{} 1} \\plus{} \\frac {\\ln a}{3}\\right)\\leq0.$??? :maybe:", "Solution_9": "[quote=\"pardesi\"][quote=\"arqady\"]Let $ f(a) \\equal{} \\frac {a \\minus{} 1}{a^2 \\plus{} a \\plus{} 1} \\minus{} \\frac {\\ln a}{3}$. Then $ f'(a) \\equal{} \\frac {(1 \\minus{} a)(a^3 \\plus{} 3a^2 \\plus{} 9a \\plus{} 1)}{2(a^2 \\plus{} a \\plus{} 1)^2}.$\nThus, $ a_{max} \\equal{} 1$ and it's done.\n[/quote]\nu said $ a_{max} \\equal{} 1$ which is wrong as a can be any real no.\nactually is suppose u meant $ f_{max}$ is at $ a \\equal{} 1$[/quote]\r\nlet $ f(x)\\equal{}\\frac{a\\minus{}1}{a^2\\plus{}a\\plus{}1}\\minus{}\\frac{\\ln a}{3}$ as he said,then we get that for $ a\\equal{}1$ the derivative of $ f$ becomse $ 0$ i.e. $ f'(1)\\equal{}0$ and also for $ a>1$ the function will become decreasing and for $ a\\leq 1$ we get that $ f(a)\\leq f(1)$...", "Solution_10": "yes i already got the proof .but was confused with my earlier interpretation due to the way the proof was presented", "Solution_11": "I am sorry. My proof is wrong.\r\n$ f'(a)\\equal{}\\frac{(1\\minus{}a)(a^3\\plus{}6a^2\\plus{}3a\\minus{}1)}{3a(a^2\\plus{}a\\plus{}1)^2}$ and it's nothing.", "Solution_12": "yes but the general idea is great and can be seen pretty often now in ML.\r\nafter Harazi Substitution i think this is a pretty good method to solve inequalities of this type.", "Solution_13": "Ok. Here is the problem in my book (we prove it, for six numbers instead of five)\r\n\r\nIs any idea for it, Arqady? :lol: You are close to the end of the solution.", "Solution_14": "Yes, hungkhtn! It's easy enough and true for six numbers. :lol:\r\nThe beginning is the same! :wink:", "Solution_15": "10p of searching, finally, I've gone to \r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?search_id=1021565391&t=152115\r\n\r\nAbove problem was once chosen for VIF problem of day. I think Zaizai also has the same solution. It is very easy if we know the technique, but very hard if not. Am I true, Arqady?", "Solution_16": "I agree with you, hungkhtn. The same technique see here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=45516" } { "Tag": [ "inequalities", "geometry", "circumcircle", "cyclic quadrilateral", "perpendicular bisector" ], "Problem": "On the circle $K$ there are given three distinct points $A,B,C.$ Construct (using only a straightedge and a compass) a fourth point $D$ on $K$ such that a circle can be inscribed in the quadrilateral thus obtained. \r\n\r\n[hide=\"Solution\"]Note that a circle can be inscribed within a quadrilateral $ABCD$ if and only if $AB+CD = AB+CD.$ Also, note that given a triangle $ABC$ and a point $D$ such that $AB+CD = AB+CD,$ the locus of $D$ is a line. So, given points $A,B,C$ on the quadrilateral, let $AB = b, BC = a.$ Now, draw a circle with centre $A$ and radius $2b+a$ and another circle with centre $C$ and radius $2a+b.$ Choose the intersection points of these circles that is farther away from $B$ and call this point $D_{1}.$ Then, draw a circle with centre $A$ with radius $3b/2+a/2$ and another circle with centre $C$ with radius $3a/2+b/2.$ Take the intersection points of these circles that is farther away from $B$ and call this point $D_{2}.$ Draw $D_{1}D_{2}$ and extend this segment infintitely. Take the intersection point between $D_{1}D_{2}$ and the circle circumscribing $ABC$ which makes $ABCD$ convex and call this point $X.$ $X$ is the desired point.[/hide]", "Solution_1": "[quote=\"Karth\"] $AB+CD = AC+BD$ [/quote]\r\nThis is incorrect. It should be $AB+CD=AD+BC$.\r\n\r\nEDIT: You still haven't corrected it. (Nothing special about $AB+CD=AB+CD$ :D ).", "Solution_2": "Ah, yes, my mistake. I shall change it. :)", "Solution_3": "Here's another solution:\r\n\r\n[hide]\nRearrange the equation $AB+CD=AD+BC$ (1) into $BC-AB=CD-AD$ (2). Construct a circle with radius AB and center B and let its intersection with BC be E (WLOG assume that BC>=AB and CD>=AD, if the inequalities are reversed then a similar situation results). Construct a circle with radius CE which intersects CD at F (CD hasn't been constructed yet) We can now reduce equation (2) to $CE=CF$. Subtracting equation (1) from equation (2), we find that $AB+DF=AD+BE$. However, $AB=BE$, so $DF=AD$. Let angle $ABC=\\theta$. Because ABCD is a cyclic quadrilateral, angle $CDA=180-\\theta$. Since triangle FDA is isosceles, angle $AFC=180-\\frac{\\theta}{2}$. Construct the perpendicular bisector of $AC$ and let its intersection above the line with the circumcircle of triangle ABC be G. Now construct a circle with center G passing through A and C. Mark its intersection with circle C as H. Note that $H=F$ because angles AHC and AFC are congruent and they both lie on circle C. In order to complete the construction, draw a line through C and F. The intersection of line CF with the circle (besides C itself) must be D, so we are done.\n[/hide]", "Solution_4": "[quote=\"knexpert\"][quote=\"Karth\"] $AB+CD = AC+BD$ [/quote]\nThis is incorrect. It should be $AB+CD=AD+BC$.\n\nEDIT: You still haven't corrected it. (Nothing special about $AB+CD=AB+CD$ :D ).[/quote]\r\n\r\nWow, that was pretty stupid :blush: I'll go back and correct it correctly now :D\r\n\r\nEDIT: Gah, it won't let me edit it... dang it. Oh well, you know what I mean, I guess..." } { "Tag": [ "inequalities", "rearrangement inequality", "inequalities unsolved" ], "Problem": "Let $a>0,b>0,c>0$. Prove the following inequality.\r\n\r\n\\[\\frac{1}{a^2(b+c)}+\\frac{1}{b^2(c+a)}+\\frac{1}{c^2(a+b)}\\geqq \\frac{3}{2abc}\\]", "Solution_1": "Rewrite the inequality like: \r\n bc/[a(b+c)] + ac/[b(a+c)] + ab/[c(a+b)] >= 3/2\r\n\r\nNow, using the rearrangement inequality, It\u2019ll be true.", "Solution_2": "Sorry, I am not accustomed to use rearrangement inequality.\r\nSo I can't confirm your solution. :(", "Solution_3": "Let $a=\\frac{1}{x},b=\\frac{1}{y},c=\\frac{1}{z}$. Then the ineq becomes \\[\\frac{x}{y+z}+\\frac{y}{z+x}+\\frac{z}{x+y}\\ge\\frac{3}{2}\\]which is exactly the Nesbitt's Inequality." } { "Tag": [], "Problem": "A somewhat interesting problem:\r\n\r\nA long rectangular conducting loop of width $ L$, resistance $ R$, and mass $ m$, is hung in a horizontal, uniform magnetic field $ \\vec{B}$ that is directed perpendicular to the plane of the rectangular loop and that exists only above a certain horizontal line parallel to the plane of the loop. The loop is then dropped and the gravitational force of the Earth acts on it with downward acceleration $ \\vec{g}$. During its fall, it accelerates unitl it reaches a certain terminal speed $ v_t$. Ignoring air drag, find an expression for $ v_t$ in terms of the given information.", "Solution_1": "the terminal velocity can only be obtained once a part of it is above that fixed horizontal line\r\nand another part below \r\nthen the current in it is $ \\frac{Bvl}{R}$ hence force is $ \\frac{B^{2}l^{2}v}{R}$.So to acheive terminal speed \r\n$ \\frac{B^{2}l^{2}v}{R}\\equal{}mg$ \r\nhence $ v\\equal{}\\frac{mgR}{B^{2}l^{2}}$" } { "Tag": [ "LaTeX" ], "Problem": "When the number $n$ is written in base b its representation is the two-digit number AB where A=b-2 and B=2. What is the representation of n in base (b-1)? (MA 1991)\r\n\r\nthis is what I've got so far:\r\n[hide] $AB=(b-2)(b)+2=b^{2}-2b+2=b^{2}+2(b-1)$ to convert to base b-1 do I divide $b^{2}-2(b-1)/b-1$? but it dosen't really get me anywhere[/hide] and how do you make a divide(straight line, not slash) in latex?\r\n\r\nthanks", "Solution_1": "To make a fraction in $\\LaTeX$, use \\frac{numerator}{denominator}. $\\frac{b^{2}-2(b-1)}{b-1}$\r\n\r\n[hide=\"hint\"]I would try using long division (or synthetic division) here. Think about what the remainder would equal and what the quotient would be (in terms or the base $b-1$ number).[/hide]", "Solution_2": "Well, to simplify $\\frac{b^{2}-2(b-1)}{b-1}$, just expand the parentheses in the numerator, then factor the quadratic.", "Solution_3": "[hide]think of how number bases are represented in base 10.\nfor example ABCDE in base n would be represented as $(A$x$n^{4})+(B$x$n^{3})+(C$x$n^{2})+(D$x$n^{1})+(E$x$n^{0})$.\nso you know that if it's in base b-1, you'll have something like \n$(A$x$(b-1))+B$ and then it'll be $AB_{b-1}$\nlalala =] what if it was $(b-1)(b-1)+$ something[/hide]", "Solution_4": "Just $AB_{b}=b^{2}-2b+2=(b-1)^{2}+1=101_{b-1}$.", "Solution_5": "The number in base 10 is:\r\n$(b-2) * b^{1}+2 * b^{0}$\r\n$(b-1)^{2}+1$\r\n$1 * (b-1)^{2}+0 * (b-1)^{1}+1 * (b-1)^{0}$\r\nwhich is $101_{b-1}$" } { "Tag": [ "trigonometry", "function", "geometry", "angle bisector", "trig identities", "Law of Cosines" ], "Problem": "i've been trying but i can't find a proof from the angle bisector theorem AD^2=AB*AC-BD*CD with AD the angle bisector.\r\n\r\ni've been using law of cosines but i can't find a way to cancel out the cosines.\r\n\r\nthanks for posting.", "Solution_1": "[quote=\"unimpossible\"]i've been trying but i can't find a proof from the angle bisector theorem AD^2=AB*AC-BD*CD with AD the angle bisector.\n\ni've been using law of cosines but i can't find a way to cancel out the cosines.\n\nthanks for posting.[/quote]\r\n\r\n[hide]Given $ \\triangle{ABC}$ with angle bisector of $ \\angle{B}$ being $ BD$, draw a line parallel to $ BD$ through $ A$. Now extend $ BC$ so that it intersects the parallel line through $ A$ to create a new $ \\triangle{ACF}$ . Now we know that $ \\angle{ABD}\\equal{}\\angle{CBD}$ and by alternalte interior angles, and corresponding angles we know that $ \\angle{AFC}\\equal{}\\angle{CBD}$ and $ \\angle{BAF}\\equal{}\\angle{ABD}$ so $ \\angle{BAF}\\equal{}\\angle{AFC}$. So its an isoscles triangle, so $ FB\\equal{}AB$. Now since we know that $ AF\\parallel BD$ $ \\frac{FB}{BC}\\equal{}\\frac{AD}{CD}$ and by substitution $ \\frac{AB}{BC}\\equal{}\\frac{AD}{CD}$[/hide][/hide]", "Solution_2": "We have cos(CDA)=-cos(BDA), pluging for those cos functions from law of cosines yields the result (using AB:AC=DB:DC).", "Solution_3": "[quote=\"unimpossible\"]i've been trying but i can't find a proof from the angle bisector theorem AD^2=AB*AC-BD*CD with AD the angle bisector.\n\ni've been using law of cosines but i can't find a way to cancel out the cosines.\n\nthanks for posting.[/quote]\r\n\r\ni need a solution to AD^2=AB*AC-BD*CD\r\n\r\nthanks.", "Solution_4": "[quote=\"unimpossible\"][quote=\"unimpossible\"]i've been trying but i can't find a proof from the angle bisector theorem AD^2=AB*AC-BD*CD with AD the angle bisector.\n\ni've been using law of cosines but i can't find a way to cancel out the cosines.\n\nthanks for posting.[/quote]\n\ni need a solution to AD^2=AB*AC-BD*CD\n\nthanks.[/quote]\r\n\r\nwhy use laws of cosines, you can just use stewarts..", "Solution_5": "[hide=\"proof\"]\nlet $ AD$ be the angel bisector .\n$ \\cos ADB \\equal{}\\minus{}\\cos ADC$\nthis gives by the cosine rule \n$ \\frac{AD^{2}\\plus{}BD^{2}\\minus{}c^{2}}{2 AD .BD}\\equal{}\\minus{}\\frac{AD^{2}\\plus{}CD^{2}\\minus{}b^{2}}{2 CD .AD}$\nbut $ \\frac{BD}{CD}\\equal{}\\frac{c}{b}$\nso $ AD^{2}\\plus{}BD^{2}\\minus{}c^{2}\\equal{}\\minus{}(AD^{2}\\plus{}CD^{2}\\minus{}b^{2})\\frac{c}{b}$ *\nalso $ \\frac{(b\\plus{}c)BD}{CD}\\equal{}\\frac{c(b\\plus{}c)}{b}\\equal{}\\frac{bc}{c^{2}}\\plus{}c\\equal{}\\frac{b BD^{2}}{CD^{2}}\\plus{}c$\nthis gives $ \\frac{bBD^{2}\\plus{}cCD^{2}}{b\\plus{}c}\\equal{}BD.CD$ **\nsimplifying * \ngives\n$ (b\\plus{}c)AD^{2}\\plus{}(b\\plus{}c)\\frac{(bBD^{2}\\plus{}cCD^{2})}{b\\plus{}c}\\equal{}bc(b\\plus{}c)$\nusing ** we get\n$ AD^{2}\\equal{}bc\\minus{}BD.CD$[/hide]" } { "Tag": [ "real analysis", "real analysis solved" ], "Problem": "Compute the limit of a(n) \r\na(n)=n+1 - \\sum_{k=2 to n}(1/2! + 2/3! +...+(k-1)/k!)\r\n-----------------------------------------------------------------\r\na(n+1)= a(n) + 1 -n/(n+1)! => a(n) is increasing, if a(n) converge to k\r\nthen k=k+1 absurd so lim a(n)=+oo", "Solution_1": "no..\r\n\r\nI think it is like so:\r\n\r\nbecause (k-1)/k!=1/(k-1)!-1/k! we have that 1/2! + 2/3! +...+(k-1)/k!=1-1/k!\r\n\r\nso a_n=n+1-sum(k=2,n) (1-1/k!) =1+1/1!+..+1/n! so lim a_n=e", "Solution_2": "[quote=\"Moubinool\"]\na(n+1)= a(n) + 1 -n/(n+1)! [/quote]\r\nOk, the correct relation is a(n+1)=a(n)+1/(n+1)! \r\nthanks Lagrangia" } { "Tag": [], "Problem": "http://kevan.org/brain.cgi?Treething", "Solution_1": "ohhh... So now I'm a zombie? But I feel normal :D", "Solution_2": "AAAAAAA I don't know how I'm typing this since I can't even think.", "Solution_3": "so what was the point of that site", "Solution_4": "[quote=\"noneoftheabove\"]so what was the point of that site[/quote]\r\n\r\nIt's a massivly multiplayer game thing.", "Solution_5": "that site was weird\r\n[url=http://kevan.org/brain.cgi?math92]my proof for fermats last theorem[/url]", "Solution_6": "Change your title to something more appealing, such as \"my proof for fermat's last theorem.\"", "Solution_7": "I have eatten 10 brains :rotfl: I'm full! :D", "Solution_8": "Since we all like posting links to sites check out this graphiti site while your having your brains eaten...\r\n\r\n[url=http://kevan.org/brain.cgi?Black_stripe*36]Graphiti[/url]\r\n\r\nYeah, I really don't know how to spell graphiti, graffiti.... It has funny jokes all the same.", "Solution_9": "I don't think I will do that :lol:!", "Solution_10": "Wow, I've eaten a lot of brains o_o; ." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Find all integer solutions of \\[ x^{4}\\plus{}y^{2}\\equal{}z^{4}\\]", "Solution_1": "Already discussed before (especially not new)...", "Solution_2": "Easy to check the equation not have solution (x,y,z) where all $ x,y,z$ different from 0\r\nCase: x=0\r\n$ y^{2}\\equal{} z^{4}$\r\nCase 2 y=0 \r\n$ x^{4}\\equal{} y^{4}$", "Solution_3": "I think it appear in DIOPHANTE EQUATION,A book of Dr.ANDREESCU", "Solution_4": "Why is this very famous problem in an Olympiad Cotest? :huh:", "Solution_5": "We will prove that the only solutions are ones where $xyz = 0$ (it's easy to solve the problem in this case). \n\nSuppose, for contradiction, that there existed a solution where $x, y, z > 0$. Take the one where $z$ is minimized.\n\nThen we clearly have that $x, y, z$ are relatively prime. We consider two cases.\n\n[b] Case 1. [/b] $z, x$ are both odd.\n\nThen we know that $z^2 - x^2 = 2a^2, z^2 + x^2 = 2b^2$ for some $a, b \\in \\mathbb{N}.$ This means that $z^2 = a^2 + b^2$ and $x^2 = b^2 - a^2$, so that $a^4 + (zx)^2 = b^4.$ However, $b < z$ and so this contradicts the minimality of $z.$\n\n[b] Case 2. [/b] $z, x$ are not both odd.\n\nThen we must have that $z$ is odd and $x$ is even. \n\nWe have two subcases here. In the first case, there are integers $a, b \\in \\mathbb{N}$ so that $a^2 + b^2 = z^2$ and $a^2 - b^2 = x^2$. Then we have $a^4 - b^4 = (xz)^2$ and $a < z$, contradicting the minimality of $z$.\n\nElse there are integers $a, b \\in \\mathbb{N}$ so that $a^2 + b^2 = z^2$ and $2ab = x^2.$ Note that $\\gcd(a, b) = 1$. \n\nThe first equation yields that there are integers $t, u \\in \\mathbb{N}$ so that $\\{a, b\\} = \\{t^2 - u^2, 2tu\\}$. This means that $x^2 = 2 \\cdot (t^2 - u^2) \\cdot 2tu$, and so $tu (t^2 - u^2)$ is a square. We have that $t, u, t^2 - u^2$ are pairwise relatively prime and so they're all squares. Hence, $(\\sqrt{t})^4 - (\\sqrt{u})^4$ is a square. As $\\sqrt{t} < z$, this contradicts the minimality of $z$.\n\nWe've derived a contradiction in all cases, and so our assumption that there is a solution where $xyz \\neq 0$ was incorrect. So all solutions are those where $xyz = 0$ (easy to find).\n\n$\\square$ " } { "Tag": [ "inequalities", "symmetry", "inequalities unsolved" ], "Problem": "Prove that\r\n\\[ \\frac {x_1^2}{x_2} \\plus{} \\frac {x_2^2}{x_3} \\plus{} \\ldots \\plus{} \\frac {x_n^2}{x_1} \\geq x_1 \\plus{} x_2 \\plus{} \\ldots \\plus{} x_n\\]\r\nfor all positive numbers $ x_1,x_2,\\ldots,x_n$", "Solution_1": "By AM_GM we have\r\n\r\n$ \\frac {x_1^2}{x_2}\\ge 2x_1 \\minus{} x_2$", "Solution_2": "OK could we say $ \\left(\\minus{}1,2,0\\right) \\succ \\left(1,0,0\\right)$ :?:", "Solution_3": "Just use Cauchy-Schwarz.", "Solution_4": "Chebycheff also works!!", "Solution_5": "[quote=\"hasan4444\"]Prove that\n\\[ \\frac {x_1^2}{x_2} \\plus{} \\frac {x_2^2}{x_3} \\plus{} \\ldots \\plus{} \\frac {x_n^2}{x_1} \\geq x_1 \\plus{} x_2 \\plus{} \\ldots \\plus{} x_n\\]\nfor all positive numbers $ x_1,x_2,\\ldots,x_n$[/quote]\r\n\r\nNobody seemed to solve it as the title says. $ x^{2}_{1}\\frac {1}{x_{2}} \\plus{} x^{2}_{2}\\frac {1}{x_{3}} \\plus{} ... \\plus{} x^{2}_{n}\\frac {1}{x_{1}}\\geq x^{2}_{1}\\frac {1}{x_{1}} \\plus{} x^{2}_{2}\\frac {1}{x_{2}} \\plus{} ... \\plus{} x^{2}_{n}\\frac {1}{x_{n}} \\equal{} x_{1} \\plus{} x_{2} \\plus{} ... \\plus{} x_{n}$\r\n\r\nTo Hasan->> How do you come to this conclusion? $ \\left( \\minus{} 1,2,0\\right)\\succ\\left(1,0,0\\right)$. It's false but...", "Solution_6": "Thanks \"\u0393\u03b9\u03ce\u03c1\u03b3\u03bf\u03c2\" I think my post was completely false I forget about that symmetry thing." } { "Tag": [ "calculus", "integration", "trigonometry", "calculus computations" ], "Problem": "Calculate\r\n\r\n\\[\\int^{\\frac{3\\pi}{2}}_{\\frac{\\pi}{2}} \\left|\\left(\\frac{2}{x^3}+\\frac{1}{x}\\right)\\sin x\\right|dx\\]", "Solution_1": "$\\left(-\\frac1{x^2}\\sin x-\\frac 1x\\cos x\\right)'=\\left(\\frac 2{x^3}+\\frac 1x\\right)\\sin x$. The rest is routine." } { "Tag": [ "invariant", "geometry", "linear algebra" ], "Problem": "Let $ \\left(\\begin{array}{c} U\\\\ V\\\\ W\\end{array}\\right)=\\left(\\begin{array}{ccc}a&b&c\\\\\r\nb&c&a\\\\\r\nc&a&b\\end{array}\\right)\\left(\\begin{array}{c}x\\\\\r\ny\\\\\r\nz\\end{array}\\right) $ and denote $ E(\\alpha,\\beta,\\gamma)=\\alpha^3+\\beta^3+\\gamma^3-3\\alpha\\beta\\gamma. $\r\nProve that $ E(U,V,W)=E(a,b,c)E(x,y,z)\\; . $\r\n [Apply this equality in order to obtain some identities/inequalities in the geometry of triangle.]", "Solution_1": "$U=ax+by+cz$\r\n$V=bx+cy+az$\r\n$W=cx+ay+bz$\r\n\r\n\r\n$E(U,V,W)=a^3x^3+b^3y^3+c^3z^3+b^3x^3+c^3y^3+a^3z^3+c^3x^3+\r\na^3y^3+b^3z^3+9abcxyz-3abcx^3-3abcy^3-3abcz^3-3a^3xyz-3b^3xyz-3c^3xyz$\r\n\r\n\r\nExpand $E(a,b,c)E(x,y,z)$ gives the same result\r\n\r\n------------------------------------\r\nTaking the determinant of \r\n\r\n$\\left(\\begin{array}{ccc}U&V&W\\\\V&W&U\\\\W&U&V\\end{array}\\right)=\r\n\\left(\\begin{array}{ccc}a&b&c\\\\b&c&a\\\\c&a&b\\end{array}\\right)\\left(\\begin{array}{ccc}x&y&z\\\\y&z&x\\\\z&x&y\\end{array}\\right)$\r\n\r\nGives also the result" } { "Tag": [ "algebra", "polynomial" ], "Problem": "hello \r\nplease help explaining it step by step\r\nthanx\r\n\r\n\r\nSolve the recurrence relation\r\n\r\nan = an\u22121 + 20an\u22122\r\n\r\nwith a0 = \u22121 and a1 = 13.", "Solution_1": "You could google the string \"recurrence relations\" if you want. I'll write my own solution down. [hide]Let $b_{n}= a_{n}+4a_{n-1}$. Then $b_{n}= 5b_{n-1}$ and $b_{1}= a_{1}+4a_{0}= 9$, so $b_{n}= 5b_{n-1}= 5^{2}b_{n-2}= 5^{3}b_{n-3}= ... = 5^{n-1}b_{1}= 9\\cdot 5^{n-1}$. So we know that $a_{n}+4a_{n-1}= 9\\cdot 5^{n-1}$. \nNow let $c_{n}= \\frac{5}{9}\\cdot\\frac{(-1)^{n}a_{n}}{4^{n}}$. Then $a_{n}= \\frac{9}{5}(-1)^{n}4^{n}c_{n}$, so $9\\cdot 5^{n-1}= \\frac{9}{5}(-1)^{n}4^{n}c_{n}+\\frac{9}{5}(-1)^{n-1}4^{n}c_{n-1}= \\frac{9}{5}(-1)^{n}4^{n}(c_{n}-c_{n-1})$. Therefore, $c_{n}-c_{n-1}= \\left(\\frac{-5}{4}\\right)^{n}$. It follows immediately that \\[\\begin{aligned}c_{n}& = \\left(\\frac{-5}{4}\\right)^{n}+c_{n-1}= \\left(\\frac{-5}{4}\\right)^{n}+\\left(\\frac{-5}{4}\\right)^{n-1}+c_{n-2}= ... \\\\ & = c_{0}+\\sum_{i=1}^{n}\\left(\\frac{-5}{4}\\right)^{i}=-\\frac{5}{9}+\\frac{-5}{4}\\cdot\\frac{\\left(\\frac{-5}{4}\\right)^{n}-1}{\\frac{-5}{4}-1}= \\frac{5}{9}\\left(\\left(\\frac{-5}{4}\\right)^{n}-2\\right)\\end{aligned}\\]So $a_{n}= \\frac{9}{5}(-1)^{n}4^{n}c_{n}= 5^{n}-2(-4)^{n}$.[/hide]", "Solution_2": "Woah, how did you see that you were supposed to make that substitution?", "Solution_3": "[hide=\"Or....\"]\nYou can use the [u]characteristic polynomial[/u]. (Look it up; I don't have the time to write up an explanation right now.) :)\n\n\\[a_{n+2}-a_{n+1}-20 a_{n}= 0 \\implies x^{2}-x-20=0 \\implies x_{1}=5, \\ x_{2}=-4 \\]\n\\[\\implies a_{n}= A \\cdot 5^{n}+B \\cdot (-4)^{n}, \\ \\{ A,B \\}\\subset \\mathbb{R}, \\ n \\in \\mathbb{Z^{*}}\\]\nNow, the initial terms give a linear system in $A$ and $B$, which can be solved to obtain the explicit formula for the recurrence, starting with $a_{0}$: \\[\\boxed{ a_{n}= 5^{n}-2 \\cdot (-4)^{n}}\\] [/hide]", "Solution_4": "[quote=\"cincodemayo5590\"][hide=\"Or....\"]\nYou can use the [u]characteristic polynomial[/u]. (Look it up; I don't have the time to write up an explanation right now.) :)\n\\[a_{n+2}-a_{n+1}-20 a_{n}= 0 \\implies x^{2}-x-20=0 \\implies x_{1}=5, \\ x_{2}=-4 \\]\n\n\\[\\implies a_{n}= A \\cdot 5^{n}+B \\cdot (-4)^{n}, \\ \\{ A,B \\}\\subset \\mathbb{R}, \\ n \\in \\mathbb{Z^{*}}\\]\nNow, the initial terms give a linear system in $A$ and $B$, which can be solved to obtain the explicit formula for the recurrence, starting with $a_{0}$:\n\\[\\boxed{ a_{n}= 5^{n}-2 \\cdot (-4)^{n}}\\]\n[/hide][/quote]\r\n\r\n\r\nHmmm Interesting. I looked up Characteristic Polynomial and I read about it and understood the gist of it. I also understand Eigenvalue and Eigenvectors etc.\r\n\r\nCan you please explain how this method helps to find the solution for this sequence? I sort of have an idea of the reasoning but not sure of it.", "Solution_5": "[quote=\"cincodemayo5590\"][hide=\"Or....\"]\nYou can use the [u]characteristic polynomial[/u]. (Look it up; I don't have the time to write up an explanation right now.) :)\n\\[a_{n+2}-a_{n+1}-20 a_{n}= 0 \\implies x^{2}-x-20=0 \\implies x_{1}=5, \\ x_{2}=-4 \\]\n\n\\[\\implies a_{n}= A \\cdot 5^{n}+B \\cdot (-4)^{n}, \\ \\{ A,B \\}\\subset \\mathbb{R}, \\ n \\in \\mathbb{Z^{*}}\\]\nNow, the initial terms give a linear system in $A$ and $B$, which can be solved to obtain the explicit formula for the recurrence, starting with $a_{0}$:\n\\[\\boxed{ a_{n}= 5^{n}-2 \\cdot (-4)^{n}}\\]\n[/hide][/quote]\r\n\r\nThat's exactly what Godel used, except using it to facilitate his own steps in the proof in creating an elementary understandable proof, instead of just citing it as you have :wink:", "Solution_6": "thnax for help.... however i could not understnad the characteristic polynomial. can some one explain this in steps ...please this type of questions will be in the test and i need some help thanx", "Solution_7": "[quote=\"me@home\"][quote=\"cincodemayo5590\"][hide=\"Or....\"]\nYou can use the [u]characteristic polynomial[/u]. (Look it up; I don't have the time to write up an explanation right now.) :)\n\\[a_{n+2}-a_{n+1}-20 a_{n}= 0 \\implies x^{2}-x-20=0 \\implies x_{1}=5, \\ x_{2}=-4 \\]\n\n\\[\\implies a_{n}= A \\cdot 5^{n}+B \\cdot (-4)^{n}, \\ \\{ A,B \\}\\subset \\mathbb{R}, \\ n \\in \\mathbb{Z^{*}}\\]\nNow, the initial terms give a linear system in $A$ and $B$, which can be solved to obtain the explicit formula for the recurrence, starting with $a_{0}$:\n\\[\\boxed{ a_{n}= 5^{n}-2 \\cdot (-4)^{n}}\\]\n[/hide][/quote]\n\nThat's exactly what Godel used, except using it to facilitate his own steps in the proof in creating an elementary understandable proof, instead of just citing it as you have :wink:[/quote]\r\n\r\nI understand that. I am just offering a quicker (with respect to work) solution for those who are interested, i.e. one that would give a result in less time, on a non-proof contest or the like. I actually prefer [b]Kurt G\u00f6del[/b]'s solution to my own; but I know that the method I have used will give a result in under a minute or two on the AMC. :)", "Solution_8": "Find $(\\alpha ,\\ \\beta)$ such that $a_{n}=a_{n-1}+20a_{n-2}\\Longleftrightarrow a_{n}-\\alpha a_{n-1}=\\beta (a_{n-1}-\\alpha a_{n-2}).$\r\nComparing the coefficients of $a_{n-1},\\ a_{n-2}$ gives $\\alpha+\\beta =1$ and $\\alpha \\beta =-20,$ yielding $(\\alpha ,\\ \\beta)=(-5,\\ 4),\\ (4,\\-5).$", "Solution_9": "[quote=\"hyderman\"]thnax for help.... however i could not understnad the characteristic polynomial. can some one explain this in steps ...please this type of questions will be in the test and i need some help thanx[/quote]\r\n\r\n[url=http://www.cs.cmu.edu/~adamchik/21-127/lectures/recursion_4_print.pdf]This[/url] PDF, by Victor Adamchik, will hopefully help.", "Solution_10": "hello \r\nplease check my answer and let me know if there are any errors and correct it......\r\n\r\n\r\n\r\nSolve the recurrence relation\r\n\r\nan = an\u22121 + 20an\u22122\r\n\r\nwith a0 = \u22121 and a1 = 13.\r\n\r\n\r\n\r\n\r\nStep \u2013 I : Find Characteristic equation and solve for its roots:\r\n\r\nan = an\u22121 + 20an\u22122 ------equation (0)\r\n\r\nFirst of all make corresponding characteristic equation of equation 0 and find roots of that equation\r\n\r\nlet \r\nan= r2\r\nan-1 =r\r\nan-2 =r0\r\n\r\nSubstituting above values in equation (0)\r\nr2 =r + 20\r\n\r\n\r\nr2-r-20=0 is the characteristic equation.\r\n\r\nr2-5r+4r-20=0\r\nr(r-5)+4(r-5)=0\r\n(r-5 ) (r+4)=0\r\n\r\nr1 = 5 , r2 = -4 are the roots of this characteristic equation\r\n\r\nStep\u2013 II : Find Constants\u2019 value using Initial Condition\r\n\r\nSo solution of this polynomial would be\r\n\r\nan = c1 r1n + c2 r2n ------------------equation (1)\r\n\r\n\r\n\r\nWhere c1 & c2 are some arbitrary constants .\r\n\r\nNow we have to find values of c1 and c2 using Initial Condition.\r\n\r\n\r\nAs a0 = \u22121 and a1 = 13.\r\n\r\nSo put n=0 in equation (1)\r\n\r\na0= c1 r10 + c2 r20 \r\na0= c1 + c2 \r\n-1 = c1 + c2 ---------equation (2)\r\n\r\nNow put n=1 in equation (1)\r\na1 = c1 r11 + c2 r21\r\n\r\n13= c1 (5)1 + c2 (-4)1 --------------equation (3)\r\n\r\n\r\nMultiply equation (2) by 4\r\n\r\n-4= 4c1 + 4c2 -------------------------------------------------equation (4)\r\n\r\nAdd equation (3) and (4) to get the values of c1 & c2.\r\n-4= 4c1 + 4c2\r\n13= 5c1 - 4c2 \r\n\r\n9 = 9c1\r\nc1=1\r\n\r\nput c1=1 in equation (2)\r\n\r\nc2=-2\r\n\r\nStep-III ..Substitute values of c1 , c2 , r1 and r2 to get the final solution \r\n\r\nso substituting these values in equation \r\nan = 1.5n + (-2) 4n\r\n\r\n\r\nan = 5n \u2013 (2).4n is the final solution of this recurrence relation" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Find all ordered triple of primes$ (p,q,r)$ such that\r\n\r\n$ p|q^r \\plus{} 1\\ \\ ,\\ \\ q|r^p \\plus{} 1\\ \\ ,\\ \\ r|p^q \\plus{} 1$", "Solution_1": "USA TST 2002 :P \r\n\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=357351#357351[/url]" } { "Tag": [ "percent", "puzzles" ], "Problem": "There are a lot of pirates on their island. Every pirate has the same habits which are\r\n1. They don't want to die.\r\n2. They love money.\r\n3. They like to see the others die.\r\nHabit no.1 is more important than no.2, e.g. if a pirate has to choose between getting money but being killed and getting no money but being alive, they will choose the second one. Assume that all of them are so clever that they will choose the best way according to their habits. If there are n pirates, each of them has his own number in {1,2,3,...,n} which are all different for each one. To clarify, there are P1,P2,P3,...,Pn on the island if there are n pirates. The pirate with more number has more power.\r\n\r\nWhenever they find a treasure(usually 100 coins), the one with the most power will decide how to divide 100 coins, e.g. n=3, no.3 gives 30, 30, 40 coins for number 1,2,3(himself). And everyone(including himself) will vote whether he agree with that idea or not. If there are A HALF OR MORE THAN agreement, that idea from the most powerful pirate will be accapt. But if there are LESS THAN A HALF agreement, the most powerful pirate will be killed, and the second most will be the one who decides how to divide that 100 coins. And so on until there are a half or more than agreement.\r\n\r\nThere are n pirate on the island. The question is...(Do it one by one)\r\n[hide=\"Problem1\"]n=2 How will pirate no.2 divide 100 coins? and How much will he get?[/hide][hide=\"Problem2\"]n=3 How will pirate no.3 divide 100 coins? and How much will he get?[/hide][hide=\"Problem3\"]n=4 How will pirate no.4 divide 100 coins? and How much will he get?[/hide][hide=\"Problem4\"]n=5 How will pirate no.5 divide 100 coins? and How much will he get?[/hide][hide=\"Problem5\"]n=100 How will pirate no.100 divide 100 coins? and How much will he get? You can try 6,7,8,.. first if you can't do 100.[/hide][hide=\"Problem6\"]n=200 How will pirate no.200 divide 100 coins? and How much will he get?[/hide][hide=\"Problem7\"]n=201 Will no.201 die?[/hide][hide=\"Problem8\"]n=202 Will no.202 die?[/hide][hide=\"Problem9\"]n=203 Will no.203 die?[/hide][hide=\"Problem10\"]n=204 Will no.204 die?[/hide][hide=\"Problem11\"]n=205 Will no.205 die?[/hide][hide=\"Problem12\"]n=206 Will no.206 die?[/hide][hide=\"Problem13\"]n=207 Will no.207 die?[/hide][hide=\"Problem14\"]n=208 Will no.208 die?[/hide][hide=\"Problem15\"]Who's the next one who can survive?[/hide]", "Solution_1": "[hide=\"n=2\"]The pirate #2 suggests 100 coins for himself and no coins for #1. He votes his suggestion, so he has the 50 percent[/hide]\n[hide=\"n=3\"]The pirate #2 knows that, if #3 die, he can take 100 coins. So he is not going to vote the suggestion of #3.\n\n#3 knows this, so he suggests 99 coins for himself and 1 coin for #1\n#1 prefers to get 1 instead of 0 coins, so he will vote #3's suggestion\n[/hide]\n[hide=\"n=4\"]#3 is not going to vote #4's suggestion\nSo #4 wants to get a vote from #1 or #2.\nIt's better for #4 to offer 1 coin to #2 and keep 99 coins for himself.\n\nIndeed, if he give 1 coin to #1, then:\n#1 knows that he can take (such or else) 1 coin from #3, so he won't vote #4's suggestion, because he wants to see #4 die\nSo, if #4 wants the vote of #1 he must give to him #2 coins. [/hide]\n[hide=\"n=5\"]#4 is not going to vote #5's suggestion.\n#5 wants 2 votes (except of his vote). It's better for him to offer coins to the pirates who took less money in the previous step.\n#1,#3 took nothing, so he offers 1 coin to each and keeps 98 for himself[/hide]\n[hide=\"n>5\"]If we continue this procedure we'll see that \nthe pirate with number $2k+1$ gives 1 coin to each the k odd numbered pirates $1,3,5, \\cdots , 2k-1$. He keeps $100-k$ for himself\nthe pirate with number $2k+2$ gives 1 coin to each the k even numbered pirates $2,4,6, \\cdots , 2k$. He keeps $100-k$ for himself\n\nThe above can be done if $k\\leq 100 \\Rightarrow $ \n$2k+1\\leq 201$ and\n$2k+2\\leq 202$\n\nNotice that the pirate #201 keeps no money for himself, but he stays alive.\nThe same for #202\n\nNow, the pirate #203 wants the vote of the odd-numbered pirates $1,3, \\cdots ,199,201$. \nThey are 101 persons, but he can offer to them only 100 coins.\n\nSo, I think that every pirate with number $n\\geq 203$ will die[/hide]", "Solution_2": "Nope, $204$ will not die.", "Solution_3": "[quote=\"pontios\"][hide=\"n=2\"]The pirate #2 suggests 100 coins for himself and no coins for #1. He votes his suggestion, so he has the 50 percent[/hide]\n[hide=\"n=3\"]The pirate #2 knows that, if #3 die, he can take 100 coins. So he is not going to vote the suggestion of #3.\n\n#3 knows this, so he suggests 99 coins for himself and 1 coin for #1\n#1 prefers to get 1 instead of 0 coins, so he will vote #3's suggestion\n[/hide]\n[hide=\"n=4\"]#3 is not going to vote #4's suggestion\nSo #4 wants to get a vote from #1 or #2.\nIt's better for #4 to offer 1 coin to #2 and keep 99 coins for himself.\n\nIndeed, if he give 1 coin to #1, then:\n#1 knows that he can take (such or else) 1 coin from #3, so he won't vote #4's suggestion, because he wants to see #4 die\nSo, if #4 wants the vote of #1 he must give to him #2 coins. [/hide]\n[hide=\"n=5\"]#4 is not going to vote #5's suggestion.\n#5 wants 2 votes (except of his vote). It's better for him to offer coins to the pirates who took less money in the previous step.\n#1,#3 took nothing, so he offers 1 coin to each and keeps 98 for himself[/hide]\n[hide=\"n>5\"]If we continue this procedure we'll see that \nthe pirate with number $2k+1$ gives 1 coin to each the k odd numbered pirates $1,3,5, \\cdots , 2k-1$. He keeps $100-k$ for himself\nthe pirate with number $2k+2$ gives 1 coin to each the k even numbered pirates $2,4,6, \\cdots , 2k$. He keeps $100-k$ for himself\n\nThe above can be done if $k\\leq 100 \\Rightarrow$ \n$2k+1\\leq 201$ and\n$2k+2\\leq 202$\n\nNotice that the pirate #201 keeps no money for himself, but he stays alive.\nThe same for #202\n\nNow, the pirate #203 wants the vote of the odd-numbered pirates $1,3, \\cdots ,199,201$. \nThey are 101 persons, but he can offer to them only 100 coins.\n\nSo, I think that every pirate with number $n\\geq 203$ will die[/hide][/quote][hide]All are correct except one. $204$ will not die like ZataX said.[/hide]", "Solution_4": "Could someone explain why? And what happens with the next pirates?", "Solution_5": "Note that number $203$ will vote for $204$ even if he gives him nothing, because if $204$ is killed, $203$ knows he will be killed, too.", "Solution_6": "Nice.\r\n\r\nBut will the pirates 2,4,6,... vote for 204?\r\n\r\nThey know that, if $204$ is killed $\\Rightarrow 203$ is killed $\\Rightarrow 202$ gives them 1 coin to each.\r\n\r\nand they like very much to see 203,204 been killed!!", "Solution_7": "He will give the coins to number $1,3,5,7,...$.", "Solution_8": "Yes! :) \r\nWho's the next one who survives?", "Solution_9": "[quote=\"ZetaX\"]He will give the coins to number $1,3,5,7,...$.[/quote]\r\n\r\nI see now :)\r\n\r\nThe pirates $1,3,5,..$ know that, if don't vote for him, then $202$ will give the coins to the pirates $2,4,6,...$\r\n\r\nIt's too complicated after 200 :maybe:\r\n\r\nI can't find who's the next :(", "Solution_10": "[hide=\"Hint\"]$208$[/hide]", "Solution_11": "[quote=\"ZetaX\"][hide=\"Hint\"]$208$[/hide][/quote][hide=\"Hint2(Give away)\"]$216$[/hide]", "Solution_12": "Could it possibly be $200+2^{x}$?", "Solution_13": "Yes, it is.", "Solution_14": "I think I got it now!\r\n\r\n204 will survive. He gives coins to 1,3,5,...,199.\r\n\r\n205,206,207 will die (3 persons)\r\n208 will survive \r\nhe gets 100 votes (from those who get the coins) + 1 (from himself) + 3 (from those who don't want to die)\r\nBut he gives coins to 2,4,...,200\r\n\r\n\r\n209,...,215 will die (7 persons)\r\n216 will survive \r\nhe gets 100 votes (from those who get the coins) + 1 (from himself) + 7 (from those who don't want to die)\r\nHe gives coins to 1,3,...,199\r\n\r\n\r\n217,...,231 will die (15 persons)\r\n232 will survive \r\nhe gets 100 votes (from those who get the coins) + 1 (from himself) + 15 (from those who don't want to die)\r\nHe gives coins to 2,4,...,200\r\n\r\nand so on.." } { "Tag": [ "calculus", "integration", "algebra", "partial fractions", "calculus computations" ], "Problem": "$ \\text{(i)}\\quad \\int\\;\\sqrt {\\; \\sqrt {\\; x \\;} \\plus{}x \\;} \\;\\textbf dx$\r\n\r\n$ \\text{(ii)}\\quad \\int\\;\\sqrt {\\; \\sqrt {\\; \\sqrt {\\; x \\;} \\plus{}x \\;} \\plus{}x \\;} \\;\\textbf dx$\r\n\r\n :D", "Solution_1": "[quote=\"misan\"]$ \\text{(i)}\\quad \\int\\;\\sqrt {\\; \\sqrt {\\; x \\;} + x \\;} \\;\\textbf dx$[/quote]\r\n\r\n\\begin{eqnarray*}\r\n\\int \\sqrt{\\sqrt{x}+x}dx\r\n& = & \\int \\left( \\frac{1}{16 \\sqrt{x} \\sqrt{\\sqrt{x}+x}} + \\frac{-1 + 16x + 16x^{3/2}}{16 \\sqrt{x} \\sqrt{\\sqrt{x}+x}} \\right) dx \\\\\r\n& = & \\frac{\\sqrt{\\sqrt{x}+x} \\ \\text{Arc}\\sinh{x^{1/4}}}{4\\sqrt{1+\\sqrt{x}} \\ x^{1/4}} - \\frac{1}{4} \\sqrt{\\sqrt{x}+x} + \\frac{1}{6} \\sqrt{x} \\sqrt{\\sqrt{x}+x} + \\frac{2}{3} x \\sqrt{\\sqrt{x}+x} \\\\\r\n\\end{eqnarray*}", "Solution_2": "[b]Kouichi Nakagawa[/b], how did you get this \"partial fractions\" :huh: :?:\r\n\r\nThe first one I would do that way:\r\n$ \\int \\sqrt {\\sqrt {x} \\plus{} x} \\, \\textbf d x \\equal{} \\int x^{1/4} \\left( 1 \\plus{} x^{1/2} \\right)^{1/2} \\, \\textbf d x$\r\nand substitute $ t \\equal{} \\sqrt {\\frac {1 \\plus{} \\sqrt {x}}{\\sqrt {x}}} \\, \\ldots$", "Solution_3": "[quote=\"@petko\"][b]Kouichi Nakagawa[/b], how did you get this \"partial fractions\" :huh: :?:[/quote]\r\n\r\n$ (\\text{Arc}\\sinh{x})' = \\frac{1}{\\sqrt{1+x^2}}$\r\n\\begin{eqnarray*}\r\n(\\text{Arc}\\sinh{x^{1/4}})'\r\n& = & (x^{1/4})' \\frac{1}{\\sqrt{1+(x^{1/4})^2}} \\\\\r\n& = & \\frac{1}{4 \\sqrt{1+\\sqrt{x}} \\ x^{3/4}} \\\\\r\n& = & \\frac{1}{4 \\sqrt{x} \\sqrt{x+\\sqrt{x}}}\r\n\\end{eqnarray*}\r\n$ (\\frac{\\text{Arc}\\sinh{x^{1/4}}}{4})' = \\frac{1}{16 \\sqrt{x} \\sqrt{x+\\sqrt{x}}}$\r\n$ \\sqrt{\\sqrt{x}+x} - \\frac{1}{16 \\sqrt{x} \\sqrt{x+\\sqrt{x}}} = \\frac{-1 + 16x + 16x^{3/2}}{16 \\sqrt{x} \\sqrt{x+\\sqrt{x}}}$" } { "Tag": [ "geometry", "circumcircle", "rectangle", "trapezoid", "rhombus", "trigonometry", "perimeter" ], "Problem": "I'm sorry if i can't explain this problem better in English.\r\nSource: Indonesian province selection\r\n\r\nA quadrilateral $ABCD$ with $a$ as the largest side, and the circumradius of the triangle $ABD$ is 1. Determine the smallest value of $a$ and when it is occur!", "Solution_1": "The smallest value of $a$ is when $ABCD$ is a square and so all sides are equally long.\r\nIn this case $\\Delta ABD$ is rectangle and $a = \\sqrt{2}$\r\n\r\nmmh... it should be wrong :?", "Solution_2": "Well, shouldn't it be $2$, all the sides can't be equal and the largest side can't be equal to another side based on his wording, thus, the smallest it can be is a trapezoid like figure with the largest base being the diameter of the circle.", "Solution_3": "I'm thinking we can take a to be infinitely small.\r\nHere's how I'd construct it:\r\n\r\nDraw a circle with radius 1.\r\n\r\nWe pick 3 points on the circle such that the \"middle\" point is the same distance from both other points. This is triangle ABD. It's an isosceles. To complete the quadrilateral, we just complete the rhombus, with all sides \"a\".\r\nSince these 3 points can be indefinitely close together on the circle, a is indefinitely small?", "Solution_4": "[quote=\"krustyteklown\"]I'm thinking we can take a to be infinitely small.\nHere's how I'd construct it:\n\nDraw a circle with radius 1.\n\nWe pick 3 points on the circle such that the \"middle\" point is the same distance from both other points. This is triangle ABD. It's an isosceles. To complete the quadrilateral, we just complete the rhombus, with all sides \"a\".\nSince these 3 points can be indefinitely close together on the circle, a is indefinitely small?[/quote]\r\n\r\nActually, I think you are right because as I said before that if you take a trapezoid for example, you can keep making the top base smaller, and the sides farther and farther, thus the bottom base (largest side) can keep getting smaller, so you can't calculate it's length.", "Solution_5": "if you draw a triangle in a circle, you can have an obtuse triangle where all the sides approach 0, the one side that gets bigger while the others get smaller can be the diagonal BD, say $\\angle A = 180 - 10^{-15}$, them BD = is essentially 0, then the other sides would be tiny as well", "Solution_6": "the answer is $a=\\sqrt{2}$ and it occurs when all sides are equal.\r\nI use cosine law. Here is my effort:\r\n$a^2=2-2cos(A)$\r\n$b^2=2-2cos(B)$\r\n$c^2=2-2cos(C)$\r\n$d^2=2-2cos(D)$ \r\nto minimize $a$ , which are the largest side, is equivalen to minimize the Perimeter of quadrilateral or minimize $a+b+c+d$ ....\r\nand so we have to minimize \r\n$\\sqrt{2-2cos(A)}+\\sqrt{2-2cos(B)}+\\sqrt{2-2cos(C)}+\\sqrt{2-2cos(D)}$", "Solution_7": "Hm, maybe the question was worded wierd then, cuz I'm pretty sure my construction works, same with Altheman's" } { "Tag": [ "linear algebra" ], "Problem": "[url=http://imageshack.us][img]http://img474.imageshack.us/img474/6008/determinantub8.gif[/img][/url]\r\nShot at 2007-08-14", "Solution_1": "Can you help solving that questions?", "Solution_2": "This is just straightforward calculation. You (probably) can't reduce it to a form where the answers come out nicely. You can tell because the expected results are not given as arithmetic combinations, which all determinants are." } { "Tag": [ "Euler", "vector", "induction" ], "Problem": "There are two paticles with equal mass and charge .they are thrown with an horizontal speed $ v$ from heights $ h_{1}$ and $ h_{2}$ from the earth along the same vertical .The particle one reaches ground at a distance $ l$ from the vertical.Find the height $ H_{2}$ at which the second particle is when the first one reaches the ground.Neglect air drag and any charge induced on the ground.", "Solution_1": "The system is conservative :)\r\n\r\n$ m g h_1 \\plus{} \\frac{1}{2} m v^2 \\plus{} K \\frac{q^2}{h_2\\minus{}h_1} \\equal{}\r\n m g 0 \\plus{} \\frac{1}{2} m {V_1}^2 \\plus{} K \\frac{q^2}{H_2}$\r\n\r\n$ m g h_2 \\plus{} \\frac{1}{2} m v^2 \\plus{} K \\frac{q^2}{h_2\\minus{}h_1} \\equal{}\r\n m g H_2 \\plus{} \\frac{1}{2} m {V_2}^2 \\plus{} K \\frac{q^2}{H_2}$\r\n\r\nSolving for $ H_2,v$\r\n\r\n$ H_2 \\equal{} \\frac{(V_1\\minus{}V_2)(V_1\\plus{}V_2)}{2 g}\\plus{}h_2\\minus{}h_1$\r\n\r\nI don't know any good way to determine the final velocity of each particle. To Solve the euler equation for this system will be a little bit difficult.", "Solution_2": "[hide=\"hint\"] \nconsider the motion of the centre of mass [/hide]", "Solution_3": "This isn't a hint, is the solution.\r\n\r\nNewton is really a good boy\r\n\r\nThe trajectory of CM is a parabola.\r\n\r\nThe equation of the movement is:\r\n\r\n$ y_1\\plus{}y_2 \\equal{} \\frac{1}{2} g t^2 \\plus{} \\frac{1}{2}( h_1\\plus{}h_2)$\r\n$ x \\equal{} v t$\r\n\r\nFor the problem $ x \\equal{} l$ and $ y_1 \\equal{} 0$\r\n\r\nAnd doone. I was trying to use Lagrange for this problem, :rotfl:", "Solution_4": "everything right except that it will be $ \\frac{\\minus{}gt^{2}}{2}$", "Solution_5": "$ g < 0$ :)\r\n\r\nI'm not the type of physician that write $ \\minus{}g$ :D \r\n\r\nFor me is a retrocesse use only \"positive variables\" 'cause we have the negative ones.\r\n\r\nSo don't say it is wrong, I never assumed $ g>0$, in the contex is clear that $ g<0$", "Solution_6": "if u aren't that type of [b]physicist[/b] (not [b]physician[/b] which means a doctor) then please use $ \\vec{g}$ instead :wink:", "Solution_7": "$ \\mathbf{g}$ is a more fashioned way to express this :) (but take time, and is just a retrocesse\r\n\r\nperhaps, you are wrong gallant pardersi, you said that my answer was wrong, when it was just a misunderstand of yourself.\r\n\r\njust apologize and don\u2019t make a fuss about that", "Solution_8": "Eu acho que o erro e seu", "Solution_9": "[quote=\"EulerIntegral\"]\nperhaps, you are wrong gallant pardersi, you said that my answer was wrong, when it was just a misunderstand of yourself.[/quote]\ni never said the word wrong. :huh: \n[quote=\"pardesi\"]everything right except that it will be $ \\frac { \\minus{} gt^{2}}{2}$[/quote]\n\n[quote=\"EulerIntegral\"]just apologize and don\u2019t make a fuss about that[/quote]\napologize for what????? pointing out a mistake.\nEither u should have written in the vector form or since u think ur a ...physiciAN :rotfl: u should have explained the meaning of the \"new\" symbols u r using and please stop asking people to do what they are supposed not to :ninja:\n\n\n[quote=\"Immanuel Bonfils\"]Eu acho que o erro e seu[/quote]\r\nimmanuel i beleive most people don't understand that please can u translate that", "Solution_10": "a thing that inst right, by induction (fermat) is wrong, QED\r\n\r\nI used $ g<0$ and any kid can see that $ g<0$.\r\n\r\nIs a OBVIOUSLY thing, the error was your to think wrongly that I used $ g>0$\r\n\r\nJust like when i put the electric field $ E \\equal{} \\frac{q^2}{r^2}$ it's clear that $ E$ is normalized.", "Solution_11": "First - One can't just put g in a vector form on an energy expression (unless in a scalar product or something of this kind...); it'll mess out scalars with vectors.\r\n\r\nSecond - The \"retrocession using only positive variables\" doesn't applies to g, a physical constant; it is definitely positive.\r\n\r\nThird - It shouldn't be $ \\;\\;{{y_1 \\plus{} y_2}\\over 2}\\;\\;$ ?\r\n\r\nNote: I've used Portuguese cause it seems EulerIntegral could understand better...", "Solution_12": "[quote=\"EulerIntegral\"]a thing that inst right, by induction (fermat) is wrong, QED[/quote]\nnice proof :P \n\n[quote=\"EulerIntegral\"]I used $ g < 0$ and any kid can see that $ g < 0$.\n\nIs a OBVIOUSLY thing, the error was your to think wrongly that I used $ g > 0$[/quote]\nsorry i am not a kid nor am i kidding . :huh: i haven't seen a book where $ g < 0$ and probably everywhere in this world in physics $ g \\equal{} 9.8 ms^{ \\minus{} 2}$ and nothing else.just to have ur answer right right u cannot change meanings. :noo: \n\n[quote=\"Immanuel Bonfils\"]First - One can't just put g in a vector form on an energy expression (unless in a scalar product or something of this kind...); it'll mess out scalars with vectors.\n[/quote]\nthat wasn't an energy conservation equation.it was the trajectory of the centre of mass.so no problem in putting vector sign through out\n[quote=\"Immanuel Bonfils\"]Note: I've used Portuguese cause it seems EulerIntegral could understand better...\n[/quote]\r\nok", "Solution_13": "Let's Prove That $ g>0$ or $ g<$\r\n\r\n$ \\minus{}9.8 \\equal{} (\\minus{}1)(9.8)$\r\n\r\nSo the minus signal can be outside the constant or inside.\r\n\r\n$ a\\equal{}7$ and you anwser is $ \\minus{}a$\r\n\r\nthis is the same that $ b\\equal{}\\minus{}7$ and you answer $ b$\r\n\r\nMinus operator is a magical operator.\r\n\r\nAnd for immanuel \"O erro \u00e9 meu \u00e9 a ... Eu nunca assume que eu usei uma outra configura\u00e7\u00e3o" } { "Tag": [ "function" ], "Problem": "consider the function f(x)=x/3-1/2. if we were to graph this function, on what point would the x and y value be the same?", "Solution_1": "[hide]$y = \\frac{x}{3}-\\frac{1}{2}$\n\nIts inverse is:\n$y=3x+\\frac{3}{2}$\n\nSince we want x to equal y, \n\n$\\frac{x}{3}-\\frac{1}{2}=3x+\\frac{3}{2}$\n\n$\\frac{8}{3}x=-2$\n\n$x=-\\frac{3}{4}$ \n\n [/hide]", "Solution_2": "this is like the tenth time today that you are correct. :D", "Solution_3": "It helps for both of us to be online from 8am this morning. :D", "Solution_4": "good point. :D", "Solution_5": "wow, don't you guys have a camp or something? I am gone from 7:45 to 5:00.\r\nthat's why 236! has a trillion posts i the last couple of days.", "Solution_6": "no, i don't go to a camp. :D", "Solution_7": "[quote=\"math92\"]wow, don't you guys have a camp or something? I am gone from 7:45 to 5:00.\nthat's why 236! has a trillion posts i the last couple of days.[/quote]\r\n\r\nI had 93 posts yesterday :rotfl:" } { "Tag": [ "LaTeX" ], "Problem": "\u5728\u8fd9\u4e2aforum\u4e0a\u662f\u5426\u8981\u7528WORD\u516c\u5f0f\u7f16\u8f91\u6765\u5199\u516c\u5f0f?\u5728\u4e0a\u9762\u80fd\u591f\u4e0d\u7528\u6253\u5f00WORD\u5c31\u5199\u7684\u65b9\u6cd5\u5417?\u6bd4\u5982\u5185\u90e8\u63d0\u4f9b\u516c\u5f0f\u7f16\u8f91\u7684\u56fe\u7247?", "Solution_1": ":huh: \u5144\u5f1f\uff0c\u5feb\u53bb\u5b66\u5b66LaTex\u7f16\u8f91\u8f6f\u4ef6\u5427\uff01\u597d\u5904\u591a\u591a\u3002", "Solution_2": "\u53ef\u662f\u6211\u4f9d\u7136\u6ca1\u6709\u5f04\u592a\u660e\u767d\u600e\u4e48\u7528!", "Solution_3": "\u6709\u4e00\u5e74\u591a\u6ca1\u53d1\u5e16\u4e86\u2026\u2026\r\n\r\n\u7f16\u8f91\u516c\u5f0f\u5f88\u5bb9\u6613\uff0c\u53ea\u9700\u8981\u5728\u516c\u5f0f\u4ee3\u7801\u4e24\u7aef\u52a0\u4e0a $ \\$$ \u5c31\u884c\u4e86\u3002\u6bd4\u5982\u5728f(x)=ax^2+bx+c\u4e24\u8fb9\u52a0\u4e0aDollar Sign, \u5c31\u6210\u4e86 $ f(x)\\equal{}ax^2\\plus{}bx\\plus{}c$\r\n\r\n\u5269\u4e0b\u7684\u5c31\u662f\u591a\u5b66\u4e60\u591a\u7ec3\u4e60\u5566\uff01", "Solution_4": "\u53bb\u8fd9\u4e2a\u7248\u5757:\r\nhttp://www.mathlinks.ro/index.php?f=123\r\n\u5b66\u4e60\u4e0b\u54af...", "Solution_5": "\u7528MathType\u6253\u51fa\u6765\u7136\u540e\u8f6c\u6362\u6210\u4ee3\u7801\u5c31\u884c\u4e86~", "Solution_6": "[quote=\"RichardGao\"]\u7528MathType\u6253\u51fa\u6765\u7136\u540e\u8f6c\u6362\u6210\u4ee3\u7801\u5c31\u884c\u4e86~[/quote]\n\n\u8f6c\u51fa\u6765\u7684\u5168\u662f\u201c\u5927\u201d\u516c\u5f0f\uff0c\u5373\u5c45\u4e2d\u7684\u3002\u5f88\u591a\u65f6\u5019\u8981\u5c06\u524d\u540e\u4e24\u4e2a\u62ec\u6539\u6210\u7f8e\u5143\u7b26\u53f7", "Solution_7": "\u8c8c\u4f3c\u00b7\u00b7\u00b7\u6211\u76f4\u63a5\u4e0a\u622a\u56fe\u7684\u00b7\u00b7\u00b7", "Solution_8": "[quote=\"DoctorRoyal\"]\u8c8c\u4f3c\u00b7\u00b7\u00b7\u6211\u76f4\u63a5\u4e0a\u622a\u56fe\u7684\u00b7\u00b7\u00b7[/quote]\n\u4e0a\u4f20\u622a\u56fe\u53ef\u4ee5\uff0c\u4e0d\u8fc7\u8fd9\u4e2a\u4e0d\u65b9\u4fbf\u5f15\u7528\u3002\u5f53\u7136\u51e0\u4f55\u56fe\u5f62\u5c31\u53e6\u5f53\u522b\u8bba\u4e86\uff0c\u8fd9\u91cc\u662f\u6307\u516c\u5f0f\u3002", "Solution_9": "\u8c8c\u4f3c\u6211\u5f97\u82b1\u4e0a\u4e00\u6bb5\u65f6\u95f4\u6765\u7814\u7a76\u4e0b\u516c\u5f0f\u7684\u8bed\u6cd5\u00b7\u00b7\u00b7", "Solution_10": "\u8fd9\u5e16\u5b50\u5c45\u7136\u8fd8\u5728\u9996\u9875\u4e0a\uff0c\u8ddd\u79bb\u6211\u4e0a\u6b21\u53d1\u5e16\u90fd4\u5e74\u4e86\u554a\n\n[quote=\"shobber\"]\u6709\u4e00\u5e74\u591a\u6ca1\u53d1\u5e16\u4e86\u2026\u2026\n\n\u7f16\u8f91\u516c\u5f0f\u5f88\u5bb9\u6613\uff0c\u53ea\u9700\u8981\u5728\u516c\u5f0f\u4ee3\u7801\u4e24\u7aef\u52a0\u4e0a $ \\$$ \u5c31\u884c\u4e86\u3002\u6bd4\u5982\u5728f(x)=ax^2+bx+c\u4e24\u8fb9\u52a0\u4e0aDollar Sign, \u5c31\u6210\u4e86 $ f(x)\\equal{}ax^2\\plus{}bx\\plus{}c$\n\n\u5269\u4e0b\u7684\u5c31\u662f\u591a\u5b66\u4e60\u591a\u7ec3\u4e60\u5566\uff01[/quote]", "Solution_11": "LS\u80af\u5b9a\u662f\u53c3\u52a0\u904e\u4ec0\u9ebd\u806f\u8cfd\u51ac\u590f\u4ee4\u71df\u96c6\u8a13\u968a\u9019\u985e", "Solution_12": "\u4f60\u8bf4\u6211\u4e48\uff1f\u6211\u6ca1\u6709\u554a\n\n[quote=\"colorfuldreams\"]Ls\u80af\u5b9a\u662f\u53c3\u52a0\u904e\u4ec0\u9ebd\u806f\u8cfd\u51ac\u590f\u4ee4\u71df\u96c6\u8a13\u968a\u9019\u985e[/quote]", "Solution_13": "For learn about latex, see [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:About]LaTeX Guide[/url]. And, for upload the image use Attachment.", "Solution_14": "[quote=\"shobber\"]\u4f60\u8bf4\u6211\u4e48\uff1f\u6211\u6ca1\u6709\u554a\n[/quote]\n\n\n\u96e3\u5f97\u5c0d\u6578\u5b78\u90a3\u9ebd\u6709\u8208\u8da3\u3002\u8a71\u8aaa\u6211\u898b\u904e\u4e0d\u5c11\u5927\u5b78\u751f\u4f86\u73a9\u9ad8\u4e2d\u7af6\u8cfd\u7684\u3002\n\nps , \u9019\u500b\u8ad6\u58c7\u7684 Olympiad Section \u5340\u7684\u554f\u984c\u611f\u89ba\u8ddf\u7af6\u8cfd\u9019\u985e\u554f\u984c\u5f88\u4e0d\u540c\u5730\u8aaa\u3002\u3002\u3002\u6216\u8005\u8aac\u6bd4\u7af6\u8cfd\u96e3\u4e14\u8a08\u7b97\u91cf\u6d77\u91cf\uff0c\u7af6\u8cfd\u4e00\u822c\u4e0d\u6703\u90a3\u9ebd\u8003\u3002\u611f\u89ba\u6709\u96c6\u8a13\u968a\u8a13\u7df4\u984c\u7684\u5473\u9053\u3002\n\nps . High School Pre-Olympiad (Ages 14+) \u9019\u500b\u5340\u7684\u554f\u984c\u5f88\u8cbc\u8fd1\u7af6\u8cfd\u5340\uff0c\u4e0d\u904e\u6709\u4e0d\u5c11\u7c21\u55ae\u7684\u554f\u984c\u5076\u723e\u53c8\u6703\u51fa\u73fe\u975e\u5e38\u96e3\u7684\u554f\u984c\u3002\u6709\u9ede\u50cf\u806f\u8cfd\u3002\n\nps , \u9664\u4e86\u4ee5\u4e0a2\u500b\u5340\uff0c\u5225\u7684\u5340\u6211\u90fd\u6c92\u53bb\u904e\u3002\n\nps . youarebad \u662f\u4e00\u500b\u597d\u4eba\u3002(\u53ef\u60dc\u6a13\u4e3b\u4e0d\u77e5\u9053\u5df2\u7d93\u9084\u5728\u4e0d\u5728\u4e86\u3002)", "Solution_15": "Thx, so the writer not online now ?? Sorry, i am use google translate to translate your word.", "Solution_16": "[quote=\"youarebad\"]Thx, so the writer not online now ?? Sorry, i am use google translate to translate your word.[/quote]\n\nbecuse the post is at Oct 12, 2007, 11:02 am, 4 years ago. I think if he was a high school student , and today he becomes to a university student", "Solution_17": "\u5174\u8da3\u867d\u7136\u8fd8\u6709\uff0c\u53ef\u662f\u5df2\u7ecf\u5b8c\u5168\u4e0d\u4f1a\u505a\u9898\u4e86\u2026\u2026\u5927\u5b66\u5ff5\u4e86\u8ddf\u6570\u5b66\u5b8c\u5168\u4e0d\u642d\u7684\u4e13\u4e1a\uff0c\u4ee5\u524d\u5b66\u8fc7\u7684\u4e1c\u897f\u5fd8\u5f97\u4e00\u5e72\u4e8c\u51c0\u4e86\n\n[quote=\"colorfuldreams\"][quote=\"shobber\"]\u4f60\u8bf4\u6211\u4e48\uff1f\u6211\u6ca1\u6709\u554a\n[/quote]\n\n\n\u96e3\u5f97\u5c0d\u6578\u5b78\u90a3\u9ebd\u6709\u8208\u8da3\u3002\u8a71\u8aaa\u6211\u898b\u904e\u4e0d\u5c11\u5927\u5b78\u751f\u4f86\u73a9\u9ad8\u4e2d\u7af6\u8cfd\u7684\u3002\n\nps , \u9019\u500b\u8ad6\u58c7\u7684 Olympiad Section \u5340\u7684\u554f\u984c\u611f\u89ba\u8ddf\u7af6\u8cfd\u9019\u985e\u554f\u984c\u5f88\u4e0d\u540c\u5730\u8aaa\u3002\u3002\u3002\u6216\u8005\u8aac\u6bd4\u7af6\u8cfd\u96e3\u4e14\u8a08\u7b97\u91cf\u6d77\u91cf\uff0c\u7af6\u8cfd\u4e00\u822c\u4e0d\u6703\u90a3\u9ebd\u8003\u3002\u611f\u89ba\u6709\u96c6\u8a13\u968a\u8a13\u7df4\u984c\u7684\u5473\u9053\u3002\n\nps . High School Pre-Olympiad (Ages 14+) \u9019\u500b\u5340\u7684\u554f\u984c\u5f88\u8cbc\u8fd1\u7af6\u8cfd\u5340\uff0c\u4e0d\u904e\u6709\u4e0d\u5c11\u7c21\u55ae\u7684\u554f\u984c\u5076\u723e\u53c8\u6703\u51fa\u73fe\u975e\u5e38\u96e3\u7684\u554f\u984c\u3002\u6709\u9ede\u50cf\u806f\u8cfd\u3002\n\nps , \u9664\u4e86\u4ee5\u4e0a2\u500b\u5340\uff0c\u5225\u7684\u5340\u6211\u90fd\u6c92\u53bb\u904e\u3002\n\nps . youarebad \u662f\u4e00\u500b\u597d\u4eba\u3002(\u53ef\u60dc\u6a13\u4e3b\u4e0d\u77e5\u9053\u5df2\u7d93\u9084\u5728\u4e0d\u5728\u4e86\u3002)[/quote]", "Solution_18": ":D :D :D \n\u54c8\u54c8\uff01shobber\uff0c \u53c8\u89c1\u5230\u4f60\u4e86\uff0c\u4f60\u597d\u50cf\u5df2\u7ecf\u5f88\u957f\u65f6\u95f4\u6ca1\u6765Mathlinkers \u4e86\uff0c\u4f60\u5e94\u8be5\u662f\u8fd9\u91cc\u6700\u6709\u540d\u7684Chinese mathlinker :) \u4e86\uff01", "Solution_19": "[quote=\"lssl\"]:D :D :D \n\u54c8\u54c8\uff01shobber\uff0c \u53c8\u89c1\u5230\u4f60\u4e86\uff0c\u4f60\u597d\u50cf\u5df2\u7ecf\u5f88\u957f\u65f6\u95f4\u6ca1\u6765Mathlinkers \u4e86\uff0c\u4f60\u5e94\u8be5\u662f\u8fd9\u91cc\u6700\u6709\u540d\u7684Chinese mathlinker :) \u4e86\uff01[/quote]\n\n\u563f\uff0c\u786e\u5b9e\u5f88\u4e45\u6ca1\u6765\u4e86\uff01\u4e0d\u8fc7\u73b0\u5728\u8fd9\u91cc\u662f\u4e0d\u662f\u5f7b\u5e95\u6539\u6210Art of Problem Solving\u4e86\uff1fMathlinks\u5f7b\u5e95\u88ab\u541e\u5e76\u4e86\uff1f\u5230\u5e95\u53d1\u751f\u4e86\u4ec0\u4e48\u5440" } { "Tag": [ "MATHCOUNTS", "geometry", "percent", "ratio", "3D geometry", "parallelogram", "LaTeX" ], "Problem": "I am thinking of starting the math chats back up. If anyone would like to participate in one of these post here. AIM seems to be the easiest way to do this since it seems that most ppl use it. I am starting to put together some problems to go over. The level of the problems I am getting are mathcounts to AMC 10. The level is aimed at true beginners. Other people can participate but don't expect this to go much past AMC 10 level. Right now I am thinking about having it go for about an hour (it doesn't have to be exact but I am going to try to get enough problems for one hour). AS always you can stay after and go over different problems etc. As for the date and time I don't have an exact one yet but I am looking at later this week or on the weekend. I wont make it during one of the Mock AMC's so don't worry about that. I would like to hear some of your thoughts on time availability.", "Solution_1": "Sure, AIM screen name [u]Feiqi J1989[/u], anytime", "Solution_2": "I'd love to do one. It would be especially helpful when I take the AMC 10 this year, and for my MATHCOUNTS team. The only problem is that I don't have AIM, and my dad is crazy about me downloading stuff, but hopefully, I can get it.\r\n\r\nMy screenname will probably h_s_potter2002, it's usually like that for everything.", "Solution_3": "I can participate/help out! My IM is whitehorseking88.", "Solution_4": "Bringing them back sounds pretty good. My SN is caocoprez", "Solution_5": "I'd be interested aim sceen name- BHorseMath.", "Solution_6": "Oh yeah I forgot to give my sn. It is BaseballJoe2002.", "Solution_7": "I should be able to participate...should fun. Screen name is lonelytraveler8.", "Solution_8": "Man, am I the only person on here who doesn't have AIM? I feel so alone right now.", "Solution_9": "I don't think that you have to download anything to use AIM express [url]http://www.aim.com/get_aim/express/aim_expr.adp?aolp=[/url]\r\n\r\nFor the time I'm thinking either friday, saturday, or sunday night. If it is friday it would have to end before 7:30 because I have a class at that time. I think I ll make a poll to see what is good.", "Solution_10": "joml88 is right, you don't need anything to use AIM express.\r\nI'm not sure when I can do it this weekend because one of the days I'm going to be going to six flags and it's still undecided exactly what day that is. But it'll be one of the three choices posted.", "Solution_11": "I'd really be happy to join a math chat! s/n = KilliingInSleep \r\n\r\nNote: I voted for Friday night, but now that I think about it, Saturday night is better since there isn't MC Counting then.", "Solution_12": "I'd be glad to help, reaperofthegrave is my sn.", "Solution_13": "i'd like to join! i have 3 sn's. first try suenified. if that doesn't work, try pinkksunsets or xX caro0 Xx", "Solution_14": "I'd be willing to help/participate\r\n\r\nAIM ID: hezgotzbeenz", "Solution_15": "so math is hard im a dumn person, and i really need help, what is the answer to 888z*59x-45y=78 ;) ;) ;) ;) :blush: :maybe:", "Solution_16": "Depends on what x, y, and z are :P . Although I seriously doubt you did anything other than type random numbers o_O ...", "Solution_17": "I might b able to do it... depending on what time it is... I'll give u my sn and if I'm on at that sec, I can probably to it. My sn is: luver of rain\r\nAnd also, if some of my friends wants to do it, can they come to ?", "Solution_18": "no answer to I hate genius2154 problem. You can only set the answer up as an inequality and you should be able to set it up as a conjunction. (I think please correct me if I'm wrong) and yes it does depend on what x, y, and z are.", "Solution_19": "I'll do it for sure, but it's just that my dad kida blocked the pop-up thing for aim, and ummm.. i just can't seem to find a way to get the thing unblocked. \r\nSoo.. umm if anyone would like to help mua out here, then pleaseee do! \r\nAnyways my AIM sn is Kristy24seven, and my e-mail's me4everfun@yahoo.com.", "Solution_20": "If anyone wants to do a \"freestyle\" math chat this friday from about 5:30-6:30 EST (of course this will probably go later for those who want to stay) I ll be willing to do it. What I mean by freestyle is that there won't really be much of a certain topic covered. So if you want I can just get some old ARML, Mandelbrot, AHSME or AIME problems to discuss. Of course, if anyone has any problems they would like to discuss that is fine also.", "Solution_21": "SN: MTrance12\r\n\r\nAnytime", "Solution_22": "I apologize but I don't think I ll be able to do a mathchat tomorrow.", "Solution_23": "joml88, are you still too busy to be hosting math chats?\r\n\r\nIf you are, would someone start hosting math chats.\r\n\r\nI really enjoyed them.", "Solution_24": "Silverfalcon is hosting one this Friday at 6 eastern - look at AMC 10 mathchat.", "Solution_25": "EST = GMT - 5:00\r\n\r\nAt least, I think so...", "Solution_26": "depends on when.... sn: kristy24seven", "Solution_27": "Math Chat Tonight!!!!", "Solution_28": "Me too. I would also like to have a math chat.", "Solution_29": "I won't be here for a week. So, I'm sure Silverfalcon can manage things on his own for now. I'm skiing in Vermont. :D Happy Holidays!" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Find all functions $f: \\mathbb Z \\to \\mathbb Z$ such that \\[f(x + y + f(y))=f(x) + 2y ,\\] for all integers $x$ and $y$.", "Solution_1": "[quote=\"Vova-LFML\"]Find all f (x + y + f(y))=f(x) + 2y , f : Z:Z, x,y : Z[/quote]\r\n\r\n$ f(x\\plus{}y\\plus{}f(y))\\equal{}f(x)\\plus{}2y$ $ \\implies$ $ Q(x,y,p)$ : $ f(x\\plus{}p(y\\plus{}f(y))\\equal{}f(x)\\plus{}2py$ $ \\forall x, y, p\\in\\mathbb Z$\r\n\r\n$ Q(0,x,1\\plus{}f(1))$ : $ f((1\\plus{}f(1))(x\\plus{}f(x))\\equal{}f(0)\\plus{}2(1\\plus{}f(1))x$\r\n$ Q(0,1,x\\plus{}f(x))$ : $ f((x\\plus{}f(x))(1\\plus{}f(1))\\equal{}f(0)\\plus{}2(x\\plus{}f(x))$\r\n\r\nAnd so $ 2(1\\plus{}f(1))x\\equal{}2(x\\plus{}f(x))$\r\n\r\nAnd so $ f(x)\\equal{}f(1)x$\r\n\r\nPutting back this value in the original equation, we get $ f(1)(f(1)\\plus{}1)x\\equal{}2x$ and so $ f(1)\\equal{}1$ or $ f(1)\\equal{}\\minus{}2$\r\n\r\nHence the two solutions :\r\n$ f(x)\\equal{}x$\r\n$ f(x)\\equal{}\\minus{}2x$" } { "Tag": [ "calculus", "integration", "geometry", "rectangle", "function", "complex analysis" ], "Problem": "First of all, in what situations is it valid to evaluate an integral of the form given by Perron's formula by instead considering a rectangular contour which has its rightmost side along the line of integration? The line integral along the other three sides of the rectangle must evaluate to zero, right? So how does this work with the zeta function and the explicit formula for Chebyshev's psi function given by Riemann?\r\n\r\nSecondly, if you have a complex function F(x) = g(x)/h(x), and you want to use the residue theorem on F(x) for a contour integral, I know you must consider the zeros of h(x); but, should you consider the singularities of h(x) at all? Technically, the function is not defined there, right? But would its residue there be 0 anyway, or what?", "Solution_1": "A (nonessential) singularity of $ h$ means that $ h$ maps the point to $ \\infty$ with some well-defined multiplicity. So taking the reciprocal means you're going to have a zero of some multiplicity, which you don't need to worry about when considering residues, unless, of course, you have a singularity of $ g$ of a higher order at the same exact spot.\r\n\r\nUnfortunately I don't know Perron's formula so I can't help with that. :(", "Solution_2": "I think I understand now; if the singularity is a removable one, the residue there is zero anyway, so it may be ignored.\r\nThanks, Mystic.\r\n\r\nAs for Perron's formula...\r\n\r\nWell, the way I use it is this; if you have $ g(s) \\equal{} \\sum_{n\\equal{}1}^{\\infty} {{f(n)} \\over {n^s}}$, then Perron's formula can be used to find an explicit formula for a finite sum of f(n) values for n not exceeding x. The only catch is that the answer you get is at a \"halfway point\" for integers x because it's trying to act like a step function.\r\n\r\nFor example...\r\n\r\nLet $ A(x) \\equal{} \\sum_{n\\equal{}1}^{x} f(n)$.\r\n\r\nLet $ A^{\\star}(x)$ be the function with \"halfway points\" at the integers to make it better behaved.\r\n\r\nThen Perron's formula implies (because we chose s = 0) $ A^{\\star}(x) \\equal{} {1 \\over {2\\pi i}}\\int_{c\\minus{}i\\infty}^{c\\plus{}i\\infty} g(z){{x^z} \\over z}\\, dz$.\r\n\r\nYou can choose the value of c you use, but you have to look at where the function is convergent.\r\nhttp://en.wikipedia.org/wiki/Perron%27s_formula" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Determine all rational u, v satisfying the following equation:\r\n\r\n[tex]u^3-v^2+1 = 0[/tex]", "Solution_1": "I think it's already been proven that (2,3) is the only pair of consecutive square and cubic powers (in either direction)... :D", "Solution_2": "$a^3+1=b^2$\r\n$(1+a)(1-a+a^2)=b^2$\r\nObviously $a,b>1$ (except the trivial solution $(a,b)=(0,1)$ so $a^2>a^2-a+1>(a-1)^2$ so $a^2-a+1$ can't be a perfect square. Results that $(a+1,a^2-a+1)=d>1$. From $d|a+1$ and $d|a^2-a+1$ results that $d|3$ which leads to $d=3$.\r\nSo $a\\equiv 1(\\mod 3)$ and $b\\equiv 0(\\mod 3)$.\r\nLet $a=3x+1$ and $b=3y$.\r\n The equation is equivalent to\r\n$x(3x^2-3x+1)=y^2$. Since $(x,3x^2-3x+1)$ results that both $x$ and $3x^2-3x+1$ are perfect squares.\r\nLet $x=q^2\\ge1$. If $q=1$ we get the solution $(a,b)=(2,3)$. If $q>1$ then $3q^4-3q^2+1=t^2>1$.\r\n$3q^2(q^2-1)=t^2-1$ which leads to $3q^2|t^2-1$ (1).\r\n$(3q^2-1)(q^2-1)=t^2-q^2$ which leads to $3q^2-1|3t^1-1$ (2).\r\nFrom (1) and (2) we deduce that there exist $m,n>0$ s.t. \r\n$\\left\\{\\begin{array}{l}\r\nt^2-1=3mq^2\\\\\r\n3t^2-1=n(3q^2-1)\\end{array}\\right.\\Leftrightarrow\\left\\{\r\n\\begin{array}{l}\r\n3t^2-1=9mq^2+2\\\\\r\n3t^2-1=3nq^2-n\r\n\\end{array}\\right.\\Rightarrow 9mq^2+2=3nq^2-n\\Rightarrow n+2=3q^2(n+3m)$\r\n but $q>1$ and $n+3m>n+2$ (because $n,m>0$ so we can't have a solution for $q>1$.\r\n\r\nThe solutions remain $(a,b)\\in\\{(0,1),(2,3)\\}$", "Solution_3": "Did I miss something or both answers only deal with integers instead of rational numbers? :? \r\n\r\nPierre.", "Solution_4": "[quote=\"pbornsztein\"]Did I miss something or both answers only deal with integers instead of rational numbers? :? [/quote]\r\nIt is unimportant thing...\r\n :lol: :lol: :lol: :lol: :lol:", "Solution_5": "Did Ulysses realy meant rationals, or should it be replaced by integers?", "Solution_6": "I think, Ulysses meant rational numbers!", "Solution_7": "In fact, it doesn't change anything about the solutions, even if they omit $(-1,0)$.\r\nBut, it seems to be a bit harder :\r\nLet $u = \\frac p q, v = \\frac a b$ where $\\gcd(p,q)= \\gcd (a,b) = 1$, and $b,q > 0$.\r\nClearing the denominators, it is easy from the equation to deduce that $b^2$ divides $q^3$ and that $q^3$ divides $b^2$, so that $q^3 = b^2 = r^6$ for some positive integer $r$, coprime with $a$ and $p$.\r\nThus, the equation becomes :\r\n$a^2 + (-p)^3 = r^6$.\r\n\r\nThis last equation is quite well-known, but it would be too lengthy to solve it here, so that I give the appropriate link :\r\nhttp://www.mathpages.com/home/kmath213.htm\r\n\r\nNote that they have omitted the case $x = +/- k^2, y = 0, z = +/- k$ in their solutions.\r\nUsing that $r$ is coprime with $a$ and $b$ and is positive, it leads to the solutions $(-1,0),(0,1),(2,3)$.\r\n\r\nPierre." } { "Tag": [ "AwesomeMath", "summer program" ], "Problem": "In the math jam, Zuming said something about writing about a problem that you solved in the past, but I can't find anything about that in the application information on the website. Can someone give me some more information?\r\n\r\nThanks", "Solution_1": "Zuming was referring to the second prompt on the personal information form.\r\n\r\nComplete application information is here: [url]http://www.awesomemath.org/application.shtml[/url].\r\n\r\nThe form in question is here: [url]http://www.awesomemath.org/application/Personal_Information_Form_electronic.doc[/url].\r\n\r\nHope this helps!\r\n\r\nZac Cox\r\nProgram Coordinator, AwesomeMath", "Solution_2": "Thank you very much." } { "Tag": [], "Problem": "Let $A=\\{1,2,\\ldots, 2006\\}$. Find the maximal number of subsets of $A$ that can be chosen such that the intersection of any 2 such distinct subsets has 2004 elements.", "Solution_1": "it's ok that every subset has at least 2004 elements, and at most 2006 elemnts.\r\n\r\nIf we have subset with 2006 elemts, then the maximal number of subsets is 2.\r\n\r\nIf we don't have subset with 2006 elemnts:\r\n\r\n1) If we don't have subset with 2004 elemnts maximal number is 2006. Subsets with 2005 elements.\r\n2) we have subset with 2004 which doesn;t contain numbers $a$ and $b$. Now we can at most 2004 subsets with 2005 because it can't contain $a$ and $b$.\r\n\r\nHence 2006 is max" } { "Tag": [ "geometry", "trapezoid", "power of a point", "radical axis", "similar triangles", "geometry proposed" ], "Problem": "Suppose that the circles $C_1$ and $C_2$ of different radii intersect in the points A and B . The tangents to these circles from A intersect $C_1$ and $C_2$ at the points P and Q, respectively. Extend PB and QB until they intersect $C_2$ and $C_1$ in M and N respectivly. prove that PM=QN.", "Solution_1": "Nice problem. :) \r\n\r\nFirst, we'll prove a lemma:\r\n\r\nLet $A, B$ be fixed points, and let $r_1, r_2$ be variable reals such that $r_1+r_2 \\geq AB$. Let $C_1, C_2$ be circles with centers at $A, B$ and radii $r_1, r_2$ respectively. Let $P$ be one of the points of intersection of the circles, and let $X$ be a fixed angle. We prove that if the line through $P$ such that its angle with line $AB$ is $X$ intersects the circles at points $M, N$, then $MN$ is fixed.\r\n\r\nTo prove it, fix $r_2$, take $r_1'$ not equal to $r_1$ and take the circle with center $A$ and radius $r_1'$. Let $P', M', N'$ be defined analogously. Then it's easy to show that $MPP'M'$ is an isosceles trapezoid and the same for $PNN'P'$, and thus $MNN'M'$ is a parallelogram. That's it.\r\n\r\nNow, back to the problem. Let the circle centers be $O_1, O_2$. The angle $PAQ$ is semi-inscribed in both circles, so it is equal to both $AMQ$ and $ANP$. Then the angles $AMQ, ANP$ are equal. Thus angles $PBA, QBA$ are equal, because they are supplementary of $AMQ, ANP$. Then substracting equal angles $MBQ, NBP$, $NBA=MBA$. So the bisector of angle $MBN$ is the radical axis of $C_1, C_2$.\r\n\r\nNow, by applying the lemma, $PM, QN$ are equal to the corresponding segments we get when we take $r_1=O_1O_2-r_2$. But then since angles $NBA=MBA$, the segments are symmetric and thus equal.", "Solution_2": "It's an easy problem :D \r\nIt's easy to see that triangles $ APB$ and $ ABQ$ are similar\r\nthen : $ PA/QA=PB/AB=AB/BQ$\r\nso $PA^2/QA^2=PB/QB$\r\nBut $PB.PM=PA^2$ and $QB.QN=QA^2$\r\nSo$PM=QN$ QED", "Solution_3": "[quote=\"marko avila\"]Suppose that the circles $C_1$ and $C_2$ of different radii intersect in the points A and B . The tangents to these circles from A intersect $C_1$ and $C_2$ at the points P and Q, respectively. Extend PB and QB until they intersect $C_2$ and $C_1$ in M and N respectivly. prove that PM=QN.[/quote]\r\n\r\nThe problem is really straightforward:\r\n\r\nUsing directed angles modulo 180\u00b0, we have < AQB = < AMB, since the points A, B, Q and M lie on one circle (namely, on the circle $C_2$). In other words, < AQN = < AMP. Similarly, < ANQ = < APM. Thus, the triangles AQN and AMP are similar.\r\n\r\nSince the line AP is the tangent to the circle $C_2$ at the point A, we have < (AP; AB) = < AQB by the tangent-chordal angle theorem. Similarly, < (AQ; AB) = < APB. Thus,\r\n\r\n < ABP = < (AB; BP) = < (AP; BP) - < (AP; AB) = < APB - < (AP; AB)\r\n = < (AQ; AB) - < AQB = < (AQ; AB) - < (AQ; BQ) = < (BQ; AB) = < NBA.\r\n\r\nSince the points A, B, P and N lie on one circle (namely, on the circle $C_1$), we have < ABP = < ANP and < NBA = < NPA; thus, this becomes < ANP = < NPA. Thus, the triangle NAP is isosceles with AP = AN. Hence, the similar triangles AQN and AMP must be congruent. Thus, PM = QN. Problem solved.\r\n\r\n darij", "Solution_4": "i just have a question. i solved this problem but im not sure if its ok. what i proved is the following:\r\n\r\ntriangles APN and AQM are similar \r\ntriangle APN is isoceles\r\nhence triangle AQM is also isoceles\r\nangleNAQ=anglePAM \r\nhence since having two corresponding sides equal and the angle between also equal we have \r\nAPM and ANQ are congruent so were done. im correct right darij??? its just that in your proof you avoid proving the isoceles and similar triangles APN and AQM SO IM HAVING SOME DOUBTS :blush:", "Solution_5": "[quote=\"marko avila\"]i just have a question. i solved this problem but im not sure if its ok. what i proved is the following:\n\ntriangles APN and AQM are similar \ntriangle APN is isoceles\nhence triangle AQM is also isoceles\nangleNAQ=anglePAM \nhence since having two corresponding sides equal and the angle between also equal we have \nAPM and ANQ are congruent so were done. im correct right darij??? its just that in your proof you avoid proving the isoceles and similar triangles APN and AQM SO IM HAVING SOME DOUBTS :blush:[/quote]\r\n\r\nHmm, how exactly do you prove that the triangles APN and AQM are similar, and, more importantly, that the triangle APN is isosceles? I mean, I don't see any really [i]obvious[/i] reason for that, but maybe there is one and I am just blind. However, if you prove this using an angle chase, then your solution does not differ really much from mine.\r\n\r\n Darij", "Solution_6": "well here is my proof :\r\nconstruct lines AB, AM and AN . so we have anglePBN=angleQBM=angleQAM=anglePAN= :theta: . also angleAQB=anglePAB= :varepsilon: because AP is a tangent to C2. analogously angleBAM=angleANB= :beta: because AQ is tangent to C1. now by pure angle chase calculated with the angles already given we hace anglePBA= :beta: + :varepsilon: now angleANP=anglePBA= :beta: + :varepsilon:. angleAPN=angleAPB+angleBPN= :beta: + :varepsilon: because they subtend the same arcs. also angleAQN= :beta: + :varepsilon: analogously. so angle APN=angleAQM and angleQAM=anglePAN APN and AQM are similar but angleAPN=ANP= :beta: + :varepsilon: so APN is isoceles and hence AQM to now angleNAQ=angleMAP so since we have AQ=AM and AP=AN we have that triangles QAN and MAP ARE CONGRUENT so were done. by the way i think ypu proved that APN so i know im right im pretty sure AQM IS TOO :blush:" } { "Tag": [ "calculus", "integration", "logarithms", "calculus computations" ], "Problem": "\\[ \\int\\limits_{0}^{4}{\\frac {x + 1}{(x^2 + 2x + 2)^{2}} dx\r\n}\\]\r\n\r\nEDIT: Problem has been fixed.", "Solution_1": "$ I \\equal{} \\frac{1}{2}\\left[ {\\ln \\left( {{x^2} \\plus{} 2x \\plus{} 2} \\right)} \\right]_0^4 \\equal{} \\frac{1}{2}\\ln 13$", "Solution_2": "Sorry, I copied the problem incorrectly. It has been fixed.", "Solution_3": "$ \\int_0^4 \\frac{x\\plus{}1}{(x^2\\plus{}2x\\plus{}2)^2}\\,dx \\equal{} \\frac12 \\int_0^4 (2x\\plus{}2) (x^2\\plus{}2x\\plus{}2)^{\\minus{}2}\\,dx$\r\n\r\n$ \\equal{} \\minus{} \\frac12 \\left[\\frac{1}{x^2\\plus{}2x\\plus{}2}\\right]_0^4 \\equal{} \\frac{3}{13}$." } { "Tag": [ "function", "combinatorics unsolved", "combinatorics" ], "Problem": "\"In how many ways can a total of 1001 be made from numbers 1-100 without repeating any number\"\r\nremember-using all numbers is not necessary but no number should be repeated...", "Solution_1": "plz anyone reply...\r\nthis is very imp for me...", "Solution_2": "this same problem was posted a couple of days ago by someone else. where is it from?", "Solution_3": "i dont know...just encountered it while solving another question....\r\nif it was solved earlier can u provide the link as i cant find it...", "Solution_4": "anyone????", "Solution_5": "there's no nice way to calculate the number of partitions of a number... but this may help you a bit:\r\n \r\nhttp://en.wikipedia.org/wiki/Partition_of_a_number", "Solution_6": "thanks...\r\n :lol:", "Solution_7": "Hi arshad;\r\n\r\nThat problem requires generating functions and the use of a computer. The number of unrestricted partitions of 1001 is 25032297938763929621013218349796.\r\n\r\nSince your problem is a restricted partition problem the number of partitions of 1001 will be way too large.\r\n\r\nYou start with the linear diophantine equation. \r\n\r\na1 + a2 + a3 + a4 + ... a100 = 1001\r\n\r\nThe ordinary generating functions for this are:\r\n\r\n$ (1 \\plus{} x ) (1 \\plus{} x^2) (1 \\plus{} x^3 ) \\dots (1 \\plus{} x^{100})$\r\n\r\nor another one:\r\n\r\n$ \\frac {(1 \\minus{} x^2)(1 \\minus{} x^4)(1 \\minus{} x^6)(1 \\minus{} x^8)(1 \\minus{} x^{10})\\dots(1 \\minus{} x^{200})}{(1 \\minus{} x)(1 \\minus{} x^2)(1 \\minus{} x^3)(1 \\minus{} x^4)(1 \\minus{} x^5)\\dots(1 \\minus{} x^{100})}$\r\n\r\nYou must find the coefficient of x^1001 in either of those expansions and that is the answer.\r\n\r\nNormally it can be done by binomial tricks or residues but finding the coefficient of x^1001 here turned out to be difficult so I used a computer. The answer was found to be.\r\n\r\n646972680324577041212 ways without repeating any number.", "Solution_8": "thanx for the nice solution...\r\n:)", "Solution_9": "Hi arshad;\r\n\r\nHope it helps." } { "Tag": [ "geometry theorems", "geometry" ], "Problem": "[hide=\"A proof of the Butterfly theorem.\"][b]The Butterfly theorem.[/b] Given a chord $PQ$ of a circle $w=C(O,r)$ draw any other two chords $AB$ and $CD$ passing through its midpoint $M$. Call the points $X$, $Y$ where $AD$ and $BC$ respectively meet $PQ$. Then $M$ is also the midpoint of $XY$.\n\n[u]Proof.[/u] Call the projections $U$, $V$ of the center $O$ on the chords $AD$, $BC$ respectively. \n$\\triangle ADM\\sim\\triangle CBM\\Longrightarrow \\widehat {AUM}\\equiv \\widehat {CVM}\\ .$ The quadrilaterals $OMXU$, $OMYV$\nare cyclically (are inscribed in the circles with the diameters $[OX]$, $[OY]$ respectively).\nTherefore, $\\widehat {AUM}\\equiv \\widehat {XOM},\\ \\widehat {CVM}\\equiv \\widehat {YOM}\\Longrightarrow \\widehat {XOM}\\equiv \\widehat {YOM}\\Longrightarrow MX=MY\\ .$[/hide]", "Solution_1": "Using butterfly theorem, prove this problem:\nProblem: Let triangle $ABC$, $BK\\perp{AC}$, $CL\\perp{AB}$, intersect in $H$. A line through $H$ cut $AB$, $AC$ in $P$, $Q$ respectively. Prove that: $HP=HQ$ iff $MP=MQ$.", "Solution_2": "Dear Mathlinkers,\nfor the last problem, you can see\nHttp://perso.orange.fr/jl.ayme vol. 7 Butterfly p. 50-53\n\nYou can also see at the end of the article references and historical notes...\nSincerely\nJean-Louis", "Solution_3": "But I can not find the solution to this whole", "Solution_4": "[quote=\"AnhIsGod\"]Using butterfly theorem, prove this problem:\nProblem: Let triangle $ABC$, $BK\\perp{AC}$, $CL\\perp{AB}$, intersect in $H$. A line through $H$ cut $AB$, $AC$ in $P$, $Q$ respectively. Prove that: $HP=HQ$ iff $MP=MQ$.[/quote]\n\nwhat's $M$?", "Solution_5": "[quote=\"XmL\"][quote=\"AnhIsGod\"]Using butterfly theorem, prove this problem:\nProblem: Let triangle $ABC$, $BK\\perp{AC}$, $CL\\perp{AB}$, intersect in $H$. A line through $H$ cut $AB$, $AC$ in $P$, $Q$ respectively. Prove that: $HP=HQ$ iff $MP=MQ$.[/quote]\n\nwhat's $M$?[/quote]\nSorry, M is midpoint BC", "Solution_6": "I saw this problem in Turkish olympiad P-5(think it TST). :wink:", "Solution_7": "Dear Mathlinkers,\nhave a look tp\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=192515\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=260415\nSincerely\nJean-Louis", "Solution_8": "It is clear that $ LBCK $ is cyclic, and also observe that ,$ M $ is the center of the circle through the points $ L,B,C,K $ , now we draw the circle with these point, applying butterfly theorem we get, $ H $ is the midpoint of $ PQ $ iff $ H $ is the mid point of $ XY $ , which means $ PH=HQ $ iff $ HM \\perp XY $ , because $ M $ is the center of the circle, so, finally the desired condition will be satisfied iff the perpendicular condition situates , and if $ HM \\perp XY $ proves $ MP=MQ $ as $ \\triangle PXM $ and $ \\triangle YMQ $ will be congruent..", "Solution_9": "[quote=\"SayanRoychowdhuri\"]It is clear that $ LBCK $ is cyclic, and also observe that ,$ M $ is the center of the circle through the points $ L,B,C,K $ , now we draw the circle with these point, applying butterfly theorem we get, $ H $ is the midpoint of $ PQ $ iff $ H $ is the mid point of $ XY $ , which means $ PH=HQ $ iff $ HM \\perp XY $ , because $ M $ is the center of the circle, so, finally the desired condition will be satisfied iff the perpendicular condition situates , and if $ HM \\perp XY $ proves $ MP=MQ $ as $ \\triangle PXM $ and $ \\triangle YMQ $ will be congruent..[/quote]\n\nWhat is $X,Y$? Intersections $PQ$ with $k(LBCK)$?" } { "Tag": [ "Putnam", "IMO" ], "Problem": "Hi. If all the outstanding contestants at IMO from past to present would compete in a harder than IMO competition , who would be the ultimate winner? Give your comments. \r\n THANKS", "Solution_1": "probably Ciprian Manolescu :D", "Solution_2": "christian reiher!", "Solution_3": "I also think that Christian Reiher is one of the favourits of this fantasy competition!!! :D\r\n\r\nBut we will never know who will win...cause I think that there will never be a thing like that( never say never :D). \r\n\r\ncheers!", "Solution_4": "Probably there would be an acerbic fight between Reid Barton and Ciprian Manolescu :)", "Solution_5": "I think that such a competition would be hard to organize.\r\nCuz some problems are of such type (Fermats TH. or others) that may require years even for them, but other problems would be easy for all of them. I don't thnk that there wud be a problem, such that one of thm wud solve it, but the other wudnt. Maybe only if it were very specific on some topic. ( I mean R. Barton is the best at informatics, so maybe in combinatorics he would now something that the others dont know ) \r\nHope u didn't get tired reading this.\r\n:)\r\n:)", "Solution_6": "Laurent Lafforgue got 42/42 two times in 1990 and 1991 I think. (his brother got the Fields Medal!).\r\nThere's also a norvegian who did 6 or 7 times the IMO !!! (1st place with Olivier Benoist at the Concours General 2003)", "Solution_7": "This norvegian is David Kunszenti - Kovacs I think.\r\n\r\nI thought the \"concours gnral\" was only for French students ?\r\n\r\nYou know, Olivier (26 p) beats David K - K (23 p) at the IMO. :D", "Solution_8": "Yes, but David Kunszenti - Kovacs was in a french school, so he could do it.", "Solution_9": "yes, and he isn't even french, nor novegian... I think he was from eastern europe, but i can't remember from where.", "Solution_10": "Well, I don't think that the best is the contestant that got perfect score on the maximum number of IMOs. For me the best is :maybe: Grigorij Perelman. He participated only in one olympiad (IMO 1982) and got the perfect score there. Now he seems to have proved the Poincare Conjecture.", "Solution_11": "David Kunszenti-Kovacs is from Hungary.\r\n\r\nThe greatest mathematician for the moment and who did extremely well at IMOs is Laszlo Lovasz (from Hungary). He took part at IMO from 1963 to 1966 winning three gold medals and a silver medal.\r\n\r\nKantor", "Solution_12": "Now we are playing fantasy IMO now? My fantasy basketball team is enough to manage. . . =)\r\n\r\nBut to actually stay on the topic. . . it could be YOU! So work hard =D\r\n\r\nTo be serious, these things are hard to judge, for a variety of reasons:\r\n\r\n1) IMOs are getting harder, but the rate at which this is happening is hard to determine, so how do you know how much to handicap them?\r\n2) People know more nowadays going to the IMO (correlated to #1) thanks to information transmission, books, great sites like this one (go Valentin), peers, and the evolution of the art of olympiad problem solving. So people on average are much better - but it doesn't mean that someone from 10 years ago, given the same situations, might not be as good as others are now if they were born later.\r\n3) depends on the problems, styles, etc. so much. Judges too.\r\n4) in more competitive countries, even the best people get knocked down sometimes just because the pool is so high (i.e. China) - you might see people that would otherwise go to the IMO 5 times or something only be on one Chinese team since the selection process is so rigorous and all. . .\r\n\r\nDang that's a long first post.\r\n\r\nLater,\r\n\r\n-Y.Z.\r\n\r\nP.S. Haha this site makes me feel warm and fuzzy sorta", "Solution_13": "I would wish that the winner will be Christian Reiher. But I don't belive that. And I think participantes from IMO 10 years ago would have a chanche too. Because they had 10 years more to improve their skills in Problem Solving. I think the young 11year-old (at that time) Australian who participated in the IMO 1986-1988 (I don't know his name anymore) would be one of the favourites.", "Solution_14": "Alas. . . 10 more years to improve, yes. But olympiad problem solving skill is not something that necessarily goes up with skill in mathematics - so once someone is done with the IMO and such they (probably) won't do their training at nearly an intense level as they did before - unless their job is to train IMO kids or something =P if anything, I think this skill deteriorates in college/grad school when you are crunching away at more real yet less nice math problems.\r\n\r\nAs a side note, a big portion (a majority by now???) of kids near the top of IMO level seem to end up doing something BESIDES pure math - this phenomenon is also seen with the Putnam exam. Hmm. . . that probably calls for a different philosophical topic.\r\n\r\nOne last muse - this topic is iffy since it might just become a combat of nationalism ^_^\r\n\r\nBut I'll leave it at that.\r\n\r\n-Yan", "Solution_15": "L\u00ea H\u00f9ng Vi\u1ec7t B\u1ea3o !", "Solution_16": "[quote=\"Igor\"]Laurent Lafforgue got 42/42 two times in 1990 and 1991 I think. (his brother got the Fields Medal!).\n[/quote]\r\n\r\nI do believe this is Vincent who got the gold medal in 1990 and 1991.\r\nLaurent attended the IMO contest in 1984 and 1985.\r\n\r\n-- \r\nFabrice", "Solution_17": "Yes, you are right, sorry :blush:", "Solution_18": "Manolescu X Barton...?\r\n\r\nActually, I will be happy to participate in the IMO once time in my life...", "Solution_19": "[quote=\"spiritrock\"]L\u00ea H\u00f9ng Vi\u1ec7t B\u1ea3o ![/quote]\r\n\r\nHey, I suppose that Le Hung Viet Bao is brilliant, and possibly very brilliant among the Vietnamese IMO prize winners. Could you please explain why you choose Le Hung Viet Bao ?\r\nThanks", "Solution_20": "[quote=\"imam12\"]Hi. If all the outstanding contestants at IMO from past to present would compete in a harder than IMO competition , who would be the ultimate winner? Give your comments. \n THANKS[/quote]\r\n\r\nPhan Thi Ha Duong\r\n\r\nShe is the most brilliant person I have ever met.\r\nAnd the way she outstands is that she decides to come back to Vietnam to help Vietnamese young talents to win.", "Solution_21": "to me I bel\u0131eve the chinese are better in terms of skills and flair in mathematics....but unfourtunately,most of the contestants from ch\u0131na only participate once and they go away w\u0131th gold medal.....pls guys,take this into consideration", "Solution_22": "Lisa Sauermann.", "Solution_23": "They say the questions are getting harder but that's because more people get good training which makes a difference.\r\nWhen I went in 1977 and 1978 as part of the British team we had no training at all. So I must make a special mention for\r\ntwo of my team mates, John Rickard who had three golds and Richard Borcherds, 1 gold + 1 silver who is a Fields medalist. \r\nI am sure they would do very well against more recent winners, but you would have to hold the contest in another platonic \r\nrealm because John has sadly left this one.", "Solution_24": "Iurie Boreico with no doubt.", "Solution_25": "I kinda of read an article about Artur Avila's excentric life (http://www.imo-official.org/participant_r.aspx?id=3801) and he seems a monster.\r\n\r\nHe was young when he got gold on his first attempt on IMO and he refused to represent Brazil again because it seems he already conquered what was to be conquered in a competition of this level. He refused to go again saying: \"Problems that have a certain solution don't interest me. The uncertain is much more fun\" (or something like that, the article was in portuguese and was too long so I read bits of it) and to follow the path of a mathematician. Apparently what shocked the most was the young boy strong resolution about his future.\r\n\r\nWhat I find most interesting about him was that when he was 6th grade and he first participated an mathematic olimpyad he got bronze in the national competition (sounds like he is human after all), this turned out gold medal the following years needless to say...", "Solution_26": "Terence Tao I guess? The youngest bronze, silver, and gold medalist in IMO history. Won Fields Prize in 2006. ", "Solution_27": "Either Terence Tao or Reid Barton or Iurie Boreico.", "Solution_28": "[quote=RunyangWang]Terence Tao I guess? The youngest bronze, silver, and gold medalist in IMO history. Won Fields Prize in 2006.[/quote]\n\nagree." } { "Tag": [ "integration", "trigonometry", "analytic geometry", "geometry", "calculus", "graphing lines", "slope" ], "Problem": "Given that, for non-negative integers $ n$,\r\n\r\n$ I_n \\equal{} \\int_0^\\frac{\\pi}{2} \\sin ^{n} \\theta\\, d\\theta$,\r\n\r\nshow that, for $ n\\ge2$, $ nI_n \\equal{} (n\\minus{}1)I_{n\\minus{}2}$.\r\n\r\nThe equation of a curve, in polar coordinates, is\r\n\r\n$ r\\equal{}\\sin^{3}\\theta,$ for $ 0\\le\\theta\\le \\pi$.\r\n\r\nFind the equations of the tangents at the pole and sketch the curve.\r\n\r\nFind the exact area of the region enclosed by the curve.", "Solution_1": "For the first one, use integration by parts.", "Solution_2": "That's the easy part:\r\n\r\n$ I_{n}\\equal{}\\int_{0}^\\frac{\\pi}{2}\\sin^{n}\\theta\\, d\\theta$\r\n\r\n$ \\equal{} \\int_{0}^\\frac{\\pi}{2}\\sin^{n\\minus{}1}\\theta\\sin\\theta\\,d\\theta$\r\n\r\nIntegrating by parts, we have\r\n\r\n$ (n\\minus{}1)\\int_{0}^\\frac{\\pi}{2}\\sin^{n\\minus{}2}\\theta\\cos^{2}\\theta\\,d\\theta$\r\n\r\n$ \\equal{}(n\\minus{}1)\\int_{0}^\\frac{\\pi}{2}(\\sin^{n\\minus{}2}\\theta \\minus{} \\sin^{n}\\theta)\\, d\\theta$\r\n\r\n$ \\Rightarrow nI_{n} \\equal{} (n\\minus{}1)I_{n\\minus{}2}$.", "Solution_3": "The area of the region bounded by the curve is given by $ A\\equal{}\\int_{\\theta_{1}}^{\\theta_{2}}\\frac{1}{2}(r(\\theta))^{2}\\,d\\theta$.", "Solution_4": "Thanks, JRav. I know that, but do I assume the limits $ \\theta_1$ and $ \\theta_2$ to be $ 0$ and $ \\pi$ respectively or do I need to find the tangents at the pole first? Are these tangents $ \\theta \\equal{} 0$ and $ \\theta \\equal{} \\pi$?\r\n\r\nFor the graph I have a single loop.\r\n\r\nI have seen the following (are there other ways of finding tangents at the pole?):\r\n\r\nIf $ f(\\alpha) \\equal{} 0$ but $ f(\\theta) > 0$ in an interval $ \\alpha < \\theta < ...$ or $ ... < \\theta < \\alpha$, then the line $ \\theta \\equal{} \\alpha$ is a tangent to the graph of $ r \\equal{} f(\\theta)$ at the origin (pole).\r\n\r\nIf so, I have \r\n\r\n$ \\frac {1}{2}\\int_{0}^{\\pi} \\sin ^{6}\\theta\\,d\\theta$\r\n\r\n$ \\equal{} \\int_{0}^\\frac {\\pi}{2} \\sin^{6}\\theta\\, d\\theta$\r\n\r\nwhich by Wallis' formula is\r\n\r\n$ \\frac {5}{6}\\times\\frac {3}{4}\\times\\frac {1}{2}\\times\\frac {\\pi}{2}$\r\n\r\n$ \\equal{} \\frac {5\\pi}{32}$.\r\n\r\nIs this correct?", "Solution_5": "Can someone post a sketch of the curve and/or tell me how to post graphs?\r\n\r\nI would also like to learn more about finding the equations of tangents at the pole (and to curves given in polar form in general).", "Solution_6": "I'm fairly sure my sketch is correct, but I'm still bothered about tangents at the pole. There must be a number of different ways of approaching this. For instance, can we change to Cartesian coordinates and differentiate implicitly and then substitute $ x\\equal{}0$ and $ y\\equal{}0$? \r\n\r\nI get (and I'm rushing again, so I've probably made some mistakes)\r\n\r\n$ (x^{2} \\plus{} y^{2})^{2} \\equal{} y^{3}$\r\n\r\n$ \\Rightarrow\\, y' \\equal{} \\frac{\\minus{}4x(x^{2}\\plus{}y^{2})}{4x^{2}y \\plus{} 4y^{3} \\minus{}3y^{2}}$\r\n\r\n$ \\Rightarrow$ the slope of the tangents at the origin and at $ (0, 1)$ are $ 0$ and so the tangents at the pole are $ \\theta \\equal{} 0$ and $ \\theta \\equal{} \\pi$.\r\n\r\nIs this correct? \r\n\r\nHow else could one approach this problem?", "Solution_7": "Does anyone have any ideas on this?", "Solution_8": "I'm hoping that different countries have different ways of approching the problem of tangents at the pole." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Supposing $ a,b,c > 0$,prove that:\r\n$ (a \\plus{} 1 \\minus{} \\frac {1}{b})(b \\plus{} c \\minus{} \\frac {1}{c}) \\plus{} (b \\plus{} 1 \\minus{} \\frac {1}{c})(c \\plus{} 1 \\minus{} \\frac {1}{a}) \\plus{} (c \\plus{} 1 \\minus{} \\frac {1}{a})(a \\plus{} 1 \\minus{} \\frac {1}{b})$\u2265$ 3$", "Solution_1": "[quote=\"SUPERMAN2\"]Supposing $ a,b,c > 0$,prove that:\n$ (a \\plus{} 1 \\minus{} \\frac {1}{b})(b \\plus{} c \\minus{} \\frac {1}{c}) \\plus{} (b \\plus{} 1 \\minus{} \\frac {1}{c})(c \\plus{} 1 \\minus{} \\frac {1}{a}) \\plus{} (c \\plus{} 1 \\minus{} \\frac {1}{a})(a \\plus{} 1 \\minus{} \\frac {1}{b})$\u2265$ 3$[/quote]\r\n\r\nThat's wrong: try $ a\\equal{}b\\equal{}c\\equal{}\\frac{1}{2}$\r\nI think you mean:\r\n$ (a \\plus{} 1 \\minus{} \\frac {1}{b})(b \\plus{} 1 \\minus{} \\frac {1}{c}) \\plus{} (b \\plus{} 1 \\minus{} \\frac {1}{c})(c \\plus{} 1 \\minus{} \\frac {1}{a}) \\plus{} (c \\plus{} 1 \\minus{} \\frac {1}{a})(a \\plus{} 1 \\minus{} \\frac {1}{b})$\u2265$ 3$\r\n\r\nbut it still wrong, try the same...$ a\\equal{}b\\equal{}c\\equal{}\\frac{1}{2}$", "Solution_2": "[quote=\"SUPERMAN2\"]Supposing $ a,b,c > 0$,prove that:\n$ (a \\plus{} 1 \\minus{} \\frac {1}{b})(b \\plus{} c \\minus{} \\frac {1}{c}) \\plus{} (b \\plus{} 1 \\minus{} \\frac {1}{c})(c \\plus{} 1 \\minus{} \\frac {1}{a}) \\plus{} (c \\plus{} 1 \\minus{} \\frac {1}{a})(a \\plus{} 1 \\minus{} \\frac {1}{b})$\u2265$ 3$[/quote]\r\nI am really sorry.This problem is from \"Secrets in ineq\" written by Pham Kim Hung(I don't know how to solve it).He must make some mistakes in his book.", "Solution_3": "[quote=\"SUPERMAN2\"][quote=\"SUPERMAN2\"]Supposing $ a,b,c > 0$,prove that:\n$ (a \\plus{} 1 \\minus{} \\frac {1}{b})(b \\plus{} c \\minus{} \\frac {1}{c}) \\plus{} (b \\plus{} 1 \\minus{} \\frac {1}{c})(c \\plus{} 1 \\minus{} \\frac {1}{a}) \\plus{} (c \\plus{} 1 \\minus{} \\frac {1}{a})(a \\plus{} 1 \\minus{} \\frac {1}{b})$\u2265$ 3$[/quote]\nI am really sorry.This problem is from \"Secrets in ineq\" written by Pham Kim Hung(I don't know how to solve it).He must make some mistakes in his book.[/quote]\r\nThis inequality 's true with condition $ abc\\equal{}1$ :)", "Solution_4": "[quote=\"quykhtn-qa1\"][quote=\"SUPERMAN2\"][quote=\"SUPERMAN2\"]Supposing $ a,b,c > 0$,prove that:\n$ (a \\plus{} 1 \\minus{} \\frac {1}{b})(b \\plus{} c \\minus{} \\frac {1}{c}) \\plus{} (b \\plus{} 1 \\minus{} \\frac {1}{c})(c \\plus{} 1 \\minus{} \\frac {1}{a}) \\plus{} (c \\plus{} 1 \\minus{} \\frac {1}{a})(a \\plus{} 1 \\minus{} \\frac {1}{b})$\u2265$ 3$[/quote]\nI am really sorry.This problem is from \"Secrets in ineq\" written by Pham Kim Hung(I don't know how to solve it).He must make some mistakes in his book.[/quote]\nThis inequality 's true with condition $ abc \\equal{} 1$ :)[/quote]\r\nIf $ abc\\equal{}1$, so, is my solution right?\r\n\r\n$ \\sum (a\\plus{}1\\minus{}\\frac 1b)(b\\plus{}c\\minus{}\\frac 1c) \\ge 3$,\r\n\r\n$ \\sum abc(a\\plus{}1\\minus{}\\frac 1b)(b\\plus{}c\\minus{}\\frac 1c) \\ge 3$,\r\n\r\n$ \\sum a(ab\\plus{}b\\minus{}1)(bc\\plus{}c^2\\minus{}1) \\ge 3$,\r\n\r\n$ \\sum (a^2b^2c \\plus{} a^2bc^2 \\minus{} a^2b \\plus{} ab^2c \\plus{} abc^2 \\minus{} ab \\minus{}abc \\minus{} ac^2 \\plus{} a)\\ge 3$,\r\n\r\n$ \\sum (a\\plus{}b\\plus{}c \\plus{} ac \\plus{} \\frac ac \\plus{} \\frac cb) \\ge \\sum (3\\sqrt[3]{abc} \\plus{} ac \\plus{} \\frac ac \\plus{} \\frac cb) \\ge 6$.\r\n\r\n :wink:", "Solution_5": "[quote=\"SUPERMAN2\"]Supposing $ a,b,c > 0$,prove that:\n$ (a \\plus{} 1 \\minus{} \\frac {1}{b})(b \\plus{} c \\minus{} \\frac {1}{c}) \\plus{} (b \\plus{} 1 \\minus{} \\frac {1}{c})(c \\plus{} 1 \\minus{} \\frac {1}{a}) \\plus{} (c \\plus{} 1 \\minus{} \\frac {1}{a})(a \\plus{} 1 \\minus{} \\frac {1}{b})$\u2265$ 3$[/quote]\r\n\r\nCan you tell me the number of page of this problem in the book?" } { "Tag": [ "geometry" ], "Problem": "Medians $ BD$ and $ CE$ of triangle $ ABC$ are perpendicular, $ BD \\equal{} 8$, and $ CE \\equal{} 12$. The area of triangle $ ABC$ is\r\n[asy]defaultpen(linewidth(.8pt));\ndotfactor=4;\n\npair A = origin;\npair B = (1.25,1);\npair C = (2,0);\npair D = midpoint(A--C);\npair E = midpoint(A--B);\npair G = intersectionpoint(E--C,B--D);\n\ndot(A);dot(B);dot(C);dot(D);dot(E);dot(G);\nlabel(\"$A$\",A,S);label(\"$B$\",B,N);label(\"$C$\",C,S);label(\"$D$\",D,S);label(\"$E$\",E,NW);label(\"$G$\",G,NE);\ndraw(A--B--C--cycle);\ndraw(B--D);\ndraw(E--C);\ndraw(rightanglemark(C,G,D,3));[/asy]$ \\textbf{(A)}\\ 24\\qquad \\textbf{(B)}\\ 32\\qquad \\textbf{(C)}\\ 48\\qquad \\textbf{(D)}\\ 64\\qquad \\textbf{(E)}\\ 96$", "Solution_1": "[hide=\"Click for solution\"]\nWe have $ CG\\equal{}\\frac{2}{3}CE\\equal{}8$ and $ [BCD]\\equal{}\\frac{1}{2}(BD)(CG)\\equal{}\\frac{1}{2} \\times 8 \\times 8\\equal{}32$. Because $ BD$ is a median, $ [ABD]\\equal{}[DBC]$, so the area of the triangle is $ 64$, or $ \\boxed{\\textbf{(D)}}$.\n[/hide]" } { "Tag": [ "geometry", "rectangle" ], "Problem": "Here's a very delightful problem (from MOP) that I enjoyed.\r\nLet $ABCD$ be a rectangle with an interior point $P$. If $\\angle APD+\\angle BPC=180^{\\circ}$, what is the sum of $\\angle DAP$ and $\\angle BCP$?\r\n\r\n[hide=\"Solution\"]\nMove $\\triangle DPC$ on top of $AB$ and call the new triangle $AQB$. $QP$, $AD$, and $BC$ are parallel. It is clear that $AQBP$ is cyclic due to the fact that $\\angle AQB+\\angle APB=180^{\\circ}$. $\\angle DAP=\\angle APQ$ (PLT-Z) and $\\angle BCP=\\angle PQB$. $AB$ and $PQ$ intersect at $O$ at $90^{\\circ}$; hence, $\\angle POB=90^{\\circ}$. $\\angle DAP+\\angle BCP=\\angle APQ+\\angle PQB$ is equal to half the sum of the arcs of $AQ$ and $PB$. $\\angle POB=90^{\\circ}$ is also equal to half the sum of the arcs of $AQ$ and $PB$, thus $\\angle DAP+\\angle BCP=\\angle APQ+\\angle PQB=90^{\\circ}$.\n[/hide]\r\nMasoud Zargar", "Solution_1": "[hide]\nYou can also note that since \n$a + b = c + d = a + c = b + d = 180$, letting $a = APB, b = BPC, c = CPD, d = DPA$, we see $a = b = c = d = 90$, and the result follows.\n[/hide]" } { "Tag": [ "algebra", "polynomial", "induction", "LaTeX" ], "Problem": "Prove that any polynomial with real coefficients that takes only nonnegative values can be written as the sum of the squares of two polynomials.", "Solution_1": "[hide]Let the polynomial be $ Z(x)$. Clearly we can assume that $ Z(x)$ and all of its factors are monic. The given conditions imply that $ Z(x) \\equal{} P(x)^2Q(x)$ where $ Q(x)$ has no real roots but all real coefficients. Note that every real root of $ Z(x)$ must be a double root otherwise $ Z(x)$ would change signs. Because $ Q(x)$ has real coefficients, every root must have a corresponding conjugate root. Therefore $ Q(x) \\equal{} \\left(\\prod^n_{i \\equal{} 0}(x \\minus{} \\alpha_i)\\right)\\left(\\prod^n_{i \\equal{} 0}(x \\minus{} \\bar{\\alpha}_i)\\right)$. Through induction we can prove that if $ \\left(\\prod^n_{i \\equal{} 0}(x \\minus{} \\alpha_i)\\right) \\equal{} a(x) \\plus{} ib(x)$, then $ \\left(\\prod^n_{i \\equal{} 0}(x \\minus{} \\bar{\\alpha}_i)\\right) \\equal{} a(x) \\minus{} ib(x)$ where $ a(x)$ and $ b(x)$ are polynomials with real coefficients. I would actually show this but I can't seem to format the latex nicely. Do it yourself, it's not that hard. Once we have this, we have $ Q(x) \\equal{} a(x)^2 \\plus{} b(x)^2$ so $ Z(x) \\equal{} P(x)^2Q(x) \\equal{} (P(x)a(x))^2 \\plus{} (P(x)b(x))^2$.\r\n\r\nEDIT: I should at least acknowledge the case where $ Z(x)$ is a positive constant for which the statement is true. Also, my first edit was pointless. [\\hide]" } { "Tag": [ "modular arithmetic", "ARML" ], "Problem": "S28\r\n\r\nA [i]lucky number[/i] is a positive integer which has a value 19 times the sum of its digits. How many different [i]lucky numbers[/i] are there?", "Solution_1": "Not many. I'm not sure (maybe I made mistake in my calculations) but it's just 399, 266 and 133.\r\n\r\nEDIT: So It's little bit more. My solution was same as DiscreetFourierTransform's\r\nbut I wrote the last two digits in wrong order and didn't release one thing :blush: .", "Solution_2": "The answer :maybe: is 11 lucky numbers (I found 190 , 171, 133 , 152 , 114 , 285 , 266 , 247 , 228 , 209 and 399 ). Am i correct?", "Solution_3": "Two digit:\r\n$ 19(a+b) = 10a+b$\r\n$ 19a+19b = 10a+b$\r\n$ 9a =-18b$\r\n\r\nmeaning either a or b is negative, which is impossible (digits themselves are never negative, only the number they represent.\r\nNo Two Digit perfect numbers.\r\n\r\nThree Digit: (a not equal to zero\r\n$ 19(a+b+c) = 100a+10b+c$\r\n$ 81a-9b-18c = 0$\r\n$ 9a-b-2c=0$\r\n$ 9a = b+2c$\r\n\r\nCases:\r\n$ b+2c \\equiv 0 \\pmod 9$\r\n\r\n$ b+2c = 9$ (9, 0) ; (7, 1) ; (5, 2) ; (3, 3) ; (1, 4) --> Five Solutions (for a = 1)\r\n----------------------\r\n$ b+2c = 18$ \r\nB must be even for $ 2c$ to divide $ 18-B$.\r\n(0, 9) ; (2, 8) ; (4, 7) ; (6, 6) ; (8, 5) --> Five Solutions. (for a = 2)\r\n---------------------\r\n$ b+2c$ = 27\r\nB must be odd for $ 2c$ to divide $ 27-B$.\r\n2c is maximally 18, so B must be minimally 9. However, this limits us to once case, because B is also maximally 9.\r\n(9, 9) --> One Solution (for a = 3)\r\n---------------------\r\nWe can't have any more solutions:\r\nBecause $ 2c$ is at most 18, and B is at most 9, the highest digital sum they can produce is 27. Therefore, there are\r\n\r\n5 + 5 + 1 solutions = 11.\r\n\r\nEDIT: Oops, forgot to consider four digits:\r\n$ 19(a+b+c+d) = 1000a+100b+10c+d$\r\n$ 981a+81b-9c-18d = 0.$\r\n$ 981a+81b = 9c+18d$\r\n$ 109a+9b = c+2d$\r\nAnd it's clear that for maximal c and d, we have $ 109a+9b = 27$ which is clearly unreachable: Therefore there are no four digit lucky numbers.", "Solution_4": "shoot this is an ARML 2007 Team Question T2", "Solution_5": "Looks like the ARML is a good source of problems.. also there was a problem on the American MC on how many lucky numbers were 13 times the sum of their digits i think" } { "Tag": [ "MATHCOUNTS", "trigonometry", "AMC", "AIME", "calculus", "integration", "geometry" ], "Problem": "I know trig/log is a must, but anyone have a list of useful formulas/theorems for the olympiad? (especially ones involving circles)", "Solution_1": "i dont think this test is about how many theorems you know. I'll bet all the formulas you learned for mathcounts would suffice.\r\n\r\nBut while we're on this topic, how long is the test and how many points for each correct and unanswered question?", "Solution_2": "Ugh, now that you brought MAML up again, I have found out that our school may not be doing it since we don't have classes next Tuesday (Its Parents' Weekend, we had Parent's Day yesterday).", "Solution_3": "[quote=\"cats...\"]i dont think this test is about how many theorems you know. I'll bet all the formulas you learned for mathcounts would suffice.\n\nBut while we're on this topic, how long is the test and how many points for each correct and unanswered question?[/quote]\r\nDetails [url=http://www.artofproblemsolving.com/wforum/viewtopic.php?t=1773]here[/url]\r\n\r\nUm, sorry to contradict you but judging from the fact that most problems need some kind of formula whether it be n(n+1)/2 or power of point...I really need to know this stuff [b]fast[/b] you guys, the comp is in three days and I must make it, or else! (and I don't even want to think about the \"or else\"... :( )", "Solution_4": "You can't learn to use things quickly in three days. It's a skill acquired with time, as a result of doing many different problems. Anyhow, calm down.", "Solution_5": "Mmm, maybe you need to know about the \"or else\"...\r\n\r\nSorry to act jitterish, but the \"or else\" is scaring the heck out of me...\r\n\r\n\"or else\": (military) camp", "Solution_6": "sorry, but i really agree with joel about not trying to cram new stuff in. I don't think it'd be worth it.\r\n\r\nI really don't think it's about how many theorems/formulas you know, anyways. Like I said, I wouldnt be surprised if you learned all the necessary formulas in mathcounts. You cited the sum of an arithmetic progression and the power of a point theorem... that is stuff (i think) that you probablly already learned in your preparation for mathcounts. The theorems/formulas on this test don't get much more advanced than that. It's more about how well you can think, so you'd be better off doing past tests.", "Solution_7": "*sigh* trig formulas and circle stuff (as in problem 19 for the last year's ([b]2002[/b]-2003) test) we don't learn for MATHCOUNTS...I'm doomed...", "Solution_8": "not to be mean, but if you don't do well its not because you didnt know enough math, but because you were too nervous. Seriously, calm down... if you're too nervous, regardless if you know how to do every question on there, you won't be able to do too well. Im sure you have all the tools necessary to do well. And you will do well if you calm down.\r\n\r\nFor trig identities, you really only need to know the sum and difference formulas: sin(x+y)=sinx*cosy+siny*cosx. cos(x+y) = cosx*cosy-sinx*siny. tan(x+y) = (tanx+tany)/(1-tanx*tany) .\r\n\r\nAnd this might actually matter for 1 question. Maybe 2...\r\nBut that really it for trig...\r\n\r\nThen as far as that circle question goes, all you needed to know is that when 2 chords intersect in a circle, ... im not sure how to articulate this exactly, but the products of the segments up until their intersection points are equal to each other. Going by the diagram given in the question, this means that AE*EB=CG*GD. The rest of that question can be solved just by knowing that, and then by thinking your way to the solution.", "Solution_9": "Actually, power of a point comes from knowing other things that you should know. It's just similar triangles. You have to know that the inscribed angle of any arc of a circle is constant. And that gives you similar triangles (when you draw two additional chords), and then you're done. Besides which, I think that was fair game for Mathcounts. And I agree that you may not have been expected to know trig yet, but it is unlikely to feature heavily at all.", "Solution_10": "I must say though, that not knowing the math needed to solve some problems is a major hit on one's confidence. Sometimes, when I try to solve some problems on national olympiads, I give up (too easily) just because I think I do not know the math needed to solve a problem, even though I may in fact know enough. On the other hand, when I do AMC12's, I am completely confident that I know all the math needed to solve the problems on the exams, and so I do well.", "Solution_11": "that happens to me, but for a slighly different reason. i quickly get discouraged and if i dont solve a problem in a matter of minutes, i give up and say its too hard for me. when i practice aimes or more recently olmypiads, i tend to have a larger score when i take it under the official test conditions than at home, which is funny. but the key thing is not too get discouraged, at home , or at the contest. tare, probably you already know enough math to solve a majority of the problems. if you cram too much, you might actually become less confident in your knowledge than you are now. and if you go in too nervous (like me before the amc 12 last year), you will make a myriad of stupid mistakes. go in confidently (but not overly so) and do your best on each problem, knowing that you can solve all of them if you keep trying.", "Solution_12": "Okay, thanks everyone, guess as long as i think I'm doomed I'm doomed and if I think not then I'm not :)", "Solution_13": "I dont know if you guys are allowed to keep your tests or not, but i dont think my teacher lets us. Can you guys keep track of your answers somehow (write it on a piece of scrap) so that we can have some idea of our scores afterwards?", "Solution_14": "so, what did you guys think?", "Solution_15": "It wasn't too bad, although I made a very stupid mistake on one of them. I answered 26, at least 1 of which was incorrect.", "Solution_16": "Hmm. It was alright. I got 22 right, 2 wrong, and omitted the rest. I don't know this officially of course, but I compared with all the other people. The last couple questions were, for me, pretty impossible (the hexagon with the squares thing? Distance? GRR!) \r\nWhat was everyone else's score and experience with it?\r\nAlso, when do they announce the Level 2 qualifiers?", "Solution_17": "what was the one with ax^2+bx+c=0 and b^4-a^2*c^2=65", "Solution_18": "I got E - Can't determine anything about it, but I'm not sure if I got that right. A lot of people got different answers.", "Solution_19": "Hold on, everyone is done taking the test right? If they aren't, then I'll have to take my answer down.", "Solution_20": "[quote=\"cats...\"]what was the one with ax^2+bx+c=0 and b^4-a^2*c^2=65[/quote] The key to this problem is that a, b, and c are integers.\r\n\r\nWe have b^4-a^2*c^2=65.\r\nSo (b^2-4ac)(b^2+4ac)=65. So one of these factors is 13 and the other is 5 or 1 is -5 and the other is -13. Suppose that both are negative. Than b^2-4ac<0 and b^2+4ac<0, so adding yields 2b^2<0, a contradiction. So b^2-4ac is either 5 or 13, and the solutions are therefore irrational.", "Solution_21": "NO!!! It could be 65 and 1!!!!!", "Solution_22": "NO!!! It could be 65 and 1!!!!!", "Solution_23": "oops. my mistake, you're right. i'm wrong. :(\r\n\r\nEDIT: what integral a, b, and c values produce 65 and 1? I can't think of any.\r\nEDIT 2: now i'm pretty sure there aren't any. if b^2-4ac=65 and b^2+4ac=1 (or the other way around), 2b^2=66 and b=sqrt 33. \r\nSo my logic was bad, but i think i'm right now.", "Solution_24": "Now it's officially over. I got 18 and guesstimated 1...should've guessed more :( \r\nThe hexagon thing I guessed since it must be smaller than the side of the square+side of the hexagon+diagonal of the square=x+x+x:rt2:=2x+x:rt2: (I don't remeber the side but the side is x), so two of the answers didn't make sense. Thus I guesstimated it was the second largest answer...am I right?\r\n\r\nP.S. You do need to memorize all the formulas...the half-angle theorem was a must on this test and such.", "Solution_25": "Is it perhaps possible to get some of these questions posted, as opposed to commontary on them? Are they online somewhere?", "Solution_26": "they're not online and i dont have them. But i think this is the question, though i dont remember the answer choices.\r\n\r\nWe have a regular hexagon of side 4. The midpoints of each side are connected to form a smaller hexagon, and this process is repeated. Find the sum of all areas of these hexagons. \r\n\r\nTare, if you're referring to the trig one where you had to simplify, that was just 1 question. Also, that circle question w/ the chords was essentially the same as the one in 2002.\r\n\r\nEdit: and the question above that i posted was ax^2+bx+c=0 and b^4-a^2*c^2=65. The question asked what can you determine about the roots: i dont remember all the choices, but some of them are a)they're imaginary b) they're irrational c)they're the same d)the roots can't be determined.\r\n\r\ni remember some more now, so i'll post them. z_1=1-ai, z_2=1+ai, z_3 = 2+ai. Given that z_1*z_2*z_3 is a real number, find (z_1*z_2*z_3)^10.\r\n\r\nHow many ways can you make 2 dollars out of nickels, quarters, and dimes, given that the absolute difference between the number of quarters and nickels is 2.\r\n\r\nlog_a 21 = x, log_a 49 = y , find an expression for log_a 27\r\n\r\nIf you put $500 in your bank account today, and every 3 days after today, you take out $12, and every 17th day after today you take out $25, and every 51st day you put in $100, on what day will you have a negative amt. of money.", "Solution_27": "These sound like difficult questions. How much time do you have to solve them?", "Solution_28": "90 minutes, 30 questions. See my post about the official rules somewhere else...you know isn't it annoying when you type up this really detailed thing and then you can't find/remember where you posted it? It's so much \"real life\"ish it's giving me the left-over creeps from Halloween :lol:" } { "Tag": [ "LaTeX", "logarithms" ], "Problem": "Can anyone give me a [u]hint[/u] on this problem. \r\n\r\nThe formula $N=8\\times10^{8}\\times x^{-3/2}$ gives, for a certain group, the number of individuals whose income exceeds $x$ dollars. What is the smallest possible value of the lowest income of the wealthiest 800 individuals? (AHSME 1960) \r\n\r\n thanks\r\n\r\n[size=75][color=red]b-flat: Check your $\\LaTeX$. I fixed it for you. Use \\times for *, and when your exponents have more than one character, surround them all by the braces.[/color][/size]", "Solution_1": "[hide=\"Hint\"] Well, just set $N=800$ and solve. Do you need any hints beyond that? [/hide]", "Solution_2": "[quote=\"myc\"]Can anyone give me a [u]hint[/u] on this problem. \n\nThe formula $N=8\\times10^{8}\\times x^{-3/2}$ gives, for a certain group, the number of individuals whose income exceeds $x$ dollars. What is the smallest possible value of the lowest income of the wealthiest 800 individuals? (AHSME 1960) \n\n thanks\n\n[size=75][color=red]b-flat: Check your $\\LaTeX$. I fixed it for you. Use \\times for *, and when your exponents have more than one character, surround them all by the braces.[/color][/size][/quote]\r\n\r\nI set n=800...", "Solution_3": "[quote=\"ch1n353ch3s54a1l\"][quote=\"myc\"]Can anyone give me a [u]hint[/u] on this problem. \n\nThe formula $N=8\\times10^{8}\\times x^{-3/2}$ gives, for a certain group, the number of individuals whose income exceeds $x$ dollars. What is the smallest possible value of the lowest income of the wealthiest 800 individuals? (AHSME 1960) \n\n thanks\n\n[size=75][color=red]b-flat: Check your $\\LaTeX$. I fixed it for you. Use \\times for *, and when your exponents have more than one character, surround them all by the braces.[/color][/size][/quote]\n\nI set n=800 to get [hide]$10000$[/hide][/quote]\r\n\r\nHe asked for a hint. Please only give him a hint, not the answer. Thanks. :)", "Solution_4": "Sorry. That's why I hid the answer, in case he might eventually want it...", "Solution_5": "o yea it works out, subsituting 800 for N , how come I didn't think of that. Thanks :lol:", "Solution_6": "[quote=\"myc\"]Can anyone give me a [u]hint[/u] on this problem. \n\nThe formula $N=8\\times10^{8}\\times x^{-3/2}$ gives, for a certain group, the number of individuals whose income exceeds $x$ dollars. What is the smallest possible value of the lowest income of the wealthiest 800 individuals? (AHSME 1960) \n\n thanks\n\n[size=75][color=red]b-flat: Check your $\\text{\\LaTeX}$. I fixed it for you. Use \\times for *, and when your exponents have more than one character, surround them all by the braces.[/color][/size][/quote]\r\n[hide] \\begin{eqnarray*}N & = & 8\\times10^{8}\\times x^{-3/2}\\\\ 800 & = & 8\\times10^{8}\\times x^{-3/2}\\\\ x^{3/2}& = & 10^{6}\\\\ \\log_{10}{x}& = & \\frac{6}{3/2}\\\\ \\log_{10}{x}& = & 4 \\\\ x & = & 10000 \\end{eqnarray*} [/hide]" } { "Tag": [ "limit", "calculus", "integration", "real analysis", "calculus computations" ], "Problem": "The question is as follows (pardon my formatting):\r\n\r\nEvaluate:\r\n\r\nlim n-> infinity n(sum)i=1 [root(1+(8i/n)) * 8/n]", "Solution_1": "Perhaps this?\r\n\r\n$\\lim_{n\\to\\infty}\\sum_{i=1}^n\\sqrt{1+\\frac{8i}n}\\cdot\\frac8n$\r\n\r\n[hide=\"Two word hint.\"]Riemann sum.[/hide]", "Solution_2": "Yup, that's the question.... and I figured it was a Riemann Sum, but I wasn't sure how to \"evaluate\" it exactly... Do I simply express it as an integral? I'm just not sure where to go with it... thanks.", "Solution_3": "You're looking for this to be of the form $\\sum f(x_i)\\Delta x.$\r\n\r\nIn looking at this, you see the square root, which looks like a candidate for $f(x),$ and you see the extra factor of $\\frac8n.$\r\n\r\nSo let's start by trying to pin down $\\Delta x.$ Try the possibility $\\Delta x=\\frac8n.$\r\n\r\nThen as $i$ runs from $1$ to $n,$ the net increase in $x$ will be $n\\Delta x=8.$ We're looking for $x$ to increase by $8.$\r\n\r\nInside the square root, you see $\\frac{8i}n.$ Try letting that be $x.$ As $i$ ranges from $1$, to $n,$ this would range from close to $0$ to $8.$ That looks like our interval.\r\n\r\nPutting the pieces together we get that this is a Riemann sum for\r\n\r\n$\\int_0^8\\sqrt{1+x}\\,dx.$\r\n\r\nThis answer is by no means unique, and by no means the only reasonable answer. We could just as reasonably have written\r\n\r\n$\\int_1^9\\sqrt{x}\\,dx$ or $\\int_0^1\\sqrt{1+8x}\\cdot8\\,dx$\r\n\r\nOf course, these various forms of the integral are interconnected by substitutions and represent the same number.", "Solution_4": "Beautiful...thanks so much for breaking it down so well!" } { "Tag": [], "Problem": "Consideram un sir $\\ s_{n}=1,9,7,7,...$ in care numarul urmator este suma celorlalte patru din fata luate modulo $\\ 10$.\r\n$\\ (a)$Poate aparea secventa $\\ 1977$ dupa un numar finit de miscari?\r\n$\\ (b)$Daca sirul incepe cu patru numere pare, demonstrati ca sirul e periodic si aflati perioada sa.\r\n$\\ (c)$Daca sirul incepe cu patru numere impare, demonstrati ca sirul e periodic si aflati perioada sa.\r\n\r\n[i]OMUS[/i]", "Solution_1": "Fie $\\left\\{ s_{n}\\right\\}_{n \\geq 1}$ un \u015fir astfel \u00eenc\u00e2t $s_{n}\\in \\overline{0,9}, \\forall n \\geq 1$ \u015fi $s_{n+4}\\equiv s_{n+3}+s_{n+2}+s_{n+1}+s_{n}\\left( \\bmod \\, 10 \\right), \\, \\forall n \\geq 1$.\r\n\r\nFie $a_{n}= \\left( s_{n}, s_{n+1}, s_{n+2}, s_{n+3}\\right)$. Exista un numar finit de posibilitati pentru valorile lui $a_{n}$, $10^{4}$. Dar cum sirul $\\left\\{ a_{n}\\right\\}_{n \\geq 1}$ e infinit, exista $u0$ then $ \\frac{\\sqrt{a}\\plus{}\\sqrt{b}\\plus{}\\sqrt{c}}{2\\sqrt{abc}}\\ge\\frac{ab}{a^3\\plus{}b^3}\\plus{}\\frac{bc}{b^3\\plus{}c^3}\\plus{}\\frac{ca}{c^3\\plus{}a^3}$", "Solution_1": "[quote=\"moldovan\"]If $ a,b,c > 0$ then $ \\frac {\\sqrt {a} \\plus{} \\sqrt {b} \\plus{} \\sqrt {c}}{2\\sqrt {abc}}\\ge\\frac {ab}{a^3 \\plus{} b^3} \\plus{} \\frac {bc}{b^3 \\plus{} c^3} \\plus{} \\frac {ca}{c^3 \\plus{} a^3}$[/quote]\r\n\r\nobvious by AM-GM: $ \\sum\\frac {bc}{b^3 \\plus{} c^3} \\le\\sum\\frac{bc}{2bc\\sqrt {bc} }\\equal{}\\frac{\\sqrt a}{2\\sqrt {abc}}$ :)" } { "Tag": [ "geometry", "3D geometry", "combinatorics unsolved", "combinatorics" ], "Problem": "Eight singers participate in an art festival where $ m$ songs are performed. Each song is performed by $ 4$ singers, and each pair of singers performs together in the same number of songs. Find the smallest $ m$ for which this is possible.", "Solution_1": "[quote=\"moldovan\"]Eight singers participate in an art festival where $ m$ songs are performed. Each song is performed by $ 4$ singers, and each pair of singers performs together in the same number of songs. Find the smallest $ m$ for which this is possible.[/quote]\r\n\r\n[hide=\"here is my solution\"]Each song brings together $ 6$ pairs of singers so the $ m$ songs together bring together $ 6m$ (not necessarily distinct) pairs of singers. There are a total of $ \\binom{8}{2} \\equal{} 28$ distinct pairs of singers. So given that each distinct pair collaborate an equal number of times it follows that $ 28|6m$ or $ 14|3m$ or $ 14|m$. Hence in particular $ m\\ge 14$ and if we can find a specific example for which $ m \\equal{} 14$ we will be done. To do this, imagine the $ 8$ singers to correspond to the $ 8$ vertices of a cube. The $ 6$ faces of the cube give us $ 6$ sets of $ 4$ singers. The $ 6$ face diagonals give us another $ 6$ sets of $ 4$ singers. To get the remaining $ 2$ sets of $ 4$ singers, color the vertices alternatively black and white, and pick the $ 4$ black vertices as one set and the $ 4$ white vertices as the other set. This gives us a total of $ 14$ sets of $ 4$ vertices among the $ 8$ vertices and it is fairly easy to show (I won't provide the details) that each pair of vertices is represented together in exactly $ 3$ such sets.\n\nSo the smallest possible $ m$ meeting the problem conditions is $ m \\equal{} 14$.[/hide]" } { "Tag": [ "probability", "search", "geometry", "MATHCOUNTS", "number theory" ], "Problem": "I am in 8th grade and this is my 1st year participating in Math Counts (long story - forget about it..)\r\n\r\nI've always been interested and pretty good at Math (aced my classes, blah blah)\r\n\r\nHowever, Math Counts is at a whole new level... and I want to do good this year since it's my last chance. (I got selected in my school team)\r\n \r\nI ordered AoPS volumes 1 and 2. \r\nI read most of one (spent a long time especially on number theory, how to count, and probability)\r\n\r\nTwo seems really tough- kinda beyond me, and my Chapter competition is on February 24th. (I am in Cincinnati)\r\nI have a 4 day weekend now- I can spend 6-7 hours each day studying and also spend 1-2 hrs on weekdays until next Saturday.\r\n\r\nMy team is ok.. not amazing or anything, so I don;t think I can get into State by team, I'll need to do it individually.\r\n\r\nQuestion 1: How many people go to the State in my Chapter? (Cincinnati)\r\n\r\nI've been doing Sprints and targets at the chapter level, from 2001-2006, besides the ones that my teacher at school gives me... I seem to only get between 22-25 on sprints and 6-7 on targets.\r\n\r\nQuestion 2: I know this is kind of hard to answer but what would be the average score I'd need to get on my sprint and target to go to state?\r\n\r\nQuestion 3: Also, what can I do to prepare for Math Counts. As said earlier, my chapter competition is next weekend (not this one) and I have a 4 day weekend in which I can do a solid 6-7 hrs of study everyday.\r\n\r\nPlease help me out. My main goal is to first get to state, and get a decent score at the state level, (not necessarily go to Nationals)", "Solution_1": "1. it depends on the size of the chapter, from 4 individuals to top 12. i'm guessing its top 12 because it's cincinatti.\r\n\r\n2.probably about a total of 35 at least\r\n\r\n3. problems usually get harder and harder as years pass, so it's a good idea to do state problems to practice for chapter", "Solution_2": "1. it really depends. you might want to search up on the internet for past results of cincy chapter. my chapter is top 4 teams and 5 ind. while a chapter as big as jhredsox's allow 7 teams and 12 ind., i think.\r\n\r\n2. this also depends on how tough your chapter is. for example, in my chapter last year, a 44 was needed to advance individually. :!: other chapters, a 31 could let you advance...\r\n\r\n3. figure out you weakness area(s) and do more problems in that specific area. don't cram the last few days before chapters, instead let you mind cool off...", "Solution_3": "[quote=\".:Logic:.\"]I am in 8th grade and this is my 1st year participating in Math Counts (long story - forget about it..)\n...\nPlease help me out. My main goal is to first get to state, and get a decent score at the state level, (not necessarily go to Nationals)[/quote]\r\n\r\nOne Word: Relax. ( I've been told that for another test coming up)\r\n\r\nand by do good, don't you mean you want to do well?\r\n\r\nI'm sure doing tests would help; and rest? bleh this doesn't sound like me. \r\n\r\nand at least top 4 get in, maybe more through district.", "Solution_4": "Ask Gyan for questions regarding Cincinatti. He's like the only active poster I know from there. Otherwise, post this in the Ohio local forum and see what they have to say.", "Solution_5": "I'm from Cleveland, but I can probably say that a [b]40[/b] would get you to States.\r\n\r\nFor the chapter competition, I would prepare with past chapter competitions under a 30 minute time limit for Sprint (not 40) and a 10 minute time limit for all 8 target round questions (pretty fast).\r\n\r\nGood Luck and Hopefully I'll see you at states. :)", "Solution_6": "remember it depneds on the year, but just try your best\r\n\r\nmy chapter's big. we advance 8 teams and 20-30 individuals\r\n\r\nmaking for 50-60 that go to states \r\n\r\njorian", "Solution_7": "Okay.. I can confidently get atleast a 25 on old chapter sprints. (taking full 40 mins)\r\n\r\nHowever on state sprints, I get 16-20 :( (taking full 40 mins)\r\n\r\n:(\r\n\r\nI don't have much of a chance of making the state do I?? :|\r\n\r\nHmm... Will the chapter sprint this year, be as hard as state sprints from previous years?\r\n\r\nI can guarantee myself ATLEAST 5 right on the target. That means I'll need a 27-28 to get a good chance at the state... :maybe: \r\n\r\n :stretcher:", "Solution_8": "[quote=\".:Logic:.\"]\nHmm... Will the chapter sprint this year, be as hard as state sprints from previous years?[/quote]\r\n\r\nThere's really no way to predict that. But remember, if it's hard for you, it's hard for other people too, and conversely, if it's easy for you, it's easy for other people too. What matters is how good you are relative to the other people in your chapter, not the number you get for a score.", "Solution_9": "[quote=\"mathmom\"][quote=\".:Logic:.\"]\nHmm... Will the chapter sprint this year, be as hard as state sprints from previous years?[/quote]\n\nThere's really no way to predict that. But remember, if it's hard for you, it's hard for other people too, and conversely, if it's easy for you, it's easy for other people too. What matters is how good you are relative to the other people in your chapter, not the number you get for a score.[/quote]\r\nAlso, the more prone you are to careless errors the harder you want the test to be.\r\nI'm less prone than most people, so a nationals similar to last years, this year would benefit me.", "Solution_10": "[quote=\".:Logic:.\"]Hmm... Will the chapter sprint this year, be as hard as state sprints from previous years? [/quote]\r\n\r\nDo not discuss the difficulty of the 2007 Chapter Competition yet.", "Solution_11": "[quote=\"vishalarul\"][quote=\".:Logic:.\"]Hmm... Will the chapter sprint this year, be as hard as state sprints from previous years? [/quote]\n\nDo not discuss the difficulty of the 2007 Chapter Competition yet.[/quote]\r\nHe is asking, will it be?", "Solution_12": "Can we say the cutoff this year for our chapter?", "Solution_13": "[quote=\"#H34N1\"]Can we say the cutoff this year for our chapter?[/quote]\r\n\r\nNo, that would give quite a lot of information about how hard the Chapter contest is. Not giving that kind of information is the point of not talking about scores before everyone is done.", "Solution_14": "Okay.\r\n", "Solution_15": "What I reccommend in the way of studying is to do 5 problems, take a 30 minute break, and repeat. If you cram too hard, it won't help. You'll get sick and tired of it and lose your motivation. I also reccomend you get as much sleep as possible.\r\n\r\nIt worked for me. :D", "Solution_16": "Hmm I don't mind doing 6-7 hours a day.... :) \r\nI still have a bit of today and tomorrow left for quality studying without school stuff..\r\n\r\n[b]Are there any specific chapters in AoPS: Volume One that I should focus on?? I found Learning to Count and Probability since there's always 2-3 of those problems in sprint, any other chapters like that?[/b]", "Solution_17": "Sorry to double post.. but could someone please just answer the above question :?:", "Solution_18": "Ok last time I'm asking.. I'm desperate :P\r\n\r\nSomeone suggest chapters from AoPS Volume 1 that will greatly help for the Chapter Round- its this Saturday :(", "Solution_19": "[quote=\".:Logic:.\"]Ok last time I'm asking.. I'm desperate :P\n\nSomeone suggest chapters from AoPS Volume 1 that will greatly help for the Chapter Round- its this Saturday :([/quote]\r\n\r\n[hide=\"Here you go\"]Sych! Keep going\nYou that desperate?\nGo Go Go\nYou really want to know?\nFine\n\nI read chapters XX, XX, XX, XX, XX, XX, XX, XX, XX, XX, XX, XX and XX. These chapters are usually what many tests in mathcounts have. I'm sure there are more but these are what I read before my competition.\n\n[color=red][size=75]Specific chapter number edited by moderator rcv[/size][/color][/hide]", "Solution_20": "[quote=\".:Logic:.\"]Are there any specific chapters in AoPS: Volume One that I should focus on?? I found Learning to Count and Probability since there's always 2-3 of those problems in sprint, any other chapters like that?[/quote]\r\nLogic: You should not ask such a specific question while the Chapter or State rounds are in progress. While your question might be innocent, it might also result in discovery of some of the content areas of this year's Chapter test. That is absolutely forbidden.\r\n\r\nStudy all of AoPS Volume I. Try all old contests you can get your hands on. Try all of this year's and previous year's handbook problems you can get your hands on.\r\n\r\nYou certainly should practice. Practice will increase your speed, your confidence, and your score. But don't expect to become a MATHCOUNTS Champion with just one week of cramming.\r\n\r\nI am locking this thread, because of the high risk that further responses will reveal areas of emphasis on the current Chapter competition." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Please give me a proof of Muirhead! Or tell me where I can find it in this forum. Thanks! :wink:", "Solution_1": "Already posted. Click [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15896]here[/url] and scroll down.", "Solution_2": "Except of it, a proof of very similar theorem which can be transformed into proof of the exact Muirhead can be found [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=17288]here[/url]. I heared that it also can be proved by [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14975]Karamata inequality[/url]. A very short proof is in [url=http://www.artofproblemsolving.com/Resources/Papers/KedlayaInequalities.pdf]this pdf by Dr. Kiran Kedlaya[/url] at the beginning of the 13-th page (12-th numbered page). \r\nI hope it will help." } { "Tag": [ "geometry", "3D geometry", "sphere", "circumcircle", "tetrahedron", "geometric transformation", "reflection" ], "Problem": "Given any five distinct points on the surface of a sphere, show that we can find a closed hemisphere which contains at least 4 of them.", "Solution_1": "Not sure about this, but here goes nothing.\r\n\r\n[hide]\n[b]Case 1:[/b] Let us first take any three points. If the circumcircle of these points passes through the center, then adding the fourth one defines a hemisphere, and we are done.\n\n[b]Case 2:[/b] Otherwise, we add a fourth point. If the tetrahedron formed by the four points does not pass through the center of the sphere, then we clearly have such a hemisphere (cut the sphere with a plane passing through the center that is parallel to the face closest to the center).\n\n[b]Case 3:[/b] Otherwise, we add a fifth point and repeat the above procedure with each of the four faces of the tetrahedron we have.\n\n[b]Aside:[/b] [i]For points $ A$, $ B$, and $ C$ on a sphere, their respective reflections across the center $ A'$, $ B'$ and $ C'$, and a point $ D$, the tetrahedron $ ABCD$ contains the center iff. $ D \\in$ spherical triangle $ A'B'C'$.[/i]\n\nThe four spherical triangles formed by the points are non-overlapping, thus the four spherical triangles formed by the projections of the points are also non-overlapping. Wherever we place the fifth point on the sphere, there is at least one such region that it is not in (it is either in none of them, or in only one of them).\n\nHence, upon adding the fifth point (and creating 4 new tetrahedrons), there is at least one tetrahedron that does not pass through the center, which implies a group of four points with a common hemisphere.\n[/hide]" } { "Tag": [ "trigonometry", "algebra", "polynomial", "sum of roots" ], "Problem": "Prove that sin(2pi/7) + sin(4pi/7) + sin(8pi/7)= (root of 7)/2\r\nPlease help me out. \r\nThanks.", "Solution_1": "Consider $ \\sin 7x\\equal{}0$, so that $ 7x\\equal{}n\\pi$ where $ n$ is an integer.\r\n\r\n$ \\sin 7x \\equal{} Im (c\\plus{}is)^{7}$, where $ c\\equal{}\\cos x, s\\equal{} \\sin x$.\r\n\r\nExpand by the binomial and write $ c^{2}\\equal{}1\\minus{}s^{2}$, to get $ s(64s^{6}\\minus{}112s^{4}\\plus{}56s^{2} \\minus{}7)\\equal{}0$\r\n\r\nSo the sextic has roots $ \\pm sin \\frac{n\\pi}{7}, n\\equal{} \\pm1, \\pm2, \\pm3$\r\n\r\nLet $ s^{2} \\equal{} t$, so $ \\sin^{2}(\\frac{n\\pi}{7}), n\\equal{}1, 2, 3$ are the roots of $ 64t^{3} \\minus{}112t^{2} \\plus{}56t \\minus{}7\\equal{}0$.\r\n\r\nThen sum of roots $ \\equal{}\\frac{112}{64} \\equal{} \\frac{7}{4}$.\r\n\r\nThen square of the sum of the roots of the sextic = sum of roots squared + twice cross products gives required sum $ \\equal{}\\sqrt{\\frac{7}{4}} \\equal{} \\frac{\\sqrt{7}}{2}$.\r\n\r\nSorry, a bit rushed at the end.", "Solution_2": "[quote=\"kushal.sharma09\"]Prove that sin(2pi/7) + sin(4pi/7) + sin(8pi/7)= (root of 7)/2\nPlease help me out. \nThanks.[/quote]\r\n\r\nIs this $ \\sin \\frac{2\\pi}{7}+\\sin\\frac{4\\pi}{7}+\\sin\\frac{8\\pi}{7}=\\frac{\\sqrt_7}{2}$ what you mean ?" } { "Tag": [ "modular arithmetic" ], "Problem": "Provez que 8/ (13^23)*(27^41)+1. :lol:", "Solution_1": "[quote=\"stancioiu sorin\"]Pro[color=red]u[/color]vez que $8|13^{23} \\cdot 27^{41} + 1$. :lol:[/quote]\r\n\r\n:)\r\n\r\nEDIT: Et une r\u00e9ponse rapide :\r\n$13^{23} \\equiv 5 \\pmod 8$ (puisque congru \u00e0 5 mod 8 si puissance impaire et \u00e0 1 mod 8 si puissance paire)\r\nEnsuite : $27^{41}=3^{123} \\equiv 3 \\pmod 8$ (car congru \u00e0 3 mod 8 si puissance impaire et \u00e0 1 mod 8 si puissance paire)\r\nDonc, le produit est congru \u00e0 15 mod 8, d'o\u00f9, en ajoutant 1, \u00e0 16 mod 8, soit 0 mod 8, ce qui conclut. :)" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "the sequence (x_n) is defined by x_0 =4, x_1 = 0, x_2 = 2c, x_3= 3b, and\r\n x_{n+4} = a*x_{n} + b*x_{n+1} + c*x_{n+2} ; where a , b and c are integers with b odd. prove that p^m|x_{p^m} for all primes p and positive integer m.", "Solution_1": "http://www.mathlinks.ro/viewtopic.php?t=5452" } { "Tag": [], "Problem": "I noticed one day that \r\n\r\n$\\frac{3}{4}\\cdot\\frac{8}{9}\\cdot\\frac{15}{16}\\cdot..\\cdot\\frac{k^{2}-1}{k^{2}}\\cdot....=\\frac{1}{2}$. \r\n\r\nI can't find a very elegant proof for this. Does anyone know a good proof?", "Solution_1": "The numerator is $\\prod_{k=2}^{n}(k^{2}-1)=\\prod_{k=2}^{n}(k-1)(k+1)=\\prod_{k=2}^{n}(k-1)\\prod_{k=2}^{n}(k+1)=(n-1)!\\frac{(n+1)!}{2}$, and the denominator is $\\prod_{k=2}^{n}k^{2}=(n!)^{2}$. Hence\r\n\r\n$P(n)={1\\over 2}\\frac{(n-1)!(n+1)!}{(n!)^{2}}={1\\over 2}\\frac{(n-1)!\\cdot n!(n+1)}{(n-1)!n\\cdot n!}={1\\over 2}\\cdot\\frac{n+1}{n}$, and that tends to ${1\\over 2}$ when $n$ tends to infinity." } { "Tag": [], "Problem": "Hi everyone,\r\nSorry for the delay. Technical difficulties have forced us to temporarily use an alternate server for hosting JHMT materials. The online registration form, old problems/solutions, results from previous years, and other general information about the contest can now be found online at:\r\n\r\nhttp://webhost5.nts.jhu.edu/~liondance/jhmt/index.php\r\n\r\nThere are still glitches with the website. Please do not register until the newest registration form (\"eApplication for JHMT 2007\") is posted online (probably tonight).\r\n\r\nOnce again, the tournament will be held on November 11, 2006. Thanks for your patience.\r\n\r\nRegards,\r\nJHU Math Club", "Solution_1": "And it has just come to my attention that the original server is now back in place.\r\n\r\nhttp://www.tripleintegral.org/jhmt\r\n\r\nThe finalized registration form will be up shortly. Go register :)." } { "Tag": [ "inequalities open", "inequalities" ], "Problem": "Prove or disprove that \r\n\\[ \\frac1{3a^3\\plus{}4}\\plus{}\\frac1{3b^3\\plus{}4}\\plus{}\\frac1{3c^3\\plus{}4}\\le\\frac37,\r\n\\]\r\nwhere $ a,b,$ and $ c$ are positive integers whose product is 1.", "Solution_1": "positive integers!!!\r\n $ R^ \\plus{} \\rightarrow$ wrong for $ a,b \\rightarrow 0$", "Solution_2": "[quote=\"Johan Gunardi\"]Prove or disprove that\n\\[ \\frac1{3a^3 \\plus{} 4} \\plus{} \\frac1{3b^3 \\plus{} 4} \\plus{} \\frac1{3c^3 \\plus{} 4}\\le\\frac37,\n\\]\nwhere $ a,b,$ and $ c$ are positive integers whose product is 1.[/quote]\r\n\r\nIf $ (a,b,c)\\in\\mathbb{N}^3$ (positive integers) such that $ abc\\equal{}1$ (whose product is 1) then $ a\\equal{}b\\equal{}c\\equal{}1$ and \r\n$ \\frac{1}{3a^3\\plus{}4}\\plus{}\\frac{1}{3b^3\\plus{}4}\\plus{}\\frac{1}{3c^3\\plus{}4}\\equal{}\\frac{3}{7}\\le\\frac{3}{7}.$" } { "Tag": [ "calculus", "integration", "trigonometry", "geometry" ], "Problem": "A triangle has sides of 13, 14, and 15. The altitude to one of these sides is integral. What is that altitude?", "Solution_1": "[hide]Let A be opposite the 13 side. By the cosine rule, cos A = 0.6 so sin A = 0.8. BE = c sinA = 15x0.8=12[/hide]", "Solution_2": "[hide=\"or\"]By Herons, the triangle's area is $\\sqrt{(21)(6)(7)(8)}=84$. The area is also $\\frac12bh$. The only value of $b$ that is a side length that makes $\\frac{168}{b}$ an integer is $14$, so the altitude is of length $12$.[/hide]" } { "Tag": [ "vector", "geometry", "3D geometry", "combinatorics proposed", "combinatorics" ], "Problem": "Given 30 non-zero vectors in space, show that there must be two which have the angle between them less than $ 45^\\circ$. \r\n\r\nKalva provides a kind of heavy-machinery solution. Does anyone have an elementary one?", "Solution_1": "See [url=http://www.animath.fr/tutorat/dossier_06075sol.pdf]here[/url].", "Solution_2": "I don't understand. Any English?", "Solution_3": "Oh, sorry, \r\n \r\nEach vector ~ u, join a cone with a circular base with axis of revolution is extended \r\nby ~ u, whose length of a generator is 1 which is the angle between the axis and the generator is \r\nequal to = 22, 5. The design was cut following the situation: \r\n ~ u \r\n1 \r\nOn this figure, the cone is the triangle shaded and obtained by gure tridimentionnelle \r\nby turning the design around the axis ~ u. \r\nLet the absurd assumption that every angle between vectors considered \r\nexceed 2. This meant that all e interiors cones associated with vectors are separated. \r\nThe volume of a cone is the third of the proceeds from the surface of the disk base with the length \r\nof the arrivals. It is therefore \r\n3 sin2 cos. At their thirties, cones are therefore an equal volume \r\nto: \r\n10 sin2 cos> 4, 25. \r\nBut this is impossible because the volume of the ball radius 1, which contains all that is \r\n3 \r\n4 6 4 19.\r\n\r\nP.S.\r\n\r\nI can't link to kalva. What should I do? :(", "Solution_4": "[quote=\"kunny\"]\nI can't link to kalva. What should I do? :([/quote]Go to [url=http://web.archive.org/web/20040211000554/www.kalva.demon.co.uk/index.html]web archive[/url].", "Solution_5": "Thank you, Johan Gunardi. :)" } { "Tag": [ "algorithm", "number theory", "Diophantine equation", "algebra unsolved", "algebra" ], "Problem": "give [a^2/b]+[b^2/a]=[(a^2+b^2)/ab]+ab\r\n [x] is a integer y so that y \\leq x\r\n find integer a,b :D :?", "Solution_1": "The only solutions of the Diophantine equation\n\n$(1) \\;\\; {\\textstyle [\\frac{a^2}{b}] + [\\frac{b^2}{a}] = [\\frac{a^2+b^2}{ab}] + ab}$,\n\nare $(a,b)=(n^2+1,n)$ and $(a,b)=(n,n^2+1)$, where $n \\in \\mathbb{Z} \\setminus \\{-1,0\\}$.\n\n\n[u][b]Proof:[/b][/u]\nFirst of all, we may WLOG assume $a \\geq b$ since $a$ and $b$ are symmetric in (1). Secondly $a,b \\neq 0$ by (1). \nLet us consider the following three cases depending on the signs of $a$ and $b$: \n\n[b]Case 1: [/b] $a,b<0$. Then in (1), $LHS < 0$ and $RHS>0$. Hence there is no solution of (1) in this case.\n\n[b]Case 2: [/b] $a>0>b$. Set $c=-b>0$. According to (1)\n\n$(2) \\;\\; {\\textstyle ac + [-\\frac{a^2}{c}] = -[\\frac{c^2}{a}] + [-\\frac{a^2+c^2}{ac}]}$.\n\nIn (2), $LHS<0$ since $RHS<0$. Therefore \n\n$(3) \\;\\; {\\textstyle ac +1 \\leq -[-\\frac{a^2}{c}]}$.\n\nThe fact $-[-x] < x + 1$ for all $x$, meaning\n\n$(4) \\;\\; {\\textstyle -[-\\frac{a^2}{c}] < \\frac{a^2}{c} + 1}$.\n\nCombining (3) and (4) yields ${\\textstyle ac < \\frac{a^2}{c}}$, i.e. $a>c^2$. Consequently $0 < {\\textstyle \\frac{c^2}{a} < 1}$, implying ${\\textstyle [\\frac{c^2}{a}]=0}$, which inserted in (2) gives \n\n${\\textstyle ac = - [-\\frac{a^2}{c}] + [-\\frac{a^2+c^2}{ac}]}$.\n\nNow $-[-x] \\geq x$ and $[x] > x-1$ for all $x$, which means\n\n${\\textstyle ac > \\frac{a^2}{c} - \\frac{a^2+c^2}{ac} - 1}$,\n\ni.e.\n\n$(5) \\;\\; a^2(a-c^2) < a^2 + ac + c^2$.\n\nDividing both side of (5) by $a^2$ we obtain\n\n${\\textstyle (a-c^2) < 1 + \\frac{c}{a} + (\\frac{c}{a})^2 < 3}$\n\nsince ${\\textstyle 1 > \\frac{c^2}{a} \\geq \\frac{c}{a}}$. Hence $0 < a-c^2 < 3$, i.e. $a - c^2 \\in \\{1,2\\}$.\n\nAssume $a - c^2 = 2$. Then by (5)\n\n$0 > a^2 - ac - c^2 > (c^2)^2 - c^2 \\cdot c - c^2$\n\nsince $a = c^2 + 2 > c^2$, i.e.\n\n$c^2(c^2 - c - 1) \\leq -2$, \n\nyielding $c<0$. This contradiction implies $a \\neq c^2 + 2$. \n\nConsequently $a = c^2 + 1$. Then in (2)\n\n$ {\\textstyle LHS = ac + [-\\frac{a^2}{c}] = c(c^2+1) + [-\\frac{(c^2+1)^2}{c}] = c^3 + c - (c^3 + 2c) + [-\\frac{1}{c}] = -c-1}$\n\nand\n\n$ {\\textstyle RHS = -[\\frac{c^2}{a}] + [-\\frac{a^2+c^2}{ac}] = [\\frac{c^2}{c^2+1}] + [-\\frac{c^2+1}{c} - \\frac{c}{c^2+1})] = -c + [-\\frac{1}{c} - \\frac{c}{c^2+1}}]$.\n\nIf $c=1$, then $LHS = -2 \\neq -3=RHS$. Thus $c > 1$, which means\n\n${\\textstyle 0 > -\\frac{1}{c} - \\frac{c}{c^2+1} > -\\frac{1}{c} - \\frac{c}{c^2} = -\\frac{1}{c} - \\frac{1}{c} = -\\frac{2}{c} \\geq -1}$.\n\nTherefore ${\\textstyle [-\\frac{1}{c} - \\frac{c}{c^2+1}] = -1}$, giving $RHS = -c-1 = LHS$. \n\nSo in this case the solutions of (1) are $(a,b)=(b^2+1,b)$, where $b \\geq -2$. \n\n[b]Case 3: [/b] $a,b>0$. Combining (1) and the facts $a^2+b^2 \\geq 2ab$ for all $a,b$ and $x-1 < [x] \\leq x$ for all $x$, yield\n\n${\\textstyle \\frac{a^2}{b} + \\frac{b^2}{a} \\geq [\\frac{a^2}{b}] + [\\frac{b^2}{a}] = [\\frac{a^2+b^2}{ab}] + ab > \\frac{a^2+b^2}{ab} + ab - 1 \\geq 2 + ab - 1 = ab + 1}$,\n\nhence \n\n$(6) \\;\\; {\\textstyle \\frac{a^2}{b} + \\frac{b^2}{a} > ab}$.\n\nMultiplying (6) by ${\\textstyle \\frac{b}{a^2}}$ we obtain\n\n${\\textstyle \\frac{b^2}{a} < 1 + (\\frac{b}{a})^3}$,\n\nimplying ${\\textstyle \\frac{b^2}{a}<2}$ since $a \\geq b > 0$.\n\nFirst assume ${\\textstyle 0 < \\frac{b^2}{a} <1}$. Then ${\\textstyle [\\frac{b^2}{a}]=0}$, which inserted in (1) gives \n\n$(7) \\;\\; {\\textstyle [\\frac{a^2}{b}] = [\\frac{a^2+b^2}{ab}] + ab}$.\n\nAccording to the Division algorithm there exist integers $q>0$ and $0 \\leq r < b$ s.t. $a = qb + r$. \nNow $b^2 < a=qb+r < b(q+1)$ since $0 \\leq r < b$, implying $q \\geq b$ and \n\n${\\textstyle \\frac{a^2+b^2}{ab} = \\frac{a}{b} + \\frac{b}{a} =\\frac{qb+r}{b} + \\frac{b}{qb+r} = q + \\frac{r}{b} + \\frac{b}{qb+r}\\leq q + \\frac{r}{b} + \\frac{1}{q} \\leq q + \\frac{r}{b} + \\frac{1}{b} \\leq q + \\frac{r+1}{b} \\leq q+1}$,\n\nso ${\\textstyle \\frac{a^2+b^2}{ab} \\leq q+1}$ with equality when $r=0, q=b, r=b-1$, i.e. $q=b=1$ giving $a=1=b^2$. This contradiction of the fact $a>b^2$ means ${\\textstyle q \\leq \\frac{a^2+b^2}{ab} < q+1}$. Hence ${\\textstyle [\\frac{a^2+b^2}{ab}] = q}$, which inserted in (7) with $a=qb+r$ gives\n\n${\\textstyle [\\frac{(qb+r)^2}{b}] = q + b(qb + r)}$, \n\ni.e. \n\n${\\textstyle q^2b + 2qr + [\\frac{r^2}{b}] = q + qb^2 + br}$,\n\nor alternatively \n\n$(8) \\;\\; {\\textstyle [\\frac{r^2}{b}] = qb(b-q) + r(b-2q) + q}$.\n\nAssume $q \\neq b$, i.e. $q>b$. \nLet $r=0$. Then by (8)\n\n$0 = qb(b-q) + q = q(b(b-q)+1)$,\n\nimplying\n\n$b(q-b) = 1,$\n\nsince $q>b>0$, which also implies $b=1$ and $q=2$, yielding $a = qb+r = 2 \\cdot 1 + 0 = 2$. So the only solution when $r=0$ is $(a,b) = (2,1)$. \n \nNext assume $r>0$. Now $q>b$, so $b-q \\leq -1$ and $b-2q < -q$, which according to (8) means\n\n$(9) \\;\\; {\\textstyle [\\frac{r^2}{b}] < -qb - qr + q = q(1-b-r)}$.\n\nEvidently ${\\textstyle [\\frac{r^2}{b}] \\geq 0}$, so $1-b-r>0$ by (9). Thus $b+r<1$, which is impossible since $b > r > 0$. Therefore the only solution of (1) when $q \\neq b$ is $(a,b)=(2,1)$. \n\nFinally let $q=b$. Inserted in (8) we obtain\n\n$(10) \\;\\; {\\textstyle [\\frac{r^2}{b}] = b(1-r)}$.\n\nNow ${\\textstyle 0 \\leq [\\frac{r^2}{b}] < [\\frac{b^2}{b}] = b}$ since $0 \\leq r 1=r$. Observe that $b=1$ gives the solution $(a,b)=(2,1)$. In other words, the only solutions of (1) in this case are $(a,b)=(n^2+1,n)$, where $n>0$. \n\nSumma summarum, the only solutions of (1) with $a \\geq b$ are $(a,b)=(n^2+1,n)$, where $n \\neq -1,0$. $\\;\\;$ [b]q.e.d.[/b]", "Solution_2": "I did that...but my solution is almost same as http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1218612&sid=030abdcaaf3c9ef2979fb264df0eba70#p1218612" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Given a natural number, one may repeatedly\r\n\r\na) multiply it by any other natural number, and\r\nb) delete zeros in its decimal representations.\r\n\r\nProve that, starting with any natural number, one can perform a sequence of these operations that transforms this number into a one-digit number.", "Solution_1": "http://www.mathlinks.ro/Forum/viewtopic.php?t=47945" } { "Tag": [ "inequalities", "Cauchy Inequality", "inequalities unsolved" ], "Problem": "[quote=\"Vasc\"]For $a,b,c$ positive real numbers and $p \\ge 7/8$ prove that\n\n$\\displaystyle \\frac{1}{a^2+b^2+pab}+\\frac{1}{a^2+c^2+pac}+\\frac{1}{b^2+c^2+pac}\\ge \\frac{27}{(2+p)(a+b+c)^2}$[/quote]", "Solution_1": "hmmm.... :lol: \r\nthat is a generalisation of the inequality : \r\n$\\frac{1}{a^{2}+b^{2}+ab}+\\frac{1}{b^{2}+c^{2}+bc}+\\frac{1}{c^{2}+a^{2}+ca}\\geq \\frac{9}{\\left( a+b+c\\right) ^{2}}$\r\nbut both of them are hard !!! :lol:", "Solution_2": "i just found a solution in case $p\\geq2$\r\n\r\nand here my solution:\r\n\r\nthe ineq is equivalent to:\r\n$\\sum\\frac{(a+b+c)^2}{(a+b)^2+(p-2)ab}\\geq\\frac{27}{2+p}$\r\napplying CAUCHY we get:\r\n$\\sum\\frac{(a+b+c)^2}{(a+b)^2(1+\\frac{p-2}{4})}\\geq\\frac{27}{2+p}$\r\n\r\nit reduces to:\r\n\r\n$\\sum(1+\\frac{c}{a+b})^2\\geq\\frac{27}{4}$\r\n\r\nlet $\\frac{c}{a+b}=x$\r\n\r\n$f(c)=(1+c)^2$ is concave\r\n\r\nappying jensen and nesbit ;) \r\n\r\nthis tell that they are very weak :D \r\n\r\nonly in case p>=2", "Solution_3": "to erdos:\r\n\r\nyour ineq can prove by brute force and :\r\n\r\n$a^6+b^6+c^6+3b^5c+3ab^5+3ac^5+3bc^5+3a^5b+3a^5c\\geq3a^4c^2+3a^4b^2+3b^4a^2+3b^4c^2+3c^4a^2+3c^4b^2+a^3b^3+b^3c^3+c^3a^3$\r\n\r\nwhich is true by CAUCHY :D \r\n\r\nsomething tell me that : brute force for this one may get the result to the generalized ineq ;)", "Solution_4": "i have another proof for $p \\ge 2$\r\nBy cauchy schwarz and iran1996,\r\n\\[\r\n\\begin{array}{l}\r\n \\left( {2\\left( {a^2 + b^2 + c^2 } \\right) + p(ab + bc + ca)} \\right)\\left( {\\sum {\\frac{1}{{a^2 + b^2 + pab}}} } \\right) \\ge 9 \\\\ \r\n (4 - p)(ab + bc + ca)\\left( {\\sum {\\frac{1}{{a^2 + b^2 + pab}}} } \\right) \\\\ \r\n \\ge (4 - p)(ab + bc + ca)\\left( {\\sum {\\frac{1}{{\\frac{{p + 2}}{4}(a + b)^2 }}} } \\right) \\\\ \r\n = \\frac{{4(4 - p)}}{{p + 2}}\\left( {\\frac{9}{4}} \\right) \\\\ \r\n = \\frac{{9(4 - p)}}{{p + 2}} \\\\ \r\n \\end{array}\r\n\\]\r\nSum them up we have\r\n\\[\r\n\\begin{array}{l}\r\n 2\\left( {a + b + c} \\right)^2 \\left( {\\sum {\\frac{1}{{a^2 + b^2 + pab}}} } \\right) \\ge 9\\left( {1 + \\frac{{4 - p}}{{p + 2}}} \\right) = \\frac{{54}}{{p + 2}} \\\\ \r\n \\left( {a + b + c} \\right)^2 \\left( {\\sum {\\frac{1}{{a^2 + b^2 + pab}}} } \\right) \\ge \\frac{{27}}{{p + 2}} \\\\ \r\n \\end{array}\r\n\\]", "Solution_5": "There is a mistake where you write\r\n\r\n[quote=\"siuhochung\"] \n $(4 - p)(ab + bc + ca)( {\\sum {\\frac{1}{{a^2 + b^2 + pab}}} } \n \\ge (4 - p)(ab + bc + ca)( {\\sum {\\frac{1}{{\\frac{{p + 2}}{4}(a + b)^2 }}} }$ \n [/quote]\r\n\r\nas $a^2+b^2 \\geq \\frac{(a+b)^2}{2}$", "Solution_6": "Can you please give a hint for a proof, Vasc?\r\n\r\nThank you.", "Solution_7": "This ineq can be obtained adding up the inequalities\r\n$\\displaystyle \\sum \\frac{2a^2+(1+2p)bc}{b^2+pbc +c^2} \\ge \\frac{3(3+2p)}{2+p}$,\r\n$\\displaystyle 4\\sum \\frac{a(b+c)+(p-1)bc}{b^2+pbc+c^2} \\ge \\frac{12(1+p)}{2+p}$.", "Solution_8": "Thanks, Vasc.\r\n\r\nBut these two inequalities seem harder than original inequality. :? \r\n\r\n\r\nHowever this is really a nice inequality to solve. ;)", "Solution_9": "[quote=\"Vasc\"]For $ a,b,c$ positive real numbers prove that\n\n$ \\frac {23}{8b^{2} \\plus{} 7bc \\plus{} 8c^{2}} \\plus{} \\frac {23}{8c^{2} \\plus{} 7ca \\plus{} 8a^{2}} \\plus{} \\frac {23}{8a^{2} \\plus{} 7ab \\plus{} 8b^{2}}\\ge\\frac {27}{(a \\plus{} b \\plus{} c)^2}.$[/quote]Using Cauchy inequality,\r\n\r\n$ \\left(\\sum{\\frac {1}{8b^2 \\plus{} 7bc \\plus{} 8c^2}}\\right)\\left[\\sum{(19a \\plus{} 4b \\plus{} 4c)^2(8b^2 \\plus{} 7bc \\plus{} 8c^2)}\\right]$\r\n\r\n$ \\geq\\left[\\sum{(19a \\plus{} 4b \\plus{} 4c)}\\right]^2 \\equal{} 729\\left(\\sum{a}\\right)^2,$\r\n\r\nwe must prove that\r\n\r\n$ 621\\left(\\sum{a}\\right)^4\\geq \\sum{(19a \\plus{} 4b \\plus{} 4c)^2(8b^2 \\plus{} 7bc \\plus{} 8c^2)}.$\r\n\r\nThis comes down to \r\n\r\n$ \\frac {621\\left(\\sum{a}\\right)^4 \\minus{} \\sum{(19a \\plus{} 4b \\plus{} 4c)^2(8b^2 \\plus{} 7bc \\plus{} 8c^2)}}{5}$\r\n\r\n$ \\equal{} 73\\sum{a^4} \\plus{} 180\\sum{a^3(b \\plus{} c)} \\minus{} 506\\sum{b^2c^2} \\plus{} 73\\sum{a^2bc}$\r\n\r\n$ \\equal{} 73\\sum{a^2(a \\minus{} b)(a \\minus{} c)} \\plus{} 253\\sum{bc(b \\minus{} c)^2}\\geq 0,$\r\n\r\nwhich is true by [url=http://www.mathlinks.ro/viewtopic.php?t=1166]Schur inequality[/url].\r\n\r\nSee also here : http://www.mathlinks.ro/viewtopic.php?t=123636" } { "Tag": [ "geometry", "trigonometry", "trig identities", "Law of Sines", "Law of Cosines", "perpendicular bisector", "Pythagorean Theorem" ], "Problem": "Someone help me on this problem:\r\n\r\nConsider two concentric circles. The circumference of the larger circle is one foot longer than the circumference of the smaller circle. Find the [positive] difference between the two radii of the circles.", "Solution_1": "[hide=\"Long Way\"]\nLet the circumference of the smaller circle be $ x$ and the larger circle's circumference is then $ x \\plus{} 1$. The radii of the circles are then $ \\frac x{2\\pi}$ and $ \\frac {x \\plus{} 1}{2\\pi} \\equal{} \\frac x{2\\pi} \\plus{} \\frac 1{2\\pi}$, respectively. Thus the difference between the radii is simply $ \\frac 1{2\\pi}$.[/hide]\n\n[hide=\"The Short Way\"]\nWe notice that the initial radius doesn't matter. Just divide 1 by $ 2\\pi$ to get $ \\frac 1{2\\pi}$ to get the answer.[/hide]", "Solution_2": "[hide]\r\nLet the radius of the smaller circle by a and the larger one b. So, 2b\\pi - 1 = 2a\\pi. Divide by 2\\pi to get b-1/(2\\pi) = a. So, b is 1/(2\\pi) larger than a.\r\n[\\hide]\r\n\r\nEdit: Woops. Yongyi beat me.", "Solution_3": "Ok, \r\n\r\nC circle I= $ 2\\pi$\r\nC circle II: $ 2\\pi\\plus{}1$\r\n\r\nBecause C=$ \\pi$d, we divide each by $ \\pi$ to get the diameter, and divide that by 2 to get the radius.\r\n$ \\frac {2\\pi} {2\\pi}$ $ \\equal{}$ $ 1$\r\n$ \\frac {2\\pi\\plus{}1} {2\\pi}$ $ \\equal{}$ ... no simplifying to do so its just $ \\frac {2\\pi\\plus{}1} {2\\pi}$.\r\nThe positive difference is $ \\frac {2\\pi\\plus{}1} {2\\pi} \\minus{}1$. We want to make a common denominator so....\r\n$ \\frac {2\\pi\\plus{}1} {2\\pi} \\minus{}\\frac {2\\pi} {2\\pi}$ $ \\equal{}$ $ \\frac {1} {2\\pi}$\r\n\r\nDang lots of people beat me...", "Solution_4": "Um...What are concentric circles? Are they circles who share the same center?", "Solution_5": "Yes - con = same, centric = well, obviously, center", "Solution_6": "I get it! Thanks! \r\nNOTE: I will probably be asking any other Geometry questions I have here.", "Solution_7": "Note that proofs do not belong here though... :P", "Solution_8": "Are they really proofs, though, not just simply answering questions?", "Solution_9": "I know that this was just answering a question, I just wanted to remind everyone that no proofs should really be posted here.\r\n\r\ni.e. For this question, if the problem had been: \"Prove that the difference between the lengths of the radii is...\", then it would probably not belong here. (As simple as it would be.)", "Solution_10": "Let hexagon $ ABCDEF$ have a side length of 2. What is the distance between $ C$ and $ E$?", "Solution_11": "Um...I just completed Alg II Honors, so I used Law of Sines/Cosines...\r\n ----------------------------------------------------------------------------------------------------------------\r\nRelabel triangle CDE as follows:\r\n\r\nSide CD is now a.\r\nSide DE is now b.\r\nSide CE is now c.\r\n\r\n$ \\gamma\\equal{}120$ (obviously)\r\n\r\n[u][b]Law of Sines Method:[/b][/u]\r\n\r\n$ \\alpha\\equal{}\\beta\\equal{}30$ (Because this is isosceles.)\r\n\r\n$ \\frac{\\sin30}{2}\\equal{}\\frac{\\sin120}{c}$\r\n\r\n$ c\\equal{}\\frac{2\\sin120}{\\sin30}$\r\n\r\n$ \\equal{}\\frac{2\\left(\\frac{\\sqrt{3}}{2}\\right)}{\\frac{1}{2}}$\r\n\r\n$ \\equal{}2\\sqrt{3}$\r\n\r\n[u][b]Law of Cosines Method:[/b][/u]\r\n\r\n$ c^2\\equal{}2^2\\plus{}2^2\\minus{}2(2)(2)\\cos120$\r\n\r\n$ c^2\\equal{}4\\plus{}4\\minus{}8\\left(\\minus{}\\frac{1}{2}\\right)$\r\n\r\n$ c^2\\equal{}8\\plus{}4$\r\n\r\n$ c^2\\equal{}12$\r\n\r\n$ c\\equal{}\\sqrt{12}$\r\n\r\n$ c\\equal{}2\\sqrt{3}$", "Solution_12": "Yes, but you're assuming it's a regular hexagon! Maybe you're supposed to... :D \r\n\r\nNon-Law of Sines or Cosines Method:\r\n\r\nNotice that $ CDE$ is a 120-30-30 triangle, and if you know enough about 120-30-30 triangles, the base is the leg times $ \\sqrt 3$. Therefore, the answer is $ 2\\sqrt 3$.\r\n\r\nProof (not rigurous, trying to be understood clearly):\r\n\r\nDrawing the perpendicular bisector of $ CE$ creates two 30-60-90 triangles, both with hypotenuse 2. Therefore, the desired length is twice the longer leg of one of the triangles, so it follows that the length of $ CE$ is $ 2\\sqrt 3$.", "Solution_13": "Darn it! I knew there was some way to do this using Pythagorean Theorem, except it just didn't occur to me at the time. (At least I have 3 months of summer vacation to get Alg II out of my head... :lol: )\r\n\r\nYeah, I guess it could be a non-regular hexagon, but that wouldn't be solvable. Maybe the problem should have been clearer.", "Solution_14": "Oh, I get it! But I don't get why this was moved to High School math. Probably because it is.", "Solution_15": "Sorry. That was probably my fault. I used Law of Sines/Cosines which is Alg II/trig.", "Solution_16": "[quote=\"qwertythecucumber\"]Let hexagon $ ABCDEF$ have a side length of 2. What is the distance between $ C$ and $ E$?[/quote]\r\n\r\nPlease create new a thread when asking a seperate question. We're not trying to start a marathon here. :wink:", "Solution_17": "Yeah, marathons do seem to end up locked... :D", "Solution_18": "[quote=\"Yongyi781\"]Yes, but you're assuming it's a regular hexagon! Maybe you're supposed to... :D [/quote]\r\n\r\nHes saying it has [b]side[/b] length of 2. That means all sides are equal. Doesn't that mean that it is a regular hexagon?", "Solution_19": "No, because the angles could still be different.", "Solution_20": "OK, I forgot to say it was a regular hexagon." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Natural numbers x,y,k,n where n>1 is odd. Prime number p>2 satisfied:\r\n\r\nx^n + y^n = p^k\r\n\r\nprove that n=p^m for some natural m.\r\n\r\n :help:", "Solution_1": "it's appeared on the forum before" } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Fin all functions $f:\\mathbb{N}\\to \\mathbb{N}$ such that :\n\\[f(x^2+y^2+z^2+t^2) =(f(x))^2 +(f(y))^2+(f(z))^2+(f(t))^2 \\]", "Solution_1": "If you want to practice in posting replies than you should do it in the Test Section of this forum. ;)", "Solution_2": "[quote=\"bellahsayn\"]FIND ALL FONCTIONS SATIS[/quote]\r\n\r\nare u mad?", "Solution_3": "Please, be polite!" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let $ a,b,c>0$ such that $ a\\plus{}b\\plus{}c\\equal{}\\frac{1}{2}$\r\nprove that\r\n$ \\frac{1}{(a\\plus{}b\\plus{}1)^4}\\plus{}\\frac{1}{(b\\plus{}c\\plus{}1)^4}\\plus{}\\frac{1}{(c\\plus{}a\\plus{}1)^4} \\le \\frac{9}{2^{11}abc}$\r\n :!:", "Solution_1": "[quote=\"devil_maths\"]Let $ a,b,c > 0$ such that $ a \\plus{} b \\plus{} c \\equal{} \\frac {1}{2}$\nprove that\n$ \\frac {1}{(a \\plus{} b \\plus{} 1)^4} \\plus{} \\frac {1}{(b \\plus{} c \\plus{} 1)^4} \\plus{} \\frac {1}{(c \\plus{} a \\plus{} 1)^4} \\le \\frac {9}{2^{11}abc}$\n :!:[/quote]\r\n\r\n\r\n$ \\frac {1}{(a \\plus{} b \\plus{} 1)^4} \\plus{} \\frac {1}{(b \\plus{} c \\plus{} 1)^4} \\plus{} \\frac {1}{(c \\plus{} a \\plus{} 1)^4} \\equal{} \\sum \\frac {1}{(a \\plus{} b \\plus{} 6* \\frac {1}{6})^4}$ $ \\le \\sum \\frac {1}{(8\\sqrt [8]{(\\frac {1}{6})^6ab})^4} \\equal{} \\sum \\frac {27}{512\\sqrt {ab}}$\r\n\r\nwe only prove \r\n\r\n$ \\sum \\frac {27}{512\\sqrt {ab}}\\le \\frac {9}{2048 abc}................(*)$\r\n\r\nsubstitution $ a \\equal{} \\frac {x^2}{2(x^2 \\plus{} y^2 \\plus{} z^2)},a \\equal{} \\frac {y^2}{2(x^2 \\plus{} y^2 \\plus{} z^2)},a \\equal{} \\frac {z^2}{2(x^2 \\plus{} y^2 \\plus{} z^2)}$\r\n\r\n$ (*) \\Longleftrightarrow \\frac {9(x^2 \\plus{} y^2 \\plus{} z^2)^3}{256x^2y^2z^2} \\minus{} \\frac {27(x^2 \\plus{} y^2 \\plus{} z^2)}{256}$ $ (\\frac {1}{xy} \\plus{} \\frac {1}{yz} \\plus{}$ $ \\frac {1}{zx})\\geq 0$\r\n\r\n$ \\Longleftrightarrow 9(x^2 \\plus{} y^2 \\plus{} z^2)(x^4 \\plus{} 2y^2x^2 \\plus{} 2x^2z^2 \\plus{} y^4 \\plus{} 2y^2z^2$ $ \\plus{} z^4 \\minus{} 3yz^2x \\minus{} 3yzx^2 \\minus{} 3y^2zx)\\geq 0$\r\n\r\n$ \\Longleftrightarrow 9(x^2 \\plus{} y^2 \\plus{} z^2)(\\sum x^2(x \\minus{} y)(x \\minus{} z) \\plus{} \\sum x(y \\plus{} z)(x \\minus{} y)(x \\minus{} z) \\plus{}$ $ 4\\sum yz(x \\minus{} y)(x \\minus{} z))\\geq 0$\r\n\r\n\r\n\r\n :lol:" } { "Tag": [ "limit" ], "Problem": "Again, this isn't all that difficult, but it has the word \"converge\" in it so I'll put it in advanced.\r\n\r\nProve that for non-negative [tex]a_n[/tex], the infinite product [tex]\\displaystyle \\prod_{n=1}^\\infty(1+a_n)[/tex] converges if and only if the series [tex]\\displaystyle \\sum_{n=1}^\\infty a_n[/tex] converges.\r\n\r\n(Edited to say \"non-negative\" instead of \"real,\" although I don't actually know if that change is necessary or not.)", "Solution_1": "This is a standard theorem in mathematical analysis, is it?", "Solution_2": "Probably, but it's nice to prove it anyway.", "Solution_3": "how is this true? let ai = -1 for instance. Product is 0. sum is clearly -infinity", "Solution_4": "[quote=\"polymorphic\"]how is this true? let ai = -1 for instance. Product is 0. sum is clearly -infinity[/quote]\r\n\r\nCheck out the definition of convergence/divergence of infinite product carefully, you might find a surprising answer.\r\n\r\nIn the original statement, we probably need to assume absolutely convergence on the :Sigma: a_n part. [b]zableman[/b]? (Again, I am not sure about this, I don't have an analysis reference book here to check upon this).\r\n\r\nBy the way, this is what I found (of course, on the web again):\r\n\r\nhttp://www.math.ucsb.edu/~mckernan/Teaching/02-03/Autumn/202A/l_26.pdf", "Solution_5": "I'm not entirely sure, so I edited it.", "Solution_6": "interesting. I think that the other way is true though without any other restrictions. Namely, for any reals an, convergence of the sum implies convergence of the product. The product clearly cannot diverge since log(1+x) < x. THen we just have to show that it can't oscillate, but i don't have a very rigorous way for showing that.", "Solution_7": "Let [tex]\\displaystyle a_{2k-1}= -\\frac{1}{\\sqrt{k+1}}[/tex] and [tex]\\displaystyle a_{2k}= \\frac{1}{\\sqrt{k+1}}[/tex] for [tex]k\\ge 1[/tex].\r\n\r\nDefine [tex]\\displaystyle S_n = \\sum_{k=1}^n a_k[/tex] and [tex]\\displaystyle P_n = \\prod_{k=1}^n (1+a_k)[/tex].\r\n\r\nThen [tex]S_{2n} = 0[/tex] and [tex]S_{2n-1} = a_{2n-1}[/tex], So [tex]\\displaystyle \\lim_{n\\rightarrow\\infty}S_n = 0[/tex]. That is, [tex]\\displaystyle \\sum_{k=1}^{\\infty} a_k[/tex] converges (but not absolutely converges).\r\n\r\nOn the other hand [tex]\\displaystyle 0 < P_{2n-1}\\le P_{2n} = \\frac{1}{n+1}[/tex], So [tex]\\displaystyle \\lim_{n\\rightarrow\\infty}P_n = 0[/tex]. That is, [tex]\\displaystyle \\prod_{k=1}^{\\infty} a_k[/tex] diverges.", "Solution_8": "There is a very interesting fact (I think it's due to Riemann if I remember it correctly):\r\n\r\nIf the infinite series :Sigma: a_n converges conditionally (which means :Sigma: |a_n| diverges, but :Sigma: a_n converges), then for any given real x, we can rearrange the terms so that the series converges to x.", "Solution_9": "I took a second look at the place where I got this fact, and their proof is unsatisfactory (and incomplete) the way it was originally given. But I'm pretty sure that replacing the 'real' with 'non-negative' makes it work, (hint in spoiler) [hide]since then both the sum and the product are monotinically increasing and we just have to show that if one has a finite value than the other does as well.[/hide]", "Solution_10": "[quote=\"zabelman\"]... But I'm pretty sure that replacing the 'real' with 'non-negative' makes it work, ...[/quote]\r\n\r\nSure it works if all a_n :ge: 0, since absolutely convergence is the same as convergence in this case.\r\n\r\nHowever, if you limit a_n :ge: 0, then your criterion can not be applied to infinite product such as [tex]\\displaystyle \\prod_{n=2}^{\\infty}\\left(1+\\frac{(-1)^n}{n^{1+\\alpha}}\\right)[/tex], where [tex]\\alpha > 0[/tex].", "Solution_11": "A tweak on fuzzylogic's example, put to a slightly different (but related) purpose:\r\n\r\nLet a_(2k-1) = -1/sqrt(k)\r\nLet a_(2k) = (1-1/sqrt(k))^(-1) - 1\r\n\r\nThen (1+a_(2k-1))*(1+a_(2k)) = 1 and the infinite product of (1+a_n) converges to 1.\r\n\r\nBut if you expant out a_(2k) by the binomial series, you'll see that a_(2k-1) + a_(2k) is comparable to 1/k for large k and hence the infinite sum of a_n diverges to infinity.\r\n\r\nBut these are the only two directions you can go. You can have the sum converge and the product diverge to zero (as in the first example) and you can have the product converge and the sum diverge to + :inf: . That's it for bad examples.", "Solution_12": "[quote=\"Kent Merryfield\"]But these are the only two directions you can go. You can have the sum converge and the product diverge to zero (as in the first example) and you can have the product converge and the sum diverge to + :inf: . That's it for bad examples.[/quote]\r\nSo the Pro painted the whole picture. :-) That's nice to know.", "Solution_13": "hm. diverge to zero? that seems like a strange concept.", "Solution_14": "[quote=\"polymorphic\"]hm. diverge to zero? that seems like a strange concept.[/quote]\r\n\r\nLoosely speaking, Prod (1+a_k) diverges to 0 is same as saying :Sigma: log(1+a_k) diverges to -:inf: since Prod_(k: 1 to n) (1+a_k) = exp [:Sigma:_(k: 1 to n) log(1+a_k)].", "Solution_15": "This is some of my thoughts when a_i :ge: 0\n\n\n\n[hide]Case I: Does pro(a_i) converge if sum(a_i) converges?\n\nSince a_n are nonnegative, then we can apply AM-GM on pro(1+a_i). We have\n\n\n\npro(1+a_i) = ((pro(1+a_i))^(1/n))^n :le: (1+sum(a_i)/n)^((n/sum(a_i))*sum(a_i)) :le: e^(sum(a_i))\n\n\n\nSo if sum(a_i) converges, e^(sum(a_i)) converges, and pro(a_i+1) converges.\n\n\n\nCase II: Does sum(a_i) converges if pro(1+a_i) converges?\n\nWell, since a_i :ge: 0. It is obvious that pro(1+a_i) :ge: sum(a_i). Thus, sum(a_i) does converge if pro(a_i +1) converges.[/hide]\n\n\n\nThus, if one of pro(a_i +1) and sum(a_i) converges, so does the other." } { "Tag": [ "topology", "advanced fields", "advanced fields unsolved" ], "Problem": "The problem is as follows:\r\n\r\nLet P be the set of all lines in $ R^{2}$. We establish the following notion of convergence in P: a sequence $ L_{n}$ converges to the line $ L_{0}$ if and only if $ \\forall x\\in{L_{0}}$ $ d(x, L_{0})$ converges to 0, where d is the standard Euclidean distance between a point and a line. Is there a metric D on P such that convergence in terms of metric D is exactly the convergence in the sense explained above?\r\n\r\nI've tried a metric described as follows:\r\n\r\nLet r(L) be the distance between a line L and the point (0, 0) taken with sign, i.e. it's positive if the orthogonal projection of (0, 0) onto the line lies below y=x and it's negative otherwise, and let a(L) be the angle between L and the line y=0 taken modulo $ \\pi$. The distance D between L and K is\r\n\r\n d[(r(L), a(L)), (r(K), a(K))], \r\n\r\nwhere d is the standard Euclidean metric on $ R^{2}$. Does this metric satisfy conditions of the problem?\r\n\r\nIf my metric is incorrect, please don't spoil the correct answer to problem - just show me why it's wrong.", "Solution_1": "That should be $ d(x,L_n)$ in the definition- it doesn't make sense as stated.\r\n\r\nIt's almost right, but you have to be more careful about \"taken with sign\". Look at the family of lines $ L_k\\equal{}y\\plus{}kx\\equal{}1$, as $ k\\to 1$. If $ k<1$, we get $ r(L_k)\\equal{}\\frac1{\\sqrt{k^2\\plus{}1}}$ by your definition, and if $ k>1$, we get $ r(L_k)\\equal{}\\minus{}\\frac1{\\sqrt{k^2\\plus{}1}}$.\r\n\r\nThis space of lines is homeomorphic to a Mobius strip (with open edges)." } { "Tag": [ "floor function", "number theory proposed", "number theory" ], "Problem": "Let $f(x)=\\lfloor x^2 \\rfloor + \\{x\\}$ for all positive reals $x$.\r\nProve: There exists an arithmetical progression $ = $ of positive rational numbers for which all (canceled) denominators are $3$ and $a_n$ is not in $f(x)$ for all $n$.", "Solution_1": "=<19/3 + n*24/3> works." } { "Tag": [ "function", "calculus", "integration", "algebra unsolved", "algebra" ], "Problem": "Suppose function $f: N\\times Z \\to N$ satisfies the conditions:\r\n (i) $f(0,0) =5^{2003}$ , $f(0,n)=0$, for all number integer $n \\neq 0$\r\n (ii) $f(m,n) = f(m-1,n)-2\\left[\\frac{1}{2}f(m-1,n)\\right]+\\left[\\frac{1}{2}f(m-1,n+1)\\right]$ for all natural $m > 0$ and for all integer $n$.\r\n\r\nProve that exist $M \\in N$ such that :\r\n1) $f(M,n)=1$ for all $n \\in Z$ that $|n| \\leq \\frac{5^{2003}-1}{2}$\r\n2) $f(M,n)=0$ for all $n \\in Z$ that $|n| > \\frac{5^{2003}-1}{2}$.\r\n\r\nP.S. $[x]$ denotes the integral part of $x$.\r\n\r\n[i]Edited by Myth[/i]", "Solution_1": "Hm... It seems the statement is not correct, since $f(m,n)=0$ for all $m$ and $n>0$." } { "Tag": [ "LaTeX" ], "Problem": "I'm trying to put pictures in my pdf but whenever I do, it makes the text after it go down farther than it should\r\n\r\nso I try to use vskip with the negative component, and the only lengths that will bring the text back up is \\vskip -4 in+ and when I do this it brings the text back up too far\r\n\r\n\r\nthe picture is made by geogebra\r\nhelp?", "Solution_1": "There could be blank space in the image. Could you post your code plus the image here (or a different one with the same problems if you don't wish to make it public)?", "Solution_2": "here it is", "Solution_3": "Thanks, but could you post the 3 pngs as I suspect it's a problem of extra space in those that is causing the problem? I tried a different image and it was fine.", "Solution_4": "the second image was causing the problem\r\n\r\nhere they are", "Solution_5": "I think you can see from your post the problem. Each png consists of a diagram with loads of blank space below them and this is being faithfully reproduced in your pdf.\r\n\r\nIf you can tell GeoGebra not to do this that will be best (GeoGebra has a [url=http://www.geogebra.org/forum/]forum[/url] if you need help) but meanwhile you could open the images in an image program and cut out and resave the part of the image you need.", "Solution_6": "yes but it's only the second picture that's giving the major problem with the other pictures \\vskip takes care of that space", "Solution_7": "The problem is that the (A) (B) (C) etc line has gone on to the next page so overriding the negative vertical space. You can use \\newpage after the (A) (B) (C) etc line which will work, though I still think it best to use the right size image.", "Solution_8": "what image program could I use to wipe out that extra space?", "Solution_9": "Which operating system are you using? In Windows, Paint will do the job (cut and paste into a small canvas). You may also find you have an image or photo editing software on your computer eg Microsoft Photo Editor that you can use. \r\n\r\nIf not then [url=http://www.openoffice.org/]OpenOffice[/url] Draw program has a crop command which will work. [url=http://www.gimp.org/]GIMP[/url] is a powerful image program. Both are free and work on most operating systems.", "Solution_10": "ok it works now. thank you for your help" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "where p+q+r=3", "Solution_1": "See here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=35839\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=78010" } { "Tag": [ "induction", "number theory solved", "number theory" ], "Problem": "Let p_n denote the nth prime number and let pi(m) denote the number of primes \\leq m, e.g. pi(12)=5. Prove for n \\geq 6 that pi((p_1p_2*...*p_n)^0.5) > 2n.", "Solution_1": "Use induction on n. \r\n\r\nIt is not difficult to verify that it holds for n = 6.\r\n\r\nSuppose it holds for n :\r\nSince n \\geq 6, we have p_(n+1) \\geq 17.\r\nIt follows that \\sqrt (p_1....p_(n+1)) \\geq 4 \\sqrt (p_1....p_n) = 4x.\r\nNow, from Bertrand's postulate :\r\n(pi(4x) - pi(2x)) + (pi(2x) - pi(x)) \\geq 2\r\nThen pi(4x) \\geq 2 + pi(x). And we are done.\r\n\r\nPierre." } { "Tag": [], "Problem": "In a sequence, each term is obtained by calculating the sum of the preceding two terms. the eighth term is 81, and the sixth term is 31. What is the fourth term?", "Solution_1": "It's a Fibonacci-like sequence. To calculate the seventh term, we have $ 31 \\plus{} x \\equal{} 81 \\implies x \\equal{} 50$. Working backwards, we find that the fifth term is $ 19$, and the fourth term is $ 31 \\minus{} 19 \\equal{} \\boxed{12}$." } { "Tag": [ "USAMTS" ], "Problem": "I heard there is a math jam after every round to dicuss how to solve those problems. Do you have to do USAMT to have access to those math jams.", "Solution_1": "In general Math Jams are free to anyone with an AoPS account.", "Solution_2": "Wait Im looking at the upcoming math jams and I dont see a USAMTS math jam anywhere . Where did you hear that? It would be a cool idea though.", "Solution_3": "Well, the deadline for Round 1 submissions is still more than a month away, so I assume they'll post the date for the Math Jam later.", "Solution_4": "Because (starting last year) we now post full solutions to every problem shortly after the conclusion of each round, we have decided to not do the USAMTS Math Jams anymore.", "Solution_5": "On a slightly related question, are commendations to solutions still begin given out? I was under the impression that commendations were abolished last year, but on my grading page it says \"Note: If your number is [b]bold[/b], then your solution to that problem has been commended as particuarly excellent\"", "Solution_6": "no, they said that last year on my profile too, and again this year. They will most likely not commend anymore, or at least this year." } { "Tag": [ "inequalities", "limit", "real analysis", "real analysis unsolved" ], "Problem": "Show that f is Lipschitz if one of its Dini derivates is bounded.\r\n\r\nI tried this with $ D^{\\plus{}}f$ bounded and I got something like for each $ x$, $ \\frac{f(y)\\minus{}f(x)}{y\\minus{}x} \\leq M\\plus{}1$ for $ 0x$.", "Solution_2": "I would have thought that from the definition of $ D^{\\plus{}}$ as a $ \\limsup$, the lower bound for the expression isn't for all $ y \\in (x,x\\plus{}\\delta_{x})$ but rather for each $ \\delta$, there is a $ y \\in (x,x\\plus{}\\delta)$ such that it holds.", "Solution_3": "[quote] you actually have $ \\minus{}M\\minus{}1 \\leq \\frac{f(y)\\minus{}f(x)}{y\\minus{}x} \\leq M\\plus{}1$ for $ 0x$, is some compactness argument required?[/quote]", "Solution_4": "I think I may need more help with this problem. I haven't been able to solve it." } { "Tag": [], "Problem": "I don't know if anyone has seen the playing cards with politicians or terrorists on them. However, I am wondering if similar playing cards exist except with mathematical formulas or theorems on them?", "Solution_1": "I don't think I've ever seen those kinds of cards, but I think they'd be cool!\r\n[url=http://imageshack.us][img]http://img146.imageshack.us/img146/2949/6d2nm.png[/img][/url]", "Solution_2": "Well, one of my friends has a deck of CERN cards; that's almost as cool." } { "Tag": [ "number theory", "greatest common divisor", "relatively prime" ], "Problem": "Show that if $ c|a$ and $ c|b$, then $ c|\\gcd (a,b)$.", "Solution_1": "let gcd(a,b)=d\r\ndx=a, dy=b where x and y are relatively prime\r\n(x,y)=1: so there is always mx+ny=1\r\nc|dx, c|dy, \r\nc|dmx, c|dny\r\nthus c|dmx+dny\r\nc|d(mx+ny)\r\nc|d\r\n\r\nI'm Pretty sure I overcompicated it...", "Solution_2": "$ c|a\\implies a\\equal{}ck$\r\n\r\n$ c|b\\implies b\\equal{}cj$\r\n\r\n$ \\text{gcd}(ck,cj)\\equal{}c\\cdot\\text{gcd}(k,j)$, which is clearly divisible by $ c$." } { "Tag": [ "geometry", "circumcircle" ], "Problem": "Let $ (O)$ be a circle.Let $ A$ be a fixed point outside the circle .$ BC$ be a changleful diameter of $ (O)$.$ AB;AC$ cut $ (O)$ at $ D;E$,respectively.Prove that the line $ DE$ has a fixed point.", "Solution_1": "the problem is equivalent to the following one:\r\n[b]show that the circumcircle of $ \\triangle ABC$ passes thruogh a fixed point (except $ A$ of course...)[/b] :D" } { "Tag": [], "Problem": "On a recent trip, Ellie drove 273 km in the same length of time Carol took to drive 198 km. Ellie's speed was 17 km/h greater than Carol's speed. How many km/h was Ellie driving?", "Solution_1": "[quote=\"ckck\"]On a recent trip, Ellie drove 273 km in the same length of time Carol took to drive 198 km. Ellie's speed was 17 km/h greater than Carol's speed. How many km/h was Ellie driving?[/quote]\r\n[hide]\n$\\frac{273}{x}=\\frac{198}{x-17}$[/hide]", "Solution_2": "[hide=\"solution\"]$273=xt$\n$198=yt$\n$t=273/x$\n$t=198/y$\n$y=x-17$\n$t=198/(x-17)$\n$198/(x-17)=273/x$\nTherefore $x=61.88$[/hide]" } { "Tag": [ "symmetry", "geometry", "inequalities", "number theory proposed", "number theory" ], "Problem": "Find all triples $ (x,y,z)$ of integers which satisfy\r\n\r\n$ x(y \\plus{} z) \\equal{} y^2 \\plus{} z^2 \\minus{} 2$\r\n\r\n$ y(z \\plus{} x) \\equal{} z^2 \\plus{} x^2 \\minus{} 2$\r\n\r\n$ z(x \\plus{} y) \\equal{} x^2 \\plus{} y^2 \\minus{} 2$.", "Solution_1": "we have $ (y^2 \\plus{} z^2 \\minus{} 2) \\minus{} (z^2 \\plus{} x^2 \\minus{} 2) \\equal{} z(x \\minus{} y)$ so $ (x \\minus{} y)(x \\plus{} y \\plus{} z) \\equal{} 0$ and with the same way $ (y \\minus{} z)(x \\plus{} y \\plus{} z) \\equal{} 0$ and $ (z \\minus{} x)(x \\plus{} y \\plus{} z) \\equal{} 0$.\r\nif $ x \\plus{} y \\plus{} z \\equal{} 0$ then $ y \\plus{} z \\equal{} \\minus{} x$ so the first equation gives $ x^2 \\plus{} y^2 \\plus{} z^2 \\equal{} 2$ so solution are $ (\\plus{}\\minus{}1,\\plus{}\\minus{}1,0) (\\plus{}\\minus{}1,0,\\plus{}\\minus{}1)$ and $ (0,\\plus{}\\minus{}1,\\plus{}\\minus{}1)$.\r\nif $ x \\plus{} y \\plus{} z$ is non zero then two of these integers are equal if $ x \\equal{} y$ then the first and the last equation give $ z^2 \\equal{} xz \\plus{} 2$ and $ xz \\equal{} x^2 \\minus{} 1$ so $ x^2 \\minus{} 1 \\equal{} z^2 \\minus{} 2$ so $ (z \\minus{} x)(z \\plus{} x) \\equal{} 1$ so $ (0,,0,1)$ and $ (0,0, \\minus{} 1)$ are solution in this case.", "Solution_2": "[quote=\"Johan Gunardi\"]Find all triples $ (x,y,z)$ of integers which satisfy\n\n$ x(y \\plus{} z) \\equal{} y^2 \\plus{} z^2 \\minus{} 2$\n\n$ y(z \\plus{} x) \\equal{} z^2 \\plus{} x^2 \\minus{} 2$\n\n$ z(x \\plus{} y) \\equal{} x^2 \\plus{} y^2 \\minus{} 2$.[/quote]\r\nAdd them all together, subtract $ 2(xy \\plus{} yz \\plus{} zx)$ and add $ 6$. Then we have $ (x \\minus{} y)^2 \\plus{} (y \\minus{} z)^2 \\plus{} (z \\minus{} x)^2 \\equal{} 6$. With the only solution $ {|x \\minus{} y|,|y \\minus{} z|,|z \\minus{} x|} \\equal{} {1,1,2}$ or a permutation.\r\nBecause of symmetry assume $ x \\ge y \\ge z$. Then we must have $ x \\minus{} z \\equal{} 2$ and $ x \\minus{} y \\equal{} y \\minus{} z \\equal{} 1$. So we have all possible solutions $ (x,y,z) \\equal{} (k \\plus{} 2,k \\plus{} 1,k)$ or a permutation.\r\n\r\nInserting gives: $ (k \\plus{} 2)(2k \\plus{} 1) \\equal{} (k \\plus{} 1)^2 \\plus{} k^2 \\minus{} 2$ $ \\iff$ $ 2k^2 \\plus{} 5k \\plus{} 2 \\equal{} 2k^2 \\plus{} 2k \\minus{} 1$ $ \\iff$ $ k \\equal{} 1$. So we just have to check the solutopn $ (x,y,z) \\equal{} (1,2,3)$. Which gives:\r\n$ 1(2 \\plus{} 3) \\equal{} 2^2 \\plus{} 3^2 \\minus{} 2$ But this is false. Hence: No solutions!\r\n\r\nEdit: Wooops! Of course $ (k \\plus{} 2)(2k \\plus{} 1) \\equal{} (k \\plus{} 1)^2 \\plus{} k^2 \\minus{} 2$ $ \\iff$ $ k \\equal{} \\minus{}1$ :) This gives the solution $ (\\minus{}1,0,1)$ or any permutation, which all works.", "Solution_3": "Add them, divide by two, subtract each of them, you get\r\n$ yz \\equal{} x^2 \\minus{} 1$\r\n$ xz \\equal{} y^2 \\minus{} 1$\r\n$ xy \\equal{} z^2 \\minus{} 1$\r\nMultiply them:\r\n$ x^2y^2z^2 \\equal{} (x^2 \\minus{} 1)(y^2 \\minus{} 1)(z^2 \\minus{} 1)$\r\nBut $ x^2 > x^2 \\minus{} 1$, etc. since x,y,z are integers.\r\nThus, one of them must be 0, another 1.\r\nThis gives $ x \\equal{} 0$ $ y \\equal{} 1$ $ z \\equal{} \\minus{} 1$ and permutations, which does indeed work. This is the only solution, some given by tchebytchev don't work.", "Solution_4": "adding $ x^2$ to the first equation,\r\n$ x(x\\plus{}y\\plus{}z)\\equal{}x^2\\plus{}y^2\\plus{}z^2\\minus{}2$\r\nthe same can be done to the others showing that $ x\\equal{}y\\equal{}z$ or $ x\\plus{}y\\plus{}z\\equal{}x^2\\plus{}y^2\\plus{}z^2\\minus{}2\\equal{}0$\r\nIf $ x\\equal{}y\\equal{}z$ then $ 2n^2\\equal{}2n^2\\minus{}2$ so no solution.\r\nIn the other case, $ x^2\\plus{}y^2\\plus{}z^2\\equal{}2$ means one of the variables is zero, and the other two have norm 1. Thus all solutions are permutations of $ (0,1,\\minus{}1)$", "Solution_5": "[quote=\"Johan Gunardi\"]Find all triples $ (x,y,z)$ of integers which satisfy\n\n$ x(y \\plus{} z) \\equal{} y^2 \\plus{} z^2 \\minus{} 2$ . . . (1)\n\n$ y(z \\plus{} x) \\equal{} z^2 \\plus{} x^2 \\minus{} 2$ . . . (2)\n\n$ z(x \\plus{} y) \\equal{} x^2 \\plus{} y^2 \\minus{} 2$. . . (3)[/quote]\r\n\r\nfrom (1) we have : xy+xz=y^2+z^2 -2 /+2yz\r\n xy+yz + xz+yz=(y+z)^2 -2 \r\n y(x+z) + z(x+y)=(y+z)^2 -2 ; from (2) and (3) we have : \r\n z^2+x^2 -2 + x^2+y^2 -2 = y^2+2yz+z^2 -2 \r\n 2x^2 -2 =2yz /:2\r\n x^2 -1=yz \r\n (x-1)(x+1)=yz\r\n \r\nFrom here we have 3 cases \r\n 1*. x-1=1 and x+1=yz =>x=2 and 3=yz => (x,y,z)={(2,1,3),(2,3,1)}\r\n 2*. x-1=y and x+1=z (from using these equations at we find x^2=4 =>x=+-2 => (x,y,z)={(2,1,3),(-2,-3,1)}\r\n 3*. x-1=yz and x+1=1 => x=0 => (y,z)={(-1,1),(1,-1)}\r\n\r\nSo all solutions are :(x,y,z)={(0,-1,1),(0,1,-1),(2,1,3),(-2,-3,1),(2,3,1)}\r\n\r\n \r\nP.S- I did not check out the solution, some of results may be incorrect !", "Solution_6": "let us write the equation $ x(y \\plus{} z) \\equal{} y^2 \\plus{} z^2 \\minus{} 2$ on the form $ y^2 \\plus{} z^2 \\minus{} \\frac {1}{2}\\cos(120) yz \\equal{} xy \\plus{} yz \\plus{} zx \\plus{} 2$. The $ LHS$ of the given equation can be seen as the side of a triangle. Let us put:\r\n\r\n$ x^2 \\plus{} y^2 \\minus{} \\frac {1}{2}\\cos(120)xy \\equal{} a^2$\r\n$ y^2 \\plus{} z^2 \\minus{} \\frac {1}{2}\\cos(120)yz \\equal{} b^2$\r\n$ z^2 \\plus{} x^2 \\minus{} \\frac {1}{2}\\cos(120)zx \\equal{} c^2$\r\n\r\nso we can organize a, b, c, x, y, z as in the figure.\r\nAdding the equations\r\n$ a \\equal{} xy \\plus{} yz \\plus{} zx \\plus{} 2$\r\n$ b \\equal{} xy \\plus{} yz \\plus{} zx \\plus{} 2$\r\n$ c \\equal{} xy \\plus{} yz \\plus{} zx \\plus{} 2$\r\n\r\nwe get: $ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} 3(xy \\plus{} yz \\plus{} zx) \\plus{} 6$ (1)\r\nBut the area of the triangle in the picture is: $ \\frac {1}{2}xy\\sin(120) \\plus{} \\frac {1}{2}yz\\sin(120) \\plus{} \\frac {1}{2}zx\\sin(120) \\equal{} \\frac {\\sqrt {3}}{4}(xy \\plus{} yz \\plus{} zx)$\r\nAnd so if we call $ A$ this area we have $ xy \\plus{} yz \\plus{} zx \\equal{} \\frac {4}{\\sqrt {3}}A$\r\nAnd so (1) becomes: $ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} 4\\sqrt {3}A \\plus{} 6$\r\nbut a well known inequality states that $ a^2 \\plus{} b^2 \\plus{} c^2 \\leq 4\\sqrt {3}A$, so we conclude that at least one of a, b, c must be zero, and so at least one of x, y, z must be zero. From here the solution is easy" } { "Tag": [ "geometry", "cyclic quadrilateral" ], "Problem": "[b]1.[/b] Find the area of a quadrilateral ABCD inscribed in a circle with sides AB, BC, CD, and DA of 15, 20, 24, and 7 respectively.\r\n\r\n[b]2.[/b] In the harmonic sequence 6, 3, 2, 3/2, 6/5, . . . , what will the eighth term be?\r\n\r\n[b]3.[/b] Find the real values of x if x :^4: + 8x :^3: + 24x :^2: + 32x - 713 = 0\r\n\r\n[b]4.[/b] Let x = :sqrt:(3 + :sqrt:3) - :sqrt:(3 - :sqrt:3). Find 192x :^2: - x :^4:.\r\n\r\n[b]5.[/b] For x, y, z, and w:ge:0, compute the smallest value of x satisfying the following system:\r\n\r\ny = x - 2002\r\nz = 2y - 2002\r\nw = 3z - 2002", "Solution_1": "2. challenge. finding the nth term of a harmonic sequence (or indeed of any sequence where each term is the jth mean* of the two surrounding terms) is not very difficult, once this hint is used: [hide]considering taking all terms of that sequence to the jth power and seeing what you get[/hide]. However, try to find a formula for the sum of a jth mean series. I am not sure one exists, and it in fact seems rather unlikely for |j|:ge:2 (:sqrt:2+:sqrt:5+:sqrt:8+:sqrt:11+:sqrt:14, for example, doesn't seem likely to come out of a nice simple formula), but see if you can find one for j=-1 (the harmonic sequence).\n\n\n\n*the jth mean of two numbers a and b is defined as ((a^n+b^n)/2)^(1/n)", "Solution_2": "just bringing this up to the top.", "Solution_3": "[i][b]Why isn't anybody trying these problems?????[/b][/i]", "Solution_4": "Okay, I'll play. For #1, [hide]we have a cyclic quadrilateral, so we can use Brahmagupta's formula, which says that\n\nA = sqrt((s-a)(s-b)(s-c)(s-d)), where s=(a+b+c+d)/2\n\n\n\n(15+20+24+7)/2=66/2 = 33\n\n\n\nA = sqrt((33-15)(33-20)(33-24)(33-7)) = sqrt(18*13*9*26) = sqrt(2*9*13*9*2*13) = 2*9*13 = 234[/hide]", "Solution_5": "Maybe I'll have this in contest format.", "Solution_6": "Now try number 1 with only theorems you know how to prove ;)", "Solution_7": "And in so doing, see if you can use your approach to prove the cyclic version of Brahmagupta that you cited above.", "Solution_8": "Ok I'll try 4:\n\n[hide]\n\nI'm assuming you mean that each one of those is under a bigger radical, so here it goes:\n\nx = :sqrt: (3 + :sqrt: 3) - :sqrt: (3 - :sqrt: 3)\n\nx :^2: = 3+ :rt3: - 2 :sqrt: ((3 + :sqrt: 3)(3 - :sqrt: 3) + 3 - :rt3: \n\n= 6 - 2 :sqrt: 6\n\n\n\n192x :^2: - x :^4: = x :^2: (192 - x :^2:)\n\n=(6 - 2 :sqrt: 6)(192-6+2 :sqrt: 6)\n\n=(6 - 2 :sqrt: 6)(186+2 :sqrt: 6)\n\n=1092-360 :sqrt: 6[/hide]", "Solution_9": "On number 1:\n\n[hide]\n\nI drew a picture and recognized the adjacent sides as pythagorean triples 7, 24, 25 and 15, 20, 25, so they are 2 right triangles that share a hypotenuse. The area is then (15)(10)+7(12)=234. A bit easier than Brahmagupta \n\n[/hide]", "Solution_10": "1) Cheap solution:\n\n\n\n[hide]Notice that 72 + 242 = 252, and 152 + 202 = 252. Thus ADC and ABC could be two right triangles, with o. If AC is any longer than 25, both o, thus ABCD is not cyclic; similarly, if AC is any shorter than 25, both o, and ABCD won't be cyclic. Therefore 4 = 729 = 36. Thus x + 2 = :pm: 33/2, and x = 33/2 - 2 or x = -33/2 - 2.[/hide]", "Solution_11": "Yes, you said what I wanted to say on number one." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Eleven Teachers run a conference. Every hour, one or more teachers give a one-hour presentation, while all of the other teachers observe the presentations. Find the least amount of time during which it is possible for each teacher to observe all otherpresentations at least once.", "Solution_1": "What is this problem asking?? I think you need to polish the problem statement up a bit.", "Solution_2": "http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1670656#1670656" } { "Tag": [ "superior algebra", "superior algebra unsolved" ], "Problem": "Find infinitely many pairwise distinct extensions $M$ such that $F_{p}(x^{p},y^{p}) \\subset M \\subset F_{p}(x,y)$.", "Solution_1": "Consider $M_{c}= \\mathbb{F}_{p}(x^{p},y^{p},x+cy)$, for $c\\in\\mathbb{F}_{p}(x^{p},y^{p})$." } { "Tag": [ "limit", "geometry", "3D geometry", "combinatorics solved", "combinatorics" ], "Problem": "Some of the vertices of the unit squares of a n x n chessboard are colored such that any k x k square formed by these unit squares has a colored point on at least one of its sides. If l(n) denotes the minimum number of colored points require to ensure the above condition, prove that \\lim n -> 00 l(n)/n 2 = 2/7.", "Solution_1": "This one has been proposed earlier in Math Magazine (pb. 1296, proposed in 1988. Solved in 1989).\r\n\r\nLet's say that we color the points in red. And suppose that we have colored l(n) points to have the desired property.\r\n\r\nLet P be a red point. Let S be a unit square with P as one of its vertices. Let n(P,S) be the number of red points which are on the boundary of S (in fact its vertices), including P, so that n(P,S) > 0.\r\nLet N(P) = \\sum 1/n(P,S) , where the sum is over all the unit squares with P as vertex.\r\nWe have :\r\n \\sum {P red} N(P) = \\sum {P red} ( \\sum {S such that P vertex of S} 1/n(P,S) )\r\n= \\sum {S unit square} ( \\sum {P red and vertex of S} 1/n(P,S) )\r\n= \\sum {S unit square} 1\r\n= n 2 . (1)\r\n\r\nIn another hand :\r\n- If P is a corner of the chessboard, then clearly N(P) \\leq 1 (there is only one unit square having P as vertex)\r\n- If P is on the boundary of the chessboard, but not a vertex, then there are two unit squares having P as vertex, so that N(P) \\leq 2.\r\n- If P is an interior point of the chessboard, then there are 4 unit squares having P as vertex. But the 2x2 square centered at P must have at least one red point on its boundary, say Q. If P and Q are adjacent vertices of a unit square, then they are for two, so that N(P) \\leq 3. If P and Q are opposite vertices of an unit square, then N(P) \\leq 7/2.\r\nThus, for each red point P, we have N(P) \\leq 7/2.\r\n\r\nIt follows that \\sum {P red} N(P) \\leq 7*l(n)/2, and using (1), we deduce that l(n) \\geq 2n 2 /7. (2)\r\n\r\nConversely, let's consider the following 7x7 square, where x stands for a non colored point, and r for a red point :\r\n\r\nxxxrxrx\r\nxrxxxxr\r\nxxrxrxx\r\nrxxxxrx\r\nxrxrxxx\r\nxxxxrxr\r\nrxrxxxx\r\n\r\nNote that is has the desired property with 14 red points.\r\n\r\nAny nxn chessboard can be considered as a corner of a mxm chessboard with m = 0 mod[7] and n \\leq m+6. Let m = 7p.\r\nWe use p 2 copies of the 7x7 chessboard above to tile the mxm chessboard (by translated copies...), so we have a total of 14p 2 = 2m 2/7 red points, and it is easy to verify that any kxk square on the chessboard has at least a red point on its boundary. Thus the mxm chessboard has the desired property, so that the nxn chessboard has too even if we delete m-n rows or columns from the top and from the right, respectively.\r\nThus l(n) \\leq 2m 2 /7 \\leq 2(n+6) 2 /7. (3)\r\n\r\nFrom (2) and (3), we then deduce that lim l(n)/n 2 = 2/7.\r\n\r\nNote that it can be proved that in higher dimensions (square boundaries are replaced by surfaces of cubes), the general limit is 2/(2 (d+1) - 1), where d is the dimension of the space.\r\n\r\nPierre." } { "Tag": [ "email", "USAMTS" ], "Problem": "Our office will be closed on Thursday and Friday this week due to Thanksgiving. \r\n\r\nWe will try to process as many of the Round 2 submissions as we can by Wednesday. But please do not panic if you do not see the checkmark when you log in at http://www.usamts.org; we may not be able to process them all on Wednesday. And, needless to say, if you don't have the checkmark on Wednesday, then it will not show up until next Monday (at the earliest).", "Solution_1": "Update: all of the Round 2 papers have been checked in.\r\n\r\nIf you submitted Round 2, but do not have a green checkmark next to your name, then there was some sort of problem with your submission. Please email usamts@usamts.org (or in many cases we have already emailed you) to find out what the status is. When doing so, please include your USAMTS ID#, name, and the date, time, and method that you submitted your solutions." } { "Tag": [ "group theory", "abstract algebra", "modular arithmetic", "linear algebra", "matrix", "algebra", "polynomial" ], "Problem": "What is struture of Sylow 2-group of simple group L_2(q) with q odd?", "Solution_1": "Set $ Z: \\equal{} \\{\\pm I_2\\}$ the centre of $ SL_2(q)$.\r\n\r\nAssume $ q\\equiv 1\\pmod{4}$ and take $ \\langle \\zeta\\rangle$ the Sylow 2-subgroup of the cyclic group $ \\mathbb{F}_q^\\times$.\r\nThen $ |\\langle \\zeta\\rangle | \\equal{} 2^n$ with $ n\\ge 2$ and $ i: \\equal{} \\zeta^{2^{n \\minus{} 2}}$ satisfies $ i^2 \\equal{} \\minus{} 1$.\r\nThen $ r: \\equal{} {\\zeta \\plus{} \\zeta^{ \\minus{} 1}\\over 2},s: \\equal{} {\\zeta \\minus{} \\zeta^{ \\minus{} 1}\\over 2i}\\in\\mathbb{F}_q$ satisfy $ r \\plus{} is \\equal{} \\zeta,r \\minus{} is \\equal{} \\zeta^{ \\minus{} 1}$ and $ r^2 \\plus{} s^2 \\equal{} (r \\plus{} is)(r \\minus{} is) \\equal{} 1$.\r\nSo $ A: \\equal{} \\begin{pmatrix}r & s \\\\\r\n\\minus{} s & r\\end{pmatrix}\\in SL_2(q)$ has characteristic polynomial $ X^2 \\minus{} 2rX \\plus{} 1 \\equal{} (X \\minus{} \\zeta)(X \\minus{} \\zeta^{ \\minus{} 1})$ and order $ 2^n$.\r\nThis gives $ a: \\equal{} AZ\\in PSL_2(q)$ of order $ 2^{n \\minus{} 1}$ and $ b: \\equal{} \\begin{pmatrix}i & \\\\\r\n& \\minus{} i\\end{pmatrix}Z\\in PSL_2(q)$ of order $ 2$ with $ b^{ \\minus{} 1}ab \\equal{} a^{ \\minus{} 1}$.\r\nSo we find the [url=http://en.wikipedia.org/wiki/Dihedral_group]dihedral group[/url] $ \\langle a\\rangle\\rtimes \\langle b\\rangle$ of order $ 2^n$ as Sylow 2-subgroup of $ PSL_2(q)$ with central involution $ \\begin{pmatrix} & 1 \\\\\r\n\\minus{} 1 & \\end{pmatrix}Z$.\r\n\r\nAnalog for $ q\\equiv \\minus{} 1\\pmod{4}$ we take $ \\langle \\zeta\\rangle$ the Sylow 2-subgroup of $ \\mathbb{F}_{q^2}^\\times$ of order $ 2^n,n\\ge 3$\r\nand $ i: \\equal{} \\zeta^{2^{n \\minus{} 2}}$ with $ i^2 \\equal{} \\minus{} 1$.\r\nBut $ \\mathbb{F}_{q^2}/\\mathbb{F}_q$ is galois with the only nontrivial automorphism $ x\\mapsto x^q$,\r\nwhich fixes $ r: \\equal{} {\\zeta^2 \\plus{} \\zeta^{ \\minus{} 2}\\over 2},s: \\equal{} {\\zeta^2 \\minus{} \\zeta^{ \\minus{} 2}\\over 2i},t: \\equal{} {\\zeta \\minus{} \\zeta^{ \\minus{} 1}\\over 2},u: \\equal{} {\\zeta \\plus{} \\zeta^{ \\minus{} 1}\\over 2i}\\in\\mathbb{F}_{q^2}$.\r\nSo $ r,s,t,u\\in\\mathbb{F}_q$ with $ r \\plus{} is \\equal{} \\zeta^2,r \\minus{} is \\equal{} \\zeta^{ \\minus{} 2},t \\plus{} iu \\equal{} \\zeta,t \\minus{} iu \\equal{} \\minus{} \\zeta^{ \\minus{} 1}$ \r\nand $ r^2 \\plus{} s^2 \\equal{} (r \\plus{} is)(r \\minus{} is) \\equal{} 1$ and $ t^2 \\plus{} u^2 \\equal{} (t \\plus{} iu)(t \\minus{} iu) \\equal{} \\minus{} 1$.\r\nSo $ A: \\equal{} \\begin{pmatrix}r & s \\\\\r\n\\minus{} s & r\\end{pmatrix},B: \\equal{} \\begin{pmatrix}t & u \\\\\r\nu & \\minus{} t\\end{pmatrix}\\in SL_2(q)$ have characteristic polynomials \r\n$ X^2 \\minus{} 2rX \\plus{} 1 \\equal{} (X \\minus{} \\zeta^2)(X \\minus{} \\zeta^{ \\minus{} 2})$ and $ X^2 \\plus{} 1 \\equal{} (X \\minus{} i)(X \\plus{} i)$.\r\nThis gives $ a: \\equal{} AZ,b: \\equal{} BZ\\in PSL_2(q)$ of orders $ 2^{n \\minus{} 2},2$\r\nand $ AB \\equal{} BA^{ \\minus{} 1}\\equal{}\\begin{pmatrix}rt\\plus{}su & ru\\minus{}st \\\\\r\nru\\minus{} st & \\minus{}rt\\minus{}su\\end{pmatrix}$ gives $ b^{ \\minus{} 1}ab \\equal{} a^{ \\minus{} 1}$.\r\nSo again we get the [url=http://en.wikipedia.org/wiki/Dihedral_group]dihedral group[/url] $ \\langle a\\rangle\\rtimes \\langle b\\rangle$ of order $ 2^{n \\minus{} 1}$ as Sylow 2-subgroup of $ PSL_2(q)$ \r\nwith central involution $ \\begin{pmatrix} & 1 \\\\\r\n\\minus{} 1 & \\end{pmatrix}Z$." } { "Tag": [ "logarithms", "complex numbers", "imaginary numbers" ], "Problem": "[b]How can I arrive from the proper way of solving this kind of logarithmic equation: $log_2(4-x)-log_4(x-1)=1$. What is x in this case? I use trial and error and I get x=2 but how can I solve it in proper way because the case is they have different base?[/b]", "Solution_1": "by the properties of logs, $log_ab=\\frac{log_cb}{log_ca}$. So $\\displaystyle log_4(x-1)=\\frac{log_2(x-1)}{log_24}$", "Solution_2": "Hydron, I get x=10 and x=2....What is the valid solution?", "Solution_3": "$\\log_4 x = \\frac{ \\log_2 x }{\\log_2 4 } = \\frac 1 2 \\log_2 x$\r\n$\\implies \\frac{(x-4)^2}{x-1} = 4$\r\n$x^2 -12x +20 = 0 \\implies x = 10,2$", "Solution_4": "10 is an extraneous solution because the first logarithm wil then be $\\log_2 -6$, which is an imaginary number that does not satisfy the requirements. 2 is the only valid solution.", "Solution_5": "[quote=\"E^(pi*i)=-1\"]10 is an extraneous solution because the first logarithm wil then be $\\log_2 -6$, which is an imaginary number that does not satisfy the requirements. 2 is the only valid solution.[/quote]\r\n\r\nYea, you can't have negatives there" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "$ x,y,z$ are positive reals and $ \\sum x=1$\r\n\r\nProof: $ \\sum \\frac{xy}{\\sqrt {xy+yz}} \\le \\frac{3 \\sqrt{3}}{4} \\sqrt{(x+y)(y+z)(z+x)}$", "Solution_1": "[url]http://www.mathlinks.ro/viewtopic.php?t=216138[/url]" } { "Tag": [ "modular arithmetic", "number theory proposed", "number theory" ], "Problem": "Does there exist a multiple of $ 103$, say $ n$, such that $ 2^{2n \\plus{} 1} \\equiv 2 \\pmod{n}$?", "Solution_1": "If $ n$ is even, then $ n\\equal{}2m$, were m odd. Let $ m\\equal{}n$ if n is odd.\r\nThen condition equavalent to $ T_m|2n, \\ 103|m$, were $ T_k$ mean minimal period for 2 mod k. Becouse $ 51\\equal{}T_{103}|T_m|2n$, we get $ 51|m$. It give $ 8\\equal{}T_{17}|T_m$ or $ 8|4m$ contradition with odd m." } { "Tag": [ "linear algebra", "matrix" ], "Problem": "I am planning to take the October SAT II in Mathematics 2. \r\n\r\nI want to know how much detail and rigor is need for the exam. I've seen the questions - they are easy. I'm asking about the topics - how many different, which ones and in what detail should they be done. Which topics are covered (the official syllabus is a bit vague)Is the Barron's book perfect. Where can I get more practice??\r\n\r\nPlease help! I can clarify my question if you want.", "Solution_1": "I don't think you'll have much of a problem with the test. Most people on this forum would consider it easy, I believe. Although I haven't taken it yet and i haven't prepared very much for it, from what I've seen it doesn't look too bad.\r\n\r\nAs for the test, it's all multiple choice and probably grid-ins. Therefore, since there aren't any proofs, short answers, or whatever, you don't have to be rigorous at all. The official syllabus should be enough to tell you the topics. And Barron's should be adequate.", "Solution_2": "I was very surprised to find a matrix question on my test. It is not covered anywhere in my high school curriculum, and it's not explicitly listed on the syllabus that I found online. I made a lucky guess and got it right. It was only one question, though, and considering how the raw scores are curved, I think I would have gotten 800 even if I got it wrong.", "Solution_3": "The Math 2C has had one matrix question on about half of the most recent administrations. If you know how to compute the determinant of a 2x2 matrix, and know that multiplying a 2x3 matrix with a 3x4 matrix gives a 2x4 matrix, then you should be okay.\r\n\r\nAnother fairly recent topic is standard deviations. You don't need to know any quantitative formulas for it. Just know that standard deviation measures how far apart numbers are. In particular, numbers bunched together have small standard deviation, and numbers spread apart have large standard deviation.", "Solution_4": "http://augustusmath.hypermart.net/tofchtml.html\r\n\r\nDo these topics cover EVERYTHING tested on the SATII Math 2c exam?", "Solution_5": "Any one precalculus book may or may not cover all the topics that are fair game on the SAT II Math Level 2 test. But don't worry about that too much. It is possible to get the highest scaled score, 800, even if you miss a few of the items on the test. That's how the scaled scores work on that test. \r\n\r\nI'm sure that you want to get a good review of precalculus math to learn that math, and you have found some good sources. They will help you learn more math, and that is important. Don't worry too much about the test--that will just spoil the fun of learning more math. You should do well on the test if you focus on the math ideas and how interesting they are while you are studying. \r\n\r\nBest wishes for your math studies. Good luck on the test.", "Solution_6": "Any online resources for practice???", "Solution_7": "I wrote the SAT II Math Test today..\r\noverall, it wasn't very hard\r\nI was glad that I read this post. :D \r\nThere was a question on STANDARD DEVIATION..", "Solution_8": "so, when i took the Math SAT IIs i first thought i shoudl take the IC exam, thinking it would be easy and i would do well- but no... i got a 750... (that was because i did something like 6+6=18, but that's not the point- i think i missed like 3 questions) then i decided that being on math team with a 750 on math SAT II is a bad thing, so i took the IIC- i finished pretty fast, took a 10 minute bathroom break (because they wouldn't let me go!) and then a short nap- that was good relaxation after writing and physics ;)" } { "Tag": [ "inequalities", "algebra", "binomial theorem", "inequalities open" ], "Problem": "I need a solution to this problem as soon as possible. I think there's a trivial proof, but I can't find it. Can you?\r\n\r\nProve or disprove that ${a \\choose b} \\leq (a/b)^{2b}$, when $b \\leq a/2$.", "Solution_1": "Here is one approach. By the binomial theorem, we know that ${a \\choose b} p^b (1-p)^{a-b} \\leq 1$ for every p between 0 and 1. Choose $p = b/a$ to optimize. We get the inequality\r\n\\[{a \\choose b} \\leq \\frac{a^a}{b^b (a-b)^{a-b}} .\\]\r\nThat bound turns out to be better than yours. Do you see how to compare the two bounds?", "Solution_2": "The inequality between the two bounds is equivalent to $(x+1)^{x-1} \\leq x^x$, for $x \\geq 1$. I suppose this is trivial...\r\nThanks a lot!", "Solution_3": "It could also be proved by a two-fold induction..." } { "Tag": [ "calculus", "integration" ], "Problem": "Balance the following unbalanced equations: $ \\ce{S2Cl2} \\plus{} \\ce{NH3} \\rightarrow \\ce{S4N4} \\plus{} \\ce{S8} \\plus{} \\ce{NH4Cl}$, and $ \\ce{KCl} \\plus{} \\ce{HNO3} \\plus{} \\ce{O2} \\rightarrow \\ce{KNO3} \\plus{} \\ce{Cl2} \\plus{} \\ce{H2O}$ (state symbols ommitted).\r\n\r\nAlso, I don't understand this:\r\n\r\n[hide=\"question\"]Balance the unbalanced chemical equation: $ \\ce{NH3}\\plus{}\\ce{O2} \\rightarrow \\ce{NO}\\plus{}\\ce{H2O}$.[hide=\"solution\"]Firstly, I balanced the hydrogen and nitrogen to get $ \\ce{2NH3}\\plus{}\\ce{O2} \\rightarrow \\ce{2NO}\\plus{}\\ce{3H2O}$. I then added a stoichiometric coefficient of $ \\frac{1}{2}$ to the $ O2$ on the LHS, and then I multiplied the whole equation by 2 to get integral stoichiometric coefficients. Thus the answer is $ \\ce{4NH3}\\plus{}\\ce{O2} \\rightarrow \\ce{2NO}\\plus{}\\ce{3H2O}$. Why can we simply add $ \\frac{1}{2}$ and multiply the equation. Also, in what circumstances do I need to add non-integer real values of $ x$ as stoichiometric coefficients? Please give some practice problems. [/hide][/hide]\r\n\r\nDoes anyone know where to get the answers for the problems in the textbook Chemistry: Principles, Patterns and Applications (volume 1) by Bruce Averill and Patricia Eldredge?", "Solution_1": "for second\r\n$ 4\\ce{KCl} \\plus{} 4\\ce{HNO3} \\plus{} \\ce{O2} \\rightarrow 4\\ce{KNO3} \\plus{} 2\\ce{Cl2} \\plus{} 2\\ce{H2O}$\r\nfor first\r\n$ 6\\ce{S2Cl2} \\plus{} 16\\ce{NH3} \\rightarrow \\ce{S4N4} \\plus{} \\ce{S8} \\plus{} 12\\ce{NH4Cl}$\r\n\r\nfor doing these, [u]STEP I[/u] first see which element is having most atoms unbalanced and balance it.\r\n[u]STEP II [/u] repeat step I.", "Solution_2": "Which element did you first balance? Any advice on which elements to balance first (generally) in a given unbalanced chemical equation?", "Solution_3": "[hide]$ 6S_2Cl_2 \\plus{} 16NH_3 \\minus{}> S_4N_4\\plus{}S_8\\plus{}12NH_4Cl$[/hide]", "Solution_4": "theta pi, u can READ THE EDITED POST.for wut u wanted 2 know.\r\n\r\ni have first heard the word stoichiometric coefficients :oops: . i thought u had great knowledge as it is a powerful word.but when i googled it i came 2 know that it was nothing. :rotfl: i jus got decieved by a powerful sounding word. not criticizing y're knowledge though :|", "Solution_5": "My chemistry teacher taught us that. It's no use really. That's why I don't really like Chemistry: too many unnecessary definitions, like Avogadro's number equalling a mole etc., formula mass, molecular mass, atomic mass, blah. Oh please! They are all masses!", "Solution_6": "[quote=\"ThetaPi\"]\nAlso, I don't understand this:\n\nBalance the unbalanced chemical equation: $ \\ce{NH3} \\plus{} \\ce{O2} \\rightarrow \\ce{NO} \\plus{} \\ce{H2O}$.[hide=\"solution\"]Firstly, I balanced the hydrogen and nitrogen to get $ \\ce{2NH3} \\plus{} \\ce{O2} \\rightarrow \\ce{2NO} \\plus{} \\ce{3H2O}$. I then added a stoichiometric coefficient of $ \\frac {1}{2}$ to the $ O2$ on the LHS, and then I multiplied the whole equation by 2 to get integral stoichiometric coefficients. Thus the answer is $ \\ce{4NH3} \\plus{} \\ce{O2} \\rightarrow \\ce{2NO} \\plus{} \\ce{3H2O}$. Why can we simply add $ \\frac {1}{2}$ and multiply the equation. Also, in what circumstances do I need to add non-integer real values of $ x$ as stoichiometric coefficients? Please give some practice problems. [/hide]\n\n[/quote]\r\n\r\nerr after you balanced H and N there, you can see that there are 5 Os in the right hand side and 2 on the left. So now you want to make it a 5,so multiply it by $ 5\\frac{1}{2}$ and then to get integers multiply the whole thing by 2.", "Solution_7": "theta pi, that happens with everyone. u have learn those confusing things. but they are basics. and wud be applied afterwards. however , i enjoy chemistry now, learning about carbon and all, \r\n\r\nshadysaysurspammed pointed a good thing there that whenever u have a element unbalanced (isolated on one side i.e. either only element or homoatomic compound) and on one side the element is a fraction of that of other side, multiply by whole equation by the reciprocal of that fraction.", "Solution_8": "[quote=\"ThetaPi\"]My chemistry teacher taught us that. It's no use really. That's why I don't really like Chemistry: too many unnecessary definitions, like Avogadro's number equalling a mole etc., formula mass, molecular mass, atomic mass, blah. Oh please! They are all masses![/quote]\r\n\r\nLet's be fair if they are all exactly the same thing then people won't bother with different names :D .\r\n\r\nAnyway, Chemistry is my favourite (science wise) and it becomes really interesting when you delve into deep theories of bonding and organic chemistry.", "Solution_9": "Mathematics is the best, admit it." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "a,b,c are real number, they are not equal and a,b,c>o\r\nprove:\r\n(m^2)/(mp+mq-pq)+(p^2)/(mp+pq-mq)+(q^2)/(qm+pq-mp)<3\r\n :maybe:", "Solution_1": "I will assume m=a, b=p, c=q. Then we want to prove\r\n\r\n\\[ \\sum_{\\text{cyc}}\\frac{a^{2}}{ab+ac-bc}<3\\]\r\n\r\nOkay, try $ a=b=1, c=\\frac{1}{2}+\\epsilon$. The expression goes to infinity.\r\nUnless m, p, and q are some totally different numbers, it appears the inequailty doesn't hold (even if \"not equal\" meant \"no two are equal\", just let $ b$ be really close to $ a$)." } { "Tag": [ "geometry", "parallelogram" ], "Problem": "Here is another one:\r\n The sides of AB and BC of \\[ \\triangle\\]ABC", "Solution_1": "hey wats the ques????", "Solution_2": "Please tell me how I can draw an image of a geometric figure and insert it here.Then I can post the sum and also what i have done .Sorry for the incomplete sum.", "Solution_3": "see the \"Geogebra\" forum--and u'll understand better :D", "Solution_4": "Here is the correct sum: The sides AB and Ac of $ \\triangle$ ABC are respectively equal and parallel to the sides XY and ZX of the $ \\triangle$ XYZ.Show that AC and XY are equal and parallel.\r\n\r\nThis is what I have done. I have joined Z and B,Z and A,A and X,X and C,C and Y and B and B and Y.\r\nXZ $ \\|$ BC and XZ = BC.\r\nSo, XZBC is a parallelogram.\r\nZB=XC\r\n\r\nAB $ \\|$ XY and AB=XY\r\nSo, ABYX is a parallelogram.\r\nAX=BY\r\nWhat next?\r\nAm I going in the right direction?\r\nIf yes, how should this be solved?In fact, I am trying to prove that $ \\triangle$ ZBY $ \\cong$ $ \\triangle$ XYC and that $ \\triangle$ AZB $ \\cong$ $ \\triangle$ XCY.\r\nThank you.\r\n\r\n [geogebra]562df5d3508734afa260e4d19a39ebcc95bbd847[/geogebra]", "Solution_5": "The problem isn't true. What they required to be proven should be $ BC$ and $ YZ$ are equal and parallel.\r\n\r\nAs a general note on geometry, never draw a special case. It is sometimes misleading as it shows special properties which is not true in general, but provokes you to try to prove them. Your figure has the triangles equilateral, which is kind of a special case. And also the other triangle is kind of symmetric to the main triangle. That is why it seemed $ \\Delta ZBY\\cong \\Delta XYC$. But unfortunately this doesn't hold in general. So you can't prove it. \r\n\r\nTry this picture:\r\n\r\n [geogebra]f18ee2ab6d2671a38c77fd7a524046f1ede33ca4[/geogebra] \r\n\r\nNote that you can move around the blue points and still the conditions of the problem hold (that is because I defined the points precisely in geogebra) . Though this luxury is not available with pencil-paper, it's best if you draw a figure that is not a special case. For example draw triangles with no 2 sides equal, no right angles etc. like my figure.\r\n\r\nYou were in the right direction. Try to prove the two parallelograms again. But there was some mistake in there before. You stated at first that $ XZ\\parallel{}BC$ but that isn't given.\r\n\r\nI've joined the points which should be joined (red lines), that is what you too wanted to, but messed up a bit because of the misleading diagram.", "Solution_6": "another advice.. muks.. \r\ni hope that u have actually tried the problems for some time b4 asking for solutions... \r\nsome time = quite some time...", "Solution_7": "I am sorry.I unnecessarily made this complex while the sum was quite easy.In fact, having made the sum complex, I worked on it for quite some time. \r\nThank you.(I will be very careful next time before posting a sum)", "Solution_8": "[quote=\"muks\"]I will be very careful next time before posting a sum[/quote]\r\nNo really. You need not be all that careful. Though Keshav is right you should spend some time yourself, but that was meant for your own improvement. Don't take this as wasting someone's time. You are always welcome to post any problem in here. :)", "Solution_9": "muks make good use of Akashnil's experience here. Frankly to say, I have become better at math than what i was in the beginning of 9th std only after being inspired by Akashnil.", "Solution_10": "It seems people are here to help! Thank you. If I am not wrong, I will have to complete the class 12 syllabus before May-June while I have completed $ \\frac {2}{3}$ of class 9 syllabus and $ \\frac {3}{11}$ of Class 10 syllabus.\r\n\r\nThank you again.I will make it a point to really elevate my standard,which is very poor at the moment.", "Solution_11": "[quote=\"muks\"]I will have to complete the class 12 syllabus before May-June [/quote]\r\nFor what? If you think that you'll have to complete the class 12 syllabus before May-June for Math Olympiads, I can assure you that you are wrong :o :P :wink: .", "Solution_12": "Well muks, u have a completely wrong understanding of olympiad maths......olympiad maths is not about theory at all.....though u need to know some basics in all fields...its about problem solving...", "Solution_13": "Then I will give myself another year after May-June. In fact, I do not have any idea of this RMO and such olympiads.", "Solution_14": "[quote=\"muks\"]Then I will give myself another year after May-June.[/quote]\r\nNo. What I meant to say is that you need not even complete upto class 12 syllabus at all for math olympiad.", "Solution_15": "Akashnil, why do u go offline in just 1 min or something and again come online and go offline? \r\nBTW, ur soln for my $ \\pm1\\pm2....$ problem can be made more rigorous by a simple statement which is very very very very obvious but still to be written for rigority. Guess urself......\r\nThere is only 1 thing tht prevents u from the IMO team : 100% rigority in proof writing. Please take it in the right sense. I am just saying as a friend, though I am not even fit to comment about someone who know 1000...0000{100 zeros} times the theory I know and is the same no. of times better than me.", "Solution_16": "[quote=\"Aravind Srinivas L\"]Akashnil, why do u go offline in just 1 min or something and again come online and go offline? [/quote]\nDunno. I think it is some bug on the site. I never do that. It just displays wrong things to you. I've got the web-browser to remember my passwords and stuff. So it auto logs in when I open the site. o that can't happen right?\n\n[quote=\"Aravind Srinivas L\"]BTW, ur soln for my $ \\pm1\\pm2....$ problem can be made more rigorous by a simple statement which is very very very very obvious but still to be written for rigority. Guess urself......[/quote]\r\nCan't. :(" } { "Tag": [ "LaTeX", "irrational number", "number theory unsolved", "number theory" ], "Problem": "This guy insists that he found much simpler way to prove Fermat's Last Theorem, not only one method, but two, and he says he also proved 4 Color Theorem without using computer. Please find some errors so that this guy can live usual life!! This guy is demonstrating alone for 18 weeks for these proof to be accepted!!\r\n\r\n\r\n\r\n\r\n[b]4 Color Theorem proof of the regions on global surface [/b] \r\n3 inter relationship of the regions on global surface \r\nsharing a common boundary line. sharing a common point. not adjacent regions. \r\n[1] 4 colors suffice for the distinguishing all regions about one region and the regions those share a common one region`s boundary line. \r\n[Proof] The reason is this. 3 colors suffice for the distinguishing all regions those share a common one region`s boundary line. \r\n[Example1] 3 colors suffice for the distinguishing all hexagonal shape regions. The reason is this. 2 colors suffice for the distinguishing 6 hexagonal shape regions those share a common one hexagonal shape region`s boundary line. \r\n[Example2] 2 colors suffice for the distinguishing all square shape regions. The reason is this. One color suffice for the distinguishing 4 square shape regions those share a common one square shape region`s boundary line. \r\n[2] 3 colors suffice for the distinguishing all regions those share a common one region`s boundary line. \r\n[Proof] The reason is this. When there are lined from a region`s inner one point to the points on one region`s boundary line those are met the adjacent regions boundary lines, all extension regions are the regions those share one common point. 3 colors suffice for the distinguishing all regions those share one common point. \r\n[3] 3 colors suffice for the distinguishing all regions those share one common point. \r\n[Proof] The reason is this. When one region is selected, 2 colors suffice for the distinguishing all regions those share a common one selected region`s boundary line. \r\n\r\n\r\n[b]2 methods of FLT proof and Pythagorean triples [/b]\r\n$X^{n}+Y^{n}=Z^{n}$\r\nFermat had made a proof that the equation cannot have nonzero natural number solution in the even number n that is greater or equal 4. Therefore we need to make a proof that the equation cannot have nonzero natural number solution in the odd and prime number n. \r\n$Y+A=X+B=Z, A=Z-Y, B=Z-X X-A=Y-B=Z-A-B=X+Y-Z G=(X-A)/(AB)^{(}1/n)=(Y-B)/(AB)^{(}1/n)=(Z-A-B)/(AB)^{(}1/n)=(X+Y-Z)/(AB)^{(}1/n) X=G(AB)^{(}1/n)+A, Y=G(AB)^{(}1/n)+B, Z=G(AB)^{(}1/n)+A+B{G(AB)^{(}1/n)+A}^{n}+{G(AB)^{(}1/n)+B}^{n}={G(AB)^{(}1/n)+A+B}^{n}$\r\nWhen n=1, G=0. When n=2, G=2^(1/2). When n>2, G=function(A,B) is the positive real number. \r\n$X=(2AB)^{(}1/2)+A, Y=(2AB)^{(}1/2)+B, Z=(2AB)^{(}1/2)+A+B$\r\n(X,Y,Z) are the irrational numbers or all Pythagorean triples in all natural number (A,B). \r\nWe can translate the upper form into this. \r\n$AB=2k^{2}, B=2k^{2}/A X=2k+A, Y=2k(k+A)/A, Z=2k+A+2k^{2}/A XY=2k(2k+A)(k+A)/A$\r\nWhen A is the odd number, $k=hA, XY=2A^{2}h(2h+1)(h+1)$and $hk=A, XY=2k^{2}(2+h)(1+h)/h$\r\nWhen A is the even number, $2k=hA, XY=A^{2}h(h+1)(h+2)/2$and $2hk=A, XY=2k^{2}(1+h)(1+2h)/h$\r\nTherefore XY cannot be the power numbers in all Pythagorean triples. \r\n[b]* * * * * 1st method of FLT proof * * * * * [/b]\r\n$G(AB)^{(}1/n)$ is the irrational number in all natural number (A,B), so (X,Y,Z) are the irrational numbers. \r\n${G(AB)^{(}1/n)+A}^{n}+{G(AB)^{(}1/n)+B}^{n}={G(AB)^{(}1/n)+A+B}^{n}$\r\nWhen $A=B, 2{GA^{(}2/n)+A}^{n}={GA^{(}2/n)+2A}^{n}{2^{(}1/n)-1}GA^{(}2/n)={2-2^{(}1/n)}A G=[2^{(n-1)/n}+...+2^{(}2/n)+2^{(}1/n)]A^{(n-2)/n}$\r\nWe make new form with $[2^{(n-1)/n}+...+2^{(}2/n)+2^{(}1/n)]. [2^{(n-1)/n}+...+2^{(}2/n)+2^{(}1/n)][{A^{(}n-1)B}^{(}1/n)+{AB^{(}n-1)}^{(}1/n)]/2$\r\nThis form is the irrational number in all natural number (A,B). \r\n$G(AB)^{(}1/n)$ divide and multiply by $[2^{(n-1)/n}+...+2^{(}2/n)+2^{(}1/n)][{A^{(}n-1)B}^{(}1/n)+{AB^{(}n-1)}^{(}1/n)]/2$, and now we can get two forms. And when $A=B, q=1. G(AB)^{(}1/n)=q[2^{(n-1)/n}+...+2^{(}2/n)+2^{(}1/n)][{A^{(}n-1)B}^{(}1/n)+{AB^{(}n-1)}^{(}1/n)]/2 q=2G(AB)^{(}1/n)/[2^{(n-1)/n}+...+2^{(}2/n)+2^{(}1/n)][{A^{(}n-1)B}^{(}1/n)+{AB^{(}n-1)}^{(}1/n)]$\r\nIf G(AB)^(1/n) is the natural number (N) in some (a,b), $G(ab)^{(}1/n)=N$ can not have $[2^{(n-1)/n}+...+2^{(}2/n)+2^{(}1/n)]$, and $G(AB)^{(}1/n)$can not have $[2^{(n-1)/n}+...+2^{(}2/n)+2^{(}1/n)]$. \r\nSo when A=B, q cannot be 1. That is an apparent contradiction. \r\nTherefore $G(AB)^{(}1/n)$ is the irrational number in all natural number (A,B). 1st method. end. \r\n[b]* * * * * 2nd method of FLT proof * * * * * [/b]\r\n$X^{n}+Y^{n}=Z^{n}{X^{(}n/2)}^{2}+{Y^{(}n/2)}^{2}={Z^{(}n/2)}^{2}$\r\nWhen n=2, we can display ${X^{(}n/2),Y^{(}n/2),Z^{(}n/2)}$ with (a,b). \r\n$a=Z^{(}n/2)-Y^{(}n/2), b=Z^{(}n/2)-X^{(}n/2) X^{(}n/2)=(2ab)^{(}1/2)+a, Y^{(}n/2)=(2ab)^{(}1/2)+b, Z^{(}n/2)=(2ab)^{(}1/2)+a+b$\r\nWhen n is the prime number and (X,Y,Z) is co prime, the ab is the irrational number. \r\n$ab=Z^{n}-(YZ)^{(}n/2)-(XZ)^{(}n/2)+(XY)^{(}n/2)$\r\nWe multiply $X^{(}n/2)$ and $Y^{(}n/2)$. \r\n$(XY)^{n}=2a^{3}b+2ab^{3}+13(ab)^{2}+6ab(a+b)(2ab)^{(}1/2)$\r\nIf (X,Y,Z) is the natural number, $(XY)^{n}$ is the natural number, but $2a^{3}b+2ab^{3}+13(ab)^{2}+6ab(a+b)(2ab)^{(}1/2)$ is the irrational number. \r\nThat is an apparent contradiction. \r\nTherefore (X,Y,Z) must be the irrational number. 2nd method. end.\r\n\r\n\r\nThere is some problem with latex..So..\r\n\r\n\r\n\r\n\r\nOriginal text:\r\n\r\n4 Color Theorem proof of the regions on global surface \r\n3 inter relationship of the regions on global surface \r\nsharing a common boundary line. sharing a common point. not adjacent regions. \r\n[1] 4 colors suffice for the distinguishing all regions about one region and the regions those share a common one region`s boundary line. \r\n[Proof] The reason is this. 3 colors suffice for the distinguishing all regions those share a common one region`s boundary line. \r\n[Example1] 3 colors suffice for the distinguishing all hexagonal shape regions. The reason is this. 2 colors suffice for the distinguishing 6 hexagonal shape regions those share a common one hexagonal shape region`s boundary line. \r\n[Example2] 2 colors suffice for the distinguishing all square shape regions. The reason is this. One color suffice for the distinguishing 4 square shape regions those share a common one square shape region`s boundary line. \r\n[2] 3 colors suffice for the distinguishing all regions those share a common one region`s boundary line. \r\n[Proof] The reason is this. When there are lined from a region`s inner one point to the points on one region`s boundary line those are met the adjacent regions boundary lines, all extension regions are the regions those share one common point. 3 colors suffice for the distinguishing all regions those share one common point. \r\n[3] 3 colors suffice for the distinguishing all regions those share one common point. \r\n[Proof] The reason is this. When one region is selected, 2 colors suffice for the distinguishing all regions those share a common one selected region`s boundary line. \r\n\r\n\r\n2 methods of FLT proof and Pythagorean triples \r\nX^n+Y^n=Z^n \r\nFermat had made a proof that the equation cannot have nonzero natural number solution in the even number n that is greater or equal 4. Therefore we need to make a proof that the equation cannot have nonzero natural number solution in the odd and prime number n. \r\nY+A=X+B=Z, A=Z-Y, B=Z-X \r\nX-A=Y-B=Z-A-B=X+Y-Z \r\nG=(X-A)/(AB)^(1/n)=(Y-B)/(AB)^(1/n)=(Z-A-B)/(AB)^(1/n)=(X+Y-Z)/(AB)^(1/n) \r\nX=G(AB)^(1/n)+A, Y=G(AB)^(1/n)+B, Z=G(AB)^(1/n)+A+B \r\n{G(AB)^(1/n)+A}^n+{G(AB)^(1/n)+B}^n={G(AB)^(1/n)+A+B}^n \r\nWhen n=1, G=0. When n=2, G=2^(1/2). When n>2, G=function(A,B) is the positive real number. \r\nX=(2AB)^(1/2)+A, Y=(2AB)^(1/2)+B, Z=(2AB)^(1/2)+A+B \r\n(X,Y,Z) are the irrational numbers or all Pythagorean triples in all natural number (A,B). \r\nWe can translate the upper form into this. \r\nAB=2k^2, B=2k^2/A \r\nX=2k+A, Y=2k(k+A)/A, Z=2k+A+2k^2/A \r\nXY=2k(2k+A)(k+A)/A \r\nWhen A is the odd number, k=hA, XY=2A^2h(2h+1)(h+1) and hk=A, XY=2k^2(2+h)(1+h)/h \r\nWhen A is the even number, 2k=hA, XY=A^2h(h+1)(h+2)/2 and 2hk=A, XY=2k^2(1+h)(1+2h)/h \r\nTherefore XY cannot be the power numbers in all Pythagorean triples. \r\n* * * * * 1st method of FLT proof * * * * * \r\nG(AB)^(1/n) is the irrational number in all natural number (A,B), so (X,Y,Z) are the irrational numbers. \r\n{G(AB)^(1/n)+A}^n+{G(AB)^(1/n)+B}^n={G(AB)^(1/n)+A+B}^n \r\nWhen A=B, \r\n2{GA^(2/n)+A}^n={GA^(2/n)+2A}^n \r\n{2^(1/n)-1}GA^(2/n)={2-2^(1/n)}A \r\nG=[2^{(n-1)/n}+\u2026+2^(2/n)+2^(1/n)]A^{(n-2)/n} \r\nWe make new form with [2^{(n-1)/n}+\u2026+2^(2/n)+2^(1/n)]. \r\n[2^{(n-1)/n}+\u2026+2^(2/n)+2^(1/n)][{A^(n-1)B}^(1/n)+{AB^(n-1)}^(1/n)]/2 \r\nThis form is the irrational number in all natural number (A,B). \r\nG(AB)^(1/n) divide and multiply by [2^{(n-1)/n}+\u2026+2^(2/n)+2^(1/n)][{A^(n-1)B}^(1/n)+{AB^(n-1)}^(1/n)]/2, and now we can get two forms. And when A=B, q=1. \r\nG(AB)^(1/n)=q[2^{(n-1)/n}+\u2026+2^(2/n)+2^(1/n)][{A^(n-1)B}^(1/n)+{AB^(n-1)}^(1/n)]/2 \r\nq=2G(AB)^(1/n)/[2^{(n-1)/n}+\u2026+2^(2/n)+2^(1/n)][{A^(n-1)B}^(1/n)+{AB^(n-1)}^(1/n)] \r\nIf G(AB)^(1/n) is the natural number (N) in some (a,b), G(ab)^(1/n)=N can not have [2^{(n-1)/n}+\u2026+2^(2/n)+2^(1/n)], and G(AB)^(1/n) can not have [2^{(n-1)/n}+\u2026+2^(2/n)+2^(1/n)]. \r\nSo when A=B, q cannot be 1. That is an apparent contradiction. \r\nTherefore G(AB)^(1/n) is the irrational number in all natural number (A,B). 1st method. end. \r\n* * * * * 2nd method of FLT proof * * * * * \r\nX^n+Y^n=Z^n \r\n{X^(n/2)}^2+{Y^(n/2)}^2={Z^(n/2)}^2 \r\nWhen n=2, we can display {X^(n/2),Y^(n/2),Z^(n/2)} with (a,b). \r\na=Z^(n/2)-Y^(n/2), b=Z^(n/2)-X^(n/2) \r\nX^(n/2)=(2ab)^(1/2)+a, Y^(n/2)=(2ab)^(1/2)+b, Z^(n/2)=(2ab)^(1/2)+a+b \r\nWhen n is the prime number and (X,Y,Z) is co prime, the ab is the irrational number. \r\nab=Z^n-(YZ)^(n/2)-(XZ)^(n/2)+(XY)^(n/2) \r\nWe multiply X^(n/2) and Y^(n/2). \r\n(XY)^n=2a^3b+2ab^3+13(ab)^2+6ab(a+b)(2ab)^(1/2) \r\nIf (X,Y,Z) is the natural number, (XY)^n is the natural number, but 2a^3b+2ab^3+13(ab)^2+6ab(a+b)(2ab)^(1/2) is the irrational number. \r\nThat is an apparent contradiction. \r\nTherefore (X,Y,Z) must be the irrational number. 2nd method. end.", "Solution_1": "I have totally no idea about what the \"4 Color Theorem proof\" are trying to express. Seems that it only considers parts of the whole graph separately, but when the regions are put together, the claim may not be true.", "Solution_2": "You should be very doubtful if someone not belonging to the world's best mathemticians claims to have proven some old open problem (or one that was some years ago).\r\nYou should be even more doubtful if the proof is short, simple, and elementary (by what the author says).\r\nThe doubts should grow to $\\infty$ if the author claims to have more than one proof or to have proven more than one of these hard problems.\r\n\r\n[quote]This guy is demonstrating alone for 18 weeks for these proof to be accepted!! [/quote]\r\nSo what\u00bf", "Solution_3": "[quote]\n\n[quote]\nThis guy is demonstrating alone for 18 weeks for these proof to be accepted!! [/quote]\n\n\n\nSo what\u00bf\n[/quote]\r\n\r\nI think that guy is so pitiful, believing that he's right, but he's not and is fighting for nothing.\r\n\r\nWell... if you say \"who cares?\" or something..I don't know.\r\n\r\nAnyway Thanks.", "Solution_4": "All that he's asking for is to find a fault in the proofs; we know there is one, of course, ZetaX.\r\n\r\nView it as the opposite of a proof problem :P", "Solution_5": "Its kind of necessary to fix the latex, so:\r\n[quote=\"car202\"]\n[b]2 methods of FLT proof and Pythagorean triples [/b]\n[hide=\"Preliminary Part\"]\n$x^{n}+y^{n}=z^{n}$\nParaphrased: Since FLT proof for $n=4$ is well known, it remains to prove it for any odd prime exponent.\nDefine $a,b$ as follows: \\[y+a=x+b=z\\implies a=z-y, \\ b=z-x, \\ x-a=y-b=z-a-b=x+y-z\\]\n\nDefine $g$ also: \\[g=\\frac{x-a}{\\sqrt[n]{ab}}=\\frac{y-b}{\\sqrt[n]{ab}}=\\frac{z-a-b}{\\sqrt[n]{ab}}=\\frac{x+y-z}{\\sqrt[n]{ab}}\\] Thus,\n\\[x=g\\sqrt[n]{ab}+a, \\ y=g\\sqrt[n]{ab}+b, \\ z=g\\sqrt[n]{ab}+a+b\\]\nso that \\[\\left(g\\sqrt[n]{ab}+a\\right)^{n}+\\left(g\\sqrt[n]{ab}+b\\right)^{n}=\\left(g\\sqrt[n]{ab}+a+b\\right)^{n}\\]\nNotice that $n=1\\implies g=0; \\text{ and }n=2\\implies g=\\sqrt{2}$. When $n>2, \\ g=f(a,b)$ is a positive real number. \n\\[x=\\sqrt{2ab}+a, \\ y=\\sqrt{2ab}+b, \\ z=\\sqrt{2ab}+a+b\\]\nWhere $(x,y,z)$ are irrational numbers [forming?] all Pythagorean triples in all natural number [pairs] $(a,b)$.\nLet $k=\\sqrt{\\frac{ab}{2}}$; now solve for $x,y,z$ in terms of $a, \\ k$ (not sure why this is done)\n\\[x=2k+a, \\ y=\\frac{2k(k+a)}{a}, \\ z=2k+a+\\frac{2k^{2}}{a}\\implies xy=\\frac{2k(2k+a)(k+a)}{a}\\]\nWhen $a$ is odd, $k=ha, \\ xy=2a^{2}h(2h+1)(h+1)$ and $hk=a, \\ xy=\\frac{2k^{2}(2+h)(1+h)}{h}$\nWhen $a$ is even, $2k=ha, \\ xy=\\frac{a^{2}h(h+1)(h+2)}{2}$ and $2hk=a, \\ xy=\\frac{2k^{2}(1+h)(1+2h)}{h}$\nTherefore $xy$ cannot be the power numbers in all Pythagorean triples. \n(I'm not sure what that last sentence actually means)\n[/hide]\n[hide=\" $1^{st}$ method of FLT proof\"]\n$g\\sqrt[n]{ab}$ is irrational in all naturals number $(a,b)$ (?), so $(x,y,z)$ are irrational.\n\\[\\left(g\\sqrt[n]{ab}+a\\right)^{n}+\\left(g\\sqrt[n]{ab}+b\\right)^{n}=\\left(g\\sqrt[n]{ab}+a+b\\right)^{n}\\]\nWhen $a=b$, \\[2\\left(g \\cdot \\sqrt[n]{a^{2}}+a\\right)^{n}= \\left(g\\cdot \\sqrt[n]{a^{2}}+2a\\right)^{n}\\]\n[Added these steps, using fact that each side is in the reals and that $n$ is odd] \\[\\iff \\sqrt[n]{2}\\left(g \\cdot \\sqrt[n]{a^{2}}+a\\right)=g\\cdot \\sqrt[n]{a^{2}}+2a \\iff \\]\n\\[\\left(\\sqrt[n]{2}-1\\right) g\\cdot \\sqrt[n]{a^{2}}= \\left(2-\\sqrt[n]{2}\\right) a \\implies g = \\frac{\\left(2-\\sqrt[n]{2}\\right) a}{\\left(\\sqrt[n]{2}-1\\right) \\cdot \\sqrt[n]{a^{2}}}\\] \\[\\iff g = \\left(\\sum_{i=1}^{n-1}\\sqrt[n]{2^{i}}\\right) \\sqrt[n]{a^{n-2}}\\]\nConsider the expression \\[\\left(\\sum_{i=1}^{n-1}\\sqrt[n]{2^{i}}\\right)\\]\nAnd use it in: \\[\\left(\\sum_{i=1}^{n-1}\\sqrt[n]{2^{i}}\\right)\\left(a^{n-1}b^{1/n}+(ab)^{\\frac{n-1}{n}}\\right)/2\\]\nThis expression is irrational for all naturals $(a,b)$. \nConsider the quantity $g\\sqrt[n]{ab}$ divided and multiplied by \\[\\left(\\sum_{i=1}^{n-1}\\sqrt[n]{2^{i}}\\right)\\left(a^{n-1}b^{1/n}+(ab)^{\\frac{n-1}{n}}\\right)/2\\]\n, which yields two expressions. When $a=b, q=1$ we have that: \\[g\\sqrt[n]{ab}= q\\left(\\sum_{i=1}^{n-1}\\sqrt[n]{2^{i}}\\right)\\left(a^{n-1}b^{1/n}+(ab)^{\\frac{n-1}{n}}\\right)/2 \\] \\[q=\\frac{2g\\sqrt[n]{ab}}{\\left(\\sum_{i=1}^{n-1}\\sqrt[n]{2^{i}}\\right)\\left(a^{n-1}b^{1/n}+(ab)^{\\frac{n-1}{n}}\\right)\\]\nIf $g\\sqrt[n]{ab}=N$ is natural for some $(a,b)$, then $g\\sqrt[n]{ab}=N$ can not have $\\left(\\sum_{i=1}^{n-1}\\sqrt[n]{2^{i}}\\right)$ (what does that mean), and $g\\sqrt[n]{ab}$ can not have $\\left(\\sum_{i=1}^{n-1}\\sqrt[n]{2^{i}}\\right)$ (oh I think it means divides. \nSo when $A=B, q \\not = 1$; contradiction.\nTherefore $g\\sqrt[n]{ab}$ is an irrational number for all naturals $(A,B)$.\n[/hide]\n[hide=\" $2^{nd}$ method of FLT proof\"]\n$X^{n}+Y^{n}=Z^{n}{X^{(}n/2)}^{2}+{Y^{(}n/2)}^{2}={Z^{(}n/2)}^{2}$\nWhen n=2, we can display ${X^{(}n/2),Y^{(}n/2),Z^{(}n/2)}$ with (a,b). \n$a=Z^{(}n/2)-Y^{(}n/2), b=Z^{(}n/2)-X^{(}n/2) X^{(}n/2)=(2ab)^{(}1/2)+a, Y^{(}n/2)=(2ab)^{(}1/2)+b, Z^{(}n/2)=(2ab)^{(}1/2)+a+b$\nWhen n is the prime number and (X,Y,Z) is co prime, the ab is the irrational number. \n$ab=Z^{n}-(YZ)^{(}n/2)-(XZ)^{(}n/2)+(XY)^{(}n/2)$\nWe multiply $X^{(}n/2)$ and $Y^{(}n/2)$. \n$(XY)^{n}=2a^{3}b+2ab^{3}+13(ab)^{2}+6ab(a+b)(2ab)^{(}1/2)$\nIf (X,Y,Z) is the natural number, $(XY)^{n}$ is the natural number, but $2a^{3}b+2ab^{3}+13(ab)^{2}+6ab(a+b)(2ab)^{(}1/2)$ is the irrational number. \nThat is an apparent contradiction. \nTherefore (X,Y,Z) must be the irrational number. 2nd method. end.\n[/hide][/quote]\r\n\r\nI'm not really understanding what the original proof was, because in key places the messy writing is leading to misunderstandings etc.", "Solution_6": "Anybody else want to try to decipher the \"proofs\"... ?\r\n(It will be a good lesson in writing convincing fallacious proofs)", "Solution_7": "My point is:\r\n- proofreading is mostly boring\r\n- this stuff is very badly written\r\n- I doubt that the author could be called mathematician, looks like a swank\r\n\r\nIf he really want's to know if his reasoning is true and not to show off, he could always try to read is stuff himself, write it more clearly and especially pay someone to proofread.\r\n\r\nI will never read that kind of stuff and hope others don't so, either. That's like feeding trolls.", "Solution_8": "[quote=\"ZetaX\"]My point is:\n- proofreading is mostly boring\n- this stuff is very badly written\n- I doubt that the author could be called mathematician, looks like a swank\n\nIf he really want's to know if his reasoning is true and not to show off, he could always try to read is stuff himself, write it more clearly and especially pay someone to proofread.\n\nI will never read that kind of stuff and hope others don't so, either. That's like feeding trolls.[/quote]\r\nI agree with ZetaX.\r\nI seen SO many of these stupid claims.\r\n\r\nYou know who that guy is. Some person why knows barely any math, and understood the Last theorem and got interested in it, because it is so elementary and then made up some argument that he found it.", "Solution_9": "I think this guy's name is Lee Jae Yuel (South Korea)\r\nHe's the most stubborn(and the most famous) of people who claim to have :) proven Fermat's last theorem and the four color theorem. \r\n He enjoys terrorizing the Korean Mathematic Society Homepage,\r\nalong with another of his friends who claims to have proven the Goldbach conjecture.\r\n like Zetax said, your time is better spent not reading this proof.", "Solution_10": "At a glance, you can already tell that he doesn't have a definite concept with numbers (like introducing the quantity g, which is a real number in the first place) and working with integers at the same time with rationals and irrationals??? His variables are badly defined. Aww, he just wasted 2 minutes of my time. :furious:" } { "Tag": [ "limit", "calculus" ], "Problem": "$a$ $b$ $c$ $d$ are reals \r\nfind\r\n$\\lim_{x \\to 0}\\frac{1-cos(ax)\\time cos(bx)\\time cos(cx) \\time cos(dx)}{x^{2}}$", "Solution_1": "Maybe L'Hospital's might help, even if it is Calculus.", "Solution_2": "don't use L'Hospital", "Solution_3": "[hide]$cos(x)=1-x^{2}/2+...$\n\n$=\\frac{1-(1-\\frac{a^{2}x^{2}+...}{2})(1-\\frac{b^{2}x^{2}+...}{2})(1-\\frac{c^{2}x^{2}+...}{2})(1-\\frac{d^{2}x^{2}+...}{2})}{x^{2}}$\n$=\\frac{1-(1-(\\frac{a^{2}x^{2}}{2}+\\frac{b^{2}x^{2}}{2}+\\frac{c^{2}x^{2}}{2}+\\frac{d^{2}x^{2}}{2})+...)}{x^{2}}$\n$=\\frac{\\frac{a^{2}+b^{2}+c^{2}+d^{2}}{2}x^{2}+...}{x^{2}}$\n\nThe $...$ represents higher powers of $x$, which are omitted because we know they will reduce to zero when we divide through by $x^{2}$ at the end.\n\nDividing through by $x^{2}$ and substituting $x=0$ gives:\n$\\boxed{L=\\frac{a^{2}+b^{2}+c^{2}+d^{2}}{2}}$[/hide]", "Solution_4": "Limits = calculus = belongs in the calculus forum.", "Solution_5": "limits=in my precalculus book and in AoPS vol 2. :wink: \r\n\r\nSo I think it's okay here." } { "Tag": [ "inequalities" ], "Problem": "If $a , b$ are positive integers, prove that\r\n\r\n $\\sqrt[3]{( \\frac{a^{4}+b^{4}}{a+b})^{a+b}}\\geq{a^{a}b^{b}}$\r\n\r\n Note that $a+b$ is an exponent on the fraction.", "Solution_1": "I think this problem is true for positive reals, and not only for integers.\r\n\r\nWeighted AM-GM:\r\n\r\n\\[\\sqrt[3]{\\left(\\frac{a.a^{3}+b.b^{3}}{a+b}\\right)^{a+b}}\\geq \\sqrt[3]{\\left(\\sqrt[a+b]{a^{3a}b^{3b}}\\right)^{a+b}}=a^{a}b^{b}\\]", "Solution_2": "[quote=\"Jan\"]I think this problem is true for positive reals, and not only for integers.\n\nWeighted AM-GM:\n\\[\\sqrt[3]{\\left(\\frac{a.a^{3}+b.b^{3}}{a+b}\\right)^{a+b}}\\geq \\sqrt[3]{\\left(\\sqrt[a+b]{a^{3a}b^{3b}}\\right)^{a+b}}=a^{a}b^{b}\\]\n[/quote]\r\n \r\nYES !!! Your solution and generalization for positive reals are good . Just a friendly note : usually for positive reals we do not use radicals but rational fractions. So , the above solution [color=red]is perfect [/color]for positive integers ! A little modification is necessary for the other case(positive reals).\r\n\r\n Babis", "Solution_3": "Thanks for the comment.\r\nI'm not sure I understand, though...\r\n\r\nDo you mean I should have written:\r\n\\[\\sqrt[3]{\\left(\\frac{a.a^{3}+b.b^{3}}{a+b}\\right)^{a+b}}\\geq \\sqrt[3]{\\left((a^{3a}b^{3b})^{\\frac{1}{a+b}}\\right)^{a+b}}=a^{a}b^{b}\\]", "Solution_4": "Exactly ! Now your solution is OK for all positive reals $a , b$ .\r\n\r\n This is the power of '' weighted means '' !!!\r\n\r\n babis" } { "Tag": [], "Problem": "Na lythei i parakato eksisosi stous thetikous akeraious:\r\n\r\n $ x^2\\plus{}y^2\\equal{}1997(x\\minus{}y)$", "Solution_1": "\u0391\u03c5\u03c4\u03bf \u03c0\u03bf\u03c5 \u03b8\u03b1 \u03c1\u03c9\u03c4\u03b7\u03c3\u03c9 \u03b9\u03c3\u03c9\u03c2 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b2\u03bb\u03b1\u03ba\u03b5\u03b9\u03b1 :blush: \r\n\u039c\u03c0\u03bf\u03c1\u03bf\u03c5\u03bc\u03b5 \u03bd\u03b1 \u03be\u03b5\u03c7\u03c9\u03c1\u03b9\u03c3\u03bf\u03c5\u03bc\u03b5,\u03c0\u03bf\u03c4\u03b5 \u03bf\u03b9 \u03bb\u03c5\u03c3\u03b5\u03b9\u03c2 \u03c0\u03bf\u03c5 \u03b8\u03b1 \u03b4\u03bf\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c3\u03b5 \u03bc\u03b9\u03b1 \u03b4\u03b9\u03b1\u03c6\u03b1\u03bd\u03c4\u03b9\u03ba\u03b7 \u03b8\u03b1 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c3\u03c5\u03b3\u03ba\u03b5\u03ba\u03c1\u03b9\u03bc\u03b5\u03bd\u03bf\u03b9 \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03b9 \u03b7 \u03b8\u03b1 \u03b5\u03c7\u03bf\u03c5\u03bd \u03b3\u03b5\u03bd\u03b9\u03ba\u03bf\u03c4\u03b5\u03c1\u03b7 \u03bc\u03bf\u03c1\u03c6\u03b7 :?: (\u03c0.\u03c7 $ x \\equal{} t^2 \\plus{} 1$ \u03ba\u03b1\u03b9 $ y \\equal{} 3t \\plus{} 5$)", "Solution_2": "[quote=\"Dimitris X\"]\u0391\u03c5\u03c4\u03bf \u03c0\u03bf\u03c5 \u03b8\u03b1 \u03c1\u03c9\u03c4\u03b7\u03c3\u03c9 \u03b9\u03c3\u03c9\u03c2 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b2\u03bb\u03b1\u03ba\u03b5\u03b9\u03b1 :blush: [/quote]\r\n\r\n\u039a\u03b1\u03b8\u03cc\u03bb\u03bf\u03c5 \u03b2\u03bb\u03b1\u03ba\u03b5\u03af\u03b1 \u03b4\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9. (\u0391\u03bb\u03bb\u03ac \u03ba\u03b1\u03b9 \u03bd\u03b1 \u03ae\u03c4\u03b1\u03bd, \u03b5\u03b4\u03ce \u03b5\u03c0\u03b9\u03c4\u03c1\u03ad\u03c0\u03b5\u03c4\u03b1\u03b9 \u03bd\u03b1 \u03c1\u03c9\u03c4\u03ac\u03bc\u03b5 \u03b2\u03bb\u03b1\u03ba\u03b5\u03af\u03b5\u03c2.)\r\n\r\n\u039b\u03bf\u03b9\u03c0\u03cc\u03bd, \u03b3\u03b5\u03bd\u03b9\u03ba\u03ac \u03b4\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bd\u03b1 \u03be\u03ad\u03c1\u03bf\u03c5\u03bc\u03b5. \u0392\u03bb\u03ad\u03c0\u03b5 \u03b3\u03b9\u03b1 \u03c0\u03b1\u03c1\u03ac\u03b4\u03b5\u03b9\u03b3\u03bc\u03b1 \u03c4\u03bf \u03c4\u03b5\u03bb\u03b5\u03c5\u03c4\u03b1\u03af\u03bf \u03b8\u03b5\u03ce\u03c1\u03b7\u03bc\u03b1 \u03c4\u03bf\u03c5 Fermat \u03c0\u03bf\u03c5 \u03b4\u03b5\u03bd \u03b3\u03bd\u03c9\u03c1\u03af\u03b6\u03b1\u03bc\u03b5 \u03ba\u03b1\u03bd \u03b1\u03bd \u03c5\u03c0\u03ac\u03c1\u03c7\u03bf\u03c5\u03bd \u03bb\u03cd\u03c3\u03b5\u03b9\u03c2 \u03ae \u03cc\u03c7\u03b9. (\u039c\u03ad\u03c7\u03c1\u03b9 \u03c4\u03bf 1994 \u03c0\u03bf\u03c5 \u03bf Wiles \u03b1\u03c0\u03ad\u03b4\u03b5\u03b9\u03be\u03b5 \u03c0\u03c9\u03c2 \u03b4\u03b5\u03bd \u03c5\u03c0\u03ac\u03c1\u03c7\u03bf\u03c5\u03bd.) \u0391\u03ba\u03cc\u03bc\u03b7 \u03c7\u03b5\u03b9\u03c1\u03cc\u03c4\u03b5\u03c1\u03b1, \u03ad\u03c7\u03b5\u03b9 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03c7\u03b8\u03b5\u03af \u03c0\u03c9\u03c2 \u03b4\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03c5\u03c0\u03ac\u03c1\u03be\u03b5\u03b9 \u03b1\u03bb\u03b3\u03cc\u03c1\u03b9\u03b8\u03bc\u03bf\u03c2 \u03c0\u03bf\u03c5 \u03bd\u03b1 \u03c4\u03bf\u03c5 \u03b4\u03af\u03bd\u03bf\u03c5\u03bc\u03b5 \u03bc\u03b9\u03b1 \u03b4\u03b9\u03bf\u03c6\u03b1\u03bd\u03c4\u03b9\u03ba\u03ae \u03b5\u03be\u03af\u03c3\u03c9\u03c3\u03b7 \u03ba\u03b1\u03b9 \u03bd\u03b1 \u03bc\u03b1\u03c2 \u03b1\u03c0\u03b1\u03bd\u03c4\u03ac\u03b5\u03b9 \u03b1\u03bd \u03ad\u03c7\u03b5\u03b9 \u03bb\u03cd\u03c3\u03b5\u03b9\u03c2 \u03ae \u03cc\u03c7\u03b9. (\u039a\u03ac\u03bd\u03bf\u03bd\u03c4\u03b1\u03c2 \u03c0\u03b5\u03c0\u03b5\u03c1\u03b1\u03c3\u03bc\u03ad\u03bd\u03bf \u03b1\u03c1\u03b9\u03b8\u03bc\u03cc \u03b2\u03b7\u03bc\u03ac\u03c4\u03c9\u03bd.)\r\n\r\n\u03a5\u03c0\u03ac\u03c1\u03c7\u03bf\u03c5\u03bd \u03ba\u03ac\u03c0\u03bf\u03b9\u03b5\u03c2 \u03bc\u03bf\u03c1\u03c6\u03ad\u03c2 \u03b4\u03b9\u03bf\u03c6\u03b1\u03bd\u03c4\u03b9\u03ba\u03ce\u03bd \u03b5\u03be\u03b9\u03c3\u03ce\u03c3\u03b5\u03c9\u03bd \u03c0\u03bf\u03c5 \u03be\u03ad\u03c1\u03bf\u03c5\u03bc\u03b5 \u03c0\u03c9\u03c2 \u03bd\u03b1 \u03c4\u03b9\u03c2 \u03bb\u03cd\u03bd\u03bf\u03c5\u03bc\u03b5 \u03ba\u03b1\u03b9 \u03b1\u03c0\u03cc \u03b5\u03ba\u03b5\u03af \u03ba\u03b1\u03b9 \u03c0\u03ad\u03c1\u03b1 \u03b1\u03bd \u03bc\u03bf\u03b9\u03ac\u03b6\u03b5\u03b9 \u03ba\u03ac\u03c0\u03c9\u03c2 \u03bc\u03b5 \u03ba\u03ac\u03c0\u03bf\u03b9\u03b1 \u03b4\u03b9\u03bf\u03c6\u03b1\u03bd\u03c4\u03b9\u03ba\u03ae \u03c0\u03bf\u03c5 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03b4\u03b5\u03b9, \u03af\u03c3\u03c9\u03c2 \u03bc\u03c0\u03bf\u03c1\u03ad\u03c3\u03bf\u03c5\u03bc\u03b5 \u03bd\u03b1 \u03bc\u03b1\u03bd\u03c4\u03ad\u03c8\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b9 \u03bc\u03bf\u03c1\u03c6\u03ae \u03b8\u03b1 \u03ad\u03c7\u03bf\u03c5\u03bd \u03bf\u03b9 \u03bb\u03cd\u03c3\u03b5\u03b9\u03c2. \r\n\r\n\u03a3\u03b5 \u03b1\u03c5\u03c4\u03ae\u03bd \u03c4\u03b7\u03bd \u03c0\u03b5\u03c1\u03af\u03c0\u03c4\u03c9\u03c3\u03b7 \u03b7 \u03c0\u03c1\u03ce\u03c4\u03b7 \u03bc\u03bf\u03c5 \u03c3\u03ba\u03ad\u03c8\u03b7 \u03ae\u03c4\u03b1\u03bd \u03bd\u03b1 \u03c3\u03c5\u03bc\u03c0\u03bb\u03b7\u03c1\u03ce\u03c3\u03c9 \u03c4\u03bf \u03c4\u03b5\u03c4\u03c1\u03ac\u03b3\u03c9\u03bd\u03bf. \u0388\u03c7\u03bf\u03c5\u03bc\u03b5 $ (x \\minus{} 1997/2)^2 \\plus{} (y \\plus{} 1997/2)^2 \\equal{} 1997^2/2$. \u03a0\u03bf\u03bb\u03bb\u03b1\u03c0\u03bb\u03b1\u03c3\u03b9\u03ac\u03b6\u03c9\u03bd\u03c4\u03b1\u03c2 \u03bc\u03b5 \u03c4\u03bf 4 \u03c0\u03b1\u03af\u03c1\u03bd\u03bf\u03c5\u03bc\u03b5 $ (2x \\minus{} 1997)^2 \\plus{} (2y \\plus{} 1997)^2 \\equal{} 2 \\cdot 1997^2$. \u0391\u03c0\u03cc \u03b5\u03b4\u03ce \u03b2\u03bb\u03ad\u03c0\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03b7 \u03b5\u03be\u03af\u03c3\u03c9\u03c3\u03b7 \u03bd\u03b1 \u03ad\u03c7\u03b5\u03b9 \u03c0\u03b5\u03c0\u03b5\u03c1\u03b1\u03c3\u03bc\u03ad\u03bd\u03bf \u03b1\u03c1\u03b9\u03b8\u03bc\u03cc \u03bb\u03cd\u03c3\u03b5\u03c9\u03bd. \r\n\r\n\u0391\u03c0\u03cc \u03b5\u03b4\u03ce \u03ba\u03b1\u03b9 \u03c0\u03ad\u03c1\u03b1 \u03c0\u03b1\u03af\u03b6\u03b5\u03b9 \u03c1\u03cc\u03bb\u03bf \u03ba\u03b1\u03b9 \u03b7 \u03b5\u03bc\u03c0\u03b5\u03b9\u03c1\u03af\u03b1 \u03ba\u03b1\u03b9 \u03c4\u03b9 \u03c0\u03b1\u03c1\u03cc\u03bc\u03bf\u03b9\u03b5\u03c2 \u03b4\u03b9\u03bf\u03c6\u03b1\u03bd\u03c4\u03b9\u03ba\u03ad\u03c2 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03b4\u03b5\u03b9. \u0388\u03bd\u03b1\u03c2 \u03c0\u03c1\u03bf\u03b3\u03c1\u03b1\u03bc\u03bc\u03b1\u03c4\u03b9\u03c3\u03c4\u03ae\u03c2 \u03b8\u03b1 \u03ad\u03c6\u03c4\u03b9\u03b1\u03c7\u03bd\u03b5 \u03ad\u03bd\u03b1 \u03c0\u03c1\u03cc\u03b3\u03c1\u03b1\u03bc\u03bc\u03b1 \u03c0\u03bf\u03c5 \u03b8\u03b1 \u03ad\u03b2\u03c1\u03b9\u03c3\u03ba\u03b5 \u03cc\u03bb\u03bf\u03c5\u03c2 \u03c4\u03bf\u03c5\u03c2 \u03c4\u03c1\u03cc\u03c0\u03bf\u03c5\u03c2 \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03b3\u03c1\u03b1\u03c6\u03c4\u03b5\u03af \u03c4\u03bf $ 2 \\cdot 1997^2$ \u03c3\u03b1\u03bd \u03ac\u03b8\u03c1\u03bf\u03b9\u03c3\u03bc\u03b1 \u03b4\u03cd\u03bf \u03c4\u03b5\u03c4\u03c1\u03b1\u03b3\u03ce\u03bd\u03c9\u03bd \u03ba\u03b1\u03b9 \u03b3\u03b9\u03b1 \u03ba\u03ac\u03b8\u03b5 \u03c4\u03ad\u03c4\u03bf\u03b9\u03bf \u03ac\u03b8\u03c1\u03bf\u03b9\u03c3\u03bc\u03b1 \u03b8\u03b1 \u03c5\u03c0\u03bf\u03bb\u03cc\u03b3\u03b9\u03b6\u03b5 \u03b3\u03c1\u03ae\u03b3\u03bf\u03c1\u03b1 \u03c4\u03b1 \u03c7 \u03ba\u03b1\u03b9 \u03c8.\r\n\r\n\u039a\u03ac\u03c0\u03bf\u03b9\u03bf\u03c2 \u03c0\u03b9\u03bf \u03ad\u03bc\u03c0\u03b5\u03b9\u03c1\u03bf\u03c2 \u03b8\u03b5 \u03b3\u03bd\u03ce\u03c1\u03b9\u03b6\u03b5 \u03cc\u03c4\u03b9 \u03b5\u03c0\u03b5\u03b9\u03b4\u03ae \u03bf 1997 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03c1\u03ce\u03c4\u03bf\u03c2 \u03c4\u03b7\u03c2 \u03bc\u03bf\u03c1\u03c6\u03ae\u03c2 1mod4 \u03c4\u03bf $ 2 \\cdot 1997^2$ \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03b3\u03c1\u03b1\u03c6\u03c4\u03b5\u03af \u03c3\u03b1\u03bd \u03ac\u03b8\u03c1\u03bf\u03b9\u03c3\u03bc\u03b1 \u03b4\u03cd\u03bf \u03c4\u03b5\u03c4\u03c1\u03b1\u03b3\u03ce\u03bd\u03c9\u03bd \u03bc\u03b5 \u03b1\u03ba\u03c1\u03b9\u03b2\u03ce\u03c2 3 \u03c4\u03c1\u03cc\u03c0\u03bf\u03c5\u03c2 (\u03ba\u03b1\u03b9 \u03c0\u03c9\u03c2 \u03bd\u03b1 \u03c4\u03bf \u03b3\u03c1\u03ac\u03c8\u03b5\u03b9) \u03ba\u03b1\u03b9 \u03ad\u03c4\u03c3\u03b9 \u03b8\u03b1 \u03b3\u03bb\u03cd\u03c4\u03c9\u03bd\u03b5 \u03b1\u03c1\u03ba\u03b5\u03c4\u03cc \u03c7\u03c1\u03cc\u03bd\u03bf.\r\n\r\n\u038a\u03c3\u03c9\u03c2 \u03b5\u03be\u03b7\u03b3\u03ae\u03c3\u03c9 \u03b1\u03c1\u03b3\u03cc\u03c4\u03b5\u03c1\u03b1 \u03c0\u03c9\u03c2 \u03b3\u03c1\u03ac\u03c6\u03bf\u03c5\u03bc\u03b5 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03bd \u03b1\u03c1\u03b9\u03b8\u03bc\u03cc \u03c3\u03b1\u03bd \u03ac\u03b8\u03c1\u03bf\u03b9\u03c3\u03bc\u03b1 \u03b4\u03cd\u03bf \u03c4\u03b5\u03c4\u03c1\u03b1\u03b3\u03ce\u03bd\u03c9\u03bd. (\u0391\u03bd \u03ad\u03c7\u03b5\u03b9 \u03be\u03b1\u03bd\u03b1\u03c3\u03c5\u03b6\u03b7\u03c4\u03b7\u03b8\u03b5\u03af \u03c3\u03c4\u03bf \u03c0\u03b1\u03c1\u03b5\u03bb\u03b8\u03cc\u03bd \u03b1\u03c2 \u03b2\u03ac\u03bb\u03b5\u03b9 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03c2 \u03ba\u03b1\u03bd\u03ad\u03bd\u03b1 link.)", "Solution_3": "\u039c\u03b9\u03b1 \u03c0\u03b9\u03bf \u03b2\u03bf\u03bb\u03b9\u03ba\u03ae \u03bc\u03bf\u03c1\u03c6\u03ae \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03c0\u03b1\u03c1\u03b1\u03ba\u03ac\u03c4\u03c9 . \u0397 \u03b5\u03be\u03af\u03c3\u03c9\u03c3\u03b7 \u03c0\u03bf\u03bb\u03bb\u03b1\u03c0\u03bb\u03b1\u03c3\u03b9\u03ac\u03b6\u03bf\u03bd\u03c4\u03b1\u03c2 \u03bc\u03b5 \u03c4\u03bf 2 \u03ba\u03b1\u03b9 \u03c0\u03c1\u03bf\u03c3\u03b8\u03ad\u03c4\u03bf\u03bd\u03c4\u03b1\u03c2 \u03c4\u03bf $ 1997^2$ \u03c3\u03c4\u03b1 \u03b4\u03cd\u03bf \u03bc\u03ad\u03bb\u03b7 , \u03b3\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 :\r\n\r\n$ (x\\plus{}y)^2\\plus{}(1997\\minus{}x\\plus{}y)^2\\equal{}1997^2$ \u03ba\u03b1\u03b9 \u03b1\u03c6\u03bf\u03cd \u03c4\u03bf 1997 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03c1\u03ce\u03c4\u03bf\u03c2 \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bd\u03b1 \u03b4\u03bf\u03c5\u03bb\u03ad\u03c8\u03bf\u03c5\u03bc\u03b5 \u03bc\u03b5 \u03c4\u03b9\u03c2 primitive solutions \u03c4\u03c9\u03bd \u03c0\u03c5\u03b8\u03b1\u03b3\u03cc\u03c1\u03b5\u03b9\u03c9\u03bd \u03c4\u03c1\u03b9\u03ac\u03b4\u03c9\u03bd .\u0394\u03b7\u03bb\u03b1\u03b4\u03ae \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 $ 1997\\equal{}m^2\\plus{}n^2$ , $ x\\plus{}y\\equal{}2mn$ \u03ba\u03b1\u03b9 $ 1997\\minus{}x\\plus{}y\\equal{}m^2\\minus{}n^2$ . \u0395\u03c0\u03af\u03c3\u03b7\u03c2 \u03c3\u03c4\u03b7 \u03c3\u03c5n\u03ad\u03c7\u03b5\u03b9\u03b1 \u03c0\u03b1\u03af\u03c1\u03bd\u03bf\u03bd\u03c4\u03b1\u03c2 mod 3 kai mod 5 \u03bc\u03b5\u03b9\u03ce\u03bd\u03bf\u03c5\u03bc\u03b5 \u03ba\u03b1\u03c4\u03ac \u03c0\u03bf\u03bb\u03cd \u03c4\u03bf \u03c0\u03bb\u03ae\u03b8\u03bf\u03c2 \u03c4\u03c9\u03bd \u03b1\u03c1\u03b9\u03b8\u03bc\u03ce\u03bd \u03bf\u03c5 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03c4\u03c3\u03b5\u03ba\u03ac\u03c1\u03bf\u03c5\u03bc\u03b5 . \u03a4\u03b5\u03bb\u03b9\u03ba\u03ac \u03bc\u03ad\u03bd\u03b5\u03b9 \u03bd\u03b1 \u03c4\u03c3\u03b5\u03ba\u03ac\u03c1\u03bf\u03c5\u03bc\u03b5 \u03bc\u03cc\u03bd\u03bf \u03c4\u03b1 $ m\\equal{}34,41,44$ \u03b1\u03c0\u03cc \u03c4\u03b1 \u03bf\u03c0\u03bf\u03af\u03b1 \r\n$ (m,n)\\equal{}(34,29)$ \u03b7 \u03bf\u03c0\u03bf\u03af\u03b1 \u03c4\u03b5\u03bb\u03b9\u03ba\u03ac \u03bc\u03b1\u03c2 \u03bf\u03b4\u03b7\u03b3\u03b5\u03af \u03c3\u03b5 \u03b4\u03cd\u03bf \u03bb\u03cd\u03c3\u03b5\u03b9\u03c2 . $ (x,y)\\equal{}(170,145)$ \u03ba\u03b1\u03b9 \r\n$ (x,y)\\equal{}(1827,145)$ \r\n\r\n :)", "Solution_4": "\u03c9\u03c1\u03b1\u03b9\u03b1 \u03b1\u03c3\u03ba\u03b7\u03c3\u03b7. \u03b5\u03bb\u03c0\u03b9\u03b6\u03c9 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03b5\u03ba\u03b1\u03bd\u03b1 \u03ba\u03b1\u03c0\u03bf\u03c5 \u03bb\u03b1\u03b8\u03bf\u03c2.\r\n\u03b1\u03c1\u03c7\u03b9\u03ba\u03b1 \u03b3\u03c1\u03b1\u03c6\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b7\u03bd \u03b5\u03be\u03b9\u03c3\u03c9\u03c3\u03b7 \u03bc\u03b1\u03c2 \u03b9\u03c3\u03bf\u03b4\u03c5\u03bd\u03b1\u03bc\u03b1,\r\n$ (x \\plus{} y)^2 \\plus{} (x \\minus{} y)^2 \\minus{} 2*1997(x \\minus{} y) \\plus{} 1997^2 \\equal{} 1997^2$\r\n$ \\leftrightarrow (x \\plus{} y)^2 \\plus{} ( x \\minus{} y \\minus{} 1997)^2 \\equal{} 1997^2$\r\n\u03b1\u03bd \u03b8\u03b5\u03c3\u03bf\u03c5\u03bc\u03b5,\r\n$ k \\equal{} x \\plus{} y , l \\equal{} 1997 \\minus{} x \\plus{} y$ \u03ae \u03b1\u03bd\u03c4\u03b9\u03c3\u03c4\u03c1\u03bf\u03c6\u03b1 $ k \\equal{} 1997 \\minus{} x \\plus{} y , l \\equal{} x \\plus{} y$\r\n\u03c4\u03bf\u03c4\u03b5 \u03b5\u03c7\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b7\u03bd \u03b5\u03be\u03b9\u03c3\u03c9\u03c3\u03b7,\u03b4\u03b5\u03b4\u03bf\u03bc\u03b5\u03bd\u03bf\u03c5 \u03bf\u03c4\u03b9 $ 0 < k,l < 1997$\r\n$ k^2 \\plus{} l^2 \\equal{} 1997^2$ , \r\n\u03c0\u03b1\u03c1\u03b1\u03c4\u03b7\u03c1\u03bf\u03c5\u03bc\u03b5 \u03bf\u03c4\u03b9 $ gcd(k,l) \\equal{} 1$ \u03b1\u03c6\u03bf\u03c5 \u03bf 1997 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c0\u03c1\u03c9\u03c4\u03bf\u03c2.\r\n\u03b5\u03c0\u03bf\u03bc\u03b5\u03bd\u03c9\u03c2 \u03b1\u03c0\u03bf \u03c4\u03b9\u03c2 \u03c0\u03c1\u03c9\u03c4\u03b1\u03c1\u03c7\u03b9\u03ba\u03b5\u03c2 \u03c0\u03c5\u03b8\u03b1\u03b3\u03bf\u03c1\u03b5\u03b9\u03b5\u03c2 \u03c4\u03c1\u03b9\u03b1\u03b4\u03b5\u03c2 \u03c0\u03c1\u03b5\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03c5\u03c0\u03b1\u03c1\u03c7\u03bf\u03c5\u03bd $ a,b$ \u03bc\u03b5 $ a > b$ \u03c0\u03bf\u03c5 \u03bd\u03b1 \u03b9\u03ba\u03b1\u03bd\u03bf\u03c0\u03bf\u03b9\u03bf\u03c5\u03bd \u03c4\u03b9\u03c2 \u03c3\u03c7\u03b5\u03c3\u03b5\u03b9\u03c2,\r\n\r\n$ a^2 \\plus{} b^2 \\equal{} 1997$(3)\r\n$ 2ab \\equal{} l$\r\n$ a^2 \\minus{} b^2 \\equal{} k$\r\n\r\n\u03bf\u03bc\u03c9\u03c2 \u03b9\u03c3\u03c7\u03c5\u03b5\u03b9,\r\n\r\n$ 1997 \\equiv 2 (mod7) , 1997 \\equiv 2 (mod5)$\r\n\u03b1\u03c1\u03b1 \r\n$ a,b \\equiv \\plus{} \\minus{} 1 (mod7)$ \u03ba\u03b1\u03b9 $ a,b \\equiv \\plus{} \\minus{} 1 (mod5)$\r\n\u03b4\u03bb\u03b4\r\n\r\n$ a,b \\equiv 6,24,31,34 (mod35)$\r\n\r\n\u03bf\u03bc\u03c9\u03c2 \u03c0\u03c1\u03bf\u03c6\u03b1\u03bd\u03c9\u03c2 $ 1997 > a > \\frac {1997}{2}$\r\n\r\n\u03b1\u03c1\u03b1 \u03b5\u03bb\u03b5\u03b3\u03c7\u03bf\u03bd\u03c4\u03b1\u03c2 \u03bc\u03b5 \u03c4\u03b9\u03c2 \u03c3\u03c7\u03b5\u03c4\u03b9\u03ba\u03b5\u03c2 \u03c0\u03c1\u03b1\u03be\u03b5\u03b9\u03c2 \u03c0\u03bf\u03b9\u03b1 \u03c4\u03b5\u03c4\u03c1\u03b1\u03b3\u03c9\u03bd\u03b1 \u03b1\u03c1\u03b9\u03b8\u03bc\u03c9\u03bd \u03b9\u03ba\u03b1\u03bd\u03bf\u03c0\u03bf\u03b9\u03bf\u03c5\u03bd \u03c4\u03b9\u03c2 (1),(2) \u03c0\u03b1\u03b9\u03c1\u03bd\u03bf\u03c5\u03bc\u03b5,\r\n$ a \\equal{} 41$ \u03ae $ a \\equal{} 34$\r\n\r\n\u03bf\u03bc\u03c9\u03c2 \u03b9\u03ba\u03b1\u03bd\u03bf\u03c0\u03bf\u03b9\u03b5\u03b9 \u03c4\u03b7\u03bd (3) \u03bc\u03bf\u03bd\u03bf \u03c4 \u03b6\u03b5\u03c5\u03b3\u03bf\u03c2 $ (a,b) \\equal{} (34,29)$\r\n\u03b1\u03c1\u03b1 \u03b5\u03c7\u03bf\u03c5\u03bc\u03b5\r\n$ k \\equal{} 315 , l \\equal{} 1972$\r\n\u0391\u03c1\u03ba\u03b5\u03b9 \u03bb\u03bf\u03b9\u03c0\u03bf\u03bd \u03bd\u03b1 \u03bb\u03c5\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b1 \u03c3\u03c5\u03c3\u03c4\u03b7\u03bc\u03b1\u03c4\u03b1,\r\n\r\n$ x \\plus{} y \\equal{} 315$\r\n$ y \\minus{} x \\plus{} 1997 \\equal{} 1972$\r\n\r\nkai\r\n\r\n$ x \\plus{} y \\equal{} 1972$\r\n$ y \\minus{} x \\plus{} 1997 \\equal{} 315$\r\n\r\n\u03bb\u03c5\u03c3\u03b5\u03b9\u03c2 \u03c4\u03c9\u03bd \u03bf\u03c0\u03bf\u03b9\u03c9\u03bd \u03b5\u03b9\u03bd\u03b1\u03b9 \u03bf\u03b9, \r\n$ (x,y) \\equal{} ( 170,145), ( 1827,145)$\r\nQED\r\nEDIT 2 \u03bb\u03b5\u03c0\u03c4\u03b1 \u03c0\u03b9\u03bf \u03c0\u03c1\u03b9\u03bd \u03c1\u03b5 silouan! !! \u03c0\u03bf\u03c5 \u03b8\u03b1 \u03bc\u03bf\u03c5 \u03c0\u03b1\u03b5\u03b9 \u03bf\u03bc\u03c9\u03c2 ! \u03b8\u03b1 \u03b5\u03b9\u03bc\u03b1\u03b9 \u03c3\u03b5 \u03b5\u03c4\u03bf\u03b9\u03bc\u03bf\u03c4\u03b7\u03c4\u03b1 next time :P", "Solution_5": "\u03a3\u03b9\u03bb\u03bf\u03c5\u03b1\u03bd\u03b5 \u03ba\u03b1\u03b9 \u0397\u03bb\u03af\u03b1 \u03c9\u03c1\u03b1\u03af\u03b5\u03c2 \u03bf\u03b9 \u03bb\u03cd\u03c3\u03b5\u03b9\u03c2 \u03c3\u03b1\u03c2.\r\n\r\n\u0392\u03c1\u03ae\u03ba\u03b1 \u03c3\u03ae\u03bc\u03b5\u03c1\u03b1 \u03bb\u03af\u03b3\u03bf \u03c7\u03c1\u03cc\u03bd\u03bf \u03ba\u03b1\u03b9 \u03ad\u03b3\u03c1\u03b1\u03c8\u03b1 [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1443992#1443992]\u03b5\u03b4\u03ce[/url] \u03ba\u03ac\u03c0\u03bf\u03b9\u03b1 \u03b8\u03b5\u03c9\u03c1\u03af\u03b1 \u03ba\u03b1\u03b9 \u03c0\u03c9\u03c2 \u03b1\u03c5\u03c4\u03ae \u03b2\u03bf\u03b7\u03b8\u03b1\u03b5\u03b9 \u03c3\u03c4\u03bf\u03bd \u03c4\u03c1\u03cc\u03c0\u03bf \u03c0\u03bf\u03c5 \u03c0\u03c1\u03cc\u03c4\u03b5\u03b9\u03bd\u03b1 \u03c3\u03c4\u03bf \u03b1\u03c1\u03c7\u03b9\u03ba\u03cc \u03bc\u03bf\u03c5 post." } { "Tag": [ "LaTeX", "\\/closed" ], "Problem": "For someone reason AoPS and Mathlinks have been really slow to load on my computer for the past two months or so. I haven't noticed any other sites that do this. Anyone have any ideas?", "Solution_1": "maybe it's because of all the posts and pics", "Solution_2": "I'm having this problem too. In particular, it takes forever for the latex to display. Weirdly enough, by quoting someone, I can scroll down and view the latex faster :huh:", "Solution_3": "Weird. Do you have hard disk cache enabled?", "Solution_4": "Yes, but disabling it changes nothing.\r\n\r\nEdit: Latency isn't too bad, but it seems like throughput is the problem. The things that take longest to load are the images... I'm going to go do a test on downloading a large-ish file.\r\n\r\nEdit 2: A 1MiB file downloads pretty quickly, it seems. Can't upload anything larger, though." } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "a continuous map $ f : (X,d) \\to (X,d)$ is called a contraction if there is $ 0 < \\alpha < 1$ st : $ d(f(x),f(y))\\leq \\alpha d(x,y)$\r\n\r\nlet $ C$ be the union of line segments in $ \\mathbb{R}^2$ from $ (0,1)$ to $ (1/n, 0)$ ,$ n \\equal{} 1,2,3,\\ldots$ and the line segment from $ (0,1)$ to $ (0,0)$,\r\n prove that there is no contraction map $ f : C\\to C$ s.t : $ f(0,0) \\equal{} (0,0)$\r\n\r\nthanks so much", "Solution_1": "How about $ f(x)\\equal{}(0,0)$ for all $ x$? :?" } { "Tag": [ "algebra", "polynomial", "quadratics" ], "Problem": "Define the polynomials $P_0, P_1, P_2 \\cdots$ by:\r\n\\[ P_0(x)=x^3+213x^2-67x-2000 \\]\r\n\\[ P_n(x)=P_{n-1}(x-n), n \\in N \\]\r\nFind the coefficient of $x$ in $P_{21}(x)$.", "Solution_1": "BUMP\r\nHow has this gone unanswered for two years!!! Should it be moved to a tougher forum? Right now I can't think of anything but its probably tractable.", "Solution_2": "what? how is this in the least bit difficult, you just need the x coefficient in: $P(x-\\frac{21*22}{2})=P_{21}(x)$", "Solution_3": "What? THat was obvious. I just read it wrong... must be a pretty weak olympiad :P", "Solution_4": "so is answer 61744?", "Solution_5": "[quote=\"me@home\"]I just read it wrong... [/quote]\r\n\r\nMe too. Personally, I read the recursion as $P_{n}(x) = P_{n-1}(x) \\cdot (x-n)$, which is a little more interesting, I think, although still not hard. For one thing, quadratic and higher terms can be ignored entirely.", "Solution_6": "[quote=\"t0rajir0u\"][quote=\"me@home\"]I just read it wrong... [/quote]\n\nMe too. Personally, I read the recursion as $P_{n}(x) = P_{n-1}(x) \\cdot (x-n)$, which is a little more interesting, I think, although still not hard. For one thing, quadratic and higher terms can be ignored entirely.[/quote]\r\nif it was $p_{n}(x)=p_{n-1}(x)(x-n)$\r\n$p_{n}(x)=p_{0}(x)\\prod_{i=1}^{n}(x-i)=\\sum_{k=0}^{n+3}c_{k}x^{k}$\r\nthe coefficient of $x$ is $c_{1}=(-1)^{n}(-67+2000\\sum_{k=1}^{n}\\frac{1}{k})n!$", "Solution_7": "this seems to be the same problem:\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=80691", "Solution_8": "$P_{21} (x) = (x-(1+2+...+21))^3 +213 (x- (1+2+..+21))^2 -67(x-(1+2+...+21)) +2000$\n\nThe coefficient of x is $3(231)^2 -213(231)(2)-67=61610$", "Solution_9": "[hide=induction]claim:- $P_{n}(x)=P_{0}\\left( x-\\sum_{i=1}^{n} i \\right)$\\\\\npf:- base case:\\\\\n$n=1$ ( trivial)\\\\\nassume the claim is true for $n$\\\\\nfrom definition we have $P_{n+1}=P_{n}(x-(n+1)) \\qquad \\qquad (\\spadesuit)$\\\\\nwe also have $P_{n}(x)=P_{0}\\left( x-\\sum_{i=1}^{n} i \\right)\\implies P_{n}(x-(n+1))=P_{0}\\left(x-(n+1)-\\sum_{i=1}^{n} i \\right)$\\\\\nfrom $(\\spadesuit)$ we conclude $P_{n+1}=P_{0}\\left( x-\\sum_{i=1}^{n+1} i \\right)$, which makes the claim true $\\blacksquare$\\\\\nso from claim it follows $P_{21}(x)=P_{0}(x-231)$\\\\\ncoefficient of $x$ in $P_{21}(x)= 3(231)^2 -213(231)(2)-67=\\boxed{61610}$ $\\blacksquare$[/hide]" } { "Tag": [ "function", "calculus", "calculus computations" ], "Problem": "The rate at which water flows out of a pipe, in gallons per hour, is given by a differentiable function R of time t. The table below shows the rate as measured every 3 hours for a 24-hour period.\r\n\r\nt (hours) R(t) (gallons per hour)\r\n0 9.6 \r\n3 10.4\r\n6 10.8\r\n9 11.2\r\n12 11.4\r\n15 11.3\r\n18 10.7\r\n21 10.2\r\n24 9.6\r\n\r\nIs there some time t, 03 there are infinitely many solutions but i found somewhere demonstrated that the equation has only a finite number of solutions!", "Solution_8": "Grobber,you are right!\r\nIf for n we have the solutions x_1,x_2,...,x_n then for n+1 x_1,x_2,...,x_n,x_n+1 where x_n+1=x1*x2*...*xn works(Just make the calculations!)\r\nSo solutions exist for every n.\r\nThe question is now if these solutions are unique.\r\nCan someone prove this?", "Solution_9": "But are we positive that these are the ONLY solutions? This seems somewhat nasty to prove. I've given it some thought,but still nothing.", "Solution_10": "[quote]But are we positive that these are the ONLY solutions?[/quote]\r\n\r\nI doubt it. The discussion in UPINT3 at D28 mentions an extensive analysis by Judith Longyear for finding all solutions, and reviews of some of the cited references refer to parametric solutions." } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Actually, I'm even going to provide the answer and just ask for proof. \r\n\r\nShow that\r\n\r\n$\\sum_{n=0}^\\infty \\frac{(-1)^n(4n+3)\\prod_{m=1}^n(2m-1)}{2^n(n+1)!} = 2$\r\n\r\nThe notation is a little more compact if I use the gamma function:\r\n\r\n$\\sum_{n=0}^\\infty \\frac{(-1)^n(4n+3)\\Gamma(n + \\frac{1}{2})}{\\sqrt{\\pi}\\Gamma(n+2)} = 2$\r\n\r\nEnjoy!", "Solution_1": "My friend points out that it is nicer to write the sum as \r\n\r\n$\\sum_{n=0}^\\infty \\frac{(-1)^n(4n+3)(2n)!}{2^{2n}n!(n+1)!\\sqrt{\\pi}}$\r\n\r\nwhich is then just \r\n\r\n$\\sum_{n=0}^\\infty \\frac{(-1)^n(4n+3)}{2^{2n}(n+1)\\sqrt{\\pi}}\\binom{2n}{n}$\r\n\r\nAny thoughts?" } { "Tag": [ "conics", "ellipse" ], "Problem": "31. A divided highway into the city passes under a number of bridges. The arch over each lane is in the form of a semi-ellipse, where the height is equal to the width. What is the lowest bridge under which a truck 6 feet wide and 12 feet high can theoretically pass?\r\n\r\nNOTE: This is multiple choice, but I did not include that part :D", "Solution_1": "[hide]\n\nlet the equation of this ellipse be\n$\\frac{x^{2}}{\\left(\\frac{h}{2}\\right)^{2}}+\\frac{y^{2}}{h^{2}}= 1$\n\n$\\frac{4x^{2}}{h^{2}}+\\frac{y^{2}}{h^{2}}= 1$\nwhere h is the height\n\nsince all the possible bridges are similar, the lowest bridge must pass through the point (3,12)\n\n$h^{2}= 4(3^{2})+12^{2}= 2^{2}3^{2}+2^{4}3^{2}= 2^{2}3^{2}(1+2^{2})$\n$h = 6\\sqrt{5}$\n[/hide]", "Solution_2": "[hide]\nEquation of ellipse is \n$x^{2}+(\\frac{y}{2})^{2}= r^{2}$\nThe point (3, 12) lies on the ellipse.\nSubbing it in gives $r = 3\\sqrt{5}$.\nSet $x = 0$ which gives a height of $\\boxed{6\\sqrt{5}}$.\n[/hide]" } { "Tag": [ "quadratics", "algebra" ], "Problem": "I just get little confused about this I see it easy but I'm just confused :huh:. If $ \\begin{array}{l}a \\plus{} b \\equal{} 1 \\\\ a^2 \\plus{} b^2 \\equal{} 2 \\\\ \\end{array}$ Find $ a$ and $ b$", "Solution_1": "hello, rewriting the equation $ a\\plus{}b\\equal{}1$ in the form $ b\\equal{}1\\minus{}a$ and plugging this in the second equation we have $ a^2\\plus{}(1\\minus{}a)^2\\equal{}2$ or $ a^2\\minus{}a\\minus{}\\frac{1}{2}\\equal{}0$.\r\nSolving this quadratic equation we get $ a_1\\equal{}\\frac{1\\plus{}\\sqrt{3}}{2}$ or $ a_2\\equal{}\\frac{1\\minus{}\\sqrt{3}}{2}$ and we find $ b_1\\equal{}1\\minus{}a_1$ and $ b_2\\equal{}1\\minus{}a_2$.\r\nSonnhard.", "Solution_2": "Thanks a lot Dr Sonnhard Graubner" } { "Tag": [ "calculus", "integration", "function", "real analysis", "real analysis unsolved" ], "Problem": "determine the following integral\r\n\\[ \\int_0^\\infty\\frac{dx}{1+x^n} \\]\r\nwhere n is of course positive integer greater than 1.", "Solution_1": "For a complex variable approach, see [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=39543[/url].", "Solution_2": "That's not the only place; there's also a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21316]thread I posted almost a year ago[/url] and one it was inspired by.\r\n\r\nMy problem was: Calculate $\\sum_{n=-\\infty}^\\infty\\frac{(-1)^n}{1+nk}$. The answer shouldn't depend on $k$ being an integer, although $k>1$ is useful.", "Solution_3": "[quote=\"jmerry\"]That's not the only place; there's also a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21316]thread I posted almost a year ago[/url] and one it was inspired by.\n\nMy problem was: Calculate $\\sum_{n=-\\infty}^\\infty\\frac{(-1)^n}{1+nk}$. The answer shouldn't depend on $k$ being an integer, although $k>1$ is useful.[/quote]\r\nyou can reduce the series to very nice result with the sinus function", "Solution_4": "Of course I know the answer- I proposed that problem in the first place because I thought it was nice. You could at least follow the link..." } { "Tag": [ "function", "algebra proposed", "algebra" ], "Problem": "If $ f: \\mathbb{R} \\rightarrow \\mathbb{R}$ is a continuous increasing function such that $ f^2(x) \\equal{} x$ $ \\forall$ $ x \\in \\mathbb{R}$, prove that $ f(x) \\equal{} x$ $ \\forall$ $ x \\in \\mathbb{R}$.", "Solution_1": "[quote=\"bakerbakura\"]If $ f: \\mathbb{R} \\rightarrow \\mathbb{R}$ is a continuous increasing function such that $ f^2(x) \\equal{} x$ $ \\forall$ $ x \\in \\mathbb{R}$, prove that $ f(x) \\equal{} x$ $ \\forall$ $ x \\in \\mathbb{R}$.[/quote]\r\n\r\nYou should modify your problem : We cant have both $ f^2(x)\\equal{}x$ and $ f(x)\\equal{}x$ $ \\forall x$", "Solution_2": "Do you mean $ f\\left( {f\\left( x \\right)} \\right) \\equal{} x$ implies $ f(x) \\equal{} x$ ? So we can prove it easily by reductio ad absurdum. If $ f(x) > x$ then $ f(f(x)) > f(x) > x$, the contradiction. The case $ f(x) < x$ is similar. So $ f(x) \\equal{} x$" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "prove that the equation x^{3}+y^{3}=z^{3} has no solutions.", "Solution_1": "Consider either Fermat's last theorem for $ n\\equal{}3$ or\r\n\r\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=177026[/url]" } { "Tag": [ "geometry", "rhombus", "ratio", "LaTeX", "angle bisector" ], "Problem": "Consider rhombus $ABCD$ and point $E$ which lies on $AC$, the longest diagonal fo the rhombus. If angle $BCD=60$ degrees, and $CE=CD$, then what is the ratio of the area of quadrilateral $ABED$ to the area of quadrilateral $BCDE$. Hopefully, y'all won't need a diagram :P", "Solution_1": "Ohh, this one. :lol: I've solved it before, but I'll redo it. ;)\r\n\r\n[hide=\"Answer\"]Setting each side of the rhombus equal to $1$, we see that diagonal $AC=\\sqrt{3}$, and $BD=1$, since the rhombus consists of two equilateral triangles. If $F$ is the intersection of the diagonals, we see that $CF=\\frac{\\sqrt{3}}{2}$, so $EF=1-\\frac{\\sqrt{3}}{2}$. Since $BF=\\frac{1}{2}$, the area of $\\triangle BDE=\\frac{1}{2}\\cdot (1-\\frac{\\sqrt{3}}{2})=\\frac{1}{2}-\\frac{\\sqrt{3}}{4}$. Therefore, the ratio between the two quadrilaterals is $\\frac{\\frac{\\sqrt{3}}{4}-(\\frac{1}{2}-\\frac{\\sqrt{3}}{4})}{\\frac{\\sqrt{3}}{4}+(\\frac{1}{2}-\\frac{\\sqrt{3}}{4})}= \\frac{\\frac{\\sqrt{3}}{2}-\\frac{1}{2}}{\\frac{1}{2}}=\\sqrt{3}-1$.[/hide]", "Solution_2": "[hide]By setting CE=CD, we are making two isosceles 75-75-30 triangles. We can put the midpoint of AC as well and connect and see the two 15-75-90 triangles which are formed. By the angle bisector theorem, we can find that tan15 degrees=2+sqrt[3]. Thus, setting a side of the rhombus as x, we can find that the sum of the areas of those two triangles is (x/2)(x/(4+2sqrt[3]))=x^2/8+4sqrt[3]=x^2(8-4sqrt[3])/(16)=x^2(2-sqrt[3])/(4). The area of BDC is sqrt[3](x^2)/4. Thus, we can add and get the area of BCDE as (x^2)/2. We know that the original area of the rhombus is (x^2)(sqrt[3])/2. Thus, by subtracting, we get the ratio between ABED and BCDE is $\\sqrt{3}-1$.[/hide]\r\nSorry if it's a bit unclear, pretty void of latex and hard to follow, it's late over here and i'm sleepy...i might fix it in the morning....", "Solution_3": "Wasnt this on the AMC daily contest thing going on right now????(I misunderstood that time and got it wrong :blush: )", "Solution_4": "AMC daily contest? Where is that? Is this problem over yet, or should i edit my post?", "Solution_5": "The problem was over around a week ago, probably. It's on the AMC forums, but it's kind of on hold for now, since Silverfalcon, who's doing the practice, has computer problems.", "Solution_6": "[quote=\"xxreddevilzxx\"]Wasnt this on the AMC daily contest thing going on right now????(I misunderstood that time and got it wrong :blush: )[/quote]\r\n\r\nIt was? :? I got it from a South Carolina Math competition", "Solution_7": "same answer but different method. :D" } { "Tag": [ "trigonometry", "inequalities" ], "Problem": "Slove this equation: $ tanx\\plus{}tan^2x\\plus{}tan^3x\\plus{}cotx\\plus{}cot^2x\\plus{}cot^3x\\equal{}6$", "Solution_1": "[hide]\\begin{eqnarray*}\\tan^3 x + \\cot^3 x & = & (\\tan x + \\cot x)^3 - 3\\tan x \\cot x (\\tan x + \\cot x) \\\\\n& = & (\\tan x + \\cot x)^3 - 3(\\tan x + \\cot x) \\\\\n\\tan^2 x + \\cot^2 x & = & (\\tan x + \\cot x)^2 - 2 \\tan x \\cot x \\\\\n& = & (\\tan x + \\cot x)^2 - 2\\end{eqnarray*}\nSo the original equation is equivalent to\n$ 0 = (\\tan x + \\cot x)^3 + (\\tan x + \\cot x)^2 - 2(\\tan x + \\cot x) - 8$\nLet $ y = \\tan x + \\cot x$, then\n$ 0 = y^3 + y^2 - 2y - 8 = (y - 2)(y^2 + 3y + 4) \\Rightarrow y = 2 \\iff \\tan x + \\cot x = 2$\nBoth $ \\tan x$ and $ \\cot x$ must be positive for this to be true, so we can use AM-GM to find\n$ \\frac {\\tan x + \\cot x}{2} \\geq \\sqrt {\\tan x \\cot x} = 1 \\Rightarrow \\tan x + \\cot x \\geq 2$\nSince equality holds, $ \\tan x = \\cot x = 1$, and $ x = \\frac {\\pi}{4} + n\\pi , n\\in \\mathbb{Z}$\n[/hide]", "Solution_2": "[quote=\"Brut3Forc3\"][hide]\\begin{eqnarray*}\\tan^3 x + \\cot^3 x & = & (\\tan x + \\cot x)^3 - 3\\tan x \\cot x (\\tan x + \\cot x) \\\\\n& = & (\\tan x + \\cot x)^3 - 3(\\tan x + \\cot x) \\\\\n\\tan^2 x + \\cot^2 x & = & (\\tan x + \\cot x)^2 - 2 \\tan x \\cot x \\\\\n& = & (\\tan x + \\cot x)^2 - 2\\end{eqnarray*}\nSo the original equation is equivalent to\n$ 0 = (\\tan x + \\cot x)^3 + (\\tan x + \\cot x)^2 - 2(\\tan x + \\cot x) - 8$\nLet $ y = \\tan x + \\cot x$, then\n$ 0 = y^3 + y^2 - 2y - 8 = (y - 2)(y^2 + 3y + 4) \\Rightarrow y = 2 \\iff \\tan x + \\cot x = 2$\nBoth $ \\tan x$ and $ \\cot x$ must be positive for this to be true, so we can use AM-GM to find\n$ \\frac {\\tan x + \\cot x}{2} \\geq \\sqrt {\\tan x \\cot x} = 1 \\Rightarrow \\tan x + \\cot x \\geq 2$\nSince equality holds, $ \\tan x = \\cot x = 1$, and $ x = \\frac {\\pi}{4} + n\\pi , n\\in \\mathbb{Z}$\n[/hide][/quote]\r\nThank you very much. :P", "Solution_3": "in this case, i think AM-GM inequality with 6 variables works better, again this is my opinion\r\n\r\nthe 6 variables would be the 6 terms on the left side in the beginning equation\r\nahh... brut3forc3 just used brut3forc3\r\n:rotfl:", "Solution_4": "[quote=\"stevenmeow\"]in this case, i think AM-GM inequality with 6 variables works better, again this is my opinion\n\nthe 6 variables would be the 6 terms on the left side in the beginning equation[/quote]\r\n......\r\n*slaps self for taking long solution*" } { "Tag": [ "inequalities", "trigonometry", "inequalities proposed" ], "Problem": "I make \"rude way\"(That is, you do not like the method) to prove that:\r\nLet $ x,y,z>0$,then\r\n\r\n$ 6\\,zy \\plus{} 6\\,xz \\plus{} 6\\,yx\\leq 8\\,{\\frac {\\left( x \\plus{} y \\plus{} z \\right) xyz}{\\left( y \\plus{} z \\right) \\left( z \\plus{} 2\\,x \\plus{} y \\right) }} \\plus{} 8\\,{\\frac {\\left( x \\plus{} y \\plus{} z \\right) xyz}{\\left( z \\plus{} x \\right) \\left( x \\plus{} 2\\,y \\plus{} z \\right) }} \\plus{} 8\\,{ \\frac {\\left( x \\plus{} y \\plus{} z \\right) xyz}{\\left( x \\plus{} y \\right) \\left( y \\plus{} 2\\,z \\plus{} x \\right) }} \\plus{} 3\\,{x}^{2} \\plus{} 3\\,{y}^{2} \\plus{} 3\\,{z}^{2}.$\r\n\r\n\r\n[size=150]Tell me a wonderful solution,[/size]\r\n\r\nand\r\n\r\n[size=200]I also you a wonderful background[/size]", "Solution_1": "It's obviously trues by AM-GM and the well-known:\r\n$ \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca} \\plus{} \\frac {8abc}{(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)} \\ge 2$\r\n :)\r\nBy this way, we can also prove the general problem:\r\n\\[ 6(xy\\plus{}yz\\plus{}zx) \\le 2(k\\plus{}2)xyz(x\\plus{}y\\plus{}z) \\sum \\frac{1}{(y\\plus{}z)(kx\\plus{}y\\plus{}z)}\\plus{}3(x^2\\plus{}y^2\\plus{}z^2)\\]\r\nWith $ a, b, c, k > 0$", "Solution_2": "[quote=\"xzlbq\"]I make \"rude way\"(That is, you do not like the method) to prove that:\nLet $ x,y,z > 0$,then\n\n$ 6\\,zy \\plus{} 6\\,xz \\plus{} 6\\,yx\\leq 8\\,{\\frac {\\left( x \\plus{} y \\plus{} z \\right) xyz}{\\left( y \\plus{} z \\right) \\left( z \\plus{} 2\\,x \\plus{} y \\right) }} \\plus{} 8\\,{\\frac {\\left( x \\plus{} y \\plus{} z \\right) xyz}{\\left( z \\plus{} x \\right) \\left( x \\plus{} 2\\,y \\plus{} z \\right) }} \\plus{} 8\\,{\\frac {\\left( x \\plus{} y \\plus{} z \\right) xyz}{\\left( x \\plus{} y \\right) \\left( y \\plus{} 2\\,z \\plus{} x \\right) }} \\plus{} 3\\,{x}^{2} \\plus{} 3\\,{y}^{2} \\plus{} 3\\,{z}^{2}.$\n\n\n[size=150]Tell me a wonderful solution,[/size]\n\nand\n\n[size=200]I also you a wonderful background[/size][/quote]\r\n\r\n\r\n$ 6\\,zy \\plus{} 6\\,xz \\plus{} 6\\,yx\\leq 8\\,{\\frac {\\left( x \\plus{} y \\plus{} z \\right) xyz}{\\left( y \\plus{} z \\right) \\left( z \\plus{} 2\\,x \\plus{} y \\right) }} \\plus{} 8\\,{\\frac {\\left( x \\plus{} y \\plus{} z \\right) xyz}{\\left( z \\plus{} x \\right) \\left( x \\plus{} 2\\,y \\plus{} z \\right) }} \\plus{}$ $ 8\\,{\\frac {\\left( x \\plus{} y \\plus{} z \\right) xyz}{\\left( x \\plus{} y \\right) \\left( y \\plus{} 2\\,z \\plus{} x \\right) }} \\plus{} 3\\,{x}^{2} \\plus{} 3\\,{y}^{2} \\plus{} 3\\,{z}^{2}.$\r\n\r\n$ \\Longleftrightarrow \\sum_{sym}(6x^7z \\plus{} 15x^6y^2 \\plus{} 19x^6yz \\minus{} 6x^5y^3$ $ \\plus{} 42x^5y^2z \\minus{} 15x^4y^4 \\minus{} 38x^4y^3z \\minus{} 23x^3y^3z^2)\\geq 0$\r\n\r\n\r\ntrue due to Muirhead theorem.\r\n\r\n :lol:", "Solution_3": "[quote=\"nguoivn\"]It's obviously trues by AM-GM and the well-known:\n$ \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca} \\plus{} \\frac {8abc}{(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)} \\ge 2$\n :)\nBy this way, we can also prove the general problem:\n\\[ 6(xy \\plus{} yz \\plus{} zx) \\le 2(k \\plus{} 2)xyz(x \\plus{} y \\plus{} z) \\sum \\frac {1}{(y \\plus{} z)(kx \\plus{} y \\plus{} z)} \\plus{} 3(x^2 \\plus{} y^2 \\plus{} z^2)\\]\nWith $ a, b, c, k > 0$[/quote]\r\n\r\nwait a lot,I go home,copy background ,\r\nthank you.\r\nBQ", "Solution_4": "Let $ ABC$ is a triangle,\r\n\r\nNote:\r\n$ \\cos\\frac {A}{2}\\le \\frac {2h_a}{b \\plus{} c} ,$\r\n\r\n$ 4\\sqrt {3}S \\plus{} 3\\sum{(b \\minus{} c)^2} \\geq a^2 \\plus{} b^2 \\plus{} c^2$(F-H ineq)\r\n<=\r\n\r\n$ 4(\\frac {2}{3}\\sum{\\cos\\frac {A}{2}})S \\plus{} 3\\sum{(b \\minus{} c)^2} \\geq a^2 \\plus{} b^2 \\plus{} c^2$\r\n\r\n<=\r\n$ 4(\\frac {2}{3}\\sum{\\frac {2h_a}{b \\plus{} c}})S \\plus{} 3\\sum{(b \\minus{} c)^2} \\geq a^2 \\plus{} b^2 \\plus{} c^2$\r\n\r\n(a=y+z,b=z+x,c=x+y)\r\n<=>\r\n\r\n$ 6\\,yz \\plus{} 6\\,zx \\plus{} 6\\,xy\\leq 8\\,{\\frac {x \\left( x \\plus{} y \\plus{} z \\right) yz}{\\left( y \\plus{} z \\right) \\left( 2\\,x \\plus{} z \\plus{} y \\right) }} \\plus{} 8\\,{\\frac {x \\left( x \\plus{} y \\plus{} z \\right) yz}{\\left( z \\plus{} x \\right) \\left( 2\\,y \\plus{} x \\plus{} z \\right) }} \\plus{} 8\\,{ \\frac {x \\left( x \\plus{} y \\plus{} z \\right) yz}{\\left( x \\plus{} y \\right) \\left( 2\\,z \\plus{} y \\plus{} x \\right) }} \\plus{} 3\\,{x}^{2} \\plus{} 3\\,{y}^{2} \\plus{} 3\\,{z}^{2} .$\r\nBQ", "Solution_5": "$ 4\\sqrt {3}S \\plus{} 3\\sum{(b \\minus{} c)^2} \\geq a^2 \\plus{} b^2 \\plus{} c^2$(F-H ineq)----------------(*)\r\n<=\r\n\r\n$ 4(\\frac {h_a \\plus{} h_b \\plus{} h_c}{s})S \\plus{} 3\\sum{(b \\minus{} c)^2} \\geq a^2 \\plus{} b^2 \\plus{} c^2$\r\n\r\n(a=y+z,b=z+x,c=x+y)\r\n<=>\r\n$ 2xyz(\\frac {1}{y \\plus{} z } \\plus{} \\frac {1}{z \\plus{} x} \\plus{} \\frac {1}{x \\plus{} y}) \\plus{} x^2 \\plus{} y^2 \\plus{} z^2 \\minus{} 2xy \\minus{} 2zx \\minus{} 2yz\\geq 0.$-----------------\uff08**\uff09\r\n\r\n<=>\r\n$ 0\\leq {x}^{4}y \\minus{} {x}^{3}{y}^{2} \\minus{} {x}^{3}{z}^{2} \\plus{} {x}^{4}z \\plus{} {y}^{4}z \\minus{} {y}^{3} {x}^{2} \\plus{} {y}^{4}x \\minus{} {y}^{3}{z}^{2} \\minus{} {z}^{3}{y}^{2} \\plus{} {z}^{4}x \\plus{} {z}^{4}y \\minus{} {z}^{ 3}{x}^{2}$\r\n\r\n<=>\r\n$ {x}^{2} \\left( y \\plus{} z \\right) \\left( x \\minus{} y \\right) \\left( x \\minus{} z \\right) \\plus{} {y}^{2} \\left( z \\plus{} x \\right) \\left( y \\minus{} z \\right) \\left( y \\minus{} x \\right) \\plus{} {z}^{2} \\left( x \\plus{} y \\right) \\left( z \\minus{} x \\right) \\left( z \\minus{} y \\right) \\plus{} 2\\,xyz \\left( {x}^{2} \\plus{} {y}^{2} \\plus{} {z}^{2} \\minus{} xy \\minus{} yz \\minus{} zx \\right)\\geq 0 .$\r\n\r\nClear from the above,\r\nIneq (*) Is not too strong\u3002\r\nBQ\r\n\r\n[size=200]Last\uff0cfor nice,Ineq(**),by CS\uff1f[/size]BQ", "Solution_6": "[size=150]The above facts give us an good idea:\nIf you feel empty in Algebra,\nWent to the triangle to find\u3002\nBQ[/size]" } { "Tag": [ "function", "algebra", "polynomial", "number theory", "relatively prime", "algebra unsolved" ], "Problem": "(i) Are there such rational functions $ f$ and $ g$ that $ (f(x))^3 \\plus{} (g(x))^3 \\equal{} x$\r\n\r\n(ii) Are there such rational functions $ f,g,h$ that $ (f(x))^3 \\plus{} (g(x))^3 \\plus{} (h(x))^3\\equal{} x$", "Solution_1": "i) Equivalently, are there polynomials $ p, q, r$ such that $ p(x)^3 \\plus{} q(x)^3 \\equal{} x r(x)^3$? Such polynomials can be taken to be pairwise relatively prime without loss of generality. Now, by the [url=http://mathworld.wolfram.com/MasonsTheorem.html]Mason-Stothers theorem[/url], the number of distinct roots among $ p(x)^3, q(x)^3, x r(x)^3$ is at least $ 1 \\plus{} 3 \\text{max}( \\text{deg}(p), \\text{deg}(q), \\text{deg}(r) \\plus{} \\frac {1}{3} )$, but the number of distinct roots is at most $ \\deg{p} \\plus{} \\deg{q} \\plus{} \\deg{r} \\plus{} 1$; contradiction.\r\n\r\nii) Over what field?" } { "Tag": [ "geometry", "trigonometry", "ARML", "AMC", "AIME", "USA(J)MO", "USAMO" ], "Problem": "Does anyone know a good site that has good information on theorems/techniques that are good for amc/aime geometry? I know that the AoPS are good but until I get those I would like to know if there are any good sites to learn from.\r\n\r\nOr you guys could just post stuff here. Either way I would appreciate it because I really need to get better with geometry.", "Solution_1": "AMC/AIME geometry usually uses the same geometry as any normal geometry class. I don't think there's any one topic in geometry that is most important for AMC/AIME. You might need to know some trig though. Just basically try to label a few lengths/angles/whatever as variables, and using your knowledge, express other lengths/angles/whatever using previously established variables.", "Solution_2": "Well, I just need a ton of practice then.", "Solution_3": "joml88 you should try the formula sheet that was made by Jackpo and others. It doesn't give any proofs or motivations for the results, but it gives you an idea of some things that are good to know. You should see if you can try to prove some of these results (some are easier to prove than others). We put a copy of it on our ARML website here: http://155.225.48.46/statemat/stuf2004/formulas.pdf\r\n\r\nAnd then you can try these Geometry Problems for practice. Most of them are former AIME and ARML problems that were discussed here. http://155.225.48.46/statemat/stuf2004/Geoprobs.htm\r\n\r\nBring a copy of these on the bus to ARML and we can talk about the ones that you have questions about on the way up there.", "Solution_4": "We will eventually offer a course in AMC/AIME level geometry.", "Solution_5": "Get this book: [url=http://www.amazon.com/exec/obidos/tg/detail/-/0486691543/qid=1084731527/sr=8-1/ref=sr_8_xs_ap_i1_xgl14/104-0048912-6303167?v=glance&s=books&n=507846]Challenging Problems in Geometry[/url] (best value math book, only $10) and do all the problems in the book. \r\n\r\nA friend of mine started working on the book when he is in 7th grade. After a year, He solved all the geometry problems (left out some easy algebra problems) in AMC12B/AIME and made USAMO. And he got 7 points on the USAMO by essentially solving the first (geometry) problem!" } { "Tag": [ "calculus", "integration", "Putnam", "trigonometry", "calculus computations" ], "Problem": "starting from the LHS, show [u][b]how to arrive[/b][/u] at the integrals shown:\r\n\r\n\r\n$\\textbf{(i)}\\;\\;\\;\\quad \\frac{22}{7}-\\pi\\;=\\;\\int_{0}^{1}\\;\\frac{\\left[ x\\;(1-x)\\right]^{4}}{1+x^{2}}\\;dx$\r\n\r\n$\\textbf{(ii)}\\;\\;\\;\\quad \\frac{355}{113}-\\pi\\;=\\;\\int_{0}^{1}\\;\\frac{(25+816\\;x^{2}) \\;\\left[ x\\;(1-x)\\right]^{8}}{3164\\; (1+x^{2})}\\;dx$", "Solution_1": "$\\int_{0}^{1}\\frac{x^{4}(1-x)^{4}}{1+x^{2}}\\ dx=\\frac{22}{7}-\\pi$. Putnam A-1 1968.\r\n\r\nProof: Let $x=\\tan \\theta$ , we have \r\n\r\n$\\int_{0}^{1}\\frac{x^{4}(1-x^{4})}{1+x^{2}}\\ dx$\r\n\r\n$=\\int_{0}^{\\frac{\\pi}{4}}\\frac{\\tan^{4}\\theta (1-\\tan \\theta)^{4}}{1+\\tan^{2}\\theta }\\cdot \\frac{1}{\\cos^{2}\\theta}\\ d\\theta$\r\n\r\n$=\\int_{0}^{\\frac{\\pi}{4}}(\\tan^{4}\\theta-5\\tan^{5}\\theta+6\\tan^{6}\\theta-4\\tan^{7}\\theta+\\tan^{8}\\theta )\\ d\\theta$\r\n\r\nLet $I_{n}=\\int_{0}^{\\frac{\\pi}{4}}\\tan^{n}\\theta \\ d\\theta (n=0,\\ 1, 2, \\cdots)$, we have \r\n\r\n$I_{n+2}=\\int_{0}^{\\frac{\\pi}{4}}\\tan^{n+2}\\theta =\\int_{0}^{\\frac{\\pi}{4}}\\tan^{n}\\theta \\tan^{2}\\theta d\\theta=\\int_{0}^{\\frac{\\pi}{4}}\\tan^{n}\\theta \\left(\\frac{1}{\\cos^{2}\\theta }-1\\right)d\\theta$\r\n\r\n$=\\left[\\frac{\\tan^{n+1}\\theta}{n+1}\\right]_{0}^{\\frac{\\pi}{4}}=\\frac{1}{n+1}-I_{n}$, yielding $I_{n+2}+I_{n}=\\frac{1}{n+1}\\ (n=0,\\ 1,\\ 2,\\ \\cdots)$.\r\n\r\nThus $I_{6}=\\frac{1}{5}-I_{4}=\\frac{1}{5}-\\left(\\frac{1}{3}-I_{2}\\right)=I_{2}-\\frac{2}{15}=\\int_{0}^{\\frac{\\pi}{4}}\\left(\\frac{1}{\\cos^{2}\\theta }-1\\right) d\\theta-\\frac{2}{15}=\\frac{13}{15}-\\frac{\\pi}{4}$,\r\n\r\n$\\int_{0}^{1}\\frac{x^{4}(1-x)^{4}}{1+x^{2}}\\ dx=I_{4}-4I_{5}+6I_{6}-4I_{7}+I_{8}$\r\n\r\n$=(I_{4}+I_{6})+(I_{6}+I_{8})-4(I_{5}+I_{7})+4I_{6}$\r\n\r\n$=\\frac{1}{5}+\\frac{1}{7}-4\\cdot \\frac{1}{6}+4I_{6}$\r\n\r\n$=4I_{6}-\\frac{34}{105}=4\\left(\\frac{13}{15}-\\frac{\\pi}{4}\\right)-\\frac{34}{105}=\\frac{22}{7}-\\pi$. Q.E.D.\r\n\r\n[url=http://www.mathlinks.ro/viewtopic.php?t=18726]Kent Merryfield's Solution[/url]", "Solution_2": "sweet !! :thumbup: \r\n\r\ndon't worry, i am not going to even try and attempt $\\textbf{(ii)}$ ! LOL !! :D \r\n\r\nyesterday, my colleague asked me to find one similar integral for\r\n\r\n$\\boxed{20\\;+\\;\\pi-e^{\\pi}}$ and i already wasted half of my day today without any significant lead. :(\r\n\r\n\r\n :rotfl:", "Solution_3": "misan, can't you just do\r\n\r\n$\\int_{e^{\\pi}}^{20+\\pi}dx$ ?\r\n\r\n :rotfl:" } { "Tag": [ "inequalities", "triangle inequality" ], "Problem": "[b]Three sides of a quadrilateral are 1,2, and 5. The fourth side is an integer, $ x$,. What is the sum of all possible values of $ x$?[/b]", "Solution_1": "[hide]\n\nlet's see it has to be less than 1+2+5=8, but more or equal to than 2. Hence, 2+3+4+5+6+7=27\n\nSo, the answer is 27?\n\n\n[/hide]", "Solution_2": "[quote=\"Nexmus\"][hide]let's see it has to be less than 1+2+5=8, but more or equal to than 2. Hence, 2+3+4+5+6+7=27\n\nSo, the answer is 27?[/hide][/quote]\n[hide]I dont think 2 would work\n\nIts sides are 1,2,5,x then 8>x and 1+2+x>5, so x<8 and x>2 so [3,7] work so we have 3+4+5+6+7=25[/hide]\r\n\r\n[color=darkred][size=75]Don't forget hide tags.\n-nebula42[/size][/color]", "Solution_3": "[hide] Yes, two won't work, since then it would be $ 2+1+2=5$, which would result in a straight line. Also, please hide your answers. [/hide]", "Solution_4": "can you guys explain your reasoning? im completely lost in this problem. the first thing that i thought when i encountered it was triangle inequality which is stupid because its a quadrilateral. :blush:", "Solution_5": "It's the same logic as triangle inequality, If it is too little, it will be a straight line, so basically sum of 3 sides<1 side.", "Solution_6": "well, if we can find the longest possible side and the shortest possible side. if we add them together we get 8. we cant have 8 because then it would just be a straight line. now we take 5 and subtract the other two and we get 2. once again we have a straight line. so it must be the numbers between 3 and 7 or $ \\sum_{k=3}^{k=7}{k}$" } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "Let $ ABC$ be a triangle with $ \\angle ABC$ being acute. Suppose that $ K$ be a point on the side $ AB$, and $ H$ be its orthogonal projection on the line $ BC$. A ray $ Bx$ cuts the segment $ KH$ at $ E$ and meets the line passing through $ K$ parallel to $ BC$ at $ F$. Prove that $ \\angle ABC \\equal{} 3\\angle CBF$ if and only if $ EF \\equal{} 2BK$.", "Solution_1": "For History, we must say that this problem ( easy to prove ), is the [b][size=100]Pappos method[/size][/b], for the trisection of arbitrary acute angle.\r\n\r\nBut, the point $ F$ on the line through $ K$ and parallel to $ BC$ such that $ EF \\equal{} 2\\cdot BK,$ it is not constuctable geometrically (with ruler and compass).\r\n\r\nKostas Vittas.", "Solution_2": "Can you show the solution of the problem as well?", "Solution_3": "[quote=\"outback\"]Can you show the solution of the problem as well?[/quote]\r\nLet $ F$ be, the point on the line through $ K$ and parallel to $ BC,$ such that $ EF \\equal{} 2\\cdot BK,$ where $ E\\equiv KH\\cap BF$ and let $ M$ be, the midpoint of the segment $ EF.$\r\n\r\nSo, from $ BK \\equal{} KM \\equal{} MF$ and $ KF\\parallel BC,$ we have $ \\angle ABF \\equal{} \\angle KMB \\equal{} \\angle 2\\angle KFB \\equal{} 2\\angle CBF$ $ ,(1)$\r\n\r\nFrom $ (1)$ $ \\Longrightarrow$ $ \\angle ABC \\equal{} 3\\angle CBF$ and we are done.\r\n\r\nKostas Vittas." } { "Tag": [ "geometry", "3D geometry", "probability" ], "Problem": "A cube 4 units on each side is composed of 64 unit-cubes. 2 faces of the larger cube that share an edge are painted blue and then disassembled into the 64 cubes. Two of the cubes are selected randomly. What is the probability that exactly 1 cube will have 2 painted faces and the other cube has none?", "Solution_1": "A total of $ 28$ faces are painted, and only $ 2$ have $ 2$ painted faces. There are also $ 8$ cubes which have no paint. The required probability is $ \\frac{2}{64}\\times\\frac{8}{63}\\equal{}\\boxed{\\frac{1}{252}}$", "Solution_2": "ok, first of all, way more than 8 cubes have no paint, second of all, 2 faces of the large cube are painted, that means that there are 16*2= 32 small faces painted....\r\nthe number of cubes that arent painted is 4*3*3= 36\r\nok so the answer is $ 2*\\frac{2*36}{64*63}$ we have to multiply by 2 because the one with no faces painted can be the first cube, or the second one...\r\n\r\nthe answer is $ \\frac{1}{28}$", "Solution_3": "Wouldn't four unit cubes have two faces painted?", "Solution_4": "Yeah, that's what I was thinking, four unit cubes have two faces painted, those cubes are on the edge that the blue faces share.\n\nSo like RoFlLoLcOpT said, there are $16 + 16 - 4 = 28$ cubes which are painted. Then there are $36$ cubes which do not have any paint on them. So $\\frac{36 \\cdot 4}{\\binom{64}{2}}$ would be the answer. $\\boxed{\\frac{1}{14}}$.", "Solution_5": "Right. RoFlLoLcOpT missed a factor of 2.", "Solution_6": "Sorry for revive but shouldn't there be only 8 with no colored sides and 24 with 2 colored sides?\n\nEDIT: Nvm I didn't read the question I'm stupid :P", "Solution_7": "[hide=Solution]Since there are four unit cubes with exactly two painted faces and 64-28=36 unit cubes with no painted faces, our answer is $$2\\cdot \\frac{4}{64} \\cdot \\frac{36}{63}= \\boxed{\\frac{1}{14}}.$$[/hide]", "Solution_8": "HEY! I PUT $\\dfrac{\\dbinom{4}{1} \\cdot \\dbinom{36}{1}}{\\dbinom{64}{2}}$ and it didn't work even though it was right!!! :mad:", "Solution_9": "......did you simplify", "Solution_10": "it's kind of common sense to simplify lmao calculators are allowed in c&p calculations like this tho", "Solution_11": "[quote=Natsuki]NOO i put that because i was too lazy to compute and ALCUMUS SAID IT WAS WRONG! \n\nIt could've atleast given me a warning that I HAD TO SIMPLIFY, right????? But it didn't!!![/quote]\n\nStop blaming the system for your mistake. You forgot to simplify, so the blame goes totally to you. (The formatting tips says to simplify everything)", "Solution_12": "ouch lol\n\ntbh I think natsuki trying to act like natsuki, angry at everything", "Solution_13": "[quote=ilovepizza2020][quote=Natsuki]NOO i put that because i was too lazy to compute and ALCUMUS SAID IT WAS WRONG! \n\nIt could've atleast given me a warning that I HAD TO SIMPLIFY, right????? But it didn't!!![/quote]\n\nStop blaming the system for your mistake. You forgot to simplify, so the blame goes totally to you. (The formatting tips says to simplify everything)[/quote]\n\nLet's all be aware of other's feelings.", "Solution_14": "[quote=HIA2020][quote=ilovepizza2020][quote=Natsuki]NOO i put that because i was too lazy to compute and ALCUMUS SAID IT WAS WRONG! \n\nIt could've atleast given me a warning that I HAD TO SIMPLIFY, right????? But it didn't!!![/quote]\n\nStop blaming the system for your mistake. You forgot to simplify, so the blame goes totally to you. (The formatting tips says to simplify everything)[/quote]\n\nLet's all be aware of other's feelings.[/quote]\n\nuh ok then", "Solution_15": "noooooooo i did 16/64*4/64=9/256 like an idiotttttgttgtgdfsgre", "Solution_16": "yea it 1/14 dude" } { "Tag": [ "parameterization", "vector", "advanced fields", "advanced fields unsolved" ], "Problem": "Let $ \\gamma: (0, 1) \\longrightarrow M$ be a natural parametrization of curve $ \\Gamma$ which is on a smooth surface $ M \\subset \\mathbb{R}^3$ with the orientation given by $ N: M \\longrightarrow \\mathbb{R}^3$ (i. e. $ N$ is an unit vector such that $ N(p) \\perp T_pM$). Prove that if $ N(\\gamma(s)) \\equal{} b(s)$, for $ s \\in (0, 1)$, then the torsion $ \\tau$ of $ \\Gamma$ and the Gaussian curvature $ K$ of $ M$ satisfy the equation $ K(\\gamma(s)) \\equal{} \\minus{}\\tau(s)^2$, $ s \\in (0, 1)$ ($ b(s)$ denotes the binormal vector of $ \\Gamma$)", "Solution_1": "We have \\[ \\dot{\\overrightarrow{N}}(s)=\\pm{\\dot{\\overrightarrow{b}}(s)}\\] then using frenet \\[ \\dot{\\overrightarrow{N}}(s)=\\pm{(-\\tau(s)\\overrightarrow{n}(s))}\\] \\[ (\\dot{\\overrightarrow{N}}(s))^{2}=(\\pm{(-\\tau(s)\\overrightarrow{n}(s)})^{2}\\] but \\[ (\\dot{\\overrightarrow{N}}(s))^{2}=(\\overrightarrow{N}_{u}\\dot{u}(s)+\\overrightarrow{N}_{v}\\dot{v}(s))^{2}=III(du,dv)\\] So \\[ III(du,dv)={\\tau}^{2}(s)\\] because $ (\\overrightarrow{n}(s))^{2}=1$. Using the identity ${ III-2HII+KI=0}$ where $ 2H$ the mean curvature and $ K$ the gaussian curvature we have \\[ 2HII-KI={\\tau}^{2}(s)\\] but $ 2H=0$ because we have $ K_{N}=0$ and $ K_{N}=\\frac{II}{I}$ where $ K_{N}$ the normal curvature. So we have \\[ -KI={\\tau}^{2}(s)\\] but $ I=(\\dot{x}(s))^{2}=1$ because $ s$ natural parameter and we have \\[ -K(s)={\\tau}^{2}(s)\\]" } { "Tag": [ "quadratics", "algebra" ], "Problem": "$ a,b,c,d\\in R$\r\n\r\n$ a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} d^2 \\minus{} ab \\minus{} bc \\minus{} cd \\minus{} d \\plus{} \\frac {2}{5} \\equal{} 0$ \r\n\r\n$ a \\equal{} ?$", "Solution_1": "hello, writing your equation in the form\r\n$ a^2\\minus{}ab\\plus{}b^2\\plus{}c^2\\plus{}d^2\\minus{}bc\\minus{}cd\\minus{}d\\plus{}\\frac{2}{5}\\equal{}0$ and applying the formula for a quadratic equation we got\r\n$ a{_{1_2}}\\equal{}\\frac{b}{2}\\pm\\sqrt{\\frac{b^2}{4}\\minus{}b^2\\minus{}c^2\\minus{}d^2\\plus{}bc\\plus{}cd\\plus{}d\\minus{}\\frac{2}{5}}$\r\n$ a_{1_2}\\equal{}\\frac{b}{2}\\pm\\sqrt{\\minus{}\\frac{3}{4}b^2\\minus{}c^2\\minus{}d^2\\plus{}bc\\plus{}cd\\plus{}d\\minus{}\\frac{2}{5}}$\r\nSonnhard.", "Solution_2": "[hide=\"solution\"]\nsee the equation for $ a$,its discriminant $ D$ must be non-negative since $ a$ is real,\nthus $ D \\equal{} b^2 \\minus{} 4(b^2 \\plus{} c^2 \\plus{} d^2 \\minus{} bc \\minus{} ca \\minus{} d \\minus{} \\frac {2}{5}) \\equal{} \\minus{} 3(b \\minus{} \\frac {2c}{3})^2 \\minus{} \\frac {8}{3}(c \\minus{} \\frac {3d}{4})^2 \\minus{} \\frac {5}{2}(d \\minus{} \\frac {4}{5})^2\\ge 0$,then because $ b,c,d$ are reals,\n$ b \\minus{} \\frac {2c}{3} \\equal{} c \\minus{} \\frac {3d}{4} \\equal{} d \\minus{} \\frac {4}{5} \\equal{} 0$ must hold,i.e.\n\n$ b \\equal{} \\frac {2}{5},c \\equal{} \\frac {3}{5},d \\equal{} \\frac {4}{5}$.\n\nthen the solution is $ a \\equal{} \\frac {b}{2} \\equal{} \\frac {1}{5}$.\n[/hide]" } { "Tag": [ "AMC", "AMC 10", "AMC 12" ], "Problem": "In the teacher's manual, it says the 10th or under student can choose which test to take - the AMC 10 or AMC 12, but I see several students here who have goals for both exams. Do you have to take one on the first administration (a) and the other at the B time? Is that OK?", "Solution_1": "You can take both the AMC 10 and 12. You are correct in that you would probably have to take one Test A and one Test B, as the AMC 10 and 12 are administered on the same day.\r\n\r\n[quote=\"AMC website\"]\nAMC 12 Eligibility \u2013 A student in a program\nleading to a high school diploma, and under 19.5\nyears of age on the day of the contest.\n\nAMC 10 Eligibility \u2013 A student in a program\nleading to a high school diploma, under 17.5\nyears of age on the day of the contest, and not\nenrolled in grades 11 or 12 or equivalent.\n[/quote]\r\n\r\nAlso, I'm not sure if this belongs in Middle School Contests.", "Solution_2": "Anyone in or below 12th grade can take two of the tests, as long as he/she can find a place to take it.\r\n\r\nFor 10th graders or below there are 4 options for two dates:\r\n1) AMC10 on the A and B dates\r\n2) AMC10 on the A date and AMC12 on the B date\r\n3) AMC12 on the A date and AMC10 on the B date\r\n4) AMC12 on the A and B dates\r\nIf you choose to take the test on only one date, it can either be AMC10 or AMC12, but not both.\r\n\r\nFor 12th graders and 11th graders, there is only 3 options:\r\n1) AMC12 on both A and B dates\r\n2) AMC12 on only A date\r\n3) AMC12 on only B date\r\n\r\nFor more information, see [url=http://www.unl.edu/amc/f-miscellaneous/faq.shtml]here[/url] (near the bottom, where it says Q, AMC 10/12 #12). Also, this might better belong in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=133]AMC Forum[/url].", "Solution_3": "Thank you for all the thorough info. I'm sorry I posted in the wrong place - the student I was inquiring about (has an interest in AMC 10 and maybe 12) is in middle school - that's how I ended up here. I will try to pay better attention in the future! Thanks again." } { "Tag": [ "exterior angle" ], "Problem": "In circle $ O$ chord $ AB$ is produced so that $ BC$ equals a radius of the circle. $ CO$ is drawn and extended to $ D$. $ AO$ is drawn. Which of the following expresses the relationship between $ x$ and $ y$?\n\n[asy]size(200);defaultpen(linewidth(0.7)+fontsize(10));\npair O=origin, D=dir(195), A=dir(150), B=dir(30), C=B+1*dir(0);\ndraw(O--A--C--D);\ndot(A^^B^^C^^D^^O);\npair point=O;\nlabel(\"$A$\", A, dir(point--A));\nlabel(\"$B$\", B, dir(point--B));\nlabel(\"$C$\", C, dir(point--C));\nlabel(\"$D$\", D, dir(point--D));\nlabel(\"$O$\", O, dir(285));\nlabel(\"$x$\", O+0.1*dir(172.5), dir(172.5));\nlabel(\"$y$\", C+0.4*dir(187.5), dir(187.5));\ndraw(Circle(O,1));\n[/asy]\n\n\n$ \\textbf{(A)}\\ x\\equal{}3y \\\\\n\\textbf{(B)}\\ x\\equal{}2y \\\\\n\\textbf{(C)}\\ x\\equal{}60^\\circ \\\\\n\\textbf{(D)}\\ \\text{there is no special relationship between }x\\text{ and }y \\\\\n\\textbf{(E)}\\ x\\equal{}2y \\text{ or }x\\equal{}3y\\text{, depending upon the length of }AB$", "Solution_1": "[hide=\"sketching strongly advised\"]\nFirst let $ E$ be intersection of the circle and $ CD$.\n\n$ \\triangle AOD$ is isosceles $ (|OA|\\equal{}|OD|)$ $ \\implies \\angle DAO\\equal{}90^\\circ\\minus{}\\frac{x}{2}, \\ \\angle ODA\\equal{}90^\\circ\\minus{}\\frac{x}{2}$\n$ (90^\\circ\\minus{}\\frac{x}{2}\\plus{}90^\\circ\\minus{}\\frac{x}{2}\\plus{}x\\equal{}180^\\circ)$\n\n$ \\angle DAE\\equal{}90^\\circ$ (Thales' theorem) $ \\implies \\angle OAE\\equal{} \\angle DAE \\minus{} \\angle DAO \\implies \\angle OAE\\equal{}\\frac{x}{2}$\n\n$ \\triangle CBO$ is isosceles $ (|BO|\\equal{}|BC|) \\implies \\angle BOE\\equal{} \\angle OCB \\equal{} y$\n\n$ \\angle BAE\\equal{} \\frac{\\angle BOE}{2}$ (because of the common chord $ BE$) $ \\implies \\angle BAE\\equal{}\\frac{y}{2}$\n\n$ \\angle AOE \\plus{} \\angle AOD\\equal{}180^\\circ \\implies \\angle AOE\\equal{}180^\\circ \\minus{} x$\n\nNow we have all the angles of $ \\triangle AOC$:\n\n$ \\angle OAE \\plus{}\\angle BAE \\plus{} \\angle OCB \\plus{} \\angle AOE\\equal{}\\frac{x}{2}\\plus{}\\frac{y}{2}\\plus{}y\\plus{}(180^\\circ \\minus{} x)\\equal{}180^\\circ \\implies \\boxed{x\\equal{}3y}$ or $ \\textbf{(A)}$[/hide]", "Solution_2": "[hide=\"No sketching solution\"]Just use the Exterior Angle Theorem (EAT) twice:\n$ OB\\equal{}BC$ implies that $ \\angle BOC\\equal{}\\angle BCO\\equal{}y$ (Angles of Isosceles Triangle Theorem) and so by EAT, $ \\angle OBA\\equal{}\\angle BOC\\plus{}\\angle BCO\\equal{}y\\plus{}y\\equal{}2y$. \n$ OB\\equal{}OA$ implies that $ \\angle OBA\\equal{}\\angle OAB\\equal{}2y$ (AITT) and so again EAT gives $ x\\equal{}\\angle OAC\\plus{}\\angle OCA\\equal{}2y\\plus{}y\\equal{}3y$.[/hide]" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Given $ a, b, c \\ge 0$. Prove that: $ \\frac {1}{(a \\plus{} b)^2} \\plus{} \\frac {1}{(b \\plus{} c)^2} \\plus{} \\frac {1}{(c \\plus{} a)^2} \\plus{} \\frac {8(ab \\plus{} bc \\plus{} ca)}{(a \\plus{} b \\plus{} c)^4} \\ge \\frac {11}{4(ab \\plus{} bc \\plus{} ca)}$ \r\n:)", "Solution_1": "with very ugly way :blush: I have proved a stronger one:\r\n\r\n$ \\sum{ \\frac{1}{(a\\plus{}b)^2}} \\geq \\frac{ a^4\\plus{}b^4\\plus{}c^4\\plus{}7(ab\\plus{}ac\\plus{}bc)^2}{(a\\plus{}b)^2(a\\plus{}c)^2(b\\plus{}c)^2}$ one can prove that is stronger with pqr :wink:", "Solution_2": "This inequality is in my opinion very nice but also hard as we have equality only for $ a\\equal{}0,b\\equal{}c$ and cyclic. I tried to solve it by full expansion and p,q,r method but it is very ugly, but I cannot solve it by a nice trick :blush: so I am waiting for a nice solution. :) \r\n\r\nThank you very much to anyone will find a solution. :roll:", "Solution_3": "[quote=\"nguoivn\"]Given $ a, b, c \\ge 0$. Prove that: $ \\frac {1}{(a \\plus{} b)^2} \\plus{} \\frac {1}{(b \\plus{} c)^2} \\plus{} \\frac {1}{(c \\plus{} a)^2} \\plus{} \\frac {8(ab \\plus{} bc \\plus{} ca)}{(a \\plus{} b \\plus{} c)^4} \\ge \\frac {11}{4(ab \\plus{} bc \\plus{} ca)}$ \n:)[/quote]\n[quote=\"anas\"]with very ugly way :blush: I have proved a stronger one:\n\n$ \\sum{\\frac {1}{(a \\plus{} b)^2}} \\geq \\frac { a^4 \\plus{} b^4 \\plus{} c^4 \\plus{} 7(ab \\plus{} ac \\plus{} bc)^2}{(a \\plus{} b)^2(a \\plus{} c)^2(b \\plus{} c)^2}$ one can prove that is stronger with pqr :wink:[/quote]\r\nIn the both inequalities enough check only two cases:\r\n1) $ b\\equal{}1,$ $ c\\equal{}0$;\r\n2) $ b\\equal{}c\\equal{}1.$", "Solution_4": "[quote=\"anas\"]with very ugly way :blush: I have proved a stronger one:\n\n$ \\sum{\\frac {1}{(a \\plus{} b)^2}} \\geq \\frac { a^4 \\plus{} b^4 \\plus{} c^4 \\plus{} 7(ab \\plus{} ac \\plus{} bc)^2}{(a \\plus{} b)^2(a \\plus{} c)^2(b \\plus{} c)^2}$ one can prove that is stronger with pqr :wink:[/quote]\r\nThis is not true; try $ a\\equal{}5, b\\equal{}6, c\\equal{}7.$", "Solution_5": "[quote=\"nguoivn\"]Given $ a, b, c \\ge 0$. Prove that: $ \\frac {1}{(a \\plus{} b)^2} \\plus{} \\frac {1}{(b \\plus{} c)^2} \\plus{} \\frac {1}{(c \\plus{} a)^2} \\plus{} \\frac {8(ab \\plus{} bc \\plus{} ca)}{(a \\plus{} b \\plus{} c)^4} \\ge \\frac {11}{4(ab \\plus{} bc \\plus{} ca)}$ \n:)[/quote]\r\nWe can prove it by only Am-Gm and Cauchy Schwarts :)", "Solution_6": "Can you please post your proof, nguoivn?\r\n\r\nThank you very much. :)", "Solution_7": "[quote=\"manlio\"]Can you please post your proof, nguoivn?\n\nThank you very much. :)[/quote]\r\nI am now very ill, but I will help you...\r\n\r\nWrite the inequality as\r\n$ \\sum \\frac{ab\\plus{}bc\\plus{}ca}{(b\\plus{}c)^2}\\plus{}\\frac{8(ab\\plus{}bc\\plus{}ca)^2}{(a\\plus{}b\\plus{}c)^4} \\ge \\frac{11}{4},$\r\nor\r\n$ \\sum \\frac{a}{b\\plus{}c}\\plus{}\\sum \\frac{bc}{(b\\plus{}c)^2} \\plus{}\\frac{8(ab\\plus{}bc\\plus{}ca)^2}{(a\\plus{}b\\plus{}c)^4} \\ge \\frac{11}{4}.$\r\nIt is easy to check that (by applying the Cauchy-Schwarz Inequality)\r\n$ \\sum \\frac{a}{b\\plus{}c} \\ge \\frac{(a\\plus{}b\\plus{}c)^2}{2(ab\\plus{}bc\\plus{}ca)}.$\r\nIn addition, one has\r\n$ \\sum \\frac{bc}{(b\\plus{}c)^2} \\ge \\frac{1}{2}\\sum \\frac{bc}{b^2\\plus{}c^2} \\equal{}\\frac{1}{4}\\sum \\frac{(b\\plus{}c)^2}{b^2\\plus{}c^2}\\minus{}\\frac{3}{4} \\ge \\frac{(a\\plus{}b\\plus{}c)^2}{2(a^2\\plus{}b^2\\plus{}c^2)}\\minus{}\\frac{3}{4}.$\r\nBy using this and the above inequality, we may reduce our inequality to the following\r\n$ \\frac{(a\\plus{}b\\plus{}c)^2}{2(ab\\plus{}bc\\plus{}ca)}\\plus{}\\frac{(a\\plus{}b\\plus{}c)^2}{2(a^2\\plus{}b^2\\plus{}c^2)}\\plus{}\\frac{8(ab\\plus{}bc\\plus{}ca)^2}{(a\\plus{}b\\plus{}c)^4} \\ge \\frac{7}{2}.$\r\nNote that \r\n$ \\frac{(a\\plus{}b\\plus{}c)^2}{2(ab\\plus{}bc\\plus{}ca)} \\equal{}\\frac{(a\\plus{}b\\plus{}c)^2}{8(ab\\plus{}bc\\plus{}ca)}\\plus{}\\frac{(a\\plus{}b\\plus{}c)^2}{8(ab\\plus{}bc\\plus{}ca)}\\plus{}\\frac{a^2\\plus{}b^2\\plus{}c^2}{4(ab\\plus{}bc\\plus{}ca)}\\plus{}\\frac{1}{2},$\r\nand\r\n$ \\frac{(a\\plus{}b\\plus{}c)^2}{2(a^2\\plus{}b^2\\plus{}c^2)}\\equal{}\\frac{ab\\plus{}bc\\plus{}ca}{a^2\\plus{}b^2\\plus{}c^2}\\plus{}\\frac{1}{2}.$\r\nTherefore, the last inequality can be written as\r\n$ \\frac{(a\\plus{}b\\plus{}c)^2}{8(ab\\plus{}bc\\plus{}ca)}\\plus{}\\frac{(a\\plus{}b\\plus{}c)^2}{8(ab\\plus{}bc\\plus{}ca)}\\plus{}\\frac{a^2\\plus{}b^2\\plus{}c^2}{4(ab\\plus{}bc\\plus{}ca)} \\plus{}\\frac{ab\\plus{}bc\\plus{}ca}{a^2\\plus{}b^2\\plus{}c^2}\\plus{}\\frac{8(ab\\plus{}bc\\plus{}ca)^2}{(a\\plus{}b\\plus{}c)^4} \\ge \\frac{5}{2}.$\r\nIn this form, we can see immediately that it is a direct consequence of the AM-GM Inequality.", "Solution_8": "My dear friend can_hang2007,\r\n\r\nI regret that you are ill. :( \r\n\r\nYour proof is really beautiful. :) \r\n\r\nThank you very much. :)" } { "Tag": [ "geometry", "rhombus", "angle bisector", "perpendicular bisector" ], "Problem": "Hello everyone,\r\n\r\n I would appreciate it greatly again if you could lend me a hand on this problem: without the solution please! \r\n\r\n---\r\n\r\nQuestion: Consider the triangle ABC with AC = 6, AB = 12 and BC = 8. A rhombus is inscribed in triangle ABC in such a way that one of its vertices is A and two of its sides lie along AB and AC. Find the length of the side of the rhombus. \r\n\r\n(Hint: denote the side of the rhombus x and write similarity ratios.)\r\n\r\n---\r\n\r\nHere is my diagram of how I visually interpreted the question:\r\n\r\n[img]http://img95.imageshack.us/img95/6540/problem5mv8.jpg[/img]\r\n\r\n\r\nThank you very much!", "Solution_1": "[hide]Here's how you construct the rhombus:\n\nThe diagonals of a rhombus bisect the angles, so construct the angle bisector of $\\angle A.$ The intersection of the the angle bisector and side $BC$ is one of the vertices of the rhombus. Then, construct the perpendicular bisector of the diagonal. The intersections of the perpendicular bisector with the sides of the triangle are the other two vertices.[/hide]", "Solution_2": "[hide=\"hint\"]One of the sides of the rhombus makes a similar triangle to the bigger triangle.[/hide]\n\n[hide=\"bigger hint\"]The side that does this is parallel to AC.[/hide]", "Solution_3": "[hide=\"solution\"]Let the other 3 vertices, D, E, and F be on the sides AB, BC, and AC respectively. Since opposite sides of a rhombus are parallel, DE is parallel to AF and therefore, DE is also parallel to AC. Since triangle BED is similar to triangle BCA, we have:\n$\\frac{DE}{AC}= \\frac{BD}{BA}$\nSince AD = x, BD = 12 - x. AC = 6, DE = x, AB = 12...\n$\\frac{x}{6}= \\frac{12-x}{12}$\nMultiplying by 12 on both sides,\n$2x = 12-x$\n$3x = 12$\n$x = 4$[/hide]", "Solution_4": "He asked for people to NOT post a solution.\r\n\r\nPlease honor his wishes :)" } { "Tag": [ "function", "inequalities", "integration", "real analysis", "real analysis unsolved" ], "Problem": "Given this question, but I'm stuck need some help. Thanks. \r\n\r\n(1). Show that if $x_1, ...x_n$ are positive real numbers, $\\eta_1, ... \\eta_n \\in [0,1]$, $\\sum_{i=1}^{n} \\eta_i = 1$ and $0 3$ is a prime number then $ p \\equal{} 6*k \\plus{} 1$ or $ p \\equal{} 6*k \\minus{} 1$ for some positive integer k. In the first case:\n\\[ {7}^{p} \\minus{} {6}^{p} \\minus{} 1 \\equal{} {7}^{6*k \\plus{} 1} \\minus{} {6}^{6*k \\plus{} 1} \\minus{} 1 \\equiv 7*({7}^{6})^{k} \\minus{} 6*({6}^{6})^{k} \\minus{} 1 \\equiv 7 \\minus{} 6 \\minus{} 1 \\equiv 0\\pmod{43}\n\\]\nusing : $ 7^{6}\\equiv 6^{6}\\equiv 1\\pmod{43}$.\nIn the second case:\n\\[ {7}^{p} \\minus{} {6}^{p} \\minus{} 1 \\equal{} {7}^{6*k \\minus{} 1} \\minus{} {6}^{6*k \\minus{} 1} \\minus{} 1 \\equiv{7}^{5}*({7}^{6})^{k \\minus{} 1} \\minus{} {6}^{5}*({6}^{6})^{k \\minus{} 1} \\minus{} 1 \\equiv 37 \\minus{} 36 \\minus{} 1 \\equiv 0\\pmod{43}\n\\]\nThis proves the statment.[/quote]", "Solution_2": "By the well-known fact that $ x^2 \\plus{} x \\plus{} 1|x^{2k} \\plus{} x^k \\plus{} 1$ for every $ k\\equiv 1,2 \\pmod{3}$ we have that $ 6^2 \\plus{} 6 \\plus{} 1|6^{2p} \\plus{} 6^p \\plus{} 1$ because 3 does not divide p. So $ 6^{2p} \\plus{} 6^p \\plus{} 1\\equiv 0 \\pmod{43}$.\r\nAlso we have that $ 6\\cdot 7\\equiv \\minus{} 1\\pmod{43}$ so $ (6\\cdot 7)^p\\equiv \\minus{} 1\\pmod{43}$ because p is odd.\r\n$ 6^{2p} \\plus{} 6^p \\plus{} 1\\equiv6^{2p} \\plus{} 6^p \\minus{} (6*7)^p\\equiv6^p*(6^p \\plus{} 1 \\minus{} 7^p)\\equiv0\\pmod{43}$\r\nBecause $ \\gcd(6,43) \\equal{} 1$ we have that $ 43|7^p \\minus{} 6^p \\minus{} 1$", "Solution_3": "It is spesialiast problem about $ A_n\\equal{}a^n\\plus{}b^n\\plus{}c^n, a\\equal{}7, b\\equal{}\\minus{}6,c\\equal{}\\minus{}1,a\\plus{}b\\plus{}c\\equal{}0$.", "Solution_4": "Can you be more specific, Rust? Do you have a generalisation (with proof)?", "Solution_5": "[quote=\"Peter\"]Can you be more specific, Rust? Do you have a generalisation (with proof)?[/quote]\r\nLet $ x_1,x_2$ are roots $ x^2 \\plus{} ax \\plus{} b \\equal{} 0,a,b\\in Z$ and ab -even (if $ x_1,x_2$ are integers, then always $ ab \\equal{} \\minus{} (x_1 \\plus{} x_2)x_1x_2$ even). Let $ A_n \\equal{} x_1^n \\plus{} x_2^n \\plus{} a^n, P \\equal{} a^2 \\minus{} b$. Then $ (2P)|A_{3k \\minus{} 1}, \\ (2P^2)|A_{3k \\plus{} 1}.$ \r\nIn your case $ P \\equal{} 43$ and prime $ p > 3$ is $ p \\equal{} \\pm 1\\mod 3$.\r\nIf $ ab$ odd we can prove only $ P|A_{3k \\minus{} 1}, \\ P^2|A_{3k \\plus{} 1}$.\r\nGeneralized. Let $ x_1,x_2,...,x_k$ roots of $ x^k\\equal{}ax\\plus{}b, \\ a,b\\in Z$ and $ A_n\\equal{}x_1^n\\plus{}x_2^n\\plus{}...\\plus{}x_k^n$.\r\nThen\r\n$ A_{n\\plus{}k}\\equal{}aA_{n\\plus{}1}\\plus{}bA_n,A_0\\equal{}k,A_1\\equal{}A_2\\equal{}...\\equal{}A_{k\\minus{}2}\\equal{}0,$ and\r\n$ A_{k\\minus{}1}\\equal{}(k\\minus{}1)!(\\minus{}1)^{k}a, A_k\\equal{}bk.$\r\nTherefore \r\n$ A_{nk}\\equal{}kc_nb^n\\plus{}k!c_{n\\plus{}1\\minus{}k}b^{n\\plus{}1\\minus{}k}a^k\\plus{}k!c_{n\\plus{}2\\minus{}2k}b^{n\\plus{}2\\minus{}2k}a^{2k}\\plus{}...$\r\n$ A_{nk\\minus{}i}\\equal{}k!c_{n\\minus{}i}b^{n\\minus{}i}a^i\\plus{}k!c_{n\\plus{}1\\minus{}i\\minus{}k}b^{n\\plus{}1\\minus{}i\\minus{}k}a^{i\\plus{}k}\\plus{}...$\r\nTherefore $ a^i|A_{nk\\minus{}i}$.", "Solution_6": "Rust could you make you generalization clearer?I get confused with the expressions you name $ A_n$.. Thank you.", "Solution_7": "If $ A_n = x_1^n + x_2^n + .. + x_k^n$, then it may be defineed by recurent\r\n$ A_{n + k} = \\sigma_1 A_{n + k - 1} - \\sigma_2 A_{n + k - 2} + ... + ( - 1)^{k - 1}\\sigma_kA_n$, were $ \\sigma_1 = x_1 + x_2 + ... + x_k,\\sigma_2 = \\sum_{i < j}x_ix_j,...$ \r\n$ x_1,x_2,..,x_k$ roots of $ x^k - \\sigma_1x^{k - 1} + \\sigma_2x^{k - 2} - ... + ( - 1)^k\\sigma_k = 0$.\r\nIf all $ \\sigma_i$ are integer, then $ A_n = P_n(\\sigma_1,...,\\sigma_k )$ are integer for all natural n.\r\nConsider case, when ${ \\sigma_1 = \\sigma_2 = \\sigma_{k - 2} = 0,\\sigma_{k - 1}( - 1)^k = a,\\sigma_k ( - 1)^k - 1} = b$.\r\nThen $ A_{n + k} = aA_{n + 1} + bA_n$ and $ A_1 = ... = A_{k - 2} = 0, A_{k - 1} = aA_0 + bA_{ - 1} = (k - 1)a, A_k = bk.$\r\nIn these case $ A_n = P_n(a,b)$, were $ P_n$ is polinom with common degree n for all members, $ deg(a) = k - 1,deg(b) = k$.\r\n$ P_n(a,b) = aP_{n - k + 1}(a,b) + P_{n - k}b,P_{k - 1}(a,b) = (k - 1)a,P_k(a,b) = kb$.\r\nTherefore $ A_{n} = P_n(a,b) = \\sum_{i,ik + j(k - 1) = n} B(n,i,j)b^ia^j$. \r\nFor coefficients $ B(n,i,j)$ we had $ B((k - 1)j,0,j) = k - 1,B(ki,i,0) = k,$\r\n$ B(ki + (k - 1)j,i,j) = k\\binom{i - 1 + j}{j} + (k - 1)\\binom{i + j - 1}{i} = \\frac {n(i + j - 1)!}{i!j!}$.\r\nFor case $ k = 3,x_1 = 7,x_2 = - 6,x_3 = - 1,\\sigma_1 = 0,\\sigma_2 = - 43,a = 43,\\sigma_3 = 42 = b$ we had\r\n$ 7^n + ( - 6)^n + ( - 1)^n = n\\sum_{3i + 2j = n}\\frac {(i + j - 1)!}{i!j!}a^ib^j$\r\nIf $ n$ odd, then $ i$ -odd, therefore $ a|P_n(a,b)$.", "Solution_8": "Such heavy machinery is not needed.\n\nThe powers of 7 mod 43 follow the pattern $7, 6, -1, -7, -6, 1, 7$, ... The powers of 6 mod 43 follow the pattern $6, -7, 1, 6$, ... All primes greater than 3 are 1 or 5 mod 6. Thus the expression $7^p - 6^p - 1$ must be either $(7 - 6 - 1)$ or $(-6 -(-7) - 1)$ mod 43.", "Solution_9": "Since $7 \\equiv -6^{-1} \\pmod{43}$, hence it suffices to show $6^{p}+6^{-p}+1 \\equiv 0 \\pmod{43} \\Leftrightarrow 6^{2p}+6^p+1 \\equiv 0 \\pmod{43}$ . \nNow since $ord_{43}(6)=3$ and $p>3$, hence $6^{p} \\not\\equiv 1 \\pmod{43}$. \nHence $(6^p-1)\\left(6^{2p}+6^p+1\\right)=6^{3p}-1=216^p-1 \\equiv 0 \\pmod{43}$ yields the result. $\\square$" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Find all pairs of prime numbers $ (p,q)$ satisfying $ 2p^q\\minus{}q^p\\equal{}7$", "Solution_1": "[quote=\"thaithuan_GC\"]Find all pairs of prime numbers $ (p,q)$ satisfying $ 2p^q \\minus{} q^p \\equal{} 7$[/quote]\r\n[hide=\" I think it can be helped you\"]http://www.mathlinks.ro/Forum/viewtopic.php?t=217641[/hide][/hide]", "Solution_2": "[hide=\"Solution\"][quote=\"thaithuan_GC\"]Find all pairs of prime numbers $ (p,q)$ satisfying $ 2p^q \\minus{} q^p \\equal{} 7$[/quote]\nAccording Feramt's theorem, we easily show that 2p-q=7\nSo, $ 2p^q\\minus{}q^p\\equal{}2p\\minus{}q$\nWe know that $ x^y>y^x$ for every $ y>x>3$, it help us to solve this problem.[/hide]", "Solution_3": "[quote=\"mr.danh\"]According Feramt's theorem, we easily show that 2p-q=7[/quote]\r\nI don't think so :blush:", "Solution_4": "Wow, if 2p-7 >0 then 2p-q=7. If 2p-7<0, we get p=2,q=3\r\nAnswer: two solutions (p,q) = (2,3), (5,3)", "Solution_5": "[quote=\"thaithuan_GC\"]Find all pairs of prime numbers $ (p,q)$ satisfying $ 2p^q \\minus{} q^p \\equal{} 7$[/quote]\r\nSee that $ q \\mid 2p\\minus{}7$ and $ p \\mid q\\plus{}7$ This implies $ pq \\mid (2p\\minus{}7)(q\\plus{}7) \\Rightarrow pq \\mid 14p \\minus{}7q \\minus{} 49$\r\nIf $ p \\equal{} 7$ we have $ q \\mid 14\\minus{}7 \\equal{} 7$ And so $ q \\equal{} 7$ But $ (7,7)$ is not a solution.\r\nIf $ q \\equal{} 7$ we have $ p \\mid 7\\plus{}7 \\equal{} 7$ but neither $ (7,7)$ or $ (2,7)$ are solutions.\r\n\r\nThis means that $ pq \\mid 2p \\minus{} q \\minus{} 7$ and therefore $ pq \\le | 2p\\minus{}q\\minus{}7 |$ or $ 2p\\minus{}q\\minus{}7\\equal{}0$\r\nSo $ pq \\le 2p\\minus{}q\\minus{}7$ or $ pq \\le q\\plus{}7\\minus{}2p$ or $ 2p\\minus{}q\\minus{}7 \\equal{} 0$\r\n\r\n$ pq \\le 2p\\minus{}q\\minus{}7 \\iff (p\\plus{}1)(q\\minus{}2) \\le \\minus{}9$ which is obviously false.\r\nSo $ pq \\le q\\plus{}7\\minus{}2p \\iff (p\\minus{}1)(q\\plus{}2) \\le 5 \\Rightarrow q \\le 3, p \\equal{} 2$\r\n\r\nObviously $ p \\equal{} q$ is not a solution, So from $ pq \\le q\\plus{}7\\minus{}2p$ we only get $ (p,q) \\equal{} (2,3)$\r\n\r\nMr.Danh have taken care of $ 2p\\minus{}q \\equal{} 2p^q\\minus{}q^p \\equal{} 7$.\r\n\r\nThe key is to first see that $ q \\mid 2p\\minus{}7$ and $ p \\mid q\\plus{}7$ and after that multiply them. This means we have product is at most a sum or a sum is zero, which leaves very few opportunities for $ p$ and $ q$ :idea:", "Solution_6": "posted before:[url]http://www.mathlinks.ro/viewtopic.php?t=209652[/url]" } { "Tag": [], "Problem": "What product is obtained when allyl vinyl sulfide goes through the following steps?\r\n\r\n1. $ sec\\minus{}BuLi$\r\n\r\n2. $ \\ce{PhCH2Br}$\r\n\r\n3. reflux\r\n\r\n4. $ \\ce{H3O\\plus{}}$", "Solution_1": "isnt allyl vinyl sulfide unreactive????\r\n\r\nIt is nothing but a thio ether isnt it????\r\n\r\nSO why shloud it be reactive???? :maybe: :maybe: :maybe:", "Solution_2": "Of course it is reactive.\r\n\r\n[quote=\"Valeriummaximum\"]SO why shloud it be reactive????[/quote]\r\n\r\nIt is your job to explain that.", "Solution_3": "the hydrogen alpha to sulphur is quite acidic, so first step is an acid base reaction\r\nsecond probably sn\r\nthird, god knows\r\nfourth, hydrolysis\r\nsorry for being so brief, no time to think abt this problem as im having a tight schedule rite now :D", "Solution_4": "So far, so good.", "Solution_5": "the third step probably involves claisen rearrangent and in the fourth(assuming third to be correct) you get ketone.\r\nthe intermediate for the third step is probably this(forgot to add benzene.)", "Solution_6": "Very good. This is called the thio-Claisen rearrangement." } { "Tag": [ "category theory", "advanced fields", "advanced fields unsolved" ], "Problem": "let $ A,B$ be rings (commutative, unital) and $ f,g: A \\to B$ homomorphisms with $ Spec(f) \\equal{} Spec(g)$, i.e. for every $ x \\in A$ the prime ideals containing $ f(x)$ coincide with the ones containing $ g(x)$. how does it follow, that $ f \\equal{} g$?\r\n \r\nI just can show the existence of natural numbers $ n_x , m_x$, depending on $ x$, with $ f(x)\\equal{}g(x)^{n_x}$ and $ g(x)\\equal{}f(x)^{m_x}$.", "Solution_1": "hm, probably it's wrong. spec is faithful as a functor rings -> locally ringed spaces, but not -> topological spaces.", "Solution_2": "I think that a field and two different automorphism will do for a counterexample.", "Solution_3": "yes, thank you." } { "Tag": [ "LaTeX", "trigonometry" ], "Problem": "[i]A London telephone number consisted of three letters followed by four figures. How many telephone numbers could be made by \narranging the symbols in [b]BYR[/b] [b]2556[/b]?[/i]\r\n\r\nWhat i think is that you would do the $ 3!$ for the letters and $ 4!$ for the numbers and multiply them, this gives $ 144$ but the the answer is ment to be $ 72$. What did i do wrong?", "Solution_1": "There are two $ 5$'s, so you must account for over-counting by dividing the result by $ 2!$.", "Solution_2": "[quote=\"i_like_pie\"]There are two $ 5$'s, so you must account for over-counting by dividing the result by $ 2!$.[/quote]\r\n\r\nI noticed the two $ 5$'s, but should not they be counted as two separate entities?", "Solution_3": "There are many ways to look at this.\r\n\r\nmy advice: when in doubt, try some examples!\r\n\r\nyou start out with BYR 2556. \r\nNow switch the two 5's around. You get: BYR 2556, the same thing!\r\n\r\nTry YBR 2655. \r\nNow switch the two 5's around. You get: YBR 2655, the same thing!\r\n\r\nIn permutations, the order does matter. But I think that you are getting confused here. The question asks you to find how many telephone numbers are possible. So the one way is to set it up as a permutation question. But when you get 144, you are really getting the number of telephone numbers if either one of the 5's was not 5, 2, or 6. So there would be 144 telephone numbers for a number like BYR 2956, and there are six more possibilities. But then you could switch the second 5, and that would give you another 7 different possibilities. But I'm getting off topic - you should understand my point. \r\n\r\nAt this point, if you're really good, you could almost start to conjecture that you would have half of the possibilities if you switched one of the 5's to something else rather than 2, 5, or 6. Or, you could use the formula. \"If a set of n elements has n1 elements of one kind alike [here, alike means the same], n2 of another kind alike, and so on, then the number of permuations, P, of the n elements taken n at a time is given by P = n!/(n1!n2!...) [here, n1, n2, ... the one and two after the n is simply the subscript but I'm not very good at latex]\" (Algebra and Trigonometry Structure and Method Book 2 by McDougal Littell Houghton Mifflin). So you would divide 144 by 2!, as i_like_pie pointed out. \r\n\r\nOr, you could set up the permutation \"procedure\" - as I like to call it - and say: _*_*_*_*_*_*_ (where * denotes multiplication). Then fill in the blanks to get 3*2*1*4*2*2*1. Note the first 2 in the pair of twos and why it was 2 instead of 3. This is very important!\r\n\r\nAnother way of looking at it would be to think: How many telephone numbers could be made by arranging the symbols in 555 - 5555. It's obviously 1 (this matches what you would get if you used the formula, too. Try it!). This example makes it clear. (You could also use the Reflexive Axiom and possibly Closure Axiom, too). \r\n\r\nHope this was helpful.", "Solution_4": "BYR can be arranged in 3! ways, like you said.\r\n\r\n2556, however, can only be arranged in 4!/2 ways, because\r\ngiven every single combination, you are overcounting twice.\r\nIn other words, you are counting 2[u]5[/u][b]5[/b]6 and\r\n2[b]5[/b][u]5[/u]6 as different things, as if 5 and 5 were\r\nseparate numbers. But they are, obviously, not.\r\n\r\nSo, the answer is simply:\r\n3!*4!/2=[b]72[/b], as the book said.", "Solution_5": "thanks everyone for your time and help, i think i am getting the hang of things." } { "Tag": [ "algebra", "polynomial", "quadratics", "complex numbers", "quadratic formula" ], "Problem": "At how many points do the graphs of $ y\\equal{}x^{3}$ and $ y\\equal{}x^{2}\\plus{}4x$ intersect?\r\n\r\nPlease submit explanation and example.", "Solution_1": "At how many points do the graphs\r\nof the equations y = x3 and y = x2 +\r\n4x intersect? They intersect when\r\nthe value of y is the same for both.\r\n\r\n[hide=\"Solution\"] x3 = x2 + 4x\nx3 \u2013 x2 \u2013 4x = 0\nx(x2 \u2013 x \u2013 4) = 0\nNow we set each part equal to zero\nand solve.\nx = 0 is one solution\nThere are two more solutions which\noccur when x2 \u2013 x \u2013 4 = 0\nx2 = x + 4\nx = x + 4 is one solution.\nx = - x + 4 is the other solution.\n1 + 2 = 3 Ans. [/hide]\n\n\n[hide=\"Click this\"] :ddr: \nAwesome, isn't it![/hide]", "Solution_2": "[quote=\"gogators\"]At how many points do the graphs of $ y \\equal{} x^{3}$ and $ y \\equal{} x^{2} \\plus{} 4x$ intersect?\n\nPlease submit explanation and example.[/quote]\r\n\r\nThe graphs intersect when the y values are the same. Thus, we have $ x^3\\equal{}x^2\\plus{}4x$.\r\n\r\n$ x^3\\minus{}x^2\\minus{}4x\\equal{}0 \\implies x(x^2\\minus{}x\\minus{}4)\\equal{}0$. Thus there are 3 roots.", "Solution_3": "Once we have $ x^3\\minus{}x^2\\minus{}4x\\equal{}0$ we know there are $ 3$ points because all roots are real and this polynomial has $ \\text{deg3}$", "Solution_4": "To check to see if none of the roots are the same, the discriminant ($ b^2\\minus{}4ac$) of the quadratic BOGTRO gave must be positive (negative gives complex numbers and zero gives three equal roots).", "Solution_5": "[quote=\"gogators\"]At how many points do the graphs of $ y \\equal{} x^{3}$ and $ y \\equal{} x^{2} \\plus{} 4x$ intersect?\n\nPlease submit explanation and example.[/quote]\r\n\r\n$ x^3 \\equal{} x^2 \\plus{} 4x$ \r\n\r\nThere will be a maximum of three intersection points, because if there were any more then both of the graphs would have to be the same, and obviously they aren't. A trivial solution to this equation is $ x \\equal{} 0$. We can find more points by dividing both sides by x; \r\n\r\n$ x^2 \\equal{} x \\plus{} 4$ \r\n$ x^2 \\minus{} x \\minus{} 4 \\equal{} 0$ \r\n\r\n\r\nUsing the quadratic formula we find that the solutions to this equation are; \r\n\r\n$ \\frac{1 \\plus{} \\sqrt{17}}{2}$ and $ \\frac{1 \\minus{} \\sqrt{17}}{2}$\r\n\r\nSo there are three intersection points." } { "Tag": [ "ratio", "geometry", "circumcircle", "geometry proposed" ], "Problem": "Let $ ABCD$ be a bicentric quadrilateral, Diagonals $ AC$ and $ BD$ meet at $ T$. Let $ EF$ be the diameter of its circumcircle such that $ EF$ is perpendicular to $ BD,BD$ intersects $ EF$ at $ S$. $ A,E$ are on the same side of $ BD$. Prove that $ \\frac {ES}{SF} \\equal{} \\frac {AT}{TC}$", "Solution_1": "See \r\n[url]http://www.mathlinks.ro/viewtopic.php?t=40994[/url]\r\n[url]http://www.mathlinks.ro/viewtopic.php?t=156535[/url]" } { "Tag": [ "number theory", "number theory proposed" ], "Problem": "find all natural nos. a,b,c\r\n s.t.\r\n 2^n*a+b divides c^n+1 for all natural nos. n.", "Solution_1": "Please, stop posting problems from the new issue of Mathematical Reflections.", "Solution_2": "anybody with nontrivial sol?", "Solution_3": "Don't you understand that it is from a current issue? :mad: \r\nSomeone should lock or delete this topic.", "Solution_4": "@who: explain why you posted this!\r\n@all others: please tell me if there are more to be locked in Number Theory or if it is better to delete them.\r\n(and locked now)\r\n\r\n\r\nNow unlocked again..." } { "Tag": [ "geometry", "3D geometry", "algebra unsolved", "algebra" ], "Problem": "Find the sum of 100 numbers after the comma of this number{(2)^1/2 + 1 }^2001 \r\n*x^y = y times x", "Solution_1": "Hi;\r\n\r\nIf you mean after the decimal point, then the sum of the first hundred digits after the decimal point is 0.", "Solution_2": "Hello ,,, yes i mean after the decimal point .. but why can you explain me this .?", "Solution_3": "Hi;\r\n\r\nHere is how I did it. Take your number and convert it to decimal.\r\n\r\n$ \\sqrt {2} \\plus{} 1 \\approx 2.414213562$\r\n\r\nNow cube it:\r\n\r\n$ (\\sqrt {2} \\plus{} 1)^3 \\approx 14.071067811$\r\n\r\nYou are interested in the part after the decimal. when you raise a number that has an integer part and a fractional part ( digits after the decimal ) the number grows in a particular way when being raised to a power. To the left of the decimal point grows to the left and to the right of the decimal point grows to the right by adding leading zeros. We only have to raise the right side up to some power to deduce the answer. Here is how. \r\n\r\nTake the fractional part of the number cubed and raise it to the 19 th power.\r\n\r\n$ .071067811^{19} \\equal{} 1.51979788* 10^{ \\minus{} 22}$\r\n\r\nWe already have a number with 21 leading zeros. Raise the new number to the 5 th power again.\r\n\r\n$ ( 1.51979788* 10^{ \\minus{} 22})^{5} \\equal{} 8.10828811*10^{ \\minus{} 110}$\r\n\r\nSince ((x^3)^19)^5) is equal to x^285 we have just raised the fractional part of your number to the 285 th power. As you can see their is already 109 leading zeros, That means their are 109 zeros after the decimal point. The number of leading zeros is only going to get larger than this as we raise the number up to 2001. So you already know that the first 100 digits after the decimal point of:\r\n\r\n$ (\\sqrt {2} \\plus{} 1)^{285}$ \r\n\r\nare zeros, so the sum of them is 0.", "Solution_4": "There is another way :\r\n\r\nConsider the sequence $ a_n$ defined as $ a_0 \\equal{} a_1 \\equal{} 2$ and $ a_{n \\plus{} 2} \\equal{} 2a_{n \\plus{} 1} \\plus{} a_n$. It's immediate to establish $ a_n \\equal{} (\\sqrt 2 \\plus{} 1)^n \\plus{} (1 \\minus{} \\sqrt 2)^n$\r\n\r\nSo $ (\\sqrt 2 \\plus{} 1)^{2001} \\equal{} a_{2001} \\plus{} (\\sqrt 2 \\minus{} 1)^{2001}$ and, since $ a_n$ is integer and $ \\sqrt 2 \\minus{} 1 < \\frac 12$, we get $ \\{(\\sqrt 2 \\plus{} 1)^{2001}\\} \\equal{} (\\sqrt 2 \\minus{} 1)^{2001} < \\frac 1{2^{2001}} < 10^{ \\minus{} 600}$\r\n\r\nHence the result $ 0$" } { "Tag": [ "function", "integration", "calculus", "linear algebra", "matrix", "inequalities", "derivative" ], "Problem": "a nice (and hard) one:\r\nlet $f$ be a $C^1$ function $f:R^n -> R^n$ such that\r\n$df(x)$ always belongs to the orthogonal group of $R^n$ with the usual scalar product.\r\nProve that $f$ is an isometry.", "Solution_1": "We have first of all: $f(x)-f(y)=\\int_{0}^{1}Df(x+t(y-x))dt(x-y)$ for all $x,y\\in\\Bbb R^n$ where the integration of the matrix means integration of the elements of it. Now we have to show that $\\int_{0}^{1}Df(x+t(y-x))dt$ is orthogonal which can be proven by standard computations (I will complete it later when I have more time).\r\n\r\n\r\nMisha", "Solution_2": "Alekk are you sure $f$ is $C^{1}$ ?\r\n\r\nThere is a version with $f$ $C^{2}$ using the trilinear linear form\r\n\r\n$g(h,k,z)=(f''(x).h.k|f'(x)z)$", "Solution_3": "$C^1$ works", "Solution_4": "I was randomly browsing the forum for interesting problems, and I found this. I think this should work:\r\n\r\nLet $x,y\\in\\mathbb R^n$ be two vectors. Lagrange's inequality gives us $\\|f(x)-f(y)\\|\\le\\sup_{[x,y]}\\|f'\\|\\cdot\\|x-y\\|$. Since $\\|f'\\|=1$ everywhere, we find that a function $f$ defined on a convex open subset $D$ of $\\mathbb R^n$ with values in $\\mathbb R^n$ for which $f'$ is always orthogonal decreases the distances: $\\|f(x)-f(y)\\|\\le\\|x-y\\|,\\ \\forall x,y\\in D$. \r\n\r\nNow, the Inverse Function Theorem says that for every $x\\in\\mathbb R^n$ there are open subsets $x\\in V\\subset\\mathbb R^n,W\\in\\mathbb R^n$ s.t. $f|_V: V\\to W$ is invertible, and its inverse is differentiable, it's derivative in $f(a)$ being the inverse of the derivative of $f$ in $a,\\ \\forall a\\in V$, thus also orthogonal. This means (from the observations in the paragraph above) that the restriction of $(f|_V)^{-1}$ to some open ball in $W$ centered around $f(x)$ decreases the distances, and, since its inverse does the same, $f$ is a local isometry, i.e. for every $x$ we can find a neighbourhood $V_x$ of $x$ s.t. the restriction of $f$ to $V_x$ is an isometry. The conclusion that $f$ is an isometry is easy to draw from here.", "Solution_5": "I remember I did the same, and said to my teacher the conclusion was straighforward from here. But it is not very straightforward. You have to write a few lines to conclude if my memories are correct ...", "Solution_6": "Oh, I know it's not [b]that[/b] obvious :). Anyway, here's what I did (I don't know if it works, as I haven't checked it properly; I was walking around my university waiting for someone and just sort of mentally playing around with this :)):\r\n\r\nTake a hypercube having two prescribed points as opposite vertices, and cover each point of this hypercube with an open neighbourhood for which the restriction of $f$ is an isometry. By Lebesgue's Number Lemma, if we divide this hypercube into small enough hypercubes (in the obvious manner), then any two adjacent hypercubes will lie in some common open set belonging to the described covering. Now go from one of the corners of the hypercube to the opposite corner passing through the vertices of adjacent small hypercubes one by one. Each time, the structure is preserved isometrically, because at each step, the distances from the next point we take into consideration to the vertices of the previous small hypercube are exactly what we want them to be, and the position of a point in $n$-space is determined by its distances to the vertices of an $n$-hypercube (actually, it's determined by a lot fewer distances). This means that the whole hypercube is transformed isometrically by $f$ so, in particular, the distance between the two opposite corners we considered is not changed by $f$.\r\n\r\nI'm terribly sorry if it sounds very unclear.", "Solution_7": "nice :)", "Solution_8": "Even more, brilliant! :10:" } { "Tag": [ "linear algebra", "matrix", "algebra", "polynomial", "quadratics", "complex numbers", "linear algebra unsolved" ], "Problem": "[b]a)[/b] Let $ A$ be a skew-symmetric matrix (this means $ A^T \\equal{} \\minus{} A$, where $ A^T$ denotes the transpose of $ A$) with real entries.\r\nShow that its determinant $ \\det A$ is nonnegative.\r\n[b]b)[/b] If $ A$ is a skew-symmetric matrix with integer entries, then show that its determinant $ \\det A$ is the square of an integer!", "Solution_1": "You mean \"nonnegative\", not \"positive\", since zero matrix is skew symmetric and has det=0 :).\r\n[Moderator edit: I have just edited Arline's post #1 according to this.]", "Solution_2": "I always like to see just how much mileage I can get out of Schur's lemma: every complex square matrix can be unitarily upper triangularized. That is, there exists $U$ with $U^{*}U=I$ such that $U^{*}AU=T$ with $T$ upper triangular.\r\n\r\nTake the $*$ of that: $U^{*}A^{*}U=T^{*}.$ Since $A^{*}=A^{T}=-A,$ it follows that $T^{*}=-T.$ This in turn implies that all of the off-diagonal elements of $T$ are zero and all of the elements of $T$ on the main diagonal (the eigenvalues of $A$) are pure imaginary.\r\n\r\nThe eigenvalues of $A$ are pure imaginary. But since $A$ is real, the eigevalues come in complex conjugate pairs. The eigenvalues of $A$ (and hence the diagonal entries of $T$) are $i\\alpha,-i\\alpha, i\\beta,-i\\beta,\\dots i\\omega,-i\\omega,0,\\dots,0.$\r\n\r\nNote that if $A$ is of odd size, there must be at least one zero in that list of eigenvalues and $\\det A=0.$ To make the rest of the problem interesting, assume that $A$ is of even size and nonsingular. The characteristic polynomial of $A$ is $(x^{2}+\\alpha^{2})(x^{2}+\\beta^{2})\\cdots(x^{2}+\\omega^{2}).$ If $A$ has only integer entries, this characteristic polynomal has only integer coefficients.\r\n\r\nI'd like to get from that polynomial having integer coefficients to $\\alpha,\\beta,\\dots$ being themselves integers, but I don't have that argument yet.", "Solution_3": "The determinant of a skew-symmetric matrix (for even n) is equal to square of its Pfaffian, thus nonnegative. See [url]http://en.wikipedia.org/wiki/Pfaffian[/url]. You can also try here: [url]http://www.math.harvard.edu/~elkies/M55a.99/pfaff.html[/url].", "Solution_4": "[quote=\"I\"]I'd like to get from that polynomial having integer coefficients to $\\alpha,\\beta,\\dots$ being themselves integers, but I don't have that argument yet.[/quote]\r\nTwo points: first, we can get the determinant to be the square of an integer without this. Second, this is false anyway.\r\n\r\nFor a cheap example, if \r\n\r\n$A=\\begin{bmatrix}0&1&0&0\\\\-1&0&1&0\\\\ 0&-1&0&1\\\\ 0&0&-1&0\\end{bmatrix}$\r\n\r\nThen the characteristic polynomial of $A$ is $x^{4}+3x^{2}+1,$ which cannot be factored into integer-coefficient quadratic polynomials." } { "Tag": [ "algorithm", "combinatorics unsolved", "combinatorics" ], "Problem": "Arutyun and Amayak show another effective trick. A spectator writes down on a board a sequence of $N$ (decimal) digits. Amayak closes two adjacent digits by a black disc. Then Arutyun comes and says both closed digits (and their order). For which minimal $N$ they may show such a trick?\n[i]K. Knop, O. Leontieva[/i]", "Solution_1": "It's clear that when Arutyun sees the sequence with the closed digits, he has to identify it in an unique way. So the possibilities of covering the digits by Amayak in the initial sequence has to be at least equal to the 100 possibilities that the sequence can be for Arutyun. The possibilities of covering is $N-1$, so $N-1\\geq 100$. So the minimum is $101$. In this case a bijection can be constructed between the \"complete sequences\" and the \"covered sequences\". This is of course a silly game which demands a huge memory :P , and let's not forget that N can be anything larger then 101 so as it tends to infinite the capacity of our memory tends to infinite. This problem could be solved if we found an algorithm which can be applicable for larger N-s.", "Solution_2": "[quote=\"N.T.TUAN\"][i]K. Knop, O. Leontieva[/i]\nArutyun and Amayak show another effective trick.\nA spectator writes down on a board a sequence of $N$ (decimal) digits.\nAmayak closes two adjacent digits by a black disc. Then Arutyun comes\nand says both closed digits (and their order). For which minimal $N$\nthey may show such a trick?[/quote]\r\n\r\nVery nice problem. I searched for a solution a long time.\r\n\r\n1) $N$ must be $\\geq 101$\r\nArutyun may have in front of him $(N-1)10^{N-2}$ different configurations ($N-1$ positions for the disc and $10^{N-2}$ for the reminder).\r\nThis number must be great enough to be mapped on any of the $10^{N}$ configurations the spectator may have choosen.\r\nSo $(N-1)10^{N-2}\\geq 10^{N}$ and $N\\geq 101$\r\n\r\n2) there exists a $101$ strategy :\r\n\r\nLet $a_{0},a_{1}, \\cdots,a_{99},a_{100}$ be the first $101$ ordered numbers written by the spectator.\r\n\r\nAmayak computes $S_{1}=10a_{0}+a_{1}+10a_{2}+a_{3}+\\cdots+10a_{100}[100]$ (where $[100]$ means modulo $100$).\r\nAmayak put then the black disc on numbers $a_{S_{1}}$ and $a_{(S_{1}+1)}$ \r\n\r\nAruyun first looks at the position of the black disc and knows $S_{1}$\r\nAruyun then computes $S_{2}=10a_{0}+a_{1}+10a_{2}+a_{3}+\\cdots+10a_{100}[100]$, using $0$ and $0$ for the two hidden positions.\r\nAruyun computes $S_{3}=S_{1}-S_{2}[100]$ and the final step is elementary :\r\n$S_{3}$ is a number between $0$ and $99$ and so $S_{3}=10a+b$\r\n\r\nThen, if $S_{1}$ is even, the ordered two hidden numbers are $(a,b)$\r\nAnd if $S_{1}$ is odd, the ordered two hidden numbers are $(b,a)$\r\n\r\nAnd minimal $N$ is $101$", "Solution_3": "Very nice strategy. I'd been trying to find it too, but i gave it up and thought there isn't a good strategy.", "Solution_4": "[quote=\"pco\"]\n\nThen, if $S_{1}$ is even, the ordered two hidden numbers are $(a,b)$\nAnd if $S_{1}$ is odd, the ordered two hidden numbers are $(b,a)$\n\nAnd minimal $N$ is $101$[/quote]\r\nShouldn't it be $S_{3}$?", "Solution_5": "if N=101 there's also strategy by marriage theorem" } { "Tag": [ "geometry", "algebra", "function", "domain", "geometry solved" ], "Problem": "Let ABC be a triangle. On it's sides are constructed 3 regular n - gons.\r\nLet O1,O2,O3 be their centers, find all n for which AO1, BO2, CO3 are collinear.\r\n It is clear for n = 3 - Torricelli and for n = 4 - Vecten.\r\nIf this problem is anywhere appeared, i am sorry for giving it, but i don't know \r\nthe solution.", "Solution_1": "This is called Kiepert's theorem: if we construct similar isosceles triangles having their bases on the sides of $ABC$, then the lines joining the vertices of $ABC$ with the respective vertices of the triangles are concurrent, so, to answer your question, the concurrence occurs for all $n$ as a particular case of Kiepert's theorem.", "Solution_2": "Have you got a prove for this theorem, where can it be found?", "Solution_3": "You can find something on this subject [url=http://www.cut-the-knot.org/Curriculum/Geometry/Kiepert.shtml]here[/url]. There are more links on [url=http://www.google.com/custom?domains=cut-the-knot.org&q=kiepert&sitesearch=cut-the-knot.org&sa=Search&client=pub-9118218894256794&forid=1&ie=ISO-8859-1&oe=ISO-8859-1&cof=GALT%3A%23003324%3BGL%3A1%3BDIV%3A%2366CC99%3BVLC%3AFF6600%3BAH%3Acenter%3BBGC%3AC5DBCF%3BLBGC%3A73B59C%3BALC%3A000000%3BLC%3A000000%3BT%3A330033%3BGFNT%3A333300%3BGIMP%3A333300%3BFORID%3A1%3B&hl=en]this[/url] page, but they are all links to articles on the same site: Cut the Knot. Try Goolge-ing it :).\r\n\r\nI never took the time to actually write the proof, but it can be proven with the aid of projective geometry.", "Solution_4": "Thanks for that." } { "Tag": [ "geometry", "perimeter", "calculus", "integration", "inequalities", "triangle inequality" ], "Problem": "There is a triangle with side's length expressed by integer . One side is 3 time biger than the other side and the third side is 6 .Find the highest value of the perimeter .", "Solution_1": "Let me restate the question as I understand it.\r\n\r\nThere is a triangle with integral side lengths. One side of the triangle is 6, and the other two sides are in a 3:1 ratio. What is the greatest possible perimeter of this triangle?\r\n\r\nIs this a correct interpretation of the problem?\r\n\r\n[hide=\"If it is, then...\"]I believe the answer is 14, which would be a triangle with sides of 2, 6, and 6.\n\nAssume that a triangle exists that meets the requirements and has a perimeter of greater than 14. Since the triangle has integral side lengths, the next larger triangle will be a 3-6-9 triangle. However, such a triangle is not possible because the length of the third side is equal to the lengths of the other two sides.[/hide]", "Solution_2": "p=a+3a+6=4a+6. Because of triangle inequality we have a+6>3a--> a<3 and 4a>6--> a>1 so a=2--> p=14\r\n\r\nedited for shorter answer.", "Solution_3": "[b]Follow-up:[/b] [i]Same setup, different concept[/i]\r\nSuppose the side-lengths don't have to be integers, but the sides are still $ 6,x,3x$. What is the maximum area of this triangle?", "Solution_4": "From Heron's Theorem\r\n\r\n$ A = \\sqrt {(2x + 3)(2x - 3)(3 + x)(3 - x)} = \\sqrt {(4x^{2} - 9)(9 - x^{2}) = \\sqrt { - (2x^{2} - \\frac {45}{4}) + ((\\frac {45}{4})^{2} - 81)}}$\r\n\r\nThus the maximum is $ \\sqrt{(\\frac {45}{4})^{2} - 81} = \\frac {27}{4}$" } { "Tag": [], "Problem": "If the three-digit number $ 2d2$ is divisible by 7, what is $ d$?", "Solution_1": "The rule for divisibility of $ 7$ is twice the units digit subtracted from the number with the units digit dropped is divisible by $ 7$. Meaning, for a number $ 100a \\plus{} 10b \\plus{} c$ that is divisible by $ 7$, $ 10a \\plus{} b \\minus{} 2c$ is divisible by $ 7$. So, this means that $ 2(10) \\plus{} d \\minus{} 2(2)$ is divisible by $ 7$, so we have that $ 16 \\plus{} d$ is divisible by $ 7$. The closest number that is divisible by $ 7$ is $ 21$, so $ d \\equal{} \\boxed{5}$.", "Solution_2": "[hide=\"Silly way\"]Or you could just try out numbers one by one (202,212,222...252) I think that would be just slightly faster.[/hide]", "Solution_3": "Okay, just notice that 42 is divisible by 7, and it ends with a 2. So ${d=\\boxed[5}$ ." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Find the biggest number of positive integers which we can choose among the integers $1,2,...,100$ such that there isn't $a,b,c,d$ with $a+b+c=d.$", "Solution_1": "67: 34,35,36,...,100.\r\nLet's suppose we have n>67 numbers, say $x_{1} 0$, show that\n\\[ \\sqrt {\\frac {a^2 + bc}{b(c + a)}} + \\sqrt {\\frac {b^2 + ca}{c(b + a)}} + \\sqrt {\\frac {c^2 + ab}{a(b + c)}}\\geq 3\\]", "Solution_13": "[quote=sqing][b]Potla inequality[/b]\nFor all $ a,b,c > 0$, show that\n\\[ \\sqrt {\\frac {a^2 + bc}{b(c + a)}} + \\sqrt {\\frac {b^2 + ca}{c(b + a)}} + \\sqrt {\\frac {c^2 + ab}{a(b + c)}}\\geq 3\\][/quote]\nIt's just AM-GM.\n\n", "Solution_14": "[quote=ductrung]For any positive real numbers $a,b$ and $c$,\n\\[\\sqrt{\\frac{a(b+c)}{a^{2}+bc}}+\\sqrt{\\frac{b(c+a)}{b^{2}+ca}}+\\sqrt{\\frac{c(a+b)}{c^{2}+ab}}\\le \\sqrt{ \\left(\\sqrt{a}+\\sqrt{b}+\\sqrt{c}\\right)\\left(\\frac{1}{\\sqrt{a}}+\\frac{1}{\\sqrt{b}}+\\frac{1}{\\sqrt{c}}\\right) }\\][/quote]\nFor any positive real numbers $a,b$ and $c$,\n$$ \\sqrt{\\frac{a(b+c)}{a^2+bc}}+\\sqrt{\\frac{b(c+a)}{b^2+ca}}+\\sqrt{\\frac{c(a+b)}{c^2+ab}}\\le \\sqrt{2(a+b+c)(\\frac{1}{a+b}+\\frac{1}{b+c}+\\frac{1}{c+a})}$$\nCrux,Problems 3507 ", "Solution_15": "[quote=sqing][quote=ductrung]For any positive real numbers $a,b$ and $c$,\n\\[\\sqrt{\\frac{a(b+c)}{a^{2}+bc}}+\\sqrt{\\frac{b(c+a)}{b^{2}+ca}}+\\sqrt{\\frac{c(a+b)}{c^{2}+ab}}\\le \\sqrt{ \\left(\\sqrt{a}+\\sqrt{b}+\\sqrt{c}\\right)\\left(\\frac{1}{\\sqrt{a}}+\\frac{1}{\\sqrt{b}}+\\frac{1}{\\sqrt{c}}\\right) }\\][/quote]\nFor any positive real numbers $a,b$ and $c$,\n$$ \\sqrt{\\frac{a(b+c)}{a^2+bc}}+\\sqrt{\\frac{b(c+a)}{b^2+ca}}+\\sqrt{\\frac{c(a+b)}{c^2+ab}}\\le \\sqrt{2(a+b+c)(\\frac{1}{a+b}+\\frac{1}{b+c}+\\frac{1}{c+a})}$$\nCrux,Problems 3507[/quote]\n\n[img]http://s4.sinaimg.cn/middle/006ptkjAzy75Kc0TuRd23&690[/img]", "Solution_16": "[quote=ductrung]For any positive real numbers $a,b$ and $c$,\n\\[\\sqrt{\\frac{a(b+c)}{a^{2}+bc}}+\\sqrt{\\frac{b(c+a)}{b^{2}+ca}}+\\sqrt{\\frac{c(a+b)}{c^{2}+ab}}\\le \\sqrt{ \\left(\\sqrt{a}+\\sqrt{b}+\\sqrt{c}\\right)\\left(\\frac{1}{\\sqrt{a}}+\\frac{1}{\\sqrt{b}}+\\frac{1}{\\sqrt{c}}\\right) }\\][/quote]\n\n[img]http://s2.sinaimg.cn/middle/006ptkjAzy75KgXRPQ571&690[/img]", "Solution_17": "$x,y,z>0$,prove that\n\n\n\\[\\sqrt {{\\frac {x \\left( y+z \\right) }{{x}^{2}+yz}}}+\\sqrt {{\\frac {y\n \\left( z+x \\right) }{{y}^{2}+zx}}}+\\sqrt {{\\frac {z \\left( x+y\n \\right) }{{z}^{2}+xy}}}\\geq 2\\,\\sqrt {1+20\\,\\sqrt {2} \\left( {\\frac {xyz}\n{ \\left( y+z \\right) \\left( z+x \\right) \\left( x+y \\right) }}\n \\right) ^{3/2}}\\]\n", "Solution_18": "[quote=luofangxiang][quote=ductrung]For any positive real numbers $a,b$ and $c$,\n\\[\\sqrt{\\frac{a(b+c)}{a^{2}+bc}}+\\sqrt{\\frac{b(c+a)}{b^{2}+ca}}+\\sqrt{\\frac{c(a+b)}{c^{2}+ab}}\\le \\sqrt{ \\left(\\sqrt{a}+\\sqrt{b}+\\sqrt{c}\\right)\\left(\\frac{1}{\\sqrt{a}}+\\frac{1}{\\sqrt{b}}+\\frac{1}{\\sqrt{c}}\\right) }\\][/quote]\n\n[img]http://s2.sinaimg.cn/middle/006ptkjAzy75KgXRPQ571&690[/img][/quote]\n\n\u8fd9\u4e2a\u4e0d\u7b49\u5f0f\u5df2\u7ecf\u975e\u5e38\u5f3a\u4e86\uff0c\u4e0d\u8fc7\uff0c\u8fd8\u53ef\u4ee5\u8fdb\u4e00\u6b65\u52a0\u5f3a\uff0c\u4e24\u4e2a\u4e0d\u7b49\u5f0f\u90fd\u53ef\u4ee5\u5305\u529b\u5c55\u6740\u3002\n$\\sum{\\frac{2\\sqrt{a\\left( b+c \\right)}}{a+\\sqrt{bc}}}\\le \\sqrt{9+\\left( \\sum{\\sqrt{a}} \\right)\\left( \\sum{\\frac{1}{\\sqrt{a}}} \\right)}$", "Solution_19": "[quote=zdyzhj][quote=luofangxiang][quote=ductrung]For any positive real numbers $a,b$ and $c$,\n\\[\\sqrt{\\frac{a(b+c)}{a^{2}+bc}}+\\sqrt{\\frac{b(c+a)}{b^{2}+ca}}+\\sqrt{\\frac{c(a+b)}{c^{2}+ab}}\\le \\sqrt{ \\left(\\sqrt{a}+\\sqrt{b}+\\sqrt{c}\\right)\\left(\\frac{1}{\\sqrt{a}}+\\frac{1}{\\sqrt{b}}+\\frac{1}{\\sqrt{c}}\\right) }\\][/quote]\n\n[img]http://s2.sinaimg.cn/middle/006ptkjAzy75KgXRPQ571&690[/img][/quote]\n\n\u8fd9\u4e2a\u4e0d\u7b49\u5f0f\u5df2\u7ecf\u975e\u5e38\u5f3a\u4e86\uff0c\u4e0d\u8fc7\uff0c\u8fd8\u53ef\u4ee5\u8fdb\u4e00\u6b65\u52a0\u5f3a\uff0c\u4e24\u4e2a\u4e0d\u7b49\u5f0f\u90fd\u53ef\u4ee5\u5305\u529b\u5c55\u6740\u3002\n$\\sum{\\frac{2\\sqrt{a\\left( b+c \\right)}}{a+\\sqrt{bc}}}\\le \\sqrt{9+\\left( \\sum{\\sqrt{a}} \\right)\\left( \\sum{\\frac{1}{\\sqrt{a}}} \\right)}$[/quote]\n\ncould explain in English?", "Solution_20": "the two can be proved by CS and then BW" } { "Tag": [ "function", "algebra", "functional equation", "algebra unsolved" ], "Problem": "I found a solution to this functional equation, but was wondering whether I made any flaws?\r\n\r\nFind all Functions f from the Real Numbers to the Real Numbers such that\r\n\r\n$ f(x\\minus{}f(y))\\equal{}f(f(y))\\plus{}xf(y)\\plus{}f(x)\\minus{}1$ for all x,y in the real numbers.\r\n\r\nSolution\r\n\r\n[hide]First, let $ x\\equal{}0$, then we have $ f(\\minus{}f(y))\\equal{}f(f(y))\\plus{}f(0)\\minus{}1$.\n\nThen set $ f(y)\\equal{}g(y^{2})$, to get $ f(\\minus{}g(y^{2}))\\equal{}f(g(y^{2}))\\plus{}g(0)\\minus{}1$, which implies that $ g((g(y^{2}))^{2})\\equal{}g((g(y^{2}))^{2})\\plus{}g(0)\\minus{}1$, so $ g(0)\\equal{}1$, and then $ f(0)\\equal{}1$.\n\nNext, we observe that plugging $ y\\equal{}0$, we get that\n$ f(x\\minus{}1)\\equal{}f(1)\\plus{}x\\plus{}f(x)\\minus{}1$. Upon plugging $ x\\equal{}1$, we get that $ f(1)\\equal{}\\frac{1}{2}$.\n\nFurthermore, we see that $ f(x\\minus{}1)\\equal{}f(1)\\plus{}x\\plus{}f(x)\\minus{}1$(*) implies that $ g(x^{2}\\minus{}2x\\plus{}1)\\equal{}x\\minus{}1/2\\plus{}g(x^{2})$ (**).\n\nHowever, $ \\frac{g(x^{2}\\minus{}2x\\plus{}1)\\minus{}g(x^{2})}{\\minus{}2x\\plus{}1}\\equal{}\\minus{}\\frac{1}{2}$, for all values of x, except $ x\\equal{}\\frac{1}{2}$. So, for right now, for all values of x, except for $ x\\equal{}\\frac{1}{2}$, $ g(x)\\equal{}1\\minus{}\\frac{x}{2}$ (***), for we know that $ g(0)\\equal{}1$.\n\nNow, we set $ x\\equal{}\\frac{\\sqrt{2}}{2}$ into (**).\n\nThen we get that $ g(\\frac{3}{2}\\minus{}\\sqrt{2})\\equal{}\\frac{\\sqrt{2}}{2}\\minus{}\\frac{1}{2}\\plus{}g(\\frac{1}{2})$.\n\nThis implies that $ g(\\frac{1}{2})\\equal{}\\frac{3}{4}$ from knowing (***) for all $ x \\neq \\frac{1}{2}$.\n\nLikewise, we plug in $ x\\equal{}\\minus{}\\frac{\\sqrt{2}}{2}$ into (**), and we see that we get $ g(\\frac{1}{2})\\equal{}\\frac{3}{4}$.\n\nSo, in either case, we get that $ g(x)\\equal{}1\\minus{}\\frac{x}{2}$, for all $ x$.\n\nTherefore, $ f(x)\\equal{}1\\minus{}\\frac{x^{2}}{2}$ is the only function that works in this problem, which can be checked easily.[/hide]", "Solution_1": "[quote=\"mathgeniuse^ln(x)\"]I found a solution to this functional equation, but was wondering whether I made any flaws?[/quote]\n\nI think you made some.\n\n[quote=\"mathgeniuse^ln(x)\"] \n...\nThen set $ f(y) \\equal{} g(y^{2})$, \n...\n[/quote]\n\nYou can't set $ f(y) \\equal{} g(y^{2})$ except if you prove first that $ f(\\minus{}x)\\equal{}f(x)$\n\n[quote=\"mathgeniuse^ln(x)\"] \n...\n$ \\frac {g(x^{2} \\minus{} 2x \\plus{} 1) \\minus{} g(x^{2})}{ \\minus{} 2x \\plus{} 1} \\equal{} \\minus{} \\frac {1}{2}$, for all values of x, except $ x \\equal{} \\frac {1}{2}$. So, for right now, for all values of x, except for $ x \\equal{} \\frac {1}{2}$, $ g(x) \\equal{} 1 \\minus{} \\frac {x}{2}$ (***)[/quote]\r\n\r\nHow could you conclude from $ \\frac {g(x^{2} \\minus{} 2x \\plus{} 1) \\minus{} g(x^{2})}{ \\minus{} 2x \\plus{} 1} \\equal{} \\minus{} \\frac {1}{2}$ that $ g(x) \\equal{} 1 \\minus{} \\frac {x}{2}$.\r\nIt can't be immediate.", "Solution_2": "[quote=\"mathgeniuse^ln(x)\"]I found a solution to this functional equation, but was wondering whether I made any flaws?\n\nFind all Functions f from the Real Numbers to the Real Numbers such that\n\n$ f(x \\minus{} f(y)) \\equal{} f(f(y)) \\plus{} xf(y) \\plus{} f(x) \\minus{} 1$ for all x,y in the real numbers.\n\n[/quote]\r\n\r\nIt's a problem I searched for a long time on another forum without finding general solution. I have a solution with one constraint more :\r\n\r\n\"The set of discontinuity points of $ f(x)$ is included in an interval $ (a,b)$\" (for example, $ f(x)$ has a finite set of discontinuity points).\r\n\r\nWith this condition, the only solution is $ f(x)\\equal{}1\\minus{}\\frac{x^2}{2}$. Here is the solution I gave on this other forum :\r\n\r\nLet $ P(x,y)$ the property $ f(x\\minus{}f(y)) \\equal{} f(f(y)) \\plus{} xf(y) \\plus{} f(x) \\minus{} 1$.\r\n\r\n$ 1)$ $ f(x)$ is not bounded (it does not exist $ u,v$ such that $ u>10440<<[/hide]\r\n\r\nOh, and a formula you forgot:\r\n\r\nThe sum of the first n cubes:\r\n[nth triangular number]^2\r\n-in other words-\r\n{[n(n+1)]/2}^2", "Solution_8": "Of course, it's easiest just to know the formula and it's proper applications rather than adding out... but similar to what basooner said, it's good to know that its derivation is from averaging and then multiplying.", "Solution_9": "well... i learned it as the first plus the last times how many there are divided by 2. because the first number plus the last number is 145 and as u go in, the sum stays the same unless the number of numbers is odd. u divide by two cause u cut the amount the number of number u have in half. i hope i explained well. im usually bad at that. which is y i hate masters round. how do u people do that?", "Solution_10": "[quote=\"mathlover\"]well... i learned it as the first plus the last times how many there are divided by 2. because the first number plus the last number is 145 and as u go in, the sum stays the same unless the number of numbers is odd. u divide by two cause u cut the amount the number of number u have in half. i hope i explained well. im usually bad at that. which is y i hate masters round. how do u people do that?[/quote]\r\n...which can be written as $\\frac{n(a+z)}{2}$, where n is the number of terms ($n=a-z+1$), a is the greatest term, and z is the least term" } { "Tag": [], "Problem": "Find a closed form for $1+22+333+....+n(11\\ldots1)$ where in the last term 1 is written $n$ times.", "Solution_1": "[hide=\"Answer\"]$S_{n}=\\frac{2(9n-1)10^{n+1}-81n(n+1)+20}{1458}$[/hide]\r\n\r\n(Will post the solution if no one else does.)", "Solution_2": "I put it into series form, but I'm not sure how to evaluate the double series:\r\n\r\n$1+22+333+\\cdots+n\\left(11\\dots 11\\right)=\\sum_{k=0}^{n}\\sum_{i=0}^{k}\\left(k+1\\right)10^{i}$", "Solution_3": "Try to find single-series form.", "Solution_4": "[quote=\"amcavoy\"]I put it into series form, but I'm not sure how to evaluate the double series:\n\n$1+22+333+\\cdots+n\\left(11\\dots 11\\right)=\\sum_{k=0}^{n}\\sum_{i=0}^{k}\\left(k+1\\right)10^{i}$[/quote]\r\n\r\nEvaluate the series on the inside first, of course! Didn't you learn the order of operations in elementary school? :P", "Solution_5": "For some reason I was thinking that I couldn't. A similar problem came up with a double sum where I had to use the identity \\[\\sum_{n=0}^{\\infty}\\sum_{k=0}^{n}a_{k,n-k}=\\sum_{i=0}^{\\infty}\\sum_{j=0}^{\\infty}a_{i,k}\\] to evaluate the double series and I guess I just assumed I had to for this as well. Anyways, thanks for the comment." } { "Tag": [ "algebra", "polynomial", "algebra proposed" ], "Problem": "Let $a_1,a_2,....,a_n\\in Z$ be different to each other.Prove that the polynomial \r\n$f(x)=(x-a_1)(x-a_2)....(x-a_n)-1$ doesn't write as a product of polynomials $f(x)$ and $g(x)$ with integer coefficients and with degree $\\geq 1$", "Solution_1": "[hide]By contradiction and use the theorem which says that every polynomial of $n$ degree which has $n+1$ roots it is the polynomial of $0$ degree.[/hide]\r\nCould anyone now finish it?", "Solution_2": "Let's suppose $f(x) = g(x)h(x)$ with $g, h \\in Z[X]$ with d\u00b0g and d\u00b0h > 0\r\n\r\n$g(a_i)h(a_i) = -1$ so $g(a_i) + h(a_i) = 0$ \r\nBut f and g have degree at most n-1 so we have $g+h = 0$\r\n$f(x) = -g(x)^2$ is not possible since in f coefficient of $x^n$ is > 0.\r\n\r\n\r\n\r\nWhat about $f(x)=(x-a_1)(x-a_2)....(x-a_n) + 1$ ?\r\nIs there a general result ? x(x-1)+1 is irreducible but not x(x-2)+1 :huh:", "Solution_3": "[quote=\"t\u00b5t\u00b5\"]\nBut f and g have degree at most n-1 so we have $g+h = 0$\n$f(x) = -g(x)^2$ is not possible since in f coefficient of $x^n$ is > 0.\n[/quote]\r\nHow do you find this? Could you explain it with all the details?", "Solution_4": "Hi,\r\n\r\n$g+h$ has degree at most n-1 (d\u00b0g, dh $\\leq$ n-1, by hypothesis) and is null at n different places, hence it's null.\r\n\r\nf(x) can be written $f(x) = x^n + ...$ so coefficient of $x^n$ is 1 but the main coefficient of $-g(x)^2$ can't be positive.\r\n\r\nPerhaps I'm wrong it seems to make sense.", "Solution_5": "yes but my questions are:\r\nWhy $g+h=0$?\r\nAnd the second why $f(x)=-g(x)^2$?", "Solution_6": "Silouan, did you READ the above post? :huh:\r\n\r\nI don't understand. Don't you ever try to figure something out yourself?\r\n\r\nAnyway...\r\n\r\nSuppose that $f(x) = g(x)h(x)$ where $g(x),h(x) \\in \\mathbb{Z}[x]$ and $\\deg{g},\\deg{h} \\geq 1.$ Then, $g(x)h(x) = -1$ for $x = a_1,a_2,\\dots,a_n$. Hence for all of these $a_i$ we have $g(a_i) = 1$ and $h(a_i) = -1$ (or the other way around). Anyway, we get $g(a_i) + h(a_i) = 0$ for all $a_i$. Then $g(x) + h(x)$ is a polynomial in $x$ and has $a_1$, $a_2$, ..., $a_n$ as zeroes. So, that's $n$ zeroes. The degree of $g(x) + h(x)$ is at most $n - 1$, since $f(x) = g(x)h(x)$ is a polynomial of degree $n$. Hence, $g(x) + h(x)$ must vanish everywhere (which means that $g(x) + h(x) \\equiv 0$ for all $x$). So, $g(x) = -h(x)$ for all $x$. Since we have $f(x) = g(x)h(x)$ by assumption, we get $f(x) = -g(x)^2$, contradiction (because the leading coefficient of $f$ is 1).", "Solution_7": "Yes but I cannot understand. Could you explain it to me Arne with details. Please, Please. (My English are awful and i cannot understand what is <>,etc.)Please help", "Solution_8": "I typed a detailed explanation in my above post. Read it carefully.\r\nPlease use a dictionary if you don't understand.", "Solution_9": "Thank you very much Arne I catch it", "Solution_10": "[quote=\"t\u00b5t\u00b5\"]\nWhat about $f(x)=(x-a_1)(x-a_2)....(x-a_n) + 1$ ?\nIs there a general result ? x(x-1)+1 is irreducible but not x(x-2)+1 :huh:[/quote]\r\n\r\nI think that this is noy always true but this it is:\r\n$f(x)=[(x-a_1)(x-a_2)....(x-a_n)]^2+1$ with the conditions from my first post. Try to solve it", "Solution_11": "Silouan, I'm just wondering now - you named this problem \"easy\" and you posted it in the proposed problems section, but at first you didn't understand the solution. Does that mean that you posted it in the wrong section? Or did you find a different solution? If you found a different one, I'd like to see it.", "Solution_12": "I had a different one. And ok if I have time i will post it. Ok? But now try to solve the new problem." } { "Tag": [ "calculus", "integration", "logarithms", "function", "algebra", "binomial theorem", "real analysis" ], "Problem": "find an asymptotic formula for $f(x)=\\int_2^x\\sqrt{\\frac{x-t}{t^2-4}}\\;dt$ as $x\\to+\\infty$.", "Solution_1": "[quote=\"cimabue\"]find an asymptotic formula for $f(x)=\\int_2^x\\sqrt{\\frac{x-t}{t^2-4}}\\;dt$ as $x\\to+\\infty$.[/quote]Write $f(x) = \\sqrt{x} \\int_h^1 \\sqrt{ \\frac{1-s}{s^2-h^2}} ds$ with $h = \\frac{2}{x}$. \r\n\r\nSplit the integral into $\\int_h^{1/2} + \\int_{1/2}^1$. \r\nIn the first integral, use the binomial theorem to write ${\\int_h^{1/2}\\sqrt{ \\frac{1-s}{s^2-h^2}} ds = \\int_h^{1/2} \\frac{1}{\\sqrt{s^2-h^2}} ds + \\sum_{k \\ge 1}}\\int_h^{1/2} \\frac{c_ks^k}{\\sqrt{s^2-h^2}} ds$, where the $c_k$ are binomial coefficients. The first term can be integrated explicitly and behaves like $\\ln{\\frac{2}{h}} = \\ln x$, and all other terms converge to constants and are differentiable in $h$.\r\n\r\nThe second integral converges to $\\int_{1/2}^1 \\frac{\\sqrt{1-s}}{s}ds = 2 atanh(\\frac{1}{\\sqrt{2}}) - \\sqrt{2}$ and is differentiable in $h$. \r\n\r\n\r\nAsymptotically therefore, $f(x) = \\sqrt{x} (\\ln x + const. + O(x^{-1})$. Were you interested in higher order terms?", "Solution_2": "nice! that's what i was looking for. my guess is that the function itself is related to some special functions but i couldnt get maple to do the work. i was hoping this would come up." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Find all positive integers $n$ such that \\[ \\frac{3^n - 2^n}{n} \\in \\mathbb Z \\]", "Solution_1": "Only $1$. Indeed, suppose that $n>1$ and let $p$ be the smallest prime divisor of $n$. Then:\r\n$(\\frac{3}{2})^n \\equiv 1 \\mod{p}$\r\nLet $k=ord(\\frac{3}{2})$. We have $k|p-1$ and $k|n$. But $k \\leq p-1 < p$ which is impossible because of the minimality of $p$, unless $k=1$. But $k=1$ implies that $\\frac{3}{2} \\equiv 1 \\mod{p}$, so $1 \\equiv 0 \\mod{p}$, contradiction.", "Solution_2": "Thanks TomciO", "Solution_3": "This problem has been discussed here http://www.mathlinks.ro/Forum/viewtopic.php?p=6799#p6799 ;)" } { "Tag": [ "geometry", "geometric transformation", "reflection", "circumcircle", "trapezoid", "geometry unsolved" ], "Problem": "Let $ ABC$ be an acute triangle.Let $ M$ be a point in the interior of $ \\triangle ABC$.\r\nLet $ M_1$ be the reflection of $ M$ wrt $ AB$, $ M_2$ be the reflection of $ M_1$ wrt $ BC$, $ M_3$ be the reflection of $ M_2$ wrt $ AC$.\r\nLet $ K$ be the midpoint of $ MM_3$.\r\nFind the locus of $ K$.\r\n\r\nI have some difficulties with this problem.Please help.Thanks.", "Solution_1": "It is easy to show that M,M2,M1 lie on the circle which have B as the circumcenter\r\nThe same as C\r\nDenote E,F are the perpendicular foot of C and B , B\u2019,C\u2019 are the reflection of B,C wrt AB,AC respectively\r\nWe have BB\u2019M3M2 is an isosceles trapezoid => BM2=B\u2019M3 => KL=LH=LF , where L is the midpoint of BM3 => L is the circumcenter of the circumcircle of \u22bfFKL \r\n\u2220HFK=1/2\u2220KLH=1/2\u2220MBM2=\u2220B=\u2220EFA => K,F,E are collinear \r\nThus K moves on the line EF \r\nWhen M is coincide with B then K=F, when M is coicide with C K=G\r\nThe locus of K is segment FG", "Solution_2": "This is very classical, actualy we get $M_3$ from $M$ by isometry which is composition of 3 reflections thus it is indirect isometry, so by Hjelmslev's theorem $K$ describe (part of) line." } { "Tag": [ "linear algebra", "matrix", "Putnam" ], "Problem": "I don't know the correct terminology in english so sorry if I make any mistakes\r\n\r\nGive then matrices $ n\\times n$ , such that $ A^3\\equal{}B^3$ and $ A^2B\\equal{}B^2A$ , prove that $ A^2\\plus{}B^2$ is non-inversible.\r\n\r\nThanks in advance for any help.", "Solution_1": "I'm pretty sure this has been posted before, but it's been quite a while.\r\n\r\nHint: what happens when you try multiplying out the product\r\n\\[ (A^2\\plus{}B^2)(A\\minus{}B)\\ ?\\]", "Solution_2": "If $ A\\equal{}B$ ?", "Solution_3": "Ah, yes. \r\n\r\ntriplebig: did you by any chance leave out a condition in the statement of the problem - the condition that $ A\\ne B\\,?$", "Solution_4": "I didn't, I checked and I copied it as it is in the book. The problem is from a WL Putnam Competition exam I think, but am not sure.", "Solution_5": "As you wrote it, your problem is false in a fairly trivial way. Suppose $ A\\equal{}B\\equal{}I$ (over a field of characteristic other than 2). Then $ A^3\\equal{}B^3\\equal{}A^2B\\equal{}B^2A\\equal{}I$ and $ A^2\\plus{}B^2\\equal{}2I$ is invertible. You may think you copied it correctly but you didn't. Maybe the condition is verbal: $ A$ and $ B$ are distinct matrices, or some wording to that effect.\r\n\r\nJust assume it does have the condition $ A\\ne B.$ Can you follow my hint?", "Solution_6": "I copied it word for word as it is in the list! Perhaps the author was the one that copied it wrong.\r\n\r\nSuppose the condition that they are distinct, how would I solve this problem?", "Solution_7": "I checked with my teacher today and he confirmed that it is indeed missing the condition $ A\\neq B$ . I have solved it using your first post. Thank you very much for your help. \r\n\r\nThe conclusion can be generalized right?\r\n\r\n$ XY\\equal{}O_n\\,;\\,Y\\neq O_n\\,\\Rightarrow\\,\\nexists X^{\\minus{}1}$", "Solution_8": "Right: any two nxn matrices A and B such that $ AB \\equal{} O_n$ and $ A,B \\neq O_n$ are singular. Proof: Suppose A is non-singular: therefore $ A^{\\minus{}1}$ exists and $ A^{\\minus{}1}AB \\equal{} A^{\\minus{}1} O_n <\\equal{}> B \\equal{} O_n$, contrary to the hypothesis that $ B \\neq O_n$. Therefore, A is singular. In a similar fashion, suppose B is non-singular: then $ B^{\\minus{}1}$ exists and $ ABB^{\\minus{}1} \\equal{} O_n B^{\\minus{}1} <\\equal{}> A \\equal{} O_n$. Therefore, B is also singular." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,b,c$ are real numbers such that $ a^2\\plus{}2b^2\\plus{}3c^2\\equal{}6$. Find the min of $ A\\equal{}a\\plus{}b\\plus{}c$", "Solution_1": "We use BCS inequality, we obtain:\r\n$ (a^2\\plus{}2b^2\\plus{}3c^2)(1\\plus{}\\frac{1}{2}\\plus{}\\frac{1}{3}) \\geq (a\\plus{}b\\plus{}c)^2$\r\nNow, we get the result: \r\n$ a\\plus{}b\\plus{}c \\leq \\sqrt{11}$" } { "Tag": [], "Problem": "How to do the different arithmetical operations in other number bases? could anyone post links to documents that explain that?", "Solution_1": "[url]http://www.cut-the-knot.org/blue/SysTable.shtml[/url]", "Solution_2": "Thanks, but it doesn't show, explain a lot. It is not complete. It discusses only addtion and multiplication. How about division, and subtraction?", "Solution_3": "another one:\r\n\r\nhttp://mathforum.org/library/drmath/view/55727.html" } { "Tag": [ "probability", "expected value" ], "Problem": "A penny, nickel, dime, and quarter are simultaneously flipped. What is the expected value of the amount of the coins which come up heads? Express your answer in cents, rounded to the nearest tenth of a cent.", "Solution_1": "Each coin has a value of whatever it's worth (obviously) and a probability of $ \\frac12$ of showing that value (otherwise, it shows a $ 0$, so we can discount those possibilities), for a total expected value of $ \\frac12(1\\plus{}5\\plus{}10\\plus{}25)\\equal{}\\frac{41}2\\equal{}\\boxed{20.5}$.", "Solution_2": "Perhaps the question could list the value in cents of how much these coins are worth.", "Solution_3": "A penny is worth 1 cent, a nickel is worth 5 cents, a dime is worth 10 cents, and a quarter is worth 25 cents.\n\nP.S. Can you teach me how to Dougy? :)", "Solution_4": "Can't two or three or four coins come up heads?\n\nWhat can only 1 coin come up heads?", "Solution_5": "^Yeah;\nI did all of the casework of 1,2,3,4 heads but why doesnt that work.\n\nAdditionally, if this wasnt casework, there is not (1/2) probability of getting heads, there is (1/2)^4 probability of getting all heads................", "Solution_6": "If you do the casework you get the same answer. \n\nIt follows a binomial distribution for the number of heads. You get 4 heads with a value of 0.41 cents (occurring 1/16), 3 heads with a summated value of 1.23 cents (consider the value of each possible 3 coin configuration and add all the individual configurations), 2 heads with a summated value of 1.23, and 1 head with a summated value of 0.41. After factoring in the probability of each outcome, you get 20.5 cents. \nNote - If you add the configuration values together, you do not need to multiply by the number of configurations there are by this solution method. ", "Solution_7": "[hide=Solution]Each coin has a probability $\\tfrac{1}{2}$ to come up heads. Thus,\n$$\\frac{1}{2} \\cdot \\left(1+5+10+25\\right)=\\boxed{\\frac{41}{2}}$$", "Solution_8": "oh my god i calculated every single possibility" } { "Tag": [ "quadratics", "inequalities proposed", "algebra", "polynomial", "three variable inequality", "Inequality" ], "Problem": "let$a,b,c >0$ and $abc=1$.\r\n prove that : \r\n$a^2+b^2+c^2 +3 \\geq 2(ab+bc+ca)$", "Solution_1": "[quote=\"tranthanhnam\"]let$a,b,c >0$ and $abc=1$.\n prove that : \n$a^2+b^2+c^2 +3 \\geq 2(ab+bc+ca)$[/quote]\r\nToo easy. Just use Schur.\r\n$\\sum a^2+3\\sqrt[3]{a^2b^2c^2}\\ge \\sum(\\sqrt[3]{a^4b^2}+\\sqrt[3]{a^4c^2}\\geq 2(ab+bc+ca)$", "Solution_2": "See also\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=18430\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=19076\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=50745\r\n\r\nand a more general version in\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=49049\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=19666 (beginning with post #2)\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=40979 (beginning with post #3)\r\n\r\n Darij", "Solution_3": "Darij Grinberg \r\n\r\nI have some questions for you: why dont you post a problem but\r\n everytime solve them send the problems you had enjoyed ...\r\nPlease ...\r\n - one more question as i remember before the exam in the IMO 2005 you and your teamates were saying womething in your lanquage with a loud voice can you say what were you saying please ...\r\n\r\nDavron", "Solution_4": "[quote=\"tranthanhnam\"]Let $ a,b,c > 0$ and $ abc \\equal{} 1,$ prove that\n \n$ a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} 3 \\geq 2( bc \\plus{} ca \\plus{} ab).$[/quote]Mr. Qing Song pointed out that this problem is [b]Example 1[/b] in :\r\n\r\nYu-Zhong Weng, Structuring quadratic function to prove inequalities (Chinese), High School Science References (Nanning), 1998, No. 6, page 14. (ISSN: 1002-6363)\r\n\r\nBy the way, \r\n\r\nthe polynomial $ 2(1 \\minus{} bc \\minus{} ca \\minus{} ab) \\plus{} a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} a^2b^2c^2$ is nonnegetive for real numbers $ a,b,c,$ \r\n\r\nbut cannot be represented as a sum of squares of other polynomials with real coefficients. \r\n\r\nSee also : http://www.mathlinks.ro/viewtopic.php?t=24093", "Solution_5": "[quote=\"Ji Chen\"][quote=\"tranthanhnam\"]Let $ a,b,c > 0$ and $ abc \\equal{} 1,$ prove that\n \n$ a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} 3 \\geq 2( bc \\plus{} ca \\plus{} ab).$[/quote]Mr. Qing Song pointed out that this problem is [b]Example 1[/b] in :\n\nYu-Zhong Weng, Structuring quadratic function to prove inequalities (Chinese), High School Science References (Nanning), 1998, No. 6, page 14. (ISSN: 1002-6363)\n\nBy the way, \n\nthe polynomial $ 2(1 \\minus{} bc \\minus{} ca \\minus{} ab) \\plus{} a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} a^2b^2c^2$ is nonnegetive for real numbers $ a,b,c,$ \n\nbut cannot be represented as a sum of squares of other polynomials with real coefficients. \n\nSee also : http://www.artofproblemsolving.com/viewtopic.php?t=24093[/quote]\nthank Ji Chen.\nI come.\nOur dialogue \uff1a\n\nhttp://blog.sina.com.cn/s/blog_4c1131020100yscg.html\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=483075]here[/url]", "Solution_6": "The following inequality is also true.\nLet $a,b,c >0$ , prove that\\[a^2+b^2+c^2 +2abc+1 \\geq 2(ab+bc+ca).\\]\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=483075]here[/url]\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&p=3559544]here[/url]\nLet $ {a,b,c}\\ge{0}$, prove that\\[{a^{2}+b^{2}+c^{2}+3\\sqrt[3]{a^{2}b^{2}c^{2}}\\geq 2(ab+bc+ca)}.\\][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=359138]here[/url] [url=http://math.stackexchange.com/questions/510104/how-prove-this-abc3-sqrt3abc-ge-2-sqrtab-sqrtbc-sqrtac?rq=1]here[/url]\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=607569&p=3611330#p3611330\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=612340&p=3642543#p3642543\nLet$a,b,c >0$ and $abc=1$.prove that $a^3+b^3+c^3 +3 \\geq 2(a^2+b^2+c^2).$", "Solution_7": "[quote=tranthanhnam]let$a,b,c >0$ and $abc=1$.\n prove that : $a^2+b^2+c^2 +3 \\geq 2(ab+bc+ca)$[/quote]\nThe following inequality is also true.\n\nLet$a,b,c >0$ and $a+b+c+2=abc$. Prove that : \n$$a^2+b^2+c^2 +12 \\geq 2(ab+bc+ca).$$\n", "Solution_8": "[quote=sqing]\nLet$a,b,c >0$ and $a+b+c+2=abc$. Prove that : \n$$a^2+b^2+c^2 +12 \\geq 2(ab+bc+ca).$$[/quote]\n[url=http://www.artofproblemsolving.com/community/c6h278791]uvw[/url] kills it immediately.\n\n", "Solution_9": "Is this problem true?\nLet $a, b, c > 0$ such that $a^2 + b^2 + c^2 = 3$. Prove that:\n$$2 + abc \\ge a + b + c$$\nIf this problem wrong, which $0 < k \\le 3$ satisfying:\n$$k + (3 - k)abc \\ge a + b + c$$", "Solution_10": "[hide=*]Let $a,b,c>0$ and $abc=1.$ For the positive integer $n$ , [url=https://artofproblemsolving.com/community/c6h1598026p9930634]prove or disprove [/url]\\[a^{n+1}+b^{n+1}+c^{n+1}+3 \\ge 2(a^n+b^n+c^n).\\][/hide]\nFor $a,b,c\\geq 0 .$ [url=http://artofproblemsolving.com/community/c6h457354p2569858]Prove that[/url]$$a^3+b^3+c^3+3abc \\ge 2\\sqrt[3]{abc}(a^2+b^2+c^2).$$\n[url=https://artofproblemsolving.com/community/c6h612340p3641832]here\n[/url]", "Solution_11": "[quote=tranthanhnam]let$a,b,c >0$ and $abc=1$.\n prove that : \n$a^2+b^2+c^2 +3 \\geq 2(ab+bc+ca)$[/quote]\n\nLet $x,y,z>0$, prove\n\n$\\sum_{cyc}x^4z^2+3x^2y^2z^2\\geq 2\\sum_{cyc}x^3y^2z$\n\nTakeya.O", "Solution_12": "[quote=sqing]Let $a,b,c>0$ and $abc=1.$ For the positive integer $n$ , [url=https://artofproblemsolving.com/community/c6h1598026p9930634]prove or disprove [/url]\\[a^{n+1}+b^{n+1}+c^{n+1}+3 \\ge 2(a^n+b^n+c^n).\\][/quote]\nCounterExample\n$n=3, a=\\frac{3}{2}, b=\\frac{2}{3}, c=1$", "Solution_13": "Let $a,b,c,d >0.$ Prove that$$a^2+b^2+c^2 +2abc+1 \\geq 2(ab+bc+ca).$$\n$$a^2+b^2+c^2 +d^2+abcd+1\\geq a^2+b^2+c^2+d^2+2\\sqrt{abcd} \\geq ab+ac+ad+bc+bd+cd.$$\n[url=https://artofproblemsolving.com/community/c6h1709871p11020577]Turcevichi[/url]", "Solution_14": "[quote=sqing]The following inequality is also true.\nLet $a,b,c >0$ , prove that\\[a^2+b^2+c^2 +2abc+1 \\geq 2(ab+bc+ca).\\]\n[hide=*][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=483075]here[/url]\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&p=3559544]here[/url]\nLet $ {a,b,c}\\ge{0}$, prove that\\[{a^{2}+b^{2}+c^{2}+3\\sqrt[3]{a^{2}b^{2}c^{2}}\\geq 2(ab+bc+ca)}.\\][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=359138]here[/url] [url=http://math.stackexchange.com/questions/510104/how-prove-this-abc3-sqrt3abc-ge-2-sqrtab-sqrtbc-sqrtac?rq=1]here[/url]\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=607569&p=3611330#p3611330\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=612340&p=3642543#p3642543\nLet$a,b,c >0$ and $abc=1$.prove that $a^3+b^3+c^3 +3 \\geq 2(a^2+b^2+c^2).$[/hide][/quote]\nProve that for all positive real numbers $x, y, z,$ the inequality holds\n$$x^2 + y^2 + z^2 + 2xyz + 1\\geq2 (xy + yz + zx).$$\n([b]Training to the Baltic Way 2018 team , Estonian[/b])\nProof:\nThe net is symmetric with respect to $x, y, z,$ so we can assume that $x> y> z.$\nIf $y \\geq 1, $ then the inequality is equivalent to the equation\n$(x-y)^ 2+ (z-1)^2 + 2z (x-1) (y-1) \\geq 0.$\nIf $y <1, $ then the inequality is equivalent to the equality\n$(x-y)^2+ (x-1)^2 + 2x (1-y) (1-z) \\geq 0.$\nAnother solution. [b]The difference can be written in the form[/b]\n$(x-1)^2+ (y-1)^2+ (z-1)^2 + 2 (x-1) (y-1) (z-1) \\geq 0.$\nIf the expressions $x-1, y-1, z-1$ are exactly $0$ or $2$, then the equation is valid.\nSo let's assume that there are $1$ or $3$ negative ones. \nLet's not narrow the universe $x-1 $ is negative, in this case $(y-1) (z-1) \\geq 0. $ \nSince $x$ is positive, then $-1 (x-1)^2+(y-1)^2+ (z-1)^2-2 (y-1) (z-1) =(x-1)^2 + ((y-1) - (z-1))^2\\geq0.$$", "Solution_15": "[quote=sqing]The following inequality is also true.\nLet $a,b,c >0$ , prove that\\[a^2+b^2+c^2 +2abc+1 \\geq 2(ab+bc+ca).\\]\n[hide=*][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=483075]here[/url]\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&p=3559544]here[/url]\nLet $ {a,b,c}\\ge{0}$, prove that\\[{a^{2}+b^{2}+c^{2}+3\\sqrt[3]{a^{2}b^{2}c^{2}}\\geq 2(ab+bc+ca)}.\\][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=359138]here[/url] [url=http://math.stackexchange.com/questions/510104/how-prove-this-abc3-sqrt3abc-ge-2-sqrtab-sqrtbc-sqrtac?rq=1]here[/url]\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=607569&p=3611330#p3611330\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=612340&p=3642543#p3642543\nLet$a,b,c >0$ and $abc=1$.prove that $a^3+b^3+c^3 +3 \\geq 2(a^2+b^2+c^2).$[/hide][/quote][hide=Yu-Zhong Weng]Let $a,b,c$ be positive real numbers such that $abc=1 .$ Prove that$$a^2+b^2+c^2+3\\geq 2(ab+bc+ca).$$\n[b]Proof A:[/b]\nWLOG $(a-1)(b-1)\\geq 0,$ so $$a^2+b^2+c^2+3- 2(ab+bc+ca)=(a-b)^2+(c-1)^2+2c(a-1)(b-1)\\geq 0.$$\n[b]Proof B:[/b]\n$$a^2+b^2+c^2+3- 2(ab+bc+ca)=\\frac{1}{a^2(b^2+1)}\\left(2b^2c^2(b-1)^2+(bc^2-1)^2+b^2(bc^2+1-c-b^2c)^2\\right)\\geq 0.$$\nhttps://artofproblemsolving.com/community/c6h1657181p11607206[/hide]", "Solution_16": "[quote=sqing]\nLet $ {a,b,c}\\ge{0}$, prove that\\[{a^{2}+b^{2}+c^{2}+3\\sqrt[3]{a^{2}b^{2}c^{2}}\\geq 2(ab+bc+ca)}.\\][/quote]\nLet $a, b, c \\ge 0$. For any positive integer n prove that [url=https://artofproblemsolving.com/community/c6h1880582p12793518]the following inequality exist[/url]\n$$(2n-1)(a^2+b^2+c^2)+3\\sqrt[3]{a^2b^2c^2} \\geq 2n(ab+bc+ca).$$\n[hide=P]\\[2(n-1)(a^2+b^2+c^2-ab-bc-ca)+\\left[a^2+b^2+c^2+3\\sqrt[3]{a^2b^2c^2} - 2(ab+bc+ca)\\right] \\geqslant 0.\\][/hide]\n", "Solution_17": "[quote=sqing]The following inequality is also true.\nLet $a,b,c >0$ , prove that\\[a^2+b^2+c^2 +2abc+1 \\geq 2(ab+bc+ca).\\]\n[hide=*][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=483075]here[/url]\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&p=3559544]here[/url][/hide]\nLet $ {a,b,c}\\ge{0}$, prove that\\[{a^{2}+b^{2}+c^{2}+3\\sqrt[3]{a^{2}b^{2}c^{2}}\\geq 2(ab+bc+ca)}.\\][hide=**][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=359138]here[/url] [url=http://math.stackexchange.com/questions/510104/how-prove-this-abc3-sqrt3abc-ge-2-sqrtab-sqrtbc-sqrtac?rq=1]here[/url]\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=607569&p=3611330#p3611330\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=612340&p=3642543#p3642543\nLet$a,b,c >0$ and $abc=1$.prove that $a^3+b^3+c^3 +3 \\geq 2(a^2+b^2+c^2).$[/hide][/quote]\n[url]https://artofproblemsolving.com/community/c6h1827390p12225661[/url]\nFor $a,b,c>0$. [url=https://artofproblemsolving.com/community/c6h1827390p15189416]Prove that [/url]$$a^2+b^2+c^2+abc+2\\ge ab+bc+ca+ a+b+c$$\n[hide=Wlog]assume that $(a-1)(b-1) \\ge 0$ then\n$$a^2+b^2+c^2+abc+2-\\left(a+b+c+ab+bc+ca\\right)= \\left(a-b\\right)^2 + \\left(c-1\\right)^2 + \\left(a-1\\right)\\left(b-1\\right)\\left(c+1\\right) \\ge 0$$[/hide]", "Solution_18": "redacted", "Solution_19": "[quote=tranthanhnam]let$a,b,c >0$ and $abc=1$. Prove that : \n$$a^2+b^2+c^2 +3 \\geq 2(ab+bc+ca)$$[/quote]\nLet$a,b,c >0$ and $abc=1.$ Prove that \n$$a^2+b^2+c^2 +3 \\geq a^ 2 +b^ 2 +c^ 2 +\\frac{{9}}{{a+b+c}} \\geq 2(ab+bc+ca)$$\n[quote=sqing]The following inequality is also true.\nLet $a,b,c >0$ , prove that\\[a^2+b^2+c^2 +2abc+1 \\geq 2(ab+bc+ca).\\][/quote]\nLet $a,b,c$ be three positive real numbers . [url=https://artofproblemsolving.com/community/c1068820h2234630p17094843]Prove that[/url]\n$$a^2 +b^ 2+c^ 2+abc+ 4 \\geq 2 (ab +bc+ca)$$", "Solution_20": "Let $a_1,a_2,\\cdots,a_n (n\\ge 3)$ be positive numbers such that $a_1a_2\\cdots a_n=1.$ Prove that$$a_1^2+a_2^2+\\cdots+a_n^2+n\\geq2\\left(\\frac{1}{a_1}+\\frac{1}{a_2}+\\cdots+\\frac{1}{a_n}\\right).$$\n[url=https://artofproblemsolving.com/community/c6h1064117p11637470]h[/url]", "Solution_21": "Let $a,b,c,d\\geq 0.$ Prove that$$a^2+b^2+c^2+d^2+a+b+c+d+abc+bcd+cda+dab\\geq 2(ab+ac+ad+bc+bd+cd)$$\n[quote=sqing]For $a,b,c>0$. [url=https://artofproblemsolving.com/community/c6h1827390p15189416]Prove that [/url]$$a^2+b^2+c^2+abc+2\\ge ab+bc+ca+ a+b+c$$[/quote]\n[url]https://artofproblemsolving.com/community/c6h518938p12694099[/url]\n[url]https://artofproblemsolving.com/community/c6h1827390p12225661[/url]\nLet $a_1,a_2,\\cdots,a_n (n\\ge 3)$ be positive numbers such that $a_1a_2\\cdots a_n=1.$ Prove that$$a_1+a_2+\\cdots+a_n- n\\geq\\min\\{a_1,a_2,\\cdots,a_n\\}(\\frac{1}{a_1}+\\frac{1}{a_2}+\\cdots+\\frac{1}{a_n}-n)$$\n[size=50](p/7303459116)[/size]\n[quote=sqing]Let $ {a,b,c}\\ge{0}$, prove that\\[{a^{2}+b^{2}+c^{2}+3\\sqrt[3]{a^{2}b^{2}c^{2}}\\geq 2(ab+bc+ca)}.\\][/quote]\n[url=https://math.stackexchange.com/questions/510104/how-prove-this-abc3-sqrt3abc-ge-2-sqrtab-sqrtbc-sqrtac?rq=1]Solution[/url]", "Solution_22": "Let $a,b,c >0$ . Prove that\\[a^3+b^3+c^3 +abc+2 \\geq 2(ab+bc+ca)\\]\n\\[a^5+b^5+c^5 +abc+5 \\geq 3(ab+bc+ca)\\]\n\\[a^3+b^3+c^3 +5abc \\geq (a+b)(b+c)(c+a)\\]\n\\[a^3+b^3+c^3 +abc+4\\geq (a+1)(b+1)(c+1)\\]" } { "Tag": [ "vector", "function", "integration", "Functional Analysis", "advanced fields", "advanced fields unsolved" ], "Problem": "Let $ \\Omega$ be a bounded, open subset of $ \\mathbf R^{n}$ and assume that $ f\\colon\\mathbf R\\to\\mathbf R$ is strictly decreasing with $ f(0)=0$. Let $ u \\in C^{2}(\\Omega) \\cap C^{0}(\\bar\\Omega)$ be a solution of the following problem\r\n\\[ \\begin{cases}-\\Delta u = f(u) & \\text{ in }\\Omega, \\\\ u = 0 & \\text{ on }\\partial\\Omega. \\end{cases}\\]\r\nShow that $ u=0$.", "Solution_1": "Any $ u$ satisfying the equation $ -\\Delta u=f(u)$ cannot have a positive local maximum or a negative local minimum; $ u$ is always increasing in some direction whenever $ u$ is positive since $ \\Delta u$ is also positive, and always decreasing in some direction whenever $ u$ is negative since $ \\Delta u$ is also negative.\r\n\r\nNow, consider the absolute maximum and minimum of $ u$; if the maximum is positive or the minimum is negative, this point must lie in the interior, where it is a local maximum or minimum. This is impossible by the above, and both the maximum and minimum of $ u$ are zero.", "Solution_2": "Why do you write gradient $ \\nabla$ instead of Laplace operator $ \\Delta$? As you wrote, the left hand side is a vector function while the right hand side is a scalar function... :maybe: I guess you mean the following solution (I slightly changed your approach): :maybe: \r\n[hide=\"Solution\"] Note that $ u$ cannot have positive local maximum and negavite local minimum inside $ \\Omega$. Indeed, if $ x_{*}$ is a point of positive maximum, then $ u''(x_{*})$ is negative-semidefinite, yielding that $ -\\Delta u(x_{*}) \\ge 0$ while $ f(u(x_{*})) < 0$, a contradiction. The situation with negative local minimum is analoguous.\n\nNow, it follows from above reasonings that $ u \\le 0$ (because there is no positive abolute maxumum of $ u$ in $ \\bar\\Omega$). Analoguously, $ u \\ge 0$ (because there is no negative absolute minumum of $ u$ in $ \\bar\\Omega$). Thus $ u=0$. [/hide]", "Solution_3": "It seems that $ f$ must not be [i]strictly[/i] decreasing, if it were just decreasing, all would be OK. Here're my two solutions not using strict monotonicity of $ f$, and I wonder are there any mistakes?\r\n[hide=\"Solution 1\"] Since $ f(s) \\le f(0) = 0$ for $ s\\ge0$, one has\n\t\\[ \\int_\\Omega |\\nabla u^{+}|^{2}= \\int_\\Omega f(u)u^{+}\\le 0,\\]\n\tyielding that $ u \\le 0$. Now, since $ f(s) \\ge f(0) = 0$ for $ s\\le0$ and since $ u\\le0$, $ -\\Delta u \\ge 0$, yielding that $ u\\ge0$. Thus, $ u=0$.\n[/hide]\n[hide=\"Solution 2\"]Set $ X : = H_{0}^{1}(\\Omega)$ and $ Y : = H^{-1}(\\Omega)$. Consider the map $ G \\colon X \\to Y$ defined by $ G(u) =-\\Delta u-f(u)$, then $ G'(u) =-\\Delta-f'(u)$. By data, $ f'(0) \\le 0$, yielding that $ G'(0) \\colon X \\to Y$ is a Banach space isomorphism. Since $ G(0)=0$, by Implicit Function Theorem, $ G$ is a diffeomorphism of some neighbourhood $ U_{X}$ of 0 in $ X$ onto some neighbourhood $ U_{Y}$ of 0 in $ Y$. Thus $ u=0$ is the unique solution to our problem.\n[/hide]", "Solution_4": "Mistake fixed.\r\n\r\nAlso, there's no problem if it's not strictly decreasing; harmonic functions don't have local maxima/minima unless they're constant." } { "Tag": [ "algebra", "polynomial" ], "Problem": "What is the general factorization of $x^{n}-1$ ?", "Solution_1": "[hide]We know by Bezout's Lemma that since $x = 1 \\Rightarrow x^{n}-1 = 0$, $x-1 | x^{n}-1.$ As a result, we have:\n\n$\\frac{x^{n}-1}{x-1}= x^{n-1}+x^{n-2}+\\cdots+1$\n\nThis follows by the formula for the sum of a geometric series. As a result, we can factor this polynomial into:\n\n$(x-1)(x^{n-1}+x^{n-2}+\\cdots+1)$[/hide]", "Solution_2": "What about $x^{n}+1$? :)", "Solution_3": "[quote=\"Arvind_sn\"]What about $x^{n}+1$? :)[/quote]\r\nWell it depends on whether x is odd or even I think.", "Solution_4": "That is true, if n is odd, then the factorization is $x^{n}+1=(x+1)(x^{n-1}-x^{n-2}+x^{n-3}...-x+1)$. However, if n is then there is no [b]real[/b] factorization.", "Solution_5": "[quote=\"The QuattoMaster 6000\"]That is true, if n is odd, then the factorization is $x^{n}+1=(x+1)(x^{n-1}-x^{n-2}+x^{n-3}...-x+1)$. However, if n is [i][even][/i] then there is no [b]real[/b] factorization.[/quote]\r\n\r\nno\r\n\r\n$4x^{4}+1=4x^{4}+4x^{2}+1-(2x)^{2}=(2x^{2}-2x+1)(2x^{2}+2x+1)$, take $y=x\\sqrt{2}$, to get\r\n\r\n$y^{4}+1=(y^{2}-\\sqrt{2}y+1)(y^{2}+\\sqrt{2}y+1)$\r\n\r\n@Karth...that is not Bezout's lemma", "Solution_6": "[quote=\"Karth\"]$x = 1 \\Rightarrow x^{n}-1 = 0$, $x-1 | x^{n}-1.$ [/quote]\n\nThey call this Bezout's Lemma? :huh: \n\nhttp://planetmath.org/encyclopedia/BezoutsLemma.html\n\n[quote=\"The QuattoMaster 6000\"]That is true, if n is odd, then the factorization is $x^{n}+1=(x+1)(x^{n-1}-x^{n-2}+x^{n-3}...-x+1)$. However, if n is then there is no [b]real[/b] factorization.[/quote]\r\n\r\nPresumably you mean \"if $n$ is even.\" But what about\r\n\r\n$x^{6}+1 = (x^{2}+1)(x^{4}-x^{2}+1)$\r\n\r\nAnyway, the general factorization (over the rationals) takes the form\r\n\r\n$x^{n}-1 = \\prod_{d | n}\\Phi_{d}(x)$\r\n\r\nhttp://mathworld.wolfram.com/CyclotomicPolynomial.html\r\n\r\nEdit: And, of course,\r\n\r\n$x^{n}+1 = \\frac{x^{2n}-1}{x^{n}-1}= \\prod_{d | 2n, d \\not | n}\\Phi_{d}(x)$", "Solution_7": "Whoops my bad... :blush: It won't let me edit it though." } { "Tag": [ "inequalities", "induction", "inequalities proposed" ], "Problem": "Let $x_1 ,x_2 ,x_3 ,...,x_n >0$ and $n\\geq 3$ such that their product is equal with $1$. Prove that:\r\n\r\n $\\frac{1}{\\sqrt{1+2x_1 }}+\\frac{1}{\\sqrt{1+2x_2 }}+...+\\frac{1}{\\sqrt{1+2x_n }} 0$ and $n\\geq 3$ such that their product is equal with $1$. Prove that:\n\n $\\frac{1}{\\sqrt{1+2x_1 }}+\\frac{1}{\\sqrt{1+2x_2 }}+...+\\frac{1}{\\sqrt{1+2x_n }} n-1\\]\r\nWe will now prove the claim by induction. We arrange all $a_i$ in increasing order, starting induction with the lowest 3. Suppose it was true for k. Then it is sufficient to prove that for some $a_i,a_{i+1}$, we can put a new variable $a_{i+1}$ in between them, labeling the old $a_{i+1}$ as $a_{i+2}$, so that the ineq is still true. This is obvious, since we can prove that\r\n\\[\\sqrt{\\frac{a_i}{a_i+2a_{i+2}}} + 1 \\ge \\sqrt{\\frac{a_i}{a_i+2a_{i+1}}} + \\sqrt{\\frac{a_{i+1}}{a_{i+1}+2a_{i+2}}}\\]\r\nand the other terms are invariant. Proof: $1>\\sqrt{\\frac{a_{i+1}}{a_{i+1}+2a_{i+2}}}$ since $a_{i+2}>0$ and $\\sqrt{\\frac{a_i}{a_i+2a_{i+2}}}\\ge \\sqrt{\\frac{a_i}{a_i+2a_{i+1}}}$ since $a_{i+1}$ the largest. Which proves the claim. $\\Box$\r\n\r\nWe can remark that for all $x_i$ going to 0, except for $x_1$ going to infinity, we can approximate $n-1$ arbitrarily close.", "Solution_2": "This inequality can be improved for $n\\ge4$ as follows:\r\n\r\nLet $x_1 ,x_2 ,x_3 ,...,x_n >0$ and $n\\geq 4$ such that their product is equal with $1$. Prove that:\r\n $\\frac{1}{\\sqrt{1+x_1 }}+\\frac{1}{\\sqrt{1+x_2 }}+...+\\frac{1}{\\sqrt{1+x_n }} 0 \r\nxyz(x+y+z)=1 find the smallest possible value of (x+y)(y+z). Russia 1989 - solved in mathlinks)", "Solution_6": "[quote=\"borislav_mirchev\"][b]N. T. THUAN[/b], I think there are some very intelligent people on mathlinks. I saw the ideas on the link. It is far better than before, but I'm just wondering if is there simpler solution to this problem?\n[/quote]\r\nYes, but i am N.T.TUAN. :P", "Solution_7": "Suppose contrary that $ S$ is rational. Consider the following set of polynomials $ \\{P_{n}\\}$, $ 1\\leq{n}\\leq1000$.\r\n\r\nLet $ P_{1}(x) \\equal{} x^{2}\\minus{}(1001^{2}\\plus{}1)$. We also recursively define \r\n\\[ P_{k\\plus{}1}(x) \\equal{} P_{k}(x\\minus{}\\sqrt{(1000\\plus{}k)^{2}\\plus{}1})\\cdot{P_{k}(x\\plus{}\\sqrt{(1000\\plus{}k)^{2}\\plus{}1})}\\,,\\] \r\nfor all $ 1\\leq{k}\\leq999$. It is easy to see that $ S$ is a zero of $ P_{1000}$. Since the leading coefficient of $ P_{1000}$ is $ 1$ and all of its coefficients are integers, then the fact that $ S$ is a rational number implies that $ S$ is an integer.\r\n\r\nIt is clear that $ S > 1001\\plus{}1002\\plus{}\\cdots\\plus{}2000$. Also, $ \\sqrt{N^{2}\\plus{}1}\\equal{} N\\sqrt{1\\plus{}N^{\\minus{}2}}< N\\left(1\\plus{}N^{\\minus{}2}/2\\right)$. Thus,\r\n\\[ \\sqrt{N^{2}\\plus{}1}< N\\plus{}1/2N\\,,\\]\r\nfor all $ N \\equal{} 1001, 1002,\\dots, 2000$. Consequently,\r\n\\[ S < (1001\\plus{}\\plus{}1002\\plus{}\\cdots\\plus{}2000)\\plus{}\\frac{(1/1001\\plus{}1/1002\\plus{}\\cdots\\plus{}1/2000)}{2}\\,.\\]\r\nHence,\r\n\\[ S < (1001\\plus{}\\plus{}1002\\plus{}\\cdots\\plus{}2000)\\plus{}\\frac{(1/1000\\plus{}1/1000\\plus{}\\cdots\\plus{}1/1000)}{2}\\,.\\] \r\nThus,\r\n\\[ S < (1001\\plus{}\\plus{}1002\\plus{}\\cdots\\plus{}2000)\\plus{}1/2\\,.\\]\r\nTherefore, \r\n\\[ 1001\\plus{}1002\\plus{}\\cdots\\plus{}2000 < S < (1001\\plus{}\\plus{}1002\\plus{}\\cdots\\plus{}2000)\\plus{}1/2\\,,\\]\r\nwhich is a contradiction!", "Solution_8": "It's very good! It seems you have lots of experience." } { "Tag": [ "trigonometry", "inequalities unsolved", "inequalities" ], "Problem": "let x, y, z be real #, suppose that x^2+y^2+z^2+2xyz=1.\r\nProve that\r\nx^2+y^2+z^2>=3/4", "Solution_1": "[hide=\"hint\"]Exist a triangle $ABC$ such that $x=\\cos{A},y=\\cos{B},z=\\cos{C}$[/hide]", "Solution_2": "[quote=\"N.T.TUAN\"]Exist a triangle $ABC$ such that $x=\\cos{A},y=\\cos{B},z=\\cos{C}$[/quote]\n[quote=\"davemiller\"]\nlet x, y, z be real :wink: [/quote]", "Solution_3": "[quote=\"arqady\"][quote=\"N.T.TUAN\"]Exist a triangle $ABC$ such that $x=\\cos{A},y=\\cos{B},z=\\cos{C}$[/quote]\n[quote=\"davemiller\"]\nlet x, y, z be real :wink: [/quote][/quote]\r\nIf $|x|,|y| \\text{or}|z| >1$ then trivial :D", "Solution_4": "[quote=\"davemiller\"]let x, y, z be real #, suppose that $x^{2}+y^{2}+z^{2}+2xyz=1.$\nProve that\n$x^{2}+y^{2}+z^{2}\\geq \\frac{3}{4}$[/quote]" } { "Tag": [], "Problem": "Compute: $ \\frac{6!}{5!}\\minus{}\\frac{4!}{3!}$.", "Solution_1": "$ 6!/5!\\equal{}6*5!/5!\\equal{}6$\r\n$ 4!/3!\\equal{}4*3!/3!\\equal{}4$", "Solution_2": "hello, we have $ \\frac{6!}{5!}\\minus{}\\frac{4!}{3!}\\equal{}\\frac{5!\\cdot6}{5!}\\minus{}\\frac{3!\\cdot4}{3!}\\equal{}6\\minus{}4\\equal{}2$.\r\nSonnhard." } { "Tag": [ "trigonometry", "geometry unsolved", "geometry" ], "Problem": "An acute triangle $ ABC$. Prove that: $ \\sum(\\sin2B+\\sin2C)^{2}.\\sinh \\leq 12.\\sinh.\\sinh.\\sinh$", "Solution_1": "[quote=\"Nbach\"]An acute triangle $ ABC$. Prove that: $ \\sum(\\sin2B+\\sin2C)^{2}.\\sinh \\leq 12.\\sinh.\\sinh.\\sinh$[/quote]\r\n$ \\sum(\\sin2B+\\sin2C)^{2}.\\sin A\\leq 12\\sin A\\sin B\\sin C$" } { "Tag": [], "Problem": "\u039d\u03b1 \u03c0\u03c1\u03bf\u03c3\u03b4\u03b9\u03bf\u03c1\u03af\u03c3\u03b5\u03c4\u03b1\u03b9 \u03cc\u03bb\u03b5\u03c2 \u03c4\u03b9\u03c2 \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03ad\u03c2 \u03b8\u03b5\u03c4\u03b9\u03ba\u03ad\u03c2 \u03c4\u03c1\u03b9\u03ac\u03b4\u03b5\u03c2 x,y \u03ba\u03b1\u03b9 z \u03bf\u03b9 \u03bf\u03c0\u03bf\u03af\u03b5\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bb\u03cd\u03c3\u03b5\u03b9\u03c2 \u03c4\u03bf\u03c5 \u03c3\u03c5\u03c3\u03c4\u03ae\u03bc\u03b1\u03c4\u03bf\u03c2 \r\n\r\n $x+y+z = 6$\r\n\r\n ${\\frac{1}^{x}}+{\\frac{1}^{y}}+{ \\frac{1}^{z}}= 2-{\\frac{4}^{xyz}}$", "Solution_1": "\u0391\u03c5\u03c4\u03ac \u03b1\u03c1\u03ad\u03c3\u03bf\u03c5\u03bd \u03c3\u03c4\u03bf\u03bd \u03b1\u03b4\u03b5\u03c1\u03c6\u03cc \u03bc\u03bf\u03c5 :D .\r\n\u0386\u03c3\u03b5 \u03c0\u03bf\u03c5 \u03ac\u03bc\u03b1 \u03c4\u03b7\u03bd \u03ad\u03b2\u03b1\u03b6\u03b5\u03c2 \u03c3\u03c4\u03bf \u03c3\u03c7\u03bf\u03bb\u03b5\u03af\u03bf \u03b8\u03b1 \u03c3\u03bf\u03c5 \u03ad\u03bb\u03b5\u03b3\u03b1\u03bd 2 \u03b5\u03be\u03b9\u03c3\u03ce\u03c3\u03b5\u03b9\u03c2 \u03bc\u03b5 \u03c4\u03c1\u03b5\u03b9\u03c2 \u03b1\u03b3\u03bd\u03ce\u03c3\u03c4\u03bf\u03c5\u03c2 \u03b3\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03b1\u03c5\u03c4\u03cc \u03bd\u03b1 \u03bb\u03c5\u03b8\u03b5\u03af ? \u0386\u03bb\u03c5\u03c4\u03bf \u03c0\u03c1\u03bf\u03b2\u03bb\u03b7\u03bc\u03b1 \u03b3\u03b9\u03b1 \u03b1\u03c5\u03c4\u03bf\u03cd\u03c2 . :) :P \r\n\u0392\u03ac\u03b6\u03c9 \u03bb\u03cd\u03c3\u03b7 \u03bc\u03cc\u03bd\u03bf \u03ba\u03b1\u03b9 \u03bc\u03cc\u03bd\u03bf \u03b5\u03c0\u03b5\u03b9\u03b4\u03ae \u03c4\u03b7\u03bd \u03c0\u03c1\u03cc\u03c4\u03b5\u03b9\u03bd\u03b5\u03c2 \u03b5\u03c3\u03cd \u0393\u03b9\u03ce\u03c1\u03b3\u03b7 . :wink: \r\nHere is my solution \r\n\u0391\u03c0\u03cc Cauchy-Swartz $(x+y+z)(\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z})\\geq 9$ \r\n\r\n\u03ac\u03c1\u03b1 $\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}\\geq \\frac{3}{2}$ \r\n\r\n\u03ba\u03b1\u03b9 \u03b7 \u03b4\u03b5\u03cd\u03c4\u03b5\u03c1\u03b7 \u03b5\u03be\u03af\u03c3\u03c9\u03c3\u03b7 \u03b4\u03af\u03bd\u03b5\u03b9 \r\n$\\frac{3}{2}\\leq \\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=2-\\frac{4}{xyz}$ \r\n\r\n\u03b1\u03c0\u03cc \u03cc\u03c0\u03bf\u03c5 $xyz\\geq 8$ \r\n\r\n\u039c\u03b5 AM-Gm \u03cc\u03bc\u03c9\u03c2 \u03b7 \u03c0\u03c1\u03ce\u03c4\u03b7 \u03b4\u03af\u03bd\u03b5\u03b9 $6=x+y+z\\geq 3\\sqrt[3]{xyz}$ \u03ac\u03c1\u03b1 \r\n$xyz\\leq 8$ \r\n\u0386\u03c1\u03b1 $xyz=8$ kai x=y=z ara $x=y=z=2$", "Solution_2": "\u0391\u03b4\u03b5\u03c1\u03c6\u03ad \u03bc\u03c0\u03c1\u03ac\u03b2\u03bf :o \u03ba\u03b1\u03bb\u03ac \u03b5\u03c3\u03cd \u03ad\u03c7\u03b5\u03b9\u03c2 \u03b5\u03bb\u03b1\u03c7\u03b9\u03c3\u03c4\u03bf\u03c0\u03bf\u03b9\u03ae\u03c3\u03b5\u03b9 \u03c4\u03bf\u03bd \u03c7\u03c1\u03cc\u03bd\u03bf . \u0391\u03c5\u03c4\u03ae\u03bd \u03c4\u03b7\u03bd \u03bb\u03cd\u03c3\u03b7 \u03b5\u03af\u03c7\u03b1 \u03ba\u03b1\u03b9 \u03b5\u03b3\u03ce \u03b2\u03c1\u03b5\u03af", "Solution_3": "[quote=\"gdeskos\"]\u03ba\u03b1\u03bb\u03ac \u03b5\u03c3\u03cd \u03ad\u03c7\u03b5\u03b9\u03c2 \u03b5\u03bb\u03b1\u03c7\u03b9\u03c3\u03c4\u03bf\u03c0\u03bf\u03b9\u03ae\u03c3\u03b5\u03b9 \u03c4\u03bf\u03bd \u03c7\u03c1\u03cc\u03bd\u03bf.[/quote]\r\n\r\n\u0388\u03bb\u03b1 \u03c1\u03b5 \u03c6\u03af\u03bb\u03b5 \u03b5\u03bd\u03c4\u03ac\u03be\u03b5\u03b9 \u03c5\u03c0\u03b5\u03c1\u03b2\u03ac\u03bb\u03bb\u03b5\u03b9\u03c2 :) .\r\n\u03a0\u03ac\u03bd\u03c4\u03c9\u03c2 \u03b5\u03bd\u03b4\u03b9\u03b1\u03c6\u03ad\u03c1\u03bf\u03bd \u03ad\u03c7\u03b5\u03b9 \u03b7 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03c3\u03c4\u03bf\u03c5\u03c2 \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03bf\u03cd\u03c2 . :wink: \r\n\u0398\u03b1 \u03b2\u03ac\u03bb\u03bb\u03c9 \u03ba\u03b1\u03b9 \u03b5\u03b3\u03ce \u03bc\u03b9\u03b1 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03c4\u03ce\u03c1\u03b1 \u03b1\u03bd \u03b8\u03b5\u03c2 \u03ba\u03bf\u03af\u03c4\u03b1 \u03c4\u03b7\u03bd ." } { "Tag": [ "linear algebra", "matrix" ], "Problem": "Suppose $ A$ and $ B$ are $ n\\times n$ matrices satisfying $ A^2 \\plus{} A\\equal{}O$, $ B^2 \\plus{} B\\equal{}O$ and $ AB\\plus{}BA\\equal{}O$. Determine $ \\det (A\\minus{}B)$, $ \\det(A\\plus{}B)$.", "Solution_1": "[hide]\n$ (A \\plus{} B)^2 \\equal{} A^2 \\plus{} AB \\plus{} BA \\plus{} B^2 \\equal{} \\minus{} A \\minus{} B$\n$ det(A \\plus{} B)^2 \\equal{} ( \\minus{} 1)^(n).det(A \\plus{} B)$\nIf $ n$ is even : (Write $ det(A \\plus{} B) \\equal{} x$)\n$ x^2 \\equal{} x$, then $ x \\equal{} 0$ or $ x \\equal{} 1$, the example for $ x \\equal{} 0$ is $ A \\equal{} B \\equal{} 0$\nand the example for $ x \\equal{} 1$ is $ A \\equal{} 0$ and $ B \\equal{} \\minus{} I$ \nIf $ n$ is odd : \n$ x^2 \\equal{} \\minus{} x$,$ x \\equal{} 0$ or $ x \\equal{} \\minus{} 1$, the example for $ x \\equal{} \\minus{} 1$ is $ A \\equal{} 0$ and $ B \\equal{} \\minus{} I$\nBecause $ AB\\plus{}BA\\equal{}0$, then $ BAB\\plus{}B^2A\\equal{}BAB\\minus{}BA\\equal{}0$\n$ AB^2\\plus{}BAB\\equal{}0$, then $ \\minus{}AB\\plus{}BAB\\equal{}0$, $ BAB\\minus{}BA\\minus{}(\\minus{}AB\\plus{}BAB)\\equal{}0$,\n$ AB\\minus{}BA\\equal{}0$, then : $ AB\\minus{}BA\\plus{}(AB\\plus{}BA)\\equal{}0$, $ 2AB\\equal{}0$, if $ A$ and $ B$ is not a matrix over a field with characteristics $ 2$, we obtain that : $ AB\\equal{}0$,$ BA\\equal{}0$, then :\n$ (A\\minus{}B)(A\\plus{}B)\\equal{}A^2\\minus{}B^2\\equal{}\\minus{}(A\\minus{}B)$, $ det(A\\minus{}B).det(A\\plus{}B)\\equal{}(\\minus{}1)^n.det(A\\minus{}B)$\nIf $ det(A\\plus{}B)\\equal{}0$ then $ det(A\\minus{}B)$ must be $ 0$ (The example is $ A\\equal{}B\\equal{}0$\nIf $ det(A\\plus{}B)\\equal{}1$ and $ n$ is even then we have many infinitely $ A$ and $ B$ that statisfies the condition such as $ A\\equal{}kI$ and $ B\\equal{}(1\\minus{}k)I$, $ A\\minus{}B\\equal{}(1\\minus{}2k)I$, $ det(A\\minus{}B)$ can approach infinity and negative infinity\nIf $ det(A\\plus{}B)\\equal{}\\minus{}1$ and $ n$ is odd, then we have many infinitely $ A$ and $ B$ that statisfies the condition such as $ A\\equal{}\\minus{}kI$ and $ B\\equal{}(\\minus{}1\\plus{}k)I$, $ A\\minus{}B\\equal{}(1\\minus{}2k)I$, $ det(A\\minus{}B)$ can approach infinity and also can approach negative infinity.\n[/hide]", "Solution_2": "Hi Kruni,\r\nthe result about $ det(A\\minus{}B)$ is false.\r\nYou have shown that if $ charac(K)\\not\\equal{}2$ then $ AB\\equal{}BA\\equal{}0$.\r\n$ A,B$ are simultaneously diagonalizable. We may assume that $ A\\equal{}diag(a_1,\\cdots,a_n),B\\equal{}diag(b_1,\\cdots,b_n)$ where $ a_ib_i\\equal{}0,a_i,b_i\\in\\{0,\\minus{}1\\}$. Then $ a_i\\equal{}b_i\\equal{}0$ or $ a_i\\equal{}0,b_i\\equal{}\\minus{}1$ or $ a_i\\equal{}\\minus{}1,b_i\\equal{}0$. Thus $ a_i\\plus{}b_i\\equal{}0$ or $ \\minus{}1$ and $ a_i\\minus{}b_i\\equal{}0,1$ or $ \\minus{}1$.\r\nConclusion: $ det(A\\plus{}B)\\equal{}0$ or $ 1$ with $ n$ even or $ \\minus{}1$ with $ n$ odd.\r\n$ det(A\\minus{}B)\\equal{}0,1,$ or $ \\minus{}1$.\r\nIf charac(K)=2 then we have only $ AB\\equal{}BA$.\r\n$ A,B$ are simultaneously diagonalizable. We may assume that $ A\\equal{}diag(a_1,\\cdots,a_n),B\\equal{}diag(b_1,\\cdots,b_n)$ where $ a_i,b_i\\in\\{0,1\\}$.\r\nConclusion: $ det(A\\plus{}B)\\equal{}det(A\\minus{}B)\\in\\{0,1\\}$." } { "Tag": [ "LaTeX", "function", "algebra unsolved", "algebra" ], "Problem": "Find all pair of numbers (x,y) that satifies the following equation:\r\n[code](3+2cos(x-y))= \\sqrt {3+2*x-x*2}.(1+cos(x-y))+(1-cos(x-y)[/code]", "Solution_1": "plz write in $LaTeX$ form", "Solution_2": "Yes, or at least use standard forms of functions and operators.\r\n\r\nsqroot([i]x[/i]) is the square root of [i]x[/i]\r\n[i]x[/i]^[i]y[/i] is the [i]x[/i] raised to the [i]y[/i]th power\r\n[i]x[/i]*[i]y[/i] is the product of [i]x[/i] and [i]y[/i]\r\n\r\nOnly use {} for sets, etc.", "Solution_3": "[quote=\"hien\"]Find all pair of numbers (x,y) that satifies the following equation:\n[code](3+2cos(x-y))= \\sqrt {3+2*x-x*2}.(1+cos(x-y))+(1-cos(x-y)[/code][/quote]\r\n\r\nI think the correct question is:\r\n$3+2cos(x-y)= \\sqrt{3+2x-x^2} (1+cos(x-y)) + (1-cos(x-y))$\r\nIt's an easy equation... :lol:" } { "Tag": [ "function", "inequalities", "Cauchy Inequality", "algebra proposed", "algebra" ], "Problem": "For two given positive integers $ m,n > 1$, let $ a_{ij} (i = 1,2,\\cdots,n, \\; j = 1,2,\\cdots,m)$ be nonnegative real numbers, not all zero, find the maximum and the minimum values of $ f$, where\n\\[ f = \\frac {n\\sum_{i = 1}^{n}(\\sum_{j = 1}^{m}a_{ij})^2 + m\\sum_{j = 1}^{m}(\\sum_{i= 1}^{n}a_{ij})^2}{(\\sum_{i = 1}^{n}\\sum_{j = 1}^{m}a_{ij})^2 + mn\\sum_{i = 1}^{n}\\sum_{j=1}^{m}a_{ij}^2}. \\]", "Solution_1": "[quote=\"Fang-jh\"]For two given positive integers $ m,n > 1$, let $ a_{ij} (i \\equal{} 1,2,\\cdots,n,j \\equal{} 1,2,\\cdots,m)$ be nonnegative real numbers, not all zero, find the maximum and the minimum values of $ f$, where $ f \\equal{} \\frac {n\\sum_{i \\equal{} 1}^{n}(\\sum_{j \\equal{} 1}^{m}a_{ij})^2 \\plus{} m\\sum_{j \\equal{} 1}^{m}(\\sum_{i \\equal{} 1}^{n}a_{ij})^2}{(\\sum_{i \\equal{} 1}^{n}\\sum_{j \\equal{} 1}^{m}a_{ij})^2 \\plus{} mn\\sum_{i \\equal{} 1}^{n}\\sum_{i \\equal{} j}^{m}a_{ij}^2}$.[/quote]\r\n\r\nDear fang-jh, \r\nI was wondering if there was a typo in the denominator in this term: $ \\sum_{i \\equal{} 1}^{n}\\sum_{i \\equal{} j}^{m}a_{ij}^2$ .. i guess its a valid sum, but then the (inner) sum should be written from $ j\\equal{}i$ to $ j\\equal{}m$, not from $ i \\equal{} j$ to $ i\\equal{}m$ ; and the (outer) sum from $ i \\equal{} 1$ to $ m$, not $ i \\equal{} 1$ to $ n$ (since if $ n \\geq i \\geq m$, the second sum is empty), and it is quite a strange sum. Perhaps it may be this: \r\n$ \\sum_{i \\equal{} 1}^{n}\\sum_{j \\equal{} 1}^{m}a_{ij}^2$?", "Solution_2": "[quote=\"epitomy01\"][quote=\"Fang-jh\"]For two given positive integers $ m,n > 1$, let $ a_{ij} (i \\equal{} 1,2,\\cdots,n,j \\equal{} 1,2,\\cdots,m)$ be nonnegative real numbers, not all zero, find the maximum and the minimum values of $ f$, where $ f \\equal{} \\frac {n\\sum_{i \\equal{} 1}^{n}(\\sum_{j \\equal{} 1}^{m}a_{ij})^2 \\plus{} m\\sum_{j \\equal{} 1}^{m}(\\sum_{i \\equal{} 1}^{n}a_{ij})^2}{(\\sum_{i \\equal{} 1}^{n}\\sum_{j \\equal{} 1}^{m}a_{ij})^2 \\plus{} mn\\sum_{i \\equal{} 1}^{n}\\sum_{i \\equal{} j}^{m}a_{ij}^2}$.[/quote]\n\nDear fang-jh, \nI was wondering if there was a typo in the denominator in this term: $ \\sum_{i \\equal{} 1}^{n}\\sum_{i \\equal{} j}^{m}a_{ij}^2$ .. i guess its a valid sum, but then the (inner) sum should be written from $ j \\equal{} i$ to $ j \\equal{} m$, not from $ i \\equal{} j$ to $ i \\equal{} m$ ; and the (outer) sum from $ i \\equal{} 1$ to $ m$, not $ i \\equal{} 1$ to $ n$ (since if $ n \\geq i \\geq m$, the second sum is empty), and it is quite a strange sum. Perhaps it may be this: \n$ \\sum_{i \\equal{} 1}^{n}\\sum_{j \\equal{} 1}^{m}a_{ij}^2$?[/quote]\r\nDear epitomy01,\r\nI check and compare with the Chinese version carefully, find no typo.", "Solution_3": "[quote=\"Fang-jh\"]For two given positive integers $ m,n > 1$, let $ a_{ij} (i \\equal{} 1,2,\\cdots,n,j \\equal{} 1,2,\\cdots,m)$ be nonnegative real numbers, not all zero, find the maximum and the minimum values of $ f$, where $ f \\equal{} \\frac {n\\sum_{i \\equal{} 1}^{n}(\\sum_{j \\equal{} 1}^{m}a_{ij})^2 \\plus{} m\\sum_{j \\equal{} 1}^{m}(\\sum_{i \\equal{} 1}^{n}a_{ij})^2}{(\\sum_{i \\equal{} 1}^{n}\\sum_{j \\equal{} 1}^{m}a_{ij})^2 \\plus{} mn\\sum_{i \\equal{} 1}^{n}\\sum_{i \\equal{} j}^{m}a_{ij}^2}$.[/quote]\r\n\r\nA typo exists. It should be\r\n\r\n$ f \\equal{} \\frac {n\\sum_{i \\equal{} 1}^{n}(\\sum_{j \\equal{} 1}^{m}a_{ij})^2 \\plus{} m\\sum_{j \\equal{} 1}^{m}(\\sum_{i \\equal{} 1}^{n}a_{ij})^2}{(\\sum_{i \\equal{} 1}^{n}\\sum_{j \\equal{} 1}^{m}a_{ij})^2 \\plus{} mn\\sum_{i \\equal{} 1}^{n}\\sum_{j \\equal{} 1}^{m}a_{ij}^2}$\r\n\r\nSee", "Solution_4": "shalex is right, please moderator correct it.\r\nthank a lot!", "Solution_5": "The minimum value is $ \\frac{m\\plus{}n}{1\\plus{}mn}$ while the maximum value is $ 1$.\r\n\r\nCorrect ?", "Solution_6": "The maximum value is $ 1,$ the minimum value is $ \\frac {m\\plus{}n}{mn\\plus{}min\\{m,n\\}}$.", "Solution_7": "Ahh, I realize it now. In my proof, the equality for the minimum value is $ a_{ij}\\equal{}0 \\forall i \\not\\equal{} j$ and $ a_{ii}\\equal{}\\text{const} \\forall i\\equal{}1, 2, ..., min(m,n)$.", "Solution_8": "couyld anyone post his proof please :)", "Solution_9": "yes! Can you post it?", "Solution_10": "For sake of convenience, denote\r\n$ \\sum_{j \\equal{} 1}^{m} a_{ij} \\equal{} x_i$\r\n$ \\sum_{i \\equal{} 1}^{n} a_{ij} \\equal{} y_j$\r\n\r\nWhat we have is $ f \\equal{} \\frac {n \\sum ( x_i )^2 \\plus{} m \\sum ( y_i )^2 }{(\\sum \\sum a_{ij} )^2 \\plus{} mn \\sum \\sum a_{ij}^2}$\r\n\r\n(1) Maximum of $ f$\r\nWe shall prove that the maximum is 1.\r\nIt suffices to show that\r\n$ n \\sum (x_i )^2 \\plus{} m \\sum ( y_i )^2 \\leq (\\sum \\sum a_{ij} )^2 \\plus{} mn \\sum \\sum a_{ij}^2$\r\n\r\nConsider the two following sequence of $ m^2 \\plus{} n^2$ nonnegative real numbers;\r\n\r\n$ (x_1 , x_1 , \\cdots , x_1 , x_2 , \\cdots , x_n , y_1 , y_1 , \\cdots , y_1 , y_2 , \\cdots , y_m )$\r\n(Where each $ x_i$ appears $ n$ times and each $ y_i$ appears $ m$ times)\r\n$ ( \\sum \\sum a_{ij} , a_{11}, a_{12} , \\cdots , a_{mn} , 0 , 0 , \\cdots , 0 )$\r\n(the 0's exist to make the number of terms equal)\r\n\r\nIt is obvious that the latter sequence majorizes the former sequence (The largest is the largest, and any other term in the latter sequence is smaller in size to an appropriate term in the former sequence)\r\nSince the function $ g(x) \\equal{} x^2$ is convex in the nonnegative reals, applying Karamata inequality, we arrive at the result.\r\n\r\n(2) Minimum of $ f$\r\nWe wish to show that $ f \\geq \\frac {m \\plus{} n}{mn \\plus{} min(m,n)}$.\r\nAssume $ m \\geq n$, so that $ min(m,n) \\equal{} n$\r\nFor sake of convenience, denote\r\n$ \\sum (x_i )^2 \\equal{} A$\r\n$ \\sum (y_i )^2 \\equal{} B$\r\n$ \\sum \\sum a_{ij}^2 \\equal{} C$\r\n$ (\\sum \\sum a_{ij})^2 \\equal{} D$\r\n\r\nWhat we wish to prove is that\r\n$ \\frac {nA \\plus{} mB}{mnC \\plus{} D} \\geq \\frac {m \\plus{} n}{mn \\plus{} n}$ .. (*)\r\n\r\n$ (nA \\plus{} mB)(mn \\plus{} n) \\geq (mnC \\plus{} D)(m \\plus{} n)$\r\n\r\nBy the Cauchy Inequality, we have $ nA \\geq D$, or $ (m \\plus{} n)nA \\geq (m \\plus{} n)D$ .. (1)\r\nSince $ \\sum (x_i )^2 \\equal{} \\sum (\\sum a_{ij} )^2 \\geq \\sum \\sum a_{ij}^2$, we have $ A \\geq C$\r\nlikewise, we have $ B \\geq C$\r\nTherefore, we have that\r\n$ n(mn \\plus{} n)A \\geq n(mn \\plus{} n)C$ .. (2) \r\n$ (m^2 n \\minus{} n^2 )B \\geq (m^2 n \\minus{} n^2 )C$ .. (3)\r\nAdding (1), (2), (3) yields (*), the desired result.", "Solution_11": "My solution for (1) is incorrect - the upper bound solution is nonsensical. I was drowzy when I wrote it =[\r\n\r\nHere's a modified version, please edit the post (I don't know why, but I lost my right to do so):\r\n\r\nUpper bound - it suffices to show, as mentioned,\r\n$ n \\sum x_i^2 \\plus{} m \\sum y_i^2 \\leq (\\sum \\sum a_{ij} )^2 \\plus{} mn \\sum \\sum a_{ij}^2$\r\nOr, equivalently,\r\n$ n \\sum x_i^2 \\minus{} (\\sum \\sum a_{ij})^2 \\leq m \\sum ( n \\sum a_{ij}^2 \\minus{} \\sum y_i^2 )$\r\nOr, equivalently,\r\n$ n \\sum x_i^2 \\minus{} (\\sum x_i )^2 \\leq m \\sum (n \\sum a_{ij}^2 \\minus{} \\sum y_i^2 )$ (because $ \\sum \\sum a_{ij} \\equal{} \\sum x_i$)\r\nNow, recall the well-known identity $ n \\sum a_i^2 \\minus{} (\\sum a_i )^2 \\equal{} \\sum (a_i \\minus{} a_j )^2$\r\nThe inequality becomes\r\n$ \\sum (x_i \\minus{} x_j )^2 \\leq m \\sum ( \\sum (a_{ik} \\minus{} a_{jk})^2 )$\r\nApplying the cauchy inequality in the form\r\n$ (x_i \\minus{} x_j )^2 \\equal{} (\\sum a_{ik} \\minus{} \\sum a_{jk})^2 \\leq m \\sum (a_{ik} \\minus{} a_{jk})^2$ and summing them up yields the result.", "Solution_12": "WI UM DDDD", "Solution_13": "\\[ f = \\frac {n\\sum_{i = 1}^{n}(\\sum_{j = 1}^{m}a_{ij})^2 + m\\sum_{j = 1}^{m}(\\sum_{i= 1}^{n}a_{ij})^2}{(\\sum_{i = 1}^{n}\\sum_{j = 1}^{m}a_{ij})^2 + mn\\sum_{i = 1}^{n}\\sum_{j=1}^{m}a_{ij}^2}. \\]\nWe write these 2 identities:\n\\[n\\sum_{i=1}^{n}\\sum_{j=1}^{m}a_{ij}^2-\\sum_{j=1}^{m}(\\sum_{i=1}^{n}a_{ij})^2=\\sum_{j=1}^{n}(n\\sum_{i=1}^{n}a_{ij}^2-(\\sum_{i=1}^{n}a_{ij})^2)=\\sum_{j=1}^{m}((n-1)\\sum_{i=1}^{n}a_{ij}^2-2*\\sum_{1\\leq k= 25/4 because\r\na/(b+c) + b/(a+c) + c/(a+ b) + 1 = (a+b+c)(1/(b+c) + 1/(a+c) + 1/(a+b)) - 2\r\n(a+b+c)(1/(b+c) + 1/(a+c) + 1/(a+b)) >= (a+b+c)*9/(2a+2b+2c) = 9/2\r\n\r\nso inequality is when is k>=(a^2 + b^2 + c^2)*k/(ab+ac+bc)\r\nbut (a^2 + b^2 +c^2)/(ab+ac+bc) >= 1\r\nso k has to be less than or equal to 0?", "Solution_2": ".You don't know this type of ineq ;) ?. Try $ a \\equal{} b;c \\equal{} 0$ ,we have :$ \\frac {11}{4} \\ge k$ ;)\r\nTo prove :\r\nif $ a,b,c \\ge 0$ then :\r\n$ ( \\sum \\frac {a}{b \\plus{} c} \\plus{} 1)^2 \\ge \\frac {11}{4}.\\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca} \\minus{} \\frac {11}{4} \\plus{} \\frac {25}{4}$\r\nWe use Mixing Varibles \r\n[b]P/s[/b] I don't like to use it ^^", "Solution_3": "why would the earlier solution not work and why do we have to use mixing\r\nI assumed that the inequality has to be true for all a,b,c", "Solution_4": "But we find the maximum value of $ k$ ;) .\r\nIf don't use mixing ;) ,we use pqr ^^ .I can prove it :)", "Solution_5": "[quote=\"mitdac123\"]But we find the maximum value of $ k$ ;) .\nIf don't use mixing ;) ,we use pqr ^^ .I can prove it :)[/quote]\r\ncould you please post your solution??mitdac123", "Solution_6": "$ a,b,c \\ge 0$ and no two of which are zero .\r\nProve that :\r\n$ ( \\frac{a}{b\\plus{}c}\\plus{}\\frac{b}{c\\plus{}a}\\plus{}\\frac{c}{a\\plus{}b}\\plus{}1)^2 \\ge \\frac{11(a^2\\plus{}b^2\\plus{}c^2)}{4(ab\\plus{}bc\\plus{}ca)} \\plus{}\\frac{7}{2}$\r\n[b]My solution[/b]\r\nAssume that :$ a\\plus{}b\\plus{}c\\equal{}1$\r\n<=>$ ( \\frac{1\\minus{}q\\plus{}2r}{q\\minus{}r})^2 \\ge \\frac{11}{4q}\\minus{}2$\r\n$ \\bigstar$ .$ 4q \\le 1$ .\r\nWe prove :\r\n$ \\frac{(1\\minus{}q)^2}{q^2} \\ge \\frac{11}{4q}\\minus{}2$\r\n$ \\leftrightarrow (4 q \\minus{} 1) (3 q \\minus{} 4) \\ge 0$ (Right because $ q \\le \\frac{1}{4}$)\r\n$ \\bigstar$.$ 4q \\ge 1$\r\nUse schur ,we have :$ r \\ge \\frac{4q\\minus{}1}{9}\\longrightarrow LHS \\ge (\\frac{q\\minus{}7}{5q\\plus{}1})^2$\r\nWe prove that :\r\n$ (\\frac{q\\minus{}7}{5q\\plus{}1})^2 \\ge\\frac{11}{4q}\\minus{}2$\r\n$ \\leftrightarrow (3q\\minus{}1)(17q\\minus{}11)(4q\\minus{}1) \\ge 0$\r\nWhich is obvious true because $ \\frac{1}{3} \\ge q \\ge \\frac{1}{4}$\r\nDone ;)" } { "Tag": [ "algebra", "polynomial", "Galois Theory", "superior algebra", "superior algebra theorems" ], "Problem": "hi\r\n\r\ncan anyone give me an example of a four degree irreducible polynomial over Q with a splitting field of degree 24 (so the galois group is isomorphic to S4)\r\n\r\nit would help me finding a number with degree 4 over Q without being constructible\r\n\r\n\r\nas usual, any more general ideas to find galois groups are welcome\r\n\r\nfred", "Solution_1": "theorem:\r\nLet be $f \\in K \\subset \\mathbb{R}$ ($K$ field) a polynomial of degree four and $g$ it's cubic resolvent.\r\nIf $f,g$ are both irreducible (over $K[x]$) and $f$ has no real roots, then you have as Galois group of $f(x)=0$:\r\n- $A_4$ if the discriminant of $f$ is a perfect square (in $K$)\r\n- $S_4$ otherwise\r\n\r\nfor example take $f(x)=x^4+x^2+x+1$ in $\\mathbb{Q}$" } { "Tag": [ "geometry", "rectangle", "analytic geometry", "algebra unsolved", "algebra" ], "Problem": "Hello everyone\r\n\r\nau = area units\r\n\r\nThe question is, what are the coordinates of the point P.\r\n\r\nxcosx = 0,5\r\n\r\nHow do you calculate x?\r\n\r\n[b]The graph ==> y = cosx[/b]", "Solution_1": "I am sure, it is impossible to solve this transcendental equation." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "For $ a>0,\\ b>0,\\ c>0$, prove the following inequality.\r\n\r\n$ \\frac{2}{3}(a\\plus{}b\\plus{}c)\\leq \\frac{a^3\\plus{}b^3}{a^2\\plus{}ab\\plus{}b^2}\\plus{}\\frac{b^3\\plus{}c^3}{b^2\\plus{}bc\\plus{}c^2}\\plus{}\\frac{c^3\\plus{}a^3}{c^2\\plus{}ca\\plus{}a^2}\\leq 6(a\\plus{}b\\plus{}c)$", "Solution_1": "[quote=\"kunny\"]For $ a > 0,\\ b > 0,\\ c > 0$, prove the following inequality.\n\n$ \\frac {2}{3}(a \\plus{} b \\plus{} c)\\leq \\frac {a^3 \\plus{} b^3}{a^2 \\plus{} ab \\plus{} b^2} \\plus{} \\frac {b^3 \\plus{} c^3}{b^2 \\plus{} bc \\plus{} c^2} \\plus{} \\frac {c^3 \\plus{} a^3}{c^2 \\plus{} ca \\plus{} a^2}\\leq 6(a \\plus{} b \\plus{} c)$[/quote]\r\nThe inequality follows from $ \\frac{1}{3}\\leq\\frac {a^2 \\minus{} ab \\plus{} b^2}{a^2 \\plus{} ab \\plus{} b^2}\\leq1.$", "Solution_2": "That's incorrect in part.", "Solution_3": "[quote=\"kunny\"]That's incorrect in part.[/quote]\r\nWhere? Explain please. Thank you.", "Solution_4": "$ \\frac{a^2\\minus{}ab\\plus{}b^2}{a^2\\plus{}ab\\plus{}b^2}\\leq 1$. :lol:", "Solution_5": "It IS true, since $ a,b,c$ are positive, so $ 2ab>0$, then $ a^2 \\minus{} ab \\plus{} b^2 < a^2 \\plus{} ab \\plus{} b^2$\r\n\r\nUnless you're saying it is incorrect because the inequality is strict" } { "Tag": [ "algebra", "polynomial", "induction", "calculus", "integration", "abstract algebra", "geometry" ], "Problem": "Let $P(x)\\in \\mathbb{Z}[x]$ be a monic polynomial with even degree. Prove that, if for infinitely many integers $x$, the number $P(x)$ is a square of a positive integer, then there exists a polynomial $Q(x)\\in\\mathbb{Z}[x]$ such that $P(x)=Q(x)^2$.", "Solution_1": "It is not true, because there are infinetely many solutions $y^{2}-2x^{2}=1$. It mean $P(x)=2x^{2}+1=y^{2}$ for infinetely many x.", "Solution_2": "That is not a monic polynomial.", "Solution_3": "Yes. I forgot about monic.\r\nIf $P(x)=x^{2n}+a_{2n-1}x^{2n-1}+...+a_{0}$, then\r\n $4P(x)=Q(x)^{2}+R(x), \\ Q(x)=2x^{n}+b_{n-1}x^{n-1}+...+b_{0})^{2}, \\ R(x),$ were $deg(R(x))\\le n-1.$\r\nTherefore for big x $|R(x)|<2|Q(x)|-1$. Because $4P(x)$ and $Q(x)^{2}$ are perfect square, it mean $R(x)=0$ for infinetely many x, or $R(x)\\equiv 0.$\r\nBecause all coefficients $Q(x)^{2}$ divide by 4, by induction we get all coefficients Q(x) are even.", "Solution_4": "These [equations of Pell type] are, essentially, the only exceptions. A special case of Siegel's theorem on integral points on algebraic curves (\"there are at most finitely many integral points on an affine algebraic curve $C$ with at least three points at infinity, i.e. such that $\\# \\tilde{C}\\setminus C \\geq 3$, where $\\tilde{C}$ is the projective closure of $C$ [relative to the considered embedding in affine space]\") implies: if $P(x) \\in \\mathbb{Z}[x]$ is an integer polynomial with at least three distinct roots, then the equation $y^{2}= P(x)$ has only finitely many solutions in integers $x,y$. In particular, any irreducible polynomial of degree $\\geq 3$ takes at most finitely many perfect square values. \r\n\r\nA proof of this is possible via Roth's theorem of diophantine approximation. (And, as well, other results of diophantine approximations, such as the $2$-variable $S$-unit equation, which can be solved effectively). The general version of Siegel's theorem, valid for arbitrary curves, is proven standardly by embedding a curve in its Jacobian variety and then using Roth's theorem and the arithmetic of abelian varieties. But more elegant (and short, granted \"the subspace theorem\" of W. Schmidt, which is a far-reaching extension of Roth's theorem) is a relatively recent proof, by Corvaja and Zannier, which doesn't at all use the arithmetic of abelian varieties! \r\n\r\nBut of course, for the special case you have proposed, no algebraic geometry is needed; it suffices to know what is the \"2-variable $S$-unit equation\" for number fields (\"if $K$ is a number field and $S \\subset M_{K}$ any finite set of places, then the equation $x+y = 1$ has only finitely many solutions in $S$-units, i.e. with $|x|_{v}, |y|_{v}\\leq 1$ for all $v \\notin S$\"); it is then straightforward to deduce that $y^{2}= P(x)$ has only finitely many solutions, provided that $P(\\cdot)$ has at least three complex roots. \r\n\r\nIt is a much deeper theorem that if $P(x)$ is an irreducible polynomial of degree at least $5$, then $y^{2}= P(x)$ has at most finitely many solutions in [i] rational[/i] numbers $x,y$; this was proven only in 1983 by G. Faltings (other proofs are known since then, but they are also extremely deep). \r\n\r\nFor the problem that you propose, you want the degree to be even and $>2$ in order to bound a square between consecutive squares.", "Solution_5": "[quote=\"Rust\"]Because all coefficients $ Q(x)^{2}$ divide by 4, by induction we get all coefficients Q(x) are even.[/quote]\r\n\r\nCan you show this induction?", "Solution_6": "[quote=\"Rust\"] then\n $ 4P(x) \\equal{} Q(x)^{2} \\plus{} R(x), \\ Q(x) \\equal{} 2x^{n} \\plus{} b_{n \\minus{} 1}x^{n \\minus{} 1} \\plus{} ... \\plus{} b_{0})^{2}, \\ R(x),$ were $ deg(R(x))\\le n \\minus{} 1.$\n[/quote]\r\n\r\nActually I don't agree with this! \r\n\r\nConsidering the degree $ x_{2n\\minus{}1}$, you get $ 4b_{n\\minus{}1}\\equal{}4a_{2n\\minus{}1}$, so $ b_{n\\minus{}1}\\equal{}a_{2n\\minus{}1}$.\r\n\r\nConsidering the degree $ x_{2n\\minus{}2}$, you get $ 4b_{n\\minus{}2}\\plus{}b_{n\\minus{}1}^2\\equal{}4a_{2n\\minus{}2}$, so if $ b_{n\\minus{}1}\\equal{}a_{2n\\minus{}1}$ is odd, you cannot choose $ b_{n\\minus{}2}$ to be an integer...", "Solution_7": "Let me certify Rust's correct argument.\r\n\r\nAssume $ P(x)\\in\\mathbb{Z}[x]$ with degree $ 2n$ and leading coefficient $ c^2$ for some $ n,c\\in\\mathbb{N}$,\r\nand $ P(m)$ is a perfect square for infinitely many $ m\\in\\mathbb{Z}$.\r\n\r\nThen $ p(x) \\equal{} x^{2n}P({1\\over x})\\in\\mathbb{Z}[x]$ of degree $ \\le 2n$ and $ p(0) \\equal{} c^2$.\r\nSo we can write $ p(x) \\equal{} c^2 \\plus{} xt(x)$ for some $ t(x)\\in\\mathbb{Z}[x]$.\r\nThen using [url=http://en.wikipedia.org/wiki/Binomial_theorem]Newton's generalized binomial theorem[/url] we get\r\n$ p(x) \\equal{} \\left(\\sum_{k\\ge 0}\\binom{1/2}{k}c^{1 \\minus{} 2k}x^kt(x)^k\\right)^2$ with $ \\sum_{k\\ge 0}\\binom{1/2}{k}c^{1 \\minus{} 2k}x^kt(x)^k\\in\\mathbb{Q}[ [x] ]$.\r\nSo $ p(x)\\equiv q(x)^2\\pmod{x^{n \\plus{} 1}}$ for some $ q(x)\\in\\mathbb{Q}[x]$ of degree $ \\le n$.\r\nThen $ p(x) \\equal{} q(x)^2 \\plus{} x^{n \\plus{} 1}r(x)$ for some $ r(x)\\in\\mathbb{Q}[x]$, \r\nand $ x^{n \\plus{} 1}r(x) \\equal{} p(x) \\minus{} q(x)^2$ has degree $ \\le 2n$, so $ \\deg r(x)\\le n \\minus{} 1$.\r\n\r\nAnd now $ P(x) \\equal{} x^{2n}p({1\\over x}) \\equal{} x^{2n}(q({1\\over x})^2 \\plus{} {1\\over x^{n \\plus{} 1}}r({1\\over x})) \\equal{} Q(x)^2 \\plus{} R(x)$\r\nwith $ Q(x): \\equal{} x^nq({1\\over x})\\in\\mathbb{Q}[x]$ of degree $ \\le n$ and $ R(x): \\equal{} x^{n \\minus{} 1}r({1\\over x})\\in\\mathbb{Q}[x]$ of degree $ \\le n \\minus{} 1$.\r\n\r\nLet $ d\\in\\mathbb{N}$ be minimal with $ dQ(x)\\in\\mathbb{Z}[x]$.\r\nThen $ \\deg P(x) \\equal{} 2n$ gives $ \\deg Q(x) \\equal{} n > \\deg R(x)$, \r\nand therefore $ 2|dQ(x)| \\minus{} 1 > |d^2R(x)|$ for sufficiently large $ |x|$.\r\nThen $ d^2P(x) \\equal{} |dQ(x)|^2 \\plus{} d^2R(x) > |dQ(x)|^2 \\minus{} 2|dQ(x)| \\plus{} 1 \\equal{} (|dQ(x)| \\minus{} 1)^2$\r\nand $ d^2P(x) \\equal{} |dQ(x)|^2 \\plus{} d^2R(x) < |dQ(x)|^2 \\plus{} 2|dQ(x)| \\minus{} 1 < (|dQ(x)| \\plus{} 1)^2$\r\nfor sufficiently large $ |x|$.\r\n\r\nSo if $ P(m)$ is a perfect square for sufficiently large $ |m|$ with $ m\\in\\mathbb{Z}$, \r\nthen so is $ d^2P(m)$, which gives $ d^2P(m) \\equal{} (|dQ(m)|)^2$ and $ P(m) \\equal{} Q(m)^2$.\r\nBut there are infinitely many such $ m$, which gives $ P(x) \\equal{} Q(x)^2$.\r\n\r\nAnd now if $ d > 1$ we take $ p\\mid d$ prime, and then $ (dQ(x))^2 \\equal{} d^2P(x)\\equiv 0 \\pmod{p}$.\r\nBut $ \\mathbb{F}_p \\equal{} \\mathbb{Z}/p\\mathbb{Z}$ is a field, so $ \\mathbb{F}_{p}[x]$ is an [url=http://en.wikipedia.org/wiki/Euclidean_domain]Euclidean domain[/url], \r\nand therefore has no [url=http://en.wikipedia.org/wiki/Zero_divisors]zero divisors[/url].\r\nThis gives $ dQ(x)\\equiv 0 \\pmod{p}$ and $ {d\\over p}Q(x)\\in\\mathbb{Z}[x]$, which contradicts the minimal choice of $ d$.\r\n\r\nSo $ d \\equal{} 1$ and $ Q(x)\\in\\mathbb{Z}[x]$.", "Solution_8": "We can generalize this problem:\nLet $P(x)\\in \\mathbb{Z}[x]$ be a monic polynomial with degree divisible by an integer $d$. Prove that, if for infinitely many integers $x$, the number $P(x)$ is a $d$th power of a positive integer, then there exists a polynomial $Q(x)\\in\\mathbb{Z}[x]$ such that $P(x)=Q(x)^d$.", "Solution_9": "[quote=bilarev]Let $P(x)\\in \\mathbb{Z}[x]$ be a monic polynomial with even degree. Prove that, if for infinitely many integers $x$, the number $P(x)$ is a square of a positive integer, then there exists a polynomial $Q(x)\\in\\mathbb{Z}[x]$ such that $P(x)=Q(x)^2$.[/quote]\n\nIt would be a nice problem:\nProve that if $P(x)\\in Z[x]$ is a perfect square for all $n\\in N$ then there exists $Q(x)\\in Z[x]$ such that $P(x)=Q(x)^2$\nUse this problem to prove the above problem ", "Solution_10": "[quote=K.I.M][quote=bilarev]Let $P(x)\\in \\mathbb{Z}[x]$ be a monic polynomial with even degree. Prove that, if for infinitely many integers $x$, the number $P(x)$ is a square of a positive integer, then there exists a polynomial $Q(x)\\in\\mathbb{Z}[x]$ such that $P(x)=Q(x)^2$.[/quote]\n\nIt would be a nice problem:\nProve that if $P(x)\\in Z[x]$ is a perfect square for all $n\\in N$ then there exists $Q(x)\\in Z[x]$ such that $P(x)=Q(x)^2$\nUse this problem to prove the above problem[/quote]\n\nwell in that case we need to prove the following statement\n\nfor $\\exists n \\in \\mathbb{N}, \\exists x \\in \\mathbb{Z}$, $P(x)=n^2$ then prove that $\\forall x \\in \\mathbb{Z}$, $P(x)=n^2$ for proper $n$s", "Solution_11": "[quote=Skravin][quote=K.I.M][quote=bilarev]Let $P(x)\\in \\mathbb{Z}[x]$ be a monic polynomial with even degree. Prove that, if for infinitely many integers $x$, the number $P(x)$ is a square of a positive integer, then there exists a polynomial $Q(x)\\in\\mathbb{Z}[x]$ such that $P(x)=Q(x)^2$.[/quote]\n\nIt would be a nice problem:\nProve that if $P(x)\\in Z[x]$ is a perfect square for all $n\\in N$ then there exists $Q(x)\\in Z[x]$ such that $P(x)=Q(x)^2$\nUse this problem to prove the above problem[/quote]\n\nwell in that case we need to prove the following statement\n\nfor $\\exists n \\in \\mathbb{N}, \\exists x \\in \\mathbb{Z}$, $P(x)=n^2$ then prove that $\\forall x \\in \\mathbb{Z}$, $P(x)=n^2$ for proper $n$s[/quote]\n\nCould you please explain more?\nI don't get what you say", "Solution_12": "[quote=K.I.M]Could you please explain more?\nI don't get what you say[/quote]\n\nI used some letters to lessen the statement but it had rather distracted\n\n\"For infinitely many(not could be all) positive integers, $P(x)$ becomes the square number. Then prove that $P(x)$ becomes the square number for all $x \\in \\mathbb{Z^+}$", "Solution_13": "i dont think its this hard. actually we need to prove that there exists a an positive integer m such that (x+m)^2 < P(x) < (x+m+1)^2 ( for all x>N0) . ofcourse if we assume that p(x) is not Q(x)^2 . and if we assume that there exists infinitely many POSITIVE integer r such that p(r) is square then things will be trivial . if not then there exists \ninfinitely many negative integers and so we may need to prove there is a m which (x-m)^2 < p(x) < (x-m+1)^2 . which is not hard to prove . its the summary of my solution . but i can write it completely . just inform me", "Solution_14": "[quote=olorin]\n[...]\nAnd now if $ d > 1$ we take $ p\\mid d$ prime, and then $ (dQ(x))^2 \\equal{} d^2P(x)\\equiv 0 \\pmod{p}$.\nBut $ \\mathbb{F}_p \\equal{} \\mathbb{Z}/p\\mathbb{Z}$ is a field, so $ \\mathbb{F}_{p}[x]$ is an [url=http://en.wikipedia.org/wiki/Euclidean_domain]Euclidean domain[/url], \nand therefore has no [url=http://en.wikipedia.org/wiki/Zero_divisors]zero divisors[/url].\nThis gives $ dQ(x)\\equiv 0 \\pmod{p}$ and $ {d\\over p}Q(x)\\in\\mathbb{Z}[x]$, which contradicts the minimal choice of $ d$.\n\nSo $ d \\equal{} 1$ and $ Q(x)\\in\\mathbb{Z}[x]$.[/quote]\nCan anybody deal with this part without using field or domain, i.e in more elementary way? Or you have another more simple solution? Thanks very much!\n" } { "Tag": [ "trigonometry", "inequalities", "inequalities proposed" ], "Problem": "Prove that in any triangle\r\n\r\n$\\sum \\sin\\alpha\\sin\\beta \\leq\\left(\\sum\\cos\\alpha \\right)^2$\r\n\r\n\r\nPlease, a trigonometric solution.\r\n\r\nThank you very much.", "Solution_1": "This inequality is equivalant to \r\n (1-cosA)(1-cosB)(1-cosC)>=cosAcosBcosC. :cool:", "Solution_2": "Also my solution is based on Treegoner's inequality.\r\n\r\nMultiplying the inequality (see the topic: \"Only trigonometric solution, please !!\")\r\n\r\n$\\prod\\tan\\frac{\\alpha}{2}\\geq\\prod\\cot\\alpha$ $ (\\clubsuit)$\r\n\r\nby $\\prod\\sin\\alpha$ we obtain\r\n\r\n$\\prod(1-cos\\alpha)\\geq\\prod\\cos\\alpha$\r\n\r\nI am curious to know if there is a more direct trigonometric solution not based on inequality $ (\\clubsuit)$" } { "Tag": [], "Problem": "Hi,\r\n\r\nI was wondering what is the best Chinese University (or Japanese) for Mathematics? How do they compare to American Universities within the mathematics that is taught? For example, does the Chinese Universities teach the equivalent of 3rd year mathematics in their first year?\r\n\r\nThanks everyone for your comments/help?", "Solution_1": "Hm.. Funny. I haven't heard of much Chinese Universities before. I live in the U.S. so I guess that's the problem. \r\n\r\nBut there are a lot of good colleges in U.K. and the U.S.", "Solution_2": "Disclaimer: this is purely my personal knowledge.\r\n\r\nBeijing University and Qinghua University (both in Beijing) are the two most prestigious tertiary institutes (in pretty much all aspects) in China; some other good math&science schools I've heard of include Zhejiang University, Wuhan University of Science & Technology, and Jiaotong University (in Shanghai - I'm not sure about math, but they seem to be beastly in computer). \r\n\r\nIt's generally easier for foreign students to get in top Chinese schools than nationals. However, most ranking systems - mostly based on reserach - place Chinese schools in the lower 100 or so, so you should consider carefully if you really want to go there. Also note that typically the best students in these universities vie for the scarce opportunities to transfer to top U.S. schools like MIT." } { "Tag": [], "Problem": "Hi, I have another chemistry question:\r\n\r\n$ \\text{In an experiment it was found that the total charge on an oil drop was } 5.93\\times 10^{\\minus{}18} \\text{ C. How many negative charges does the drop contain?}$\r\n\r\nI was wondering if anyone could explain this question to me? The answer is 37.\r\n\r\nThank you!", "Solution_1": "Well the charge on an electron is 1.6 x 10 ^ -19 . total charge is 5.93 x 10^ -18 , so the number of charged particles is total charge / charge on each particle = 37", "Solution_2": "Okay, I understand!\r\n\r\nMy textbook didn't tell me the charge of an individual electron. :mad: haha" } { "Tag": [ "geometry", "3D geometry", "sphere", "trigonometry", "integration", "probability and stats" ], "Problem": "We are given a sphere of radius 1 and a tangent plane. We select a point on the surface of the sphere uniformly. How is distributed the distance between this point and the plane? Calculate it's variance.", "Solution_1": "Let $ \\theta$ be the angle between the radius of the sphere to a given point on that sphere and the radius to the point that contacts the plane. This angle is a random variable which is distributed on $ [0,\\pi]$ with density proportional to the circumference of the circle of points lying at that angle. That is, the density of the distribution of $ \\theta$ is $ C\\sin\\theta$ for some $ C.$ Since $ \\int_0^{\\pi}\\sin\\theta\\,d\\theta \\equal{} 2,$ we have $ C \\equal{} \\frac12,$\r\n\r\nThe quantity we are interested in is $ X \\equal{} 1 \\minus{} \\cos\\theta.$ That's the distance to the tangent plane.\r\n\r\n$ E(X) \\equal{} E(1 \\minus{} \\cos\\theta) \\equal{} \\frac12\\int_0^{\\pi}(1 \\minus{} \\cos\\theta)\\sin\\theta\\,d\\theta \\equal{} 1.$\r\n\r\n(This result is exactly what we should have expected - on average, the distance in question is the distance to the equatorial circle.)\r\n\r\n$ E(X^2) \\equal{} \\frac12\\int_0^{\\pi}(1 \\minus{} \\cos\\theta)^2\\sin\\theta\\,d\\theta \\equal{} \\frac43.$\r\n\r\n$ \\text{Var}(X) \\equal{} E(X^2) \\minus{} (E(X))^2 \\equal{} \\frac13,$ so $ \\sigma \\equal{} \\frac1{\\sqrt {3}}.$\r\n\r\n[hide=\"Hidden because it is wrong.\"]This is exactly half of the standard deviation of a uniform $ [0,2]$ distribution.[/hide]", "Solution_2": "I think it's exactly uniform [0;2] (not \"half\")\r\nRecently i've met an old but very beautiful problem using this fact:\r\nhttp://rand.org/pubs/research_memoranda/2006/RM408.pdf", "Solution_3": "My incorrect comment is still there, but I just hid it.\r\n\r\nYou're right. The distribution on $ X$ is exactly uniform on $ [0,2].$ This is effectively the same fact as the following: if a spherical loaf of bread is sliced into $ n$ slices of equal thickness, then each slice contains the same amount of crust. This fact was, in essence, known by Archimedes.", "Solution_4": "Maybe this is quite funny:\r\n\r\nThere is a possibility to make a sum of $ n$ uniform rv to be uniform.\r\nKent's post http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1132376#1132376\r\n\r\nBut there is a possiblity to make a sum of $ n$ uniform rv to be constant! How?\r\n[hide=\"hint\"]With sphere and $ n$ planes [/hide]", "Solution_5": "I think the question of whether $ n$ is odd or even makes a difference.", "Solution_6": "I think there are many ways to do this, here is one:\r\n\r\nAs i remember the sum of all heights in regular $ n$-gon is constant.\r\nSo for $ n>2$ we draw regular $ n$-gon, inscribe a circumpherence.\r\nWe draw a sphere (this circumpherence should be the equator of the sphere).\r\nWe select uniformly a point on the surface of the sphere. \r\nWe project this point to the plane of $ n$-gon.\r\nWe have $ n$ heights which are uniformly disributed but their sum is constant.\r\n\r\nThe case of $ n\\le 2$ is easy." } { "Tag": [ "conics", "ellipse", "geometry", "hyperbola", "parabola", "projective geometry", "advanced fields" ], "Problem": "A variable tangent to an ellipse intersects the tangents from vertices A and A' in T and T'. Find put the locus of the intersection point of TA' and T'A.", "Solution_1": "It's also an ellipse tangent to the 2 tangents through A and A' in the same points (I'm assuming A and A' are on the elipse, right?).\r\n\r\nI don't want to get into details right now, but it's true for projective geometry reasons: the map T->T' is a projective transformation from the tangent through A (call it a) and the tangent trough A' (call it a') (this is a well-known theorem in projective geometry). This means that the map AT'->A'T is a projective transformation from the pencil of lines through A to the pencil of lines through A'. It's also well-known that in this case the locus of the intersection pts of 2 analogous lines (i.e. lines s.t. one of them is the image of the other one in the mentioned projective map) is a conic tangent in A and A' to the lines a and a' s.t. a->AA' and AA'->a'. \r\n\r\nThis shows that our locus is a conic, and since the point of intersection is always inside the ellipse, the conic can't be a hyperbola, a parabola or degenerate (line(s)), so it must be an ellipse tangent in A and A' to a and a' respectively.\r\n\r\nI don't know that much on the subject myself, but I got the impression that \"De la matematica elementara spre matematica superioara\" by Constantin Avadanei, Neculai Avadanei, Constantin Bors and Cristina Ciurea is a very good introduction to projective geometry (everything I know is taken from the chapter in that book :)).\r\n\r\nAnyway, I hope I'm not wrong about this.." } { "Tag": [ "trigonometry", "logarithms", "calculus", "calculus computations" ], "Problem": "Find the following derivative.\r\n\r\n[1] $\\frac{d}{dx} e^{2x}\\cos 3x$\r\n\r\n[2] $\\frac{d}{dx} (\\ln x)^2 $\r\n\r\n[3] $\\frac{d}{dx} a^{2x}\\ (a>0,a\\neq 1)$\r\n\r\n[4] $\\frac{d}{dx} \\cos 2x$\r\n\r\n[5] $\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1$\r\n\r\n[6] $\\frac{d}{dx} (x+2)^2 (x+3)^3 (x+4)^4$\r\n\r\n[7] $\\frac{d}{dx} \\tan ^ 3 x$\r\n\r\n[8] $\\frac{d}{dx} \\sqrt{x+3}$", "Solution_1": "By a cursory glance it appears that.....\r\n[hide]\n[4] $-2\\sin 2x$\n\n[2] $\\frac{2\\ln x}{x}$\n\n[5] $\\frac{-b^{2}x}{a^{2}y}=\\frac{dy}{dx}$\n\n[1] $e^{2x}(2\\cos 3x - 3\\sin 3x)$\n\n[7] $3\\tan^{2}x\\sec^{2}x$\n\n[3] $2a^{2x}\\ln a$\n\n[8] $\\frac{1}{2\\sqrt {x+3}}$[/hide]", "Solution_2": "O.K. adidasty :) How about for [6]?", "Solution_3": "im lazy.....lol jk ill get to it tomorrow its late", "Solution_4": "6\r\n$(x+2)(x+3)^2(x+4)^3[2(x+4)(x+3)+3(x+2)(x+4)+4(x+2)(x+3)]$", "Solution_5": "[quote=\"amirhtlusa\"]6\n$(x+2)(x+3)^2(x+4)^3[2(x+4)(x+3)+3(x+2)(x+4)+4(x+2)(x+3)]$ :) [/quote]\r\n$=(x+2)(x+3)^2(x+4)^3(9x^2+52x+72)$\r\n\r\nedited by kunny" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "solve the equation:\r\n$ (2\\plus{} \\sqrt2)^{log_2x} \\plus{} x^2(2\\minus{}\\sqrt2)^{log_2x} \\equal{}1\\plus{}x^2$", "Solution_1": "Let $ y\\equal{}log_2(x),2\\plus{}\\sqrt 2\\equal{}4^a$, then equation equavalent to $ f(y)\\equal{}2^{(2a\\minus{}1)y}\\plus{}2^{(1\\minus{}2a)y}\\equal{}2^{\\minus{}y}\\plus{}2^y$. If $ y\\equal{}zln(2)$, then it equavalent to\r\n$ ch((1\\minus{}2a)z)\\equal{}ch(z)$, It give unique solution $ z\\equal{}0$ or $ x\\equal{}1$." } { "Tag": [ "induction", "geometry", "Gauss", "LaTeX", "AMC", "USA(J)MO", "USAMO" ], "Problem": "Umm...can someone please explain this to me? As if I was four years old please...", "Solution_1": "Gauss202 explained it very nicely in this post... check it out [url]http://artofproblemsolving.com/Forum/viewtopic.php?t=5966[/url]", "Solution_2": "Okay, mathematical induction is a method of proof. It's probably the easiest way to prove things that have to do w/sequences and series. I'll show you an example while explaining the steps:\r\n\r\nProve that the sum of the first n odd numbers is [tex]n^2[/tex].\r\n\r\n(1) Show that it is true for n=1 .\r\n\r\nThe sum of the first 1 odd numbers = [tex]1^2=1[/tex]. 1 is the first odd number, so it =1, so you know this is true. (duh, but it's part of the process)\r\n\r\nSince n=1 seemed pointless, I like to try another like n=3. I asked AoPS people before and some said it was not necessary, but I like it anyway, so here:\r\n\r\n1+3+5 = sum of first 3 odd numbers = 9\r\n[tex]3^2[/tex] = sum of first 3 odd numbers = 9\r\n\r\nSo it works for n=3.\r\n\r\n(2) If it is true for some integer [tex]k[/tex], then it must be true for some integer [tex](k+1)[/tex]. So, show that it is true for [tex](k+1)[/tex] assuming that it is true for [tex]k[/tex]. \r\n\r\nThe sum of the first [tex]k[/tex] odd numbers is:\r\n\r\n1 + 3 + 5 + 7 + ... + [tex](2k-1)[/tex] = [tex]k^2[/tex]\r\n\r\nThe sum of the first [tex](k+1)[/tex] odd numbers is:\r\n\r\n1 + 3 + 5 + 7 + ... + [tex](2k-1)[/tex] + [tex](2k+1)[/tex] = [tex](k+1)^2[/tex]\r\n\r\nWe need to show that this is true, so, we want to make the LHS (left hand side) of the equation look exactly the same as the RHS of the equation.\r\n\r\n 1 + 3 + 5 + 7 + ... + [tex](2k-1)[/tex] + [tex](2k+1)[/tex] = [tex](k+1)^2[/tex]\r\n\r\nWe already assumed that the bold part of the LHS = [tex]k^2[/tex], so substitute that in:\r\n\r\n[tex]k^2[/tex]+[tex](2k+1)[/tex] = [tex](k+1)^2[/tex]\r\n\r\nFoil out the RHS:\r\n\r\n[tex]k^2[/tex]+[tex](2k+1)[/tex] = [tex]k^2+2k+1[/tex]\r\n\r\nSo:\r\n\r\n[tex]k^2+2k+1 = k^2+2k+1[/tex]\r\n\r\nCool, huh? =)\r\n\r\nOkay, now one thing: inductive and deductive reasoning are different. When you reason inductivley, you have to assume some things. When you reason deductivley, you use only stuff you already know. So the two column proofs you do in geometry employ deductive reasoning. (math whiz people - correct me if this is some sort of egregious error)\r\n\r\nAlright, these AoPS threads you will find helpful:\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=12557\r\n(from gauss)\r\n\r\n\r\n? I can't find it but I know he posted it in beginner's forum.\r\n(from Osiris)\r\n\r\nOkay, now here are some problems to practice with. Guys, please let SingTheSorrow do them first. In increasing difficulty:\r\n\r\n(1) Prove: [tex]1^3 + 2^3 + 3^3 + 4^3 + ... + n^3 = \\frac{n^2(n+1)^2}{4}[/tex].\r\n\r\n(2) Prove: [tex]\\displaystyle(1+\\frac{1}{1})(1+\\frac{1}{2})(1+\\frac{1}{3})...(1+\\frac{1}{n})=n+1[/tex].\r\n\r\n(3) Prove: [tex]\\frac{1}{\\sqrt{1}} + \\frac{1}{\\sqrt{2}} + \\frac{1}{\\sqrt{3}} + ... + \\frac{1}{\\sqrt{n}} > \\sqrt{n}[/tex] for [tex]n \\ge 2[/tex].\r\n\r\n(4) Prove: [tex]n! > 2^n[/tex] for [tex]n \\ge 4[/tex].\r\n\r\n(5) Prove: [tex]\\displaystyle(\\frac{x}{y})^{n+1} < \\displaystyle(\\frac{x}{y})^{n}[/tex] for [tex]n \\ge 1[/tex] and[tex]0 :rt2: \r\n\r\nI wasn't too sure about this...but I just put it in my calculator and it verfied the statement. Can I do something different rather than rounding :rt2: to 1.41? \r\n\r\nNow, I'm not too sure about how do do the next part either...just assume that I showed all my work, but now how do I show sqrt (k) + 1/ sqrt (k+1) > sqrt (k + 1)?\r\n\r\nI attempted the others but I'm sleeping now.", "Solution_10": "You brought up a very valid point. I am not satisfied with the answer that I come up w/for this one, and the solutions manual does not give a great answer. I will let someone else answer this one.\r\n\r\n#2 looks good.", "Solution_11": "3) It obviously works for n = 2.\r\n\r\nStart with the obviously true statement [tex]2 > \\sqrt{2}[/tex]. Divide 2 on both sides to get [tex]1 > {{\\sqrt{2}}\\over{2}}[/tex]. Now add [tex]{{\\sqrt{2}}\\over{2}}[/tex] to both sides to get [tex]1 + {{\\sqrt{2}}\\over{2}} > \\sqrt{2}[/tex]. Divide the numerator and the denominator of the fraction by [tex]\\sqrt{2}[/tex]. You should arrive at the desired inequality.\r\n\r\nSuppose it works for n = k; that is,\r\n\r\n[tex]\\displaystyle {{1}\\over{\\sqrt{1}}} + {{1}\\over{\\sqrt{2}}} + \\ldots + {{1}\\over{\\sqrt{k}}} > \\sqrt{k}[/tex].\r\n\r\nAdd [tex]{{1}\\over{\\sqrt{k+1}}}[/tex] to both sides. We obtain a RS of [tex]\\sqrt{k} + {{1}\\over{\\sqrt{k+1}}}[/tex]. We then wish to prove the inequality\r\n\r\n[tex]\\displaystyle \\sqrt{k}+{{1}\\over{\\sqrt{k+1}}} > \\sqrt{k+1}[/tex].\r\n\r\nCan you prove this inequality?", "Solution_12": "[quote=\"SingTheSorrow\"]3)\n\nI'm confused now with the part b of matematical induction...help? But here's what I have:\nN=2\n\nSo. 1/ sqrt (1) + 1/ :rt2: > :rt2: \n\nI wasn't too sure about this...but I just put it in my calculator and it verfied the statement. Can I do something different rather than rounding :rt2: to 1.41? \n\nNow, I'm not too sure about how do do the next part either...just assume that I showed all my work, but now how do I show sqrt (1): + 1/ sqrt (k+1) > sqrt (k + 1)?\n\nI attempted the others but I'm sleeping now.[/quote]\r\n\r\nUmm... if you don't want to use a calculator to prove that [tex]1+\\frac{1}{\\sqrt{2}}>\\sqrt{2}[/tex] you can do this:\r\nMake the left side one fraction\r\n[tex]\\frac{\\sqrt{2}+1}{\\sqrt{2}}>\\sqrt{2}[/tex]\r\nAssume it is true and multiple by sides by :rt2: .\r\n\r\n[tex]\\sqrt{2}+1>2[/tex]\r\n[edited]\r\n[tex]\\sqrt{2}>1[/tex] \r\nThe last equation is obviously true and so your assumption is true. I think that you can do this but I am not sure. That is the best way that I can think of.", "Solution_13": "[quote=\"joml88\"]\nUmm... if you don't want to use a calculator to prove that [tex]1+\\frac{1}{\\sqrt{2}}>\\sqrt{2}[/tex] you can do this:\nMake the left side one fraction\n[tex]\\frac{\\sqrt{2}+1}{\\sqrt{2}}>\\sqrt{2}[/tex]\nAssume it is true and multiple by sides by :rt2: .\n\n[tex]\\sqrt{2}+1>1[/tex]\n\n[tex]\\sqrt{2}>0[/tex] \nThe last equation is obviously true and so your assumption is true. I think that you can do this but I am not sure. That is the best way that I can think of.[/quote]\r\n\r\nYou made a tiny mistake: :rt2: * :rt2: = 2 not 1.", "Solution_14": "Oops... sry bout that. I edited that.", "Solution_15": "That's what I was having trouble with, I got it into a bunch of different forms, and none of them shouted \"I am bigger than [tex]\\sqrt{k}[/tex]!\" I don't really know how to prove innequalities...Only AM-GM, if that is what you are referring to...\r\n\r\nHowever this should follow w/what you did for n=2.\r\n\r\nStart w/ the statement [tex]k>\\sqrt{k}[/tex].\r\n\r\nEDITED: Bunch of unhelpful blahblah...\r\n\r\nNot exactly what we want...maybe SingTheSorrow can answer this...", "Solution_16": "JS1527 wrote:That's what I was having trouble with, I got it into a bunch of different forms, and none of them shouted \"I am bigger than [tex]\\sqrt{k}[/tex]!\"\n\n\n\nI haven't been following this topic much, but I assume you are trying to prove that k >= sqrt[k]\n\nSolution:\n\n[hide]\n\nIf k = 0 or 1, then there is equality. If k is greater than 1, then\n\n\n\nk(k-1) >=0\n\nk^2 - k >=0\n\nk^2 >=k\n\nk >=sqrt[k]\n\nQED\n\n[/hide]", "Solution_17": "OK:\r\n\r\nWe start with\r\n\r\n[tex]\\displaystyle 1 > {{1}\\over{k + 1}}[/tex]. This is true for any k :ge: 1. Add k to both sides, and move 1 to the right:\r\n\r\n[tex]\\displaystyle k > k - 1 + {{1}\\over{k + 1}}[/tex]. Rewrite (- 1) as (+ 1 - 2):\r\n\r\n[tex]\\displaystyle k > (k + 1) + {{1}\\over{k + 1}} - 2[/tex]. The RS can be written as a square:\r\n\r\n[tex]\\displaystyle (k + 1) - 2 + {{1}\\over{k + 1}}[/tex]\r\n[tex]= \\left(\\sqrt{k + 1} - {{1}\\over{\\sqrt{k + 1}}}\\right)^2[/tex]. Notice that the terms inside the brackets have a positive value. Therefore,\r\n\r\n[tex]\\displaystyle \\left(\\sqrt{k}\\right)^2 >\\left(\\sqrt{k + 1} - {{1}\\over{\\sqrt{k + 1}}}\\right)^2[/tex]. Taking square root of both sides, and moving terms around, we obtain the desired inequality.", "Solution_18": "nr, we are trying to prove that:\r\n\r\n[tex]\\sqrt{k}+\\frac{1}{\\sqrt{k+1}} > \\sqrt{k+1}[/tex]\r\n\r\nFollowing what Joe did:\r\n\r\n[tex]\\frac{(\\sqrt{k+1})(\\sqrt{k})}{\\sqrt{k+1}}+\\frac{1}{\\sqrt{k+1}} > \\sqrt{k+1}[/tex]\r\n\r\nMultiply both sides by [tex]\\sqrt{k+1}[/tex], assuming that it is true:\r\n\r\n[tex](\\sqrt{k+1})(\\sqrt{k}) + 1 > k+1[/tex]\r\n[tex](\\sqrt{k+1})(\\sqrt{k})> k[/tex]\r\n[tex]\\sqrt{k^2+k}>k[/tex]\r\n\r\nNow since [tex]\\sqrt{k^2} = k[/tex] for k positive, if you add a k to the inside of the square root:\r\n\r\n[tex]\\sqrt{k^2+k}>k[/tex]\r\n\r\nGuys, is this correct? Danny do you follow all this?", "Solution_19": "It's correct, but only because each step is reversible (as Joel mentioned). It is best to start with an universally true statement, and manipulate it to arrive at the target inequality, not backwards.", "Solution_20": "Osiris, I have to admit, what you did is very cool, but I would never think to do any of that...", "Solution_21": "I did follow what you did! But Osirsis is, um, wayyy too advance for me. But here I go stumbling along with matematical induction. \r\n\r\n4) \r\n\r\nn! > 2^n\r\n\r\nSubstitute n for 4.\r\n\r\n4! > 2^4\r\n24 > 16 <--- Correct.\r\n\r\nAssuming that is correct, let's prove the statement for k + 1.\r\n\r\n2^n (k+1) > 2^k+1\r\n\r\nDivide both sides by 2^n to get:\r\n\r\nn+1>2\r\n\r\nSubstract 1 from both sides.\r\n\r\nn > 1 Which is correct since n has to be 4 or greater.\r\n\r\n... Right?\r\n\r\n5) \r\n\r\nSubstitute n for 1, x for one, and y for two. \r\n\r\n(1/2)^2 < (1/2)^1\r\n1/4<1/2 <--Correct.\r\n\r\nAssuming this is correct, let's prove this statement for k+1. \r\n\r\n(x/y)^n (x/y)^n+2 < (x/y)^n+1\r\n\r\nMultiply together to get:\r\n\r\n(x/y)^2n+2 <(x/y^n+1\r\n\r\nDivide both sides by x/y^2n+2 which gives you...\r\n\r\n1<1/2...Um, where's my stupid mistake? \r\n\r\n6) \r\n\r\nLet n = 1\r\n\r\n2^2(1)-1 + 3^2(1)-1\r\n5\r\n\r\nWhich is divisable by 5. \r\n\r\nNow how do I do this part? \r\n\r\n(5 + 2^2n+2 + 3^2n-2)/5?", "Solution_22": "(4)[quote]Assuming that is correct, let's prove the statement for k + 1. \n\n2^n (k+1) > 2^k+1 \n[/quote]\r\n\r\nYou should not have both n and k in the equation.\r\n\r\nYou assume that [tex]k!>2^k[/tex] when [tex]k \\ge 4[/tex].\r\n\r\nNow you are going to show that:\r\n[tex](k+1)! > 2^{k+1}[/tex].\r\n\r\nI don't see how what you did is correct, cause [tex]2^k[/tex] is not on both sides...or am I mistaken?\r\n\r\nI would try:\r\n\r\n[tex](k+1)k! > 2^k \\cdot 2[/tex]\r\n\r\nNow we already assumed that [tex]k! > 2^k[/tex], and you are left w/ [tex]k+1 > 2[/tex], or [tex]k > 1[/tex] as you mentioned, and this is true, so the statement must be true.\r\n\r\nI am not great at proving stuff, but I think this is okay (maybe).\r\n\r\nI am not exactly sure what you were doing in the beginning steps. Did you divide the new expression [tex](k+1)! > 2^{k+1}[/tex] by the assumed expression [tex]k! > 2^k[/tex]? I think you did it pretty well regardless, good work, keep up the effort.", "Solution_23": "(5) Try this:\r\n\r\nAssume that [tex] (\\frac{x}{y})^{k+1} < (\\frac{x}{y})^{k}[/tex]\r\n\r\nWhat does multiplying both sides by the constant [tex]\\frac{x}{y}[/tex] produce?\r\n\r\nYour work:\r\n\r\n[quote](x/y)^n (x/y)^n+2 < (x/y)^n+1 \n[/quote]\r\n\r\nWhat you want to show is not this, letting [tex]n=k+1[/tex],\r\n\r\n[tex](\\frac{x}{y})^{k+2} < (\\frac{x}{y})^{k+1}[/tex]." } { "Tag": [ "function", "topology", "real analysis", "real analysis unsolved" ], "Problem": "$f: R^2 \\to R^2$ a continuous function such that $f(f(X))=X$ for any $X \\in R^2$.\r\nProve that $f$ has at least on fixed point.\r\nAnd in $R^n$", "Solution_1": "Deleted (see below why) :).", "Solution_2": "------------------------", "Solution_3": "Sorry about that. I got confused during that post. I've deleted my message. \r\n\r\nWhat I really meant was this: given a closed path $u$ in $\\mathbb R^2$ homotopic to a constant path, and a continuous function $g: \\mathbb R^2\\to\\mathbb R^2$, the closed path $g\\circ u$ is homotopic to a constant path in $g(\\mathbb R^2)$. In this sense, we have a kind of \"preservation of simple connectedness\". In the above post I attempted to do precisely this: find a nullhomotopic closed path which is no longer nullhomotopic after applying $g$, thus getting a contradiction. Anyway, since it wasn't really a solution, it's beter off deleted :). \r\n\r\nI'll repost it when/if I feel like making everything precise :).", "Solution_4": "I had read your proof and thought it worked, because $f$ is an homeomorphism (while $exp$ not).\r\nThe idea was beautiful :)", "Solution_5": "No, cimabue was right. What I said there was indeed wrong, but luckily, as I wrote in my second post, it was not what I meant :). I did not apply those considerations to $f$, but to $g$, which was defined by $g(x)=x-f(x)$, and $g$ is not necessarily a homeomorphism.", "Solution_6": "Consider a segment $[0,1]$. A point $a\\in [0,1]$ is called good, if $f(a)$ belongs to $[0,1]$. The set of good points is closed, so we may find a good $b$, for which the distance between $b$ and $f(b)$ is minimal. The segment $\\delta$ between $b$ and $f(b)$ does not intersect with its $f$-image $f(\\delta)$. The union of $\\delta$ and $f(\\delta)$ is a Jordan curve $\\gamma$. We have $f(\\gamma)=\\gamma$. The interior $D$ of $\\gamma$ is connected. If there are $x\\in D$ and $y\\in D$ such that $f(X)\\notin D,\\,f(Y)\\in D$, then by connectness we find $z\\in D$ such that $f(z)\\in \\gamma$ --- a contradiction. Analogously, the $f$-image of the exterior $D'$ of $\\gamma$ either lies inside $D$, or inside $\\D'$. Since $f$ is bounded on $D$, $f(D')$ contains some points in $D'$ (any point in $D'$ has a preimage). Finally, $f(D)=D,\\,f(D')=D'$.\r\n\r\nNow apply Brower theorem for $D\\cup \\gamma$. I do not know exactly, whether $D\\cup \\gamma$ is homeomorphic to a closed disc, but I remember there was some theorem about this.", "Solution_7": "has someone an idea in higher dimensions ?", "Solution_8": "http://groups.google.ca/group/sci.math.research/browse_frm/thread/138d813a4e4be074/f90f0e83f2483cd3" } { "Tag": [ "geometry", "trigonometry", "symmetry", "trig identities", "Law of Sines", "Law of Cosines", "angle bisector" ], "Problem": "Here's a challenge:\r\n\r\nGiven:\r\n\r\nABCD is a square\r\nPoint M is inside ABCD\r\nAngle MDC = Angle MCD = 15 degrees\r\n\r\nProve:\r\nTriangle ABM is equilateral\r\n\r\nHints:\r\n[hide]You need to construct one line[/hide]\n[hide]Uses tangent and tangent inverse[/hide]\r\n\r\nIf any one knows a different way to solve this problem that doesn't use the hints above, then please post, I would like to see it.", "Solution_1": "Though it's not necessary to use only [hide] tangent [/hide], I think the only solution is trigonometric. :D", "Solution_2": "yeah. because you can use \r\n\r\n[hide]sin \nby eliminating the possiblity of scalene and doing some other stuff, but i don't remember that way.[/hide]", "Solution_3": "Eh. I found that it was isoscoles. \r\nI assumed it was equilateral so then I found the altitude of the triangle in terms of the square's side and then I just used trig to find the length of the altitude of the 15-15-150 triangle in terms of the side of the square...adding them up the answer should be the length of the square, right...? I found that it was something different...\r\nAnd it's obviously not scalene.", "Solution_4": "its equilateral, i don't know what you did...", "Solution_5": "[hide=\"Try this :D \"]\nLet $N$ be a point inside the square such that $\\angle ADN=\\angle DAN=15^\\circ$.\n\nStep 1: Prove that $\\triangle DMN$ is equilateral.\nStep 2: Prove that $AD=AM=AB$.\nStep 3: Prove that $\\angle MAB=60^\\circ$.\n\nThen we're done! :lol: \n[/hide]", "Solution_6": "but wouldnt that mean that DN = DM = MN = AN and since AM = AD, then AN+ND = AD - which it cant be since it wouldnt be a triangle.\r\n\r\nthere is either something wrong with my logic or yours...", "Solution_7": "err ok.\r\nSo then the altitude of the triangle is $\\frac{s\\sqrt{3}}{2}$. So then with the law of sines, $\\frac{\\sin75}{\\frac{s}{2}}=\\frac{\\sin15}{x}$ where $x$ is the altitude of $\\triangle MDC$. Solving for $x$, $x=\\frac{s\\sin15}{2\\sin75}$. Then add: $\\frac{s\\sin15}{2\\sin75}+\\frac{s\\sqrt{3}}{2}$, take out $s$ and get $s\\displaystyle\\left(\\frac{\\sin15}{2\\sin75}+\\frac{\\sqrt{3}}{2}\\right)$ and then I used the trigonometric identities (and a calculator afterwards), and if $\\triangle ABM$ is equilateral, then $\\frac{\\sin15}{2\\sin75}+\\frac{\\sqrt{3}}{2}=1$...oops. It does. hehe, ABM is equilateral..", "Solution_8": "[hide=\"my trig way\"]\nok draw a line through $M$ parallel to $AD$ and $BC$. So the leg of the isoceles triangle $MCD$ will be $\\frac{s}{2\\cos{15}}$ by simple trig. Then we have $\\angle ADM = 75$ because square has right angles. then we have find $\\cos{15}$ and $\\cos{75}$ by sum-difference formula or other ways you want to, and using law of cosines and simplify heavily we have that $AM = s$, and by symmetry $CM = s$, therefore it'll be an equilateral triangle\n[/hide]", "Solution_9": "Correct me is im wrong:\r\n\r\nGiven DM=CM, angle ADM=MCB, and AD=BC. Therefore triangle ADM is congruent to BMC. So AM=BM. Now i just need to prove AB equals AM or BM.\r\n\r\nDraw Line EM, so that EM is the angle bisector of triangle AME intersecting AB at E. Note that the angle bisector, the median and the altitude of a iscoceles triangle are the same line. Draw line FM so that it is the angle bisector of triangle DMC, intersecting CD at F. Call AD, a so that a=EM+FM. FM=a/2*tan 15. Therefore EM=(1-tan15/2)a, which equals sqr3/2. Notice that AE=EB=DF=FC=a/2. So ((1/2)^2+(sqr3/2)^2)a=AM. Which proves AM=AD=BM.\r\n\r\nSorry i cant get a picture. :oops: [/list]", "Solution_10": "[quote=\"star99\"]but wouldnt that mean that DN = DM = MN = AN and since AM = AD, then AN+ND = AD - which it cant be since it wouldnt be a triangle.\n\nthere is either something wrong with my logic or yours...[/quote]\r\nHow did you get $MN=AN$? :? \r\n\r\nAlright, I'll post my solution based on the hint I've posted:\r\n[hide=\"Geometric solution\"]\nBy ASA, $\\triangle AND$ and $\\triangle DMC$ are congruent., so $DN=DM$.\nAlso, $\\angle NDM=90^\\circ-15^\\circ-15^\\circ=60^\\circ$, which implies $\\triangle MDN$ is an equilateral triangle.\n\n$\\angle AND=180^\\circ-2\\cdot 15^\\circ=150^\\circ$ and $\\angle DNM=60^\\circ$, so $\\angle ANM=360^\\circ-150^\\circ-90^\\circ=150^\\circ$\nBy SAS, $\\triangle AND$ and $\\triangle ANM$ are congruent.\nSo we have $AD=AM(=AB)$ and $\\angle NAM=15^\\circ$.\nThat gives $\\angle MAB=90^\\circ-2\\cdot 15^\\circ=60^\\circ$.\n\nSince $AM=AB$ and $\\angle MAB=60^\\circ$, $\\triangle ABM$ is equilateral.\n[/hide]\r\nHope it's clear now.", "Solution_11": "[quote=\"frt\"][hide=\"Try this :D \"]\nLet $N$ be a point inside the square such that $\\angle ADN=\\angle DAN=15^\\circ$.\n\nStep 1: Prove that $\\triangle DMN$ is equilateral.\nStep 2: Prove that $AD=AM=AB$.\nStep 3: Prove that $\\angle MAB=60^\\circ$.\n\nThen we're done! :lol: \n[/hide][/quote]\n\nThat's genius\u2014I'm kinda jealous I didn't think of it first. Anyways here's a fully written-out non-trig solution based on that.\n\n[hide]\nLet $N$ be a point inside the square such that $\\angle ADN=\\angle DAN=15^\\circ$. By ASA, triangles $ADN$ and $CDM$ are congruent. Thus, $DM = DN$. $\\angle MDN$ must equal $60^\\circ$, because $\\angle ADN + \\angle MDN + \\angle CDM = 90^\\circ$. Since triangle $MDN$ is isosceles and one of the angles is $60^\\circ$, it's clearly an equilateral triangle. So that means that $MN = DN = AN$. Also, $\\angle ANM = 150^\\circ$, because $\\angle AND + \\angle DNM + \\angle ANM = 360\\^circ \\implies 150^\\circ + \\angle ANM + 60^\\circ = 360^\\circ$. So triangles $AND$ and $ANM$ are congruent by $SAS$. Thus $AM = AD = AB$. Since triangles $ADM$ and $BCM$ are congruent by $SAS$, $AM = BM$. So our conclusion is that $AM = BM = AB$ and thus triangle $ABM$ is equilateral.\n\nNo trig.\n[/hide]", "Solution_12": "Let $N$ be the point in $ABCD$ such that $\\angle BAN=30^{\\circ}=\\angle CBN$.\r\n\r\nThen $\\angle NAB=\\angle NBA=90^{\\circ}-30^{\\circ}=60^{\\circ}=180^{\\circ}-60^{\\circ}-60^{\\circ}=\\angle ANB$, so $\\triangle ANB$ is equilateral.\r\n\r\n$\\triangle DAN=\\triangle CBN$ by SAS, so $ND=NC$, and $\\angle NDC=\\angle NCD$.\r\n\r\nThen $\\angle ADN=\\angle AND=\\frac{180^{\\circ}-30^{\\circ}}{2}=75^{\\circ}$. \r\n\r\n$\\angle NDC=90^{\\circ}-75^{\\circ}=15^{\\circ}=\\angle NCD$.\r\n\r\n$N$ has the same properties as $M$ and they are uniquely defined so $N\\equiv M$ giving: $\\triangle AMB$ is equilateral.", "Solution_13": "wow thats a lot of ways.....\r\n\r\nn e one have more geometry challenge problems?" } { "Tag": [ "trigonometry", "inequalities", "inequalities unsolved" ], "Problem": "Given triangle $ABC$ with $A\\le 120^{0}$ find minimum of\r\n$\\frac{(\\sin A\\sin B+\\sin B\\sin C+\\sin C\\sin A)^{2}}{\\sin A\\sin B\\sin C}$", "Solution_1": "what is the minimum of $\\sin A+\\sin B+\\sin C$?", "Solution_2": "anyone? *bump*", "Solution_3": "[quote=\"nayel\"]what is the minimum of $\\sin A+\\sin B+\\sin C$?[/quote]\r\nIf $A,$ $B$ and $C$ are measureangles of triangle then \r\n$\\inf(\\sin A+\\sin B+\\sin C)=0.$ :wink:", "Solution_4": "[quote=\"arqady\"][quote=\"nayel\"]what is the minimum of $\\sin A+\\sin B+\\sin C$?[/quote]\nIf $A,$ $B$ and $C$ are measureangles of triangle then \n$\\inf(\\sin A+\\sin B+\\sin C)=0.$ :wink:[/quote]\r\n\r\nthank you. but there's a tiny bit of problem: i lost my paper! :blush:", "Solution_5": "[quote=\"arqady\"][quote=\"nayel\"]what is the minimum of $\\sin A+\\sin B+\\sin C$?[/quote]\nIf $A,$ $B$ and $C$ are measureangles of triangle then \n$\\inf(\\sin A+\\sin B+\\sin C)=0.$ :wink:[/quote]\r\nWith $A\\le 120^{0}$ that express has minimum when $A=120^{0},B=C=30^{0}$, I haven't still solved that this problem.", "Solution_6": "ok i rediscovered my theory: \r\n\r\nwe have \r\n\r\n$(\\sin A\\sin B+\\sin B\\sin C+\\sin C\\sin A)^{2}$\r\n$\\ge 3\\sin A\\sin B\\sin C(\\sin A+\\sin B+\\sin C)$\r\n\r\nso we only need find minimum of $\\sin A+\\sin B+\\sin C$\r\n\r\ni think that didn't help... :(", "Solution_7": "[quote=\"nayel\"]ok i rediscovered my theory: \n\nwe have \n\n$(\\sin A\\sin B+\\sin B\\sin C+\\sin C\\sin A)^{2}$\n$\\ge 3\\sin A\\sin B\\sin C(\\sin A+\\sin B+\\sin C)$\n\nso we only need find minimum of $\\sin A+\\sin B+\\sin C$\n\ni think that didn't help... :([/quote]\r\nI think you can't do follow your way because follow me my express gets minimum iff $A=120,B=C=30$ but your inequality gets equality iff $A=B=C=60$ :maybe:", "Solution_8": "The minimum of the expression doesn't get minimum only if $A=120$ because the expression is cyclic so we can chose also $B=120$ or $C=120$.For example we can have $A=20,B=20,C=140$ so we have an angle bigger than $120$ and because the expression is cyclic we can re-denote the triangle so it has no importance that $A\\leq120$.We have no restriction.I hope you understood me.", "Solution_9": "I haven't understood you Zamfirmihai, my express won't be symmetric woth condition $A\\le 120$, by the way really I sure my problem is true minimum when $A=120,B=C=30$, it is from a problem follow me is very hard\r\nGiven triangle $ABC,A\\le 120^{0}$ prove that\r\n\\[(FA+FB+FC)(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c})\\ge 4+\\frac{4}{\\sqrt{3}}\\]\r\nWith $F$ is Fermat point Equality holds iff $A=120,B=C=30$ two problems is the same. but I am sure the second is true, symmetric isn't problem here." } { "Tag": [], "Problem": "How man seconds will it take a train 300 meters long travelling at 100 km/hr to pass a man jogging at 10 km/hr in the same direction?", "Solution_1": "[hide]12[/hide]", "Solution_2": "[hide]You want the front of the train to start at where the man is, and then travel 0.3 kilometers.\n\n\n\nx = the number of hours\n\n\n\n10x+0.3 = 100x\n\n0.3 = 90x\n\nx = 0.00333...\n\n\n\nBut x is the number of hours. We want the number of seconds.\n\n\n\n0.0333.../x = 1/3600\n\nx = 12\n\n\n\nSo it would take 12 seconds for the train to pass.\n\n\n\n[/hide]", "Solution_3": "[hide]The train is going 100-10 = 90km/h relative to the man so,\n\n\n\n 0.3km/ 90km/h= 1/300h = 3600/300s = 12 seconds[/hide]" } { "Tag": [ "summer program", "MathPath", "ceiling function", "AMC", "USA(J)MO", "USAMO", "email" ], "Problem": "Who are the freshmen who made it to MOSP? Try to include name and state, and score maybe to help those in waiting.\r\n\r\nI'm Jason Hoch, CT, 9", "Solution_1": "Carlos Dominguez, Ohio, Score: 10", "Solution_2": "uh...I think they'll send out a complete list pretty shortly", "Solution_3": "Edward Gan, MD, 12", "Solution_4": "[quote=\"EFuzzy\"]Edward Gan, MD, 12[/quote]\r\n\r\nHey, I saw you at MathPath! :D Congrats!", "Solution_5": "Alex Ryan, TX, 10", "Solution_6": "John Doe, AK, $9 \\frac{3}{4}$. I think they will be sending out a list shortly too, but I guess people are just really excited and want to share :P.", "Solution_7": "[quote=\"probability1.01\"]John Doe, AK, $9 \\frac{3}{4}$.[/quote]\r\n\r\nI assume that rounds up, right? :D", "Solution_8": "It rounds up to 41.", "Solution_9": "41!!!??? Some rounding! :D Yeah, so congrats to making red mop with $9\\frac 34 =41$ :lol:", "Solution_10": "[quote=\"pieterminate\"]41!!!??? Some rounding! :D Yeah, so congrats to making red mop with $9\\frac 34 =41$ :lol:[/quote]\r\n\r\nWell actually it's $\\left\\lceil 9\\frac 34 \\right\\rceil = 41$. So $10 = 41$, which is true if the LHS is in base $41$ and the RHS is in base $10$.\r\n\r\nNo more math until MOSP! :P", "Solution_11": "It's actually MOP, and you have homework - you have to complete every problem (approximately USAMO level difficulty; some towards the end are a bit harder) or you'll be cut from the program.", "Solution_12": "[quote=\"nr1337\"][quote=\"pieterminate\"]41!!!??? Some rounding! :D Yeah, so congrats to making red mop with $9\\frac 34 =41$ :lol:[/quote]\n\nWell actually it's $\\left\\lceil 9\\frac 34 \\right\\rceil = 41$. So $10 = 41$, which is true if the LHS is in base $41$ and the RHS is in base $10$.\n\nNo more math until MOSP! :P[/quote]\r\n\r\n10 is *congruent* to 41 mod 31! :lol:", "Solution_13": "[quote=\"MysticTerminator\"]It's actually MOP, and you have homework - you have to complete every problem (approximately USAMO level difficulty; some towards the end are a bit harder) or you'll be cut from the program.[/quote]\r\n\r\nEh, [b]m[/b]ath [b]o[/b]lympiad [b]s[/b]ummer [b]c[/b]amp \r\n\r\nBut are they actually that hard....? \r\nAnd has [i]anyone[/i] ever been cut? :roll: :? ;)", "Solution_14": "[quote=\"MysticTerminator\"]you have to complete every problem (approximately USAMO level difficulty; some towards the end are a bit harder) or you'll be cut from the program[/quote]\r\nand, I might add, you get sent home on a Metro Bus.... or ride a bike, you're pick. They will provide you with a week's worth of sacked lunches if you choose the latter option, at no expense to your family (but they will charge you for the bike :D)\r\n\r\nOh, and I completely forgot that probability1.01 counts in base $\\frac{7+\\sqrt{689}}{64}$. Then his USAMO score is 9.75 :?", "Solution_15": "[quote=\"pieterminate\"][quote=\"MysticTerminator\"]It's actually MOP, and you have homework - you have to complete every problem (approximately USAMO level difficulty; some towards the end are a bit harder) or you'll be cut from the program.[/quote]\n\nEh, [b]m[/b]ath [b]o[/b]lympiad [b]s[/b]ummer [b]c[/b]amp \n\nBut are they actually that hard....? \nAnd has [i]anyone[/i] ever been cut? :roll: :? ;)[/quote]\r\n\r\num, that would be MOSC, which is obviously wrong - it's actually Math Olympiad Program, hence the name MOP and the corresponding MOPpers (not MOPsters, MOSPers, anything else stupid)\r\n\r\nyes, if you use MOSP (ugh, feel dirty even typing it), you're automatically an outcast and get sent home forcibly with all the people who didn't do their homework", "Solution_16": "[quote=\"MysticTerminator\"]you have to complete every problem[/quote]\r\n\r\nNote how he cleverly avoids the word \"correctly...\" :)", "Solution_17": "[quote=\"nr1337\"][quote=\"MysticTerminator\"]you have to complete every problem[/quote]\n\nNote how he cleverly avoids the word \"correctly...\" :)[/quote]\r\n\r\n?! well of course - if you do ANYTHING at MOP incorrectly, ... (you don't just get sent home)", "Solution_18": "[quote=\"MysticTerminator\"][quote=\"nr1337\"][quote=\"MysticTerminator\"]you have to complete every problem[/quote]\n\nNote how he cleverly avoids the word \"correctly...\" :)[/quote]\n\n?! well of course - if you do ANYTHING at MOP incorrectly, ... (you don't just get sent home)[/quote]\r\n\r\nSomething tells me that you also get some unimaginable punishment by the [b]MOP[/b] people if you qualify but don't go, right... :D", "Solution_19": "[quote=\"pieterminate\"]\nBut are they actually that hard....? \nAnd has [i]anyone[/i] ever been cut? :roll: :? ;)[/quote]\r\n\r\nUmmm... obviously? But of course if you don't believe me you can always go there and find out for yourself ;). So yeah, don't get too excited yet.", "Solution_20": "Oh well, I was just reading from the email:\r\n[quote]\nThe Mathematical Olympiad Summer Program\n(MOSP)[/quote]\r\n\r\n\r\nYeah, you people are scaring me too much. :roll: :lol: lol\r\nOh well, that makes MOP more fun then. :D :lol:", "Solution_21": "[quote=\"pieterminate\"]Oh well, I was just reading from the email:\n[quote]\nThe Mathematical Olympiad Summer Program\n(MOSP)[/quote]\n\n\nYeah, you people are scaring me too much. :roll: :D :lol: lol[/quote]\r\n\r\nthat's b/c the AMC has changed the name from MOP to MOSP, but only losers actually call it MOSP. In fact, if you are going to use the \"S\" at all, you can't capitalize it or else you will be crucified. So it's MOP or MOsP, but preferably just plain old MOP. \r\n\r\nDon't worry; when they cut you from the program, they don't actually send you home. Instead, MOP just gives you a mop, and you work there as a janitor ;) \r\n\r\nIn all seriousness, if you don't want to go to MOP with a mop, you better complete EVERY homework problem. Otherwise, you might end up like what almost happened to Alex last year :(\r\n(sorry, alex, I know you didn't want me to mention it... I didn't give any details though :oops: )", "Solution_22": "[quote=\"bubala\"][quote=\"pieterminate\"]Oh well, I was just reading from the email:\n[quote]\nThe Mathematical Olympiad Summer Program\n(MOSP)[/quote]\n\n\nYeah, you people are scaring me too much. :roll: :D :lol: lol[/quote]\n\nor else you will be crucified. \n\nDon't worry; when they cut you from the program, they don't actually send you home. Instead, MOP just gives you a mop, and you work there as a janitor ;) \n[/quote]\r\n\r\nHow cruel! :rotfl: \r\n\r\nHow hard are the problems exactly? Stupid people like me might not be able to solve them... :D", "Solution_23": "I heard that the mops are even colour-coded: Freshman get red mops, winners get black mops, and everyone else gets blue mops. :D", "Solution_24": "[quote=\"nr1337\"]I heard that the mops are even colour-coded: Freshman get red mops, winners get black mops, and everyone else gets blue mops. :D[/quote]\r\n\r\nwait . . . how did you know? (actually, on the '04 name tags, the shirt of the guy with the mop was indeed colored that way)", "Solution_25": "Anyone know how to change the topic title.... it has the dreaded letter S in it.", "Solution_26": "Word is Edward Trout made it too. And a couple of Japanese people but I don't remember how to spell their names.\r\n\r\n[quote]How hard are the problems exactly? Stupid people like me might not be able to solve them...[/quote]\nHere's a sample problem:\n\n[quote]Prove or disprove Goldbach's conjecture.[/quote]", "Solution_27": "Oh if they're at that level i'll be all set i started toying with that a couple years ago", "Solution_28": "[quote=\"randomdragoon\"]\nHere's a sample problem:\nProve or disprove Goldbach's conjecture.[/quote]\r\n\r\nOh, okay. :D\r\nYeah, it's rather hard... you want to do it for us? I can't get anywhere, I can't even read the fourth word! :lol: Yeah, counter example: [hide]12319240932847598032643250981379132421340723906853097093405849735939274859274502394875093284590328745690382048209327569817624793208459863205986387563287456823746583274659832857623495876324985123192409328475980326432509813791324213407239068530970934058497359392748592745023948750932845903287456903820482093275698176247932084598632059863875632874568237465832746598328576234958763249851231924093284759803264325098137913242134072390685309709340584973593927485927450239487509328459032874569038204820932756981762479320845986320598638756328745682374658327465983285762349587632498512319240932847598032643250981379132421340723906853097093405849735939274859274502394875093284590328745690382048209327569817624793208459863205986387563287456823746583274659832857623495876324985123192409328475980326432509813791324213407239068530970934058497359392748592745023948750932845903287456903820482093275698176247932084598632059863875632874568237465832746598328576234958763249851231924093284759803264325098137913242134072390685309709340584973593927485927450239487509328459032874569038204820932756981762479320845986320598638756328745682374658327465983285762349587632498512319240932847598032643250981379132421340723906853097093405849735939274859274502394875093284590328745690382048209327569817624793208459863205986387563287456823746583274659832857623495876324985123192409328475980326432509813791324213407239068530970934058497359392748592745023948750932845903287456903820482093275698176247932084598632059863875632874568237465832746598328576234958763249851231924093284759803264325098137913242134072390685309709340584973593927485927450239487509328459032874569038204820932756981762479320845986320598638756328745682374658327465983285762349587632498512319240932847598032643250981379132421340723906853097093405849735939274859274502394875093284590328745690382048209327569817624793208459863205986387563287456823746583274659832857623495876324985123192409328475980326432509813791324213407239068530970934058497359392748592745023948750932845903287456903820482093275698176247932084598632059863875632874568237465832746598328576234958763249851231924093284759803264325098137913242134072390685309709340584973593927485927450239487509328459032874569038204820932756981762479320845986320598638756328745682374658327465983285762349587632498512319240932847598032643250981379132421340723906853097093405849735939274859274502394875093284590328745690382048209327569817624793208459863205986387563287456823746583274659832857623495876324985123192409328475980326432509813791324213407239068530970934058497359392748592745023948750932845903287456903820482093275698176247932084598632059863875632874568237465832746598328576234958763249851231924093284759803264325098137913242134072390685309709340584973593927485927450239487509328459032874569038204820932756981762479320845986320598638756328745682374658327465983285762349587632498512319240932847598032643250981379132421340723906853097093405849735939274859274502394875093284590328745690382048209327569817624793208459863205986387563287456823746583274659832857623495876324985123192409328475980326432509813791324213407239068530970934058497359392748592745023948750932845903287456903820482093275698176247932084598632059863875632874568237465832746598328576234958763249851231924093284759803264325098137913242134072390685309709340584973593927485927450239487509328459032874569038204820932756981762479320845986320598638756328745682374658327465983285762349587632498512319240932847598032643250981379132421340723906853097093405849735939274859274502394875093284590328745690382048209327569817624793208459863205986387563287456823746583274659832857623495876324985123192409328475980326432509813791324213407239068530970934058497359392748592745023948750932845903287456903820482093275698176247932084598632059863875632874568237465832746598328576234958763249851231924093284759803264325098137913242134072390685309709340584973593927485927450239487509328459032874569038204820932756981762479320845986320598638756328745682374658327465983285762349587632498512319240932847598032643250981379132421340723906853097093405849735939274859274502394875093284590328745690382048209327569817624793208459863205986387563287456823746583274659832857623495876324985123192409328475980326432509813791324213407239068530970934058497359392748592745023948750932845903287456903820482093275698176247932084598632059863875632874568237465832746598328576234958763249851231924093284759803264325098137913242134072390685309709340584973593927485927450239487509328459032874569038204820932756981762479320845986320598638756328745682374658327465983285762349587632498512319240932847598032643250981379132421340723906853097093405849735939274859274502394875093284590328745690382048209327569817624793208459863205986387563287456823746583274659832857623495876324985123192409328475980326432509813791324213407239068530970934058497359392748592745023948750932845903287456903820482093275698176247932084598632059863875632874568237465832746598328576234958763249851231924093284759803264325098137913242134072390685309709340584973593927485927450239487509328459032874569038204820932756981762479320845986320598638756328745682374658327465983285762349587632498512319240932847598032643250981379132421340723906853097093405849735939274859274502394875093284590328745690382048209327569817624793208459863205986387563287456823746583274659832857623495876324985123192409328475980326432509813791324213407239068530970934058497359392748592745023948750932845903287456903820482093275698176247932084598632059863875632874568237465832746598328576234958763249851231924093284759803264325098137913242134072390685309709340584973593927485927450239487509328459032874569038204820932756981762479320845986320598638756328745682374658327465983285762349587632498512319240932847598032643250981379132421340723906853097093405849735939274859274502394875093284590328745690382048209327569817624793208459863205986387563287456823746583274659832857623495876324985123192409328475980326432509813791324213407239068530970934058497359392748592745023948750932845903287456903820482093275698176247932084598632059863875632874568237465832746598328576234958763249851231924093284759803264325098137913242134072390685309709340584973593927485927450239487509328459032874569038204820932756981762479320845986320598638756328745682374658327465983285762349587632498512319240932847598032643250981379132421340723906853097093405849735939274859274502394875093284590328745690382048209327569817624793208459863205986387563287456823746583274659832857623495876324985123192409328475980326432509813791324213407239068530970934058497359392748592745023948750932845903287456903820482093275698176247932084598632059863875632874568237465832746598328576234958763249851231924093284759803264325098137913242134072390685309709340584973593927485927450239487509328459032874569038204820932756981762479320845986320598638756328745682374658327465983285762349587632498512319240932847598032643250981379132421340723906853097093405849735939274859274502394875093284590328745690382048209327569817624793208459863205986387563287456823746583274659832857623495876324985123192409328475980326432509813791324213407239068530970934058497359392748592745023948750932845903287456903820482093275698176247932084598632059863875632874568237465832746598328576234958763249851231924093284759803264325098137913242134072390685309709340584973593927485927450239487509328459032874569038204820932756981762479320845986320598638756328745682374658327465983285762349587632498512319240932847598032643250981379132421340723906853097093405849735939274859274502394875093284590328745690382048209327569817624793208459863205986387563287456823746583274659832857623495876324985123192409328475980326432509813791324213407239068530970934058497359392748592745023948750932845903287456903820482093275698176247932084598632059863875632874568237465832746598328576234958763249851231924093284759803264325098137913242134072390685309709340584973593927485927450239487509328459032874569038204820932756981762479320845986320598638756328745682374658327465983285762349587632498512319240932847598032643250981379132421340723906853097093405849735939274859274502394875093284590328745690382048209327569817624793208459863205986387563287456823746583274659832857623495876324985123192409328475980326432509813791324213407239068530970934058497359392748592745023948750932845903287456903820482093275698176247932084598632059863875632874568237465832746598328576234958763249851231924093284759803264325098137913242134072390685309709340584973593927485927450239487509328459032874569038204820932756981762479320845986320598638756328745682374658327465983285762349587632498512319240932847598032643250981379132421340723906853097093405849735939274859274502394875093284590328745690382048209327569817624793208459863205986387563287456823746583274659832857623495876324985123192409328475980326432509813791324213407239068530970934058497359392748592745023948750932845903287456903820482093275698176247932084598632059863875632874568237465832746598328576234958763249851231924093284759803264325098137913242134072390685309709340584973593927485927450239487509328459032874569038204820932756981762479320845986320598638756328745682374658327465983285762349587632498512319240932847598032643250981379132421340723906853097093405849735939274859274502394875093284590328745690382048209327569817624793208459863205986387563287456823746583274659832857623495876324985123192409328475980326432509813791324213407239068530970934058497359392748592745023948750932845903287456903820482093275698176247932084598632059863875632874568237465832746598328576234958763249851231924093284759803264325098137913242134072390685309709340584973593927485927450239487509328459032874569038204820932756981762479320845986320598638756328745682374658327465983285762349587632498512319240932847598032643250981379132421340723906853097093405849735939274859274502394875093284590328745690382048209327569817624793208459863205986387563287456823746583274659832857623495876324985123192409328475980326432509813791324213407239068530970934058497359392748592745023948750932845903287456903820482093275698176247932084598632059863875632874568237465832746598328576234958763249851231924093284759803264325098137913242134072390685309709340584973593927485927450239487509328459032874569038204820932756981762479320845986320598638756328745682374658327465983285762349587632498512319240932847598032643250981379132421340723906853097093405849735939274859274502394875093284590328745690382048209327569817624793208459863205986387563287456823746583274659832857623495876324985123192409328475980326432509813791324213407239068530970934058497359392748592745023948750932845903287456903820482093275698176247932084598632059863875632874568237465832746598328576234958763249851231924093284759803264325098137913242134072390685309709340584973593927485927450239487509328459032874569038204820932756981762479320845986320598638756328745682374658327465983285762349587632498512319240932847598032643250981379132421340723906853097093405849735939274859274502394875093284590328745690382048209327569817624793208459863205986387563287456823746583274659832857623495876324985123192409328475980326432509813791324213407239068530970934058497359392748592745023948750932845903287456903820482093275698176247932084598632059863875632874568237465832746598328576234958763249851231924093284759803264325098137913242134072390685309709340584973593927485927450239487509328459032874569038204820932756981762479320845986320598638756328745682374658327465983285762349587632498512319240932847598032643250981379132421340723906853097093405849735939274859274502394875093284590328745690382048209327569817624793208459863205986387563287456823746583274659832857623495876324985123192409328475980326432509813791324213407239068530970934058497359392748592745023948750932845903287456903820482093275698176247932084598632059863875632874568237465832746598328576234958763249851231924093284759803264325098137913242134072390685309709340584973593927485927450239487509328459032874569038204820932756981762479320845986320598638756328745682374658327465983285762349587632498512319240932847598032643250981379132421340723906853097093405849735939274859274502394875093284590328745690382048209327569817624793208459863205986387563287456823746583274659832857623495876324985123192409328475980326432509813791324213407239068530970934058497359392748592745023948750932845903287456903820482093275698176247932084598632059863875632874568237465832746598328576234958763249851231924093284759803264325098137913242134072390685309709340584973593927485927450239487509328459032874569038204820932756981762479320845986320598638756328745682374658327465983285762349587632498512319240932847598032643250981379132421340723906853097093405849735939274859274502394875093284590328745690382048209327569817624793208459863205986387563287456823746583274659832857623495876324985123192409328475980326432509813791324213407239068530970934058497359392748592745023948750932845903287456903820482093275698176247932084598632059863875632874568237465832746598328576234958763249851231924093284759803264325098137913242134072390685309709340584973593927485927450239487509328459032874569038204820932756981762479320845986320598638756328745682374658327465983285762349587632498512319240932847598032643250981379132421340723906853097093405849735939274859274502394875093284590328745690382048209327569817624793208459863205986387563287456823746583274659832857623495876324985123192409328475980326432509813791324213407239068530970934058497359392748592745023948750932845903287456903820482093275698176247932084598632059863875632874568237465832746598328576234958763249851231924093284759803264325098137913242134072390685309709340584973593927485927450239487509328459032874569038204820932756981762479320845986320598638756328745682374658327465983285762349587632498512319240932847598032643250981379132421340723906853097093405849735939274859274502394875093284590328745690382048209327569817624793208459863205986387563287456823746583274659832857623495876324985123192409328475980326432509813791324213407239068530970934058497359392748592745023948750932845903287456903820482093275698176247932084598632059863875632874568237465832746598328576234958763249851231924093284759803264325098137913242134072390685309709340584973593927485927450239487509328459032874569038204820932756981762479320845986320598638756328745682374658327465983285762349587632498512319240932847598032643250981379132421340723906853097093405849735939274859274502394875093284590328745690382048209327569817624793208459863205986387563287456823746583274659832857623495876324985123192409328475980326432509813791324213407239068530970934058497359392748592745023948750932845903287456903820482093275698176247932084598632059863875632874568237465832746598328576234958763249851231924093284759803264325098137913242134072390685309709340584973593927485927450239487509328459032874569038204820932756981762479320845986320598638756328745682374658327465983285762349587632498512319240932847598032643250981379132421340723906853097093405849735939274859274502394875093284590328745690382048209327569817624793208459863205986387563287456823746583274659832857623495876324985123192409328475980326432509813791324213407239068530970934058497359392748592745023948750932845903287456903820482093275698176247932084598632059863875632874568237465832746598328576234958763249851231924093284759803264325098137913242134072390685309709340584973593927485927450239487509328459032874569038204820932756981762479320845986320598638756328745682374658327465983285762349587632498512319240932847598032643250981379132421340723906853097093405849735939274859274502394875093284590328745690382048209327569817624793208459863205986387563287456823746583274659832857623495876324985123192409328475980326432509813791324213407239068530970934058497359392748592745023948750932845903287456903820482093275698176247932084598632059863875632874568237465832746598328576234958763249851231924093284759803264325098137913242134072390685309709340584973593927485927450239487509328459032874569038204820932756981762479320845986320598638756328745682374658327465983285762349587632498512319240932847598032643250981379132421340723906853097093405849735939274859274502394875093284590328745690382048209327569817624793208459863205986387563287456823746583274659832857623495876324986[/hide]", "Solution_29": "[quote=\"nr1337\"][quote=\"EFuzzy\"]Edward Gan, MD, 12[/quote]\n\nHey, I saw you at MathPath! :D Congrats![/quote]\r\n\r\nCongratulations to you too! Do you know anyone else from Mathpath who might come?", "Solution_30": "Look for a long-haired Georgian wearing a blue blazer and green vest.", "Solution_31": "is it possible that students outside the states get invited to the mosp?", "Solution_32": "[quote=\"jrshoch\"]Anyone know how to change the topic title.... it has the dreaded letter S in it.[/quote] \r\n\r\nThat's easy. The person who posted the first post in the thread just has to change the subject line of his post with an edit." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let x,y,z be positive numbers such that $ x^2\\plus{}y^2\\plus{}z^2\\equal{}6$\r\nand A,B,C are three angles of an acute triangle.\r\nProve that\r\n $ \\sum_{cyc}\\frac{1}{1\\plus{}yzcosA\\plus{}xyz^2cosAcosB} \\geq 1$", "Solution_1": "[quote=\"8826\"]Let x,y,z be positive numbers such that $ x^2 \\plus{} y^2 \\plus{} z^2 \\equal{} 6$\nand A,B,C are three angles of an acute triangle.\nProve that\n $ \\sum_{cyc}\\frac {1}{1 \\plus{} yzcosA \\plus{} xyz^2cosAcosB} \\geq 1$[/quote]\r\nUse: $ \\sqrt{3\\sum \\ {xyz^2cosAcosB}} \\le \\sum \\ yzcosA \\le \\frac{\\sum \\ x^2}{2}$ and Cauchy-Schwarz inequality,we have Q.E.D" } { "Tag": [ "function", "geometry", "3D geometry", "calculus", "calculus computations" ], "Problem": "$ f(A,B,C,D)$ is continuous and convex function, $ A,B,C,D \\in$ in a cube.\r\nIs it true that we have the maximum of $ f(A,B,C,D)$ when $ A,B,C,D$ lie in the vertices of the cube?", "Solution_1": "A hypercube, but yes. To see this, consider each of $ A, B, C, D$ separately.", "Solution_2": "No.\r\nThe problem is this.\r\nConsider a cube on $ R^3$, and $ A,B,C,D \\in$ cube.\r\nFind the maximum of $ f(A,B,C,D) \\equal{} AB \\plus{} BC \\plus{} CD \\plus{} AD$.", "Solution_3": "[quote=\"Goxer\"]Consider a cube on $ R^3$, and $ A,B,C,D \\in$ cube.[/quote]\r\n\r\nI'm not sure what you mean. I originally interpreted $ (A, B, C, D) \\in \\mathbb{R}^4$. Do you mean that $ A, B, C, D$ are points in $ \\mathbb{R}^3$? Does $ AB$ denote the dot product?", "Solution_4": "$ A,B,C,D$ are point in $ R^3$.\r\n$ AB$ is the euclidean distance.", "Solution_5": "For your example: the maximum occurs when $ A\\equal{}C$ and $ B\\equal{}D$, and these pairs are opposite corners of the cube.", "Solution_6": "Hmm. Well, I'll tell you what I originally meant.\r\n\r\n[b]Proposition:[/b] Let $ f : \\mathbb{R}^n \\to \\mathbb{R}$ be a continuous and convex function defined on a hypercube (or a cube, or a square, or an interval). Then $ f(\\vec{v})$ attains its maximum when $ \\vec{v}$ is some vertex of the hypercube (etc.).\r\n\r\nThis problem can be interpreted as a function $ f : \\mathbb{R}^{12} \\to \\mathbb{R}$. The vertices of the $ 12$-dimensional hypercube correspond to each of $ A, B, C, D$ being vertices of a cube in $ \\mathbb{R}^3$, so the original conclusion should still hold as long as $ f$ can be shown to be convex in all $ 12$ variables.", "Solution_7": "Then the answer initially given by [b]t0rajir0u[/b] still applies. With $ A,B,C$ fixed, you have a convex function of $ D$, which attains its maximum at an extreme point of the cube, that is, a vertex. And so on.", "Solution_8": "Thank you very much." } { "Tag": [ "geometry", "3D geometry", "probability", "number theory", "least common multiple", "function" ], "Problem": "I'm sorry to tell you that I didn't have time to grade or post any solutions yet.\r\n\r\nYou guys can discuss it though. I think Drunken Math and 236factorial were only ones who sent the answers but it'll be good discussion for you guys to have. :) \r\n\r\nI'll post the results and solutions ASAP.", "Solution_1": "It looks like i lost the original copy, so you'll have to read a PDF :D\r\n\r\n[hide=\"warning\"]some are wrong :D (like 10 i think) [/hide]", "Solution_2": "Results should be posted soon but I can't give anyone exact date. I have practices everyday and since I'm sophomore, I have more school work to do than last year (and I'm taking higher classes than my age) so be patient (I know everybody is and I appreciate this greatly) and I'll grade them ASAP.\r\n\r\nThanks.", "Solution_3": "Here are my answers (from the PM I sent Silverfalcon):\r\n\r\n[hide]1.) Figuring out the first few terms of the sequence,\n\n3, 5, 9, 17, 33, 65, 129...\n\nThere's a pattern on the units digits of the terms of this sequence; it goes 3, 5, 9, 7 repeatedly.\n\nSince 20 is divisible by 4, the units digit of 20 must be $\\boxed{7}$.\n\n2.) $h(4)=(2*4)(4+3)=8*7=\\boxed{56}$\n\n3.) To maximize the sum, we must make abc have the largest possible value that satisfies the conditions. Since there are no conditions restricting the value of b, we make the largest we can make it: 9. a and c can't be equal to each other and must be smaller than c, so the largest possible values of those are 8 and 7. It doesn't matter what order they are in, since the brother number is the number flipped. So the largest sum of $abc$ and $cba$ is $897+798=\\boxed{1695}$\n\n4.) Set T has these numbers: 66, 55, 44, 33, 22, and 11, all in base 7. Converting them into base 10 would yield Set R. Set R has these numbers: 8, 16, 24, 32, 40 and 48. Out of those 6 numbers, the only two that are a perfect square or a perfect cube are 8, and 16. So the probability would be $\\frac{2}{6}=\\boxed{\\frac{1}{3}}$\n\n5.) Counting the number of possibilities, the number of possibilitites is:\n\n$(6+5+4+3+2+1)+(5+4+3+2+1)+(4+3+2+1)+(3+2+1)+(2+1)+1=\\boxed{56}$\n\n6.) $n+n+1+n+2=3n+3=3(n+1)$ The LCM of 3 and 4 is 12, so we must make $n+1=4$. So $\\boxed{n=3}$\n\n7.) If you try out different digits for the thousands digit, you will find that there are $\\boxed{6}$ different numbers that satisfy the condition.\n\n8.) Do notice that if you divide $12^1$ by 13, the remainder is 12. If you divide $12^2$ by 13, the remainder is 1. If you proceed with casework, you find that this is a neverending pattern, in which if the exponent is odd, the remainder would be 12, and if the exponent is even, the remainder is 1. Since 12 is odd, the remainder would be $\\boxed{1}$\n\n9.)\n\n10.) The largest circle that can be inscribed in a unit square has a radius of $\\frac{1}{2}$, so the area of that circle would be $\\boxed{\\frac{\\pi}{4}}$[/hide]", "Solution_4": "[quote=\"Drunken_Math\"]10.) The largest circle that can be inscribed in a unit square has a radius of $\\frac{1}{2}$, so the area of that circle would be $\\boxed{\\frac{\\pi}{4}}$[/quote]\r\n\r\nI knew i should've put that... I got tricked :D or did I? :?", "Solution_5": "Since you guys seem to be anxious about your score, I'll post the scores first. Both of you qualified!!\r\n\r\nDrunken_Math: 9 correct\r\n263factorial: 8 correct\r\n\r\nSince the solution is pretty long and important, I'll post it on the next post.", "Solution_6": "Here is a solution for this COPSD 3rd A.\r\n\r\n[hide=\"Solution 1\"]\nThere are two methods for this one.\n\nMethod 1:\n\n$t_1 = 3$ and $t_2 = 5$. Therefore:\n\n$t_3 = 3(5) - 2(3) = 9\\\\\nt_4 = 3(9) - 2(5) = 17\\\\\nt_5 = 3(17) - 2(9) = 33\\\\\nt_6 = 3(33) - 2(17) = 65\\\\\nt_7 = 3(65) - 2(33) = 129$ \n\nThus, the last digit repeats in every 4 terms. The last digit of $t_{20}$ is then 7. \n\nMethod 2:\n\n[b]Lemma: For $n>2$ in $t_n$, $t_n = 2^n + 1$ [/b]\n\nProof: For $n = 3$ (Let's start there although $n = 1,2$ work), we get $t_3 = 2^3+1 = 9$ which is the same number from Method 1. Now, assume for some integer $k$, we have $t_k = 2^k + 1$. Hence, $t_{k+1}$ will be:\n\n\\[{t_{k+1} = 3(2^k + 1) - 2(2^{k-1} + 1)\\\\\n= 3+3(2^k} - 2^k - 2\\\\\n= 2^{k+1} + 1\\]\n\nwhich implies $t_k = 2^k + 1$ works for all natural $k$ values. Thus, $t_{20} = 2^{20} + 1 = 1048577$ and 7 is indeed the last digit. [/hide]\n\n[hide=\"Solution 2\"]\n\nFirst, put 4 into each function:\n\n$f(4) = 2 \\cdot 4 = 8\\\\\ng(4) = 4+3 = 7$\n\n$h(4)$ is simply product of those two which is 56. [/hide]\n\n[hide=\"Solution 3\"]\n\nTo maximize $abc+cba$, we note that this equals:\n\n$10^2(a+c) + 10^1(2b) + 10^0(a+c)$\n\nThus, $a+c$ has to be maximum in order for whole thing to be maximized. Since $a \\neq c$ and $b > a,c,$ we can have $a = 8$ or $c = 8$ for maximum (which means $b = 9$!). So, $10^2(a+c) + 10^1(2b) + 10^0(a+c) = 15(100) + 18(10) + 15 = 1695$. [/hide]\n\n[hide=\"Solution 4\"]\n\nLet's simply say numbers that have identical unit and tens digit as palindromes. Six palindromes in base 7, or elements of set $T$, are 11,22,33,44,55,66. Now, when we convert them to base 10:\n\n$11_7 = (1 \\cdot 7^1 + 1 \\cdot 7^0)_{10} = 8_{10}\\\\\n22_7 = (2 \\cdot 7^1 + 2 \\cdot 7^0)_{10} = 16_{10}\\\\\n33_7 = (3 \\cdot 7^1 + 3 \\cdot 7^0)_{10} = 24_{10}\\\\\n44_7 = (4 \\cdot 7^1 + 4 \\cdot 7^0)_{10} = 32_{10}\\\\\n55_7 = (5 \\cdot 7^1 + 5 \\cdot 7^0)_{10} = 40_{10}\\\\\n66_7 = (6 \\cdot 7^1 + 6 \\cdot 7^0)_{10} = 48_{10}$\n\nOnly 2 are square or cube so the probability is $\\frac{2}{6} = \\frac{1}{3}$. [/hide]\n\n[hide=\"Solution 5\"]\n\nThis problem is same as finding number of ways put $( \\star )$ in $( * )$ spaces where $( \\blacksquare )$ are apples arranged in a way such that no two bananas (or $\\star$ ) can be next to each other.\n\n$* \\blacksquare * \\blacksquare * \\blacksquare * \\blacksquare * \\blacksquare * \\blacksquare * \\blacksquare *$\n\nThere are 8 $*$'s and 3 $\\star$ we need. There are $\\binom{8}{3} = 56$. [/hide]\n\n[hide=\"Solution 6\"]\n\n$n+n+1+n+2 = 3n+3$. In base 4, this equation is $3n + 3 \\equiv 0 \\equiv 4 \\equiv 8 \\cdots \\mod 4$. Ths smallest $n$ is if $3n+3$ or $3(n+1)$ is divisible by 4, and obviously, at $n = 3$. [/hide]\n\n[hide=\"Solution 7\"]\n\nTo be divisible by 15, it must be divisible by 5 AND 3 as well. So, we have two cases:\n\nCase I: A790\n\nSince $3|A+7+9$, $A$ can be 2,5,8.\n\nCase II: A795\n\nSince $3|A+7+9+5$, $A$ can be 3 (not 0 because $A$ is the first digit!), 6, 9.\n\nHence, there are 6 $A$ values or such numbers. [/hide]\n\n[hide=\"Solution 8\"]\n\n$12^{12} \\equiv (-1)^{12} \\mod 13\\\\\n\\equiv 1^{12} \\mod 13\\\\\n\\equiv 1 \\mod 13$\n\nThus, the remainder is 1. You can also solve this by seeing that remainder alternates between 1 and 12. [/hide]\n\n[hide=\"Solution 9\"]\n\n$pq = \\frac{p}{q}\\\\\nq = \\frac{1}{q}\\\\\nq^2 = 1\\\\\nq = \\pm 1$\n\nFor $q = 1$, we get $p-1 = p$ and that implies $-1 = 0$ which is false. For $q = -1$, we get $p+1 = -p$ which implies $p = \\frac{-1}{2}$. So, $k = -1 \\cdot \\frac{-1}{2} = \\frac{1}{2}$. [/hide]\n\n[hide=\"Solution 10\"]\n\nThe largest square is one that contains the 4 midpoints of four sides of square (draw it and you'll see it!). Thus, the inscribed circle has radius of $\\frac{1}{2}$, and it will have area of $\\frac{\\pi}{4}$. [/hide]", "Solution_7": "Number 7.... forgot could end with 0... so dumb... I suck at math :D", "Solution_8": "[quote=\"236factorial\"]Number 7.... forgot could end with 0... so dumb... I suck at math :D[/quote]\r\n\r\nDon't worry, everybody makes their share of careless mistakes. :)", "Solution_9": "And you... skipped number 9?! :o \r\n\r\nIt's better to try and get it wrong than to not try at all.. Santa thinks you're naughty :rotfl: \r\n\r\nOf course, that means everyone is naughty :D", "Solution_10": "$p-q = pq =\\frac{p}{q} = k$\r\n\r\nHere was my reasoning for this problem:\r\n\r\n$pq=\\frac{p}{q}$\r\n\r\n$pq^2=p$\r\n\r\n$q^2=1$\r\n\r\n$q=1$\r\n\r\nI forgot to make it $\\pm1$\r\n\r\nAs I said, everyone makes their share of careless mistakes. :)", "Solution_11": "I'm glad to see some kind of talk is going on.\r\n\r\nI wasn't sure if all my hard work typing the solutions attracted nobody. :P" } { "Tag": [ "geometry", "3D geometry", "dodecahedron", "pigeonhole principle" ], "Problem": "prove that choosing any 5 vertexes of a regular dodecahedron three of them will form an isosceles triangle", "Solution_1": "[hide]From any 5 points, there are $ \\binom52\\equal{}10$ lengths. However, there are only 6 different lengths for diagonals possible, so thus, by the pigeonhole principle, at least two of them must be the same length and thus, an isosceles triangle can be formed by choosing the vertices used to create those two sides.[/hide]", "Solution_2": "[quote=\"b-flat\"][hide]From any 5 points, there are $ \\binom52 \\equal{} 10$ lengths. However, there are only 6 different lengths for diagonals possible, so thus, by the pigeonhole principle, at least two of them must be the same length and thus, an isosceles triangle can be formed by choosing the vertices used to create those two sides.[/hide][/quote]\r\n\r\nactually there are 7 different lengths possible, because the sides of the triangle don't have to be diagonals. but the results is the same anyway" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Solve the equation:\r\n$ \\sqrt{x \\minus{} 3} \\plus{} \\sqrt{7 \\minus{} x} \\equal{} \\minus{}x^{2} \\plus{} 6x \\minus{} 7$", "Solution_1": "here is my solution:\r\n$ \\minus{}x^2\\plus{}6x\\minus{}7<\\equal{}2$ and $ \\sqrt {x \\minus{} 3} \\plus{} \\sqrt {7 \\minus{} x} > \\equal{} \\sqrt{4} \\equal{} 2$ so x=3 is the solution", "Solution_2": "[quote=\"Nam Luu\"]Solve the equation:\n$ \\sqrt {x \\minus{} 3} \\plus{} \\sqrt {7 \\minus{} x} \\equal{} \\minus{} x^{2} \\plus{} 6x \\minus{} 7$[/quote]\r\n\r\nPut $ y\\equal{}x\\minus{}3$ the equation $ \\Leftrightarrow \\sqrt{y}\\plus{}\\sqrt{4\\minus{}y}\\equal{}2\\minus{}y^2$\r\n\r\nNow $ LHS^2 \\ge 4 \\Rightarrow LHS \\ge 2 \\ge RHS$\r\n\r\nEquality holds $ \\Rightarrow y\\equal{}0 or 4$ (rejected)\r\n\r\nSo $ y\\equal{}0 \\Rightarrow x\\equal{}3$ is the only solution after checking." } { "Tag": [ "geometry", "circumcircle", "geometry proposed" ], "Problem": "Let $XOY$ be an angle formed by lines $OX,OY$.Let$P$ be a point, inside the angle.\r\nFind the line $\\ell$ such that it passes through $P$ and meets $OX,OY$ in $A,B$ such that $AP*PB$ be minimum.", "Solution_1": "I believe it's the line passing through $P$ which makes equal angles with $OX,OY$. In other words, $OAB$ should be isosceles.\r\n\r\nTake $AB$ as above (in order to make $OAB$ isosceles), and let $A'B',\\ A'\\in OX,B'\\in OY$ be some other line through $P$. We want to show that $PA'\\cdot PB'\\ge PA\\cdot PB$. \r\n\r\nTake $A''\\in OX,B''\\in OY$ so that $\\angle A''PA=\\angle A'PA=\\angle B''PB=\\angle B'PB$ ($A'',A'$ are separated by $A$, and $B'',B'$ are separated by $B$). It's easy to see that the quadrilateral $A''A'B''B'$ is cyclic, and let its circumcircle cut $(PA,(PB$ in $A_1,B_1$ respectively. Then $PA_1\\ge PA,PB_1\\ge PB\\Rightarrow PA'\\cdot PB'=PA_1\\cdot PB_1\\ge PA\\cdot PB$, as required.", "Solution_2": "Exactly grobber,here is a another solution. which i really like: :D \r\nLet $ABO$ be an isosceles trinagle such that $P \\in AB$.\r\nLet $C$ be a circle ,inside the $XOY$ and tangent to $OX,OY$ at $A,B$.\r\nWe claim that $PA*PB$ is the minimum.(note that It is power of $P$ wrt $C$).\r\nDraw another line which intersects $OX,OY$ in $R,S$.\r\nClearly we have $RP*SP \\ge$ power of $P$ wrt $C$.\r\nProblem solved.\r\n\r\nPS.Its a reall pitty that you guys dont have goemetry in university. :(" } { "Tag": [ "number theory", "greatest common divisor", "probability", "relatively prime", "number theory proposed" ], "Problem": "What is the probability that a random 2 digit base b number will be relativly prime with its digit reversal?", "Solution_1": "By direct calculation there are 41 two-digit integers relatively prime to their reversal, counting 10. Thus, if you wish to include integers ending in 0 (whose reversal has only one digit), the answer is $ \\frac{41}{90}$. If you want to exclude these numbers, the answer is $ \\frac{40}{81}$.\r\n\r\nThe numbers which are not relatively prime to their reversals can be characterized as:\r\n1) Numbers with both digits even\r\n2) Numbers in which the first and last digit are both 5 or 0\r\n3) Palindromes\r\n4) Multiples of 11\r\n5) Multiples of 3\r\n6) 70\r\n\r\nThere is substantial overlap among these categories for 2-digit numbers (e.g. (3) and (4) are actually the same), but for higher-digit numbers each group will be distinct and this characterization may no longer be comprehensive.\r\n\r\nEdit: in the 3-digit case, excluding numbers which end in 0, the first five sets form a complete characterization with the exception of the numbers 259 and 952 which have a GCD of 7. In higher-digit cases, the number of neq categories needed will increase. In fact, each prime will have some list associated to it of pairs of integers which are reversals of each other and divisible by that prime, and it is most likely not straightforward to characterize these lists as it is for 2, 3, 5 and 11.", "Solution_2": "Thank you for the reply. I wrote a python script to find the probability for higher numbers of digits(Up to 10000), and it seems to hover around .43 . That is odd, as your characterization of numbers that are relatively prime with there reversal almost suggests something more intricate. I guess I'll have to study the problem further." } { "Tag": [ "probability", "function", "integration", "real analysis", "real analysis unsolved" ], "Problem": "Consider the set $F$ of functions defined as follows $f(0)=0,f(x)=f(i)+\\alpha_i(x-i)$ for $i \\leq x \\leq i+1, 0 \\leq i \\leq n-1 $ and $\\alpha_i$ takes values $-1$ or $1$. Find the probability that for arbitrary taken function $f$ from the set $F$\r\na).$\\int_0^n f(x)dx=0$.\r\nb). the function $f$ has $k$ roots on $(0,n]$", "Solution_1": "a) and b) simultaneously?", "Solution_2": "[quote=\"pleurestique\"]a) and b) simultaneously?[/quote]\r\nnot, a) is one problem and b) is another." } { "Tag": [], "Problem": "There is a strange island where people who can climb trees always tell\r\nthe truth, and people who couldn't be bothered always lie. You go up\r\nto four people on that island and ask the first person, \"Can you climb\r\ntrees?\" The first person nodded. \"Yes,\" he said. You go to the second\r\nperson and ask him. He answers, \"I can't climb trees.\" You go to the\r\nthird person, who says, \"As a fact, I can tell you that the first and\r\nsecond person are both liars. I can climb trees.\" The fourth person\r\ntells you, \"I can climb trees, and I can tell you that the third\r\nperson is a liar.\"\r\n\r\nOne of these people gave an impossible answer, given the information\r\nin the first sentence. Who was it?\r\n\r\n\r\nApart from the person who gave you the impossible answer, tell which\r\nperson is a truth-teller and which one is a liar.", "Solution_1": "I'm not sure this is right, but the third person gave the impossible answer because in that island you can't lie. Uh...I THINK the third person is a lier and the 4th is the truth-teller.\r\n\r\nI know I'm wrong. (I think, who knows!)", "Solution_2": "That is incorrect. People who [b]don't[/b] climb trees always lie. I've leave this up for a week or so, and if no one else replies, I'll reveal the answer.", "Solution_3": "The second person gave an impossible reply. If he was a liar, he would say he could climb a tree and if he was a truth teller, he would still say he could climb a tree.\r\n\r\nThird person is a liar and the first and fourth person are truth tellers.", "Solution_4": "That's right! :first:", "Solution_5": "Here is another problem. It's similar, but also tricky.\r\n\r\nA princess visits an island inhabited by two tribes. Members of one tribe always tell the truth, and members of the other tribe always lie.\r\n\r\nThe princess comes to a fork in the road. She needs to know which road leads to the castle so as to avoid the fire-breathing dragon and rescue the prince from the wizard holding him captive in the castle. (Although the princess doesn't know it, the south road leads to the castle and the north road leads to the dragon.)\r\n\r\nStanding at this fork in the road is a member of each tribe, but the princess can't tell which tribe each belongs to. What question should she ask to find the road to the castle?", "Solution_6": "[quote=\"isabella2296\"]Here is another problem. It's similar, but also tricky.\n\nA princess visits an island inhabited by two tribes. Members of one tribe always tell the truth, and members of the other tribe always lie.\n\nThe princess comes to a fork in the road. She needs to know which road leads to the castle so as to avoid the fire-breathing dragon and rescue the prince from the wizard holding him captive in the castle. (Although the princess doesn't know it, the south road leads to the castle and the north road leads to the dragon.)\n\nStanding at this fork in the road is a member of each tribe, but the princess can't tell which tribe each belongs to. What question should she ask to find the road to the castle?[/quote]\r\n\r\n\"Show me where you come from.\" :D", "Solution_7": "Well...I guess that could work but it's not really a question...\r\n\r\nBy the way, there are a couple answers that could work. I'll leave this up for a while and if no one else offers any answers I'll give one.", "Solution_8": "It's a classic problem. The princess should ask [hide]\"If I ask the other islander which is the right road to the castle, what will he answer?[/hide]\r\nI don't know any other possible solution, so I'm interested in the second answer.\r\n\r\nGoodbye\r\nGM", "Solution_9": "Since the question doesn't limit you to one question, you could ask something like \"Is the sky blue?\" and then ask which way to go.", "Solution_10": "[quote=\"isabella2296\"]Since the question doesn't limit you to one question, you could ask something like \"Is the sky blue?\" and then ask which way to go.[/quote]\r\n\r\nOf course, the sky is not blue, it just appears blue :lol:", "Solution_11": "Then let me rephrase that: \"Does the sky appear blue?\"\r\n\r\nBetter? :D", "Solution_12": "Of course, I'm from the liar tribe. :D", "Solution_13": ":) Ha ha. Very funny.", "Solution_14": "Usually the puzzle says \"one question only\", and in that case the \"is the sky blue\" scheme doesn't work.", "Solution_15": "I guess. Well, I'll post another logic problem.\r\n\r\nThere are three people in front of you. You know that:\r\n\r\nOne of them is God. He knows everything, and always tells the truth.\r\nOne of them is the Devil. He also knows everything, but lies. \r\nThe third person knows nothing, but answers questions as if he knows \r\nthe answers. His answers, however, are completely useless and could be \r\nright or wrong.\r\n\r\nYou can ask a total of three questions that can correctly be answered \r\nwith yes or no, each to one of the persons. You may choose whom to ask.\r\n\r\nDetermine who is who...", "Solution_16": "I'm not sure...\r\n\r\n[hide=\"answer\"]Ask: \"If i asked you yesterday if (a certain one of the three) was the third person, would you have answered yes?\"\n\nThe Devil will be forced to tell the truth because he would have said a lie yesterday; he has to lie about that.\n\nIf at least two of them say yes, ask the trademork \"Does the sky appear blue?\" and we shall get who is who now that we have narrowed it down to God and the Devil.\n\nIf at least two of them say no (With the first question, God and the Devil must agree), we repeat the question on one of the other people...\"If i asked you yesterday if (a another one of the three) was the third person, would you have answered yes?\"\nWhat the majority answers will be correct. Then we go back to the \"obvious\" question stage[/hide]", "Solution_17": "Yes, that's correct. Congrats, I didn't get that one on my own. :10:", "Solution_18": "the third person gave an impossible answer...he is a liar and the fourth person is telling the truth", "Solution_19": "2nd person lied?" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "1. Find every pair of distinct positive integers $ a,b$ such that $ b^2 \\plus{} a | a^2 \\plus{} b$ and there exists prime $ p$ and power $ k \\geq 1$ such that $ b^2 \\plus{} a \\equal{} p^k$.\r\n\r\n2. Does there exist a pair of distinct positive integers $ a,b$ such that $ (b^2 \\plus{} a \\minus{} 1) | (a^2 \\plus{} b \\minus{} 1)$ there exists prime $ p$ and power $ k \\geq 1$ such that $ b^2 \\plus{} a \\equal{} p^k$?", "Solution_1": "$ a\\equal{}p^k\\minus{}b^2$ and $ p^k|(p^k\\minus{}b^2)^2\\plus{}b\\to p^k|b(b^3\\plus{}1)$. If $ p|b$, then $ p\\not|b^3\\plus{}1\\to p^k|b$ contradition with $ p^k>b^2$.\r\nTherefore $ p^k|b^3\\plus{}1\\equal{}(b\\plus{}1)(b^2\\minus{}b\\plus{}1)$. Obviosly $ (b\\plus{}1,b^2\\minus{}b\\plus{}1)\\equal{}(b\\plus{}1,3)$. It give $ p\\equal{}3,b^2\\minus{}b\\plus{}1\\equal{}3^{k\\minus{}1}.$\r\nOnly $ b\\equal{}2,k\\equal{}2,p\\equal{}3,a\\equal{}5$ is solution.", "Solution_2": "[quote=\"Mashimaru\"]1. Find every pair of distinct positive integers $ a,b$ such that $ b^2 \\plus{} a | a^2 \\plus{} b$ and there exists prime $ p$ and power $ k \\geq 1$ such that $ b^2 \\plus{} a \\equal{} p^k$.\n\n2. Does there exist a pair of distinct positive integers $ a,b$ such that $ (b^2 \\plus{} a \\minus{} 1) | (a^2 \\plus{} b \\minus{} 1)$ there exists prime $ p$ and power $ k \\geq 1$ such that $ b^2 \\plus{} a \\equal{} p^k$?[/quote]\r\n\r\nThis is problem 3 of St. Petersburg City Mathematical Oympiad \r\nLooking for problem above, you can find solutions. :roll:", "Solution_3": "[quote=\"Rust\"]...\nTherefore $ p^k|b^3 \\plus{} 1 \\equal{} (b \\plus{} 1)(b^2 \\minus{} b \\plus{} 1)$. ...$ b^2 \\minus{} b \\plus{} 1 \\equal{} 3^{k \\minus{} 1}.$\n...[/quote]\r\n\r\nCould you please explain this step? :(\r\nI get p=3. then $ 3^{k \\minus{} 1}|b^2 \\minus{} b \\plus{} 1$, then..??", "Solution_4": "use $ b^2<3^k$ and $ 3^{k\\minus{}1}| b^2\\minus{}b\\plus{}1\\equal{}1\\mod 2\\to b^2\\minus{}b\\plus{}1\\equal{}3^{k\\minus{}1}$." } { "Tag": [ "modular arithmetic" ], "Problem": "By the expansion of a natural n, in basis 8 , find the criteria of divisibility for 7", "Solution_1": "[hide]http://www.divisibilitybyseven.mat.br/[/hide]", "Solution_2": "A number in base 8 can be written $d_0+8d_1+8^2d_2+\\cdots +8^nd_n$, where $n$ is some nonnegative integer and $0\\leq d_k\\leq 7$ for each $k$.\r\n\r\nSince $8^k\\equiv 1\\pmod{7}$ for all $k$, this number is congruent to $d_0+d_1+d_2+\\cdots +d_n$. So the number is divisible by 7 if and only if the sum of its base 8 digits is.", "Solution_3": "In general, a number is divisible by $n$ iff the sum of its digits in base $n + 1$ is divisible by $n$.", "Solution_4": "And as a sidenote, you can also use the fact that $n \\equiv -1 \\pmod{n+1}$ to find divisibility by n+1. For example, is 121 in base 10 divisible by 11? $121 \\equiv 1 \\times 10^2 + 2 \\times 10 + 1 \\equiv 1 \\times (-1)^2 + 2 \\times (-1) + 1$\r\n$\\equiv 1-2+1 \\equiv 0 \\pmod{11}$", "Solution_5": "And on the divisibility rules lists I learned in elementary/middle school, it always said there was no general method like the other numbers to ten...", "Solution_6": "[quote=\"quantum leap\"]And on the divisibility rules lists I learned in elementary/middle school, it always said there was no general method like the other numbers to ten...[/quote]\r\n\r\nWhat do you mean? In base $10$, there are the following methods.\r\n\r\n- Methods that require the last $k$ digits for some $k$: All numbers of the form $2^a 5^b$, with $a, b \\in \\mathbb{N} \\cup \\{ 0 \\}$. Most popularly, $2, 4, 5, 8, 10$.\r\n\r\n- Methods that involve an easy summation of all digits: $3, 9, 11$.\r\n\r\n- Methods that involve a difficult summation of all digits: $\\{ n | (n, 10) = 1 \\}$ (it's just a matter of how hard you want to make your job ;) )\r\n\r\nDemonstration: Everyone says that there is no divisibility rule for $7$. In fact, there is, but it is annoying to use:\r\n\r\nWe will denote by $P_p = $ the pattern of numbers by which you multiply the digits of some number to figure out if it is divisible by $p$ (starting from the ones digit). For example,\r\n\r\n$P_3 = P_9 = <1, 1, 1, ... >$\r\n$P_{11} = <1, -1, 1, -1, ... >$\r\n\r\nNow,\r\n\r\n$P_7 = <1, 3, 2, 1, 3, 2, ... >$\r\n\r\nIt is not easy to use, but it works. In fact, it is trivial to prove\r\n\r\n$P_p = <1, b \\bmod p, b^2 \\bmod p, b^3 \\bmod p, ... >$\r\n\r\nIn some base $b$. This sequence terminates to zero iff $\\exists k : p | b^k$, repeats the $1$ term iff $(b, p) = 1$, and repeats without a $1$ term otherwise." } { "Tag": [ "calculus", "derivative" ], "Problem": "The cutoff this year was a 27, with a max score of 37, so the scores were lower than last year's, although not by much. Personally, I felt that the exam was significantly harder this year, since anatomy is probably my weakest subject. Although I made the cut, I failed to even break 30 -- my score dropped by 9 points compared to last year.\r\n\r\nGood luck to all who will be taking the semifinal exam!", "Solution_1": "wat was the maximum marks?? :maybe:", "Solution_2": "have you given exam papers?", "Solution_3": "i think u meant \"r u given exam papers\" :D", "Solution_4": "i feel so guilty.\r\n\r\ni qualified with a 27.", "Solution_5": "kef ellem :D :P", "Solution_6": "[quote=\"unimpossible\"]i feel so guilty.\n\ni qualified with a 27.[/quote]\r\n\r\n:wallbash missed it with 26", "Solution_7": "Made it with a 35! Without doing eco or biodiversity...time to get to work.", "Solution_8": "what was the maximum marks??\r\ni mean how much mark exam it was?? :)", "Solution_9": "It was out of 50, the highest score in the nation being 37.", "Solution_10": "Thanks..\r\n\r\nand could u guys post some questions ,in free time?? :)", "Solution_11": "ugh sure? took it today it was INHUMAN ZOMG. fail.", "Solution_12": "I thought it was okay. I'm hoping for 120 points.", "Solution_13": "Does anyone else think this exam is flawed? Honestly, the highest score was 37 out of 50. The person with the most knowledge of biology missed 13 questions on a multiple choice test. \r\n\r\nI got a 22, literally not knowing any question, and using a little bit of reasoning.", "Solution_14": "Imho, it could be designed better, but that's just the nature of the beast. The fact that the best score was 37 just means that the test is hard, not necessarily that it is flawed. Differentiation might be somewhat of an issue, but more difficult questions simulate the multiple-choice questions from semis more closely." } { "Tag": [ "quadratics", "algorithm", "number theory proposed", "number theory" ], "Problem": "Do there exist a method to solve the following one equation without using theory of Pell's equation?\r\n\r\nSolve in positive integers : $ 3q^2\\minus{}p^2\\equal{}2$ if $ p$and$ q$ are both odd.", "Solution_1": "Define \"without the theory of Pell's equation.\" One way to derive the Pell's method is to use continued fractions, which is [i]technically[/i] an answer to your question. In particular, one computes the value of $ 3q_n^2 \\minus{} p_n^2$ for convergents $ \\frac{p_n}{q_n}$ of $ \\sqrt{3}$ and checks if they are equal to $ 2$. It is known that this sequence is always finite for any quadratic surd, so this algorithm terminates. Every solution appears in this way.", "Solution_2": "Could You explain in more details about what is Your method about? I would be really glad if anyone could show me how to kill this equation. Or maybe You know where is an article of method described above? thanks for interests", "Solution_3": "Briefly: the convergents $ \\frac{p_n}{q_n}$ have the property that $ \\left| \\sqrt{3} \\minus{} \\frac{p_n}{q_n} \\right| < \\frac{1}{q_n^2}$, which can be used to show that $ \\left| 3q_n^2 \\minus{} p_n^2 \\right| < 2 \\sqrt{3}$ (and this generalizes). Other lemmas on continued fractions can be used to show that this sequence is in fact periodic, and what's more has the property that every solution to $ 3q_n^2 \\minus{} p_n^2 \\equal{} k$ appears for any integer $ |k| < 2 \\sqrt{3}$ (this is because all such numbers form a quotient close enough to $ \\sqrt{3}$ that it must be a convergent). This is not true when $ |k| > 2 \\sqrt{3}$ - the question is much harder in this regime, but ZetaX knows more than I do here." } { "Tag": [ "FTW", "email" ], "Problem": ":) \r\nName: V V V Satyanarayana(Vcube)\r\nresidence: Hyderabad\r\n hello sir \r\n I want to intract with in maths", "Solution_1": ":) \r\nName:George W Bush(dubya)\r\nresidence: Texas\r\nhello sir\r\nI want to intract with in iraqi oil\r\n\r\n[hide]use the welcome to india thread for intro :spam: [/hide]", "Solution_2": "Fikar not V^3\r\nThey had term everything as spam :D \r\nWelcome!\r\n :welcomeani: Which class?", "Solution_3": "[quote=\"Scary math\"]They had term everything as spam :D [/quote]\r\nu forgot 2 mention \r\nSPAMMING IS ALLOWED HERE, EVERYONE SPAMS, EVEN THE MOD :rotfl:", "Solution_4": "@Sagar\r\nAbt spamming\r\nthere are sections of AoPS where spamming is strictly forbidden,like the FTW forum.Over there,even a wrong solution is considered a spam!!!! :( \r\n@V^3\r\nThis is a wried and wonderful place for lost souls.Enjoy!!!!! :thumbup:", "Solution_5": "@scary\r\ni was talking of the india section only :)", "Solution_6": "hello sir??? :rotfl: :rotfl: :rotfl:", "Solution_7": "Hi! I spam.", "Solution_8": "King of spammers is here!!! :D \r\nName:James Bond \r\nresidence:varies\r\n\r\nFound :Top secret info on ur computer.Run!!coz i m pursuing u", "Solution_9": "@ritu\r\nBond's a spammer???????????? :huh:", "Solution_10": "course he is!\r\nif he can get a royal flush in every hand in casino royale , y the heck cant he be a spammer? :P \r\nAnyway here goes \r\nName: Ajmal Amir Kasab (A2K)\r\nresidence: Arthur road Jail\r\n HELLO SIR\r\n[img]http://pics.shareordie.in/2008/12/Mohammed_Ajmal_Amir_Kasab.jpg[/img]\r\nI want to intract with in terrorism :ninja: :idea:", "Solution_11": "I think you guys scared V3 away.... :rotfl:", "Solution_12": "My intro\r\nName-Rob Bourdon\r\nResidence-Agoura hills\r\nI wish to interact in rock++", "Solution_13": "[quote=\"franklampard8\"]course he is!\nif he can get a royal flush in every hand in casino royale , y the heck cant he be a spammer? :P \nAnyway here goes \nName: Ajmal Amir Kasab (A2K)\nresidence: Arthur road Jail\n HELLO SIR\n[img]http://pics.shareordie.in/2008/12/Mohammed_Ajmal_Amir_Kasab.jpg[/img]\nI want to intract with in terrorism :ninja: :idea:[/quote]\r\n!! :roftl: \r\nA really frightening joke ... :ninja:\r\nBTW; \r\nName: Sayan Mukherjee\r\nAge : 13\r\nFrom: Kolkata\r\nNice to meet you. :P", "Solution_14": "@potla\r\nTui terrorist na rock star??? :rotfl:\r\n@franky\r\nForgot to mention,some girl in our class is a die-hard kasab fan. :rotfl: :rotfl: \r\nGuess she will believe me if I tell her he's at AoPS.", "Solution_15": "[quote=\"Scary math\"]Guess she will believe me if I tell her he's at AoPS.[/quote]\r\nYeah he has some are of 'problem solving' :P Just KILL 'em and the probs over :D", "Solution_16": "[quote=\"Scary math\"]Forgot to mention,some girl in our class is a die-hard kasab fan. :rotfl: :rotfl: \nGuess she will believe me if I tell her he's at AoPS.[/quote]\r\n\r\neither the girl is mentally ill or dis is a sick joke :wallbash_red: \r\nask her if she wud feel the same when she wud have the barrel of a fully loaded ak47 shoved up her pu*sy :ninja: \r\n\r\ni always had believed commies are anti-nationals :P", "Solution_17": "@Sagar\r\nWell,for ur info,I am the only one what u call by anti nationalist in my class, :P \r\nEarlier,her heartthrob was Shoaib Malik,till a few months ago,it was Kasab,I am not sure of the current one. :rotfl: :D", "Solution_18": "dat girl really is sick\r\nEDIT: abusive language", "Solution_19": "@sagar\r\nStrange the mod hasnt kicked u out yet :mad: \r\nBetter stop using such words,and she isnt sick,she is too childish. :furious:", "Solution_20": "@Scary math I was just viewing the thread.I totally agree with you.Moderator should take action.It can not be tolerated in mathlinks.", "Solution_21": "ok sorry :( \r\ni just lost my temper when i read that the girl has a crush on kasab\r\ni still refuse to believe it and i think scary is saying such things just to stand out in the crowd :P \r\n\r\n[hide]\nedited one post, cant edit the previous one\nmods please delete it :roll: \n[/hide]", "Solution_22": "Man,I am serious! :mad: \r\nWhy would I be makin it up?\r\nBetter appreciate the fact that some wriedos do have soft corners for these brainwashed,heartless creatures. :P :rotfl:", "Solution_23": "What is their to appreciate in it?Its actually funny that such weird creatures exist in India :rotfl:", "Solution_24": "Well,dat doesnt make her eligible for abusive language :mad: \r\n\r\nV^3 realised India forum is a spammer's paradise,he has stopped visiting his own intro!", "Solution_25": "chill guys.... \r\nsagar whats your email id?", "Solution_26": "@frank\r\nvaze.sagar at gmail dot com\r\nsagarvaze at rediff dot com" } { "Tag": [], "Problem": "SIGNUPS:\r\n\r\n1. Joe10112 \r\n2. r15s11z55y89w21 \r\n3. dragon96 \r\n4. BOGTRO \r\n5. vahalla \r\n6. ernie \r\n7. auroraak \r\n8. bluecarneal \r\n9. Bugi \r\n10.mz94\r\n\r\nRULES:\r\n\r\nYou should read each part of the story. Usually, there will be some puzzle. Your goal is to a) figure out the puzzle and b) somehow over throw the school system. You may do any action you could do with the materials provided.\r\n\r\nYour stuff: (per person)\r\n\r\n10 pencils\r\n24 color pencils\r\n1 protractor\r\n1 pair of scissors \r\n1 ruler\r\n5 rubber bands (many more on teacher's desks, hint hint).\r\n1 box of 100 paper clips\r\n\r\nSTORY:\r\n\r\n[color=red][u][b][size=150]Chapter 1: Windows[/size][/b][/u][/color]\r\n\r\nGrimly, the students file in, one by one, into their seats. As they sit down, they see the dark black lens of the security cameras focusing on them. Backs rigid, faces blank, they attempt to pay attention to the dull, flat voice of the teacher. Today, one of the students dropped off. Soft snores came from his nose. Turning around, the teacher swiftly pressed a button, causing the floor beneath the child to disappear. He fell into the hole, never to be seen again...\r\n\r\nAfter the incident, the children paid better attention. At the ring of a computerized bell, the class quietly shuffled off to their next class. As they routinely filled the classroom, Joe10112 was sent to the office to make copies for the class. As Joe10112 quickly stepped through the quiet hallway, he noticed a series of windows that he had always seen covered by shades. Today however, was different. Joe10112 cautiously peered into the window. Robots! Lots and lots of robots. They strode around the room. Some peered into monitors, others worked on a strange machine. As Joe10112 continued to stare into the window, he felt a strange presence. Something was wrong here. He turned around, and everything was black.\r\n\r\nAll he heard was muffled voices. Straining to open his eyes, Joe10112 finally regained vision. Above him were many concerned faces. r15s11z55y89w21 pulled him up. Joe10112 turned around and saw the windows, except this time, they were closed...\r\n\r\nClassroom 3A1Q, chemistry. Once again, the kids were not paying attention. Joe10112\u2019s incident led them all to think about everything except chemistry. Then, an idea sparked. Soon a note spread around the class, and on it, one word. Rebel.\r\n\r\n[color=red][u][b][size=150]Chapter 2: Textbook.[/size][/b][/u][/color]\r\n\r\nThe teacher had them crack open their thick textbooks. As dragon96 did so, a note slipped out on to his desk. \r\n\r\n[quote]Puzzle 1: \nbrmyd[/quote]\r\n\r\nThe students looked at each other, faces solemn. The question is. What do you do?", "Solution_1": "do we know who _3_ is?\r\n\r\n\r\ndragon96?", "Solution_2": "Yes, sorry. I wrote most of the story before hand and I must have forgot to include the name.", "Solution_3": "Eh vents?\r\n\r\nAre there vents?", "Solution_4": "What are you going to do about them?", "Solution_5": "Isn't that copying off the Race to survival??\r\nCrawl through them to get out, or to that room. I have no Idea what you are trying to point to.", "Solution_6": "Vents are infamous in all situations.\r\n\r\nI'm trying to point to the fact that you do something like:\r\n\r\nI use a ruler to pry of the tiled upper celing and climb through the vents.", "Solution_7": "I use a ruler to pry of the tiled upper celing and climb through the vents. :D", "Solution_8": "Looks like a copy of RtS. Out", "Solution_9": "*tells ppl wat hapnd to me*\r\n\r\nonce in a good place with NO TEACHERS snooping\r\n\r\nwhere did code drop from?", "Solution_10": "Read the story.\r\n\r\nAha! dragon96 exclaimed. Excitedly, dragon96 used a ruler to pry off the celing panels. Using a desk he reached up to climb up into the vent. As he dragged himself through, he met another code, scratched on to the side of the vent:\r\n\r\n[quote]Puzzle 2:\nUPJ ZY OPDK[/quote]", "Solution_11": "Can anyone remind me what Caesarean Cipher arrangement is? I suspect it is, because I have tried other codes, which don't seem to work. Or better yet, tell me if Caesarean works.", "Solution_12": "http://en.wikipedia.org/wiki/Caesar_cipher", "Solution_13": "Wait, so are we in the hall or the room?", "Solution_14": "Where do you want to be?\r\n(By the way: I'm going to ask frustrating rhetorical questions from now on).", "Solution_15": "I'm guessing this thread is dead?", "Solution_16": "I think this is too hard. Cause what are you supposed to do?", "Solution_17": "Decode the message, obviously...if you bothered to read anything?", "Solution_18": "*sigh*\r\n\r\ndojo, EASIER problms plz?\r\n\r\nthnks\r\n\r\nand is the thread dead????\r\n\r\ndojo, post then we know the thread is not dead", "Solution_19": "I would if people would participate." } { "Tag": [ "geometry", "calculus", "function", "probability", "algebra", "linear equation" ], "Problem": "The subjective questions of this year's ISI paper can be found at http://www.googolplexideas.com/index.php/math-olympiad/4-problems/48-isi-2009.html", "Solution_1": "link appear dead\r\n\r\nYou better post them here", "Solution_2": "Some server problem please see http://thepizone.com/gi/isi-2009.pdf", "Solution_3": "[hide=\"1\"]\n\n We have 3 linear equations:\n\n$ \\alpha x \\plus{} \\beta y \\plus{} \\gamma z \\equal{} 0$\n$ \\frac {\\alpha}{x} \\plus{} \\frac {\\beta}{y} \\plus{} \\frac {\\gamma}{z} \\equal{} 0$ , which follows from taking conjugate of the above equation and using |$ \\alpha$| =1 .\n\nand $ x \\plus{} y \\plus{} z \\equal{} 0$\n\nThis forms a homogeneous equation in x ,y and z. But since x , y , z are non zero , this system has infinite solutions. Hence the determinant of coefficients of the above 3 linear equation should be 0 which gives the required condition.[/hide]\r\n\r\nWill post others in evening.", "Solution_4": "@ Manjil : for the 6th Q, it should be [b]ad \u2013 bc not equal to zero[/b], \r\n\r\n[b]not ab \u2013 cd[/b].", "Solution_5": "Yes thanks, and question 9 should have been $ \\mid f(x)\\mid <\\frac{3}{2} \\forall x \\in [\\minus{}1,1]$ and not as mentioned.", "Solution_6": "[hide=\"The 4th one\"]\n\nAnswer was x = 5/13, y = -12/13,\nMin value of x^2 + y^2 being 1.\n\n2 methods to do this : by Geometry and Calculus.\n\nI did it by Calculus taking x = 14 cos@ - 5 and y = 14 sin@ + 12.\n\n[/hide]", "Solution_7": "[hide=\"Solution to the 5th one\"]\n$ \\frac{10^r}{p} \\minus{} \\frac{1}{p} \\equal{} a_1a_2 \\cdots a_r .0$\n\n$ 10^{r}$ [b] \u2013 1 = A.p [/b] where A = $ a_1 a_2 \\ldots a_r$ = a natural no.\n\n$ 10^r \\equiv 1 \\ (mod p)$\n\nQ.E.D ! \n[/hide]", "Solution_8": "[hide=\"the 10th one\"]\n\nRational root => D must be perfect square\n\nD = b^2 \u2013 4ac = odd { odd \u2013 even }\n\nAnd every odd perfect square is of the form $ 8n \\plus{} 1 \\ ; \\ n \\in N$\n\n This gives 8 m + 1 \u2013 4 x ac = 8 n + 1 \n\nor ac = odd = 2 ( m \u2013 n ) \nwhich is a contradiction.\n\nSo, no rational roots.\n\n[b]\nOR [/b] :idea: \n\nu can prove that the discriminant is of the form [b]8 n + 5 [/b]\nso it can\u2019t be the perfect square of an odd integer.\n\n[/hide]", "Solution_9": "[u]The correct 6th Question[/u] : \r\n\r\nIf ad - bc is a real number not equal to zero and if $ b_1$ and $ b_2$ are integer multiples of (ad - bc) \r\nthen prove that there exists two integers x and y such that ax + by = $ b_1$ and cx + dy = $ b_2$ are both simultaneously satisfied. \r\n\r\n\r\n[hide=\"Solution\"]\n\nLet $ b_1 \\equal{} n_1$ ( ad \u2013 bc )\n\nand $ b_2 \\equal{} n_2$ ( ad \u2013 bc ) \n\nwhere $ n_1$ & $ n_2 \\ \\in \\ I$\n\nax + by = $ b_1$ \u2026. (1)\n\ncx + dy = $ b_2$ \u2026. (2)\n\neqn (1) x d \u2013 eqn (2) x b gives u\n\n\n x ( ad \u2013 bc ) = d $ b_1$ \u2013 b $ b_2$\n\nor [b] x = d $ n_1$ \u2013 b $ n_2$\n \nwhich is an integer. [/b]\n\nSimilar proof for y .\n\n\n[i]But stupid me - couldn't think of this in the exam hall [/i] :( \n\n [/hide]", "Solution_10": "[quote=\"Nora.91\"][hide=\"The 4th one\"]\n\nAnswer was x = 5/13, y = -12/13,\nMin value of x^2 + y^2 being 1.\n\n2 methods to do this : by Geometry and Calculus.\n\nI did it by Calculus taking x = 14 cos@ - 5 and y = 14 sin@ + 12.\n\n[/hide][/quote\r\nisnot x^2+y^2 minimum when x=y......", "Solution_11": "[quote=\"Nora.91\"][hide=\"the 10th one\"]\n\nRational root => D must be perfect square\n\nD = b^2 \u2013 4ac = odd { odd \u2013 even }\n\nAnd every odd perfect square is of the form $ 8n \\plus{} 1 \\ ; \\ n \\in N$\n\n This gives 8 m + 1 \u2013 4 x ac = 8 n + 1 \n\nor ac = odd = 2 ( m \u2013 n ) \nwhich is a contradiction.\n\nSo, no rational roots.\n\n[b]\nOR [/b] :idea: \n\nu can prove that the discriminant is of the form [b]8 n + 5 [/b]\nso it can\u2019t be the perfect square of an odd integer.\n\n[/hide][/quote]\r\n\r\nWe can do this by assuming that there is a rational root and coming to a contradiction. This is a very old Canadian math Olympiad problem.", "Solution_12": "[quote=\"manjil\"]The subjective questions of this year's ISI paper can be found at http://www.googolplexideas.com/index.php/math-olympiad/4-problems/48-isi-2009.html[/quote]\r\n\r\nThis is now working!", "Solution_13": "[quote=\"UKO\"]\nis not x^2 + y^2 minimum when x=y......\n[/quote]\r\n\r\nyes, but this case u also have d condition $ (x\\plus{}5)^2 \\plus{} (y\\minus{}12)^2 \\equal{} 14^2$\r\n\r\nif x = y, this is not satisfied !!", "Solution_14": "[hide=\"2nd first part\"]\n\nLet us consider the function $ f(x) = x(x+1)..(x+2009)$\n\nLet $ \\alpha$ be a root to the equation in the question, then $ f(\\alpha) = c$\n\nAssume $ \\alpha$ is a root of multiplicity $ 2$. Then $ f'(\\alpha) =c$\n\nIts easy to see that $ f'(x) = f(x)(\\frac{1}{x} + \\frac{1}{x+1} + .... +\\frac{1}{x+2009})$\n\nPut $ x = \\alpha$ in the above equation, we get\n\n$ c = c(\\frac{1}{\\alpha} + \\frac{1}{\\alpha +1} ....)$\n\nHence, for the root to have multiplicity atleast $ 2$ we should have $ 1 = \\frac{1}{\\alpha} + \\frac{1}{\\alpha +1} ....+\\frac{1}{\\alpha +2009}$\n\nNow if the root is of multiplicity $ 3$, $ f''(\\alpha) = c$\n\n\n\n$ f''(x) = f'(x)(\\frac{1}{x} + \\frac{1}{x+1} + .... +\\frac{1}{x+2009}) - f(x)(\\frac{1}{x^2} + \\frac{1}{(x+1)^2} + ...... +\\frac{1}{(x+2009)^2}$\n\n\n\nNow putting $ x=\\alpha$ and using $ f(x) = f'(x) = c$ and $ 1 = ({\\frac{1}{\\alpha} + \\frac{1}{\\alpha +1} ....)}$ , we get\n\n$ f''(\\alpha) = c - c(\\frac{1}{\\alpha^2} + \\frac{1}{(\\alpha+1)^2} + ...... +\\frac{1}{(\\alpha+2009)^2})$[/hide]\r\n\r\nwhich cannot be equal to c for any real value of $ \\alpha$. Hence proved.", "Solution_15": "conjurer\r\n\r\nshudnt f'(x)=0 ????", "Solution_16": "[quote=\"preetham\"]conjurer\n\nshudnt f'(x)=0 ????[/quote]\r\n\r\nIncredibly wronged..bare with me please.hehe.\r\n\r\nStill continuing with prev solution. we have f'(a) =0\r\n\r\nf''(a) = -f(a)(1/a^2 + 1/(a+1)^2 + ... +1/(a+2009)^2) which cannot be 0.", "Solution_17": "[quote=\"Nora.91\"][quote=\"UKO\"]\nis not x^2 + y^2 minimum when x=y......\n[/quote]\n\nyes, but this case u also have d condition $ (x \\plus{} 5)^2 \\plus{} (y \\minus{} 12)^2 \\equal{} 14^2$\n\nif x = y, this is not satisfied !![/quote]\r\n\r\n\r\nSorry I couldnot understand................... :( \r\nSolving..........(x+5)^2+(x-12)^2=14^2will give us the solution.", "Solution_18": "Pardon me i wrote a wrong sttment there .... about it not being satisfied\r\n\r\nBUT\r\n\r\n[quote=\"UKO\"]\nis not x^2+y^2 minimum when x=y......\n[/quote]\r\n\r\ni don't get this \r\n\r\nconsider \r\n\r\n$ F \\equal{} x^2 \\plus{} y^2 \\\\ if \\ x \\equal{} y \\\\ then \\ F \\equal{} 2x^2 \\\\ but \\ if \\ y < x \\\\ then \\ y^2 < x^2 \\\\ \\\\ So, \\ F \\equal{} x^2 \\plus{} y^2 < x^2 \\plus{} x^2 < 2x^2 $\r\n\r\n[b]this value is less than when x = y[/b]\r\n\r\nso, Why do you say that minimum value is when x = y :?:", "Solution_19": "[url]http://targetiit.com/iit_jee_forum/posts/isi_2009_subjective_mib_8978.html[/url]\r\n\r\n\r\nall the Q have been solved here", "Solution_20": "ISI B.Stat results out ....", "Solution_21": "well,y do u never post ur own result :?: \r\n\r\nbest of luck though..", "Solution_22": "[quote=\"Nora.91\"]ISI B.Stat results out ....[/quote]\r\n\r\nplease give the link and what abt B.math?", "Solution_23": "[quote=\"manjil\"][quote=\"Nora.91\"]ISI B.Stat results out ....[/quote]\n\nplease give the link and what abt B.math?[/quote]\r\n\r\nOK I found the link myself...http://www.isical.ac.in/~deanweb/BSTATINTLIST09.pdf", "Solution_24": "[quote=\"rituraj007\"]well,y do u never post ur own result :?: \n\nbest of luck though..[/quote]\r\n\r\n\r\nThanx ...\r\n\r\nThat's becoz my course results ( B.Math ) aren't out yet ! :rotfl: \r\n\r\nand even if they r, its kinda shameful to post it in front of people of your level :( \r\n\r\n\r\n@ manjil : i don't know when B.Math will be out, i think we have to check their website everyday.", "Solution_25": "i m so much unaware of ISI stuff :P", "Solution_26": "Hi .\r\n\r\nISI BMath Results are out http://www.isibang.ac.in/~statmath/bmath09.htm", "Solution_27": "[quote=\"apratimgtr\"]Hi .\n\nISI BMath Results are out http://www.isibang.ac.in/~statmath/bmath09.htm[/quote]\r\n\r\nI got thr.....Sl.No 60\r\n\r\nnora.91 too!", "Solution_28": "Congrats!! :)", "Solution_29": "congrats to all who are selected :) \r\n\r\n\r\n[hide=\"by the way\"]what is diff between B.math and B.Stat(stat is a part of maths,i suppose)[/hide]", "Solution_30": "@skand : Statistics is a completely different course .... look at their syllabi\r\n\r\n[b]Congrats Manjil !!![/b] :thumbup:", "Solution_31": "[quote=\"Nora.91\"]\n\n[b]Congrats Manjil !!![/b] :thumbup:[/quote]\r\n\r\nThanks same to you too!!!!:thumbup:", "Solution_32": "Are the results make you elligible for interveiw or admission?", "Solution_33": "@ murgi : Interview", "Solution_34": "i got through too!!! but what worries me now is the interview! do u have any idea what sort of questions are asked in the interview?", "Solution_35": "How many are selected for CMI through the written exams?And how many are eliminated through interviews?\r\nIs it true that INMO awardees get direct admission to CMI for BSc?Reply at mahadgos@yahoo.com as soon as possible! :juggle:", "Solution_36": "[quote=\"Scary math\"]How many are selected for CMI through the written exams?And how many are eliminated through interviews?\nIs it true that INMO awardees get direct admission to CMI for BSc?Reply at mahadgos@yahoo.com as soon as possible! :juggle:[/quote]\r\n\r\nthere are no interview for CMI I guess, abt 45 are selected and yes INMO awardees get direct admsn.", "Solution_37": "Well my neighbour who appeared for CMI this year said there were interviews.", "Solution_38": "Yeah, i think there is interview fr CMI\r\n\r\nand INMO and INAO awardees get admission only after the interview, they just don't have to take the test.\r\n\r\n\r\nEDIT : The [url=http://www.isibang.ac.in/~statmath/bmath09.htm]ISI interview schedule's[/url] been put up,\r\n\r\nmine on the first day itself ! :noo:", "Solution_39": "Where could I get the ISI B-Stat intreview results?\r\n And one more request,I am a 10th grade Math Olympiad begginer therefore I seek some advice:\r\nWhat is the minnimum qualifying marks required and the minnimum no of questions one should get right in RMO? Which is more important,probability or combinatorics?From where do more questions come? What are the prime topics ?Could I skip convergency & divergency,limits,Recurring series and calculus when preparing for Regional Olympiads? \r\nPlease reply at mahadgos@yahoo.com as soon as possible \r\nYaar koi to help kaar dimag ka :idea: jalane mein! :fool: \r\n :maybe:" } { "Tag": [ "linear algebra", "matrix" ], "Problem": "Hello friends,\r\n Recently I read how dynamic programming can be applied to multiply a chain of matrices (which of course are multiplication compatible) in a reasonable amount of time. What I don\u2019t understand however is why can\u2019t we apply greedy approach for the same wherein we do the following:-\r\n\r\nFirst multiply the matrices M(i) and M(i+1) that minimize the product d(i-1).d(i).d(i+1) and continue in the same way.\r\n\r\n\r\nWe assume that we have a chain of matrices M1.M2.M3-------.Mn-1.Mn and in addition to that an array d[0\u2026n] where Mi is of dimension d(i-1).d(i).", "Solution_1": "try these cases:\r\n1 10\r\n10 3\r\n3 20\r\n20 1\r\n\r\n2 10\r\n10 3\r\n3 20\r\n20 1" } { "Tag": [], "Problem": "Some one of you can help me to find\r\n \"A Classical Introduction to Modern Number Theory\"", "Solution_1": "Yes, but what is author of it?", "Solution_2": "m.candales, you can find many good books in webpage http://gigapedia.org [url=http://www.mediafire.com/?3sxz5kg5oyy]here[/url] may a book you needed.\r\n\r\nMichael.", "Solution_3": "I think the book in concern is 'A Classical introduction to Modern Number Theory' by Ireland and Rosen. It is under copyright so please do not try and acquire it illegally." } { "Tag": [ "parameterization", "geometry", "geometric transformation", "rotation", "function", "quadratics", "ceiling function" ], "Problem": "Hi folks. \r\n\r\nI've got a problem here that I am reaching out for help on. I've gone through 3 courses of calc a few years back, and haven't used it much since. I've got this problem I'm working on. I've got an approximation I'll probably use, but my curiosity is pushing me to find a precise equation.\r\n\r\nIt's relatively simple. A cloth is wound up on a roll using a motor rotating at a constant speed. I'd like to increment the cloth by a constant length for the entire roll. I suppose the explanation of it is that simple.\r\n\r\nNot that I expect anyone to give me the answer, but in case you'd like to know, the parameters of the system are:\r\n\r\nEmpty Core Diameter: 1.95 inches\r\nFull Cloth Diameter: 4.75 inches\r\nMotor RPM: 0.89\r\nTotal Cloth Length: 75 feet\r\nDesired Increment: 0.5 inch\r\n\r\nI can calculate that:\r\nEmpty cloth speed: 0.091 inch/sec\r\nFull Cloth Speed: 0.221 inch/sec\r\nMotor On Time (Empty): 5.505 sec\r\nMotor On Time (Full): 2.26 sec\r\n\r\nI'll probably end up using a linear approximation, but I'd love to know the precise equation.\r\n\r\nCan anyone give me a kick start?", "Solution_1": "With this design, the speed at which the cloth is added is proportional to the radius of the spool. There are limits; the motor is working a lot harder with a full roll.\r\n\r\nLet $ C(t)$ be the amount of cloth, and $ r(t)$ be the radius. We have $ C(t)\\equal{}a(r(t)^2\\minus{}r_0^2)$, where $ a$ is an empirically derived constant and $ r_0$ is the initial radius; this just estimates $ C$ by the volume taken up. In addition, $ C'(t)\\equal{}br(t)$, where $ b$ is another constant (found either empirically or from the rotation rate).\r\n\r\nNow, we can also differentiate our equation for $ C$, and get $ C'(t)\\equal{}ar(t)r'(t)$. This implies that $ b\\equal{}ar'(t)$, and $ r(t)\\equal{}r_0\\plus{}\\frac{b}{a}t$.\r\nFrom that, we can find $ C(t)\\equal{}\\frac{b^2}{a}t^2\\plus{}2br_0t$.\r\n\r\nThe exact increment of $ t$ needed to increase $ C$ by a certain amount involves a square root.", "Solution_2": "Wow, great.\r\n\r\nI think that makes sense. I think what was always missing in school was teaching the concepts needed to be able to model a problem like this. By the time I got done with all the math I took I could manipulate all sorts of equations, but couldn't model the most basic of real life scenarios.\r\n\r\nSo to recap what you did. You could write an equation for the volume of the cloth on the take up roll, and know that that is proportional to the length of cloth.\r\n\r\nThen you can write an equation stating that the rate of change of the length of cloth was proportional to the take up roll radius.\r\n\r\nAfter that, it's a matter of differentiating the first, equating to the second and solving for the radius as a function of time, and plugging back into the original equation. Use algebra to solve, and end up with a quadratic equation. Simple and brilliant.\r\n\r\nSo I'm left with an equation that gives me the amount of cloth on the roll for a given amount of time the motor has been running.\r\n\r\nBut I need something slightly different. I need to calculate a preset for a timer in a PLC to turn on the cloth motor for the Nth time.\r\n\r\nSo how to take the next step..... As you stated it requires a square root. I can set C(t) to some number, then to a number 1/2\" greater, and find delta t for each iteration, but that won't work well in a PLC program...\r\n\r\nWhat I really need is an equation for t(n), if that makes any sense.\r\n\r\nThis is exactly the help I needed to get me started - thanks a lot!", "Solution_3": "I'm assuming you mean you need the time it takes for your motor to wind $ n$ partitions of some length $ l$?... I really can't go on without knowing: assuming your motor runs at constant power I think the rotational velocity of the spool can be written as a function of time. Once you have that done you can probably use rotational kinematics to find nearly anything you need.", "Solution_4": "igiul- the motor is running at constant speed, not constant power. It's simpler than that.\r\n\r\njbenderjb- It makes sense- that's the inverse function. Taking the inverse of a quadratic function is essentially solving a quadratic equation:\r\n$ \\frac {b^2}at^2 \\plus{} 2br_0t \\minus{} C \\equal{} 0$\r\n$ t \\equal{} \\frac { \\minus{} 2br_0\\pm\\sqrt {4b^2r_0^2 \\plus{} \\frac {b^2}{a}C}}{2\\frac {b^2}{a}} \\equal{} \\frac { \\minus{} ar_0\\pm a\\sqrt {r_0^2 \\plus{} \\frac1aC}}{b}$\r\n\r\nWe choose the positive sign in the square root; time runs forward. If we put in the $ n$th value of $ C$, we get the $ n$th value of $ t$ back from this formula.\r\nThis $ t$ is the total time the motor has spent running.", "Solution_5": "[quote=\"jmerry\"]igiul- the motor is running at constant speed, not constant power. It's simpler than that.[/quote]\r\n\r\nI was actually think of finding the moment of inertia as a function of time. From there, equating $ KE_R$ to the work of the engine allows us to isolate $ w$ (rotational velocity). I think your point is that the engine runs at a constant RPM whereas I took to the notion that this implied that the engine would do the same amount of work winding the cloth albeit it slows down as mass on the spool increases.\r\n\r\nCorrect me if I'm wrong; it's been a year since I took advanced physics so my terminology and fluency on the subjects at hand are a bit rusty.", "Solution_6": "It is a simple set-up - as I'd like to define it, at least.\r\n\r\nI have a motor that will spin at a constant speed when on. It is winding up a spool of cloth which cleans something. Every once in a while I turn the motor on to increment the cloth to get a fresh section of cloth. I'd like to have the same cloth increment each time over 75 feet of cloth, so I need some decrement logic to take into account the changing take up roll diameter.\r\n\r\nI am assuming constant speed for our purposes. It is, in reality, a constant power system, but through an 1800:1 gearing without all that much friction to be overcome. So I think the constant speed is a very safe estimate.\r\n\r\nThis will be programmed into a computer of sorts (PLC). Working in real world, I need to take into account the fact that my timer precision is in milliseconds, and so in the end I will probably keep track of how long the motor has run so far, and calculate how long it should run for another 1/2 inch increment. Just to make things more fun, it will be an adjustable increment by the user.\r\n\r\nThis has been huge help, and I look forward to writing some of this out.", "Solution_7": "Let the full cloth radius be $ r_f$ and the empty core radius be $ r_o$\r\n$ r_o = 0.975\\text{in}$\r\n$ r_f = 2.375\\text{in}$\r\n$ \\Delta r = 1.40\\text{in}$\r\n\r\nThen, $ r_n = r_o + n\\rho$ where $ \\rho$ is the thickness of our cloth. There exists $ n: r_f = r_o + n\\rho$ which implies $ n\\rho = \\Delta r$\r\n[hide=\"The Tricky Part\"]\n\\begin{eqnarray*}75 & = & 2\\pi\\sum_{i = 1}^n r_n \\\\\n& = & 2\\pi\\displaystyle\\left(\\sum_{i = 1}^n r_o + \\rho\\sum_{i = 1}^n i\\right) \\\\\n& = & 2\\pi\\left(r_on + \\rho\\frac {n(n + 1)}{2}\\right) \\\\\n& = & \\pi n(2r_o + \\rho n + \\rho) \\\\\n& = & 3.35\\pi n + 1.4\\pi \\\\\n\\end{eqnarray*}\n[/hide]\r\nFrom there I found $ n$, then $ \\rho$ by $ \\Delta r = n\\rho$.\r\n\r\n$ n\\approx 6.708430288$\r\n$ \\rho\\approx 0.208692636$\r\n\r\nThus, $ r_n = r_o + \\rho\\lceil 0.89t\\rceil$ where $ \\lceil x\\rceil$ denotes the least integer greater than or equal to $ x$. And finally, for $ n$ partitions of half inch strips, $ \\boxed{t(n,t_e) = .89\\pi n(r_o + \\rho\\lceil 0.89t_e\\rceil )}$ where $ t_e$ is total time elapsed. $ t(n)$ [i]should[/i] hold so long as $ r_n$ does not change during the interval $ (t_e,t_e + t(n,t_e)$.\r\n\r\nFunnily enough, I used absolutely no calculus at all to do that.\r\n\r\n[b]edit:[/b] The radius that $ t(n)$ generates is that on the outside of each layer of cloth, so you may want to alter that just a bit by subtracting $ \\frac{\\rho}{2}$ from our $ r_n$ function.", "Solution_8": "OK. That's a lot for me to think about.\r\n\r\nI can get to the point where you define the radius at any given time (accumulated motor run time)\r\n\r\nI'm lost at the point where that is multiplied by 0.89 * pi * n (Sorry I don't know TeX...)\r\nThe rotation rate times ( pi * radius ) gives us a \"cloth speed\" at a given time\r\nThough shouldn't it be multiplied by 2 for the circumference?\r\n\r\nSo then we multiply by n... Is n still the total number of rotations in a full cloth?\r\n\r\nConceptually, I can't wrap my head around that one...\r\n\r\nThe cloth length is actually 75 feet, so 900 inches, but I can easily recalculate the p and n.", "Solution_9": "I've been thinking about this all day...\r\n\r\nI can see where \"cloth rate\" is divided by the partition length, giving you a \"partition rate\", and that would account for what I thought was a missing \"2\" for the circumference.\r\n\r\nn is the \"partition count\", and I just can't figure out why we multiply by that. (Of-course I could be way off all together)\r\n\r\nBut if I have partition speed given run time, couldn't I take the inverse of that to give me a \"seconds per partition\"?", "Solution_10": "Yes, and that's the linear approximation you were talking about at the beginning. Your increments are short enough that the error in that approximation is negligible." } { "Tag": [ "factorial", "AMC", "USA(J)MO", "USAMO", "induction", "algebra proposed", "algebra" ], "Problem": "Help me find value of $ \\sum_{k\\equal{}0}^{n}k!(n\\minus{}k)!$ in terms of $ n\\in\\mathbb N$. \r\n\r\nThanks in advance for your nice solution.", "Solution_1": "Mathematica gives :\r\n$ \\minus{}i 2^{\\minus{}(1\\plus{}n)} [\\pi \\minus{}i \\beta(2,2\\plus{}n,0)] \\Gamma(2\\plus{}n)$", "Solution_2": "Since we are in \"proposed and own\" category, the OP has the solution.\r\n\r\nSo, after strange mathematica and mapple solutions, could we have J.Y.Choi's one, please ?", "Solution_3": "I posted this problem in this category because I made this problem and couldn't solve it by myself. \r\n\r\nMy insignificant approach is thinking of $ \\binom{n}{k}^{\\minus{}1}$, but I can't make an advance any more. I need mathlinkers' help.", "Solution_4": "[quote=\"J.Y.Choi\"]I posted this problem in this category because I made this problem and couldn't solve it by myself.[/quote]\r\nPlease, read the forum definitions and respect the usage. It's just a matter of politeness :\r\n\r\n[i][u]Unsolved category[/u][/i] : \"Problems you couldn't solve and to which you know that there is a solution (i.e. a problem from a contest, etc.) but you don't know it.\"\r\n\r\n[i][u]Proposed and own category[/u][/i] : \"Problems [u][b]that you have already solved [/b][/u]and you are interested in second opinions or solutions.\"\r\n\r\n[i][u]Open category[/u][/i] : \"An open question is a question that has no known solution up to this moment, and it is not known wheter the problem has one or not.\"\r\n\r\nSome users are interested in only some categories (personnaly, I generally spend more time on \"proposed and own\" than on \"Open\" since I'm sure there is a clever answer (olympiad level) and I know, when I fail finding it, that original poster will post the answer when asked.\r\n\r\nIn your case : you randomly created a problem and dont know if there is a solution : Post it in \"Open\" category and write in the beginning that you created it.", "Solution_5": "If my memory serves me right this is an adaptation of a USAMO problem. Its mentioned somewhere in Titu's book: Mathematical Olympiad Challenges", "Solution_6": "The given series is the form of $ n!\\sum_{k\\equal{}0}^{n}\\binom{n}{k}^{\\minus{}1}$. So,let's denote a sequence $ a_n$ by $ a_n\\equal{}\\sum_{k\\equal{}0}^{n}\\binom{n}{k}^{\\minus{}1}$. For $ n\\geq 2$, we have \r\n\r\n$ a_n\\equal{}\\sum_{k\\equal{}0}^{n}\\frac{1}{\\binom{n}{k}}\\equal{}\\sum_{k\\equal{}0}^{n\\minus{}1}\\frac{k!(n\\minus{}1\\minus{}k)!(n\\minus{}k)}{(n\\minus{}1)!n}\\plus{}1$ \r\n$ \\equal{}\\sum_{k\\equal{}0}^{n\\minus{}1}\\frac{1}{\\binom{n\\minus{}1}{k}}\\minus{}\\sum_{k\\equal{}0}^{n\\minus{}1}\\frac{k}{n}\\frac{(n\\minus{}1\\minus{}k)!k!}{(n\\minus{}1)!}\\plus{}1$\r\n$ \\equal{}a_{n\\minus{}1}\\minus{}\\sum_{k\\equal{}0}^{n\\minus{}1}\\frac{k}{n}\\frac{1}{\\binom{n\\minus{}1}{k}}\\plus{}1\\quad\\cdots\\boxed{1}$.\r\n\r\nMoreover, $ \\sum_{k\\equal{}0}^{n\\minus{}1}\\frac{k}{n}\\frac{1}{\\binom{n\\minus{}1}{k}}\\equal{}\\sum_{k\\equal{}0}^{n\\minus{}1}\\frac{n\\minus{}1\\minus{}k}{n}\\frac{1}{\\binom{n\\minus{}1}{k}}\\equal{}\\frac{n\\minus{}1}{n}a_{n\\minus{}1}\\minus{}\\sum_{k\\equal{}0}^{n\\minus{}1}\\frac{k}{n}\\frac{1}{\\binom{n\\minus{}1}{k}}$. \r\n\r\nTherefore $ \\sum_{k\\equal{}0}^{n\\minus{}1}\\frac{k}{n}\\frac{1}{\\binom{n\\minus{}1}{k}}\\equal{}\\frac{n\\minus{}1}{2n}a_{n\\minus{}1}$. \r\n\r\nFinally, by $ \\boxed{1}$, we get $ a_n\\equal{}a_{n\\minus{}1}\\minus{}\\frac{n\\minus{}1}{2n}\\plus{}1\\equal{}\\frac{n\\plus{}1}{2n}a_{n\\minus{}1}\\plus{}1$. Would anyone can solve this inductive form for $ a_n$? I have no idea...", "Solution_7": "I can't find it in Mathematical Olympiad Challenges, but Kurlyandchik and Lisizkiy prove in Kvant 10/1978 that $ \\sum_{k\\equal{}0}^n \\frac{1}{\\binom{n}{k}} \\equal{} \\frac{n\\plus{}1}{2^{n\\plus{}1}}\\sum_{k\\equal{}1}^{n\\plus{}1}\\frac{2^k}{k}$. Of course, the proof is trivial by induction since J.Y.Choi has done all the hard work. I doubt that the term on the right hand side can be simplified.\r\n\r\n darij" } { "Tag": [ "calculus", "integration", "function", "trigonometry", "calculus computations" ], "Problem": "Find the function $f(x)$ such that $f(x)=\\cos (2mx)+\\int_{0}^{\\pi}f(t)|\\cos t|\\ dt$ for positive inetger $m.$", "Solution_1": "The integral equality implies that \r\n\r\n$f(x)=\\cos(2mx)+C$ where $C$ is a constant.\r\n\r\nFpr calculating $C$ use the integral equality and you should get that\r\n\r\n$C=\\int_{0}^{\\pi}(C+\\cos(2mt))|\\cos(t)|dt$\r\n\r\n$C=C\\int_{0}^{\\pi}|\\cos t|dt+\\int_{0}^{\\pi}\\cos(2mt)|\\cos(t)|dt$.\r\n\r\nHence, $C=\\frac{-1}{3}\\int_{0}^{\\pi}\\cos(2mt)|\\cos t|dt$." } { "Tag": [ "algebra", "polynomial", "quadratics", "function" ], "Problem": "Find all polynomial with real-valued coefficients $P$ and $Q$ satisfying\r\n(a) $2P(x)-Q(x)=P(y)-y$ , \r\n(b) $P(x)Q(x)\\geq x+1$ ,\r\nfor all real numbers $x$ and $y$", "Solution_1": "This one is really easy... :? Why is it still unsolved?\r\n\r\nPierre.", "Solution_2": "Ok, so I will do it :\r\nSince the lhs of the first condition does not depend on $y$, it follows that the rhs is constant, so that $P(y)=y+c$ for some constant $c$.\r\nFrom the first condition, it now follows that $Q(x)=2x+c.$\r\n\r\nThus the second condition asks for $2x^2+3cx+c^2 \\geq x+1$ for all $x$\r\nwhich means that $2x^2 + (3c-1)x + c^2 - 1 \\geq 0.$\r\nThis is equivalent to ask that the discriminant of this quadratic is non-positive.\r\nBut $\\Delta = (c-3)^2$ thus $c=3$ which gives $P(x) = y+3$ and $Q(x) = 2x+3.$\r\n\r\nPierre.", "Solution_3": "Are P and Q functions there? I just don't understand your notation...", "Solution_4": "P and Q are polynomials with real coefficients, as the question says.", "Solution_5": "Ah...it appeared that P and Q were the coefficients..." } { "Tag": [ "geometry", "3D geometry", "algebra", "polynomial", "Rational Root Theorem" ], "Problem": "Rationalize the denominator. \r\n\r\n$1$ divided by the square root of $(1+\\sqrt{2})$\r\n\r\nEdit: I don't know whats wrong. Can someone help me?\r\nEdit2: okay i fixed it. kinda.", "Solution_1": "you mean $\\frac{1}{\\sqrt{1+\\sqrt{2}}}$ ??", "Solution_2": "[quote=\"anyone\"]you mean $\\frac{1}{\\sqrt{1+\\sqrt{2}}}$ ??[/quote]\r\nyea. I didn't know until now.", "Solution_3": "sooo... does anyone know how to do it? :lol:", "Solution_4": "Mm, this one gets a bit ugly.\r\nBut basically you multiply both the numerator and the denominator first by $\\sqrt{1+\\sqrt{2}}$, and then by $1-\\sqrt{2}$, and simplify.\r\nThen, you would get $\\sqrt{2+2\\sqrt{2}}-\\sqrt{1+\\sqrt{2}}.$", "Solution_5": "Let the equation = x, then take x^2, multiply num. and denom. by 1-sqrt(2), then take the square root again: \r\n\r\nsqrt(sqrt(2)-1)", "Solution_6": "lol, nice, that's definitely better than my method.", "Solution_7": "[quote=\"davidyko\"]Mm, this one gets a bit ugly.\nBut basically you multiply both the numerator and the denominator first by $\\sqrt{1+\\sqrt{2}}$, and then by $1-\\sqrt{2}$[/quote]\r\n\r\nThe book said to multiply, but I don't see how you multiply something with two brackets above it. I always thought of it as a cube root. Could someone elaborate? how to mutiply multiple brackets like that. It really confused me.", "Solution_8": "[quote=\"davidyko\"]Mm, this one gets a bit ugly.\nBut basically you multiply both the numerator and the denominator first by $\\sqrt{1+\\sqrt{2}}$, and then by $1-\\sqrt{2}$, and simplify.\nThen, you would get $\\sqrt{2+2\\sqrt{2}}-\\sqrt{1+\\sqrt{2}}.$[/quote]\r\nThats exactly what I got. I just wasn't sure.", "Solution_9": "[quote=\"now a ranger\"][quote=\"davidyko\"]Mm, this one gets a bit ugly.\nBut basically you multiply both the numerator and the denominator first by $\\sqrt{1+\\sqrt{2}}$, and then by $1-\\sqrt{2}$[/quote]\n\nThe book said to multiply, but I don't see how you multiply something with two brackets above it. I always thought of it as a cube root. Could someone elaborate? how to mutiply multiple brackets like that. It really confused me.[/quote]\r\n\r\nYou mean an embedded radical? The square root of a square root is a fourth root, by the way. (You can see this by writing the square roots as exponents.)\r\nIf you're multiplying two radicals together, and one or both have embedded radicals, don't panic; just multiply the stuff inside the outer square roots together as if it weren't there, and then just slap the square root on top when you're done.\r\n\r\nFor example,\r\n$(\\sqrt{2})({\\sqrt{1+\\sqrt{2}})=\\sqrt{(2)(1+\\sqrt{2})}=\\sqrt{2+2\\sqrt{2}}}$.\r\n\r\nOff-topic: Oh, cool, post #250. Do I have two-and-a-half stars now...? Let's see.", "Solution_10": "${\\frac{1}{\\sqrt{1+\\sqrt{2}}}=\\frac{1}{\\sqrt{1+\\sqrt{2}}}*\\frac{\\sqrt{\\sqrt{2}-1}}{\\sqrt{\\sqrt{2}-1}}=\\sqrt{\\sqrt{2}-1}}$", "Solution_11": "woo! nice method.\r\n\r\nThe bracket there kept driving me nuts.", "Solution_12": "[b]Follow-up:[/b] Show that $\\sqrt{1+\\sqrt{2}}$ cannot be written in the form\r\n\r\n$p+q \\sqrt{2}$\r\n\r\nWhere $p, q$ are rational.", "Solution_13": "Is it because the 2 brackets make it the square root of 1 + the 4th root of 2. In the p+q part, there has to be a bracket around it, or the number is the square of the one with the brackets.", "Solution_14": "[quote=\"now a ranger\"]the 2 brackets make it the square root of 1 + the 4th root of 2. [/quote]\r\n\r\nClearly false. By this logic, $\\sqrt{4+4}= \\sqrt{4}+\\sqrt{4}= 4$, but of course $\\sqrt{4+4}= 2 \\sqrt{2}$.", "Solution_15": "oh now I see what you meant by \"showing things\". It says to \"show\" that something is true or false in aops 1. Ooo :blush:", "Solution_16": "[quote=\"t0rajir0u\"][b]Follow-up:[/b] Show that $\\sqrt{1+\\sqrt{2}}$ cannot be written in the form\n\n$p+q \\sqrt{2}$\n\nWhere $p, q$ are rational.[/quote]\r\n\r\n\r\n[hide=\"Ugly Solution\"]\nProceed by contradiction:\n$\\sqrt{1+\\sqrt{2}}=p+q \\sqrt{2}$\n$1+\\sqrt{2}= p^{2}+2pq\\sqrt{2}+2q^{2}$\n\nSo equation rational and irrational parts:\n\n$1= p^{2}+2q^{2}$\nAND\n$1\\sqrt{2}=2pq\\sqrt{2}$\n\nSimplifying:\n$1= p^{2}+2q^{2}$\nAND\n$pq=\\frac{1}{2}$\n\nSo $p = \\frac{1}{2q}$\n\nSubstitute that into the first equation:\n\n$1= (\\frac{1}{2q})^{2}+2q^{2}$\n$1 = \\frac{1}{4q^{2}}+2q^{2}$\n$4q^{2}=1+8q^{4}$\n$8q^{4}-4q^{2}+1 = 0$\nBy rational root theorem, there are no rational roots of this equation. Hence, $\\sqrt{1+\\sqrt{2}}$ cannot be written in the form $p+q \\sqrt{2}$ where $p, q$ are rational.\n\nEDIT: Actually, the discriminant is negative, so it doesn't even have any REAL solutions. (I think).\n\n\n[/hide]", "Solution_17": "Contradiction is indeed a good way to go, but not quite in that form.\r\n\r\n[hide=\"Instead take this route...\"] Yes, we have $1 = p^{2}+2q^{2}$ and $1 = 2pq$. Write this as\n\n$p^{2}+2q^{2}= 2pq$\n\nNow, clear the denominators and let $p = \\frac{a}{d}, q = \\frac{b}{d}$, where $gcd(a, b, d) = 1$ are integers. Then we wish to solve\n\n$a^{2}+2b^{2}= 2ab$\n\nIn integers. Well, $2 | a$, so let $a = 2c$. But then\n\n$2c^{2}+b^{2}= 2bc$\n\nWhich implies no solutions by infinite descent. QED. [/hide]\nEdit: [hide=\"Alternately...\"] If $p^{2}+2q^{2}= 2pq$ then \n\n$(p-q)^{2}=-q^{2}$\n\nWhich clearly has no nontrivial solutions. [/hide]" } { "Tag": [ "trigonometry", "inequalities" ], "Problem": "Daca ABC este un triunghi ascutitunghic, sa se afle maximul expresiei:\r\n\\[ E \\equal{} \\frac {\\cos A\\cos B\\cos C}{\\sin ^{2} A\\sin ^{2} B\\sin ^{2} C} .\r\n\\]", "Solution_1": "$ \\begin{array}{l}\r\n tgAtgBtgC \\equal{} tgA \\plus{} tgB \\plus{} tgC; \\\\ \r\n \\frac{1}{{(tgA \\plus{} tgB \\plus{} tgC)\\sin A\\sin B\\sin C}}\\mathop \\le \\limits^{Jensen} \\frac{1}{{3\\sqrt 3 \\sin A\\sin B\\sin C}}(*); \\\\ \r\n \\sin A\\sin B\\sin C \\equal{} \\frac{1}{2}(\\cos (A \\minus{} B) \\minus{} \\cos (A \\plus{} B))\\cos C; \\\\ \r\n \\left. \\begin{array}{l}\r\n \\cos (A \\minus{} B)\\cos C \\equal{} \\frac{1}{2}(\\cos (A \\minus{} B \\plus{} C) \\plus{} \\cos (A \\minus{} B \\minus{} C))(1) \\\\ \r\n \\cos (A \\plus{} B)\\cos C \\equal{} \\frac{1}{2}(\\cos \\pi \\plus{} \\cos (A \\plus{} B \\minus{} C))(2) \\\\ \r\n \\end{array} \\right\\| \\Rightarrow \\sin A\\sin B\\sin C \\equal{} \\\\ \r\n \\frac{1}{4}((\\cos (A \\minus{} B \\plus{} C) \\plus{} \\cos (A \\minus{} B \\minus{} C)) \\minus{} (\\cos \\pi \\plus{} \\cos (A \\plus{} B \\minus{} C))) \\equal{} \\frac{1}{4}(\\cos (\\pi \\minus{} 2B) \\plus{} \\\\ \r\n \\cos (\\pi \\minus{} 2A) \\plus{} 1 \\minus{} \\cos (\\pi \\minus{} 2C)); \\\\ \r\n \\minus{} (\\cos 2B \\plus{} \\cos 2A \\plus{} \\cos (\\pi \\minus{} 2C))\\mathop \\ge \\limits^{Jensen} \\minus{} 3\\cos \\frac{{2B \\plus{} 2A \\plus{} \\pi \\minus{} 2C}}{3} \\equal{} \\minus{} 3\\cos (\\pi \\minus{} \\frac{{4C}}{3}); \\\\ \r\n \\sin A\\sin B\\sin C \\ge \\minus{} \\frac{3}{4}\\cos (\\pi \\minus{} \\frac{{4C}}{3}) \\ge \\minus{} \\frac{3}{4} \\\\ \r\n \\end{array}$ pentru $ C\\equal{}\\frac{{{\\rm 3}\\pi }}{{\\rm 4}} \\frac{1}{{3\\sqrt 3 \\sin A\\sin B\\sin C}} \\le \\frac{{ \\minus{} 4}}{{9\\sqrt 3 }}$ dar egalitate nu se obtine in inegalitatea a doua in sensul lui Jensen deoarece nu avem 2B=2A=$ \\pi \\minus{}2C$" } { "Tag": [], "Problem": "\u0388\u03c3\u03c4\u03c9 \u03c4\u03c1\u03b5\u03b9\u03c2 \u03ba\u03cd\u03ba\u03bb\u03bf\u03b9, \u03c91, \u03c92 \u03ba\u03b1\u03b9 \u03c93 \u03bc\u03b5 \u03ba\u03ad\u03bd\u03c4\u03c1\u03b1 \u039f1, \u039f2 \u03ba\u03b1\u03b9 \u039f3 \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03bf\u03b9\u03c7\u03b1, \u03c0\u03bf\u03c5 \u03b5\u03c6\u03ac\u03c0\u03c4\u03bf\u03bd\u03c4\u03b1\u03b9 \u03b1\u03bd\u03ac \u03b4\u03c5\u03bf. P1 \u03ba\u03b1\u03b9 P2 \u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u03b5\u03c0\u03b1\u03c6\u03ae\u03c2 \u03c4\u03c9\u03bd \u03c93 \u03bc\u03b5 \u03c91 \u03ba\u03b1\u03b9 \u03c93 \u03bc\u03b5 \u03c92 \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03bf\u03b9\u03c7\u03b1. \u0391\u0392 \u03bc\u03af\u03b1 \u03b4\u03b9\u03ac\u03bc\u03b5\u03c4\u03c1\u03bf\u03c2 \u03c4\u03bf\u03c5 \u03c93 \u039f\u03b9 \u0391\u03a11 \u03ba\u03b1\u03b9 \u0392\u03a12 \u03c4\u03ad\u03bc\u03bd\u03bf\u03c5\u03bd \u03c4\u03bf\u03c5\u03c2 \u03c91 \u03ba\u03b1\u03b9 \u03c92 \u03c3\u03c4\u03b1 \u03a7 \u03ba\u03b1\u03b9 \u03a5 \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03bf\u03b9\u03c7\u03b1. \u0391\u03bd \u0396 \u03c4\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03c4\u03bf\u03bc\u03ae\u03c2 \u03c4\u03c9\u03bd \u0391\u03a12 \u03ba\u03b1\u03b9 \u0392\u03a11, \u03bd.\u03b4.\u03bf. \u03a7,\u03a5,\u0396 \u03c3\u03c5\u03bd\u03b5\u03c5\u03b8\u03b5\u03b9\u03b1\u03ba\u03ac.", "Solution_1": "\u03bb\u03cd\u03c3\u03b7 \u03c3\u03c4\u03bf attachment", "Solution_2": "\u03a0\u03bf\u03bb\u03cd \u03ba\u03b1\u03bb\u03ae \u03b7 \u03bb\u03cd\u03c3\u03b7 \u03c3\u03b1\u03c2 \u03ba\u03b1\u03b9 \u03c0\u03b9\u03bf \u03b1\u03c0\u03bb\u03ae \u03b1\u03c0\u03cc \u03b1\u03c5\u03c4\u03ae \u03c0\u03bf\u03c5 \u03c3\u03ba\u03b5\u03c6\u03c4\u03cc\u03bc\u03bf\u03c5\u03bd. \u0395\u03b3\u03ce \u03b1\u03c0\u03ad\u03b4\u03b5\u03b9\u03be\u03b1 \u03cc\u03c4\u03b9 \u03c4\u03b1 $ P_1,P_2,P_3,Z$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf\u03bc\u03bf\u03ba\u03c5\u03ba\u03bb\u03b9\u03ba\u03ac \u03ba\u03b1\u03b9 \u03bc\u03b5 \u03b2\u03ac\u03c3\u03b7 \u03b1\u03c5\u03c4\u03cc \u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 $ Y,Z,P_3$ \u03ba\u03b1\u03b9 $ X,Z,P_3$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c5\u03bd\u03b5\u03c5\u03b8\u03b5\u03b9\u03b1\u03ba\u03ac \u03ac\u03c1\u03b1 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03c4\u03bf \u03b6\u03b7\u03c4\u03bf\u03cd\u03bc\u03b5\u03bd\u03bf.\r\n\r\n\u039a\u03ac\u03c4\u03b9 \u03b1\u03ba\u03cc\u03bc\u03b1. \u03a4\u03bf \u03c3\u03c7\u03ae\u03bc\u03b1 \u03b1\u03c5\u03c4\u03cc \u03c0\u03c9\u03c2 \u03c4\u03bf \u03c6\u03c4\u03b9\u03ac\u03be\u03b1\u03c4\u03b5? :)", "Solution_3": "[quote=\"kalantzis\"]\u03a0\u03bf\u03bb\u03cd \u03ba\u03b1\u03bb\u03ae \u03b7 \u03bb\u03cd\u03c3\u03b7 \u03c3\u03b1\u03c2 \u03ba\u03b1\u03b9 \u03c0\u03b9\u03bf \u03b1\u03c0\u03bb\u03ae \u03b1\u03c0\u03cc \u03b1\u03c5\u03c4\u03ae \u03c0\u03bf\u03c5 \u03c3\u03ba\u03b5\u03c6\u03c4\u03cc\u03bc\u03bf\u03c5\u03bd. \u0395\u03b3\u03ce \u03b1\u03c0\u03ad\u03b4\u03b5\u03b9\u03be\u03b1 \u03cc\u03c4\u03b9 \u03c4\u03b1 $ P_1,P_2,P_3,Z$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf\u03bc\u03bf\u03ba\u03c5\u03ba\u03bb\u03b9\u03ba\u03ac \u03ba\u03b1\u03b9 \u03bc\u03b5 \u03b2\u03ac\u03c3\u03b7 \u03b1\u03c5\u03c4\u03cc \u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 $ Y,Z,P_3$ \u03ba\u03b1\u03b9 $ X,Z,P_3$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c5\u03bd\u03b5\u03c5\u03b8\u03b5\u03b9\u03b1\u03ba\u03ac \u03ac\u03c1\u03b1 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03c4\u03bf \u03b6\u03b7\u03c4\u03bf\u03cd\u03bc\u03b5\u03bd\u03bf.\n\n\u039a\u03ac\u03c4\u03b9 \u03b1\u03ba\u03cc\u03bc\u03b1. \u03a4\u03bf \u03c3\u03c7\u03ae\u03bc\u03b1 \u03b1\u03c5\u03c4\u03cc \u03c0\u03c9\u03c2 \u03c4\u03bf \u03c6\u03c4\u03b9\u03ac\u03be\u03b1\u03c4\u03b5? :)[/quote]\r\n\u03a4\u03bf \u03c3\u03c7\u03ae\u03bc\u03b1 \u03bc\u03b5 \u03c4\u03bf Word" } { "Tag": [ "Princeton", "college" ], "Problem": "I though it would be cool if everyone who is graduating this year posts what college they plan on attending. People in college now can post their college also.", "Solution_1": "This is an excellent thread for this time of year. \r\n\r\nJust for completeness, I will note that my undergraduate university was the University of Minnesota College of Liberal Arts, the only college to which I applied. I was in class of '80 there, majoring in Chinese language.", "Solution_2": "[url=http://www.mtu.edu/]Michigan Technological University[/url], where I will be getting so many AP credits that I will likely graduate early with a double major. I'll be there for grad school as well (hey, they're going to pay me to go there for grad school. I'm not looking further than that)", "Solution_3": "I think most grad schools will pay you to go there anyway.", "Solution_4": "[quote=\"ComplexZeta\"]I think most grad schools will pay you to go there anyway.[/quote]\r\nReally? MTU offered full tuition, partial fees, and a monthly stipend (oh, and guaranteed admission if I keep a 3.0 GPA in undergrad). Can I get this deal elsewhere too?", "Solution_5": "A graduate student here told me that at UCSB, if you teach a few classes, then you get free tuition and a stipend of something like $15000 a year, if I recall correctly. I don't think you can get guaranteed admission, but if you do well, you will presumably have many options for graduate school (or at least I hope so).", "Solution_6": "Princeton University, Simon Fraser University, or the University of Waterloo.", "Solution_7": "Princeton University", "Solution_8": "Hmm, it seems Princeton is quickly becoming over-represented :)\r\n\r\n\r\nI attend Harvard.", "Solution_9": "[quote=\"JBL\"]Hmm, it seems Princeton is quickly becoming over-represented :)\n[/quote]\r\n\r\nThat's never possible! Go Tigers!" } { "Tag": [], "Problem": "$ \\cfrac{\\cfrac{3}{8}\\plus{}\\cfrac{7}{8}}{\\cfrac{4}{5}}\\equal{}$", "Solution_1": "$ \\equal{}\\frac{\\frac{10}{8}}{\\frac{4}{5}}\\equal{}\\frac{\\frac{5}{4}}{\\frac{4}{5}}\\equal{}(\\frac{5}{4})^2\\equal{}\\boxed{\\frac{25}{16}}$.", "Solution_2": "doesn't \"simplified fraction\" mean the answer it 1 3/4", "Solution_3": "[quote=mohan77]doesn't \"simplified fraction\" mean the answer it 1 3/4[/quote]\n\n$1\\dfrac{3}{4}\\neq\\dfrac{25}{16}$", "Solution_4": "[quote=gauss1181]$ \\equal{}\\frac{\\frac{10}{8}}{\\frac{4}{5}}\\equal{}\\frac{\\frac{5}{4}}{\\frac{4}{5}}\\equal{}(\\frac{5}{4})^2\\equal{}\\boxed{\\frac{25}{16}}$.[/quote]\n\nI agree", "Solution_5": "[quote=mohan77]doesn't \"simplified fraction\" mean the answer it 1 3/4[/quote]\n\n25/16 is not equal to 1 3/4. Where in the problem does it say simplified fraction anyway?", "Solution_6": "[quote=mohan77]doesn't \"simplified fraction\" mean the answer it 1 3/4[/quote]\n\n$1\\dfrac{3}{4}\\neq\\dfrac{25}{16}$ \nThe problem just says:\n$ \\cfrac{\\cfrac{3}{8}+\\cfrac{7}{8}}{\\cfrac{4}{5}}=$" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "let a,b,c are non-negative such that:$ a \\plus{} b \\plus{} c \\equal{} 1$.prove that:\r\n$ \\frac {a^2(b \\plus{} c)}{1 \\plus{} 4bc} \\plus{} \\frac {b^2(a \\plus{} c)}{1 \\plus{} 4ac} \\plus{} \\frac {c^2(a \\plus{} b)}{1 \\plus{} 4ab} \\leq \\frac {1}{4}$", "Solution_1": "[quote=\"tuantam1lan\"]let a,b,c are non-negative such that:$ a \\plus{} b \\plus{} c \\equal{} 1$.prove that:\n$ \\frac {a^2(b \\plus{} c)}{1 \\plus{} 4bc} \\plus{} \\frac {b^2(a \\plus{} c)}{1 \\plus{} 4ac} \\plus{} \\frac {c^2(a \\plus{} b)}{1 \\plus{} 4ab} \\leq \\frac {1}{4}$[/quote]\r\nHello, it is clear that\r\n$ LHS \\le \\sum a^2(b \\plus{} c) \\le \\frac {1}{4}.$\r\nSo your inequality is valid. :)", "Solution_2": "[quote=\"can_hang2007\"][quote=\"tuantam1lan\"]let a,b,c are non-negative such that:$ a \\plus{} b \\plus{} c \\equal{} 1$.prove that:\n$ \\frac {a^2(b \\plus{} c)}{1 \\plus{} 4bc} \\plus{} \\frac {b^2(a \\plus{} c)}{1 \\plus{} 4ac} \\plus{} \\frac {c^2(a \\plus{} b)}{1 \\plus{} 4ab} \\leq \\frac {1}{4}$[/quote]\nHello, it is clear that\n$ LHS \\ge \\sum a^2(b \\plus{} c) \\le \\frac {1}{4}.$\nSo your inequality is valid. :)[/quote]\r\n\r\n :D Ooops can hang! You placed the inequality sign the wrong way. It should read:\r\n\r\n$ LHS \\le \\sum a^2(b \\plus{} c) \\le \\frac {1}{4}.$", "Solution_3": "[quote=\"Q.E.D.avid\"][quote=\"can_hang2007\"][quote=\"tuantam1lan\"]let a,b,c are non-negative such that:$ a \\plus{} b \\plus{} c \\equal{} 1$.prove that:\n$ \\frac {a^2(b \\plus{} c)}{1 \\plus{} 4bc} \\plus{} \\frac {b^2(a \\plus{} c)}{1 \\plus{} 4ac} \\plus{} \\frac {c^2(a \\plus{} b)}{1 \\plus{} 4ab} \\leq \\frac {1}{4}$[/quote]\nHello, it is clear that\n$ LHS \\ge \\sum a^2(b \\plus{} c) \\le \\frac {1}{4}.$\nSo your inequality is valid. :)[/quote]\n\n :D Ooops can hang! You placed the inequality sign the wrong way. It should read:\n\n$ LHS \\le \\sum a^2(b \\plus{} c) \\le \\frac {1}{4}.$[/quote]\r\nThank you very much, Q.E.D.avid. I typed it so fast and didn't preview, so I didn't see the mistake in typing. :blush:", "Solution_4": "[quote=\"can_hang2007\"][/quote][quote=\"Q.E.D.avid\"][quote=\"can_hang2007\"][quote=\"tuantam1lan\"]let a,b,c are non-negative such that:$ a + b + c = 1$.prove that:\n$ \\frac {a^2(b + c)}{1 + 4bc} + \\frac {b^2(a + c)}{1 + 4ac} + \\frac {c^2(a + b)}{1 + 4ab} \\leq \\frac {1}{4}$[/quote]\nHello, it is clear that\n$ LHS \\le \\sum a^2(b + c) \\le \\frac {1}{4}.$\nSo your inequality is valid. :)[/quote][/quote][quote=\"can_hang2007\"][/quote]\r\nThank you very much,can-hang2007,you really are smarter people.\r\nI created it after read your proof for hungkhtn's inequality in math.vn.it is an innovative simple and not particularly.\r\nanother proof of mine:\r\n$ \\frac {1 + 4bc}{4} = \\frac {(a + b + c)^2}{4} + bc \\ge ab + bc + ca$\r\nConsequently:\r\n$ \\frac {4a^2(b + c)}{1 + 4bc} \\le \\frac {a^2(b + c)}{(ab + bc + ca)}$\r\nwe have:\r\n$ 4\\sum{\\frac {a^2(b + c)}{1 + 4bc}} \\le \\frac {\\sum{a^2(b + c)}}{ab + bc + ca} \\le \\frac {\\sum{a^2(b + c) + 3abc}}{ab + bc + ca} = 1$", "Solution_5": "please dont post the topic with names like \"easy\" or \"hard' , etc.\r\nif its really easy to you then why do you want to post it! that too in \"unsolved\" section?\r\n\r\nplease take care of that :)", "Solution_6": "[quote=\"phymax\"]please dont post the topic with names like \"easy\" or \"hard' , etc.\nif its really easy to you then why do you want to post it! that too in \"unsolved\" section?\n\nplease take care of that :)[/quote]\r\nif you have'n idea for problem then you should'n Comment like ditto!", "Solution_7": "[quote=\"can_hang2007\"][quote=\"tuantam1lan\"]let a,b,c are non-negative such that:$ a \\plus{} b \\plus{} c \\equal{} 1$.prove that:\n$ \\frac {a^2(b \\plus{} c)}{1 \\plus{} 4bc} \\plus{} \\frac {b^2(a \\plus{} c)}{1 \\plus{} 4ac} \\plus{} \\frac {c^2(a \\plus{} b)}{1 \\plus{} 4ab} \\leq \\frac {1}{4}$[/quote]\nHello, it is clear that\n$ LHS \\le \\sum a^2(b \\plus{} c) \\le \\frac {1}{4}.$\nSo your inequality is valid. :)[/quote]\r\n i think we can prove that:$ a^2(b \\plus{} c) \\plus{} b^2(c \\plus{} a) \\plus{} c^2(a \\plus{} b) \\plus{} \\frac {3}{4}abc \\le \\frac {1}{4}$\r\nequality when $ a\\equal{}b\\equal{}c\\equal{}\\frac{1}{3}$ or $ a\\equal{}b\\equal{}\\frac{1}{2},c\\equal{}0$", "Solution_8": "[quote=\"tuantam1lan\"][quote=\"can_hang2007\"][quote=\"tuantam1lan\"]let a,b,c are non-negative such that:$ a \\plus{} b \\plus{} c \\equal{} 1$.prove that:\n$ \\frac {a^2(b \\plus{} c)}{1 \\plus{} 4bc} \\plus{} \\frac {b^2(a \\plus{} c)}{1 \\plus{} 4ac} \\plus{} \\frac {c^2(a \\plus{} b)}{1 \\plus{} 4ab} \\leq \\frac {1}{4}$[/quote]\nHello, it is clear that\n$ LHS \\le \\sum a^2(b \\plus{} c) \\le \\frac {1}{4}.$\nSo your inequality is valid. :)[/quote]\n i think we can prove that:$ a^2(b \\plus{} c) \\plus{} b^2(c \\plus{} a) \\plus{} c^2(a \\plus{} b) \\plus{} \\frac {3}{4}abc \\le \\frac {1}{4}$\nequality when $ a \\equal{} b \\equal{} c \\equal{} \\frac {1}{3}$ or $ a \\equal{} b \\equal{} \\frac {1}{2},c \\equal{} 0$[/quote]\r\nWell, it is equivalent to Schur's Inequality of third degree, my friend. :)", "Solution_9": "[quote=\"can_hang2007\"][/quote][quote=\"tuantam1lan\"][quote=\"can_hang2007\"][quote=\"tuantam1lan\"]let a,b,c are non-negative such that:$ a + b + c = 1$.prove that:\n$ \\frac {a^2(b + c)}{1 + 4bc} + \\frac {b^2(a + c)}{1 + 4ac} + \\frac {c^2(a + b)}{1 + 4ab} \\leq \\frac {1}{4}$[/quote]\nHello, it is clear that\n$ LHS \\le \\sum a^2(b + c) \\le \\frac {1}{4}.$\nSo your inequality is valid. :)[/quote]\n i think we can prove that:$ a^2(b + c) + b^2(c + a) + c^2(a + b) + \\frac {3}{4}abc \\le \\frac {1}{4}$\nequality when $ a = b = c = \\frac {1}{3}$ or $ a = b = \\frac {1}{2},c = 0$[/quote][quote=\"can_hang2007\"]\nWell, it is equivalent to Schur's Inequality of third degree, my friend. :)[/quote]\r\ncan we find maximal value of the expression?\r\n$ a^2b + b^2c + c^2a + \\frac {3}{8}abc$\r\nwith $ a,b,c$ are non-negative such that:$ a + b + c = 1$", "Solution_10": "[quote=\"tuantam1lan\"][/quote][quote=\"can_hang2007\"][/quote][quote=\"tuantam1lan\"][quote=\"can_hang2007\"][quote=\"tuantam1lan\"]let a,b,c are non-negative such that:$ a + b + c = 1$.prove that:\n$ \\frac {a^2(b + c)}{1 + 4bc} + \\frac {b^2(a + c)}{1 + 4ac} + \\frac {c^2(a + b)}{1 + 4ab} \\leq \\frac {1}{4}$[/quote]\nHello, it is clear that\n$ LHS \\le \\sum a^2(b + c) \\le \\frac {1}{4}.$\nSo your inequality is valid. :)[/quote]\n i think we can prove that:$ a^2(b + c) + b^2(c + a) + c^2(a + b) + \\frac {3}{4}abc \\le \\frac {1}{4}$\nequality when $ a = b = c = \\frac {1}{3}$ or $ a = b = \\frac {1}{2},c = 0$[/quote][quote=\"can_hang2007\"]\nWell, it is equivalent to Schur's Inequality of third degree, my friend. :)[/quote][quote=\"tuantam1lan\"]\ncan we find maximal value of the expression?\n$ a^2b + b^2c + c^2a + \\frac {3}{8}abc$\nwith $ a,b,c$ are non-negative such that:$ a + b + c = 1$[/quote]\r\nI think that we can, I found thye maximal value, is $ \\frac {361}{136}$ but I'm not sur, we must wait for the best solver problem here,", "Solution_11": "[quote=\"tuantam1lan\"][/quote][quote=\"can_hang2007\"][/quote][quote=\"tuantam1lan\"][quote=\"can_hang2007\"][quote=\"tuantam1lan\"]let a,b,c are non-negative such that:$ a + b + c = 1$.prove that:\n$ \\frac {a^2(b + c)}{1 + 4bc} + \\frac {b^2(a + c)}{1 + 4ac} + \\frac {c^2(a + b)}{1 + 4ab} \\leq \\frac {1}{4}$[/quote]\nHello, it is clear that\n$ LHS \\le \\sum a^2(b + c) \\le \\frac {1}{4}.$\nSo your inequality is valid. :)[/quote]\n i think we can prove that:$ a^2(b + c) + b^2(c + a) + c^2(a + b) + \\frac {3}{4}abc \\le \\frac {1}{4}$\nequality when $ a = b = c = \\frac {1}{3}$ or $ a = b = \\frac {1}{2},c = 0$[/quote][quote=\"can_hang2007\"]\nWell, it is equivalent to Schur's Inequality of third degree, my friend. :)[/quote][quote=\"tuantam1lan\"]\ncan we find maximal value of the expression?\n$ a^2b + b^2c + c^2a + \\frac {3}{8}abc$\nwith $ a,b,c$ are non-negative such that:$ a + b + c = 1$[/quote]\r\nHello, we have the known result of Hungkhtn and Vasc is: $ a^2b + b^2c + c^2a + abc \\le \\frac {4}{27}$ with equality holds for $ a = b = c = 1/3$ and $ a = 2/3,b = 1/3,c = 0$. From this result, we can see that the maximum you asked is $ 4/27.$ :)", "Solution_12": "$ \\frac{3}{8}abc\\le abc$. Is that right?\r\nI don't think this nice inequality belongs to hungkhtn, canhang2007. As I know it appeared on Canada Mathematic olympiad 1996 .Sorry If my information source is wrong . :)", "Solution_13": "[quote=\"Allnames\"]$ \\frac {3}{8}abc\\le abc$. Is that right?\nI don't think this nice inequality belongs to hungkhtn, canhang2007. As I know it appeared on Canada Mathematic olympiad 1996 .Sorry If my information source is wrong . :)[/quote]\r\nCanada's problem is to prove that $ x^2y\\plus{}y^2z\\plus{}z^2x\\leq \\frac{4}{27}$ with $ a\\plus{}b\\plus{}c\\equal{}1$ and $ a,b,c>0$ and it's Canada 1999, Problem 5.", "Solution_14": "[quote=\"Allnames\"]$ \\frac {3}{8}abc\\le abc$. Is that right?\nI don't think this nice inequality belongs to hungkhtn, canhang2007. As I know it appeared on Canada Mathematic olympiad 1996 .Sorry If my information source is wrong . :)[/quote]\r\nIt is valid as $ abc \\ge 0$ and $ 3/8<1$. Let me explain more, because the inequality of Hungkhtn (you can see it is stronger than Canadian MO 1996 ;) ), we have\r\n$ a^2b\\plus{}b^2\\plus{}c^2a\\plus{}\\frac{3}{8}abc \\le a^2b\\plus{}b^2c\\plus{}c^2a\\plus{}abc \\le \\frac{4}{27}.$\r\nMoreover, the equality can attain when $ a\\equal{}2/3,b\\equal{}1/3,c\\equal{}0,$ so this is also our maximum value. :)" } { "Tag": [ "MATHCOUNTS" ], "Problem": "The rules are the same in Cap. Sugar's saga.\r\n\r\nProblem 1\r\n\r\nWhile I type this, I'm listening to my ipod. Due to difficulties with the computer, only 5 songs of my favorite album were put into it. There are 10 total songs in the whole album. Here are the times for each:\r\n\r\nOceans and Rain: 3 min 45 sec\r\nGhoom Charakhana: 2 min 54 sec\r\nTranscendence: 3 min 26 sec\r\nHarmon Session Special XI: 2 min 34 sec\r\nSweet Lassi Dub: 3 min 54 sec\r\nSirocco: 4 min 11 sec\r\nDer Bauch: 3 min 29 sec\r\nJust You and I: 2 min 59 sec\r\nCaravane: 3 min 1 sec\r\nShaskin: 3 min 3 sec\r\n\r\nIt so happens that the average of those songs equals one of the songs +.8 second times 2. Find the average of the minutes and the seconds of that song.", "Solution_1": "You mean (time of song+0.8) *2 right?", "Solution_2": "don't you think [b]four[/b] marathons in one forum is a little too much?", "Solution_3": "@kyyuanmathcount: Yes.\r\n@splashD: Well, I may decide to move it to another forum...", "Solution_4": "Yeah, I believe people in this forum are really overloaded with work. I will, however, move it to a forum of your choice, you just have to make the level appropriate :). Pm me.\r\n\r\n\r\nEdit: Alright. Mathcounts --> MOEMS", "Solution_5": "uh, this isn't exactly [i]active[/i]", "Solution_6": "Sorry.\r\n\r\n#2\r\n\r\nMy dad needs time to put the other 5 songs onto my ipod. He also needs to finish working on the downstairs bathroom. If he will finish in x-3 days if he works x hours per day, and if he works 8 more hours, he will finish that. He will fix my ipod in one day maximum. If he never fixes my ipod the same day he works, and never starts on my ipod until he is finished in the bathroom, and he will work x hours per day in the bathroom, what is the minimum time needed so that I can have my ten songs onto my ipod?", "Solution_7": "I have no idea what that question says. Please repeat it clearly." } { "Tag": [ "geometry", "perimeter", "3D geometry", "sphere", "incenter" ], "Problem": "A circle is inscribed in a square. The area of the square divided by its perimiter is the area of the circle divided by its circumfrence.\r\n\r\n1) How big must a triangle be?\r\n\r\n2) How big must a pentagon be?\r\n\r\n3) Prove the general rule that should be obvious from here.\r\n\r\n4) Is the rule also trule for Archimedean solids and spheres?", "Solution_1": "I'm not sure I understand. Do you mean \"how big must a triangle inscribed in the same circle be\"?", "Solution_2": "Oh, sorry. I meant a triangle that has the property that its area over its perimeter is the same as that as the square and circle. Same for a pentagon.", "Solution_3": "[hide=\"the general rule is\"]\nfor [b]any[/b] polygon that has a circle of radius $r$ inscribed in it, perimeter $2s$, and area $K$, $K=rs$. (it can be rewritten as $\\frac{K}{2s}=\\frac{\\pi r^{2}}{2\\pi r}$)\n\nTo see that this formula works, we can split an $n$-sided polygon with an incircle into $n$ triangles, each formed by the incenter and two consecutive vertices. Let the polygon be $P_{1}P_{2}P_{3}\\cdots P_{n}$ with incenter $I$. Then triangle $P_{k}P_{k+1}I$ has a base length of $P_{k}P_{k+1}$ and a height of $r$, therefore an area of $\\frac{r}{2}P_{k}P_{k+1}$.\n\nNow add up the areas of the $n$ triangles to get the area of the whole polygon, and note that\n\\[P_{1}P_{2}+P_{2}P_{3}+\\cdots+P_{n}P_{1}=2s\\]\nso that\n\\[K=\\frac{r}{2}(2s)=rs\\]\nas desired.\n[/hide]\n\n[hide=\"for three dimensions...\"]\nif a polyhedron has a sphere of radius $r$ inscribed in it, surface area $S$, and volume $V$, then\n\\[V=\\frac{1}{3}(rS)\\]\n[/hide]" } { "Tag": [ "AMC", "AIME", "USA(J)MO", "USAMO", "geometry", "induction", "pigeonhole principle" ], "Problem": "I have gotten a lot better at problem solving the past few weeks (i.e., 6's on practice NYSMLs as opposed to 2's; 5's and 6's [out of 6]on practice westchester math league contests). However, I was a wimp last year and decided to only do 12 of the AMC 12, so I got in with a 104.5 and did zilch on AIME. Should I try to work on my AMC skills and do the Art of Problem solving AMC 12 Practice Series, or should I do the AIME series?", "Solution_1": "Learning to solve AIME problems better will probably improve your AMC score anyway.", "Solution_2": "It depends on your capabilities. If you are getting under 5s on AIMEs then work on AMC 12... If you are getting over 4s then work on AIME stuff.", "Solution_3": "You should do AoPS 2. All of it. And do every single AIME/AMC12, ever.\r\n\r\nIf you do that, the classes will be a waste of time, and you can work through that faster than waiting every week for the next lesson. It also saves you $\\$100$.", "Solution_4": "So you think I can qualify for USAMO doing just practice AIME/AMC 12 and AoPS Volume 2?", "Solution_5": "[quote=\"OVB\"]So you think I can qualify for USAMO doing just practice AIME/AMC 12 and AoPS Volume 2?[/quote]\r\nYes. Most definitely. What do you think you'd be doing special in WOOT or problem series anyway?", "Solution_6": "[quote=\"PenguinIntegral\"][quote=\"OVB\"]So you think I can qualify for USAMO doing just practice AIME/AMC 12 and AoPS Volume 2?[/quote]\nYes. Most definitely. What do you think you'd be doing special in WOOT or problem series anyway?[/quote]\r\n\r\nWell, I'm not completely sure about problem series, but WOOT is completely a different story. In WOOT, you actually learn how to approach problems (through problems, of course). However, they teach you how to think through certain formulae and show you where to apply what.", "Solution_7": "[quote=\"Karth\"][quote=\"PenguinIntegral\"][quote=\"OVB\"]So you think I can qualify for USAMO doing just practice AIME/AMC 12 and AoPS Volume 2?[/quote]\nYes. Most definitely. What do you think you'd be doing special in WOOT or problem series anyway?[/quote]\n\nWell, I'm not completely sure about problem series, but WOOT is completely a different story. In WOOT, you actually learn how to approach problems (through problems, of course). However, they teach you how to think through certain formulae and show you where to apply what.[/quote]\r\nAlso, there's really no harm in trying WOOT. If you decide before the third class that you've changed your mind and don't want to take it anymore, you can drop the class and get a full refund.", "Solution_8": "Hmm... I had a question about this policy... is this after the third class? As in Building Bridges, Geometry of the Circle, and Induction & Pigeonhole Principle? Or is this Building Bridges A, Building Bridges B, and Geometry of the Circle A?", "Solution_9": "[quote=\"Karth\"]Hmm... I had a question about this policy... is this after the third class? As in Building Bridges, Geometry of the Circle, and Induction & Pigeonhole Principle? Or is this Building Bridges A, Building Bridges B, and Geometry of the Circle A?[/quote]\r\nI'm not sure... but you'd be better off asking this question in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=140]Classes Information Forum[/url]. :wink:", "Solution_10": "[quote=\"Karth\"]Hmm... I had a question about this policy... is this after the third class? As in Building Bridges, Geometry of the Circle, and Induction & Pigeonhole Principle? Or is this Building Bridges A, Building Bridges B, and Geometry of the Circle A?[/quote]\r\n\r\nI think Orientation counts as a lesson too", "Solution_11": "[quote=\"PenguinIntegral\"][quote=\"OVB\"]So you think I can qualify for USAMO doing just practice AIME/AMC 12 and AoPS Volume 2?[/quote]\nYes. Most definitely. What do you think you'd be doing special in WOOT or problem series anyway?[/quote]\r\n\r\nWell just doing problems is helpful until you can't solve one of them...then it's kind of pointless unless someone is there to explain how to [i]come up with[/i] the solution, not just solve the problem.", "Solution_12": "i agree with most comments mentioned above (except somebody talking about W00T... I don't think W00T was part of this thread's purpose :D)\r\n\r\nIf you are doing [i]okay[/i] on the AIME (around 5 or higher) than you should definitely work on the AIME (unless you're worried about not qualifying; even so, you might think about working AIME problems); by working on AIME, your brain learns more about harder problems, and the mid-range-level problems become easier (So your AMC12 score should also go up).\r\n\r\nIf you are not quite doing that well on AIME yet, then maybe you should try to do a mix of both. Figure out what any obvious weaknesses you have (for example, if you just skip [i]every[/i] problem that starts out with \"A triangle...\") and try to cap those holes. Work on AMC12 tests until you feel comfortable working 18 or more problems (just an estimate) and then move on to AIME.\r\n\r\nOf course, I shouldn't be talking as I screwed both the AMC12 and AIME last year... and still made it to MOP... dunno how that worked out...", "Solution_13": "[quote=\"Mathlete9366\"]for example, if you just skip [i]every[/i] problem that starts out with \"A triangle...\"[/quote]\r\n\r\nI'm like that the first year when I really try to study for AIME... :rotfl: (except I also skip every problem with a picture)" } { "Tag": [], "Problem": "[url=http://news.yahoo.com/s/afp/20090528/od_afp/swedeneducationoffbeatiraq_20090528124335]A student in Sweden[/url]\r\n\r\nI'm not familiar with Bernoulli's number. Is this a really a huge achievement?", "Solution_1": "That depends on the nature of the formula. The [url=http://en.wikipedia.org/wiki/Bernoulli_number]Bernoulli numbers[/url] are rather deep, but any particular formula would have to be evaluated on the basis of its explanatory power or its [url=http://en.wikipedia.org/wiki/Bernoulli_number#Efficient_computation_of_Bernoulli_numbers]complexity[/url].", "Solution_2": "http://news.yahoo.com/s/afp/20090528/wl_mideast_afp/swedeneducationoffbeatreply_20090528174251\r\n\r\nSo the formula isn't new... and according to http://de.wikipedia.org/wiki/Mohamed_Altoumaimi it's similar to the Worpitzky's formula. Now I don't understand one thing: Worpitzky's formula isn't exactly 'a formula', it's more of an algorithm.", "Solution_3": "Oh, Sweden. As if [url=http://plus.maths.org/latestnews/sep-dec03/oxenhielm/]Elin Oxenhielm[/url] wasn't bad enough.", "Solution_4": "In any case, a single achievement, even if it is real, can always be just a random stroke of luck. If you want to call someone a \"genius\" or some other word like that, you need some consistent record of many achivements over some reasonable time. The media just cannot wait that long; each reporter needs to be the first to deliver an exciting news, so they often come with premature judgements and announcements, especially when something like \"a child prodigy\" is concerned. This makes much more harm than good to both the community and the person in question. I only hope that Mohamed and his friends and relatives have enough common sense to just ignore all that media noise around his name.", "Solution_5": "What \"fedja\" has said is precisely true. I think, maybe an apt title for this thread should be -- [b] \"Another young 'fraudigy'\" [/b] :lol: though, I would be really surprised if he turned out to be a real prodigy. It really sounds pretty absurd (I really can't believe it) how a young person equipped with a limited amount of mathematical tools can solve a problem that has confounded experts for so many years.", "Solution_6": "[quote=\"t0rajir0u\"]That depends on the nature of the formula. The [url=http://en.wikipedia.org/wiki/Bernoulli_number]Bernoulli numbers[/url] are rather deep, but any particular formula would have to be evaluated on the basis of its explanatory power or its [url=http://en.wikipedia.org/wiki/Bernoulli_number#Efficient_computation_of_Bernoulli_numbers]complexity[/url].[/quote]\r\nhave you looked at his formula yet?", "Solution_7": "The only picture I could find that includes the formula itself is [url=http://www.thelocal.se/19710/20090528/]here[/url]. This is exactly what is called [url=http://en.wikipedia.org/wiki/Bernoulli_number#Connection_with_the_Worpitzky_number]Worpitzky representation[/url] except $ p$ near the end should be $ (p\\plus{}1)$ (I do not know why this error appears on the picture; it certainly results in a different answer, which is incorrect, but it may be just a stupid \"misprint\"). The Worpitzky formula is far from obvious, of course, but it is rather well-known and I do not see why one should get so excited about its being rediscovered.", "Solution_8": "It could well be that the picture is just of him in front of something that looks like a math formula. Generally, newspapers do not care much about mathematical accuracy. If the story is some boy-genius, so long as there is something in the background that looks complicated enough to a non-mathematical observer, even if it has nothing to do with what he proved (though this seems to have a bunch to do with what he proved, so it could just have been one of many things he wrote on the board explaining what he proved to someone, and the photographer chose that spot as the most photogenic...)\r\n\r\nThat said, just as I do not have much faith in photojournalists correctly representing mathematical achievements, I just barely have a little more faith in written journalists correctly representing mathematical achievements. It could be the AP ballooning something not very big to be a big deal. That's not saying this kid isn't incredibly smart, just that his proof may be an incredible achievement for someone his age, with the journalist just jumping to the conclusion that it therefore must be incredible consequential to mathematics as a whole.\r\n\r\nBut I would have to see the *actual* formula, (and probably ask someone far smarter than me) in order to figure out what is the case.", "Solution_9": "It seems that professional mathematicians in Sweden are very reluctant to contact news papers. Unfortunately, the only somewhat recent achievement I know of that should have been written about is this: http://www.claymath.org/research_award/Dencker/", "Solution_10": "The only thing that grabs the average reader's attention is that the kid is Iraqi. I assume this has something to do with why they bothered to publish such a story.", "Solution_11": "TZF this is random but are you going into pure math or applied?", "Solution_12": "[b]@ mathmonster [/b]: It would be much better if you had asked [b] TZF [/b] via a PM.", "Solution_13": "likewise\r\ntrue, most assume refugees are poor and unresourceful\r\nnationality really has nothing to add to the story" } { "Tag": [ "ratio", "geometry", "vector", "function" ], "Problem": "Two particles move along the edges of equilateral triangle $ \\triangle ABC$ in the direction\r\n\\[ A\\rightarrow B\\rightarrow C\\rightarrow A\r\n\\]starting simultaneously and moving at the same speed. One starts at $ A$, and the other starts at the midpoint of $ \\overline{BC}$. The midpoint of the line segment joining the two particles traces out a path that encloses a region $ R$. What is the ratio of the area of $ R$ to the area of $ \\triangle ABC$?\r\n\r\n$ \\textbf{(A)}\\ \\frac {1}{16}\\qquad \\textbf{(B)}\\ \\frac {1}{12}\\qquad \\textbf{(C)}\\ \\frac {1}{9}\\qquad \\textbf{(D)}\\ \\frac {1}{6}\\qquad \\textbf{(E)}\\ \\frac {1}{4}$", "Solution_1": "[hide=\"Not really a solution\"] Draw the perpendicular from $A$ to $BC$. At the beginning the midpoint of the particles is the midpoint of this perpendicular. Once they have both moved half a side, the midpoint of the particles is at the midpoint of the perpendicular from $B$. And so forth symmetrically. \n\nDivide up $\\triangle ABC$ into four congruent equilateral triangles, then divide each of those into four more congruent triangles. The path of the midpoint traces out exactly the boundary of the middle triangle, which has area $\\frac{1}{16}$ that of the original. Hence the answer is $\\boxed{ A }$. [/hide]", "Solution_2": "blah...use vectors to show that the image is a triangle", "Solution_3": "Could somebody show the method that Altheman is describing with the vectors?", "Solution_4": "The position vector as a function of time of one point can be written as $ v_1 t\\plus{}c_1$. The other is $ v_2 t\\plus{}c_2$. Then the midpoint is $ \\frac{v_1\\plus{}v_2}{2}\\cdot t\\plus{}\\frac{c_1\\plus{}c_2}{2}$. It follows that while the velocity (i.e. speed and direction are constant) that the midpoint of those two points also has a constant velocity. Thus in such an interval, the midpoint traces a line segment. The whole path is just the union of 6 line segments.", "Solution_5": "Each time the particles move one half side the midpoints form a line, because of for example what Altheman said above (you don't need to actually compute anything to see this). Two points form a line and so there are 3 lines which form a triangle that we know must have vertices at the midpoint of the medians of the original triangle. The rest follows from what t0rajir0u said above.", "Solution_6": "Can someone draw a picture for this since I am having a hard time understanding what the figure looks like.", "Solution_7": "Picture is here :)", "Solution_8": "Thus the answer is $A$, the ratio being $1/16$.\n\nBest regards,\nsunken rock", "Solution_9": "umm this one is rlly easy for a #22\na homothety centered at the centroid and scale factor $\\frac{3}{2} \\cdot (\\frac{2}{3}-\\frac{1}{2})=\\frac{1}{4}$ gets you the [hide=result]$\\boxed{\n\\textbf{(A)}\\ \\frac {1}{16}}$[/hide] immediately ", "Solution_10": "I kind of guessed that the locus was an equilateral triangle (after plotting three vertices and noting that a conic section wouldn't work, based on the answer choices), but I was looking at wiki solution #1 for this, and they used the fact that the distance between the two points was linear (at least, it's a square root of a quadratic) in terms of $x$, where $x$ is the distance traversed by the two particles, but I don't understand why that would justify the locus being an equilateral triangle." } { "Tag": [ "inequalities", "real analysis", "real analysis solved" ], "Problem": "This is from the county round of the Romanian olympiad, guess what year :D :\r\n\r\nLet b_n be a real sequence s.t. a_n=b_1+b_2+..+b_n is bounded. Let f:[0,1]->R a Riemann integrable function. Prove that \r\n\r\n1/n*(\\sum {1<=k<=n} f(k/n)*b_k) -> 0 as n->00.", "Solution_1": "(1/n) \\sum f(k/n)*b_k=(1/n)*(a_1*(f(1/n)-f(2/n))+a_2*(f(3/n)-f(2/n))+...+a_(n-1)*(f((n-1)/n)-f(n/n))+a_n*f(n/n)). [b](1)[/b] (I think this is Abel's identity but it is easy anyway)\r\n\r\nWe can assume a_k>0 for all k. That is because (a_n) is bounded so we can add a b>0 so b_1 such that a_n>0, and if we replace b_1 by b+b_1, we don't change the limit.\r\nSo assume 0(1/n)*m*(f(1/n)-f(2/n)+f(2/n)-...+f((n-1)/n)-f(1)+f(1))=(m/n)*f(1/n).\r\nBecause f is Riemann integrable, lim(m/n)f(1/n)=0, so our limit is \\geq 0.\r\nDo the same for M to get our limit \\leq 0 and we are done.", "Solution_2": "[quote=\"amfulger (1)>(1/n)*m*(f(1/n)-f(2/n)+f(2/n)-...+f((n-1)/n)-f(1)+f(1))=(m/n)*f(1/n).\r\nBecause f is Riemann integrable, lim(m/n)f(1/n)=0, so our limit is \\geq 0.\r\nDo the same for M to get our limit \\leq 0 and we are done.[/quote]\r\n\r\nNobody bothered to see if what I wrote was correct, so nobody noticed the mistake. We can have f(k/n)k| \\leq M for all k.\r\n|(1)| \\leq (1/n)*M*(|f(1/n)-f(2/n)|+|f(2/n)-f(3/n)|+...+|f(n-1/n)-f(n/n)|+|f(1)|.\r\nIt is easy to see that after removing the || (write f(a)-f(b) or f(b)-f(a)), in the RightHandSide (RHS), we have M/n*( the difference of two Riemman sums associated to f), so the RHS tends happily to 0 :D , so (1) tends to 0." } { "Tag": [ "geometry", "circumcircle", "Russia" ], "Problem": "Given a triangle $ABC$. Let a circle $\\omega$ touch the circumcircle of triangle $ABC$ at the point $A$, intersect the side $AB$ at a point $K$, and intersect the side $BC$. Let $CL$ be a tangent to the circle $\\omega$, where the point $L$ lies on $\\omega$ and the segment $KL$ intersects the side $BC$ at a point $T$. Show that the segment $BT$ has the same length as the tangent from the point $B$ to the circle $\\omega$.", "Solution_1": "Let $AL$ meet the circumcircle of $\\triangle{ABC}$ at $L'$, then obviously $KL||BL'$.\r\nSince $\\angle{CTL}=\\angle{CBL'}=\\angle{L'AC}$ we get that $ATLC$ is cyclic. \r\nTherefore, if $LC$ meets the circumcircle of $\\triangle{ABC}$ at $W$, then $\\angle{CAL}=\\angle{WAL}=\\angle{BAT}=\\angle{KTB}$. \r\nThus $BT^2=BK\\cdot{BA}$ and the conclusion follows.", "Solution_2": "More complicated:\nLet $\\{A,J\\}\\in\\omega\\cap AC$; clearly $KJ\\parallel BC$. $JAKL$ is cyclic, so $\\angle JAL=\\angle JKL=\\angle CTL\\implies CATL$ cyclic and $\\angle LCT=\\angle LAT\\ (\\ *\\ )$.\nIf $\\{M,N\\}\\in\\omega\\cap BC$, order $B-M-N-C$ and $\\{A,X\\}\\in\\omega\\cap AT$, from $(*)$ we infer $\\stackrel{\\frown}{MN}-\\stackrel{\\frown}{NL}=\\stackrel{\\frown}{LX}$, so $LX\\parallel BC\\parallel KJ$, thence $\\angle LTC=\\angle LAC=\\angle JAL=\\angle KAT$ and $BT$ is tangent to $\\odot (KAT)$. Since $AK\\equiv AB$ is radical axis of circles $\\odot (KAT)$ and $\\omega$, we are done.\n\nBest regards,\nsunken rock", "Solution_3": "Join $AT$ and $AL$. Let $\\omega\\cap AC$ = $M$. Join $KM$.\nSince, $\\omega$ and circle $ABC$ touch each other,\nso, $KM$ is parallel to $BC$. \nThus, $\\angle CTL$ = $\\angle MKL$ = $\\angle MAL$.\n\u2234 $ATLC$ is also cyclic.\nSo, $\\angle ATC$ = $\\angle ALC$ = $\\angle AKL$ [alternate segment theorem]\n\u2234$BC$ is tangent to circle $AKT$.\nBy power of a point theorem,\n$BT^2$ = $BK\\cdot BA$ = $BN^2$.\nHence, $BT$ = $BN$.", "Solution_4": "Let BS be tangent to $\\omega$. BS^2 = BK.BA so we need to prove BT^2 = BK.BA or in fact AKT is tangent to BC at T. \u2220AKT = \u2220ALC and we want to prove \u2220AKT = \u2220ATC so we will prove ATLC is cyclic. Let J be random point on Line L which is tangent to $\\omega$. \u2220ACT = \u2220BTJ = \u2220KAJ = \u2220ALT so ATLC is cyclic as wanted. we're Done.", "Solution_5": "Let $\\omega$ again meet $AC$ at $M.$ Clearly $KM\\parallel BC,$ hence $\\angle CTL=\\angle MKL=\\angle CAL$ yields that $ATLC$ is cyclic.\nThus $\\angle ATC=\\angle ALC=\\angle AKL$ and $|BT|^2=|BK|\\cdot |AB|,$ so we are done.", "Solution_6": "Harmonic bundles!\nLet $\\omega \\cap AC=N, \\omega \\cap BC = D,E$. Also let $U$ be point such a $B$ is midpoint of $UT$ and $KU \\cap \\omega = V$. From condition we have $KN \\parallel BC$.\nNote that we want to prove that $(U,T;D,E)=-1$. But $(U,T;D,E)=^K(V,L;D,E)$, so we need to prove that $CV$ is tangent to $\\omega$.\nBut $-1=(U,T;B, \\infty_{BC})=^K(V,L;A,N)$, so $VALN$ is harmonic and really $CV$ is tangent to $\\omega$. $\\blacksquare$" } { "Tag": [ "function", "inequalities", "calculus", "integration" ], "Problem": "SMO 2007, Open #14\r\n\r\nLet $N$ be the set of positive integers and let $f: N \\rightarrow N$ be such that \\[f(m+n) \\geq f(m)+f(n)-1\\]for any $m, n$ which are elements of $N$ and $f(1)>1$ and $f(8888) < 9000$. Determine the value of $f(2007)$.", "Solution_1": "if $f(0)=1$\r\n$\\forall n\\in \\mathbb{N}^{*},\\ f(n)\\ge f(n-1)+f(1)-1\\ \\Longrightarrow\\ f(n)\\ge n(f(1)-1)+1\\ge n+1\\\\ we\\ take\\ g(n)=f(n)-n-1\\\\ \\\\ i)\\ \\forall (n,m)\\in \\mathbb{N}^{2},\\ g(n+m)\\ge g(n)+g(m) \\\\ ii)\\ g(8888)<111$\r\nthere are infinity of solutions.\r\nif $g$ check i), then $\\exists (a,b,h)\\in \\mathbb{N}\\times \\mathbb{N}\\times \\mathbb{N}^{\\mathbb{N}}$\r\nsuch that:\r\n$\\forall n\\in [|0,a-1|],\\ g(n)=0$\r\n$g(a)=b$\r\n$\\forall n\\in[|a,+\\infty[,\\ g(n)=Max(\\{g(k)+g(n-k)| 00\r\nprove that: 2(a+b+C)=< (a^2)(b+C)+(b^2)(a+c)+(c^2)(a+b)", "Solution_1": "Please don't post into the \"Solved Problems\" section.\r\nAnd inequalities belong not to Algebra but to inequalities as the name says ;)", "Solution_2": "Note that your inequality is equivalent with the following:\r\n$\\sum_{sym} a^2b \\ge \\sum_{sym} a^{\\frac{5}{3}}b^{\\frac{2}{3}}c^{\\frac{2}{3}}$\r\nwhich holds by Muirhead." } { "Tag": [ "geometry", "trapezoid", "trigonometry" ], "Problem": "Trapezoid $ TRAP$ is a trapezoid inscribed in $ \\odot O$ such that $ T$, $ O$, and $ R$ are collinear, $ TR\\equal{}8$, and $ \\angle PTR\\equal{}75$ degrees. The area of trapezoid TRAP, in simplest radical form, can be written in the form $ k\\plus{}w\\sqrt{p}$. Find the value of $ k\\plus{}w\\plus{}p$.", "Solution_1": "[geogebra]9ccbaffe45fdeacbcfc86f10a2f2111b37b78db0[/geogebra] \r\n[hide=\"solution\"]\nSince $ TOR$ are collinear, they lie on the diameter of the cirlce and $ TRAP$ is perfectly symetrical.\nSo $ TP \\equal{} 8 \\cos(75) \\equal{} 8\\cos (30\\plus{}45) \\equal{} 8\\left(\\frac{\\sqrt{6}\\minus{}\\sqrt{2}}{4}\\right)\\equal{}2(\\sqrt{6}\\minus{}\\sqrt{2})$\n$ AP \\equal{} 8\\sin(30)\\cos(45)\\plus{}\\cos(30)\\sin(45) \\equal{} 2(\\sqrt{6}\\plus{}\\sqrt{2})$\n\nBy Ptolemy\n$ 8\\cdot AP \\plus{} 4(\\sqrt{6}\\minus{}\\sqrt{2})^2 \\equal{} 4(\\sqrt{6}\\plus{}\\sqrt{2})^2$\n$ \\cdot AP \\equal{} 2\\sqrt{3}$\n\n$ h\\equal{} 2(\\sqrt{6}\\minus{}\\sqrt{2})\\cos(75) \\equal{} \\frac{1}{2}\\cdot (\\sqrt{6}\\minus{}\\sqrt{2})^2 \\equal{} 4\\minus{}2\\sqrt{3}$\n\n$ [TRAP] \\equal{} \\frac{1}{2}\\cdot (4\\minus{}2\\sqrt{3})\\cdot 8 \\cdot 2\\sqrt{3} \\equal{} \\boxed{16(2\\sqrt{3}\\minus{}3)}$[/hide]", "Solution_2": "I did a lot of trig, and I got\r\n\\[ 4r^2 \\sin^3 \\theta \\cdot \\cos \\theta.\\]\r\nIs that right?", "Solution_3": "Edit: Sorry, I was being stupid.", "Solution_4": "When I evaluate my formula, I get something around 14.", "Solution_5": "Edit: Sorry, I was being stupid.", "Solution_6": "What situation must be impossible?", "Solution_7": "[hide=\"Solution\"]\nSo unless I'm being stupid again, the area is just $ \\frac12\\cdot2\\cdot(8 \\plus{} 4\\sqrt3) \\equal{} 8 \\plus{} 4\\sqrt3\\implies8 \\plus{} 4 \\plus{} 3 \\equal{} \\boxed{15}$.\n[/hide]\r\n\r\n[b]123456789:[/b] Your formula is correct." } { "Tag": [ "geometry" ], "Problem": "A vertical line divides a triangle with vertices (0, 0) , (1, 1) , and (9, 1) into two regions of equal area. What is the equation of this line?", "Solution_1": "As the line that divides the triangle is vertical line,\r\nso assume that the equation of that line is as follow:\r\n\r\nx - a = 0\r\n==> x = a \r\n\r\nAsuume triangle is ABC\r\nA=(0,0)\r\nB=(1,1)\r\nC=(9,1)\r\n\r\nBC = 8\r\n\r\nArea of the triangle : (1/2) . 8 . 1 = 4\r\n\r\nSuppose line (x - a = 0) intersects BC at D and AC at E.\r\n\r\nthen CD = 9 - a\r\n\r\nED = (1/9) . (CD) = (1/9).(9 - a)\r\n\r\nArea of triangle CDE = (1/2).(CD).(ED) = (1/2).(9 - a).(1/9).(9 - a)\r\n\r\n==> (1/2).(9 - a).(1/9).(9 - a) = 4/2 = 2\r\n==> (9 - a).(9 - a) = 36\r\n==> 9 - a = 6\r\n==> a = 3\r\nSo equation of the line is x - 3 = 0\r\n\r\n\r\nThanks,\r\nhttp://www.teacherone.com" } { "Tag": [ "MATHCOUNTS" ], "Problem": "Does anyone know the formula for the fibonnacci sequence? in algebratic terms?\r\nlike 1, 2, 3, 5, 8, 13, ....", "Solution_1": "$F(n) = \\frac{a^n - b^n}{\\sqrt{5}}$\r\n\r\nwhere $a = \\frac{1+\\sqrt{5}}{2}$ and $b = \\frac{1-\\sqrt{5}}{2}$.\r\n\r\nHope that helps.", "Solution_2": "isn't a term of the fibonnacci sequence just the sum of the two terms preceeding it?", "Solution_3": "chess64 gave you a theorem known as \"Binet's Formula\"\r\n\r\nAnother one you can use is the following:\r\n\r\n$F_n = \\displaystyle\\binom{n}{0} + \\displaystyle\\binom{n-1}{1} + \\displaystyle\\binom{n-2}{2} +...$\r\n\r\nThe sequence above will continue until $n-k$ is less than $k$.\r\n\r\nProofs of these can be found in section 16-3 of AoPS Volume II, but this is much beyond MathCounts, so it's best to remember the pattern as $F_0 = F_1 = 0, F_{n+1} = F_n + F_{n-1}$" } { "Tag": [ "LaTeX" ], "Problem": "Is there a command that makes the whole document double spaced?", "Solution_1": "[quote=\"Fierytycoon\"]\nUse: \n\n\\linespread{2} \n\nOr use the setspace package and the \\doublespacing command.\n[/quote]" } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "The letters $A$ or $B$ are assigned on the points divided equally into $2^{n}\\ (n=1,\\ 2,\\cdots)$ parts of a circumference.If you choose $n$ letters from any succesively arranging points directed clockwise, prove that there exists the way of assignning for which the line of letters are mutually distinct.", "Solution_1": "Draw a directed graph whose vertices are the sequences of length $ n\\minus{}1$,with an edge between two sequences if the last $ n\\minus{}2$ letters of the first vertex match the first $ n\\minus{}2$ letters of the second.\r\nThis graph has two edges into and out of each vertex,so there exists a (directed) path traversing each edge exactly once.We convert this into a cycle of the desired form by starting at any vertex,writting down the sequence corresponding to it,then appendinginturn thelast letter of each sequence we encounter along the path." } { "Tag": [ "search", "\\/closed" ], "Problem": "I search for *\\frac{a^2+bc}{b+c}*, then the search engine gave me a lot of irrelevant links.... How can I search normally? Thanks for your help.", "Solution_1": "See http://www.mathlinks.ro/Forum/viewtopic.php?t=97725", "Solution_2": "Yes, there is a problem with the search part... I tried $x^y=y^x$, which was post like 10 times, and it appeared no results.... are the admins trying to repear it?", "Solution_3": "Yes. Read the supplied links :)\r\n\r\nHave patience, fixes are on the way, but we are only human." } { "Tag": [ "function", "complex analysis", "complex analysis unsolved" ], "Problem": "Show that an even function f(z) can have only even powers in its Taylor expansion about z = 0, and similarly for odd functions. (Use f(z) - f(-z) = 0 and uniqueness) \r\n\r\nThank you.", "Solution_1": "So $ f(z)\\minus{}f(\\minus{}z)$ has a taylor expansion. And this is zero identically, so if \r\n\r\n$ f(z)\\equal{}\\sum a_n x^n$, we get\r\n\r\n$ (a_0\\plus{}a_1 x\\plus{}a_2 x^2\\plus{}...)\\minus{}(a_0\\minus{}a_1 x\\plus{}a_2 x^2\\plus{}...)\\equal{}0$\r\n\r\nCollecting terms on the left, we get\r\n\r\n$ 2a_1x\\plus{}2 a_3 x^3\\plus{}...\\equal{}0\\equal{}0\\plus{}0x\\plus{}0x^2\\plus{}...$\r\n\r\nso $ a_1\\equal{}a_3\\equal{}...\\equal{}0$\r\n\r\nWe can equate coefficients by uniqueness." } { "Tag": [ "geometry", "perpendicular bisector", "cyclic quadrilateral" ], "Problem": "Two points $ A$ and $ C$ are chosen on a circle $ \\Gamma$. Let $ O$ be an arbitrary point on the perpendicular bisector of $ AC$. The perpendicular bisectors of $ OA$ and $ OC$ meet $ \\Gamma$ at points $ B$ and $ D$ respectively such that $ A,B,C,D$ lie on $ \\Gamma$ in this order. $ AB$ and $ CD$ when extended meet at $ P$. Show that the segment adjoining the midpoints of $ AB$ and $ CD$ is perpendicular to $ OP$.", "Solution_1": "$ \\textbf{Lemma 1: }$ If $ ABCD$ is a cyclic quadrilateral and $ BA\\cap CD = P$ then,\r\n\\[ PB^2 - AB\\cdot PB = PC^2 - PC\\cdot DC\r\n\\]\r\n$ \\textbf{Proof: }$ From power of point, \r\n\\begin{align*} & PA \\cdot PB = PD \\cdot PC \\\\\r\n\\iff & PB(PB - AB) = PC(PC - DC) \\\\\r\n\\iff & PB^2 - AB\\cdot PB = PC^2 - PC\\cdot DC \\end{align*}\r\n\r\n$ \\textbf{Lemma 2}:$ If $ AB \\perp CD$ then, $ AC^2 - CB^2 = AD^2 - DB^2$\r\nIt is quite wellknown.\r\nNow, we need to prove that,\r\n\\[ PM^2 - MO^2 = PN^2 - ON^2 \\iff PM^2 - PN^2 = OM^2 - ON^2\r\n\\]\r\nHere,\r\n\\[ PM^2 = \\left( PB - \\frac {AB}{2}\\right)^2 = PB^2 + \\frac {AB^2}{4} - AB \\cdot PB\r\n\\]\r\nAnd,\r\n\\[ PN^2 = PC^2 + \\frac {DC^2}{4} - DC.PC\r\n\\]\r\nSo,\r\n\\[ PM^2 - PN^2 = \\frac {1}{4}(AB^2 - CD^2)\r\n\\]\r\nAgian, from the theorem of Apollonius,\r\n\\[ OM^2 = \\frac {1}{4}(2AO^2 + 2OB^2 - AB^2)\r\n\\\\\r\nON^2 = \\frac {1}{4}(2OD^2 + 2OC^2 - DC^2)\r\n\\]\r\nSubtracting we get,\r\n\\[ OM^2 - ON^2 = \\frac {1}{2}(AO^2 - OC^2 + OB^2 - OD^2) - \\frac {1}{4}(AB^2 - DC^2)\r\n\\]\r\nAs, $ AO = OC$, it is left to prove that, $ OB^2 - OD^2 = AB^2 - DC^2$\r\nWhich follows from, $ AB = OB$ and $ OD = DC$\r\n\r\nEDIT: Eh...Making a lot of typos :mad:", "Solution_2": "[quote=\"nayel\"] [color=darkred]Two points $ A$ and $ C$ are chosen on a circle $ w$ . Let $ O$ be an arbitrary point on the perpendicular bisector of the segment $ [AC]$ .\n\nThe perpendicular bisectors of the segments $ [OA]$ , $ [OC]$ meet the circle $ w$ at $ B$ , $ D$ respectively such that $ A,B,C,D$ lie on the circle $ w$ in this order.\n \nDenote the intersection $ P\\in AB\\cap CD$ and the midpoints $ M$ , $ N$ of the segments $ [AB]$ and $ [CD]$ respectively. Prove that $ OP\\perp MN$ .[/color] [/quote]\r\n\r\n[color=darkblue][b][u]Proof (metrical).[/u][/b]\n\nObserve that $ \\left\\|\\begin{array}{c}\nOA\\equal{}OC\\ \\ ;\\ \\ OB\\equal{}AB\\ \\ ;\\ \\ OD\\equal{}CD\\\\\\\\\np_w(P)\\equal{}PA\\cdot PB\\equal{}PC\\cdot PD\\\\\\\\\n2\\cdot PM\\equal{}PA\\plus{}PB\\ \\ ;\\ \\ 2\\cdot PN\\equal{}PC\\plus{}PD\\end{array}\\right\\|$ . I\"ll show (similarly to [b]Moonmathpi496[/b]) that $ PM^2\\minus{}PN^2\\equal{}OM^2\\minus{}ON^2$ . \n\n$ \\blacktriangleright\\ 4\\cdot\\left(PM^2\\minus{}PN^2\\right)\\equal{}(PA\\plus{}PB)^2\\minus{}(PC\\plus{}PD)^2\\equal{}$ $ PA^2\\plus{}PB^2\\minus{}PC^2\\minus{}PD^2\\equal{}(PA\\minus{}PB)^2\\minus{}(PC\\minus{}PD)^2\\equal{}AB^2\\minus{}CD^2$ .\n\n$ \\blacktriangleright\\ 4\\cdot\\left(OM^2\\minus{}ON^2\\right)\\equal{}$ $ \\left[2\\left(OA^2\\plus{}OB^2\\right)\\minus{}AB^2\\right]\\minus{}\\left[2\\left(OC^2\\plus{}OD^2\\right)\\minus{}CD^2\\right]\\equal{}$ $ \\left(2\\cdot OB^2\\minus{}AB^2\\right)\\minus{}\\left(2\\cdot OD^2\\minus{}CD^2\\right)\\equal{}AB^2\\minus{}CD^2$ .\n\nIn conclusion, $ PM^2\\minus{}PN^2\\equal{}OM^2\\minus{}ON^2\\equal{}\\frac 14\\cdot\\left(AB^2\\minus{}CD^2\\right)$ , i.e. $ OP\\perp MN$ .\n\n[b]Remark.[/b] I denoted the [b]power[/b] $ p_w(P)$ [b]of the point[/b] $ P$ [b]w.r.t. the circle[/b] $ w$ and I used the [b]theorem of median[/b] in $ \\triangle ABC\\ : \\ 4m_a^2\\equal{}2\\left(b^2\\plus{}c^2\\right)\\minus{}a^2$ . [/color]", "Solution_3": "Nice proof Mr. Nicula, indeed. :) \r\nBut if you just look at [url=http://www.mathlinks.ro/viewtopic.php?t=213117]the solution of yetti[/url] you are going to laugh or cry(like me) :wink:" } { "Tag": [], "Problem": "Find the value of $ K$ for which these two equations will not have a common solution:\n\\[ 6x \\plus{} 4y \\equal{} 7\\]\n\\[ Kx \\plus{} 8y \\equal{} 7\\]", "Solution_1": "hello, multiplying the first equation by $ \\minus{}2$ and adding the second equation we get the equation $ x(k\\minus{}12)\\equal{}\\minus{}7$.\r\nSo we have for $ k\\equal{}12$ a contradiction and it follows there exist no intersection point.\r\nSonnhard." } { "Tag": [ "algebra", "polynomial", "Vieta" ], "Problem": "Given that $ a, b,$ and $ c$ are the roots of $ x^3\\minus{}2x^2\\plus{}3x\\plus{}4$, find $ (a\\plus{}b)(a\\plus{}c)(b\\plus{}c)$.", "Solution_1": "Since a+b+c=2 from Vietas theorem, rewrite it as (2-a)(2-b)(2-c).\r\n\r\nThen you can use Vietas theorem once more when that term is expanded as you will get like an a+b+c thing, an ab+ac+bc thing, and abc thing, Vietas covers all three.", "Solution_2": "Even easier, $ (2\\minus{}a)(2\\minus{}b)(2\\minus{}c)$ is just $ P(2) \\equal{} 10$." } { "Tag": [ "topology", "induction", "superior algebra", "superior algebra theorems" ], "Problem": "OK. Maybe this is because I'm not into college algebra yet... but does (the infinite form of ) the axiom of choice imply a well-ordering on the reals? Is it possible to find that well-ordering? and... if not, is it possible to create a new axiom to allow that? :maybe: \r\n\r\n(and can we induct on the reals? :D )", "Solution_1": "The axiom of choice is equivalent to the statement that any set can be well-ordered. On the other hand, the rule of thumb about anything for which you invoked the axiom of choice or any equivalent just to claim it exists is something you're never going to be able to visualize in any concrete way.\r\n\r\nAs for your last question, try searching for the words \"Transfinite Induction.\" (Also, Zorn's Lemma and the Hausdorff Maximal Principle.)", "Solution_2": "by the way, a well-ordering on IR implies the existence of a IQ-basis of IR :-)", "Solution_3": "1. Yes, the axiom of choice implies a well-ordering on the reals. It also implies a well-ordering on any non-empty set. See [url=http://en.wikipedia.org/wiki/Well-ordering_theorem]here[/url].\r\n\r\n2. Yes, there is induction, sort of. However, it's a slightly different form called Transfinite Induction that involves ordinals. See [url=http://en.wikipedia.org/wiki/Transfinite_induction]here[/url]\r\n\r\n3. I don't think you need to say \"infinite form\" of the axiom of choice. The 'finite' form is in fact an obvious theorem of standard ZF axioms. The axiom of choice refers to an extension of ZF, and to be an extension it must to be the infinite form. So the distinction probably isn't necessary.\r\n\r\n4. One of the reasons for the axiom of choice being controversial is that it allows you state existence of things that can't be, or are really hard to, construct. The canonical example is using the Axiom of Choice to choose an element from each class of $ \\mathbb{R}/ \\mathbb{Q}$, which I believe is a partition of the reals where for every two sets, there do not exist elements from each with difference being a rational number, and every two elements in one set have their difference rational. (maybe not though, I don't know much about quotients) Anyways, if you try to actually construct that list of elements, you'll find yourself baffled quickly. The Axiom of Choice says you can do this anyways.\r\n\r\nThe well-ordering of the reals I believe is one of the things that can't be concisely described. Another example of something that is rather difficult to construct is the four pieces of a ball that allow the Banach-Tarski paradox.\r\n\r\nEDIT: Der, I got beaten." } { "Tag": [], "Problem": "I don't get the difference between them. If anybody could help, that would be awesome.", "Solution_1": "Should this go to chemistry forum?" } { "Tag": [ "probability" ], "Problem": "Rules: You have an egg. You choose someone to throw your egg at. Whoever's egg has a \"bomb\" in it kills the other person. We play until there is only one person left.\r\nI will be taking 14 people to start.\r\nSign up please, guys!\r\n1. spaceguy524\r\n2. meewhee009\r\n3. eqjj168\r\n4. 15!!!\r\n5. Scamper\r\n6. EggyLv.999\r\n7. Yoshi\r\n8. westiepaw\r\n9. cf249\r\n10. andrewjjiang97\r\n11. lotsofmath\r\n12. bluecarneal\r\n13. Richard_Min\r\n14.", "Solution_1": "I will join. :)\r\n\r\nAnd don't make the eggs thrown at me the bomb just because I beat you in the previous Egg Toss.", "Solution_2": "Mehz is /in.", "Solution_3": "Lol, I won't.\r\nI'll use random.org.", "Solution_4": "I'm in! :D \r\n1. spaceguy524\r\n2. meewhee009\r\n3. [b]eqjj168[/b]\r\n4.\r\n5.\r\n6.\r\n7.\r\n8.\r\n9.\r\n10.\r\n11.\r\n12.\r\n13.\r\n14.", "Solution_5": "i join\r\n1. spaceguy524 \r\n2. meewhee009 \r\n3. eqjj168 \r\n4. 15!!!\r\n5. \r\n6. \r\n7. \r\n8. \r\n9. \r\n10. \r\n11. \r\n12. \r\n13. \r\n14.", "Solution_6": "/in............", "Solution_7": "in.\r\n\r\nThe message is too small. Please make the message longer before submitting.", "Solution_8": "join. I hope I don't die early like last time.", "Solution_9": "In. Maybe one of these days, I'll win one of these games.", "Solution_10": "I finally get the rules!\r\n\r\n\r\n1. spaceguy524 \r\n2. meewhee009 \r\n3. eqjj168 \r\n4. 15!!! \r\n5. Scamper\r\n6. \r\n7. \r\n8. \r\n9. \r\n10. \r\n11. \r\n12. \r\n13. \r\n14.\r\n\r\nHow do you know if an egg has a bomb?", "Solution_11": "/in... :lol:", "Solution_12": "/in\r\n\r\n :D :D :D", "Solution_13": "[b]/in[/b]", "Solution_14": "1. spaceguy524 \r\n2. meewhee009 \r\n3. eqjj168 \r\n4. 15!!! \r\n5. Scamper \r\n6. EggyLv.999\r\n7. Yoshi\r\n8. westiepaw \r\n9. cf249\r\n10. andrewjjiang97\r\n11. lotsofmath\r\n12. bluecarneal\r\n13. \r\n14.\r\n\r\nEdit: 100th post!!! :lol: :clap2:", "Solution_15": "All right, he's inactive.\r\nJjfun1's voice comes over the speaker. \"Err... looks like spaceguy is in a coma. Everyone must throw their egg at him. Thank you.\" The speaker fell silent with an earsplitting squeak. Meewhee shrugged. She held not an egg in her hand, but a smelly sock. BAM! It hit spaceguy. Nothing happened except for the fact that he seemed to wrinkle his nose in disgust while still in his coma. Lotsofmath smashed is egg on spaceguy's head. Spaceguy snored. Yoshi threw a fail egg at lotsofmath, forgetting jjfun1's instructions. The egg seemed to stop in mid-air. It zoomed towards spaceguy and smashed into his face. He exploded. \r\n\r\n[b]Deadline: Sunday[/b]\r\n[hide=\"Player List\"]\n1.meewhee009\n2. Yoshi\n3. westiepaw\n4. lotsofmath[/hide]", "Solution_16": "Throw a Tyrannosaurus Rex Egg at westiepaw", "Solution_17": "[b]Throw a YOSHI egg at lotsofmath[/b].\r\n\r\nYay, I haven't died yet!", "Solution_18": "[b]Throw a RED Yoshi egg at lotsofmath[/b]", "Solution_19": "[b]Those who still need to throw: only meeewhee.[/b]", "Solution_20": "You guys have time zone advantage :o\r\n[b]Throws :rotfl: smiley @ lotsofmath[/b]", "Solution_21": "Awesome!\r\nJjfun1 smiled. lotsofmath swung his arm around and threw a Tyrannosaurus Rex Egg at westiepaw. Westiepaw threw a YOSHI egg at lotsofmath. Yoshi threw a RED Yoshi egg at lotsofmath. meewhee threw a :rotfl: smiley egg at lotsofmath. Everyone looked around, confused. Meewhee stammered, \"Ummm. Why is everyone still alive?\" Jjfun1 shrugged. \"I don't know,\" she replied, \"I guess this poison isn't what it said on the bottle.\" Suddenly, westiepaw fell to the ground. \"I don't feel so good...\" she whispered. Rawr.\r\n\r\n[b]Deadline: Sunday Night[/b]\r\nNote: Because there are less players, I'm going to make the deadline shorter. :P\r\n\r\n[hide=\"Player List\"]\n\n1.meewhee009\n2. Yoshi\n3. lotsofmath[/hide]", "Solution_22": "[b]Throws :ninja: egg at lotsofmath[/b]", "Solution_23": "OMG epic luck for two straight rounds!\r\n\r\nThrow a :bomb: at meewhee009", "Solution_24": "Throw a :sleeping: egg at lotsofmath", "Solution_25": "Nice activity guys. :)\r\nMeewhee threw a :ninja: egg at lotsofmath.\r\nLotsofmath hurled a :bomb: egg at meewhee.\r\nYoshi swung his arm and flung a :sleeping: egg at lotsofmath.\r\n\r\nSuddenly, jjfun1's voice came on.\r\n\"Ermm....it looks like there's a glitch with the eggs...they don't seem to be working!\" True to her words, the egg that hit lotsofmath broke open, and a bomb with a timer, frozen, was visible. \"Err...it seems as if everyone needs to throw another egg at the bomb...to activate it I guess.\" She flipped through the EGG TOSS GAME HANDBOOK! \"Heh. Yeah. So. Everyone throw another egg at the bomb.\"\r\n\r\nThe players whispered among one another. One by one, sparkly polka-dotted eggs appeared in their hands.\r\nMeewhee squeezed her eyes shut, tight, and dropped her egg on the bomb.\r\nShe opened one eye, then sighed in relief.\r\nLotsofmath snickered, and smashed his egg on top of the bomb.\r\nBOOM!\r\nAn invisible force field had appeared around Meewhee and Yoshi, but not lotsofmath.\r\nHe disappeared with a loud [i]pop![/i]\r\n\r\n[b]Ok, it's the last round, and this is how it'll work. Meewhee and Yoshi will take turns guessing a number from 1-10 inclusive. When one of them guesses the correct number, they win! Good luck guys, and great game to everyone else who played. Thanks for playing. :) Now....GO![/b]", "Solution_26": "3\r\n\r\nGood game people.", "Solution_27": "7.\r\n\r\n\r\nOoooh....", "Solution_28": "[color=white]Good luck to both of you![/color]", "Solution_29": "Wow.\r\nI'm surprised!\r\n[b]Meewhee guessed the number on her very first try. :O[/b]\r\n@Yoshi, nice guess, that's my favorite number. :)\r\nGreat game everyone. Thanks for playing!!!!" } { "Tag": [ "probability" ], "Problem": "Coupons in cereal boxes are numbered 1 to 5, and a set of one of each is required for a prize. With one coupon per box, how many boxes on the average are required to make a complete set?", "Solution_1": "Solution:\r\n[hide]\n11.42\n\nif the probability of an event occuring is 1/n, it will take n times before, chances are, u have what u want.\nSo probabilities for getting coupons:\n\n1 (1st coupon)\n4/5 (2nd coupon)\n3/5 (3rd coupon)\n2/5 (4th coupon)\n\n1/5 (5th coupon)\n\nso each of these can be put into 1/n for by making them each 1/(reciprical). \nso, its then 1+5/4+5/3+5/2+5 = 11.42 [/hide]" } { "Tag": [ "geometry" ], "Problem": "Find the area of a segment of a circle within a $60^\\circ$ sector, with a radius of 9.", "Solution_1": "[hide]$60^{\\circ}$ is $\\frac{1}{6}$ of the circle, so the area is $\\frac{81\\pi}{6}=13.5\\pi$.[/hide]", "Solution_2": "I think he means the circular segment...\r\n[hide]The area of the sector is $\\frac{81\\pi}{6}$. The area of the triangle is $\\frac{81\\sqrt{3}}{4}$. Just subtract the area of the triangle from the sector.[/hide]", "Solution_3": "[hide]$60^\\circ$ is $\\frac16$ of the entire circle, so the area is $\\frac{9^{2}\\pi}6=\\frac{27\\pi}2$[/hide]", "Solution_4": "ragnarok23 got the right answer.", "Solution_5": "Oh I see, I never heard it referred to as the circular segment before.", "Solution_6": "Well, now you know for future use." } { "Tag": [ "function", "integration", "calculus", "derivative", "real analysis", "real analysis unsolved" ], "Problem": "Let $ f$ be a measurable function of $ {\\mathbb R}^n$ to $ {\\mathbb C}$ such that $ \\sup_{x,y\\in{\\mathbb R}^n}\\frac{|f(x)\\minus{}f(y)|}{1\\plus{}|x\\minus{}y|}<\\plus{}\\infty$\r\nProve that $ f$ is sum of a bounded measurable function and of a continously differentiable function with bounded first order derivatives. \r\n\r\nNice to see your proof. :)", "Solution_1": "Convolving $ f$ with a positive function $ \\phi\\in C_c^{\\infty}(\\mathbb R^n)$ such that $ \\int \\phi(x)\\,dx \\equal{} 1$, we get a $ C^{\\infty}$ function $ g \\equal{} f*\\phi$. Clearly,\r\n\\[ |g(x) \\minus{} f(x)| \\equal{} \\left|\\int (f(y) \\minus{} f(x))\\phi(x \\minus{} y)\\,dy\\right|\\le \\int C(|y \\minus{} x| \\plus{} 1)\\phi(x \\minus{} y)\\,dy \\equal{} : C_2\r\n\\]\r\nAlso, the derivative of $ k$th order $ D^kg(x)$ can be written as $ (\\minus{}1)^k\\int (f(y) \\minus{}f(x))D^k\\phi(x\\minus{}y)dy$, which is shown to be uniformly bounded as above.", "Solution_2": "I also obtained a same proof as your. Good work mlok :)", "Solution_3": "I may be a bit jaded by now, but somehow I am more likely to call this \"a standard mollification result\" than something \"extremely beautiful\". It's that damned mechanization of math, again." } { "Tag": [ "calculus", "integration", "algebra", "function", "domain", "logarithms", "limit" ], "Problem": "I had a bunch of questions like this in my homework and I just don't get the difference. Can someone just help me out with this concept with an example question?\r\n\r\nCompute the value of the following improper integral if it converges. If it diverges, enter INF if it diverges to infinity, MINF if it diverges to minus infinity, or DIV otherwise (hint: integrate by parts).\r\n\r\n{Integral from 1 to infinity} (9lnx)/(x^6) dx \r\n\r\nDetermine whether\r\n{Summation from 1 to infinity}(9ln(n))/(n^6)\r\nis a convergent series. Enter C if the series is convergent, or D if it is divergent. \r\n\r\nAre they always the same? If one is convergent are they both and vice versa? Thanks =)", "Solution_1": "[quote=\"Beachic109\"]Are they always the same? If one is convergent are they both and vice versa? Thanks =)[/quote]\r\n\r\nThis is the [url=http://en.wikipedia.org/wiki/Integral_test_for_convergence]integral test[/url]. You should be able to find it in your textbook.", "Solution_2": "To determine if an improper integral converges (or its value), replace the problem point(s) in the domain of integration with variables and take limits, e.g.\r\n\r\n\\[ \\int_{1}^{\\infty}\\frac{9\\ln x}{x^6}dx \\equal{} \\lim_{L\\to\\infty}\\int_{1}^{L}\\frac{9\\ln x}{x^6}dx\\]\r\n\r\nnow compute the integral as normal: use integration by parts with $ u\\equal{}\\ln x$ to get\r\n\r\n\\[ \\equal{}\\lim_{L\\to\\infty} \\left[ \\minus{}\\frac{9\\ln x }{5x^5}\\right]_{1}^{L}\\plus{}\\int_{1}^{L}\\frac{9}{5x^6}dx \\equal{} \\lim_{L\\to\\infty} \\frac{9}{25}\\left[ \\frac{5\\ln x\\plus{}1}{x^5}\\right]_{L}^{1}\\]\r\n\\[ \\equal{} \\lim_{L\\to\\infty} \\frac{9}{25}\\left(1\\minus{} \\frac{5\\ln L\\plus{}1}{L^5}\\right)\\]\r\n\r\nNow if the above limit converges, then the improper integral converges and has the same value as this limit.\r\n\r\nThere is a theorem that states that if $ a_n$ is monotonically decreasing then the series $ \\sum_{n\\equal{}1}^{\\infty}a_n$ and the integral $ \\int_{1}^{\\infty}a_n\\, dn$ converge or diverge together." } { "Tag": [ "inequalities", "inequalities open" ], "Problem": "hi everyone~\r\n\r\nI am Korean...\r\n\r\nMaybe someone know the problem which proposed last 2005 in Germany TST.\r\n\r\nI try to solve it using a nesbitt's inequality but I failed.\r\n\r\nProblem is\r\n\r\n1+a/1-a + 1+b/1-b + 1+c/1-c =<2(b/a+c/b+a/c)\r\n\r\nOops! The condition is a+b+c=1, a,b,c>=0", "Solution_1": "http://www.mathlinks.ro/Forum/viewtopic.php?t=25780\r\n\r\n darij" } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra solved" ], "Problem": "any compact subgroup $G$ of $GLn(R)$ is of the following form:\r\n$G=M K M^{-1}$ where:\r\n$K$ is a subgroup of $O(n)$\r\n$M$ an invertible matrices.", "Solution_1": "Alekk, do you happen to know any normal (meant for mortal human beings) solution for this problem? I asked my teacher and he says it can be done using the Haar measure or something like this, but these tools I imagine cannot be used in a kholle or exam. And anyway, I undersood nothing from his solution.", "Solution_2": "It is a friend who got this exercice during an oral examination. He told me that he didn't manage to solve it(of course). So I ask on another forum if someone could solve it with elementary results. It turns out that the usual solution uses Haar stuff you just mentioned but someone gave me a (long) proof with elementary mathematics. I will post a pdf file with the sln.", "Solution_3": "here is a proof(in fact it is a ps-file):" } { "Tag": [ "number theory open", "number theory" ], "Problem": "who can give the solution? please!! :) :) \r\nthanks for all!!", "Solution_1": "42", "Solution_2": "thanks to you!!\r\nlet n be a positive integer, prove that :exist &>0 so that with n positive real x1,x2,...xn arbitrary always exist t>0 so that \r\n&<{tx1},{tx2}...{txn} < 1/2" } { "Tag": [ "inequalities", "inequalities solved" ], "Problem": "Let a,b,c be real positive numbers.\r\n\r\nProve that\r\n\r\n \\sqrt (a + 1/b) + \\sqrt (1/ab + 1/a) + \\sqrt (b + ab) \\geq 3 \\sqrt 2", "Solution_1": "this is a quite straight forward AM-GM application:\r\n\r\nfirst (a+1)(b+1)(ab+1) \\geq 8ab (AM-GM for each of the pairs (a,1), (b,1), (ab,1) and multiply) then \r\n\r\n\\sqrt ( (ab+1)(a+1)(b+1) / ab) \\geq 2\\sqrt 2 and extracting the cubic root we get \r\n\r\n3 ( \\sqrt ( (ab+1)(a+1)(b+1) / ab) )1/3 \\geq 3\\sqrt 2 and the LHS is exactly the GM of the three terms in LHS of the initial inequality." } { "Tag": [], "Problem": "The Selective Examination (1) for the National Team of China for IMO 2005 \r\n2005\u5e74\u4e2d\u56fd\u6570\u5b66\u5965\u6797\u5339\u514b\u56fd\u5bb6\u96c6\u8bad\u961f\u9009\u62e8\u8003\u8bd5\uff08\u4e8c\uff09", "Solution_1": "Thanks for posting this!", "Solution_2": "\u8fd8\u6709\u51e0\u6b21\u6d4b\u8bd5\u9898,\u8d39\u8001\u5e08\u80fd\u5426\u4e00\u5e76\u63d0\u4f9b ? \u591a\u8c22 !" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "For a given integer $d$, let us de\ufb01ne $S = \\{m^{2}+dn^{2}| m, n \\in\\mathbb{Z}\\}$. Suppose that $p, q$ are two elements of $S$ , where $p$ is prime and $p | q$. Prove that $r = q/p$ also belongs to $S$ .", "Solution_1": "[quote=\"N.T.TUAN\"]For a given integer $d$, let us de\ufb01ne $S = \\{m^{2}+dn^{2}| m, n \\in\\mathbb{Z}\\}$. Suppose that $p, q$ are two elements of $S$ , where $p$ is prime and $p | q$. Prove that $r = q/p$ also belongs to $S$ .[/quote]\r\nLet $p=m_{0}^{2}+dn_{0}^{2}, k=\\frac{m_{0}}{n_{0}}(mod \\ p)=\\sqrt{-d}(mod \\ p).$\r\nLet $q=pr=m^{2}+dn^{2}$. If $p|m, \\ p|n$, then $r=(\\frac{mm_{0}-nn_{0}d}{p})^{2}+d(\\frac{nm_{0}+n_{0}m}{p})^{2}$. \r\nElse $m=\\pm nk(mod \\ p)$ and $r=(\\frac{mm_{0}\\pm nn_{0}d}{p})^{2}+d(\\frac{nm_{0}\\mp n_{0}m}{p})^{2}.$", "Solution_2": "following link is easy than what Rust posted http://www.mathlinks.ro/Forum/viewtopic.php?t=66248 but same :wink: . Sorry, because I don't remember it." } { "Tag": [ "geometry", "3D geometry" ], "Problem": "ok the title says it all. i'll start off and the next poster edits it to be funny and so on and so on\r\n\r\n\"I learned how to count!\"", "Solution_1": "\"I learned female anatomy!!!\"\r\n[hide]im a guy by the way[/hide]", "Solution_2": "[quote=\"life=tennis+math\"]\"I learned female anatomy!!!\"\n[hide]im a guy by the way[/hide][/quote]\r\n\r\nI doubt there's many girls who'd say that anyways.\r\n\r\n\r\noh man i learned a lot today\r\ndon't remember what though", "Solution_3": "\"I learend about the Grangers!\"", "Solution_4": "I learned Math today!", "Solution_5": "I learned that math isn't overrated; neither is flushing the toilet in a public restroom.", "Solution_6": "i learned the value of switchblades", "Solution_7": "i learned stuff", "Solution_8": "I learned that nobody learns anything of value in school!", "Solution_9": "i learned how to make a bomb using only a can in macgyver movie", "Solution_10": "i learned how to supress the need of going to the bathroom", "Solution_11": "I learned nothing. :huh: \r\nSeriously.", "Solution_12": "I learned it's hard to kidnap fat people. jk.", "Solution_13": "i learned that fob teachers can be born in brazil while being chinese", "Solution_14": "well...um, i learned that i suck at basketball?\r\n\r\nwell in PE, i jammed my finger and now it's painful and swollen. :stretcher: \r\n\r\ni got mad skillz :roll:", "Solution_15": "[quote=\"dragongrl6\"]well...um, i learned that i suck at basketball?\n\nwell in PE, i jammed my finger and now it's painful and swollen. :stretcher: \n\ni got mad skillz :roll:[/quote]\r\n\r\nlol\r\n\r\ni learned at school that bowling is a sissy sport (we do it in pe so boring) :showoff: :sleep2: :weightlift:", "Solution_16": "i learned that substitute teachers are fun to annoy", "Solution_17": "I learned that when someone is in possession of my Rubik's cube, he/she wouldn't put it in lost and found. He/she would take it and not give it back." } { "Tag": [ "USAMTS", "email" ], "Problem": "i am trying to submit my solutions but it says the dealine was march 9th even tho today is marh 9th what do i do !!!!!!!!!!!!!!", "Solution_1": "From the problems sheet: \"Web deadline: 3 PM Eastern / Noon Pacific on the due date\"", "Solution_2": "omg i didnt see that i submitted my last three by mail its 9 o clock and i cant do anything now ><", "Solution_3": "you could have still submitted it by mail", "Solution_4": "i just said its 9 o clock i cant submit by mail anymore ><", "Solution_5": ":o [quote=\"peregrinefalcon88\"]i just said its 9 o clock i cant submit by mail anymore ><[/quote]\r\nomg what are u going to do :o \r\nget to one quick.\r\nUSAMTS staff: please accept his solns!!! he has proof he finished in time", "Solution_6": "[quote=\"mathemonster\"]USAMTS staff: please accept his solns!!! he has proof he finished in time[/quote]\r\nHow does he have proof? He posted this topic 4 hours and 48 minutes after the due time.\r\nAnd even if he did post it before, there is no proof that he finished in time. This is the internet, so you can lie (I don't mean this sarcastically). It also doesn't even matter if he finished on time provided that he didn't turn in his solutions. If you don't turn in your solutions on time, then they won't be accepted (except possibly under extraordinary circumstances, such as the situation with Round 2 solutions being lost, but even in this scenario, only the people that turned in their solutions on time were able to resend).", "Solution_7": "tough luck, read carefully, meh, i'm not being very helpful.", "Solution_8": "[quote=\"kcn2rivers\"]From the problems sheet: \"Web deadline: 3 PM Eastern / Noon Pacific on the due date\"[/quote]This was announced in at least four places: on the problem sheet, on the USAMTS website, by email, and on this forum; and on multiple occasions." } { "Tag": [ "trigonometry" ], "Problem": "suppose A and B are two angles such that\r\nsin A + sin B = 1\r\ncos A + cos B = 0\r\nfind the value of 12 cos 2A + 4 cos 2B.", "Solution_1": "Looks like your computer accidentally double-posted.", "Solution_2": "Umm...that's the computer problem if his computer accidentally double-posted...Anyhow, do not post a same question twice :D", "Solution_3": "$\\cos A =-\\cos B \\rightarrow \\sin A = \\pm \\sin B$\r\n\r\nIf $\\sin A =-\\sin B$ then $\\sin A+\\sin B = 0$, so $\\sin A = \\sin B = 1/2 \\rightarrow A=30^{o}$, $B=150^{o}$ or vice-versa. Either way, $\\cos2A = \\cos2B = \\frac{1}{2}\\Rightarrow 12\\cos2A+4\\cos2B = 8$" } { "Tag": [ "geometry", "3D geometry", "sphere", "calculus", "derivative", "calculus computations" ], "Problem": "A sphere shrinks at a rate of 1 cm^3 per second. At what rate does the diameter decrease when the diameter is 10 cm? I know it has something to with optimization, and the use of the area formula. The problem is that I do not know the diameter for which it started with and it gave too little info on this problem(at least i think).", "Solution_1": "You are given diameter (which I'll call $ q$) and rate of change of volume or $ \\frac{dV}{dt}$. You should look for a formula that has these relations in them.\r\n\r\n$ V=\\frac{4}{3}\\pi r^{3}$, but you want diameter so: $ V=\\frac{4}{3}\\pi \\left(\\frac{q}{2}\\right)^{3}=\\frac{1}{6}\\pi q^{3}$. Now take the derivative to get $ \\frac{dV}{dt}=\\frac{1}{2}\\pi q^{2}\\frac{dq}{dt}$ (which is why I used $ q$ to avoid the infamous $ \\frac{dd}{dt}$). Your problem asks for $ \\frac{dq}{dt}$ at the time of $ q=10$ and $ \\frac{dV}{dt}=-1$, so plug in and solve from here." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Prove that there are infinitely many positive integers $ a$ such that the sequence $ (z_{n})$ $ n \\ge 1$ ,\r\n$ z_{n} \\equal{} n^{4} \\plus{} a$ ,does not contain a prime number.", "Solution_1": "[hide=\"hint\"]use sophie-germain identity[/hide]", "Solution_2": "Let a = 4 into n power 4. \r\nI FORGOT THIS IN RMO 2009. GRRR..." } { "Tag": [ "MATHCOUNTS", "FTW", "geometry", "3D geometry", "LaTeX" ], "Problem": "Hello,\r\n\r\nThis is a mock mathcounts sprint round.\r\n\r\nPlease see the questions [url=http://mersenne.zapto.org/Submission/Sprint.pdf]here[/url].\r\n\r\nPlease submit your answers [url=http://mersenne.zapto.org/Submission]here[/url]. Do not PM me. PM'ed answers will not count.\r\n\r\nDo not disclose any answers via any form of communication.\r\n\r\nI will post the target round on Wednesday. I will post all scores and correct answers on Sunday.\r\n\r\nI have also attached the questions.\r\n\r\n-#H34N1\r\n\r\n[b]Clarifications[/b]\r\n[b]#3:[/b] You do not have to express it in simplest radical form.\r\n[b]#6:[/b] Integer Lengths\r\n[b]#9:[/b] $ a$, $ b$, and $ c$ are positive, real numbers. Find the least positive value.\r\n[b]#11[/b] Express your answer in terms of $ n$.\r\n[b]#12.[/b] Express your answer in simplest radical form.\r\n[b]#13[/b] Express your answer in common fraction form.\r\n[b]#19[/b] Either decimal or fraction is permissible.\r\n[b]#20[/b] You need the distance that hte ball travelled.\r\n[b]#29:[/b] There are in reality 99 cards, not 50 cards as mentioned in the contest.\r\n\r\n\r\n[i]Authentication Issue: If you are misauthenitcated by the Submission form, please wait about 4-5 minutes. This happens when two people are filling out the form. Thank you.[/i]", "Solution_1": "When is it due?", "Solution_2": "Answers will not be posted until Sunday, 9:00 PM EST, so you have until then.\r\n\r\n[b]Clarification[/b]: There was a typo, and on #3, you do not have to express it in simplest radical form.", "Solution_3": "the format is exactly like the real ones. nice dude!!", "Solution_4": "I must remind everyone that they have [b]1[/b] chance before turning in answers. There may be NO corrections.\r\n\r\nThey are all saved onto a database, and I will take your FIRST submission.", "Solution_5": "Cool, how did you edit the PDF?\r\nThat site is getting better and better!\r\nEdit: It says I am #H34N1 when I go to the submission page, but I am minicon.", "Solution_6": "[quote=\"minicon\"]Cool, how did you edit the PDF?\nThat site is getting better and better![/quote]\r\n\r\nI made the template. It is not edited off of a previously made MATHCOUNTS pdf.", "Solution_7": "#29 is messed up, u might want to check it out. 50 cards numbered 1-99???", "Solution_8": "[quote=\"cognos599\"]#29 is messed up, u might want to check it out. 50 cards numbered 1-99???[/quote]\r\n\r\nYeah I just realized that and have added it to the Clarification section. Forgot to proofread these :oops:", "Solution_9": "thanks........(fricking message too short)", "Solution_10": "i think there are parts that are missing from some of the problems. some are listed below\r\n\r\nPROBLEMS\r\n6 (not specified whether integers or not)\r\n9 (i believe -infinity works, if only one of the values is neg infinity)\r\n11 (i think it should have, \"in terms of n\" at the end)\r\n12 (\"simplest radical form\")\r\n13 (\"in common fraction form\")\r\n17 (some ppl may not know what perfect numbers are, so define for them maybe)\r\n19 (what form do you want it in? decimal or fraction or mixed numeral?)\r\n20 (clarify what distance you are looking for)", "Solution_11": "what level is this? chapter or state?", "Solution_12": "[quote=\"mchoi815\"]i think there are parts that are missing from some of the problems. some are listed below\n\nPROBLEMS\n6 (not specified whether integers or not)\n9 (i believe -infinity works, if only one of the values is neg infinity)\n11 (i think it should have, \"in terms of n\" at the end)\n12 (\"simplest radical form\")\n13 (\"in common fraction form\")\n17 (some ppl may not know what perfect numbers are, so define for them maybe)\n19 (what form do you want it in? decimal or fraction or mixed numeral?)\n20 (clarify what distance you are looking for)[/quote]\r\n\r\n6- Integer lengths\r\n9 - Positive value\r\n11 - yes\r\n12 - yes\r\n13 - yes\r\n17 - no\r\n19. when it is not specifried, you can put either\r\n20. i think its pretty clear there\r\n\r\n[b]IF YOU ARE MISAUTHENTICATED, THEN PLEASE WAIT 4 OR 5 MINUTES AND RE-AUTHENTICATE. THE AUTHENTICATION BUG HAPPENS WHEN TWO PEOPLE ARE AT THE SITE AT THE SAME TIME[/b]", "Solution_13": "The bug is caused by deadlock, correct?", "Solution_14": "[quote=\"minicon\"]The bug is caused by deadlock, correct?[/quote]\r\n\r\nIt should be fixed now. You just need to wait 5 seconds if you are misauthenticated.", "Solution_15": "This was because you clicked on Submit query when you were supposed to just click on the auto authenticate. :P", "Solution_16": "Hmm, I could have gotten another if I had actually felt like doing #30, which was the worst probblem, according to me.", "Solution_17": "[quote=\"xpmath\"]Hmm, I could have gotten another if I had actually felt like doing #30, which was the worst probblem, according to me.[/quote]\r\n\r\ni spent 10 minutes multiplying it out :rotfl:", "Solution_18": "I put $ \\frac{2}{3}\\sqrt{3}\\minus{}1, \\frac{2}{3}\\sqrt{3}\\plus{}1$ for Target Round #2, and the answer was the same thing, except it was expressed in a different form. Was that counted as incorrect?\r\nbecause i got 1 and 8 right and i only got 4 points for target, so....", "Solution_19": "I will double check that.", "Solution_20": "Hey can you post the answers? I finished my sprint", "Solution_21": "[quote=epic2004_aops]Hey can you post the answers? I finished my sprint[/quote]\n\nwhy the revive tho?", "Solution_22": "I don't know I was just surfing and found this and once I finish I found there was no key... can a key be made?", "Solution_23": "ask the OP in an pm or something like that", "Solution_24": "[quote=green_dog_7983]ask the OP in an pm or something like that[/quote]\n\nThe OP hasn't been active in 3 months and this post was made in 2008; there's a high possibility he won't respond", "Solution_25": "[quote=Kagebaka][quote=green_dog_7983]ask the OP in an pm or something like that[/quote]\n\nThe OP hasn't been active in 3 months and this post was made in 2008; there's a high possibility he won't respond[/quote]\n\ntrue tho", "Solution_26": "Can any of y'all make an answer key", "Solution_27": "Eh, I can't access any of the links at all...", "Solution_28": "Hello. How do you join?", "Solution_29": "You are about 9 years too late..." } { "Tag": [ "limit", "real analysis", "real analysis unsolved" ], "Problem": "$ f: \\mathbb{Q}\\to\\mathbb{R}$ is continuous.\r\n\r\nprove, that $ f$ has a finite limit in some (at least 1) irrational points.", "Solution_1": "[hide=\"Idea\"]\nConsider metric spaces $ (X, d), \\ (Y, \\rho)$. \nLet $ A \\subset X$, and $ g: A \\to Y$.\nFor $ x\\in X$ define ${ o_{g}(x) = \\inf\\{\\varepsilon > 0 \\ : \\exists U - \\mbox{open neighbourhood of }x \\ : \\ \\mathrm{diam}\\ g(U\\cap A}) \\leqslant \\varepsilon\\}$\n(we assume that $ \\inf \\emptyset = + \\infty$)\nand $ o(g) = \\{x \\in X \\ : \\ o_{g}(x) = 0\\}$.\nNow it's not hard to prove that $ o(g)$ is a countable intersection of open sets, and if $ g$ is continuous in $ x$ then $ o_{g}(x) = 0$. \nIt is easy to see that if $ x_{0} \\in o(f)$ then $ \\lim_{x\\to x_{0}}f(x)$ exists.\nSince $ \\mathbb{Q}$ is not a countable intersection of open sets (this follows from Baire Category Theorem), we have $ \\mathbb{Q}\\subsetneq o(f)$, and $ f$ must have limit in some irrational point.\n[/hide]", "Solution_2": "Thank you.", "Solution_3": "I didn't work through that idea fully, since my eye was drawn to the word Baire at the bottom. Doesn't Baire Category Theorem only work for complete metric spaces (and $ \\mathbb{Q}$ is not)?", "Solution_4": "That sentence really means \"$ \\mathbb{Q}$ is not an intersection of open sets in $ \\mathbb{R}$\".", "Solution_5": "http://planetmath.org/encyclopedia/LetXdBeACompleteMetricSpaceWithNoIsolatedPointsAndLetDsubsetXBeACountableDenseSetThenDIsNotAG_delta.html" } { "Tag": [ "linear algebra", "matrix" ], "Problem": "1: If the entries of a 3 by 3 matrix are chosen randomly between 0 and 1, what are the most likely dimensions of the 4 subspaces? What if the matrix is 3 by 5?\r\n\r\nThe answer is - it's most likely to be independent. But how can we tell? There are 2^9 possible combinations. But this is where I get stuck...\r\n\r\n2: If A and B share the same 4 subspaces, then A is a multiple of B. True or false? Answer is false - involving an invertible matrix. This is talking about subspaces, right? Not just the bases? If they share the same 4 subspace bases, then A MUST be a multiple of B, right? But if they do not share the same 4 subspace bases, but nonetheless span the same subspace, then A doesn't have to be a multiple of B.", "Solution_1": "I assume that \"the four subspaces\" are the spaces spanned by the first $k$ columns for $k=0,1,2,3$; I've never heard this terminology before.\r\n\r\nFor #1, I offer this count:\r\nThere are $7\\cdot 6\\cdot 4=168$ invertible matrices over the field with two elements. All of these are still invertible over the rationals, since they have odd determinant.\r\nIn addition, the matrices with determinant $\\pm2$ are invertible. Looking at the permutations, we can show that there are three matrices of determinant $2$ and three of determinant $-2$ with elements in $\\{0,1\\}$, so a total of 174 of the 512 matrices of this form are invertible.\r\nAfter that, the other possible lists are comparatively easy:\r\n0, 0, ?, ?: 1*8*8=64\r\n0, 1, ?, ?: 512-64=448\r\n0, 0, 0, ?: 1*1*8=8\r\n0, 0, 1, ?: 64-8=56\r\n0, 1, 1, ?: 7*2*8=112\r\n0, 1, 2, ?: 448-128=336\r\n0, 0, 0, 0: 1\r\n0, 0, 0, 1: 7=8-1\r\n0, 0, 1, 1: 14=1*7*2\r\n0, 0, 1, 2: 42=56-14\r\n0, 1, 1, 1: 28=1*7*2*2\r\n0, 1, 1, 2: 84=112-28\r\n0, 1, 2, 2: 162=336-174\r\n0, 1, 2, 3: 174\r\n\r\nInvertibility is most likely, but only in a very narrow sense. That's only a third of the time.\r\n\r\nFor part 2: You've got it right. If the spaces in question are the column spaces I mentioned, $A$ and $B$ share the same spaces whenever $AX=B$ for some upper triangular matrix $X$ with nonzero diagonal entries. (This makes $X$ invertible, so $B=AX^{-1}$, and $X^{-1}$ is also upper triangular)\r\nYou should be able to construct plenty of examples." } { "Tag": [ "calculus", "derivative", "analytic geometry", "graphing lines", "slope", "conics", "parabola" ], "Problem": "another one", "Solution_1": "2 ways to do it:\n\n\n\n[hide]1)the square of the distance from the point (a,1/3 a^2) to the point (15,-3) is (15-a)^2+(-3-1/3 a^2)^2. We wish to minimize this, so set it's derivative equal to 0.\n\n\n\n2) the slope of the line tangent to the parabola at the point (a,1/3 a^2) is 2/3 a. the slope of the line perpendicular to this line is -3/(2a). So, the line perpendicular to the parabola at the point (a,1/3 a^2) is (y-1/3 a^2)=-3/(2a) (x-a). We want this line to pass through (15,-3). Solve for a.[/hide]", "Solution_2": "100% Correct.", "Solution_3": "I used his first way to get the answer. Is it [hide]13.416[/hide]?", "Solution_4": "yep 6sqrt(5)", "Solution_5": "Don't use approximations! Feh. Who knows what that number is? Is it (10 :pi: - 18)? Is it :sqrt:180 (= 6 :rt5:)? Those two numbers are very close together, but the [i]reasons[/i] that it would be one as opposed to the other are very different!\r\n\r\n\r\n*Note* edited because I can't subtract.", "Solution_6": "10 :pi:-18?", "Solution_7": "remmebeer alwasy store your numbers when do the AP CALC exam. and then round at the end to three digits.", "Solution_8": "Right-o, (10 :pi: - 18). But it's okay, because now that I'm doing real math, we don't actually have to know how to subtract, we just have to acknowledge the existence of the inverse-additive operator. \r\n\r\nAnd it's worth pointing out that this website is [i]not[/i] the AP calculus exam, and that exact forms really are much nicer than decimal approximations because they contain more information.", "Solution_9": "Yeah sorry, I was too lazy to do the derivatives and stuff so I just used my calculator and graphed it :).", "Solution_10": "rlee this probelm is from a non-calculator exam that fubu almost aced.", "Solution_11": "What exam is this, anyway?", "Solution_12": "USMC 10-12", "Solution_13": "That isn't really helpful at all. It might be nice if you gave the full name of the competition and some information (or possibly a link) about it." } { "Tag": [ "topology" ], "Problem": "Proof that Every second-countable space is first-countable and separable", "Solution_1": "The first countable fact is obvious; the separability follows from the fact that the union of one point from each basis element forms a countable dense subset." } { "Tag": [ "inequalities", "algebra unsolved", "algebra" ], "Problem": "Determine the position where a parenthese should be put in an exponential tower- \r\n ${{{{{{{x^{x}}^{x}}^{x}}^{x}}^{.}}^{.}}^{.}}^{x}$there being $x$ exponents odf $x$ such that the resultant number has maximum value($x$ is a natural number)", "Solution_1": "Is the tower evaluated from the right or the left conventionally?\r\nIs only one pair of parentheses allowed to be added?", "Solution_2": "the tower is to be evaluated from the left \r\nand only one pair of parentheses is to be added?", "Solution_3": "I think I have this all clear now. Solution follows: \r\n[hide]\nI am going to invent some notation because I can't currently work Latex.\n[X,n] means a stack of n Xs not in parenthesis. \n[X, m][X, n] means a stack of height m as the base for a stack of height n adjoined upwards to it.\n\nWrite [X, X] as [X, a]([X, b])[X, X - a - b] where 1 <= a, b and a + b <= X\n (1)\nParentheses are inserted around the middle stack of height b.\n\nEvalutating from the left-hand-side: \n[X, a] = X^(X^(a-1))\n[X, b] = X^(X^(b-1)). \nThe third stack collapses down the the level of the power of the first.\n(1) can therefore be expressed as X raised to the following power, which I then simplify:\nX^(a - 1) * X^(X^(b - 1)) * X^(X - a - b). \n= X^(a - 1 + X^(b-1) + X - a - b)\n= X^(X^(b - 1) + X - b - 1)\nNotice how the first and third stacks result in two tier exponentiation whilst the middle stack results in three tier exponentiation. Note also the independence of this answer upon a!\n\nTo maximise [X, X] therefore, we maximise f(X, b) = X^(b-1) + X - b - 1 by varying b. Since X > 1, f(X, b + 1) = f(X, b) + X - 1 > f(X, b), so \nmaximise b subject to the inequalities in (1). \nThis gives a = 1, b = X - 1.\n\nTherefore the maximum value of the stack is [X:1]([X:X-1]).\n\nIt has value: X^(X^(X^(X - 2)))\n\n(N.B. a >= 1 because if a = 0 [X:X] = X^(X^(X-1)). b >= 1 because the case of a stack vanishing is already covered since the upper stack may have zero height.)\n[/hide]" } { "Tag": [ "symmetry", "geometry", "parallelogram", "geometry unsolved" ], "Problem": "$W$ is a polygon which has a center of symmetry $S$ such that if $P$ belongs to $W$, then so does $P'$, where $S$ is the midpoint of $PP'$. Show that there is a parallelogram $V$ containing $W$ such that the midpoint of each side of $V$ lies on the border of $W$.", "Solution_1": "It's useful to have seen [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=35947]this problem[/url] before :).\r\n\r\nFor compactness reasons, there is a parallelogram $ABCD$ of minimal area containing the polygon $\\cal W$ (and which has the center of symmetry $O$ of $\\cal W$ as its own center of symmetry). Let's prove that such a polygon satisfies the requirements. \r\n\r\nSuppose, for instance, that the midpoints $M,N$ of $AD,BC$ (respectively) do not belong to the border $\\partial\\cal W$ of $\\cal W$. The intersection between $\\partial\\cal W$ and $AD$ is a closed (maybe degenerate) segment $s$ which lies either in $[AM)$ or $(MD]$. Suppose it's the former. Then, the intersection $\\partial\\cal W\\cap BC$ will be a closed sgement $s'$ which lies in $(NC]$. Now keep $M,N$ and the lines $AB,CD$ fixed, and move $A$ slightly, putting $D=AM\\cap CD$, until $AD$ no longer meets $\\partial\\cal W$, but such that $ABCD$ still contains $\\cal W$ (a drawing should make all of this clear). Then $BC$ will not meet $\\partial\\cal W$ either, and the new parallelogram $ABCD$ will have the same area as the old one. We can slightly shrink it along the direction of $AB$ so that it still contains $\\cal W$, and this contradicts the minimality of the area of the initial parallelogram." } { "Tag": [ "geometry", "inradius", "number theory proposed", "number theory" ], "Problem": "Prove that there are infinite positive integer$ n$ such that $ p\\equal{}nr$ , where $ p$ and $ r$ are the semiperimeter and the inradius of a triangle $ ABC$ , where $ ABC$ has integer side lengths.", "Solution_1": "Has anyone got it? I've reduced it to finding rational solutions $ x,y$ for $ xy(x\\plus{}y\\minus{}n)\\equal{}n$. Any ideas?", "Solution_2": "[quote=\"cosinator\"]Has anyone got it? I've reduced it to finding rational solutions $ x,y$ for $ xy(x \\plus{} y \\minus{} n) \\equal{} n$. Any ideas?[/quote]\r\n\r\nI don't know english very well...I didn't know I had posted it correctly, but here's my solution.\r\n\r\nLet $ p \\equal{} 2s$. If $ s \\equal{} nr$, $ p \\equal{} 2nr$. So if we prove that there are infinite positive intergers $ n$ such that $ s \\equal{} nr$, this completes the proof.\r\n\r\n$ n \\equal{} \\frac {s}{r} \\equal{} \\frac {s^2}{S} \\equal{} \\frac {s^2}{\\sqrt {s(s \\minus{} a)(s \\minus{} b)(s \\minus{} c)}} \\equal{} \\sqrt {\\frac {s^3}{(s \\minus{} a)(s \\minus{} b)(s \\minus{} c)}}$\r\n\r\nwhere S is $ ABC$'s area and $ a, b, c$ are $ ABC$'s side length.\r\n\r\nIf we let $ s \\minus{} a \\equal{} x, s \\minus{} b \\equal{} y, s \\minus{} c \\equal{} z$, than $ x, y, z$ are any possible natural numbers, since $ a, b, c$ are triangle's side lengths.\r\n\r\nSo we can get $ n \\equal{} \\sqrt {\\frac {(x \\plus{} y \\plus{} z)^3}{xyz}}$. Now we have to prove that there are infinite many interger $ \\frac {(x \\plus{} y \\plus{} z)^3}{xyz}$ that is a perfect square.\r\n\r\nLet's put $ z \\equal{} k(x \\plus{} y)$. Then the value is $ \\frac {(k \\plus{} 1)^3(x \\plus{} y)^2}{xyk}$. Let's prove that this value can turn into $ 9(k \\plus{} 1)^2$ for infinite many interger $ k$.\r\n\r\n$ \\frac {(k \\plus{} 1)^3(x \\plus{} y)^2}{xyk} \\equal{} 9(k \\plus{} 1)^2$.\r\n\r\n$ (k \\plus{} 1)(x \\plus{} y)^2 \\equal{} 9xyk$.\r\n\r\nIf we let $ y \\equal{} xt$, $ (k \\plus{} 1)(t \\plus{} 1)^2 \\equal{} 9kt^2$.\r\n\r\nSo, we can get $ (k \\plus{} 1)t^2 \\plus{} ( \\minus{} 7k \\plus{} 2)t \\plus{} (k \\plus{} 1) \\equal{} 0$.\r\n\r\n$ D \\equal{} ( \\minus{} 7k \\plus{} 2)^2 \\minus{} 4(k \\plus{} 1)^2 \\equal{} 9k(5k \\plus{} 4)$. So we can let $ k(5k \\plus{} 4) \\equal{} u^2$.\r\n\r\nSince $ gcd(k, 5k \\plus{} 4)$ is $ 1$ or $ 2$ or $ 4$, we can get $ k \\equal{} X^2, 5k \\plus{} 4 \\equal{} Y^2$ or $ k \\equal{} 2X^2, 5k \\plus{} 4 \\equal{} 2Y^2$.\r\n\r\nBut either can produce $ Y^2 \\minus{} 5X^2 \\equal{} 4$.\r\n\r\nWith mod 8, we can easily prove that $ X, Y$ are not odd. Let's let $ X \\equal{} 2v, Y \\equal{} 2w$.\r\n\r\nSo, $ v^2 \\minus{} 5w^2 \\equal{} 1$. Since this is a pell's equation, it has infinite many solutions.\r\n\r\nThis completes the proof. :lol:" } { "Tag": [ "conics", "parabola", "Putnam", "analytic geometry", "graphing lines", "slope", "symmetry" ], "Problem": "Given a parabola, construct its focus using the ruler and the compasses.\r\n\r\nI liked it because it was so simply stated. This is A5 from Putnam 1955 (I found it on Kalva).", "Solution_1": "Choose the coordinate system with origin at the (unknown) parabola vertex and the (unknown) parabola axis identical with the y-axis. In this coordinate system, the parabola has an equation $y = ax^2$. An arbitrary line intersecting the parabola at points $P = [x_P, y_P] , Q = [x_Q, y_Q]$ has an equation\r\n\r\n$y - y_P = \\frac{y_Q - y_P}{x_Q - x_P}\\ (x - x_P)$\r\n\r\n$y - ax_P^2 = \\frac{ax_Q^2 - ax_P^2}{x_Q - x_P}\\ (x - x_P) = a(x_Q + x_P) (x - x_P)$\r\n\r\nbecause for the points $P, Q$ on the parabola, $y_P = ax_P^2, y_Q = ax_Q^2$. Similarly, an arbitrary line $RS$ parallel to $PQ$ and intersecting the parabola at points $R = [x_R, y_R] , S = [x_S, y_S]$ has an equation\r\n\r\n$y - ax_R^2 = a(x_S + x_R) (x - x_R)$\r\n\r\nSince the lines $PQ, RS$ are parallel, their slopes are equal, i.e., $a(x_Q + x_P) = a(x_S + x_R)$. This means that the x-coordinates $\\frac 1 2 (x_Q + x_P) = \\frac 1 2 (x_S + x_R)$ of the midpoints $M, N$ of the segments $PQ, RS$ are the same, i.e., the line $MN$ connecting the midpoints is parallel to the y-axis, identical with the parabola axis. A line through an arbitrary point on the parabola, for example through the point $P$, and perpandicular to the line $MN$ intersects the parabola at another point $P'$. Because of symmetry of the parabola in our coordinate system, the midpoint $K$ of the segment $PP'$ lies on the parabola axis and the line through $K$ parallel to $MN$ intersects the parabola at its vertex $O$ (the origin). The line perpendicular to $KO$ at $O$ is the tangent to the parabola at the vertex, identical with the x-axis.\r\n\r\nThe slope $m$ of the parabola at the point $P$ is the derivative of $y = ax^2$ at $x_P$, i. e., $m = 2ax_P$. Let $H$ be the foot of a normal from the point $P$ to the x-axis and $L$ the midpoint of the segment $OH$, i.e., $H = [x_P, 0]$, $L = [\\frac{x_P}{2}, 0]$. A line passing through the points $L$ and $P$ has an equation\r\n\r\n${y - y_P = \\frac{y_L - y_P}{x_L - x_P}\\ (x - x_P) = -\\frac{ax_P^2}{\\frac{x_P}{2} - x_P}}\\ (x - x_P) = 2ax_P (x - x_P)$\r\n\r\nThe line $LP$ passes through the point $P$ and it has the same slope as the parabola at this point, i.e., it is the tangent to the parabola at the point $P$. Now it is possible to proceed in 2 ways. First a parabolic mirror focuses light rays parallel to the mirror axis to the focus. Hence, it would be possible to draw a line through the point $P$ parallel to the axis $OM$ and then reflect this line in the normal to the tangent $LP$ at the point $P$. The reflected line would intersect the parabola axis at the focus $F$. Alternately, it is possible to use the fact that the pedal curve of a parabola corresponding to the focus is the tangent to the parabola at its vertex. The line through the point $L$ perpendicular to $LP$ has an equation\r\n\r\n$y - y_L = -\\frac{1}{2ax_P}\\ ( x - x_L)$\r\n\r\n$y = -\\frac{1}{2ax_P}\\ \\left( x - \\frac{x_P}{2} \\right)$\r\n\r\nThe y-coordinate of the intersection $F$ of this line with the parabola axis is obtained by putting $x = 0$:\r\n\r\n$y_F = \\frac{1}{2ax_P} \\frac{x_P}{2} = \\frac{1}{4a}$\r\n\r\nwhich does not depend on the point $P$. The curvature of the parabola at the vertex $O$ is \r\n\r\n$\\frac{|y\\\"|}{(1 + y'^2)^{\\frac 3 2}} = \\frac{2a}{(1 + 4a^2x^2)^{\\frac 3 2}} = 2a$ at $x = 0$\r\n\r\nThe kissing circle of the parabola at its vertex $O$ has the same curvature and is centered on the axis, i.e., its radius is $r = \\frac{1}{2a}$ and its center $C = [0, \\frac{1}{2a}]$. The focus of the parabola lies halfway between the points $O$ and $C$, i.e., it has coordinates $[0, \\frac{1}{4a}]$. But these are just the coordinates of the constructed point $F$.", "Solution_2": "Thank you, but I didn't read it :). I'll take your word for it :). There are easier ways (no computations), but there are some things to prove. I'll leave them as exercises :).\r\n\r\nConstruct a chord of the parabola, and another chord parallel to the first one. Now draw the chord passing through the midpoints of the two initial chords. This last chord is parallel to the axis of the parabola, so it's perpendicular to its directrice. Call its direction $d$. \r\n\r\nNow choose any point $P$ outside the parabola, construct the tangents $PT_1,PT_2$ to the parabola, using the fatc that $T_1T_2$ is the polar of $P$ wrt the parabola, and draw a line $d'$ through $T_1$ which is $\\|d$. The reflection of $d'$ in $PT_1$ should pass through the focal point. If you do this again for $T_2$, we get two lines intersecting at the focal point.\r\n\r\nThere might be shorter solutions, but this is all I can think of right now.", "Solution_3": "I put down the computations because I am not that familiar with conic sections and besides, I am trying to get used to precision. The first computation just showed that the line connecting the midpoints of 2 parallel chords to a parabola is parallel to its axis. Then I very simply constructed the axis and the tangent at the vertex, which is the pedal curve of the focus. The second computation showed a simple way to construct a tangent at an arbitrary point. A normal to this tangent at its intersection with the pedal curve cuts the axis at the focus, which is indeed very simple.", "Solution_4": "[quote=\"grobber\"]Given a parabola, construct its focus using the ruler and the compasses.[/quote]\r\n\r\nI have another solution of this problem, which is based on Grobber's idea, but uses only rational operations (so, for instance, I don't draw tangents to the parabola from a point outside the parabola):\r\n\r\nCall $\\mathcal{P}$ our parabola.\r\n\r\nFirst, consider a subroutine for our construction:\r\n\r\n[b]Subroutine 1.[/b] For any point M on the parabola $\\mathcal{P}$, construct the tangent to this parabola $\\mathcal{P}$ at the point M.\r\n\r\nIndeed, the simplest construction uses the Pascal theorem: Take four arbitrary points A, B, C, D on the parabola $\\mathcal{P}$; connect the points $AB\\cap DM$ and $CD\\cap MA$, and intersect the connecting line with the line BC. Join the point of intersection with the point M; the resulting line is the tangent to the parabola $\\mathcal{P}$ at the point M. This is a direct consequence of the fact that, after the Pascal theorem, applied to the points A, B, C, D, M, M on the parabola $\\mathcal{P}$, the points $AB\\cap DM$, $BC\\cap MM$ and $CD\\cap MA$ are collinear, where the line MM has to be interpreted as the tangent to the parabola $\\mathcal{P}$ at the point M.\r\n\r\nNow, apply Subroutine 1 to four different points on the parabola $\\mathcal{P}$. You get four tangents to the parabola $\\mathcal{P}$. These four tangents bound four triangles, each of these triangles having all its sidelines tangent to the parabola $\\mathcal{P}$. But there is a well-known theorem stating that if all sidelines of a triangle are tangent to a parabola, then the circumcircle of this triangle passes through the focus of the parabola. Thus, we have to take our four triangles, draw their circumcircles; since all these circumcircles pass through the focus of our parabola, we can immediately find this focus by intersecting the circumcircles.\r\n\r\n Darij" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Given a_i , b_i (i = 1,2,3,4,5) satisfy : $ a_i \\leq a_j$ and $ b_i \\leq b_j (i \\leq j)$\r\n$ a_1 \\plus{} a_2 \\plus{} a_3 \\plus{} a_4 \\plus{} a_5 \\equal{} b_1 \\plus{} b_2 \\plus{} b_3 \\plus{} b_4 \\plus{} b_5 \\equal{} 0$\r\n$ a_2 \\plus{} a_3 \\plus{} a_4 \\plus{} a_5 \\leq b_2 \\plus{} b_3 \\plus{} b_4 \\plus{} b_5$\r\n$ a_3 \\plus{} a_4 \\plus{} a_5 \\leq b_3 \\plus{} b_4 \\plus{} b_5$\r\nProve that :$ a_1^2 \\plus{} a_2^2 \\plus{} ... \\plus{} a_5^2 \\leq b_1^2 \\plus{} b_2^2 \\plus{} ... \\plus{} b_5^2$\r\nPlz help me >\"<", "Solution_1": "I don't think the inequality is correct\r\n\r\nPut $ a_i \\equal{} b_i$ for $ 1\\le i\\le3$\r\n\r\nand $ a_4 \\equal{} \\minus{}a_5 \\equal{} 3$ \r\n$ b_4 \\equal{} \\minus{}b_5 \\equal{} 2$", "Solution_2": "[quote=\"arpit goel\"]I don't think the inequality is correct\n\nPut $ a_i \\equal{} b_i$ for $ 1\\le i\\le3$\n\nand $ a_4 \\equal{} \\minus{} a_5 \\equal{} 3$ \n$ b_4 \\equal{} \\minus{} b_5 \\equal{} 2$[/quote]\r\noh sr :P I was so sleepy when writing this post so I missed sth ...:P edited :P", "Solution_3": "I don't understand how does this change anything\r\n\r\nPut $ a_4 \\equal{} \\minus{}a_5 \\equal{} \\minus{}3$\r\n\r\nand $ b_4 \\equal{} \\minus{}b_5 \\equal{} \\minus{}2$", "Solution_4": "[quote=\"arpit goel\"]I don't understand how does this change anything\n\nPut $ a_4 \\equal{} \\minus{} a_5 \\equal{} \\minus{} 3$\n\nand $ b_4 \\equal{} \\minus{} b_5 \\equal{} \\minus{} 2$[/quote]\r\nok ... I'll consider again this problem ...>\"< anyway, thx:D", "Solution_5": "[quote=\"arpit goel\"]I don't understand how does this change anything\n\nPut $ a_4 \\equal{} \\minus{} a_5 \\equal{} \\minus{} 3$\n\nand $ b_4 \\equal{} \\minus{} b_5 \\equal{} \\minus{} 2$[/quote]\r\nI see.... iF $ a_5 \\equal{} \\minus{}3$, then $ a_5 < a_1 , a_2 , a_3$ , which is wrong :D" } { "Tag": [ "algebra", "polynomial", "inequalities", "complex analysis", "complex numbers", "Polynomials" ], "Problem": "A prime $p$ has decimal digits $p_{n}p_{n-1} \\cdots p_0$ with $p_{n}>1$. Show that the polynomial $p_{n}x^{n} + p_{n-1}x^{n-1}+\\cdots+ p_{1}x + p_0$ cannot be represented as a product of two nonconstant polynomials with integer coefficients", "Solution_1": "Set $x=10$. If the polynomial can be factored into two other polynomials, then $p$ can be expressed as two decimal factors, and so is not prime, a contradiction.\r\n\r\nThis seems really easy. Is there a flaw?", "Solution_2": "Flaw: The decimal factors could be $\\pm 1$ or $\\pm p$ if you set $x=10$.", "Solution_3": "Indeed, that would be too easy to be true (but I admit it's deceiving). I'm afraid you'll need some complex analysis to kill this one correctly.", "Solution_4": "[quote=\"Peter\"]Indeed, that would be too easy to be true (but I admit it's deceiving). I'm afraid you'll need some complex analysis to kill this one correctly.[/quote]\r\nNope,this problem has very elementary solution,if I am not mistaken,it is enough to observe one inequality between roots and coefficients of the given polynomial...", "Solution_5": "Inequalities between complex roots is complex analysis, no? You won't need deep theorems, obviously, since it's from the BaMO.", "Solution_6": "[quote=\"Peter\"]Inequalities between complex roots is complex analysis, no? You won't need deep theorems, obviously, since it's from the BaMO.[/quote]\r\nHmm...I didn't know that.Probably I have some misconceptions in definition of complex analysis.I can't figure out what you mean by your last sentence,whatever it means I agree with you,solution is pretty simple...\r\nMaybe you would be so kind to post the proof,assuming you have solved this problem.", "Solution_7": "Well I don't even know a definition of complex analyis, but if you do analysis (inequalities) on complex numbers, I'd be inclined to call that complex analysis. Sorry if that's a wrong wording. :)\r\n\r\nFrom Kalva: [quote]If w is a (complex) root of the polynomial, then since each coefficient is at most 9, we have |pn| \u2264 9(1/|w| + 1/|w|2 + ... + 1/|w|n-1). So if |w| \u2265 9, then |pn| \u2264 9(1/9 + 1/81 + ... ) < 9/8. But pn \u2265 2, so we must have |w| < 9.\n\nSuppose the polynomial has a factor with integer coefficients (and positive degree less than n). Then Gauss' lemma tells us that it must be a product of two polynomials with integer coefficients (and each with positive degree less than n). Suppose they are f(x) and g(x). Suppose the (complex) roots of f(x) are z1, ... , zm. Then f(x) = A(x - z1) ... (x - zm) with A and integer. Hence |f(10)| = |A| |10 - z1| ... |10 - zm|. But |A| \u2265 1 and each factor |10 - zi| > 10 - 9 = 1. So |f(10)| > 1. Similarly, |g(10)| > 1. But f(10) and g(10) are integers and their product is the prime p. So we have a contradiction. So the polynomial cannot have any such factor. [/quote]\r\n\r\nBy the way, it is also in the Resources section on this site.", "Solution_8": "Isn't this Cohn's irreducibility criterion?" } { "Tag": [], "Problem": "$\\sqrt{x+1}>1+\\sqrt{\\frac{x-1}{x}}$", "Solution_1": "[quote=\"ashrafmod\"]$\\sqrt{x+1}>1+\\sqrt{\\frac{x-1}{x}}$[/quote]\r\nNice! :lol:", "Solution_2": "[hide]Obviously, x must be greater than -1. SImilarly, [0,1) is undefined. However, values from [-1,0) are clearly failures since the left hand side has an unattainable maximum of 1 while the right hand side has a minimum of $1+\\sqrt{2}$. For $[1,\\infty)$, we get $\\frac{(x^{2}-x+1)^{2}}{4x^{2}-4x}>1$. Subtracting one, we get $\\frac{x^{4}-2x^{3}-x^{2}+2x+1}{4x^{2}-4x}>0$. Now, that factors into $\\frac{(x^{2}-x-1)^{2}}{4x^{2}-4x}>0$ which is true for all values of x greater than 1 except x=$\\frac{1+\\sqrt{5}}{2}$.[/hide]" } { "Tag": [], "Problem": "23. Consider two uniform spherical planets of equal density but unequal radius. Which of the following quantities is the same for both planets?\r\n(a) The escape velocity from the planet's surface.\r\n(b) The acceleration due to gravity at the planet's surface.\r\n(c) The orbital period of a satellite in a circular orbit just above the planet's surface.\r\n(d) The orbital period of a staellite in a circular orbit at a given distance from the planet's center.\r\n(e) None of the above.", "Solution_1": "I'll give a few hints instead of answering:\r\n\r\na) $ E_{tot} \\equal{} \\frac {1}{2} mv_{esc}^2 \\minus{} \\frac {GMm}{R} \\equal{} 0$ (at infinity, where potential = 0 and kinetic = 0)\r\n\r\nb) Use the Law of Gravitation (you probably know this)\r\n\r\nc) Use the concept of centripetal force, coupled with the Law of Gravitation. Use volume density $ \\rho \\equal{} \\frac {M}{\\frac {4}{3} \\pi R^3}$. Main equation: $ T \\equal{} \\frac {2\\pi R}{v}$.\r\n\r\nd) Same as option c, except differentiate between the planets' radii, $ R_{1}$ and $ R_{2}$, and the orbit radius $ r$.", "Solution_2": "Oh yeah, I forgot to ask for hints. Ahh, thanks! I can't believe I keep on forgetting such simple concepts!\r\n\r\nHere's the solution just to make sure I'm doing it correctly (I'm just showing c here)\r\n[hide=\"solution\"]Knowing that $ T \\equal{}\\frac{2\\pi R}{v}$ and $ F_g\\equal{}F_c\\equal{}G\\frac{Mm}{f^2}\\equal{}\\frac{mv^2}{r}$, you can rearrange to get $ v\\equal{}\\sqrt{\\frac{GM}{r}}$ and plug it into the first equation to get $ T\\equal{}2\\pi r\\sqrt{\\frac{r}{GM}}$.\nSince mass $ M$ is proportional to the planet's volume, one can say that $ M\\equal{}k\\frac{4}{3}\\pi r^3\\equal{}jr^3$ (where $ k$ and $ j$ are both constants).\nTherefore, $ T\\equal{}2\\pi r\\sqrt{\\frac{r}{Gjr^3}}\\equal{}\\frac{2\\pi}{Gj}$. Thus, the period does not depend on anything except some constants, so the period has to be same for both planets.[/hide]", "Solution_3": "Yes, looks right.\r\n\r\nAnd option D doesn't work because the orbital radius doesn't cancel with the planet's radius, so $ T$ isn't independent of the two planets' radii." } { "Tag": [ "algebra", "polynomial" ], "Problem": "prove that $a(a+1)(a+2)(a+3)+1$ is perfect square.", "Solution_1": "[quote=\"mahyar_math\"]prove that $a(a+1)(a+2)(a+3)+1$ is perfect square.[/quote]\r\n\r\n[hide]$a(a+1)(a+2)(a+3)+1=(a^{2}+3a)(a^{2}+3a+2)+1=(a^{2}+3a+1)^{2}$[/hide]", "Solution_2": "Show that the polynomial $x^{5}+x+1$can be expressed as the product\r\nof two nonconstant polynomials with integer coefcients.", "Solution_3": "[quote=\"mahyar_math\"]Show that the polynomial $x^{5}+x+1$can be expressed as the product\nof two nonconstant polynomials with integer coefcients.[/quote]\r\n\r\n[hide=\" :spam: \"]$x^{5}+x+1=(x^{2}+x+1)(x^{3}-x^{2}+1)$[/hide]", "Solution_4": "[quote=\"mahyar_math\"]Show that the polynomial $x^{5}+x+1$can be expressed as the product\nof two nonconstant polynomials with integer coefcients.[/quote]\r\n\r\n[hide=\"simple factorization...\"]\n$(x^{2}+x+1)(x^{3}-x^{2}+1)$\nuse synthetic division[/hide]" } { "Tag": [ "superior algebra", "superior algebra solved" ], "Problem": "Please help me with this problem\r\n\r\nLet a and b be elements of a finite group G. Show that ab and ba have the same order.\r\n\r\nThank you.", "Solution_1": "Assume that $(ab)^n=e$, then $(ba)^n=b(ab)^nb^{-1}=b e b^{-1}=e$. Changing $a,b$ gives that they have the same orders." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "let $ m\\in\\mathbb{N}$ be a fixed positive integer and $ x_n\\equal{}1\\plus{}mn^2$. Is it true that for every $ p$ - prime there is a term of this sequence divisible by $ q^p$ for some $ q\\ge 2$ ?\r\nThe case $ p\\equal{}2$ is trivial (consider $ x_{4m\\plus{}3}$ and $ q\\equal{}4m\\plus{}1$), but I have no idea for $ p\\equal{}3,5,7,\\dots$ or for the general proof (if this conjecture is true)", "Solution_1": "Yes. If $ q$ is an odd prime not dividing $ m$ one can use [url=http://en.wikipedia.org/wiki/Hensel%27s_lemma]Hensel's lemma[/url]. There's no benefit to be gained by restricting $ p$ to primes.", "Solution_2": "obviously if it works for $ p$ primes then for other $ p$ it works too, so the main problem was the case with primes\r\nThanks :) I haven't heard about Hensel's lemma before", "Solution_3": "Oh, then you must have meant $ p^q$." } { "Tag": [ "invariant", "combinatorics unsolved", "combinatorics" ], "Problem": "There's a $ 3\\times3$ square, one is empty and the others are with numbers 1~8 in it. Move one number up or down or left or right to the empty place each time. Prove it is impossible to do such change in finite steps,\r\n$ \\boxed{ \\begin{array}{*{20}c}1 & 2 & 3 \\\\ 4 & 5 & 6 \\\\ 8 & 7 &{}\\\\ \\end{array}}\\to \\boxed{ \\begin{array}{*{20}c}1 & 2 & 3 \\\\ 4 & 5 & 6 \\\\ 7 & 8 &{}\\\\ \\end{array}}$", "Solution_1": "Loyd's game :lol: Find one invariant :)", "Solution_2": "[quote=\"N.T.TUAN\"]Loyd's game :lol: Find one invariant :)[/quote]\r\nBut...to be more specific...what is the invariant :?:" } { "Tag": [ "percent" ], "Problem": "Calculate the weight percent of Fe(III) in \r\n$ K_3[Fe(C_2O_4)]3H_2O$ (theoretical)", "Solution_1": "Isn't missing something in your formula?", "Solution_2": "I don't know. What's missing? and once I find it how can use it to give me the answer", "Solution_3": "Look at the charges. If what you meant is $ K_3[Fe(C_2O_4)] \\cdot 3H_2O$ then the formula is incorrect; if you wanted to mean $ K_3[Fe(C_2O_4)]_3 \\cdot H_2O$ then it is correct but the \"3s\" are not needed.\r\nAlso, to calculate a mass percentage of an element in a compound just divide the mass of the element present in one mole of the compound by its molar mass, and then multiply by 100." } { "Tag": [], "Problem": "In a triangle $ ABC$, where $ a \\equal{} BC$, $ b \\equal{} CA$ and $ c \\equal{} AB$, it is known that: $ a \\plus{} b \\minus{} c \\equal{} 2$ and $ 2ab \\minus{} c^2 \\equal{} 4$. Prove that $ ABC$ is an equilateral triangle.", "Solution_1": "a+b-c=2 et 2ab-c\u00b2=4 ===> (a+b-c)\u00b2=(2ab-c\u00b2) ===> a\u00b2+b\u00b2+2c\u00b2-2c(a+b)=0\r\n\r\na\u00b2+b\u00b2+2c(c-a-b)=0 ===> 2ab-4c<=a\u00b2+b\u00b2-4c=0 ===> 2ab-4c<=0\r\n\r\n2ab=4+c\u00b2 ===> 2ab-4c=4+c\u00b2-4c=(c-2)\u00b2>=0 so (c-2)=0 ==>c=2\r\n\r\n\r\ndonc a+b=4 et 2ab=8 (a+b)\u00b2=16 et 4ab=16 ==> (a+b)\u00b2=4ab ===> a=b\r\n\r\npuis en deduit que a=b=c=2\r\n\r\nso ABC is an equilateral triangle", "Solution_2": "[quote=\"mohamed_3ab9or_maths\"]$ a \\plus{} b \\minus{} c \\equal{} 2$ and $ 2ab \\minus{} c^2 \\equal{} 4\\implies (a \\plus{} b \\minus{} c)^2 \\equal{} 2ab \\minus{} c^2\\implies a^2 \\plus{} b^2 \\plus{} 2c^2 \\minus{} 2c(a \\plus{} b) \\equal{} 0$\n\n$ a^2 \\plus{} b^2 \\plus{} 2c(c \\minus{} a \\minus{} b) \\equal{} 0\\implies 2ab \\minus{} 4c\\leq a^2 \\plus{} b^2 \\minus{} 4c \\equal{} 0\\implies 2ab \\minus{} 4c\\leq 0$\n\n$ 2ab \\equal{} 4 \\plus{} c^2\\implies 2ab \\minus{} 4c \\equal{} 4 \\plus{} c^2 \\minus{} 4c \\equal{} (c \\minus{} 2)^2\\leq 0$. So $ c \\minus{} 2 \\equal{} 0\\implies c \\equal{} 2$.\n\n\nTherefore, $ a \\plus{} b \\equal{} 4$ and $ 2ab \\equal{} 8,\\ (a \\plus{} b)^2 \\equal{} 16$ and $ 4ab \\equal{} 16\\implies (a \\plus{} b)^2 \\equal{} 4ab\\implies a \\equal{} b$\n\nThen we deduce that $ a \\equal{} b \\equal{} c \\equal{} 2$\n\nSo $ \\triangle ABC$ is an equilateral triangle.[/quote]\r\n\r\nI corrected and cleaned some parts of solution." } { "Tag": [ "conics", "geometry", "angle bisector", "geometry proposed" ], "Problem": "Let a triangle ABC and a point P inside it. AP, BP, CP meet BC, CA, AB in D,E,F. The angle bisector of $ \\angle CPE$ meet the angle bisector of $ \\angle PBC$ in $ P_1$ and the angle bisector of $ \\angle PCB$ in $ P_2$, the angle bisector of $ \\angle CPD$ meet the angle bisector of $ \\angle PCA$ in $ P_3$ and the angle bisector of $ \\angle PAC$ in $ P_4$, the angle bisector of $ \\angle APE$ meet the angle bisector of $ \\angle PAB$ in $ P_5$ and the angle bisector of $ \\angle PBA$ in $ P_6$.\r\n\r\n1) Prove that $ P_1$, $ P_2$, $ P_3$, $ P_4$, $ P_5$, $ P_6$ stay on the same conic $ \\xi(P)$\r\n2) Find the locus of the point P for which A is on the interior of $ \\xi(P)$\r\n3) Can A,B,C be all on the interior of $ \\xi(P)$?\r\n\r\n[hide=\"figure\"][img]http://img442.imageshack.us/img442/6910/conicaty5.jpg[/img][/hide]", "Solution_1": "No one can solve it :?:" } { "Tag": [ "induction", "number theory unsolved", "number theory" ], "Problem": "$p$ is a prime number that is greater than $2$. Let $\\{ a_{n}\\}$ be a sequence such that $ na_{n+1}= (n+1) a_{n}-\\left( \\frac{p}{2}\\right)^{4}$. \r\n\r\nShow that if $a_{1}=5$, the $16 \\mid a_{81}$.", "Solution_1": "It is easy to prove (for example by induction), that $a_{n}=na_{1}-(\\frac{p}{2})^{4}(n-1)$. It give $a_{81}=81a_{1}-5p^{4}=5(81-p^{4})$. Because $16|(p^{4}-1)$ for any odd number p, $16|a_{81}$.", "Solution_2": "$na_{n+1}= (n+1) a_{n}-\\left( \\frac{p}{2}\\right)^{4}$\r\n\r\n$\\Longleftrightarrow \\frac{1}{n+1}\\left\\{a_{n+1}-\\left(\\frac{p}{2}\\right)^{4}\\right\\}=\\frac{1}{n}\\left\\{a_{n}-\\left(\\frac{p}{2}\\right)^{4}\\right\\}=\\cdots\\cdots =\\frac{1}{1}\\left\\{a_{1}-\\left(\\frac{p}{2}\\right)^{4}\\right\\},$ yielding Rust's result." } { "Tag": [ "geometry open", "geometry" ], "Problem": "For circumference (O1), (O2) contact ben wherewith (O) for before at P, Q and in addition to after-birth. Circumference cutting this two-way's ben general tangent line (O) at 4 points, among them choosed B, C it be 2 points to lie the same margarine half-plane O1O2. Two-way's wherewith tangential rather only general parallel proof BC (O1), (O2) \r\n :( :(", "Solution_1": "Maybe you had used an automatic way of translating, but the wording is ununderstandable, at least for me." } { "Tag": [ "percent" ], "Problem": "One-third of a 24-student class is absent today. One-half of those were also absent yesterday. What percent of the class has been absent for two straight days? Express your answer to the nearest whole number.", "Solution_1": "[hide]1/3 * 24 = 8 1/2*8 = 4 4/24 = 1/6 = 16 and 1/6 percent and rounded to the nearest whole number is 17%[/hide]", "Solution_2": "[hide]one third of the class=8\nhalf of 8 = 4\n4 / 24 = 1/6\n\n100/6%=50/3% = 16.6.........%[/hide]", "Solution_3": "[quote=\"MCrawford\"]One-third of a 24-student class is absent today. One-half of those were also absent yesterday. What percent of the class has been absent for two straight days? Express your answer to the nearest whole number.[/quote]\r\n\r\n[hide]1/3 of 24 = 8\n1/2 of 8 = 4\n8-4=4\n4/24=1/6\n=0.166...\n17%[/hide]", "Solution_4": "One piece of information in that problem could have been left out, and the problem would still be solvable. What was that piece of information?", "Solution_5": "i don't know..\r\nwait...could it be solved without knowing that there were 24 students?", "Solution_6": "the info about the 24 students can be left out cuz 1/2*1/3 = 1/6, so the answer would be 16.6... anyways.", "Solution_7": "8 are absent the firstt time \r\n8/2=4 \r\n4/24=1/6 \r\n16.6..%", "Solution_8": "[hide]\n17%\n[/hide]", "Solution_9": "Not necessarily. What if there weren't 6n students? Do you cut 'em in halves and thirds like I do?", "Solution_10": "i don't know the answer to Treething's question but the answer to the original problem is[hide]17%[/hide]", "Solution_11": "one-third of 24 is 8, and one-half of that is 4. 4/24=1/6 or $16.\\overline{6}$. The nearest whole number is 17.", "Solution_12": "[hide=\"Answer\"]$17$[/hide]", "Solution_13": "We can solve the problem without being that there are 24 students", "Solution_14": "[quote=\"shinwoo\"][hide]one third of the class=8\nhalf of 8 = 4\n4 / 24 = 1/6\n\n100/6%=50/3% = 16.6.........%[/hide][/quote]\r\nround the answer", "Solution_15": "[quote=\"Devendra Patel\"]We can solve the problem without being that there are 24 students[/quote]\r\nwhat, i don't understand what you are saying" } { "Tag": [ "function", "linear algebra", "matrix" ], "Problem": "Could someone translate in english thank you \r\n\r\n----------------------------------\r\n1/ \r\n\r\nfie f,g :R $\\rightarrow$R \r\nfunctii surjective \r\nDaca $go(g-f)$ este descrescatoare\r\nsi exista L>1, astfel incat $|g(x)-g(y)|\\geq L|x-y|$ $\\forall x,y \\in R$\r\n\r\nsa se arate ca f are un unic punit fix\r\n--------------------------------------------------------\r\n2/fie $A\\in M(n,C)$ o matrice cu coeficienti intregi inversabila si astfel incat\r\nmultinea {$A^{k}, k\\in N$} este finita\r\nsa se demonstreze ca aceasta multine are cel mult $3^{n^2}$ element", "Solution_1": "1.Let f,g be surjective functions. If ... is a nonincreasing\r\nfunction and if there exists L >1 such that ...., prove that f has\r\na unique fixed point. \r\n2. Let A... be an invertible matrix with\r\ninteger coefficients, such that the set... is finite. Prove that this\r\nset has at most ... elements." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a_1,a_2,...,a_6$ be real numbers such that: \r\n$ a_1 \\not \\equal{} 0, a_1a_6 \\plus{} a_3 \\plus{} a_4 \\equal{} 2a_2a_5 \\ \\mathrm{and}\\ a_1a_3 \\ge a_2^2$\r\nProve that $ a_4a_6\\le a_5^2$. When does equality holds?", "Solution_1": "That's false!Consider for example a1=1;a2=2;a3=4;a4=14;a5=7;a6=10" } { "Tag": [ "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "Let $M$ be a set of matrix $A_{3,3}$ that has elements from ${ { 1, 2,3,4,5,6,7,8,9} }$. Prove the fallowing:\r\nShow that, if B is from M then $B^{-1}$ is $not$ from M", "Solution_1": "If $A$ and $B$ are in $M$, each entry of $AB$ is at least $1\\cdot 1+1\\cdot 1+1\\cdot 1=3$. Therefore, $AB$ is not the identity and the two matrices are not inverses." } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "Solve in $\\mathbb Z$:\r\n$\\sqrt{x+\\sqrt{x+\\sqrt{x+...+\\sqrt{x}}}}=y$, the number of $\\sqrt$ is $2007$", "Solution_1": "[quote=\"Zamfirmihai\"]Solve in $\\mathbb Z$:\n$\\sqrt{x+\\sqrt{x+\\sqrt{x+...+\\sqrt{x}}}}=y$, the number of $\\sqrt$ is $2007$[/quote]\r\n\r\nIf I call $f_{n}(x)=\\sqrt{x+\\sqrt{x+\\sqrt{x+...+\\sqrt{x}}}}$, where the number of $\\sqrt$ is $n$, then I have to solve $f_{2007}(a)=b$\r\nSo $f_{2006}(a)=b^{2}-a \\in\\mathbb{Z}$, so $f_{n}(a)\\in \\mathbb{Z}$ $\\forall n\\leq 2007$\r\n\r\nSo $f_{1}(a)\\in\\mathbb{Z}$ and so $a=u^{2}$\r\nThen $f_{2}(a)\\in\\mathbb{Z}$ and $u^{2}+u=v^{2}$\r\n\r\nBut, if $u\\neq 0$ $u^{2}+u$ is in $((|u|-1)^{2},(|u|+1)^{2})$ and can't be $v^{2}$\r\n\r\n\r\nAnd the only solution $(x,y)$ is $(0,0)$" } { "Tag": [ "advanced fields", "advanced fields solved" ], "Problem": "How to compute $\\pi_{1}(\\mathbb{S}^{2}\\vee \\mathbb{S}^{1})$ and $\\pi_{1}(\\mathbb{S}^{2}\\vee \\mathbb{S}^{2})$?", "Solution_1": "i guess you mean $X\\vee Y$ to be $X \\sqcup Y$ where you identify two points, one for each space, right?\r\n\r\nif so, you can use van kampen's theorem, in both cases:\r\n$A = \\mathbb{S}^{2}\\vee\\mathbb{S}^{1}$: call $x_{2}\\in \\mathbb{S}^{2}, x_{1}\\in \\mathbb{S}^{1}$ the two points you identified, then you can find contractible open neighbourhoods of each point in its \"original\" space, and call them $U_{2}, U_{1}$. then $U_{1}\\cup U_{2}\\subset A$ is open and connected, which allows you to use quoted theorem.\r\nsame as before for the other thing.\r\nso they come out to be $\\mathbb{Z}$ and the trivial group.\r\n\r\nanyway, you can do it more \"by hands\", and split any path $\\gamma$ in $\\gamma_{i}$, where each of these \"sub-path\" is closed, and starts in the points you've identified, and $\\text{spt }\\gamma_{i}$ lies in one of the two \"original\" spaces. you can do it, since $\\text{spt }\\gamma$ is compact...\r\nthen you have the same result..\r\nof course, this is nothing but the proof of van kampen's theorem, a bit simplified..." } { "Tag": [ "function" ], "Problem": "If $r$ and $s$ are the roots of the equation $ax^2 + bx + c = 0$, the value of $\\frac{1}{r^2} + \\frac{1}{s^2}$ is:\r\n\\[ \\text{(A)}\\ b^2 - 4ac \\qquad \\text{(B)}\\ \\frac{b^2 - 4ac}{2a} \\qquad \\text{(C)}\\ \\frac{b^2 - 4ac}{c^2} \\qquad \\text{(D)}\\ \\frac{b^2 - 2ac}{c^2} \\qquad \\text{(E) none of these} \\]", "Solution_1": "[hide]We wish to find the value of $\\frac{1}{r^{2}}+\\frac{1}{s^{2}}$. If we can write this sum as a combination of the sums and products of the roots, we can find the value, because we know that the sum of the roots is $-\\frac{b}{a}$, and the product is $\\frac{c}{a}$.\n\nWe can do this by writing $\\frac{1}{r^{2}}+\\frac{1}{s^{2}}=\\frac{r^{2}+s^{2}}{r^{2}s^{2}}=\\frac{(r+s)^{2}-2rs}{(rs)^{2}}$. Substituting in the sums and products of the roots gives us $\\frac{\\displaystyle\\left(\\displaystyle-\\frac{b}{a}\\right)^{2}-2\\left(\\displaystyle\\frac{c}{a}\\right)}{\\left(\\displaystyle\\frac{c}{a}\\right)^{2}}$, which simplifies to:\n\n[b]D[/b]: $\\frac{b^{2}-2ac}{c^{2}}$.[/hide]", "Solution_2": "[hide=\"heres another way\"]\nThe equation:\n\n\\[a\\left(\\frac{1}{\\sqrt{x}}\\right)^2+b\\left(\\frac{1}{\\sqrt{x}}\\right)+c=0\\]\n\nhas roots $\\frac{1}{r^2}$ and $\\frac{1}{s^2}$. So let's just find the sum of the roots of this equation. First multiply through by $x$:\n\n\\[a+b\\sqrt{x}+cx=0.\\]\n\nIsolate $b\\sqrt{x}$ and square:\n\n\\[b\\sqrt{x}=-(a+cx)\\Rightarrow b^2x=a^2+2acx+c^2x^2.\\]\n\nSo then:\n\n\\[c^2x^2+(2ac-b^2)x+a^2=0\\]\n\nand it follows that:\n\n\\[\\frac{1}{r^2}+\\frac{1}{s^2}=\\frac{b^2-2ac}{c^2}.\\]\n\nQED[/hide]", "Solution_3": "yea, using transformation of the function is clever.\r\nbtw, what is QED?", "Solution_4": "[quote=\"BornForMath\"]yea, using transformation of the function is clever.\nbtw, what is QED?[/quote]I'm not sure exactly what it stands for, but I think it's used to signify that a proof or whatever is complete.", "Solution_5": "[quote=\"BornForMath\"]yea, using transformation of the function is clever.\nbtw, what is QED?[/quote]\r\n\r\nQED stands for [color=blue]Q[/color]uad [color=blue]E[/color]sta [color=blue]D[/color]emostrandom in Latin, \r\nwhich can be translated in English, [b]which is to be proved[/b].", "Solution_6": "[quote=\"kunny\"][quote=\"BornForMath\"]yea, using transformation of the function is clever.\nbtw, what is QED?[/quote]\n\nQED stands for [color=blue]Q[/color]uad [color=blue]E[/color]sta [color=blue]D[/color]emostrandom in Latin, \nwhich can be translated in English, [b]which is to be proved[/b].[/quote]\r\n\r\nActually, it stands for \"Quod Erat Demonstrandum.\"", "Solution_7": "Oh! yes, I am wrong. esta is for Spanish? Thank you. ;)" } { "Tag": [ "inequalities", "AMC", "USA(J)MO", "USAMO", "ARML", "function", "trigonometry" ], "Problem": "For those of you preparing for the USAMO, here's an inequality.\r\n\r\nGiven $x,y,z>0$, prove that:\r\n\r\n$\\frac {x}{(x+y)(x+z)}+\\frac {y}{(y+x)(y+z)}+\\frac{z}{(z+x)(z+y)}\\leq \\frac {9}{4(x+y+z)}$", "Solution_1": "My solution:\n\n\n\n[hide]\n\nAs before, homogenize to get x+y+z = 1. Get a common denominator. Now:\n\n\n\n\n\n\n\nlast ineq is clearly true, its equivalent to the first, so the first is true, QED.\n\n[/hide]\n\n\n\nwow, this ineq was easy!", "Solution_2": "In your hint, I get the part up until applying Schur. I can't seem to get it to work that way. Would someone mind explaining that last bit?\r\n\r\nHowever, simply expanding the whole thing out under a common denominator was the way I first did it and it seems to be very easy (as shown by Singular).", "Solution_3": "i dont get how you apply the homogenous rule of x+y+z = 1. isnt it just simple to multiply and factor some out and use the derivations of \r\n$\\displaystyle x^2 + y^2 + z^2 = (x+y+z)^2 - 2(xy+yz+zx)$?", "Solution_4": "oh never mind.....\r\n\r\ni misread the $(x+y)(y+z)(z+x)$ part.... \r\n\r\nbut i still dont get the aprt with x+y+z = 1 or how to apply Schur's.", "Solution_5": "xxreddevilzxx, I didn't use Schur so I don't know about that. As for the x+y+z=1, assume x+y+z=k for some real k then there exist reals a, b, c such that x=ka, y=kb, and z=kc so the condition is equivalent to a+b+c=1. So we can make this substitution since the degree is the same on both sides (the k factors out) and then proceed from there. I hope this is enough. \r\n\r\nP.S. Were you on the ARML Washinton A team last year?", "Solution_6": "I also don't see where to apply Schur here, but to be honest, I also tend to think that the function $\\frac{1}{x}$ is convex and not concave (at least on $\\mathbb{R}^{+}$), so I'm not sure whether Blahblahblah's hint is correct at all...\r\n\r\nAnyway, here is how I would solve the problem:\r\n\r\n[hide=\"Solution of the problem\"]\n\nWe have to prove the inequality\n\n$\\frac{x}{\\left(z+x\\right)\\left(x+y\\right)}+\\frac{y}{\\left(x+y\\right)\\left(y+z\\right)}+\\frac{z}{\\left(y+z\\right)\\left(z+x\\right)}\\leq\\frac{9}{4\\left(x+y+z\\right)}$\n\nfor any positive reals x, y, z.\n\nLet a = y + z, b = z + x and c = x + y. Then, since b + c = (z + x) + (x + y) = y + z + 2x > y + z = a, and similarly, c + a > b and a + b > c, the numbers a, b, c are the sides of a triangle. Let A, B and C be the corresponding angles and $s=\\frac{a+b+c}{2}$ the semiperimeter of this triangle. Then,\n\n$s=\\frac{a+b+c}{2}=\\frac{\\left(y+z\\right)+\\left(z+x\\right)+\\left(x+y\\right)}{2}=\\frac{2\\left(x+y+z\\right)}{2}=x+y+z$,\n\nand s - a = (x + y + z) - (y + z) = x, and similarly s - b = y and s - c = z. Thus, the inequality that we have to prove,\n\n$\\frac{x}{\\left(z+x\\right)\\left(x+y\\right)}+\\frac{y}{\\left(x+y\\right)\\left(y+z\\right)}+\\frac{z}{\\left(y+z\\right)\\left(z+x\\right)}\\leq\\frac{9}{4\\left(x+y+z\\right)}$,\n\nbecomes\n\n$\\frac{s-a}{bc}+\\frac{s-b}{ca}+\\frac{s-c}{ab}\\leq\\frac{9}{4s}$.\n\nMultiply both sides of this inequality with s to get:\n\n$\\frac{s\\left(s-a\\right)}{bc}+\\frac{s\\left(s-b\\right)}{ca}+\\frac{s\\left(s-c\\right)}{ab}\\leq\\frac94$.\n\nUpon multiplication with 2, this becomes:\n\n$2\\frac{s\\left(s-a\\right)}{bc}+2\\frac{s\\left(s-b\\right)}{ca}+2\\frac{s\\left(s-c\\right)}{ab}\\leq\\frac92$.\n\nAnd upon subtraction of 3 this becomes\n\n$2\\frac{s\\left(s-a\\right)}{bc}+2\\frac{s\\left(s-b\\right)}{ca}+2\\frac{s\\left(s-c\\right)}{ab}-3\\leq\\frac92-3$,\n\nwhat is equivalent to\n\n$\\left(2\\frac{s\\left(s-a\\right)}{bc}-1\\right)+\\left(2\\frac{s\\left(s-b\\right)}{ca}-1\\right)+\\left(2\\frac{s\\left(s-c\\right)}{ab}-1\\right)\\leq\\frac32$.\n\nBut after the half-angle formulas for the angles of triangle ABC, we have\n\n$\\cos^2\\frac{A}{2}=\\frac{s\\left(s-a\\right)}{bc}$.\n\nThus,\n\n$\\cos A=\\cos\\left(2\\frac{A}{2}\\right)=2\\cos^2\\frac{A}{2}-1=2\\frac{s\\left(s-a\\right)}{bc}-1$.\n\nSimilarly,\n\n$\\cos B=2\\frac{s\\left(s-b\\right)}{ca}-1$;\n$\\cos C=2\\frac{s\\left(s-c\\right)}{ab}-1$.\n\nThus, the inequality in question becomes\n\n$\\cos A+\\cos B+\\cos C\\leq\\frac32$.\n\nBut this is a well-known inequality (see, for instance, http://www.mathlinks.ro/Forum/viewtopic.php?t=21188 ). This completes the proof of the initial inequality.[/hide]\r\n\r\n[b]EDIT:[/b] As I see, this inequality has also been proven in http://www.mathlinks.ro/Forum/viewtopic.php?t=4908 . I have a bad memory!\r\n\r\n Darij", "Solution_7": "nice darij!", "Solution_8": "that's a great trigo sub!", "Solution_9": "The function is convex, $f''(x)=\\frac{2}{x^3}$ and $f''(x)>0$ for all $x>0$", "Solution_10": "[quote=\"DPopov\"]The function is convex, $f''(x)=\\frac{2}{x^3}$ and $f''(x)>0$ for all $x>0$[/quote]\r\n\r\n*sigh* duh... no wonder when i tried to do it after applying Jensen's everything was backwards :blush:", "Solution_11": "Too bad I'm not familiar with Jensen's :(", "Solution_12": "[quote=\"DPopov\"]Too bad I'm not familiar with Jensen's :([/quote]\r\n\r\n[url=http://mathworld.wolfram.com/JensensInequality.html]Here[/url] you can find it. Basically it says that if you have a continuous concave function in $[a,b]$, and $x_1,\\dots,x_n$ are points in that interval, $t_i>0,\\ t_1+\\cdots+t_n=1$, then\r\n\r\n\\[\\sum_{i=1}^{n}t_if(x_i)\\leq f\\left(\\sum_{i=1}^{n}t_ix_i\\right)\\]\r\n\r\nIf the function is convex, then the inequality goes in the other direction\r\n\r\n\\[\\sum_{i=1}^{n}t_if(x_i)\\geq f\\left(\\sum_{i=1}^{n}t_ix_i\\right)\\]\r\n\r\nI have seen some books that call it \"The Mother of all Inequalities\" (Specially some Math Olympiad Books) because most of the inequalities that you use on Olympiads are particular cases of the Jensen's Inequality (Included GM-AM and many others).\r\n\r\nBest,", "Solution_13": "thanx! :)", "Solution_14": "could someone explian or point me to schurs inequality?", "Solution_15": "Schur:\r\n\r\n$a^k(a-b)(a-c)+b^k(b-c)(b-a)+c^k(c-a)(c-a) \\ge 0$\r\n\r\nfor $a,b,c$ nonnegative reals and $k>0$.\r\n\r\nCouldn't find it on Mathworld, interestingly enough.", "Solution_16": "I still don't understand how you can assume that x+y+z=1. Does that mean if an inequality is homogenous of degree 2 you can say x^2 + y^2 + z^2=2", "Solution_17": "[quote=\"mna851\"]I still don't understand how you can assume that x+y+z=1. Does that mean if an inequality is homogenous of degree 2 you can say x^2 + y^2 + z^2=2[/quote]\r\n\r\nSo assume x+y+z=k instead. By scaling each variable down to x'=x/k, y'=y/k, z'=z/k, we see that x',y', and z' still satisfy the inequality (by canceling out all the k's) and x'+y'+z'=1.\r\n\r\nAnother way is by proving it for x+y+z=1, we can multiply the whole thing by k/k^2 to get x+y+z=k for any k.", "Solution_18": "lollerskates. I think that using Jensen's in the wrong direction has to be the most common mistake in inequalities. Anyways, it holds easily by Muirhead's, although all such solutions are inevitably very unsatisfying.", "Solution_19": "We have to prove that\r\n$\\frac{2(xy+xz+yz)}{(x+y)(x+z)(y+z)}\\leq \\frac{9}{4(x+y+z)}$\r\n\r\nor $(x+y)(x+z)(y+z)\\geq \\frac{8(xy+xz+yz)(x+y+z)}{9}$\r\nIt is easy\r\n$(x+y)(x+z)(y+z)= (x+y+z)(xy+xz+yz)-xyz\\geq (x+y+z)(xy+xz+yz)-\\frac{(x+y+z)(xy+xz+yz)}{9}=\\frac{8(xy+xz+yz)(x+y+z)}{9}$", "Solution_20": "[hide=\"brute force w/o muirhead\"]\nWe have:\n$\\sum_{cyc}\\frac{x}{(x+y)(x+z)}\\leq \\frac{9}{4(x+y+z)}$\n$\\sum_{cyc}4x(y+z)(x+y+z)\\leq \\sum_{cyc}3(x+y)(x+z)(y+z)$\n$\\sum_{cyc}4x^{2}y+4xy^{2}+4xyz+4x^{2}z+4xyz+4xz^{2}\\leq $$\\sum_{cyc}3x^{2}y+3x^{2}z+3xyz+3xz^{2}+3xy^{2}+3xyz+3y^{2}z+3yz^{2}$\n\n$6xyz\\leq \\sum_{sym}x^{2}y$ Which follows from $\\text{AM-GM}$.\n[/hide]" } { "Tag": [ "geometry", "ratio", "geometry unsolved" ], "Problem": "Inside triangle $ ABC$, point $ P$ is taken. The lines that pass through $ P$ and vertices\r\nof the triangle intersect the sides at points $ A_1$, $ B_1$ and $ C_1$. Prove that the area of the triangle\r\ndetermined by the midpoints of segments $ AA_1$, $ BB_1$ and $ CC_1$ is equal to $ 1/4$ of the area of\r\ntriangle $ A_1B_1C_1$. :wink: :huh:", "Solution_1": "A2C\u2019=A2\u2019B\u2019 B\u2019C2\u2019=C2A\u2019 A\u2019B2=B2\u2019C\u2019 \r\nDenote A\u2019B2=a A\u2019C2=b C2B\u2019=c B\u2019A2=d A2C\u2019= e C\u2019B2=f\r\nS(\u22bfA\u2019B2C2)+S(\u22bfC2B\u2019A2)+S(\u22bfA2C\u2019B2)=ab*sinA + cd*sinB + ef*sinC \uff081\uff09\r\nS(\u22bfA\u2019B2\u2019C2\u2019)+S(\u22bfC2\u2019B\u2019A2\u2019)+S(\u22bfA2\u2019C\u2019B2\u2019)=cf*sinA + be*sinB + ad*sinC \uff082\uff09\r\nNotice sinA=(e+d)/2R sinB=(a+f)/2R sinC=(d+e)/2R \uff083\uff09\r\n(1)=abd + abe + acd + cdf + ebf + efc\r\n(2)=abd + adc + aeb + ebf + fcd +efc\r\nSo (1)=(2)\r\nThus we have S(\u22bfA2B2C2)=S(\u22bfA2\u2019B2\u2019C2\u2019)=1/4S(\u22bfA1B1C1)", "Solution_2": "Let $ M, N$ and $ P$ the midpoint of sides $ [BC], [CA]$ and $ [AB]$, respectively; and let $ G$ the centroid of $ \\triangle {ABC}\\Rightarrow \\{G\\} \\equal{} AA_1\\cap BB_1 \\cap CC_1.$\r\nDenote \r\n$ \\{A_0\\} \\equal{} AA_1\\cap PN; \\ \\{A_2\\} \\equal{} A_1G\\cap PN; \\ \\{B_0\\} \\equal{} BB_1\\cap PM; \\\\\r\n\\{B_2\\} \\equal{} B_1G\\cap PM; \\ \\{C_0\\} \\equal{} CC_1\\cap MN; \\ \\{C_2\\} \\equal{} C_1G\\cap MN.$\r\n[b]1).[/b] Have: $ |A_0P| \\equal{} |A_2N|; |B_0M| \\equal{} |B_2P|; |C_0N| \\equal{} |C_2M|\\Rightarrow$\r\n$ \\Rightarrow area (\\triangle{A_0B_0C_0}) \\equal{} area(\\triangle{A_2B_2C_2}). \\ \\ (1)$\r\n[b]2). [/b]The $ \\triangle{A_2B_2C_2}$ and $ \\triangle{A_1B_1C_1}$ is inversly homotetic; the homotetic center is $ G$ and the homotetic ratio is $ \\frac {1}{2}\\Rightarrow$\r\n$ \\Rightarrow area(\\triangle{A_2B_2C_2}) \\equal{} \\frac {1}{4}.area(\\triangle{A_1B_1C_1}). \\ \\ (2).$\r\n[b]3).[/b] $ (1);(2)\\Rightarrow area(\\triangle{A_0B_0C_0}) \\equal{} \\frac {1}{4}.area(\\triangle{A_1B_1C_1}).$" } { "Tag": [ "algebra", "system of equations" ], "Problem": "[b]If $ (x,y)$ is a solution for the system of equations below, find the maximum value of $ x^{2}\\plus{}y^{2}$.\n\n$ 2x^{2}\\plus{}5xy\\plus{}3y^{2}\\equal{}2$\n$ 6x^{2}\\plus{}8xy\\plus{}4y^{2}\\equal{}3$\n[/b]", "Solution_1": "[hide=\"solution\"]\nMultiplying the first equation by -3 and the second by 2 and then adding them we obtain\n$ 6x^{2}\\plus{}xy\\minus{}y^{2}\\equal{}0$\nDivide the last equation by $ y^{2},\\ (y\\neq 0)$ and note $ \\frac{x}{y}\\equal{}t$:\n$ 6t^{2}\\plus{}t\\minus{}1\\equal{}0\\Rightarrow t_{1}\\equal{}\\minus{}\\frac{1}{2},t_{2}\\equal{}\\frac{1}{3}$\nIf $ \\frac{x}{y}\\equal{}\\minus{}\\frac{1}{2}$ then $ x^{2}\\equal{}\\frac{1}{2},y^{2}\\equal{}2\\Rightarrow x^{2}\\plus{}y^{2}\\equal{}\\frac{5}{2}$\nIf $ \\frac{x}{y}\\equal{}\\frac{1}{3}$ then $ x^{2}\\equal{}\\frac{1}{27},y^{2}\\equal{}\\frac{1}{3}\\Rightarrow x^{2}\\plus{}y^{2}\\equal{}\\frac{10}{27}$\nSo the maximum of $ x^{2}\\plus{}y^{2}$ is $ \\frac{5}{2}$\n[/hide]", "Solution_2": "OK for the solution and for the results but this :\r\n[quote=\"red_dog\"] If $ \\frac{x}{y}\\equal{}\\frac{1}{3}$ then $ x^{2}\\equal{}\\frac{1}{27},y^{2}\\equal{}\\frac{1}{3}$[/quote]\r\nis wrong.\r\n\r\nFortunately, this does not change the result.", "Solution_3": "\\[\\begin{array}{l}\n\\left. \\begin{array}{l}\n2{r^2}{\\cos ^2}\\theta + 5{r^2}\\sin \\theta \\cos \\theta + 3{r^2}{\\sin ^2}\\theta = 2\\\\\n6{r^2}{\\cos ^2}\\theta + 8{r^2}\\sin \\theta \\cos \\theta + 4{r^2}{\\sin ^2}\\theta = 3\n\\end{array} \\right\\}\\mathop \\Rightarrow \\limits^{(II) \\div (I)} \\frac{{6{{\\cos }^2}\\theta + \\sin \\theta \\cos \\theta + {{\\sin }^2}\\theta }}{{2{{\\cos }^2}\\theta + 5\\sin \\theta \\cos \\theta + 3{{\\sin }^2}\\theta }} = \\frac{3}{2}\\\\\n \\Rightarrow \\sin \\theta = \\cdots ,\\cos \\theta = \\cdots \\Rightarrow r = \\cdots \\Rightarrow {x^2} + {y^2} = {r^2} = \\cdots \n\\end{array}\\]" } { "Tag": [], "Problem": "why is genetic polymorphism important to evolution?\r\n1.individual variability provides the raw material for natural selection to act on\r\n2.genes cant mutate unless they are polymorphic\r\n3.only heterozygous individuals are selected in natural populations\r\n4. the hardy weinberg equilibrium is less likely to be disturbed in polymorphic population\r\n5. none of the above; genetic polymorphism is not important to evolution\r\n\r\nin a population of wildflowers, the frequency of the alle for the red flowers was 0.8 what was the frequency of the white allele the only other allele for flower color?\r\n\r\nit is .2?\r\n\r\nrefer to the question above what is the frequency of homozygous white flower plants in the population? \r\n1. .04\r\n2. .16\r\n3. .32\r\n4. .48\r\n5. .64\r\n\r\n\r\nrefer to the question above what is the frequency of the plants in the population that are heterozygous for flower color?\r\n1. .04\r\n2. .16\r\n3. .32\r\n4. .48\r\n5. .64\r\n\r\n\r\na virus killed most of the seals in the north sea eg. dropped the population from 8000 to 800. in an effort to help preserve the species scientists caught 20 seals and used them to start a new population in the northwest pacific ocean. what of the following factors would most likely have the least impact in this new population?\r\n1. founder principle\r\n2. random mating\r\n3.genetic drift\r\n4.bottleneck effect\r\n\r\ndoes inbreeding occur in plants?\r\n\r\n\r\n\r\nan individual of which of the following species is likely to have the highest level of heterozygous genes? \r\n1. fish\r\n2.land snail\r\n3. self pollinating plant\r\n4. bird \r\n5. mammal", "Solution_1": "a scientist measures the circumference of acorns in a population of oka trees and discovers that the most common circumference is 2 cm. what would you expect the most commom circumference to be after 10 generation of stabilizing selection? \r\n1. 2cm \r\n\r\nrefer to the question above, but this time answer what you expect after 10 genertation of disruptive selection. \r\n3. greater than 2 cm and less than 2 cm \r\n\r\nrefer to the question above, but this time answer what you expect after 10 genertation of directional selection. \r\n2. greater than 2 cm or less than 2 cm \r\n\r\nwhat is the ultimate source of genetic variability? \r\n1. mutation \r\n\r\nsry can't help u on the last one\r\n\r\nIf u look at the graphs of stabilzing, disruptive, and directional selections u will understand...", "Solution_2": "thank you so much!", "Solution_3": "Fade, genetic polymorphism is one of the most important aspects of evolution. This is the reason:\r\n\r\nMutations produce individual genetic variation (aka, polymorphism.) Polymorphism in a population means a variety of alleles and nucleotide sequences for a certain gene. These differences provide a basis for natural selection to work. For example, the genes that control eye color are polymorphic among members of [i]Homo sapiens[/i]." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Prove that the sides a,b,c of any triangle suck that $ a \\plus{} b \\plus{} c \\equal{} 3$ satisfy the inequality\r\ni/$ \\sum\\frac {a}{b\\plus{}c^2}>\\equal{}\\frac{3}{2}$", "Solution_1": "[quote=\"tranquoc\"]Prove that the sides a,b,c of any triangle suck that $ a \\plus{} b \\plus{} c \\equal{} 3$ satisfy the inequality\ni/$ \\sum\\frac {a}{b \\plus{} c^2} > \\equal{} \\frac {3}{2}$[/quote]\r\n\r\nIt is true with $ a,b,c$ are nonegative real numbers such that $ a\\plus{}b\\plus{}c\\equal{}3$." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let be $ a,b,c$ the side-lenghts of an triangle. Prove that the follow inequality holds: \\[ 4(a^5\\plus{}b^5\\plus{}c^5)>abc(a\\plus{}b\\plus{}c)^2\\][b]\nClaudiu Mindrila[/b]", "Solution_1": "$ 4\\sum a^5\\geq\\frac{4}{3^4}(\\sum a)^5\\geq\\frac{4}{3^4}\\sum a(3\\sum ab)^2\\equal{}\\frac{4}{9}\\sum a(\\sum ab)^2\\geq\\frac{4}{3}abc(\\sum a)^2$", "Solution_2": "It is easily killed by rearrangement theorem as shown\r\n$ 3\\left( a^5\\plus{}b^5\\plus{}c^5 \\right) \\geq \\left(a^2\\plus{}b^2\\plus{}c^2 \\right)\\left(a^3\\plus{}b^3\\plus{}c^3 \\right).$\r\nI wonder whether there would be a nice proof for this problem using the condition the a, b, c be lengths of the triangle.\r\nHow about your idea, M\u00eendril\u0103 Claudiu?", "Solution_3": "We know that $ \\frac{a}{b\\plus{}c}\\plus{}\\frac{b}{c\\plus{}a}\\plus{}\\frac{c}{a\\plus{}b}<2$ . Use that and the rearrangement :) .", "Solution_4": "$ f(a,b,c) \\equal{} 4 (a^5 \\plus{} b^5 \\plus{} c^5) \\minus{} a b c (a \\plus{} b \\plus{} c)^2$.\r\n\r\n$ f(a, b, c) \\minus{} f(a, \\frac {b \\plus{} c}{2},\\frac {b \\plus{} c}{2})$\r\n$ \\equal{} \\frac {1}{4} (b \\minus{} c)^2 (a^3 \\plus{} 2 a^2 b \\plus{} a b^2 \\plus{} 15 b^3 \\plus{} 2 a^2 c \\plus{} 2 a b c \\plus{} 25 b^2 c \\plus{} a c^2 \\plus{} 25 b c^2 \\plus{} 15 c^3)$.\r\n\r\nSo WLOG $ a \\plus{} b \\plus{} c \\equal{} 3$, $ b \\equal{} c \\equal{} x$, $ a \\equal{} 3 \\minus{} 2x$. WLOG $ a\\le b\\le c$ so $ 3 \\minus{} 2x\\le x\\implies x\\ge 1$. $ a\\ge 0\\implies x\\le\\frac {3}{2}$.\r\n\r\nSo it suffices to check $ f(3 \\minus{} 2x,x,x) > 0$ for $ x\\in [1,\\frac {3}{2}]$.\r\n\r\n$ f(3 \\minus{} 2x,x,x) \\equal{} 3 \\plus{} (x \\minus{} 1)^2(323 \\minus{} 434x \\plus{} 240x^2 \\minus{} 40x^3)$ which is an easy enough computation. The second factor is greater than 72." } { "Tag": [], "Problem": "A $ 3''\\times5''$ photo is enlarged proportionally such that its smaller dimension is now $ 1'3''$. How many inches are in the larger dimension?", "Solution_1": "Since it is proportional, we divide to get the proportionality constant, then multiply through to get the other dimension.\r\n $ \\frac{15''}{3''}\\equal{}5$\r\nNow, multiply $ 5$ by the other dimension.\r\n $ 5''\\times5\\equal{}\\boxed{25''}$" } { "Tag": [], "Problem": "my teacher says that atoms have a full valence of 8 electrons and i disagreed because i know the valences go \r\n2, 8, 8, 18, 18, 32, 32 but then she coutners by saying something about sub levels which i do not believe so can someone please clear up the situation on valence electrons :)", "Solution_1": "the last shell of an atom always has 8 electrons .this is called the octet rule" } { "Tag": [ "MATHCOUNTS", "blogs", "probability", "geometry", "3D geometry" ], "Problem": "1) NO SIGNING UP REQUIRED. This tournament(more like test) is open to everyone. i just made that other thread to see if anyone would be interested.\r\n\r\n2) PM me the answers. no solutions required.\r\n\r\n3) No rescources, calculators, ect. Only pencil and paper.\r\n\r\n4) All of these problems can be solved using mathcounts techniques.\r\n\r\n5) Each part of the test is due 1 week after it is posted, Not necisarily when i post the next set of problems.\r\n\r\nCombinatorics problems will be posted on my blog in a few moments...\r\n\r\nAt the end, i will make a word document of the problems", "Solution_1": "why did you revive that topic? you don't have to post another thread on the same thing.\r\n\r\nand when are the problems gonna be posted?", "Solution_2": "EDIT:\r\nI have a feeling some people dont know the problems are on my blog, so now they will be right here:\r\n\r\nCombinatorics\r\n1.\tCalculate $ 0*\\binom{10} {0} \\plus{} 1*\\binom{10} {1} \\plus{} 2*\\binom{10} {2}...10*\\binom{10} {10}$.\r\n2.\tThere are 20 candies and 7 kids. One kid will envy another if his/her amount of candies is 3 or more less than the other. For example, a kid with 2 candies will envy another kid with 5 candies, but not a kid with 4 candies. How many envy-free ways can we distribute the candies?\r\n3.\tHow many 8 digit integers with distinct digits in decreasing order when read from left to right are multiples of 3?\r\n4.\tWe fill a jar with marbles such that probability of drawing a blue marble from the jar is $ \\frac {1} {3}$, and the probability of drawing 2 blue marbles(without replacement) from the jar is $ \\frac {1} {11}$. We draw 3 marbles from this jar. Given that a blue marble is drawn, find the probability that all 3 marbles are blue.\r\n5.\tOn a circle circular track radius 100 meters, 3 people stand on the track. What is the probability that each person can run $ 50\\pi$ meters along the track in either direction and not run into someone else?\r\n\r\nDue date is extended to next sunday", "Solution_3": "For number 5, are do you mean that \"each pair of people are at least $ 50\\pi$ meters away from each other\" or like \"it's impossible for them to run in to each other\" or \"it's possible for them to run in a fashion such that no two meet\" or something I was too stupid to think of :P ?", "Solution_4": "Im not sure what you are asking. If the wording is confusing, this is the same question worded differently:\r\nWhat is the probability that for any person to run to another along the track, he/she must run atleast $ 50\\pi$ meters.", "Solution_5": "OK, that was what I thought...thanks.", "Solution_6": "Combinatorics problems are not due yet.(sunday is the due date)\r\n\r\nNumber thoery\r\n\r\n1. How many positive integers less than 1000 cannot be expressed as the sum of 2 cubes?\r\n2. How many positive integer solutions $ (a,b,c,d)$ are there to $ a^42 \\plus{} b^6 \\plus{} c^2 \\plus{} d \\equal{} 1337$?\r\n3. Let $ N$ be the largest positive integer with the following property: reading from left to right, each pair of consecutive digits of $ N$ forms a perfect square. What are the leftmost three digits of $ N$?\r\n4. How many integers less than $ 10^6$ are perfect square multiples of 24?\r\n5. How many factors does $ 111,111,111,111$ have?\r\n\r\nRemember, no calculators. These problems are due Tuesday.", "Solution_7": "#1. What does envy-free mean? Can some candies don't be given to kids? for example 2,2,2,2,2,2,2 are given and another 6 candies are not given.", "Solution_8": "I think he means that all candies must be given and if the number of candies given to each kid is $ \\{a_1,a_2,a_3,a_4,a_5,a_6,a_7\\}\\equal{}A$, then $ \\max{A}\\minus{}\\min{A}\\leq2$. This notation is probably incorrect, but I hope you know what I mean. :P", "Solution_9": "[quote=\"Bachukas\"]#1. What does envy-free mean? Can some candies don't be given to kids? for example 2,2,2,2,2,2,2 are given and another 6 candies are not given.[/quote]\r\n\r\nAll the candies must be given", "Solution_10": "[quote=\"abacadaea\"]2. How many positive integer solutions $ (a,b,c,d)$ are there to $ a^42 \\plus{} b^6 \\plus{} c^2 \\plus{} d \\equal{} 1337$? [/quote]\r\n\r\nI think you mean $ a^{42} \\plus{} b^6 \\plus{} c^2 \\plus{} d \\equal{} 1337$.\r\n\r\nEDIT: Where have I seen #3 before... hmm...", "Solution_11": "For #5 on the NT one, do I seriously have to check for primes under a four digit number?", "Solution_12": "Be clever. :P", "Solution_13": "I HAVE SENT MY ANSWERS IN.\r\n\r\nIf you have done the test:\r\nOk, I'm sorry if this is against the rules or anything, but is there a simple, applicable, practical primality test for a big number, oh [hide] that big number just below 10000 perhaps?[/hide]\r\n\r\nI'm getting off topic with this subject; can someone who finished pm me?", "Solution_14": "lol. i guess i can answer that. the answer is yes. \r\n\r\nI guess this could be considered a failure. only 3 people have submitted answers, but ill just keepn posting problems until we get through all 25", "Solution_15": "[quote=\"abacadaea\"] 1. How many positive integers less than 1000 cannot be expressed as the sum of 2 cubes? [/quote]\r\n\r\nCannot be expressed as the sum of 2 [b]positive integer[/b] cubes, or cannot be expressed as the sum of 2 [b]integer[/b] cubes?", "Solution_16": "if it was integer, you would get the same answer if it was positive integer.", "Solution_17": "[quote=\"shentang\"]if it was integer, you would get the same answer if it was positive integer.[/quote]\r\n\r\nI don't think so.\r\n\r\n19 can be expressed as $ (\\minus{}2)^3\\plus{}3^3$, but it can't be expressed as a sum of two positive integer cubes.", "Solution_18": "hmmm good point, hadn't thought about that :oops:", "Solution_19": "positive integer cubes.", "Solution_20": "I would totally do this if it was during the summer. :(", "Solution_21": "lol. its still the summer for me" } { "Tag": [ "analytic geometry", "geometry", "circumcircle" ], "Problem": "The points (18,0), (0,0) and (0,80) lie on a circle in teh coordinate plane. Find its area in terms of pi.\r\n\r\nThis is to be done without a calculator. Thanks in advance!", "Solution_1": "Hello\r\n[hide=\"hint\"]\nAs the eq. of the an arbitrary circle is $ x^2\\plus{}y^2\\plus{}2gx\\plus{}2fy\\plus{}c\\equal{}0$\nput the three points on the eq. so that u will get the value of g,f,c.\nradius of circle=$ \\sqrt{g^2\\plus{}f^2 \\minus{}c}$\nNow find the area by its radius.\n[/hide]\r\nThank u.", "Solution_2": "[quote=\"brianb\"]The points (18,0), (0,0) and (0,80) lie on a circle in teh coordinate plane. Find its area in terms of pi.\n\nThis is to be done without a calculator. Thanks in advance![/quote]\r\n\r\n[hide=\"Hint 1\"]\nDraw the triangle formed by the three points. What type of triangle is it?\n[hide=\"Hint 2\"]\nIt's a right triangle. What is special about the circumradius of a right triangle?\n[/hide]\n[/hide]" } { "Tag": [ "inequalities" ], "Problem": "Prove that $ \\sqrt{a^{2}\\plus{}b^{2}\\plus{}c^{2}}\\le |a|\\plus{}|b|\\plus{}|c|$ for a,b,c real", "Solution_1": "(lal+lbl+lcl)^2=a^2+b^2+c^2+2labl+2lacl+2lbcl>=a^2+b^2+c^2\r\n\r\nSince both a^2+b^2+c^2 and lal+lbl+lcl are both positive taking the square root of both sides of the above equation we get the desired result. \r\n\r\n\r\nIf this is hard to read, sorry, I'm in the process of getting Latex.", "Solution_2": "[hide]Consider the distance in Euclidean 3-space from the origin to $ (a,b,c)$ along the following two paths:\n\n1. A straight line -- length is $ \\sqrt {a^2 \\plus{} b^2 \\plus{} c^2}$\n\n2. The path $ O \\to (a,0,0) \\to (a,b,0) \\to (a,b,c)$ -- length is $ |a| \\plus{} |b| \\plus{} |c|$\n\nThe shortest distance between two points is via a direct line segment.[/hide]" } { "Tag": [ "number theory", "relatively prime" ], "Problem": "Find all positive integers (a,b), so that a-b is a prime, and ab is a perfect sqaure", "Solution_1": "[hide=\"What's wrong with my logic?\"]\n1. $a$ and $b$ cannot be perfect squares or $a-b$ would be factorable as a difference of squares.\n2. $a$ and $b$ cannot share any factors, or $a-b$ would be factorable by that shared factor.\n3. Since $ab$ is a perfect square, either both $a$ and $b$ are perfect square (invalidating 1) or $a$ and $b$ share one or more factors which are odd powers in both. However, that would invalidate 2, as $a$ and $b$ are relatively prime.[/hide]", "Solution_2": "[quote=\"worthawholebean\"][hide=\"What's wrong with my logic\"]\n1. $a$ and $b$ cannot be perfect squares or $a-b$ would be factorable as a difference of squares.\n2. $a$ and $b$ cannot share any factors, or $a-b$ would be factorable by that shared factor.\n3. Since $ab$ is a perfect square, either both $a$ and $b$ are perfect square (invalidating 1) or $a$ and $b$ share one or more factors which are odd powers in both. However, that would invalidate 2, as $a$ and $b$ are relatively prime.[/hide][/quote]Perhaps prove your statements first. Counterexample to statement #1: Let $a=4$, $b=1$. $4-1=3$ is prime and $4\\cdot1=4$ is a perfect square, both $a,b$ are also perfect squares...", "Solution_3": "[quote]Perhaps prove your statements first. Counterexample to statement #1: Let $a=4$, $b=1$. $4-1=3$ is prime and $4\\cdot1=4$ is a perfect square, both $a,b$ are also perfect squares...[/quote]This exception occurs when one of the factors of $a-b$ is 1, which only occurs when $\\sqrt{a}$ and $\\sqrt{b}$ are consecutive integers. As for $a-b$ being prime, this occurs infinitely many times, as the difference of two consecutive perfect squares can take on any odd number.\r\n\r\nA given odd prime, $p$ can be rewritten as $2k+1$. The two squares whose difference is $p$ are $(k+1)^{2}$ and $k^{2}$. The difference is $p$, and their product is the square of $k(k+1)$.\r\n\r\nI think that worthawholebean's second point is valid, so those are the only solutions.", "Solution_4": "wait... aren't there infinitely many solutions?\r\n\r\n[hide=\"because\"]worthawholebean's second point is valid. Therefore, a is a perfect square, and b is also a perfect square relatively prime to a. But a and b cant be of the same parity. So one is odd, and one is even.\n\nLet's say 1 and 4. that's good. 4 and 9. that's good. 9 and 16. that's good. 25 and 36. that's good. so on. Those squares differ by primes, and there are infinitely many of them![/hide]", "Solution_5": "[quote=\"1=2\"]wait... aren't there infinitely many solutions?[/quote]Yes, there are certainly infinitely many solutions, as long as $2k+1=p$ is prime and we know there are an infinite number of odd primes.", "Solution_6": "Yes there are infinite many solutions, but can someone find some way to calculate a and b?", "Solution_7": "[quote=\"Cartman\"]Yes there are infinite many solutions, but can someone find some way to calculate a and b?[/quote]Already did.\r\n\r\nIf $2k+1 = p$ (prime), then $a$ and $b$ are $k^{2}$ and $(k+1)^{2}$", "Solution_8": "[quote=\"renatzu\"][quote=\"Cartman\"]Yes there are infinite many solutions, but can someone find some way to calculate a and b?[/quote]Already did.\n\nIf $2k+1 = p$ (prime), then $a$ and $b$ are $k^{2}$ and $(k+1)^{2}$[/quote]\r\nOh... sorry. Of some reason i missed that" } { "Tag": [ "pigeonhole principle", "algebra", "system of equations" ], "Problem": "1. Find all pairs of positive integers $(x,y)$ which satisfy the equation: $x^{3}-y^{3}= xy+61.$\r\n\r\n2. Find all integer solutions to the equation: $xy+3x-5y=-3.$\r\n\r\n3. Find all solutions $(x,y,z,t)$ that satisfy: $x!+y!+z!=t!$\r\n\r\n4. Find all positive integers $(x,y)$ that satisfy: $\\sqrt{x}+\\sqrt{y}=\\sqrt{2001}$\r\n\r\n5. Let $x,y,z,a,b,c$ be integers such that\r\n\r\n$x^{2}+y^{2}=a^{2}$\r\n$x^{2}+z^{2}=b^{2}$\r\n$y^{2}+z^{2}=c^{2}$\r\n\r\nProve that the number $xyz$ is divisible by 55.\r\n\r\nHas anyone got any equations or systems of equations that they found the solution to be interesting?", "Solution_1": "Please post each problem separately. It becomes difficult to post solutions and keep track of them if more than one problem are posted all together.", "Solution_2": "[quote=\"FieryHydra\"]Please post each problem separately. It becomes difficult to post solutions and keep track of them if more than one problem are posted all together.[/quote]\r\n\r\nSorry about that. :blush:", "Solution_3": "[hide=\"1\"]$(x-y)(x^{2}+xy+y^{2})=xy+61$\n$(x-y)((x-y)^{2}+3xy)=xy+61$\n$(x-y)^{3}+3xy(x-y)=xy+61$\n$3xy(x-y)-xy=61-(x-y)^{3}$\n$xy(3(x-y)-1)=61-(x-y)^{3}$\n$xy=\\frac{61-(x-y)^{3}}{3(x-y)-1}$\n\nSince $x$ and $y$ are positive, the numerator and denominator are either both positive or both negative. If $x-y$ is not positive, then the denominator is negative but the numerator is positive. Therefore, $x-y$ is positive. But if $x-y$ is greater than $3$, then the numerator is negative while the denominator is positive. Testing $x-y=1,2,3$, we find that $xy$ is only an integer when $x-y=1$. Therefore, $x-y=1$, and $xy=30$. The only solution is $(6,5)$.[/hide]\r\n\r\nNice problem! :)", "Solution_4": "[quote=\"matt276eagles\"][hide=\"1\"]$(x-y)(x^{2}+xy+y^{2})=xy+61$\n$(x-y)((x-y)^{2}+3xy)=xy+61$\n$(x-y)^{3}+3xy(x-y)=xy+61$\n$3xy(x-y)-xy=61-(x-y)^{3}$\n$xy(3(x-y)-1)=61-(x-y)^{3}$\n$xy=\\frac{61-(x-y)^{3}}{3(x-y)-1}$\n\nSince $x$ and $y$ are positive, the numerator and denominator are either both positive or both negative. If $x-y$ is not positive, then the denominator is negative but the numerator is positive. Therefore, $x-y$ is positive. But if $x-y$ is greater than $3$, then the numerator is negative while the denominator is positive. Testing $x-y=1,2,3$, we find that $xy$ is only an integer when $x-y=1$. Therefore, $x-y=1$, and $xy=30$. The only solution is $(6,5)$.[/hide]\n\nNice problem! :)[/quote]\r\nWe may also argue by looking at the original equation that $x>y$ since $x^{3}-y^{3}= xy+61>0$ and $x,y$ are both positive.\r\n\r\nNow looking at $xy=\\frac{61-(x-y)^{3}}{3(x-y)-1},$ we note that $3(x-y)-1 > 0$ and so $61-{(x-y)}^{3}$ must be positive. So, we only have the following possibilities for the values of $(x-y): 1,2,3.$\r\n\r\nAfter this, we proceed the way [b]matt276eagles [/b]mentioned in his solution above.", "Solution_5": "[hide] 1. Let $d=x-y$ then substitute $x=d+y$ into the equation to obtain $(3d-1)y^{2}+(3d^{2}-d)y+d^{3}=61$.\n\nTherefore, $d^{3}\\leqslant 61$ or $d \\leqslant 3$. So, $d=1, d=2$ or $d=3$ and by substituting these values into the equation above, we can safely conclude that the only solution in positive integers that exists is $(x,y)=(6,5)$.\n$QED$.\n [/hide]\r\n\r\nI did these a while ago so I might have made some minor mistakes :) . If you find these hard, try doing number 4 (which is really easy).\r\n\r\nOops, I was a bit slow.", "Solution_6": "[quote=\"Franz Joseph\"]\n2. Find all integer solutions to the equation: $xy+3x-5y=-3.$\n[/quote]\r\n$xy+3x-5y=-3$\r\n$\\Rightarrow (5-x)(y+3)=2\\cdot3^{2}$\r\n\r\nAfter this, it's just a matter of evaluating individual cases, such as $5-x=1, y+3=18,$ and so on.", "Solution_7": "[hide=\"#2\"]\n$xy+3x-5y=-3$\n$(x-5)(y+3)=-18$\nthe solutions are $(6,-21), (4,15), (-13,-2), (23,-4)$,etc..\nwow alot of them....did i make a mistake?? :maybe: [/hide]\r\nFieryHydra, u beat me 3 min..", "Solution_8": "[hide]2. By rearranging the equation, we obtain \n\n$y= \\frac{3-3x}{x-5}=\\frac{15-3x-18}{x-5}=-3-\\frac{18}{x-5}$\n\nWe now observe that $x-5$ must be a factor of $18$ so in total we should have $12$ solutions all together.\n\n(Since $x-5= \\pm 1, \\pm 2, \\pm 3, \\pm 6, \\pm 9, \\pm18$).\n\n$QED$. [/hide]", "Solution_9": "[hide=\"#4\"]$\\sqrt{y}=\\sqrt{2001}-\\sqrt{x}\\&\\implies\\& y=2001-2\\sqrt{2001x}+x$\n\nSince $x$ is natural, $\\sqrt{2001x}$ can be either natural or irrational, but as $y$ is natural, it must be natural. Hence $2001x$ must be a perfect square, but since $2001=3\\cdot 23\\cdot 29$, we obtain $x=2001a^{2}$ where $a$ is some natural number.\n\nIn the same manner we find $y=2001b^{2}$.\n\nSubstituting those into the given equation, we get $a+b=1$, which has no solution in positive integers. (It DOES have a solution in non-negative integers - $(0,1),(1,0)$ - which would give $(x,y)\\in\\{(0,2001),(2001,0)\\}$)[/hide]", "Solution_10": "[hide=\"3\"]PS: Nevermind, I found a solution: $(2,2,2,3)$[/hide]My solution was faulty, still finding a proof :oops:", "Solution_11": "[quote=\"10000th User\"][hide=\"3\"]This reduces to $1+(y-x)!+(z-x)!=(t-x)!$. For integers $m>1$, $m!$ is always even. If $t-x>1$, RHS is even; but, LHS is odd. So, $t-x\\le1$ and $(t-x)!=1$. But, $(y-x)!\\ge1$ and $(z-x)!\\ge1$ and so $1=(t-x)!=1+(y-x)!+(z-x)!\\ge3$, which is impossible. Therefore, no solution.\n\nPS: Nevermind, I found a solution: $(2,2,2,3)$\n[/hide][/quote]\r\n\r\nIt does have a solution.", "Solution_12": "[hide]3. We can safely assume that $x \\leqslant y \\leqslant z \\leqslant t$. Since $z!2 /(j 2 +1)) from j=1 to \\infty , where \\pi is the product symbol as \\sum is the summation symbol\r\n\r\nQuery: Is this related to Riemann-Zeta function? Try it and find out!\r\n\r\nEdit: Hey you are right I put wrong symbol, now it is correct :D", "Solution_1": "Am I missing something obvious but it seems to me that the sum is +oo ??", "Solution_2": "Thanks for the reminder belenos", "Solution_3": "Sorry for the confusion, what i meant is the following\r\n\r\n \\pi (j 2 /(j 2 +1)) = (1 2 /(1 2 +1))(2 2 /(2 2 +1))...", "Solution_4": "or how about this?\r\n\r\nDoes the \\sum (j / (j 2 +1)) converge? If so evaluate it.", "Solution_5": "[quote=\"3X.lich\"]or how about this?\n\nDoes the \\sum (j / (j 2 +1)) converge? If so evaluate it.[/quote]\r\n\r\nYes it converges, but in IR U +oo :D \r\n\r\nj/^2+1 = 1/j + 1/^2+1 and \\sum 1/j = +oo (harmonic sum)", "Solution_6": "you mean this?\r\n\r\n(j/(j 2 +1)) = 1/j - 1/(j(j 2 +1))\r\n\r\nbecause it's not quite harmonic :(", "Solution_7": "[quote=\"3X.lich\"]\nDoes the \\sum (j / (j 2 +1)) converge? If so evaluate it.[/quote]\r\n\r\nIt is divergent since j/(j^2+1) ~ 1/j when j-->+oo \r\n \\sum (j / (j 2 +1)) has the same nature of \\sum 1/j wich \r\nis divergent numerical serie.", "Solution_8": "[quote=\"3X.lich\"]Sorry for the confusion, what i meant is the following\n\n \\pi (j 2 /(j 2 +1)) = (1 2 /(1 2 +1))(2 2 /(2 2 +1))...[/quote]\r\n\r\np(n)= \\pi _(j=1 to n) j^2/(j^2+1) \r\n\r\nlnp(n) = \\sum _(j=1 to n) ln (j^2/(j^2+1)) \r\n= - \\sum _(j=1 to n) ln((j^2+1)/j^2)\r\n= - \\sum _(j=1 to n) ln(1+ 1/j^2) \r\n\r\nwhen j-->+oo, ln(1+1/j^2) ~ 1/j^2 so \r\n- \\sum _(j=1 to n) ln(1+ 1/j^2) has the same nature of -- \\sum _(j=1 to n) 1/j^2 wich is convergent \r\np(n) is convergent", "Solution_9": "I have something to add here, for those in college, many students are familiar with the 'integral convergence test'. It's kind of overly used and way off from elementary but it helps to do a quick check for convergence in this problem.\r\n\r\nLet f(x)=x/(x 2 +1). Clearly this function is positive definite for x>0 and strictly decreasing since \r\n\r\ndf/dx = (1 - x 2 )/(1 + x 2 ) 2 \\leq 0 for all x \\geq 1.\r\n\r\nNow convergence of \\int f(x) dx <==> convergence of \\sum f(n)\r\n\r\nBut \\int {0--> \\infty } f(x) dx = (1/2) \\lim log (1+x 2 ) --> \\infty , the original sum is divergent.\r\n\r\nPS: Has anyone gotten a value for the infinite product? I know it's convergent by means of the log on the product but to find its value is not easy. I was trying some Taylor expansions around powers of trigonometric functions or exponentials but couldn't find the right pattern. This is like a method I did to find the value of the zeta function z(2) = \\pi 2 /6" } { "Tag": [ "linear algebra", "matrix", "geometry", "geometric transformation", "rotation", "linear algebra unsolved" ], "Problem": "Caracterise A matrix in M(n,R) such that tA=A^(k) where tA is the transpose of A\r\n\r\nA^(k) is k power of A, k is integer greater than 1", "Solution_1": "the complex case:\r\nsuppose that $ A^{*}= A^{k}$.\r\nSchur's Lemma shows that there exists a unitary matrix $ U$ such that $ A = U^{*}T U$ and $ T$ is upper triangular. Then $ T^{*}= T^{k}$, wich shows that $ T$ is actually diagonal: $ D=diag(d_{1}, d_{2}, \\ldots, d_{n})$ where $ d_{i}^{k}= \\overline{d_{i}}$. Hence every $ d_{i}$ is a $ k+1$-th root of the unity, or is null. is it a good description ?", "Solution_2": "The real case (following Alekk).\r\n$ d_{j}=e^{\\frac{2i\\pi\\lambda_{j}}{k+1}}$ where $ \\lambda_{j}\\in\\{0,\\cdots,k\\}$.\r\nBy associating pairwise the conjugate $ A$-eigenvalues we obtain that there exists an orthogonal matrix $ P$ s.t. ${ A=P^{T}diag[Rot(\\frac{2\\pi\\lambda_{1}}{k+1}),\\cdots,Rot(\\frac{2\\pi\\lambda_{p}}{k+1}}),0_{1},\\cdots,0_{q})]P$ where $ 2p+q=n$ and $ Rot(\\alpha)$ is the matrix of the vectorial rotation of angle $ \\alpha$." } { "Tag": [ "geometry", "geometric transformation", "reflection", "geometry proposed" ], "Problem": "Let B, C be two fixed points on the circle (O). Let M be the middle point of the smaller arc of BC. Let D be an changing point on the smaller arc of BC. DM intersects BC at H. Let H' be the reflection of H by DC. Find the locus of H'. :D", "Solution_1": "nobody can??? What a pity???", "Solution_2": "If $DH'$ cuts $(O)$ again at $N,$ then $DC$ bisects $\\angle HDH' \\equiv \\angle MDN$ $\\Longrightarrow$ arcs $CM$ and $CN$ of $(O)$ are congruent $\\Longrightarrow$ $N$ is fixed. $DM$ bisects $\\angle BDC$ $\\Longrightarrow$ $\\angle CDM=\\angle BDH$ and since $\\angle DBH=\\angle DMC,$ then it follows that $\\angle DHC=\\angle DCM$ $\\Longrightarrow$ $\\angle DNM=\\angle DCM=\\angle DHC=\\angle DH'C.$ Hence, if $U \\equiv CH' \\cap MN,$ $\\triangle UNH'$ is U-isosceles, or $UN=UH'$ $\\Longrightarrow$ $H'$ runs on an oblique strophoid with pole $C,$ cusp $N$ and construction line $NM.$", "Solution_3": "What happens when D varies all over the circle?" } { "Tag": [ "geometry" ], "Problem": "I am currently an eigth grader enrolled in Honors Geometry and am rather ignorant about how high school works. \r\n\r\nDo the counselors at high schools allow testing out of Algebra 2, you is it necessary for me to take 2 math courses simultaneously next year?", "Solution_1": "It all depends on your high school. Some may let you test out of a course, some may require you take a summer course. The best thing to do is discuss the issue with your counselor and the administration.", "Solution_2": "I am pretty sure that you have to take the class. It's an easy class but you have lots to learn (it's the basics for all the problem solving stuff here). I think Algebra II is a requirement of MY high school.. I don't know about yours though.", "Solution_3": "If I want to I can pass out of it. I'm going to study the book this summer and take Pre Cal next year." } { "Tag": [ "limit" ], "Problem": "[hide=\"solution\"]\ntaking $ x^{2}$ out we get $ \\lim_{x \\to \\minus{} \\infty} \\frac { \\sqrt {2 \\plus{} \\frac {1}{x^{2}}}}{ \\minus{} 4 \\minus{} \\frac {1}{x}} \\equal{} \\lim_{u\\to 0^{ \\minus{} }}\\frac { \\sqrt {2 \\plus{} u^{2}}}{ \\minus{} 4 \\minus{} u} \\equal{} \\frac {1}{\\minus{}2 \\sqrt {2}}$[/hide]", "Solution_1": "Final answer is wrong. Think Logically. Numerator is positve, denominator is negative.", "Solution_2": "sorry this happend when u don't solve it with pen and paper :D \r\nthe answer is $ \\frac{\\minus{}1}{2\\sqrt{2}}$", "Solution_3": "Edit your post above. The key point is that, when you take $ x^2$ out of the root, it becomes -x and not x. ;)", "Solution_4": "ye si have done that :) \r\nthe only thing i did wrong though i divided by $ \\minus{}x$ but forgot to take the minus sign from above :)" } { "Tag": [ "Euler", "function", "number theory solved", "number theory" ], "Problem": "Solve the equation \r\n\r\nphi(n) = d(n) where phi(n) is Euler function, d(n) is the number \r\n\r\nof divisor of n", "Solution_1": "I think I solved it, but my soln is so long it scares me. There's nothing smart abt it, it's just a case chase. We write n=p1^a1*..*pk^a^k where pi are primes, we write phi(n) and we use the fact that d(n)<=(a1+1)*..*(ak+1). Then we take the cases k=1, k=2 etc. and we find that k>=4 does not work (I think, I haven't checked everything). I won't post the actual soln, here's what I found: n is in the set {1, 3, 8, 10, 18, 24, 30}. Are these all?" } { "Tag": [], "Problem": "A leaky faucet can be expensive. Suppose you put a measuring cup under a leak and collect a 1/2 cup of water from 8am to 4pm. How many gallons of water would this amount to in 1 day if there are 16 cups in a gallon?", "Solution_1": "[hide]That's 1/2 per 8 hours and 3/2 per day. So 3/32 of a gallon.[/hide]", "Solution_2": "I got it.\r\n\r\nThanks!" } { "Tag": [ "geometry", "induction", "pigeonhole principle", "USAMO", "Hi" ], "Problem": "Let $n$ be an integer greater than 1. Suppose $2n$ points are given in the plane, no three of which are collinear. Suppose $n$ of the given $2n$ points are colored blue and the other $n$ colored red. A line in the plane is called a [i]balancing line[/i] if it passes through one blue and one red point and, for each side of the line, the number of blue points on that side is equal to the number of red points on the same side. \r\nProve that there exist at least two balancing lines.", "Solution_1": "Hmm... A strange problem.\r\n\r\nConsider a convex hull of the whole set. If it contains red and blue points then, obviously, we are done. Therefore suppose the convex hull consists of blue points only.\r\n\r\nTake $A,B,C$ three consecutive vertices on the convex hull. We state there is a balancing $A$ through $A$. Indeed, consider the leftmost line $k$ through $A$ s.t. the right open halfplane contains the same number of red and blue points :!:. The line $k$ exists due to discrete continuity arguments (just rotate $k$ from $AB$ towards to $AC$). We know that $k$ contains a point $D$ from our set. If $D$ is a red point then we are done. If $D$ is a blue point, then there is another line between $AB$ and $k$ satisfying the condition :!: . Contradiction.\r\nSince the convex hull contains at least three points we conclude that there are at least three balancing lines for this case.\r\n\r\nPlease, someone read it! I think I missed something :(", "Solution_2": "I solved it the same way. So, I missed the same thing as you.", "Solution_3": "I solved it almost exactly the same way, except I didn't know what a convex hull was so I had to write a page defining the convex hull. \r\n\r\nLet C be right of A. Label the red points $R_i, i=1,2,...,n $ so that $i > j \\Rightarrow \\angle CAR_i > \\angle CAR_j $ (i.e., counterclockwisely)\r\n\r\nIf we draw all the lines $AR_i $, then the number of red points on the right side of $AR_i $ is $i-1, i=1,2,...,n.$ The number of blue points on the right side of $AR_i $ increases from 1 to at most n-2 as i increases from 1 to n. \r\n\r\nSo $i-1$ starts smaller and ends larger than the number of blue points right of $AR_i $, and $i-1$ also takes on every single value between 0 and n-1, so they have to equal at some point. So this is a balancing line. Do the same thing for B.", "Solution_4": "So, it means that the problem is really \"trivial\". I.e. it is just a direct application of well known ideas: convex hull + discrete continuity.", "Solution_5": "I guess, iff you're experienced. I spent like an hour trying to do it via induction... :?", "Solution_6": "[quote=\"tekno10m\"]I guess, iff you're experienced. I spent like an hour trying to do it via induction... :?[/quote]\r\n\r\nI also tried it via induction. I didn't have much time to do it though, and ended up getting in the base case, which was to show its true for 2 reds and 2 blues.\r\n\r\nThen, obviously, if you put the red and blue point on the same point of the balancing line, that line still acts as a balancing line.\r\n\r\nIf not, that's where it gets a little wierd. Lets say you put the red on side \"A\" of the line and the blue on side \"B.\" Rotate your old balancing line about the blue point so that the red point it formerly passed through falls into side \"B.\" If the line next crosses a red point, you have found a new balancing line. If not, you cross a blue point before a red point, then you will have more blues on side B than reds. Therefore, once you hit a red point, rotate the line about that red point so as to move more blues to side A.\r\n\r\nI conjectured (but was unable to prove) that eventually you will eventually find a new balancing line by this method. Therefore, each balancing line for 2n points has a corresponding balancing line for 2n+2 points, QED.\r\n\r\nObviously I dont expect full credit, but hopefully they will give me a point or two for my base case and conjecture.\r\n\r\nJB", "Solution_7": "Just a word about balanced lines, about which there are many papers.\r\n\r\nJ. Pach and R. Pinchasi proved that, according to a conjecture of G. Baloglou, for $n$ red points and $n$ blue points in general position in the plane there are at least $n$ balanced lines.\r\nAnd this bound is sharp.\r\n\r\nJ.Pach, R. Pinchasi, 'On the number of balanced lines', Discrete and Computational Geometry, 2001, p. 611-628.\r\n\r\nTheir lemma 1-5 states that for any vertex $v$ of the convex hull of the given set, there is a balanced line passing through $v$.\r\nTheir argumentation in the proof of this lemma is Myth's one.\r\n\r\nPierre.", "Solution_8": "[quote=\"Myth\"]So, it means that the problem is really \"trivial\". I.e. it is just a direct application of well known ideas: convex hull + discrete continuity.[/quote]\r\n\r\nWell-known to you, perhaps. But not to most high school students, even olympiad-types. Furthermore, there are so many other techniques that less-experienced students might be led to try. With your level of experience, a glance at the problem suggests discrete continuity, and the convex hull comes soon after. But we don't all have your level of experience!\r\n\r\nI might also describe #3 in the same terms. #3 uses even simpler well-known ideas - parallel lines, inscribed angles, and cyclic quadrilaterals. But I wouldn't describe it as trivial.", "Solution_9": "America... What should I do that everyone see I wrote [b]\"trivial\"[/b], not trivial. \r\nIndeed, problem is not trivial, but it is absolutely straightforward in some sense. You see that Fedor did exactly the same things as me.", "Solution_10": "The students more experienced with problem solving that I know all jumped straight to the convex hull. I, for one, spent many hours trying induction, pigeonhole, extremal, contradiction, and even connecting-the-lines (kinda sorta graph theory I guess) while expending many pieces of paper on plots. So I guess this problem gives a huge advantage to experience, arguably much more so than most of the other problems.", "Solution_11": "I did this for fun. My solution is not fundamentally different than those presented here. Just a slightly different explanation.\r\n\r\n\r\nConsider the convex hull of points. If the convex hull contains both red and blue points, then as we consider the points on the hull in a counterclockwise direction, there must be at least one blue point immediately followed by a red point, and there must be at least one red point immediately followed by a blue point. Lines connecting these pairs of points will be balancing lines. There are at least two such lines.\r\n\r\nOtherwise, the convex hull is monochrome. In the following, we consider the convex hull to be monochrome blue, but we note that similar arguments would apply if the convex hull were monochrome red.\r\n\r\nIf the convex hull is monocrome blue, we choose one point on the convex hull and divide the half-plane containing all points into $n+1$ sectors, by drawing a series of rays from the chosen point through each red point. Sectors $S_1$ and $S_{n+1}$ must each contain at least one blue point. The remaining sectors may or may not contain blue points.\r\n\r\n[img]http://www.artofproblemsolving.com/Forum/files/usamo2005-5_673.gif[/img]\r\n\r\nRecall that the total number of blue points is equal to the total number of red points. If we consider $L_1$, we note there is an excess of blue points to the right of the ray. We denote the excess of blue points to the right of line $L_i$ by $K_i$. $(K_1 > 0)$ If we consider $L_n$, we note there is an excess of blue points to the left of the ray. (Or a negative excess of blue points to the right of the ray.) $(K_n < 0)$\r\n\r\nConsider each successive line, $L_i$, from $L_2$ through $L_n$. If Sector $S_i$ contains one blue point, the imbalance in $L_i$ will equal the imbalance in line $L_{i-1}$. $(K_i = K_{i-1})$ If Sector $S_i$ contains more than one blue point, the excess of blue points to the right of $L_i$ will increase, compared to the excess of blue points to the right of $L_{i-1}$. $(K_i > K_{i-1})$ However, if Sector $S_i$ contains no blue points, the excess of blue points to the right of $L_i$ will decrease by exactly one, compared to the excess of blue points to the right of $L_{i-1}$. $(K_i = K_{i-1} - 1)$\r\n\r\nThe value(s) where $K_i = 0$ correspond to a balancing line. Since the sequence, $K_1, K_2, K_3, \\ldots, K_n$ is known to go from a positive value at $K_1$ to a negative value at $K_n$, by positive integer steps, zero steps, or steps of negative one, the sequence must pass through 0 at least once. Thus, at least one of the lines is a balancing line.\r\n\r\nIf we choose another point on the convex hull, we are guaranteed to find another (distinct) balancing line.\r\n\r\nThus, in every case, there are at least two balancing lines. QED", "Solution_12": "If you've seen IMO 99 #1 you were probably more than ready to use Convex Hull on this problem.", "Solution_13": "I experimented for a while with the rotate-the-line method, then succumbed to a very seductive application of the Pigeonhole Principle and induction. And in any case, my line-rotating was restricted to an inductive method that would generate a new balancing line from those in a set of 2n-2 points. I abandoned it because it seemed messy, and I couldn't get it to work. Darn.", "Solution_14": "I didn't know about convex shells, either, so I just proved it. Then, with the line rotation, I made $ a_0, a_1, ... a_n$ the counts of blue points before, after, and in between the n red points. Then you have the sum of the $a_i$ equal to $n - 1$. Then a balancing line exists through our blue point and the $k + 1^{st}$ red point met in the rotation iff the sum of the $a_i$ for $i$ from $0$ to $k$ is equal to $k$. It does not seem hard, then, to prove that there must be at least one $k$ for which this is true. I did not have time to do it on the USAMO, but I would have liked to. It sounds like if I had scattered the words \"finite\", \"discrete\", and \"continuity\" here and there in my proof, the judges would be happier with my proof. Oh well. I liked this problem anyways. I might have done it the hard way, though. Or is the hard way the inductive proof ... ? (Which I also briefly looked at, then decided that it was too hard. Then I solved it this way)\r\nYeah. I don't see anything wrong with your proof, Myth. On the other hand, I might just be missing something ... oh wait! You forgot to prove that deterministic primality testing only takes $O(n^{O(1)})$ in the worst case! I fill in for your glaring omission here:\r\n\r\nTheorem: Deterministic primality testing for positive integer n can be computed in $O(n^{O(1)})$\r\n\r\nProof: It has been done; such algorithms exist, and I was standing within about 20 feet of the source code (and the source of the source code) for one such algorithm written in Mathematica, at the NorthWest Science Expo. I got to do that for several hours. That was fun.", "Solution_15": "Actually, my doubts were concerned only with the fact, that problem is too straightforward (for experienced olympiad students). I see now that it were baseless doubts, since many people did the same things as me.", "Solution_16": "I solved this problem the same way a [i]very[/i] similar problem is solved in [u]Math Olympiad Challenges[/u]: \r\n\r\n[quote]3.1 Arrange in Order\nExample 2: Given $2n+2$ points in the plane, no three collinear, prove that two of them determine a line that separates $n$ of the points from the other $n$.[/quote]\r\nI guess I'm just lucky I happened to read that particular section and that that problem stuck in my mind.", "Solution_17": "I tried induction and wrote a 5 page long solution that really made no sense, but that's ok. I have no idea what a convex hull is....oh well, I still have many more years left.", "Solution_18": "[quote]\n3.1 Arrange in Order\nExample 2: Given 2n+2 points in the plane, no three collinear, prove that two of them determine a line that separates n of the points from the other n. [/quote]\r\n\r\nSo basically almost the exact same problem wow.", "Solution_19": "[quote=\"rrusczyk\"]\n\nWell-known to you, perhaps. But not to most high school students, even olympiad-types. [/quote]\r\n\r\nI solved the problem very similar, using convex h. and d. continuity... I think one can easily think of convex hull because you think of it as a trivial case, when one side of the balancing line has no points and the other has the remaining 2n-2... \r\nonce you get to a convex hull you try some example for small n and see how can you get a balancing line.. making some arguments clear may be the matter of expierience, but the idea isn't.. \r\ncroatia hasn't got as good results as US but contestants are mostly familiar with those terms (hull and continuity)...", "Solution_20": "I think the proof given in the AMC website is flawed; it claims that every point on the convex hull lies on a balancing line.\r\n\r\nBut suppose we have the 2 red points (0,0) and (0,1) and the 2 blue points (-1,-1) and (1,-1).\r\nObviously the convex hull is (0,1) (-1,-1) (-1,1).\r\nBut the point (0,1) is not on any balancing lines (look at the diagram) but it is on the convex hull.", "Solution_21": "[quote=\"Philip_Leszczynski\"]I think the proof given in the AMC website is flawed; it claims that every point on the convex hull lies on a balancing line.\n\nBut suppose we have the 2 red points (0,0) and (0,1) and the 2 blue points (-1,-1) and (1,-1).\nObviously the convex hull is (0,1) (-1,-1) (-1,1).\nBut the point (0,1) is not on any balancing lines (look at the diagram) but it is on the convex hull.[/quote]\r\n\r\nErr, aren't the lines through (0,1); (-1,-1) and (0,1); (1,-1) both balancing lines? They both have one red and one blue on one side and none on the other...", "Solution_22": "[b][u]Lemma.[/u][/b]\nLet $k \\ge 3$ be an integer. Then, given any $k$ distinct points in the plane with no three points collinear, there exists a convex polygon, with vertices from the $k$ given points, such that all of the $k$ points lie within the polygon.\n[i]Proof.[/i]\nWe proceed by induction on $k.$ The base case, $k = 3$, is trivial. We simply connect the three points into a triangle. Since every triangle is convex, the claim follows. Now, for the induction hypothesis, assume the claim to be true for some $k = m$, where $m \\in \\mathbb{N}, m \\ge 3.$ We prove the claim true for $k = m + 1.$\n\nLet us label the points $X_1, X_2, ..., X_{m + 1}.$ By the induction hypothesis, there exists a convex polygon, which we will denote by $\\oslash$ with vertices from the set $\\{X_1, X_2, ..., X_m\\}$, that contains all points $X_1, X_2, ..., X_m.$ Now, if the point $X_{m + 1}$ also lies in $\\oslash$, then we are done. If $X_{m + 1}$ does not lie in $\\oslash$, then consider a line $\\gamma$ passing through $X_{m + 1}$ such that $\\gamma$ does not intersect $\\oslash.$ Note that since $X_{m + 1} \\notin \\oslash$, such a line $\\gamma$ must exist. \n\nNow, we rotate $\\gamma$ couterclockwise until it first intercepts a vertex $X_p$ of $\\oslash.$ We continue to rotate $\\gamma$ counterclockwise until it intercepts a vertex $X_q$ of $\\oslash$, such that $X_q$ is the last vertex that $\\gamma$ intercepts during this rotation (until it once again intercepts $X_p$ after rotating through $180^{\\circ}$). \nDenote by $S$ the sequence of consecutive vertices of $\\oslash$ that connect $X_p$ and $X_q$, where $S$ lies on the opposite side of line $X_pX_q$ as $X_{m + 1}.$ \nIt follows that all vertices of $\\oslash$ that are not $X_p, X_q$ or in $T$ are contained within the lines $X_{m + 1}X_p$ and $X_{m + 1}X_q.$ Therefore, $X_{m + 1}X_pTX_q$ is a convex polygon that contains all $m + 1$ points. \nTherefore we have shown the claim true for $k = m + 1.$ This completes the induction.\n\nNow, since $n \\ge 2$, there are at least $2 \\cdot 2 = 4$ given points. Therefore, by our Lemma, there must exist a convex polygon, which we will denote by $\\otimes$, with vertices from the $2n$ given points, such that all of the $2n$ points lie within $\\otimes.$ We label these vertices (in the order that they appear in $\\otimes$) $P_1, P_2, ..., P_i$, where $i \\in \\mathbb{N}, i \\le 2n.$ We now consider two cases:\n\n[u]Case 1:[/u] $P_1, P_2, ..., P_i$ are all the same color.\nWLOG, assume that these points are all red. Then the adjacent vertices $P_1$ and $P_2$ are both red. Now, consider the line $P_1P_2.$ We rotate this line about $P_1$ in the direction such that $P_2$ is rotated into $\\otimes.$ Let $Q$, where $Q$ is one of the given $2n$ points, be the first blue point that the line $P_1P_2$ intercepts during this rotation. Then the line $P_1Q$ divides the plane into two regions, one of which contains only red points (among which is $P_2$), and the other contains the remaining of the given $2n$ points (other than $P_1$ and $Q$, of course). Since the first region contains only red points, it follows that the second region, which contains the point $P_i$, contains more blue points than red points. \nNow, we continue rotating the line $P_1P_2$ in the same direction until it intercepts $Q*$, where $Q*$ is one of the $2n$ given points, and $Q*$ is the last blue point that the line $P_1P_2$ will intercept during its rotation, until $P_1P_2$ intercepts $P_i.$ Then the line $P_1Q*$ divides the plane into two regions, one of which contains only red points (among which is $P_i$), and the other contains the remaining of the given $2n$ points (other than $P_1$ and $Q*$, of course).\n\nTherefore, during this rotation, the line $P_1Q$ divides the plane into two regions. The region containing the point $P_i$ has more blue points than red points. Then, the line is further rotated to $P_1Q*.$ The line $P_1Q*$ divides the plane into two regions, where the region containing $P_i$ contains more red points than blue points.\n\nTherefore, by the principle of continuity, there must exist a blue point $R \\in \\otimes$, such that the line $P_1R$ divides $\\otimes$ into two regions, with each region containing an equal number of red and blue points. Hence, there exists a balancing line passing through $P_1.$ Similarly, we can show that there exists a balancing line passing through $P_2$, and in general, through all of the $P_k$'s. Since we have shown that there exist at least two balancing lines, this concludes Case 1.\n\n[u]Case 2:[/u] The set $\\{P_1, P_2, ..., P_i\\}$ contains both red and blue elements.\nConsider the longest consecutive sequence of red vertices on $\\otimes.$ Since the vertices of $\\otimes$ are not all red, it follows that the first and last red vertex in this sequence must be adjacent to a blue vertex. Therefore, $\\otimes$ must contain at least two pairs of consecutive vertices that are opposite in color. Suppose these pairs of adjacent vertices are $(P_a, P_b), (P_c, P_d)$ where $P_a, P_c$ are red and $P_b, P_d$ are blue. Note that these pairs are distinct, since if $P_a = P_c$ and $P_b = P_d$, then it follows that $\\otimes$ has only two vertices, absurd.\nNow, since the pairs $(P_a, P_b)$ and $(P_c, P_d)$ are distinct, the two lines $P_aP_b$ and $P_cP_d$ are also distinct. In addition, since the $P_i$'s are simply vertices of a convex polygon, $\\otimes$, that encloses the given $2n$ points, it follows that the each of the lines $P_aP_b$ and $P_cP_d$ divides the plane into two regions, one of which fully contains $\\otimes.$ Therefore each of these lines divides the plane into two regions, one of which contains the remaining $2n - 2$ points (other than the two in the line itself), and the other containing none of the given points. Clearly these $2n - 2$ points are comprised of $n - 1$ red points and $n - 1$ blue points, i.e., an equal number of red and blue points. Therefore, the lines $P_aP_b$ and $P_cP_d$ are both balancing lines. Since we have shown that there exist at least two balancing lines, this concludes Case 2.\n\nSince we have considered all cases, the proof is complete.", "Solution_23": "Actually good cg problem??\n\nLet $S$ be the set of all of the points. Also, given a line $\\overline{PQ}$ (where the order of vertices matters) and a point $X$ not on that line, say $X$ is to the \"left\" of $\\overline{PQ}$ if it is on the left half-plane from the perspective of an ant at $P$ looking towards $Q$, and say it is to the \"right\" otherwise.\nConsider the convex hull of $S$. If it contains points of both colors, then any line drawn from a red point to a blue point is clearly balancing. Since there are at least two of these, the result is true in this case.\nHence suppose that the convex hull of $S$ contains points of only one color. WLOG let this color be blue. I will now prove the following key claim:\n\n[b]Claim:[/b] For every blue point $B$ on the convex hull of $S$, there exists at least one balancing line passing through $B$.\n[b]Proof:[/b] Consider the $n$ red points, none of which are on the convex hull. Number these points $R_1,R_2,\\ldots,R_n$ such that $R_i$ is to the left of $\\overline{BR_j}$ if and only if $i0$, since there are no red points to the left of $\\overline{BR_1}$, but there must be at least one blue point (else $R_1$ lies on the convex hull), and by similar reasoning $f(n)<0$. Hence there must be some value $1 \\leq x \\leq n$ such that $f(x)=0$, otherwise $f(i)$ decreases by at least $2$ when $i$ increases by $1$. (This is known as discrete continuity/discrete IVT).\n\nTo finish, note that the number of points on the convex hull of $S$ is at least $3$, hence there are at least $2$ balancing lines for $S$, as desired. $\\blacksquare$", "Solution_24": "Consider the convex hull of the $2n$ points. It must consist of at least $3$ points such that any 3 are not collinear. First consider when it consists of both red and blue points. WLOG assume there must be at least 2 red points. Call these 2 red points $R_1$ and $R_2$. Then, the 2 balancing lines are $R_1B$ and $R_1B$, where $B$ is any blue point on the convex hull. For both $R_1B$ and $R_2B$, the other $2n-2$ points all lie on one side of the line, so $R_1B$ and $R_2B$ are indeed balancing.\n\nNow consider when all the points on the convex hull is one color. WLOG let the color be red. Label the points on the convex hull $R_1, R_2, \\dots, R_i$ $(i \\ge 3)$. Focus on $R_1$. Then, label the blue points $B_1, B_2, \\dots B_n$ such that $R_1B_j$ is to the left of $R_1B_{j+1}$ for all $j$. For all $R_1B_j$, let $X_j$ denote the number of blue points to the left of $R_1B_j$ subtracted from the number of red points to the left of $R_1B_j$. We know that $R_1B_1 \\ge 1$ and $R_1B_n \\le -1.$ Moreover, if $X_{j+1} < X_j$, $X_{j+1}$ must be at most 1 less than $X_j$ since there are no blue points between $B_j$ and $B_{j+1}$. By the intermediate value theorem, there exists an $X_{\\ell}$ such that $X_{\\ell}=0.$ Then $A_1B_{\\ell}$ is the desired balancing line. A similar process can be done for $R_2$, so we have another balancing line and we are done. (In fact, every red line on the convex hull is part of a balancing line).", "Solution_25": "POV: you spend 1 hour proving the existence of a convex hull", "Solution_26": "Consider the convex hull of the points. If both red and blue points exist on the convex hull, then we are done because taking an adjacent red/blue pair produces a balancing line.\n\nOtherwise, WLOG suppose only blue points are on the convex hull. Taking one such point, we can continuously sweep clockwise by taking red points inside the convex hull.\n\nInitially, one side of the line will have at least one blue points and no red points, and at the end, the same side of the line will contain $n-1$ red points, but at most $n-2$ blue points. Since the number of red points on this side of the line increases by exactly $1$ each time we take a new line, and the number of blue points on that side cannot decrease, one of the lines drawn will be a balancing line.\n\nThis works for all points on the boundary, and since our convex hull consists of at least $3$ points, we will have at least $2$ balancing lines, as desired.", "Solution_27": "CASE 1: One can notice That if any blue and red point have not points on one of it's sides this mean the line between them is a balancing line (as on the other side we will have $n-1$ blue points and $n-1$ red).\\\\\n\nCASE 2: Now we consider the case in which all the points are inside a polygon with all vertices being one color (assume WLOG it's blue).\nNow consider a line that is pivoting at any vertex and rotating till reaching the other vertex of the polygon (convex hull) passing through every point in side the polygon. and let $x$ be the number of red points.\\\\\nminus the number of blue points on one side of the line.\nIn other words every time the line passes through a blue point $x$ will increase by $1$, and decrease by $1$ if it passes through a red point.\\\\\nFirstly we have that $x=1$ and when the line reaches the other vertex of the polygon it will be equal to $-1$.\nSo when $x$ will be equal to zero, we have a balancing line.\nNow repeating this process on all the other vertices of the polygon (convex hull) then we will have balancing lines equal to the number of the vertices.", "Solution_28": "If the convex hull is not monochromatic, the result is true by considering any dichromatic edge of the convex hull.\n\nOtherwise, take a line $\\ell$ through two (say) consecutive blue points on the convex hull, and rotate $\\ell$ counterclockwise until it touches a red point; call this line $\\ell_0$. We keep track of the difference $D$ between the number of red and blue points on the side that originally contained the edge containing $\\ell$. In particular, $D$ must be negative to begin (as there are no red points on that side by assumption). Now, we rotate $\\ell$, stopping to compute $D$ every time $\\ell$ hits another red point.\n\nIn particular, any difference $D$ increases by at most $1$ every time we hit a new red point. In particular, the last such line $\\ell$ will have $D$ nonnegative. So by continuity, there exists a point in between where $D=0$ precisely.\n\nWe have demonstrated that there exists a balancing line through some vertex of the convex hull. As there are at least two such vertices, we are done." } { "Tag": [ "trigonometry" ], "Problem": "Solve equation $ 2sin2x\\plus{}(2\\sqrt{3}\\minus{}3)sinx\\plus{}(2\\minus{}3\\sqrt{3})cosx\\minus{}6\\plus{}\\sqrt{3}\\equal{}0$.\r\n\r\nThank.", "Solution_1": "hello, rewriting your equation in the form\r\n$ 4\\sin(x)\\cos(x)+(2\\sqrt{3}-3)\\sin(x)+(2-3\\sqrt{3})\\cos(x)-6+\\sqrt{3}=0$ expressing $ \\sin(x)$ and $ \\cos(x)$ by $ \\tan(\\frac{x}{2})$ we get\r\n$ {\\frac{8\\tan(\\frac{x}{2})(1-\\tan(\\frac{x}{2})^2)}{(1+\\tan(\\frac{x}{2})^2)^2}-\\frac{6\\tan(\\frac{x}{2})}{1+\\tan(\\frac{x}{2})^2}+\\frac{4\\sqrt{3}\\tan(\\frac{x}{2})}{1+\\tan(\\frac{x}{2})^2}+\\frac{2(1-\\tan(\\frac{x}{2})^2)}{1+\\tan(\\frac{x}{2})^2}-\\frac{3\\sqrt{3}(1-\\tan(\\frac{x}{2}))^2}{1+\\tan(\\frac{x}{2})^2}-6-\\sqrt{3}=0}$ substituting $ t=\\tan(\\frac{x}{2})$ and simplifying we have to solve\r\n$ \\frac{-2(-2+\\sqrt{3})(-2t^2+t\\sqrt{3}-1)(t+2+\\sqrt{3})^2}{(1+t^2)^2}=0$.\r\nSonnhard." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Find all k$\\in\\mathbb{N}$, k$\\ge$2 such that:\r\n $\\sqrt{a_{1}+\\sqrt[3]{a_{2}+...+\\sqrt[k+1]{a_{k}}}}\\ge\\sqrt[32]{a_{1}a_{2}...a_{k}}$, for any $a_{1},a_{2},...,a_{k}\\in[0,\\infty).$", "Solution_1": "I've already proved that's imposible for k=2, k=3 and k>4 but I don't know how to solve this:\r\n $\\sqrt{a_{1}+\\sqrt[3]{a_{2}+\\sqrt[4]{a_{3}}}}\\ge\\sqrt[32]{a_{1}a_{2}a_{3}}$ for any real numbers $a_{1},a_{2},a_{3}\\ge0$.", "Solution_2": "if not,we'll have\r\n$\\sqrt a_{1}0 then Prove that:\r\n\\[ \\sum{\\frac{a}{4b^{2}+1}}\\geq(\\sum{a\\sqrt{a}})^{2}\\]", "Solution_1": "It proved by Cauchy-Schwarz\r\n\\[ \\sum{\\frac{a}{4b^{2}+1}}=\\sum{\\frac{a^{3}}{4a^{2}b^{2}+a^{2}}}\\geq\\frac{(a\\sqrt{a}+b\\sqrt{b}+c\\sqrt{c})^{2}}{a^{2}+b^{2}+c^{2}+4(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})}\\]\r\nAnd we prove\r\n\\[ \\sum{a^{2}}+\\sum{4a^{2}b^{2}}\\leq\\ 1=(a+b+c)^{2}\\]\r\n\r\n\\[ \\Leftrightarrow\\sum{2a^{2}b^{2}}\\leq\\sum{ab}\\]\r\n\r\n\\[ \\Leftrightarrow\\sum{ab(1-2ab)}\\geq\\ 0 \\]\r\n(This is easily :a+b+c=1 => ab,bc,ca<1/2)\r\nEquation when (a,b,c)=(0,0,1)and the permutations.\r\n[color=red]Keep it simple: As simple as possible, but no simpler\n\n___Albert Einstein__[/color]\r\n[color=blue]I'm Nguy\u1ec5n T\u00e2m \u0110\u0103ng , vietnam[/color]" } { "Tag": [ "Alcumus", "Support" ], "Problem": "Sometimes alcumus wont give me two tries... is this normal?", "Solution_1": "[quote=\"Lucky_Luke\"]Sometimes alcumus wont give me two tries... is this normal?[/quote]\r\n\r\nIf you get it right, then you don't get two tries...\r\n\r\nMaybe you had tried it, closed the window, and later came back to the problem. You would be on your second try, but thought it was your first...", "Solution_2": "no, i mean when i get the problem wrong the first try, it doesn't even show the \"That is incorrect\" screen it just jumps to the solution and says i'm wrong", "Solution_3": "I kinda have the same problem. When I get a question wrong and want to try it again, it takes me to a new problem.", "Solution_4": "What browsers are you guys using?", "Solution_5": "same problem on all, I used everyone except safari.", "Solution_6": "yeah, happened on IE6 and Firefox", "Solution_7": "How many people are experiencing this problem? \r\n\r\nJust to confirm the issue:\r\n\r\nUser enters answer incorrectly\r\nUser directed directly to solution page, bypassing second chance\r\nUser gets new problem and answers incorrectly\r\nUser directed directly to solution page, bypassing second chance", "Solution_8": "well, not consecutively. it's only happened three times", "Solution_9": "Maybe it's an issue on a specific problem(s)?", "Solution_10": "Probably not, if its three different times.\r\n\r\nWell it works for me:\r\nWindows XP\r\nOpera 9.5 I think." } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "Three mutually external circles $ C_1,C_2,C_3$ with respective centers $ O_1,O_2,O_3$ are given. \r\n\r\nLet $ l_1$ be the tangent from $ O_1$ to $ C_2$, one nearer to $ C_3$.\r\nLet $ l_2$ be the tangent from $ O_3$ to $ C_2$, one nearer to $ C_1$.\r\nLet $ l_3$ be the tangent from $ O_2$ to $ C_3$, one nearer to $ C_1$.\r\nLet $ l_4$ be the tangent from $ O_1$ to $ C_3$, one nearer to $ C_2$.\r\nLet $ l_5$ be the tangent from $ O_3$ to $ C_1$, one nearer to $ C_2$.\r\nLet $ l_6$ be the tangent from $ O_2$ to $ C_1$, one nearer to $ C_3$.\r\n\r\nLet $ A\\equal{}l_1\\cap l_2,B\\equal{}l_2\\cap l_3,...,F\\equal{}l_6\\cap l_1$\r\n\r\nProve that $ AB\\cdot CD\\cdot EF\\equal{}BC\\cdot DE\\cdot FA$ if and only if $ AD,BE,CF$ are concurrent.", "Solution_1": "Anybody? :)" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Find all solutions in integers of the equation:\r\n$ y^2\\plus{}y\\equal{}x^4\\plus{}x^3\\plus{}x^2\\plus{}x$ :ninja:", "Solution_1": "Fix some $ x\\in \\mathbb{Z}$ and try to find $ y$. We have an second grade equality with variable $ y$: $ y^2 \\plus{} y \\minus{} (x^4 \\plus{} x^3 \\plus{} x^2 \\plus{} x) \\equal{} 0$.\r\nTo obtain $ y\\in \\mathbb{Z}$ we need that $ 1 \\plus{} 4(x^4 \\plus{} x^3 \\plus{} x^2 \\plus{} x)$ be an perfect square but note the following: $ 1 \\plus{} 4(x^4 \\plus{} x^3 \\plus{} x^2 \\plus{} x) \\equal{} (2x^2 \\plus{} x)^2 \\plus{} 3x^2 \\plus{} 4x \\plus{} 1$.\r\nIf $ 3x^2 \\plus{} 4x \\plus{} 1 \\equal{} 0$, the $ x \\equal{} \\minus{} 1$ and this is a solution.\r\nOtherwise $ 3x^2 \\plus{} 4x \\plus{} 1 > 0$ for $ x\\in\\mathbb{Z}$. Then $ (2x^2 \\plus{} x \\plus{} 1)^2\\le 1 \\plus{} 4(x^4 \\plus{} x^3 \\plus{} x^2 \\plus{} x) \\to 1 \\plus{} 4x^2 \\plus{} 2x\\le 3x^2 \\plus{} 4x \\plus{} 1 \\to x^2\\le 2x \\to 0\\le x(2 \\minus{} x)$.\r\nFinally we put $ x \\equal{} 0,1,2$ and get the solution, $ x \\equal{} 2, \\minus{} 1,0$. Finally the pair are (2,6), (2,-5), (-1, 0), (-1,1), (0, 0), (0,-1)." } { "Tag": [ "function", "abstract algebra", "algebra", "polynomial", "induction", "group theory", "calculus" ], "Problem": "Let $ k$ be an algebraically closed field. Prove that any $ k\\subset A\\subset k(x)$ which is a valuation ring is in fact a discrete valuation ring.\r\nIs this result true ?\r\nI was thinking that this might be a counterexample: $ A \\equal{} \\left\\{\\frac{f(x)}{g(x)}\\in k(x): g(0)\\neq0 \\right\\}$. This $ A$ satisfies the property if $ a\\in k(x)$ then either $ a$ or $ a^{\\minus{}1}$ is in $ A$ (which is an alternative definition for a valuation ring). But $ A$ is not Noetherian so it can't be a DVR.\r\nIs there something wrong ?", "Solution_1": "$ A$ is a localisation of a noetherian ring $ k[x]$, thus it is notherian.", "Solution_2": "[quote]$ A$ is a localisation of a noetherian ring $ k[x]$, thus it is notherian.[/quote]\r\n:oops: stupid mistake...", "Solution_3": "We don't need $ k$ algebraically closed.\r\nStatement: Let $ k$ be a field and let $ K/k$ be a finitely generated field extension of trdeg=1. Then any valuation ring $ k\\subset A\\subsetneq K$ is a DVR.\r\n\r\nStep 1. $ K = k(X)$. Let $ k\\subset A\\subsetneq K$ be a valuation ring. We can suppose that $ X\\in A$, otherwise $ X^{ - 1}\\in A$ and $ K = k(Y), Y = X^{ - 1}$ so it will be the same argument.\r\nObserve now that there is a unique irreducible polynomial $ P\\in k[X]$ such that $ P^{ - 1}\\not\\in A$: there must exist at least one, else $ A$ would be $ K$. If there exist $ P,Q$ irreducible with $ P^{ - 1},Q^{ - 1}\\not\\in A$ then by localisation we see that there exist $ \\frak{m}, \\mathfrak{n}\\in A$ different maximal ideals - which is a contradiction since $ A$ is local.\r\nSo our ring $ A$ must be $ S^{ - 1}k[X]$ where $ S = \\{Q\\in k[X], (P,Q) = 1\\}$ and this is clearly a discrete valuation ring.\r\n\r\nStep 2. If $ L\\supset K\\supset k$ where $ K = k(X)$ is a finite algebraic extension with $ [L: K] = r$ and if $ v: L^*\\to G$ is a valuation of $ L$ then if we denote by $ H = v(K^*)$ we have $ [G: H] \\le r$. Take $ \\alpha_i\\in G, i = 1,r + 1$. Then there exist $ x_i, i = 1,r + 1$ in $ L$ such that $ v(x_i) = \\alpha_i$. Since $ [L: K] = r$ it follows that there exist $ t_i\\in K, i = 1,r + 1$ not all $ 0$ such that $ \\sum_{i = 1}^{r + 1}t_ix_i = 0$. Using the fact that if $ v(x)\\neq v(y)$ then $ v(x + y) = min(v(x),v(y))$ we deduce by induction that there is some $ i\\neq j$ such that $ t_ix_i = t_jx_j$. So $ \\alpha_i - \\alpha_j = v(t_j) - v(t_i)\\in H$ and this is what we want to conclude that $ [G: H]\\le r$.\r\n\r\nStep 3. If $ G$ is an abelian ordered group and if $ H < G$ is an [url=http://en.wikipedia.org/wiki/Archimedean_property]archimedean[/url], finite index subgroup of $ G$ then $ G$ is archimedian. If $ G$ has a least positive element then it is isomorphic to $ \\mathbb{Z}$.\r\nLet $ n = [G: H]$ and $ x,y\\in G, x,y > 0$. Then $ nx,ny\\in H$ so there is some $ m$ such that $ mnx > mny$ and then $ mx > y$ and $ G$ is therefore archimedean. Therefore if $ H$ has a least positive element also $ G$ has a least positive element.\r\n\r\nLet $ x\\in G, x > 0$ be the least positive element. Then because $ G$ is archimedean any element of $ G$ is of the form $ nx$ with $ n\\in\\mathbb{Z}$, so $ G$ is isomorphic to $ \\mathbb{Z}$. \r\n\r\nStep 4 Let $ k\\subset L$ of trdeg=1 and $ v: L^*\\to G$ be a valuation. Then $ L/K, K = k(X)$ is a finite algebraic extension so $ v(K^*)$ is a finite index subgroup of $ G$ from Step 2. From Step 1 we have that $ v(K^*)$ is isomorphic to $ \\mathbb{Z}$ and from Step 3 we get $ G$ isomorphic to $ \\mathbb{Z}$.", "Solution_4": "Is this correct?\r\n\r\n1. Consider an element $ x \\in K$ which is transcendental over $ k$. Without loss of generality, assume that $ x \\in A$ (otherwise we have $ \\frac{1}{x} \\in A$), i.e. $ k[x] \\subset A$. Since $ A$ is integrally closed, the integral closure of $ k[x]$, say $ B$, satisfies $ B \\subset A$.\r\n\r\nOf course, since $ x$ is transcendental over $ k$, we have that $ K$ is a finite extension of $ k(x)$.\r\n\r\nTheorem: The integral closure of a Dedekind domain in a finite extension of its quotient field is also a Dedekind domain.\r\n\r\nSince $ k[x]$ is a Dedekind domain, we see that $ B$ is also a Dedekind domain by the theorem.\r\n\r\n2. Let $ m$ be the maximal ideal of $ A$, and consider $ n \\equal{} B \\cap m$. Since $ B$ has dimension 1, $ n$ is either maximal or $ (0)$. If $ n$ is maximal, then note that $ B_n \\subset A$ and A dominates $ B_n$ (as local ring). However, \r\na. $ B_n$ is a DVR since $ B$ is a Dedekind domain.\r\nb. valuation rings are maximal elements in the poset ordered by domination. Since $ A$, $ B_n$ are both valuation rings, we must have $ A \\equal{} B_n$, i.e. $ A$ is also DVR.\r\n\r\n3. Therefore it suffices to show that $ n$ is not 0. Indeed, if $ n \\equal{} (0)$, that means $ K(B) \\subset A$, where $ K(B)$ is the fractional field of $ B$. However, as $ K$ is a finite extension of $ k(x)$ and $ B$ is the integral closure of $ k[x]$, this implies that $ K(B) \\equal{} K$, which means that $ A \\equal{} K$, which contradicts the given assumption.", "Solution_5": "I don't quite understand your proof, and may have made some mistakes.\r\n\r\nI think the vital process is as following: suppose $ \\frac {P}{Q} \\in A$, since (P,Q)=1, there is such a U(x) that PU+QV=1. Since A is a ring, $ \\frac {P}{Q}U \\plus{} V \\equal{} \\frac {1}{Q} \\in A$. If all prime polynomials' reciproties belongs to A then $ A \\equal{} k(x)$, a contradiction. So suppose $ \\frac {1}{H(x)}$ is not in A, where H is prime. Then for any other prime polynomial L(x), and from the above argument we have $ \\frac {L}{H}$ is not in A and $ \\frac {H}{L}$ is in A, hence $ \\frac {1}{L} \\in A$.\r\n\r\nThus (H) is the only maximal ideal in A.", "Solution_6": "I don't think you are proving Kuba's statement in his fourth post.", "Solution_7": "oh, I see. What does feild extension trdeg=1 means? I search the google and wiki but didn't find it, Seems familiar.", "Solution_8": "Transcendence degree. The maximal number of algebraically independent elements you can find.", "Solution_9": "It seems to be very similar proof with Soarer's, explaining a little more, why discrete val ring is \"ordered\". The first step is the same, consider the int ring in A, Soarer has already showed it's Dedkind. Then it remains to prove:\r\nIf D is a Dedkind domain, $ D\\subsetneq A\\subsetneq Frac(D)$, A is valuation, then A is discrete valuation.\r\n\r\n(1).by a similar process in my #6 post, if $ \\frac {a}{b}\\in A$ then $ \\frac {k_{1}a \\plus{} k_{2}b}{b}\\in A$ thus \"pure\" fractional ideal $ \\frac {(a,b)}{b}\\in A$.\r\n(2).If all fractional ideal is in A then A=Frac(D), a contradiction.\r\n(3).If two prime ideals' reciprocal $ \\frac {1}{\\alpha},\\frac {1}{\\beta}$ are not in A, consider $ \\frac {q_{1}\\alpha}{q_{2}\\beta}$ which belongs to principle ideal class, such $ q_{1},q_{2}$ must exist by chinese remainder theorem, then from (1) either $ \\frac {1}{q_{1}\\alpha}$ or $ \\frac {1}{q_{1}\\beta}$ is in A, thus either $ \\frac {1}{\\alpha}$ or $ \\frac {1}{\\beta}$ is in A. Which means A is local, and thus, discrete valuation." } { "Tag": [ "logarithms", "inequalities", "function", "calculus", "calculus computations" ], "Problem": "Show that if $ \\sum a_n$ converges, then so does $ \\sum\\frac{\\sqrt{a_n}}{n}$.", "Solution_1": "[hide=\"Answer\"]Let $ \\sum a_n$ and $ \\sum b_n$ be to convergent series of positive terms. Then, as the sum of two convergent series also converges, then the sum whose general term is $ \\frac{a_n \\plus{} b_n}{2}$ is convergent. Therefore, $ \\sum \\sqrt{a_n b_n}$ also converges because for every n, $ \\sqrt{a_n b_n} \\leq \\frac{a_n \\plus{} b_n}{2}$.\n\nIn the present case, consider $ b_n \\equal{} \\frac{1}{n^2}$.[/hide]", "Solution_2": "And in fact Carcul's proof shows that $ \\sum_{n \\geq 1} \\frac {\\sqrt {a_n}}{n^{\\frac {1}{2} \\plus{} \\varepsilon}}$ converges for any $ \\varepsilon > 0$. However, it's not true that $ \\sum_{n \\geq 1} \\frac {\\sqrt {a_n}}{n^{\\frac {1}{2}}}$ will converge: take $ a_n \\equal{} \\frac{1}{n (\\log n)^2}$, for example.", "Solution_3": "One can use the Cauchy-Schwarz inequality:\r\n\r\n$ \\sum \\frac {\\sqrt {a_n}}n\\le\\left(\\sum a_n\\right)^{\\frac12}\\left(\\sum\\frac1{n^2}\\right)^{\\frac12}$\r\n\r\nWhat Carcul used was something that could be used in a proof of Cauchy-Schwarz, so it's a closely related argument.\r\n\r\nAnd Cauchy-Schwarz also works with everything JBL said.\r\n\r\nNow prove this:\r\n\r\nSuppose $ 0 < s < 1$ and $ a_n > 0.$\r\n\r\nIf $ \\sum a_n$ converges, then $ \\sum\\frac {a_n^s}{n}$ converges.", "Solution_4": "Don't all of these follow straight from the comparison test?", "Solution_5": "Depends on what you mean. Could you explain your argument further?", "Solution_6": "[quote=\"Kent Merryfield\"]Suppose $ 0 < s < 1$ and $ a_n > 0.$\n\nIf $ \\sum a_n$ converges, then $ \\sum\\frac {a_n^s}{n}$ converges.[/quote]\r\n\r\n[hide]If $ p \\plus{} q \\equal{} 1$, $ p, q \\geq 0$ then $ pa \\plus{} qb \\geq a^pb^q$ for all $ a, b > 0$. (This is just weighted AM-GM, or equivalently the convexity of the function $ a^xb^{1 \\minus{} x}$.) Thus $ pa_n \\plus{} qn^{ \\minus{} \\frac {1}{q}} \\geq \\frac {a_n^p}{n}$ and so $ \\sum \\frac {a_n^p}{n}$ converges by comparison. (Note $ \\minus{} \\frac {1}{q} < \\minus{} 1$, so $ \\sum n^{ \\minus{} \\frac {1}{q}}$ converges.)[/hide]", "Solution_7": "Just to give an argument along the lines of what SouthStar might have meant:\r\n[hide]Let's say that $ n$ is small if $ a_n < n^{\\minus{}1/(1 \\minus{} s)}$, and large otherwise. The small terms are bounded above by the sum of $ n^{\\minus{}1/(1\\minus{}s)}$, which converges. If $ n$ is large, then $ \\frac{a_n^s}{n} \\le a_n$, so the large terms are bounded above by the sum of $ a_n$.[/hide]" } { "Tag": [ "trigonometry", "function" ], "Problem": "Prove that if $0\\leq uu, the expression can never equal one. Also, because of that stipulation, the expression is never less than 1.\nMaximize the expression. That is done by maximizing both fractions. Incidentally, doing so still allows v>u. Set $v=\\pi/2$ and set the limit of u to 0. The expression approaches $\\pi/2$.[/hide]" } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "hi,\r\nsolve the equation \r\n\\[ x^{2n} \\plus{} a_{1}x^{2n\\minus{}1} \\plus{} ... \\plus{} a_{2n \\minus{} 2}x^{2} \\minus{} 2nx \\plus{} 1 \\equal{} 0,\\]\r\nknown that the equation has real positive roots.", "Solution_1": "Maybe you wanted to say that [b]all[/b] the roots are positive reals. If so, use Viete's relations : \r\n\r\n$ 2n\\equal{}S_{2n}\\equal{} \\sum_{k\\equal{}1}^{2n} \\frac{P_{2n}}{x_k}$ , where $ x_k$ are the roots, and $ P_{2n}\\equal{} x_1 x_2 \\cdots x_{2n}$\r\n\r\nand $ P_{2n}\\equal{}1$ \r\n\r\nBut from AM-GM we know that $ S_{2n} \\geq 2n \\sqrt[2n] P_{2n}$ , but in our case we have equality , therefore $ \\frac{1}{x_1} \\equal{} \\cdots \\frac{1}{x_{2n}}\\equal{}1$.So all roots are equal to $ 1$." } { "Tag": [], "Problem": "Let a quadrangle $ ABCD$ with $ AB\\cap CD \\equal{} E,AD\\cap BC \\equal{} F$. Prove that three circles with diameter $ AC, BD, EF$ have two common points.", "Solution_1": "Dear Mathlinkers,\nI think that this result is not true for any kind of quadrangle...Make a figure and toys with one point.\nSincerely\nJean-Louis", "Solution_2": "As far as I remember, if $ABCD$ was cyclic, then the statement is true.\n\nBest regards,\nsunken rock" } { "Tag": [ "email" ], "Problem": "So how did everyone do?", "Solution_1": "My results for Silver:\r\n\r\n[code]\n -- case number --\n 1 2 3 4 5 6 7 8 9 10 11 12\ncounfr * * * * * * t t t t\nevents * * * * * * * * * *\nwall * x * * t t * * * t t t\n[/code]\r\n\r\nMediocre compared to you guys but I thought it was okay. Events was my only iterative solution... I made quick recursive solutions for the other two (slow, but I got a couple right for them at least). I hate optimization.", "Solution_2": "I got fullscore in gold.\r\n\r\n[code] 1 2 3 4 5 6 7 8 9 10\ncfair * * * * * * * * * *\nmqueue * * * * * * * * * *\ntwohead * * * * * * * * * *[/code]", "Solution_3": "Hmm... I just changed the file names for Silver's countyfr and submitted for Gold's cfair and got the same results. Does that happen often?", "Solution_4": "[quote=\"Xuan768\"]Hmm... I just changed the file names for Silver's countyfr and submitted for Gold's cfair and got the same results. Does that happen often?[/quote]\r\n\r\nHappens occasionally.\r\n\r\nMy results:\r\n\r\ncfair * * * * * * * * * *\r\nmqueue * * x x x * t x * x\r\ntwohead * * * * * t t t t t\r\n\r\nMeh, kinda botchy. Oh well, hopefully it's still enough to make USAICO.", "Solution_5": "So anyone get camp notifications yet?", "Solution_6": "Well, I haven't. Not that I was even close to getting in.", "Solution_7": "This will be my first year at USAICO. JS, I assume you're going too? Do you know when they will publish a list of all the qualifiers?", "Solution_8": "[quote=\"PenguinMath\"]This will be my first year at USAICO. JS, I assume you're going too? Do you know when they will publish a list of all the qualifiers?[/quote]\r\n\r\nYeah, I'm also going. Qualifiers may not be published for a week or so, as people have to respond yes or no to the invitation, and that doesn't necesarilly happen immediately (though clearly I accepted immediately upon receiving the invitation, there are apparently some people who actually have to think about whether they want to miss a week of school to go to the highest level of pre-college computer science instruction in the world).\r\n\r\nHowever if you would like a probable list of people, I would say:\r\n\r\n2007 Richard McCutchen\r\n2006 Kevin Modzelewski\r\n2007 John Pardon\r\n2006 Richard Ho\r\n2006 George Boxer\r\n2007 Thomas Morgan\r\n2007 Bohua Zhan\r\n2007 Zarathustra Brady\r\n2008 Jacob Steinhardt\r\n2007 Michael Gummelt\r\n2007 Peiyu Wang\r\n2009 Thomas Wang\r\n\r\nP.S. Are you Bohua, Thomas, or Michael?", "Solution_9": "[quote=\"JSteinhardt\"](though clearly I accepted immediately upon receiving the invitation, there are apparently some people who actually have to think about whether they want to miss a week of school to go to the highest level of pre-college computer science instruction in the world).[/quote]\r\n\r\nThere are reasons, like maybe their graduation is that week, or something. You never know.", "Solution_10": "[quote=\"JSteinhardt\"][quote=\"PenguinMath\"]This will be my first year at USAICO. JS, I assume you're going too? Do you know when they will publish a list of all the qualifiers?[/quote]\n\nYeah, I'm also going. Qualifiers may not be published for a week or so, as people have to respond yes or no to the invitation, and that doesn't necesarilly happen immediately (though clearly I accepted immediately upon receiving the invitation, there are apparently some people who actually have to think about whether they want to miss a week of school to go to the highest level of pre-college computer science instruction in the world).\n\nHowever if you would like a probable list of people, I would say:\n\n2007 Richard McCutchen\n2006 Kevin Modzelewski\n2007 John Pardon\n2006 Richard Ho\n2006 George Boxer\n2007 Thomas Morgan\n2007 Bohua Zhan\n2007 Zarathustra Brady\n2008 Jacob Steinhardt\n2007 Michael Gummelt\n2007 Peiyu Wang\n2009 Thomas Wang\n\nP.S. Are you Bohua, Thomas, or Michael?[/quote]\r\n\r\nList is posted - [url]http://oldweb.uwp.edu/academic/mathematics/usaco/2006/camp06/finalists.htm[/url]", "Solution_11": "[quote=\"h_s_potter2002\"][quote=\"JSteinhardt\"][quote=\"PenguinMath\"]This will be my first year at USAICO. JS, I assume you're going too? Do you know when they will publish a list of all the qualifiers?[/quote]\n\nYeah, I'm also going. Qualifiers may not be published for a week or so, as people have to respond yes or no to the invitation, and that doesn't necesarilly happen immediately (though clearly I accepted immediately upon receiving the invitation, there are apparently some people who actually have to think about whether they want to miss a week of school to go to the highest level of pre-college computer science instruction in the world).\n\nHowever if you would like a probable list of people, I would say:\n\n2007 Richard McCutchen\n2006 Kevin Modzelewski\n2007 John Pardon\n2006 Richard Ho\n2006 George Boxer\n2007 Thomas Morgan\n2007 Bohua Zhan\n2007 Zarathustra Brady\n2008 Jacob Steinhardt\n2007 Michael Gummelt\n2007 Peiyu Wang\n2009 Thomas Wang\n\nP.S. Are you Bohua, Thomas, or Michael?[/quote]\n\nList is posted - [url]http://oldweb.uwp.edu/academic/mathematics/usaco/2006/camp06/finalists.htm[/url][/quote]\r\n\r\nThanks for the heads up.\r\n\r\nHm...so it looks like I was right for everyone except Zimmerman going instead of Thomas.\r\n\r\nAlthough I must admit that I am surprised that Filip Wolski is not coming from the international group.", "Solution_12": "I'm Michael. Congrats on being the only sophomore to qualify! We need more underclassmen or else the team isn't looking so hot for the future. Oh, and I'm not from Virginia! I'm a proud Texan! They said I was from VA on that website. I guess there's so many good math/sci students from TJHSST that if they just assume that an arbitrary student is from there then odds are they're right. \r\n\r\nOh and pray for me and my attempt to switch from Java -> C++ in the next few weeks. I was a C++ programmer back in the day, but its been a while. I have NO familiarity with the STL so that'll be the hardest part. Best of Luck!", "Solution_13": "[quote=\"PenguinMath\"]I'm Michael. Congrats on being the only sophomore to qualify! We need more underclassmen or else the team isn't looking so hot for the future. Oh, and I'm not from Virginia! I'm a proud Texan! They said I was from VA on that website. I guess there's so many good math/sci students from TJHSST that if they just assume that an arbitrary student is from there then odds are they're right. \n\nOh and pray for me and my attempt to switch from Java -> C++ in the next few weeks. I was a C++ programmer back in the day, but its been a while. I have NO familiarity with the STL so that'll be the hardest part. Best of Luck![/quote]\r\n\r\nLol yeah I'm in about the same position. Tom went to camp last year knowing only JAVA and didn't do too poorly. But preparation is never a bad thing.\r\n\r\nI agree with you about the whole lack of underclassmen. I was actually sure Kevin Li at least would have made it, if not Thomas Wang, but Kevin told me personally that he didn't make it. It seems to me to be very illogical, since he has done roughly as well as Peiyu/Jonathan, who is going to camp, but Kevin is only a Sophomore. Go figure.\r\n\r\nGood luck to you too! I guess I'll see you in June. From what I hear USAICO is awesome in that there are pretty good classes but also some nice down time as well.", "Solution_14": "Ya I've heard good things as well. I'm excited that there's so many internationals coming, that'll be interesting.", "Solution_15": "Dont worry, I'll make it next year. [\\blatent lie]", "Solution_16": "What's the best thing you can do to prepare for the USACO?", "Solution_17": "What's the best thing you can do to prepare for the USACO?", "Solution_18": "What's the best thing you can do to prepare for the USACO?", "Solution_19": "jSteinhardt said in one of his posts something like \"working through the [usaco] training pages are the best thing you can do to train for the usaco. Took me from bronze to gold in about a year.\"", "Solution_20": "jSteinhardt said in one of his posts something like \"working through the [usaco] training pages are the best thing you can do to prepare for the usaco. Took me from bronze to gold in about a year.\"", "Solution_21": "Thanks. Well, I'm in a different situation. I'm a freshman, and I have no computer science experience whatsoever though I plan on taking AP Computer Science (not sure which level my school offers, A or AB) next year. Is this enough knowledge to help me do well in USACO? Also, where can I get the USACO training pages?", "Solution_22": "[url=http://train.usaco.org/usacogate]http://train.usaco.org/usacogate[/url]\r\nGoogle stuff when you want to find things-just a click away ;)\r\nYeah, you do need to know how to program to use the pages (i kind of overlooked that when i registered before realizing i had no experience), so take whatever class they have at your school that you think will help.", "Solution_23": "I agree, training site is really great. It took me to gold in 10 months.", "Solution_24": "Did you have experience, r2rd? Also, what's an ICQ number?", "Solution_25": "I have no programming experience at all, so all this stuff looks like Chinese to me.", "Solution_26": "http://java.sun.com/docs/books/tutorial/java/TOC.html\r\n\r\nThat teaches JAVA, although personally C++ is better.\r\n\r\nYou can just go to amazon.com or some store and buy a book on C++ if you really want to, although I guess it's a lot easier with a teacher. I am guessing that the best person to contact is Rob Kolstad, the USACO head coach who is almost always willing and able to help out. Make sure all emails have USACO in their subject, his e-mail is kolstad@ace.DELOS.COM and his AIM is RobAtDelos if you prefer that. Explain your situation, he probably has some good recommendations.", "Solution_27": "Thanks a bunch, JS.\r\n\r\nWow, from what friends have told me, C++ is usually taught before Java, at least at my school." } { "Tag": [ "function", "algebra", "polynomial", "trigonometry", "IMO", "IMO Shortlist" ], "Problem": "IMO 1975 ShortList:", "Solution_1": "[b][size=150]IMO 1975 ShortListed Problems: (Posted in Forums)[/size][/b]\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2025364#p2025364]1.(FRA)[/url]There are six ports on a lake. Is it possible to organize a series of routes satisfying the following conditions ?:\n\n(i)\tEvery route includes exactly three ports;\n(ii)\tNo two routes contain the same three ports;\n(iii)\tThe series offers exactly two routes to each tourist who desires to visit two different arbitrary ports.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=367460&#p367460]2.(CZS)[/url] [b]IMO1[/b] We consider two sequences of real numbers $x_{1} \\geq x_{2} \\geq \\ldots \\geq x_{n}$ and $\\ y_{1} \\geq y_{2} \\geq \\ldots \\geq y_{n}.$ Let $z_{1}, z_{2}, .\\ldots, z_{n}$ be a permutation of the numbers $y_{1}, y_{2}, \\ldots, y_{n}.$ Prove that $\\sum \\limits_{i=1}^{n} ( x_{i} -\\ y_{i} )^{2} \\leq \\sum \\limits_{i=1}^{n}$ $( x_{i} - z_{i})^{2}.$\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2025366#p2025366]3.(USA)[/url] Find the integer represented by $\\left[ \\sum_{n=1}^{10^9} n^{-2/3} \\right] $. Here $[x]$ denotes the greatest integer less than or equal to $x.$\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2025368#p2025368]4.(SWE)[/url] Let $a_1, a_2, \\ldots , a_n, \\ldots $ be a sequence of real numbers such that $0 \\leq a_n \\leq 1$ and $a_n - 2a_{n+1} + a_{n+2} \\geq 0$ for $n = 1, 2, 3, \\ldots$. Prove that\n\\[0 \\leq (n + 1)(a_n - a_{n+1}) \\leq 2 \\qquad \\text{ for } n = 1, 2, 3, \\ldots\\]\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2025369#p2025369]5.(SWE)[/url] Let $M$ be the set of all positive integers that do not contain the digit $9$ (base $10$). If $x_1, \\ldots , x_n$ are arbitrary but distinct elements in $M$, prove that\n\\[\\sum_{j=1}^n \\frac{1}{x_j} < 80 .\\]\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=849354&#p849354]6.(USS)[/url] [b]IMO4[/b] When $4444^{4444}$ is written in decimal notation, the sum of its digits is $ A.$ Let $B$ be the sum of the digits of $A.$ Find the sum of the digits of $ B.$ ($A$ and $B$ are written in decimal notation.)\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2025371#p2025371]7.(GDR)[/url] Prove that from $x + y = 1 \\ (x, y \\in \\mathbb R)$ it follows that\n\\[x^{m+1} \\sum_{j=0}^n \\binom{m+j}{j} y^j + y^{n+1} \\sum_{i=0}^m \\binom{n+i}{i} x^i = 1 \\qquad (m, n = 0, 1, 2, \\ldots ).\\]\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=367456&#p367456]8.(NET)[/url] [b]IMO3[/b] In the plane of a triangle $ABC,$ in its exterior$,$ we draw the triangles $ABR, BCP, CAQ$ so that $\\angle PBC = \\angle CAQ = 45\\,^{\\circ},$ $\\angle BCP = \\angle QCA = 30\\,^{\\circ},$ $\\angle ABR = \\angle RAB = 15\\,^{\\circ}.$\n\nProve that \n\n[b]a.)[/b] $\\angle QRP = 90\\,^{\\circ},$ and \n\n[b]b.)[/b] $QR = RP.$\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2025374#p2025374]9.(NET)[/url] Let $f(x)$ be a continuous function defined on the closed interval $0 \\leq x \\leq 1$. Let $G(f)$ denote the graph of $f(x): G(f) = \\{(x, y) \\in \\mathbb R^2 | 0 \\leq x \\leq 1, y = f(x) \\}$. Let $G_a(f)$ denote the graph of the translated function $f(x - a)$ (translated over a distance $a$), defined by $G_a(f) = \\{(x, y) \\in \\mathbb R^2 | a \\leq x \\leq a + 1, y = f(x - a) \\}$. Is it possible to find for every $a, \\ 0 < a < 1$, a continuous function $f(x)$, defined on $0 \\leq x \\leq 1$, such that $f(0) = f(1) = 0$ and $G(f)$ and $G_a(f)$ are disjoint point sets ?\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=367453&#p367453]10.(GBR)[/url] [b]IMO6[/b] Determine the polynomials P of two variables so that:\n\n[b]a.)[/b] for any real numbers $t,x,y$ we have $P(tx,ty) = t^n P(x,y)$ where $n$ is a positive integer, the same for all $t,x,y;$\n\n[b]b.)[/b] for any real numbers $a,b,c$ we have $P(a + b,c) + P(b + c,a) + P(c + a,b) = 0;$\n\n[b]c.)[/b] $P(1,0) =1.$\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=367455&#p367455]11.(GBR)[/url] [b]IMO2[/b] Let $a_{1}, \\ldots, a_{n}$ be an infinite sequence of strictly positive integers, so that $a_{k} < a_{k+1}$ for any $k.$ Prove that there exists an infinity of terms $ a_{m},$ which can be written like $a_m = x \\cdot a_p + y \\cdot a_q$ with $x,y$ strictly positive integers and $p \\neq q.$\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2025379#p2025379]12.(GRE)[/url] Consider on the first quadrant of the trigonometric circle the arcs $AM_1 = x_1,AM_2 = x_2,AM_3 = x_3, \\ldots , AM_v = x_v$ , such that $x_1 < x_2 < x_3 < \\cdots < x_v$. Prove that\n\\[\\sum_{i=0}^{v-1} \\sin 2x_i - \\sum_{i=0}^{v-1} \\sin (x_i- x_{i+1}) < \\frac{\\pi}{2} + \\sum_{i=0}^{v-1} \\sin (x_i + x_{i+1})\\]\n\n\n13.(ROM) Let $A_0,A_1, \\ldots , A_n$ be points in a plane such that\n(i) $A_0A_1 \\leq \\frac{1}{ 2} A_1A_2 \\leq \\cdots \\leq \\frac{1}{2^{n-1} } A_{n-1}A_n$ and\n(ii) $0 < \\measuredangle A_0A_1A_2 < \\measuredangle A_1A_2A_3 < \\cdots < \\measuredangle A_{n-2}A_{n-1}A_n < 180^\\circ,$\nwhere all these angles have the same orientation. Prove that the segments $A_kA_{k+1},A_mA_{m+1}$ do not intersect for each $k$ and $n$ such that $0 \\leq k \\leq m - 2 < n- 2.$\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2025382#p2025382]14. (YUG)[/url] Let $x_0 = 5$ and $x_{n+1} = x_n + \\frac{1}{x_n} \\ (n = 0, 1, 2, \\ldots )$. Prove that\n$45 < x_{1000} < 45. 1.$\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=367461&#p367461]15. (USS)[/url] [b]IMO5[/b] Can there be drawn on a circle of radius $1$ a number of $1975$ distinct points, so that the distance (measured on the chord) between any two points (from the considered points) is a rational number?" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Find value for this sum\r\n\\[6+66+666+6666+\\ldots+\\underbrace{666\\ldots 666}_{n\\, \\rm{digit number}}\\]\r\nHow about the general problem?\r\n\\[aa+aa+aaa+aaaa+\\ldots+\\underbrace{aaa\\ldots aaa}_{n\\, \\rm{digit number}}\\]", "Solution_1": "[quote=\"color\"]Find value for this sum\n\\[6+66+666+6666+\\ldots+\\underbrace{666\\ldots 666}_{n\\, \\rm{digit number}}\\]\nHow about the general problem?\n\\[aa+aa+aaa+aaaa+\\ldots+\\underbrace{aaa\\ldots aaa}_{n\\, \\rm{digit number}}\\]\n[/quote]\r\n\r\n$S=a+aa+aaa+aaaa+\\ldots+\\underbrace{aaa\\ldots aaa}_{n\\, \\rm{digit number}})$\r\n$S=\\frac{a}{9}(9+99+999+9999+\\ldots+\\underbrace{999\\ldots 999}_{n\\, \\rm{digit number}})$\r\n$S=\\frac{a}{9}(10^{1}+10^{2}+10^{3}+\\ldots+10^{n}-n)$\r\n\r\n$S=\\frac{a}{9}(10\\frac{10^{n}-1}{9}-n)$\r\n\r\n-- \r\nPatrick" } { "Tag": [ "number theory", "prime numbers" ], "Problem": "How many duplets of prime numbers, (p,q) are possible such that\r\np^2 + 7pq + q^2 is a whole square, i.e. square of some integer?", "Solution_1": "For infinitely many couples.\r\n\r\nJust take the $ p\\equal{}q$\r\n\r\n\r\n :)", "Solution_2": "So let's make the problem harder: describe all such $ (p,q)$.\r\n\r\nI got that $ p \\neq q$ has no solutions with some nasty casework, but there's probably a much better way to determine that.\r\n\r\nEDIT: And this is why you don't solve problems with nasty casework, because I missed $ (3,11)$ as shown in the links below.", "Solution_3": "I thought this looked familiar:\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=58233[/url]\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80152[/url]\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=19090[/url]" } { "Tag": [ "geometry", "geometric transformation", "rotation", "vector", "analytic geometry", "complex numbers", "superior algebra" ], "Problem": "Hi,\r\n\r\nFirst, I would like to apologise if I have misplaced this topic - I hope superior algebra is the right forum for my question.\r\nMy question concerns quaternion rotations. I know the formulas that need to be applied to rotate a certain object around the x-,y- and z-axis, but I do not know how to do this 'practically'.\r\nFor example, one point I want to rotate is the point (1,1,1).\r\nI want to rotate this around the point (1,2,3), 90 degrees, so pi/2 radians.\r\nI believe the formula is:\r\n(I qoute from a non-math site now)\r\n*******************************\r\nIf the axis of rotation is (ax, ay, az)- must be a unit vector\r\nand the angle is theta (radians)\r\n\r\n w = cos(theta/2)\r\n x = ax * sin(theta/2)\r\n y = ay * sin(theta/2)\r\n z = az * sin(theta/2)\r\n*******************************\r\nw=cos(pi/4)= sqrt(2)/2\r\nx = 1x * sin(pi/4) = x*sqrt(2)/2\r\ny = 2y * sin(pi/4) = 2y*sqrt(2)/2 = ysqrt(2)\r\nz = 3z * sin(pi/4) = 3z*sqrt(2)/2 = 3/2zsqrt(2)\r\n\r\nBut how would I apply this to my (1,1,1) point?\r\nDo you know of any other examples?\r\n\r\nThanks,\r\nRicOh", "Solution_1": "When working in the three-dimensional space $\\mathbb R^{3}$, it is helpful to use the standard basis vectors $\\vec\\i$, $\\vec\\j$, and $\\vec k$ - these are simply vector of unit length that point in the direction Ox, Oy, Oz, respectively. So, your point $(1,1,1)$ can be identified with the vector $\\vec\\i+\\vec\\j+\\vec k$. (In general, a point $(x,y,z)$ will be represented by the vector $x\\vec\\i+y\\vec\\j+z\\vec k$ --- the [i]position vector[/i] of the point.) \r\n\r\nQuaternions are objects of the form $a+bi+cj+dk$, where $i$, $j$, and $k$ are imaginary units, and $a\\dots d$ are real numbers. There are addition/subtraction and multiplication/division rules that you can [url=http://mathworld.wolfram.com/Quaternion.html]look up[/url]. The notation suggests that you can interpret vectors as quaternions with real part $a$ equal to $0$. General quaternions can be written as pairs $(a,\\vec v)=a+v_{1}\\vec\\i+v_{2}\\vec\\j+v_{3}\\vec k$. Multiplying two vectors $\\vec v$ and $\\vec w$ [b]as quaternions[/b] you get the quaternion $(-\\vec v\\cdot\\vec w, \\vec v\\times\\vec w)$ - this multiplication contains both the dot product and the cross product. :o \r\n\r\nYou have already computed the rotation quaternion $q=w+xi+yj+zk$. The position of a point $p$ (such as $i+j+k$ in your example) after rotation by $q$ is given by the quaternion $p'=qpq^{-1}$. Since $q$ is a unit quaternion, its inverse $q^{-1}$ is the same as its conjugate, which is $\\bar q=w-xi-yj-zk$. The quaternion $p'=qp\\bar q$ has zero real part, so it indeed represents a vector in 3D.", "Solution_2": "Thank you for replying. My problem was how to 'enter' the point in the formula. By representing (1,1,1) as i+j+k, my problem is solved.\r\nAlso, I'm sorry for my inferior knowledge, I am just a Belgian hugh school student with weird interests. :roll:", "Solution_3": "No problem. I suspect, but don't know for sure, that we now use the letters i,j,k for basis vectors because Hamilton used them for quaternions. A long tim ago, quaternions were much more en vogue than they are now.", "Solution_4": "So now, step by step\r\n1) Coordinates (a,b,c) --> ai+bj+ck\r\n2) Rotation --> quaternion using the formula\r\n3) Apply the rotation to the ai+bj+ck\r\n4)Result: di+ej+fk--> (d,e,f)\r\n\r\nI do not fully understand why you still have to multiply the result with the conjugate, but that's not (very) important.", "Solution_5": "Of course, it is not obvious why $qp\\bar q$ gives the coordinates of the rotated point. I can't give a proof off the top of my head. But it is clear that $qp$ would not work - this is in general a quaternion with nonzero real part, so it does not represent a vector.", "Solution_6": "I'll Google some more then - it's the experience that counts. :D" } { "Tag": [ "function", "complex analysis", "complex analysis unsolved" ], "Problem": "Let $ D \\equal{} \\{ z \\in \\mathbb{C} : |z|<2 \\}$. Suppose that $ f: D \\to \\mathbb{C}$ is a continuous function that is analytic over $ D\\minus{}[0,1]$. Show that $ f$ is analytic in $ D$.", "Solution_1": "Use Morera's theorem (the version with triangles). Partition the triangle into a few, so that the segment lies on the boundary." } { "Tag": [], "Problem": "Find $\\sum_{r=n}^{2n} n^2$\r\n\r\nCan I say\r\n\r\n$\\sum_{r=n}^{2n} n^2$ = $\\sum_{r=1}^{2n} n^2$ - $\\sum_{r=1}^{n} n^2$\r\n\r\n?\r\n\r\nI don't get some sigma notation things sometimes. (like when you can add them or multiply them or things)", "Solution_1": "Not quite.\r\n\r\nThat statement is saying that\r\n\r\n$(2n)^2+(2n-1)^2+\\ldots+n^2=((2n)^2+(2n-1)^2+\\ldots+1)-(n^2+(n-1)^2+\\ldots+1)$\r\n\r\nCan you see why it's not true?", "Solution_2": "Also, you have another notational problem -- most of your n's are supposed to be r's.", "Solution_3": "[quote=\"blahblahblah\"]Not quite.\n\nThat statement is saying that\n\n$(2n)^2+(2n-1)^2+\\ldots+n^2=((2n)^2+(2n-1)^2+\\ldots+1)-(n^2+(n-1)^2+\\ldots+1)$\n\nCan you see why it's not true?[/quote]\r\nYes I see now, \r\nthe statement would be saying:\r\n$(2n)^2+(2n-1)^2+\\ldots+n^2=((2n)^2+(2n-1)^2+\\ldots+n^2+(n-1)^2+\\ldots+1)-(n^2+(n-1)^2+\\ldots+1)$\r\nin which the $n^2$ term occurs on the left hand side but not the right. so:\r\n\r\n$\\sum_{r=n}^{2n} {r^2} = \\sum_{r=1}^{2n} {r^2} - \\sum_{r=n}^{2n} {r^2} + n^2$\r\n\r\nthanks also, I just realised the $n^2$ should say $r^2$ after the sigma sign. \r\nneed.more.practice.", "Solution_4": "You should write $\\sum_{r=n}^{2n} {r^2} = \\sum_{r=1}^{2n} {r^2} - \\sum_{r=1}^{n-1} {r^2}$ ;)", "Solution_5": "ah I get it now. always making mistakes somehow\r\n\r\nThanks." } { "Tag": [ "function", "floor function", "symmetry" ], "Problem": "Compute the number of intersection points of the graphs of $(x-\\lfloor x\\rfloor)^2+y^2=x-\\lfloor x \\rfloor$ and $y=\\frac15x$.", "Solution_1": "[hide=\"Solution\"]\nFloor functions are usually hard to deal with directly. So just straight algebra wont cut it. Since we're looking at intersection points, let's try graphing.\n\nFirst, note that $x-\\lfloor x\\rfloor=\\{ x\\}$ (i.e. the fractional part of $x$). So our original equation is equivalent to $\\{x\\}^2 - \\{x\\}+y^2=0$ or $(\\{x\\}-1/2)^2 + y^2 = 1/4$. By definition though, the fractional part of $x$ is on $[0,1)$. This means that the graph is periodical with period 1. We also note it will consist of a bunch of tangent circles by looking at the equation. Just concentrating on quadrant I, we want to find where $y=x/5$ \"leaves\" the circles. We know that the highest point the circles achieve is at $y=.5$. It just so happens that $(2.5,.5)$ is on both graphs so this is our last point. So the line will intersect the first circle in the first quadrant once (exclude the origin for now), the second circle twice, and the third circle twice for a total of 5 times. By symmetry, there are 5 intersections in quadrant 4. Add in the origin for a total of $\\boxed{11.}$\n\n[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=760[/img][/hide]", "Solution_2": "[quote=\"4everwise\"]Compute the number of intersection points of the graphs of $(x-\\lfloor x\\rfloor)^2+y^2=x-\\lfloor x \\rfloor$ and $y=\\frac15x$.[/quote]\r\n[hide]\nGraph it!\n[url=http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=761]Link to Picture[/url]\nWe see that there are 5 intersections on each side, and then the origin.\n5+5+1=[b]11[/b]\n[/hide]" } { "Tag": [ "AMC", "AIME" ], "Problem": "The sum of AMC and AIME is the No.1 of the state of my highschool. Does it help?\r\nI know one year the winners of each state got 1000 dollars.\r\nAny awards this year for states winners?", "Solution_1": "I'm assuming that you mean your index (AMC+10xAIME) the highest in the state, not just the sum of AMC+AIME. I don't know about any monetary prize, but I think you automatically qualify for USAMO.", "Solution_2": "that's true, you do automatically qualify. i don't know abt cash prizes tho..", "Solution_3": "It is unlikely you are highest in Virginia though given that TJ probably had multiple indices that were close to 300.", "Solution_4": "Does TJ select people from the state or country like a magnet school or is it just an average high school with a really good math program?", "Solution_5": "I believe TJ draws from students all over VA?\r\nI'm not sure...", "Solution_6": "It's a county magnet school as far as I know, but leave it to one of the many AoPSers from TJ to really answer that question.", "Solution_7": "TJ is one of Virginia's \"governor's schools\" each of which draws students from a small number of surrounding counties. I'm fairly sure that in practice most of TJ is Fairfax county. The high school I went to was another such governor's school in central Virginia.", "Solution_8": "I got a plaque from WI for having highest AMC 10 B score", "Solution_9": "same for me last year, good for college i suppose", "Solution_10": "what score did u guys need to get? to get the plaque", "Solution_11": "That varies from state to state, if you're interested, AMC stats are up at their website, you can find out the top scores needed for each state there:\r\n[url]http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2009-12a/09-1-1012-StateStats.shtml[/url]" } { "Tag": [ "ratio", "geometry" ], "Problem": "Through an internal point $O$ of $\\Delta ABC$ one draws 3 lines, parallel to each of the sides, intersecting in the points shown on the picture.\r\n\r\n[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=289[/img]\r\n\r\nFind the value of $\\frac{|AF|}{|AB|}+\\frac{|BE|}{|BC|}+\\frac{|CN|}{|CA|}$.", "Solution_1": "an easy nonrigorous way is just to note that the problem statement implies that the expression is constant for any choice of O, then choose something convenient.", "Solution_2": "You can [b]make[/b] it rigorous by showing that it's invariant...", "Solution_3": "OK // AC, and MN // BC $\\Rightarrow$ ONKC is a paralellogram. \r\nThus, $CN$ = $OK$\r\n\r\n$\\angle EOK = \\angle BAC, \\angle OKE = \\angle ACB \\Rightarrow \\Delta ABC$ is similar to $\\Delta OEK$\r\n\r\n\\[ \\Rightarrow \\dfrac{CN}{CA}=\\dfrac{OK}{CA}=\\dfrac{EK}{BC} \\]\r\n\r\nBecause AF // OD, and OF // AD, AFOD is a parallellogram.\r\nTherefor, AF = OD.\r\nBecause $\\angle ODN = \\angle BAC$, and $\\angle DON = \\angle ABC, \\Delta ODN$ is similar to $\\Delta ABC$\r\n\\[ \\Rightarrow \\dfrac{AF}{AB}=\\dfrac{OD}{AB}=\\dfrac{ON}{BC}=\\dfrac{KC}{BC} \\]\r\n\r\n\\[ \\dfrac{AF}{AB}+\\dfrac{BE}{BC}+\\dfrac{CN}{CA}=\\dfrac{BE+EK+KC}{BC}=1 \\]", "Solution_4": "Through an internal point $O$ of $\\Delta ABC$ one draws $3$ lines, parallel to each of the sides, intersecting in the points shown on the picture.\n[img]https://cdn.artofproblemsolving.com/attachments/e/3/03d4d1bb61eda8b4a72ff84466d700de47c147.png[/img]\nFind the value of $\\frac{|AF|}{|AB|}+\\frac{|BE|}{|BC|}+\\frac{|CN|}{|CA|}$." } { "Tag": [ "inequalities", "calculus", "derivative", "function", "inequalities proposed" ], "Problem": "PVThuan is my friend in Viet Nam (actually, he teaches English in my school). The following problem is to him with my compliment... It's also for the other if you are interested in it.\r\n\r\nProve that if $a,b,c,d$ are non-negative real numbers and $a^2+b^2+c^2+d^2=4$ then\r\n\\[ \\frac{1}{\\sqrt{a^3+bcd}}+\\frac{1}{\\sqrt{b^3+cda}}+\\frac{1}{\\sqrt{c^3+dab}}+\\frac{1}{\\sqrt{d^3+abc}} \\ge 2\\sqrt{2} \\]\r\nAnd a similar one is that\r\n\\[ \\frac{1}{\\sqrt{a^3+abc}}+\\frac{1}{\\sqrt{b^3+bcd}}+\\frac{1}{\\sqrt{c^3+cda}}+\\frac{1}{\\sqrt{d^3+dab}} \\ge 2\\sqrt{2} \\]\r\nI want to imply a stronger one. But first try to prove the inequalities above...", "Solution_1": "for the second one we can use just AM-GM\r\n :)", "Solution_2": "[quote=\"Albanian Eagle\"]for the second one we can use just AM-GM\n :)[/quote]\r\n\r\nYes... It's my first solution... But may you post a detail solution?", "Solution_3": "$\\sum_{cyc} \\frac{1}{\\sqrt{a^3+abc}}\\geq 4 \\left(\\frac{1}{abcd \\prod_{cyc} (a^2+bc)}\\right)^{\\frac{1}{8}}$ \r\nthen we use the fact that $8\\geq \\sum a^2+bc \\geq 4\\prod (a^2+bc) ^{\\frac{1}{4}}$", "Solution_4": "Ok... It's easy. The first one can be solved by the same method, but fairly harder.", "Solution_5": "Using Jensen for the first, we need to prove a very nice inequality:\r\n\r\n$a^3+b^3+c^3+d^3+abc+bcd+cda+dab \\leq 8$\r\n\r\nfor $a^2+b^2+c^2+d^2=4$.\r\nCan someone post a solution to this nice and hard inequality?", "Solution_6": "[quote=\"Vasc\"]Using Jensen for the first, we need to prove a very nice inequality:\n\n$a^3+b^3+c^3+d^3+abc+bcd+cda+dab \\leq 8$\n\nfor $a^2+b^2+c^2+d^2=4$.\nCan someone post a solution to this nice and hard inequality?[/quote]\r\n\r\nmaybe we can use derivative\r\n\r\nWLOG let $a\\geq b\\geq c\\geq d$\r\n\r\n$f(a)=a^3+a(bc+bd+cd)+(b^3+c^3+d^3+bcd)$\r\n\r\n$f'(a)=3a^2+(bc+bd+cd)>0$ for $0\\leq a,b,c,d\\leq 2$ \r\n\r\nso $f(a)$ is increasing . If we fix $bc+bd+cd=k$ then $4=a^2+b^2+c^2+d^2\\geq a^2+(bc+bd+cd)=a^2+k$\r\n\r\nso $a\\leq \\sqrt{4-k}$ . And since $f(a)$ is increasing , the maximum is when\r\n\r\n$a=\\sqrt{4-k}$ and $bc+bd+cd=k$ $\\implies b=c=d=\\sqrt{k/3}$ ($\\because b^2+c^2+d^2=k$)\r\n\r\nwhich means we only need to check the case for $a\\geq b=c=d$\r\n\r\nand we reduce the question into proving $a^3+3ab^2+4b^3\\leq 8$ with constrain $a^2+3b^2=4$\r\n\r\nor equivalent to $(a^3+3ab^2+4b^3)^2\\leq (a^2+3b^2)^3$ $\\Longleftrightarrow (a-b)^2b^2(3a^2-2ab+11b^2)\\geq 0$\r\n\r\ninteresting to see that the equality case at $a=b=c=d=1$ or $a=2,b=c=d=0$(or its permutation)", "Solution_7": "We have\r\n$f'(a)=3a^2+(bc+bd+cd)$ \r\nif $b,c,d$ are fixed. Then, $a^2+b^2+c^2+d^2=4$ implies $a$ fixed, and $f(a)$ is not function. :o", "Solution_8": "[quote=\"Vasc\"]Using Jensen for the first, we need to prove a very nice inequality:\n\n$a^3+b^3+c^3+d^3+abc+bcd+cda+dab \\leq 8$\n\nfor $a^2+b^2+c^2+d^2=4$.\nCan someone post a solution to this nice and hard inequality?[/quote]\r\n$abcd=const$ and with $f(a)=a^3+\\frac{M}{a}$ while using (n-1)EV \r\nwe see that the maximum ocurrs when 3 variables are equal \r\nso we need to show that\r\n$a^3+4b^3+3ab^2\\le 8$ when $a^2+3b^2=4$ which is equivalent to $b^3+\\sqrt{4-3b^2}\\le 2$ which is true by AM-gM", "Solution_9": "Very nice, Vasc. Are the following true?\r\n1) \\[ a^3+b^3+c^3+d^3+(abc)^2+(bcd)^2+(cda)^2+(dab)^2\\leq8. \\]\r\n2) \\[ a^3+b^3+c^3+d^3+(abc)^3+(bcd)^3+(cda)^3+(dab)^3\\leq8. \\]", "Solution_10": "[quote=\"Vasc\"]Using Jensen for the first, we need to prove a very nice inequality:\n\n$a^3+b^3+c^3+d^3+abc+bcd+cda+dab \\leq 8$\n\nfor $a^2+b^2+c^2+d^2=4$.\nCan someone post a solution to this nice and hard inequality?[/quote]\r\n\r\nYes, VASC. This is exactly the stronger inequality which I want to mention in this topic. It was appear in chapter I in my book, one exercises. My solution is using SOS method (and truly, it's nice but not hard).\r\n\r\nI hope someone gives the solution.", "Solution_11": "[quote=\"Albanian Eagle\"] $abcd=const$ and with $f(a)=a^3+\\frac{M}{a}$ while using (n-1)EV \nwe see that the maximum ocurrs when 3 variables are equal [/quote] The function is $f(u)=u^{3/2}+M.u^{-1/2}$ if we replace $a^2,b^2,c^2,d^2$ by $a,b,c,d$, respectively. But\r\n$g(x)=f'(1/x)$ is not convex, and then the method fails.", "Solution_12": "[quote=\"hungkhtn\"] My solution is using SOS method (and truly, it's nice but not hard).\nI hope someone gives the solution.[/quote]\r\nTo use SOS method, did you square the inequality or not ?", "Solution_13": "Of course, my inital problem is \r\n\\[ (a^3+b^3+c^3+abc+bcd+cda+dab)^2 \\le (a^2+b^2+c^2+d^2)^3 \\]\r\nBut the reason why I post the first problem (in square-root form) is that L: This one has a very nice solution only by AM-GM.", "Solution_14": "If I am not wrong then\r\n$a^3+b^3+c^3+d^3+abc+bcd+cda+dab \\leq 8\\Leftrightarrow$\r\n$\\Leftrightarrow(a^3+b^3+c^3+abc+bcd+cda+dab)^2 \\le (a^2+b^2+c^2+d^2)^3\\Leftrightarrow$\r\n$\\Leftrightarrow\\sum_{sym}(a-b)^2(3c^4+c^3d+3c^2d^2+cd^3+3d^4+3a^2b^2-c^2ab-d^2ab)\\geq0,$\r\nwhich obviously true.", "Solution_15": "[quote=\"arqady\"]If I am not wrong then\n$a^3+b^3+c^3+d^3+abc+bcd+cda+dab \\leq 8\\Leftrightarrow$\n$\\Leftrightarrow(a^3+b^3+c^3+abc+bcd+cda+dab)^2 \\le (a^2+b^2+c^2+d^2)^3\\Leftrightarrow$\n$\\Leftrightarrow\\sum_{sym}(a-b)^2(3c^4+c^3d+3c^2d^2+cd^3+3d^4+3a^2b^2-c^2ab-d^2ab)\\geq0,$\nwhich obviously true.[/quote]\r\n\r\nYes, Arqady. I only remember the coefficient all $S_{a,b},S_{a,c},...,S_{c,d}$ is non-negative, so the problem is easy.", "Solution_16": "Prom this result, We get a direct corollary\r\n\\[ (a+b+c+d)^6 \\ge 2^8(1+a^2)(1+b^2)(1+c^2)(1+d^2) \\]\r\nfor all $abcd=1$. I wait for a nice solution for this nice one.", "Solution_17": "[quote=\"hungkhtn\"]Of course, my inital problem is \n\\[ (a^3+b^3+c^3+abc+bcd+cda+dab)^2 \\le (a^2+b^2+c^2+d^2)^3 \\]\nBut the reason why I post the first problem (in square-root form) is that L: This one has a very nice solution only by AM-GM.[/quote]\r\nWe choose $a^2+b^2+c^2+d^2=4$. The inequality has the form.\r\n\\[ 4(a+b)+(c^2-cd+d^2+ab)(c+d-a-b)\\leq8. \\]\r\nI have tried to use AM-GM several ways but I failed: Grouping, supposing $a\\geq b\\geq c\\geq d$..... \r\n\r\nCan you give a hint?", "Solution_18": "From the condition, we have $a+b+c+d\\leq4$. WLOG, we suppose that $a+b\\geq c+d$. Then $a+b\\leq2$. Thus,\r\n\\[ 4(a+b)+(c^2-cd+d^2+ab)(c+d-a-b)\\leq8. \\]\r\nWe are done.\r\n\r\nIs that right?", "Solution_19": "[quote=\"pvthuan\"]From the condition, we have $a+b+c+d\\leq4$. WLOG, we suppose that $a+b\\geq c+d$. Then $a+b\\leq2$. Thus,\n\\[ 4(a+b)+(c^2-cd+d^2+ab)(c+d-a-b)\\leq8. \\]\nWe are done.\n\nIs that right?[/quote]\r\n\r\nNo, why $a+b\\le 2$?", "Solution_20": "[quote=\"hungkhtn\"][quote=\"pvthuan\"]From the condition, we have $a+b+c+d\\leq4$. WLOG, we suppose that $a+b\\geq c+d$. Then $a+b\\leq2$. Thus,\n\\[ 4(a+b)+(c^2-cd+d^2+ab)(c+d-a-b)\\leq8. \\]\nWe are done.\n\nIs that right?[/quote]\n\nNo, why $a+b\\le 2$?[/quote]\r\n\r\nYes, I was wrong. :oops:", "Solution_21": "Only by $AM-GM$ inequality, we can make a very nice solution for\r\n\\[ 64(a^3+bcd)(b^3+cda)(c^3+dab)(d^3+abc) \\le (a^2+b^2+c^2+d^2)^6 \\]\r\nI hope to see this one.", "Solution_22": "I wonder if a proof of the inequality shown in the preceding post is available elsewhere. If so, please give me a reference. I would have guessed that the inequality that was required to establish the initial inequality posed by hungkhtn would have required the exponent on the sum of the squares comprising the right hand side of the preceding inequality to be 5 , not 6. Unfortunately, the result is false when the 6 is replaced by a 5." } { "Tag": [ "integration", "trigonometry", "combinatorics unsolved", "combinatorics" ], "Problem": "$ {2}^{2n}\\equal{}C(0,0).C(2n,n)\\plus{}C(2,1).C(2(n\\minus{}1),n\\minus{}1)\\plus{}C(4,2).C(2(n\\minus{}2),n\\minus{}2)\\plus{}\\cdots\\plus{}C(2n,n).C(0,0)$, where $ n$ is a positive integer. Thanks in advance.", "Solution_1": "Okay there are two ways to do this - one is to consider \\[ \\int(\\sin^2\\theta \\plus{}cos^2\\theta)^n d\\theta\\] and look at it two ways. Another is to prove it combinatorially as follows: \r\nConsider the number of paths on the integer line starting from zero of length 2n. The number of such paths is $ 2^{2n}$. We count this quantity in another way. \r\nClaim: The number of paths that 'last cross' the origin after $ 2k$ moves is $ \\binom{2k}{k}\\binom{2n\\minus{}2k}{n\\minus{}k}$. \r\nProving the above claim relies on the task of finding a bijection between the number of paths of length $ 2k$ that end at $ 0$ and the number of paths of length $ 2k$ that never cross zero. \r\n Proving the last bit requires some work so I'll put it up after I work it out again. The general idea seems clear though." } { "Tag": [ "logarithms", "real analysis", "real analysis unsolved" ], "Problem": "pro 1) \r\n$1+y'^{2}=y.y''$\r\npro 2 ) \r\n$yy''+2y'^{2}=y^{2}-\\frac{yy'}{x}$\r\npro3 )\r\n$y'^{2}=4yy''+y^{2}$\r\nb) $xyy''-xy'^{2}=yy'$\r\nc) $y'^{2}+yy''-yy'=0$\r\nd)$(x^{2}+1)(y'^{2}-yy')=xyy'$\r\ne)$xyy''+xy'^{2}=2yy'$", "Solution_1": "Eeew! Nonlinear, I do remember a technique that goes like this, though...\r\n\r\n$1+y'^{2}=y\\cdot y''$, Let $v=y'$ so $y''=\\frac{dv}{dx}= \\frac{dv}{dy}\\cdot\\frac{dy}{dx}=v\\frac{dv}{dy}$ So this transforms the D.E. to\r\n\r\n$1+v^{2}=yv\\frac{dv}{dy}$ which is separable $\\frac{dy}{y}=\\frac{v}{1+v^{2}}dv$, so $\\ln y=\\frac{1}{2}\\ln(1+v^{2})+C \\Rightarrow Cy=(1+v^{2})^{1/2}\\Rightarrow v=\\sqrt{Cy^{2}-1}$, where $C$ isn't necessarily the same constant between all those steps.\r\n\r\nNow sub back $\\frac{dy}{dx}=\\sqrt{Cy^{2}-1}$ which is separable: $\\frac{dy}{\\sqrt{Cy^{2}-1}}=dx$, which isn't all that nice, so I'll let you finish it." } { "Tag": [ "trigonometry" ], "Problem": "Find all the roots of the equation $ x\\,(3\\minus{}x)^2\\equal{}1.$", "Solution_1": "Actually, the solutions are $ 4\\sin^2\\frac{\\pi}{18}.\\ 4\\sin^2\\frac{5\\pi}{18},\\ 4\\sin^2\\frac{7\\pi}{18}.$", "Solution_2": "Yeah, Mary Barton is right.\r\n\r\nSonnhard, did you read the problem wrong? Because I checked, and those numbers were not equivalent to Mary's. (They were complex...).", "Solution_3": "hello, i have also checked may solutions. For example on can show that \r\n$ 4\\sin\\left(\\frac{\\pi}{18}\\right)^2$ is \r\n$ \\left(\\frac{1}{2}(\\minus{}1\\minus{}i\\sqrt{3})(\\frac{1}{2}(\\minus{}1\\plus{}i\\sqrt{3}))^{\\frac{1}{3}}\\plus{}\\frac{1}{2}(\\frac{1}{2}(\\minus{}1\\minus{}i\\sqrt{3}))^{\\frac{1}{3}}(\\minus{}1\\plus{}i\\sqrt{3})\\right)^2$\r\nSonnhard.", "Solution_4": "Solutions for any answers would be appreciated :) ...me wants to know :maybe:", "Solution_5": "Is it just me, or do those messy solutions look like the cubic equation?", "Solution_6": "When in doubt, brute-force it. Then try to prove it nicely now that you have the solutions :D", "Solution_7": "For this problem, there is no need for brute force. There is actually an elegant solution that is no more than three steps long. See if you can find it. ;)", "Solution_8": "If we substitute $ x$ for $ y^2$ then $ x(3-x)^2 = (3y-y^3)$ looks almost like triple angle identity for sine. So let $ x=4\\sin^2k$. Then we have $ 4(3\\sinh - 4\\sin^2k)^2$ = $ 4(\\sin3k)^2$= $ 1$\r\n\r\nSo we have $ \\sin3k = \\frac{1}{2}$ => $ k=\\frac{\\pi}{18}, \\frac{5\\pi}{18}, \\frac{7\\pi}{18}$.", "Solution_9": "Yes, you\u2019ve got the right idea. :-D\r\n\r\n[hide=\"Full three-step solution\"][b][u]Step 1[/u]:[/b] $ \\sin{3\\theta}\\ \\equal{} \\ 3\\sin{\\theta} \\minus{} 4\\sin^3{\\theta}\\ \\equal{} \\ \\sin{\\theta}\\left(3 \\minus{} 4\\sin^2{\\theta}\\right)$\n\n[b][u]Step 2[/u]:[/b] $ 4\\sin^2{3\\theta}\\ \\equal{} \\ 4\\sin^2{\\theta}\\left(3 \\minus{} 4\\sin^2{\\theta}\\right)^2\\ \\equal{} \\ x\\,(3 \\minus{} x)^2$ where $ x \\equal{} 4\\sin^2{\\theta}$\n\n[b][u]Step 3[/u]:[/b] Put $ \\theta\\ \\equal{} \\ \\frac {\\pi}{18},\\ \\frac {5\\pi}{18},\\ \\frac {7\\pi}{18}$ (these are all in the range $ \\left[0,\\tfrac{\\pi}2\\right]$ so their sines are all positive and distinct; hence the squares of their sines are all distinct)[/hide]", "Solution_10": "[quote=\"Dr Sonnhard Graubner\"]hello, i have also checked may solutions. For example on can show that \n$ 4\\sin\\left(\\frac {\\pi}{18}\\right)^2$ is \n$ \\left(\\frac {1}{2}( \\minus{} 1 \\minus{} i\\sqrt {3})(\\frac {1}{2}( \\minus{} 1 \\plus{} i\\sqrt {3}))^{\\frac {1}{3}} \\plus{} \\frac {1}{2}(\\frac {1}{2}( \\minus{} 1 \\minus{} i\\sqrt {3}))^{\\frac {1}{3}}( \\minus{} 1 \\plus{} i\\sqrt {3})\\right)^2$\nSonnhard.[/quote]\r\n\r\nHow did you get this expression?\r\nBy the way, if we can't memorize the identity $ \\sin{3\\theta}\\ \\equal{} \\ 3\\sin{\\theta} \\minus{} 4\\sin^3{\\theta}\\ \\equal{} \\ \\sin{\\theta}\\left(3 \\minus{} 4\\sin^2{\\theta}\\right)$, does it mean we can't solve this problem? :roll:", "Solution_11": "you can derive it from the sine and cosine addition formulas (i think thats what they are called)", "Solution_12": "An easier way to derive it is expanding $ \\cos{3x}\\plus{}i\\sin{3x}\\equal{}(\\cos{x}\\plus{}i\\sin{x})^3$ and equate the real parts.", "Solution_13": "I know how to derive the identity $ \\sin{3\\theta}\\ \\equal{} \\ 3\\sin{\\theta} \\minus{} 4\\sin^3{\\theta}$, but what if we couldn't recognize it in this problem (I don't like to memorize formulas!)? :( :D" } { "Tag": [ "trigonometry" ], "Problem": "I wrote this out not too long ago.. Inspiration came up while doing English paper ( :rotfl: ). I'm not quite sure if this problem is solvable but there is an answer that works so.. hopefully, it is valid. :D \r\n\r\nNo Calculator.\r\n\r\nTriangle $ ABC$ has $ \\angle ACB \\equal{} 90^\\circ$ and $ AB \\equal{} \\sqrt{21}$. If $ BC \\equal{} 4$ and $ \\angle ABC \\equal{} \\theta,$ then $ \\tan \\theta \\equal{} \\sin 2 \\alpha \\cos \\alpha$ where $ 0^\\circ < \\alpha < 60^\\circ$. Find the measure of $ \\alpha$.", "Solution_1": "[quote=\"Silverfalcon\"]I wrote this out not too long ago.. Inspiration came up while doing English paper ( :rotfl: ). I'm not quite sure if this problem is solvable but there is an answer that works so.. hopefully, it is valid. :D \n\nNo Calculator.\n\nTriangle $ ABC$ has $ \\angle ACB \\equal{} 90^\\circ$ and $ AB \\equal{} \\sqrt {21}$. If $ BC \\equal{} 4$ and $ \\angle ABC \\equal{} \\theta,$ then $ \\tan \\theta \\equal{} \\sin 2 \\alpha \\cos \\alpha$ where $ 0^\\circ < \\alpha < 60^\\circ$. Find the measure of $ \\alpha$.[/quote]\r\n\r\n[hide]\n$ \\tan{\\theta}\\equal{}\\frac{\\sqrt{5}}{4}$ by pythag\nthen the rhs of equation 2 is $ 2(\\sin{\\alpha}\\minus{}\\sin^3{\\alpha})$\ncall $ \\sin{\\alpha}\\equal{}x$\n$ x^3\\minus{}x\\plus{}\\frac{\\sqrt{5}}{8}\\equal{}0$\n\nthis is a depressed cubic\nunfortunately i dont know how to solve those\n[/hide]", "Solution_2": "i'm getting 18 degrees", "Solution_3": "[quote=\"Altheman\"]i'm getting 18 degrees[/quote]\r\n\r\nThat's correct.\r\n\r\n[hide=\"Way I Created It\"]\nI was goofing around and realize how similar that $ \\sin 36$ and $ \\cos 18$ looked.\n\n$ \\sin 36^\\circ \\equal{} \\frac{\\sqrt{5 \\minus{} \\sqrt{5}}}{2 \\sqrt{2}}$\n\n$ \\cos 18^\\circ \\equal{} \\frac{\\sqrt{5 \\plus{} \\sqrt{5}}}{2 \\sqrt{2}}$\n\nMultiplying those two simplifies to $ \\frac{\\sqrt{5}}{4}$. But, working backwards, I got the cubic that was mentioned earlier and did not know how to proceed either. Knowing these numbers help but I don't know what's other way. [/hide]\r\n\r\nCan you show how?", "Solution_4": "$ x^3\\minus{}x\\plus{}\\frac{\\sqrt{5}}{8}\\equal{}0\\implies (x^3\\minus{}x)^2\\minus{}\\frac{5}{64}\\equal{}\\frac{1}{64}\\cdot (4x^2\\minus{}5)(4x^2\\minus{}2x\\minus{}1)(4x^2\\plus{}2x\\minus{}1)\\equal{}0$" } { "Tag": [ "LaTeX", "linear algebra" ], "Problem": "Hi every one. Any one here know any thing about maple?\r\n\r\nWhat is the maple command for writing for example\r\n\r\n2*X congruent 1 mod17 i.e 2*x=1mod17\r\nI searched the net but didn't find any thing", "Solution_1": "First of all , maybe this should have been better placed in superior algebra\r\nbut then again, there isn't really any official for maple questions( i once tried in the latex help forum but got no response)\r\n\r\nwhat exactly do you want\r\ndo you actually mean you just want to write it down as text ? \r\n\r\nthen i suggest you put this between dollar signs : 2^x \\equiv 1\\bmod{17} :\r\n$2^x \\equiv 1\\bmod{17}$\r\n\r\nor you could type this in maple\r\n\r\n2^x=1 mod 17\r\n\r\nyou'll have to elaborate I guess\r\n\r\nIt's nice to see someone from afghanistan here (are there others?)\r\nhttp://en.wikipedia.org/wiki/Afghanistan\r\nwhy is your flag different from the flag here?", "Solution_2": "$2x \\equiv 1\\bmod{17}$\r\n\r\nyes thats exactly what i want. I want the command equivalent to that in maple. and then ask maple to find $x$\r\n\r\nMy county's flag has changed a bit every now and then in the past but now its permanent. The administrator of this site should change Afghanistans flag to the current one.", "Solution_3": "actually in that case I suggest you type\r\n\r\nmsolve(2*x=1,17);\r\n\r\nor am I not understand your question correctly?" } { "Tag": [ "function", "integration", "inequalities", "real analysis", "real analysis unsolved" ], "Problem": "Let $f: [a,b]\\rightarrow R$ be an integrable function having the property that\r\n\r\n$0\\leq f(x)\\leq \\int_{a}^{x}f(t)dt$ for all $x\\in [a,b]$.\r\n\r\nProve that $f=0$.", "Solution_1": "This is indeed a [url=http://en.wikipedia.org/wiki/Gr%C3%B6nwall%27s_inequality]known result[/url]." } { "Tag": [ "trigonometry", "algebra", "polynomial", "function", "vector", "calculus", "calculus computations" ], "Problem": "Find the general solutions of \r\n(i) $ y'' \\plus{} 9y \\equal{} \\cos 3x$\r\n(ii) $ y'' \\minus{} 2y' \\plus{} y \\equal{} (x \\minus{} 1)e^x$\r\n\r\nIn general how do we solve for inhomogeneous ODEs when the particular solution isn't easy to guess?", "Solution_1": "You don't ;) But in these cases it actually isn't so hard: for the second one, you want to guess something like \"polynomial times $ e^x$\" where \"polynomial\" is of degree high enough that it isn't part of the homogeneous solution. For the first, you want to guess trigonometric function, but it turns out that the natural guesses are in the homogeneous solution. In this case, you do the same thing: replace your guess of \"trigonometric thing\" with \"polynomial times trigonometric thing\" where \"polynomial\" is of the minimal degree necessary to kick you out of the homogeneous solution.\r\n\r\nSee also Kent's recent posts here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=311377", "Solution_2": "I see - I had problems guessing for the first function but then I plugged it into WolframAlpha. Thanks for the link - looks enlightening!\r\n\r\nAlso, I wanted to ask about the solution space determined by solution vectors $ {y_1 \\choose y_1'}$ and $ {y_2 \\choose y_2'}$. What is the meaning of \"solution space/solution vector\" and why are all particular solutions of linear, second-order inhomogeneous ODEs of the form $ y_p \\equal{} u(x)y_1 \\plus{} v(x)y_2$ where $ y_1$ and $ y_2$ are homogeneous solutions?", "Solution_3": "hello, for (i) i have got\r\n$ y(x)\\equal{}C_1\\cos(3x)\\plus{}C_2\\sin(3x)\\plus{}\\frac{1}{18}\\cos(3x)\\plus{}\\frac{1}{6}x\\sin(3x)$.\r\nSonnhard.", "Solution_4": "I think we can combine the $ \\frac{1}{18} \\cos 3x$ into the $ C_1 \\cos 3x$? Otherwise I get the same answer :)" } { "Tag": [ "geometry", "rectangle", "perimeter" ], "Problem": "Twelve square tiles, one inch on a side, are arranged in a rectangle,\nwithout overlapping. What is the number of inches in the sum of all\npossible distinct perimeters?", "Solution_1": "We can have \r\n\r\n1,12\r\n2,6\r\n3,4\r\n\r\nThe distinct perimeters are $ 26\\plus{}16\\plus{}14\\equal{}\\boxed{56}$." } { "Tag": [ "algebra", "polynomial" ], "Problem": "If $b^2-4ac \\leq 0$, prove that $ax^2+bx+c \\geq 0$ for all reals $x$.\r\nWhat can you tell if $b^2-4ac \\geq 0$?", "Solution_1": "i dont think that first statement is necessarily correct. if a>0 then it would be, i believe.", "Solution_2": "Arkmastermind is correct. If the discriminant is less than 0, there are no roots to the equation, thus the graph of the polynomial will never touch the x-axis. So, depending on a's sign, $ax^2+bx+c$ will always either be greater or less than 0. If the discrimant is 0, the graph will touch the x-axis at one point, so the value will always be greater or less than or equal to 0.\r\n\r\nIf the discriminant is positive or zero, then we know $ax^2+bx+c$ is greater than/less than (depending on a's sign) or equal to $\\frac{-b}{2a}$", "Solution_3": "Oops... I forgot the condition of $a$. :blush:" } { "Tag": [], "Problem": "What common fraction is $ 50\\%$ less than $ \\frac{5}{8}$?", "Solution_1": "Hi;\r\n\r\nHow about:\r\n\r\n5 / 16", "Solution_2": "$ \\frac58(1\\minus{}\\frac12)\\equal{}\\frac58(\\frac12)\\equal{}\\boxed{\\frac{5}{16}}$." } { "Tag": [ "limit", "trigonometry", "calculus", "calculus computations" ], "Problem": "$ \\lim\\limits_{x \\to 0} \\frac {\\sin (a \\plus{} 2x) \\minus{} 2 \\sin (a \\plus{} x) \\plus{} \\sin a}{x^2} \\equal{} \\lim\\limits_{x \\to 0} \\frac {\\sin a \\cos 2x \\plus{} \\sin 2x \\cos a \\minus{} 2\\sin a \\cos x \\minus{} 2 \\cos a \\sin x \\plus{} \\sin a}{x^2} \\equal{} \\lim\\limits_{x \\to 0} \\frac { 2\\cos a \\sin x (\\cos x \\minus{} 1)}{x^2} \\equal{} 0$\r\nBut the true solution: $ \\minus{} \\sin a$", "Solution_1": "Well, The factored form ($ 2\\cos a \\sin x \\ldots$) doesn't seem to be correct. Here's my factoring:\r\n\r\n$ \\sin(a \\plus{} 2x) \\minus{} 2\\sin(a \\plus{} x) \\plus{} \\sin a \\equal{} \\sin(a \\plus{} 2x) \\minus{} \\sin(a \\plus{} x) \\plus{} \\sin a \\minus{} \\sin(a \\plus{} x) \\equal{}$\r\n$ 2\\cos\\left(a \\plus{} \\frac {3x}{2}\\right)\\sin\\left(\\frac {x}{2}\\right) \\minus{} 2\\cos\\left(a \\plus{} \\frac {x}{2}\\right)\\sin\\left(\\frac {x}{2}\\right) \\equal{}$\r\n$ 2\\sin\\left(\\frac {x}{2}\\right)\\left(\\cos\\left(a \\plus{} \\frac {3x}{2}\\right) \\minus{} \\cos\\left(a \\plus{} \\frac {x}{2}\\right)\\right) \\equal{}$\r\n$ 2\\sin\\left(\\frac {x}{2}\\right)\\left( \\minus{} 2\\sin\\left(\\frac {x}{2}\\right)\\sin(a \\plus{} x)\\right) \\equal{}$\r\n$ \\minus{} 4\\sin^2\\left(\\frac {x}{2}\\right)\\sin(a \\plus{} x)$\r\n\r\nwhich gives your limit $ \\lim_{x\\to 0} \\frac { \\minus{} \\sin^2\\left(\\frac {x}{2}\\right)\\sin(a \\plus{} x)}{\\left(\\frac {x}{2}\\right)^2} \\equal{} \\minus{} \\sin a$" } { "Tag": [ "calculus", "induction", "number theory unsolved", "number theory" ], "Problem": "The sequence $ a_1\\equal{}20$,$ a_2\\equal{}30$, $ a_{n\\plus{}1}\\equal{}3a_n\\minus{}a_{n\\minus{}1}$ .\r\n Find all n so that $ 1\\plus{}5a_na_{n\\plus{}1}$ is a perfect square.", "Solution_1": "[quote=\"shortlist\"]The sequence $ a_1 \\equal{} 20$,$ a_2 \\equal{} 30$, $ a_{n \\plus{} 1} \\equal{} 3a_n \\minus{} a_{n \\minus{} 1}$ .\n Find all n so that $ 1 \\plus{} 5a_na_{n \\plus{} 1}$ is a perfect square.[/quote]\r\n\r\nI'm sure there is a smarter method, but here is one :\r\n\r\nDirect resolution gives $ a_n \\equal{} (15 \\minus{} 5\\sqrt 5)\\left(\\frac {3 \\plus{} \\sqrt 5}2\\right)^n \\plus{}$ $ (15 \\plus{} 5\\sqrt 5)\\left(\\frac {3 \\minus{} \\sqrt 5}2\\right)^n$\r\n\r\nAnd then (after some calculus) $ 1 \\plus{} 5a_na_{n \\plus{} 1} \\equal{} 501 \\plus{} u_n^2$ where $ u_n \\equal{} (25 \\minus{} 5\\sqrt 5)\\left(\\frac {3 \\plus{} \\sqrt 5}2\\right)^n \\plus{} (25 \\plus{} 5\\sqrt 5)\\left(\\frac {3 \\minus{} \\sqrt 5}2\\right)^n$\r\n\r\nWe have $ u_1 \\equal{} 50$ and $ u_2 \\equal{} 100 > u_1$ and $ u_{n \\plus{} 2} \\equal{} 3u_{n \\plus{} 1} \\minus{} u_n$ and so $ u_n$ is a strictly increasing sequence made of positive integers\r\n\r\n$ u_1 \\equal{} 50$ and $ 1 \\plus{} 5a_1a_{2} \\equal{} 501 \\plus{} u_1^2 \\equal{} 3001$ is not a perfect square\r\n$ u_2 \\equal{} 100$ and $ 1 \\plus{} 5a_2a_{3} \\equal{} 501 \\plus{} u_2^2 \\equal{} 10501$ is not a perfect square\r\n$ u_3 \\equal{} 250$ and $ 1 \\plus{} 5a_3a_{4} \\equal{} 501 \\plus{} u_3^2 \\equal{} 63001 \\equal{} 251^2$ is a perfect square\r\n$ \\forall n > 3$ : $ u_n > u_3 \\equal{} 250$ and so $ u_n^2 < 501 \\plus{} u_n^2 < (u_n \\plus{} 1)^2$ and $ 501 \\plus{} u_n^2$ can not be a perfect square.\r\n\r\nHence the answer : $ \\boxed{n \\equal{} 3}$", "Solution_2": "wonderful\r\nThanhs you very much :lol: :roll:", "Solution_3": "[quote=\"pco\"][quote=\"shortlist\"]The sequence $ a_1 \\equal{} 20$,$ a_2 \\equal{} 30$, $ a_{n \\plus{} 1} \\equal{} 3a_n \\minus{} a_{n \\minus{} 1}$ .\n Find all n so that $ 1 \\plus{} 5a_na_{n \\plus{} 1}$ is a perfect square.[/quote]\n\nI'm sure there is a smarter method, but here is one :\n\nDirect resolution gives $ a_n \\equal{} (15 \\minus{} 5\\sqrt 5)\\left(\\frac {3 \\plus{} \\sqrt 5}2\\right)^n \\plus{}$ $ (15 \\plus{} 5\\sqrt 5)\\left(\\frac {3 \\minus{} \\sqrt 5}2\\right)^n$\n\nAnd then (after some calculus) $ 1 \\plus{} 5a_na_{n \\plus{} 1} \\equal{} 501 \\plus{} u_n^2$ where $ u_n \\equal{} (25 \\minus{} 5\\sqrt 5)\\left(\\frac {3 \\plus{} \\sqrt 5}2\\right)^n \\plus{} (25 \\plus{} 5\\sqrt 5)\\left(\\frac {3 \\minus{} \\sqrt 5}2\\right)^n$\n\nWe have $ u_1 \\equal{} 50$ and $ u_2 \\equal{} 100 > u_1$ and $ u_{n \\plus{} 2} \\equal{} 3u_{n \\plus{} 1} \\minus{} u_n$ and so $ u_n$ is a strictly increasing sequence made of positive integers\n\n$ u_1 \\equal{} 50$ and $ 1 \\plus{} 5a_1a_{2} \\equal{} 501 \\plus{} u_1^2 \\equal{} 3001$ is not a perfect square\n$ u_2 \\equal{} 100$ and $ 1 \\plus{} 5a_2a_{3} \\equal{} 501 \\plus{} u_2^2 \\equal{} 10501$ is not a perfect square\n$ u_3 \\equal{} 250$ and $ 1 \\plus{} 5a_3a_{4} \\equal{} 501 \\plus{} u_3^2 \\equal{} 63001 \\equal{} 251^2$ is a perfect square\n$ \\forall n > 3$ : $ u_n > u_3 \\equal{} 250$ and so $ u_n^2 < 501 \\plus{} u_n^2 < (u_n \\plus{} 1)^2$ and $ 501 \\plus{} u_n^2$ can not be a perfect square.\n\nHence the answer : $ \\boxed{n \\equal{} 3}$[/quote]\r\nI've read your post with much interest. I'm especially interested in how you came up with the idea of letting $ 1 \\plus{} 5a_na_{n \\plus{} 1} \\equal{} 501 \\plus{} u_n^2$. That was very nice!\r\nI've found a (in my opinion) \"nicer\" formula for $ a_n$: $ a_n \\equal{} 10 \\left ( \\left(\\frac {3 \\plus{} \\sqrt 5}2\\right)^{n\\minus{}1} \\plus{} \\left(\\frac {3 \\minus{} \\sqrt 5}2\\right)^{n\\minus{}1} \\right )$\r\nOne can see that $ u_n \\equal{} a_{n \\plus{} 1} \\plus{} a_{n}$, and then $ 1 \\plus{} 5a_na_{n \\plus{} 1} \\equal{} 501 \\plus{} (a_n \\plus{} a_{n \\plus{} 1})^2 \\iff a_{n \\plus{} 1}^2 \\minus{} 3a_na_{n \\plus{} 1} \\plus{} a_{n}^2 \\equal{} \\minus{} 500$.\r\nThis is very interesting as it looks like $ x^2 \\minus{} 3x \\plus{} 1 \\equal{} 0$! And one can actually quite easily see (i did it with induction) that if $ c_{n \\plus{} 2} \\equal{} ac_{n \\plus{} 1} \\plus{} bc_n$ then $ c_{n \\plus{} 2}^2 \\minus{} ac_{n \\plus{} 2}c_{n \\plus{} 1} \\minus{} bc_{n \\plus{} 1}^2 \\equal{} ( \\minus{} b)^n (c_{2}^2 \\minus{} ac_2c_1 \\minus{} bc_1^2)$.\r\n(In this case $ b \\equal{} \\minus{} 1$ so $ a_{n \\plus{} 1}^2 \\minus{} 3a_{n \\plus{} 1}a_n \\plus{} a_n^2$ is constant)\r\n\r\nCan I ask what motivation you had for writing $ 1 \\plus{} 5a_na_{n \\plus{} 1} \\equal{} 501 \\plus{} u_n^2$ ? It was not obvious (for me at least), and i used some time to prove it, so I'm curious if you just played around with the numbers, or if you had it in mind from the start.", "Solution_4": "Just as a reference - this is the second problem of Balkan Mathematical Olympiad, 2002.", "Solution_5": "[quote=\"Mathias_DK\"] ...Can I ask what motivation you had for writing $ 1 \\plus{} 5a_na_{n \\plus{} 1} \\equal{} 501 \\plus{} u_n^2$ ? It was not obvious (for me at least), and i used some time to prove it, so I'm curious if you just played around with the numbers, or if you had it in mind from the start.[/quote]\r\n\r\nNo :blush: , I had nothing in mind. Just computed $ 1 \\plus{} a_na_{n \\plus{} 1}$ and saw that this quantity was $ 1501 \\plus{} 25(5 \\minus{} \\sqrt 5)^2r_1^{2n} \\plus{} 25(5 \\minus{} \\sqrt 5)^2r_2^{2n}$ (where $ r_1$ and $ r_2$ are the roots of $ x^2 \\minus{} 3x \\plus{} 1 \\equal{} 0$)\r\n\r\nAnd then I saw that this expression could be written as a square + something, hence the result.\r\n\r\nAnd, btw, I think your simpler formula for $ a_n$ is wrong : it gives $ a_1 \\equal{} 30$ instead of $ 20$", "Solution_6": "[quote=\"pco\"]And, btw, I think your simpler formula for $ a_n$ is wrong : it gives $ a_1 \\equal{} 30$ instead of $ 20$[/quote]\r\nThanks for correcting me! I wrote $ n$ instead of $ n\\minus{}1$ by mistake." } { "Tag": [ "search", "inequalities unsolved", "inequalities" ], "Problem": "$ a,b,c > 0$, $ a\\plus{}b\\plus{}c\\equal{}3$ then prove:\r\n\r\n$ 4(a^2\\plus{}b^2\\plus{}c^2) \\plus{}abc \\ge 13$\r\n\r\n[hide=\"HINT\"]MV[/hide]", "Solution_1": "APMO of some year.Use search function. :wink:", "Solution_2": "[quote=\"Erken\"]APMO of some year.Use search function. :wink:[/quote]\r\n\r\nOk, I changed the problem.", "Solution_3": "[hide=\"Solution\"]$ \\sigma_1\\equal{}\\minus{}(a\\plus{}b\\plus{}c) \\\\ \\sigma_2\\equal{}ab\\plus{}bc\\plus{}ca \\\\ \\sigma_3\\equal{}\\minus{}abc \\\\ 4(a^2\\plus{}b^2\\plus{}c^2) \\plus{}abc \\ge 13 \\iff 8 \\sigma_2 \\plus{} \\sigma_3 \\le 23$\n\nBut:\n$ P(t)\\equal{}(t\\minus{}a)(t\\minus{}b)(t\\minus{}c)\\equal{}t^3\\plus{} \\sigma_1 t^2 \\plus{} \\sigma_2 t \\plus{} \\sigma_3\\equal{}t^3 \\minus{}3t^2 \\plus{} \\sigma_2 t \\plus{} \\sigma_3$\n\nLet $ t\\equal{}8$:\n$ P(8)\\equal{}8^3 \\minus{} 3 \\cdot 8^2\\plus{}8 \\sigma_2 \\plus{} \\sigma_3\\equal{}320\\plus{}8\\sigma_2\\plus{}\\sigma_3\\equal{}(8\\minus{}a)(8\\minus{}b)(8\\minus{}c) \\le (\\frac{(8\\minus{}a)\\plus{}(8\\minus{}b)\\plus{}(8\\minus{}c)}{3})^3\\equal{}343$\n\nSo:\n$ 8 \\sigma_2 \\plus{} \\sigma_3 \\le 23$[/hide]\r\n\r\nMy first post :)", "Solution_4": "[quote=\"FOURRIER\"]$ a,b,c > 0$, $ a \\plus{} b \\plus{} c \\equal{} 3$ then prove:\n\n$ 4(a^2 \\plus{} b^2 \\plus{} c^2) \\plus{} abc \\ge 13$\n\n[hide=\"HINT\"]MV[/hide][/quote]\r\n$ 4(a^2 \\plus{} b^2 \\plus{} c^2) \\plus{} abc \\ge 13$\r\n$ \\Longleftrightarrow 36(a^2\\plus{}b^2\\plus{}c^2)(a\\plus{}b\\plus{}c)\\plus{}27abc \\geq 13(a\\plus{}b\\plus{}c)^3$\r\n$ \\Longleftrightarrow 23(a^3\\plus{}b^3\\plus{}c^3) \\geq 3[a^2(b\\plus{}c)\\plus{}b^2(c\\plus{}a)\\plus{}c^2(a\\plus{}b)]\\plus{}51abc$\r\nIt's right because :\r\n$ 17(a^3\\plus{}b^3\\plus{}c^3) \\geq 51abc$\r\nand $ 6(a^3\\plus{}b^3\\plus{}c^3) \\geq 3[a^2(b\\plus{}c)\\plus{}b^2(c\\plus{}a)\\plus{}c^2(a\\plus{}b)]$", "Solution_5": "[b]lemma:[/b]for $ x,y,z>0$ and $ x\\plus{}y\\plus{}z\\equal{}3$ we have $ x^2\\plus{}y^2\\plus{}z^2\\plus{}xyz\\geq 4$.\r\n\r\nnow according to this lemma,its sufficient to show that:\r\n\r\n$ x^2\\plus{}y^2\\plus{}z^2\\geq 3$\r\n\r\nwhich is true since:\r\n\r\n$ x^2\\plus{}y^2\\plus{}z^2\\geq \\frac{(x\\plus{}y\\plus{}z)^2}3\\equal{}3$", "Solution_6": "we can write A=4[(a+b+c)\u00b2-2(ab+bc+ac)]+abc;\r\n we have A=36-8(ab+bc+ac)+abc\r\n\r\nI think it is easier to prove!!" } { "Tag": [ "geometry", "rhombus", "trapezoid" ], "Problem": "Prove that any quadrilateral can be divided into at least 8 isosceles triangles.", "Solution_1": "Is this from WOOT? (and also, do you want at least 8 or for all greater than 8)?\r\n\r\nAs a hint, trying experimenting with splitting an arbitrary triangle into 4 isosceles triangles.", "Solution_2": "I think its from woot (I don't do woot but I was given the problem by a friend). It says 8 or more.\r\nGOT IT!!!! zomg this is the first problem at this level that I solved.\r\n[hide]\nThis is sorta an algorithmic process.\nWe know that any quadrilateral can be split into 2 triangles. Either of these triangles can be made into 4 isoceles triangles by drawing the altitude from one of the vertices, creating 2 right triangles which can be turned into 2 isoceles triangles by drawing the median to the hypotenuse. That's for 8.\n\nNow how do we get more than 8?\n\nWe can see that there are a few things we can do with modifying the steps we perform before and after that can change the isoceles triangle count.\n\nIf we take the original quadrilateral and draw in a line segment somewhere so that it creates another isoceles triangle, we have an isoceles triangle and a new quadrilateral left (which we can split by the previous method into 8 or more). This means we can add however many multiples of 1 that we want, and that solves the problem.\nThis is assuming however, that no sides are congruent. Special cases:\n2 sides congruent: easy to find a way to draw quadrilateral so chopping off an isoceles triangle will yield a quadrilateral instead of triangle...find a different side to cut.\n3 sides congruent: easy also, if last side is longer than the other 3, chop of isoceles triangle with congruent sides that are the same length as the 3 congruent sides. If shorter, ''''''''''''''''''' that are the same length as the non congruent side.\n4 sides: It is a rhombus, very easy to make 8 with. It might be a little annoying with the fact that its already got 2 isoceles triangles. That doesn't matter though, because if its not 60-120-60-120, we can draw an isoceles triangle with congruent sides NOT congruent to the sides of the rhombus, by drawing in a segment from one of the vertices with the greater angle to one of the opposite segments so it forms an isoceles triangle with a totally new side length, and a new irregular quadrilateral. If it is a square, its really too easy because any triangle cut from segments drawn to existing vertices will be a 45-45-90 and when you split those in half you add 1 to the count. If it is the 60 120 60 120, 8 can be made obviously. 10 or more can be made by pinching off a smaller equilateral triangle (side length must be irrational when compared to rhombus side length, go play with drawings to see why) from the rhombus to create a trapezoid, and 2 equilateral triangles.\nThe trapezoid can now be split by the original method into 8 or more, which added to the 2 equilateral triangles makes 10 or more. For 9, we can specifically pick the pinched off equilateral so that its side length is 3/5 the rhombus' side length. If you do some drawing you will see how it can be made into 7 small equilateral triangles.\n\n[/hide]\r\nI know its not THAT clear because I didn't draw in those fancy figures however you do that...but this method works for all quadrilaterals so whatever :P" } { "Tag": [ "calculus", "integration", "trigonometry", "calculus computations" ], "Problem": "Let $ f(x) \\equal{} cosx \\minus{} \\int^x_0f(t)f(x \\minus{} t)dt$\r\nFind $ f(x) \\plus{} f^\"(x)$", "Solution_1": "I think your problem is :\r\n\r\nLet $ f(x)\\equal{}\\cos x\\minus{}\\int_0^x (x\\minus{}t)f(t)\\ dt$.\r\n\r\nFind $ f(x)\\plus{}f''(x)$.\r\n\r\n[hide=\"Answer\"]-\\cos x[/hide]" } { "Tag": [ "topology" ], "Problem": "Let M be a compact metric space. Show that two point a,b $ \\in$ M belong to distinct connected components of M $ \\Longleftrightarrow$ exists a split M = A $ \\cup$ B, with a $ \\cup$ A and b $ \\cup$ B. Then, conclude that in a metric space, the connected component of a point is a intersection of open-close subsets that have one point.", "Solution_1": "Use the fact that if two connected subsets intersect then their union is connected." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Given $ a, b, c \\geq\\ 0$. Prove that:\r\n$ \\frac{a^2\\plus{}b^2\\plus{}c^2}{ab\\plus{}bc\\plus{}ca} \\geq\\ \\frac{a^2}{a^2\\plus{}2bc} \\plus{} \\frac{b^2}{b^2\\plus{}2ca} \\plus{} \\frac{c^2}{c^2\\plus{}2ab}$", "Solution_1": "[quote=\"nguoivn\"]Given $ a, b, c \\geq\\ 0$. Prove that:\n$ \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca} \\geq\\ \\frac {a^2}{a^2 \\plus{} 2bc} \\plus{} \\frac {b^2}{b^2 \\plus{} 2ca} \\plus{} \\frac {c^2}{c^2 \\plus{} 2ab}$[/quote]\r\n$ \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca} \\geq\\sum_{cyc}\\frac {a^2}{a^2 \\plus{} 2bc}\\Leftrightarrow\\sum_{cyc}\\left(\\frac{a^2}{ab\\plus{}ac\\plus{}bc}\\minus{}\\frac {a^2}{a^2 \\plus{} 2bc}\\right)\\geq0\\Leftrightarrow$\r\n$ \\Leftrightarrow\\sum_{cyc}\\frac{a^2(a\\minus{}b)(a\\minus{}c)}{a^2\\plus{}2bc}\\geq0.$\r\nLet $ a\\geq b\\geq c.$ Then $ \\frac{a^2}{a^2\\plus{}2bc}\\geq\\frac{b^2}{b^2\\plus{}2ac}$ and $ a\\minus{}c\\geq b\\minus{}c.$\r\nThus, $ \\sum_{cyc}\\frac{a^2(a\\minus{}b)(a\\minus{}c)}{a^2\\plus{}2bc}\\geq(a\\minus{}b)(b\\minus{}c)\\left(\\frac{a^2}{a^2\\plus{}2bc}\\minus{}\\frac{b^2}{b^2\\plus{}2ac}\\right)\\plus{}$\r\n$ \\plus{}\\frac{c^2(c\\minus{}a)(c\\minus{}b)}{c^2\\plus{}2ab}\\geq0.$" } { "Tag": [], "Problem": "Jane and John, with masses of 50 kg and 60 kg, respectively, stand on a frictionless surface 10 m apart. John pulls on a rope that connects him to Jane, giving Jane an acceleration of 0.92 m/s\u00b2 toward him. (a) What is John's acceleration? (b) If the pulling force is applied constantly, where will Jane and John meet?\r\n\r\nMy classmate said that $ m_{John} a_{John} \\equal{} m_{Jane} a_{Jane}$, but I don't really see why that's true. Shouldn't the acceleration of John be $ (F\\minus{}T)/a$ where F is the force he pulls Jane and T the tension? And $ T \\equal{} m_{Jane} a$ where $ a$ is the same for both Jane and John?\r\nI just don't understand the concept of tension well; could anyone explain it to me? Thanks.", "Solution_1": "[quote=\"mathwizarddude\"]My classmate said that $ m_{John} a_{John} = m_{Jane} a_{Jane}$, but I don't really see why that's true. Shouldn't the acceleration of John be $ (F - T)/a$ where F is the force he pulls Jane and T the tension? And $ T = m_{Jane} a$ where $ a$ is the same for both Jane and John?\n.[/quote]\r\nu r cent pc correct :) \r\njust tell where's the problem?? :)", "Solution_2": "[quote=\"mathwizarddude\"]My classmate said that $ m_{John} a_{John} \\equal{} m_{Jane} a_{Jane}$, but I don't really see why that's true.[/quote]\n\nNewton's 3rd Law\n\n[quote]Shouldn't the acceleration of John be $ (F \\minus{} T)/a$ where F is the force he pulls Jane and T the tension?[/quote]\n\nTypo; you probably mean $ (F\\minus{}T)/m$ ;)\n\n[quote]And $ T \\equal{} m_{Jane} a$ where $ a$ is the same for both Jane and John?[/quote]\r\n\r\nI think so. I don't understand it very well either though :(" } { "Tag": [ "AMC", "AIME", "modular arithmetic", "geometry", "3D geometry", "number theory" ], "Problem": "Here's the link if the way it copied out is confusing: http://www.artofproblemsolving.com/Wiki/index.php/1989_AIME_Problems/Problem_9\r\n\r\nThe problem is:\r\n\r\nOne of Euler's conjectures was disproved in the 1960s by three American mathematicians when they showed there was a positive integer such that 133^5+110^5+84^5+27^5=n^{5}. Find the value of n. \r\n\r\nThe solution is:\r\nBy Fermat's Little Theorem, we know {n^{5}} is congruent to n modulo 5. Hence,\r\n3 + 0 + 4 + 7 \\equiv n\\pmod{5}\r\n4 \\equiv n\\pmod{5}\r\n\r\nContinuing, we examine the equation modulo 3,\r\n-1 + 1 + 0 + 0 \\equiv n\\pmod{3}\r\n0 \\equiv n\\pmod{3}\r\n\r\nThus, n is divisible by three and leaves a remainder of four when divided by 5. It's obvious that n>133, so the only possibilities are n = 144 or n = 174. It quickly becomes apparent that 174 is much too large, so n must be 144. \r\n\r\nA couple issues:\r\n\r\n1. Where did modulo 3 come from? Ok, modulo 5 came from the theorem and the equation being in the 5th power, but where did 3 come from?\r\n\r\n2. Barring that, they said the only possibilities are 144 and 174. What about 159, 189, 204, etc?", "Solution_1": "1) mod 3 is good when dealing with squares, cubes, etc.\r\n\r\n2) $ 133^5 \\plus{} 110^5 \\plus{} 84^5 \\plus{} 27^5 \\equal{} n^5 < 133^5 * 4$ so $ n < 175.494...$\r\n\r\nand 159 is out of the question because n must be even.", "Solution_2": "Thanks, your answer to my second question made perfect sense.\r\n\r\nHowever, about the first question, does that mean mod 3 is good for all powers? Is there a theorem about this?", "Solution_3": "No, not really. Mod 4 is also really good when dealing with squares(since the residues are just 0 and 1); mod 8 too. But in this problem, I think we could've used any mods, except mod 3 was more convenient because 2 of the 4 numbers on the LHS were divisible by 3. But try using 2 other mods to see if it still works, like 4 and 7." } { "Tag": [ "limit", "real analysis", "real analysis unsolved" ], "Problem": "Let $ \\sum_{n\\equal{}0}^{\\infty} a_n z^n$ with radius of convergence $ r>0$ and $ a_n\\geq 0$ non increasing, and $ \\lim_{\\infty} a_n \\equal{}0$.\r\nClearly if $ s>r$, $ (a_n s^n)_{n \\geq 0}$ cannot be bounded, but I cannot show that $ \\lim_{\\infty} a_n s^n \\equal{} \\plus{}\\infty$, and I suspect that it is not always the case, but I cannot prove that either.\r\nAny idea?", "Solution_1": "It is indeed not always the case. That monotonicity makes it tricky, but here's an example:\r\n\r\n$ a_n \\equal{} 2^{ \\minus{} 2^k}$ for $ 2^{k \\minus{} 1} < n\\le 2^k$. The values $ a_n^{ \\minus{} 1/n}$ range from $ 2$ to $ 4$ in each block, so the radius of convergence is 2 but there are infinitely many $ n$ for which $ a_n\\cdot 3^n < 1$.", "Solution_2": "Thanks jmerry,\r\nIs there some (general) conditions we could add to $ a_n$ so that the result $ \\lim_{\\infty} a_n s^n \\equal{} \\plus{} \\infty$ becomes true?\r\n\r\nI thought about decreasing instead of non increasing but I'm afraid it is not going to work either.", "Solution_3": "No, making it strictly monotone wouldn't help; just take the antiderivative of the example I just used.\r\n\r\nThere's nothing there except to essentially assume the result." } { "Tag": [ "Harvard", "college" ], "Problem": "Since school is starting soon for most ppl, how about we post tips to make school easier? \r\n\r\n :?:", "Solution_1": "Kinda obvious, but don't procrastinate.", "Solution_2": "Relax, and don't do your homework last minute (commonly called procrastination).", "Solution_3": "relax, chill out, don't ever take school seriously\r\n\r\nwatch out for classes you choose. don't take any that you know has a reputation of doing unneccesary amounts of work or just bs work in general. if you're in high school, there shouldn't ever be a reason to have to do homework at home. occasionally you might not feel like doing an essay at school and decide to do it at home, but that's at your own discretion. most school hw cna be done in 10 minutes or so and you can easily do it in hte period directly before the class that it's due in. this way, you cna cycle all your homework so you're doign it all a period before it's due. and what do youget? no homework, easy A's, easy high school.", "Solution_4": "@ Elemennop - Commonly known as procrastination...\r\n\r\nDon't get stressed out about stuff, just take it easy but never leave work unfinished until the last minute. Once you start procrastinating, it's hard to get back into a normal cycle. If you're getting too much homework (judge by the normal amount, not by [i]your[/i] idea of a normal amount), then try to figure out how to do it faster instead of just wasting all your time on schoolwork and having no time left for other more important things.", "Solution_5": "[quote=\"i_like_pie\"]@ Elemennop - Commonly known as procrastination...\n[/quote]\r\n\r\nwhatever yo uwant to call it, it's still an extremely good method for getting schoolwork done and i don't see why people act as if it's a bad thing to do homework not at home.", "Solution_6": "What Elemenop is suggesting - do all of your homework at school and in class, and never any of it at home - may well work in high school. But habits can be hard to break in the long run, and you need to remember that it won't work in college.\r\n\r\nIt's reasonably common for a college class to cover in one semester, in 45 to 60 class hours, a subject that would have taken a whole year and 160 or more class hours in high school. No one's going to give you any in-class time for finishing your homework. And we start expecting more from you than you high school teachers did - more depth, more independence. Yes, you might actually have to read the textbook yourself.", "Solution_7": "-Prioritize. If you have a lot of homework and there is no way you can get it done that night (sports, etc.) then put do the toughest homework and leave the easier ones for that following school day (period before it is due).\r\n\r\n-Know your teachers a bit. Obviously, doing your next period's homework needs a bit of skill and a teacher that is amazing at spotting things behind his back will be a problem to your procrastinating needs. It also helps having friends in higher grades so they can warn you about teachers that may be mean.", "Solution_8": "Basically what elemmenop said, plus find out the workload before you sign up for any given class, (why take AP english in 12th grade...is it really gonna get you into Harvard??)", "Solution_9": "Here's what I'm doing: Find out what books you're going to be reading in English class this year, and then read as many as you can during the last few long days of summer vacation. This way, you won't have to spend as much time reading class books during the school year, and it will reduce the overall stress level.", "Solution_10": "Taking really good notes and reviewing them 10 minutes every day is a really good idea. \r\n\r\nStaying orgainized is important too. Do you hw and then play. \r\n\r\n Do what you know will please your teachers, getting on their good side is important.\r\n\r\nPrepare for pop quizes.", "Solution_11": "[quote=\"aanjohri\"] Do what you know will please your teachers, getting on their good side is important.[/quote]\r\n\r\n\r\nYeah umm anybody who's hung out with me enough knows that I've learned this lesson the very hard way...", "Solution_12": "[quote=\"aanjohri\"]Taking really good notes and reviewing them 10 minutes every day is a really good idea. \n\nStaying orgainized is important too. Do you hw and then play. \n\n Do what you know will please your teachers, getting on their good side is important.\n\nPrepare for pop quizes.[/quote]\r\n\r\nuh...why are u answering yr own question?\r\n\r\n-dont cram the night b4 on test\r\n-get to become friends w/ yr teacher. it can really be good help\r\n-DONT BE AFRAID TO ASK FOR HELP", "Solution_13": "Just do things as early as possible, NOT quickly.\r\n\r\nLook at your notes everyday for maybe 15 minutes. For a few days, it will start to stick in your head. Cramming the night before won't get you an A. Only as high as a B. I learned that the hard way.", "Solution_14": "[quote=\"Bicameral\"]Cramming the night before won't get you an A. Only as high as a B.[/quote]\r\n\r\ni don't really care what point you're trying to make, but lying isn't going to help you get it across.", "Solution_15": "Procrastination is fun!\r\n\r\nYou should take notes, save them, and study them.\r\n\r\nTry to enjoy the class, the teacher usually notices it (Note: You may get more work for this)\r\n\r\nAsk for help, it will never do anything wrong, except if you ask the mafia.\r\n\r\n\r\n\r\nMost of these things I never really did but procrastinate, and still got all As.", "Solution_16": "Since most of what I have to say has already been said, here's something else to think about.\r\n\r\nI respectfully disagree with those against procrastination.\r\n\r\nWhile there's the bad kind of procrastination (I'm too lazy to do it now, I'll do it \"later\"), controlled procrastination is actually very useful. For example, if you know you can easily finish some really stupid assignment in 45 minutes (especially writing), start it with 40 minutes of time left. Now why would someone do that?\r\n\r\nIt develops a lot of skills that aren't developed by blowing it off on time:\r\n1. [b]It teaches you to be more creative [/b]- you need to pull out the same creativity and ideas under time pressure\r\n2. [b]It improves your English[/b] - by being forced to express your ideas into clear sentences under time pressure, your English communication skills rise dramatically\r\n3. [b]It improves your fast decision making[/b] - so there are two kinds of fast decision making: intelligent and rash. When you're writing under pressure, you have to make fast, intelligent decisions. Over time, this will improve the quality of your fast decisions to make them less rash.\r\n4. [b]It teaches you to be more focused[/b] - because of the time pressure and the fact that the assignment will be graded, you are forced to be more focused - a valuable skill\r\n5. Finally, [b]in the long run, it saves you time[/b] because you complete your work in a compact timeframe with no distractions. This is certainly faster than doing homework while on AIM or watching TV (or AoPSing! xD)\r\n\r\nNow if you can force yourself to do the assignment like that [b]without[/b] saving it until teh last minute, then that is, of course, much better. But how many people can do that?", "Solution_17": "[quote=\"The Zuton Force\"]But how many people can do that?[/quote]\r\nI used to do that before, and now I can do homework very quickly, so your reasoning is absolutely correct.\r\n\r\nI never thought of it as procrastination, however, because usually that refers to the \"I'll do it later\" category.", "Solution_18": "[quote=\"mz94\"]\nuh...why are u answering yr own question?\n[/quote]\r\n\r\ni dunno, i had some ideas, so i posted them", "Solution_19": "-Procrastinate. Rush in everything at the last second, you'll still get a perfect on that test.\r\n-Don't study. This stresses you out for tests and lowers your grade. Only look something up if you're sure you can't remember and that it will be on the test. Then repeat that twenty times to yourself and close the book.\r\n-Don't get expelled.\r\n\r\nThis is only half serious, but i do endorse all three of these... well maybe not the first.", "Solution_20": "[quote=\"The Zuton Force\"]Since most of what I have to say has already been said, here's something else to think about.\n\nI respectfully disagree with those against procrastination.\n\nWhile there's the bad kind of procrastination (I'm too lazy to do it now, I'll do it \"later\"), controlled procrastination is actually very useful. For example, if you know you can easily finish some really stupid assignment in 45 minutes (especially writing), start it with 40 minutes of time left. Now why would someone do that?\n\nIt develops a lot of skills that aren't developed by blowing it off on time:\n1. [b]It teaches you to be more creative [/b]- you need to pull out the same creativity and ideas under time pressure\n2. [b]It improves your English[/b] - by being forced to express your ideas into clear sentences under time pressure, your English communication skills rise dramatically\n3. [b]It improves your fast decision making[/b] - so there are two kinds of fast decision making: intelligent and rash. When you're writing under pressure, you have to make fast, intelligent decisions. Over time, this will improve the quality of your fast decisions to make them less rash.\n4. [b]It teaches you to be more focused[/b] - because of the time pressure and the fact that the assignment will be graded, you are forced to be more focused - a valuable skill\n5. Finally, [b]in the long run, it saves you time[/b] because you complete your work in a compact timeframe with no distractions. This is certainly faster than doing homework while on AIM or watching TV (or AoPSing! xD)\n\nNow if you can force yourself to do the assignment like that [b]without[/b] saving it until teh last minute, then that is, of course, much better. But how many people can do that?[/quote]\r\n\r\nI think procrastination isn't that bad, as long as you can handle heavy work on some days. I did it last year, and got all A's. :P", "Solution_21": "equation: let a-z be 1-26 respectively.\r\n\r\nprocrastination = 16+18+15+3+18+1+19+20+9+14+1+20+9+15+14=192. Procrastination>hard work=98>knowledge=96.\r\n\r\nTherefore, procrastination is bigger than knowledge.\r\n\r\nDo you see the point that I'm trying to make here?\r\n\r\n\r\n[hide=\"if you don't,\"]procrastination is obviously less than knowledge.[/hide]", "Solution_22": "Shouldn't you multiply those...isn't this a math forum...sometimes i'm not so sure", "Solution_23": "Ummm.... did a moderator move this thread to games and fun factory or did i actually post this here and am completely clueless?", "Solution_24": "Lol it suddenly got spammy I think." } { "Tag": [ "function" ], "Problem": "If it is 12 o clock (noon) and we need to go to a fuction that starts at 7, but need to arrive there 10 minutes early and have to go to the library before (which is 20 minutes from your house and 10 minutes from the fuction) and you need 12 minutes at the library, what time should you leave home?", "Solution_1": "[hide]7.00pm-10minutes=6.50pm\n6.50-(20minutes+10minutes)=6.20pm\n6.20pm-12minutes=6.08pm.\nwe should leave at 6.08pm.[/hide]", "Solution_2": "[hide]6:08, cool question, im kinda noob but this is ok for me lol[/hide]", "Solution_3": "[hide]$10 minutes+ 20 minutes + 10 minutes+ 12 minutes= 52 minutes$\n$7: 00 pm - 52 minutes = 6: 08$[/hide]", "Solution_4": "Please hide your answers so that people can do the problem without looking at the answer first (temptation). It's not cool to do it for eveyone. :) :D", "Solution_5": "[hide]7:00-10-10-20-12=6:08[/hide]" } { "Tag": [ "function", "algebra proposed", "algebra" ], "Problem": "let \u03a8 is a function \u03a8: ]-infini ;-2[ ------> R\r\n\r\n such that f(x) = x^2/(x+1)\r\n\r\nis this function injective? :blush:", "Solution_1": "[quote=\"greatestmaths\"]let \u03a8 is a function \u03a8: ]-infini ;-2[ ------> R\n\n such that f(x) = x^2/(x+1)\n\nis this function injective? :blush:[/quote]\r\n\r\nYes, because $ f'(x)\\equal{}\\frac{x(x\\plus{}2)}{(x\\plus{}1)^{2}}$ shows that $ f(x)$ is strictly increasing for $ x\\in]\\minus{}\\infty,\\minus{}2[$", "Solution_2": "[hide]\n\\[ \\frac{a^{2}}{a\\plus{}1}\\equal{}\\frac{b^{2}}{b\\plus{}1}\\Rightarrow a^{2}b\\plus{}a^{2}\\equal{}ab^{2}\\plus{}b^{2}\\]\n\\[ \\Rightarrow (a\\minus{}b)(ab\\plus{}a\\plus{}b)\\equal{}0\\]\nEither $ a\\equal{}b$, or $ ab\\plus{}a\\plus{}b\\equal{}0$. In the second case, this would mean $ (a\\plus{}1)(b\\plus{}1)\\equal{}1$. Now $ a,b\\leq\\minus{}2$ imply $ a\\plus{}1,b\\plus{}1\\leq\\minus{}1$ so $ (a\\plus{}1)(b\\plus{}1)\\geq 1$ with equality only when $ a\\equal{}b\\equal{}\\minus{}2$. In this case, $ a\\equal{}b$ anyway.\n\nTherefore, $ a,b\\leq\\minus{}2$ and $ \\frac{a^{2}}{a\\plus{}1}\\equal{}\\frac{b^{2}}{b\\plus{}1}$ imply $ a\\equal{}b$, so the function is injective.\n[/hide]" } { "Tag": [ "calculus", "derivative", "function", "real analysis", "real analysis theorems" ], "Problem": "Darboux Theorem states that the derivative of a function has the Intermediate Value Property\r\n--------------------------------------------------------------------------\r\nDarboux property. \r\n\r\nLet E be an interval. It is said that the function f:E-->R has the property of Daboux on E, if for any aR has the property of Daboux on E, if for any a0 and f'(b)<0. (the similar argument works for f'(a)<0 and f'(b)>0).\r\n\r\nf(x) is differential on [a,b] implies that it is continuous. Thus it has a maximum value. f(a) is not the maximum value since f'(a)>0, f(x) is increasing at a. Simiarly, f(b) is not either since f(x) decreases at b.\r\nHence there is a \\xi \\in (a,b) s.t. f(\\xi) is the maximum value. By Fermat's theorem, f(\\xi)=0.\r\n\r\nNow back to the general cases.\r\n\r\nsuppose c is between f'(a) and f'(b).\r\nLet g(x) = f(x) - cx, g'(x) = f'(x) - c. g'(a)g'(b) < 0. we know from the lemma that there is a \\xi such that g'(\\xi) = 0, which is f'(\\xi) = c.", "Solution_4": "Can someone tell me what Fermat's theorem say (maybe we have the same in french in a different name)", "Solution_5": "FERMAT THEOREM: suppose f(x) reaches its maximum at x_0 (x_0 is an interior point of a interval) and f'(x_0) exists, then f'(x_0) = 0." } { "Tag": [ "AMC" ], "Problem": "Define $ a@b \\equal{} ab \\minus{} b^{2}$ and $ a\\#b \\equal{} a \\plus{} b \\minus{} ab^{2}$. What is $ \\frac {6@2}{6\\#2}$?\r\n\r\n$ \\textbf{(A)}\\ \\minus{} \\frac {1}{2}\\qquad \\textbf{(B)}\\ \\minus{} \\frac {1}{4}\\qquad \\textbf{(C)}\\ \\frac {1}{8}\\qquad \\textbf{(D)}\\ \\frac {1}{4}\\qquad \\textbf{(E)}\\ \\frac {1}{2}$", "Solution_1": "[hide]\n\\[a@b=ab-b^{2}=b(a-b)\\Rightarrow 6@2=2(6-2)=2(4)=8 \\]\nNow,\n\\[a\\#b=a+b-ab^{2}\\Rightarrow 6@2=6+2-6(2)(2)=6+2-24=-16 \\]\nHence,\n\\[\\frac{6@2}{6\\#2}=\\frac{8}{-16}=\\frac{-1}{2}\\]\nSo, the answer is $A$ (by the way, A was supposed to be $\\frac{-1}{2}$, not $\\frac{1}{2}$). [/hide]", "Solution_2": "$\\dfrac{6@2}{6\\#2} = \\dfrac{6*2-2^2}{6+2-6*2^2}=\\dfrac{12-4}{8-24}=\\dfrac{8}{-16}=\\boxed{-\\dfrac{1}{2}~~\\text{(A)}}$", "Solution_3": "After just substituting values you get $-\\frac{1}{2}$ Which as A", "Solution_4": "[hide=Solution]\n$\\frac{6@2}{6\\#2}=\\frac{6\\cdot2-2^2}{6+2-6\\cdot2^2}=\\frac{8}{-16}=\\boxed{\\textbf{(A)}-\\frac12}$.\n[/hide]", "Solution_5": "[hide=sol]6@2=6*2-2^2=12-4=8\n6#2=6+2-6*2^2=-16\n6@2/6#2=-1/2\nA[/hide]" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $a,b,c>0$.\r\n\r\nProve that\r\n$\\sum_{cyc}a^6 + 2 \\sum_{cyc}a^5 b + \\sum_{cyc}a^4 b^2 \\geq \\sum_{cyc} a^2 b^4 +\\sum_{cyc}a^3 b^3 +\\sum_{sym} a^3 b^2 c$.", "Solution_1": "Look like a part in Japan 1997's proof for ineq problem.\r\nTry AM-GM...", "Solution_2": "In all the sums you have <> exept for the 6th where you have <>\r\n\r\n :huh: \r\n\r\nIf all are symmetric sums then use Muirhead", "Solution_3": "Maybe cyclic not symmetric" } { "Tag": [], "Problem": "Source: AMC 10B 2005\r\n\r\nFor how many positive integers $n$ less than or equal to $24$ is $n!$ evenly divisible by $1+2+...+n$?", "Solution_1": "[hide=\"Click to reveal hidden content\"]The question is asking for how many values of n is: $\\frac{n!} {\\frac{n(n+1)} {2}}$ which can be transformed into:\n\n$\\frac{2(n-1)!} {n+1}$ It becomes obvious that the bottom will not cancel if the result is a prime, so count the number of primes under 25.\n\n2,3,5,7,11,13,17,19,23. That's 9 primes, but when the denominator is 2, it will cancel out with the 2 on the top.\n\nSo there are 8 values of n that it will not cancel out, therefore there are $\\boxed {16}$ values of n such that it will cancel out.[/hide]", "Solution_2": "[quote=\"ch1n353ch3s54a1l\"]Source: AMC 10B 2005\n\nFor how many positive integers $n$ less than or equal to $24$ is $n!$ evenly divisible by $1+2+...+n$?[/quote]\r\n\r\n[hide] $\\frac{n!}{1+2+...+n}=\\frac{2*(n-1)!}{n+1}$.\n\nWe see that if the denominator is prime, the result is not integrer. So, we have $9$ primes wich are less or equal to $25$, but one of those primes is $2$, were the result is an integrer, because both $2's$ (one in the numerator, and one in the denominator) cancel out each other. So, we have $8$ numbers that can't be, so the result is $\\boxed {16}$[/hide]" } { "Tag": [ "search" ], "Problem": "[url]http://video.on.nytimes.com/ifr_main.jsp?nsid=b26f34614:10f9606731b:-71e7&rf=bm&fr_story=d14603c1e23e6ce37920a8134a2e27b1405a4991&st=1166453285801&mp=FLV&cpf=false&fvn=9&fr=121806_094751_26f34614x10f9606731bxw71e6&rdm=175553.69351783473[/url]\r\n\r\n\r\n\r\nHAHAHAHAHAHAHAHAHAHAHA!!!!!!! If you don't laugh, nothing is funny to you.\r\n\r\n\"widgets\u2260gadgets\"", "Solution_1": "the link doesn't work for me", "Solution_2": ":D :D HAHAHA I agree. That was really funny. \"right and left are not the same thing\".", "Solution_3": "Yeah, it's stupid that Windows copies good features of other operating systems. It would be a lot better if it instead added features that were left out of other operating systems because they were bad. I mean, seriously, a desktop search? Why would anyone want that?" } { "Tag": [ "inequalities", "algebra", "polynomial", "inequalities unsolved" ], "Problem": "Let $a_i;b_i \\in [1001;2002]$ ($i=1,...n$) such that: $\\displaystyle \\sum_{i=1}^n{a_i}^2=\\sum_{i=1}^n{b_i}^2$.\r\nProve that: $\\displaystyle \\sum_{i=1}^n \\frac {{a_i}^3}{b_i} \\leq \\frac {17}{10}\\sum_{i=1}^n{a_i}^2$.", "Solution_1": "Solution was posted by Myth. But my soln is nice.\r\nI'll post may son", "Solution_2": "In Old and New inequalities there is a really nice solution for this one.", "Solution_3": "We have: $\\displaystyle \\frac {1}{2} \\leq \\frac {a_i}{b_i} \\leq 2$ then $\\displaystyle (\\frac {a_i}{b_i} -\\frac {1}{2})(\\frac {a_i}{b_i}-2) \\leq 0$.\r\n$\\displaystyle \\Leftrightarrow {a_i}^2+{b_i}^2 \\leq \\frac {5}{2}a_ib_i ; \\Rightarrow \\sum_{i=1}^na_ib_i \\geq \\frac {4}{5}\\sum_{i=1}^n{a_i}^2$;$(1)$\r\nAnd: $\\displaystyle \\frac {{a_i}^3}{b_i} + a_ib_i \\leq \\frac {5}{2}{a_i}^2 \\Rightarrow $$\\displaystyle \\sum_{i=1}^n(\\frac {{a_i}^3}{b_i} + a_ib_i) \\leq \\frac {5}{2}\\sum_{i=1}^n{a_i}^2$;$(2)$\r\nCombining $(1)$ and $(2)$; then we get: $\\displaystyle \\sum_{i=1}^n \\frac {{a_i}^3}{b_i} \\leq \\frac {17}{10}\\sum_{i=1}^n{a_i}^2$.\r\nQED", "Solution_4": "Yes, that's it!", "Solution_5": "Hey Harazi, would you please send all problem in \"Old and New inequalities\" to me? :D \r\nThank you very much!", "Solution_6": "this one is a bit old...\r\nit appeared in the Chinese Provincial Contest 1998, Part II.\r\nin that problem, $a_i,b_j\\in [1,2]$.", "Solution_7": "[quote=\"hxtung\"]Solution was posted by Myth. But my soln is nice.\nI'll post may son[/quote]\r\nbtw, what's Myth's solution?", "Solution_8": "See: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=152[/url]", "Solution_9": "Here is a slightly more compressed solution of the above argument:\r\n===========================================================================\r\nFor 1/2 <= x <= 2, the polynomial [tex] $P(x) = (x+2/5)(x-1/2)(x-2)$ [/tex] is non-positive.\r\nHence, [tex] $P(x) = x^3 -\\frac{21}{10} x^2 +\\frac{2}{5} \\le 0$ [/tex].\r\n\r\nPlugging in x= a_i/b_i, and multiplying by (b_i)^2 yields\r\n[tex]$(a_i)^3/b_i - (21/10) *(a_i)^2 +(2/5)*(b_i)^2 <= 0$.[/tex]\r\n\r\nAdding this inequality up over all i, and using sum a_i^2 = sum b_i^2\r\nyields the desired inequality." } { "Tag": [ "videos", "Support" ], "Problem": "Okay, I know this topic has been posted before on this forum, but I tried the pause thingy, but it still doesn't work for the videos. It would go to about 1 minute, then show the advertisement for every single video. It's getting really frustrating!!! I would be watching the video, then all of a sudden, this advertisement pops up and I have to start over from the beginning. I would first pause, wait like 45 sec, then play the video. Once again, it gets stuck at the same place. Please help!!!!!", "Solution_1": "Please provide information on your connection (e.g., dial-up, dsl, cable modem, and download speed). Please also provide operating system, browser, browser version, and flash version. If you are using Windows, TO obtain the flash version right click anywhere over the video and select About.\r\n\r\nUnfortunately, this is very likely due to a speed issue on your end or some sort of network issue between your ISP and our server host and is out of our control. However, by providing the information, we can look into and see if there is anything we can do on our end.\r\n\r\nAnyone else experiencing similar difficulties?", "Solution_2": "Uh, we have Time Warner Cable for internet, so cable modem? I'm not too sure about the download speed, but I tried both Firefox and IE. I normally use Firefox 3. The IE was I think version 6 or 7. My OS is Windows XP, and Adobe Flash Player.", "Solution_3": "What Flash version are you using? You can obtain the Flash version by right clicking anywhere over the video and selecting About. \r\n\r\nThis is definitely sounding like an issue somewhere between your computer and up to, but not including our hosting environment. Try doing it with all virus software, internet security software, and other running programs disabled. Also, when trying to get it all to work, hook your modem directly to your computer rather the through a router. Routers can often have trouble with video. (I have to reboot my router every couple days to keep things working well.)" } { "Tag": [ "geometry", "circumcircle", "geometry proposed" ], "Problem": "The bisectors BB1 and CC1 of a triangle ABC meet at point I. The\r\nline B1C1 intersects the circumcircle of triangle ABC at points M and N.\r\nProve that the circumradius of the triangle MIN is twice as long as the\r\ncircumradius of \u25b3ABC.", "Solution_1": "[hide]No. $ R=R_{IMN}$\nDenote $ AI\\cap MN={\\J\\}$. You can prove that $ AJ=IJ$[/hide]", "Solution_2": "u r missing something maybe because it is ARO 2006(final round grade 11 first day problem 4)", "Solution_3": "Sorry, I'm stupid :blush: \r\nHere is my solution:\r\nLet $ I_a, I_b, I_c$ be the excenters of triangle ABC then I is the orthocenter of triangle $ I_aI_bI_c$.\r\nWe get $ R_{II_bI_c}\\equal{}R_{I_aI_bI_c}\\equal{}2R$ (because $ (ABC)$ is the [b]Nine-point [/b]circle of triangle $ I_aI_bI_c$)\r\nWe will show that M, N lie on $ (II_bI_c)$\r\nWe have $ \\angle I_cAB\\equal{}\\frac{\\angle B}{2}\\plus{}\\frac{\\angle C}{2}\\equal{}\\angle I_cIB$ thus $ I_cAIB$ is cyclic quadrilateral.\r\n$ \\Rightarrow C_1I.C_1I_c\\equal{}C_1A.C_1B\\equal{}C_1M.C_1N \\Rightarrow I_cMIN$ is cyclic\r\nSimilarly, $ MINI_b$ is cyclic.\r\n$ \\Rightarrow \\angle II_cN\\equal{}\\angle IMN\\equal{}\\angle II_bN \\Rightarrow M,N$ lies on $ (II_bI_c)$\r\nWe are done.", "Solution_4": "I proved this problem in Geometry Marathon. See here:-\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=331763&start=761" } { "Tag": [ "geometry", "integration", "calculus", "calculus computations" ], "Problem": "Curve with parametric equations:\r\nx= t+sint , y =sint , 0_ \\binom{2n-2}{n} + \\binom{2n-2}{n-1} + \\binom{2n-2}{n-2} > \\dots$\r\n\r\n$\\dots > $ $\\sum_{r=0}^n {n \\choose r} = 2^n$, as desired.", "Solution_2": "Another solution to the left proceeds as follows\r\n\r\nWe want to show that $(n!)^2\\cdot 2^n< (2n!)$\r\n\r\nBut $(n!)^2\\cdot 2^n=n!\\cdot (2\\cdot 4\\cdot 6\\cdot\\ldots 2n)(time newspaper arrives)." } { "Tag": [ "MATHCOUNTS", "geometry", "geometric transformation", "reflection", "\\/closed" ], "Problem": "It's been moved?? Why??", "Solution_1": "Perhaps that the content in that forum doesn't necessarily correlate to MATHCOUNTS? Perhaps it fits better under Classroom Math?\r\n\r\nAnd does it really matter?", "Solution_2": "[quote=\"7h3.D3m0n.117\"]\n\nAnd does it really matter?[/quote]\r\n\r\nI'm just used to it being in the MATHCOUNTS forum.. Oh well.", "Solution_3": "I think that the tournaments in the \"MATHCOUNTS\" sub-forum were too hard for even MATHCOUNTS level! \r\n\r\nI don't really understand the appropriateness of the shift; although, the administrators and/or moderators might've looked at a bigger picture than I.", "Solution_4": "This is really an useless topic. Oh no, it's been moved, what now? Has anything changed that can really reflect on your abilities?\r\nUseless...", "Solution_5": "In general, either ask the moderators directly or in the thread. Or even better: don't ask at all. Just imagine if moderators need to justify every act of moving...", "Solution_6": "[quote=\"ZetaX\"]In general, either ask the moderators directly or in the thread. Or even better: don't ask at all. Just imagine if moderators need to justify every act of moving...[/quote]\r\n\r\nYeah, or if it concerns whole sub-forums, ask an admin." } { "Tag": [ "function", "floor function" ], "Problem": "Find all $x$ such that\r\n\\[ x\\lfloor x\\lfloor x\\lfloor x\\rfloor\\rfloor\\rfloor=54 \\]\r\nMasoud Zargar", "Solution_1": "$x \\in \\mathbb{R}$?", "Solution_2": "absolutely (trying to use long words that teh post can actually post :P)", "Solution_3": "[hide=\"hint\"]use the identity: $x=\\lfloor x \\rfloor+\\{x\\}$ where $\\{y\\}$ is the fractional part of $y$[/hide]", "Solution_4": "[hide=\"question\"]Why pick a number for which there is no real solution?[/hide]", "Solution_5": "[hide=\"not sure\"]$\\frac{54}{x}$ must be an integrer?????[/hide]", "Solution_6": "[hide=\"hmm\"]\nPut $a=\\lfloor x\\rfloor$ and $b=x-\\lfloor x \\rfloor$.\n\nThe equation simplifies to \\[ (a+b)\\lfloor a^3+2a^2b+ab^2\\rfloor=54, \\] unless I made a mistake. Now we take advantage of the fact that $\\lfloor z \\rfloor \\in \\mathbb{Z} \\forall z\\in \\mathbb{R}$?\n[/hide]", "Solution_7": "[hide=\"no solution\"]\nLet $n^4 = 54$, $x$ is about $2.7108$. This will be our starting number.\n\nClearly, $x$ must be bigger than this value. Note that $x=3$ overshoots by achieving a value of 81 so it is too high.\n\nAlso note that this function has the property that a higher x value will produce a higher final result.\n\nWe therefore conclude that $2.71080$ is a real number, then $\\arctan\\alpha+\\arctan\\alpha^{-1}=\\frac\\pi2$. I found three different solutions to it and at least two of them are completely basic, just trigonometry, the other needs a simple fact from calculus.\r\n\r\nEnjoy :)", "Solution_1": "isn't that obvious?? ;)", "Solution_2": "[quote=\"amirhtlusa\"]isn't that obvious?? ;)[/quote]Well, it could be for you, but given that I have seen you in the College fora pretty often I assume you are not at HS level and you are not a beginner learning from more experienced problems solvers, and this forum is thought to be! ;)", "Solution_3": "[quote=\"djimenez\"]Someone asked me this one yesterday, I thought it could be nice to post it in this forum.\n\nProve that if $\\alpha>0$ is a real number, then $\\arctan\\alpha+\\arctan\\alpha^{-1}=\\frac\\pi2$. I found three different solutions to it and at least two of them are completely basic, just trigonometry, the other needs a simple fact from calculus.\n\nEnjoy :)[/quote]\r\n\r\n[hide=\"Proof\"]\n\nSuppose $\\exists\\beta\\ni\\,\\arctan\\beta+\\arctan\\beta^{-1}\\neq\\frac{\\pi}{2}$ then\n\n\n$\\arctan\\beta^{-1}\\neq\\frac{\\pi}{2}-\\arctan\\beta\\\\\n\\\\\n\\Rightarrow \\tan(\\arctan\\beta^{-1})\\neq\\tan(\\frac{\\pi}{2}-\\arctan\\beta)\\\\\n\\\\\n\\Rightarrow \\beta^{-1}\\neq\\frac{\\sin(\\frac{\\pi}{2}-\\arctan\\beta)}{\\cos(\\frac{\\pi}{2}-\\arctan\\beta)}\\\\\n\\\\\n\\Rightarrow \\beta^{-1}\\neq\\frac{\\sin(\\frac{\\pi}{2})\\cos(\\arctan\\beta)-\\cos(\\frac{\\pi}{2})\\sin(\\arctan\\beta)}{\\cos(\\frac{\\pi}{2})\\cos(\\arctan\\beta)+\\sin(\\frac{\\pi}{2})\\sin(\\arctan\\beta)}\\\\\n\\\\\n\\Rightarrow \\beta^{-1}\\neq\\frac{\\cos(\\arctan\\beta)}{\\sin(\\arctan\\beta)}\\\\\n\\\\\n\\Rightarrow \\beta^{-1}\\neq\\frac{1}{\\tan(\\arctan\\beta)}\\\\\n\\\\\n\\Rightarrow \\beta^{-1}\\neq\\frac{1}{\\beta}\\\\\n\\\\\n\\Rightarrow \\beta\\cdot\\beta^{-1}\\neq\\beta\\cdot\\frac{1}{\\beta}\\\\\n\\\\\n\\Rightarrow 1\\neq 1 $\n\nOur supposition has lead us to a contradiction; it follows that our supposition is false.\n\nThat our supposition is false implies: $\\forall\\alpha\\in\\textbf{R}\\,\\arctan\\alpha+\\arctan\\alpha^{-1}=\\frac{\\pi}{2}\\;\\square\\\\\n\\\\$\n[/hide]", "Solution_4": "[quote=\"Dr. No\"][hide=\"Proof\"]Suppose $\\exists\\beta\\ni\\,\\arctan\\beta+\\arctan\\beta^{-1}\\neq\\frac{\\pi}{2}$ then\n\n\n$\\arctan\\beta^{-1}\\neq\\frac{\\pi}{2}-\\arctan\\beta\\\\\n\\\\\n\\Rightarrow \\tan(\\arctan\\beta^{-1})\\neq\\tan(\\frac{\\pi}{2}-\\arctan\\beta)\\\\\n\\\\\n\\Rightarrow \\beta^{-1}\\neq\\frac{\\sin(\\frac{\\pi}{2}-\\arctan\\beta)}{\\cos(\\frac{\\pi}{2}-\\arctan\\beta)}\\\\\n\\\\\n\\Rightarrow \\beta^{-1}\\neq\\frac{\\sin(\\frac{\\pi}{2})\\cos(\\arctan\\beta)-\\cos(\\frac{\\pi}{2})\\sin(\\arctan\\beta)}{\\cos(\\frac{\\pi}{2})\\cos(\\arctan\\beta)+\\sin(\\frac{\\pi}{2})\\sin(\\arctan\\beta)}\\\\\n\\\\\n\\Rightarrow \\beta^{-1}\\neq\\frac{\\cos(\\arctan\\beta)}{\\sin(\\arctan\\beta)}\\\\\n\\\\\n\\Rightarrow \\beta^{-1}\\neq\\frac{1}{\\tan(\\arctan\\beta)}\\\\\n\\\\\n\\Rightarrow \\beta^{-1}\\neq\\frac{1}{\\beta}\\\\\n\\\\\n\\Rightarrow \\beta\\cdot\\beta^{-1}\\neq\\beta\\cdot\\frac{1}{\\beta}\\\\\n\\\\\n\\Rightarrow 1\\neq 1 $\n\nOur supposition has lead us to a contradiction; it follows that our supposition is false.\n\nThat our supposition is false implies: $\\forall\\alpha\\in\\textbf{R}\\,\\arctan\\alpha+\\arctan\\alpha^{-1}=\\frac{\\pi}{2}\\;\\square\\\\\n\\\\$\n[/hide][/quote]The proof is basically right... and understandable even when... you never used the fact that $\\alpha>0$... and that $\\tan x$ is not injective!\r\n\r\nWe can say that you used it ($\\alpha>0$) but you didn't explicitely mention it. Can you tell in what step you used it strongly without mention it? ;)", "Solution_5": "[quote=\"djimenez\"]The proof is basically right... and understandable even when... you never used the fact that $\\alpha>0$... and that $\\tan x$ is not injective!\n\nWe can say that you used it ($\\alpha>0$) but you didn't explicitely mention it. Can you tell in what step you used it strongly without mention it? ;)[/quote]\r\n\r\nDavid,\r\n\r\nI don't believe the constriant $\\alpha>0$ is necessary.\r\n\r\nBoth $\\sin x$ and $\\tan x$ are odd functions; as such they have the same parity as their arguments. Therefore, the proof is valid regardless of the parity of $\\alpha$, or rather $\\beta$, in this case.", "Solution_6": "[quote=\"Dr. No\"]I don't believe the constriant $\\alpha>0$ is necessary.[/quote] Of course it is... note that\r\n\r\n$(-1)^{-1}=-1$ and $\\arctan(-1)+\\arctan(-1)=-\\frac\\pi2\\neq\\frac\\pi2$ ;)", "Solution_7": "[quote=\"djimenez\"][quote=\"Dr. No\"]I don't believe the constriant $\\alpha>0$ is necessary.[/quote] Of course it is... note that\n\n$(-1)^{-1}=-1$ and $\\arctan(-1)+\\arctan(-1)=-\\frac\\pi2\\neq\\frac\\pi2$ ;)[/quote]\r\n\r\nAh, but $-\\frac{\\pi}{2}\\equiv \\frac{\\pi}{2}\\bmod{\\pi}$.\r\n\r\nFurther, $\\pi$ is the period of $\\tan x$\r\n\r\nErgo, any answer resulting from not supplying the constraint will be likewise congruent to $\\frac{\\pi}{2} \\bmod{\\pi}$. This suggests a more general statement of the problem is:\r\n\r\n$\\forall\\alpha\\in\\textbf{R}\\;\\arctan\\alpha + \\arctan\\alpha^{-1}\\equiv \\frac{\\pi}{2}\\bmod{\\pi}$\r\n\r\n :lol:", "Solution_8": "[quote=\"djimenez\"]Someone asked me this one yesterday, I thought it could be nice to post it in this forum.\n\nProve that if $\\alpha>0$ is a real number, then $\\arctan\\alpha+\\arctan\\alpha^{-1}=\\frac\\pi2$. I found three different solutions to it and at least two of them are completely basic, just trigonometry, the other needs a simple fact from calculus.\n\nEnjoy :)[/quote]\r\n\r\nI had this problem in the semestrial math paper. We had to calculate for some particularily $\\alpha$'s. But I told one of my classmates that this is true for any $\\alpha>0$.\r\n\r\n[color=blue]The problem\nIf $x\\in (0,+\\infty)$ show that: $\\arctan x+\\arctan\\frac{1}{x}=\\frac{\\pi}{2}.$[/color]\r\n\r\n[i]The proof[/i]\r\nIt's clearly that $\\tan(\\arctan x)=x$. Because $\\tan(\\frac{\\pi}{2}-m)=\\cot m$ results that ${\\tan(\\frac{\\pi}{2}- \\arctan\\frac{1}{x})=\\cot(\\arctan\\frac{1}{x})=\\frac{1}{\\tan(\\arctan\\frac{1}{x}})}=\\frac{1}{\\frac{1}{x}}=x$\r\nSo we found out that $\\arctan x=\\frac{\\pi}{2}-\\arctan\\frac{1}{x}+k\\pi$ with $k\\in Z$ or else written it's:\r\n$\\arctan x+\\arctan\\frac{1}{x}=\\frac{\\pi}{2}+k\\pi, k\\in Z$. \r\nBut it's said that $x>0$ so $\\arctan x\\in(0,\\frac{\\pi}{2})$ and $\\arctan\\frac{1}{x}\\in(0,\\frac{\\pi}{2})$. It fallows that $\\arctan x+\\arctan\\frac{1}{x}\\in (0,\\pi)$. Finally that's possible when $k=0$. And that's all :D", "Solution_9": "[hide=\"what is needed to make this sufficiently rigorous..\"]Suppose $\\alpha$ is the tangent of the angle $\\theta$. Then $\\tan\\theta=\\alpha\\Rightarrow\\frac{\\sin\\theta}{\\cos\\theta}=\\alpha\\Rightarrow\\alpha^{-1}=\\frac{\\cos\\theta}{\\sin\\theta}=\\frac{\\sin\\left(\\frac{\\pi}{2}-\\theta\\right)}{\\cos\\left(\\frac{\\pi}{2}-\\theta\\right)}=\\tan\\left(\\frac{\\pi}{2}-\\theta\\right),$ so $\\arctan\\alpha+\\arctan\\alpha^{-1}=\\theta+\\frac{\\pi}{2}-\\theta=\\frac{\\pi}{2}.$[/hide]", "Solution_10": "[quote=\"djimenez\"][quote=\"amirhtlusa\"]isn't that obvious?? ;)[/quote]Well, it could be for you, but given that I have seen you in the College fora pretty often I assume you are not at HS level and you are not a beginner learning from more experienced problems solvers, and this forum is thought to be! ;)[/quote]\r\ni'm a Junior in HS :D", "Solution_11": ", my proof is more like \"AntonioMainenti\"'s", "Solution_12": "[quote=\"AntonioMainenti\"][hide=\"what is needed to make this sufficiently rigorous..\"]Suppose $\\alpha$ is the tangent of the angle $\\theta$. Then $\\tan\\theta=\\alpha\\Rightarrow\\frac{\\sin\\theta}{\\cos\\theta}=\\alpha\\Rightarrow\\alpha^{-1}=\\frac{\\cos\\theta}{\\sin\\theta}=\\frac{\\sin\\left(\\frac{\\pi}{2}-\\theta\\right)}{\\cos\\left(\\frac{\\pi}{2}-\\theta\\right)}=\\tan\\left(\\frac{\\pi}{2}-\\theta\\right),$ so $\\arctan\\alpha+\\arctan\\alpha^{-1}=\\theta+\\frac{\\pi}{2}-\\theta=\\frac{\\pi}{2}.$[/hide][/quote]\r\n\r\nAntonio,\r\n\r\nThe problem is sufficiently rigorous as stated. A proof wherein $\\alpha > 0$ was assumed has already been given.\r\n\r\nThe question at hand is whether this restriction is necessary. David has proved that it is by counterexample -- I shant quibble with the obvious. For me, the thing that remains is whether this is the most general form of the problem that can be given. If it is then the resulting theorem is of little value since it is merely an abstruse restatement of an obvious fact pointed out by another poster earlier in the thread: $\\tan (\\frac{\\pi}{2}-m) =\\cot m$.", "Solution_13": "[quote]Antonio,\n\nThe problem is sufficiently rigorous as stated. A proof wherein $\\alpha > 0$ was assumed has already been given.\n\nThe question at hand is whether this restriction is necessary. David has proved that it is by counterexample -- I shant quibble with the obvious. For me, the thing that remains is whether this is the most general form of the problem that can be given. If it is then the resulting theorem is of little value since it is merely an abstruse restatement of an obvious fact pointed out by another poster earlier in the thread: $\\tan (\\frac{\\pi}{2}-m) =\\cot m$.[/quote]Ah I see. It seems like the sum would be $\\frac{\\pi}{2}$ if $\\alpha>0$, undefined for $\\alpha=0$, and $-\\frac{\\pi}{2}$ if $\\alpha<0$. I'm sure you already realize this though (assuming it's right). As for generalizing further.. I have no idea." } { "Tag": [ "vector", "function", "group theory", "abstract algebra", "induction", "superior algebra", "superior algebra solved" ], "Problem": "let (G.*) a croupe such that for every element X X*X=e prove that exist k \\ card(G)=2^k :?:", "Solution_1": "Let $x,y\\in G$. Then $e=x^{2}y^{2}=(xy)^{2}$ which leads to $xy=yx$, so $G$ is abelian. From now on, denote $xy$ by $x+y$.\r\nWith this in view, it should be easy to see that $G$ has a natural structure of $\\mathbb Z/2\\mathbb Z$ vector space. \r\nChoose $B$ a basis for $G$ as a vector space. Then we can identify any element of $G$ with a finite subset of $B$, because all the coefficients that appear are either 0 or 1 and there are only finitely many 1's, so we can look at this as of a characteristic function of a subset of $B$ with finite support.\r\n\r\nIf $B$ is finite, which I think is what you care for, then if $|B|=k$, then $|G|=|\\mathcal P(B)|=2^{k}$.\r\nIf $k$ is not a finite, then the cardinal of finite subsets of $B$ is the same as the cardinal of $B$, so $|G|=|B|$.\r\n\r\nBut maybe by \"croupe\" you mean \"finite group\" :D", "Solution_2": "alternative: the assumption tells us that every element has order $1$ or $2$. so by cauchy's theorem, the order of $G$ has no odd prime divisor.", "Solution_3": "[quote=\"-oo-\"]alternative: the assumption tells us that every element has order $1$ or $2$. so by cauchy's theorem, the order of $G$ has no odd prime divisor.[/quote]\r\n\r\nAlternative to the finite setting you mean.", "Solution_4": "we can use simply lacrange :D", "Solution_5": "the problem - as stated - is false.\r\nyou actually need finiteness hypothesis, since there are at least two ways to find a counterexample: either $(\\overline{\\mathbb{F}}_{2},+)$ and $(\\mathbb{Z}/2\\mathbb{Z})^\\omega_{0}$ (i.e. definitively null sequences in $(\\mathbb{Z}/2\\mathbb{Z})^\\omega$) are countable.", "Solution_6": "ma_go the problem is not false becouse we have card(G)=card(H)card(G\\H).H={e.x} is subgroup of G. and we prove that G\\H is abelien .and done by induction in n. :ninja:", "Solution_7": "mine was just an answer to amfulger.\r\n\r\nas i said, the statement needs finiteness hypothesis: you can't drop it, since i found two [isomorphic] infinite counterexamples.\r\n\r\nof course, if you ask for your group to be finite, there are at least three different proofs of it.", "Solution_8": "Seems I forgot to end my argument there. I proved that if $G$ is infinite, then $|G|$ is the same as the cardinal of a basis $B$ for the $\\mathbb Z/2\\mathbb Z$ vector space $G$. \r\n\r\nWhat I forgot to mention and what ma-ga completed, is that there exist cardinals that are not $2^{\\mbox{some other cardinal numer}}$, for example $\\aleph_{0}=|\\mathbb N|$." } { "Tag": [ "algebra", "polynomial" ], "Problem": "Hi all, \r\n\r\nI had a question about the rule of signs that wasn't covered in my algebra 2 or precalc course.\r\n\r\nHow do we determine the number of zeros there are if we are missing a term, such as x^3+6x+20 or x^5-4x^4+2x^2-4", "Solution_1": "You can ignore that term since 0 is not considered a sign change.", "Solution_2": "Do you mean that, for example: in the polynomial, x^2-4, we can just skip the x term and look at the sign change between x^2 and -4?", "Solution_3": "[quote=\"SuperMathBoy\"]Do you mean that, for example: in the polynomial, x^2-4, we can just skip the x term and look at the sign change between x^2 and -4?[/quote]That's right. As you can see, there is only 1 sign change so you should expect one 1 positive root and 1 negative root or two complex roots (if the roots are not real).", "Solution_4": "Ok... Thanks" } { "Tag": [ "\\/closed" ], "Problem": "I use UTC+0 (= GMT) timezone and it is showing as an hour behind. I am posting this at 23:57 UTC but the forum is showing the time as 22:57. When the clocks go forward in a week's time to UTC+1 (= BST, British Summer Time) it will be still be an hour behind if I use the summer time option.", "Solution_1": "This seems global. My UTC-8+DST (Pacific time zone in the USA) is displaying one hour too early, and I just saw the same complaint from someone two hours east of me.", "Solution_2": "That would be me I guess. I made a topic a few hours earlier but put in in the wrong forum. It's here: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=195533[/url], and I also experimented with turning DST off.", "Solution_3": "same here too the clocks are slow by an hour", "Solution_4": "Yeah, we know. It should be fixed by tomorrow.", "Solution_5": "It actually seems fixed right now..." } { "Tag": [ "calculus", "integration", "trigonometry", "logarithms", "geometry", "geometric transformation", "reflection" ], "Problem": "The problem was didn't make sense, so that was deleted.", "Solution_1": "but when $x=1$ , the expression inside the integral is undefined(reaching infinity)? So we cannot perfome the integration ?", "Solution_2": "Sorry for that , you are right. I will edit.\r\n\r\nkunny", "Solution_3": "Could you give a hint please? I've tried all the tricks in my book with no success.", "Solution_4": "Kalle, sorry for my delaying reply to you.This problem is very trciky. Remark that $\\sin^{2}\\theta+\\cos^{2}\\theta =1.$\r\n\r\nkunny", "Solution_5": "Kunny do you know solution. I tried very much but i can not solve . As if it is a mount.", "Solution_6": "Hello, everyone.\r\n\r\nI'm very sorry that this problem is mistaken and taking troubles you to take much time. I should have examined in detail. I see to it that afterwards.\r\n\r\nKunny", "Solution_7": "[quote]but when x=1 , the expression inside the integral is undefined(reaching infinity)? So we cannot perfome the integration ?[/quote]\r\n\r\nI don't get it: $1 \\notin [e^\\frac{\\pi}{4},e^\\frac{\\pi}{3}]$\r\nIs that the correct version of the problem(after that remark)?", "Solution_8": "[quote=\"anih\"][quote]but when x=1 , the expression inside the integral is undefined(reaching infinity)? So we cannot perfome the integration ?[/quote]\n\nI don't get it: $1 \\notin [e^\\frac{\\pi}{4},e^\\frac{\\pi}{3}]$\nIs that the correct version of the problem(after that remark)?[/quote]\r\n\r\nAs kunny said, the original post was edited to reflect shyong's complaint. Mathematica (and also kunny, from his last post) thinks there is no elementary expression for this integral.", "Solution_9": "[quote=\"JBL\"]Mathematica thinks there is no elementary expression for this integral.[/quote]\r\nHow do you tell Mathematica to investigate this?", "Solution_10": "Mathematica has built-in simplification algorithms in its integrator, so it usually doesn't spit out things like a hypergeometric function (as it does in this case) unless it is unable to find an elementary form. There are also the functions Simplify and FullSimplify, which respectively try and try harder to make things less ugly. It's not flawless, of course, (I'm always annoyed because it doesn't simplify $\\sqrt{x^{2}}$) but for the vast majority of familiar integrals it will give you a nice answer when a nice answer exists." } { "Tag": [ "vector" ], "Problem": "[hide=\"hint\"]vectors[/hide]\nany other way?\n[hide]it's actually easy to see that this is right if you look at this a,b c,d e,f g,h and put some of them negative, but what about a generalization a,b,c d,e,f..[/hide]", "Solution_1": "The sums naturally hint a vector solution.\r\n\r\nConsider the plane vectors (a,b),(c,d),(e,f),(g,h).\r\n\r\nThe dot product between some two of these must form an angle less than 90 degrees." } { "Tag": [], "Problem": "Weather right or not..I wrote a math paper(theme is secret for now :)) At least, I want to confirm it. Is there anyway?\r\n\r\nOne more thing..flag is S.Korea's, but I live in USA.", "Solution_1": "[quote=\"physicsholic\"]Weather right or not..I wrote a math paper(theme is secret for now :)) At least, I want to confirm it. Is there anyway?\n\nOne more thing..flag is S.Korea's, but I live in USA.[/quote]\r\n\r\nTalk to some college professor, he'll know", "Solution_2": "[quote=\"Jos\u00e9\"]\nTalk to some college professor, he'll know[/quote]\r\n\r\nOr she'll know, as the case may be.\r\n\r\nWhether you're a high-school student, college student, graduate student, or faculty member, the process of publishing original research is the same. (Expository papers are different.) You need to find an appropriate journal, and follow the submission instructions there. Most journals require (or at least request) that papers be typeset in LaTeX. What journals would be appropriate depends on the subfield of mathematics and on how significant the result is. So, in order to get specific advice, you'll have to share your results.\r\n\r\nIn general, researchers do not keep their results secret. If you're worried about having your results stolen, the thing to do is to submit your paper to the arXiv. Then the results appear with your name on them and the date of submission, so you have evidence of when you produced the results.", "Solution_3": "[quote=\"physicsholic\"]\nOne more thing..flag is S.Korea's, but I live in USA.[/quote]\r\n\r\nHm, you're the one who choose the flag when you created your account...identity crisis perhaps :D (just kidding with you). Go to profile > country flag and change it to the usa flag. Then just enter your password and hit submit.", "Solution_4": "[quote=\"physicsholic\"]Weather right or not..I wrote a math paper(theme is secret for now :)) At least, I want to confirm it. Is there anyway?\n\nOne more thing..flag is S.Korea's, but I live in USA.[/quote]\r\n\r\nWell, what have you done finally???" } { "Tag": [ "trigonometry", "geometry unsolved", "geometry" ], "Problem": "I feel this question is easy but I post it in High School forum and I feel it's too hard there. So I just get confused:\r\nThe question is:\r\n\r\nProve that:\r\n$ \\tan 35 \\left(\\sin 10 \\plus{} \\cos 10\\right) \\equal{} \\cos 10 \\minus{} \\sin 10$\r\n\r\nThis is my first post in this hard forum be merciful in the solutions please don't put harsh stuff :D \r\n\r\nThanking you", "Solution_1": "Take a $ \\cos 10$ outside from both the sides. so we need to prove $ \\tan 35 ( 1 \\plus{} \\tan 10 ) \\equal{} 1 \\minus{} \\tan 10$ \r\n\r\n$ \\tan 35 \\equal{} \\tan ( 45 \\minus{} 10 ) \\equal{} \\frac { 1 \\minus{} \\tan 10 }{ 1 \\plus{} \\tan 10 }$ as $ \\tan 45 \\equal{} 1$\r\n\r\nTherefore $ \\tan 35 ( 1 \\plus{} \\tan 10 ) \\equal{} 1 \\minus{} \\tan 10$ which is the result we wanted", "Solution_2": "$ \\tan 35(\\sin10 \\plus{} \\cos 10) \\equal{} \\cos 10 \\minus{} \\sin 10$ \r\n$ \\Leftrightarrow \\tan 35 \\equal{} \\frac{1 \\minus{} \\sin 20}{\\cos20}$ \r\n$ \\Leftrightarrow \\sin70\\tan35 \\plus{} \\cos70 \\equal{} 1$ \r\n$ \\Leftrightarrow \\sin^2(35) \\plus{} \\cos^2 (35) \\equal{} 1.$\r\n\r\n\r\ndone :P", "Solution_3": "Please guys could you explain your computaions :?: :(", "Solution_4": "$ \\tan 35 \\equal{} (\\cos 10 \\minus{} \\sin 10)\\div (\\cos 10 \\plus{} \\sin 10) \\equal{} (\\cos 10 \\minus{} \\sin 10)^2 \\div (\\cos^2 10 \\minus{} \\sin^2 10) \\equal{} (1 \\minus{} \\sin 20) \\div (\\cos 20)$\r\n\r\nuse the fact that $ \\cos 20 \\equal{} \\sin 70$ and $ \\sin 70 \\equal{} 2 sin 35 cos 35$ and $ \\cos 70 \\equal{} \\cos^2 35 \\minus{} sin^2 35$ then we get the desired result.", "Solution_5": "To prove $ \\tan 35(\\sin 10+\\cos 10) =\\cos 10-\\sin 10$\r\n\r\nNow take out $ \\cos 10$ on both sides so we have to prove $ \\cos 10 \\tan 35(\\tan 10+1) =\\cos 10(1-\\tan 10$\r\n\r\nCancel the two $ \\cos 10$ on both side leaving us to prove $ \\tan 35(\\tan (10) +1 ) = 1-\\tan 10$\r\n\r\nnow $ \\tan 35 =\\tan ( 45-10 ) =\\frac{ 1-\\tan 10 }{ 1+\\tan 10 }$ \r\n\r\n $ [ \\tan ( A-B) = \\frac{ \\tan A - \\tan B }{ \\tan A + \\tan B }$ and $ \\tan 45 = 1 ]$ \r\n \r\nTherefore $ \\tan 35(\\tan (10) +1 ) = 1-\\tan 10$\r\n\r\nHence proved", "Solution_6": "Thanks a lot guys \r\n\r\nThe last two questions :oops: \r\n\r\n@zool007\r\nJust explain this only :) $ \\tan 35 \\equal{} (\\cos 10\\minus{}\\sin 10)\\div (\\cos 10\\plus{}\\sin 10)$\r\n@\"vijaymenon\"\r\nOK :| where is the $ \\sin$", "Solution_7": "When iam taking a $ \\cos 10$ outside then it becomes $ \\frac{ \\sin 10 }{\\cos 10 } \\equal{} \\tan 10$\r\n\r\nIf u have any doubt jus write $ \\tan 10 \\equal{} \\frac{ \\sin 10 }{\\cos 10 }$ and we will get back the original expression.\r\n\r\nWhat do u do ?? in high school ??", "Solution_8": "it's something called working backwards." } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "Let $\\left\\{U_{1},....,U_{1985}\\right\\}$ and $\\left\\{K_{1},....,K_{1985}\\right\\}$ two partitios of the set $A=\\left\\{1,2,...,n\\right\\}$\r\n,(i.e $U_{i}\\cap U_{j}=\\oslash$ for all $i\\neq j$ and $U_{1}\\cup ...U_{1985}=A$\r\nand $K_{i}\\cap K_{j}=\\oslash$ for all $i\\neq j$ and $K_{1}\\cup ...K_{1985}=A$)\r\n\r\nsuch that for all $1\\leq i,j\\leq 1985$ if $U_{i}\\cap K_{j}=\\oslash$ , then \r\n$\\left|U_{i}\\right|+\\left|K_{j}\\right|\\geq 1985$. Find the minimum value of n for which such partitions exist .", "Solution_1": "Hey I wonder about the source of this problem :blush:( [color=red]Armenia TST[/color] :roll: ). From which TST is this problem? If it is from IMO team selection test, I can't figure out where did you find this problem?As far as I know there is no any book with Armenian TST,is there?\r\nPlease, don't leave me in the dark :D .", "Solution_2": "[quote=\"KMN\"]Let $\\left\\{U_{1},....,U_{1985}\\right\\}$ and $\\left\\{K_{1},....,K_{1985}\\right\\}$ two partitios of the set\nsuch that for all $1\\leq i,j\\leq 1985$ if $U_{i}\\cap K_{j}=\\oslash$ [/quote]\r\nthere existe $1\\leq j\\leq 1985$ such that $1\\in k_{j}$\r\nthere existe $1\\leq i\\leq 1985$ such that $1\\in U_{i}$\r\nthen $U_{i}\\cap K_{j}\\neq \\oslash$\r\nso, ((((((((((for all $1\\leq i,j\\leq 1985$,$U_{i}\\cap K_{j}=\\oslash$)))))))) is false :huh:", "Solution_3": "The assumption on $U, K$ is $\\forall i\\forall j (A(i,j) \\Longrightarrow B(i, j))$ \r\nwhere $A(i, j)$ is $U_{i}\\cap K_{j}= \\emptyset$ and $B(i, j)$ is $|U_{i}|+|K_{j}| \\geq 1985$.", "Solution_4": "thank you,\r\nsorry becaue i am bad in englesh :blush:" } { "Tag": [ "function", "algebra", "domain", "induction", "algebra solved" ], "Problem": "Find all functions f : Z (integers) \\ {0} -> Q (rationals) such that \r\nf((x+y)/3) = 1/2*(f(x)+f(y)) for all x,y in the domain of the function.", "Solution_1": "First of all we can observe that by taking x=1 and y=2 we get f(1)=f(2). We then take x=y=3 and we get f(2)=f(3). We now take x=2, y=4 to get f(1)=f(2)=f(3)=f(4). The next step is x=1, y=5 and we get f(1)=f(2)=..=f(5). \r\n\r\nFrom now on we can use induction: Let's assume we've proven f(1)=f(2)=f(3)=..=f(3k-1) (let's call this P_k). We make x=3 and y=3k to get f(k+1)=(f(3)+f(3k))/2, meaning that f(3k)=f(1). Then x=2, y=3k+1 gives us f(3k+1)=f(1). The next step is x=1,y=3k+2. This gives f(3k+2)=f(1) so f(3(k+1)-1)=f(1), which is, in fact, P_(k+1), so induction works: P_k->P_(k+1).\r\n\r\nThis means that f is constant for the positive numbers (call its value a). By the same method we prove that it's constant for the negative numbers (call its value b). x=-1,y=4 gives us f(1)=(f(-1)+f(4))/2, so a=(a+b)/2, so a=b, so f is constant on Z\\{0}." } { "Tag": [ "calculus", "calculus computations" ], "Problem": "How do you find the equation of the line tangent to the graph of $ f$ at $ (1,1)$ where $ f$ is given by $ f(x)\\equal{}3x^3\\minus{}3x^2\\plus{}1$.", "Solution_1": "$ y\\minus{}y_{0}\\equal{} (y_{0})^{\\prime}(x\\minus{}x_{0}),$ where $ (x_{0}, y_{0})\\equal{}(1,1)$ and $ (y_{0})^{\\prime}\\equal{}\\frac{dy}{dx}(x_{0}).$\r\n\r\nJust express $ y$ in terms of $ x$ in the above equation.", "Solution_2": "Hmm... I am a little confused about how you solve this part: $ (y_{0})^{\\prime} \\equal{} \\frac {dy}{dx}(x_{0})$\r\n\r\nIs it $ {(f(x)}'\\equal{}9x^2\\minus{}6x)*(1)$?\r\n\r\nMaking it $ y\\minus{}1\\equal{}(9x^2\\minus{}6x)(x\\minus{}1)$ which would make $ y(x)\\equal{}9x^3\\minus{}15x^2\\plus{}6x\\plus{}1$?\r\n\r\nOr am I way off and just confusing myself with my limited skills?", "Solution_3": "You should calculate $ f^{\\prime}(x_{0})\\equal{}f^{\\prime}(1)\\equal{}3$.\r\n\r\nNow you get $ y\\minus{}1\\equal{}3(x\\minus{}1)$. So $ y\\equal{}3x\\minus{}2.$" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let $ ABC$ be an acute triangle and $ H$ its orthocentre.\r\n\r\nShow that: $ \\sqrt{2}< \\frac{AB\\plus{}BC\\plus{}CA}{AH\\plus{}BH\\plus{}CH}<2.$", "Solution_1": "Actually, I think your inequality is not very strong.\r\nTry that proof:\r\n$ \\frac{AB\\plus{}BC\\plus{}CA}{AH\\plus{}BH\\plus{}CH}<2\r\n\\Leftrightarrow AB\\plus{}BC\\plus{}CA<2AH\\plus{}2BH\\plus{}2CH\r\n\\equal{}(AH\\plus{}BH)\\plus{}(BH\\plus{}CH)\\plus{}(CH\\plus{}AH)$\r\nWhich is obviously true (triangle basic inequality)\r\nOne inequality's done\r\nNow another:\r\nLet $ N\\in AB, M\\in AC$ such that $ \\widehat{NHB}\\equal{}\\widehat{MHC} \\equal{}90^o$\r\n $ AM\\plus{}AN\\equal{}AN\\plus{}NH>AH$\r\n $ NB>BH$\r\n$ MC>HC$\r\n$ \\rightarrow AB\\plus{}AC>HA\\plus{}HB\\plus{}HC$\r\nDo the same work we have:\r\n$ AB\\plus{}AC>HA\\plus{}HB\\plus{}HC$\r\n$ AB\\plus{}BC>HA\\plus{}HB\\plus{}HC$\r\n$ AC\\plus{}BC>HA\\plus{}HB\\plus{}HC$\r\n$ \\rightarrow AB\\plus{}BC\\plus{}CA>\\frac{3}{2}(HA\\plus{}HB\\plus{}HC)>\\sqrt{2}(HA\\plus{}HB\\plus{}HC)$", "Solution_2": "[quote=\"Mateescu Constantin\"]Let $ ABC$ be an acute triangle and $ H$ its orthocentre.\n\nShow that: $ \\sqrt {2} < \\frac {AB \\plus{} BC \\plus{} CA}{AH \\plus{} BH \\plus{} CH} < 2.$[/quote]\r\n\r\nthe inequality is equivalent to \r\n\r\n$ \\sqrt {2}(R\\plus{}r) < p < 2(R \\plus{} r)$ Which is true and kind of weak" } { "Tag": [ "quadratics", "LaTeX", "algebra", "linear equation" ], "Problem": "How to solve this?\r\n\r\n2x + sqr{2x^2} = 16 + 16sqr{2} ?\r\n\r\nThanks.", "Solution_1": "[hide=\"Hint\"]Show that $ x$ has to be positive, and then take the root (making $ \\sqrt 2x^2\\equal{}x\\sqrt2$) and solve the linear equation regularly.\n\n[/hide]\n\n[hide=\"Answer\"]$ x\\equal{}8 \\sqrt 2$[/hide]", "Solution_2": "I dont know how to get rid of LHS root, could you please show me the procedure? Thanks.", "Solution_3": "2x + sqr{2x^2} = 16 + 16sqr{2}\r\n\r\nin the first term 2x, look at 2 as (sqr{2})^2\r\nand second term sqr{2x^2} = x(sqr{2})\r\n\r\nthen move everything to the right , and we have a quadratic equation\r\n\r\nx (sqr{2})^2 + (x-16)(sqr{2}) -16 = 0\r\nfactor it out\r\n\r\n( x(sqr{2}) - 16 )*(sqr{2} + 1 ) = 0\r\nso\r\nx(sqr{2}) - 16 must = 0\r\n\r\nx(sqr{2}) - 16 =0\r\nx(sqr{2})=16\r\nx=16/sqr{2} = 16(sqr{2})/2 = 8(sqr{2})\r\n\r\n(sorry, i dont know how to use LaTex, so this is kind of messy :oops: )\r\n\r\n\r\n[/b]", "Solution_4": "I'll LaTeXify it for you. :) \r\n\r\n$ 2x \\plus{} \\sqrt{2x^2} \\equal{} 16 \\plus{} 16\\sqrt{2}$\r\n\r\nIn the first ter $ 2x$, look at $ 2$ as $ \\sqrt{2}^2$\r\nand second term $ \\sqrt{2x^2} \\equal{} x\\sqrt{2}$.\r\n\r\nThen move everything to the right, and we have a quadratic equation.\r\n\r\n$ x\\sqrt{2}^2 \\plus{} (x \\minus{} 16)(\\sqrt{2}) \\minus{} 16 \\equal{} 0$\r\n\r\nFactor it out.\r\n\r\n$ (x\\sqrt{2} \\minus{} 16)(\\sqrt{2} \\plus{} 1) \\equal{} 0$\r\n\r\nso\r\n\r\n$ x\\sqrt{2} \\minus{} 16$ must be equal to $ 0$.\r\n\r\n$ x\\sqrt{2} \\minus{} 16 \\equal{} 0$\r\n\r\n$ x\\sqrt{2} \\equal{} 16$\r\n\r\n$ x \\equal{} \\frac{16}{\\sqrt{2}} \\equal{} \\frac{16\\sqrt{2}}{2} \\equal{} \\boxed{8\\sqrt{2}}$", "Solution_5": "No real need for complicating it (btw, generally speaking, keep in mind that $ \\sqrt x^2 \\neq x$, it is $ |x|$)\r\n\r\nso our equation could have been rewritten as\r\n\r\n$ 2x \\plus{} |x|\\sqrt 2 \\equal{} 16 \\plus{} 16\\sqrt 2$\r\n\r\nNow, if we look at two cases, we have:\r\n\r\n1. $ x < 0$, LHS is obviously negative (it becomes $ 2x \\minus{} x\\sqrt 2 \\equal{} x\\sqrt 2 (\\sqrt 2 \\minus{} 1))$, and as $ x$ is negative, it is all negative), and RHS is positive, so it is a contradicting case. No solutions.\r\n\r\n2.$ x\\geq 0$, our equation is now $ 2x \\plus{} x\\sqrt 2 \\equal{} 16 \\plus{} 16\\sqrt 2$, or $ x\\sqrt 2(\\sqrt 2 \\plus{} 1) \\equal{} 16(\\sqrt 2 \\plus{} 1)$, i.e. $ x \\equal{} 16/\\sqrt 2 \\equal{} 8\\sqrt 2$." } { "Tag": [], "Problem": "Let $ z$ be a complex number and $ \\bar{z}$ be its conjugate. What are the values of $ z$ for which $ z\\bar{z}=5$ and $ z^{2}+\\bar{z}^{2}=6$?", "Solution_1": "[quote=\"mountain.dew\"]Let $ z$ be a complex number and $ \\bar{z}$ be its conjugate. What are the values of $ z$ for which $ z\\bar{z}=5$ and $ z^{2}+\\bar{z}^{2}=6$?[/quote]\r\n\r\nIf $ z\\bar{z}=5$ and $ z^{2}+\\bar{z}^{2}=6$ then $ (z+\\bar{z})^{2}=6+2\\cdot5=16$ and $ (z-\\bar{z})^{2}=6-2\\cdot5=-4$. Therefore $ z=\\pm2\\pm i$." } { "Tag": [], "Problem": "Let be $ a\\in \\mathbb{R}$ . Find $ x,y,z\\in \\mathbb{R}$ that satisfies $ x\\plus{}y\\plus{}z\\equal{}4a$ and $ xy\\plus{}yz\\plus{}zx\\equal{}7a^2\\minus{}az$ .", "Solution_1": "[quote=\"alex2008\"]Let be $ a\\in \\mathbb{R}$ . Find $ x,y,z\\in \\mathbb{R}$ that satisfies $ x \\plus{} y \\plus{} z \\equal{} 4a$ and $ xy \\plus{} yz \\plus{} zx \\equal{} 7a^2 \\minus{} az$ .[/quote]\r\n$ \\Leftrightarrow 16xy \\plus{}16 yz \\plus{} 16zx \\equal{} 7*16a^2 \\minus{} 16az$\r\n$ \\Leftrightarrow 16xy \\plus{}16 yz \\plus{} 16zx \\equal{} 7(x\\plus{}y\\plus{}z)^2 \\minus{} 4(x\\plus{}y\\plus{}z)z$\r\n$ \\Leftrightarrow 16xy \\plus{}16 yz \\plus{} 16zx \\equal{} 7(x\\plus{}y\\plus{}z)^2 \\minus{} 4(x\\plus{}y\\plus{}z)z$\r\n$ \\Leftrightarrow 7x^2\\plus{}7y^2\\plus{}3z^2\\minus{}2xy\\minus{}6yz\\minus{}6xz\\equal{}0$\r\n$ \\Leftrightarrow (x\\minus{}y)^2\\plus{}\\frac{3}{2}(z\\minus{}2x)^2\\plus{}\\frac{3}{2}(z\\minus{}2y)^2\\equal{}0$\r\n$ \\Leftrightarrow \\left\\{\\begin{array}{c} x\\equal{}y\\\\z\\equal{}2x\\\\z\\equal{}2y\\end{array}\\right.$\r\nConsider $ x\\equal{}y\\equal{}t \\Rightarrow z\\equal{}2t$\r\nSince $ x\\plus{}y\\plus{}z\\equal{}4a \\Rightarrow 4t\\equal{}4a \\Rightarrow t\\equal{}a \\Rightarrow \\left\\{\\begin{array}{c} x\\equal{}y\\equal{}a\\\\z\\equal{}2a\\end{array}\\right.$" } { "Tag": [ "trigonometry" ], "Problem": "Can anyone post a solution to the following:\r\nIf $ x \\plus{} y \\equal{} 270$, $ 0 < x,y < 180$, and $ x,y \\neq 90$, find the maximum value of $ 4sinx \\plus{} 3siny$. \r\n\r\n\r\nWhat I did was very non-legit. I said that since 4 and 3 are relatively close, just set both x,y =135 to get the answer of 4.95 when the actual answer is 5 (this wasnt a proof question, so it didnt really matter that I made this assumption since I rounded up to 5 anyways). I was wondering what the actual legit solution to this is.\r\n\r\n\r\n\r\nThanks.", "Solution_1": "[hide=\"solution\"]$ 4\\sin x \\plus{} 3\\sin y \\equal{} 4\\sin x \\plus{} 3\\cos (90 \\minus{} y) \\equal{} 4\\sin x \\plus{} 3\\cos(x \\minus{} 180) \\equal{} 4\\sin x \\minus{} 3\\cos x$\n\n$ \\equal{} 5\\left(\\frac {4}{5}\\sin x \\minus{} \\frac {3}{5}\\cos x\\right) \\equal{} 5\\sin \\left(x \\minus{} \\sin^{ \\minus{} 1}\\frac {3}{5}\\right)\\leq 5$.\n\nThus the answer is 5, for $ x \\equal{} 90 \\plus{} \\sin^{ \\minus{} 1} \\frac {3}{5}$ and $ y \\equal{} 270 \\minus{} x$.[/hide]" } { "Tag": [ "geometry", "similar triangles", "arithmetic sequence", "perpendicular bisector" ], "Problem": "1. [hide] z=5\n\nsolution- x=2 y=1. 2*2+2*1=6 1+2+5=8[/hide]", "Solution_1": "Note to BHorseMath: [hide]What you did was okay because it worked for this problem. However you cannot concluded that x=2 and y=1. Try doing it a little more algebra-ish by simplifying 2x + 2y = 6 so that the coefficents are relativley prime. [/hide]", "Solution_2": "[hide]2) the triangle is a 5-12-13, the altitude of the hypotenuse is 5 units long\n\n\n\n[/hide]", "Solution_3": "Be more careful mathfiend -- look what it is asking you for and then check if that's what you found.", "Solution_4": "I believe it would depend on how one would draw the triangle. When I drew it, the altitude was [hide]5[/hide]. When drawing it another way, the altitude would be [hide]12[/hide].", "Solution_5": "[quote=\"Rep123max\"]\n2. A right triangle has an area of 30. If the length of the hypotenuse is 13, what is the length of the altitude [b]to the hypotenuse[/b]?\n[/quote]\r\n\r\nThat means the one perpendicular to the hypotenuse.", "Solution_6": "???", "Solution_7": "An altitude is the line at right angles to the given side of a triangle through the opposite vertex.. its not actually one of the sides of a triangle. Maybe you are mixing it up with the word \"adjacent\"?", "Solution_8": "On number 3, why do you even need to know that [tex]1+3+5+7+9+ \\ldots +n=(\\frac{n+1}{2})^2[/tex]?", "Solution_9": "Mathfiend -- what is the hypotenuse of a right triangle?", "Solution_10": "By the way Robert, knowing both those formulas would be good as some of the terms are similar in both. Suppose you subtracted them... what would you get?", "Solution_11": "13", "Solution_12": "Rep123max wrote:Exactly. Instead of just telling the formula for the sum of all even numbers, I told the formulas for the sum of all natural numbers and the sum of all the odd numbers.\n\n\n\nI think Robert was trying to insinuate that:\n\n\n\n[hide]The sum of the first n natural numbers multiplied by two gives you some even numbers [/hide]", "Solution_13": "So the answer for number 3 is [hide]2(1 + 2 + 3 ... + 60) = (60)(61) = 3660.[/hide]", "Solution_14": "[hide] 1. x+y+z=8, 2x+2y=6, this means the x+y=3, so z=5.\n\n2. If it has an area of 30, then 13*altitude/2=30, so 13*a/2=30, 13*a=60, a=60/13=4 and 8/13.\n\n3. This is just double n(n+1)/2, so it is n(n+1), where is half the last number in the sequence, so n=60. 60(60+1)=3600+60=3660.\n\n4. Using a:^2:+b:^2:=c:^2:, then we know that the sum of the areas of the other two semicircles should equal the area of the semicircle drawn on the hypoteneuse, therefore the answer is 29.7. [/hide]", "Solution_15": "[quote=\"Ragingg\"]By the way Robert, knowing both those formulas would be good as some of the terms are similar in both. Suppose you subtracted them... what would you get?[/quote]\r\n\r\ncouldnt you do the formula 1+2+3... and multiply it by 2?", "Solution_16": "1. [hide]2x+2y=6, so x+y=3 and substitute that into x+y+z=8 to get z=5.[/hide]\n\n\n\n2. [hide]It's a 5-12-13 triangle. Draw the altitude to the hypotenuse to form two triangles that are similar to the big one. The altitude to the hypotenuse is one of the sides of the small triangles. Using the similar triangles, you can then solve for the altitude. (it's 60/13?)[/hide]\n\n\n\n3. [hide]2+4+6+...+120 has 60 terms, sum of arithmetic sequence is n(t1+tn)/2=60(2+120)/2=3660.[/hide]\n\n\n\n4. [hide]The radius of each semicircle is 1/2 the length of the side it's on: the area of a semicircle is :pi:/2*r^2. Multiply the Pythagorean Thm. by :pi:, you can see that the area of the semicircle on the hypotenuse is equal to the sum of the areas of the other two semicircles.[/hide]", "Solution_17": "1.[hide]x+y = 8-z\n\n2(x+y) = 2x+2y = 6 = 2(8-z)\n\n\n\n6 = 16-2z\n\n-10 = -2z\n\nz = 5[/hide]\n\n\n\n2.[hide]\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nI don't think this is right, since it can't be factored.\n\n[/hide]\n\n\n\n3.[hide]You just add 1 to (n+1) since 1+1 = 2, 3+1=4, and so on.\n\n\n\n(120+2)2/22 \n\n1242/4\n\n622\n\n3844[/hide]\n\n\n\n4.[hide]Let's first try to find out the length of the hypontenuse.\n\n\n\n29.7*2 = 59.4\n\n59.4/3.14 is about 18.9\n\nThe radius is :sqrt: 18.9\n\nThe diameter, or hypotenuse is 2 :sqrt: 18.9\n\nSquared, that is 75.6, or the other two sides' added together after being suqared.\n\n\n\nI don't know where to go after here, since the number isn't really a number I fell confortably working with. Did something go wrong here?[/hide]", "Solution_18": "Am I supossed to notice that the triangle in number 2 is a [hide]5-12-13[/hide] triangle? I didn't think I should have, and I complicated it I guess.", "Solution_19": "h_s_potter2002 wrote:Am I supossed to notice that the triangle in number 2 is a [hide]5-12-13[/hide] triangle? I didn't think I should have, and I complicated it I guess.\n\nYOu don't have to notice that it's a 5-12-13 triangle. Hint: [hide]Hypotenuse [tex]\\times[/tex] Altitude to the hypotenuse/2=the area of the triangle[/hide]", "Solution_20": "Help with 2.\n\n[hide]It's a 5-12-13 triangle. Doesn't the altitude to the hypotenuse cut the hypotenuse in half?\n\nIf it does, that makes the hypotenuse 6.5, and so 6.5 2 +x 2 =12 2 . So then we get x 2 =101.75, and x=root101.75 \n\nIs this correct?[/hide]", "Solution_21": "[quote=\"NightFlarer\"]Help with 2.\n[spoiler]It's a 5-12-13 triangle. Doesn't the altitude to the hypotenuse cut the hypotenuse in half?\n[/quote]\r\n\r\nDefinitely no! A point lies on the perpendicular bisector of a segment iff it's equidistance from the endpoints. Since 5 is not equal to 12, it's impossible.", "Solution_22": "[hide]\n\n1. 2x+2y+2z = 16\n\n2x+2y = 6\n\n2z = 10\n\nz = 5\n\n2. Let h be the hypotenuse and a be the altitude to the hypotenuse.\n\n\n\nha = 30\n\n\n\nThe hypotenuse is 13, so\n\n\n\n13a = 30\n\na = 30/13.\n\n3.\n\n2+4+6+...+120 = 2(1+2+3+...+60) = 2/2 * (60*(60+1)) = 3660.\n\n[/hide]", "Solution_23": "oh my god...that are problems we are doing in school...except that it could be that we are not given the formula for number 3 but have to derive them on our own...\r\n\r\nPeter", "Solution_24": "Oh yeah, Peter, you really hit the point... and actually, even [i]my[/i] maths teacher, after I returned from the IMO, organized a little geometry contest in my class, and the problems given in that contest were much more \"challenging\" (oh dear, I hate this word) problems than the ones proposed in this thread...\r\n\r\nBut I can understand Rep123max's intention on posing those problems; he just wanted \"[i]the people who only visit the non math forums[/i]\" to post less spam and to learn the tiniest bit of maths... :D \r\n\r\n Darij (yes I know I'm spamming again, I'll try to post some more mathematical comments the next days)", "Solution_25": "Do you guys have any respect for people no as smart, or as old as you? There is a reason this part of the board is called Getting Started. The people here aren't as smart as IMO patricipants. A lot of people haven't even gone into Algebra yet, they can probably barely do the American qualifying rounds for their team. \r\n\r\nIf you don't want to try easy problems, then don't come to this part of the board, go to the Advanced or Undergraduate boards. Not everyone can get into IMO, or even do well in the qualifying rounds since we're barely in our teens and haven't learned much maths yet. \r\n\r\nSo basically, don't make fun of other people just becuase they don't know as much maths as you do.", "Solution_26": "Maybe you didn't notice that these problems are posted in the Getting Started area. Perhaps you should look up \"getting started\" in the dictionary. That should help. Or perhaps your English is poor and you think \"Getting Started\" means \"Advanced.\" Need I remind you that both of you were once beginners? It's sad how so many people become cocky once they make it to the IMO. \r\n\r\nAnd if I have misunderstood your messages, excuse me.", "Solution_27": "It appears to be the German thing over again:\r\n\r\n \" Deuchland Deuchland uber alles \".\r\n\r\n You guys had to do much better at IMO to make your statements.\r\n\r\n\r\n Pestich.", "Solution_28": "I feel I'm getting mocked... :blush: \r\n\r\nI agree with hspotter and nightflarer but Rep says this is for everyone to do. Maybe you IMO people should just keep it low a little bit.", "Solution_29": "Actually, maybe it was about a misinterpretation of the Getting Started topic. As far as I understand, it is a topic aiming at school students who are already good at maths in school but want to learn some more maths. At least, this is the usual aim of such fores. Helping students who are not even on the school level in maths is usually the task of pedagogists and, on the internet, the task of some other fores, which are devoted to helping people with their homework. Maybe I just misunderstand the intention of this forum - then I will be glad if you clear me up about these things. But if I am right thinking that this forum is devoted for school students already being at school niveau and now wishing to increase their level beyound what is usually studied in school - then the \"challenge\" problems posted in these thread clearly miss the point.\r\n\r\nConcerning the ages, both of us - Peter and me - are teens, too. Peter is 16, and I am 15. I am actually opposed to thinking that the mathematical abilities of someone are tied to his age - but even if you think otherwise, you will still agree that the main question is not the age.\r\n\r\nSo much concerning the posting of h_s_potter2002. Actually I am not sure about some of the points since I really have not seen any official and precise information what kind of students the Getting Started forum aims at, beyond the very laconic subtitle \"For beginners to learn from more experienced problems solvers.\" leaving the question whether the writer meant \"beginners\" in mathematical olympiads (that is the way I understand the title) or \"beginners\" in school mathematics. I admit that I hadn't read the http://artofproblemsolving.com/Forum/viewtopic.php?t=10857 thread before, but now I have read it and it didn't answer most of my questions, but all discussions, including this one, make me think that the users of the Getting Started forum are participants of some mathematical olympiads, although not the IMO, and not just school students seeking help with their homework.\r\n\r\nWhat concerns NightFlarer's post, the arguments were quite the same so I refer to above.\r\n\r\nFinally, about Pestich's reply, I think he gives just one reasonable argument or at least something that can be understood as an argument in the wider sense of the word. By this I mean the fact that we (Peter and me) were not the best scorers in this year's IMO. But actually we didn't claim anything like this. We just claimed that the problems posted in this thread were faaaar too easy, and in order to understand this it is not of much importance whether one has 42 points on the IMO or a honorable mention (many people who don't even have a HM on the IMO are far above this level of problems). This, BTW, has nothing to do with Germany. None of us spoke in the name of Germany, neither in the name of the German IMO team. Each of us spoke in his and only his name. Incidentally, I have emigrated from Russia, now living in Germany, my roots are Jewish and my mother was born in the Ukraine. I don't identify with any nationality just 'cuz I don't find this necessary (more to this point, I find this contraproductive)! I don't know what is Peter's opinion about this, but I can hardly imagine he spoke out of proudness for Germany.\r\n\r\n Darij[/i]", "Solution_30": "Ok. I will try to outline what kinds of problems should and should not go in the getting started forum:\r\n\r\nThese problems should be 1-3 step problems. By this I mean that all you have to do is apply a couple of formulas or draw in a few lines, etc. to finish the problem. \r\n\r\nAnything with the word prove in it is 99% of the time not supposed to be in this forum. \r\n\r\nAny kind of Olympiad problem regardless of country or how stupidly easy it was should go in another forum. \r\n\r\nThese problems should be from contests such as Mathcounts/AMC 8/AMC 10/ and easier AMC 12 problems. If you are not familiar with these contests here is a link for where you can find AMC 12 problems. \r\n\r\n[url]http://www.math.ksu.edu/main/events/hscomp/samples/amc12/sample.htm[/url]\r\n\r\nI also advise some of you newer people to look at old topics. You should get a sense of what this forum used to be. \r\n\r\nFinally, if you are not sure what level a problem is put it in Intermediate Forum. If it is too easy for that forum they can put it in here... if it is too hard they can advance it to a higher forum. \r\n\r\nYou IMO guys didn't just know how to do beginner stuff when you are born. You shouldn't expect everyone on this forum to know as much as you(or even half as much for that matter). This is the forum where people can acquire such knowledge. It seems like you are just trying to do away with the \"Getting Started\" forum and create another forum for olympiad problems. Maybe I should ve mentioned this earlier but this is not Getting Started to Olympiad problems it is Getting Started to problem solving in general. I am sorry if you were confused about that. You guys already have more than 10+ forums/subforums for Olympiad level problems. I can't imagine that would want more.\r\n\r\nI am sorry if I came across rude but I think that we really need to reach an understanding on what difficulty this forum should be. \r\n\r\nThank you,\r\n\r\nJoseph L.", "Solution_31": "Okay, looking at the problems posted here I find you are right. Thanks for giving a very precise definition of the aim of this forum, now it's clear.\r\n\r\n Darij", "Solution_32": "[hide]\n\n1. If 2x + 2y = 6, (x+y) = 3, and so z + 3 = 8, so z is 5.\n\n\n\n2. Since it is a right triangle, the altitude must be half the hypotenuse, but (13*6.5)/2 =/= 30. I don't think it's possible for a right angled triangle to have an area dependant on anything else but the hypotenuse or the adjacent sides.\n\n\n\n4. As the area of the semicircles are proportional to the squares of the lengths=, pythagoras will work here (hopefully). So the sum of the two semicircles on the legs will also be 29.7[/hide]", "Solution_33": "[quote=\"darij grinberg\"]Actually, maybe it was about a misinterpretation of the Getting Started topic. As far as I understand, it is a topic aiming at school students who are already good at maths in school but want to learn some more maths.[/quote]\n\nThere has been a misinterpretation here about what \"Getting Started\" means. The Getting Started forum is for AoPS-Mathlinks site participants who have little or NO experience with mathematics competitions at any level. Anyone who has been to IMO, even once, is far beyond the intended level of the Getting Started forum and shouldn't be answering challenge problems in this forum. \n\nIn the United States, and in China (at least), mathematics competitions start at the late elementary school level and include problems that involve NO knowledge of algebra. Problems at that level are very appropriate for this forum, even if they look easy to people who have more experience in mathematics competitions. The solution for finding harder problems is to look for them in the higher level forums (Intermediate, Advanced, and College Playground). The persistent problem on most forums on AoPS has been problems that are too hard for their stated level, not problems that are too easy, so I as moderator often \"promote\" problems to a higher level if they are posted here but not really at a getting started level. \n\n[quote=\"darij grinberg\"]Helping students who are not even on the school level in maths is usually the task of pedagogists and, on the internet, the task of some other fores, which are devoted to helping people with their homework. Maybe I just misunderstand the intention of this forum - then I will be glad if you clear me up about these things.[/quote]\n\nYes, you are misunderstanding the intent of this forum. Thank you for asking to be cleared up. On this Getting Started forum, anyone is welcome to ask informational questions about mathematics at even the most elementary level. It is best to post questions with the title \"Question: \" so that everyone is clear what is a question and what is a challenge problem. Anyone is also welcome to post challenge problems here, but we moderators ask that only the BEGINNING problem-solvers post answers to challenge problems, until the true beginners have had a chance to answer. The experienced problem-solvers may then comment (preferably in spoiler text) on the solutions posted by the beginners. Please give the beginners a chance to practice answering easy challenge problems, and please remember that everyone has to start at the beginning. \n\n[quote=\"darij grinberg\"]Concerning the ages, both of us - Peter and me - are teens, too. Peter is 16, and I am 15.[/quote]\n\nSeveral of the participants on this forum are eleven years old. In any case, some of the older participants have had very limited opportunities to learn mathematics well at school (not all schools in all places are very good) and have no math competition experience. This forum is designed to help the beginners, even if they are older than you are. \n\n[quote=\"darij grinberg\"]Actually I am not sure about some of the points since I really have not seen any official and precise information what kind of students the Getting Started forum aims at, beyond the very laconic subtitle \"For beginners to learn from more experienced problems solvers.\" leaving the question whether the writer meant \"beginners\" in mathematical olympiads (that is the way I understand the title) or \"beginners\" in school mathematics.[/quote]\r\n\r\nYou admitted that you didn't read earlier threads on the intended level of this forum until AFTER you posted in this thread the first time. What you still need to think about is that the structure of mathematics competitions in different countries is not always the same. In China, for example, the term \"olympiad\" is used for mathematics competitions at all levels, but many of those competitions involved very elementary mathematics (although they are given to pupils of impressively young ages, by comparison to math contests in the United States). Much of the English-speaking world has math competitions for young people who are completely ignorant of algebra, and that is expecially the level that this Getting Started forum is intended for. If you have the experience level you indicate for yourself (and at the AGE you indicate for yourself, in any event), then you should be posting answers to challenge problems AT LEAST in the Intermediate forum, and probably mostly in the Advanced forum. The Getting Started forum is for younger people with less competition experience. \r\n\r\nI hope that clears everything up. Different countries have different systems, but on this international Web site the Getting Started is consciously intended to serve up EASY problems for absolute beginners in any form of math competition.", "Solution_34": "Thanks a lot for explaining. I usually thought that \"Getting Started\" meant something like the former \"Beginner's Corner\" on MathLinks, but now I see it's a big difference...\r\n\r\n Darij", "Solution_35": "yeah, i agree with darij in any point. i thought as well this \"getting started\" would already deal with easy olympiad level problems, sorry for misunderstanding. i won't do such a post in the future.\r\nand though i am actually a german(from the former gdr) and not immigrated like darij, i don't hold much onto the fact i am one and words like \"deutschland deutschland ueber alles\" would never come out of my mouth.\r\nso, again sorry for that post.\r\n\r\nPeter", "Solution_36": "some problems like that you are doing at school Peter? Lucky you. I'm as old as you're but I started in school first when I was seven.\r\nI have to say some of that problems I would have to think some time but I would be able to solve them. And I'm a real unexperienced problem solver. But maybe it's just because of my maths teacher. Once he wasn't able to solve a problem into our math book 10 minutes later I solved it for him (and these are real easy ones). Later I told it to my boyfriend who was a member of the German IMO team 2003 and who had the same math teacher he just asked: What do you expect from a basketball player?\r\nSo maybe I have just really bad luck with this stupid teacher." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "geometry", "similar triangles", "perpendicular bisector" ], "Problem": "[quote]Let ${\\cal C}_1$ and ${\\cal C}_2$ be concentric circles, with ${\\cal C}_2$ in the interior of ${\\cal C}_1$. From a point $A$ on ${\\cal C}_1$ one draws the tangent AB to ${\\cal C}_2 (B\\in {\\cal C}_2)$. Let $C$ be the second point of intersection of $AB$ and ${\\cal C}_1$, and let $D$ be the midpoint of $AB$. A line passing through $A$ intersects ${\\cal C}_2$ at $E$ and $F$ in such a way that the perpendicular bisectors of $DE$ and $CF$ intersect at a point $M$ on $AB$. Find, with proof, the ratio $AM/MC$.[/quote]\r\nI like this problem. You should too. It is very simple :lol:", "Solution_1": "Yeah. 5/3.", "Solution_2": "[quote=\"probability1.01\"]Yeah. 5/3.[/quote]\r\nwhat part of \"with proof\" do you not understand, the \"with\" or the \"proof\"?? JK, lol. proof, please :D", "Solution_3": "[hide]WLOG, assume that length AB = 1\nLet x = length AE\n\nAE * AF = AB * AB by a certain geometry theorem with a name I don't remember\nAF = 1/x\n\nAfter this point, I assume E is between A and F rather than F between A and E. If this assumption is not true, just switch all references to E and F past this point.\nAE/AD = AC/AF, since x/(1/2) = 2/(1/x) = 2x\ntriangle ADE ~ triangle AFC since sides are proportional, and angle DAE is equal to itself\nangle ADE = angle AFC by similar triangles\nangle CDE is supplementary to ADE, and therefore supplementary to AFC\ntherefore C, F, E, and D are cyclic.\nSince M is on the perpendicular bisector of CF and DE, it must be the center of this circle.\n\nCM = MD = 1/2 * 3/4 = 3/8\nAM = 3/8 + 1/4 = 5/8\nAM/MC = 5/3[/hide]\r\nMy geometry is a bit rusty.", "Solution_4": "[quote=\"TMill\"][hide]WLOG, assume that length AB = 1\nLet x = length AE\n\nAE * AF = AB * AB by a certain geometry theorem with a name I don't remember\nAF = 1/x\n\nAfter this point, I assume E is between A and F rather than F between A and E. If this assumption is not true, just switch all references to E and F past this point.\nAE/AD = AC/AF, since x/(1/2) = 2/(1/x) = 2x\ntriangle ADE ~ triangle AFC since sides are proportional, and angle DAE is equal to itself\nangle ADE = angle AFC by similar triangles\nangle CDE is supplementary to ADE, and therefore supplementary to AFC\ntherefore C, F, E, and D are cyclic.\nSince M is on the perpendicular bisector of CF and DE, it must be the center of this circle.\n\nCM = MD = 1/2 * 3/4 = 3/8\nAM = 3/8 + 1/4 = 5/8\nAM/MC = 5/3[/hide]\nMy geometry is a bit rusty.[/quote]\r\nvery good :lol: and welcome to AoPS" } { "Tag": [], "Problem": "As you know, Cantor's diagonalization can be used to prove that EXPTIME problems exist, by listing out all the PTIME programs, and then listing out the decision problems each one solves, selecting all the values on the diagonal and flipping them, thus creating a new decision problem that is not solved by any program in PTIME. However, why can't we use the same proof, but list out all possible turing machines instead of just the PTIME ones? Then when we pick out the values along the diagonal and flip them, we get a decision problem that cannot be solved by any turing machine, thus its unsolvable. But it can't be unsolvable because you can always generate the nth turing machine, calculate the output and flip it to output the answer.\r\n\r\nThere's a problem in the logic somewhere here, but I'm not sure where it is.", "Solution_1": "[i] But it can't be unsolvable because you can always generate the nth turing machine, calculate the output and flip it to output the answer.[/i]\r\n\r\nThe problem is in the \"calculate the output\" part. For an arbitrary Turing machine you can not calculate the output AND THEN flip it, because a Turing machine can reject a word by making an infinite computation that never reaches a final (accepting) state.\r\n\r\nIn other words, what you found is just a proof that there really are problems that are \"not solvable\", or, better phrased, you constructed a language that is not recursively enumerable. (Please note the difference between solvable=recursive and partially solvable=recursively enumerable problems/languages.)", "Solution_2": "ahh okay thanks for the clarification." } { "Tag": [ "algebra", "polynomial", "inequalities", "complex numbers", "advanced fields", "advanced fields unsolved" ], "Problem": "This problem is difficult for me . Can anyone help me ? :) \r\n\r\nLet $A$ be a commutative Banach Algebra with Unit . Suppose $x\\in A$ and the set of all polynomials in $x$ is dense in $A$ .\r\nProve that the completion of the spectrum of $x$ (in $\\mathbb{C}$) is connected .", "Solution_1": "What do you mean by \"completion of the spectrum\"? :? My first guess was that you meant \"closure of the spectrum\", but that would be pointless, since the spectrum is already closed (compact, actually).", "Solution_2": "Oh , I'm sorry .My English is terrible :( \r\nDenote $e$ the unit of $A$\r\nThe spectrum of $x$ is the set of all complex numbers $\\lambda$ such that $x-\\lambda e$ is not invertible . We denote it $\\sigma (x)$ .\r\nThe complement of $\\sigma (x)$ is $\\mathbb{C}\\backslash \\sigma (x)$ :)", "Solution_3": "You didn't need to actually define the terms, jusy mention that instead of \"completion\" you meant \"complement\" :).\r\n\r\nSuppose $\\mathbb C\\setminus\\sigma(x)$ has at least one bounded connected component $U$, and let $\\lambda\\in U$. Take some $\\varepsilon>0$ (we'll determine later how small it should be), and a complex polynomial $P$ such that $\\|P(x)-(x-\\lambda e)^{-1}\\|<\\varepsilon\\ (*)$. From the fact that every complex homomorphism of $A$ has norm at most $1$ and $(*)$ we deduce that $|P(z)-(z-\\lambda)^{-1}|<\\varepsilon,\\ \\forall z\\in\\sigma(x)$ ($\\sigma(x)$ is the set of all images of $x$ through complex homomorphisms of $A$), which means that $|(z-\\lambda)P(z)-1|<\\varepsilon M,\\ \\forall z\\in\\sigma(x)\\ (**)$, where $M=\\max_{\\omega\\in\\sigma(x)}|(\\omega-\\lambda)^{-1}|$. By the Maximum Modulus Principle, the inequality $(**)$ holds for all $z\\in U$ as well, and by putting $z=\\lambda$ we get $1<\\varepsilon M$. Taking $\\varepsilon<\\frac 1M$ in the beginning gives the desired contradiction." } { "Tag": [ "probability", "search" ], "Problem": "If someone could explain how to do the following problems, I'd greatly appreciate it.\r\n\r\n$ \\bold{1.}$ A town contains $ 4$ people who repair televisions. If $ 4$ sets break down, what is the probability that exactly $ i$ of the repairers are called? Solve the problem for $ i\\equal{}1,2,3,4$. What assumptions are you making?\r\n\r\n[hide=\"Answers\"]$ P(i\\equal{}1)\\equal{}\\frac{1}{64}$\n$ P(i\\equal{}2)\\equal{}\\frac{21}{64}$\n$ P(i\\equal{}3)\\equal{}\\frac{36}{64}$\n$ P(i\\equal{}4)\\equal{}\\frac{6}{64}$[/hide]\n\n$ \\bold{2.}$ The second Earl of Yarborough is reported to have bet at odds of $ 1000$ to $ 1$ that a bridge hand of $ 13$ cards would contain at least one card that is ten or higher. (By [i]ten or higher[/i] we mean that it is either a ten, a jack, a queen, a king, or an ace.) Nowadays, we call a hand that has no cards higher than $ 9$ a [i]Yarborough[/i]. What is the probability that a randomly selected bridge hand is a Yarborough?\n\n[hide=\"Note\"]I found a solution (trying to search to avoid posting on here) [url]http://www.physicsforums.com/showthread.php?t=199138[/url] at that link, but I cannot make sense of what their final answer is and how they arrived at it.[/hide]", "Solution_1": "No one? I also have another problem I'm interested in:\r\n\r\n$ \\bold{3.}$ An urn contains $ n$ red and $ m$ blue balls. They are withdrawn one at a time until a total of $ r$, $ r \\leq n$, red balls have been withdrawn. Find the probability that a total of $ k$ balls are withdrawn.\r\n[i]Hint:[/i] A total of $ k$ balls with be withdrawn if there are $ r \\minus{} 1$ red balls in the first $ k \\minus{} 1$ withdrawals and the $ k^{\\text{th}}$ withdrawal is a red ball.", "Solution_2": "[quote=\"skimnc\"]No one?[/quote]\r\n\r\nBe patient -- the members of this forum are not an on-call service to help you with your work on your schedule :)\r\n\r\nFor your second question, post 2 is the only worthwhile portion of it.", "Solution_3": "bump\r\n\r\nI figured out $ \\bold{1.}$ and $ \\bold{2.}$, but if anyone could help with $ \\bold{3.}$, I would greatly appreciate it." } { "Tag": [ "probability", "geometry", "3D geometry", "ratio" ], "Problem": "1. So far a student's math exam grades are 90, 96, 82, and 88. It takes a grade of 70 or better to pass and all grades are whole numbers. If the student receives a passing grade on each of the next two exams, what is the probability that the student will have an average of at least 90? Assume that any score is equally likely to occur and that all exams have 100 possible points.\r\n\r\n2. A tank in the form of a rectangular solid with base area of 12 square feet has 8 inches of water in it. If a 1 foot gold cube is placed in the tank, by how many inches will the water rise?\r\n\r\n[hide]I did these problems and got 153/961 and 1 respectively. But appearantly the answers are 133/961 and 8/11. Can someone point out where (if) I might be going wrong?[/hide]", "Solution_1": "Okay, I'm kinda new at this so try and bear with me (my Algebra is a little rusty). In order to find the first problem I put X in for the two unknown grades, and solved it so that the answer was 90(the minumum grade). X comes out to 92 then. That means there are 9 different grades that can make the average grade at least 90. That also means that there are 22 grades that would create an average of below 90%. So there fore the ratio is 9/22. \r\n I Thought the second problem was easy, but I came up with the same answer( 1 inch).\r\n!!!!![b]Can someone pleaase help me [/b]!!!!", "Solution_2": "In the second problem, the cube is 12x12x12 so it has a volume\r\n$12*12*12 = 1728\\mbox{ in^3}$\r\nSo when the cube is placed in the water, it displaces that much water. The volume of one inch of water is $12\\mbox{ in^3}$ so the water rises $\\frac{1728}{12} = 144\\mbox{ in.}$", "Solution_3": "note that for number 2, not all of the cube is submerged in water..since the cube is 12 inches high and the water is only 8. \r\nalso, as more of the cube goes into the tank, the higher the water level is, thus the more of the cube is in the water, and thus the water level gets higher still...\r\nthis looks like calc to me....", "Solution_4": "ah oops!\r\nyes it does look like calc now that i think about it.. i also didnt see that the area of the base is 12 sq ft for some reason", "Solution_5": "No, it's not really calc, but it has a trick. Have in mind that the total volume of water is the same, but without the cube its base is the rectangular of the pool, and with the cube its the rectangular minus square of the cube... now it should be easy.", "Solution_6": "No, it's not really calc, but it has a trick. Have in mind that the total volume of water is the same, but without the cube its base is the rectangular of the pool, and with the cube its the rectangular minus square of the cube... now it should be easy.", "Solution_7": "Ok so its\r\n$1728(8) = (1728 - 144)x$\r\n$13824 = 1584x$\r\n$x = \\frac{96}{11}$\r\n$\\frac{96}{11} - 8 = \\frac{8}{11}$", "Solution_8": "the first one i got 483/1010 which i used my calculator... but for the second one i got 12... notice it says inches while a 144 in^3 cube is dropped. so basically,\r\n\r\n8*12=96 and 12*12=144\r\nso the water is 96 in^3 and the cube is 144 in^3\r\nand since it is already 8 inches high, it rises 144/12 inches or 12 inches", "Solution_9": "remember that gold is denser then water... so it [u]will[/u] sink... and the base of the cube dropped epuals the base of the container (tank) sorry :lol: :lol:" } { "Tag": [ "AMC" ], "Problem": "Can somebody explain a few approaches and differences for the following problems?\r\n\r\nLet a, b, c, d, and e be integers whose sum is 150. How many possible combinations are there for the set {a,b,c,d,e}?\r\n\r\nLet a,b,c,d and e be integers whose sum is 150 where a10$ ;)", "Solution_4": "For the second proof, I have two questions.\r\n\r\nFirst, is there a way to find a proof without using $ \\mod$?\r\n\r\nSecondly, what's FLT?\r\n\r\nThanks.", "Solution_5": "[quote=\"314.math\"]For the second proof, I have two questions.\n\nFirst, is there a way to find a proof without using $ \\mod$?\n\nSecondly, what's FLT?\n\nThanks.[/quote]There is a proof without mod, I'm sure someone will post that. FLT stands for \"Fermat's Little Theorem\" in here.", "Solution_6": "Actually i think the stricter divisibility is that $ n^5 \\minus{} n \\equiv 0 \\mod 30$", "Solution_7": "[quote=\"madness\"]Actually i think the stricter divisibility is that $ n^5 \\minus{} n \\equiv 0 \\mod 30$[/quote]Is that the largest positive integer answering my question posted above? Prove it :P", "Solution_8": "that can be easily proved by factorisation with cases :)", "Solution_9": "[hide=\"1\"]\nIf all odd numbers are across from another, then they come in pairs, so there's an even number of odd numbers.\n1 to 2n-1 are the odds, there being n of them. From above 2 divides n, so n=2k, k integer, then number of integers = 2n = 2(2k) = 4k.\n[/hide]\n\n[hide=\"2\"]\nn^5 - n = n(n-1)(n+1)(n^2+2), one of n and n-1 are even, one of n, n-1, n+1 is divisible by 3, by FLT, 5|n^5-n, so 30|n^5-n for all n, and for n=2, n^5-n=30, so 30 is the largest number that divides all n^5-n\n[/hide]", "Solution_10": "I think this might be the solution for second problem, \r\n\r\n10 | n^5 - n , that means n^5 - n = 10*k\r\n\r\nthat is true if we prove that the last digit of n^5 = n for all integers, \r\nthe last digit in n^x , repeates periodically, so all that is left is to show that \r\nlast digit of n^5 = n, for numbers from 1 to 9. \r\n\r\ni am sure there must be more elegant way but i did it with putting everything on paper(i have written just the last digits to show that they occur periodically):\r\n\r\n1 1 1 1 1 \r\n2 4 8 6 2\r\n3 9 7 1 3\r\n4 6 4 6 4\r\n5 5 5 5 5\r\n6 6 6 6 6\r\n7 9 3 1 7\r\n8 4 2 6 8\r\n9 1 9 1 9\r\n\r\nCan please someone comment that solution tell if it's even suitable?", "Solution_11": "I got the first problem now, but am still confused about the second one. Could someone help me, please? Thanks!", "Solution_12": "[hide=\"Painless solution to number 2\"]\nNote that $ s(n) \\equal{} n^5 \\minus{} n \\equal{} (n \\minus{} 1)n(n \\plus{} 1)(n^2 \\plus{} 1)$, and it suffices to prove divisibility by 2 and 5:\n\n$ 2|n(n \\plus{} 1) \\implies s(n) \\equiv 0 \\pmod{2}$\n\n$ n^2 \\plus{} 1 \\equiv (n \\minus{} 2)(n \\plus{} 2) \\pmod{5}$\n$ \\implies s(n) \\equiv (n \\minus{} 2)(n \\minus{} 1)n(n \\plus{} 1)(n \\plus{} 2) \\equiv 0 \\pmod{5}$[/hide]\r\n\r\n[quote=\"Gonky\"]...that is true if we prove that the last digit of n^5 = n for all integers, \nthe last digit in n^x , repeates periodically, so all that is left is to show that \nlast digit of n^5 = n, for numbers from 1 to 9.\n...\nCan please someone comment that solution tell if it's even suitable?[/quote]\r\n\r\nIt sure is.", "Solution_13": "Is there anything wrong with my proof? I know it is proves the statements but it is strong enough? Can it be taken as less \"acceptable\" than the others?", "Solution_14": "The inductive proof is fine, although you could write it up more cleanly.", "Solution_15": "Thanks t0rajir0u. :) \r\nHow could I present it more cleanly?", "Solution_16": "You don't make it clear that $ k^5 \\equiv k \\bmod 10$ is your inductive step; it looks like you're assuming the result to prove it.", "Solution_17": "i am really quite new to this thing ... but what is mod ? I have some clue what it does, but what is the definition?", "Solution_18": "[quote=\"t0rajir0u\"]You don't make it clear that $ k^5 \\equiv k \\bmod 10$ is your inductive step; it looks like you're assuming the result to prove it.[/quote]\r\n\r\nThanks. :)", "Solution_19": "[quote=\"Gonky\"]i am really quite new to this thing ... but what is mod ? I have some clue what it does, but what is the definition?[/quote]\r\n\r\n[b]Definition:[/b] We say that $ a \\equiv b \\bmod m$ (read: $ a$ is congruent to $ b$ modulo $ m$) if and only if $ m | a \\minus{} b$. In other words, $ a, b$ give the same remainder upon division by $ m$.\r\n\r\n[b]Examples:[/b]\r\n\r\n$ 12 \\equiv 5 \\bmod 7$\r\n$ 22 \\equiv \\minus{}11 \\bmod 11$\r\n$ 87 \\equiv \\minus{}13 \\bmod 25$\r\n\r\n[b]Properties:[/b]\r\n\r\n$ a \\equiv a \\bmod m$\r\n$ a \\equiv b \\bmod m \\Leftrightarrow b \\equiv a \\bmod m$\r\n$ a \\equiv b \\bmod m, b \\equiv c \\bmod m \\implies a \\equiv c \\bmod m$\r\n$ a \\equiv b \\bmod m \\implies a \\plus{} c \\equiv b \\plus{} c \\bmod m$\r\n$ a \\equiv b \\bmod m \\implies ac \\equiv bc \\bmod m$\r\n\r\nOne useful way to think about mods (especially $ \\bmod 10$) is that the last digit in base $ m$ must be equal. So $ k^5 \\equiv k \\bmod 10$ mean that $ k^5, k$ have the same last digit in base $ 10$. For example,\r\n\r\n$ 2^5 \\equal{} 32$\r\n$ 3^5 \\equal{} 243$\r\n$ 4^5 \\equal{} 1024$\r\n$ 5^5 \\equal{} 3125$\r\n\r\nand so forth." } { "Tag": [ "\\/closed" ], "Problem": "I keep looking, and I can't find a way.... Is there a way to display all the threads to which one is subscribed, and/or all the subscribed threads with new posts?\r\n\r\nThanks!", "Solution_1": "One thing you can do instead of subscribing is 'add the topic to favorites' (there's a little link for it near the top when you are looking at the topic). You can then click the favorites button up top to see all your favorites.", "Solution_2": "There will be a watched topics list again (it used to be on the old ml board), but I need to debug it a little bit, since some of the structural matters of the forum have changed. Until then you can use Favorite topics :)", "Solution_3": "That is one option, but I get automatically subscribed to threads to which I post, so using the \"subscribed\" threads is a more appealing option, if it is possible. I'm surprised this feature isn't build into the bb code, as it's one I've seen on many other boards.", "Solution_4": "Thanks for the update, Valentin, I'll watch for it :)", "Solution_5": "[quote=\"mathmom\"]I'm surprised this feature isn't build into the bb code, as it's one I've seen on many other boards.[/quote]You shouldn't be. First of all this option is available default only on paid boards (such as invision/vbulletin). Second of all, as you might have seen this is not a regular bboard, it is very \"customized\" :D", "Solution_6": "[quote=\"Valentin Vornicu\"] Second of all, as you might have seen this is not a regular bboard, it is very \"customized\" :D[/quote]\r\n\r\nYes, but it seems \"better\" than most other boards where I have seen that feature, so that is why I was surprised. :)", "Solution_7": "[quote=\"mathmom\"]Yes, but it seems \"better\" than most other boards where I have seen that feature, so that is why I was surprised. :)[/quote]Thanks! :D" } { "Tag": [ "function", "induction", "algebra solved", "algebra" ], "Problem": "Let $f$ be a function defined on the positive integers such that \\[ f(1) = 2,\\;f(2) = 1,\\;f(3n) = 3f(n), \\\\ f(3n + 1) = 3f(n) + 2,\\;f(3n + 2) = 3f(n) + 1.\\]\r\nHow many $n$ satisfy $n \\leq 2004$ and $f(n) = 2n$?", "Solution_1": "It looks to me (it doesn't have to be correct, though :D) that $f(n)$ is obtained from $n$ by writing it in base $3$ and keeping all its $0$'s while changing $1$'s to $2$'s and vice-versa. I also believe that this means that the numbers for which $f(n)=2n$ are precisely those which have no $2$'s in their ternary representation.\r\n\r\nIs this even remotely accurate,Arne? :)", "Solution_2": "Well it's quite correct Grobber ;) Of course there is still a proof to be given (induction :D)\r\nThe answer should be 127 :)", "Solution_3": "You don't really need an induction of any kind.\r\n\r\nSince for $n \\geq 2$, $\\{f(i)\\}_{i=1}^n$ uniquely determines $f(n+1)$, according to the given relations, it is sufficient to find a single function $f$ satisfying these conditions.\r\nAla grobber, we observe that $f(n)=$ $n$'s ternary representation with 1's and 2's switched trivially satisfies the given equations implying that it is valid and, by our above remarks, is the unique $f$ satisfying these conditions.\r\nThen obviously $f(n) = 2n$ iff $n$ contains no $2$'s in its ternary representation. The largest such $n \\leq 2004$ is $1093=1111111_3$. Reading this number in binary implies that there are indeed the $127$ different $n$ meeting the requirements." } { "Tag": [ "linear algebra", "matrix" ], "Problem": "Find $ 2 \\times 2$ matrices $ A, B$ such that $ AB \\equal{} 0$ but $ BA \\neq 0.$", "Solution_1": "$ A\\equal{}(a_{i,j})$ with $ a_{1,1}\\equal{}a_{2,2}\\equal{}\\minus{}1$ and $ a_{1,2}\\equal{}a_{2,1}\\equal{}1$ for $ B \\equal{}(b_{i,j})$ with $ b_{1,1}\\equal{}b_{2,1}\\equal{}3$ and $ b_{1,2}\\equal{}b_{2,2}\\equal{}2$.", "Solution_2": "How did you construct this, and is this the only one?", "Solution_3": "it's not the only one , to construct such matrix , I fix a matrix $ A$ like in my previous post , for $ B$ set $ B \\equal{} (b_{i,j})$ with $ b_{1,1} \\equal{}x, b_{2,1} \\equal{} z , b_{1,2} \\equal{} y$ and $ b_{2,2} \\equal{} t$.solve the system $ AB\\equal{}0$ to find that we must have $ x\\equal{}z$ and $ y\\equal{}t$ , and for $ BA \\neq 0$ we choose $ x\\minus{}y \\neq0$.", "Solution_4": "General solution:\r\n$ AB\\equal{}0$ implies that $ A,B$ are simultaneously triangularizable.\r\nIn this case $ A\\equal{}\\begin{pmatrix}a&b\\\\0&c\\end{pmatrix},B\\equal{}\\begin{pmatrix}d&e\\\\0&f\\end{pmatrix}$\r\nwith $ ad\\equal{}cf\\equal{}0,ae\\plus{}bf\\equal{}0,db\\plus{}ec\\not\\equal{}0$.\r\nThen the general solution is:\r\n$ A\\equal{}P\\begin{pmatrix}0&b\\\\0&c\\end{pmatrix}P^{\\minus{}1},B\\equal{}P\\begin{pmatrix}d&e\\\\0&0\\end{pmatrix}P^{\\minus{}1}$ with $ db\\plus{}ec\\not\\equal{}0$ and $ P$ invertible.", "Solution_5": "Actually,a more general result holds: We can find two matrices $ A,B \\in\\mathcal{M}_n(\\mathbb{R})$ such that $ AB\\equal{}0$ and $ BA\\neq 0$, where $ n\\geq 2$." } { "Tag": [ "trigonometry", "analytic geometry", "graphing lines", "slope" ], "Problem": "How many points of intersection do the graphs of $ y \\equal{} cos(\\frac {\\pi}{x})$ and $ y \\equal{} 100x \\plus{} 2$ have?", "Solution_1": "Looks interesting. Can you give us a hint of how big the number will be?", "Solution_2": "[quote=\"chetjan\"]How many points of intersection do the graphs of $ y \\equal{} cos(\\frac {\\pi}{x})$ and $ y \\equal{} 100x \\plus{} 2$ have?[/quote]\r\n[hide=\"Hint on what x can be\"]$ \\cos(\\frac{\\pi}{x})\\equal{}100x\\plus{}2$. cosine is between $ \\minus{}1$ and $ 1$. \n$ \\minus{}1\\le 100x\\plus{}2\\le 1$\n$ \\minus{}\\frac{3}{100}\\le x\\le \\minus{}\\frac{1}{100}$[/hide]", "Solution_3": "[quote=\"thejackchaner\"]Looks interesting. Can you give us a hint of how big the number will be?[/quote]\r\n[hide=\"General Hint\"]The number of intersection will be less than 100\n[hide=\"Specific Hint\"]The number of intersections will either be 67 or 68[/hide][/hide]", "Solution_4": "[hide=\"Answer\"]67\n[hide=\"Solution\"] all you need to know is y(x) and y ' (x) for the given x.\n\nWhen x=-0.03, the wavy line is at y=-0.5 and heading upwards steeply. The straight line is at y=-1.\nThe wavy line will then hit 1 when x=-1/34. It hasn't intersected yet.\nBy the time x reaches -1/36, when the wavy line is again at 1, it will have intersected the straight line twice.\nBy the time x reaches -1/38, it will have intersected it twice more.\nAnd so on...\nBetween -1/98 and -1/100, it will intersect twice, and then intersect once more at the endpoint (-1/100, 1).\n\nThere are 33 even numbers from 36 to 100 inclusive. Double that to get 66. And add the last intersection to get 67.\n\nTo see that there are two intersections between -1/98 and -1/100, remember that, at (-1/100,1) the slope of the straight line is 100, and the slope of the wavy line is 0. So the straight line is under the wavy line for a brief moment beforehand.[/hide][/hide]" } { "Tag": [], "Problem": "Mr, Mendez awards extra credit on quizzes to his students with quiz grades that exceed the class mean. Given that 107 students take the same quiz, what is the largest number of students who can be awarded extra credit?", "Solution_1": "Let's say one person gets 0 and everyone else (106 people) gets 100. The mean will then be less than 100. Then the total number of people who can get extra credit is $ \\boxed{106}$." } { "Tag": [ "AMC", "AIME", "trigonometry", "geometry", "geometric transformation", "reflection", "AoPSwiki" ], "Problem": "Hello. I was just wondering if you guys wouldn't mind helping me come up with as many non-trig solutions to this problem as possible. I pressed the link to the old thread and there was only one solution not involving trigonometry (construction of equilateral triangle $ CMN$ such that $ N$ is the reflection of $ M$ across median $ CD$). Anyways, I have included the problem as well as my solution with cyclic quadrilaterals. Thanks for your help.\r\n\r\nTriangle $ ABC$ is isosceles with $ AC\\equal{}BC$ and $ \\angle{ACB}\\equal{}106^{\\circ}$. Point $ M$ is in the interior of the triangle so that $ \\angle{MAC}\\equal{}7^{\\circ}$ and $ \\angle{MCA}\\equal{}23^{\\circ}$. Find $ \\angle{CMB}$.\r\n\r\n[hide=\"Cyclic Quadrilateral Solution\"]\nLet $ D$ be the midpoint of $ AB$. Let $ E$ be the intersection of $ AB$ and $ CM$, and let $ H$ be the intersection of $ AM$ and $ CD$. By angle chasing one can determine that $ \\angle{AEC}\\equal{}120$, $ \\angle{AHC}\\equal{}120$, $ \\angle{CED}\\equal{}60$ and $ \\angle{DAH}\\equal{}30$. Because $ DH$ is the perpendicular bisector of $ AB$, then $ AH\\equal{}BH$ and $ \\angle{HAB}\\equal{}\\angle{HBA}\\equal{}30$. Therefore, $ \\angle{AHB}\\equal{}120$ and since $ \\angle{MEB}\\plus{}\\angle{BHA}\\equal{}60\\plus{}120\\equal{}180$, $ MHBA$ is cyclic. Also, because $ \\angle{AHC}\\equal{}\\angle{AEC}\\equal{}120$, $ HEAC$ is cyclic. Therefore, by arc-angle theorem $ \\angle{ACE}\\equal{}\\angle{AHE}\\equal{}\\angle{MBE}\\equal{}23$. By pure angle chasing, one can obtain that $ \\angle{CMB}\\equal{}83$\n[/hide]", "Solution_1": "You don't really need trigonometry - you are correct.\r\n\r\nHere is a non trigonometric solution from the AoPSWiki:\r\n\r\n[img]http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/0/a/c/0acadfa5d916afcaa6d423b731b121aa2a373bf2.png[/img]\r\n[img]http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/a/9/3/a93a41f2e73a496bb66b6bc1a1372ef1bb7ae0bb.png[/img]\r\nTake point $ N$ inside $ \\triangle ABC$ such that $ \\angle CBN \\equal{} 7^\\circ$ and $ \\angle BCN \\equal{} 23^\\circ.$\r\n\r\n$ \\angle MCN \\equal{} 106^\\circ \\minus{} 2\\cdot 23^\\circ \\equal{} 60^\\circ.$ Also, since $ \\triangle AMC$ and $ \\triangle BNC$ are congruent (by ASA), $ CM \\equal{} CN.$ Hence $ \\triangle CMN$ is an equilateral triangle, so $ \\angle CNM \\equal{} 60^\\circ.$\r\n\r\nThen $ \\angle MNB \\equal{} 360^\\circ \\minus{} \\angle CNM \\minus{} \\angle CNB \\equal{} 360^\\circ \\minus{} 60^\\circ \\minus{} 150^\\circ \\equal{} 150^\\circ.$ We now see that $ \\triangle MNB$ and $ \\triangle CNB$ are congruent. Therefore, $ CB \\equal{} MB,$ so $ \\angle CMB \\equal{} \\angle MCB \\equal{} \\boxed{083^\\circ}.$" } { "Tag": [ "Asymptote", "LaTeX" ], "Problem": "I'm wondering if it's about time we split the two help forums. There are many users asking questions about Asymptote so it would be nicer to have it in a separate forum.\r\n\r\nWhat do you guys think?", "Solution_1": "The vote was 8-1 so topic closed, Asymptote forum created. Please place all your Asymptote related questions in that forum." } { "Tag": [ "counting", "derangement", "factorial" ], "Problem": "What is $1!+2!!+!3+4!+5!!+!6$?\r\n\r\n[b]NOTE:[/b]$x!!\\ne (x!)!$ (It's called a double factorial...)\r\n\r\n\r\nEdited: Before, it ended up being an insanely large number, so I changed it so it would be smaller.\r\nEdited again: Too big to find out", "Solution_1": "whats $!3$? :maybe:", "Solution_2": "derangements\r\n\r\nif you take $(1,2,3)$ it is all the possible ways such that none of the numbers are in their original place like $(2,3,1)$ or $(3,1,2)$", "Solution_3": "Then isn't $!n$ just $n!-1$? :maybe:", "Solution_4": "[quote=\"stupidityismygam\"]\t\nderangements\n\nif you take $(1,2,3)$ it is all the possible ways such that [b]none [/b]of the numbers are in their original place like $(2,3,1)$ or $(3,1,2)$[/quote]\n[quote=\"i_like_pie\"]Then isn't $!n$ just $n!-1$? :maybe:[/quote]", "Solution_5": "[quote=\"i_like_pie\"]Then isn't $!n$ just $n!-1$? :maybe:[/quote]\r\nNo.\r\n\r\nFor example, $!4=9\\ne24-1$", "Solution_6": "[quote=\"i_like_pie\"]Then isn't $!n$ just $n!-1$? :maybe:[/quote]\r\n$!n=n!\\cdot\\sum_{k=0}^{n}\\frac{(-1)^{k}}{k!}$" } { "Tag": [ "geometry", "rectangle", "probability", "analytic geometry" ], "Problem": "A point $ (x,y)$ is randomly picked from inside the rectangle with vertices $ (0,0)$, $ (4,0)$, $ (4,1)$, and $ (0,1)$. What is the probability that $ x y. Now we just need to consider the square formed by the points (0, 0), (0, 1), (1, 0), and (1, 1). Exactly 1/2 satisfies x > y, since it is random and no one has higher probability than the other. Or, you can see that for a point (x, y) within the square to have x < y, the x coordinate must be on the line formed by f(x) = y, where y is the y - coordinate, but behind f(x) = x, where x is the x - coordinate. If the length of the square increases, the width increases the same amount, so the possible region is a triangle formed by the points (0, 0), (1, 1), and (0, 1). This square is 1/4 of the total figure, so the probability is 1/4 * 1/2 = 1/8.\n\n$ \\boxed{A}$[/hide]\r\nSorry if I made a careless mistake if I wasn't clear.", "Solution_2": "[hide] We can solve this by considering the intersection of the line y=x with the rectangle, then taking only the region in the rectangle above the line. The intersection points are (0,0) and (1, 1), so our desired region is a triangle with vertices (0, 0), (0, 1), and (1, 1). The area is hence $ \\frac{1}{2}$. The area of the rectangle is 4, so our answer is $ \\textbf{(A)}\\ \\frac{1}{8}$.[/hide]" } { "Tag": [ "probability", "ARML", "geometric sequence", "arithmetic sequence" ], "Problem": "One tries to pick 3 distinct numbers from {1,...34} so that their sum is divisible by 3. how many ways can this be done.\r\n\r\nWhen a certain biased coin is flipped 5 times the probability of getting heads exactly once is the same as of getting heads twice. What is the probability that the coin comes up heads 3 times out of 5.\r\n\r\nArml How many ways can you pick 3 numbers from the set {2^1,...2^2002} so that they form a geometric progression?", "Solution_1": "[hide=\"1\"]\nThis was the tie-breaker for last year's ARML, wasn't it?\n\nThere are 11 multiples of 3, 11 numbers 1 less than a multiple of 3, and 12 numbers 1 more than a multiple of 3.\n\nPossibilities:\nAll 3 are multiples of 3: $(11)(10)(9)/3!=165$\nAll 3 are 1 less than multiples of 3: $(11)(10)(9)/3!=165$\nAll 3 are 1 more than multiples of 3: $(12)(11)(10)/3!=220$\nOne of each: $(12)(11)(11)=1452$\n\nTotal: $165+165+220+1452=\\boxed{2002}$\n[/hide]\n\n[hide=\"2\"]\nLet $h$ be the probability of flipping a heads.\nThen, $1-h$ is the probability of flipping a tails.\n\nWe have $C_{1}^{5}h(1-h)^{4}=C_{2}^{5}h^{2}(1-h)^{3}$\n$(1-h)=2h$\n$h=\\frac13$\n\nThe probability of flipping 3 heads is:\n$C_{3}^{5}h^{3}(1-h)^{2}=(10)\\frac{1}{3^{3}}\\frac{2^{2}}{3^{2}}=\\boxed{\\frac{40}{243}}$\n\n[/hide]\n\n[hide=\"3\"]\nI will assume that this implies 3 different numbers.\n\nThis is the same as asking how many ways you can pick an arithmetic progression of lenth 3 out of the set {1,2,3,...,2002}.\n\nThere is one such arithmetic progression for any pair of starting and ending values that differ by a mulitple of 2.\n\nFor first term 1, there are (2001-1)/2 = 1000 such values for the third term.\nFor 2, there are (2002-2)/2 = 1000 such values.\nFor 3, there is 1 less value than there is for 1.\nFor 4, there is 1 less value than there is for 2.\netc.\nFor 1999, there is (2001-1999)/2 = 1 value\nFor 2000, there is (2002-2000)/2 = 1 value.\n\nThus there are a total of $\\frac{(1000)(1001)}{2}+\\frac{(1000)(1001)}{2}=\\boxed{1001000}$ geometric progressions in this set.\n[/hide]" } { "Tag": [ "symmetry" ], "Problem": "I need help with these:\r\n\r\n1) standard deviation = 15.6, 10% are above 17.2, find the mean\r\n\r\n2) mean = 88, 2.5% are below 50, find the standard deviation.\r\n\r\n3) While only 5% of babies have learned to walk by the age of 10 months, 75% are walking by 13 months of age. If the age at which babies develop the ability to walk can be described by a normal model, find the mean and standard deviation", "Solution_1": "Assuming these are all normal distributions:\r\n[hide=\"1.\"]\nThe Z score is given by $Z = \\frac{x-\\mu}{\\sigma}$. A table of Z-scores gives the probabilities of values less than or equal to Z. We want the value greater than Z so we have\n$1-\\frac{17.2-\\mu}{15.6}= .1$ The answer follows. [/hide]\n\n[hide=\"2.\"]\nAssuming normal distribution, this uses the [url=http://www.isixsigma.com/dictionary/Empirical_Rule-467.htm]Empirical Rule[/url]. So that $95\\%$ of the values are in $88 \\pm 2\\sigma$ or $5\\%$ of the values are greater OR less than $88\\pm2\\sigma$. Symmetry gives that $2.5\\%$ are below $2\\sigma$ so that $88-2\\sigma = 50 \\implies \\sigma = 19$. [/hide]\n\n[hide=\"3.\"]Use the Z-scores again to get a system of equations.\n$\\frac{10-\\mu}{\\sigma}=.05$\n$\\frac{13-\\mu}{\\sigma}=.75$\nSolve the first for $\\mu$ and then substitute into the second:\n$\\frac{13+.05\\sigma-10}{\\sigma}=.75\\sigma$ Solving for $\\sigma$ gives $\\sigma \\approx 4.286$. Then go back and solve for $\\mu$.[/hide]" } { "Tag": [ "calculus", "integration", "logarithms", "calculus computations" ], "Problem": "Prove the following inequality.\r\n\r\n$ \\int_0^1 2008^{x^{2008}}\\ dx\\geq \\left(1\\minus{}\\frac{1}{2008^{2008}}\\right)\\frac{1}{\\ln 2008}$.", "Solution_1": "$ \\int_0^1 2008^{x^{2008}}\\ dx >\\int_0^1 1 dx \\equal{}1 > \\left(1 \\minus{} \\frac {1}{2008^{2008}}\\right)\\frac {1}{\\ln 2008}$.", "Solution_2": "The problem seemed to be a stupid question. :oops: \r\n\r\nHere is my solution:\r\n\r\n$ \\int_0^1 2008^{x^{2008}}\\ dx\\geq \\int_0^1 2008^{2008x\\minus{}2007}\\ dx$\r\n\r\n$ \\equal{}\\left[\\frac{2008^{2008x\\minus{}{2007}}}{2008\\ln 2008}\\right]_0^1\\equal{}\\left(1\\minus{}\\frac{1}{2008^{2008}}\\right)\\frac{1}{\\ln 2008}$." } { "Tag": [ "geometry", "incenter", "inradius", "Pythagorean Theorem", "geometry proposed" ], "Problem": "Let $\\triangle ABC$ be a right triangle with hypotenuse $[BC]$. Let $[AH]$ be the altitude. Incircle of $\\triangle ABH$ touches $[AB]$ at $D$. Incircle of $\\triangle ACH$ touches $[AC]$ at $E$. Let $E'$ be a point on $[AB]$ such that $|AE|=|AE'|$. If $M$ is the midpoint of $[AH]$, and $I$ is the incenter $\\triangle ABH$, prove that $\\widehat{EE'M}= \\widehat{IHD}$.", "Solution_1": "i apologize to everyone for this question. I had forgotten the point $E'$ in the problem. forgive me!", "Solution_2": "AE' = AE, $\\triangle AEE'$ is isosceles right, $\\angle AE'E = 45^\\circ.$ HI bisects the right angle $\\angle AHB,$ $\\angle AHI = 45^\\circ.$ $\\angle DHI = \\angle AHI-\\angle AHD = 45^\\circ-\\angle AHD$ and $\\angle ME'E = \\angle AE'E-\\angle AE'M = 45^\\circ-\\angle AE'M.$ To show $\\angle DHI = \\angle ME'E,$ it is sufficient to show that $\\triangle AHD \\sim \\triangle AE'M$ with the common angle $\\angle DAH = \\angle MAE'$ are similar. Denote a = BC, b = CA, c = AB, h = AH and $s_{b},\\ s_{c}$ the semiperimeters of $\\triangle ABH,\\ \\triangle ACH.$ As $\\angle A$ is right and AH the A-altitude\r\n\r\n$h = AH = \\frac{bc}{a},\\ BH = \\frac{c^{2}}{a},\\ CH = \\frac{b^{2}}{a}$\r\n\r\n$AD = s_{b}-BH = \\frac{h+c-BH}{2}= \\frac{c(a+b-c)}{2a}$\r\n \r\n$AE' = AE = s_{c}-CH = \\frac{h+b-CH}{2}= \\frac{b(c+a-b)}{2a}$\r\n\r\n$\\frac{AD}{AH}= \\frac{AM}{AE'}\\ \\Longleftrightarrow\\ AD \\cdot AE' = AM \\cdot AH\\ \\Longleftrightarrow$\r\n\r\n$\\frac{bc(a+b-c)(c+a-b)}{4a^{2}}= \\frac{h^{2}}{2}= \\frac{b^{2}c^{2}}{2a^{2}}\\ \\Longleftrightarrow\\ a^{2}-(b-c)^{2}= 2bc$\r\n\r\nwhich is true because of Pythagorean theorem $a^{2}= b^{2}+c^{2}.$", "Solution_3": "in my solution, the first part is same\r\n[b]lemma:[/b]\r\nLet $\\triangle ABC$ be a right triangle with $\\angle B = 90^\\circ$. The incircle touches the hypotenuse at $D$. The area of the triangle is equal to $AD\\cdot CD$.\r\n[hide=\"Proof:\"]\nLet $s$ be the semiperimeter, $r$ be the inradius and $a,b,c$ be sides ($c$ is the hypotenuse). We have $r = s-b$. So\n$[ABC]=s\\cdot r= s(s-b)= \\sqrt{s(s-a)(s-b)(s-c)}\\Rightarrow$ $s(s-b)=(s-a)(s-c)=[ABC]$\n[/hide]\r\n\r\nFrom the lemma we know that $\\frac{AH\\cdot BH}{2}=AD\\cdot BD$. From the similarity $\\frac{BD}{AE}=\\frac{BH}{AH}$. Multiply both equation. So we get $\\frac{AH^{2}}{2}=AD\\cdot AE$. We have $\\frac{AH^{2}}{2}=AM\\cdot AH$. So $AE\\cdot AD=AE'\\cdot AD=AM\\cdot AH$ implies $E',M,H,D$ are concyclic. $\\angle EE'M = |\\angle AE'E-\\angle AE'M|$. Similarly $\\angle DHI = |\\angle IHA-\\angle DHA|$. We know $\\angle AE'M = \\angle AHD$ and $\\angle AE'E = \\angle IHA = 45^\\circ$. So $\\angle EE'M = \\angle IHD$. Q.E.D" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Prove that\r\n\r\n$ \\frac{a}{a^{2}\\plus{}4w^{2}_{a}}\\plus{}\\frac{b}{b^{2}\\plus{}4w^{2}_{b}}\\plus{}\\frac{c}{c^{2}\\plus{}4w^{2}_{c}}\\geq\\frac{9}{8s}$", "Solution_1": "[quote=\"Ligouras\"]Prove that\n\n$ \\frac {a}{a^{2} \\plus{} 4w^{2}_{a}} \\plus{} \\frac {b}{b^{2} \\plus{} 4w^{2}_{b}} \\plus{} \\frac {c}{c^{2} \\plus{} 4w^{2}_{c}}\\geq\\frac {9}{8s}$[/quote]\r\n\r\n\r\n\r\nAlgebraization \r\n\r\nCan be shown using the Muirhead theorem.\r\n\r\n :lol:", "Solution_2": "[quote=\"Ligouras\"]Prove that\n\n$ \\frac {a}{a^{2} \\plus{} 4w^{2}_{a}} \\plus{} \\frac {b}{b^{2} \\plus{} 4w^{2}_{b}} \\plus{} \\frac {c}{c^{2} \\plus{} 4w^{2}_{c}}\\geq\\frac {9}{8s}$[/quote]\r\n\r\n-------------------------------\r\n\r\n$ \\frac{1}{a^2\\plus{}4h^2_a}\\geq\\frac{1}{a^2\\plus{}2w^2_a\\plus{}2h^2_a}\\geq\\frac{1}{a^2\\plus{}4w^2_a}\\geq\\frac{1}{a^2\\plus{}s(s\\minus{}a)}\\equal{}\\frac{1}{(b\\plus{}c)^2}$\r\n\r\nIRAN 1996 :)", "Solution_3": "[quote=\"Ligouras\"]Prove that\n\n$ \\frac {a}{a^{2} \\plus{} 4w^{2}_{a}} \\plus{} \\frac {b}{b^{2} \\plus{} 4w^{2}_{b}} \\plus{} \\frac {c}{c^{2} \\plus{} 4w^{2}_{c}}\\geq\\frac {9}{8s}$[/quote]\r\n\r\n---------------------------\r\n\r\nFind the best $ k>0$, such that\r\n\r\n$ \\frac{a}{a^{2}\\plus{}4w^{2}_{a}}\\plus{}\\frac{b}{b^{2}\\plus{}4w^{2}_{b}}\\plus{}\\frac{c}{c^{2}\\plus{}4w^{2}_{c}}\\geq k\\frac{1}{s}$", "Solution_4": "[quote=\"Ligouras\"][quote=\"Ligouras\"]Prove that\n\n$ \\frac {a}{a^{2} \\plus{} 4w^{2}_{a}} \\plus{} \\frac {b}{b^{2} \\plus{} 4w^{2}_{b}} \\plus{} \\frac {c}{c^{2} \\plus{} 4w^{2}_{c}}\\geq\\frac {9}{8s}$[/quote]\n\n---------------------------\n\nFind the best $ k > 0$, such that\n\n$ \\frac {a}{a^{2} \\plus{} 4w^{2}_{a}} \\plus{} \\frac {b}{b^{2} \\plus{} 4w^{2}_{b}} \\plus{} \\frac {c}{c^{2} \\plus{} 4w^{2}_{c}}\\geq k\\frac {1}{s}$[/quote]\r\n\r\n\r\n$ k_{max}\\equal{}\\frac{9}{8}$." } { "Tag": [ "algebra", "polynomial", "function", "geometry", "geometric transformation", "number theory", "number theory unsolved" ], "Problem": "Let $ P(x) \\in Z[x]$. Suppose that $ P(n)$ is a kth power for all $ n \\in N$. Then, there exists a polynomial $ Q(X) \\in Z[x]$ such that $ P(x)\\equal{}Q(x)^k$.", "Solution_1": "CHECK OUT \r\nGeorge P\u00b4olya and G\u00b4abor Szeg\u02ddo, Problems and Theorems in Analysis, vol. 2; Theory of functions, zeros,polynomials, determinants, number theory, geometry; revised and enlarged translation by C. E. Billigheimer of the fourth German edition, Springer-Verlag, New York, 1976. PROBLEM: VIII-114. \r\nto prove the case whwn k=2, check http://www.mathlinks.ro/viewtopic.php?t=241684", "Solution_2": "[quote=\"conan_naruto236\"]Let $ P(x) \\in Z[x]$. Suppose that $ P(n)$ is a kth power for all $ n \\in N$. Then, there exists a polynomial $ Q(X) \\in Z[x]$ such that $ P(x) \\equal{} Q(x)^k$.[/quote]\r\nPlease help me :(" } { "Tag": [], "Problem": "Sabet konid agar $ (xy\\plus{}1)(xz\\plus{}1)(yz\\plus{}1)$ Morabbae kamel bashad angah tamame adade xy+1,xz+1,yz+1 morabbae kameland.", "Solution_1": "kasai ke mikhan hallesho bebinan mitoonan be [url=http://www.mathlinks.ro/viewtopic.php?t=150369]inja[/url] moraje'e konan." } { "Tag": [ "geometry" ], "Problem": "a,b, and c are the lengths of the three sides of a right triangle. a+b+c=22 and a^2 + b^2 + c^2 = 200. what is the number of square units in the area of the triangle. please help", "Solution_1": "Is there a theorem that you could relate to this problem that involves squares of sides of a right triangle?", "Solution_2": ".....? i have no idea!", "Solution_3": "Hint: It begins with a P...was a name of a person...he had a cult...golden ratio...ring a bell? :lol:", "Solution_4": "oh hehehe, i misread the reply as involving the shape of squares and triangles. i got it...PYTHAGOREAN!!! so could you explain further because i tried using it.", "Solution_5": "Um, just substitute the a^2 + b^2 with c^2 and go from there.\r\n\r\n3 variables, 3 equations, one destiny :lol:", "Solution_6": "use the pythagorean theorem, and a simple manipulation will yield the result.", "Solution_7": "i did substitute so i got a=6, b=8, and c=10 but it doesnt add up to a+b+c=22. but it works for a^2+b^2+c^2=200. where did i go wrong?", "Solution_8": "[quote=\"eqn1190\"]i did substitute so i got a=6, b=8, and c=10 but it doesnt add up to a+b+c=22. but it works for a^2+b^2+c^2=200. where did i go wrong?[/quote]\r\n\r\nwhere did you get a=6, b=8 from?", "Solution_9": "you dont even have to solve for a and b", "Solution_10": "am i correct about c=10? then i can explain further.", "Solution_11": "c=10 is correct, but nothing in the question said a and b had to be integers.\r\n\r\n(You really don't have to solve for a and b. Some simple algebraic manipulations can easily yield a relationship between a and b that would give away the area of the triangle.)", "Solution_12": "how do we make the third equation, and how do we solve when are equations arent equal (^2 and ^1)", "Solution_13": "Okay, full solution then...\n\n\n\n[hide]WLOG assume a :le: b < c. Then, a2 + b2 = c2. Substituting that into a2 + b2 + c2 = 200 yields c2 = 100 and thus c = 10 because c > 0.\n\n\n\nTherefore a + b = 22 - 10 = 12. Squaring this yields a2 + 2ab + b2 = 144; subtract a2 + b2 = c2 = 100 to get 2ab = 44, so ab/2 = 11 is the desired area.[/hide]\n\n\n\nFor the curious, a = [hide]6 - :sqrt:14[/hide] and b = [hide]6 + :sqrt:14[/hide].", "Solution_14": "yup that's what I got. remember that you don't always have to solve for specific variables, and that you can square expressions, etc. (in general, manipulate them) to get what you want.", "Solution_15": "Correcto...except what's \"WLOG\"?", "Solution_16": "\"without loss of generality\"\r\n\r\nI don't think Osiris needed to put that in there, but it looks a whole lot cooler now.", "Solution_17": "Well you do need it actually.. nothing in the question said that c had to be the longest side. So you can't just assume that without saying WLOG :)", "Solution_18": "thanks a lot, i understand now. im not really good at simple manipulations like that because i dont really see it. are there any more problems i can try to solve that are like the one i posted?", "Solution_19": "4a + b = 1/(2a) + 2/b. If a and b are positive, compute the value of ab.\r\n\r\nIf r = 20/(r + 5) and s = -5 + 20/s, where r and s are two distinct real numbers, compute r + s.\r\n\r\nIf r:^2: + br + 4 = 0 and s + r = -b, what is the value of s:^2: + bs?" } { "Tag": [ "calculus", "integration", "least common multiple", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $ x_i, y_i$ be integers for $ i \\equal{} 1,2,\\cdots, s$. Is there a solution of the following equation system on $ l$?\r\n\r\n\\[ \\left\\{\\begin{array}{ll} \\frac {x_1}{y_1}[\\frac {y_1}{x_1},\\frac {y_2}{x_2},\\cdots,\\frac {y_s}{x_s}]\\cdot l\\equiv 1(\\bmod x_1) \\\\\r\n\\frac {x_2}{y_2}[\\frac {y_1}{x_1},\\frac {y_2}{x_2},\\cdots,\\frac {y_s}{x_s}]\\cdot l\\equiv 1(\\bmod x_2) \\\\\r\n\\cdots\\cdots\\cdots \\\\\r\n\\frac {x_s}{y_s}[\\frac {y_1}{x_1},\\frac {y_2}{x_2},\\cdots,\\frac {y_s}{x_s}]\\cdot l\\equiv 1(\\bmod x_s) \\\\\r\n\\end{array}\\right.\\]", "Solution_1": "Not before you explain what the $ \\left[...\\right]$ are supposed to mean.\r\n\r\n darij", "Solution_2": "$ [n_1,n_2,\\cdots,n_s]$ means least common multiple of numbers $ n_1,n_2,\\cdots,n_s$.", "Solution_3": "What is the least common multiple of rational numbers supposed to mean? Maximal power of each prime? Is $ l$ to be choosen in a way that makes the left hand sides integral?\r\n\r\n darij", "Solution_4": "sorry, where $ \\frac{y_i}{x_i}$ with $ i\\equal{}1,2,\\cdots,s$ and $ l$ are all integers.", "Solution_5": "The question still doesn't make much sense to me: take $ s\\equal{}2$, $ x_1\\equal{}2$, $ y_1\\equal{}4$, $ x_2\\equal{}1$, $ y_2\\equal{}8$. Then your first equation spells $ \\frac12\\cdot 8\\cdot l\\equiv 1\\mod 2$, impossible. On the other hand, there are many cases where a solution exists. I doubt they can be classified.\r\n\r\n darij", "Solution_6": "thanks, how to classify?" } { "Tag": [ "geometry", "3D geometry", "tetrahedron", "analytic geometry", "IMO", "IMO 1964" ], "Problem": "In tetrahedron $ABCD$, vertex $D$ is connected with $D_0$, the centrod if $\\triangle ABC$. Line parallel to $DD_0$ are drawn through $A,B$ and $C$. These lines intersect the planes $BCD, CAD$ and $ABD$ in points $A_2, B_1,$ and $C_1$, respectively. Prove that the volume of $ABCD$ is one third the volume of $A_1B_1C_1D_0$. Is the result if point $D_o$ is selected anywhere within $\\triangle ABC$?", "Solution_1": "In tetrahedron $ ABCD $, vertex $ D $ is connected with $ D_{0} $, the centrod in $ \\triangle ABC $. Line parallel to $ DD_{0} $ are drawn through $ A,B $ and $C$. These lines intersect the planes $ BCD, CAD $ and $ ABD $ in points $ A_{1}, B_{1} $ and $ C_{1} $, respectively. Prove that the volume of $ ABCD $ is one third the volume of $ A_{1}B_{1}C_{1}D_{0} $. Is the result if point $ D_{o} $ is selected anywhere within $ \\triangle ABC $?\n\n [size=150]Part 1: Construction of the points $ A_{1}, B_{1}, C_{1} $[/size]\n\nGiven a tetrahedron $ ABCD $ with vertex $ D $,see picture.\nLines $d1$ and $d2$ are bisectors of $ \\triangle ABC $. The crossing of $d1$ and $d2$ is the point $ D_{0} $.We draw the line $ D D_{0} $ and also the line $z$, parallel to $ D D_{0}$.\nIn the plane $ (ABC) $, we draw the line $ p $, parallel to $ BC $.\nIn this plane $ (ABC) $, the line $ p $ is crossing the 2 bisectors $d1$ and $d2$ in 2 points, where $b1$ and $b2$ are parallel to $ D D_{0}$.\nIn the plane $ (b1, d1,D D_{0},C) $, the lines $ d1 $ and $ CD $ are crossing in the point $A1$.\nIn the plane $ (b2, d2,D D_{0},B) $, the lines $ d2 $ and $ BD $ are crossing in the point $A2$.\nIn the plane $ (p, b1, b2,z,A1, A2) $, the lines $ A1A2 $ and $ z $ are crossing in the point $A'$; plane $ (BCD) \\cap z $ = $A'$.\nThat point $A'$ is the asked point $ A_{1}$.\nThe same construction can be used for the points $ B_{1} $ and $ C_{1} $.\nIf the tetrahedron is regular, then the points $ A_{1}, B_{1}, C_{1} $ form a plane, parallel to the plane $ (ABC)$, see part 2.\n\n[size=150]Part 2: Calculation of the coordinates of the points $ A_{1}, B_{1}, C_{1} $[/size]\n\nGiven a regular tetrahedron $ ABCD $ with vertex $ D $; each side has length 1.\n\t\\[A(0,0,0), B(\\frac{\\sqrt{3}}{2},\\frac{1}{2},0), C(0,1,0)\\]\n\t\\[D_{0}(\\frac{\\sqrt{3}}{6},\\frac{1}{2},0), D(\\frac{\\sqrt{3}}{6},\\frac{1}{2},\\frac{\\sqrt{6}}{3})\\]\nIn this regular tetrahedron: $ D D_{0} \\bot (ABC)$, hence $ D D_{0} \\bot (x\\;y)$.\nThe equation of a plane $(PQR)$ with \n\t\\[P(x_{1},y_{1},z_{1}),Q(x_{2},y_{2},z_{2}),R((x_{3},y_{3},z_{3})\\]\nis given by:\t\n\\[ \\left|\n\\begin{array}{ccc}\nx- x_{1}& y- y_{1}& z- z_{1}\\\\\nx_{2}- x_{1}& y_{2}- y_{1}& z_{2}- z_{1} \\\\\nx_{3}- x_{1}& y_{3}- y_{1}& z_{3}- z_{1} \\end{array}\n\\right|=0\\]\na) equation of the plane $(DBC)$:\n\\[ \\left |\n\\begin{array}{ccc}\nx& y- 1& z\\\\\n\\frac{\\sqrt{3}}{2}& - \\frac{1}{2}& 0 \\\\\n\\frac{\\sqrt{3}}{6}& - \\frac{1}{2}& \\frac{\\sqrt{6}}{3} \\end{array}\n\\right|=0\\]\n\t\\[\\frac{\\sqrt{6}}{6}x+\\frac{\\sqrt{2}}{2}(y-1)+\\frac{\\sqrt{3}}{6}z=0\\]\nThrough point $A$, we are drawing a parallel with $D D_{0}$, i.e. this parallel $\\bot (x\\;y)$\nThe equation of this parallel is: \n\t\\[ \\left\\{\\begin{array}{ll} x=0\\\\ y=0\\end{array}\\right.\\]\nThis parallel through $A$ $ \\cap (DBC) = A'$:\n\t\\[ \\left\\{\\begin{array}{lll} \\frac{\\sqrt{6}}{6}x+\\frac{\\sqrt{2}}{2}(y-1)+\\frac{\\sqrt{3}}{6}z=0\\\\x=0\\\\ y=0\\end{array}\\right.\\]\n\t\\[A'(0,0,\\sqrt{6})=A_{1}(0,0,\\sqrt{6})\\]\nb) equation of the plane $(ADB)$:\n\\[ \\left|\n\\begin{array}{ccc}\nx& y& z\\\\\n\\frac{\\sqrt{3}}{2}& \\frac{1}{2}& 0 \\\\\n\\frac{\\sqrt{3}}{6}& \\frac{1}{2}& \\frac{\\sqrt{6}}{3} \\end{array}\n\\right|=0\\]\n\t\\[\\frac{\\sqrt{6}}{6}x-\\frac{\\sqrt{2}}{2}y+\\frac{\\sqrt{3}}{6}z=0\\]\nThrough point $C$, we are drawing a parallel with $D D_{0}$, i.e. this parallel $\\bot (x\\;y)$\nThe equation of this parallel is: \n\t\\[ \\left\\{\\begin{array}{ll} x=0\\\\ y=1\\end{array}\\right.\\]\nThis parallel through $C$ $ \\cap (ADB) = C'$:\n\t\\[ \\left\\{\\begin{array}{lll} \\frac{\\sqrt{6}}{6}x-\\frac{\\sqrt{2}}{2}y+\\frac{\\sqrt{3}}{6}z=0\\\\x=0\\\\ y=1\\end{array}\\right.\\]\n\t\\[C'(0,1,\\sqrt{6})\\]\nc) equation of the plane $(ADC)$:\n\\[ \\left|\n\\begin{array}{ccc}\nx& y& z\\\\\n0& 1& 0\\\\\n\\frac{\\sqrt{3}}{6}& \\frac{1}{2}& \\frac{\\sqrt{6}}{3} \\end{array}\n\\right|=0\\]\n\t\\[\\frac{\\sqrt{6}}{3}x-\\frac{\\sqrt{3}}{6}z=0\\]\nThrough point $B$, we are drawing a parallel with $D D_{0}$, i.e. this parallel $\\bot (x\\;y)$\nThe equation of this parallel is: \n\t\\[ \\left\\{\\begin{array}{ll} x=\\frac{\\sqrt{3}}{2}\\\\ y=\\frac{1}{2}\\end{array}\\right.\\]\nThis parallel through $B$ $ \\cap (ADC) = B'$:\n\t\\[ \\left\\{\\begin{array}{lll} \\frac{\\sqrt{6}}{3}x-\\frac{\\sqrt{3}}{6}z=0\\\\x=\\frac{\\sqrt{3}}{2}\\\\ y=\\frac{1}{2}\\end{array}\\right.\\]\n\t\\[B'(\\frac{\\sqrt{3}}{2},\\frac{1}{2},\\sqrt{6})\\]\nBecause the value of the z-coordinate of $A',B'$ and $C'$ is $\\sqrt{6}$, are the surface of $\\triangle ABC$ and the surface of $\\triangle A'B'C'$ equal.\nBecause the height of the tetrahedron $ ABCD $ equals $\\frac{\\sqrt{6}}{3}$ (the z-coordinate of the point $D$) and the height of the tetrahedron $ A'B'C'D_{0} $ equals $\\sqrt{6}$, is the volume of $ ABCD $ indeed one third the volume of $ A_{1}B_{1}C_{1}D_{0} $.\n\n[size=150]Part 3: Calculation of another tetrahedron[/size]\n\na) Given a arbitrary tetrahedron $ ABCD $ with vertex $ D $.\n\t\\[A(0,0,0), B(a,b,0), C(0,c,0)\\]\nPoint $D_{0}$ is the crossing point of the bisectors of $\\triangle ABC$.\n$D D_{0}$ is perpendicular on the plane $(ABC)$, points $D$ and $ D_{0}$ have the same x-coordinate and y-coordinate; the z-coordinate of $D$ is 1.\nThe same calculation as in Part 2 is made with result: the volume of $ ABCD $ is one third the volume of $ A_{1}B_{1}C_{1}D_{0} $.\nb) Given a regular tetrahedron $ ABCD $ with vertex $ D $; each side has length 1.\n\t\\[A(0,0,0), B(\\frac{\\sqrt{3}}{2},\\frac{1}{2},0), C(0,1,0), D(\\frac{\\sqrt{3}}{6},\\frac{1}{2},\\frac{\\sqrt{6}}{3})\\]\nThe point $ D_{o} $ is selected anywhere within $ \\triangle ABC $:\n\\[D_{0}(k,l,0)\\] with $k \\neq \\frac{\\sqrt{3}}{6}$ \\ and/or \\ $l \\neq \\frac{1}{2}$.\nThe same calculation as in Part 2 is made with result: the volume of $ ABCD $ is NEVER one third the volume of $ A_{1}B_{1}C_{1}D_{0} $." } { "Tag": [ "LaTeX" ], "Problem": "Is there another way to align equations other than eqnarray*?\r\nThanks!\r\nminicon", "Solution_1": "Eqnarray is obsolete - see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]Latex FAQ 5[/url]\r\nMake sure you have loaded amsmath package in your document and use align\r\n\\begin{align}\r\n2x+y&=3 \\\\\r\nx+2y&=5\r\n\\end{align}\r\n[code]\\begin{align}\n2x+y&=3 \\\\\nx+2y&=5\n\\end{align}[/code]", "Solution_2": "Thanks stevem!\r\n:)" } { "Tag": [ "LaTeX" ], "Problem": "Is it possible to use \\input on a file that's not in the same directory as your .tex file? I'm trying it now by including the full pathname in single quotation marks but it's not sure it this will work...", "Solution_1": "Difficult to diagnose the problem without the details. But remember\r\n1. the path must be [i]relative[/i] to the directory of the tex file;\r\n2. always avoid directories with spaces in them; you [i]may[/i] be able to use the DOS version of windows paths with spaces in them;\r\n3. the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=import]import package[/url] may be useful." } { "Tag": [ "trigonometry", "conics", "ellipse" ], "Problem": "[code][(x - x1) / a]^2 + [(y - y1) / b]^2 = 1 \n[(x - x2) / c]^2 + [(y - y2) / d]^2 = 1[/code]\r\n\r\nThose are my two equations.\r\n\r\nAll I need is for someone to simplify these into a quartic. I don't have access to Mathematica right now, but I think this can easily be solved. I just need a quartic, I don't need the steps involved (I'm pretty sure that simplifies to a quartic).\r\n\r\nEliminating y from both equations gives this (if it helps) :\r\n[b]sqrt {b^2 - b^2[(x - x1) / a]^2} + y1 = sqrt {d^2 - d^2[(x - x2) / c]^2} + y2[/b]\r\n\r\nObviously, if it's easier to solve for x than that would be just as good.\r\n\r\nAnything is really appreciated and thank you for reading!!", "Solution_1": "I'll put what you posted into latex. I'm not going to try solving it now, as it's late and I need to do something else. Maybe tomorrow.\r\n\r\n$\\left(\\frac{x-x_{1}}{a}\\right)^{2}+\\left(\\frac{y-y_{1}}{b}\\right)^{2}= 1$\r\n$\\left(\\frac{x-x_{2}}{c}\\right)^{2}+\\left(\\frac{y-y_{2}}{d}\\right)^{2}= 1$\r\n\r\nAnd you think it becomes\r\n$\\sqrt{b^{2}-b^{2}\\left(\\frac{x-x_{1}}{a}\\right)^{2}}+y_{1}= \\sqrt{d^{2}-d^{2}\\left(\\frac{x-x2}{c}\\right)^{2}}+y_{2}$\r\n\r\nIs this right?\r\n----------------------------------\r\nThe only way to get a quartic is to multiply the first equation by the second, and that would be a lot of messy algebra. It's doable, but, as you said, mathematica would be nicer.\r\n\r\nI have a copy at school, I'll burn this out for you tomorrow if nodoby answers sooner.\r\n\r\nBy the way, what are you solving for, and why do you want a quartic? That adds to the complexity. As it is, those equations are fairly nice looking and simple.", "Solution_2": "[quote=\"DiscreetFourierTransform\"]The only way to get a quartic is to multiply the first equation by the second[/quote]\n\nI beg to differ.\n\n[hide=\"The method that I am too lazy to finish\"] Let $x = a \\cos \\theta+x_{1}, y = b \\sin \\theta+y_{1}$. Then let $t = \\tan \\frac{\\theta}{2}$. \n\nThe first equation is automatically satisfied; and because we have the relations\n\n$\\cos \\theta = \\frac{1-t^{2}}{1+t^{2}}$\n$\\sin \\theta = \\frac{2t}{1+t^{2}}$\n\nThe second equation gives us a quartic in $t$, which we can solve to calculate $x, y$. [/hide]\nEdit:\n\n[quote=\"DiscreetFourierTransform\"]By the way, what are you solving for[/quote]\r\n\r\nLooks like the intersections of two ellipses to me.", "Solution_3": "Yeah, I saw the elipses thingy too.\r\nHowever, I think that other quartic method is nice too.\r\nTor, you pwn me on every math post I make :oops: :P \r\nHow long have you been doing comp. math? I just started this year, and I'm still kind of intimidated by the stuff in Pre-Olympiad...", "Solution_4": "Since my join date, more or less. Read other people's solutions - you'll get a better flavor for the types of logic you need to solve interesting, non-school kinds of problems.", "Solution_5": "Thanks for the replies!! I actually found my solution. Someone used Maple from another math forum to get it.\r\n\r\n[quote]Looks like the intersections of two ellipses to me.[/quote]\r\nIndeed, you are right! :wink:\r\n\r\nIf anyone is interested in the horrible quartic that results, [url=http://mymathforum.com/viewtopic.php?p=2800#2800]here[/url] you go!" } { "Tag": [ "function", "search" ], "Problem": "How do you arrive at\r\n$1^{-1} + 2^{-1} + 3^{-1} +4^{-1} +... =-1/12$?", "Solution_1": "I think this is one of the things like $\\infty!=\\sqrt{2\\pi}$.\r\nIt's clear to see it's not technically true since the sequence contains only positive numbers. :)", "Solution_2": "well I know that it is a special thing and not exactly equal. One person put a little a above the equals sig to show that they weren't really equal. I was just wodnering where that came from.", "Solution_3": "It's the Riemann Zeta Function at -1. You can search here or look it up on mathworld for more.", "Solution_4": "What I find most fascinating about this summation is its applications in the real world. The computation of certain physical quantities often involves summing this (divergent) series to give a small, negative number. Quite interesting. :)" } { "Tag": [ "national olympiad" ], "Problem": "Released today.\r\n\r\nFIRST PRIZE\r\n[b]Yufei Zhao[/b], Don Mills Collegiate Institute, Don Mills, Ontario. :)\r\n\r\nSECOND PRIZE\r\n[b]Jacob Tsimerman,[/b] University of Toronto Schools, Toronto, Ontario.\r\n\r\nTHIRD PRIZE\r\n[b]David Rhee[/b], McNally School, Edmonton, Alberta. \r\n\r\nHONOURABLE MENTIONS \r\n[b]Boris Braverman[/b], Simon Fraser Junior High, Calgary, Alberta\r\n[b]Dennis Chuang[/b], Strathcona-Tweedsmuir School, Okotoks, Alberta\r\n[b]Gabriel Gauthier-Shalom[/b], Marianopolis College, Montreal, Quebec\r\n[b]Oleg Ivrii[/b], Don Mills Collegiate Institute, Don Mills, Ontario\r\n[b]Jnos Kramr[/b], University of Toronto Schools, Toronto, Ontario\r\n[b]Andrew Mao[/b], A.B. Lucas Secondary School, London, Ontario\r\n[b]Richard Peng[/b], Vaughan Road Academy, Toronto, Ontario\r\n[b]Peng Shi[/b], Sir John A. MacDonald Collegiate Institute, Agincourt, Ontario. \r\n\r\nSource: [url]http://www.cms.math.ca/MediaReleases/2004/cmo.html[/url]", "Solution_1": "Can someone please post the problems?\r\nThe CanMO website doesn't give them.", "Solution_2": "Yey, congrats Billzhao! :D I suppose you are admitted in the Canada IMO Team? :D\r\n\r\nCould you please post the questions also? :)", "Solution_3": "The problems are posted [url=http://www.mathlinks.ro/viewtopic.php?t=5547]here[/url].\r\n\r\nBtw, on the front page, it should spell Can[b]a[/b]da, not Canda", "Solution_4": "[quote=\"billzhao\"]The problems are posted [url=http://www.mathlinks.ro/viewtopic.php?t=5547]here[/url].\n\nBtw, on the front page, it should spell Can[b]a[/b]da, not Canda[/quote] sorry, I will fix it :blush: :blush: :blush:" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Let f(x)= x^2 + x, prove that there doesn't exist such integers a,b for which 4f(a)=f(b).\r\n :blush:", "Solution_1": "[quote=\"Klaas\"]Let f(x)= x^2 + x, prove that there doesn't exist such integers a,b for which 4f(a)=f(b).\n :blush:[/quote]\r\n\r\nThat's wrong : \r\n$ 4f(0)\\equal{}f(0)$\r\n$ 4f(0)\\equal{}f(\\minus{}1)$\r\n$ 4f(\\minus{}1)\\equal{}f(0)$\r\n$ 4f(\\minus{}1)\\equal{}f(\\minus{}1)$\r\n\r\nIn fact :\r\n\r\n$ 4f(a)\\equal{}f(b)$ $ \\Leftrightarrow$ $ 4a^2\\plus{}4a\\equal{}b^2\\plus{}b$ $ \\Leftrightarrow$ $ (2a\\plus{}1)^2\\equal{}b^2\\plus{}b\\plus{}1$\r\n\r\nThen, if $ b>0$, we have $ b^2misunderestimating Bush. Oh wait, no.[/quote]\r\n\r\nAnd you say BUSH has communication issues?", "Solution_7": "[quote=\"Rep123max\"]Well, I guess I might be misunderestimating Bush. Oh wait, no.[/quote]\r\n\r\n :lol: :lol: :lol: \r\n\r\nAlthough the 42% does include Camp David, they still calculated that he spent about 16-17% at his home, which I believe is still more than any previous president. However, he might well have been working when at his ranch, so we don't really know whether the 42% is a relavent statistic or not. However, what really bothers me is the [i]month-long[/i] vacation he took in August 2001. He'd only been in office for five and a half months! I know that he may have done some work over that vacation, but I doubt that he would be nearly as productive at home as in the White House.\r\n\r\nAlso, confuted, I don't think calling Rep's posts pathetic is going to stop him from posting what he believes nor further your argument. I would like to hear the rationalization for Bush's month-long vacation, so if you could provide this I would appreciate it.", "Solution_8": "Come on Rep- Do you really think a president has any vacation time? When he is at the Crawford Ranch every day he is briefed and has work to do- calls to make. I think that should be obvious to anyone who really thinks about the job that there is very little if any down time when president! I suppose what Clinton did in the Oval Office was down time (the blue dress does prove he had some).\r\n\r\nAs for staying in the classroom- even democrats have defended that. They said that it was better than rushing out and scaring a group of little kids. Also, he wasn't reading the book- he was probably thinking about what he had to do next. They would also never have shot the plane down- the military hadn't gotten to the plane before it crashed (PA) and everything else happened fast. Remember before this our country would most likely never have shot down a plane with passengers. \r\n\r\nNow I'm sure that judging by what you say everything in the world is Bush's fault- does that mean the economy is his fault? Or could it perhaps be because idiots crashed into our buildings and caused people to stop flying and if people don't fly then they don't need hotels and if they don't need hotels then they don't need restaurants etc....Yep, if Bush would just go out to eat more the economy would be thriving!", "Solution_9": "[quote=\"soilent green\"]As for staying in the classroom- even democrats have defended that. They said that it was better than rushing out and scaring a group of little kids. Also, he wasn't reading the book- he was probably thinking about what he had to do next. They would also never have shot the plane down- the military hadn't gotten to the plane before it crashed (PA) and everything else happened fast. Remember before this our country would most likely never have shot down a plane with passengers.[/quote]\r\n\r\nFirst of all, I'd like to know what democrats have defended this. Second, I don't think it's unreasonable to say that it's OK to scare a few kids when [i]national security is at stake![/i] Whether or not he was thinking of what he needed to do, what he should have done was left the room and gotten briefed by someone who knew the situation so that at least he could be in a position to decide what to do. Sitting silently in a room full of schoolchildren in a time of crisis accomplished nothing.", "Solution_10": "One Democrat that defended Bush was George Stephanopolis when interviewed on Good Morning America. He was one of Clintons Cronies- if you remember from the first term (press secretary). It was also stated that never before had a sitting president been informed about a huge tragedy while being filmed. Presidents should be given this time so they don't make huge errors like using nuclear weapons. Even Roosevelt when Japan bombed Pearl Harbor took time making a decision before commiting this country to war in the Pacific.\r\n\r\nPeople want to see leaders calm under stress.", "Solution_11": "[quote=\"JSRosen3\"]First of all, I'd like to know what democrats have defended this. Second, I don't think it's unreasonable to say that it's OK to scare a few kids when [i]national security is at stake![/i] Whether or not he was thinking of what he needed to do, what he should have done was left the room and gotten briefed by someone who knew the situation so that at least he could be in a position to decide what to do. Sitting silently in a room full of schoolchildren in a time of crisis accomplished nothing.[/quote]\r\nNobody knew the situation better than 'a plane just flew into the WTC' and then 'another!'. The briefing you mentioned easily could have been (and I'm sure was) whispered into his ear. We all know that very very little information was available in the first few minutes - and how could it have been?", "Solution_12": "Uh... Rep, when information, as confuted said, isn't available, doesn't that mean you [b]can't[/b] go out and get the information? What confuted said doesn't support your point at all, however much you may want it to.", "Solution_13": "[quote=\"Rep123max\"]It means you don't wait in a classroom for it to come for you. You leave the classroom, get all your top aids, discuss what they know/don't know. Get on Air Force 1, and take it from there.[/quote]\r\n\r\nExactly. No matter how little information there was, the president needed to [i]put himself in a position[/i] to make a decision. That means leaving the classroom and taking charge of the situation.", "Solution_14": "Or maybe George Bush considered his image in front of the children to be more important than the security of United States.", "Solution_15": "20/1 hindsight, I say. Even if he had been briefed, what possibly decisions could he have possibly made to react to this? The argument put forth is that he should have been briefed so that he could make an informed decision. A decision on what? The terrorists' intentions weren't known until it would have been far too late to make a decision. Sure, the President and his team could work on their reaction time, but it's just Bush's bum luck that 9/11 happened while he was trying to do something charitable for the American people. How much experience with crisis has Bush had, anyway? To expect for him to get it right on the first try without faltering really seems like asking a bit much.", "Solution_16": "[quote=\"Rep123max\"]And can somebody tell me what Bush did pre 9/11 to fight terrorism?[/quote]\r\n\r\nRep, no one really thought there was a serious threat to U.S. untill 9/11. The only reason there is this much hype for fighting terrorism is because of 9/11.", "Solution_17": "im glad rep isnt our president. to be a president, you have to be calm in the line of fire, not hotheaded." } { "Tag": [ "abstract algebra" ], "Problem": "Solve $ 5^{x}-3^{y}=2$ in $ \\mathbb{N}^{+}$. I didn't find any topic regarding this diophantine. I tried some combinations of modules, but didn't find any contraditction. Also, I'd like to know if there is any elegant solution.", "Solution_1": "EDIT: see ZetaX two posts below. :oops: \r\n\r\n[hide=\"solution\"]Look at it $ \\mod(5)$\n\n$ \\rightarrow 3^{y}= 2 \\mod(5)$\nBy casework and FLT then $ y=1 \\mod(4)$\nAll we need is that $ y$ is odd.\n\nNow $ \\mod(3)$\n$ \\rightarrow 5^{x}= 2 \\mod(3)$\nSame argument we get $ x$ is odd.\n\nBut then $ 5^{x}-3^{y}=(5-3)(5^{x-1}+3.5^{x-2}+\\cdots+3^{y-1})$\n\nFor $ x \\text{ or }y >1$ the second bracket is $ >1$ thus $ x=y=1$ only solution.[/hide]", "Solution_2": "I find another solution :) \r\nwe can easily prove both $ x,y$ are odd . now if $ y=1$ then $ x=1$ . \r\nnow assume that $ y\\geq2$ we have $ 5^{x}\\equiv 2\\mod[9]$ and we give $ x=6k+5$\r\nknow use $ mod7$ , we give $ 5^{6k+5}\\equiv 3\\mod[7]$ and it results $ 3^{y}\\equiv 1\\mod[7]$ and we give $ y=6m$ which is contradiction because $ y$ is odd .", "Solution_3": "[quote=\"seamusoboyle\"]But then $ 5^{x}-3^{y}=(5-3)(5^{x-1}+3.5^{x-2}+\\cdots+3^{y-1})$[/quote]\r\nNo, that's wrong (the second bracket makes in fact no sense if $ x \\neq y$)." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let $ x_{1},x_{2},....$ be a sequence of positive real numbers .If $ a_{k}$ and $ g_{k}$ are the arithmetic and geometric means , respectively, of $ x_{1},....,x_{k}$, prove that\r\n\r\n1) $ \\frac{a_{n}^{n}}{g_{k}^{n}}$ $ \\ge$ $ \\frac{a_{n\\minus{}1}^{n\\minus{}1}}{g_{k}^{n\\minus{}1}}$\r\n\r\n2) $ n(a_{n}\\minus{}g_{n})$ $ \\ge$ $ (n\\minus{}1) (a_{n\\minus{}1}\\minus{}g_{n\\minus{}1})$", "Solution_1": "2)\r\n$ \\dfrac{(n\\minus{}1)g_{n\\minus{}1}\\plus{}x_n}{n}\\ge\\sqrt[n]{g_{n\\minus{}1}^{n\\minus{}1}x_n}\\equal{}g_n$ (by AM-GM)\r\nand $ x_n\\equal{}na_n\\minus{}(n\\minus{}1)a_{n\\minus{}1}$\r\n$ \\Rightarrow n(a_n\\minus{}g_n)\\ge(n\\minus{}1)(a_{n\\minus{}1}\\minus{}g_{n\\minus{}1})$", "Solution_2": "hey can anyone answer the first inequality??", "Solution_3": "Is the question wrong or very difficult??\r\n[u]can no one solve it??[/u]", "Solution_4": "I assume you want to prove $ \\frac{a_n^n}{g_n^n}\\ge\\frac{a_{n\\minus{}1}^{n\\minus{}1}}{g_{n\\minus{}1}^{n\\minus{}1}}$.\r\n\r\n$ a_n^n\\equal{}(\\frac{(n\\minus{}1)a_{n\\minus{}1}\\plus{}x_n}n)^n\\ge a_{n\\minus{}1}^{n\\minus{}1}x_n\\equal{}a_{n\\minus{}1}^{n\\minus{}1}\\frac{g_n^n}{g_{n\\minus{}1}^{n\\minus{}1}}$." } { "Tag": [ "percent" ], "Problem": "Matt scored $ 70\\%$ on the first 60 questions of the 100-question math final. If each question has the same value, what percent of the remaining questions must he answer correctly to receive a total score of $ 80\\%$?", "Solution_1": "Since he scored $ 70\\%$ on the first $ 60$ questions, that means he answered $ .7\\times{60} \\equal{} 42$ of those questions right. In order to get $ 80\\%$, he must have answered $ 80$ questions correctly. So, of the last $ 40$ questions, he must answer $ 80 \\minus{} 42 \\equal{} 38$ of those questions correctly. His percentage on those questions must be $ \\frac {38}{40} \\equal{} \\boxed{95\\%}$." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Prove that for $ \\triangle ABC;A,B,C<\\dfrac{\\pi}{2} m\\geq 2; m \\in N$ we have\r\n$ cos^{m}A \\plus{} cos^{m}B \\plus{} cos^{m}C \\geq \\dfrac{3}{2^{m}}$", "Solution_1": "nobody have the solution?", "Solution_2": "[quote=\"desmrhai\"]Prove that for $ \\triangle ABC;A,B,C < \\dfrac{\\pi}{2} m\\geq 2; m \\in N$ we have\n$ cos^{m}A \\plus{} cos^{m}B \\plus{} cos^{m}C \\geq \\dfrac{3}{2^{m}}$[/quote]\r\n\r\n$ LSH\\equal{}\\sum (cos^2A)^{\\dfrac m{2}} \\ge 3\\left(\\dfrac{\\sum cos^2A}3\\right)^{\\dfrac m{2}}\\ge 3\\left(\\dfrac{1}4\\right)^{\\dfrac m{2}}\\equal{}\\dfrac{3}{2^{m}}$ :)", "Solution_3": "[quote=\"dduclam\"][quote=\"desmrhai\"]Prove that for $ \\triangle ABC;A,B,C < \\dfrac{\\pi}{2} m\\geq 2; m \\in N$ we have\n$ cos^{m}A \\plus{} cos^{m}B \\plus{} cos^{m}C \\geq \\dfrac{3}{2^{m}}$[/quote]\n\n$ LSH \\equal{} \\sum (cos^2A)^{\\dfrac m{2}} \\ge 3\\left(\\dfrac{\\sum cos^2A}3\\right)^{\\dfrac m{2}}\\ge 3\\left(\\dfrac{1}4\\right)^{\\dfrac m{2}} \\equal{} \\dfrac{3}{2^{m}}$ :)[/quote]\r\ncan you explain $ \\sum (cos^2A)^{\\dfrac m{2}} \\ge 3\\left(\\dfrac{\\sum cos^2A}3\\right)^{\\dfrac m{2}}$", "Solution_4": "[quote=\"desmrhai\"][quote=\"dduclam\"][quote=\"desmrhai\"]Prove that for $ \\triangle ABC;A,B,C < \\dfrac{\\pi}{2} m\\geq 2; m \\in N$ we have\n$ cos^{m}A \\plus{} cos^{m}B \\plus{} cos^{m}C \\geq \\dfrac{3}{2^{m}}$[/quote]\n\n$ LSH \\equal{} \\sum (cos^2A)^{\\dfrac m{2}} \\ge 3\\left(\\dfrac{\\sum cos^2A}3\\right)^{\\dfrac m{2}}\\ge 3\\left(\\dfrac{1}4\\right)^{\\dfrac m{2}} \\equal{} \\dfrac{3}{2^{m}}$ :)[/quote]\ncan you explain $ \\sum (cos^2A)^{\\dfrac m{2}} \\ge 3\\left(\\dfrac{\\sum cos^2A}3\\right)^{\\dfrac m{2}}$[/quote]\r\n\r\nDear friend,\r\nIt's the lemma: $ \\frac{a^n\\plus{}b^n\\plus{}c^n}3\\ge \\left(\\frac{a\\plus{}b\\plus{}c}3\\right)^n$\r\nEasy to prove this lemma by using AM-GM inequality. :)\r\n\r\nInfact, your inequality's also true for all $ 2\\le m\\in R^*$. :)" } { "Tag": [ "MATHCOUNTS" ], "Problem": "I was just wondering if I have a chance of getting into Countdown or the top 4 in Virginia. Right now, I'm averaging about a 37 or a 38 on practice tests I take. Is this a decent score? I don't know too much about the whole thing because even though I qualified for states the last two years, I had a scheduling conflict and couldn't make it. Anyway, thanks for any help.", "Solution_1": "Are you the type of person who does better at home than at the actual competition?", "Solution_2": "It all comes down to which state you live in. You live in Virginia so I assume that state is EXTREMELY competitive. I live in OK so I have pretty much no REAL competition. Since you live in the more populated states, I think the cutoff for state in Virginia would be above 40. :lol:", "Solution_3": "[quote=\"Ihatepie\"]Are you the type of person who does better at home than at the actual competition?[/quote]\r\nMy first year I was, but after that I pretty much got used to it and scored about the same score that I got on practice.", "Solution_4": "Are you in 7th or 8th? I'd guess that if you can get a couple more points, you're basically guaranteed for nats.", "Solution_5": "I'm in 8th :(, but thanks for giving me some info. At least now I know I have a chance.", "Solution_6": "I actually think you have a good chance of getting in with your current score. But keep on studying.\r\n\r\nI could only do one year of Mathcounts (learned about it in 8th grade) and didn't study enough. I got a decent 76th (top third :lol:) at nationals, but I could've easily cracked 50 with more practice.\r\n\r\n[color=blue] [size=59]and if one kid from my school hadn't been hiding mathcounts in 6th and 7th I could've easily made the top 30 and maybe even made countdown[/color][/size]", "Solution_7": "Yea a 37 or 38 will get you to countdown. Anything mid 30's should get you there. A 37 will not guarantee you top 4 but you should be around the middle. It also depends how hard this years test is.", "Solution_8": "Actually I was lying; you're not close to the cutoff.\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\njk :D", "Solution_9": "I'm pretty sure Virginia is VERY competitive, so that score would probably get you into countdown, maybe. Study really hard!\r\nIf you were in my state, atleast last year, you would have won by a lot. My state is SOOO much better this year though. :(", "Solution_10": "I doubt there is any state not counting Texas and California that 38 won't get you to countdown.\r\n\r\nIs countdown official in your state? And how is it held?", "Solution_11": "I think you would be secure with a 37 in state written.\r\nMy state OK is horrible..last year number 1 got like a 28.", "Solution_12": "omg omg, i'm going to Va states too, i usually get around 20-24 in sprint, and usually a perfect on target. This is my first year going to states, and i'm in 8th grade :( . so, vamathlete, i guess i'll see u there. Keep practicing.", "Solution_13": "[quote=\"Ihatepie\"]I doubt there is any state not counting Texas and California that 38 won't get you to countdown.\n\nIs countdown official in your state? And how is it held?[/quote]\r\n\r\nnew york?\r\nand yeah texas...i have a feeling 42/43+ required 4 countdown", "Solution_14": "[quote=\"theprodigy\"][quote=\"Ihatepie\"]I doubt there is any state not counting Texas and California that 38 won't get you to countdown.\n\nIs countdown official in your state? And how is it held?[/quote]\n\nnew york?\nand yeah texas...i have a feeling 42/43+ required 4 countdown[/quote]\r\n\r\nI believe the cutoff last year for going to nats last year in texas was 41 actually. Getting to CD would probably require at least a 34 there, but not 42/43+. If that were the case, they'd be out of tiebreakers before the top 4 was decided :D", "Solution_15": "I think I heard from someone that the cutoff for countdown in Texas was around a 35.\r\n\r\nAs for your question, try for a 46. I'm not sure if that'll get you in though. :wink:" } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "\\/closed" ], "Problem": "Just wondering, who were the two non-WOOT alumni that were USAMO winners this year? I think Sergei was one, don't know about the other. According to the WOOT Information page 10 of the 12 winners were alumni.", "Solution_1": "Everyone who makes mop is invited to WOOT next year, so Sergei might actually count as an alumn (don't know why he wouldn't enroll if it was free...) Brian and Sherry, on the other hand, probably didn't enroll." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Define the sequence of positive integers $\\{x_{n}\\}$ by $x_{1}=10^{999}+1$ and for all $n \\geq 2$ the number $x_{n}$ is obtained from $11x_{n-1}$ by deleting the first digit.Is the sequence bounded?Why?", "Solution_1": "well, the sequence of the number of digits of $x_{i}$ is nonincreasing..." } { "Tag": [ "geometry", "perpendicular bisector" ], "Problem": "Right triangle $ABC$ is made of paper. Hypotenuse $\\overline{AB} = 36$. Vertex $C$ is folded onto the midpoint of $\\overline{AB}$, forming crease $\\overline{XY}$ (where points $X$ and $Y$ are on the legs of the triangle). Given that $XY = 20$, compute the area of triangle $XYC$.", "Solution_1": "[hide]\nLet midpt of AB = M. CM= 18\nDraw in X and Y. Quadrilateral CXMY is made up of two congruent right triangles, XCY and XMY.\nXY=20 and is perpendicular bisector of CM. So area of triangle CXY = 9 x 20/2 = 90.[/hide]", "Solution_2": "Nice. I don't know why it took me so long to figure out the distance from C to the midpoint of the hypotenuse." } { "Tag": [ "analytic geometry", "geometry", "rectangle", "combinatorics proposed", "combinatorics" ], "Problem": "Consider an 8x8 board divided in 64 unit squares. We call [i]diagonal[/i] in this board a set of 8 squares with the property that on each of the rows and the columns of the board there is exactly one square of the [i]diagonal[/i]. Some of the squares of this board are coloured such that in every [i]diagonal[/i] there are exactly two coloured squares. Prove that there exist two rows or two columns whose squares are all coloured.", "Solution_1": "Ok, it id easier than I first thought :\r\n\r\n1) Consider any diagonal and then its two coloured cells in different rows and columns. Assume that they are the ones with coordinates $(i,j)$ and $(r,s)$. Then consider the diagonal obtained from the one above by replacing the two coloured cells by the cells $(i,s)$ and $(r,j)$. Since this new diagonal has two coloured cells and that the initial one has no other coloured cells, it follows that these two new cells are coloured too.\r\n\r\n2) There is no coloured cells which does not belong to row $j$ or $s$ and to columns $i$ or $r$, otherwise with the two first coloured cells above, we may find a diagonal with at least three coloured cells.\r\n\r\n3) Wlog, assume that the four coloured cells in 1) are the ones in the upper left $2\\times 2$ square, that are $(1,8),(2,8),(1,),(2,7).$\r\nNow, assume that there exists a non-coloured cell in the upper row, say $(x,8)$ with $x>2.$.\r\nBy considering this cell and $(2,7)$ and $(1,y)$ with $y <7$, we may complete them to form a diagonal and, using 2), we deduce that $(1,y)$ is coloured. Thus all the first column is coloured.\r\nWe prove in the same way that the second column is coloured by considering $(2,y)$ instead of $(1,y).$\r\nTherefore there are two coloured columns.\r\n\r\nIf all the upper row is coloured, then as in 1), we deduce that all the second row from above is coloured, and we are done.\r\n\r\nPierre.", "Solution_2": "[hide=\"Solution\"]Label the columns $a,b,c\\ldots$ and the rows $1,2,3\\ldots$. Begin by considering the main diagonal $a1-h8$. It must have two colored squares. WLOG rearrange the rows and columns so that they are on $a1$ and $b2$. Then diagonals containing $a1$ and $b2$ can contain any square within the rectangle defined by $c3$ and $h8$, and since the diagonal must have exactly two colored squares ($a1$ and $b2$), none of the squares in that rectangle can be colored.\n\nNow consider the diagonal containing $a3, b2, c1$ and any other 5 squares we know are not colored. We know $b2$ is colored, so either $a3$ or $c1$ is colored. \n\nAssume $a3$ is colored. Then the diagonals containing $b1,a3$ and $b2,a3$ show that no square on columns $c-h$ is colored. And then by necessity the entire $a$ and $b$ columns must be colored to satisfy the conditions. If $c1$ is colored, the entire $1$ and $2$ rows are colored by a similar argument. This completes the proof.[/hide]" } { "Tag": [ "calculus", "ratio", "algorithm" ], "Problem": "Ok, the time is rolling around to start picking a new Science Fair Topic that is required by the school. I'd really love to do a math topic but can't really think of any good ones. Any of you guys have ideas? \r\n(Note: I come from a ghetto town and will be completely on my own in this endeavor so try not to post any really advanced/hardcore topics. Also, I will not plagiarize any ideas posted unless given explicit permission. I'm just looking for ideas to jump start my own.)", "Solution_1": "I was thinking about doing a research in math too...\r\n\r\nbut what exactly can we do???", "Solution_2": "What level of math do you know? Do you know calculus yet?", "Solution_3": "for me, I'm taking AB this year (junior)", "Solution_4": "I've taken BC already. Well, I've already come up with one I think. I'm going to look at the different definitions/approximations of the golden ratio and do a precision analysis, determine which one approaches the value of phi the \"fastest\". I'm going to try to come up with a few on my own and compare those against the standard ones out there.... (good luck to me, heh). It's still tentative but I may settle for that one.", "Solution_5": "I did a project applying the Collatz algorithm to different subsets of numbers and observing the results. It was a really EASY project, but everyone else thinks it was complicated. :lol: I'm going to the citywide fair this weekend." } { "Tag": [ "algebra", "polynomial", "algebra unsolved" ], "Problem": "Find all polynomials $ P(x)\\in \\mathbb R[x]$ such that \n$$ P(x^2)=P(x)P(x+2),$$\nfor all $x\\in \\mathbb R$.", "Solution_1": "let $ a$ be a complex root of $ P$. so $ P(a)\\equal{}0$\r\ntherefore $ a^2,a^4,a^8$ etc. have to be roots of $ P$ also. This leeds to contradiction because we get infinitely many roots and polinomial should be of infinite degree except if $ |a|\\equal{}1;0$ not hard to exclude the $ 0$ part\r\n\r\nbut if we put $ a\\minus{}2$ into the expression we get $ P((a\\minus{}2)^2)\\equal{}P(a\\minus{}2)P(a)$ meaning $ P((a\\minus{}2)^2)\\equal{}0$ which means again infinitely many roots with the same reason as before. except if $ |a\\minus{}2|\\equal{}1$\r\n\r\nthe only $ a$ such that $ |a\\minus{}2|\\equal{}|a|\\equal{}1$ is $ a\\equal{}1$ and it is the only theoretical root of the polynomial\r\n\r\nnow we put $ a_n(x\\minus{}1)^n$ into starting expression and conclude that all polinomials of form $ (x\\minus{}1)^n$ for any natural $ n$ are solutions plus trivial solutions $ P(x)\\equal{}0$ and $ P(x)\\equal{}1$", "Solution_2": "[quote=\"radio\"]Find all $ P(x)\\in R[x]$ such that \n$ P(x^2) \\equal{} P(x)P(x \\plus{} 2)$[/quote]\r\n\r\n$ P(x)\\equal{}0$ and $ P(x)\\equal{}1$ are solution.\r\nIf $ P(x)$ is different from constant polynomial, then :\r\nIf $ z$ is a complex root, then so is $ z^2$ and so, since we have finite number of roots, all non zero roots have $ |z|\\equal{}1$.\r\nIf $ 0$ is a root, $ P((\\minus{}2)^2)\\equal{}P(\\minus{}2)P(0)$ and $ 4$ is a root, but $ |4|\\neq 1$ and so $ 0$ is not a root.\r\nIf $ z$ is a complex root, then so is $ (z\\minus{}2)^2$ and so all roots have $ |z|\\equal{}1$ and $ |z\\minus{}2|\\equal{}1$\r\n\r\nSo the only complex root is $ 1$.\r\n\r\nSo $ P(x)\\equal{}a(x\\minus{}1)^n$ and it is easy to check that $ a\\equal{}1$\r\n\r\nSo the solutions are :\r\n$ P(x)\\equal{}0$\r\n$ P(x)\\equal{}1$\r\n$ P(x)\\equal{}(x\\minus{}1)^n$" } { "Tag": [ "inequalities", "algebra", "function", "domain" ], "Problem": "Solve $ 2x\\minus{}1\\le\\sqrt{x^2\\plus{}3x}.$", "Solution_1": "$ \\sqrt{x^2\\plus{}3x} \\geq 0 \\\\\r\n\\Rightarrow x^2\\plus{}3x\\geq 0\\\\\r\n\\Rightarrow x\\geq 0 \\ or \\ x \\leq \\minus{}3........(1)$\r\n\r\nnow squaring we get \r\n$ 4x^2\\minus{}4x\\plus{}1 \\leq \\ x^2\\plus{}3x\\\\\r\n\\Rightarrow 3x^2\\minus{}7x\\plus{}1\\leq 0\\\\\r\n\\Rightarrow x^2\\minus{}7x/3\\plus{}1/3 \\leq 0\\\\\r\n\\Rightarrow \\frac{7\\minus{}\\sqrt{37}}{6} \\leq x \\leq \\frac{7\\plus{}\\sqrt{37}}{6}.....(2)$\r\ncombining 1 and 2\r\n$ \\frac{7\\minus{}\\sqrt{37}}{6} \\leq x \\leq \\frac{7\\plus{}\\sqrt{37}}{6}$", "Solution_2": "[quote=\"implex\"]\ncombining 1 and 2\n$ \\frac {7 \\minus{} \\sqrt {37}}{6} \\leq x \\leq \\frac {7 \\plus{} \\sqrt {37}}{6}$[/quote]\r\nIs this the final answer? It's not correct then. :o", "Solution_3": "[hide=\"my solution\"]\n1) $ 0\\le 2x\\minus{}1 \\Rightarrow x\\ge \\frac{1}{2} \\dots (1)\\\\\n2x\\minus{}1\\le \\sqrt{x^2\\plus{}3x}\\\\\n4x^2\\minus{}4x\\plus{}1\\le x^2\\plus{}3x\\\\\n3x^2\\minus{}7x\\plus{}1\\le 0\\\\\n\\frac{7\\minus{}\\sqrt{37}}{6}\\le x \\le \\frac{7\\plus{}\\sqrt{37}}{6}$\nfrom (1):\n$ \\frac{1}{2}\\le x \\le \\frac{7\\plus{}\\sqrt{37}}{6}$\n2) $ 0 > 2x\\minus{}1 \\Rightarrow x<\\frac{1}{2}\\dots (2)\\\\\n\\minus{}\\sqrt{x^2\\plus{}3x}\\le 2x\\minus{}1\\\\\nx^2\\plus{}3x \\le 4x^2\\minus{}4x\\plus{}1\\\\\n0 \\le 3x^2\\minus{}7x\\plus{}1\\\\\n\\frac{7\\minus{}\\sqrt{37}}{6} \\le x \\le \\frac{7\\plus{}\\sqrt{37}}{6}$\nfrom (2):\n$ \\Phi \\text{ No Solution}$\nthe final solution:\n$ \\frac{1}{2}\\le x \\le \\frac{7\\plus{}\\sqrt{37}}{6}$\n\n[/hide]", "Solution_4": "Put $ x\\equal{}0$. It works!\r\n\r\nPut $ x\\equal{}\\minus{}100$. It works too!\r\n\r\nSo... what's going on here? Look closely at the domains and ranges.", "Solution_5": "ok : \r\n$ \\left[0,\\frac {7 \\plus{} \\sqrt {37}}{6}\\right]\\cup ( \\minus{} \\infty,\\minus{}3]$", "Solution_6": "[quote=\"mathson\"]ok : \n$ \\left[0,\\frac {7 \\plus{} \\sqrt {37}}{6}\\right]\\cup ( \\minus{} \\infty,3]$[/quote]A small typo, it should be $ \\minus{}3$ ;)", "Solution_7": "[quote=\"10000th User\"][quote=\"mathson\"]ok : \n$ \\left[0,\\frac {7 \\plus{} \\sqrt {37}}{6}\\right]\\cup ( \\minus{} \\infty,3]$[/quote]A small typo, it should be $ \\minus{} 3$ ;)[/quote]\r\n\r\n$ \\text{thanks }10^4 \\text{th User}$" } { "Tag": [ "inequalities", "Cauchy Inequality", "algebra unsolved", "algebra" ], "Problem": "3829. Let $a_1$, $a_2$, ..., $a_n$ be positive reals\r\n\r\n\\[\\sum_{i = 1}^n\\frac{a_i^2}{a_{i + 1} + a_{i + 2}} \\geq \\frac{1}{2}\\sum_{i = 1}^na_i.\\]", "Solution_1": "First, I can only guess the correct formulation of the problem. Here is my guess:\r\n\r\n[color=blue][b]Problem.[/b] Let $a_1$, $a_2$, ..., $a_n$ denote positive numbers. Prove that $\\sum_{i=1}^n\\frac{a_i^2}{a_i+a_{i+1}}\\geq\\frac12\\sum_{i=1}^n a_i$.[/color]\r\n\r\n[i]Solution.[/i] First,\r\n\r\n$2\\sum_{i=1}^n\\frac{a_i^2}{a_i+a_{i+1}}-\\sum_{i=1}^n\\frac{a_i^2+a_{i+1}^2}{a_i+a_{i+1}}=\\sum_{i=1}^n\\frac{2a_i^2-\\left(a_i^2+a_{i+1}^2\\right)}{a_i+a_{i+1}}=\\sum_{i=1}^n\\frac{a_i^2-a_{i+1}^2}{a_i+a_{i+1}}$\r\n$=\\sum_{i=1}^n\\left(a_i-a_{i+1}\\right)=0$.\r\n\r\nThus, $2\\sum_{i=1}^n\\frac{a_i^2}{a_i+a_{i+1}}=\\sum_{i=1}^n\\frac{a_i^2+a_{i+1}^2}{a_i+a_{i+1}}$. Now, for any two positive reals x and y, we have $\\frac{x^2+y^2}{x+y}\\geq\\frac{x+y}{2}$, since $2\\left(x^2+y^2\\right)\\geq\\left(x+y\\right)^2$ (because $2\\left(x^2+y^2\\right)-\\left(x+y\\right)^2=\\left(x-y\\right)^2\\geq 0$). Thus,\r\n\r\n$2\\sum_{i=1}^n\\frac{a_i^2}{a_i+a_{i+1}}=\\sum_{i=1}^n\\frac{a_i^2+a_{i+1}^2}{a_i+a_{i+1}}\\geq\\sum_{i=1}^n\\frac{a_i+a_{i+1}}{2}=\\sum_{i=1}^n a_i$,\r\n\r\nso that $\\sum_{i=1}^n\\frac{a_i^2}{a_i+a_{i+1}}\\geq\\frac12\\sum_{i=1}^n a_i$, and the problem is solved.\r\n\r\n darij", "Solution_2": "Using Cauchy inequality,\r\n$\\left ( \\sum_{i=1}^{n} (a_{i+1}+a_{i+2}) \\right ) \\cdot \\left (\\sum_{i = 1}^n\\frac{a_i^2}{a_{i + 1} + a_{i + 2}} \\right ) \\geq (\\sum_{i = 1}^n a_i)^2$.\r\nWhile $\\sum_{i=1}^{n} (a_{i+1}+a_{i+2}) = 2\\sum_{i=1}^{n}a_i$, the result follows.", "Solution_3": "Another proof.\r\nUsing AM-GM inequality.\r\nSince $\\frac{a_i^2}{a_{i + 1} + a_{i + 2}} + \\frac{a_{i + 1} + a_{i + 2} }{4} \\geq a_i,$ $i = 1, 2, ..., n.$\r\nAdding such $n$ inequlities, and notice that $\\sum_{i=1}^{n}\\frac{a_{i + 1} + a_{i + 2}}{4} = \\frac{1}{2}\\sum_{i=1}^{n}a_i.$\r\nThe result follows.\r\n\r\nFrom these two proofs, we see that the denominator can be selected freely." } { "Tag": [ "calculus", "calculus computations" ], "Problem": "sum(n=1 to infinity){1/((2^(2n+1))*n*(2n-1))}", "Solution_1": "Try the more general sum $ \\sum_{n\\equal{}1}^{\\infty}\\frac{x^{2n}}{(2n)(2n\\minus{}1)}$. (Your series evaluates this at $ x\\equal{}\\frac{1}2$)\r\n\r\nWhat happens when you differentiate this series a couple of times?", "Solution_2": "Thank you for the help.\r\n\r\nI differentiated the series twice, cancelled the common terms, and obtained a geometric series.\r\nThen I evaluted the series and integrated to recover the original sum :)" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Determine all rational numbers $ r$ such that all solutions of the equation:\r\n\r\n$ rx^2\\plus{}(r\\plus{}1)x\\plus{}(r\\minus{}1)\\equal{}0$ are integers.", "Solution_1": "[quote=\"moldovan\"]Determine all rational numbers $ r$ such that all solutions of the equation:\n\n$ rx^2 \\plus{} (r \\plus{} 1)x \\plus{} (r \\minus{} 1) \\equal{} 0$ are integers.[/quote]\r\n\r\nClearly $ r\\equal{}0$ is a solution, so for $ r \\neq 0$, let $ r\\equal{}\\frac{q}{p}$, where $ \\gcd(p,q)\\equal{}1$, $ p$ is integer and $ q$ is positive integer, then the equation\r\n\r\n$ \\Leftrightarrow qx^2\\plus{}(q\\plus{}p)x\\plus{}(q\\minus{}p)\\equal{}0$\r\n\r\nNow let the solutions of the equation be integers $ a, b$.\r\n\r\nThen $ q(x\\minus{}a)(x\\minus{}b) \\equiv qx^2\\plus{}(q\\plus{}p)x\\plus{}(q\\minus{}p)$ gives $ q|q\\plus{}p \\Rightarrow q\\equal{}1$\r\n\r\nNow we are going to find $ p$ so that $ x^2\\plus{}(p\\plus{}1)x\\plus{}(1\\minus{}p)$ has only integer roots.\r\n\r\nRewrite the equation like this: $ x^2\\plus{}x\\plus{}1\\equal{}p(1\\minus{}x)$.\r\nClearly for $ x$ is integer, $ (1\\minus{}x) \\neq 0$\r\nAnd $ (1\\minus{}x)|\\gcd(x^2\\plus{}x\\plus{}1,1\\minus{}x)\\equal{}\\gcd(3,1\\minus{}x)$\r\n$ \\Rightarrow 1\\minus{}x\\equal{}1,\\minus{}1,3,\\minus{}3$\r\n\r\n$ 1\\minus{}x\\equal{}1 \\Rightarrow p\\equal{}1 \\Rightarrow r\\equal{}1$ as solution after checking.\r\n$ 1\\minus{}x\\equal{}\\minus{}1 \\Rightarrow p\\equal{}\\minus{}7 \\Rightarrow r\\equal{}\\minus{}\\frac{1}{7}$ as solution after checking.\r\n$ 1\\minus{}x\\equal{}\\minus{}3 \\Rightarrow p\\equal{}\\minus{}7$ again.\r\n$ 1\\minus{}x\\equal{}3 \\Rightarrow p\\equal{}1$ again.\r\n\r\nSo $ r\\equal{}0$ or $ r\\equal{}1$ or $ r\\equal{}\\minus{}\\frac{1}{7}$ are the only solutions." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Find the maximum value of $A$ for all $x,y,z>0$:\r\n\r\n$\\frac{x}{\\sqrt{y^2+z^2}} + \\frac{y}{\\sqrt{z^2+x^2}} + \\frac{z}{\\sqrt{x^2+y^2}} \\geq A$", "Solution_1": "max{A}=3sin45", "Solution_2": "No, A=2 ,see $x=y,z\\to 0$", "Solution_3": "can you please post whole solution?", "Solution_4": "It's a very old problem,I think it should be in mathlink.", "Solution_5": "[quote=\"Adalbert\"]Find the maximum value of $A$ for all $x,y,z>0$:\n\n$\\frac{x}{\\sqrt{y^2+z^2}} + \\frac{y}{\\sqrt{z^2+x^2}} + \\frac{z}{\\sqrt{x^2+y^2}} \\geq A$[/quote]\r\nIbelieve that it should be:\r\nFind the maximum value of $A$ for all $x,y,z>0$:\r\n\r\n$\\frac{x}{\\sqrt{y^2+z^2}} + \\frac{y}{\\sqrt{z^2+x^2}} + \\frac{z}{\\sqrt{x^2+y^2}} > A$\r\n\r\nif so then using $a\\geq \\frac{2a^2}{a^2+1}$\r\n$\\frac{x}{\\sqrt{y^2+z^2}} + \\frac{y}{\\sqrt{z^2+x^2}} + \\frac{z}{\\sqrt{x^2+y^2}}\\geq \\frac{2x^2}{x^2+y^2+z^2}+\\frac{2y^2}{x^2+y^2+z^2}+\\frac{2z^2}{x^2+y^2+z^2}=2$ but we can not have equality ;)", "Solution_6": "i saw it $\\geq$ but it could be $>$ i guess...nice solution, thanks", "Solution_7": "We can think of this inequality. $x^2+y^2+z^2 \\ge 2x\\sqrt{y^2+z^2}, \\frac{x}{\\sqrt{y^2+z^2}} \\ge \\frac{2x^2}{x^2+y^2+z^2}$, then we have $\\sum{\\frac{x}{\\sqrt{y^2+z^2}} \\ge 2}$ but here the inequality isn't equal to 2." } { "Tag": [ "function", "algebra", "polynomial", "trigonometry", "domain" ], "Problem": "A function \"slides\" if f(x) = f(x+1) for all x in [0,1]. Give one non-polynomial function which slides and prove that a non-constant polynomial cannot slide.", "Solution_1": "[hide=\"Hint for the first part\"] This is pretty obvious; sliding is a weaker condition than periodicity. [/hide]\n[hide=\"Hint for the second part\"] How many roots can the function $ g(x) \\equal{} f(x) \\minus{} f(x\\plus{}1)$ can have? [/hide]", "Solution_2": "Can you elaborate on that? I would be pleased to see your solution.", "Solution_3": "The problems you've just posted look like exercises out of a textbook, so I'd prefer it if you showed us your progress so far instead. What's confusing about the hints?", "Solution_4": "$ g(x) \\equal{} f(x) \\minus{} f(x\\plus{}1)$ can't have infinite amount of roots unless it is a constant polynomial. One function that works could be $ f(x) \\equal{} \\minus{}2$ if $ x<0$, $ f(x) \\equal{} 0$ if $ 0\\leq x \\leq2$ and $ f(x) \\equal{} 2$ if $ 2\\leq x$", "Solution_5": "Thanks, aznlord1337.\r\nt0rajir0u, your assumptions are wrong: the questions are not from a textbook (someone was asked them at a university interview, which is why I thought they might be interesting); I have solved them (the first example I thought of is Sin (2pi x); I am not confused by your hints. Thanks anyway.", "Solution_6": "[hide=\"first part\"]Since the sine function repeats ever $ 2\\pi$, we could multiply $ x$ by $ 2\\pi$ and it repeats every 1. Thus $ \\sin{2\\pi x}$ slides. So does $ \\cos{2\\pi x}$. You can also add any constant to a sliding function and it will still be a sliding function.[/hide]\n\n[hide=\"second part\"]Let $ e(x)\\equal{}f(x)\\minus{}f(x\\plus{}1)$. Thus $ g(x)\\equal{}0$ for all $ x\\in [0,1]$. Thus $ g(x)$ has infinitely many roots, and $ g(x)$ is not a polynomial, and hence $ f(x)$ isn't either.[/hide]", "Solution_7": "[quote=\"1=2\"]Let $ e(x) \\equal{} f(x) \\minus{} f(x \\plus{} 1)$. Thus $ g(x) \\equal{} 0$ for all $ x\\in [0,1]$. Thus $ g(x)$ has infinitely many roots, and $ g(x)$ is not a polynomial[/quote]\r\nThis is not quite correct; $ g(x)$ could be the zero polynomial. You then need to show that it is not possible for $ f(x \\plus{} 1)$ to equal $ f(x)$ identically (if $ f$ is a nonconstant polynomial).", "Solution_8": "oh....\r\n\r\n[hide=\"well then\"]Note that the constant terms of $ f(x)$ and $ f(x\\plus{}1)$ are the same. When we plug in $ x\\plus{}1$ into $ f(x)$, we must not have it affect the constant term. But $ (x\\plus{}1)^n$ contains a constant term for all $ n$. Thus we may not have a $ x^n$ term in $ f(x)$ for all $ n$. Thus $ f(x)$ isn't a polynomial.[/hide]\r\n\r\nIs that right?", "Solution_9": "Alternatively, for 2 you could\r\n\r\n\r\n[hide]\nPolynomial functions are continuous, so f(x) is bounded on [0,1) which implies that it is bounded, which implies that it is the 0 function[/hide]", "Solution_10": "[quote=\"1=2\"]Note that the constant terms of $ f(x)$ and $ f(x \\plus{} 1)$ are the same. When we plug in $ x \\plus{} 1$ into $ f(x)$, we must not have it affect the constant term. [/quote]\r\nThere's no contradiction here if $ f(0) \\equal{} f(1)$.\r\n\r\nOne easy way to finish from here is to show that if $ f(x)$ is a polynomial of degree exactly $ n$ then $ f(x \\plus{} 1) \\minus{} f(x)$ is a polynomial of degree exactly $ n\\minus{}1$, which is just done by comparing coefficients. You could also do what pythag011 did.\r\n\r\nAnd for a third alternative, $ f(x) \\minus{} f(c)$ for any constant $ c$ has infinitely many roots.", "Solution_11": "Polynomial Case: If $ f(x)\\equal{}f(x\\plus{}1)$, then $ f(x)\\equal{}f(x\\plus{}n)$ for all $ n\\in\\mathbb{N}$. If we look at this algebraically, it is clear that we can extend the domain to $ \\mathbb{C}$. If $ f$ constant, it obviously works. If $ f$ is non-constant, then it has a root, $ r$. Let $ deg(f)\\equal{}m$. Observe that $ r$, $ r\\plus{}1$, ..., $ r\\plus{}m$ are $ m\\plus{}1$ distinct roots of a non-constant polynomial. Thus $ deg(f)\\ge m\\plus{}1$ which is a contradiction.", "Solution_12": "The message here is that a lot of things go wrong if you require a nonconstant polynomial to also be periodic. Here's another: since polynomials are smooth, not only is it necessary that $ f(0) \\equal{} f(1)$, but we must also have $ f'(0) \\equal{} f'(1), f''(0) \\equal{} f''(1)$, and so on until you get to a linear polynomial - which must then be constant." } { "Tag": [ "probability", "videos", "Pascal\\u0027s Triangle" ], "Problem": "Probability: Count *Equally Likely* Events!\n\nWe do two probability problems that teach us what it is important to count equally likely outcomes when counting successful and possible outcomes in a probability problem.", "Solution_1": "This video made it easy for me to see how to deal with probability", "Solution_2": ":lol: this is so great! i was horrible at this site before! :P", "Solution_3": "Probability of 0 yellows is $ \\frac{1}{64}$\r\nProbability of 1 yellow is $ \\frac{3}{32}$\r\nProbability of 2 yellows: $ \\frac {\\binom{6}{2}}{2^6} \\equal{} \\frac{12}{64} \\equal{} \\frac{3}{16}$\r\nProbability of 3 yellows:$ \\frac {\\binom{6}{3}}{2^6} \\equal{} \\frac{20}{64} \\equal{} \\frac{5}{16}$\r\nProbability of 4 yellows is equal to the probability of 2 blues so $ \\frac {3}{16}$\r\nProbability of 5 yellows is 1 blue, so $ \\frac{3}{32}$.\r\nProbability of 6 yellows is just $ \\frac{1}{64}$\r\nBut summing gives$ \\frac{29}{32}$, what did I do wrong?", "Solution_4": "[quote=\"brightzhu\"]Probability of 0 yellows is $ \\frac {1}{64}$\nProbability of 1 yellow is $ \\frac {3}{32}$\nProbability of 2 yellows: $ \\frac {\\binom{6}{2}}{2^6} \\equal{} \\frac {12}{64} \\equal{} \\frac {3}{16}$\nProbability of 3 yellows:$ \\frac {\\binom{6}{3}}{2^6} \\equal{} \\frac {20}{64} \\equal{} \\frac {5}{16}$\nProbability of 4 yellows is equal to the probability of 2 blues so $ \\frac {3}{16}$\nProbability of 5 yellows is 1 blue, so $ \\frac {3}{32}$.\nProbability of 6 yellows is just $ \\frac {1}{64}$\nBut summing gives$ \\frac {29}{32}$, what did I do wrong?[/quote]\r\n\r\nYou mean pulling 6 balls from an infinite supply of yellow, blue balls?\r\n\r\nEDIT: $ \\binom{6}{2}\\equal{}12$? :)", "Solution_5": "oh i see, those numbers form the sixth row of pascal's triangle! :D", "Solution_6": "This video made it very easy for me to understand probability. It was explained very well and his humor made it fun to watch.", "Solution_7": "These are the simplest of probability problems. These are usually found on the AMC 8.", "Solution_8": "[hide=\"My solution\"]0 yellow faces:$\\binom{6}{0}$\n1 yellow face: $\\binom{6}{1}$\n2 yellow faces: $\\binom{6}{2}$\n3 yellow faces: $\\binom{6}{3}$\n4 yellow faces: $\\binom{6}{4}$\n5 yellow faces: $\\binom{6}{5}$\n6 yellow faces: $\\binom{6}{6}$\n\nWe know the fact that:\n$\\binom{n}{0}+\\binom{n}{1}+.....\\binom{n}{n-1}+\\binom{n}{n}=2^n$\nWe have $\\binom{6}{0}+\\binom{6}{1}+\\binom{6}{2}+\\binom{6}{3}+\\binom{6}{4}+\\binom{6}{5}+\\binom{6}{6}\\implies 2^6=64$\n\n$\\mathbb{P}=\\frac{Success's}{\\text{Total possibilities}}\\implies \\frac{Success's}{2^6}\\implies \\frac{Success's}{64}$\n\n0 yellow faces: $\\frac{\\binom{6}{0}}{64}\\implies \\frac{1}{64}$\n1 yellow face: $\\frac{\\binom{6}{1}}{64}\\implies \\frac{6}{64}\\implies \\frac{3}{32}$\n2 yellow faces: $\\frac{\\binom{6}{2}}{64}\\implies \\frac{\\frac{6*5}{2}}{64}\\implies \\frac{15}{64}$\n3 yellow faces: $\\frac{\\binom{6}{3}}{64}\\implies \\frac{\\frac{6*5*4}{3*2}}{64}\\implies \\frac{20}{64}\\implies \\frac{10}{32}\\implies \\frac{5}{16}$\n4 yellow faces: $\\frac{\\binom{6}{4}}{64}\\implies \\frac{\\binom{6}{2}}{64}\\implies \\frac{\\frac{6*5}{2}}{64}\\implies \\frac{15}{64}$\n5 yellow faces: ${\\frac{\\binom{6}{5}}{64}\\implies \\frac{6}{1}}{64}\\implies \\frac{6}{64}\\implies \\frac{3}{32}$\n6 yellow faces: $\\frac{\\binom{6}{6}}{64}\\implies \\frac{\\binom{6}{0}}{64}\\implies \\frac{1}{64}$\n\nAlso, as Fersolve pointed out, that those numbers:\n$\\in{1,6, 15, 20, 15, 6, 1}$ are the sixth row of Pascal's triangle, and we know that the nth row of pascal's triangle is $2^n$ so we can find that the denominator is easily $2^6$[/hide]\n\nEdit: YAY! 100th math post! Click on the link in my signature to see what I am at now =)\nEdit 2: [quote=\"brightzhu\"]Probability of 0 yellows is $ \\frac {1}{64}$\nProbability of 1 yellow is $ \\frac {3}{32}$\nProbability of 2 yellows: $ \\frac {\\binom{6}{2}}{2^6} = \\frac {12}{64} = \\frac {3}{16}$\nProbability of 3 yellows:$ \\frac {\\binom{6}{3}}{2^6} = \\frac {20}{64} = \\frac {5}{16}$\nProbability of 4 yellows is equal to the probability of 2 blues so $ \\frac {3}{16}$\nProbability of 5 yellows is 1 blue, so $ \\frac {3}{32}$.\nProbability of 6 yellows is just $ \\frac {1}{64}$\nBut summing gives$ \\frac {29}{32}$, what did I do wrong?[/quote]\nProbability of 2 yellow is $\\frac{\\binom{6}{2}}{2^6}$\n$\\binom{6}{2}=\\frac{6*5}{2}\\implies \\frac{\\cancel{6}*5}{\\cancel{2}}\\implies 3*5=\\boxed{15}$, so the probability of 2 yellows= $\\frac{15}{64}$ not $\\frac{12}{64}$\nYou made that mistake in $\\binom{6}{4}$ also.\n\nAlso: \n$\\frac{1}{64}+\\frac{6}{64}+\\frac{12}{64}+\\frac{20}{64}+\\frac{12}{64}+\\frac{1}{64}$ using your method is:\n$\\frac{1+6+12+20+12+6+1}{64}\\implies \\frac{7+12+20+12+6+1}{64}\\implies \\frac{19+20+6+12+1}{64}\\implies \\frac{39+6+12+1}{64}\\implies \\frac{45+12+1}{64}\\implies \\frac{57+1}{64}\\implies \\frac{58}{64}\\implies \\frac{\\frac{58}{2}}{\\frac{64}{2}}\\implies \\frac{29}{32}$ which is correct with your mistake, if your reasoning for $\\binom{6}{2}=12$" } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Find all functions $ f: R\\minus{}>R$ such that \r\n\r\n$ f(x\\minus{}f(y))\\equal{}f(x\\plus{}y^{2008})\\plus{}f(f(y)\\plus{}y^{2008})\\plus{}1$ for all $ x,y \\in R$\r\n\r\n--------------------------------------------------------------------------------------------------------------- :lol: :)", "Solution_1": "http://www.mathlinks.ro/viewtopic.php?t=251508" } { "Tag": [ "trigonometry", "calculus", "calculus computations" ], "Problem": "The region bounded by $ y\\equal{}\\cos x$, $ y\\equal{}x$, and $ x\\equal{}0$ is revolved around the line $ y\\equal{}\\minus{}2$. Find the volume of the solid of revolution.\r\n~~~\r\n\r\nI have sketched the curves. From the origin to the point of intersection, $ y\\equal{}\\cos x$ is always above $ y\\equal{}x$, so I'm thinking that the washer method can be used for this problem.\r\n\r\nThe problem is, I'm not sure of an exact answer to $ x\\equal{}\\cos x$. If I knew that, then I could solve the definite integral.", "Solution_1": "The solution to $ x\\equal{}\\cos{x}\\approx 0.739085$. It's not particularly easy to show that by hand, though. Use a calculator.", "Solution_2": "$ x^2\\plus{}2x\\minus{}2\\equal{}0$ gives an O.K. approximation in case you can't (don't want to) use a calculator.", "Solution_3": "Or we could use Taylor Expansion. :lol:" } { "Tag": [ "trigonometry", "LaTeX" ], "Problem": "Simple problem but it's driving me crazy.\r\n\r\n[code]\n\n\\begin{array}{|l|l|}\n\\hline\n\\hspace{10pt} \\sin(\\theta) = \\frac{O}{H} & \\hspace{20pt} \\cos(\\theta) = \\frac{A}{H} & {} \\\\[10pt] \\hline\n\\hspace{10pt} \\tan(\\theta) = \\frac{O}{A} & \\hspace{20pt} \\csc(\\theta) = \\frac{H}{O} & {} \\\\[10pt] \\hline\n\\hspace{10pt} \\sec(\\theta) = \\frac{H}{A} & \\hspace{20pt} \\cot(\\theta) = \\frac{A}{O} & {}\n\\end{array}\n\n[/code]\r\n\r\nI keep getting 2 error messages for no reason. What's going on?", "Solution_1": "There are 2 errors here:\r\n1. array must be in maths mode so you need dollar signs as in $ \\$$\\begin{array} .... \\end{array}$ \\$$\r\n2. You have 3 columns (the last one blank) yet you have only 2 columns setup in \\begin{array}{|l|l|} \r\n\r\nI'm not sure of the purpose of \\hspace and there's no need for \\\\[10pt] as the way to stretch the distance between rows is to use \\renewcommand{\\arraystretch}{[i]factor[/i]} where factor=1 is normal, factor=1.5 is 1.5 times normal etc.\r\n\r\nThis looks OK to me:\r\n[code]$\\begin{array}{|c|c|c|} \n\\hline \n\\sin(\\theta) = \\frac{O}{H} & \\cos(\\theta) = \\frac{A}{H} & \\\\ \\hline \n\\tan(\\theta) = \\frac{O}{A} & \\csc(\\theta) = \\frac{H}{O} & \\\\ \\hline \n\\sec(\\theta) = \\frac{H}{A} & \\cot(\\theta) = \\frac{A}{O} & \\\\ \\hline \n\\end{array}$[/code]", "Solution_2": "Now the spacing of the array is being a pain in the rear end.\r\n\r\n[code] \\begin{mcen}$\\begin{array}{|c|c|c|c|}\n\\hline\n{} & {} & {} & {} \\\\\n\\sin(0) = 0 & \\sin(\\frac{\\pi}{2}) = 1 & \\hspace{-7.5pt} \\sin(\\pi) = 0 & \\sin(\\frac{3\\pi}{2}) = -1 \\\\\n\\hline\n{} & {} & {} & {} \\\\\n\\cos(0) = 1 & \\cos(\\frac{\\pi}{2}) = 0 & \\cos(\\pi) = -1 & \\cos(\\frac{3\\pi}{2}) = 0 \\\\\n\\hline\n{} & {} & {} & {} \\\\\n\\tan(0) = 0 & \\tan(\\frac{\\pi}{2}) = * & \\hspace{-7.5pt} \\tan(\\pi) = 0 & \\tan(\\frac{3\\pi}{2}) = * \\\\\n\\hline\n{} & {} & {} & {} \\\\\n\\csc(0) = * & \\csc(\\frac{\\pi}{2}) = 1 & \\csc(\\pi) = * & \\csc(\\frac{3\\pi}{2}) = -1 \\\\\n\\hline\n{} & {} & {} & {} \\\\\n\\sec(0) = 1 & \\sec(\\frac{\\pi}{2}) = * & \\hspace{-7.5pt} \\sec(\\pi) = -1 & \\sec(\\frac{3\\pi}{2}) = * \\\\\n\\hline\n{} & {} & {} & {} \\\\\n\\cot(0) = * & \\cot(\\frac{\\pi}{2}) = 0 & \\cot(\\pi) = * & \\cot(\\frac{3\\pi}{2}) = 0 \\\\\n\\hline\n\\end{array}$ \\\\[10pt] \\end{mcen}\n[/code]\r\n\r\nwhere mcen stands for the \"mini-center\" command, which centers something without any other auto-spacing annoyance to worry about.\r\n\r\nHow can I manipulate the \"white space\" above and below the text in each box?", "Solution_3": "I don't really understand what you are trying to do. What is wrong with\r\n[code]\\renewcommand{\\arraystretch}{2}\n$\\begin{array}{|l|l|l|l|} \n\\hline \n\\sin(0) = 0 & \\sin(\\frac{\\pi}{2}) = 1 & \\sin(\\pi) = 0 & \\sin(\\frac{3\\pi}{2}) = -1 \\\\ \n\\hline \n\\cos(0) = 1 & \\cos(\\frac{\\pi}{2}) = 0 & \\cos(\\pi) = -1 & \\cos(\\frac{3\\pi}{2}) = 0 \\\\ \n\\hline \n\\tan(0) = 0 & \\tan(\\frac{\\pi}{2}) = * & \\tan(\\pi) = 0 & \\tan(\\frac{3\\pi}{2}) = * \\\\ \n\\hline \n\\csc(0) = * & \\csc(\\frac{\\pi}{2}) = 1 & \\csc(\\pi) = * & \\csc(\\frac{3\\pi}{2}) = -1 \\\\ \n\\hline \n\\sec(0) = 1 & \\sec(\\frac{\\pi}{2}) = * & \\sec(\\pi) = -1 & \\sec(\\frac{3\\pi}{2}) = * \\\\ \n\\hline \n\\cot(0) = * & \\cot(\\frac{\\pi}{2}) = 0 & \\cot(\\pi) = * & \\cot(\\frac{3\\pi}{2}) = 0 \\\\ \n\\hline \n\\end{array}$[/code]" } { "Tag": [ "AMC", "USA(J)MO", "USAMO" ], "Problem": "THE USAMO list is really up. I am the first person to post this topic!!!\r\nI make it!!!\r\nSo far the index is XXX :D\r\n\r\nhttp://www.unl.edu/amc", "Solution_1": "where? :rotfl:", "Solution_2": "[quote=\"pieterminate\"]where? :rotfl:[/quote]\r\n\r\nYou are fooled so many times???\r\nThis time is real, and you don't believe???", "Solution_3": "What a character! A true jest.", "Solution_4": "[quote=\"Go Beyond\"][quote=\"pieterminate\"]where? :rotfl:[/quote]\n\nYou are fooled so many times???\nThis time is real, and you don't believe???[/quote]\r\n\r\nWell.....i didn't make it anyways, so.....", "Solution_5": "http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2006-ua/06usamoqualstate.html\r\n\r\n\r\n7th grade: 2\r\n8th grade: 15\r\n9th grade: 47 <--All of you freshmen MOP aspirers, note this number carefully!\r\n10th grade: 98\r\n11th grade: 89\r\n12th grade: 116\r\n\r\nI might be off by 1 for some but these should be correct.", "Solution_6": "Good job Randy Jia! Amazing. One of 2 7th graders." } { "Tag": [ "algebra", "polynomial", "calculus", "integration", "induction", "number theory solved", "number theory" ], "Problem": "Let [tex] P(x) [/tex] is a polynomial with integral coefficient . [tex] P(x) \\geq 0 [/tex] with every x . [tex] a_0 = 0 , a_1 = f(a_0) , a_2 = f(a_1) ,...., a_n= f(a_{n-1}) , .... [/tex] . Prove that [tex] (a_m , a_n ) = a_{(m,n)}[/tex] with m , n are natural numbers , m is different from n .", "Solution_1": "What does P ?\r\nMaybe f=P ?", "Solution_2": "Let $P_k = PoPo \\cdots oP$ $k$ times.\r\nThen $a_{i+k} = P_k(a_i)$ for all $i,k$.\r\n\r\nLet $P(x) = \\sum^{i=0}^q \\alpha_ix^i$.\r\n\r\nLet $m \\geq n$, and $d = gcd(a_m,a_n)$. Let $m=n+k$.\r\nWe have $a_m = P_k(a_n) = a_n(\\sum^{i=1}^q \\alpha{}_ia{}_n^{i-1} ) + P_k(0)$\r\nIt folows that $d$ divides $P_k(0) = P_k(a_0)=a_k=a_{m-n}$.\r\nThus, an easy induction leads to $d$ divides $a_{gcd(m,n)}$.\r\n\r\nConversely, for all $i$ and all $r \\geq 1$, $a_i$ divides $a_{ri}$. This follows from another easy induction because if $a_i$ divides $a_{ri}$ then it divides $P_i(a_{ri} )= a_{(r+1)i}$ (since, from above, the coefficient of $x^0$ in $P_i$ is $a_i$).\r\nThus $a_{gcd(m,n)}$ divides both $a_m$ and $a_n$ so that it divides $d$.\r\nSince everything is non-negative here, we deduce that $a_{gcd(m,n)}= d$, and we are done.\r\n\r\nPierre." } { "Tag": [ "limit", "logarithms", "trigonometry", "calculus", "calculus computations" ], "Problem": "$ \\lim_{x \\to 0_{+}}\\sqrt{x}\\ln x \\sin \\frac{1}{x^{2}}-\\frac{\\sin \\sqrt{x}}{x}$", "Solution_1": "$ \\lim_{x \\to 0_{+}}\\sqrt{x}\\ln x \\sin \\frac{1}{x^{2}}-\\frac{\\sin \\sqrt{x}}{x}=-\\infty?$", "Solution_2": "why is that ?", "Solution_3": "Because\r\n\r\n$ \\lim_{x \\to 0_{+}}\\sqrt{x}\\ln x \\sin \\frac{1}{x^{2}}= 0$\r\n\r\nand\r\n\r\n$ \\lim_{x \\to 0_{+}}\\frac{\\sin \\sqrt{x}}{\\sqrt{x}}= 1$" } { "Tag": [ "function", "real analysis", "real analysis unsolved" ], "Problem": "Prove the Intermediate Value Theorem using uniform continuity :)", "Solution_1": "Why? :maybe:", "Solution_2": "Because I'm going through problems in an Intro to Analysis book, and I'm stuck :(", "Solution_3": "I meant: why would anyone want to use uniform continuity here? In its general form the IVT is a purely topological statement: the image of a [url=http://en.wikipedia.org/wiki/Connected_space]connected set[/url] under a continuous map is connected. Whether the map is uniformly continuous is irrelevant. \r\n\r\nIf you are interested in the proof for continuous functions on real line, it's [url=http://en.wikipedia.org/wiki/Intermediate_value_theorem]here[/url]." } { "Tag": [ "geometry", "angle bisector" ], "Problem": "What is angle chasing? :oops:", "Solution_1": "That's ch.14 in vol.1. Angle chasing is basically trying to find the measure of a given angle. You can use many different tools for angle chasing,i.e, angle bisector, parallel/ perpendicular lines, sum of angles in a triangle,etc. To learn angle chasing, do lots of problems and then it will come easier. :)" } { "Tag": [ "AMC", "AIME", "geometry", "3D geometry", "AMC 10" ], "Problem": "Can we use paper, tape, and scissors to cut out and create 3D geometry figures to make it easier to visualize? For example when I had a picture of the 1985 AIME problem #15 I was able to solve it in nearly 2 seconds.", "Solution_1": "i believe you are allowed to use any eucliedean tools... ;)", "Solution_2": "I'm not entirely sure, but cutting out figures might be viewed as distracting to the other test-takers (this happened to me when I tried to make a paper cube on the AMC10 my freshman year).", "Solution_3": "From the front cover of the 2005 AIME:\r\n\r\n[quote]No aids other than scratch paper, ruler, compass, and protractors are permitted. In particular, [b][u]calculators and computers are not permitted[/u][/b][/quote]", "Solution_4": "[quote=\"joml88\"]From the front cover of the 2005 AIME:\n\n[quote]No aids other than scratch paper, ruler, compass, and protractors are permitted. In particular, [b][u]calculators and computers are not permitted[/u][/b][/quote][/quote]Hmmm...then it doesn't seem they ban ripping the paper...", "Solution_5": "[quote=\"Elemennop\"]I'm not entirely sure, but cutting out figures might be viewed as distracting to the other test-takers (this happened to me when I tried to make a paper cube on the AMC10 my freshman year).[/quote]\r\n\r\no_0\r\nwouldn't that have been time consuming?\r\nI don't think its against the rules or anything, i mean, the compass's point is quite sharp and can be used as a cutting tool : |", "Solution_6": "Hahahaha well tape and scissors are technically not permitted there ;-). But I guess if your folding/compass-cutting skills were good enough, you could get by without them. Still, it doesn't seem like it would help you that much (especially if you had to take the trouble to fold it and stuff). The figures shouldn't ever be too hard to visualize.", "Solution_7": "Hehe, this is where my advanced origami skills come into play. :P" } { "Tag": [], "Problem": "A pitcher contains $ 4\\frac{1}{2}$ cups of water. A $ \\frac{1}{3}$-cup scoop is used to\ntransfer water from the pitcher to a bowl. How many scoops are\nneeded until the bowl has more water than the pitcher?", "Solution_1": "Basically, we just need to do:\r\n\r\n$ 4\\frac {1}{2} \\div \\frac {1}{3}$\r\n\r\nFirst we change $ 4\\frac {1}{2}$ into an improper fraction, to get $ \\frac {9}{2}$. \r\n\r\nSecond, we take the reciprocal of the other fraction, $ \\frac {3}{1}$ and multiply it by $ \\frac {9}{2}$, to get:\r\n\r\n$ \\frac {9}{2} \\times \\frac {3}{1} \\implies \\frac {9 \\cdot 3}{2 \\cdot 1} \\implies 13.5$\r\n\r\nHowever, we want the bowl to have more water than the pitcher, so our answer is $ \\boxed{14}$.\r\n\r\nEDIT: Oops, I see.", "Solution_2": "@izzy, the problem asked for something different\r\n\r\nAfter 6 scoops, there $ 2\\frac {1}{2}$ cups of water and 2 cups in the bowl.\r\nWhen we scoop again, there is $ 2\\frac {1}{6}$ cup left in the pitcher and $ 2\\frac {1}{3}$ cups in the bowl. At this point there is more water in the bowl than in the pitcher, so the answer is 7.\r\n\r\nWe can also used izzy's solution and said that since she rounded up the number of scoops, the answer is the ceiling of the number of scoops divided by 2, or 7." } { "Tag": [ "algebra solved", "algebra" ], "Problem": "Let a_1, a_2,...,a_1999 be non-negative real numbers satisfying the following two conditions:\r\n\r\n(a) \\sum^{1999}_{k=1} a_k = 2,\r\n\r\n(b) a_1*a_2 + a_2*a_3 + ... + a_1998*a_1999 + a_1991*a_1 = 1\r\n\r\nLet S = \\sum^{1999}_{k=1} (a_k)^2. Find the maximum and minimum possible values of S.\r\n\r\nAlso consider the ILL problem 3 at: http://www.kalva.demon.co.uk/short/sh82.html", "Solution_1": "The following fact is very interesting:\r\n We have already discussed the fact that (x_1+...+x_n))^2>=4(x_1*x_2+...+x_n*x_1) for all reals x_1,...,x_n. But the sequence in the problem verifies the above relation with equality. So, we should think about the cases of equality, which I believe it's not an easy job.", "Solution_2": "I think the condition (b) is : a_1*a_2+...+a_1999*a_1=1\r\n[img]http://www.geocities.com/duyduca1/MAth1.gif[/img]", "Solution_3": "Yes, of course. I've been thinking about this problem, but I really don't know how to deduce the case of equality." } { "Tag": [ "LaTeX" ], "Problem": "The only thing I know is that you can put signs around something.\r\n\r\nHow do you do basics? Like drawing a triangle. making 20 rt 4 look real. and like showing squared. and stuff like that\r\n\r\nand labeling things", "Solution_1": "Please have a look at the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]LaTeX FAQ[/url]" } { "Tag": [ "calculus", "function", "geometry", "analytic geometry", "integration", "derivative", "topology" ], "Problem": "I came upon this question while browsing the Cornell site. It is quite funny.\r\nSo do you like mathematical analysis?\r\nIf you don't and you are somewhat forced to study it, do you think that \"is there life after calculus?\"", "Solution_1": "Yes Analysis IS fun. And it's often intuitive.", "Solution_2": "in my opinion the correct statement would sound \"some analysis is fun, but not all of it\" :D", "Solution_3": "I agree with Valentin. But this is a general statement. I don't know a serious field from math that is all fun.\r\n\r\nI don't have a completely formed, life lasting opinion about analysis. Until now I didn't like it. But my future research projects include Analytical Number Theory, so analysis will have to be one of my tools.", "Solution_4": "Which part of Calculus do you hate most ??", "Solution_5": "the part about double intregrals, analitic functions, etc. :D", "Solution_6": "it is said that the most difficult part in analysis (not a part of mathematical analysis) is Real Variable Functions Analysis which was developed greatly by Lesbegue.", "Solution_7": "[quote=\"Valentin Vornicu\"]the part about double intregrals, analitic functions, etc. :D[/quote]\r\n\r\nDon't tell me you prefer Analytic Geometry, which is so ugly - to my opinion - that is shouldn't be allowed to be part of our beloved Maths :D", "Solution_8": "I also hate analytic geom, and I'm a big fan of synthetic geometry, but sometimes you really can't do without coordinates and calculations and ugly stuff like that :D \r\n\r\nAs for calculus, I like it, but I don't think my mind is built for it ( :D ). In my opinion, to do well in calculus you have to be good at manipulating lots of abstract structures, but I'm more of a visual thinker. Would you believe I make drawings for most of the calculus problems I solve ? This isn't entirely accurate, since my interests include integrals now and it's harder to make drawings which suit this sort of problems, but back in 11'th grade, when my main interest was derivatives, I drew lots of tangents to curves and things like that. Despite writing a 'standard' solution, using \\varepsilon-delta definitions and stuff, my first step would be to draw something, this kind of thinking might be harmful when I start studying more advanced things which can't be solved by looking at a drawing :D", "Solution_9": "i agree with df/dx\r\ni also hate that problems which u don't have to think... u have to make ONLY calculus... a lot of calculus", "Solution_10": "For df/dx or d 2 f/d 2 x:\r\nWhat's intuitive about this:\r\nfind a primitive for \\sqrt (2+cos(x))?", "Solution_11": "[quote=\"df/dx\"]\nDon't tell me you prefer Analytic Geometry, which is so ugly - to my opinion - that is shouldn't be allowed to be part of our beloved Maths :D[/quote]\r\nFirst of all I think that you ment Differential Geometry, cuz Analytic Geometry is in fact algebra. I don't like both of them. \r\n\r\nas amfulger pointed out, analysis can be a real pain in the ass :) it most certainly has its beauty, but I for one tend to see its uglyness :)", "Solution_12": "It might look intuitive at the beginning, but when you get to more advanced stuff I think it loses all its intuitiveness (is that a word ? :D ).", "Solution_13": "So everybody seems to hate Analysis :( ...\r\n\r\nNot me :D !", "Solution_14": "it's not that we hate it. we just don't like most of it. :) \r\nand now that you've mentioned it, who are you? :D", "Solution_15": "Yes. Analytic function seems to be appeared in Complex Variable number theory.\r\n\r\nI don't like tensor analysis....", "Solution_16": "i like very much analysis :D", "Solution_17": "I think you all have a misinterpretation of \"analysis\" and what it really is. There are many beautiful and unintuitive things in analysis. The invention of calculus is one of the great inventions of mankind!! It forever changed the world and opened up whole new worlds of mathematics.\r\n\r\nProblems in analysis can range from everything from foundational - the study of infinite sets and the construction of the real numbers (a truly beautiful subject). To the theory of measure and integration (which lays the foundation for proper understanding of probability). To the absolute gorgeousness of complex analysis (perhaps the most beautiful of all subjects). To fourier analysis (which can be used to solve a surprising range of problems from physics to number theory). To differential geometry and differential topology (deceptively subtle and powerful subjects).\r\n\r\nI don't think it is good to cordon off any areas of mathematics as \"good\" or \"bad\", because a great mathematician should be able to recognize connections between seemingly separate areas and integrate them (no pun intended).", "Solution_18": "[quote=\"gauss202\"]There are many beautiful and unintuitive things in analysis.[/quote]\r\n\r\nI agree, how intuitive is it that the p-adic rationals (and integers) are homeomorphic to the Cantor sets?", "Solution_19": "[quote=\"gauss202\"]I think you all have a misinterpretation of \"analysis\" [/quote]\r\nDon't think I do.\r\nFor me not liking anlysis just means I will not chose it as a research field. You can't do research in all mathematics. You chose a NT problem. You realize you need some analysis. But it is just a tool for you. You apply a statement that fits the problem conditions. You don't even have to know a proof for what you use. You even might not have known what you have used until you grabbed an analysis book and found ouy what you where looking for (sound like a U2 song).\r\nI saw a proof for Dirichlet's theorem (the one about the distribution of primes) that uses only real analysis. I was able of understanding it and I am not good in analysis (I don't solve very hard problemsin analysis). \r\n\r\nAnalysis was created for engineering purposes. They wanted to measure areas, so they invented the integrals.", "Solution_20": "[quote=\"amfulger\"]\nAnalysis was created for engineering purposes. They wanted to measure areas, so they invented the integrals.[/quote]\r\n\r\nNo this is not true. I mean, Newton was inspired by physical problems in his discovery of calculus but that is not all it is for! Look at Cantor and Riemann, would you consider them \"egineers\"?\r\n\r\nAnalysis is really the study of the infinite, to put it broadly. Anything that involves infinite sets could be said to fit under the realm of analysis. That covers a tremendously broad range of things. It is not enough to just know the results of theorems in analysis. The more deeply you understand them, the greater your insight into mathematics as a whole. Math is not separated into boxes of different fields. It is tied together.\r\n\r\nLook at Alaine Connes, the great french mathematician, for instance. He has proposed some very interesting lines of attack on the Riemann Hypothesis that are rooted in deep ideas from analysis. He could not have come up with these ideas without very good understanding and subtle insights into why things were true in analysis. And this is becoming increasingly true in Number Theory as a whole.\r\n\r\nI'm not saying that all of analysis interest me, I'm just saying that there is more to the field than just mundane integrations and obscure inequalities.", "Solution_21": "[quote=\"gauss202\"][quote=\"amfulger\"]\nAnalysis was created for engineering purposes. They wanted to measure areas, so they invented the integrals.[/quote]\nNo this is not true. I mean, Newton was inspired by physical problems in his discovery of calculus but that is not all it is for! Look at Cantor and Riemann, would you consider them \"egineers\"?\nI'm not saying that all of analysis interest me, I'm just saying that there is more to the field than just mundane integrations and obscure inequalities.[/quote]\r\n\r\nI only wanted to say analysis started as a computational tool.\r\nI know that there is more to this field than just computations. I verry much like the study of cardinal numbers.\r\nI just say that I can't be an analyst because most of the analysis does not suite my representative way of thinking. I know mathematics is a whole. I've heard that until now it is impossible to avoid calculus in any proof for the Fundamental Theorem of Algebra.", "Solution_22": "analysis can nearly be seen everywhere in mathematics.\r\nmany things referring to functions are met with analysis.\r\n\r\nproofs of the fundamental theorem of algebra can be roughly divided into three categories. first there are the topological proofs, which are vased on topological considerations such as the winding number of a curve in R^2 around 0. then, there are analytical proofs, which are related to Liouville's theorem: a nonconstant entire function on C is unbounded. finally, there are the algebraic proofs and these proofs use only the fact that every polynomial of odd degree with real coefficients has a real root and that every complex number has a square root. and recently, a proof based on linear algebra is published. the paper is 'The Fundamental Theorem of Algebra and Linear Algebra', MAA, Monthly 110.", "Solution_23": "But even all of those proofs rely on analysis on a subtle level. The Fundamental Theorem of Algebra is an analysis result. It relies essentially on the fact that the real numbers have no \"gaps\" in them, i.e. that they have the least upper bound property. This is, for example, what you need to prove that every odd degree polynomial has a real root. All proofs of the FTA rely on some result of that kind. A good discussion of the Fundamental Theorem of Algebra can be found in a book called [u]Numbers[/u] by Ebbinghaus, et. al. It is a very fun read.", "Solution_24": "In my opinion it doesn't exist life after calculus, and math is so great, it makes you 'think' :D ....and may sometimes be fun, or sometimes can be hard, but always will be interesting :)", "Solution_25": "Analysis is useful for differential equation wich is use in economy. \r\nanalysis is useful for probability modelisation of stock exchange (bank...)\r\nAlso in applied mathematics for planes, rockets, ... \r\nOne day I have seen a book of Debreu nobel prize of economy, in its book there was a lot topology, metric space, point fixed theorem.\r\nJohn Nashe used analysis \r\nYeah could have great life after calculus.", "Solution_26": "I think that analysis being developed as a tool fpr physics in the beginning has no importance, and I dare say that it would have been discovered and used in all those fields anyway. This is because people are just too complex to rely only on finite quantities. We were bound to tackle infinity-related problems sooner or later, and it just so happenes that the man who first used analysis was a physicist and discovered it for purposes related to physics.", "Solution_27": "It's not entirely true that calculus was \"invented\" by any one man. It really arose from a lot of contributions over a long period of time. Newton and Leibniz just unified a lot of these concepts and were the first to be able to solve a wide range of problems with them. Nor is it true that they used calculus only for physics problems. They were both concerned with Calculus from a pure mathematical standpoint.\r\n\r\nThere is a false perception of calculus from high school students that it's somehow the \"end\" of mathematics. It may be the end of high school mathematics, but it's just the beginning of what you might call modern mathematics. If you imagine all of the math topics that you study in high school - you really have only studied topics that were developed before the 18th century. In University you cover mainly topics developed during the 18th and 19th century, and maybe a few topics from the 20th century. Then, only when you get to graduate school, do you start to study really modern topics from the last century.\r\n\r\nSo you can think of calculus as your first taste of mathematics from the 18th century and beyond. Hardly an \"end\" to mathematics.", "Solution_28": "You know, I don't think \"invented\" is a good word. There's a difference between invented and discovered things. My opinion is that inventions are those things which probablt wouldn't have come to existence if it weren't for the inventors who first thought of them. \r\n\r\nOn the other hand, I think calculus is something we have discovered: someone would have found it sooner or later because, as I said in my previous post, it's only natural for beings as complex as we are to be concerned with the concept of infinity.\r\n\r\nEven the fact that both Newton and Leibniz found nearly the same results in roughly the same period of time shows us that calculus was bound to \"happen\".\r\n\r\nI don't think there is such a thing as \"the end of mathematics\", or \"the end of physics\" or anything like that. We' way too far from a comprehensive truth which explains what we see around us, and until we reach something like that, all sciences will/must continue to \"grow\" and develop. We may never reach such a unifying truth, but we can try getting closer and closer (asymptotically :D).", "Solution_29": "[quote=\"Anonymous\"]In my opinion it doesn't exist life after calculus, and math is so great, it makes you 'think' :D ....and may sometimes be fun, or sometimes can be hard, but always will be interesting :)[/quote]\r\n\r\nexactly my opinoin (kind of)... :D", "Solution_30": "df/dx, you forgot to answer a question asked by a very important person: our admin Valentin. He asked: WHO ARE YOU? Pls respond.", "Solution_31": "I just wanted to say I have nothing against calculus, but I have something against application of calculus to elementary theorems (especially when this application is unnecessary)...\r\n\r\n dg", "Solution_32": "[quote=\"darij grinberg\"]I just wanted to say I have nothing against calculus, but I have something against application of calculus to elementary theorems (especially when this application is unnecessary)...\n\n dg[/quote]\r\n\r\nBut why is that? In some cases, calculus may provide a more elegant proof of a theorem. For example, the Prime Number Theorem has a proof using only \"elementary\" mathematics, but it has a more elegant proof using Calculus.\r\n\r\nPlus, many facts that are taken for granted in elementary mathematics are really built on a hidden calculus foundation. For example, the definition of e and natural logarithms, and the area of a circle.\r\n\r\nI don't think that such a distinction should be made between calculus and the rest of mathematics; no more than a distinction should be made between geometry and algebra. The only reason it is done for competition purposes is to make ensure that results aren't needed that some high school students wouldn't have been exposed to yet. But there is nothing impure, or inelegant about calculus.", "Solution_33": "[quote=\"gauss202\"]But why is that? In some cases, calculus may provide a more elegant proof of a theorem. For example, the Prime Number Theorem has a proof using only \"elementary\" mathematics, but it has a more elegant proof using Calculus.[/quote]\n\nWell, I don't know what you mean by \"prime number theorem\" (if it is about density, then its formulation already uses calculus, since it depends on the notion of a limit). Yes, calculus often simplifies a proof or shows the \"real\" reason behind a fact. But calculus is different from elementary mathematics, it uses more axioms and often approaches things with very different arguments. So why don't also search for an elementary, non-calculus proof? Any alternative proof of a theorem already proven can be useful for the development of mathematics. Especially, there are lots of cases when calculus provides a straightforward but ugly proof of something (think about the use of differentiation to kill inequalities!). \n\n[quote=\"gauss202\"]Plus, many facts that are taken for granted in elementary mathematics are really built on a hidden calculus foundation. For example, the definition of e and natural logarithms, and the area of a circle.[/quote]\r\n\r\nAs for pi and e, I don't call them elementary mathematics. In my opinion, for instance, elementary geometry is the part of geometry where [b]R[/b] can be replaced by the set of all algebraic numbers. I know well that there are parts of calculus \"hidden\" in the high school curriculum, but the curriculum is not my criterion of distinction between elementary and advanced mathematics.\r\n\r\n Darij", "Solution_34": "Maybe but idk and irdc calc\n\nXdddddddddd", "Solution_35": "#bumpoldthreads" } { "Tag": [ "Vieta", "algebra unsolved", "algebra" ], "Problem": "if a, b, c are the roots of $3x^{3}-7x^{2}+5x-32=0$, find the value of $a^{2}+b^{2}+c^{2}$.", "Solution_1": "please u have to know where to pose each exercices.be carful\r\n$a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(ab+bc+ca)$ .then use Vieta's formula.", "Solution_2": "$a^{2}+b^{2}+c^{2}=\\frac{19}{9}$" } { "Tag": [ "calculus", "integration", "modular arithmetic", "symmetry" ], "Problem": "Suppose $ xz\\minus{}2yt\\equal{}3$ and $ xt\\plus{}yz\\equal{}1$, find the integral solutions to $ (x,y,t,z)$.", "Solution_1": "first, note that $ xz \\equal{} 3\\plus{}2yt$ so $ xz \\equiv 1 \\pmod{2}$, which tells us that both $ x$ and $ y$ are odd. From the second equation, we can also see that $ gcd(x,y) \\equal{} gcd(x,z) \\equal{} gcd(y,t) \\equal{} gcd(z,t) \\equal{} 1$.\r\n\r\nstuck.", "Solution_2": "Big Hint:\r\n[hide]$ (xz \\minus{} 2yt)^2 \\plus{} 2(xt \\plus{} yz)^2 \\equal{} (x^2 \\plus{} 2y^2)(z^2 \\plus{} 2t^2)$.[/hide]", "Solution_3": "ok so $ 11 \\equal{} (x^2\\plus{}2y^2)(z^2\\plus{}2t^2)$, and we can only factor $ 11$ as $ 1$ and $ 11$, so let, WLOG (b/c of symmetry)\r\n$ x^2\\plus{}2y^2 \\equal{} 1$ which implies $ x\\equal{}1$ and $ y\\equal{}0$\r\n\r\nwhich in our original equations gives $ xt\\plus{}yz \\equal{} 1 \\implies t\\equal{}1$ and $ xz\\minus{}2yt\\equal{}3 \\implies z\\equal{}3$\r\n\r\nSo our solutions are $ (1,0,1,3)$ and $ (3,1,0,1)$" } { "Tag": [ "geometry", "circumcircle", "geometry proposed" ], "Problem": "The vertices of triangle ABC are lattice points and there is no smaller triangle similar to triangle BAC with its vertices at lattice points. Prove that the circumcenter of ABC is not a lattice point.", "Solution_1": "http://www.artofproblemsolving.com/community/c6h597717p3547050" } { "Tag": [ "algorithm" ], "Problem": "What is the maximum number of consecutive odd positive integers \r\nthat can be added together before the sum exceeds 401?", "Solution_1": "$ \\sum_{k\\equal{}1}^n(2k\\minus{}1)\\equal{}n^2$\r\nAnd $ 400\\equal{}20^2$.\r\nSo, $ 400\\equal{}\\sum_{k\\equal{}1}^{20}(2k\\minus{}1)\\equal{}1\\plus{}3\\plus{}\\cdots\\plus{}37\\plus{}39$.\r\nThus $ \\boxed{39}$.", "Solution_2": "Isn't the answer 20?", "Solution_3": "[quote=\"mentalgenius\"]Isn't the answer 20?[/quote]\n\nYep, an ancient case of \"answer the question that's asked\".\nNote that you could also replace the word \"consecutive\" with \"distinct\", and you'd still get the same answer by using an anti-greedy algorithm starting with $1$ and packing in the smallest positive odd integers that you can.", "Solution_4": "[hide=\"Solution\"]\nQ: What is the maximum number of consecutive odd positive integers that can be added together before the sum exceeds 401?\nA: The formula for the sum of the first n perfect squares is $n^2$. $20^2$ is 400, so the sum of the first 20 odd numbers is 400. For the first 21 odd numbers, the sum is 441, which is greater than 401. Therefore, the maximum number of consecutive odd positive numbers is $20$.\n[/hide]" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Prove that there are infinitely many twin prime pairs iff there are infinitely many positive integers $ n$ that cannot be written in the form $ n \\equal{} 6|ab|\\plus{}a\\plus{}b$ for any choice of integers $ a$ and $ b$.", "Solution_1": "See http://www.mathlinks.ro/Forum/viewtopic.php?t=150559 ." } { "Tag": [ "abstract algebra", "algebra", "polynomial", "analytic geometry", "Ring Theory", "superior algebra", "superior algebra unsolved" ], "Problem": "Let A = k[x,y,z] / (xy - z^2) (k is a field).\r\n\r\nLet X, Y, Z denote the images of x, y, z in A. Show that p = (X, Z) is a prime ideal of A.\r\n\r\n\r\nNow, I want to show that A / p is an ID. It looks like A / p is isomorphic to k[y].\r\n\r\n...not 100% sure how to show that. \r\n\r\nCan we consider a homomorphism from A to k[y] by sending X to 0 and Z to 0?? it's kernel is p = (X, Z) and it's onto...", "Solution_1": "your approach is correct. but to avoid a nasty calculation with elements to show that the kernel equals p, use the universal property of polynomial algebras and quotient algebras:\r\n\r\nfor every $ k$-algebra $ B$, we have natural bijections\r\n \r\n$ Hom(A/p , B) \\equal{} \\{f \\in Hom(A,B) : f(X) \\equal{} f(Z) \\equal{} 0\\}$\r\n$ \\equal{} \\{f \\in Hom(k[x,y,z],B) : f(xy \\minus{} z^2) \\equal{} f(x) \\equal{} f(z) \\equal{} 0\\}$\r\n$ \\equal{} \\{f \\in Hom(k[x,y,z],B) : f(x) \\equal{} f(z) \\equal{} 0\\}$\r\n$ \\equal{} \\{(a,b,c) \\in B^3 : a \\equal{} c \\equal{} 0\\} \\equal{} B \\equal{} Hom(k[y],B)$,\r\n \r\nso the yoneda-lemma implies $ A/p \\cong k[y]$. if you don't want to use the yoneda-lemma, you have to repeat its proof here (which is done very often, I wonder why ..): verify with the universal properties, that $ A \\to k[y], X \\to 0, Z \\to 0, Y \\mapsto y$ yields a well-defined $ k$-algebra homomorphism. the kernel contains $ p$, thus we get $ A/p \\to k[y]$. now define $ k[y] \\to A/p$ by $ y \\mapsto Y$. now it's clear that $ k[y] \\to A/p \\to k[y]$ and $ A/p \\to k[y] \\to A/p$ are identity homomorphism for this can be checked on the variables $ y$ resp. $ X,Y,Z$.\r\n\r\nas you can see, the proof with the yoneda-lemma uses exactly the same arguments, but it's more easy. besides it's very geometric: you intersect the variety $ xy \\minus{} z^2 \\equal{} 0$ with the variety $ x \\equal{} z \\equal{} 0$ and then you just get the affine line with coordinate $ y$. with this approach you can determine quotients of finitely generated $ k$-algebras without really thinking about it. :)", "Solution_2": "hm, that's slick.....didn't actually know about yoneda lemma. Thanks!", "Solution_3": "One could also apply the fact that quotients \"commute.\" Let $R$ be ring and $I_0,I_1,\\dots\\subseteq R$ be (a perhaps infinite set of) ideals. Take $\\mathbb N$ to be the natural number object, i.e. the category generated by $$\\textbf{0}\\overset{f_0}{\\longrightarrow} \\textbf{1}\\overset{f_1}{\\longrightarrow} \\textbf{2}\\overset{f_2}{\\longrightarrow}\\cdots$$ Then the diagram $\\mathcal Q:\\mathbb N\\to \\text{RING}$ with $$\\mathcal Q(\\textbf{0})=R$$ $$\\mathcal Q(\\textbf{n+1})=\\mathcal Q(\\textbf{n})/\\mathcal Q(f_{n-1})\\circ \\cdots \\circ \\mathcal Q(f_{0})``(I_n)$$ $$\\text{And } \\mathcal{Q}(f_n):\\mathcal Q(\\textbf{n})\\to \\mathcal Q(\\textbf{n+1})\\text{ the induced quotient morphism}$$ Has colimit $R/(I_1+I_2+\\cdots)$. In the original problem, take $R=k[x,y,z]$, $I_0=(xy-z^2)$, and $I_1=I_2=\\cdots=(x,z)$. Functor $\\mathcal Q$ has image $$k[x,y,z]\\overset{\\eta_0}{\\longrightarrow} k[x,y,z]/(xy-z^2)\\overset{\\eta_1}{\\longrightarrow} k[x,y,z]/(xy-z^2)/\\eta_0``((x,z))\\overset{\\text{id}}{\\longrightarrow}k[x,y,z]/(xy-z^2)/\\eta_0``((x,z))\\overset{\\text{id}}{\\longrightarrow}\\cdots$$ Which is equivalent to the diagram $$k[x,y,z]\\overset{\\eta_0}{\\longrightarrow} k[x,y,z]/(xy-z^2)\\overset{\\eta_1}{\\longrightarrow} k[x,y,z]/(xy-z^2)/\\eta_0``((x,z))$$ And thus has colimit $A/p=k[x,y,z]/(xy-z^2)/\\eta_0``((x,z))$ for purely categorical reasons. Appealing, however, to the above result, $\\text{colim}(\\mathcal Q)$ may also be written as $k[x,y,z]/(xy-z^2,x,z)= k[x,y,z]/(x,z)\\cong k[y]$, and it follows that $$A/p\\cong \\text{colim}(\\mathcal Q)\\cong k[y]$$ At which point we may conclude that $p\\subseteq A$ is prime, as noted in the OP $\\blacksquare$" } { "Tag": [ "conics", "ellipse" ], "Problem": "An ellipse is inscribed in triangle $ABC$. From a focus $F$, lines are drawn to vertices $A$, $B$, $C$ and to tangent point $D$ on side $AB$. Prove that $\\angle AFD + \\angle BFC = 180^{\\circ}$\r\n\r\na hint woud be nice, or a proof.", "Solution_1": "[hide]By Poncelet we have:\n$\\angle AFD = \\angle AFG = \\alpha$\n$\\angle CFD = \\angle CFE = \\beta$\n$\\angle BFE = \\angle BFG = \\gamma$\n\nThen...\n\n$2\\alpha + 2\\beta + 2\\gamma = 360^{\\circ}$\n$\\alpha + \\beta + \\gamma = 180^{\\circ}$\n\nBut...\n\n$\\angle BFC = \\angle BFE + \\angle CFE = \\gamma + \\beta$\nand\n$\\angle AFD = \\alpha$\n\nHence...\n\n$\\angle AFD + \\angle BFC = \\alpha + \\beta + \\gamma = 180^{\\circ}$[/hide]", "Solution_2": "What is Poncelet?", "Solution_3": "About Poncelet: http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Poncelet.html" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let $ a,b,c\\geq0$ s.t. $ abc\\equal{}1$.Prove that :\r\n$ (\\dfrac{a\\plus{}b\\plus{}c}{3})^5\\geq\\dfrac{a^2\\plus{}b^2\\plus{}c^2}{3}$", "Solution_1": "@@ .So old \r\n$ 3abc(a^2\\plus{}b^2\\plus{}c^2)\\equal{}\\frac{3abc(a\\plus{}b\\plus{}c)(a^2\\plus{}b^2\\plus{}c^2)}{a\\plus{}b\\plus{}c} \\le \\frac{(ab\\plus{}bc\\plus{}ca)^2(a^2\\plus{}b^2\\plus{}c^2)}{a\\plus{}b\\plus{}c} \\le (a\\plus{}b\\plus{}c)^5/27$\r\nDone ;)" } { "Tag": [ "logarithms", "number theory", "prime numbers" ], "Problem": "If the sum of all solutions to the equation $ (x^{log 3})^{2}\\minus{}3^{log x}\\minus{}2\\equal{}0$ is $ a^{log_{b}c}$, find $ a\\plus{}b\\plus{}c$, where $ b$ and $ c$ are prime numbers and $ a$, $ b$, and $ c$ are all positive integers.", "Solution_1": "Since $ \\log{(x^{\\log 3})}\\equal{}\\log 3\\log x\\equal{}\\log{(3^{\\log x})}$, we see that $ x^{\\log 3}\\equal{}3^{\\log x}.$ So you're equation becomes $ (3^{\\log x})^2\\minus{}3^{\\log x}\\minus{}2\\equal{}0$. Solve this to obtain $ 3^{\\log x}\\equal{}2$ or $ \\minus{}1$. $ \\minus{}1$ is extraneous since $ 3^{\\log x}$ is positive, so it must be $ 3^{\\log x}\\equal{}2$. Solve for $ x$ to get $ x\\equal{}10^{\\log_32}$. This is the same as $ 2^{\\log_3{10}}$, but the problem says $ b$ and $ c$ are prime, and either way, the answer to you're problem is $ 15$." } { "Tag": [ "geometry", "3D geometry", "algebra", "polynomial", "arithmetic sequence", "More Sequences" ], "Problem": "Show that if an infinite arithmetic progression of positive integers contains a square and a cube, it must contain a sixth power.", "Solution_1": "[quote=\"Peter\"]Show that if an infinite arithmetic progression of positive integers contains a square and a cube, it must contain a sixth power.[/quote]\r\nhttp://www.kalva.demon.co.uk/short/soln/sh9715.html", "Solution_2": "Let $ r,s$ be coprime. If an arithmetic progression contains a $ r$-th power and a $ s$-th power, then it contains a $ rs$-th power.\r\n\r\n\r\nProof:\r\nThis is equivalent to: Let $ a,d$ be integers. If there are integers $ x,y$ such that $ a \\equiv x^r \\equiv y^s \\mod d$, then there is an integer $ z$ such that $ a \\equiv z^{rs} \\mod d$.\r\n\r\nFirst observation: we can assume that $ d\\equal{}p^q$ is a prime power, $ p$ prime.\r\nIndeed, by the chinese remainder theorem, an integer polynomial equation can be solved $ \\mod n$ iff it can be solved $ \\mod$ all prime powers dividing $ n$. Here, the equations are $ x^r\\minus{}a , y^s\\minus{}a, z^{rs}\\minus{}a$.\r\n\r\nSecond observation: let $ d\\equal{}p^q$ as said, then we can assume $ \\gcd(a,p^q)\\equal{}1$.\r\nIndeed, if $ p|a$, we write $ x\\equal{} p^u x', y\\equal{} p^v y'$ with $ x', y'$ both not divisible by $ p$. Then $ x^r \\equiv y^s \\mod p^q$ gives $ a \\equiv p^{ur} x'^r \\equiv p^{vs} y'^s \\mod p^q$.\r\nThus either $ a \\equiv 0 \\mod p^q$, in this case we can take $ z\\equal{}0$, or we get that $ vs\\equal{}v_p(x^r) \\equal{} v_p(a) \\equal{} v_p(y^s) \\equal{} ur 0 [(1+h)^4-1]/h How to find the answer? my teacher said since it looked like the derivative form lim f(a+h)-f(a)/h, but then I didn't know how to do it.\r\nThanks for help.", "Solution_1": "2)\r\n\r\nYou mean, $ \\lim_{h\\to 0}\\frac{(1+h)^{4}-1}{h}$?\r\n\r\nIf so, let $ f(x) = x^{4}$. By definition we have that\r\n\\begin{eqnarray*}f^{\\prime}(a) & = &\\lim_{h\\to 0}\\frac{f(a+h)-f(a)}{h}\\\\ & = &\\lim_{h\\to 0}\\frac{(a+h)^{4}-a}{h}.\\end{eqnarray*} \r\n\r\nNow, let $ a=1$ and we are done, aren't we?", "Solution_2": "0 slopes and positive/negative slopes are a good start. Let's call the original function whose graph is given $ f(x)$. First plot the points where the derivative is 0 (that is, where $ f^{\\prime}(x)\\equal{}0$). Draw a few tangent lines on the graph of $ f(x)$ (eyeball them) and guess the values of the slopes at these points (you know what slopes of $ \\pm 1,\\pm 2$ look like, so compare and guess,) then plot these points. Also you can tell if the derivative, $ f^{\\prime}(x)$, is increasing or decreasing by checking the concavity of $ f(x)$ (by looking at the graph): \r\n\r\n$ f(x)$ is concave up => $ f^{\\prime}(x)$ is increasing , $ f(x)$ is concave down => $ f^{\\prime}(x)$ is decreasing \r\n\r\nOn that note, the inflection points of $ f(x)$ (a point where the concavity of $ f(x)$ changes; such points occur where $ f^{\\prime\\prime}(x)\\equal{}0$) are the critical points (critical point = a maximum, minimum, or saddle point) of $ f^{\\prime}(x)$, in particular:\r\n\r\nIf $ x_{0}$ is an inflection point of $ f(x)$, and\r\n\r\n[b]i.[/b] $ f(x)$ changes from concave up to concave down at $ x_{0}$, then $ f^{\\prime}(x)$ has a max at $ x_{0}$\r\n\r\n[b]ii.[/b] $ f(x)$ changes from concave down to concave up at $ x_{0}$, then $ f^{\\prime}(x)$ has a min at $ x_{0}$\r\n\r\n[b]iii.[/b] $ f(x)$ does not change concavity at $ x_{0}$ (even though $ f^{\\prime\\prime}(x_{0})\\equal{}0$), then $ f^{\\prime}(x)$ has a saddle point at $ x_{0}$\r\n\r\nNote: An example of case [b]iii[/b] is $ f(x)\\equal{}x^{4}$ with $ x_{0}\\equal{}0$.\r\n\r\nThis will only give you the locations of the extrema/saddle points of $ f^{\\prime}(x)$, for the values of $ f^{\\prime}(x)$ at these points you will need to draw the tangent lines and guess the slopes (as above)." } { "Tag": [ "probability", "analytic geometry", "geometry", "trapezoid", "inequalities", "graphing lines", "slope" ], "Problem": "(a) Two mathematicians take a lunch break seperately sometime between 12:00 and 1:00 and stay for $\\frac{m}{n}$ minutes. If the probability of them meeting each other is $.4$, compute $m+n$.\r\n\r\n(b) Three mathematicians take a lunch break seperately sometime between 12:00 and 1:00 and stay for $\\frac{m}{n}$ minutes. If the probability that two of them meet but the third doesn't meet with either of them is $.4$, compute $m+n$.", "Solution_1": "[hide=\"a\"]\nSay the first person enters at time $x$ and the second at time $y$, where $0 \\leq x,y \\leq 60$ is the amount of time elapsed since 12:00. Then, in order to have the mathematicians meeting each other, we need that\n\\[|x-y| \\leq \\frac{m}{n}=t \\]\nGraphing this on a coordinate axes, we see that it divides the square of side $60$ (i.e. the time frame of the problem) into two congruent isosceles triangles of side length $60-t$ and a trapezoid. We want that the area of the trapezoid to be $\\frac{2}{5}\\cdot 3600 = 1440$. This means that:\n\\[3600-2 \\cdot (\\frac{1}{2}\\cdot (60-t)^{2}) =1440 \\implies \\boxed{ t= 60-12 \\sqrt{5}}\\]\nBut that doesn't fit with the answer format... Did I interpret the problem wrong? Must each mathematician stay for the whole $t$ minutes [u]within[/u] the time frame? [/hide]\n\n[hide=\"b\"] Similar method. Take the inequality for the first two, then subtract the intersection of the inequalities with each of the first two and the third. Can we do Shoelace in 3-D ? :D [/hide]", "Solution_2": "[quote=\"cincodemayo5590\"][hide=\"a\"]\nSay the first person enters at time $x$ and the second at time $y$, where $0 \\leq x,y \\leq 60$ is the amount of time elapsed since 12:00. Then, in order to have the mathematicians meeting each other, we need that\n\\[|x-y| \\leq \\frac{m}{n}=t \\]\nGraphing this on a coordinate axes, we see that it divides the square of side $60$ (i.e. the time frame of the problem) into two congruent isosceles triangles of side length $60-t$ and a trapezoid. We want that the area of the trapezoid to be $\\frac{2}{5}\\cdot 3600 = 1440$. This means that:\n\\[3600-2 \\cdot (\\frac{1}{2}\\cdot (60-t)^{2}) =1440 \\implies \\boxed{ t= 60-12 \\sqrt{5}}\\]\nBut that doesn't fit with the answer format... Did I interpret the problem wrong? Must each mathematician stay for the whole $t$ minutes [u]within[/u] the time frame? [/hide]\n\n[hide=\"b\"] Similar method. Take the inequality for the first two, then subtract the intersection of the inequalities with each of the first two and the third. Can we do Shoelace in 3-D ? :D [/hide][/quote]\n\nI got something else for (a):\n\n[hide=\"Possibly wrong solution\"]\nP(t) where t is time spent and P is probability of meeting is LINEAR, is it not? Whic means that you can pick 2 times such as 10 and 20 to get the slope and then you can solve for .4. This was my original thought, but now I think I'm wrong. Is P(t) linear?\n[/hide]", "Solution_3": "[quote=\"miyomiyo\"][quote=\"cincodemayo5590\"][hide=\"a\"]\nSay the first person enters at time $x$ and the second at time $y$, where $0 \\leq x,y \\leq 60$ is the amount of time elapsed since 12:00. Then, in order to have the mathematicians meeting each other, we need that\n\\[|x-y| \\leq \\frac{m}{n}=t \\]\nGraphing this on a coordinate axes, we see that it divides the square of side $60$ (i.e. the time frame of the problem) into two congruent isosceles triangles of side length $60-t$ and a trapezoid. We want that the area of the trapezoid to be $\\frac{2}{5}\\cdot 3600 = 1440$. This means that:\n\\[3600-2 \\cdot (\\frac{1}{2}\\cdot (60-t)^{2}) =1440 \\implies \\boxed{ t= 60-12 \\sqrt{5}}\\]\nBut that doesn't fit with the answer format... Did I interpret the problem wrong? Must each mathematician stay for the whole $t$ minutes [u]within[/u] the time frame? [/hide]\n\n[hide=\"b\"] Similar method. Take the inequality for the first two, then subtract the intersection of the inequalities with each of the first two and the third. Can we do Shoelace in 3-D ? :D [/hide][/quote]\n\nI got something else for (a):\n\n[hide=\"Possibly wrong solution\"]\nP(t) where t is time spent and P is probability of meeting is LINEAR, is it not? Whic means that you can pick 2 times such as 10 and 20 to get the slope and then you can solve for .4. This was my original thought, but now I think I'm wrong. Is P(t) linear?\n[/hide][/quote]\r\n\r\nI think it's quadratic, since it's based on the area of the probability.", "Solution_4": "Hmmm\r\n\r\nYeah, you're right." } { "Tag": [ "calculus", "integration", "trigonometry", "calculus computations" ], "Problem": "Calculate \r\n\r\n\\[\\int^{\\frac{\\pi}{2}}_{0}2x\\sin ^2 x dx\\]", "Solution_1": "Use $2\\sin^2x=1-\\cos (2x)$ and integration by parts" } { "Tag": [ "geometry", "ratio", "search", "MATHCOUNTS", "analytic geometry" ], "Problem": "I need some help on this first on, but I want to check if I got the second one. :yup: \r\n\r\n23. A right triangle has side lengths 3,4 and 5 units. Another \r\ntriangle has sides of length 4, x and y units and has the same\r\narea as the right triangle. If x0$ prove:\r\n\r\n$ \\sqrt[3]{\\frac{a}{b\\plus{}c}}\\plus{}\\sqrt[3]{\\frac{b}{a\\plus{}c}}\\plus{}\\sqrt[3]{\\frac{c}{a\\plus{}b}}>2$", "Solution_1": "[quote=\"zaya_yc\"]$ a,b,c > 0$ prove:\n\n$ \\sqrt [3]{\\frac {a}{b \\plus{} c}} \\plus{} \\sqrt [3]{\\frac {b}{a \\plus{} c}} \\plus{} \\sqrt [3]{\\frac {c}{a \\plus{} b}} > 2$[/quote]\r\n\r\nsee here\r\nhttp://www.mathlinks.ro/viewtopic.php?t=301485", "Solution_2": "[b]Inequalities Marathon: Problem 428[/b] : For nonnegative numbers a, b and c prove\n$ \\sqrt\\frac{a}{b+c}+\\sqrt\\frac{b}{c+a}+\\sqrt\\frac{c}{a+b} \\ge 2 $\n[b]Remark on problem 428[/b]With same conditions$\\sum_{cyc}\\sqrt[n]{\\frac{a}{b+c}} \\ge 2 $ $(n \\ge 2 )$\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=299899&start=1140\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=397993&p=3476910#p3476910", "Solution_3": "[url=http://www.artofproblemsolving.com/community/c6h299899p2129359]Remark on problem 428[/url]", "Solution_4": "Let $a,b,c$ be non negative real number .Prove that$$ \\sqrt[3]{\\frac{a}{b+c}}+\\sqrt{\\frac{b}{c+a}}+\\frac{c}{a+b}>2.$$\n\n[hide=h]Let $a,b,c$ be non negative real number and $c(a+b)>0$ .[url=https://artofproblemsolving.com/community/c4h2041239p14439858]Find minimum of [/url] $ \\sqrt{\\frac{a}{b+c}}+\\sqrt{\\frac{b}{c+a}}+\\frac{c}{a+b}$[/hide]", "Solution_5": "[quote=sqing]Let $a,b,c$ be non negative real number .Prove that$$ \\sqrt[3]{\\frac{a}{b+c}}+\\sqrt{\\frac{b}{c+a}}+\\frac{c}{a+b}>2.$$\n\n[/quote]\nIt's wrong. Try $a=0$, $b=\\sqrt[3]4$ and $c=1$.", "Solution_6": "[hide=Thanks.][quote=arqady][quote=sqing]Let $a,b,c$ be non negative real number .Prove that$$ \\sqrt[3]{\\frac{a}{b+c}}+\\sqrt{\\frac{b}{c+a}}+\\frac{c}{a+b}>2.$$\n\n[/quote]\nIt's wrong. Try $a=0$, $b=\\sqrt[3]4$ and $c=1$.[/quote]\n[/hide]\n", "Solution_7": "[quote=sqing]Let $a,b,c$ be non negative real number .Prove that$$ \\sqrt[3]{\\frac{a}{b+c}}+\\sqrt{\\frac{b}{c+a}}+\\frac{c}{a+b}>2.$$\n\n[hide=h]Let $a,b,c$ be non negative real number and $c(a+b)>0$ .[url=https://artofproblemsolving.com/community/c4h2041239p14439858]Find minimum of [/url] $ \\sqrt{\\frac{a}{b+c}}+\\sqrt{\\frac{b}{c+a}}+\\frac{c}{a+b}$[/hide][/quote]\n\nMay be \n\nLet $a,b,c$ be non-negative real numbers. Prove or Disprove:\n$$\\sqrt[3]{\\frac{a}{b+c}}+\\sqrt{\\frac{b}{c+a}}+\\frac{c}{a+b}\\ge\\frac{4\\sqrt[4]{3}}{3}$$\n\n[hide=Note]Equality occurs when $a=\\frac{c\\sqrt[4]{3}}{3},b=0$[/hide]", "Solution_8": "[hide=Thanks.][quote=thuanz123][quote=sqing]Let $a,b,c$ be non negative real number .Prove that$$ \\sqrt[3]{\\frac{a}{b+c}}+\\sqrt{\\frac{b}{c+a}}+\\frac{c}{a+b}>2.$$\n\n[hide=h]Let $a,b,c$ be non negative real number and $c(a+b)>0$ .[url=https://artofproblemsolving.com/community/c4h2041239p14439858]Find minimum of [/url] $ \\sqrt{\\frac{a}{b+c}}+\\sqrt{\\frac{b}{c+a}}+\\frac{c}{a+b}$[/hide][/quote]\n\nMay be \n\nLet $a,b,c$ be non-negative real numbers. Prove or Disprove:\n$$\\sqrt[3]{\\frac{a}{b+c}}+\\sqrt{\\frac{b}{c+a}}+\\frac{c}{a+b}\\ge\\frac{4\\sqrt[4]{3}}{3}$$\n\n[hide=Note]Equality occurs when $a=\\frac{c\\sqrt[4]{3}}{3},b=0$[/hide][/quote][/hide]\n$$\\frac{4\\sqrt[4]{3}}{3}=1.7547...$$\n" } { "Tag": [ "function", "logarithms", "trigonometry", "geometric series", "real analysis", "real analysis unsolved" ], "Problem": "Consider the function $ f(x)=\\sum_{n\\geq 1}{\\frac{1}{n}sin(\\frac{x}{4^n})}$. Prove that there is $c$ not depending on $x$ such that $ |f(x)|< c ln ln x$ for all $x>e$.", "Solution_1": "It seems fairly easy to me. Split the sum into two parts: $n \\le \\log_4 x$ and $n > \\log_4 x$. The first part we can bound by a finite harmonic series, giving a bound of $O(\\log\\log x)$. The second part we can bound by an infinite geometric series, using $|\\sin x| \\le |x|$, giving a bound of $O(1)$.", "Solution_2": "It is, but this was the first question. The second one was to prove that there is a sequence of real numbers $x_n$ which goes to infinity such that $inf_{n>0}{\\frac{ |f(x_n)|}{lnln x_n}>0}$.", "Solution_3": "Okay, that's a little more interesting. How about $x_k = 4^k \\cdot 120^\\circ$? For the first $k$ terms of the series, the sine term will be $\\sin 120^\\circ = \\sqrt{3}/2$, the same positive constant. So again we can approximate the first part of the sum by a finite harmonic series. The remaining part of the series we can again bound by $O(1)$." } { "Tag": [ "analytic geometry" ], "Problem": "How can one locate any known stars in any constellation[b]?[/b] How can one identify[b] any stars in the heavens? Is there any way of counting[/b] them? If so,how? Is there any truth that in counting stars,one must count[b] each[/b] stars[b] with an INTERVAL[/b] of[b] second[/b] each[b]?[/b]", "Solution_1": "There are several ways to locate and identify or to 'categorise' stars. \r\nThe oldest and simplest method is obviously actual names (like Vega or Rigel) and using memory for location. Johann Bayer made the first systematic catalogue now known as [url=http://en.wikipedia.org/wiki/Bayer_designation]Bayer's designation[/url], in which visible stars of a constellation are named by Greek letters in order of brightness (like $ \\alpha$ Lyrae or $ \\beta$ Orionis). Another commonly used catalogue is the [url=http://en.wikipedia.org/wiki/Flamsteed_designation]Flamsteed's catalogue[/url] in which stars are numbered starting from the lowest RA (like 3 Lyrae or 19 Orionis).The most faint stars are catalogued in advanced surveys like Henry Draper Catalogue (HD/HDE) made with large telescopes (like HD 172167 or HD 34085)\r\n\r\nAs to locating stars, Astronomers use the most common-sense idea: [url=http://en.wikipedia.org/wiki/Celestial_coordinate_system]Coordinate system[/url], the most widely used among them being the [url=http://en.wikipedia.org/wiki/Equatorial_coordinate_system]\"Celestial Coordinates\"[/url]: or [b] RA-Dec[/b] system.\r\n\r\nI don't rightly know what you mean by \"counting with interval of a second\" but if we have detailed stellar catalogues, there is hardly any need to count stars seperately. For the record, depending on your eye-sight and location on earth, about 6000 stars in the sky are visible to the naked eye and about 1000 of them have a Bayer designation.", "Solution_2": "Nice post. :)\r\n\r\nA current project that is attempting to image the night sky is SLOAN Digital Sky Survey. Some of the mappings are used in planetarium software, uncluding GoogleSky, the GoogleEarth addition.", "Solution_3": "Try this... :wink: \r\nhttp://www.rssd.esa.int/SA-general/Projects/Hipparcos/apps/SelectArea.html" } { "Tag": [ "geometry", "3D geometry", "quadratics", "algebra", "quadratic formula", "complex numbers", "AMC" ], "Problem": "Source: Columbus State University Math Tournament, 2001 Ciphering Round\r\nDifficulty: Very Easy (Medium AMC10/12)\r\n\r\ni) Find the sum of the solutions, both real and imaginary (potentially complex), of x:^3: + 8 = 0. \r\n\r\nii) On the interval 0 :le: x :le: 5:pi:, how many solutions does the equation tan(x) = -7 have?\r\n\r\niii) Find the exact value of (2x + 3y)/(5x + 7y) if y/x = 3.\r\n\r\niv) In how many distinct ways can 16 be written as the sum of four positive odd integers?\r\n\r\nAgain, show work for full credit.", "Solution_1": "i, ii, and iv are indeed all very easy. The easiest way to do iii is probably to multiply the fraction by (1/x)/(1/x), so that all you have is constants or (y/x)'s.", "Solution_2": "Since no one wants to try my challenges, I think I will. \n\n\n\nQuote:i) Find the sum of the solutions, both real and imaginary (potentially complex), of x:^3: + 8 = 0. \n\n[hide]\nx:^3:+8 is the sum of two cubes, and therefore factors to (x+2)(x^2+2x+4). So one solution is x=-2. To find the other two solutions, we will use the quadratic formula:\nx = [-2 +- sqrt(4-4*1*4)]/2 = [-2 +- sqrt(-12)]/2 = -2 +- sqrt(-3) = -2 +- i:rt3:. So the other solutions are -2 + i:rt3; and -2 - i:rt3:. The sum of these three is -6.[/hide]\n\nQuote:ii) On the interval 0 :le: x :le: 5:pi:, how many solutions does the equation tan(x) = -7 have? \n\n[hide]tan(x) will be equal to -7 in the intervals (-pi/2, 0), (pi/2, pi), (3pi/2, 2pi), etc. The intervals from ..., pi) to ..., 5pi) are within (0, 5pi). Thus there are five x-values.[/hide]\n\nQuote:iii) Find the exact value of (2x + 3y)/(5x + 7y) if y/x = 3.\n\n[hide]y/x=3\ny=3x\n\n(2x + 3y)/(5x + 7y) = (2x + 3(3x))/(5x + 7(3x)) = 11x/26x = 11/26, which is a fairly easy method, too.[/hide]\n\nQuote:iv) In how many distinct ways can 16 be written as the sum of four positive odd integers? \n\n\n\n[hide]List (cannot use any number bigger than first number)\n\n13 1 1 1\n\n11 1 1 3\n\n9 1 1 5 or 9 1 3 3\n\n7 1 1 7 or 7 1 3 5 or 7 3 3 3 \n\n5 5 3 3 \n\n\n\n8 ways.\n\n[/hide]", "Solution_3": "mathfanatic wrote:Since no one wants to try my challenges, I think I will. \n\nQuote:i) Find the sum of the solutions, both real and imaginary (potentially complex), of x:^3: + 8 = 0. \n\n[hide]\nx:^3:+8 is the sum of two cubes, and therefore factors to (x+2)(x^2+2x+4). So one solution is x=-2. To find the other two solutions, we will use the quadratic formula:\nx = [-2 +- sqrt(4-4*1*4)]/2 = [-2 +- sqrt(-12)]/2 = -2 +- sqrt(-3) = -2 +- i:rt3:. So the other solutions are -2 + i:rt3; and -2 - i:rt3:. The sum of these three is -6.[/hide]\n\nyour factorization is incorrect; the second factor should be [hide]x:^2:-2x+4[/hide]. also, while using the quadratic formula, you made an algebraic error, forgetting to [hide]divide the real part by 2.[/hide] Once you take these into account, you get the answer of [hide]0, which you get also get simply by noting that the three roots will be the vertices of an equilateral triangle (symmmetric with respect to the origin) in the complex plane.[/hide] You could also get that answer by [hide]using Viete's formula, onse you ensure there are no double/triple roots.[/hide]", "Solution_4": "alternate solution to ii) (which is very similar to matfan's) [hide]the period of tangent is :pi:, as there are 5 periods in the interval (0,5:pi:), any real value will be assumed exactly 5 times on that interval.[/hide]", "Solution_5": "MysticTerminator wrote:mathfanatic wrote:Since no one wants to try my challenges, I think I will. \n\nQuote:i) Find the sum of the solutions, both real and imaginary (potentially complex), of x:^3: + 8 = 0. \n\n[hide]\nx:^3:+8 is the sum of two cubes, and therefore factors to (x+2)(x^2+2x+4). So one solution is x=-2. To find the other two solutions, we will use the quadratic formula:\nx = [-2 +- sqrt(4-4*1*4)]/2 = [-2 +- sqrt(-12)]/2 = -2 +- sqrt(-3) = -2 +- i:rt3:. So the other solutions are -2 + i:rt3; and -2 - i:rt3:. The sum of these three is -6.[/hide]\nyour factorization is incorrect; the second factor should be [hide]x:^2:-2x+4[/hide]. also, while using the quadratic formula, you made an algebraic error, forgetting to [hide]divide the real part by 2.[/hide] Once you take these into account, you get the answer of [hide]0, which you get also get simply by noting that the three roots will be the vertices of an equilateral triangle (symmmetric with respect to the origin) in the complex plane.[/hide] You could also get that answer by [hide]using Viete's formula, onse you ensure there are no double/triple roots.[/hide]\n\n\n\nOh, duh. I feel stupid now.\n\n\n\nOkay, so corrected, we have\n\n[hide]x:^3:+8 is the sum of two cubes, and therefore factors to (x+2)(x^2-2x+4). So one solution is x=-2. To find the other two solutions, we will use the quadratic formula:\n\nx = [2 +- sqrt(4-4*1*4)]/2 = [2 +- sqrt(-12)]/2 = 1 +- sqrt(-3) = 1 +- i:rt3:. So the other solutions are 1 + i:rt3; and 1 - i:rt3:. The sum of these three is 0.[/hide]", "Solution_6": "Actually, Mystic, \"The sum of the roots\" implies that you take multiplicities counted multiple times. That is, the sum of the roots of \r\n(x - 1)^753 = 0 is 753, not 1.", "Solution_7": "[quote=\"JBL\"]Actually, Mystic, \"The sum of the roots\" implies that you take multiplicities counted multiple times. That is, the sum of the roots of \n(x - 1)^753 = 0 is 753, not 1.[/quote]\r\n\r\noh sorry didn't know that. hm. that seems a bit...um...nonintuitive, but it sure helps for using viete!" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "there are n points on the plane and we know that all of them are not on a line.\r\nprove that there is a line that only two points are on it.", "Solution_1": "It's the classic \"Sylvester's probem\" :D \r\n\r\nDomi", "Solution_2": "[url=http://en.wikipedia.org/wiki/Sylvester-Gallai_theorem#Proof_of_the_Sylvester.E2.80.93Gallai_theorem]Wikipedia's proof[/url]" } { "Tag": [ "LaTeX", "geometry" ], "Problem": "I just have six problems that I don't understand in PSAT Writing Section. I wish I was good at Writing/Vocabulary like Math (I got perfect score without really trying! :D ) but that's not the case.. They were supposed to be label as A,B,C,D, and E but since I don't know how to do without using Latex, I'll put them in different color.\r\n\r\n1. Because Roberto wants [color=blue]to help preserve [/color] marine life, he intends [color=darkblue]on declaring[/color] marine biology [color=indigo]as[/color] his major when he [color=green]begins college[/color] next fall. [color=violet]No error[/color]\r\n\r\n2. This year's benefit game [color=blue]drew[/color] [color=darkblue]nearly twice[/color] [color=indigo]as many[/color] spectators [color=green]as last year[/color]. [color=violet]No error[/color]\r\n\r\n3. [color=blue]Opposite to[/color] the opinion of several respected literary critics, Jane Austen [color=darkblue]does not make[/color] good taste or manners in themselves sure [color=indigo]signs of[/color] virtue [color=green]in her characters[/color]. [color=violet]No error[/color] \r\n\r\n4. [color=blue]If not[/color] in [color=darkblue]good[/color] physical condition, a bicycle excursion through the countryside [color=indigo]can be[/color] [color=green]painfully exhausting[/color]. [color=violet]No error[/color]\r\n\r\n5. [color=blue]Setting herself[/color] up [color=darkblue]as[/color] a preserver of the Classical tradition, Clara Schumann was an [color=indigo]outspoken[/color] [color=green]critic toward[/color] Romantic composer Franz Liszt. [color=violet]No error[/color]\r\n\r\n6. [color=blue]When looking[/color] at modern photograph of that area of Indonesia, the effects of the 1883 [color=darkblue]eruption of[/color] the volcano Krakatau [color=indigo]are[/color] [color=green]still evident[/color]. [color=violet]No error[/color]\r\n\r\nI hope this is OK to post.. I'm taking PSAT next week and I want to make sure I understand everything in the packet.. :lol: \r\n\r\n[hide=\"Answers\"]\n1. on declaring\n2. as last year\n3. Opposite to\n4. If not\n5. critic toward\n6. When looking [/hide]", "Solution_1": "1. [hide][i]to help[/i] is an infinitive, so [i]on declaring[/i] should also be an infinitive - [i]to declare[/i].[/hide]\n2. [hide][i]This game[/i] is being compared to something - it should be [i]last year's game[/i], or, more casually, [i]last year's[/i].[/hide]3. [hide][i]Opposite to[/i] is not a gramatically correct phrase, in general. The context here suggests [i]Contrary to[/i] or [i]In contrast to[/i].[/hide]\n4. [hide][i]If not[/i] makes the good condition refer to the bicycle excursion. Perhaps, [i]If one is not[/i].[/hide]5. [hide]One is not a [i]critic to[/i] - most appropriate might be [i]critic of[/i].[/hide]\n6. [hide]I'm guessing you missed a letter or a word or something around \"modern photograph\"? The lack of a subject at the beginning of the sentence makes \"the effects\" the subject. It might be replaced by [i]When one looks[/i].[/hide]\r\n\r\nHope that helps.", "Solution_2": "Thanks. That helps a lot.\r\n\r\nI just few questions though.\r\n\r\n1. Why is phrase \"Oppsite to\" grammatically incorrect? I know phrase like \"Contrary to..\" or \"In contrast\" is used more often but still, :? .\r\n\r\n2. For #5, what's the problem with using \"critic toward\" instead of \"critic of\"? I'm not quite sure when do you use word \"toward.\"\r\n\r\nAnd last question (irrelevant to the above questions):\r\n\r\nI know I have no way to prepare for PSAT this year but I want to be prepared for next year because I'll be junior (taking both PSAT and SAT!! And I heard PSAT picks national scholarships too!) and wondering how I can prove my critical reading/writing skill. What texts do you recommend? :) \r\n\r\nI know some people in this site got perfect score or close to on the SAT so I hope I get really good responses.. :)", "Solution_3": "[quote=\"Silverfalcon\"]Thanks. That helps a lot.\n\nI just few questions though.\n\n1. Why is phrase \"Oppsite to\" grammatically incorrect? I know phrase like \"Contrary to..\" or \"In contrast\" is used more often but still, :? .\n\n2. For #5, what's the problem with using \"critic toward\" instead of \"critic of\"? I'm not quite sure when do you use word \"toward.\"\n\nAnd last question (irrelevant to the above questions):\n\nI know I have no way to prepare for PSAT this year but I want to be prepared for next year because I'll be junior (taking both PSAT and SAT!! And I heard PSAT picks national scholarships too!) and wondering how I can prove my critical reading/writing skill. What texts do you recommend? :) \n\nI know some people in this site got perfect score or close to on the SAT so I hope I get really good responses.. :)[/quote]\r\n\r\n1. \"Opposite to\" is just not used, and thus gramatically incorrect. The correct phrase would be \"opposite of.\"\r\n\r\n2. I believe \"toward\" indicates some sort of direction, so it wouldn't work here. I'm not completely sure about that, though.", "Solution_4": "\"Opposite of\" doesn't fit in the sentence, though. It's used in the context of, \"One is the opposite of the other.\" But it is correct that \"opposite to\" is never correct. \r\n\r\n\"Toward\" isn't used only for physical direction though - \"I feel derision towards his plan.\" However, it is used with a verb in all cases. Therefore, \"a critic towards\" is not correct. Note that it is not always correct to use \"towards\" with a verb, either.\r\n\r\nHope that helps.", "Solution_5": "[quote=\"inteluser\"]\"Opposite of\" doesn't fit in the sentence, though. It's used in the context of, \"One is the opposite of the other.\" But it is correct that \"opposite to\" is never correct. \n\n\"Toward\" isn't used only for physical direction though - \"I feel derision towards his plan.\" However, it is used with a verb in all cases. Therefore, \"a critic towards\" is not correct. Note that it is not always correct to use \"towards\" with a verb, either.\n\nHope that helps.[/quote]\r\n\r\nThanks! :)", "Solution_6": "Anybody here taking the PSAT at their school on Wednesday?", "Solution_7": "[quote=\"elston\"]Anybody here taking the PSAT at their school on Wednesday?[/quote]\r\n\r\nYup. BTW, there's another thread on this." } { "Tag": [ "function", "calculus", "derivative", "limit", "logarithms", "real analysis", "real analysis unsolved" ], "Problem": "Let M be a large real number. Explain briefly why there must be exactly one root w of the equation Mx=e^x with w > 1. Why is log M a reasonable approximation to w? Write w = log M + y. Can you give an approximation to y, and hence improve on log M as an approximaton to w?", "Solution_1": "$ f(x)\\equal{}\\frac{e^x}{x}$ is a strictly increasing function on $ [1,\\infty)$ (easily shown after differentiation) and $ \\lim_{x \\to \\infty} f(x) \\equal{} \\infty$.\r\nSo $ Mx\\equal{}e^x$ has exactly one solution for large $ M$.\r\nNow $ x \\equal{} \\log(M) \\plus{} \\log(x)$.\r\nIf $ x>1$ then $ x > \\log(M)>2$ for large $ M$.\r\nThus from now on we suppose that $ x>2$.\r\nThen $ \\log(x) < \\frac12 x$.\r\nand $ x \\equal{} \\log(M) \\plus{} \\log(x) < \\log(M) \\plus{} \\frac12 x$.\r\nThen $ \\frac12 x < \\log(M)$ and $ \\log(x) < \\log(2) \\plus{} \\log(\\log(M))$\r\nNow we have $ x \\equal{} \\log(M) \\plus{} \\log(x) \\equal{} \\log(M) \\plus{} O(\\log(\\log(M))$.\r\nTaking on both sites logarithms:\r\n$ \\log(x) \\equal{} \\log( \\log(M) \\plus{} O(\\log(\\log(M))) \\equal{} \\log(\\log(M)) \\plus{} \\log(1\\plus{}O(\\frac{\\log(\\log(M))}{\\log(M)}))$.\r\n$ \\equal{}\\log(\\log(M)) \\plus{} O(\\frac{\\log(\\log(M))}{\\log(M)})$.\r\nSubstitute this result in $ x \\equal{} \\log(M) \\plus{} \\log(x)$.\r\n$ x \\equal{} \\log(M) \\plus{} \\log(x) \\equal{} \\log(M) \\plus{} \\log(\\log(M)) \\plus{} O(\\frac{\\log(\\log(M))}{\\log(M)})$..", "Solution_2": "Thanks. \r\n\r\nThere must be an easier way of doing this as the question is aimed at students in their final year of school who probably would not have met O-notation.", "Solution_3": "Can anyone supply an easier (to understand) solution?", "Solution_4": "Can anyone show me a simpler way of doing this?", "Solution_5": "Still nothing?", "Solution_6": "Someone must be able to do this.", "Solution_7": "Consider the function $ g(x)\\equal{}\\frac{e^x}{x}$. Now, $ g(1)\\equal{}e$ and $ g'(x)\\equal{}x^2e^x(x\\minus{}1)$ for $ x>1$, so $ g$ is increasing. This means that for every large $ M$, there is an $ x$ such that $ g(x)\\equal{}M$, hence $ e^x\\equal{}Mx$.\r\n\r\nTo see why $ \\log M$ is a good approximation to $ x$, take log on both sides of the equation $ e^x\\equal{}Mx$, and get $ x\\equal{}\\log M \\minus{} \\log x$. The idea is now that $ x>>\\log x$ for large $ x$, and $ \\log M$ is a good approximation.\r\n\r\nThe precise argument for the 3rd question is in Peter's post. \r\n\r\nAn please: One post is enough, 6 posts nagging for another solution is annpoying..", "Solution_8": "Thanks for your help." } { "Tag": [ "inequalities", "trigonometry", "inequalities unsolved" ], "Problem": "In two Triangle $ABC$ and $A'B'C'$ ,the following inequality is true(or not)?\r\n$ \\sin\\frac{A}{2} \\sin\\frac{A'}{2} +\\sin\\frac{B}{2}\\sin\\frac{B'}{2}+\\sin\\frac{C}{2}\\sin\\frac{C'}{2} \\leq \\frac{3}{4}(\\tan\\frac{A}{2}\\tan\\frac{A'}{2}+\\tan\\frac{B}{2}\\tan\\frac{B'}{2}+\\tan\\frac{C}{2}\\tan\\frac{C'}{2})$", "Solution_1": "Wrong. Take $A=B'=0, C=C'=\\epsilon\\to 0$.\r\n\r\nEdit: I was just told that it is not at all clear what I mean, so if you want a real counterexample, take $A=B' = 0.02^{o}$, $C=C'=50^{o}$ and$A'=B=129.98^{o}$." } { "Tag": [ "modular arithmetic" ], "Problem": "Find all positive integers (x,y) such that \r\n$7^{x}-3^{y}= 4$", "Solution_1": "Grrr, this is annoying...\r\n\r\nthe only significant thing I've come up with so far is that $x \\equiv y \\pmod 2$...", "Solution_2": "You can add something: $x\\equiv y\\equiv1\\pmod2$ :)\r\n\r\nRe write it as $7(7^{x-1}-1)=3(3^{y-1}-1)$", "Solution_3": "[quote=\"manuel\"]You can add something: $x\\equiv y\\equiv1\\pmod2$ :)\n\nRe write it as $7(7^{y-1}-1)=3(3^{y-1}-1)$[/quote]Do you mean $7(7^{x-1}-1)$?", "Solution_4": "yes, typo sorry" } { "Tag": [ "\\/closed" ], "Problem": "I have a problem. Usually whenever I try to log in, the only way I can log in is when I click the \"log in automaticly button\". But Iv found out that the whenevr I do click that button and log in, when i try to reply to posts, i get the message \"invalid session please resubmit the form\" which prevents me from posting. Whenever I wouldn't click that button and acually am able to log in(very rarly), I would not get the \"invalid session\" notice when posting. Iv tried this on 3 different computers each on opera,firefox,and IE , and they all work the same way. Recently, Iv tried deleting all Private data and I have been able to log in without clicking the button, which has allowed me to post on the forum. Im not very good with computers at all, but im pretty sure this is a cookie issue. But the problem is that even when my browser is on \"accept all cookies,\" it still wont log me in unless i click the \"log in automaticly button\", which therefore wont allow me to post on the forum since its a \"invalid session\". Its a pain deleting all my private data every time i go to AoPS (which is about everyday) so does somone know about this bug( if it is a bug)? If this problem has already been solved, you can delete this topic.", "Solution_1": "This is very strange. Where were the 3 computers you tried this on, and when did you test this, first of all.", "Solution_2": "where? well these computers are all at home and use the same WLAN connection, and i tested it like a week ago on all computers, and i still test it every day on my laptop. any ideas whats wrong?", "Solution_3": "Seems like something is wrong with either your connection setup or your account settings. To make it completely sure which one is the case, one can try to login from some other place using your username and password (you can try it yourself from school or local library; any admin can also do it from his PC) and to try to login from one of your computers to a different account. If you have somebody like your sibling or neighbor or schoolmate who has an account on AoPS, just ask him to try it; the duplicate accounts are generally a strict No-No, but if no admin or somebody else who knows better answers you soon, I guess you'll not be really penalized for setting one more account and trying it for login (just do not use it for posting except, maybe, in the test forum!). If you do it, choose a shorter user name as different from your current one as possible (though I doubt this is the problem). If that other account has no problem with it on your computer, it will indicate that the problem is most likely on the AoPS side and you'll have to wait for Valentin to figure out what it is and fix it. If you have the same problem with another account (and/or if you have no problem with your account on the computers outside your home), it will almost certainly mean that the problem is on your end and you'll need to take a close look at your internet connection settings.", "Solution_4": "Definitely the problem is not on our side, otherwise a lot more people would be complaining. This is a local issue with your cookies. Try completely erasing your cookies." } { "Tag": [], "Problem": "Find the length of $ AC$ in the right triangle.\n\n[asy]draw((0,0)--(7,0)--(7,sqrt(2))--cycle);\nlabel(\"7\",(0,0)--(7,0),S);\nlabel(\"$\\sqrt{2}$\",(7,0)--(7,sqrt(2)),E);\nlabel(\"A\",(0,0),W);\nlabel(\"B\",(7,0),SE);\nlabel(\"C\",(7,sqrt(2)),NE);[/asy]", "Solution_1": "By Pythagorean's Theorem, $ AC\\equal{}\\sqrt{AB^2\\plus{}AC^2}\\equal{}\\sqrt{7^2\\plus{}(\\sqrt2)^2}\\equal{}\\sqrt{51}$" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Here are the problems and their proof (Vietnamese version). \r\n\r\nWe haven't translated it to English version. I think it will be not easy to read, especially for the Geometrical Problems. :(\r\n\r\nAnyway, here are they. :)", "Solution_1": "Thank you very much, Honey_S, for this very nice document :)" } { "Tag": [ "invariant", "induction", "combinatorics proposed", "combinatorics" ], "Problem": "Prove that each finite set of integers can be arranged without intersection.", "Solution_1": "orl, maybe you could restate some of the problems you've recently posted? For instance, I have no idea what this one means :?. First of all, what does \"arranging a finite set of integers\" mean? Second of all, what does \"intersection\" mean in this case? \r\n\r\nSorry if this is common language, it's not common to me.. :)\r\n\r\nThanks!", "Solution_2": "[quote=\"grobber\"]\n\nSorry if this is common language, it's not common to me.. :)\n\nThanks![/quote]\r\n\r\nMaybe I should define it first what I want to know. :lol: \r\n\r\nIn the sequence $\\{a_1,a_2,a_3,a_4\\} = \\{1,3,6,5\\}$ and $a_2 = \\frac{a_1 + a_4}{2}$ is the arithmetic mean of $a_1$ and $a_4$. A sequence $\\{a_1,a_2,a_3,a_4\\}$ of integers is called free of intersections if there are no indices $i,j,k$ with $1 \\leq i < j < k \\leq n$ and $a_j = \\frac{a_i + a_k}{2}$, e.g. that there is no element which is the arithmetic mean of its left and right neighboring element. \r\n\r\nHint: Find some operations which remains a the sequence invariant.", "Solution_3": "Ah, ok. Thus, I still do not see where there must be an invariant (unless you are talking about induction) but here it goes :\r\nLet $n$ be a positive integer. And $U_n = a_1,..., a_n $ be a sequence of positive integers.\r\nWe will say that $U_n$ is good if no term $a$ is the arithmetical mean of two other terms of the sequence one on each side from $a$.\r\nThus both $U_3 = -1,1,0$ and $U_6 = -2,2,0,-3,1,-1$ are good.\r\n\r\nNow assume that $U_n$ is good.\r\nLet $U_{2n} = 2a_1,2a_2, \\cdots , 2a_n, 2a_1-1,2a_2-1, \\cdots, 2a_n - 1 = b_1, \\cdots, b_{2n}.$\r\n\r\nWe will verify that $U_{2n}$ is good too :\r\nLet $1 \\leq i2n$ then $b_i$ and $b_k$ have opposite parities so that their sum is not equal to $2b_j$.\r\nIf $k \\leq 2n$, then $b_i = 2a_i,b_j = 2a_j,b_k = 2a_k$, and the induction hypothesis ensures us that $b_i+b_k \\neq 2b_j.$\r\nIf $i > 2n$ then $b_i = 2a_{i-2n} - 1,b_j = 2a_{j-2n}-1,b_k = 2a_{k-2n}-1$, and again $b_i+b_k \\neq 2b_j.$\r\n\r\nThis proves that $U_{2n}$ is good, as claimed.\r\n\r\nNow, it is easy from $U_3$ and $U_6$ given above to verify that $U_{3 \\cdot 2^n}$ is a permutation of $E_n = \\{-2^{n+1} , -2^{n+1} + 1, \\cdots, 0, 1 ,2, \\cdots, 2^n \\}$.\r\nThen, since any finite set of integers is contained in such set $E_n$ for sufficientely large $n$ and since to obtain a good permutation of our given set it suffices to delete the members of $E_n$ which do not belong to our given set (so that the sequence remains good), we are done.\r\n\r\nPierre.", "Solution_4": "More or less the same problem was used by the Bundeswettbewerb Mathematik 2005 as the problem 4 in the 1st round:\r\n\r\nFor which positive integers n is it possible to arrange the n numbers 1, 2, 3, ..., n in a sequence such, that for any two numbers the arithmetic mean of these two numbers doesn't stand somewhere between them?\r\n\r\nOf course, the answer is: This is possible for all positive integers n.\r\n\r\nAnother thread with a discussion of the same topic: http://www.mathlinks.ro/Forum/viewtopic.php?t=14758 ." } { "Tag": [ "symmetry", "superior algebra", "superior algebra solved" ], "Problem": "Show that $\\mathfrak{S}_{4}$ can be embedded in $\\mbox{SL}_{3}(\\mathbb{Z})$.\r\n\r\nCan we generalize?", "Solution_1": "So, how could you embed $S_{4}$ ? I've searched for ten minutes or so (\"randomly\") and couldn't grasp the thing. Is there a general approach or do you have to \"guess\" the form of the image by this embedding ?", "Solution_2": "Note that $S_{4}$ is the symmetry group of the cube.\r\nNow it should be easy.", "Solution_3": "Actually, that symmetry group is $S_{4}\\times \\mathbb{Z}_{2}$. You want just the rotations.", "Solution_4": "Thank you. I wouldn't have thought about that :lol:" } { "Tag": [ "percent", "probability" ], "Problem": "Battery lifetime is normally distributed for large samples. The mean lifetime is 500 days and the standard deviation is 61 days. What percent, to the nearest whole number, of batteries have lifetimes longer than 561 days?", "Solution_1": "[hide]Normalcdf (561, 1E99, 500, 61)[/hide]", "Solution_2": "Usually calculator commands don't suffice for answers... in any case, it's simply the probability that you're at least one standard deviation greater than the mean... around 16% I believe.", "Solution_3": "I'm just wondering if we can use Chebyshev's inequality...\r\n\r\n[EDIT]Oh, it [u]is[/u] Chebyshev's theorem :blush:" } { "Tag": [ "algebra", "polynomial", "function" ], "Problem": "Devise a pair of dice with exactly the same outcomes as ordinary dice (the sum 2 comes out once, etc.), but which are not ordinary dice.", "Solution_1": "[hide=\"Hint\"] Factor the polynomial $ (x \\plus{} ... \\plus{} x^6)^2$ into two factors that take the value $ 6$ at $ x \\equal{} 1$ with non-negative integer coefficients. There is exactly one other way to do this. [/hide]", "Solution_2": "[hide=\"Solution\"][b]t0rajir0u[/b] is basically saying that the generating functions corresponding to the two individual dice must multiply to the same generating function as multiplying those for two normal dice would. Additionally, if each dice must have six sides (we're just putting a number on each face), then the coefficients of each polynomial must add to 6.\n\nIt turns out that $ (x \\plus{} \\ldots x^6)^2 \\equal{} x^{2} (x^2 \\minus{} x \\plus{} 1)^2 (x \\plus{} 1) (x^2 \\plus{} x \\plus{} 1)^2$\n\nIf we don't limit ourselves to non-negative face values, then we should write it as follows:\n$ x^{1 \\minus{} n} x^{1\\plus{}n} (x^2 \\minus{} x \\plus{} 1)^2 (x \\plus{} 1) (x^2 \\plus{} x \\plus{} 1)^2$\n\nFor now, I'll pretend we want positive face values (we can ditch this assumption by adding any constant to one die's faces and subtracting it from the other die). Thus we need the leading coefficients to be 0 in each case. We know the sum of the coefficients of a polynomial is equal to that polynomial evaluated at $ 1$. We find the following contributions:\n$ x \\to 1$\n$ x^2 \\minus{} x \\plus{} 1 \\to 1$\n$ x \\plus{} 1 \\to 2$\n$ x^2 \\plus{} x \\plus{} 1 \\to 3$\n\nWe see that each dice must get one factor of $ (x \\plus{} 1)(x^2 \\minus{} x \\plus{} 1)$, which contribute a factor of 6. We additionally need a factor of $ x$ in each so that the leading coefficients are $ 0$. Now we need to figure out how we want to distribute the two $ (x^2 \\minus{} x \\plus{} 1)$'s.\n\nWe can do:\n$ x(x^2 \\minus{} x \\plus{} 1)(x \\plus{} 1)(x^2 \\plus{} x \\plus{} 1) \\; ; \\qquad x (x^2 \\minus{} x \\plus{} 1) (x \\plus{} 1)(x^2 \\plus{} x \\plus{} 1)$\n$ x(x \\plus{} 1)(x^2 \\plus{} x \\plus{} 1) \\; ; \\qquad x (x^2 \\minus{} x \\plus{} 1)^2 (x \\plus{} 1)(x^2 \\plus{} x \\plus{} 1)$\n\nExpanding the above, we get the following pairs:\n$ x \\plus{} \\ldots x^6 \\; ; \\qquad x \\plus{} \\ldots x^6$\n$ x \\plus{} 2x^2 \\plus{} 2x^3 \\plus{} x^4 \\; ; \\qquad x \\plus{} x^3 \\plus{} x^4 \\plus{} x^5 \\plus{} x^6 \\plus{} x^8$\n\nThese correspond to dice with face values:\n$ (1,2,3,4,5,6) \\; ; \\qquad (1,2,3,4,5,6)$\n$ (1,2,2,3,3,4) \\; ; \\qquad (1,3,4,5,6,8)$\n\nThe first is just a standard pair. The only other possibility is $ (1,2,2,3,3,4) \\qquad (1,3,4,5,6,8)$[/hide]" } { "Tag": [ "algebra", "polynomial", "algebra proposed" ], "Problem": "1-) Find all polynomials P(x) with (for all real x):\r\n\r\nP(1) = 1\r\n\r\nP(x\u00b2 + 1) = [P(x)]\u00b2 + 1\r\n\r\n2-) Find all polynomials P(x) with (for all real x):\r\n\r\nP(x\u00b2) + P(x).P(x + 1) = 0\r\n\r\nThanks a lot!!! :)", "Solution_1": "I think the first one is easy. $ P(1)\\equal{}1 , P(2)\\equal{}2 , P(5)\\equal{}5 , P(26)\\equal{}26 , ...$ So $ P(n)\\equal{}n$ for infinitely many $ n \\Longrightarrow$ \r\n\r\n$ P(x)\\equal{}x$ for all $ x$ real numbers. :wink:", "Solution_2": "The second one is a little harder. If $ P(x)$ is constant so $ P(x) \\equal{} 0, \\minus{} 1$ \r\n\r\n$ u$ is a complex root of $ P \\Longrightarrow$\r\n\r\n$ P(u^{2^k}) \\equal{} 0$ for all positive integer $ k$. There exists integers $ m$ and $ n$ so $ u^{2^m} \\equal{} u^{2^n} \\Longrightarrow$\r\n\r\n$ |u| \\equal{} 0,1$ \r\n\r\n$ P((u \\minus{} 1)^{2^k}) \\equal{} 0$ for all positive integer $ k$. So by same reason $ |u \\minus{} 1| \\equal{} 0,1$ \r\n\r\n$ u \\equal{} 0, 1 , \\frac {1 \\plus{} \\sqrt {3}i}{2}, \\frac {1 \\minus{} \\sqrt {3}i}{2}$ \r\n\r\n$ P(x) \\equal{} x^{\\alpha}(x \\minus{} 1)^{\\beta}(x \\minus{} \\frac {1 \\plus{} \\sqrt {3}i}{2})^{\\theta}(x \\minus{} \\frac {1 \\minus{} \\sqrt {3}i}{2})^{\\gamma}$\r\n\r\n$ x^{2\\alpha}(x^2 \\minus{} 1)^{\\beta}(x^2 \\minus{} \\frac {1 \\plus{} \\sqrt {3}i}{2})^{\\theta}(x^2 \\minus{} \\frac {1 \\minus{} \\sqrt {3}i}{2})^{\\gamma} \\equal{}$\r\n\r\n$ \\minus{}x^{\\alpha}(x \\minus{} 1)^{\\beta}(x \\minus{} \\frac {1 \\plus{} \\sqrt {3}i}{2})^{\\theta}(x \\minus{} \\frac {1 \\minus{} \\sqrt {3}i}{2})^{\\gamma} (x \\plus{} 1)^{\\alpha}x^{\\beta}(x \\plus{} \\frac {1 \\minus{} \\sqrt {3}i}{2})^{\\theta}(x \\plus{} \\frac {1 \\plus{} \\sqrt {3}i}{2})^{\\gamma}$\r\n\r\nBut there isn't any non constant polynomial.\r\n\r\nRemark: If $ P(x^2) \\equal{} P(x).P(x \\plus{} 1)$ and $ P$ is non constant so $ P(x) \\equal{} (x^2 \\minus{} x)^{\\theta}$.", "Solution_3": "First of all, thanks for the answers! :) \r\n\r\nBut still about the first problem, i also found that answer (which is the right one, by the way), although i couldn't prove it's unic.\r\n\r\nI tried the following:\r\n\r\nQ(x) = P(x\u00b2 + 1)\r\nT(x) = [P(x)]\u00b2 + 1\r\n\r\nx = P(x):\r\n\r\nQ(P(x)) = P([P(x)]\u00b2 + 1)\r\nQ(P(x)) = P(T(x))\r\n\r\nSince T(x) = Q(x):\r\n\r\nQ(P(x)) = P(Q(x))\r\n\r\nIdeed, P(x) = x is a valid answer.\r\nBut does it prove anything about the uniticity?\r\n\r\nAbout the second question, i just got to thank you once again.", "Solution_4": "[quote=\"Valle\"] Ideed, P(x) = x is a valid answer.\nBut does it prove anything about the uniticity?[/quote]\r\n\r\nAs crazyfehmy said : $ P(n)\\equal{}n$ for infinitely many $ n$, so $ P(x)\\minus{}x$ is a polynomial with infinitely many roots, so is zero. So $ P(x)\\equal{}x$ $ \\forall x$" } { "Tag": [ "geometry", "3D geometry", "pyramid", "induction", "ratio", "geometry unsolved" ], "Problem": "Consider a pyramid $VA_1A_2...A_n$ with $n\\geq 5$. On the side $VA_i$ consider the point $B_i$.\r\nLet $M=\\left\\{P_{ij}\\right\\}$, where $P_{ij}=A_iB_j \\cap A_jB_i$. Prove that, if at least $\\binom{n-1}{2}+1$ points of $M$ are in the same plane, then the points $B_1$, $B_2$, ..., $B_n$ are in the same plane.", "Solution_1": "[quote=\"S_Alex\"]Consider a pyramid VA1A2...An. (n>=5) On the side VAi consider the point Bi. \nLet M={Pij / Pij=AiBj \\cap AjBi. Prove that if at least C(n-1,2) +1 points of M are in the same plane then the points B1, B2, ..., Bn are in the same plane.[/quote]\r\nAnyone any ideeas ?.......... ?.............?............. No ? ........... :D :D :D", "Solution_2": "[quote=\"S_Alex\"]Consider a pyramid VA1A2...An. (n>=5) On the side VAi consider the point Bi. \nLet M={Pij / Pij=AiBj \\cap AjBi. Prove that if at least C(n-1,2) +1 points of M are in the same plane then the points B1, B2, ..., Bn are in the same plane.[/quote]\r\nIs it really that hard ????..... :D :D", "Solution_3": "[quote=\"S_Alex\"]Consider a pyramid VA1A2...An. (n>=5) On the side VAi consider the point Bi. \nLet M={Pij / Pij=AiBj \\cap AjBi. Prove that if at least C(n-1,2) +1 points of M are in the same plane then the points B1, B2, ..., Bn are in the same plane.[/quote]\r\nGuess so. :D :D", "Solution_4": "i have seen this problem earlier and dint solve it back then and cant solve it now :D and if i remember correctly its arad 1999 county olympiad", "Solution_5": "[quote=\"galois\"]i have seen this problem earlier and dint solve it back then and cant solve it now :D and if i remember correctly its arad 1999 county olympiad[/quote]\r\nYou're right galois but that doesn't help me very much :D :D :D because the book which contains that county olympiad does not provide any proofs whatsoever. :D", "Solution_6": "I was asking myself this: if we know that all the Bi's are coplanar, then does it follow that all the Pij's are coplanar?", "Solution_7": "[quote=\"grobber\"]I was asking myself this: if we know that all the Bi's are coplanar, then does it follow that all the Pij's are coplanar?[/quote]\r\nGood question. But I don't believe that the reverse statement it's true. Why is that I can't tell (I could be wrong).", "Solution_8": "If we can prove it for n=5, induction can continue the argument.", "Solution_9": "[quote=\"grobber\"]If we can prove it for n=5, induction can continue the argument.[/quote]\r\nYes I know that but how do you prove it for n=5. It's hard even to draw it. \r\n :D (joking).", "Solution_10": "First of all we can assume that $P_{12}, P_{13}$ and $P_{23}$ lie in that plane (associated graph necesserily contains triangle). :!:\r\nSuppose $C_{n-2}^2+1$ points lie in the plane $\\pi$ and points $A_1,...,A_n$ lie in the plane $\\alpha$, $(B_1B_2B_3)=\\beta$. Suppose $\\alpha$ and $\\beta$ intersect. Then we can apply the projective transformation, which maps $\\alpha\\cap \\beta$ to the line of infinity. \r\nSo we have now that $\\alpha ||\\beta$. It immediately gives that $\\alpha||(P_{12}P_{23}P_{31})$, i.e. $\\alpha||\\pi$. Since our graph is connected it immediately follows that all ratios $VB_:VA_i$ are equal, so all $B_i$ lies in $\\beta$. Hence the result." } { "Tag": [ "quadratics", "function", "absolute value" ], "Problem": "We may say concerning the solution of\r\n\\[ |x|^2 \\plus{} |x| \\minus{} 6 \\equal{} 0\r\n\\]\r\nthat:\r\n\r\n$ \\textbf{(A)}\\ \\text{there is only one root}\\qquad \r\n\\textbf{(B)}\\ \\text{the sum of the roots is }{\\plus{}1}\\qquad \r\n\\textbf{(C)}\\ \\text{the sum of the roots is }{0}\\qquad \\\\\r\n\\textbf{(D)}\\ \\text{the product of the roots is }{\\plus{}4}\\qquad \r\n\\textbf{(E)}\\ \\text{the product of the roots is }{\\minus{}6}$", "Solution_1": "EDIT: got it now.\r\n\r\n[hide=\"POTENTIALLY DANGEROUS CONTENT\"]$ |x|^2 \\equal{} x^2$. Always.\n\nSo the equation can be rewritten as $ x^2 \\pm x \\minus{} 6$ if we get rid of the abs. value.\n\nThe sum of the roots in $ x^2 \\plus{} x \\minus{} 6$ is 1 by Veita's.\n\nThe sum of the roots in $ x^2 \\minus{} x \\minus{} 6$ is -1 by Veita's.\n\nThe total sum is $ 1 \\plus{} ( \\minus{} 1) \\equal{} \\boxed 0$\n\nOur answer is C[/hide][/hide]", "Solution_2": "[hide=\"Alternate Solution\"]$ |x|^2 \\plus{} |x| \\minus{} 6 \\equal{} 0$\n\nLet $ u \\equal{} |x|$\n\nThen,\n\n$ u^2 \\plus{} u \\minus{} 6 \\equal{} 0$\n\n$ \\left(u \\minus{} 2\\right)\\left(u \\plus{} 3\\right) \\equal{} 0$\n\n$ u \\equal{} 2, \\minus{} 3$\n\nBut $ u\\geq0$\n\n$ \\therefore u \\equal{} 2$\n\nie $ |x| \\equal{} 2$\n\n$ \\therefore x \\equal{} \\pm 2$\n\nHence the sum of the roots is 0 \n\nie $ \\boxed{C}$[/hide]", "Solution_3": "[hide=\"Another Solution\"]\nWe can note that the function $ f(x)\\equal{}|x|^2\\plus{}|x|\\minus{}6$ is even. Thus, if $ r$ is a root then $ \\minus{}r$ is also a root, so the sum of the roots must be $ \\boxed{0}$.\n[/hide]", "Solution_4": "This problem has been solved, but I want to comment on [b]abcak[/b]'s solution:\r\n\r\nYour solution implies that there are four roots, two from each of the two quadratics you ended up with. However, the actual function has only two roots. The problem is that absolute value asks you to pick + or - in a very particular way -- it does not give you the freedom to choose. By changing $ |x|$ to $ \\pm x$, you are creating a multivalued function.\r\n\r\nIn actuality, only the positive root of $ x^2\\plus{}x\\minus{}6$ and the negative root of $ x^2\\minus{}x\\minus{}6$ are valid." } { "Tag": [ "Putnam", "inequalities", "function", "logarithms", "inequalities unsolved" ], "Problem": "a1, a2, ... , an, b1, ... , bn are non-negative reals. Show that (\u220f ai)1/n + (\u220fbi)1/n \u2264 (\u220f(ai+bi))1/n.[/tex][/list]", "Solution_1": "I guess that (\u220f ai)1/n is a root of n -th power of (\u220f ai). So \r\n \u2211 ai/(ai+bi) >= n* [ \u220fai/(ai+bi)]1/n\r\n \u2211 bi/(ai+bi) >= n* [ \u220fbi/(ai+bi)]1/n\r\nSumming:\r\n n =n* [ \u220fai/(ai+bi)]1/n + n* [ \u220fbi/(ai+bi)]1/n or \r\n (\u220f ai)1/n + (\u220fbi)1/n \u2264 (\u220f(ai+bi))1/n", "Solution_2": "This is Putnam 2003 A2, and prowler's solution is the nicest and cleanest such one. Another nice solution is to write k_i = a_i/b_i and divide through by (\u220fbi)^(1/n) so that the inequality is equivalent to:\r\n\r\n1 + \u220f(k_i)^(1/n) <= \u220f(1+k_i)^(1/n)\r\n\r\nNow, letting f(x) = ln(1+e^x), this is interpreted as:\r\n\r\n\u2211 f(k_i) / n >= f( \u2211 f(k_i) / n ), but this is exactly Jensen's inequality for the convex function f(x).", "Solution_3": "You are right [b]blahblahblah[/b],it is easy to solve the problem using that the function $f(x)=\\ln{(1+e^{x})}$ is convex everywhere on $R$ and\r\nmoreover we can prove more general inequality(Hyugens inequality(sorry if I do mistake)).\r\n${a_{1}}^{\\gamma_{1}}\\*{a_{2}}^{\\gamma_{2}}\\cdots{a_{n}}^{\\gamma_{n}}+{b_{1}}^{\\gamma_{1}}\\*{b_{2}}^{\\gamma_{2}}\\cdots{b_{n}}^{\\gamma_{n}}\\leq{(a_{1}+b_{1})}^{\\gamma_{1}}\\*{(a_{2}+b_{2})}^{\\gamma_{2}}\\cdots{(a_{n}+b_{n})}^{\\gamma_{n}}$\r\nwhere $a_{i},b_{i},\\gamma_{i}>0(i=1,2,...,n)$ and $\\sum_{i=1}^{n}{\\gamma_{i}}=1 (*)$.\r\nTo prove it we have to write the Jensen inequality $f(\\sum_{i=1}^{n}{\\gamma_{i}\\*x_{i}})\\leq\\sum_{i=1}^{n}{\\gamma_{i}\\*f(x_{i})}$,for \r\nthe function $f(x)=\\ln{(1+e^{x})}$ and for $x_{i}=\\ln{\\frac{a_{i}}{b_{i}}}(i=1,2,...,n)$.\r\nBut there is another approach which has already used [b]prowler[/b]:\r\nthere is another nice inequality:\r\n${x_{1}}^{\\gamma_{1}}\\*{x_{2}}^{\\gamma_{2}}\\cdots{x_{n}}^{\\gamma_{n}}\\leq{\\gamma_{1}}\\*x_{1}+{\\gamma_{2}}\\*x_{2}+\\cdots+{\\gamma_{n}}\\*x_{n}$,where \r\n$x_{i},\\gamma_{i}>0(i=1,2,...,n) and \\sum_{i=1}^{n}{\\gamma_{i}}=1 (**)$\r\n $(**)$ is a simple corolarry of the following simple inequality\r\n$a^{\\alpha}\\*b^{\\beta}\\leq\\alpha\\*a+\\beta\\*b$,for $a,b,\\alpha,\\beta>0$ and $\\alpha+\\beta=1$.\r\nUsing $(**)$ we can write \r\n$\\prod_{i=1}^{n}{(\\frac{a_{i}}{a_{i}+b_{i}})^{\\gamma_{i}}}\\leq\\sum_{i=1}^{n}{\\gamma_{i}\\*\\frac{a_{i}}{a_{i}+b_{i}}}$\r\nand\r\n$\\prod_{i=1}^{n}{(\\frac{b_{i}}{a_{i}+b_{i}})^{\\gamma_{i}}}\\leq\\sum_{i=1}^{n}{\\gamma_{i}\\*\\frac{b_{i}}{a_{i}+b_{i}}}$.\r\nNow if we add last to inequalities we get $(*)$." } { "Tag": [ "AMC", "AIME", "AIME I", "AIME II" ], "Problem": "Hi,\r\n\r\nI was just wondering if AIME (alternate AIME in particular) must be written at the same time on March 28 or it just has to be written some time that day (like 1 pm or something).\r\nIt's just that AIME I is during March Break and I am pretty sure my school won't write it at that time. AIME II is on March 28, which is the same day for Canadian Math Olympiad, and CMO has to be written at 9 am local time.\r\n\r\nThank you for your help.", "Solution_1": "Last year, it did not have to be administered at a specific time. I do not believe this has changed, but I don't know for sure." } { "Tag": [ "linear algebra" ], "Problem": "Let $ \\mathrm{T}: \\mathbb{R}^{3}\\rightarrow\\mathbb{R}$ be linear. Show that there exists scalars $ a$, $ b$, and $ c$ such that $ \\mathrm{T}(x,y,z)\\equal{}ax\\plus{}by\\plus{}cz$ for all $ (x,y,z)\\in\\mathbb{R}^{3}$. Can you generalize this result for $ \\mathrm{T}: F^{3}\\rightarrow F$? State and prove an analogous result for $ \\mathrm{T}: F^{n}\\rightarrow F^{m}$.", "Solution_1": "Uh, by linear you mean also respects multiplication, right? Otherwise this is not true.\r\n\r\n$ a \\equal{} T(1,0,0), b \\equal{} T(0,1,0), c \\equal{} T(0,0,1)$, uniquely determined", "Solution_2": "Linear as in $ T(cx\\plus{}y)\\equal{}cT(x)\\plus{}T(y)$." } { "Tag": [ "trigonometry", "trig identities" ], "Problem": "Find the amplitude of the graph of $ 3\\sin x\\plus{}4\\cos x$.", "Solution_1": "$ 3\\sin x\\plus{}4\\cos x\\equal{}5\\left(\\sin x \\cdot \\frac{3}{5}\\plus{}\\cos x \\cdot\\frac{4}{5}\\right)$", "Solution_2": "in genl .\r\n$ \\minus{}\\ce{sqrt(a^2\\plus{}b^2)} <\\equal{} a.sinx\\plus{}b.cosx <\\equal{} \\ce{sqrt(a^2\\plus{}b^2)}$", "Solution_3": "[quote=\"kunny\"]$ 3\\sin x \\plus{} 4\\cos x \\equal{} 5\\left(\\sin x \\cdot \\frac {3}{5} \\plus{} \\cos x \\cdot\\frac {4}{5}\\right)$[/quote]\r\nerr... I still don't get it :(", "Solution_4": "[quote=\"AnonAmoz\"][quote=\"kunny\"]$ 3\\sin x \\plus{} 4\\cos x \\equal{} 5\\left(\\sin x \\cdot \\frac {3}{5} \\plus{} \\cos x \\cdot\\frac {4}{5}\\right)$[/quote]\nerr... I still don't get it :([/quote]\r\n\r\nHe just used standard trig identities to rearrange the expression into something where the amplitude was obvious.", "Solution_5": "[quote=\"AnonAmoz\"][quote=\"kunny\"]$ 3\\sin x \\plus{} 4\\cos x \\equal{} 5\\left(\\sin x \\cdot \\frac {3}{5} \\plus{} \\cos x \\cdot\\frac {4}{5}\\right)$[/quote]\nerr... I still don't get it :([/quote]\r\n\r\n$ 3\\sin x \\plus{} 4\\cos x \\equal{} 5\\left(\\sin x \\cdot \\frac {3}{5} \\plus{} \\cos x \\cdot\\frac {4}{5}\\right)$\r\n\r\n$ \\equal{}5(\\sin x\\cos \\alpha \\plus{}\\cos x\\sin \\alpha)$\r\n\r\n$ \\equal{}5\\sin (x\\plus{}\\alpha)$, where $ \\cos \\alpha \\equal{}\\frac{3}{5},\\ \\sin \\alpha \\equal{}\\frac{4}{5}$", "Solution_6": "I don't understand how to get from here:\r\n[quote=\"kunny\"]\n$ 3\\sin x \\plus{} 4\\cos x \\equal{} 5\\left(\\sin x \\cdot \\frac {3}{5} \\plus{} \\cos x \\cdot\\frac {4}{5}\\right)$\n[/quote]\nto here: \n[quote=\"kunny\"]\n$ \\equal{} 5(\\sin x\\cos \\alpha \\plus{} \\cos x\\sin \\alpha)$\n[/quote]\r\n :maybe:", "Solution_7": "$ \\alpha$ is defined as an angle such that $ \\cos \\alpha \\equal{} \\frac {3}{5}, \\sin \\alpha \\equal{} \\frac {4}{5}$. This is possible because $ \\left( \\frac {3}{5} \\right)^2 \\plus{} \\left( \\frac {4}{5} \\right)^2 \\equal{} 1$. (In particular, $ \\alpha$ is one of the angles in a $ 3 \\minus{} 4 \\minus{} 5$ triangle.)" } { "Tag": [ "calculus", "integration", "rotation", "calculus computations" ], "Problem": "I need help setting this up as an integral:\r\n\r\nFind the volume that are within these bounds:\r\ny=e^(-x), y=0, x=-1, x=0\r\n\r\nRotate about x=1\r\n\r\nThanks!", "Solution_1": "what does the \"about x = 1\" mean?", "Solution_2": "I think that might indicate we're actually talking about the volume of a solid of revolution.\r\n\r\nxboxlive2020, could you clarify your question? Use words, not just shorthand.", "Solution_3": "Fixed the question. Sorry.\r\n\r\nCan someone please move this to the unsolved and proposed problems? Thanks!" } { "Tag": [ "FTW", "algebra unsolved", "algebra" ], "Problem": "Represent the number $ 2008$ as a sum of natural number such that the addition of the reciprocals of the summands yield 1.", "Solution_1": "$ 8\\plus{}10*80\\plus{}30*40\\equal{}2008$\r\n\r\n$ 1/8 \\plus{} (1/80)10 \\plus{} (1/40)30\\equal{}1$\r\n\r\nTrain of Thought:\r\n\r\nTo figure this out, I figured that we know perfect squares work instead of 2008 (very obviously), so the idea was to take a perfect square close to 2008 and manipulate the fractions to give the sum 2008 instead of the perfect square. Then, trying many squares, I came to the realization that $ 40^2\\equal{}1600$ would work out nicely, the fractions would combine nicely since 40 has many factors. So I had $ (1/40)40$ to start with, which gave a sum of 1600. But I needed 2008. So I needed to increase the 1's digit by 8. This came by taking $ 1/40\\plus{}1/40\\plus{}1/40\\plus{}1/40\\plus{}1/40\\equal{}1/8$ which decreased 1600 by 200, then increased by 8. So now I had 1408 and needed to get to 2008. But that's a difference of 600, and every time you split up $ 1/40\\equal{}1/80\\plus{}1/80$ it increases by 120. Since 120 |600, I needed to do this 5 times. And the result then follows.\r\n\r\nWishful Thinking FTW", "Solution_2": "$ 1*(1*4) \\plus{} 4*(4*4)\\plus{} 14*(14*4)\\plus{}17*(17*4)\\equal{}2008$\r\n$ 1/4\\plus{}4/16\\plus{}14/56\\plus{}17/68\\equal{}1$\r\n\r\nRational: $ 2008\\equal{}a*b$ and $ b\\equal{}$ sum of $ a$ squares.\r\n\r\nMore interesting, Which positive integers have such partition (clearly not for 2 or 3)? for a given one (Example: 2008) which partition has minimum number of terms?", "Solution_3": "Every integer greater than 77 has such a partition. It's one of those facts that is often quoted as a property of sufficiently large integers where \"sufficiently large\" has a peculiar precise bound. I remember reading about this some years ago but Googling quickly provides some references, like this one:\r\n\r\nhttp://www.math.uiuc.edu/Bulletin/Abstracts/October/oct31-97analnotheory.html", "Solution_4": "The 77 result is given in Engel, and as such, this is not a very good competition problem.", "Solution_5": "Can anyone show if $ 23$ is or is not a lucky number? Thanks.", "Solution_6": "Finally found the paper that proves that all integers greater than 77 can be partition into distinct positive integers whose reciprocals sum to 1.\r\n[url]http://www.math.ucsd.edu/~fan/ron/papers/63_02_partitions.pdf[/url]" } { "Tag": [], "Problem": "Find exact solutions to $ y^3\\plus{}y^2\\equal{}2y\\plus{}1$", "Solution_1": "Just because I was not lazy and actually looked for an algebraic solution for the 7th roots of unity doesn't mean you have to abuse your privileges! Give credit to the correct people at the times of positive rectitude!\r\n\r\nThe equation $ y^3\\plus{}2y^2\\minus{}y\\minus{}1\\equal{}0$ is totally mine!!!! I claim it!!!!", "Solution_2": "Huh wait what? This is just random, i used random number genereator on my calc for coefficients. Cubic formula mayhaps?", "Solution_3": "I am not in the mood to spam about random topics.\r\n\r\nStupid Problem:\r\n\r\nGiven that $ a,b,c$ are positive real numbers and $ a\\plus{}b\\plus{}c\\equal{}3$, find the minimum value of $ \\sum_{\\text{cyc}}\\frac{a^2}{b^2\\plus{}c^2\\minus{}bc}$.\r\n\r\nEasier version:\r\n\r\nGiven that $ a,b,c$ are positive real numbers and $ a\\plus{}b\\plus{}c\\equal{}3$, find the minimum value of $ \\sum_{\\text{cyc}}\\frac{a^2}{b^2\\plus{}c^2\\minus{}1}$.", "Solution_4": "Whats the cyc at the bottom?", "Solution_5": "Basically, it means you cycle the set $ (a,b,c)\\equal{}(a,b,c)\\equiv(b,c,a)\\equiv(c,a,b)$ and substitute the variables in." } { "Tag": [ "limit", "induction", "blogs", "function", "inequalities", "triangle inequality" ], "Problem": "I'm not sure whether this problem belongs in this section, a mod should move it if necessary.\r\n\r\n$ a_1\\equal{}a_2\\equal{}1, a_n\\equal{} \\frac{1}{a_{n\\minus{}1}} \\plus{} \\frac{1}{a_{n\\minus{}2}}, n \\ge 2$. Find the $ \\lim_{n\\to\\infty}a_n$ and the convergence rate.", "Solution_1": "Substituting the limit $ l$ for $ a_{n}, a_{n\\minus{}1}$ and $ a_{n\\minus{}2}$, we get $ l \\equal{} \\frac{1}{l} \\plus{} \\frac{1}{l}$ so that $ l^{2} \\equal{} 2$ and therefore $ \\lim_{n \\to \\infty} a_{n} \\equal{} \\sqrt{2}$.", "Solution_2": "[quote=\"AndrewTom\"]Substituting the limit $ l$ for $ a_{n}, a_{n \\minus{} 1}$ and $ a_{n \\minus{} 2}$, we get $ l \\equal{} \\frac {1}{l} \\plus{} \\frac {1}{l}$ so that $ l^{2} \\equal{} 2$ and therefore $ \\lim_{n \\to \\infty} a_{n} \\equal{} \\sqrt {2}$.[/quote]\r\n\r\nthat doesn't prove anything, unless you can show that the sequence converges..\r\n\r\n[hide=\" a start\"]\nWe will show that the sequence is bounded by induction\nBase: $ 1 \\le a_1, a_2 \\le 2$\nInductive step: \n$ 1 \\le a_k, a_{k \\minus{} 1} \\le 2 \\Longrightarrow 1 \\ge \\frac {1}{a_k}, \\frac {1}{a_{k \\minus{} 1}} \\ge \\frac {1}{2} \\Longrightarrow 2 \\ge \\frac {1}{a_k} \\plus{} \\frac {1}{a_{k \\minus{} 1}} \\ge 1 \\Longrightarrow 2 \\ge a_{k \\plus{} 1} \\ge 1$\n\nSo by induction $ 1 \\le a_n \\le 2$ for all $ n \\in \\mathbb{N}$\n\nNote: this may not help because the sequence doesn't seem to be monotonic decreasing or incresing (i think it is an oscilating convergance)\n[/hide]", "Solution_3": "I am a bit doubtful but is the convergence rate [hide]$ \\frac {1}{\\sqrt {2}}$?[/hide] :blush:", "Solution_4": "$ a_1\\equal{}a_2\\equal{}1 ,a_3\\equal{}\\frac{1}{1} \\plus{}\\frac{1}{1}\\equal{}2$ and $ a_4\\equal{}\\frac{1}{1} \\plus{}\\frac{1}{2}\\equal{}\\frac{3}{2}$ \r\n\r\nWe see that $ a_2a_4$\r\n\r\nOcha was right , this is oscilating sequence,so it is not convergent (no limit)", "Solution_5": "No, this does converge. Oscillation doesn't imply it has no limits; imagine a sequence that's powers $ \\frac{1}{2^n}$ for even $ n$ and $ \\frac{\\minus{}1}{2^n}$ for odd $ n$.\r\n\r\nI have a [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=297241]blog post[/url] on this. It's not that easy. I got this problem from Problem-Solving Strategies, and Engel happens to omit the hardest part too. :P\r\n\r\nI posted a variant of this problem in the calc forum, too. \r\n\r\nI wouldn't mind if anyone could improve/correct my solution in my blog.", "Solution_6": "Unfortunately, clicking on blue text (within someone's post) does not work on my computer so I don't know what ocha, Potla or Zhero have posted. However, I can offer the following idea:\r\n\r\n$ a_{i}, a_{i \\plus{} 1}$ in $ [a, \\frac {2}{a}]$ (for $ a \\leq \\sqrt {2}) \\implies a_{j}$ in $ [a, \\frac {2}{a}]$ for all $ j \\geq i$ (by substituting in the recurrence).\r\n\r\nI would now like to be able to prove that \r\n\r\n$ a_{i}$ in $ [a, \\frac {2}{a}]$ and $ a_{i \\plus{} 1}$ in $ [a \\plus{} c, \\frac {2}{a} \\minus{} c] \\implies a_{j}$ in $ [a \\plus{} d, \\frac {2}{a} \\minus{} d]$ for some $ d$ a function of $ c$.\r\n\r\nHowever, I get $ [a \\plus{} d', \\frac {2}{a} \\minus{} d'']$ where $ d' \\equal{} \\frac {a^{2}c}{4 \\minus{} 2ac}$ and $ d'' \\equal{} \\frac {c}{a^{2} \\plus{} ac}$\r\n\r\nI hoped that $ d'$ and $ d''$ would be greater than $ c$ to ensure convergence but (unless I've made some errors) they don't seem to be.\r\n\r\nI guess this doesn't work because the sequence oscillates about $ \\sqrt {2}$ and perhaps I should look at $ a_{i \\plus{} 2}$ as well as $ a_{i}$ and $ a_{i \\plus{} 1}$ before considering $ a_{j}$ above as the interval $ [a\\plus{}d', \\frac{2}{a} \\minus{} d'']$ will only get narrower from one end at a time.", "Solution_7": "Yes; the fact that this sequence oscillates so much makes it harder to prove that it converges. \r\n\r\nI'll just copy and paste what I wrote in my blog: \r\n\r\n[hide]\nWith the help of leshik from the calc forums (and from fwolth's random methods) I believe I have found a solution to this following limit: \n\nSuppose $ a_0, a_1 > 0$, and $ a_{n \\plus{} 1} \\equal{} \\frac {1}{a_n} \\plus{} \\frac {1}{a_{n \\minus{} 1}}$. \n\n[hide=\"Step 1\"]If the limit exists, the limit is clearly $ \\sqrt {2}$. So we let $ b_i \\equal{} \\frac {a_i}{\\sqrt {2}}$. Then our recurrence relation becomes $ b_{n \\plus{} 1} \\equal{} \\frac {\\frac {1}{b_n} \\plus{} \\frac {1}{b_{n \\minus{} 1}}}{2}$. Let $ c \\equal{} \\frac {1}{b_i}$ now; we thus have $ c_0, c_1 > 1$, $ c_{i \\plus{} 1} \\equal{} \\frac {2}{c_i \\plus{} c_{i \\minus{} 1}}$. We now wish to show that $ \\lim_{n \\to \\infty} c_n \\equal{} 1$. [/hide]\n\n[hide=\"The hard part...\"]Let $ x$ be the inferior limit of $ c_i$, and $ y$ be the superior limit of $ a_i$. We wish to show that $ x \\geq 1$ and $ y \\leq 1$; then $ 1 \\geq y \\geq x \\geq 1$, so $ x \\equal{} y \\equal{} 1$, as desired. \n\nBecause of fail notation, I will here denote $ a_n$ by $ c_n$. \n\nWe will only show that $ x \\geq 1$; the proof that $ y \\leq 1$ will be similar. By definition, we can find a subsequence $ a_{n_1}$, $ a_{n_2}$, ... that converges to $ x$. By our recurrence definition, we have that $ a_{n_k \\minus{} 1} \\plus{} a_{n_k \\minus{} 2}$ approaches $ \\frac {2}{x}$. \n\nAssume now for the sake of contradiction that $ a_{n_k \\minus{} 1}$ does not converge to $ \\frac {1}{x}$. We claim now that there exists a subseqnece of $ a_{n}$ that converges to a value greater than $ \\frac {1}{x}$. We now consider an arbitrary convergent subsequence of $ a_{n_k \\minus{} 1}$, $ a_{n_{m_k} \\minus{} 1}$ that converges to some real number $ c$. If $ c \\geq \\frac {1}{x}$, we are done (since then $ c > \\frac {1}{x}$, since $ c \\neq \\frac {1}{x}$.) Otherwise, we have $ c < \\frac {1}{x}$. Consider now $ a_{n_{m_k} \\minus{} 2}$; by our recurrence relation, we know this converges to $ \\frac {2}{x} \\minus{} c$ (because $ a_{n_{m_k} \\minus{} 1} \\plus{} a_{n_{m_k} \\minus{} 2}$ converges to $ \\frac {2}{x}$, which is greater than $ \\frac {1}{x}$ since $ c < \\frac {1}{x}$.)\n\nSo now, let $ a_{p_n}$ be the subsequence that converges to $ \\frac {1}{t} > \\frac {1}{x}$. Then we get that $ a_{p_n \\minus{} 1} \\plus{} a_{p_n \\minus{} 2}$ converges to $ 2t$, where $ t < x$. We claim now that there exists a subsequence of $ a_n$ that converges to a value less than or equal to $ t$. \n\nWe have two cases: there are infintely many terms in $ a_{p_n \\minus{} 1}$ and $ a_{p_n \\minus{} 2}$ such that both terms are greater than or equal to $ t$, or that there are only finitely many such terms. In the first case, we simply pick the terms greater than $ t$ in $ a_{p_n \\minus{} 1}$ as a subsequence; we easily see that it is necessary that both sequences must converge to $ t$ (they are all positive, they are always greater than $ t$, and their sum is bounded above by something that converges to $ 2t$.) Otherwise, we consider a subsequence that takes terms from both $ a_{p_n \\minus{} 1}$ and $ a_{p_n \\minus{} 2}$, such that every term in that subsequence is less than $ t$ (there are infinitely many such terms in the union of these sequences, by assumption.) Thus, all convergent subsequences of this sequence converge to a value less than $ t$; we can find such a subsequence, so we are done. \n\nBut now, we have that some subsequence converges to $ t < x$, which contradicts our assumption that $ x$ is the infimum. It thus follows that $ a_{n_k \\minus{} 1}$ must converge to $ \\frac {1}{x}$, and thus $ a_{n_k \\minus{} 2}$ must also converge to $ \\frac {1}{x}$. But then, by our recurrence relation, we see that $ a_{n_k \\minus{} 3}$ converges to $ 2x \\minus{} \\frac {1}{x}$. Since $ x$ is the infimum, we have $ 2x \\minus{} \\frac {1}{x} \\geq x$, so $ x \\geq 1$, as desired.\n\nMeh this is probably a bit more complicated than it needs to be...\n[/hide]\n[/hide]", "Solution_8": "Thanks Zhero. Unfortunately the only blue text on this page that does anything on my computer is blog post. Once there, clicking on (post your comment) works but clicking on Step 1 or The hard part ... has no effect. I've thought of modifying what I posted but the fact that the sequence doesn't alternate but oscillates in a way I can't predict makes it difficult for me.\r\n\r\nP.S. I have a copy of Arthur Engel's Problem-Solving Strategies and have looked at pages 225 and 231. I had earlier considered the sequence $ \\frac{1}{a_{n}}$ but got nowhere with it.", "Solution_9": "EDIT: I've cleaned up the solution\r\n\r\n[hide=\"solution\"]\n[b]Lemma 1[/b] if $ a_n,a_{n - 1} \\in \\left[\\frac {2}{c_n},c_n\\right]$ then $ a_m \\in \\left[\\frac {2}{c_n},c_n\\right]$ for all $ m \\ge n$ \n[hide=\"proof\"] From the condition \n$ \\frac {2}{c_n} \\le a_{k}, a_{k - 1}\\le c_n\\Longrightarrow \\frac {c_n}{2}\\ge\\frac {1}{a_{k}},\\frac {1}{a_{k - 1}}\\ge\\frac {1}{c_n}\\Longrightarrow c_n\\ge\\frac {1}{a_{k}} + \\frac {1}{a_{k - 1}}\\ge \\frac {2}{c_n}\\Longrightarrow c_n\\ge a_{k + 1}\\ge \\frac {2}{c_n}$ \nSo by the inductive hypothesis, the statement is true for all $ n\\in \\mathbb{N}$ [/hide]\n\nWe wish to find a decreasing family $ \\left[\\frac {2}{c_n},c_n\\right]$ that traps the higher order elements of $ a_n$.\nSo let $ c_n = \\max\\left(a_n,a_{n - 1},\\frac {2}{a_n},\\frac {2}{a_{n - 1}}\\right)$\n\nThe choice of $ c_n$ assures that $ a_{n - 1},a_{n} \\in \\left[\\frac {2}{c_n},c_n\\right]$, and hence by our lemma $ a_m \\in \\left[\\frac {2}{c_n},c_n\\right]$ for all $ m\\ge n$. It follows that $ c_n$ is a monotonicly decreasing sequence. \n\nIn addition $ c_n$ is bounded below by $ \\sqrt {2}$ because clearly $ \\max\\left(a_n,\\frac {2}{a_n}\\right) \\ge \\sqrt {2}$ so by the [b]monotonic convergence theorem[/b] the sequence $ \\{c_n\\}$ converges $ (*)$.\n\n\nNow we use inequalities to show how $ |c_{n + 2} - c_{n}|$ varies with $ a_n$\n\n[hide=\"one\"]$ \\max(a_n,a_{n - 1}) \\ge \\frac {a_n + a_{n - 1}}{2} \\ge \\frac {2}{\\frac {1}{a_n} + \\frac {1}{a_{n - 1}}} = \\frac {2}{a_{n + 1}}$\n\n$ \\therefore \\max(a_n,a_{n - 1}) - \\frac {2}{a_{n + 1}} \\ge \\max(a_n,a_{n - 1}) - \\frac {a_n + a_{n - 1}}{2} = \\left|\\frac {a_n - a_{n - 1}}{2}\\right| \\qquad(1)$ [/hide]\n\n[hide=\"two\"]$ \\max\\left(\\frac {2}{a_n},\\frac {2}{a_{n - 1}}\\right) \\ge \\frac {1}{a_n} + \\frac {1}{a_{n - 1}} = a_{n + 1}$\n\n$ \\therefore \\max\\left(\\frac {2}{a_n},\\frac {2}{a_{n - 1}}\\right) - a_{n + 1} \\ge \\left|\\frac {1}{a_n} - \\frac {1}{a_{n - 1}}\\right|\\qquad(2)$ [/hide]\n\n\nIt follows from $ (1)$ and $ (2)$ that $ c_n - \\max(a_{n + 1},\\frac {2}{a_{n + 1}}) \\ge \\min\\left(\\frac {|a_n - a_{n - 1}|}{2}, \\left|\\frac {1}{a_n} - \\frac {1}{a_{n - 1}}\\right|\\right)$\n\nReplacing $ a_n$ with $ a_{n + 1}$ in the above and putting everything in terms of $ c_n$ gives\n\n$ c_n - c_{n + 2} \\ge \\min\\left(\\frac {|a_n - a_{n - 1}|}{2}, \\left|\\frac {1}{a_n} - \\frac {1}{a_{n - 1}}\\right|,\\frac {|a_{n + 1} - a_{n}|}{2}, \\left|\\frac {1}{a_{n + 1}} - \\frac {1}{a_{n}}\\right|\\right) \\qquad(**)$\n\n\nWe know from $ (*)$ that $ c_n$ converges so by Cauchy's Lemma, $ |c_{n + 2} - c_n| \\to 0$ as $ n\\to \\infty$. So from $ (**)$ we have ${ \\min (|a_n - a_{n - 1}|,|a_{n + 1} - a_n}|)\\to 0$ has $ n\\to \\infty$\n\nIt is now easy to show using the triangle inequality that $ |a_n - a_{n - 1}|\\to 0$ as $ n\\to \\infty$ and it's been proven that $ a_n$ converges!\n\nNow let $ \\lim_{n\\to \\infty} a_n \\to a$ and $ a = \\frac {1}{a} + \\frac {1}{a} \\Rightarrow a = \\sqrt {2}$\n\n$ \\lim_{n\\to \\infty} a_n \\to \\sqrt {2}$ \n[/hide]" } { "Tag": [ "superior algebra", "superior algebra unsolved" ], "Problem": "G finite with p| |G| (p=prime) Prove that\r\n\r\np| |{g in G: g^p =1}|\r\n\r\nand\r\n\r\n|{H 0$ then wlog, $x,y > 1$ but then $x^2+y^2 > 5 + (x-1)*(y-1)$" } { "Tag": [], "Problem": "$ x \\plus{} {1\\over y} \\equal{} 4 ,\\quad y \\plus{} {1\\over z} \\equal{} 1 ,\\quad z \\plus{} {1\\over x} \\equal{} {3 \\over 7}$\r\n\r\nFind $ xyz \\equal{}$?", "Solution_1": "$ (x\\plus{}\\frac{1}{y})(y\\plus{}\\frac{1}{z})(z\\plus{}\\frac{1}{x})\\equal{}(x\\plus{}\\frac{1}{y})\\plus{}(y\\plus{}\\frac{1}{z})\\plus{}(z\\plus{}\\frac{1}{x})\\plus{}xyz\\plus{}\\frac{1}{xyz}$.\r\nSo, $ 4\\cdot1\\cdot\\frac{3}{7}\\equal{}4\\plus{}1\\plus{}\\frac{3}{7}\\plus{}xyz\\plus{}\\frac{1}{xyz} \\iff \\frac{12}{7}\\equal{}\\frac{38}{7}\\plus{}xyz\\plus{}\\frac{1}{xyz} \\iff xyz\\plus{}\\frac{1}{xyz}\\equal{}\\minus{}\\frac{26}{7}$.\r\nThus, $ xyz\\plus{}\\frac{1}{xyz}\\equal{}\\minus{}\\frac{26}{7} \\Rightarrow (xyz)^2\\plus{}\\frac{26}{7}xyz\\plus{}1\\equal{}0 \\iff xyz\\equal{}\\boxed{\\frac{\\minus{}13\\pm2\\sqrt{30}}{7}}$.", "Solution_2": "[quote=\"Kouichi Nakagawa\"]$ (x \\plus{} \\frac {1}{y})(y \\plus{} \\frac {1}{z})(z \\plus{} \\frac {1}{x}) \\equal{} (x \\plus{} \\frac {1}{y}) \\plus{} (y \\plus{} \\frac {1}{z}) \\plus{} (z \\plus{} \\frac {1}{x}) \\plus{} xyz \\plus{} \\frac {1}{xyz}$.\nSo, $ 4\\cdot1\\cdot\\frac {3}{7} \\equal{} 4 \\plus{} 1 \\plus{} \\frac {3}{7} \\plus{} xyz \\plus{} \\frac {1}{xyz} \\iff \\frac {12}{7} \\equal{} \\frac {38}{7} \\plus{} xyz \\plus{} \\frac {1}{xyz} \\iff xyz \\plus{} \\frac {1}{xyz} \\equal{} \\minus{} \\frac {26}{7}$.\nThus, $ xyz \\plus{} \\frac {1}{xyz} \\equal{} \\minus{} \\frac {26}{7} \\Rightarrow (xyz)^2 \\plus{} \\frac {26}{7}xyz \\plus{} 1 \\equal{} 0 \\iff xyz \\equal{} \\boxed{\\frac { \\minus{} 13\\pm2\\sqrt {30}}{7}}$.[/quote]\r\n\r\nYou have to show that both answers work, i.e. there are two distinct answers .", "Solution_3": "From 1st, 2nd equation, we have $ x\\equal{}\\frac{4y\\minus{}1}{y},\\ z\\equal{}\\frac{1}{1\\minus{}y}$.\r\n\r\n$ \\therefore z\\plus{}\\frac{1}{x}\\equal{}\\frac{3}{7}\\Longleftrightarrow \\frac{1}{1\\minus{}y}\\plus{}\\frac{y}{4y\\minus{}1}\\equal{}\\frac{3}{7}\\ \\cdots (*)$\r\n\r\n$ \\Longleftrightarrow 5y^2\\plus{}20y\\minus{}4\\equal{}0$\r\n\r\n$ \\Longleftrightarrow (5y)^2\\plus{}20\\cdot (5y)\\minus{}4\\equal{}0$\r\n\r\n$ \\Longleftrightarrow 5y\\equal{}\\minus{}10\\pm 2\\sqrt{30}$, thus from $ (*)$, we have $ xyz\\equal{}\\frac{4y\\minus{}1}{y}\\cdot y\\cdot \\frac{1}{1\\minus{}y}$\r\n\r\n$ \\equal{}\\frac{4y\\minus{}1}{1\\minus{}y}\\equal{}\\frac{3}{7}(4y\\minus{}1)\\minus{}y\\equal{}\\frac{5y\\minus{}3}{7}\\equal{}\\boxed{\\frac{\\minus{}13\\pm 2\\sqrt{30}}{7}}.$" } { "Tag": [ "geometry", "similar triangles" ], "Problem": "In a right triangle with sides $ a$ and $ b$, and hypotenuse $ c$, the altitude drawn on the hypotenuse is $ x$. Then:\r\n\r\n$ \\textbf{(A)}\\ ab \\equal{} x^2 \\qquad\\textbf{(B)}\\ \\frac {1}{a} \\plus{} \\frac {1}{b} \\equal{} \\frac {1}{x} \\qquad\\textbf{(C)}\\ a^2 \\plus{} b^2 \\equal{} 2x^2$\r\n$ \\textbf{(D)}\\ \\frac {1}{x^2} \\equal{} \\frac {1}{a^2} \\plus{} \\frac {1}{b^2} \\qquad\\textbf{(E)}\\ \\frac {1}{x} \\equal{} \\frac {b}{a}$", "Solution_1": "hello, after a well-known fact we have $ c \\cdot x \\equal{} a\\cdot b$, from here we get $ x^2 \\equal{} \\frac {a^2b^2}{c^2} \\equal{} \\frac {a^2b^2}{a^2 \\plus{} b^2} \\equal{} \\frac {1}{\\frac {1}{a^2} \\plus{} \\frac {1}{b^2}}$ and $ \\frac {1}{x^2} \\equal{} \\frac {1}{a^2} \\plus{} \\frac {1}{b^2}$. So the answer is $ \\text{D}$.\r\nSonnhard.", "Solution_2": "[quote=\"Dr Sonnhard Graubner\"]after a well-known fact we have $ c\\cdot x \\equal{} a\\cdot b$[/quote]What is that well known fact?\r\nOh yeah areas.\r\nBut please don't just say that it's a well known fact without stating what fact it is.\r\n[hide=\"outline\"]similar triangles destroys this problem[/hide]", "Solution_3": "hello, yes, we have $ \\frac{c \\cdot x}{2}\\equal{}\\frac{a\\cdot b}{2}$, sorry for my well-known.\r\nSonnhard." } { "Tag": [ "inequalities", "trigonometry", "inequalities proposed" ], "Problem": "Prove that if $ A, B,C$ are angles of a triangle then\r\n\\[ \\cos^3A\\plus{}\\cos^3B\\plus{}\\cos^3C\\plus{}5\\cos A\\cos B\\cos C\\leq1.\\]", "Solution_1": "[quote=\"hitek\"]Prove that if $ A, B,C$ are angles of a triangle then\n\\[ \\cos^3A\\plus{}\\cos^3B\\plus{}\\cos^3C\\plus{}5\\cos A\\cos B\\cos C\\leq1.\\][/quote]\n\n\nProve or disprove the following inequality:\nIf $ a>0,c>0,b>0, a ^ 2 + b ^ 2 + c ^ 2 + abc = 4 $,\n then $ a ^ 3 + b ^ 3 + c ^ 3 +5 abc\\le\\ 8 $\n_______________________________________\nSandu Marin, Bucuresti, Romania", "Solution_2": "[quote=\"hitek\"]Prove that if $ A, B,C$ are angles of a triangle then\n\\[ \\cos^3A\\plus{}\\cos^3B\\plus{}\\cos^3C\\plus{}5\\cos A\\cos B\\cos C\\leq1.\\][/quote]\n\n\n\\[1-\\cos^3{A}-\\cos^3{B}-\\cos^3{C}-5\\cos{A}\\cos{B}\\cos{C} = \\frac{1}{4}\\frac{Q}{R^3}.\\]\n\n$Q=(3r-5R)s^2-Rr^2+20R^3+8R^2r-r^3.$\n\n\n$y=4R^2+4Rr+3r^2-s^2-r^2\\frac{R-2r}{R-r}\\geq 0.$ is yang xue zhi INEQ.\n\nNote this:\n\n\\[r^2(R+r)\\frac{R-2r}{R-r}+y(-3r+5R)=Q.\\]\n\nand done.", "Solution_3": "[quote=sandumarin][quote=\"hitek\"]Prove that if $ A, B,C$ are angles of a triangle then\n\\[ \\cos^3A\\plus{}\\cos^3B\\plus{}\\cos^3C\\plus{}5\\cos A\\cos B\\cos C\\leq1.\\][/quote]\n\n\nProve or disprove the following inequality:\nIf $ a>0,c>0,b>0, a ^ 2 + b ^ 2 + c ^ 2 + abc = 4 $,\n then $ a ^ 3 + b ^ 3 + c ^ 3 +5 abc\\le\\ 8 $\n[/quote]\nGiven inequality is true for all triangle. ;) \nBy the way, the inequality $\\sum_{cyc}\\frac{(a^2+b^2-c^2)^3}{8a^3b^3}+5\\prod_{cyc}\\frac{a^2+b^2-c^2}{2ab}\\leq1$ is true for all positives $a$, $b$ and $c$.\n\n", "Solution_4": "[quote=hitek]Prove that if $ A, B,C$ are angles of a triangle then\n\\[ \\cos^3A\\plus{}\\cos^3B\\plus{}\\cos^3C\\plus{}5\\cos A\\cos B\\cos C\\leq1.\\][/quote]\n\nLet $\\cos\\alpha=\\frac{x-2}{2}$, $\\cos\\beta=\\frac{y-2}{2}$ and $\\cos\\gamma=\\frac{z-2}{2}$, $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.\nHence, $\\{x,y,z\\}\\subset(0,4)$.\nSince $\\cos\\alpha+\\cos\\beta+\\cos\\gamma\\geq1$, we obtain $u\\geq\\frac{8}{3}$.\nIn another hand, $\\cos^2\\alpha+\\cos^2\\beta+\\cos^2\\gamma+2\\cos\\alpha\\cos\\beta\\cos\\gamma=1$ gives $2(xy+xz+yz)-x^2-y^2-z^2=xyz$ \nor $w^3=3(4v^2-3u^2)$ and we need to prove that $\\sum_{cyc}\\frac{(x-2)^3}{8}+\\frac{5(x-2)(y-2)(z-2)}{8}\\leq1$ or\n$\\sum_{cyc}(x^3-6x^2-10xy+32x)+5xyz\\leq72$ or $\\sum_{cyc}(x^3-11x^2+32x)\\leq72$ or $f(w^3)\\geq0$, where\n$f(w^3)=72\\left(\\frac{w^3}{3(4v^2-3u^2)}\\right)^3-96u\\left(\\frac{w^3}{3(4v^2-3u^2)}\\right)^2+11(9u^2-6v^2)\\cdot\\frac{w^3}{3(4v^2-3u^2)}-27u^3+27uv^2-3w^3$.\nBut $f''(w^3)=\\frac{16w^3}{(4v^2-3u^2)^3}-\\frac{64u}{3(4v^2-3u^2)^2}=\\frac{48}{(4v^2-3u^2)^2}-\\frac{64u}{3(4v^2-3u^2)^2}\\leq\\frac{48}{(4v^2-3u^2)^2}-\\frac{512}{9(4v^2-3u^2)^2}<0$.\nId est, $f$ is a concave function.\nHence, $f$ gets a minimal value for an extremal value of $w^3$, which happens in the following cases:\n1. $w^3\\rightarrow0^+$. \nLet $x\\rightarrow0$. Hence, $\\alpha\\rightarrow180^{\\circ}$ and $\\beta=\\gamma\\rightarrow0^{\\circ}$. We see that in this case our inequality is true;\n2. $y=x$, which gives $\\beta=\\alpha$ and $\\gamma=180^{\\circ}-2\\alpha$, which gives $(2\\cos\\alpha-1)^2(2\\cos^2\\alpha+2\\cos\\alpha+1)\\cos^2\\alpha\\geq0$.\n Done! :)" } { "Tag": [ "calculus", "calculus computations" ], "Problem": "if i have $\\alpha_{n}\\to \\alpha$ as $n\\to \\infty$\r\nthen why is\r\n$ [-\\infty,\\alpha]=\\cup_{n=1}^{\\infty}[-\\infty,\\alpha_{n}]$\r\n\r\nshouldn't it be intersection ?", "Solution_1": "Please post such problems in the calculus forum next time, it's not complex analysis.\r\n\r\nIt's union if the sequence $\\alpha_{n}$ is always $\\leq \\alpha$, it's intersection if always $\\geq \\alpha$. So it depends, and none is correct if the sequence is arbitrary.", "Solution_2": "forgot to mention\r\n$\\alpha_{n}< \\alpha$\r\n\r\nzetax you're right, sorry,it was obvious" } { "Tag": [], "Problem": "a car travels 30 miles per gallon of gas. ever gallon of gas is $ \\$$4.(yes thats where its headed). a certain toll road is 120 miles long and costs 6 dollars to drive on it. another free road is x miles long.if both ways require the same amount of money(including gas), what is x?\r\n\r\ni made this up on the spot so it may not make sense but im kinda out of it so i cant really tell...", "Solution_1": "Is it\r\n[hide] 165? [/hide]", "Solution_2": "[hide]\nTaking the toll road means that you spend 6+16=22 dollars\nso 22 dollars is what you spend on gas meaning you use 22/4 = 5.5 gallons of gas\n5.5*30=165[/hide]", "Solution_3": "[quote=\"abacadaea\"]a car travels 30 miles per gallon of gas. ever gallon of gas is $ \\$$4.(yes thats where its headed). a certain toll road is 120 miles long and costs 6 dollars to drive on it. another free road is x miles long.if both ways require the same amount of money(including gas), what is x?\n\ni made this up on the spot so it may not make sense but im kinda out of it so i cant really tell...[/quote]\r\nHere's my solution:\r\n[hide]The 120 mile road consumes 4 gallons of gas, so it would take $ \\$$16 for gas mileage. It also costs $ \\$$6 more to drive on it, so the cost of driving on that road is $ \\$$16 + $ \\$$6=$ \\$$22. So the $ x$-mile road, which is free to drive on, requires $ 22/4$ = $ 5.5$ gallons of gas, so $ x$ = $ 30 * 5.5$ = $ 165$ miles.[/hide]", "Solution_4": "[hide]\n\ncosts 6+120/30*4=22 dollars so\n\n22/4*30=165 =)\n\n[/hide]", "Solution_5": "I got [hide]165[/hide], just to confirm your answer." } { "Tag": [ "superior algebra", "superior algebra unsolved" ], "Problem": "Let $G$ be a finite group and $\\phi: G\\to C^\\times$ be a non trivial homomorphism. Show that\r\n\\[\\sum_{g\\in G}\\phi(g)=0\\]", "Solution_1": "Also note the dual result:\r\n\\[\\sum_{\\phi \\in G^{*}}\\phi(g) = 0 \\]\r\nwith $G^{*}$ being the dual group of $G$ (homomorphisms from $G$ to $\\mathbb C^{*}$) and $g \\in G$, $g$ nontrivial.\r\n\r\nBoth are proven by multiplying with $\\phi(g) \\neq 1$ (here each is one fixed).", "Solution_2": "[quote=\"ZetaX\"]A\nBoth are proven by multiplying with $\\phi(g) \\neq 0$ (here each is one fixed).[/quote]\r\n\r\nCan you elaborate more, please?", "Solution_3": "Sorry, it's $\\neq 1$, not $\\neq 0$:\r\n\r\nFor the first, take a $h \\in G$ with $\\phi(h) \\neq 1$.\r\nThen we get $\\sum_{g\\in G}\\phi(g)=\\sum_{g\\in G}\\phi(hg)=\\sum_{g\\in G}\\phi(h)\\phi(g)=\\phi(h) \\cdot \\sum_{g\\in G}\\phi(g)$, thus $(1-\\phi(h)) \\cdot \\sum_{g\\in G}\\phi(g) =0$, giving the result.\r\nFor the second, take a homomorphism $\\psi$ with $\\psi(g) \\neq 1$ and use the same idea.", "Solution_4": "Thanks, Zetax" } { "Tag": [ "calculus", "derivative", "calculus computations" ], "Problem": "we just started calculus this year, and we done chain, product, quotient, power rules, implicit diff, second derivative...and i cant get some of the questions assigned to cover all these\r\n\r\nFind the second derivative d^2y / dx^2\r\n\r\nx^2-3x+y^2=0\r\n\r\nx^3+y^3=3", "Solution_1": "I'll do the second one only, because it's quicker.\r\n\r\nLet $y'=\\frac{dy}{dx}$.\r\n\r\nWe have $x^3+y^3=3$, so $3x^2+3y^2y'=0$, or $y'=-\\left(\\frac xy\\right)^2$\r\n\r\nDifferentiating again, we have \r\n\r\n$6x+3y^2y''+6y(y'^2)=0$. We already know what $y'$ is - substitute in. Then solve for $y''$ to finish.", "Solution_2": "hm, i seemed to forgot to substitute y', lol...damn, wtf was i thinking...anyway, thx a lot! i just got the answer for the second derivative, one of the signs is opposite than the answer in the back of the book, but im sure i'll figure something out...THX a lot" } { "Tag": [ "LaTeX", "quadratics", "algebra", "polynomial", "Rational Root Theorem" ], "Problem": "Once I found this on a math site: A man goes into a shop and buys 4 things when the shopkeeper calculates the bill he stars to multiply the prices and obtains 19.7$. The customer tells him that he calculated wrong so he added the prices this time and he obtained 19.7$. So the question is which were the prices?\r\n The problem resumes to calculating the sistem:\r\na+b+c+d=19.7\r\na*b*c*d=19.7 \r\n Is there a special method to solve it or one must ''guess\" the answer?\r\n Can a computer programe be written to solve it, and other similar problems?", "Solution_1": "[quote=\"Alicea\"]Once I found this on a math site: A man goes into a shop and buys 4 things when the shopkeeper calculates the bill he stars to multiply the prices and obtains $19.7$. The customer tells him that he calculated wrong so he added the prices this time and he obtained $19.7$. So the question is which were the prices?\n The problem resumes to calculating the system:\n$a+b+c+d=19.7$\n$a \\cdot b \\cdot c \\cdot d=19.7$ \n Is there a special method to solve it or one must ''guess\" the answer?\n Can a computer programe be written to solve it, and other similar problems?[/quote]\r\n\r\ncareful w/ your dollar signs :wink:", "Solution_2": "Well, they're prices, so they all have to take the form of an integer divided by $100$.", "Solution_3": "Forgive me I'm new around so I didn't know what signs from the keyboard cause problems :blush: .", "Solution_4": "If you know how to use them, they don't cause problems.\r\n\r\nGenerally, to make a dollar sign, type:\r\n[code]$ \\$ $[/code]\r\n\r\nThis will make:\r\n$\\$$\r\n\r\nDollar signs represent the start of an equation, expression, etc. written in $\\LaTeX$.", "Solution_5": "[hide=\"not quite a solution\"]Let a, b, c, and d be roots of a quartic.\n\n$x^{4}-19.7x^{3}+dx^{2}+ex+19.7=0$\n\n\n$d=ab+ac+ad+bc+bd+cd$\n\n$e=abc+abd+acd+bcd$\n\nFrom the rational root theorem, the roots must be \"factors\" of +-19.7. But 197 is prime, so I think I did something wrong somewhere....[/hide]", "Solution_6": "You can't apply the rational root theorem with non-integer coefficients. In fact, even after you've multiplied by $10$ you don't know whether your other two coefficients (don't call one of them $d$ when you've used that letter already!) are integer. \r\n\r\nReally, the rational root theorem doesn't tell you anything you didn't already know.", "Solution_7": "oooooooooooooooops! :oops: \r\n\r\n\r\nWell, 19.7 is the sum and product of the roots of the quadratic, so I got that part right." } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "Compute the sums: $ \\displaystyle\\sum_{i\\equal{}0}^{n}\\frac{\\binom{k\\plus{}i}{k}}{k\\plus{}i}$ and $ \\displaystyle\\sum_{i\\equal{}0}^{n} \\frac{\\binom{k\\plus{}i}{k}}{k\\plus{}i\\minus{}1}, n,k \\in \\mathbb{N}, k \\ge 2.$", "Solution_1": "For the first, write $ \\frac {\\binom{k \\plus{} i}{k}}{k \\plus{} i} \\equal{} \\frac {1}{k} \\cdot \\binom{k \\plus{} i \\minus{} 1}{k \\minus{} 1}$; for the second, write $ \\frac {\\binom{k \\plus{} i}{k}}{k \\plus{} i \\minus{} 1} \\equal{} \\frac {1}{k} \\binom{k \\plus{} i \\minus{} 1}{k \\minus{} 1} \\plus{} \\frac {1}{k} \\cdot \\frac {\\binom{k \\plus{} i \\minus{} 1}{k \\minus{} 1}}{k \\plus{} i \\minus{} 1}$ and then repeat the same idea again with the second summand.\r\n\r\nIn both cases, finish with $ \\sum_{i \\equal{} 0}^n \\binom{m \\plus{} i}{i} \\equal{} \\binom{m \\plus{} n \\plus{} 1}{n}$." } { "Tag": [], "Problem": "I just came up with this a little bit ago as I was playing around with some infinite sequences.\r\n\r\nGiven: $y=x+x+x+x+x+x+x+x+x+x+x+x+x+x+x+x+x+x...$\r\n\r\nSolve for x.\r\n\r\n$y=x+x+x+x+x+x+x+x+x+x+x+x+x+x+x+x+x+x...$\r\nGiven\r\n$y=x+y$\r\nSubsitution\r\n$0=x$\r\nSubtraction.\r\n\r\nThus any number added infinately is equal to zero, or even more simply, any number is equal to zero. Could this be an extraneous solution?\r\n\r\nIt doesn't seem right, and it kind of reminds me of Zeno's Paradox... What are your thoughts on this?", "Solution_1": "Hmm\r\nWont y+x always be x greater than y so they cannot be equal even if it's an infinite sum?\r\nSo the only time it will work is when x = 0.", "Solution_2": "True, but thats the thing. If you are to look at it that way, other problems that use this message of substitution are similarly flawed.", "Solution_3": "When $x=0$, then what you have is correct. When $x\\ne0$, you have basically (using symbolism liberally) $\\infty=x+\\infty,$ where $x$ is a real number, but this does not imply $x=0$. You have to be careful when you're dealing with infinite sums.", "Solution_4": "But I want the reasoning for why this is false.\r\n\r\nBecause you can put almost any number as x and get the same exact thing.\r\n\r\n$y=5+5+5+5+5+...$\r\n$y=5+y$\r\n$0=5$", "Solution_5": "[quote=\"Farazy\"]But I want the reasoning for why this is false.\n\nBecause you can put almost any number as x and get the same exact thing.\n\n$y=5+5+5+5+5+...$\n$y=5+y$\n$0=5$[/quote]$y$ is not number that actually has a value. You're attempting to do arithmetic with infinitely large numbers. \r\n\r\nThe algebraic principles that you're using to derive $x=0$ (namely subtracting $y$ from both sides) do not apply to infinite numbers. For example, if $\\aleph_0$ is the cardinality of the set of all integers, then $\\aleph_0=\\aleph_0+C$, where $C$ is any real number, but certainly this does not imply $C=0$. The algebraic rule that you may use to derive $C=0$ simply does not apply.", "Solution_6": "I also came up with something else that proves what I said wrong...\r\n\r\n$y=x+x+x...$\r\n\r\nWe can subsitute y in like before.\r\n\r\n$y=x+y$ but we notice that y is x more then that...So:\r\n\r\n$y=x+(y-x)$\r\n\r\nWhich goes back to\r\n\r\n$y=y$", "Solution_7": "This \"paradox\" is also closely related to the Infinite Hotel paradox.\r\n\r\nA hotel has a (countably) infinite number of rooms, numbered 1, 2, 3, etc. The hotel is currently full -- every room has an occupant.\r\n\r\nA man shows up at the front desk asking for a room. The manager regretfully informs the man that the hotel is full. The man then says, \"Well, you can free up a room for me by just telling everybody to move to the room that's one higher!\" So the person in Room #1 moves to Room #2, the person in Room #2 moves to Room #3, etc., and then Room #1 is empty for the man to move into.\r\n\r\nIs this a paradox?", "Solution_8": "what does \"countably infinite\" mean? I thought you can't count infinity?", "Solution_9": "[quote=\"DPatrick\"]This \"paradox\" is also closely related to the Infinite Hotel paradox.\n\nA hotel has a (countably) infinite number of rooms, numbered 1, 2, 3, etc. The hotel is currently full -- every room has an occupant.\n\nA man shows up at the front desk asking for a room. The manager regretfully informs the man that the hotel is full. The man then says, \"Well, you can free up a room for me by just telling everybody to move to the room that's one higher!\" So the person in Room #1 moves to Room #2, the person in Room #2 moves to Room #3, etc., and then Room #1 is empty for the man to move into.\n\nIs this a paradox?[/quote]\r\n\r\nWait, but where does that leave the person in the highest numbered room...where does he move into?", "Solution_10": "What highest number room?\r\n\r\nThere are an infinite number of rooms.", "Solution_11": "So, how is that countable? The hotel can't claim to be full ever - actually, practically the hotel would never know if it was full or not.", "Solution_12": "\"Countably infinite\" alludes to the fact that there are different sizes of infinite sets.\r\n\r\nBasically, an infinite set is \"countably infinite\" (or just \"countable\") if we can count it 1, 2, 3, 4, ..., or more precisely, it can be put into 1-to-1 correspondence with the positive integers.\r\n\r\nSo in the problem I posted above, the hotel has a Room #n for every positive integer n. So indeed it has an infinite number of rooms.\r\n\r\nOther infinite sets, like the set of all points on a line segment, are \"uncountably infinite\" (or just \"uncountable\").\r\n\r\nAll of this is explained quite nicely in Section 17.7 of AoPS Volume 2 (which is inside the chapter titled \"Counting in the Twilight Zone\").", "Solution_13": "[quote=\"Farazy\"]But I want the reasoning for why this is false.\n\nBecause you can put almost any number as x and get the same exact thing.\n\n$y=5+5+5+5+5+...$\n$y=5+y$\n$0=5$[/quote]\r\n\r\nIt dosnt tend to a value, it just get bigger, so you cant use that technique with these kind of problems." } { "Tag": [ "integration", "calculus", "algebra", "polynomial", "real analysis", "real analysis unsolved" ], "Problem": "$P_{1}(x),P_{2}(x),P_{3}(x)$ and $P_{4}(x)$ are polynomials. Prove that \r\n\r\n$\\int^{x}_{-1}P_{1}(t)P_{3}(t)dt\\int^{x}_{-1}P_{2}(t)P_{4}(t)dt-\\int^{x}_{-1}P_{1}(t)P_{4}(t)dt\\int^{x}_{-1}P_{2}(t)P_{3}(t)dt$\r\n\r\nis divisible by $(x+1)^{4}$.", "Solution_1": "Denote\r\n$Q(x)=\\int^{x}_{-1}P_{1}(t)P_{3}(t)dt\\int^{x}_{-1}P_{2}(t)P_{4}(t)dt-\\int^{x}_{-1}P_{1}(t)P_{4}(t)dt\\int^{x}_{-1}P_{2}(t)P_{3}(t)dt$\r\n\r\nIt's very easy to check that $Q(-1)=Q'(-1)=Q''(-1)=Q'''(-1)=0$", "Solution_2": "I liked the problem. Easy to check but still a neat construction. \r\n\r\nNew problem:\r\nConstruct an integral equation in terms of n general polynomials divisible by (x + a)^n.", "Solution_3": "Any solution of the question of googleplex ?", "Solution_4": "15 year bump\n\nwow" } { "Tag": [], "Problem": "Three girls and two boys are on the Math Team. [Note: Names do not indicate gender.]\r\n\r\nTwo students wear white shirts; three wear black shirts.\r\nMiken and Carry wear shirts of different colors.\r\nBari and Jamie wear shirts of the same color.\r\nPita and Carry are the same gender.\r\nJamie and Miken are of different genders.\r\n\r\nThe boy in the white shirt scored the most points. What is his name?\r\n\r\na. Bari b. Carry c. Jamie d. Miken e. Pita", "Solution_1": "[hide=\" click to reveal hidden content\"]\n\nI think that Bari and Miken are boys and that Carry, Jamie and Pita are girls; Miken and Pita wear white, Bari, Carry and Jamies wear black. So Miken scored the most points.\n\n[/hide]" } { "Tag": [ "induction", "modular arithmetic", "algebra", "binomial theorem", "algebra unsolved" ], "Problem": "Let $ \\left(x_{n}\\right)$ be a real sequence satisfying $ x_{0}=0$, $ x_{2}=\\sqrt[3]{2}x_{1}$, and $ x_{n+1}=\\frac{1}{\\sqrt[3]{4}}x_{n}+\\sqrt[3]{4}x_{n-1}+\\frac{1}{2}x_{n-2}$ for every integer $ n\\geq 2$, and such that $ x_{3}$ is a positive integer. Find the minimal number of integers belonging to this sequence.", "Solution_1": "First, let's simplify everything a little by writing $a = \\sqrt[3]{2}$ so we keep in mind that $a^3 = 2$.\r\n\r\n$x_{n+1} = \\frac{ ax_n + 2a^2 x_{n-1} + x_{n-2} }{2}$\r\n\r\nSo we can find that\r\n\r\n$x_3 = \\frac{ a^2 x_1 + 2a^2 x_1 + 0}{2} = \\frac{3a^2}{2} x_1$\r\n$x_1 = \\frac{2}{3a^2} x_3 = \\frac{a}{3} x_3$ for our integer $x_3$.\r\n\r\nWrite the sequence in terms of $x_3$.\r\n\r\n$x_0 = 0$\r\n$x_1 = \\frac{a}{3} x_3$\r\n$x_2 = \\frac{a^2}{3} x_3$\r\n$x_3 = x_3$\r\n\r\nAt this point, let's rewrite the recursion to make calculation easier.\r\n\r\n$x_{n+1} = \\frac{ 3ax_n + 6a^2 x_{n-1} + 3x_{n-2} }{6}$\r\n\r\nIt should be clear now that the exponent of $a$ in $x_n$ is $n \\bmod 3$, which means for $3 | n$ we have a rational coefficient of $x_3$. \r\n\r\nNow, let's calculate a few more terms.\r\n\r\n$x_4 = \\frac{3a + 2a^4 + a}{6} x_3 = \\frac{4a}{3} x_3$\r\n$x_5 = \\frac{4a^2 + 6a^2 + a^2}{6} x_3 = \\frac{11a^2}{6} x_3$\r\n$x_6 = \\frac{5.5a^3 + 8a^3 + 3}{6} x_3 = 5x_3$, an integer. \r\n$x_7 = \\frac{15a + 11a^4 + 4a}{6} x_3 = \\frac{41a}{6} x_3$\r\n$x_8 = \\frac{20.5a^2 + 30a^2 + 5.5a^2}{6} x_3 = \\frac{56a^2}{6} x_3$\r\n$x_9 = \\frac{28a^3 + 41a^3 + 15}{6} = \\frac{153}{6} x_3$, not an integer.\r\n\r\nConjecture: All $x_{3k}$ are non-integer for $k \\ge 3$. This would mean the only integers in the squence are $x_0, x_3, x_6$. However, I am unsure if I have made a calculation mistake, and even if I haven't I lack the brainpower to complete the proof :(", "Solution_2": "The offical solution is very very long and I am crazy to see it.The answer is 5,and $x_0,x_3,x_6,x_{12},x_{24}$ must be integers. :wacko:\r\nThe problem is created by a university professor and he's famous to create such crazy problem for MO and luckily he has retired. :bye:", "Solution_3": ":censored: Calculation up to $x_{24}$ would be very tedious, and proving non-integer for all $x_{3k}, k > 8$ more tedious still - why write a problem that does not have elegant solution?!", "Solution_4": "Solution:Suppose $n\\geq2$,so\r\n\\begin{eqnarray*} x_{n+1}-\\sqrt[3]{2}x_n-\\frac{1}{\\sqrt[3]{2}}x_{n-1}&=&\\frac{1}{\\sqrt[3]{4}}x_n-\\sqrt[3]{2}x_n+\\sqrt[3]{4}x_{n-1}-\\frac{1}{\\sqrt[3]{2}}x_{n-1}+\\frac{1}{2}x_{n-2}\\\\ &=&-\\frac{\\sqrt[3]{2}}{2}x_n+\\frac{\\sqrt[3]{4}}{2}x_{n-1}+\\frac{1}{2}x_{n-2}\\\\ &=&-\\frac{\\sqrt[3]{2}}{2}(x_n-\\sqrt[3]{2}x_{n-1}-\\frac{1}{\\sqrt[3]{2}}x_{n-2}) \\end{eqnarray*}\r\nSince $x_2-\\sqrt[3]{2}x_1-\\frac{1}{\\sqrt[3]{2}}x_0=0$,\r\n\\[ x_{n+1}=\\sqrt[3]{2}x_n+\\frac{1}{\\sqrt[3]{2}}x_{n-1}(n\\geq1)\\eqno (1) \\]\r\nThe character equation of (1) is $x^2=\\sqrt[3]{2}x+\\frac{1}{\\sqrt[3]{2}}$ and the root is\r\n$x=\\frac{\\sqrt[3]{2}}{2}\\pm\\sqrt{\\frac{\\sqrt[3]{4}}{4}+\\frac{1}{\\sqrt[3]{2}}}=\\frac{\\sqrt[3]{2}}{2}(1\\pm\\sqrt{3})$\r\nSo by $x_0=0$,$x_n=A(\\frac{\\sqrt[3]{2}}{2})^n[(1+\\sqrt{3})^n-(1-\\sqrt{3})^n]$.\r\n$x_3=\\frac{A}{4}[(1+\\sqrt{3})^3-(1-\\sqrt{3})^3]=3\\sqrt{3}A$,\r\nSo $A=\\frac{x_3}{3\\sqrt{3}}$.\r\n$x_n=\\frac{x_3}{3\\sqrt{3}}(\\frac{\\sqrt[3]{2}}{2})^n[(1+\\sqrt{3})^n-(1-\\sqrt{3})^n]$.\r\n\r\n\r\nLet $a_n=\\frac{1}{\\sqrt{3}}[(1+\\sqrt{3})^n-(1-\\sqrt{3})^n]$,\r\n$b_n=(1+\\sqrt{3})^n+(1-\\sqrt{3})^n$\r\nIt's east to show that $a_n$ is even and $b_n$ is also even.\r\nBy $x_3$ is a positive integer,so the necessary condition of that $x_n$ is integer is $3|n$.\r\nAnd we also have for any nonnegative integer $m,n$\r\n$a_{n+m}=\\frac{1}{2}(a_nb_m+a_mb_n)$\r\n$b_{n+m}=\\frac{1}{2}(b_nb_m+3a_na_m)$\r\nLet $m=n$,\r\n$a_{2n}=a_nb_n$\r\n$b_{2n}=\\frac{1}{2}(b_n^2+3a_n^2)$\r\n\r\nAssume $a_n=2^{k_n}p_n,b_n=2^{l_n}q_n$,where $n,k_n,l_n$ is positive integer and $p_n,q_n$ are odd.\r\nWe can get $a_1=b_1=2$,that means $k_1=l_1=1$.Then we can get \r\n\r\n$k_2=2,l_2=3,k_4=5,l_4=3,k_8=8,l_8=5$\r\nAnd by induction \r\n\\[ k_{2^m}=\\left\\{\\begin{array}{cc} 1,&m=0\\\\ 2,&m=1\\\\ 2^{m-1}+m+1,&m\\geq2 \\end{array}\\right.\\\\ l_{2^m}=\\left\\{\\begin{array}{cc} 1,&m=0\\\\ 2,&m=1\\\\ 2^{m-1}+1,&m\\geq2 \\end{array}\\right. \\]\r\n\r\nFor any $m_1>m_2\\geq2$,\r\n$a_{2^{m_1}+2^{m_2}}=\\frac{1}{2}(a_{2^{m_1}}b_{2^{m_2}}+a_{2^{m_2}}b_{2^{m_1}})$\r\n$b_{2^{m_1}+2^{m_2}}=\\frac{1}{2}(b_{2^{m_1}}b_{2^{m_2}}+3a_{2^{m_1}}a_{2^{m_2}})$\r\n$k_{2^{m_1}+2^{m_2}}=2^{m_1-1}+2^{m_2-1}+m_2+1$\r\n$l_{2^{m_1}+2^{m_2}}=2^{m_1-1}+2^{m_2-1}+1$\r\nBy induction,for $m_1>m_2>\\ldots>m_r\\geq2$\r\n$k_{2^{m_1}+2^{m_2}+\\ldots+2^{m_r}}=2^{m_1-1}+2^{m_2-1}+\\ldots+2^{m_r-1}+m_r+1$\r\n$l_{2^{m_1}+2^{m_2}+\\ldots+2^{m_r}}=2^{m_1-1}+2^{m_2-1}+\\ldots+2^{m_r-1}+1$\r\n\r\nSo if $n=2^rp$,where positive integer $r\\geq2$,$p$ is odd,\r\n$k_n=\\frac{n}{2}+r+1$\r\n$l_n=\\frac{n}{2}+1$\r\n\r\nIf $n=4m+1$,\r\n$a_{4m+1}=\\frac{1}{2}(a_{4m}b_1+a_1b_{4m})=a_{4m}+b_{4m}$\r\nSo $k_{4m+1}=2m+1$.\r\nLikewise,$k_{4m+2}=k_{4m+3}=2m+2$\r\nSo \\[ k_n=\\left\\{\\begin{array}{cc} \\frac{n+1}{2},&n \\,\\,\\mbox{is odd}\\\\ \\frac{n}{2}+1,&n\\equiv2\\pmod{4}\\\\ \\frac{n}{2}+r+1,&n=2^rp,r\\geq2,p \\,\\,\\mbox{is odd} \\end{array}\\right. \\]\r\n\r\nWhen $3|n$,$x_n=\\frac{x_3}{3}2^{-\\frac{2}{3}n}a_n=\\frac{x_3}{3}2^{k_n-\\frac{2}{3}n}p_n$\r\nSince $a_3=12$,so $3|a_3$ and $3|a_{3\\times2^r}$ where $r$ is nonnegative integer.\r\n$k_3=2=\\frac{2}{3}\\cdot3$\r\n$k_6=4=\\frac{2}{3}\\cdot6$\r\n$k_{12}=9>\\frac{2}{3}\\cdot12$\r\n$k_{24}=16=\\frac{2}{3}\\cdot24$\r\nSo $x_3,x_6,x_{12},x_{24}$ are all integers.\r\n\r\nLet $x_3=3$,$x_n=2^{k_n-\\frac{2}{3}n}p_n$.\r\nIf $n\\not\\equiv0\\pmod{4}$,then $k_n\\leq\\frac{n}{2}+1$.\r\nSo for any $n>6$,$k_n-\\frac{2}{3}n\\leq1-\\frac{n}{6}<0$.\r\n\r\nIf $4|n$,suppose that $n=2^r3^kq$,where $r\\geq2,k\\geq1$,$3|q,2|q$.\r\nSo $k_n=2^{r-1}3^kq+r+1$\r\n$k_n-\\frac{2}{3}n=2^{r-1}3^kq+r+1-2^{r+1}3^{k-1}q=r+1-2^{r-1}3^{k-1}q\\leq r+1-2^{r-1}$\r\nThe equation holds iff $k=q=1$,\r\nSo if $r>3$,$2^{r-1}=(1+1)^{r-1}>r+1$\r\nTherefore when $r>3$,or$2\\leq r\\leq3$ and either $k$ or $q$ is larger than 1,we have $k_n-\\frac{2}{3}n<0$.\r\nSo if $x_3=3$,there is only five integer $x_0,x_3,x_6,x_{12},x_{24}$ in $\\{x_n\\}$.\r\nSo the answer is 5. \r\n\r\nCrazy,isn't it?", "Solution_5": "I've a friend who took part in this contest and failed in this problem,so he can't take part in IMO.And he hated the problem very much.", "Solution_6": "It is easy to show $x_{n}=\\frac{2^{-2n/3}((1+\\sqrt{3})^{n}-(1-\\sqrt{3})^{n})a}{3\\sqrt{3}}$ when $x_{3}=a$.\r\nOf course, if xn is integer, then 3|n, and when number of natural integer in this sequence is minimum, then a=1\r\nLet $y_{n}=x_{3n}, a=1$.\r\nthen $y_{n+2}=5y_{n+1}+\\frac{1}{2}y_{n}$\r\nbinomial theorem and power of 2 will give you a answer.\r\nonly $y_{0},y_{1},y_{2},y_{4},y_{8}$ are integer.\r\n\r\nIs it hard?" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,b,c>0$ . Prove that :\r\n\r\n\\[ \\frac{\\sqrt{a\\plus{}b\\plus{}c}\\plus{}\\sqrt{a}}{b\\plus{}c}\\plus{}\\frac{\\sqrt{a\\plus{}b\\plus{}c}\\plus{}\\sqrt{b}}{c\\plus{}a}\\plus{}\\frac{\\sqrt{a\\plus{}b\\plus{}c}\\plus{}\\sqrt{c}}{a\\plus{}b}\\ge \\frac{9\\plus{}3\\sqrt{3}}{2\\sqrt{a\\plus{}b\\plus{}c}}\\]", "Solution_1": "[quote=\"Pain rinnegan\"]Let $ a,b,c > 0$ . Prove that :\n\\[ \\frac {\\sqrt {a \\plus{} b \\plus{} c} \\plus{} \\sqrt {a}}{b \\plus{} c} \\plus{} \\frac {\\sqrt {a \\plus{} b \\plus{} c} \\plus{} \\sqrt {b}}{c \\plus{} a} \\plus{} \\frac {\\sqrt {a \\plus{} b \\plus{} c} \\plus{} \\sqrt {c}}{a \\plus{} b}\\ge \\frac {9 \\plus{} 3\\sqrt {3}}{2\\sqrt {a \\plus{} b \\plus{} c}}\\]\n[/quote]\r\n\r\nwithout loss of generality, assume that $ a \\plus{} b \\plus{} c \\equal{} 1$, so\r\n\r\n\\[ \\frac {\\sqrt {a \\plus{} b \\plus{} c} \\plus{} \\sqrt {a}}{b \\plus{} c} \\plus{} \\frac {\\sqrt {a \\plus{} b \\plus{} c} \\plus{} \\sqrt {b}}{c \\plus{} a} \\plus{} \\frac {\\sqrt {a \\plus{} b \\plus{} c} \\plus{} \\sqrt {c}}{a \\plus{} b}\\ge \\frac {9 \\plus{} 3\\sqrt {3}}{2\\sqrt {a \\plus{} b \\plus{} c}}\\]\\[ \\Leftrightarrow \\sum \\frac {1 \\plus{} \\sqrt {a}}{1 \\minus{} a} \\ge \\frac {9 \\plus{} 3\\sqrt {3}}{2}\\]\r\nwe can prove that $ \\frac {1 \\plus{} \\sqrt {a}}{1 \\minus{} a}\\ge\\frac {3}{4} \\plus{} \\frac {9 \\plus{} 6\\sqrt {3}}{4}a$, thus\r\n\\[ \\sum \\frac {1 \\plus{} \\sqrt {a}}{1 \\minus{} a} \\ge \\sum\\left(\\frac {3}{4} \\plus{} \\frac {9 \\plus{} 6\\sqrt {3}}{4}a\\right) \\equal{} \\frac {9 \\plus{} 3\\sqrt {3}}{2}\\]", "Solution_2": "my solution\nit is equivalent to $\\sum_{cyc}\\frac{a}{b+c}+\\sum_{cyc}\\frac{\\sqrt{a(a+b+c)}}{b+c}\\ge\\frac{3+3\\sqrt{3}}{2}$\nwe only need to prove that \n$\\sum_{cyc}\\frac{\\sqrt{a(a+b+c)}}{b+c}\\ge\\frac{3\\sqrt{3}}{2}$\nby Holder inequality,we have $(\\sum_{cyc}\\frac{\\sqrt{a}}{b+c})^{2}(\\sum_{cyc}(b+c)^{2}a^{2})\\ge\\ (a+b+c)^{3}$\nso we only need to prove that\n$4(a+b+c)^{4}\\ge\\ 27(2\\sum_{cyc}a^{2}b^{2}+2\\sum_{cyc}a^{2}bc)$\nhences to $4\\sum_{cyc}a^{4}+16\\sum_{sym}a^{3}b-30\\sum_{cyc}a^{2}b^{2}-6\\sum_{cyc}a^{2}bc\\ge\\ 0$ and that's trivial by Muirhead" } { "Tag": [], "Problem": "\u039f\u03c1\u03b8\u03bf\u03b3\u03ce\u03bd\u03b9\u03bf \u03c3\u03c5\u03bc\u03c0\u03bb\u03ae\u03c1\u03c9\u03bc\u03b1 \u03c4\u03bf\u03c5 \u03c5\u03c0\u03cc\u03c7\u03c9\u03c1\u03bf\u03c5 U \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 \u03ad\u03bd\u03b1\u03bd \u03c0\u03b5\u03c0\u03b5\u03c1\u03b1\u03c3\u03bc\u03ad\u03bd\u03b7\u03c2 \u03b4\u03b9\u03ac\u03c3\u03c4\u03b1\u03c3\u03b7\u03c2 \u03b4\u03b9\u03b1\u03bd\u03c5\u03c3\u03bc\u03b1\u03c4\u03b9\u03ba\u03cc \u03c7\u03ce\u03c1\u03bf V, \u03bf\u03bd\u03bf\u03bc\u03ac\u03b6\u03b5\u03c4\u03b1\u03b9 o \u03c5\u03c0\u03cc\u03c7\u03c9\u03c1\u03bf\u03c2 W, \u03bc\u03b5 $ W\\subset V$ ,\r\n\u03ad\u03c4\u03c3\u03b9 \u03ce\u03c3\u03c4\u03b5 o V \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf \u03b5\u03c5\u03b8\u03cd \u03ac\u03b8\u03c1\u03bf\u03b9\u03c3\u03bc\u03b1 \u03c4\u03c9\u03bd W \u03ba\u03b1\u03b9 U \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae $ V\\equal{} U\\plus{}W$ \u03ba\u03b1\u03b9 $ U\\cap W\\equal{}\\emptyset$.\r\n\u0391\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03c4\u03b5 \u03cc\u03c4\u03b9 \u03b3\u03b9\u03b1 \u03bf\u03c0\u03bf\u03b9\u03bf\u03bd\u03b4\u03ae\u03c0\u03bf\u03c4\u03b5 \u03c5\u03c0\u03cc\u03c7\u03c9\u03c1\u03bf U \u03c4\u03bf\u03c5 V,\u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03ad\u03bd\u03b1 \u03bf\u03c1\u03b8\u03bf\u03b3\u03ce\u03bd\u03b9\u03bf \u03c3\u03c5\u03bc\u03c0\u03bb\u03ae\u03c1\u03c9\u03bc\u03b1 \u03c4\u03bf\u03c5 ,\u03ba\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03bf\u03b9 \u03b4\u03b9\u03b1\u03c3\u03c4\u03ac\u03c3\u03b5\u03b9\u03c2 \u03b4\u03cd\u03bf \u03bf\u03c0\u03bf\u03b9\u03bf\u03bd\u03b4\u03ae\u03c0\u03bf\u03c4\u03b5 \u03bf\u03c1\u03b8\u03bf\u03b3\u03ce\u03bd\u03b9\u03c9\u03bd \u03c3\u03c5\u03bc\u03c0\u03bb\u03b7\u03c1\u03c9\u03bc\u03ac\u03c4\u03c9\u03bd \u03c4\u03bf\u03c5 U \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 \u03c4\u03bf\u03bd V, \u03b5\u03af\u03bd\u03b1\u03b9 \u03af\u03b4\u03b9\u03b5\u03c2.", "Solution_1": "[quote=\"spinos\"]\u039f\u03c1\u03b8\u03bf\u03b3\u03ce\u03bd\u03b9\u03bf \u03c3\u03c5\u03bc\u03c0\u03bb\u03ae\u03c1\u03c9\u03bc\u03b1 \u03c4\u03bf\u03c5 \u03c5\u03c0\u03cc\u03c7\u03c9\u03c1\u03bf\u03c5 U \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 \u03ad\u03bd\u03b1\u03bd \u03c0\u03b5\u03c0\u03b5\u03c1\u03b1\u03c3\u03bc\u03ad\u03bd\u03b7\u03c2 \u03b4\u03b9\u03ac\u03c3\u03c4\u03b1\u03c3\u03b7\u03c2 \u03b4\u03b9\u03b1\u03bd\u03c5\u03c3\u03bc\u03b1\u03c4\u03b9\u03ba\u03cc \u03c7\u03ce\u03c1\u03bf V, \u03bf\u03bd\u03bf\u03bc\u03ac\u03b6\u03b5\u03c4\u03b1\u03b9 o \u03c5\u03c0\u03cc\u03c7\u03c9\u03c1\u03bf\u03c2 W, \u03bc\u03b5 $ W\\subset V$ ,\n\u03ad\u03c4\u03c3\u03b9 \u03ce\u03c3\u03c4\u03b5 o V \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf \u03b5\u03c5\u03b8\u03cd \u03ac\u03b8\u03c1\u03bf\u03b9\u03c3\u03bc\u03b1 \u03c4\u03c9\u03bd W \u03ba\u03b1\u03b9 U \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae $ V \\equal{} U \\plus{} W$ \u03ba\u03b1\u03b9 $ U\\cap W \\equal{} \\emptyset$. \u0391\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03c4\u03b5 \u03cc\u03c4\u03b9 \u03b3\u03b9\u03b1 \u03bf\u03c0\u03bf\u03b9\u03bf\u03bd\u03b4\u03ae\u03c0\u03bf\u03c4\u03b5 \u03c5\u03c0\u03cc\u03c7\u03c9\u03c1\u03bf U \u03c4\u03bf\u03c5 V,\u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03ad\u03bd\u03b1 \u03bf\u03c1\u03b8\u03bf\u03b3\u03ce\u03bd\u03b9\u03bf \u03c3\u03c5\u03bc\u03c0\u03bb\u03ae\u03c1\u03c9\u03bc\u03b1 \u03c4\u03bf\u03c5 ,\u03ba\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03bf\u03b9 \u03b4\u03b9\u03b1\u03c3\u03c4\u03ac\u03c3\u03b5\u03b9\u03c2 \u03b4\u03cd\u03bf \u03bf\u03c0\u03bf\u03b9\u03bf\u03bd\u03b4\u03ae\u03c0\u03bf\u03c4\u03b5 \u03bf\u03c1\u03b8\u03bf\u03b3\u03ce\u03bd\u03b9\u03c9\u03bd \u03c3\u03c5\u03bc\u03c0\u03bb\u03b7\u03c1\u03c9\u03bc\u03ac\u03c4\u03c9\u03bd \u03c4\u03bf\u03c5 U \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 \u03c4\u03bf\u03bd V, \u03b5\u03af\u03bd\u03b1\u03b9 \u03af\u03b4\u03b9\u03b5\u03c2.[/quote]\r\n\r\n\r\n\u0391\u03c5\u03c4\u03cc \u03b4\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c0\u03bf\u03bb\u03cd\u03c4\u03c9\u03c2 \u03c4\u03b5\u03c4\u03c1\u03b9\u03bc\u03bc\u03ad\u03bd\u03bf \u03b3\u03b9\u03b1 \u03b1\u03c0\u03b5\u03b9\u03c1\u03bf\u03b4\u03b9\u03ac\u03c3\u03c4\u03b1\u03c4\u03bf\u03c5\u03c2 \u03c7\u03ce\u03c1\u03bf\u03c5\u03c2, \u03b1\u03bb\u03bb\u03ac \u03b3\u03b9\u03b1 \u03c7\u03ce\u03c1\u03bf\u03c5\u03c2 \u03c0\u03b5\u03c0\u03b5\u03c1\u03b1\u03c3\u03bc\u03ad\u03bd\u03b7\u03c2 \u03b4\u03b9\u03ac\u03c3\u03c4\u03b1\u03c3\u03b7\u03c2 \u03c4\u03b1 \u03c0\u03c1\u03ac\u03b3\u03bc\u03b1\u03c4\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c1\u03ba\u03b5\u03c4\u03ac \u03b1\u03c0\u03bb\u03ac. \u0391\u03c1\u03ba\u03b5\u03af \u03bd\u03b1 \u03b5\u03c0\u03b5\u03ba\u03c4\u03b5\u03af\u03bd\u03b5\u03b9\u03c2 \u03bc\u03af\u03b1 \u03b2\u03ac\u03c3\u03b7 \u03c4\u03bf\u03c5 $ U$ \u03bc\u03ad\u03c7\u03c1\u03b9 \u03bd\u03b1 \u03b5\u03be\u03b1\u03bd\u03c4\u03bb\u03ae\u03c3\u03b5\u03b9\u03c2 \u03cc\u03bb\u03bf\u03bd \u03c4\u03bf \u03c7\u03ce\u03c1\u03bf.\r\n\r\n\u0397 \u03bb\u03bf\u03b3\u03b9\u03ba\u03ae \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03b5\u03be\u03ae\u03c2: \u03b1\u03bd $ U \\equal{} V$, \u03c4\u03b5\u03bb\u03b5\u03af\u03c9\u03c3\u03b5\u03c2. \u0391\u03bb\u03bb\u03b9\u03ce\u03c2, \u03b8\u03b1 \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03c4\u03bf\u03c5\u03bb\u03ac\u03c7\u03b9\u03c3\u03c4\u03bf\u03bd \u03ad\u03bd\u03b1 \u03b4\u03b9\u03ac\u03bd\u03c5\u03c3\u03bc\u03b1 \u03c3\u03c4\u03bf\u03bd $ V\\setminus U$, \u03c4\u03bf \u03bf\u03c0\u03bf\u03af\u03bf \u03bc\u03c0\u03bf\u03c1\u03b5\u03af\u03c2 \u03bd\u03b1 \u03c0\u03c1\u03bf\u03c3\u03b8\u03ad\u03c3\u03b5\u03b9\u03c2 \u03c3\u03c4\u03b7 \u03b2\u03ac\u03c3\u03b7 \u03c4\u03bf\u03c5 $ U$ \u03ba\u03b1\u03b9 \u03bd\u03b1 \u03c0\u03ac\u03c1\u03b5\u03b9\u03c2 \u03ad\u03bd\u03b1\u03bd \u03ba\u03b1\u03b9\u03bd\u03bf\u03cd\u03c1\u03b9\u03bf \u03c7\u03ce\u03c1\u03bf $ U_1$. \u039e\u03b1\u03bd\u03b1\u03ba\u03ac\u03bd\u03b5\u03b9\u03c2 \u03c4\u03b7\u03bd \u03af\u03b4\u03b9\u03b1 \u03b4\u03bf\u03c5\u03bb\u03b5\u03b9\u03ac \u03bc\u03b5 \u03c4\u03bf\u03bd $ U_1$ \u03ba\u03b1\u03b9 \u03c6\u03c4\u03b9\u03ac\u03c7\u03bd\u03b5\u03b9\u03c2 \u03ad\u03bd\u03b1\u03bd $ U_2$ \u03ba.\u03bf.\u03ba. \u03bc\u03ad\u03c7\u03c1\u03b9 \u03bd\u03b1 \u03c6\u03c4\u03ac\u03c3\u03b5\u03b9\u03c2 \u03c3\u03c4\u03bf\u03bd $ V$. \u03a4\u03bf \u03c3\u03c5\u03bc\u03c0\u03bb\u03ae\u03c1\u03c9\u03bc\u03b1 \u03c4\u03bf\u03c5 $ U$ \u03b8\u03b1 \u03c0\u03b1\u03c1\u03ac\u03b3\u03b5\u03c4\u03b1\u03b9 \u03b1\u03ba\u03c1\u03b9\u03b2\u03ce\u03c2 \u03b1\u03c0\u03cc \u03b1\u03c5\u03c4\u03ac \u03c4\u03b1 \u03b4\u03b9\u03b1\u03bd\u03cd\u03c3\u03bc\u03b1\u03c4\u03b1 \u03c0\u03bf\u03c5 \u03c0\u03c1\u03bf\u03c3\u03ad\u03b8\u03b5\u03c3\u03b5\u03c2.\r\n\r\n\u0391\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c0\u03b5\u03b9\u03c1\u03bf\u03b4\u03b9\u03ac\u03c3\u03c4\u03b1\u03c4\u03bf\u03b9 \u03bf\u03b9 \u03c7\u03ce\u03c1\u03bf\u03b9, \u03c7\u03c1\u03b5\u03b9\u03ac\u03b6\u03b5\u03c3\u03b1\u03b9 \u03c4\u03bf \u03b1\u03be\u03af\u03c9\u03bc\u03b1 \u03b5\u03c0\u03b9\u03bb\u03bf\u03b3\u03ae\u03c2 \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03ad\u03c7\u03b5\u03b9\u03c2 \u03b2\u03ac\u03c3\u03b7 \u03b3\u03b9\u03b1 \u03ba\u03ac\u03b8\u03b5 \u03b4/\u03ba\u03cc \u03c7\u03ce\u03c1\u03bf... \u03b2\u03b1\u03c3\u03b9\u03ba\u03ac, \u03bc\u03c0\u03bf\u03c1\u03b5\u03af\u03c2 \u03bd\u03b1 \u03c4\u03bf \u03ba\u03ac\u03bd\u03b5\u03b9\u03c2 \u03c0\u03ac\u03bd\u03c9 \u03ba\u03ac\u03c4\u03c9 \u03cc\u03c0\u03c9\u03c2 \u03c0\u03b1\u03c1\u03b1\u03c0\u03ac\u03bd\u03c9, \u03b1\u03bb\u03bb\u03ac \u03bf Dem \u03c3\u03af\u03b3\u03bf\u03c5\u03c1\u03b1 \u03b8\u03b1 \u03be\u03ad\u03c1\u03b5\u03b9 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03bd \u03ba\u03b1\u03bb\u03cd\u03c4\u03b5\u03c1\u03bf \u03c4\u03c1\u03cc\u03c0\u03bf ;)\r\n\r\nCheerio,\r\n\r\nDurandal 1707" } { "Tag": [ "analytic geometry", "graphing lines", "slope", "trigonometry" ], "Problem": "Find the acute angle between two lines with slopes -2 and -1/2 to the nearest degree.", "Solution_1": "[quote=\"THECHAMP\"]Find the acute angle between two lines with slopes -2 and -1/2 to the nearest degree.[/quote]\r\n[hide]Call the first line $ l_1$ and the second $ l_2$. The slope of $ l_1$ is $ \\minus{}2$ or $ \\tan\\theta\\equal{}\\minus{}2$, thus it makes a $ \\tan^{\\minus{}1}(\\minus{}2)$ angle with the x-axis. Similarly with the second line, it has the angle $ \\tan^{\\minus{}1}(\\minus{}\\frac{1}{2})$. Now we want the angle between these two lines so we have that $ \\alpha\\equal{}\\tan^{\\minus{}1}(\\minus{}2)\\minus{}\\tan^{\\minus{}1}(\\minus{}\\frac{1}{2})$. Taking the tangent of both sides and using the tangent subtraction formula we have $ \\tan\\alpha\\equal{}\\frac{\\minus{}2\\minus{}\\frac{1}{2}}{1\\plus{}(\\minus{}2)(\\minus{}\\frac{1}{2})}\\equal{}\\frac{5}{4}$. Thus the angle between these two angles is $ \\tan^{\\minus{}1}(\\frac{5}{4})$ which approximated to the nearest degree is $ 51^{\\circ}$[/hide]" } { "Tag": [ "geometry theorems", "geometry" ], "Problem": "Given two lines $(k: l : m), (p : q: r)$ in trilinears, what is the formula for the (smaller) angle between them?", "Solution_1": "Given $\\triangle ABC$ and two lines $l_{i}=[p_{i}: q_{i}: r_{i}]$ where $i=1,2$ in baricentric coordinates. \\[\\cos(l_{1},l_{2})=\\frac{ \\sum_{\\text{c\\'iclica}}S_{A}(q_{1}-r_{1})(q_{2}-r_{2})}{\\sqrt[]{\\sum_{\\text{c\\'iclica}}{S_{A}(q_{2}-r_{2})^{2}}}\\sqrt[]{\\sum_{\\text{c\\'iclica}}S_{A}(q_{1}-r_{1})^{2}}}.\\]" } { "Tag": [ "counting", "distinguishability", "derangement" ], "Problem": "8 girls and 16 boys are standing in a line. How many of the permutations of their ordering are there such that exactly 5 of the girls are not in the spot they were before, while 3 are. (The positions of the boys don't matter, and everyone is distinguishable)", "Solution_1": "The 16 boys permute in $ 16!$ ways. There are $ 8!/5!3!$ ways to categorize the girls. Then use the derangement formula to find the number of derangements of the 5 girls. (which I believe to be $ \\sum_{k\\equal{}0}^5 (\\minus{}1)^k/k!$)", "Solution_2": "Even if we derange the set of 5 girls, what if some of those girls take spots which were previously occupied by boys?", "Solution_3": "I see... this problem seems much more difficult than I thought it would be (I didn't think carefully.... last period of school day and I was tooo tired... needa take a nap when got home... See? I even got the derangement formula wrong :oops: :P )", "Solution_4": "First, let's choose the three girls that stay in their place. There 56 ways to do this.\r\n\r\nWithout regard to restrictions, the remaining 21 people may be arranged in 21! ways. There are 16! ways to place all 5 girls in the spot they were in before. There are 17! ways to place any 4 of those girls in the spot they were in before.\r\n\r\nUsing the Principle of Inclusion-Exclusion, there are $ 5(20!) \\minus{} 10(19!) \\plus{} 10(18!) \\minus{} 5(17!) \\plus{} 16!$ ways to place them so that atleast one of them are in their previous position. Simplifying, we get $ 526236(16!)$ Because we want the number of ways to place them in a position where none of them are in their previous position, we need to subtract $ 526236(16!)$ from $ 21!$ \r\n$ 21! \\minus{} 526236(16!) \\equal{} 2441880(16!) \\minus{} 526236(16!) \\equal{} 1915644(16!)$\r\n\r\nMultiplying this by our previous answer of 56, and we get $ 107276064(16!)$\r\nor, if you prefer $ 2,244,514,547,083,640,832,000$ ways." } { "Tag": [ "geometry", "circumcircle", "trigonometry", "inequalities", "function", "geometry proposed" ], "Problem": "In a triangle $ABC$ we have $\\angle C>10^0$ and $\\angle B=\\angle C+10^0.$We consider point $E$ on side $AB$ such that $\\angle ACE=10^0,$ and point $D$ on side $AC$ such that $\\angle DBA=15^0.$ Let $Z\\neq A$ be a point of interection of the circumcircles of the triangles $ABD$ and $AEC.$Prove that $\\angle ZBA>\\angle ZCA.$", "Solution_1": "[quote=\"nickolas\"]In a triangle $ABC$ we have $\\angle C>10^0$ and $\\angle B=\\angle C+10^0.$We consider point $E$ on side $AB$ such that $\\angle ACE=10^0,$ and point $D$ on side $AC$ such that $\\angle DBA=15^0.$ Let $Z\\neq A$ be a point of interection of the circumcircles of the triangles $ABD$ and $AEC.$Prove that $\\angle ZBA>\\angle ZCA.$[/quote]\r\n\r\nHere what I did...\r\nFirst put $\\angle BCA=10^0+x$ then $\\angle ECB=x,\\angle ABE=x+20^0,\\angle DBC=x+5^0$\r\nHowever the cord $AZ$ is the same for both circles we will just prove that circumradius of ABD is smaller than the circumradius of AEC (this follows by the sine law) or the inequality $\\frac{AB}{\\sin{\\angle ADB}} < \\frac{AC}{\\sin{\\angle AEC}}(*)$\r\nBy the sine law it also follows that \r\n$\\frac{AB}{\\sin{(10^0+x)}}=\\frac{AC}{\\sin{(20^0+x)}}(**)$ now dividing (*) and (**) side by side gives:\r\n$\\sin({\\angle AEC})\\cdot\\sin{(x+10^0)}<\\sin({\\angle ADB})\\cdot\\sin{(x+20^0)}$ because of $\\angle AEC=2x+20^0$ and $\\sin{\\angle ADB}=2x+15^0$ it is enough to prove that $\\sin(2x+20^0)\\cdot\\sin{(x+10^0)}<\\sin(2x+15^0)\\cdot\\sin{(x+20^0)}(***)$\r\nNow I proceeded as follows:\r\nTake $f(y)=\\frac{\\sin(y+x)}{\\sin(y+2x)},x<90^0,f'(y)=\\frac{\\sin(x)}{(\\sin(y+2x))^2}\\ge 0$ so $f$ is nondecreasing...Hence $\\frac{\\sin(20^0+x)}{\\sin(20^0+2x)}\\ge \\frac{\\sin(15^0+x)}{\\sin(15^0+2x)} >\\frac{\\sin(10^0+x)}{\\sin(15^0+2x)}$ and (***) follow...\r\nIn this proof I didn't bothered about the sign of the sine functions because from the condition of the problem we have $2x+20^0<90^0$ or $x<35^0$" } { "Tag": [ "analytic geometry", "geometry", "geometric transformation", "perimeter" ], "Problem": "The circle is tangent to both coordinate axes and contains the point(6,3). Find the center of the circle and the radius.", "Solution_1": "Too easy i think [hide]\nLet the center of the circle is $O(x,y)$ as it touches both coordinate axis then $x=y$. Then if $A(6,3)$ $OA^{2}=(x-6)^{2}+(x-3)^{2}=2x^{2}-18x+45$\nNow if $A$ is in the circle $OA \\le x^{2}$ giving $x \\in [3,15]$.[/hide]", "Solution_2": "[quote=\"delta\"]Too easy i think [hide]\nLet the center of the circle is $O(x,y)$ as it touches both coordinate axis then $x=y$. Then if $A(6,3)$ $OA^{2}=(x-6)^{2}+(x-3)^{2}=2x^{2}-18x+45$\nNow if $A$ is in the circle $OA \\le x^{2}$ giving $x \\in [3,15]$.[/hide][/quote]\r\nTechnically a circle is only the points that are a constant distance from a point, and not the points in the interior. A [i]disk[/i] contains the interior points.", "Solution_3": "[quote=\"randomdragoon\"]Technically a circle is only the points that are a constant distance from a point, and not the points in the interior. A [i]disk[/i] contains the interior points.[/quote]\r\nIt could be a language problem.\r\n\r\nTake a look at the following translations English/Spanish:\r\n\r\nA disk is bounded by a circle, and its perimeter is called circumference.\r\nDisk $\\to$ C\u00edrculo\r\nCircle $\\to$ Circunferencia\r\nCircumference $\\to$ Longitud de la circunferencia\r\n\r\nI've been told the Spanish meanings are more faithful to the original latin words and English messed it up. But that's maybe because I'm Spanish. Judge yourself:\r\n\r\n'Deus est circulus cuius centrum ubique, circunferentia vero nusquam'" } { "Tag": [ "articles", "geometry", "LaTeX" ], "Problem": "Is the code for page margin \r\n[code]\\setlength\\leftmargin{1in}[/code]\ncorrect?\nBasically, my document is otherwise perfect, but it seems to be shifted to the right, so much so that if I put textwidth as more than 7, the text goes off the page on the right side. For 6, it is noticeably right aligned and for 5, it seems correct. \nMy preamble is this\n[code]\\documentclass[11pt]{article}\n\\usepackage{amsmath,amssymb}\n\\setlength\\topmargin{-1in}\n\\setlength\\textwidth{5in}\n\n\\pdfpagewidth 8.26in\n\\pdfpageheight 11.5in[/code]\r\nThis gives correct view, but there is too much (but symmetric) left and right margin which I can't reduce.", "Solution_1": "Page layout is very complicated with lots of variables so use the geometry package which makes it very easy indeed. For example [code]\\usepackage[left=1in, right=1in]{geometry}[/code] gives you left and right margins of 1in (you can use cm if you wish)", "Solution_2": "It worked. Thanks for the help.", "Solution_3": "If you use the textwidth command and want a centered page, you should set the side margins to zero. Anything else moves the page off center. For example, suppose you use a page that's 8.5 in wide and 6.5 in of text. If you set the left margin to 1 in, the page is shifted 1 in to the right of center, so its right edge is justified to the right edge of the page.\r\nI use 6.5 in textwidth, with left and right margins set to zero; this gives me a centered page, with actual margins of 1 inch on each side.\r\n\r\nIf you're using a negative top margin, you're probably putting a lot of space into some other commands; there are at least three commands I know that add to the top margin on a standard page. From my experience, you probably want the sum slightly positive; the page looks better if the text is a little below center.", "Solution_4": "Thanks for the help. When I put all margins 0 (in the geometry package) the text is horrribly aligned towards the top-left corner. Problem is that I'm using a previous document as a 'template' for my document, i.e. I've kept the preamble same and have only added my text between \\begin{document} and \\end{document}. So I don't really know what settings were used by the previous author.", "Solution_5": "Zero margins is my advice without the geometry package; that package follows different rules.\r\n\r\nActually, now that I look at it: that leftmargin command seems to do nothing at all, and the default without it is about two inches of margin on the left.\r\n\r\nWhat does work for me is the \\oddsidemargin command; that gives about one inch of margin on the left if you set it to zero, plus whatever you add. It doesn't default to center.\r\n\r\nFor the margins at the top, look for \\headheight and \\headsep. They add to the top margin, and (at least in my system) they default to slightly positive if you don't define them. The total margin at the top comes out to about 1 inch plus the sum of the \\topmargin, \\headheight, and \\headsep values.", "Solution_6": "I would advise deleting all the page layout settings in the preamble and only using the geometry package. The [url=http://tug.ctan.org/tex-archive/macros/latex/contrib/geometry/manual.pdf]Users' manual[/url] will give you all the control you want." } { "Tag": [], "Problem": "Find the only integer n such that $ \\frac{(n\\plus{}1)^2}{n\\plus{}12}$ is an integer.", "Solution_1": "'m not sure if there is only 1, but if the denominator evaluates to 1, and the numerator evaluates to integer *which it would be if n=-11*, then the fraction is an integer\r\n\r\n$ n \\equal{} \\minus{} 11$\r\n\r\nEDIT:\r\nwait what about denominator evaulates to -1 and n=-13?\r\n\r\n$ n\\equal{}\\minus{}13$ also", "Solution_2": "sorry I meant to say find the only positive integer n. :blush:", "Solution_3": "UNLESS you mean positive in which it is 109 as follows:\r\n\r\n$ \\frac{(n\\plus{}1)^2}{(n\\plus{}12)}$\r\n$ \\frac{n^2\\plus{}2n\\plus{}1}{n\\plus{}12}$\r\nsynthetic division...\r\n\r\n$ n\\minus{}10\\plus{}\\frac{121}{n\\plus{}12}$\r\n$ n\\minus{}10$ is always an integer given n is an integer so the only question is\r\nis $ \\frac{121}{n\\plus{}12}$ an integer?\r\nor\r\nwhat integer when added to 12 is a factor of 121?\r\nthe factors are -121, -11, -1, 1, 11, and 121\r\nof these, only 121 is positive when 12 is subtracted from it\r\n$ 121\\minus{}12\\equal{}\\boxed{109}$" } { "Tag": [ "geometry", "rectangle", "geometry solved" ], "Problem": "Hi, \r\nI am stucked on a question on my homework. Here is the problem: \r\nThe diagonals of a rectangle are 8 units long and intersect at a 60 degree angle. Find the dimensions of the rectangle. \r\n\r\nWhat I dont get is , what is \"dimension\"? This is supposed to be about the \"Special Right Triangle\" (30-60-90) (45-45-90) Theorems. \r\n\r\nAny help will be appreciated. Thank you!", "Solution_1": "I am pretty sure that \"dimensions\" means sidelengths here.\r\n\r\nAnyway, this is neither Olympiad Section level, nor an open question.\r\n\r\n darij", "Solution_2": "Thank you for the help. \r\nSorry about posting in the wrong forum :S It is my first time here, I don't know where else I should post. Sorry about that. Thanks again." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "$\\boxed{\\ \\left\\{\\begin{array}{c}\\{x,y,z\\}\\subset (1,\\infty )\\\\\\\\ x+y+z+2=xyz\\end{array}\\right\\|\\Longrightarrow \\frac{3+x+y+z}{6}\\ge\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}\\ }\\ .$", "Solution_1": "Like a remark, with the substitutions: $x=\\frac{a+b}{c}$, $y=\\frac{b+c}{a}$, $z=\\frac{c+a}{b}$ respecting the condition : $x+y+z+2=xyz$\r\n\r\nthe inequality is $\\Longleftrightarrow \\sum{\\frac{a+b}{c}}+3\\geq 6\\sum{\\frac{a}{b+c}}$ $(\\star)$ , with $a,b,c$ sides of a triangle (because the other condition : $x,y,z > 1$ ) \r\n\r\nbut $(\\star) \\Longleftrightarrow (a+b+c)(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c})\\geq 6\\sum{\\frac{a}{b+c}}$, which is the almost famous now inequality given in Vietnam TST 2006.", "Solution_2": "Put $a=x-1$ ,etc and it is equivalent to \r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=122198\r\n\r\n :wink:" } { "Tag": [ "geometry", "parallelogram", "number theory", "greatest common divisor", "geometry unsolved" ], "Problem": "An acute triangle $ABC$ is given. Let $C'$ be the foot of the altitude on $AB$, and let $D,E$ be different points of segments $CC'$. Denote by $F,G$ be orthogonal projections of the point $D$ onto the sides $AC,BC$. Prove that if $DGEF$ is a parallelogram, then $ABC$ is isosceles.", "Solution_1": "$\\triangle CDF$ and $\\triangle CDG$ can be inscribed in the same semicircle because of the equal hypotenuses. But the perpendicular distances from $F$ and $G$ to segment $CC'$ are equal, so these triangles are congruent. If $\\angle GDC=\\angle FCD$, $\\angle C$ would be a right angle. But $\\triangle ABC$ is acute, therefore, $\\angle GCD=\\angle FCD$. We know that $CC'$ is the altitude so $\\angle A=\\angle B$." } { "Tag": [], "Problem": "The positive integers $ A$, $ B$, $ A \\minus{} B$, and $ A \\plus{} B$ are all prime numbers. The sum of these four primes is\r\n\r\n$ \\textbf{(A)}\\ \\text{even} \\qquad \\textbf{(B)}\\ \\text{divisible by }3 \\qquad \\textbf{(C)}\\ \\text{divisible by }5 \\qquad \\textbf{(D)}\\ \\text{divisible by }7 \\\\ \\textbf{(E)}\\ \\text{prime}$", "Solution_1": "A cannot be 2, since there is no prime below it for A-B to equal. Therefore, A is an odd prime. B must be 2, because otherwise A-B and A+B would both be even and not prime. The only 3 consecutive odd numbers that are prime are 3, 5, and 7, so A = 5. These four primes sum up to 17, so the answer is E.\r\nEDIT: Oops, messed up a bit in writing. Fixed.", "Solution_2": "[quote=\"Brut3Forc3\"]A cannot be 2, since there is no prime below it for A-B to equal. Therefore, A is an odd prime. B must be 2, because otherwise A-B and A+B would both be even and not prime. The only 3 consecutive odd numbers that are prime are 3, 5, and 7, so B = 5. These four primes sum up to 17, so the answer is E.[/quote]\r\n\r\nNot true, there may be more triple primes.\r\nBut you know that they are of the form a, a-2, and a+2. Add those up and you get 3a, so the answer is B\r\n\r\nEDIT: Oh snaps, forgot to include 2. Ehehe. Still though, I don't think it will always be prime.", "Solution_3": "[quote=\"CircleSquared\"]\nBut you know that they are of the form a, a-2, and a+2. Add those up and you get 3a, so the answer is B[/quote]\r\nFirst of all, you would have to add 2 to that sum, so it wouldn't be divisible by 3.\r\nAlso, there can't be any more triple primes because of any 3 consecutive odd numbers, there will be one that is divisible by 3.", "Solution_4": "[quote=\"Brut3Forc3\"][quote=\"CircleSquared\"]\nBut you know that they are of the form a, a-2, and a+2. Add those up and you get 3a, so the answer is B[/quote]\nFirst of all, you would have to add 2 to that sum, so it wouldn't be divisible by 3.\nAlso, there can't be any more triple primes because of any 3 consecutive odd numbers, there will be one that is divisible by 3.[/quote]\r\n\r\ni see. i knew i was missing something :rotfl:", "Solution_5": "[quote]B must be 2, because otherwise A-B and A+B would both be even and not prime. The only 3 consecutive odd numbers that are prime are 3, 5, and 7, so B = 5.[/quote]\nYou said that B must be 2, and then you said that it is 5.\nDid you mean that A is 5? that's the only thing that fits.[/quote]", "Solution_6": "[quote=\"alexk\"][quote]B must be 2, because otherwise A-B and A+B would both be even and not prime. The only 3 consecutive odd numbers that are prime are 3, 5, and 7, so B = 5.[/quote]\nYou said that B must be 2, and then you said that it is 5.\nDid you mean that A is 5? that's the only thing that fits.[/quote][/quote]\r\nYup. Sorry. :oops:" } { "Tag": [], "Problem": "IN THE NAME OF GOD\r\n\r\nFind all points (4, y) that are 10 units from the point (\u20132, \u20131). \r\n\r\nTHANKS\r\nSALAH", "Solution_1": "Hi salahcool.\r\n\r\nThe points which are $ 10$ units from $ (\\minus{}2, \\minus{}1)$ lie on the circle, centre $ (\\minus{}2, \\minus{}1)$ and radius $ 10$, that is on the circle whose equation is $ (x\\plus{}2)^{2} \\plus{} (y\\plus{}1)^{2} \\equal{} 100$. The line $ x\\equal{}4$ cuts this circle where $ (4\\plus{}2)^{2} \\plus{} (y\\plus{}1)^{2} \\equal{} 100$, that is where $ (y\\plus{}1)^{2} \\equal{} 64$, so $ y\\equal{}\\minus{}9$ or $ 7$ and so the points are $ (4, \\minus{}9)$ and $ (4, 7)$.", "Solution_2": "hello, the distance between the two points is given by\r\n$ \\sqrt{(4\\plus{}2)^2\\plus{}(y\\plus{}1)^2}$ and this must be $ 10$, hence we have\r\n$ \\sqrt{(4\\plus{}2)^2\\plus{}(y\\plus{}1)^2}\\equal{}10$ from here we get\r\n$ y_1\\equal{}\\minus{}9$ or $ y\\equal{}7$\r\nSonnhard.", "Solution_3": "ok,thank you very much\r\n\r\nWell done\r\n\r\nDr Sonnhard Graubner\r\n\r\nAndrewTom" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Prove that if $ x \\plus{} y$ is a square-free number then the equation $ x^{2009} \\plus{} y^{2009} \\equal{} z^{2009}$ has no solutions in natural numbers.", "Solution_1": "$ x \\plus{} y|x^{2009} \\plus{} y^{2009} \\equal{} z^{2009}$\r\nSo $ (x \\plus{} y)^{2009}|z^{2009} < (x \\plus{} y)^{2009}$, impossible.Done." } { "Tag": [ "analytic geometry", "vector", "linear algebra", "matrix" ], "Problem": "Does there exist a linear map T : R^6-------->R^2 such that 2rank(T) = nul(T)? \r\n\r\n\r\nYes. rank(T)+null(T)=dimm(R^6)=6 \r\n3rank(T)=6 so rank(T)=2 \r\n\r\ncan you please give me an example ?\r\n\r\nThanks", "Solution_1": "How about $ T(x_1,x_2,x_3,x_4,x_5,x_6) \\equal{} (x_1,x_2)$, or similar.", "Solution_2": "Wow Thanks for superfast reply.\r\n\r\n\r\n$ x_{1}$ has six components? i.e (1,2,3,4,5,6)", "Solution_3": "No, $ x_1$ is just the \"first coordinate\" of a vector $ x \\equal{} (x_1,x_2,x_3,x_4,x_5,x_6) \\in \\mathbb{R}^6$. So, if you define $ y \\equal{} (x_1,x_2) \\in \\mathbb{R}^2$, the linear mapping is $ T(x) \\equal{} y$.", "Solution_4": "Written as a matrix, mavropnevma's example is $ \\begin{bmatrix}1&0&0&0&0&0\\\\0&1&0&0&0&0\\end{bmatrix}.$\r\n\r\nBut $ \\begin{bmatrix}3&0&\\minus{}7&21&9\\\\1&4&32&\\sqrt{2}&11\\end{bmatrix}$ would also work.\r\n\r\nAnd if you write down just any $ 2\\times 6$ matrix (which is exactly what I just did) it would be a staggeringly unlikely coincidence for it to have any rank other than $ 2.$", "Solution_5": "Thanks Kent Merryfield.\r\n\r\nwhere do i plug this matrix and what do i get?", "Solution_6": "What do you mean by that? Those matrices are simply representations of your linear transformation (you are familiar with matrix representations of linear transformations, right?). Basically, for this case just write down any 2x6 matrix such that (for example) its 2x2 submatrix made up of rows 1,2 and columns 1,2 is invertible." } { "Tag": [ "LaTeX" ], "Problem": "How do you put a strikethrough, cross, or line through an equation (to cancel it, in a fraction)? I have tried \\not{equation}, but it only works on single digits, and then, not well. \\cancel{equation} does not work in the AoPS Classroom or Forum. \\sout{equation} does not work either. How would I do this? I have seen it done in the Classroom before, so I know it can be done. Please help! Soon! Thank you!", "Solution_1": "Since the cancel package is not available on the forum you need to be creative eg using \\not\\! as in \\[ \\frac{32}{64}\r\n\\equal{}\\frac{{2}\\times\\not\\!4\\times\\not\\!4}{4\\times\\not\\!4\\times\\not\\!4}\r\n\\equal{}\\frac{2}{4}\r\n\\equal{}\\color{black}\\boxed{\\color{blue} \\frac{1}{2}}\\]\r\nFor cancelling out more complicated stuff you will just have to use lots of \\not\\!\r\n\\[ \\frac{x^2\\minus{}1}{x\\minus{}1}\\equal{}\\frac{(x\\plus{}1)(\\not\\!x\\not\\!\\minus{}\\not\\!1)}{\\not\\!x\\not\\!\\minus{}\\not\\!1}\\equal{}x\\plus{}1\\]\r\nIt's not ideal but will have to do.", "Solution_2": "Thank you!" } { "Tag": [ "puzzles" ], "Problem": "this is kind of a brain teaser. a easy one.\r\n\r\n$(x-a)(x-b)(x-c)...(x-z)=?$\r\n\r\n :D", "Solution_1": "[hide=\"Click to reveal hidden text\"]It has $(x-x)$, so $\\boxed{0}$.[/hide]", "Solution_2": "[quote=\"Zenith Xan\"][hide=\"Click to reveal hidden text\"]It has $(x-x)$, so $\\boxed{0}$.[/hide][/quote]\r\n\r\nSimple and nice!!!!!!!!!!!!" } { "Tag": [ "superior algebra", "superior algebra unsolved" ], "Problem": "We introduce the following operators mapping classes of algebra to classes of algebras (all of the same type):\r\n$ A\\in S(K)$ iff is a subalgebra of some member of K\r\n$ A\\in H(K)$ iff is a homomorphic image of some member of K\r\n$ A\\in P(K)$ iff is a direct product of family of algebras in K\r\n\r\nWe know that:$ SH\\subset HS, PS\\subset SP, PH\\subset HP$\r\nShow:\r\n$ HP \\neq PH$\r\n$ SP \\neq PS$", "Solution_1": "So I don't know much about universal algebra, but hopefully these examples are easily modified to your purposes:\r\n\r\n\r\nFirstly notice that HP contains central products of groups, but PH has very few indecomposable members.\r\n\r\nTake K to be the class of dihedral groups G of order 8. Denote their central involution by z.\r\n\r\nH(K) is the class containing { 1, 2, 2 x 2, G }\r\n\r\nThe only directly indecomposable members of P(H(K)) are already in H(K)\r\n\r\nP(K) contains G x G, and H(P(K)) contains $ G \\times G / \\langle (z,z) \\rangle$, a directly indecomposable extra-special group of order 32, sometimes denoted $ 2^{2\\cdot2\\plus{}1}$ or $ G \\circ G$.\r\n\r\n\r\nSecondly notice that SP contains subdirect products, but PS has very few indecomposable members.\r\n\r\nTake K to be the class of nonabelian groups G of order 6.\r\n\r\nThe only directly indecomposable members of P(S(K)) are already in S(K) and so have order at most 6.\r\n\r\nHowever P(K) contains a group isomorphic to $ \\langle (1,2,3), (1,2) \\rangle \\times \\langle (4,5,6), (4,5) \\rangle$, and so S(P(K)) contains a group isomorphic to the indecomposable group $ \\langle (1,2,3), (4,5,6), (1,2)(4,5) \\rangle$." } { "Tag": [ "AMC" ], "Problem": "Let $P$ be the product of the first 100 positive odd integers. Find the largest integer $k$ such that $P$ is divisible by $3^k$.", "Solution_1": "[hide]There are 33 multiples of 3, 11 multiples of 9, 4 multiples of 27 and 1 multiple of 81 for a total of 49 factors of 3.[/hide]", "Solution_2": "A slightly nicer way:\r\n\r\n[hide]\n\n1*3*5*...*199 = 200!/2*(100)!\n\nHighest power of 3 that divides 200! can be found with the usual method:\n\n[200/3] + [200/9] + [200/27] + [200/81] = 66 + 22 + 7 + 2 = 97\n\nHighest power of 3 can be found with the same idea:\n\n[100/3] + [100/9] + [100/27] + [100/81] = 33 + 11 + 3 + 1 = 48\n\nNow subtract the two: 97 - 48 = 49.\n\n049\n[/hide]", "Solution_3": "[quote=\"JSteinhardt\"]A slightly nicer way:\n\n1*3*5*...*199 = 200!/2*(100)!\n\n...[/quote]\r\n\r\nNot that it changes the answer any, but $2\\cdot 6\\cdot 8\\cdots 200\\neq 2\\cdot 100!$, it equals $2^{100}\\cdot 100!$...", "Solution_4": "[quote=\"E^(pi*i)=-1\"][hide]There are 33 multiples of 3, 11 multiples of 9, 4 multiples of 27 and 1 multiple of 81 for a total of 49 factors of 3.[/hide][/quote]\r\n\r\nAren't there only 3 multiples of 27? I got 48 as my answer.", "Solution_5": "[quote=\"ktm\"][quote=\"E^(pi*i)=-1\"][hide]There are 33 multiples of 3, 11 multiples of 9, 4 multiples of 27 and 1 multiple of 81 for a total of 49 factors of 3.[/hide][/quote]\n\nAren't there only 3 multiples of 27? I got 48 as my answer.[/quote]\r\n27/81/135/189", "Solution_6": "Question: In the above solution it is mentioned:\r\n[quote] There are 33 multiples of 3, 11 multiples of 9, 4 multiples of 27 and 1 multiple of 81 for a total of 49 factors of 3. [/quote]\r\n\r\nAren't the 11 multiples of 9 (or the 4 multiples of 27 and so on...) counted with the 33 multiples of 3 (since they are multiples of this number as well) and, thus, they should not be re-counded;", "Solution_7": "[quote=\"technox\"]Question: In the above solution it is mentioned:\n[quote] There are 33 multiples of 3, 11 multiples of 9, 4 multiples of 27 and 1 multiple of 81 for a total of 49 factors of 3. [/quote]\n\nAren't the 11 multiples of 9 (or the 4 multiples of 27 and so on...) counted with the 33 multiples of 3 (since they are multiples of this number as well) and, thus, they should not be re-counded;[/quote]'\r\n\r\nWe count the multiples of 3 and then add 1 more for each multiple of 9 (so we have counted the multiples of nine twice and the multiples of ONLY three once). This is repeated with 27 and 81.", "Solution_8": "[quote=\"bpms\"][quote=\"ktm\"][quote=\"E^(pi*i)=-1\"][hide]There are 33 multiples of 3, 11 multiples of 9, 4 multiples of 27 and 1 multiple of 81 for a total of 49 factors of 3.[/hide][/quote]\n\nAren't there only 3 multiples of 27? I got 48 as my answer.[/quote]\n27/81/135/189[/quote]\r\n\r\nOh, I read the question as the first 100 positive integers... woops.", "Solution_9": "What is the \"usual method?\"", "Solution_10": "there's a formula for finding the highest power of a some number k such that it divides n! ", "Solution_11": "[quote=solafidefarms]Let $P$ be the product of the first 100 positive odd integers. Find the largest integer $k$ such that $P$ is divisible by $3^k$.[/quote]\n\nthis is inclusion-exclusion principle.", "Solution_12": "[quote=Andrew_maybe]there's a formula for finding the highest power of a some number k such that it divides n![/quote]\n\nYes, but we are talking about the odd integers thus it doesn't work\n\n[hide=Solution]\nThe first odds are $1\\cdot 3\\cdot 5\\cdot 7\\cdots 195\\cdot 197\\cdot 199$. So, we want to see how many $3'$s are in the first $200$ integers.\n$$\\left\\lfloor \\frac{200}{3}\\right\\rfloor - \\left\\lfloor \\frac{200}{6}\\right\\rfloor +\\left\\lfloor\\frac{200}{9}\\right\\rfloor - \\left\\lfloor \\frac{200}{18}\\right\\rfloor +\\left\\lfloor \\frac{200}{27}\\right\\rfloor - \\left\\lfloor \\frac{200}{54}\\right\\rfloor+\\left\\lfloor\\frac{200}{81}\\right\\rfloor - \\left\\lfloor \\frac{200}{162}\\right\\rfloor= 66 - 33 + 22 - 11 + 7 - 3 + 2 - 1 = 49.$$", "Solution_13": "$v_3(P)=v_3(200!)-v_3(2^{100}\\cdot 100!)=v_3(200!)-v_3(100!)=97-48=\\boxed{049}$", "Solution_14": "aime p8\n\nThere are $33$ multiples of $3$, $11$ multiples of $9$, $4$ multiples of $27$, $1$ multiple of $81$, $0$ multiples of $243$, so answer is $33+11+4+1=\\boxed{049}$." } { "Tag": [ "calculus", "integration", "complex analysis" ], "Problem": "Find\r\n\\[\\int_{C}{\\frac{1}{{z+2}}dz}\\]\r\nwhere C:\r\n\\[z = 2e^{i\\theta },-\\pi \\leqslant \\theta \\leqslant 0 \\]\r\nDoes anyone knows how to solve this complex integral when we have a singularity at the begining of the curve?\r\n\r\nthanks!", "Solution_1": "The integral diverges." } { "Tag": [ "linear algebra", "matrix", "superior algebra", "superior algebra unsolved" ], "Problem": "Following up from http://www.artofproblemsolving.com/Forum/topic-36105.html where it was shown by Leva1980 that $\\mathrm{SL}_{n}(\\mathbb{Z})$ is generated by only two elements, $S, A_{1,2}$.\r\nShow that there exist an integer $K_{n}$, s.t. every matrix of $\\mathrm{SL}_{n}(\\mathbb{Z})$ is the product of at most $K_{n}$ elements belonging to $\\{S, A_{1,2}\\}$.\r\n\r\nP.S.: I don't ask to find the smallest possible value for $K_{n}$ but that would be a very nice surprise (although I think it is extremelly difficult).", "Solution_1": "It seems to me that if you only allow finite products, you will only get a finite subset of $SL_{n}$", "Solution_2": "amfulger,\r\nI think what I wrote is true, even if it seems a bit conter-intuitive. It's a nice result :wink:", "Solution_3": "As amfulger said, with at most $K_{n}$ factors from the set $\\{S, A_{12}\\}$ you can get at most $2^{K_{n}+1}-1$ different products. Since $SL_{n}(\\mathbb Z)$ is infinite (the matrices $I+k N_{12}, k \\in \\mathbb Z$ are pw. different), you cannnot generate $SL_{n}(\\mathbb Z)$ in this way.\r\n\r\nDid you mean what we understand by what you wrote? :)\r\n\r\nLiMa", "Solution_4": "Amfulger, LiMa,\r\nYou are absolutely right. I jumped to quickly from \"boundly generation against a set of elementary matrices\".\r\nI should have written:\r\n\r\nShow that there exist an integer $K_{n}$, s.t. every matrix of $\\mathrm{SL}_{n}(\\mathbb{Z})$ is the product of at most $K_{n}$ elementary matrices, where an elementary matrix is a power of $I_{n}+E_{i,j}$ ($i\\not=j$).\r\nSorry for the inconvenience, still a nice problem I hope.", "Solution_5": "read a proof of the fact that these matrices generate $SL_{n}(\\mathbb{Z})$ (I think it has been done on the forum) and you will immediately get such a $K_{n}$, but maybe not a tight bound :)" } { "Tag": [ "logarithms", "modular arithmetic", "Recursive Sequences" ], "Problem": "For a given positive integer $k$ denote the square of the sum of its digits by $f_{1}(k)$ and let $f_{n+1}(k)=f_{1}(f_{n}(k))$. Determine the value of $f_{1991}(2^{1990})$.", "Solution_1": "It is clear that $ f_1(k) \\leq (9(\\log_{10}k\\plus{}1))^2$ and, in particular, $ f_1(k) hy then obviously (hy + i + j)/(y + j) < 2.\r\n(Since (hy + i + j)/(y + j) < 2 <=> hy + i + j < 2y + 2j <=> (h - 2)y + i < j.\r\nBut we chose j > hy so (h - 2)y + i < (h - 2)y + y = (h - 1)y < hy < j.)\r\nSo we must have x + j = y + j since (x + j)/(y + j) < 2 is an integer.\r\nHence x = y.\r\n\r\nFor (a), you have to come up with a clever construction but I don't see how to do it.\r\n(I am very bad at constructing solutions, see for example the post about xx = y3 + z3.)", "Solution_1": "Oops. It's easier than I thought.\r\n\r\nTake y = 1 and x = (n + 1)! + 1.\r\n\r\nNow j + 1 divides (n + 1)! since j \\leq n. Hence k(j + 1) = (n + 1)!.\r\n\r\nSo (x + j)/(y + j) = ((n + 1)! + j + 1)/(j + 1) = k + 1.\r\n\r\nProblem solved. Sorry :?" } { "Tag": [ "probability", "calculus", "conics", "hyperbola", "geometry", "integration", "logarithms" ], "Problem": "My professor isn't very good so I'm not getting this class at all...\r\n\r\n1. Let X and Y have the pdf f(x,y) = 1, 0u)\\equal{}\\int_u^1\\int_{\\frac{u}x}^11\\,dy\\,dx\\equal{} \\int_u^1\\left(1\\minus{}\\frac{u}x\\right)\\,dx$\r\n\r\n$ \\equal{}1\\minus{}u\\minus{}u\\ln 1\\plus{}u\\ln u\\equal{}1\\minus{}u\\plus{}u\\ln u.$\r\n\r\nThat's the reverse cdf; we subtract it from $ 1$ to get the usual cdf:\r\n\r\n$ F(u)\\equal{}P(U\\le u)\\equal{}u\\minus{}u\\ln u$ for $ u\\in(0,1).$\r\n\r\n(We have $ F(u)\\equal{}1$ for $ u\\ge 1$ and $ F(u)\\equal{}0$ for $ u\\le 0.$)\r\n\r\nDifferentiate this to get the density:\r\n\r\n$ f(u)\\equal{}\\begin{cases}\\minus{}\\ln u&0 n/2*n = 1/2\r\n\r\nSo H(2^n) > n/2", "Solution_2": "Thanks. I was wayyyy off. :P \r\n\r\nI'll need to teach myself calc to understand most of that resource, but it was interesting nonetheless." } { "Tag": [ "induction", "inequalities", "combinatorics proposed", "combinatorics" ], "Problem": "Let T be a tree having vertex {1,2,...,n} and edge set denote by $E(T)$ . If $A_1,...,A_n\\subset X$ ,prove that\r\n$|A_1\\cup...\\cup A_n|\\leq \\sum_{i=1}^{n} |A_i| - \\sum_{(i,j)\\in E(T)}|A_i\\cap A_j|$", "Solution_1": "Cute one :).\r\n\r\nWe can prove this by induction on $n$. Suppose WLOG that $1$ is a terminal vertex in the tree, and $2$ is its unique neighbour. Replace the LHS in the inequality with $|A_2\\cup\\ldots\\cup A_n|$, and eliminate $|A_1|-|A_1\\cap A_2|$ from the RHS. By the induction hypothesis, the inequality we end up with holds. Now, when we modify this expression back to the original one, we add $|A_1\\setminus (A_2\\cup\\ldots\\cup A_n)|$ to the LHS, and $|A_1\\setminus A_2|$ to the RHS, i.e. the RHS increases at least as much as the LHS does, and we're done." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Gicen $ 0 < a \\le b \\le c \\le 3 ;;2c \\plus{} 3b \\ge 3bc;;bc \\plus{} 2ca \\plus{} 3ab \\ge 3abc$.Prove that:\r\n$ a^5 \\plus{} b^5 \\plus{} c^5 \\le 276$", "Solution_1": "[quote=\"quykhtn-qa1\"]Gicen $ 0 < a \\le b \\le c \\le 3 ;;2c \\plus{} 3b \\ge 3ab;;bc \\plus{} 2ca \\plus{} 3ab \\ge 3abc$.Prove that:\n$ a^5 \\plus{} b^5 \\plus{} c^5 \\le 276$[/quote]\r\nI think it's must be liked this:\r\n$ 0 < a \\le b \\le c \\le 3 ; 2c \\plus{} 3b \\ge 2bc;;bc \\plus{} 2ca \\plus{} 3ab \\ge 3abc$.Prove that:\r\n$ a^5 \\plus{} b^5 \\plus{} c^5 \\le 276$", "Solution_2": "[quote=\"tuandokim\"][quote=\"quykhtn-qa1\"]Gicen $ 0 < a \\le b \\le c \\le 3 ;;2c \\plus{} 3b \\ge 3ab;;bc \\plus{} 2ca \\plus{} 3ab \\ge 3abc$.Prove that:\n$ a^5 \\plus{} b^5 \\plus{} c^5 \\le 276$[/quote]\nI think it's must be liked this:\n$ 0 < a \\le b \\le c \\le 3 ; 2c \\plus{} 3b \\ge 2bc;;bc \\plus{} 2ca \\plus{} 3ab \\ge 3abc$.Prove that:\n$ a^5 \\plus{} b^5 \\plus{} c^5 \\le 276$[/quote]\r\nYes,you are right.I edited it.Thanks.Now,try it :)", "Solution_3": "$ 1^5\\plus{}2^5\\plus{}3^5\\equal{}a^5(\\frac{1^5}{a^5}\\plus{}\\frac{2^5}{b^5}\\plus{}\\frac{3^5}{c^5})\\plus{}(b^5\\minus{}a^5)(\\frac{2^5}{b^5}\\plus{}\\frac{3^5}{c^5})\\plus{}(c^5\\minus{}b^5)(\\frac{3^5}{c^5})\\geq \\frac{a^5}{81}(\\frac{1}{a}\\plus{}\\frac{2}{b}\\plus{}\\frac{3}{c})^5\\plus{}\\frac{b^5\\minus{}a^5}{16}(\\frac{2}{b}\\plus{}\\frac{3}{c})^5\\plus{}(c^5\\minus{}b^5)\\frac{3}{c^5}\\geq \\frac{a^5}{81}*81\\plus{}\\frac{b^5\\minus{}c^5}{16}*32\\plus{}c^5\\minus{}b^5\\equal{}a^5\\plus{}b^5\\plus{}c^5$\r\nQ>E>D", "Solution_4": "[quote=\"tuandokim\"]$ 1^5 \\plus{} 2^5 \\plus{} 3^5 \\equal{} a^5(\\frac {1^5}{a^5} \\plus{} \\frac {2^5}{b^5} \\plus{} \\frac {3^5}{c^5}) \\plus{} (b^5 \\minus{} a^5)(\\frac {2^5}{b^5} \\plus{} \\frac {3^5}{c^5}) \\plus{} (c^5 \\minus{} b^5)(\\frac {3^5}{c^5})\\geq \\frac {a^5}{81}(\\frac {1}{a} \\plus{} \\frac {2}{b} \\plus{} \\frac {3}{c})^5 \\plus{} \\frac {b^5 \\minus{} a^5}{16}(\\frac {2}{b} \\plus{} \\frac {3}{c})^5 \\plus{} (c^5 \\minus{} b^5)\\frac {3}{c^5}\\geq \\frac {a^5}{81}*243 \\plus{} \\frac {b^5 \\minus{} c^5}{16}*32 \\plus{} c^5 \\minus{} b^5 \\equal{} a^5 \\plus{} b^5 \\plus{} c^5$\nQ>E>D[/quote]\r\nNice solution :)" } { "Tag": [ "inequalities", "function", "logarithms", "geometry", "3D geometry", "sphere", "inequalities proposed" ], "Problem": "Let $a,b,c$ be non-negative numbers such that $a^2+b^2+c^2=1$. Prove that\r\n\r\n$\\sqrt{1-ab}+\\sqrt{1-bc}+\\sqrt{1-ca} \\geq \\sqrt{6}$.", "Solution_1": "Use $ \\sqrt {1 \\minus{} ab} \\equal{} \\sqrt {\\frac {1}{\\frac {1}{1 \\minus{} ab}}}$ and use Jensen inequality for convex function $ \\sqrt {\\frac {1}{x}}$ you will get $ LHS\\geq 3\\sqrt {\\frac {3}{\\frac {1}{1 \\minus{} ab} \\plus{} \\frac {1}{1 \\minus{} bc} \\plus{} \\frac {1}{1 \\minus{} ca}}}$. It should be greater that $ \\sqrt {6}$, but it is equivalent to show that $ \\frac {1}{1 \\minus{} ab} \\plus{} \\frac {1}{1 \\minus{} bc} \\plus{} \\frac {1}{1 \\minus{} ca}\\leq \\frac {9}{2}$ and this problem was already solved on mathlinks ;) :D, here: http://www.mathlinks.ro/Forum/viewtopic.php?t=36873", "Solution_2": "Indeed, it is right.\r\nBut what you say about the following:\r\n\r\n$(1-ab)(1-bc)(1-ca) \\geq \\frac{8}{27}$ ? :|", "Solution_3": "We can use logarithm and get equivalent:\r\n$\\ln(1-ab)+\\ln(1-bc)+\\ln(1-ca)\\geq \\ln\\left(\\frac{8}{27}\\right)$\r\nNow using Jensen inequality on $\\ln{\\frac{1}{x}}$ we get:\r\n$\\ln(1-ab)+\\ln(1-bc)+\\ln(1-ca)\\geq 3\\ln\\left(\\frac{3}{\\frac{1}{1-ab}+\\frac{1}{1-bc}+\\frac{1}{1-ca}}\\right)\\geq \\ln\\left(\\frac{8}{27}\\right)$\r\nThe last is one more time equivalent to the one once proved at mathlinks, I think, but I also doubt cause by the way of your post, Vasc, I think that it should not be solvable the same way as ineq in first post ;)\r\nSo wheres the miestake in my thoughts? :D", "Solution_4": "Let $x=\\sqrt{1-ab},y=\\sqrt{1-bc},z=\\sqrt{1-ca,}$ we have $x^2+y^2+z^2=3-(ab+bc+ca)\\geq 3-(a^2+b^2+c^2)\\geq 2.\\left(x\\geq \\frac{1}{\\sqrt{2}},y\\geq\\frac{1}{\\sqrt{2}},z\\geq\\frac{1}{\\sqrt{2}}\\right)\\cdots [*].$\r\n\r\nFor $(x.y,z)$ satisfying $[*],$ in $xyz$-plane, seeking the condition that the plane $x+y+z=k$ have the intersection points with $[*],$ we have \r\n\r\n$\\frac{|0+0+0-k|}{\\sqrt{1^2+1^2+1^2}}\\geq \\sqrt{2}$ which is the radius of sphere $x^2+y^2+z^2=2$\r\n\r\n$\\Longleftrightarrow |-k|\\geq \\sqrt{6}.$ Since $k\\geq 0,$ we obtain $x+y+z\\geq \\sqrt{6},$ which completes the proof.", "Solution_5": "Your method, Ondrob, works again. :lol: \r\nBut the last step follows, for the sharper inequality:\r\n$(2-3ab)(2-3bc)(2-3ca) \\geq 1$.\r\nWho give a solution?", "Solution_6": "Of course, THis one is still true :\r\n\r\n$\\sqrt{3-abc}+\\sqrt{3-bcd}+\\sqrt{3-cda}+\\sqrt{3-dab} \\ge 4\\sqrt{2}$\r\n\r\nIf $a^2+b^2+c^2+d^2=4$.\r\n\r\n:)\r\n\r\nFor your problem, Vasc. I think Expanding is the best way! :)", "Solution_7": "Actually, we have\r\n\r\n$(3-abc)(3-bcd)(3-cda)(3-dab) \\geq 16$\r\n\r\nand, better,\r\n\r\n$(2-abc)(2-bcd)(2-cda)(2-dab) \\geq 1$.\r\n\r\nThe last is nice. ;)", "Solution_8": "[quote=\"kunny\"]Let $x=\\sqrt{1-ab},y=\\sqrt{1-bc},z=\\sqrt{1-ca,}$ we have $x^2+y^2+z^2=3-(ab+bc+ca)\\geq 3-(a^2+b^2+c^2)\\geq 2.\\left(x\\geq \\frac{1}{\\sqrt{2}},y\\geq\\frac{1}{\\sqrt{2}},z\\geq\\frac{1}{\\sqrt{2}}\\right)\\cdots [*].$\n\nFor $(x.y,z)$ satisfying $[*],$ in $xyz$-plane, seeking the condition that the plane $x+y+z=k$ have the intersection points with $[*],$ we have \n\n$\\frac{|0+0+0-k|}{\\sqrt{1^2+1^2+1^2}}\\geq \\sqrt{2}$ which is the radius of sphere $x^2+y^2+z^2=2$\n\n$\\Longleftrightarrow |-k|\\geq \\sqrt{6}.$ Since $k\\geq 0,$ we obtain $x+y+z\\geq \\sqrt{6},$ which completes the proof.[/quote]\r\nsorry,I don't think this \"proof\" is right.\r\nif it is right then you can prove \r\n$x+y+z\\geq \\sqrt{6},$ if $x^2+y^2+z^2=2$.\r\nbut is certainly wrong :( \r\nsorry,if I am wrong", "Solution_9": "Could anyone tell me if we could prove the following ineq with [color=blue]ONDROB[/color]'s method ?\r\n\r\n$\\frac{1}{1-ab}+\\frac{1}{1-bc}+\\frac{1}{1-ca}\\leq \\frac{9}{2}$", "Solution_10": "[quote=\"zhaobin\"][quote=\"kunny\"]Let $x=\\sqrt{1-ab},y=\\sqrt{1-bc},z=\\sqrt{1-ca,}$ we have $x^2+y^2+z^2=3-(ab+bc+ca)\\geq 3-(a^2+b^2+c^2)\\geq 2.\\left(x\\geq \\frac{1}{\\sqrt{2}},y\\geq\\frac{1}{\\sqrt{2}},z\\geq\\frac{1}{\\sqrt{2}}\\right)\\cdots [*].$\n\nFor $(x.y,z)$ satisfying $[*],$ in $xyz$-plane, seeking the condition that the plane $x+y+z=k$ have the intersection points with $[*],$ we have \n\n$\\frac{|0+0+0-k|}{\\sqrt{1^2+1^2+1^2}}\\geq \\sqrt{2}$ which is the radius of sphere $x^2+y^2+z^2=2$\n\n$\\Longleftrightarrow |-k|\\geq \\sqrt{6}.$ Since $k\\geq 0,$ we obtain $x+y+z\\geq \\sqrt{6},$ which completes the proof.[/quote]\nsorry,I don't think this \"proof\" is right.\nif it is right then you can prove \n$x+y+z\\geq \\sqrt{6},$ if $x^2+y^2+z^2=2$.\nbut is certainly wrong :( \nsorry,if I am wrong[/quote]\r\n\r\nI should have shown when the equality holds.\r\n\r\nThe equality holds when the sphere $x^2+y^2+z^2=2$ contact with the plane $x+y+z=\\sqrt{6}\\Longleftrightarrow x=y=z=\\frac{\\sqrt{6}}{3}.$ I wonder my solution is wrong basically.Could someone point that? \r\n\r\nkunny", "Solution_11": "[quote=\"Vasc\"]Your method, Ondrob, works again. :lol: \nBut the last step follows, for the sharper inequality:\n$(2-3ab)(2-3bc)(2-3ca) \\geq 1$.\nWho give a solution?[/quote]\r\n\r\nIt's easy but nice with AM-GM :\r\n\r\nLet $t=\\frac{xy+yz+zx}{3} \\le 1$ we need to prove :\r\n$(2-x)(2-y)(2-z) \\geq 1$\r\n\r\nNotice that : $(2-xy)(2-yz)(2-zx)+(2+xy)(2+yz)(2+zx) =16+12t$\r\n\r\nBy AM-GM : $(2+xy)(2+yz)(2+zx) \\le (2+t)^3$ \r\n\r\nBecause $t \\le 1$, then $16+12t-(2+t)^3 \\ge 1$. We have done.\r\n\r\nNotice that $x=\\sqrt{3}a,y=\\sqrt{3}{b, z=\\sqrt{3}c}$.", "Solution_12": "[quote=\"hungkhtn\"] \nLet $t=\\frac{xy+yz+zx}{3} \\le 1$ .\nNotice that : $(2-xy)(2-yz)(2-zx)+(2+xy)(2+yz)(2+zx) =16+12t$[/quote]\r\nThere is a mistake in your calculation. ;)", "Solution_13": ":( A big mistake, I'll try it again...", "Solution_14": "[quote=\"Vasc\"]Let $a,b,c$ be non-negative numbers such that $a^2+b^2+c^2=1$. Prove that\n\n$\\sqrt{1-ab}+\\sqrt{1-bc}+\\sqrt{1-ca} \\geq \\sqrt{6}$.[/quote]\r\nVacs, I have a very simple solution to this inequality in my future book. The solution is due to a Mathematics Team in my country. It's involved in just AG and Cauchy.", "Solution_15": "And the harder problem is in my book, to. :lol: It's problem of VASC.", "Solution_16": "Back to Vasc's original inequality:\r\n\r\n\r\nSquaring both sides of the inequality gives\r\n\\[ 3-(ab+bc+ca)+2\\sqrt{(1-ab)(1-bc)}+2\\sqrt{(1-ca)(1-ab)}+2\\sqrt{(1-bc)(1-ac)}\\geq6. \\]\r\nNote that we have\r\n\\[ 1-ab=\\frac{1+a^2+b^2+c^2}2-ab=\\frac12+\\frac{a^2+b^2}2-ab+\\frac{c^2}2\\geq\\frac12+\\frac{c^2}2. \\]\r\nBy Cauchy,\r\n\\[ (1+u^2)(1+v^2)\\geq(1+uv)^2. \\]\r\nNow, the desired inequality follows.\r\n\r\n :P", "Solution_17": "[quote=\"Vasc\"]Indeed, it is right.\nBut what you say about the following:\n\n$(1-ab)(1-bc)(1-ca) \\geq \\frac{8}{27}$ ? :|[/quote]\r\n\r\nYes, it's right Vasc. It was solved by $p,q,r$ substitution \\[p=a+b+c,\\; q=ab+bc+ca,; r=abc.\\]", "Solution_18": "[quote=\"Vasc\"] \nBut the last step follows, for the sharper inequality:\n$(2-3ab)(2-3bc)(2-3ca) \\geq 1$.\nWho give a solution?[/quote]\r\n$\\sum_{cyc}\\frac{1}{2-3ab}\\leq3$ is not true: $a=b=\\frac{12}{17},$ $c=\\frac{1}{17}.$\r\nBut $(2-3ab)(2-3ac)(2-3bc)\\geq1\\Leftrightarrow$\r\n$\\Leftrightarrow\\sum_{cyc}(a-b)^{2}(7a^{4}-10a^{3}b+15a^{2}b^{2}-10ab^{3}+7b^{4}+$\r\n$+6a^{2}bc+6b^{2}ac-5c^{2}bc+c^{3}a+c^{3}b)\\geq0$:D, which obviously true.", "Solution_19": "In \"Algebraic Inequalities\" there are other two nice solutions to the equivalent problem:\r\n[i]If $a,b,c$ are non-negative numbers such that $a^{2}+b^{2}+c^{2}=3$, then\n$(2-ab)(2-bc)(2-ca) \\geq 1$.[/i]", "Solution_20": "A.Gladki\u0441h promised to bring to me this book. :P :)", "Solution_21": "[quote=\"arqady\"]A.Gladki\u0441h promised to bring to me this book. :P :)[/quote]\r\nBut what a pity that no one can bring it to me :(", "Solution_22": "[quote=\"zhaobin\"][quote=\"arqady\"]A.Gladki\u0441h promised to bring to me this book. :P :)[/quote]\nBut what a pity that no one can bring it to me :([/quote]\r\nit is too expensive for me... I can't afford it at all :(", "Solution_23": "[quote=\"Vasc\"]Actually, we have\n\n$ (2 \\minus{} abc)(2 \\minus{} bcd)(2 \\minus{} cda)(2 \\minus{} dab) \\geq 1$.\nThe last is nice. ;)[/quote]\r\n\r\nWhat is your solution, Vasc?", "Solution_24": "I think EV-Theorem works here." } { "Tag": [ "geometry", "incenter", "circumcircle", "geometry unsolved" ], "Problem": "Dear Mathlinkers,\r\nlet ABC be a triangle, I the incenter of ABC, UVW the cevian triangle of I wrt ABC,\r\nO the center of the circumcircle of ABC, \r\nX the meetpoint of the parallel to AI with BC, \r\nY, Z symmetrically\u2026\r\nU' the symmetric of U wrt X, \r\nV', Z' symmetrically\u2026\r\nA*B*C* the tangential triangle of ABC.\r\nProve that the lines A*U', B*V', C*W' are concurrent.\r\nSincerely\r\nJean-Louis", "Solution_1": "I can't understand your problem.\r\nI'm intersted in but what is definition of X?\r\nAnd $ AU$and $ AU'$ is isotomic line wrt $ angle BAC$ right?", "Solution_2": "Dear Yang,\r\nyes I understand your observation. Sorry I missed to specifie that\r\n\"X the meetpoint of the parallel to AI pasing through O with BC\".\r\nSorry again.\r\nSincerely\r\nJean-Louis", "Solution_3": "I will redifine points.\r\n$ A^{*}, B^{*}, C^{*}$ to $ P, Q, R$\r\nThen,\r\nBy cevian nest theorem, \r\nIf we want to show $ PU', QV', RW'$ are concurrent,\r\n$ AP, BQ, CR$ are concurrent and $ AU', BV', CW'$ are concurrent.\r\nBut $ AP, BQ, CR$ are $ A \\minus{} symmedian, B \\minus{} symmedian, C \\minus{} symmedian$ of triangle $ ABC$ respectively.\r\nand with simple calculating, by ceva theorem, we can get $ AU', BV', CW'$ are concurrent.\r\n\r\nSo $ PU', QV', RW'$ are concurrent :lol:\r\n\r\nif you want to know what is cevian nest theorem\r\nThen see here\r\n\r\nhttp://www.mathlinks.ro/viewtopic.php?t=6579&search_id=1788733684\r\n\r\nIt's sometimes very useful theorem.", "Solution_4": "Dear Mathlinkers,\r\ntwo synthetic proofs of this known result originated from Clark Kimberling has been put on my website\r\nhttp://perso.orange.fr/jl.ayme vol. 5 The second mid-arc point\u2026\r\nSincerely\r\nJean-Louis" } { "Tag": [ "topology", "algebra", "function", "domain", "real analysis", "real analysis theorems" ], "Problem": "$ k \\le n$\r\n\r\nLet $ S \\subset R^n$ be a smooth k-dimensional surface and let $ \\varphi : I^k \\toU$ be a local chart of S.\r\n\r\nDoes $ \\partial \\varphi (I^k ) \\cap S$ have k-dimensional measure zero in $ R^n$*?\r\n\r\n*In one sense, this means if $ f: I^k \\to T$ is some (any) local chart of S, the set theoretic preimage $ f^{ - 1} (\\partial \\varphi (I^k ))$ has k-dimensional measure zero in $ R^k$.\r\n\r\n\r\nI'm trying to investigate whether $ \\partial \\varphi (I^k )$ is 'negibligible' in $ R^n$ in a 'k-dimensional sense'.\r\n\r\nIf not, counterexample?", "Solution_1": "I don't think the question in your title is the same as the one in your post. The answer to the question in your title is yes- but that involves a measure-zero subset of the domain. Extending to the closure of your domain is not something that can be guaranteed.", "Solution_2": "Yes, the question in the title has nothing to do with what I'm asking here. :blush: \r\n\r\nWhat are the counterexamples then?" } { "Tag": [], "Problem": "So, I have to find all $ x,y,z \\in \\mathbb{N}$ s.t. $ \\frac {xy}{z} \\plus{} \\frac {yz}{x} \\plus{} \\frac {zx}{y} \\equal{} 3$. I've seen some similiar problems but I can't get it right. I tried to multiply by $ xyz$ and then to factor everything on the left side. Any hints?", "Solution_1": "[quote=\"neuro\"]So, I have to find all $ x,y,z \\in \\mathbb{N}$ s.t. $ \\frac {xy}{z} \\plus{} \\frac {yz}{x} \\plus{} \\frac {zx}{y} \\equal{} 3$. I've seen some similiar problems but I can't get it right. I tried to multiply by $ xyz$ and then to factor everything on the left side. Any hints?[/quote]\r\n\r\n\r\n\r\n$ a^{2}\\plus{}b^{2}\\plus{}c^{2}\\geq ab\\plus{}ac\\plus{}bc$ -----> \r\n\r\n$ \\frac{xy}{z}\\plus{}\\frac{yz}{x}\\plus{}\\frac{zx}{y}\\equal{}3$ <-> $ 3xyz\\equal{}(xy)^{2}\\plus{}(yz)^{2}\\plus{}(zx)^{2}\\geq (xy)(yz)\\plus{}(xy)(xz)\\plus{}(yz)(zx)\\equal{}(x\\plus{}y\\plus{}z)xyz$ ----> \r\n\r\n\r\n$ 3xyz\\geq (x\\plus{}y\\plus{}z)xyz$ ----> $ 3\\geq x\\plus{}y\\plus{}z$ ---> $ x\\equal{}y\\equal{}z\\equal{}1$", "Solution_2": "Thank you very much, can you point me to some similar problems?", "Solution_3": "I think you could use AM-GM at the beginning.\r\n\\[ \\frac {xy}{z} \\plus{} \\frac {yz}{x} >\\equal{}2y\\]\r\n\\[ \\equal{}>\\frac {xy}{z} \\plus{} \\frac {yz}{x} \\plus{} \\frac {zx}{y}>\\equal{}x\\plus{}y\\plus{}z\\]\r\n\\[ \\frac {xy}{z} \\plus{} \\frac {yz}{x} \\plus{} \\frac {zx}{y} \\equal{} 3\\]\r\n\\[ <\\equal{}>3>\\equal{}x\\plus{}y\\plus{}z\\]\r\n.....", "Solution_4": "Let $ x \\leq y \\leq z$\r\n\r\n$ \\frac {yz}{x} < 3 \\implies z < \\frac {3x}{y} \\leq 3$\r\n\r\n$ \\therefore 1 \\leq x \\leq y \\leq z \\leq 2$\r\n\r\nThere are four possibilities:\r\n$ (1,1,1) \\qquad (1,1,2)\\qquad (1,2,2) \\qquad (2,2,2)$\r\n\r\nThe solution is $ (x,y,z) \\equal{} (1,1,1)$", "Solution_5": "By AM-GM\r\n\r\n$ \\dfrac{xy}{z} \\plus{} \\dfrac{yz}{x} \\plus{} \\dfrac{xz}{y} \\ge 3xyz$\r\n\r\n$ \\dfrac{1}{z^2} \\plus{} \\dfrac{1}{y^2} \\plus{} \\dfrac{1}{x^2} \\ge 3$\r\n\r\nClearly the only solution to is $ x\\equal{}y\\equal{}z\\equal{}1$", "Solution_6": "Shouldn't that be $ 3 \\sqrt[3]{xyz}$?\r\n\r\nRegardless, you'd still have $ 3\\equal{} \\dfrac{xy}{z} \\plus{} \\dfrac{yz}{x} \\plus{} \\dfrac{xz}{y} \\ge 3 \\sqrt[3]{xyz} \\implies xyz \\leq 1$" } { "Tag": [ "algebra", "polynomial", "algebra unsolved" ], "Problem": "Let polynomial f(x) in Z(x) ,deg f(x)=n and\r\n f(2^i)= 1/2^i ( i=1,2,..n)\r\nFind value f(0)", "Solution_1": "Let f(x) one is polinoms (may be find by Lagrange approximation), g(x)=(x-2)(x-4)...(x-2^n). Then for all whole m\r\nf(x)+mg(x) satisfied your conditions. Therefore f(0) can not be find from your conditions, because g(0) is not equal 0.", "Solution_2": "If f(x) belongs Z(x), then f(x) is whole for all whole x. I think f(x) belongs Q[x], and f(2^i)=2^(-i) for all i=0,1,...,n. Then \r\nf(x) can be find by Lagrange approximation.", "Solution_3": "Sorry for my topo\r\n f(x) must be in R(x)\r\nNow help me", "Solution_4": "TRUE PROBLEM\r\n\r\n[quote=\"bui_haiha2000\"]Let polynomial f(x) in R(x) ,deg f(x)=n and\n f(2^i)= 1/2^i ( i=1,2,..n)\nFind value f(0)[/quote]", "Solution_5": "No one solve my problem???", "Solution_6": "[quote=\"bui_haiha2000\"]No one solve my problem???[/quote]\r\nNo one wait more than one day\u00bf\u00bf\u00bf", "Solution_7": "f(2^i)= 1/2^i ( i=0,1,2,..n)\r\n\r\nLet $g(x)=\\prod_{i=0}^n (x-2^i) +(-1)^n.2^{\\binom{n}{2}}$\r\ng(0)=0\r\n\r\n${f(x)=\\frac{g(x)}{x.(-1)^n.2^{\\binom{n}{2}}}}$\r\nIf k is the coefficient of x in g(x) then $f(0)=\\frac{k}{(-1)^n.2^{\\binom{n}{2}}}$\r\n\r\n$k=(-1)^{n}.2^{\\binom{n}{2}}.(1+2^{-1}+2^{-2}+\\ldots+2^{-n})$\r\ntherefore $f(0)=2-\\frac{1}{2^n}$" } { "Tag": [ "quadratics", "calculus", "derivative", "algebra" ], "Problem": "The real numbers $s,t$ varies being satisfied with $s^2+t^2=1,s\\geqq 0,t\\geqq 0$.\r\nFind the range of the value of the root for the following equation can be valued.\r\n\\[x^4-2(s+t)x^2+(s-t)^2=0\\]", "Solution_1": "does that mean between some minimum and infinity, or...?", "Solution_2": "Sorry for my poor English, probability1.01.\r\nI have just edited.Please rewrite this problem's sentense into plain English for my brushing my English up.\r\n\r\nkunny", "Solution_3": "[hide] \\[x^2=s+t \\pm 2^{1/2}. Since... s+t \\leq \\sqrt{2(s^2+t^2)}=2^{1/2}, x^2=s+t+2^{1/2} \\]\n So \\[2^{3/4} \\leq x_1=\\sqrt{s+t+2^{1/2} }\\geq 2^{1/4} , and -2^{1/4}\\geq x_2= -\\sqrt{s+t+2^{1/2} }\\geq -2^{3/4}\\][/hide][/hide]", "Solution_4": "Oooops I made a little mistake. I hope this is true :? \r\n[hide] \\[x^2=s+t \\pm 2^{1/2}. Since... s+t \\leq \\sqrt{2(s^2+t^2)}=2^{1/2}, x^2=s+t+2^{1/2} \\]\n So \\[2^{3/4} \\geq x_1=\\sqrt{s+t+2^{1/2} }\\geq 2^{1/4} , and -2^{1/4}\\geq x_2= -\\sqrt{s+t+2^{1/2} }\\geq -2^{3/4}\\][/hide]", "Solution_5": "Regrettably you are wrong, Mamat.", "Solution_6": "[hide]Using the quadratic equation:\n$x^2=s+t\\pm\\sqrt{4st}$\nThen substituting $s=sin(\\theta)$ and $t=cos(\\theta)$ for $0<\\theta<\\pi/2$ (I couldn't get less than or equal to or the greater to render correctly)\n\n$x^2=sin(\\theta)+cos(\\theta)\\pm\\sqrt{2sin(2\\theta)}$\nThe derivative of the right side is $cos( \\theta)-sin( \\theta)+\\sqrt{2}\\frac{cos(2 \\theta)}{\\sqrt{sin(2\\theta)}}$ which = 0 at $x=\\pi/4$ on the necessary interval.\n\nSo maximum $x^2=\\frac{\\sqrt(2)}{2}+\\frac{\\sqrt(2)}{2}+\\sqrt{2}=2\\sqrt{2}=2^{\\frac{3}{2}}$\nTherefore x max=$2^{\\frac{3}{4}}$ and x min is negative that.[/hide][/hide]", "Solution_7": "Yes.you are right, Wumbate. \r\n\r\nHere is my solution.\r\n\r\n$s\\geqq 0,t\\geqq 0,s^2+t^2=1\\Longleftrightarrow (s+t)^2=2(s^2+t^2)-(s-t)^2\\leqq 2 \\Longleftrightarrow 1\\leqq s+t\\leqq \\sqrt{2}$,\r\n\r\nequality occurs when $(s,t)=(0,1),(1,0),\\left(\\frac{1}{\\sqrt{2}},\\frac{1}{\\sqrt{2}}\\right)$.\r\n\r\nSolving for quadratic equation to $x^2$, we have $x^2=s+t\\pm\\sqrt{4st}=s+t\\pm\\sqrt{(s+t)^2-(s-t)^2}$.\r\n\r\nFrom $A.M.\\geqq G.M.\\ 0\\leqq s+t-\\sqrt{4st},s+t+\\sqrt{4st}\\leqq 2(s+t)\\leqq 2\\sqrt{2}$.\r\n\r\nTherefore $0\\leqq x^2\\leqq 2\\sqrt{2}$, the desired answer is $-2^{\\frac{3}{4}}\\leqq x\\leqq 2^{\\frac{3}{4}}$." } { "Tag": [ "trigonometry", "inequalities", "inequalities proposed" ], "Problem": "Let real numbers $ a,b,c\\in (0;1)$ satisfying $ (a\\plus{}b\\plus{}c\\minus{}2)^2\\equal{}2abc$. Prove that $ \\sqrt{\\frac{a}{2\\minus{}a}}\\plus{}\\sqrt{\\frac{b}{2\\minus{}b}}\\plus{}\\sqrt{\\frac{c}{2\\minus{}c}}\\ge \\sqrt{3}$", "Solution_1": "[quote=\"thaithuan_GC\"]Let real numbers $ a,b,c\\in (0;1)$ satisfying $ (a \\plus{} b \\plus{} c \\minus{} 2)^2 \\equal{} 2abc$. Prove that $ \\sqrt {\\frac {a}{2 \\minus{} a}} \\plus{} \\sqrt {\\frac {b}{2 \\minus{} b}} \\plus{} \\sqrt {\\frac {c}{2 \\minus{} c}}\\ge \\sqrt {3}$[/quote]\r\nIf $ a\\plus{}b\\plus{}c \\ge 2$, then we have\r\n\\[ \\sqrt{\\frac{a}{2\\minus{}a}} \\plus{}\\sqrt{\\frac{b}{2\\minus{}b}} \\plus{}\\sqrt{\\frac{c}{2\\minus{}c}} \\ge \\sqrt{\\frac{a}{b\\plus{}c}} \\plus{}\\sqrt{\\frac{b}{c\\plus{}a}}\\plus{}\\sqrt{\\frac{c}{a\\plus{}b}} \\ge 2>\\sqrt{3}\\]\r\nIf $ a\\plus{}b\\plus{}c \\le 2$, the given condition can be rewriten as: $ a\\plus{}b\\plus{}c\\plus{}\\sqrt{2abc} \\equal{}2$. This identity suggests us to put $ a\\equal{}2\\cos^2 A,b\\equal{}2\\cos^2B,b\\equal{}2\\cos^2C$ where $ A,B,C \\in (0,\\frac{\\pi}{2})$, then we have $ \\cos^2A\\plus{}\\cos^2B\\plus{}\\cos^2 C\\plus{}2\\cos A\\cos B \\cos C\\equal{}1$, it shows that $ A,B,C$ are the angles of a triangle. Moreover, the above inequality becomes\r\n\\[ \\cot A \\plus{}\\cot B \\plus{}\\cot C \\ge \\sqrt{3}\\]\r\nwhich is just a known inequality in triangle. \r\n:)", "Solution_2": "Nice can hang.\r\n\r\nBy the way whatz ur real name?", "Solution_3": "How you get this?\r\n[quote=\"can_hang2007\"]\n then we have $ \\cos^2A \\plus{} \\cos^2B \\plus{} \\cos^2 C \\plus{} 2\\cos A\\cos B \\cos C \\equal{} 1$, it shows that $ A,B,C$ are the angles of a triangle. :)[/quote]", "Solution_4": "[quote=\"befuddlers\"]Nice can hang.\n\nBy the way whatz ur real name?[/quote]\r\n\r\nMy real name is Vo Quoc Ba Can. You can call me Can. Thanks. :blush:", "Solution_5": "Can_Hang, could u please explain ur solution further ? Especially the first few lines. PLease." } { "Tag": [ "calculus", "geometry", "geometric transformation", "reflection", "vector" ], "Problem": "I didnt know where to post this so excuse me if its the wrong forum.\r\nAnyway, I have not heard anything about ap statistics, on this website. Is anybody taking the class or exam, what book did you use, are you really preapred for the exam?\r\nDid anybody study it by them selves? How long do you think it would take to study all of it?\r\nIf you used a review study guide book which did you use?", "Solution_1": "I think some old threads that have sunk out of view on the first page described experiences taking the AP statistics exam. The consensus here seems to be that the math on that exam is easy and that the concepts are common-sense enough that this is a good exam to self-study for.", "Solution_2": "For what it's worth I (somewhat) disagree with the general sentiment. I think some of the nuances of statistics can be hard to pick up by studying on your own, especially from a prep book alone. While some people might be able to do well on the exam by studying, it doesn't mean they got a full understanding of the subject. If someone does self-study for the exam I definitely advise them to get a good textbook. A lot of the more commonly used ones can be found here: http://apcentral.collegeboard.com/apc/public/courses/teachers_corner/51924.html\r\n\r\nBut in general, self-studying for any AP exam is not for everyone, and people who give that advice are sometimes doing a disservice by assuming that what worked for them will work for everyone else. Just for comparison:\r\n\r\nAP Calculus AB (2006)\r\nGrade - Number - % At\r\n 5 43,925 22.3\r\n 4 40,353 20.5\r\n 3 36,627 18.6\r\n 2 30,486 15.5\r\n 1 45,790 23.2\r\n\r\n3 or Higher = 61.3\r\nMean Score = 3.03\r\nStand. Dev. = 1.48\r\n\r\nAP Statistics (2006)\r\nGrade - Number - % At\r\n 5 11,154 12.6\r\n 4 19,568 22.2\r\n 3 22,357 25.3\r\n 2 16,119 18.3\r\n 1 19,039 21.6\r\n\r\n3 or Higher = 60.2\r\nMean Score = 2.86\r\nStand. Dev. = 1.32\r\n\r\nSo there are a lot of people who found the AP Stats exam pretty hard, and there are a lot of people who really do need a teacher to explain this stuff to them. You might assume that none of those people are here on AoPS, but that may or may not be true.\r\n\r\nI think each individual person has to give a self-assessment of whether self-studying for an AP Exam is right for them. Maybe it is, and maybe it isn't.", "Solution_3": "[quote=\"gauss202\"]While some people might be able to do well on the exam by studying, it doesn't mean they got a full understanding of the subject. If someone does self-study for the exam I definitely advise them to get a good textbook.[/quote] \r\n\r\nI agree with this. I see conceptual errors in thinking about statistics almost every day from the keyboards of students who claim to have taken an AP statistics class. The human brain has built-in cognitive illusions that tend to result in misunderstanding fairly straightforward concepts of statistics, so it takes practice and careful guidance to learn to think statistically.", "Solution_4": "The inferential reasoning used in statistics doesn't come naturally to all highly skilled in mathematics and deductive reasoning. I wonder how many of those who self-studied \"5's\" last year could explain the topic to interested students this year.\r\n\r\nWhen thinking about taking AP courses and/or AP exams, look carefully at the corresponding college courses. For BC Calculus, the courses typically form the base for substantial further study in mathematics. \r\n\r\nCollege courses that you might gain exemptions from by taking the AP Statistics exam are designed for majors in the social sciences, business, nursing, etc. At most schools, these majors typically take just one course in statistics. \r\n\r\nThe AP statistics course is algebra-based. Statistics courses designed for majors in statistics or mathematics will be calculus-based. The credits/exemptions that you might earn taking AP Statistics might be of little value if you need a calculus-based Statistics course in college.\r\n\r\nSimilarly, introductory college science courses such as Physics are sometimes typically taught in two flavors, calculus-based and algebra-based. If you major in the physical sciences (or math) you will probably start with the calculus-based courses. The algebra-based courses are for students who are knocking out a science requirement and plan no further study of the topic. AP credits for the algebra-based Physics courses won't mean much for such students.\r\n\r\nEdited: As Kent Merryfield points out below, General Chemistry is typically taught with an algebra prereq, not calculus. The AP Chemistry exam is useful for most folks. The above post has been edited to reflect this.", "Solution_5": "College level General Chemistry is not a calculus-based course anywhere that I know of. The AP Chemistry exam does widely \"count,\" especially for biological science majors - in fact, it's the AP course I'd suggest has the most value for a prospective biological science major.\r\n\r\nAP Physics does come in the two \"flavors\" mentioned here; AP Physics B won't \"count\" for a physical science or engineering major.", "Solution_6": "I am self-studying Statistics this year, although in a different way than most people on these forums. During my vector calculus class I can choose to either stay in class, listen to lecture, or go outside and work on Stats on the computer. It appears on my high school transcript that I took the class \"Early Bird\", which means before school, and it is an AP credit and all of that.\r\n\r\nHowever, the coordinator of the online course (from APEX) didn't get me the textbook which was explicitly stated as necessary, and I only realized this recently. I had been studying the online lectures and something felt like it was missing (they say in places \"refer to resources section\" which is a blank page). They also quiz over materail not lectured about.\r\n\r\nFor this reason, I got the school coordinator to order me a textbook and its due to come in sometime next week. \r\n\r\nAll in all, I will essentially be learning the whole course from next week to April or so.\r\n\r\n\r\nFor a very busy person, is this enough time to master the course? (I know that people here get 5's and do it in 2 weeks, so,....)", "Solution_7": "Check this really good [url=http://statland.org/MAAFIXED.PDF]review of statistics textbooks for mathematicians[/url] for the rationale of the AP statistics syllabus and some issues to look out for when studying statistics.", "Solution_8": "[quote=\"me@home\"]\nFor a very busy person, is this enough time to master the course? (I know that people here get 5's and do it in 2 weeks, so,....)[/quote]\r\n\r\nI'm going to go ahead and say no. Even though people sometimes rave about Statistics being easy, mastering the course definitely takes a lot more time input. The people you're talking about did simply that, \"get 5's\" and I'm pretty sure they don't quite remember much statistics now :D", "Solution_9": "There is definitely a difference between \"master the course,\" and \"gain a 5 on the AP test.\" You can see that my first answer in this thread related mostly to passing the test, and my second answer mostly to gaining an understanding of statistics. The important thing about any AP course is not to think that getting a 5 on the test means that you are done learning the subject.", "Solution_10": "Does anybody have or no where there might be good\r\nprograms for the graphing calculator either the ti-83 or 89, that could be use full for Ap statistics", "Solution_11": "The TI 83/84 calculators come with lots of useful standard features for AP Stats. Normal, binomial, Poisson, etc. distributions are pre-programmed, eliminating the need for tables (2nd VARS - Distr.) Univariate statistics and some rudimentary analysis of variance and regression output are all available. There might be a package or two that might help with data entry or that expand the typical output, but I think virtually everything that you really need for AP Stats and much more is already on your calculator.", "Solution_12": "Yes, the latest versions of the 89 also have built-in stats software much like that that has long been on the 83/84 series.", "Solution_13": "I assure you that AP Statistics was many times easier than AP Calculus. It does not require much understanding as Calculus and most of the material consists of definitions and basic understandings. For example, the Free Response parts ask to do a statistical test. Instead of just finding the $t$ value, you have to state the null hypothesis, conditions, and a conclusion to get full points.", "Solution_14": "I think many of the AoPS-type people will find calculus easier to learn than stats. Maybe I'm just biased because I found calculus to be more fun although more work. However, I'm comfortable calling things easier than others even if the thing that I consider easier seems that way only because I like it more...", "Solution_15": "AP stats is generally speaking not a hard AP, especially since the cutoff for 5 is low. However, I found it incredibly lame because you don't really get enough theory to actually understand 90% of what you learn. Also, it's not connected with most other math, so you will very likely forget most of the material if you're going into math or hard sciences. That's why I don't really consider stats to be actual math. If you want to do well on the AP, just memorize all the tests and their stipulations, understanding the reasons behind them when possible, and you should be fine.", "Solution_16": "[quote=\"probability1.01\"]AP stats is generally speaking not a hard AP, especially since the cutoff for 5 is low. However, I found it incredibly lame because you don't really get enough theory to actually understand 90% of what you learn. Also, it's not connected with most other math, so you will very likely forget most of the material if you're going into math or hard sciences. That's why I don't really consider stats to be actual math. If you want to do well on the AP, just memorize all the tests and their stipulations, understanding the reasons behind them when possible, and you should be fine.[/quote]\r\n\r\nYeah that's why I don't particularly like AP stats.", "Solution_17": "Statistics ISN'T math. See \r\n\r\nhttp://statland.org/MAAFIXED.PDF \r\n\r\nto get a better idea of what it is. Physics isn't math either, but that doesn't mean physics is a bad subject.", "Solution_18": "As a senior in high school i took AP statistics concurrently with Precalculus. Right now i am a college freshman in what is equivalent to AP Calculus BC. So far i find Calculus much easier than statistics, but maybe i am the oddball like that. Statistics isn't mathematically rigorous. It is all just simple computations that your calculator does for you anyways. The concepts in Statistics may seem easy at the start, but they are hard and it can be very difficult remembering all the procedures for all the inference tests and stuff. The AP test isn't too hard. I had \"Senioritis\" so i didn't study at all for the test and i got a 4.", "Solution_19": "It's unfortunate that good math students get the impression from AP Statistics that the field of statistics is not \"mathematically rigorous.\" Take a look at some of the leading journals in the field: The Annals of Statistics, The Journal of the American Statistical Association, Biometrika, Technometrics, The Journal of the Royal Statistical Society, etc. You will find that new ideas in the field are typically well-defined and \"proven.\" \r\n\r\nBefore AP Statistics, most good math/stat students were introduced to statistics through courses in \"Mathematical Statistics.\" In these classes, not much effort was spent on the various statistical models you see in AP Stats. However, basic concepts and properties of estimators were derived in a \"rigorous\" fashion. After these classes, you will appreciate the difference between \"least squares,\" \"maximum likelihood\" and other types of estimators. You should be able to engage in a conversation about the properties of an innovative estimation technique or method.\r\n\r\nMathematical Statistics courses still exist at most universities. I'd recommend taking a couple of these before giving up on the field of statistics. \r\n\r\nI'm beginning to think that it's a real mistake to introduce top math students to the statistics field with the \"AP Stats\" course. Most math/physical science/stat students will never be required to take the AP Stats equivalent in college. It's a course designed for majors in business, life and social sciences. \r\n\r\nSome statisticians are highly mathematical and spend their careers developing methodology or statistical theory (i.e. theorems and proofs.) Many of these folks are in math or statistics departments at universities and a few work in government agencies. \r\n\r\nOther statisticians are more applied and get their hands dirty designing experiments or surveys and analyzing real data. Many applications point toward a need for new theory and techniques. Those working on more demanding applications will develop the theory and methods required and publish in both theoretical and applied journals. You'll find many of these statisticians in medical research and the pharmaceutical industry.", "Solution_20": "I still think that there is a good place in the high school curriculum for a course that emphasizes--as an AP statistics course should--that statistics is all about data, and mathematical analysis of data has to be connected to the real-world context in which the data were gathered.", "Solution_21": "In response to gt59\r\n\r\nI apologize if i was unclear. I didn't mean that statistics as a whole was not mathematically rigorous, i just meant that the AP class specifically doesn't require much math beyond intermediate algebra, so the math isn't \"hard\" so to speak. I am aware that more advanced studies in statistics use much more \"advanced\" methods of math such as calculus and whatnot." } { "Tag": [ "probability" ], "Problem": "It has come to a policeman's ears that 5 gangsters (all of different height) are meeting, one of them is the clan leader, he's the tallest of the 5. He knows the members will leave the building one by one, with a 10-minute break between them, and too bad for him Belgium has not enough policemen to follow all gangsters, so he's on his own to spot the clanleader, and he can only follow one member.\r\n\r\nSo he decides to let go the first 2 people, and then follow the first one that is taller than those two. What's the chance he actually catches the clan leader like this?", "Solution_1": "Let ABCDE be the gangsters with A the tallest.\r\n\r\nThe ways that the first 2 gangstars can come out are AB AC AD AE BC BD BE CD CE DE and the reverse, all with equal probability. Then A is \"marked\" 40% of the time and caught 0%; B is marked 30%, C 20%, and D 10%.\r\n\r\nThen when B is marked, A is caught 100%. When C is marked, either of A or B could come first with equal chances, so 50% prob. When D is marked, A comes first 33% of the time. So in total .40(0) + .30(1) + .20(.5) + .10(1/3) = 43.33... %" } { "Tag": [ "quadratics", "algebra", "quadratic formula" ], "Problem": "problem 1\r\n\\[ {\\rm{Solve using quadratic formula}},{\\rm{ for }}x{\\rm{ }}: {\\rm{ }}a\\left( {x{\\rm{ }} \\plus{} {\\rm{ }}1} \\right){\\rm{ }} \\equal{} {\\rm{ }}x\\left( {a{\\rm{ }} \\plus{} {\\rm{ }}1} \\right)\r\n\\]\r\n\r\nproblem2\r\n\r\n\\[ {\\rm{Solve for }}x: {\\rm{ }}\\left( {x \\minus{} 3} \\right)\\left( {x \\plus{} 9} \\right)\\left( {x \\minus{} 7} \\right)\\left( {x \\plus{} 5} \\right){\\rm{ }} \\equal{} 1680\r\n\\]\r\n\r\n\r\n\r\n\r\nmembers please submit more sums of same type with solutions\r\nSTUDENTS FROM INDIA\r\nONLINE COACHING FOR SSC CBSE ICSE BOARD MATHS SITTING AT YOUR HOUSE\r\nFOR DETAILS MAIL sanbittu@hotmail.com", "Solution_1": "[quote=\"sanbittu\"]problem 1\n\\[ {\\rm{Solve using quadratic formula}},{\\rm{ for }}x{\\rm{ }}: {\\rm{ }}a\\left( {x{\\rm{ }} \\plus{} {\\rm{ }}1} \\right){\\rm{ }} \\equal{} {\\rm{ }}x\\left( {a{\\rm{ }} \\plus{} {\\rm{ }}1} \\right)\n\\]\nproblem2\n\\[ {\\rm{Solve for }}x: {\\rm{ }}\\left( {x \\minus{} 3} \\right)\\left( {x \\plus{} 9} \\right)\\left( {x \\minus{} 7} \\right)\\left( {x \\plus{} 5} \\right){\\rm{ }} \\equal{} 1680\n\\]\nmembers please submit more sums of same type with solutions\nSTUDENTS FROM INDIA\nONLINE COACHING FOR SSC CBSE ICSE BOARD MATHS SITTING AT YOUR HOUSE\nFOR DETAILS MAIL sanbittu@hotmail.com[/quote]\r\n1. $ a(x\\plus{}1)\\equal{}x(a\\plus{}1) \\Leftrightarrow x\\equal{}a$\r\n2. $ (x\\minus{}3)(x\\plus{}9)(x\\minus{}7)(x\\plus{}5)\\equal{}1680 \\Leftrightarrow [(x\\minus{}3)(x\\plus{}5)][(x\\plus{}9)(x\\minus{}7)]\\equal{}1680$\r\n$ (x^2\\plus{}2x\\minus{}15)(x^2\\plus{}2x\\minus{}63)\\equal{}1680$\r\nSet $ t\\equal{}x^2\\plus{}2x\\minus{}39 \\ge \\minus{}40$\r\nThe equation becomes:\r\n$ (t\\minus{}24)(t\\plus{}24)\\equal{}1680 \\Leftrightarrow t^2\\equal{}2256 \\Leftrightarrow t\\equal{}4\\sqrt{141}$ ( because $ t \\ge \\minus{}40$)\r\n$ \\Leftrightarrow (x\\plus{}1)^2\\equal{}4(10\\plus{}\\sqrt{141})$\r\n$ \\Leftrightarrow x\\equal{}\\minus{}1\\minus{}2\\sqrt{10\\plus{}\\sqrt{141}}$ or $ x\\equal{}\\minus{}1\\plus{}2\\sqrt{10\\plus{}\\sqrt{141}}$" } { "Tag": [ "geometry", "advanced fields", "advanced fields unsolved" ], "Problem": "I have to submit them by Monday and they are crazy....errr!!!!\r\n\r\n1. One in seven mathematicians is a philosopher, while one in nine philosophers\r\n is a mathematician. Are there more philosophers than mathematicians? \r\n \r\n 2. A study shows that the percentage of left-handed people among mathematicians\r\n is greater than the percentage of left-handed people among \r\n non-mathematicians. Prove that it follows that the percentage of \r\n mathematicians among left-handed people is greater than the percentage of\r\n mathematicians among right-handed people. \r\n \r\n 3. Three sailors were shipwrecked on an island. To make sure that they would \r\n have enough to eat, they spent the day gathering bananas. They put the \r\n bananas in a large pile and decided to divide the bananas equally among \r\n themselves the next morning. However, each sailor distrusted the other \r\n two, so after they were asleep, one of the sailors awoke and divided the \r\n pile of bananas into three equal shares. When he did so, he found that he \r\n had one banana left over; he fed it to a monkey. He then hid his share and\r\n went back to sleep. A second sailor awoke and divided the remaining bananas\r\n into three equal shares. He too found that he had one banana left over and \r\n fed it to the monkey. He then hid his share and went back to sleep. The \r\n third sailor awoke and divided the remaining bananas into three equal shares.\r\n Again, one banana was left over, so he fed it to the monkey. He then hid his\r\n share and went back to sleep. When the sailors got up the next morning, the\r\n pile was noticeably smaller, but since they all felt guilty, none of them \r\n said anything and they all agreed to divide the bananas. When they did so, \r\n one banana was left over and they again gave it to the monkey.\r\n What was the minimal possible number of bananas at the beginning? \r\n \r\n 4. Alice, Betty, and Carol took the same series of examinations. For each \r\n examination there was one mark of x, one mark of y, and one mark of z, where \r\n x, y, z are distinct positive integers. After all the examinations, Alice \r\n had a total score of 20, Betty a total score 10, and Carol a total score \r\n of 9. \r\n If Brabara was placed first in Algebra, who was placed second in Geometry? \r\n (Terence Tao, \"Solving mathematical problems: a personal perspective\", \r\n pages 86-89, books.google.com) \r\n \r\n 5. A party was held at the house of the Schlobodkins. There were four other\r\n couples present (besides Mr. and Mrs. Schlobodkins), and many, but not all, \r\n pairs of people shook hands. Nobody shook hands with anyone twice, and \r\n nobody shook hands with his/her spouse. Both the host and the hostess shook \r\n some hands. At the end of the party, Mr. Schlobodkins polls every person \r\n present to see how many hands each person (other than himself) shook. Each \r\n person gives a different answer. \r\n Determine how many hands Mrs. Schlobodkins must have shaken. \r\n (Paul R. Halmos, \"The Thrills of Abstraction\",\r\n The Two-Year College Mathematics Journal\", Vol. 13, No. 4 (1982) JSTOR)\r\n \r\n 6. In general, a magic square is an arrangenment of the integers from 1 to n*n, \r\n in cells of an n-by-n square such that the numbers in each row, column, and \r\n diagonal give the same sum, the magic sum. \r\n (a) What is the magic sum for a 3-by-3 magic square? \r\n (b) What is the magic sum for an n-by-n magic square? \r\n (c) Place the numbers 1 through 9 into the nine cells of a 3-by-3 square \r\n to make it a magic square.\r\n\r\nI have been busy with meetings and all and didn't get time to think about it....help!!!!!!!!!", "Solution_1": "- Problem 3 is similar to this one:\r\n\r\nhttp://www.mathlinks.ro/viewtopic.php?t=185766\r\n\r\nThe minimal number of bananas is 79, according to my count.\r\n\r\n- Problem 6 is the easiest one.\r\n\r\nSolve part (b) first. The rows of a $ n \\times n$ magic square partition the set $ \\{1, 2, ..., n(n \\minus{} 1), n^{2}\\}$ into $ n$ disjoint subsets whose elements sum to the same number, say $ N$. Since $ 1 \\plus{} 2 \\plus{} ... \\plus{} n^{2} \\equal{} \\frac {n^{2}(n^{2} \\plus{} 1)}{2}$, it follows that $ N \\equal{} \\frac {n(n^{2} \\plus{} 1)}{2}$." } { "Tag": [ "summer program", "MathPath" ], "Problem": "A few questions on the format of the quiz answers:\r\n\r\nIn the qualifying quiz it says to \"show all steps in your reasoning and in your computation\". If, for example, you did a massive long division, do they want to see it, or does it assume use of calculators?\r\n\r\nAlso, do they accept typed quizzes?\r\n\r\nThanks.", "Solution_1": "I don't really know, but since you can use any resource except another person (textbooks, calculators, etc.) I don't think you need to show really long long division." } { "Tag": [ "geometry", "rectangle", "ratio", "symmetry" ], "Problem": "If someone could help me on these questions and explain the answers, I would be very appreciative... :)\r\n\r\n1. In rectangle $ ABCD$, points $ F$ and $ G$ lie on segment $ AB$ so that $ AF \\equal{} FG \\equal{} GB$ and $ E$ is the midpoint of segment $ DC$. Also, segment $ AC$ intersects segment $ EF$ at $ H$ and segment $ EG$ at $ J$. The area of rectangle $ ABCD$ is 70. Find the area of triangle $ AHF$.\r\n[img]http://usera.ImageCave.com/jnma/untitled3.bmp.jpg[/img]\r\n\r\n2. $ ABCD$ is a square. Parallel lines $ m$, $ n$, and $ p$ pass through vertices $ A$, $ B$, and $ C$, respectively. The distance between $ m$ and $ n$ is 7 units, and the distance between $ n$ and $ p$ is 9 units. Find the number of square units in the area of square $ ABCD$.", "Solution_1": "[hide=\"1\"] DC is parallel to AB, so triangles EHC and FHA are similar. The ratio of their bases is $ \\frac {x}{2}$ to $ \\frac {x}{3}$, or 3 to 2 (let the base of the rectangle be x and you see that).\n\nThat means the ratio of their heights is also 3 to 2. Since the sum of their heights is the height of the rectangle, triangle FHA has a height that is 2/5 of the rectangle's and a base that is 1/3 of the rectangles. The ratio of it's area to the rectangle's is thus $ \\frac {1}{15}$, so our answer is $ \\frac {14}{3}$.[/hide]\n\n\n[hide=\"2\"] Utilize symmetry to construct more of these lines:\n [geogebra]afe69729a2fb205b74c8bd2c539ea39a5d66a0d3[/geogebra] \n\nNow label a bunch of things with 7s and 9s, and you get four right triangles with legs 7 and 9, so the area from all four of those is 126. You also get a square with side (9-7)=2, which has an area of 4.\n\nThe total area is 130. Sorry for my bad drawing, I'm in a hurry.[/hide]", "Solution_2": "Wow xpmath! That was a terrific explanation. :D\r\n\r\nThanks again!" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "prove that among the numbers of the arithmetic progressions 3,7,11,15,19,23,... and 5,11,17,23,29,35,... there are an infinite nuber of primes.", "Solution_1": "It means that there exist infinite prime of the form $ 4k \\minus{} 1$ \r\nYou can induction. \r\nCall $ p_1,..,p_n$ is the fisrt n prime of the form $ 4k \\minus{} 1$\r\nConsider \r\n$ t \\equal{} 4\\prod_{i \\equal{} 1}^np_i \\minus{} 1$\r\nbecause $ t\\equiv 3(\\mod 4)$ so it has a prime divisor in the form $ 4k \\minus{} 1$ .Call it is m\r\nBut from $ \\gcd(t,p_i) \\equal{} 1$ so $ m\\not\\in\\{p_i\\}$\r\nThis problem claim.\r\nTo prove there exist infinite prime in the form consider \r\n$ x^2 \\plus{} 1$" } { "Tag": [ "LaTeX", "Pascal\\u0027s Triangle" ], "Problem": "I know you can use pascal's triangle to solve this, but you wouldn't want to generate down to the 12th row in a countdown round. Is there any other trick?", "Solution_1": "It's the same as $ 12\\choose{6}\\equal{}\\frac{12!}{6!6!}\\equal{}924$ Erm, what's wrong with my latex??\r\n\r\n$ 12\\choose{6}$$ \\equal{}\\frac{12!}{6!6!}\\equal{}924$" } { "Tag": [ "LaTeX" ], "Problem": "I want to display \"R \\ A\" in LaTeX, but I cannot use a backslash as this has a different function. What do I use for this?\r\nAlso, when I use \\mathbb{R} as in the Guide, the compiler says this is an undefined control sequence. Do I need to include a package?", "Solution_1": "Try \"\\backslash\".\r\n\r\nThis gives $\\backslash$ ;)", "Solution_2": "[quote=\"jmadsen\"]I want to display \"R \\ A\" in LaTeX, but I cannot use a backslash as this has a different function. What do I use for this?\nAlso, when I use \\mathbb{R} as in the Guide, the compiler says this is an undefined control sequence. Do I need to include a package?[/quote]\r\n\r\nfor displaying mathblackboard, you shoud include the amslatex package. Just to make sure that it will working just include everything here:\r\n\r\n\\usepackage{amsmath,amssymb,amsfonts}" } { "Tag": [], "Problem": "Solve this Maths Problems within 60 Secs with Tricks\r\n\r\nPlz provide your solution alongwith\r\n\r\n[b][color=#FF0000]\n1) 68% of 595 - 43% of 372 = ?\n\n\n2) (96)\u00b2 + (63)\u00b2 = (?)\u00b2 - (111)\u00b2 - 8350\n\n\n3) [(135)\u00b2 \u00f7 15 * 32] \u00f7 ? = 45 * 24\n\n\n4) (739% of 383) \u00f7 628 = ?\n\n\n5) 814296 * 36 = ? * 96324\n\n\n6) 628.306 + 6.1325 * 44.0268 = ?\n\n\n7) (935.82)\u00b2 = ?\n\n\n8 ) 92.5% of 550 = ?\n\n[/color][/b]\r\n\r\n\r\n\r\n[color=#0000FF][b]Solve above problems within 60 secs..Without use of calculator (calc)\n\nUse Paper, Pen & Brain only\n\nPlz provide ur solution alongwith time u consumed[/b] [/color]", "Solution_1": "Not sure why you've posted this.\r\n\r\nI did the first three, having skipped trying to spot the square root of $ 33856$ as question three was easy and cancelled nicely. Those three took me longer than $ 60$ seconds, so having accepted the challenge and failed, I look forward to seeing the tricks." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Let $ p$ \u0430 prime such that $ p\\equal{}4k\\plus{}1$ for some $ k\\in\\mathbb{N}$. Prove that $ k^{2k}\\equiv1\\left(\\mod p\\right)$.", "Solution_1": "You can make this problem better (and is an well-known exercise):\r\nhttp://www.mathlinks.ro/viewtopic.php?t=251936" } { "Tag": [], "Problem": "What is the sum of the real values of $ x$ for which the following expression is undefined? Express your answer as a common fraction.\n\\[ \\frac{8}{\\frac5x\\minus{}\\frac23}\\]", "Solution_1": "x could be 0, and if it's not, then $ 5/x\\equal{}2/3$ so x can be $ 15/2$\r\nSum: $ 15/2$" } { "Tag": [], "Problem": "How many of the natural numbers from 1 to 800, inclusive,\ncontain the digit 6 at least twice? (The numbers 266 and 663\nare two natural numbers that contain the digit 6 at least twice,\nbut 430 or 16 are not.)", "Solution_1": "There are $ 3$ choices for the non-six digit and $ 10$ choices for that digit, for a total of $ 27$.\r\n\r\nHowever, we have missed $ 666$ and have erroneously counted $ 866$ and $ 966$, so we need to subtract $ 1$.\r\n\r\nOur answer is $ \\boxed{26}$.\r\n\r\nEdit: Sorry, misread.", "Solution_2": "See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1056895860&t=217824]here[/url], post number 4. The answer is $ \\boxed{26}$.", "Solution_3": "Why wouldn't the answer be 25? There are only 7 numbers with a 6 in the tens and units place. 866 is not allowed.", "Solution_4": "[hide=\"Casework\"]There is 1 2-digit number that works: 66\nWe divide the 3 digit cases into four parts:\nCase A) If the number is in the form of $66A$, where the digit A is not 6.\nThere are 9 values of A.\nCase B) If the number is in the form of $A66$, where digit A is not 6.\nThere are 6 values of A.\nCase C) If the number is in the form of $6A6$, where digit A is not 6.\nThere are 9 values of A.\nCase C) All digits are 6- there is only one- $666$.\nTherefore we have: $1+9+6+9+1=\\boxed{26}$[/hide]", "Solution_5": "Here is what I did.\n\n[hide=\"My Solution\"]\n2 digit: 66 That's 1.\n3 digit:\n166 606 661\n266 616 662\n366 626 663\n466 636 664\n566 646 665\n666 656 666\n766 676 667\n866 686 668\n966 696 669 \nThere are 28 numbers here. But the numbers have to between 1 and 800 so 866 and 966 are eliminated making it $\\boxed{26}$.\n[/hide]", "Solution_6": "800 factoral?", "Solution_7": "Unless you were being serious, you shouldn't revive to say a troll answer.", "Solution_8": "I DIDN'T KNOW WE CAN COUNT 2 DIDGIT NUMBERS", "Solution_9": "The solution seems wrong because it should be 66_ has 10 numbers. 6_6 has 10. _66 has 8 numbers, but then there should also be 66, so it should be 27. no?", "Solution_10": "@BudgieBricks I think you are counting $666$ twice, once in the 66_ case and once in the 6_6 case. I think there are only nine possibilities for 6_6 after doing the 66_ case.", "Solution_11": "[quote=roderickhuang]Here is what I did.\n\n[hide=\"My Solution\"]\n2 digit: 66 That's 1.\n3 digit:\n166 606 661\n266 616 662\n366 626 663\n466 636 664\n566 646 665\n666 656 666\n766 676 667\n866 686 668\n966 696 669 \nThere are 28 numbers here. But the numbers have to between 1 and 800 so 866 and 966 are eliminated making it $\\boxed{26}$.\n[/hide][/quote]\n\nNo offense, but that's a really sad way of doing it. lol", "Solution_12": "[hide=very dumb solution]\n166 616 2\n266 626 4\n366 636 6\n466 646 8\n566 656 10\n666 676 12\n766 686 14\n661 696 16\n662 17\n663 18\n664 19\n665 20 \n667 21\n668 22\n669 23\n66 24\n660 25\n606 26\n[/hide]", "Solution_13": "[hide=\"PIE\"]\nIf the first two digits are 6: $10$ cases\nIf the first and last are 6: $10$ cases\nIf the last two digits are 6: $8$ cases\nIf all digits are 6: $1$ case\n\nThis results in a total of $10 + 10 + 8 - 1 - 1 - 1 + 1 = \\boxed{26}$ cases by PIE." } { "Tag": [ "quadratics", "modular arithmetic", "USAMTS", "ceiling function", "summer program", "PROMYS", "number theory" ], "Problem": "Ok, so I understand what quadratic residues are now, but can someone please explain to me how you would go about finding the quadratic residues of a given $\\pmod{n}$?", "Solution_1": "[quote=\"DanK\"]Ok, so I understand what quadratic residues are now, but can someone please explain to me how you would go about finding the quadratic residues of a given $\\pmod{n}$?[/quote]\r\nI can give you an example, and that will make it very simple:\r\nLets say we are asked to find the quadratic residues in mod 7. We will look at the squares of the numbers from 0 to 6 in mod 7. So the distinct ones would be known as the [b]quadratic residues[/b] in mod 7. We have\r\n$0^{2}=0,1,4,9\\equiv 2, 16\\equiv 2, 25\\equiv 4, 36\\equiv 1$. Thus, the quadratic residues are 0, 1, 4 and 2.\r\nhope that helps.", "Solution_2": "Yeah, I saw that you could figure out the individual square in mod $n$ like that, but I didn't know when to stop. So I guess I understand how to do it now, but I don't understand one thing about why it works: how do we know all of the possible quadratic residues will occur within the squares of numbers less than $n$?", "Solution_3": "$(kn+r)^{2}\\equiv r^{2}\\pmod{n}$", "Solution_4": "[quote=\"Elemennop\"]$(kn+r)^{2}\\equiv r^{2}\\pmod{n}$[/quote] :wallbash_red: :wallbash: Sometimes I really wonder how I can be so stupid...", "Solution_5": "[quote=\"DanK\"][quote=\"Elemennop\"]$(kn+r)^{2}\\equiv r^{2}\\pmod{n}$[/quote] :wallbash_red: :wallbash: Sometimes I really wonder how I can be so stupid...[/quote]\r\nI doubt you're stupid, after all you helped me. Sometimes it takes time for the idea to reach your head, it does for me.", "Solution_6": "Yeah, I just get annoyed with myself sometimes...\r\n\r\nAt least I understand it now. They kept talking about in in one of the solutons to a USAMTS problem from last year, and it was driving me crazy...", "Solution_7": "You only have to check the first $\\lceil\\frac{n}{2}\\rceil$ squares. The residues repeat, but backwards. For example, $0,1,4,2$ and $2,4,1$. This is because in $\\mod 7$, numbers can be written as $0,1,2,3,-3,-2,-1$. $(-k)^{2}=k^{2}$, giving the same residue.", "Solution_8": "[quote=\"lotrgreengrapes7926\"]You only have to check the first $\\lceil\\frac{n}{2}\\rceil$ squares. The residues repeat, but backwards. For example, $0,1,4,2$ and $2,4,1$. This is because in $\\mod 7$, numbers can be written as $0,1,2,3,-3,-2,-1$. $(-k)^{2}=k^{2}$, giving the same residue.[/quote]\r\n\r\nNice. I always forget to check negative results from mods...", "Solution_9": "Well, if $n=p_{1}^{a_{1}}p_{2}^{a_{2}}...p_{k}^{a_{k}}$, then \\[x^{2}\\equiv r\\mod{n}\\iff x^{2}\\equiv r\\mod{p_{i}^{a_{i}}}\\] for $1 \\le i \\le k$. Then, we use a simple divisibility argument to show that there are exactly zero or two solutions to that congruence mod $p_{i}$, and then, using an inductive method, prove there are exactly zero or two solutions mod $p_{i}^{a_{i}}$. Then, given a bunch of congruences in each $p_{i}^{a_{i}}$, there are two possible residues in each, so it seems there are either zero or $2^{k}$ solutions mod $n$ for each possible $r$, so then we have $\\frac{\\phi(n)}{2^{k}}$ quadratic residues...\r\n\r\nI don't know if this is making any sense if you haven't done some number theory, but it's an interesting chain of reasoning (woot for PROMYS!)...start off with proving $x^{2}\\equiv r\\mod{p}$ has either zero or two solutions, and if two, they are additive inverses of each other mod p, and go from there.", "Solution_10": "How do you use quadratic residues in problem solving?", "Solution_11": "For example last post from [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=102730[/url].", "Solution_12": "Or: there exists infinitely many primes $\\equiv 1 \\mod 3$:\r\nAssume there are only finitely many, call their product $P$. Then consider the number $P^{2}+3$ and let $q$ be a prime divisor of it. Then $P^{2}\\equiv-3 \\mod q$, thus $-3$ is a quadratic residue $\\mod q$. By quadratic reciprocity we get that $q$ is a quadratic residue $\\mod 3$. Thus $q \\equiv 1 \\mod 3$ ($q \\neq 3$ should be clear). But $q$ divides $P$ by definition, which gives $q|3$, acontradiction." } { "Tag": [ "percent" ], "Problem": "A baby goes to sleep at 8 p.m., wakes up after two hours, stays awake for 15 minutes, and then goes back to sleep. If the baby repeats this sleep pattern until she is awakened at 6 a.m., what percent of the time from 8 p.m. to 6 a.m. was the baby asleep?", "Solution_1": "In the first $ 9$ hours from $ 8$ p.m. to $ 5$ a.m., there are $ 8$ hours of sleep, as four complete cycles have been completed. Then, there is $ 1$ more hour of sleep until she is awakened, so the answer is $ \\frac{8\\plus{}1}{9\\plus{}1}\\equal{}\\boxed{90\\%}$." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Find all $ x,y,z \\in N$ that $ 3^x\\plus{}4^y\\equal{}5^z$", "Solution_1": "don't you think the topic should be more specific ?\r\n\r\n\r\n\r\n\r\nit was posted some time ago\r\n\r\nthe sketch of solution:\r\n\r\n1. $ 2|x$\r\n\r\n2. $ 2|z$ (cong. mod 3)\r\n\r\n3. then $ 4^y\\equal{}5^z\\minus{}3^x\\equal{}5^{2Z}\\minus{}3^{2X}\\equal{}(5^Z\\minus{}3^X)(5^Z\\plus{}3^X)$ so both of them are powers of 2.\r\nThe rest isn't difficult.\r\n\r\nanswer: $ 2,2,2$ (only)" } { "Tag": [ "geometry", "rectangle" ], "Problem": "the centers of fence posts are 30 inches apart. they enclose a triangular region 20 feet by 20 feet by 10 feet. How many posts will be needed?", "Solution_1": "[hide]22\nThe sides are 240in, 240in, 120in \nDivide each by 30 and you get:\n8,8,4\n8+8+4= [b]22 ANS[/b][/hide]", "Solution_2": "i think that equals [b]20[/b]. [hide]Wouldn't the corners of the fence be shared by two sides? I think it would then equal [b]17[/b][/hide][/hide]", "Solution_3": "[hide]22 \nThe sides are 240in, 240in, 120in \nDivide each by 30 and you get: \n8,8,4 \n8+8+4= 22 ANS [/hide]", "Solution_4": "[quote=\"Xantos C. Guin\"][hide]22\nThe sides are 240in, 240in, 120in \nDivide each by 30 and you get:\n8,8,4\n8+8+4= [b]22 ANS[/b][/hide][/quote]\n[quote=\"4everwise\"][hide]22 \nThe sides are 240in, 240in, 120in \nDivide each by 30 and you get: \n8,8,4 \n8+8+4= 22 ANS [/hide][/quote]\n\nWOW!\nthose 2 posts are exactly identical \nexept 4everwise didn't put the [hide]22 ANS[/hide] in bold.\r\nI guess great minds do think alike.", "Solution_5": "8+8+4=$20$\r\n2 people made a careless error!\r\n[b]2 people!!!!!!!!!!!!!!!!!!!!!!!!![/b]\r\nand it was the SAME error!", "Solution_6": "Notes to penguin:\r\nStop making careless errors.", "Solution_7": "lol :D", "Solution_8": "Yeah, if the answer was 22, then there will be two posts stacked on top of another.", "Solution_9": "oh well what if you have 60 yards of fence and a built in wall that expands 2 miles. What is the max area? Make a rectangle using the wall as one side and using the fence for 3 sides." } { "Tag": [ "trigonometry", "function", "symmetry", "inequalities", "quadratics", "algebra", "system of equations" ], "Problem": "Please help with these problems:\r\n\r\n1.\r\n\r\nBy choosing different real values for $ a$ in the system of equations:\\[ x^{2}\\minus{}y^{2}\\equal{}0\\]\\[ (x\\minus{}a)^{2}\\plus{}y^{2}\\equal{}1\\]the number of real-valued solutions can be:\r\n\r\na) 0, 1, 2, 3, 4, or 5 b) 0, 1, 2, or 4 c) 0, 2, 3, or 4 m d) 0, 2, or 4 e) 2 or 4", "Solution_1": "2.\r\n\r\nThe graph of the equation $ y\\plus{}\\mid y\\mid \\equal{} x\\plus{}\\mid x\\mid$, restricted to $ x < 0$, is:\r\n\r\na) the empty set b) a line c) shaped like a V d) quadrant of a plane e) two quadrants of a plane", "Solution_2": "3.\r\n\r\nHow many real number solutions does $ \\left(\\frac{x\\minus{}21}{2x}\\right)^{x^{2}\\minus{}x\\minus{}12}\\equal{}1$ have?\r\n\r\na) 0 b) 1 c) 2 d) 3 e) 4\r\n\r\nI seem to get 3 for this problem but the answers say 4 can someone explain", "Solution_3": "4.\r\n\r\nIf $ \\sin\\theta\\plus{}\\cos\\theta\\equal{}\\frac{1}{5}$ and $ 0\\le\\theta<\\pi$, then $ \\tan\\theta$ is:\r\n\r\na) $ \\minus{}\\frac{4}{3}$ b) $ \\minus{}\\frac{3}{4}$ c) $ \\frac{3}{4}$ d) $ \\frac{4}{3}$ e) not completely determined by the given information", "Solution_4": "5.\r\n\r\nWhat is the set of values for $ k$ for which the function $ f(x) \\equal{}\\cos^{4}x\\minus{}k^{2}\\cos^{2}(2x)\\plus{}sin^{4}x$ is constant?\r\n\r\na) $ {\\frac{1}{\\sqrt{2}},\\minus{}\\frac{1}{\\sqrt{2}}}$ b) $ \\{\\frac{1}{2},\\minus{}\\frac{1}{2}\\}$ c) $ \\{\\frac{1}{2}\\}$ d) $ \\{\\frac{1}{\\sqrt{2}}\\}$ e) cannot be determined", "Solution_5": "[hide=\"(1)\"]\nThe answer is (c). \n\nGraphically, the first equation is the pair of lines $ x \\equal{} y$, $ x \\equal{}\\minus{}y$. The second equation is a circle with center $ (a,0)$ and radius 1. When we place a circle of radius one along the x-axis, we find that when the center is near the origin we can intersect at four points, when the center is at $ (1,0)$ it intersects at 3 points, when the center is at $ (\\sqrt{2},0)$ its tangent to the two lines and so there it intersects at 2 points. If we drag the circle away from the lines there are 0 intersections. However, we cannot get 1 point because of symmetry across the x-axis.\n\nNot sure if it is possible algebraically. (also there's an \"m\" next to choice (c), though I'm assuming that's a typo)[/hide]\n\n[hide=\"(2)\"]\nThe answer is (d).\n\nLet's break up the absolute value signs:\n\n$ y\\plus{}|y|\\minus{}x \\equal{} |x|$\n\nWhen $ x < 0$, we have \n\n$ y\\plus{}|y|\\minus{}x \\equal{}\\minus{}x$\n$ y\\plus{}|y| \\equal{} 0$\n\nThis will be true when $ y$ is negative; so we have the inequalities $ x < 0, y < 0$ which is the third quadrant of the plane.\n[/hide]\n\n[hide=\"(3)\"]\nThe solution you are probably missing is $ x \\equal{} 7$.\n\nWe find that $ x^{y}\\equal{} 1$ is true when $ x \\equal{}\\boxed{\\pm}1$ (under real numbers; the negative is true when $ y$ is even) or when $ y \\equal{} 0$.\n\nCase 1: $ x \\equal{}\\pm 1$\n\nCheck the parity of $ x^{2}\\minus{}x\\minus{}12$: $ x$ being even causes $ x^{2}\\minus{}x\\minus{}12$ to be even, and $ x$ being odd also causes that to be even. I don't think this is rigorous under real numbers ... oh well. \n\nThen $ \\frac{x\\minus{}21}{2x}\\equal{}\\pm 1\\Longrightarrow x\\minus{}21 \\equal{}\\pm 2x\\Longrightarrow x \\equal{} 7,\\minus{}21$. \n\nCase 2: $ x^{2}\\minus{}x\\minus{}12 \\equal{} 0$\n\nFactors into $ (x\\minus{}4)(x\\plus{}3) \\equal{} 0\\Longrightarrow x \\equal{} 4,\\minus{}3$.\n\nSo solutions: $ \\boxed{\\minus{}21,\\minus{}3,4,7}$\n[/hide]\n\n[hide=\"(4)\"]\nJust by looking at the solutions (which imply that $ \\sin\\theta \\equal{}\\frac{4}{5}$ and $ \\cos\\theta \\equal{}\\minus{}\\frac{3}{5}$ is a possibility), guess and check quickly tells us that the answer is either (a) or (e). On a contest I would just guess (a) and continue, but here I'll be rigorous.\n\nNotice that $ \\theta$ must be in the second quadrant; in the first, both $ \\sin\\theta$ and $ \\cos\\theta$ are positive and at least one is $ \\ge\\frac{1}{\\sqrt{2}}$. Hence $ \\tan\\theta < 0$.\n\nSquare the equation and we get $ \\sin^{2}\\theta\\plus{}\\cos^{2}\\theta\\plus{}2\\sin\\theta\\cos\\theta \\equal{}\\frac{1}{25}\\Longrightarrow\\sin 2\\theta \\equal{}\\frac{\\minus{}24}{25}$. Hence $ \\theta \\equal{}\\frac{\\sin^{\\minus{}1}\\left(\\frac{\\minus{}24}{25}\\right)}{2}$.\n\nUse the tangent half-angle formula:\n\n$ \\tan\\left(\\frac{\\theta}{2}\\right) \\equal{}\\pm\\sqrt{\\frac{1\\minus{}\\cos (\\sin^{\\minus{}1}\\frac{\\minus{}24}{25})}{1\\plus{}\\cos (\\sin^{\\minus{}1}\\frac{\\minus{}24}{25})}}$\n\n$ \\cos (\\sin^{\\minus{}1}\\frac{\\minus{}24}{25}) \\equal{}\\frac{7}{25}$ (just draw it out, you get a 7-24-25 triangle in the second quadrant), so we get \n\n$ \\tan\\left(\\frac{\\theta}{2}\\right) \\equal{}\\pm\\sqrt{\\frac{1\\minus{}\\frac{7}{25}}{1\\plus{}\\frac{7}{25}}}\\equal{}\\pm{\\frac{32}{18}}\\equal{}\\pm\\frac{4}{3}$. It's in the second quadrant, take the negative root, and our answer is (a).\n\n[/hide]\n\n[hide=\"(5)\"]\nThe answer is (a). \n\n$ f(x) \\equal{} (\\cos^{2}x\\plus{}\\sin^{2}x)^{2}\\minus{}2\\sin^{2}x\\cos^{2}x\\minus{}k^{2}\\cos^{2}2x$\n$ f(x) \\equal{} 1\\minus{}\\frac{1}{2}\\sin^{2}2x\\minus{}k^{2}\\cos^{2}2x$\n\nSince we want it to be constant, we want the last two terms to be constant; we will use the identity $ \\sin^{2}\\theta\\plus{}\\cos^{2}\\theta \\equal{} 1$. Then $ k^{2}\\equal{}\\frac{1}{2}\\Longrightarrow k \\equal{}\\pm\\frac{1}{\\sqrt{2}}$\n\n\n[/hide]", "Solution_6": "[quote=\"azjps\"]Not sure if it is possible algebraically. (also there's an \"m\" next to choice (c), though I'm assuming that's a typo)[/quote]\r\n\r\n[hide=\"Algebra\"] The first equation gives $ y^{2}\\equal{} x^{2}$, so a substitution into the second produces $ (x\\minus{}a)^{2}\\plus{}x^{2}\\equal{} 1$. A quadratic can have $ 0, 1, 2$ solutions (varying $ a$ appropriately). \n\nGenerally, each value of $ x$ corresponds to two values of $ y$; however, when $ y \\equal{} 0$ this does not occur, so there can be $ 0, 2, 3, 4$ solutions. [/hide]" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Let $x, y$ be natural numbers such that\r\n\r\n$2x^{2}-1=y^{15}$\r\n\r\nShow that if $x>1$, then $5\\mid x$.", "Solution_1": "Russia 2005\r\nhttp://www.mathlinks.ro/viewtopic.php?p=220224#220224", "Solution_2": "Thanks a lot... I was wondering where it had come from..." } { "Tag": [ "function", "calculus", "calculus computations" ], "Problem": "can you show that :D\r\n\r\n$ \\sum_{n\\equal{}1}^\\infty \\; \\frac{1}{(1\\plus{}\\pi^2\\,n^2)^3}\\;\\equal{}\\;\\boxed{\\frac{11\\minus{}31e^2\\plus{}41e^4\\minus{}5e^6}{16\\;(e^2\\minus{}1)^3}}$\r\n\r\nstrange thing is we have $ \\pi$ on the thing that we are trying to sum, but the sum itself actually involves $ e$ :|", "Solution_1": "it's simple exercises for complx analysis and it's not important there will be $ \\pi^{2}$ or not . we can find\r\n$ \\sum \\frac {1}{(1 \\plus{} d^{2}n^{2k})^{m}}$", "Solution_2": "OK, let me make it easier, you [b]cannot[/b] use complex analysis.", "Solution_3": "this question [u]is[/u] a complex analysis question", "Solution_4": "[quote=\"kenn4000\"]this question [u]is[/u] a complex analysis question[/quote]\r\nYes, but I think there is also some way by real analysis. \r\nFor example when Ramanujan proved many interesting results, He did not know complex analysis.", "Solution_5": "Note that\r\n\r\n\\begin{eqnarray*}\r\n\\sum_{k=1}^{\\infty} \\frac{1}{(1 + \\pi^2 k^2)^3}\r\n& = & \\sum_{k=1}^{\\infty} \\sum_{n=0}^{\\infty} \\frac{(n+1)(n+2)}{2} \\frac{(-1)^n}{(\\pi k)^{2(n+3)}}\\\\\r\n& = & \\sum_{n=0}^{\\infty} \\frac{(n+1)(n+2)}{2} (-1)^n \\frac{\\zeta(2n+6)}{\\pi^{2(n+3)}}\\\\\r\n& = & \\sum_{n=1}^{\\infty} \\frac{(n-1)(n-2)}{2} (-1)^{n-1} \\frac{\\zeta(2n)}{\\pi^{2n}} \\quad \\quad \\cdots (1)\r\n\\end{eqnarray*}\r\n\r\nUsing the identity $ (-1)^{n-1} \\frac{\\zeta(2n)}{\\pi^{2n}} = \\frac{B_{2n}}{(2n)!} 2^{2n-1}$ and $ \\frac{x}{e^x - 1} = \\sum_{n=0}^{\\infty} \\frac{B_n}{n!} x^n,$ we have\r\n\r\n$ \\sum_{n=1}^{\\infty} (-1)^{n-1} \\frac{\\zeta(2n)}{\\pi^{2n}} x^{2n-1} = \\frac{1}{2} \\left( \\frac{e^{2x} + 1}{e^{2x} - 1} - \\frac{1}{x} \\right). \\quad \\quad \\cdots (2)$\r\n\r\nDifferentiate $ (2)$ once and twice, put $ x = 1$ to obtain\r\n\r\n$ \\sum_{n=1}^{\\infty} (-1)^{n-1} \\frac{\\zeta(2n)}{\\pi^{2n}} = \\frac{1}{e^2 - 1}$\r\n\r\n$ \\sum_{n=1}^{\\infty} (2n-1) (-1)^{n-1} \\frac{\\zeta(2n)}{\\pi^{2n}} = \\frac{1}{2} - \\frac{2e^2}{(e^2 - 1)^2}$\r\n\r\n$ \\sum_{n=1}^{\\infty} (2n-1)(2n-2) (-1)^{n-1} \\frac{\\zeta(2n)}{\\pi^{2n}} = \\frac{1+e^2+7 e^4-e^6}{(e^2 - 1)^3}$\r\n\r\nFinally, choose a suitable linear combination of these three series to obtain $ (1)$. Then we have the result.\r\n\r\n\r\nBut I think it is not a direct way to obtain the value of the given series. There may be an easier, direct way.", "Solution_6": "[quote=\"sos440\"]\n... Using the identity $ ( \\minus{} 1)^{n \\minus{} 1} \\frac {\\zeta(2n)}{\\pi^{2n}} \\equal{} \\frac {B_{2n}}{(2n)!} 2^{2n \\minus{} 1}$ ...[/quote]\r\nI only note that this identity :lol: can be proved without using complex analysis.", "Solution_7": "touche. although bernoulli numbers and zeta functions might be hard to find in a noncomplex textbook =P", "Solution_8": "There is a book called \"Proofs from the Book\" that gives famous theorems and elegant proofs. In the chapter \"Cotangent and the Herglotz Trick\" they derive that zeta formula using only real analysis. Anyone who wants an interesting read on mathematics and beautiful proofs should pick up a copy.", "Solution_9": "here is elementary proof by using [b]Papadimitriou's[/b] idea for $ \\zeta(2)$. \r\n\r\nExcept Papadimitriou's I any more did not hear about modern mathematicians of Greece :(" } { "Tag": [], "Problem": "The product of any two of the positive integers 30, 72, and $N$ is divisible by the third. What is the smallest possible value of $N$?", "Solution_1": "[hide]The least is $\\displaystyle 60$.\n\nThis was done using prime factorization.\n\n$\\displaystyle 30=2\\cdot 3\\cdot 5$\n$\\displaystyle 72=2^3 \\cdot 3^2$\nTherefore $\\displaystyle N$ must be a factor of $\\displaystyle 2^4 \\cdot 3^3 \\cdot 5$.\n\n$\\displaystyle 2^3 \\cdot 3^2 | 2 \\cdot 3 \\cdot 5 \\cdot N$\nSo $\\displaystyle N$ is divisible by $\\displaystyle 2^2 \\cdot 3$\n\n$\\displaystyle 2 \\cdot 3 \\cdot 5| 2^3 \\cdot 3 ^2 \\cdot 5 \\cdot N$\nSo $\\displaystyle N$ is divisible by $\\displaystyle 5$\n\n$\\displaystyle 2^2 \\cdot 3 \\cdot 5=60$\n[/hide]" } { "Tag": [ "geometry", "parallelogram", "circumcircle", "trigonometry", "function" ], "Problem": "Let be given a parallelogram $ ABCD$ and two points $ A_1$, $ C_1$ on its sides $ AB$, $ BC$, respectively. Lines $ AC_1$ and $ CA_1$ meet at $ P$. Assume that the circumcircles of triangles $ AA_1P$ and $ CC_1P$ intersect at the second point $ Q$ inside triangle $ ACD$. Prove that $ \\angle PDA \\equal{} \\angle QBA$.", "Solution_1": "Let $ (X) \\equiv \\odot(AA_1P)$ cut $ DA$ again at $ A_2$ and let $ (Y) \\equiv \\odot (CC_1P)$ cut $ CD$ again at $ C_2.$\r\n\r\n$ \\angle CQP \\equal{} \\angle PC_1B \\equal{} \\angle PAA_2 \\equal{} \\pi \\minus{} \\angle A_2QP$\r\n\r\n$ \\Longrightarrow$ $ C, Q, A_2$ are collinear and similarly, $ A, Q, C_2$ are collinear. From $ (X)$, $ A_1Q$ is antiparallel of $ AA_2 \\parallel BC$ WRT angle $ \\angle (AB, A_2C)$ $ \\Longrightarrow$ $ A_1BCQ$ is cyclic, and $ A_2P$ is antiparallel of $ AA_1 \\parallel DC$ WRT angle $ \\angle (DA, CA_1)$ $ \\Longrightarrow$ $ CDA_2P$ is also cyclic. From $ \\odot (A_1BCQ), \\odot(CDA_2P),$\r\n\r\n$ \\angle QBA \\equal{} \\angle QBA_1 \\equal{} \\angle QCA_1 \\equal{} \\angle QCP \\equal{} \\angle A_2CP \\equal{} \\angle A_2DP \\equal{} \\angle ADP.$", "Solution_2": "[color=darkblue]From [url=http://www.mathlinks.ro/viewtopic.php?p=1486021#1486021][b]Virgil Nicula[/b]'s extension[/url] we have\n\n$ \\frac {\\sin \\angle ABQ}{\\sin \\angle CBQ} \\equal{} \\frac {AA_1}{CC_1}$\n\nDenote by $ E$ the intersection of the lines $ AD$ and $ CP$. Then\n\n$ \\dfrac{\\sin \\angle ADP}{\\sin \\angle CDP} \\equal{} \\dfrac{EP}{PC} \\cdot \\dfrac{CD}{ED}$\n\nBut\n\n$ \\dfrac{EP}{PC} \\equal{} \\dfrac{EA}{CC_1}$\n\nand\n\n$ \\dfrac{AD}{ED} \\equal{} 1 \\minus{} \\dfrac{EA}{ED} \\equal{} 1 \\minus{} \\dfrac{AA_1}{CD} \\equal{} \\dfrac{CD \\minus{} AA_1}{CD}$\n\n$ ED \\equal{} \\dfrac{CD \\cdot AD}{CD \\minus{} AA_1} \\equal{} \\dfrac{CD \\cdot AD}{AB \\minus{} AA_1} \\equal{} \\dfrac{CD \\cdot AD}{BA_1}$\n\nSo\n\n$ \\dfrac{\\sin \\angle ADP}{\\sin \\angle CDP} \\equal{} \\dfrac{EA}{CC_1} \\cdot \\dfrac{CD}{\\dfrac{CD \\cdot AD}{BA_1}} \\equal{} \\dfrac{EA}{BC} \\cdot \\dfrac{BA_1}{CC_1} \\equal{} \\dfrac{AA_1}{BA_1} \\cdot \\dfrac{BA_1}{CC_1} \\equal{} \\frac {AA_1}{CC_1} \\equal{} \\frac {\\sin \\angle ABQ}{\\sin \\angle CBQ}$\n\nAs function $ f(x) \\equal{} \\dfrac{\\sin (\\alpha \\minus{} x)}{\\sin x}$ is strictly decreasing on $ (0,\\pi)$ we get that $ \\angle PDA \\equal{} \\angle QBA$.[/color]", "Solution_3": "Sorry to revive this old thread, But I have an alternative solution to this beautifull problem.\n[hide=\" My solution\"]\nFrom Miquel's theorem, we have $Q\\in \\odot(A_1BC)$. Then $\\angle{QCB}=\\angle{AA_1Q}=\\angle{APQ}$. But also $\\angle{PAQ}=\\angle{PA_1Q}=\\angle{CBQ}$ it follow that $\\triangle{PAQ}\\sim \\triangle{CBQ}$ therefore $\\frac{AP}{PQ}=\\frac{BC}{QC}=\\frac{AD}{QC}$, then $\\triangle{PAD}\\sim \\triangle{PQC}$ because $\\angle{PQC}=\\angle{PAD}$ it follow that $\\angle{PDA}=\\angle{PCQ}=\\angle{ABQ}$ Which end the proof.\n[/hide]", "Solution_4": "Use complex numbers. Let $Q=0$. Note that in angle measure we have $QAP=QA_1P=QA_1C=QBC$, $QPA=QA_1A=QCC_1=QCB$ since $Q$ takes $AC_1$ to $A_1C$. So $Q$ takes $AP$ to $BC$ and $pb=ac$. So $p(a+c-d)=ac$, $-pd=-ap-cp+ac$, and $p(p-d)=(p-a)(p-c)$. Thus, $P$ takes $AD$ to $QC$. So now $PDA=PCA=A_1CQ=A_1BQ=QBA$, so we're done.", "Solution_5": "Nice angle chasing exercise! :)\n\n[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */\nimport graph; size(14.182cm); \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 = -4.3, xmax = 15.96, ymin = -5.24, ymax = 6.3; /* image dimensions */\npen xdxdff = rgb(0.49019607843137253,0.49019607843137253,1.); pen ffxfqq = rgb(1.,0.4980392156862745,0.); pen bfffqq = rgb(0.7490196078431373,1.,0.); pen ffqqff = rgb(1.,0.,1.); pen xfqqff = rgb(0.4980392156862745,0.,1.); \n /* draw figures */\ndraw((4.32,3.2)--(5.36,0.), xdxdff); \ndraw((2.66,0.)--(5.36,0.), xdxdff); \ndraw((1.62,3.2)--(4.32,3.2), xdxdff); \ndraw((1.62,3.2)--(2.66,0.), xdxdff); \ndraw((4.32,3.2)--(4.1,0.)); \ndraw((4.859378365410247,1.5403742602761632)--(2.66,0.)); \ndraw(circle((3.38,0.5817777444167372), 0.9256702133582054), linetype(\"2 2\") + ffxfqq); \ndraw(circle((3.919179933099226,2.152271624016165), 1.1217804936225562), linetype(\"2 2\") + ffxfqq); \ndraw(circle((4.73,1.56425), 1.686350515907057), linetype(\"4 4\") + bfffqq); \ndraw((3.518359866198457,3.2)--(2.66,0.), linewidth(1.2) + linetype(\"4 4\") + ffqqff); \ndraw(circle((2.569179933099227,1.739483478257249), 1.7418527652134657), dotted + xfqqff); \n /* dots and labels */\ndot((2.66,0.),dotstyle); \nlabel(\"$A$\", (2.38,-0.26), NE * labelscalefactor); \ndot((5.36,0.),dotstyle); \nlabel(\"$B$\", (5.46,-0.38), NE * labelscalefactor); \ndot((4.32,3.2),dotstyle); \nlabel(\"$C$\", (4.38,3.36), NE * labelscalefactor); \ndot((1.62,3.2),linewidth(3.pt) + dotstyle); \nlabel(\"$D$\", (1.36,3.04), NE * labelscalefactor); \ndot((4.1,0.),dotstyle); \nlabel(\"$A_1$\", (4.14,-0.44), NE * labelscalefactor); \ndot((4.859378365410247,1.5403742602761632),dotstyle); \nlabel(\"$C_1$\", (5.02,1.52), NE * labelscalefactor); \ndot((4.172843888664785,1.059547471487769),linewidth(3.pt) + dotstyle); \nlabel(\"$P$\", (4.34,0.78), NE * labelscalefactor); \ndot((4.172843888664785,1.059547471487769),linewidth(3.pt) + dotstyle); \ndot((3.0478152622528927,1.4457908484299304),linewidth(3.pt) + dotstyle); \nlabel(\"$Q$\", (3.12,1.56), NE * labelscalefactor); \ndot((4.32,3.2),linewidth(3.pt) + dotstyle); \ndot((3.518359866198457,3.2),linewidth(3.pt) + dotstyle); \nlabel(\"$A_2$\", (3.6,3.32), NE * labelscalefactor); \nclip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); \n /* end of picture */\n\n[/asy]\n\n[b]Lemma 1.[/b] Points $A_1, B, C, Q$ lie on a circle.\n\n[i]Proof.[/i] We have $$\\angle A_1QC=\\angle A_1QP+\\angle PQC=\\angle A_1AP+\\angle PC_1B=180^{\\circ}-\\angle A_1BC_1 \\Longrightarrow A_1, B, C, Q \\, \\text{are concyclic}. \\, \\square$$\n\n[b]Lemma 2.[/b] Points $A, Q, A_2$ are collinear where $A_2$ is the second intersection of $(CC_1P)$ with $CD$.\n\n[i]Proof.[/i] Note that from $A_2C \\parallel A_1B$ we have $$\\angle AQP+\\angle A_2QP=\\angle PA_1B+\\left(180^{\\circ}-\\angle PCA_2\\right)=180^{\\circ} \\Longrightarrow A, Q, A_2 \\, \\text{are collinear}. \\, \\square$$\n\n[b]Lemma 3.[/b] Points $D, A_2, P, A$ are concyclic.\n\n[i]Proof.[/i] We have $$\\angle APA_2=\\angle A_2CC_1=180^{\\circ}-\\angle ADA_1 \\Longrightarrow A, D, A_2, P \\, \\text{are concyclic}. \\, \\square$$\n\nAs $$\\angle PDA=\\angle PA_2Q=\\angle QCA_1=\\angle QBA,$$ we are done.", "Solution_6": "Let $APQ$ interrsect $AD$ at $X$ and let $CPQ$ intersect $CD$ at $Y$. \n\nObserve that $\\angle XAP = \\angle PC_1B = \\angle PQC$. Thus $X, Q, C$ are collinear. Similarly, $A, Q, Y$ are collinear. \n\nThis mean that $DAPY$ is cyclic and $BA_1QC$ is cyclic (properties of the Miquel Point). This gives us\n\n$\\angle PDA = \\angle PYA = \\angle PYQ = \\angle PCQ = \\angle A_1CQ = \\angle ABQ$ as required.", "Solution_7": "Pick $A_2 \\in \\overline{AD}, C_2 \\in \\overline{CD}$ such that $AA_1A_2PQ$, $CC_1C_2PQ$ are concylic. Observe that $Q$ is the Miquel point of $PA_1BC_1$. \n\n$\\angle PQA_2 + \\angle PQC = 180 - \\angle PAA_2 + 180 - \\angle PC_1C = 180$, and hence $A_2, Q, C$ are colinear. Similarly $C_2, A, Q$ are colinear. Thus $P$ is the Miquel point of $QC_2DA_2$. \n\nNow we're done since $\\angle QBA = \\angle QC_1A = \\angle PC_1Q = \\angle PC_2Q = \\angle AC_2Q = \\angle ADP$", "Solution_8": "Let circle $AA_1P$ and $CC_1P$ meet $AD$ and $CD$ at $S,K$.\nClaim $: BA_1QC$ is cyclic.\nProof $:$ Note that $\\angle CQA_1 = \\angle CQP + \\angle PQA_1 = \\angle BC_1A + \\angle PAB = \\angle 180 - \\angle A_1BC$.\nClaim $: A,Q,K$ and $C,Q,S$ are collinear.\nProof $:$ Note that $\\angle PQA = \\angle PA_1B = \\angle PCK = \\angle 180 - \\angle PQK$. we prove the other one with same approach.\nClaim $: CPSD$ is cyclic.\nProof $:$ Note that $\\angle PSA = \\angle PQA = \\angle PCK = \\angle PCD$.\nNow Note that $\\angle QBA = \\angle QCA_1 = \\angle QCP = \\angle SCP = \\angle SDP = \\angle PDA$.\nwe're Done.", "Solution_9": "Denote by $R$ reflection of $P$ wrt center of parallelogram. In $A_1BC_1P$ Miquel point $Q$ is the isogonal conjugate of $\\infty_{BR},$ since $BR$ is homothetic to it's Gauss line with center $P$ and coefficient $2,$ therefore $$\\angle PDA\\stackrel{\\text{symmetry}}{=}\\angle RBC=\\angle QBA.$$", "Solution_10": "[hide = solution]\nRemark that $Q$ is the centre of the spiral similarity sending $AA_1 \\mapsto CC_1$. Making use of the complete quadrilateral configuration, we deduce that $Q$ also maps $AP \\mapsto BC$. Therefore $QB/BC = BC/AP$. Observe that $DA = BC$ and thus we may rewrite the condition earlier as $QB/AD = QA/AP$. Moreover, observe that $\\angle BQA = \\angle BC_1A = \\angle C_1AD = \\angle DAP$ since $DA \\parallel BC$. This along with the previous length condition implies that $\\triangle BAQ \\simeq \\triangle DPA$. In particular, $\\angle PDA = \\angle QBA$, as desired.\n[/hide]" } { "Tag": [ "limit", "probability", "expected value", "probability and stats" ], "Problem": "There are $ n$ cards labbeled 1 to $ n$ in a box. You take one card from the box and take it back $ k$ times repeatedly.\r\nDenote by $ X$ the number written in the card took out.\r\n\r\n(1) For $ n \\equal{} 4,\\ k \\equal{} 3$, find $ P(X \\equal{} 2)$ and the expected value of $ X$.\r\n\r\n(2) For $ k \\equal{} 3$, express the expected value of $ X$ in terms of $ n$.\r\n\r\n(3) Two persons $ A,\\ B$ make the procedure as follows. \r\n\r\n$ A$ conducts the procedure for $ k \\equal{} 1$, $ B$ conducts the procedure $ k \\equal{} a > 1$. Find $ \\lim_{n\\to\\infty} P(X_A\\leq X_B)$.", "Solution_1": "hi kunny could you explain this differently.. im not sure i understand what X is..\r\n\r\nif you take more than one card out shouldnt you get multiple values for X?", "Solution_2": "[quote=\"kunny\"]There are $ n$ cards labbeled 1 to $ n$ in a box. You take one card from the box and take it back $ k$ times repeatedly.\nDenote by $ X$ the number written in the card took out.\n\n(1) For $ n \\equal{} 4,\\ k \\equal{} 3$, find $ P(X \\equal{} 2)$ and the expected value of $ X$.\n\n(2) For $ k \\equal{} 3$, express the expected value of $ X$ in terms of $ n$.\n\n(3) Two persons $ A,\\ B$ make the procedure as follows. \n\n$ A$ conducts the procedure for $ k \\equal{} 1$, $ B$ conducts the procedure $ k \\equal{} a > 1$. Find $ \\lim_{n\\to\\infty} P(X_A\\leq X_B)$.[/quote]\r\n\r\nTo Moderators:\r\n\r\nCould you correct the context of thar part ''Denote by $ X$ the number written in '' into ''Denote by $ X$ the [color=red]maximum[/color] number written [color=red]on[/color]?\r\n\r\nThank you for your troubles.\r\n\r\nkunny", "Solution_3": "let $ Y_j$ be the number drawn on the jth try\r\n$ P(Y_j\\equal{}m)\\equal{}\\frac{1}{n}$ for each $ 1\\le m \\le n$\r\n\r\nand $ Z_k\\equal{}\\max\\{Y_1,...,Y_k\\}$ so\r\n$ p_k(m)\\equal{}P(Z_k\\equal{}m)\\equal{}P(Z_{k\\minus{}1}\\equal{}m \\cap Y_k\\le m)\\plus{}P(Z_{k\\minus{}1}X_B))\\equal{}1\\minus{}\\frac{1}{n}\\sum_{m\\equal{}1}^n \\left(\\frac{m\\minus{}1}{n}\\right)^a$\r\n$ \\equal{}1\\minus{}\\frac{1}{n^{a\\plus{}1}}\\sum_{m\\equal{}0}^{n\\minus{}1} m^a\\to 1\\minus{}\\frac{1}{a\\plus{}1}$", "Solution_4": "That's correct answer. I have nothing to add more than this, but\r\n\r\nLet me say only one thing, most students those who can solve use usually use the fact:\r\n\r\n$ \\boxed{P(X\\equal{}k)\\equal{}P(X\\leq k)\\minus{}P(X\\leq k\\minus{}1) \\ (k\\geq 2)}$" } { "Tag": [ "inequalities" ], "Problem": "Fie x,y,z numere reale strict pozitive astefel incat x+y+z=xyz. Sa se demonstreze ca: 4(xy+xz+yz) <= 9+x^2*y^2*z^2 (inegalitatea \"mai mic sau egal\").", "Solution_1": "Facem substitu\u0163ia $a = \\frac1{yz}, \\, b = \\frac1{zx}, \\, c = \\frac1{xy}$.\r\n\r\nCondi\u0163ia devine $a+b+c=1$, iar inegalitatea de demonstrat este $4(bc+ca+ab) \\leq 9abc+1$:\r\n$ \\, \\Longleftrightarrow \\, \\, 4 \\left( a+b+c \\right) \\sum_\\textrm{cyc}bc \\leq 9abc+\\left( a+b+c \\right)^{3}$\r\n$ \\, \\Longleftrightarrow \\, \\, \\sum_\\textrm{cyc}\\left( 4abc+4 b^{2}c+4 b c^{2}\\right) \\leq 9abc+\\sum_\\textrm{cyc}\\left( a^{3}+3 a^{2}b+3 a b^{2}\\right)+6abc$\r\n$ \\, \\Longleftrightarrow \\, \\, \\sum_\\textrm{cyc}\\left( b^{2}c+b c^{2}\\right) \\leq \\sum_\\textrm{cyc}a^{3}+3abc$\r\n$ \\, \\Longleftrightarrow \\, \\, 0 \\leq \\sum_\\textrm{cyc}a(a-b)(a-c)$, care-i Schur.", "Solution_2": "[color=darkblue][b]Putina geometrie ![/b] Pentru triunghiul $ABC$ notam : semiperimetrul $p$ , $N$- punctul lui Nagel , $C(I,r)$- cercul inscris , $C(O,R)$- cercul circumscris.\n\nAtunci exista identitatile evidente : $\\{\\begin{array}{c}(p-a)+(p-b)+(p-c)=p\\\\\\\\ (p-a)(p-b)+(p-b)(p-c)+(p-a)(p-b)=r(4R+r)\\\\\\\\ (p-a)(p-b)(p-c)=pr^{2}\\end{array}$\n\nadica ecuatia $t^{3}-pt^{2}+r(4R+r)t-pr^{2}=0$ are radacinile $x=p-a$ , $y=p-b$ , $z=p-c\\ .$\n\nSe stie ca $NI^{2}=p^{2}+5r^{2}-16Rr\\ .$ Asadar, $\\boxed{\\ p^{2}+5r^{2}\\ge 16Rr\\ }\\ \\ (*)\\ .$ Acum vom transforma echivalent inegalitatea $(*)$ :\n\n$(*)\\Longleftrightarrow$ $p^{2}+9r^{2}\\ge 4r(4R+r)$ $\\Longleftrightarrow$ $p^{3}+9pr^{2}\\ge 4p\\cdot r(4R+r)$ $\\Longleftrightarrow$ $\\boxed{\\ (x+y+z)^{3}+9xyz\\ge 4(x+y+z)(xy+yz+zx)\\ }\\ \\ (**)\\ .$\n\nSe stie ca inegalitatea $(**)$ este echivalenta cu [b][u]inegalitatea Schur[/u][/b] (asa cum a aratat si Perfect_radio) : $\\boxed{\\ \\sum [x(x-y)(x-z)]\\ge 0\\ }\\ \\ (***)\\ .$[hide]\n$\\{\\begin{array}{c}(\\sum x)^{3}=\\sum x^{3}+3\\prod (y+z)\\\\\\\\ \\prod (y+z)=\\sum [yz(y+z)]+2xyz\\\\\\\\ \\sum [yz(y+z)]=\\sum [x^{2}(y+z)]\\end{array}$ $\\Longrightarrow$ $\\{\\begin{array}{ccc}\\boxed{\\ \\sum [x(x-y)(x-z)]\\ }& = & \\blacktriangleleft S\\blacktriangleright\\\\\\\\ \\sum x^{3}-\\sum [x^{2}(y+z)]+3xyz & = & \\blacktriangleleft C\\blacktriangleright\\\\\\\\ (\\sum x\\right)^{3}-4\\sum [yz(y+z)]-3xyz & = & \\blacktriangleleft H\\blacktriangleright\\\\\\\\ (\\sum x)^{3}+9xyz-4\\{\\sum [yz(y+z)]+\\sum (xyz)\\}& = & \\blacktriangleleft U\\blacktriangleright\\\\\\\\ (\\sum x)^{3}+9xyz-4\\sum [yz(x+y+z)]] & = & \\blacktriangleleft R\\blacktriangleright\\\\\\\\ \\boxed{\\ (\\sum x)^{3}+9xyz-4\\cdot\\sum x\\cdot\\sum (yz)\\ge 0\\ }& \\mathrm{\\ }& \\mathrm{O.K.}\\end{array}$\n\nAsadar , $\\boxed{\\ x(x-y)(x-z)+y(y-z)(y-x)+z(z-x)(z-y)\\ge 0\\Longleftrightarrow (x+y+z)^{3}+9xyz\\ge 4(x+y+z)(xy+yz+zx)\\ }\\ .$[/hide]\n[b]Concluzie.[/b] Cele trei inegalitati (dintre care una geometrica) sunt echivalente. Cu aceasta ocazie am obtinut si [b][u]o demonstratie geometrica a inegalitatii Schur ![/u][/b]\n\n[b]Observatie.[/b] Inegalitatea propusa este inegalitatea $(**)$ transformata prin conditia $x+y+z=1$ si substitutia $x: =\\frac{1}{yz}$ etc. [/color]", "Solution_3": "[quote=\"Virgil Nicula\"][color=darkblue][b]Putina geometrie ![/b] Pentru triunghiul $ABC$ notam : semiperimetrul $p$ , $N$- punctul lui Nagel , $C(I,r)$- cercul inscris , $C(O,R)$- cercul circumscris.\n\nAtunci exista identitatile evidente : $\\{\\begin{array}{c}(p-a)+(p-b)+(p-c)=p\\\\\\\\ (p-a)(p-b)+(p-b)(p-c)+(p-a)(p-b)=r(4R+r)\\\\\\\\ (p-a)(p-b)(p-c)=pr^{2}\\end{array}$\n\nadica ecuatia $t^{3}-pt^{2}+r(4R+r)t-pr^{2}=0$ are radacinile $x=p-a$ , $y=p-b$ , $z=p-c\\ .$\n\nSe stie ca $NI^{2}=p^{2}+5r^{2}-16Rr\\ .$ Asadar, $\\boxed{\\ p^{2}+5r^{2}\\ge 16Rr\\ }\\ \\ (*)$\n[/quote]\r\n\r\nsuperba demonstratia `geometrica` pt inegalitatea lui gerretsen :blush:" } { "Tag": [ "geometry", "trigonometry", "trig identities", "Law of Sines" ], "Problem": "Let $ a$, $ b$, and $ c$ be the side lengths of $ \\triangle ABC$. Let $ R$ be the circum-radius of $ \\triangle ABC$. Prove that\r\n\r\n\\[ [ABC]\\equal{}\\frac{abc}{4R}\\]\r\n\r\nwhere $ [ABC]$ denotes the area of $ \\triangle ABC$.", "Solution_1": "[hide=\"hint\"]Prove that the area is equal to the $ (ab \\sin C)/2$, and then use the Extended Law of Sines to get rid of the angle.[/hide]", "Solution_2": "Since $ A\\equal{}\\dfrac{bh}{2}$, and $ h\\equal{}b\\sin C$, the area of the triangle is $ \\dfrac{ab\\sin C}{2}$. Also $ \\dfrac{2\\sin C}{c}\\equal{}\\dfrac{1}{4R}$. Need some help there :dry:", "Solution_3": "[\u25b3ABC] = 1/2*sinA*bc = 1/2*sinB*ac = 1/2*sinC*ab -------(1)\r\n\r\nby 'sine rule'\r\n2R= a/sinA = b/sinB = c/sinC \r\n\r\nso you can change the (1) to [size=150][\u25b3ABC] = abc/4R [/size](because, [b]sinA is a/2R[/b])", "Solution_4": "[quote=\"#H34N1\"]Also $ \\dfrac{2\\sin C}{c} \\equal{} \\dfrac{1}{4R}$. Need some help there :dry:[/quote]That's not correct. It should be $ \\dfrac{\\sin C}{2c} \\equal{} \\dfrac{1}{4R}$." } { "Tag": [ "geometry", "perpendicular bisector", "congruent triangles" ], "Problem": "Derive the equation of the locus of all points that are equidistant from points A and B if \r\na. A= (-3,5) and B= (3,-5)\r\nb. A= (-4,-1) and B= (10,5)\r\n\r\nHINT: [hide][How is the locus related to the line segmant AB?][/hide][/hide]", "Solution_1": "HINT Define a line that goes through $A$ and $B$ and find a perpendicular line such that the intersecting of the two lines is at the center of segment $\\overline{AB}$.", "Solution_2": "[hide=\"Hint\"]\n$\\cdot \\text{It is a trace of a point in an equidistance from two fixed point A, B}\\iff \\text{A perpendicular bisector of AB}$\n$\\cdot z,\\alpha,\\beta\\in\\mathbb{C}\\ s.t. \\ |z-\\alpha|=|z-\\beta| \\ (\\alpha \\not= \\beta) \\iff \\text{Point z is a point on a perpendicular bisector of segment of a line}\\ \\alpha \\beta$[/hide]", "Solution_3": "[hide=\"A solution\"]\n1. Draw the line $AB$ (Construction)\n2. Draw a point $P$ in the locus (Construction)\n3. Drop a perpendicular from $P$ to $AB$ at $M$ (Construction)\n4. $AP = BP$ (Given)\n5. $MP = MP$ (Reflexive property)\n6. $\\Delta AMP \\cong \\Delta BMP$ ($HL$ from steps 4 and 5)\n7. $AM = MB$ (Corresponding parts of congruent triangles are congruent)\n8. $P$ lies on the perpendicular bisector of $AB$ (Steps 3 and 7)\n[/hide]", "Solution_4": "would the solution be 3/5x=y for a \r\nand y=3/7x+5/7 for b\r\n??" } { "Tag": [ "trigonometry", "function" ], "Problem": "Which number equals the sum of all numbers [i]a [/i]between 0 and 2 :pi: that satisfy : tan [i]a [/i]- 2003tan [i]a [/i]+ 1 = 0", "Solution_1": "I know this is offtopic but where I can see the Flanders Math Olympiad problems? I don't think the site offers it.", "Solution_2": "It's zero\r\n[hide]\nConsider for what tan(a) it holds and then consider graph of function tan(x)\n[/hide]", "Solution_3": "[quote=\"Silverfalcon\"]I know this is offtopic but where I can see the Flanders Math Olympiad problems? I don't think the site offers it.[/quote]Seach for Flanders on the forum, and you will find some of the problems proposed in the past years." } { "Tag": [ "geometry", "3D geometry", "tetrahedron", "counting", "distinguishability", "rotation", "symmetry" ], "Problem": "Each face of a regular tetrahedron is painted either red, white or blue. Two colorings are considered indistinguishable if two congruent tetrahedra with those colorings can be rotated so that their appearances are identical. How many distinguishable colorings are possible?\r\n\r\n$ \\textbf{(A)}\\ 15 \\qquad \\textbf{(B)}\\ 18 \\qquad \\textbf{(C)}\\ 27 \\qquad \\textbf{(D)}\\ 54 \\qquad \\textbf{(E)}\\ 81$", "Solution_1": "We can do casework.\r\n[hide]\nIf the tetrahedron is all one color, there are $3$ ways to do this.\n\nIf the tetrahedron has three faces of one color and one of another, there are $3*2=6$ ways to do this.\n\nIf the tetrahedron is two and two, there are $3$ ways to do this that are distinguishable.\n\nIf the tetrahedron is two, one, and one, there are $3$ ways to do this because many end up being reflections. Thus there are only three distinguishable types, and you get to choose only what color is used for two faces.\n\nThe total is $15$, or $A$.[/hide]", "Solution_2": "#red faces=r, #blue faces=b, #white faces=w\r\n\r\nr+b+w=4,\r\nthere are $6C2=15$ different tetrahedrons :D", "Solution_3": "[quote=\"Allan Z\"]#red faces=r, #blue faces=b, #white faces=w\n\nr+b+w=4,\nthere are $6C2=15$ different tetrahedrons :D[/quote]\r\n\r\nWow, why didn't I think of that? :D Nice solution. Lol, I love how balls and urns can be applied to almost ANYTHING. :) \r\n\r\n(Oh, and you can use [code] \\dbinom{6}{2} [/code] to get it to show up as $\\dbinom{6}{2}$ if you want.)", "Solution_4": "[quote=\"Allan Z\"]#red faces=r, #blue faces=b, #white faces=w\n\nr+b+w=4,\nthere are $6C2=15$ different tetrahedrons :D[/quote]\r\nare you sure that's valid?", "Solution_5": "[quote=\"Allan Z\"]#red faces=r, #blue faces=b, #white faces=w\n\nr+b+w=4,\nthere are $6C2=15$ different tetrahedrons :D[/quote]\r\n\r\nhah, great.", "Solution_6": "[quote=\"iamagenius\"][quote=\"Allan Z\"]#red faces=r, #blue faces=b, #white faces=w\n\nr+b+w=4,\nthere are $6C2=15$ different tetrahedrons :D[/quote]\nare you sure that's valid?[/quote]\r\n\r\nI think it is, because of all the symmetry and rotations. And on an AMC, nothing needs to be rigorous, as long as it works (quickly). :)", "Solution_7": "[quote=\"iamagenius\"][quote=\"Allan Z\"]#red faces=r, #blue faces=b, #white faces=w\n\nr+b+w=4,\nthere are $6C2=15$ different tetrahedrons :D[/quote]\nare you sure that's valid?[/quote]\r\n\r\nYes, I'm pretty sure the rotations make the individual triangle sides indistinguishable.", "Solution_8": "Not quite: the symmetry group of a tetrahedron up to rotations has 12 elements, while the symmetric group on four elements has 24 elements. (For those of you who don't speak group theory, this just means that there are 12 different ways you can rotate the four faces of the tetrahedron, but 24 permutations of four objects. This means that there are some permutations of the faces that aren't achieved only by rotations. In fact, you can get these extra arrangments if you also allow reflection.) However, it takes 4 colors of paint (one for each side) in order for you to tell a tetrahedron apart from its reflection. With only three colors, each tetrahedron looks the same as its reflection and so this calculation is valid. I wonder if this solution occured to the problem writers, and whether they thought about what I just said.", "Solution_9": "[quote=\"JBL\"]With only three colors, each tetrahedron looks the same as its reflection and so this calculation is valid.[/quote]\r\n\r\nPity I only realized this immediately after I'd finished all my casework. :oops:", "Solution_10": "yes, it worked because its just a tetrahedron, with 4 faces. So it's just basically asking how many nonnegative integer set (r,b,w) to make r+b+w=4.. If you have the same (r, b, w), you can rotate them and see that they are indistinguisable", "Solution_11": "[quote=iamagenius][quote=\"Allan Z\"]#red faces=r, #blue faces=b, #white faces=w\n\nr+b+w=4,\nthere are $6C2=15$ different tetrahedrons :D[/quote]\nare you sure that's valid?[/quote]\n\nIf it isn't, then the answer is at most $15$. Seeing that there are no other answer choices less than $15$, it's the answer." } { "Tag": [], "Problem": "mines only dialup! :( :(", "Solution_1": "Am using dialup right now, but by tonight should have DSL.", "Solution_2": "w00t im using DSL! Leaving all you snails in my dust!\r\n\r\nlol jp", "Solution_3": "I have DSL :) (SBC yahoo!)", "Solution_4": "No option for T1, T2, T3, or Internet 2???\r\n\r\nLOL jk, I highly doubt anyone here has that in their home, unless you are some one with too much money for you own good.", "Solution_5": "Cable. *nods*\r\n\r\nBah I need to get rid of Time Warner. Any suggestions?", "Solution_6": "Dialup......30 kb/ sec......So sometimes I am afraid to open the topics which have been replied by darij or Yetti.", "Solution_7": "DSL! Used to be cable...but DSL is cheaper. :D", "Solution_8": "aha... the power of cable!", "Solution_9": "dial up!\r\nthe sad thing is... before i got my new laptop the computer i was using was too slow for the dial-up data transfer speed anyway...", "Solution_10": "cable! more expensive, but faster than dsl =D", "Solution_11": "[quote=\"shobber\"]Dialup......30 kb/ sec......So sometimes I am afraid to open the topics which have been replied by darij or Yetti.[/quote]\r\n\r\nWhy? \r\n\r\nI have cable too, getting expensive...", "Solution_12": "optimum online!!! :D", "Solution_13": "[quote=\"236factorial\"][quote=\"shobber\"]Dialup......30 kb/ sec......So sometimes I am afraid to open the topics which have been replied by darij or Yetti.[/quote]\n\nWhy? [/quote]\r\nI think that's because they often post very long (and complete!) solutions, as well as some large geometric figures.\r\n\r\nI use dialup (AOL). I've got a new version of AOL recently, but it's still kinda slow :(", "Solution_14": "USE COMCAST!!!", "Solution_15": "Go SBC Yahoo DSL. That one's best cause of all the acronyms.", "Solution_16": "i have cable :lol:", "Solution_17": "Yay for DSL!", "Solution_18": "what does DSL stand for?\r\n[hide]i don't know, i'm asking you!![/hide]", "Solution_19": "[quote=\"math92\"]what does DSL stand for?\n[hide]i don't know, i'm asking you!![/hide][/quote]\r\nDigital Subscriber Lines", "Solution_20": "dial up!!! *grumble grumble* ...no matter, i'll be off to college in a year...", "Solution_21": "w00t cable...altho i live in the middle of nowhere and it's kinda spotty everynow and then...by which i mean all of the bloody time\r\n\r\nbut hey it's cable :P", "Solution_22": "[quote=\"math92\"]what does DSL stand for?\n[hide]i don't know, i'm asking you!![/hide][/quote]\r\n\r\ndigital subscriber line. Also, what about SDSL and ADSL? You know, SDSL is faster than DSL, and ADSL is slower. :?:", "Solution_23": "i have Internet Explorer, rarely use AOL", "Solution_24": "But SDSL is over 10MBPS true connection speed. :)" } { "Tag": [ "geometry", "combinatorics unsolved", "combinatorics" ], "Problem": "A tramp has a coat of area 1 with 5 patches; each patch has area 1/2. Prove that two patches exist with common area>=1/5", "Solution_1": "anyone? :)", "Solution_2": "It has almost certainly appeared repeatedly ... do a search." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let $b^n=b^{n-1}+1$ ($n\\in N^*, b\\geq 1, n\\geq 3$)\r\nProve that $b^n-nb+1<0$", "Solution_1": "[b]EDIT:[/b] I did not see the first condition :oops:", "Solution_2": "Are you sure that $5^5=5^4+1$ :roll:" } { "Tag": [ "MATHCOUNTS" ], "Problem": "Okay, I just got very confused by a very ambiguous site from mathworld (http://mathworld.wolfram.com/WholeNumber.html)[url=http://mathworld.wolfram.com/WholeNumber.html]mathworld.wolfram.com/WholeNumber.html[/url]\r\nFor the purposes of mathcompetitions, can someone provide me with a definition for natural number, whole number, and counting number? Thanks.", "Solution_1": "The natural and counting numbers are the same.\r\nThey are 1,2,3,4,5...\r\nWhole numbers are the same except they include 0\r\nThey are\r\n0,1,2,3,4,5,6,7...\r\nYou rarely hear the term natural or counting numbers on MATHCOUNTS, though. It is almost always \"positive integers\"", "Solution_2": "Think natrual/counting numbers as the numbers they taught you in preschool.\r\n1,2,3,4,5... (The idea of zero and negetive numbers is coufusing to preschool kids)\r\n\r\nThink whole numbers as the numbers they taught you in kindergarden/1st grade.\r\n0,1,2,3,4... (The idea of zero should be understandable by kindergarden/1st grade kids, but not negetive numbers)\r\n\r\nI don't know if that helps. That's what they made us do in preschool and kindergarden/1st grade. But I knew about the integers when i was 3.\r\n\r\nLife was so easy back then.", "Solution_3": "I should add, always look for Non-negative (0 included) and Positive (no 0). \r\n\r\n\r\nSo, Counting/Natural=positive\r\n\r\nWhole number=non-negative.", "Solution_4": "natural=counting numbers: so 1,2,3,4,5,6,7,8 well you get the idea\r\nwhole numbers just have 0 and all the natural numbers." } { "Tag": [ "function", "calculus", "integration", "induction", "real analysis", "real analysis solved" ], "Problem": "Note R+ = the non negative real numbers set and R = the real numbers set.\r\nLet A be a positive real number.\r\nFind all the continuous functions f such that :\r\n- f : R+ ---> R\r\n- f(x) = Integral(0 =< t =< A*x){f(t).dt} , for all x of R+\r\n\r\nSee you", "Solution_1": "Let x in a compact [0;b] of R+\r\n\r\nf'(x) = af(ax) by induction f^(n)(x) = a^(n(n+1)/2)f(a^nx)\r\n\r\nf^(n)(0) = 0 since f(0) = 0\r\n\r\nTaylor formula gives \r\n\r\nf(x) = sum_{k= 0 to n}f^(k)(0)/k! + int_[0;x](x-t)^n/n!.f^(n+1)(t)dt\r\n\r\nf(x) =a^((n+1)(n+2))/n!.int_[0;x](x-t)^nf(a^(n+1)t)dt\r\n\r\n0<= f(x) <= .M.x^(n+1)/(n+1)! where \r\n\r\nM = sup |f| on [0;1]\r\n\r\nlim x^(n+1)/(n+1)! = 0 \r\n\r\nso f(x) = 0 for any x in [0;b]\r\n----------------------------------------------------------------------------------\r\nThis exercice is well known in preparatory classes (math-sup; math-sp)\r\n----------------------------------------------------------------------------------\r\nDiogene where yu from ?", "Solution_2": "Sorry Moubi, I dont agree your proof .. It's Ok if A =< 1 ..\r\nBut could you explain the case : A > 1 ?! I suppose it's very easy for the famous \"preparatory classes\" ... !", "Solution_3": "I didn't it is easy, i considered the case 0<=A<=1\r\n------------------------------------------------------------\r\n\r\nAnother way:\r\n\r\nThe function g:R+--->R, \r\n\r\ng(A) = int_[0;Ax]f(t)dt - f(x) is derivable as\r\n\r\na function of A variable, g(A) = 0 by the hypothesis \r\n\r\ng'(A) = x.f(Ax) - 0 = 0 \r\n\r\nfor x>0 =>for any A >= 0 f(Ax) = 0", "Solution_4": "sorry the guest was me, i forgot log in", "Solution_5": "Sorry Guest , I dont agree your proof .. Dont forget : f might depend on A ..\r\nSee you ..", "Solution_6": "For 0 \\leq A \\leq 1 see the post Fri sept 05, 2002 11:46 pm\r\n\r\n\r\nFor A >1 it seems more interesting, have a look on Fabius function\r\n\r\nhttp://www.math.ohio-state.edu/~edgar/selfdiff\r\n\r\nThank you Diogene for this problem i never studied the case A>1\r\n\r\noubi", "Solution_7": "Let F distribution function of \r\nX= \\sum_{k \\geq 1}2 -nR_k , R_k are independant random\r\nvariables, uniformly distributed on [0;1]\r\nConsider f the unique function on [0;+\\infty[ such that \r\n\r\nf(x) = F(x) on [0;1]\r\n\r\nf(x) = 1- f(1-x) on ]1,2]\r\n\r\nf(x) = - f(x-2 n) on ]2 n;2 n+1]\r\n\r\nExercice : Check f(x) = \\int_[0;2x] f(t) dt for any x \\in R+\r\n\r\n\r\nreference:\r\nJ.Fabius, \"A probalistics example of nowhere analytic C \\infty function\", Z.Wahrsch.Verw.Geb.5 (1966) 173-174\r\n\r\n(in math research library)", "Solution_8": "\"This exercice is well known in preparatory classes (math-sup; math-sp)\" -- What classes are that?", "Solution_9": "Great Moubi .. I didnt know these Fabius functions , and I was looking for a such function !!! Thank you very much .. \r\nJust one remark : if F is the Fabius function , from X = Sum[k>=1]{R(k)/(a^k)} with a > 1, \r\nI think its more convenient to define f such that \r\nf(x) = F(x) , for 0 =< x =< min((a-1)/a ,1/a)\r\nf(x) = f(x/a)/a , for min((a-1)/a ,1/a) =< x \r\nThank you again for the reference. Im still looking for the book" } { "Tag": [ "geometry", "circumcircle", "radical axis", "geometry proposed" ], "Problem": "Let $ABC$ be a triangle. Let $D,E,F$ be points on $\\overline{BC},\\overline{CA}$ and $\\overline{AB}$ respectively, such that $AD,BE$ and $CF$ meet in a single point. Suppose that $ACDF$ and $BCEF$ are cyclic quadrilaterals. Prove that $AD\\perp BC, BE\\perp AC$ and $CF\\perp AB$.", "Solution_1": "someone post a solution please? i know it's an easy problem... :maybe:", "Solution_2": "Some hints by Menelaus theorem we can show $AEBD$ is cyclic too, we well know Miquel theorem circumcircle of triangl $AEF,CDE,BFD$ are concurrent at a point $M$ call concurrent point of $AD,BE,CF$ be $M'$ we will show $M\\equiv M'$ indeed we have $A$ is radical center of the circles $(BFD),(CDE),(BCEF)\\Rightarrow\\ A,M',D$ are colinear similarly we get $M\\equiv M'$ thus the quadrilaterals $AEMF,CDME,BFMD$ are cyclic, too.By calculate the angles we can show $M$ is othorcenter.", "Solution_3": "Suppose cevians$AD,BE,CF$ meet in $H$\r\n\r\nbecause $ACBF$ and $CEBF$ cyclic so $\\angle ADC=\\angle AFC$,$\\angle FBE=\\angle FCE$ and $\\angle CEB=\\angle CFB=\\pi-\\angle AFC$\r\n\r\nso $EHDC$ is cyclic too..\r\n\r\nbecause $\\angle FBE=\\angle FCE=\\angle HDE$ so $ABED$ cyclic or $\\angle AEB=\\angle ADE \\rightarrow \\angle ADB =\\frac{\\pi}{2}$\r\n\r\nthe conclusion follows", "Solution_4": "here's my solution: first verify that if one of $AD\\perp BC, BE\\perp CA, CF\\perp AB$ holds then all three holds(easy). so we need only show that $AD\\perp BC$. let $AD,BE,CF$ intersect at $O$. then $\\angle OEC=\\angle BFC=180-\\angle AFC=180-\\angle ODC\\quad (1)$. So $ODCE$ is cyclic $\\iff AO\\cdot AD=AE\\cdot AC$. but as $BCEF$ is cyclic we have $AE\\cdot AC=AF\\cdot AB$. so $AO\\cdot AD=AF\\cdot AB\\iff ODBF$ is cyclic hence $\\angle ODB+\\angle OFB=180\\quad (2)$. but $\\angle OFB=\\angle OEC$ so from $(1)$ and $(2)$, we get $\\angle ODB=\\angle ODC$ and the conclusion follows immediately." } { "Tag": [ "function", "algebra proposed", "algebra" ], "Problem": "Find all nondecreasing functions $ f : R\\rightarrow R$ such that $ f(0) \\equal{} 0, f(1) \\equal{} 1$ and $ f(a) \\plus{} f(b) \\equal{} f(a)f(b) \\plus{} f(a \\plus{} b \\minus{} ab)$ for all real numbers $ a,b$ such that $ a < 1 < b$", "Solution_1": "Answer:\r\n[hide]\nI.\n\\[ f(t) \\equal{} \\begin{cases} 1 \\minus{} (1 \\minus{} t)^p, & \\text{ if } t \\leq 1; \\\\\n1 \\plus{} C |t \\minus{} 1|^p , & \\text{ if } t \\geq 1. \\end{cases}\n\\]\nwhere $ p > 0$, $ C > 0$.\n\nII. $ f(t) \\equal{} 1$ when $ t\\geq 1$; $ f(t)$ is an arbitrary non-decreasing function when $ t\\leq 1$ s.t. $ f(0) \\equal{} 0$, $ f(1) \\equal{} 1$.\n\n[/hide]\nHint:\n[hide]\nConsider the function $ g(s) \\equal{} 1 \\minus{} f(1 \\minus{} s)$. Observe that $ g(s) g(t) \\equal{} g (st)$ when $ s < 0$ and $ t > 0$. \n[/hide]", "Solution_2": "Can you solve clearly , yury?", "Solution_3": "Consider the function $ g(s) \\equal{} 1 \\minus{} f(1 \\minus{} s)$. For every $ s < 0$ and $ t > 0$, we have\r\n\\[ g(s) g(t) \\equal{} (1 \\minus{} f(1 \\minus{} s)) (1 \\minus{} f(1 \\minus{} t)) \\equal{} 1 \\minus{} f(1 \\minus{} s) \\minus{} f(1 \\minus{} t) \\plus{} f(1 \\minus{} s) f(1 \\minus{} t) \\equal{} 1 \\minus{} f((1 \\minus{} s) \\plus{} (1 \\minus{} t) \\minus{} (1 \\minus{} s) (1 \\minus{} t)) \\equal{} 1 \\minus{} f(1 \\minus{} st) \\equal{} g(st).\r\n\\]\r\n\r\n\\[ g(0) \\equal{} 1 \\minus{} f(1) \\equal{} 0.\r\n\\]\r\n\r\n\\[ g(1) \\equal{} 1 \\minus{} f(0) \\equal{} 1.\r\n\\]\r\nDenote $ C \\equal{} \\minus{} f( \\minus{} 1)$. For every $ t > 0$, we have $ f( \\minus{} t) \\equal{} f( \\minus{} 1) f(t) \\equal{} \\minus{} C f(t)$. Now we consider two cases: $ C \\neq 0$ and $ C \\equal{} 0$.\r\nI. First, let us assume that $ C \\neq 0$. Then for every $ s, t > 0$,\r\n\\[ f(s) f(t) \\equal{} \\minus{} f( \\minus{} s) f(t) /C \\equal{} \\minus{} f( \\minus{} st) / C \\equal{} f(st).\r\n\\]\r\nIt is well known (and it was discussed at this forum many times) that every monotone function $ f(s)\\not\\equiv 0$ that satisfies this identity equals $ s^p$ for some $ p$ when $ s > 0$. Now, $ f( \\minus{} s) \\equal{} \\minus{} C f(s) \\equal{} \\minus{} C s^p$ (when $ s > 0$). We conclude that\r\n\\[ f(s) \\equal{} \\begin{cases} s^p, & \\text{ if } s \\geq 0; \\\\\r\n\\minus{} C |s|^p, & \\text{ if } s < 0. \\end{cases}\r\n\\]\r\nPlugging this expression into the formula $ f(a) \\equal{} 1 \\minus{} f(1 \\minus{} a)$, we get the answer (see my previous post). It is easy to check that it satisfies the condition of the problem.\r\n\r\nII. Assume $ C \\equal{} 0$. Then $ f(s) \\equal{} 0$ when $ s < 0$. It is easy to see that when $ s > 0$ the function $ f(s)$ can be an arbitrary non-decreasing function s.t. $ f(1) \\equal{} 1$." } { "Tag": [], "Problem": "I can't say this in spanish so I'll just say in english.\r\n\r\nAnyone compete in spanish olympiad?\r\n\r\nYes o No", "Solution_1": "\u00a1 Eso es un buen ejercicio de traducci\u00f3n !", "Solution_2": "[quote]Eso es un buen ejercicio de traducci\u00f3n ![/quote]\n\n=\n\n[quote=\"Translator\"]That's a good exercise of translation![/quote]\r\n\r\nGracias.\r\n\r\nP.S. Does that mean you compete for the high school math competition? If so, can you tell what kind and some rules?\r\n\r\nI'll try to take the idea by translator if I don't get it.\r\n\r\nThnx.", "Solution_3": "No !\r\nI'm French (as you can see with the flag)...\r\nSo...\r\n\r\nI don't even participate in French Olympiad...", "Solution_4": "Yo tengo amigos que participan, que quiere saber? igual creo que existe una buena pagina web, busque: \"olimpiada espa\u00f1ola de matematicas\"", "Solution_5": "[url=http://www.e-thug.net/sounds/aww.wav]http://www.e-thug.net/sounds/aww.wav[/url]" } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "The circles $C_{1}$ and $C_{2}$ touch externally at $M$ and the radius of $C_{2}$ is larger than that of $C_{1}$. $A$ is any point on $C_{2}$ which does not lie on the line joining the centers of the circles. $B$ and $C$ are points on $C_{1}$ such that $AB$ and $AC$ are tangent to $C_{1}$. The lines $BM, CM$ intersect $C_{2}$ again at $E, F$ respectively. $D$ is the intersection of the tangent at $A$ and the line $EF$. Show that the locus of $D$ as $A$ varies is a straight line.", "Solution_1": "It has already been posted before at http://www.mathlinks.ro/Forum/viewtopic.php?t=16007\r\n\r\nKostas Vittas." } { "Tag": [ "email", "MATHCOUNTS", "\\/closed" ], "Problem": "I have been at the classroom as of about 7:55. The gray little button that says connect is grayed out, however. I'm pretty sure I'm logged in, so why can't I connect? BTW, it is 8:07 EST right now.", "Solution_1": "Sorry I missed this earlier, Mystic!\r\n\r\nWhen the little button is grayed out, that means the class is in the process of loading for you. (The button is grayed out for everyone when they go there logged in and class is open.) It sometimes takes several minutes for the classroom to load, but it shouldn't take 12 minutes.\r\n\r\nAre you sure your computer is Java enabled? You should see an interactive Mandelbrot set on the Virtual Classroom page (even now, when class is closed).\r\n\r\nAgain, sorry I didn't see this and answer it sooner. If you like, you can send me an email (rusczyk@artofproblemsolving.com) and we can set up a time to test your connection (I'll create a page for you to go see and try to connect - if it fails again, we'll get together & figure it out.).", "Solution_2": "ok, I went over there and checked. I was able to see the Mandelbrot, and zoom up on it with the mouse, if that's what you mean.", "Solution_3": "I'll try again at the upcoming mathcounts mathjam. If that doesn't work, I'll send an e-mail.", "Solution_4": "I made a little test page for you; I sent you an email with details. Please email me back and let me know what happens...\r\n\r\nThanks.", "Solution_5": "I am having a problem with joining the Math Jams. It seems that everytime I click on the Math Jams link (leave the Forum) I get logged off. When I go back to the Forum, I have to log in again.", "Solution_6": "Peculiar error. What browser/platform are you using? I saw you (I think) in class today towards the end - did you solve your problem?", "Solution_7": "Are you sure you saw me? I didn't go in the classroom. I'm using Microsoft Internet Explorer.", "Solution_8": "I sent you an email yesterday about what to try - please send me an email back after you try it out." } { "Tag": [ "induction", "superior algebra", "superior algebra solved" ], "Problem": "Let $S$ be finite commutative semigroup wrt multiplication consisting of integer-valued non-invertable matrices of order 2. Prove that if $|S|>5$ then $AB=0$ for all $A,B\\in S$.", "Solution_1": "[quote=\"Myth\"]Let $S$ be finite commutative semigroup wrt multiplication consisting of integer-valued non-invertable matrices of order 2. Prove that if $|S|>5$ then $AB=0$ for all $A,B\\in S$.[/quote]\r\n\r\nMyth: Nice problem, it seems trivial when you look at it, but it is not, actually, I haven't been able to find a counterexample, but I am not 100% is sure. the condition $AB=0$ is for $A\\neq B$ or in general? If it is in general, I think it should be enough to prove that the matrices should be nilpotent. Any way, I have to run, need to give a class in five minutes, I'll try it later!", "Solution_2": "We can assume that $S$ is such that $0\\in S$ and $X\\in S\\Rightarrow -X\\in S$. This is because adding $0$ and $-S$ to $S$ does not change the property of $S$ of being a finite semigroup.\r\n\r\nLet $A\\in S$. Because $A,A^2,A^3,\\ldots$ is a finite set, so is $\\det A, \\det A^2,\\ldots$. But $\\{n^k|k\\in\\mathbb{N}\\}$ is finite iff $n\\in\\{-1,0,1\\}$.\r\nIf $\\det A=\\pm 1$ then $A$ is invertible which is false by hypothesis. So $\\det A=0\\ \\forall\\ A\\in S$.\r\n$A^2=aA$ (Hamilton-Cayley), where $\\mathbb{Z}\\ni a=Tr(A)=Trace(A)$.\r\nBy induction, $A^n=a^{n-1}A\\ \\forall n\\in\\mathbb{N}^*$. For the set $\\{a^nA|n\\in\\mathbb{N}\\}$ to be finite we must have $a\\in\\{-1,0,1\\}$ or $A=0$(which in turn leads to $a=0$). So $Tr(A)\\in\\{-1,0,1\\}\\ \\forall\\ A\\in S$.\r\n\r\nAssume there exists $A\\in S$ such that $A^2\\neq 0$. Then $a=Tr(A)\\neq 0$, so $a=\\pm 1$.\r\nLet $X\\in S$. Because $AX=XA$ and $A\\not\\in\\mathbb{Z}\\cdot I_2$, there exist $f(X),g(X)\\in\\mathbb{Q}$ such that $X=f(X)A+g(X)I_2$.\r\nBecause the eigenvalues of $X$ are $0$ and $Tr(X)$, we get $\\{0,Tr(X)\\}=\\{f(X)\\cdot a+g(X), g(X)\\}$.\r\nIf $g(X)=0$ then $f(X)=aTr(X)=\\pm Tr(X)$, so $X=\\pm A$ or $f(X)=0\\Rightarrow X=0_2$.\r\nIf $a\\cdot f(X)+g(X)=0$ and $g(X)=Tr(X)=\\pm 1$. Then $f(X)=-a\\cdot g(X)$ and $g(X)=\\pm 1$, leading to the solutions $X=\\pm (I_2-a\\cdot A)$.\r\n\r\nSo if $a=1$ then $S\\subset\\{0_2,A,-A,I_2-A,A-I_2\\}$. If $a=-1$ then $S\\subset\\{0_2,A,-A,I_2+A,-I_2-A\\}$. Both cases contradict $|S|>5$.\r\nSo $A^2=0_2\\ \\forall\\A\\in S$.\r\nChoose $0_2\\neq A\\in S$. For any $X\\in S$ consider $f(X),g(X)$ as above. Then $X=f(X)A+g(X)I_2$.\r\nEigenvalues reasoning leads to $g(X)=0$, so $X=f(X)A$. Multiply by $X$ to obtain $f(X)\\cdot (AX)=0_2\\Rightarrow$ $(f(X)=0\\Rightarrow X=0_2)$ or $AX=0$.\r\nThis immediately implies $AX=0_2\\ \\forall\\ A,X\\in S$.\r\n\r\n\r\nAs an example of such $S$, take $S=\\{f\\cdot A|f\\in F\\}$, where $F$ is a finite set containing $0$ and $A$ is a nilpotent matrix." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "If $a,b,c>0$ then prove that $(a^2+b^2+c^2)^2\\geq 3(ab^3+bc^3+ca^3)$", "Solution_1": "Is there any difference between this one and vasc's one?! :?", "Solution_2": "[quote=\"Soarer\"]Is there any difference between this one and vasc's one?! :?[/quote]\r\n\r\nYes .Vasc's one is $(a^2+b^2+c^2)^2\\geq 3(a^3b+b^3c+c^3a)$", "Solution_3": "...so how is that different? You're just switching c and b...", "Solution_4": "[quote=\"K81o7\"] You're just switching c and b...[/quote]\r\n\r\nOh yes....It is easy....." } { "Tag": [ "geometry", "AMC", "USA(J)MO", "USAMO", "search", "articles", "induction" ], "Problem": "Can someone tell me how to get good at proofs?\r\n\r\nI can't write a good proof if my life depended on it... I need some dire help...\r\n\r\n :noo: :help: :ewpu:", "Solution_1": "Take Geometry. When you are proving something, you really can't use the theorem that you are proving inside your proof. Geometry is the main thing with proofs. You prove SAS, ASA, SSS, and a lot more congruences. You'll learn. Don't worry. If you're worrying about it because you are proving stuff for USAMO and stuff, then you need to seriously search up how to prove things.", "Solution_2": "Proofs is mostly about geometry in middle school and stuff, but it can get much harder. I started learning proofs this year, and I'm understanding a bit since they teach you geometric proofs in geometry. \r\n\r\nThere are some easy proofs like proving that any odd number is a difference between 2 squares. To prove this i would say that any odd number is in form 2a+1. And $ 2a \\plus{} 1$=$ (a \\plus{} 1)^2$-$ a^2$ because $ (a \\plus{} 1)^2$ is $ a^2$+$ 2a \\plus{} 1$ and subtract $ a^2$ from that and you get the equation 2a+1=2a+1. \r\n\r\nHowever, there are very complicated proofs. My suggestion would be start from the very basic and work your way up. Here is a great site to start learning proofs: [url]http://zimmer.csufresno.edu/~larryc/proofs/proofs.html[/url]\r\n\r\nGood luck!!", "Solution_3": "eh... I guess I'll take intro to geometry!", "Solution_4": "Wait! They might not have proofs in online classes. It's actually better to take it in real person with a teacher that you can see and a class that is actually the full length. Never skip Geometry. Only skip Algebra 1 or Algebra 2. As I am skipping Algebra 2.", "Solution_5": "read this [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_HowWrite.php]article[/url] on writing proofs.\r\n\r\nThe more advanced AoPS books cover proof by induction, contradiction, pigeonhole, and direct proof", "Solution_6": "[quote=\"jonathanchou711\"]Wait! They might not have proofs in online classes. It's actually better to take it in real person with a teacher that you can see and a class that is actually the full length. Never skip Geometry. Only skip Algebra 1 or Algebra 2. As I am skipping Algebra 2.[/quote]\r\n\r\nEr, yes they do. Also, proofs you learn at school are just two-column trivial proofs where you have to show about 30 steps in a 1-line proof. AoPS Intro to Geometry is a much better course than school geometry (took both).", "Solution_7": "eh... AIME15 is the [i]expert[/i]... I'll agree with him", "Solution_8": "[quote=\"Isabella2296\"]-- Before posting, think hard whether your post is necessary or not. Remember, you do not have to push the \"Submit\" button. [/quote]\r\n\r\nThink about your post. Was it really necessary?\r\n\r\n---EDIT---\r\nI just realized that my post was spam, too. Sorry! :blush:" } { "Tag": [], "Problem": "Explain why Isoborneol ([url]http://en.wikipedia.org/wiki/Isoborneol[/url]) on treatment with 50% sulphuric acid gives Camphene ([url]http://en.wikipedia.org/wiki/Camphene[/url]) and not Bornylene ([url]http://www.pherobase.com/database/compound/compounds-detail-bornylene.php[/url])\r\n\r\nGive a mechanism if possible.\r\n\r\n[hide=\"My Opinion\"]Of course the reason should be that the double bond would rather be outside the ring than within it due to high angle strain. In bornylene, 2 carbon atoms are under angle strain, whereas in camphene, only one atom is under angle strain.\nBut who can give the mechanism? It seems very different, to me atleast. :maybe: [/hide]", "Solution_1": "Mechanism given by my friend:\r\n\r\n :)", "Solution_2": "nice mechanism! :)" } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "AMC 12", "AIME", "USAJMO", "AMC 10" ], "Problem": "OK, so I am kind of confused on how they do the USAMO Index. I understand that they will take your [AMC12]*10[AIME] scores, but I don't know where to see the floor scores from past years. What score should I be shooting for? I've heard 200, but will it increase or decrease because of the USAJMO? Does anyone know where I can find the floor scores or the Index scores by state?", "Solution_1": "Well, they shouldn't have any floor scores this year for either USAMO or USAJMO because you can only make USAMO by taking the AMC12 and make USAJMO by taking the AMC10 (the only exception is if you solely take the AMC10 and get an 11+ on the AIME)\r\n\r\nPrevious approximates were at 210 because that is how it's been in the past when there were fewer qualifiers. Before there was Red MOP qualification through USAJMO for 10th graders as well, a better estimate would have been 220, since many of the ambitious underclassmen who took the AMC10 last year will likely take the AMC12 this year to get to USAMO, and they are not factored in.\r\n\r\nOf course, with Red MOP qualification, there will likely be a decrease in AMC12 takers. I'd say you are probably safe with a 215 index.", "Solution_2": "lol i hope this helps\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=321488", "Solution_3": "Is there any site that has records of the indices of the USAMO participants from past years?", "Solution_4": "http://en.wikipedia.org/wiki/USAMO#Recent_qualification_indices\r\n\r\nNot a very useful list anymore, considering all the changes that have been put in place between this year and last year.", "Solution_5": "If you take the average of the USAMO invitation indices listed, the result is 213. If you take the average of the USAMO invitation indices for 2000, 2001, 2003, 2004, 2005 when we invited about 250-260 participants the result is 217. Using the average like this is not perfect, but year-by-year values and the overall average are consistent with my estimate based on 2008 and 2009 results that an index of 210 will be the approximate invitation level for 2010. Of course, the precise value will depend on the variability in difficulty of the 2010 AMC 10 and AMC 12 and the AIME compared to previous years. The committees do an excellent job of calibrating the difficulty level given that there are 6 contests each year, each with new problems, but the contests do vary in overall difficulty with a standard deviation which is about 1 problem from an overall average.\r\n\r\n-- Steve Dunbar\r\nMAA Director, American Mathematics Competitions", "Solution_6": "To Steve,\r\n\r\nIs it true that the average AIME score in 2008 and 2009 are higher than that of 2000 - 2005? If true, what is the impact of higher AIME score on the USAMO index estimate?", "Solution_7": "The information is at:\r\nGeneral Statistics\r\non\r\nhttp://www.unl.edu/amc/e-exams/e7-aime/archiveaime.shtml\r\n\r\nSteve Dunbar\r\nMAA Director, American Mathematics Competitions", "Solution_8": "everyone in usajmo will have an AIME score 10 or less right?", "Solution_9": "Oh yeah, I didn't notice that. It's still going to be 120 to qualify for AIME if you take the amc 10 right?", "Solution_10": "its going to be 120 or the minimum score of the top 1%, whichever is lower\r\ni remember 2007 AMC10B the cutoff for AIME was 117, so occasionsly it is less than 120", "Solution_11": "What is the USAJMO index probably going to be then?", "Solution_12": "it will be at most 250 lol\r\n\r\nbut im guessing usajmo will be around 210." } { "Tag": [ "search", "number theory", "prime numbers", "number theory proposed" ], "Problem": "Prove that there exists an infinity of pairs of primes $ (p, q)$ such that $ p|2^{q\\minus{}1}\\minus{}1$ and $ q|2^{p\\minus{}1}\\minus{}1$", "Solution_1": "[quote=\"andyciup\"]Prove that there exists an infinity of pairs of primes $ (p, q)$ such that $ p|2^{q \\minus{} 1} \\minus{} 1$ and $ q|2^{p \\minus{} 1} \\minus{} 1$[/quote]\r\n\r\nI would suggest $ (p,p)$ for any odd prime $ p$\r\n\r\n:)", "Solution_2": "It should be ''different prime numbers '' and have a look here :\r\n http://www.mathlinks.ro/viewtopic.php?search_id=1054974708&t=204055" } { "Tag": [ "number theory", "prime numbers" ], "Problem": "Let $p_n$ be the $n$-th prime. ($p_1=2$)\r\nDefine the sequence $(f_j)$ as follows: \r\n- $f_1=1, f_2=2$\r\n- $\\forall j\\ge 2$: if $f_j = kp_n$ for $k p_y$. Then $p_x$ divides into $b$ implying b is a multiple of $p_x$, contradiction $b < p_y$. Similarily for $p_y > p_x$. For $p_x = p_y$ we get $a = b$, so that they are the same term.\r\n\r\n(b) When we get to the prime number 101 (26th prime), we will have used 1 + the sum of the prime numbers up to 101. Afterwards, we always have a factor > 100, so we are guaranteed. We are also guaranteed previous up to 2 * 97, where before that we have 97, which is 2 digit. So we go back 96 numbers.\r\n\r\n(c) any number where dividing by the largest prime yields a number larger than that prime.\r\n\r\n(d) We go up to 9*11; then 7*13; 5* 17,19 ; 4 * 23; 3 * 29, 31; 2 * 37, 41, 43, 47; 1 * 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. This can be counted as 1 + the sum of the primes up to 11 exclusive, plus 9 + 7 + 5*2 + 4 + 3*2 + 2*4 + 10.", "Solution_2": "[quote=\"Singular\"](a) Suppose two f_i are the same, say $ap_x$ and $bp_y$. (a < p_x ; b < p_y). Suppose $p_x > p_y$. Then $p_x$ divides into $b$ implying b is a multiple of $p_x$, contradiction $b < p_y$. Similarily for $p_y > p_x$. For $p_x = p_y$ we get $a = b$, so that they are the same term.[/quote]Correct. :)\n\n[quote=\"Singular\"](b) When we get to the prime number 101 (26th prime), we will have used 1 + the sum of the prime numbers up to 101. Afterwards, we always have a factor > 100, so we are guaranteed. We are also guaranteed previous up to 2 * 97, where before that we have 97, which is 2 digit. So we go back 96 numbers.[/quote]So the explicit index number? ...\n\n[quote=\"Singular\"](c) any number where dividing by the largest prime yields a number larger than that prime.[/quote]Correct. :)\n\n[quote=\"Singular\"](d) We go up to 9*11; then 7*13; 5* 17,19 ; 4 * 23; 3 * 29, 31; 2 * 37, 41, 43, 47; 1 * 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. This can be counted as 1 + the sum of the primes up to 11 exclusive, plus 9 + 7 + 5*2 + 4 + 3*2 + 2*4 + 10.[/quote]So... you say 2+3+5+7+9+7+5*2+4+3*2+2*4+10 = 69? Not correct in that case." } { "Tag": [ "function", "real analysis", "real analysis solved" ], "Problem": "grobber wrote:\r\n4) \r\nf:[0,1]->[0,1] is continuous and bijective. Determine the set A={f(x)-f(y) | x, y are in [0,1]\\Q}. \r\n\r\n(Students may use the result: there are no injective functions from [a,]\\Q to Q)\r\n----------------------------------------------------------------------\r\n\r\nVery good job, Harazi", "Solution_1": "If I'm not wrong, my solution was the following:\r\n Clearly, f must bestrtly monotonuous. Since changing f to -f doesn't moify anythind, let us assume that f is strictly increasing. Then f(0)=0 and f(1)=1 and also for each x and y we have 1 \\geq f(x)-f(y) \\geq -1. Thus, A is in (-1,1). Now let us take t in (-1,0). Suppose t is not in A.Let us take the interval I=[0, f^(-1)(t+1)] . For each x in I we have 0 \\leq f(x)-t \\leq 1. THus, we can define the function g from I\\Q to Q, g(x)=f^(-1)(f(x)-t). This is well defined because if there is x such that g(x) belongs to R\\Q, then t=f(x)-f(g(x)). And g is clearly injective, contradiction. So, we have that (-1,0) is in A. But if x is in A, so is -x. So, (-1,1) is in A and A=(-1,1)." } { "Tag": [ "inequalities", "function", "inequalities solved" ], "Problem": "Prove, for x1,x2,....,xn>0 , that\r\n\r\n (x1^x1*x2^x2*...*xn^xn)^(1/(x1+x2+....+xn)) >=(x1+x2+...+xn)/n", "Solution_1": "Write the inequality in the form \\sum xi*ln(xi) \\geq (\\sum xi)*ln(\\sum xi/n) and then apply Jensen's inequality for the convex function f(x)=x*ln(x)." } { "Tag": [ "geometry" ], "Problem": "Wich point $ P$ inside a triangle $ ABC$ minimizes the sum $ PA\\plus{}PB\\plus{}PC$\r\n\r\n :)", "Solution_1": "That is Fermat point.", "Solution_2": "[quote=\"AoPSWiki\"]A method to find the point is to construct three equilateral triangles out of the three sides from $ \\triangle ABC$, then connect each new vertex to each opposite vertex, as these three lines will concur at first Fermat point.[/quote]\r\n\r\n\r\nTake directly from [[Fermat point]]." } { "Tag": [ "function", "number theory", "algebra unsolved", "algebra" ], "Problem": "Let a_1, a_2, ..., a_n be a string of symbols from the set {a,b,...,j}. Prove that there is a bijective function f : {a,b,...,j} -> {0,1,...,9} such that the string f(a_1)f(a_2)...f(a_n) represents an n-digit integer in decimal expansion which is divisible by 3.", "Solution_1": "Let symbols be $f_{0}$, $f_{1}$, ..., $f_{9}$, and they counts in the initial string are $b_{0}$, ..., $b_{9}$ respectively. We may assume $b_{i}\\in\\{0,1,2\\}$.\r\n\r\nCase 1. All $b_{i}$ are equal modulo 3. Then any bijection satisfy the condition.\r\n\r\nCase 2. There are $b_{i}=0$, $b_{j}=1$ and $b_{k}=1$. Then consider arbitrary bijection where $f_{i}=0$, $f_{j}=1$ and $f_{k}=2$. If it doesn't satisfy the condition then consider the same bijection but $f_{i}=1$ and $f_{j}=0$ OR $f_{i}=2$ and $f_{k}=0$. One of these two bijections will satisfy the condition.\r\n\r\nCase 3. All other cases are analoguosly to Case 2.\r\n\r\nP.S. Please, prove the general statement: $3\\to p$, $9\\to p^{2}$.\r\n\r\nP.P.S. Clearly, it is a Number Theory problem :)" } { "Tag": [ "algebra", "polynomial", "function", "calculus", "integration" ], "Problem": "Let $f \\in \\mathbb{C} [X]$ and degree $f=n$. We know that $f(0),f(1),\\ldots,f(n) \\in \\mathbb{Z}$.\r\n\r\nProve that $f(a) \\in \\mathbb{Z}$, $\\forall a \\in \\mathbb{Z}$", "Solution_1": "n+1 points uniquely determines a polynomial of degree n?\r\n\r\n(I dont understand the notation $grad f = n$ for a function of one variable.)", "Solution_2": "Yeah, I think by grad he means degree.", "Solution_3": "It follows from difference tables and what Singular said.", "Solution_4": "and what are difference tables?", "Solution_5": "For any polynomial $f$ of degree $n$ on the integers, the function $\\delta f(n) = f(n + 1) - f(n)$ is a polynomial of degree $n - 1$. You can make this into a pretty-looking table, which is why I said \"difference table,\" but I probably should have said \"finite differences.\" In any event, it follows that we can keep doing this, $n$ times, until we get a bottom row that will be constant. This row, and every other entry in the table at that point, will be integral. Thus, building back up from the bottom, the function always takes integral values. (This was not a good description -- can someone do a better job?)\r\n\r\n\r\nHere's a little picture, though, for a known cubic with 4 integral values:\r\n\r\n[code]f: 2 9 28 65\ndf: 7 19 37\nddf: 12 18\ndddf: 6[/code]\n[code]f: 2 9 28 65\ndf: 7 19 37\nddf: 12 18\ndddf: 6 6 6 6 ...[/code]\n[code]f: 2 9 28 65 126 217 ...\ndf: 7 19 37 61 91 ...\nddf: 12 18 24 30 ...\ndddf: 6 6 6 6 ...[/code]", "Solution_6": "i think i got the idea pretty much" } { "Tag": [ "function" ], "Problem": "could nebdy please give a brief review on the use of NBS in the formation of allyl bromides.\r\ni mean, for example, if we have 2,4 dimethyl hex-3-ene, to which carbon does the bromine radikl get added? y?\r\nif possibl, pls suggest some references.\r\nthanks in advance", "Solution_1": "$ \\text{NBS}$ is a very useful comound for replacing an allylic $ \\text{H}$ by $ \\text{Br}$.\r\nit is used with high temperatures.In high temepratures the alyylic $ \\text{H}$ is captured by $ \\text{Br}^.$ which may come from $ \\text{NBS}$ or otherwise.The important function of $ \\text{NBS}$ is to react with so formed $ \\text{HBr}$ and convert to succinimide nad step up the forward reaction\r\nReferences-Morrison&Boyd.\r\nIn $ \\text{2,4 - Dimethyl Hex - 3 - ene}$ there are two allylic $ H$ but the $ 4 - \\text{H}$ is resonance , hyperconjugation and inductively stabilized so the $ \\text{H}$ is captured from there\r\n\r\n[b]NOTE[/b]:the name of the compound is $ \\text{N - Bromo Succinimide}$ and not otherwise", "Solution_2": "thanks pardesi for the soln, reference as well as correction ( truly spkin, i was actually confused whether its succinate or succinamide)", "Solution_3": "The allylic bromination of compounds (specially alkenes, but also alkynes, carbonyls, and benzylic positions) with NBS is commonly called the Wohl-Ziegler bromination. The reaction is often performed in a non polar solvent such as CCl4, and some heating is always involved (refluxing for some time) but not very high temperatures. Regarding the regioselectivity, it is known that a secondary position is more readily substituted than a primary one; however, the relative reactivity of tertiary carbons is not clear, with sometimes they being substituted faster than secondary, but quite often secondary is substituted faster than tertiary. So, regarding your compound, 2,4-dimethylhex-3-ene, there are three allylic positions - carbon 2 (tertiary), the 4-methyl group (primary), and carbon 5 (secondary) -, so I would perhaps vote on carbon 5 to be the more readily substituted.\r\nAn important issue in this reaction is its mechanism. The first step is formation of some bromine atoms from NBS via the cleavage of the weak nitrogen-bromine bond - mediated by heat or by some radical initiator present in the mixture:\r\n\r\n$ (NBS)N\\minus{}Br \\longrightarrow (NBS)N \\cdot \\plus{} \\cdot Br$\r\n\r\nNow the bromine atom will abstract a hydrogen atom from an alkene molecule:\r\n\r\n$ R\\minus{}H \\plus{} \\cdot Br \\longrightarrow HBr \\plus{} R \\cdot$\r\n\r\nNow there is a fast ionic reaction between NBS and the HBr liberated in the last step:\r\n\r\n$ NBS \\plus{} HBr \\longrightarrow (NBS)\\minus{}H \\plus{} Br_2$,\r\n\r\nand this step ilustrates that the main function of NBS is to consume the HBr liberated and provide a low but constant concentration of Br2. Now, in the next step the allyl radical reacts with bromine molecule to form the allyl bromide,\r\n\r\n$ R \\cdot \\plus{} Br_2 \\longrightarrow R\\minus{}Br \\plus{} Br \\cdot$\r\n\r\nand the chain repeats. Now, an important question: if bromine (Br2) is one of the reacting species why it does not add to the double bond?", "Solution_4": "1. NBS gives a low concentration of Br.\r\n2. at high temperatures, radical reaction competes`successfully with the addition reaction\r\nand moreover,\r\n3. in the presence of a non polar solvent, since there aren't any polar molecules to solvate and thus stabilise the bromide ion formed in the first step of the addition reaction, the bromide ion uses a bromine atom as a substitute. So, in a non polar solvent the rate equation is second order w.r.t. bromine, so the low concentration (point 1.) has an even more pronounced effect in slowing the reaction.", "Solution_5": "[quote=\"Cyprusad\"]at high temperatures, radical reaction competes`successfully with the addition reaction[/quote]\r\n\r\nAnd that's because the first step of the addition reaction is reversible." } { "Tag": [], "Problem": "How many different combinations of $ \\$5$ bills and $ \\$2$ bills can\nbe used to make a total of $ \\$17$? Order does not matter in this\nproblem.", "Solution_1": "two ways: 3 5s and one 2\r\nor 1 5 and 6 2s" } { "Tag": [ "algebra", "polynomial", "LaTeX", "probability", "linear algebra", "matrix" ], "Problem": "Prove that determinant\r\n\r\ncol 1 col2 col3\r\n\r\n2bc-a^2 , c^2 , b^2\r\n\r\nc^2 , 2ac-b^2 , a^2 ==a^3+b^3+c^3-3abc\r\n\r\n b^2 , a^2 , 2ab-c^2\r\nplease refine your language .this does not look nice for a person, intelligent like you. apology if my words hurts you\r\n\r\n\r\n\r\n,", "Solution_1": "This is complete nonsense. I have no idea what you even intend that statement to mean.", "Solution_2": "Presumably that\r\n\r\n\\[ \\det\\left( \\begin{array} {ccc}\r\n2bc \\minus{} a^2 & c^2 & b^2 \\\\\r\nc^2 & 2ac \\minus{} b^2 & a^2 \\\\\r\nb^2 & a^2 & 2ab \\minus{} c^2\r\n\\end{array} \\right) \\equal{} a^3\\plus{}b^3\\plus{}c^3\\minus{}3abc\r\n\\]\r\n\r\nOf course, this isn't true since the LHS is a homogeneous polynomial of degree $ 6$ and the RHs is a polynomial of degree $ 3$.", "Solution_3": "[b] ayush_2008 [/b]\r\n\r\nI think it would be much better if you learned LATEX (I really can't offer you such an advice because I myself am learning LATEX). If you managed to learn LATEX, it would be much better because it would make your technical posts more legible. It also increases the probability of you getting substantive replies/solutions.", "Solution_4": "The original version of this didn't include the word \"determinant\". I saw it anyway, and dismissed it as obviously wrong for the same reasons as blahblahblah.\r\nOn the $ \\text{\\LaTeX}$, both this thread and your previous thread contain examples of how to build a matrix. Use them; I learned a great deal of what I know of the language by looking at other people's constructions on this forum.\r\n\r\nOn my language- \"nonsense\" is a dangerous word, and I'm aware that I shouldn't use it around my students. Still, some statements truly are nonsense, and this isn't a classroom. Stating that a matrix is equal to a number rises to that level, especially when the obvious choices of taking a trace or a determinant make the statement false.", "Solution_5": "Anyways, it appears that the correct problem is the 'obvious' correction;\r\n\r\n\\[ \\det\\left(\\begin{array}{ccc}2bc\\minus{}a^{2}& c^{2}& b^{2}\\\\ c^{2}& 2ac\\minus{}b^{2}& a^{2}\\\\ b^{2}& a^{2}& 2ab\\minus{}c^{2}\\end{array}\\right) \\equal{} (a^{3}\\plus{}b^{3}\\plus{}c^{3}\\minus{}3abc)^2 \\]\r\n\r\nPerhaps the identity $ a^3\\plus{}b^3\\plus{}c^3\\minus{}3abc \\equal{} (a\\plus{}b\\plus{}c)(a^2\\plus{}b^2\\plus{}c^2\\minus{}ab\\minus{}bc\\minus{}ca)$ is useful here?", "Solution_6": "[quote=\"blahblahblah\"]Anyways, it appears that the correct problem is the 'obvious' correction;\n\\[ \\det\\left(\\begin{array}{ccc}2bc \\minus{} a^{2} & c^{2} & b^{2} \\\\\nc^{2} & 2ac \\minus{} b^{2} & a^{2} \\\\\nb^{2} & a^{2} & 2ab \\minus{} c^{2}\\end{array}\\right) \\equal{} (a^{3} \\plus{} b^{3} \\plus{} c^{3} \\minus{} 3abc)^2\n\\]\nPerhaps the identity $ a^3 \\plus{} b^3 \\plus{} c^3 \\minus{} 3abc \\equal{} (a \\plus{} b \\plus{} c)(a^2 \\plus{} b^2 \\plus{} c^2 \\minus{} ab \\minus{} bc \\minus{} ca)$ is useful here?[/quote]\r\n\r\nblahblahblah is right, I just have not found an elegant way yet to prove the result, just develop the determinant and identify the result as the square...", "Solution_7": "It is well-known that \r\n\r\n$ \\det \\left( \\begin{array}{ccc} a & b & c \\\\\r\nc & a & b \\\\\r\nb & c & a \\end{array} \\right) \\equal{} a^3 \\plus{} b^3 \\plus{} c^3 \\minus{} 3abc$\r\n\r\ndue to the properties of the eigenvalues of a circulant matrix. I'm reasonably certain there's a nice way to express the given matrix as a product of two matrices similar to the above. Alternately, it's a symmetric matrix, so it should have three real eigenvalues (for real $ a, b, c$), which might be something like\r\n\r\n$ (a \\plus{} b \\plus{} c)^2, (a^2 \\plus{} b^2 \\plus{} c^2 \\minus{} ab \\minus{} bc \\minus{} ca), (a^2 \\plus{} b^2 \\plus{} c^2 \\minus{} ab \\minus{} bc \\minus{} ca)$.\r\n\r\n(but this choice isn't consistent with the trace.) I'll try to see if either of these calculations work out.\r\n\r\n[b]Edit:[/b] Got it. The given matrix is equal to\r\n\r\n$ \\left( \\begin{array}{ccc} a & c & b \\\\\r\nb & a & c \\\\\r\nc & b & a \\end{array} \\right) \\left( \\begin{array}{ccc} \\minus{} a & \\minus{} b & \\minus{} c \\\\\r\nb & c & a \\\\\r\nc & a & b \\end{array} \\right).$\r\n\r\n[b]Edit #2:[/b] If it's not clear why the above works (in particular, why the sign is correct), multiply on the right by $ \\left( \\begin{array}{ccc} 0 & 0 & 1 \\\\\r\n0 & 1 & 0 \\\\\r\n1 & 0 & 0 \\end{array} \\right)$. Anyway, I'm wondering what the source of this problem is. The matrix wants to be a [url=http://en.wikipedia.org/wiki/Gramian_matrix#Gram_determinant]Gram matrix[/url], but isn't.", "Solution_8": "After some experimenting, this seems to be the \"right\" solution:\r\n\r\n$ \\left(\r\n\\begin{array}\r\n[c]{ccc}\r\n2bc\\minus{}a^{2} & c^{2} & b^{2}\\\\\r\nc^{2} & 2ca\\minus{}b^{2} & a^{2}\\\\\r\nb^{2} & a^{2} & 2ab\\minus{}c^{2}\r\n\\end{array}\r\n\\right) \\equal{}\\left(\r\n\\begin{array}\r\n[c]{ccc}\r\n1 & 0 & 0\\\\\r\n0 & 0 & 1\\\\\r\n0 & 1 & 0\r\n\\end{array}\r\n\\right) \\left(\r\n\\begin{array}\r\n[c]{ccc}\r\na & b & c\\\\\r\nc & a & b\\\\\r\nb & c & a\r\n\\end{array}\r\n\\right) \\left(\r\n\\begin{array}\r\n[c]{ccc}\r\n\\minus{}1 & 0 & 0\\\\\r\n0 & 1 & 0\\\\\r\n0 & 0 & 1\r\n\\end{array}\r\n\\right) \\left(\r\n\\begin{array}\r\n[c]{ccc}\r\na & b & c\\\\\r\nc & a & b\\\\\r\nb & c & a\r\n\\end{array}\r\n\\right)$,\r\n\r\nand\r\n\r\n$ \\det\\left(\r\n\\begin{array}\r\n[c]{ccc}\r\n1 & 0 & 0\\\\\r\n0 & 0 & 1\\\\\r\n0 & 1 & 0\r\n\\end{array}\r\n\\right) \\equal{}\\minus{}1$,\r\n\r\n$ \\det\\left(\r\n\\begin{array}\r\n[c]{ccc}\r\n\\minus{}1 & 0 & 0\\\\\r\n0 & 1 & 0\\\\\r\n0 & 0 & 1\r\n\\end{array}\r\n\\right) \\equal{}\\minus{}1$ and\r\n\r\n$ \\det\\left(\r\n\\begin{array}\r\n[c]{ccc}\r\na & b & c\\\\\r\nc & a & b\\\\\r\nb & c & a\r\n\\end{array}\r\n\\right) \\equal{}\\left( a\\plus{}b\\plus{}c\\right) \\left( a\\plus{}\\zeta b\\plus{}\\zeta^{2}c\\right) \\left(\r\na\\plus{}\\zeta^{2}b\\plus{}\\zeta c\\right)$ (by the circulant formula, where $ \\zeta$ is a primitive $ 3$-rd root of unity)\r\n$ \\equal{}a^{3}\\plus{}b^{3}\\plus{}c^{3}\\minus{}3abc$.\r\n\r\n darij" } { "Tag": [], "Problem": "From 6:00pm until midnight the temperature dropped $ 2^\\circ$ per hour. The temperature was $ 5^\\circ$ at 6:00pm. In degrees, what was the temperature at midnight?", "Solution_1": "There are $ 6$ hours from $ 6: 00$pm to midnight, so the temperature is $ 5\\minus{}2(6)\\equal{}\\boxed{\\minus{}7^\\circ}$." } { "Tag": [ "topology", "limit" ], "Problem": "Precisely three of the following ten collections of subsets of $ \\mathbb R$ are topologies. Identify these and justify your answer.\r\n\r\n(1) $ \\tau_1$ consists of $ \\mathbb R$, $ \\emptyset$, and every $ (a,b)$ for $ a,b\\in\\mathbb R$ with $ a < b$.\r\n\r\n(2) $ \\tau_2$ consists of $ \\mathbb R$, $ \\emptyset$, and every $ ( \\minus{} r,r)$ for $ r\\in\\mathbb R^ \\plus{}$.\r\n\r\n(3) $ \\tau_3$ consists of $ \\mathbb R$, $ \\emptyset$, and every $ ( \\minus{} r,r)$ for $ r\\in\\mathbb Q^ \\plus{}$.\r\n\r\n(4) $ \\tau_4$ consists of $ \\mathbb R$, $ \\emptyset$, and every $ [ \\minus{} r,r]$ for $ r\\in\\mathbb Q^ \\plus{}$.\r\n\r\n(5) $ \\tau_5$ consists of $ \\mathbb R$, $ \\emptyset$, and every $ ( \\minus{} r,r)$ for $ r\\in\\mathbb R^ \\plus{} \\backslash \\mathbb Q$.\r\n\r\n(6) $ \\tau_6$ consists of $ \\mathbb R$, $ \\emptyset$, and every $ [ \\minus{} r,r]$ for $ r\\in\\mathbb R^ \\plus{} \\backslash\\mathbb Q$.\r\n\r\n(7) $ \\tau_7$ consists of $ \\mathbb R$, $ \\emptyset$, and every $ [ \\minus{} r,r)$ for $ r\\in\\mathbb R^ \\plus{}$.\r\n\r\n(8) $ \\tau_8$ consists of $ \\mathbb R$, $ \\emptyset$, and every $ ( \\minus{} r,r]$ for $ r\\in\\mathbb R^ \\plus{}$.\r\n\r\n(9) $ \\tau_9$ consists of $ \\mathbb R$, $ \\emptyset$, every $ [ \\minus{} r,r]$, and every $ ( \\minus{} r,r)$ for $ r\\in\\mathbb R^ \\plus{}$.\r\n\r\n(10) $ \\tau_{10}$ consists of $ \\mathbb R$, $ \\emptyset$, every $ [ \\minus{} n,n]$, and every $ ( \\minus{} r,r)$ for $ n\\in\\mathbb Z^ \\plus{}$ and $ r\\in\\mathbb R^ \\plus{}$.\r\n\r\n[hide]For (1) - (6), I think only $ \\tau_2$ is a topology. But for (7) - (10) I am not sure.[/hide]", "Solution_1": "Isn't (1) the standard topology? Why don't you think that counts?", "Solution_2": "[quote=\"JRav\"]Isn't (1) the standard topology? Why don't you think that counts?[/quote]\r\n\r\n$ \\tau_1$ is a [b]base[/b] for the standard topology. But It's not a topology. The union of $ (1,2)$ and $ (3,4)$ is not contained...\r\n\r\nIn (1)-(6) only $ \\tau_2$ is a topology. $ \\tau_9$ and $ \\tau_{10}$ are topology too...\r\n\r\nIf $ \\tau_7$ is a topology $ \\tau_8$ must be a topology too. So I have simply to show that $ \\tau_7$ is not a topology... Consider an infinite series with limits $ \\lim_{n\\to\\infty} r_n \\equal{} l$ then the union of the sets in the form $ [r_n,r_n)$ is $ (l,l)$ that it's not in $ \\tau_7$.", "Solution_3": "[quote=\"vict85\"]Consider an infinite series with limits $ \\lim_{n\\to\\infty} r_n \\equal{} l$ then the union of the sets in the form $ [\\minus{}r_n,r_n)$ is $ (\\minus{}l,l)$ that it's not in $ \\tau_7$.[/quote]\r\nI also think that $ \\tau_7$ and $ \\tau_8$ are not topologies. But for your reason, could you please explain $ \\bigcup_{n\\equal{}1}^{\\infty} [\\minus{}r_n,r_n)\\equal{}(\\minus{}l,l)$ in detail?", "Solution_4": "[quote=\"ifai\"][quote=\"vict85\"]Consider an infinite series with limits $ \\lim_{n\\to\\infty} r_n \\equal{} l$ then the union of the sets in the form $ [ \\minus{} r_n,r_n)$ is $ ( \\minus{} l,l)$ that it's not in $ \\tau_7$.[/quote]\nI also think that $ \\tau_7$ and $ \\tau_8$ are not topologies. But for your reason, could you please explain $ \\bigcup_{n \\equal{} 1}^{\\infty} [ \\minus{} r_n,r_n) \\equal{} ( \\minus{} l,l)$ in detail?[/quote]\r\n\r\nOk, I forgot to write that $ r_n < l$ for any $ n$ (it's a necessary condiction). With this condiction $ \\bigcup_{n \\equal{} 1}^{\\infty} [ \\minus{} r_n,r_n) \\subseteq [ \\minus{} l,l)$.\r\n\r\nThe points $ \\minus{} l$ and $ l$ are not in any of the $ [ \\minus{} r_n,r_n)$ and so $ \\bigcup_{n \\equal{} 1}^{\\infty} [ \\minus{} r_n,r_n) \\subseteq ( \\minus{} l,l) \\subset [ \\minus{} l,l)$. Now consider a real number $ \\epsilon > 0$, then you know that $ \\bigcup_{n \\equal{} 1}^{\\infty} [ \\minus{} r_n,r_n) \\supseteq [ \\minus{} l \\plus{} \\epsilon,l \\minus{} \\epsilon)$ for any $ \\epsilon$ and so $ \\bigcup_{n \\equal{} 1}^{\\infty} [ \\minus{} r_n,r_n) \\equal{} ( \\minus{} l,l)$\r\n\r\n\r\nP.S: the sequences $ r_n \\equal{} r \\minus{} \\frac {1}{n}$ and $ r_n \\equal{} r \\minus{} \\frac {1}{2^n}$ are common choices...", "Solution_5": "Oh, I see. Thank you very much, vict85. Originally I was confused since $ \\lim_{n\\to\\infty} r_n\\equal{}l$ even if $ r_n\\notin l$ for all $ n$. Hence I thought $ \\minus{}l$ should be in $ \\bigcup_{n\\equal{}1}^{\\infty}[\\minus{}r_n,r_n)$.\r\n\r\nNow I know I was wrong. Note that $ \\bigcup_{S\\in\\mathcal F} S\\equal{}\\{x: x\\in S \\text{ for some } S\\in\\mathcal F\\}$. But $ \\minus{}l\\notin [\\minus{}r_n,r_n)$ for all $ n$, thus we have $ \\minus{}l\\notin\\bigcup_{n\\equal{}1}^{\\infty} [\\minus{}r_n,r_n)$.", "Solution_6": "I agree that testing topologies can be tricky (as you saw, I forgot the definition above), and I got an A in my topology course! Perhaps I'm rusty... :)", "Solution_7": "[quote=\"JRav\"]I agree that testing topologies can be tricky (as you saw, I forgot the definition above), and I got an A in my topology course! Perhaps I'm rusty... :)[/quote]\r\n\r\nWell, I simply gave more attention... At a first look it's normal to see in the first one the standard topology...", "Solution_8": "Also there are some books which define topological space as what in other books is the fundamental neighborhoods system, and the more common definition using open sets is regarded as an alternative definition which does not alters the subject of study (together with the other alternatives definitions using closed sets, closure operation, etc...). \r\n\r\nSo I guess it is kind of a justified \"mistake\"...\r\n\r\n:)" } { "Tag": [ "number theory", "relatively prime" ], "Problem": "Prove that if p is a prime then\r\n(p-1)!\u2261p-1(mod p(p-1)/2)", "Solution_1": "Hi,\n\n\n\n[hide]\n\nUsing the fact that p is relatively prime to (p-1)/2 and Wilson's theorem:\n\n\n\np | (p-1)!-(p-1)\n\n(1/2)(p-1) | (p-1)!-(p-1)\n\n\n\nimplies\n\n\n\n(1/2)p(p-1) | (p-1)!-(p-1)\n\n[/hide]", "Solution_2": "I don't understand something in your solution.\r\nWhy (1/2)(p-1) | (p-1)!-(p-1).\r\nMaybe it's trivial, but...", "Solution_3": "(1/2)(p-1) * 2[(p-2)!-1] = (p-1)! - (p-1)" } { "Tag": [], "Problem": "How many $ (x,y)$ solution in positive integers of the given equation :\r\n$ \\sqrt{xy}\\minus{}71\\sqrt{x}\\plus{}30\\equal{}0$", "Solution_1": "[hide=\"is there something wrong with this?\"]\n$ \\sqrt{xy}\\minus{}71\\sqrt{x}\\plus{}30\\equal{}0 \\rightarrow \\sqrt{x}*(\\sqrt{y}\\minus{}71)\\equal{}\\minus{}30$.\nThis suggests that we should pair up factors of $ 30$.\n$ \\sqrt{x}>0 \\rightarrow \\sqrt{x}\\equal{}1,2,3,5,6,10,15,30$.\nThis suggests there are 8 solutions.\n\n\n[/hide]", "Solution_2": "does $ \\sqrt{x}$ have to be integer", "Solution_3": "Wouldn't it have to be?\r\n\r\nAssume that $ \\sqrt{x}$ is not an integer, which suggests it is irrational. Then we still have $ \\sqrt{x}(\\sqrt {y} \\minus{}71)\\equal{}\\minus{}30$. $ \\sqrt y\\equal{}\\frac{\\minus{}30}{\\sqrt x}\\plus{}71$, we will get a rational and irrational portion. But y is an integer, contradiction.", "Solution_4": "[quote=\"azn6021023\"]Wouldn't it have to be?\n\nAssume that $ \\sqrt {x}$ is not an integer, which suggests it is irrational. Then we still have $ \\sqrt {x}(\\sqrt {y} \\minus{} 71) \\equal{} \\minus{} 30$. $ \\sqrt y \\equal{} \\frac { \\minus{} 30}{\\sqrt x} \\plus{} 71$, we will get a rational and irrational portion. But y is an integer, contradiction.[/quote]\r\nIs $ \\frac { \\minus{} 30}{\\sqrt x} \\plus{} 71$ an integer if $ \\sqrt x$ is irrational?" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Prove that $n$ is prime if and only if:\r\n$\\phi(n) + \\sigma(n) = n \\cdot d(n)$,\r\nwhere $\\phi(n)$ is the number of numbers $m\\leq n$ : $gcd(m,n) = 1$, $\\sigma(n)$ is the sum of the divisors of $n$, \r\n$d(n)$ is the number of divisors of $n$.", "Solution_1": "If $n$ is prime then $\\phi(n) + \\sigma(n) = n \\cdot d(n)$ takes a trivial form $(n-1)+(n+1)=n\\cdot 2$.\r\n\r\nSuppose that $\\phi(n) + \\sigma(n) = n \\cdot d(n)$. Then $\\phi(n)\\leq n-1$ and \r\n\\[ \\phi(n) = n d(n) - \\sigma(n) = n \\sum_{q\\mid n} 1 - \\sum_{q\\mid n} q = \\sum_{q\\mid n} (n - q) \\geq n-1. \\]\r\nTherefore, $\\phi(n)=n-1$ meanning that $n$ is prime." } { "Tag": [], "Problem": "For a positive integer $ n$, find \r\n\r\n$ \\binom{2n}{0}^2\\plus{}\\binom{2n}{2}^2\\plus{}\\binom{2n}{4}^2\\plus{}...\\plus{}\\binom{2n}{2n}^2$", "Solution_1": "Nobody? ...", "Solution_2": "Upon introspection, the same sum without the squared terms seems to be $ 2^{2n \\minus{} 1}$. With the squared terms, it goes like 2, 38, 452, ...", "Solution_3": "[quote=\"Yongyi781\"]Upon introspection, [/quote] You looked deep into your soul, and discovered a classic problem on binomial series? \r\n\r\nThe answer is [url=http://www.research.att.com/~njas/sequences/A098772]A098772[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url], with a nice simple closed form, but I'm still looking for the combinatorial proof.", "Solution_4": "A related sum:\r\n$ \\sum_{k\\equal{}0}^{n} \\ {n\\choose k}^{2}\\equal{}{2n\\choose n}$\r\nThis one has a very easy combinatorial proof :wink:", "Solution_5": "Well, I think there is a combinatorial proof to be had, but I don't really want do the work necessary to see if the outline of an idea I have can be turned into an actual proof. So, instead, I give you a generating functional proof:\r\n\r\n[hide]$ 2\\sum_{k \\equal{} 0}^n \\binom{2n}{2k} x^{2k} \\equal{} (1 \\plus{} x)^{2n} \\plus{} (1 \\minus{} x)^{2n}$. The expression we want is $ \\sum_{k \\equal{} 0}^n \\binom{2n}{2k} \\binom{2n}{2n \\minus{} 2k}$, and this is the coefficient of $ x^{2n}$ in the product $ \\left(\\sum_{k \\equal{} 0}^n \\binom{2n}{2k} x^{2k}\\right)^2$. So we want one quarter of the coefficient of $ x^{2n}$ in $ \\left((1 \\plus{} x)^{2n} \\plus{} (1 \\minus{} x)^{2n}\\right)^2 \\equal{} (1 \\plus{} x)^{4n} \\plus{} (1 \\minus{} x)^{4n} \\plus{} 2(1 \\minus{} x^2)^{2n}$, and this is $ \\frac{1}{4}\\left(\\binom{4n}{2n} \\plus{} \\binom{4n}{2n} \\plus{} 2(\\minus{}1)^n \\binom{2n}{n}\\right) \\equal{} \\frac{\\binom{4n}{2n} \\plus{} (\\minus{}1)^n \\binom{2n}{n}}{2}$.[/hide]", "Solution_6": "Consider all sequences of $ n$ $ A$'s and $ n$ $ B$'s. Obviously there are $ \\binom{2n}{n}$ ways. Now lets count this differently. Suppose there are $ k$ $ A$'s in the first half of the sequence. Then there are $ \\binom{n}{k}$ ways for the first half. The the other half has $ n \\minus{} k$ $ A$'s, and $ k$ $ B$'s so there are $ \\binom{n}{k}$ ways for that two. Summing from $ k \\equal{} 0$ to $ k \\equal{} n$ establishes the identity posted by [polskimisiek]." } { "Tag": [ "AMC 10", "AMC" ], "Problem": "The newly published book:\r\nFirst Steps for Math Olympians: Using the American Mathematics Competitions, by J. Dougles Faires \r\nis now available. You can purchase it through the MAA on-line bookstore (http://www.maa.org, then choose Bookstore) or by calling 1-800-331-1622. The catalog code is PSC/SF06, the ISBN is 0-88385-824-X.\r\n\r\nYou can also find it listed in the Art of Problem Solving Bookstore, and at the major web bookstores.\r\n \r\nIt is 320 pages and covers mostly problem solving techniques for the AMC 10, some for the AMC 12. Doug Faires has been the AMC 10 SubCommittee Chair since the inception of the AMC 10. Faires is a very talented writer and teacher. This would be an excellent book to use to prepare for the AMC 10, and the AMC 12.\r\n\r\nSteve Dunbar\r\nAMC Director", "Solution_1": "[url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=52]Click here for information about the book on the AoPS site[/url]." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,b,c,d$ are non-negative real numbers. Prove that\r\n\\[ a^{4}+b^{4}+c^{4}+d^{4}+8abcd \\ge \\sum_{cyclic}abc(a+b+c)+\\frac{1}{6}\\sum_{cyclic}ab(a-b)^{2}\\]", "Solution_1": "We have\r\n\r\n$ a^{4}+b^{4}+c^{4}+d^{4}+8abcd\\geq \\sum abc(a+b+c))+2/3\\sum(ab(a-b)^{2})$ which is equivalent to $ \\frac13g_{1}+\\frac16g_{4}\\geq0$ where\r\n\r\n\\[ g_{1}: =a^{2}(a-b)(a-c)+a^{2}(a-b)(a-d)+a^{2}(a-c)(a-d)+b^{2}(b-c)(b-d)+b^{2}(b-a)(b-c)+b^{2}(b-a)(b-d)+c^{2}(c-b)(c-d)+c^{2}(c-a)(c-d)+c^{2}(c-a)(c-b)+d^{2}(d-b)(d-c)+d^{2}(d-a)(d-c)+d^{2}(d-a)(d-b)\\geq 0;\\]\r\nand\r\n\\[ g_{4}: =(b+c)(b+c-2d)(a-b)(a-c)+(b+d)(b+d-2c)(a-b)(a-d)+(d+c)(c+d-2b)(a-c)(a-d)+(a+c)(a+c-2d)(b-a)(b-c)+(a+d)(a+d-2c)(b-a)(b-d)+(d+c)(c+d-2a)(b-c)(b-d)+(a+b)(a+b-2d)(c-a)(c-b)+(a+d)(a+d-2b)(c-a)(c-d)+(b+d)(b+d-2a)(c-b)(c-d)+(a+b)(a+b-2c)(d-a)(d-b)+(a+c)(a+c-2b)(d-a)(d-c)+(b+c)(b+c-2a)(d-b)(d-c)\\geq0.\\]", "Solution_2": "Nice! Thank you, fjwxcsl!", "Solution_3": "[quote=\"fjwxcsl\"]We have\n\n$ a^{4}+b^{4}+c^{4}+d^{4}+8abcd\\geq \\sum(abc(a+b+c))+\\frac{2}{3}\\sum(ab(a-b)^{2})$\n$ \\iff \\frac{1}{3}g_{1}+\\frac{1}{6}g_{4}\\geq 0$\n\nin which\n\n$ g_{1}: =a^{2}(a-b)(a-c)+a^{2}(a-b)(a-d)+a^{2}(a-c)(a-d)$\n$ +b^{2}(b-c)(b-d)+b^{2}(b-a)(b-c)+b^{2}(b-a)(b-d)$\n$ +c^{2}(c-b)(c-d)+c^{2}(c-a)(c-d)+c^{2}(c-a)(c-b)$\n$ +d^{2}(d-b)(d-c)+d^{2}(d-a)(d-c)+d^{2}(d-a)(d-b)\\geq 0$\n\n$ g_{4}: =(b+c)(b+c-2d)(a-b)(a-c)+(b+d)(b+d-2c)(a-b)(a-d)+(d+c)(c+d-2b)(a-c)(a-d)$\n$ +(a+c)(a+c-2d)(b-a)(b-c)+(a+d)(a+d-2c)(b-a)(b-d)$\n$ +(d+c)(c+d-2a)(b-c)(b-d)+(a+b)(a+b-2d)(c-a)(c-b)$\n$ +(a+d)(a+d-2b)(c-a)(c-d)+(b+d)(b+d-2a)(c-b)(c-d)$\n$ +(a+b)(a+b-2c)(d-a)(d-b)+(a+c)(a+c-2b)(d-a)(d-c)$\n$ +(b+c)(b+c-2a)(d-b)(d-c)\\geq 0$.[/quote]" } { "Tag": [ "integration" ], "Problem": "Se aleg trei numere la intamplare din intervaul (o,1). Care este probabilitatea ca diferenta dintre cel mai mare dintre ele si cel mai mic sa fie mai mica decat 1/3? \r\n\r\nProblema este propusa [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=614198#614198]aici[/url]. Eu am postat acolo o rezolvare geometrica, m-ar interesa daca stie cineva si alta rezolvare.", "Solution_1": "am incercat eu sa fac ceva aici, nu stiu cat de bine e, macar rezultatu e la fel :)\r\n[hide=\"solutie\"]fie numerele $a,b,c$. \nil aleg pe $c$ sa fie cel din mijloc.\napoi, pentru $a$, ma uit in ce interval trebuie sa fie $b$ :\npt $a\\in[0,1/3]$, rezulta ca $b\\in[0,a+1/3]$. (i)\npt $a\\in[1/3,2/3]$, rezulta ca $b\\in[a-1/3,a+1/3]$. (ii)\npt $a\\in[2/3,1]$, rezulta ca $b\\in[a-1/3,1]$. (iii)\nevident, $c\\in[a,b]$ sau $c\\in[b,a]$.\npentru cazul (i) :\npentru un $a$ fix, probabilitatea ca $c$ sa fie intre $a$ si $b$ este :\n$\\int_{0}^{a+1/3}|x-a|\\,dx=\\int_{0}^{a}(a-x)\\,dx+\\int_{a}^{a+1/3}(x-a)dx=\\frac{a^{2}}{2}+\\frac{1}{18}$.\nacum integrez si dupa $a$ :\n$\\int_{0}^{1/3}\\left(\\frac{a^{2}}{2}+\\frac{1}{18}\\right)\\,da=\\frac{1}{27\\cdot 6}+\\frac{1}{27\\cdot 2}=\\frac{2}{27\\cdot 3}$.\ndeci probabilitatea cazului (i), cu conditia din ipoteza si presupunerile facute, este $\\frac{2}{27\\cdot 3}$.\ncazul (iii) este analog, deoarece este \"simetric\". (se obtine aceeasi probabilitate).\npentru cazul (ii) :\nil fixez pe $a$. probabilitatea ca $c$ sa fie intre $a$ si $b$ este :\n$\\int_{a-1/3}^{a+1/3}|x-a|\\,dx=2\\int_{a}^{a+1/3}(x-a)\\, dx=\\frac{1}{9}$.\napoi integrez dupa $a$ :\n$\\int_{1/3}^{2/3}\\frac{1}{18}\\, da=\\frac{1}{27}$.\ndeoarece evenimentele (i), (ii), (iii) sunt independente (nu se poate simultan $a\\in[0,1/3]$ si $a\\in[1/3,2/3]$), voi aduna probabilitatile, si voi obtine rezultatul :\n$p_{c\\ mijloc}=\\frac{2}{27\\cdot 3}+\\frac{1}{27}+\\frac{2}{27\\cdot 3}=\\frac{7}{27\\cdot 3}$.\nam scris indicele \"c mijloc\" deoarece asa am presupus de la inceput. cum combinatiile cu $c$ in \"mijloc\" sunt $1/3$ din total, rezulta ca probabilitatea dorita este chiar :\n$p=3p_{c\\ mijloc}=\\frac{7}{27}$.[/hide]", "Solution_2": "Great! :) \r\n\r\nMi se pare ok rezolvarea ta si cred ca, spre deosebire de mea, poate fi generalizata pentru n numere (vad ca a ridicat cineva intre timp problema asta pe topicul la care am dat link mai sus).\r\n\r\nImi mirosea si mie a integrale, da' nu stiam cum sa le aranjez.", "Solution_3": "solutia ta are avantajul elegantei. felicitari :)\r\nacum ma gandesc la generalizare, sper sa nu fie prea mult de caculat :)\r\n\r\nedit :\r\n\r\n[hide=\"generalizare\"]procedez exact la fel :\n$a$ cel mai mare, $b$ cel mai mic, restul sunt $c_{1},c_{2},\\ldots,c_{n-2}$.\nam trei cazuri : \n$a\\in\\left[0,\\frac{1}{n}\\right]$.\n$a\\in\\left[\\frac{1}{n},\\frac{n-1}{n}\\right]$.\n$a\\in\\left[\\frac{n-1}{n},1\\right]$.\nde aici analog ca pentru cazul $n=3$.\nbla bla am gresit la calcule :)[/hide]", "Solution_4": "E o greseala pe undeva, fiindca nu se potriveste raspunsul pentru n=3 :huh: (nici pentru n=2, de altfel).\r\n\r\nNu stiu ce functie ai integrat tu.\r\n\r\nPentru $c$ si $d$ fixate, probabilitatea ca alte $n-2$ numere sa fie intre $c$ si $d$ este $|c-d|^{n-2}$ (sunt $n-2$ evenimente independente cu aceeasi probabilitate), deci cred ca trebuie integrata initial functia $|x-a|^{n-2}$, in loc de $|x-a|$.\r\n\r\nProcedand exact cum ai procedat tu in cazul $n=3$, dar cu functia asta in loc de cea veche, mi-a dat in final:\r\n\r\n$p_{n}=\\frac{n^{2}-n+1}{n^{n}}$\r\n\r\nPentru $n=2$ obtinem $3/4$, iar pentru $n=3$ obtinem $7/27$, adica exact cat trebuia sa dea.", "Solution_5": "da, si eu cred ca am gresit la calcule :) \r\neu sunt mai neatent de obicei, eh..macar am gasit ideea, dar n`am avut rabdare sa le fac cum trebe.\r\nte cred pe cuvant ca ai dreptate :)" } { "Tag": [], "Problem": "Prove that the fraction $ \\dfrac{21n \\plus{} 4}{14n \\plus{} 3}$ is irreducible for every natural number $ n$.", "Solution_1": "Let $k$ be a common factor, greater than zero, of both the numerator and the denominator. Let\r\n\\[ 21n+4=Ak \\]\r\nand\r\n\\[ 13n+3=Bk \\]\r\nThen\r\n\\[ k(3B-2A)=3Bk-2Ak=(42n+9)-(42n+8)=1 \\]\r\nThe first expression is divisible by $k$, so $k$ must be a divisor of 1. Thus the greatest common factor of the numerator and the denominator is one, and the fraction is not reducible.", "Solution_2": "Using Euclid's algorithm for determining the greatest common divisor of the two numbers we obtain\r\n\r\n\\[ 21n+4=1 \\cdot (14n+3)+(7n+1) \\]\r\n\\[ 14n+3=2 \\cdot (7n+1)+1 \\]\r\n\\[ 7n+1=1 \\cdot (2n+1)+0 \\]\r\n\r\nThe greatest common divisor of the numerator and denominator is one. Therefore the given fraction is irreducible.", "Solution_3": "Easiest IMO prob. ever. :D \r\nWe have 2(21n+4)-3(14n+3)=-1, o we are done...\r\n\r\nBomb", "Solution_4": "[quote=\"bomb\"]Easiest IMO prob. ever. :D \n[/quote]Well it was the [b]first[/b] imo problem .... [i]ever[/i] :)", "Solution_5": "It was :) I never realised..\r\n\r\nBomb", "Solution_6": "Of course it seems easy to you since everyone knows how to use Euclid's algorithm, but 49 years ago it was a new method, probably.", "Solution_7": "Probably, it was new 2300 years ago :wink:", "Solution_8": "I mean that an ordinary kid didn't learn this method at school. I'm not ignorant - I know that Euclid lived in Ancient Greece. :lol:", "Solution_9": "[quote=\"apiarist1\"]I mean that an ordinary kid didn't learn this method at school. I'm not ignorant - I know that Euclid lived in Ancient Greece. :lol:[/quote]\r\n\r\n1) Ordinary kids don't go to IMO's.\r\n2) 49 years ago people DID learn the Euclidean algorithm at school; it's rather nowadays that they don't. You will be amazed at what people learnt at school 49 years ago; the textbooks contained things that are meanwhile forgotten enough that you can pose them as IMO problems...\r\n\r\n dg", "Solution_10": "[quote=\"darij grinberg\"]49 years ago people DID learn the Euclidean algorithm at school; it's rather nowadays that they don't. You will be amazed at what people learnt at school 49 years ago; the textbooks contained things that are meanwhile forgotten enough that you can pose them as IMO problems...[/quote]\r\n\r\nCould you provide some examples for such IMO problems?", "Solution_11": "Hi, this is my first ever post here so please forgive me for being a little stupid (perhaps), but all I did was:\r\n\r\n$ \\dfrac{21n \\plus{} 4}{14n \\plus{} 3} \\equal{} \\dfrac{14n \\plus{} 3 \\plus{} 7n \\plus{} 1}{14n \\plus{} 3} \\equal{} 1 \\plus{} \\dfrac{7n \\plus{} 1}{14n \\plus{} 3} \\equal{} 1 \\plus{} \\dfrac{7n \\plus{} 1}{2(7n \\plus{} 1) \\plus{} 1}$\r\n\r\nand so we see that the part by which the fraction exceeds 1 can never be reduced because the numerator does not divide the denominator.\r\n\r\nEDIT: Sorry, what I mean is that nothing divides both the numerator and the denominator because of the extra \"1\" hanging around.", "Solution_12": "I'm not really sure if this is a solution. I'm not too smart so I might be wrong. What I did was assume that the fraction is reducible. Thus\r\n\r\n(21n+4)/(14n+3)=k for some positive integer k.\r\n\r\nThus 21n+4=14nk+3k\r\n 7n(3-2k)=3k-4\r\n 7n =(3k-4)/(3-2k)\r\nSince n is a natural number (3k-4)/(3-2k) must be a positive integer for some k. Unfortunately, (3k-4)/(3-2k)>0 only in the interval (4/3,3/2). We see that for (3k-4)/(3-2k) to be an integer k must be between 3/2 and 4/3. But then, k isn't an integer and we reach a contradiction. Thus (21n+4)/(14n+3) is irreducible.\r\nI apologize for the lack of LaTeX in advance. I still have to learn.", "Solution_13": "[quote=\"darij grinberg\"][quote=\"apiarist1\"]I mean that an ordinary kid didn't learn this method at school. I'm not ignorant - I know that Euclid lived in Ancient Greece. :lol:[/quote]\n\n1) Ordinary kids don't go to IMO's.\n2) 49 years ago people DID learn the Euclidean algorithm at school; it's rather nowadays that they don't. You will be amazed at what people learnt at school 49 years ago; the textbooks contained things that are meanwhile forgotten enough that you can pose them as IMO problems...\n\n dg[/quote]\n\n\nPeople do realize this was in the soviet Union, america didn't enter till 1970's, with Paul Zeitz being first IMO member for the US (author of \"Arts and Crafts of Problem Solving\").", "Solution_14": "[moderator says: do not quote the entire post before yours]\n\n... and your point is?__________________________", "Solution_15": "........", "Solution_16": "[b]Solution 1.[/b] Note that\n\n\\[3(14n+3)-2(21n+4)=1,\\]\n\nso by B\u00e9zout's lemma, we have $\\gcd(21n+4,14n+3)=1,$ meaning that $21n+4$ and $14n+3$ are coprime, so the fraction $\\frac{21n + 4}{14n + 3}$ is irreducible.\n\n[b]Solution 2.[/b] Repeated applications of the Euclidean algorithm yield,\n\n$\\gcd(21n+4,14n+3)=\\gcd(14n+3,7n+1)=\\gcd(7n+2,7n+1)=\\gcd(7n+1,1)=1,$\n\nand thus the fraction $\\frac{21n + 4}{14n + 3}$ is irreducible.", "Solution_17": "It's not like the solutions here are actually different from one another, so I might as well post mine.\n\nObserve that$$3(14n+3)-2(21n+4)=1,$$so by B\u00e9zout's Identity we have $\\gcd (21n+4,14n+3)=1$ (alternatively, any common divisor of $21n+4$ and $14n+3$ divides any integer linear combination of them, so it divides $1$), hence $\\frac{21n+4}{14n+3}$ is irreducible.", "Solution_18": "First Post and first IMO problem solved by me\n[hide=Solution] Observe that when we have a fraction $\\frac{m}{f}$, if the greatest common divisor of $m$ and $f$ is 1 then the fraction can't be reduced.\n$$gcd(14n+3, 21n+4)=gcd(7n+1, 14n+3)=gcd(1, 7n+1)= 1$$\nBy the Euclidean Algorithm $\\frac{21n+4}{14n+3}$ is irreducible for every natural number $n$.\n\n[b]Remark[/b]: Notice how most of the solutions to this problem is very similar.\n[/hide]\n", "Solution_19": "Because $-2(21n+4)+3(14n+3)=1$, $21n+1$ and $14n+3$ are prime (bezout)\nSo the fraction is not reducible", "Solution_20": "This is so easy. We need to use Euclid's algorithm. So $\\gcd(21n+4,14n+3)=\\gcd(7n+1,14n+3)=\\gcd(7n+1,7n+2)=1.$ So since the gcd is 1 the fraction is irreducible.", "Solution_21": "[hide=solution] Note that $$-2(21n+4) + 3(14n+3)=1,$$ so by Bezout's we have $$\\gcd(21n+4, 14n+3)=1,$$ as desired. $\\square$[/hide]", "Solution_22": "[hide=solution]Note that the fraction is irreducible if and only if gcd(21n+4,14n+3) = 1. Using the Euclidean algorithm, we get gcd(21n+4,14n+3) = gcd(7n+1,14n+3) = gcd(7n+1,1) which is obviously 1. Thus, the fraction is irreducible over all natural numbers n.[/hide]", "Solution_23": "oh look its the imo problem of all time\n\nEuclidean Algorithm suffices. $\\blacksquare$", "Solution_24": "[quote=apiarist1]I mean that an ordinary kid didn't learn this method at school. I'm not ignorant - I know that Euclid lived in Ancient Greece. :lol:[/quote]\n\nNo one learns this at school nowdays though", "Solution_25": "By Euclidean Algorithm, $$\\text{gcd}(21n + 4, 14n + 3) = \\text{gcd}(14n + 3, 7n + 1) = \\text{gcd}(7n + 1, 1) = 1.$$ Hence, $ \\dfrac{21n + 4}{14n + 3}$ is irreducible for every natural number $ n$.", "Solution_26": "Simple one:\n\n[hide]\nhttps://youtu.be/hORB1K2F1sY[/hide]", "Solution_27": "By Euclidean division Algorithm,\n\n$$gcd(21n+4,14n+3) = gcd(7n+1,14n+3) = gcd(7n+1,7n+2) = gcd(7n+1,1) = 1.$$\n\n Hence, $\\frac{21n+4}{14n+3}$ is irreducible for natural ${n}$.", "Solution_28": "Assuming that the fraction $\\frac{21n+4}{14n+3}$ is reducible for some $n$, we get that $(21n+4,14n+3)\\neq1$. $\\Rightarrow \\exists$ $d\\neq1 \\in \\mathbb{N}$ such that $d \\vert 21n+4$, $d\\vert 14n+3$.\n\nNow, $d\\vert21n+4$ $\\Rightarrow$ $d\\vert2(21n+4)$ $\\Rightarrow d\\vert 42n+8$.\n\nAlso, $d\\vert14n+3 \\Rightarrow d\\vert3(14n+3) \\Rightarrow d\\vert42n+9$\n\n$\\Rightarrow d\\vert(42n+9)-(42n+8) \\Rightarrow d\\vert1$\n\nAs $d \\in \\mathbb{N}$, $d=1$, which implies that $21n+4$ and $14n+3$ have no common factor other than 1.\n$(21n+4,14n+3)=1$, which contradicts our assumption. $(\\Rightarrow\\Leftarrow)$. Proved.", "Solution_29": "you can just Euclidean Algorithm this\n\nalternatively\n\nLet $k = 7n + \\frac{3}{2}$.\n\nThis transforms our fraction into $\\frac{3k - \\frac{1}{2}}{2k}$ and since $2k \\nmid 3k - \\frac{1}{2}$, this fraction is irreducible. $\\blacksquare$" } { "Tag": [], "Problem": "I just wanted to point out something.In the next few years, a majority 49% of students passing out of institutes of higher education in India will be underserving candidates.The fact is that 49% of the seats are reserved for students of some \"backward classes\". This decision is based on a data of 1930.Wow! That's called rubbing salt on an already-wounded India. This is politically motivated(that's what I think).", "Solution_1": "You think? Obviously there isn't a single advantage to the system, and the sole reason is to gain votes.\r\n\r\n1. Non-SC/ST/OBC (from hereon, SSO) students will have a harder time gaining access to these prestigious universities. When they do make it there, there won't be as much synergy. Unlike American affirmative action, where the goal is to bring variform viewpoints and skills to the table, a non-ST/ST/OBC student will probably have little to gain from someone who is allowed in because of class-status.\r\n\r\n2. SSO students might not be prepared for the education ahead of them. There is a good reason why non-SSO students have to grind it out to get in. They might also be the targets of discrimination among their peers, or otherwise, just might not fit in. Finally, employers will know that they got in because of class-status.\r\n\r\nAlso consider that the reservation allows many students to get in because of their ancestry, even though they themselves might not necessarily be disadvantaged.\r\n\r\n3. The institution loses out because of a diminished student body caliber.\r\n\r\n\r\nAnd wait... that's not all... what's this I hear about a law to institute reservations among the [b]faculty [/b]in IIT's? Talk about shooting their institutions in the foot! I was planning to study abroad at IIT Kanpur for a semester... obviously I'm not going to do that anymore.", "Solution_2": "Same thinking here.\r\n\r\nPoliticians never understand what IIT is for general indian students.\r\nThe dream for which students are more than willing to sacrifice 2 years.\r\n\r\nthey are diluting standards of IITs just for votes thats wrong.\r\n\r\nReservatin is not going to help the deserving people.Most of students who'll get admission in good streams(eg,CSE,IIT mumbai,etc)will be students who would have got ranks below the requirement but they'll be selected just because of the fact that they were lucky to be born in OBC family.\r\n\r\nAlso the definition of creamy layer has gone up to Rs.450,000.(do you think people with family incomes like Rs.400,000 deserve to get reservations?????)\r\n\r\nabout standard there is not any second thought,STANDARD WILL FALL DOWN.\r\nEarlier good IITs had students of high caliber,it was something that made IIT famous(even today infrastructure,etc of IITs is not that good as many institutions),one can jus assume that IIT studen will be certainly good(it proved true in majority of cases)\r\nbut now before deciding about caliber of student beforehand,you'll have to ask his caste(it divides the society) \r\n\r\n@TZF, you seem to have good idea about India as well as these things", "Solution_3": "That is why,the craze for IIT is now a little less. :( \r\n\r\n\r\n@TZF: You are a very motivating person,with very good thoughts.Kudos to you man :)", "Solution_4": "another discussion!! :) \r\n\r\nwell i think u can go to the other thread--dont remember the name...", "Solution_5": "[quote=\"TZF\"] I was planning to study abroad at IIT Kanpur for a semester... obviously I'm not going to do that anymore.[/quote]\r\nI'd advise you not to be hasty :maybe: Even though there is some reservation among students, there are still bright students in IIT's and after all, everything depends upon you and the type of people you hang with wherever you study. And i can ascertain you that there are still bright and focussed people in the IIT's (not to mention that [url=http://indiaedunews.net/IIT/IIT-JEE_topper_bags_gold_at_International_Physics_Olympiad_5413/]people like this[/url] are at IITK too :P)", "Solution_6": "but these are only exceptions to the pool of applicant's out of which some who just get in IIT but don't deserve to get in(This is basically about the reservation system//////)\r\nAccording to me,a lot of politics is engulfing IIT(not just IITK..............)which should be stopped immediately", "Solution_7": "[quote=\"TZF\"]\nI was planning to study abroad at IIT Kanpur for a semester... obviously I'm not going to do that anymore.[/quote]\r\n\r\nDon't even think about it. It is a breeding ground for hindu extremist policies. Some professors there are core hindu extremists, who hate islam and muslims., and are recruiting hindu students into their fray.\r\n\r\nThis is a true story and has appeared in many major newspapers." } { "Tag": [ "induction", "inequalities", "combinatorics unsolved", "combinatorics" ], "Problem": "In a country there are several cities; some of these cities are connected by airlines, so that an airline connects exactly two cities in each case and both flight directions are possible. Each airline belongs to one of $k$ flight companies; two airlines of the same flight company have always a common final point. Show that one can partition all cities in $k+2$ groups in such a way that two cities from exactly the same group are never connected by an airline with each other.", "Solution_1": "Observe that airlines of each company compose a triangle or a star. (Triangle is a graph $(\\{u,v,w\\},\\{uv,vw,wu\\})$ and star is a graph $(\\{c,u_1,\\ldots,u_l\\},\\{cu_1,\\ldots,cu_l\\})$).\r\n\r\nInduction by k.\r\n\r\nThe statement of the problem is obvious for $k=1$.\r\n\r\n[b]Inductive step.[/b]If for some company airlines compose a star then delete the center vertex $u$ of the star (with all incident edges). Then for the rest cities we may apply inductive conjecture and put $u$ in new group. We are done in this case.\r\n\r\nNow consider a case when all companies are triangles. Consider any company composing a triangle $uvw$. For the rest cities we may apply inductive conjecture and put them into $k+1$ groups in a proper way. And now we have to put two vertices from $\\{u,v,w\\}$ into some old groups. Suppose we can't do it. It means that there exists $x,y \\in \\{u,v,w\\}$ such that there are edges from $x$ to each group and from $y$ to each group. Totally $2(k+1)$ edges incident to $u$, $v$ or $w$. But we have $k-1$ companies exept $uvw$. Each company is a triangle, and it is easy to see that no company can have all 3 edges incident to $u$, $v$ or $w$. There are at most two such edges. Totally we may have $2(k-1)$ edges incident to $u$, $v$ or $w$. Since $2(k+1)>2(k-1)$, we have a contradiction. So, we can put some two vertices from $\\{u,v,w\\}$ to old groups. The rest vertex we will put into new group. In this case we are also done.\r\n\r\nThe problem is solved.", "Solution_2": "[quote=\"Remike\"] it is easy to see that no company can have all 3 edges incident to $u$, $v$ or $w$. [/quote]\r\n\r\nCan anyone tell me why this is?", "Solution_3": "[quote=\"barasawala\"][quote=\"Remike\"] it is easy to see that no company can have all 3 edges incident to $u$, $v$ or $w$. [/quote]\n\nCan anyone tell me why this is?[/quote]\r\n\r\nBecause all the companies have to be triangles?", "Solution_4": "OK, I misunderstood something. Sorry.\r\n\r\nThanks for your reply anyway, weiquan.", "Solution_5": "For the case where everything is a triangle, we can do it with the following inequality: $\\binom{\\chi(G)}{2} \\le |E|$. $\\chi(G)$ = minimum number of colors you need to color the vertices such that no adjacent vertices have the same color (aka chromatic number).\n\nProof: Color it with the minimum amount of color. Assume that $\\binom{\\chi(G)}{2} > |E|$. Then some pair of colors is not connected by an edge. (Say 1, 2). Then just change all the 1's to 2's, and since they aren't adjacent, this new coloring works, contradicts minimality. $\\blacksquare$\n\nSo assume $\\chi(G) \\ge k+3$. Then $\\binom{k+3}{2} \\le |E| = 3k$, a contradiction for any $k \\ge 1$.", "Solution_6": "Note that the set of edges of the same airline forms either a star or a triangle.\n\nLet $x$ be the number of stars. For each star, let the central vertex be the vertex of degree $\\ge 2$. Then I color this vertex with a color that is different from all other vertices. This takes at most $x$ colors.\n\nLet $y$ be the number of remaining triangles, which is at most the number of original triangles. There are $3y$ edges. \n\nClaim: $E\\ge \\binom{\\chi(G)}{2}$.\n\nProof: There exists an edge between two vertices of any pair of colors $(i,j)$; if $(i,j)$ is a counterexample, $i,j$ can be merged to be the same color.\n\nIf follows that if the graph needs $y+3$ colors, then we need at least $\\binom{y+3}{2}$ edges, but $\\binom{y+3}{2} > 3y$ for all $y\\ge 1$, contradiction.", "Solution_7": "We do a little trolling\n\n\nI assume that no two flight companies can operate a flight between the same two cities (this seems to be what the wording indicates). We are essentially being asked to show that it is possible to $k+2$-color such a graph.\n\nIt is clear that the set of edges of some flight company is either a triangle or a star. We now use induction, with the case of $k=1$ being obvious. If a flight company has edges forming a star, then temporarily delete the company and color the resulting graph with $k+1$ colors. Then add it back, and and recolor the \"center\" of the star with the $k+2$-nd color, which clearly works. Therefore, suppose that all of the flight companies are triangles.\n\nIf some non-isolated vertex $u$ has degree $d \\leq k+1$, then temporarily delete the $\\tfrac{d}{2}$ flight companies with an edge attached to the city and color the remaining graph. When we add the $\\tfrac{d}{2}$ flight companies back, assign a unique one of our new $\\tfrac{d}{2}$ colors to each flight company containing $u$. If a flight company on vertices $\\{u,v,w\\}$ is assigned a color, then change the color of vertex $u$ to the assigned color. Finally, since $d \\leq k+1$, one of the $k+2$ available colors is not present among the neighbors of $u$, so we can change the color of $u$ to that. It is clear that the resulting graph coloring is valid.\n\nTherefore, suppose that every vertex has degree at least $k+2$. On the other hand, we have $3k$ edges, hence there are at most $\\frac{6k}{k+2}$ vertices, i.e. at most $5$ vertices. Since we supposed $k \\geq 2$, we actually need $5$ vertices and for the graph to be a $K_5$, but this is clearly impossible. Hence we are done. $\\blacksquare$\n\n[b]Remark:[/b] [url]https://en.wikipedia.org/wiki/Spoke%E2%80%93hub_distribution_paradigm[/url]", "Solution_8": "funny\n[hide = solution]\nWe know that the graph can be edge decomposed into a bunch of triangles and stars, so that there are exactly $k$ of them together. We've to show that $\\chi(G) \\leq k+2$. To prove this, we induct on $n$. Bases cases are easy to deal with. For the inductive step, if there's a star, we delete the central vertex and all the edges it is part of, induct down to the left over graph and then add in the central vertex as a new group all by itself. If there's only triangles, there are at most $3k$ edges. Since this implies ${\\chi(G) \\choose 2} \\leq 3k$ the result follows quickly now.\n[/hide]" } { "Tag": [ "number theory", "least common multiple", "greatest common divisor" ], "Problem": "three numbers a, b, and c have a LCM of 60. numbers a and c have a GCF of 6. numbers a and b have a GCF of 2. numbers b and c have a GCF of 4. \r\n\r\n1. what is the smallest possible sum of a, b, and c?\r\n\r\n2. what is the largest possible sum of a, b, and c? \r\n\r\n3. how many possible sums are there?", "Solution_1": "to do the first in 5 minutes:\r\nthree numbers a, b, and c have a LCM of 60. numbers a and c have a GCF of 6. numbers a and b have a GCF of 2. numbers b and c have a GCF of 4.\r\n\r\n1. what is the smallest possible sum of a, b, and c? \r\n[hide]6*X \n\n4*x\n\n12*X\n\n60=2^2*3*5\n\nso we already have 2^2, and we already have 3, but we don't have 5 so we add this to b(for smallest value)\n\n6x\n20x\n12x\nand for the smallest value x=1\n\n38?!?[/hide]", "Solution_2": "true and correct.\r\nnow try the other 2", "Solution_3": "[hide]The smallest sum is $6+20+12=38$\n\nThe largest sum is $6+4+60$. \n\nThere are three possible sums. [/hide]", "Solution_4": "davidli is correct." } { "Tag": [ "calculus", "integration", "trigonometry", "function", "calculus computations" ], "Problem": "Hi,\r\nI need help to solve these examples:\r\n\r\n1) $ \\int \\frac{cosx \\, dx}{2sin^2x\\plus{}5}$\r\n\r\n2)$ \\int(x^2\\plus{}3)sin 2x\\,dx$\r\n\r\n3)$ \\int(4x^3\\plus{}2x)arctgx\\, dx$\r\n\r\nThanks in advance.", "Solution_1": "First one\r\n\r\nsubstitute sinx as t\r\n\r\nthis gives\r\n\r\n$ \\int\\frac{dt}{2t^2\\plus{}5} \\equal{} \\frac{1}{2} \\int\\frac{dt}{t^2\\plus{}(\\sqrt{\\frac{5}{2}})^2}$\r\n\r\n\r\n$ \\equal{} \\frac{1}{2.\\sqrt{\\frac{5}{2}}} tan^{\\minus{}1}(\\frac{t\\sqrt{2}}{\\sqrt{5}}) \\plus{} C$\r\n\r\n\r\n\r\n$ \\equal{} \\frac{1}{\\sqrt{10}} \\ tan^{\\minus{}1}(\\sqrt{\\frac{2}{5}}sinx) \\plus{} C$\r\n\r\n\r\nSecond one, try Integration by Parts ...", "Solution_2": "hello, for 1), use the substitution $ z\\equal{}\\tan(\\frac{x}{2})$ with $ dx\\equal{}\\frac{2dz}{1\\plus{}z^2}$, $ \\cos(x)\\equal{}\\frac{1\\minus{}z^2}{1\\plus{}z^2}$, $ \\sin(x)\\equal{}\\frac{2z}{1\\plus{}z^2}$.\r\nSonnhard.", "Solution_3": "[quote=\"Ligeia\"]Hi,\nI need help to solve these examples:\n\n1) $ \\int \\frac {cosx \\, dx}{2sin^2x \\plus{} 5}$\n\n2)$ \\int(x^2 \\plus{} 3)sin 2x\\,dx$\n\n3)$ \\int(4x^3 \\plus{} 2x)arctgx\\, dx$\n\nThanks in advance.[/quote]\r\n1)\r\n $ \\int \\frac {cosx \\, dx}{2sin^2x \\plus{} 5}\\equal{}\\int \\frac {d\\sin x}{2\\sin^2x \\plus{} 5}\\equal{}\\sqrt {\\frac 25}\\int \\frac {d\\sqrt {\\frac 25}\\sin x}{2\\sin^2x \\plus{} 5}\\equal{}\\dots$\r\n2)\r\n$ \\int(x^2 \\plus{} 3)sin 2x\\,dx\\equal{}\\int x^2 \\sin 2x\\,dx\\plus{}\\int 3\\sin 2x\\,dx\\equal{}\\frac 12\\int x^2 d\\cos 2x\\plus{}3\\int \\sin 2x\\,dx\\equal{}\\dots$\r\n3)\r\n$ \\int(4x^3 \\plus{} 2x)\\arctan x\\, dx\\equal{}\\int 4x^3 \\arctan x\\, dx\\plus{}\\int 2x \\arctan x dx\\equal{}\\int \\arctan x dx^4\\plus{}\\int \\arctan x dx^2\\equal{}\\dots$", "Solution_4": "hello, in the second case write the integral in the form\r\n$ \\int x^2\\sin(2x)\\,dx\\plus{}3\\int\\sin(3x)\\,dx$ and use integration by parts for the first one.\r\nSonnhard.", "Solution_5": "the Third one\r\n\r\nu can write x = tan z\r\n\r\nIntegral reduces to\r\n\r\n$ 2 \\int tanz(2tan^2z \\plus{} 1)sec^2z.z.dz$\r\n\r\nwhich can be integarted by parts taking z as first function ..." } { "Tag": [], "Problem": "Last year Mr. John Q. Public received an inheritance. He paid $ 20\\%$ in federal taxes on the inheritance, and paid $ 10\\%$ of what he had left in state taxes. He paid a total of $ \\$10,500$ for both taxes. How many dollars was \tthe inheritance?\r\n\t\r\n$ \\textbf{(A)}\\ 30,000 \\qquad \\textbf{(B)}\\ 32,500 \\qquad \\textbf{(C)}\\ 35,000 \\qquad \\textbf{(D)}\\ 37,500 \\qquad \\textbf{(E)}\\ 40,000$", "Solution_1": "[hide] Say that John got 100x dollars of inheritance. We know that 20% of this, or 20x of it went to federal taxes. Thus, he has 80x dollars left. 10% of this, or 8x went to state taxes. Thus, a total of 28x was what he paid. Hence, 28x=10500, so x=375. Therefore, John got 375*100=37500 dollars from inheritance, so the answer is $D$. [/hide]", "Solution_2": "[quote=\"4everwise\"]Last year Mr. John Q. Public received an inheritance. He paid $20\\%$ in federal taxes on the inheritance, and paid $10\\%$ of what he had left in state taxes. He paid a total of $\\$10,500$ for both taxes. How many dollars was \tthe inheritance?\n\t\n$\\text{(A)}\\ 30,000 \\qquad \\text{(B)}\\ 32,500 \\qquad \\text{(C)}\\ 35,000 \\qquad \\text{(D)}\\ 37,500 \\qquad \\text{(E)}\\ 40,000$[/quote]\r\nMe remember this problem, I did it the guess and check way, but I know the real way.\r\n[hide=\"The real way\"]So he has x amount of money. So he pays .2x for federal taxes. he has .8x left. Then he pays .08x to pay for state taxes. So in total he pays 10500 so the equation is .28x=10500. Solving this you get that x=37500.[/hide]", "Solution_3": "[quote=\"4everwise\"]Last year Mr. John Q. Public received an inheritance. He paid $20\\%$ in federal taxes on the inheritance, and paid $10\\%$ of what he had left in state taxes. He paid a total of $\\$10,500$ for both taxes. How many dollars was \tthe inheritance?\n\t\n$\\text{(A)}\\ 30,000 \\qquad \\text{(B)}\\ 32,500 \\qquad \\text{(C)}\\ 35,000 \\qquad \\text{(D)}\\ 37,500 \\qquad \\text{(E)}\\ 40,000$[/quote]\r\n\r\n[hide=\"solution\"]Let the inheritance be $x$. \n\n$.2x+.8\\cdot .1x=10500$.\n\n$x=\\boxed{37500}$.[/hide]", "Solution_4": "Also on the 10A\r\n The answer is [b]3500[/b]", "Solution_5": "[quote=\"King_James23\"]Also on the 10A\n The answer is [b]3500[/b][/quote]Answer is wrong, see above solutions, and we don't care what you got.", "Solution_6": "[quote=\"4everwise\"]Last year Mr. John Q. Public received an inheritance. He paid $20\\%$ in federal taxes on the inheritance, and paid $10\\%$ of what he had left in state taxes. He paid a total of $\\$10,500$ for both taxes. How many dollars was \tthe inheritance?\n\t\n$\\text{(A)}\\ 30,000 \\qquad \\text{(B)}\\ 32,500 \\qquad \\text{(C)}\\ 35,000 \\qquad \\text{(D)}\\ 37,500 \\qquad \\text{(E)}\\ 40,000$[/quote]\r\n[hide=\"Answer\"]20% off 100% is 80%. 10% off 80% is 72%. 100% -72% = 28%. Let the inheritance be x. Then, 10500 = 28% of x. Thus, x equals 37500, or D.[/hide]" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let a,b,c are lengths sides of triangel. Prove that\r\n$ \\sum \\frac {a}{\\sqrt{5a^2\\plus{}13bc}}\\ge \\frac {1}{\\sqrt {2}}$\r\nFind the best constant for this inequality and prove it. :)", "Solution_1": "No anyone prove this inequality??? :( argady, can-hang, where are you?", "Solution_2": "We can use Holder inquality. Like \r\n\\[ \\begin{array}{l}\r\n (a^3 \\plus{} b^3 \\plus{} c^3 )(m^3 \\plus{} n^3 \\plus{} p^3 )(x^3 \\plus{} y^3 \\plus{} z^3 ) \\ge (amx \\plus{} bny \\plus{} cpz)^3 \\\\ \r\n With \\\\ \r\n a^3 \\plus{} b^3 \\plus{} c^3 \\equal{} m^3 \\plus{} n^3 \\plus{} p^3 \\equal{} \\sum\\limits_{cyc} {(\\frac{{a^{\\frac{1}{3}} }}{{(5a^2 \\plus{} 13bc)^{\\frac{1}{6}} }}} )^3 \\\\ \r\n x^3 \\plus{} y^3 \\plus{} z^3 \\equal{} \\sum\\limits_{cyc} {(a^{\\frac{1}{3}} } (5a^2 \\plus{} 13bc)^{\\frac{1}{3}} )^3 \\\\ \r\n \\end{array}\r\n\\]\r\n :wink:", "Solution_3": "Yes, but it is not finish. :D \r\nWho can finish it?? :lol:", "Solution_4": "[quote=\"thegod277\"]Let a,b,c are lengths sides of triangel. Prove that\n$ \\sum \\frac {a}{\\sqrt {5a^2 \\plus{} 13bc}}\\ge \\frac {1}{\\sqrt {2}}$\nFind the best constant for this inequality and prove it. :)[/quote]\r\n\r\nBy Holder inequality we have\r\n\\[ \\left( {\\sum {\\frac {a}{{\\sqrt {5a^2 \\plus{} 13bc} }}} } \\right)^2 \\left( {\\sum {a\\left( {5a^2 \\plus{} 13bc} \\right)} } \\right) \\ge \\left( {\\sum a } \\right)^3\r\n\\]\r\nThen it suffices to show that\r\n\\[ \\Leftrightarrow a^3 \\plus{} b^3 \\plus{} c^3 \\plus{} 6abc \\le 2\\left( {ab\\left( {a \\plus{} b} \\right) \\plus{} bc\\left( {b \\plus{} c} \\right) \\plus{} ca\\left( {c \\plus{} a} \\right)} \\right)\r\n\\]\r\nwhich is an old problem of nhocnhoc. :)\r\n\r\nI think the followed holds for $ a,b,c$ be nonnegative numbers \r\n\\[ \\frac{a}{{\\sqrt {4a^2 \\plus{} 5bc} }} \\plus{} \\frac{b}{{\\sqrt {4b^2 \\plus{} 5ca} }} \\plus{} \\frac{c}{{\\sqrt {4c^2 \\plus{} 5ab} }} \\le 1\r\n\\]", "Solution_5": ":D oh, yes. A finish inequality is\r\n$ \\sum (a \\plus{} b \\minus{} c)(a \\minus{} c)(b \\minus{} c)\\ge 0$\r\nwhich is true by vornicur schur. :) \r\nAnd the best constan for this inequality is 2,53...", "Solution_6": "[quote=\"thegod277\"]:D oh, yes. A finish inequality is\n$ \\sum (a \\plus{} b \\minus{} c)(a \\minus{} c)(b \\minus{} c)\\ge 0$\nwhich is true by vornicur schur. :) \nAnd the best constan for this inequality is 2,53...[/quote]\r\n\r\nShow your solution if you can, the god." } { "Tag": [ "trigonometry", "complex analysis", "complex analysis unsolved" ], "Problem": "Another problem from Conway:\r\n\r\nI'm trying to show that the image of $ \\frac{\\cos z}{z}$ outside of any ball around the origin is $ \\mathbb{C}$. This is of course very easy to do using Picard's Theorem, but surely there is a more elementary way to see it?", "Solution_1": "Or did he just put it in his book to make us realize how hard it is without Picard? :P\r\n\r\nI wouldn't put it past him..." } { "Tag": [ "algebra", "polynomial", "search", "number theory", "prime numbers", "arithmetic sequence", "number theory proposed" ], "Problem": "Prove that there are an infinite number of primes $ p$ such that neither $ p \\minus{} 2$ nor $ p \\plus{} 2$ is a prime, i.e., there is an infinite number of prime numbers not belonging to any pair of twin primes.", "Solution_1": "$ p \\equiv 2 \\bmod 3, p \\equiv \\minus{} 2 \\bmod 5$ is an arithmetic progression with infinitely many primes by Dirichlet's theorem. (CRT lets you generalize as follows: there are infinitely many primes $ p$ such that $ p \\minus{} 2$ and $ p \\plus{} 2$ both have at least $ n$ prime factors for any $ n$.)\r\n\r\n(The strongest result I know about elementary proofs of special cases of Dirichlet's theorem couldn't cover this case as far as I know, but I hesitate to claim that there isn't a simpler proof.)", "Solution_2": "It would be sufficient to prove there are infinitely many primes of the form $ 15n\\plus{}8$, by elementary methods.", "Solution_3": "I don't see how you are doing $ 15n\\plus{}8$ on elementary ways (note that $ 8^2 \\not \\equiv 1 \\mod 15$).", "Solution_4": "I was not stating that it can be done for sure (only a suggestion). Does there exist a solution without the use of Dirichlet's Theorem?", "Solution_5": "Not that I know of. As ZetaX has hinted, the strongest result (at least to my knowledge) is that there exists a \"polynomial\" proof (like the proof for $ 4n \\plus{} 1$) about primes congruent to $ a \\bmod m$ if and only if $ a^2 \\equiv 1 \\bmod m$. See [url=http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/dirichleteuclid.pdf]this paper[/url]. \r\n\r\nThis is the only elementary method to prove [i]any[/i] special cases of Dirichlet's method I know of. Therefore if there's an elementary solution to this problem one might have to prove compositeness without exhibiting an explicit divisor.", "Solution_6": "That doesn't rule out other approaches entirely, such as showing that there aren't very many twin primes compared to primes.", "Solution_7": "Asuming the Schinzel hypothesis, one can even prove that there is no \"euclid-style\" proof if $ a^2 \\not\\equiv 1 \\mod n$. The paper was posted here about a year ago, I will take a search if someone wants it." } { "Tag": [ "quadratics", "algebra" ], "Problem": "Solve x and y\r\n\r\nProblem 1\r\n$\\displaystyle x^2y+xy^2=70$\r\n$\\displaystyle (x+y)(x^2+y^2)=203$\r\n\r\nProblem 2\r\n$\\displaystyle (x^2+y^2)x/y=6$\r\n$\\displaystyle (x^2-y^2)y/x=1$\r\n\r\nProblem 3\r\n$\\displaystyle x^4+y^4-x^2y^2=13$\r\n$\\displaystyle x^2-y^2+2xy=1$\r\n\r\nProblem 4\r\n$x^3-y^3=19(x-y)$\r\n$x^3+y^3=7(x+y)$\r\n\r\nProblem 5\r\n$x^2/y+y^2/x=12$\r\n$1/x+1/y=1/3$", "Solution_1": "[hide] Let $xy=a$\n$x+y=b$\nand $x^2+y^2=b^2-2a$\nthen we have\n$ab=70$\n$b(b^2-2a)=203$\nor\n$b^3-2ab=203$\nsubstituting $70$ for $ab$ gives $b^3=343$ or $b=7$\nso $xy=10$\nplugging in some numbers, we see that\n$x=5$, $y=2$ [/hide]", "Solution_2": "[hide] \n#4\n$x^3-y^3\\Rightarrow(x-y)(x^2+xy+y^2)=19(x-y)\nalso\nx^3+y^3\\Rightarrow(x+y)(x^2-xy+y^2)=7(x+y)$\n\nwe can cross out $(x-y)$ and $(x+y)$ from the equations respectively to get...\n\n$(x^2+xy+y^2)=19\nand\n(x^2-xy+y^2)=7$\n\nsubtract the equations to get...\n\n$2xy=12\\Rightarrow xy=6$\n\nsubstituting xy=6 to either equation, we get $x^2+y^2=13$\n\nwe can add and subtract 2xy into $x^2+y^2=13$ to eventually get\n\n$(x+y)^2=25 and (x-y)^2=1$\n\ntherefore $x+y=\\pm5 and x-y=\\pm1$\n\nthen we get $x=\\pm3 and y=\\pm2$\n\n#5\n\nwe can multiply the second equation by 36 and set it equal to equation 1 to get...\n\n\\[\\frac{36}{x}+\\frac{36}{y}=\\frac{x}{y}(x+y)\\]\n\nthis is also equal to \\[\\frac{36}{xy}=\\frac{x}{y}\\]\n\ncancelling, we get $x=\\pm6$\n\nplugging this into equation 2, we get $x,y=6$ or $x=-6, y=2$\n\nwe plug these into equation 1 and see that only $x,y=6$ works so that is our answer.\n[/hide]", "Solution_3": "The easiest way to do all of these is to use this\r\n[hide=\"Problem5\"]\nUsing $s=x+y$ and $p=xy$\nSo substituting in the equation, $\\frac{s(s^2-3p)}{p}=12$-$\\boxed2$\nand then the $2nd$ we get $3s=p$ -$\\boxed1$\nso substituting $\\boxed1$ in $\\boxed2$ we get the quadratic equation,\n$s^2-9s-36=0$ which gives the roots $s=12$ and $s=-3$\n[/hide]", "Solution_4": "[quote=\"felipesa\"]Problem 2\n$\\displaystyle (x^2+y^2)x/y=6$\n$\\displaystyle (x^2-y^2)y/x=1$\n[/quote]\r\nDivide them, we get\r\n$\\frac{x^2+y^2}{x^2-y^2}\\frac{x^2}{y^2}=6$\r\nor\r\n$\\frac{\\frac{x^2}{y^2}+1}{\\frac{x^2}{y^2}-1}\\frac{x^2}{y^2}=6$\r\n\r\nLet $t=x^2/y^2$,\r\n$\\frac{t+1}{t-1}t=6$, we get $t=2$ or $t=3$.\r\n\r\nI think you can finish it from this.", "Solution_5": "[hide=\"partial solution for problem 3\"]\n$x^2-y^2=1-2xy$\nsquare both sides\n$x^4+y^4+2x^2y^2=1-4xy+4x^2y^2$\nsubtract $3x^2y^2$ from both sides, and now you can use make it fit into the other equation\n$x^2y^2-4xy+1=13$\n$(xy-6)(xy+2)=0$\nsubstitution and manipulation will probably get to the answer from here :) \n[/hide]", "Solution_6": "[quote=\"felipesa\"]\nProblem 3\n$\\displaystyle x^4+y^4-x^2y^2=13$\n$\\displaystyle x^2-y^2+2xy=1$\n[/quote]\r\nthe first equation is $(x^2-y^2)^2+x^2y^2=13$\r\nand from the second equation we have $x^2-y^2 = 1-2xy$\r\nhence\r\n$(1-2xy)^2+x^2y^2=13$\r\nLet $t=xy$ and it becomes $(1-2t)^2+t^2=13$ and we get $t=-\\frac{6}{5}$ or $t=2$.\r\n\r\nFor $t=2$, we have\r\n$x^4+y^4=17$ and $x^2-y^2=-3$\r\nNotice that $x^4+y^4 = (x^2+y^2)^2-2x^2y^2=(x^2+y^2)^2-8=17$\r\nhence $x^2+y^2=5$.\r\nNow we can solve out $x=1,y=2$, or $x=-1,y=-2$. (Since $xy>0$, $x$ and $y$ have the same sign)\r\n\r\nYou can do it in a similar way for $t=-\\frac{6}{5}$ but the answers are complicated.", "Solution_7": "I solved 2 in a different way:\r\n\r\nMultiply the two equations, we get $(x^2+y^2)(x^2-y^2)=x^4-y^4=6$\r\n(1)+(2), we get: \r\n\\[\\frac{x^4-y^4+2x^2y^2}{xy}=7\\]\r\n\r\nLet $xy=t$, we have: $2t^2-7t+6=0$. Hence $t=2$,$t=\\frac{3}{2}$.\r\n\r\n(1)-(2): $\\frac{x^4+y^4}{xy}=5$. \r\n\r\nSo we have: $x^4+y^4=10$ or $x^4+y^4=\\frac{15}{2}$\r\nWe already know that $x^4-y^4=6$. Solve this system, get:\r\n$x=\\frac{2}{\\sqrt[4]{2}}, y=\\sqrt[4]{2}$\r\n\r\n$x=\\frac{3}{\\sqrt[4]{12}}, y=\\frac{\\sqrt[4]{12}}{2}$" } { "Tag": [ "function", "vector", "trigonometry", "calculus", "calculus computations" ], "Problem": "Find all continuous functions f:R->R, if $ f\\left( x \\right) \\plus{} f\\left( {x \\plus{} 1} \\right) \\plus{} f\\left( {x \\plus{} 2} \\right) \\equal{} 6x \\plus{} 6$\r\nfor every real number x", "Solution_1": "There's far too many to give \"all\". It's a translate of an infinite-dimensional vector space.\r\n\r\nFirst, let $ g(t)\\equal{}f(t)\\minus{}2t$. We have $ g(x)\\plus{}g(x\\plus{}1)\\plus{}g(x\\plus{}2)\\equal{}0$, hence $ g(x\\plus{}3)\\equal{}g(x)$. Write its Fourier series $ g(t)\\equal{}\\sum_n a_n e^{2\\pi i n t/3}$. From $ g(x)\\plus{}g(x\\plus{}1)\\plus{}g(x\\plus{}2)\\equal{}0$, $ a_n\\equal{}0$ for each $ n$ divisible by 3. From $ f$ being real-valued, $ a_n\\equal{}\\overline{a_{\\minus{}n}}$ (or we could do the whole thing with sines and cosines).\r\nThat's basically all we can say.", "Solution_2": "Thank you very much jmerry" } { "Tag": [ "geometry", "3D geometry", "analytic geometry" ], "Problem": "A cube of cheese $ C\\equal{}${$ (x,y,z)|0\\leq x,y,z\\leq 1$} is cut along the planes $ x\\equal{}y$, $ y\\equal{}z$, and $ z\\equal{}x$. How many pieces are there? (No cheese is moved until all three cuts are made.)\r\n\r\n(a) 5\r\n(b) 6\r\n(c) 7\r\n(d) 8\r\n(e) 9\r\n\r\nI don't get what $ C\\equal{}${$ (x,y,z)|0\\leq x,y,z\\leq 1$} and \"planes $ x\\equal{}y$, $ y\\equal{}z$, and $ z\\equal{}x$\" means. What does the graph look like? Could someone please help me? Thanks!", "Solution_1": "The problem is talking about a 3-dimensional coordinate system with perpendicular axes $ x, y, z$ intersecting at $ (0,0,0)$, the origin. The graph of an equation like $ x\\equal{}y$ would be the set of all points with coordinates satisfying the equation. It will take some imagination to get the picture in your head if you aren't familiar with the 3D system yet.\r\n\r\n$ x\\equal{}y$ is a plane that includes the entire $ z$-axis.\r\n\r\nI'll give you another opportunity to try to solve the problem. :)", "Solution_2": "I still don't get it. Could someone please provide me with a solution? Thanks!", "Solution_3": "[hide=\"Approach A\"]$ x \\equal{} y$ partitions space into two sections $ x > y$ and $ x < y$.\nSimilarly for $ x \\equal{} z$ and $ y \\equal{} z$.\n\nSo, the three cuts together partition the space into sections $ x < y < z$ and permutations. There are $ 6$ such possibilities, which correspond to the permutations of $ (x,y,z)$, so the answer is $ \\textbf{(B)}$[/hide]\n\n[hide=\"Approach B\"]All three planes contain the line $ x \\equal{} y \\equal{} z$. When three different planes all intersect in a common line, they will divide space into 6 parts.\n\n[url=http://www.josechu.com/planes_in_3d/pl6_gif_p.gif]Here is an image[/url] showing three planes with a common line dividing space into 6 sections.[/hide]" } { "Tag": [ "linear algebra", "matrix", "probability", "IMO Shortlist", "algebra", "combinatorics" ], "Problem": "Consider a matrix of size $n\\times n$ whose entries are real numbers of absolute value not exceeding $1$. The sum of all entries of the matrix is $0$. Let $n$ be an even positive integer. Determine the least number $C$ such that every such matrix necessarily has a row or a column with the sum of its entries not exceeding $C$ in absolute value.\n\n[i]Proposed by Marcin Kuczma, Poland[/i]", "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": ":D ,ans. n/2,construction is easy,and to prove ans i use ideas of sub. desciption", "Solution_3": "Let $ n \\equal{} 2k$, and let $ a_{ij}$ be the entry in the $ i$th row and $ j$th column. Then we claim that $ C \\equal{} k$. We know that $ C \\ge k$, since we can set $ a_{ij}$ to be 1 if $ i, j \\le k$, -1 if $ i, j > k$, and 0 otherwise; this construction has every row and column sum having absolute value $ k$.\r\n\r\nWe will now show that $ C \\equal{} k$ works. Suppose for sake of contradiction that we have a matrix whose row and column sums all have absolute value more than $ k$. WLOG we can assume that the first $ a$ row sums and first $ b$ column sums are positive, while the last $ 2k \\minus{} a$ and $ 2k \\minus{} b$ row and column sums are non-positive (this can be achieved simply by a permutation of the rows and columns). Furthermore, we can assume that $ a \\plus{} b \\ge 2k$, or else we may exchange $ a$ and $ b$ with $ 2k \\minus{} a$ and $ 2k \\minus{} b$ and flip all signs.\r\n\r\nDefine\r\n\r\n$ P \\equal{} \\sum_{i \\le a, j \\le b} a_{ij}$;\r\n\r\n$ N \\equal{} \\sum_{i > a, j > b} a_{ij}$;\r\n\r\n$ X \\equal{} \\sum_{i \\le a, j > b} a_{ij}$;\r\n\r\n$ Y \\equal{} \\sum_{i > a, j \\le b} a_{ij}$.\r\n\r\nSince $ P \\plus{} X$ is the sum of the first $ a$ row sums, it is strictly greater than $ ka$. Since $ P \\plus{} Y$ is the sum of the first $ b$ column sums, it is strictly greater than $ kb$. Therefore,\r\n\r\n$ P \\plus{} X \\plus{} P \\plus{} Y > k(a \\plus{} b)$.\r\n\r\nNoting that $ P \\plus{} X \\plus{} Y \\plus{} N \\equal{} 0$, we have\r\n\r\n$ P \\minus{} N > k(a \\plus{} b)$.\r\n\r\nThere are only $ ab$ entries in the sum by which we calculate $ P$, and there are only $ (2k \\minus{} a)(2k \\minus{} b)$ entries for the sum for $ N$. Thus,\r\n\r\n$ k(a \\plus{} b) < P \\minus{} N \\le ab \\plus{} (2k \\minus{} a)(2k \\minus{} b)$\r\n$ k(a \\plus{} b) < 4k^2 \\minus{} 2k(a \\plus{} b) \\plus{} 2ab$\r\n$ 0 < 4k^2 \\minus{} 3k(a \\plus{} b) \\plus{} 2ab$\r\n$ k^2 < 9k^2 \\minus{} 6k(a \\plus{} b) \\plus{} 4ab \\equal{} (3k \\minus{} 2a)(3k \\minus{} 2b)$\r\n$ 4k^2 < (6k \\minus{} 2a \\minus{} 2b)^2 \\le (2k)^2 \\equal{} 4k^2$,\r\n\r\na contradiction. (Recall that $ 4k \\ge a \\plus{} b \\ge 2k$).", "Solution_4": "Let $n=2k$, then $C=k$. Construction is easy, so let's prove it.\n\nIn this solution, include 0 in positive numbers.\nWhen $k=1$, there exists a row or a column where the two entries have opposite sign. If not, all entries must have the same sign and they must all be 0, which is a trivial case. Sum of two opposite numbers where each has absolute value not bigger than 1 yields a number also having absolute value not bigger than 1.\nHere, notice that if we have one column not bigger than 1 in absolute value, the other column also satisfies this.\n\nIf $k>1$\nlet the matrix be $M$.\nfirst, since switching rows or columns does not affect the problem, switch the order of rows and the order of columns so that both row sums and column sums are arranged biggest to smallest. (row sum : sum of entries in the row, column sum : sum of entries in the column)\nNow let\n$A= \\sum_{1 \\leq i,j \\leq k} a_{ij}, B= \\sum_{1 \\leq i \\leq k < j} a_{ij}, C= \\sum_{1 \\leq j \\leq kk} a_{ij}$,\nand make a 2-by-2 matrix $M'$ with entries $A, B, C, D$, i.e., merge the four k-by-k squares in $M$.\nNote that $|A|, |B|, |C|, |D| \\leq k^2$ and $A+B+C+D=0$, so considering the $\\frac{1}{k^2} M'$, by the result of case when $k=1$, we can assume $|A+C| \\leq k^2$ and $|B+D| \\leq k^2$, i.e. the sum of the column sums of first half columns and second half columns of $M'$ does not exceed $k^2$.\nSince the column sums were arranged biggest to smallest, either all column sums are positive in the first half or all column sums are negative in the second half. Let it be the first half since we can consider $-M$ if it was not the case.\nk column sums have sum smaller than $k^2$, so one of them must be smaller than $k$, and since it's positive, it's absolute value is smaller than k.", "Solution_5": "[quote=orl]Consider a matrix of size $n\\times n$ whose entries are real numbers of absolute value not exceeding $1$. The sum of all entries of the matrix is $0$. Let $n$ be an even positive integer. Determine the least number $C$ such that every such matrix necessarily has a row or a column with the sum of its entries not exceeding $C$ in absolute value.\n[/quote]\nDivide the matrix into four $n/2 \\times n/2$ quarters. Fill the top left with $1$s, the top right and bottom left with $0$s and the bottom right with $-1$s. Thus we get $C \\ge n/2.$\nWe now show that $C=n/2.$ Assume on the contrary that there exists a matrix $A=(a_{ij})_{i,j=1}^n$ such that every row and column sum is either greater than $n/2$ or less than $-n/2.$ Let $R_i$ represent the $i$ th row sum and $C_i$ the $i$ th column sum. Permute the rows so that all the \"positive\" rows are on the top of the \"negative\" rows. Represent the positive ones by $R^+$ and negative by $R^-.$ Similarly permute the columns. Also, let $|R^+|$ denote the number of positive rows while $|R^-|$ the negative. \nNote that the sum $=0$ condition implies that the sum of the positive rows equals the magnitude of the sum of the negative rows.\n\n[b]Claim:[/b] $|R^+|=|R^-|=n/2.$ Similarly $|C^+|=|C^-|=n/2$.\n[i]Proof:[/i] If $|R^+|>n/2,$ then $|R^-|n/2$ and $-n \\le R^- <-n/2$ by our assumption. Hence\n$$\\left |\\sum R^- \\right | \\le \\frac{n}{2}|R^-|<\\frac{n^2}{4} < \\frac{n}{2} |R^+|<\\sum R^+$$\na contradiction. The other case is similar. $\\square$\n\nNote that sum of all the elements is given by the sum of the top $n/2$ rows and left $n/2$ colums, minus the top left $n/2 \\times n/2$ square (which is double counted before) plus the bottom right $n/2 \\times n/2$ square. These top rows and left colums have a positive sum by the claim, while the latter two have the maximum, minimum value when all the member are $1, -1$ respectively. Hence \nThus, \n\\begin{align*}\n\\sum_{i,j} a_{ij} =\\sum_{in/2} a_{ij} -\\sum_{i,j\\left( \\frac{n}{2} \\right) \\left( \\frac{n}{2} \\right)+\\left( \\frac{n}{2} \\right) \\left( \\frac{n}{2} \\right)-\\frac{n^2}{4}-\\frac{n^2}{4}=0 \n\\end{align*}\na contradiction. $\\blacksquare$\n\n\n[b]Edit:[/b] As pointed out below, this solution has a fault; the lemma is wrong. I believe that it can be fixed, but I am yet to try it.", "Solution_6": "[quote=Wizard_32][quote=orl]Consider a matrix of size $n\\times n$ whose entries are real numbers of absolute value not exceeding $1$. The sum of all entries of the matrix is $0$. Let $n$ be an even positive integer. Determine the least number $C$ such that every such matrix necessarily has a row or a column with the sum of its entries not exceeding $C$ in absolute value.\n[/quote]\nDivide the matrix into four $n/2 \\times n/2$ quarters. Fill the top left with $1$s, the top right and bottom left with $0$s and the bottom right with $-1$s. Thus we get $C \\ge n/2.$\nWe now show that $C=n/2.$ Assume on the contrary that there exists a matrix $A=(a_{ij})_{i,j=1}^n$ such that every row and column sum is either greater than $n/2$ or less than $-n/2.$ Let $R_i$ represent the $i$ th row sum and $C_i$ the $i$ th column sum. Permute the rows so that all the \"positive\" rows are on the top of the \"negative\" rows. Represent the positive ones by $R^+$ and negative by $R^-.$ Similarly permute the columns. Also, let $|R^+|$ denote the number of positive rows while $|R^-|$ the negative. \nNote that the sum $=0$ condition implies that the sum of the positive rows equals the magnitude of the sum of the negative rows.\n\n[b]Claim:[/b] $|R^+|=|R^-|=n/2.$ Similarly $|C^+|=|C^-|=n/2$.\n[i]Proof:[/i] If $|R^+|>n/2,$ then $|R^-|n/2$ and $-n \\le R^- <-n/2$ by our assumption. Hence\n$$\\left |\\sum R^- \\right | \\le \\frac{n}{2}|R^-|<\\frac{n^2}{4} < \\frac{n}{2} |R^+|<\\sum R^+$$\na contradiction. The other case is similar. $\\square$\n\nNote that sum of all the elements is given by the sum of the top $n/2$ rows and left $n/2$ colums, minus the top left $n/2 \\times n/2$ square (which is double counted before) plus the bottom right $n/2 \\times n/2$ square. These top rows and left colums have a positive sum by the claim, while the latter two have the maximum, minimum value when all the member are $1, -1$ respectively. Hence \nThus, \n\\begin{align*}\n\\sum_{i,j} a_{ij} =\\sum_{in/2} a_{ij} -\\sum_{i,j\\left( \\frac{n}{2} \\right) \\left( \\frac{n}{2} \\right)+\\left( \\frac{n}{2} \\right) \\left( \\frac{n}{2} \\right)-\\frac{n^2}{4}-\\frac{n^2}{4}=0 \n\\end{align*}\na contradiction. $\\blacksquare$[/quote]\n\ni feel there is a flaw in ur solution.why is |R+| and |R-| same. i feel that this is not always true\n", "Solution_7": "The [i]number[/i] of rows with positive sum is equal to the [i]number[/i] of rows with negative sum. I guess you got confused about the notation.\n\nIf you find a flaw or a counterexample, then kindly let me know.", "Solution_8": "yes this is not always true .the number of rows with positive sum is not always equal to number of rows with negative sum.here is counterexample for n=10.\n 0 0 0 0 1 1 1 1 1 1 ---(sum 6>10/2=5)\n 0 0 0 0 1 1 1 1 1 1\n 0 0 0 0 1 1 1 1 1 1\n 0 0 0 0 1 1 1 1 1 1\n 0 0 0 0 1 1 1 1 1 1\n 0 0 0 0 1 1 1 1 1 1\n 0 -1 -1 -1 -1 -1 -1 -1 -1 -1\n 0 -1 -1 -1 -1 -1 -1 -1 -1 -1 \n 0 -1 -1 -1 -1 -1 -1 -1 -1 -1\n 0 -1 -1 -1 -1 -1 -1 -1 -1 -1\nso this counterexample is valid.\n", "Solution_9": "[quote=Wizard_32]The [i]number[/i] of rows with positive sum is equal to the [i]number[/i] of rows with negative sum. I guess you got confused about the notation.\n\nIf you find a flaw or a counterexample, then kindly let me know.[/quote]\n\ni feel u messed up with the inequality of sigma|R-| to prove that the sum of matrix is not zero\nactually the max value of mod of the sum negative row is n but u have took it as n/2 so the inequality condition goes wrong", "Solution_10": "We claim that $C=\\tfrac{N}{2},$ achieved by splitting the matrix into the four equal square submatrices $\\begin{pmatrix}0 & 1 \\\\ -1 & 0\\end{pmatrix}.$\n\nNote that we can permute the rows and columns such that only the first $w$ columns and the first $y$ columns have positive sum. Let $x=N-w,z=N-y.$ Let the corresponding sums of the entries in each region be $W,X,Y,Z$ as shown:\n\n[asy]\nsize(6cm);\ndraw((0,0)--(5,0)--(5,5)--(0,5)--(0,0));\ndraw((2.7,0)--(2.7,5));\ndraw((0,2.1)--(5,2.1));\nlabel(\"$w$\",(0,5)--(2.7,5),N);\nlabel(\"$x$\",(2.7,5)--(5,5),N);\nlabel(\"$y$\",(0,5)--(0,2.1),W);\nlabel(\"$z$\",(0,2.1)--(0,0),W);\nlabel(\"$W$\",(1.35,3.55));\nlabel(\"$X$\",(3.85,3.55));\nlabel(\"$Y$\",(1.35,1.05));\nlabel(\"$Z$\",(3.85,1.05));\n[/asy]\nSo we easily get \n\\[W+Y\\geq wC,\\ X+Z\\leq -xC,\\ W+X\\geq yC,\\ Y+Z\\leq -zC.\\]\nBut for instance $X+Z\\leq -xC\\implies W+Y\\geq xC,$ and similar inequalities give\n\\[W+Y\\geq \\max(w,x)C,\\ W+X\\geq \\max(y,z)C,\\ Y+Z\\leq -NC+\\min(y,z)C,\\ X+Z\\leq -NC+\\min(w,x)C.\\]\nSumming in pairs, we get\n\\[2W+X+Y\\geq C(\\max(w,x)+\\max(y,z))\\geq C(\\max(w,x)-\\min(w,x)+\\max(y,z)-\\min(y,z))+2NC+X+Y+2Z\\]\nand the extreme inequalities gives \n\\[2(W-Z)\\geq C(|w-x|+|y-z|+2N).\\]\nBut $W-Z\\leq wy+xz=wy+(N-w)(N-y)$ and $|w-x|+|y-z|=|2w-N|+|2y-N|,$ so we get \n\\[wy+(N-w)(N-y)\\geq C(|2w-N|+|2y-N|+2N).\\]\nThis rearranges to \n\\[(2w-N)(2y-N)+N^2\\geq C(|2w-N|+|2y-N|+2N).\\]\nFinally, noting that $(2w-N)(2y-N)\\leq |(2w-N)(2y-N)|$ and letting $a=|2w-N|,b=|2y-N|,$ we get \n\\[C\\leq \\frac{ab+N^2}{a+b+2N}\\qquad (\\spadesuit).\\]\nWe show that $\\tfrac{ab+N^2}{a+b+2N}\\leq \\tfrac{N}{2}.$ This is equivalent to $(2a-N)(2b-N)\\leq N^2,$ which is true since \n\\[-N\\leq 2w-N,2y-N\\leq N\\implies 0\\leq a,b\\leq N\\implies -N\\leq 2a-N,2b-N\\leq N.\\]\nThus we conclude $C\\leq \\tfrac{N}{2}$ from $(\\spadesuit).$", "Solution_11": "let $n=2k$\nwe claim thet the answer $k$ \nthe construction is easy\nwe will prove that $C=k$ work\nWLOG let the first $x$ row are positive the next $y$ are negative and do the same for columns as the picture\n[asy]\nsize(6cm);\ndraw((0,0)--(5,0)--(5,5)--(0,5)--(0,0));\ndraw((2.7,0)--(2.7,5));\ndraw((0,2.1)--(5,2.1));\nlabel(\"$w$\",(0,5)--(2.7,5),N);\nlabel(\"$z$\",(2.7,5)--(5,5),N);\nlabel(\"$x$\",(0,5)--(0,2.1),W);\nlabel(\"$y$\",(0,2.1)--(0,0),W);\n[/asy]\n[color=#f00][u][b]lemma: [/b][/u][/color]if we have a matrix $2 \\times 2$ and $|a_{ij}| \\le m$ and the sum of all entries is zero thene there exists a row or column with the sum of its entries not exceeding $m$ in absolute value.\nit's obvious\n$\\blacksquare$\nnow split the matrix into four identical matrices\n[asy]\nsize(6cm);\ndraw((0,0)--(5,0)--(5,5)--(0,5)--(0,0));\ndraw((2.5,0)--(2.5,5));\ndraw((0,2.5)--(5,2.5));\nlabel(\"$W$\",(1.35,3.55));\nlabel(\"$X$\",(3.85,3.55));\nlabel(\"$Y$\",(1.35,1.05));\nlabel(\"$Z$\",(3.85,1.05));\n[/asy]\nfrom the lemma suppose $|W+X| \\le k^2$ but $W+X=-Y-Z \\implies |Y+Z| \\le k^2$\nif $x>k$ then we're done from the PHP becuase $r_1+r_2....r_{k} \\le k^2$\nif $x < k$ then $ y >k$ just do the same\nand we win :D", "Solution_12": "What if $n$ is odd?", "Solution_13": "The answer is $\\tfrac{n}2$ with every in the top left quadrant $1$, all bottom right quadrant $-1$ and the rest are zero.\n\nAssume on the contrary that we can have every row/column sum greater than $\\tfrac{n}2$ or less than $-\\tfrac{n}{2}$. We can rearrange rows and columns so arrange them in descending order from left to right, from top to bottom. Let $A$ be sum of top left quadrant, $B$ be top right, $C$ be bottom left, $D$ be bottom right.\n\nWLOG suppose there are as many positive-summed rows as negative-summed ones, so the region of $A+B$ has all positive rows sums, so $A+B\\ge \\tfrac{n^2}{4}$. Since each column in the region $A+B$ sums to at most $\\tfrac{n}2$, there are at least $\\tfrac{n}2$ nonnegative sums. If the sum is nonnegative no amount of negative numbers can make the entire column a negative enough sum. Thus, $A+C\\ge \\tfrac{n^2}{4}$.\n\nWe have\n\\[n^2=A+B+C+D=(A+B)+(A+C)-(A-D)> \\frac{n^2}{4}+\\frac{n^2}{4}-(A-D)\\]\nso $A-D> \\tfrac{n^2}{2}$, absurd.", "Solution_14": "[quote=Ali3085]let $n=2k$\nwe claim thet the answer $k$ \nthe construction is easy\nwe will prove that $C=k$ work\nWLOG let the first $x$ row are positive the next $y$ are negative and do the same for columns as the picture\n[asy]\nsize(6cm);\ndraw((0,0)--(5,0)--(5,5)--(0,5)--(0,0));\ndraw((2.7,0)--(2.7,5));\ndraw((0,2.1)--(5,2.1));\nlabel(\"$w$\",(0,5)--(2.7,5),N);\nlabel(\"$z$\",(2.7,5)--(5,5),N);\nlabel(\"$x$\",(0,5)--(0,2.1),W);\nlabel(\"$y$\",(0,2.1)--(0,0),W);\n[/asy]\n[color=#f00][u][b]lemma: [/b][/u][/color]if we have a matrix $2 \\times 2$ and $|a_{ij}| \\le m$ and the sum of all entries is zero thene there exists a row or column with the sum of its entries not exceeding $m$ in absolute value.\nit's obvious\n$\\blacksquare$\nnow split the matrix into four identical matrices\n[asy]\nsize(6cm);\ndraw((0,0)--(5,0)--(5,5)--(0,5)--(0,0));\ndraw((2.5,0)--(2.5,5));\ndraw((0,2.5)--(5,2.5));\nlabel(\"$W$\",(1.35,3.55));\nlabel(\"$X$\",(3.85,3.55));\nlabel(\"$Y$\",(1.35,1.05));\nlabel(\"$Z$\",(3.85,1.05));\n[/asy]\nfrom the lemma suppose $|W+X| \\le k^2$ but $W+X=-Y-Z \\implies |Y+Z| \\le k^2$\nif $x>k$ then we're done from the PHP becuase $r_1+r_2....r_{k} \\le k^2$\nif $x < k$ then $ y >k$ just do the same\nand we win :D[/quote]\n\nI believe that it is not true that all quadrants have sum less than k^2. ", "Solution_15": "your brain on smoothing\n\n\nThe answer is $n/2$, achieved by splitting the matrix into four $n/2 \\times n/2$ submatrices and filling the top left with $1$ and the bottom right with $-1$. We now show that $C>n/2$ is impossible.\n\nSuppose that there was some $n \\times n$ matrix with every row/column having absolute value greater than $n/2$. Since every row/column also has absolute value at most $n$, it follows that there are at least $n/3$ rows with positive sum, and at least $n/3$ rows with negative sum (same for columns).\n\nNow WLOG permute the matrix so that the rows are sorted in decreasing order from top to bottom, and the columns are sorted in decreasing order from left to right. Consider the top-left region, formed by the intersection of the positive rows and positive columns. If some entry in this region is not $1$, but we can find some other entry in its row lying in the top-right region, formed by the intersection of the positive rows and the negative columns, which is not $-1$, then we can increase the first entry and decrease the second at the same rate until some entry's absolute value reaches $1$: this clearly does not affect the minimum absolute value of any row/column nor change the signs of any row/column, and the sum of matrix entries should be the same, so we may apply this smoothing operation freely until we cannot (clearly this will terminate).\n\nWhen this process terminates, I claim that the entire top-left region is filled with $1$s. Indeed, if there is still some term in the top-left region which is less than $1$, then all the terms in its row which are in the top-right region must be $-1$. Since there are at least $n/3$ columns in this top-right region, it follows that the sum of the term's row is at most $-n/3+2n/3\\frac{n}{2} \\implies x>\\frac{na}{2}-ab \\text{ and } a+\\frac{y}{b}>\\frac{n}{2} \\implies y>\\frac{nb}{2}-ab.$$\nOn the other hand, since the sum of the terms in the matrix is $0$, we should have\n$$x+y=(n-a)(n-b)-ab=n^2-na-nb \\implies n^2-na-nb>\\frac{na}{2}+\\frac{nb}{2}-4ab\\geq \\frac{n(a+b)}{2}-(a+b)^2 \\implies (2n-(a+b))(n-(a+b))>0,$$\nwhere we use $(a+b)^2 \\geq 4ab$ (AM-GM), but this is clearly impossible for $2n\\geq a+b \\geq n$: contradiction. $\\blacksquare$", "Solution_16": "Let $n=2k$(why is the problem not phrased this way lol). Then the answer is $\\boxed{k}$, achieved by the top left quarter grid filled with $1$s and bottom right quarter grid filled with $-1$s.\n\nNow suppose that we could achieve $C>k$. Then at least $k$ of the rows must have the same sum(by pigeonhole). WLOG assume these are positive and the first $k$ rows. Since the sum of these rows is greater than $k^2$, at least $k$ of these columns have their first $k$ rows sum up to a positive number(since each is at most $k$). But since each column has absolute value at least $C$, this means that those $k$ columns have a positive sum(they don't have eonugh cells to get to $-k$). Assume WLOG that these columns are the first $k$ columns.\n\nNow consider the sum of the entire grid. The top half and the left half of the board are both greater than $k^2$, but by PIE their intersection and the outside can change the sum by at most $k^2$, meaning that the sum of the board is positive, contradiction. $\\blacksquare$" } { "Tag": [], "Problem": "you all remember the old boys vs girls game. click here is you dont:[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=293724[/url]\r\n\r\n\r\nanyway, this is the boys vs girls game, but instead of adding 1 or subtracting 1, you are going to subtract how many stars you have (dont know what they are called), like the new members have 0 stars, yang-mills theory have 3, etc. if you are a girl, you subtract how many stars you have from the total. if you are a boy, you add how many stars you have. \r\n\r\nnew members can add 1 star. \r\n\r\nif you have half stars, round up.\r\n\r\ni will start from 25\r\n\r\n\r\n25", "Solution_1": "20. :D", "Solution_2": "Wow, the girls have isabella. 23", "Solution_3": "Hehe. 18.", "Solution_4": "this iz so unfair. 20.5", "Solution_5": "[b]24[/b]", "Solution_6": "wait, i guess i have 2.5 stars?\r\n\r\n26.5\r\n\r\nill edit my first post", "Solution_7": "You round up hjoon.\r\n\r\n30", "Solution_8": "oh, thnx blazer.\r\n\r\n33", "Solution_9": "[b]36[/b]", "Solution_10": "$ 39$", "Solution_11": "[b]42[/b]", "Solution_12": "$ 45$", "Solution_13": "[b]48[/b]", "Solution_14": "46\r\n\r\nThe message is too small. Please make the message longer before submitting.", "Solution_15": "$ 49$\r\n\r\nomg so close", "Solution_16": "[b]52[/b] WE WIN!!! :D", "Solution_17": "once again, an unfair game...\r\n\r\nmore girls than boys! :( \r\n\r\nLet's try again", "Solution_18": "u mean more boys than girls. ok, i'll be starting a new topic. lets play to 5. boys 1, girls 0", "Solution_19": "whoops... yeah, thats wut i meant! :oops:" } { "Tag": [ "abstract algebra", "function", "Support", "inequalities", "group theory", "linear algebra", "triangle inequality" ], "Problem": "Let $G$ be a finite abelian group, and let $\\hat{G}$ denote the dual group of $G$ consisting of the linear complex characters of $G$ (i.e. the group homomorphisms $\\chi : G \\to \\mathbb{C}^{\\times}$). If $f : G \\to \\mathbb{C}$ is any complex-valued function on $G$, we may define the Fourier transform $\\hat{f} : \\hat{G} \\to \\mathbb{C}$ by \r\n\\[\r\n\\hat{f} (\\chi ) := \\frac{1}{|G|} \\sum_{g \\in G} f(g) \\chi (-g).\r\n\\]\r\nDenote by $\\mathrm{supp} \\, (f) := \\{x \\in G \\mid f(x) \\neq 0\\}$ the support of $f$. \r\n\r\nProblem. If $f \\neq 0$ is not identically zero, show that\r\n\\[\r\n|\\mathrm{supp} \\, (f)| \\cdot |\\mathrm{supp} \\, (\\hat{f})| \\geq |G|.\r\n\\]\r\n\r\nEnjoy! ;) \r\n\r\n--Vesselin", "Solution_1": "By the way, Vesselin, I'm the one who moved it over here. Since I was planning to answer it with analysis arguments, I didn't think it belonged in algebra. I'd also like to see \"advanced fields\" get more use, anyway.\r\n\r\nA little background information: The set of all characters is itself a group. Call the group of characters $\\hat{G}.$ The characters are orthogonal and span the set of functions on $G,$ so as a matter of linear algebra, $|G|=|\\hat{G}|.$\r\n\r\nIn his definition of the Fourier transform, Vesselin wrote $\\chi(-g).$ This could have been said in several other ways: $\\chi(-g)=(\\chi(g))^{-1}=\\overline{\\chi(g)} =\\chi^{-1}(g)$ where by $\\chi^{-1}$ I mean the group inverse in $\\hat{G}.$\r\n\r\nThe orthogonality of the characters leads to the Plancherel theorem:\r\n\r\n$\\frac1{|G|}\\sum_{g\\in G}|f(g)|^2=\\sum_{\\chi\\in \\hat{G}}|\\hat{f}(\\chi)|^2.$\r\n\r\nFuthermore, we can say just as a matter of the triangle inequality that\r\n\r\n$\\forall\\chi\\in \\hat{G},$ $|\\hat{f}(\\chi)|\\le\\frac1{|G|}\\sum_{g\\in G}|f(g)|.$\r\n\r\nWrite $||f||_p=\\left(\\frac1{|G|}\\sum_{g\\in G}|f(g)|^p\\right)^\\frac1p$ for functions on $G.$ Out of deference to the different normalization, I'll refrain from this notation on $\\hat{G},$ except that I will use $||\\hat{f}||_{\\infty}$ on $\\hat{G}.$ The two theorems above can then be expressed as:\r\n\r\n$||f||_2^2=\\sum_{\\chi\\in \\hat{G}}|\\hat{f}(\\chi)|^2$ and $||\\hat{f}||_{\\infty}\\le||f||_1.$\r\n\r\n(Continued in next post)", "Solution_2": "Proof of the problem:\r\n\r\nAssume WLOG that $||f||_2=1.$ (Once we know the function is not the zero function, we may multiply by any nonzero constant without changing any of the supports.)\r\n\r\n$||f||_1^2=\\frac1{|G|^2}\\left(\\sum_{g\\in\\, \\mathrm{supp}\\,(f)}|f(g)|\\right)^2 $\r\n\r\n$\\le\\frac1{|G|^2}\\cdot |\\mathrm{supp}\\,(f)|\\cdot\\sum_{g\\in \\,\\mathrm{supp}\\,(f)}|f(g)|^2$\r\n\r\n(using the Cauchy-Schwarz inequality)\r\n\r\nBut $\\frac1{|G|}\\sum_{g\\in \\,\\mathrm{supp}\\,(f)}|f(g)|^2=||f||_2^2=1,$ so we have\r\n\r\n$||f||_1^2\\le\\frac{|\\mathrm{supp}\\,(f)|}{|G|}.$\r\n\r\nNow work with the other side of Plancherel's Theorem:\r\n\r\n$1=\\sum_{\\chi\\in \\hat{G}}|\\hat{f}(\\chi)|^2= \\sum_{\\chi\\in \\,\\mathrm{supp}\\,(\\hat{f})}|\\hat{f}(\\chi)|^2\\le |\\mathrm{supp}\\,(\\hat{f})|\\cdot ||\\hat{f}||_{\\infty}^2$\r\n\r\n$\\le |\\mathrm{supp}\\,(\\hat{f})|\\cdot ||f||_1^2\\le \\frac{|\\mathrm{supp}\\,(\\hat{f})|\\cdot|\\mathrm{supp}\\,(f)|}{|G|}.$\r\n\r\nHence, $|\\mathrm{supp}\\,(\\hat{f})|\\cdot |\\mathrm{supp}\\,(f)|\\ge |G|.$", "Solution_3": "We appear to have equality in this inequality if and only if $f$ is a scalar multiple of the characteristic function of a coset of a subgroup of $G.$", "Solution_4": "[quote=\"Kent Merryfield\"]We appear to have equality in this inequality if and only if $f$ is a scalar multiple of the characteristic function of a coset of a subgroup of $G.$[/quote]\r\n\r\nYes.\r\n\r\nActually, as recently discovered by Terence Tao, the inequality can be significally improved in the special case when $G = \\mathbb{Z}_p$ is a cyclic group of prime order $p$:\r\n\r\nProblem 2. If $G = \\mathbb{Z} / p\\mathbb{Z}$ with $p$ prime, and if $f : G \\to \\mathbb{C}$ is not identically zero, show that\r\n\\[\r\n|\\mathrm{supp} \\, (f)| + |\\mathrm{supp} \\, (\\hat{f})| \\geq p+1.\r\n\\]\r\nAmazingly, the celebrated Cauchy-Davenport theorem (which was so much discussed on this forum :lol: ) is a fairly straightforward corollary of this result:\r\n\r\nProblem 3. Deduce the Cauchy-Davenport theorem (that $|A+B| \\geq \\min \\{p,|A|+|B| - 1\\}$ for $A,B \\subset \\mathrm{Z}_p$) from Problem 2.\r\n\r\n--Vesselin" } { "Tag": [ "inequalities", "rearrangement inequality", "inequalities unsolved" ], "Problem": "Help me to prove this inequality for $a,b,c$ are the sides of the triangle ,then:\r\n$9(\\frac{a^2}{b}+\\frac{b^2}{c}+\\frac{c^2}{a})\\geq (a+b+c)^2(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c})$.", "Solution_1": "Hey, I managed to get the problem, or so i think, but required me to DA(dumbass it. Anyway, here is a hint\r\n[hide]\nif a,b,c are sides of triangle, then you can make the substitutions a = y+x, b = z+x, c = x+y, for x,y,z>0\n\nHope this helps\n[/hide]\n\nUh-oh, didn't help, here's the solution\n[hide]\n\nafter you clear the denominators, u can subtract a^3bc, b^3a, and c^3b from both sides, and you can rewrite that expression to: 8a^3c + 8b^3a + 8c^3b >= 5a^2bc + 5ab^2c + 5abc^2 + a^3b + b^3c + c^3a + 2a^2b^2 + 2a^2c^2 + 2b^2c^2. Now, doing the egregious substituion as suggested in the hint, we can tragically deduce the expressiion to be:\n8x^4 + 32x^3y + 16x^3z + 24x^2y^2 + 48x^2yz + 24x^2z^2 + 16xy^3 + 48xy^2z + 48xyz^2 + 32xz^3 + 8y^4 + 32y^3z + 24y^2z^2 + 16yz^3 + 8z^4 >= 3x^4 + 16x^3y + 18x^3z + 29x^2y^2 + 62x^2yz + 29x^2z^2 + 18xy^3 + 62xy^2z + 62xyz^2 + 16xz^3 + 3y^4 + 16y^3z + 29y^2z^2 + 18yz^3 + 3z^4.\n(I dare you to verify this :P )\n\nWe can subtract the RHS from both sides to obtain something-simpler-but-still-complicated-...-unfortunately, to get:\n5x^4 + 16x^3y + 16xz^3 + 5y^4 + 16y^3z + 5z^4>= 2x^3z+5x^2y^2+14x^2yz + 5x^2x^2 + 2xy^3 + 14xy^2z + 14xyz^2 + 5y^2z^2 + 2yz^3.\n\nNow we show the following is true, and adding these up will give the inequality derived above:\n\n2x^4 + 2y^4 + 2z^4 >= 2x^3z + 2xy^3 + 2yz^3 (is true by rearrangement inequality)\n\n3x^4 + 3y^4 + 3z^4 >= 3x^2y^2 + 3x^2z^2 + 3y^2z^2 (is true by rearrangement inequality, or break it down into symmetric AM-GM, whichever you prefer, or Bunchning)\n\n2x^3y + 2y^3z + 2z^3x >= 2x^2y^2 + 2x^2z^2 + 2y^2z^2 (holds by bunching)\n\n14x^3y + 14y^3z + 14z^3x >= 14x^2yz + 14xy^2z + 14xyz^2 (holds by bunching) \n\n[/hide]\r\n\r\nToodles", "Solution_2": "[quote=\"RealZeta\"]\n2x^3y + 2y^3z + 2z^3x >= 2x^2y^2 + 2x^2z^2 + 2y^2z^2 (holds by bunching)[/quote]\r\nBut this ineq is not true for all positive x, y, z. :P", "Solution_3": "Touche Vasc :P\r\n\r\n(reworking solution)\r\n\r\nToodles,", "Solution_4": "When I'm beat, I'm beat, the problem is a little beyond me, yet it seems that I can get it. The expression is partially symettric, but the other half is missing. So I have no clue (dont think Bunching will get anywhere)...maybe a clever work on Schur? Or just coming up with something deviously clever.\r\nIt is one of th hardest inequalities i've dealt with.\r\n\r\nIm assuming its a chinese olympiad problem?", "Solution_5": "If I understand you right, you want to show:\r\n\r\n$3x^4 + 3y^4 + 3z^4 \\geq 3x^2y^2 + 3x^2z^2 + 3y^2z^2$\r\n$2x^3y + 2y^3z + 2z^3x \\geq 2x^2y^2 + 2x^2z^2 + 2y^2z^2$\r\n\r\nBut these are not true on their own. But we can show the following:\r\n$x^4 + y^4 + z^4 \\geq x^2y^2 + x^2z^2 + y^2z^2$\r\n$2x^4 + 2y^4 + 2z^4 \\geq 2xy^3 + 2yz^3 + 2zx^3$ (rearrangement)\r\n$2x^3y + 2y^3z + 2z^3x + 2xy^3 + 2yz^3 + 2zx^3 \\geq 4x^2y^2 + 4x^2z^2 + 4y^2z^2$ (Now it is symmetric and true!)\r\n\r\nAnd add them up ;)", "Solution_6": "Clever..." } { "Tag": [ "search" ], "Problem": "whats your favorite website?", "Solution_1": "Google :D so useful", "Solution_2": "Meh, how about besides search engines.\r\n\r\nHmm... I think, AOPS!!! :D", "Solution_3": "AoPS!!! most definately!", "Solution_4": "me too!", "Solution_5": "besides AoPS, my favorite site would have to be [url=http://www.dragongoserver.net]DragonGoServer[/url].", "Solution_6": "I enjoy reading political blogs. Two of my favorites are [url=http://atrios.blogspot.com]Eschaton[/url] and [url=http://plumer.blogspot.com]Brad Plumer[/url]. I also read the [url=http://www.ucomics.com/boondocks/]Boondocks[/url] comic strip daily.", "Solution_7": "http://www.fark.com", "Solution_8": "Well, since the questions is on AoPS, definently AoPS." } { "Tag": [ "inequalities", "geometry proposed", "geometry" ], "Problem": "Let $G$ the centroid of a triangle $ABC$ and let a line which passes through $G$ and intersects the lines $AB,\\ AC$ at the points $K,L$ respectively .Prove that \r\n$(\\frac{BK}{KA})^{4}+(\\frac{CL}{LA})^{4}\\geq\\frac{1}{8}$", "Solution_1": "$(\\frac{BK}{KA})^{4}+(\\frac{CL}{LA})^{4}\\ge 2(\\frac{\\frac{BK}{KA}+\\frac{CL}{LA}}{2})^{4}=\\frac{1}{8}$", "Solution_2": "What's your proof for $\\frac{BK}{KA}+\\frac{CL}{LA}=1$ ? :wink:", "Solution_3": "I known that I need to explain more :D :D \r\nAbove a forgot the notation of algebraic lengths, here we always use algebraic lengths!\r\nNow let $\\frac{\\overline{BK}}{\\overline{KA}}=x,\\frac{\\overline{CL}}{\\overline{LA}}=y$ then\r\n$x\\vec{KA}+\\vec{KB}=\\vec{0}\\Rightarrow \\vec{AB}=(x+1)\\vec{AK}$\r\nSimilar $\\vec{AC}=(y+1)\\vec{AL}$ but\r\n$\\vec{AG}=\\frac{1}{3}(\\vec{AB}+\\vec{AC})=\\frac{1}{3}((x+1)\\vec{AK}+(y+1)\\vec{AL})$\r\nbut $G,K,L$ are colinear$\\Rightarrow \\frac{1}{3}(x+1)+\\frac{1}{3}(y+1)=1\\Rightarrow x+y=1$\r\nCan you show your solution? :)" } { "Tag": [ "Ring Theory", "superior algebra", "superior algebra unsolved" ], "Problem": "let (R, m, k) be a commutative noether local ring.\r\n$ p_i$ is prime ideal, not m.(i=1,2,...n)\r\nI is an ideal, not m .\r\nthen m$ \\nsubseteq$ $ I\\cup (\\cup_{i\\equal{}1}^n p_i)$ ???", "Solution_1": "By union, do you mean the set-theoretic union or the ideal union, i.e. sum, of two ideals?", "Solution_2": "of course set-theoretic union is meant.\r\n\r\nhttp://www.math.uiuc.edu/~r-ash/ComAlg/ComAlg0.pdf" } { "Tag": [ "search", "limit", "calculus", "integration", "logarithms", "trigonometry", "calculus computations" ], "Problem": "Sometimes someone appears on this forum asking how to make some computation that have already been done before in an old or recent post. Also, people that often post interesting problems forget or don't know if he/she have already asked that before. It is not also very easy (I think) to make a search in order to find some particular post.\r\nTherefore I think it could be an usefull idea to make a sticky post having some kind of index of computations that have already been done before, together with the link(s) to the corresponding post(s), in order for people interested to find more easily what he/she is looking for. For example, I am thinking in something like the following:\r\n\r\n\r\n1. Limit Computations\r\n\r\n1.1 $ \\lim_{n\\to\\infty}\\frac{1^{k}+2^{k}+3^{k}+...+n^{k}}{n^{k+1}}$, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=118632]Post1[/url]\r\n\r\n1.2 ...\r\n\r\n...\r\n\r\n2. Integral Calculations\r\n\r\n1.1 ...\r\n\r\n1.2 ...\r\n\r\n...\r\n\r\n1.10 $ \\int^{\\pi /2}_{0}[\\ln(\\sin x)]^{2}\\,dx$, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=154194]Post1[/url]\r\n\r\n...\r\n\r\n3. Series Computations\r\n\r\n1.1 ...\r\n\r\n...\r\n\r\n1.6 $ \\sum_{n=1}^{+\\infty}\\frac{\\sin n}{2^{n}}$, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=120999]Post1[/url]\r\n\r\n...\r\n\r\n\r\nWhat is your opinion about this?", "Solution_1": "Unfortunately, the list of such computations will be at least 20 pages long (and that is an optimistic estimate), so, despite it may be ordered a bit better than the forum itself, not many people will have enough patience to search through it :(. Besides, proper maintainance of such a list will be a full time job and I do not know anyone who would wish to spend his/her free time copying the contents of the new posts to that sticky every day. So, my opinion is that this suggestion, despite being quite attractive, is not very practical.", "Solution_2": "I was only referring to the most common/important computations, not all of them. I also noted that in this forum, contrary to others, we cannot edit a post after some time, which, you are right, its not pratical for that purpose. Anyway, thanks for your opinion." } { "Tag": [ "probability" ], "Problem": "A driver approaches a toll booth and randomly selects two coins from his pocket. If the pocket contains 2 quarters, 2 dimes, and 2 nickels, what is the probability that the value of the two coins he selects will be at least enough to pay the 30-cent toll? Express your answer as a common fraction.", "Solution_1": "[hide]$\\frac{9}{\\binom{6}{2}}=\\boxed{\\frac{3}{5}}$[/hide]", "Solution_2": "[hide]\nTo have 30 cents, the driver needs at least one quarter.\nThere are ${6 \\choose 2}$ ways to pick two coins randomly.\nTo get two quarters, there is one way.\nTo get a quarter and a dime, there are $2\\cdot2=4$ ways.\nTo get a quarter and a nickel, there are $2\\cdot2=4$ ways.\n\n$\\frac{4+4+1}{{6 \\choose 2}}=\\frac{9}{15}=\\boxed{\\frac{3}{5}}$[/hide]" } { "Tag": [ "Columbia" ], "Problem": "Este topic lo cree para quienes esten interesados en la Ibero y quieran saber algo acerca del lugar donde se va a llevar a cabo: Cartagena, Colombia! Ademas de lo que creen pasara alli! Chao!", "Solution_1": "[quote=\"Pascual2005\"]Este topic lo cree para quienes esten interesados en la Ibero y quieran saber algo acerca del lugar donde se va a llevar a cabo: Cartagena, Colombia! Ademas de lo que creen pasara alli! Chao![/quote]\r\n\r\nBueno, yo ando bastante desconectado de todo lo que respecta a la Ibero desde que estoy por aqui en el norte, pero... hasta ahora me entero que va a ser en Colombia :D . En fin, esta bien, veinte aniversario, pues queda genial, la primera (1985, cuando ninguno de los actuales participantes habia tan siquiera nacido) fue en Colombia tambien.\r\n\r\nEspero que quienes vayan la disfruten!", "Solution_2": "Si, este a\u00f1o es el 20 cumplea\u00f1os de la ibero y por eso va a estar llena de sorpresas: que se esperan de esta ibero?", "Solution_3": "Bueno, no se que esperar, pero espero pasarla bien, porque seria mi ultima olimpiada a la que puedo ir :D.", "Solution_4": "Por cierto, no estara pensando Colombia en tratar de llevar a cabo alguna IMO alli? Digo, en el 2001 la Centro, en el 2005 la Ibero, que tal si en el 2009 la IMO.\r\n\r\nBueno, solo una idea al aire. Ciao,", "Solution_5": "Pascual es cierto que Cartagena es la ciudad donde mas se festeja en colombia???? eso me dijo una colombiana ahi pero no sabia si era cierto...", "Solution_6": "Manuel: Si es cierto!\r\n\r\nJimenez: Creo que en el 2001 colombia iba a ser la sede de la IMO, pero problemas legales no lo permitieron, dudo que halla una dentro de poco aca, aunque a mi modo de verlo junto con Brazil son los 2 paises latinos mas probables en hacerla.\r\n\r\nLeo: Creame que se va a divertir mucho, y ojala ganarse una dorada para venezuela.", "Solution_7": "Eso espero yo tambien ;) darle una segunda presea de oro a venezuela en la ibero. Pero eso lo veremos halla, :D", "Solution_8": "Hola yo solo quisiera saber un poco m\u00e1s sobre el SEMI2005 y si conoce los nombres de los gu\u00edas le agradeceria q me dijera el de El Salvador! Adios :)", "Solution_9": "Los nombres de los gu\u00edas est\u00e1n en la p\u00e1gina de la Ibero (pero sin pa\u00edses) quiz\u00e1 alguien sepa algo m\u00e1s.\r\n[quote]Por cierto, no estara pensando Colombia en tratar de llevar a cabo alguna IMO alli? Digo, en el 2001 la Centro, en el 2005 la Ibero, que tal si en el 2009 la IMO. \n[/quote]\r\nY por cierto me parece que la IMO del 2009 ya tiene sede: Frankfurt, Alemania. Creo que la que todav\u00eda no se sabe ni qu\u00e9 es la del 2008 pero no estoy seguro.\r\nSale nos vemos en Cartagena y por cierto: \u00bfQu\u00e9 se hace exactamente en el SEMI?\r\nNos preguntaron si queriamos ir y dijimos que s\u00ed. Sabemos que es un seminario de resoluci\u00f3n de problemas o algo as\u00ed pero exactamente \u00bfc\u00f3mo se maneja?", "Solution_10": "Creo que al SEMI 2005 vienen Titu Andrescu Y Andy Liu a dictar algunas charlas....\r\n\r\nNo se todavia quien es guia de quien pero cuando sepa lo confirmo!\r\n\r\nSuerte", "Solution_11": "Pascual que es el SEMI?\r\neso es el cmpamento de verano en bogota?", "Solution_12": "Semi no es el entrenamiento de colombia es un evento paralelo a la olimpiada que se hara en cartagena donde se hablara de resolucion de problemas", "Solution_13": "ahhh bacano,\r\nque mal que me lo pierdo :(" } { "Tag": [ "Olimpiada de matematicas", "geometry" ], "Problem": "$ D$ es un punto variable en el lado $ \\overline{BC}$ de un $ \\triangle ABC.$ Por $ D$ se trazan las paralelas a los lados $ AB$ y $ AC$ que intersectan a $ AC$ y $ AB$ en $ X,Y,$ respectivamente. Demostrar que el lugar geom\u00e9trico del circuncentro de $ \\triangle AXY$ es una recta que es perpendicular a la A-simediana del $\\triangle ABC.$", "Solution_1": "Si $ M$ y $ N$ sin puntos medios de $ AB$ y $ AC,$ con semejanza de tri\u00e1ngulos es facil probar que $ \\frac {_{MX}}{^{NY}} \\equal{} \\frac {_c}{^b}.$ Las circunferencias $ \\odot(AMN)$ y $ \\odot(AXY)$ se cortan pues en el centro de la semejanza espiral que lleva $ \\overrightarrow{MX}$ a $ \\overrightarrow{NY}.$ Entonces $ \\frac {_{PM}}{^{PN}} \\equal{} \\frac {_{AM}}{^{AN}} \\equal{} \\frac {_c}{^b}$ $ \\Longrightarrow$ $ P$ es fijo y pertenece entonces a la $ A$-circunferencia de Apolonio del tri\u00e1ngulo $ \\triangle AMN.$\n\nExpandiendo con la homotecia de centro en $ A$ y coeficiente $ 2,$ se ve que $ P$ es punto medio de $ AQ,$ donde $ Q$ es el punto de corte de la A-circunferencia de Apolonio de $ \\triangle ABC$ con su circunferencia circunscrita, la cual no es mas que la recta de la A-simediana de $\\triangle ABC.$ En definitiva, el lugar del circuncentro de $ \\triangle AXY$ es la mediatriz de $ AP.$" } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "AIME" ], "Problem": "I purposely put this topic in this forum.\r\n\r\ni suppose that most people who post in the high school basics forum are not \"advanced\" problem solvers (yet?)\r\n\r\nbut i was wondering what are your goals? do you think you will ever be a top 3 USAMO winner? do you work toward that?", "Solution_1": "over the course of the last three years I ve realized I sucked at competition math in middle school. so over this summer, Im going to work extra hard for AIME and possibly (but unlikely) USAMO. along with that, Im going to try and hit the Intermediate forums :P .", "Solution_2": "for example, do you think you will ever go to IMO?", "Solution_3": "[quote=\"iamagenius\"]for example, do you think you will ever go to IMO?[/quote]\r\n\r\nwell unless i get unnaturally lucky next year, no... if i reach USAMO id be pretty happy but right now looking at the pre-olympiad problems make my head spin :D . \r\n\r\nive never taken AMC before so next year what i think i'll do is take it (and try AIME too) and see how far I can get. The summer after that I will study what needs work on and then try for USAMO.", "Solution_4": "My aims right now are similar to 7h3.D3m0n.117's. Really, I'll be ecstatic if I get into USAMO, and happy if I make a good score on AIME. I haven't taken the AMC before (in a competition at least, I've taken a few from the resources section, and of those I'm scoring well enough to qualify into AIME), but between now and next year I plan on learning a lot. The only problem is I wish I got an earlier start, as I'm going to be in 11th grade after summer is over.", "Solution_5": "well\r\ni don't really study much\r\nand i'm not really that interested in math\r\nso i guess i'd be satisfied with blue mop one day", "Solution_6": "My goals are just like thedemon's and stuck's goals. I've never taken an AMC before but hope to do well next year, and at least make it to AIME." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Problem 1(Solved):\r\n\r\nPositive Reals $a,b$ and $k=\\sqrt{ab}\\geq 1$. Prove:\r\n\r\n$\\frac{1}{1+a}+\\frac{1}{1+b}\\geq \\frac{2}{1+k}$\r\n\r\nSolution 1:\r\n\r\nBy Cauchy ineq. we get:\r\n\r\n$\\frac{1}{1+a}+\\frac{1}{1+b}\\geq$\r\n$\\frac{M^{2}}{b(1+a)+a(1+b)}$\r\n$=\\frac{2M^{2}}{2M^{2}+(k-1)4\\sqrt{ab}}$\r\n$\\geq\\frac{2M^{2}}{2M^{2}+(k-1)M^{2}}$\r\n$=\\frac{2}{1+k}$\r\n\r\nHere, $M=\\sqrt{a}+\\sqrt{b}$\r\n\r\nProblem 2:(Unsolved)\r\n\r\nPositive Reals $a,b,c$ and $k=(abc)^{1/3}\\geq 1$. Prove:\r\n\r\n$\\frac{1}{1+a+a^{2}}+\\frac{1}{1+b+b^{2}}+\\frac{1}{1+c+c^{2}}\\geq\\frac{3}{1+k+k^{2}}$", "Solution_1": "[b]Any idea on Problem 2?[/b] :maybe:", "Solution_2": "Just an idea... try mixing variables on it, and possibly mildorf's method of finding a tight bounding inequality to sandwich between...\r\n\r\nSorry I have little more than ideas but its what you asked for :wink:", "Solution_3": "See here\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?p=202141#p202141 \r\n(Vasc's post about two thirds down the first page) :)", "Solution_4": "[quote=\"Vasc\"][b]Proposition[/b]. Let $k,r$ and $x_{1},x_{2},...,x_{n}$ be positive numbers such that $\\sqrt[n]{x_{1}x_{2}...x_{n}}=r$.\n\n(a) If $n \\ge k+1$ and $r\\ge n^{1/k}-1$, then\n$\\frac{1}{(1+x_{1})^{k}}+\\frac{1}{(1+x_{2})^{k}}+...+\\frac{1}{(1+x_{n})^{k}}\\ge \\frac{n}{(1+r)^{k}}$;\n\n (b) If $n\\ge 1+1/k$ and $r\\le (\\frac{n}{n-1})^{1/k}-1$, then\n$\\frac{1}{(1+x_{1})^{k}}+\\frac{1}{(1+x_{2})^{k}}+...+\\frac{1}{(1+x_{n})^{k}}\\le \\frac{n}{(1+r)^{k}}$;\n\n(c)\tIf $k \\ge n-1$ is a positive integer and ${r\\ge1}$, then\n$\\frac{1}{1+x_{1}+...+x_{1}^{k}}+\\frac{1}{1+x_{2}+...+x_{2}^{k}}+...+\\frac{1}{1+x_{n}+...+x_{n}^{k}}\\ge \\frac{n}{1+r+...+r^{k}}$. :lol:[/quote]\r\n\r\nThanks, spanferkel, I found it, but i didn't find how to prove them.", "Solution_5": "[quote=\"libra_gold\"][quote=\"Vasc\"][b]Proposition[/b]. Let $k,r$ and $x_{1},x_{2},...,x_{n}$ be positive numbers such that $\\sqrt[n]{x_{1}x_{2}...x_{n}}=r$.\n\n(a) If $n \\ge k+1$ and $r\\ge n^{1/k}-1$, then\n$\\frac{1}{(1+x_{1})^{k}}+\\frac{1}{(1+x_{2})^{k}}+...+\\frac{1}{(1+x_{n})^{k}}\\ge \\frac{n}{(1+r)^{k}}$;\n\n (b) If $n\\ge 1+1/k$ and $r\\le (\\frac{n}{n-1})^{1/k}-1$, then\n$\\frac{1}{(1+x_{1})^{k}}+\\frac{1}{(1+x_{2})^{k}}+...+\\frac{1}{(1+x_{n})^{k}}\\le \\frac{n}{(1+r)^{k}}$;\n\n(c)\tIf $k \\ge n-1$ is a positive integer and ${r\\ge1}$, then\n$\\frac{1}{1+x_{1}+...+x_{1}^{k}}+\\frac{1}{1+x_{2}+...+x_{2}^{k}}+...+\\frac{1}{1+x_{n}+...+x_{n}^{k}}\\ge \\frac{n}{1+r+...+r^{k}}$. :lol:[/quote]\n\nThanks, spanferkel, I found it, but i didn't find how to prove them.[/quote]\r\n\r\nYou can use HOLDER", "Solution_6": "[quote=\"mysterious\"]\n\nYou can use HOLDER[/quote]\r\nPlease show how :roll: :?:" } { "Tag": [ "algebra", "binomial theorem", "AMC" ], "Problem": "Find the last two digits of 11^10-1", "Solution_1": "[hide]11^1=...11\n\n11^2=...21\n\n11^3=...31\n\n...\n\n11^9=...91\n\n11^10=...01\n\n11^10-1=...00\n\n00[/hide]", "Solution_2": "ya", "Solution_3": "[quote=\"darkquantum\"]Find the last two digits of 11^10-1[/quote]\r\n11^10-1=(10+1)^10-1=(binomial theorem) 1+C(10,1)*10*1+bunch of terms divisible by 100 -1 :equiv:1+100-1:equiv:0, where i calculate modulus 100.", "Solution_4": "err\r\nwat did u do mystic??/\r\ndd u asnwer the question?", "Solution_5": "He did answer the question.", "Solution_6": "He just expressed 11^10-1 = (10+1)^10 - 1. Expanding (10+1)^10 by binomial theorem, all the terms cancel out mod 100 except for 1 and 100, so 1 + 100 - 1 = 100, and the last two digits are 00.", "Solution_7": "ahh\r\ni cic" } { "Tag": [ "vector" ], "Problem": "[b]Problem.[/b] Let $M,N,P,Q,R,S$ be the midpoints of the sides $AB,BC,CD,DE,EF,FA$ of a hexagon, respectively. Prove that\r\n\\[RN^{2}=MQ^{2}+PS^{2}\\]\r\nif and only if $MQ\\perp PS$.", "Solution_1": "[hide]\n[b]VECTORS!!!!!!![/b] :rotfl: :P \n\nThe condition $RN^{2}=MQ^{2}+PS^{2}$ is equivalent to\n\\[|{\\bf b+c-e-f}|^{2}=|{\\bf a+b-d-e}|^{2}+|{\\bf c+d-a-f}|^{2}\\]\nwhile the condition $MQ\\perp PS$ is equivalent to\n\\[({\\bf a+b-d-e})\\cdot ({\\bf c+d-a-f})=0\\]\nNow just observe that ${\\bf b+c-e-f}={\\bf a+b-d-e}+{\\bf c+d-a-f}$. For all ${\\bf u}$ and ${\\bf v}$, $|{\\bf u+v}|^{2}=|{\\bf u}|^{2}+|{\\bf v}|^{2}+2{\\bf u}\\cdot{\\bf v}$, so that\n\\[|{\\bf b+c-e-f}|^{2}\\]\n\\[=|{\\bf a+b-d-e}|^{2}+|{\\bf c+d-a-f}|^{2}+2({\\bf a+b-d-e})\\cdot ({\\bf c+d-a-f})\\]\nAnd that's it!\n[/hide]", "Solution_2": "My solution was similar to yours, but I used Complex Numbers. Is there a solution without using such powerful tools?:)" } { "Tag": [ "modular arithmetic", "Congruences" ], "Problem": "Let $a_{1}$, $\\cdots$, $a_{k}$ and $m_{1}$, $\\cdots$, $m_{k}$ be integers with $2 \\le m_{1}$ and $2m_{i}\\le m_{i+1}$ for $1 \\le i \\le k-1$. Show that there are infinitely many integers $x$ which do not satisfy any of congruences \\[x \\equiv a_{1}\\; \\pmod{m_{1}}, x \\equiv a_{2}\\; \\pmod{m_{2}}, \\cdots, x \\equiv a_{k}\\; \\pmod{m_{k}}.\\]", "Solution_1": "We look at residues $a \\mod m= m_1 m_2 ... m_k$ and call $a$ \"forbidden\" iff $a \\equiv a_1 \\mod m_i$ for some $i$.\r\n\r\nWe clearly have $m_i \\geq 2^{i-1} m_1$ for all $i$.\r\nThe total number of forbidden residues $\\mod m$ by some fixed $i$ is $\\frac{m}{m_i}$.\r\nThus the total number of forbidden residues $\\mod m$ is at most $\\sum_{i=1}^k \\frac{m}{m_i} < \\frac{m}{m_1} \\cdot \\sum_{i=1}^k \\frac{1}{2^{i-1}} < \\frac{2m}{m_1} \\leq m$.\r\nSo there is some residue class that is not forbidden, giving that all it's infinitely many members don't satisfy any of these congruences." } { "Tag": [ "logarithms", "inequalities" ], "Problem": "\"Find the greatest integer x for which $3^{20} > 32^x$.\"\r\n\r\nIs there anyone you can solve this without the use of calculator?", "Solution_1": "$\\frac {20}{\\log_3 32} > x$\r\n\r\nYou can kinda estimate at that.", "Solution_2": "Oh yeah.\r\n\r\nI am just worried that the estimation would come out wrong.", "Solution_3": "well, $\\log_{3} 32$ is a number between $3$ and $4$, so $x=5$ ;)", "Solution_4": "Actually you're right.\r\n:)", "Solution_5": "[quote=\"amirhtlusa\"]well, $\\log_{3} 32$ is a number between $3$ and $4$, so $x=5$ ;)[/quote]\r\n\r\nCan't $x=6$ work as well from this approximation? And btw, $x=6$ does work because $3^{20} > 2^{30}$.\r\n\r\nTo show that $\\frac{20}{\\log_3{32}} > 6$ we write $\\log_3{32} = 5\\log_3{2}$ so $\\frac{20}{\\log_3{32}} = \\frac{4}{\\log_3{2}} > \\frac{4}{\\frac{2}{3}} = 6$ because $\\log_3{2} < \\frac{2}{3}$.", "Solution_6": "The way I solved is that since $ 2^5=32$, we have $(3^4)^5 > (2^x)^5$. If you take the fifth root on both sides, the inequality still holds and we get $ 3^4 > 2^x$ => $81>2^x$. The smallest power of two less than 81 is 64, which equals $2^6$. Thus, the largest value of x is 6.", "Solution_7": "[quote=\"frodo\"]The way I solved is that since $ 2^5=32$, we have $(3^4)^5 > (2^x)^5$. If you take the fifth root on both sides, the inequality still holds and we get $ 3^4 > 2^x$ => $81>2^x$. The smallest power of two less than 81 is 64, which equals $2^6$. Thus, the largest value of x is 6.[/quote]\r\n\r\nYeah. That's a much more efficient/slick way. I got caught up in the approximation frenzy :dry: :play_ball:", "Solution_8": "yes, $x=6$ is the correct answer,\r\ni made a mistake. :P", "Solution_9": "Thanks :D", "Solution_10": "Or, you could notice that log base 3 32 is right above log base 3 27..." } { "Tag": [ "factorial", "inequalities" ], "Problem": "Find the smallest positive integer $n$ such that $1! \\cdot 2! \\cdot 3! \\cdot\\cdot\\cdot n!$ is divisible by 1998^1998.", "Solution_1": "[hide]First, factor 1998:\n\n$1998 = 2^1 * 3^3 * 37^1$\n\nSo, we have to find an $n$ such that the factorial has the following factors:\n\n$1998^{1998} = (2^1 * 3^3 * 37^1)^{1998} = 2^{1998} * 3^{5994} * 37^{1998}$\n\nWe will have an abundance of 2s and 3s, so we don't need to worry about those. We only have to count how many times 37 appears in the product. The amount of factors in each factorial that are divisible by 37 clearly goes up by 1 every 37 factorials. This means from 0-36 factorials, there are 0 factors divisible by 37, there is 1 per factorial from 37-73, 2 for 74-110, etc. So, in the first group we gain 0 of our 1998 37s, 37 in the next, then 74, etc. So, we have to minimize the LHS of this inequality:\n\n$37(0) + 37(1) + 37(2) + 37(3) + \\cdots + 37(x) + (x + 1)y \\ge 1998$\n\nwhere x is positive and $0 \\le y < 37$. To do this, the first x terms must be less than 1998, so the largest possible x is 9, which yields\n\n$(10)y \\ge 333$\n\nThe greatest y must then be 34. Now, to calculate n, we first multiply x + 1 by 37 and subtract 1 (this counts the factorials up to $37(x+1)$, but we are also counting 0, which we don't want). Then, we add y. So, our answer is\n\n$(9 + 1)37 - 1 + 34 = 370 + 34 = 403$[/hide]" } { "Tag": [ "algebra", "polynomial", "LaTeX", "combinatorics unsolved", "combinatorics" ], "Problem": "Find the number of $ P_n(x)$($ n$ is even) satisfying:\r\ni)$ degP_n(x)\\equal{}n$\r\nii)The coefficients of $ P_n(x)\\in$ {$ \\minus{}1,0,1$}\r\niii)There exists a polynomial $ Q(x)$ whose all the coefficients $ \\in$ {$ \\minus{}1,0,1$} satisfying $ P_n(x)\\equal{}(x^2\\minus{}1)Q(x)$", "Solution_1": "It's essentially the same as http://www.artofproblemsolving.com/Forum/viewtopic.php?t=276784 -- if we split the sequence of coefficients of such a polynomial into two sequences by taking every other term, each of the sequences we get must satisfy the property in the link (though one of them must begin with either 1 or -1). But this means we can easily write the answer to this problem as (essentially) the product of two answers to the other problem.\r\n\r\nA couple LaTeX suggestions: \\deg exists. To get curly brackets, use \\{ and \\}." } { "Tag": [], "Problem": "Given $ P(x)\\equal{}a_{2008} x^{2008} \\plus{} a_{2007} x^{2007} \\plus{} ... \\plus{} a_{1} x \\plus{} a_{0}$ and $ Q(x)\\equal{}x^{2} \\plus{} 2x \\plus{} 2008$. Let $ P(x)\\equal{}0$ has $ 2008$ real roots and $ P(2008) \\le 1$. Prove that there exists real number $ x$ such that $ P(Q(x))\\equal{}0$.", "Solution_1": "Are you sure the problem statement is correct? I don't think the current statement is true. :| \r\n\r\n[hide]$ Q(x) \\equal{} (x \\plus{} 1)^{2} \\plus{} 2007$, so the range of $ Q$ is $ [2007,\\infty)$. We need to prove that $ P$ has a root greater than or equal to $ 2007$.\n\nBut we could just make $ P(x) \\equal{} a(x \\plus{} 1)x(x \\minus{} 1)(x \\minus{} 2)\\cdots (x \\minus{} 2006)$ and make $ a$ sufficiently small so that $ P(2008)\\le 1$ still holds.[/hide]" } { "Tag": [ "pigeonhole principle", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $ A$ be a finite ring.Prove that there exists two positive integers $ m,p,m \\geq p \\geq 1$ so that $ a^{m}\\equal{}a^{p}, \\forall a \\in A$.", "Solution_1": "If $ A\\equal{}\\{x_i\\}_{i\\equal{}1}^n$, let $ S\\equal{}\\{(x_{i}^k)_{i\\equal{}1}^n | k \\equal{} 1, 2, ...\\}$ be a set of n-tuples of powers of elements of $ A$. Since $ S$ contains at most $ n^n$ elements, it is finite, so we must have $ (x_{i}^m)_{i\\equal{}1}^n \\equal{} (x_{i}^p)_{i\\equal{}1}^n$ for some $ m>\\equal{}p>\\equal{}1$.", "Solution_2": "Oh, I have a feeling that we both meant: $ m>p>\\equal{}1$. :)", "Solution_3": "I think is obvious the case of equality :D...But can you be more specific with the demonstration? :blush:", "Solution_4": "[quote=\"bogdanl_yex\"]I think is obvious the case of equality :D...But can you be more specific with the demonstration? :blush:[/quote]\r\n\r\nHow much more specific? It's an application of the pigeonhole principle. We have an infinite sequence of n-tuples:\r\n\r\n$ (x_{1}, x_{2},..x_{n})$\r\n$ (x_{1}^2, x_{2}^2,..x_{n}^2)$\r\n$ (x_{1}^3, x_{2}^3,..x_{n}^3)$\r\n...etc.\r\n\r\n but there are only $ n^n$ possible, so two of them must be equal. Say it's the m-th and p-th ones. We have:\r\n\r\n$ (x_{1}^m, x_{2}^m,..x_{n}^m) = (x_{1}^p, x_{2}^p,..x_{n}^p)$\r\n\r\nBut that's the same as ${ x_{1}^m=x_{1}^p, x_{2}^m=x_{2}^p}$, etc. Which is exactly what we were looking for.", "Solution_5": "thank's a lot :lol:" } { "Tag": [ "linear algebra", "matrix", "algebra", "polynomial", "linear algebra unsolved" ], "Problem": "Solve in $M_{3}(R)$, $X^3=-X$", "Solution_1": "We geometrize, let $f$ be the endomorphism associated to $X$.\r\nIts characteristic polynomial is $x(x^{2}+1)$, then its eigenvalues are 0, i and -i, in $\\mathbb{C}$, $f$ is diagonalizable.\r\nThe rest is simple." } { "Tag": [ "Yale", "college", "Harvard", "MIT", "AMC", "AIME", "USA(J)MO" ], "Problem": "Hello AoPSers. I am beginning a new game that anybody can join. Basically, once we have enough people, we will begin. \r\n\r\n AoPS City, a game that I made up myself is one that can be set up many different ways. The people that sign up will be the townspeople that live in AoPS City. I am the mayor of AoPS City, and I make up all the rules that go on in AoPS. Basically you have to follow the rules to stay in the game. The last person in AoPS City is the winner, and they will learn what happens later. It is a game that is hard to explain, and you just have to begin, and you learn the idea of it as you play it. There are many different things that can happen in AoPS City. The mayor runs the only store in AoPS City. \r\n\r\nHere are the basic rules; \r\n\r\n1.) You need to be alive, or the mayor (me) will get rid of your body, and you will no longer be in AoPS City. \r\n2.) You need to eat a proper amount of food and water to live each day. \r\n3.) You can not live in AoPS City, unless you have a house. \r\n4.) Sometimes there will be random events each day that either complicate the game, or make it easier. \r\n5.) These rules will be described in better detail later. \r\n6.) You must be active in AoPS City. (you have to post on the thread)\r\n\r\n\r\n\r\nI will explain the game more when it begins. \r\nSign-ups begin now!!! \r\nIf you want to play then post it on the thread. Don't pm me!!!! \r\n\r\nPeople Playing;\r\nAIME15 \r\nernie \r\ndragon96\r\nerabel \r\nAIME15\r\nBugi \r\nPoincare \r\npythag011\r\nchenhsi \r\npacman2812 \r\njames41\r\nzapi2007\r\nmewto55555\r\nisabella2296", "Solution_1": "/me is IN!!!!!!!!!", "Solution_2": "I might drop out, but currently [b]in[/b].", "Solution_3": "in\r\n\r\n[color=white]The message is too small. Please make the message longer before submitting.[/color]", "Solution_4": "Sure dude. Sounds fun.", "Solution_5": "[quote=\"dragon96\"]in\n\n[color=white]The message is too small. Please make the message longer before submitting.[/color][/quote]\r\n\r\nI can see your white text against the gray background.\r\n\r\nWhen will we start? Also, describe the rules in better detail now.", "Solution_6": "In, might drop out.", "Solution_7": "\\in! sounds fun...", "Solution_8": "in\r\n\r\ntoo short", "Solution_9": "/in to funness...", "Solution_10": "*travels to AoPS City*\r\n[size=0]means /in[/size]", "Solution_11": "ok sure why not", "Solution_12": "I'm glad that a lot of actives are joining.", "Solution_13": "I might die on random but I'm in!", "Solution_14": "MEWTO%%%%% IS IN!", "Solution_15": "This is boring, [b]drop out[/b]", "Solution_16": "[b]Chip in 15 dollars[/b] :D", "Solution_17": "Stats - \r\n\r\nlotsofmath - 83 dollars - house, car, food and water \r\nerabel - 95 dollars - house, car, food and water \r\nPoincare - 100 dollars - house, car, food and water \r\npythag011 - 74 dollars- house, car, food and water \r\nchenhsi - 85 dollars - house, car, food and water \r\npacman2812 - 90 dollars - house, car, food and water \r\njames41 - 65 dollars - house, car, food and water \r\nzapi2007 - 80 dollars - house, car, food and water \r\nmewto55555 - 95 dollars - house, car, food and water \r\nisabella2296 - 85 dollars - house, car, food and water \r\nDojo - 84 dollars - house, car, food and water \r\ncrazychinesemaniac - 85 dollars - house, car, food and water \r\npcstaa - 85 dollars - house, car, food and water \r\nicemathematician7 - 85 dollars - house, car, food and water \r\nTOWN - 10,000 dollars \r\n\r\nOkay, we only need five more dollars.", "Solution_18": "pays 10 dollars, gets an upgrade on arcade :)", "Solution_19": "Stats - \r\n\r\nlotsofmath - 83 dollars - house, car, food and water \r\nPoincare - 100 dollars - house, car, food and water \r\npythag011 - 74 dollars- house, car, food and water \r\njames41 - 65 dollars - house, car, food and water \r\nzapi2007 - 70 dollars - house, car, food and water \r\nDojo - 84 dollars - house, car, food and water \r\nicemathematician7 - 85 dollars - house, car, food and water \r\nTOWN - 10,000 dollars\r\n\r\n\r\nOh, yes, and zapi2007 is also the richest person in AoPS City, what a coincidence. :) \r\n\r\n\r\nNow we have paid 35 dollars for the arcade. \r\nOh, and lotsofmath has graduated from college. I think we should take mewto out of this considering he has been banned from posting and pming. I am also getting rid of pcstaa and CCM for inactivity and not replying to the pm I sent, but opening it. There are now seven people left in AoPS City. \r\n\r\n\r\nOkay, now as for the jobs. Each person makes thirty dollars a week, and they must also pay 25 dollars a week for food, water, their car, and their house. Starting at day 4. \r\n\r\n\r\n[size=200][b]Day 4[/b][/size]\r\n\r\nAlright, if you are first on the list above, then you live in house 1, and you live in house 2 if you are second on the list above, and third on the list is in house three. (understand the pattern? :) ) \r\n\r\nFortune Cookie: Listen for hints for an advantage in the game. \r\n\r\nAlright, good morning to all seven of you. You have been playing for a long time, but now somebody is going to die. The murderer has dropped a bomb, but for some reason it is falling really slowly. It is about to fall on lotsofmath's house unfortunately. lotsofmath has the power to change houses with somebody else. All he has to do is say \"I switch houses with *****\" [b]in bold. [/b] If the person who has been offered to switch houses doesn't say anything in three days, then I will assume that they will accept to change houses with the person living in the house about to explode. Although, if the person does reply in three days, then they can decline the offer of changing houses. If the person declines the offer, then lotsofmath can offer the house to another person. If three people decline to change houses with the person living in house 1, then the person living in house 1 will have to stay (and die due to the bomb). The person living in house 1 three weeks from now will die. Let the moving begin. \r\n\r\nHint: There is a way for everybody to survive, so pay attention to this thread.", "Solution_20": "Plan A: Build a new house and move to it.\r\nIf not,\r\nPlan B: [b]I switch houses with icemathematician7.[/b]", "Solution_21": "way to switch houses with the person most likely to be inactive", "Solution_22": "I'll notify him on aim...", "Solution_23": "Way to play it safe lotsofmath. :lol:\r\n\r\nI also forgot to tell you that you can switch houses with a person even if you are not living in the house that is about to blow up. If you guys are all going to survive, it will require some generosity. (Hint ?)", "Solution_24": "hey can i be in!!!!!!!!!!!!!!", "Solution_25": "Hmm... Would you like to replace icemathematician? \r\n\r\nRemember if you guys all want to survive it is going to require some generosity. \r\n\r\nOkay icemathematician and lotsofmath left their houses at Jan 31st, 2009 2:43 pm. They waited outside their houses while the packers took their things and put them in their new houses. It took about an hour, but once it had been an hour they were able to go into their new houses at Jan 31st, 2009 3:43 pm. \r\n\r\nYa know, there are two hints in this post for everybody surviving. The first one was in the previous post so you guys should know what it is. The other one is the most important.", "Solution_26": "Oh.\r\n\r\nIt takes an hour to switch houses. THat will be very important.", "Solution_27": "The bomb will land on house 1 at Feb 18, 2009 12:33 pm. \r\n\r\nHmm... That's interesting.", "Solution_28": "especially the fact that nothing will happen now.", "Solution_29": "Well, like I said. You don't have to be living in house 1 to switch houses with somebody. If a person declines the offer in three days however, then they remain in their current homes. You just have to offer the house at just the right time." } { "Tag": [ "ratio", "ceiling function", "logarithms", "integration", "calculus", "derivative", "geometric sequence" ], "Problem": "Consider the sum:\r\n\r\n$S=(x)+ (xx) + (xxx)+(xxxx)+...+(xx...n \\ times...xx)$,\r\n\r\nwhere [b]x[/b] is a natural number and [b]a[/b] is the number of algarisms of x. Find a general formula for $S$.\r\n\r\n\r\n\r\nps: (xxx) is not (x\u00b3), (xxx) is just the juxtaposition of x. For instance, for x equal to 31, (xxx) will be 313131.", "Solution_1": "[quote=\"F u r u y \u00e1\"]Consider the sum:\n\n$S=(x)+ (xx) + (xxx)+(xxxx)+...+(xx...n \\ times...xx)$,\n\nwhere [b]x[/b] is a natural number and [b]a[/b] is the number of algarisms of x. Find a general formula for $S$.\n\n\n\nps: (xxx) is not (x\u00b3), (xxx) is just the juxtaposition of x. For instance, for x equal to 31, (xxx) will be 313131.[/quote]\r\n\r\nHmm ... I just got something like $S=\\frac{x}{9}\\left(\\frac{10^n-1}{9}\\times 10-n\\right)$\r\n\r\nIs it true ? :?", "Solution_2": "I don't think that's correct.\r\n\r\nPlugging in x=1 and n=3 should yield 111+11+1=123, but the formula returns something else.", "Solution_3": "[quote=\"fourierseries\"]I don't think that's correct.\n\nPlugging in x=1 and n=3 should yield 111+11+1=123, but the formula returns something else.[/quote]\r\n\r\nwhen $x=1$ , $n=3$ ;\r\n\r\n$S=\\frac{1}{9}\\left(\\frac{999}{9}\\times 10 -3\\right)=\\frac{1}{9}\\times 1107=123$\r\n\r\n :ninja:", "Solution_4": "You have to find a way to put the variable [b]a[/b] into the formula, so it will work with any natural x. I tried x=23, n=4:\r\n\r\n$S=\\frac{23}{9}\\left(\\frac{10^4-1}{9}\\times 10-4\\right)=28382$\r\n$23+2323+232323+23232323=23466992$", "Solution_5": "considering that [b]x[/b] have [b]a[/b] digits, we can construct:\r\n\r\n$x=x$\r\n\r\n$xx=x.10^a + x$\r\n\r\n$xxx=x.10^{2a} + x.10^a + x$\r\n\r\n$xxxx=x.10^{3a} + x.10^{2a} + x.10^a + x$\r\n\r\n$xxxxx=x.10^{4a} + x.10^{3a} + x.10^{2a} + x.10^a + x$\r\n.\r\n.\r\n$(xx...n times...xx)=x.10^{(n-1)a} + x.10^{(n-2)a} ... + x.10^a + x$ \r\n\r\nwe observe that the number [b]x[/b] appear [b]n[/b] times in the set of sums consequently:\r\n\r\n$S=n.x + (n-1).10^a.x + (n-2).10^{2a}.x + (n-3).10^{3a}.x.. + [n-(n-1)].10^{(n-1)a}.x$ \r\n,to make the expression easy, $S=x.(N + M)$ such that:\r\n\r\n$N=n + n.10^a + n.10^{2a}+ n.10^{3a}... + n.10^{(n-1)a}$, we have a geometric sequence of ration $10^a$ then:\r\n\r\n$N=\\frac{n.(1 - 10^{na})}{(1 - 10^a)}$\r\n\r\n\r\n$M=-1.10^a - 2.10^{2a} - 3.10^{3a}...- (n-1)10^{(n-1).a}$\r\n\r\n$M= -(10^a) - (10^{2a}) - (10^{2a}) - (10^{3a}) - (10^{3a}) - (10^{3a}) - (10^{4a}) - (10^{4a}) - (10^{4a}) - (10^{4a})...$\r\n\r\n$M=\\frac{-(10^a)[1 - 10^{(n-1)a}]}{1 - 10^a} - \\frac{(10^{2a})[1 - 10^{(n-2)a}]}{1 - 10^a} - \\frac{(10^{3a})[1 - 10^{(n-3)a}]}{1 - 10^a}...\\frac{10^{(n-1)a}(1 - 10^{[n-(n-1)]a})}{1 - 10^a} $\r\n\r\n$M=\\frac{ -10^a - 10^{2a} - 10^{3a}...-10^{(n-1)a} + (n-1).10^{an}}{1 - 10^a}=\\frac{ 10^{an}.[(1-10^a)(n-1) +1] - 10^a}{(1 - 10^a)^2}$ thus:\r\n\r\n$S=x(N + M)=x.[\\frac{n -10^a(n + 1) + 10^{a(n+1)}(1 -2n)}{(1 - 10^a)^2}]$", "Solution_6": "I think something is wrong, your formula is returning a negative decimal number.\r\n\r\n[hide][img]http://img.photobucket.com/albums/v280/Furuyah/soma.png[/img][/hide]", "Solution_7": "let $r = \\lceil \\log{x} \\rceil$\r\n\r\n$\\sum_{k=0}^{n-1} ( (10^{r})^{(n-k)} * (n-k) * x$", "Solution_8": "[quote=\"Altheman\"]let $r = \\lceil \\log{x} \\rceil$\n\n$\\sum_{k=0}^{n-1} ( (10^{r})^{(n-k)} * (n-k) * x$[/quote]\r\ni get the sam thing, but can it get simplified easily?", "Solution_9": "you can move over the x since it is a constant, but i am not sure if that can be simplified...i have a feeling that it can\r\nthere might be a general simplification of\r\n\r\n$x* \\sum^{n}_{k=0} ( y^{n-k} * (n-k) )$ could be\r\n$x* \\sum^{n}_{k=1} (k* y^{k})$ since this generates the same terms\r\nso having $y = 10^{\\lceil \\log{x} \\rceil}$\r\n$x* \\sum^{n}_{k=1} ((k+1)* y^{k}) - x* \\sum^{n}_{k=1} (y^{k})$\r\nconsider $\\int (n+1)* y^{n} = y^{n+1} + C$\r\nso $D_y(y^{n+1} + y^{n} + ... + y^{3} + y^{2})$ equals the first summation symbol\r\nthe inside is a geometric sequence so\r\n$x * D_y(\\frac{y^{2}*(1- y^{n})}{1-y}) + x* \\frac{y*(1- y^{n})}{1-y}$\r\ni bet that this is completely unnecessary, but i don't see the other way to do it...\r\ni don't feel like doing the differentiation now...it is not that difficult (think quotient rule)\r\nsorry about the calculus if i made a mistake, i am not that good at calculus, i just recognized that pattern with the derivitives and the geometric sequence" } { "Tag": [ "linear algebra", "matrix", "vector", "linear algebra unsolved" ], "Problem": "Each entry of the matrix $A=(a_{ij})\\in M_{n}(\\mathbb{R})$ is equal to $0$ or $1$ and moreover $a_{ii}=0,\\,a_{ij}+a_{ji}=1\\,(1\\leq i1$ be a fixed integer. For a prime p that does not divide a, let $ f(p)$ be the order of a modulo p. Prove that $\\frac{p}{f(p)}$ is unbounded.", "Solution_1": "oh thanks, i did it with a more general result as showed below:\r\n${\\bf Problem.}$ Let be given $a>1$ is integer and $h$ is a polynomial with integer coefficients($h\\not\\equiv0$). For each prime $p$ which $(p,a)=1$ we put $f(p)$ is the order of $p$ respectively to $a$. Then $\\frac{p}{h(f(p))}$ is unbounded.\r\n\r\nProof. If $h$ is constant , it's trivial. Assume that $deg(h)\\ge1$. It's enough to show that for each positive integer $d$ then \r\n\\[\\frac{p}{(f(p))^d}\\]\r\nis unbounded. Okie, \r\nAssume that \\[\\frac{p}{(f(p))^d}\\] is bounded, ie... there exist $n$ positive integer such that $p<(f(p))^d\\cdot n$. Now by Dirichlet Theorem there exist infinitely many $q$ is prime and $q>a$ and $q\\equiv1\\pmod{(n+1)!}$. Take $p$ is a prime number which\r\n\\[p\\mid \\frac{a^{q^d}-1}{a^{q^{d-1}}-1}\\]\r\n1st) At first let us show $p\\nmid a^{q^{d-1}}-1$\r\nIn fact , if not then \r\n\\[a^{q^{d-1}}\\equiv1\\pmod{p}\\]\r\nso\r\n\\[\\frac{a^{q^d}-1}{a^{q^{d-1}}-1}\\equiv q\\pmod{p}\\]\r\nthen $p=q$. We now get $q\\mid a^{q^{d-1}}-1$ ,then $q\\mid (a^{q-1}-1,a^{q^{d-1}}-1)=a-1$ this contracts to the definition of $q$.\r\n\r\n2nd) The next step, we show $f(p)=q^{d}$. \r\nwe have $p\\mid (a^{f(p)}-1,a^{q^d-1}-1)=a^{q^t}-1$ where $t\\le d$. By the fist step $t>d-1$. So $t=d$ ie... $f(p)=q^d$ (note $f(p)$ min)\r\nNow we have \r\n$p-1=uf(p)=uq^d$ with $1\\le u< n$. Then\r\n\\[p\\equiv1+u\\pmod{(n+1)!}\\]\r\nso $p=1+u$ then $q^d=1$ contracts!\r\n\r\nSo we have \\[\\frac{p}{(f(p))^d}\\] \r\nis unbounded. We are done!!!!\r\n\r\nI am wondering that if \\[\\frac{p}{e^{f(p)}}\\] \r\nis bounded or unbounded?", "Solution_2": "What a pity that when i were home i found that there is something wrong in my proof. In fact the proof is only true for $h$ with $deg(h)\\le 1$ and you can do as we said!" } { "Tag": [ "geometry", "conics", "parabola", "analytic geometry", "symmetry", "LaTeX" ], "Problem": "A bridge is to be built in the shape of a parabolic arch and is to have a span of 100 feet. The height of the arch a distance of 40 feet from the center is to be 10 feet. Find the height of the arch at its center.", "Solution_1": "[hide]Is it 250/9? If thats right i can post a solution-all I did was anylytic geometry, lable the highest point (0,h)[/hide]", "Solution_2": "I'm almost certain my answer is right this time:\r\n\r\n[hide=\"My Answer\"]250/9[/hide]\n\nit's the same thing Art of Owna got, but i think the way i got it is rather different.\n\n[hide=\"My Rather Lengthy Solution\"]First, i drew the parabola on a cartesian coordinate grid such that the axis of symmetry was on the y axis and the zeros were 50 and -50. And you also know that in the equation of the parabola ($ax^{2}+bx+c = y$), a has to be negative (since it opens downward). You also know that $b = 0$, because $\\frac{-b}{2a}= 0$ (using the fact that the axis of symm. is zero). So, then you start writing the equation (lets start by using the point (40,10) given to us in the problem):\n\n$-ax^{2}+c = y$ becomes $(-a)(1600)+c = 10$. So you get $c = 10+1600a$.\n\nSub that back into the original equation and you get $-ax^{2}+10+1600a = y$.\n\nNow you apply the formula for the point (50,0). You should get $(-a)(2500)+1600a+10 = 0$. Simplify it to get $a = \\frac{1}{90}$.\n\nRemember that we took a to be negative in the original, so when you plug it back in, you get $\\frac{-1}{90}x^{2}+\\frac{1600}{90}+10 = y$\n\nThe problem is asking for the y value (height) when $x = 0$, so you plug in $0$ for $x$ and the final equation is $\\frac{1600}{90}+10 = y$\n\nSolve for $y$ and you get $y = \\boxed{\\frac{250}{9}}$[/hide]", "Solution_3": "Well, my equation turned out to be x^2=-90(y-250/9), plugging in (50,0) (-50,0) and (40,10) seem to work out..", "Solution_4": "is the bridge supposed open upward (valley) or downward (hill)? i thought it was like a hill but my dad was reading over my solution and he kept thinking that it was like a valley... so which one is it?", "Solution_5": "O I just had a general equation of x^2=a(y-h), then had two points (50,0) and (40,10) plug them in for x and y, two equations for two variables a and h. I found a, then found h, and h is the vertex, which is also the height.\r\n\r\n\r\nAnd, wouldnt it have to be a hill (I mean a bridge isnt usually valley shaped),", "Solution_6": "[quote=\"Art of Owna\"]\nAnd, wouldnt it have to be a hill (I mean a bridge isnt usually valley shaped),[/quote]\r\n\r\nthat's exactly what i thought", "Solution_7": "[hide=\"Answer and my solution\"]\nArt of Owna got it, the answer is $\\frac{250}{9}$.\n\na parabola: $y = a(x-h)^{2}+k$\n\nfrom this, one can conclude:\n$10 = a(40-0)^{2}+k$\n$10 = 1600a+k$\n$0 = 1600a+k-10$ ... 1\n\n$0 = a(50-0)^{2}+k$\n$0 = 2500a+k$ ... 2\n\nnow set the two equations equal to each other\n\n$1600a+k-10 = 2500a+k$\n$-10 = 900a$\n$a =-\\frac{1}{90}$\n\nthen plug in $a$ in equation 2:\n$0 = 2500 \\cdot-\\frac{1}{90}+k$\n$\\frac{2500}{90}= k$\n$k = \\frac{250}{9}$\n[/hide]\r\n\r\nand yes, it's a hill, not a valley\r\n\r\n-----------------\r\n\r\nGAHH i hate LaTeX errors.. spent like 20 minutes to figure out i had a parenthesis instead of a curly bracket :rotfl:", "Solution_8": "hey! i got it too!! and i had a couple LaTeX errors, also. just hit preview and see what ur missing." } { "Tag": [ "inequalities" ], "Problem": "Let $ a,b,c$ be positive real numbers that satisfy $ ab+bc+ca=1$. Show that $ a+b+c-3abc\\ge \\frac{1}{3}(\\sqrt{a^{2}+1}+\\sqrt{b^{2}+1}+\\sqrt{c^{2}+1})$.", "Solution_1": "We have\r\n\\begin{eqnarray*}\\frac{1}{3}\\cdot\\sum\\sqrt{a^{2}+1}&\\le&\\sum a-3\\cdot abc\\\\ \\iff\\frac{1}{3}\\cdot\\sum\\sqrt{(a+b)\\cdot (a+c)}&\\le&\\sum a-3\\cdot abc\\\\ \\iff\\frac{1}{3}\\cdot\\sum\\frac{2a+b+c}{2}&\\le&\\sum a-3\\cdot abc\\\\ \\iff a+b+c&\\ge& 9\\cdot abc \\end{eqnarray*} What is truly because \\[ \\left\\{\\begin{array}{c}1=ab+bc+ca\\ge 3\\cdot\\sqrt[3]{a^{2}b^{2}c^{2}}\\implies abc\\le\\frac{1}{3\\sqrt 3}\\\\ a+b+c\\ge \\sqrt{3\\cdot\\left(ab+bc+ca\\right)}=\\sqrt 3 \\end{array}\\right\\|\\\\ \\implies a+b+c\\ge 9\\cdot abc \\]" } { "Tag": [ "geometry", "circumcircle", "geometry unsolved" ], "Problem": "Let triangle ABC be inscribed in the circle C(O).\r\na) Build the triangle MNP such that MN, NP, MP are perpendicular on OC, OA and OB and M, N, P lie on the circle C(O).\r\nb) Let MS be perpendicular on BC. OS intersect the arc BC in R ( the arc BC which doesn't contain the point A). D lies on (OS such that DR=RS. The points E and F are for B and C what is D for A ( the analogous points). Prove that the circumscribed circles of the triangles OMD, ONE and OPF are meeting in a point different by O. \r\n :gathering: :diablo: ( I'm from USA)", "Solution_1": "Just a hint for a) MN is perendicular to OC iff $\\angle MOC=\\angle CON$ and similar equivalence holds for the othe perpendiculars.\r\nSo we denote \r\n$a=|\\angle NOA|=|\\angle AOP|$,\r\n$b=|\\angle POB|=|\\angle BOM|$,\r\n$c=|\\angle MOC|=|\\angle CON|$.\r\nNow we have $a+b=|\\angle AOB|$,$b+c=|\\angle BOC|$,$c+a=|\\angle COA|$, solving this system\r\n$a=\\frac{|\\angle AOB|+|\\angle COA|-|\\angle BOC|}{2}$. \r\nAfter constructing angle $a=|\\angle NOA|=|\\angle AOP|$ we can draw the lines OP and ON, then the perpendicular to OB lead via P meets the circle in M.", "Solution_2": "Thank you for your answer. :D \r\nAny ideas for b) ? :help: \r\nI'm waiting ...", "Solution_3": "Is it imposible ? Help me, please. I have to know it until next week. :help: :help: :help:", "Solution_4": "a) \u2013 This is exactly the configuration of a given triangle ABC, with it\u2019s altitudes AS, BT,\r\nCU, which intersect the circumcircle of ABC, at points M, N, P and the lines MN, NP, MP, are perpendicular to OC, OA, OB respectively ( well known ).\r\n\r\nb) \u2013 If now, R, Q, L, are the intersection points of circumcircle of ABC, from the lines OS, OT, OU respectively and we denote as D, E, F, the symmetrical points of S, T, U, with respect to the points R, Q, L respectively, then the circumcircles of the triangles OMD, ONE, OPF, have not any common point, other than O ( they are not coaxal circles ). \r\n\r\n Kostas Vittas.", "Solution_5": "Thank you :D" } { "Tag": [], "Problem": "Find all numbers $x$ such that $3^{2x}+3^{x}=90$", "Solution_1": "[hide=\"hint\"]\nMake a substitution. [hide=\":)\"]Why do you randomly click smileys? :huh: Well, since you have such weird habits, here's another hint for you. :D $y=3^x$[/hide]\n[/hide]", "Solution_2": "if you consider 3^x as t and 3^2x as t^2, then you get this equation: t^2 + t-90=0. Then you solve it, and the right solution is 9. so, as 3^x is 9, x must be 2 because 3^2 =9, and that's your solution.\r\nAnyway, you can solve it without doing this.", "Solution_3": "[hide]\n$3^{2x}+3^x=(3^x)^2+3^x$, so $0=(3^x)^2+3^x-90$\nFactoring, $0=(3^x+10)(3^x-9)$\n$3^x= -10$ which gives no solution, and $3^x=9$, which gives $\\boxed{x=2}$[/hide]", "Solution_4": "[hide]Let 3^x = y \ny^2 + y = 90\ny^2 + y - 90 = 0\n(y - 9)(y + 10) = 0\ny = 9 or -10\n3^x = 9 or -10\n3^x = 9 (-10 is impossible)\nx = 2 \n[/hide]" } { "Tag": [ "modular arithmetic", "calculus", "integration", "function", "MATHCOUNTS", "number theory", "totient function" ], "Problem": "I keep seeing things about 'mods'. I might've done something with them at one point or another, but I'm really not sure. i keep seeing them used, and have also heard that some problems with them in them don't necessary require them...there are alternate methods?\r\nI would check the AoPS wiki, but it's down for maintenance?\r\nThanks for all your help. :lol:", "Solution_1": "Example: $ 42 \\equiv 6 \\mod 9$. So if you take $ \\mod x$, where $ x$ is the divisor, then $ 6$ is the remainder when $ 42$ is divided by $ 9$.", "Solution_2": "Hmm. I thought it was something like that. Thanks. :lol:", "Solution_3": "Basically...\r\n$ x \\equiv y(mod z)$\r\nMeans that..\r\n$ x \\equal{} y \\plus{} kz$\r\nWhere $ x, y, z$ and $ k$ are integers. They do not need to be positive.", "Solution_4": "Basically, if $ x\\equiv y$ (mod m), then $ x\\minus{}y$ is divisible by $ m$\r\n\r\nwait, $ x$, $ y$, $ k$, and $ k$???", "Solution_5": "and remember that adding kz mod z is just like adding 0", "Solution_6": "Oops.\r\n:blush:\r\nAnyways, I fixed it.", "Solution_7": "[quote=\"ZhangPeijin\"]Basically...\n$ x \\equiv y(mod z)$\nMeans that..\n$ x \\equal{} y \\plus{} kz$\nWhere $ x, y, z$ and $ k$ are integers. They do not need to be positive.[/quote]\r\n\r\ni thought just k and z had to be integers\r\n\r\n$ 3.5 \\equiv 9.5 \\pmod{6}$", "Solution_8": "no, I don't think so\r\n\r\nusing the idea of remainders, how would you define remainders for non-integral values?", "Solution_9": "Yeah, like in fractions, you don't have remainders.", "Solution_10": "know important mod formulas (but know how to derive them first!!!) such as Fermat's little theorem, Euler's Totient Formula...", "Solution_11": "Yeah, they basically mean:\r\nIf p is prime and a is a natural number,\r\n$ a^p\\equiv a (\\mod {p})$\r\n\r\nAnd\r\nWhen p and a are relatively prime.\r\n$ a^{p \\minus{} 1} \\equal{} 1 (\\mod {p})$", "Solution_12": "Oh no, I don't know any of those. Better start studying...[/b]", "Solution_13": "[quote=\"ZhangPeijin\"]Yeah, they basically mean:\n$ a^p\\equiv a (\\mod {p})$\nAnd\n$ a^{p \\minus{} 1} \\equal{} 1 (\\mod {p})$\nWhere p and a are relatively prime.[/quote]\r\n\r\nthose 2 are equivalent, and it's still the wrong formula\r\n\r\nhere, $ a$ is any integer, and $ p$ is a prime. Try $ a\\equal{}5$ and $ p\\equal{}12$, you'll see that formula doesn't work.", "Solution_14": "What?\r\nIt was on Wolfram...", "Solution_15": "don't see it there, I think you misread it", "Solution_16": "No, you forgot to specify that p is prime.", "Solution_17": "If p is a prime number and a a natural number, then\r\n$ a^p \\equal{} a (mod p)$. \t\r\n(1)\r\n\r\nFurthermore, if p\u2224a (p does not divide a), then there exists some smallest exponent d such that\r\n$ a^d \\minus{} 1 \\equal{} 0 (mod p)$ \t\r\n(2)\r\n\r\nand d divides p-1. Hence,\r\n$ a^{(p \\minus{} 1)} \\minus{} 1 \\equal{} 0 (mod p)$. \t\r\n(3)\r\n\r\nThe theorem is sometimes also simply known as \"Fermat's theorem\" (Hardy and Wright 1979, p. 63). \r\n\r\n\r\n\r\n\r\nTaken exactly from Wolfram.", "Solution_18": "[quote=\"ZhangPeijin\"]If p is a prime number and a a natural number, then\n$ a^p \\equal{} a (mod p)$. \t\n(1)\n\nFurthermore, if p|a (p does not divide a), then there exists some smallest exponent d such that\n$ a^d \\minus{} 1 \\equal{} 0 (mod p)$ \t\n(2)\n\nand d divides p-1. Hence,\n$ a^(p \\minus{} 1) \\minus{} 1 \\equal{} 0 (mod p)$. \t\n(3)\n\nThe theorem is sometimes also simply known as \"Fermat's theorem\" (Hardy and Wright 1979, p. 63). \n\n\n\n\nTaken exactly from Wolfram.[/quote]\r\n\r\nyou forgot to specify that in the second formula, $ p$ is prime. \r\n\r\ndo NOT treat them as separate theorems though, one easily follows from the other", "Solution_19": "Yes, $ a^p\\equiv{a}\\pmod{p}$ is just a corallary of Fermat's Little Theorem, taken by multiplying by a. I just don't remember either of them and just go with the totient formula.", "Solution_20": "lol... $ a^{\\phi(p)}\\equiv1\\mod p$\r\n\r\nWhere $ \\phi$ represents the totient function... and $ p$ is any natural number so that $ (a,p)\\equal{}1$", "Solution_21": "Hmm...\r\n...and where did you learn this?\r\n\r\nBut, thanks! :D\r\nI would've never learned this on my own...:mad:", "Solution_22": "What's the totient function?", "Solution_23": "[quote=\"ZhangPeijin\"]What's the totient function?[/quote]\r\nBasically a generalization of FLT. Look it up.\r\n\r\nI think we should stop talking about this, it's not exactly Mathcounts level.", "Solution_24": "Actually, totient and fermat's can be applied to lots of mathcounts problems... I think. \r\n\r\nTotient function isn't a generalization of FLT, totient theorem is. The totient function is how many numbers less than a number are relatively prime to it.", "Solution_25": "Well, that's what I meant, considering the totient function appears in the theorem.\r\nYeah, I've used it for countless problems, such as target #7 on state last year." } { "Tag": [], "Problem": "(a) Compute the greatest value of $ x \\plus{} y$ for integers $ x,y$, such that\r\n\r\n$ x^2 \\plus{} y^2 \\equal{} 884$\r\n\r\n\r\n\r\n(b) Compute the greatest value of $ x \\plus{} y$ for integers $ x,y$, such that\r\n\r\n$ x^2 \\minus{} y^2 \\equal{} 884$", "Solution_1": "I don't really know whether this will work,but \r\n\r\n$ x^2 \\plus{} y^2$=884\r\n$ \\implies$ $ 2(x^2 \\plus{} y^2)$=1768\r\n\r\nor, $ (x \\plus{} y)^2 \\plus{} (x \\minus{} y)^2 \\equal{} 1768$\r\nor $ (x \\plus{} y)^2 \\plus{} (x \\minus{} y)^2 \\equal{} 1764 \\plus{} 4$\r\nor,$ (x \\plus{} y)^2 \\plus{} (x \\minus{} y)^2 \\equal{} 42^2 \\plus{} 2^2$\r\nor when $ (x \\minus{} y)^2 \\equal{} 4$, $ (x \\plus{} y)^2 \\equal{} 42^2$\r\nor, max. value of (x+y)=42.", "Solution_2": "[quote=\"bbgun34\"](a) Compute the greatest value of $ x \\plus{} y$ for integers $ x,y$, such that\n\n$ x^2 \\plus{} y^2 \\equal{} 884$\n\n\n\n(b) Compute the greatest value of $ x \\plus{} y$ for integers $ x,y$, such that\n\n$ x^2 \\minus{} y^2 \\equal{} 884$[/quote]\r\n\r\n[hide=\"a\"]We want $ x$ and $ y$ to be as close as possible to obtain the maximum possible value. The closest square to 442 is 441. So after a bit of brute force, the two integers whose squares add up to 884 are 20 and 22, which give a sum of 42.[/hide]\n\n[hide=\"b\"]This factors into $ (x\\minus{}y)(x\\plus{}y)\\equal{}884$. Since we want to maximize the $ (x\\plus{}y)$ term, we set $ x\\minus{}y\\equal{}1$ and $ x\\plus{}y\\equal{}884$. But this doesn't give us integer values for x and y, thus the next factors are $ x\\minus{}y\\equal{}2$ and $ x\\plus{}y\\equal{}442$. This does give us integer values $ (222,220)$, thus the max is $ 442$.[/hide]", "Solution_3": "(b) We can factor $ x^2\\minus{}y^2 \\equal{} 884$ into $ (x\\plus{}y)(x\\minus{}y) \\equal{} 884$. We now simply find the factors of 884 and plug them in. If we use 884 and 1 as the factors, $ x$ would not be an integer. If we go to the next set of factors, 2 and 442, we see $ x\\equal{}222$ which is an integer and therefore $ y$ must be an integer too. The maximum value that can be attained for $ (x\\plus{}y)$ is then $ \\boxed{442}$.\r\n\r\nI hate when someone replies while someone else is posting..." } { "Tag": [ "LaTeX" ], "Problem": "Hello\r\nI have previous never had any problems with references, but now for some reason they are not properly reference in the text. The caption of a figure or table is correct, i.e \"Table 1 shows...\" \"Table 2 is ...\". But in the text the \\ref{} shows only the section number; \"Table 4.2.1 shows\" \"Table 4.2.1 is....\"! I am using winedt, and maybe I have toggled the wrong button, but not that I know of. Can anyone help? \r\n\r\nRegards\r\nJT", "Solution_1": "Double-check that in each table, your \\label command comes after the \\caption command, but before the \\end{table} command. Having the \\label command in the wrong place often produces spurious references.", "Solution_2": "Thank you. I had my \\label before \\caption. Now it works if I compile twice. (I though WinEdt would compile the necessary number of times with just 1 click on the texify button..)\r\n\r\nJT" } { "Tag": [ "probability" ], "Problem": "How many different 4-digit numbers can be formed using the digits 2, 4, 5, 6, and 7 such that no digits repeat and the number is divisible by four?", "Solution_1": "A number is divisible by 4 only if the last two digits form a number divisible by four(e.g., 12, 24, etc.) Out of the five given digits, the possible last two digits are 24, 52, 56, and 64. \r\n\r\nNo matter what we choose as the last two digits, there are $ 3 \\cdot 2 = 6$ ways to choose the other two digits, so there are $ 6 \\cdot 4 = 24$ ways, out of $ 5 \\cdot 4 \\cdot 3 \\cdot 2 = 120$ total ways, giving us a probability of 1/5.", "Solution_2": "When I asked this question, this page was pulled up. The correct answer is given as 36.", "Solution_3": "1) The question did not ask for probability\r\n2) You missed the last two digits being 72 or 76\r\n\r\nso $ 6\\cdot 3!\\equal{}\\boxed{36}$", "Solution_4": "How did you get $6 \\cdot 3!$? I think the 6 is from the possible last digit combinations, but where does the $3!$ come from?", "Solution_5": "[quote=\"mathgenie\"]How did you get $6 \\cdot 3!$? I think the 6 is from the possible last digit combinations, but where does the $3!$ come from?[/quote]\n\nI think that the 3! is actually 3 chose 2 (cause there are 3 numbers left and and you want only 2 of them). Which equals 6.\n6*6=36", "Solution_6": "[b]Solution:[/b]\n\nThere are [u]6[/u] different types of endings:\n$24$\n$52$\n$72$\n$64$\n$56$\n$76$\n\nSince we want 4 digit numbers, [u]with no digit repeating[/u], there are $3\\cdot 2$ ways to place the first 2 numbers, nomatter the ending.\n\nSo, in total we have $6\\cdot 6=36$ different numbers." } { "Tag": [], "Problem": "images........", "Solution_1": "", "Solution_2": "" } { "Tag": [ "inequalities theorems", "inequalities" ], "Problem": "Does anyone has a proof for the Vasile's RCF LCF Theorem?\r\nI can't find it, Who has the proof please post a link here, please.\r\nThank you. :)", "Solution_1": "[quote=\"epsilonist\"]Does anyone has a proof for the Vasile's RCF LCF Theorem?\nI can't find it, Who has the proof please post a link here, please.\nThank you. :)[/quote]\r\nYou can find very nice proof in Vasc's book:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=100399" } { "Tag": [ "Harvard", "college", "MIT", "geometry", "ratio" ], "Problem": "For people who are going to private high schools next year(like myself), which are you considering greatly? I'm going to either Belmont Hill, Rox. Lat., or B.U.A.", "Solution_1": "I'm going to Belmont Hill. Hope to see you there!\r\n\r\n-jorian", "Solution_2": "Umm, why go to a private school when you could go to a beter Public school. Boston Latin and Lexington are way more 1337 than BUA. \r\nLexington more so for math. \r\nBLS more so for getting into Harvard :P w00t 16 people got in early, expecting 35 to make it overall.", "Solution_3": "[quote=\"jhredsox\"]I'm going to Belmont Hill. Hope to see you there!\n\n-jorian[/quote]\r\n\r\nOh no...", "Solution_4": "I'm getting the impression you don't like me :maybe: \r\n-jorian", "Solution_5": "Your powers of observation never cease to [b]ASTOUND[/b] me.", "Solution_6": "[quote=\"yunorman\"]Umm, why go to a private school when you could go to a beter Public school. Boston Latin and Lexington are way more 1337 than BUA. \nLexington more so for math. \nBLS more so for getting into Harvard :P w00t 16 people got in early, expecting 35 to make it overall.[/quote]\r\n\r\nYeah but...so many people go to Boston Latin, and so many are children of Harvard alumni. I think we still have a better ratio.\r\n\r\nAnyway, s private schools are not necessarily better for getting into colleges if you'd be a star in a public school and just top 30 or so in a private school.", "Solution_7": "Yes Lexington is pretty 1337. Well I would consider applying to PEA if my parents wanted to pay all the money, but I don't think private schools are a whole lot more of an advantage. \r\n\r\nI think Belmont Hill is overrated. A friend of mine started there in 7th grade and he told me it was a jock school and the kids there weren't that much smarter than the people at our middle school (then again, Lexington's schools are 1337)", "Solution_8": "dude... most BLS kids are not harvard alumni children. None of us who got into harvard were alumni children this year. and there were only 2 last year.", "Solution_9": "Lol my b. I just assumed it because of the Boston connection. There must be some legacy going on though?", "Solution_10": "Philips Exeter, which is a stronger and more selective high school than Boston Latin, has fewer of its students (both absolutely and per capita) attend Harvard than the admissions numbers from BLS. Assuming the huge yield rate for Harvard is at least as true at BLS as it is nationwide, the same should be true for attendance as well. I believe that the reverse is true for MIT admissions -- P.E. sends 10 or so people per year and I don't think BLS has anywhere near that success rate. This is evidence of a Harvard-BLS special relationship.\r\n\r\nhttp://www.exeter.edu/documents/Profile_2006_final.pdf\r\nhttp://www.bls.org/cfml/l3tmpl.cfm?page=Demographics", "Solution_11": "[quote=\"yunorman\"]dude... most BLS kids are not harvard alumni children. None of us who got into harvard were alumni children this year. and there were only 2 last year.[/quote]\r\n\r\nHarvard alumni tend not to live in neighborhoods served by BLS, except for Beacon Hill, some medical personnel living in areas near the hospitals (JP, etc), and a sprinkling of BU/Harvard/MIT academics living in Brighton. BLS is not as minority-heavy as it used to be before the anti-discrimination lawsuit, but as far as I know it is still more or less a representation of the city of Boston, a mostly middle-class cross-section of it with some upper- and working-class (largely immigrants) mixed in.", "Solution_12": "Boston Latin is considered a public school, correct?", "Solution_13": "Statiscal,\r\n\r\nBoston Latin school does have many students matriculate to MIT every year. Though we have far fewer applicants ap[plying to MIT than P.E. Pretty much most of the elite students get into Harvard Early and only those who have MIT as their top choice will bother with the MIT app after getting into Harvard. There is a \"connection\" between BLS and Harvard, but from first hand experience, 95% of the people who are admitted are absolutely amazing. This is probably true for P.E. [Some kid aren't so amazing... but get in through their own connections]", "Solution_14": "[quote=\"yunorman\"]Boston Latin school does have many students matriculate to MIT every year. [/quote]\n\nFor acceptance to every prestigious school *other than Harvard*, BLS is strongly outperformed by PE and PA. This is in line with the much greater objective strength of PE/PA in any measurable category such as SAT scores, number of college-level courses offered, level of faculty, size of library, teacher/student ratio, geographic reach, budget and endowment, athletics. For Harvard, BLS suddenly outperforms Philips. This reversal with Harvard indicates that it has some form of special preference for BLS that other universities do not. \n\nFor example, for MIT the BLS web site reports 12 matriculants over 4 years 2003-6 (3 per year) and 7 accepted last year. PE website shows 7 matriculated per year during the same time. One also expects more acceptances per matriculant at Exeter, because it is not a Boston-specific school, and as the web site data indicate, students at PE/PA have other comparable options even if accepted to Harvard/MIT, which is not (statistically) the case at BLS. In other words, one expects yield to be higher at BLS.\n\nFor all schools other than Harvard and MIT, PE/PA completely dominate compared to BLS and the discrepancy is clearer.\n\n[quote]There is a \"connection\" between BLS and Harvard, but from first hand experience, 95% of the people who are admitted are absolutely amazing. [/quote]\r\n\r\nBut are they amazing by BLS standards or Harvard standards? It can be argued whether Harvard's preferential admission of Boston Latin students is objectively reasonable (let's say, if BLS schooling were better adapted for success at Harvard than the schooling elsewhere), or a lowered admission standard due to the history and proximity." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let $a,b,c >0$ .Prove that :\r\n$\\frac{a^3}{ab+1}+\\frac{b^3}{bc+1}+\\frac{c^3}{ca+1}\\geq \\frac{(ab+bc+ca)^2(a+b+c)}{3(abc(a+b+c)+ab+bc+ca)}$.", "Solution_1": "This is a good one :D :D", "Solution_2": "Have you got a proof ,RaduSorici ? Can you post it ?", "Solution_3": "yes , a nice ineq :D \r\n\r\ni spent 2 hours solving it and found that cauchy is a good idea\r\n\r\nusing $\\sqrt{(ab+1)a}$\r\n\r\nor some number like this, i will complete it this afternoon :)", "Solution_4": "ok just prove this ineq: ;) \r\n\r\n$\\frac{3((xz)^2+(yz)^2+(xy)^2)^2}{(xyz)^2(3+xz+zy+xy)}\\geq\\frac{(yz+zx+xy)(x+y+z)^2}{xyz(x+y+z)+yz+zx+xy}$ :D", "Solution_5": "can you use caughy-schwarz theorm directly? :)", "Solution_6": "no , let $a=\\frac{1}{x}$..... :D \r\n\r\nafter that using CAUCHY_SCHWART ;) \r\n\r\nTHE REST 'S MAYBE EASY :blush:" } { "Tag": [ "factorial", "ratio", "geometric sequence", "AMC" ], "Problem": "The fifth and eighth terms of a geometric sequence of real numbers are $ 7!$ and $ 8!$ respectively. What is the first term?\r\n\r\n$ \\textbf{(A)}\\ 60\\qquad \r\n\\textbf{(B)}\\ 75\\qquad \r\n\\textbf{(C)}\\ 120\\qquad \r\n\\textbf{(D)}\\ 225\\qquad \r\n\\textbf{(E)}\\ 315$", "Solution_1": "[hide=\"Answer\"]\n$ 8!\\equal{}8\\cdot7!$ You had to multiply by three common ratios to get there, therefore $ r^3\\equal{}8 \\implies r\\equal{}2$ Since $ 7!$ is the fifth term, you have to divide $ r$ four times to get the first term, giving $ \\frac{7!}{2^4}\\equal{}315 \\implies \\boxed{(E)}$\n[/hide]" } { "Tag": [ "conics", "parabola", "algebra", "polynomial", "geometry", "projective geometry" ], "Problem": "Prove or disprove that there is a way to \"fit\" a circle in a parabola by having it be tangent at at least 3 points without any intersections other than the tangents. Intuition tells me it can't be possible because those three points define the circle, and can never define a parabola...", "Solution_1": "You can have a circle intersecting a parabola at exactly 3 points, for example the two equations $x^2+(y-1)^2=1$ and $x^2=y$ meet at (-1,1), (0,0), and (1,1). However, I think you may have meant the circle fits inside the parabola and the parabola doesn't ever go inside the circle, which would be impossible with 3 points of intersection. Not sure of the proof, however.", "Solution_2": "Yes, that's what I meant about no other intersections...except the tangents, wumbate. I want the circle to \"stay inside the parabola\".", "Solution_3": "Of course not. to have $n \\geq 3$ tangent points your polynomial needs to be of grade $n \\geq 3$ :D\r\n\r\n\r\nAnd anyway you could've tried projective geometry if all else fails :P" } { "Tag": [ "group theory", "abstract algebra", "topology", "linear algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "I'm struggling with the following:\r\n\r\na) determine the automorphisms Aut(Q) of the additive group\r\nb) determine the automorphisms Aut(Q*) of the multiplicative group\r\n\r\n\r\nas for a) the only thing I have managed to find so far is that there must be a subgroup H in Aut(Q) which is isomorphic to Q*, namely the automorphisms that multiply by a nonzero rational constant.\r\nAre there others?\r\n\r\nas for b) I have no idea altogether. x -> 1/x seems to be one. But there are certainly many, many more, aren't there.\r\n\r\nHow does one attack such a problem in general?", "Solution_1": "a)\r\nThere are no others, since the whole automorphism is defined by the mapping of any non-zero element.\r\nproof: WLOG assume that this element is $1$ and set $f(1) = a$. Then you get that also all multiplies (the integers) are defined by this as $f(n)=an$ and now you get $y\\cdot f(\\frac{x}{y}) = f(y\\cdot \\frac{x}{y})= x$ and so also all rational numbers are defined by this.\r\n\r\nb)\r\nThe map is exactly defined by the values the primes take. But it looks complicated to get all automorphisms...\r\nFor example all values taken on at primes must be 'squarefree' in the sense that every prime factor in a value occurs with power $0,1$ or $-1$. Also they must be 'multiplicativly independet' in the sense that every rational occurs only once to be bijective...", "Solution_2": "That's not exactly the answer on the multiplicative automorphisms- the images of primes can have square factors. For example, we can send 3 to 12 while preserving all other primes.\r\n\r\nAn equivalent form which might be easier to think about: Consider the $\\mathbb{Z}$-module of finitely supported integer sequences on countably many elements. The automorphisms we're looking for are the invertible linear transformations: each is a change of basis transformation.\r\n\r\nIf you care about the topology, the identity and inversion are the only continuous automorphisms.", "Solution_3": "thank you!\r\n\r\na) thanks, ZetaX. I had already known this idea for the integers, but I didn't know that it can be extended to all the rational numbers so easily.\r\n\r\nb) the automorphisms need not be continuous for my purposes. They just have to be automorphisms w.r.t. multiplication... I've understood that an automorphism is determined by its values for the prime integers...\r\nNow, what numbers exactly is it allowed to map prime integers to in an automorphism?\r\nDid I get your point right, jmerry, that there must be some kind of 'unique factorization property' for all integers together with the images of the primes?\r\nIsn't it that we CANNOT send 3 to 12 because then we inevitably lose the integers that are powers of 3 without having equally many powers of 4, for instance? As, to repair this, we would have to send another prime to 3 (and analogously some prime to 2) and doing this will destroy uniqueness of factorization...\r\n\r\nmore help still appreciated... :)", "Solution_4": "[quote=\"ido2005\"]Isn't it that we CANNOT send 3 to 12 because then we inevitably lose the integers that are powers of 3 without having equally many powers of 4, for instance?[/quote]\r\nRationals, not integers. $\\frac34$ is sent to 3. I could have sent 2 to $\\frac{-4954}{909}$ if I wanted to, fixing 3 and all other primes.\r\n\r\nOn the first pass, I neglected the negatives: you are free to choose $+$ or $-$ signs randomly for the images of primes. Formally, $\\mathbb{Q}^*$ is the direct product of $\\mathbb{Z}_2$ and a free abelian group on countably many variables. There is no simple description of the automorphisms of the second structure, although you can translate to the group of finitely supported integer sequences and use linear algebra language." } { "Tag": [ "algebra", "polynomial", "search", "algebra unsolved" ], "Problem": "Let $ P(x) \\in \\mathbb{R}[X]$ a polynomial such that $ P(t) \\in \\mathbb{Q} \\leftrightarrow t \\in \\mathbb{Q}, \\forall t \\in \\mathbb{R}$.\r\n\r\nShow that $ \\exists (a,b) \\in \\mathbb{Q}^2$ such that $ P(x) \\equal{} ax \\plus{} b$.", "Solution_1": "http://www.mathlinks.ro/viewtopic.php?search_id=1506724380&t=237911" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let $ a,b,c$ be length of edges of a triangle.Prove that\r\n$ 10(a^2b\\plus{}b^2c\\plus{}c^2a)\\plus{}4abc \\ge 6(ab^2\\plus{}bc^2\\plus{}ca^2)\\plus{}(a^2\\plus{}b^2\\plus{}c^2)(a\\plus{}b)(b\\plus{}c)(c\\plus{}a)$", "Solution_1": "There is something wrong here.", "Solution_2": "Sorry,but $ ab \\plus{} bc \\plus{} ca \\equal{} 1$ :( I cann't edited my post. Thank you,hungkhtn!" } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Find all functions $ f: \\mathbb R^ \\plus{} \\rightarrow \\mathbb R^ \\plus{}$ satisfying $ f(x^y) \\equal{} (f(x))^{f(y)}$ for every $ x,y\\in\\mathbb R^ \\plus{}$", "Solution_1": "[hide]Replace $ y$ by $ mn$ and compute $ f(x^y)$ in two different ways.\n\nWhat do you conclude?\n\nThe used property here is $ a^{(bc)} \\equal{} (a^b)^c$. What other properties of multiplication can you use?\n\nWhat do you conclude with these properties?[/hide]", "Solution_2": "[hide=\"Solution!\"]$ f(1) \\equal{} f(1^1) \\equal{} f(1)^{f(1)}$, so $ f(1) \\equal{} 1$.\n\n$ f(x) \\equal{} 1$ is clearly a solution, so from now on assume $ f(x_0) \\ne 1$ for some $ x_0$. Then $ f(x_0)^{f(yz)} \\equal{} f(x_0^{yz}) \\equal{} f(x_0^y)^{f(z)} \\equal{} f(x_0)^{f(y)f(z)}$, so $ f(yz) \\equal{} f(y)f(z)$.\n\nAlso, $ f(x_0)^{f(y \\plus{} z)} \\equal{} f(x_0^{y \\plus{} z}) \\equal{} f(x_0^y x_0^z) \\equal{} f(x_0^y) f(x_0^z)$ $ \\equal{} f(x_0)^{f(y)} f(x_0)^{f(z)} \\equal{} f(x_0)^{f(y) \\plus{} f(z)}$, so $ f(y \\plus{} z) \\equal{} f(y) \\plus{} f(z)$.\n\nIf $ x > y$, we can write $ x \\equal{} y \\plus{} z$ for some $ z > 0$, and then $ f(x) \\equal{} f(y) \\plus{} f(z) > f(y)$, so $ f(x)$ is increasing. Thus $ f(x) \\equal{} x^c$ for some constant $ c$ (because if $ f(x) \\equal{} x^c$, $ f(y) \\equal{} y^d$, then for any integers $ a, b$ with $ x^a < y^b$, $ x^{ac} \\equal{} f(x^a) < f(y^b) \\equal{} y^{bd}$ since $ f(x)$ is multiplicative, so $ c \\le d$, and similarly $ c \\ge d$ so $ c \\equal{} d$).\n\nFrom $ f(2)^2 \\equal{} f(2*2) \\equal{} f(2^2) \\equal{} f(2)^{f(2)}$, we get $ f(2) \\equal{} 1 \\equal{} 2^0$ or $ f(2) \\equal{} 2 \\equal{} 2^1$, so $ c$ is either $ 0$ or $ 1$, and $ f(x)$ is either $ 1$ or $ x$.[/hide]\r\n\r\nEdit: Oh, ninja'd.\r\n\r\nAlso, hide tags look like a good idea.", "Solution_3": "[quote=\"zeb\"]\nThus $ f(x) \\equal{} x^c$ for some constant $ c$ (because if $ f(x) \\equal{} x^c$, $ f(y) \\equal{} y^d$, then for any integers $ a, b$ with $ x^a < y^b$, $ x^{ac} \\equal{} f(x^a) < f(y^b) \\equal{} y^{bd}$ so $ c \\le d$\n[/quote]\r\nWhy?", "Solution_4": "[quote=\"thaithuan_GC\"][quote=\"zeb\"]\nThus $ f(x) \\equal{} x^c$ for some constant $ c$ (because if $ f(x) \\equal{} x^c$, $ f(y) \\equal{} y^d$, then for any integers $ a, b$ with $ x^a < y^b$, $ x^{ac} \\equal{} f(x^a) < f(y^b) \\equal{} y^{bd}$ so $ c \\le d$\n[/quote]\nWhy?[/quote]\r\n\r\nThis is easier to explain if we first assume $ x,y > 1$...\r\n\r\nLet $ \\frac {ln(y)}{ln(x)} \\minus{} \\epsilon < \\frac {a}{b} < \\frac {ln(y)}{ln(x)}$, then from $ x^a < y^b$ we get $ x^{ac} < y^{bd}$ so $ \\frac {ac}{bd} < \\frac {ln(y)}{ln(x)}$ so $ \\frac {c}{d} < \\frac {\\frac {ln(y)}{ln(x)}}{\\frac {a}{b}} < \\frac {\\frac {ln(y)}{ln(x)}}{\\frac {ln(y)}{ln(x)} \\minus{} \\epsilon}$, true for all $ \\epsilon > 0$, so $ \\frac {c}{d} \\le 1$.\r\n\r\nEdit: Ok, there's a simpler way to do it once we know it's multiplicative, additive (unless $ f \\equal{} 1$), and increasing (unless $ f \\equal{} 1$) - we know that $ f$ acts like the identity on the rationals (unless of course $ f \\equal{} 1$), and it's increasing. Whatever. I thought of this way first.", "Solution_5": "OK, thanks for your solution." } { "Tag": [ "geometry", "trapezoid", "ratio", "geometric transformation", "dilation", "rotation", "geometry proposed" ], "Problem": "Let ABCD be an orthodiagonal trapezoid (AB || CD), O the common point of its diagonals, M and N the projections of O onto AB, CD respectively, and Q the orthocenter of the triangle ABN.\r\nProve that the perpendiculars from D onto AN and from C onto BN concur at P on MN and NP = OQ.\r\n\r\nBest regards,\r\nsunken rock", "Solution_1": "what is orthodiagonal", "Solution_2": "To: md45\r\n\r\nOrthodiagonal = having perpendicular diagonals.\r\n\r\nBest regards,\r\nsunken rock", "Solution_3": "It is easy, based on the below Lemma (well known and easy to prove).\r\n\r\n[b][size=100][color=DarkBlue]LEMMA. - A triangle $ \\bigtriangleup ABC$ is given and let $ AD$ be, its altitude. We draw the cemicircle $ (K),$ with diameter the side-segment $ BC,$ which intersects the line segment $ AD,$ at one point so be it $ Z.$ Prove that $ (AD)\\cdot (HD) \\equal{} (ZD)^{2},$ where $ H$ is the orthocenter of $ \\bigtriangleup ABC$[/color][/size][/b].\r\n\r\nKostas Vittas.", "Solution_4": "[hide]We draw a line through the point $ O$ and parallel to the bases $ AB,\\ CD,$ of the given orthodiagonal trapezoid $ ABCD,$ which intersects the line segments $ BQ,\\ AQ,$ at points $ E,\\ F,$ respectively.\n\nIt is enough to prove that $ OE \\equal{} ND$ and similarly $ OF \\equal{} NC,$ which is equivalent to prove, by [b][size=100]Thales theorem[/size][/b], that $ \\frac {MQ}{QO} \\equal{} \\frac {MO}{ON}$ $ \\Longrightarrow$ $ (MQ)\\cdot (MN) \\equal{} (MO)^{2}$ $ ,(1)$\n\nBut $ (1)$ is true, because of the above [b][size=100]Lemma[/size][/b] $ ($ $ Q$ is the orthocenter of the triangle $ \\bigtriangleup NAB$ and the point $ O,$ lies on the semicircle $ (K),$ with diameter its side-segment $ AB$ $ ).$ [/hide]\r\nKostas Vittas.", "Solution_5": "The aim of this problem was to emphasyze the following\r\n[b]Lemma[/b]\r\n[i]Let D be the projection of A onto BC, E the projection of D onto AC and P a point on (DE) such as PE/PD=BD/DC ( * ), ABC being a triangle. Then AP and BE are perpendicular[/i]\r\n(It appeared on this forum some time ago for and isosceles triangle, this being a sort of extention).\r\n\r\nProof: Let R be the projection of B onto AC. Then triangles ADE and BCR are similar, E and P divide similar sides to same ratio, therefore they are similar, and so are AP and BE. As the subject triangles have the correspondent sides perpendicular, it follows that AP and BE are perpendicular too.\r\nFrom the proof we can see that the said lines are perpendicular iff equality (*) holds.\r\n\r\nNow the problem: O is the projection of D onto AC, N the one of O onto CD, so the perpendicular from D onto AN divides ON at the ratio NP'/OP'=OA/OC ( according to (*) ), P' being the intersection of this perpendicular with ON; analogously, the perpendicular from C onto BN divides ON at the ratio NP\"/OP\"=OB/OD, so P\"=P'=P and NP/PO=OA/OC ( 1 ). Then see that the triangles OCP and OAQ are similar, consequently OQ/OP=OA/OC ( 2 ), from (1) and (2) getting OQ = NP.\r\n\r\nBest regards,\r\nsunken rock\r\n\r\n[/b]", "Solution_6": "[quote=\"sunken rock\"]The aim of this problem was to emphasyze the following\n[b]Lemma[/b]\n[i]Let D be the projection of A onto BC, E the projection of D onto AC and P a point on (DE) such as PE/PD=BD/DC ( * ), ABC being a triangle. Then AP and BE are perpendicular[/i]\n(It appeared on this forum some time ago for and isosceles triangle, this being a sort of extention).\n\nProof: Let R be the projection of B onto AC. Then triangles ADE and BCR are similar, E and P divide similar sides to same ratio, therefore they are similar, and so are AP and BE. As the subject triangles have the correspondent sides perpendicular, it follows that AP and BE are perpendicular too.\nFrom the proof we can see that the said lines are perpendicular iff equality (*) holds.\n\nNow the problem: O is the projection of D onto AC, N the one of O onto CD, so the perpendicular from D onto AN divides ON at the ratio NP'/OP'=OA/OC ( according to (*) ), P' being the intersection of this perpendicular with ON; analogously, the perpendicular from C onto BN divides ON at the ratio NP\"/OP\"=OB/OD, so P\"=P'=P and NP/PO=OA/OC ( 1 ). Then see that the triangles OCP and OAQ are similar, consequently OQ/OP=OA/OC ( 2 ), from (1) and (2) getting OQ = NP.\n\nBest regards,\nsunken rock\n\n[/b][/quote]\r\nDear [b]sunken rock[/b] and dear [b]vittasko[/b]: nice solutions! I have a slightly different solution for the concurrency, however:\r\n[hide=\"Solution\"]\nLet the feet of the perpindiculars of $ D$ and $ C$ on $ AN$ and $ BN$ be $ X$ and $ Y$ respectively and let $ DX$ and $ CY$ meet at $ R$. It follows that $ \\angle ABO \\equal{} \\angle BDC$ since $ AB\\parallel DC$. Also, $ DX\\perp AN$ and $ BQ\\perp AN$, so $ BQ\\parallel DX\\implies \\angle OBQ \\equal{} \\angle ODX$. We thus find that $ \\angle ODC$ and $ \\angle ABO$ are equal and are partitioned into the same angles as each other by $ DX$ and $ BQ$ respectively. Similarly, $ \\angle OCD$ and $ \\angle OAB$ are equal and they are partitioned into the same angles by $ CY$ and $ BQ$ respectively. Thus, after a dilation around $ O$, with $ 180$ degrees, with a scale factor of $ \\frac {OC}{AO}$ maps $ \\triangle AOB$ to $ \\triangle COD$ since $ \\triangle AOB\\sim \\triangle COD$ since $ AB\\parallel CD$. Thus, line $ AQ$ maps to line $ CY$ and line $ BQ$ maps to line $ DX$, so after a $ 180$ degree rotation about $ O$, $ Q$ maps to $ R$. This means that $ Q$, $ O$, and $ R$ are collinear, so $ R$ lies on $ MN$, which means that $ R \\equal{} P$, as desired. [/hide]", "Solution_7": "[quote=\"sunken rock\"][color=darkred]Let $ ABCD$ be an orthodiagonal trapezoid ( $ AB\\parallel CD$ ), the common point $ O$ of its diagonals, the projections $ M$ , $ N$ of $ O$ onto $ AB$ , $ CD$ respectively and the orthocenter $ Q$ of the triangle $ ABN$ . Prove that the perpendiculars from $ D$ onto $ AN$ and from $ C$ onto $ BN$ concur at $ P$ on $ MN$ and $ NP = OQ$ . [/color][/quote]\n[color=darkblue][b][u]Proof.[/u][/b] \n\n$ \\blacktriangleright$ I\"ll apply the [b]Carnot's lemma[/b] to $ \\triangle ABN$ and the points $ M$ , $ C$ , $ D$ :\n\n$ (MA^2 - MB^2) + (CB^2 - CN^2) + (DN^2 - DA^2) =$ $ (OA^2 - OB^2) + CB^2 + (ND^2 - NC^2) - AD^2 =$\n\n$ (OA^2 - OB^2) + (OB^2 + OC^2) + (OD^2 - OC^2) - (OA^2 + OD^2) = 0$ . In conclusion,\n\nthe perpendiculars from the points $ M$ , $ C$ , $ D$ onto the lines $ AB$ , $ BN$ , $ NA$ are concurrently ( in $ P$ ).\n\n$ \\blacktriangleright$ $ \\triangle AQB\\ \\sim\\ \\triangle CPD\\ \\implies\\ \\frac {PN}{MQ} = \\frac {CD}{AB} = \\frac {ON}{OM}$ . Thus, $ MO^2 = MA\\cdot MB = MQ\\cdot MN$ $ \\implies$\n\n$ \\frac {MN}{OM} = \\frac {OM}{QM}$ $ \\implies$ $ 1 + \\frac {ON}{OM} = 1 + \\frac {OQ}{QM}$ $ \\implies$ $ \\frac {ON}{OM} = \\frac {OQ}{QM}$ $ \\implies$ $ \\frac {PN}{QM} = \\frac {OQ}{QM}$ $ \\implies$ $ \\boxed {\\ PN = OQ\\ }$ .[/color]\n\n[quote=\"Virgil Nicula\"][color=darkred][b][u]An equivalent enunciation.[/u][/b] Let $ ABC$ be an acute triangle with orthocenter $ H$ . Denote the point $ L\\in (AH)$ \n\nand which belongs to the circle with diameter $ [BC]$ . Construct the points $ \\{\\begin{array}{c} M\\in BL\\ ,\\ MA\\parallel BC \\\\\n \\\\\nN\\in CL\\ ,\\ NA\\parallel BC\\end{array}$ \n\nand the intersection $ K$ of the perpendiculars from $ M$ , $ N$ onto $ AC$ , $ AB$ . Prove that $ K\\in AH$ and $ AK = LH$ .[/color][/quote]\r\n[b][u]Proof.[/u][/b] \r\n\r\n$ \\blacktriangleright\\ \\{\\begin{array}{ccc} NK\\perp AB & \\implies & NA^2 - NB^2 = KA^2 - KB^2 \\\\\r\n \\\\\r\nMK\\perp AC & \\implies & MC^2 - MA^2 = KC^2 - KA^2 \\\\\r\n \\\\\r\nAL\\perp MN & \\implies & AM^2 - AN^2 = LM^2 - LN^2 \\\\\r\n \\\\\r\nLM\\perp LC & \\implies & LM^2 + LC^2 = MC^2 \\\\\r\n \\\\\r\nLN\\perp LB & \\implies & NB^2 = LN^2 + LB^2\\end{array}\\|\\ \\bigoplus$ $ \\implies$ \r\n\r\n$ KC^2 - KB^2 = LC^2 - LB^2$ $ \\implies$ $ KL\\perp BC$ $ \\implies$ $ \\boxed {\\ K\\in AH\\ }$ . Denote $ D\\in BC$ for which $ HD\\perp BC$ .\r\n\r\n$ \\blacktriangleright\\ \\{\\begin{array}{ccccc} \\triangle KAM\\sim\\triangle HDB & \\implies & \\frac {KA}{HD} = \\frac {AM}{DB} & \\implies & \\frac {KA}{HD} = \\frac {LA}{LD} \\\\\r\n \\\\\r\nLD^2 = DB\\cdot DC = DH\\cdot DA & \\implies & \\frac {DA}{LD} = \\frac {LD}{DH} & \\implies & \\frac {LA}{LD} = \\frac {HL}{DH}\\end{array}\\|$ $ \\implies$\r\n\r\n$ \\frac {KA}{HD} = \\frac {HL}{HD}$ $ \\implies$ $ \\boxed {\\ KA = HL\\ }$ ." } { "Tag": [ "inequalities", "algebra", "polynomial", "calculus", "calculus computations" ], "Problem": "$ \\parallel{}p(x)^k\\parallel{}\\ \\parallel{}p(x)\\parallel{} \\leq \\parallel{}p(x)^{k\\plus{}1}\\parallel{}$\r\nIs this true?", "Solution_1": "Any ideas whatsoever?", "Solution_2": "What is \"height\" in this case?", "Solution_3": "Hey!\r\n\r\nIn this case: http://mathworld.wolfram.com/PolynomialHeight.html\r\n\r\nEdit: Maybe it is just true in the real case and not for complex coefficients.", "Solution_4": "This is just false: take $ k\\equal{}1$ and $ p(x)\\equal{}x^2\\minus{}10x\\minus{}1$." } { "Tag": [ "inequalities", "calculus", "algebra unsolved", "\\/closed" ], "Problem": "At some point tonight, a whole bunch of forums started showing 0 threads. The threads are still there, and in some cases still accessible; [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=313109]this thread[/url] is reachable as the newest post in its section, even though the unsolved and proposed subforum shows no posts. The test forum shows no posts from the outside, but shows 50 topics when you actually go there. \r\n\r\nForums definitely affected: MS Classroom, MS Contests, HS Basics, HS Intermediate, HS Pre-Olympiad, Algebra Unsolved, Inequalities Proposed, NT Proposed, Calculus U&P, Calculus T&F, Linear Alg U&P, Superior Alg U&P, Intro. Physics, Test forum.", "Solution_1": "I thought it only happened to me..\r\n\r\nPlease fix this soon..", "Solution_2": "Should be fixed now." } { "Tag": [ "geometry", "parallelogram", "power of a point" ], "Problem": "[i][color=red]Problem 1[/color][/i]\r\n\r\nIn the first shape $KB ,KF$ are perpedicular to $CB,CF$ respectively. $EB//FD$ .Proove that $A$ is the midpoint of $FD$\r\n\r\n[i][color=red]Problem 2[/color][/i]\r\n\r\nIn the second shape ABCD is a parallelogram with $AB=2BC$\r\n$BE$ is perpedicular to $AD$.Proove that $EMC=3DEM$(angel)", "Solution_1": "Nice.\r\n\r\n[u][b]First problem: [/b][/u]Call $\\left\\{S\\right\\}=BF \\cap DE$.\r\n$BF$ is the polar of $D$ wrt the circle, so $E$ and $C$ are harmonically divided by $P$ and $D$.\r\n$\\Rightarrow BE, BF,BC,BD$ is a harmonic bundle. \r\n\r\nBecause $DF \\parallel BE$, their intersection point lies at infinity.\r\nHence it has to be true that $\\left\\{A\\right\\}=BC\\cap DF$ is the midpoint of $\\left[DF\\right]$.", "Solution_2": "A question about prob 1: Isn't $K$ on the circle, because $C$ is on the circle? Then, because of the similarity of $\\triangle{BCE}$ and $\\triangle{ACD}$, and Power of a Point, it reduces to showing that $2BC=EC$, which clearly isn't always true...", "Solution_3": "[u][b]Second Problem:[/b][/u] \r\nWe can easiley see that $\\vec{AM}= \\vec{AD}+\\vec{DM}$ and $\\vec{MB}=\\vec{MC}+\\vec{CB}$\r\nThis leads us to $\\vec{AM}.\\vec{MB}=\\left(\\vec{AD}+\\vec{DM}\\right).\\left(\\vec{MC}+\\vec{CB}\\right)$\r\n\r\nNow because $\\vec{CB}=-\\vec{AD}$ and $\\vec{DM}=\\vec{MC}$, we can see that the above expression equals $\\vec{DM}^{2}-\\vec{AD}^{2}=0$\r\nHence $AM \\bot MB$.\r\n\r\nNow, clearly $BM$ bisects $\\angle ABC$ (Proof: $\\left|MC\\right|=\\left|CB\\right|$ so $\\angle MBC = \\frac{\\pi-\\angle BCM}{2}= \\frac{1}{2}\\angle ABC$)\r\nThis gives us $\\angle AEM = \\angle ABM$.\r\n\r\nNow, construct point $G$ on $\\left[AB\\right]$ such that $GM \\parallel BC$.\r\nWe see that $\\angle EMC = \\angle GMC+\\angle GME = 2\\angle AEM+\\angle AEM$\r\n\r\nDone!", "Solution_4": "Done! :wink:", "Solution_5": "Amen.\r\n\r\nK81o7: I think giannis meant that K is the center of the circle, I don't understand your comment. Please look at the diagram", "Solution_6": "[quote=\"Jan\"]Amen.\n\nK81o7: I think giannis meant that K is the center of the circle, I don't understand your comment. Please look at the diagram[/quote]\r\n\r\nI have already looked at the diagram: that's where the confusion arose. The information given implies KBCF is cyclic. So if B, C, F are all on the circle, K is too.", "Solution_7": "I think there is a typo in the question, but anyway, it's clear what he means. CP,CF should be replaced with DP,DF. :wink: \r\nHe just means that DP,DF are tangent to the circle. If my interpretation is incorrect, please let me know giannis and I'll try to find a solution to the new problem, but I think it makes sense this way and it is a nice and easy problem.", "Solution_8": "Very nice solutions Jan , and yours notes are correctly. \r\nA have another solution using similarity. :)" } { "Tag": [ "calculus", "integration", "function", "trigonometry", "calculus computations" ], "Problem": "Find the constants $ a,\\ b,\\ c$ such that a function $ f(x)\\equal{}a\\sin x\\plus{}b\\cos x\\plus{}c$ satisfies the following equation for any real numbers $ x$.\r\n\r\n\\[ 5\\sin x\\plus{}3\\cos x\\plus{}1\\plus{}\\int_0^{\\frac{\\pi}{2}} (\\sin x\\plus{}\\cos t)f(t)\\ dt\\equal{}f(x).\\]", "Solution_1": "hello kunny, do you mean\n$5\\sin(x)+3\\cos(x)+1+\\int_{0}^{\\frac{\\pi}{2}}(\\sin(t)+\\cos(t))f(t)\\,dt=f(x)$?\ninstead of\n$ 5\\sin x+3\\cos x+1+\\int_{0}^{\\frac{\\pi}{2}}(\\sin x+\\cos t)f(t)\\ dt=f(x).$\nSonnhard.", "Solution_2": "One solution\n$5\\sin x + 3\\cos x + 1 + \\int_0^{\\frac{\\pi }{2}} ( \\sin x + \\cos t)f(t)\\;dt = f(x) \\Leftrightarrow $\n\n$5\\sin x + 3\\cos x + 1 + \\sin x\\int_0^{\\frac{\\pi }{2}} {f\\left( t \\right)} dt + \\int_0^{\\frac{\\pi }{2}} {\\cos t} f(t)\\;dt = f(x) \\Leftrightarrow $\n\n$\\left( {5 + \\int_0^{\\frac{\\pi }{2}} {f\\left( t \\right)} dt} \\right)\\sin x + 3\\cos x + 1 + \\int_0^{\\frac{\\pi }{2}} {\\cos t} f(t)\\;dt = a\\sin x + b\\cos x + c$\n\nBut\n\n\\[\\int_0^{\\frac{\\pi }{2}} {\\cos t} f(t)\\;dt = \\int_0^{\\frac{\\pi }{2}} {\\cos t} \\left( {a\\sin t + b\\cos t + c} \\right)\\;dt = \\]\n\n\\[c + \\frac{a}{2}\\int_0^{\\frac{\\pi }{2}} {\\sin 2tdt} + \\frac{b}{2}\\int_0^{\\frac{\\pi }{2}} {\\left( {1 + \\cos 2t} \\right)dt} = c + \\frac{a}{2} + \\frac{{b\\pi }}{4}\\]\n\nand\n\\[\\int_0^{\\frac{\\pi }{2}} {f\\left( t \\right)} dt = \\int_0^{\\frac{\\pi }{2}} {\\left( {a\\sin x + b\\cos x + c} \\right)} dt = a + b + \\frac{{\\pi c}}{2}\\]\n\nSo\n$\\left( {5 + a + b + \\frac{{\\pi c}}{2}} \\right)\\sin x + 3\\cos x + 1 + \\frac{a}{2} + \\frac{{b\\pi }}{4} + c = a\\sin x + b\\cos x + c \\Leftrightarrow $\n\n$\\left( {5 + b + \\frac{{\\pi c}}{2}} \\right)\\sin x + 3\\cos x + 1 + \\frac{a}{2} + \\frac{{b\\pi }}{4} = b\\cos x \\Leftrightarrow $ , for every real x\n\nFor \n$\\left. {\\begin{array}{*{20}{c}}\n {x = 0:4 + \\frac{a}{2} + \\frac{{b\\pi }}{4} = b} \\\\\n {x = \\pi : - 2 + \\frac{a}{2} + \\frac{{b\\pi }}{4} = - b} \\\\\n {x = \\frac{\\pi }{2}:6 + b + \\frac{{\\pi c}}{2} + \\frac{a}{2} + \\frac{{b\\pi }}{4} = 0} \\\\\n\\end{array}} \\right\\} \\Leftrightarrow \\begin{array}{*{20}{c}}\n {b = 3} \\\\\n {c = - \\frac{{16}}{\\pi }} \\\\\n {a = - \\frac{{3\\pi }}{2} - 2} \\\\\n\\end{array}$\n\nSo we have\n$f\\left( x \\right) = - \\left( {\\frac{{3\\pi }}{2} + 2} \\right)\\sin x + 3\\cos x - \\frac{{16}}{\\pi }$\nthe only thing left is to check if it satisfies the ypothesis" } { "Tag": [ "Gamebot" ], "Problem": "My classroom has 11 rows of chairs, with 11 chairs in each row. The chairs in each row are numbered from 1 to 11. How many chairs have odd numbers?", "Solution_1": "Each row has $ 6$ odd numbered chairs (1,3,5,7,9,11), and there are $ 11$ rows, so $ 6\\cdot 11\\equal{}\\boxed{66}$", "Solution_2": "how do you know its 6 odd numbered chairs.\n Thank you", "Solution_3": "... Why is this in the multiplication principle in counting?", "Solution_4": "[quote=Tacy]how do you know its 6 odd numbered chairs.\n Thank you[/quote]\n\nbecause you count 1,3,5,7,9,and 11 :-D ", "Solution_5": "[quote=AlexLikeMath]... Why is this in the multiplication principle in counting?[/quote]\n\nI had the same question. \n\n[hide=Answer to @AlexLikeMath]Maybe because you could realize that there are $11$ rows of $6$ odd numbers so you could multiply them to get $\\boxed{66}$. [/hide]\n\nThis was a relatively easy question.\n\n[hide=Problem Vitals]\nLevel 11 PA\nLevel 3 C&P\n[/hide]", "Solution_6": "[hide=Sol]From 1 to 11, the odd numbers are 1, 3, 5, 7, 9, and 11. Since there are 11 rows, 6x11 is 66.[/hide]", "Solution_7": "Here is[hide=my solution]There are 11 rows, each with 11 chairs. But only 6 of those chairs are odd numbers. So it is 6 $\\cdot$ 11 = $\\boxed{66}$ chairs.[/hide]. Here is a [hide=mistake]There are 11 rows, 11 columns, and 11 numbers, but only 6 of those are odd. So we do 11 $\\cdot$ 11 $\\cdot$ = $\\boxed{726}$.[/hide]. Here is why it[hide=does not work] This does not work, because the 11 chairs in each row [i]are[/i] the chairs numbered 1 to 11. So we do 11 $\\cdot$ 6, not 11 $\\cdot$ 11 $\\cdot$ 6.[/hide].", "Solution_8": "[hide=My reasoning][tip=Hover plz][hide=Just to be safe *thumbsup*]So since there are $6$ odd numbers from $1$ through $11$, there are $6$ odd numbers per row. Which means there should be $6\\cdot11$ odd seats. Therefore, the answer is $\\boxed{66}$. :D[/hide][/tip][/hide]", "Solution_9": "when you forget that 9 is odd because you just had a problem about the first 5 odd primes :(" } { "Tag": [ "probability", "function", "probability and stats" ], "Problem": "NO1.Let X,Y be a random variable defined as follows: \r\nX and Y are real valued functions on a sample space W ={(x,y): 0<=x<=1 & 0<=y<=1}, where a probability is given by width, X satisfies \r\nX((x,y))=8x for all (x,y) in W. \r\nand Y satisfies \r\nY((x,y))=7(x+y) for all (x,y) in W. \r\nIf F is a joint probability distribution of X,Y, anser the value F3,2 , down to two places of decimals. \r\n\r\nNo.\uff12\u3000Let X,Y be a random variable defined as follows: \r\nX and Y are real valued functions on a sample space W ={(x,y): 0<=x<=1 & 0<=y<=1}, where a probability is given by width, X satisfies \r\nX((x,y))=9x for all (x,y) in W. \r\nand Y satisfies \r\nY((x,y))=9(x+y) for all (x,y) in W. \r\nIf p(x,y) is a joint probability density of X,Y, anser the value p(1,2), down to two places of decimals. \r\n\r\nNO\uff13\r\nLet Y be a random variable defined as follows: \r\nY is a real valued function on a sample space W ={(x,y): 0<=x<=1 & 0<=y<=1}, where a probability is given by width, and Y satisfies \r\nY((x,y))=10(x+y) for all (x,y) in W. \r\nanser the variance V(Y), down to two places of decimals. \r\n\r\nplease tell me how to sove.", "Solution_1": "These all seem like pretty much the same problem. Recall that if a joint density function is given by $ f_{X,Y}(x,y)$, then the joint probability is $ \\iint_{D}f_{X,Y}(x,y)\\,dx\\,dy$, where $ D$ is the region you're considering." } { "Tag": [ "function", "articles" ], "Problem": "http://www.math.upenn.edu/~wilf/DownldGF.html\r\ni read a few pages of this. i didnt undrestand it completly. can someone please explain the example in page 4 and give some easier examples than that ?", "Solution_1": "Here's a clearer solution to the problem.\r\n\r\nLet us take the 1 and put it into the series, with:\r\nA(n) +1 = 2(A(n-1)+1)\r\nThen let's define the sequence B, so that\r\nB(n) = A(n) +1\r\nand notice that B(n) = 2(B(n-1))\r\nand the sequence is 1,2,4,8,..., so we can give the solution immediately.\r\n\r\nThere is a good article about Generating Functions at\r\n[url]http://en.wikibooks.org/wiki/High_School_Mathematics_Extensions/Counting_and_Generating_functions[/url]" } { "Tag": [ "MATHCOUNTS", "probability", "AMC", "AIME", "FTW" ], "Problem": "discuss your individual and team rankings and scores here.\r\n\r\nI got a 22/14=36, putting me in 20th place. Dang I was just 1 point off from cd qualification :( just b/c of one stupid mistake on the sprint round :mad:\r\n\r\nMy team took 14th with a team score of 45. Apparently we screwed up team round and got an 8.\r\n\r\nIndividual team members placed 62nd, 100th, and 102nd.", "Solution_1": "where did u get results?", "Solution_2": "Well I think they tell you. But they tell you the team this afternoon.", "Solution_3": "The written results were all released this morning. My coach received them.", "Solution_4": "silly mistakes are costly.", "Solution_5": "could u plz tell me the al individual and team results?", "Solution_6": "The coaches already got them this morning!? \r\nThe last 3 years I thought they were picked up a couple hours before the ceremony, after touring all day?", "Solution_7": "who won masters?", "Solution_8": "I've somehow heard Max Schindler won?", "Solution_9": "they don't announce it until the awards.", "Solution_10": "Right, that's the only one that's not already known (though after watching there are usually 2 clear favorites)", "Solution_11": "[quote=\"gauss1181\"]I've somehow heard Max Schindler won?[/quote]\r\n\r\nCan you confirm all the other results that jmerry has posted in the other thread? Thanks.", "Solution_12": "So apparently, MathCounts was around the same difficulty as last year but without so much COUNTING AND PROBABILITY problems? My 3/8 on the target last year was deadly :rotfl:", "Solution_13": "haha i got 69th place individual with a 20/8=28 i think\r\n\r\nour team got 8th overall and others placed 4th, 36th (i think), and 52nd", "Solution_14": "ugh. 14/12=26, putting me at 94th. I'm so pissed because on target, i KNOW i could have gotten a perfect. it was so easy. i misread one of them and needed like 30 more seconds for another. i probably could have done better at sprint too, but oh well.\r\n\r\nhaha our team got 30th (delaware), but we could have done soo much better. cause my other teammates also made stupid mistakes (like a lot of them) and on team round for the 1.499 one, i rounded it to the nearest tenth :mad: i know i'm stupid haha.", "Solution_15": "Hmmm... got 28/14... darn, if I had not misread #5 target I would be first written.... darn that just sucks...\r\n\r\nAlso don't watch the webcast for 2 reasons:\r\n\r\n* pythag011 ruins it\r\n* I failed....\r\n\r\n :( :( :(", "Solution_16": "[quote=\"dnkywin\"]\n* pythag011 ruins it\n[/quote]\r\n\r\nwell i just watched it haha. maybe his parents should drop him in a public school or something, get some social skills. it's too bad - some of those things he did we just completely inappropriate. he's going to get really ragged on aops.\r\n\r\nand btw - you didn't fail (how does that make the rest of us feel? lol)", "Solution_17": "Yeah seriously if anyone failed countdown it's I. I didn't even buzz in... :| \r\n\r\nBut the test was not legitimate in my opinion. I got really lucky and somehow made countdown with 26/14=40.\r\n\r\nOh well, based on written I got 3-1-3-7 throughout this year.\r\n\r\nIt can be interpreted also as 1(number of points missed on school, or the number of people who got my score)-3(-way tie for first)-3-7. :P", "Solution_18": "really you didn't fail.\r\n\r\nyou still got the IMO who knows you'll make it.\r\n\r\nI totally failed,\r\nI didn't make aime.", "Solution_19": "is this what serious mathy kids do? discuss how fail-y they are?\r\ni mean i never even heard of aime before this year and in public school world i'm already considered smart.\r\nbut i must say the nats competition itself was very interesting this year.", "Solution_20": "[quote=\"the.bookblogger\"]is this what serious mathy kids do? discuss how fail-y they are? [/quote]\n\nyes.\n\n[quote=\"the.bookblogger\"]i mean i never even heard of aime before this year and in public school world i'm already considered smart.[/quote]\n\nSame here and that's one of the reasons I epic fail, and not in a pythag011 way of fail.\n\n[quote=\"the.bookblogger\"]but i must say the nats competition itself was very interesting this year.[/quote]\r\nDid you make nats this year?", "Solution_21": "[quote]Did you make nats this year?[/quote]\n\nHA no. didnt even get countdown at states. (dont laugh) well it was also my first year so i have excuses.[/quote]", "Solution_22": "Mhm. No comment.", "Solution_23": "I made 13th with 25+12=37", "Solution_24": "Dont think anyone failed as bad as I. Made 13 points of stupid mistakes (5 sprint, 4 target) (copying, adding, incorrect reading, not dividing by 2, wrong form). So in my practices, I got like about 32, and my goal was to get atleast 80th. I should have gotten 50th w/o 13 pts of mistakes, but that dropped me so bad. I am very disappointed my MC career had to end this way. I have gotten so much better, yet I scored lower than last year.\r\n\r\nAnd yay LAST on the team. LOL\r\n\r\nWE did okay on team, 9/10, if that is a consolation......\r\n\r\n :( :( :(", "Solution_25": "[quote=\"pinkmuskrat\"]. Made 13 points of stupid mistakes (5 sprint, 4 target) (copying, adding, incorrect reading, not dividing by 2, wrong form). So in my practices, I got like about 32, and my goal was to get atleast 80th. I should have gotten 50th w/o 13 pts of mistakes, but that dropped me so bad. I am very disappointed my MC career had to end this way. I have gotten so much better, yet I scored lower than last year.[/quote]\r\n\r\nSo what did you get? 32?", "Solution_26": "How hard was this national test for you? Rating: 1 being super easy to 10 being insanely tough?", "Solution_27": "Eh, in my opinion it was harder than last year but not that much harder. I'll give it a 4.\r\n\r\nAlthough I've noticed that my difficulty ratings are often off :(", "Solution_28": "I'd rank it like a 4ish [High expectations for the test? Low expectations for me?], but I totally blanked on the sprint round (putting stupid random answers for the whole last page, expect for the one where the answer was 6, and not checking). The sprint round made me focus for the target round though, which was easy and I did well on.\r\n\r\nObviously, it wasn't 2005 hard. The back page was surprisingly easy. Euh, now, it seems so.\r\n\r\nTeam round was just fun. #5 listing FTW :).\r\n\r\nDang... So Hawaii still never had anyone make CD.\r\nI got (21/16?=37?), putting me in 17th place. What a fun number.\r\nThe CD cutoff was so close...\r\n\r\nRandom Question: Do the graders ever make mistakes?", "Solution_29": "ok, disregard my other thread where i said that i got a 37, i failed, and got a 35...\r\n\r\n21 on sprint, and 14 on target... i failed.\r\ni ended up getting 28th, and my team got 15th (michigan) because we epic failed the team round and got 7 out of 10." } { "Tag": [ "number theory open", "number theory" ], "Problem": "Solve it\r\n$ y^{p}\\equal{}x^{2}\\plus{}3$\r\nwhere $ x,y,p$ are positive integers.\r\nAny elementary solutions?", "Solution_1": "The only solutions I could find are:\r\n1) y=2, p=2, x=1\r\n2) p=1 and y=x^2+3\r\nHere is a partial proof that there are no other possibilities:\r\n[hide]First note that if x is even then y is odd and vice versa. If y is even, let y=2n and you have (2^p)(n^p)=x^2+3. If p>2, then x^2+3 must be a multiple of 8, an impossible scenario (x^2=0,1,4 mod 8). If p=2, then the only solution is y=2, p=2, x=1. After that, the gaps between two consecutive perfect squares grow monotonically larger. If p=1, then there is the solution y=x^2+3.\nNow let's try y is odd and x is even. Let x=2q and you have y^p=4q+3. Since y is odd, this can be true only if p is odd (y^2=1 mod 4 if y is odd). Sorry, this is as far as I could get. If there are any other solutions, they must satisfy the conditions that p is odd, y is odd, and x is even.[/hide]", "Solution_2": "Well , I think that algebraic number theory works here but I'm not pretty sure... :maybe:\r\nLEFTERIS" } { "Tag": [], "Problem": "Hi everyone,\r\n\r\nI was wondering if any of you knew any good volunteering opportunities (either in the US or abroad) that either you've had personal experience with or have heard of from reliable sources.\r\n\r\nThanks", "Solution_1": "Well, there should always be opportunites where you live. For example, at a library, nursing home, local food bank, etc. I'm sure they would be happy to accomodate volunteers.\r\n\r\nI myself have volunteered at the local public library, a nursing home, and a (public) school for disabled children.", "Solution_2": "if you just want a good way to spend a summer, u might also want to consider a math summer program- some are discussed in [url=http://www.artofproblemsolving.com/Forum/index.php?f=136]Other US contests and programs[/url]", "Solution_3": "[quote=\"banblink182\"]Hi everyone,\n\nI was wondering if any of you knew any good volunteering opportunities (either in the US or abroad) that either you've had personal experience with or have heard of from reliable sources.\n\nThanks[/quote]\r\n\r\nIt's usually quite gratifying giving back to the community that you live in, that way, you can continue the relations that you build during the school year.", "Solution_4": "tutor math! It's always fun", "Solution_5": "I was a counselor at a week-long day camp over the summer. I'd definitely recommend doing something like that, it was definitely worth the time." } { "Tag": [], "Problem": "Violet, her brother and her parents are going to take the Metrolink from Santa Clarita to Los Angeles Union Station. This trip covers three zones, and they need round-trip tickets. They will save $ 30\\%$ on the two adult tickets and $ 50\\%$ on the two youth tickets since they are traveling during off-peak hours. Using the fare chart, what is the total cost for Violet's family?\t\t\n\n\n \\begin{tabular}{|c|c|c|}\n \\hline No. of & One-Way & Round-Trip \\\\\n Zones & Peak & Peak \\\\ \\hline\n $ 1$ & $ \\$4.25$ & $ \\$6.75$ \\\\ \\hline\n $ 2$ & $ \\$5.25$ & $ \\$9.00$ \\\\ \\hline\n $ 3$ & $ \\$6.25$ & $ \\$11.50$ \\\\\\hline\n \\end{tabular}", "Solution_1": "$ \\begin{tabular}{|c|c|c|}\r\n\\hline No. of & One\\minus{}Way & Round\\minus{}Trip \\\\\r\nZones & Peak & Peak \\\\ \\hline\r\n1&\\$4.25 &\\$6.75 \\\\ \\hline\r\n2&\\$5.25 &\\$9.00 \\\\ \\hline\r\n3&\\$6.26 &\\$11.50 \\\\\\hline\r\n\\end{tabular}$\r\n\r\nThey pay 11.50 for a normal peak round-trip at three zones. However they pay (2*0.7+2*1)=2.4 times this amount. 2.4*11.5=$ \\$27.60$", "Solution_2": "[quote=\"james4l\"]$ \\begin{tabular}{|c|c|c|}\n\\hline No. of & One\\minus{}Way & Round\\minus{}Trip \\\\\nZones & Peak & Peak \\\\ \\hline\n1&\\$4.25 &\\$6.75 \\\\ \\hline\n2&\\$5.25 &\\$9.00 \\\\ \\hline\n3&\\$6.26 &\\$11.50 \\\\\\hline\n\\end{tabular}$\n\nThey pay 11.50 for a normal peak round-trip at three zones. However they pay (2*0.7+2*1)=2.4 times this amount. 2.4*11.5=$ \\$27.60$[/quote]\n\n\n\ninstead of (2*0.7+2*1) it's (2*0.7+2*0.5) because its 50% off the two youth tickets, than it will add up to 2.4 times! :)" } { "Tag": [ "superior algebra", "superior algebra unsolved" ], "Problem": "We know that([b]Z[/b][sqrt(2)],N) is an Euclidean domain. And 2 is not a atom in [b]Z[/b][sqrt(2)]. However, 2= (root 2).(root 2)= (2+ sqrt2).(2-sqrt2) would seem to contradict this. Explain this with justification?\r\n\r\nI dont know how to explain it, can you show me or give me some hints please?\r\n\r\nThank you in advanced for your time on this question", "Solution_1": "One has $ 2 \\pm \\sqrt {2} \\equal{} \\sqrt {2}(\\sqrt {2} \\pm 1)$. What are the numbers $ \\sqrt {2} \\pm 1$ ?\r\nOne is the inverse of the other, so they are invertible (also called units).\r\nUnits (like $ \\pm 1$ in $ \\mathbb{Z}$, or them and $ \\pm i$ in $ \\mathbb{Z}[i]$) do not matter in factorizations." } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Prove that for any subset A of the complex plane, the three sets int(A), bd(A), and int(A complement) partition the complex plane.", "Solution_1": "That is a trivial consequence of the definitions in any topological space." } { "Tag": [ "geometry", "3D geometry", "prism" ], "Problem": "Here's an original problem that I created. I just want to verify if my answer.\r\n\r\nWhat is the maximum number of cylindrical logs with length 2m and a diameter of 0.8 m ca be put in a covered rectangular container with length, width and height of 3.25 m by 4.05 m, by 3 m respectively.\r\n\r\n\r\nNote: I put the 0.05 allowance because giving and exact number is unrealistic, the logs will not fit.\r\n\r\n[hide=\"My Answer\"]28[/hide]", "Solution_1": "[hide]I didn't give any allowance or anything, and I got 30.[/hide]", "Solution_2": "How did you get your answre? Would you like to show us your solution?", "Solution_3": "I get 28 as well. The logs do not go all in one direction. \r\n\r\nI think surge assumed the logs to be rectangular prisms (0.8x0.8x2). Then he assumed the box to be 3.2x4x3. If you divide those, you get 30.", "Solution_4": "Why can't the logs go in one direction?", "Solution_5": "The logs should be packed hexagonally with the $2$-meter dimension of the log aligned with the $4.05$-meter dimension of the box. I'll leave the actual calculation to someone else.", "Solution_6": "Yeah, hexagonal packing is most efficient for circles, and with that you can get 30 logs in it...", "Solution_7": "[quote=\"surge\"]Why can't the logs go in one direction?[/quote]\r\n\r\nAt least not for the one I got 28 on...", "Solution_8": "You put two logs along the 4.05m edge, and then it's a simple matter of hexagonal packing using the circles of radius .8 and the face with dimensions 3.25m x 3m...\r\n\r\nYou should find that you can get 15 logs in that way, giving a total of 30, because it is 2 deep..." } { "Tag": [ "probability" ], "Problem": "A standard, fair, 6-sided die is rolled 8 times. Given that the number 3 appears exactly three times, what is the probability that no two 3's appear on consecutive rolls?\r\n\r\nThe solution to this problem given to me was long. Is there a shorter way?", "Solution_1": "Hmm... this gets very complicated. I tried splitting it into cases, but then you need to split again, and again...\r\n\r\nAnd I don't think you can just write out the combinations that work, either, as there would be way too many of those.", "Solution_2": "perhaps we should think just of the combinations of 3's. Maybe we should count what we don't want first, and subtract that from all possible outcomes then dividing to get the probability.\r\n\r\nEDIT: ok, i'll keep thinking...", "Solution_3": "[quote=\"plokoon51\"]I wouldn't want to post a wrong solution, so could someone (236factorial) tell me if my answer is right? (i doubt it is)\n\n[hide]29/56???[/hide][/quote]\r\n\r\nNo.", "Solution_4": "[hide=\"Maybe, just maybe, this works\"]Let's count the ones that DO have 2 or more together: \n\nWe simply count the 2 together as 1 and find how many ways you can arrange these 2 parts in the 7 places to put them. So c(2,7) ways that it doesnt' work out of the c(3,8) possible ways fail, so $\\frac{c(3,8)-c(2,7)}{c(3,8)}=\\frac{35}{56}$[/hide]", "Solution_5": "[hide]How about\n3-3-???? 4 spots\n3--3-??? 3 spots\n...\n=10\n\n-3-3-??? 3 spots\n-3--3-?? 2 spots\n...\n=6\n\n--3-3-?? 2 spots\n--3--3-? 1 spot\n=3\n\n---3-3-? 1 spot\n=1\n\n1+3+6+10=20\ndivide by 8 cohoose 3 =56 to get 5/14\n\n\n\n\nOR\nyou know you must have\n3-3-3 and you have 3 -s to put where you want there are 4 spots. so partitions of three with 4 numbers\n3000 4 permutations\n2100 12 permutations\n1110 4 permutations\n\n4+12+4=20[/hide]", "Solution_6": "[quote=\"solafidefarms\"][hide=\"Maybe, just maybe, this works\"]Let's count the ones that DO have 2 or more together: \n\nWe simply count the 2 together as 1 and find how many ways you can arrange these 2 parts in the 7 places to put them. So c(2,7) ways that it doesnt' work out of the c(3,8) possible ways fail, so $\\frac{c(3,8)-c(2,7)}{c(3,8)}=\\frac{35}{56}$[/hide][/quote]\r\n\r\noops...i thought that 3 3's in a row was a separate case when all of those possibilities would be counted in c(7,2). silly me.\r\n\r\nis that answer correct, 236factorial?\r\n\r\nEDIT: hmmm...i guess not", "Solution_7": "[quote=\"solafidefarms\"][hide=\"Maybe, just maybe, this works\"]Let's count the ones that DO have 2 or more together: \n\nWe simply count the 2 together as 1 and find how many ways you can arrange these 2 parts in the 7 places to put them. So c(2,7) ways that it doesnt' work out of the c(3,8) possible ways fail, so $\\frac{c(3,8)-c(2,7)}{c(3,8)}=\\frac{35}{56}$[/hide][/quote]\r\nbut you could also put 3 together so you also have to subtract c(1,7) which will get 29/56, but I don't see why mine wouldn' work", "Solution_8": "the answer isn't 35/56. Should I give you the answer?" } { "Tag": [ "analytic geometry", "vector", "real analysis", "real analysis unsolved" ], "Problem": "show that the following pairs of families of curves are orthogonal (a,b are arbitrary constants). That is, prove that where any member of the first family intersects a member of the second family, the curves are orthogonal.\r\n\r\na) r=a cos(t) ; r=b sin(t)\r\nb) x(t) = a-ln(t), y(t)=t, t>0 ; x(s)=(s-b)/2, y(s)=root(s) s>0", "Solution_1": "I'll do b). I am still confused about a): are those equations in polar coordinates or something else?\r\n\r\nWe need to show that at the intersection point, the tangent vectors to our curves are orthogonal. The tangent vector to the first curve is $(-\\frac1t,1)$ (just differentiate with respect to the parameter). The tangent vector to the second curve is $(\\frac12,\\frac{1}{2\\sqrt s})$. Since the $y$-coordinates coincide, we have $t=\\sqrt s$, so the second vector is the same as $(\\frac 12,\\frac1{2t})$. Now, the scalar product is $-\\frac1t\\cdot\\frac12 +1\\cdot \\frac1{2t}=0$ and we are done." } { "Tag": [ "probability", "geometry", "3D geometry", "tetrahedron", "algebra", "function", "domain" ], "Problem": "Here are some moderate to hard ciphering questions\r\nComments are appreciated\r\n\r\n1. The expression (1+i)^k is a real number when k is a multiple of what number? (i is defined as the square root of \u20131)\r\n\r\n2. x, y, and z are distinct real numbers that sum to 0. What is the maximum possible value of \r\n\t\t(xy + xz + yz)/(x^.5 + y^.5 + z^.5)\r\n\r\n3. You are given a 4ft board. You go to the shop and receive 2 boards, each of which has a maximum possible length of 6ft. What is the probability that you can construct a triangle from the three boards?\r\n\r\n4. xyz=1 , x+y+z=0 , and xy+yz+xz=0. Exactly one of x, y, and z is a real number. Find this value.\r\n\r\n5. Two men are betting money on a game of tic-tac-toe. A man pays $1 dollar to randomly pick 4 spaces. If he gets tic-tac-toe, he wins$3. On average, how much can the player expect to win or lose in dollars?\r\n\r\n6. Find sqrt(3 + sqrt(3 + sqrt(3 + \u2026.))). \r\n\r\n7. In triangle ABC, D lies on side BC and AB=6, BD=3, DC=7 and AC=5. The length of the segment AD can be written in the form sqrt(a/b) where a/b is a fraction in lowest terms. What is a+b?\r\n\r\n8. The sum of the coefficients of the expansion (a+b+c+d)^9 can be written in the form 4^k where k is an integer. Find k.\r\n\r\n9. Find the ordered 5-tuple (a,b,c,d,e) that satisfies the following system of equations:\r\n4a + b + c + d + e = 8\r\na + 4b + c + d + e = 14\r\na + b + 4c + d + e = 24\r\na + b + c + 4d + e = 34\r\na + b + c + d + 4e = 40\r\n\r\n10. Find the volume of a regular tetrahedron with side length 1\r\n\r\n11. The rational numbers a,b,c, and d have the property that (ax+b)/(cx+d)=1 has no solution in x. What is the value of (a^2)/(a^2 +c^2)?\r\n\r\n12. The number X20191817\u2026321 is divisible by 11 if and only if X equals what positive integer? (note that X is a digit of the number; this is not multiplication)\r\n\r\n13. x/(y-6)=y/(z-8)=z/(x-10)=3. What is the value of x+y+z?\r\n\r\n14. Find the sum\r\n\r\n(sqrt(1+1) \u2013 sqrt(1))/(sqrt(1^2 +1) + (sqrt(2+1) \u2013 sqrt(2))/(sqrt(2^2 +2) +\r\n\t(sqrt(3+1) \u2013 sqrt(3))/(sqrt(3^2 +3) + \u2026 + (sqrt(15+1) \u2013 sqrt(15))/(sqrt(15^2 +15)\r\n\r\n15. Find the largest n such that (2004!)! is divisible by ((n!)!)!\r\n\r\n16. Compute\r\n\r\n(1*1! + 2*2! +3*3! + \u2026 +7*7! +1)/(6!)\r\n\r\n17. Simplify\r\n\r\n1(nC1) + 2(nC2) + 3(nC3) + \u2026 + n(nCn)\r\n\r\n18. A teacher wants to divide un-graded papers into a collection of several stacks, each consisting of the same number of papers. He knows that there is somewhere between 1000 and 2000 papers. He tries dividing the stack into groups of 2, 3, 4, 5, 6,7, and 8, but repeatedly gets 1 paper left over. How many papers are there?\r\n\r\n19. Find the sum of the series\r\n\r\n1+ 2(1/5) + 3(1/5)^2 + 4(1/5)^3 + \u2026\r\n\r\n20. What is the smallest positive integer greater than 1 that leaves a remainder of 1 when divided by 2^2^2 and 2^3^3.(note: a^b^c=(a^b)^c )\r\n\r\n21. What is the remainder when 2^100 is divide by 15?\r\n\r\n22. Find one value of a for which the triangle with vertices (20,30), (30,60), and (a,50) has area 20.\r\n\r\n23. In triangle ADC, angle bisector DB is drawn. If AB=3, AD=6, and CD=8, find DB.\r\n\r\n24. If f(x/(x-1)) for all x not equal to 0 or 1, and 0moon", "Solution_1": "[hide]280 socks in total 20 are to be picked out....\nworst case scenario\n1 1 1 1\n1\nfirst pair\n1111\n1\n1\nnot yet\n1111\n1\n1\n1\n.\n.\n.\nis the answer 11/70?[/hide]", "Solution_2": "What do u wana mean by 11/70 i dont think ur ans correct.Ok a hint try to use box principle\r\n>moon", "Solution_3": "[quote=\"Moonmathpi496\"]What do u wana mean by 11/70 i dont think ur ans correct.Ok a hint try to use box principle\n>moon[/quote]\r\n[hide]i don't see how the box principle would help-\ni just calculated the largest amount of socks you'd have to take(i am rather sure but this is probably where you want me to apply box principle :( ) and put that over the amount of socks and simplified [/hide]\r\nsorry that i am not getting this", "Solution_4": "Ok use the fact if want 1 pair of socks u have to take at least 5 socks.\r\nAnd i told u to find the smallest number of socks to get 10 pair.\r\n>moon", "Solution_5": "well what i am saying is\r\n[hide]you already have a sock from each other than the one you have 2 of so you should add those instead for more steps....[/hide]" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "prove $3^n-2^n$ is not divisible by $n$", "Solution_1": "Just like you prove the famous $n\\not |2^n-1$: let $p$ be the smallest prime divisor of $n$. We have $p\\ne 2,3$, so $p|(3^{p-1}-2^{p-1},3^n-2^n)=3^{(p-1,n)}-2^{(p-1,n)}=3-2=1$, contradiction.\r\n\r\nP.S. I assumed there is an $n$ s.t. $n|3^n-2^n$.", "Solution_2": "[quote=\"grobber\"]Just like you prove the famous $n\\not |2^n-1$.[/quote]\r\nCan we make the same thing for $n|2^{n+1}-1$ ?", "Solution_3": "[quote=\"Collins\"][quote=\"grobber\"]Just like you prove the famous $n\\not |2^n-1$.[/quote]\nCan we make the same thing for $n|2^{n+1}-1$ ?[/quote]\r\nBut $n\\mid 2^{n+1}-1$ for $n=3,15$, for example.", "Solution_4": "[quote=\"Myth\"][quote=\"Collins\"][quote=\"grobber\"]Just like you prove the famous $n\\not |2^n-1$.[/quote]\nCan we make the same thing for $n|2^{n+1}-1$ ?[/quote]\nBut $n\\mid 2^{n+1}-1$ for $n=3,15$, for example.[/quote]\r\nOk, but my question is: can we find all $n$ such that: $n|2^{n+1}-1$?", "Solution_5": "For example, all $n=2^{2^{k}}-1$ are a good choice. :)\r\nI think, it can be all possible values for $n$ :?", "Solution_6": "I think it's not that easy,how about n=35", "Solution_7": "i think this pro very very hard\r\nand we have:\r\n[b]SIMILAR-PRO[/b]: find all possitive integer $n$ such that: $2^n-1|3^n-1$\r\nand new pro very hard :D", "Solution_8": "[quote=\"nguyenquockhanh\"]\nsimilar problem: find all possitive integer $n$ such that: $2^n-1|3^n-1$\nand new pro very hard [/quote]\r\nWhy do you think it is similar? :? Your problem was posted by me before.Look at:\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?p=134641#134641" } { "Tag": [ "inequalities", "function", "integration", "inequalities proposed" ], "Problem": "Suppose $-\\infty\\left(\\frac{f(a)^r+f(b)^r}{2}\\right)^{\\frac{1}{r}}[/tex]. But by construction, [tex]f[/tex] is a convex function, whence [tex]M_r (f,[a,b])\\leq \\left(\\frac{f(a)^p+f(b)^p}{2}\\right)^{\\frac{1}{p}}=A_p([f(a),f(b)])[/tex]. Putting together these two inequalities, we get [tex]A_{r}([f(a),f(b)]) p[/tex].\r\n\r\nI don't know if it's right or useful, but maybe somebody's inspired by this ;)\r\n\r\n\r\nHanno", "Solution_2": "[quote=\"Hanno\"]Hi all!\n[quote]then prove/or disprove that the point (p,q) belongs to a subset of D which is not convex. [/quote]\nWhat's meant by this? Hanno[/quote] Thanks for remark ! \r\nLet us denote by $\\; \\Omega \\; , \\; \\Omega \\subset D\\; ,\\; $ [b] the set of all pairs[/b] $ (r,p) $ for which the inequalities are satisfied for all positive convex functions on $ [a,b] $ and for all $ (x,y) ,x\\ne y, a\\le x1).", "Solution_1": "Here is my solution. Define the series of sequences by putting $ (x_n)^{0}=\\frac{b^n+1}{a^n+1}, (x_n)^{i+1}=b\\cdot (x_n)^{i}-a^i\\cdot (x_n)^{i} $. It is clear that the terms of these sequences are integers. Moreover, $ (x_n)^{0} $ behaves like $(\\frac{b}{a})^n $, $ (x_n)^{1} $ behaves like $(\\frac{b}{a^2})^n $ and so on, an easy induction shows that $ (x_n)^{k} $ behaves like $ (\\frac{b}{a^{k+1}})^n $. Thus, one of these sequences becomes 0 from a certain point. Let j minimal such that $ (x_n)^{j} $ becomes 0 from a certain point. Now, we can compute the terms of the sequence $ (x_n)^{j-1} $ from that certain point and it is immediate to see that if b is not a multiple of a then $ (x_n)^{j-1} $ also becomes 0 from a certain point, contradicting the minimality of j. Thus, b is a multiple of a. Let $ b=ka $. Then $ a^n+1|k^n-1$ and we define $ (y_n)^{0}=\\frac{k^n-1}{a^n+1} $ and so on (just like in the argument showing that a|b) and with the same reasoning we find that $ a|k $. Thus, $ k=ua $ and $ a^n+1| u^n+1 $. Now, you know what you have to do: a|u, we write u=sa, a|s, The conclusion follows.", "Solution_2": "Hi harazi,\r\n\r\ncould you please explain a little more your solution ? there is some points I don't understand .. thanxx :)", "Solution_3": "Today I had a very bad day. Nothing worked. Ok, I will try to explain it more. Too bad its too long. So, define $ x_n^1=\\frac{b^n+1}{a^n+1} $ and $ x_n^{i}=b\\cdot x_n^{i-1}-a^i\\cdot x_{n+1}^{i-1} $. By induction we can prove the following statement:\r\n For any $j\\geq 1$ there exist constants $ c_0,c_1,, c_j $ which depend on $j$ such that $x_n^j=\\frac{ c_j\\cdot b^n+c_{j-1}\\cdot a^{(i-1)n}++c_1\\cdot a^n+c_0}{(a^n+1)(a^{n+1}+1)\\cdot\\cdot (a^{n+i-1}+1) } $. This is not difficult. This shows that $ x_n^j $ behaves like $(\\frac{b}{a^j})^n $. Thus, for $ a^k>b $ the sequence $ x_n^k $ tends to 0, so becomes constant from a certain point. Let j minimal such that the sequence $ x_n^j $ becomes constant from a certain point $n_0$. Thus, we have $ x_n^{j-1}=(\\frac{b}{a^j})^{n-n_0}\\cdot x_{n_0}^{j-1} $. Suppose that $ b$ is not a multiple of $ a$. Then since the number $(\\frac{b}{a^j})^{n-n_0}\\cdot x_{n_0}^{j-1} $ is an integer for an infinite numbers n, we must have $ x_{n_0}=0 $ and so the sequence $ x_n^{j-1} $ becomes 0 from a certain point. Thus, j is not minimal, false. Therefore, b is a multiple of a and from now on I think it is clear." } { "Tag": [ "articles", "inequalities", "AMC", "USA(J)MO", "USAMO", "trigonometry", "geometry" ], "Problem": "Just athought, maybe some aopser's that are also mopper could share their notes, especially the people that are in red.", "Solution_1": "Try Mildorf's notes, they're in the articles section of AoPS, I believe. I think he was in blue.", "Solution_2": "[quote=\"13375P34K43V312\"]Try Mildorf's notes, they're in the articles section of AoPS, I believe. I think he was in blue.[/quote]\r\nMildorf was an IMO gold medalist, so he was probably in black...", "Solution_3": "I believe MOPpers are told to not distribute notes, especially not on a hugely popular open forum such as this one. However, some lecturers have posted theirs online, or even more polished versions of notes. (reid barton, kiran among others)", "Solution_4": "Mildorf hasn't posted any MOP notes (the ones under articles weren't from MOP).", "Solution_5": "[quote=\"krustyteklown\"][b]I believe MOPpers are told to not distribute notes, especially not on a hugely popular open forum such as this one. [/b]However, some lecturers have posted theirs online, or even more polished versions of notes. (reid barton, kiran among others)[/quote]\r\nWhy :huh:", "Solution_6": "for the same reason as if you take an AoPS class, you're not supposed to give the materials to anyone else\r\n\r\nif they're not public, you shouldn't really ask individuals in order to obtain them", "Solution_7": "[quote=\"krustyteklown\"]for the same reason as if you take an AoPS class, you're not supposed to give the materials to anyone else\n\nif they're not public, you shouldn't really ask individuals in order to obtain them[/quote]\r\n\r\nI belive DPatrick said in the Classes Information forum that distributing NOTES was acceptable, as long as they are not like copied directly from the transcripts.", "Solution_8": "What puzzles me is why the AMC foundation doesn't make outlines or notes from MOP available. What's the point in keeping them proprietary? AoPS sells them as classes, so I understand wanting those kept secret. But I ask again, if the AMC foundation truely wants to promote nontrivial mathematics, why don't they make MOP notes available freely, or at least sell them? I can see no reason for keeping them from the public.", "Solution_9": "[quote=\"PenguinIntegral\"]I can see no reason for keeping them from the public.[/quote]\r\n\r\nThey don't want those dirty, seedy Russians getting those documents. The Russians will do anything to win the IMO.", "Solution_10": "[quote=\"13375P34K43V312\"][quote=\"krustyteklown\"]for the same reason as if you take an AoPS class, you're not supposed to give the materials to anyone else\n\nif they're not public, you shouldn't really ask individuals in order to obtain them[/quote]\n\nI belive DPatrick said in the Classes Information forum that distributing NOTES was acceptable, as long as they are not like copied directly from the transcripts.[/quote]\r\n\r\nAs long as they are your own notes, and have just about nothing to do with material from the class. If you write down the problems and solutions from the transcripts without class conversations, I would still imagine that you can't distribute those. Basically, you shouldn't be sharing anything that was obtained from the class and might decrease someone else's incentive to take it. \r\n\r\n@PenguinIntegral: some notes are availalbe (note Kiran Kedlaya's). I imagine that most of the notes belong to the professors, who might not want their materials flying all over the internet :wink:", "Solution_11": "[quote=\"Elemennop\"][quote=\"PenguinIntegral\"]I can see no reason for keeping them from the public.[/quote]\n\nThey don't want those dirty, seedy Russians getting those documents. The Russians will do anything to win the IMO.[/quote]\r\n\r\n....says the Russian.", "Solution_12": "[quote=\"Elemennop\"][quote=\"PenguinIntegral\"]I can see no reason for keeping them from the public.[/quote]\n\nThey don't want those dirty, seedy Russians getting those documents. The Russians will do anything to win the IMO.[/quote]\nI'd imagine that the Russians and other major countries already have programs of equal or greater value. I thought the IMO was for friendly competition, not information hording.\n\n\n[quote=\"Phelpedo\"]\n@PenguinIntegral: some notes are availalbe (note Kiran Kedlaya's). I imagine that most of the notes belong to the professors, who might not want their materials flying all over the internet :wink:[/quote]\r\nWhy wouldn't they want their materials flying over the internet, or being bought for a minor sum? Isn't the whole goal of the AMC foundation and it's affiliates to spread mathematical knowledge and interest?", "Solution_13": "[quote=\"PenguinIntegral\"][quote=\"Elemennop\"][quote=\"PenguinIntegral\"]I can see no reason for keeping them from the public.[/quote]\n\nThey don't want those dirty, seedy Russians getting those documents. The Russians will do anything to win the IMO.[/quote]\nI'd imagine that the Russians and other major countries already have programs of equal or greater value. I thought the IMO was for friendly competition, not information hording.\n\n\n[quote=\"Phelpedo\"]\n@PenguinIntegral: some notes are availalbe (note Kiran Kedlaya's). I imagine that most of the notes belong to the professors, who might not want their materials flying all over the internet :wink:[/quote]\nWhy wouldn't they want their materials flying over the internet, or being bought for a minor sum? Isn't the whole goal of the AMC foundation and it's affiliates to spread mathematical knowledge and interest?[/quote]\r\nThat's what they want you to think...", "Solution_14": "Well, despite what rumors I have read on this thread, I think the real reason that most MOPpers (not [i]mopers[/i] :wink: ) don't post their notes on the internet is that it would be a lot of tedious work to typeset more than 30 sets of notes. Besides, there are plenty of good materials already out there (e.g., Engel, and Kiran's packet for inequalities) to give you the prerequisites for solving olympiad problems. Indeed, much, if not most, of the lecture time at MOP is spent working [b]problems[/b], not learning material. And this site is testimony to the fact that there are already massive quanitities of problems available for free on the internet, many of them from the AMC site itself. If you want advice for where to start, I think the first few IMOs and (to a lesser extent) USAMOs are quite accessible. Besides all this, there's the fact that I'm pretty sure they recycle problems (at the very least, they borrow from other sources), so it wouldn't necessarily be in the best interests of the program to have to come up with a zillion new red problems each year because everybody had already read last year's notes, now would it?", "Solution_15": "Indeed, MOP notes are not some kind of holy grail for getting better at olympiad problems. Old IMO shortlists are just as good as most MOP sheets. Anyways, if you're interested in becoming a MOPer, then you could just ignore math completely and start walking around with your head down.", "Solution_16": "[quote=\"bpms\"]\nThat's what they want you to think...[/quote]\r\n\r\nJust what are you implying?", "Solution_17": "[quote=\"Phelpedo\"][quote=\"bpms\"]\nThat's what they want you to think...[/quote]\n\nJust what are you implying?[/quote]\r\nI was joking...\r\nPerhaps it wasn't too obvious...\r\nSorry if no one understood that I wasn't serious...\r\nNext time I will use :D...", "Solution_18": "There are three reasons that the notes from MOSP are not available from the AMC.\r\n\r\nFirst and foremost, the notes belong to the graders and instructors who create them. They create them, and they own them, in the sense that any instructor owns his or her own intellectual property. So the AMC is not free to use the material as if the AMC created or owned it. We'd need to get permission. Not hard, but it would take effort which would divert us form other activities. Furthermore, some of the MOSP material comes from sources which may be copyrighted. Using copyrighted mateiral in the classroom falls under the Fair Use category, but not if we tried to sell it or use it for more than classroom purposes.\r\n\r\nSecond, the notes are not edited. Some may contain mistakes, impossible problems, trivial problems, old well-known problems , copyrighted material and the like. The notes are not in a uniform format, so it would take us some effort to gather them and edit them into a uniform and useable format that would be more than the web version of a pile of papers.\r\n\r\nThird, good versions of some of the notes do exist. Most of the books by Titu Andreescu contain material used at MOSP in one way or another. His books on trig and combinatorics come immediately to mind. There is also the new book published by Springer that has every IMO short-listed problem. Likewise for the volumes called Olympiad Problems from Around the World, published by MAA. Also look at the books Mathematical Gems, volums I-IV by Honsberger, published by MAA. Many MOSP topics are covered in there, in more detail and with exposition.\r\n\r\nSteve Dunbar\r\nAMC Director", "Solution_19": "[quote=\"Boy Soprano II\"] Indeed, much, if not most, of the lecture time at MOP is spent working [b]problems[/b], not learning material. Besides all this, there's the fact that I'm pretty sure they recycle problems (at the very least, they borrow from other sources), so it wouldn't necessarily be in the best interests of the program to have to come up with a zillion new red problems each year because everybody had already read last year's notes, now would it?[/quote] First as described by mellowmelon in http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=222065\r\n\"Le had a geometry class for us almost every day. We did start off with the basics, but later on he got into the theory of polar and projective transformations.\" So MOSP is rather heavy theory contrary to what you said. Also I think a lot of us that have no chance getting into MOSP or are beyond high school would really like to learn about all the cool problem solving concepts and the theories behind them. Like a lot of Titu's books don't contain that much theory and I cannot find a book that discusses pole and polar transformations. Like a lot of the problems may be on the web but without the notes they would be rather unapproachable.", "Solution_20": "please don't revive old threads, a mod should lock this", "Solution_21": "Actually, those classes were by far the heaviest on theory; that was atypical, which is one reason I noted it was my favorite lecture at MOP. Boy Soprano is right when he says most of them emphasize problem solving." } { "Tag": [ "Recursive Sequences" ], "Problem": "An integer sequence satisfies $a_{n+1}={a_n}^3 +1999$. Show that it contains at most one square.", "Solution_1": "$ x^3(\\mod 7) \\equal{} \\{ \\minus{} 1,1,0\\}$\r\nwe prove that $ n > 2$ then \r\n$ a_n$ can't be a square.\r\nWe has $ 7\\not |a_n$\r\nConsider the equation:\r\n$ x^2 \\equal{} y^3 \\plus{} 1999$\r\n$ x^2(\\mod 7)\\in \\{0,1,4,2\\}$\r\n$ y^3 \\plus{} 1999 (\\mod 7)\\in \\{3,5\\}$\r\nSo it has no solution when $ n > 2$\r\nNow we prove that at most one of $ (a_1,a_2)$ can be a perfect square.\r\n[u]Case 1[/u]$ a_1$ is a perfect square $ a_1 \\equal{} m^2$\r\nSuppose $ a_2$ is a perfect square then\r\n$ m^6 \\plus{} 1999 \\equal{} x^2$ (contradiction!)\r\nSo if $ x_1$ is a perfect square then $ x_2$ is not a perfect square .\r\nSo at most one of term in this sequence can be a perfect square.", "Solution_2": "doesn't '$ m^{6} \\plus{} 1999 \\equal{} x^{2}$ (contradiction!)' need a little more explanation? Anyway, my solution using a different modulus: \r\n\r\nIf $ {a_{n}} \\equal{} 1 (mod 4),$ then $ {a_{n \\plus{} 1}} \\equal{} 0 (mod 4)$\r\nIf $ {a_{n}} \\equal{} 0 (mod 4),$ then $ {a_{n \\plus{} 1}} \\equal{} 3 (mod 4)$\r\nIf $ {a_{n}} \\equal{} 3 (mod 4),$ then $ {a_{n \\plus{} 1}} \\equal{} 2 (mod 4)$\r\nIf $ {a_{n}} \\equal{} 2 (mod 4),$ then $ {a_{n \\plus{} 1}} \\equal{} 3 (mod 4)$\r\n\r\nSince squares are never 2,3 (mod 4), we must have that both $ {a_{1}}$ and $ {a_{2}}$ are squares to contradict our theorem. Assume $ {a_{1}} \\equal{} x^{2}$ and $ {a_{2}} \\equal{} y^{2}.$ Then: $ (y \\plus{} x^{3})*(y \\minus{} x^{3}) \\equal{} y^{2} \\minus{} (x^{2})^{3} \\equal{} 1999$ . Since 1999 is prime, $ y \\minus{} x^{3}$ must be 1. So:\r\n$ 1999 \\equal{} (y \\plus{} x^{3})*(y \\minus{} x^{3}) \\equal{} 2*x^{3} \\plus{} 1$\r\n$ 999 \\equal{} x^{3}$ ; impossible for an integervalued x" } { "Tag": [ "MIT", "college", "AMC", "USA(J)MO", "USAMO", "AIME", "calculus" ], "Problem": "I'm just an average sophomore who happens to like math. However, these days (actually a long before) the issue of college admission starts dominating my mind and I question myself, should math be my hook.\r\n\r\nDo you think I should find another passion of mine and start working on that? Or build up on my current love; math. I really hate to do something just for the sake of college resume and I don't want deceive myself about my true passion. However math happens to be a subject where one's talent can be measured so easily through a single test paper. And partly for this reason, it is generalised that it is the outcome (implying success in competitions) that shows how much time you've spent doing maths. All of the winners tend to be people who've developed interest in math since earlier age. I just don't think I'll stand a chance and compete with them. I sound very paranoid about college and yes, I think I am. I'm perfectly aware that not going to MIT means you've failed your life and that there are other exceptional colleges out there. But lets not mind about college and focus on developing my passion.\r\n\r\nI've thought about ways to show my interest in math and came up with the following ways: \r\n1. Succeed in competitions eg: make it into USAMO (yes, very obvious) \r\n2. Research: compete in Intel STS or Siemens.\r\n3. Tutoring in math and help others in a form of comm. service\r\n\r\nThe road of number 1 is a possibility but I fear whether I will be able to successfully do it. The nature of competition is that if you feel bad on the test date and screw up, tough luck. I'm a current sophomore and got a 4 on AIME with a very low AMC score due to random guessing. 4 is a promising score and shows potential to qualify for USAMO after a year but since this years AIME was far easier than the rest, its not too encouraging. Can anyone share their 'success stories'? I think inspiration from the stories is what I need now to stop my habit of spending hours on CC rather than solving some problems. \r\n\r\nI'm also focusing on number 2 and found that many of the winners for their math research aren't prodigies who went to MOSP or people who did RSI. But I have no idea where to begin and get my initial idea; I'm just surfing thru mathworld in a glimpse of hope that I would encounter an 'Eureka' moment. Any suggestions?\r\n\r\n \r\nThank you for reading my very boring, incredibly random, narrow-minded, stubborn, rude and grammatically awkward post. Thanks.", "Solution_1": "and are there any resources I can use for the initial stage of math research? I've read the brief descriptions of the math researches the Intel winners have done and I really don't know how in-depth you have to go. For example, it simply says either \"work on number derivatives and its special properties\" or \"work on abstract graphs\". They don't happen to publish their work on the internet do they? (I've read Gabriel Carrol's poset research (came 3rd in Intel STS and he was the producer of full mark IMO paper in Washington) he posted on his website and I was baffled by all the alien language he was using.) \r\n\r\nI've read a very interesting post about the nature of Intel at CC (just search up for \"intel research\" and you'll find it) and it says most of the winners work with world-class labs or professors and produce undergraduate lvl researches. Does anyone have a full paper of any of the winners?", "Solution_2": "If you have Word Perfect, I could e-mail you my Intel paper. (I was a finalist two years ago.) If you only have word, I might be able to translate it, but no promises. Send me a PM if you're interested.", "Solution_3": "Do math if you like it. If you don't and just try to be good at it without any passion for it, its unlikely you'll get the desired results.\r\n\r\nBoth competition and research will require more problem solving skills than (I assume) you currently have, so you should try to remedy this. Solve more problems, buy problem solving books, solve more problems, repeat.\r\n\r\nIf its research you want to do, you should also try and learn more 'advanced' math - if you're not sufficiently proficient with calculus, review it some more. Same with linear algebra, etc, etc.", "Solution_4": "[quote=\"nodrog2002\"]Research: compete in Intel STS or Siemens. [/quote]\r\n\r\nWhat exactly is the Intel STS? I have heard of it, but what age level is it for? What exactly is it? Is it hard?", "Solution_5": "that is mainly for junior i believe...\r\n\r\ni could be wrong though anyone else mind to correct me(if i am mistaken) feel free...", "Solution_6": "[url=http://www.sciserv.org/sts/]Intel STS[/url] -- basically, you submit a research project the beginning of your Senior year and compete for money.", "Solution_7": "I would guess that it would be good to get really good at some math-related skill (cryptography, computer science, physics, economics, etc.) because I don't know how you get money from being a mathematician, unless you are Euler or something.", "Solution_8": "nodrog2002, some thoughts:\r\n\r\nWriting a research paper is a great idea but you should do it mostly for the math you'll learn, not with hopes of winning the Intel or Siemens Competition. A mere handful of math papers win each year so your chances are very, very slim. If you're interested in math research, you should find a mentor. Are there any math professors or mathematicians in your community? If not, try contacting one by e-mail. You shouldn't expect to come up with a topic on your own. Even graduate students don't usually choose their own dissertation topics. Read the recent Siemens Competition Math Jam. PoLing Loh said her RSI mentor gave her her topic.\r\n\r\nI'm a firm believer in work. If you decide you want to qualify for the USAMO next year, I'm sure you can. Just set aside time every week (preferably every day) to work on problems. If you work on problems regularly, it will be impossible for you not to improve. If you miss the USAMO next year, there's always the year after. In the meantime, you'll learn a lot of math, which can't hurt.\r\n\r\nThe college admissions scene is a complicated one. I wouldn't worry too much about it right now. The most selective colleges are so hard to get into, even for the very best students. I wouldn't count on getting into any particular one (or four or five) but here are my thoughts on what they're looking for:\r\n\r\nThe selective colleges receive so many applications from bright students, they can afford to be very choosy. They want students who will be the leaders of tomorrow, who will be active participants in campus life. They do not want students who just show up for class and do their homework. They would prefer someone who has started a software consulting business while still in high school to someone with 1600 SAT and no work experience. Excelling at math competitions certainly wouldn't hurt your chances but you should show you have other interests too.\r\n\r\nHope this helps. Relax and good luck.", "Solution_9": "if this makes you feel better, i didn't start my involvement in math till 9th grade and participation in serious competitions till 10th grade. my scores are not the greatest, but i'm very thankful for my participoation, the learning i gained, and the great people i got to meet. i have never made USAMO, though i'm very satisfied with my results overall... you dont' have to make usamo to make the college of your choice or to succeed in life!\r\nif you do want to do an intel paper, i suggest you begin early... i left my planning to the summer after junior year and procrastinated till i never got one done. if you want to do it, you must have the initiative to begin and find someone willing to mentor you. also, there are many summer math programs that you can participate in. they look good for college because you're spending your summer doing math AND you learn many interesting things (and you can come out of them with an idea for a paper :) )\r\ni have found tutoring to be very fun and rewarding experience. alot of understanding takes place when you're trying to explain something to others. tutoring, however, doesnt' really help you build up those skills you need for the USAMO.\r\nall these ideas are great, hope this helps you decide what to do... best of luck in high school :)" } { "Tag": [ "percent" ], "Problem": "When its digits are reversed, a particular positive two-digit\ninteger is increased by $ 20\\%$. What is the original number?", "Solution_1": "$ \\frac{100\\plus{}20}{100}(10a\\plus{}b)\\equal{}(10b\\plus{}a) \\iff 5a\\equal{}4b$.\r\nSo, $ (a, b)\\equal{}(4, 5)$.\r\n\r\nThus $ \\boxed{45}$.", "Solution_2": "[hide]Note that the original number must be divisible by 5. Also, since the difference of a two-digit number and its reverse is 9 times the difference of the two digits, and since the percent isn't relevant to a 9 whatsoever, the original number is also divisible by 9. Guessing $9 \\times 5 = \\boxed{45}$ works, since 90 obviously is too large.[/hide]" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Let $ H$ a big set if $ \\exists M \\subset H$ measurable such that $ \\lambda(M)>0$.Is true that if $ H_1,H_2,H_3,...,H_n,...$ are big then $ \\exists$ infinetly many $ (H_j)$ such that $ \\bigcap_{j}H_j$ is big?", "Solution_1": "[quote=\"msaid\"]Consider $ H_n \\equal{} [n \\minus{} 1,n]$ Then $ H_i$ is a Big Set for all $ i$ but $ \\cap {H_i}$ is empty.[/quote]\r\n[color=green][I think msaid intended to post that as a reply here and accidentally made a new thread instead.][/color]" } { "Tag": [], "Problem": "Contrast Ascomycota and Basidiomycota: how are they \"closely related\"?", "Solution_1": "Ascomycota: http://en.wikipedia.org/wiki/Ascomycota\r\n\r\nBasidiomycota: http://en.wikipedia.org/wiki/Basidiomycota", "Solution_2": "both show sexual reproduction of similar type--i'll tell u more if u want to kno", "Solution_3": "Ok. What is the similarity between their reproduction?" } { "Tag": [], "Problem": "Hi does anyone know who proposes the problems for the RMO and the INMO?", "Solution_1": "some names\r\n\r\nDr.BJ Venkatachala\r\nCR pranesachar\r\nRB Bapat,etc \r\nor in short,all Indian authors who write olympiads book", "Solution_2": "[quote=\"skand\"]some names\n\nDr.BJ Venkatachala\nCR pranesachar\nRB Bapat,etc \nor in short,all Indian authors who write olympiads book[/quote]\r\n\r\nThanks!" } { "Tag": [ "algorithm", "number theory", "Euclidean algorithm", "greatest common divisor" ], "Problem": "What is the greatest common divisor of 693 and 882?", "Solution_1": "We use the Euclidean algorithm here. $ 882 \\minus{} 693 \\equal{} 189$, and then $ 693 \\minus{} 189 \\equal{} 504$. $ 504 \\minus{} 189 \\equal{} 315$, $ 315 \\minus{} 189 \\equal{} 126$, $ 189 \\minus{} 126 \\equal{} 63$, $ 126 \\minus{} 63 \\equal{} 63$, and then we can easily see that the greatest common divisor is gcd(63,63) = 63.", "Solution_2": "Another method is to prime factorize each number:\n\n$693=\\boxed{2^0}\\cdot \\boxed{3^2}\\cdot \\boxed{7^1}\\cdot 11^1\n\\\\882=2^1\\cdot 3^2\\cdot7^2\\cdot \\boxed{11^0}$\n\nThen we do the multiplication of the boxed terms:\n\n$2^0\\cdot 3^2\\cdot 7^1 \\cdot 11^0=1\\cdot 9\\cdot 7\\cdot 1=\\boxed{\\boxed{63}}$" } { "Tag": [ "trigonometry", "geometry", "perpendicular bisector", "AMC" ], "Problem": "Triangle ABC has AC = BC and \u2220ACB = 106o. M is a point inside the triangle such that \u2220MAC = 7o and \u2220MCA = 23o. Find \u2220CMB. \r\n\r\nAs I went over this problem using Kalva's solution: [color=white]Take X on the line AM so that angle XBC = 7o. Since CA = CB, X must lie on the angular bisector of \u2220C. So \u2220BCX = 53o. Hence \u2220XCM = 53o-\u2220ACM = 30o. Also \u2220XMC = 7o+23o = 30o. Now XB is perpendicular to MC. But XM = XC, so it must be the perpendicular bisector of MC. Hence \u2220CMB = \u2220MCB = 83o. [/color]I have trouble figuring out why XB is perpendicular to MC? Can somebody please explain it to me, thanks.", "Solution_1": "saturnlife wrote:Triangle ABC has AC = BC and \u2220ACB = 106o. M is a point inside the triangle such that \u2220MAC = 7o and \u2220MCA = 23o. Find \u2220CMB. \n\n\n\nKalva's solutions can be very confusing sometimes. If you don't mind, might I propose an alternate one? (A while back someone on MathLinks solved this problem, and recalling the crucial first step, it's not too bad.)\n\n\n\n[hide]Pick a point N in triangle ABC such that :ang: NCB = 23 degrees and :ang: NBC = 7 degrees. (N is like the equivalent of M for the side BC.) By ASA, triangles AMC and BNC are congruent. Therefore MC = NC, since the two are corresponding parts. Therefore MCN is isosceles.\n\n\n\nNow, since ACM + MCN + NCB = 106 degrees, and ACM = NCB = 23 degrees, MCN = 106 - 23 - 23 = 60 degrees. Since MCN is isosceles, the other two angles of the triangle (CMN and CNM) are congruent and are therefore both equal to 60 degrees as well. Thus CMN is actually equilateral, so MN = NC. Finally, since MNC, CNB, and BNM comprise a full 360 degrres together, and MNC = 60 and CNB = 150, BNM = 360 - 60 - 150 = 150. But CNB also equals 150, so BNM = CNB.\n\n\n\nNow we look back at triangles MNB and CNB. We found that BNM = CNB and MN = NC, and they share a side NB. So by SAS, MNB is congruent to CNB. Therefore MB = CB because they are corresponding parts.\n\n\n\nAlmost finished! Since MB = CB, triangle MBC is isosceles. Since angle MBC = 7 + 7 = 14, the other two angles of the triangle (BMC, the desired angle, and BCM) both have measure (180-14)/2 = 83. Answer.\n\n\n\nLooking back, the crucial step was constructing that congruent triangle using side BC. I'll have a look at Kalva's solution, but I think this one is much more insightful and understandable.[/hide]", "Solution_2": "The easiest way to do this is to apply trig ceva... You do this and notice that 2sin7cos7 = sin14 and ur done.", "Solution_3": "Thanks for the solutions guys, they helped me alot!" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "In every turn it can be taken (c+5b,3c-5a,2b-3a) or (2a+3b,b+3c,4c+2a) instead of (a,b,c).At the beginning if (a,b,c)=(1,2,4) , is it possible that get (2003,2004,2005) using such turns?", "Solution_1": "No, it is impossible. To show this we analyze the congruence modulo 2:\r\n\r\nterns of start terns after first move terns after second move\r\n\r\n0,0,0 0,0,0 0,0,0\r\n0,0,1 1,1,0 0,1,0\r\n0,1,0 1,0,0 1,1,0\r\n0,1,1 0,1,0 1,0,0\r\n1,0,0 0,1,1 0,0,0\r\n1,0,1 1,0,1 0,1,0\r\n1,1,0 1,1,1 1,0,0\r\n1,1,1 0,0,1 1,0,0\r\n\r\nso we want obtain the tern (2003,2004,2005)=(1,0,1) and it can build only from another tern (1,0,1) but we start from a (1,0,0) so it is impossible.", "Solution_2": "Quote my last message to read the correct layout." } { "Tag": [ "induction" ], "Problem": "On a large, flat field there are n people positioned so that for each person the distances to all the other people are different. Each person holds a water pistol and at a given signal fires at the person who is closest. When n is odd\r\nprove that at least one person remains dry. Is this always true when n is even?", "Solution_1": "[quote=\"mdk\"]On a large, flat field there are n people positioned so that for each person the distances to all the other people are different. Each person holds a water pistol and at a given signal fires at the person who is closest. When n is odd\nprove that at least one person remains dry. Is this always true when n is even?[/quote]\r\n\r\n[hide=\"Solution\"]\nEVEN: Fails because we can pair people together.\n\nODD: If $ n=3$ it is trivial. So now we proceede with induction. Say there are $ n+2$ people and we want to show at least one remains dry. Use the Extreme Principle and look at the [i]smallest distance[/i]. Then those two people must shoot at each other. Now there are $ n$ other people. If any one of those other people shoots one of those two minimized people the proof is compelte because there are more remaining people then the people shooting. Otherwise there are $ n$ people shooting amongst eachother and by induction one is dry.\n[/hide]" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ (x,y,z)\\in{R_{+}^3},k\\in{R_{ + }},$prove that:\r\n\r\n$ A.\\sum{x^k[}$(x-y)$ +$(x-z)${ ]}\\ge0;$\r\n\r\n$ B.\\sum{x^k(x^2 - yz)}\\ge0.$", "Solution_1": "[quote=\"fjwxcsl\"]Let $ (x,y,z)\\in{R_{ + }^3},k\\in{R_{ + }},$prove that:\n\n$ A.\\sum{x^k[}$(x-y)$ +$(x-z)${ ]}\\ge0;$\n\n$ B.\\sum{x^k(x^2 - yz)}\\ge0.$[/quote]\r\n\r\njust ues Muirhead\r\n\r\n$ A.\\sum{x^k[(x - y) + (x - z)]}\\ge0 \\iff [k + 1,0,0] \\ge [k,1,0]$;\r\n\r\n$ B.\\sum{x^k(x^2 - yz)}\\ge0 \\iff [k + 2,0,0] \\ge [k,1,1]$.\r\n\r\nclear true.", "Solution_2": "[quote=\"fjwxcsl\"]Let $ (x,y,z)\\in{R_{ + }^3},k\\in{R_{ + }},$prove that:\n\n$ A.\\sum{x^k[}$(x-y)$ +$(x-z)${ ]}\\ge0;$\n\n$ B.\\sum{x^k(x^2 - yz)}\\ge0.$[/quote]\r\n\r\nor, \r\n\r\n$ A.\\sum{x^k[(x-y)+(x-z)]}\\ge0 \\iff \\sum{(x^k-y^k)(x-y)} \\ge 0$, \r\n\r\n$ B.\\sum{x^k(x^2 - yz)}\\ge0 \\iff \\sum{[(x^k-y^k)(x - y)(x + y) + z^{k}(x - y)^2]} \\ge 0$,\r\n\r\neasy to prove that when $ k>0$ have $ (x^k-y^k)(x-y) \\ge 0$, so both inequalitys holds;", "Solution_3": "[quote=\"kuing\"][quote=\"fjwxcsl\"]Let $ (x,y,z)\\in{R_{ + }^3},k\\in{R_{ + }},$prove that:\n\n$ A.\\sum{x^k[}$(x-y)$ +$(x-z)${ ]}\\ge0;$\n\n$ B.\\sum{x^k(x^2 - yz)}\\ge0.$[/quote]\n\nor, \n\n$ A.\\sum{x^k[(x - y) + (x - z)]}\\ge0 \\iff \\sum{(x^k - y^k)(x - y)} \\ge 0$, \n\n$ B.\\sum{x^k(x^2 - yz)}\\ge0 \\iff \\sum{[(x^k - y^k)(x - y)(x + y) + z^{k}(x - y)^2]} \\ge 0$,\n\neasy to prove that when $ k > 0$ have $ (x^k - y^k)(x - y) \\ge 0$, so both inequalitys holds;[/quote]\r\n\r\n Simple and nice ,kuing!\r\n\r\nFurther,we have\r\n\r\n $ A.\\sum_{x_1\\to{x_2}\\to{...}\\to{x_n}}{x_1^k[(x_1 - x_2) + (x_1 - x_3) + ... + (x_1 - x_n)]}\\ge0 ;$ \r\n\r\n $ B.\\sum_{x_1\\to{x_2}\\to{...}\\to{x_n}}{x_1^k(x_1^{n - 1} - x_2x_3...x_n)}\\ge0 ;$\r\n\r\n $ C.\\sum_{x_1\\to{x_2}\\to{...}\\to{x_n}}{x_1^k[(n - 1)x_1 - x_2 - x_3 - ... - x_n](x_1^{n - 1} - x_2x_3...x_n)}\\ge0 .$" } { "Tag": [ "modular arithmetic", "number theory unsolved", "number theory" ], "Problem": "Suppose that $ p$ is an odd prime, and write $ \\frac {1} {1}\\minus{}\\frac {1} {2}\\plus{}\\frac {1} {3}\\minus{}\\ldots\\minus{}\\frac {1} {p\\minus{}1}\\equal{}\\frac {a} {(p\\minus{}1)!}$. Show that $ a \\equiv \\frac {2\\minus{}2^p} {p} \\pmod p$.", "Solution_1": "By $ 2 \\minus{} 2^p \\equal{} 2 \\minus{} (\\sum_{i \\equal{} 0}^pC_p^i) \\equal{} \\minus{} (C_p^1 \\plus{} \\cdots \\plus{} C_p^{p \\minus{} 1})$ we have $ \\frac {2 \\minus{} 2^p}{p} \\equal{} \\minus{} \\sum_{i \\equal{} 1}^{p \\minus{} 1}\\dfrac{C_p^i}{p}$. Now in $ \\mathbb{F}_p$ : $ \\dfrac{C_p^i}{p} \\equal{} \\dfrac{(p \\minus{} i \\plus{} 1)\\cdots (p \\minus{} 1)}{i!} \\equal{} ( \\minus{} 1)^{i \\minus{} 1}\\cdot\\frac {1}{i}$ and we're done!.", "Solution_2": "See also http://www.mathlinks.ro/viewtopic.php?t=37352 .\r\n\r\n darij" } { "Tag": [ "geometry", "angle bisector" ], "Problem": "[color=darkred][b]18.[/b] Let $ ABC$ have an inscribed semicircle with a diameter on $ [BC]$ which touches $ AB$ and $ AC$ at $ F$ and $ E$ respectively. \nProvided that $ BF=1$ and $ CE=9$ and $ AB+AC=40$, find the area of $ \\triangle ABC$.[/color]\r\n\r\n[hide=\"Proof.\"][color=darkblue]Show easily that $ AE=AF=15$, $ AB=16$, $ AC=24$ and $ OC^{2}-OB^{2}=CE^{2}-BF^{2}=80$.\n\nFrom the relation $ \\frac{OB}{OC}=\\frac{[OAB]}{[OAC]}$ obtain $ \\frac{OB}{OC}=\\frac{AB}{AC}=\\frac{16}{24}=\\frac{2}{3}$. Thus, $ \\frac{OC^{2}}{9}=\\frac{OB^{2}}{4}=\\frac{OC^{2}-OB^{2}}{9-4}=\\frac{80}{5}=16$, i.e.\n\n$ OB=8$, $ OC=12$, $ BC=20$ and $ \\rho=OE=OF=3\\sqrt{7}$. In conclusion, $ [ABC]=\\frac{1}{2}\\cdot\\rho\\cdot (AB+AC)$ $ \\implies$ $ \\boxed{[ABC]=60\\sqrt 7}$.\n\n[b]Remark.[/b] The triangle $ ABC$ is similarly with the triangle $ A'B'C'$, where $ A'B'=4$, $ B'C'=5$, $ C'A'=6$. [b]Easy and nice problem, thank you ![/b][/color][/hide]", "Solution_1": "This is a nice solution. Here is a slightly different proposed problem that uses highlights his idea. \r\n\r\nGiven triangle $ ABC$, define $ A'\\in BC$ as the point such that there is a semicircle with center $ A'$ and diameter on $ BC$ that is tangent to $ AB$ and $ AC$. Define $ B'$ and $ C'$ similarly. Prove that $ AA'$, $ BB'$, and $ CC'$ are concurrent.", "Solution_2": "[quote=\"Altheman\"]This is a nice solution. Here is a slightly different proposed problem that uses highlights his idea. \n\nGiven triangle $ ABC$, define $ A'\\in BC$ as the point such that there is a semicircle with center $ A'$ and diameter on $ BC$ that is tangent to $ AB$ and $ AC$. Define $ B'$ and $ C'$ similarly. Prove that $ AA'$, $ BB'$, and $ CC'$ are concurrent.[/quote]\r\n\r\n[hide]That is, $ A'$ is on the angle bisector of A? (internal if $ B$ and $ C$ are acute, external otherwise).[/hide]", "Solution_3": "I never said that problem was difficult. I guess that I should have specified that it is acute.\r\n\r\n$ \\frac{BA'}{A'C}=\\frac{[BAA']}{[CAA']}=\\frac{AB}{AC}$ which implies that $ AA'$ bisects angle $ A$.", "Solution_4": "[quote=\"Altheman\"]I never said that problem was difficult. I guess that I should have specified that it is acute.\n\n$ \\frac{BA'}{A'C}=\\frac{[BAA']}{[CAA']}=\\frac{AB}{AC}$ which implies that $ AA'$ bisects angle $ A$.[/quote] It's just based on the fact that $ A'$ is equidistant from $ AB$ and $ AC$. I thought I'd give that way of seeing it: from a locus standpoint.\r\n\r\nThen again, it would still use Virgil Nicula's method on the original problem." } { "Tag": [ "superior algebra", "superior algebra solved" ], "Problem": "Find, with proof, the number of automorphisms of the group $Z_{n}$ with itself under modular addition.", "Solution_1": "That's easy: Every endomorphism of $\\mathbb{Z}_{n}$ is uniquely determined by the image of 1. Hence there are exactly n endomorphims which can be seen as multiplication in the ring $\\mathbb{Z}_{n}$. The multiplication with a is an automorphism iff a is invertible mod n iff gcd(a,n)=1.\r\nHence there are $\\phi(n)$ automorphisms of $\\mathbb{Z}_{n}$." } { "Tag": [ "inequalities", "inequalities solved" ], "Problem": "In a book written by J. I. Diaz, namely, Nonlinear partial differential equations and free boundaries, Lemma 4.10, there are 2 inequalities as follows\r\n\r\n\\[ \\begin{gathered} \\left| {a - b} \\right|^p \\leqslant C\\left( {\\left| a \\right|^{p - 2} a - \\left| b \\right|^{p - 2} b} \\right)\\left( {a - b} \\right) {\\rm \\quad for \\quad } p \\geqslant 2, \\hfill \\\\ \\left| {a - b} \\right|^2 \\leqslant C\\left( {\\left| a \\right|^{p - 2} a - \\left| b \\right|^{p - 2} b} \\right)\\left( {a - b} \\right)\\left( {\\left| a \\right| + \\left| b \\right|} \\right)^{2 - p} {\\rm \\quad for \\quad }1 < p < 2, \\end{gathered} \\]\r\nwhere $a$ and $b$ are in $\\mathbb{R}^n$.\r\n\r\nBecause I have not this book, so, I want to ask all of us for which conditions on C, these above inequalities hold. Thanks for each comment(s)...", "Solution_1": "$C=\\frac{1}{p-1}$ :)" } { "Tag": [ "\\/closed" ], "Problem": "[quote]Could not query weblogs information\n\nDEBUG MODE\n\nSQL Error : 1030 Got error -1 from storage engine\n\nSELECT w.*, u.*, e.* FROM phpbb_users u, phpbb_weblogs w LEFT JOIN phpbb_weblog_entries e ON e.weblog_id = w.weblog_id WHERE u.user_weblog = w.weblog_id AND w.deleted = 0 AND e.entry_deleted = 0 AND e.entry_time < 1110337280 ORDER BY e.entry_time DESC\n\nLine : 736\nFile : functions_weblog.php [/quote]\r\n\r\n*blinks* Well, it works now...Funny thing is i was nowhere near the weblogs...", "Solution_1": "Bet it didn't keep working for long. We had a little database problem tonight, but we've fixed it. Please let me know if you see any odd database behavior on the site.", "Solution_2": "Yea. It crashed again before I hit the next screen. Everything looks fine now. :-)" } { "Tag": [ "algebra", "polynomial", "linear algebra", "linear algebra unsolved" ], "Problem": "$A_{k} \\in M_{n}(R)$ such that $A_{i}A{j}=A_{j}A_{i}$ for $i\\neq j$ \r\n\r\nconsider the polynomial $P(x_{1},...x_{m})=det(A_{1}x_{1}+...+A_{m}x_{m})$ \r\n\r\nLet $M_{k}\\in M_{n}(R)$ matrices such that $A_{1}M_{1}+...+A_{m}M_{m}=0$\r\n\r\nProve that $P(M_{1},...,M_{m})=0$", "Solution_1": "The same idea as the one I used to solve the problem posted by you for $m=2$ (as problem from training agregation) works here as well. Of course, we have to place ourselves in the algebra generated by $A_i$, which is commutative." } { "Tag": [ "inequalities" ], "Problem": "Let k be a positive integer. $\\sum_{i=1}^{n}x_i=k$. What are the possible values for $\\sum{x_i^2}$", "Solution_1": "from what i remember, the original problem didn't require $k$ to be an integer", "Solution_2": "For the lower limit we apply the power mean - arithmetic mean inequality:\r\n\r\n$\\sqrt{\\frac{\\sum x_i^2}{n}} \\geq \\frac{\\sum x_i}{n}$\r\n\r\n$\\Rightarrow \\sum x_i^2 \\geq \\frac{k^2}{n}$\r\n\r\nwith equality iff $x_1 = x_2 = ... = x_n$. Note that this implies a lower limit of 0 as $n \\rightarrow \\infty$, with an arbitrarily small non-zero sum able to be constructed by setting equality of all variables and adjusting $n$ accordingly.\r\n\r\nFor the upper limit we note that\r\n\r\n$\\sum x_i^2 \\leq \\left(\\sum x_i \\right)^2 = k^2$\r\n\r\nwith equality iff $n = 1$ (we note that for $n > 1$, all $x_i$ are greater than zero, so the inequality is strict).\r\n\r\nWe therefore see that\r\n\r\n$0 < \\sum x_i^2 \\leq k^2$\r\n\r\nwhere we are able to tend toward the lower limit as $n$ increases, and toward the upper limit as $n$ decreases (with equality iff $n = 1$).", "Solution_3": "$k$ can be any positive real. Sorry about that.\r\n\r\nCan you also show how to get the value $k$ by construction?", "Solution_4": "If the problem doesn't specify $x_i$ to be positive, the expression can go to infinity...", "Solution_5": "[quote=\"paladin8\"]If the problem doesn't specify $x_i$ to be positive, the expression can go to infinity...[/quote]\r\nI think the problem did specify that (looking at Kalva's version).", "Solution_6": "The problem here is stated incorrectly (we're dealing with strictly infinite sums). The best way of approaching it is probably to deal with geometric series.", "Solution_7": "Yes. The original problem (paraphrased):\r\nLet $x_1,x_2,\\dots$ be a sequence of positive real numbers with $\\sum_{n=1}^\\infty x_n=A$. What are all possible values for $\\sum_{n=1}^\\infty x_n^2$?\r\n\r\nIt was a fairly easy problem, but it's easy to miss part of it if you're not careful." } { "Tag": [ "Euler", "email" ], "Problem": "As you probably know, a lot of terms in math are not pronounced how they would seem. In this thread, please post any terms you have heard pronounced wrong or once did pronounce wrong, and how to pronounce it correctly.\r\n\r\nI'll start with a few:\r\n\r\nEuler - OI ler [like an \"oiler\"]\r\nPtolemy - tal e mi [p is silent]\r\nPoisson - PWA so\r\nPoussin - Pu-(nasal sen)\r\n\r\nAnd I have a question on \"cyclic\":\r\nCyclic - sick-lick? sigh-clik? are both acceptable?", "Solution_1": "Its definitely sigh-click.", "Solution_2": "[url]http://www.merriam-webster.com/cgi-bin/dictionary?book=Dictionary&va=cyclic[/url]\r\n\r\nBoth are acceptable. I use both strangely enough.", "Solution_3": "I've only ever heard sick-lick.\r\n\r\nHow do you pronounce Godel?", "Solution_4": "I believe it is approximately \"guur-del.\" Anyhow, most of these aren't unusual pronunciations, they are just foreign pronunciations.", "Solution_5": "I've always pronounced it like the English word \"girdle\", which I think is close to what Joel said.", "Solution_6": "[url=http://en.wikipedia.org/wiki/Close-mid_front_rounded_vowel]Wikipedia[/url] has a quicktime sound file for the first vowel in his name -- it's close to what Ravi and I said, but with perhaps a little bit less \"r\" sound. It is, technically, the \"close-mid front rounded vowel.\" Go figure.", "Solution_7": "The one I haven't been able to figure out is Sylow.", "Solution_8": "Yes, I would also like to be definite on how to pronounce Sylow. I've heard it both like \"silo\" and like \"see-low\", and am under the impression it's the latter, but am not sure.", "Solution_9": "I remember my professor, Prof. Wolmer Vasconcelos, pronounce Sylow as \"see-low\". Perhaps that helps.", "Solution_10": "I have heard two pronunciations: Zee-lof and See-low. Do we have any Norwegian speakers who can clarify the matter for us?", "Solution_11": "[quote=\"ComplexZeta\"]I have heard two pronunciations: Zee-lof and See-low. Do we have any Norwegian speakers who can clarify the matter for us?[/quote]There are a couple of Norwegians on the site, however I'm not sure that they are active in this part :?", "Solution_12": "[quote=\"ComplexZeta\"]I have heard two pronunciations: Zee-lof and See-low. Do we have any Norwegian speakers who can clarify the matter for us?[/quote]\r\n\r\nThe first pronunciation would be much more plausible for a Norwegian pronunciation of \"Sylow.\" Alas, Sylow is not in the edition of Webster's New Biographical Dictionary I have at hand, which would provide a definitive answer for speakers of American English. Maybe you could send an email to the math department of St. Olaf College, which would surely know the answer to this question. ;) \r\n\r\nPersonally, I always thought \"Cauchy\" was a fairly tricky name, but it is easy to look up, and it has lots of oral tradition transmitting the preferred pronunciation to most math students.", "Solution_13": "Cauchy is pronounced co-she, right?\r\n\r\nHow about Heron as in Heron's formula?\r\nOr phi?", "Solution_14": "phi can be pronounced either fee or fie, but not fo or fum.", "Solution_15": "For the pronunciation of Cauchy, is it \r\n\r\n\"KO she\" \r\n\r\nor \r\n\r\n\"ko SHE\"\r\n\r\n? \r\n\r\nNever could figure this one out..." } { "Tag": [ "calculus", "integration", "analytic geometry", "trigonometry", "calculus computations" ], "Problem": "I just started learning how to evaluate evaluating double integrals by converting to polar coordinates. my question is how do you set the limits on each integrand?\r\n\r\nfor example i know that to evaluate $\\int_{0}^{\\infty}e^{x^{2}}$ you need to use $I^{2}=\\int_{0}^{\\infty}\\int_{0}^{\\infty}e^{-(x^{2}+y^{2})}dxdy$\r\nand then when u convert to polar you have \r\n$I^{2}=\\int\\int re^{-r^{2}}drd\\theta$ but how do i find the limits?", "Solution_1": "It's $e^{-x^{2}}$ in the first integral. $I^{2}$ is taken over the first quadrant, so you need a description of this quadrant in terms of $r$ and $\\theta$. With a very simple sketch you can convince yourself that $0\\le\\theta\\le\\pi/2$ while $r$ is unrestricted ($0\\le r<\\infty$).", "Solution_2": "thanks. but in general how do you go about determining the limits?", "Solution_3": "(1) draw the region of integration; \r\n(2) determine the range of $\\theta$ within the region;\r\n(3) find the smallest and largest value of $r$ along the half-line that makes angle $\\theta$ with the positive $x$-axis.\r\n\r\nTwo examples from my midterm exams:\r\n[quote]Evaluate the integral $\\int_{-1}^{1}\\int_{0}^{\\sqrt{1-x^{2}}}(x^{2}+y^{2})^{5/2}dy\\,dx$ using polar coordinates.[/quote]\n(1) the region of integration is the upper half of the disk of radius $1$.\n(2) $0\\le\\theta\\le\\pi$\n(3) for each $\\theta$ as in (2), the polar distance $r$ is between $0$ and $1$.\nObtain $\\int_{0}^{\\pi}\\int_{0}^{1}r^{6}\\,dr\\,d\\theta$, which is $\\pi/7$.\n\n[quote]Convert the integral $\\int_{-1}^{1}\\int_{|x|}^{1}xy \\,dy\\,dx$ into polar coordinates.[/quote]\r\n(1) the region of integration is the triangle with vertices $(0,0)$, $(1,1)$, and $(-1,1)$.\r\n(2) $\\frac{\\pi}{4}\\le\\theta\\le\\frac{3\\pi}{4}$\r\n(3) for each $\\theta$ as in (2), the smallest value of $r$ is $0$, while the largest occurs on the line $y=1$ (hypotenuse of the triangle). In polar coordinates this line has equation $r\\sin\\theta=1$, hence $r=\\csc\\theta$.\r\nAnswer: $\\int_{\\pi/4}^{3\\pi/4}\\int_{0}^{\\csc\\theta}r^{3}\\cos\\theta\\sin\\theta\\,dr\\,d\\theta$." } { "Tag": [ "function", "Euler", "calculus", "integration", "number theory" ], "Problem": "this person i know wants to find out what the Riemann Zeta Function is, and its uses. He does have an username on aops, but seems to be hibernating. so, i'm asking for him.\r\n\r\nif this is too easy or hard then please can a mod move it?", "Solution_1": "Dzeta function is defined for Re(s)>1 as Dzeta(s)=1/1^s+1/2^s+1/3^s+...; By unique factorization theorem a so called Euler formula holds: Dzeta(s)=Product(1/(1-1/p_n^s)), where the product is taken over p_n prime; Then it can be proved that it has meromorphic continuation to the whole plane (if I'm not mistaken it's done by the use of formula Dzeta(s)=1/Gamma(s)*integral from 0 to infinity of x^(s-1)/(e^x-1) dx; ) It can be then shown that it is regular everywhere in the plane, except a simple pole at s=1;", "Solution_2": "This function (because of existance of Euler formula) has connection to distribution of primes, and therefore is used in number theory; But it is not elementary; The famous Riemann conjecture asserts that all nontrivial zeros of Dzeta lie on the line Im(s)=1/2; (trivial zeroes are zeroes at -2,-4,-6...)", "Solution_3": "Another facts can be found in [url=http://mathworld.wolfram.com/RiemannZetaFunction.html]Mathworld[/url] site :lol:", "Solution_4": "Where are the zeros of zeta of s?\r\nto the tune of \"Sweet Betsy from Pike\"; words by Tom Apostol\r\n(Without any comments - Saw it some where...)\r\n\r\n\r\nWhere are the zeros of zeta of s?\r\nG.F.B. Riemann has made a good guess,\r\nThey're all on the critical line, sai he,\r\nAnd their density's one over 2pi log t.\r\n\r\nThis statement of Riemann's has been like trigger\r\nAnd many good men, with vim and with vigor,\r\nHave attempte to find, with mathematical rigor,\r\nWhat happens to zeta as mod t gets bigger.\r\n\r\nThe efforts of Landau and Bohr and Cramer,\r\nAnd Littlewood, Hardy and Titchmarsh are there,\r\nIn spite of their efforts and skill and finesse,\r\n(In) locating the zeros there's been no success.\r\n\r\nIn 1914 G.H. Hardy did find,\r\nAn infinite number that lay on the line,\r\nHis theorem however won't rule out the case,\r\nThere might be a zero at some other place.\r\n\r\nLet P be the function pi minus li,\r\nThe order of P is not known for x high,\r\nIf square root of x times log x we could show,\r\nThen Riemann's conjecture would surely be so.\r\n\r\n\r\nRelated to this is another enigma,\r\nConcerning the Lindelof function mu(sigma)\r\nWhich measures the growth in the critical strip,\r\nOn the number of zeros it gives us a grip.\r\n\r\nBut nobody knows how this function behaves,\r\nConvexity tells us it can have no waves,\r\nLindelof said that the shape of its graph,\r\nIs constant when sigma is more than one-half.\r\n\r\nOh, where are the zeros of zeta of s?\r\nWe must know exactly, we cannot just guess,\r\nIn orer to strengthen the prime number theorem,\r\nThe integral's contour must not get too near 'em." } { "Tag": [ "percent" ], "Problem": "hi, \r\nim stuck with this problem please help,\r\n\r\nAnne borrows 3200. Loan is paid 7 months later by a lump sum that includes 168 in interest. What was the simple rate of interest on the loan?\r\n\r\nIm not sure which formulate to use? its not the simple interest formula right?i dont think it is\r\n\r\n\r\nthanks", "Solution_1": "ans. (168/7)multiplied by 12 gives you amt. of int. that accrues for a year\r\n\r\nrate of int={(168/7)*12}/3200 best of luck.........", "Solution_2": "Believe it or not, it IS the simple interest thm. The percentage of the interest over the borrowed amount is 5.25 percent over the whole time period. If you want to calculate the monthly interest, you divide that by 7, to get 0.75 percent interest. If you want the yearly interest, you will multiply the monthly interest by 12 to get 9 percent simple interest. Hopefully this helps =D" } { "Tag": [], "Problem": "\\noindent The numbers 1, 2 and 3 are written in these nine unit squares. \n\\begin{itemize}\n\\item Each of the numbers appears three times, and \n there is only one number placed in each of the \n nine unit squares. \n\\item Each number is in a unit square horizontally \n or vertically adjacent to a unit square with the same number. \n\\item The sum of the numbers in the leftmost column and the sum \n of the numbers in the top row are each 7.\n\\end{itemize}\nWhat is the sum of the numbers in the four shaded squares?\n\n[asy]draw((0,0)--(30,0)--(30,30)--(0,30)--cycle);\ndraw((10,0)--(10,30));\ndraw((20,0)--(20,30));\ndraw((0,10)--(30,10));\ndraw((0,20)--(30,20));\nfill((0,0)--(10,0)--(10,10)--(0,10)--cycle,gray(0.7));\nfill((0,20)--(10,20)--(10,30)--(0,30)--cycle,gray(0.7));\nfill((20,0)--(30,0)--(30,10)--(20,10)--cycle,gray(0.7));\nfill((20,20)--(20,30)--(30,30)--(30,20)--cycle,gray(0.7));[/asy]", "Solution_1": "One way to fill this box is:\r\n\r\n$ 3$ $ 2$ $ 2$\r\n$ 3$ $ 3$ $ 2$\r\n$ 1$ $ 1$ $ 1$ (i believe this is legal.)\r\n\r\nIf this is so, then the corners add up to $ 7$.", "Solution_2": "Also, since you know the sum of the numbers in the left most column and the sum of the numbers in the top column are equal to 7, you can tell that the corners are equal to 7. I think. I'm not quite sure about this." } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "How to show there is no simple group of order $ 210$ , without using the odd-test of simple group .\r\n\r\nI always stuck in the techniques I know.\r\n\r\n1) Caley's Generalized theorem.\r\n\r\n2) the method of counting elements\r\n\r\n3) playing by center of the group , and normalizer .", "Solution_1": "Please ignore", "Solution_2": "I really don't know .\r\n\r\nthe odd test say :\r\n\r\nthere is no simple group of order $ 2n$ , where $ n$ is odd.\r\n\r\n :)", "Solution_3": "Done, Simon :wink: \r\n\r\nI just thinking about to get a contradiction from Sylow $ 7$-subgroup and Sylow $ 5$-Subgroup .\r\n\r\nSince If we assume $ n_7\\ne 1 \\Rightarrow n_7\\equal{}15$($ H_1,\\cdots, H_{15}$), and I assume there more than one Sylow $ 5$-subgroup say $ K$\r\n\r\nCould we show there is no other than identity element common in $ H_iK, \\forall i\\equal{}1,2\\cdots , 15$.?\r\n\r\n :)" } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Determine function $ f : R^{\\plus{}} \\minus{}\\minus{}>R^{\\plus{}}$ such that $ f(x\\plus{}y)\\plus{}f(x\\minus{}y)\\equal{}4 (f(x)f(y))^{1/2}$ for every $ x>y>0$.", "Solution_1": "There is no such function.\r\n\r\n1) $ f$ is bounded\r\nLet $ a\\ne1$ be positive, we know \r\n$ f(1\\plus{}a)<4\\sqrt{f(1)f(a)}$,\r\n$ P(1\\plus{}a,1): f(a)<4\\sqrt{f(1\\plus{}a)f(1)}$.\r\nHence, \r\n$ f(a)^2<16f(1\\plus{}a)f(1)<64f(1)^{3/2}f(a)^{1/2}\\Rightarrow f(a)<16f(1)$.\r\nSince $ a$ can be arbitrary, $ f$ is bounded.\r\n\r\nNow let $ A\\equal{}\\sup_{x>0}f(x)<\\infty$.\r\nCase 1)\r\nIf $ A$ is not attainable, we can find $ x>y$ such that $ f(x),f(y)$ are arbitrarily close to $ A$. \r\nThen $ 4\\sqrt{A^2}\\le 2A$. Contradiction!\r\nCase 2)\r\nIf $ f(b)\\equal{}A$. Then \r\n$ 4\\sqrt{f(b)f(3b)}\\le 2A$ and $ 4\\sqrt{f(b)f(2b)}\\le 2A$ imply \r\n$ 4\\sqrt{f(3b)f(2b)}\\le A$.\r\nBut\r\n$ 4\\sqrt{f(3b)f(2b)}\\equal{}f(5b)\\plus{}f(b)>A$. Contradiction!", "Solution_2": "maybe you made mistke :maybe: \r\n\r\n$ f(x)\\equal{}kx^2$ can be a solution", "Solution_3": "Maybe you made mistake.\r\n\r\n$ LHS \\equal{} 2k(x^2 \\plus{} y^2)$\r\n$ RHS \\equal{} 4kxy$.\r\n\r\nIf the original problem gives\r\n$ f(x\\plus{}y)\\minus{}f(x\\minus{}y)\\equal{}4\\sqrt{f(x)f(y)}$, not\r\n$ f(x\\plus{}y)\\plus{}f(x\\minus{}y)\\equal{}4\\sqrt{f(x)f(y)}$,\r\nthen it should be easy to derive the only solution $ f(x)\\equal{}kx^2$.", "Solution_4": "Sorry plz kill me :wallbash_red:" } { "Tag": [ "algebra", "polynomial", "induction", "binomial theorem", "algebra unsolved" ], "Problem": "A polynomial $ f(x)\\equal{}ax^4\\plus{}bx^3\\plus{}cx^2\\plus{}dx$ with $ a,b,c,d>0$ is such that $ f(x)$ is an integer for $ x \\in \\{ \\minus{}2,\\minus{}1,0,1,2 \\}$ and $ f(1)\\equal{}1$ and $ f(5)\\equal{}70$. \r\n$ (a)$ Show that $ a\\equal{}\\frac{1}{24}, b\\equal{}\\frac{1}{4},c\\equal{}\\frac{11}{24},d\\equal{}\\frac{1}{4}$. \r\n$ (b)$ Prove that $ f(x)$ is an integer for all $ x \\in \\mathbb{Z}$.", "Solution_1": "b) is a standard result in finite differences; any five consecutive integer values determines the finite difference table of the polynomial, from which all the other values of $ f(x), x \\in \\mathbb{Z}$ can be easily computed.", "Solution_2": "Interesting problem, I think. (Sorry for the length; I wanted to make my work more clear, and included a lot of space...)\r\n\r\n[hide=\"Solution\"](a).From the givens, we have the following equations in $ a$, $ b$, $ c$, $ d$:\n[list]\n[*]$ 16a\\minus{}8b\\plus{}4c\\minus{}2d\\in \\mathbb{Z}$ (1)\n[*]$ a\\minus{}b\\plus{}c\\minus{}d\\in \\mathbb{Z}$ (2)\n[*]$ a\\plus{}b\\plus{}c\\plus{}d\\equal{}1\\in \\mathbb{Z}$ (3)\n[*]$ 16a\\plus{}8b\\plus{}4c\\plus{}2d\\in \\mathbb{Z}$ (4)\n[*]$ 625a\\plus{}125b\\plus{}25c\\plus{}5d\\equal{}70$ (5)[/list]\nNote that sums and differences of integer multiples of integers remain integral. Thus, we can add and subtract (2) and (3) and do the same for (1) and (4):\n[list]\n[*]$ 2a\\plus{}2c\\in \\mathbb{Z}$ (6)\n[*]$ 2b\\plus{}2d\\in \\mathbb{Z}$ (7)\n[*]$ 32a\\plus{}8c\\in \\mathbb{Z}$ (8)\n[*]$ 16b\\plus{}4d\\in \\mathbb{Z}$ (9)[/list]\nSubtracting multiples of (6) from (8) and doing likewise for (7) and (9) yields the following:\n[list]\n[*]$ 24a\\in \\mathbb{Z} \\Leftrightarrow 24c\\in \\mathbb{Z}$ (10)\n[*]$ 12b\\in \\mathbb{Z} \\Leftrightarrow 12d\\in \\mathbb{Z}$ (11)[/list]\nNote that if $ 2a\\plus{}2c>2$, then $ a\\plus{}b\\plus{}c\\plus{}d>1$, contradiction. Also if $ 2a\\plus{}2c<0$, then at least one of $ a$ or $ c$ is negative, contradiction. Similarly, if $ 2a\\plus{}2c\\equal{}0$, then we cannot have $ a,c>0$, contradiction. In addition, if $ 2a\\plus{}2c\\equal{}2$, then $ b\\plus{}d\\equal{}0$ via (3). We conclude that because $ 2a\\plus{}2c\\in \\mathbb{Z}$, we have $ 2a\\plus{}2c\\equal{}1\\Leftrightarrow 2b\\plus{}2d\\equal{}1$ (via (3)). We can substitute $ c\\equal{}1/2\\minus{}a$ and $ d\\equal{}1/2\\minus{}b$ into (5) to obtain (simplifying) $ 120a\\plus{}24b\\equal{}11$. If $ a\\geq 3/24$, $ b<0$, which is a contradiction. If $ a\\equal{}1/12$, $ b\\equal{}1/24$, contradicting (11). Therefore, as $ a>0$ and $ 24a\\in \\mathbb{Z}$, we must have $ a\\equal{}1/24\\Leftrightarrow c\\equal{}11/24$ and $ b\\equal{}1/4\\Leftrightarrow d\\equal{}1/4$, as desired. We can check again that the properties are satisfied by this definition of $ f(x)$.\n\n(Note: It is not necessary to have $ a,b,c,d>0$; at least $ a,b,c,d\\geq 0$ or perhaps something even sharper suffices with small modifications to this proof.)\n\n(b). t0rajir0u is correct. However, we can also prove this by writing $ f(x)$ as $ \\frac{(x)(x\\plus{}1)(x\\plus{}2)(x\\plus{}3)}{24}$ and noting that among the four consecutive numbers $ x$, $ x\\plus{}1$, $ x\\plus{}2$, and $ x\\plus{}3$, there must be at least $ 1$ multiple of $ 3$, exactly $ 2$ multiples of $ 2$, and exactly $ 1$ multiple of $ 4$ (which is one of the multiples of $ 2$). Therefore, the numerator of $ f(x)$ is divisible by $ 3\\cdot 2\\cdot 4\\equal{}24$, as desired. We can also notice that $ f(x)$ equals the binomial coefficient $ \\binom{x\\plus{}3}{4}$ which must be an integer (we might have to make special considerations if we want to avoid $ a$ being negative in the definition of $ \\binom{a}{b}$.[/hide]", "Solution_3": "Here's an elementary proof of the fact that given a polynomial $ \\deg P(x)\\equal{}n$ so that $ P(t)\\in\\mathbb{Z}$ for a set of $ n\\plus{}1$ consecutive integers, $ P(t)\\in\\mathbb{Z}$ for all integers $ t$.\r\n\r\nWe proceed by induction on $ n$; consider $ n\\equal{}0\\Rightarrow P(x)\\equal{}c$. If we have that $ c\\equal{}t$ for an integer $ t$, then clearly $ P$ is an integer everywhere, so our base case is established.\r\n\r\nNow assume we have this result for $ 1,2,\\ldots,k\\minus{}1$; we show that it's true for $ n$. To establish this, let the integers so that $ P(x)$ is an integer be $ a_i\\equal{}a\\plus{}i,0\\le i \\le n,a\\in \\mathbb{Z}$. Observing that $ j(b\\plus{}1)^{k}\\minus{}jb^{k}$ is a $ k\\minus{}1$-degree polynomial as a trivial result of the binomial theorem, we define the polynomial $ Q(x)\\equal{}P(x)\\minus{}P(x\\minus{}1)$; from our result above $ \\deg Q(x)\\equal{}k\\minus{}1$. Setting $ x\\equal{}a_1,\\ldots,a_n$, we have that $ Q(x)$ is an integer (as the difference of two other integers) for $ k$ consecutive integers, so our inductive step is complete and the result is true.", "Solution_4": "Aren't we looking for the converse?" } { "Tag": [], "Problem": "Find the last three digits of the number 2005^2005.", "Solution_1": "[quote=\"cezar lupu\"]Find the last three digits of the number 2005^2005.[/quote]\r\n\r\n(2000 + 5)^2005 == x mod (1000);\r\n5^2005 == x mod (1000);\r\ncycle of 125 and 625 repeat for odd and even; hence last 3 digits are 125." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ x,y,z$ be real numbers, such that $ x\\plus{}y\\plus{}z\\equal{}1$.\r\nProve that the following inequality holds\r\n\r\n$ \\left.\\frac{x^2}{3x^2\\minus{}2x\\plus{}3}\\plus{}\\frac{y^2}{3y^2\\minus{}2y\\plus{}3}\\plus{}\\frac{z^2}{3z^2\\minus{}2z\\plus{}3}\\geq \\frac{1}{8}\\right.$", "Solution_1": "[quote=\"marin.bancos\"]Let $ x,y,z$ be real numbers, such that $ x + y + z = 1$.\nProve that the following inequality holds\n\n$ \\left.\\frac {x^2}{3x^2 - 2x + 3} + \\frac {y^2}{3y^2 - 2y + 3} + \\frac {z^2}{3z^2 - 2z + 3}\\geq \\frac {1}{8}\\right.$[/quote]\r\n\r\nNote that : $ 3x^2 - 2x + 3 = (x - 1)^2 + 2x^2 + 2 = (y + z)^2 + 2x^2 + 2 \\leq 2(x^2 + y^2 + z^2 + 1)$,\r\n\r\nHence : $ \\frac {x^2}{3x^2 - 2x + 3} \\geq \\frac {x^2}{2(x^2 + y^2 + z^2 + 1)}$ ,\r\n\r\nSimilarly,we get :\r\n\r\n$ \\frac {y^2}{3y^2 - 2y + 3} \\geq \\frac {y^2}{2(x^2 + y^2 + z^2 + 1)}$ and $ \\frac {z^2}{3z^2 - 2z + 3} \\geq \\frac {z^2}{2(x^2 + y^2 + z^2 + 1)}$ ,\r\n\r\nHence :\r\n\r\n$ \\left.\\frac {x^2}{3x^2 - 2x + 3} + \\frac {y^2}{3y^2 - 2y + 3} + \\frac {z^2}{3z^2 - 2z + 3}\\geq \\frac {x^2 + y^2 + z^2^}{2(x^2 + y^2 + z^2 + 1)} \\geq \\frac {1}{8}$\r\n\r\nBecause $ x^2 + y^2 + z^2 \\geq \\frac {(x + y + z)^2^}{3} = \\frac {1}{3}$\r\n\r\nThis ends the proof .", "Solution_2": "[quote=\"rachid\"]\n $ \\frac {x^2}{3x^2 \\minus{} 2x \\plus{} 3} \\geq \\frac {x^2}{2(x^2 \\plus{} y^2 \\plus{} z^2 \\plus{} 1)}$ \n[/quote]\r\nWithout this remark, written for $ x,y,z$, it's really difficult to prove the inequality! In fact, I don't think that it's possible. I invite you to find new solutions.\r\nThanks, [b]rachid[/b]! Your solution is also my solution!", "Solution_3": "[size=150]But Would is it easy,please?\n\n$ {\\frac {{x}^{2}}{3\\,{x}^{2} \\minus{} 2\\,x \\plus{} 3}} \\plus{} {\\frac {{y}^{2}}{3\\,{y}^{2} \\minus{} 2\\,y \\plus{} 3}} \\plus{} {\\frac {{z}^{2}}{3\\,{z}^{2} \\minus{} 2\\,z \\plus{} 3}}\\leq \\frac{1}{4}$[/size]", "Solution_4": "Could you be more specific, Mr. xzlbq ? Is it mentained the initial condition $ x\\plus{}y\\plus{}z\\equal{}1$? What do you mean? :mad:", "Solution_5": "I think the new condition should be $ |x| \\plus{} |y| \\plus{} |z| \\equal{} 1$ so that the inequality of xzlbq is true. :)\r\n\r\nLet's try to analyze this a bit further.\r\n\r\nIf I make $ x \\rightarrow \\infty$ , $ y \\rightarrow \\minus{}\\infty$ and $ z \\rightarrow 1$ then the expression will come to a value that exceeds $ \\frac{1}{4}$ . Hence, the inequality happens (not holds) when $ |x| \\plus{} |y| \\plus{} |z| \\equal{} 1$", "Solution_6": "[quote=\"marin.bancos\"]Let $ x,y,z$ be real numbers, such that $ x \\plus{} y \\plus{} z \\equal{} 1$.\nProve that the following inequality holds\n\n$ \\left.\\frac {x^2}{3x^2 \\minus{} 2x \\plus{} 3} \\plus{} \\frac {y^2}{3y^2 \\minus{} 2y \\plus{} 3} \\plus{} \\frac {z^2}{3z^2 \\minus{} 2z \\plus{} 3}\\geq \\frac {1}{8}\\right.$[/quote]\r\n\r\n\r\n$ \\frac {x^2}{3x^2 \\minus{} 2x \\plus{} 3} \\ge \\frac{6x\\minus{}1}{24}$\r\n\r\n$ \\iff \\frac{(3x\\minus{}1)^2(3\\minus{}2x)}{24(3x^2\\minus{}2x\\plus{}3)} \\ge 0,$\r\n\r\nso\r\n\r\n$ \\frac {x^2}{3x^2 \\minus{} 2x \\plus{} 3} \\plus{} \\frac {y^2}{3y^2 \\minus{} 2y \\plus{} 3} \\plus{} \\frac {z^2}{3z^2 \\minus{} 2z \\plus{} 3}$\r\n\r\n$ \\geq \\frac{6x\\minus{}1}{24}\\plus{}\\frac{6y\\minus{}1}{24}\\plus{}\\frac{6z\\minus{}1}{24}$\r\n\r\n$ \\equal{}\\frac18.$", "Solution_7": "Kuing, the inequality states that[b][u][i] $ x,y,z$ are reals [/i][/u][/b]", "Solution_8": "[quote=\"great math\"]Kuing, the inequality states that[b][u][i] $ x,y,z$ are reals [/i][/u][/b][/quote]\r\n\r\noh...sorry for my mistake... :( \r\n\r\nthank you for reminding me... :blush:", "Solution_9": "[size=150][quote=\"marin.bancos\"]Could you be more specific, Mr. xzlbq ? Is it mentained the initial condition $ x \\plus{} y \\plus{} z \\equal{} 1$? What do you mean? :mad:[/quote]\nno positive,Thanks.\nBQ[/size]", "Solution_10": "[quote=\"kuing\"]\n\n$ \\frac {x^2}{3x^2 \\minus{} 2x \\plus{} 3} \\ge \\frac {6x \\minus{} 1}{24}$\n\n$ \\iff \\frac {(3x \\minus{} 1)^2(3 \\minus{} 2x)}{24(3x^2 \\minus{} 2x \\plus{} 3)} \\ge 0,$\n\nso\n\n$ \\frac {x^2}{3x^2 \\minus{} 2x \\plus{} 3} \\plus{} \\frac {y^2}{3y^2 \\minus{} 2y \\plus{} 3} \\plus{} \\frac {z^2}{3z^2 \\minus{} 2z \\plus{} 3}$\n\n$ \\geq \\frac {6x \\minus{} 1}{24} \\plus{} \\frac {6y \\minus{} 1}{24} \\plus{} \\frac {6z \\minus{} 1}{24}$\n\n$ \\equal{} \\frac18.$[/quote]\r\nThis way still usefull, kuing. We only need more sentences.\r\nIf $ x, y, z$ are in $ [0, \\frac{3}{2}]$, the way above effect.\r\nIf in $ x, y, z$ at least have a number $ \\ge \\frac{3}{2}$, example assume $ 3\\minus{}2x \\le 0$\r\nNow, easily to see that $ \\frac{x^2}{3x^2\\minus{}2x\\plus{}3} \\ge \\frac{x^2}{3x^2}\\equal{}\\frac{1}{3}$\r\nSo, $ \\left.\\frac{x^{2}}{3x^{2}\\minus{}2x\\plus{}3}\\plus{}\\frac{y^{2}}{3y^{2}\\minus{}2y\\plus{}3}\\plus{}\\frac{z^{2}}{3z^{2}\\minus{}2z\\plus{}3}\\ge \\frac{1}{3}\\plus{}0\\plus{}0 \\ge \\frac{1}{8}\\right.$\r\nThe proof completed :)", "Solution_11": "[quote=\"xzlbq\"][size=150]But Would is it easy,please?\n\n$ {\\frac {{x}^{2}}{3\\,{x}^{2} \\minus{} 2\\,x \\plus{} 3}} \\plus{} {\\frac {{y}^{2}}{3\\,{y}^{2} \\minus{} 2\\,y \\plus{} 3}} \\plus{} {\\frac {{z}^{2}}{3\\,{z}^{2} \\minus{} 2\\,z \\plus{} 3}}\\leq \\frac {1}{4}$[/size][/quote]\r\n\r\nthen,\r\n[size=150]\nx,y,z are real numbers,prove that:\n$ {\\frac {{x}^{2}}{3\\,{x}^{2} \\minus{} 2\\,x \\plus{} 3}} \\plus{} {\\frac {{y}^{2}}{3\\,{y}^{2} \\minus{} 2\\,y \\plus{} 3}} \\plus{} {\\frac {{z}^{2}}{3\\,{z}^{2} \\minus{} 2\\,z \\plus{} 3}}\\leq \\frac {13}{12}$\nplease?[/size]", "Solution_12": "[quote=\"nguoivn\"][quote=\"kuing\"]\n\n$ \\frac {x^2}{3x^2 \\minus{} 2x \\plus{} 3} \\ge \\frac {6x \\minus{} 1}{24}$\n\n$ \\iff \\frac {(3x \\minus{} 1)^2(3 \\minus{} 2x)}{24(3x^2 \\minus{} 2x \\plus{} 3)} \\ge 0,$\n\nso\n\n$ \\frac {x^2}{3x^2 \\minus{} 2x \\plus{} 3} \\plus{} \\frac {y^2}{3y^2 \\minus{} 2y \\plus{} 3} \\plus{} \\frac {z^2}{3z^2 \\minus{} 2z \\plus{} 3}$\n\n$ \\geq \\frac {6x \\minus{} 1}{24} \\plus{} \\frac {6y \\minus{} 1}{24} \\plus{} \\frac {6z \\minus{} 1}{24}$\n\n$ \\equal{} \\frac18.$[/quote]\nThis way still usefull, kuing. We only need more sentences.\nIf $ x, y, z$ are in $ [0, \\frac {3}{2}]$, the way above effect.\nIf in $ x, y, z$ at least have a number $ \\ge \\frac {3}{2}$, example assume $ 3 \\minus{} 2x \\le 0$\nNow, easily to see that $ \\frac {x^2}{3x^2 \\minus{} 2x \\plus{} 3} \\ge \\frac {x^2}{3x^2} \\equal{} \\frac {1}{3}$\nSo, $ \\left.\\frac {x^{2}}{3x^{2} \\minus{} 2x \\plus{} 3} \\plus{} \\frac {y^{2}}{3y^{2} \\minus{} 2y \\plus{} 3} \\plus{} \\frac {z^{2}}{3z^{2} \\minus{} 2z \\plus{} 3}\\ge \\frac {1}{3} \\plus{} 0 \\plus{} 0 \\ge \\frac {1}{8}\\right.$\nThe proof completed :)[/quote]\r\n\r\nthank you... :)", "Solution_13": "Very interesting your remarks, dear friends! Thanks for them. I'd like also to thank for the alternative solution to this inequality.", "Solution_14": "Another Solution :\r\n\r\nWe can rewrite given inequality as:\r\n\\[ 1 \\minus{} \\frac {1}{8} \\geq \\sum_{cyc}{\\frac {1 \\minus{} \\frac {2x}{3}}{3x^2 \\minus{} 2x \\plus{} 3}}\\]\r\nSince $ 3x^2 \\minus{} 2x \\plus{} 3 \\geq \\frac {8}{3}$ It is obvious that :\r\n\\[ 1 \\minus{} \\frac {1}{8} \\equal{} \\sum_{cyc}{\\frac {1 \\minus{} \\frac {2x}{3}}{\\frac {8}{3}}}\\geq \\sum_{cyc}{\\frac {1 \\minus{} \\frac {2x}{3}}{3x^2 \\minus{} 2x \\plus{} 3}}\\]\r\nDone !", "Solution_15": "[quote=\"enndb0x\"]Another Solution :\n\nWe can rewrite given inequality as:\n\\[ 1 \\minus{} \\frac {1}{8} \\geq \\sum_{cyc}{\\frac {1 \\minus{} \\frac {2x}{3}}{3x^2 \\minus{} 2x \\plus{} 3}}\\]\nSince $ 3x^2 \\minus{} 2x \\plus{} 3 \\geq \\frac {8}{3}$ It is obvious that :\n\\[ 1 \\minus{} \\frac {1}{8} \\equal{} \\sum_{cyc}{\\frac {1 \\minus{} \\frac {2x}{3}}{\\frac {8}{3}}}\\geq \\sum_{cyc}{\\frac {1 \\minus{} \\frac {2x}{3}}{3x^2 \\minus{} 2x \\plus{} 3}}\\]\nDone ![/quote]\r\n\r\nYour solution is similar with kuing's and you still need nguoivn's remark.", "Solution_16": "Using the substitutions $ x\\equal{}\\frac{a\\plus{}1}{3}$, $ y\\equal{}\\frac{b\\plus{}1}{3}$, $ z\\equal{}\\frac{c\\plus{}1}{3}$, where $ a\\plus{}b\\plus{}c\\equal{}0$, we can write the inequality as\r\n$ \\sum \\frac {(a\\plus{} 1)^2}{a^2 \\plus{} 8} \\ge \\frac 3{8}$.\r\nWithout loss of generality, assume that $ bc\\ge 0$. By the Cauchy-Schwarz inequality, we have\r\n$ \\frac {(b\\plus{} 1)^2}{b^2 \\plus{} 8} \\plus{} \\frac {(c\\plus{} 1)^2}{c^2 \\plus{} 8}\\ge \\frac {(b\\plus{}c\\plus{}2)^2}{b^2 \\plus{} c^2 \\plus{} 16} \\equal{} \\frac {(a\\minus{}2)^2}{a^2 \\plus{} 16 \\minus{} 2bc}\\ge \\frac {(a\\minus{}2)^2}{a^2 \\plus{} 16}$.\r\nThus, it suffices to show that\r\n$ \\frac {(a\\plus{} 1)^2}{a^2 \\plus{} 8} \\plus{} \\frac {(a\\minus{}2)^2}{a^2 \\plus{} 16}\\ge \\frac 3{8}$,\r\nthat is\r\n$ a^2(13a^2\\minus{}16a\\plus{}160)\\ge 0$." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "AwesomeMath", "summer program" ], "Problem": "Hello,\r\nI am a girl who is interested in the China Girls Math Olympiad. I have been unable to find any information regarding how the team(s?) will be selected or if the US is even participating in it this year. I have already sent an e-mail to rjensen@maa.org more than 3 weeks ago but have not received any reply (although I realize that I may simply need to wait longer). It seems that last year's participants were already informed before the 2007 USAMO. Is this also the case this year, or is the 2008 USAMO factored in for team selection? Does anyone on this forum have information regarding the CGMO 2008?\r\n\r\nThank you.", "Solution_1": "CGMO is being run differently this year. I think USAMO 2008 scores will be used... so no one has been notified yet.", "Solution_2": "Thank you for your information!\r\nIs there any place I could find more information on it? Any additional details that you might know would also be great since I can find nothing whatsoever online regarding CGMO 2008.", "Solution_3": "Tell me how this goes! \r\n\r\nI've taken the Chinese Math Olympiad before and I have a stash of girl's olympiad problems, so PM me if you'd like some practice!", "Solution_4": "My understanding -- and this is unofficial, since I don't know for sure -- is that they will use a combination of 2007 and 2008 USAMO results to select the team. Last year the team trained at AwesomeMath before going to China, and I assume it's the same this year (but again, I don't know for sure)." } { "Tag": [ "number theory solved", "number theory" ], "Problem": "Prove that for a prime $p$ there is another prime $q$ for which $n^p-p$ is never divisible by $q$ for any integer value of $n$.", "Solution_1": "IMO 03\\6.", "Solution_2": "It's problem $6$ from IMO 2003. It's been discussed before, and there's even a generalization around here somewhere. I think the generalization states that there are infinitely many such primes $q$.", "Solution_3": "http://www.mathlinks.ro/Forum/viewtopic.php?t=98\r\n\r\n And there is also something about the density of the primes satisfiyng that condition.. now i don't remember exactly", "Solution_4": "Oh! thanks every one! i really did n't know that it is IMO problem :blush: !\r\n\r\nAnyhow thanks a lot!" } { "Tag": [ "algebra", "polynomial" ], "Problem": "Solve of the system of equations \r\n$y+xy^{2}=6 x^{2}$\r\n$1+x^{2}y^{2}=5x^{2}$", "Solution_1": "We have $6x^{2}-xy^{2}-y = 0$ and $(5-y^{2})x^{2}-1 = 0$. Solving for $x$ in terms of $y$,\r\n\r\n$x = \\frac{y^{2}+-\\sqrt{y^{4}+24y}}{12}$ \r\nand\r\n$x = \\frac{0+-\\sqrt{20-y^{2}}}{10-2y^{2}}$\r\n\r\nThus $20-y^{2}\\geq 0$ and $y(y^{3}+24) \\geq 0$. \r\n\r\nHmm.. right track so far?", "Solution_2": "[quote=\"chien than\"]Solve of the system of equations \n$y+xy^{2}=6 x^{2}$\n$1+x^{2}y^{2}=5x^{2}$[/quote]\r\n\r\nRahh...\r\n[hide]\n$x=0$ yields no solution, so $xy+x^{2}y^{2}=6x^{3}$. Then $6x^{3}-5x^{2}=xy-1$, $y=6x^{2}-5x+\\frac{1}{x}$, and by the second original equation, $y=5-\\frac{1}{x^{2}}$, so $6x^{2}-5x+\\frac{1}{x}=5-\\frac{1}{x^{2}}$, $6x^{4}-5x^{3}-5x^{2}+x+1=0$, $x=\\frac{1}{2}$, $y=1$?[/hide]\r\n\r\nEDIT: Wrong", "Solution_3": "[quote=\"ch1n353ch3s54a1l\"][quote=\"chien than\"]Solve of the system of equations \n$y+xy^{2}=6 x^{2}$\n$1+x^{2}y^{2}=5x^{2}$[/quote]\nby the second original equation, $y=5-\\frac{1}{x^{2}}$[/quote]\r\n\r\nErm, how did you get that?", "Solution_4": "[quote=\"Zellex\"]Erm, how did you get that?[/quote]\r\n\r\nPah. The answer works though >_>\r\n\r\n[hide]$xy=\\sqrt{5x^{2}-1}$, $6x^{3}=(xy)^{2}+xy$, $5x^{2}-1+\\sqrt{5x^{2}-1}=6x^{3}$, which...well, conveniently has the solution $x=\\frac{1}{2}$...I guess you can just test the reciprocals of the factors of 18...[/hide]", "Solution_5": "$a=xy$\r\n\r\n$a+a^{2}=6x^{3}$\r\n\r\n$1+a^{2}=5x^{2}$\r\n\r\n$[(a+a^{2})/6]^{2}=[(1+a^{2})/5]^{3}$\r\n\r\nnow we have an awesome 6th degree polynomial to solve...\r\n\r\nat which point...i bust out the calculator, $a=2,1/2, 18a^{4}+45a^{3}+86a^{2}+45a+18=0$\r\n\r\nthat factor happens to have nonreal solutions...which you can solve for...make the substiution $u=a+1/a$, etc\r\n\r\nthe problem is now solved conceptually...too much computation to finish" } { "Tag": [ "geometry", "3D geometry", "trigonometry", "integration", "calculus" ], "Problem": "We have a cone shaped glass dish. We put the open end of it in a another dish full of mercury. The distance between the mercury in the big dish and the tip of the cone is $H$. There is some mercury in the cone too. In the closed part above the mercury there is nytrogen of mass $m$. The outside pressure is equal to the hidrostatic pressure of a mercury pillar of height $H$. how much heat does the nytrogen pick up if the temperature rises slowly by $\\Delta T$ ? (The change of the mercury level in the big dish is negligible, and the level of mercury in the cone doenst go below the level of the big dish)", "Solution_1": "Is the coefficient of thermal expansion of mercury given?", "Solution_2": "$\\Delta T$ is small , so heat expansion is negligible.", "Solution_3": "So then, does that mean that the nitrogen's temperature rises by $\\Delta T$, and there are no other effects such as the thermal expansion of mercury?", "Solution_4": "Yes. all heat expansion of mercury is negligibe.", "Solution_5": "If nytrogen molecule is considered with 5 degrees of freedom, the heat should be\r\n\r\n $Q\\, ={13 \\over 4}{m \\over M}R\\Delta T$ where R stands for Clapeyron's constant and M for nytrogen's mol.\r\n I enjoyed it: quite interesting. Could you tell us where did you find it? ", "Solution_6": "Could you post your answer?", "Solution_7": "Yes Immanuel bonfils's answer is correct. you can post the solution if you want. :D", "Solution_8": "Can someone please post the solution to this problem?\r\n\r\nThanks.", "Solution_9": "Here is a solution:\r\nlet the height of the gase cone be $h$.\r\nSo the volume of the gas:\r\n$v=\\frac{\\pi h^{3}\\tan^{2}\\theta}{3}$\r\nWhere $\\theta$ is the half opening angle of the cone.\r\nSo the height of the gas cone:\r\n$h=V^{\\frac13}\\sqrt[3]{\\frac{3}{\\pi \\tan^{2}\\theta}}$\r\nnow lets find the pressure of the gas.\r\nThe outside pressure:\r\n$p_{0}=\\rho g H$ (\\rho is the mass density of mercury)\r\nIn the tub the pressure is equal throughout the surface:\r\n$p_{0}=p+\\rho g (H-h)$\r\nSo the pressure of the gas:\r\n$p=\\rho gh$\r\nUsing the formula we got for $h$:\r\n$p=\\rho g V^{\\frac13}\\sqrt[3]{\\frac{3}{\\pi \\tan^{2}\\theta}}$\r\nnow lets make the following substitution:\r\n$\\psi =\\sqrt[3]{\\frac{3}{\\pi \\tan^{2}\\theta}}$\r\nso the pressure:\r\n$p=\\psi V^{\\frac13}$\r\nThe work of the gas:\r\n$W=\\int_{V_{1}}^{V_{2}}pdV =\\psi \\int_{V_{1}}^{V_{2}}V^{\\frac13}dV$\r\nAfter evaluating the integral:\r\n$W=\\frac34 (\\psi V_{2}^{\\frac13}V_{2}-\\psi V_{1}^{\\frac13}V_{1})=\\frac34 (p_{2}V_{2}-p_{1}V_{1})=\\frac34 nR\\Delta T$\r\nAccording to the first law of thermodynamics, the heat taken up by the gas (nytrogen)\r\n$Q=\\Delta E+W = \\frac{5}{2}nR\\Delta T+\\frac34 nR \\Delta T =\\frac{13}{4}nR \\Delta T$" } { "Tag": [ "vector", "geometry", "function", "complex numbers", "real analysis", "real analysis unsolved" ], "Problem": "This is, actually, a geometry question that came up during an attempt to understand something about random analytic functions but, since there is no geometry subforum on the College Playground and since it can be asked without mentioning complex numbers at all, I'm placing it here.\r\n\r\nLet $n$ be a large number and $\\delta$ be a very small number (you may think that $\\delta=e^{-n}$). Suppose that $v_1,\\dots,v_n$ and $w_1,\\dots,w_n$ are unit vectors in the Euclidean space such that all the scalar products $(v_i,w_j)$ are less than $\\delta$ in absolute value. Is it true that then there exist unit vectors $v'_1,\\dots,v'_n$ and $w'_1,\\dots,w'_n$ such that $|v_i-v'_i|$ and $|w_j-w'_j|$ are all very small (say, do not exceed $n^{100}\\sqrt[10]\\delta$ or something similar) and each vector $v'_i$ is orthogonal to each vector $w'_j$?", "Solution_1": "[quote=\"fedja\"]and each vector $ v_i $ is orthogonal to each vector $ w_j $?[/quote]\r\n\r\nYou mean $ v_i' $ is orthogonal to $ w_j' $?", "Solution_2": "Yes. I've edited. Sorry for the misprint. :blush:", "Solution_3": "Gm. Consider a compact set $K=(S_{k})^{2n}$ of $2n$-uples of unit $n$-dimensional vectors, lying in a subspace generated by $v_i$ and $w_i$ ($k$ is a dimension of the span of these vectors). Define a function $f$ on $K$ as $\\sum_{i,j=1}^n |(x_i,y_j)|$, where $(x_1,x_2,\\dots,x_n,y_1,y_2,\\dots,y_n)$ is our $2n$-uple. Note that this function is continious, so the preimage of 0 is closed, denote it by $K_0$ and its open $\\epsilon$-neighborhood by $K_{\\varepsilon}$. Then the complement of $K_{\\epsilon}$ is closed, so $f$ is bounded by below on it. It means that if $v_1,\\dots,v_n,w_1,\\dots,w_n$ are far from any system of vectors with $(v_i',w_j')=0$, then the scalar product of some two $(v_i,w_j)$ is not too small. \r\n\r\nYou need some concrete estimates, yes?", "Solution_4": "Yes, I need the distances $|v_j-v'_j|$ and $|w_j-w'_j|$ to be less than $e^{-n/10}$ when $\\delta=e^{-n}$.\r\n($10$ can be replaced by any other absolute constant, but the estimate should be exponential in $n$)." } { "Tag": [ "number theory", "prime numbers", "number theory theorems" ], "Problem": "when i was learning prime numbers, this question came into my mind but i was unable to find any satisfactory answer.....so plz help me to clear my doubt.\r\n\r\nthe question is:\r\n\r\ncan we explain the reason of the existence of prime numbers in the number system?i mean can we give any logic against their existence or just they do?\r\n\r\nplease reply.......thank you.", "Solution_1": "so,what does this silence mean?isnt there any explanation of the existence of the primes?", "Solution_2": "They obviously exist ;)\r\nPlease state the question better or tell us why the existence of $ 2$ doesn't satisfy you.", "Solution_3": "i am not telling that i am not satisfied with the existence of primes e.g. 2.....i was just wondering why they exist necessarily?\r\nthe question just came into my mind.i know it sounds foolish....still i asked.\r\nthank you..... :|", "Solution_4": "[quote=\"hzbrl\"]when i was learning prime numbers, this question came into my mind but i was unable to find any satisfactory answer.....so plz help me to clear my doubt.\n\nthe question is:\n\ncan we explain the reason of the existence of prime numbers in the number system?i mean can we give any logic against their existence or just they do?\n\nplease reply.......thank you.[/quote]\r\n\r\nIf you have a set $ S$ and an operator \"*\" : $ S X S \\rightarrow S$ which is associative, commutative and with an identity element e, you can call \"prime element\" an element $ p\\neq e$ such that the only couples $ (x,y)\\in S^{2}$ such that $ p=x*y$ are $ (e,p)$ and $ (p,e)$\r\n\r\nWith such a definition :\r\nPrime numbers exist in $ (%Error. \"MathBB\" is a bad command.\n{N},multiplication)$. For example : $ 2,3,5,7,11,\\ldots$\r\nPrime numbers don't exist in $ (%Error. \"MathBB\" is a bad command.\n{Z},addition)$. \r\nPrime numbers exist in $ (\\{0\\}\\cup%Error. \"MathBB\" is a bad command.\n{N},addition)$. For example : $ 1$ (and it is the only prime number here).\r\nPrime numbers exist in $ (\\{1\\}\\cup\\{EvenPositiveIntegers\\},multiplication)$. For example : $ 6,10,14,\\ldots$\r\nPrime numbers don't exist in $ (%Error. \"MathBB\" is a bad command.\n{Q},multiplication)$.\r\n...\r\nSo there is no need for existtence of prime numbers. It just depend on the set and the operator (and on ... the definition :wink:) .", "Solution_5": "A good question. Prime numbers exist in all bases and I don't think the question can be answered easily. Primality/compositeness are not properties of numbers, rather they are properties of [i]relation[/i] by way of multiplication and division. Numbers by themselves do not constitute mathematics and as primitive objects they have no properties. They simply exist. These properties only come into existence with the idea of relation and from relation comes mathematics. Music is an analogy. Notes in themselves are not music. Music only comes into being when the notes are played in relation to each other. Relation is the essence of music. Notes are its 'atoms'. You might read Plotinus on Number and quantity.", "Solution_6": "Just a note: pco's definition of primes coincedes with the known ones on $ \\mathbb N$, but not for a lot of other sets (especially the ring theoretic defintion).", "Solution_7": "There is no logical reason why one number is prime and another is not. Numbers are primitive. They [i]precede[/i] logic. They simply are what they are." } { "Tag": [ "geometry", "geometric transformation", "reflection", "symmetry", "circumcircle", "trapezoid", "perpendicular bisector" ], "Problem": "Let $ABC$ be an acute triangle. Points $P,U$ are on segment $BC$, $Q,S$ on segment $CA$, $R,T$ on segment $AB$, such that\r\n\r\n$PR,US$ is perpendicular to $BC$\r\n$PQ,ST$ is perpendicular to $AC$\r\n$QR,TU$ is perpendicular to $BA$\r\n\r\n\r\nProve that triangles $PQR$,$STU$ are congruent.", "Solution_1": "you can prove that $QU//AB , PT//AC$ and$RS//BC$\r\nI think it easy", "Solution_2": "You can easly prove that these 6 points lie on one circle and then the thesis is obvious. :D", "Solution_3": "I don't think that's so easy...", "Solution_4": "[quote=\"TomciO\"]I don't think that's so easy...[/quote]\r\nReally?\r\n :idea: \r\nI think that when you know that these points are concyclic, it isn't difficult, just prove that $S$ is the reflection of point $P$ in the symmetry through the circumcenter of $\\triangle PQR$, then use some inequalities...", "Solution_5": "Of course. But I don't think that's easy to prove that these points are concyclic...", "Solution_6": "Oh, I solved this problem (because I take part in the Polish MO) and my proof is very simillar to what nxxx said. Especially, it's really easy to prove that these points are concyclic... :lol:", "Solution_7": "So, I would be glad to see the proof.", "Solution_8": "Let $O$ be the circumcenter of $\\triangle PQR$. Denote $P', Q', R'$ the points symmetric by $O$ to $P, Q, R$, respectively. Since $O$ lies on the perpendicular bisector of segment $PQ$, using Thales Theorem we obtain that $P'\\in AC$. Analogously, $Q'\\in AB$ and $R'\\in BC$. Triangle $P'Q'R'$ is congruent to triangle $PQR$ and has it's sides parallel to respective sides of $\\triangle PQR$ so $P'Q'$ is perpendicular to $AC$, simillarly for other sides.\r\n\r\nSuppose that $Q'$ is different from $T$, for example $AQ'CS$, then $CR'>CU$, $BU>BR'$, $BT>BQ'$, $AQ'>AT$. Contradiction.\r\n\r\nHence $Q'=T$, $P'=S$, $R'=U$ which means that triangles $PQR$ and $STU$ are congruent so we are done. :)", "Solution_9": "Well, this is the proof which I know, but actually I had something different on mind... When nxxx said that you can easily prove that these points are concyclic I thought that he means that it can be showed directly (without reflecting the points at first). And I think that's not that easy!\r\nBy the way, it's a nice proof.", "Solution_10": "It is clear that TRSQ, SQUP, and UPTR are cyclic quadrilaterals. If the circles around any two of these are equal than the 6 points are concyclic. Suppose then that all three circles are distinct. It is easy to see that the radical axes of pairs of these circles are the sides of the triangle ABC. But the sides of the triangle do not concur.", "Solution_11": "My proof uses a pretty strange lemma:\n\n[color=blue][b]Third theorem.[/b] Let ABC be a non-degenerate triangle. Let $A_{1}$, $A_{2}$, $B_{1}$, $B_{2}$, $C_{1}$, $C_{2}$ be points on the lines BC, BC, CA, CA, AB, AB, respectively. Assume that\n\n- the points $B_{1}$, $B_{2}$, $C_{1}$, $C_{2}$ lie on one circle,\n- the points $C_{1}$, $C_{2}$, $A_{1}$, $A_{2}$ lie on one circle,\n- the points $A_{1}$, $A_{2}$, $B_{1}$, $B_{2}$ lie on one circle.\n\nThen, all six points $A_{1}$, $A_{2}$, $B_{1}$, $B_{2}$, $C_{1}$, $C_{2}$ lie on one circle.[/color]\n\nHere, if the points $B_{1}$ and $B_{2}$ coincide, then an assertion like \"the points $B_{1}$, $B_{2}$, $C_{1}$, $C_{2}$ lie on one circle\" has to be understood in the following way: \"the circle through the points $B_{1}$, $C_{1}$, $C_{2}$ touches the line CA at the point $B_{1}$\". Similar conventions apply if the points $C_{1}$ and $C_{2}$ coincide or the points $A_{1}$ and $A_{2}$ coincide.\n\n[i]Proof of the Third theorem.[/i] Consider the following three circles:\n- the circle $k_{a}$ passing through the points $B_{1}$, $B_{2}$, $C_{1}$, $C_{2}$;\n- the circle $k_{b}$ passing through the points $C_{1}$, $C_{2}$, $A_{1}$, $A_{2}$;\n- the circle $k_{c}$ passing through the points $A_{1}$, $A_{2}$, $B_{1}$, $B_{2}$.\n\nIf at least two of these circles $k_{a}$, $k_{b}$, $k_{c}$ coincide, then they form one circle through the six points $A_{1}$, $A_{2}$, $B_{1}$, $B_{2}$, $C_{1}$, $C_{2}$, so it follows that the six points $A_{1}$, $A_{2}$, $B_{1}$, $B_{2}$, $C_{1}$, $C_{2}$ lie on one circle, and the Third theorem is proven.\n\nThus, it remains to prove that at least two of the circles $k_{a}$, $k_{b}$, $k_{c}$ coincide. In order to prove this, we assume the contrary, i. e. assume that no two of these circles $k_{a}$, $k_{b}$, $k_{c}$ coincide. Then, consider the radical axis of the circles $k_{b}$ and $k_{c}$. This radical axis must pass through the points of intersection of the circles $k_{b}$ and $k_{c}$, i. e. through the points $A_{1}$ and $A_{2}$; thus, this radical axis is the line BC (since the line passing through the points $A_{1}$ and $A_{2}$ is the line BC). So we see that the radical axis of the circles $k_{b}$ and $k_{c}$ is the line BC. Similarly, the radical axis of the circles $k_{c}$ and $k_{a}$ is the line CA, and the radical axis of the circles $k_{a}$ and $k_{b}$ is the line AB. Now, the pairwise radical axes of three circles always concur; applying this fact to the circles $k_{a}$, $k_{b}$, $k_{c}$, we thus obtain that the lines BC, CA, AB concur. But this is obviously wrong, since the triangle ABC is non-degenerate. So we get a contradiction, and thus our assumption (that no two of the circles $k_{a}$, $k_{b}$, $k_{c}$ coincide) was wrong. Hence, we have shown that at least two of the circles $k_{a}$, $k_{b}$, $k_{c}$ coincide, and therefore the Third theorem is proven.\n\nNow, a slightly extended statement of the problem:\n\n[color=blue][b]Problem.[/b] Let ABC be a non-degenerate triangle. Let $A_{1}$, $A_{2}$, $B_{1}$, $B_{2}$, $C_{1}$, $C_{2}$ be points on the lines BC, BC, CA, CA, AB, AB, respectively, such that $A_{1}B_{1}\\perp BC$, $B_{1}C_{1}\\perp CA$, $C_{1}A_{1}\\perp AB$, $C_{2}A_{2}\\perp BC$, $A_{2}B_{2}\\perp CA$, $B_{2}C_{2}\\perp AB$. Then, prove that:\n\n[b](a)[/b] The points $A_{1}$, $A_{2}$, $B_{1}$, $B_{2}$, $C_{1}$, $C_{2}$ lie on one circle.\n[b](b)[/b] We have $B_{1}C_{2}\\parallel BC$, $C_{1}A_{2}\\parallel CA$, $A_{1}B_{2}\\parallel AB$.\n[b](c)[/b] The triangles $C_{1}A_{1}B_{1}$ and $B_{2}C_{2}A_{2}$ are congruent.[/color]\n\n[i]Solution.[/i] Since $B_{1}C_{1}\\perp CA$ and $B_{2}C_{2}\\perp AB$, we have $\\measuredangle B_{2}B_{1}C_{1}=90^{\\circ}$ and $\\measuredangle B_{2}C_{2}C_{1}=90^{\\circ}$. Thus, the points $B_{1}$ and $C_{2}$ lie on the circle with diameter $B_{2}C_{1}$. This shows that the points $B_{1}$, $B_{2}$, $C_{1}$, $C_{2}$ lie on one circle. Similarly, the points $C_{1}$, $C_{2}$, $A_{1}$, $A_{2}$ lie on one circle, and the points $A_{1}$, $A_{2}$, $B_{1}$, $B_{2}$ lie on one circle. Thus, by the Third theorem, all six points $A_{1}$, $A_{2}$, $B_{1}$, $B_{2}$, $C_{1}$, $C_{2}$ lie on one circle, and problem [b](a)[/b] is solved.\n\nSince the points $A_{2}$, $B_{1}$, $B_{2}$, $C_{1}$ lie on one circle, we have $\\measuredangle B_{2}A_{2}C_{1}=\\measuredangle B_{2}B_{1}C_{1}$. Together with $\\measuredangle B_{2}B_{1}C_{1}=90^{\\circ}$, this yields $\\measuredangle B_{2}A_{2}C_{1}=90^{\\circ}$, so that $C_{1}A_{2}\\perp A_{2}B_{2}$. Since $A_{2}B_{2}\\perp CA$, this becomes $C_{1}A_{2}\\parallel CA$. Similarly, $A_{1}B_{2}\\parallel AB$ and $B_{1}C_{2}\\parallel BC$. Thus, part [b](b)[/b] is solved.\n\nThe fact that $C_{1}A_{2}\\parallel CA$ rewrites as $C_{1}A_{2}\\parallel B_{1}B_{2}$. Hence, the quadrilateral $C_{1}A_{2}B_{2}B_{1}$ is a trapezoid. Since this trapezoid has a circumcircle (the circle through the points $A_{1}$, $A_{2}$, $B_{1}$, $B_{2}$, $C_{1}$, $C_{2}$), it must be isosceles, so that $B_{1}C_{1}=A_{2}B_{2}$. Similarly, $C_{1}A_{1}=B_{2}C_{2}$ and $A_{1}B_{1}=C_{2}A_{2}$. Thus, the triangles $C_{1}A_{1}B_{1}$ and $B_{2}C_{2}A_{2}$ are congruent, and problem [b](c)[/b] is solved.\n\n Darij" } { "Tag": [ "geometry", "circumcircle", "geometry solved" ], "Problem": "Let points $A,B,C,D$ be on circle $O$, and a line $l$ pass through $O$. $AB,AC,CD,BD$ cut $l$ at $P,Q,R,S$ respectively. Show that $OP=OR\\leftrightarrow OQ=OS$.", "Solution_1": "If $O$ is unit circle of complex plane and $l$ intersect $O$ at $x$, we easly get that both conditions are equivalent to $x^{2}=\\frac{abc+bcd+cda+dab}{a+b+c+d}$.", "Solution_2": "[quote=\"malinger\"][color=darkred]Let points $A,\\,B,\\,C,\\,D$ be on circle $O$, and a line $l$ pass through $O$. $AB,\\,AC,\\,CD,\\,BD$ cut $l$ at $H,\\,J,\\,K,\\,I$ respectively. \nShow that $OH=OK\\Longleftrightarrow OI=OJ$[/color].[/quote]\r\n[color=indigo][b]Lemma.[/b] Let be given a quadrilateral $ABCD$ incribed in the circle $(O).$ A mobile point $M$ lies on arc $CD$ ( not contain $A,\\,B$). \nDefined $X=MA\\cap CD,\\,Y=MB\\cap CD$, then we have: $\\frac{XD\\cdot YC}{XY}=const$\n[b]Proof.[/b] Take point $E$ on the opposite to $CD$ so that $\\angle BEC=\\angle AMB$\n\nSo the quadrilateral $XMEB$ is cyclic\n $\\Longrightarrow XY\\cdot YE=YM\\cdot YB=YC\\cdot YD$ $\\Longrightarrow XY\\cdot CE=XD\\cdot YC\\Longrightarrow\\frac{XD\\cdot YC}{XY}=CE (const)$\n\n[b]Proof for the proposed problem.[/b] Denote $E,\\,F=l\\cap (O)$\n$\\bullet\\quad OH=OK$ \nApplying this [b]Lemma.[/b] for quadrilateral $BCFE,$ we have: $\\frac{EH\\cdot JF}{HJ}=\\frac{EI\\cdot KF}{IK}$\n\nOn the otherhand, $OH=OK\\Longrightarrow HE=KF\\Longrightarrow HF=EK$\n\nHence: $\\frac{JF}{HJ}=\\frac{EI}{IK}\\Longrightarrow\\frac{HF}{HJ}=\\frac{EK}{IK}\\Longrightarrow HJ=IK\\Longrightarrow OI=OJ$\n\n$\\bullet\\quad OI=OJ$ Similar, we have: $OH=OK$ [/color]", "Solution_3": "Please show the proof to your lemma, April. :)", "Solution_4": "How come $XY\\cdot CE=XD\\cdot YC$?\r\n\r\nMoreover, I do not see which points you applied the lemma to.", "Solution_5": "[color=indigo]We have: $XY\\cdot YE=YC\\cdot YD\\Longleftrightarrow XY\\cdot (YC+CE)=YC\\cdot (XY+XD)$ $\\Longleftrightarrow XY\\cdot CE= XD\\cdot YC$\n\nAnd I applying this [b]Lemma.[/b] for quadrilateral $BCFE$ and two points $A,\\,B$ lie on it's circumcircl, then we have $\\frac{EH\\cdot JF}{HJ}=\\frac{EI\\cdot KF}{IK}=const$.[/color]", "Solution_6": "[quote=\"malinger\"]Let points $ A,B,C,D$ be on circle $ O$, and a line $ l$ pass through $ O$. $ AB,AC,CD,BD$ cut $ l$ at $ P,Q,R,S$ respectively. Show that $ OP \\equal{} OR\\leftrightarrow OQ \\equal{} OS$.[/quote]\r\n\r\nIn other words, you want to prove that O is the midpoint of PR if and only if O is the midpoint of QS. This follows from Theorem 3 in the note \"On cyclic quadrilaterals and the butterfly theorem\" on [url=http://www.cip.ifi.lmu.de/~grinberg/]my website[/url] (in fact, the point O is its own orthogonal projection on the line $ l$, because O lies on $ l$).\r\n\r\n darij" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Let m,n be the positive integers.Show that 4mn-m-n can never be a square.", "Solution_1": "[hide=\"Solution\"]\nWe prove by contradiction.\nSuppose $ 4mn \\minus{} m \\minus{} n \\equal{} x^2$. We factor the left side (a usual trick) like this:\n$ 16mn \\minus{} 4m \\minus{} 4n \\equal{} 4x^2$\n$ (16mn \\minus{} 4m) \\minus{} (4n \\minus{} 1) \\equal{} 4x^2 \\plus{} 1$\n$ (4m \\minus{} 1)(4n \\minus{} 1) \\equal{} 4x^2 \\plus{} 1$\nWe see that the LHS is divisible by at least one prime in form of $ 4a \\minus{} 1$, so the RHS is also divisible by $ 4a \\minus{} 1$. But, that is impossible by Fermat's Little Theorem. So, it's never a square. [/hide]" } { "Tag": [ "trigonometry", "geometry", "power of a point", "trig identities", "Law of Cosines", "Pythagorean Theorem" ], "Problem": "I need to find out the angle of a satellite whos coverage is [u]+[/u]1482 Km. I have included a diagram. The only other information I have is its altitude 833 Km. See if you can solve.\r\n[img]http://img.photobucket.com/albums/v145/sealcock6/sat_swath_angle.gif[/img]", "Solution_1": "Use law of cosines and law of sines to establish 2 equations with two unknowns and then you can find the angle.", "Solution_2": "That doesn't work because of the curved side, DPopov.\r\n\r\nI think I have a method:\r\nIf you know the circumference of the earth, then you can find the fraction of the total circumference 1482 is. You can make it a degree measurement as a fraction of 360.\r\nPower of a Point is then appliable.\r\n\r\nOr something else with Power of a Point - you likely use that 833 somehow.\r\nHope that helps. :)", "Solution_3": "Draw $AE$, $AC$, and $EC$ and extend $BE$ to intersect $C$. Now it works. I wasn't reffering to use the curved side?? Of course make a triangle. You can express $AC$ using the Law of Cosines with $\\cos(2x)$ and also $\\frac{AC}{2}$ using the Law of Cosines with $\\cos(x)$.", "Solution_4": "How can you use those methods with the curved measurements of AE and EC?", "Solution_5": "Uhh..they are not curved....EA and AC are segments and NOT arcs.", "Solution_6": "It seems to me they are. Look at the problem title, and kilometers would probably only be surface measurement.", "Solution_7": "Oh, wait, does he mean the arcs....i thought he meant the segments. Then you are correct, I apologize.", "Solution_8": "so no one knows how to get the angle of ABC?", "Solution_9": "If the satellite's coverage is $\\pm 1482 \\text{ km}$ then $A$ and $C$ are points of tangency. Perpendicular radii, etc. solve for r, then use trig to find angle.\r\n\r\n$(AB)^2 + r^2 = (BE + r)^2$\r\n\r\n$\\angle ABC = 2 \\angle ABE = 2 \\arctan \\frac{r}{AB}$", "Solution_10": "Not nescerraly tangents. Bu if not, then the radius of the earth would probably be needed", "Solution_11": "Then again, there will always be some error due the earth not being perfectly spherical.", "Solution_12": "[quote=\"seamusoboyle\"] Bu if not, then the radius of the earth would probably be needed[/quote]\r\nIndeed, you're thinking about eratosthene method, right?", "Solution_13": "[quote=\"mathmanman\"][quote=\"seamusoboyle\"] Bu if not, then the radius of the earth would probably be needed[/quote]\nIndeed, you're thinking about eratosthene method, right?[/quote]\r\nYou can help me to solve equation with perfect part, thks", "Solution_14": "I solved the problem. Connecting A and C to the center of the earth R, then connecting A to C and connecting E to R. \r\nI then used 1482/ 40076 (circum of the earth at the equator). from the internet. Then I multiplied it by 360 and got 0.23 rad. This gave me ARE. \r\nThen I used the radius of the earth 6374 for AR.\r\nI made a new point D, where BR and AC intersected.\r\nUsing sin(13.31)x6374 = 1467.71 = AD\r\nand cos(13.31)x6374 = 6202.72 = DR\r\nER-DR = ED = 6374-6202.72 = 171.28\r\nED+BE = BD = 833 + 171.21 = 1004.28\r\nPythagorean Theorem AD^2 + BD^2 = AB^2\r\n1467.42^2 +1004.21^2 = AB^2, 1778.42 = AB\r\nUsing law of cosines\r\nAD^2 = BD^2 + BA^2 \u2013 2(BD*BA)cos(ABC)\r\ncos(ABC) = 0.56, acos(0.56) = ABC\r\nThen multiplied by 180/pi\r\nI got 55.62 degrees" } { "Tag": [ "inequalities open", "inequalities" ], "Problem": "Find the biggest $ k$ such that for all $ x,y,z\\ge0$ with $ x \\plus{} y \\plus{} z \\equal{} 3$, we have $ x^ky\\plus{}y^kz\\plus{}z^kx\\le 3$.", "Solution_1": "[quote=\"spanferkel\"]Find the biggest $ k$ such that for all $ x,y,z\\ge0$ with $ x \\plus{} y \\plus{} z \\equal{} 3$, we have $ x^ky \\plus{} y^kz \\plus{} z^kx\\le 3$.[/quote]\r\nI mentioned it here (a hungkhtn's problem),with his result we only need find k to $ 3\\ge \\frac {3^{k \\plus{} 1}k^k}{(k \\plus{} 1)^{k \\plus{} 1}}$,but how to solve it (hungkhtn 'problem) is very hard for me\r\n\r\n\r\nhttp://www.mathlinks.ro/viewtopic.php?t=219968" } { "Tag": [], "Problem": "$ \\frac{7x \\minus{} 14}{2x^2 \\minus{} 9x \\plus{}4} \\equal{} \\frac{A}{x\\minus{}4} \\plus{} \\frac{B}{2x \\minus{} 1}$\r\n\r\nWhat is A + B?", "Solution_1": "Yay partial fractions.\r\n[hide=\"outline\"]Clear denominators, sub in a value that makes a term 0, solve, blah.[/hide]", "Solution_2": "[hide=\"Answer\"]Do as XPMath said to get $ \\boxed{5}$.[/hide]", "Solution_3": "[hide=\"Heaviside Method\"]\n\n$7x - 14 = A(2x - 1) + B(x - 4)$\n\nLet $x = \\frac{1}{2}$:\n\n$7 \\cdot \\frac{1}{2} - 14 = A(2(\\frac{1}{2} - 1) + B(\\frac{1}{2} - 4) \\Rightarrow$\n$\\frac{7}{2} - \\frac{28}{2} = -\\frac{7}{2}B \\Rightarrow$\n$1 - 4 = -B \\Rightarrow$\n$B = 3$\n\nNow let $x = 4$:\n\n$7(4) - 14 = A(2(4) - 1) + B(4 - 4) \\Rightarrow$\n$14 = 7A \\Rightarrow$\n$A = 2$\n\nAnd $A + B = 2 + 3 = 5$[/hide]" } { "Tag": [ "geometry", "trigonometry" ], "Problem": "Find the area of triangle ABC if a=40ft b=25ft and 0$ then $ \\frac{x}{ay+bz} + \\frac{y}{az+bx} + \\frac{z}{ax+by} \\geq \\frac{3}{a+b}$.\r\n\r\nThe solution goes like so:\r\nusing this inequality :\r\n\r\n$ \\frac{a_{1}^2}{x_{1}}+ \\dots + \\frac{a_{n}^2}{x_{n}} \\geq \\frac{(a_{1}+ \\ldots +a_{n})^2}{x_{1}+ \\ldots +x_{n}}$ for $n=3 $\r\n\r\nwe get that:\r\n\r\n$ \\sum \\frac{x^2}{axy+bxz} \\geq \\frac{(x+y+z)^2}{(a+b)(xy+yz+zx)} \\geq \\frac{3}{a+b}$.\r\n\r\nfrom the fact that $( \\sum(x))^2 \\geq 3 \\sum(xy)$\r\n\r\nDoes anyone see another approach? Thank you very much!", "Solution_1": "very nicesoluton", "Solution_2": "The same problem is as an exercise on Titu's Lemma at [url=http://www.mathlinks.ro/weblog.php?w=1396]Potla's blog[/url] as problem 5." } { "Tag": [ "search" ], "Problem": "Consider a paper punch that can be centered at any point of the plane and that,\r\nwhen operated, removes from the plane precisely those points whose distance from the\r\ncenter is irrational. How many punches are needed to remove every point?", "Solution_1": "http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1314245630&t=90918" } { "Tag": [ "trigonometry", "function" ], "Problem": "If i want to solve for example $\\cos{\\frac{x}{2}}=\\sqrt{\\frac{1+\\cos x}{2}}$ does it mean that i have to draw a triangle, choose the angle $x$ and give it the value of $\\frac{adjacent}{hypotenuse}$ ?\r\n\r\nFor example $\\frac{a}{c}\\cdot \\frac{1}{2}= \\sqrt{\\frac{1+\\frac{a}{c}}{2}}$ where $a$ is the adjacent side and $c$ is the hypotenuse. Is it the way to solve this type of problems?", "Solution_1": "What do you mean by \"solve\"? If you want to \"solve an equation\", you can't, because that's an identity. If you want to prove the identity, you're off the wrong start, because $\\cos{x\\over 2}\\neq{a\\over c}\\cdot{1\\over 2}$", "Solution_2": ":blush: Indeed i made a terrible mistake and it seems that i lack some basic math vocabulary..\r\n\r\nIn other words, what are the steps to prove an identity which has some trig functions? What about solving an equation with a trig function?\r\n\r\nHope i made myself clear.", "Solution_3": "To prove the half-angles you need to have basic mastery of the double angles, which you can derive from the addition identities. For cosine, the double identity is: \r\n\r\n$\\cos(2a)=\\cos^{2}a-\\sin^{2}a = 2\\cos^{2}a-1$ \r\n\r\nExpressing $\\cos a$ in terms of $\\cos 2a$, we get\r\n\r\n$\\cos(2a)+1=2\\cos^{2}a$\r\n\r\n$\\cos^{2}a=\\frac{\\cos(2a)+1}{2}$\r\n\r\n$\\cos a=\\sqrt{\\frac{\\cos(2a)+1}{2}}$\r\n\r\n$\\cos \\left(\\frac{a}{2}\\right) = \\sqrt{\\frac{\\cos a+1}{2}}$ \r\n\r\nYou can do similar things for all the other identities." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Provethat there are many infinitely n s.t $n!$ is begun by 2005.", "Solution_1": "Have a look here :\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?highlight=langlois&t=23417\r\n\r\nPierre." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let positive real numbers $ a,b,c$ satisfying $ abc\\equal{}1$. Prove that $ \\frac{a}{b^2\\plus{}c^2}\\plus{}\\frac{b}{c^2\\plus{}a^2}\\plus{}\\frac{c}{a^2\\plus{}b^2}\\ge \\frac{3}{2}$", "Solution_1": "[quote=\"thaithuan_GC\"]Let positive real numbers $ a,b,c$ satisfying $ abc \\equal{} 1$. Prove that $ \\frac {a}{b^2 \\plus{} c^2} \\plus{} \\frac {b}{c^2 \\plus{} a^2} \\plus{} \\frac {c}{a^2 \\plus{} b^2}\\ge \\frac {3}{2}$[/quote]\r\n\r\ntry a=1/4, b=c=2" } { "Tag": [ "invariant" ], "Problem": "This is a favorite problem of mine.\r\n\r\nWhen 44444444 is written in decimal notation, the sum of its digits is A. Let B be the sum of the digits of A. Find the sum of the digits of B. (A and B are written in decimal notation).\r\n\r\nIf you've seen it before, hold off on responding for a while to give others a chance to work on it.", "Solution_1": "I love this problem. :)", "Solution_2": "It's a nice problem. I have seen it before right here on AoPS.", "Solution_3": "Yes, it's a cute problem, from an IMO I believe. I showed the problem to my (then) 9-year-old daughter, and she \"solved\" it -- with my generous hints, of course. She loved the fact that we could reason about such large numbers.", "Solution_4": "I have never seen it and I don't know how to solve it... could you provide some hints so that I might get started?", "Solution_5": "Spoon wrote:some hints so that I might get started?\n\n\n\n[hide]mod 9 + rough upper bound[/hide]", "Solution_6": "I'm curious... WHy did you choose mod 9?", "Solution_7": "[hide]Working modulo 9 is often a good thing to do for such a problem because it is invariant under sum of digits.[/hide]" } { "Tag": [ "limit", "real analysis", "real analysis unsolved" ], "Problem": "hello\r\n1/ Prove that :$ : (\\forall n\\in \\mathbb{N}^{*}) (\\forall \\alpha \\in [0,1[) 1\\leq (1 \\plus{} \\frac {\\alpha}{n})^{n}\\leq \\frac {1}{1 \\minus{} \\alpha}$\r\n\r\n2/let $ (a_n)$ a sequence , and $ (\\forall n \\in N) a_n > \\equal{} 0$ \r\nprove : if $ \\lim_{n\\rightarrow \\plus{} \\infty}n\\alpha_{n} \\equal{} 0$\r\nthan $ \\lim_{n\\rightarrow \\plus{} \\infty}(1 \\plus{} \\alpha_{n})^{n} \\equal{} 1$\r\n\r\nthx", "Solution_1": "1.) Write the problem in the following form:\r\nShow that the graph of $ f$ is below the graph of $ g$ with:\r\n$ f(x) \\equal{} 1\\plus{}\\frac{x}{n}$ and $ g(x) \\equal{} \\frac{1}{\\sqrt[n]{1\\minus{}x}}$\r\n\r\n2.) Let $ \\epsilon > 0$.\r\nThen there is an $ N$ so that $ |n\\alpha_n| < \\epsilon$ for $ n > N$.\r\nFor those n: $ 1 \\le (1 \\plus{} \\frac {n\\alpha_n}{n})^n < (1 \\plus{} \\frac {\\epsilon}{n})^n$\r\netc." } { "Tag": [ "symmetry" ], "Problem": "How many lines of symmetry does the symbol $\\Omega$ have?", "Solution_1": "[hide] 1 (down middle)[/hide]", "Solution_2": "[quote=\"math92\"]How many lines of symmetry does the symbol $\\Omega$ have?[/quote]\r\n[hide]yes absolutely it has only one symmetry line and it is the down middle. :D [/hide]" } { "Tag": [ "function", "probability and stats" ], "Problem": "Let $ X,Y: \\Omega \\to \\mathbb{R}$ be two random variables such that $ Y$ is $ \\sigma (X)$ measurable. I'm trying to proove that $ Y\\equal{}\\Phi (X)$ where $ \\Phi: \\mathbb{R} \\to \\mathbb{R}$ is Borel measurable. Our professor told us to take the $ \\Phi$ defined by $ \\Phi (X (Y^{\\minus{}1}(y)))\\equal{}y$ and prove that it's a well defined function. I couldnt manage to show that $ \\Phi$ is Borel measurable... My attempts to do so seem to depend on whether $ Im( X )$ is a Borel set or not... Any suggestions?", "Solution_1": "try first with simple functions $ Y_n \\equal{} \\sum 1_{A_i}$ and remember that any measurable function is a pointwise limit of such functions." } { "Tag": [ "geometry", "circumcircle", "incenter", "geometry proposed" ], "Problem": "[color=darkblue] Let $DEF$ be the orthic triangle of the [u]acute[/u] triangle $ABC\\ ,$ where $D\\in BC$ , $E\\in CA$ , $F\\in AB\\ .$ For a interior point $P$ denote the intersection $R\\in AP\\cap BC\\ .$ Prove that the circumcenter of the triangle $ABC$ coincides with the incenter of the triangle $CPR$ if and only if $\\triangle DEF\\sim \\triangle CPR\\ .$[/color]", "Solution_1": "Suppose that the circumcenter of the triangle $ABC$ coincides with the incenter of the triangle $CPR$.\r\n\r\nWe need to prove that the angles of triangle $CPR$ are the equal to $\\hat P,\\hat Q,\\hat R$ which are $180^\\circ-2A,180^\\circ-2B,180^\\circ-2C$ respectively.\r\n\r\nLet $X,Y,Z$ be foots of perpendicular from $O$(the circumcenter of $ABC$) to the sides $CP,RP,PC$ respectively. So we get $OX=OY=OZ$. And because $OA=OC$, $\\triangle OXC\\cong \\triangle OYA$. Because $\\widehat{BOC}=2A$, so $\\widehat{XOC}=A$. Therefore, $\\widehat{YOC}=A$. From $\\widehat{AOC}=2B$, we get $\\widehat{XOY}=2C$. Because $\\Box OXAY$ is cyclic, $\\widehat{PRC}=180^\\circ-2C$. $\\widehat{OCX}=90^\\circ-\\widehat{COX}=90^\\circ-A$. So $\\widehat{PCR}=180^\\circ-2C$.", "Solution_2": "Should it be[color=darkblue] the circumcenter of the triangle $ABC$ coincides with the incenter of the triangle $CPR$ if and only if $D=C,E=P,F=R$[/color]?", "Solution_3": "Yes ! Two similar triangles (in the mentioned notation $\\rightarrow$) : $XYZ\\sim MNP\\Longleftrightarrow X=M\\ ,\\ Y=N\\ ,\\ Z=P .$", "Solution_4": "O.K. then\r\n\r\nSuppose that $\\hat C,\\hat P,\\hat R$ are $180^\\circ-2A,180^\\circ-2B,180^\\circ-2C$\r\n\r\nLet $X,Y,Z$ be foots of perpendicular from $I$(the incenter of $CPR$) to the sides $CP,RP,PC$ respectively. Because $\\widehat{PRC}=180^\\circ-2C$, $\\widehat{RAC}=C$. So $RC=RA$. From $RX=RY$, $CX=AY$. And because $IX=IY$, so $\\triangle IXC\\cong \\triangle IYA$. Because $\\widehat{PCR}=180^\\circ-2A$, $\\widehat{XIC}=A$ also $\\widehat{YIC}=A$. Then, because $\\widehat{XIY}=2C$, we get $\\widehat{CIA}=2B$. From $\\widehat{CIA}=2B$ and because $IA=IC$, $I$ is the circumcenter of $ABC$." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "induction", "search" ], "Problem": "A lucky number is a positive integer that can be written as the sum of positive integers whose reciprocals sum to $ 1$. A number is unlucky if not lucky. How many unlucky numbers are there? Prove it.", "Solution_1": "Check [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=215882[/url]", "Solution_2": "Thank you for the related post, but I still have proof for my answer..", "Solution_3": "Infinitely many ;)", "Solution_4": "Certainly, but how do you prove so?", "Solution_5": "A USAMO problem asked you to prove that all integers greater than 32 are lucky.", "Solution_6": "Cool! But I'm not a great prover, so could you please give a hint of how to prove this? Thanks a bunch", "Solution_7": "It said, and I quote: \"Given that the integers between 33 and 73 inclusive are lucky,\"\r\nI don't know how to prove it either, but I'm working on it!", "Solution_8": "I've been working on a formula for lucky numbers composed of $ x$ different numbers (so far only successful with $ x\\equal{}2$) and finding a periodic number of lucky occurrences when the lucky number is made up of multiples of a certain number (e.g. lucky numbers made up of multiples of $ 2$ occur every $ 6$ numbers ($ 4\\equal{}2*2 ; 10\\equal{}2\\plus{}2*4 ; 16\\equal{}4*4 ; 22\\equal{}2\\plus{}4\\plus{}2*8$). Numbers composed of multiples of $ 3$ occur every $ 9$ ($ 9\\equal{}3*3 ; 18\\equal{}2*3\\plus{}2*6 ; 27\\equal{}3\\plus{}4*6$). Numbers composed of multiples of $ 5$ occur every $ 15$ ($ 25\\equal{}5*5 ; 40\\equal{}4*5\\plus{}2*10 ; 55\\equal{}3*5\\plus{}4*10$). Numbers composed of multiples of $ 7$ occur every $ 21$ ($ 49\\equal{}7*7 ; 70\\equal{}7*6\\plus{}2*14$; Etc.). From here it is easy to see (I still need to show algebraically) that when based solely on multiples of one number $ n$, there is a periodic occurrence of lucky numbers every $ 3n$. Now I feel I oughta find the relation of period of occurrences when composed of multiples of more than one number (e.g. $ 11\\equal{}2\\plus{}3\\plus{}6$) From there, I feel that induction wouldn't be bad. Anyone got an algebraic relation for the $ 3n$ of an idea for numbers composed of multiple multiples? Hopefully my rambling thoughts on this problem are of use..", "Solution_9": "$ \\frac{1}{a}\\plus{}\\frac{1}{b}\\equal{}\\frac{a\\plus{}b}{ab}$, so $ a\\plus{}b\\equal{}ab$. Solve for $ a$: $ a\\equal{}ab\\minus{}b\\equal{}b(a\\minus{}1)$. Solve for $ b$: $ b\\equal{}ab\\minus{}a\\equal{}a(b\\minus{}1)$. Substituting $ a(b\\minus{}1)$ for $ b$, we have $ a\\equal{}a(a\\minus{}1)(b\\minus{}1)$. Dividing by $ a$, we get $ (a\\minus{}1)(b\\minus{}1)\\equal{}1$. Since $ (\\minus{}1)(\\minus{}1)$ does not yeild positive integers for $ a$ and $ b$, we have to use $ (1)(1)$. Now we have $ a\\minus{}1\\equal{}b\\minus{}1\\equal{}1$, so $ a\\equal{}b\\equal{}2$. Thus, $ a\\plus{}b\\equal{}4$, so $ 4$ is the only lucky number. There are infinite unlucky numbers.", "Solution_10": "You do not only have to use two positive integers. For example, $ 11$ is lucky because $ 11 \\equal{} 2 \\plus{} 3 \\plus{} 6$ and $ \\frac {1}{2} \\plus{} \\frac {1}{3} \\plus{} \\frac {1}{6} \\equal{} 1$.\r\n\r\naahmadi: that seems like an unnecessarily complicated approach. I do not, however, seem to be able to find this problem in Engel. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1958492651&t=215882]this thread[/url].", "Solution_11": "@mathelete: There can be more than two variables.", "Solution_12": "Check these links:\r\n[url]http://www.research.att.com/~njas/sequences/A028229[/url]\r\n\r\n[url]http://www.research.att.com/~njas/sequences/A052428[/url]", "Solution_13": "[quote=\"1=2\"]It said, and I quote: \"Given that the integers between 33 and 73 inclusive are lucky,\"\nI don't know how to prove it either, but I'm working on it![/quote]\r\n\r\nUsing the ideas from [url]http://www.math.ucsd.edu/~fan/ron/papers/63_02_partitions.pdf[/url]\r\n\r\nthe transformations: \r\n(a) $ 1 \\equal{} \\frac {1}{d_1} \\plus{} \\cdots \\plus{} \\frac {1}{d_n} \\equal{} \\frac {1}{2} \\plus{} \\frac {1}{2d_1} \\plus{} \\cdots \\plus{} \\frac {1}{2d_n}$ transforms $ d_1 \\plus{} \\cdots \\plus{} d_n\\equal{}m$ to $ 2m \\plus{} 2$\r\n\r\n(b) $ 1 \\equal{} \\frac {1}{d_1} \\plus{} \\cdots \\plus{} \\frac {1}{d_n} \\equal{} \\frac {1}{3} \\plus{} \\frac {1}{6} \\plus{} \\frac {1}{2d_1} \\plus{} \\cdots \\plus{} \\frac {1}{2d_n}$ transforms $ m$ to $ 2m \\plus{} 9$" } { "Tag": [], "Problem": "For a concert, 100 more balcony tickets than main-floor tickets were sold. The balcony tickets cost $4 and the main-floor tickets cost$12. If the total sales for BOTH types of tickets was $3056, find the number of balcony tickets that were sold. My work: Let x = number of balcony tickets$4(x + 100) + $12 =$3056 \r\n\r\nIs this the correct equation to use? \r\n\r\nOr do I use this one: \r\n\r\n4X + 100 + 12X = 3056?", "Solution_1": "[hide]let $x$=balcony tickets and $y$=floortickets\n\n$x=y+100$\n$4x+12y=3056$\n\nNow just solve the system and find x+y\n[/hide]", "Solution_2": "[quote=\"anyone\"][hide]let $x$=balcony tickets and $y$=floortickets\n\n$x=y+100$\n$4x+12y=3056$\n\nNow just solve the system and find x+y\n[/hide][/quote]\n\nEdited\n[hide=\"Finishing up the solution\"]Substitute equation 1 into equation 2:\n\\[4(y+100)+12y = 3056 \\]\n\n\\[16y=2656 \\]\n\n\\[y=166 \\]\n\n\\[x=y+100=266 \\][/hide]", "Solution_3": "The book's answer is 266 balcony tickets.\r\n\r\nHow do I get 266 tickets?", "Solution_4": "[quote=\"vishalarul\"]\n\\[4(y+100)+12y=3056\\]\n\\[16y=2056\\]\n[/quote]\r\nThere's your mistake, that should be a $2656$, thus making $y=166$. Thus, we find that $x=166+100=\\boxed{266}$." } { "Tag": [ "function", "calculus", "derivative", "real analysis", "real analysis solved" ], "Problem": "Let f:R->R a function wich has a primitive F:R->(R+)*.\r\nShow that for every \\varepsilon >0 there exists a real x such that\r\nIf(x)I< \\varepsilon", "Solution_1": "Let's assume there is a c>0 s.t. |f(x)|>=c for all x \\in R. Since f is a derivative, it has Darboux's property, so it can't change signs, so we may assume f(x)>=c for all x \\in R.\r\n\r\nThe property stated above means that the function G(x)=F(x)-cx is increasing. G(x)->00 as x->-00 (because F(x)>0 for all x \\in R), so we have a contradiction (we can find, for example, a small enough x<0 s.t. G(x)>G(0), but G(x)<=G(0) because G is supposd to be increasing).\r\n\r\nThe samt thing happens if we assume f(x)<-c for all x\\in R.\r\n\r\nIt's obvious that all of this works for all f which have a bounded primitive F." } { "Tag": [ "calculus", "real analysis", "linear algebra" ], "Problem": "So I finally got around to buying this book recently. I hadn't seen it since I was in high school, when it was in it's second edition. I mostly liked it because it had proofs of some things that I hadn't seen before. Like the proof that $\\pi$ is irrational and $e$ is transcendental. Upon seeing it again though, I have a couple of comments.\r\n\r\nFirst of all, its a really well written and thought out textbook. I like Spivak's writing style and the problems lead you to some interesting results (as is typical of all of his textboks). But the book seems to fall into an awkward place in terms of curriculum.\r\n\r\nIts too technical in my opinion to be used as into a \"Calculus\" textbook, and at the same time not quite modern or thorough enough to be used for an \"Introduction to Analysis\" textbook. So I'm not sure that it has much of an audience aside from being a cool book to have on the shelves of people who've already studied single variable calculus.\r\n\r\nThe reason why I say its too technical for a \"Calculus\" book, is that I believe students should get aquinted with the techniques and applications of Calculus before they get into the proofs of its theorems. Of course the two pedagogies aren't mutually exclusive, but this book lies heavily on the side of the latter. Nearly all of the exercises are of an advanced nature and very few of them deal with applications or purely conceptual understanding.\r\n\r\nFor intro to Calculus, I like the books [u]Thomas' Calculus[/u] by Thomas, Weir, Hass, and Giorgano or [u]Calculus[/u] by Stewart. They are for the most part what people call standard \"plug and chug\" textbooks, but they have some good problems in them (especially Thomas' applications and Stewarts Concept Check and Problem Corner problems).\r\n\r\nThe reason why I say its not modern or thorough enough to be used for an \"Analysis\" book, is that well.. it isn't. Just compare it to one of the standard Intro to Analysis books, Rudin's [u]Principles of Mathematical Analysis[/u], and you will see that the coverage of topics is vastly different. I don't think this book alone would prepare someone for upperclass work in Mathematics.\r\n\r\nOne way that I could see this book being used is for motivated students who finish the typical BC/IB Calculus curriculum early in high school and want to go back and learn more technical details of WHY the theorems are true. This is something that can be done in place of, or in addition to, the more typical Multivariable/Diff. Eqn's/Lin. Algebra courses that high schoolers do after BC Calculus. But I doubt there are many high school teachers who could help you with the problems in this book.", "Solution_1": "[quote=\"gauss202\"]\nOne way that I could see this book being used is for motivated students who finish the typical BC/IB Calculus curriculum early in high school and want to go back and learn more technical details of WHY the theorems are true. This is something that can be done in place of, or in addition to, the more typical Multivariable/Diff. Eqn's/Lin. Algebra courses that high schoolers do after BC Calculus. But I doubt there are many high school teachers who could help you with the problems in this book.[/quote]\r\n\r\nThat sounds like something I might do. But what do you think would be the best route for someone just taking a BC class? I am going to try to change my schedule next year so I can take Linear Algebra and/or the next level of Calculus at a local college.\r\n\r\nFrom your last paragraph I get the impression that Spivak's book is a [i]good[/i] book for high school students under those circumstances but are there books that are better for the same types of students, i.e. do you know of any books that are better than Spivak's?\r\n\r\n[i][edit: I just read the thread \"Sequence of Learning Advanced Math\" and it answered a bunch of my questons][/i]", "Solution_2": "Well, it is a great source of calculus problems, regardless of whether or not its used in a curricular setting.", "Solution_3": "Rudin has a proof of irrationality of $e$ (it's a standard exercise in all analysis texts; e.g, it's in all four introductory analysis texts I own). As I recall, the one for $\\pi$ isn't much different, expand taylor series, blah blah blah blah.", "Solution_4": "[quote=\"blahblahblah\"]Rudin has a proof of irrationality of $e$ (it's a standard exercise in all analysis texts; e.g, it's in all four introductory analysis texts I own). As I recall, the one for $\\pi$ isn't much different, expand taylor series, blah blah blah blah.[/quote]\r\n\r\nThe proof for $e$ is an exercise, the proof for $\\pi$ is incomparably harder.", "Solution_5": "The proof of irrationality of $e$ is on p.65 of the third edition.\r\n\r\nAs for the proof of $\\pi$, yes, it is harder - I was mistaken in what I said. They both end up with proving that if $\\pi$ or $e$ were rational, there would be an integer between $0$ and $1$, which is probably why I remembered them as similar.", "Solution_6": "Re: the use of Spivak\r\n\r\nWhen I was a senior in high school, I had the standard high school calculus course, that is, all mechanics, no theory. I can't even remember which book we used.\r\n\r\nIn freshman year of college, the math department grouped the smart math majors (or were we self-selected?) into a two-year honors sequence. The first year was Calculus, using Spivak's book. I thought it fit the bill nicely -- I loved the book.", "Solution_7": "[quote=\"blahblahblah\"]The proof of irrationality of $e$ is on p.65 of the third edition.\n\nAs for the proof of $\\pi$, yes, it is harder - I was mistaken in what I said. They both end up with proving that if $\\pi$ or $e$ were rational, there would be an integer between $0$ and $1$, which is probably why I remembered them as similar.[/quote]\r\n\r\nActually when I wrote the comments above I was thinking of the proofs that $\\pi$ is transcendental, not irrational.\r\nStill, proving $e$ is irrational is at the level of an olympiad problem (there is only one move you can make, so make it...) whereas it would be astonishing if anybody were to come up with the irrationality proof for $\\pi$ within a few hours of learning that the result is true.\r\n\r\nre: Spivak's book, I made several possibly-relevant postings about it in the thread \"Best Calculus Textbook\", on this board around Dec 2004." } { "Tag": [ "real analysis", "calculus", "integration", "function", "trigonometry", "limit", "abstract algebra" ], "Problem": "If $ f: \\mathbb R\\rightarrow \\mathbb R$ ,such that :\r\n\r\n\\[ f(x)\\equal{}\\left\\{ \\begin{array}{crl} x\\plus{}1 &,&\\mathrm{ x \\ is \\ rational } \\\\ 2x\\minus{}1 &,& \\mathrm{x \\ is \\ irrational } \\end{array}\\right.\\]\r\n\r\n find the Leb. $ \\int_{\\mathbb R} f$", "Solution_1": "It doesn't exist. The \"rational\" part is irrelevant (sets of measure zero never matter) and the function $ x\\mapsto 2x\\minus{}1$ is not Lebesgue integrable over $ \\mathbb R$.", "Solution_2": "[quote=\"fedja\"]It doesn't exist. The \"rational\" part is irrelevant (sets of measure zero never matter) and the function $ x\\mapsto 2x \\minus{} 1$ is not Lebesgue integrable over $ \\mathbb R$.[/quote]\r\n\r\nWhat about to put $ \\cos(x)$ instead of $ 2x\\minus{}1$.\r\n\r\nThx", "Solution_3": "Indeed , I'm warping about to find the value of it .", "Solution_4": "The integral of $ \\cos x$ over $ \\mathbb R$ does not exist either in Lebesgue or improper-Riemann sense. One could argue for it being equal to $ 0$ in some Pickwickian sense. (Like in summability methods for series)", "Solution_5": "[quote=\"mlok\"]The integral of $ \\cos x$ over $ \\mathbb R$ does not exist either in Lebesgue or improper-Riemann sense. One could argue for it being equal to $ 0$ in some Pickwickian sense. (Like in summability methods for series)[/quote]\r\n\r\nwhat is that ?Pickwickian sense? Sorry :blush:", "Solution_6": "For example, I can define a generalization of Lebesgue integral of $ f$ over $ \\mathbb R$ as \r\n$ \\lim_{\\epsilon\\to 0}\\int_{\\mathbb R}\\exp(\\minus{}\\epsilon^2 x^2) f(x)\\,dx$ when the limit exists. Note that the factor $ \\exp(\\minus{}\\epsilon^2 x^2)$ converges to $ 1$ uniformly on compact subsets as $ \\epsilon\\to 0$. It is also uniformly bounded. One can show that if $ f$ is Lebesgue integrable on $ \\mathbb R$, the above limit is equal to the integral. \r\n\r\nNow take $ f(x)\\equal{}\\cos x$: \r\n$ \\int_{\\mathbb R}\\exp(\\minus{}\\epsilon^2 x^2) \\cos (x)\\,dx\\equal{}\\frac{\\sqrt{\\pi}}{\\epsilon} \\exp(\\minus{}1/(4\\epsilon^2))$. This converges to $ 0$ as $ \\epsilon\\to 0$. Hence the Pickwickian integral of $ \\cos x$ over $ \\mathbb R$ is $ 0$.\r\n\r\n(It may have a real name, unrelated to Dickens, but I don't know it.)", "Solution_7": "That's a summability method, of course. It's not exactly Abel summability, but it's related.\r\n\r\nTwo-sided Abel summability would be $ \\lim_{t\\to 0^\\plus{}}\\int_{\\minus{}\\infty}^{\\infty}e^{\\minus{}t|x|}f(x)\\,dx.$\r\n\r\nThe Abel sum of this particular integral would be $ \\lim\\frac{2t}{1\\plus{}t^2}\\equal{}0,$ also.\r\n\r\nOr you could try Cesaro means: $ \\lim_{N\\to\\infty}\\frac1N\\int_0^N\\int_{\\minus{}y}^yf(x)\\,dx\\,dy$\r\n\r\nIn the case of this particular integral, that would also be $ \\lim_{N\\to\\infty}\\frac2N(1\\minus{}\\cos N)\\equal{}0.$\r\n\r\nFor all of these methods, including your no-name method, it shouldn't be too difficult to prove the \"abelian\" theorem: if the integral exists as an improper integral, then it is summable to the same number. There should also be an array of \"tauberian\" theorems: if $ f$ satisfies such-and-such a condition (perhaps a size condition at $ \\infty$), and the integral is summable, then it converges as an improper integral. Of course, $ f(x)\\equal{}\\cos x$ isn't going to satisfy any of those tauberian conditions." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Prove that for all nonnegative real numbers $ a,b,c$ we have\r\n\r\n$ a\\sqrt {a^2 \\plus{} bc} \\plus{} b\\sqrt {b^2 \\plus{} ca} \\plus{} c\\sqrt {c^2 \\plus{} ab} \\ge \\sqrt {2} (ab \\plus{} bc \\plus{} ca).$\r\n\r\n[hide=\"Solution 1\"] By Minkowski Inequality we have\n\n$ \\quad a\\sqrt {a^2 \\plus{} bc} \\plus{} b\\sqrt {b^2 \\plus{} ca} \\plus{} c\\sqrt {c^2 \\plus{} ab} \\\\\n\\equal{} \\sqrt {(a^2)^2 \\plus{} \\left(a\\sqrt {bc}\\right)^2} \\plus{} \\sqrt {(b^2)^2 \\plus{} \\left(b\\sqrt {ca}\\right)^2} \\plus{} \\sqrt {(c^2)^2 \\plus{} \\left(c\\sqrt {ab}\\right)^2} \\\\\n\\ge \\sqrt {\\left(a^2 \\plus{} b^2 \\plus{} c^2\\right)^2 \\plus{} \\left(a\\sqrt {bc} \\plus{} b\\sqrt {ca} \\plus{} c\\sqrt {ab}\\right)^2}.$\n\nHence, it suffices to show\n\n$ \\left(a^2 \\plus{} b^2 \\plus{} c^2\\right)^2 \\plus{} \\left(a\\sqrt {bc} \\plus{} b\\sqrt {ca} \\plus{} c\\sqrt {ab}\\right)^2 \\ge 2(ab \\plus{} bc \\plus{} ca)^2.$\n\nWhich is equivalent to\n\n$ a^4 \\plus{} b^4 \\plus{} c^4 \\plus{} 2abc\\left( \\sqrt {ab} \\plus{} \\sqrt {bc} \\plus{} \\sqrt {ca} \\right) \\ge 3abc(a \\plus{} b \\plus{} c).$\n\nBy Schur Inequality\n\n$ a^4 \\plus{} b^4 \\plus{} c^4 \\ge a^3(b \\plus{} c) \\plus{} b^3(c \\plus{} a) \\plus{} c^3(a \\plus{} b) \\minus{} abc(a \\plus{} b \\plus{} c).$\n\nTherefore we just need to prove\n\n$ \\quad a^3(b \\plus{} c) \\plus{} b^3(c \\plus{} a) \\plus{} c^3(a \\plus{} b) \\plus{} 2abc\\left( \\sqrt {ab} \\plus{} \\sqrt {bc} \\plus{} \\sqrt {ca} \\right) \\\\\n\\ge 4abc(a \\plus{} b \\plus{} c).$\n\nUsing AM-GM Inequality we have\n\n$ a^3 b \\plus{} a^3 c \\plus{} abc\\sqrt {bc} \\plus{} abc\\sqrt {bc} \\ge 4a^2 bc$\n$ b^3 c \\plus{} b^3 a \\plus{} abc\\sqrt {ca} \\plus{} abc\\sqrt {ca} \\ge 4b^2 ca$\n$ c^3 a \\plus{} c^3 b \\plus{} abc\\sqrt {ab} \\plus{} abc\\sqrt {ab} \\ge 4c^2 ab.$\n\nAdding them up we get the result.[/hide]\n\n[hide=\"Solution 2\"]Applying Schur Inequality we have\n\n$ a\\sqrt{a^2\\plus{}bc}\\plus{} b\\sqrt{b^2\\plus{}ca}\\plus{} c\\sqrt{c^2\\plus{}ab} \\ge \\frac{1}{\\sqrt{2}}( a(a\\plus{}\\sqrt{bc})\\plus{} b(b\\plus{}\\sqrt{ca})\\plus{} c(c\\plus{}\\sqrt{ab}))\\\\\n\\ge \\sqrt{2}(\\sqrt{ab}(a\\plus{}b)\\plus{}\\sqrt{bc}(b\\plus{}c)\\plus{}\\sqrt{ca}(c\\plus{}a)) \\ge \\sqrt{2}(ab\\plus{}bc\\plus{}ca)$[/hide]\r\n\r\nSolution 2 is from Zheng Fan from Shanghai High School.", "Solution_1": "[quote=\"shalex\"]Prove that for all nonnegative real numbers $ a,b,c$ we have\n\n$ a\\sqrt {a^2 \\plus{} bc} \\plus{} b\\sqrt {b^2 \\plus{} ca} \\plus{} c\\sqrt {c^2 \\plus{} ab} \\ge \\sqrt {2} (ab \\plus{} bc \\plus{} ca).$\n\n[hide=\"Solution 1\"] By Minkowski Inequality we have\n\n$ \\quad a\\sqrt {a^2 \\plus{} bc} \\plus{} b\\sqrt {b^2 \\plus{} ca} \\plus{} c\\sqrt {c^2 \\plus{} ab} \\\\\n\\equal{} \\sqrt {(a^2)^2 \\plus{} \\left(a\\sqrt {bc}\\right)^2} \\plus{} \\sqrt {(b^2)^2 \\plus{} \\left(b\\sqrt {ca}\\right)^2} \\plus{} \\sqrt {(c^2)^2 \\plus{} \\left(c\\sqrt {ab}\\right)^2} \\\\\n\\ge \\sqrt {\\left(a^2 \\plus{} b^2 \\plus{} c^2\\right)^2 \\plus{} \\left(a\\sqrt {bc} \\plus{} b\\sqrt {ca} \\plus{} c\\sqrt {ab}\\right)^2}.$\n\nHence, it suffices to show\n\n$ \\left(a^2 \\plus{} b^2 \\plus{} c^2\\right)^2 \\plus{} \\left(a\\sqrt {bc} \\plus{} b\\sqrt {ca} \\plus{} c\\sqrt {ab}\\right)^2 \\ge 2(ab \\plus{} bc \\plus{} ca)^2.$\n\nWhich is equivalent to\n\n$ a^4 \\plus{} b^4 \\plus{} c^4 \\plus{} 2abc\\left( \\sqrt {ab} \\plus{} \\sqrt {bc} \\plus{} \\sqrt {ca} \\right) \\ge 3abc(a \\plus{} b \\plus{} c).$\n\nBy Schur Inequality\n\n$ a^4 \\plus{} b^4 \\plus{} c^4 \\ge a^3(b \\plus{} c) \\plus{} b^3(c \\plus{} a) \\plus{} c^3(a \\plus{} b) \\minus{} abc(a \\plus{} b \\plus{} c).$\n\nTherefore we just need to prove\n\n$ \\quad a^3(b \\plus{} c) \\plus{} b^3(c \\plus{} a) \\plus{} c^3(a \\plus{} b) \\plus{} 2abc\\left( \\sqrt {ab} \\plus{} \\sqrt {bc} \\plus{} \\sqrt {ca} \\right) \\\\\n\\ge 4abc(a \\plus{} b \\plus{} c).$\n\nUsing AM-GM Inequality we have\n\n$ a^3 b \\plus{} a^3 c \\plus{} abc\\sqrt {bc} \\plus{} abc\\sqrt {bc} \\ge 4a^2 bc$\n$ b^3 c \\plus{} b^3 a \\plus{} abc\\sqrt {ca} \\plus{} abc\\sqrt {ca} \\ge 4b^2 ca$\n$ c^3 a \\plus{} c^3 b \\plus{} abc\\sqrt {ab} \\plus{} abc\\sqrt {ab} \\ge 4c^2 ab.$\n\nAdding them up we get the result.[/hide]\n\n[hide=\"Solution 2\"]Applying Schur Inequality we have\n\n$ a\\sqrt {a^2 \\plus{} bc} \\plus{} b\\sqrt {b^2 \\plus{} ca} \\plus{} c\\sqrt {c^2 \\plus{} ab} \\ge \\frac {1}{\\sqrt {2}}( a(a \\plus{} \\sqrt {bc}) \\plus{} b(b \\plus{} \\sqrt {ca}) \\plus{} c(c \\plus{} \\sqrt {ab})) \\\\\n\\ge \\sqrt {2}(\\sqrt {ab}(a \\plus{} b) \\plus{} \\sqrt {bc}(b \\plus{} c) \\plus{} \\sqrt {ca}(c \\plus{} a)) \\ge \\sqrt {2}(ab \\plus{} bc \\plus{} ca)$[/hide]\n\nSolution 2 is from Zheng Fan from Shanghai High School.[/quote]\r\nNice solution, shalex.\r\n\r\nMy solution:\r\n$ a\\sqrt {a^2 \\plus{} bc} \\plus{} b\\sqrt {b^2 \\plus{} ca} \\plus{} c\\sqrt {c^2 \\plus{} ab} \\ge \\sqrt {2} (ab \\plus{} bc \\plus{} ca).$\r\n\r\n$ \\Leftrightarrow \\sum \\sqrt{\\left(\\frac{a}{bc}\\right)^2\\plus{}\\frac{1}{bc}} \\ge 2\\sum \\frac{1}{a}$\r\n\r\nBy Minkowski inequality:\r\n$ \\sum \\sqrt{\\left(\\frac{a}{bc}\\right)^2\\plus{}\\frac{1}{bc}} \\ge \\sqrt{\\left(\\sum \\frac{a}{bc}\\right)^2\\plus{}\\left(\\sum \\frac{1}{\\sqrt{bc}}\\right)^2}$\r\n\r\nWe have:\r\n$ A\\equal{}\\left(\\sum \\frac{a}{bc}\\right)^2\\plus{}\\left(\\sum \\frac{1}{\\sqrt{bc}}\\right)^2\\equal{}\\sum \\frac{a^2}{b^2c^2}\\plus{}2\\sum \\frac{1}{a\\sqrt{bc}}\\plus{}2\\sum \\frac{1}{a^2}\\plus{}\\sum \\frac{1}{ab}$\r\n\r\nBy AM-GM inequality:\r\n$ \\frac{a^2}{b^2c^2}\\plus{}\\frac{2}{a\\sqrt{bc}} \\ge \\frac{3}{bc}$\r\n$ \\Rightarrow \\sum \\frac{a^2}{b^2c^2}\\plus{}2\\sum \\frac{1}{a\\sqrt{bc}} \\ge 3\\sum \\frac{1}{bc}$\r\n\r\nSo $ \\sqrt{A} \\ge \\sqrt{2\\left(\\sum \\frac{1}{ab}\\right)^2}\\equal{}\\sqrt{2}(\\sum \\frac{1}{a})$", "Solution_2": "[quote=\"shalex\"]Prove that for all nonnegative real numbers $ a,b,c$ we have\n\n$ a\\sqrt {a^2 \\plus{} bc} \\plus{} b\\sqrt {b^2 \\plus{} ca} \\plus{} c\\sqrt {c^2 \\plus{} ab} \\ge \\sqrt {2} (ab \\plus{} bc \\plus{} ca).$\n\n[hide=\"Solution 1\"] By Minkowski Inequality we have\n\n$ \\quad a\\sqrt {a^2 \\plus{} bc} \\plus{} b\\sqrt {b^2 \\plus{} ca} \\plus{} c\\sqrt {c^2 \\plus{} ab} \\\\\n\\equal{} \\sqrt {(a^2)^2 \\plus{} \\left(a\\sqrt {bc}\\right)^2} \\plus{} \\sqrt {(b^2)^2 \\plus{} \\left(b\\sqrt {ca}\\right)^2} \\plus{} \\sqrt {(c^2)^2 \\plus{} \\left(c\\sqrt {ab}\\right)^2} \\\\\n\\ge \\sqrt {\\left(a^2 \\plus{} b^2 \\plus{} c^2\\right)^2 \\plus{} \\left(a\\sqrt {bc} \\plus{} b\\sqrt {ca} \\plus{} c\\sqrt {ab}\\right)^2}.$\n\nHence, it suffices to show\n\n$ \\left(a^2 \\plus{} b^2 \\plus{} c^2\\right)^2 \\plus{} \\left(a\\sqrt {bc} \\plus{} b\\sqrt {ca} \\plus{} c\\sqrt {ab}\\right)^2 \\ge 2(ab \\plus{} bc \\plus{} ca)^2.$\n\nWhich is equivalent to\n\n$ a^4 \\plus{} b^4 \\plus{} c^4 \\plus{} 2abc\\left( \\sqrt {ab} \\plus{} \\sqrt {bc} \\plus{} \\sqrt {ca} \\right) \\ge 3abc(a \\plus{} b \\plus{} c).$\n\nBy Schur Inequality\n\n$ a^4 \\plus{} b^4 \\plus{} c^4 \\ge a^3(b \\plus{} c) \\plus{} b^3(c \\plus{} a) \\plus{} c^3(a \\plus{} b) \\minus{} abc(a \\plus{} b \\plus{} c).$\n\nTherefore we just need to prove\n\n$ \\quad a^3(b \\plus{} c) \\plus{} b^3(c \\plus{} a) \\plus{} c^3(a \\plus{} b) \\plus{} 2abc\\left( \\sqrt {ab} \\plus{} \\sqrt {bc} \\plus{} \\sqrt {ca} \\right) \\\\\n\\ge 4abc(a \\plus{} b \\plus{} c).$\n\nUsing AM-GM Inequality we have\n\n$ a^3 b \\plus{} a^3 c \\plus{} abc\\sqrt {bc} \\plus{} abc\\sqrt {bc} \\ge 4a^2 bc$\n$ b^3 c \\plus{} b^3 a \\plus{} abc\\sqrt {ca} \\plus{} abc\\sqrt {ca} \\ge 4b^2 ca$\n$ c^3 a \\plus{} c^3 b \\plus{} abc\\sqrt {ab} \\plus{} abc\\sqrt {ab} \\ge 4c^2 ab.$\n\nAdding them up we get the result.[/hide]\n\n[hide=\"Solution 2\"]Applying Schur Inequality we have\n\n$ a\\sqrt {a^2 \\plus{} bc} \\plus{} b\\sqrt {b^2 \\plus{} ca} \\plus{} c\\sqrt {c^2 \\plus{} ab} \\ge \\frac {1}{\\sqrt {2}}( a(a \\plus{} \\sqrt {bc}) \\plus{} b(b \\plus{} \\sqrt {ca}) \\plus{} c(c \\plus{} \\sqrt {ab})) \\\\\n\\ge \\sqrt {2}(\\sqrt {ab}(a \\plus{} b) \\plus{} \\sqrt {bc}(b \\plus{} c) \\plus{} \\sqrt {ca}(c \\plus{} a)) \\ge \\sqrt {2}(ab \\plus{} bc \\plus{} ca)$[/hide]\n\nSolution 2 is from Zheng Fan from Shanghai High School.[/quote]\r\nOne more solution is\r\nSquaring both sides, we can rewrite the inequality as\r\n\\[ \\sum a^4\\plus{}2\\sum ab\\sqrt{(a^2\\plus{}bc)(b^2\\plus{}ca)} \\ge 2\\sum a^2b^2\\plus{}3\\sum a^2bc\\]\r\nBy the Cauchy Schwarz, we have\r\n\\[ \\sum ab\\sqrt{(a^2\\plus{}bc)(b^2\\plus{}ca)} \\ge \\sum ab(ab\\plus{}c\\sqrt{ab})\\]\r\nIt suffices to show that\r\n\\[ \\sum a^4\\plus{} 2\\sum ab(ab\\plus{}c\\sqrt{ab}) \\ge 2\\sum a^2b^2\\plus{}3\\sum a^2bc\\]\r\n\\[ \\Leftrightarrow \\sum (a^4\\plus{}2abc\\sqrt{bc} \\minus{}3a^2bc) \\ge 0\\]\r\nBy the AM-GM Inequality, we get\r\n\\[ a^4\\plus{}2abc\\sqrt{bc} \\ge 3\\sqrt[3]{a^6b^3c^3}\\equal{}3a^2bc\\]\r\nThe inequality is proved. :)\r\n\r\nActually, the stronger result is also true\r\n\\[ \\sum a\\sqrt{a^2\\plus{}bc} \\ge (2\\minus{}\\sqrt{2})(\\sum a^2\\plus{}\\sqrt{2}\\sum ab)\\]\r\n:)", "Solution_3": "[quote=\"NguyenDungTN\"][quote=\"shalex\"]Prove that for all nonnegative real numbers $ a,b,c$ we have\n\n$ a\\sqrt {a^2 \\plus{} bc} \\plus{} b\\sqrt {b^2 \\plus{} ca} \\plus{} c\\sqrt {c^2 \\plus{} ab} \\ge \\sqrt {2} (ab \\plus{} bc \\plus{} ca).$\n\n[hide=\"Solution 1\"] By Minkowski Inequality we have\n\n$ \\quad a\\sqrt {a^2 \\plus{} bc} \\plus{} b\\sqrt {b^2 \\plus{} ca} \\plus{} c\\sqrt {c^2 \\plus{} ab} \\\\\n\\equal{} \\sqrt {(a^2)^2 \\plus{} \\left(a\\sqrt {bc}\\right)^2} \\plus{} \\sqrt {(b^2)^2 \\plus{} \\left(b\\sqrt {ca}\\right)^2} \\plus{} \\sqrt {(c^2)^2 \\plus{} \\left(c\\sqrt {ab}\\right)^2} \\\\\n\\ge \\sqrt {\\left(a^2 \\plus{} b^2 \\plus{} c^2\\right)^2 \\plus{} \\left(a\\sqrt {bc} \\plus{} b\\sqrt {ca} \\plus{} c\\sqrt {ab}\\right)^2}.$\n\nHence, it suffices to show\n\n$ \\left(a^2 \\plus{} b^2 \\plus{} c^2\\right)^2 \\plus{} \\left(a\\sqrt {bc} \\plus{} b\\sqrt {ca} \\plus{} c\\sqrt {ab}\\right)^2 \\ge 2(ab \\plus{} bc \\plus{} ca)^2.$\n\nWhich is equivalent to\n\n$ a^4 \\plus{} b^4 \\plus{} c^4 \\plus{} 2abc\\left( \\sqrt {ab} \\plus{} \\sqrt {bc} \\plus{} \\sqrt {ca} \\right) \\ge 3abc(a \\plus{} b \\plus{} c).$\n\nBy Schur Inequality\n\n$ a^4 \\plus{} b^4 \\plus{} c^4 \\ge a^3(b \\plus{} c) \\plus{} b^3(c \\plus{} a) \\plus{} c^3(a \\plus{} b) \\minus{} abc(a \\plus{} b \\plus{} c).$\n\nTherefore we just need to prove\n\n$ \\quad a^3(b \\plus{} c) \\plus{} b^3(c \\plus{} a) \\plus{} c^3(a \\plus{} b) \\plus{} 2abc\\left( \\sqrt {ab} \\plus{} \\sqrt {bc} \\plus{} \\sqrt {ca} \\right) \\\\\n\\ge 4abc(a \\plus{} b \\plus{} c).$\n\nUsing AM-GM Inequality we have\n\n$ a^3 b \\plus{} a^3 c \\plus{} abc\\sqrt {bc} \\plus{} abc\\sqrt {bc} \\ge 4a^2 bc$\n$ b^3 c \\plus{} b^3 a \\plus{} abc\\sqrt {ca} \\plus{} abc\\sqrt {ca} \\ge 4b^2 ca$\n$ c^3 a \\plus{} c^3 b \\plus{} abc\\sqrt {ab} \\plus{} abc\\sqrt {ab} \\ge 4c^2 ab.$\n\nAdding them up we get the result.[/hide]\n\n[hide=\"Solution 2\"]Applying Schur Inequality we have\n\n$ a\\sqrt {a^2 \\plus{} bc} \\plus{} b\\sqrt {b^2 \\plus{} ca} \\plus{} c\\sqrt {c^2 \\plus{} ab} \\ge \\frac {1}{\\sqrt {2}}( a(a \\plus{} \\sqrt {bc}) \\plus{} b(b \\plus{} \\sqrt {ca}) \\plus{} c(c \\plus{} \\sqrt {ab})) \\\\\n\\ge \\sqrt {2}(\\sqrt {ab}(a \\plus{} b) \\plus{} \\sqrt {bc}(b \\plus{} c) \\plus{} \\sqrt {ca}(c \\plus{} a)) \\ge \\sqrt {2}(ab \\plus{} bc \\plus{} ca)$[/hide]\n\nSolution 2 is from Zheng Fan from Shanghai High School.[/quote]\nNice solution, shalex.\n\nMy solution:\n$ a\\sqrt {a^2 \\plus{} bc} \\plus{} b\\sqrt {b^2 \\plus{} ca} \\plus{} c\\sqrt {c^2 \\plus{} ab} \\ge \\sqrt {2} (ab \\plus{} bc \\plus{} ca).$\n\n$ \\Leftrightarrow \\sum \\sqrt {\\left(\\frac {a}{bc}\\right)^2 \\plus{} \\frac {1}{bc}} \\ge 2\\sum \\frac {1}{a}$\n\nBy Minkowski inequality:\n$ \\sum \\sqrt {\\left(\\frac {a}{bc}\\right)^2 \\plus{} \\frac {1}{bc}} \\ge \\sqrt {\\left(\\sum \\frac {a}{bc}\\right)^2 \\plus{} \\left(\\sum \\frac {1}{\\sqrt {bc}}\\right)^2}$\n\nWe have:\n$ A \\equal{} \\left(\\sum \\frac {a}{bc}\\right)^2 \\plus{} \\left(\\sum \\frac {1}{\\sqrt {bc}}\\right)^2 \\equal{} \\sum \\frac {a^2}{b^2c^2} \\plus{} 2\\sum \\frac {1}{a\\sqrt {bc}} \\plus{} 2\\sum \\frac {1}{a^2} \\plus{} \\sum \\frac {1}{ab}$\n\nBy AM-GM inequality:\n$ \\frac {a^2}{b^2c^2} \\plus{} \\frac {2}{a\\sqrt {bc}} \\ge \\frac {3}{bc}$\n$ \\Rightarrow \\sum \\frac {a^2}{b^2c^2} \\plus{} 2\\sum \\frac {1}{a\\sqrt {bc}} \\ge 3\\sum \\frac {1}{bc}$\n\nSo $ \\sqrt {A} \\ge \\sqrt {2\\left(\\sum \\frac {1}{ab}\\right)^2} \\equal{} \\sqrt {2}(\\sum \\frac {1}{a})$[/quote]\r\n\r\nA typo at last; it should be\r\n\r\n$ \\sqrt {A} \\ge \\sqrt {2\\left(\\sum \\frac {1}{a}\\right)^2}$", "Solution_4": "[quote=\"shalex\"][quote=\"NguyenDungTN\"][quote=\"shalex\"]Prove that for all nonnegative real numbers $ a,b,c$ we have\n\n$ a\\sqrt {a^2 \\plus{} bc} \\plus{} b\\sqrt {b^2 \\plus{} ca} \\plus{} c\\sqrt {c^2 \\plus{} ab} \\ge \\sqrt {2} (ab \\plus{} bc \\plus{} ca).$\n\n[hide=\"Solution 1\"] By Minkowski Inequality we have\n\n$ \\quad a\\sqrt {a^2 \\plus{} bc} \\plus{} b\\sqrt {b^2 \\plus{} ca} \\plus{} c\\sqrt {c^2 \\plus{} ab} \\\\\n\\equal{} \\sqrt {(a^2)^2 \\plus{} \\left(a\\sqrt {bc}\\right)^2} \\plus{} \\sqrt {(b^2)^2 \\plus{} \\left(b\\sqrt {ca}\\right)^2} \\plus{} \\sqrt {(c^2)^2 \\plus{} \\left(c\\sqrt {ab}\\right)^2} \\\\\n\\ge \\sqrt {\\left(a^2 \\plus{} b^2 \\plus{} c^2\\right)^2 \\plus{} \\left(a\\sqrt {bc} \\plus{} b\\sqrt {ca} \\plus{} c\\sqrt {ab}\\right)^2}.$\n\nHence, it suffices to show\n\n$ \\left(a^2 \\plus{} b^2 \\plus{} c^2\\right)^2 \\plus{} \\left(a\\sqrt {bc} \\plus{} b\\sqrt {ca} \\plus{} c\\sqrt {ab}\\right)^2 \\ge 2(ab \\plus{} bc \\plus{} ca)^2.$\n\nWhich is equivalent to\n\n$ a^4 \\plus{} b^4 \\plus{} c^4 \\plus{} 2abc\\left( \\sqrt {ab} \\plus{} \\sqrt {bc} \\plus{} \\sqrt {ca} \\right) \\ge 3abc(a \\plus{} b \\plus{} c).$\n\nBy Schur Inequality\n\n$ a^4 \\plus{} b^4 \\plus{} c^4 \\ge a^3(b \\plus{} c) \\plus{} b^3(c \\plus{} a) \\plus{} c^3(a \\plus{} b) \\minus{} abc(a \\plus{} b \\plus{} c).$\n\nTherefore we just need to prove\n\n$ \\quad a^3(b \\plus{} c) \\plus{} b^3(c \\plus{} a) \\plus{} c^3(a \\plus{} b) \\plus{} 2abc\\left( \\sqrt {ab} \\plus{} \\sqrt {bc} \\plus{} \\sqrt {ca} \\right) \\\\\n\\ge 4abc(a \\plus{} b \\plus{} c).$\n\nUsing AM-GM Inequality we have\n\n$ a^3 b \\plus{} a^3 c \\plus{} abc\\sqrt {bc} \\plus{} abc\\sqrt {bc} \\ge 4a^2 bc$\n$ b^3 c \\plus{} b^3 a \\plus{} abc\\sqrt {ca} \\plus{} abc\\sqrt {ca} \\ge 4b^2 ca$\n$ c^3 a \\plus{} c^3 b \\plus{} abc\\sqrt {ab} \\plus{} abc\\sqrt {ab} \\ge 4c^2 ab.$\n\nAdding them up we get the result.[/hide]\n\n[hide=\"Solution 2\"]Applying Schur Inequality we have\n\n$ a\\sqrt {a^2 \\plus{} bc} \\plus{} b\\sqrt {b^2 \\plus{} ca} \\plus{} c\\sqrt {c^2 \\plus{} ab} \\ge \\frac {1}{\\sqrt {2}}( a(a \\plus{} \\sqrt {bc}) \\plus{} b(b \\plus{} \\sqrt {ca}) \\plus{} c(c \\plus{} \\sqrt {ab})) \\\\\n\\ge \\sqrt {2}(\\sqrt {ab}(a \\plus{} b) \\plus{} \\sqrt {bc}(b \\plus{} c) \\plus{} \\sqrt {ca}(c \\plus{} a)) \\ge \\sqrt {2}(ab \\plus{} bc \\plus{} ca)$[/hide]\n\nSolution 2 is from Zheng Fan from Shanghai High School.[/quote]\nNice solution, shalex.\n\nMy solution:\n$ a\\sqrt {a^2 \\plus{} bc} \\plus{} b\\sqrt {b^2 \\plus{} ca} \\plus{} c\\sqrt {c^2 \\plus{} ab} \\ge \\sqrt {2} (ab \\plus{} bc \\plus{} ca).$\n\n$ \\Leftrightarrow \\sum \\sqrt {\\left(\\frac {a}{bc}\\right)^2 \\plus{} \\frac {1}{bc}} \\ge 2\\sum \\frac {1}{a}$\n\nBy Minkowski inequality:\n$ \\sum \\sqrt {\\left(\\frac {a}{bc}\\right)^2 \\plus{} \\frac {1}{bc}} \\ge \\sqrt {\\left(\\sum \\frac {a}{bc}\\right)^2 \\plus{} \\left(\\sum \\frac {1}{\\sqrt {bc}}\\right)^2}$\n\nWe have:\n$ A \\equal{} \\left(\\sum \\frac {a}{bc}\\right)^2 \\plus{} \\left(\\sum \\frac {1}{\\sqrt {bc}}\\right)^2 \\equal{} \\sum \\frac {a^2}{b^2c^2} \\plus{} 2\\sum \\frac {1}{a\\sqrt {bc}} \\plus{} 2\\sum \\frac {1}{a^2} \\plus{} \\sum \\frac {1}{ab}$\n\nBy AM-GM inequality:\n$ \\frac {a^2}{b^2c^2} \\plus{} \\frac {2}{a\\sqrt {bc}} \\ge \\frac {3}{bc}$\n$ \\Rightarrow \\sum \\frac {a^2}{b^2c^2} \\plus{} 2\\sum \\frac {1}{a\\sqrt {bc}} \\ge 3\\sum \\frac {1}{bc}$\n\nSo $ \\sqrt {A} \\ge \\sqrt {2\\left(\\sum \\frac {1}{ab}\\right)^2} \\equal{} \\sqrt {2}(\\sum \\frac {1}{a})$[/quote]\n\nA typo at last; it should be\n\n$ \\sqrt {A} \\ge \\sqrt {2\\left(\\sum \\frac {1}{a}\\right)^2}$[/quote]\r\n :oops: You're right.", "Solution_5": "[quote=\"can_hang2007\"]Actually, the stronger result is also true\n\\[ \\sum a\\sqrt {a^2 \\plus{} bc} \\ge (2 \\minus{} \\sqrt {2})(\\sum a^2 \\plus{} \\sqrt {2}\\sum ab)\n\\]\n:)[/quote]\r\nHere is my solution for the weaker one :)", "Solution_6": "[quote=\"can_hang2007\"][quote=\"shalex\"]Prove that for all nonnegative real numbers $ a,b,c$ we have\n\n$ a\\sqrt {a^2 \\plus{} bc} \\plus{} b\\sqrt {b^2 \\plus{} ca} \\plus{} c\\sqrt {c^2 \\plus{} ab} \\ge \\sqrt {2} (ab \\plus{} bc \\plus{} ca).$\n\n[hide=\"Solution 1\"] By Minkowski Inequality we have\n\n$ \\quad a\\sqrt {a^2 \\plus{} bc} \\plus{} b\\sqrt {b^2 \\plus{} ca} \\plus{} c\\sqrt {c^2 \\plus{} ab} \\\\\n\\equal{} \\sqrt {(a^2)^2 \\plus{} \\left(a\\sqrt {bc}\\right)^2} \\plus{} \\sqrt {(b^2)^2 \\plus{} \\left(b\\sqrt {ca}\\right)^2} \\plus{} \\sqrt {(c^2)^2 \\plus{} \\left(c\\sqrt {ab}\\right)^2} \\\\\n\\ge \\sqrt {\\left(a^2 \\plus{} b^2 \\plus{} c^2\\right)^2 \\plus{} \\left(a\\sqrt {bc} \\plus{} b\\sqrt {ca} \\plus{} c\\sqrt {ab}\\right)^2}.$\n\nHence, it suffices to show\n\n$ \\left(a^2 \\plus{} b^2 \\plus{} c^2\\right)^2 \\plus{} \\left(a\\sqrt {bc} \\plus{} b\\sqrt {ca} \\plus{} c\\sqrt {ab}\\right)^2 \\ge 2(ab \\plus{} bc \\plus{} ca)^2.$\n\nWhich is equivalent to\n\n$ a^4 \\plus{} b^4 \\plus{} c^4 \\plus{} 2abc\\left( \\sqrt {ab} \\plus{} \\sqrt {bc} \\plus{} \\sqrt {ca} \\right) \\ge 3abc(a \\plus{} b \\plus{} c).$\n\nBy Schur Inequality\n\n$ a^4 \\plus{} b^4 \\plus{} c^4 \\ge a^3(b \\plus{} c) \\plus{} b^3(c \\plus{} a) \\plus{} c^3(a \\plus{} b) \\minus{} abc(a \\plus{} b \\plus{} c).$\n\nTherefore we just need to prove\n\n$ \\quad a^3(b \\plus{} c) \\plus{} b^3(c \\plus{} a) \\plus{} c^3(a \\plus{} b) \\plus{} 2abc\\left( \\sqrt {ab} \\plus{} \\sqrt {bc} \\plus{} \\sqrt {ca} \\right) \\\\\n\\ge 4abc(a \\plus{} b \\plus{} c).$\n\nUsing AM-GM Inequality we have\n\n$ a^3 b \\plus{} a^3 c \\plus{} abc\\sqrt {bc} \\plus{} abc\\sqrt {bc} \\ge 4a^2 bc$\n$ b^3 c \\plus{} b^3 a \\plus{} abc\\sqrt {ca} \\plus{} abc\\sqrt {ca} \\ge 4b^2 ca$\n$ c^3 a \\plus{} c^3 b \\plus{} abc\\sqrt {ab} \\plus{} abc\\sqrt {ab} \\ge 4c^2 ab.$\n\nAdding them up we get the result.[/hide]\n\n[hide=\"Solution 2\"]Applying Schur Inequality we have\n\n$ a\\sqrt {a^2 \\plus{} bc} \\plus{} b\\sqrt {b^2 \\plus{} ca} \\plus{} c\\sqrt {c^2 \\plus{} ab} \\ge \\frac {1}{\\sqrt {2}}( a(a \\plus{} \\sqrt {bc}) \\plus{} b(b \\plus{} \\sqrt {ca}) \\plus{} c(c \\plus{} \\sqrt {ab})) \\\\\n\\ge \\sqrt {2}(\\sqrt {ab}(a \\plus{} b) \\plus{} \\sqrt {bc}(b \\plus{} c) \\plus{} \\sqrt {ca}(c \\plus{} a)) \\ge \\sqrt {2}(ab \\plus{} bc \\plus{} ca)$[/hide]\n\nSolution 2 is from Zheng Fan from Shanghai High School.[/quote]\nOne more solution is\nSquaring both sides, we can rewrite the inequality as\n\\[ \\sum a^4 \\plus{} 2\\sum ab\\sqrt {(a^2 \\plus{} bc)(b^2 \\plus{} ca)} \\ge 2\\sum a^2b^2 \\plus{} 3\\sum a^2bc\n\\]\nBy the Cauchy Schwarz, we have\n\\[ \\sum ab\\sqrt {(a^2 \\plus{} bc)(b^2 \\plus{} ca)} \\ge \\sum ab(ab \\plus{} c\\sqrt {ab})\n\\]\nIt suffices to show that\n\\[ \\sum a^4 \\plus{} 2\\sum ab(ab \\plus{} c\\sqrt {ab}) \\ge 2\\sum a^2b^2 \\plus{} 3\\sum a^2bc\n\\]\n\n\\[ \\Leftrightarrow \\sum (a^4 \\plus{} 2abc\\sqrt {bc} \\minus{} 3a^2bc) \\ge 0\n\\]\nBy the AM-GM Inequality, we get\n\\[ a^4 \\plus{} 2abc\\sqrt {bc} \\ge 3\\sqrt [3]{a^6b^3c^3} \\equal{} 3a^2bc\n\\]\nThe inequality is proved. :)\n\nActually, the stronger result is also true\n\\[ \\sum a\\sqrt {a^2 \\plus{} bc} \\ge (2 \\minus{} \\sqrt {2})(\\sum a^2 \\plus{} \\sqrt {2}\\sum ab)\n\\]\n:)[/quote]\r\n\r\nDo you have the solution for this one? :)", "Solution_7": "[quote=\"shalex\"]\nDo you have the solution for this one? :)[/quote]\r\nYes, I have. :)", "Solution_8": "[quote=\"shalex\"]\nApplying Schur Inequality we have\n\n$ a\\sqrt {a^2 \\plus{} bc} \\plus{} b\\sqrt {b^2 \\plus{} ca} \\plus{} c\\sqrt {c^2 \\plus{} ab} \\ge \\frac {1}{\\sqrt {2}}( a(a \\plus{} \\sqrt {bc}) \\plus{} b(b \\plus{} \\sqrt {ca}) \\plus{} c(c \\plus{} \\sqrt {ab})) \\\\\n\\ge \\sqrt {2}(\\sqrt {ab}(a \\plus{} b) \\plus{} \\sqrt {bc}(b \\plus{} c) \\plus{} \\sqrt {ca}(c \\plus{} a)) \\ge \\sqrt {2}(ab \\plus{} bc \\plus{} ca)$\n\nSolution 2 is from Zheng Fan from Shanghai High School.[/quote]\r\n\r\nAnyway, a typo here,\r\n\r\n$ \\sqrt {2}(\\sqrt {ab}(a \\plus{} b) \\plus{} \\sqrt {bc}(b \\plus{} c) \\plus{} \\sqrt {ca}(c \\plus{} a))$\r\n\r\nshould be\r\n\r\n$ \\frac {1}{\\sqrt {2}}(\\sqrt {ab}(a \\plus{} b) \\plus{} \\sqrt {bc}(b \\plus{} c) \\plus{} \\sqrt {ca}(c \\plus{} a))$", "Solution_9": "[quote=\"shalex\"]Prove that for all nonnegative real numbers $ a,b,c$ we have\n\n$ a\\sqrt {a^2 \\plus{} bc} \\plus{} b\\sqrt {b^2 \\plus{} ca} \\plus{} c\\sqrt {c^2 \\plus{} ab} \\ge \\sqrt {2} (ab \\plus{} bc \\plus{} ca).$\n\n[/quote]\r\n\r\nThis one is from me, http://hi.baidu.com/shfdfzhjj/blog/item/991a61d9b8cbf9ed38012fd2.html :)" } { "Tag": [ "modular arithmetic", "search", "number theory unsolved", "number theory" ], "Problem": "Prove that there exist such number $x$ that $x$ is not divisable with any $n^2,n\\in N$ and numbers \\[x+1, x+2,..., x+2004\\] are all divisable with squares of some numbers.", "Solution_1": "If we use C.R.T . it is trivial", "Solution_2": "Yes Chinese Riminder Theorem ;)", "Solution_3": "CRT gives you the divisibility of $x+1$ etc by any squares you please --- but you need to ensure that $x$ is square-free: that's not a congruence condition.", "Solution_4": "You can even find prime $x$'s with the required property, but you need to employ Dirichlet's Theorem as well:\r\n\r\nPick any $2004$ primes $p_1,\\ldots,p_{2004}$, none of them not dividing any of the numbers $1,2,\\ldots,2004$, and determine (by the Chinese Remainder Theorem) $k$ s.t. $x\\equiv k\\pmod{p_1^2\\cdot\\ldots\\cdot p_{2004}^2}\\Rightarrow x\\equiv -i\\pmod{p_i^2},\\ \\forall i\\in\\overline{1,2004}$, and then take $x$ to be a prime with the property $x\\equiv k\\pmod{p_1^2\\cdot\\ldots\\cdot p_{2004}^2}$ (such a prime number exists, by Dirichlet's Theorem, since our choices ensure that $(k,p_1^2\\cdot\\ldots\\cdot p_{2004}^2)=1$).", "Solution_5": "Well, I don't know $C.R.T$ but now i am going to learn it, thanx a lot (can you post some nice link, from where to learn it :lol: :lol:)...", "Solution_6": "you can search it ;)", "Solution_7": "[quote=\"grobber\"]You can even find prime $x$'s with the required property, but you need to employ Dirichlet's Theorem as well:\n\nPick any $2004$ primes $p_1,\\ldots,p_{2004}$, none of them not dividing any of the numbers $1,2,\\ldots,2004$, and determine (by the Chinese Remainder Theorem) $k$ s.t. $x\\equiv k\\pmod{p_1^2\\cdot\\ldots\\cdot p_{2004}^2}\\Rightarrow x\\equiv -i\\pmod{p_i^2},\\ \\forall i\\in\\overline{1,2004}$, and then take $x$ to be a prime with the property $x\\equiv k\\pmod{p_1^2\\cdot\\ldots\\cdot p_{2004}^2}$ (such a prime number exists, by Dirichlet's Theorem, since our choices ensure that $(k,p_1^2\\cdot\\ldots\\cdot p_{2004}^2)=1$).[/quote] \r\n\r\nWhat do you mean by Dirichlet's theorem: In infinit aritmetic sequence there are infinit prime numbers...?\r\nI never saw some problem where that can be used, and I am not sure that you can use that theorem in competition....\r\nNice solution, but I am looking for some more elementary, CRM OK but dirichlet theorem ;)", "Solution_8": "Yes, that's the theorem I was referring to. Where did you find the problem? I'd like to know that before I even attempt to solve it in an elementary manner. :)", "Solution_9": "Let $S$ be the set of positive integers $x$ such that $x + 1$, $x + 2$, ..., $x + 2004$ are not square-free. By the Chinese Remainder Theorem, we know that $S$ is not empty. Let $x_0$ be the smallest element of $S$. Because $x_0 - 1$ is not in $S$ but $x_0$ is, we know that $x_0$ must be square-free. Thus $x_0$ is the required value of $x$.", "Solution_10": "Thanx Ravi B nice sol.\r\n\r\nGrobber, I found this problem in some papers from some trening, but they are for 9th grade....", "Solution_11": "As a follow-up, is there an elementary proof that every AP contains infinitely many square-free numbers if it contains two?", "Solution_12": "AP-????", "Solution_13": "(bump)\r\n[b]A[/b]rithmetic [b]P[/b]rogression. :)" } { "Tag": [ "calculus", "integration", "trigonometry", "logarithms", "algebra", "partial fractions", "calculus computations" ], "Problem": "Compute this integral: \r\n\\[\\int{\\sqrt{\\tan (x)}dx}\\]", "Solution_1": "[quote=\"mofidy\"]Compute this integral:\n\\[\\int{\\sqrt{\\tan (x)}dx}\\]\n[/quote]\r\nIt was already here. there was many many methods.", "Solution_2": "Yep.\r\n\r\nJust set $u^{2}=\\tan x$, and let the imagination blow up your mind.", "Solution_3": "Start by making the substitution $\\tan x = \\frac{1}{u^{2}}, x' = \\frac{-2u}{1+u^{4}}$, and we get\r\n\r\n[hide=\"Click to reveal hidden content\"]$\\int \\sqrt{\\tan x}dx =-2\\int \\frac{1}{1+u^{4}}du$.[/hide]\n\nFinding this last primitive is a nice exercise, and you should get\n\n[hide=\"Click to reveal hidden content\"]$\\int \\frac{1}{1+u^{4}}du = \\frac{1}{4\\sqrt{2}}\\ln\\left(\\frac{u^{2}+\\sqrt{2}u+1}{u^{2}-\\sqrt{2}u+1}\\right)+\\frac{1}{2\\sqrt{2}}\\arctan\\left(\\frac{\\sqrt{2}u}{1-u^{2}}\\right)+C$.[/hide]\n\nNow you just have to multiply this by -2 and make the reverse substitution:\n\n[hide=\"Final Answer\"]$\\int \\sqrt{\\tan x}dx =-\\frac{1}{2\\sqrt{2}}\\ln\\left(\\frac{1+\\sqrt{2 \\tan x}+\\tan x}{1-\\sqrt{2 \\tan x}+\\tan x}\\right)-\\frac{1}{\\sqrt{2}}\\arctan\\left(\\frac{\\sqrt{2 \\tan x}}{\\tan x-1}\\right)+C$[/hide].", "Solution_4": "[quote=\"mofidy\"]Compute this integral:\n\\[ I=\\int{\\sqrt{\\tan (x)}\\,dx}\\]\n[/quote]\r\nFirst set a change of variables according to $ u =\\sqrt{\\tan x}$, the integral becomes to\r\n\r\n$ \\int\\frac{2u^{2}}{u^{4}+1}\\,du$\r\n\r\nNasty expression, but applyin' partial fractions yields\r\n\r\n$ I =\\int\\frac{1}{2\\sqrt2}\\left(\\frac{2u}{u^{2}-\\sqrt 2u+1}-\\frac{2u}{u^{2}+\\sqrt 2u+1}\\right)\\,du$\r\n\r\nFinally\r\n\r\n\\begin{eqnarray*}I & = &\\frac{1}{{2\\sqrt 2 }}\\int{\\left({\\frac{{2u-\\sqrt 2+\\sqrt 2 }}{{u^{2}-\\sqrt 2 u+1}}-\\frac{{2u+\\sqrt 2-\\sqrt 2 }}{{u^{2}+\\sqrt 2 u+1}}}\\right)\\,du}\\\\ & = &\\frac{1}{{2\\sqrt 2 }}\\left({\\ln\\left|{\\frac{{u^{2}-\\sqrt 2 u+1}}{{u^{2}+\\sqrt 2 u+1}}}\\right|+\\int{\\frac{{\\sqrt 2 }}{{\\left({u-\\frac{1}{{\\sqrt 2 }}}\\right)^{2}+\\left({\\frac{1}{{\\sqrt 2 }}}\\right)^{2}}}\\,du}+\\int{\\frac{{\\sqrt 2 }}{{\\left({u+\\frac{1}{{\\sqrt 2 }}}\\right)^{2}+\\left({\\frac{1}{{\\sqrt 2 }}}\\right)^{2}}}\\,du}}\\right)\\\\ & = &\\frac{1}{{2\\sqrt 2 }}\\left({\\ln\\left|{\\frac{{u^{2}-\\sqrt 2 u+1}}{{u^{2}+\\sqrt 2 u+1}}}\\right|+2\\left[{\\arctan\\left({\\sqrt 2 u-1}\\right)+\\arctan\\left({\\sqrt 2 u+1}\\right)}\\right]}\\right)+k,\\,k\\in\\mathbb{R}\\\\ \\end{eqnarray*}\r\n\r\nBack substitute and we have the desired $ \\blacksquare$" } { "Tag": [ "percent", "geometry", "rectangle", "congruent triangles" ], "Problem": "1. A store charges $28 for a certain type of sweater. This price is 40 percent more than the amount it costs the store to buy one of these sweaters/ At an end of season sale, store employees can urchase any remaining sweaters at 30 percent off the store's cost. How much would it cost an employee to purchase a sweater of this type at this sale?\r\n\r\n2. In rectangle ABCD, point E is the midpoint of line BC. If the area of quadrilateral ABED is 2/3, what is the area of rectangle ABCD?", "Solution_1": "[hide=\"1\"]Isn't it just 28*0.7=19.60? [/hide]\n\n[hide=\"2\"]Draw lines ED, AE, and EF, where F is the midpoint of AD. There are four congruent triangles created. ABED is three of them. The are fo the rectangle is 4/3*2/3=8/9. [/hide]", "Solution_2": "I'm not sure I understand the wording on #1. When it says the store's cost, I assume it's referring to the 28 dollar price? If so, why does it bother to include the 40% increase in what the store pays for it? And if it's referring to the price that the store bought the sweater for, it isn't very clear...\r\n\r\n#2. (Note: $[ABC...]$ denotes the area of polygon $ABC...$.) $E$ is the midpoint of rectangle $ABCD$, and $E$ is the midpoint of $BC$. If we take the midpoint of $AD$ and label it $F$, then $[ABEF]=[FCDE]=\\frac{1}{2}\\cdot [ABCD]$. $ED$ divides rectangle $ECDF$ in half, since it is a diagonal, which means that $[EDF]=[ECD]=\\frac{1}{2}\\cdot [ECDF]$. Therefore, we have $\\frac{1}{2}\\cdot [ABCD]+\\frac{1}{2}\\cdot \\frac{1}{2}\\cdot [ABCD]=\\frac{3}{4}\\cdot [ABCD]=\\frac{2}{3}$. Therefore, $[ABCD]=\\frac{8}{9}$.\r\n\r\nI hope you understood that. If not, please tell me what needs clarification.", "Solution_3": "I think the 40% was just a trick? They do that a lot on tests, but not usually on SAT's.", "Solution_4": "but the answer is $14.00", "Solution_5": "Q1: let x be the cost of a sweater for the store.\r\n\r\nthen (28 - x)/x =0.4\r\n\r\nat the end, you will find out x = 20\r\n\r\nthe answer: x * 70% = 14", "Solution_6": "Cheerful Coffin, welcome to AoPS! And your solution is correct.", "Solution_7": "[quote=\"woohooxd\"]but the answer is 14.00[/quote]\r\n\r\nOkay, so they mean that the employees can buy a sweater for 30% off the price that the store pays to buy the sweaters. You can find the price that the store pays, although it isn't necessary, but I'll explain it that way first. The store sells sweaters for 28 dollars, which is 40% more than they buy it for. Therefore, $1.4x=28$, where $x$ is the price the store pays. Solving that gives $x=20$, so the store pays 20 dollars. The employees get 30% off of that price, meaning that $y=.7\\cdot x \\Rightarrow y=14$, so the employees pay 14 dollars.\r\n\r\nThe other, quicker way of solving it is the following: $1.4x=28$, $.7x=y \\Rightarrow 1.4x=2y \\Rightarrow 2y=28 \\Rightarrow y=14$. :)" } { "Tag": [ "ratio", "geometry" ], "Problem": "In the accompanying figure $ \\overline{CE}$ and $ \\overline{DE}$ are equal chords of a circle with center $ O$. Arc $ AB$ is a quarter-circle. Then the ratio of the area of triangle $ CED$ to the area of triangle $ AOB$ is:\r\n[asy]defaultpen(linewidth(.8pt));\nunitsize(2cm);\n\npair O = origin;\npair C = (-1,0);\npair D = (1,0);\npair E = (0,1);\npair A = dir(-135);\npair B = dir(-60);\n\ndraw(Circle(O,1));\ndraw(C--E--D--cycle);\ndraw(A--O--B--cycle);\n\nlabel(\"$A$\",A,SW);\nlabel(\"$C$\",C,W);\nlabel(\"$E$\",E,N);\nlabel(\"$D$\",D,NE);\nlabel(\"$B$\",B,SE);\nlabel(\"$O$\",O,N);[/asy]\r\n$ \\textbf{(A)}\\ \\sqrt {2} : 1\\qquad \\textbf{(B)}\\ \\sqrt {3} : 1\\qquad \\textbf{(C)}\\ 4 : 1\\qquad \\textbf{(D)}\\ 3 : 1\\qquad \\textbf{(E)}\\ 2 : 1$", "Solution_1": "Since arc $ AB$ is a quarter circle, arc $ AB$ measures $ 90$ degrees. Since $ \\overline {CD}$ passes through the center, $ \\overline {CD}$ is a diameter, making the measure of arc $ CD$ equal to $ 180$ degrees. Thus, the ratio is $ 2: 1$.", "Solution_2": "the triangle CED is clearly equivalent to two quarters of circle, thus the answer is (E)", "Solution_3": "We know that angle $ AOB\\equal{}90^o$ because the arc $ AB$ is a quarter circle. Say the radius of the circle is $ r$. The area of $ AOB$ is $ \\frac{r^2}{2}$. We know that $ CD$ is $ 2r$ because it is the diameter. We also know that angle $ CED\\equal{}90^o$ \r\n\r\nbecause we have a triangel inscribed in a circle with one of the sides being the diameter. The height of triangle $ CED$ is $ r$ and the base is $ 2r$. The area of triangle $ \\frac{2r^2}{2}\\equal{}r^2$. Our answer is $ r^2: \\frac{r^2}{2}\\equal{}2: 1$. That is answer choice $ \\mathbf{(E)}$." } { "Tag": [ "calculus", "college", "MIT", "Stanford", "geometry" ], "Problem": "Can anyone tell me,how many questions are there in an one hour SAT physics test? :maybe: \r\nThanks beforehand :) !", "Solution_1": "There are 75 questions on the SAT Subject Test in Physics.", "Solution_2": "All of them are concept questions. I don't think formulas helped me, except to give me an idea of what happens to one variable when another goes through [insert change]. Relativity is asked. I took the subject test last May right before the Physics B exam, and relativity was really the only difference between what was covered, c'est \u00e0 dire, if you've only had physics C, there's going to be more than E&M covered, and definitely no calculus is required. There are some practice questions on collegeboard's website to give you an idea of how the subject test is.", "Solution_3": "How many questions did you manage to answer during 60 minutes? :maybe: \r\nI think the time is too short to answer 75 questions :maybe: .", "Solution_4": "I think I did all of them when I took it. But I know I didn't get all of them right. I think you can miss a significant number of questions and still get an 800. (I got 800.) \r\n\r\nPart of the reason it's fast is that (as quantum leap said) a lot of the questions are concept questions, you don't have to do any calculations. A good strategy for a test where there is not enough time is to skip the problems that you know will take a lot of time and do the quicker ones first, so that you are sure to get all of the easy ones. Then go back and work on the ones that will take more time.", "Solution_5": "Because the designers of the SAT II tests can't be sure what high school curriculum test-takers have had, the tests are designed to have a lot of questions about very many topics. And then the scoring is such that most test-takers can miss several questions and still get the highest standard score of 800. \r\n\r\nI took an old SAT II United States history test recently, for my own practice. (My son is taking an AP United States history course this year.) I missed about ten questions out of ninety, but still got the peak score of 800. This just shows that you don't have to be perfect to obtain the highest possible score. \r\n\r\nDefinitely most SAT I or SAT II tests are easier for persons who read English rapidly. I think you are a second-language reader of English, and that will be taken into account by the colleges to which you apply. If your specific test scores of English proficiency are high enough, and your other test scores are high, even if not the highest possible, that will be quite impressive. After all, it's difficult to take a test in a second language. I think you know that it is very competitive to get into the most famous United States universities if you are an international student, because so many international students apply to those universities, but you have some prospect of being admitted, so don't worry too much. \r\n\r\nGood luck in your applications.", "Solution_6": "With what Tokenadult said, I took the SAT 2 Physics in October, and walking out of that test (never taken a physics course, just studied from a prep book) I was sure I had got at least 12 wrong, not counting miscalculations and careless mistakes. I ended up with a 780, so my best guess at the curve is around 12 raw score.", "Solution_7": "Sorry for reviving this very old thread. But since I intend to take a Physics Test on October 4th, I'd like to ask some questions. First, I just learn all of the background knowledge by myself throughout three main books: KAPLAN, BARRON and Schaum's College Physics Outline. I see that all of them have the same number of chapters which cover similar topics. But I don't know whether I just need to learn from those stuffs or need other materials? I really want to get the high score on this test but I seem to can't remember all of the abstract knowledge in such topics: light and sound motion, interference and diffraction, color and light,electric potential and potential difference. Wow! I know that's a big hole but I'm curious how many questions are asked about those topics?(just approximation). Does anybody take the test tell me how to fill up those holes?I can't remember all of the phenomenons.Tell me what should I do now is the best for preparation outside as well as inside the room test.", "Solution_8": "I guess it is not the best place,but I just don't want to create another topic,so I will ask right here:\r\nIs there any electronic material,which can help to prepare for the SAT Physics Subject test?I mean e-book,some flash application or anything else.Also what is more preferable for you: to take a test in math 1 or math 2?I have never taken this test before,that's why I have no representation which of them is better and what are advantages of each of them.\r\nThanks in advance...", "Solution_9": "Try this: http://www.sparknotes.com/testprep/books/sat2/physics/", "Solution_10": "Math level 2 has precalc/trig, math level 1 just has algebra 2 and below. If you've taken precalc, take level 2, or if not, wait until you've taken precalc, because I've heard it's actually easier to get a very high score on level 2 if you've taken a high enough level of math. For the University of California (my state, so I know those requirements best), math level 1 doesn't count as one of your required subject tests, but I think it does for most other colleges.\r\n\r\nThe sparknotes website linked in the previous post also has stuff for the math subject tests.", "Solution_11": "[quote=\"mechanicalpencil\"]Math level 2 has precalc/trig, math level 1 just has algebra 2 and below. If you've taken precalc, take level 2, or if not, wait until you've taken precalc, because I've heard it's actually easier to get a very high score on level 2 if you've taken a high enough level of math. For the University of California (my state, so I know those requirements best), math level 1 doesn't count as one of your required subject tests, but I think it does for most other colleges.\n\nThe sparknotes website linked in the previous post also has stuff for the math subject tests.[/quote]\r\nThanks,this web-site is really awesome.\r\nCould you explain a bit clearer what math level 1 contains besides algebra?\r\n\r\nWhat is the University of California,you have mentioned in the text?Los Angeles,Berkeley,Davis or something else...\r\n\r\nMaybe you or someone else know about requirements of other universities,like CalTech,Berkeley,Stanford.\r\nI know that MIT requires either math 1 or math 2,but what's more preferable for their admission?", "Solution_12": "[url=http://en.wikipedia.org/wiki/University_of_California]University of California[/url] is a collection of the [url=http://www.universityofcalifornia.edu/campuses/welcome.html]following state schools in California.[/url] Stanford is a private school in Palo Alto, California (part of Silicon Valley). Mostly they recquire the basic tests as SAT/ACT, subject specific tests, essays, and then references of people who know you academically/socially for admission.\r\n\r\nThere will also be a [url=http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php]MIT Math Jam[/url] on 23rd of Octobre if you consider this school.\r\n\r\nCaltech is a private university close to Los Angeles. You may want to consult the mathlinker [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=9911]Gjergji Zaimi[/url] for that matter. I think Caltech and MIT are at the same level, and actual choice is rather due to subject preference. I think this forum is a bit MIT-biased for some reason! :D Of course, it depends on how much fin aid you will get, type of area you prefer etc. I mean discussing weather matters gives a hint where you should study as you may not want to study the whole day. :wink:", "Solution_13": "[quote=\"orl\"][url=http://en.wikipedia.org/wiki/University_of_California]University of California[/url] is a collection of the [url=http://www.universityofcalifornia.edu/campuses/welcome.html]following state schools in California.[/url] Stanford is a private school in Palo Alto, California (part of Silicon Valley). Mostly they recquire the basic tests as SAT/ACT, subject specific tests, essays, and then references of people who know you academically/socially for admission.\n\nThere will also be a [url=http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php]MIT Math Jam[/url] on 23rd of Octobre if you consider this school.[/quote]\r\nThanks,it was really helpful info.\r\nWhat about CalTech?Do they have some specific requirements or preferences in SAT?What is the best choice among the mentioned universities,if I am going to have math major.Also please have in mind that I am international applicant.\r\nIt seems strange that I haven't heard anything about these classes yet?But it sounds really helpful in any case.\r\nWe are straying away from the subject,so I think,it makes sense to change title,what do you think mods?\r\nHow do you feel about \"several advices conerning passing SAT and some info about American Universities\"?", "Solution_14": "Well...probably nobody saw my last reply,so I will repeat:\r\nCould you suggest me some top US universities,if I am international applicant,I would like to have math major and it is preferable for me to pass TOEFL,not SAT...\r\nThanks in advance.\r\n[i]Added after reading [b]orl's[/b] reply above:[/i]\r\nActually,I mainly consider MIT,but as an alternative option I will apply for other universities as well.As I said I would better choose math major and it is quite preferable for me to pass TOEFL,and MIT is the best place due to all these factors.\r\nBut I do really wonder if there exists any other such universities?", "Solution_15": "Regarding the test: don't worry about answering all of the questions. Before I took the test (last year), I took one practice test for the Physics subject test and found a comfortable number of questions I knew I could answer in the time without sacrificing accuracy -- I left about 6 questions blank and still got the score." } { "Tag": [ "linear algebra", "matrix", "IMC", "college contests" ], "Problem": "The 16th IMC 2009 will be in Budapest, Hungary on 25th - 30th July, 2009.\r\nIt is being co-organized by University College London and Eotvos Lorand University. \r\n\r\nhttp://www.imc-math.org.uk/\r\n\r\nSchedule:\r\n\r\nFirst Day - 25th July - Arriving and Accommodation, Registration \r\nSecond Day - 26th July - Opening Ceremony, Additional Registration, Meeting of the Jury\r\nThird Day - 27th July - First Exam Day \r\nFourth Day - 28th July - Second Exam Day \r\nFifth Day - 29th July - Closing Ceremony, Final Dinner\r\nSixth Day - 30th July - Departure \r\n\r\nSo it seems there will be no excursion day and therefore there will be a time for one coordination session only (instead of two in 2008, i.e. the first in the evening on excursion day and the second in the morning on closing day). :maybe:", "Solution_1": "do u know anything about the accommodation?", "Solution_2": "In our invitation letter is written this:\r\n\r\nThe participants will be accommodated in the student\r\nresidences BME K\u00e1rm\u00e1n T\u00f3dor Koll\u00e9gium (Budapest, Irinyi J. u. 911, H1111)\r\nand BME Sch\u00f6nherz Zolt\u00e1n Koll\u00e9gium (Budapest, Irinyi J. u. 42, H1111).\r\n\r\n :)", "Solution_3": "[quote=\"hana\"]In our invitation letter is written this:\n\nThe participants will be accommodated in the student\nresidences BME K\u00e1rm\u00e1n T\u00f3dor Koll\u00e9gium (Budapest, Irinyi J. u. 911, H1111)\nand BME Sch\u00f6nherz Zolt\u00e1n Koll\u00e9gium (Budapest, Irinyi J. u. 42, H1111).\n\n :)[/quote]\r\n\r\nWell, you seem to be better informed than me. :) \r\nAnyway, here are some photos I found of the rooms:\r\nK\u00e1rm\u00e1n dormitory: http://ghk.ktk.bme.hu/demoszoba.php\r\nSch\u00f6nherz dormitory: http://picasaweb.google.com/SaTi85/SchSzoba# (this one's a student's photos about his room)\r\n\r\nFurther information is available here: http://www.imc-math.org.uk/index.php?year=2009&item=travel", "Solution_4": "[b]kdano,[/b] would you be stayin in the dormitory too? :)\r\n\r\nBtw, I've found clear ncie and big map of Budapest ~ 6mb, 4880x3710\r\nDirect link: http://www.orangesmile.com/destinations/img/budapest-map-big.jpg\r\nedit: it's was just a part of the city... noway you can find the XI district there =)\r\nhttp://www.fsz.bme.hu/hungary/budapest/cgi-bin/search_tkp is bigger map and the Irinyi str is in the upper right corner of dictrict XI :wink:", "Solution_5": "[quote=\"rustam\"][b]kdano,[/b] would you be stayin in the dormitory too? :)\n\nBtw, I've found clear ncie and big map of Budapest ~ 6mb, 4880x3710\nDirect link: http://www.orangesmile.com/destinations/img/budapest-map-big.jpg[/quote]\r\n\r\nNo, I'm gonna sleep at home :)\r\nIf you want to use this map for sightseeing, it's quite alright, though it has some inconveniences:\r\nFirst of all, none of the IMC-related buildings (the dormitories, the campus) are on this map :) they're simply more to the south.\r\nAnother one is that there's been a major reorganization in the public transport system. Couple of routes got a new number (like bus 78 is now 178 and 178A) , some got extended (like bus 5 now goes all the way through R\u00e1k\u00f3czi \u00fat), but eg. bus 7A no more exists, while others, like trams 4, 6 didn't change at all (neither did undergrounds :P). So don't expect the routes to be exactly like this :D", "Solution_6": "can someone attach the problems of IMC 2009 ?", "Solution_7": "Is there any results from IMC 2009?", "Solution_8": "http://imc2009.elte.hu/", "Solution_9": "Thanks a lot, kdano.", "Solution_10": "it's funny how problem 3 from the second day changes from the competition paper to the solution paper ;)\r\nwhich one is the real problem 3?", "Solution_11": "can someone tell me this year cut off? any mathlinkers there?.. so we only have 10 problems this year.\r\n\r\ni think, 2nd day problem 3 in the solution paper was already purposed before > :maybe: \r\n\r\nthanks", "Solution_12": "[quote=\"ma_go\"]it's funny how problem 3 from the second day changes from the competition paper to the solution paper ;)\nwhich one is the real problem 3?[/quote]\r\n\r\nThat [i]is[/i] funny :P\r\nThe actual second day problems are that of the \"solutions\" file.", "Solution_13": "Another funny (?) thing noted by my friend:\r\n\r\nThe 2nd day Problem No 3:\r\n\r\nThe condition can be written in this way:\r\n\r\n$ A(AB\\minus{}BA)\\equal{}(AB\\minus{}BA)A.$\r\n\r\nThen the statement of the problem is the special case ($ n\\equal{}1$) of the Linear Algebra No 2\r\nproblem of Iran 2009 olympiad (15-17 July), see\r\n\r\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=291882[/url]", "Solution_14": "I also noticed that. \r\nA very similar problem appears as an exercise in Matrix Analysis Book by Horn and Johnson.\r\nAlso the ISOM problem is a problem in V.V Prasolov, Problems and Theorems in Linear Algebra.\r\n\r\nEven the idea has already discussed here in mathlinks a few times, one thread is (posted not long before IMC and ISOM )\r\nhttp://www.mathlinks.ro/viewtopic.php?t=280822\r\n\r\nand also the percentage of participants who solved this problem in IMC is relatively low.. :maybe:", "Solution_15": "During the competition I tried to solve first of all this problem because I have a feeling that I've seen it somewhere before. After 2 hours of struggling, I got 10 points worth solution :lol: \r\n\r\nThanx a lot for mathlinks again!", "Solution_16": "Please: Mathematical Olympiad Challenges, 2.9, problem 11.", "Solution_17": "Out of curiosity: anybody knows where will next IMC be held? :D", "Solution_18": "[quote=\"EUCLA\"]Out of curiosity: anybody knows where will next IMC be held? :D[/quote]\r\n\r\nProbably Bulgaria (but this time not Blagoevgrad, somewhere also not too far from Sofia). At least in Budapest J.J. said that he thought about.\r\n\r\nBy the way in 2010 three olympiads for university students are supposed to be in Bulgaria - [url=http://www.seemous.eu/]SEEMOUS[/url] in Plovdiv (March 2010), Open National Olympiad [url=http://translate.google.com/translate?prev=hp&hl=en&js=y&u=http%3A%2F%2Fnsom.ru.acad.bg%2F&sl=bg&tl=en&history_state0=]\"NSOM'[/url] in Veliko Tarnovo (May 2010) and probably IMC (July 2010?)." } { "Tag": [ "geometry", "3D geometry" ], "Problem": "The volume of a cube is 12.5 cubic centimeters. If each side is doubled, what is the number of cubic centimeters in the new volume?", "Solution_1": "[hide=\"kind of messy but it is the solution\"]If you notice, we have a cubic measurment of 12.5, and we are introducing a an element of 2. So we need to cube the 2 and multiply it by 12.5 giving us 100.\n\nIf you dont see that, consider the following:\n\nLet x be the side of the cube.\n\n$x^3=volume=12.5$\n\nIf side is doubled we know have side = 2x\n\nSo:\n\n$2^3 x^3=8x^3=100$[/hide]", "Solution_2": "[quote=\"MCrawford\"]The volume of a cube is 12.5 cubic centimeters. If each side is doubled, what is the number of cubic centimeters in the new volume?[/quote]\r\n\r\n[hide=\"Quick and easy\"]\nSince a cube has three dimensions, doubling the side length multiplies the volums by 2:^3:, or 8.\n\n12.5 x 8 = [b]100[/b]\n[/hide]", "Solution_3": "[hide]\n100\n[/hide]", "Solution_4": "[hide][color=indigo]\nIf the side is x, then the volume is x*x*x.\nThat means if the side is doubled the volume is 2x*2x*2x.\nSo the volume is increased by 8.\n8*12.5=[/color][color=green]100[/color][/hide]" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let $a,b,c,d\\in R^+$ and $abcd=1$.\r\nProve that\r\n$(\\frac{1+ab}{1+a})^4+(\\frac{1+bc}{1+b})^4+(\\frac{1+cd}{1+c})^4+(\\frac{1+da}{1+d})^4\\geq 4$", "Solution_1": "Posted before. To prove thi is equivalent to proving without the fourth powers I.E.\r\n\r\n(ab+1)/(a+1)+.......>=4, in which case a triple aplication of harmonic mean finishes it off. \r\n\r\nYOu first homogenixe by letting a=x/y b=y/z c=z/w d=w/x and the apply AMHM.\r\n\r\nBomb" } { "Tag": [ "AMC", "AIME" ], "Problem": "anyone think the two most recent olymon (june july august) are really hard?\r\n\r\ni beasted the 1998 aime and the 1984 ahsme with near perfect scores just now (timed)... then i tried olymon and got owned on the hard questions", "Solution_1": "Good, then its not just me. Yes, it seems harder than I remembered it being before.", "Solution_2": "The questions aren't in increasing diffuculty, right?" } { "Tag": [ "modular arithmetic", "algorithm", "number theory", "algebra unsolved", "algebra" ], "Problem": "7a \u2261 1 (mod 30) I have the resolution of the problem but i dont understand the first step.\r\n\r\n7a \u2261 1 (mod 30) <--> [color=red]7a \u2261 91 (mod 30)[/color] <--> 7a \u2261 7.13 (mod 30)\r\n\r\nAnd using that (7:30)=1 its like saying a \u2261 13 (mod 30), then \r\n\r\na=30q + 13\r\n\r\n\r\nSo i dont understand where the 91 comes from, i dont see it.\r\n\r\n[color=red]7a \u2261 91 (mod 30)[/color]", "Solution_1": "Because $91\\equiv 1\\pmod{30}$.", "Solution_2": "Thanks N.T.TUAN, so have then to look at the properties of cngruence and because:\r\n\r\na \u2261 b (m) --> b \u2261 a (m)\r\n\r\nand trnasitivity\r\n\r\na \u2261 b (m) and b \u2261 c (m) --> a \u2261 c (m)\r\n\r\nThat would be the theory to explain right?\r\n\r\nWhat can i use to find such a number as this 91 but for other exercises, is there any algorithm or just by \"eye\" ?", "Solution_3": "For $ax\\equiv b\\pmod{m}$, where $\\gcd{(a,m)}=1$, we can slove same as before. Find $c$ such that $ca\\equiv 1\\pmod{m}$.", "Solution_4": "Thank you very much! :D", "Solution_5": "[quote=\"sanchez45\"]Thank you very much! :D[/quote]\r\nNo problem, I think you need a text book on Number Theory ( rank Olympiad, if you want to *MO ) . Example I.Niven or D.Burton.", "Solution_6": "Actually i have a book in spanish from a profesor of my university but i am studying this by myself." } { "Tag": [ "number theory", "greatest common divisor", "least common multiple" ], "Problem": "I suddenly realise that the scope of problems based on number system is much larger than i thought initially... now what should i know to strengthen my fundamentals....", "Solution_1": "What are number systems? Could you give an example? Is it like number theory? :maybe:\r\n\r\nEDIT: Or are you refering to bases?", "Solution_2": "Number Theorey and all... lol.... Number system just sounds weird... I just realised it! GCD, LCM,Digits, and so on..." } { "Tag": [ "geometry", "geometric transformation", "rotation", "Pythagorean Theorem" ], "Problem": "Let A, B fixed in the Euclidean plane. find the geometric locus of all points M in the plane satisfying the equality MA^2 - MB^2 = k, where k is some real constant\r\n\r\nPlease Help me i don't know where to start", "Solution_1": "Put a line through the points $A, B$, pick an arbitrary 3rd point $P$ on this line and erect a normal at this point. Now, pick an arbitrary point $M$ on the normal and imagine 2 squares constructed on the segments $MA$ and $MB$. By Pythagorean theorem, the 1st square has the side $MA = \\sqrt{PA^2 + MP^2}$ and the area $MA^2 = PA^2 + MP^2$. Likewise, the 2nd square has the side $MB = \\sqrt{PB^2 + MP^2}$ and the area $MB^2 = PB^2 + MP^2$. The difference of the areas of these 2 squares is\r\n\r\n$MB^2 - MA^2 = (PB^2 + MP^2) - (PA^2 + MP^2) = PB^2 - PA^2$\r\n\r\nDoes this difference depend on the position of the point $M$ on the normal at $P$ ? So, what is the locus ? What remains to be done is to find the position of the point $P$ (the foot of the normal), given the segment $AB$ and some real number $k$. Obviously,\r\n\r\n$k = PB^2 - PA^2 = (PA + AB)^2 - PA^2 =$\r\n\r\n$= PA^2 + 2 \\PA \\cdot AB + AB^2 - PA^2 = AB \\cdot (2PA + AB)$ \r\n\r\n$PA = \\frac 1 2 \\left(\\frac{k}{AB} - AB\\right)$", "Solution_2": "Start by assuming that A and B are both points on the x-axis. Let $A = (-a,0)$ and $B = (a,0)$ for some number $a$. So you are looking for the set of points $(x,y)$ such that $[(x+a)^2 + y^2] - [(x-a)^2 + y^2] = k$. Try and solve this for x and y.\r\n\r\nOnce you have that, you pretty much have your solution - because every other placement of A and B will just be translations and rotations of that solution." } { "Tag": [ "ceiling function", "number theory unsolved", "number theory" ], "Problem": "Find the number of elements that a set $B$ can have, contained in $(1, 2, ... , n)$, according to the following property: For any elements $a$ and $b$ on $B$ ($a \\ne b$), $(a-b) \\not| (a+b)$.", "Solution_1": "Obviously we cannot choose any numbers whose difference is 1 or 2, so an upper bound is $\\lceil{\\frac{n}{3}}\\rceil$. This is also attainable; just choose all numbers that are 1 mod 3." } { "Tag": [ "induction", "modular arithmetic", "number theory proposed", "number theory" ], "Problem": "Show that there are an infinity of positive integers $n$ such that $2^{n}+3^{n}$ is divisible by $n^{2}$.", "Solution_1": "i) $a^{2}|2^{a}+3^{a}$ and $b^{2}|2^{b}+3^{b}\\Longrightarrow (ab)^{2}|2^{ab}+3^{ab}$\r\n\r\nii)5 is 'good' and 55 is 'good'\r\n\r\nusing this 2 facts, we have one solution, because we easily see that the numbers $5^{k}55^{l}$ are 'good's'.", "Solution_2": "I don't get why i) is true for $a, b$ not being coprime. If it would be true for all $a, b$ you could use just $5$ and you would have $5^{2k}|2^{5^{k}}+3^{5^{k}}$ but that's false. Am I wrong?", "Solution_3": "[quote=\"TomciO\"]I don't get why i) is true for $a, b$ not being coprime. If it would be true for all $a, b$ you could use just $5$ and you would have $5^{2k}|2^{5^{k}}+3^{5^{k}}$ but that's false. Am I wrong?[/quote]\r\nYou are right $5^{2k}\\not |2^{5^{k}}+3^{5^{k}}$ for $k>1$, but $55^{2k}|2^{55^{k}}+3^{55^{k}}$.", "Solution_4": "So it probably follows from the induction easily (I'm too lazy to check) and we are done, but it doesn't follow from i), I think.", "Solution_5": "[quote=\"Rust\"]but $55^{2k}|2^{55^{k}}+3^{55^{k}}$.[/quote]\r\n\r\nIt seems to me that this is false too. Look,\r\n\r\nwe know that if $\\parallel a-b\\parallel _{p}=k \\geq1$ then $\\parallel a^{n}-b^{n}\\parallel _{p}=k+\\parallel n\\parallel _{p}$; but assuming that your clame is correct we should have that $2k \\leq \\parallel 2^{55^{k}}-(-3)^{55^{k}}\\parallel _{5}=\\parallel 2+3\\parallel _{5}+\\parallel 55^{k}\\parallel _{5}=1+k$ so $k=1$????But you clamed that $55^{2k}|2^{55^{k}}+3^{55^{k}}$ for all positive integer $k$s,don't you?", "Solution_6": "[hide]\nWe know there exists a positive integer solution, since 1 clearly works. Suppose there are finitely many, then there exists a largest integer $M$ for which this property holds.\n\nFor all positive integers $M$, we have $2^{M}+3^{M}>M^{2}$. (In fact $3^{M}>M^{2}$ which is easliy verifiable through induction and $3M^{2}>(M+1)^{2}$ for $M\\geq 2$.). Therefore, there exists a prime $p$ such that $2^{M}+3^{M}$ is divisible by $pM^{2}$. We will now show that $2^{pM}+3^{pM}$ is divisible by $(pM)^{2}$, which will give us the contradiction we need.\n\nTo prove the above assertion, it's clear that $p,M$ are both odd and not divisible by 3 in order to have $2^{M}+3^{M}$, an odd number not divisible by 3, divisible by $pM^{2}$. We now have\n\\[2^{pM}+3^{pM}=(2^{M}+3^{M})\\left(\\sum_{k=0}^{p-1}(-1)^{k}2^{kM}3^{(p-1-k)M}\\right) \\]\nIt remains to prove that the second factor in the right hand side is divisible by $p$. Observe that $2^{M}\\equiv-3^{M}\\pmod{p}$, so that\n\\[(-1)^{k}2^{kM}3^{(p-1-k)M}\\equiv 3^{(p-1)M}\\equiv 1\\pmod{p}\\]\nThen the right hand sum, consisting of $p$ terms all congruent to 1 modulo $p$, is divisible by $p$, as desired.\n[/hide]", "Solution_7": "We will construct the infinite sequence $ {n_{k}}$ such that for any member $ n_{k}$ of this sequence satisfies the condition:\r\n$ 2^{n_{k}} + 3^{n_{k}}$ is divisible by $ n_{k}^{2}$. Easy to verify that $ n_{1} = 1$ and $ n_{2} = 5$. Assume we constructed it until $ n_{k}$. We will easily find $ n_{k + 1}$. Let $ 2^{n_{k}} + 3^{n_{k}} = an_{k}^{2}$,where $ a$ is odd positive integer(as $ n_{k} > = n_{2} = 5$, easy to show that $ 2^{n_{k}} + 3^{n_{k}} > n_{k}^{2}$) and $ a > 1$. Take $ p$ the odd prime divisor of $ a$. Substitute $ 2^{n_{K}}$ by $ x$ and $ 3^{n_{k}}$ by $ y$. Then, obviously ${ x^{p} + y^p} = (x + y)(x^{p - 1} - yx^{p - 2} + ... + y^{p - 1})$. As $ x + y$ is divisible by $ pn_{k}^{2}$,\r\nif we define $ r$ to be the reminder of $ x mod p$, than the reminder $ y mod p$ will be $ - r$. So, $ x^{p - 1}, - yx^{p - 2},...,y^{p - 1}$ numbers have the same reminder if we divide them by $ p$, but the number of them is $ p$. So, $ x^{p - 1} - x^{p - 2} + ... + y^{p - 1}$ is divisible by $ p$. It follows \r\n$ x^{p} + y^{p}$ is divisible by $ (pn_{k})^{2}$. Now we take $ n_{k + 1} = pn_{k}$. As $ p > 1$, $ n_{k + 1} > n_{k}$ and \r\n$ 2^{n_{k + 1}} + 3^{n_{k + 1}}$ is divisible by $ n_{k + 1}^{2}$, which completes our proof.\r\nP.S. That's not my own solution. I have read this solution before.", "Solution_8": "The official solution was(i'm sorry,if it was posted before):\r\nIf $ n_k$ is the number,such that $ n_k^2|3^{n_k}\\plus{}2^{n_k}$ then $ n_{k\\plus{}1}\\equal{}\\frac{3^{n_k}\\plus{}2^{n_k}}{n_k}$ also satisfies to the given condition.\r\nSo by taken $ n_1\\equal{}1$ it follows that $ n_2\\equal{}5,n_3\\equal{}55\\dots$ and we are done.", "Solution_9": "Obviously $5$ satisfy $n^2 | 2^n + 3^n$.\nNow, we'll prove that for each odd number $a$ satisfy, we could construct another larger odd number satisfy.\nTake an odd prime $q | 2^a + 3^a$ that hasn't been used before. This must exists due to [b]Zsigmondy Theorem[/b].\nWe'll now prove $aq$ satisfy.\nNow, notice that our hypothesis gives us \n\\[ a^2 | 2^a + 3^a | (2^a)^q + (3^a)^q \\]\nMoreover, $v_q (2^{aq} + 3^{aq}) = v_q (2^a + 3^a) + v_q (q) \\ge 2 $.\n\nUsing this method, one could construct arbitrarily long sequence $a_n$ starting with $a_1 = 5$, and applying the strategy above to find $a_n$ by using $a_{n - 1}$ we're then finished." } { "Tag": [ "function", "inequalities", "algebra unsolved", "algebra" ], "Problem": "let function $f$:[0,1]-R such that:\r\n\r\n$f(1/n)$=$(-1)^{n}$, :forall $n$=1,2..\r\n\r\nProve that: no exists 2 functions creasing $h(x)$,$g(x)$:[0,1]-$R$ such that:\r\n$f(x)$=$g(x)$-$h(x)$ :forall $x$ :in [0,1]", "Solution_1": "$var(f(x))=var(g(x))+var(h(x))=\\infty$, therefore or $var(h(x))=\\infty$ or $var(h(x))=\\infty .$ It give contradition for increase or decrease function.", "Solution_2": "Official solution:\r\nSuppose that there exist such functions $g$ and $h$. Then for any $k\\in\\mathbb{N}$ we have \\[g\\left(\\frac1{2k+1}\\right)= f\\left(\\frac1{2k+1}\\right)+h\\left(\\frac1{2k+1}\\right)=-1+h\\left(\\frac1{2k+1}\\right)<-1+h\\left(\\frac1{2k}\\right)= 1+h\\left(\\frac1{2k}\\right)-2= f\\left(\\frac1{2k}\\right)+h\\left(\\frac1{2k}\\right)-2= g\\left(\\frac1{2k}\\right)-2< g\\left(\\frac1{2k-1}\\right)-2.\\] Applying this inequality we get \\[g(0)< g\\left(\\frac1{2k+1}\\right)< g\\left(\\frac1{2k-1}\\right)-2< g\\left(\\frac1{2k-3}\\right)-4< \\ldots< g\\left(\\frac13\\right)-2(k-1)< g(1)-2k.\\] Thus $g(1)-g(0)>2k$ for any $k$, which is impossible. \r\n@[b]Rust[/b]: maybe it's not a clever question, but what is $var$?" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "consider this problem,\r\nproblem:there are unknown number of doors for a secret place.Each door has 4 different locks.There are 8 people posessing keys to locks in such a way that any combination of 4 people can open all keys of all doors.\r\n what is the minimum number of doors? and how many keys?", "Solution_1": "so for any combination of 3 of them they can't open the door .\r\nso we have $\\binom{8}{3}=56$ locks so we have $\\frac{56}{4}=14$ doors. ;)", "Solution_2": "Thank you a_vakilian.The solution is incomplete in two aspects,so let me complete the formalities.\r\n(1)First let me prove that \"for every group of three people there exists a unique lock which cant be opened\".\r\n On contrary,let there be a lock that cant be opened by atleast two groups of three people.This means if a person from one of the two groups joins the other group , the four cant still open the lock. This is contradiction.\r\n(2)The number of locks is exhaustive.\r\n Let us say there may exist some more lock(s).If it is not the lock already enumerated then it must be such that three can open it.Again contradiction.", "Solution_3": "I consider it obvious. ;) \r\nbut thank" }