{ "Tag": [ "function", "induction", "linear algebra", "matrix", "quadratics", "algebra", "polynomial" ], "Problem": "The sequence $a_{n}$ satisfies the recursion $a_{n}=3a_{n-1}+28a_{n-2}$ for integers $n>2$. If $a_{1}=3$ and $a_{2}=65$, write an explicit formula for $a_{n}$ in terms of $n$.", "Solution_1": "I used generating functions to obtain that [hide]$a_{n}= 7^{n}+(-4)^{n}$[/hide]Perhaps if one conjectures it, it's easy to prove by induction?", "Solution_2": "Yeah. \r\n\r\nI'm not very good at generating functions..could you post your solution?", "Solution_3": "[hide=\"Solution\"] We have the characteristic equation $x^{2}-3x-28 = 0$ with roots $-4, 7$. Then\n\n$a_{n}= \\alpha \\cdot 7^{n}+\\beta \\cdot (-4)^{n}$\n\nAnd we calculate $\\alpha = \\beta = 1$. [/hide]", "Solution_4": "Yeah, I'm not good at characteristic equations :( I'd like to learn though. Is there a good explanation of it on the internet?\r\n\r\nI'll post my step-by-step solution with generating functions. [hide]To keep it readable, let's use $S_{k}(\\psi(x)) = \\sum_{i=k}^{\\infty}\\psi(x)$. Let $f(x) = S_{1}\\left(a_{i}x^{i}\\right)$. We have that \\[\\begin{aligned}f(x) & = S_{1}\\left(a_{i}x^{i}\\right) = a_{1}x^{1}+a_{2}x^{2}+S_{3}\\left(a_{i}x^{i}\\right) \\\\ & = 3x+65x^{2}+S_{3}\\left(3a_{i-1}x^{i}+28a_{i-2}x^{i}\\right) \\\\ & = 3x+65x^{2}+3xS_{3}\\left(a_{i-1}x^{i-1}\\right)+28x^{2}S_{3}\\left(a_{i-2}x^{i-2}\\right) \\\\ & = 3x+65x^{2}+3xS_{2}\\left(a_{i}x^{i}\\right)+28x^{2}S_{1}\\left(a_{i}x^{i}\\right) \\\\ & = 3x+65x^{2}+3x\\left(S_{1}\\left(a_{i}x^{i}\\right)-a_{1}x^{1}\\right)+28x^{2}S_{1}\\left(a_{i}x^{i}\\right) \\\\ & = 3x+65x^{2}+3x\\left(f(x)-3x\\right)+28x^{2}f(x) \\\\ & = 3x+56x^{2}+f(x)\\left(3x+28x^{2}\\right).\\end{aligned}\\] Solving for $f(x)$, we get \\[f(x) = \\frac{3x+56x^{2}}{1-3x-28x^{2}}=-2-\\frac{3x-2}{1-3x-28x^{2}}.\\] But $1-3x-28x^{2}= (1-7x)(1+4x)$, so \\[f(x) =-2-\\frac{3x-2}{(1-7x)(1+4x)}.\\] Split that last one into partial fractions: \\[\\frac{3x-2}{(1-7x)(1+4x)}=-\\frac{1}{1-7x}-\\frac{1}{1+4x}.\\] We obtain that \\[\\begin{aligned}f(x) & =-2+\\frac{1}{1-7x}+\\frac{1}{1-(-4x)}\\buildrel{\\text{Sum of geo. series}}\\over=-2+S_{0}\\left((7x)^{i}\\right)+S_{0}\\left((-4x)^{i}\\right) \\\\ & = S_{1}\\left((7x)^{i}+(-4x)^{i}\\right) = S_{1}\\left((7^{i}+(-4)^{i})x^{i}\\right).\\end{aligned}\\] From which, $a_{n}= 7^{n}+(-4)^{n}$. Tadaaaaaaaaaa :P [/hide]", "Solution_5": "[quote=\"t0rajir0u\"][hide=\"Solution\"] We have the characteristic equation $x^{2}-3x-28 = 0$ with roots $-4, 7$. Then\n\n$a_{n}= \\alpha \\cdot 7^{n}+\\beta \\cdot (-4)^{n}$\n\nAnd we calculate $\\alpha = \\beta = 1$. [/hide][/quote]\r\n\r\nWHy do you get that characteristic equation...? then, why are the roots expressed as $r^{n}$ in $a_{n}=\\alpha \\cdot 7^{n}+\\beta \\cdot (-4)^{n}$? then why $\\alpha=\\beta=1$? :blush:", "Solution_6": "Yeah... I like the characteristic equation solution a bit more than that generating function madness...\r\n\r\nOnly when I google characteristic equation, I get matrix stuff. Explanation/link t0r?", "Solution_7": "How about this one?\r\n\r\nHosei University entrance exam/management 1987\r\n\r\nFind the $n$ th term of the sequence $\\{a_{n}\\}$ such that $a_{1}=1,\\ a_{2}=2,\\ a_{n+2}-a_{n+1}-2a_{n}=2.$", "Solution_8": "Substitute $a_{n}=b_{n}-1$ and then you can use characteristic equations", "Solution_9": "We have that \\[\\begin{aligned}a_{n+2}+a_{n+1}& = 2+2(a_{n+1}+a_{n}) = 2+2(2+2(a_{n}+a_{n-1})) \\\\ & = 2+4+4(a_{n}+a_{n-1}) = 2+4+8+8(a_{n-1}+a_{n-2}) = \\cdots \\\\ & = \\sum_{i=1}^{n}2^{i}+2^{n}(a_{2}+a_{1}) = 2(2^{n}-1)+3\\cdot2^{n}= 5\\cdot 2^{n}-2\\end{aligned}\\] and \\[\\begin{aligned}a_{n+2}-2a_{n+1}-1 & =-(a_{n+1}-2a_{n}-1) = a_{n}-2a_{n-1}-1 = \\cdots \\\\ & = (-1)^{n}(a_{2}-2a_{1}-1) = (-1)^{n+1}\\end{aligned}\\] So we have \\[\\begin{aligned}3a_{n+2}& = 2(a_{n+2}+a_{n+1})+(a_{n+2}-2_{n+2}) \\\\ & = 5\\cdot 2^{n+1}-4+(-1)^{n+1}+1 = 5\\cdot 2^{n+1}+(-1)^{n+1}-3.\\end{aligned}\\] This means that $a_{n}= \\frac{1}{3}\\left(5\\cdot 2^{n-1}+(-1)^{n+1}-3\\right)$.\r\n\r\nIs this good, kunny? :)", "Solution_10": "[quote=\"Hokkage\"]Yeah... I like the characteristic equation solution a bit more than that generating function madness...\n\nOnly when I google characteristic equation, I get matrix stuff. Explanation/link t0r?[/quote]\r\n\r\n[hide=\"A theorem\"] [b]Theorem:[/b] Every sequence of the form $a_{n}= \\alpha \\gamma^{n}+\\beta \\delta^{n}$ is a second-order linear recurrence of the form\n\n$a_{n}= A a_{n-1}+B a_{n-2}$\n\nProof: The proof is actually very simple. Let $A = \\alpha+\\beta, B =-\\alpha \\beta$. Then\n\n$\\alpha \\gamma^{n}+\\beta \\delta^{n}= A \\left( \\alpha \\gamma^{n-1}+\\beta \\delta^{n-1}\\right)+B \\left( \\alpha \\gamma^{n-2}+\\beta \\delta^{n-2}\\right)$\n\nAnd we see that the quadratic\n\n$x^{2}= Ax+B$\n\nHas roots $\\alpha, \\beta$. \n\nA very nonrigorous way of understanding this: Let $a_{n}$ be a second-order linear recurrence. We assume that $a_{n}$ is \"approximately\" exponential, that is, $a_{n}= \\epsilon \\zeta^{n}$ and then we simply divide out to get\n\n$\\zeta^{2}= A \\zeta+B$\n\nThis is called the characteristic equation of the recurrence.\n\nI claim that every recurrence such that its characteristic equation has distinct roots can be written in the form I gave above. The proof uses generating functions and is very long, so I will not give it here.\n\nFrom here it is a simple matter of calculation. :) [/hide]", "Solution_11": "Something on recurrence equations and characteristic polynomials:\r\n\r\n[url]http://en.wikipedia.org/wiki/Recurrence_relation[/url]\r\n[url]http://www2.ics.hawaii.edu/~sugihara/course/ics141s98/note/04-23n24[/url]", "Solution_12": "[quote=\"Kurt G\u00f6del\"]\n\n$a_{n}= \\frac{1}{3}\\left(5\\cdot 2^{n-1}+(-1)^{n+1}-3\\right)$.\n\nIs this good, kunny? :)[/quote]\r\n\r\nYes, your answer is correct. :)", "Solution_13": "I got $a_{n}=-\\frac13 (-1)^{n}+\\frac56 (2)^{n}-1$ using Elemennop's idea.", "Solution_14": "[quote=\"SplashD\"]I got $a_{n}=-\\frac13 (-1)^{n}+\\frac56 (2)^{n}-1$ using Elemennop's idea.[/quote]\r\n\r\nThat's the same as [b]Kurt G\u00f6del[/b]'s answer, just with some exponents and common factors juggled around." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Solve for x\r\n$ \\sqrt{x\\plus{}\\sqrt{x^2\\minus{}1}}\\equal{}\\frac{9\\sqrt{2}}{4}(x\\minus{}1)\\sqrt{x\\minus{}1}$", "Solution_1": "By using Maple, we get the result: $ x \\equal{} \\frac {5}{3}$. (Plugging in $ x \\equal{} \\frac {5}{3}$ in the equation $ \\sqrt{x\\plus{}\\sqrt{x^2\\minus{}1}}\\equal{}\\frac{9\\sqrt{2}}{4}(x\\minus{}1)\\sqrt{x\\minus{}1}$, it becomes $ \\sqrt3\\equal{}\\sqrt3$.)", "Solution_2": "this is the problem from the newest Mathematics and Youth Magazine (7/2008). The dealine for submitting solutions is 15 September, 2008, so don't answer.See detail here:\r\n[url]http://www.mathlinks.ro/viewtopic.php?p=1195403#1195403[/url]", "Solution_3": "[quote=\"greenvert\"]this is the problem from the newest Mathematics and Youth Magazine (7/2008). The dealine for submitting solutions is 15 September, 2008, so don't answer.See detail here:\n[url]http://www.mathlinks.ro/viewtopic.php?p=1195403#1195403[/url][/quote]\r\nThank you. I didn't know that. Sorry for posting my comments. Moderators, please lock the topic.\r\nAnd Doctor Graubner: Please don't post a solution in the meantime." } { "Tag": [], "Problem": "Three non-collinear points $A,B,C$ exist in the same plane. $P$ internally divides $AB$ into segments of $t : (1-t)$ and $Q$ divides the $BC$ into the same proportion internally as well. $R$ divides $PQ$ internally into segments of $t: (1-t)$. Find the locus of $R$ as $t$ varies if $A= (1,1)$, $B=(2,3)$ and $C=(3,1)$.", "Solution_1": "question:\r\n\r\nwhen the segment $AB$ is divided, there are 2 possible points for $P$. This shouldn't matter, since t varies.\r\n\r\nBut what does matter is whether $Q$ divides $BC$ on the same side as $P$ or at the other point. Is the bigger side closer to B or closer to C?", "Solution_2": "I said that segment $AB$ is divided internally so for every $t$ there is only one possible $P$ (I think, correct me if I am wrong).\r\n\r\n$BQ: QC = t: (1-t)$. Everything is divided internally (for simplicity) but you can try to prove the general case for if the points divide internally or externally.", "Solution_3": "[hide=\"what I was trying to differentiate between\"]\n\nI hope I'm not missing something obvious in your question\n\n[img]http://img110.imageshack.us/img110/1310/mathvz6.jpg[/img]\n\n[/hide]", "Solution_4": "The first one." } { "Tag": [], "Problem": "It's easy to show that there are arbitrarily many composite consecutine positive integers but, \r\n\r\nIs there a set of 1000 positive consecutive integers, with exactly 5 of them prime?", "Solution_1": "Well, consider the first 1000 numbers, which contains more than 5 primes, and a string of 1000 composite numbers, which contains 0 primes.\r\nEvery time we shift our string over 1, we either change the number of primes by 1,0, or -1, depending whether the next number is a prime, and whether the number you just left when you shifted is a prime.\r\nHence, starting with a number > 5, we end up with 0 after finitely many operations of +1,-1, and 0(do nothing). Therefore, logic says that every number between this number greater than 5 and 0 must have been reached, specifically, at some point 5 was reached. Hence we're done.", "Solution_2": "A question: Is there a number $ n > 1$ such that there exists a set of $ n$ consecutive numbers with more than $ \\pi(n)\\plus{}1$ primes? If so, what numbers? Are there infinitely many?\r\n\r\nHeuristically the answer should be no, since primes get rarer as you move up. Note: The reason for the $ \\plus{}1$ is that for any prime $ p$, the $ p\\minus{}1$ numbers $ 2,3,\\ldots,p$ will contain $ \\pi(p\\minus{}1)\\plus{}1$ primes.", "Solution_3": "What is $ \\pi (n)$?", "Solution_4": "$ \\pi \\left( n \\right)$ is the number of primes $ \\leq n$.", "Solution_5": "[quote=\"Hamster1800\"]A question: Is there a number $ n > 1$ such that there exists a set of $ n$ consecutive numbers with more than $ \\pi(n) \\plus{} 1$ primes? If so, what numbers? Are there infinitely many?\n[/quote]\r\n\r\nThe number of primes between $ n\\plus{}1$ and $ n\\plus{}k$ is $ \\pi(n\\plus{}k)\\minus{}\\pi(n)$. The question of whether this is always no more that $ \\pi(k)$, or in other words whether $ \\pi$ is subadditive\r\n\r\n$ \\pi(n\\plus{}k) \\leq \\pi(n) \\plus{} \\pi(k)$\r\n\r\nis the (second) Hardy-Littlewood conjecture, which is unsettled but probably false, since it was shown to contradict other widely believed conjectures." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Find all positive real a,b,c such that\r\n4(ab+bc+ac)-1>=a^2+b^2+c^2>=3(a^3+b^3+c^3)", "Solution_1": "$ 4(ab \\plus{} bc \\plus{} ac) \\minus{} 1 \\ge a^2 \\plus{} b^2 \\plus{} c^2 \\ge ab \\plus{} bc \\plus{} ca$, so $ ab \\plus{} bc \\plus{} ca \\ge \\frac {1}{3}$\r\n$ a^2 \\plus{} b^2 \\plus{} c^2 \\ge 3(a^3 \\plus{} b^3 \\plus{} c^3) \\ge (a^2 \\plus{} b^2 \\plus{} c^2)(a \\plus{} b \\plus{} c)$, so $ a \\plus{} b \\plus{} c \\le 1$, $ ab \\plus{} bc \\plus{} ca \\le \\frac {(a \\plus{} b \\plus{} c)^2}{3} \\le \\frac {1}{3}$,\r\nso $ \\begin{cases}ab \\plus{} bc \\plus{} ca \\equal{} \\frac {1}{3} \\\\\r\na \\equal{} b \\equal{} c \\end{cases}$\r\n$ a\\equal{}b\\equal{}c\\equal{}\\frac{1}{3}$ are the only solution." } { "Tag": [ "LaTeX", "function", "blogs", "\\/closed" ], "Problem": "In this topic you will see the current confirmed bugs\r\n\r\n:arrow: LaTeX pop-up window doesn't show the code (IE, Firefox, Opera) (Valentin)\r\n\r\n:arrow: Some lang entries need filling (lang) (Valentin, Dave)\r\n\r\n:arrow: Highlight function doesn't work with LaTeX properly (Valentin)\r\n\r\n:arrow: Smilies and other spam limits (Valentin)\r\n\r\n:arrow: Topic shadow can't be erased (Valentin)\r\n\r\n:arrow: Blogs page needs some more patching up (attachments box not displaying properly) (Valentin) reported @ http://www.mathlinks.ro/Forum/viewtopic.php?t=97878", "Solution_1": "Blogs auths now revised. Users can now post in blogs :)", "Solution_2": "Also, could the PDF in resource center bug also be added? Thanks.\r\n\r\nIt used to be soo awesome :)", "Solution_3": "[quote=\"anirudh\"]Also, could the PDF in resource center bug also be added?[/quote]\r\n\r\nI think it's not a bug but a decision because of copyright morons who don't want a considerable amount of problems of one competition posted at one place.\r\n\r\nWhat would be much more awesome (and should not violate copyright): http://www.mathlinks.ro/Forum/viewtopic.php?t=125784 .\r\n\r\n darij", "Solution_4": "We need a printer-friendly thing" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let \\[ a,b,c>0 \\]Prove the inequality:\r\n\\[ \\frac{a^2}{b}+\\frac{b^2}{c}+\\frac{c^2}{a}\\geq a+b+c+\\frac{4(a-b)^2}{a+b+c} \\]", "Solution_1": "Man, i'm sick of posting solutions to this one. Anyway, here it goes: \r\n\r\nLet us observe that $\\frac{a^{2}}{b}=2a-b+\\frac{(a-b)^{2}}{b}$. Writing the analogous and applying Cauchy-Schwarz inequality we have: \r\n\r\n \\[ \\frac{(a-b)^{2}}{b}+\\frac{(b-c)^{2}}{c}+\\frac{(c-a)^{2}}{a}\\geq\\frac{(2a-2b)^{2}}{a+b+c} \\].\r\n\r\n So from this inequality it follows that the inequality is done! ;)", "Solution_2": "I do not know how you get to:\r\n\r\n\\[ \\frac{(a-b)^{2}}{b}+\\frac{(b-c)^{2}}{c}+\\frac{(c-a)^{2}}{a}\\geq\\frac{(2a-2b)^{2}}{a+b+c} \\]", "Solution_3": "[quote=\"yassin88\"]I do not know how you get to:\n\n\\[ \\frac{(a-b)^{2}}{b}+\\frac{(b-c)^{2}}{c}+\\frac{(c-a)^{2}}{a}\\geq\\frac{(2a-2b)^{2}}{a+b+c} \\][/quote]\r\ncezar lupu has already stated that this is from Cauchy-Schwarz.\r\n$(a+b+c)(\\frac{(a-b)^{2}}{b}+\\frac{(b-c)^{2}}{c}+\\frac{(c-a)^{2}}{a})=(a+b+c)(\\frac{(a-b)^{2}}{b}+\\frac{(c-b)^{2}}{c}+\\frac{(a-c)^{2}}{a})\\geq (a-b+c-b+a-c)^2=(2a-2b)^2$", "Solution_4": "Thanks, this is exactly what I wanted to say. ;)", "Solution_5": "This was given in BMO 2005(problem 1) if i am not wrong...\r\n\r\nnice! :D :lol:", "Solution_6": "[quote=\"socrates\"]This was given in BMO 2005(problem 1) if i am not wrong...\n\nnice! :D :lol:[/quote]\r\n\r\nIt was also given in Morocco National Olympiads 2005. :D" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let a,b,c>0 and $ a\\plus{}b\\plus{}c\\equal{}1$. Prove that:\r\n$ \\frac{ab}{\\sqrt{ab\\plus{}c}}\\plus{}\\frac{bc}{\\sqrt{bc\\plus{}a}}\\plus{}\\frac{ca}{\\sqrt{ca\\plus{}b}} \\le \\frac{1}{2}$", "Solution_1": "I have just realized this ineq is very easy by AM-GM. :blush: \r\nWe continiuos with another old lemma:\r\nLet a,b,c>0. Prove that:\r\n$ \\frac {1}{a^2 \\plus{} ab \\plus{} b^2} \\plus{} \\frac {1}{b^2 \\plus{} bc \\plus{} c^2} \\plus{} \\frac {1}{c^2 \\plus{} ca \\plus{} a^2} \\ge \\frac {9}{(a \\plus{} b \\plus{} c)^2}$", "Solution_2": "[quote=\"mathstarofvn\"]Let a,b,c>0 and $ a \\plus{} b \\plus{} c \\equal{} 1$. Prove that:\n$ \\frac {ab}{\\sqrt {ab \\plus{} c}} \\plus{} \\frac {bc}{\\sqrt {bc \\plus{} a}} \\plus{} \\frac {ca}{\\sqrt {ca \\plus{} b}} \\le \\frac {1}{2}$[/quote]\r\n $ ab\\plus{}c\\equal{}ab\\plus{}c(a\\plus{}b\\plus{}c)\\equal{}(c\\plus{}a)(c\\plus{}b)$\r\nwe have \r\n$ LHS \\le \\frac{1}{2}(\\sum \\frac{ab}{c\\plus{}b}\\plus{}\\sum \\frac{ab}{c\\plus{}a})\\equal{}\\frac{1}{2}$" } { "Tag": [ "calculus", "integration", "geometry", "function", "factorial", "calculus computations" ], "Problem": "calculate the area betewen the graphe of the function f(x)=x^6/(1+x^4)^1/2\r\nand x=0 and x=1", "Solution_1": "since the function $f(x)$ is always positive between $0$ and $1$ .so the problem is equivalent to this \r\n\r\n $\\int_{0}^{1}\\frac{x^{6}}{\\sqrt{1+x^{4}}}$ $dx$\r\n\r\nsubstitute $x^{2}$ $=$ $tan$ $\\theta$\r\n\r\nthe inegral becomes \r\n $\\int_{0}^{\\frac{\\pi}{4}}sin^{\\frac{5}{2}}\\theta cos^{-\\frac{7}{2}}\\theta$ $d\\theta$\r\n\r\nwhich can be solved using $gamma$ $functions$", "Solution_2": "waht is gamma function??", "Solution_3": "[quote=\"drapt@\"]waht is gamma function??[/quote]\r\n\r\n\r\n\r\n$gamma$ $function$ is defined $\\Gamma(x)$\r\n\r\n\r\n$\\Gamma(x)$ $=$ $\\int_{0}^{\\infty}e^{-t}t^{x-1}dt$\r\n\r\nalso it is the extension of factorial into the field of real no.s :)", "Solution_4": "is there an other way to fin the result???" } { "Tag": [ "number theory", "number theory theorems" ], "Problem": "from number theory excercises:\r\nafter solving Pellian, I get solutions given by\r\n\\[ w_n\\plus{}y_n\\sqrt{15} \\equal{} \\left(5\\plus{}\\sqrt{15}\\right)\\left(4\\plus{}\\sqrt{15}\\right)^n\\]\r\nthen $ w_0\\equal{}5$.\r\nhow can I prove that $ w_0$ is the only power of 5 in it ?", "Solution_1": "no idea ?\r\n\r\n :(" } { "Tag": [ "function", "floor function", "ceiling function" ], "Problem": "$\\lfloor{x}\\rfloor+\\lfloor{2x}\\rfloor+\\lfloor{4x}\\rfloor+\\lfloor{8x}\\rfloor=2005$", "Solution_1": "lets say that $\\lfloor{x}\\rfloor=k$ if $y$ is the decimal part of x then. $\\lfloor{2x}\\rfloor=2k+\\lfloor{2y}\\rfloor$ also for the others then $17k\\leq2005=17k+\\lfloor{2y}\\rfloor+\\lfloor{4y}\\rfloor+\\lfloor{8y}\\rfloor\\leq17k+1+3+7=17k+11$ but 2005 is 17x+16 then that isnt true for any x.", "Solution_2": "You need to replace 17 with a 15. And a solution does exist. Try x=133.8", "Solution_3": "Stemming from joshua_mex we know\r\n\\[15k\\leq 2005\\leq 15k+\\lfloor{2y}\\rfloor+\\lfloor{4y}\\rfloor+\\lfloor{8y}\\rfloor\\]\r\nThis implies k=133.\r\nOur problem is then reduced to solving\r\n\\[\\lfloor{2y}\\rfloor+\\lfloor{4y}\\rfloor+\\lfloor{8y}\\rfloor=10\\]\r\nSubstitute $y=1-p, p>0$ and using the identity $\\lceil{u}\\rceil=-\\lfloor{-u}\\rfloor$, we now must solve\r\n\\[\\lceil{2p}\\rceil +\\lceil{4p}\\rceil+\\lceil{8p}\\rceil=4\\]\r\nbecause $p>0$ all terms on the LHS will be at least 1 implying that the LHS is at least 3. So we need for $\\lceil{4p}\\rceil=1$ and $\\lceil{8p}\\rceil=2$. For all real $u$ it is known $u\\leq\\lceil{u}\\rceil3$\r\n$L=-\\frac{1}{6}$ for $n=3$\r\n$L=0$ for $n<3$ (your case)", "Solution_3": "Perhaps a geometric argument would qualify? \r\n\r\nIt's enough to consider $03$ and $n=3$" } { "Tag": [ "function", "integration", "limit", "induction", "real analysis", "real analysis solved" ], "Problem": "$\\{f_n(x)\\}$ are defined on $[0,1]$. $f_0(x)$ is Riemann-integrable, and $f_0(x) > 0$. The sequence of functions satisfies the following recurrence:\r\n\\[f_n(x) = \\sqrt{\\int_0^x f_{n-1}(x) dx}\\]\r\nFind $\\lim\\limits_{n\\to\\infty} f_n(x)$.", "Solution_1": "$x/2$", "Solution_2": "That's what I found in all the cases I've tried. It seems amazing, doesn't it?", "Solution_3": "I can prove it. We just need to show that for arbitrary $a>0$ and sufficiently large $n$ we have \r\n\\[(1+a)\\frac{x}{2}\\geq f_n(x)\\geq \\frac{x-a}{2}.\\]\r\nIndeed, consider $f_1(x)$. It is increasing positive function. Let $M=\\max f_1(x)=f_1(1)$. One can easy prove by induction that for all $x\\in[0,1]$: \r\n\\[f_{n+1}(x)\\leq M^{1/2^n}\\alpha_n\\cdot x^{\\frac{2^n-1}{2^n}},\\]\r\nwhere $\\alpha_n\\to\\frac{1}{2}$. So upper estimate is true.\r\nFor second estimate take $m=\\min_{x\\geq\\delta} f_1(x)=f_1(\\delta)>0$, then for all $x\\in[\\delta,1]$:\r\n\\[f_{n+1}(x)\\geq m^{1/2^n}\\alpha_n\\cdot (x-\\delta)^{\\frac{2^n-1}{2^n}}.\\]\r\nAnd from here we obtain lower estimate.", "Solution_4": "[quote=\"Myth\"]Indeed, consider $f_1(x)$. It is increasing positive function. Let $M=\\max f_1(x)=f_1(1)$. One can easy prove by induction that for all $x\\in[0,1]$: \n\\[f_{n+1}(x)\\leq M^{1/2^n}\\alpha_n\\cdot x^{\\frac{2^n-1}{2^n}},\\]\nwhere $\\alpha_n\\to\\frac{1}{2}$. So upper estimate is true.\n[/quote]\r\nThe difficulty for me is to prove that $\\alpha_n\\to\\frac{1}{2}$ :(", "Solution_5": "Ok!\r\nLet's reconstruct $\\alpha_n$. What we have?\r\nDenote $u_k=\\frac{2^{k-1}}{2^k-1}$, then\r\n\\[\\alpha_n=u_n^{1/2}u_{n-1}^{1/2^2}...u_1^{1/2^n}\\]\r\n(am I right?)\r\nMy statement follows from the fact that $u_k\\to \\frac{1}{2}$ if $k\\to\\infty$. Indeed, it easy to prove that $\\alpha_n$ is decreasing bounded sequence, so it has limit, and \r\n\\[\\alpha_n=\\sqrt{\\alpha_{n-1}}\\cdot\\sqrt{u_n},\\]\r\ntherefore $a=\\sqrt{a}\\cdot\\sqrt{\\frac{1}{2}}$, where $a=\\lim_{n\\to\\infty}a_n$. Thus $a=\\frac{1}{2}$.", "Solution_6": "Assuming that the sequence converges and the limit exists, why not do it this way:\r\n\r\nlet [tex]f(x) = \\lim\\limits_{n\\to\\infty} f_n(x)[/tex]\r\n\r\nso [tex]f(x)^2 = \\int_0^x f(x) dx[/tex]\r\n\r\nand after differentiating both sides [tex]2*f(x)*\\frac{d f(x)}{dx} = f(x)[/tex]\r\n\r\nthis leads to [tex]\\frac{d f(x)}{dx} = 1/2[/tex] or [tex]0[/tex], therefore [tex]f(x) = \\frac{x}{2}[/tex], since f(x) cannot converge to 0.", "Solution_7": "And what? It gives nothing, but only possible answer. I am sure that Moubinool found it in the same way. Main part of the problem is convergence itself.", "Solution_8": "Ok now we know that $f_n$ converge simply to$x/2$.\r\n\r\nDo we have uniform convergence on $[0,1]$ ?", "Solution_9": "Hmm... Moubinool is kidding :lol: :lol: :lol: \r\nUniform convergence is a direct consequence of my proof, isn't it?", "Solution_10": ":huh: looks like the problem i had to do in my talk at the 3rd of BWM last year...", "Solution_11": "And what? :? \r\nDo you have another proof?", "Solution_12": "For uniform convergence we need to prove that \r\n\r\n$||f_n - f||_{\\infty}= \\sup_{x\\in [0,1]}|f_n(x)- x/2|$ tends to $0$", "Solution_13": "Moubinool!\r\n[quote=\"Myth\"]Let $M=\\max f_1(x)=f_1(1)$. One can easy prove by induction that for all $x\\in[0,1]$: \n\\[f_{n+1}(x)\\leq M^{1/2^n}\\alpha_n\\cdot x^{\\frac{2^n-1}{2^n}},\\]\nwhere $\\alpha_n\\to\\frac{1}{2}$.[/quote]\n[quote]$m=\\min_{x\\geq\\delta} f_1(x)=f_1(\\delta)>0$, then for all $x\\in[\\delta,1]$:\n\\[f_{n+1}(x)\\geq m^{1/2^n}\\alpha_n\\cdot (x-\\delta)^{\\frac{2^n-1}{2^n}}.\\][/quote]\r\nFix $\\epsilon>0$. Take $\\delta=\\epsilon/2$ then\r\n\\[\\sup |f_{n+1}(x)-x/2|\\leq \\sup \\left|M^{1/2^n}\\alpha_n\\cdot x^{\\frac{2^n-1}{2^n}}-m^{1/2^n}\\alpha_n\\cdot (x-\\delta)\\right|< \\epsilon\\]\r\nfor all sufficiently large $n$." } { "Tag": [ "LaTeX", "trigonometry", "induction", "algebra unsolved", "algebra" ], "Problem": "yay i learned latex! its so much easier now..\r\n\r\nLet $x_1, x_2, ..., x_n$ be $n$ positive real numbers such that\r\n\r\n$x_1=\\frac{1}{x_1}+x_2=\\frac{1}{x_2}+x_3=...=\\frac{1}{x_{n-1}}+x_n=\\frac{1}{x_n}$.\r\n\r\nProve that $x_1=2 \\cos \\left( \\frac{\\pi}{n+2} \\right)$.\r\n\r\ni finding this very hard and pleas if you could help me i would be very glad. note that this problem work in radians not in degrees", "Solution_1": "[hide=\"hint\"]notice that since the equation is read towards the left hand we have that changing $x_i$ with $\\frac{1}{x_{n+1-i}}$ the equation is still true\n$x_1=\\frac{1}{x_n}$ or\n$\\frac{1}{x_1}+x_2=x_2+x_n=\\frac{1}{x_{n-1}}+x_n\\to x_2=\\frac{1}{x_{n-1}}$\nBy induction $x_v=x_{n+1-v}$\nLet's take the median term(s) $x_m$ and find their value using that fact...maybe use complex numbers... :D [/hide]", "Solution_2": "i be sorry but i still can not do it so if you would show me the solusion i would be very glad. thank you very much, in advance" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "$a,b,c$ are positive numbers such as ${a b c = 1}$\r\n\r\nProove that\r\n${\\frac{1}{c+\\frac{1}{a+\\frac{1}{b}}}+\\frac{1}{a+\\frac{1}{b+\\frac{1}{c}}}+\\frac{1}{b+\\frac{1}{c+\\frac{1}{a}}}\\geq 2}$\r\n\r\nwithout \"brute force\"", "Solution_1": "[quote=\"yagaron\"]$a,b,c$ are positive numbers such as ${a b c = 1}$\n\nProove that\n${\\frac{1}{c+\\frac{1}{a+\\frac{1}{b}}}+\\frac{1}{a+\\frac{1}{b+\\frac{1}{c}}}+\\frac{1}{b+\\frac{1}{c+\\frac{1}{a}}}\\geq 2}$\n[/quote]\r\nMy ugly proof: ${\\frac{1}{c+\\frac{1}{a+\\frac{1}{b}}}+\\frac{1}{a+\\frac{1}{b+\\frac{1}{c}}}+\\frac{1}{b+\\frac{1}{c+\\frac{1}{a}}}\\geq 2}\\Leftrightarrow\\sum_{cyc}\\frac{ab+1}{b+c+1}\\geq2.$\r\nLet $a=\\frac{x}{y},$ $b=\\frac{y}{z}$ and $c=\\frac{z}{x},$ where $x,$ $y$ and $z$ are positive numbers. \r\nHence, $\\sum_{cyc}\\frac{ab+1}{b+c+1}=\\sum_{cyc}\\frac{x^{2}+xz}{z^{2}+xy+xz}=\\sum_{cyc}\\frac{x}{z+\\frac{xy}{x+z}}=\\sum_{cyc}\\frac{x^{2}}{xz+\\frac{x^{2}y}{x+z}}\\geq$\r\n$\\geq\\frac{(x+y+z)^{2}}{xy+xz+yz+\\frac{x^{2}y}{x+z}+\\frac{y^{2}z}{x+y}+\\frac{z^{2}x}{y+z}}.$\r\nId est, it remains to prove that ( my old inequality :D )\r\n $x^{2}+y^{2}+z^{2}\\geq2\\left(\\frac{x^{2}y}{x+z}+\\frac{y^{2}z}{x+y}+\\frac{z^{2}x}{y+z}\\right),$\r\nwhich equivalent to the following inequality: $\\sum_{sym}(x^{4}y-x^{3}y^{2})\\geq0.$ :)" } { "Tag": [], "Problem": "$64x^{3}+8\\sqrt{3}x^{2}+52x-49=0$", "Solution_1": "[hide]TI-89 gives an approximation of $\\boxed{x \\approx .511312}$ but I don't know how to solve it long hand really.[/hide]", "Solution_2": "Well, obviously RRT won't work, seeing as there won't be any rational roots.\r\nHmm... :|" } { "Tag": [ "trigonometry", "real analysis", "real analysis unsolved" ], "Problem": "Determine whether $ \\Sigma _{n\\equal{}1} ^\\infty \\frac{\\tan n}{2^n}$ is convergent or not.", "Solution_1": "You can use the command \\sum instead of \\Sigma to get the nice looking $ \\sum$." } { "Tag": [ "linear algebra", "matrix", "induction", "vector", "Gauss", "algorithm", "superior algebra" ], "Problem": "This problem came up in another forum, and I can't make any progress on it.\r\n\r\nShow that the natural quotient map from $ \\textsf{SL}_n(\\mathbb{Z})$ to $ \\textsf{SL}_n(\\mathbb{Z}_m)$ is surjective.\r\n\r\nJust to be clear about notation, $ \\textsf{SL}_n(R)$ is the group of all n\u00d7n matrices over the (commutative, unital) ring R that have determinant 1, and $ \\mathbb{Z}_m$ is the quotient ring $ \\mathbb{Z}/m\\mathbb{Z}$.\r\n\r\nEven in the very simplest possible case, where n=2, m is prime, and the matrix to be lifted is diagonal, this result is less than obvious. For example, the natural lifting of the matrix $ \\begin{bmatrix}4&0\\\\0&2\\end{bmatrix}\\in\\textsf{SL}_2(\\mathbb{Z}_7)$ to a 2\u00d72 matrix over $ \\mathbb{Z}$ has determinant 8. But there is a lifting of this matrix to $ \\textsf{SL}_2(\\mathbb{Z}),$ namely $ \\begin{bmatrix}25&7\\\\7&2\\end{bmatrix}.$\r\n\r\nAs far as I can tell by looking at examples, the claimed result is true for all positive integers m and n. But I have no idea how to prove it.", "Solution_1": "[quote=\"Opalg\"]Show that the natural quotient map from $ \\textsf{SL}_n(\\mathbb{Z})$ to $ \\textsf{SL}_n(\\mathbb{Z}_m)$ is surjective.[/quote][hide=\"Hint\"]Use induction on n. Given some matrix $ M\\in\\textsf{SL}_n(\\mathbb{Z}_m)$ lift the first column of $ M$ to a vector $ x_0\\in\\mathbb Z^n$ whose components generate the unit ideal in $ \\mathbb Z$. Then $ \\mathbb Z^n/\\mathbb Zx_0$ is torsion free, hence free of rank $ n\\minus{}1$. This implies that there is a matrix $ M_0\\in \\textsf{SL}_n(\\mathbb{Z})$ such that $ M_0e_1\\equal{}x_0\\equiv Me_1 \\mod m$. The first column of the matrix $ N\\equal{}M_0^{\\minus{}1}M \\in \\textsf{SL}_n(\\mathbb{Z}_m)$ is the first unit vector. Now build appropriate blocks in $ N$ and lift them. ...[/hide]", "Solution_2": "Thanks! That works very nicely. :10:", "Solution_3": "This can (even in a more general setting) be reduced to $ A$ diagonal and then $ n \\equal{} 2$. Then, we have $ \\begin{pmatrix}a & 0 \\\\\r\n0 & a^{ \\minus{} 1}\\end{pmatrix} \\equal{} \\begin{pmatrix}1 & a \\\\\r\n0 & 1\\end{pmatrix}\\begin{pmatrix}1 & 0 \\\\\r\n\\minus{} a^{ \\minus{} 1} & 1\\end{pmatrix}\\begin{pmatrix}1 & a \\\\\r\n0 & 1\\end{pmatrix}\\begin{pmatrix}1 & \\minus{} 1 \\\\\r\n0 & 1\\end{pmatrix}\\begin{pmatrix}1 & 0 \\\\\r\n1 & 1\\end{pmatrix}\\begin{pmatrix}1 & \\minus{} 1 \\\\\r\n0 & 1\\end{pmatrix}$.", "Solution_4": "[quote=\"myan\"]This can (even in a more general setting) be reduced to $ A$ diagonal and then $ n \\equal{} 2$.[/quote]\r\nhow does the reduction work?", "Solution_5": "We want to show that $ SL_n(R) \\equal{} E_n(R)$ ($ E_n$: generated by elementary matrices) for a ring $ R$ with finitely many maximal ideals.\r\n\r\nReduce to\r\n1. upper triangular\r\n2. diagonal\r\n3. $ n \\equal{} 2$\r\n\r\nFor the first steps use something like the Gauss algorithm together with prime avoidance (here we use that $ R$ has finitely many maximal ideals) in the refinement given e.g. in Exercise 3.19b of Eisenbud, Commutative Algebra.", "Solution_6": "There is also a generalisation of the form \"some condition on $ R,n$ $ \\implies$ $ SL_m(R) \\equal{} E_m(R)$ for $ m \\geq n$\" (which is satisfied for $ n \\equal{} 2$, $ R$ finitely many maximal ideals), but the proof is quite lengthy." } { "Tag": [ "geometry", "3D geometry", "pyramid", "Support", "calculus", "trigonometry", "quadratics" ], "Problem": "You're trying to build a bigger pyramid than your father, Aphluhent III, but you're on a tight budget. The pyramid will cost you $20 per cubic meter of interior. However, each of the sides will cost you $10 per square meter to finish. The greatest incline your architects can support is 45 degrees. If your budget is $10000, what is the tallest pyramid you can build?\r\n\r\n(The pyramid has a square base and the surface area of the base is not included in the cost.)", "Solution_1": "Is this required to be a square-based pyramid?", "Solution_2": "Yes, the base is square.", "Solution_3": "Can anyone solve this?", "Solution_4": "Is it ok to use calculus?", "Solution_5": "Yes, though it should be solvable without calculus.", "Solution_6": "I believe the greatest height possible is approximately 8.774 feet. All you really need is right triangle trig and algebra.", "Solution_7": "The volume of the interior of a 45-degree angled square pyramid is\r\n[hide] $\\frac{4x^{3}}{3}$ - that was easy (x is the height, also half the length of the base) [/hide]\nand the surface area (not counting the base) is\n[hide] $8x^{2}$ - 8 45-45-90 triangles with side length $x\\sqrt2$[/hide]\nso our price formula is\n[hide] $\\frac{80x^{3}}{3}+80x^{2}=(80x^{2})(\\frac{x}{3}+1)$[/hide]\nour max price is 10000, so\n[hide] $(80x^{2})(\\frac{x}{3}+1)=10000$\n$(x^{2})(\\frac{x}{3}+1)=125$\n$(x^{2})(x+3)=375$\n$x^{3}+3x^{2}-375=0$\n$x=\\frac{\\sqrt{1509}-3}{2}=17.923$ $feet$[/hide]\r\n\r\nEDIT: I screwed that up - quadratic in the wrong place\r\nI dont feel like solving the cubic" } { "Tag": [], "Problem": "Hi guys...\r\nI \"stumbled\" upon during one of my school's math team meetings.\r\n\r\nThe problems states something along these lines:\r\n\r\n9*9=81\r\n9+9=18\r\n\r\n3*24=72\r\n3+24=27\r\nFind another pair where their product is the reverse of their sum.\r\n\r\nMy mom said that the digits in the product/sum should add up to 9, but even then, I wasn't able to find another pair.\r\n\r\nHelp...please?\r\n\r\n :huh:", "Solution_1": "Guess and check? The algebra seems far too tedious...\r\n\r\nx + y = 10a + b\r\nxy = 10b + a\r\n\r\n[hide=\"One answer\"]\nZero and zero =D .\n[/hide]", "Solution_2": "I don't know if 0 and 0 would count. I think the problem really wants another solution like 9 and 9. Besides, does reversing a one digit number work. That would have to be dancingblob's say because s/he posted the problem.", "Solution_3": "No, I don't think two one-digit numbers would work.\r\nAnd I'm a she. :D \r\n\r\nThis is proving to be quite a problem.", "Solution_4": "What about 2 and 2? 4 is the reverse of 4.", "Solution_5": "[quote=\"JEngel\"]What about 2 and 2? 4 is the reverse of 4.[/quote]\r\nI think that dancingblob said that two one-digit numbers don't work. ;)", "Solution_6": "other pair of number are \r\n\r\n[hide]2 and 47[/hide]", "Solution_7": "[quote=\"joseluis\"]other pair of number are \n\n[hide]2 and 47[/hide][/quote]\r\nNice! :D How did you find it? I can't seem to find any... :(", "Solution_8": "$xy + x + y = 11(a+b)$\r\n$(x+1)(y+1) = 11(a+b) + 1$\r\n\r\n$xy - x - y = 9(a-b)$\r\n$(x-1)(y-1) = 9(a-b) + 1$\r\n\r\nThis is about as simple as I can get it. The two cases you propose have $a + b = 9$, which (and I have checked) has no other solutions.\r\n\r\nI've been picking different values of $a+b$ and $a-b$ and have still not found another two-digit solution. :huh:\r\n\r\nEdit: Joselius has found one! Plugging in $a-b = 5$ gives $(x-1)(y-1) = 46$. Choosing $x = 3, y = 24$ gives a solution already given, but choosing $x = 2, y = 47$ produces $x+y = 49, xy = 94$ :o (I admit I had not bothered to check this one :| )\r\n\r\nAlternately, from the top equation, $a+b = 13$ gives $(x+1)(y+1) = 144$ (which, again, I had checked and not found solutions for :| ), and here again, choosing $x = 2, y = 47$ produces a valid solution :o", "Solution_9": "really the number came to me keep traing but i think that there is not other number with that characteristic in thet the product of the number is less tha 100.", "Solution_10": "you guys never cease to amaze me.\r\nthanks for your help.\r\n :lol:" } { "Tag": [ "trigonometry", "inequalities", "inequalities proposed" ], "Problem": "Prove if a,b,c>0 and ab+bc+ca=3 then\r\n$ \\sum\\sqrt{a^2\\plus{}3} >\\equal{}a\\plus{}b\\plus{}c\\plus{}3$", "Solution_1": "[quote=\"tranquoc\"]Prove if $ a,b,c > 0$ and $ ab \\plus{} bc \\plus{} ca \\equal{} 3$ then\n$ \\sum\\sqrt {a^2 \\plus{} 3} \\geq a \\plus{} b \\plus{} c \\plus{} 3$[/quote]\r\nThe idea $ a \\equal{} \\sqrt3\\tan\\frac {\\alpha}{2},$ $ b \\equal{} \\sqrt3\\tan\\frac {\\beta}{2}$ and $ c \\equal{} \\sqrt3\\tan\\frac {\\gamma}{2}$ works here.", "Solution_2": "Here is my proof:\r\nLet $ a \\equal{} \\sqrt3\\tan\\frac {\\alpha}{2},$ $ b \\equal{} \\sqrt3\\tan\\frac {\\beta}{2}$ and $ c \\equal{} \\sqrt3\\tan\\frac {\\gamma}{2},$ where $ \\{\\alpha,\\beta,\\gamma\\}\\subset[1,\\pi).$ \r\nThen $ \\alpha\\plus{}\\beta\\plus{}\\gamma\\equal{}\\pi$ and our inequality equivalent to $ \\sum_{cyc}\\left(\\frac{1}{\\cos\\frac{\\alpha}{2}}\\minus{}\\tan\\frac{\\alpha}{2}\\right)\\geq\\sqrt3.$\r\nLet $ f(x)\\equal{}\\frac{1}{\\cos\\frac{x}{2}}\\minus{}\\tan\\frac{x}{2}.$ Then $ f'(x)\\equal{}\\frac{\\sin\\frac{x}{2}}{2\\cos^2\\frac{x}{2}}\\minus{}\\frac{1}{2\\cos^2\\frac{x}{2}}\\equal{}\\minus{}\\frac{1}{2(1\\plus{}\\sin\\frac{x}{2})}.$\r\n$ f''(x)\\equal{}\\frac{\\cos\\frac{x}{2}}{4(1\\plus{}\\sin\\frac{x}{2})^2}\\geq0.$\r\nThus, $ \\sum_{cyc}f(\\alpha)\\geq3f\\left(\\frac{\\alpha\\plus{}\\beta\\plus{}\\gamma}{3}\\right)\\equal{}\\sqrt3.$\r\nI don't see another way. :(", "Solution_3": "but I don't let a=..;....... AND you?????????????????/", "Solution_4": "$ a^2\\plus{}3\\equal{}(a\\plus{}b)(a\\plus{}c)$......-> done", "Solution_5": "[quote=\"tranquoc\"]$ a^2 \\plus{} 3 \\equal{} (a \\plus{} b)(a \\plus{} c)$......-> done[/quote]\r\nWhat do you mean? Why it helps? Explain pleas. :wink: Thank you!", "Solution_6": "[quote=\"arqady\"][quote=\"tranquoc\"]$ a^2 \\plus{} 3 \\equal{} (a \\plus{} b)(a \\plus{} c)$......-> done[/quote]\nWhat do you mean? Why it helps? Explain pleas. :wink: Thank you![/quote]\r\n\r\nIt helps because it reduces the problem to http://www.mathlinks.ro/viewtopic.php?t=133953 .\r\n\r\n darij", "Solution_7": "Thank you, darij! :D", "Solution_8": "Prove if $a,b,c>0$ and $a+b+c\\leq abc$ . then\n$$\\sqrt{a^2\\plus{}1} +\\sqrt{b^2\\plus{}1} +\\sqrt{c^2\\plus{}1} +3\\leq ab+bc+ca$$ \n" } { "Tag": [ "geometry", "calculus", "AMC", "AIME" ], "Problem": "Does anyone here use books from the REA series for math? I have REA geometry and pre-calculus and i wanted to know if they were a good preparation for any AMC's(AMC 12, AIME, etc.). the problems seem easier than the ones in AOPS volume 1, at least in geometry, yet volume 1 geometry is challenging for me. are the REA books a good prep for volume 1 geometry? as for algebra, volume 1 is okay for me, not much of a challenge, but its not fantastically simple either. however, volume 2 algebra is way beyond my level. what books should i use for algebra in that case?", "Solution_1": "i have the geometry and algebra/trigonometry books. i guess it depends what your goals are, I used the algebra/trigonometry book to prepare for class, and i think it might have helped for contests too. overall i thought it was really good. i never used the geometry book though so i cant say much about it. I think i flipped through it and it looked pretty easy.\r\n\r\nthe REA algebra/trig book might help for AMC, but the best thing is just to stick with AOPS. if volume 1 is too hard, maybe you can use the REA books to reinforce basic ideas and then go back to aops.", "Solution_2": "thanks for your feedback! :)" } { "Tag": [], "Problem": "the number of permutations of letters of the word PARALLEL taken four at a time are ...................... \r\n\r\npl suggest a shorter way \r\n\r\n\r\n1 .A music academy has to send 10 members to 5 shows ,two in each show. two of the shows are in city and the 3 are outside .two of the five members prefer to work in city while 3 prefer to work outside city..the number of ways in which five shows can be held if preference are to be satisfied . \r\noptions are 5200, 5400,5472,5250 \r\n\r\nQ2 There are two students A and B having a and b numbers of books respectively.the number of ways in which they can exchange their their books so that after exchange they have same number of books with them but not he same books is ............. \r\n\r\noptions are a+b^(C)a ,(a+b^C a)-1,a+b^(P)a,( a+b^P a)-1", "Solution_1": "[quote=\"ayush_2008\"]\n1 .A music academy has to send 10 members to 5 shows ,two in each show. two of the shows are in city and the 3 are outside .two of the five members prefer to work in city while 3 prefer to work outside city..the number of ways in which five shows can be held if preference are to be satisfied . \noptions are 5200, 5400,5472,5250 \n[/quote]\r\n\r\n\r\nSince 2 members of internal show are fixed, the other 2 can be selected from 5 in $ {5\\choose 2}$ ways. This automaticaly fixes the selection of the external show members. Now, the internal show guys have to be distributed into 2 distinct groups of 2 each. This is done in $ {4\\choose 2} \\cdot {2\\choose 2}$ ways. The external show guys have to be distributed into 3 distinct groups of 2 each. This is done in $ {6\\choose 2} \\cdot {4\\choose 2} \\cdot {2\\choose 2}\\equal{}\\frac {6!}{2!2!2!}$ ways.\r\n\r\nSo, total=$ {5\\choose 2} \\cdot {4\\choose 2}\\cdot {2\\choose 2} \\cdot \\frac {6!}{2!2!2!}\\equal{}5400$ ways." } { "Tag": [ "inequalities", "geometry", "circumcircle", "incenter", "geometry proposed" ], "Problem": "[color=purple] Let $ O$and $ I$ are circumcenter and incenter of triangle $ ABC$ which has area $ S$ . Prove that\n $ 2\\sqrt {2}OI^2\\ge S[|\\frac {b \\minus{} c}{a}| \\plus{} |\\frac {c \\minus{} a}{b} \\plus{} |\\frac {a \\minus{} b}{c}|]$\n :) [/color]", "Solution_1": "[color=purple] I have a mistake when I created the problem in previous post :| but any one can show me the truly of it ? \n The following is true : $ 2OI^2\\ge S[|\\frac {b \\minus{} c}{a}\\plus{}\\frac {c \\minus{} a}{b}\\plus{}\\frac {a \\minus{} b}{c}|]$\n :) :) [/color]" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ x_1,x_2,x_3,x_4\\geq0$,prove that:\r\n\r\n$ (x_1^2x_2\\plus{}x_2^2x_3\\plus{}x_3^2x_4\\plus{}x_4^2x_1)(x_1x_2^2\\plus{}x_2x_3^2\\plus{}x_3x_4^2\\plus{}x_4x1^2) \\geq x_1x_2x_3x_4(x_1\\plus{}x_2\\plus{}x_3\\plus{}x_4)^2.$\r\nBQ", "Solution_1": "[quote=\"xzlbq\"]Let $ x_1,x_2,x_3,x_4\\geq0$,prove that:\n\n$ (x_1^2x_2\\plus{}x_2^2x_3\\plus{}x_3^2x_4\\plus{}x_4^2x_1)(x_1x_2^2\\plus{}x_2x_3^2\\plus{}x_3x_4^2\\plus{}x_4x1^2) \\geq x_1x_2x_3x_4(x_1\\plus{}x_2\\plus{}x_3\\plus{}x_4)^2.$\nBQ[/quote]\nI'll prove that\n \\[(a^2b+b^2c+c^2d+d^2a)(ab^2+bc^2+cd^2+da^2)\\geq abcd(a+b+c+d)^2\\]\nfor non-negatives $a$, $b$, $c$ and $d$.\n$(a^2b+b^2c+c^2d+d^2a)(ab^2+bc^2+cd^2+da^2)-abcd(a+b+c+d)^2=$\n$=bd(a^2+ac+c^2)(a-c)^2+ac(b^2+bd+d^2)(b-d)^2+a^2c^2(b-d)^2+$\n$+b^2d^2(a-c)^2+(a+c)(b+d)((b^2-bd+d^2)(a^2-ac+c^2)-abcd)\\geq0$.", "Solution_2": "By Cauchy Schwarz Inequality:\n$\\left(\\sum a^2b\\right)\\left(\\sum ab^2\\right)\\ge\\left(\\sum abc\\right)^2$,\n$\\left(\\sum a^2b\\right)\\left(\\sum acd\\right)\\ge abcd\\left(\\sum a\\right)^2$,\n$\\left(\\sum ab^2\\right)\\left(\\sum bcd\\right)\\ge abcd\\left(\\sum b\\right)^2$." } { "Tag": [], "Problem": "Determine the sum of all the positive integers from $1$ to $2006$, inclusive, that do not contain the digit $7$.", "Solution_1": "[hide=\"flawed first attempt\"]\nFirst consider the sum of integers from $1$ to $999$ without any $7$s.\n\nThe sum of units digits $0$ to $9$ without $7$ is $\\frac{9(9+1)}2-7=38$\nAnd this sum of units digits appears 100 times.\n\nThe sum of tens digits $00$ to $90$ without $70$ is $38\\cdot10=380$\nAnd this sum of tens digits appears 10 times.\n\nThe sum of hundreds digits $000$ to $900$ without $700$ is $38\\cdot100=3800$\nAnd this sum of hundreds digits appears once.\n\nThe sum from $1000$ to $1999$ is the above sum plus $1000\\cdot1000$\n\nTherefore the sum is\n$2\\cdot(3\\cdot3800)+1000\\cdot1000+6\\cdot(\\frac{2001+2006}2)=\\boxed{1,034,821}$[/hide]\n\n[hide=\"Correct Solution\"]\nFirst consider the sum of integers from $1$ to $999$ without any $7$s.\n\nThe sum of units digits $0$ to $9$ without $7$ is $\\frac{9(9+1)}2-7=38$\nThe sum of tens digits $00$ to $90$ without $70$ is $38\\cdot10=380$\nThe sum of hundreds digits $000$ to $900$ without $700$ is $38\\cdot100=3800$\nSo the total sum is $3800+380+38=4218$ and this set of sums appears $\\frac{1000}{10}=100$ times from $1$ to $999$.\n\n\nNow let's consider the sum from $1000$ to $1999$.\n\nBy PIE, the # of numbers with a $7$ in them from $1000$ to $1999$ is equal to $100+100+100-10-10-10+1=271$. So the # of numbers that don't have $7$ is $1000-271=729$. So the sum is equal to the sum from $1$ to $999$ plus $729\\cdot1000$\n\nTherefore, the total sum is:\n\n$2(100\\cdot4218)+729\\cdot1000+6\\left(\\frac{2001+2006}2\\right)=\\boxed{1,584,621}$[/hide]", "Solution_2": "Whoops, I just realized my solution was totally flawed! :blush:", "Solution_3": "What is PIE?", "Solution_4": "[quote=\"surge\"]What is PIE?[/quote]\r\n\r\nPIE stands for Principle of Inclusion-Exclusion.", "Solution_5": "My reasoning for the problem is follows, if you can understand it...I'm really bad at explaining things (sorry :blush:)\r\n\r\n[hide]We'll take care of the numbers 0-9 first. They sum up to 38 without 7. Now, we take care of the two digit numbers. If we break them up into tens and units (such as 17 = 10 + 7), for each tens digit x, we have the sum of numbers 10x + 0 through 10x + 9 as 9(10x) + 38. If we add them all up for x's 0 - 9 excluding 7, we have (9)(10 + 20 + 30 + ... + 90 - 70) + (9)(38) = (9)(418) = 3762, so the sum of the numbers between 0-99 without a 7 is 3762.\n\nNow, we take care of the three digit numbers. If we split them up as a hundreds digit and a two digit part, we can call the hundreds digit x. Then the sum of numbers 100x + 0 through 100x + 99 is (81)(100x) + 3762 where we use 81 because there are (9 numbers w/o 7 in units digit)(9 numbers w/o 7 in tens digit) = 81 numbers between (0-99) w/o 7. For all numbers 0-999 ignoring the 700s, we have (81)(3800) + (9)(3762) = 341658.\n\nFinally, we work on the thousands digit. We split the four digit numbers up into thousands digit, x, and a three digit number. Once again, the sum of all the thousands digits together for the numbers 1000x to 1000x + 999 is (9)(9)(9)(1000x) = (729)(1000x), with 729 numbers total since there are (9 # w/o 7 in units digit)(9 # w/o 7 in tens digit)(9 # w/o 7 in hundreds digit) = 729 numbers for every thousand without a 7. To add all the numbers for 0-999 and 1000-1999 we then have to do (729)(1000) + (2)(341658) = 1412316. Then, just simply add 2000 through 2006 = 14021 and the final answer is:\n\n$\\boxed{1426337}$[/hide]" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,b,c,d$ are non-negative real numbers such that $ a+b+c+d=1$. Prove that\r\n\t\\[ \\sqrt{a+\\frac{33(b+c-2d)^{2}}{64}}+\\sqrt{b+\\frac{33(c+d-2a)^{2}}{64}}+\\sqrt{c+\\frac{33(d+a-2b)^{2}}{64}}+\\sqrt{d+\\frac{33(a+b-2c)^{2}}{64}}\\ge 2 \\]", "Solution_1": "Another one but easier\r\nLet $ a,b,c,d$ are real numbers such that $ a+b+c+d=4$. Find the best constant $ k$ such that\r\n$ \\sqrt{a+b+k(a+b-c-d)^{2}}+\\sqrt{b+c+k(b+c-d-a)^{2}}+\\sqrt{c+d+k(c+d-a-b)^{2}}+\\sqrt{d+a+k(d+a-b-c)^{2}}\\ge 4\\sqrt{2}$", "Solution_2": "Another one and may be the hint for the second problem\r\nLet $ a,b,c,d$ are real numbers such that $ a\\plus{}b\\plus{}c\\plus{}d \\equal{} 4$. Find the best constant $ k$ such that\r\n$ \\sqrt{a\\plus{}k(a\\minus{}b\\minus{}c\\minus{}d)^{2}}\\plus{}\\sqrt{b\\plus{}k(b\\minus{}c\\minus{}d\\minus{}a)^{2}}\\plus{}\\sqrt{c\\plus{}k(c\\minus{}d\\minus{}a\\minus{}b)^{2}}\\plus{}\\sqrt{d\\plus{}k(d\\minus{}a\\minus{}b\\minus{}c)^{2}}\\ge 3\\sqrt2$", "Solution_3": "The last hint\r\nLet $ x_{1},x_{2},...,x_{n}$ are real numbers such that $ \\sum_{i \\equal{} 1}^{n}x_{i}\\equal{} m$. Find the best constant $ k$ such that\r\n$ \\sum_{i \\equal{} 1}^{n}\\sqrt{x_{i}\\plus{}k\\left(\\sum_{j\\neq i}^{n}x_{j}\\minus{}x_{i}\\right)^{2}}\\ge (n\\plus{}2)\\sqrt{\\frac{m}{8}}$" } { "Tag": [ "geometry", "3D geometry", "logarithms", "algebra unsolved", "algebra" ], "Problem": "For all positive integer $n$ prove the equality \r\n$[\\sqrt{n}]+[\\sqrt[3]{n}]+...+[\\sqrt[n]{n}]=[log_2n]+[log_3n]+...+[log_nn]$", "Solution_1": "LHS counts the number os squares, cubes, ..., $n$-th powers $\\leq n$.\r\nRHS counts the number of powers of $2$, $3$, ..., $n$ $\\leq n$.\r\n\r\nThey are clearly the same.", "Solution_2": "huh..clever...\r\nBut then again, aren't the first powers counted on the right as well? For example, doesn't $\\log_{[2]}{n}$ count 2 as well?", "Solution_3": "Yes, I forgot some of them to count...\r\n\r\nIt's important to note that some powers are counted more than one time, e.g. $64$ is counted as $2^6, 4^3 , 8^2$ and $64^1$.\r\n\r\nNow the full and correct proof:\r\nOn the LHS, also $1$ is counted everytime, thus $n-1$ times in total. On the RHS, also the first powers $\\neq 1$ are counted, also $n-1$ of them. Thus we can now neglect $1$ and all first powers.\r\nA number of type $s^t$ (by the above $s,t>1$) with $s$ being no power of degree $>1$ is counted as $\\frac{t}{d}$-th power of $s^d$ for any divisor $d \\neq t$ of $t$ on the LHS and as power of $s^d$ for all such $d \\neq 1$ on the RHS. Thus both count the same.\r\nBy \"changing\" the counted $1$es into counted first powers, the argument posted before (\"#(squares, cubes,...) = #(powers of 2,3,...)\") already works and gives a similar proof." } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "Let $ C_1$ be a circle centered at $ O_1$. A circle $ C_2$, centered at $ O_2$, passing through $ O_1$ and meets $ C_1$ at $ A$ and $ B$. Let $ l$ be a line passing through $ O_1$. Points $ P$ and $ Q$ are on $ l$ such that $ AP$ and $ BQ$ are perpendicular to $ l$.\r\nProve that if $ R$ is the midpoint of $ AB$, then $ PQR$ is an isoceles triangle.", "Solution_1": "Because $ O_1O_2 \\perp AB$ and $ R$ is the intersection of $ O_1O_2$ and $ AB$, so $ O_1, R, A, P$ and $ O_1, R, B, Q$ are concyclic. So:\r\n\r\n$ \\angle{RPQ}\\equal{}\\angle{RAO_1}\\equal{}\\angle{RBO_1}\\equal{}\\angle{RQP}$\r\n\r\nHence $ \\triangle{PQR}$ is an isosceles triangle.", "Solution_2": "PQ is the projection of AB onto l, therefore the middle of AB is equally apart of P and Q (the projection of R will be the middle of PQ).\r\n\r\nBest regards,\r\nsunken rock" } { "Tag": [ "factorial", "number theory", "prime factorization" ], "Problem": "How many perfect squares are divisors of the product (1!)(2!)(3!)(4!).....(10!)?", "Solution_1": "It is well known that the exponent of a prime $ p$ in the prime factorization of $ n!$ equals $ \\sum_{k=1}^{\\infty} \\left[{\\frac{n}{p^k}\\right]}$, where $ [x]$ denotes the integer part of $ x$.\r\n\r\nUsing this formula we can get the exponent of the primes $ 2$, $ 3$, $ 5$, and $ 7$ (they are the only ones to appear) in each of $ 1!, 2!, ..., 10!$, and then we add them up to get the prime factorization of $ P=(1!)(2!)...(10!)$.\r\n\r\nIndeed, $ P=2^{38} \\cdot 3^{17} \\cdot 5^{7} \\cdot 7^{4}$.\r\n\r\nThe perfect squares divisors of $ P$ are of the form $ 2^{\\alpha} \\cdot 3^{\\beta} \\cdot 5^{\\gamma} \\cdot 7^{\\delta}$, with $ \\alpha, \\beta, \\gamma, \\delta$ even nonnegative integers less or equal than $ 38, 17, 7$ and $ 4$ respectively.\r\n\r\nWe have $ 20 \\times 9 \\times 4 \\times 3 = 2160$ ways to choose $ \\alpha, \\beta, \\gamma, \\delta$, so there are $ 2160$ divisors of $ P$ which are perfect squares.", "Solution_2": "@uglysolutions, your answer can't be correct.\r\n\r\n10 factorial= 3628800 and the highest square below or equal to this is 1904^2=3625216\r\n\r\nThis means there are AT MOST 1904 perfect squares which divide 10 factorial, so your answer is clearly wrong.\r\n\r\n[b]EDIT: Ignore this post, I misread the question.[/b]", "Solution_3": "He's not counting perfect squares dividing 10!, he's counting perfect squares dividing the (substantially larger) number the question asked about ;)", "Solution_4": "Oops, misread the question. Oh well, lesson learnt." } { "Tag": [ "function", "analytic geometry", "calculus", "calculus computations" ], "Problem": "Sketch the level curves in the xy-plane of f(x,y)=3-y-x^2 for k=0,2,4.\r\n\r\n\r\n\r\nAnyone mind helping me how to do these type of problems? Thank you. :)", "Solution_1": "A level curve is formed by all points with coordinates $ (x,y)$ such that $ f(x,y)=k$. Write this equation as $ y=3-k-x^{2}$: do you recognize the curve?" } { "Tag": [ "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Assume that $ L$ is a finite extension of a field $ K$. Let $ E_{1},\\cdots,E_{n}$ (resp. $ F_{1},\\cdots,F_{n}$) be linearly disjoint field extensions of $ K$ (resp. $ L$). Let $ \\sigma_{i}: E_{i}\\to F_{i}$ be isomorphism which coincide on $ K$ and for which $ \\sigma_{i}(K)=L$ for every $ i=1,\\cdots,n$. Then\r\n\\[ \\sigma_{1}\\otimes\\cdots\\otimes \\sigma_{n}: E_{1}\\otimes_{K}\\cdots\\otimes_{K}E_{n}\\to F_{1}\\otimes_{L}\\cdots\\otimes_{L}F_{n}\\]\r\nis a well defined isomorphism.", "Solution_1": "$ \\sigma_{i}(K) \\subseteq L$ implies that the map $ (x_{1},...,x_{n}) \\mapsto \\sigma_{1}(x_{1}) \\otimes ... \\otimes \\sigma_{n}(x_{n})$ is $ K$-linear in every variable, so it extends to a unique $ K$-linear map of the tensor products. since the $ \\sigma_{i}$ are multiplicative, their tensor product is also multiplicative. the $ E_{1},...,E_{n}$ are linearly disjoint, i.e. their tensor product is a field. thus the map is injective. since the $ \\sigma_{i}$ are surjective, it can be easily seen, that the map is also surjective." } { "Tag": [ "FTW", "AMC", "AIME" ], "Problem": "Sign up for Mewto55555 and Ernie's co-hosted [size=150][b]64 person[/b][/size] tournament. Will be double elimination. Please include your highest rating when signing up!\r\n\r\n\r\nParticipants\r\n\r\n1. Mewto55555 (1527)\r\n2. FantasyLover (1502)\r\n3. Ernie (1455)\r\n4. RomanianGenius (1438)\r\n5. Bogtro (1437)\r\n6. Lotsofmath (1422)\r\n7. Skaldjfhrulez (1414)\r\n8. Budi713 (1412)\r\n9. Violin321 (1367)\r\n10. Ash-Pokemon (1351)\r\n11. Isabella2296 (1349) (won tiebreaker vs. thdanh90 4-1)\r\n12. Thdanh90 (1349)\r\n13. Dragonlaird7(1342)\r\n14. Nikeballa96 (1325)\r\n15. Pinkpiglet (1323)\r\n16. Meta Knight (1262)\r\n17. Jjx1 (1223)", "Solution_1": "reserved...\r\n\r\n1 mewto55555 (1529)\r\n2 ernie (1455)\r\n3 RomanianGenius (1438)\r\n4 violin321 (1367)\r\n5 thdanh90 (1349)\r\n6 nikeballa96 (1325)", "Solution_2": "reserved...", "Solution_3": "I will join. highest=1367", "Solution_4": "You got me.\r\n1349 :oops:", "Solution_5": "nikeballa313\r\nhighest: 1325\r\n\r\nill get killed, why not...=]", "Solution_6": "why not? 1414 highest.", "Solution_7": "bunch of tourneys... :rotfl: \r\n\r\nill join highest-1502", "Solution_8": "Put \"izzy\" on your sign-up list, please. :) \r\n\r\n\r\nHighest rating -- 1349", "Solution_9": "I guess I'll join.\r\n\r\nhighest: 1422", "Solution_10": "join happy face tourney!", "Solution_11": "good luck with the bracket, mewto and ernie :D", "Solution_12": "i sign up\r\n\r\n1412", "Solution_13": "Join Happy Face Tourney! :lol:", "Solution_14": "I'll sign up\r\n1342 :(", "Solution_15": "COME ON PPLS JOIN ALREADY!!!!! I'M GOING 2 FORGET ABOUT THIS TOURNEY SOON!! :D", "Solution_16": "ill join highest rating 1398", "Solution_17": "this might take a while", "Solution_18": "ernie are u cancelling this or not?\r\n(i hope not)", "Solution_19": "The status of this tourney is undecided. (kinda sounds like the FBI :ninja:, wh00ps, thats a ninja :D)", "Solution_20": "BTW, my highest rating is now 1464\r\nno change in seeding, but still", "Solution_21": "Yep. Highest=1200", "Solution_22": "I'll join.\r\nHighest rating=1414", "Solution_23": "I'm sorry to disappoint everyone, but this tournament is now [b]canceled[/b]", "Solution_24": "NOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO! :(\r\nWhy? Just give the top couple of people byes to a later round (or just give randomly selected people byes if you prefer that). There are lots of other solutions, too. There's lots of people signed up, and it can still be a really successful tourney.", "Solution_25": "THIS HAS NOTHING TO DO WITH POEPLE SIGNED UP....", "Solution_26": "WHAT?!?!??!?!?! NOW U TELL US MEWTO!!?!?!? does ctc have do with this. =]", "Solution_27": "Oh... :oops_sign: Sorry, then. What's the problem, though?", "Solution_28": "TOURNEY MANIA!", "Solution_29": "But you know what I find stupid? Mewto can certify other tourneys, but not his own.\r\n\r\nI find that kinda stupid, like this guy.\r\n\r\n[hide=\"Stupid Guy\"]:yankchain:[/hide]" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Find all x,y be a positive numbers such that\r\n$ m^n\\minus{}n^m\\equal{}3$ :?: \r\n :huh:", "Solution_1": "Anybody? :huh:", "Solution_2": "I suppose you meant \"find $ m$ and $ n$\".\r\n\r\nClearly $ m$ is odd or $ n$ is odd, and $ m \\not\\equiv n mod 2$. If $ m > 1$ is odd, $ m^n \\equiv 1 mod 4$ (every odd square is 1 mod 4) and $ 4|n^m$, thus $ 3 \\equal{} m^n \\minus{} n^m \\equiv 1 mod 4$, contradiction. Note that $ m \\equal{} 1$ is obviously absurd.\r\n\r\nIf $ n$ is odd and $ n > 1$, $ 8|m^n$ and $ n^m \\equiv 1 mod 8$ (every odd square is $ 1 mod 8$), thus $ m^n \\minus{} n^m \\equiv 7 mod 8$, then $ 3 \\equiv 7 mod 8$, contradiction. Then $ n \\equal{} 1 \\Rightarrow m \\equal{} 4$." } { "Tag": [ "calculus", "derivative", "calculus computations" ], "Problem": "[url=http://imageshack.us][img]http://img49.imageshack.us/img49/3323/question1aq4.gif[/img][/url]", "Solution_1": "After substitution you should work out the derivatives using the product rule. You will get an overall factor $ e^{\\minus{}(bx\\plus{}ay)}$ on the left hand side which you can drop.\r\n\r\n[hide=\"Hint\"]\n\\[ a \\frac{\\partial u}{\\partial x} \\equal{} a \\left( \\frac{\\partial w}{\\partial x} \\right) e^{\\minus{}(bx\\plus{}ay)} \\plus{} a w \\frac{\\partial}{\\partial x} e^{\\minus{}(bx\\plus{}ay)} \\equal{} a \\left(\\frac{\\partial w}{\\partial x} \\minus{} b w \\right) e^{\\minus{}(bx\\plus{}ay)}\n\\][/hide]" } { "Tag": [], "Problem": "Outline a general method for making the homologation of nitro alkanes:\r\n\r\n$ R\\minus{}CH_2NO_2 \\longrightarrow R\\minus{}NO_2$.", "Solution_1": "R-CH2-NO2 + H3O+->RCOOH\r\nFOLL BY SCHMIDT REACTION TO GIVE RNH2 FOLL BY OXIDATION.", "Solution_2": "[quote=\"VARUNARASU\"]R-CH2-NO2 + H3O+->RCOOH[/quote]\r\n\r\nI don't think so.", "Solution_3": "why not. Hydrolysis of nitro alkanes in presence of acid gives the corr acid. not sure abt the mech but i think it involves tautomerisation to acinitro form. this is applicable to only primary compounds tho", "Solution_4": "[quote=\"Carcul\"][quote=\"VARUNARASU\"]R-CH2-NO2 + H3O+->RCOOH[/quote]\n\nI don't think so.[/quote]\r\n\r\nIs VARUNARASU referring to this:\r\n\r\nhttp://en.wikipedia.org/wiki/Nef_reaction", "Solution_5": "not exactly. i know what nef reaction is but my book says acid catalysed hydrolysis of primary nitro compounds gives acids. there is no base involved. also aldehydes are not formed", "Solution_6": "[quote=\"VARUNARASU\"]not exactly.[/quote]\r\n\r\nYes you are. The Nef reaction is the name given to the conversion of nitro compounds into aldehydes, ketones, or carboxilic acids. However, the conversion to carboxilic acids with acid requires the use of [i]concentrated mineral acids[/i], which, because often leads to side reactions, is a method seldom used.", "Solution_7": "Just wondering Carcul. How do you know which ones are often use in synthesis and which ones are seldom/never used? Is it just experience?", "Solution_8": "It's just experience, but any reaction with a strong mineral acid is a reaction in harsh conditions, and so side reactions are always to be expected.", "Solution_9": "experience :w00tb: :wow: :wow:", "Solution_10": "Yes, Pardesi, are you trying to say something?", "Solution_11": "no just wondering about ur experience . :)", "Solution_12": "i don know if this is write....correct me\r\n\r\nfirst reduce it with Lithium Aluminium Hydride to get the primary amine. Treat with Caro's acid,it undergoes a rearrangement called LOSSEN REARRANGEMENT and we get the primary amine with 1 carbon less. Oxidise with CF3COOOH-triflouro peracetic acid", "Solution_13": "The Lossen rearrangement is the rearrangement of hydroxamic esthers, which are prepared from hydroxamic acids. So, how does a hydroxamic acid/esther forms from a primary amine and Caro's acid? Also, the treatment of aliphatic primary amines with Caro's acid yields nitroso compounds or oximes.", "Solution_14": "sorry\r\nanother method of mine...........see if it's right\r\nfirst reduce with Lithium Aluminium Hydride.treat with nitrous acid to get the alcohol.oxidise using Jones reagent to get the respective acid.Heat with ammonia to get the respective amide.\r\nAmide undergoes Hoffmann Bromamide rearrangement to give you the amine.Amine with triflouro peracetic acid gives you the product.\r\n\r\n\r\n\r\nIs this Correct??????? :)", "Solution_15": "[quote=\"valeriummaximum\"]sorry\nanother method of mine...........see if it's right\nfirst reduce with Lithium Aluminium Hydride.treat with nitrous acid to get the alcohol.oxidise using Jones reagent to get the respective acid.Heat with ammonia to get the respective amide.\nAmide undergoes Hoffmann Bromamide rearrangement to give you the amine.Amine with triflouro peracetic acid gives you the product.\n\n\n\nIs this Correct??????? :)[/quote]\r\nbut when u add excess NH3 to the acid you mainly get RCOONH4+ AND not RCONH2 so what i think is that to get that RCONH2 we can convert RCOOH to RCOCl and then add NH3 to convert it into RCONH2?? am i right carcul coz this is the correction you made in one of my previous conversions????", "Solution_16": "i heat with ammonia srinitrs.\r\nheating with ammonia eliminates a mole of water to give you the amide.............. :lol:", "Solution_17": "[quote=\"Valeriummaximum\"]treat with nitrous acid to get the alcohol[/quote]\n\nWrong: the treatment of primary aliphatic amines with nitrous acid is not a synthetically usefull reaction, because the diazonium salt is not stable enough to survive in solution, and so it forms a carbocation which then undergoes several reactions: elimination to the alkene, nucleophilic addition to the alcohol and/or alkyl halide, etc.\n\n[quote=\"Valeriummaximum\"]Heat with ammonia to get the respective amide.[/quote]\r\n\r\nThat's not a good way of making an amide from an acid, because of the reasons pointed out by Srinitrs: the use of the acid chloride is preferred. However, if you want to convert directly the acid into the amide you can use ammonia and DCC.", "Solution_18": "What is DCC?\r\n\r\nAnyway my method:-\r\n\r\n[list]1.Nef reaction (to covert it into aldehyde)\n2.Tollen's reagent (to convert it into acid)\n3.Smidt reaction (Reaction of HN3) to get primary amine\n4.CF3COOOH(to oxidise -NH2 --> -NO2[/list]", "Solution_19": "Yes, that's right. Here's another method:\r\n\r\n1) $ R\\minus{}CH_2NO_2 \\plus{} NaNO_2/DMSO \\longrightarrow R\\minus{}CO_2H$\r\n\r\n2) $ R\\minus{}CO_2H \\plus{} HN_3/heat \\longrightarrow R\\minus{}NH_2$\r\n\r\n3) $ R\\minus{}NH_2 \\plus{} oxidation \\longrightarrow R\\minus{}NO_2$", "Solution_20": "[quote]\nWhat is DCC? \n[/quote]\r\nDCC is dicyclohexyl carbodiimide." } { "Tag": [ "trigonometry", "limit", "AMC", "AIME" ], "Problem": "Could someone please post a solution to this problem?\r\n\r\nFind $\\sum_{n=0}^{\\infty}\\cot^{-1}(n^{2}+n+1)$, where $\\cot^{-1}(t)$ for $t\\geq 0$ denotes the number $\\theta$ in the interval $0\\leq\\theta\\leq\\frac{\\pi}{2}$ with $\\cot(\\theta)=t$.\r\n\r\nIt is in the \"Do you need this\" PDF on trig. I currently can do all of it except this one. Any help would be appreciated.", "Solution_1": "[hide=\"Solution\"]$S=\\sum_{n=0}^{\\infty}\\cot^{-1}\\left(n^{2}+n+1\\right)=\\sum_{n=0}^{\\infty}\\tan^{-1}\\left(\\frac{1}{n^{2}+n+1}\\right)$\n\nNote that $\\frac{1}{n^{2}+n+1}=\\frac{\\left(n+1\\right)-n}{1+n\\left(n+1\\right)}$.\n\nSo you have $S=\\sum_{n=0}^{\\infty}\\tan^{-1}\\left(n+1\\right)-\\sum_{n=0}^{\\infty}\\tan^{-1}\\left(n\\right)=\\lim_{n\\rightarrow\\infty}\\tan^{-1}\\left(n\\right)=\\boxed{\\frac{\\pi}{2}}$.[/hide]\r\nWhat is this \"Do you need this?\" that you speak of?", "Solution_2": "[quote=\"amcavoy\"]What is this \"Do you need this?\" that you speak of?[/quote]\r\n\r\ni think it's something like a pretest for the online class of this site...", "Solution_3": "I would just like to note that lately, a bunch of people have been posting these problems from the pretest things (a trend started by me!). That's probably good news (financially) for AoPS, because it means more people are considering taking classes.", "Solution_4": "[quote=\"mysmartmouth\"]I would just like to note that lately, a bunch of people have been posting these problems from the pretest things (a trend started by me!). That's probably good news (financially) for AoPS, because it means more people are considering taking classes.[/quote]\r\nBy the way, where do you find pretests? Or do you have to enroll in the class to get them?", "Solution_5": "http://www.artofproblemsolving.com/Classes/AoPS_C_ClassesS.php", "Solution_6": "[quote=\"towersfreak2006\"]http://www.artofproblemsolving.com/Classes/AoPS_C_ClassesS.php[/quote]\r\nSo... I guess WOOT doesn't have a pretest? Or is it on another page? :| \r\nWow, my 200th post! :) :P", "Solution_7": "WOOT! does not have any of those. I believe its Week 0 class (Orientation) is designed to place you into the correct level though.", "Solution_8": "Note that with WOOT, you can do it, and if you find it too hard, you can drop out and get your money back within x number of days. The pretests are designed to see if you are ready for the class. And Neal, I think you are ready for WOOT, despite that drastically low AIME score *sarcasm*. :rotfl:", "Solution_9": "I can remember seeing this problem in a GRE practice book..." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "1.Let $ a,b,c \\in [3,4].$\r\nProve that\r\n$ \\sum \\sqrt{a^2\\plus{}b^2\\minus{}c^2} \\leq a\\plus{}b\\plus{}c$\r\n2.Let $ a,b,c \\in [3,4].$\r\nProve that:\r\n $ \\sqrt[3]{a(a^2\\plus{}b^2\\minus{}c^2)}\\plus{}\\sqrt[3]{b(b^2\\plus{}c^2\\minus{}a^2)}\\plus{}\\sqrt[3]{c(c^2\\plus{}a^2\\minus{}b^2)}\\leq a\\plus{}b\\plus{}c$", "Solution_1": "My idea is let $ a,b,c$ be sides of a triangle :) \r\n\r\n(Since $ a,b,c \\in [3,4]$, it is possible)\r\n\r\nThen $ \\sum \\sqrt{a^2\\plus{}b^2\\minus{}c^2} \\equal{} \\sum \\sqrt{2abcosA} \\leq \\sqrt{(\\sum 2ab)(\\sum cosA)}$\r\n\r\nBut we have $ \\sum cosA \\leq \\frac{3}{2}$, $ 3 \\sum ab \\leq (a\\plus{}b\\plus{}c)^2$.\r\n\r\nAlso, $ \\sum \\sqrt[3]{a(a^2\\plus{}b^2\\minus{}c^2)} \\equal{} \\sum \\sqrt[3]{2a^2bcosA} \\leq \\sqrt[3]{(\\sum a)(\\sum 2ab)(\\sum cosA)}$.\r\n\r\nThe remaining parts should be trivial :P" } { "Tag": [ "Mafia", "puzzles" ], "Problem": "A group of jealous professors are locked in a room. There is nothing else in the room but pencils and one tiny scrap of paper per person. The professors want to determine their average (mean, not median) salary so that each one can gloat or griece over their personal situation compared to their peers. However, they are secretive people and do not want to give away any personal salary information to anyone else. Can they determine the average salary in such a way that no professor can discover [i]any[/i] fact about the salary of anyone but herself? For example, even facts such as \"3 people earn more than $\\$$40,000\" or \"no one earns more than $\\$$90,000\" are not allowed.\r\n\r\n\r\nAnyone? I decided that they *could* have the pencil represent like $\\$$20,000 or some other base amount, then rip up the pieces of paper to represent other increments and toss it all into a pot, but you would need to be able to rip the paper into some pretty small pieces...", "Solution_1": "Every professor can do the following process for himself or herself. \r\n\r\nProf#1 writes any arbitary salary figure and passes that to Prof # 2. Prof # 2 adds his salary to the received figure; forwards the total to the 3rd Prof after tearing off the value received from the 1st professor. This process continues and every one can know just the mean salary without any further details. \r\n\r\nSuppose, there are 5 professors. \r\n\r\nProf#1 writes 100 (dummy value) and passes it to Prof # 2. Prof # 2 adds his own salary, say 300, and passes just the total value of 400 to Prof#3. Prof # 3 adds his salary (say 500) and lets Prof # 4 know the total as 900. Prof # 4 passes a total of 1600 to Prof # 5. Prof # 5 returns back a total of 2500 to Prof # 1. \r\n\r\nNow, prof # 1 can determine the average salary. Say, his actual salary was 350.\r\n\r\nAverage salary = {(2500 - 100) + 350}/5 = 550. All other professors can also follow a similar process.", "Solution_2": "[quote]\nProf#1 writes any arbitary salary figure and passes that to Prof # 2. Prof # 2 adds his salary to the received figure; forwards the total to the 3rd Prof after tearing off the value received from the 1st professor. This process continues and every one can know just the mean salary without any further details.\n[/quote]\r\n\r\nProf #3 now knows that that value passed to him is the upper limit of what prof#2 makes. Isn't this one of the \"facts\" that the problem statement doesn't allow? Moreover, if the group consists of only 2 profs (the problem doesn't specify the group's minimum size), the first one can easily determine what the second prof makes. \r\n\r\nI don't think there's a solution. Here's why:\r\n\r\n[hide]\nConsider a group that consists of only two profs. Suppose there is some elaborate protocol that allows them to determine their average salary without revealing any personal salary info to the other prof. At the end of this, each will know the average salary. If prof #1 makes the same as the average salary, he's also just determined that prof #2 must make exactly the same amount. If prof #1 makes _more_ than the average salary, he's just determined prof #2 makes _less_ than the average salary. And if prof #1 makes _less_ than the average, then prof #2 must make more. I assume these are the type of \"facts\" that the problem statement doesn't want you to be able to determine.\n\nIf you claim that \"group\" implies 3 or more people, then consider the case with 3 people. Once everyone knows the average, they can simply compare it to their own salary to conclude some facts about the salaries of the others: If a prof makes the same as the average, then either everyone makes exactly the same or there's one person who makes more and one person who makes less. If a prof makes more than the average, then there must be at least one prof who makes less than the average. \n[/hide]", "Solution_3": "[quote=\"gubd\"]\nProf #3 now knows that that value passed to him is the upper limit of what prof#2 makes. Isn't this one of the \"facts\" that the problem statement doesn't allow? Moreover, if the group consists of only 2 profs (the problem doesn't specify the group's minimum size), the first one can easily determine what the second prof makes. \n\nI don't think there's a solution. Here's why:\n\n[hide]\nConsider a group that consists of only two profs. Suppose there is some elaborate protocol that allows them to determine their average salary without revealing any personal salary info to the other prof. At the end of this, each will know the average salary. If prof #1 makes the same as the average salary, he's also just determined that prof #2 must make exactly the same amount. If prof #1 makes _more_ than the average salary, he's just determined prof #2 makes _less_ than the average salary. And if prof #1 makes _less_ than the average, then prof #2 must make more. I assume these are the type of \"facts\" that the problem statement doesn't want you to be able to determine.\n\nIf you claim that \"group\" implies 3 or more people, then consider the case with 3 people. Once everyone knows the average, they can simply compare it to their own salary to conclude some facts about the salaries of the others: If a prof makes the same as the average, then either everyone makes exactly the same or there's one person who makes more and one person who makes less. If a prof makes more than the average, then there must be at least one prof who makes less than the average. \n[/hide][/quote]\r\n\r\nFirst of all, Shamik's procedure does work. If the initial number prof#1 writes down is large enough, the maximum value prof#3 recieves will say almost nothing about prof#2's salary. For example, if prof#1 writes down 1,002, 345 and prof#2's salary is 3,000, then the maximum value that prof#3 gets is 1, 005, 345, and isn't very revealing at all about prof#2's salary.\r\n\r\nAlso, I agree that such a procedure would be impossible for a group of 2 professors because knowing the mean and their own salary they can figure out the other's salary. However the argument for 3 or more professors is faulty. Knowing the mean and their own salary they can figure out a fact about [b]someone's[/b] salary, but not a fact about the salary of [b]someone in particular[/b].", "Solution_4": "[quote]\nHowever the argument for 3 or more professors is faulty. Knowing the mean and their own salary they can figure out a fact about [b]someone's salary[/b], but not a fact about the salary of [b]someone in particular[/b].\n[/quote]\r\n\r\nFrom the problem statement: [i]even facts such as \"3 people earn more than \\$40,000\" ... are not allowed[/i]. This is a fact about [b]someone's salary[/b], not the salary of [b]someone in particular[/b], as you put it. My interpretation of this is that learning either type of fact is ruled out by the problem statement. \r\n\r\nIf the problem statement didn't rule out learning such general facts, then I agree, Shamik's solution would work in a group of 3 or more people.", "Solution_5": "Gubd's right. I guess I didn't read the problem very carefully :blush: .\r\n\r\nHalluci gave me an idea though:\r\n\r\n[hide]They could rip their scraps into pieces and write on each scrap a portion of their salary in different handwritings. For example, if prof#1 earned 500 dollars, he could rip his scrap into three pieces and write 400, 200, and -100 on different scraps. Then all pieces would be dumped into the middle of the room , and they would then find the mean. They need to write down the portions in different handwritings so that no one could piece the scraps together.[/hide]\r\n\r\nThis still gives away information after the pofessors know the mean though, as gubd said. Perhaps the problem wasn't worded correctly? :huh: :huh:", "Solution_6": "I was thinking, could gubd's problem be solved if professor #1 is randomly determined? Then they can't make any assumptions about how much anyone makes. And making professor #1 randomly determined can be done if they setup a system like in mafia games(shuffling cards).", "Solution_7": "I don't think I quite understand what you mean. After any professor knows the mean and his own salary, he can compare his salary to the mean and find out a fact about someone else in the group. (For example, if the mean was 500 and prof#1 made 100, he would know that at least on person in the group made more than 500.) Also, I have just noticed that knowing the mean, a professor can also find the total of everyone else. I still think the problem was misworded.", "Solution_8": "[quote=\"gubd\"]Prof #3 now knows that that value passed to him is the upper limit of what prof#2 makes. Isn't this one of the \"facts\" that the problem statement doesn't allow?[/quote]\r\n\r\nSeconding laughinghead: this objection is a good objection to the statement of the problem, not to the given solution. If a professor is to know the mean salary, he will also learn the total salary, and thus an upper bound on all salaries and in particular an upper bound on the salary of any particular professor. So any solution to the problem is a violation of that restriction. What this suggests is that the restriction in the problem is flawed and should be re-written. Under a reasonable re-writing of the problem, the given solution would be correct.", "Solution_9": "I agree. :lol:", "Solution_10": "In any case, the original objection about the third professor knowing upper limits is easily removed if the first arbitrary value passed is allowed to be negative.", "Solution_11": "[quote=\"Shamik Banerjee\"]Every professor can do the following process for himself or herself. \n\nProf#1 writes any arbitary salary figure and passes that to Prof # 2. Prof # 2 adds his salary to the received figure; forwards the total to the 3rd Prof after tearing off the value received from the 1st professor. This process continues and every one can know just the mean salary without any further details. \n[/quote]\r\n\r\nThere is a problem with that solution. If there are five professors, the fifth one can memorize the last number, he can figure out what the first professor made.", "Solution_12": "Prof#1 writes any arbitary salary figure and passes that to Prof # 2. Prof # 2 adds his arbitary salary to the received figure; forwards the total to the 3rd Prof after tearing off the value received from the 1st professor. This process continues \r\n\r\nand then they start again and again and finally after n times they will find the mean...", "Solution_13": "[quote=\"mustafa\"]There is a problem with that solution. If there are five professors, the fifth one can memorize the last number, he can figure out what the first professor made.[/quote]\r\n\r\nNo, he can't. It goes around to all professors. Then the first professor takes the result back, subtracts the arbitrary number he began with, and adds his own salary. (He has to not let the second professor see the final total before he does this. In fact, there must be strictly more than 2 professors, since knowing the mean of 2 numbers and one of the numbers tells you the other number.) Then he divides by the number of professors, and the result is the mean. The arbitrary initial value is highly relevant.", "Solution_14": "I have a feeling Shamik has seen this type of puzzle before and therefore has the expected and correct solution. Or he's just a genius with great insight :lol:", "Solution_15": "Similar: Ten people has to select one from two candidates. However they do not want to make it openly and there is no poll. Assume that they will not cheat.", "Solution_16": "Make it so that the number the first professor wrote can be negative", "Solution_17": "Why is this in the games forum? " } { "Tag": [], "Problem": "A passenger's train speed is 60 mph, and freight train speed is 40 mph. The passenger train travels the same distance in 1.5 h less time than the freight train. how long does each train take to make the trip?\r\n\r\nDoes this problem use d = rt? \r\n\r\nif i use (60 * t-1.5)=40t would that be right?", "Solution_1": "um.. as far as I know, it uses the d=rt.\r\n\r\nI think your equation looks right, but just remeber to plug it back in for passenger train.", "Solution_2": "[quote=\"DonkeyKong\"]um.. as far as I know, it uses the d=rt.\n\nI think your equation looks right, but just remeber to plug it back in for passenger train.[/quote]Ya I agree with DonkeyKong it should use the d=rt", "Solution_3": "I agree w/ DK also, it does use d=r x t\r\nwhen it says they both travel the same distance, you should use d=r x t because you already know the rate and you can just substitute r x t for distance. \r\nYour equation looks fine to me! :)", "Solution_4": "[quote=\"epatjn\"]A passenger's train speed is 60 mph, and freight train speed is 40 mph. The passenger train travels the same distance in 1.5 h less time than the freight train. how long does each train take to make the trip?\n\nDoes this problem use d = rt? \n\nif i use (60 * t-1.5)=40t would that be right?[/quote]\r\n\r\nUh, technically, it should be $ 60\\times (t\\minus{}1.5)\\equal{}40t$. The parentheses are very important.\r\n\r\nAnd then make sure you answer the question. :wink:", "Solution_5": "you guys are all correct\r\ni mean it looks right\r\n\r\n\r\n :play_ball: \r\n :coolspeak:" } { "Tag": [ "geometry", "calculus", "integration", "real analysis", "real analysis theorems" ], "Problem": "I was wondering how the Jacobian was derived. It seems that dealing with projections would lead me there (similar to the surface area integral), but I want to see the full thing.\r\n\r\nSo far I have:\r\n\r\narea(F1)=area(F2)cos(theta)\r\n\r\nThen I can replace the left side of the equation with dxdy, but I'm not sure what to do from there.\r\n\r\nAny ideas?\r\n\r\nThanks.", "Solution_1": "Actually, I asked the same question a while ago, here is the discussion: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32350[/url]" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Find $ x \\in R$ of this equation: $ 4\\sqrt { 1\\plus{} x} \\minus{} 1 \\equal{} 3x \\plus{} 2\\sqrt {1 \\minus{} x} \\plus{} \\sqrt {1 \\minus{} x^2}$", "Solution_1": "We can rewrite the equation as\r\n$ (\\sqrt{1\\minus{}x}\\plus{}\\sqrt{1\\plus{}x}\\minus{}2)(\\sqrt{1\\minus{}x}\\minus{}2\\sqrt{1\\plus{}x})\\equal{}0$ \r\nThe solution is $ 0$ or $ \\minus{}\\frac35$." } { "Tag": [], "Problem": "Let $m$ be a 4-digit number such that $m^2-n^2$ is a square number, where $n$ is the 4-digit number formed by reversing the digits of $m$. Find $m$.", "Solution_1": "This is similar to one of the 2005 AMC 12B problems, except that $m$ is a 4-digit number, not 2-digit...\r\n\r\n[hide=\"my work so far\"](EDITED) Let $m=1000a+100b+10c+d$ and $n=a+10b+100c+1000d$.\n\n$m^2-n^2=(m+n)(m-n)$\n$=(1001a+110b+110c+1001d)(999a+90b-90c-999d)$\n$=99(91(a+d)+10(b+c))(111(a-d)+10(b-c))=k^2$\n\n$(91(a+d)+10(b+c))(111(a-d)+10(b-c))$ must be in the form of $11p^2$.\nBut then, it's a little hard to check the values, and I don't know if this is the right approach :( [/hide]", "Solution_2": "[quote=\"frt\"]This is similar to one of the 2005 AMC 12B problems, except that $m$ is a 4-digit number, not 2-digit...\n\n[hide=\"my work so far\"](EDITED) Let $m=1000a+100b+10c+d$ and $n=a+10b+100c+1000d$.\n\n$m^2-n^2=(m+n)(m-n)$\n$=(1001a+110b+110c+1001d)(999a+90b+90c+999d)$\n$=99(91(a+d)+10(b+c))(111(a+d)+10(b+c))=k^2$\n\nLet $a+d=x$ and $b+c=y$, then $(91x+10y)(111x+10y)$ must be in the form of $11p^2$.\nBut then, it's a little hard to check the values for $x$ and $y$, and I don't know if this is the right approach :( [/hide][/quote]\r\n\r\nI thought it should be \r\n$m^2-n^2=(m+n)(m-n)$\r\n$=(1001a+110b+110c+1001d)(999a+90b-90c-999d)$.....", "Solution_3": "[quote=\"jelly119\"]\nI thought it should be \n$m^2-n^2=(m+n)(m-n)$\n$=(1001a+110b+110c+1001d)(999a+90b-90c-999d)$.....[/quote]\r\nOops :oops: Edited.", "Solution_4": "Notice the three numbers form a pythagorean triple....", "Solution_5": "[quote=\"seamusoboyle\"]Notice the three numbers form a pythagorean triple....[/quote]\r\nbut I cant understand how that would help.", "Solution_6": "It puts severe limits on what the numbers can be.", "Solution_7": "I understand I think.....", "Solution_8": "I cant think how you can do this using logical hit and trial way or using frt's way.", "Solution_9": "i have two numbers \r\nfor $m=6565$\r\n$6565^2-5656^2=3333^2$\r\nfor $m=5625$\r\n$5625^2-5265^2=1980^2$\r\n\r\ni found them with microsoft excel", "Solution_10": "[quote=\"gunes\"]i have two numbers \nfor $m=6565$\n$6565^2-5656^2=3333^2$\nfor $m=5625$\n$5625^2-5265^2=1980^2$\n\ni found them with microsoft excel[/quote]\r\nWow... power of technology :lol: \r\nI should have thought of $6565^2-5656^2=3333^2$ since I knew $65^2-56^2=33^2$...", "Solution_11": "also \r\n$656565^2-565656^2=333333^2$\r\n$656565656565^2-565656565656^2=333333333333^2$ and going on so...", "Solution_12": "that is soo weird!", "Solution_13": "What about a 3-digit number? Think about it!", "Solution_14": "[quote=\"jelly119\"]What about a 3-digit number? Think about it![/quote]\r\nI don't think there's any 3-digit number that satisfies the given condition.\r\nHere's what I did.\r\n[hide]Let $m=100a+10b+c$\nThen $(100a+10b+c)^2-(a+10b+100c)^2$\n$=(101a+20b+101c)(99a-99c)=99(101a+20b+101c)(a-c)=k^2$\nThis means $(101a+20b+101c)(a-c)$ has to be in the form of $11s^2$.\nNote that $a-c\\ne 11$\nWe then have to think about some cases.\n[hide=\"case 1\"]$a-c=p^2$ and $101(a+c)+20b=11q^2$\n\nFrom the second equation, we obtain $b=\\frac{11q^2-101(a+c)}{20}$ ($2\\le a+c\\le 18$ and $0\\le b\\le 9$), but there doesn't exist any solution.[/hide]\n[hide=\"case 2\"]$a-c=p$ and $101(a+c)+20b=11p$\n\nCombine two equations to get $101(a+c)+20b=11(a-c)$ or $90a+20b+112c=0$, but there's no solution for $a,c\\ge 1$ and $b\\ge 0$.[/hide]\n[hide=\"case 3\"]$a-c=p$ and $101(a+c)+20b=11p^3$\n\nAgain, we obtain $b=\\frac{11p^3-101(a+c)}{20}$ from the second equation, but there's no solution for $2\\le a+c\\le 18$ and $0\\le b\\le 9$.[/hide]\nThere are still some minor cases like $a-c=p$ and $101(a+c)+20b=11p^5$ etc., so I'm not completely sure, but probably there is no such 3-digit integer.[/hide]\r\nSince it's late at night, I hope I didn't do my arithmetic wrong...", "Solution_15": "[quote=\"frt\"][quote=\"jelly119\"]What about a 3-digit number? Think about it![/quote]\nI don't think there's any 3-digit number that satisfies the given condition.\nHere's what I did.\n[hide]Let $m=100a+10b+c$\nThen $(100a+10b+c)^2-(a+10b+100c)^2$\n$=(101a+20b+101c)(99a-99c)=99(101a+20b+101c)(a-c)=k^2$\nThis means $(101a+20b+101c)(a-c)$ has to be in the form of $11s^2$.\nNote that $a-c\\ne 11$\nWe then have to think about some cases.\n[hide=\"case 1\"]$a-c=p^2$ and $101(a+c)+20b=11q^2$\n\nFrom the second equation, we obtain $b=\\frac{11q^2-101(a+c)}{20}$ ($2\\le a+c\\le 18$ and $0\\le b\\le 9$), but there doesn't exist any solution.[/hide]\n[hide=\"case 2\"]$a-c=p$ and $101(a+c)+20b=11p$\n\nYou've got the correct answer!! :) \nCombine two equations to get $101(a+c)+20b=11(a-c)$ or $90a+20b+112c=0$, but there's no solution for $a,c\\ge 1$ and $b\\ge 0$.[/hide]\n[hide=\"case 3\"]$a-c=p$ and $101(a+c)+20b=11p^3$\n\nAgain, we obtain $b=\\frac{11p^3-101(a+c)}{20}$ from the second equation, but there's no solution for $2\\le a+c\\le 18$ and $0\\le b\\le 9$.[/hide]\nThere are still some minor cases like $a-c=p$ and $101(a+c)+20b=11p^5$ etc., so I'm not completely sure, but probably there is no such 3-digit integer.[/hide]\nSince it's late at night, I hope I didn't do my arithmetic wrong...[/quote]" } { "Tag": [ "inequalities", "real analysis", "real analysis solved" ], "Problem": "$f(x)\\geq 0$ is Riemann-integrable on $[a,b]$.\r\nShow that $\\sqrt{f(x)}$ is also Riemann-integrable on $[a,b]$ using $\\sum_i \\omega_i \\Delta x_i<\\varepsilon$.\r\n\r\n========================================\r\n\r\nMy puzzle is: I don't know how to use $\\sum_i \\omega_i \\Delta x_i<\\varepsilon$ directly. I use $\\sum_{\\omega_i\\geq \\epsilon} \\Delta x_i<\\sigma$.", "Solution_1": "[quote=\"liyi\"]$f(x)\\geq 0$ is Riemann-integrable on $[a,b]$.\nShow that $\\sqrt{f(x)}$ is also Riemann-integrable on $[a,b]$ using $\\sum_i \\omega_i \\Delta x_i<\\varepsilon$.\n\n========================================\n\nMy puzzle is: I don't know how to use $\\sum_i \\omega_i \\Delta x_i<\\varepsilon$ directly. I use $\\sum_{\\omega_i\\geq \\epsilon} \\Delta x_i<\\sigma$.[/quote]\r\n\r\n We have $\\sum_i{\\omega_i}\\Delta x_i <\\varepsilon$\r\n A fundamental inequality gives : \\[\\sum_i \\left( \\sqrt{\\omega_i}\\Delta x_i \\right)^2 \\leq \\left(\\sum_i{\\omega_i}\\Delta x_i \\right) \\cdot \\left( \\sum_i \\Delta x_i \\right ) <\\varepsilon \\cdot (b-a) \\] so \\[\\sum_i \\left( \\sqrt{\\omega_i}\\Delta x_i \\right) < \\sqrt{(b-a)\\varepsilon} \\]", "Solution_2": "I worked it out later. My proof is exactly the same. Thank you all the same.\r\nThe inequality is called Cauchy's Inequality.", "Solution_3": "[quote=\"dickchimney\"]\n A fundamental inequality gives : \\[\\sum_i \\left( \\sqrt{\\omega_i}\\Delta x_i \\right)^2 \\leq \\left(\\sum_i{\\omega_i}\\Delta x_i \\right) \\cdot \\left( \\sum_i \\Delta x_i \\right ) \\][/quote]\r\nA typo there.\r\nIt must be $\\left(\\sum \\sqrt{\\omega_i}\\Delta x_i \\right)^2 \\leq \\left(\\sum{\\omega_i}\\Delta x_i \\right) \\cdot \\left( \\sum \\Delta x_i \\right )$" } { "Tag": [], "Problem": "If $ x \\plus{} 5 < 8$ and $ x$ is a prime number, what is the value of $ x$?", "Solution_1": "$ x\\plus{}5<8$\r\n$ x<3$\r\nx is prime. Only prime number under 3 is 2." } { "Tag": [ "complex analysis", "complex analysis unsolved" ], "Problem": "Let $ z_1,z_2,z_3\\in C$ and $ |z_1|\\equal{}|z_2|\\equal{}|z_3|$. Prove that\r\n\\[ \\text{arg}\\,\\frac{z_3\\minus{}z_2}{z_3\\minus{}z_1}\\equal{} \\frac{1}{2}\\,\\text{arg}\\,\\frac{z_2}{z_1}\\]", "Solution_1": "Interpret this geometrically. Draw the circle centered at the origin with radius $ r\\equal{}|z_1|$, and some segments as appropriate. What are these angles in the picture?", "Solution_2": "Yes, I see. Perifirical angle is $ 1/2$ of central." } { "Tag": [ "calculus" ], "Problem": "What are some good books for introductory physics?", "Solution_1": "What level? [url=http://www.amazon.com/University-Physics-Modern-Mastering-MasteringPhysics/dp/080538684X]University Physics (Young & Freedman)[/url] is a widely used calculus based introduction, first year college. Another one is [url=http://www.amazon.com/Physics-Scientists-Engineers-Douglas-Giancoli/dp/0132431068/ref=pd_sim_b_36]Physics for Scientists and Engineers (Giancoli)[/url], and [url=http://www.amazon.com/Fundamentals-Physics-Extended-David-Halliday/dp/0471758019/ref=cm_cr_pr_pb_t]Funamentals of Physics (Halliday & Resnick)[/url] is also known.\r\nAll these books are very big, covering lots of material (all the basic stuff: mechanics, thermodyamics, electricity+magnetism, waves, optics, etc.), but of course also on a very basic level.\r\n\r\nIf you're self studying, and can follow your own plan, I think there are better ways to learn physics. For example, first get a good working knowledge of calculus/linear algebra (at a computational level), and then directly go to some deeper, advanced introductory texts, like [url=http://www.amazon.com/Introduction-Mechanics-Daniel-Kleppner/dp/0070350485/ref=sr_1_4?ie=UTF8&s=books&qid=1234976789&sr=1-4]An Introduction to Mechanics (Kleppner)[/url], [url=http://www.amazon.com/Introduction-Electrodynamics-3rd-David-Griffiths/dp/013805326X/ref=pd_sim_b_3]Introduction to Electrodynamics (Griffiths)[/url] and [url=http://www.amazon.com/Special-Relativity-M-I-T-Introductory-Physics/dp/0393097935/ref=pd_sim_b_10]Special Relativity (French)[/url]", "Solution_2": "I think Halliday & Resnick is the best at an introductory level. it explains things very clearly and there are thought-provoking questions and problems behind every chapter. It is great :omighty: .", "Solution_3": "French's Speical Relativity is awesome! I'd highly recommmend it. Also Wheeler's and Taylor's [i]Spacetime Physics [/i]is pretty for starters." } { "Tag": [ "inequalities", "geometry", "circumcircle", "inradius", "geometry solved" ], "Problem": "Let ABC be a triangle with circumradius R and inradius r. Let [tex]m_a, h_a[/tex] be the lengths of the median and altitude, respectively, through A. Prove that\r\n\r\n[tex]\\frac{R}{2r} \\geq \\frac{m_a}{h_a}[/tex]", "Solution_1": "This was definitely posted before. I don't know whether it was posted in the \"Geometry\" or \"Inequalities\" section, however..\r\n\r\n[color=red][Moderator edit: Indeed. http://www.mathlinks.ro/Forum/viewtopic.php?t=2558 ][/color]" } { "Tag": [ "inequalities", "linear algebra", "matrix" ], "Problem": "[b]Let [/b]$ A \\in M_{n} \\mathbb{(R)}$,[b]I is identity nxn matrix .[/b]\r\n [b]Such that :[/b] $ 19A^{4}-5A^{3}+1909A^{2}-5A+1890I=0$\r\n[b]Prove that :[/b]\r\n $ det^{2}(A+xI)+det^{2}(A+yI) \\geq 2det^{2}(A+\\sqrt{xy}I)$, $ \\forall x,y \\in R^{+}$\r\n[b]Here x, y are positive real number .[/b]\r\n[b]\"..Do math to eat very sweet ...\"[/b]", "Solution_1": "[b]You can prove from inequality :[/b]\r\n $ (ax^{2}\\plus{}bx\\plus{}c)(ay^{2}\\plus{}by\\plus{}c) \\geq (axy\\plus{}b \\sqrt {xy}\\plus{}c)^{2}$ , $ a>0,b<0,c>0 ,b^{2}\\minus{}4ac < 0$\r\n $ (x^{2}\\plus{}1)(y^{2}\\plus{}1) \\geq (xy\\plus{}1)^{2}$ ,$ x ,y \\geq 0$\r\n[b]\" .. You can real Jean Marie Monier professor to find the nice problem of math ...\"[/b]\r\n[b]\"..I can copy all books to understand but I will creat to find nice problem of math ...\"[/b][/b]\r\nA", "Solution_2": "$ b>0$,[b](I think (-5))[/b]\r\n[b]I am sorry !!![/b]" } { "Tag": [ "analytic geometry", "graphing lines", "slope", "calculus", "derivative", "function", "symmetry" ], "Problem": "I recently begun study regression, for personal interest, but now i've hit the wall. So, I have a little problem, that i think should be fairly easy to you guys.\r\n\r\nI need the slope, of a curvilinear (non-linear) regression. I will use this on a time series study, and I can get the value correct (intercept) but not the slope. I think you somehow shoul be able to extract the slope from the formula of non-linear regression, but i can't pull it off.\r\n\r\nAnyone willing to show me what formula to use for the non- linear regressino slope, whit the dependent variable x = timeserie?\r\n\r\nThanks guys.", "Solution_1": "What do you mean by slope? Any non-linear curve has variable slope... it's differentiation in calculus. Unless I misunderstood your question.", "Solution_2": "[quote=\"Flirrrt\"]I recently begun study regression, for personal interest, but now i've hit the wall. So, I have a little problem, that i think should be fairly easy to you guys.\n\nI need the slope, of a curvilinear (non-linear) regression. I will use this on a time series study, and I can get the value correct (intercept) but not the slope. I think you somehow shoul be able to extract the slope from the formula of non-linear regression, but i can't pull it off.\n\nAnyone willing to show me what formula to use for the non- linear regressino slope, whit the dependent variable x = timeserie?\n\nThanks guys.[/quote]\r\n\r\nYou need calculus (differentiation to find the slope). \r\nHowever, if you haven't studied calculus yet, here's an approach:\r\n\r\nLet your regression function be $ f(x)$, and suppose you would like to find the \"slope\" at $ x_{0}$. Pick a very small number $ \\Delta x > 0$, and calculate\r\n$ m = \\frac{f(x+\\Delta x)-f(x)}{\\Delta x}$\r\n\r\nOf course, this is only an approximation. But as you decrease the size of $ \\Delta x$, then you can get really close to the actual value.", "Solution_3": "Well, what power is the function closest to? 2? 3? 4? if greater than 2, I would second vish. Otherwise, calculate the distance from the point on the function $ p_{o}$ to the approximate focus. The extend another line from the focus, along the line of symmetry, that has that same length $ d$, and ends at point $ p_{f}$. The lines that connects these is tangent to point $ p_{o}$.\r\n\r\nIf I'm not mistaken this is a more rudimentary form of calculus differentiation.\r\n\r\nEDIT: whew, gotta pay attention in english :(" } { "Tag": [ "inequalities solved", "inequalities" ], "Problem": "Let a,b,c,n>0 such that a^n+b^n=c^n. Prove that a^(n(n+1))*c>[n(c-b)]^(n+1)*b^(n^2) and that (c/b-1)(c/a-1)<[1/n^2]*(c^2/ab)^1/(n+1).", "Solution_1": "A=c^(n-1)+c^(n-2)*a+...+a^(n-1)\r\nB=c^(n-1)+c^(n-2)*b+...+b^(n-1)\r\n\r\na.) Pro\r\n<=>[(c^n-b^n)]^(n+1)*c > n^(n+1)*b^(n^2)*[(c-b)]^(n+1)\r\n<=>B^(n+1)*c>n^(n+1)*b^(n^2)\r\nWe have LHS > c * [n * b^(n-1)]^(n+1) > RHS\r\n\r\nb.)Pro\r\n<=>[(c-b)(c-a)]^(n+1)*n^(2n+2) < c^2*(ab)^n\r\n<=>[(c^n-b^n)(c^n-a^n)]^(n+1)*n^(2n+2) < c^2*(ab)^n*(AB)^(n+1)\r\n<=>(a^n*b^n)^(n+1)*n^(2n+2) < c^2*(ab)^n*(AB)^(n+1)\r\nWe have RHS\r\n>c^2 * (ab)^n * [n^2 * (ab)^(n-1)]^(n+1)\r\n>c^2 * (ab)^n * n^(2n+2) * (ab)^(n^2-1)\r\n>c^2 * n^(2n+2) * (ab)^(n^2+n-1)\r\n>n^(2n+2) * (ab)^(n^2+n) (c^2>ab)\r\n=LHS\r\n.....................................................\r\nQ.E.D" } { "Tag": [ "calculus", "calculus computations" ], "Problem": "Write the parametric equations of the line that goes through the point (6,-2,1) and its perpendicular to both:\r\n[x,y,z]=[1,4,-2]+t[3,-1,1] and [x,y,z]=[9,5,-3]+s[1,-3,7]", "Solution_1": "And what have you thought to try? (A hint: in what context have you heard the phrase \"perpendicular to both\" before?)" } { "Tag": [ "symmetry", "geometry", "circumcircle", "incenter", "ratio", "trigonometry", "number theory" ], "Problem": "Today, 5 November the IMAR contest was held at National College Sf. \r\n Sava between 10-13. There were given 3 problems. The results will be available on Tuesday. \r\n\r\n [b] Problem 1. [/b]\r\n\r\nConsider the equation\r\n\\[\\frac{xy-C}{x+y}= k, \\]\r\nwhere all symbols used are positive integers.\r\n\r\n1. Show that, for any (fixed) values $C, k$, this equation has at least a solution $x, y$;\r\n\r\n2. Show that, for any (fixed) values $C, k$, this equation has at most a finite number of solutions $x, y$;\r\n\r\n3. Show that, for any $C, n$, there exists $k = k(C,n)$ such that the equation has more than $n$ solutions $x, y$.\r\n\r\n [hide=\"Solution\"]\n\n The equation can be written $(x-k)(y-k) = k^{2}+C$. In order to have solution(s) we\nneed $x-k$ and $y-k$ of same sign. Now, if both $x \\leq k$ and $y \\leq k$, then \n$(x-k)(y-k) = (k-x)(k-y) < k\\cdot k = k^{2}< k^{2}+C$, therefore we need have \n$x > k$ and $y > k$. But if either of them is greater than $k^{2}+k+C$, again there is no\nsolution, so the number of possible solutions is finite. As $k^{2}+k+C$ and $k+1$ are a solution, we have \nanswered affirmatively to the stated assertion of the problem. In fact, the number of distinct positive integer\nsolutions is equal to the number of positive divisors of $k^{2}+C$. $\\square$\n\n\nIn order to prove 3., we shall use the {\\it lemma} that\n$m^{3}+1 = (m+1)[(m+1)(m-2)+3]$, so a common divisor of the two factors can be at most 3, but this \ncannot happen when $m$ is itself divisible by 3, therefore the two factors are relatively prime. \nMoreover, for $m > 1$, both factors are greater than 1. By repeated application, for $m_{0}= m$, $m_{i+1}= m_{i}^{3}$,\nit follows that $m_{n}+1$ has more than $n$ relatively prime factors.\n\nLet us now take\n\\[p = C\\cdot 3^{3^{n}}, \\quad{ }m = C\\cdot 3^{3^{n}+1}, \\quad{ }\\textrm{and}\\quad{ }k = p^{\\frac{3^{n}+1}{2}}, \\]\ntherefore $k^{2}+C = C(m^{3^{n}}+1)$, where $m$ is seen to be divisible with 3. According to the above lemma,\nthis will have more than $n$ relatively prime divisors. \\hfill $\\blacksquare$\n\n[/hide]\n\n [b] Problem 2 [/b]\n\n A number of $n > m \\geq 1$ soccer teams play a full tournament, each team meeting (once) each other.\nPoints are awarded: 2 for a victory, 1 for a tie and 0 for a loss. At the end, each team has won half of its points\nagainst the $m$ teams placed last (including each of these teams, who won half of its points against the other $m-1$).\n\nFind all possible values for $n$ and $m$, supported with examples of such tournaments.\n\n [hide=\"Solution.\"]\n\n Let us call the last $m$ teams {\\it the losers}, and the other $n-m$ teams {\\it the winners}. \nIt is clear that the losers together have won 2(2${m}\\choose{2}$) points, while\nthe winners together have won 2(2${n-m}\\choose{2}$) points; but in all there are 2${n}\\choose{2}$ \npoints awarded, hence we have the equation\n2(2${m}\\choose{2}$) + 2(2${n-m}\\choose{2}$) = 2${n}\\choose{2}$\nwhich yields $n^{2}-(4m+1)n+4m^{2}= 0$, or $(n-2m)^{2}= n$, so we need have $n = (k+1)^{2}$, with $k \\geq 1$.\nBut at least one loser won not less than $2(m-1)$ points, while at least one winner has won at most $2(n-m-1)$\npoints; this implies $n \\geq 2m$, therefore $n-2m = k+1$, so $m = k(k+1)/2$ and $n-m = (k+1)(k+2)/2$.\n\nWe will give an example of high symmetry: assume all losers tied the matches between themselves, and similarly,\nall winners tied the matches between themselves; this gives that each loser won $2(m-1) = (k-1)(k+2)$ points,\nwhile each winner won $2(n-m-1) = k(k+3)$ points. We shall now consider the possible cases:\n\n{\\bf k even}: arrange the winners in a rectangular array of $(k+1)$ rows and $(k+2)/2$ columns; there will be\n${k+1}\\choose{k-1}$ = ${k+1}\\choose{2}$ = $k(k+1)/2 = m$ groups of $(k-1)$ rows. Establish a one-to-one\nmapping between the losers and these groups, and have each loser tie the winners in the corresponding group, and lose to the others.\n\n{\\bf k odd}: arrange the winners in a rectangular array of $(k+1)/2$ rows and $(k+2)$ columns; there will be\n${(k+1)/2}\\choose{(k-1)/2}$ = ${(k+1)/2}\\choose{1}$ = $(k+1)/2 = m/k$ groups of $(k-1)/2$ rows. Establish a one-to-one\nmapping between the partition of the losers in $m/k$ classes of $k$ losers each and these groups, and have each loser tie \nthe winners in the corresponding group, and lose to the others.\nThus, each loser ties $(k-1)(k+2)/2$ and loses the other $(k+2)$ of these cross-matches, accounting for\n$(k-1)(k+2)/2$ points, while each winner ties\n$k(k-1)/2$ and wins the other $k$ of these cross-matches, accounting for $k(k+3)/2$ points, as needed. \\hfill $\\blacksquare$\n\n {\\bf Remarks.} This problem is based on a Swiss problem, where $m = 10$ was given (therefore $k = 4$);\nthus $n$ came out to be 25; but the examples given there lacked the grouping idea behind. Our implementation both\ngeneralizes the problem and offers a generic way of constructing the required examples.\n \n [/hide]\n\n [b] Problem 3. [/b]\n\n Consider the isosceles triangle $ABC$ with $AB = AC$, and $M$ the midpoint of $BC$.\nFind the locus of the points $P$ interior to the triangle, for which $\\angle BPM+\\angle CPA = \\pi$.\n\n [hide=\"Solution\"]\n \n When asked during the contest, the problem started with the following helping preliminary question\n(omitted in this published version), which we will use as a {\\it lemma:}\n\n{\\it Prove that if, for a point $P$ interior to the triangle, $\\angle ABP = \\angle BCP$, then $\\angle BPM+\\angle CPA = \\pi$.}\n\nThe proof starts by noticing that the configuration is fully symmetrical:\n\\[\\angle ABP = \\angle BCP \\quad{ }\\textrm{iff}\\quad{ }\\angle ACP = \\angle CBP\\]\n\\[\\angle BPM+\\angle CPA = \\pi \\quad{ }\\textrm{iff}\\quad{ }\\angle CPM+\\angle BPA = \\pi\\]\nand also that $P \\in AM$ guarantees the result, therefore we may assume wlog (for construction's sake)\nthat $P$ is interior to the triangle $ABM$. The given angle equality is readily seen to be equivalent with $P \\in \\Gamma$, \nwhere $\\Gamma$ is the arc interior to $ABC$ of the circumcircle $\\mathcal K$ of triangle $BCI$, with $I$ the incenter of $ABC$.\nIt is also immediately seen that $AB$ and $AC$ are tangent to $\\mathcal K$. \n\nConsider the Apollonius circle $\\mathcal A$ for points $A$, $M$ and ratio $BA/BM$; \ndenote by $U$ and $V$ the points where $AM$ meets $\\mathcal A$. Clearly, $B, C \\in \\mathcal A$. Then $BA/BM = UA/UM = VA/VM$, \nfrom which follows $BA^{2}/BM^{2}= UA\\cdot VA/UM\\cdot VM = UA\\cdot VA/BM^{2}$, by symmetry and the power of a point relation, so\n$BA^{2}= UA\\cdot VA$, hence $AB$ is tangent to $\\mathcal A$, again by the power of a point relation. Therefore the two\ncircles $\\mathcal K$ and $\\mathcal A$ coincide. Now, prolong $AP$ until it meets $\\mathcal K$ again at $Q$, and prolong\n$PM$ until it meets $\\mathcal K$ again at $R$. Then $BA/BM = PA/PM = QA/QM$, therefore $BA^{2}/BM^{2}= PA\\cdot QA/PM\\cdot QM = BA^{2}/PM\\cdot QM$, by the power of a point relation, so $BM^{2}= PM\\cdot QM$. But $BM^{2}= PM\\cdot RM$, again by the power\nof a point relation, so $QM = RM$, hence $BQ = CR$. It follows that $\\angle BPQ = \\angle CPR$, and this is enough\nto yield $\\angle BPM+\\angle CPA = \\pi$. \\hfill $\\square$\n\n\nAlternatively, one can calculate ratios using the symmetrical points $P'$ and $Q'$ with respect to $AM$, and \ndenoting by $N$ the meeting point of $PQ'$ and $P'Q$. It can be obtained that $AM = AN$, therefore $M \\equiv N$\nand the rest easily follows as above. \n\n\\bigskip\n\nWe claim the locus is the arc $\\Gamma$ (defined in the above), together with the (open) segment $(AM)$. \nClearly $P \\in (AM)$ fills the bill, so from now on we will assume $P \\notin (AM)$. Also, as above, we will\nassume $P$ interior to the triangle $ABM$ (otherwise we work with the symmetrical relations).\n\nAssume $P \\notin \\Gamma$, equivalent to $\\angle BPC \\neq \\pi-\\angle ABC$; then $AB$ and $AC$ are not tangent to $\\mathcal K$. \nTake $B'$ and $C'$ to be the tangency points on $\\mathcal K$ from $A$, and $M'$ to be the midpoint of $B'C'$.\nWe are now under the conditions from the lemma (for triangle $AB'C'$), therefore $\\angle B'PM'+\\angle C'PA = \\pi$.\nBut $\\angle B'PM' = \\angle BPM+\\delta(\\angle B'PB+\\angle MPM')$ and $\\angle C'PA = \\angle CPA-\\delta\\angle C'PC$,\nwhere $\\delta = 1$ if $\\angle BPC < \\pi-\\angle ABC$, respectively $\\delta =-1$ if $\\angle BPC > \\pi-\\angle ABC$.\nWe have $\\angle B'PB = \\angle C'PC$ from the symmetry of the configuration, and $\\angle BPM+\\angle CPA = \\pi$ given;\nthese relations therefore imply $\\angle MPM' = 0$, which can only happen when $M$, $M'$ and $P$ are collinear, \ni.e. $P \\in AM$, which was ruled out from the start in this part of the proof. The contradiction thus reached \nconfirms our claim. $\\blacksquare$\n\nAlternatively, when $\\angle BPC > \\pi/2$, construct the point $A'$ as the apex of the isosceles triangle of base $BC$ and \nequal angles of measure $\\angle PBC+\\angle PCB < \\pi/2$, on the same side of $BC$ with $A$; \nthen $P$ is like in the lemma (for triangle $A'BC$), \ntherefore $\\angle BPM+\\angle CPA' = \\pi$, hence $\\angle CPA = \\angle CPA'$ and so $A \\equiv A'$, as $P$, $A$ and $A'$\nare not collinear. But this is clearly indicating that $P \\in \\Gamma$.\n\nShowing that if $\\angle BPC \\leq \\pi/2$, then $\\angle BPM+\\angle CPA < \\pi$, will finish the proof; this follows\nfrom noticing that, considering $\\Gamma_{M}$ the circumcircle of triangle $BPM$, $\\Gamma_{M}$ will meet $(AM)$ at a point $X$\n(in fact we even have $X \\in (AH)$, with $H$ the orthocenter of $ABC$). \nAlso, considering the (open) hemicircle $\\gamma$ of diameter $BC$ and on the same side of $BC$ with $A$, \n$\\Gamma_{M}$ will meet $\\gamma$ at a point $Y$ ($X$ may even coincide with $Y$, when $CX$ is tangent to $\\Gamma_{M}$).\n\nBut the points $C$, $X$ and $Y$ are collinear, so the point $P$, being on $\\Gamma_{M}$ and interior to $ABM$, will be such that\n$X$ is within the triangle $CPA$, therefore $\\angle CXA > \\angle CPA$. But $X$ and $P$ being on $\\Gamma_{M}$\nyields $\\angle BPM = \\angle BXM$, while $X$ being on $(AM)$ yields $\\angle BXM+\\angle CXA = \\pi$. Together\nthese add up to $\\angle BPM+\\angle CPA < \\pi$, as needed. \n\n\n[b] Remarks. [/b] This problem is based on a Moldavian problem (the preliminary question), to which the locus\nquestion has been added (as kind of a converse proposition).\n\n\n [/hide]", "Solution_1": "i`ll give now another solution to problem 3 :)\r\n\r\nso..let`s take $\\angle BPM=\\alpha$ and $\\angle CPM = \\beta$ so $\\angle APC=180-\\alpha$ and $\\angle APB=180-\\beta$\r\n\r\nwe consider $\\angle CBP = u$ and $\\angle BCP = v$\r\n\r\napplying the law of sin in triangles $ABP$ and $ACP$ we have:\r\n\r\n$\\frac{l}{\\sin({180-\\beta})}=\\frac{AP}{\\sin({B-u})}$ and\r\n\r\n$\\frac{l}{\\sin({180-\\alpha})}=\\frac{AP}{\\sin({C-v})}$ , where $l=AB=AC$\r\n\r\nby dividing the two relations we obtain that :\r\n\r\n$\\frac{\\sin{\\alpha}}{\\sin{\\beta}}=\\frac{\\sin({C-v})}{\\sin({B-u})}$ $(\\star)$ (we used the fact that $\\sin({180-x})=\\sin x$)\r\n\r\n\r\nnow let`s apply the law of sin in triangles $BPM$ and $CPM$:\r\n\r\n$\\frac{PM}{\\sin{u}}=\\frac{BM}{\\sin{\\alpha}}$\r\n\r\n$\\frac{PM}{\\sin{v}}=\\frac{CM}{\\sin{\\beta}}$\r\n\r\nby dividing the two relations we obtain that:\r\n\r\n$\\frac{\\sin{\\alpha}}{\\sin{\\beta}}=\\frac{\\sin{u}}{\\sin{v}}$ $(\\star\\star)$\r\n\r\nby $(\\star)$ and $(\\star\\star)$ we have that : $\\frac{\\sin{u}}{\\sin{v}}=\\frac{\\sin({C-v})}{\\sin({B-u})}$\r\n\r\n$\\Rightarrow \\sin{u}\\cdot \\sin(B-u) = \\sin{v}\\cdot \\sin(C-v)$\r\n\r\nit is well known that we have the following relation : $\\sin{x}\\sin{y}=\\frac{\\cos(x-y)-\\cos(x+y)}{2}$\r\n\r\nso : $\\cos(B-2u)-\\cos{B}= \\cos(C-2v)-\\cos{C}$, but $B=C$\r\n\r\n$\\Rightarrow \\cos(B-2u) = \\cos(B-2v)$\r\n\r\n$1.$ $B-2u = B-2v$ $\\Rightarrow u=v$ so $P \\in (AM)$\r\n\r\n$2.$ $B-2u = 2v-B$ $\\Rightarrow B=u+v$ so $v=\\angle ABP \\Rightarrow \\angle ABP+\\angle ACP = \\angle PBC+\\angle PCB \\Rightarrow P$ is on arc $BIC$ \r\n\r\nthe end :oops:" } { "Tag": [ "algebra", "binomial theorem", "number theory unsolved", "number theory" ], "Problem": "Assume that the equation x^2 - d*y^2 = -1 is solvable, with least positive solution x_0 + y_o*sqrt(d) > 1.\r\n\r\nLet also x_1 + y_1*sqrt(d) be a solution to x^2 - d*y^2 = 1.\r\n\r\nShow that (x_1 + y_1*sqrt(d)) = (x_0 + y_0*sqrt(d))^2.", "Solution_1": "Ok, let me rephrase the question.\r\nSuppose $d$ is not a perfect square. Let $(x, y) = (x_1, y_1)$ be the pair of smallest positive integers satisfying $x^2 - dy^2 = -1$. Let $(a,b) = (a_1, b_1)$ be the pair of smallest positive integers satisfying $a^2 - db^2 = 1$. Show that $a_1 + b_1 \\sqrt{d} = {(x_1 + y_1 \\sqrt{d})}^2$.\r\n\r\nLet $x_2 + y_2 \\sqrt{d} = {(x_1 + y_1 \\sqrt{d})}^2$. Note that by Binomial Theorem and by grouping the rational and irrational terms together, we also have $x_2 - y_2 \\sqrt{d} = {(x_1 - y_1 \\sqrt{d})}^2$. Thus ${x_2}^2 - d{y_2}^2 = {(x_1 + y_1 \\sqrt{d})}^2 {(x_1 - y_1 \\sqrt{d})}^2 = ({x_1}^2 - d{y_1}^2)^2 = 1$. Thus $(a,b) = (x_2, y_2)$ is a solution to $a^2 - db^2 = 1$.\r\n\r\nNow note that ${a_1}^2 - d{b_1}^2 = 1$ by definition, thus $a_1 \\leq x_2, b_1 \\leq y_2$. Suppose $a_1 < x_2, b_1 < y_2$, then since $a_1 + b_1 \\sqrt{d} > 1$, $\\exists$ some $m \\in \\mathbb{Z}^+$ such that \\[ {(a_1 + b_1 \\sqrt{d})}^m \\leq x_2 + y_2 \\sqrt{d} < {(a_1 + b_1 \\sqrt{d})}^{m+1}. \\] Multiplying throughout by ${(a_1 - b_1 \\sqrt{d})}^m$, we get \\[ 1 \\leq (x_2 + y_2 \\sqrt{d}){(a_1 - b_1 \\sqrt{d})}^m < a_1 + b_1 \\sqrt{d}. \\] Let $(x_2 + y_2 \\sqrt{d}){(a_1 - b_1 \\sqrt{d})}^m = a_0 + b_0 \\sqrt{d}$ $\\Rightarrow (x_2 - y_2 \\sqrt{d}){(a_1 + b_1 \\sqrt{d})}^m = a_0 - b_0 \\sqrt{d}$. Thus we have ${a_0}^2 - d{b_0}^2 = ({x_2}^2 - d{y_2}^2){({a_1}^2 - d{b_1}^2)}^m = 1$.\r\n\r\nSo $(a,b) = (a_0, b_0)$ is a solution of $a^2 - db^2 = 1$ such that $1 \\leq a_0 + b_0 \\sqrt{d} < a_1 + b_1 \\sqrt{d}$, which contradicts the minimality of $(a_1, b_1)$.\r\n\r\n$\\therefore a_1 + b_1 \\sqrt{d} = {(x_1 + y_1 \\sqrt{d})}^2$." } { "Tag": [], "Problem": "[b]Find roots of equation ![/b]\r\nx^4 -5x^3+4x^2-2x+2=0", "Solution_1": "By the rational roots theorem, we see that the possible roots are $ \\pm1, \\pm2$.\r\n\r\nBy plugging in $ 1$ in synthetic division, we see that it's a root. So we can factor it to:\r\n\r\n$ (x\\minus{}1)(x^3 \\minus{}4x^2 \\minus{}2) \\equal{} 0$\r\n\r\nNow for $ x^3 \\minus{}4x^2 \\minus{}2$, there is 1 real root and 2 complex roots. I don't know if you are solving for reals or complex...", "Solution_2": "$ x^3 \\minus{} 4x^{^2 } \\minus{} 2 \\equal{} 0$ \r\nMaybe, we will find the real root with formula [b]Cadano.[/b]/", "Solution_3": "It's irrational and somewhere between 4 and 5. You can use a graphing calculator or Wolfram Alpha, but not by hand." } { "Tag": [], "Problem": "since long that i haven't been here.\r\njust had exams.\r\n\r\nhere is the question.\r\nA man have head 40 Cows and birds.\r\nhe counts the legs of them 148 totallly.\r\n\r\nhow much cows and birds does he have ?", "Solution_1": "[quote=\"most-wanted\"]since long that i haven't been here.\njust had exams.\nhere is the question.\nA man have head 40 Cows and birds.\nhe counts the legs of them 148 totallly.\nhow much cows and birds does he have ?[/quote]\r\nI used to do questions like this all the time. :) \r\nLet x be the no. of cows, and y be the no. of birds. So,\r\n$4x+2y=148$-->$2x+y=74$\r\n$x+y=40$\r\nSo, sub the first in the second to get $x=34$\r\nAnd then $y=6$.\r\nSo, 6 birds and 34 cows. :)", "Solution_2": "i am not sure how you got that.\r\ncan you please explain it in details.\r\nhow you got 2x + y = 74 ?", "Solution_3": "[quote=\"most-wanted\"]i am not sure how you got that.\ncan you please explain it in details.\nhow you got 2x + y = 74 ?[/quote]\r\nWell, a cow has 4 legs, and a bird 2 (unless they are mutated...), so we get the equation\r\n4x+2y=148, where x is the no. of cows and y the no. of bids. The common factor is 2, so we divide both sides of the equation by it, to get\r\n2x+y=74. Which, is the simplfied form. :)", "Solution_4": "thanks i would remember that always by now :lol:", "Solution_5": "no problem. :)", "Solution_6": "[hide]using elimination, $\\boxed{34}$ cows and $\\boxed{6}$ birds[/hide]", "Solution_7": "[hide]Call cows $x$ and birds $y$. We can set up the equations $4x+2y=148$ and $x+y=40$. Solving we find $(x,y)=(34,6)$.[/hide]", "Solution_8": "[hide]$4x+2y=148$\n$x+y=40$\n\n$x=34$\n$y=6$\n\n$(34,6)$[/hide]", "Solution_9": "[quote=\"most-wanted\"]since long that i haven't been here.\njust had exams.\n\nhere is the question.\nA man have head 40 Cows and birds.\nhe counts the legs of them 148 totallly.\n\nhow much cows and birds does he have ?[/quote]\r\n[hide=\"for those who do not get systems\"]If they all are cows there would be 160 legs. There are two less legs per member. Twelve legs means six birds and 40-6=34 cows.[/hide]", "Solution_10": "[hide]\nusing systems of eqations and then using elimination i get 34 and 6[/hide]", "Solution_11": "[quote=\"most-wanted\"]since long that i haven't been here.\njust had exams.\n\nhere is the question.\nA man have head 40 Cows and birds.\nhe counts the legs of them 148 totallly.\n\nhow much cows and birds does he have ?[/quote]\r\n\r\n[hide]\nc+b=40\n4c+2b=148\n\n4c+4b=160\n4c+2b=148\n\n2b=12\nb=6\nc=34[/hide]" } { "Tag": [], "Problem": "When a positive three-digit dividend with a units digit of 2 is divided by a positive one-digit divisor, the result is a whole number quotient with a remainder of 1. How many distinct values are possible for this three-digit dividend?", "Solution_1": "[hide=\"Hint\"]\nThis is the same as if the units digit was 1 and the remainder was 0. Then immediately, you know that the divisors cannot be some values, like 2, 4, 5, 6, 8.[/hide]", "Solution_2": "[hide]there are 30 3-digit with lats digit as 1 divisible by 3\n7*23=161, ...7*133=931\nso there are 12 divisible by 7\n\n30+12=42\n\nbu the answer should be 38\n\nwhat did I miss?[/hide]", "Solution_3": "how did you conclude that their are 30 three digit numbers that are divisible by 3? Also since there are numbers like 231 which are divisible by three and seven their could be some overlaps.", "Solution_4": "I got it \r\n\r\n[hide]111,141,171\n201,231,261,291\n321,351,381\n411,441,171\n501,531,561,591\n621,651,681\n711,741,771\n801,831,861,891\n921,951,981\n\n7*23=161, ... 7*133=931\nyou are right, there are 4 overlap (33,63,93,123)\n\nso 30+12 -4=38 :P[/hide]" } { "Tag": [ "inequalities", "trigonometry", "inequalities proposed" ], "Problem": "Consider $ a_{1}$, $ a_{2}$, $ \\ldots$, $ a_{10}$ pozitive real numbers such that $ a_{1}\\plus{}a_{2}\\plus{}\\ldots\\plus{}a_{10}\\equal{}1$. Prove that\r\n\\[ \\sqrt{a_{1}a_{2}}\\plus{}\\sqrt{a_{2}a_{3}}\\plus{}\\ldots\\plus{}\\sqrt{a_{9}a_{10}} \\leq \\cos{\\frac{\\pi}{11}}.\\]", "Solution_1": "I think if you use mixing variables, you can get maximum occurs when $ a_1 \\equal{} a_3 \\equal{} ... \\equal{} a_9$ and $ a_2 \\equal{} a_4 \\equal{} ... \\equal{} a_10$. From this I think the maximum occurs when all the variables are equal, so the value is 0.9 < cos pi/11... this looks strange. Is there any case where the max is > 0.9?" } { "Tag": [ "geometry", "incenter", "geometry unsolved" ], "Problem": "Let a tangential quadrangle $ ABCD$; $ E,F$ in $ BC$; $ AE\\cap DF \\equal{} G$; $ H,I,J,K$ are incenters of triangles $ ABE,EFG,CDF,DAG$. Prove that: $ HIJK$ is a cyclic quadrangle.", "Solution_1": "I also see that: If $ AE\\cap HK \\equal{} M,DF\\cap JK \\equal{} N$ then $ AMND$ is a tangential quadrangle, but I can prove it. Anybody can?", "Solution_2": "Two non-intersecting directed circles (with no common interior points) have only 2 (directed) common tangents. For example, if both circles are directed counter-clockwise, their directed common tangents are the common external tangents of undirected circles. If one circle is directed counter-clockwise and the other one clockwise, their directed common tangents are the common internal tangents of undirected circles. Two intersecting directed circles have common tangents only if they have the same direction. \r\n\r\n[color=red]Theorem: [/color] Let $ \\mathcal A, \\mathcal B, \\mathcal C, \\mathcal D$ be four directed circles. Let $ a, a'$ be the directed common tangents of $ \\mathcal A, \\mathcal B,$ let $ b, b'$ be the directed common tangents of $ \\mathcal B, \\mathcal C,$ let $ c, c'$ be the directed common tangents of $ \\mathcal C, \\mathcal D,$ and let $ d, d'$ be the directed common tangents of $ \\mathcal D, \\mathcal A.$ Suppose the directed lines $ a, b, c, d$ are tangent to a single directed circle $ \\mathcal I.$ Then the directed lines $ a', b', c', d'$ are also tangent to a single directed circle $ \\mathcal I'.$\r\n\r\n[hide=\"See the well known dual theorem.\"] Let $ \\mathcal X, \\mathcal Y, \\mathcal Z, \\mathcal T$ be four (undirected) circles. Let $ X, X'$ be intersections of $ \\mathcal X, \\mathcal Y,$ let $ Y, Y'$ be intersections of $ \\mathcal Y, \\mathcal Z,$ let $ Z, Z'$ be intersections of $ \\mathcal Z, \\mathcal T,$ and let $ T, T'$ be intersections of $ \\mathcal T, \\mathcal X.$ Suppose the points $ X, Y, Z, T$ are concyclic. Then the points $ X', Y', Z', T'$ are also concyclic.[/hide]\r\n\r\nThe proof is a bit tedious without reference to a figure. So, assume $ a, b, c, d$ form a convex quadrilateral $ ABCD,$ where $ A \\equiv d \\cap a,$ $ B \\equiv a \\cap b,$ $ C \\equiv b \\cap c,$ $ D \\equiv c \\cap d,$ so that $ a \\equiv \\overline{AB},$ $ b \\equiv \\overline{BC},$ $ c \\equiv \\overline{CD},$ $ d \\equiv \\overline{DA}$ and the circles $ \\mathcal A, \\mathcal B, \\mathcal C, \\mathcal D$ are all inside this quadrilateral and directed counter-clockwise, same as the directed circle $ \\mathcal I.$ Denote intersections of the other directed common tangents $ A' \\equiv d' \\cap a',$ $ B' \\equiv a' \\cap b',$ $ C' \\equiv b' \\cap c',$ $ D' \\equiv c' \\cap d',$ so that $ a' \\equiv \\overline{B'A'},$ $ b' \\equiv \\overline{C'B'},$ $ c' \\equiv \\overline{D'C'},$ $ d \\equiv \\overline{A'D'}.$ Let $ \\mathcal A$ be tangent to $ d, a$ at $ P_1, P_2$ and to $ d', a'$ at $ P_1', P_2'.$ Similarly, let $ \\mathcal B$ be tangent to $ a, b$ at $ Q_1, Q_2$ and to $ a', b'$ at $ Q_1', Q_2'.$ Cyclically, define the tangency points $ R_1, R_2, R_1', R_2'$ of $ \\mathcal C$ and $ S_1, S_2, S_1', S_2'$ of $ \\mathcal D.$\r\n\r\nSuppose that $ ABCD$ is tangential, i.e., the directed lines $ a, b, c, d$ touch a single directed circle $ \\mathcal I.$ Then\r\n\r\n$ \\overline{AB} \\plus{} \\overline{CD} \\equal{} \\overline{BC} \\plus{} \\overline{DA}.$\r\n\r\nSubtracting $ \\overline{P_1A} \\equal{} \\overline{AP_2}, \\overline{Q_1B} \\equal{} \\overline{BQ_2},$ etc., it follows that\r\n\r\n$ \\overline{P_2Q_1} \\plus{} \\overline{R_2S_1} \\equal{} \\overline{Q_2R_1} \\plus{} \\overline{S_2P_1}.$\r\n\r\nThese are tangent lengths between tangency points, therefore\r\n\r\n$ \\overline{Q_1'P_2'} \\plus{} \\overline{S_1'R_2'} \\equal{} \\overline{R_1'Q_2'} \\plus{} \\overline{P_1'S_2'}.$\r\n\r\nSubtracting $ \\overline{A'P_1'} \\equal{} \\overline{P_2'A'}, \\overline{B'Q_1'} \\equal{} \\overline{Q_2'B'},$ etc., yields\r\n\r\n$ \\overline{B'A'} \\plus{} \\overline{D'C'} \\equal{} \\overline{C'B'} \\plus{} \\overline{A'D'},$\r\n\r\nso that $ A'B'C'D'$ is also tangential, directed lines $ a', b', c', d'$ are tangent to a single directed circle. QED.\r\n\r\nAs to the posted problem: The circles $ \\mathcal D, \\mathcal A$ are of zero radius, degenerated to points $ D, A.$ Their common tangents $ d \\equiv \\overline{DA}, d' \\equiv \\overline {AD}$ coincide in position, but they still have opposite directions. The incircles $ (H), (J)$ of $ \\triangle ABE, \\triangle CDF$ are the circles $ \\mathcal B, \\mathcal C.$ Let $ b'$ be the common external tangent of the incircles $ (H), (J)$ other that $ b \\equiv \\overline{BC}.$ Let this tangent intersect $ \\overline{EA}, \\overline{DF}$ at $ M, N.$ According to the theorem, quadrilateral $ AMND$ formed by the lines $ d' \\equiv \\overline{AD}, a' \\equiv \\overline{EA}, b' \\equiv \\overline{NM}, c' \\equiv \\overline{DF}$ is tangential. Its incircle is the incircle $ (K)$ of the $ \\triangle DAG.$\r\n\r\nConsider now the general quadrilateral $ MEFN.$ (If $ G$ is inside of $ ABCD,$ $ MEFN$ is reflex.) Let $ k_m, k_e, k_f, k_n$ be the external bisectors of its angles at $ M, E, F, N.$ The incenters of $ \\triangle ABE, \\triangle EFG, \\triangle CDF, \\triangle DAG$ are intersections $ H \\equiv k_e \\cap k_m,$ $ I \\equiv k_e \\cap k_f,$ $ J \\equiv k_f \\cap k_n,$ $ K \\equiv k_n \\cap k_m.$ It follows that $ HIJK$ is cyclic. (This is well known and a simple angle chase.)" } { "Tag": [], "Problem": "My organization wants to make the US more immigrant friendly; so, we are asking you guys/girls what you think should be done to make the US a better place for Korean nationals?", "Solution_1": "hmmm, i can think of one really good way\r\nIMPEACH BUSH GD.", "Solution_2": "lol, get bush out of office" } { "Tag": [ "function", "trigonometry" ], "Problem": "Maybe It is silly math question, But I really want to know\r\nIf I have cos(x)=1/5 or cos(x)=2/3\r\nIs it possible to find x ?", "Solution_1": "Yes. If you want to do it on a calculator, use the $sin^{-1}$ function.", "Solution_2": "Well, in your case, it'd rather be $cos^{-1}$ ;)", "Solution_3": "When using inverse functions on a calculator, keep in mind that the inverse trig functions have limited range, so if you were looking for a really big angle that's well out of the inverse trig function's range, it's not going to give you the right angle.", "Solution_4": "so $cos^{-1}$ is same as arccos ?", "Solution_5": "Yes, i think so.", "Solution_6": "[quote=\"invisal\"]so $cos^{-1}$ is same as arccos ?[/quote]\r\nYes, you can also type [url=http://www.google.com/search?sourceid=navclient&ie=UTF-8&rls=HPIB,HPIB:2005-19,HPIB:en&q=arccos%281%2F3%29]This in google.[/url] (note the answer would be given in radians." } { "Tag": [], "Problem": "Recently I read TPOTO and I found it quite nice but the movie is a bit different. That's usual of course but I think that the unique characteristic is that both the book and the movie are... :!: ...brilliant and marvelus. The plots are also different but they are both unique. The movie is not a copy or a copy that a few pieces are changed or cut but an evolution of the book imo. What do you think?", "Solution_1": "I didn't read the book but I happened to see the movie (which I didn't like thaaaaaaaat much).\r\nHowever, when I was in New York last year, I happened to see the show on Broadway (my professor there told me it has been playing on Broadway forever) and it was really good. When I return to France next year, I'll try to go watch it in Paris.", "Solution_2": "Haven't read the book or seen the movie, but we played the song in band and IT IS AWESOME!!!!!!!! :lol:", "Solution_3": "Do you see the musical?", "Solution_4": "YAY I'M NOT THE ONLY PHANTOM OF THE OPERA GEEK! :lol: \r\n\r\nI personally prefer the movie, but that's just me. Both are very good. The book makes you feel more for Raoul's point of view. The movie... well, not only does it make you almost hate Raoul, but it really lets you see the human side of the Phantom. Plus, with Gerry Butler as the Phantom... there's just no comparison! \r\n\r\nWhat do you think of Emmy Rossum (Christine Daae) in the movie? Some people say her voice was not operatic enough for the role, while others (me included) like her voice because it conveys a gentle, innocent persona." } { "Tag": [ "complex numbers" ], "Problem": "What is $\\sqrt{-2}^{2}$?\r\nIs it $\\sqrt{-2}\\times \\sqrt{-2}=\\sqrt{4}=2$\r\n OR\r\n$\\sqrt{-2}^{2}=(i \\sqrt{2})^{2}=-2$?", "Solution_1": "It would be the latter.\r\n\r\nAlways express in terms of a real number times the imaginary unit.", "Solution_2": "Yes, another way to see it is if $\\sqrt{x}^2, x|x<0$\r\n\r\nThen $\\sqrt{x}^2 = \\sqrt{x}\\sqrt{x} = -\\sqrt{x^2}$", "Solution_3": "the convension is that you cannot do that with negative numbers, i.e. $\\sqrt{a}*\\sqrt{b} \\ne \\sqrt{ab}$ $\\forall a,b<0$, the reason that we run into a problem is that with square roots we always take the positive one, but with negative roots, it is not real there are 2 roots, and you cannot really say that one is larger than the other, so you have to convert them into positive numbers before you can do the radical thing", "Solution_4": "Actually, the point is this:\r\n\r\nEvery complex number (and real numbers are just the special case of complex numbers) have TWO values of the square root. In exponential form, we have $-2=2e^{i(\\pi+2n\\pi)}$, hence $\\sqrt{-2}$ can be either $\\sqrt{2}e^{i\\pi/2}$ or $\\sqrt{2}e^{-i\\pi/2}$.\r\n\r\nIf you multiply SAME values of the root (which is understood as the proper way of squaring), you'll get $2e^{i\\pi}=-2$, and if you multiply DIFFERENT values (which is possible, but then it's not squaring), you'll get $2e^{i0}=2.$\r\n\r\nHence, in the \"identity\" $\\sqrt{-2}^2=\\underline{\\sqrt{-2}\\cdot\\sqrt{-2}}=\\sqrt{(-2)\\cdot(-2)}=2$, the two underlined roots are actually different numbers, though written in the same way, and that's similar to the \"identity\" $\\sqrt{4}^2=\\sqrt{4}\\cdot\\sqrt{4}=2\\cdot(-2)=-4.$", "Solution_5": "[quote=\"jnsoccer66\"]What is $\\sqrt{-2}^{2}$?\nIs it $\\sqrt{-2}\\times \\sqrt{-2}=\\sqrt{4}=2$\n OR\n$\\sqrt{-2}^{2}=(i \\sqrt{2})^{2}=-2$?[/quote]\r\n\r\n\u221a(-2)\u00b2 =2 or (-2).", "Solution_6": "What I mean is that both could be.", "Solution_7": "What I mean is that both could be." } { "Tag": [ "integration", "calculus", "trigonometry", "calculus computations" ], "Problem": "I know it's easy question but can you tell me how do you compute $ \\int xdx\\plus{}ydy\\plus{}zdz$ when $ x\\equal{}acost, y\\equal{}a sint, z\\equal{}t$ i don't know the definition of double integral of type 2 but i guess it's not just puting $ dx\\equal{}acost(\\minus{}sint)$ and so on and all becoming $ \\int t^2/2$ doesn't it??", "Solution_1": "$ x \\equal{} a\\cos t$ so $ dx \\equal{} \\minus{}a\\sin t \\,dt$,\r\n$ y \\equal{} a\\sin t$ so $ dy \\equal{} a\\cos t \\,dt$,\r\nand $ z \\equal{} t$ so $ dz \\equal{} dt$\r\n\r\nTherefore $ \\int_{\\sigma(t)} xdx \\plus{} ydy \\plus{} zdz$\r\n$ \\equal{} \\int_{t_1}^{t_2} (a\\cos t)(\\minus{}a\\sin t \\,dt) \\plus{} (a\\sin t)(a\\cos t \\,dt) \\plus{} (t)(dt)$\r\n$ \\equal{} \\int_{t_1}^{t_2} t \\,dt$" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "let $0\\leq a_{1}\\leq b_{1}$ and sequence $(a_{n})$and$(b_{n})$ defined by $a_{n+1}=\\frac{a_{n}+b_{n}}{2},,b_{n+1}=\\sqrt{a_{n}.b_{n}}$\r\nFind what limit of $a_{n}$ and$b_{n}$ when n---> $\\inf$", "Solution_1": "That's the arithmetic-geometric mean, see some web resources for it.", "Solution_2": "Here's a reference:\r\n[url]http://mathworld.wolfram.com/Arithmetic-GeometricMean.html[/url]", "Solution_3": "thanks you \r\nnsato \r\nZetax" } { "Tag": [ "induction", "number theory proposed", "number theory" ], "Problem": "Prove among $k+m-1$ arbitraty integer number $m \\geq k \\geq 2$ and $m$ divisible $k$ then exist $m$ numbers have sum divisible $k$.\r\nGoodluck", "Solution_1": "Let $m=nk$. We perform induction on $n$. For $n=1$ it's famous theorem of Erdos (look for example here http://www.mathlinks.ro/Forum/viewtopic.php?highlight=combinatorial&t=19496 at Vesselin's post - it is corollary 2). Suposse we have prove it for $n$ -then let's prove it for $n+1$. Due to induction hypothesis from the set of cardinality $(n+1)k-1$ we can take $nk$ numbers with sum divisible by $k$. So also we can take $nk$ numbers with sum divisible by $k$ from the set of cardinality $(n+2)k-1$ and we are left with $2k-1$. But from this set we can take $k$ numbers with sum divisible by $k$ according to base of induction. Hence, all in all, we have chosen $(n+1)k$ with sum divisible by $k$. QED" } { "Tag": [ "rotation" ], "Problem": "Hi everyone.\r\nI want to practice to draw a circle with hand, without any help from any tools. It is very difficluts to draw a complete and smooth one. So do you guys have any hint or idea to help me to do that? :? \r\nThanks in advance.", "Solution_1": "Try putting one of your fingers (maybe your thumb, maybe some other finger) down on the paper like a compass point, hold your pencil comfortably, and then rotate the piece of paper with your other hand. (I'm just thinking this up; I haven't tried the experiment.) \r\n\r\nDoes anyone else have a suggestion?", "Solution_2": "[quote=\"tokenadult\"]Try putting one of your fingers (maybe your thumb, maybe some other finger) down on the paper like a compass point, hold your pencil comfortably, and then rotate the piece of paper with your other hand. (I'm just thinking this up; I haven't tried the experiment.) \n\nDoes anyone else have a suggestion?[/quote]\r\n\r\nI tried that and got more of an oval shape but i guess it sorta works.", "Solution_3": "Whoa. I tried tokenadult's idea and if you do it carefully, it works almost perfectly!\r\nThe only problem I see is if your workspace is fairly limited. That will result in the paper constantly bumping into other things and jiggling and messing things up. And it only works for fairly large circles due to the size of your fingers. Still, it's pretty cool.", "Solution_4": "if you hold your arm about the length of your fingers above the paper and rotate your wrist, some sort of circle is drawn... you need to be more careful with this method if you want a good circle though", "Solution_5": "Maybe it is a kind of practise and experience and feel...My physics teacher can draw a perfect circle without any help on the blackborad. He just stares at the blackboard and then raise his hand and ....complete!", "Solution_6": "[quote=\"shobber\"]Maybe it is a kind of practise and experience and feel...My physics teacher can draw a perfect circle without any help on the blackborad. He just stares at the blackboard and then raise his hand and ....complete![/quote]\r\n\r\nMaybe he's a robot... or a jedi! :starwars:", "Solution_7": "what is Jedi...?", "Solution_8": "i refer you to paladin's comical graphic", "Solution_9": "[quote=\"shobber\"]what is Jedi...?[/quote]\r\n :rotfl: :rotfl: :rotfl: shobber must watch Star Wars movies!!!", "Solution_10": "Replying to Shobber's original question, I think the key to drawing an accurate circle on a blackboard is to THINK about an ideal circle before starting to move the chalk. Don't look at what your hand is doing; just see in your mind a picture of a perfect circle, and what your hand will do will look better than if you think too much like \"Now I have to move here, now I have to move here,\" etc. \r\n\r\nTo answer Shobber's second question, a jedi is a fictional outer-space \"knight\" that strongly influences the story action in a movie in the Star Wars series. No one is really a jedi, but some actors portray jedi in movies.", "Solution_11": "[quote=\"shobber\"]what is Jedi...?[/quote]\r\nhaven't you ever seen star wars???\r\nand also this topic is really random.\r\ni think that if you practice drawing cricles you will get better at it. Draw 50 a day and you will be good at it. ;)", "Solution_12": "Thanks for all your help! :D \r\nI shall follow your suggestions. First: Use my mind as tokenadult suggested. Second: Practice makes progress as math92 suggested. \r\nI hope I shall be able to draw at least smooth circle after practice.", "Solution_13": "[quote=\"shobber\"]Maybe it is a kind of practise and experience and feel...My physics teacher can draw a perfect circle without any help on the blackborad. He just stares at the blackboard and then raise his hand and ....complete![/quote]\r\n\r\nHe suffers from a rare disease that we in America call \"being artistic\". A circle is like any form of art, you can get better with practice but you will reach a limit. Some people can naturally draw a circle, that's just how it is.", "Solution_14": "[quote=\"shobber\"]Thanks for all your help! :D \nI shall follow your suggestions. First: Use my mind as tokenadult suggested. Second: Practice makes progress as math92 suggested. \nI hope I shall be able to draw at least smooth circle after practice.[/quote]\r\nYes, shobber use mind! Be Jedi and use [u]force[/u] to guide drawing technique :D" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Are there $n\\in\\mathbb{N}$ and $p\\geq99$ a prime number so that $n^{10}+p^{n+1}$ is a perfect square?", "Solution_1": "If $n^{10}+p^{n+1}=m^{2}$, then n is even (else $n^{10}+p^{n+1}=2(mod \\ 8)$). It give $m-n^{5}=p^{k},m+n^{5}=p^{n+1-k}$ It give $n^{5}=\\frac{p^{n+1-k}-p^{k}}{2}, 2k\\le n$.\r\n It give $n^{5}\\ge \\frac{p-1}{2}p^{n/2}\\ge 50p^{n/2}$ - contradition." } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Find the integer part of \r\n\r\n$\\frac{1}{\\sqrt{1}+\\sqrt{2}}+\\frac{1}{\\sqrt{3}+\\sqrt{4}}+...+\\frac{1}{\\sqrt{99}+\\sqrt{100}}$", "Solution_1": "the answer is 4 if I am right.\r\njust notice $\\frac 1 {\\sqrt{k}+\\sqrt{k+1}} > \\frac 1 {\\sqrt{k+1}+\\sqrt{k+2}},k \\ge 0$\r\nso we have:\r\n$2\\left(\\frac{1}{\\sqrt{1}+\\sqrt{2}}+\\frac{1}{\\sqrt{3}+\\sqrt{4}}+...+\\frac{1}{\\sqrt{99}+\\sqrt{100}}\\right) > \\frac{1}{\\sqrt{1}+\\sqrt{2}}+\\frac{1}{\\sqrt{2}+\\sqrt{3}}+\\frac{1}{\\sqrt{3}+\\sqrt{4}}+...+\\frac{1}{\\sqrt{99}+\\sqrt{100}}+\\frac{1}{\\sqrt{100}+\\sqrt{101}}= \\sqrt{101} -1>9$\r\nand \r\n$2\\left(\\frac{1}{\\sqrt{1}+\\sqrt{2}}+\\frac{1}{\\sqrt{3}+\\sqrt{4}}+...+\\frac{1}{\\sqrt{99}+\\sqrt{100}}\\right) < \\frac{1}{\\sqrt{1}+\\sqrt{0}}+\\frac{1}{\\sqrt{1}+\\sqrt{2}}+\\frac{1}{\\sqrt{2}+\\sqrt{3}}+\\frac{1}{\\sqrt{3}+\\sqrt{4}}+...+\\frac{1}{\\sqrt{99}+\\sqrt{100}}= \\sqrt{100} =10$\r\nso we have \r\n$4.5<\\frac{1}{\\sqrt{1}+\\sqrt{2}}+\\frac{1}{\\sqrt{3}+\\sqrt{4}}+...+\\frac{1}{\\sqrt{99}+\\sqrt{100}}<5$", "Solution_2": "$\\left. \\begin{array}{l} S_1 = \\frac{1}{{1 + \\sqrt 2 }} + \\frac{1}{{\\sqrt 3 + \\sqrt 4 }} + \\cdots + \\frac{1}{{\\sqrt{99} + \\sqrt{100} }} = \\sum\\limits_{k = 1}^{50} {\\frac{1}{{\\sqrt{2k - 1} + \\sqrt{2k} }}} \\\\ S_2 = \\frac{1}{{\\sqrt 2 + \\sqrt 3 }} + \\frac{1}{{\\sqrt 4 + \\sqrt 5 }} + \\cdots + \\frac{1}{{\\sqrt{98} + \\sqrt{99} }} = \\sum\\limits_{k = 1}^{49} {\\frac{1}{{\\sqrt{2k} + \\sqrt{2k + 1} }}} \\\\ \\end{array} \\right\\} \\Rightarrow$\r\n\r\n$\\Rightarrow S_1 + S_2 = \\frac{1}{{1 + \\sqrt 2 }} + \\frac{1}{{\\sqrt 2 + \\sqrt 3 }} + \\cdots + \\frac{1}{{\\sqrt{99} + \\sqrt{100} }} = 9$\r\n\r\nBut:\r\n$\\left. \\begin{array}{l} \\frac{1}{{\\sqrt 0 + \\sqrt 1 }} > \\frac{1}{{\\sqrt 1 + \\sqrt 2 }} > \\frac{1}{{\\sqrt 2 + \\sqrt 3 }} \\\\ \\frac{1}{{\\sqrt 2 + \\sqrt 3 }} > \\frac{1}{{\\sqrt 3 + \\sqrt 4 }} > \\frac{1}{{\\sqrt 4 + \\sqrt 5 }} \\\\ \\frac{1}{{\\sqrt 4 + \\sqrt 5 }} > \\frac{1}{{\\sqrt 5 + \\sqrt 6 }} > \\frac{1}{{\\sqrt 6 + \\sqrt 7 }} \\\\ \\cdots \\cdots \\cdots \\cdots \\cdots \\cdots \\cdots \\cdots \\cdots \\cdots \\cdots \\\\ \\frac{1}{{\\sqrt{98} + \\sqrt{99} }} > \\frac{1}{{\\sqrt{99} + \\sqrt{100} }} > \\frac{1}{{\\sqrt{100} + \\sqrt{101} }} \\\\ \\end{array} \\right\\} \\Rightarrow$\r\n\r\n$\\Rightarrow 1 + S_2 > S_1 > S_2 + \\frac{1}{{\\sqrt{100} + \\sqrt{101} }} \\Rightarrow$\r\n\r\n$\\Rightarrow 1 + S_1 + S_2 > 2S_1 > S_1 + S_2 + \\frac{1}{{\\sqrt{100} + \\sqrt{101} }} \\Rightarrow$\r\n\r\n$\\Rightarrow 1 + 9 > 2S_1 > 9 + \\frac{1}{{\\sqrt{100} + \\sqrt{101} }} \\Rightarrow$\r\n\r\n$\\Rightarrow 5 > S_1 > 4.5 + 0.04 \\Rightarrow \\left[ {S_1 } \\right] = 4$" } { "Tag": [ "IMO", "IMO 2004" ], "Problem": "Are someone know the results of brazilian team?", "Solution_1": "Well, after many hours with no news, we finally have confirmation of 2 silver and 4 bronze medals for Brazil. Were very happy with this result! Congrats to all!", "Solution_2": "and dont oyu know anything about Argentina? Is there any gold medal from latinamerica", "Solution_3": "ok.......congratulations for your performance........ on which position did you finish?\r\n\r\nPD: Do you know Davi (imo 2001/2002/2003) because i lost his email.....", "Solution_4": "I still dont have new information, except that Brazil made 132 points total, ending up in 21st position, together with Canada. Argentina made 92 points, 39th position. \r\n(see the other post in this forum with the complete country list).\r\n\r\nP.S. Conozco a Davi, te envo un mensaje privado a respecto.", "Solution_5": "BRA 1 - Fbio Moreira - Bronze ( 6 + 7 + 1 + 5 + 1 + 1= 21)\r\n[b]BRA 2 - Gabriel Bujokas - Silver (3 + 6 + 1 + 7 + 5 + 7 = 29) [/b]\r\nBRA 3 - Henry Hsu - Bronze (6 + 1 + 1 + 7 + 3 + 1 = 19) \r\n[b]BRA 4 - Rafael Hirama - Silver (6 + 2 + 2 + 7 + 7 + 2 = 26) [/b]\r\nBRA 5 - Rafael Marini - Bronze (6 + 1 + 1 + 1 + 0 + 7 = 16) \r\nBRA 6 - Thiago Santos - Bronze (7 + 3 + 3 + 6 + 1 + 1 = 21)", "Solution_6": "Hi all!\r\nI'am Luis Hernandez, spanish imo contestant in 01, 02 , 03. \r\nI've got some news about the argentinian team. I met spanish team this morning and they told me yue yang (arg) got a GOLD! 32, just over the cutoff, exactly as Carlos did last year didn't he? I hope a gold for iberoamerica wll become a tradition into the future imos. Maybe one for spain in 2008, when we host the imo!! \r\nLuis", "Solution_7": "[quote=\"luishhh\"]Hi all!\nI'am Luis Hernandez, spanish imo contestant in 01, 02 , 03. \nI've got some news about the argentinian team. I met spanish team this morning and they told me yue yang (arg) got a GOLD! 32, just over the cutoff, exactly as Carlos did last year didn't he? I hope a gold for iberoamerica wll become a tradition into the future imos. Maybe one for spain in 2008, when we host the imo!! \nLuis[/quote] wow, you will host the imo in 2008? I didn't knew that ... it's great news :)", "Solution_8": "Hi Luis... Im Carlos, as far as i know Yue got a better gold than mine...... he got 34.... 751777.... \r\n (es patetico tener que responder en ingles....... como andas?)" } { "Tag": [ "vector", "combinatorics proposed", "combinatorics" ], "Problem": "Suppose $V= \\mathbb{Z}_2^n$ and for a vector $x=(x_1,..x_n)$ in $V$ and permutation $\\sigma$.We have $x_{\\sigma}=(x_{\\sigma(1)},...,x_{\\sigma(n)})$\nSuppose $ n=4k+2,4k+3$ and $f:V \\to V$ is injective and if $x$ and $y$ differ in more than $n/2$ places then $f(x)$ and $f(y)$ differ in more than $n/2$ places.\nProve there exist permutaion $\\sigma$ and vector $v$ that $f(x)=x_{\\sigma}+v$", "Solution_1": "First you should write the other two parts of the question!\r\n1) If $D(x,y)$ be the number of differences between $x$ and $y\\in V$ then the number of $z\\in V$ such that $D(z,x)>{n\\over 2}$ and $D(z,y)>{n\\over 2}$ only depends on $D(x,y)$.\r\nProof is easy!Only you should change the members of vectors.\r\n2) Suppose the number of shuch $z$'s in (1) is $l_k$.($k=D(x,y)$) Then \r\n$l_1 =l_2 >l_3 =l_4 >l_5 =l_6>\\ldots$\r\nProof of this one is also easy.By only transforming each $z$ into another $z$!\r\nNow that we know these we come to prove the main problem.\r\nBecause $n=4k+2$ or $n=4k+3$ , $[{n\\over 2}]=2k+1$ so $[{n\\over 2}]+1=2k$.\r\nNow suppose that $D(x,y)=k$. The number of $z$'s that are good (like the part 2) to $x$ and $y$ are $l_k$. Suppose $D(f(x),f(y))=r$. We have that $f(z)$ is also good to $f(x)$ and $f(y)$ and also the inverse. By part 2 we have $|r-k|<2$.\r\nNow suppose that $D(x,y)=[{n\\over 2}]+1$. Because it's an even number then $D(f(x),f(y))=[{n\\over 2}]$ or $[{n\\over 2}]+1$. But its more than $n\\over 2$. So it is exactly $[{n\\over 2}]+1$. So $D(f(x),f(y))=D(x,y)$\r\nBy the same proof we have if $D(x,y)=[{n\\over 2}]$ then $D(x,y)=D(f(x),f(y))$.\r\nNow suppouse $x\\in V$ and $i\\in \\{1,\\ldots,n\\}$.There exist a $y\\in V$ such that $D(x,y)=[{n\\over2}]+1$ and and differs with $x$ in the $i$'th member.\r\nAlso there exist a $y'$ such that it only differs with $y$ in $i$'th member.\r\nWe have $D(x,y)=D(f(x),f(y))$ and $D(x,y')=D(f(x),f(y'))$.\r\nNow suppose that $I$ is the set of indexes that $f(x)$ and $f(y)$ differ in.\r\nConstruct $I'$ like the $I$ for $f(x)$ and $f(y')$.\r\nWe want to prove that $I'\\subset I$. If it is not then it is easily seen that $I$ and $I'$ differ in more than 2 members. But $D(y,y')=1$ so $D(f(y),f(y'))<3$.\r\nSo $I\\subset I'$.\r\nBy changing $x$ we can make the $y$ to become all the members of $V$.\r\nSo for every $y\\in V$ changing one member causes one change to $f(y)$.\r\nSuppose that $x\\in V$. And suppose that changing the $i$'th member causes the $g(i)$'th member of $f(x)$ to change. We have $g$ is a permutation because $f$ is injective.\r\nWe shall now prove that $g$ is uniqe for all $x\\in V$.\r\nSuppose that it is not.\r\nSo we have $x,y\\in V$ and $i\\in \\{1,\\ldots,n\\}$ such that a change to $x$ in $i$ causes the $j$'th member of $f(x)$ to change, and a change to $y$ causes the $k$'th member.\r\nSuppose that the changed $x$ is $x'$ and the changed $y$ is $y'$.\r\n$D(x,y)=D(x',y')=t$\r\n$|D(f(x'),f(y'))-t|<2$\r\nBut |D(f(x'),f(y'))-t| is even because we have two changes in $f(x)$ and $f(y)$ to obtain $f(x')$ and $f(y')$.\r\nSo $D(f(x'),f(y'))=t$\r\nSo $D(f(x'),f(y))-D(f(x),f(y))=1$ xor $D(f(x),f(y'))-D(f(x),f(y))=1$ and the other one is $-1$.\r\nSuppose the first one is $-1$.\r\nThen it is easily seen that $D(f(x),f(y))-D(x,y)=1$.(Because in other ways we have $|D(f(x'),f(y))-D(x',y)|>1$)\r\nBut from the $D(f(x),f(y'))-D(f(x),f(y))=1$ we get $D(x,y)=D(f(x),f(y))$.(Like the last part)\r\nSo $g$ is uniqe for all $x\\in V$.\r\nNow because that we can change every $x$ to every $y$ by changing members, we have that the problem is proved!", "Solution_2": "Of course Nima you got the best mark from this question.Well done. ;) ;)" } { "Tag": [], "Problem": "I'm just curious.\r\n\r\ni'm pretty puny for an 8th grader.... :(", "Solution_1": "5'11\" and 135\r\n\r\nhowever i can bench press my weight\r\nand dead lift like 225\r\n\r\nnot something many underweight people can do", "Solution_2": "[quote=\"The Herald of Doom\"]I'm just curious.\n\ni'm pretty puny for an 8th grader.... :([/quote]\r\n\r\nArnav, how many grades did you fail?!", "Solution_3": "six\r\npoint nine\r\n\r\ngoing on seven\r\ngod knows", "Solution_4": "DOOM DOOM DOOM", "Solution_5": "size of what? there are many sizes you could give", "Solution_6": "i'm pretty sure he's referring to weight and stuff (unless you're a girl, then you can read between his lines)", "Solution_7": "[quote=\"pianoforte\"]size of what? there are many sizes you could give[/quote]\r\n\r\nIT IS HOW BIG YOU ARE DUH", "Solution_8": "I don't think The Herald of Doom is ArnAv.", "Solution_9": "He might be an even bigger spammer than Arnav. :rotfl:", "Solution_10": "No, I'm pretty sure the moderators agree that HoD is Arnav.\r\n\r\nIn any case pretty soon his rating will be less than Treething's :|...", "Solution_11": ":ninja: :ninja: :ninja: ", "Solution_12": "WHAT DO YOU MEAN", "Solution_13": "6'4'' 190lbs squat 450 lbs but I can only bench press 185lbs twice.", "Solution_14": "[quote=\"solafidefarms\"]No, I'm pretty sure the moderators agree that HoD is Arnav.\n\nIn any case pretty soon his rating will be less than Treething's :|...[/quote]\r\n\r\nIs this because of IP address, or just behavior and stuff?", "Solution_15": "guys guys\r\n\r\nThe Herald of Doom is not ArnAv.", "Solution_16": "3\r\n[color=white]_[/color]" } { "Tag": [ "summer program", "PROMYS", "geometry", "AMC", "USA(J)MO", "USAMO", "inequalities" ], "Problem": "I have a hard time to choose between PROMYS or AoPS Olympiad classes. Right now I solved/wrote up solutions to every PROMYS problems. PROMYS is a fun camp and I want to meet some good math people out there. However PROMYS only focus on number theory(which isn't my favorite area in math). I have heard that AoPS Olympiad Classes are excellent preparations for USAMO. My main goal this summer is to get better at proof-writing, which I am horrible at according to my USAMO score this year. Taking AoPS classes later is an option but I think I should concentrate more on school during school time since my grades this year are horrible. \r\n\r\nSuggestions are appreciated!", "Solution_1": "PROMYS - will pound rigor into you like no other. Very good for proof writing, fun, meet cool people, play games, have an awesome time, etc. However, it does only focus on NT and usually this NT goes \"beyond\" in a sense what is required of the USAMO. To get the full package from PROMYS, all I can say is DO YOUR PROBLEM SETS EVERY DAY. Last year, some people just wasted their time away and probably left with a lot less than most of us. It's a lot of work, but it pays off both in proof writing and pure mathematical dedication.\r\n\r\nAoPS Olympiad Classes - very useful for the USAMO, almost specific to it. Make sure to take Problem Solving and Geometry, though I would say Inequalities is the coolest (it didn't show up on USAMO, though :(). But really these don't take as much time as you might think, especially if you have no other plans over the summer. Even if you take them together it's 5 hours max instruction per week, whereas PROMYS is 6 weeks straight.\r\n\r\nOverall, my opinion would be to go to PROMYS, build up the mentality to take on the USAMO and then, when you get back, take the Olympiad classes in the Fall/Winter/Spring. Oh, that's what I did, come to think of it... and I didn't do so hot on the USAMO, so maybe you should get another opinion.", "Solution_2": "Go to PROMYS. You'll have all year to USAMO prep (and AoPS class if you choose to). You won't have all year to be exposed in person all day to the caliber of students you'll be around at PROMYS. \r\n\r\nIf it's possible to do Olympiad Geom while at PROMYS, you could potentially do both, but if you had to pick one, I'd pick virtually any of the top math camps if you want to be immersed in interesting math and spend lots of time in person with cool people." } { "Tag": [ "geometry", "incenter", "geometry unsolved" ], "Problem": "Let $ ABC$ be a triangle with I as its incenter.Circle $ k$ is centerd at $ I$ and lies inside triangle $ ABC$.Point $ A_1$ lies on $ k$ such that $ IA_1$ is perpendicular to $ BC$.Points $ B_1$ and $ C_1$ are defined anlogously.Prove that $ AA_1,BB_1,CC_1$ are concurrent.", "Solution_1": "Dear Mathlinkers,\r\nthis is the Kariya's theorem.\r\nSincerely\r\nJean-Louis", "Solution_2": "This is an application of isogonal theorem.\r\nSee here for the general: \r\nhttp://www.mathlinks.ro/viewtopic.php?t=284298" } { "Tag": [], "Problem": "A pond is enclosed by a wooden deck that is 3 feet wide. The fence surrounding the deck is 100 feet long. \r\n\r\n(a) If the pond is square, what are its dimensions? \r\n\r\n(b) If the pond is rectangular and the length of the pond is three times its width, what are the dimensions of the pond?", "Solution_1": "[quote=\"sharkman\"]A pond is enclosed by a wooden deck that is 3 feet wide. The fence surrounding the deck is 100 feet long. \n\n(a) If the pond is square, what are its dimensions? \n\n(b) If the pond is rectangular and the length of the pond is three times its width, what are the dimensions of the pond?[/quote]\r\ntry using equations", "Solution_2": "bpms,\r\n\r\nI know you mean well but your hints are not helpful at all. \r\n\r\nWhat equations?\r\n\r\nWhy not set up the equations for me?\r\n\r\nI can take it from there.", "Solution_3": "[quote=\"sharkman\"]bpms,\n\nI know you mean well but your hints are not helpful at all. \n\nWhat equations?\n\nWhy not set up the equations for me?\n\nI can take it from there.[/quote]\r\n\r\n[hide=\"Equations\"](a) Let the side of the square be $x$. Then the length of the fence is $4 x+24$, which is $100$. So $4x+12 = 100$.\n\n(b) Let the dimensions be $y\\times 3y$. Then the length of the fence is $8 y+24$, which is $100$. So $8 y+12 = 100$[/hide]", "Solution_4": "Telling us to set up equations for you is completely useless on your part, sure you can solve equations, but that's not the essence of the problem and you don't learn anything if we just tell you the equations to solve. You might as well do a worksheet of 100 problems such as\r\n\r\n$4x+30=110$\r\n$3x+7=16$\r\netc etc" } { "Tag": [], "Problem": "If you write out the natural numbers one, two, three, four, five, \u2026 , and stop when you get to sixty, how many times have you written the letter \u201cf\u201d?\r\n\r\nA. 32\tB. 33\tC. 41\tD. 42\tE. 43", "Solution_1": "[hide] $ 4$ and $ 5$ are the only digits with an 'f'. Thus, just by considering numbers which end in either $ 4$ or $ 5$, we have $ 13$ 'f's (because $ 15$ has two 'f's). also, every number in the forties will contribute an extra 'f' and every number in the fifties will contribute two extra 'f's. Thus, $ 13\\plus{}10\\plus{}2(10)\\equal{}\\boxed{43}$. Thus, the answer is E.[/hide]" } { "Tag": [], "Problem": "John attended a three-country multicultural festival. At the first exhibit, John paid a $3 admission fee. He then spent half the money he had left on souvenirs. Finally, he spent $3 on food and left. At each of the second and third exhibits, he again paid a $3 admission fee, then spent half of the money he had left for souvenirs, then spent $3 on food, and left. When he left the third exhibit, he had no money remaining. How many dollars did he spend at the festival?\r\n\r\n\r\nI solved this problem by setting up an equation to solve for the original amount of money he had, but it took way too long and I made a lot of silly mistakes. Is there a better way to do this problem?", "Solution_1": "Just read the question \"backwards\". For example, when it says spent $3 on food, just add $3 to the amount you have. When it says used half the money, jsut multiply the amount by 2." } { "Tag": [], "Problem": "Suppose $a,b,c$ are digits. Find $[abbc]$ (where $[ \\hskip .1in ]$ means the concatenated number as opposed to the product) if\r\n\r\n$[abbc] = a-[bb]+[bc]^{2}$.\r\n\r\n[hide=\"hint\"] It hacks. [/hide]", "Solution_1": "[hide=\"Solution\"]\\begin{eqnarray*}&& 1000a+110b+c=a-11b+\\overline{bc}^{2}\\\\ &\\iff& 999a+121b+c=\\overline{bc}^{2}\\\\ &\\iff& 999a+111b=\\overline{bc}^{2}-\\overline{bc}=\\overline{bc}(\\overline{bc}-1)\\\\ &\\iff& 3\\cdot 37(9a+b)=\\overline{bc}(\\overline{bc}-1)\\end{eqnarray*}\n\nHence $\\overline{bc}\\in\\{37,38,74,75\\}$. Out of those, only $37$ and $75$ give RHS which is divisible by $3$:\n\n1. $\\overline{bc}=37\\implies 3(9a+3)=36\\implies a=1$, hence $\\overline{abbc}=1337$\n\n2. $\\overline{bc}=75\\implies 3(9a+7)=150\\implies a\\not\\in\\mathbb{Z}$.\n\nHence the only solution is $\\overline{abbc}=1337$\n\nNOTE: Degenerate cases like $\\overline{bc}=0\\lor\\overline{bc}=1$ (giving $\\overline{abbc}=0000\\lor\\overline{abbc}=0001$) are dismissed. [/hide]" } { "Tag": [ "calculus", "integration", "trigonometry", "calculus computations" ], "Problem": "Hello,\r\n\r\nwho can you integrate this? who can you simplify before you integrate?\r\n\r\n[img]http://img296.imageshack.us/img296/6746/inte3th.jpg[/img]\r\n\r\nThanks.", "Solution_1": "$\\left( \\frac{ds}{dt} \\right)^2 = (2 \\sin t + 2 \\sin 2t)^2 + (2 \\cos t + 2 \\cos 2t)^2 = 4 \\sin^2 t + 8 \\sin t \\sin 2t + 4 \\sin^2 2t + 4 \\cos^2 t + 8 \\cos t \\cos 2t + 4 \\cos^2 2t = 8 + 8 \\left( \\sin t \\sin 2t + \\cos t \\cos 2t \\right) = 8 + 8 \\left( \\cos (2t - t) \\right) = 16 \\frac{1 + \\cos t}{2} = 16 \\cos^2 \\frac{t}{2}$\r\n\r\n$\\frac{ds}{dt} = 4 \\cos \\frac{t}{2}$\r\n\r\nThen $\\int ds = 8 \\sin \\frac{t}{2} + C$. :)", "Solution_2": "Thank you very well now I put in my upper limit 2 pi and my lower limit 0 \r\n\r\nThen I get o for my definte integral but the solution in my book is 16 were do I miss ?", "Solution_3": "Maybe this problem is finding the arc length of Deltoid.", "Solution_4": "yes that is the problem but is the solution in my book not correct?", "Solution_5": "$\\int_0^{2\\pi} \\sqrt{16\\cos ^ 2 \\frac{t}{2}}dt=4\\int_0^{2\\pi} \\left|\\cos \\frac{t}{2}\\right| dt=....$", "Solution_6": "yes but what do you find at the finale solution in other words what is the lenght of the arc.\r\n\r\nThanks.", "Solution_7": "The answer is 16.", "Solution_8": "[img]http://img95.imageshack.us/img95/5157/mo5mh.jpg[/img]\r\n\r\ngreets.", "Solution_9": "As kunny noted (and I forgot to specify), the integrand is not $4 \\cos \\frac{t}{2}$, but rather $4 | \\cos \\frac{t}{2} |$.\r\n\r\nHence the correct integral is $8 \\sin \\frac{t}{2} |_{0}^{\\pi} - 8 \\sin \\frac{t}{2} |_{\\pi}^{2\\pi} = 8 \\sin \\frac{\\pi}{2} + 8 \\sin \\frac{\\pi}{2} = \\boxed{16}$.", "Solution_10": "thank you verry well do someone know what kind of software (freeware) you can use to plot with parametric.\r\n\r\nJust to have some idee of the function.\r\n\r\nGreets." } { "Tag": [], "Problem": "If $ S \\equal{} i^n \\plus{} i^{\\minus{}n}$, where $ i \\equal{} \\sqrt{\\minus{}1}$ and $ n$ is an integer, then the total number of possible distinct values for $ S$ is:\r\n\r\n$ \\textbf{(A)}\\ 1\\qquad \r\n\\textbf{(B)}\\ 2\\qquad \r\n\\textbf{(C)}\\ 3\\qquad \r\n\\textbf{(D)}\\ 4\\qquad \r\n\\textbf{(E)}\\ \\text{more than 4}$", "Solution_1": "If n=0: $ i^0 \\plus{} i^0 \\equal{} 2$.\r\nIf n=1: $ i \\plus{} \\frac {1}{i} \\equal{} i \\minus{} i \\equal{} 0$\r\nIf n=2: $ i^2 \\plus{} \\frac {1}{i^2} \\equal{} \\minus{} 1 \\minus{} 1 \\equal{} \\minus{} 2$\r\nIf n=3: $ i^3 \\plus{} \\frac {1}{i^3}\\equal{}\\minus{}i\\minus{}\\frac{1}{i}\\equal{}\\minus{}i\\plus{}i \\equal{} 0$\r\n\r\nEverything just repeats from n=4 and so forth. This gives a total of $ \\boxed{3}$ distinct values for $ n$.\r\n\r\nAnswer: C", "Solution_2": "[hide=\"Solution\"]$ n\\equal{}0: 2$\n$ n\\equal{}1: 0$\n$ n\\equal{}2: \\minus{}2$\n$ n\\equal{}3: 0$\n\n$ n\\equal{}4: 2$\n$ n\\equal{}5: 0$\n$ n\\equal{}6: \\minus{}2$\n$ n\\equal{}7: 0$\n\nand so on.\n\nTherefore, $ \\boxed{\\textbf{(C)}\\ 3}$.[/hide]\r\n\r\nEdit: Aw, too late." } { "Tag": [ "trigonometry", "geometry", "geometric transformation", "rotation", "analytic geometry" ], "Problem": "A man is standing in a courtyard of [b]square shape[/b]. \r\nThe man then marks his original position in the courtyard, and measures the distances between the mark and three corners of the courtyard one by one, and obtained the measurements of 30 feet, 40 feet and 50 feet respectively.\r\n\r\nQuestion:\r\nWhat is the length of the [b]diagonal [/b]of the courtyard? \r\nNote: The answer should be [b]exact[/b],and no approximation.\r\n\r\nHint:\r\n[hide]You may tackle the problem with any one of the following techniques:\n\n1. Trigonometry, with the knowledge of Cosine Formula and also a basic Identity \n$ sin^2 \\theta \\plus{} cos^2 \\theta \\equal{} 1$. \n\n2. Geometry, with the application of a rotation of figure.\n\n3. Coordinates Geometry, with the knowledge of \"Equation of Circle\".[/hide]", "Solution_1": "Ah I think I get it.\r\n\r\nIf you set this in coordinate geometry you can solve this.\r\nDraw a circle of radius $ 3$,$ 4$,$ 5$ respectively on the vertexes.\r\nFind the unique intersection, that gives you the point.\r\nThen you can use trig to find the angles and then calculate the length of a side of the square.\r\nThe diagonal is $ \\sqrt2$ times the side.\r\n\r\nIs my approach correct?", "Solution_2": "Wait.. I think this is really simple..\r\nDot is where man is standing. \r\nBy the picture, you can tell that diagonal is $ \\boxed{80}$\r\nAm I right??\r\n[asy]size(150); pair A,B,C,D,E; A= (0,10); B= (0,40); C= (-40,0); D= (40,0); E= (0,-40); draw(A--B); draw(A--C); draw(A--D); draw(A--E); draw(B--C); draw(B--D); draw(C--E); draw(D--E); label(\"30\",(5,20)); label(\"40\",(-20,0)); label(\"50\",(5,-20)); dot(A);[/asy]", "Solution_3": "If it was $ 80$ then the edge with length $ 40$ would be on it's midpoint, which is a contradiction that half a diagonal is $ 30$ and $ 50$", "Solution_4": "[quote=\"Fraenkel\"]If it was $ 80$ then the edge with length $ 40$ would be on it's midpoint, which is a contradiction that half a diagonal is $ 30$ and $ 50$[/quote]\r\nOh.. Yeah.. You are right..\r\n\r\nNever mind, then.", "Solution_5": "Fraenkel, you've got the correct procedure.\r\nFind the length of the square first, then multiply it by root 2.\r\nHave you tried to solve it by yourself?\r\nI suggest you should try other methods too. :)" } { "Tag": [ "trigonometry", "inequalities", "trig identities", "Law of Cosines", "AMC" ], "Problem": "A ship sails $ 10$ miles in a straight line from $ A$ to $ B$, turns through an angle between $ 45^{\\circ}$ and $ 60^{\\circ}$, and then sails another $ 20$ miles to $ C$. Let $ AC$ be measured in miles. Which of the following intervals contains $ AC^2$?\r\n[asy]unitsize(2mm);\ndefaultpen(linewidth(.8pt)+fontsize(10pt));\ndotfactor=4;\n\npair B=(0,0), A=(-10,0), C=20*dir(50);\n\ndraw(A--B--C);\ndraw(A--C,linetype(\"4 4\"));\n\ndot(A);\ndot(B);\ndot(C);\nlabel(\"$10$\",midpoint(A--B),S);\nlabel(\"$20$\",midpoint(B--C),SE);\nlabel(\"$A$\",A,SW);\nlabel(\"$B$\",B,SE);\nlabel(\"$C$\",C,NE);[/asy]$ \\textbf{(A)}\\ [400,500] \\qquad \\textbf{(B)}\\ [500,600] \\qquad \\textbf{(C)}\\ [600,700] \\qquad \\textbf{(D)}\\ [700,800]$\r\n$ \\textbf{(E)}\\ [800,900]$", "Solution_1": "[hide]From the law of cosines, we have:\n\n$ AC^2 \\equal{} 100 \\plus{} 400 \\plus{} 2(10)(20)\\cos{\\theta} \\equal{} 500 \\plus{} 400\\cos{\\theta}$\n\nNote that we're adding the cosine term instead of subtracting, because the actual angle between AB and BC is $ 180 \\minus{} \\theta$, so $ \\cos{180 \\minus{} \\theta} \\equal{} \\minus{}\\cos{\\theta}$\n\nWe know that $ 45 \\leq \\theta \\leq 60$, so:\n\n$ \\frac{\\sqrt{2}}{2} \\geq \\cos{\\theta} \\geq \\frac{1}{2}$\n\n$ 200\\sqrt{2} \\geq 400\\cos{\\theta} \\geq 200$\n\n$ 500 \\plus{} 200\\sqrt{2} \\geq 500 \\plus{} 400\\cos{\\theta} \\geq 700$\n\nIf you feel the need to confirm, you can estimate the LHS of the inequality, which is about 782. So our answer is $ \\boxed{D}$\n\nAlso, you could probably just evaluate $ AC^2$ at the extremes, it comes to essentially the same thing.[/hide]" } { "Tag": [ "geometry", "trigonometry" ], "Problem": "A farmer has four straight pieces of fencing: 1, 2, 3, and 4 metres in length. What is the maximum area, in metres, that he can enclose by connecting the pieces? Assume the land is flat\r\n\r\nPlease help me with this question.", "Solution_1": "[hide]\nby Brahmagupta's formula\n\n$A = \\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\\cos^{2}\\theta}$\n\nwhere $\\theta$ is half the sum of the opposite angles and $s$ is the semiperimeter\n\n$s = \\frac{1+2+3+4}{2}= 5$\n\n$A^{2}= (5-1)(5-2)(5-3)(5-4)-(1)(2)(3)(4)\\cos^{2}\\theta$\n$A^{2}= 24-24\\cos^{2}\\theta$\n\n$A$ is maximized when $A^{2}$ is maximized\n$A^{2}$ is maximized when $\\cos^{2}\\theta$ is minimized\n\n$\\cos^{2}\\theta$ is minimized when $\\cos\\theta = 0$ or when $\\theta=90$\n\n$\\theta = 90$ is a valid angle choice because that would mean that the sum of the opposite angles would be $180$, which is possible in a quadrilateral.\n\nplugging in $\\theta=90$ yields\n\n$A^{2}= 24$\n$A = 2\\sqrt{6}$\n[/hide]" } { "Tag": [ "geometry open", "geometry" ], "Problem": "There are given a quadrilateral A1B1C1D1 inscribed in a circle with radii R1 and quadrilateral A2B2C2D2 inscribed in a circle with radii R2.\r\nProve that if \r\n\\sqrt[3]{ab}$.", "Solution_1": "http://www.mathlinks.ro/Forum/viewtopic.php?t=292\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=18698\r\n\r\nNote that using the [url=http://www.mathlinks.ro/Forum/search.php]search function[/url] with, as a keyword, \"russia\" and restricting your search to the NT subforum, you'd have found those quite quickly." } { "Tag": [ "logarithms", "calculus", "calculus computations" ], "Problem": "Say $ \\sum_{n\\equal{}1}^\\infty a_n$ converges. Is it necessarily the case that $ \\sum_{n\\equal{}1}^\\infty a_n \\log (a_n)$ converges?", "Solution_1": "This should be in the calculus computations and tutorials forum. (But don't re-post it -- a moderator will probably move it.) The answer is no. We need an example where convergence rests on something log-size in the terms. So a natural thought is something like $ a_n \\equal{} \\frac {1}{n (\\log n)^2}$. Then $ a_n \\log a_n \\equal{} \\minus{} \\frac {\\log n \\plus{} 2\\log \\log n}{n (\\log n)^2}$, and the associated series does not converge.", "Solution_2": "what does converging mean in math?", "Solution_3": "Let me Wikipedia that for you: http://en.wikipedia.org/wiki/Convergence .", "Solution_4": "A sequence $ (a_{n})$ converges to a point $ x$ if $ \\forall\\varepsilon>0$, $ \\exists N>0$ such that $ \\left|a_{n}\\minus{}x\\right|<\\varepsilon$ $ \\forall n\\geq N$. Basically, you can always get \"most\" of your sequence within any $ \\varepsilon$ of your point of convergence." } { "Tag": [ "geometry", "circumcircle", "trigonometry", "power of a point", "radical axis", "perpendicular bisector", "geometry proposed" ], "Problem": "On the sides $AB$, $BC$, $CA$ of the triangle $ABC$ consider the points $C'$, $A'$, $M$ such that $MA' = MC$ and $MC'=MA$.\r\n Find the geometric locus of the center of the circumcircle of the triangle $A'BC'$.", "Solution_1": "Let a = BC, b = CA, c = AB the sides of the triangle $\\triangle ABC$, H be the orthocenter and R the circumradius. Let K be the intersection of the line MA' with the A-altitude AH and L the intersection of the line MC' with C-altitude CH. Since $\\angle MC'A = \\angle MAC' = \\angle A$, we have $\\angle C'BH = \\angle C'LH = 90^\\circ - \\angle A$, which means that the quadrilateral C'BLH is cyclic. Likewise, since $\\angle MA'C = \\angle MC'A = \\angle C$, we have $\\angle A'BH = \\angle A'KH = 90^\\circ - \\angle C$, which means that the quadrilateral A'BKH is also cyclic. Let (P), (Q) be the circumcircles of the quadrilaterals A'BKH and C'BLH and assume that these 2 circles are different. Then their radical axis is the B-altitude BH. The triangles $\\triangle AMK, \\triangle CML$ are both isosceles with MA = MK, MC = ML: For example, $\\angle MAK = 90^\\circ - \\angle C$ and $\\angle AMK = \\angle MCA' + \\angle MA'C = 2 \\angle C$, hence,\r\n\r\n$\\angle MKA = 180^\\circ - (90^\\circ - \\angle C) - 2 \\angle C = 90^\\circ - \\angle C = \\angle MAK$\r\n\r\nLikewise, $\\angle MCL = 90^\\circ - \\angle A$ and $\\angle CML = \\angle MAC' + \\angle MC'A = 2 \\angle A$, hence,\r\n\r\n$\\angle MLC = 180^\\circ - (90^\\circ - \\angle A) - 2 \\angle A = 90^\\circ - \\angle A = \\angle MCL$\r\n\r\nThe power of the vertex C to the circumcircle (P) of the quadrilateral A'BKH is equal to\r\n\r\n$CA' \\cdot CB = 2\\ CM \\cos \\widehat C \\cdot a = 4\\ CM \\cdot R\\ \\sin \\widehat A \\cos \\widehat C$ \r\n\r\nThe power of the vertex C to the circumcircle (Q) of the quadrilateral C'BLH is equal to\r\n\r\n$CQ \\cdot CH = c\\ \\cot \\widehat C \\cdot 2\\ CM \\cos (90^\\circ - \\widehat A) =$\r\n\r\n$= 4\\ CM \\cdot R\\ \\cot \\widehat C \\sin \\widehat C \\sin \\widehat A = 4\\ CM \\cdot R\\ \\cos \\widehat C \\sin \\widehat A$\r\n\r\nThus the power of the vertex C to the circles (P), (Q) is the same, which means that the vertex C lies on their radical axis. But their radical axis is the B-altitude BH, which does not pass through the vertex C (unless the angle $\\angle C = 90^\\circ$ is right, but then the isosceles triangle $\\triangle MCA'$ would not exist). Consequently, the 2 circles must be identical. Let's call this common circumcircle (P). Since it passes through the points A', B, C', it is the circumcircle of the triangle $\\triangle A'BC'$. Since it passes through 2 fixed points BH, the circumcircles of all triangles $\\triangle A'BC'$ form a pencil with the base points B, H and common radical axis BH. Their centers all lie on the perpendicular bisector of their common chord BH.", "Solution_2": "in http://www.mathlinks.ro/Forum/viewtopic.php?t=59510 \r\n\r\nwas proven by Darij that $A'$, $C'$, $B$ and $H$ are concyclic, where $H$ is the ortocenter of $ABC$.\r\n\r\nso, the circuncircle of $A'C'B$ always pass trought $B, H$.\r\n\r\nthis implies that the circuncenter lies in the bisector of $BH$. \r\n\r\n\r\nE.L" } { "Tag": [], "Problem": "Determine the amount of times the mass of the molecule of water (H2O) is contained in the mass of the molecule of glucose (C6H12O6)\r\n:\r\n\r\nAtomic mass: H = 1 ; O = 16 ; C = 12\r\n\r\na) 1\r\nb)5\r\nc)10\r\nd)15\r\ne) 20", "Solution_1": "Mass of glucose $ C_6 H_{12} O _6$ = $ (12*6) \\plus{} (12*1) \\plus{} (16*6) \\equal{} 180amu$\r\n\r\nMass of water $ H_2 O$ = $ (1*2) \\plus{} 16*1 \\equal{} 18amu$\r\n\r\nSo, 10 times the mass of one molecule of water is contained in one molecule of glucose.\r\nAnswer is c", "Solution_2": "Thank you very much ;)\r\nNow I understand." } { "Tag": [ "geometry", "3D geometry" ], "Problem": "The difference between the cube of an integer and the square of that same integer is 100. What is the integer?\r\n\r\nanyone have a quick and easy method? i could only solve it by brute force and it's an easy problem to take that long...", "Solution_1": "If you remember that all powers of 5 above 1 end in 25, then you can quickly realize that this will always leave a multiple of 100 as the difference.[/hide]", "Solution_2": "I would call this wimp force.\r\n\r\n[hide]x^3 - x^2 = 100\n(x^2)(x - 1) = 100\nperfect square factors of 100 are 4, 10, 25.\nthen by just looking at the numbers it is obvious that x^2 = 25 and x = 5[/hide]", "Solution_3": "It's not really wimpy... it's really the best way to solve the problem without resorting to \"drastic measures,\" don't you think?", "Solution_4": "When you look at this problem, you need to understand that the last two digits have to be the same. If you were more lenient and would only go for the last digit, all the multiples of 5 would occur, nothing else. So try 10 because it is an easy number. Too high? Yes, so it has to be lower, so only 5 works. Doing this problem took all of 10 seconds for me.", "Solution_5": "[quote=\"archimedes1\"]I would call this wimp force.\n\n[hide]x^3 - x^2 = 100\n(x^2)(x - 1) = 100\nperfect square factors of 100 are 4, 10, 25.\nthen by just looking at the numbers it is obvious that x^2 = 25 and x = 5[/hide][/quote]\r\n10 isn't a perfect square... :P" } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Let $ f: R \\mapsto R$ and $ |f(x)| \\leq 1$ for all $ x \\in R$ and $ f(x) \\plus{} f(x\\plus{} \\frac{13}{42})\\equal{} f(x\\plus{} \\frac{1}{6}) \\plus{} f(x\\plus{} \\frac{1}{7})$. Prove that: f is periodic function", "Solution_1": "Put x + 1/6 , x + 2/6 + ... + x + 5/6 instead of x and sum the equations that you have we obtain f(x) + f(x+1+1/7) = f(x+1) + f(x+1/7). Now if we apply the similar process for this equation (Put x + 1/7, x + 2/7, ... , x + 6/7 instead of x) we have the following equation: f(x+2) - f(x+1) = f(x+1) - f(x). If we put a, instead of f(x+1) - f(x); summing gives that f(x+k) = f(x) + k.a . Here, if a=0, we arrive at f(x+1) = f(x). This shows that f is a periodic function. If a doesn't equal to 0 , turn back to the equation. f(x+k) = f(x) + k.a , since f is bounded, f(x+k) is bounded, too. But when k reaches infinity, the right of the equation reaches infinity, too. That is a contradiction. So a must be 0. Therefore f(x+1) = f(x) and it is periodic. :P" } { "Tag": [ "function", "inequalities", "calculus", "derivative", "integration", "limit", "logarithms" ], "Problem": "Let $f\\colon [0,+\\infty) \\to [0,+\\infty)$ a crescent and bijective function. Prove that the sum $\\sum_{n=1}^\\infty{1\\over f(n)}$ converges if and only if the sum $\\sum_{n=1}^\\infty{f^{-1}(n)\\over n^{2}}$ converges, where $f^{-1}$ is the inverse function of $f$.", "Solution_1": "What is the meaining of the word \"crescent\" here? Would it be \"concave upwards,\" or convex in the Jensen sense? (That is, assuming it has a second derivative, that $f''>0?$) Or does it just mean \"increasing\"?", "Solution_2": "I think the word \"crescent\" means \"(strict) increase\".", "Solution_3": "Sorry, I meant \"strictly increasing\". I only noticed that \"crescent\" thing after posting.", "Solution_4": "Hi,\r\nI think I have a method for solving ~ the Brazilian problem~.\r\n\r\nFirst we observe that since $f$ is an increasing function we get that \r\n\r\n$\\frac{1}{f(k+1)}<\\int_{k}^{k+1}\\frac{1}{f(x)}dx=\\frac{1}{f(c_{k})}<\\frac{1}{f(k)}$\r\nThus $\\sum_{n=1}^{\\infty}\\frac{1}{f(n)}$ has the same nature as the integral $\\int_{1}^{\\infty}\\frac{1}{f(x)}dx$.\r\n\r\nA couple of observations:\r\n$f$ is increasing thus, f is differentiable almost everywhere, so without loosing generality we assume that $f$ is differentiable everywhere, since the Riemann integral does not depend on a set of Lebesque measure zero.\r\n\r\nNow, via the substitution $f(x)=y$, $x=f^{-1}(y)$, we get that \r\n$A=\\int_{1}^{\\infty}\\frac{1}{f(x)}dx=\\int_{f(1)}^{\\infty}\\frac{1}{yf'(f^{-1}(y)}dy$.\r\nWe compare A to $\\int_{1}^{\\infty}\\frac{f^{-1}(y)}{y^{2}}dy$ via the limit test.\r\nThus we calculate\r\n\r\n${{L=\\lim_{x\\rightarrow \\infty}\\frac{\\frac{1}{yf'(f^{-1})(y)}}{\\frac{f^{-1}(y)}{y^{2}}}}}=\\lim_{x\\rightarrow \\infty}\\frac{y}{f'(f^{-1}(y)f^{-1}(y)}$.\r\nNow we make the substitution $f^{-1}(y)=u$, observe that when $u$ converges to $\\infty$ then $y$ also converges to $\\infty$.\r\nThus we get that \r\n$L=\\lim_{u\\rightarrow \\infty}\\frac{f(u)}{uf'(u)}$.\r\nWe distinquish here some cases:\r\n1) Definitely that if $L\\neq 0, \\infty$ then we are done, via the limit test.(of course the limit exists)\r\n\r\n2) $L=\\lim_{x\\rightarrow \\infty}\\frac{f(x)}{xf'(x)}=\\infty$ then $\\frac{f}{xf'}\\geq \\frac{1}{A}$ for $x>a$ , where $A<1$ is such taken, this follows via the limit definition\r\nit follows that $\\frac{f'}{f}\\leq \\frac{A}{x}$ from which it follows that \r\n$\\left(\\ln f-\\ln x^{A}\\right)'<0$ thus we get that the function $x\\rightarrow \\ln\\frac{f}{x^{A}}$ is decreasing and after simple calculations we get that\r\n$fa$.\r\nthus $\\frac{1}{f}>\\frac{1}{Cx^{A}}$ and since $A<1$ we get that \r\n$\\int_{a}^{\\infty}\\frac{1}{f}dx>\\int_{a}^{\\infty}\\frac{1}{Cx^{A}}dx=\\infty$.\r\nNow observe that the inequality $f(x)\\leq Cx^{A}$ implies via $x=f^{-1}(y)$ that $y\\leq C(f^{-1}(y))^{A}$ or $f^{-1}(y)>\\frac{y^{1/A}}{C^{1/A}}$.\r\nI will drop the constant, $\\int_{b}^{\\infty}\\frac{f^{-1}(y)}{y^{2}}dy>\\int_{b}^{\\infty}\\frac{y^{1/A}}{y^{2}}dy=\\int_{b}^{\\infty}\\frac{1}{y^{2-1/A}}dy=\\infty$ since $A<1$. thus the two integrals have the same nature, namely both divergent.\r\nthe third case when $L=0$ is handled similarly, namely we get that the two integrals are convergent, we use the same technique as in the second case.\r\n\r\nRemark:\r\nThis problem implies the following integral test\r\n\r\nLet $f: (0,\\infty)\\rightarrow (0,\\infty)$ be a strictly increasing function. Then $\\int_{0}^{\\infty}\\frac{1}{f}dx$ converges if and only if \r\n$\\int_{0}^{\\infty}\\frac{f^{-1}}{x^{2}}dx$ converges.", "Solution_5": "I think the following observation can simplify the argument, although I didn't actually push to completion.\r\n\r\nYour argument studies the integral\r\n\\[\\int \\frac{f^{-1}(x)}{x^{2}}\\, dx \\, . \\]\r\nLet $y = f^{-1}(x)$. By the substition rule, the integral changes to\r\n\\[\\int \\frac{y f'(y)}{f(y)^{2}}\\, dy \\, . \\]\r\nIntegration by parts shows that this integral is equal to\r\n\\[\\int \\frac{dy}{f(y)}-\\frac{y}{f(y)}\\, . \\]\r\nNotice that the final integral is the other integral you are interested in. So we have related the two integrals of interest." } { "Tag": [], "Problem": "How many positive integer divisors does 72 have?", "Solution_1": "Prime factorizing it, we get $ 2^3\\times 3^2$. We then do $ (3\\plus{}1)(2\\plus{}1)\\equal{}\\boxed {12}$" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Let $a;b$ be real numbers satisfying $(a-\\sqrt{a^{2}+1})(b-\\sqrt{b^{2}+1})=2$\r\nCaculate $a+b=?$", "Solution_1": "$|a+b|\\ge \\frac{\\sqrt 2}{2}$.\r\nIf z=a+b, then $a,b=\\frac{z\\pm 3\\sqrt{z^{2}-\\frac{1}{2}}}{2}.$", "Solution_2": "I am surely that i have seen the problem has posted before, but now i don't know where is it, exactly. If i don't find the link, i will post again solution. The problem is fairly easy", "Solution_3": "Can you write you solution?", "Solution_4": "[quote=\"chien than\"]Caculate $a+b=?$[/quote]\r\nObviously $a+b=b+a$ AND WE ARE DONE :rotfl:", "Solution_5": "hey chien tian, can you post the answer? I can only deduce the fact that the product of the larger roots of the equations x^2-2ax-1=0 and x^2-2bx-1=0 as 2..., but it seems not sufficient enough to find the sum of the four roots in the two equation, which is 2(a+b)" } { "Tag": [ "vector", "calculus", "geometry", "search", "abstract algebra" ], "Problem": "Can anyone suggest me some books on General Relativity which explains things in more \"physics\" way than a more mathematical way . Like more emphasis on Conceptual explanations without advanced high tech complex mathematical proofs .", "Solution_1": "How much math have you taken? Do you know special relativity well? If you have taken math through vector calculus and differential equations, and have a strong grasp of special relativity I would recommend Schutz's [i]A First Course in General Relativity[/i]. However, if you don't yet completely understand special relativity then I would recommend A.P French's [i]Special Relatvity[/i] (all you need is basic calculus). Lastly, if you don't have the prerequisites in math then I would recommend Kip Thorne's [i]Black Holes and Time Warps[/i], which is good conceptual, but doesn't allow you to fully understand general relativity, because it doesn't include the maths.", "Solution_2": "Schutz's [i]A First Course in General Relativity[/i], as mentioned above, is a popular book and I would recommend it as well. If you're interested in black holes and cosmology, Hobson, Efstathiou, and Lasenby's [i]General Relativity: An Introduction for Physicists[/i] covers these in more detail than Schutz, while keeping the text very mathematically light. I would say the bare minimum of mathematical experience needed for these books is vector calculus, although a stronger background would definitely be useful.\r\n\r\nOn the other end of the spectrum, if you someday happen to be looking for a more mathematical text to read, perhaps once you have gone through a book geared towards physical intuition, I would recommend Wald's [i]General Relativity[/i]. I probably wouldn't recommend this as a first text even to someone with a good math background, but if you are really interested in general relativity you should definitely read it at some point.", "Solution_3": "Thanks for the info friends ,\r\ner.......I am just a 10th grade student and I know some basic calculus ...I only know special relativity conceptually not with mathematics rigour......can you suggest me books which cover all the pre-requisites of general relativity ... Vector calculus ,Differential Geometry all should be learnt by me .\r\nPlease also mention the pre requisite of the pre requisite for general relativity .Ya I am also interested in the black holes book.....I feel it would be first better to understand it in a more conceptual way before going to mathematics part of it ...sometimes too much math hurts the brain (atleast for me ) :rotfl:. \r\nThe book by Robert Wald is greek and latin for me .....I feel like fainting :D\r\n[quote=\"zhyllilom\"]\nIf you're interested in black holes and cosmology, Hobson, Efstathiou, and Lasenby's General Relativity: An Introduction for Physicists covers these in more detail than Schutz, while keeping the text very mathematically light. I would say the bare minimum of mathematical experience needed for these books is vector calculus, although a stronger background would definitely be useful.\n[/quote]\r\nFriend ,I have never even seen vector calculus in my life ,please dont scoff that I plan to learn general relativity . Can you suggest the pre requisites for Vector calculus etc...\r\nAlso I am very much interested in cosmology ,black holes ...It would be nice if I first read books which treat stuffs in a more conceptual and humorous way .....like the dead famous etc......\r\nOk humour is not that mandatory.....", "Solution_4": "Okay, so are you currently taking calculus? Vector Calculus can be taken as soon as you complete single variable calculus. Differential Geometry is another story, as its usually taught for advanced undergraduates or graduate students. Also, a course in differential equations could help - again you can take it as soon as you finish single variable calculus. However, don't underestimate the physics you must understand, especially special relativity, in order to grasp general relativity. Again I would highly recommend you go through a special relativity textbook (like A.P French's [i]Special Relativity[/i]) before attempting to learn general relativity. While you're waiting on it to learn the math you could also read Schutz's [i]Gravity From the Ground Up[/i] which only uses basic high school math (no calculus).\r\n\r\nIf you just want pop science books try searching for:\r\n\r\nLeonard Susskind\r\nBrian Greene\r\nStephen Hawking\r\nKip Thorne \r\n\r\nThey all have written popular books... Thorne's [i]Black Holes & Time Warps[/i] is probably the best \"laymen\" explanation of black holes that I've read.", "Solution_5": "Is Differential Geometry a must for the general theory of relativity ? \r\nAnyway can you suggest me books for the Vector Calculus etc......\r\n\r\nThe black holes and time warps book is cool my friend ,thank you very much .", "Solution_6": "Well in order to [b]completely[/b] understand general relativity you need differential geometry. However, you can get a good understanding of it without differential geometry. (general relativity isn't usually studied until grad school anyways). \r\n\r\nAs for vector calculus there are several textbooks you can use (search on amazon) - I would also highly recommend MIT's vector calculus lectures (for free) on iTunes!", "Solution_7": "What are your views regarding the book on GR by James.B.Hartle ?\r\nAlso I need to know tensor algebra and some advanced classical mechanics right ?", "Solution_8": "I'm not familiar with Hartle's book - however you could look on amazon for reviews as to the difficulty level.", "Solution_9": "I would say, with your preparation, there is absolutely no rush. If you just want to toy with some ideas for fun, that's fine, but you won't get out much if you don't have the math prerequisites.\r\n\r\nLearn the cool things that are withing grasp first (trust me, there are a LOT of those). There's not reason to rush things up, you only hurt yourself.\r\n\r\nFor example, after learning single variable really well, start learning vector calculus (multivariable) with a decent textbook, and practice your skills on a book like Griffiths Electrodynamics. Learn at the same time Special Relativity, and you are well on your way on building an understanding of relativistic electrodynamics, sth I find very cool, and well within grasp of a motivated high school student (within 1-2 yrs of hard work, you'll get there).", "Solution_10": "thanks for the advice friend :lol: ,I am not in a rush anyway. I want to learn all the math pre-requisite . After Vector calculus what is the math I require ?" } { "Tag": [ "geometry", "geometry unsolved" ], "Problem": "Given 4 lines in general position, construct a quadrilateral with internal \r\n angle bisectors on these lines.\r\n\r\n\r\n\r\n Comd. Pestich", "Solution_1": "I'm more interested in the problem source than the problem itself.\r\n\r\n\"Politbureau Entrance exam\" :D What is it? I've never heard that entering Politburo needs a math test :blush:", "Solution_2": "It shows one more time that the political leadership liked to keep \r\n population in the dark about it's practices, and it will continue\r\n with this policy, especially towards comrades and their\r\n families who send in their census forms late.\r\n\r\n\r\n Lt. Col. Pestich\r\n\r\n\r\n\r\n P.S. What about the geometry problem at hand?" } { "Tag": [ "Euler", "modular arithmetic", "search" ], "Problem": "Find the remainder when $ 37^{100}$ is divided by $ 29$.\r\n\r\nPlease explain it :)", "Solution_1": "[hide=\"Solution\"]By [url=http://en.wikipedia.org/wiki/Euler%27s_theorem]Euler's Theorem[/url], $ 37^{28}\\equiv 1 \\mod 29$.\n\n$ 37^{100} \\equal{} (37^{28})^3\\cdot37^{16}\\equiv 37^{16}\\equiv23\\mod 29$\n\nAnswer: 23[/hide]", "Solution_2": "Or use Fermat's little theorem (which Euler generalizes)", "Solution_3": "How did you find that $ 37^{16}\\equiv23\\mod 29$?", "Solution_4": "Lol, calculator bash... :blush: \r\n\r\nWell, it's easier than $ 37^{100}$, right?", "Solution_5": "You could also do\r\n\r\n$ 37^{16} \\equiv 2^{48} \\equiv 2^{20} \\equiv 32^{4} \\equiv 3^4 \\equiv 81 \\equiv \\minus{}6 \\equiv 23 \\pmod{29}.$", "Solution_6": "Yeah, here's what I did:\r\n$ 37^{16}\\equiv 6^{8}\\equiv 7^{4}\\equiv 20^{2}\\equiv 400\\equiv 23\\pmod{29}.$", "Solution_7": "Sorry for being late. I understand this part $ 37^{28}\\equiv 1\\mod 29$ because I know that Fermat Little Theorem is $ a^{p\\minus{}1}\\equiv 1\\mod p$.\r\nBut I didn't understand how you did the second part. :(", "Solution_8": "You repeatedly have to apply the properties of modular arithmetic. (Search on the wiki if you don't know them.)", "Solution_9": "People are leaving out the steps of their modular reductions, so you can't tell what calculations they did or whether they did them by hand or by calculator.\r\n\r\nIn detail: $ 37^{100} \\equiv 8^{100} \\bmod {29}$, because $ 37 \\equiv 8 \\bmod {29}$.\r\nAlso $ 2^{28} \\equiv 1 \\bmod {29}$, by Fermat's Little Theorem.\r\nSo $ 8^{100} \\equal{} 2^{300} \\equal{} (2^{28})^{10} \\cdot 2^{20} \\equiv 2^{20} \\bmod {29}$\r\n\r\n$ 2^5 \\equal{} 32 \\equiv 3 \\bmod {29}$\r\nSo $ 2^{20} \\equal{} (2^5)^4 \\equiv 3^4 \\equiv 23 \\bmod{29}$\r\n\r\nTherefore, $ 37^{100} \\equiv 23 \\bmod {29}$" } { "Tag": [ "geometry proposed", "geometry", "U sviti matematyky" ], "Problem": "Points $ C_1,$ $ A_1$ and $ B_1$ are chosen at sides $ AB,$ $ BC$ and $ AC$ of triangle $ ABC$ in such a way that the straight lines $ AA_1,$ $ BB_1$ and $ CC_1$ are concurrent. Points $ C_2,$ $ A_2$ and $ B_2$ are chosen at sides $ A_1B_1,$ $ B_1C_1$ and $ A_1C_1$ of triangle $ A_1B_1C_1$ in such a way that the straight lines $ A_1A_2,$ $ B_1B_2$ and $ C_1C_2$ are concurrent. Prove that the straight lines $ AA_2,$ $ BB_2$ and $ CC_2$ are concurrent.", "Solution_1": "The [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=6579]Cevian Nests Theorem[/url], again.\r\n\r\nKostas Vittas." } { "Tag": [ "videos", "Ring Theory", "Support" ], "Problem": "I can't really find any topics about gaming (although I only looked in the first 4 pages), so discuss gaming here.\r\n\r\nI have a 360, I'm camping out for a Wii, and I'm not getting within 30 feet of a PS3.", "Solution_1": "I dont like Xbox \r\nthe controlers are bulky and i dont give a :cursing: bout Halo\r\nAny ways i would camp out for a Wii the games are so fun!\r\nIf you want a preview watch On the Spot at gamestop.com\r\nAlso 360_Fan why do u hate the PS3 so much it will have an amazing varity of features", "Solution_2": "I might get a Wii. Some of my friends think that the Wii will do much better than others expect. We were talking about that earlier today.", "Solution_3": "I agree. I think the Wii is going to establish a good base in every gamers home while the second choice of system will be between 360 and PS3.\r\n\r\nHere is the link to see the demo of the all-new kick- :cursing: Legend of Zelda:Twilight Princess For Wii. \r\n[url]http://www.gameinformer.com/NR/exeres/4093da30-d529-4dfa-8689-973d2bda4268.htm[/url]\r\nThis is a great site! GameInformer was the nations best game magazine these past 2 years.\r\n\r\nThat's just the description of the game. Just imagine doing all that with the nunchuck and remote!", "Solution_4": "its amazing how [size=200]BAD[/size] at advertising Nintendo is!\r\n-jorian", "Solution_5": "Sorry, that was the wrong link. This goes directly to the videos;\r\n[url]http://www.gameinformer.com/News/Media/Media.htm?CS_pid={814A5054-B8CB-4D70-BFC0-7F5E7EBA0E51}[/url]\r\n\r\nThere we go. :lol:", "Solution_6": "The X-Box 360 will probably lose its popularity in the next few days.", "Solution_7": "If you're getting a Wii then you're going to be within 30 feet of a PS3 obviously.\r\n\r\nAlso, Wii is overpriced.\r\n\r\nPS3's on the other hand, are a much better deal for the price you're playing.\r\n\r\n\r\nXBox360's are unnecessary, I already have some portable heating units.", "Solution_8": "Halo RULZ!!!!!!! the Wii sounds cool but the first PS3 i see im lighting a match...\r\n\r\ni mean, what are you gonna get for PS3 (new games)? Assassin's Creed is now for 360 too (go 360!)....yes you can play metal gear solid or something but that can be for the 360 also.......and isn't the online play on the 360 better than the ps3?", "Solution_9": "[quote].......and isn't the online play on the 360 better than the ps3?[/quote]\nExcept online play is free on the PS3 and Wii, I think.\n\n[quote]Also, Wii is overpriced. \n\nPS3's on the other hand, are a much better deal for the price you're playing. [/quote]\r\n\r\nHmmm, 250 bucks for a Wii, it's funner to play (although I desperately hope they have some third-party support lined up), and 600 bucks for a PS3, and they just think that they rule the gaming world and can dictate that the future world of gaming is going to use Blu-ray.\r\n\r\nI think I'll take the Wii.\r\n\r\nOh, I'm not trying to spark an argument on this new topic, I'm just a devout Nintendo fan and newly converted Microsoft believer.", "Solution_10": "Several of my friends are going to be camping out to buy a Wii.\r\n\r\nOn the PS3 vs. Wii debate: [url=http://www.vgcats.com/comics/?strip_id=212]Vegenomics[/url]", "Solution_11": "Sega Genisis.\r\n\r\nGames: Toy Sory, Sonic the Hedgehog, Terminator II, PacMan, Tiny Toons: Bugs Bunny.\r\n\r\nI got half-way in Tiny Toons, practically finished PacMan, got to level 4 in Terminator II, got to the last level in Sonic, got to Sid's room in Toy Story.", "Solution_12": "Yeah, I personally would go for the PS3 if it weren't for the price and fact that it's already sold out (except on ebay, where I saw one auction go up to $\\$$99,999,999.99 (seriously, though I dunno if it was real))\r\n\r\nSo in that case, It's probably Wii for now, and PS3/next Microsoft console later.", "Solution_13": "$\\$$ 600 for a PS3....when the 360 came out it was only 400!", "Solution_14": "I don't hate the PS3... I just don't like it. :P \r\n\r\nI also hear that it's made poorly and when you're actually playing it it looks and feels low quality.", "Solution_15": "stuff i read comparing the three new systems:\r\n\r\nthe PS3 has dissappoiting release titles\r\nthe 360 was released earlier and therefore the \"technology\" might be not up to par compared to the PS3\r\nthe WII looks like the design was made in 1950........", "Solution_16": "i've heard about the PS3 having technical problems. i saw a video of this guy that was trying to turn the PS3 on and this red light would start blinking, then it would shut off.", "Solution_17": "I'm planning to learn how to play Halo off my friend's 360 >.> . That's all I'm doing for video games.", "Solution_18": "[quote=\"Treething\"]I'm planning to learn how to play Halo off my friend's 360 >.> . That's all I'm doing for video games.[/quote]\r\n\r\nyou can buy a 360 just for the halo series alone :lol: ...\r\n\r\nif u need help with halo and stuff let me know :starwars:", "Solution_19": "I hate video games. I just don't see why people would pay hundreds of dollars to waste their time :D \r\nI was addicted to video games once. wasted a few years of my life. :(", "Solution_20": "Well, the Wii came out in North America today. Unfortunately, I couldn't get one... :( \r\n\r\nMeh, I'll get one maybe next week or something.", "Solution_21": "[quote=\"daermon\"]I hate video games. I just don't see why people would pay hundreds of dollars to waste their time :D \nI was addicted to video games once. wasted a few years of my life. :([/quote]\r\n\r\nlol well it depends...are you out of school or something?\r\n\r\nif you are it is a waste\r\n\r\nbut if you are not out then it can be fun to get rid of stress after a school year or something" } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "As the figure below shows,let $ ABCD$ be a square and $ P$ be a movable point on the segment $ CD$.The projection points of $ A$ on $ PB$ and $ B$ on $ PA$ are $ F$ and $ E$ respectively.$ CF$ meets $ DE$ at $ Q$.\r\n\r\n(1)Find the locus of $ Q$;\r\n(2)Let $ G$ be the projection point of $ P$ on $ AB$.Show that $ QG$ is the bisector of $ \\angle AQB$.\r\n(3)Let $ H$ be the orthocentre of $ \\triangle PAB$.Show that there exists a fixed point on $ QH$.", "Solution_1": "I think you mean the locus of Q . Let\u2019s draw a circle by center O the middle of AB and radius OA .This circle clearly passes through E and F .Is this circle the intended locus and is these facts useful?", "Solution_2": "[quote=\"christ\"]I think you mean the locus of Q . Let\u2019s draw a circle by center O the middle of AB and radius OA .This circle clearly passes through E and F .Is this circle the intended locus and is these facts useful?[/quote]\r\n\r\nThat's OK. The locus of $ Q$, exactly." } { "Tag": [], "Problem": "The diving pool shown is in the shape of a trapezoidal right prism. How many cubic feet are in its volume?\n\n[asy]import three;\ncurrentprojection=orthographic(3,-2,1);\ndraw((0,0,0)--(0,0,14)--(0,20,14)--(20,20,14)--(20,0,14)--(20,0,0)--cycle);\ndraw((20,0,14)--(0,0,14));\ndraw((20,20,14)--(20,20,4)--(20,0,0));\ndraw((0,0,0)--(0,20,4),dashed);\ndraw((20,20,4)--(0,20,4),dashed);\ndraw((0,20,14)--(0,20,4),dashed);\ndraw((20,18,14)--(20,18,12)--(20,20,12));\nlabel(\"14'\",(0,0,7),W);\nlabel(\"20'\",(10,0,0),SW);\nlabel(\"14'\",(20,0,7),E);\nlabel(\"20'\",(0,10,14),NNW);\nlabel(\"20'\",(10,20,14),NE);\nlabel(\"10'\",(20,20,9),E);[/asy]", "Solution_1": "[quote=\"GameBot\"]The diving pool shown is in the shape of a trapezoidal right prism. How many cubic feet are in its volume?\n\n[asy]import three;\ncurrentprojection=orthographic(3,-2,1);\ndraw((0,0,0)--(0,0,14)--(0,20,14)--(20,20,14)--(20,0,14)--(20,0,0)--cycle);\ndraw((20,0,14)--(0,0,14));\ndraw((20,20,14)--(20,20,4)--(20,0,0));\ndraw((0,0,0)--(0,20,4),dashed);\ndraw((20,20,4)--(0,20,4),dashed);\ndraw((0,20,14)--(0,20,4),dashed);\ndraw((20,18,14)--(20,18,12)--(20,20,12));\nlabel(\"14'\",(0,0,7),W);\nlabel(\"20'\",(10,0,0),SW);\nlabel(\"14'\",(20,0,7),E);\nlabel(\"20'\",(0,10,14),NNW);\nlabel(\"20'\",(10,20,14),NE);\nlabel(\"10'\",(20,20,9),E);[/asy][/quote]\r\n\r\nThe average height is 12. $ 20 \\cdot 20 \\cdot 12\\equal{}4800$." } { "Tag": [ "calculus", "integration", "logarithms", "calculus computations" ], "Problem": "Study the convergence of the following integral:\r\n\r\n$\\int_{t}^{\\infty}\\ln \\frac{b+(x\\ln x)^{p}}{a+(x\\ln x)^{p}}dx$,\r\n\r\nwhere $p>0$, $b>a>0$ and $t$ is a positive real number large enough such that the integrand makes sense.", "Solution_1": "We can write this as \r\n\r\n$\\int_{t}^{\\infty}\\ln\\left(1+\\frac{b}{(x\\ln x)^{p}}\\right)-\\ln\\left(1+\\frac{a}{(x\\ln x)^{p}}\\right)\\,dx$\r\n\r\nFor large $x,$ the integrand is comparable to $\\frac{b-a}{(x\\ln x)^{p}}.$ It converges for $p>1$ and diverges for $p\\le 1.$", "Solution_2": "That is right. There is a conection between these two problems since I belive that the limit of that product is $0$ if $p\\leq 1$ and that the product converges to some real number, which I think it cannot be calculated explicitely, for $p>1$." } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "Let be given n lines satisfying that no 2 lines parallel with each other, no 3 lines intersect together at one point. Prove that it is possible to write in each part of the plane bounded by the above lines a integer number x satisfying x <> 0, |x |AC$\n\nLet $ \\phi = \\frac{A}{2}$\n\nThe ray $ AD$ intersects the circumcircle of $ \\triangle ABC$ at $ O$. Then $ \\angle OBC = \\angle OAC = \\phi$\n\nLet $ L$ be the projection of $ O$ to the line $ AB$. Then $ MOBL$ is inscribed in the circle with diameter $ AB \\Rightarrow$\n$ \\angle OLM = \\angle OBM = \\phi$\n\nNotice here that, $ L,B$ are on the same side of $ OM$. \n[hide=\"Indeed\"]Let $ Q$ is the antidiametric point of $ O$ and $ L'$ be the point where the ray $ OL$ intersects again (if it does) the circle.\n$ A$ must be on the arc $ BC$ not containing $ O$.\nIf we assume directed angles, we could say that $ \\angle COA\\geq 0$ and from $ \\angle AOL'=90-\\phi$ we get $ \\angle COL'\\geq 90-\\phi\\Rightarrow \\angle QOL'\\geq 0$ which means that $ L'$ can't be in the arc $ CQ$ and consequently $ L,L',B$ are in the same side of $ OM$[/hide]\nLet $ O'$ be the projection of $ M$ to $ AD$. Then from the right triangles $ OLA,LO'A$ we get $ AO'=AO\\cdot \\cos^{2}\\phi$\n\n[img]9545[/img]\n$ \\hline$\n\nLet $ M' \\in AM$ such that $ O'M'\\parallel OM$. The parallel line through $ M'$ to $ BC$ intersects $ AB,AC,AD$ at $ B',C',D'$ respectively.\nThen the triangles $ ABC,AB'C'$ are similar\nParticularly, for any point $ X \\in \\{B,C,M,D,O\\}$ we have $ \\boxed{AX' = AX \\cos^{2}\\phi}$\n\n[img]9548[/img]\n$ \\hline$\n\nThrough $ M'$ we bring a perpendicular to $ AM'$ which intersects $ AD$ at $ F$. \nLet $ w$ be the circle with diameter $ FM$. \nThen $ M',O' \\in w$, so $ AF \\cdot AO' = AM \\cdot AM' \\Rightarrow \\boxed{AF \\cdot AO' = AM^{2}\\cos^{2}\\phi}$\n\nWe draw the rectangle $ FO'MH$. Then $ FO'\\parallel=HM$\n$ O'H$ is a diameter of $ w$, so $ O'M' \\perp M' H$. But $ O'M' \\perp B'C'\\Rightarrow O'M' \\perp D'M'$, so we get that $ D'M',H$ are collinear\n$ D'D\\parallel HM, \\ D'H\\parallel DM\\Rightarrow D'DMH$ is parallelogram $ \\Rightarrow D'D=HM=FO'$\nBut $ AD' = AD \\cos^{2}\\phi \\Rightarrow DD' = AD(1-\\cos^{2}\\phi) \\Rightarrow \\boxed{DD' =AD\\sin^{2}\\phi}$\n\n[img]9550[/img]\n$ \\hline$\n\nNow it's finished! We need to construct two points $ F,O'\\in AD$, such that \n$ \\begin{array}{l}AF \\cdot AO' = AM^{2}\\cos^{2}\\phi\\\\ FO'=AD\\sin^{2}\\phi \\end{array}\\}$\n\nIn other words the points $ F,O'$ are inverse wrt the circle centered and $ A$ with radius $ AM\\cos\\phi$\n\n[color=red][size=150]Theorem[/size]\nPoints $ X,Y$ are inverse wrt the circle $ (O,R)$ and $ XY=2d$. If $ X$ is out of the circle then $ OX=d+\\sqrt{d^{2}+R^{2}}$[/color]\n\n$ 2d = OX-OY = OX-\\frac{R^{2}}{OX}= \\frac{OX^{2}-R^{2}}{OX}\\Rightarrow$\n\n$ OX^{2}-R^{2}=2d\\cdot OX \\Rightarrow$\n\n$ OX^{2}-(2d)\\cdot OX-R^{2}=0\\Rightarrow OX = \\frac{2d+\\sqrt{4d^{2}+4R^{2}}}{2}=d+\\sqrt{d^{2}+R^{2}}$\n$ \\hline$\n\nApplying the theorem with $ R=AM\\cos\\phi$ and $ d=\\frac{AD\\sin^{2}\\phi}{2}$ we get the exact position of $ O'$, which is \n$ AO' = \\frac{AD\\sin^{2}\\phi}{2}+\\sqrt{(\\frac{AD\\sin^{2}\\phi}{2})^{2}+(AM\\cos\\phi)^{2}}$\n\n\n(Of course, $ R$ and $ d$ are constructible with straightedge and compass)\n\nFinally, we construct the circle $ s=(A,m_{a})$ and we bring a perpendicular through $ O'$ to $ AD$. \nWe'll find $ M$ on the intersection between the line and the circle.\n\n[hide=\"We will have solution iff \"]$ AO'\\leq AM \\iff d+\\sqrt{d^{2}+R^{2}}\\leq AM \\iff$\n$ \\sqrt{d^{2}+R^{2}}\\leq AM-d \\ \\ (1)\\ \\Longrightarrow^\\star$\n$ d^{2}+R^{2}\\leq (AM-d)^{2}\\ \\ (2) \\iff$\n$ d^{2}+R^{2}\\leq AM^{2}-2AM\\cdot d+d^{2}\\iff$\n$ R^{2}\\leq AM^{2}-2AM\\cdot d \\iff$\n$ 2AM\\cdot d\\leq AM^{2}-R^{2}\\iff$\n$ 2AM\\cdot d\\leq AM^{2}(1-\\cos^{2}\\phi) \\iff$\n$ 2 d\\leq AM\\sin^{2}\\phi \\iff$\n\n$ AD\\sin^{2}\\phi\\leq AM\\sin^{2}\\phi \\iff$\n$ AD\\leq AM$\n\n$ (\\star)$ If $ AD\\leq AM$ then $ (2)$ holds\n\nDoes $ (2)$ imply $ (1)$ :?: \n\n$ AM-d=AM-\\frac{AD\\sin^{2}\\phi}{2}>AM-\\frac{AD}{2}>AM-AD\\geq 0$, so $ (2) \\implies (1)$\n\n\n[b]So we have a solution if and only iff $ AD\\leq AM$[/b][/hide][/hide]" } { "Tag": [], "Problem": "I don't know where I should put this, but feel free to move it. Please don't delete this...\r\n\r\nA game of trivial pursuit. Someone posts a question(serious question), and someone else answers it.\r\n\r\nExample:\r\n\r\n{A wrote:\r\nWhat puppet was featured on the August, 1980, cover of [i]Life[/i]?\r\n\r\nB wrote:\r\nMiss Piggy}\r\n\r\nI'll start.\r\n\r\nWhat year did Mexico City host the Summer Olympics?\r\n\r\nIf there is more than one date, the first would be accepted, not the latest.", "Solution_1": "Sounds like fun. :) Can we look on the Internet, or does this need to be from our own knowledge?", "Solution_2": "uhhh..1968?", "Solution_3": "We need a next question. Let's keep these easy:\r\n\r\nWhat shape has the same number of lines as a limerick?", "Solution_4": "A pentagon.", "Solution_5": "According to 1=2's rules, you're supposed to post the next problem.", "Solution_6": "What was the name of the second [i]Star Trek[/i] series originally going to be?", "Solution_7": "Classic Trek?\r\n\r\nI actually don't know this one.", "Solution_8": "Nice try, 1=2. :) \r\n\r\nIt's been quite a while. I don't think anyone knows it. I'll just post the answer and come up with a different question, if that's all right.\r\n\r\n[hide=\"Answer\"][i]Star Trek: Phase II[/i][/hide]\r\n\r\nNew question: Who is the famous singer who was once married to actress Carrie Fisher (Leia from [i]Star Wars[/i])?" } { "Tag": [ "group theory", "abstract algebra", "linear algebra", "matrix", "modular arithmetic", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $ G$ be a group of order $ pq$, where $ p$ and $ q$ are both primes and $ q0 is a prime number(for example 5) is -p (in my example -5) also a prime number?\r\nIn other words, when you say prime number do you only refer to positive integers(as wikipedia says) or has the term been extended.\r\nIt would be logical that also negative numbers should be prime (-5 and 5 have the same divisors: -5, -1, 1 and 5).", "Solution_1": "Yes prime must be positive .\r\nIf we call $ \\minus{}p$ is prime then the fundamental law of number theory not true:''Every natural number can represent unique in the form product of prime numbers'' .And 1 is also not a prime.\r\nIf the fundamental is not true then it is quite hard to solve problem in number theory.", "Solution_2": "How would it contradict that fundamental truth?\r\nEvery natural number is represented unique as a product of natural prime numbers.\r\nIf you extend that to integers, how do you represent -60 in a product of prime numbers?", "Solution_3": "$ \\minus{} 5$ is prime in $ \\mathbb Z$. Along with $ \\minus{} 2, \\minus{} 3, \\minus{} 7, ...$ .\r\n\r\nTo clarify this, we consider arbitrary integral domains (commutative rings with $ 1$ such that $ ab \\equal{} 0$ implies $ a \\equal{} 0$or $ b \\equal{} 0$). At first some simple notions for those:\r\n$ x|y$ (spoken: $ x$ [i]divides[/i] $ y$) iff there is a $ z$ with $ y \\equal{} xz$ as usual.\r\nA [i]unit[/i] $ u$ is an element such that $ u|1$, in other words: there is an inverse of multiplication.\r\nA [i]prime element[/i] is a (nonzero nonunit) $ p$ such that if $ p|ab$, then $ p|a$ or $ p|b$. \r\nA similar notion is that of an [i]irreducible element[/i]: a nonzero nonunit $ p$ is called irreducible iff $ p \\equal{} ab$ implies that $ a$ or $ b$ is a unit.\r\n\r\nNow prime elements $ p,q$ (and similar irreducible elements) are seen to be equivalent iff they just differe by a unit, thus if $ p \\equal{} qu$ for some unit $ u$.\r\n\r\nJust a small and easy to prove remark:\r\nLemma:\r\nEvery prime element is an irreducible one.\r\n\r\n[u]Theorem[/u]:\r\nThe following are equivalent:\r\na) every nonzero element can be represented as a product of prime elements and units.\r\nb) the representation from a) exists and is unique up to multiplication by units and reordering.\r\nc) every nonzero element can be written as a product of irreducible elements and units, and this is unique as in b).\r\nIf this holds, the ring is called UFD (unique factorisation domain).\r\n\r\n\r\nLooking back, all notions make sense if we look at the theorem. Applying the deifnitions on $ \\mathbb Z$ we see: $ \\minus{} 2, \\minus{} 3, \\minus{} 5, \\minus{} 7,...$ are primes.\r\n\r\nI often consider the \"essentially different\" (rational) primes to be $ \\bar{\\mathbb P} \\equal{} \\{ 0,2,3,5,7,..., \\infty \\}$ :D(yes, this really makes sense, e.g. as [i]places[/i]).", "Solution_4": "Thank you very much for the documented clarification ZetaX. Yet, I don't really understand the last sentence.[/img][/quote]", "Solution_5": "Another way of viewing primes is that of a (real, multiplicative) valuation, this means a function $ | \\cdot | : K \\to \\mathbb R_0^\\plus{}$ (here $ K\\equal{}\\mathbb Q$) such that:\r\n$ |x| \\equal{} 0 \\iff x\\equal{}0$\r\n$ |a \\cdot b| \\equal{} |a| \\cdot |b|$\r\n$ |a\\plus{}b| \\leq |a| \\plus{} |b|$\r\n\r\nWe call $ | \\cdot |$ und $ | \\cdot |'$ equivalent if $ |x|' \\equal{} |x|^r$ for some $ r>0$.\r\n\r\nThen up to equivalence there are the following valuations of $ \\mathbb Q$:\r\n\r\n- $ | \\cdot |_\\infty$, the standard modulus.\r\n- $ | \\cdot |_0$, the trivial valuation (given by $ |x|\\equal{}1$ for $ x \\neq 0$).\r\n- for every positive natural prime $ p$ we define $ | \\cdot |_p$ as follows: write a rational $ x$ as $ x\\equal{} p^v \\cdot \\frac bc$ with $ p \\nmid b,c$ and set $ |x|_p : \\equal{} p^\\minus{}v$.\r\n\r\nThe reasoning for choosing the indices $ 0$ (see it as a prime in the sense $ 0|ab$ implies $ 0|a$ or $ 0|b$, too) and $ \\infty$ (similar as $ 0$) is a bit more complicated to explain, but this should already give the idea why to consider them prime." } { "Tag": [], "Problem": "The complex number $z$ satisfies $z + |z| = 2 + 8i$. What is $|z|^2$? Note: if $z = a + bi$, then $|z| = \\sqrt{a^2+b^2}$.\r\n\r\n(A) 68\r\n(B) 100\r\n(C) 169\r\n(D) 208\r\n(E) 289", "Solution_1": "[hide]\n$z$ = $a + bi$\n$|z| = \\sqrt{a^2 + b^2}$\n$z + |z| = 2 + 8i$\n$a + bi + \\sqrt{a^2 + b^2} = 2 + 8i$\n$b = 8$\n\n$a + \\sqrt{a^2 + 64} = 2$\n$\\sqrt{a^2 + 64} = 2 - a$\n$a^2 + 64 = 4 - 4a + a^2$\n$60 = -4a$\n$a = -15$\n\n$|z| = \\sqrt{(-15)^2 + 8^2}$\n$|z|^2 = (-15)^2 + 8^2$\n$|z|^2 = 289$[/hide]", "Solution_2": "[hide=\"Answer\"]$a+bi+\\sqrt{a^2+b^2}=2+8i$\nThe only $i$ is found in $bi$, so $b=8$. Therefore, $a+\\sqrt{a^2+64}=2\\Rightarrow (a-2)^2=a^2+64\\Rightarrow a=-15$. Hence, $a^2+b^2=225+64=289\\Rightarrow \\boxed{E}$.[/hide]" } { "Tag": [], "Problem": "please factorize-\r\n 2x^4+3x^3+x^2+x+4\r\nhow can the expression be resolved into factors???", "Solution_1": "hello, we have $ 2x^4\\plus{}3x^3\\plus{}x^2\\plus{}x\\plus{}4\\equal{}(x^2\\minus{}x\\plus{}1)(2x^2\\plus{}5x\\plus{}4)$\r\nSonnhard.", "Solution_2": "can you please work out yhe factorization step-by-step???", "Solution_3": "Mos of it is just guess and check because it is hard to use the extreme and mode which is used in simple algebraic factorization. Like: 2x^2+3x+1.", "Solution_4": "we can do the factorization by taking x^3=-1\r\n then x^3+1=0 \r\n or (x+1)(x^2-x+1)=0\r\n as any of the factors could be 0.so,\r\n 2x^4+3x^3+x^2+x+4\r\n = 2x^2(x^2-x+1)+5x(x^2-x+1)+4(x^2-x+1) \r\n = (x^2-x+1)(2x^2+5x+4)\r\nis it good enough??? please advice on its process or show other likely ways of doing it.please...", "Solution_5": "The factorization would be in the form (2x^2+ax+b)(x^2+cx+d), where a,b,c,d integers. Then when you expand, you get a system of four equations and four variables. You can solve this.", "Solution_6": "Well, the first (but hard-to-see) step is plugging in $ \\omega$ where $ \\omega^3\\equal{}\\minus{}1$ and $ \\omega\\neq\\minus{}1$. Observe that $ \\omega$ satisfies $ \\omega^2\\minus{}\\omega\\plus{}1\\equal{}0$.\r\n\r\nThen $ 2\\omega^4\\equal{}\\minus{}2\\omega$ and $ 3\\omega^3\\equal{}\\minus{}3$, so the expression is equivalent to $ \\omega^2\\minus{}\\omega\\plus{}1$, which is 0, so $ x^2\\minus{}x\\plus{}1$ is a factor.", "Solution_7": "[quote=\"earth\"]please factorize-\n 2x^4+3x^3+x^2+x+4\nhow can the expression be resolved into factors???[/quote]\r\n$ 2x^4\\plus{}3x^3\\plus{}x^2\\plus{}x\\plus{}4\\equal{}2x^4\\plus{}2x\\plus{}3(x^3\\plus{}1)\\plus{}(x^2\\minus{}x\\plus{}1)\\equal{}2x(x^3\\plus{}1)\\plus{}3(x^3\\plus{}1)\\plus{}(x^2\\minus{}x\\plus{}1)\\equal{}(2x\\plus{}3)(x^3\\plus{}1)\\plus{}(x^2\\minus{}x\\plus{}1)\\equal{}(2x\\plus{}3)(x\\plus{}1)(x^2\\minus{}x\\plus{}1)\\plus{}(x^2\\minus{}x\\plus{}1)\\equal{}(x^2\\minus{}x\\plus{}1)(2x^2\\plus{}5x\\plus{}4)$\r\n :)" } { "Tag": [ "ratio" ], "Problem": "Consider points A, B, C, and D in that order on a line, where AC: BD = 3:7 and BC:AD = 1:4. Determine AB:CD.", "Solution_1": "There's probably a much faster and cleaner solution, but here's one:\r\n\r\n[hide=\"Solution\"]First, we transform the given ratios into equations:\n\\begin{align}\\frac {AC}{BD} = \\frac {3}{7}\\quad & \\Longrightarrow\\quad7AC = 3BD \\\\\n\\frac {BC}{AD} = \\frac {1}{4}\\quad & \\Longrightarrow\\quad4BC = AD\\end{align}\nThen, we simplify $ (\\text{equation}\\,1) - 3\\cdot(\\text{equation}\\,2)$ to determine $ \\frac {AB}{BC}$:\n\\begin{align*}7AC - 3(4BC) & = 3BD - 3(AD) \\\\\n7(AB + BC) - 12BC & = 3BD - 3(AB + BD) \\\\\n7AB - 5BC & = - 3AB \\\\\n10AB & = 5BC \\\\\n\\frac {AB}{BC} & = \\frac {1}{2}\\end{align*}\nNext, we simplify $ (\\text{equation}\\,1) + 7\\cdot(\\text{equation}\\,2)$ to obtain $ \\frac {BC}{CD}$:\n\\begin{align*}7AC + 7(4BC) & = 3BD + 7(AD) \\\\\n28BC - 3(BC + CD) & = 7(AC + CD) - 7AC \\\\\n25BC - 3CD & = 7CD \\\\\n25BC & = 10CD \\\\\n\\frac {BC}{CD} & = \\frac {2}{5}\\end{align*}\nFinally, $ \\frac {AB}{CD} = \\frac {AB}{BC}\\cdot\\frac {BC}{CD} = \\frac {1}{2}\\cdot\\frac {2}{5} = \\boxed{\\frac {1}{5}}$.[/hide]", "Solution_2": "Let $ x \\equal{} AB$, $ n \\equal{} BC$, and $ y \\equal{} CD$, so we want $ \\frac xy$.\r\n\r\nWe already know that\r\n\r\n$ \\frac {x \\plus{} n}{y \\plus{} n} \\equal{} \\frac37$ and $ \\frac n{x \\plus{} n \\plus{} y} \\equal{} \\frac14\\implies4n\\equal{}x\\plus{}n\\plus{}y\\implies n \\equal{} \\frac {x \\plus{} y}3$.\r\n\r\nPlugging this into our first equation, we have $ \\frac {x \\plus{} \\frac {x \\plus{} y}3}{y \\plus{} \\frac {x \\plus{} y}3} \\equal{} \\frac {4x \\plus{} y}{x \\plus{} 4y} \\equal{} \\frac37\\implies28x \\plus{} 7y \\equal{} 3x \\plus{} 12y\\implies25x \\equal{} 5y\\implies\\frac xy \\equal{} \\boxed{\\frac15}$." } { "Tag": [], "Problem": "Find the integer solutions of the equation \\[x^{3}+y^{3}= 8^{30}\\] (without FLT).", "Solution_1": "no solutions by fermat. :wink:", "Solution_2": "hmmm. im pretty sure he said without FLT. i could be wrong, but im not.", "Solution_3": "There are solutions to this equation since it did not mention positive integers.", "Solution_4": "[quote=\"seamusoboyle\"]hmmm. im pretty sure he said without FLT. i could be wrong, but im not.[/quote]\r\n\r\nsorry i didn`t see. the equations doesn`t have sols, boxedexe :wink:", "Solution_5": "What about $(2^{30},0),(0,2^{30})$? As I said, it did not positive integers.", "Solution_6": "$(x+y)(x^{2}-xy+y^{2})=8^{30}$\r\nBy parity, the LHS will have an odd factor apart from 1 unless $x$ and $y$ are both even.\r\nLet $x=2x_{1}$ and $y=2y_{1}$\r\n$(x_{1}+y_{1})(x_{1}^{2}-x_{1}y_{1}+y_{1}^{2})=8^{29}$\r\n\r\nAnd again...\r\nLet $x_{1}=2x_{2}$ and $y_{1}=2y_{2}$\r\n$(x_{2}+y_{2})(x_{2}^{2}-x_{2}y_{2}+y_{2}^{2})=8^{28}$\r\n\r\nAnd so forth...\r\nLet $x_{n}=2x_{n+1}$ and $y_{n}=2y_{n+1}$\r\n$(x_{n}+y_{n})(x_{n}^{2}-x_{n}y_{n}+y_{n}^{2})=8^{30-n}$\r\n\r\nUntil...\r\n$(x_{30}+y_{30})(x_{30}^{2}-x_{30}y_{30}+y_{30}^{2})=1$\r\n$x_{30}^{3}+y_{30}^{3}=1$\r\n\r\nThe only integer solutions to this are $(x_{30},y_{30})=(1,0)\\implies \\boxed{(x,y)=(2^{30},0)}$ and $(x_{30},y_{30})=(0,1)\\implies \\boxed{(x,y)=(0,2^{30})}$", "Solution_7": "[quote=\"Andreas\"]Find the integer solutions of the equation\n\\[x^{3}+y^{3}= 8^{30}\\]\n(without FLT).[/quote]\r\n\r\none of $x$ or $y$ must be positive, wlog $x$, then 3 cases\r\n\r\n$y$ is negative, then $x^{3}=(2^{30})^{3}+(-y)^{3}$ has no solutions by FLT\r\n$y$ is positive, no solutions again by flt\r\n$y=0$ so $x=2^{30}$\r\nthen switching, we get the other solution, for a solution set of:\r\n\r\n$(x,y)=(2^{30},0)$, $(0,2^{30})$", "Solution_8": "Nice Solution, but the problem states \"without FLT.\"" } { "Tag": [ "linear algebra", "matrix", "group theory", "superior algebra", "superior algebra unsolved" ], "Problem": "Sorry, I'm new to abstract algebra. I don't know if this is the right forum. Anyway here's the problem:\r\n\r\nLet $G : ={\\left[\\begin{array}{cccc}1 & n \\\\0 & 1\\end{array}\\right]}$ under matrix multiplication.\r\n\r\nShow $f: G \\mapsto Z$ where $f{\\left[\\begin{array}{cccc}1 & n \\\\0 & 1\\end{array}\\right]}=n$ is an isomorphism", "Solution_1": "I have only studied group theory for a few hours, so I only know the very elementary things, more like notation. I have never read about matrices in group sand such, so bear with me. Tell me if I am correct.\r\n\r\n1) Prove that $G$ forms a group with the identity $e_{G}=\\left[\\begin{matrix}1&0\\\\0&1\\end{matrix}\\right]$, which is very easy.\r\n2) Prove that $f$ is a homomorphism by proving that $f\\left[\\begin{matrix}1&n\\\\0&1\\end{matrix}\\right]=f\\left(\\left[\\begin{matrix}1&p\\\\0&1\\end{matrix}\\right]\\left[\\begin{matrix}1&q\\\\0&1\\end{matrix}\\right]\\right)=f\\left[\\begin{matrix}1&p\\\\0&1\\end{matrix}\\right]f\\left[\\begin{matrix}1&q\\\\0&1\\end{matrix}\\right]$\r\n3) Prove that $f$ is both injective and surjective.\r\n\r\nAre these steps correct/complete?", "Solution_2": "i think they are. thanks.\r\n[quote=\"boxedexe\"]I have only studied group theory for a few hours, so I only know the very elementary things, more like notation. I have never read about matrices in group sand such, so bear with me. Tell me if I am correct.\n\n1) Prove that $G$ forms a group with the identity $e_{G}=\\left[\\begin{matrix}1&0\\\\0&1\\end{matrix}\\right]$, which is very easy.\n2) Prove that $f$ is a homomorphism by proving that $f\\left[\\begin{matrix}1&n\\\\0&1\\end{matrix}\\right]=f\\left(\\left[\\begin{matrix}1&p\\\\0&1\\end{matrix}\\right]\\left[\\begin{matrix}1&q\\\\0&1\\end{matrix}\\right]\\right)=f\\left[\\begin{matrix}1&p\\\\0&1\\end{matrix}\\right]f\\left[\\begin{matrix}1&q\\\\0&1\\end{matrix}\\right]$\n3) Prove that $f$ is both injective and surjective.\n\nAre these steps correct/complete?[/quote]", "Solution_3": "Assuming you want $G=\\left\\{\\left[\\begin{matrix}1&n\\\\ 0&1\\end{matrix}\\right] : n\\in\\mathbb{Z}\\right\\}$, then you can define your isomorphism in the natural way that you want to. Let $f: G\\rightarrow \\mathbb{Z}$ by $\\left[\\begin{matrix}1&n\\\\ 0&1\\end{matrix}\\right]\\mapsto n$. Identity goes to identity since $f\\left(\\left[\\begin{matrix}1&0\\\\ 0&1\\end{matrix}\\right]\\right)=0$.\r\n\r\nLet $\\left[\\begin{matrix}1&l\\\\ 0&1\\end{matrix}\\right]$, $\\left[\\begin{matrix}1&k\\\\ 0&1\\end{matrix}\\right]\\in G$. Then $f\\left(\\left[\\begin{matrix}1&l\\\\ 0&1\\end{matrix}\\right]\\left[\\begin{matrix}1&k\\\\ 0&1\\end{matrix}\\right]\\right)= f\\left(\\left[\\begin{matrix}1&l+k\\\\ 0&1\\end{matrix}\\right]\\right)=l+k= f\\left(\\left[\\begin{matrix}1&l\\\\ 0&1\\end{matrix}\\right]\\right)+f\\left(\\left[\\begin{matrix}1&k\\\\ 0&1\\end{matrix}\\right]\\right)$\r\n\r\nNow you need to check 1-1 and onto. Both of which should be pretty straightforward from the definition. I can do it, though, if you'd like.", "Solution_4": "Oops, I have a typo. In my post, it should say $f\\left[\\begin{matrix}1&n\\\\0&1\\end{matrix}\\right]=f\\left(\\left[\\begin{matrix}1&p\\\\0&1\\end{matrix}\\right]\\left[\\begin{matrix}1&q\\\\0&1\\end{matrix}\\right]\\right)=f\\left[\\begin{matrix}1&p\\\\0&1\\end{matrix}\\right]+f\\left[\\begin{matrix}1&q\\\\0&1\\end{matrix}\\right]$", "Solution_5": "[quote=\"boxedexe\"]Oops, I have a typo. In my post, it should say $f\\left[\\begin{matrix}1&n\\\\0&1\\end{matrix}\\right]=f\\left(\\left[\\begin{matrix}1&p\\\\0&1\\end{matrix}\\right]\\left[\\begin{matrix}1&q\\\\0&1\\end{matrix}\\right]\\right)=f\\left[\\begin{matrix}1&p\\\\0&1\\end{matrix}\\right]+f\\left[\\begin{matrix}1&q\\\\0&1\\end{matrix}\\right]$[/quote] \n\n\nBut shouldn't $f\\left[\\begin{matrix}1&n\\\\0&1\\end{matrix}\\right]=f\\left(\\left[\\begin{matrix}1&p\\\\0&1\\end{matrix}\\right]\\left[\\begin{matrix}1&q\\\\0&1\\end{matrix}\\right]\\right)=f\\left[\\begin{matrix}1&p\\\\0&1\\end{matrix}\\right]f\\left[\\begin{matrix}1&q\\\\0&1\\end{matrix}\\right]$[/quote]\r\n\r\nWhy did f(ab) = f(a) + (b). I thought its matrix multiplication, where did the $+$ sign came? (Assuming that $a$ is the matrix with the $p$ and $b$ be the matrix with the $q$) Can you please explain, I'm quite poor at this. Thanks.", "Solution_6": "No. $f\\left(\\left[\\begin{matrix}1&p\\\\0&1\\end{matrix}\\right]\\left[\\begin{matrix}1&q\\\\0&1\\end{matrix}\\right]\\right)=f\\left[\\begin{matrix}1&p+q\\\\0&1\\end{matrix}\\right]=f\\left[\\begin{matrix}1&p\\\\0&1\\end{matrix}\\right]+f\\left[\\begin{matrix}1&q\\\\0&1\\end{matrix}\\right]$", "Solution_7": "[quote=\"boxedexe\"]No. $f\\left(\\left[\\begin{matrix}1&p\\\\0&1\\end{matrix}\\right]\\left[\\begin{matrix}1&q\\\\0&1\\end{matrix}\\right]\\right)=f\\left[\\begin{matrix}1&p+q\\\\0&1\\end{matrix}\\right]=f\\left[\\begin{matrix}1&p\\\\0&1\\end{matrix}\\right]+f\\left[\\begin{matrix}1&q\\\\0&1\\end{matrix}\\right]$[/quote]\r\n\r\nCan you please explain how did that happen?\r\n\r\nEDIT:\r\nSorry,i think i got it." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $a,b,c,d$ be nonegative real numbers. Prove that\r\n\\[\\underset{sym}{\\sum}\\frac{1}{a^{3}+b^{3}}\\ge \\frac{243}{2(a+b+c+d)^{3}}\\]\r\n:)\r\n\r\n\r\n@Mr.Thuan: I have solved many problems you sent to me, but i can't send them to you by mail because i have lost my USB yesterday. :(", "Solution_1": "Good. Come on.", "Solution_2": "The similar.\r\nLet a,b,c,d be nonegative real numbers. Prove that :\r\n$\\sum_{cyc}\\frac{1}{ a^{3}+b^{3}+c^{3}}\\geq \\frac{24}{(a+b+c+d)^{3}}$\r\nAnd it's very easy.", "Solution_3": "Yes, that is correct inequality and can be solved in a similar manner proposed by Nguyen Quoc Ba Ca. Even a more general result is possible. \r\n\r\nBut I think that it's not very easy.", "Solution_4": "[quote=\"VIETNAM\"]The similar.\nLet a,b,c,d be nonegative real numbers. Prove that :\n$\\sum_{cyc}\\frac{1}{ a^{3}+b^{3}+c^{3}}\\geq \\frac{24}{(a+b+c+d)^{3}}$\nAnd it's very easy.[/quote]\r\nNo one like this? :(", "Solution_5": "Let $a\\geq b\\geq c\\geq d$\r\nWe have\r\n$(a+\\frac{c+d}{2})^{3}+(b+\\frac{c+d}{2})^{3}\\geq a^{3}+b^{3}+c^{3}\\geq a^{3}+b^{3}+d^{3}$ \r\n\r\nand \r\n$(a+\\frac{c+d}{2})^{3}\\geq a^{3}+c^{3}+d^{3}$\r\n\r\nand \r\n$(b+\\frac{c+d}{2})^{3}\\geq b^{3}+c^{3}+d^{3}$\r\n\r\nso we have to prove that \r\n$\\frac{2}{x^{3}+y^{3}}+\\frac{1}{x^{3}}+\\frac{1}{y^{3}}\\geq \\frac{24}{(x+y)^{3}}$ \r\n\r\nwhich is true since it is equivalent to\r\n$\\frac{(x-y)^{2}(x^{6}+4x^{5}y+8x^{4}y^{2}-8x^{3}y^{3}+8x^{2}y^{4}+4xy^{5}+y^{6})}{(x^{3}y^{3})(x+y)^{3}(x^{2}-xy+y^{2})}\\geq 0$ \r\n\r\nwhich holds", "Solution_6": "[quote=\"silouan\"]Let $a\\geq b\\geq c\\geq d$\nWe have\n$(a+\\frac{c+d}{2})^{3}+(b+\\frac{c+d}{2})^{3}\\geq a^{3}+b^{3}+c^{3}\\geq a^{3}+b^{3}+d^{3}$ \n\nand \n$(a+\\frac{c+d}{2})^{3}\\geq a^{3}+c^{3}+d^{3}$\n\nand \n$(b+\\frac{c+d}{2})^{3}\\geq b^{3}+c^{3}+d^{3}$\n\nso we have to prove that \n$\\frac{2}{x^{3}+y^{3}}+\\frac{1}{x^{3}}+\\frac{1}{y^{3}}\\geq \\frac{24}{(x+y)^{3}}$ \n\nwhich is true since it is equivalent to\n$\\frac{(x-y)^{2}(x^{6}+4x^{5}y+8x^{4}y^{2}-8x^{3}y^{3}+8x^{2}y^{4}+4xy^{5}+y^{6})}{(x^{3}y^{3})(x+y)^{3}(x^{2}-xy+y^{2})}\\geq 0$ \n\nwhich holds[/quote]\r\nIt 's true,but it's not nice,silouan! :blush:", "Solution_7": "Very nice proof, silouan! :lol:", "Solution_8": "That is true and NICE solution. Come on Silouan.", "Solution_9": "Thank you all . I feel happy that you liked it . :)" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Prove that there exists an integer $ n$ such that $ 2^n\\plus{}3^n$ has exactly $ 23$ distinct prime divisors.", "Solution_1": "[quote=\"mr.danh\"]Prove that there exists an integer $ n$ such that $ 2^n\\plus{}3^n$ has exactly $ 23$ distinct prime divisors.[/quote]\n\nwho can post solution?", "Solution_2": "Any solution\n" } { "Tag": [ "induction", "combinatorics open", "combinatorics" ], "Problem": "Henry plays a game where he has to guess if a card picked from a deck is red or black:\r\nHe makes a guess and picks up a card.\r\nIf his guess was correct he gets a point.\r\nHe makes his next guess; if there are more black cards left in the deck he chooses black and vice versa and picks up a card.\r\nIf there are equally many black and red cards he always chooses black.\r\n\r\nOn the average, how many points does he get when all the 52 cards have been picked up?\r\n\r\n\r\nI have no idea of the difficulty of this so I posted it here.", "Solution_1": "[quote=\"Iltanen\"]Henry plays a game where he has to guess if a card picked from a deck is red or black:\nHe makes a guess and picks up a card.\nIf his guess was correct he gets a point.\nHe makes his next guess; if there are more black cards left in the deck he chooses black and vice versa and picks up a card.\nIf there are equally many black and red cards he always chooses black.\n\nOn the average, how many points does he get when all the 52 cards have been picked up?\n\n\nI have no idea of the difficulty of this so I posted it here.[/quote]\r\nAn idea to how to solve it:\r\n\r\nLet $ a_{r,b}$ be the average number of points he gets with $ r$ red cards and $ b$ black cards.\r\nWe know that $ a_{n,m} \\equal{} a_{m,n}$ and $ a_{n,0} \\equal{} n$\r\nWe can easily see that $ a_{n,n} \\equal{} \\frac{1}{2} \\plus{} a_{n,n\\minus{}1}$\r\nAnd for $ m > n$ we get $ a_{m,n} \\equal{} \\frac{m}{m\\plus{}n} (1 \\plus{} a_{m\\minus{}1,n}) \\plus{} \\frac{n}{m\\plus{}n} a_{m,n\\minus{}1}$\r\n\r\nSo we \"just\" have to calculate $ a_{26,26}$. It's almost impossible to do without a computer, so if we could derive a general formula for $ a_{n,n}$ it would help a lot.\r\n\r\nLets instead try to calculate $ a_{n,k}$ for $ n>k$.\r\n$ k \\equal{} 0$: $ a_{n,0} \\equal{} n$ and $ a_{0,0} \\equal{} 0$\r\n$ k \\equal{} 1$: $ a_{n,1} \\equal{} \\frac{n}{n\\plus{}1}(1\\plus{}a_{n\\minus{}1,1})\\plus{}\\frac{1}{n\\plus{}1}a_{n,0} \\equal{} \\frac{2n}{n\\plus{}1} \\plus{} \\frac{n}{n\\plus{}1}a_{n\\minus{}1,1}$ and we get $ a_{n,1} \\equal{} \\frac{n^2 \\plus{} n \\plus{} 1}{n\\plus{}1} \\equal{} \\frac{n^3\\minus{}1}{n^2\\minus{}1}$ by induction, and $ a_{1,1} \\equal{} \\frac{3}{2}$\r\n\r\nI might complete it later, but I've gotta go :)\r\n\r\nIf i should guess i would say $ 26 < a_{26,26} < 27$" } { "Tag": [ "calculus", "integration", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Hi all\r\n\r\nI'm just working through problems in Marcus on my own time. I am stuck on problem 30 from chapter 1. \r\n\r\nLet $ R$ be an integral domain. Show that two ideals of $ R$ are isomorphic as $ R$-modules $ iff$ they are in the same ideal class.\r\n\r\nNow I can prove that if they are in the same ideal class they are isomorphic. But I have been unable to prove the other direction. Any help would be appreciated, but please don't use any high level machinery if possible, as it is the first chapter and none of the previous problems relied on any high-powered theorems at all.\r\n\r\nThanks for the help.", "Solution_1": "Let the two ideals be I and J, and let f be an isomorphism between them. Now fix an element x in I and look at its image in J. Basically, we want to identify x and its image. So, what are two easy a,b to multiply by so that ax=bf(x)?", "Solution_2": "marvelous thanks very much" } { "Tag": [ "geometry" ], "Problem": "Are there any books that are on specific subjects like algebra or combinatorics or geometry that are as easy to read as AoPS. I am trying to look for one that goes more in depth but still easy to understand.", "Solution_1": "You should do AoPS first because it gives you a good foundation on all of the subjects. Then youi can go into more specific books like the ones stated in the Resources section on this site." } { "Tag": [ "geometry" ], "Problem": "Given a circle of center A and let CD be the diameter of that circle. Let CF be a chord of the circle. Show that the triangle CFD is a right triangle.\r\n\r\n**Geogebra picture included to illustrate the above problem. (F=F1)\r\n[geogebra]c39b98579f820f5cab1c0c97bf5536aa01419e11[/geogebra]", "Solution_1": "This is pretty standard\r\n\r\nLine $ AF$ splits the triangle into two isoscelese triangles: $ \\bigtriangleup CAF$ and $ \\bigtriangleup DAF$\r\nIn each triangle there are two angles the same let them be $ x$ and $ y$.\r\n\r\n$ x\\plus{}x\\plus{}y\\plus{}y \\equal{} 180^o$\r\n$ 2(x\\plus{}y) \\equal{} 180$\r\n$ x\\plus{}y \\equal{} 90$\r\n\r\nSo all such triangles are right angled triangles", "Solution_2": "Just to clarify, the measure of angle CDF = y, the measure of angle DCF = x, and the measure of angle CFD = x+y. And then by using the property that all inner angles of a triangle must add up to 180 degrees, you can say that for triangle CFD: x+(x+y)+y = 180.\r\n\r\nCorrect?", "Solution_3": "[quote=\"kratsg\"]Just to clarify, the measure of angle CDF = y, the measure of angle DCF = x, and the measure of angle CFD = x+y. And then by using the property that all inner angles of a triangle must add up to 180 degrees, you can say that for triangle CFD: x+(x+y)+y = 180.\n\nCorrect?[/quote]\r\n\r\nyeah that's it" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Define $\\phi_{n}\\in (l^\\infty)^{*}$ by $\\phi(f)=n^{-1}\\sum_{j=1}^{n}f(j).$Then the sequence $\\{\\phi_{n}\\}$ \r\nhas a weak* cluster point $\\phi$, and $\\phi$ is an elemnet of $(l^\\infty)^{*}$\r\nthat does not arise from an element of $l^{1}.$\r\n\r\n( I think since $\\parallel \\phi_{n}\\parallel \\leq 1$, so by Alaoglu Thm, this sequence is \r\nrelatively weak* compact, so must have a cluster point, but \r\nwhat is it? And why it is not from $l^{1}$?)", "Solution_1": "$\\phi\\notin\\ell_{1}$ because it vanishes on all sequences with finite support." } { "Tag": [], "Problem": "Prove the Ceva Theorem using the Menelaus Theorem.", "Solution_1": "I am sure that this was already discussed in this forum.But to prove Menelaus using [b]only [/b] Ceva is much difficulter.Lets ask [b]darij[/b] to give us a link .:)", "Solution_2": "Oh well, the proof of the Ceva theorem using the Menelaos theorem is quite well-known and easy: See http://www.cut-the-knot.org/Generalization/Menelaus.shtml#CfromM .\r\n\r\n Darij" } { "Tag": [ "trigonometry", "calculus", "derivative", "function", "calculus computations" ], "Problem": "Find the derivative of $ y\\equal{}\\tan{^{\\minus{}1}(x\\minus{}\\sqrt{1\\plus{}x^{2}})}$ and simply where possible.\r\n\r\nAll I get is one big mess with nothing significant that will cancel, except for a $ 2$ but I can't seem to simplify it.", "Solution_1": "Please can you post how far you've got. I've just done that now and it all seems to cancel fairly quickly for me so I think you may have made a mistake somewhere", "Solution_2": "I got\r\n\\[ \\frac {dy}{dx} = \\frac {1}{1+(x - \\sqrt {1 - x^{2}})^{2}}\\frac {d}{dx}(x - \\sqrt {1 + x^{2}})\r\n\\]\r\n\r\n\\[ = \\frac {1}{1 + (x - \\sqrt {1 + x^{2}})^{2}}\\cdot{(1} - \\frac {2x}{2\\sqrt {1 + x^{2}}})\r\n\\]\r\n\r\n\\[ = \\frac {2\\sqrt {1 + x^{2}} - 2x}{(1 + (x - \\sqrt {1 + x^{2}})^{2})(2\\sqrt {1 + x^{2}})}\r\n\\]\r\nI have a nasty feeling that I messed up somewhere but I checked over a ton of times. Unless I'm just not seeing some really simple cancellation...", "Solution_3": "It is correct. But you could try to transform it to the form: $ \\frac{1}{2(x^2\\plus{}1)}$. I think that's the final form you are after...", "Solution_4": "From the original equation we have:\r\n1) $ \\tan(y) \\equal{} x \\minus{} \\sqrt{1 \\plus{} x^2}$\r\n\r\n2) $ \\tan^{2}(y) \\plus{} 1 \\equal{} 2(x^2 \\minus{} x\\sqrt{1 \\plus{} x^2} \\plus{}1)$\r\n\r\n3) Differentiating 1) we get,\r\n\r\n $ (\\tan^{2}(y) \\plus{} 1)dy \\equal{} (1 \\minus{} \\frac{x}{\\sqrt{1 \\plus{} x^2}})dx \\equal{} (\\frac{x^2 \\minus{}x\\sqrt{1 \\plus{}x^2} \\plus{} 1}{1 \\plus{} x^2})dx$.\r\n\r\nHence, $ \\frac{dy}{dx} \\equal{} \\frac{(\\frac{x^2 \\minus{}x\\sqrt{1 \\plus{}x^2} \\plus{} 1}{1 \\plus{} x^2})}{(\\tan^2(y) \\plus{} 1)} \\equal{} \\frac{(\\frac{x^2 \\minus{}x\\sqrt{1 \\plus{}x^2} \\plus{} 1}{1 \\plus{} x^2})}{2(x^2 \\minus{} x\\sqrt{1 \\plus{} x^2} \\plus{}1)} \\equal{} \\frac{1}{2(1 \\plus{}x^2)}$", "Solution_5": "So, in hindsight, how could the original function be simplified? If $ x\\equal{}\\tan\\theta$, we're looking at $ \\arctan(\\tan\\theta\\minus{}\\sec\\theta)\\equal{}\\arctan\\frac{\\sin\\theta\\minus{}1}{\\cos\\theta}$. That's a version of a half-angle formula- $ \\tan\\frac{\\alpha}{2}\\equal{}\\frac{1\\minus{}\\cos\\alpha}{\\sin\\alpha} \\equal{}\\frac{\\sin\\alpha}{1\\plus{}\\cos\\alpha}$. To convert this into what we need, let $ \\theta\\equal{}\\alpha\\plus{}\\frac{\\pi}{2}$; that means $ \\sin\\theta\\equal{}\\cos\\alpha$ and $ \\cos\\theta\\equal{}\\minus{}\\sin\\alpha$. Our original function is $ \\arctan\\left(\\tan\\frac{\\theta\\minus{}\\frac{\\pi}{2}}{2}\\right)\\equal{} \\frac{\\theta\\minus{}\\frac{\\pi}{2}}{2}\\equal{}\\frac{\\arctan x}{2}\\minus{}\\frac{\\pi}{4}$." } { "Tag": [ "inequalities" ], "Problem": "Is the \"Trivial Inequality\" referred to in AoPS as x^2 >= 0 really called the \"Trivial Inequality\" (the name I would use in an actual olympiad), or is it just an inequality that seems trivial?\r\n\r\nFierytycoon", "Solution_1": "It's actually called the Trivial Inequality.", "Solution_2": "thx for the info" } { "Tag": [], "Problem": "101 ones + 101 tens + 101 hundreds is equal to what number?", "Solution_1": "[quote=\"math92\"]101 ones + 101 tens + 101 hundreds is equal to what number?[/quote]\r\n\r\n[hide=\"answer\"] \n101 *1= 101\n101 * 10= 1010\n101*100=10100\n\n101 + 1010 + 10100= [b]11211[/b] [/hide]", "Solution_2": "[hide]101 + 1010 + 10100 =11211[/hide]", "Solution_3": "[hide]$101 + 1010 + 10100= 11211 $[/hide]", "Solution_4": "[hide=\"Answer\"]11211[/hide]", "Solution_5": "[hide]\n101 ones=101\n101 tens=1010\n101 hundreds=10100\nso, 101 + 1010 + 10100 =11211 [/hide]", "Solution_6": "101 ones=101\r\n101tens=1010\r\n101hundreds=10100\r\nso it would be 101=1010=10100=11211" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "1)Find all n such that \r\n$ 2^{\\phi(n)\\minus{}1}|n^{n}$\r\nSome problem :\r\nhttp://diendantoanhoc.net/forum/index.php?showtopic=33495\r\nNote :So nguyen to :Prime\r\n So nguyen duong :Natural number\r\n Tim :Find all", "Solution_1": "First I think you intend $ 2^{\\phi (n)}\\minus{}1|n^{n}$ :wink: \r\n\r\nLet $ n\\equal{}2^{k}p_{1}^{k_{1}}\\cdots p_{r}^{k_{r}}$ suppose for a moment that $ k\\geq 1$ then $ 2^{k\\plus{}r\\minus{}1}|\\phi(n)$\r\nso if $ \\phi(n)\\equal{}2^{s}\\cdot m$ then $ (2^{m}\\minus{}1)(2^{m}\\plus{}1)\\cdots(2^{2^{s\\minus{}1}m}\\plus{}1)|n^{n}$\r\nbut $ n$ has $ r$ odd prime factors and so $ k\\plus{}r\\minus{}1\\le s\\le r$ so we conclude that the solutions have $ k\\equal{}0$ or $ k\\equal{}1$ but in both cases $ s\\equal{}r$ so each $ 2^{2^{i}m}\\plus{}1$ is a prime power. From here we can easily find solutions $ n\\equal{}2,3,6$", "Solution_2": "[quote=\"TTsphn\"]1)Find all n such that \n$ 2^{\\phi(n)\\minus{}1}|n^{n}$\nSome problem :\nhttp://diendantoanhoc.net/forum/index.php?showtopic=33495\nNote :So nguyen to :Prime\n So nguyen duong :Natural number\n Tim :Find all[/quote]\r\nObviosly all solutions are even numbers and n=1.", "Solution_3": "The way I wrote in my post the problem is from Balkan MO shortlist 2005 (Proposed by Albania, E.Pisha)" } { "Tag": [ "analytic geometry", "geometry" ], "Problem": "The vertices of triangle $ PQR$ have coordinates as follows: $ P(0,a),\\,Q(b,0),\\,R(c,d),$ where $ a,\\,b,\\,c$ and $ d$ are positive. The origin and point $ R$ lie on opposite sides of $ PQ$. The area of triangle $ PQR$ may be found from the expression:\r\n\r\n$ \\textbf{(A)}\\ \\frac{ab \\plus{} ac \\plus{} bc \\plus{} cd}{2} \\qquad \r\n\\textbf{(B)}\\ \\frac{ac \\plus{} bd \\minus{} ab}{2}\\qquad \r\n\\textbf{(C)}\\ \\frac{ab \\minus{} ac \\minus{} bd}{2}\\qquad \r\n\\textbf{(D)}\\ \\frac{ac \\plus{} bd \\plus{} ab}{2}\\qquad \r\n\\textbf{(E)}\\ \\frac{ac \\plus{} bd \\minus{} ab \\minus{} cd}{2}$", "Solution_1": "By using shoelace you get C.", "Solution_2": "I think you reversed the order for shoelace, because the actual answer is $ \\boxed{(B)}$.", "Solution_3": "[quote=\"sunehra\"]I think you reversed the order for shoelace, because the actual answer is $ \\boxed{(B)}$.[/quote]\r\nI did it the wrong way :blush:" } { "Tag": [ "group theory", "abstract algebra", "number theory", "relatively prime", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $ p$ be the smallest prime number such that $ p\\mid |G|$ , where $ G$ is finite group , if $ H\\leq G$ and $ [G: H] \\equal{} p$ , show that $ H\\triangleleft G$.", "Solution_1": "Very classical.\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=215134\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=62685\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=22672\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=234462\r\n\r\n darij", "Solution_2": "Thank you :)", "Solution_3": "A very [b]elementary[/b] proof of this result follows from this observation:\r\n\r\n-- Let p be the smallest prime dividing |G| and let $ g \\in G \\minus{} H$. Then, \r\n\r\n$ H, gH, \\ldots, g^{p \\minus{} 1}H$\r\n\r\nare pairwise disjoint and\r\n\r\n$ G \\equal{} \\bigcup_{k \\equal{} 0}^{p \\minus{} 1} g^{k}H.$\r\n\r\nNone of the proofs mentioned by [b]Darij[/b] had proceeded along these lines. :wink:", "Solution_4": "[quote=\"coquitao\"]A very [b]elementary[/b] proof of this result follows from this observation:\n\n-- Let p be the smallest prime dividing |G| and let $ g \\in G \\minus{} H$. Then, \n\n$ H, gH, \\ldots, g^{p \\minus{} 1}H$\n\nare pairwise disjoint and\n\n$ G \\equal{} \\bigcup_{k \\equal{} 0}^{p \\minus{} 1} g^{k}H.$\n\n[/quote]\r\n\r\nHow do u get the normality of $ H$, from here , also how do u assure that $ g^i \\ne g^j, i\\ne j, 1\\leq i,j i > 0$, it might not be the original $ g$, am I right?. But in this case, because he choose $ g$ to identify $ mnm^{ \\minus{} 1}$, his proof went smoothly.\r\n\r\nNow if $ m n m^{ \\minus{} 1} \\equal{} g^i h'$ and $ m \\equal{} g^j h$, then you cannot deduce anything.", "Solution_18": "Let's be clear: there is no [i]original[/i] $ g$. You assume that $ g\\equal{}mnm^{\\minus{}1} \\not\\in H$, then you look at $ H, gH,\\ldots g^{p\\minus{}1} H$ and so forth. Mathematical arguments do not always occur in the order in which they are described. :)", "Solution_19": "You are absolutely right, LydianRain! :wink:", "Solution_20": "[quote=\"Hidden Scofield\"]That I got from his solution ...\n\nNote that $ |g| \\equal{} p$.So, $ g,g^2,\\cdots,g^{p \\minus{} 1}$ are distinct .\n\n[/quote]\r\n\r\nI hate to revive this thread this way, but I thought I had to do it. First off: it's not valid to suppose that [b]g[/b] is always going to have order [i]p[/i]. Wonder why it was that Scofield assumed that in the first place. I have to admit that I overlooked this important point on the first perusal of his post. Needless to say, the actual remark renders meaningless most of the comments in our discussion, except for the first four of 'em and the two previous-ones... :wallbash_red:" } { "Tag": [ "function", "calculus", "derivative", "complex analysis", "absolute value", "topology" ], "Problem": "I would like to ask a stupid question: What is the definition of an analytic function? Some text consider an analytic function $ f$ on $ S$ as a differentiable function on $ S$, some regard it as a continuously differentiable on $ S$ (i.e. $ f\\in\\mathcal C^1$), and others regard it as a function which can be represented by a power series. So, which one is more popular?", "Solution_1": "You have to be careful with your choice of words. Complex functions can be differentiable (at points) without being analytic. For example, the absolute value function is differentiable almost everywhere but it is not analytic.\r\n\r\nFormally, a function is analytic on an open set $ D$ if for any $ x_{0}\\in D$, $ f(x)\\equal{}\\sum_{n\\equal{}0}^{\\infty}a_{n}(x\\minus{}x_{0})^{n}$ and the series is convergent for $ x$ in a neighborhood of $ x_{0}$.", "Solution_2": "If by differentiable in $ S$ you mean complex differentiable in an open set containing $ S$, then all three definitions are equivalent. It's completely different from the real variable setting, where differentiable, continuously differentiable and real analytic are three distinct concepts.\r\n\r\nJRav, what do you mean? $ f(x \\plus{} iy) \\equal{} x^2 \\plus{} y^2$ satisfies the C-R equations only at 0, and there the absolute value fails to be real differentiable, so the absolute value function has a complex derivative nowhere.", "Solution_3": "All three properties turn out to be equivalent, but most people use different words to emphasize different aspects:\r\n-differentiable is used to mean just that (it holds meaning for reals too, in which case it is not equivalent to the others)\r\n-holomorphic applies to complex functions only, implying that they are continuously differentiable \r\n-analytic implies that it converges to its power series in the neighborhood of every point" } { "Tag": [ "quadratics", "Euler", "function", "number theory", "prime numbers", "arithmetic series", "algebra" ], "Problem": "I may have proof here.\r\n\r\n-y^2 = 4 - x^5\r\ny^2 = x^5 - 4\r\ny = sqrt(x^5 - 4)\r\nsqrt(x^5 - 4) must be an integer.\r\n \r\nIn order for a number to have an integer square root, each of prime factors must have an even number (i.e. 2^2 * 3^2 * 5^6) . That is because the primes must be able to be split in half.\r\n\r\nSo I don't have to keep on typing x^5, let z = x^5.\r\n\r\nLet's assume that sqrt(z) is an integer.\r\n\r\nTherefore, if sqrt(z-4) were to still to have an integer square root, the subtraction of 4 must have taken out two prime numbers, p. // Edit: I'm not sure, but this logic might not be sound.\r\n\r\nEdit 2: Nope. Drats. There goes my proof.\r\n\r\nEdit 3: Well, I think I may have another solution. \r\nz and z-4 must both be perfect squares. \r\n1, 4, 9, 25, etc. are perfect squares.\r\nThe differences between these squares is an arithmetic series with a cd of 2, starting at 3. Therefore, z and z-4 can never both be perfect squares.\r\n\r\n\r\nIgnore\r\n-----------------\r\n\r\nz-4 = z/(p^2)\r\nz = z/(p^2) + 4\r\nzp^2 = z + 4p^2\r\nzp^2 - 4p^2 - z = 0\r\n(z-4)p^2 - z = 0\r\nTherefore, it is a quadratic equation where a = (z-4), b = 0, and c = -z\r\n\r\nIn order for the quadratic equation to have a real solution, sqrt(b^2 - 4*a*c) must be >= 0.\r\n\r\nsqrt(b^2 - 4*a*c) here is sqrt(0 - 4*(z-4)(-z))\r\nsqrt(0 - (4z-16)(-z))\r\nsqrt(0 - (-4z^2 + 16z))\r\nsqrt(4z^2 - 16z)\r\nIf 4z^2 - 16z must be greater or equal to 0, \r\n4z^2 >= 16z.\r\nz^2 >= 4z\r\nz >= 4\r\n\r\nThe whole quadratic equation for (z-4)p^2 - z would be \r\n0 + (can't be minus) sqrt(4z^2 - 16z)/(2*(z-4))\r\nsqrt(4z^2 - 16z)/(2z-8)\r\nSquare both the numerator and denominator:\r\n(4z^2 - 16z)/(4z^2 - 32z + 64)\r\n4z^2 - 16z > 4z^2 - 32z + 64\r\n-16z > -32z + 64\r\n-16z > 64 - 32z\r\n16z > 32z - 64\r\n0 > 16z - 64\r\n64 > 16z\r\n4 > z\r\n\r\nProof by contradiction.\r\n\r\n\r\nI'm a freshman in highschool, so please forgive any problems I have with presenting a proof.\r\n\r\nI also hope I didn't make a silly arithmetic error, screwing my whole proof up. :?", "Solution_1": "The only residues of x^5 (mod 11) are 0, 1, -1.\r\n\r\nThe quadratic residues modulo 11, or the possible values of y^2 (mod 11), are 0, 1, 4, 9, 5, 3.\r\nSo the possible values of y^2 + 4 (mod 11) are 4, 5, 8, 2, 9, 7.\r\n\r\nSo x^5 cannot equal y^2 + 4 (mod 11), and x^5 - y^2 = 4 (mod 11) has no integer solutions. Hence, x^5 - y^2 = 4 has no integer solutions.\r\n\r\n\r\nI found 11 through a brute force search. Is there a general way to figure out what you can mod x^n by so that the only values are {0, 1, -1}?", "Solution_2": "Why must z-4 be z/p^2? \r\n\r\nIt might be some perfect square that has factors other than those of z. For example, suppose 4 were 40 instead: 7^2-40=3^2.", "Solution_3": "Right. I realize that both of my solutions are wrong.", "Solution_4": "Nice solutions, yoyo.\r\n\r\nHow did you think to use mods? \r\n\r\nAre there any budding number theorists out there who want to take a swing at yoyo's general question?", "Solution_5": "If p > 2 is prime, a^(p-1) = 1 (mod p) if (a,p) = 1 (Fermat little theorem)\r\n\r\nSo (a^(p-1)/2)^2 = 1 (mod p), so (Z/pZ is a field) a^(p-1)/2 in {1,-1}\r\n\r\n\r\nConclusion :\r\n\r\nif p is prime > 2, the values of a^(p-1)/2 are in {0, 1, -1}\r\n\r\n\r\nSet p =11", "Solution_6": "[quote=\"rrusczyk\"]How did you think to use mods? \n[/quote]\r\n\r\nMods seemed to be useful on problems of a similar form, where there are no integer solutions, which I've seen before. There are also some similar ideas (and problems) in the last part of the Diophantine Equations chapter in [u]AoPS[/u] book 2. I think the advantage of using mods is that you only have to check finitely many values of x and y, instead of infinitely many. Also, if two numbers are equal, then they're equal modulo a for all a, so the contrapositive is useful (not equal (mod a) => not equal).", "Solution_7": "I am interested in yoyo's (Tim's?) question...is there a way to find a mod n where the solutions of x^n are only {-1,0,1} ???", "Solution_8": "[quote=\"akatookey\"]I am interested in yoyo's (Tim's?) question...is there a way to find a mod n where the solutions of x^n are only {-1,0,1} ???[/quote]\r\n\r\nTake a look at i/3's post above. In case you're a little unfamiliar with the notation, (a,p) is the greatest common divisor of a & p. \r\n\r\nIf you're not convinced, look at n^2 (mod 3), n^4 (mod 5), n^6 (mod 7), n^10 (mod 11), and so on.\r\n\r\nThen, when you're convinced... It's time to try proving it.", "Solution_9": "[quote=\"yoyo\"][quote=\"rrusczyk\"]How did you think to use mods? \n[/quote]\n\nMods seemed to be useful on problems of a similar form, where there are no integer solutions, which I've seen before. There are also some similar ideas (and problems) in the last part of the Diophantine Equations chapter in [u]AoPS[/u] book 2. I think the advantage of using mods is that you only have to check finitely many values of x and y, instead of infinitely many. Also, if two numbers are equal, then they're equal modulo a for all a, so the contrapositive is useful (not equal (mod a) => not equal).[/quote]\r\n\r\nYeah, these sort of questions which involve too many variables to just solve, almost always use mods. Even if it does not solve the equation outright, it will often give some useful information. For example, it is 'wellknown' that odd squares are 1 mod 8 (you can check this if you like.) Considering the equation mod 8, if x and y are even we get -1 == 4 mod 8, not true, and if they are both odd then x^5 must be 5 mod 8, and so x must be 5 mod 8. Usually finding the right mod (or mods) will solve the question.", "Solution_10": "Heres another problem with exactly the same idea as the above one. Prove that the equation x^3 + y^3 + z^3 = 2003 has no integer solutions. (Hint : think mods!!)", "Solution_11": "2003 mod 9 = 5\r\n\r\nx^3 mod 9 in {0,1,-1} so x^3 + y^3 + z^3 always different from 5 mod 9.\r\n\r\n\r\nWhy 9 ???\r\n\r\n\r\nEuler theorem :\r\n\r\na^phi(n) = 1 mod n if (a,n) = 1 (n prime or not).\r\nphi being the Euler function\r\n\r\nphi(9) = 6\r\n\r\nso a^6 = (a^3)^2 = 1 if (a,9) = 1.", "Solution_12": "I don't have anything definite yet, but I was messing around with this in math class. Since sqrt(x^5-4) has to be an integer, you need to find an integer^5 that is 4 greater than a square. I tested this for x'es of 2 to 10 and this is what I got...\r\nX----x^5-4---Nearest square---------Value of-------Difference between\r\n------------number below x^5-4----that square----x^5-4 and the square\r\n2___28________5^2_________________25_______________3\r\n3___239_______15^2________________225_____________14\r\n4___1020______31^2________________961_____________59\r\n5___3121______55^2________________3025____________96\r\n6___7772______88^2________________7749____________23\r\n7___16803_____129^2_______________16641__________162\r\n8___32764_____181^2_______________32761__________3\r\n9___59045_____242^2_______________58564__________481\r\n10__99996_____316^2______________99956___________140\r\n\r\nI tested this for only 1,2,3, and 4 at first. I was really excited because I noticed that the differences kept on increasing. That theory was blown away when I continued the pattern. I put this up here because someone might be able to use it, perhaps there is a pattern in the differences or the squares? \r\nSorry about the messy underscores- the posting program deletes additional spaces. Hope that you can read this okay.\r\n\r\n\r\n-Isaac" } { "Tag": [ "analytic geometry", "combinatorics solved", "combinatorics" ], "Problem": "Is the graph K_5 planar?", "Solution_1": "It is a very very well-known problem....\r\nYou may find a proof that the answer is negative, in french, at :\r\nhttp://www.animath.fr\r\n\r\ngo to 'Les cours de l'olympiade franaise de mathmatiques'\r\nthen :\r\n'Les graphes'\r\n\r\nPierre.", "Solution_2": "Kuratowsky theorem: \r\nGraph F is planar if and only if it does not contain subgraph isomorphic to K_5 or K_(3,3).\r\n\r\nNamdung", "Solution_3": "You are obviously right, Namdung. But, in our case, we do not have to deal with the general case of Kuratowski theorem, which is difficult to prove.\r\nTo prove that K_5 and K_3,3 are not planar, you only need Euler's formula for planar graphs.\r\n\r\nPierre.", "Solution_4": "I hadn't heard of Kuratowsky's thm before. That's impressive.\r\n\r\nTry this (I haven't tried it, so I don't know the answer :D ): Is K_(3,3) planar on the surface of a thorus?", "Solution_5": "Yes, both K_5 and K_3,3, are planar on the torus. It may be shown that even K_7 is, but it is too boring to describe with a keyboard. Let's try for the two others :\r\n \r\nThe torus may be seen as a square whose sides are vertically and horizontally, and for which, when going outside from one vertical side you are coming inside from the opposite side, at the same y-coordinate, and the same is for the horizontal sides (for the x-coordinate). So :\r\n\r\n- For K_5 :\r\nInto a big square (the torus), draw a little square ABCD where the line (AC) goes through the midpoints of opposite sides of the big square, and the same for (BD). Note that on the torus, these two midpoints represent the same point.\r\nLet O be the center of ABCD.\r\nWe may draw each sides of ABCD and each segments connecting O to the A,B,C,D. From the orientation of ABCD, we may connect A to C 'from the outside'. And we do the same for B and D, with respect to the other direction of the torus.\r\n\r\n- For K_3,3 :\r\nChoose the same little square ABCD. And choose the common midpoints of pair of opposite sides, say E and F (see above), where E is collinear (for the big square) to A and C, and F with B and D.\r\nThen draw ABCD and connect both A and C to E, and both Band D to F. It remains to connect E and F, which is easy.\r\n\r\nPierre.", "Solution_6": "I saw this puzzle-type question on a website: \r\n\r\nA physicist wants to determine whether he lives in a plane or on the surface on the torus. How can he do that having only a rope? \r\n\r\nI don't know what the answer is, but maybe they're somehow related." } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "Let $ a_{0}=29,a_{1}=105,a_{2}=381$ and $ a_{n+3}=3a_{n+2}+2a_{n+1}+a_{n}\\forall n=0,1,2,...$\r\n\r\nProve that for each positive integer $ m$, there exists a number $ n$ such that $ a_{n},a_{n+1}-1,a_{n+2}-2$ are all divisible by $ m$.", "Solution_1": "It is seminar a problem in Romani Olympiad on 1998.\r\nConsider the sequence\r\n$ r_0 \\equal{} 0,r_1 \\equal{} 1,r_{2} \\equal{} 3$ \r\n$ r_{n \\plus{} 3} \\equal{} 3r_{n \\plus{} 2} \\plus{} 2r_{n \\plus{} 1} \\plus{} r_n$\r\n$ r_4 \\equal{} a_0,r_{5} \\equal{} a_1,r_{6} \\equal{} a_2$ so $ r_{n \\plus{} 4} \\equal{} a_n$\r\nBecause $ \\{r_n\\}(\\mod m)$ is periodic so has infinite n such that $ m|a_n,a_{n \\plus{} 1} \\minus{} 1,a_{n \\plus{} 2} \\minus{} 2$" } { "Tag": [ "modular arithmetic", "algebra", "polynomial" ], "Problem": "The natural number $ n$ has the property that $ n^{2}\\plus{}1$ is divisible by $ n\\plus{}1$. Find $ n$.", "Solution_1": "When $ n \\equal{} 1$, $ n^2 \\plus{} 1 \\equal{} 2$ and $ n \\plus{} 1 \\equal{} 2$. Clearly 2 divides itself, so the natural number is $ \\boxed{n \\equal{} 1}$.\r\n\r\nAlso, note that this value is the only satisfactory $ n$. Here's the brief proof:\r\n\r\nTake $ n^2 \\minus{}1$, which we can factor as $ (n \\plus{} 1)(n \\minus{} 1)$. We see that $ n \\plus{} 1$ is a factor of $ (n^2 \\plus{} 1) \\minus{} 2$.\r\n\r\nThus, the only case for which $ n^2 \\plus{} 1 \\equiv [(n^2 \\plus{} 1) \\minus{} 2] \\pmod {n \\plus{} 1}$ is if $ 2 \\equiv 0 \\pmod {n \\plus{} 1}$, so 2 is a multiple of $ n \\plus{} 1$.\r\n\r\nThe only solution for $ n \\in N$ is $ \\boxed{n \\equal{} 1}$.", "Solution_2": "Thanks goldenboy1.618. \r\n\r\n$ n^{2}\\minus{}1\\equal{}(n\\minus{}1)(n\\plus{}1) \\implies (n\\plus{}1) \\mid (n^{2} \\minus{}1)$.\r\n\r\n$ (n\\plus{}1) \\mid (n^{2} \\plus{}1)$ and $ (n\\plus{}1) \\mid (n^{2} \\minus{}1) \\implies (n\\plus{}1)$ divides their difference, that is, $ (n\\plus{}1) \\mid 2$.\r\n\r\nBut $ n$ is a natural number, so $ n\\plus{}1 \\ge 2$. So, $ n\\plus{}1 \\equal{} 2$, that is, $ n\\equal{}1$.", "Solution_3": "[hide=\"Alternate Solution\"]\nIf $ \\dfrac{n^2 \\plus{} 1}{n \\plus{} 1}\\in \\mathbb{Z}\\implies \\dfrac{n^2 \\plus{} 1}{n \\plus{} 1} \\minus{} (n \\minus{} 1) \\equal{} \\dfrac{2}{n \\plus{} 1}\\in \\mathbb{Z}$, from where $ n \\equal{} 1$ is the only solution.\n[/hide]\r\n\r\nEdit: Beaten..", "Solution_4": "Thanks tonypr.", "Solution_5": "Usually for those sorts of problems a technique that (almost)always works is polynomial division. \r\n\r\nSo in the case of this problem, we have $ \\frac {n^2 \\plus{} 1}{n \\plus{} 1} \\in \\mathbb{N}.$ Using synthetic division, we get $ \\frac {n^2 \\plus{} 1}{n \\plus{} 1} \\equal{} n \\minus{} 1 \\plus{} \\frac {2}{n \\plus{} 1}.$ Since the RHS must be an integer, $ (n \\plus{} 1) \\ | \\ 2,$ so the only natural number that satisfies that is $ n \\equal{} 1.$" } { "Tag": [], "Problem": "What is the hundreds digit in the following product:\n$ 5 \\times 6 \\times 7 \\times 8 \\times 9 \\times 10$ ?", "Solution_1": "alright: multiply the first 4 numbers, and find the tens digit. that wil be the hundreds digit b/c when you multiply by 10 you add a 0 and the 10s become hundreds. we only care abt the 10s place\r\n\r\n5*6=3\r\n3*7=1\r\n1*8=8\r\n8*9=2\r\n\r\nso 2 is the hundreds place" } { "Tag": [], "Problem": "I stumbled upon this recently. To my understanding, they are opening in 2009-2010 school year, enrolling 40 juniors each year for a 2 year program. Their site has an application.\r\n\r\nI'm pretty sure I'm a year too old to apply, but I thought I'd share the word.\r\n\r\nhttp://kams.fhsu.edu/", "Solution_1": "Not sure if this will be useful for me but thanks for the info!" } { "Tag": [ "function", "calculus", "derivative", "calculus computations" ], "Problem": "I am having a bit of trouble understanding how to prove the basic identities of the Poisson Brackets... could someone help me out?\r\n\r\nFor two given sets of $ n$ points $ P: {p_1, p_2, \\cdots, p_n}$ and $ Q: {q_1, q_2, \\cdots, q_n}$ and two functions $ f, g$ which are functions of $ p_i, q_i,$ the Poisson Bracket is defined to be\r\n\r\n\\[ [f, g] \\equal{} \\sum{\\frac{\\partial f}{\\partial q_i} \\frac{\\partial g}{\\partial p_i} \\minus{} \\frac{\\partial g}{\\partial q_i} \\frac{\\partial f}{\\partial p_i}}\r\n\\]\r\n\r\nThe identity I'm having trouble with proving is\r\n\r\n\\[ [f, q_i] \\equal{} \\minus{}\\frac{\\partial f}{\\partial p_i} \\equal{} \\frac{\\partial f}{\\partial q_i}\r\n\\]\r\n\r\nGenerally, I only know the definitions and minimal knowledge of partial differentiation, but no fancy manipulations. I am much better with single variable calculus. Could someone post a proof of the above identity? Thanks. :)", "Solution_1": "Substitute $ g \\equal{} q_k$ into the definition and you get\r\n\\[ [f, q_k] \\equal{} \\sum_i{\\frac {\\partial f}{\\partial q_i} \\frac {\\partial q_k}{\\partial p_i} \\minus{} \\frac {\\partial q_k}{\\partial q_i} \\frac {\\partial f}{\\partial p_i}} \\equal{}\r\n\\]\r\nNow all the $ q_i$ and $ p_i$ are independent variables, so $ \\frac {\\partial q_k}{\\partial p_i} \\equal{} 0$ for all $ i$ and $ \\frac {\\partial q_k}{\\partial q_i} \\equal{} \\begin{cases} 1 & k \\equal{} i \\\\\r\n0 & k\\neq i \\end{cases}$. Hence,\r\n\\[ [f,q_k] \\equal{} \\minus{} \\frac {\\partial f}{\\partial p_k}\r\n\\]", "Solution_2": "Ah, okay... I was just making sure that we could set $ \\partial q / \\partial p \\equal{} 0$. As a general idea, is this allowed?:\r\n\r\n\\[ \\frac{\\partial f}{\\partial x} \\frac{\\partial x}{\\partial y} \\equal{} \\frac{\\partial f}{\\partial y}\r\n\\]" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Solve in $x,y,z$:-\r\n\t$a = (2x + y + z)^2 -4x^2$\r\n\t$b = (2y + x + z)^2 -4y^2$\r\n\t$c = (2z + y + x)^2 -4z^2$\r\nUnder what co ndition on $a,b,c$ we get positive solutions?", "Solution_1": "[quote=\"rohitsingh0812\"]Solve in $x,y,z$:-\n\t$a = (2x + y + z)^2 -4x^2$\n\t$b = (2y + x + z)^2 -4y^2$\n\t[b]$c = (2z + y + z)^2 -4z^2$[/b]\nUnder what co ndition on $a,b,c$ we get positive solutions?[/quote]\r\nIt's supposed to be $c = (2z + y + x)^2 -4z^2$, right?" } { "Tag": [ "geometry", "3D geometry", "probability" ], "Problem": "Two cubes, each number with the integers 4 through 9, are tossed. What is the probability that the product of the two numbers rolled is greater than 29 and less than 61? Express your answer as a common fraction.", "Solution_1": "I'm pretty positive that my answer is correct so if you want to look at the answer before posting, look at mine.\r\n\r\nBut don't look at it before you solved it :D \r\n\r\n[hide=\"Solution\"]\n\nThere are 6 choices on the first time and another 6 chocies on the second time.\n\nThe pairs that are LESS than 29 or GREATER than 61 are:\n\n4-4\n4-5\n4-6\n4-7\n5-4\n6-4\n7-4\n\n9-9\n9-8\n9-7\n8-9\n7-9\n\nOops.. There are 5*5. \n\nBut we want the ones that are in that range so subtract the 13 from total. The total, we found on the beginning is 36, so the amount on the top is 36-13 = 23.\n\nSimplify the fraction:\n\n\\[\\frac {23}{36}$[/hide]", "Solution_2": "This casework is not yet complete.", "Solution_3": "[hide]Find what we don't want\n\n\nBelow 29, there are:\n16\n20\n24\n25\n28\n\nThere are 5*2 below 29\n\nNow above 61:\n63\n64\n72\n81\n\nThere are 4 above 61\n\nThere are 6*6/2 choices total, so there are 18-9=9 \nthat work so the probability is 9/18=1/2[/hide]", "Solution_4": "Solafidefarms,\r\n\r\nTake another look at the method you've used to count everything.", "Solution_5": "[hide=\"I think this is it...\"]11/18\n\nI tried them all..\n\n(4*8) , (4*9)\n(5*6) , (5*7) , (5*8) , (5*9)\n(6*5) , (6*6) , (6*7) , (6*8) , (6*9)\n(7*5) , (7*6) , (7*7) , (7*8) \n(8*4) , (8*5) , (8*6) , (8*7)\n(9*4) , (9*5) , (9*6)\n\nI don't really know how to solve it in a neat way, but I did it anyway because I was bored yesterday..[/hide]", "Solution_6": "[color=green]\n\n[hide]11/18 :D[/color][/hide]", "Solution_7": "[quote=\"MCrawford\"]Two cubes, each number with the integers 4 through 9, are tossed. What is the probability that the product of the two numbers rolled is greater than 29 and less than 61? Express your answer as a common fraction.[/quote]\r\n\r\n[hide]\nWe know that the product of the two numbers must be 29-61. The pairs that would satisfy that are:\n\n4, 8\n4, 9\n\n5, 6\n5, 7\n5, 8\n5, 9\n\n6, 5\n6, 6\n6, 7\n6, 8\n6, 9\n\n7, 5\n7, 6\n7, 7\n7, 8\n\n8, 4\n8, 5\n8, 6\n8, 7\n\n9, 4\n9, 5\n\nThat's 22 possibilities to satisfy the requirement. There are 36 combinations in total, because 6*6 is 36.\n\n22/36 = 11/18\n[/hide]", "Solution_8": "I hate probability problems, i usually miss them by a very little number...is the only way to do this problem listing them?", "Solution_9": "[quote=\"LuCy4EvA\"]I hate probability problems, i usually miss them by a very little number...is the only way to do this problem listing them?[/quote]\r\n\r\nI don't think so. Ii usually make a 6x6 multiplication table. And just do all the multiplication to find it all the possibilities.", "Solution_10": "[quote=\"SoccerBrainy40\"]\nWe know that the product of the two numbers must be 29-61. The pairs that would satisfy that are: \n\n4, 8 \n4, 9 \n\n5, 6 \n5, 7 \n5, 8 \n5, 9 \n\n6, 5 \n6, 6 \n6, 7 \n6, 8 \n6, 9 \n\n7, 5 \n7, 6 \n7, 7 \n7, 8 \n\n8, 4 \n8, 5 \n8, 6 \n8, 7 \n\n9, 4 \n9, 5 \n\nThat's 22 possibilities to satisfy the requirement. There are 36 combinations in total, because 6*6 is 36. \n\n22/36 = 11/18 \n[/quote]\r\nbtw, you forgot 9,6... which leads to my answer... instead of yours.\r\n\r\n[hide=\"solution\"]find the multiple that are less than 29 or greater than 61. \n(4,4)\n(4,5)(5,4)\n(4,6)(6,4)\n(4,7)(7,4)\n\n(8,8)\n(8,9)(9,8)\n(9,9)\n(7,9)(9,7)\n\ntotal = 13.\n\nsubtract from ALL possibilities = 36:\n\n36-13 = 23, $\\frac{23}{36}$[/hide]", "Solution_11": "oops... i understood the problem wrong... \r\n\r\n[hide=\"solution\"]\nsince they are the same kind of dice, a 4 on one of them and 6 on the other is not different from 6 on one and 4 ont he other...\n\nmeaning, there are only 21 possibilities in total.\n\n4,4 \n4,5\n4,6\n4,7\n4,8\n4,9\n5,5\n5,6\n5,7\n5,8\n5,9\n6,6\n6,7\n6,8\n6,9\n7,7\n7,8\n7,9\n8,8\n8,9\n9,9\n\nout of those, there are 9 which does not satisfy the given property. \n\nso, $1 - \\frac{9}{21} = \\frac{12}{21} = \\frac{4}{7}$\n[/hide]\r\n\r\ni hope that is the right way to interpret the probelm..." } { "Tag": [ "geometry", "circumcircle", "geometry unsolved" ], "Problem": "Let in quadrilateral $ABCD$ be $\\angle CAB=40, \\angle CAD=30, \\angle DBA=75, \\angle DBC=25$. Find $\\angle BDC$", "Solution_1": "[quote=\"omjfv\"]Let in quadrilateral $ABCD$ be $\\angle CAB=40, \\angle CAD=30, \\angle DBA=75, \\angle DBC=25$. Find $\\angle BDC$[/quote]\r\n\r\nYou'll find that you can determine inmediately all angles in the quadrangle except $\\angle BDC$ and $\\angle ACD$. You'll find\r\n1) Triangle $ABC$ is isosceles, base $AC$, so $AB = BC$\r\n2) Take point $E$ on $AD$ such that $\\angle BED = 110$, then $\\triangle BED$ isosceles, base $BD$ hence $BE = ED$\r\n3) Also triangle $ABE$ isosceles, base $AE$, hence $AB=BE$. Hence using 1) and 3) $B$ circumcenter of triangle $AEC$ and as $\\angle EAC = 30$, you have $\\angle EBC = 60 \\Rightarrow \\triangle BEC$ equilateral\r\n4) $BE = BC$ and using 2) $= ED$, hence $E$ circumcenter of triangle $BCD$ with $\\angle BEC = 60$ hence $\\angle BDC = 30$\r\n\r\nDaniel" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Prove that if $ a,b,c>0$ and $ a\\plus{}b\\plus{}c\\equal{}1$ then:\r\n\r\n$ \\frac {a^3\\plus{}1} {a^2\\plus{}1}\\plus{}\\frac {b^3\\plus{}1} {b^2\\plus{}1}\\plus{}\\frac {c^3\\plus{}1} {c^2\\plus{}1} \\ge 2$\r\n\r\n$ \\frac {a^{n\\plus{}1}\\plus{}1} {a^n\\plus{}1}\\plus{}\\frac {b^{n\\plus{}1}\\plus{}1} {b^n\\plus{}1}\\plus{}\\frac {c^{n\\plus{}1}\\plus{}1} {c^n\\plus{}1} \\ge 2$, where $ n\\in \\mathbb{N}^*$", "Solution_1": "$ \\sum \\frac{a^3\\plus{}1}{a^2\\plus{}1}\\equal{}\\sum a\\plus{}\\sum \\frac{1\\minus{}a}{a^2\\plus{}1}\\equal{}1\\plus{}\\sum \\frac{1\\minus{}a}{a^2\\plus{}1}$ hence it suffices to prove that\r\n$ \\sum \\frac{1\\minus{}a}{a^2\\plus{}1}\\geq 1$\r\nBy Cauchy-Schwarz it's enough to prove that\r\n$ 2\\geq \\minus{}\\sum a^3\\plus{}\\sum a^2\\iff \\sum a^3\\plus{}2\\sum a\\geq \\sum a^2$ which is true.", "Solution_2": "Using $ 2(a^{n \\plus{} 1} \\plus{} 1)\\ge (a \\plus{} 1)(a^n \\plus{} 1)$ which follows from $ (a \\minus{} 1)(a^n \\minus{} 1)\\ge 0$, we get\r\n\\[ \\sum\\frac {a^{n \\plus{} 1} \\plus{} 1}{a^n \\plus{} 1}\\ge\\sum\\frac {a \\plus{} 1}{2} \\equal{} 2\r\n\\]\r\n\r\nActually the following inequality is true for $ a\\plus{}b\\plus{}c\\equal{}1$:\r\n\\[ 3>\\sum\\frac {a^{n \\plus{} 1} \\plus{} 1}{a^n \\plus{} 1}>2\\]" } { "Tag": [ "function", "topology", "algebra", "domain", "limit", "vector", "complex analysis" ], "Problem": "This's very known one but I could find its solution: Prove that for all continous function $F:{\\mathbb R}^{n}\\to{\\mathbb R}^{n}$ such that $dist(Fx,Fy)\\geq Ldist(x,y)$ with $L>1$ then $F$ has a fixed point.\r\n\r\nI also guess that in the infinitely dimesion space, the problem is quite hard, isn't it?", "Solution_1": "In an infinitely-dimensional Banach space the statement may be false: just consider twice the forward shift in $\\ell^{2}$. On the other hand, in this case it may be natural to ask whether there exists an infinite-dimensional Banach space with this property.", "Solution_2": "Here's a proof for the initial problem (the finite-dimensional case):\r\n\r\nThe idea is to show that $f$ is onto (that is, $f(\\mathbb R^{n})=\\mathbb R^{n}$). Once we do that, it will follow that $f$ is actually a homeomorphism of $\\mathbb R^{n}$ onto itself, and we'll be able to apply the Contraction Principle to $f^{-1}$ to conclude that we can find $x\\in\\mathbb R^{n}$ such that $f^{-1}(x)=x\\Rightarrow f(x)=x$. \r\n\r\nIn order to prove that $f$ is onto, notice that, on the one hand, it's a one-to-one continuous map of $\\mathbb R^{n}$ into itself and hence it's an open map, by Brouwer's [url=http://en.wikipedia.org/wiki/Invariance_of_domain]Domain Invariance Theorem[/url]. In particular, $f(\\mathbb R^{n})$ is an open subset of $\\mathbb R^{n}$. On the other hand, notice that the range of $f$ closed: if, say, $f(x_{n})\\to y$, then $f(x_{n})$ is a Cauchy sequence and hence $x_{n}$ is a Cauchy sequence, meaning that we can find $x=\\lim_{n}x_{n}$ such that $f(x)=y$. This means that the non-empty subset $f(\\mathbb R^{n})$ of the connected set $\\mathbb R^{n}$ is both open and closed, so it must be the whole of $\\mathbb R^{n}$.", "Solution_3": "Where would grobber's proof break down in an infinite dimensional case? Note that he very much needed that $f(\\mathbb{R}^{n})$ be open, and he brought in his biggest and fanciest weapon (the Brouwer domain invariance theorem) just to show that. In fedja's infinite-dimensional example of twice the right shift, $f(\\ell^{2})$ is not open - it is a proper closed vector subspace, and contains no balls. However, that example does trivially contain a fixed point: zero. We'll have to make some adjustments to have an example.\r\n\r\nIn fact, I haven't yet found a counterexample on $\\ell^{2}.$", "Solution_4": "There's a trivial tweak to generate a counterexample: add a constant. If $f(x_{1},x_{2},\\dots)=(1,2x_{1},2x_{2},\\dots)$, $f$ has no fixed points in $\\ell^{2}$: a fixed point would have $x_{1}=1,x_{2}=2,x_{3}=4,\\dots$, and that sequence isn't in $\\ell^{2}$.\r\n\r\nThinking topologically, we can generalize to maps from a connected (finite-dimensional) manifold to itself; the theorem grobber quoted still works locally, since points have neighborhoods homeomorphic to $\\mathbb{R}^{n}$. The distance can be any distance which generates the right topology, although there are no such maps if the space is bounded with this distance.", "Solution_5": "Ahh - somehow I'd convinced myself that if $R$ is the right shift, then $(I-2R)$ was invertible. It's not, so $Tx=2Rx+b$ does not have a fixed point as long as $b$ is not in the range of $(I-2R).$", "Solution_6": "Correcting my manifold statement- we need some extra metric/topological conditions. I think the condition that closed and bounded sets are compact does it.", "Solution_7": "Thanks friends for useful discussings. Really it's a cool one. Agree with comments on manifolds by jmerry, but how the extra conditions play a role in here, for example let me take $H(\\Omega)$ space of all holomophic function on $\\Omega$. I guess this space staisfies your condition.", "Solution_8": "The original hypothesis is concerns metric properties. What's your metric on $H(\\Omega)?$\r\n\r\nThen again, if $\\Delta$ is the unit disc, and we map $H(\\Delta)$ to $H(\\Delta)$ by $T(f)(z)=1+2zf(z),$ what do we have? (Look earlier on this thread, and you'll understand the inspiration for the example.)", "Solution_9": "The space of holomorphic functions is infinite-dimensional; we should have counterexamples in any sensible metric. My statement about manifolds did not leave the finite-dimensional world.", "Solution_10": "ah I just don't understand some English words. In fact i mean $H(\\Omega)$ with metric generate from semi-norm squences (as usual). This topo on $H(\\Omega)$ satisfying the condition which jmerry want.\r\nBy the way, there is a same question: If $X$ is Banach space; $f: X\\to X$ with the above condition, ie... $|f(x)-f(y)|\\geq L|x-y|$ with $L>1$. Assume that $f$ has a fixed point. Does $f$ must be $1-1$ from $X$ onto $X$?", "Solution_11": "Certainly not. The counterexamples already given in this thread are in Banach spaces. I don't think anything infinite-dimensional works.\r\n\r\nIt doesn't make sense to ask the question unless our space is a complete metric space; if we require that it's a vector space as well and that the metric is a norm, we have a Banach space. That isn't enough.\r\n\r\nThe correct manifold extension: If $X$ is a topological manifold (locally homeomorphic to $\\mathbb{R}^{n}$ for some fixed $n$) which is connected and complete in some metric $d$, then any $f$ which satisfies $d(f(x),f(y))\\ge L(d(x,y))$ with $L>1$ is a homeomorphism from $X$ to itself, and has a unique fixed point.", "Solution_12": "[quote=\"goldhp\"]ah I just don't understand some English words. In fact i mean $H(\\Omega)$ with metric generate from semi-norm squences (as usual). This topo on $H(\\Omega)$ satisfying the condition which jmerry want.[/quote]\nThen jmerry and I did both understand you correctly, and the answer is no, because it's infinite-dimensional. jmerry's conditions started with the set being a finite-dimensional manifold. The example I gave, or something like it, should be a counterexample.\n\n[quote]By the way, there is a same question: If $X$ is Banach space; $f: X\\to X$ with the above condition, ie... $|f(x)-f(y)|\\geq L|x-y|$ with $L>1$. Assume that $f$ has a fixed point. Does $f$ must be $1-1$ from $X$ onto $X$?[/quote]\r\nFalse. Just go back to the first thing fedja said, twice the right shift on $\\ell^{2}.$ This mapping satisfies your expansion condition, and it has a fixed point: zero. It's 1-1, as any mapping satisfying the expansion condition trivially must be, but it does not map $\\ell^{2}$ onto $\\ell^{2}.$", "Solution_13": "hi. I am interested in the original problem proposed by goldhp. I went through grobber's solution , is there any other way to solve this problem? ( i dont know brouwer's domain invariance theorem) also when i first saw the problem, i thought that the problem is false. the distance between 2 points keeps increasing on applying f , so how come there is a fixed point?", "Solution_14": "Here's how you construct a fixed point: apply $f^{-1}$ repeatedly. We need powerful theorems to show that $f^{-1}$ is defined everywhere, but it is certainly a contraction if it is.", "Solution_15": "[quote=\"kouboy\"] is there any other way to solve this problem? ( i dont know brouwer's domain invariance theorem) [/quote] Something like Brouwer's theorem should be used if your function is assumed to be just continuous. On the other hand, if you know that it is differentiable, everything is easy: just consider the point where $|f(x)-x|$ attains its minimum. I played for a while with the idea of approximating continuous expansions by differentiable ones but couldn't figure out how to do that :(. If somebody can do such approximation by elementary means, it will result in another solution." } { "Tag": [ "inequalities", "function", "integration", "real analysis", "real analysis unsolved" ], "Problem": "Assume that $f(x)$ is an increasing function and $n\\in \\mathbb{N}$. \r\nProve the following inequality.\r\n\r\n\\[0\\leqq \\frac{1}{n}\\sum_{k=1}^n f\\left(\\frac{k}{n}\\right) -\\int_0^1 f(x)dx\\leqq \\frac{1}{n}\\{f(1)-f(0)\\}\\]", "Solution_1": "Since $f$ is increasing,\r\n\\[\\frac{1}{n}f(\\frac{k-1}{n})\\int_\\frac{k-1}{n}^\\frac{k}{n} f(x)dx \\leq \\frac{1}{n}f(\\frac{k}{n}),\\]\r\nthus\r\n\\[0 \\leq \\frac{1}{n}f(\\frac{k}{n}) - \\int_\\frac{k-1}{n}^\\frac{k}{n} f(x)dx \\leq \\frac{1}{n}\\left(f(\\frac{k}{n})-f(\\frac{k-1}{n})\\right)\\]\r\nSum them up for $k=1,...,n$ we obtain the inequality as desired.", "Solution_2": "A monotonous function on [0,1] is always Riemann-integrable (because it is continuous except at (at most) a countable number of points); for the two inequalities, draw a graph and see what happens (it's basic majoration / minoration).", "Solution_3": "You are right, liyi." } { "Tag": [ "geometry", "3D geometry", "sphere" ], "Problem": "[img]http://i297.photobucket.com/albums/mm223/theeater820/State28.png[/img]", "Solution_1": "This is actually a somewhat tricky problem so here's a hint: The only thing keeping you from solving the problem is that we don't know the height of the cone. However, we can find it. Since the sphere has radius 5, from the center of the sphere to one of its edges is 5 (the center of the sphere is somewhere along the center axis of the cone). And we know the radius of the cone is 3 so we can set up a triangle and solve for the distance from the center of the sphere to the base of the cone. Then we can add this to the radius of the sphere which would give us the total height." } { "Tag": [ "calculus", "integration", "function", "real analysis", "real analysis theorems" ], "Problem": "Let $ f$ and $ g$ be summable over the interval [a,b], and suppose $ F(x) \\equal{} F(a) \\plus{} \\int_a^x f(y)dy$, and $ G(x) \\equal{} G(a) \\plus{} \\int_a^x g(y)dy$, then\r\n\r\n$ \\int_a^b F(x)g(x)dx \\equal{} F(b)G(b) \\minus{} F(a)G(a) \\minus{} \\int_a^b G(x)f(x)dx$.\r\n\r\n-For the sake of example, I want to come up with a concrete $ F$ and $ g$ and carry out the integration. In fact, i'm having difficulty finding functions which are Lebesgue-integrable but not Riemann integrable (I know that statement must sound really funny since they are plenty of them). My text doesn't provide any concrete examples :| Any good webpages that can be recommended?\r\n\r\nLos", "Solution_1": "You mean that you'd like to see [i]non Riemann integrables[/i] f and g for which integration by part applies?\r\n\r\nAs for a non Riemann integrable but Lebesgue integrable function, the typical example is the characteristic function of the rationals in [0,1], $ \\mathbb{I}_{\\mathbb{Q}\\cap[0,1]}$.", "Solution_2": "What is the question? Anyway, for the formula to be true it is sufficient (and nearly necessary) that $ g$ be lebesgue integrable and $ F$ to be absolutely continuous. If $ F$ is not absolutely continuous then it has a singular part which is detected in the LHS, but not the RHS." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Did you install MathPlayer?\r\n\r\nHere's a question:\r\nIf a>0,b>0,c>0,a+b+c=1\r\nProve that:(1+a)/(1-a)+(1+b)/(1-b)+(1+c)/(1-c)\u22642(a/b+b/c+c/a)\u3002\r\n\r\n$ (1 \\plus{} a)/(1 \\minus{} a) \\plus{} (1 \\plus{} b)/(1 \\minus{} b) \\plus{} (1 \\plus{} c)/(1 \\minus{} c) < \\equal{} 2(a/b \\plus{} b/c \\plus{} c/a)$\r\n\r\n\r\nUnluckily,I can't solve it.\r\n\r\nYou're clever,can you solve it?", "Solution_1": "For $ a,b,c \\ge 0$ and $ a\\plus{}b\\plus{}c\\equal{}1$ then prove that \r\n\\[ \\frac{{1 \\plus{} a}}{{1 \\minus{} a}} \\plus{} \\frac{{1 \\plus{} b}}{{1 \\minus{} b}} \\plus{} \\frac{{1 \\plus{} c}}{{1 \\minus{} c}} \\le 2\\left( {\\frac{a}{b} \\plus{} \\frac{b}{c} \\plus{} \\frac{c}{a}} \\right)\\]\r\n\\[ \\Leftrightarrow \\frac{{2a \\plus{} b \\plus{} c}}{{b \\plus{} c}} \\plus{} \\frac{{a \\plus{} 2b \\plus{} c}}{{a \\plus{} c}} \\plus{} \\frac{{a \\plus{} b \\plus{} 2c}}{{a \\plus{} b}} \\le 2\\left( {\\frac{a}{b} \\plus{} \\frac{b}{c} \\plus{} \\frac{c}{a}} \\right)\r\n\\]\r\n\\[ \\Leftrightarrow 3 \\plus{} 2\\left( {\\frac{a}{{b \\plus{} c}} \\plus{} \\frac{b}{{c \\plus{} a}} \\plus{} \\frac{c}{{a \\plus{} b}}} \\right) \\le 2\\left( {\\frac{a}{b} \\plus{} \\frac{b}{c} \\plus{} \\frac{c}{a}} \\right)\r\n\\]\r\n\r\nwhich is the inequality of mathlinks contest \r\n\\[ \\frac{a}{b} \\plus{} \\frac{b}{c} \\plus{} \\frac{c}{a} \\minus{} \\frac{3}{2} \\ge \\frac{a}{{b \\plus{} c}} \\plus{} \\frac{b}{{c \\plus{} a}} \\plus{} \\frac{c}{{a \\plus{} b}}\r\n\\]\r\n:)", "Solution_2": "[quote=\"frankvista\"]Did you install MathPlayer?\n\nHere's a question:\nIf a>0,b>0,c>0,a+b+c=1\nProve that:(1+a)/(1-a)+(1+b)/(1-b)+(1+c)/(1-c)\u22642(a/b+b/c+c/a)\u3002\n\n$ (1 \\plus{} a)/(1 \\minus{} a) \\plus{} (1 \\plus{} b)/(1 \\minus{} b) \\plus{} (1 \\plus{} c)/(1 \\minus{} c) < \\equal{} 2(a/b \\plus{} b/c \\plus{} c/a)$\n\n\nUnluckily,I can't solve it.\n\nYou're clever,can you solve it?[/quote]\r\nLet $ c \\equal{} \\min\\{a,b,c\\}.$ We obtain:\r\n$ 2\\left(\\frac {a}{b} \\plus{} \\frac {b}{c} \\plus{} \\frac {c}{a}\\right)\\geq\\frac {1 \\plus{} a}{1 \\minus{} a} \\plus{} \\frac {1 \\plus{} b}{1 \\minus{} b} \\plus{} \\frac {1 \\plus{} c}{1 \\minus{} c}\\Leftrightarrow$\r\n$ \\Leftrightarrow2\\left(\\frac {a}{b} \\plus{} \\frac {b}{c} \\plus{} \\frac {c}{a} \\minus{} 3\\right)\\geq\\sum_{cyc}\\left(\\frac {2a}{b \\plus{} c} \\minus{} 1\\right)\\Leftrightarrow$\r\n$ \\Leftrightarrow2\\left(\\frac {a}{b} \\plus{} \\frac {b}{a} \\minus{} 2 \\plus{} \\frac {b}{c} \\minus{} \\frac {b}{a} \\plus{} \\frac {c}{a} \\minus{} 1\\right)\\geq\\frac {2a^3 \\plus{} 2b^3 \\plus{} 2c^3 \\minus{} a^2b \\minus{} a^2c \\minus{} b^2a \\minus{} b^2c \\minus{} c^2a \\minus{} c^2b}{(a \\plus{} b)(a \\plus{} c)(b \\plus{} c)}\\Leftrightarrow$\r\n$ \\Leftrightarrow2\\left(\\frac {(a \\minus{} b)^2}{ab} \\plus{} \\frac {(c \\minus{} a)(c \\minus{} b)}{ac}\\right)\\geq\\frac {2(a \\plus{} b)(a \\minus{} b)^2 \\plus{} (a \\plus{} b \\plus{} 2c)(c \\minus{} a)(c \\minus{} b)}{(a \\plus{} b)(a \\plus{} c)(b \\plus{} c)}\\Leftrightarrow$\r\n$ \\Leftrightarrow2(a \\minus{} b)^2\\left(\\frac {1}{ab} \\minus{} \\frac {1}{(a \\plus{} c)(b \\plus{} c)}\\right) \\plus{}$\r\n$ \\plus{} (c \\minus{} a)(c \\minus{} b)\\left(\\frac {2}{ac} \\minus{} \\frac {a \\plus{} b \\plus{} 2c}{(a \\plus{} b)(a \\plus{} c)(b \\plus{} c)}\\right)\\geq0,$ which is obviously true.", "Solution_3": "[quote]\nwhich is the inequality of mathlinks contest\n\\[ \\frac {a}{b} \\plus{} \\frac {b}{c} \\plus{} \\frac {c}{a} \\minus{} \\frac {3}{2} \\ge \\frac {a}{{b \\plus{} c}} \\plus{} \\frac {b}{{c \\plus{} a}} \\plus{} \\frac {c}{{a \\plus{} b}}\n\\]\n:)[/quote]\r\n\r\nCan you tell me the details of the proof?", "Solution_4": "[quote=\"frankvista\"][quote]\nwhich is the inequality of mathlinks contest\n\\[ \\frac {a}{b} \\plus{} \\frac {b}{c} \\plus{} \\frac {c}{a} \\minus{} \\frac {3}{2} \\ge \\frac {a}{{b \\plus{} c}} \\plus{} \\frac {b}{{c \\plus{} a}} \\plus{} \\frac {c}{{a \\plus{} b}}\n\\]\n:)[/quote]\n\nCan you tell me the details of the proof?[/quote]\r\n\r\nSee here frankvista, http://www.mathlinks.ro/viewtopic.php?t=206917 :)", "Solution_5": "[quote=\"frankvista\"][quote]\nwhich is the inequality of mathlinks contest\n\\[ \\frac {a}{b} \\plus{} \\frac {b}{c} \\plus{} \\frac {c}{a} \\minus{} \\frac {3}{2} \\ge \\frac {a}{{b \\plus{} c}} \\plus{} \\frac {b}{{c \\plus{} a}} \\plus{} \\frac {c}{{a \\plus{} b}}\n\\]\n:)[/quote]\n\nCan you tell me the details of the proof?[/quote]\r\n\r\nInequality <=> $ \\sum a(\\frac1{b}\\minus{}\\frac1{b\\plus{}c})\\ge\\frac3{2}$ <=> $ \\sum\\frac{ca}{b(b\\plus{}c)}\\ge\\frac3{2}$\r\nBy Cauchy-Schwarz and AM-GM we have \r\n$ \\sum\\frac{ca}{b(b\\plus{}c)}\\ge\\frac{(ab\\plus{}bc\\plus{}ca)^2}{2abc(a\\plus{}b\\plus{}c)}\\ge \\frac{3abc(a\\plus{}b\\plus{}c)}{2abc(a\\plus{}b\\plus{}c)}\\equal{}\\frac3{2}$\r\n\r\nWe have done! :roll:", "Solution_6": "Nice proof, akai! :lol:", "Solution_7": "[quote=\"arqady\"]Nice proof, akai! :lol:[/quote]\r\nThanks,arqady :lol: Nother proof (dduclam tell me):\r\n\r\nBecause $ \\sum\\frac {a}b\\ge\\frac {(a \\plus{} b \\plus{} c)^2}{ab \\plus{} bc \\plus{} ca}$\r\nthus we need prove only $ \\frac {(a \\plus{} b \\plus{} c)^2}{ab \\plus{} bc \\plus{} ca} \\minus{} 3\\ge \\frac a{b \\plus{} c} \\minus{} \\frac3{2}$\r\n<=>$ \\sum\\frac {(b \\minus{} c)^2}{2(ab \\plus{} bc \\plus{} ca)}\\ge\\sum\\frac {(b \\minus{} c)^2}{2(a \\plus{} b)(a \\plus{} c)}$\r\n<=> $ \\sum\\frac {a^2(b \\minus{} c)^2}{(a \\plus{} b)(a \\plus{} c)}\\ge0$\r\nso we are done :roll:\r\n\r\n@arqady: Your proof also nice! :lol:\r\n\r\nPS: It's Japan TST 2004 :)" } { "Tag": [ "algebra", "polynomial", "group theory", "abstract algebra", "topology", "vector", "function" ], "Problem": "Mikl\u00f3s Schweitzer Competition 2009 will be on 30 October - 9 November 2009.\r\nThe problems will be announced in the competition web-sites on this Friday, 30 October 2009 at 12:00.\r\nThe guidelines of 2009 Competition: http://www.cs.elte.hu/students/schweitzer/kiiras2009.pdf (in Hungarian)\r\n\r\nhttp://www.cs.elte.hu/students/schweitzer/\r\nhttp://www.math.klte.hu/schw09/\r\nhttp://www.bolyai.hu/schweitzer.html", "Solution_1": "Problems (in Hungarian): [url=http://www.math.klte.hu/schw09/schweitzerfeladatok2009.pdf]PDF[/url] [url=http://www.bolyai.hu/down/2009/schweitzerfeladatok2009.pdf]PDF[/url] [url=http://www.bolyai.hu/down/2009/schweitzerfeladatok2009.tex]TeX[/url]", "Solution_2": "Problems [b]#3,4,5,8,9[/b] are translated into English.\r\n(Disclaimer. This is unofficial translation. Moreover at the moment it has not been checked by native Hungarian speaker yet.)\r\n\r\n[i]Attachment deleted.[/i]", "Solution_3": "Current version of the translation. (5 and 8 are fixed, 12 added.)\r\n\r\nPlease watch the updates at Mikl\u00f3s Mar\u00f3ti's (aka [b]mmaroti[/b]) site:\r\nhttp://www.math.u-szeged.hu/~mmaroti/schweitzer/", "Solution_4": "[quote=\"dmitin\"]Current version of the translation. (5 and 8 are fixed, 12 added.)\n\nPlease watch the updates at Mikl\u00f3s Mar\u00f3ti's (aka [b]mmaroti[/b]) site:\nhttp://www.math.u-szeged.hu/~mmaroti/schweitzer/[/quote]\r\n\r\nAll problems are now translated. Please let me know if you find any typos of problems.\r\n\r\nBest,\r\nMiklos", "Solution_5": "[size=150][b]Problems of the Mikl\u00f3s Schweitzer Memorial Competition 2009[/b][/size]\r\n\r\n[url=http://www.mathlinks.ro/viewtopic.php?t=311746][b]1.[/b][/url] On every card of a deck of cards a regular 17-gon is displayed with all sides and diagonals, and the vertices are numbered from 1 through 17. On every card all edges (sides and diagonals) are colored with a color 1,2,...,105 such that the following property holds: for every 15 vertices of the 17-gon the 105 edges connecting these vertices are colored with different colors on at least one of the cards. What is the minimum number of cards in the deck?\r\n\r\n[url=http://www.mathlinks.ro/viewtopic.php?t=311748][b]2.[/b][/url] Let $ p_1,\\dots,p_k$ be prime numbers, and let $ S$ be the set of those integers whose all prime divisors are among $ p_1,\\dots,p_k$. For a finite subset $ A$ of the integers let us denote by $ \\mathcal G(A)$ the graph whose vertices are the elements of $ A$, and the edges are those pairs $ a,b\\in A$ for which $ a \\minus{} b\\in S$. Does there exist for all $ m\\geq 3$ an $ m$-element subset $ A$ of the integers such that\r\n(i) $ \\mathcal G(A)$ is complete?\r\n(ii) $ \\mathcal G(A)$ is connected, but all vertices have degree at most 2?\r\n\r\n[url=http://www.mathlinks.ro/viewtopic.php?t=311749][b]3.[/b][/url] Prove that there exist positive constants $ c$ and $ n_0$ with the following property. If $ A$ is a finite set of integers, $ |A| \\equal{} n > n_0$, then\r\n\\[ |A \\minus{} A| \\minus{} |A \\plus{} A| \\leq n^2 \\minus{} c n^{8/5}.\\]\r\n[url=http://www.mathlinks.ro/viewtopic.php?t=311750][b]4.[/b][/url] Prove that the polynomial\r\n\\[ f(x) \\equal{} \\frac {x^n \\plus{} x^m \\minus{} 2}{x^{\\gcd(m,n)} \\minus{} 1}\\]\r\nis irreducible over $ \\mathbb{Q}$ for all integers $ n > m > 0$.\r\n\r\n[url=http://www.mathlinks.ro/viewtopic.php?t=311751][b]5.[/b][/url] Let $ G$ be a finite non-commutative group of order $ t \\equal{} 2^nm$, where $ n, m$ are positive and $ m$ is odd. Prove, that if the group contains an element of order $ 2^n$, then\r\n(i) $ G$ is not simple;\r\n(ii) $ G$ contains a normal subgroup of order $ m$.\r\n\r\n[url=http://www.mathlinks.ro/viewtopic.php?t=311753][b]6.[/b][/url] A set system $ (S,L)$ is called a Steiner triple system, if $ L\\neq\\emptyset$, any pair $ x,y\\in S$, $ x\\neq y$ of points lie on a unique line $ \\ell\\in L$, and every line $ \\ell\\in L$ contains exactly three points. Let $ (S,L)$ be a Steiner triple system, and let us denote by $ xy$ the thrid point on a line determined by the points $ x\\neq y$. Let $ A$ be a group whose factor by its center $ C(A)$ is of prime power order. Let $ f,h: S\\to A$ be maps, such that $ C(A)$ contains the range of $ f$, and the range of $ h$ generates $ A$.\r\nShow, that if\r\n\\[ f(x) \\equal{} h(x)h(y)h(x)h(xy)\\]\r\nholds for all pairs $ x\\neq y$ of points, then $ A$ is commutative, and there exists an element $ k\\in A$, such that $ f(x) \\equal{} kh(x)$ for all $ x\\in S$.\r\n\r\n[url=http://www.mathlinks.ro/viewtopic.php?t=311754][b]7.[/b][/url] Let $ H$ be an arbitrary subgroup of the diffeomorphism group $ \\mathsf{Diff}^\\infty(M)$ of a differentiable manifold $ M$. We say that an $ \\mathcal C^\\infty$-vector field $ X$ is [i]weakly tangent[/i] to the group $ H$, if there exists a positive integer $ k$ and a $ \\mathcal C^\\infty$-differentiable map $ \\varphi \\mathrel{: } \\mathord{]} \\minus{} \\varepsilon,\\varepsilon\\mathord{[}^k\\times M\\to M$ such that \r\n(i) for fixed $ t_1,\\dots,t_k$ the map\r\n\\[ \\varphi_{t_1,\\dots,t_k} : x\\in M\\mapsto \\varphi(t_1,\\dots,t_k,x)\\]\r\nis a diffeomorphism of $ M$, and $ \\varphi_{t_1,\\dots,t_k}\\in H$;\r\n(ii) $ \\varphi_{t_1,\\dots,t_k}\\in H \\equal{} \\mathsf{Id}$ whenever $ t_j \\equal{} 0$ for some $ 1\\leq j\\leq k$;\r\n(iii) for any $ \\mathcal C^\\infty$-function $ f: M\\to \\mathbb R$\r\n\\[ X f \\equal{} \\left.\\frac {\\partial^k(f\\circ\\varphi_{t_1,\\dots,t_k})}{\\partial t_1\\dots\\partial t_k}\\right|_{(t_1,\\dots,t_k) \\equal{} (0,\\dots,0)}.\\]\r\nProve, that the commutators of $ \\mathcal C^\\infty$-vector fields that are weakly tangent to $ H\\subset \\textsf{Diff}^\\infty(M)$ are also weakly tangent to $ H$.\r\n\r\n[url=http://www.mathlinks.ro/viewtopic.php?t=311755][b]8.[/b][/url] Let $ \\{A_n\\}_{n \\in \\mathbb{N}}$ be a sequence of measurable subsets of the real line which covers almost every point infinitely often. Prove, that there exists a set $ B \\subset \\mathbb{N}$ of zero density, such that $ \\{A_n\\}_{n \\in B}$ also covers almost every point infinitely often. (The set $ B \\subset \\mathbb{N}$ is of zero density if $ \\lim_{n \\to \\infty} \\frac {\\#\\{B \\cap \\{0, \\dots, n \\minus{} 1\\}\\}}{n} \\equal{} 0$.)\r\n\r\n[url=http://www.mathlinks.ro/viewtopic.php?t=311756][b]9.[/b][/url] Let $ P\\subseteq \\mathbb{R}^m$ be a non-empty compact convex set and $ f: P\\rightarrow \\mathbb{R}_{ \\plus{} }$ be a concave function. Prove, that for every $ \\xi\\in \\mathbb{R}^m$\r\n\\[ \\int_{P}\\langle \\xi,x \\rangle f(x)dx\\leq \\left[\\frac {m \\plus{} 1}{m \\plus{} 2}\\sup_{x\\in P}{\\langle\\xi,x\\rangle} \\plus{} \\frac {1}{m \\plus{} 2}\\inf_{x\\in P}{\\langle\\xi,x\\rangle}\\right] \\cdot\\int_{P}f(x)dx.\\]\r\n[url=http://www.mathlinks.ro/viewtopic.php?t=311757][b]10.[/b][/url] Let $ U\\subset\\mathbb R^n$ be an open set, and let $ L: U\\times\\mathbb R^n\\to\\mathbb R$ be a continuous, in its second variable first order positive homogeneous, positive over $ U\\times (\\mathbb R^n\\setminus\\{0\\})$ and of $ C^2$-class Langrange function, such that for all $ p\\in U$ the Gauss-curvature of the hyper surface\r\n\\[ \\{ v\\in\\mathbb R^n \\mid L(p,v) \\equal{} 1 \\}\\]\r\nis nowhere zero. Determine the extremals of $ L$ if it satisfies the following system\r\n\\[ \\sum_{k \\equal{} 1}^n y^k\\partial_k\\partial_{n \\plus{} i}L \\equal{} \\sum_{k \\equal{} 1}^n y^k\\partial_i\\partial_{n \\plus{} k} L \\qquad (i\\in\\{1,\\dots,n\\})\\]\r\nof partial differetial equations, where $ y^k(u,v) : \\equal{} v^k$ for $ (u,v)\\in U\\times\\mathbb R^k$, $ v \\equal{} (v^1,\\dots,v^k)$.\r\n\r\n[url=http://www.mathlinks.ro/viewtopic.php?t=311758][b]11.[/b][/url] Denote by $ H_n$ the linear space of $ n\\times n$ self-adjoint complex matrices, and by $ P_n$ the cone of positive-semidefinite matrices in this space. Let us consider the usual inner product on $ H_n$\r\n\\[ \\langle A,B\\rangle \\equal{} {\\rm tr} AB\\qquad (A,B\\in H_n)\\]\r\nand its derived metric. Show that every $ \\phi: P_n\\to P_n$ isometry (that is a not necessarily surjective, distance preserving map with respect to the above metric) can be expressed as\r\n\\[ \\phi(A) \\equal{} UAU^* \\plus{} X\\qquad (A\\in H_n)\\]\r\nor\r\n\\[ \\phi(A) \\equal{} UA^TU^* \\plus{} X\\qquad (A\\in H_n)\\]\r\nwhere $ U$ is an $ n\\times n$ unitary matrix, $ X$ is a positive-semidefinite matrix, and $ ^T$ and $ ^*$ denote taking the transpose and the adjoint, respectively.\r\n\r\n[url=http://www.mathlinks.ro/viewtopic.php?t=311760][b]12.[/b][/url] Let $ Z_1,\\,Z_2\\dots,\\,Z_n$ be $ d$-dimensional independent random (column) vectors with standard normal distribution, $ n \\minus{} 1 > d$. Furthermore let\r\n\\[ \\overline Z \\equal{} \\frac {1}{n}\\sum_{i \\equal{} 1}^n Z_i,\\quad S_n \\equal{} \\frac {1}{n \\minus{} 1}\\sum_{i \\equal{} 1}^n(Z_i \\minus{} \\overline Z)(Z_i \\minus{} \\overline Z)^\\top\\]\r\nbe the sample mean and corrected empirical covariance matrix. Consider the standardized samples $ Y_i \\equal{} S_n^{ \\minus{} 1/2}(Z_i \\minus{} \\overline Z)$, $ i \\equal{} 1,2,\\dots,n$. Show that\r\n\\[ \\frac {E|Y_1 \\minus{} Y_2|}{E|Z_1 \\minus{} Z_2|} > 1,\\]\r\nand that the ratio does not depend on $ d$, only on $ n$.\r\n\r\nhttp://www.math.u-szeged.hu/~mmaroti/schweitzer/schweitzer-2009-eng.pdf", "Solution_6": "Can anyone post solution of Mikl\u00f3s Schweitzer Memorial Competition 2000- 2009 ( in Hungarian)?\r\nMany thank" } { "Tag": [], "Problem": "What is the conjugate of a real number $a$? of a pure imaginary number $bi$? \r\n\r\nPlease explain. :lol:", "Solution_1": "[hide=\"first question\"]let $a=a+0i$. you can see $\\overline{a}=\\overline{a+0i}=a-0i=a$[/hide]\n[hide=\"second question\"]$bi=0+bi$ so $\\overline{bi}=\\overline{0+bi}=0-bi=-bi$ [/hide]", "Solution_2": "oh snap.\r\n\r\nI guess I was thinking too much." } { "Tag": [], "Problem": "How do we derive the formula tan(x/2)=tan(x)sin(x)/(tan(x)+sin(x))", "Solution_1": "multiply numerator and denominator by cot x and we get sin/(1+cos) which we get by taking the normal tan half-angle and multplying top and bottom by sqrt(1+cos)", "Solution_2": "Ah, I see, thanks." } { "Tag": [ "linear algebra", "matrix", "abstract algebra", "group theory", "linear algebra unsolved" ], "Problem": "I will admit that this question is related to one of my previous investigations, but I have cleaned the problem up into a much nicer version. The basic idea is this.\r\n\r\nTHE CAYLEY TABLE MATRIX\r\nConstruct the Cayley table for the multiplicative group (mod m).\r\nAssign to each residue class a real positive finite value according to a specific formula.\r\nSubstitute these values into the Cayley table in their corresponding positions.\r\nNow treat the table as a matrix.\r\n\r\nTHE PROBLEM\r\nGiven knowledge about the distribution of the variables assigned to the residue classes, what is the easiest way to prove that the determinant of this matrix is nonzero (in a situation where this fact is true)?\r\n\r\nUPDATE: I have an idea. I wonder if squarefree factorization helps.", "Solution_1": "The \"group determinant\" of a finite group turns out to be a product over terms that have to do with its irreducible finite-dimensional representations; this was in fact Frobenius' original motivation for studying representations of finite groups ([url=http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.22.3825]reference[/url]). For abelian groups the story is a little simpler and the determinant splits into linear factors, so it will be nonzero if and only if all of these factors are nonzero. See:\r\n\r\n[url=http://en.wikipedia.org/wiki/Discrete_Fourier_transform]Discrete Fourier transform[/url]\r\n[url=http://en.wikipedia.org/wiki/Pontryagin_duality]Pontryagin duality[/url]\r\n[url=http://en.wikipedia.org/wiki/Regular_representation]Regular representation[/url]\r\n\r\nIn particular, for a finite cyclic group one recovers a classical formula for the eigenvalues and determinant of a [url=http://en.wikipedia.org/wiki/Circulant_matrix]circulant matrix[/url]. The general case for finite abelian groups is essentially repeated tensor products of this case.", "Solution_2": "t0rajir0u, your response has helped me a great deal. The information about the group determinant being expressible as a product of linear factors has turned out to be especially valuable, allowing me to see certain connections in a whole new way. Thank you! If anything unusual comes from this, I'll update the topic for anyone interested." } { "Tag": [ "\\/closed" ], "Problem": "The pages on the site have been coming up really slow today....please help!", "Solution_1": "Possibilities (from the most likely to least likely):\r\na1.) Server Trouble\r\na2.) Server Maintainance\r\na3.) Switching server\r\nb1.) Switching to independent server\r\nb2.) Forest fire disrupted the server :) (Although I'm not sure if there [b]is[/b] any forest fire left :lol:)\r\nb3.) Construction in/around server\r\nc.) major change somewhere on AoPS\r\nd.) Human error on the site/in the server whether it's turning the server off accidentally or tripping over the cord :) or whatever\r\ne.) phpBB programming problem\r\n\r\nj.) Problems on the user side\r\n\r\nPick your reason :)", "Solution_2": "It could also be that the large number of users are chewing up the server's bandwidth, or something to that effect. Each time you view or refresh a page, you're using some of the file transfer capacity of the site up. Multiply that by 688, factor in Tare, and you have a big number.", "Solution_3": "The server went a little batty last night for some reason. If all goes as planned, we're changing servers in 2-4 weeks and hopefully a lot of these little problems will go away.", "Solution_4": "[quote=\"mathfanatic\"]It could also be that the large number of users are chewing up the server's bandwidth, or something to that effect. Each time you view or refresh a page, you're using some of the file transfer capacity of the site up. Multiply that by 688, factor in Tare, and you have a big number.[/quote]\r\nWhat??? My computer's no better/faster than yours so it probably won't matter who the user is, that's why I put this at j instead of f.\r\n\r\nRemember file transfer capacity has little to do with the available memory in the server, and correct me if I'm wrong.\r\n\r\nAlso since this problem also happens when there's only a couple of people here, I doubt that's the problem.\r\n\r\nThanks for your input Mr. Rusczyk.", "Solution_5": "But that's the thing. I don't think server memory is necessarily a problem; rather, it's with many computers trying to access the information in the server at the same time. \r\n\r\nAnd I meant \"factor in Tare\" because you spend a lot of time on AoPS, therefore you view a lot of pages. Therefore your computer requests more pages from the server than most people's computers do. The number of users only affects the number that are likely to be requesting pages from the server.", "Solution_6": "Mmm, although I spend a lot of time here not all the time I'm logged on I'm \"active\"...I wonder if you just being there helps to clog up the connection...I mean if you are in artofproblemsolving.com but are doing absolutely nothing I doubt that takes up any memory or bandwidth.\r\n\r\nYou do have a point there, hmm...", "Solution_7": "[quote=\"Tare\"]Mmm, although I spend a lot of time here not all the time I'm logged on I'm \"active\"...I wonder if you just being there helps to clog up the connection...I mean if you are in artofproblemsolving.com but are doing absolutely nothing I doubt that takes up any memory or bandwidth.[/quote]\r\n\r\nI don't think it does. Your computer only transmits/recieves signals to/from AoPS when you're submitting something, trying to view a page, or in a virtual classroom. If the page sits idle, nothing." } { "Tag": [ "vector", "analytic geometry", "calculus", "absolute value" ], "Problem": "I was trying to get what vectors are in the Encarta Encyclopedia but I can't. I don't think they're AP. The Encarta has a strange way to explain them(to me.) Does anyone have a simpler way to explain? Or do they have the Encarta and think that is the simplest way to explain?\r\n\r\nFuncia :o", "Solution_1": "Certain quantities in physics, such as the velocity of an object, or the acceleration of an object, or the force that is being exerted on an object, have not only a magnitude but also a direction. Such quantities are called vectors. Intuitively, a vector is a \"quantity that has both a magnitude and a direction.\"\r\n\r\nCan you visualize a vector? Well... A vector $\\vec{v}$ can be \"represented\" by a \"directed line segment\" (in other words, an arrow) whose length is the magnitude of $\\vec{v}$ and whose direction is the direction of $\\vec{v}$. However, there are many (infinitely many) such directed line segments, and any one of them \"represents\" the same vector $\\vec{v}$. When I think of a vector, I visualize one of these directed line segments... But I remind myself that there are other directed line segments (parallel to the one I'm visualizing, and having the same length) which also represent the same vector.\r\n\r\nVectors can be added. For example, suppose that a cruise ship is traveling 3 meters/sec to the North. And suppose that on board that cruise ship, a boy is riding a skateboard 4 meters/sec to the West. (That's his velocity with respect to the cruise ship.) Can you find his velocity relative to the ocean surface? That's a little puzzle. Turns out the boy's velocity relative to the ocean surface is 5 meters/sec Northwest. Therefore, the sum of the vectors \"3 m/s to the North\" and \"4 m/s to the West\" is the vector \"5 m/s Northwest\". You could draw a little picture, with arrows to represent each of the vectors, to illustrate this fact.\r\n\r\nVectors can also be multiplied by real numbers. For example, if your velocity is 5 m/s Northwest, and my velocity is five times your velocity, then what's my velocity? 25 m/s Northwest.\r\n\r\nIf you introduce a coordinate system, then you can associate with each vector $\\vec{v}$ an ordered triple of real numbers $(x,y,z)$. To do that, visualize the directed line segment that represents $\\vec{v}$ and which is based at the origin. Look at the point at the tip of that directed line segment. The coordinates of that point are $x$, $y$, and $z$. The numbers $x$, $y$ and $z$ are called the \"components\" of $\\vec{v}$ with respect to the coordinate system you are using. There's a certain simple formula for adding vectors in terms of the components of those vectors. There's a similar formula for multiplying a vector by a real number. And there's a formula for the magnitude of a vector in terms of that vector's components.\r\n\r\nSince every vector can be associated in this way with an ordered triple of real numbers $(x,y,z)$ (assuming you have introduced a coordinate system), an ordered triple of real numbers may itself also be called a \"vector\". But I believe it's important to remember that there's actually a difference between the geometric/physical vector $\\vec{v}$ and the ordered triple $(x,y,z)$ that tells you the components of $\\vec{v}$, just as there is a difference between a geometric point in space and the ordered triple that tells you the coordinates of that point.\r\n\r\nThat's a very short introduction to vectors... I'm sure there are plenty of much more thorough explanations of vectors out there on the internet. Also, freshman level college physics books, or even high school physics books, tend to have decent explanations of vectors. Lots of math books do too; for example, a Pre-Calculus book, or perhaps an Algebra II book, might have a chapter or a section on vectors. You could also try Calculus books or Vector Calculus books.", "Solution_2": "Wow! Thanks! I guess I'll go research... :coolspeak:", "Solution_3": "The person above me provided a good mathematical explanation of what vectors are, but i will explain it in layman's terms just so you can have several ways to think about it. Like in the post above, a vector is a quantity with both magnitude and direction. Now, some people might be lost already at this point. What does it mean for a quantity to have \"direction\"? Well, it is quite simple actually. Here is an example to show this concept of \"direction\":\r\n\r\nFirst of all, note that distance is a scalar quantity (no direction is involved), but it's vectoral counterpart, displacement, does have direction. Now, to the example.\r\n\r\nSay if you are standing at a given point. You walk 2 meters to the right, then 2 meters down, then 2 meters back, then 2 meters up (it might help to draw this situation, since i can't type out a diagram). Anyways, the person just ends up back where they started. Now, the person traveled a total DISTANCE of 8 meters (scalar addtion = 2 + 2 + 2 + 2 = 8, since which way they traveled is negligible). HOWEVER, the displacement of this person is ZERO, since they ended up where they started. To understand why this is in a more mathematical way, let us go back an analyze each step.\r\n\r\nFirst of all, lets call traveling \"right\" (or forward) and \"back\" (or left) the X direction and call traveling up and down the Y direction. Also let's call going forward/right the positive X direction and going up the positive Y direction. The opposites of these are the negative direction of each respective axis or \"direction\".\r\n\r\nThe person traveled 2 meters to the right (+2x), then 2 meters down (-2y), then 2 meters to the left (-2x), and then up 2 meters (+2y). Notice that we have employed positive and negative signs to change direction. Doing vector addition yields:\r\n\r\n(+2x) + (-2y) + (-2x) + (+2y)\r\n\r\nnow i am just going to rearrange the terms to match up the x's and y's to make it easier to see that it equals zero.\r\n\r\n(+2x) + (-2x) + (2y) + (-2y) = 0x + 0y = 0\r\n\r\nThus, the displacement is zero.\r\n\r\nHowever, notice that if you do vector addition but take the absolute values of the quantities, you still end up with 8 meters. This is how scalars and vectors are mathematically related. A scalar is the absolute value of it's respective vector. This is because, by taking the absolute value, you make all the quantities \"positive\", thus it doesn't matter what their initial sign or \"direction\" was. This mathematically shows that sign or \"direction\" is negligible for scalar quantities.\r\n\r\nOK, so there was a basic idea for you to help you differentiate between, yet relate both scalars and vectors. I hope this helps :)" } { "Tag": [ "geometry", "Putnam", "complex numbers", "real analysis", "real analysis solved" ], "Problem": "hi all,\r\n\r\n here is a nice pb:\r\nLet z_n be a sequence of non nul complex numbers so that\r\n|z_i - z_j|>=1 for every i and j. Let a>=3 a real.\r\nProve (or disprove) that the serie \r\n Sn=sum( (1/z_i)^a, i=1..n) converges.\r\n\r\nCould you improve the bound p>=a ??", "Solution_1": "Such a nice problem! The series is absolutely convergent for any $ a>2 $. To see this, think about the points having affixes $ z_k$, let them $ P_k$. Then the sum of the terms $\\frac{1}{(OP_i)^a} $ where $ n\\leq OP_i2$.", "Solution_2": "Compare to problem A3 on the 1949 Putnam:\r\n[url]http://www.kalva.demon.co.uk/putnam/putn49.html[/url]" } { "Tag": [ "modular arithmetic" ], "Problem": "Just wondering if someone could help me out with this question. It has me completely stumped. the question is\r\n\r\nProve that the equation x^3 + 2(y^3) = 4(z^3) has no solution in positive integers x, y and z.\r\n\r\nOr you can find the proper question at:\r\n\r\nhttp://members.optusnet.com.au/gujju_style/maths2.JPG\r\n\r\nThanks.", "Solution_1": "If we look at the equation, we note $2|x^{3}\\Rightarrow 2|x \\Rightarrow x=2x_{1}$ for some positive integer $x_{1}.$\r\n\r\nSo, we have $4x_{1}^{3}+y^{3}= 2z^{3}$. Again, we note $2|y^{3}\\Rightarrow 2|y \\Rightarrow y = 2y_{1}$ for some positive integer $y_{1}$.\r\n\r\nSo, we have $2x_{1}^{3}+4y_{1}^{3}= z^{3}.$ Note again $2|z^{3}\\Rightarrow 2|z \\Rightarrow z=2z_{1}$ for some positive integer $z_{1}.$\r\n\r\nAnd, now we get $x_{1}^{3}+2y_{1}^{3}= 4z_{1}^{3}.$ This obviously has the same form as our orginal equation. And, now we can repeat our argument again, leading to an infinite descent. Thus, there are no solutions in positive integers $x,y$ and $z.$", "Solution_2": "I don't know if this is a very good argument, but... :maybe: \r\n[hide=\"Here it is\"]Take mod two of the entire equation:\n$x^{3}\\equiv0\\pmod2$\n$x\\equiv0\\pmod2$\nLet $x=2p$\n$8p^{3}+2y^{2}=4z^{2}$\n$4p^{3}+y^{3}=2z^{3}$\nSimilarly, $y=2q$:\n$4p^{3}+8q^{3}=2z^{3}$\n$2p^{3}+4q^{3}=z^{3}$\nSimilarly, $z=2r$:\n$2p^{3}+4q^{3}=8r^{3}$\n$p^{3}+2q^{3}=4r^{3}$\nSimilarly, $p=2{p}_{1}$\n\nWe can continue like this forever. $x$, $y$, and $z$ will have to have an infinite factors of $2$. This means that everything will be infinity. So there are no solutions. [/hide]\r\n\r\nDid I oversimplify the problem?", "Solution_3": "[quote=\"FieryHydra\"]And, now we can repeat our argument again, leading to an infinite descent.[/quote]\r\nOr just note that if $\\gcd(x,y,z)>1$ we can divide the given equation by $\\gcd(x,y,z)$, so we can assume that $\\gcd(x,y,z)=1$, which will lead to a contradiction, as we proved that $x\\equiv y\\equiv z\\mod{2}$.", "Solution_4": "Thanks for the reply guys, very much appreciated. I'm just a little confused as to which one to look at?\r\nAre both solutions correct?", "Solution_5": "Look at either one. Each of them has basically the same type of argument. :P", "Solution_6": "[quote=\"Kurt G\u00f6del\"][quote=\"FieryHydra\"]And, now we can repeat our argument again, leading to an infinite descent.[/quote]\nOr just note that if $\\gcd(x,y,z)>1$ we can divide the given equation by $\\gcd(x,y,z)$, so we can assume that $\\gcd(x,y,z)=1$, which will lead to a contradiction, as we proved that $x\\equiv y\\equiv z\\mod{2}$.[/quote]\r\nIn the last sentence, $x\\equiv y \\pmod 2$ is not the same as $x\\equiv 0 \\pmod 2.$", "Solution_7": "[quote=\"FieryHydra\"]And, now we can repeat our argument again, leading to an [b]infinite descent[/b]. Thus, there are no solutions in positive integers $ x,y$ and $ z.$[/quote]\r\nCan someone explain this? How can I really show by infinite descent that there are no solutions?\r\n\r\nSo how can I prove this implication (\"And, now we can repeat our argument again, leading to an infinite descent\") $ \\rightarrow$ (\"Thus, there are no solutions in positive integers $ x,y$ and $ z.$\") mathematically?", "Solution_8": "[quote=\"Ikarus\"][quote=\"FieryHydra\"]And, now we can repeat our argument again, leading to an [b]infinite descent[/b]. Thus, there are no solutions in positive integers $ x,y$ and $ z.$[/quote]\nCan someone explain this? How can I really show by infinite descent that there are no solutions?\n\nSo how can I prove this implication (\"And, now we can repeat our argument again, leading to an infinite descent\") $ \\rightarrow$ (\"Thus, there are no solutions in positive integers $ x,y$ and $ z.$\") mathematically?[/quote]\r\n\r\nThe most rigorous way to do it is using the Well Ordering Principle. You agree that the set of solutions is a subset of the set of ordered triplets of positive integers, and so by the Well Ordering Principle, there exists a solution which has a smallest value of x (WOP states that any subset of the naturals has a smallest element). However, we can use any solution to produce a smaller solution, so a smallest solution cannot exist, and we have a contradiction." } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "hi,\r\nprove the identity\r\n\\[ 1^{2}\\binom{n}{1} \\plus{} 3^{2}\\binom{n}{3} \\plus{} ... \\equal{} n(n\\plus{}1)2^{n\\minus{}3}.\\]", "Solution_1": "The LHS counts the number of ways to form a committee consisting of an odd number of people out of $ n$ people along with a president and vice president of that committee (who are allowed to be the same). Pick the president and vice president first: if they're the same person (which occurs in $ n$ cases) then select an even subset of the other $ n \\minus{} 1$ people, which can be done in $ 2^{n \\minus{} 2}$ ways. Otherwise (which occurs in $ n(n \\minus{} 1)$ cases) select an odd subset of the other $ n \\minus{} 2$ people, which can be done in $ 2^{n \\minus{} 3}$ ways. This gives $ (2n \\plus{} n(n \\minus{} 1)) 2^{n \\minus{} 3}$ as desired." } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Find $ f$ where $ f(x)\\cdot f(y) \\equal{} f(x\\plus{}y) \\plus{}xy$ and $ x, y \\in \\mathbb{R}$", "Solution_1": "[quote=\"AndrewTom\"]Find $ f$ where $ f(x)\\cdot f(y) \\equal{} f(x \\plus{} y) \\plus{} xy$ and $ x, y \\in \\mathbb{R}$[/quote]\r\n$ f(0)\\cdot f(y) \\equal{} f(y)$ so $ f(0) \\equal{} 1$\r\n\r\n$ f(1)\\cdot f(\\minus{}1) \\equal{} f(0) \\minus{} 1 \\equal{} 0$ so $ f(1) \\equal{} 0$ or $ f(\\minus{}1) \\equal{} 0$\r\n\r\nIf $ f(1) \\equal{} 0$ then $ f(x\\minus{}1)\\cdot f(1) \\equal{} f(x) \\plus{} x \\minus{} 1$ so $ f(x) \\equal{} 1 \\minus{} x$\r\n\r\nIf $ f(\\minus{}1) \\equal{} 0$ then $ f(x\\plus{}1)\\cdot f(\\minus{}1) \\equal{} f(x) \\minus{} x \\minus{} 1$ so $ f(x) \\equal{} 1 \\plus{} x$", "Solution_2": "[quote=\"AndrewTom\"]Find $ f$ where $ f(x)\\cdot f(y) \\equal{} f(x \\plus{} y) \\plus{} xy$ and $ x, y \\in \\mathbb{R}$[/quote]\r\n\r\nLet $ P(x,y)$ the equation $ f(x)\\cdot f(y) \\equal{} f(x \\plus{} y) \\plus{} xy$ with $ x, y \\in \\mathbb{R}$\r\n\r\n$ P(0,0)$ $ \\implies f^2(0) \\equal{} f(0)$ and so either $ f(0)\\equal{}0$ or $ f(0)\\equal{}1.$\r\n\r\n$ f(0)\\equal{}0$ $ \\implies f(x)\\equal{}0$ $ \\forall x \\in \\mathbb{R}$ and it's easy to check that this is not a solution.\r\n\r\nSo $ f(0)\\equal{}1.$ \r\n\r\n$ P(x,\\minus{}x)$ $ \\implies f(x)f(\\minus{}x)\\equal{}1\\minus{}x^2\\equal{}(1\\minus{}x)(1\\plus{}x)$.\r\n\r\nSo the first solution is $ f_1(x)\\equal{}(1\\minus{}x)$ and $ f_1(\\minus{}x)\\equal{}(1\\plus{}x)$ $ \\forall x \\in \\mathbb{R}.$\r\n\r\nThe second solution is $ f_2(x)\\equal{}(1\\plus{}x)$ and $ f_2(\\minus{}x)\\equal{}(1\\minus{}x)$ $ \\forall x \\in \\mathbb{R}.$\r\n\r\nIt is simple to check that both functions are solution of the original equation.\r\n\r\nHence the solution \r\n$ \\boxed{\r\n\\begin{array}{ll}\r\nf(x)\\equal{}1\\minus{}x & \\forall x \\in \\mathbb{R} \\\\ \r\nf(x)\\equal{}1\\plus{}x & \\forall x \\in \\mathbb{R} \\end{array}}$" } { "Tag": [], "Problem": "=================================\r\nProve that the set $ \\{1, 2, \\ldots , 1989\\}$ can be expressed as the disjoint union of subsets \r\n$ A_1, A_2, \\ldots , A_{117}$ in such a way that each $ A_i$ contains 17 elements and the sum of \r\nthe elements in each $ A_i$ is the same.\r\n=================================\r\n\r\nThis was the first problem posed at IMO'1989, and it has been proposed by the Philippines.\r\n\r\nDoes anybody in this forum know the name of the composer of this problem?\r\nIt should be added to\r\nhttp://www.artofproblemsolving.com/Wiki/index.php/IMO_Problems_and_Solutions", "Solution_1": "I believe it's Jose Marasigan.\r\n\r\nI'll confirm with him next time I see him." } { "Tag": [ "geometry", "trigonometry", "angle bisector", "geometry proposed" ], "Problem": "Given triangle $ ABC$ of area 1. Let $ BM$ be the perpendicular from $ B$ to the bisector of angle $ C$. Determine the area of triangle $ AMC$.", "Solution_1": "$ S_{ABC} \\equal{} \\frac {a \\cdot b \\sin C} {2} \\equal{} 1$ ( 1 )\r\n$ CM \\equal{} a \\cdot \\cos \\frac {C} {2}$ ( 2 )\r\n$ S_{ACM} \\equal{} \\frac {AC \\cdot CM \\cdot \\sin \\frac {C} {2}} {2}$ ( 3 )\r\nAfter substituting (2) in (3), taking into account (1), we get $ S_{ACM} \\equal{} \\frac {1} {2}$.\r\n\r\n(Here a and b means the lengths of BC and AC respectively), while $ S_{XYZ}$ means the area of $ \\Delta XYZ$).\r\n\r\nBest regards,\r\nsunken rock", "Solution_2": "Extend $ BM$ to meet $ CA$ at $ D$. Then as $ CM$ is both an angle bisector and an altitude, $ DM \\equal{} MB$. So the perpendicular from $ M$ to $ DC$ will be half the perpendicular from $ B$ to $ DC$, so $ [AMC] \\equal{} \\frac {1}{2}$." } { "Tag": [ "AMC", "USA(J)MO", "USAJMO", "USAMO", "AMC 10", "AIME" ], "Problem": "Can someone give me more information on MOP, like who qualifies and how, what happens there, where is it, and anything else I should know. \r\n\r\nThis is all I know about MOP:\r\n\r\n- Also called MOSP\r\n- Stands for Mathematical Olympiad (Summer) Program\r\n- There are three levels, red, blue and black (I don't know the difference, though)\r\n\r\nThanks!", "Solution_1": "[url]http://en.wikipedia.org/wiki/Mathematical_Olympiad_Program[/url]\r\n\r\n[url]http://www.artofproblemsolving.com/Wiki/index.php/Mop[/url]", "Solution_2": "this site also has a lot of good information\r\nhttp://uncyclopedia.wikia.com/wiki/MOP", "Solution_3": "It isn't Math Olympiad Program...", "Solution_4": "no one cares about the s", "Solution_5": "MOP=MOSP=MOsP\r\n\r\nAnd uncyclopedia is a joke...", "Solution_6": "[quote=\"abacadaea\"]this site also has a lot of good information\nhttp://uncyclopedia.wikia.com/wiki/MOP[/quote]\r\n\r\nWow, really?", "Solution_7": "[quote=\"AIME15USAMO\"][quote=\"abacadaea\"]this site also has a lot of good information\nhttp://uncyclopedia.wikia.com/wiki/MOP[/quote]\n\nWow, really?[/quote]\r\n\r\nuncyclopedia has given me the clearest definition of MOP I have ever read/heard. Do not criticize it. :rotfl:", "Solution_8": "[quote=\"PowerOfPi\"]Can someone give me more information on MOP, like who qualifies and how, what happens there, where is it, and anything else I should know. \n\nThis is all I know about MOP:\n\n- Also called MOSP\n- Stands for Mathematical Olympiad (Summer) Program\n- There are three levels, red, blue and black (I don't know the difference, though)\n\nThanks![/quote]\r\n\r\nBlack MOP is top 12 USAMOers.\r\nBlue MOP is top 12 that are not in Black MOP, are not seniors\r\nRed MOP is top 12 freshman not in Black or Blue MOP", "Solution_9": "[quote=\"topofmath\"][quote=\"PowerOfPi\"]Can someone give me more information on MOP, like who qualifies and how, what happens there, where is it, and anything else I should know. \n\nThis is all I know about MOP:\n\n- Also called MOSP\n- Stands for Mathematical Olympiad (Summer) Program\n- There are three levels, red, blue and black (I don't know the difference, though)\n\nThanks![/quote]\n\nBlack MOP is top 12 USAMOers.\nBlue MOP is top 12 that are not in Black MOP, are not seniors\nRed MOP is top 12 freshman not in Black or Blue MOP[/quote]\r\n\r\nActually, I believe there is a qualifying score range for each of the respectful \"MOPs\".", "Solution_10": "Yes, and this qualifying score range is chosen such that there are a certain number of MOPpers in each group.\r\n\r\nEverything in topofmath's post is true except that the number of red MOPpers is much greater than 12 - usually around 25, and that the numbers are approximate - there were more than 12 people in Blue in 2009, for example.", "Solution_11": "Blue is supposed to be as close to 18 without exceeding it, I believe.", "Solution_12": "[quote=\"Brut3Forc3\"]Blue is supposed to be as close to 18 without exceeding it, I believe.[/quote]\r\n\r\nSomething like this.\r\n\r\nBlack: Top 12 overall.\r\nBlue: Next ~12-18 non-seniors.\r\nRed: Top 20-30 freshmen. Note that USAJMO is now an option for qualifying for this...\r\n\r\nIn recent years, potential qualifiers for the China Girls' Math Olympiad (CGMO) have also trained at MOP (though not necessarily every year). The top-scoring ~8-10 girls on the USAMO [USAJMO may allow one to qualify, but not sure??] qualify.\r\n\r\nIt's 3 weeks of math and getting pwned by Zuming. Pretty awesome.", "Solution_13": "[quote=\"not_trig\"][quote=\"Brut3Forc3\"]Blue is supposed to be as close to 18 without exceeding it, I believe.[/quote]\n\nSomething like this.\n\nBlack: Top 12 overall.\nBlue: Next ~12-18 non-seniors.\nRed: Top 20-30 freshmen. Note that USAJMO is now an option for qualifying for this...\n\nIn recent years, potential qualifiers for the China Girls' Math Olympiad (CGMO) have also trained at MOP (though not necessarily every year). The top-scoring ~8-10 girls on the USAMO [USAJMO may allow one to qualify, but not sure??] qualify.\n\nIt's 3 weeks of math and getting pwned by Zuming. Pretty awesome.[/quote]", "Solution_14": "In the AoPS Wiki page it said that exceptional seventh and eighth graders have sometimes been accepted.", "Solution_15": "What's usually the average scores for red Moppers?", "Solution_16": "On the name, I believe it was once officially MOP- but if so, that was before my time. The silent S has lasted over ten years, pretty impressive when you consider how much turnover there is in everybody involved.", "Solution_17": "[quote=\"PowerOfPi\"]In the AoPS Wiki page it said that exceptional seventh and eighth graders have sometimes been accepted.[/quote] In Blue (and now Black MOP.) Red MOP is for 9th graders only\r\n@marc I think it's somewhere around 10.", "Solution_18": "[quote=\"PowerOfPi\"]In the AoPS Wiki page it said that exceptional seventh and eighth graders have sometimes been accepted.[/quote]\r\n\r\nI think thats for blue mop, not red mop. As far as I know, nobody below 9th grade has ever been accepted for red mop.", "Solution_19": "Wait, a score of 10 on what? USAMO I guess?", "Solution_20": "Yup on USAMO. And it's reaaaallly hard.", "Solution_21": "Are you being sarcastic? So, if I get a 150 on the AMC10, 8 on the AIME, qualify for USAMO because I am first in the state, and get a 12 (2 points per problem), I will make Red MOP?", "Solution_22": "[quote=\"PowerOfPi\"]Are you being sarcastic? So, if I get a 150 on the AMC10, 8 on the AIME, qualify for USAMO because I am first in the state, and get a 12 (2 points per problem), I will make Red MOP?[/quote]\r\n\r\n...And getting 2 points per problem is obviously very easy.\r\n\r\nBTW, that was sarcastic.\r\n\r\nGetting 2 points per problem is impossible almost. Its very, very hard to do.\r\n\r\nLook at some RECENT usamo problems to see why they are considered hard and why getting 2 points on each problem might be... not easy.\r\n\r\nMode and Median score in 2009: 4 on usamo.", "Solution_23": "[quote=\"PowerOfPi\"]Are you being sarcastic? So, if I get a 150 on the AMC10, 8 on the AIME, qualify for USAMO because I am first in the state, and get a 12 (2 points per problem), I will make Red MOP?[/quote]\r\n\r\nWhat pythag said. I think that getting a 2 on each problem is almost as hard as getting a 20+ (3 complete/nearly complete solutions + maybe a bit inroad on others) The most common way of making Red MOP is either two complete solutions or 1 with a few more semi-completed.", "Solution_24": "[quote=\"pythag011\"]Mode and Median score in 2009: 4 on usamo.[/quote]\r\n\r\nIf I'm not mistaken, the mode score was actually a $ 2$.", "Solution_25": "[quote=\"dysfunctionalequations\"][quote=\"pythag011\"]Mode and Median score in 2009: 4 on usamo.[/quote]\n\nIf I'm not mistaken, the mode score was actually a $ 2$.[/quote]\r\nIt was. But you could get 2 points for giving the (very obvious) answer to number 2 without proof, or so I have heard." } { "Tag": [ "geometry", "geometry unsolved" ], "Problem": "The triangle $\\triangle ABC.$ The bisectrix $AA ', BB', CC '.$ Circles $(I), (O), (E)$ to turn be inscribed circle, circumsribed circle of the triangle $\\triangle ABC$ and escribed circle to the angle $\\angle A.$ Since $I,$ draw $IH$ perpendicular $B'C '.$ Prove that:\n$R \\cdot r = OE \\cdot IH$", "Solution_1": "Let $ (I)$ touch $ BC, CA, AB$ at $ U, V, W.$ $ V$ is pole of $ CA$ and direction $ \\perp BIB'$ is pole of $ BIB'$ WRT $ (I)$ $ \\Longrightarrow$ $ b' \\perp BIB'$ through $ V$ is polar of $ B'$ and likewise, $ c' \\perp CIC'$ through $ W$ is polar of $ C'$ WRT $ (I)$ $ \\Longrightarrow$ $ U' \\equiv b' \\cap c'$ is pole of $ B'C'$ WRT $ (I).$ Therefore $ \\overline{IH} \\cdot \\overline {IU'} \\equal{} r^2$ (1). Let $ V', W'$ be poles of $ C'A', A'B'$ WRT $ (I),$ obtained in a similar way. $ \\triangle U'V'W'$ is anticomplementary to $ \\triangle UVW.$ Let $ E, F, G$ be excenters of $ \\triangle ABC$ against $ A, B, C,$ respectively. $ \\triangle EFG \\sim \\triangle U'V'W'$ are similar, having parallel sides, $ \\perp AIA', BIB', CIC'.$ $ (O), (I)$ are their respective 9-point circles $ \\Longrightarrow$ $ \\frac {\\overline{OE}}{\\overline{IU'}} \\equal{} \\minus{} \\frac {R}{r}$ (2). Combining (1) & (2), $ \\overline{OE} \\cdot \\overline{IH} \\equal{} \\minus{}R \\cdot r.$" } { "Tag": [ "trigonometry", "trig identities", "Law of Cosines" ], "Problem": "Equilateral triangles $ ABF$ and $ CAG$ are erected externally on the side $ CA$ and hypotenuse $ AB$ of right triangle $ ABC$. $ M$ is the midpoint of $ [BC]$. $ |MF|\\equal{}11$ $ |MG|\\equal{}7$ then find $ |BC|$", "Solution_1": "Does anybody know the solution ?, please share it :(", "Solution_2": "By law of cosines on CGM and AC=GC,\r\n$ AC^2 \\plus{} MC^2 \\plus{} \\sqrt {3}AC \\cdot CM \\equal{} 49$\r\nBy law of cosines on MBF and MB=MC,\r\n$ CM^2 \\plus{} AB^2 \\minus{} 2 \\cdot CM \\cdot AB \\cdot cos (60 \\plus{} ABC) \\equal{} 121$\r\n$ CM^2 \\plus{} AB^2 \\minus{} 2 \\cdot CM \\cdot AB(\\frac {1}{2} \\cdot \\frac {2CM}{AB} \\minus{} \\frac {\\sqrt {3}}{2} \\cdot \\frac {AC}{AB})$\r\n$ AB^2 \\minus{} CM^2 \\plus{} \\sqrt {3}AC\\cdot CM \\equal{} 121$\r\n$ AC^2 \\plus{} 3MC^2 \\plus{} \\sqrt {3}AC \\cdot CM \\equal{} 121$\r\nSubtracting from the first\r\n$ 2MC^2 \\equal{} 72$\r\n$ MC \\equal{} 6$\r\n$ BC \\equal{} 12$" } { "Tag": [ "calculus", "derivative", "complex analysis" ], "Problem": "Is it true that if there is a $ c \\in \\mathbb{C}$ such that $ D_wf(z)\\equal{}cw$ for each $ w$ such that $ |w| \\equal{} 1$ then $ f$ is $ \\mathbb{C}$-differentiable at $ z$?\r\nThe above $ D$ means the directional derivative in the $ w$ direction.", "Solution_1": "Yes, because $ f$ is $ \\mathbb C$-differentiable at $ z_0$ iff the total differential of $ f$ at $ z_0$ (in the sense of [b]real[/b] calculus) is $ \\mathbb C$-linear." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "I posted this at \"Unsolved Problems\" but got no reply. :( \r\n\r\n If $ a_1,a_2,...,a_n$ are positive real numbers then prove \r\n\r\n$ \\frac {a_1}{a_2 \\plus{} a_3} \\plus{} \\frac {a_2}{a_3 \\plus{} a_4} \\plus{} ... \\plus{} \\frac {a_{n \\minus{} 2}}{a_{n \\minus{} 1} \\plus{} a_n } \\plus{} \\frac {a_{n \\minus{} 1}}{a_n \\plus{} a_1} \\plus{} \\frac {a_n}{a_1 \\plus{} a_2} \\geq \\frac {n}{4}$", "Solution_1": "Let $ b_1\\equal{}max a_i$; $ b_2$ is largest of two numbers in denominator of fraction, where $ b_1$ is numerator; ... ; $ b_k$ is largest of two numbers in denominator of fraction, where $ b_{k\\minus{}1}$ is numerator; $ b_1$ is largest of two numbers in denominator of fraction, where $ b_k$ is numerator. obviously. $ k \\geq \\frac{n}{2}$", "Solution_2": "https://en.wikipedia.org/wiki/Shapiro_inequality" } { "Tag": [], "Problem": "[color=olive]Show that if a and b are integers with (a,b)=1, then (a+b,a-b)=1 or 2 [/color]\r\n\r\nShould I use the fact that (a+cb,b)=(a,b)? I've also tried to start with (a,b)=1 => 1=xa+yb, but then I was stuck, don't know how to begin. Thanks for helping me out.", "Solution_1": "http://www.mathlinks.ro/Forum/viewtopic.php?t=111291\r\n\r\nIt wasn't necessary to repost in the correct forum: the original thread was moved here." } { "Tag": [ "number theory", "prime numbers" ], "Problem": "Prime number is known by me to be defined as a number whose factors are only itself and one. The lowest prime number i previously have been tought was 2. But with a spark of random thinking on my part, i thought that [b]-1[/b]'s factors are only -1 and one. By the definition of prime numbers above, all negative numbers are composite. Example: -7's factors are 1,7,-7,-1. But negative one's factors are 1,-1. No others. Please explain.", "Solution_1": "A prime number is defined to be natural. This stuff doesn't work well with negatives.", "Solution_2": "So a negative number can't be prime? The onlynegative number that might possibly be prime is -1.", "Solution_3": "It really depends on your definition of prime, since primes are not well defined in the negatives.", "Solution_4": "hmm. We need to ask a mod. They would know", "Solution_5": "Look [url=http://en.wikipedia.org/wiki/Prime_number]here[/url].\r\n\r\nA factor is defined to be positive, and prime numbers have to be greater than $ 1$." } { "Tag": [ "geometry", "3D geometry", "calculus", "MATHCOUNTS" ], "Problem": "I found a piece of baumkuchen, about 1/8 of the original. The outer part of the piece was about 5 and a half cm. long and the inner part was about 1 and a half cm. long. It's thickness (height) was 5 cm. If my sister ate the rest of the baumkuchen, how much did she eat? (just round off the radius)\r\n\r\n(Note: baumkuchen is a German dessert/pastry. It looks like a donut except [b]all sides are flat[/b] Reference:\r\n[img]http://www.ja-inaba.or.jp/tokusan/img/t_27.jpg[/img]\r\n[img]http://www.yosanet.com/gensen/shop/003/img/item003-l.jpg[/img][img]http://www.okashi.co.jp/suzuya/suzuya08.gif[/img]\r\n[img]http://www.shunkado.co.jp/images/bamukuheni.jpg[/img]\r\n\r\nTo all the German people: Yes, I know that traditional Baumkuchen differ from the Baumkuchen I'm talking about (My version of baumkuchen is one of the Japanese versions of baumkuchen, not the traditional which can have a round outside like a donut, shaped slightly like a cone instead of a perfect circle, has creme in the middle, etc.), but I don't think calculus is in MATHCOUNTS :)", "Solution_1": "My answer might be diffrent from yours, since I rounded diffrently. The volume for a roll of toilet paper, or Baumkuchen, is V= :pi: R^2H- :pi: r^2H. R=radius of outer circle, r=radius of inner circle, H=height. Since the problem gives 1/8 of the circumferences,\r\n1/8C=5.5\r\nC=44\r\n44=2 :pi: R\r\n22= :pi: R\r\n7=R\r\nThis also gives 2 for r. So\r\nV=7^2 :pi: 5-2^2 :pi: 5\r\nV=245 :pi: -20 :pi: \r\nV=225 :pi: \r\nV=707\r\nSo the 7/8 the sister ate's volume would be 619 cm^3.", "Solution_2": "Yup yup...er, toilet paper? :lol: \r\n\r\n :cry: So happy someone solved this one, all the picture-finding didn't go to waste after all...it would really be great if you could solve all (50 :lol: ) unsolved problems I made...", "Solution_3": "My geometry teacher called this kind of problem a \"toilet paper problem\".", "Solution_4": "There were 50 unsolvable problems? Where?\r\n\r\nOh, wait. Unsolved. Slightly different. Oh, well. Let the fanatic at them!", "Solution_5": "[quote=\"topper\"]My geometry teacher called this kind of problem a \"toilet paper problem\".[/quote]\r\n :lol:, I understand where he/she's getting it from. They kind of look the same too except for the shape and color (they both have layers)\r\n\r\nI doubt I can think up of a problem where it'll be unsolvable to you, mathfanatic :)", "Solution_6": "That's not true.. Just ask whether there are infinitely many primes of a certain form. Chances are it will be unsolvable to everyone in the world.", "Solution_7": "Yeah, I guess so, but then again I would've never thought of that :)" } { "Tag": [ "analytic geometry", "articles" ], "Problem": "I saw the phrase \"mass points\" in relation to triangles and looked it up on the Wiki, but there was nothing other than a link to a nonexistent article. What is it?", "Solution_1": "Mass points are also known as [url=http://www.cut-the-knot.org/triangle/barycenter.shtml]barycentric coordinates[/url]. If you learn to use them properly, they can be very useful.", "Solution_2": "a semi-old topic but still...\r\nThanks for your reply, but I don't know how to find the barycenter of two points, which is involved in the link description of how to find the barycenter of three points. Hopefully that info. will help in understanding the rest of the article, because I'm pretty confused about the rest of it, too. :o", "Solution_3": "The barycenter is also known as the [url=http://en.wikipedia.org/wiki/Center_of_mass]center of mass[/url]." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "modular arithmetic", "geometry", "algorithm", "induction" ], "Problem": "This is where you may talk about the problems and solutions and where the results will be posted in 4 days time.\r\n\r\n2 graders gave a score to each solution, without knowing who the solution was from, and then the two scores were averaged. That is why half-points are given to some solutions.\r\n\r\nBest solutions in the near future.\r\n\r\nOverall Winners:\r\n[hide]31 - rem\n39 - JSteinhart\n18 - SplashD\n19 - pkerichang\n25 - 101101101[/hide]\n\nDay 1 Winners:\n[hide]18-SplashD\n19-pkerichang\n31 - rem\n37 - Xantos C. Guin\n39 - JSteinhart\n[/hide]\nMost Perfects: (tie)SplashD and Xantos C. Guin\n\nScores:\n[hide]\nThe average is the average score of the solutions, not the average score of the participants. \\begin{tabular}{c|l|l|l|r\\parallel l|l|l|r}ID&P1& P2&P3&Day 1&P4&P5&P6&Total\\\\ \\hline Average&5.08&5.85&3.69&12.25&2.88&1.67&3.5&14.43\\\\ 1&3.5&5.5&0&9&&&1&10\\\\ 8&3&5.5&4&12.5&&&&12.5\\\\ 14&&6.5&&6.5&.5&&&7\\\\ 15&6&5&0&11&&&&11\\\\ 18&7&6&7&20&&&&20\\\\ 19&6&7&6.5&19.5&&&&19.5\\\\ 22&5&&&5&&&&5\\\\ 25&4.5&5&1.5&11&3&.5&4.5&19\\\\ 28&5.5&6&2&13.5&&&&13.5\\\\ 31&7&6&4.5&17.5&3&3&4&27.5\\\\ 33&3.5&6.5&&10&&&&10\\\\ 37&7&7&&14&1.5&&&15.5\\\\ 39&4&5.5&4&13.5&3.5&1.5&4.5&23\\\\ junggi&4&4.5&&9.5&&&&9.5\\\\ \\end{tabular} [/hide]", "Solution_1": "#1 was the only one I had time to attempt, and it was just Euclidean algorithm. The answer was all $x$ such that $x\\equiv 3\\pmod{5}$, $x\\equiv 4\\pmod{17}$, or $x\\equiv 38\\pmod{85}$.", "Solution_2": "I managed to solve everything but maybe made some stupid mistakes, but I don't remember my answers...\r\nI really liked the problems :)", "Solution_3": "I spent a while grading and I have a few comments:\r\n\r\nOf the 12 sets of 3 problems that I graded, only one set had adequate spacing. All you have to do is press enter a few times between each line. \r\n\r\nFor the third problem...if you say that the orthric triangles are homothetic wrt the orthocenter of ABC, that does not mean that the larger triangles are as well. You have to justify that. There was only one person who gave a satisfactory argument to follow. I am also suprised that only one other person recognized the 9 point circle, which was my intent of the problem.", "Solution_4": "[hide=\"2\"] For 2, I got 279. :maybe: Is that right? [/hide]", "Solution_5": "[quote=\"lotrgreengrapes7926\"][hide=\"2\"] For 2, I got 279. :maybe: Is that right? [/hide][/quote]\r\nI think so.\r\n\r\nOMG I made an arithmetic error on this quetsion. I am so dumb! :wallbash_red: I solved the question but messed up in my calculations. hopefully the graders will take only a few marks off...", "Solution_6": "[quote=\"chess64\"]#1 was the only one I had time to attempt, and it was just Euclidean algorithm. The answer was all $x$ such that $x\\equiv 3\\pmod{5}$, $x\\equiv 4\\pmod{17}$, or $x\\equiv 38\\pmod{85}$.[/quote]\r\n\r\n$x\\equiv 38\\pmod{85}$ is already covered by both $x\\equiv 3\\pmod{5}$ and $x\\equiv 4\\pmod{17}$, so I don't think it needs to be listed again. \r\n\r\nFor number 2:\r\n[hide]Since $5$ rocks in a box is equal to $7$ rocks in the previous box, the arrangement of rocks is the base $\\frac{7}{5}$ representation of the number of rocks in base $10$. So the answer is the largest integer that has $10$ digits in base $\\frac{7}{5}$ which after using a formula is $279_{10}=5362324266_{\\frac{7}{5}}$. \n[/hide]\r\nThis is why it is solafidefarms' favorite problem.", "Solution_7": "What's the criteria for getting a 7? Does it have to be brilliant and ingenious :idea: ? Or just correct and well-explained?", "Solution_8": "[quote=\"Xantos C. Guin\"]For number 2:\n[hide]Since $5$ rocks in a box is equal to $7$ rocks in the previous box, the arrangement of rocks is the base $\\frac{7}{5}$ representation of the number of rocks in base $10$. So the answer is the largest integer that has $10$ digits in base $\\frac{7}{5}$ which after using a formula is $279_{10}=5362324266_{\\frac{7}{5}}$. \n[/hide]\nThis is why it is solafidefarms' favorite problem.[/quote]\r\n\r\nWow, and me with my brute force...", "Solution_9": "[quote=\"Altheman\"]I spent a while grading and I have a few comments:\n\nOf the 12 sets of 3 problems that I graded, only one set had adequate spacing. All you have to do is press enter a few times between each line. \n\nFor the third problem...if you say that the orthric triangles are homothetic wrt the orthocenter of ABC, that does not mean that the larger triangles are as well. You have to justify that. There was only one person who gave a satisfactory argument to follow. I am also suprised that only one other person recognized the 9 point circle, which was my intent of the problem.[/quote]\r\n\r\nI don't remember personally saying anything about orthic triangles, but it seems to me that showing that the altitudes of the two triangles are homothetic is sufficient, as a triangle is uniquely determined by its altitudes. (In fact, it is also uniquely determined by its orthic triangle, as the excenters of the orthic triangle are the vertices of the original triangle.) It therefore follows that if the orthic triangles are homothetic, so are the original triangles.\r\n\r\nI also seem to remember hearing at one point that showing that any three corresponding points on a triangle are homothetic about a point is sufficient...can anyone confirm this?\r\n\r\nThis is not an attempt to get the graders to change anything, in fact I hope they won't, as otherwise the integrity of the contest would be ruined, I am just seeking to make sure that what I think is true is actually true, as I am personally not so good at geometry.\r\n\r\nBtw, did you guys get $\\binom{n}{2}(n-1)^{4}$ for problem 4?", "Solution_10": "[quote=\"JSteinhardt\"]Btw, did you guys get $\\binom{n}{2}(n-1)^{4}$ for problem 4?[/quote]\r\n\r\nThat's probably right. I messed that one up. My answer was an integer $\\text{ iff }n\\not\\equiv 3 \\bmod{4}$", "Solution_11": "[quote=\"Xantos C. Guin\"][quote=\"JSteinhardt\"]Btw, did you guys get $\\binom{n}{2}(n-1)^{4}$ for problem 4?[/quote]\n\nThat's probably right. I messed that one up. My answer was an integer $\\text{ iff }n\\not\\equiv 3 \\bmod{4}$[/quote]\r\n\r\nI used lots of casework :roll: and it was an integer $\\text{ iff }n\\equiv0\\mod2$. :blush:", "Solution_12": "[quote=\"lotrgreengrapes7926\"]What's the criteria for getting a 7? Does it have to be brilliant and ingenious :idea: ? Or just correct and well-explained?[/quote]\r\n\r\nYes. Most people got the right answer for 1 and 2, but the majority of people got less than a 5 on the first and no more than a 5.5 on the second. 7 was for well-explained pretty solution.", "Solution_13": "Btw Day 1 seemed significantly easier than a normal USAMO (I would be ecstatic if a USAMO number 3 were that easy, on a geometry problem no less), whereas Day 2 seemed about on par with normal difficulty, (question 4 was a bit easier than most question 4's, question 5 was a bit harder than most question 5's, question 6 was about as hard as most question 6's.", "Solution_14": "[quote=\"chess64\"]#1 was the only one I had time to attempt, and it was just Euclidean algorithm. The answer was all $x$ such that $x\\equiv 3\\pmod{5}$, $x\\equiv 4\\pmod{17}$, or $x\\equiv 38\\pmod{85}$.[/quote]\r\n\r\nIf only I had waited a bit to take the first part of the test....\r\nI learned Euclidean algorithm in interm. num theory class the day after :lol:", "Solution_15": "Part of it was you were late :P. Secondly I thought it wasn't as pretty as I liked. Thirdly, I was the only grader for your solution, because the other #1s had already been graded. So maybe it should have gotten a 6.5.", "Solution_16": "\\gcd(x^{2}+1,2x+9) by the Euclidean Algorithm. Applying the algorithm again, we can say that\r\n\r\n2(x^{2}+1)-x(2x+9)=-9x+2 \\implies \\gcd(f(x),g(x))=\\gcd(2x+9,-9x+2).\r\n\r\n\r\n\r\nIt seems like you claim gcd(a, b) = gcd(b, 2a - kb). Try this with b = 4, a = 2, k = 1. We get 2 = gcd(2, 4) = gcd(4, 4 - 4) = 4 OH GOD HAVE WE DISPROVEN MATH", "Solution_17": "[quote=\"solafidefarms\"][b]Secondly I thought it wasn't as pretty as I liked.[/b][/quote]\r\n\r\nI believe this has been said three times in this thread, but Alex and I are being ignored.\r\n\r\n[b]Style does not affect your score.[/b]", "Solution_18": "[quote=\"JSteinhardt\"][quote=\"solafidefarms\"][b]Secondly I thought it wasn't as pretty as I liked.[/b][/quote]\n\nI believe this has been said three times in this thread, but Alex and I are being ignored.\n\n[b]Style does not affect your score.[/b][/quote]\r\n\r\nEven if you overly repeat and write tons of pages when it can be done with only one?", "Solution_19": "Yes, even in this case style does not affect your score. If someone writes 25 pages, but they are 25 pages of rigorous statements that eventually prove the desired result, they should get 7 points. This is how it works on the actual USAMO.\r\n\r\nIt should also be noted the chess64's solution was less than 1 page...\r\n\r\nEdit: Just so things don't get confused, I'm not complaining or anything, I'm just attempting to make this point clear for the purpose of future Mock USAMO's.", "Solution_20": "[quote=\"MysticTerminator\"][quote]$\\gcd(x^{2}+1,2x+9)$ by the Euclidean Algorithm. Applying the algorithm again, we can say that \\[2(x^{2}+1)-x(2x+9)=-9x+2 \\implies \\gcd(f(x),g(x))=\\gcd(2x+9,-9x+2).\\] [/quote]\n\nIt seems like you claim gcd(a, b) = gcd(b, 2a - kb). Try this with b = 4, a = 2, k = 1. We get 2 = gcd(2, 4) = gcd(4, 4 - 4) = 4 OH GOD HAVE WE DISPROVEN MATH[/quote]\r\n\r\nUmm Arnav...\r\n\r\nTHEY'RE POLYNOMIALS", "Solution_21": "[quote=\"chess64\"][quote=\"MysticTerminator\"][quote]$\\gcd(x^{2}+1,2x+9)$ by the Euclidean Algorithm. Applying the algorithm again, we can say that \\[2(x^{2}+1)-x(2x+9)=-9x+2 \\implies \\gcd(f(x),g(x))=\\gcd(2x+9,-9x+2).\\] [/quote]\n\nIt seems like you claim gcd(a, b) = gcd(b, 2a - kb). Try this with b = 4, a = 2, k = 1. We get 2 = gcd(2, 4) = gcd(4, 4 - 4) = 4 OH GOD HAVE WE DISPROVEN MATH[/quote]\n\nUmm Arnav...\n\nTHEY'RE POLYNOMIALS[/quote]\r\n\r\n...ok? so let's do this with polynomials\r\n\r\n2x = gcd(2x, 4x) = gcd(4x, 4x - 4x) = gcd(4x, 0) = 4x", "Solution_22": "hmm try using the euclidean algorithm arnav\r\n\r\n$\\frac{x^{2}+1}{2x+9}=\\frac{1}{2}x-\\frac{9}{4}+\\frac{85/4}{2x+9}$\r\n\r\n$\\iff 4(x^{2}+1)=(2x+9)(2x-9)+85$\r\n\r\n$\\therefore \\gcd(x^{2}+1,2x+9)=\\gcd(2x+9,85)$\r\n\r\nwhich is actually faster than my first way\r\n\r\nbut whatever\r\n\r\narnav is still wrong :P", "Solution_23": "[quote=\"JSteinhardt\"][quote=\"solafidefarms\"][b]Secondly I thought it wasn't as pretty as I liked.[/b][/quote]\n\nI believe this has been said three times in this thread, but Alex and I are being ignored.\n\n[b]Style does not affect your score.[/b][/quote]\r\n\r\nStyle did affect your score in this case, because I think it [i]should[/i] affect score. Maybe it doesn't matter on the real thing, but I'd rather consider my work nearly a Book proof or, if I had to bruteforce it, as elegant bruteforce as possible, then just to hack out a correct but not-wonderfully-readable solution. Also, in How to Write a Proof it puts a bit of emphasis on niceness. Thus it did matter, even if it doesn't usually matter. When you do a mock USAMO you don't have to grade by style, and, judging from the outcry, I may not grade on style next time either.", "Solution_24": "[quote=\"solafidefarms\"][quote=\"JSteinhardt\"][quote=\"solafidefarms\"][b]Secondly I thought it wasn't as pretty as I liked.[/b][/quote]\n\nI believe this has been said three times in this thread, but Alex and I are being ignored.\n\n[b]Style does not affect your score.[/b][/quote]\n\nStyle did affect your score in this case, because I think it [i]should[/i] affect score. Maybe it doesn't matter on the real thing, but I'd rather consider my work nearly a Book proof or, if I had to bruteforce it, as elegant bruteforce as possible, then just to hack out a correct but not-wonderfully-readable solution. Also, in How to Write a Proof it puts a bit of emphasis on niceness. Thus it did matter, even if it doesn't usually matter. When you do a mock USAMO you don't have to grade by style, and, judging from the outcry, I may not grade on style next time either.[/quote]\n\neh, it's cool either way (as USAMO is indeed independent of style while MOP tests have style), just give a heads up beforehand so people know what to expect\n\nin fact, I think the mathlinks contest on this site had a rubric that gave 6 for correct proof that was not very innovative, so it's up to the individual contest-maker's discretion\n\n[quote=\"chess64\"]hmm try using the euclidean algorithm arnav\n\n$\\frac{x^{2}+1}{2x+9}=\\frac{1}{2}x-\\frac{9}{4}+\\frac{85/4}{2x+9}$\n\n$\\iff 4(x^{2}+1)=(2x+9)(2x-9)+85$\n\n$\\therefore \\gcd(x^{2}+1,2x+9)=\\gcd(2x+9,85)$\n\nwhich is actually faster than my first way\n\nbut whatever\n\nhaha I'm an idiot :P[/quote]\r\n\r\nerm this is definitely not what you did the first time. I'm not disputing that every step in your line of equations is correct. I'm just saying your reasoning is flawed. Try justifying each step.", "Solution_25": "Yeah I agree with Arnav, if you want to grade on style you can, but the title Mock USAMO indicates by default that it would be graded identically to a USAMO. So next time it would b a good idea to let people know beforehand. In fact, if you want to do it the way it works at MOP, you could assign a score solely based on Math, 1-7, and another based solely on Style, 0.0-1.0. At MOP, a well-written proof gets 0.8, sloppy proof or \"dumbass/brute force\" solution gets 0.7, and stuff like Lagrange multipliers or mindless coordinates gets 0.5-0.6; 0.9 is reserved for elegant solutions, and 1.0 for extremely elegant, etc. etc.; also proofs with non-trivial generalizations might get higher style. You don't have to do that, it's just an idea. I think that might be easier to grade that way because you could concentrate just on whether it's correct, then just on whether it's nice. (Btw, your \"score\" would then be style * math.) It would also prevent people from asking why they lost points, because they would know exactly why -> if they lose math points, it's because they did something that wasn't rigorous or just plain wrong. If they lose style points, it's because they had bad style.\r\n\r\nI'm glad to see that the organizers are keeping an open mind, and I apologize if I was a bit harsh before, basically I wanted to get a point across for future reference and it seemed that everyone was ignoring it.", "Solution_26": "[quote=\"MysticTerminator\"]\n\n[quote=\"chess64\"]hmm try using the euclidean algorithm arnav\n\n$\\frac{x^{2}+1}{2x+9}=\\frac{1}{2}x-\\frac{9}{4}+\\frac{85/4}{2x+9}$\n\n$\\iff 4(x^{2}+1)=(2x+9)(2x-9)+85$\n\n$\\therefore \\gcd(x^{2}+1,2x+9)=\\gcd(2x+9,85)$\n\nwhich is actually faster than my first way\n\nbut whatever\n\nhaha I'm an idiot :P[/quote]\n\nerm this is definitely not what you did the first time. I'm not disputing that every step in your line of equations is correct. I'm just saying your reasoning is flawed. Try justifying each step.[/quote]\r\n\r\nhuh huh any response :P", "Solution_27": "Shouldn't multiplying by 2 be allowed here, since the other term (2x + 9) is never divisible by 2?", "Solution_28": "dude, don't revive.", "Solution_29": "no arnav\r\n\r\nno response" } { "Tag": [], "Problem": "[quote=\"KBabe\"]I'm 11, about to be 12 in two days. \n\nand i have a question, are homeschooled ppl usually smarter than public school ppl?[/quote]That mean you're gonna receive two big presents in the same day, one for your birthday and another one from Santa :) \r\n\r\nHomeschooled vs School should be discussed (if it hasn't already been) in another topic ... so go ahead and make one. Sorry for being off-topic :blush:", "Solution_1": "[quote=\"Valentin Vornicu\"][quote=\"KBabe\"]I'm 11, about to be 12 in two days. \n\nand i have a question, are homeschooled ppl usually smarter than public school ppl?[/quote]That mean you're gonna receive two big presents in the same day, one for your birthday and another one from Santa :) \n\nHomeschooled vs School should be discussed (if it hasn't already been) in another topic ... so go ahead and make one. Sorry for being off-topic :blush:[/quote]\r\n\r\nIsn't Christmas in 4, not 2 days?", "Solution_2": "[quote=\"h_s_potter2002\"]Isn't Christmas in 4, not 2 days?[/quote]Since it's 22nd (well at least in Romania :D) and usually Christmas presents are offered on the late 24th night (again at least in Romania :) ) my post was (Romanian-ly speaking) correct :)", "Solution_3": "it is fun to get double presents. what's your favorite holiday? (anyone)\r\n\r\nanyone want to talk about..maybe their fantasies, or what they wish their life was like?", "Solution_4": "[quote=\"KBabe\"]it is fun to get double presents. what's your favorite holiday? (anyone)\n\nanyone want to talk about..maybe their fantasies, or what they wish their life was like?[/quote]\r\n\r\nmy birthday is 11 days after Christmas, so it's like this one big present fest and then you have to wait another whole year. i would prefer to have my b-day and Christmas spread out, but i usually get the day off cuz of Christmas break, but not this year! :( lol. my favorite holidy is, in fact, Christmas because it is a very important holiday in regards to my faith, and the presents are simply an added bonus. :lol:", "Solution_5": "What I really want for christmas is a new calculator but thats not happening since I got a new one last year for christmas that a TI 83+. I'd love anything thats horse related and I know that I am getting my show clothes for Class A horse shows since I had to come with my trainer and my mother to try them on and pick them out. I would say I wanted a horse like I always have but I got my horse a weeks before my b-day so I suppose I should be content lol, but with all the things horses need I'll alwaysbe getting supplies." } { "Tag": [ "integration", "calculus", "analytic geometry", "function", "inequalities", "linear algebra", "matrix" ], "Problem": "Calculate :\r\n $ V_n(R) \\equal{} \\int\\int\\int_{x_1^2 \\plus{} x_2^2 \\plus{} ... \\plus{} x_n^2 \\leq R^2}...\\int{dx_1dx_2...dx_n} \\equal{} ?$ Here ($ R > 0$)\r\n[hide=\"Answer\"] \n $ V_{n}(R) \\equal{} \\frac {2(2\\pi)^{\\frac {n \\minus{} 1}{2}}}{1.3.5....n}R^n$ If $ n$ is odd\n $ V_n{(R)} \\equal{} \\frac {\\pi^{\\frac{n}{2}}}{(\\frac {n}{2})!}R^n$ If $ n$ is even \n(True ? or false ?)[/hide]", "Solution_1": "Key words: n-volume of n-ball.\r\n\r\nHere are my two favorite methods. One thing to note: by similarity and scaling, $ V_n(R)\\equal{}V_n(1)R^n.$ (I've probably posted both of these before on this site but I'm too lazy to look.)\r\n\r\n1. Assume $ n\\ge 2$ and slice off two dimensions:\r\n\r\n$ V_n(R)\\equal{}\\iint_{x_1^2\\plus{}x_2^2\\le R^2}\\left(\\underset{x_3^2\\plus{}\\cdots\\plus{}x_n^2\\le R^2\\minus{}x_1^2\\minus{}x_2^2}{\\int\\cdots\\int}dx_3\\dots dx_n\\right)dx_1\\,dx_2$\r\n\r\n$ \\equal{}\\iint_{x_1^2\\plus{}x_2^2\\le R^2}V_{n\\minus{}2}\\left(\\sqrt{R^2\\minus{}x_1^2\\minus{}x_2^2}\\right)dx_1\\,dx_2$\r\n\r\nNow do that integral in polar coordinates:\r\n\r\n$ \\equal{}\\int_0^R\\int_0^{2\\pi}V_{n\\minus{}2}\\left(\\sqrt{R^2\\minus{}r^2}\\right)r\\,d\\theta\\,dr$\r\n\r\n$ \\equal{}2\\pi V_{n\\minus{}2}(1)\\int_0^R\\left(R^2\\minus{}r^2\\right)^{\\frac n2\\minus{}1}r\\,dr$\r\n\r\nYou can do that integral by the standard first-year calculus substitution:\r\n\r\n$ V_n(R)\\equal{}2\\pi V_{n\\minus{}2}(1)\\left.\\left(\\minus{}\\frac12\\cdot\\frac2n\\left(R^2\\minus{}r^2\\right)^{\\frac n2}\\right)\\right|_{r\\equal{}0}^R$\r\n\r\n$ \\equal{}\\frac{2\\pi}{n}V_{n\\minus{}2}(1)R^n$\r\n\r\nIn particular, $ V_n(1)\\equal{}\\frac{2\\pi}{n}V_{n\\minus{}2}(1).$\r\n\r\nFrom $ V_1(R)\\equal{}2R,$ we get $ V_3(R)\\equal{} \\frac{4\\pi R^3}3,\\ V_5(R)\\equal{}\\frac{8\\pi^2R^5}{3\\cdot 5},$ and so on.\r\n\r\nAssuming $ V_0(R)\\equal{}1$ is consistent with the correct $ V_2(R)\\equal{}\\pi R^2.$ Continuing, $ V_4(R)\\equal{}\\frac{\\pi^2R^4}{2}$ and in general, $ V_{2k}(R)\\equal{}\\frac{\\pi^kR^{2k}}{k!}.$\r\n\r\n2. Use polar coordinates (the $ n$-dimensional analogue of spherical coordinate) to derive that $ V_n(R)\\equal{}\\frac1nA_nR^2,$ where $ A_n$ is the $ (n\\minus{}1)$-dimensional measure of the unit sphere. Integrate a Gaussian in both rectangular and polar coordinates. Find a gamma function in the resulting radial integral. (Supply the details yourself.) The result should be that $ V_n(R)\\equal{}\\frac{\\pi^{n/2}R^n}{\\Gamma\\left(\\frac n2\\plus{}1\\right)}.$\r\n\r\nI learned method 2 a long time ago (call it \"folklore\"). I don't think I ever saw method 1 anywhere - and I like it because it's more elementary. Yes, the answers in the form QuyBac wrote them above are correct.", "Solution_2": "May be a harder question : \r\n\r\nOn $ \\forall x\\in\\mathbb{R}^{n}$ define $ \\parallel{}x\\parallel{}_{p} = (|x_{1}|^{p} + \\ldots + |x_{n}|^{p})^{\\frac {1}{p}}$\r\nthen we know minkowski inequality : $ \\parallel{}x + y\\parallel{}_{p}\\leq \\parallel{}x\\parallel{}_{p} + \\parallel{}y\\parallel{}_{p}$ then we have $ (\\mathbb{R}^{n},\\parallel{}.\\parallel{}_{p})$ is a normed space. Find the Volume of the \"unit ball\" $ B_{n} = \\{x : \\parallel{}x\\parallel{}_{p} < 1\\}$?", "Solution_3": "[quote=\"Kent Merryfield\"]Key words: n-volume of n-ball.\n\nHere are my two favorite methods. One thing to note: by similarity and scaling, $ V_n(R) \\equal{} V_n(1)R^n.$ (I've probably posted both of these before on this site but I'm too lazy to look.)\n\n1. Assume $ n\\ge 2$ and slice off two dimensions:\n\n$ V_n(R) \\equal{} \\iint_{x_1^2 \\plus{} x_2^2\\le R^2}\\left(\\underset{x_3^2 \\plus{} \\cdots \\plus{} x_n^2\\le R^2 \\minus{} x_1^2 \\minus{} x_2^2}{\\int\\cdots\\int}dx_3\\dots dx_n\\right)dx_1\\,dx_2$\n\n$ \\equal{} \\iint_{x_1^2 \\plus{} x_2^2\\le R^2}V_{n \\minus{} 2}\\left(\\sqrt {R^2 \\minus{} x_1^2 \\minus{} x_2^2}\\right)dx_1\\,dx_2$\n\nNow do that integral in polar coordinates:\n\n$ \\equal{} \\int_0^R\\int_0^{2\\pi}V_{n \\minus{} 2}\\left(\\sqrt {R^2 \\minus{} r^2}\\right)r\\,d\\theta\\,dr$\n\n$ \\equal{} 2\\pi V_{n \\minus{} 2}(1)\\int_0^R\\left(R^2 \\minus{} r^2\\right)^{\\frac n2 \\minus{} 1}r\\,dr$\n\nYou can do that integral by the standard first-year calculus substitution:\n\n$ V_n(R) \\equal{} 2\\pi V_{n \\minus{} 2}(1)\\left.\\left( \\minus{} \\frac12\\cdot\\frac2n\\left(R^2 \\minus{} r^2\\right)^{\\frac n2}\\right)\\right|_{r \\equal{} 0}^R$\n\n$ \\equal{} \\frac {2\\pi}{n}V_{n \\minus{} 2}(1)R^n$\n\nIn particular, $ V_n(1) \\equal{} \\frac {2\\pi}{n}V_{n \\minus{} 2}(1).$\n\nFrom $ V_1(R) \\equal{} 2R,$ we get $ V_3(R) \\equal{} \\frac {4\\pi R^3}3,\\ V_5(R) \\equal{} \\frac {8\\pi^2R^5}{3\\cdot 5},$ and so on.\n\nAssuming $ V_0(R) \\equal{} 1$ is consistent with the correct $ V_2(R) \\equal{} \\pi R^2.$ Continuing, $ V_4(R) \\equal{} \\frac {\\pi^2R^4}{2}$ and in general, $ V_{2k}(R) \\equal{} \\frac {\\pi^kR^{2k}}{k!}.$\n\n2. Use polar coordinates (the $ n$-dimensional analogue of spherical coordinate) to derive that $ V_n(R) \\equal{} \\frac1nA_nR^2,$ where $ A_n$ is the $ (n \\minus{} 1)$-dimensional measure of the unit sphere. Integrate a Gaussian in both rectangular and polar coordinates. Find a gamma function in the resulting radial integral. (Supply the details yourself.) The result should be that $ V_n(R) \\equal{} \\frac {\\pi^{n/2}R^n}{\\Gamma\\left(\\frac n2 \\plus{} 1\\right)}.$\n\nI learned method 2 a long time ago (call it \"folklore\"). I don't think I ever saw method 1 anywhere - and I like it because it's more elementary. Yes, the answers in the form QuyBac wrote them above are correct.[/quote]\r\n Thanks you very much [b]Pro. Kent Merryfield[/b] !\r\n Now If can please you for a the [b]answer [/b] \r\n Let $ A \\in M_n{\\mathbb{(R)}}$ , $ A \\equal{} A^T$ and $ 0 < x^TAx \\leq 1,\\forall x \\in M_{n1}{\\mathbb{(R)}} \\minus{} {O_{n1}}$ (or $ 0 < A \\leq I_n$)\r\n Here $ x \\equal{} \\begin{bmatrix}x_1 \\\\\r\nx_2 \\\\\r\n.. \\\\\r\nx_n\\end{bmatrix},x_k \\in \\mathbb{R},k \\equal{} 1,2,..,n$\r\nCalculate : $ I \\equal{} \\int\\int_{0 < x^TAx \\leq 1}...\\int{e^{x^TAx}dx_1dx_2..dx_n}$\r\n(thanks ! I hope can learned method 1 of Pro to have answer in $ S_n^{\\plus{}\\plus{}}(\\mathbb{R})$ :) )", "Solution_4": "Hint: find positive definite $ S$ such that $ S^2\\equal{}A.$ Make the substitution $ u\\equal{}Sx,$ also written as $ x\\equal{}S^{\\minus{}1}u.$ What happens if you try that?", "Solution_5": "[quote=\"Kent Merryfield\"] What happens if you try that?[/quote]\r\n Oke I want a result for calculate ?\r\n $ A \\in S_n^{ \\plus{} \\plus{} }{\\mathbb{(R)}}$ the exist $ \\omega \\in O_n(\\mathbb{R})$ such that : $ A \\equal{} \\omega^Tdiag(\\lambda_1,\\lambda_2,..,\\lambda_n)\\omega$ , $ \\lambda_1,\\lambda_2,..,\\lambda_n$ are positive eigenvalues of $ A$\r\nLet $ y \\equal{} \\omega x$ then $ Jacobi (J)$ : $ |J| \\equal{} |\\det(\\omega)| \\equal{} 1$ and :\r\n $ x^TAx \\equal{} y^Tdiag(\\lambda_1,...,\\lambda_n)y \\equal{} \\lambda_1y_1^2 \\plus{} ... \\plus{} \\lambda_ny_n^2$ and :\r\n $ I \\equal{} \\int\\int_{0 < \\lambda_1y_1^2 \\plus{} ... \\plus{} \\lambda_ny_n^2 \\leq 1}...\\int{e^{\\lambda_1y_1^2 \\plus{} ... \\plus{} \\lambda_ny_n^2}dy_1dy_2...dy_n}$\r\n Put $ z_k \\equal{} \\sqrt {\\lambda_k}y_k,1 \\leq k \\leq n$ ...... then use method 1 of Pro to calculate .....\r\n Are you Agree ? :)", "Solution_6": "1)Edit : $ |J| \\equal{} |\\det(\\omega)| \\equal{} 1$ are $ |J| \\equal{} |\\det(\\omega^T)| \\equal{} 1$\r\n2) $ A\\equal{}S^2$ are clear but If $ A \\geq O$\r\n3) A nice result for ($ \\mathbb{R}^n$ , $ n \\geq 2$) eliptic : $ I \\equal{}\\int\\int_{0 < x^{T}Ax\\leq 1}...\\int{dx_{1}dx_{2}..dx_{n}}$ \r\n4) Thanks you very much If you are for a result calculate in $ \\mathbb{R}^n$, :roll:", "Solution_7": "[quote=\"1234567a\"]May be a harder question : \n\nOn $ \\forall x\\in\\mathbb{R}^{n}$ define $ \\parallel{}x\\parallel{}_{p} = (|x_{1}|^{p} + \\ldots + |x_{n}|^{p})^{\\frac {1}{p}}$\nthen we know minkowski inequality : $ \\parallel{}x + y\\parallel{}_{p}\\leq \\parallel{}x\\parallel{}_{p} + \\parallel{}y\\parallel{}_{p}$ then we have $ (\\mathbb{R}^{n},\\parallel{}.\\parallel{}_{p})$ is a normed space. Find the Volume of the \"unit ball\" $ B_{n} = \\{x : \\parallel{}x\\parallel{}_{p} < 1\\}$?[/quote]\n\nWe use the following formula: If the function f is homogeneous of degree k, i.e, $f(tx)=t^kf(x)$ with $x\\in \\R^n$ and $t>0$ then \n\\[\\int_{R^n}f(x)e^{-||x||_p^p}dx=C_{n,p}\\int_{B_p}f(x)dx\\]\nwhere $B_p$ is the unit ball of norm $||.||_p$, and \n\\[C_{n,p}=\\Gamma(1+\\frac{n+k}{p})\\]\nProof of this formula: We have\n\\[e^{-||x||_p^p}=\\int_{||x||_p^p}^{\\infty}e^{-t}dt\\]\napply the Fubini theorem, we get\n\\[\\int_{R^n}f(x)e^{-||x||_p^p}dx=\\int_{R^n}f(x)\\int_{||x||_p^p}^{\\infty}e^{-t}dt=\\int_{R^n}\\int_0^{\\infty}f(x)e^{-t}1_{||x||_p^p\\leq t}dtdx.\\]\nChange of variable $x=t^{1/p}y$, because of the homogeneous of f we have\n\\[\\int_{R^n}f(x)e^{-||x||_p^p}dx=\\int_{B_p}f(y)\\int_{0}^{\\infty}t^{\\frac{n+k}{p}}e^{-t}dtdy=\\Gamma(1+\\frac{n+k}{p})\\int_{B_p}f(y)dy\\]\nApply this result for the function $f=1$ with the note that \n\\[\\int_Re^{-|t|^p}dt=2\\int_0^{\\infty}e^{-t^p}dt=\\frac{2}{p}\\int_0^{\\infty}e^{-u}u^{1/p-1}du=\\frac{2}{p}\\Gamma(\\frac{1}{p})=2\\Gamma(1+\\frac{1}{p})\\]\nHence\n\\[V(B_p)=2^n\\frac{(\\Gamma(1+\\frac{1}{p}))^n}{\\Gamma(1+\\frac{n}{p})}\\]" } { "Tag": [], "Problem": "\u03a4\u03b1 \u03c0\u03c1\u03bf\u03b2\u03bb\u03ae\u03bc\u03b1\u03c4\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9:\r\n 1)\u03a3\u03b5 \u03ad\u03bd\u03b1 \u03bf\u03c1\u03b8\u03bf\u03b3\u03ce\u03bd\u03b9 \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03bf \u0391\u0392\u0393 \u03bc\u03b5 \u03bc\u03ae\u03ba\u03b7 \u03ba\u03ac\u03b8\u03b5\u03c4\u03c9\u03bd \u03c0\u03bb\u03b5\u03c5\u03c1\u03ce\u03bd \u03b1=2 \u03ba\u03b1\u03b9 \u03b2=1 \u03c6\u03ad\u03c1\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b7\u03bd \u03b5\u03c9\u03c4\u03b5\u03c1\u03b9\u03ba\u03ae \u03ba\u03b1\u03b9 \u03b5\u03be\u03c9\u03c4\u03b5\u03c1\u03b9\u03ba\u03ae \u03b4\u03b9\u03c7\u03bf\u03c4\u03cc\u03bc\u03bf \u03c4\u03b7\u03c2 \u03b3\u03c9\u03bd\u03af\u03b1\u03c2 \u0391. \u039f\u03b9 \u03b4\u03b9\u03c7\u03bf\u03c4\u03cc\u03bc\u03bf\u03b9 \u03c4\u03ad\u03bc\u03bd\u03bf\u03c5\u03bd \u03c4\u03b7\u03bd \u03b5\u03c5\u03b8\u03b5\u03af\u03b1 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ae \u0392\u0393 \u03b5\u03c3\u03c9\u03c4\u03b5\u03c1\u03b9\u03ba\u03ac \u03c3\u03c4\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03a0 \u03ba\u03b1\u03b9 \u03b5\u03c3\u03c9\u03c4\u03b5\u03c1\u03ba\u03ac \u03c3\u03c4\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03a3. \u03a0\u03bf\u03b9\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03b1 \u03bc\u03ae\u03ba\u03b7 \u03c4\u03c9\u03bd \u0393\u03a0 \u03ba\u03b1\u03b9 \u0393\u03a3?\r\n 2)\u03a3\u03b5 \u03ad\u03bd\u03b1 \u03c4\u03b5\u03c4\u03c1\u03ac\u03b3\u03c9\u03bd\u03bf \u03bc\u03b5 \u03c0\u03bb\u03b5\u03c5\u03c1\u03ac \u03bc\u03ae\u03ba\u03bf\u03c5\u03c3 2 \u03b5\u03b3\u03b3\u03c1\u03ac\u03c6\u03bf\u03c5\u03bc\u03b5 \u03ad\u03bd\u03b1 \u03b9\u03c3\u03bf\u03c3\u03ba\u03b5\u03bb\u03ad\u03c2 \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03bf. \u03a0\u03bf\u03b9\u03bf \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf \u03bc\u03ae\u03ba\u03bf\u03c2 \u03c4\u03b7\u03c2 \u03b1\u03ba\u03c4\u03af\u03bd\u03b1\u03c2 \u03c4\u03bf\u03c5 \u03b5\u03b3\u03b3\u03b5\u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03bd\u03bf\u03c5 \u03c3\u03c4\u03bf \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03bf \u03ba\u03cd\u03ba\u03bb\u03bf\u03c5? (\u039c\u03c0\u03bf\u03cd\u03bc\u03b5 \u03b1\u03bd \u03b8\u03ad\u03bb\u03bf\u03c5\u03bc\u03b5 \u03bd\u03b1 \u03c7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03ae\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf \u03b1\u03c0\u03bf\u03c4\u03ad\u03bb\u03b5\u03c3\u03bc\u03b1 \u03c4\u03bf\u03c5 \u03c0\u03c1\u03bf\u03b7\u03b3\u03bf\u03cd\u03bc\u03b5\u03bd\u03bf\u03c5 \u03c0\u03c1\u03bf\u03b2\u03bb\u03ae\u03bc\u03b1\u03c4\u03bf\u03c2).\r\n-\u0393\u03b9\u03b1 \u03c4\u03b7\u03bd \u03b5\u03c0\u03af\u03bb\u03c5\u03c3\u03b7 \u03c4\u03c9\u03bd \u03c0\u03c1\u03bf\u03b2\u03bb\u03b7\u03bc\u03ac\u03c4\u03c9\u03bd \u03bd\u03b1 \u03bc\u03b7 \u03c7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03b7\u03b8\u03bf\u03cd\u03bd \u03c4\u03c1\u03b9\u03b3\u03c9\u03bd\u03bf\u03bc\u03b5\u03c4\u03c1\u03b9\u03ba\u03ad\u03c2 \u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03ae\u03c3\u03b5\u03b9\u03c2\r\n\u038c\u03c0\u03bf\u03b9\u03bf\u03c2 \u03b4\u03cd\u03bd\u03b1\u03c4\u03b1\u03b9 \u03bd\u03b1 \u03b2\u03bf\u03b7\u03b8\u03ae\u03c3\u03b5\u03b9 \u03b1\u03c2 \u03bc\u03bf\u03c5 \u03c3\u03c4\u03b5\u03af\u03bb\u03b5\u03b9.\r\n\u0395\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03ce.", "Solution_1": "\u03c3\u03c4\u03bf 1\u03bf \u03c0\u03c1\u03bf\u03b2\u03bb\u03b7\u03bc\u03b1, \u03bf\u03c4\u03b1\u03bd \u03bb\u03b5\u03c2 \u03c6\u03b5\u03c1\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b7 \u03b4\u03b9\u03c7\u03bf\u03c4\u03bf\u03bc\u03bf \u03c4\u03b7\u03c2 \u03b3\u03c9\u03bd\u03b9\u03b1\u03c2 \u0391, \u03b5\u03bd\u03bf\u03b5\u03b9\u03c2 \u03c4\u03b7 \u03b4\u03b9\u03c7\u03bf\u03c4\u03bf\u03bc\u03bf \u03c4\u03b7\u03c2 90\u03b1\u03c1\u03b1\u03c2 \u03b3\u03c9\u03bd\u03b9\u03b1\u03c2;", "Solution_2": "\u038c\u03c7\u03b9 \u03b1\u03bd \u03b4\u03b5\u03b9\u03c2 \u03bf\u03b9 \u03ba\u03ac8\u03b5\u03c4\u03b5\u03c2 \u03c0\u03bb\u03b5\u03c5\u03c1\u03ad\u03c2 \u03b5\u03bd\u03b9\u03b1 \u03bf\u03b9 \u03b1 \u03ba\u03b1\u03b9 \u03b2 \u03bf\u03c0\u03cc\u03c4\u03b5 \u03ba\u03b1\u03c4\u03b1\u03bb\u03b1\u03b2\u03b1\u03af\u03bd\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b7 \u0393 \u03bf\u03c1\u03b8\u03ae \u03b3\u03c9\u03bd\u03af\u03b1", "Solution_3": "1)\u03b5\u03c3\u03c4\u03c9 \u03b5\u03c3\u03c9\u03c4\u03b5\u03c1\u03b9\u03ba\u03bf \u03c4\u03bf P,\u03b5\u03be\u03c9\u03c4\u03b5\u03c1\u03b9\u03ba\u03bf \u03c4\u03bf S.\u03b8\u03b1 \u03b5\u03c6\u03b1\u03c1\u03bc\u03bf\u03c3\u03bf\u03c5\u03bc\u03b5 \u03b8\u03b5\u03c9\u03c1\u03b7\u03bc\u03b1\u03c4\u03b1 \u03b4\u03b9\u03c7\u03bf\u03c4\u03bf\u03bc\u03c9\u03bd.\r\n\u03a0\u03b9\u03bf \u03c3\u03c5\u03b3\u03ba\u03b5\u03ba\u03c1\u03b9\u03bc\u03b5\u03bd\u03b1,\r\n$ \\frac{BP}{PC}\\equal{}\\frac{AB}{b} < \\equal{} > \\frac{a}{PC}\\equal{}\\frac{AB \\plus{} b}{b}$\r\n$ < \\equal{} >$\r\n$ \\frac{2}{PC}\\equal{}\\sqrt{5} \\plus{}1< \\equal{} >PC\\equal{}\\frac{2}{\\sqrt{5} \\plus{}1}$,\u03b1\u03c6\u03bf\u03c5\r\n$ AB^2\\equal{}1 \\plus{} 4\\equal{}5$\r\n\u03b1\u03bb\u03bb\u03b1 \u03ba\u03b1\u03b9\r\n$ \\frac{SC}{SB}\\equal{}\\frac{b}{AB} <\\equal{} >\\frac{SC}{a}\\equal{}\\frac{1}{\\sqrt{5}\\minus{}1}<\\equal{}>SC\\equal{}\\frac{2}{\\sqrt{5}\\minus{}1}$\r\n\u03b5\u03bb\u03c0\u03b9\u03b6\u03c9 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03b5\u03ba\u03b1\u03bd\u03b1 \u03ba\u03b1\u03c0\u03bf\u03b9\u03bf \u03bb\u03b1\u03b8\u03bf\u03c2 \u03c3\u03c4\u03b9\u03c2 \u03c0\u03c1\u03b1\u03be\u03b5\u03b9\u03c2. :)", "Solution_4": "[quote=\"theo_1\"]\u038c\u03c7\u03b9 \u03b1\u03bd \u03b4\u03b5\u03b9\u03c2 \u03bf\u03b9 \u03ba\u03ac8\u03b5\u03c4\u03b5\u03c2 \u03c0\u03bb\u03b5\u03c5\u03c1\u03ad\u03c2 \u03b5\u03bd\u03b9\u03b1 \u03bf\u03b9 \u03b1 \u03ba\u03b1\u03b9 \u03b2 \u03bf\u03c0\u03cc\u03c4\u03b5 \u03ba\u03b1\u03c4\u03b1\u03bb\u03b1\u03b2\u03b1\u03af\u03bd\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b7 \u0393 \u03bf\u03c1\u03b8\u03ae \u03b3\u03c9\u03bd\u03af\u03b1[/quote]\r\n\r\n\u03b1 \u03bf\u03ba, \u03bf\u03c0\u03c9\u03c2 \u03ba\u03b1\u03b9 \u03bd\u03b1 \u03b5\u03c7\u03b5\u03b9 \u03c0\u03b1\u03bd\u03c4\u03bf\u03c2 \u03c4\u03bf 1\u03bf \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b1\u03c0\u03bb\u03b7 \u03b5\u03c6\u03b1\u03c1\u03bc\u03bf\u03b3\u03b7 \u03c4\u03bf\u03c5 \u0398\u03b5\u03c9\u03c1\u03b7\u03bc\u03b1\u03c4\u03bf\u03c2 \u03b4\u03b9\u03c7\u03bf\u03c4\u03bf\u03bc\u03c9\u03bd, \u03c4\u03bf 2\u03bf \u03b5\u03b9\u03bd\u03b1\u03b9 \u03bb\u03b9\u03b3\u03bf \u03b6\u03bf\u03c1\u03b9\u03ba\u03bf, \u03b8\u03b1 \u03c4\u03bf \u03c0\u03c1\u03bf\u03c3\u03c0\u03b1\u03b8\u03b7\u03c3\u03c9 \u03b1\u03c5\u03c1\u03b9\u03bf \u03b3\u03b9\u03b1\u03c4\u03b9 \u03c4\u03c9\u03c1\u03b1 \u03c0\u03c1\u03b5\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03b4\u03b9\u03b1\u03b2\u03b1\u03c3\u03c9 \u0391\u039f\u0394\u0395 \u03b3\u03b9\u03b1 \u03c4\u03b7\u03c2 \u03a0\u03b1\u03bd\u03b5\u03bb\u03bb\u03b7\u03bd\u03b9\u03b5\u03c2, \u03b5\u03c7\u03c9 20 \u03bc\u03b5\u03c1\u03b5\u03c2 \u03bd\u03b1 \u03bc\u03b1\u03b8\u03c9 \u03b1\u03c0 \u03b5\u03be\u03c9 27 \u03c3\u03b5\u03bb\u03b9\u03b4\u03b5\u03c2...", "Solution_5": "\u039b\u03bf\u03b9\u03c0\u03bf\u03bd, \u03c4\u03b5\u03bb\u03b9\u03ba\u03b1 \u03b2\u03c1\u03b7\u03ba\u03b1 \u03bb\u03c5\u03c3\u03b7 \u03b3\u03b9\u03b1 \u03c4\u03bf 2\u03bf \u03c0\u03c1\u03bf\u03b2\u03bb\u03b7\u03bc\u03b1 \u03b7 \u03bf\u03c0\u03bf\u03b9\u03b1 \u03c3\u03c4\u03b7\u03c1\u03b9\u03b6\u03b5\u03c4\u03b1\u03b9 \u03c3\u03c4\u03b7 \u03bb\u03c5\u03c3\u03b7 \u03c4\u03bf\u03c5 \u03c0\u03c1\u03c9\u03c4\u03bf\u03c5 (\u03b4\u03b7\u03bb\u03b1\u03b4\u03b7 \u03b5\u03b3\u03b1\u03c1\u03bc\u03bf\u03b3\u03b7 \u0398\u03b5\u03c9\u03c1\u03b7\u03bc\u03b1\u03c4\u03bf\u03c2 \u0394\u03b9\u03c7\u03bf\u03c4\u03bf\u03bc\u03c9\u03bd), \u03bf\u03bc\u03c9\u03c2 \u03c4\u03bf \u03b1\u03c0\u03bf\u03c4\u03b5\u03bb\u03b5\u03c3\u03bc\u03b1 \u03c4\u03b7\u03c2 \u03bb\u03c5\u03c3\u03b7\u03c2 \u03b5\u03be\u03b1\u03c1\u03c4\u03b1\u03c4\u03b1\u03b9 \u03b1\u03c0\u03bf \u03c4\u03b1 \u03c3\u03c4\u03bf\u03b9\u03c7\u03b5\u03b9\u03b1 \u03c4\u03bf\u03c5 \u03c4\u03c1\u03b9\u03b3\u03c9\u03bd\u03bf\u03c5 \u03c0\u03bf\u03c5 \u03b5\u03b3\u03b3\u03c1\u03b1\u03c6\u03bf\u03c5\u03bc\u03b5, \u03ba\u03b1\u03b9 \u03b5\u03c0\u03b7\u03b4\u03b7 \u03c5\u03c0\u03b1\u03c1\u03c7\u03bf\u03c5\u03bd \u03c0\u03b1\u03c1\u03b1 \u03c0\u03bf\u03bb\u03bb\u03b1 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03b1 \u03bc\u03b5\u03c4\u03b1\u03be\u03c5 \u03c4\u03bf\u03c5\u03c2 \u03b9\u03c3\u03bf\u03c3\u03ba\u03b5\u03bb\u03b7 \u03c4\u03c1\u03b9\u03b3\u03c9\u03bd\u03b1 \u03b5\u03b3\u03c1\u03b1\u03c8\u03b9\u03bc\u03b1 \u03c3\u03b5 \u03b5\u03bd\u03b1 \u03c3\u03c5\u03b3\u03ba\u03b5\u03ba\u03c1\u03b9\u03bc\u03b5\u03bd\u03bf \u03c4\u03b5\u03c4\u03c1\u03b1\u03b3\u03c9\u03bd\u03bf, \u03b4\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03c9 \u03bd\u03b1 \u03b4\u03c9\u03c3\u03c9 \u03bc\u03b9\u03b1 \u03c3\u03c5\u03b3\u03ba\u03b5\u03ba\u03c1\u03b9\u03bc\u03b5\u03bd\u03b7 \u03bb\u03c5\u03c3\u03b7 \u03c3\u03c4\u03bf \u03c0\u03c1\u03bf\u03b2\u03bb\u03b7\u03bc\u03b1 \u03b1\u03bd \u03b4\u03b5\u03bd \u03b4\u03c9\u03b8\u03b5\u03b9 \u03c0\u03c1\u03c9\u03c4\u03b1 \u03ba\u03b1\u03c0\u03bf\u03b9\u03b1 \u03c0\u03bb\u03b7\u03c1\u03bf\u03c6\u03bf\u03c1\u03b9\u03b1 \u03b3\u03b9\u03b1 \u03c4\u03bf \u03b5\u03b3\u03b3\u03b5\u03b3\u03c1\u03b1\u03bc\u03b5\u03bd\u03bf \u03c4\u03c1\u03b9\u03b3\u03c9\u03bd\u03bf. \u0395\u03b9\u03bd\u03b1\u03b9 \u03b3\u03c1\u03b1\u03bc\u03bc\u03b5\u03bd\u03b7 \u03c3\u03b5 \u03ba\u03b1\u03bd\u03b5\u03bd\u03b1 \u03b2\u03b9\u03b2\u03bb\u03b9\u03bf \u03b1\u03c5\u03c4\u03b7 \u03b7 \u03b1\u03c3\u03ba\u03b7\u03c3\u03b7;", "Solution_6": "\u038c\u03c7\u03b9 \u03b1\u03c5\u03c4\u03ae \u03b7 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bc\u03b9\u03b1 \u03b5\u03c1\u03b3\u03b1\u03c3\u03af\u03b1 \u03b7 \u03bf\u03c0\u03bf\u03af\u03b1 \u03b4\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c0\u03cc \u03c4\u03bf \u03b2\u03b9\u03b2\u03bb\u03af\u03bf.\r\n\u0397 \u03b5\u03ba\u03c6\u03ce\u03bd\u03b7\u03c3\u03b7 \u03b4\u03b5\u03bd \u03bb\u03ad\u03b5\u03b9 \u03c4\u03af\u03c0\u03bf\u03c4\u03b1 \u03c0\u03b5\u03c1\u03b9\u03c3\u03c3\u03cc\u03c4\u03b5\u03c1\u03bf \u03ba\u03b1\u03b9 \u03c4\u03b9\u03c0\u03bf\u03c4\u03b1 \u03bb\u03b9\u03b3\u03cc\u03c4\u03b5\u03c1\u03bf \u03b1\u03c0\u03cc \u03b1\u03c5\u03c4\u03cc \u03c0\u03bf\u03c5 \u03ae\u03b4\u03b7 \u03c3\u03b1\u03c2 \u03b5\u03af\u03c0\u03b1.", "Solution_7": "[quote=\"theo_1\"]\u038c\u03c7\u03b9 \u03b1\u03c5\u03c4\u03ae \u03b7 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bc\u03b9\u03b1 \u03b5\u03c1\u03b3\u03b1\u03c3\u03af\u03b1 \u03b7 \u03bf\u03c0\u03bf\u03af\u03b1 \u03b4\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c0\u03cc \u03c4\u03bf \u03b2\u03b9\u03b2\u03bb\u03af\u03bf.\n\u0397 \u03b5\u03ba\u03c6\u03ce\u03bd\u03b7\u03c3\u03b7 \u03b4\u03b5\u03bd \u03bb\u03ad\u03b5\u03b9 \u03c4\u03af\u03c0\u03bf\u03c4\u03b1 \u03c0\u03b5\u03c1\u03b9\u03c3\u03c3\u03cc\u03c4\u03b5\u03c1\u03bf \u03ba\u03b1\u03b9 \u03c4\u03b9\u03c0\u03bf\u03c4\u03b1 \u03bb\u03b9\u03b3\u03cc\u03c4\u03b5\u03c1\u03bf \u03b1\u03c0\u03cc \u03b1\u03c5\u03c4\u03cc \u03c0\u03bf\u03c5 \u03ae\u03b4\u03b7 \u03c3\u03b1\u03c2 \u03b5\u03af\u03c0\u03b1.[/quote]\r\n\r\n\u03a4\u03bf\u03c4\u03b5 \u03bc\u03b1\u03bb\u03bf\u03bd \u03c5\u03c0\u03b1\u03c1\u03c7\u03b5\u03b9 \u03b1\u03c3\u03b1\u03c6\u03b5\u03b9\u03b1 \u03c3\u03c4\u03b7 \u03b4\u03b9\u03b1\u03c4\u03c5\u03c0\u03c9\u03c3\u03b7, \u03b4\u03b9\u03bf\u03c4\u03b9 \u03c5\u03c0\u03b1\u03c1\u03c7\u03bf\u03c5\u03bd \u03b1\u03bc\u03b5\u03c4\u03c1\u03b9\u03c4\u03b1 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03b1 \u03b5\u03b3\u03c1\u03b1\u03c8\u03b9\u03bc\u03b1 \u03c4\u03c1\u03b9\u03b3\u03c9\u03bd\u03b1 \u03c3\u03c4\u03bf \u03c3\u03c5\u03b3\u03ba\u03b5\u03ba\u03c1\u03b9\u03bc\u03b5\u03bd\u03bf \u03c4\u03b5\u03c4\u03c1\u03b1\u03b3\u03c9\u03bd\u03bf \u03bc\u03b5 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03b1 \u03c5\u03c8\u03b7, \u03b2\u03b1\u03c3\u03b5\u03b9\u03c2 \u03ba\u03b1\u03b9 \u03c0\u03bb\u03b5\u03c5\u03c1\u03b5\u03c2, \u03ba\u03b1\u03b9 \u03b4\u03b5\u03bd \u03bd\u03bf\u03bc\u03b9\u03b6\u03c9 \u03bf\u03bb\u03b1 \u03bd\u03b1 \u03b5\u03c7\u03bf\u03c5\u03bd \u03c3\u03c4\u03b1\u03b8\u03b5\u03c1\u03b7 \u03b1\u03ba\u03c4\u03b9\u03bd\u03b1 \u03b5\u03b3\u03b3\u03b5\u03c1\u03b3\u03b1\u03bc\u03b5\u03bd\u03bf\u03c5 \u03ba\u03c5\u03ba\u03bb\u03bf\u03c5", "Solution_8": "[quote=\"NickNafplio\"]....\n\u03a4\u03cc\u03c4\u03b5 \u03bc\u03ac\u03bb\u03bb\u03bf\u03bd \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03b1\u03c3\u03ac\u03c6\u03b5\u03b9\u03b1 \u03c3\u03c4\u03b7 \u03b4\u03b9\u03b1\u03c4\u03cd\u03c0\u03c9\u03c3\u03b7, \u03b4\u03b9\u03cc\u03c4\u03b9 \u03c5\u03c0\u03ac\u03c1\u03c7\u03bf\u03c5\u03bd \u03b1\u03bc\u03ad\u03c4\u03c1\u03b7\u03c4\u03b1 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03ac \u03b5\u03b3\u03b3\u03c1\u03ac\u03c8\u03b9\u03bc\u03b1 \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03b1 \u03c3\u03c4\u03bf \u03c3\u03c5\u03b3\u03ba\u03b5\u03ba\u03c1\u03b9\u03bc\u03ad\u03bd\u03bf \u03c4\u03b5\u03c4\u03c1\u03ac\u03b3\u03c9\u03bd\u03bf \u03bc\u03b5 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03ac \u03cd\u03c8\u03b7, \u03b2\u03ac\u03c3\u03b5\u03b9\u03c2 \u03ba\u03b1\u03b9 \u03c0\u03bb\u03b5\u03c5\u03c1\u03ad\u03c2, \u03ba\u03b1\u03b9 \u03b4\u03b5\u03bd \u03bd\u03bf\u03bc\u03af\u03b6\u03c9 \u03cc\u03bb\u03b1 \u03bd\u03b1 \u03ad\u03c7\u03bf\u03c5\u03bd \u03c3\u03c4\u03b1\u03b8\u03b5\u03c1\u03ae \u03b1\u03ba\u03c4\u03af\u03bd\u03b1 \u03b5\u03b3\u03b3\u03b5\u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03bd\u03bf\u03c5 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c5[/quote]\r\n\u0398\u03b1 \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03c3\u03b5 \u03af\u03c3\u03c9\u03c2 \u03c4\u03bf \u03b6\u03b7\u03c4\u03bf\u03cd\u03bc\u03b5\u03bd\u03bf \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03bc\u03ad\u03b3\u03b9\u03c3\u03c4\u03b7 \u03c4\u03b9\u03bc\u03ae \u03c4\u03b7\u03c2 \u03b1\u03ba\u03c4\u03af\u03bd\u03b1\u03c2 \u03c4\u03bf\u03c5 \u03b5\u03b3\u03b3\u03b5\u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03bd\u03bf\u03c5 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c5 \u03b5\u03bd\u03cc\u03c2 \u03bc\u03b5\u03c4\u03b1\u03b2\u03b1\u03bb\u03bb\u03cc\u03bc\u03b5\u03bd\u03bf\u03c5 \u03b9\u03c3\u03bf\u03c3\u03ba\u03b5\u03bb\u03bf\u03cd\u03c2 \u03c4\u03c1\u03b9\u03b3\u03ce\u03bd\u03bf\u03c5, \u03bc\u03b5 \u03c4\u03b7\u03bd \u03ba\u03bf\u03c1\u03c5\u03c6\u03ae \u03c4\u03bf\u03c5 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf \u03c3\u03c4\u03b1\u03b8\u03b5\u03c1\u03cc \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03c3\u03c4\u03b7\u03bd \u03c0\u03b5\u03c1\u03af\u03bc\u03b5\u03c4\u03c1\u03bf \u03c4\u03bf\u03c5 \u03b4\u03bf\u03c3\u03bc\u03ad\u03bd\u03bf\u03c5 \u03c4\u03b5\u03c4\u03c1\u03b1\u03b3\u03ce\u03bd\u03bf\u03c5.\r\n\r\n$ \\bullet$ \u0388\u03c3\u03c4\u03c9 $ ABCD$ \u03c4\u03bf \u03b4\u03bf\u03c3\u03bc\u03ad\u03bd\u03bf \u03c4\u03b5\u03c4\u03c1\u03ac\u03b3\u03c9\u03bd\u03bf \u03ba\u03b1\u03b9 $ \\bigtriangleup KLM,$ \u03ad\u03bd\u03b1 \u03b9\u03c3\u03bf\u03c3\u03ba\u03b5\u03bb\u03ad\u03c2 \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03bf \u03bc\u03b5 $ KL \\equal{} KM$ \u03ba\u03b1\u03b9 \u03c4\u03b7\u03bd \u03ba\u03bf\u03c1\u03c5\u03c6\u03ae \u03c4\u03bf\u03c5 $ K,$ \u03c4\u03c5\u03c7\u03cc\u03bd \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03b5\u03c0\u03af \u03c4\u03b7\u03c2 \u03c0\u03bb\u03b5\u03c5\u03c1\u03ac\u03c2 $ AB.$\r\n\r\n\u039c\u03b5 \u03b4\u03b5\u03b4\u03bf\u03bc\u03ad\u03bd\u03bf \u03cc\u03c4\u03b9 $ a \\equal{} 2,$ \u03cc\u03c0\u03bf\u03c5 $ a$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03c0\u03bb\u03b5\u03c5\u03c1\u03ac \u03c4\u03bf\u03c5 $ ABCD,$ \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03ba\u03ac\u03c0\u03bf\u03b9\u03b1 \u03bc\u03ad\u03b3\u03b9\u03c3\u03c4\u03b7 \u03c4\u03b9\u03bc\u03ae \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03b1\u03ba\u03c4\u03af\u03bd\u03b1 $ r$ \u03c4\u03bf\u03c5 \u03b5\u03b3\u03b3\u03b5\u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03bd\u03bf\u03c5 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c5 \u03c3\u03c4\u03bf $ \\bigtriangleup KLM,$ \u03bc\u03b5 \u03b1\u03c6\u03b5\u03c4\u03b7\u03c1\u03af\u03b1 \u03c4\u03b7\u03bd \u03c4\u03b9\u03bc\u03ae $ r \\equal{} 0$ $ ($ \u03cc\u03c4\u03b1\u03bd $ K\\equiv C\\equiv L$ $ ),$ \u03c0\u03bf\u03c5 \u03b1\u03c6\u03bf\u03c1\u03ac \u03c3\u03c4\u03bf \u03c3\u03c5\u03b3\u03ba\u03b5\u03ba\u03c1\u03b9\u03bc\u03ad\u03bd\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf $ K.$\r\n\r\n\u039c\u03c0\u03bf\u03c1\u03b5\u03af \u03c4\u03b5\u03bb\u03b9\u03ba\u03ac \u03c4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03bd\u03b1 \u03bc\u03b1\u03c2 \u03b6\u03b7\u03c4\u03ac\u03b5\u03b9 \u03bd\u03b1 \u03c0\u03c1\u03bf\u03c3\u03b4\u03b9\u03bf\u03c1\u03af\u03c3\u03bf\u03c5\u03bc\u03b5 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf $ K$ \u03c3\u03c4\u03b7\u03bd \u03c0\u03b5\u03c1\u03af\u03bc\u03b5\u03c4\u03c1\u03bf \u03c4\u03bf\u03c5 \u03b4\u03bf\u03c3\u03bc\u03ad\u03bd\u03bf\u03c5 \u03c4\u03b5\u03c4\u03c1\u03b1\u03b3\u03ce\u03bd\u03bf\u03c5, \u03ae \u03b1\u03bd \u03c5\u03c0\u03ac\u03c1\u03c7\u03bf\u03c5\u03bd \u03c0\u03b5\u03c1\u03b9\u03c3\u03c3\u03cc\u03c4\u03b5\u03c1\u03b1 \u03c4\u03bf\u03c5 \u03b5\u03bd\u03cc\u03c2 \u03bd\u03b1 \u03b2\u03c1\u03bf\u03cd\u03bc\u03b5 \u03c0\u03bf\u03b9\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9, \u03b3\u03b9\u03b1 \u03c4\u03b1 \u03bf\u03c0\u03bf\u03af\u03b1 \u03c0\u03c1\u03bf\u03ba\u03cd\u03c0\u03c4\u03b5\u03b9 \u03c4\u03bf \u03bc\u03ad\u03b3\u03b9\u03c3\u03c4\u03bf \u03b4\u03c5\u03bd\u03b1\u03c4\u03cc \u03b5\u03cd\u03c1\u03bf\u03c2 \u03c4\u03b9\u03bc\u03ce\u03bd, \u03bc\u03b9\u03b1\u03c2 \u03ba\u03b1\u03b9 \u03ba\u03ac\u03b8\u03b5 \u03c6\u03bf\u03c1\u03ac \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03b1\u03c6\u03b5\u03c4\u03b7\u03c1\u03af\u03b1 \u03c4\u03b7\u03bd \u03c4\u03b9\u03bc\u03ae $ r \\equal{} 0.$\r\n\r\n\u0394\u03b5\u03bd \u03be\u03ad\u03c1\u03c9 \u03ac\u03bb\u03bb\u03bf \u03b1\u03bd\u03ac\u03bb\u03bf\u03b3\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03ba\u03b1\u03b9 \u03c0\u03c1\u03bf\u03c6\u03b1\u03bd\u03ce\u03c2 \u03b4\u03b5\u03bd \u03ad\u03c7\u03c9 \u03ba\u03ac\u03c0\u03bf\u03b9\u03b1 \u03bb\u03cd\u03c3\u03b7 \u03c5\u03c0\u03cc\u03c8\u03b7 \u03bc\u03bf\u03c5. \u03a0\u03c1\u03bf\u03c3\u03c0\u03b1\u03b8\u03ce \u03bc\u03cc\u03bd\u03bf \u03bd\u03b1 \u03ba\u03b1\u03c4\u03b1\u03bb\u03ac\u03b2\u03c9 \u03c4\u03bf \u03b6\u03b7\u03c4\u03bf\u03cd\u03bc\u03b5\u03bd\u03bf. \r\n\r\n\u039a\u03ce\u03c3\u03c4\u03b1\u03c2 \u0392\u03ae\u03c4\u03c4\u03b1\u03c2." } { "Tag": [ "function", "calculus", "integration", "limit", "real analysis", "Support", "calculus computations" ], "Problem": "Please help!!! Thanks. \r\n\r\n$f: R\\rightarrow$ $R$ be a continuous function and suppose that\r\n$\\int_{-\\infty}^{\\infty}$ $|f|$ exists. \r\nShow that $\\lim_{h\\to0} \\int_{-\\infty}^{\\infty}$ $|f(x+h)-f(x)|dx= 0$", "Solution_1": "$f$ is Lebesgue integrable on $\\mathbb{R}$. Now use the density of the continuous functions with compact support in $L_{1}(\\mathbb{R})$ and the fact that a fonction continuous on a compact set is in particular uniformly continuous.", "Solution_2": "Could anyone explain in a simple way or others since the \"Lebesgue integrable\" by \"julien_santini\" is sort of dehind of my knowledge? Sorry!!\r\n\r\nThank.", "Solution_3": "Since $\\lim_{K\\to\\infty}\\int_{-K}^K|f(x)|\\,dx=\\int_{-\\infty}^{\\infty}|f(x)|\\,dx,$ \r\n\r\ngiven $\\epsilon>0$ we can find $K$ such that $\\int_{|x|>K}|f(x)|\\,dx<\\frac{\\epsilon}4.$\r\n\r\nCreate a function $g$ as follows:\r\n\r\nFor $-K\\le x\\le K,\\,g(x)=f(x).$\r\n\r\nOn the intervals $[-K-1,-K]$ and $[K,K+1],$ define $g(x)=f(x)(K+1-|x|).$ Then let $g(x)=0$ for $|x|K}|f(x)|\\,dx<\\frac{\\epsilon}4.$\r\n\r\n$\\int_{\\mathbb{R}}|f(x+h)-f(x)|\\,dx\\le\\int_{\\mathbb{R}}|g(x+h)-g(x)|\\,dx +\\int_{\\mathbb{R}}|h(x+h)-h(x)|\\,dx$\r\n\r\n$\\int_{\\mathbb{R}}|h(x+h)-h(x)|\\,dx\\le2\\int_{\\mathbb{R}}|h(x)|\\,dx<\\frac{\\epsilon}2.$\r\n\r\nWe have that $g$ is uniformly continuous. Find $\\delta<1$ such that if $|h|<\\delta$ then $|g(x+h)-g(x)|<\\frac{\\epsilon}{4(K+1)}$\r\n\r\nThen $\\int_{\\mathbb{R}}|g(x+h)-g(x)|\\,dx=\\int_{-K-1}^{K+1}|f(x+h)-f(x)|\\,dx$\r\n\r\n$<\\int_{-K-1}^{K+1}\\frac{\\epsilon}{4(K+1)}\\,dx=\\frac{\\epsilon}2.$\r\n\r\nSo, for $|h|<\\delta,\\,\\int_{\\mathbb{R}}|f(x+h)-f(x)|\\,dx<\\epsilon.$\r\n\r\nThis is actually the same proof that julien_santini proposed, with two modifications:\r\n\r\n1. I don't mention Lebesgue integration because I don't have to. This should give Disco Very fewer reasons to panic.\r\n\r\n2. julien_santini used the abstract assertion of the density of continuous compactly supported functions in $L^1.$ I replaced that by explicitly constructing the approximating function.", "Solution_4": "A more elementary(?) proof, with lower standpoint. We don't need \"compact support\" here. Recall the following statement, which is a standard exercise of Riemann integral:\r\nLemma. $f$ is integrable on $[a-d,b+d] (d>0)$. Show that $\\int_{h\\to 0} \\int_a^b |f(x+h) - f(x)| dx = 0$.\r\n\r\nProof. \r\n[hide]\nWe are going to prove that, for $\\epsilon > 0$, there exists $\\delta>0$ such that $\\int_a^b |f(x+h) - f(x)| dx < \\epsilon$ for all $0 < h < \\delta$. In order to use the condition of integrability, we construct a partition of $[a,b]$ by lettting $x_i = a + i\\frac{b-a}{n}$. Thus\n$\\int_a^b |f(x+h) - f(x)| dx = \\sum \\int_{x_i}^{x_{i+1}} |f(x+h)-f(x)|dx$\nLet $0 < h <\\frac{b-a}{n}$. Suppose $x$ is in the $i$-th interval of the partition, then $|f(x+h)-f(x)|$ is in $i$-th or $i+1$-th interval. If $x+h$ is in the $i$-th one, then\n$|f(x+h)-f(x)| < \\omega_i$\nIf it is in the $i+1$-th one, \n$|f(x+h)-f(x)| < |f(x+h)-f(x_i)| + |f(x_i)-f(x)| < \\omega_{i+1} + \\omega_i$\nSo we always have\n$|f(x+h)-f(x)| < |f(x+h)-f(x_i)| + |f(x_i)-f(x)| < \\omega_{i+1} + \\omega_i$\n\nSince $f$ is integrable on $[a-d,b+d]$, it is bounded, suppose $|f|\\leq M$. Then we take $\\omega_{n+1}=2M$.\nHence\n$\\sum \\int_{x_i}^{x_{i+1}} |f(x+h)-f(x)|dx < (\\omega_0 + 2(\\omega_1 + \\cdots + \\omega_n)\\frac{b-a}{n} + \\omega_{n+1}\\frac{b-a}{n}$\n\n$f$ is integrable, we can choose $n$ large enough such that\n$\\sum \\omega_i \\frac{b-a}{n} < \\epsilon / 4$\nand\n$2M\\frac{b-a}{n} < \\epsilon / 2$\n\nThen\n$\\int_a^b |f(x+h) - f(x)| dx = \\sum \\int_{x_i}^{x_{i+1}} |f(x+h)-f(x)|dx < \\epsilon/2 + \\epsilon/2 < \\epsilon$.\n[/hide]\r\n\r\nNow we can apply the lemma to $[-K-1,K+1]$ in the Kent Merryfield's proof.", "Solution_5": "Thank you all that halps me to understand alot." } { "Tag": [ "Ring Theory", "superior algebra", "superior algebra unsolved" ], "Problem": "show that if R is a PID and S is a multiplication subset of R, then the localization S^(-1)R is also a PID.\r\n\r\nI am thinking of defining some Dedekind-Hasse norm. But I can only define it as non-integer norm. Is it not allowed????", "Solution_1": "Perhaps try to look at lifting of ideals directly through the localization? I don't have a lot of experience with localization but that just seems like a logical place to start." } { "Tag": [ "calculus", "LaTeX" ], "Problem": "Kindly note the following rules for posting in this forum. \r\n\r\n1) This Forum is for [color=red]High School (non-AP) , Non- Calculus problems and simple College level[/color] discussion. Homework problems at College level can be discussed here. For more advanced theoretical, Calculus-based as well as Olympiad problems, kindly use the [url=http://www.mathlinks.ro/Forum/index.php?f=332]Advanced Physics Forum[/url]. \r\n\r\n2) Problems and exercises of simple nature ( 'cookbook' problems) mainly requiring application of formulae and [color=orange]homework problems [/color] can be posted here. Recreational and 'fun' problems should also be posted here. [b]However note that no problem shall be fully solved for you and you should try to explain what you didnt understand or were unable to do while solving the problem.[/b] \r\n\r\n3)[color=darkblue] Clarification of doubts [/color]on specific High school or Freshman College level Physics Theory are welcome here. You can also seek help on course material. Advanced theoretical discussions should be done at the Advanced Physics Forum. \r\n\r\nFor more information, kindly contact the Moderator. Suggestions are welcome.\r\n\r\nThank you.", "Solution_1": "all of u befor posting please go thru these rules by Rushil...specifically that u shud show ur work :)", "Solution_2": "What do i do if i don't know LaTex?", "Solution_3": "Read:\r\nhttp://www.artofproblemsolving.com/Wiki/index.php/LaTeX", "Solution_4": "I've began to post problems not as homework problems, but as problems I've created and wanted to share. What should I do?", "Solution_5": "[quote]Recreational and 'fun' problems should also be posted here.[/quote]", "Solution_6": "Okay. Thank you. But what if they are high-level and original?", "Solution_7": "[quote=\"Rushil\"]For more advanced theoretical, Calculus-based as well as Olympiad problems, kindly use the [url=http://www.artofproblemsolving.com/Forum/index.php?f=332]Advanced Physics Forum[/url]. [/quote]\n\nDid you read the first post?", "Solution_8": "It seems as if there are contradicting directions. The part you just quoted indicates that those problems should go in the Advanced forum. However, your other quote suggests that they should be posted in this forum. Therefore, I feel that there is a contradiction.", "Solution_9": "As far as I can tell, you haven't posted anything that uses calculus, has advanced theoretical material, or is Olympiad level.\n\nSo far, you've been posting them in the right place. I see no contradiction.\n\nHigh-level (\"Olympiad\", or \"advanced theoretical\") problems should be posted in the Advanced Physics forum.", "Solution_10": "[quote=\"Rushil\"]Kindly note the following rules for posting in this forum. \n\n\n\n1) This Forum is for [color=red]High School (non-AP) , Non- Calculus problems and simple College level[/color] discussion. Homework problems at College level can be discussed here. For more advanced theoretical, Calculus-based as well as Olympiad problems, kindly use the [url=http://www.artofproblemsolving.com/Forum/index.php?f=332]Advanced Physics Forum[/url]. \n\n\n\n2) Problems and exercises of simple nature ( 'cookbook' problems) mainly requiring application of formulae and [color=orange]homework problems [/color] can be posted here. Recreational and 'fun' problems should also be posted here. [b]However note that no problem shall be fully solved for you and you should try to explain what you didnt understand or were unable to do while solving the problem.[/b] \n\n\n\n3)[color=darkblue] Clarification of doubts [/color]on specific High school or Freshman College level Physics Theory are welcome here. You can also seek help on course material. Advanced theoretical discussions should be done at the Advanced Physics Forum. \n\n\n\nFor more information, kindly contact the Moderator. Suggestions are welcome.\n\n\n\nThank you.[/quote]\nBy non-AP, that means no discussion of the AP practice problems or real exam questions, right?", "Solution_11": "Given that the person you quoted has not touched AoPS for almost 2.5 years, I don't think you'll be getting a response from him.\n\nJust do as the rule says. Any AP stuff goes in Advanced Physics." } { "Tag": [ "geometry", "analytic geometry", "geometry proposed" ], "Problem": "Let $ ABC$ be an isosceles triangle with $ AB \\equal{} AC$, and let $ D$ be the midpoint of $ BC$. Let $ E$ be the foot of the perpendicular from $ D$ to $ AB$ and $ F$ the midpoint of $ DE$. Prove that $ AF$ is perpendicular to $ CE$.\r\n\r\nI tried coordinate bashing this and it got really ugly. Hints are appreciated.", "Solution_1": "[hide]Let M be the midpoint of EB. As $ MF \\parallel{} BC \\Rightarrow MF\\perp AD$. So F is orthocentre in $ \\Delta DMA$ (as $ DE\\perp AM$). Then $ AF \\perp DM$. But $ DM\\parallel{}EC \\Rightarrow AF\\perp EC$.[/hide]", "Solution_2": "This used to be my favorite geometry problem, I have seen the solution delta posted and I think it is the best. Here is another solution:\r\nLet $ G$ be the midpoint of $ DC$. Note that $ \\triangle AED \\sim \\triangle ADC$, and thus $ \\triangle AEF \\sim \\triangle ADG$. Thus, since $ \\angle DAG\\equal{}\\angle EAF$, $ \\angle FAG\\equal{}\\angle DAE$. Also, we have $ \\frac {AG}{AF} \\equal{} \\frac {AD}{AE}$ from the similarity, so $ \\triangle AFG \\sim \\triangle AED$ since they have a same angle and the corresponding sides are in the same ratio. Thus, $ \\angle AFG \\equal{} 90$, but $ FG \\parallel EC$, so $ AF \\perp CE$.", "Solution_3": "lisi argotera", "Solution_4": "For a generalisation you may see [url]https://artofproblemsolving.com/community/q1h202893p1119923[/url] , post #6.\n\nBest regards,\nsunken rock" } { "Tag": [ "calculus", "integration", "geometry", "function", "limit", "algebra", "domain" ], "Problem": "\\[ \\int_{\\minus{}\\frac{\\pi}{2}}^{\\frac{\\pi}{2}} csc(x) dx\\]\r\n\r\nDoes it converge? I fell it is symmetry. it should be zero , if integral is the area of the function including.\r\n\r\nthanks", "Solution_1": "Ah. Every time I teach improper integrals in Calculus II, someone in the class wants to argue this point with me.\r\n\r\nWith every improper integral, I ask the question, \"for how many different reasons is this improper?\" Or perhaps the question should be asked, at how many different places is this improper. The integral you give is improper for two reasons: the right hand side of $ 0$ and the left hand side of zero.\r\n\r\nIf an integral is improper for more than one reason, we [i]must[/i] cut it into pieces, where each piece is only improper for one reason. Then, the only way we can accept the integral as convergent is if each piece converges independently and separately.\r\n\r\nIn the particular case, $ \\int_{\\minus{}\\frac{\\pi}2}^0\\csc(x)\\,dx\\equal{}\\minus{}\\infty$ and $ \\int_0^{\\frac{\\pi}2}\\csc(x)\\,dx\\equal{}\\infty.$\r\n\r\nSince they both diverge, we must say that this improper integral diverges. We don't get to combine them. You use the (incomplete) phrase, \"the integral is the area of the function including ...\" Well, in this case, that notion of area is undefined.\r\n\r\nBut you want to combine them. Students always do that - that's why it's an argument.\r\n\r\nIt is true that $ \\lim_{a\\to 0^\\plus{}}\\left(\\int_{\\minus{}\\frac{\\pi}2}^{\\minus{}a}\\csc(x)\\,dx\\plus{}\\int_a^{\\frac{\\pi}2}\\csc(x)\\,dx\\right)\\equal{}0.$\r\n\r\nThere's your symmetry. But what I just did isn't the improper integral. It's something called a \"principal value\". Any attempt to balance the bad parts of an integral improper for two reasons by making the domain converge \"at the same rate\" is called a principal value. And to show how dependent this is on the balance, let me give you some other possibilities:\r\n\r\n$ \\lim_{a\\to 0^\\plus{}}\\left(\\int_{\\minus{}\\frac{\\pi}2}^{\\minus{}a}\\csc(x)\\,dx\\plus{}\\int_{a^2}^{\\frac{\\pi}2}\\csc(x)\\,dx\\right)\\equal{}\\plus{}\\infty.$\r\n\r\n$ \\lim_{a\\to 0^\\plus{}}\\left(\\int_{\\minus{}\\frac{\\pi}2}^{\\minus{}a^2}\\csc(x)\\,dx\\plus{}\\int_a^{\\frac{\\pi}2}\\csc(x)\\,dx\\right)\\equal{}\\minus{}\\infty.$\r\n\r\n$ \\lim_{a\\to 0^\\plus{}}\\left(\\int_{\\minus{}\\frac{\\pi}2}^{\\minus{}a}\\csc(x)\\,dx\\plus{}\\int_{2a}^{\\frac{\\pi}2}\\csc(x)\\,dx\\right)\\equal{}\\minus{}\\ln 2.$\r\n\r\nAnd so on.\r\n\r\nLet me ask you this: do you think that $ \\int_{\\minus{}\\infty}^{\\infty}x^3\\,dx$ ought to be zero?\r\n\r\nOn the other hand, $ \\int_{\\minus{}\\frac{\\pi}2}^{\\frac{\\pi}2}\\sqrt[3]{\\csc x}\\,dx$ is a convergent integral, and does equal zero.", "Solution_2": "[quote=\"Kent Merryfield\"]Ah. Every time I teach improper integrals in Calculus II, someone in the class wants to argue this point with me.\n\nWith every improper integral, I ask the question, \"for how many different reasons is this improper?\" Or perhaps the question should be asked, at how many different places is this improper. The integral you give is improper for two reasons: the right hand side of $ 0$ and the left hand side of zero.\n\nIf an integral is improper for more than one reason, we [i]must[/i] cut it into pieces, where each piece is only improper for one reason. Then, the only way we can accept the integral as convergent is if each piece converges independently and separately.\n\nIn the particular case, $ \\int_{ \\minus{} \\frac {\\pi}2}^0\\csc(x)\\,dx \\equal{} \\minus{} \\infty$ and $ \\int_0^{\\frac {\\pi}2}\\csc(x)\\,dx \\equal{} \\infty.$\n\nSince they both diverge, we must say that this improper integral diverges. We don't get to combine them. You use the (incomplete) phrase, \"the integral is the area of the function including ...\" Well, in this case, that notion of area is undefined.\n\nBut you want to combine them. Students always do that - that's why it's an argument.\n\nIt is true that $ \\lim_{a\\to 0^ \\plus{} }\\left(\\int_{ \\minus{} \\frac {\\pi}2}^{ \\minus{} a}\\csc(x)\\,dx \\plus{} \\int_a^{\\frac {\\pi}2}\\csc(x)\\,dx\\right) \\equal{} 0.$\n\nThere's your symmetry. But what I just did isn't the improper integral. It's something called a \"principal value\". Any attempt to balance the bad parts of an integral improper for two reasons by making the domain converge \"at the same rate\" is called a principal value. And to show how dependent this is on the balance, let me give you some other possibilities:\n\n$ \\lim_{a\\to 0^ \\plus{} }\\left(\\int_{ \\minus{} \\frac {\\pi}2}^{ \\minus{} a}\\csc(x)\\,dx \\plus{} \\int_{a^2}^{\\frac {\\pi}2}\\csc(x)\\,dx\\right) \\equal{} \\plus{} \\infty.$\n\n$ \\lim_{a\\to 0^ \\plus{} }\\left(\\int_{ \\minus{} \\frac {\\pi}2}^{ \\minus{} a^2}\\csc(x)\\,dx \\plus{} \\int_a^{\\frac {\\pi}2}\\csc(x)\\,dx\\right) \\equal{} \\minus{} \\infty.$\n\n$ \\lim_{a\\to 0^ \\plus{} }\\left(\\int_{ \\minus{} \\frac {\\pi}2}^{ \\minus{} a}\\csc(x)\\,dx \\plus{} \\int_{2a}^{\\frac {\\pi}2}\\csc(x)\\,dx\\right) \\equal{} \\minus{} \\ln 2.$\n\nAnd so on.\n\nLet me ask you this: do you think that $ \\int_{ \\minus{} \\infty}^{\\infty}x^3\\,dx$ ought to be zero?\n\nOn the other hand, $ \\int_{ \\minus{} \\frac {\\pi}2}^{\\frac {\\pi}2}\\sqrt [3]{\\csc x}\\,dx$ is a convergent integral, and does equal zero.[/quote]\r\n\r\nThanks, it is very detailed." } { "Tag": [ "integration" ], "Problem": "Intergrala din x^3 * sqrt (x^2 + 4)", "Solution_1": "Voi calcula doar $J=\\int x\\sqrt{x^{2}+4}dx$\r\n$J=\\int x\\frac{x^{2}+4}{\\sqrt{x^{2}+4}}dx$\r\n$J=\\int\\frac{x^{3}}{\\sqrt{x^{2}+4}}dx+4\\int\\frac{x}{x^{2}+4}dx$\r\n$J=\\int x^{2}(\\sqrt{x^{2}+4})'dx+4\\sqrt{x^{2}+4}$\r\n$J=x^{2}\\sqrt{x^{2}+4}+4\\sqrt{x^{2}+4}-2\\int x\\sqrt{x^{2}+4}dx$\r\n$3J=x^{2}\\sqrt{x^{2}+4}+4\\sqrt{x^{2}+4}$\r\nDe aici concluzia rezulta rapid. Pt. varianta ta, cu $x^{3}$, faci analog, si vei ajunge la J", "Solution_2": "$\\boxed{\\ I\\equiv\\int x^{3}\\sqrt{4+x^{2}}\\ dx\\ }=\\frac{1}{2}\\cdot\\int x^{2}\\sqrt{4+x^{2}}\\cdot (x^{2})'dx$ $\\Longrightarrow$ $\\boxed{\\ I=\\frac{1}{2}\\cdot F(x^{2})+C\\ }$,\r\nunde $F\\equiv\\int t\\sqrt{4+t}\\ dt=\\int [(t+4)-4]\\sqrt{t+4}\\ dt=\\int \\left[(t+4)^{\\frac{3}{2}}-(t+4)^{\\frac{1}{2}}\\right]\\ dt=$\r\n$\\frac{2}{5}(t+4)^{2}\\sqrt{t+4}-\\frac{2}{3}(t+4)\\sqrt{t+4}+C$ $\\Longrightarrow$ $\\boxed{\\ F=\\frac{2}{15}(3t+7)(t+4)\\sqrt{t+4}+C\\ }\\ .$\r\nIn concluzie, $\\boxed{\\ I=\\frac{1}{15}(3x^{2}+7)(x^{2}+4)\\sqrt{x^{2}+4}+C\\ }\\ .$" } { "Tag": [ "calculus", "derivative", "calculus computations" ], "Problem": "use implicit differentiation to find an equation of the tangent line\r\n\r\n(x^2 + y^2)^3=8(r^2)(y^2) at (1,-1)", "Solution_1": "$r^{2}$ ? should that be an $x^{2}$?", "Solution_2": "it's r^2 and y^2" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Prove that if x,y are real numbers therefore:\r\n\r\nx^4 + y^4 + (x^2 +1)(y^2 +1) >= (y+1)*x^3 + (x+1)*y^3 + x +y\r\n\r\n :ninja:", "Solution_1": "first let's prove a lemma:\r\n\r\n[b]lemma:for real numbers $ x,y$ we have:\n\n$ (|y|\\plus{}1)|x|^3\\plus{}(|x|\\plus{}1)|y|^3\\plus{}|x|\\plus{}|y|\\geq (y\\plus{}1)x^3\\plus{}(x\\plus{}1)y^3\\plus{}x\\plus{}y$[/b]\r\n\r\n[hide=\"proof of the lemma\"]\nif $ x\\equal{}0$ or $ y\\equal{}0$ then it's easy to see,now we have two particular cases:\n\n$ \\boxed{\\textrm{Case 1}: x<0\\textrm{ and }y>0}$\n\nin this case we have $ |x|\\equal{}\\minus{}x,|y|\\equal{}y$ so our lemma becomes:\n\n$ \\minus{}(y\\plus{}1)x^3\\plus{}(1\\minus{}x)y^3\\minus{}x\\plus{}y\\geq (y\\plus{}1)x^3\\plus{}(x\\plus{}1)y^3\\plus{}x\\plus{}y$\n\n$ \\iff 0\\geq 2(y\\plus{}1)x^3\\plus{}2xy^3\\plus{}2x$\n\nwhich is true because $ x<0,y>0$\n\n--------------------\n\n$ \\boxed{\\textrm{Case 2}: x<0\\textrm{ and }y<0}$\n\nin this case we have $ |x|\\equal{}\\minus{}x,|y|\\equal{}\\minus{}y$ and our lemma is equivalent to:\n\n$ \\minus{}(1\\minus{}y)x^3\\minus{}(1\\minus{}x)y^3\\minus{}x\\minus{}y\\geq (y\\plus{}1)x^3\\plus{}(x\\plus{}1)y^3\\plus{}x\\plus{}y$\n\n$ \\iff 0\\geq 2x^3\\plus{}2y^3\\plus{}2x\\plus{}2y$\n\nwhich is true because $ x<0,y<0$\n----------------------------------------------------------[/hide]\r\n\r\nnow according to this lemma its sufficinet to show that:\r\n\r\n$ x^4\\plus{}y^4\\plus{}(x^2\\plus{}1)(y^2\\plus{}1)\\geq(|y|\\plus{}1)|x|^3\\plus{}(|x|\\plus{}1)|y|^3\\plus{}|x|\\plus{}|y|$\r\n\r\nnow we know that $ x^4\\equal{}(\\minus{}x)^4,y^4\\equal{}(\\minus{}y)^4,x^2\\equal{}(\\minus{}x)^2,y^2\\equal{}(\\minus{}y)^2$ so its sufficient to prove this problem for $ x>0,y>0$...\r\n\r\nnow by AM-GM we have:\r\n\r\n$ x^4\\plus{}x^2y^2\\geq 2x^3y$\r\n\r\n$ y^4\\plus{}x^2y^2\\geq 2y^3x$\r\n\r\n$ x^4\\plus{}x^2\\geq 2x^3$\r\n\r\n$ y^4\\plus{}y^2\\geq 2y^3$\r\n\r\n$ x^2\\plus{}1\\geq 2x$\r\n\r\n$ y^2\\plus{}1\\geq 2y$\r\n\r\nnow sum up all these inequalities and devide bothe sides by $ 2$ to derive at your problem...", "Solution_2": "or you can use the fact that:\r\n\r\n$ x^4 \\plus{} y^4 \\plus{} (x^2 \\plus{} 1)(y^2 \\plus{} 1)\\geq (y \\plus{} 1)x^3 \\plus{} (x \\plus{} 1)y^3 \\plus{} x \\plus{} y$\r\n\r\n$ \\iff (x^2 \\minus{} x)^2 \\plus{} (y^2 \\minus{} y)^2 \\plus{} (x^2 \\minus{} xy)^2 \\plus{} (y^2 \\minus{} xy)^2 \\plus{} (x^2 \\minus{} 1)^2 \\plus{} (y^2 \\minus{} 1)^2 \\geq 0$\r\n\r\nwhich is obviously true..." } { "Tag": [ "trigonometry", "number theory proposed", "number theory" ], "Problem": "Prove that the equation $4x^3-3x+1=2y^2$ has at least $31$ solutions in positive integers $x$ and $y$ with $x\\leq 2005$.", "Solution_1": "suppose that $x = 2k+1$. ($k \\leq 1002$)\r\n\r\n$(x+1)(2x-1)^2 = 2y^2 \\Longrightarrow (k+1)(4k+1)^2 = y^2$.\r\n\r\nso, if $(k+1)$ is a square, our equation has a solution. now, we easily see that we \r\nhave 31 perfect squares $\\leq 1002$", "Solution_2": "This was given at an Australian training in 2004. As well as the straightforward solution posted by e.lopes, there's a weird one which uses the fact that substituting $x=\\cos \\alpha$ and $y=\\cos\\beta$ into the equation yields $\\cos 3\\alpha=\\cos 2\\beta$. I've just tried to reconstruct it but I can't remember exactly how it goes.", "Solution_3": "I hope this is the solution Agrippina had in mind.\n\nNotice that $\\cos (ix)=\\frac{e^x+e^{-x}}{2}$.\n\nThus from the previous post, it is sufficient to prove that there are at least $31$ values of $x$ such that $\\frac{e^{2x}+e^{-2x}}{2}\\le 2005$ and $b=\\frac{e^{3x}+e^{-3x}}{2}$ and $a=\\frac{e^{2x}+e^{-2x}}{2}$ are both integers.\nLet $c=\\frac{e^x+e^{-x}}{2}\\in \\mathbb Z$. Thus, $a=2c^2-1\\in \\mathbb Z$ and $b=4c^3-3c\\in \\mathbb Z$.\nNotice that $1\\le c\\le 31 \\le \\sqrt{1003}$. Thus, there are at least $31$ such values of $x$, as desired.", "Solution_4": "oh yes!the key step is to notice $LHS=(x+1)(2x-1)^2$.\nbesides,what's 'Donova Olympiad'?" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let $ a,b,c,d > 0$ satisfy $ abc \\plus{} bcd \\plus{} cda \\plus{} dab \\equal{} a \\plus{} b \\plus{} c \\plus{} d$. Prove that :\r\n$ (\\sqrt {a^2 \\plus{} 1} \\plus{} \\sqrt {b^2\\plus{} 1})^2 \\plus{} (\\sqrt {c^2 \\plus{} 1} \\plus{} \\sqrt {d^2 \\plus{} 1})^2\\leq (a \\plus{} b \\plus{} c \\plus{} d)^2$", "Solution_1": "[quote=\"conan_naruto236\"]Let $ a,b,c,d > 0$ satisfy $ abc \\plus{} bcd \\plus{} cda \\plus{} dab \\equal{} a \\plus{} b \\plus{} c \\plus{} d$. Prove that :\n$ (\\sqrt {a^2 \\plus{} 1} \\plus{} \\sqrt {b^2 \\plus{} 1})^2 \\plus{} (\\sqrt {c^2 \\plus{} 1} \\plus{} \\sqrt {d^2 \\plus{} 1})^2\\leq (a \\plus{} b \\plus{} c \\plus{} d)^2$[/quote]\r\nNobody... :(" } { "Tag": [ "LaTeX", "algebra unsolved", "algebra" ], "Problem": "hi. can you help me ? solve this equation:\r\n\r\n (3+sqrt(3))^x+(3-sqrt(3))^x=(3*sqrt(2)+2*sqrt(3))^x+(3*sqrt(2)-2*sqrt(3))^x", "Solution_1": "[quote=\"Svejk\"]hi. can you help me ? solve this equation:\n\n (3+sqrt(3))^x+(3-sqrt(3))^x=(3*sqrt(2)+2*sqrt(3))^x+(3*sqrt(2)-2*sqrt(3))^x[/quote]\r\nI think the only solution is x=0:\r\nlet $a=3+\\sqrt{3},b=3\\sqrt{2}+2\\sqrt{3}$ it give:\r\n$3-\\sqrt{3}=\\frac{6}{a},3\\sqrt{2}-2\\sqrt{3}=\\frac{6}{b}$\r\ntherefore:\r\n$a^{x}+\\frac{6^{x}}{a^{x}}=b^{x}+\\frac{6^{x}}{b^{x}}\\Rightarrow (a^{x}-b^{x})(a^{x}b^{x}-6^{x})=0$but ab not equal 6 and a not equal b therefore x=0", "Solution_2": "Thanks :laugh: \r\n\r\nPS:Could you tell me how did you write the symbols ?", "Solution_3": "use latex \r\nsee[url]http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php[/url]" } { "Tag": [ "induction", "number theory unsolved", "number theory" ], "Problem": "a real number call interesting if it can be written to $ \\sqrt{n}$ $ \\plus{}$ $ \\sqrt{n\\plus{}1}$ $ (n \\in N)$.prove that if X be an interesting number, then $ X^r$ is an interesting number for all natural r.", "Solution_1": "[quote=\"behdad.math.math\"]a real number call interesting if it can be written to $ \\sqrt {n}$ $ \\plus{}$ $ \\sqrt {n \\plus{} 1}$ $ (n \\in N)$.prove that if X be an interesting number, then $ X^r$ is an interesting number for all natural r.[/quote]\r\n\r\nIf $ x\\equal{}\\sqrt n\\plus{}\\sqrt{n\\plus{}1}$, then $ n\\equal{}\\left(\\frac{x^2\\minus{}1}{2x}\\right)^2$. So the question is to show that $ M\\equal{}\\left(\\frac{x^{2p}\\minus{}1}{2x^p}\\right)^2$ is an integer too.\r\n\r\n$ M\\equal{}\\left(\\frac{x^{2p}\\minus{}1}{2x^p}\\right)^2$ $ \\equal{}\\left(\\frac 12(x^p\\minus{}\\frac 1{x^p})\\right)^2$ $ \\equal{}\\left(\\frac 12((\\sqrt{n\\plus{}1}\\plus{}\\sqrt n)^p\\minus{}(\\sqrt{n\\plus{}1}\\minus{}\\sqrt n)^p)\\right)^2$ and so :\r\n\r\n$ M\\equal{}\\left(\\sum_{\\text{odd } k\\in[0,p]}{p \\choose k} (\\sqrt{n\\plus{}1})^{p\\minus{}k}(\\sqrt n)^k\\right)^2$\r\n\r\nThe quantity inside the parenthesis is either $ A\\sqrt n$ if $ p$ is odd, either $ A\\sqrt n\\sqrt{n\\plus{}1}$ if $ p$ even, for some integer $ A$.\r\n\r\nAnd so $ M$ is either $ A^2n$, either $ A^2n(n\\plus{}1)$ and so is an integer.\r\n\r\nQ.E.D.", "Solution_2": "Thanks dear Pco.\r\nIs it possible to solve it using induction,too?", "Solution_3": "if $ A \\equal{} \\sqrt {n \\plus{} 1} \\plus{} \\sqrt {n}$ and $ B \\equal{} \\sqrt {n \\plus{} 1} \\minus{} \\sqrt {n}$\r\n\r\nand if a sequence $ a_r$ is given by $ 4a_r \\plus{} 2 \\equal{} A^{2r} \\plus{} B^{2r}$\r\n\r\nuse the fact that $ AB \\equal{} 1$ so $ A^{2r} B^{2r} \\equal{} 1$\r\n\r\nso we can solve to get $ A^r \\equal{} \\sqrt {a_r \\plus{} 1} \\plus{} \\sqrt {a_r}$\r\n\r\n[ $ X \\equal{} A^{2r}$ and $ Y \\equal{} B^{2r}$ then $ X \\plus{} Y \\equal{} 4a_r \\plus{} 2$ and $ XY \\equal{} 1$\r\nthen solve to get $ X \\equal{} 2a_r \\plus{} 1 \\plus{} \\sqrt { (2a_r \\plus{} 1)^2 \\minus{} 1}$ and $ Y \\equal{} 2a_r \\plus{} 1 \\minus{} \\sqrt { (2a_r \\plus{} 1)^2 \\minus{} 1}$\r\nso $ \\sqrt {X} \\equal{} \\sqrt {a_r \\plus{} 1} \\plus{} \\sqrt {a_r}$ ]\r\n\r\nalso $ A^2$ and $ B^2$ are roots of $ x^2 \\minus{} 2(2n \\plus{} 1)x \\plus{} 1 \\equal{} 0$ [ because AB = 1 and $ A^2 \\plus{} B^2 \\equal{} 2(2n \\plus{} 1)$] \r\n\r\nso a_r satisfies a recursion relation which we can check by induction\r\n$ 4a_r \\plus{} 2 \\minus{} 2(2n\\plus{}1)(4a_{r\\minus{}1} \\plus{} 2) \\plus{} (4a_{r\\minus{}2} \\plus{} 2) \\equal{} 0$\r\nwhich simplifies to $ a_r \\equal{} 2(2n \\plus{} 1)a_{r \\minus{} 1} \\minus{} a_{r \\minus{} 2} \\plus{} 2n$ with $ a_0 \\equal{} 0$ and $ a_1 \\equal{} n$\r\nso a_r is always an integer" } { "Tag": [ "inequalities", "LaTeX" ], "Problem": "[color=green]\nprove that\n\nif a,b,c are positive reals with sum 1\n\n((a^3)/(a^2+b^2)) + ((b^3)/(b^2+c^2)) + ((c^3)/(c^2+a^2)) \n\nis >= 1/2\n\n[/color]", "Solution_1": "[hide=\"hint\"]substitute a with 1-b-c[/hide]", "Solution_2": "Please use $\\LaTeX$.\r\n\r\nIf $a,b,c \\in \\mathbb{R}$ such that $a + b + c = 1$.\r\n\r\n$\\frac{a^3}{a^2+b^2} + \\frac{b^3}{b^2+c^2} + \\frac{c^3}{c^2+a^2} \\geq 1/2$.", "Solution_3": "[hide=\"(Ugly) solution\"]\nLet us write it in a more promising manner,\n\n$\\sum_{cyc} \\frac{2a^3}{a^2 + b^2} \\ge \\sum_{cyc} a = \\sum_{cyc} \\frac{a^3 + ab^2}{a^2 + b^2}$\n$\\sum_{cyc} a \\frac{a^2 - b^2}{a^2 + b^2} \\ge 0$\n$\\sum_{cyc} a \\frac{ ( \\frac{a}{b} )^2 - 1}{ ( \\frac{a}{b} )^2 + 1} \\ge 0$\n\nJensen's on $f(x) = \\frac{x^2 - 1}{x^2 + 1}$ yields\n\n$\\sum_{cyc} a \\frac{ ( \\frac{a}{b} )^2 - 1}{ ( \\frac{a}{b} )^2 + 1} \\ge \\frac{ u^2 - 1}{u^2 + 1}$\n\nWhere\n\n$u = \\frac{a^2}{b} + \\frac{b^2}{c} + \\frac{c^2}{a}$\n\nBy Rearrangement,\n\n$u \\ge a + b + c = 1$,\n\nCompleting the proof. QED.\n[/hide]", "Solution_4": "Nice Solution:\r\n\r\n Write a^3/(a^2 + b^2) = a - (a.b^2)/(a^2 + b^2) and similar expressions for the other two terms on the left side .\r\n\r\nThe inequality to be proved now takes the form\r\n\r\na+b+c - [ (a.b^2)/(a^2+b^2) + (b.c^2)/(b^2 +c^2) + (c.a^2)/(c^2 +a^2) ] >= 1/2, or equivalently\r\n\r\n[ (a.b^2)/(a^2 +b^2) + (b.c^2)/(b^2 +c^2) + (c.a^2)/(c^2 +a^2) ] <= 1/2 which is true because a^2 +b^2 >=2ab,..... ...and a+b+c =1\r\n\r\nSorry this isn't in Latex.This was not typed on my PC.", "Solution_5": "[quote=\"Centy\"]\n\nIf $a,b,c \\in \\mathbb{R^+}$ such that $a + b + c = 1$.\n\n$\\frac{a^3}{a^2+b^2} + \\frac{b^3}{b^2+c^2} + \\frac{c^3}{c^2+a^2} \\geq 1/2$.[/quote]\r\n\r\nUse:\r\n$\\frac{a^3}{a^2+b^2} \\geq a-\\frac{b}{2}$ (It is equivalent to $\\frac12 b (a-b)^2 \\geq 0$\r\nSum up all term and use $a+b+c=1$" } { "Tag": [ "function" ], "Problem": "What exactly are they and where can they applied in mathematics?\r\n\r\nThanks in advance", "Solution_1": "[url]http://en.wikipedia.org/wiki/Logistic_function[/url]\r\n\r\nHowever, I'm too lazy and intimidated to actually read the article and summarize it, so someone else would have to do that.", "Solution_2": "A logistic function models situations where a quantity (y) grows (or decreases) at a rate jointly proportional to its size (y), and proportional to the difference of its size from some fixed value (L-y).\r\n\r\nIn other words, the rate of change of y equals $ k \\cdot y(L\\minus{}y)$ for some constant of proportionality k.\r\n\r\nFor example, some simple diseases grow logistically in closed populations. They grow proportional to the number of people who have the disease because the number of carriers (y) directly affects the rate at which it spreads. But, if there are only L people in the population, then it also spreads proportional to (L-y), the number of people in the population who are not yet infected.\r\n\r\nSome other examples of things that grow logistically are species in closed ecosystems (L is called the carrying capacity of the ecosystem), and rumors in closed populations.\r\n\r\nA neat way to test logistic growth is to take a room full of people and give everyone in the room a card. Put red X's on a few of the people's cards and have them randomly mingle around, talking to different people for 2 minutes. Every time someone with a red X talks to someone without a red X, they put a red X on the other person's card. If the person they talk to already has a red X, they make no change. Every 6 minutes or so have everyone close their eyes and take a poll of how many people have red X's. The number of people with red X's as a function of time will be roughly logistic.", "Solution_3": "Another such example is Newton's rate of cooling, where the rate that an object cools is proportional to the difference in its temperature and the temperature of its surroudings.\r\nThat is\r\n$ \\dfrac{dT}{dt} \\equal{} k(T_0 \\minus{} T)$ where $ T_0$ is the temp of the room", "Solution_4": "That is not logistic, because the rate at which the temperature changes is only proportional to the difference between the temperature and the surrounding atmosphere - not jointly proportional as is required of a logistic function.\r\n\r\nTemperature cooling/heating is a stricly exponential (decay) function with respect to the ambient temperature." } { "Tag": [ "search", "Support", "\\/closed" ], "Problem": "This is going to be a very vague question because I do not know many specifics.\r\n\r\nWhere do I learn programming (CS I think)? I need to learn it for free; preferably online. Furthermore, I don't know where to ask this.\r\n\r\nUm... any answers will be appreciated.", "Solution_1": "This might better belong in the computers forum.\r\n\r\nWhen you say CS, I think you mean CSS. A good site for learning HTML, CSS, etc. is http://www.w3schools.com . You can also get various books from the library. The SAMS Teach Yourself books are pretty good, as are the \"for the absolute beginner\".", "Solution_2": "[quote=\"isabella2296\"]When you say CS, I think you mean CSS.[quote]\n\nProbably Computer Science, not Cascading Style Sheets.\n\n[quote=\"isabella2296\"]The SAMS Teach Yourself books are pretty good, as are the \"for the absolute beginner\".[/quote][/quote][/quote]\r\n\r\nI highly doubt it. Anyway, you haven't really given us any information to work on. What purpose do you expect your programming to serve (i.e. what do you expect to get out of it)?", "Solution_3": "Well, I don't know much about it except that you can use it to solve math problems. Basically, I would prefer my entire focus on learning programming to be related to math because I already much about computers and I don't to spend more time learning about them.", "Solution_4": "You can use almost any computer language to perform mathematical calculations. A general answer to a general question. :lol: \r\nThen there is the question of efficiency, and you run into easy and inefficient languages like basic compared to C and C++ which might be a bit harder to learn (but much more efficient and robust).\r\n(by efficiency, I mean how they make use of computer resources).\r\n\r\nIt also depends on what level of mathematics you will be computing (ie, how complex it is). The more complex it is, the better it will be for you to use a more efficient language.\r\nI suggest you search online for computer languages such as c, c++, c#, java, basic, etc. Then you weigh them yourself wrt what you want to accomplish, the amount of time you have, the complexity of the language (the learning curve), etc.\r\nOn a side note,You could also use math cad to perform you're mathematical calculations. But it is geared more towards engineering, which is largely the reason why I mentioned it. :D \r\nHope that answers atleast part of your question. :)", "Solution_5": "I suggest you learn MATLAB. You can pick up a free copy by winning a local competition.\r\n\r\n[quote=\"Inspired By Nature\"]Then there is the question of efficiency, and you run into easy and inefficient languages like basic compared to C and C++ which might be a bit harder to learn (but much more efficient and robust).\n(by efficiency, I mean how they make use of computer resources).[/quote]\r\n\r\nI would disregard this as the advice of someone who has not touched either BASIC, C, or C++.", "Solution_6": "[quote=\"Differ\"]I suggest you learn MATLAB. You can pick up a free copy by winning a local competition.\n\n[quote=\"Inspired By Nature\"]Then there is the question of efficiency, and you run into easy and inefficient languages like basic compared to C and C++ which might be a bit harder to learn (but much more efficient and robust).\n(by efficiency, I mean how they make use of computer resources).[/quote]\n\nI would disregard this as the advice of someone who has not touched either BASIC, C, or C++.[/quote]\r\n\r\nI agree, matlab is indeed an excellent suggestion. It completely slipped my mind. :) \r\n\r\nAs for the efficiency statement, I stand by it. Higher level programming languages such as BASIC are generally less efficient and consume more resources (more mem consumption, slower exec speed, etc) than low level programming languages such as C and C++. \r\nAs for my level of competence in programming, I do believe I have 'touched' all three of the aforementioned languages amongst others.", "Solution_7": "So... since I need something free, will C++ be useful?", "Solution_8": "[quote=\"pytheagle\"]So... since I need something free, will C++ be useful?[/quote]\r\n\r\nIf you can't win MATLAB, I would recommend Java or C#.", "Solution_9": "[quote=\"Differ\"][quote=\"pytheagle\"]So... since I need something free, will C++ be useful?[/quote]\n\nIf you can't win MATLAB, I would recommend Java or C#.[/quote]\r\n\r\nI'm getting tired of Differing from you Differ, But I believe C and C++ have a much wider base (resources, industry use, support, IDE's, etc) than Java or C#. Interestingly enough, they are all quite similar. \r\nAlso, with all due respect, 'winning MATLAB' is not an immediate or relavent solution to the issue provided by pytheagle. \r\n\r\n[b]@pytheagle[/b]: Yes, there are many different sources you can get a IDE (integrated development environment) from. C++ is an excellent language to start of with. \r\nIDE's are basically a complete package (all-in-one) including editor (for writing code), compiler (converts your code to something the computer can understand), and debugger (for catching your errors).\r\nMy current personal favourite is codeblocks,\r\nhttp://www.codeblocks.org/\r\n\r\nBloodshed is a close second,\r\nhttp://www.bloodshed.net/download.html\r\nBut I do believe they do not update it anymore.\r\n\r\nThere is also Visual C++ express,\r\nhttp://www.microsoft.com/Express/VC/\r\n\r\nAlso, if you just want to view code of different languages, an excellent little program is notepad++,\r\nhttp://notepad-plus.sourceforge.net/uk/site.htm\r\n(pick the binary files if you are interested)\r\n\r\nNote, that the initial learning curve may be quite steep so be patient. There are many tutorials online to help you start off. Here's an excellent site to start you off,\r\nhttp://www.cplusplus.com/doc/tutorial/\r\nBest of luck. :)" } { "Tag": [ "modular arithmetic" ], "Problem": "Show that given any 17 integers, it is possible to choose 5 whose sum is divisible by 5.", "Solution_1": "[quote=\"mihail911\"]Show that given any 17 integers, it is possible to choose 5 whose sum is divisible by 5.[/quote]\r\n\r\nIf there are $ 5$ of the given integers congruent $ \\pmod{5}$ then choose the five integers and we are done.\r\n\r\nIf not, then by pigeonholeprinciple we can find an integer $ k \\in \\{0,1,2,3,4 \\}$ such that there are $ 4$ of the given integers congruent to $ k \\pmod{5}$. And we can definitely find another two integers which are congruent to $ k\\minus{}1$ and $ k\\plus{}1$ respectively (because $ \\frac{17\\minus{}4}{4}>3$). Now choose $ 3$ integers $ \\equiv k \\pmod{5}$ and the two integers $ \\equiv k\\minus{}1,k\\plus{}1 \\pmod{5}$ and we are done." } { "Tag": [ "LaTeX" ], "Problem": "Can anyone plz verfity if these two lines have a common spot (or whatever you call it in english) in (1+(3/16);(31/32))\r\n\r\nl: x=5+4y ^ m: 8y=(1/2)x-4", "Solution_1": "Do you mean that they intersect there?\r\n\r\nIf so, I believe that the solution is $(4,-\\displaystyle\\frac{1}{4})$. If we substitute the value for x into the second equation, we can simplify to find $y=-\\displaystyle\\frac{1}{4}$. Then this value gives x=4 in the first equation.", "Solution_2": "When I solve the equation, I get the same thing as E^(pi*i)=-1 got. DId you do an incorrect operation?", "Solution_3": "Hum. Could you (just to make stupid thing on a wednesday) show me how you would find x if you didn't know y, plz...", "Solution_4": "Multiply the first equation by 2, move the 8y to the left side and the 2x to the right side. Now does it look obvious?", "Solution_5": "Sorry i've been away for some time. Sometimes i just have other things to attend to (a lot of work lately moved my math study to a 2. priority).\r\n\r\nCould you show it using latex, I just don't get it." } { "Tag": [ "geometry", "circumcircle", "incenter", "geometry proposed" ], "Problem": "Given are acute triangle $ABC$ with circumcenter $O,BO\\cap CA={B'},CO\\ca AB={C'}$ prove that $ABC$ is isoceles equivalent $BB'=CC'$ :)", "Solution_1": "Gemath, verify the case $A=90^{\\circ}$ and $B\\ne C\\ !$", "Solution_2": "ok Virgil It needs an acute triangle! I have fixed it ! Sorry :)", "Solution_3": "[color=darkblue][b][u]An easy extension.[/u][/b] Given are a triangle $ABC$ and a point $P\\not\\in BC$ for which $\\widehat{PBC}\\equiv\\widehat{PCB}\\ .$\nDenote the intersections $B'\\in BP\\cap AC$ and $C'\\in CP\\cap AB\\ .$ \nProve that $BB'=CC'$ $\\Longleftrightarrow$ the triangle $ABC$ is isosceles .\n[b]Proof.[/b] Prove easily that $B<90^{\\circ}$ and $C<90^{\\circ}\\ .$ Therefore, \n$BB'=CC'$ $\\Longrightarrow$ $BCC'\\equiv CBB'$ $\\Longrightarrow$ $B=C\\ .$ Reciprocally is obviously.[/color]", "Solution_4": "ok Virgil Nicula that is a good solution and extension! Really I solve it by very long proof - a algebraic menthod, The solution by synthetic geometry is too nice and short but algebraic menthod can answer the question: which do points inside triangle have a property $AA'=BB'\\Rightarrow \\triangle ABC$ is isosceles some known point: $H$(orthorcenter),$I$(incenter Steiner-Lehmus),$N$(Nagel point -a paper in forum geometricorum),$O$(circumcenter),$G$(centroid)..et cetera\r\nI think $L$(Lemoine point) is too, hope you are interested in! :)" } { "Tag": [ "geometry", "geometric transformation", "reflection", "power of a point", "radical axis" ], "Problem": "Here's another relatively easy problem:\r\n\r\nSuppose that circles $S_1$ and $S_2$, with centers $O_1$ and $O_2$ respectively, meet at points $A$ and $B$. Extend $O_2B$ to meet $S_1$ at $C$ and $S_2$ at $F$. Extend $O_1B$ to meet $S_2$ at $D$ and $S_1$ at $E$. Prove that $AB, CE, DF$ are concurrent.", "Solution_1": "Isn't it just symmetric?", "Solution_2": "The circles can be different sizes. If the circles are congruent, then you can just show that they're all reflections over the line AB. But if they aren't...you first have to assume that both of the lines OB touch both circles. This is an interesting problem...i'll look into it a little bit...", "Solution_3": "But (or so I thought when I last posted in this thread) they are reflections across $O_1O_2$...", "Solution_4": "Let $P$ be the intersection of $CE, DF$. Since $BE, BF$ are diameters of the circles $(O_1), (O_2)$, the angles $\\angle BCE = \\angle BDF = 90^o$ are right. Hence, the angles $\\angle PDE = \\angle PCF = 90^o$ are also right. Consequently, the 4 right angle triangles\r\n\r\n$\\triangle BCE \\sim \\triangle PDE \\sim \\triangle PCF \\sim \\triangle BDF$\r\n\r\nare all similar, each pair in this sequence having one acute angle in common. In particular, from similarity of the triangles $ \\triangle PDE \\sim \\triangle PCF$, we get\r\n\r\n$\\frac{PD}{PE} = \\frac{PC}{PF},\\ \\ \\ PD \\cdot PF = PC \\cdot PE$\r\n\r\nThus the power of the point $P$ to the circles $(O_1), (O_2)$ is the same, which means that this point is on the radical axis $AB$ of these 2 circles." } { "Tag": [ "calculus", "integration", "search", "vector", "function", "real analysis", "real analysis unsolved" ], "Problem": "Four flies sit at the corners of a square card table, side a, facing inward. They start simultaneously walking at the same rate, each directing its motion steadily toward the fly on its right. Find\r\n\r\n(a) the equation of the path traced by one of the flies.\r\n(b) the total distance traced by each fly (without calculus).", "Solution_1": "im pretty sure they all walk the logarithmic spiral towards the middle,\r\n\r\n(mostly because i remember seeing a picture of it)", "Solution_2": "They indeed spiral in towards the centre. kenn4000 is right.", "Solution_3": "Someone should answer part (b), which is an old-time classic. I've been holding off because I don't want to spoil the fun.\r\n\r\n[hide=\"Note:\"]\"Without calculus\" might be a slight overstatement as arguments about what happens instantaneously or in infinitesimal increments may play a role. But no explicit integral needs to be computed.[/hide]", "Solution_4": "The last time it was posted was here: http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=48756274&t=214248", "Solution_5": "I've always felt a bit uneasy about the solution to this classical problem. The way its worded in this case especially, \"each directing its motion steadily toward the fly on its right\", seems most naturally to ascribe to each fly an acceleration vector pointing at the next fly, not a velocity vector as is assumed in the classical solutions. If each fly were actually moving (rather than merely 'directing its motion') toward the next fly, it seems the resulting implied acceleration vector would be some function requiring knowledge on part of the fly of not only the position, but also the dynamics, of the other flies. Does anyone else feel the usual solution/formulation of this problem is unnatural, or am I just being pedantic?", "Solution_6": "Would you prefer the following wording?\r\n\r\nFour flies walk on a table. Each travels at a constant speed along a certain smooth path. A passing observer notes that at each instant of their walk, every fly is facing directly towards the next fly in clockwise order around the center of the table." } { "Tag": [ "ratio", "geometry", "number theory", "relatively prime" ], "Problem": "A chord of a circle is perpendicular to a radius at the midpoint of the radius. The ratio of the area of the larger of the two regions into which the chord divides the circle to the smaller can be expressed in the form $\\frac{a\\pi+b\\sqrt{c}}{d\\pi-e\\sqrt{f}}.$, where $a, b, c, d, e,$ and $f$ are positive integers, $a$ and $e$ are relatively prime, and neither $c$ nor $f$ is divisible by the square of any prime. Find the remainder when the product $abcdef$ is divided by $1000$.\r\n\r\n[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=403[/img]", "Solution_1": "[hide=\"Solution\"]\nLet the radius of the circle be $r$. We know that $M$ is the midpoint of the radius $\\overline{AB}$. So $AM=\\frac r2.$ Additionally, $AD=r$. Thus, right triangle $AMD$ has its hypotenuse equal to twice one of its legs and is thus a 30-60-90 right triangle. This gives us $m\\angle MAD=m\\angle MAC=60^\\circ.$ Therefore the smaller sector bounded by arc $CD$ is one third the area of the circle.\n\nThe area of the smaller section formed by the cord is one third the area of the circle minus $[ACD].$ This gives $\\frac 13\\pi r^2-\\frac{r^2\\sqrt{3}}4.$ The area of the large section formed by the chord is $\\pi r^2$ minus the area of the smaller section. This comes out to $\\frac 23 \\pi 4^2+\\frac{r^2\\sqrt{3}}4.$\n\nThe ratio of these two is\n\n\\[ \\frac{\\frac 23\\pi r^2+\\frac{r^2\\sqrt{3}}4}{\\frac 13\\pi r^2-\\frac{r^2\\sqrt{3}}4}=\\frac{8\\pi+3\\sqrt{3}}{4\\pi-3\\sqrt{3}}. \\]\n\nThus, $abcdef=2592$ which gives $\\boxed{592}$ for our answer.[/hide]", "Solution_2": "[hide]Let's call the radius $2$, now if you know that $M$ cuts the radius in half the then it cuts chord $DC$ in half too since $DC$ is perpendicular to $AE$, so now we know that $AM=1$ and $DA=2$ and so we now have $60,30,90$ triangles, and we know that the circular section $ACED$ is $\\frac{1}{3}$ of the circle, so $DCE=\\frac{1}{3}circleA-DAC \\Longrightarrow DCE=\\frac{4}{3}\\pi-\\sqrt{3}$ so then the rest of the circle is $\\frac{8}{3}\\pi+\\sqrt{3}$. Thus the ratio is $\\boxed{\\frac{8\\pi+3\\sqrt{3}}{4\\pi-3\\sqrt{3}}}$[/hide]" } { "Tag": [ "algebra", "polynomial", "geometry", "3D geometry", "quadratics", "search", "calculus" ], "Problem": "How do you factor cubic polynomials such as this:\r\n\r\n $ x^3\\minus{}21x\\plus{}20\\equal{}0$, where grouping doesn't work ?", "Solution_1": "Unless I've mistaken your interpretation by factoring a cubic polynomial:\r\nOne way includes applying the [url=http://www.artofproblemsolving.com/Wiki/index.php/Rational_Root_Theorem]rational[/url] [url=http://en.wikipedia.org/wiki/Rational_root_theorem]roots[/url] theorem and testing the possible answers, which quickly deduces this cubic polynomial into\r\n[hide]\n$ x^3 \\minus{} 21x \\plus{} 20 \\equal{} (x \\minus{} 1)(x^2 \\plus{} x \\minus{} 20) \\equal{} \\boxed{(x \\minus{} 1)(x \\plus{} 5)(x \\minus{} 4)} \\equal{} 0$[/hide]\r\nIt sometimes may be tedious and time consuming (depending on the polynomial) but it's definitely worth to try out depending on the polynomial (it becomes more easier to use when working with lesser degree polynomials like this cube, but it doesn't always work).\r\n\r\n\r\nI have one question, what did you mean by grouping?", "Solution_2": "$ x^3\\minus{}21x\\plus{}20\\equal{}0$\r\n[hide=\"rational root theorem\"]\nUsing [url=http://mathworld.wolfram.com/RationalZeroTheorem.html]rational root theorem[/url] we know to test $ \\pm$ versions of the divisors of $ 20\\equal{}2^25^1$, of which there are $ 12$. After we find one, we can use synthetic division to finish the problem in a more familiar format. Luckily, it didn't take long.\n\n$ f(1)\\equal{}(1)^3\\minus{}21\\plus{}20\\equal{}0$\n\nthen...\n$ \\frac{x^3\\minus{}21x\\plus{}20\\equal{}0}{x\\minus{}1}\\equal{}x^2\\plus{}x\\minus{}20$\n\n$ x^2\\plus{}x\\minus{}20\\equal{}0$\n$ (x\\minus{}4)(x\\plus{}5)\\equal{}0\\Longrightarrow x\\equal{}\\minus{}5,4$\n\nSo now we have the three solutions: $ \\boxed{x\\equal{}\\minus{}5,1,4}$\n[/hide]", "Solution_3": "Well there is another complex method called Cardons method for solving a cubic eqn \r\nRefer http://en.wikipedia.org/wiki/Cubic_equation\r\n :)", "Solution_4": "x^3-21x+20=0\r\nx^3+x^2-21x+20=x^2\r\nx^3+(x-20)*(x-1)=x^2\r\nX^3-x^2+(x-20)*(x-1)=0\r\nX^2(x-1)+(x-20)*(x-1)=0\r\n(x^2+x-20)*(x-1)=0\r\n(x-4)*(x+5)*(x-1)=0", "Solution_5": "[quote=\"erdogankerem123\"]How do you factor cubic polynomials such as this:\n\n $ x^3 \\minus{} 21x \\plus{} 20 \\equal{} 0$, where grouping doesn't work ?[/quote]\r\nA really useful thing is that, if the coefficients add to 0, x=1 is a solution, and then it becaomes a quadratic.", "Solution_6": "[quote=\"xpmath\"][quote=\"erdogankerem123\"]How do you factor cubic polynomials such as this:\n\n $ x^3 \\minus{} 21x \\plus{} 20 \\equal{} 0$, where grouping doesn't work ?[/quote]\nA really useful thing is that, if the coefficients add to 0, x=1 is a solution, and then it becaomes a quadratic.[/quote]\r\n\r\nDoes this hold for all polynomials?", "Solution_7": "[quote=\"igiul\"][quote=\"xpmath\"][quote=\"erdogankerem123\"]How do you factor cubic polynomials such as this:\n\n $ x^3 \\minus{} 21x \\plus{} 20 \\equal{} 0$, where grouping doesn't work ?[/quote]\nA really useful thing is that, if the coefficients add to 0, x=1 is a solution, and then it becaomes a quadratic.[/quote]\n\nDoes this hold for all polynomials?[/quote]\r\n\r\nThe bit about it become quadratic isn't, but $ x\\equal{}1$ being a solution for polynomials who's coefficients sum to 0 is.", "Solution_8": "Yeah, I meant that in this problem, after factoring out x-1, it becomes a quadratic.", "Solution_9": "Usually you should try the rational root theorem.. but to narrow the search a little, you could use Descartes' rule of signs.", "Solution_10": "But Descartes rule of signs can only give you the interval of the roots and if there demand a situation to use it then the roots cant be found easily .\r\n :wink:", "Solution_11": "Well, most of the time you won't [i]need[/i] to use it, but when you have a lot of possible rational roots it might help a little.", "Solution_12": "Thanks everyone for the replies! They help a lot.\r\n\r\n[quote=\"Patterns_34\"]\nI have one question, what did you mean by grouping?[/quote]\r\n\r\nAccording to my Pre-Cal and Alg II teachers, grouping is when you take a cubic polynomial such as $ x^3\\plus{}x^2\\minus{}4x\\minus{}4$ and you \"group\" the terms that are easily seen to have a common factor, in this case $ x^3$ and $ x^2$, and $ \\minus{}4x$ and $ \\minus{}4$, then factor out the common terms so that it becomes $ x^2(x\\plus{}1)\\minus{}4(x\\plus{}1)$, which can obviously be factored further.", "Solution_13": "Who ever gave you the idea that grouping doesn't work here?\r\n\r\n$ (x^3 \\minus{} x) \\minus{} (20x \\minus{} 20) \\equal{} 0 \\Leftrightarrow$\r\n$ (x^2 \\plus{} x)(x \\minus{} 1) \\minus{} 20(x \\minus{} 1) \\equal{} 0 \\Leftrightarrow$\r\n$ (x \\minus{} 1)(x^2 \\plus{} x \\minus{} 20) \\equal{} 0 \\Leftrightarrow$\r\n$ (x \\minus{} 1)(x \\plus{} 5)(x \\minus{} 4) \\equal{} 0$\r\n\r\nJust because it's harder to see doesn't mean it's not there. :)", "Solution_14": "[quote=\"t0rajir0u\"]Who ever gave you the idea that grouping doesn't work here?\n\n$ (x^3 \\minus{} x) \\minus{} (20x \\minus{} 20) \\equal{} 0 \\Leftrightarrow$\n$ (x^2 \\plus{} x)(x \\minus{} 1) \\minus{} 20(x \\minus{} 1) \\equal{} 0 \\Leftrightarrow$\n$ (x \\minus{} 1)(x^2 \\plus{} x \\minus{} 20) \\equal{} 0 \\Leftrightarrow$\n$ (x \\minus{} 1)(x \\plus{} 5)(x \\minus{} 4) \\equal{} 0$\n\nJust because it's harder to see doesn't mean it's not there. :)[/quote]\r\nahh... didn't see that, thanks!\r\n\r\nso, can all polynomials be factored by grouping if rearranged?", "Solution_15": "[quote=\"xpmath\"][quote=\"erdogankerem123\"]How do you factor cubic polynomials such as this:\n\n $ x^3 \\minus{} 21x \\plus{} 20 \\equal{} 0$, where grouping doesn't work ?[/quote]\nA really useful thing is that, if the coefficients add to 0, x=1 is a solution, and then it becaomes a quadratic.[/quote]\r\n\r\nwhat exactly is the proof of that?\r\n\r\n@erdogankerem i think you can factor all polynomials by grouping only if it has rational (maybe even integral) roots. im not exactly sure how you would factor a polynomial if it only has irrational and/or complex roots. although if, in the course of your factoring, you get $ x^2\\plus{}1$ as a factor, you know that $ \\pm i$ are two roots.", "Solution_16": "[quote=\"mihail911\"][quote=\"xpmath\"][quote=\"erdogankerem123\"]How do you factor cubic polynomials such as this:\n\n $ x^3 \\minus{} 21x \\plus{} 20 \\equal{} 0$, where grouping doesn't work ?[/quote]\nA really useful thing is that, if the coefficients add to 0, x=1 is a solution, and then it becaomes a quadratic.[/quote]\n\nwhat exactly is the proof of that?[/quote]\n\nThe sum of the coefficients of a polynomial $ P(x)$ is $ P(1)$. If $ P(1) \\equal{} 0$, then $ 1$ is a root by definition.\n\n[quote=\"mihail911\"]@erdogankerem i think you can factor all polynomials by grouping only if it has rational (maybe even integral) roots. im not exactly sure how you would factor a polynomial if it only has irrational and/or complex roots. although if, in the course of your factoring, you get $ x^2 \\plus{} 1$ as a factor, you know that $ \\pm i$ are two roots.[/quote]\r\n\r\nMy previous post was somewhat misleading. Here is the issue:\r\n\r\nGrouping is a [i]strategy[/i] for making the factorization of certain kinds of polynomials more obvious. [b]Any[/b] polynomial can be \"factored by grouping\" [i]if you already know the roots in advance[/i], but if the roots aren't \"nice\" (generally, integer), then the groups themselves will be almost impossible to find without knowing the roots in advance.", "Solution_17": "erdogan -- is this a problem from your calculus textbook by any chance?\r\nI seem to remember it, and if youre trying to integrate, you dont need to factor.\r\n\r\nIf its not a calc problem, then assume i am insane and ignore me :)", "Solution_18": "[quote=\"erdogankerem123\"]How do you factor cubic polynomials such as this:\n\n $ x^3 \\minus{} 21x \\plus{} 20 \\equal{} 0$, where grouping doesn't work ?[/quote]\r\n\r\n\r\n[hide=\"Inspection\"](NYSML has spoiled me in that aspect. :P) gives that the sum of the coefficients is 0, therefore 1 is a root.\n\nThe product of the other two roots is $ \\frac{\\minus{}20}{1}\\equal{}\\minus{}20$, and the sum is $ 0\\minus{}1\\equal{}\\minus{}1$. We now have the quadratic $ x^2\\plus{}x\\minus{}20$. Then you can bash with the [[Quadratic Formula]], or inspect some more and factor:\n\n$ x^3\\minus{}21x\\plus{}20\\equal{}(x\\minus{}1)(x\\plus{}5)(x\\minus{}4)$[/hide]" } { "Tag": [], "Problem": "Given $n$ a natural number greater than $1$ and $p$ a prime, where $n|p-1$ and $p|n^{3}-1$, show that $4p-3$ is a square number.", "Solution_1": "This is not true for $n=1, p=2,$ or $n=1,p=5,$ and so on. I guess $n > 1$ is a necessary condition.", "Solution_2": "[quote=\"FieryHydra\"]This is not true for $n=1, p=2.$[/quote]\r\n\r\nSurely it is NOT true!! :P Now we assume that $n>1$.", "Solution_3": "So $p | (n-1)(n^{2}+n+1) \\implies p | n^{2}+n+1$, and we'll let $p-1 = kn$. Then $0 \\equiv kn^{2}+kn+k \\equiv-n+k-1 \\mod p \\implies k = n+1$. So we have $p = kn+1 = n^{2}+n+1 \\implies 4p-3 = 4n^{2}+4n+1 = (2n+1)^{2}$.", "Solution_4": "[quote=\"probability1.01\"]$0 \\equiv kn^{2}+kn+k \\equiv-n+k-1 \\mod p \\implies k = n+1$.[/quote]\r\n\r\nWouldn't it have the case that $k=n+1+np$?" } { "Tag": [ "inequalities", "geometry", "circumcircle", "inequalities unsolved" ], "Problem": "Given x, y, z are positive reals such that x + y + z = 1, show that:\r\n\r\n32.xyz.(1 - xy - yz - zx) <= [ 3.(1 - x).(1 - y).(1 - z) ]^2 \r\n\r\n :wink:", "Solution_1": "[quote=\"Jaswinder\"]Given x, y, z are positive reals such that x + y + z = 1, show that:\n\n32.xyz.(1 - xy - yz - zx) <= [ 3.(1 - x).(1 - y).(1 - z) ]^2 \n\n :wink:[/quote]\r\nId est,we need to prove that $ 32xyz(x\\plus{}y\\plus{}z)(x^2\\plus{}y^2\\plus{}z^2\\plus{}xy\\plus{}xz\\plus{}yz)\\leq9(x\\plus{}y)^2(x\\plus{}z)^2(y\\plus{}z)^2,$ which is equivalent to\r\n$ \\sum_{sym}(9x^4y^2\\minus{}7x^4yz\\plus{}9x^3y^3\\minus{}10x^3y^2z\\minus{}x^2y^2z^2)\\geq0,$ which is obviously true..", "Solution_2": "My proof: :)", "Solution_3": "arquady could u plz tell how the last expression is true i am not able to understand :P \r\n\r\nplz give the complete soln that shows that the symmetric expression that u have written is greater than or equal to zero......", "Solution_4": "here is my soln..........\r\n\r\n\r\nHomonogise and write x+y=a etc and it'll come down to a^2 + b^2 + c^2\r\n<= 9R^2 as a,b,c are sides of a triangle and R the circumradius. I\r\nproved this result in a couple of ways. First setting a=2R sinA which\r\nthen becums a trigo inequality and also is true by jensens but can be\r\nproved other wise easily as well. Another proof is that OH^2 = 9R^2 -\r\na^2 - b^2 - c^2 >= 0 where OH is distance btween circumcenter n\r\northocenter. QED.", "Solution_5": "[quote=\"arqady\"]Id est,we need to prove that $ 32xyz(x \\plus{} y \\plus{} z)(x^2 \\plus{} y^2 \\plus{} z^2 \\plus{} xy \\plus{} xz \\plus{} yz)\\leq9(x \\plus{} y)^2(x \\plus{} z)^2(y \\plus{} z)^2,$ [/quote]\r\nThis inequality holds for any real numbers $ x,y,z.$ ;) :)" } { "Tag": [ "probability", "expected value", "probability and stats" ], "Problem": "You have two decks of 26 cards. Each card in each of the two decks has a different letter of the alphabet on it. You pick at random one card from each of the two decks. A vowel is worth 3 points and a consonant is worth 0 points. Let X be the sum of the values of the two cards picked. Find E(X), V(X), and the standard deviation of X.\r\n\r\nNote:\r\nE(X) stands for expected value\r\nV(X) stands for variance", "Solution_1": "The variate X has a discrete probability distribution: (note: I take the vowels to be A,E,I,O,U, and Y.)\r\nThere are then 20 consonants and 6 vowels in each deck of cards. The probabilities are\r\n\r\n$ P(X\\equal{}0)\\equal{}P(CC)\\equal{}\\frac{20}{26}\\cdot\\frac{20}{26}\\equal{}\\frac{100}{169}$\r\n\r\n$ P(X\\equal{}3)\\equal{}P(CV\\mbox{ or }VC)\\equal{}\\frac{20}{26}\\cdot\\frac{6}{26}\\plus{}\\frac{6}{26}\\cdot\\frac{20}{26}\\equal{}\\frac{60}{169}$\r\n\r\n$ P(X\\equal{}6)\\equal{}P(VV)\\equal{}\\frac{6}{26}\\cdot\\frac{6}{26}\\equal{}\\frac{9}{169}$\r\n\r\ngiving\r\n\r\n\\[ \\begin{array}{c|c|c|c}X&P(X)&X\\cdot P(X)&X^2\\cdot P(X)\\\\\\hline0&{\\scriptstyle \\frac{100}{169}}&0&0\\\\\\hline 3&{\\scriptstyle \\frac{60}{169}} & {\\scriptstyle \\frac{120}{169}}&{\\scriptstyle \\frac{360}{169}}\\\\\\hline 6&{\\scriptstyle \\frac{9}{169}}&{\\scriptstyle \\frac{54}{169}}&{\\scriptstyle \\frac{324}{169}}\\end{array}\\]\r\n\r\nso that $ \\boxed{E(X)\\equal{}\\sum_X X\\cdot P(X) \\equal{} \\frac{174}{169}}$ and $ \\boxed{V(X)\\equal{}\\sum_X X^2\\cdot P(X)\\minus{}E(X)^2 \\equal{} \\frac{85320}{28561}}$." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Find all integers h such that there exist infintely many positive integers n for which $[\\sqrt{h^2+1}n]$ is a perfect square.", "Solution_1": "Well i dont know if the following is truth, but anyway...\r\n\r\nI think that the set ${m^2/\\sqrt{h^2+1}}$ has fractional part wich is dense in $(0,1)$. So there are infinitely many $m$, such that the interval $(m^2/\\sqrt{h^2+1},(m^2+1)/\\sqrt{h^2+1})$ contains an integer, also this integer can be expressed as $\\sqrt{n}/\\sqrt{h^2+1}$ with $n$ between $m^4$ and $m^4+2m^2+1$ so we get that $n=t^2(h^2+1)$, so this $t$ gives a value for which $t\\sqrt{h^2+1}$ has it integer part equal to $m^2$.\r\n\r\nSow e only need to prove that the set is dense, or at least it aproaches as we would like to integers, I think it is a known fact, at least I saw it in a problem from Iran or something like that!" } { "Tag": [ "probability" ], "Problem": "Huey, Dewey, and Louie are competing in the diving competitions. There are three diving events, but only two of the three divers are allowed to compete in each of the events and the diver's names are drawn randomly. What is the probability that Louie's name is drawn to compete in at least two of the three diving events? \r\n\r\nI think I'm missing something obvious but I just can't seem to get the answer from an algebraic approach using combinations/permutations etc...", "Solution_1": "We can think of it as \"one diver is chosen to not be in each event.\"\r\n\r\nWe want louie's name to be drawn at most once. It is drawn 0 times with probability $(2/3)^3$. It is drawn 1 times with probability $3(1/3)(2/3)^2$. So total probability 20/27.", "Solution_2": "[quote=\"Tharinor\"]Huey, Dewey, and Louie are competing in the diving competitions. There are three diving events, but only two of the three divers are allowed to compete in each of the events and the diver's names are drawn randomly. What is the probability that Louie's name is drawn to compete in at least two of the three diving events? \n\nI think I'm missing something obvious but I just can't seem to get the answer from an algebraic approach using combinations/permutations etc...[/quote]The probability of n successes in N trials is a binomial distribution: $P(n|N) = \\binom{N}{n} p^n(1-p)^{N-n}$\r\n\r\nThe probability that Louie's name is chosen for at least two event is the probability that his name will be chosen for either two or three of the events.\r\n\r\n$P(n \\ge 2|N =3) = P(n =2|N =3) + P(n =3 |N =3) \\\\\r\n\\\\\r\n= \\binom{3}{2}(\\frac{1}{3})^2(1-\\frac{1}{3})^1 + \\binom{3}{3}(\\frac{1}{3})^3(1-\\frac{1}{3})^0 \\\\\r\n\\\\\r\n= 3(\\frac{1}{9})(\\frac{2}{3}) + 1(\\frac{1}{27}) \\\\\r\n\\\\\r\n= \\frac{7}{27}$", "Solution_3": "well think of it this way, if we want the times that he is chosen 2 times or more, then he can be chosen 2 or 3 times\r\nthe probability that he will be chosen 2 times is $\\frac{2}{3}*\\frac{2}{3}*\\frac{1}{3}*3$ because (win for chosen, loose for not chosen) win, win, lose, and then the order of those has 3 ways (wwl wlw lww)\r\nthen the chances that he wins is $\\frac{2}{3}*\\frac{2}{3}*\\frac{2}{3}$\r\nand then adding those, you get $\\boxed{\\frac{20}{27}}$" } { "Tag": [ "number theory" ], "Problem": "Find all solution to the equation $(a^{2}+2b^{2}+2ab)(a^{2}+2b^{2}-2ab)=3b^{2}(2(a^{2}-b^{2})^{2}(a^{2}+b^{2}))^{\\frac{1}{3}}$", "Solution_1": "if $a,b\\in\\mathbb{Z}$ then $(a,b)=(0,0)$\na and b are even. infinite descent." } { "Tag": [ "puzzles" ], "Problem": "What comes next in this sequence: \r\nfawn bear lion mole ==?== \r\n\r\nSelect from: seal duck stag vole wolf.", "Solution_1": "[hide]duck, a-e-i-o-u[/hide]" } { "Tag": [ "FTW", "AMC", "AIME", "Support" ], "Problem": "its just like the top admin election BUT with a slight differnece. this time you can take away one point OR! add one point.I only know 3 mods on FTW though.Oh and all the mods start at 0.Remeber you can only choose one choice. take away or add a point. I'll take the first vote. I will vote for izzy\r\n\r\nisabella:1\r\n\r\nvallon22:0\r\n\r\niin77:0", "Solution_1": "I vote for isabella2296.", "Solution_2": "I vote for izzy. vallons a mod?", "Solution_3": "i vote izzy. yes, vallons a mod, dragon96.", "Solution_4": "I would vote all three if I could.\r\n\r\nHowever, I cannot, so I will remain neutral and not vote.", "Solution_5": "I vote iin77\r\n\r\nisabella2296: 4\r\nvallon22: 0\r\niin77: 1", "Solution_6": "I'm gonna vote for the math1337 vallon.", "Solution_7": "Umm.. why don't you type the scores...\r\nthat way everyone can see it.\r\n\r\nisabella2296: 4\r\nvallon22: 1\r\niin77: 1", "Solution_8": "isabella2296: 5\r\nvallon22: 1 \r\niin77: 1\r\n\r\nThe only one I THINK has never muted me in FTW. (Before the update I can't talk now :( )", "Solution_9": "isabella2296: 5\r\nvallon22: 1\r\niin77: 2", "Solution_10": "isabella2296: 5.5\r\niin77: 2\r\nvallon22: 1.5\r\n\r\n:P :P :P", "Solution_11": "isabella2296: 5.5 \r\niin77: 3 \r\nvallon22: 1.5", "Solution_12": "I will take a point away for once. Sorry Izzy.\r\n\r\nIzzy:4.5\r\n\r\niin77:3\r\n\r\nvallon22:1.5", "Solution_13": "gah\r\ni'm losing...\r\n\r\ni vote myself :P \r\n\r\nIzzy:4.5 \r\n\r\niin77:3 \r\n\r\nvallon22:2.5\r\n\r\ngah\r\ni'm still losing :D", "Solution_14": "izzy- 4.5\r\niin77- 4\r\nvallon22-2.5\r\n\r\nyay! go iin! sorry izzy =[", "Solution_15": "izzy and vallon: 0\r\niin: -9", "Solution_16": "izzy and vallon: 0\r\niin: -10", "Solution_17": "Getting lazier:\r\n\r\niin:-11", "Solution_18": "even lazier\r\n\r\n-12", "Solution_19": "okay gonna subtract your votes again. if you keep doing this i'm gonna ask a mod to lock this and the competition is over.\r\n\r\nIzzy:9\r\n\r\niin:6\r\n\r\nvallon:4", "Solution_20": "Well you might as well ask a mod to lock it because earlier you said in 24 hours its over. The scores are 0,0, and -12. No winner.", "Solution_21": "oka w/e mod please lock this. THE WINNER IS IZZY. this is the final score\r\n\r\n\r\nIzzy:9\r\n\r\niin:6\r\n\r\nvallon:4", "Solution_22": "izzy: 0\r\nvallon: 0\r\niin77: -13", "Solution_23": "[quote=\"jjx1\"]oka w/e mod please lock this. THE WINNER IS IZZY. this is the final score\n\n\nIzzy:9\n\niin:6\n\nvallon:4[/quote]\r\nyeah, this topic is locked down i can't even post :rotfl:", "Solution_24": "izzy: 0 \r\nvallon22: 0 \r\niin77: -14", "Solution_25": "ERNIE THE GAME IS OEVER. IZZY WON. I DID NOT COUNT YOUR VOTES OR JAMES OR DRAGONS. ALSO JOIN FFC!!!!!!!", "Solution_26": "guys\r\nit's just a game\r\nstop yelling at each other about it\r\nplus this is mostly a popularity vote\r\nand for now, i really don't care about my popularity on AoPS\r\n\r\nalso, there is no mod for FF, so no one can lock it", "Solution_27": "izzy: 0 \r\nvallon22: 0 \r\niin77: -15", "Solution_28": "vallon you're right. guys you can do whatever you want to this forum. But i thought there were two mods for ff.", "Solution_29": "Oh well... IDC.\r\n\r\nizzy: 0 \r\nvallon22: 0 \r\niin77: -16" } { "Tag": [ "function", "real analysis", "real analysis unsolved" ], "Problem": "Given that $ A$ and $ \\mathbb R\\setminus A$ are both dense subsets of $ \\mathbb R$ does it follow that there exists either a function that is continuous only on $ A$ or a function that is discontinuous only on $ A$.\r\nP.S. I have no idea whether or not this is true - I just came across it thinking about a problem that mlok posed in response to an (easy) problem I posted. (The answer is yes for countable sets.)", "Solution_1": "The set of continuity of a function must be a $ G_{\\delta}$ set (countable intersection of open sets). This is a more or less direct consequence of the $ \\epsilon$-$ \\delta$ definition. If I remember correctly, any $ G_{\\delta}$ set is the set of continuity for some function.", "Solution_2": "So this amounts to showing that there are no dense sets (with dense compliment) that are neither $ F_{\\sigma}$ or $ G_{\\delta}$? This seems unlikely to be true and even if it is it may be quite hard.\r\nAnyway, the reason I did this was to see if such a set $ A$ could satisfy the hypothesis of that previous problem by invoking a theorem of Volterra that there do not exist two pointwise discontinuous (continuous on a dense set) functions with one continuous only where the other is discontinuous.\r\n(Supposing the existence of an $ F$ as in the previous problem we could take $ f$ a function continuous only on $ A$ and then achieve a contradiction by considering $ g=f(F)$.)", "Solution_3": "Of course density has little to do with being $ G_{\\delta}$. Consider the set formed by all positive rationals and negative irrationals: it is dense, has dense complement, and is neither $ G_{\\delta}$ nor $ F_{\\sigma}$. Thus, the answer to your question is no.\r\n\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=162163]Link to the earlier thread that motivated this one[/url]." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Given $ a, b, c \\geq\\ 0$ satisfy $ ab \\plus{} bc \\plus{} ca \\equal{} 1.$\r\nProve that: $ \\sqrt []{a^2b \\plus{} b^2c \\plus{} c^2a} \\plus{} \\sqrt []{ab^2 \\plus{} bc^2 \\plus{} ca^2} \\plus{} 3 \\sqrt []{abc} \\geq\\ 2$\r\n :)\r\n\r\n@mitdac123: : Good luck in love.", "Solution_1": "I haven't solved the general problem:\r\nFind the best constant k satisfy:\r\n$ \\sqrt []{a^2b \\plus{} b^2c \\plus{} c^2a} \\plus{} \\sqrt []{ab^2 \\plus{} bc^2 \\plus{} ca^2} \\plus{} k \\sqrt []{abc} \\geq\\ 2$\r\n(with the same condition)\r\nI only have proof for $ k \\geq\\ 3$\r\n :wink:", "Solution_2": "[quote=\"nguoivn\"]I haven't solved the general problem:\nFind the best constant k satisfy:\n$ \\sqrt []{a^2b \\plus{} b^2c \\plus{} c^2a} \\plus{} \\sqrt []{ab^2 \\plus{} bc^2 \\plus{} ca^2} \\plus{} k \\sqrt []{abc} \\geq\\ 2$\n(with the same condition)\nI only have proof for $ k \\geq\\ 3$\n :wink:[/quote]\r\nHere is the hint: $ (a^2b\\plus{}b^2c\\plus{}c^2a)(ab^2\\plus{}bc^2\\plus{}ca^2) \\ge (a^2b^2\\plus{}b^2c^2\\plus{}c^2a^2)(ab\\plus{}bc\\plus{}ca)$. ;) :)", "Solution_3": "Nice hint, can_hang. But after using this hint, we'' ll have to prove:\r\n$ 2 \\sqrt [4]{3(9 \\minus{} 2pr)} \\plus{} k \\sqrt []{r} \\geq\\ 2$\r\nI think we must to use calculator :wink: \r\nCan you show me the best constant $ k \\equal{} ?$\r\nBesides, with $ k \\geq\\ 3,$ I have a nice and short proof by Am-Gm :)", "Solution_4": "[quote=\"nguoivn\"]Nice hint, can_hang. But after using this hint, we'' ll have to prove:\n$ 2 \\sqrt [4]{3(9 \\minus{} 2pr)} \\plus{} k \\sqrt []{r} \\geq\\ 2$\nI think we must to use calculator :wink: \nCan you show me the best constant $ k \\equal{} ?$\nBesides, with $ k \\geq\\ 3,$ I have a nice and short proof by Am-Gm :)[/quote]\r\nNo, I didn't mean to prove it by that way. ;)\r\nMy hint used for another way. ;) Try it!", "Solution_5": "I want to see the solution by AM-GM.Thanks you." } { "Tag": [ "inequalities", "algebra", "polynomial", "inequalities proposed" ], "Problem": "The real numbers $a_1,a_2,...,a_{100}$ satisfy the relationship :\r\n$a_1^2+ a_2^2 + \\cdots +a_{100}^2 + ( a_1+a_2 + \\cdots + a_{100})^2 = 101$\r\nProve that $|a_k| \\leq 10$ for all $k \\in \\{1,2,...,100\\}$", "Solution_1": "we have $a_2^2 + \\cdots +a_{100}^2 \\geq \\frac{(a_2 + \\cdots +a_{100})^2}{99}$\r\nso $101=a_1^2+ a_2^2 + \\cdots +a_{100}^2 + ( a_1+a_2 + \\cdots + a_{100})^2 \\geq 2a^2_1+\\frac{(a_2 + \\cdots +a_{100})^2}{99}+(a_2 + \\cdots +a_{100})^2+2a_1(a_2 + \\cdots +a_{100})$\r\nconsider it for a polynomial whose degree is 2.\r\nwe will easily get,", "Solution_2": "This is inequality was proposed by Dinu Serbanescu at jbmo test selection in 2004." } { "Tag": [ "inequalities", "number theory unsolved", "number theory" ], "Problem": "I just calculated this . It says that the error between the actual and the computed value of pi(N) using sieve is atleast in the order of ( \u221aN/log(N) ) i.e O(\u221aN/logN).\r\nThis is the proof for the above the statement .\r\nI have taken take a specific case and proved it . However it can be generalised .\r\nConsider 25 infinity . for example one knows how close 9.99 is to 10.\r\nBut how close is 10^(10^10^10^10^10) close to 10^(10^10^10^10^9.99) ???\r\nI am sorry for replying to the above post . Though i am myself not very interested in discussing these kinds of things." } { "Tag": [ "\\/closed" ], "Problem": "I was viewing the online people and I noticed that several guests were listed as \"posting a message.\" Screenshot attached for clarification if need be.", "Solution_1": "That looks like another side effect of the server issues last night; see also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=135229]here[/url]. The posts from 2013 are a dead giveaway." } { "Tag": [ "function", "quadratics", "algebra", "quadratic formula", "floor function" ], "Problem": "Greatest Integer Function Problem...\r\n\r\nSolve for all positive integer $ x$ : $ [x]^2 \\equal{} xh$ where $ [x]$ represents GIF, $ x$ is the number and $ h$ is the fractional part.\r\n\r\nI've reached this part of the solution: since $ x \\equal{} h\\plus{}[x]$, $ [x]^2 \\minus{} [x]h \\minus{} h^2\\equal{}0$. using quadratic formula, $ [x] \\equal{} \\frac{h\\plus{}h\\sqrt{5}}{2}$. what next?", "Solution_1": "[hide=\"Partial solution?\"]\nWe know $ [x] \\equal{} x\\minus{}h$ so $ (x\\minus{}h)^2 \\equal{} xh$.\nIt follows that $ x^2 \\minus{} 3hx \\plus{} h^2 \\equal{} 0$.\nThen $ D \\equal{} 5h^2$.\n\nto be continued...\n[/hide]", "Solution_2": "[hide](You meant Solve for all positive reals, not integers (it doesn't make sense for ints)).\n\n\"I've reached this part of the solution: since $ x \\equal{} h \\plus{} [x]$, $ [x]^2 \\minus{} [x]h \\minus{} h^2 \\equal{} 0$. using quadratic formula, $ [x] \\equal{} \\frac {h \\plus{} h\\sqrt {5}}{2}$. what next?[/quote]\"\n\nWell, as $ h<1$, $ [x]\\leq 2$.\n\nCase 1. $ [x]\\equal{}0$. Simply putting it into the equation, we have $ x\\equal{}0$ (not quite positive...).\nCase 2. $ [x]\\equal{}1$. Put it in the last equation you've got and voila, you have one solution.\nCase 3. $ [x]\\equal{}2$. Same as above.\n[/hide]", "Solution_3": "Ok, I was going to post a really long and ugly solution (was planning to finish it now because I was interrupted last time). I guess yours is like a hundred times better.", "Solution_4": "[quote=\"hsiljak\"][hide](You meant Solve for all positive reals, not integers (it doesn't make sense for ints)).\n\n\"I've reached this part of the solution: since $ x \\equal{} h \\plus{} [x]$, $ [x]^2 \\minus{} [x]h \\minus{} h^2 \\equal{} 0$. using quadratic formula, $ [x] \\equal{} \\frac {h \\plus{} h\\sqrt {5}}{2}$. what next?[/quote]\"\n\nWell, as $ h < 1$, $ [x]\\leq 2$.\n\nCase 1. $ [x] \\equal{} 0$. Simply putting it into the equation, we have $ x \\equal{} 0$ (not quite positive...).\nCase 2. $ [x] \\equal{} 1$. Put it in the last equation you've got and voila, you have one solution.\nCase 3. $ [x] \\equal{} 2$. Same as above.\n[/hide][/quote]\r\n\r\noh yeah....that's what i lacked... thanks for the help and info:P" } { "Tag": [ "induction" ], "Problem": "$ \\frac {1}{x^{1}\\minus{}1}\\minus{}\\frac {1}{x^{1}\\plus{}1}\\minus{}\\frac {2}{x^{2}\\plus{}1}\\minus{}\\frac {4}{x^{4}\\plus{}1}\\minus{}\\frac {8}{x^{8}\\plus{}1}$\r\n\r\n[hide=\"Stuck?\"]\nanswer is $ \\frac {16}{x^{16}\\minus{}1}$\n[/hide]", "Solution_1": "[hide=\"General Solution\"]We can prove by induction that $ \\frac{1}{x\\minus{}1} \\minus{} \\displaystyle\\sum_{i\\equal{}0}^{n}{\\displaystyle\\frac{2^i}{x^{2^i}\\plus{}1}}\\equal{}\\displaystyle\\frac{2^{n\\plus{}1}}{x^{2^{n\\plus{}1}}\\minus{}1}$.\n\nBase case: $ n \\equal{} 0$. $ \\frac{1}{x\\minus{}1} \\minus{} \\frac{1}{x\\plus{}1} \\equal{} \\frac{(x\\plus{}1)\\minus{}(x\\minus{}1)}{x^2\\minus{}1} \\equal{} \\frac{2}{x^2\\minus{}1}$\n\nWe make the assumption that the claim is true for some $ k$, and seek to prove it for $ k\\plus{}1$.\n\n$ \\frac{1}{x\\minus{}1} \\minus{} \\displaystyle\\sum_{i\\equal{}0}^{k\\plus{}1}{\\displaystyle\\frac{2^i}{x^{2^i}\\plus{}1}} \\equal{} \\frac{2^{k\\plus{}1}}{x^{2^{k\\plus{}1}}\\minus{}1} \\minus{} \\frac{2^{k\\plus{}1}}{x^{2^{k\\plus{}1}}\\plus{}1} \\equal{} \\frac{2^{k\\plus{}2}}{x^{2^{k\\plus{}2}}\\minus{}1}$.\n\nTherefore, $ \\frac{1}{x\\minus{}1} \\minus{} \\displaystyle\\sum_{i\\equal{}0}^{n}{\\displaystyle\\frac{2^i}{x^{2^i}\\plus{}1}}\\equal{}\\displaystyle\\frac{2^{n\\plus{}1}}{x^{2^{n\\plus{}1}}\\minus{}1}$.\n\nLet $ n\\equal{}3$, and the result is $ \\boxed{\\frac{16}{x^{16}\\minus{}1}}$.\n\n[/hide]", "Solution_2": "[hide=\"Or\"]\nWe could simply see that if we subtracted the first two equations, then it would become $ \\frac {2}{x^{2}\\minus{}2}$\nAnd following this we get the same answer.[/hide]", "Solution_3": "correct guys." } { "Tag": [ "abstract algebra", "group theory", "permutations", "combinatorics", "IMO Shortlist" ], "Problem": "Suppose that $ a_1$, $ a_2$, $ \\ldots$, $ a_n$ are integers such that $ n\\mid a_1 \\plus{} a_2 \\plus{} \\ldots \\plus{} a_n$.\nProve that there exist two permutations $ \\left(b_1,b_2,\\ldots,b_n\\right)$ and $ \\left(c_1,c_2,\\ldots,c_n\\right)$ of $ \\left(1,2,\\ldots,n\\right)$ such that for each integer $ i$ with $ 1\\leq i\\leq n$, we have\n\\[ n\\mid a_i \\minus{} b_i \\minus{} c_i\n\\]\n\n[i]Proposed by Ricky Liu & Zuming Feng, USA[/i]", "Solution_1": "This problem and the following generalisation appeared 1979 in Ars Combinatoria (thanks to Darij who found it):\r\n\r\nLet $ (G, \\plus{} )$ be a finite abelian group of order $ n$.\r\nLet also $ a_1,a_2,...,a_{n \\minus{} 1} \\in G$ be arbitrary.\r\nThen there exist pairwise distinct $ b_1,b_2,...,b_{n \\minus{} 1} \\in G$ and pairwise distinct $ c_1,c_2,...,c_{n \\minus{} 1} \\in G$ such that $ a_k \\equal{} b_k \\plus{} c_k$ for $ k \\equal{} 1,2,...,n \\minus{} 1$.\r\n\r\n[Moderator edit: The Ars Combinatoria paper is:\r\nF. Salzborn, G. Szekeres, [i]A problem in Combinatorial Group Theory[/i], Ars Combinatoria 7 (1979), pp. 3-5.]", "Solution_2": "so could someone post a proof of either the problem or its generalization?", "Solution_3": "you can find the proof in file shortlist 2005 which has been posted by orl\r\nThe main idea is given two sequence $a_{1}...a_{n}$ and $b_{1}...b_{n}$ s.t $\\sum a_{i}\\equiv 0(mod n)$ and $\\sum b_{i}\\equiv 0(mod n)$ and there are exactly two i;j s.t $a_{i}\\neq\\ b_{i}(modn)$ and $a_{j}\\neq\\ b_{j}(modn)$.Then if we know two permutation good for the sequence (a_1...a_n) then we can build two permuttionm good for (b_1...b_n)\r\ni will come back with detail if you need", "Solution_4": "Well, I will post the solution from Ars Combinatoria if a re-find that two sheets of paper...\r\nIt's a bit different from the ISL one.", "Solution_5": "I think you didnt keep promise,Zetax :lol: Please post it here and now!", "Solution_6": "to spdf: that was also my idea when i first saw the problem but i can't find a good way to contruct those 2 permutations for $ b_j$.you said you can post details.please do so :)", "Solution_7": "Sorry for not repsonding (I merely forgot...). But I just saw that problem again: In a slightly different manner (but being equivalent to this one here) it is solved in \"The Mathematics of Juggling\", called the \"Converse of the Average Theorem\".\r\n\r\nMain ideas:\r\nYou show that this property (being a sum of two permutations) is invariant under the operations $ a_{i,j,d}$ that add $ d$ to $ a_i$ and subtract $ d$ from $ a_j$.\r\nFor this, you need to do it algorithmically (but describing it is a bit hard without that graphics...).", "Solution_8": "[quote=ZetaX]Sorry for not repsonding (I merely forgot...). But I just saw that problem again: In a slightly different manner (but being equivalent to this one here) it is solved in \"The Mathematics of Juggling\", called the \"Converse of the Average Theorem\".\n\nMain ideas:\nYou show that this property (being a sum of two permutations) is invariant under the operations $ a_{i,j,d}$ that add $ d$ to $ a_i$ and subtract $ d$ from $ a_j$.\nFor this, you need to do it algorithmically (but describing it is a bit hard without that graphics...).[/quote]\n\nthat is what I tried to do, but I can not prove that we can do it for the general case... can someone please help?", "Solution_9": "Since it's almost twelve years without complete solution, here's the official solution:\n\nSuppose there exists permutations $\\sigma$ and $\\tau$ of $[n]$ for some sequence $\\{ a_i\\}_{i\\in [n]}$ so that $a_i\\equiv_n \\sigma (i)+\\tau (i)$ for all $i\\in [n]$.\nGiven a sequence $\\{ b_i\\}_{i\\in [n]}$ with sum divisible by $n$ that differ, in modulo $n$, from $\\{ a_i\\}_{i\\in [n]}$ in only two positions, say $i_1$ and $i_2$.\nWe want to construct permutations $\\sigma'$ and $\\tau'$ of $[n]$ so that $b_i\\equiv_n \\sigma' (i) +\\tau' (i)$ for all $i\\in [n]$.\nRecall that $b_i\\equiv a_i\\pmod{n}$ for all $i\\in [n]$ that $i\\neq i_1,i_2$.\nConstruct a sequence $i_1,i_2,i_3,...$ by, for each integer $k\\geq 2$, define $i_{k+1}\\in [n]$ to be the unique integer satisfy $\\sigma (i_{k-1})+\\tau (i_{k+1})\\equiv_n b_{i_k}$.\nLet (clearly exists) $p2$. This means $i_j\\not\\in \\{ i_1,i_2\\} \\implies \\sigma (i_j) +\\tau (i_j) \\equiv_n b_{i_j}$ for all $j\\in \\{ p,p+1,...,q\\}$.\nSumming the equation $\\sigma (i_{k-1})+\\tau (i_{k+1})\\equiv_n b_{i_k}$ for $k\\in \\{ p,p+1,...,q-1\\}$ gives us\n$$\\sum_{j=p-1}^{q-2}{\\sigma (i_j) } +\\sum_{j=p+1}^{q}{\\tau (i_j)} \\equiv_n\\sum_{j=p}^{q-1}{b_{i_j}} \\implies \\sigma (i_{p-1}) +\\sigma (i_p) +\\tau (i_{q-1}) +\\tau (i_q) \\equiv_nb_{i_p}+b_{i_{q-1}}.$$\nPlugging $i_p=i_q$ and use $\\sigma (i_p) +\\tau (i_p)\\equiv_n b_{i_p}$ gives us $\\sigma (i_{p-1}) +\\tau (i_{q-1})\\equiv_n b_{i_{q-1}} \\equiv_n \\sigma (i_{q-1})+\\tau (i_{q-1})$.\nHence, $\\sigma (i_{p-1}) \\equiv_n \\sigma (i_{q-1})\\implies i_{p-1}=i_{q-1}$, contradiction to the definition of $p,q$.\n\nSo, we've $p\\in \\{ 1,2\\}$. Let $p'=3-p$. Define the desired permutations $\\sigma'$ and $\\tau'$ as follows:\n$$\\sigma' (i_l)=\\begin{cases} \n\\sigma (i_{l-1}), & \\text{ if } l\\in \\{ 2,3,...,q-1\\} \\\\\n\\sigma (i_{q-1}), & \\text{ if } l=1\n\\end{cases} ,\\tau' (i_l)= \\begin{cases} \n\\tau (i_{l+1}), & \\text{ if } l\\in \\{ 2,3,...,q-1\\} \\\\\n\\tau (i_{p'}), & \\text{ if } l=1\n\\end{cases} $$ \nand $\\sigma' (i) =\\sigma (i),\\tau' (i)=\\tau (i)$ for the rest $i\\in [n]$ that $i\\not\\in \\{ i_1,i_2,...,i_{q-1}\\}$.\nNote that the reason we choose $\\tau (i_{p'})$ is just to not use $\\tau (i_p)=\\tau (i_{(q-1)+1})$ more than one time.\nThis construction gives us $\\sigma' (i)+\\tau' (i)\\equiv_n b_i$ for all $i\\in [n]$ except when $i=i_1$.\nBut since both $\\sigma'$ and $\\tau'$ are permutations of $[n]$, we've $\\sum_{i\\in [n]}{(\\sigma' (i)+\\tau' (i))} \\equiv_n 2\\times \\frac{n(n-1)}{2}\\equiv_n 0\\equiv_n \\sum_{i\\in [n]}{b_i}$.\nThis guarantee that $\\sigma' (i) +\\tau' (i)\\equiv_n b_i$ when $i=i_1$ too. This prove the validity of permutations we constructed.", "Solution_10": "The case with $n$ prime is also resolved in [url=https://services.math.duke.edu/~dasgupta/papers/finallatin.pdf]this paper[/url]", "Solution_11": "any other solutions ? \n" } { "Tag": [ "inequalities", "inequalities solved" ], "Problem": "If $a,b,c>0$ and $a^2+b^2+c^2=1$ then we have:\r\n$ \\sum \\frac{a^5+b^5}{ab(a+b)} \\geq 3(ab+bc+ca)-2$.", "Solution_1": "note that a^5+b^5=(a+b)(a^4+b^4-a^2*b^3-a^3*b^2+a^2*b^2)\r\ni'll prove \r\n(a^4+b^4-a^2*b^3-a^3*b^2+a^2*b^2) \\geq ab(3ab-a^2-b^2)\r\nwhich is equivalent (a^2-b^2) \\geq 0 .thus\r\n(a^5+b^5)/ab(a+b) \\geq (3ab-a^2-b^2)\r\nadd two similar inequalitites to get the result\r\ncheers", "Solution_2": "A shorter solution :\r\n\r\n a^5 +b^5 >= a^4.b +a.b^4=ab(a+b)(a^2 -ab +b^2)\r\n So \r\n L.H.S >= 2(a^2 +b^2 +c^2) -(ab+ac+bc) \r\n We have\r\n 2(a^2 +b^2 +c^2) -(ab+ac+bc) >= 3(ab+ac+bc) -2 (a^2 +b^2 +c^2)\r\n <-->\r\n a^2 +b^2 +c^2 \\geq ab+ac+bc" } { "Tag": [ "trigonometry" ], "Problem": "Prove that this is true :\r\n[img]http://img200.imageshack.us/img200/6218/cotj.jpg[/img]", "Solution_1": "[hide=\"Solution\"]\n$ \\frac {\\tan x \\minus{} \\frac {1}{\\tan x}}{\\tan x \\plus{} \\frac {1}{\\tan x}} \\equal{} \\frac {\\tan^2 x \\minus{} 1}{\\tan ^2 x \\plus{} 1}$.\n\nBut $ \\sec^2 x \\minus{} \\tan^2 x \\equal{} 1\\implies \\tan^2 x \\plus{} 1 \\equal{} \\sec^2 x$\n\n$ \\frac {\\tan^2 x \\minus{} 1}{\\sec^2 x} \\equal{} \\sin^2 x \\minus{} \\cos^2 x \\equal{} \\minus{} \\cos 2x$.\n\nTherefore we have reduced it to\n\n$ 1 \\minus{} \\cos 2x \\equal{} 2\\sin^2 x$, which is clearly true.[/hide]", "Solution_2": "$ \\frac{tgx\\minus{}cotgx}{tgx\\plus{}cotgx}\\plus{}1\\equal{}2.\\frac{tgx}{tgx\\plus{}cotgx}\\equal{}2.\\frac{tg^2x}{tg^2x\\plus{}1}\\equal{}2.\\frac{\\frac{sin^2x}{cos^2x}}{\\frac{sin^2x}{cos^2x}\\plus{}1}\\equal{}2.\\frac{sin^2x}{cos^2x\\plus{}sin^2x}\\equal{}2.sin^2x$", "Solution_3": "Sorry, grn_trtle, I didn't see your post. :read:" } { "Tag": [], "Problem": "Let $BE$ and $CF$ be the altitudes of an acute triangle $ABC$ with $E$ on $AC$ and $F$ on $AB$. Let $O$ be the point of intersection of $BE$ and $CF$. Take any line $KL$ through $O$ with $K$ on $AB$ and $L$ on $AC$. Suppose $M$ and $N$ are located on $BE$ and $CF$ respectively. such that $KM$ is perpendicular to $BE$ and $LN$ is perpendicular to $CF$. Prove that $FM$ is parallel to $EN$.", "Solution_1": "[hide]Since $AF || LN$ and $AE || KM$ we have $\\angle CAF = \\angle CLN = BKM$.\n\nThen $\\angle MKF = NLE$.\n\nSince $OFKM$ and $LNOE$ are cyclic we have $\\angle FOK = \\angle FMK$ and $\\angle LON = \\angle LEN$.\n\nSince $\\angle FOK = \\angle LON$ we have $\\angle FMK = \\angle LEN$.\n\nNow $\\angle OMF = 90^o - \\angle FMK = 90^o - \\angle LEN = \\angle NEO$. So $EN$ is parallel to $MK$.[/hide]", "Solution_2": "[i]I add one more solution to Andreas nice solution![/i]\r\n\r\nWe \u2018 ll use similar triangles.\r\n\r\n a) Triangles $OKM$ and $OEL$ are similar so :\r\n $\\frac{OM }{OE} = \\frac{OK}{OL}$ \r\n \r\nb) Triangles $OKF$ and $OLN$ are similar so: \r\n $\\frac{OK}{OL} = \\frac{OF}{ON}$\r\n\r\nFrom (1) and (2) we take\r\n\r\n $\\frac{OM}{OE} = \\frac{OF}{ON}$\r\n\r\nwhich means that $FM // EN$\r\n\r\n [u]Babis[/u]", "Solution_3": "Since, $\\angle AFC =\\angle LNC=90^{\\circ} \\implies AF||LN \\implies \\angle AKL =\\angle KLN$, but since, $FKMO$ and $ELNO$ are cyclic quadrilaterals\nHence,$\\angle AKL=\\angle FMO=\\angle KLN=\\angle BEN \\implies \\boxed{FM||EN}$" } { "Tag": [], "Problem": "I'm going to do a test to see how many people view this topic.", "Solution_1": "Hehehe I'm too curious", "Solution_2": "Couldn't resist :roll:", "Solution_3": ":blush: hahahahaha", "Solution_4": "Hehe, had to see what it was. :blush: :D", "Solution_5": "Curiosity did not kill the cat. :?", "Solution_6": "Hehe. To make sure a lot of people will view your thread, warn them not to view your thread.", "Solution_7": "Oh, curiosity killed the cat. But satisfaction brought it back. :D", "Solution_8": "\"Buuuutttt..... the cat came back, the very next day...\"", "Solution_9": "Funny. :blush:", "Solution_10": "Basic example of reverse psychology :lol:", "Solution_11": "[quote=\"Bictor717\"]\"Buuuutttt..... the cat came back, the very next day...\"[/quote]\r\n\r\nHe just couldn't stay away! :D", "Solution_12": "Uhhh.......I wasn't able to resist. :D", "Solution_13": "[quote=\"plokoon51\"]Basic example of reverse psychology :lol:[/quote]\r\n\r\nsomehow.. i thought reverse psychology was a canadian thing.. :D \r\n(we always joke around with reverse psychology in math camps)", "Solution_14": "I guess I couldn't resist... but if someone made a \"Please view this topic\" thread, I would probably look at that too...", "Solution_15": "er...i'm sure it says somewhere in the Moderator Bill of Rights that I have the right to look at a possibly inappropriate topic. *nods to self* Thus, I'm sure I didn't fall for your trick. :-)", "Solution_16": "As always, the best way to make many people to see something, is ask loudly enough not to do it. :D", "Solution_17": "This is the first topic I entered out of all I saw.\r\n\r\n\r\n :D", "Solution_18": "the easiest way to get a lot of views? (=", "Solution_19": "Cats are too smart to be killed by curiosity. That's why i'm not a cat..... eh.....", "Solution_20": "not my cat......... :D", "Solution_21": "[quote=\"mcalderbank\"]not my cat......... :D[/quote]\r\n\r\nBTW, I don't know if it would work with all the cats, but with the cat of a friend, we always play to point a laser to the floor, and move it around, and the cat always tries to catch it. It is very funny. :rotfl: \r\n\r\nBest regards,", "Solution_22": "That's why cats have nine lives. By the way, there's a poem called Curiosity by Alistair Reid:\r\n\r\nCuriosity \r\nmay have killed the cat; more likely \r\nthe cat was just unlucky, or else curious \r\nto see what death was like, having no cause \r\nto go on licking paws, or fathering \r\nlitter on litter of kittens, predictably.\r\n\r\nNevertheless, to be curious \r\nis dangerous enough. To distrust \r\nwhat is always said, what seems, \r\nto ask odd questions, interfere in dreams, \r\nleave home, smell rats, have hunches \r\ndo not endear cats to those doggy circles \r\nwhere well-smelt baskets, suitable wives, good lunches \r\nare the order of things, and where prevails \r\nmuch wagging of incurious heads and tails.\r\n\r\nFace it. Curiosity \r\nwill not cause us to die-- \r\nonly lack of it will. \r\nNever to want to see \r\nthe other side of the hill \r\nor that improbable country \r\nwhere living is an idyll \r\n(although a probable h*ll) \r\nwould kill us all\r\n\r\nOnly the curious have, if they live, a tale \r\nworth telling at all.\r\n\r\nDogs say cats change too much, are irresponsible, \r\nare changeable, marry too many wives, \r\ndesert their children, chill all dinner tables \r\nwith tales of their nine lives. \r\nWell, they are lucky. Let them be \r\nnine-lived and contradictory, \r\ncurious enough to change, prepared to pay \r\nthe cat price, which is to die \r\nand die again and again, \r\neach time with no less pain. \r\nA cat minority of one \r\nis all that can be counted on \r\nto tell the truth. And what cats have to tell \r\non each return from h*ll \r\nis this: that dying is what the living do, \r\nthat dying is what the loving do, \r\nand that dead dogs are those who do not know \r\nthat dying is what, to live, each has to do.", "Solution_23": "[quote=\"jpark\"]\n\nsomehow.. i thought reverse psychology was a canadian thing.. :D \n(we always joke around with reverse psychology in math camps)[/quote]\r\n\r\nHehehe, don't forget Jenn, there's no such thing as reverse psychology. :D", "Solution_24": "The point of this post wasn't to trick people. I'm not [i]that[/i] mean(almost). I just wanted to see how people deal with curiosity.", "Solution_25": "well at least we could use this for consensus purposes\r\nout of 25 posts, 17 are boys 5 are girls, i didnt bother to subtract the overlap (like someone posting twice)" } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Let $ f: [0 ,1]\\minus{}>[0, 1]$ an increasing function.\r\n$ f(0)\\equal{}0$\r\n$ f(x) \\plus{} f(1\\minus{}x) \\equal{} 1$\r\n$ 2f(x) \\equal{} f(3x)$\r\nCalculate $ f(18/1991)$.", "Solution_1": "I found that $ f(1\\minus{}\\frac{1}{3^n}) \\equal{} 1\\minus{}\\frac{1}{2^n}$\r\nI don't know if it helps.", "Solution_2": "Well if what you found is true, letting $ c \\equal{} \\frac{18}{1991}$\r\nwe get $ f(c) \\equal{} 1\\minus{}\\frac{1}{2^{\\frac{\\minus{}\\log(1\\minus{}c)}{3}}}$", "Solution_3": "You mean: $ f(c) \\equal{} 1\\minus{}\\frac{1}{2^{\\frac{\\minus{}\\log(1\\minus{}c)}{log{3}}}}$", "Solution_4": "$ f(x) \\equal{} \\frac12$ for $ x\\in[\\frac13,\\frac23]$.\r\n$ \\frac {1176}{1991}\\in[\\frac13,\\frac23]$\r\n$ f(\\frac {392}{1991}) \\equal{} f(\\frac {1176}{1991})/2 \\equal{} \\frac14$\r\n$ f(\\frac {1599}{1991}) \\equal{} 1 \\minus{} f(\\frac {392}{1991}) \\equal{} \\frac34$\r\n$ f(\\frac {533}{1991}) \\equal{} f(\\frac {1599}{1991})/2 \\equal{} \\frac38$\r\n$ f(\\frac {1458}{1991}) \\equal{} 1 \\minus{} f(\\frac {533}{1991}) \\equal{} \\frac58$\r\n$ f(\\frac {18}{1991}) \\equal{} f(\\frac {1458}{1991})/16 \\equal{} \\frac5{128}$\r\n\r\nAnother way to find $ f(x)$.\r\n1) Write $ x$ in base $ 3$. \r\n2) If $ 1$ shows up, then remove all the number after the first $ 1$. \r\n3) replace $ 2$ with $ 1$.\r\n4) treat the new number as $ 2\\minus{}$ base, that will be $ f(x)$.\r\n\r\n$ \\frac{18}{1991}\\equal{}0.0000201abc...$ in $ 3\\minus{}$ base. So $ f(\\frac{18}{1991})\\equal{}2^{\\minus{}5}\\plus{}2^{\\minus{}7}$.", "Solution_5": "I didn't understand this second method you used to find $ f(x)$. Could you explain it in details?", "Solution_6": "[quote=\"danilorj\"]You mean: $ f(c) \\equal{} 1 \\minus{} \\frac {1}{2^{\\frac { \\minus{} \\log(1 \\minus{} c)}{log{3}}}}$[/quote]\r\nBut it'll not work. n is integer.", "Solution_7": "Pls forget about the second method.", "Solution_8": "Man, how can you get to the solution if you didn't use the conditions of the problem?\r\n$ f(0)\\equal{}0$ Where did you use it?\r\n$ f(x)\\plus{}f(1\\minus{}x)\\equal{}1$ The same\r\n$ 2f(x)\\equal{}f(3x)$ and the same.\r\nAnd returning to your solution, why did you remove the number after the first one and replaced every two number by one ? Then you treat it as a binary expansion ?", "Solution_9": "[quote=\"danilorj\"]Man, how can you get to the solution if you didn't use the conditions of the problem?\n$ f(0) \\equal{} 0$ Where did you use it?\n$ f(x) \\plus{} f(1 \\minus{} x) \\equal{} 1$ The same\n$ 2f(x) \\equal{} f(3x)$ and the same.\n[/quote]\r\n\r\n$ f(0) \\equal{} 0$ and $ f(0) \\plus{} f(1) \\equal{} 1 \\implies f(1) \\equal{} 1$ -> first condition and second condition\r\n$ 2f(\\frac13) \\equal{} f(1) \\implies f(\\frac13) \\equal{} \\frac12$ -> third condition\r\n$ f(\\frac13) \\plus{} f(\\frac23) \\equal{} 1 \\implies f(\\frac23) \\equal{} \\frac12$ -> second condition\r\n\"f is an increasing function\" implies $ f(x) \\equal{} \\frac12 \\forall x \\in [\\frac13, \\frac23]$ -> condition that f is increasing (you forgot about that one).\r\n\r\nAnd that's only in the first line of his solution! ;)", "Solution_10": "[quote=\"xxp2000\"]$ f(x) \\equal{} \\frac12$ for $ x\\in[\\frac13,\\frac23]$.\n$ \\frac {1176}{1991}\\in[\\frac13,\\frac23]$\n$ f(\\frac {392}{1991}) \\equal{} f(\\frac {1176}{1991})/2 \\equal{} \\frac14$\n$ f(\\frac {1599}{1991}) \\equal{} 1 \\minus{} f(\\frac {392}{1991}) \\equal{} \\frac34$\n$ f(\\frac {533}{1991}) \\equal{} f(\\frac {1599}{1991})/2 \\equal{} \\frac38$\n$ f(\\frac {1458}{1991}) \\equal{} 1 \\minus{} f(\\frac {533}{1991}) \\equal{} \\frac58$\n$ f(\\frac {18}{1991}) \\equal{} f(\\frac {1458}{1991})/16 \\equal{} \\frac5{128}$\n\nAnother way to find $ f(x)$.\n1) Write $ x$ in base $ 3$. \n2) If $ 1$ shows up, then remove all the number after the first $ 1$. \n3) replace $ 2$ with $ 1$.\n4) treat the new number as $ 2 \\minus{}$ base, that will be $ f(x)$.\n\n$ \\frac {18}{1991} \\equal{} 0.0000201abc...$ in $ 3 \\minus{}$ base. So $ f(\\frac {18}{1991}) \\equal{} 2^{ \\minus{} 5} \\plus{} 2^{ \\minus{} 7}$.[/quote]\r\n\r\n(3) Notice that from $ 2f(x) \\equal{} f(3x)$\r\n$ f(x)\\equal{}\\frac{f(3x)}{2}$\r\n\r\n(4) $ f(x) \\equal{} \\frac12 \\forall x \\in [\\frac13, \\frac23]$ \r\n\r\n(5) Notice that for $ 01", "Solution_1": "[quote=\"yaa\"]${ \\frac {a - 1}{ca} + \\frac {b - 1}{ab} + \\frac {c - 1}{bc}}\\leq \\frac {3}{4}$with a,b,c>2[/quote]\r\nLet $ a=2+x,$ $ b=2+y$ and $ c=2+z.$\r\nHence, your inequality is equivalent to the following:\r\n$ \\sum_{cyc}(2xy+xyz)\\geq0.$", "Solution_2": "Sorry my problem is: a,b,c>1", "Solution_3": "[quote=\"yaa\"]Sorry my problem is: a,b,c>1[/quote]\r\nIf so it's wrong: $ c\\rightarrow1^{ \\plus{} }$ and $ a \\rightarrow \\plus{} \\infty.$ :wink:", "Solution_4": "......\r\n\r\n$ 2$ is the best constant..." } { "Tag": [ "MATHCOUNTS" ], "Problem": "I just did. How about you?", "Solution_1": "1> I have never chamged my sig.\r\n2> Mmmmmm ... chocolate. was mcuh better.\r\n3> Stop creating so many topics.", "Solution_2": "1: I only created three!\r\n2: Ok, I'll change my avatar.\r\n3: Really? uh... ok.", "Solution_3": "i changed my sig recently", "Solution_4": "I changed my signature three times. However i had two out of four of my signatures for less than three days.", "Solution_5": "I saved my old signature, check my \"new\" one out.", "Solution_6": "rather often", "Solution_7": "My first sig was MATHCOUNTS related. \r\nMy second sig was \"Penguin!\". \r\nMy third sig had a link to the MIMC thread. \r\nMy fourth sig is \"Penguin! / (:>)[O]{\".", "Solution_8": "half of them about penguins...", "Solution_9": "i change my sig every two minutes", "Solution_10": "[quote=\"Iversonfan2007\"]i change my sig every two minutes[/quote]\r\n\r\nyou dont have a sig right now :|", "Solution_11": "check his pro\r\n\r\nhe has a sig there" } { "Tag": [ "function", "algebra", "polynomial", "complex analysis", "complex analysis unsolved" ], "Problem": "Show that any proper entire function is a polynomial.\r\n\r\n// A mapping between two topological spaces $X$ and $Y$ is [i]proper[/i] iff the preimage of every compact subset of $Y$ is a compact subset of $X$.", "Solution_1": "Since the preimage of $\\{z\\colon |z|\\le 1\\}$ is compact, the function $1/f(z)$ is bounded near infinity. Therefore, $\\infty$ is a removable singularity for $1/f(z)$. From this it follows that $f$ has at most polynomial growth at $\\infty$, which by the Cauchy estimates implies that $f$ is a polynomial." } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Find all function continous \r\n$ f (0;\\infty ) \\minus{} > (0;\\infty )$\r\nsuch that:$ f(2x) \\equal{} 2f(x)$\r\n$ f((f(x))^2) \\equal{} xf(x)$\r\n$ f(x)\\in N^*$ if $ x\\in N^*$", "Solution_1": "[quote=\"Allnames\"]Find all function continous \n$ f (0;\\infty ) \\minus{} > (0;\\infty )$\nsuch that:$ f(2x) \\equal{} 2f(x)$\n$ f((f(x))^2) \\equal{} xf(x)$\n$ f(x)\\in N^*$ if $ x\\in N^*$[/quote]\r\nThe two first conditions are enough to show that $ f(x)\\equal{}x$ :\r\n\r\nLet $ f(x)$ a continuous function $ (0,\\plus{}\\infty)\\to(0,\\plus{}\\infty)$ such that :\r\n$ f(f(x)^2)\\equal{}xf(x)$\r\n$ f(2x)\\equal{}2f(x)$\r\n\r\nNotice that $ f(f^2(x))\\equal{}xf(x)$ implies $ x\\equal{}\\frac{f(f^2(x))}{f(x)}$ and so $ f(x)$ is a bijection, and so monotonous, and so strictly increasing (since $ f(2x)>f(x)$).\r\n\r\nLet $ A\\equal{}\\{x$ such that $ f(x)\\equal{}x\\}$\r\n\r\nIf $ A$ is dense in $ (0,\\plus{}\\infty)$, continuity implies $ f(x)\\equal{}x$ $ \\forall x\\in(0,\\plus{}\\infty)$\r\nIf $ A$ is not dense in $ (0,\\plus{}\\infty)$, let $ 0x_0$, then $ f^2(x_0)>x_0f(x_0)\\equal{}f(f^2(x_0))$ and so $ f(y_0)y_0$ for some $ y_0\\equal{}f^2(x_0)$ and so $ \\exists z$ such that $ f(z1)\\equal{}z$ (since $ f(x)$ is continuous).\r\n\r\nThen $ f(2x)\\equal{}2f(x)$ implies that $ f(2^kz)\\equal{}2^kz$ $ \\forall k\\in\\mathbb Z$\r\nSo $ A$ is not empty and contains elements as little as we want ($ k\\to \\minus{}\\infty$) and as great as we want ($ k\\to \\plus{}\\infty$).\r\n\r\nLet then $ a\\equal{}\\sup\\{x\\leq u$ such that $ f(x)\\equal{}x\\}$. \"$ a$\" exists since we can have in A elements as little as we want.\r\nLet then $ b\\equal{}\\inf\\{x\\geq v$ such that $ f(x)\\equal{}x\\}$. \"$ b$\" exists since we can have in A elements as great as we want.\r\nSince $ f(x)$ is continuous, $ f(a)\\equal{}a$ and $ f(v)\\equal{}v$\r\n\r\nSo we now have an interval $ [a,b]$ such that $ a\\leq ux$ in $ (a,b)$ or $ f(x)>x$ in $ (a^2,b^2)$ if $ f(x)x$ in $ (a,b)$ (else consider $ a'\\equal{}a^2$ and $ b'\\equal{}b^2$).\r\n\r\nWe then know that $ f(x)x$ $ \\forall x\\in(a^4,b^4)$, so $ f(x)c$ and so that $ f(2^nc)>2^nc$. So, if we find two integers $ n$ and $ k$ such that $ a^{2^{2k\\plus{}1}}<2^nc 1$\r\n\r\nSo, as soon as $ \\frac{2^{2k\\plus{}1}(\\ln(b)\\minus{}\\ln(a))}{\\ln(2)}> 1$ and we just have to take $ k$ great enough.\r\n\r\nSo, $ A$ not dense in $ (0,\\plus{}\\infty)$ implies a contradiction.\r\n\r\nSo the only solution is $ f(x)\\equal{}x$ $ \\forall x\\in(0,\\plus{}\\infty)$\r\n\r\nAnd it's immediate to check that this function matches all the requirements of the initial problem." } { "Tag": [ "modular arithmetic" ], "Problem": "Find all positive integers $ n$ such that both $ n$ and $ n^2\\plus{}8$ are primes.", "Solution_1": "$ x^2$ is either congruent to $ 0$ or $ 1$(mod $ 3$). If it is congruent to $ 1$(mod $ 3$), $ x^2\\plus{}8$ is divisible by $ 3$, which is impossible. Hence $ x$ is divisible by $ 3$, and since $ x$ is prime, $ x\\equal{}3$, $ x^2\\plus{}8\\equal{}17$. So $ x\\equal{}3$ is the only solution.", "Solution_2": "what about the case $ x^{2}\\equiv 2(mod3)$", "Solution_3": "$ x^2\\equiv 0,1\\pmod 3$ for all $ x\\in\\mathbb{Z}$.\r\n\r\nProof:\r\nWorking in $ \\mathbb{Z}/3\\mathbb{Z}$:\r\n$ x\\equal{}1,2$.\r\nIf $ x\\equal{}1$, $ x^2\\equal{}1$.\r\nIf $ x\\equal{}2$, $ x^2\\equal{}4$.\r\n$ 4\\equal{}1$, so there's no case where $ x^2\\equal{}2$.", "Solution_4": "Hey guys,\r\n\r\nCan n=11? If you plug that in, you get 11 and 129. I know 11's prime, but I'm pretty sure 129 is also prime. Therefore, x=3 is NOT the only solution. Can someone write a complete solution for this which finds ALL numbers, because 11 works also.\r\n\r\nThanks,\r\n\r\nrts2007", "Solution_5": "Um... 1+2+9=12. 129 is divisible by 3.", "Solution_6": "If $ 3 \\nmid n$ then $ 3 \\mid n^2\\plus{}8$ $ \\Rightarrow 3 \\mid n \\Rightarrow n\\equal{}3$", "Solution_7": "OH,\r\n\r\nThanks grn_turtle. Oh wow. I can't believe I made such a stupid mistake like that, overlooking 3. \r\n\r\n-rts2007" } { "Tag": [ "ratio" ], "Problem": "If the equal sides of an isosceles right triangle are $2^x$ and the hypoteneuse is $2^{2005}$, then $x=?$", "Solution_1": "[hide]By definition, an isosceles right triangle (45-45-90) has a ratio of side lengths of: $1: 1: \\sqrt{2}$. Since the hypoteneuse has length $2^{2005}$, the equal side lengths must be:\n\n$\\frac{2^{2005}}{\\sqrt2}=\\frac{2^{2005}}{2^{0.5}}=2^{2005-0.5}=2^{2004.5}$\n\nTherefore, if expressed as: $2^x$, $x$ must equal: $\\boxed {2004.5}$ or $\\frac{4009}{2}$[/hide]" } { "Tag": [ "function", "LaTeX", "algebra unsolved", "algebra" ], "Problem": "Determine all the strictly increasing functions $ f\\colon \\mathbb{R}\\to \\mathbb{R}$ which satisfy\r\n\r\n$ f(x\\plus{}f(y))\\equal{}f(x\\plus{}y)\\plus{}2009$,\r\n for any reals $ x,y$.", "Solution_1": "put in x such that x = -f(y), we get f(0) - 2009 = const = f(-f(y) + y), f is injection since f if strictly monotonic, so there exist a constant a such that a = -f(y) + y (*)\r\n\r\nnow we get f(y) = y - a\r\n\r\nputting in x = y = 0, we get f(f(0)) = f(0) + 2009, now using (*) , -2a = -a + 2009, a = -2009,\r\nso f(x) = x + 2009, putting back in starting relation we see that this function is the soluton", "Solution_2": "Sorry, can you write this in LATEX?", "Solution_3": "[quote=\"felixx\"]put in x such that x = -f(y) [/quote]\r\n\r\nThis step throws out any potentially non-surjective solutions. In other words you've only proven that $ f(x) \\equal{} x \\plus{} 2009$ for $ x\\in f(\\mathbb{R})$. In order to make the argument solid, you'd need to show $ f$ is surjective.", "Solution_4": "[quote=\"JoeBlow\"][quote=\"felixx\"]put in x such that x = -f(y) [/quote]\n\nThis step throws out any potentially non-surjective solutions.[/quote]\r\n\r\nCan you explain it? I think putting that into an equation does not need surjectivity, because here x is arbitrary. :huh:", "Solution_5": "[quote=\"Inequalities Master\"]Determine all the strictly increasing functions $ f\\colon \\mathbb{R}\\to \\mathbb{R}$ which satisfy\n\n$ f(x \\plus{} f(y)) \\equal{} f(x \\plus{} y) \\plus{} 2009$,\n for any reals $ x,y$.[/quote]\r\nFrom hypothesis we have $ f(x\\plus{}f(y))\\equal{}f(y\\plus{}f(x))\\forall x,y\\in\\mathbb{R}$ therefore $ x\\plus{}f(y)\\equal{}y\\plus{}f(x)\\forall x,y\\in\\mathbb{R}$ because $ f$ is strictly increasing, so $ f(x)\\equal{}x\\plus{}C\\forall x\\in\\mathbb{R}$. Now, check original identity. :)", "Solution_6": "[quote=\"polskimisiek\"][quote=\"JoeBlow\"][quote=\"felixx\"]put in x such that x = -f(y) [/quote]\n\nThis step throws out any potentially non-surjective solutions.[/quote]\n\nCan you explain it? I think putting that into an equation does not need surjectivity, because here x is arbitrary. :huh:[/quote]\r\nYes, he is wrong! I think so, too. :lol:" } { "Tag": [ "geometry" ], "Problem": "Juanita called her friend long-distance and forgot the number, except for the area code. She did remember that when she dialed the last seven digits, she dialed the digits successively from columns #$ 1$, #$ 2$, #$ 3$, #$ 1$, #$ 2$, #$ 3$, and #$ 1$. Find the number of possible phone numbers Juanita could have dialed with this dialing pattern.\n[asy]draw((0,0)--(2.6,0)--(2.6,4)--(0,4)--(0,0));\nlabel(\" 1\",(0,3.5),E);\nlabel(\" 2\",(1,3.5),E);\nlabel(\" 3\",(2,3.5),E);\nlabel(\" 4\",(0,2.5),E);\nlabel(\" 5\",(1,2.5),E);\nlabel(\" 6\",(2,2.5),E);\nlabel(\" 7\",(0,1.5),E);\nlabel(\" 8\",(1,1.5),E);\nlabel(\" 9\",(2,1.5),E);\nlabel(\" 0\",(1,0.5),E);[/asy]", "Solution_1": "There are 3 ways to pick the first number,4 the 2nd,3,3,4,3,3\r\n\r\n3x4x3x3x4x3x3=36x36x3=3x1296=3888" } { "Tag": [ "calculus", "integration", "search", "number theory unsolved", "number theory" ], "Problem": "Consquaring the equation $x^2+y^2+z^2+t^2-Nxyzt-N=0(*)$ where $N$ is known natural number .\r\na)Show that:there exists many infinitly natural number $N$ such that (*) has root. \r\nb)Supposing $N=4^k(8m+7)$ where $k,m$ be non-negative integers.Show that: for which (*) has no positive integeral root .", "Solution_1": "Markov style again yea \r\nHmmm you should search for this, it's post by Mr [b]namdung[/b]" } { "Tag": [ "pigeonhole principle", "combinatorics unsolved", "combinatorics" ], "Problem": "Prove that if there are 101 people of different heights standing in a line, it is possible to find 11 (not necessarily consecutive) people in the order they are standing in the line with heights that are either increasing or decreasing.", "Solution_1": "This is Erdos-Szekeres with $ m \\equal{} n \\equal{} 10$.", "Solution_2": "I have found this page on the theorem you mentioned\r\n\r\nhttp://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Szekeres_theorem\r\n\r\nbut how did you determine that r = s = 10 (or m = n = 10)? Thank you.", "Solution_3": "Actually, we have $ r \\equal{} s \\equal{} 11$. Note that $ rs \\minus{} r \\minus{} s \\plus{} 2 \\equal{} (r \\minus{} 1)(s \\minus{} 1) \\plus{} 1 \\equal{} 101$ in this case.\r\n\r\n(The statement that JBL is thinking of has $ m \\equal{} r \\minus{} 1, n \\equal{} s \\minus{} 1$, in which case it can be stated suggestively as follows: a sequence of length $ mn \\plus{} 1$ contains either a monotonically increasing subsequence of length $ m \\plus{} 1$ or a monotonically decreasing subsequence of length $ n \\plus{} 1$. I say suggestive because the connection to the Pigeonhole principle is more obvious this way.)", "Solution_4": "As t0r noted, I use the formulation at [url=http://mathworld.wolfram.com/Erdos-SzekeresTheorem.html]mathworld[/url], not the wikipedia one, and that explains the off-by-one error." } { "Tag": [ "function", "calculus", "derivative", "calculus computations" ], "Problem": "The coefficient $ \\beta$ of thermal expansion of a liquid relates the change in the volume V (in $ m^3$) of a fixed quantity of a liquid to an increase in its temperature T (in \u00b0 C) as shown below.\r\n\r\n$ dV \\equal{} \\beta V dT$\r\n\r\nLet $ \\rho$ be the density (in kg/$ m^3$) of water as a function of temperature. Write an expression for $ d\\rho$ in terms of $ \\rho$ and dT", "Solution_1": "hello\r\n$ \\rho \\equal{} \\frac {M}{V}$\r\n$ \\frac {d\\rho}{dt} \\equal{} \\minus{} \\frac {M}{V^2}\\beta V$\r\n$ \\frac {d\\rho}{dt} \\equal{} \\minus{} \\rho\\beta$\r\n$ {d\\rho} \\equal{} \\minus{} \\rho\\beta dt$\r\nthank u", "Solution_2": "Ok thanks for the reply. What if I have a graph that shows the density of water as a function of temperature. \r\n\r\nHow can I use it to estimate $ \\beta$ when T = 40", "Solution_3": "hello\r\nif u have graph i.e. u have the derivative .\r\nfrom the relation u can find $ \\rho$ at the given $ T$.\r\nput the values u will get $ \\beta$\r\nthank u", "Solution_4": "[quote=\"kabi\"]hello\nif u have graph i.e. u have the derivative .\nfrom the relation u can find $ \\rho$ at the given $ T$.\nput the values u will get $ \\beta$\nthank u[/quote]\r\n\r\nOk at T = 40 I got $ \\rho$ to be somewhere near 992, so what do I do now?\r\n\r\nyou said to put the value, but to which equation?" } { "Tag": [ "linear algebra", "matrix" ], "Problem": "We are given all the eigenvalues and eigenvectors of a $n\\times n$ matrix $A$ , and we are also given another $n\\times n$ matrix $Q$ . If matrix $B$ is defined by \r\n\r\n$B=QAQ^{-1}$ \r\n\r\nhow can we find the eigenvalues of $B$ without evaluating the matrix $B$ ?", "Solution_1": "Eigenvalues of of similiar matrices are the same. :)", "Solution_2": "[quote=\"Omid Hatami\"]Eigenvalues of of similiar matrices are the same. :)[/quote]\r\n\r\nso how we can prove that similar matrices have same eigenvalue ?", "Solution_3": "First of all look at [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=87254]this topic[/url].\r\nThen you see that $\\det(QAQ^{-1}-xI)=\\det(QAQ^{-1}-Q(xI)Q^{-1})$ that is equal to $\\det(Q)\\det(Q^{-1})\\det(A-xI)=\\det(A-xI)$", "Solution_4": "More directly, if $Av=\\lambda v$, $QAQ^{-1}(Qv)=QAv=\\lambda (Qv)$. That matches every eigenvector of $A$ with an eigenvector of $QAQ^{-1}$ with the same eigenvalue.", "Solution_5": "Yes but it does not prove that the multiciplity of eigenvalues are equal." } { "Tag": [ "MATHCOUNTS" ], "Problem": "I'm taking the state mathcount test next week and I would like to know the scores from previous years to get a feel on what I need to attain to try and reach nationals. Thanks.", "Solution_1": "[quote=\"Snailboto\"]I'm taking the state mathcount test next week and I would like to know the scores from previous years to get a feel on what I need to attain to try and reach nationals. Thanks.[/quote]\r\n\r\nokay, there are already like 5 forums asking about scores. look around before you post. and maybe you should post this in the CT forum.\r\n\r\ncan a mod merge some of these or something." } { "Tag": [ "geometry", "circumcircle", "trigonometry", "LaTeX", "angle bisector" ], "Problem": "(8) AB and CD are chords of a circle radius 4 on opposite sides of the centre. CB and DA produced meet at P outside the circle. THe length of arc CD is four times the length of arc AB. THe length of arc CD is 8(pi)/5. Find GIF compiler doesn't work today, so a formula can be viewed only if it was already compiled into a GIF before.\r\n\r\n Darij", "Solution_6": "For number 8 I got a different answer:\n\n[hide]The question asked for CB and AD to be produced at P right? That means to extend them to where they intersect, which is P right? Well, if that's true, the answer would be 27 degrees I think. We can find arc CD has measure of 72 degrees, and arc AB is 18 degrees, so (72-18)/2 gives 27 degrees for < APB[/hide]", "Solution_7": "NightFlarer wrote:For number 8 I got a different answer:\n[hide]The question asked for CB and AD to be produced at P right? That means to extend them to where they intersect, which is P right? Well, if that's true, the answer would be 27 degrees I think. We can find arc CD has measure of 72 degrees, and arc AB is 18 degrees, so (72-18)/2 gives 27 degrees for < APB[/hide]\n\n\n\n[tex]\\frac{3}{20}\\pi[/tex] is 27 degrees" } { "Tag": [ "function", "algebra", "polynomial", "ratio" ], "Problem": "\u6570\u5217{bn}\u4e2d\uff0cb1=0\uff0cb2=2\uff0cb3=3\uff0cb n+1=b n-1+b n-2\uff0c\u6c42\u8bc1\uff1a\u5f53p\u4e3a\u8d28\u6570\u65f6\uff0cp|bp.\u800c\u4e14\uff0c\u4e2a\u4eba\u610f\u89c1\u89c9\u5f97\uff0c\u5f53p|bp\u65f6\uff0cp\u4e3a\u8d28\u6570\u3002", "Solution_1": "\u600e\u4e48\u6c42\u901a\u9879? :?", "Solution_2": "\u6570\u5217${b_n}$\u4e2d,$b_1=0,b_2=2,b_3=3,b_{n+1}=b_{n-1}+b_{n-2}$,\u6c42\u8bc1:\u5f53$p$\u4e3a\u8d28\u6570\u65f6,$p|b_p$. \u800c\u4e14,\u4e2a\u4eba\u610f\u89c1\u89c9\u5f97,\u5f53$p|b_p$\u65f6,$p$\u4e3a\u8d28\u6570.", "Solution_3": "\u6211\u4eec\u7528\u6bcd\u51fd\u6570 (generating functions) \u6765\u89e3\u6b64\u9898.\r\n\r\n\r\n\u6211\u58f0\u660e\u5bf9\u4e8e $n>3$, \\[b_n=\\sum_{k\\geq 0} 2\\binom{k}{3k+2-n}+3\\binom{k}{3k+3-n}\\qquad (*).\\]\r\n\r\n\u8fd9\u91cc\u6211\u4eec\u7528\u5e38\u89c4 \\[s>r\\quad\\Rightarrow\\quad\\binom{r}{s}=0, \\qquad s<03)\r\n\r\n(So if you want to find a factor of 1111111 you need to try only primes like 29,..43 ..71.. .... (only 14k+1 types)..).. (before you reach 239 which indeed is a factor of 1111111)\r\n\r\nProof is not hard .", "Solution_6": "Another slight variation of the problem:\r\n\r\nProve that (1111.....1 ( 2003 times) base 16 is composite.", "Solution_7": "[quote=\"Gyan\"]Okay here is a little more chalanging problem:\nProve that any prime factor of 1111111111111 (=((10^13-1)/9) is 1 mod 26.\n(or in general a prime factor of ( 111.... p times) where p is prime is 1 mod 2p.\n[/quote]\r\n\r\nHi,\r\n\r\nI think your statement doesn't hold true for all primes $p$.\r\nFor example, \r\n$p = 2$, $11 \\neq 1 \\pmod{4}$, and\r\n$p = 3$, $111 \\neq 1 \\pmod{6}$.\r\n\r\nBut for other primes $p$, this holds.\r\n\r\nProof for primes $p \\neq 2,3$ :\r\n\r\n$10^p - 1 = (10 - 1) (10^{p-1} + 10^{p-2} + ........ + 10^2 + 10 + 1)$ ---------- (1)\r\n\r\nWe know, for prime $p$, $10^p \\equiv 10 \\pmod{p}$, and so, $10^p - 1 \\equiv 9 \\pmod{p}$.\r\n\r\nAnd in RHS of (1), (10-1)=9, so (1) reduced modulo prime $p$ gives\r\n$9 \\equiv 9 * x \\pmod{p}$, ----- (2)\r\nwhere $(10^{p-1} + 10^{p-2} + ........ + 10^2 + 10 + 1) \\equiv x \\pmod{p}$.\r\n\r\nThe only solution to (2) is $x = 1$ ($x 3.[/b]\r\n\r\nBut I could not follow your proof. I don't see that you have proved it for [b]all[/b] factors of the number. \r\n\r\nthat is, for example 11111 = 41*271 .. both 41 and 271 are 1 mod 10 (=2*5)\r\nand factors of 1111111 = 239*4649 both 239 and 4649 (and of course 1111111) are 1 mod 14.\r\n\r\nFor the second part a hint: All such numbers (1111.. 'n' times) based 16 , except n= 2, ( the number 17) in decimal notation) are not prime... :D", "Solution_10": "Hi Gyan,\n\n\n\nSorry about the carelessness. I guess I didn't read the question carefully, and hence thought that its asked to prove \"1111...11(p times) is 1 mod 2p\".\n\n\n\nNow, I will try to prove that all prime factors of 111...11(p times) are 1 mod 2p.\n\n[hide]\n\nLet q be a prime factor of \n\n111...11(p times) = (10^p - 1) / 9 = 10^(p-1) + 10^(p-2) + ....... + 10 + 1\n\n\n\nSince, RHS is not divisible by 2 or 5, so q isn't 2 or 5. Hence, gcd(q,10)=1.\n\nFurther, p isn't 3 as 111 is not 1 (mod 6); this implies 111..11(p times) is not a multiple of 3, and hence q is not 3.\n\n\n\nNow, q divides (10^p - 1). But we know 10^(q-1) = 1 (mod q) (Fermat's Little Theorem).\n\nSo, the order of 10 mod q is a common factor of both p and (q-1). Since, only factors of p are 1 and p, so order of 10 mod q is either 1 or p.\n\n\n\nNow, 10^1 = 1 ( mod q) implies q = 3 (which is disproved before)\n\nSo, order of 10 mod q = p, implies p divides (q-1).\n\n\n\nSo, we conclude that p divides (q-1). Since, q isn't 2, so, 2 divides (q-1) too.\n\nSince, p and 2 are coprime, and as both divide (q-1), \n\ntherefore 2p divides (q-1), or in other words, q is 1 mod 2p.\n\n[/hide]\n\n******************\n\nSambit", "Solution_11": "Gyan wrote:Another slight variation of the problem:\n\nProve that (1111.....1 ( 2003 times) base 16 is composite.\n\n\n\nHint if any one is interested \n\n\n\n[hide] 16 is 4^2. \n\nand 16^n -1 = (4^n+1)(4^n-1) \n\n\n\nand if n>2, the required number (10^n-1)/15 will prduce a number >1, which is not prime . [/hide]", "Solution_12": "gyan, that doesn't matter.\r\n\r\n16^2004 = 1 mod 15, that solves the problem on it's own: we got (15k+1-1)/15 is always a natural number...\r\n\r\nit doesn't even matter base what it is, if it's just sufficiently high. (so that k is not 0 or 1)", "Solution_13": "Huh????\r\n\r\n(Are you sure you are seeing the problem correctly?) \r\n\r\nit is not about (16^n-1) but about (16^n-1)/15 )\r\n\r\nPoint is to prove [ (16^n-1)/15], itself, is composite as long as n>2\r\n\r\nIn other words, what the above hint showed was: \r\n the number ((16^n-1)/15) is divisible by 17 (if n is even) or ((4^n+1)/5 ) (if n is odd) - note that ((4^n+1)/5 is an iteger if n is odd)). \r\n\r\nFor other bases, one can not prove similar results for ALL n's...\r\n(If base is a not a perfect square (eg 16) .. you may find prime numbers of the type 1111111.... Trick was to realize that 16 was a perfect square)\r\n\r\n[ eg in base 10, not only 11, but (10^23)/9 is also prime)\r\n\r\nHope that helps." } { "Tag": [ "calculus", "calculus computations" ], "Problem": "If $ \\displaystyle\\sum_{n\\equal{}1}^\\infty 1/n^2 \\equal{} s$. ($ s\\equal{}\\frac{\\pi^2}{6}$), prove that \r\n$ \\displaystyle\\sum_{n\\equal{}1}^\\infty (\\minus{}1)^{n\\minus{}1}/n^2 \\equal{}\\frac{1}{2} s$", "Solution_1": "$ \\displaystyle\\sum_{n\\equal{}1}^\\infty (\\minus{}1)^{n\\minus{}1}/n^2 \\equal{} \\sum_{n\\equal{}1}^\\infty 1/n^2 \\minus{} 2\\sum_{n\\equal{}1}^\\infty 1/(2n)^2 \\equal{} \\sum_{n\\equal{}1}^\\infty 1/n^2 \\minus{} \\frac{1}{2}\\sum_{n\\equal{}1}^\\infty 1/n^2 \\equal{} s \\minus{}\\frac{1}{2} s \\equal{} s/2$" } { "Tag": [ "inequalities" ], "Problem": "Let $a,b,c$ be sides of a triangle. Prove the following inequality:\r\n\r\n$(a+b+c)(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}) \\geq 6(\\frac{a}{b+c}+\\frac{b}{c+a}+\\frac{c}{a+b})$", "Solution_1": "[quote=\"pohoatza\"]Let $a,b,c$ be sides of a triangle. Prove the following inequality:\n\n$(a+b+c)(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}) \\geq 6(\\frac{a}{b+c}+\\frac{b}{c+a}+\\frac{c}{a+b})$[/quote]\r\nNice! :lol: \r\nIt's not true for all $a>0,$ $b>0$ and $c>0.$ :wink:", "Solution_2": "It works when $a=b=c=1$\r\nthe sign should be opposite", "Solution_3": "Whoa what's wrong with it" } { "Tag": [ "search", "\\/closed" ], "Problem": "The cache seems to have corrupted again, or whatever the cause was. :o", "Solution_1": "Hmm, I don't see it.", "Solution_2": "It's gone now, but it was there nevertheless.", "Solution_3": "[quote=\"Temperal\"]It's gone now, but it was there nevertheless.[/quote]We usually fix it when we see it, but sometimes none of us (me or Richard) are online so it takes a while to get fixed. Nevertheless please report it every time you see it.", "Solution_4": "OH GOD, IT'S BACK.\r\n\r\nCan't you permanently do something about it? It's quite annoying not being able to use the \"view new posts\" button.", "Solution_5": "I didn't see it at the time of the first post, but I do now.", "Solution_6": "[quote=\"i_like_pie\"]OH GOD, IT'S BACK.\n\nCan't you permanently do something about it? It's quite annoying not being able to use the \"view new posts\" button.[/quote]So it appears when you guys click on the view new posts / ego search buttons? Might be a clue as where the bug starts occurring if that is the case.", "Solution_7": "[quote=\"Valentin Vornicu\"][quote=\"i_like_pie\"]OH GOD, IT'S BACK.\n\nCan't you permanently do something about it? It's quite annoying not being able to use the \"view new posts\" button.[/quote]So it appears when you guys click on the view new posts / ego search buttons? Might be a clue as where the bug starts occurring if that is the case.[/quote]\r\n\r\nYes; it's a blank page with nothing but it on it, and then when we return to Mathlinks, then we find we're logged out.", "Solution_8": "Back again...\r\n(just opened http://www.mathlinks.ro/Forum/index.php )", "Solution_9": "I see it too. AoPS skin, Firefox 2, first noticed at 3:21 PM EST.\r\n\r\nedit: by the way, it appears at the top of the Topic Review when I'm posting replies, as well.", "Solution_10": "I've managed to trace back the problem to a single point, however it's still not clear what is causing it. A chain of events that I have yet been able to reproduce. Keep reporting when you see it.", "Solution_11": "[quote=\"Xevarion\"]\nedit: by the way, it appears at the top of the Topic Review when I'm posting replies, as well.[/quote]\r\n\r\nThis is because topic review is just an iframe of another page on the AoPS forums server. To be specific, posting.php?mode=topicreview&t=TOPIC_NUMBER" } { "Tag": [ "function", "topology", "real analysis", "real analysis unsolved" ], "Problem": "Let $U$ be an open subset of $C$. Prove that there exists a closed discrete subset $S$ of $U$ whose closure in $C$ covers the frontier of $U$. From here it is easy to see that there is an analytic map on $U$ which cannot be extended in a larger open set. Is there any other approach?", "Solution_1": "Consider all sets $E\\subset U$ with the following property: $|x-y|\\ge \\max\\{\\mathrm{dist}(x,\\partial U),\\mathrm{dist}(y,\\partial U)\\}$ for all $x,y\\in E$. Order them by inclusion. Let $S$ be any maximal element with respect to this order. \r\n\r\nI don't know of any other approach. We need an analytic function on $U$ which behaves in some \"non-analytic\" way near every point of the boundary. Accumulation of singularities of some kind (like zeros) is the most natural way to achieve this.", "Solution_2": "[quote=\"harazi\"] From here it is easy to see that there is an analytic map on $U$ which cannot be extended in a larger open set. Is there any other approach?[/quote] Actually, there are a few alternative approaches but most of them require more effort. For instance, you can consider a dense countable set $\\{z_{j}\\}$ in $\\mathbb C\\setminus U$ and take $f(z)=\\sum_{j}\\frac{c_{j}}{z-z_{j}}$. This, obviously, will be a nice analytic function in $U$, provided that $\\sum_{j}|c_{j}|<+\\infty$. What is much less obvious is that you can always choose the coefficients $c_{k}$ in such a way that $f$ will not be extendable to any larger open set. Can you see how to do it or why not every choice of $c_{j}\\ne 0$ with $\\sum_{j}|c_{j}|<+\\infty$ will work? ;)", "Solution_3": "I didn't make any progress with the construction of such a sequence, even though I guess that $c_{n}=1/5^{n}$ should work. Can you give some details, Fedja? I tried to use Landau's trick for series with nonnegative coefficients, but it didn't work.", "Solution_4": "Here are two possible approaches that you may want to explore:\r\na) Random choice (the idea is that, for fixed $c_{k}$ with $k\\ne j$, there are not too many values of $c_{j}$ such that your function is analytically extendable to an open set containing $c_{j}$)\r\nb) Very fast decaying coefficients (the idea is that the sum of the series will belong to some quasi-analytic class on almost every line and, therefore, if an analytic expansion exists, it must coincide with the sum of the series almost everywhere where it is defined)\r\nI believe that the idea of taking all coefficients positive should work too but I do not see right away how to proceed with it. If I come up with something in this line, I'll post it." } { "Tag": [], "Problem": "i am too much confused betwwen fundamental principal of counting ,permutation and combination.WHAT IS MAIN DIFFERENCCE BETWWEN THEM. \r\npl explain a logical way.", "Solution_1": "In a permutation, order matters. In a combination, order doesn't matter. For example, $ _5C_2$ is different from $ _5P_2.$ The first is $ 10$ while the second is $ 20.$" } { "Tag": [ "algebra", "polynomial" ], "Problem": "Is there anyway to determine when sqrt of a polynomial will give you a perfect square?\r\nFor example, for what values does $\\sqrt{32x^2+16x+1}$ give an integer?\r\nBy trial and error the largest I could find was $x=7$", "Solution_1": "$x=42$ gives $\\sqrt{32x^2+16x+1}=239$ also, according to my calculator :P \r\n\r\nAnyway, I've seen a couple of methods to approach a problem like this, but I don't think there's any method that works for any polynomial...\r\n\r\nI'll show you one way that is often useful (but it doesn't work for this probem.)\r\n[hide]\n[b]Problem[/b]: Find all positive integers $n$ such that $n^2+n+34$ is a perfect square.\n\n[b]Solution[/b]:\nLet $n^2+n+34=(n+k)^2$.\nSolving for $n$, we obtain $n=\\frac{34-k^2}{2k-1}$. ($n^2$ is gone! :) )\n\nNow the problem becomes an integer problem!\n\nSince $n$ has to be positive, $n=\\frac{34-k^2}{2k-1}>0$\nFor positive values of $k$, $2k-1>0$ and thus $34-k^2>0$\nSo $k=1,2,3,4,5$\n\nBy trial and error, we see $k=1,2,3,5$ work, and these give us $n=1,5,10,33$\n\n\nThis method is only useful when the coefficient of $n^2$ is a perfect square. Also, if $n$ can be negative, it's often hard to restrict the possible values of $n$ and $k$.\n[/hide]" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "porve Goldbach's theorem:\r\nl=1/3+1/7+1/8+1/15+1/24+1/26+1/31+1/35+...= \\sum 1/k-1\r\nk \\in p\r\nwhere P is the set of \"perfect power\" defined recursively as follows:\r\np={m ^3 |m \\geq 2, ^3 \\geq ^3 }\r\nm is not in p.", "Solution_1": "I think, author meant following:\r\n\r\nProve Goldbach's equality\r\n\\[1=\\frac{1}{3}+\\frac{1}{7}+\\frac{1}{8}+\\frac{1}{15}+\\frac{1}{24}+\\frac{1}{31}+\\frac{1}{36}+...=\\sum_{k\\in PP}\\frac{1}{k-1},\\]\r\nwhere $PP$ is a set of perfect powers $\\{a^b\\mid a,b\\geq 2\\}$.", "Solution_2": "Ok! Since problem was unsolved for 9 months, I will write solution.\r\n\r\nSuppose $k\\in PP$, then\r\n\\[\\frac{1}{k-1}=\\frac{1}{k}+\\frac{1}{k^2}+\\frac{1}{k^3}+...,\\]\r\nso\r\n\\[\\sum_{k\\in PP}\\frac{1}{k-1}=\\sum_{k\\in PP}\\left( \\frac{1}{k}+\\frac{1}{k^2}+\\frac{1}{k^3}+...\\right).\\]\r\nNote that summand $\\frac{1}{a}$ appears in this sum iff $a\\in PP$, moreover we will meet $\\frac{1}{a}$ so much times how much times $a$ can be represented as a perfect power. Fox example, $\\frac{1}{16}$ appears in sum twice, since $16=2^4=4^2$.\r\nIt follows that whole sum is\r\n\\[\\sum_{n=2}^{\\infty}\\left(\\frac{1}{n^2}+\\frac{1}{n^3}+...\\right)=\\sum_{n=2}^{\\infty}\\frac{1}{n(n-1)}=1.\\]\r\nQED", "Solution_3": "fantastic Myth,now i read the proof it is beautiful :)" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "$\\bigstar$\r\nLet $a,b,c$ be non-negative real numbers with sum $3$. Prove that\r\n\\[a^{2}+b^{2}+c^{2}-\\frac{8}{3}\\max\\{|a-b|,|b-c|,|c-a|\\}\\le 1+2abc.\\]", "Solution_1": "This is a very tight inequality, Hung. Equality occur for $a=b=c=1$ and $a=3,b=c=0$, we can prove it by the same method with the following\r\nProblem (Vo Quoc Ba Can).\r\n\\[\\sum\\sqrt{a+\\frac{(b-c)^{2}}{4}}\\le \\sqrt{3}+\\left( \\frac{2}{\\sqrt{3}}-1\\right) (|a-b|+|b-c|+|c-a|) \\]\r\nif $a,b,c \\ge 0$ and $a+b+c=1$. \r\n:)", "Solution_2": "Yes, you are really clever to realize the case of the equalities. However, my inequality can be proved with a very nice solution.", "Solution_3": "[quote=\"toanhocmuonmau\"]This is a very tight inequality, Hung. Equality occur for $a=b=c=1$ and $a=3,b=c=0$, we can prove it by the same method with the following\nProblem (Vo Quoc Ba Can).\n\\[\\sum\\sqrt{a+\\frac{(b-c)^{2}}{4}}\\le \\sqrt{3}+\\left( \\frac{2}{\\sqrt{3}}-1\\right) (|a-b|+|b-c|+|c-a|) \\]\nif $a,b,c \\ge 0$ and $a+b+c=1$. \n:)[/quote]\r\n\r\nHow do you prove it? Can? I think it is mixing variable?" } { "Tag": [ "function", "limit", "real analysis", "real analysis unsolved" ], "Problem": "Prove that if $ f: \\mathbb{R} - > \\mathbb{R}$ is non-decreasing function, continuous from the right, and if $ x > 0$ it's $ f(x) < x$, then for every ${ x > 0 \\lim_{n\\to\\infty}\\ {f^{n}(x)}} = 0$\r\nwhere we define $ f^{n+1}$ as $ f(f^{n})$", "Solution_1": "What about $ f(x)\\equal{}x\\minus{}1$? \r\n\r\nBut if you assume in addition $ f(0)\\equal{}0$, then the proof goes like this: any decreasing sequence of positive numbers has a limit; the limit $ L$ satisfies $ f(L)\\equal{}L$ by right-continuity; hence $ L\\equal{}0$." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let a,b,c>0. Prove that:\r\n$ \\sum\\frac{x\\plus{}y}{\\sqrt{2z}}\\geq \\sum \\sqrt{x\\plus{}y}$", "Solution_1": "We have \r\n$ \\sum\\frac{x+y}{\\sqrt{2z}}\\geq \\frac{(\\sqrt{a+b}+\\sqrt{b+c}+\\sqrt{a+c})^2}{\\sqrt{2a}+\\sqrt{2b}+\\sqrt{2c}}$\r\nand we should to prove \r\n$ \\sqrt{a+b}+\\sqrt{b+c}+\\sqrt{a+c})\\geq \\sqrt{2a}+\\sqrt{2b}+\\sqrt{2c}$\r\n to finish solution.\r\nSo we have to prove:\r\n $ (\\sqrt{a+b}+\\sqrt{b+c}+\\sqrt{a+c})^2\\geq (\\sqrt{2a}+\\sqrt{2b}+\\sqrt{2c})^2$\r\nor \r\n $ \\sqrt{(a+b)(a+c)}+\\sqrt{(b+c)(a+c)}+\\sqrt{(a+b)(b+c)}\\geq 2\\sqrt{ab}+2\\sqrt{bc}+2\\sqrt{ac}$\r\n which is right, becouse \r\n ${ \\sqrt{(a+b)(a+c)}\\geq 2\\sqrt{ab}}+2\\sqrt{ac}$\r\n ${ \\sqrt{(b+c)(a+c)}\\geq 2\\sqrt{bc}}+2\\sqrt{ac}$\r\n ${ \\sqrt{(a+b)(b+c)}\\geq 2\\sqrt{ab}}+2\\sqrt{bc}$\r\nis right." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Let $p\\ge 7$ be a prime number and $a\\neq 0, b, c$ are natural numbers from $1$ to $9$. Prove that the number $\\overline{\\underbrace{aa\\cdot\\cdot\\cdot a}_{p-1}\\underbrace{bb\\cdot\\cdot\\cdot b}_{p-1}\\underbrace{cc\\cdot\\cdot\\cdot c}_{p-1}}$ is devisible by $p\\cdot\\overline{a\\underbrace{00\\cdot\\cdot\\cdot 0}_{p-2}b\\underbrace{00\\cdot\\cdot\\cdot 0}_{p-2}}$.", "Solution_1": "huh? if 1<=c<=9, how can this happen? The first number is not even divisible by 10...huh???" } { "Tag": [ "inequalities", "geometry", "incenter", "circumcircle", "rectangle", "trapezoid", "3D geometry" ], "Problem": "Does somebody have any JBMO shortlists from past years? If somebody has any, please share them.", "Solution_1": "Yes, I want also to see the Shortlist pf JBMO 2007", "Solution_2": "If nobody has got any shortlists, can anyone tell me which books are good for prepairing for the JBMO and where can i find them?", "Solution_3": "[size=150][color=darkblue]Shortlist[/color] [color=darkred][b]JBMO 2002[/b][/color][/size]\r\n\r\n[color=darkred][b]1.[/b][/color] [color=darkblue]A student is playing computer. Computer shows randomly $ 2002$ positive numbers. Game's rules let do the following operations\n- to take $ 2$ numbers from these, to double first one, to add the second one and to save the sum.\n- to take another $ 2$ numbers from the remainder numbers, to double the first one, to add the second one, to multiply this sum with previous and to save the result.\n- to repeat this procedure, until all the $ 2002$ numbers won't be used.\nStudent wins the game if final product is maximum possible.\nFind the winning strategy and prove it.[/color]\r\n\r\n\r\n[b][color=darkred]2.[/color][/b] [color=darkblue]Positive real numbers are arranged in the form:\n$ 1 \\ \\ \\ 3 \\ \\ \\ 6 \\ \\ \\ 10 \\ \\ \\ 15 \\ \\ \\ ...$\n$ 2 \\ \\ \\ 5 \\ \\ \\ 9 \\ \\ \\ 14 \\ \\ \\ ...$\n$ 4 \\ \\ \\ 8 \\ \\ 13 \\ \\ \\ ...$\n$ 7 \\ \\ 12 \\ \\ \\ ...$\n$ 11 \\ \\ \\ ...$\nFind the number of the line and column where the number $ 2002$ stays.[/color]\r\n\r\n[b][color=darkred]3.[/color][/b] [color=darkblue]Let $ a,b,c$ be positive real numbers such that $ abc \\equal{} \\frac {9}{4}$. Prove the inequality:\n\\[ a^3 \\plus{} b^3 \\plus{} c^3 > a\\sqrt {b \\plus{} c} \\plus{} b\\sqrt {c \\plus{} a} \\plus{} c\\sqrt {a \\plus{} b}\n\\]\n[/color]\r\n\r\n[b][color=darkred]4.[/color][/b] [color=darkblue](Jury's variant) Let $ a,b,c$ be positive real numbers such that $ abc \\equal{} 2$. Prove the inequality:\n\\[ a^3 \\plus{} b^3 \\plus{} c^3 > a\\sqrt {b \\plus{} c} \\plus{} b\\sqrt {c \\plus{} a} \\plus{} c\\sqrt {a \\plus{} b}\n\\]\n[/color]\r\n\r\n[b][color=darkred]5.[/color][/b] [color=darkblue]Let $ a,b,c$ be positive real numbers. Prove the inequality:\n\\[ \\frac {a^3}{b^2} \\plus{} \\frac {b^3}{c^2} \\plus{} \\frac {c^3}{a^2}\\ge \\frac {a^2}{b} \\plus{} \\frac {b^2}{c} \\plus{} \\frac {c^2}{a}\n\\]\n[/color]\r\n\r\n[b][color=darkred]6.[/color][/b] [color=darkblue]Let $ a_1,a_2,...,a_6$ be real numbers such that:\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\n\\]\nProve that $ a_4a_6\\le a_5^2$. When does equality holds?[/color]\r\n\r\n[b][color=darkred]7.[/color][/b] [color=darkblue]Consider integers $ a_i, i \\equal{} \\overline{1,2002}$ such that\n\\[ a_1^{ \\minus{} 3} \\plus{} a_2^{ \\minus{} 3} \\plus{} \\ldots \\plus{} a_{2002}^{ \\minus{} 3} \\equal{} \\frac {1}{2}\n\\]\n[/color]\r\n\r\n[b][color=darkred]8.[/color][/b] [color=darkblue]Let $ ABC$ be a triangle with centroid $ G$ and $ A_1,B_1,C_1$ midpoints of the sides $ BC,CA,AB$. A paralel through $ A_1$ to $ BB_1$ intersects $ B_1C_1$ at $ F$ prove that triangles $ ABC$ and $ FA_1A$ are similiar if and only if quadrilateral $ AB_1GC_1$ is cyclic.[/color]\r\n\r\n[b][color=darkred]9.[/color][/b] [color=darkblue]In triangle $ ABC$ $ H,I,O$ are orthocenter, incenter and circumcenter, respectively. $ CI$ cuts circumcircle at $ L$. IF $ AB \\equal{} IL$ and $ AH \\equal{} OH$, find angles of triangle $ ABC$.[/color]\r\n\r\nI'll finish later...", "Solution_4": "[quote=\"Ahiles\"]I'll finish later...[/quote]\r\n\r\nIt's been a month past since you said this. Can you please finish?", "Solution_5": "[quote=\"Bugi\"]\nIt's been a month past since you said this. Can you please finish?[/quote]\r\nOK..\r\n [b][color=darkred]10.[/color][/b] [color=darkblue]Let $ ABC$ be a triangle with area $ S$ and points $ D,E,F$ points on the sides $ BC,CA,AB$. Perpendiculars at points $ D,E,F$ to the $ BC,CA,AB$ cut circumcircle of the triangle $ ABC$ at points $ (D_1,D_2), (E_1,E_2), (F_1,F_2)$. Prove that:\n\\[ |D_1B\\cdot D_1C \\minus{} D_2B\\cdot D_2C| \\plus{} |E_1A\\cdot E_1C \\minus{} E_2A\\cdot E_2C| \\plus{} |F_1B\\cdot F_1A \\minus{} F_2B\\cdot F_2A| > 4S\n\\]\n[/color] [b][color=darkred]11.[/color][/b] [color=darkblue]Let $ ABC$ be an isosceles triangle with $ AB \\equal{} AC$ and $ \\angle{A} \\equal{} 20^\\circ$. On the side $ AC$ consider point $ D$ such that $ AD \\equal{} BC$. Find $ \\angle{BDC}$.[/color]\r\n [b][color=darkred]12.[/color][/b] [color=darkblue]Let $ ABCD$ be a convex quadrilateral with $ AB \\equal{} AD$ and $ BC \\equal{} CD$. On the sides $ AB,BC,CD,CA$ we consider points $ K,L,L_1,K_1$ such that quadrilateral $ KLL_1K$ is rectangle. Then consider rectangles $ MNPQ$ inscribed in the triangle $ BLK$, where $ M\\in KB, N \\in BL, P,Q \\in LK$ and $ M_1N_1P_1Q_1$ inscribed in triangle $ DK_1L_1$ where $ P_1$ and $ Q_1$ are situated on the $ L_1K_1$, $ M_1$ on the $ DK_1$ and $ N_1$ on the $ DL_1$. Let $ S,S_1,S_2,S_3$ be the areas of the $ ABCD,KLL_1K_1,MNPQ,M_1N_1P_1Q_1$ respectively. Find the maximum possible valu of the expression:\n\\[ \\frac {S_1 \\plus{} S_2 \\plus{} S_3}{S}\n\\]\n[/color] [b][color=darkred]13.[/color][/b] [color=darkblue]Let $ A_1,A_2,...A_{2002}$ be the arbitrary points in the plane. Prove that for every circle of the radius $ 1$ and for every rectangle inscribed in this circle, exist 3 vertexes $ M,N,P$ of the rectangle such that:\n\\[ MA_1 \\plus{} ... \\plus{} MA_{2002} \\plus{} NA_1 \\plus{} ... \\plus{} NA_{2002} \\plus{} PA_1 \\plus{} ... \\plus{} PA_{2002}\\ge 6006\n\\]\n[/color]\r\n\r\nThat's all... :)", "Solution_6": "Problem 9 has appeared at a Brazilian's MO second round. :P", "Solution_7": "Dear Ahiles,\r\n\r\nWhat is the question for problem 7?\r\n\r\nIs it JBMO Shortlist 2002 or 2007. BTW, could you plz post another jbmo shortlists ?", "Solution_8": "It's the shortlist for 2002, Ahiles wrote that. I have seen problem 7 and the question is: Prove that at least 3 of the numbers are equal.I would also like to see some more shortlists.", "Solution_9": "Okey thank you, and here is PDF file of JBMO 2002 :).", "Solution_10": "[size=150][color=darkred]JBMO Shortlist 2006[/color][/size]\r\n\r\n [b][color=darkred]1.[/color][/b] [color=darkblue] For an acute triangle $ ABC$ prove the inequality\n\\[ \\frac {m_a^2}{ - a^2 + b^2 + c^2} + \\frac {m_b^2}{a^2 - b^2 + c^2} + \\frac {m_c^2}{a^2 + b^2 - c^2}\\ge \\frac {9}{4}\n\\]\nwhere $ m_a,m_b,m_c$ represent the lengths of the corresponding medians.\n[/color] \r\n\r\n [b][color=darkred]2.[/color][/b] [color=darkblue] Let $ x,y,z$ be positive real numbers such that $ x + 2y + 3z = \\frac {11}{12}$. Prove the inequality\n\\[ 6(3xy + 4xz + 2yz) + 6x + 3y + 4z + 72xyz\\le \\frac {107}{18}\n\\]\nWhen does equality occur?\n[/color] \r\n\r\n [b][color=darkred]3.[/color][/b] [color=darkblue] Let $ n\\ge3$ be a natural number. A set of real numbers $ \\{x_1,x_2,...x_n\\}$ is called $ summable$ if\n\\[ \\frac {1}{x_1} + \\frac {1}{x_2} + ... + \\frac {1}{x_n} = 1.\n\\]\nProve that for every $ n\\ge 3$ there always exists a $ summable$ set, which consits of $ n$ elements, such that the biggest element is \n\n$ a)$ bigger that $ 2{}^2{}^{n - 2}$;\n$ b)$ smaller that $ n^2$.\n[/color] \r\n\r\n\r\n[b][color=darkred]4.[/color][/b] [color=darkblue] Determine the biggest possible value of $ m$ for which equation $ 2005x + 2007y = m$ has unique solution in matural numbers.\n[/color] \r\n\r\n[b][color=darkred]5.[/color][/b] [color=darkblue] Determine all pairs $ (m,n)$ of real numbers for which $ m^2 = nk + 2$, where $ k = \\overline{n1}$.\n[/color] \r\n\r\n[b][color=darkred]6.[/color][/b] [color=darkblue] Prove that for every composite number $ n > 4$, numbers $ kn$ divides $ (n - 1)!$ for every integer $ k$, such that $ 1\\le k \\le [\\sqrt {n - 1}]$.\n[/color] \r\n\r\n[b][color=darkred]7.[/color][/b] [color=darkblue] Determine all numbers $ \\overline{abcd}$, such that $ \\overline{abcd} = 11(a + b + c + d)^2$.\n[/color] \r\n\r\n[b][color=darkred]8.[/color][/b] [color=darkblue] Prove that there do not exist natural numbers $ n\\ge 10$, such that every $ n$'s digit is non-zero, and all numbers which are obtained by permutating of its digits are perfect squares\n[/color] \r\n\r\n[b][color=darkred]9.[/color][/b] [color=darkblue] Let $ ABCD$ be a trapezoid with $ AB\\parallel CD, AB > CD$ and $ \\angle{A} + \\angle{B} = 90^\\circ$. Prove that the distance between the midpoints of the bases is equal to the semidifference of the bases.[/color] \r\n\r\n[b][color=darkred]10.[/color][/b] [color=darkblue] Let $ ABCD$ be an isosceles trapezoid inscribed in a circle $ \\mathcal{C}$ with $ AB\\parallel CD, AB = 2CD$. Let $ Q \\in AD \\cap BC$ and let $ P$ be intersection of the tangents to $ \\mathcal{C}$ at $ B$ and $ D$. Calculate the area of the quadrilateral $ ABPQ$ in terms of the area of the triangle $ PDQ$.\n[/color] \r\n\r\n[b][color=darkred]10 (bis).[/color][/b] [color=darkblue] Let $ ABCD$ be atrapezoid inscribed in a circle $ \\mathcal{C}$ of diameter $ AB$ with $ AB\\parallel CD, AB = 2CD$. Let $ Q \\in AD \\cap BC$ and let $ P$ be intersection of the tangents to $ \\mathcal{C}$ at $ B$ and $ D$. Calculate the area of the quadrilateral $ ABPQ$ in terms of the area of the triangle $ PDQ$.\n[/color] \r\n\r\n[b][color=darkred]11.[/color][/b] [color=darkblue] Circles $ \\mathcal{C}_1$ and $ \\mathcal{C}_2$ intersect at $ A$ and $ B$. Let $ M \\in AB$. A line through $ M$ (different form $ AB$) cuts circles $ \\mathcal{C}_1$ and $ \\mathcal{C}_2$ in $ Z,D$ and $ E,C$ respectively, such that $ D,E \\in ZC$. Perpendiculars at $ B$ to the lines $ EB$, $ ZB$ and $ AD$ respectively, cut circle $ \\mathcal{C}_2$ at points $ F,K$ and $ N$ respectively. Prove that $ KF = NC$.[/color]\r\n\r\n[b][color=darkred]12.[/color] [/b] [color=darkblue] Let $ ABC$ be an equilateral triangle of center $ O$, and $ M \\in BC$. Let $ K,L$ be proiections of $ M$ onto the sides $ AB$ and $ AC$ respectively. Prove that line $ OM$ passes through the midpoint of the segment $ KL$.\n.[/color]\r\n\r\n[b][color=darkred]13.[/color][/b] [color=darkblue] Let $ A$ be a subset of the set $ \\{1,2,3,..,2006\\}$, consisting of $ 1004$ elements. Prove that exists three distinct numbers $ a,b,c \\in A$ such that $ gcd(a,b)$ \n$ a)$ divides $ c$.\n$ b)$ doesn't divide $ c$.\n[/color] \r\n\r\n[b][color=darkred]14.[/color][/b] [color=darkblue] Let $ n\\ge 5$ be a positive integer. Prove that the set $ \\{1,2,3,...,n\\}$ can not be partitioned into two non-zero subsets $ S_n$ and $ P_n$ such that the sum of the elements of $ S_n$ is equal to the product of the elements of $ P_n$.\n[/color]", "Solution_11": "Thank you!\r\nHow did you find this (if it's internet please tell us!)?\r\nDo you have 2007 shortlist?\r\nAnd what's 10. bis, and are you sure that you've written 9. well?\r\nI've made it in PDF:", "Solution_12": "I've got a book by Dan Branzei. But it is in romanian. There are shortlist 2001-2006. When I have time, I will post others.", "Solution_13": "[quote=\"Ahiles\"]I've got a book by Dan Branzei. But it is in romanian. There are shortlist 2001-2006. When I have time, I will post others.[/quote]\r\n\r\nCan you tell us what's in that book (just shortlists or more)?", "Solution_14": "[quote=\"Bugi\"][quote=\"Ahiles\"]I've got a book by Dan Branzei. But it is in romanian. There are shortlist 2001-2006. When I have time, I will post others.[/quote]\n\nCan you tell us what's in that book (just shortlists or more)?[/quote]\r\n\r\nWell, I have this book!\r\n It contains :\r\n [b]a) [/b]The problems for all JBMO 1997-2006\r\n [b]b)[/b] The SL - problems from 2000-2006\r\n[b] c)[/b] The tst for Romany\r\n[b] d) [/b]Solutions to these problems ! \r\n \r\n A co-autor is a friend of me in Romany and he has sent it to me.\r\n I have the SL - 2006 in pdf format. But please find SL 2007 (and 2008, if possible) and we 'll give the list to each other with e mails.\r\n Babis Sterigou - Greece", "Solution_15": "SL 2008 is revealed at next JBMO, because participating countries can take these problems to the competitions in those countries (JBkTST-s).\r\nAhiles, can you tell us the full name of the book?", "Solution_16": "It's called \"10 ani de Olimpiade Balcanice ale Juniorilor\" and authors are Dan Br\u00e2nzei, Dinu \u0218erb\u0103nescu, Gabriel Popa, Ioan \u0218erdean, Dan \u0218tefan Marinescu, Vasile \u0218erdean.\r\nPite\u0219ti : Paralela 45, 2007.", "Solution_17": "Thanks. I've posted the most of these problems in Olympiad section, but are you sure about 9. and 14? In 14, someone proved that it always can be partitioned.\r\nAnd, can you write the name of that book in English (exact please)?", "Solution_18": "[quote=\"Ahiles\"]\n[b][color=darkred]14.[/color][/b] [color=darkblue] Let $ n\\ge 5$ be a positive integer. Prove that the set $ \\{1,2,3,...,n\\}$ can not be partitioned into two non-zero subsets $ S_n$ and $ P_n$ such that the sum of the elements of $ S_n$ is equal to the product of the elements of $ P_n$.\n[/color][/quote]\r\nFor future reference, it's actually the other way around. See http://www.mathlinks.ro/viewtopic.php?t=238634.", "Solution_19": "Yep, I did a mistake there, but I can't edit it already.", "Solution_20": "Can you write the name of that Romanian book in English?", "Solution_21": "[size=150][color=darkred]JBMO Shortlist 2000[/color][/size]\r\n\r\n [b][color=darkred]1.[/color][/b] [color=darkblue] Let $ M$ be the set of positive integers which consist of $ 2006$ digits, one of which is $ 7$, but others are $ 1$. Prove that there are at least 666 composite numbers in $ M$. \n[/color] \r\n\r\n [b][color=darkred]2.[/color][/b] [color=darkblue] Find all natural numbers, not divisible by 10, which are perfect cubes, and after deleting the last three digits, they are also nonzero perfect cubes.\n[/color] \r\n\r\n [b][color=darkred]3.[/color][/b] [color=darkblue] Find all numbers which are written as $ \\overline{abcd}$ in the base $ 10$ and $ \\overline{dcba}$ in the base 7.\n[/color] \r\n\r\n\r\n[b][color=darkred]4.[/color][/b] [color=darkblue] Find all pairs of natural numbers $ (m,n)$, for which numbers\n$ A=3m^2+2mn+n^2+3n$\n$ B=m^2+3mn+2n^2$\n$ C=2m^2+mn+3n^2$\nare consecutive in some order.\n[/color] \r\n\r\n[b][color=darkred]5.[/color][/b] [color=darkblue] Find all numbers $ n=\\overline{abcd}$, $ n=\\prod_{i=1}^k p_i^{\\alpha_i}$, such that $ \\sum p_i=\\sum \\alpha_i$.\n[/color] \r\n\r\n[b][color=darkred]6.[/color][/b] [color=darkblue] Find all positive integers $ a$ and $ b$, for which $ a^4+4b^4$ is prime.\n[/color] \r\n\r\n[b][color=darkred]7.[/color][/b] [color=darkblue] Solve in $ \\mathbb{N}^3$ the equation\n \\[ xy+yz+zy-xyz=2 \\][/color] \r\n[b][color=darkred]8.[/color][/b] [color=darkblue] Prove that there are no $ x,y,z\\in \\mathbb{Z}$ which satisfy the equation\n\\[ x^4+y^4+z^4-2x^2y^2-2y^2z^2-2y^2z^2=2000\\][/color]\r\n[b][color=darkred]9.[/color][/b] [color=darkblue]Prove that for every $ n \\in \\mathbb{Z}$ we can find $ a,b \\in \\mathbb{Z}$, such that\n\\[ n=[a\\sqrt{2}]+[b\\sqrt{3}]\\][/color] \r\n\r\n[b][color=darkred]10.[/color][/b] [color=darkblue] Consider the sequence $ (x_n)_{n \\in \\mathbb{N}}$ of natural numbers which satisfy\n$ (i)$ $ x_{2n+1}=4x_n+2n+2$, $ \\forall n \\in \\mathbb{N}$\n$ (ii)$ $ x_{3n+2}=3x_{n+1}+6x_n$, $ \\forall n \\in \\mathbb{N}$\nIf $ x_0=0$, prove that $ x_{3n-1}=x_{n+2}-2x_{n+1}+10x_n$, $ \\forall n \\in \\mathbb{N}^*$\n[/color] \r\n\r\n[b][color=darkred]11.[/color][/b] [color=darkblue] Let $ m,n \\in \\mathbb{N}$, $ m\\le 2000$, and $ k=3-\\dfrac{m}{n}$. Determine the minimum possible value for $ k$[/color]\r\n\r\n[b][color=darkred]12.[/color] [/b] [color=darkblue] Let $ x,y,a,b\\in \\mathbb{R}_+$, $ x \\not= y$, $ x \\not = 2y$, $ y \\not =2x$, $ a \\not= 3b$ and $ \\dfrac{2x-y}{2y-x}=\\dfrac{a+3b}{a-3b}$. Prove that $ \\dfrac{x^2+y^2}{x^2-y^2}\\ge 1$[/color]\r\n\r\n[b][color=darkred]13.[/color][/b] [color=darkblue] Find all $ (x,y,z) \\in R^3$, for which\n\\[ 2x\\sqrt{y-1}+2y\\sqrt{z-1}+2z\\sqrt{x-1}\\ge xy+xz+yz\\][/color] \r\n[b][color=darkred]14.[/color][/b] [color=darkblue] Let $ ABC$ be a triangle. Find all segments $ XY$, which are situated in the interior of the triangle, such that $ XY$ and five from the segments $ XA,XB,XC$, $ YA,$ $ YB$, $ YC$ divide area of $ ABC$ in five equivalent parts. Prove that all found segments are concurrent.\n[/color]\r\n\r\n[b][color=darkred]14.[/color][/b] [color=darkblue] Let $ ABC$ be a triangle and $ a,b,c$ lengths of the corresponding sides. Consider another triangle $ DEF$ with sides $ EF=\\sqrt{au}$, $ DF=\\sqrt{bu}$, $ DE=\\sqrt{cu}$. Denote by $ A,B,C,D,E,F$ measures of the angles of this triangles. Suppose that $ A>B>C$. Prove that $ A>D>E>F>C$.\n[/color]\r\n\r\n[b][color=darkred]15.[/color][/b] [color=darkblue] Let $ ABCDE$ be a hexagon with all angles equal. Prove that $ AB-DE=EF-BC=CD-FA$.\n[/color]\r\n\r\n[b][color=darkred]15.[/color][/b] [color=darkblue] Let $ ABCD$ be a quadrilateral with $ \\angle{DAB}=60^\\circ$, $ \\angle{ABC}=90^\\circ$, $ \\angle{BCD}=120^\\circ$. Diagonals of the quadrilateral intersect at $ M$, such that $ MB=1$, $ MD=2$. Calculate tha area of $ ABCD$.\n[/color]", "Solution_22": "You have to post each problem in its own thread, otherwise this is not very useful to us.", "Solution_23": "OK.. I did it..", "Solution_24": "Could someone please post the solution threads to these problems (if any)? Thanks!", "Solution_25": "I posted 2002 and 2006 a long time ago, Ahiles posted 2000. Search for ''JBMO Shortlist''...", "Solution_26": "[url=http://www.artofproblemsolving.com/community/c6h502939p2825631]JBMO Shortlist 2010.pdf (119kb)[/url]\n[url=http://www.artofproblemsolving.com/community/c3240_jbmo_shortlists]JBMO Shortlist[/url]", "Solution_27": "Revive much?" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "find all integer n and k satisfaies \r\n\r\n $n|k^2+1 ,n^2|k^3+1 ,n^3|k^4+1$", "Solution_1": "Let $n \\neq \\pm 1$ and $p \\in \\mathfrak{P}$ such that $p \\mid n$. It has to be $k^3 + 1 \\equiv k^2 + 1 \\equiv 0 \\bmod p$. It follows $k^2(k-1) \\equiv -k(k+1) \\equiv 0 \\bmod p$, and subsequently $k - 1 \\equiv k + 1 \\equiv 0 \\bmod p$, since obviously $\\gcd(k,p) = 1$. So $p = 2$, $k \\equiv 1 \\bmod 2$ and $n = 2^t$, for a suitable $t \\in \\mathbb{Z}^+$. Anyway $k^2 + 1 \\equiv 2 \\bmod 4$, if $k \\equiv 1 \\bmod 2$. Therefore $n = 2$ and you need $k^4 + 1 \\equiv 0 \\bmod 8$, which is impossibile, because $k^4 +1 \\equiv 2 \\bmod 4$, if $k \\equiv 1 \\bmod 2$. So necessarily $n = \\pm 1$ and any $k\\in\\mathbb{Z}$ sounds good." } { "Tag": [ "algebra", "polynomial", "quadratics", "algebra unsolved" ], "Problem": "Let p(x) be the polynomial $x^3 + 14x^2 - 2x + 1$. Let $p^n(x)$ denote $p(p^(n-1)(x))$. Show that there is an integer N such that $p^N(x) - x$ is divisible by 101 for all integers x.", "Solution_1": "Cute...isn't it ;)\r\nFirst of all $101$ is a prime number. This will be very,very usefull. Now for every polynomial $Q$ in $Z[x]$ ,value $Q(x) \\; mod \\; 101$ depends only on residue of $x$ modulo $101$ thus we have to prove our statement only for $0\\leq x \\leq 100$.\r\nAll the time we will operate on $Z_{101}$ so every equality will be taken modulo $101$\r\n It's easy to see that all we have to prove now is that \\[ p(0),p(1),...,p(100) \\] is a permutation of numbers \\[ 0,1,2,3,...100 \\] With it, our statement will be obvious, cos' then for every $n$ \\[ p^n(0),p^n(1),...,p^n(100) \\] will be such permutation. As the number of permutation is finite, the same permutation must appear again somewhere. So there exist $n_10$ and $ a\\plus{}b\\plus{}c\\equal{}3$, then\r\n\r\n\r\n$ \\frac{a\\minus{}1}{a^2\\plus{}2bc}\\plus{}\\frac{b\\minus{}1}{b^2\\plus{}2ca}\\plus{}\\frac{c\\minus{}1}{c^2\\plus{}2ab}\\leq\\frac{1}{ab\\plus{}bc\\plus{}ca}$", "Solution_1": "\\[ \\frac {1}{a^2 \\plus{} 2bc} \\plus{} \\frac {1}{b^2 \\plus{} 2ca} \\plus{} \\frac {1}{c^2 \\plus{} 2ab}\\geq \\frac {2}{ab \\plus{} bc \\plus{} ca}\\]\r\nit's enough to show\r\n\\[ \\frac {a}{a^2 \\plus{} 2bc} \\plus{} \\frac {1}{b^2 \\plus{} 2ca} \\plus{} \\frac {1}{c^2 \\plus{} 2ab}\\leq \\frac {3}{ab \\plus{} bc \\plus{} ca} \\equal{} \\frac {a \\plus{} b \\plus{} c}{ab \\plus{} bc \\plus{} ca}\\]\r\n\r\n\\[ \\Longleftrightarrow \\frac {a}{a^2 \\plus{} 2bc}(a \\minus{} b)(a \\minus{} c) \\plus{} \\frac {b}{b^2 \\plus{} 2ca}(b \\minus{} c)(b \\minus{} a) \\plus{} \\frac {c}{c^2 \\plus{} 2ca}(c \\minus{} a)(c \\minus{} b)\\geq 0\\]\r\nwhich is true by V-V Schur. :lol:" } { "Tag": [], "Problem": "$AB$ is a fixed diameter of a circle whose center is $O$. From $C$, any point on the circle, a chord $CD$ is drawn perpendicular to $AB$. Then, as $C$ moves over a semicircle, the bisector of angle $OCD$ cuts the circle in a point that always:\r\n\\[ \\text{(A) bisects the arc}\\ AB \\qquad \\text{(B) trisects the arc}\\ AB \\qquad \\text{(C) varies} \\]\r\n\\[{ \\text{(D) is as far from}\\ AB\\ \\text{as from}\\ D} \\qquad \\text{(E) is equidistant from}\\ B\\ \\text{and}\\ C \\]", "Solution_1": "[hide=\"Answer\"] (A) bisects the arc AB.[/hide]\n\n[hide=\"Justification\"]\nFor any segment $CD$, extend the segment so that it intersects $AB$ at $Q$. Let $\\angle QCO = y$ and $\\angle QOC = x$.\n\nLet the bisector cut the circle at point $P$. \n\nThe external angle $\\angle DCO = x + 90$. Thus, $\\angle DCP = \\frac{x+90}{2}$. Since $CO$ and $PO$ are radii, $\\triangle CPO$ is isosceles. So, the angles $\\angle OCP$ and $\\angle OPC$ must sum to $x + 90$ and $\\angle COP=y$.\n\nHere, we see that $\\angle AOP = x+y = 90^o$. Thus, the bisector must bisect arc $AB$.\n\n :lol: \n\n[/hide]" } { "Tag": [ "geometry", "ratio", "similar triangles" ], "Problem": "What is the area in square centimeters of the shaded region below?\r\n\r\n[img]7208[/img]", "Solution_1": "[quote=\"#H34N1\"]What is the area in square centimeters of the shaded region below?\n\n[img]7208[/img][/quote]\r\n[hide]So call the square thats 3x3 ABCD starting from the top left. Then call the little square DEFG starting form the the bottom left and going to the counter clockwise. So we have a 3-4-5 triangle BCE. We want to find the the area of quadrilateral BCDG. To do this we need to subtract the area of triangle DGE from triangle BCE. Triangle DGE is similiar to triangle CBE. And the ratio of the sides are 1:4. So the side length of DG will be 3/4. SO the area of triangle DGE is 3/8. Subtracting we get that the area of quafrilateral BCDG is 45/8. Then subtracting this from the area of the square which is 9 we get that the area of the shaded region is 27/8.[/hide]\r\nHope this isn't too confusing.", "Solution_2": "[hide]Using similar triangles, it is easily found that the length of the segment shared by the two squares (minus the part which touches the shaded region) is $\\frac34$. This means the second side of the shaded region is $\\frac94$.\n\n$\\frac{3\\cdot\\frac94}2=\\boxed{\\frac{27}8}$[/hide]" } { "Tag": [ "quadratics", "algebra" ], "Problem": "What is the sum of the numbers that are two more than their reciprocals?", "Solution_1": "[quote]What is the sum of the numbers that are two more than their reciprocals?[/quote][hide]\nLet $x$ be such a number.\n\nThen $x = \\frac{1}{x} + 2$\n\nThen $x$ is a root of the quadratic: $x^2 - 2x - 1 = 0$\n\nFor a quadratic equation: $x^2 + bx + c = 0$, the sum of the roots is $-b$.\n\nTherefore, the sum is $2$.[/hide]", "Solution_2": "[hide]$x = \\frac{1}{x} + 2\\\\\nx^2 = 1+2x\\\\\nx^2-1-2x = 0\n(x-2)(x+1) = 0\\\\\n{x = 2} or {x = -1}$\n\nThe answer is $1$.[/hide]", "Solution_3": "[quote=\"h_s_potter2002\"][hide]$x = \\frac{1}{x} + 2\\\\\nx^2 = 1+2x\\\\\nx^2-1-2x = 0\n(x-2)(x+1) = 0\\\\\n{x = 2} or {x = -1}$\n\nThe answer is $1$.[/hide][/quote]\r\n\r\nYou made a little mistake after $x^2-2x-1=0$. It wasn't factored correctly. Close, though.", "Solution_4": "[hide]What is the sum of the numbers that are two more than their reciprocals?\n\nx=(1/x)+2\nx2=2x+1\nx2-2x-1=0 \n\nif ax2+bx+c, the roots of this equation is -b[/hide]\n\n[hide][b]therefore the solution is 2[/b][/hide]", "Solution_5": "x = 1/x + 2\r\nx2 = 1 + 2x\r\n2 - 2x - 1 = 0\r\n\r\n(2 +/- sqrt (4 + 4) )/ 2\r\n1 +/- sqrt(2)\r\n\r\n(1 + sqrt(2)) + (1 - sqrt(2))\r\n2" } { "Tag": [ "trigonometry", "Pythagorean Theorem", "geometry", "AMC 10", "power of a point" ], "Problem": "Square $ ABCD$ has side length $ 2$. A semicircle with diameter $ \\overline{AB}$ is constructed inside the square, and the tangent to the semicricle from $ C$ intersects side $ \\overline{AD}$ at $ E$. What is the length of $ \\overline{CE}$?\n\n[asy]\ndefaultpen(linewidth(0.8));\npair A=origin, B=(1,0), C=(1,1), D=(0,1), X=tangent(C, (0.5,0), 0.5, 1), F=C+2*dir(C--X), E=intersectionpoint(C--F, A--D);\ndraw(C--D--A--B--C--E);\ndraw(Arc((0.5,0), 0.5, 0, 180));\npair point=(0.5,0.5);\nlabel(\"$A$\", A, dir(point--A));\nlabel(\"$B$\", B, dir(point--B));\nlabel(\"$C$\", C, dir(point--C));\nlabel(\"$D$\", D, dir(point--D));\nlabel(\"$E$\", E, dir(point--E));[/asy]\n\n$ \\textbf{(A)}\\ \\frac {2 \\plus{} \\sqrt5}{2} \\qquad \\textbf{(B)}\\ \\sqrt 5 \\qquad \\textbf{(C)}\\ \\sqrt 6 \\qquad \\textbf{(D)}\\ \\frac52 \\qquad \\textbf{(E)}\\ 5 \\minus{} \\sqrt5$", "Solution_1": "[hide] Call the point of tangency of CE to the semicircle R. By POP Theorem, CR=CB=2. Also ER=EA=x. By Pythagorean Theorem on triangle EDC, 2^2+(2-x)^2=(2+x)^2, so x=1/2, so CE=5/2. D. [/hide]", "Solution_2": "[hide=\"Here is an another solution.\"][b]Denote:[/b] the middlepoint $O$ of the side $AB,$\n\nthe tangent point $T\\in CE\\cap C(O,1)$, i.e. $TC=BC=2$\n\nand $TE=x$, i.e. $CE=2+x$. Therefore,\n\n$OT\\perp CE,\\ OC\\perp OE\\Longrightarrow OT^2=TC\\cdot TE\\Longrightarrow$\n\n$x=\\frac 12\\Longrightarrow \\boxed {\\ CE=\\frac 52\\ }\\ ,$ i.e. the answer is $\\boxed {\\ D\\ }$.[/hide]", "Solution_3": "Probably not rigorous, but who cares.\r\n\r\nLet O be the midpoint of AB\r\n$ \\angle BOC \\plus{} \\angle AOE \\equal{} 90$\r\n$ tan ^{ \\minus{} 1} {2} \\plus{} tan^{ \\minus{} 1} {x} \\equal{} 90$\r\n$ \\frac {2 \\plus{} x}{1 \\minus{} 2x} \\equal{} tan 90$\r\nSine $ tan 90$ is undefined, we have $ 1 \\minus{} 2x \\equal{} 0$, so $ x \\equal{} \\frac {1}{2}$\r\nNow Pythag finishes it.", "Solution_4": "So simply put you first use pythagoras to get 2 and then power of a point to find the other part is 1/2. Is this from an AMC 10?", "Solution_5": "[quote=\"mathemagician1729\"]So simply put you first use pythagoras to get 2 and then power of a point to find the other part is 1/2. Is this from an AMC 10?[/quote]\r\nYes, specifically [url=http://www.artofproblemsolving.com/Wiki/index.php/2004_AMC_12A_Problems/Problem_18]here.[/url]", "Solution_6": "alternate solution: let AE=x, then ED=2-2, CD=2, CE=(2+x) by the tangent to a circle theorem. Pythagorean theorem on CDE gives x=1/2, hence the answer is 5/2 (D).", "Solution_7": "I go like : let tangent point be T. let center of circle be O. Draw CO. Note that triangle CBO is congruent to triangle CTO by HL congruence(radius length). So CT = 2 and TO =1. Draw EO. See how OT is the altitude of this triangle and forms 3 similar triangles. TOE is COT scaled down by 1/2. Then find ET and add to get D\n", "Solution_8": "There is a much easier solution. \nMark the point of contact of the tangent with the semicircle at $X$\n$BC=CX$(TANGENTS FROM AN EXTERNAL POINT). \n$CX=2$\nALSO $EA=EX=x$\n$DE=2-x$\n$CE=2+x$\n$DC=2$.Apply Pythagoras in $DCE$ to get $2+x$", "Solution_9": "Dear Mathlinkers,\n\nhttp://jl.ayme.pagesperso-orange.fr/Docs/Miniatures%20Geometriques%20addendum%20V.pdf p. 46...\n\nSincerely\nJean-Louis", "Solution_10": "Why did people post in this thread after 6 years? ", "Solution_11": "[b]Solution:[/b] Let $\\overline{CE}$ be tangent to the semicircle at $P$. Notice that $PC=CB=2$. Now, let $EA=EP=x$. This makes $DE=2-x$ and we already know that $DC=2$. Now we can do the Pythagorean theorem to get that $(2-x)^2 + 2^2 = (2+x)^2 \\Rightarrow x = \\frac{1}{2}$. This makes $CE = 2+x = \\frac{5}{2} \\Rightarrow \\boxed{D}$." } { "Tag": [ "algebra solved", "algebra" ], "Problem": "Determine a closed expression for S_(m,n) with m,n \\in N:\r\n\r\nS_(m,n) = \\sum^{n}_{k=0} ((k+1)^2*m^k).", "Solution_1": "Let fn(x)=1+x+...+xn.\r\n\r\nS(X,n)=(X*(X*fn(X))')'", "Solution_2": "I do not call that a closed expression. :)", "Solution_3": "I think it is much shoerter than your \"closed\" expresion ;)", "Solution_4": "I would like to see a fraction involving polynomials. Probably I have a different idea of closed expression than you seem to have... :D", "Solution_5": "[quote=\"orl\"]I would like to see a fraction involving polynomials. Probably I have a different idea of closed expression than you seem to have... :D[/quote]\r\n\r\nNo, I don't have a different idea of closed expresion. I've got smth much more fearsome. I call it a laziness attack :D. I think that the battle may be long lost.\r\n---------------------------------------------------------------------------------\r\nHere is a shot:\r\n[(n+1)2mn+3-(2n 2 +6n+3)mn+2+(n+2) 2 mn+1-m-1]/(m-1)3.\r\n-------------------------------------------------------------------------------\r\nHappy? \r\nThis was an awful waste of an A4 sheet.", "Solution_6": "Hey well done, amfulger. :) I am proud you got it. But please do not forget to consider the case \r\n\r\nm = 1: 1/6((n+1)(n+2)(2n+3))." } { "Tag": [ "quadratics", "algebra", "quadratic formula" ], "Problem": "you are probably familiar with the equation for the surface are of a closed right cylinder.\r\n\r\nS=2 pi r^2 + 2 pi r h\r\n\r\nI need to isolate r on one side of the equation. i have been staring at this for hours and just cant figure it out. please help me! i just want to move on to the next section in by book!!!!", "Solution_1": "[hide=\"Hint\"]$ ax^2\\plus{}bx\\plus{}c\\equal{}0 \\Rightarrow x\\equal{}\\frac{\\minus{}b\\pm\\sqrt{b^2\\minus{}4ac}}{2a}$[/hide]", "Solution_2": "It's just a quadratic equation in $ r$. Solve it using the quadratic formula and disregard the negative root because $ r>0$.", "Solution_3": "i dont know the quadratic formula yet, havnt got that far...\r\nwhy did my book give me a problem that it has not yet taught me the solution to yet....\r\nperhaps i left out important information....\r\nfind r when S=208m^2 and h= 4m\r\ncan you solve this without the quadratic formula now?\r\n\r\ni can isolate S..... \r\nS=2pi r^2+2pi r h ......\r\nh= [u]S-2pi r^2[/u]\r\n ------ 2pi r", "Solution_4": "If $ \\pi \\approx \\frac {13}{4}$...\r\n\r\n$ S \\equal{} 2\\pi r^2 \\plus{} 2\\pi r h$\r\nSo, $ 208 \\equal{} 2 \\times \\frac {13}{4} \\times r^2 \\plus{} 2 \\times \\frac {13}{4} \\times r \\times 4$\r\n$ 208 \\times {1}{2} \\times \\frac {13}{4} \\equal{} r^2 \\plus{} 4r$\r\n$ 32 \\equal{} r^2 \\plus{} 4r$\r\n\r\nSo, $ 32 \\plus{} 4 \\equal{} r^2 \\plus{} 4r \\plus{} 4 \\equal{} (r \\plus{} 2)^2$ i.e. $ (r \\plus{} 2)^2 \\equal{} 6^2$ i.e. $ r \\plus{} 2 \\equal{} \\pm6$\r\nThus, $ r \\equal{} 6 \\minus{} 2 \\equal{} \\boxed{4}$. (Because $ r > 0$)\r\n\r\nEDIT:I fixed a typo.", "Solution_5": "oh, ooooopppss!!! im so sorry, but i wrote it wrong ......\r\nS is actually 208 Pi m^2\r\n i just cant figure out how to isolate r.......\r\n\r\nand.... i though the fraction for pi was 22/7....\r\nplease forgive me for being such a newbie!!!!!", "Solution_6": "[quote=\"davisdandrew\"]oh, ooooopppss!!! im so sorry, but i wrote it wrong ......\nS is actually 208 Pi m^2\n i just cant figure out how to isolate r.......\n\nand.... i though the fraction for pi was 22/7....\nplease forgive me for being such a newbie!!!!![/quote]\r\n\r\nOk. So similarly my post,\r\n\r\n$ 208 \\pi \\equal{} 2 \\pi r^2 \\plus{} 2 \\pi r \\times 4$\r\n$ 208 \\times \\frac{1}{2} \\equal{} r^2 \\plus{} 4r$\r\n$ 104 \\equal{} r^2 \\plus{} 4r$\r\n\r\nSo, $ 104 \\plus{} 4 \\equal{} r^2 \\plus{} 4r \\plus{} 4 \\equal{} (r \\plus{} 2)^2$ i.e. $ (r \\plus{} 2)^2 \\equal{} 6^2 \\times3$ i.e. $ r \\plus{} 2 \\equal{} \\pm 6\\sqrt {3}$\r\nThus, $ r \\equal{} 6\\sqrt {3} \\minus{} 2 \\equal{} \\boxed{6\\sqrt {3} \\minus{} 2}$. (Because $ r > 0$)", "Solution_7": "hmmmm, the answer in the book says r= 8.....\r\nim so lost....", "Solution_8": "If $ r \\equal{} 8$,\r\n$ 2\\times8^2 \\times \\pi \\plus{} (2\\times8 \\times \\pi)\\times4 \\equal{} 128\\pi \\plus{} 64\\pi \\equal{} 192\\pi \\neq 208\\pi$.", "Solution_9": "is there a way you can simply put the equation in r= form?" } { "Tag": [ "function", "logarithms" ], "Problem": "hi friends,\r\n\r\nI cam across through a problem of this kind.\r\n\r\nSolve for $n$ such that $n=2^{\\frac{n}{3}}$\r\n\r\nregards\r\n\r\nbhupala", "Solution_1": "Considering the growth, you will have only a finite number of cases to check ...", "Solution_2": "If $n$ means any integer, then I suggest this problem be moved to the Pre-Olympiad section where, I am sure, there are plenty of people who will be able solve it.", "Solution_3": "$n=2^{\\frac{n}{3}}\\Longleftrightarrow n^{3}=2^{n}.$", "Solution_4": "They're two monotonically increasing functions, so they should intersect at most once. Actually, I'm not sure that they do.", "Solution_5": "[quote=\"kunny\"]$n=2^{\\frac{n}{3}}\\Longleftrightarrow n^{3}=2^{n}.$[/quote]\r\n\r\nIf $n \\in \\mathbb{Z}$, it has to be in $\\mathbb{N}$, cause if $n<0 \\implies n^{3}<0$ but $2^{n}$ remains positive. So $n$ has to be a power of 2, call it $n=2^{k}$ for some positive integer $k$. So $2^{3k}=2^{2^{k}}\\Longleftrightarrow 3k=2^{k}$, that is clearly impossible in $\\mathbb{N}$", "Solution_6": "If $n$ is allowed to be any real:\r\n\r\nThey will intersect between $n=1$ and $n=2$.\r\n$2^{1}>1^{3}$ but $2^{2}<2^{3}$. The actual value at which they intersect is probably some really ugly number, which you could maybe find using logs.\r\nThey will also intersect at some other value $910^{3}$.\r\n\r\nBut clearly no solutions for $n \\in \\mathbb{Z}$", "Solution_7": "$x=2^{x/3}\\Leftrightarrow xe^{-x\\frac{\\ln{2}}{3}}=1\\Leftrightarrow-\\frac{\\ln{2}}{3}xe^{-x\\frac{\\ln{2}}{3}}=-\\frac{\\ln{2}}{3}$\r\n$\\Leftrightarrow-x\\frac{\\ln{2}}{3}=W\\left(-\\frac{\\ln{2}}{3}\\right)\\Leftrightarrow x=-W\\left(-\\frac{\\ln{2}}{3}\\right)\\cdot\\frac{3}{\\ln{2}}$\r\nHere W is the Lambert function (see http://en.wikipedia.org/wiki/Lambert's_W_function).\r\nI get (with http://functions.wolfram.com) $x=1.373465876$. This is the value mentioned above that is between 1 and 2. The other value (almost 10) is due to the fact that the Lambert W-function is not single valued in the interval [$-\\frac{1}{e}$,0[ and can be computed in e.g. Mathematica using the ProductLog[k,n] function which allows one to caluculate the values on other branches than the principle branch." } { "Tag": [ "inequalities", "function", "romania" ], "Problem": "Let $a,b,c$ be three positive real numbers such that $abc=1$. Prove that: \\[ 1+\\frac{3}{a+b+c}\\ge{\\frac{6}{ab+bc+ca}} . \\]", "Solution_1": "$\\displaystyle 1+\\frac{3}{a+b+c}\\geq \\frac{6}{ab+bc+ca}$ is equivalent to \r\n\r\n$\\displaystyle \\frac{ab+bc+ca}{3} + \\frac{ab+bc+ca}{a+b+c} \\geq 2$.\r\n\r\nBy AM-GM, it suffices to prove $\\displaystyle \\frac{ab+bc+ca}{3} \\cdot \\frac{ab+bc+ca}{a+b+c} \\geq 1$.\r\n\r\nThat is, $\\displaystyle (ab+bc+ca)^2 \\geq 3(a+b+c)$.\r\n\r\nLet $x=1/a, y=1/b, z=1/c$, then $xyz=1$ and it reduces to prove $(x+y+z)^2\\geq 3(xy+yz+zx)$ which is obvious.", "Solution_2": "$ 1 \\plus{} \\frac {3}{a \\plus{} b \\plus{} c}\\ge{\\frac {6}{ab \\plus{} bc \\plus{} ca}}$ is equivalent to $ (3\\plus{}a \\plus{} b \\plus{} c)(ab \\plus{} bc \\plus{} ca)\\ge 6(a \\plus{} b \\plus{} c)$.\r\n\r\nNow lets consider two cases. First $ ab \\plus{} bc \\plus{} ca\\ge a \\plus{} b \\plus{} c$ and then $ ab \\plus{} bc \\plus{} ca\\le a \\plus{} b \\plus{} c$.\r\n\r\nIf $ ab \\plus{} bc \\plus{} ca\\ge a \\plus{} b \\plus{} c$ then using AM-GM $ (3\\plus{}a \\plus{} b \\plus{} c)(ab \\plus{} bc \\plus{} ca)\\ge (3\\plus{}3)(ab \\plus{} bc \\plus{} ca))\\ge 6(ab \\plus{} bc \\plus{} ca)\\ge 6(a \\plus{} b \\plus{} c)$.\r\n\r\nIf $ ab \\plus{} bc \\plus{} ca\\le a \\plus{} b \\plus{} c$ then let $ x\\equal{} a \\plus{} b \\plus{} c$.\r\n\r\n$ (3\\plus{}a \\plus{} b \\plus{} c)(ab \\plus{} bc \\plus{} ca)\\ge (3\\plus{}a \\plus{} b \\plus{} c)(a \\plus{} b \\plus{} c) \\equal{} (3\\plus{}x)x$. Now we net that $ (3\\plus{}x)x\\ge 6x$ but this equivalent to $ x(x\\minus{}3)\\ge 0$ and this is true becouse $ a\\plus{}b\\plus{}c\\ge 3$ by AM-GM.", "Solution_3": "[quote=\"Maverick\"]Let $ a,b,c$ be three positive real numbers such that $ abc \\equal{} 1$. Prove that:\n\\[ 1 \\plus{} \\frac {3}{a \\plus{} b \\plus{} c}\\ge{\\frac {6}{ab \\plus{} bc \\plus{} ca}} .\n\\]\n[/quote]\r\nWe set \\[ a\\equal{}\\frac{1}{x} ; b\\equal{}\\frac{1}{y} , c\\equal{}\\frac{1}{z}\\] and observe xyz=1. The inequality is equivalent to:\r\n\\[ 1\\plus{}\\frac{3}{xy\\plus{}yz\\plus{}xz}\\geq \\frac{6}{x\\plus{}y\\plus{}z}\\]\r\nClearly \\[ 1\\plus{}\\frac{3}{xy\\plus{}yz\\plus{}xz}\\geq 1\\plus{}\\frac{9}{(x\\plus{}y\\plus{}z)^{2}}\\geq \\frac{6}{x\\plus{}y\\plus{}z}\\]", "Solution_4": "[quote=\"Maverick\"]Let $ a,b,c$ be three positive real numbers such that $ abc = 1$. Prove that:\n\\[ 1 + \\frac {3}{a + b + c}\\ge{\\frac {6}{ab + bc + ca}} .\n\\]\n[/quote]\r\nLet $ p = a + b + c$, $ q = ab + bc + ca$ and $ r = abc$.\r\nThen we have $ 1 + \\frac {3}{p} \\ge \\frac {6}{q}$. By AM-GM:\r\n$ 1 + \\frac {3}{p} \\ge 2\\sqrt {\\frac {3}{p}}$. So we have to prove:\r\n$ 2\\sqrt {\\frac {3}{p}} \\ge \\frac {6}{q}$, which is equivalent to $ q^2 \\ge 3pr$. This is obviously true upon expanding.\r\nEdit:\r\nAn one liner:\r\n$ 1 + \\frac {3}{a + b + c} - {\\frac {6}{ab + bc + ca}} =$ $ \\left(1 - \\sqrt {\\frac {3}{a + b + c} \\right)^2 +}$ $ \\frac {\\sqrt {3}}{\\sqrt {a + b + c}(ab + bc + ca)(ab + bc + ca + \\sqrt {3a + 3b + 3c})} \\cdot$ $ ( (ab - bc)^2 + (ab - ac)^2 + (ca - cb)^2 ) \\ge 0$\r\n(If it can fit in one line :P)", "Solution_5": "first ,we use $ {abc \\equal{} 1}$,so the inequality $ {\\Leftrightarrow}$ $ {\\frac {6}{\\frac {1}{a} \\plus{} \\frac {1}{b} \\plus{} \\frac {1}{c}} < \\equal{} 1 \\plus{} \\frac {3}{a \\plus{} b \\plus{} c}}$;we know that :$ {\\frac {3}{\\frac {1}{a} \\plus{} \\frac {1}{b} \\plus{} \\frac {1}{c}} \\le \\frac {a \\plus{} b \\plus{} c}{3}}$ so we only need to prove that :$ {\\frac {2}{3}(a\\plus{}b\\plus{}c) \\le 1 \\plus{} \\frac {3}{a \\plus{} b \\plus{} c}}$_______[#];let $ {a \\plus{} b \\plus{} c \\equal{} x}$ by $ {a \\plus{} b \\plus{} c \\le 3*(abc)^{\\frac {1}{3}}}$ ,we know $ {x \\in (0,3] }$; so [#] $ {\\Longleftrightarrow}$ $ {\\frac {2}{3}x^2 \\minus{} x \\minus{} 3 > \\equal{} 0}$ ; then consider the function: $ {\\phi (x) \\equal{} \\frac {2}{3}(x \\minus{} \\frac {3}{4})^2 \\minus{} \\frac {27}{8}}$ ; we know that : $ {\\phi (x) \\ge 0}$ on $ {x \\in (0,3]}$. then # is true ,so we have completed!", "Solution_6": "[quote=\"Maverick\"]Let $ a,b,c$ be three positive real numbers such that $ abc \\equal{} 1$. Prove that:\n\\[ 1 \\plus{} \\frac {3}{a \\plus{} b \\plus{} c}\\ge{\\frac {6}{ab \\plus{} bc \\plus{} ca}} .\\]\n[/quote]\r\n$ LHS \\equal{} 1 \\plus{} \\frac{9}{{3abc(a \\plus{} b \\plus{} c)}} \\ge 1 \\plus{} \\frac{9}{{{{(ab \\plus{} bc \\plus{} ca)}^2}}} \\ge 2\\sqrt {\\frac{9}{{{{(ab \\plus{} bc \\plus{} ca)}^2}}}} \\equal{} \\frac{6}{{ab \\plus{} bc \\plus{} ca}}$\r\n^^!", "Solution_7": "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=288827]Maranna SortList - 2009 - Italy:[/url]\nLet $ a, b, c$ be positive numbers and $ abc\\equal{}1$. Prove that\\[ \\frac{2}{a\\plus{}b\\plus{}c}\\plus{}\\frac{4}{3}\\geq1\\plus{}\\frac{3}{ab\\plus{}bc\\plus{}ca}\\geq\\frac{6}{a\\plus{}b\\plus{}c}.\\]", "Solution_8": "[quote=sqing][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=288827]Maranna SortList - 2009 - Italy:[/url]\nLet $ a, b, c$ be positive numbers and $ abc\\equal{}1$. Prove that\\[ \\frac{2}{a\\plus{}b\\plus{}c}\\plus{}\\frac{4}{3}\\geq1\\plus{}\\frac{3}{ab\\plus{}bc\\plus{}ca}\\geq\\frac{6}{a\\plus{}b\\plus{}c}.\\][/quote]\n\nI think Buffalo Way helps us here (but looking for something quicker). Any ideas arqady sir ?", "Solution_9": "Let $ a , b, c$ be positive real numbers such that $abc= 1 .$ Show that\n$$ ab+bc+ca+\\frac{9}{10(a^2+b^2+c^2)}\\geq \\frac{33}{10}. $$", "Solution_10": "Let $a_1,a_2,\\cdots,a_n (n\\ge 2)$ be positive real numbers . Prove that\n$$1+\\frac{n(n-1)}{2\\sum_{1\\le i1$ is an integer.\r\nFind all matrices $M$ such that $f$ is locally bijective around $M$\r\n(i.e the differential is an isomorphism )", "Solution_1": "Here for $k=2$ \r\n\r\n$f(M)=M^2$ \r\n\r\nthe differential at $M$ is $d_{M}f(H)=HM+MH$ \r\n\r\nIf all eigenvalues of $M$ has real parts $<0$ the \r\n$HM+MH=0$ has a unique solution $H=0$\r\nThen the differential is bijective in this case\r\n\r\n\r\nIf $M=aI_n$ an homotetie $a\\neq 0$ gives also a differential bijective" } { "Tag": [ "Duke", "college", "AMC", "USA(J)MO", "USAMO", "geometry", "summer program" ], "Problem": "So, I realize it's a bit early for some to know whether or not they are going to camp this year, but for those that do, let's get to meet each other!\r\n\r\nPost your name, age/grade, location, interests, and anything else you're are comfortable sharing.\r\n\r\nMy name is Leonid Grinberg from Boston, MA (a suburb, technically, of Boston, but whatever). I was born in Russia, and speak Russian fluently. I am 15 and an incoming sophomore (though I am in the class of 2009, for which there is somewhat of a long and complicated explanation). I like math (obviously), computers (IT and programming), physics, logic, economics, psychology, law and language.", "Solution_1": "Palmer Mebane, from near Richmond, Virginia (born there also), going to be a senior next year. I'm probably the person most disconnected from immediate reality you'll ever meet. My biggest interests are math and logic-oriented puzzles, though I detest word puzzles and riddles. On the side, I play lots of classic and old-school electronic games, program some, and dabble into personality theories.", "Solution_2": "Jeremiah Siochi, from Iowa (Iowa City - home of University of Iowa), I am 15 and next year will be a sophomore. I was born in Ohio and my dad is half Filipino and half Brazilian. I am also homeschooled. My interests are math, Latin, logic, computer programming, piano and double bass, classical music (especially Brahms), and jazz. I spent the past two summers at Duke University's TIP summer studies.", "Solution_3": "I am Ofir, born in Israel, but living in Worcester, MA. I am 15 yrs old. I, too, enjoy classical music, and even play the clarinet (and the tenor sax but I'm not good at it).", "Solution_4": "I am Mark Chen (currently a junior) from Hsinchu, Taiwan. My interests include math competitions, mathematics research, piano (Chopin, Liszt, Prokoffief), debate (National Debate Champion), British literature, chess, and Chinese. I'm also resonably good at basketball, swimming, and ping-pong. I'm not particularly good at chemistry, sudoku, go, organizing my room, and waking up for school.\r\n\r\nI attend National Experimental High School, a superbly competitive high school where it's normal to study through lunchtimes in silence (so my GPA is nowhere near my post rating). Needless to say, I can't wait for a summer back in the states :P .\r\n\r\nI've taken the USAMO twice (from 12:30 AM-5:00 AM), won a silver medal in the Chinese Mathematical Olympiad, and trained as an alternate for the Taiwanese IMO team. My AMC scores are indicated in my profile. My school has a highly active Olympiad culture; the top IPhO scorer (absolute winner) in 2005 went to my school, as did a 2005 IChO gold medallist. \r\n\r\nI'm looking forward to meeting everyone! To all pingpong tournament contestants: bring it on.\r\n\r\nMark Chen (MC 04 05 x 07)", "Solution_5": "Laura Zehender, also from near Richmond, VA (I have lived here my whole life). I am 15 and will be a sophomore next year. I've been homeschooled since 2nd grade. My interests include math (especially combinatorics and geometry), dance (ballet, tap, jazz, pointe, and have done some other kinds of dancing), Latin, piano (especially Tschaikowsky), listening to country music, ancient Greek literature and myths, and art (especially modeling clay and drawing with pencil). I also am a bookworm and chocoholic. This is my first year at Mathcamp, but last summer I went to MathPath, and my older brother went to Mathcamp twice.", "Solution_6": "I'm David Corwin, from Acton, MA. I go to Acton-Boxborough Regional High School, one of the best public high schools in MA. My interests include math, foreign languages (French/Latin so far), science (I guess?), politics, classical music, tennis, school newspaper, something like that.", "Solution_7": "[quote=\"calc rulz\"]I'm David Corwin, from Acton, MA. I go to Acton-Boxborough Regional High School, one of the best public high schools in MA. My interests include math, foreign languages (French/Latin so far), science (I guess?), politics, classical music, tennis, something like that.[/quote]\r\n\r\nAlgonquin Regional High School is way better.\r\n\r\n(even though you beat us at the state meet :mad: )", "Solution_8": "I'm Tom James from Louisville, KY, I'm 17 years old and will be a senior next year. I'm a pretty cynical fellow with a dry sense of humor. I like to make fun of myself and I don't mind criticism (based on logic, of course :-)).\r\n\r\nAt school, I play three sports: cross country, wrestling (my favorite), and track. I'm active in yearbook, theatre, quick recall, and other assorted extra-curriculars. I also really like French.\r\n\r\nOutside of school, I'm very interested in philosophy and politics and love a good discussion or debate. I also am interested in personality theory such as Myers-Briggs (I'm an INTJ). I do a fair bit of independent reading and programming too.\r\n\r\nMy favorite area of mathematics would probably be logic/foundations such as Godel's theorems and theoretical computer science such as algorithmic information theory. I think these fields penetrate into the deepest truths of our existence. I hope to be a pure mathematician one day.\r\n\r\nAlthough I'm normally slow to get comfortable around new people, I'm a good deal more extroverted at camps for kids like us (I went to Duke TIP for three years prior to Mathcamp). I hope that I can get to know most kids during the five weeks because I'm normally such an introvert :P \r\n\r\nI am looking forward to this very, very, very much :D", "Solution_9": "[quote=\"tjames\"]I'm a pretty cynical fellow with a dry sense of humor.[/quote]\r\n\r\nTom, I think you and I are going to get along! :)\r\n\r\nI guess I should also mention that I am taking Latin, and am enjoying it very much. I cannot play music, but I enjoy listening to it. I play Ultimate Frisbee for a sport, and participate in the Computer Club and Math Team as general extra-curriculars. I greatly enjoy debate and discussion in general, but have never participated in any team, and have no idea how competitive debating works. I go to Belmont High School, which, although far from the best school, is quite a nice, relaxing place to attend.\r\n\r\nI am also something of a grammar stickler.", "Solution_10": "@tjames: Those are both extremely fascinating subjects! (Information theory + logical foundations) In fact, I have tried to delve into them but have found that my mathematical talents apparently lie elsewhere as I cannot seem to get the hang of the many bizarre and counterintuitive results that creep up. Perhaps at Mathcamp I'll have the proper time and environment to understand it better, as I'm sure you will too.\r\n\r\nThough I may eschew that area entirely, as (since we're supposed to talk about ourselves) my primary interest is combinatorics, whatever that is supposed to mean. In the same vein, I do a lot of algorithmic style programming and also play chess, soccer, and frisbee, though I am not so great at the last of these (maybe some of you that are good can teach me while we are there).", "Solution_11": "Hey, I'm an INTJ too!\r\n\r\nAnyway, might as well explain that I'm going to MathCamp as well.\r\n\r\nBrian Hamrick, rising sophomore, from VA. I go to TJHSST. In math, I enjoy number theory, as well as anything else decently far away from geometry. This will be my second MathCamp of (hopefully) many. When we get to camp. You'll probably find me playing bughouse in the lounge during free time. I look forward to meeting all of you!", "Solution_12": "After discovering I've recieved a full ride, I'm most definitely going. Thanks MATHCAMP.\n\nMy name is Adrian from Alhambra, CA located in the LA Valley. I picked up Ultimate [Frisbee] from HCSSiM '06 and was like the worst player there, but when I returned home everyone was praising me, but I'm ready to be the worst again because it's so awesome to play.\n\nI really like music and dabble in a bunch of instruments. My main instrument is guitar and I used to practice very technical music (you know those crazy driving rock songs with fast guitar solos), but now I became more mellow and acoustic. I'd LOVE to start a technical (or nontechnical) rock band with anyone, if interested... that is... if Colby college happens to have an electric guitar and basses and drums and keyboards and trumpets for ska and amps sitting somewhere nearby. I hate 4/4 time and I love playing around with time signatures. Dream Theater is my favorite band.\n\nI've recently feel in love with singing, even though I'm not good, but I swear I soon will be. I think RENT is absolutely the best musical ever created (even though the only other one I've watched is Phantom of the Opera -_-). If you see some weirdo randomly bursting into song, there's a chance it'd be me.\n\nAnd then I tend to be overly romantic and overly analyze things and dream of transcendence and connect my inner soul through the fleeting, tranquil depths of nature and thus I love hiking and things of the sort.\n\nHmn, CARPE DIEM is my life philosophy. Tchao.", "Solution_13": "[quote=\"Eternica\"]\nI've recently feel in love with singing, even though I'm not good, but I swear I soon will be. I think RENT is absolutely the best musical ever created (even though the only other one I've watched is Phantom of the Opera -_-). If you see some retard randomly bursting into song, there's a chance it'd be me.[/quote]\r\n\r\nThere's also a very good chance it won't be you. A lot of people sing :).", "Solution_14": "Wait. Then that'd be awesome. We can turn MATHCAMP '07 into a musical!", "Solution_15": "Thanks, I did it, and now I have the email!", "Solution_16": "[quote=\"hahafaha\"]\nSo, I have obviously never been to Mathcamp before (can't wait though), but I do have friends who have. I am also not a Mormon, but I also have friends (one friend, actually) who is. And from everything I know about the two, I'd probably think that Mathcamp would be far stranger to the random person than Mormonism would be.[/quote]\r\n\r\nIt wasn't so much mormonism itself that we found strange, it was more the idea of mormon camp (and their crazy clap-chant-thingies). Then again, it's true that with \"games\" like bunny-bunny-toki-toki (which I'm sure you'll all learn by the end of camp), we were probably the stranger of the two.\r\n\r\nAnd, while we had the most amusing inter-camp relations with the mormons (barring, perhaps, the football players that Greg B. (who will be a Junior Counselor this year) explained the P vs NP problem to), I think the strangest camp there was, without a doubt, Yearbook camp.\r\n\r\nAnyway, a bit about me. I'm Youlian Simidjiyski, I enjoy math (surprise!) and philosophy (another surprise!). I'm originally from Bulgaria, and I probably spend more time trying to convince random people that Math is awesome than I spend actually doing Math. This will be my second and final MathCamp (since I'm a senior), and, as much as I look forward to seeing all of my friends from MC '06 again (*waves at Alex, Brian, and Dan*), I'm much more excited about meeting all of the new campers, and seeing all of your reactions to the craziness ;). Simply put, people aren't lying when they say MathCamp is like a whole new world. My two pieces of advice to new Campers:\r\n\r\n1. Go to TAU (Time: Academic, Un...scheduled?), do your homework (whoa! Amazing, right? Homework... a good thing... *headpop*), watch for signs of burnout, and resist the urge to take mid-day naps. You're going to have the other 47 weeks of the year to do math at a sluggish pace, make sure to make the most of it while you're there.\r\n\r\n2. Balance your math time and your social time carefully. It'll probably surprise most of you to hear that camp is a very social place. It's generally full of amazing people, almost all of whom would be among my best friends were they to attend my school. Thus, even though plenty of us are introverts and shun normal social behavior, at camp, the social life starts to detract first from sleep, and then, obviously, from Math. Find a good balance.\r\n\r\nAnyway, can't wait to meet all of you, and political discussions at camp would be quite fun, though, knowing how MC discussions end up, it's likely that two people originally debating free market vs communistic systems would end up talking about which one is better in a community with countably infinite, uncountably infinite, or negative amounts of capital...", "Solution_17": "Im going to have to agree with what the alumni are saying here. Mathcamp is pure awesomeness, and here are a few points:\r\n\r\nA) Political discussion is good. Other stuff at Mathcamp always seems to seem more fun, so political discussion is not very common.\r\n\r\nB) Mathcamp is VERY social. Someone once told me a story of how they did not take a single class their second year at Mathcamp because he was too busy enjoying the social life. Now, that does not mean you should not take any classes. In fact, you should take many classes, but Mathcamp is very fun socially as well.\r\n\r\nC) I am also a news Junkie outside of Mathcamp. I read digg ~5 times a day, Google News ~2, Slashdot~1. Last year at Mathcamp, throughout the five weeks, I read digg once, Slashdot once, and Google News never. Mathcamp is really too much of its own awesome world to think about this \"news\" stuff.\r\n\r\nNow, a little about me: I am Sergei Bernstein from Belmont, MA. This is my third year, and hopefully my best year. I enjoy math, frisbee, piano, video games, non-video games, living impulsively, and trying to make every problem a math problem. I am Xtremely excited about meeting new campers. In fact, I have already met some, and I can already tell that this year will be a great one. Also, I have already made some plans for Mathcamp. Let's see... what else... I believe Dan still owes me some pizza or equivalent. Here is an awesome quote: \"Your induction is weak, old man\"", "Solution_18": "[quote=\"TheBernstinator\"]\"Your induction is weak, old man\"[/quote]\r\nOh lord, this is probably going to be all over AoPS now that MOP is over. For those attending Mathcamp, a Star Wars parody is already in the works; most of the progress is a lot of spoofed quotes like this one.\r\n\r\nOther good ones:\r\n\"His diagram's gone. Luke, are you okay?\"\r\n\"Leave your compass and straightedge. You will not need them here.\"\r\n\"If you invert me now I shall become more powerful than you can possibly imagine.\" (think power of a point)\r\n(as you can see most of these were thought up during a Geometry lecture)\r\n\r\nEvidently Mathcamp has already produced a matrix parody, so this is nothing too new.", "Solution_19": "[quote=\"MellowMelon\"][quote=\"TheBernstinator\"]\"Your induction is weak, old man\"[/quote]\nOh lord, this is probably going to be all over AoPS now that MOP is over. For those attending Mathcamp, a Star Wars parody is already in the works; most of the progress is a lot of spoofed quotes like this one.\n\nOther good ones:\n\"His diagram's gone. Luke, are you okay?\"\n\"Leave your compass and straightedge. You will not need them here.\"\n\"If you invert me now I shall become more powerful than you can possibly imagine.\" (think power of a point)\n(as you can see most of these were thought up during a Geometry lecture)\n\nEvidently Mathcamp has already produced a matrix parody, so this is nothing too new.[/quote]\r\n\r\nAlso:\r\n\r\n\"We shall invert him to the dark side of the circle.\"\r\nKeep in mind the compass and straightedge part was pertinent to one's entrance into the complex cave: straightedge and compass are unneeded when one complex-bashes a geo problem.", "Solution_20": "[quote]Oh lord, this is probably going to be all over AoPS now that MOP is over. For those attending Mathcamp, a Star Wars parody is already in the works; most of the progress is a lot of spoofed quotes like this one.[/quote]\r\nActually, Serge and I have several chunks of scenes written by now.\r\n\r\nEDIT: To respond to the original subject. I'm \"definitely going\". 4 more days!!! Yay! :D", "Solution_21": "In response to the original topic, I might end up at mathcamp [b]next year[/b], so not me...", "Solution_22": "hey, i'm Prateek from Toronto, Canada.. I seem to be outnumbered by all you left-wing americans\r\nanyhow, i'm 16, going into grade 11, and among other things i'm interested in video games, wrestling, food, sleep, and math. this is my first year, and i hope i don't get lost on my way to camp.", "Solution_23": "[quote=\"Eternica\"]\nI've recently feel in love with singing, even though I'm not good, but I swear I soon will be. I think RENT is absolutely the best musical ever created (even though the only other one I've watched is Phantom of the Opera -_-). If you see some retard randomly bursting into song, there's a chance it'd be me.[/quote]\r\n\r\nHahaha. I'm a few weeks too late to warn you about Phantom of the Lecture Hall.\r\n\r\nIs anyone doing another math musical this year?\r\n\r\n(MC'06, not '07)", "Solution_24": "i love how this thread is literally from 2007 but we're all staring at it bc awesomeguy posted a link", "Solution_25": "Lol, yeah, cuz we don't know if Eternica is a boy or girl.", "Solution_26": "[quote=hahafaha]So, I realize it's a bit early for some to know whether or not they are going to camp this year, but for those that do, let's get to meet each other!\n\nPost your name, age/grade, location, interests, and anything else you're are comfortable sharing.\n\nMy name is Leonid Grinberg from Boston, MA (a suburb, technically, of Boston, but whatever). I was born in Russia, and speak Russian fluently. I am 15 and an incoming sophomore (though I am in the class of 2009, for which there is somewhat of a long and complicated explanation). I like math (obviously), computers (IT and programming), physics, logic, economics, psychology, law and language.[/quote]\n\n\nYou\u2019re 29 now ", "Solution_27": "I was like eternica is a 49% boy 51% girl and now I know. Anyways these are his most personal posts that I have ever seen. ", "Solution_28": "...maybe not talk about it?", "Solution_29": "[quote=Jwenslawski]I was like eternica is a 49% boy 51% girl and now I know. Anyways these are his most personal posts that I have ever seen.[/quote]\n\nI thought he was a girl until up to now o.o" } { "Tag": [ "trigonometry" ], "Problem": "This problem has been bugging me. I found a couple very different answers. But none worked. I really need this. :wallbash: But I don't know if there is a solution. :lol: If there isn't I at least want a method to approximate it.\r\n\r\nI have $ 5$ input values: $ Y_{1}$, $ Y_{2}$, $ Y_{3}$, $ C$, and $ M$.\r\nAnd I need $ 1$ output value: $ O$\r\n\r\n$ Y_{1}$, $ Y_{2}$, and $ Y_{3}$ are three $ y$ values on the sine curve $ y\\equal{}a \\cdot sin(b\\cdot x)$\r\nThe corresponding $ x$ values ($ x_{1}$, $ x_{2}$, and $ x_{3}$) have a distance of $ d$. (This means $ x_{1}\\equal{}x_{2}\\plus{}d\\equal{}x_{3}\\plus{}2\\cdot d$)\r\n\r\n$ C$, and $ O$ are two $ y$ values on the sine curve $ y\\equal{}a \\cdot sin(M\\cdot b\\cdot x)$\r\nThe corresponding $ x$ values ($ x_{c}$, and $ x_{o}$) have a distance of $ d$. (This means $ x_{c}\\equal{}x_{o}\\plus{}d$)\r\n\r\nSo pretty much solve for O:\r\n$ Y_{1}\\equal{}a \\cdot sin(b\\cdot x_{1})$\r\n$ Y_{2}\\equal{}a \\cdot sin(b\\cdot x_{2})$\r\n$ Y_{3}\\equal{}a \\cdot sin(b\\cdot x_{3})$\r\n$ C\\equal{}a \\cdot sin(M\\cdot b\\cdot x_{c}))$\r\n$ O\\equal{}a \\cdot sin(M\\cdot b\\cdot x_{o}))$\r\n$ x_{1}\\equal{}x_{2}\\plus{}d\\equal{}x_{3}\\plus{}2\\cdot d$\r\nand,\r\n$ x_{c}\\equal{}x_{o}\\plus{}d$\r\n\r\nIf this isn't the correct place to put this topic then move it.", "Solution_1": "Offhand, it looks like generically you need to know the value of $ a$ in advance; that lets you work with angle addition formulas and the Pythagorean theorem. And that may not even be enough; consider what happens if $ d \\equal{} \\frac{2 \\pi k}{b}$ for some unknown integer $ k$; then the value of $ O$ depends on the value of $ Mk \\bmod 2 \\pi$, which you can't compute.", "Solution_2": "Well, the value of $ b$, I think, can vary, it can be any multiple of a certain number.\r\nAlso, there will also be special cases where there is no solution. Such as where $ Y1$,$ Y2$,and $ Y3$ form a line, but I can consider those separatly.", "Solution_3": "Note that $ C$ and $ O$ correspond to points on the curve $ y \\equal{} a \\sin (bx)$, as well, if you just adjust the $ x$ components appropriately.\r\n\r\nThe question now becomes: can four points determine a sinusoid?", "Solution_4": "Can't three points determine a and b?\r\n$ y \\equal{} a \\sin (bx)$\r\nThat's what (x1,y1),(x2,y2),(x3,y3) can do, right?\r\nIf that can't happen I can always add another point (x4,y4)." } { "Tag": [ "calculus", "integration", "trigonometry", "logarithms", "complex numbers", "calculus computations" ], "Problem": "1)$ {\\int_{0}^{\\frac{\\pi}{2}}\\frac{x}{\\sin x\\plus{}\\cos x\\plus{}1}dx}$\r\n2)$ {\\int_{0}^{\\frac{\\pi}{2}}\\frac{x^{2}}{\\sin x\\plus{}\\cos x\\plus{}1}dx}$\r\nI can solve only the first one. :(", "Solution_1": "$ P\\equal{}\\int_{0}^{\\frac{\\pi }{2}}\\frac{x}{\\cos (x)\\plus{}\\sin (x)\\plus{}1}\\, dx$\r\n\r\nusing $ x\\to\\frac{\\pi }{2}\\minus{}x$\r\n\r\n$ P\\equal{}\\int_{0}^{\\frac{\\pi }{2}}\\frac{\\pi }{2 (\\cos (x)\\plus{}\\sin (x)\\plus{}1)}\\, dx\\minus{}P$\r\n\r\n$ x\\equal{}i\\log\\left(\\left(\\frac{1}{2}\\minus{}\\frac{i}{2}\\right) (y\\minus{}1)\\right)$\r\n\r\nI will use complex numbers here.\r\n\r\n$ \\int_{0}^{\\frac{\\pi }{2}}\\frac{\\pi }{2 (\\cos (x)\\plus{}\\sin (x)\\plus{}1)}\\, dx\\equal{}\\frac{1}{2}\\pi\\int_{2\\plus{}i}^{2\\minus{}i}\\left(\\frac{1}{\\minus{}i\\plus{}y}\\minus{}\\frac{1}{i\\plus{}y}\\right)\\, dy\\equal{}\\frac{1}{2}\\pi\\log (2)$\r\n\r\nSo $ P\\equal{}\\frac{1}{4}\\pi\\log (2)$" } { "Tag": [ "algebra", "polynomial", "algebra solved" ], "Problem": "If P(x) is a polynomial of degree n such that P(x) = 2^x for x = 1,2,3,...,n+1, find P(n+2).", "Solution_1": "There is an unique polynomial P with degree n and such that P(k) = 2 k for k=1,2,...,n+1..\r\n\r\nUsing that C(k,0)+C(k,1)+...+C(k,k) = 2 k , it is easy to verify that P(x) = 2[C(x-1,0)+C(x-1,1)+...+C(x-1,n)], where C(x,j) = x(x-1)...(x-j+1)/j! .\r\n\r\nIt follows that \r\nP(n+2) = 2[C(n+1,0)+...+C(n+1,n)] = 2[2 n+1 -1] = 2 n+2 - 2.\r\n\r\nPierre.", "Solution_2": "An other solution without having to guess the explicit form of Pn :\r\n\r\nLet P(n) such defined polynom.\r\n\r\nQ(n)(x)=P(n)(x+1) - P(n)(x)\r\nd(Q(n)) = n-1 and Q(n)(k) = 2^k (for k=1..n) so Q(n) = P(n-1) and P(n)(x+1) = P(n)(x) + P(n-1)(x) :) \r\n\r\nAfter that we just have to sum : Pn(n+2) = 2^n+2 + 2^(n+1) + .... + 2^3 + P(1)(3) = 2^(n+2) - 2 (since P(1)(x) = 2x)" } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $D_{n}$ denote the dihedral group of order $2n$. Count the total number of homomorphisms from $S_{4}$ to $D_{6}$.\r\n\r\nIs there any general method to count homomorphisms of finite groups?", "Solution_1": "Hi.\r\n\r\nI don't think that there is a general method for counting homomorphisms, but in this case it's relative easy:\r\n$D_{6}$ is isomorphic to $D_{3}\\times\\mathbb{Z}_{2}\\cong S_{3}\\times\\mathbb{Z}_{2}$ (little exercise within the exercise ;-) )\r\nNow $Hom(S_{4},S_{3}\\times\\mathbb{Z}_{2})$ is isomorphic to $Hom(S_{4},S_{3})\\times\\ Hom(S_{4},\\mathbb{Z}_{2})$. The right-hand factor is easy since there is only one normal subgroup of index 2. The other factor can be calculated similarly: $S_{4}/V_{4}$ is isomorphic to $S_{3}$ which yields 6 homomorphisms $S_{4}\\to S_{3}$ (the one from this isomorphism times the six automorphisms of $S_{3}$). Since there is no normal subgroup of index 3, there can't be an homomorphism from $S_{4}$ onto $A_{3}$. There are three elements of order 2 in $S_{3}$ so there are 3 additional homomorphisms onto the 2-elements-subgroups. This means that there are 10 Homomorphism $S_{4}\\to S_{3}$ (6 onto + 3 with smaller image +1 trivial)" } { "Tag": [ "advanced fields", "advanced fields unsolved" ], "Problem": "Can anyone help me with the problem found here - http://zgall1.t35.com/question2.jpg? \r\n\r\nI am having a lot of trouble even getting started on it.\r\n\r\nThanks\r\nZack", "Solution_1": "I have managed to solve most of the question but I still need some help proving that X(jack) = X(bar). Can anyone help me out with that?" } { "Tag": [ "calculus", "integration", "function", "limit", "real analysis", "real analysis unsolved" ], "Problem": "Given a sequence of measurable positive functions $(f_{n})$ defined on a measure space $(X,m,\\Sigma)$ which converges pointwise to a function $f$, such that $lim \\int_{X}{f_{n}dm}=\\int_{X}{fdm}<+\\infty$, show that $lim \\int_{E}{f_{n}dm}=\\int_{E}{fdm}$ for every measurable subset $E \\subseteq X$.\r\n\r\nN.B: this does not hold if we drop the integrability of $f$. So far, I could find a solution only if assuming $m$ to be $\\sigma$-additive (and using some \"complicated\" stuff, like Egoroff's theorem), so I feel I must be missing something easy.", "Solution_1": "Would you please check your assertion. Aren't there some superfluous $\\lim$'s?\r\nAnd what do you mean by \"converges to a function $f$\"? Pointwise convergence? Or convergence in some weak sense?", "Solution_2": "Yep, pointwise convergence. I edited the lim's, thank you for pointing it out.", "Solution_3": "Fatou's Lemma states \\[\\int_{X}\\liminf f_{n}dm\\le\\liminf\\int_{X}f_{n}dm.\\] We apply this for $E$ and $X\\setminus E$ in the place of $X$ and notice that in the present situation $\\liminf f_{n}=\\lim f_{n}=f$. So \\[\\int_{X}f dm=\\int_{E}f dm+\\int_{X\\setminus E}fdm\\le\\liminf\\int_{E}f_{n}dm+\\liminf\\int_{X\\setminus E}f_{n}dm\\le\\liminf\\int_{X}f_{n}dm=\\int_{X}fdm<\\infty.\\] Hence every $\\le$ must be $=$ and we get $\\int_{E}f dm=\\liminf\\int_{E}f_{n}dm,\\ \\int_{X\\setminus E}fdm=\\liminf\\int_{X\\setminus E}f_{n}dm.$\r\nBut since $\\limsup\\int_{E}f_{n}dm=\\int_{X}fdm-\\liminf\\int_{X\\setminus E}f_{n}dm=\\int_{E}fdm,$ we conclude \\[\\int_{E}f dm=\\lim\\int_{E}f_{n}dm.\\]", "Solution_4": "[quote=\"Kolumbus\"]we get $\\int_{E}f dm=\\liminf\\int_{E}f_{n}dm,\\ \\int_{X\\setminus E}fdm=\\liminf\\int_{X\\setminus E}f_{n}dm.$\n[/quote]\r\n\r\nEverything's looking fine but I cannot catch immediately how you infer the two independent equalities above.", "Solution_5": "If $a\\le b$ and $c\\le d$ and $a+c=b+d$, then \"every $\\le$ must be $=$\", that is $a=b,c=d$.\r\nHere we have $a=\\int_{E}fdm, b=\\liminf\\int_{E}f_{n}dm, c=\\int_{X\\setminus E}fdm, d=\\liminf\\int_{X\\setminus E}f_{n}dm$.\r\nNote that this need not be true, if $a,b,c,d$ could be $\\infty$.", "Solution_6": "Ok, I missed the fact that $a \\leq b$ by Fatou's lemma. Thank you." } { "Tag": [ "algebra", "polynomial" ], "Problem": "Let $S = (x-1)^4 + 4(x-1)^3 + 6(x-1)^2 + 4(x-1) + 1$. Then $S$ equals\r\n(A) $(x-2)^4$\r\n(B) $(x-1)^4$\r\n(C) $x^4$\r\n(D) $(x+1)^4$\r\n(E) $x^4+1$", "Solution_1": "[hide](a+1)^4 can be expanded to a^4 + 4a^3 + 6a^2 + 4a + 1\n\n\n\nIn this case the a is (x-1), so we can replace a with (x-1) to get:\n\n\n\n(a+1)^4\n\n= [(x-1) + 1)]^4\n\n= x^4\n\n\n\nSo the answer is C) x^4[/hide]", "Solution_2": "[hide](x+1)^4=x^4+4x^3+6x^2+4x+1. In this case the x is x-1, so subsitute it in. [(x-1)+1]^4=x^4. Therefore, the answer is (C) c^4[/hide]", "Solution_3": "[quote=\"MCrawford\"]Let $S = (x-1)^4 + 4(x-1)^3 + 6(x-1)^2 + 4(x-1) + 1$. Then $S$ equals\n(A) $(x-2)^4$\n(B) $(x-1)^4$\n(C) $x^4$\n(D) $(x+1)^4$\n(E) $x^4+1$[/quote]\r\n\r\nThe answer is C. But I dont know how to reply. Am I doing it correctly?", "Solution_4": "[quote=\"balloo\"][quote=\"MCrawford\"]Let $S = (x-1)^4 + 4(x-1)^3 + 6(x-1)^2 + 4(x-1) + 1$. Then $S$ equals\n(A) $(x-2)^4$\n(B) $(x-1)^4$\n(C) $x^4$\n(D) $(x+1)^4$\n(E) $x^4+1$[/quote]\n\nThe answer is C. But I dont know how to reply. Am I doing it correctly?[/quote]\r\n\r\nThat is the right answer but I can't tell you if you are doing it correctly unless you post what you are doing.", "Solution_5": "[hide]\nJust let $a=x-1$\nThen u will get the answer in a minute.[/hide]", "Solution_6": "If you have already tried it, and want to look at something which might be new, here is something which may be interesting ...\r\n[hide]\nFor a moment, assume that you did not recognize that it is the bionomial expansion , or right away saw that it is equal to $x^4$ .. Here is an interesting[b] fast[/b] method:\n\n[size=150]Check when $x=1$ !!!\n[/size]\nWhat happens when $x = 1$ , you see right away that $S=1$ and option C can be correct. \nYou can also have chosen some other values such as $x=0$ (which makes the calculation easy) to rule out some or all of the wrong options.\n\nNow can you be sure that $x^4$ is correct, if you checked it for say a few specific values.\nThe answer is Yes, if you checked it for 5 (or more different) values of $x$ . \n\nFor example, if particular values of , say $x$ for $0 , -1, +1, -2,+2$ $S$ gives values (here) (0,1,1,16,16) you can be sure that the identity will hold for all values for x.\n\nTrick is find the degree of the polynomial - obviously it is 4 here. and just try if for one more.\nThis one of not that well known trick, but often very easy to use! \n[/hide]", "Solution_7": "[quote=\"MCrawford\"]Let $S = (x-1)^4 + 4(x-1)^3 + 6(x-1)^2 + 4(x-1) + 1$. Then $S$ equals\n(A) $(x-2)^4$\n(B) $(x-1)^4$\n(C) $x^4$\n(D) $(x+1)^4$\n(E) $x^4+1$[/quote]\r\n[hide=\"solution\"]\nu = x - 1\n\nS = u^4 + 4u^3 + 6u^2 + 4u + 1\nS = (u + 1)^4\nS = (x)^4\n[b]C[/b]\n[/hide]", "Solution_8": "[quote=\"MCrawford\"]Let $S = (x-1)^4 + 4(x-1)^3 + 6(x-1)^2 + 4(x-1) + 1$. Then $S$ equals\n(A) $(x-2)^4$\n(B) $(x-1)^4$\n(C) $x^4$\n(D) $(x+1)^4$\n(E) $x^4+1$[/quote]\r\n\r\n[hide=\"solution\"]let x-1 be y\n\n$y^4+4y^3+6y^2+4y+1=(y+1)^4=((x-1)+1)^4=x^4$\n\n(C)[/hide]", "Solution_9": "[hide]\nI thought multiping it out would be a lot of work, so I tried to plug in a number.\nI tried to plug in 1.\nYou get:\n0+0+0+0+1=1\nThe answer is very obviously \"C\".\n[/hide]", "Solution_10": "Note to Gyan and asinha:\r\n[hide] Answer A would also not be eliminated after substituting x=1 - you'll need to try at least one more value to eliminate A.\n[/hide]", "Solution_11": "[quote=\"TripleM\"]Note to Gyan and asinha:\n[hide] Answer A would also not be eliminated after substituting x=1 - you'll need to try at least one more value to eliminate A.\n[/hide][/quote]\r\nYes ... but the point was to give a tool to students to use it to solve such problems quickly.. and the main point (that if the substitution works for sufficient number of values,- eg 5 in this case) may ne new to many of the sutdents here.", "Solution_12": "Yep, I know that, but you both apparently made the same mistake which could have been costly :)", "Solution_13": "Not to nit-pick but there is much more than \"just pluggig for x=1 in my original post ... see in the original: \r\n[quote][b]x=0 [/b](which makes the calculation easy) to rule out [b]some or all of the wrong [/b]options. [/quote] you are not depending on x=1 only. Yes option A might have been overlooked (BTW I did not even look that carefully at all the options except x^4, as the expression, to put it mildly, was found to be extremely simple) ..Okay so option A could be a possibility for x=1 but not when x=0 and y[b]ou need only one value to rule it out. [/b] ... Again the important point here, you can afford to be sloppy (as to the concern as \" the error could be costly\") if you are going to check it for another number where it is easy to see :) - \r\n\r\nBTW, what you pointed out is really appriciated - and yes there was an error of adding extra \"only\" - and yes I am VERY glad that we have bright people like you who would see that and take time to point it out ..That is very helpful for the forum. (I have edited out the extra \"only\" from the original post :)\r\n\r\nAnyway the takeaway for the people here should be:\r\n\r\nTo rule out wrong option - Only one exception/contradiction is needed. \r\nTo 'prove' the option is correct: - You need (n+1) cases where n is the degree of the polynomial .\r\n\r\nFor others : Read the original message by me, if you have not done so, and you may have another method in your tool box. Enjoy.:)", "Solution_14": "Just a little remark for starters who read this topic, except TripleM and Gyan, which seems to be in an argument/concern type of debate: no matter what method/technique/formula you use, as long as you can get the answer fast and be done with it quick is what matters more because those tests/competitions are timed and there are no time remaining for you to think which method works faster." } { "Tag": [ "integration", "logarithms", "trigonometry" ], "Problem": "$I=\\int^{1}_{0}\\frac{ln(1+x)}{1+x^{2}}dx$", "Solution_1": "[quote=\"radac_mail\"]$I=\\int^{1}_{0}\\frac{ln(1+x)}{1+x^{2}}dx$[/quote]\r\nLet $t = \\arctan (x)$\r\n\r\nthen\r\n$I = \\int_{0}^{1}{\\frac{{\\ln (1+x)}}{{x^{2}+1}}}dx = \\int_{0}^{\\frac{\\pi }{4}}{\\ln (1+\\tan (t))dt}$\r\n\r\nwe have\r\n$1+\\tan (t) = \\frac{{\\cos (x)+\\sin (x)}}{{\\cos (x)}}= \\frac{{\\sqrt 2 \\,\\left({\\cos \\left({\\frac{\\pi }{4}-t}\\right)}\\right)}}{{\\cos (t)}}$\r\n\r\nlet:\r\n$u = \\,\\frac{\\pi }{4}-t\\$$I\\= \\int_0^{\\frac{\\pi }{4}} {\\ln \\left( {\\frac{{\\sqrt 2 \\,\\left( {\\cos \\left( {\\frac{\\pi }{4} - t} \\right)} \\right)}}{{\\cos (t)}}} \\right)dt} \\\\$ \r\n$I\\, =-\\int_{\\frac{\\pi }{4}}^{0}{\\ln \\left({\\frac{{\\sqrt 2 \\,\\left({\\cos \\left( u \\right)}\\right)}}{{\\cos (\\frac{\\pi }{4}-u)}}}\\right)}du \\\\$\r\n $I\\,\\, = \\int_{0}^{\\frac{\\pi }{4}}{\\ln \\left({\\frac{{\\,2\\left({\\cos \\left( u \\right)}\\right)}}{{\\sqrt 2 \\cos (\\frac{\\pi }{4}-u)}}}\\right)}du \\\\$\r\n$I = \\int_{0}^{\\frac{\\pi }{4}}{\\ln (2)du+\\int_{0}^{\\frac{\\pi }{4}}{\\ln \\left({\\frac{{\\,\\left({\\cos \\left( u \\right)}\\right)}}{{\\sqrt 2 \\cos (\\frac{\\pi }{4}-u)}}}\\right)du}}\\\\$\r\n$I\\, = \\frac{\\pi }{4}\\ln (2)-\\int_{0}^{\\frac{\\pi }{4}}{\\ln \\left({\\frac{{\\sqrt 2 \\,\\left({\\cos \\left({\\frac{\\pi }{4}-u}\\right)}\\right)}}{{\\cos (u)}}}\\right)du}\\\\$\r\n $I = \\frac{\\pi }{4}\\ln (2)-I \\\\$\r\n $2I = \\frac{\\pi }{4}\\ln (2) \\\\$\r\n$I = \\frac{\\pi }{8}\\ln (2) \\\\$\r\n \\", "Solution_2": "$I=\\int^{\\frac{\\pi}{4}}_{0}ln(1+tg t)dt$\r\nNow with $u=\\frac{\\pi}{4}-t$ we have $I=\\int^{\\frac{\\pi}{4}}_{0}ln(1+\\frac{1-tg\\: u}{1+tg\\: u})du$\r\n=$\\int^{\\frac{\\pi}{4}}_{0}ln 2 \\: du-\\int^{\\frac{\\pi}{4}}_{0}ln(1+tg u)du$=$\\frac{\\pi}{4}ln2-I \\rightarrow I$\r\n\r\n...but i appreciate your hard-working :)" } { "Tag": [ "limit", "integration", "calculus", "calculus computations" ], "Problem": "Let $ a_n\\equal{}\\frac{1}{1 \\cdot n}\\plus{}\\frac{1}{2 \\cdot (n\\minus{}1)}\\plus{}\\cdots\\plus{}\\frac{1}{n \\cdot 1}$\r\nfind the limit:\r\n$ \\lim_{n \\to \\plus{} \\infty}a_n$ :)", "Solution_1": "zhaobin, \r\nWe'll use $ \\dfrac{a\\plus{}b}{ab}\\equal{}\\dfrac{1}{a}\\plus{}\\dfrac{1}{b}$, and that limit is equal to $ 0$ . :P", "Solution_2": "Hey,Thank you for your answer.That's also my way :lol: But maybe the following will be harder:\r\nLet $ a_n \\equal{} \\frac {1}{\\sqrt {1} \\cdot \\sqrt { n}} \\plus{} \\frac {1}{\\sqrt {2} \\cdot \\sqrt {n \\minus{} 1}} \\plus{} \\cdots \\plus{} \\frac {1}{\\sqrt {n} \\cdot {\\sqrt 1}}$\r\nfind the limit:\r\n$ \\lim_{n \\to \\plus{} \\infty}a_n$ :P \r\n\r\nYou can try, :)", "Solution_3": "That equals $ \\int_0^1 \\frac1{\\sqrt {x(1 \\minus{} x)}} \\, dx \\equal{} \\pi$. [Well, some arguments are needed to justify the equality between the \"Riemann sums\" and the integral, but I reckon they are not too hard.]\r\n\r\nPS: I got your PM and I'm currently thinking about it.", "Solution_4": "Yes,That's correct.I think it is nice limit.\r\nAlthough we can generalize it,but not nice.\r\n\r\n\r\nPS(Only to perfect_radio):Thank you,please reply as you have make clear of the problem.\r\nI don't also how harizi apply that identity to the original problem." } { "Tag": [ "modular arithmetic", "number theory", "relatively prime" ], "Problem": "Prove or disprove that there is an $ n$ such that $ 1988n\\equiv\\minus{}1\\pmod{3^{100}}$.", "Solution_1": "In the title, don't you mean $ 3^{100}$ divides 1988n+1?", "Solution_2": "Yeah, I'm sorry. Just ignore the title. :D", "Solution_3": "[hide]Since $ 1988$ is relatively prime to $ 3^100$, there will be a multiplicative inverse of $ 1988 \\pmod {3^100}$. Take the negative of this inverse and put that as $ n$.[/hide]", "Solution_4": "I would guess $ n\\equal{}1988^{3^{99} \\minus{}1}$", "Solution_5": "[quote=\"Johan Gunardi\"]Prove or disprove that there is an $ n$ such that $ 1988n\\equiv\\minus{}1\\pmod{3^{100}}$.[/quote]\n\nWe have $\\gcd(1988,3^{100})=1$, so $1988^{\\phi( 3^{100})} \\equiv 1 \\pmod{3^{100}}$.\n\nAlso $\\phi(p^k)=p^{k-1}(p-1)$ for every prime $p$ and positive integer $k$.\n\nSo $\\phi( 3^{100})=2 \\times 3^{99}$ and you can easily find $n=1988^{2 \\times 3^{ 99} -1}$.", "Solution_6": "[quote=\"amparvardi\"]...\nSo $\\phi( 3^{100})=2 \\times 3^{99}$ and you can easily find $n=1988^{2 \\times 3^{ 99} -1}$.[/quote]\n(Not that it matters - as it is easy to adjust n) but wouldn't $1988 n$ would be $\\equiv +1 \\pmod {3^{100}}$\n\n(Question is little more interesting if it was to find $n$ such that $1988^n \\equiv -1 \\pmod {3^{100}}$" } { "Tag": [ "integration", "function", "calculus", "calculus computations" ], "Problem": "Suppose that a sequence $ \\{a_n\\}$ is defined as $ a_n\\equal{}\\int^{n\\plus{}1}_nf(x)dx$ ,$ n\\in\\mathbb{N}$ . If we wish to show that $ a_n$ is monotonic, can we just show that $ f(x)$ is monotonic?\r\n\r\n[hide=\"original question\"]let $ a_n\\equal{}\\int^{n\\plus{}1}_n\\frac{1}{xe^x}dx$, for $ n\\geq 1$ . Show that $ a_n$ is decreasing [/hide]", "Solution_1": "Well, yes. $ a_{n\\plus{}1} \\minus{} a_n \\equal{} \\int_n^{n\\plus{}1} \\left( f(x \\plus{} 1) \\minus{} f(x) \\right) \\, dx$.", "Solution_2": "(Although of course the converse is not true -- there are non-monotonic functions for which that sequence is monotonic.)" } { "Tag": [ "induction", "combinatorics unsolved", "combinatorics" ], "Problem": "There are $37$ political leaders. They all know that some of them are being spied. They make a deal that every day have a lunch all together and talk about spying. Each one of them talks with each of the others.\r\n\r\nRules:\r\n1) All of them will tell the truth (what they know) about who's being spied, but not who spies;\r\n2) If one of the two knew that the other one was being spied, he would [b]not[/b] say it to him (so, this is an exception to rule 1);\r\n3) As soon as somebody find himself being spied, he immediately moves to some other office.\r\n\r\nAfter the first and the second lunch nothing happened. But after the third lunch some of the leaders has moved.\r\n\r\nQuestion: how many leaders has moved?", "Solution_1": "I've seen this kind of idea before. In general, if nobody leaves on the first k days and somebody leaves on the k+1 th day, then k+1 people leave then (so 3 people in our case). We can prove by induction the assertion: if a person has been told that m people are spied, and nobody leaves after the first m days, he will realise that he is spied and will leave on the m+1 th day.\r\n\r\nFor 0 it is trivial; now assume it true for m-1 and let us prove it for m. Person X has been told that m people are spied; then it can either be that m others are spied, or m other and himself is spied. But if it were that only m others are spied, then one of those (say Y) would be told that m-1 others are spied. So by the induction hypothesis, it would mean that he would get up by the m-th day. But our assumption is that nobody leaves in the first m days, so our X must also be spied. He will thus leave on the m+1 th day (he cannot leave before that, because if no one leaves, he cannot discern in the first m days whether there are m other people spied or m others and himself)\r\n\r\nSo if no one leaves in the first k days, and people leave on the k+1 th, then those people who leave must be the ones who are told that k people are spied. So in truth, there are k+1 people spied, and all those are told that k people are spied. So these k+1 must leave.", "Solution_2": "Your solution looks nice, but now I see that there is a problem. I'll try to prove that somebody must move after the first lunch.\r\n\r\nLet person $A_1$ be one of the leaders.\r\n\r\nSuppose that he heard that nobody is spied. That means that he should move.\r\n\r\nSuppose that he heard that person $A_2$ is spied. If that person doesn't move it means that person $A_2$ heard that somebody is spied, and that must be person $A_1$, so he should move.\r\n\r\nSuppose that he heard that persons $A_2$ and $A_3$ are spied. If nobody of them moves that means that persons $A_2$ and $A_3$ heard that two persons are spied, so $A_1$ is spied and he will move.\r\n\r\nContinuing this discussion, I don't see that there is a possibility that nobody will move after the first lunch.\r\n\r\nIf you (or somebody else) knew this problem, could you please post the exact statement, cause I just heard it and there may be a mistake.", "Solution_3": "Bojan, The problem seems to be in rule 3). Let suppose that two men are spied - A_1 and A_2. A_1(he is 'faster' thinker than A_2) sees that A_2 hasn't left, so he will leave. When A_2 sees that A_1 is leaving he won't know what to do :blush: \r\nIf rule 3) means:\r\nWhen somebody understand that he is being spied he won't appear on next lunch.\r\nThen if on fourth(this is after third) lunch somebody is missing(these men had realized that they are spied during third lunch) than it would be 3 people.\r\nAnd the solution of RHS is OK." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Given a positive integer $ n$. Prove that there exists an infinite number of intergers $ x,y,z,t$ satisfying $ x^3 \\plus{} y^3 \\plus{} z^3 \\plus{} t^3 \\equal{} 2(\\frac {n(n \\plus{} 1)}{2})^3 \\minus{} 1$.", "Solution_1": "[quote=\"thaithuan_GC\"]Given a positive integer $ n$. Prove that there exists an infinite number of intergers $ x,y,z,t$ satisfying $ x^3 \\plus{} y^3 \\plus{} z^3 \\plus{} t^3 \\equal{} 2(\\frac {n(n \\plus{} 1)}{2})^3 \\minus{} 1$.[/quote]\r\nThis problem has some ways to approach , my solution is only one way . Try to find out some other methods . Let consider : \r\n$ x \\equal{} \\frac {a \\plus{} b}{2},y \\equal{} \\frac {a \\minus{} b}{2},z \\equal{} \\frac {u \\plus{} v}{2},t \\equal{} \\frac {u \\minus{} v}{2}$ \r\nThen the equation becomes :\r\n\\[ a^3 \\plus{} 3ab^2 \\plus{} u^3 \\plus{} 3uv^2 \\equal{} 8m^3 \\minus{} 4\r\n\\]\r\nwhere $ m \\equal{} \\frac {n(n \\plus{} 1)}{2}$\r\nChoose $ a \\equal{} 2m,u \\equal{} \\minus{} 1$ and consider equation :\r\n\\[ 6mb^2 \\minus{} 3u^2 \\equal{} \\minus{} 3\r\n\\]\r\nIt is equivalent to equation :\r\n\\[ v^2\\minus{}2mb^2\\equal{}1\r\n\\]\r\nIt is clear $ 2m$ is not a perfect square and this equation has infinite solution $ (v_n,b_n)$ . Because $ b_0 \\equal{} 0$ so the sequence $ \\{b_n\\}$ contains infinite even number , and it clear that all term of sequence $ \\{v_n\\}$ is odd ,thus $ x,y,z,t$ are integer . Proof completes ." } { "Tag": [ "AMC", "AIME", "geometry", "AIME I" ], "Problem": "So here are the problems, answers are due by Sunday, December 11th, 2005, 8:00 PM EST. Have fun!\r\n\r\nPM or IM me your answers.", "Solution_1": "I love the poll distribution so far :D", "Solution_2": "Okay, so far only 2 people have submitted answers. Yeah I realized this is a horrible time in the year for something like this. Therefore, the deadline will be extended to [b]Sunday, December 11th, 2005, 8:00 PM EST[/b].", "Solution_3": "I'm definitely going to take the exam - I just haven't gotten time yet; probably I'll do it tomorrow.", "Solution_4": "I will do it when I am next free. :)", "Solution_5": "i want to take it, but i usually don't have a 3 hour block of free time", "Solution_6": "[quote=\"Altheman\"]i want to take it, but i usually don't have a 3 hour block of free time[/quote]\r\nThen take it in smaller blocks, and just make sure that you only work on it for a total of three hours.", "Solution_7": "[quote=\"zanttrang\"][quote=\"Altheman\"]i want to take it, but i usually don't have a 3 hour block of free time[/quote]\nThen take it in smaller blocks, and just make sure that you only work on it for a total of three hours.[/quote]\r\n\r\nYeah do that if it works better for you, right now I just need people sending in answers, :lol: .", "Solution_8": "Dec 11, now, right?", "Solution_9": "yes, it is december 11th.", "Solution_10": "Nevermind, I'm not taking it. I looked over the first few problems, and it's way harder than the standard AIME, not to mention a huge overabundance of geometry problems. I'm not going to spend my time doing something like this. It took me 20 minutes to get #1. :dry:", "Solution_11": "Who was the liar who voted easy? :rotfl:", "Solution_12": "[quote=\"JesusFreak197\"]Nevermind, I'm not taking it. I looked over the first few problems, and it's way harder than the standard AIME, not to mention a huge overabundance of geometry problems. I'm not going to spend my time doing something like this. It took me 20 minutes to get #1. :dry:[/quote]\n\nYeah but 6 geometry problems is not exactly too many, check out 2004 AIME1\n [url]http://www.kalva.demon.co.uk/aime/aime04a.html[/url]. I agree that the first few problems are much harder than something on a regular AIME, sorry about that :( .\n\nBut what's wrong with it being too difficult (wouldn't it be a good practice then)?\n\n[quote=\"Mildorf\"]My philosophy is that you should train, in part, for problems more difficult than the ones you will encounter, so I have made these mock contests extremely difficult. The idea is that, once you become acclimated to them, the real AIMEs will seem easier, and you will approach them with justifiable confidence[/quote]", "Solution_13": "[quote=\"beta\"]\n\nYeah but 6 geometry problems is not exactly too many[/quote]\nWell, I was mostly looking at the first half or so of the test, and was a little frustrated by the abundance of geometry, which is by far my worst subject.\n\n[quote=\"beta\"]But what's wrong with it being too difficult (wouldn't it be a good practice then)?\n\n[quote=\"Mildorf\"]My philosophy is that you should train, in part, for problems more difficult than the ones you will encounter, so I have made these mock contests extremely difficult. The idea is that, once you become acclimated to them, the real AIMEs will seem easier, and you will approach them with justifiable confidence[/quote][/quote]\r\n\r\nI know I should practice the harder problems; I might do some off-and-on at some point. I don't know if I'm an unusual mathy person here, but I don't like spending a long time on problems when I'm not actually in a competition. On the previous mock AIME, I only spent one hour and fifty minutes on it because three hours is just too long to spend straight. I wasn't even going to do it all straight, but I'm not a big fan of individual problems in the first five that take 20 minutes to solve. :| Maybe I'll take a look at some of the later ones, except those are probably going to be even harder, anyway... =\\", "Solution_14": "Thomas made an important point that it was important to work on the hard problems so the actual one will be a lot easier. This is how Barron's books are for SAT (I'm pretty positive that they're much harder than real SAT) but does that mean everyone who reads Barron's books get good grades on SAT? Not necessarily. This is same with Mock AMC. Although harder problems can benefit people, it may not benefit [b]everyone[/b]. \r\n\r\nJust for sidenote, don't worry too much about beta's geometry problems. They are tough problems but that's because he's good at geometry. If you can manage to solve all of them by yourself (which I can't, sadly), I'm pretty sure that you will be able to do most of the geometry problems on AIME by yourself. \r\n\r\nbeta, nice set of problems by the way! :)", "Solution_15": "you should want to challenge yourself jesusfreak, and also, you should practice for the real thing by working the full 3 hours", "Solution_16": "I very rarely have three free hours straight, and when I do, I'd rather do something other than math.", "Solution_17": "[quote=\"JesusFreak197\"]I very rarely have three free hours straight, and when I do, I'd rather do something other than math.[/quote]\r\n\r\nHmmm then why do you do math competitions if you don't like doing math?", "Solution_18": "I never said that. I enjoy doing math, but I don't like spending long periods of time straight on it.", "Solution_19": "[quote=\"JesusFreak197\"]...I'd rather do something other than math.[/quote]\r\n :what?: like what? :P", "Solution_20": "Yay! Now I can actually finish them! (too much homework this week....)", "Solution_21": "wow, i just finished this, and it was really difficult, and on number 6, i kept getting an answer that did not fit the form of a,b,c, but oh well, and number 1 was a lot like the number 14 on the 2005 aime A? talk about difficult...", "Solution_22": "[quote=\"Altheman\"]wow, i just finished this, and it was really difficult, and on number 6, i kept getting an answer that did not fit the form of a,b,c, but oh well, and number 1 was a lot like the number 14 on the 2005 aime A? talk about difficult...[/quote]\r\n\r\nI don't want to give too much away but #1 is much simpler than 2005 AIME #14...", "Solution_23": "All answers are due today!", "Solution_24": "I know I'm bumping a thread after 17 years, but this mock hasn't been listed anywhere or talked about, so I think it is severely underappreciated.\n\nDo not be fooled by this test's age. The difficulty is still on par with the harder real AIMEs from recent years. If you don't believe me, just try out the first 7-8 problems and you'll see.\n\nAnd yeah, I wrote solutions, since I was in the mood to and I thoroughly enjoyed solving these:\n[url]https://www.dropbox.com/s/n434y3tiytfrzki/beta_Mock_AIME_Solutions.pdf?dl=0[/url]\n\n", "Solution_25": "wow p13 :blush: \n\nonly thing bad abt this test is that 7/15 are geo", "Solution_26": "just did a couple of problems\nand def recommend", "Solution_27": "[quote=MockSolutionsCompiler]I know I'm bumping a thread after 17 years, but this mock hasn't been listed anywhere or talked about, so I think it is severely underappreciated.\n\nDo not be fooled by this test's age. The difficulty is still on par with the harder real AIMEs from recent years. If you don't believe me, just try out the first 7-8 problems and you'll see.\n\nAnd yeah, I wrote solutions, since I was in the mood to and I thoroughly enjoyed solving these:\n[url]https://www.dropbox.com/s/n434y3tiytfrzki/beta_Mock_AIME_Solutions.pdf?dl=0[/url][/quote]\n\nattached those solutions \n\n(problems are to be posted in separate threads in HSM in a while)", "Solution_28": "all problems have been posted in separate threads in HSM, I just created the related post collection\n\n[url=https://artofproblemsolving.com/community/c3741459_2005_beta_mock_aime]here[/url] \n\nenjoy / start solving\n\nUSA mocks [url=https://artofproblemsolving.com/community/c2439870_usa_mocks]here [/url] (also mentioned [url=https://artofproblemsolving.com/community/c5h3200761p29225908]here[/url])\nmore AIME mocks [url=https://artofproblemsolving.com/community/c2439872_aime_mocks]here[/url]\nmore User-created contests [url=https://artofproblemsolving.com/community/c2435719_user_created_contests]here[/url] (contains all above)", "Solution_29": "19 years, and @beta\u2019s legacy lives. \n\nGO AIME people!!!!!!", "Solution_30": "Gosh this thread was made when I was less than two months old. ", "Solution_31": "this thread was made when i was -4 years old", "Solution_32": "By the way, there is a results thread, which contains answers and solve rates:\n\n[url]https://artofproblemsolving.com/community/c5h65552[/url]" } { "Tag": [ "USAMTS" ], "Problem": "For the USAMTS problems, can we use programming knowledge to write a program to solve the problem for us? If so, are there any rules or guidelines in doing so?", "Solution_1": "Yes. Please see [url=http://www.usamts.org/TipsFAQ/U_FAQ.php#computer]this item in our FAQ list[/url]." } { "Tag": [ "function", "real analysis", "real analysis solved" ], "Problem": "Suppose $E\\subset \\mathbb{R}^1$ and the cardinality of $E$ is smaller than $c$. Show that there exists real number $a$ such that $E+\\{a\\} = \\{x+a: x\\in E\\}\\subset \\mathbb{R}^1\\setminus \\mathbb{Q}$.", "Solution_1": "If not, then there are functions f and g, both from the reals into the rationals, respective E such that $ f(x)=x+g(x) $. Since R is not numerable, there exists u such that $ f^{-1}(u) $ is not numerable. But then $ u-f^{-1}{u} $ is in E, and hence E is not numerable. I assumed here that any set with cardinality smaller than c is at most numerable.", "Solution_2": "I was thinking of taking a Hamel basis of $R$ (a basis of $R$ over $Q$) which includes $1$. Let $a_i,\\ i\\in I$ be the members of this basis. $I$ must have continuum cardinality $c$ and the difference of each two different $a_i$'s is not rational because otherwise it wouldn't be a basis (because th basis contains $1$ and then the $a_i$'s wouldn't be independent). Then if we assume that for each $a_i$ there is an $x_i$ s.t. $x_i+a_i$ is rational, the $x_i$'s must be distinct, so the cardinality of $E$ would be at least the cardinality of $(a_i)_{i\\in I}$, which is $c$, and we have a contradiction.\r\n\r\n[edited :D: Hammel$\\rightarrow$ Hamel]", "Solution_3": "[quote=\"grobber\"]I was thinking of taking a Hammel basis of $R$ (a basis of $R$ over $Q$) ... [/quote]\r\nHamel basis ;)", "Solution_4": "it is obviously that $a\\not\\in \\mathbb{Q}-E$, so we need only to prove that the cardinality of $\\mathbb{Q}-E$ is $